Fundamentals of
E n g i n ee r i n g E c o n o m i c
A n a ly s i s
White
Grasman
Ca s e
L a S c o l a Nee d y
P r at t
Formula
Formula
F 5 Ac
i
d
A 5Fc
11 1 i 2 n 2 1
• Future Worth of
Uniform Series
• Uniform Series
from Future Value
i
11 1 i 2 n 2 1
11 1 i 2 n 2 1
i 11 1 i 2 n
d
d
A5Pc
d
• Uniform Series
from Present Value
i 11 1 i 2
n
P 5 Ac
11 1 i 2 n 2 1
1 An 11 1 i 2 2n
1 An21 11 1 i 2 21n212
1 A 3 11 1 i 2 23 1 p
P 5 A1 11 1 i 2 21 1 A 2 11 1 i 2 22
F 5 P 11 1 i 2 n
P 5 F 11 1 i 2
2n
• Present Worth of
Uniform Series
Uniform Series
Cash Flows
Irregular Series
Cash Flows
Series Cash Flows
Future Worth
of a Present
Payment
Present Worth
of a Future
Payment
Single Cash Flows
A 5 F1A Z F i%,n2
F 5 A1F Z A i%,n2
A 5 P1A Z P i%,n2
P 5 A1P Z A i%,n2
t51
P 5 a At 1P Z F i%,t 2
n
Factor Notation
F 5 P 1F Z P i%,n2
P 5 F 1P Z F i%,n2
Factor Notation
ⴝPMT(i %,n,,ⴚF )
ⴝFV(i %,n,ⴚA)
ⴝPMT(i %,n,ⴚP )
ⴝPV(i %,n,ⴚA)
ⴝNPV(i %,A1,A2,A3, . . . ,An )
Excel Notation
ⴝFV(i %,n,,ⴚP)
ⴝPV(i %,n,,ⴚF )
Excel Notation
P
0
P
0
P
0
0
1
A
1
1
A
1
A1
n
End of Period
A
2
...
An – 1
A
n–1
n–1
A
A
n
F
n
A
n–1
3
End of Period
2
End of Period
A
A2
2
A3
Cash Flow Diagram
2
n–1
End of Period
F
Cash Flow Diagram
An
n
• Future Worth of a
Geometric Series
• Uniform Series
Equivalent of a
Geometric Series
• Present Worth of a
Geometric Series
d
nA1 y11 1 i 2
i2j
1 2 11 1 j 2 n 11 1 i 2 2n
i
2
11 1 i 2 n 2 11 1 ni 2
d
d
i5j
iZj
F 5 µ
n21
i2j
11 1 i 2 n 2 11 1 j 2 n
nA1 11 1 i 2
A1 c
A1 c
d
i5j
i?j
1 2 11 1 j 2 n 11 1 i 2 2n 2
i 11 1 i 2 n
d
d c
i2j
11 1 i 2 n 2 1
A5 µ
n21
i 11 1 i 2
nA1 a
b
11 1 i 2 n 2 1
P 5 µ
F 5 Gc
• Future Worth of a
Gradient Series
n
1
2 1A Z F i%,n2 d
i
i
i2
2n
Formula
1 2 11 1 ni 211 1 i 2
A1 c
A 5 Gc
• Uniform Series
Equivalent of a
Gradient Series
Geometric Series
Cash Flows
P 5 Gc
• Present Worth of a
Gradient Series
Gradient Series
Cash Flows
Series Cash Flows
i5j
iZj
ⴝFV(i %,n,,ⴚNPV(i %,0,G,
2G, . . . ,(n ⴚ 1)G))
ⴝPMT(i %,n,ⴚNPV(i %,0,G,
2G, . . . ,(n ⴚ 1)G))
ⴝNPV(i %,0,G,2G, . . . ,(n ⴚ 1)G)
Excel Notation
F 5 A1 1F Z A1 i%,j%,n2
A 5 A1(A Z A1 i%,j%,n)
ⴝFV(i %,n,,ⴚNPV(i %,A,(1 ⴙ j )A,
(1 ⴙ j )2A, . . . ,(1 ⴙ j )nⴚ1A))
ⴝPMT(i %,n,ⴚNPV(i %,A,
(1 ⴙ j )A,(1 ⴙ j )2A, . . . ,
(1 ⴙ j )nⴚ1A))
P 5 A1 1P Z A1 i%,j%,n2 ⴝNPV(i %,A,(1 ⴙ j )A,
(1 ⴙ j )2A, . . . ,(1 ⴙ j )nⴚ1A)
F 5 G 1F Z G i%,n2
A 5 G 1A Z G i%,n2
P 5 G 1P Z G i%,n2
Factor Notation
P
0
P
0
2
1
2
2G
...
...
3
End of Period
A1(1 + j)2
n
(n – 1)G
n–1
n
A1(1 + j)(n – 1)
n–1
A1(1 + j)n – 2
3
End of Period
A1 A1(1 + j)
1
G
(n – 2)G
Cash Flow Diagram
FUNDAMENTALS
OF ENGINEERING
ECONOMIC ANALYSIS
First Edition
JOHN A. WHITE
University of Arkansas
KELLIE S. GRASMAN
Missouri University of Science and Technology
KENNETH E. CASE
Oklahoma State University
K I M L a S C O L A N E E DY
University of Arkansas
D AV I D B . P R AT T
Oklahoma State University
VICE PRESIDENT & EXECUTIVE PUBLISHER Don Fowley
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Library of Congress Cataloging in Publication Data
White, John A., 1939Fundamentals of engineering economic analysis/John A. White, Universisty of Arkansas, Kellie Grasman,
Missouri University of Science and Technology, Kenneth E. Case, Oklahoma State University,
Kim LaScola Needy, University of Arkansas, David B. Pratt, Oklahoma State University.
pages cm
Includes index. cover
ISBN 978-1-118-41470-5 (hardback)
1. Engineering economy. I. Title.
TA177.4.W47 2013
658.15—dc23
2013023515
The inside back cover will contain printing identification and country of origin if omitted from this page. In
addition, if the ISBN on the back cover differs from the ISBN on this page, the one on the back cover is correct.
Printed in the United States of America
10 9 8 7 6 5 4 3 2 1
CONTENTS
CHAPTER 1
An Overview of Engineering Economic Analysis
01
3
Learning Objectives 4
Introduction 4
1-1 Time Value of Money 4
1-2 Engineering Economy Principles 7
1-3 Economic Justification of Capital Investments 10
1.3.1 SEAT Step 1 of 7: Identify the Investment Alternatives 11
1.3.2 SEAT Step 2 of 7: Defining the Planning Horizon 11
1.3.3 SEAT Step 3 of 7: Specifying the Minimum Attractive Rate
of Return 13
1.3.4 SEAT Step 4 of 7: Estimate the Cash Flows 14
1.3.5 SEAT Step 5 of 7: Comparing Alternatives 14
1.3.6 SEAT Step 6 of 7: Performing Supplementary Analyses 14
1.3.7 SEAT Step 7 of 7: Select the Preferred Investment 15
Summary 15
Key Terms 16
FE-Like Problems 17
Problems 18
CHAPTER 2
Time Value of Money Calculations
02
23
Learning Objectives 23
Introduction 23
2-1 Cash Flow Diagrams 24
2-2 Single Cash Flows 26
2.2.1 Future Worth Calculations (F |P) 26
2.2.2 Present Worth Calculations (P |F ) 33
2-3 Multiple Cash Flows 35
2.3.1 Irregular Cash Flows 35
2.3.2 Uniform Series of Cash Flows 39
2.3.3 Gradient Series of Cash Flows 46
2.3.4 Geometric Series of Cash Flows 51
iii
iv
Contents
2-4
Compounding Frequency 55
2.4.1 Period Interest Rate Approach 56
2.4.2 Effective Annual Interest Rate 58
2.4.3 When Compounding and Cash Flow Frequencies Differ
Summary 62
Key Terms 66
FE-Like Problems 67
Problems 69
CHAPTER 3
Equivalence, Loans, and Bonds
03
04
61
93
Learning Objectives 93
Introduction 93
3-1 Equivalence 94
3-2 Interest Payments and Principal Payments 100
3.2.1 Immediate Payment Loans 101
3.2.2 Deferred Payment Loans 104
3-3 Bond Investments 106
3-4 Variable Interest Rates 110
3-5 Annual Percentage Rate 111
Summary 113
Key Terms 114
FE-Like Problems 114
Problems 116
CHAPTER 4
Present Worth 125
Learning Objectives 125
Introduction 126
4-1 Comparing Alternatives 126
4.1.1 Methods of Comparing Economic Worth 127
4.1.2 Ranking and Incremental Methods of Economic Worth 127
4.1.3 Equivalence of Methods 128
4.1.4 Before-Tax versus After-Tax Analysis 128
4.1.5 Equal versus Unequal Lives 129
4.1.6 A Single Alternative 129
4-2 Present Worth Calculations 130
4.2.1 Present Worth of a Single Alternative 130
4.2.2 Present Worth of Multiple Alternatives 132
4.2.3 Present Worth of One-Shot Investments 135
Contents
4-3
Benefit-Cost Analysis 137
4.3.1 Benefit-Cost Calculations for a Single Alternative 137
4.3.2 Benefit-Cost Calculations for Multiple Alternatives 139
4-4 Discounted Payback Period 144
4.4.1 Discounted Payback Period for a Single
Alternative 144
4.4.2 Discounted Payback Period for Multiple
Alternatives 149
4-5 Capitalized Worth 152
4.5.1 Capitalized Worth for a Single Alternative 153
4.5.2 Capitalized Worth for Multiple Alternatives 156
Summary 157
Key Terms 159
FE-Like Problems 159
Problems 162
CHAPTER 5
Annual Worth and Future Worth
05
06
179
Learning Objectives 179
Introduction 179
5-1 Annual Worth 180
5.1.1 Annual Worth of a Single Alternative 180
5.1.2 Annual Worth of Multiple Alternatives 183
5-2 Future Worth 185
5.2.1 Future Worth of a Single Alternative 185
5.2.2 Future Worth of a Multiple Alternatives 189
5.2.3 Portfolio Analysis 192
Summary 194
Key Terms 194
FE-Like Problems 195
Problems 196
CHAPTER 6
Rate of Return 213
Learning Objectives 213
Introduction 214
6-1 Internal Rate of Return Calculations 215
6.1.1 Single Alternative 215
6.1.2 Multiple Roots 217
6.1.3 Multiple Alternatives 220
v
vi
Contents
6-2
External Rate of Return Calculations 224
6.2.1 Single Alternative 224
6.2.2 Multiple Alternatives 228
Summary 231
Key Terms 232
FE-Like Problems 232
Problems 235
CHAPTER 7
Replacement Analysis
07
08
257
Learning Objectives 258
Introduction 258
7-1 Fundamentals of Replacement Analysis 258
7-2 Cash Flow and Opportunity Cost Approaches 261
7-3 Optimum Replacement Interval 264
Summary 268
Key Terms 269
FE-Like Problems 269
Problems 271
CHAPTER 8
Depreciation 285
Learning Objectives 286
Introduction 286
8-1 The Role of Depreciation in Economic Analysis 287
8-2 Language of Depreciation 288
8-3 Straight-Line and Declining Balance Depreciation
Methods 289
8.3.1 Straight-Line Depreciation (SLN) 289
8.3.2 Declining Balance and Double Declining Balance
Depreciation (DB and DDB) 291
8.3.3 Switching from DDB to SLN with the Excel® VDB
Function 294
8-4 Modified Accelerated Cost Recovery System (MACRS) 296
8.4.1 MACRS-GDS 297
8.4.2 MACRS-ADS 301
Summary 301
Key Terms 303
FE-Like Problems 303
Problems 305
Contents
CHAPTER 9
Income Taxes 317
09
10
11
Learning Objectives 317
Introduction 317
9-1 Corporate Income Tax Rates 319
9-2 After-Tax Analysis Using Retained Earnings (No Borrowing) 322
9.2.1 Single Alternative 323
9.2.2 Multiple Alternatives 328
9-3 After-Tax Analysis Using Borrowed Capital 330
9-4 Leasing versus Purchasing Equipment 336
Summary 338
Key Terms 340
FE-Like Problems 340
Problems 342
CHAPTER 10
Inflation 359
Learning Objectives 359
Introduction 360
10-1 The Meaning and Measure of Inflation 360
10-2 Before-Tax Analysis 364
10-3 After-Tax Analysis 367
10-4 After-Tax Analysis with Borrowed Capital 369
Summary 373
Key Terms 375
FE-Like Problems 375
Problems 377
C H A P T E R 11
Break-Even, Sensitivity, and Risk Analysis
Learning Objectives 391
Introduction 392
11-1 Break-Even Analysis 393
11-2 Sensitivity Analysis 398
11-3 Risk Analysis 405
11.3.1 Analytical Solutions 407
11.3.2 Simulation Solutions 412
Summary 418
Key Terms 419
FE-Like Problems 419
Problems 421
391
vii
viii
Contents
CHAPTER 12
Capital Budgeting
12
437
Learning Objectives 438
Introduction 438
12-1 The Classical Capital Budgeting Problem 438
12-2 Capital Budgeting Problem with Indivisible Investments 440
12-3 Capital Budgeting Problem with Divisible Investments 446
12-4 Practical Considerations in Capital Budgeting 449
Summary 450
Key Terms 451
FE-Like Problems 451
Problems 453
APPENDIX A
a Single Sums, Uniform Series, and Gradient Series Interest Factors
b Geometric Series Present Worth Interest Factors 496
c Geometric Series Future Worth Interest Factors 501
APPENDIX B
Obtaining and Estimating Cash Flows
507
B-1 Introduction 506
B-2 Cost Terminology 508
B.2.1 Life Cycle Viewpoint 508
B.2.2 Past/Future Viewpoint 510
B.2.3 Manufacturing Cost Structure Viewpoint 514
B.2.4 Fixed and Variable Viewpoint 517
B.2.5 Average and Marginal Viewpoint 527
B-3 Cost Estimation 530
B.3.1 Project Estimation 532
B.3.2 Estimating Methodologies 535
B.3.3 General Sources of Data 536
B-4 General Accounting Principles 537
B.4.1 Balance Sheet 538
B.4.2 Income Statement 542
B.4.3 Ratio Analysis 545
B-5 Cost Accounting Principles 550
B.5.1 Traditional Cost Allocation Methods 551
B.5.2 Activity Based Costing 556
B.5.3 Standard Costs 557
B.5.4 Economic Value Added 558
Summary 561
FE-Like Problems 562
Problems 563
ANSWERS TO SELECTED
EVEN NUMBERED PROBLEMS
INDEX
599
577
471
PREFACE
INTRODUCTION
We are pleased to present Fundamentals of Engineering Economic Analysis,
a new text to serve the undergraduate engineering economics course. Three
overarching goals informed the development of this book and its corresponding WileyPLUS course:
•
•
•
Streamlined: This text was carefully optimized to serve a 1semester, 1–3 credit-hour course without sacrificing rigor or essential
content. As engineering programs are increasingly squeezed to cover
more content without extending the 4-year curriculum, Engineering
Economics courses are being reduced from two semesters to one semester and/or from 3 credit hours down to 2 or even 1 credit hour.
Fundamentals of Engineering Economic Analysis is built to meet the
needs of the course as it is today, tailored down to a content set that
can be covered within the allotted timeframe.
Accessible: Students who enter this course rarely have any background
in accounting or finance. This text has been designed to support a broad
spectrum of engineering students with the best-integrated pedagogy, in
both print and electronic formats. It also provides real-world vignettes
to motivate and deepen students’ engagement with the course topics.
Unique digital solution: The WileyPLUS course that accompanies
this text provides a wealth of homework problems that are automatically graded to provide instant feedback to students, along with secure
solutions. The course also provides students with an abundance of
learning support features that are available 2437, such as videos that
show an instructor walking through the solution of a typical problem.
The core content and approach of Fundamentals of Engineering Economic Analysis are built on the strong foundation of Principles of Engineering Economic Analysis, now in its sixth edition, by John A. White,
Kenneth E. Case, and David B. Pratt. As such, the content has been thoroughly and successfully class-tested, and reflects decades’ worth of accuracy checking. But the coverage has been carefully streamlined to serve the
always time-challenged one-semester course. In addition, through the
efforts of new co-authors Kellie Grasman and Kim LaScola Needy, Fundamentals of Engineering Economic Analysis provides thoughtful and wellintegrated pedagogical support to help a wide range of students better
grasp—and retain—this critical material.
ix
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Preface
We believe this synthesis of proven content and consistent pedagogy
is ideally suited to help students master a course that will inform the rest
of their lives as working engineers.
APPROACH OF THE TEXT
Fundamentals of Engineering Economic Analysis provides a rich and interesting learning experience for students as well as a strong teaching tool for
instructors. This exciting new approach was carefully crafted to support
different learning and teaching styles. An engaging visual design reinforces
the key pedagogical aids throughout the text, such as learning objectives
and key terms. Visual learning is also supported with numerous and carefully crafted illustrations, such as cash flow diagrams, that reinforce student
understanding of the underlying concepts. Each chapter begins with a realworld vignette to demonstrate why the concepts to be addressed are important to a practicing engineer. A detailed summary at the end of each chapter
succinctly captures the most important ideas to reinforce the students’
learning. Finally, each chapter ends with a set of FE-Like Problems to help
students who want to prepare for the FE Exam, as well as an abundant variety
of homework problems at various levels of difficulty.
Within the topical coverage, Fundamentals of Engineering Economic
Analysis proceeds logically from time value of money concepts, to methods
of comparing economic alternatives, to additional complexities that pervade
real-world engineering economic analysis, such as taxes, depreciation, and
inflation.
A seven-step Systematic Evaluation and Analysis Technique (SEAT) is
introduced in Chapter 1 and revisited in the following chapters. This approach
serves to remind students of the central role of comparing economic alternatives within the larger task of a full-blown engineering economic analysis.
It also emphasizes the logical and methodical nature of such analyses.
The book takes a cash flow approach throughout. We want students to
recognize that a correct engineering economic analysis depends on the
correct amount and timing of the relevant cash flows.
Use of Traditional Solution Methods and Excel Spreadsheets
Another important facet of our approach concerns the balance of traditional solution methods and Microsoft® Excel. Throughout, the text supports the use of the basic formulas and interest factors for single sums,
uniform series, gradient series, and geometric series. These have formed
the heart of time value of money evaluations by engineers over the past
60 years. Traditional solution methods are followed by discussion of Excel
tools and formulas, where they exist, to provide the balanced coverage that
Preface
today’s students need. Indeed, the complexity of problems students will
face in the real world practically demands that they develop some fluency
with Excel formulas and tools. But we introduce them only after careful
presentation of the underlying math and traditional techniques.
Fundamentals of Engineering Economic Analysis has been developed
to support the different teaching styles of instructors and learning styles of
students, including a range of approaches to the use of Excel spreadsheets.
For example, instructors and students can, if they choose, rely on hand
calculations and ignore the material on spreadsheets. With a very few
exceptions, the worked Examples in the text focus first on complete hand
solutions, followed by the Excel approaches where relevant. The exceptions to this approach are a handful of Examples that are sufficiently complex to require use of spreadsheets, and those Examples have titles that
clearly indicate that requirement. Depending on the needs of the instructor,
students, and/or the specific topic, a mixture of approaches can also be
used.
Use of Solved Examples
Solved Examples in the book feature several components that reinforce
our approach to teaching engineering economics:
•
•
•
•
Many of the Examples are linked with respect to context. That is, the same
scenario is revisited in multiple Examples to implement progressive
concepts and techniques. See, for instance, Examples 4.2, 5.3, and 6.4.
This approach also facilitates comparison of results under different
situations, such as no tax versus tax and various loan plans.
Most Examples are formatted with specific steps to aid students in
developing a methodical problem-solving approach.
Examples are first solved by traditional formula or factor techniques,
followed by Excel approaches, where relevant. The Excel versions
of solutions are distinguished by blue/boldface type. A few complex
Examples are addressed with Excel only.
Examples reflect a mix of business/industry and personal finance contexts, to optimally motivate students’ interest and attention.
CONTENT SUMMARY
Topical coverage is focused on core and essential topics for those who will
perform engineering economics analysis, including:
•
•
Principles and methodology for engineering economic analyses considering the time value of money.
Time value of money including all commonly used cash flow profiles.
xi
xii
Preface
•
•
•
•
•
•
•
•
•
Borrowing, lending, and investing with business and personal
applications.
Measures of economic worth including present worth, capitalized
worth, future worth, annual worth, internal rate of return, external rate of
return, and benefit-cost.
Depreciation including MACRS and its foundation methods—double
declining balance and straight line.
After-tax economic analysis, with and without borrowing.
Before- and after-tax replacement analysis and optimal replacement
interval.
Inflation, including before- and after-tax analyses.
Supplementary analyses, including breakeven, sensitivity analysis,
and risk analysis, as well as use of @RISK simulation software.
Capital budgeting to make sound overall investment under monetary
constraints.
Costs, estimation, accounting, and Economic Value Added concepts.
IMPORTANT STUDENT SUPPORT FEATURES
Pedagogical features in both the print text and WileyPLUS are designed to
support and reinforce student learning.
In the print text:
•
•
•
•
•
•
Learning objectives set student expectations at the beginning of the
chapter, are restated at the appropriate point in the body of the chapter,
and are reviewed in the chapter Summary.
Key terms are defined in margins and listed and cross-referenced in the
chapter summary for easy review.
Real-world vignettes and discussion questions begin each chapter,
placing the content in a practical perspective and opening it up for
further exploration.
Summary sections at ends of chapters highlight key concepts, equations, learning objectives, and key terms in an easily accessible format.
816 end-of-chapter problems are organized by text section to facilitate
assignment and self-study. Answers to even-numbered problems are
provided at the end of the book.
138 FE-like problems at ends of chapters give students an opportunity
to practice for the FE exam, as well as check their understanding of
chapter contents.
Preface
In WileyPLUS:
•
•
•
•
Video Lessons are brief lectures explaining concepts from the text, so
that students have multiple means of better understanding core material.
Video Solutions to selected problems give students an online “office hours”
experience for problems they may struggle with, and model a solution
strategy for other homework problems.
Algorithmic problems are extensions of print problems, but with
different quantitative information. This gives instructors additional
exercises to assign and students more opportunity to apply their
understanding.
GO Tutorial problems provide students with a step-by-step guide
on how to approach a problem, so they can review it (at instructor’s
discretion) and then go back and try the problem again on their own.
On the companion Web site:
•
Useful Excel spreadsheet utilities that assist in performing specific
engineering economy analyses.
INSTRUCTOR SUPPORT
The following instructor support materials are available on the Web site for this
text: http://www.wiley.com/WileyCDA/WileyTitle/productCd-1118414705.
html
•
•
•
PowerPoint lecture slides for each chapter facilitate teaching and learning from the book. They can be used for lecture and presentation purposes or serve as templates for instructors developing their own slides.
The Instructor’s Solutions Manual includes solutions in Excel spreadsheet form to make it easy to see the solutions, select particular cells
to see the formula used, or convert the entire worksheet to see the
formulas used in all cells. The Instructor’s Solutions Manual is
password-protected, and accessible only to instructors adopting this
text. Please visit the instructor section of the website at http://bcs.wiley.
com/he-bcs/Books?action=index&itemId=1118414705&bcsId=8241
to register for a password.
Useful Excel spreadsheet utilities that assist in performing specific
engineering economy analyses.
WILEYPLUS
WileyPLUS combines the complete, dynamic online text with all of the teaching and learning resources you need, in one easy-to-use system. WileyPLUS
offers today’s engineering students the iterative and visual learning materials
xiii
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Preface
they need to help them grasp difficult concepts—and apply what they’ve
learned to solve problems. A robust variety of examples and exercises enables
students to work problems, see their results, and obtain instant feedback
including hints and reading references linked directly to the online text.
For instructors, WileyPLUS:
•
•
Makes the course more manageable—automatic grading and real-time
reporting saves instructors time and energy
Allows instructors to focus on maximizing their impact on students
WileyPLUS provides an abundance of resources, both for students and
instructors:
•
•
•
A complete online version of the textbook
A wealth of practice problems with automatic grading and immediate
feedback for students
° Secure solutions make homework meaningful—problems are
algorithmic so students receive unique values in each problem
° Integration with Blackboard Course Management Systems enables the ability to have all course materials in one place
2437 support for students to ensure positive learning outcomes
° Video Solutions demonstrate how to solve a problem, both by hand
and with Excel
° Mini-lecture videos by co-author Kellie Grasman explain concepts
from the text, so that students may have multiple means of better
understanding the topic
Contact your local Wiley representative or visit www.wileyplus.com for
more information about using WileyPLUS in your course.
ACKNOWLEDGMENTS
We would like to thank the following instructors who reviewed the manuscript at various stages, or reviewed or contributed to the WileyPLUS course:
Kamran Abedini, California State Polytechnic University—Pomona
Virgil Adumitroaie, University of Southern California
Baabak Ashuri, Georgia Institute of Technology
Swaminathan Balachandran, University of Wisconsin—Platteville
Geza Paul Bottlik, University of Southern California
Stanley F. Bullington, Mississippi State University
Richard Burke, SUNY Maritime
Karen M. Bursic, University of Pittsburgh
Mark Calabrese, University of Central Florida
John R. Callister, Cornell University
Preface
Viviana I. Cesani, University of Puerto Rico
Xin Chen, Southern Illinois University Edwardsville
Oswald Chong, University of Kansas
Donald Coduto, California State Polytechnic University—Pomona
William Foley, Rennselaer Polytechnic University
Charles R. Glagola, University of Florida
Craig M. Harvey, Louisiana State University
Edgar A. Hollingsworth, Pennsylvania College of Technology
Sung-Hee Kim, Southern Polytechnic State University
Krishna K. Krishnan, Wichita State University
Leonard R. Lamberson, Western Michigan University
John Lee, University of Wisconsin—Madison
Alberto Marquez, Lamar University
Gary Maul, Ohio State University
R. Eugene McGinnis, Christian Brothers University
Mamunur Rashid, Georgia Southern University
Matthew S. Sanders, Kettering University
Alex Savachkin, University of South Florida
Dana Sherman, University of Southern California
Surendra Singh, University of Tulsa
Alice E. Smith, Auburn University
John M. Usher, Mississippi State University
Tao Yang, California State Polytechnic University—San Luis Obispo
Mehmet Bayram Yildirim, Wichita State University
Obtaining input, counsel, examples, and data has been made much easier
through the exceptional cooperation of the following individuals and others
in their organizations (listed alphabetically by company):
•
•
•
•
•
•
Abbot Laboratories: Miles D. White, Chairman and Chief Executive
Officer.
Intel: Dr. Craig R. Barrett, Chairman of the Board.
J. B. Hunt Transport Services, Inc: Jerry W. Walton, Executive Vice
President and Chief Financial Officer, Donald G. Cope, Sr. Vice President, Controller, and Chief Accounting Officer, and David G. Mee,
Executive Vice President and Chief Financial Officer.
Motorola Solutions: Dr. Thomas F. Davis, Corporate Vice President
and Chief Economist (retired).
Schneider National, Inc.: Dr. Christopher B. Lofgren, President and
Chief Executive Officer.
Wal-Mart Stores: Michael T. Duke, President and Chief Executive
Officer.
xv
A WALKTHROUGH
OF THE TEXT
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Real-world
vignettes
show how practicing engineers use
the concepts
E N G I N E E R I N G E C O N O M I C S I N P R AC T I C E :
K E L L I E S C H N E I D E R A N A LY Z E S
H E R N E T WO RT H
Immediately after receiving her engineering degree, Kellie Schneider began
employment with a multinational company. Her initial salary was $60,000
per year. Kellie decided she would invest 10 percent of her gross salary each
month. Based on discussions with the company’s personnel and with a person
in human resources, she believed her salary would increase annually at a
rate ranging from 3 percent to 15 percent, depending on her performance.
After analyzing various investment opportunities, she anticipated she would
be able to earn between 5 percent and 10 percent annually on her investment portfolio of mutual funds, stocks, bonds, U.S. Treasury notes, and certificates of deposit. Finally, Kellie believed annual inflation would vary from
2 percent to 5 percent over her professional career. With this information in
hand, she calculated the range of possible values for her net worth, first after
30 years of employment, and second after 40 years of employment. She was
amazed and pleased at what she learned.
Discussion
questions
engage students in a
thought-provoking
discussion
DISCUSSION QUESTIONS:
1. What do you suppose Kellie is attempting to do with her investment
portfolio by selecting multiple forms of investments?
2. Although retirement may seem like a long time away, why is it important to start saving early for your “golden years?”
3. If interest rates end up being lower than Kellie assumes, what must
she do to build her goal “nest egg?”
4. What impacts will a change in your family status (such as getting married or having a child) have on your investment decisions? c02TimeValueOfMoneyCalculations.indd Page 23
5/3/13 9:27 PM f-481
22
TIME VALUE OF MONEY
CALCULATIONS
Learning
objectives
help students
measure their
progress
LEARNING OBJECTIVES
When you have finished studying this chapter, you should be able to:
1. Construct a cash flow diagram (CFD) depicting the cash inflows and
outflows for an investment alternative. (Section 2.1)
2. Perform time value of money calculations for single cash flows with
annual compounding. (Section 2.2)
3. Distinguish between uniform, irregular, gradient and geometric series
of cash flows. (Section 2.3)
4. Perform time value of money calculations for multiple cash flows including:
• Irregular series of cash flows with annual compounding. (Section 2.3.1)
• Uniform series of cash flows with annual compounding. (Section 2.3.2)
• Gradient series of cash flows with annual compounding. (Section 2.3.3)
• Geometric series of cash flows with annual compounding. (Section 2.3.4)
xvi
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A Walkthrough of the Text
xvii
WileyPLUS feature icons
are called out in the text—for example,
a video lesson on Cash Flow Diagrams
is available in WileyPLUS
2-1
CASH FLOW DIAGRAMS
LEA R N IN G O B JEC TIV E: Construct a cash flow diagram (CFD) depicting the
Video Lesson:
Cash Flow Diagrams
Key terms and
definitions are
highlighted
Cash Flow Diagram
(CFD) A diagram
depicting the magnitude
and timing of cash flowing
in and out of the
investment alternative.
cash inflows and outflows for an investment alternative.
It is helpful to use cash flow diagrams (CFDs) when analyzing cash flows
that occur over several time periods. As shown in Figure 2.1, a CFD is
constructed using a segmented horizontal line as a time scale, with vertical
arrows indicating cash flows. An upward arrow indicates a cash inflow or
positive-valued cash flow, and a downward arrow indicates a cash outflow,
or negative-valued cash flow. The arrows are placed along the time scale to
correspond with the timing of the cash flows. The lengths of the arrows can
be used to suggest the magnitudes of the corresponding cash flows, but in
most cases, little is gained by precise scaling of the arrows.
Worked Examples
Many examples are linked to others in context. That is, the/208/WB00898/XXXXXXXXXXXX/ch02/text_s
same scenario is re-visited in multiple Examples to implement
progressively complex concepts and techniques. This approach also
facilitates comparison of results under different situations, such as no
tax versus tax and various loan plans.
c02TimeValueOfMoneyCalculations.indd Page 46 5/3/13 9:27 PM f-481
Who Wants to Be a Millionaire?
EXAMPLE
Examples follow a
consistent problemsolving methodology
Suppose Crystal Wilson wants to accumulate $1,000,000 by the time she
retires in 40 years. If she earns 10 percent on her investments, how much
must she invest each year in order to realize her goal?
KEY DATA
SOLUTION
Shows the handcalculation
Given: F 5 $1,000,000; i 5 10%; n 5 40
Find: A
Applying Equation 2.27,
A 5 $1,000,0001A Z F,10%,402
5 $1,000,00010.00225942
5 $2,259.40/year
Using the Excel® PMT worksheet function gives
A 5PMT110%,40,,210000002 5 $2,259.41/year
Shows how to solve with Excel
xviii A Walkthrough of the Text
Chapter Summaries
SUMMARY
KEY CONCEPTS
Brief summary of
key concepts,
linked to Learning
Objectives
1. Learning Objective: Construct a cash flow diagram (CFD) depicting the
cash inflows and outflows for an investment alternative. (Section 2.1)
The CFD is a powerful visual tool used to depict an investment alternative.
The graphic depicts the timing, magnitude and direction of the cash flow.
Cash flows can represent present values, future values, uniform series, gradient series or geometric series. Cash flows are positive or negative depending upon the perspective (such as borrower versus lender) from which they
are drawn.
KEY TERMS
Reviews
Key Terms
c02TimeValueOfMoneyCalculations.indd Page 67 5/3/13 9:28 PM f-481
Capital Recovery Factor, p. 42
Cash Flow Diagram (CFD), p. 24
Compounding, p. 26
Interest Rate, p. 26
Nominal Annual Interest Rate, p. 55
Period Interest Rate, p. 56
Effective Annual Interest Rate, p. 58
Future Value, p. 28
Present Value, p. 27
Sinking Fund Factor, p. 45
Geometric Series, p. 51
Gradient Series, p. 46
Uniform Series, p. 26
/208/WB00898/XXXXXXXXXXXX/ch02/text_s
Problem available in WileyPLUS
GO Tutorial Tutoring Problem available in WileyPLUS
Video Solution Video Solution available in WileyPLUS
FE-LIKE PROBLEMS
1.
If you want to triple your money at an interest rate of 6% per year compounded annually, for how many years would you have to leave the money in
the account?
a. 12 years
c. 32 years
b. 19 years
d. Cannot be determined without knowing the
amount invested.
FE-Like Problems
for exam preparation
c02TimeValueOfMoneyCalculations.indd
2. Let F be the accumulated sum, P the
principal invested, i the annual Page
com-69 5/3/13 9:28 PM f-481
pound interest rate, and n the number of years. Which of the following
correctly relates these quantities?
a. F 5 P (1 1 in)
c. F 5 P (1 1 n)i
b. F 5 P (1 1 i)n
d. F 5 P (1 1 ni)n 2 1
PROBLEMS
Introduction
1.
Abundant end-of-chapter
problems at various
difficulty levels and keyed
to chapter sections
You are offered $200 now plus $100 a year from now for your used
computer. Since the sum of those two amounts is $300, the buyer suggests
simply waiting and giving you $300 a year from now. You know and trust the
buyer, and you typically earn 5.0% per year on your money. So, is the offer
fair and equitable?
2. You are offered $500 now plus $500 one year from now. You can earn
6% per year on your money.
a. It is suggested that a single fair amount be paid now. What do you con-
sider fair?
b. It is suggested that a single fair amount be paid one year from now. What
do you consider fair?
3. What words comprise the abbreviation “DCF”? Tell/describe/define what it
means in 10 words or less.
4. State the four DCF rules.
Section 2.1
Cash-Flow Diagrams
5. Pooi Phan needs $2,000 to pay off her bills. She borrows this amount from a
bank with plans to pay it back over the next four years at $X per year. Draw
a cash flow diagram from the bank’s perspective.
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xx
Automatically graded homework
MEANINGFUL HOMEWORK—
ALGORITHMIC PROBLEMS GIVE EACH STUDENT UNIQUE VALUES
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ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E : WAL M A RT
In fiscal year (FY) 2011, Walmart Stores, Inc. (WMT) employed more than
2.1 million people worldwide, including 1.4 million in the United States. Its
annual sales increased 3.47 percent from the previous year to a record
$419 billion, and income increased 6.3 percent to a record $15.4 billion. As of
January 31, 2011, Walmart had a total of 8,907 retail outlets worldwide,
including 3,804 Supercenters, discount stores, and Neighborhood Markets
as well as 609 Sam’s Clubs in the United States and 4,557 retail outlets in
stores in 14 countries outside the United States. Walmart imports billions of
dollars of products from 70 countries. Its imports are greater than those of
most countries in the world. In a 15-year period (from 1990 to 2005) the
number of Walmart stores in the United States increased by more than
2,000; in the next 6-year period (from 2005 to 2011), the number of stores
increased by 3,800.
A major factor in Walmart’s success has been the efficiency of its supply
chain. Walmart was among the first to adopt electronic data interchange
(EDI), allowing it to pay vendors for products purchased by customers in its
stores. The real-time management of money across its thousands of stores
has contributed significantly to Walmart’s economic performance.
Walmart is an international leader in logistics, and its logistics operations employ more than 85,000 associates. Its distribution network includes
approximately 150 distribution centers and 50 transportation offices. A major
user of transportation services from external sources, Walmart has the
fourth-largest private fleet of trucks in the United States, including 7,200 tractors, 53,000 trailers, and 8,000 drivers. Highly regarded for its rapid response
to the September 11, 2001 attacks on the World Trade Center, to Hurricane
Katrina in 2005, and to the 2010 oil spill in the Gulf of Mexico, Walmart has
9 disaster distribution centers, strategically located across the United States
and stocked with relief supplies to assist communities in distress.
Walmart was a pioneer in the use of radio frequency identification (RFID)
to do real-time tracking of products throughout the supply chain. In recent
2
AN OVERVIEW OF
ENGINEERING
ECONOMIC ANALYSIS
years Walmart has targeted sustainability and worked with its vendor community to dramatically reduce the amount of waste arriving at and leaving
from its stores in the form of packing and packaging materials. Its engineers
work with vendors’ engineers to significantly reduce energy consumption in
distributing the products it sells and to increase energy efficiency in its
stores. Walmart gives priority attention to reducing life-cycle costs, resulting
in a 19.2 percent return on investment (ROI) in FY 2011.
ENGINEER ING EC ONOM I C S I N P R AC T I C E :
ROB ERT THOM PS ON’S FI N AN C I AL P L AN N I N G
Robert Thompson is a recent engineering graduate who desires to invest a
portion of his annual income each year. He is unsure about the kind of investments he should make—stocks, bonds, mutual funds, U.S. Treasury notes,
certificates of deposit, rental property, land, and so on. Also, he is undecided
about how much of his annual income he should set aside for investment.
Further, he does not know what annual return he should expect to earn on
his investments. Finally, he wonders how long it will take for him to achieve
his financial goals.
DISCUSSION QUESTIONS:
1. What do the preceding two examples have in common?
2. What role does engineering economic analysis play in these scenarios?
3. Is an engineering economic analysis just as relevant for the complicated business transactions at Walmart as for Robert Thompson’s
much simpler transactions?
4. Compare the length of the financial planning horizon that each investor
(Walmart and Robert) is considering. Are they the same or different?
3
4
Chapter 1
An Overview of Engineering Economic Analysis
LEARNING OBJECTIVES
When you have finished studying this chapter, you should be able to:
1. Apply the four discounted cash flow (DCF) rules to simple time value
of money (TVOM) situations (Section 1.1).
2. Identify the 10 principles of engineering economic analysis that can be
used by all engineers in analyzing the economic performance of the
products, processes, and systems they design (Section 1.2).
3. Identify the seven steps of the systematic economic analysis technique
(SEAT) used to perform engineering economic analyses (Section 1.3).
INTRODUCTION
The subject matter of this text is variously referred to as economic analysis, engineering economy, economic justification, capital investment
analysis, or economic decision analysis. Traditionally, the application of
economic analysis techniques in the comparison of engineering design
alternatives has been referred to as engineering economy. The emergence of a widespread interest in economic analysis in public-sector
decision making, however, has brought about greater use of the more
general term economic analysis. We define engineering economic analysis
as using a combination of quantitative and qualitative techniques to analyze economic differences among engineering design alternatives in selecting the preferred design.
In this text we use a cash flow approach to the subject. A cash flow
occurs when money actually changes hands from one individual to another or from one organization to another. Thus, money received and
money dispersed (spent or paid) constitutes a cash flow.
It is always tempting to skip the first chapter of an engineering textbook. It would be a big mistake to do so here, however. You will build on
the foundation concepts introduced in this chapter throughout the rest of
your study of engineering economics.
1-1
TIME VALUE OF MONEY
LEARN I N G O B JEC T I V E : Apply the four discounted cash flow (DCF) rules to
Video Lesson:
Time Value of Money
simple time value of money (TVOM) situations.
A fundamental concept underlies much of the material covered in the text:
Money has a time value. This means that the value of a given sum of
money depends on both the amount of money and the point in time when
1-1 Time Value of Money
the money is received or paid. Just as in construction the placement of
forces along a beam matters in designing structures, so the placement in
time of money received or paid matters when evaluating the economic
worth of an investment.
The Time Value of Money Illustrated
Time Value of Money
(TVOM) The value of a
given sum of money
depends on both the
amount of money and
the point in time when the
money is received or paid.
EXAMPLE
To illustrate the time value of money (TVOM) concept, suppose a wealthy
individual approaches you and says, “Because of your outstanding ability
to manage money, I am prepared to present you with a tax-free gift of
$1,000. If you prefer, however, I will postpone the presentation for a year,
at which time I will guarantee that you will receive a tax-free gift of $X.”
(For purposes of this example, assume that the guarantee is risk-free.) In
other words, you can choose to receive $1,000 today or receive $X 1 year
from today. Which would you choose if X equals (1) $1,000, (2) $1,050,
(3) $1,100, (4) $1,500, (5) $2,000, (6) $5,000, (7) $10,000, (8) $100,000?
In presenting this situation to students in our classes, none preferred to
receive $1,000 a year from now instead of receiving $1,000 today. Also,
none preferred to receive $1,050 a year from now instead of $1,000 today.
Gradually, as the value of X increased, more and more students switched
from preferring $1,000 today to preferring $X a year from now. Not surprisingly, every student preferred to receive $100,000 a year from now to
receiving $1,000 today. Also, it was no surprise that all students preferred
to receive $10,000 a year from now to receiving $1,000 today. The greatest
debate and uncertainty among the students concerning which amount to
choose occurred when X was between $1,100 and $2,000.
For each student, some value (or range of values) of $X exists for
which the student is indifferent—that is, has no preference—about receiving $1,000 today versus receiving $X a year from today. If a student is
indifferent when X equals $1,200, then we would conclude that $1,200
received 1 year from now has a present value or present worth of $1,000
for that particular student in his or her current circumstances. We would
conclude that this student’s TVOM is 20 percent.
From students’ responses to the questions posed in the preceding example,
patterns have been noted in the choices they make. Several indicate a very
strong need for money now, not later. Their personal circumstances are
5
SOLUTION
6
Chapter 1
An Overview of Engineering Economic Analysis
Discounted Cash Flow
(DCF) The movement of
money forward or backward
in time.
such that they do not believe they can wait a year to receive the money, even
if they will receive significantly more at that time. Others are skeptical
regarding the guarantee of the money being available a year later—they
resort to the ‘‘bird in the hand, versus many birds in the bush’’ philosophy.
Such responses occur in industry as well. Corporate executives exhibit
similar tendencies when faced with current versus deferred choices.
Occasionally, students claim that inflation will make the future amount
of money worth less to them. While it is certainly true that inflation, which
represents a decrease in the purchasing power of money, will diminish the
present worth of a future sum of money, money has time value even in the
absence of inflation. (Chapter 10 will examine the effects of inflation on
engineering economic decisions.)
Why do we claim that money has time value in the absence of inflation?
Because money has “earning power.” If you own money and someone else
temporarily needs it, you can loan it to them and charge them interest. The
interest rate you charge should be based on your TVOM. After all, if you
loan it to someone, then you can no longer invest it; hence, you forego the
opportunity to earn a return on your money. The lost opportunity should factor into how much you charge someone for using your money. Because of
this, the TVOM is sometimes referred to as the opportunity cost of money.
Other terms used to express the TVOM are interest rate, discount rate,
hurdle rate, minimum attractive rate of return, and cost of capital. These
terms will be used somewhat interchangeably throughout the text. In the
first few chapters, however, interest rate and discount rate are used most
frequently.
Another term used in financial circles and throughout the text is
discounted cash flows, often referred to as DCF. Originally, DCF referred
to the process of using the TVOM or discount rate to convert all future
cash flows to a present single sum equivalent.1 Today, DCF tends to refer
to any movement of money backward or forward in time.
The fact that money has a time value changes how mathematical operations involving money should be performed. Simply stated, because money
has a time value, one should not add or subtract money unless it occurs at
the same point in time.
Having introduced the notion of new rules for dealing with money, we
can now summarize the four DCF rules:
Money has a time value.
2. Quantities of money cannot be added or subtracted unless they occur
at the same point in time.
1.
1
Strictly speaking, compounded cash flow means moving money forward in time; that is, using
the TVOM to calculate a future value. This text, however, follows the generally accepted
practice of using the term DCF to refer to the movement of money forward or backward in time.
1-2
Engineering Economy Principles
3. To move money forward one time unit, multiply by 1 plus the discount
or interest rate.
4. To move money backward one time unit, divide by 1 plus the discount
or interest rate.
Applying the Four DCF Rules
EXAMPLE
Recall the previous example, where the student’s TVOM was 20 percent.
Suppose the student is guaranteed to receive $1,100 one year from today,
and nothing thereafter, if $1,000 is invested today in a particular venture.
What is the return on the student’s investment?
It would be a mistake for the student to subtract the $1,000 investment from
the $1,100 return and conclude that the investment yielded a net positive
return of $100. Why? Rule 1 establishes that money has a time value; for
this student, it can be represented by a 20 percent annual rate. Rule 2 establishes that the $1,000 investment cannot be subtracted from the $1,100
return, because they occur at different points in time.
So, what should the student do? Apply either Rule 3 or Rule 4. Using
Rule 3, the student would conclude that the future value or future worth of
the $1,000 investment, based on a 20 percent TVOM, equals $1,000 (1.20)
or $1,200 one year later. Because the $1,000 was an expenditure or investment, it is a negative cash flow, whereas the $1,100 return on the investment
was a positive cash flow. Hence, the net future worth of the investment is
2$1,200 1 $1,100, or 2$100. Because the future worth is negative, the
investment would not be considered a good one by the student.
Using Rule 4, the student would conclude that the present value or
present worth of $1,100 a year from now equals $1,100/1.20, or $916.67.
Therefore, the $1,000 investment yields a negative net present value of
$83.33. Again, the student should conclude that the investment was not a
good one. Note that different conclusions would have occurred if the student’s TVOM had been 8 percent instead of 20 percent.
By the way, if you did not understand the mathematics used here,
don’t worry. The mathematics of TVOM operations are covered in Chapters 2 and 3.
1-2
ENGINEERING ECONOMY PRINCIPLES
LEARNING O BJECTI VE: Identify the 10 principles of engineering economic
analysis that can be used by all engineers in analyzing the economic performance of the products, processes, and systems they design.
SOLUTION
7
8
Chapter 1
An Overview of Engineering Economic Analysis
Throughout this text, basic principles are presented that all engineers can
use in analyzing the economic performance of the products, processes, and
systems they design. No matter how impressive or how sophisticated an
engineering design might be, if it fails to ‘‘measure up’’ economically, it
usually is doomed to failure.
The following 10 principles of engineering economic analysis provide
a foundation for this text:
1. Money has a time value. This principle underlies almost everything we
2.
3.
4.
5.
cover in the text. Due to the TVOM, we prefer to receive a fixed sum
of money sooner rather than later; likewise, we prefer to pay a fixed
sum of money later, rather than sooner. (Notice how many of the following principles are corollaries of the TVOM principle.)
Make investments that are economically justified. The second principle is captured in a succinct statement attributed to Henry Ford in the
early 1900s: ‘‘If you need a new machine and don’t buy it, you pay for
it without ever getting it.’’ The key to his quote is the word need; need
indicates justification. The need can manifest itself in terms of cost
reductions that will occur if a new machine is purchased, or the need
can reflect the added business that will result from adding new manufacturing capacity or capability. Hence, if savings or revenues that will
easily offset the purchase price of a new machine are foregone, then
the new machine’s price is paid by continuing to incur higher costs
than will occur with the new machine or by passing up the profits that
will result from increased capacity or added capability.
Choose the mutually exclusive investment alternative that maximizes
economic worth. This is a corollary of the first and second principles.
The third principle addresses the situation when multiple investment
alternatives exist and only one can be chosen. We refer to such investments as mutually exclusive. Note that the third principle considers
only monetary aspects of the alternatives. Nonmonetary considerations may cause an alternative to be chosen that does not maximize
economic worth. (The third principle is discussed in Chapters 4, 5,
and 6.)
Two investment alternatives are equivalent if they have the same economic worth. The fourth principle is an extension of the third, which
states that for well-behaved cash flow profiles, the equivalence holds
only for the TVOM that equates their economic worths. (Chapters 2
and 6 examine this principle more closely.)
Marginal revenue must exceed marginal cost. The fifth principle
comes from a first course in economics. Based on this principle, one
should not make an investment unless the added revenues are greater
than the added costs. Based on the first principle, the TVOM must be
1-2
6.
7.
8.
9.
10.
Engineering Economy Principles
used in comparing marginal revenues and marginal costs if they occur
at different points in time.
Continue to invest as long as each additional increment of investment
yields a return that is greater than the investor’s TVOM. The sixth
principle, a corollary of the fifth principle, was verbalized in a statement made in 1924 by General Motors’ chief financial officer, Donald
Brown: ‘‘The object of management is not necessarily the highest rate
of return on capital, but . . . to assure profit with each increment of
volume that will at least equal the economic cost of additional capital
required.’’ The sixth principle can also be stated as follows: Use someone else’s money if you can earn more by investing it than you have
to pay to obtain it. Of course, one must consider the risks involved in
borrowing money in order to make good investments. (More will be
said about this principle in Chapters 6, 9, 10, and 12.)
Consider only differences in cash flows among investment alternatives.
In performing engineering economic analyses, decisions are made
between alternatives; hence, costs and revenues common to all investment alternatives can be ignored in choosing the preferred investment. (The seventh principle, which is also a corollary of the
fifth principle, is at the heart of Chapters 4 through 11.)
Compare investment alternatives over a common period of time. The
eighth principle is often violated. When alternatives have useful lives
that differ in duration, there is a temptation to compare the life cycle
of one investment with the unequal life cycle of another. As you will
learn, it is important to compare the alternatives over the same length
of time. (We address this principle in this chapter, Chapters 4 through
6, and in Chapter 12.)
Risks and returns tend to be positively correlated. The higher the risks
associated with an investment, the greater the anticipated returns must
be to justify the investment.
Past costs are irrelevant in engineering economic analyses, unless
they impact future costs. The tenth principle relates to past costs or
investments made previously. Past costs that have no carryover effect
in the future, also called sunk costs, must be ignored when performing
an engineering economic analysis. (Applications of this principle will
occur throughout the text.)
This text emphasizes the use of these 10 principles in choosing the best
investment to make and addresses specific kinds of engineering investments. What kind of investments do we consider? When an existing production machine or process must be replaced, many alternatives are usually available as replacement candidates; but which one is best? Alternative
candidates for replacement also exist when faced with replacing bridges,
9
10
Chapter 1
An Overview of Engineering Economic Analysis
transformers, telecommunications base stations, computers, road surfaces,
sewers, chemical mixers, furnaces, and so forth. Likewise, investments in
existing equipment—such as overhauling the equipment or adding new
features to extend its useful life or to add new production capability—are
included in the text.
In designing a new product, many design decisions involve choosing
from among alternatives. Some examples of choices include whether to
use standardized parts that can be purchased or specially designed parts
that must be produced; whether to enclose the product in a molded plastic
case or a formed metal case; whether to use standard, replaceable batteries
or a specially designed rechargeable battery; and whether to perform the
manufacturing and assembly in-house versus outsourcing the manufacturing and assembly. Decisions must also be made regarding materials to use
in construction and repair activities, as well as transportation alternatives
for moving people and materials. Discounted cash flow methods also play
a critical role in decisions regarding mergers, acquisitions, and disposition
of manufacturing plants. Regardless of which branch of engineering is
involved, numerous choices occur in designing and improving products,
processes, and systems.
Finally, while this text emphasizes engineering applications, the material it presents can be of great personal value. The principles provided can
be used to identify the best engineering design, product, process, or system, and to assist in personal investing. This is particularly true for the
material presented in Chapters 2 and 3.
1-3
ECONOMIC JUSTIFICATION
OF CAPITAL INVESTMENTS
LEARN I N G O B JEC T I V E : Identify the seven steps of the systematic economic
Video Lesson:
Systematic Economic
Analysis Technique
(SEAT)
analysis technique (SEAT) used to perform engineering economic analyses.
In performing engineering economic analyses, it is helpful to follow a consistent methodology. The following seven-step systematic economic analysis technique (SEAT) is recommended:
1.
2.
3.
4.
5.
6.
7.
Identify the investment alternatives.
Define the planning horizon.
Specify the discount rate.
Estimate the cash flows.
Compare the alternatives.
Perform supplementary analyses.
Select the preferred investment.
1-3
Economic Justification of Capital Investments
11
This text focuses particularly on step 5—comparing alternative investments
in order to identify the best one—although several of the others also will be
mentioned. Although typically not every step of this method is explicitly
performed in the examples and problems in this text, you will need to understand the important concepts behind the steps. With this end in mind, the
following sections take a closer look at each step.
1.3.1
SEAT Step 1 of 7: Identify the Investment Alternatives
Generally, the aim is to select the best investment from a feasible set of
mutually exclusive and collectively exhaustive investment alternatives.
“Mutually exclusive” as used here means ‘‘either/or but not both.’’
“Collectively exhaustive” means that no other investment alternatives are
available—all possible investments are considered.
The collectively exhaustive assumption is critical. Care must be taken
when forming the alternatives to ensure that all available alternatives are
being considered.
A ‘‘do nothing’’ alternative frequently is included in the set of feasible
investment alternatives to be compared. Such an alternative is intended to
represent ‘‘business as usual’’ or ‘‘maintaining the status quo.’’ Business conditions rarely stand still, however. Doing nothing does not mean that nothing
will be done; rather, it could mean that management has opted to forego the
opportunity to influence future events. The ‘‘do nothing’’ alternative often is
used as a baseline against which other investment alternatives are compared.
In this text, as a matter of convenience, zero incremental costs often are
associated to doing nothing. In practice, when the ‘‘do nothing’’ alternative
is feasible, extreme care must be taken not to underestimate the cost of doing
nothing. For many firms, business as usual is the most expensive alternative;
‘‘standing pat’’ for too long can be a disastrous course, because while the
firm is doing nothing, its competitors are generally doing something.
1.3.2
SEAT Step 2 of 7: Defining the Planning Horizon
As noted in the eighth principle of engineering economic analysis, it is
important to compare investment alternatives over a common period of time.
In this text, this period of time is referred to as the planning horizon. For
investments in, say, equipment to perform a required service, the period of
time over which the service is required might be used as the planning horizon. Likewise, with one-shot investment alternatives, the period of time over
which receipts continue to occur might define the planning horizon.
In a sense, the planning horizon defines the width of the ‘‘window’’
through which the economic performance of each investment alternative
will be viewed. Using too short a planning horizon could preclude an investment that would yield very sizeable returns in the long run. Conversely, too
Planning Horizon The
period of time or width of
the “window” over which
the economic performance
of each investment alternative will be viewed.
12
Chapter 1
An Overview of Engineering Economic Analysis
long a planning horizon can result in the firm going out of business before
realizing the promised long-term benefits.
When the lives of investment alternatives differ, five general approaches
are used to determine the planning horizon’s length:
1.
2.
3.
4.
5.
Set the planning horizon equal to the shortest life among the alternatives.
Set the planning horizon equal to the longest life among the alternatives.
Set the planning horizon equal to the least common multiple of the
lives of the various alternatives.
Use a standard length horizon equal to the period of time that best fits
the organization’s need, such as 10 years.
Use an infinitely long planning horizon.
Approach #3, the least common multiple of lives, is a popular choice. When
it is used, each alternative’s cash flow profile is assumed to repeat in the
future until all investment alternatives under consideration conclude at the
same time. Such an assumption might be practical for some applications,
but it can prove untenable for others. For example, assume Alternative A
has a useful life of 4 years, Alternative B has a useful life of 5 years, and
Alternative C has a useful life of 6 years. What is the least common multiple of 4, 5, and 6? The answer is 60. It does not seem realistic to assume that
identical cash flow profiles will repeat over a period of 60 years.
Using the shortest life approach (#1) would establish the planning horizon’s length for Alternatives A, B, and C as 4 years. With a 4-year planning
horizon, one must estimate the value of the one remaining year of useful
service for Alternative B. Similarly, one must estimate the value of the 2 years
of useful service available with Alternative C.
The longest life approach (#2) would establish 6 years as the planning
horizon for Alternatives A, B, and C. In this case, decisions must be made
about the 2-year gap for Alternative A and the 1-year gap for Alternative B.
Will identical replacements be made? If so, what will be their monetary
values at the end of the planning horizon?
Next, we consider the impact of a standard planning horizon of, say,
10 years (approach #4). As before, it is assumed that identical replacements will occur. For Alternative A, 10 years represents two complete life
cycles and one-half a life cycle. For the latter, a salvage value would have to
be estimated. For Alternative B, 10 years represents two complete life cycles.
For Alternative C, 10 years represents one full life cycle and one partial life
cycle; thus, requiring another salvage value estimate.
Finally, we consider the infinitely long planning horizon (#5). Obviously,
for many investments, it makes no sense to assume an indefinite planning
horizon. When the least common multiple of lives is quite long, however, an
infinitely long planning horizon will yield a reasonably good approximation.
1-3
Economic Justification of Capital Investments
13
1.3.3 SEAT Step 3 of 7: Specifying the Minimum Attractive
Rate of Return
The chapters that follow will examine many investment opportunities,
always using an interest rate to compound (move forward in time) or discount (move backward in time) cash flows. This interest rate is commonly
referred to as the minimum attractive rate of return or MARR. The
value used for the minimum attractive rate of return matters—a lot!
Because any investment will consume some portion of a firm’s scarce
resources, it is important for the investment to earn more than it costs to obtain
the investment capital. In addition, the MARR should reflect the opportunity
cost associated with investing in the candidate alternative as opposed to
investing in other available alternatives. In fact, it is assumed that investment
capital not committed to the candidate alternative is earning a return at least
equal to the MARR. Generally, a company has multiple sources of capital:
loans, bonds, stocks, and so on. Each has a different cost associated with it.
Because firms have multiple sources of capital, they typically calculate the weighted average cost of capital (WACC) and use it to establish
a lower bound on the MARR. It is only a lower bound because certain
unprofitable investments often are required. For example, investments
made to ensure compliance with governmental regulations, to enhance
employee morale, to protect lives, and to prevent injuries have positive
returns, particularly if social costs and social returns are included. It is difficult to quantify such returns, however. Even though the returns are less
than the WACC, these investments still will be made. For this reason,
optional investments must earn more than enough to cover the WACC.
Capital available to a corporation can be categorized as either debt
capital or equity capital. Examples of debt capital are bonds, loans, mortgages, and accounts payable; examples of equity capital are preferred
stock, common stock, and retained earnings. Typically, capital for a particular investment consists of a mixture of debt and equity capital.
Although many variations are available, a widely accepted WACC
formula is
WACC 5 (E/V ) ie 1 (D/ V) id (1 2 itr)
(1.1)
where E 5 a firm’s total equity, expressed in dollars
D 5 a firm’s total debt and leases, expressed in dollars
V 5 E 1 D, a firm’s total invested capital
ie 5 cost of equity or expected rate of return on equity, expressed
in percent
id 5 cost of debt or expected rate of return on borrowing, expressed
in percent
itr 5 corporate tax rate
Minimum Attractive Rate
of Return (MARR) Also
referred to as the hurdle
rate or discount rate, the
MARR is the minimum
rate of return on an
investment that a decision
maker is willing to accept
given the associated risk
and the opportunity cost of
other forgone investments.
Weighted Average Cost
of Capital (WACC)
The WACC is a value
calculated to establish the
lower bound on the MARR
and to take into account
that most firms have
multiple sources of capital.
14
Chapter 1
An Overview of Engineering Economic Analysis
The costs associated with debt capital are deductible from taxable income;
however, the costs associated with equity capital are not deductible. As a
result, (1 2 itr) is associated with debt capital but not equity capital.
1.3.4
SEAT Step 4 of 7: Estimate the Cash Flows
Once the planning horizon is determined, cash flow estimates are needed
for each investment alternative for each year of the planning horizon. The
Association for the Advancement of Cost Engineering International (www.
aacei.org) defines cost estimating as
a predictive process used to quantify, cost, and price the resources
required by the scope of an asset investment option, activity or
project. As a predictive process, estimating must address risks and
uncertainties. The outputs of estimating are used primarily as
inputs for budgeting, cost or value analysis, decision making in
business, asset and project planning, or for project cost and schedule
control processes.
Cost estimating is not an exact science. Rather, it is an approximation that
involves the availability and relevancy of appropriate historical data, personal judgments based on the estimator’s experience, and the time frame
available for completing the estimating activity.
1.3.5
SEAT Step 5 of 7: Comparing Alternatives
After the investment alternatives are identified, the planning horizon is
defined, the discount rate is specified, and the cash flows are estimated, it
is time to evaluate the alternatives in terms of their economic performance.
When doing the comparison, it is necessary to select a criterion to use.
Many options exist. In fact, two already have been presented. In Example
1.2, the investment was evaluated on the basis of its present value or present
worth, as well as its future value or future worth. Later chapters will consider present worth analysis in more detail, as well as benefit cost, payback period, discounted payback period, capitalized worth, future worth,
annual worth, and rate of return analyses.
Depending on the particular type of investment as well as the country
in which the investment is made, depreciation, income taxes, replacement,
and inflation may need to be considered in comparing the alternatives.
These topics also are covered in more detail in later chapters.
1.3.6
SEAT Step 6 of 7: Performing Supplementary Analyses
The sixth step in comparing investment alternatives is performing supplementary analyses. The intent of this step is to answer as many ‘‘what if’’
questions as possible. Up to this point, it has been assumed that the cash
flow estimates, the length of the planning horizon, and the TVOM used
Summary 15
were error free. Obviously, that will not always be the case. Conditions
change, errors are made, and risks and uncertainties exist. In this step, risk
and uncertainty are explicitly considered.
1.3.7 SEAT Step 7 of 7: Select the Preferred Investment
Selecting the preferred investment is the final step in a systematic engineering analysis. Because many factors must be considered in making the
selection, the preferred investment may not be the one that performs best
when considered using only the economic criteria.
Typically, multiple criteria exist rather than a single criterion of maximizing, say, present worth. The presence of multiple criteria coupled with
the risks and uncertainties associated with estimating future outcomes
makes the selection process quite complicated. To make this process easier,
the engineer is encouraged to address as many of management’s concerns
as possible in comparing the investment alternatives. If management’s
concerns are adequately addressed, the selection decision will agree with
the engineer’s recommendations.
The text concentrates on economic factors throughout. Keep in mind,
however, that management’s ultimate choice may be based on a host of
criteria rather than a single monetary criterion. Despite attempts to quantify all benefits in economic terms, some intangible factors or attributes
probably will not be reduced to dollars. Consider, for example, such factors as improved safety, reduced cycle times, improved quality, increased
flexibility, increased customer service, improved employee morale, being
the first in the industry to use a particular technology, and increased market visibility. Clearly, some of these factors are more readily measured in
economic terms than are others.
KEY CONCEPTS
1. Learning Objective: Apply the four discounted cash flow (DCF) rules to
simple time value of money (TVOM) situations. (Section 1.1)
The four DCF rules state that
1. Money has a time value.
2. Quantities of money cannot be added or subtracted unless they occur
at the same point in time.
3. To move money forward one time unit, multiply by 1 plus the discount
or interest rate.
4. To move money backward one time unit, divide by 1 plus the discount
or interest rate.
SUMMARY
16
Chapter 1
An Overview of Engineering Economic Analysis
2. Learning Objective: Identify the 10 principles of engineering economic
analysis that can be used by all engineers in analyzing the economic performance of the products, processes, and systems they design. (Section 1.2)
The 10 principles of engineering economic analysis can be summarized as
follows:
1. Money has a time value.
2. Make investments that are economically justified.
3. Choose the mutually exclusive investment alternative that maximizes
economic worth.
4. Two investment alternatives are equivalent if they have the same economic worth.
5. Marginal revenue must exceed marginal cost.
6. Continue to invest as long as each additional increment of investment
yields a return that is greater than the investor’s TVOM.
7. Consider only differences in cash flows among investment alternatives.
8. Compare investment alternatives over a common period of time.
9. Risks and returns tend to be positively correlated.
10. Past costs are irrelevant in engineering economic analyses, unless they
impact future costs.
3. Learning Objective: Identify the seven steps of the systematic economic
analysis technique (SEAT) used to perform engineering economic analyses.
(Section 1.3)
The seven-step systematic economic analysis technique (SEAT) used to
perform engineering economic analysis can be summarized as follows:
1. Identify the investment alternatives.
2. Define the planning horizon.
3. Specify the minimum attractive rate of return (MARR), also known as
the discount rate.
4. Estimate the cash flows.
5. Compare the alternatives.
6. Perform supplementary analyses.
7. Select the preferred investment.
KEY TERMS
Discounted Cash Flow (DCF), p. 6
Time Value of Money (TVOM), p. 5
Minimum Attractive Rate of Return, p. 13 Weighted Average Cost of Capital
(WACC), p. 13
Planning Horizon, p. 11
Summary
Problem available in WileyPLUS
GO Tutorial Tutoring Problem available in WileyPLUS
Video Solution Video Solution available in WileyPLUS
FE-LIKE PROBLEMS
1.
The fact that one should not add or subtract money unless it occurs at the
same point in time is an illustration of what concept?
a. Time value of money
c. Economy of scale
b. Marginal return
d. Pareto principle
2.
If a set of investment alternatives contains all possible choices that can
be made, then the set is said to be which of the following?
a. Coherent
c. Independent
b. Collectively exhaustive
d. Mutually exclusive
3.
Risks and returns are generally __________ correlated.
a. inversely
c. not
b. negatively
d. positively
4.
The “discounting” in a discounted cash flow approach requires the use of
which of the following?
a. An interest rate
b. The economic value added
c. The gross margin
d. The incremental cost
5.
Answering “what if” questions with respect to an economic analysis is
an example of which step in the systematic economic analysis technique?
a. Identifying the investment alternatives
b. Defining the planning horizon
c. Comparing the alternatives
d. Performing supplementary analysis
6.
If a student’s time value of money rate is 30 percent, then the student
would be indifferent between $100 today and how much in 1 year?
a. $30
c. $103
b. $100
d. $130
7.
Which of the following best represents the relationship between the
weighted average cost of capital (WACC) and the minimum attractive rate
of return (MARR)?
a. WACC and MARR are unrelated.
b. WACC is a lower bound for MARR.
c. WACC is an upper bound for MARR.
d. MARR # WACC.
17
18
Chapter 1
An Overview of Engineering Economic Analysis
PROBLEMS
Section 1.1 Time Value of Money
1.
Wylie has been offered the choice of receiving $5,000 today or an agreedupon amount in 1 year. While negotiating the future amount, Wylie notes that
he would be willing to take no less than $5,700 if he has to wait a year. What
is his TVOM in percent?
2. RT is about to loan his granddaughter Cynthia $20,000 for 1 year. RT’s
TVOM, based upon his current investment earnings, is 8 percent. Cynthia’s
TVOM, based upon earnings on investments, is 12 percent.
a. Should they be able to successfully negotiate the terms of this loan?
b. If so, what range of paybacks would be mutually satisfactory? If not, how far
off is each person from an agreement?
3.
RT is about to loan his granddaughter Cynthia $20,000 for 1 year. RT’s
TVOM, based upon his current investment earnings, is 12 percent, and he
has no desire to loan money for a lower rate. Cynthia is currently earning
8 percent on her investments, but they are not easily available to her, and she
is willing to pay up to $2,000 interest for the 1-year loan.
a. Should they be able to successfully negotiate the terms of this loan?
b. If so, what range of paybacks would be mutually satisfactory? If not, how
many dollars off is each person from reaching an agreement?
4. If your TVOM is 15 percent and your friend’s is 20 percent, can the two of
you work out mutually satisfactory terms for a 1-year, $3,000 loan? Assume
the lender has the money available and neither of you wants to go outside
their acceptable TVOM range. Be explicit about who is lending and what is
the acceptable range of money paid back on the loan.
Sections 1.2 and 1.3 Engineering Economy Principles and Economic
Justification of Capital Investments
5.
The following stages of a project are each contingent upon the preceding
stage. If the preceding stage is not performed (accepted), then none of the
subsequent stages may be performed.
Stage
1
2
3
4
5
Investment
A small FCC-licensed
commercial radio station
New antenna and hardline from transmitter
New amplifier to boost
from 5,000 watts to
30,000 watts
New lightningprotection equipment
New control console
Cost in
Today’s $
Revenues in
Today’s $
$180,000
$270,000
$30,000
$50,000
$40,000
$30,000
$30,000
$60,000
$15,000
$5,000
Summary 19
a. Remembering that no stages can be skipped, which set of the five stages
do you recommend be purchased?
b. Of the ten principles, which one(s) is well illustrated by this problem?
c. Of the systematic economic analysis technique’s seven steps, which one(s)
is well illustrated by this problem?
6. Barbara and Fred have decided to put in an automatic sprinkler system at
their cabin. They have requested bids, and the lowest price received is $5,500
from Water Systems Inc (WSI). They decide to do the job themselves and
obtain a set of materials (plastic pipe, nozzles, fittings, and regulators) from
an all-sales-are-final discount house for $1,100. They begin the installation
and rent a trencher at $80 per day. Unfortunately, they quickly hit sandstone
in many places of the yard and require a jackhammer and air compressor at
another $80 per day. They keep all the rental equipment for 5 days. By this
time, Fred has hurt his knee, and Barbara is sick of the project. They again
contact WSI, who tells them that they can use only some of the materials,
reducing the cost by $500, and only some of the trenching, reducing the cost
by another $500, bringing the total to $4,500, finished and ready to go.
a. How much have they already spent?
b. How much will they have spent when the project is over if they accept the
new offer from WSI?
c. A different contractor, Sprinkler Systems (SS), who heard of their situa-
tion approaches Barbara and Fred and recommends a design for a sprinkler system that would require a different set of materials and a new
routing of the trenches. They offer to (1) backfill all existing trenches,
(2) cut new trenches with their rock-impervious Ditch Witch, and (3) install
the system. Their charge is $6,000 for a finished-and-ready-to-go project,
and they correctly note that this is less than the total that will have been
spent if Barbara and Fred go with WSI. Should Barbara and Fred go with
WSI or SS? Why?
d. Of the ten principles, which one(s) is well illustrated by this problem?
e. Of the systematic economic analysis technique’s seven steps, which one(s)
is well illustrated by this problem?
7. List some nonmonetary factors in the alternative decision process that you
should be prepared to address when presenting a proposal to management.
Let your mind run free and think this out on your own, rather than trying to
find words that fit from the text.
a. Come up with 10 or more items.
b. Of the ten principles, which one(s) is well illustrated by this problem?
c. Of the systematic economic analysis technique’s seven steps, which one(s)
is well illustrated by this problem?
8. Four proposals (A, B, C, and D) are available for investment. Proposals A and
C cannot both be accepted; Proposal B is contingent upon the acceptance of
either Proposal C or D; and Proposal A is contingent on D.
a. List all possible combinations of proposals and clearly show which are
feasible.
20
Chapter 1
An Overview of Engineering Economic Analysis
b. Of the ten principles, which one(s) is well illustrated by this problem?
c. Of the systematic economic analysis technique’s seven steps, which one(s)
is well illustrated by this problem?
9.
Five proposals (V, W, X, Y, and Z) are available for investment. At least two
and no more than four must be chosen. Proposals X and Y are mutually exclusive. Proposal Z is contingent on either Proposal X or Y being funded. Proposal
V cannot be pursued if either W, X, Y, or any combination of the three is pursued.
a. List all feasible mutually exclusive investment alternatives.
b. Of the ten principles, which one(s) is well illustrated by this problem?
c. Of the systematic economic analysis technique’s seven steps, which
one(s) is well illustrated by this problem?
10. Three proposals (P, Q, and R) are available for investment. Exactly one or
two proposals must be chosen; Proposals P and Q are mutually exclusive.
Proposal R is contingent on Proposal P being funded. List all feasible mutually exclusive investment alternatives.
11.
AutoFoundry has contacted Centrifugal Casting Company about the
purchase of machines for the production of (1) Babbitt bearings, and (2)
diesel cylinder liners (engine sleeves). The cost of the machines prohibits
AutoFoundry from purchasing both, so they decide to base their selection on
which machine will provide the greatest net income on a “today” basis (also
known later as a present worth basis). The bearing machine has a life of 5 years,
after which it is expected to be replaced. It has “today” costs (considering
first cost and 5 years of operating cost, maintenance, etc.) of $460,000 and
provides new revenues over the 5 years of $730,000 in today’s dollars. The
cylinder liner machine has a life of 9 years, with “today” costs and new revenues over the 9-year life of $650,000 and $990,000, respectively.
a. What would you recommend that AutoFoundry do?
b. Of the ten principles, which one(s) is well illustrated by this problem?
c. Of the systematic economic analysis technique’s seven steps, which one(s)
is well illustrated by this problem?
12. Suppose you have been out of school and gainfully employed for 5 years.
You have three alternatives available for investment with your own money.
Each has some element of risk, although some are safer than others. Following is a summary of the alternatives, the risks, and the returns:
You Invest, $
Chance of
Success
A
$100,000
95%
$110,789.47
5%
$95,000
B
$100,000
60%
$150,000.00
40%
$50,000
C
$100,000
20%
$510,000.01
80%
$10,000
Alternative
a.
b.
c.
d.
Returned to You Chance of Returned to You
If Success, $
Failure
If Failure, $
Which would you select?
Why would you make this selection?
Of the ten principles, which one(s) is well illustrated by this problem?
Of the systematic economic analysis technique’s seven steps, which one(s)
is well illustrated by this problem?
Summary 21
13. A Payne County commissioner has $20,000 remaining in the budget to spend
on one of three worthy projects. Each is a one-time investment, and there
would be no follow-on investment, regardless of which project is chosen.
Project A involves the placement of gravel on a rough and often muddy road
leading to a public observatory, providing net benefits (consider this as net
revenue-in-kind) of $8,000 per year for 4 years, at which time the road will
again be in disrepair. Project B involves the building of a water-retention dam
to hold water during big rains, thereby lessening damage due to flash flooding; the benefits are expected to be worth $6,000 for each of 6 years, after
which silt will have made the pond ineffective. Project C is to provide water,
sewer, and electrical hookups for recreational vehicles at the fairgrounds; net
benefits of $4,000 per year would be realized for 10 years, after which the
system would need to be replaced. No matter which alternative is selected,
once its useful life is over, there will be no renewal.
a. What planning horizon should be used in evaluating these three projects?
b. Of the ten principles, which one(s) is well illustrated by this problem?
c. Of the systematic economic analysis technique’s seven steps, which one(s)
is well illustrated by this problem?
14.
Reconsider the county commissioner’s evaluation of three projects in
Problem 13. Take the facts as given, except now suppose the commissioner
can commit the county to renewing these investments, even if a different
commissioner is elected. So, after 4 years in project A, the road would be
renewed with gravel or perhaps even paved. After 6 years, the water-retention
dam could be dredged and renewed, or a new dam could be built. After 10 years,
the RV hookups could be modernized and replaced.
a. What are the considerations in selecting the appropriate planning horizon
in this case?
b. Of the ten principles, which one(s) is well illustrated by this problem?
c. Of the systematic economic analysis technique’s seven steps, which one(s)
is well illustrated by this problem?
ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E :
K ELLIE S C HN E I DE R AN ALY Z E S
HER NE T WO RT H
Immediately after receiving her engineering degree, Kellie Schneider began
employment with a multinational company. Her initial salary was $60,000
per year. Kellie decided she would invest 10 percent of her gross salary each
month. Based on discussions with the company’s personnel and with a person
in human resources, she believed her salary would increase annually at a
rate ranging from 3 percent to 15 percent, depending on her performance.
After analyzing various investment opportunities, she anticipated she would
be able to earn between 5 percent and 10 percent annually on her investment portfolio of mutual funds, stocks, bonds, U.S. Treasury notes, and certificates of deposit. Finally, Kellie believed annual inflation would vary from
2 percent to 5 percent over her professional career. With this information in
hand, she calculated the range of possible values for her net worth, first after
30 years of employment, and second after 40 years of employment. She was
amazed and pleased at what she learned.
DISCUSSION QUESTIONS:
1. What do you suppose Kellie is attempting to do with her investment
portfolio by selecting multiple forms of investments?
2. Although retirement may seem like a long time away, why is it important to start saving early for your “golden years?”
3. If interest rates end up being lower than Kellie assumes, what must
she do to build her goal “nest egg?”
4. What impacts will a change in your family status (such as getting married or having a child) have on your investment decisions?
22
TIME VALUE OF MONEY
CALCULATIONS
LEARNING OBJECTIVES
When you have finished studying this chapter, you should be able to:
1. Construct a cash flow diagram (CFD) depicting the cash inflows and
outflows for an investment alternative. (Section 2.1)
2. Perform time value of money calculations for single cash flows with
annual compounding. (Section 2.2)
3. Distinguish between uniform, irregular, gradient and geometric series
of cash flows. (Section 2.3)
4. Perform time value of money calculations for multiple cash flows including:
• Irregular series of cash flows with annual compounding. (Section 2.3.1)
• Uniform series of cash flows with annual compounding. (Section 2.3.2)
• Gradient series of cash flows with annual compounding. (Section 2.3.3)
• Geometric series of cash flows with annual compounding. (Section 2.3.4)
5. Perform time value of money calculations for multiple compounding
periods per year using the period interest rate and the effective annual
interest rate. (Section 2.4)
INTRODUCTION
In the previous chapter, we identified 10 principles of engineering economic analysis.
Principle #1 is the subject of this chapter: Money has a time value. As
noted in Chapter 1, TVOM considerations apply when moving money forward or backward in time. Recall, we referred to the movement of money
forward or backward in time as discounted cash flow or DCF. Also, recall
the following four DCF rules:
1. Money has a time value.
2. Quantities of money cannot be added or subtracted unless they occur
at the same points in time.
23
24
Chapter 2
Time Value of Money Calculations
3. To move money forward one time unit, multiply by 1 plus the discount
or interest rate.
4. To move money backward one time unit, divide by 1 plus the discount
or interest rate.
In this chapter, we present the mathematics and basic operations
needed to perform engineering economic analyses incorporating the DCF
rules. These are the same techniques that Kellie Schneider used to analyze
her future net worth. The only thing we will not cover in this chapter is
how to incorporate the effects of inflation in such an analysis; we save that
for Chapter 10. You will also learn how to determine the present worth and
the future worth for three particular types of cash flow series.
The material in this chapter serves as a foundation for the remainder
of the text. Hence, a solid understanding of the mathematics and concepts
contained in this chapter is essential.
2-1
CASH FLOW DIAGRAMS
LEARN I N G O B JEC T I V E : Construct a cash flow diagram (CFD) depicting the
Video Lesson:
Cash Flow Diagrams
Cash Flow Diagram
(CFD) A diagram
depicting the magnitude
and timing of cash flowing
in and out of the
investment alternative.
cash inflows and outflows for an investment alternative.
It is helpful to use cash flow diagrams (CFDs) when analyzing cash flows
that occur over several time periods. As shown in Figure 2.1, a CFD is
constructed using a segmented horizontal line as a time scale, with vertical
arrows indicating cash flows. An upward arrow indicates a cash inflow or
positive-valued cash flow, and a downward arrow indicates a cash outflow,
or negative-valued cash flow. The arrows are placed along the time scale to
correspond with the timing of the cash flows. The lengths of the arrows can
be used to suggest the magnitudes of the corresponding cash flows, but in
most cases, little is gained by precise scaling of the arrows.
$5,000
$5,000
$5,000
(+)
0
(–)
1
2
3
4
Time
$2,000
$3,000
$4,000
FIGURE 2 . 1 A Cash Flow Diagram (CFD)
5
2-1
Cash Flow Diagrams
The CFD in Figure 2.1 depicts an expenditure of $4,000, followed by the
receipt of $5,000, followed by an expenditure of $3,000, followed by a second receipt of $5,000, followed by a final expenditure of $2,000, followed by
a final receipt of $5,000. This CFD is drawn from the investor’s perspective:
Downward arrows denote expenditures, and upward arrows denote receipts.
A Simple Illustration of the Time Value of Money
EXAMPLE
As an illustration of how TVOM can impact the preference between investment alternatives, consider investment Alternatives A and B, having the cash
flow profiles depicted in Figure 2.2. The CFDs indicate that the positive cash
flows for Alternative A are identical to those for Alternative B, except that
the former occurs 1 year sooner; both alternatives require an investment of
$6,000. If exactly one of the alternatives must be selected, then Alternative
A would be preferred to Alternative B, based on the time value of money.
$3,000
(+)
0
1
(–)
$3,000
2
3
End of Year
4
$3,000
5
6
7
Alternative A
$6,000
$3,000
(+)
0
(–)
1
2
3
End of Year
$3,000
4
5
$3,000
6
7
Alternative B
$6,000
FIGURE 2.2
CFDs for Alternatives A and B
When faced with cash flows of equal magnitude occurring at different
points in time, a corollary to the four DCF rules in Chapter 1 is when
receiving a given sum of money, we prefer to receive it sooner rather than
later, and when paying a given sum of money, we prefer to pay it later
rather than sooner. Since we prefer to receive the $3,000 sooner, Alternative A is preferred to Alternative B.
25
26
Chapter 2
Time Value of Money Calculations
Compounding The
practice of calculating
interest on both the original
principal and any interest
accumulated to date.
Uniform Series A series
characterized by cash flows
of equal (or uniform) value
for each period.
In this chapter, we emphasize end-of-period cash flows and end-ofperiod compounding. Depending on the financial institution involved,
in personal finance transactions, savings accounts might not pay interest
on deposits made in ‘‘the middle of a compounding period.’’ Consequently, answers obtained using the methods we describe may not be
exactly the same as those provided by the financial institution.
Beginning-of-period cash flows can be handled easily by noting that
the end of period t is the beginning of period t 1 1. For example, one can
think of a payment made at the beginning of, say, March as having been
made at the end of February. In this and subsequent chapters, end-of-year
cash flows are assumed unless otherwise noted.
Drawing CFDs for economic transactions is important for at least two
reasons. First and foremost, CFDs are powerful communication tools. A
CFD presents a clear, concise, and unambiguous description of the amount
and timing of all cash flows associated with an economic analysis. Usually,
a well-drawn CFD can be readily understood by all parties of an economic
transaction regardless of whether they have had any formal training in
economic analysis.
A second important reason for drawing CFDs is that they frequently
aid in the identification of significant cash flow patterns that might exist
within an economic transaction. One such pattern, for example, is a uniform series where all of the cash flows are equally spaced and have the
same magnitude over a certain period of years.
2-2
Video Lesson:
Transformations—Single
Cash Flow
LEARN I N G O B JEC T I V E : Perform time value of money calculations for single
cash flows with annual compounding.
Our analysis of the time value of money begins with the simplest scenario:
a single cash flow.
2.2.1
Interest Rate The rate of
change in the accumulated
value of money.
SINGLE CASH FLOWS
Future Worth Calculations (F |P)
In nearly all business and lending situations, compound interest is used.
For that reason, interest should be charged (or earned) against both the
principal and accumulated interest to date. Such a process is called compounding. It is at the heart of everything else we do in the text.
When compound interest is used, the interest rate (i) is interpreted as
the rate of change in the accumulated value of money, and In is the accumulated interest over n years, given by
n
In 5 a iFt21
t51
(2.1)
2-2
where t increments the years from 1 to n, F0 5 P where P is the current or
present value of a single sum of money, Fn is the accumulated value of P
over n years, and
Fn 5 Fn21 11 1 i2
Single Cash Flows
Present Value The value
of money at a present
point in time.
(2.2)
Compounding Interest for 5 Years
EXAMPLE
Suppose you loan $10,000 for 1 year to an individual who agrees to pay
you interest at a compound rate of 10 percent/year. At the end of 1 year, the
individual asks to extend the loan period an additional year. The borrower
repeats the process several more times. Five years after loaning the person
the $10,000, how much would the individual owe you?
Video Lesson:
Simple & Compound
Interest
Given: Loan amount 5 $10,000; interest rate 5 10%/year; length of loan 5
5 years
Find: Value of the loan amount after 5 years
KEY DATA
As shown in Table 2.1, the $10,000 owed, compounded over a 5-year
period at 10 percent annual compound interest, totals $16,105.10.
SOLUTION
TABLE 2.1
Tabular Solution to Example 2.2
Unpaid Balance at the
Beginning of the Year
Annual
Interest
$10,000.00
$1,000.00
2
$11,000.00
3
$12,100.00
4
5
Year
1
Payment
Unpaid Balance at
the End of the Year
$0.00
$11,000.00
$1,100.00
$0.00
$12,100.00
$1,210.00
$0.00
$13,310.00
$13,310.00
$1,331.00
$0.00
$14,641.00
$14,641.00
$1,464.10
$16,105.10
$0.00
The previous example involved two cash flows: an amount borrowed and
an amount repaid. We can generalize the loan example and develop an
equation to determine the amount owed after n periods, based on a compound interest rate of i%/period, if P is borrowed. As shown in Table 2.2,
the future amount, F, owed is related to P, i, and n as follows:
F 5 P11 1 i2 n
(2.3)
where i is expressed as a decimal amount or as an equivalent percentage. As
a convenience in computing values of F (the future worth) when given values
of P (the present worth), the quantity (1 1 i)n is tabulated in Appendix A for
27
28
Chapter 2
Time Value of Money Calculations
TABLE 2.2
Derivation of Equation 2.3
(A)
(B)
(C) 5 (A) 1 (B)
End of
Period
Amount
Owed
Interest for
Next Period
Amount Owed for
Next Period*
0
P
Pi
1
P 11 1 i 2
P 11 1 i 2i
2
P 11 1 i 2 2
3
P 11 1 i 2 3
P 11 1 i 2 2i
A
A
A
n21
n
P 11 1 i 2
n21
P 11 1 i 2 n
P 11 1 i 2 1 P 11 1 i 2i 5 P 11 1 i 2 2
P 11 1 i 2 2 1 P 11 1 i 2 2i 5 P 11 1 i 2 3
P 11 1 i 2 3i
P 11 1 i 2
P 1 Pi 5 P 11 1 i 2
P 11 1 i 2 3 1 P 11 1 i 2 3i 5 P 11 1 i 2 4
A
n21
i
P 11 1 i 2
n21
1 P 11 1 i 2 n21i 5 P 11 1 i 2 n
*Notice, the value in column (C) for the end of period (n 2 1) provides the value in column (A) for
the end of period n.
various values of i and n. The quantity (1 1 i)n is referred to as the single
sum, future worth factor. It is denoted (F Z P i%,n) and reads ‘‘the F, given P
factor at i% for n periods.’’ The above discussion is summarized as follows.
Let
P 5 the equivalent value of an amount of money at a present point
in time, or present worth.
F 5 the equivalent value of an amount of money at a future point
in time, or future worth.
i 5 the interest rate per interest period.
n 5 the number of interest periods.
Thus, the mathematical relationship between the future and present worth
is given by Equation 2.3; the mnemonic representation is given by1
F 5 P1F Z P i%,n2
Future Value The value
of money at some future
point in time.
(2.4)
A CFD depicting the relationship between F and P is given in Figure 2.3.
Remember, F occurs n periods after P.
In addition to solving numerically for the value of (1 1 i)n and using
tabulated values in Appendix A of the single sum, future worth factor, (F Z P
i%,n), an Excel® financial function2 can be used to solve for the future
worth. Specifically, the FV, or future value, function can be used. The
parameters for the FV function are, in order of appearance, interest rate (i),
A comma may be placed between the factor identifier and the interest rate. Thus, (F Z P, i%,n)
and (F Z P i%,n) are equivalent representations of (1 1 i)n.
2
Microsoft’s Excel® software is used throughout the text. Where a computer is used to generate
a solution, software functions are shown in blue boldface type in the text.
1
2-2
Single Cash Flows
29
F
0
1
2
End of Period
n–1
n
F occurs n periods after P
P
F = P (F|P i%,n)
P = F (P|F i%,n)
number of periods (n), equal-sized cash flow per period (A), present
amount (P), and type, which denotes either end-of-period cash flows (0 or
omitted) or beginning-of-period cash flows (1).
To solve for F when given i, n, and P, the answer can be obtained by
entering the following in any cell in an Excel® spreadsheet: 5FV(i,n,,2P).
Notice there are no spaces between the equal sign and the closing parenthesis; also, two commas are placed between n and P in the function, since
no equal cash flow per period applies, and type is not included, since endof-period cash flows apply. Finally, notice that a negative value is entered
for P, since the sign of the value obtained for F by using the FV function
will be opposite the sign used for P. The reason for the sign change in
Excel® is simple: The FV function was developed for a loan situation
where $P are loaned (negative cash flow) in order to receive $F (positive
cash flow) n periods in the future when the money is loaned at i% interest
per period.
Recall Example 2.2, in which $10,000 was loaned for 5 years at
10 percent annual compound interest. An Excel® solution to the example is
shown in Figure 2.4. As indicated, the value of i can be entered as a decimal
The Excel® FV
Financial Functions Used with a
Single Sum of Money
F I GU R E 2 . 4
CFD of the
Time Relationship Between
P and F
F I G U RE 2 . 3
30
Chapter 2
Time Value of Money Calculations
or as a percentage, and the value of P can be entered as either a positive or
a negative amount, since the sign for the future value obtained will be opposite that used for P. Finally, notice that dollar signs and commas are not
used in denoting the value of P.
Repaying a 5-Year Loan with a Single Payment
EXAMPLE
Dia St. John borrows $1,000 at 12 percent compounded annually. The loan
is to be paid back after 5 years. How much should she repay?
KEY DATA
Given: Loan amount 5 $1,000; interest rate 5 12% compounded annually;
length of loan 5 5 years
Find: Value of the loan amount after 5 years
SOLUTION
Using the compound interest tables in Appendix A for 12 percent and
5 periods, the value of the single sum, future worth factor 1F Z P 12%,52 is
shown to be 1.76234. Thus,
F 5 P1F Z P 12%,52
5 $1,00011.762342
5 $1,762.34
Using the Excel® FV worksheet function,
F 5FV112%, 5,,210002 5 1762.34
A familiar application of a future worth analysis for a single cash flow is
to wonder how long it would take for an investment to double in value if
invested at i% compounded per period. It turns out there are several ways
of answering this question:
1.
2.
3.
4.
5.
6.
Obtain an approximation by using the Rule of 72.
Consult the tables in Appendix A for the stated interest rate and find
the value of n that makes the (F Z P i%,n) factor equal 2, then interpolate as necessary.
Solve mathematically for the value of n that makes (1 1 i)n equal 2.
Use the Excel® NPER worksheet function.
Use the Excel® GOAL SEEK tool.
Use the Excel® SOLVER tool.
Example 2.4 explores each of these methods.
2-2
Doubling Your Money
Single Cash Flows
EXAMPLE
How long does it take for an investment to double in value if it earns
(a) 2 percent, (b) 4 percent, or (c) 12 percent annual compound interest?
The first approach is to apply a rule of thumb called the Rule of 72. Specifically, the quotient of 72 and the interest rate provides a reasonably good
approximation of the number of interest periods required to double the
value of an investment:
a. n < 72/2 5 36 yrs
b. n < 72/4 5 18 yrs
c. n < 72/12 5 6 yrs
The second approach is to consult Appendix A and, for each value of i,
determine the value of n for which 1F Z P i%,n2 5 2.
a. For i 5 2%, n is between 30 and 36; interpolating gives
n < 30 1 612.0000 2 1.811362/ 12.03989 2 1.811362 5 34.953 yrs.
b. n < 17 1 12.0000 2 1.947902/ 12.02582 2 1.947902 5 17.669 yrs;
c. n < 6 1 12.0000 2 1.973822/ 12.21068 2 1.973822 5 6.111 yrs.
With the third approach, we solve mathematically for n such that (1 1 i)n 5
2 gives n 5 log 2/log11 1 i2. Therefore, the correct values of n (to 3
decimal places) are
a. n 5 log 2/log 1.02 5 35.003 yrs;
b. n 5 log 2/log 1.04 5 17.673 yrs; and
c. n 5 log 2/log 1.12 5 6.116 yrs.
The parameters of the Excel® NPER worksheet function are, in order of
placement, interest rate, equal-sized cash flow per period, present amount,
future amount, and type. As before, type refers to end-of-period (0 or omitted)
versus beginning-of-period (1) cash flows. Letting F equal 2 and P equal −1,
the NPER function yields identical results to those obtained mathematically:
a. n 5 NPER12%,,21,22 5 35.003 yrs
b. n 5 NPER14%,, 21,22 5 17.673 yrs and
c. n 5 NPER112%,, 21,22 5 6.116 yrs
To use the Excel® GOAL SEEK tool requires a spreadsheet, as shown in
Figure 2.5. Letting x denote the row number in the spreadsheet, the parameters for GOAL SEEK are the following:
Set cell: Cx
To value: 2
By changing cell: Bx
SOLUTION
31
32
Chapter 2
Time Value of Money Calculations
Any number is entered in cell Bx, and the future value of $1, invested at
interest rate Ax for Bx number of years, is calculated using the Excel®
FV worksheet function. Then, the Excel® GOAL SEEK tool is used to
determine the value of Bx that makes Cx 5 2. The results obtained by
GOAL SEEK are shown in Figure 2.5. Namely,
a. n 5 34.999 yrs
b. n 5 17.672 yrs
c.
n 5 6.116 yrs
FIGURE 2 . 5
The Excel® Goal Seek Tools Used to Solve Example 2.4
The same spreadsheet used with GOAL SEEK can be used with the
Excel® SOLVER tool, as shown in Figure 2.6. The SOLVER parameters
are the following:
Set Target Cell: Cx
Equal To: s Max s Min d Value of: 2
By Changing Cells:
Bx
As with GOAL SEEK, any number is entered in cell Bx, and the future
value of $1, invested at interest rate Ax for Bx number of years, is calculated using the Excel® FV worksheet function. Then, the Excel® SOLVER
tool is used to determine the value of Bx that makes Cx 5 2, where x
denotes the row number in the spreadsheet. The results obtained by
SOLVER are shown in Figure 2.6. Namely,
a. n 5 35.003 yrs
b. n 5 17.673 yrs
c.
n 5 6.116 yrs
2-2
FIGURE 2.6
Single Cash Flows
The Excel® Solver Tool Used to Solve Example 2.4
The answers obtained using GOAL SEEK and SOLVER differ from each
other and those obtained using the Excel® NPER worksheet function,
especially if the calculation is carried out to 8 or 10 decimal places. GOAL
SEEK and SOLVER use a search procedure that can end prematurely
(i.e., before obtaining an exact solution). Of the six approaches to solving
this example, only the mathematical one using logarithms, and the Excel®
NPER function yielded exact solutions.
2.2.2 Present Worth Calculations (P |F)
Since we can determine values of F when given values of P, i, and n, it is
a simple matter to determine the values of P when given values of F, i, and
n. In particular, from Equation 2.3,
F 5 P11 1 i2 n
(2.5)
Dividing both sides of Equation 2.5 by (1 1 i)n, we have the relation,
P 5 F11 1 i2 2n
(2.6)
EXPLORING THE
SOLUTION
33
34
Chapter 2
Time Value of Money Calculations
or
P 5 F1P ƒ F i%,n2
(2.7)
where (1 1 i)–n and (P Z F i%,n) are referred to as the single sum, present
worth factor.
In addition to solving numerically for the value of (1 1 i)–n and using
tabulated values of the single sum, present worth factor, (P Z F i%,n), provided in Appendix A, an Excel® financial function can be used to solve for
the present worth—specifically, the PV, or present value function. The
parameters of the PV function, in order, are interest rate (i), number of
periods (n), equal-sized cash flow per period (A), future amount (F), and
type, which denotes either end-of-period cash flows (0 or omitted) or
beginning-of-period cash flows (1).
To solve for P when given i, n, and F, the following can be entered in
any cell in an Excel® spreadsheet: 5PV(i,n,,2F). Notice, as with the FV
function, there are no spaces between the equal sign and the closing parenthesis; also, as before, since no equal-sized cash flow per period occurs,
two commas are placed between n and F. Again, the sign of the value
obtained for P when using the PV function will be opposite the sign of the
value of F that is entered in the cell.
Saving Money
EXAMPLE
To illustrate the computation of P given F, i, and n, suppose you wish
to accumulate $10,000 in a savings account 4 years from now, and the
account pays interest at a rate of 5 percent compounded annually. How
much must be deposited today?
KEY DATA
SOLUTION
Given: F 5 $10,000; i 5 5% compounded annually; n 5 4 years
Find: P
Using the compound interest tables in Appendix A for 5 percent and
4 periods, the value of the single sum, present worth factor, 1P Z F 5%,42, is
shown to be 0.82270. Thus,
P 5 F1P Z F 5%,42
5 $10,00010.822702
5 $8,227.00
As shown in Figure 2.7, using the Excel® PV worksheet function,
P 5PV(5%,4,,210000) 5 $8227.02
2-3 Multiple Cash Flows
35
The Excel® PV Financial
Function Used with a Single Sum of
Money
FIGURE 2.7
2-3
MULTIPLE CASH FLOWS
LEARNING OBJECTIVES: Distinguish between uniform, irregular, gradient
and geometric series of cash flows. Perform time value of money calculations
for multiple cash flows with annual compounding.
So far we have considered present and future values of single cash flows. In
this section we will extend our analysis to multiple cash flows. We begin with
multiple cash flows that do not exhibit a pattern from one to another. We follow with cash flow series that form a pattern, allowing the use of shortcuts in
determining the present worth and future worth. We will look at three such
series as well: the uniform series, the gradient series, and the geometric series.
2.3.1 Irregular Cash Flows
LEARNING O BJECTI VE: Perform time value of money calculations for irreg-
ular cash flows with annual compounding.
Most engineering economic analyses involve more than a single return occurring after an investment is made. In such cases, the present worth equivalent
of the future cash flows can be determined by adding the present worth of the
individual cash flows. Similarly, the future worth of multiple cash flows can
be determined by adding the future worth of the individual cash flows.
If we let At denote the magnitude of a cash flow (receipt or disbursement) at the end of time period t, then
P 5 A1 11 1 i2 21 1 A2 11 1 i2 22 1 A3 11 1 i2 23 1 p 1 An21 11 1 i2 21n212
1 An 11 1 i2 2n
(2.8)
or, using summation notation,
n
P 5 a At 11 1 i2 2t
t51
(2.9)
Video Lesson:
Transformations—Multiple
Cash Flows
36
Chapter 2
Time Value of Money Calculations
or, equivalently,
n
P 5 a At 1P Z F i%,t2
(2.10)
t51
Computing the Present Worth of a Series of Cash Flows
EXAMPLE
Video Example
KEY DATA
SOLUTION
Consider the series of cash flows depicted by the CFD given in Figure 2.8.
Using an interest rate of 6 percent per interest period, what is the present
worth equivalent of cash flows?
Given: Cash flow series in Figure 2.8; i 5 6% per interest period.
Find: P
The present worth equivalent is given by
P 5 $3001P Z F 6%,12 2 $3001P Z F 6%,32 1 $2001P Z F 6%,42
1 $4001P Z F 6%,62 1 $2001P Z F 6%,82
5 $30010.943402 2 $30010.839622 1 $20010.792092
1 $40010.704962 1 $20010.627412
5 $597.02
$400
$300
(+)
$200
0
1
2
3
4
5
End of Period
$200
6
7
8
(–)
$300
FIGURE 2.8
CFD of Multiple Cash Flows
The future worth equivalent of a cash flow series is equal to the sum of the
future worth equivalents for the individual cash flows. Thus,
F 5 A1 11 1 i2 n21 1 A2 11 1 i2 n22 1 A3 11 1 i2 n23 1 p 1 An22 11 1 i2 2
1 An21 11 1 i2 1 An
(2.11)
2-3 Multiple Cash Flows
or, using summation notation,
n
F 5 a At 11 1 i2 n2t
(2.12)
t51
or, equivalently,
n
F 5 a At 1F Z P i%,n 2 t2
(2.13)
t51
Notice, in Equations 2.11 and 2.12, the exponent of the interest factor
counts the number of periods between the cash flow and the time period
where F is located. By convention, the future worth coincides in time with
the nth or last cash flow; as such, it does not draw interest, as shown by An
in Equation 2.11.
Alternately, since we know the value of future worth is given by
F 5 P11 1 i2 n
(2.14)
substituting Equation 2.9 into Equation 2.14 yields
n
F 5 11 1 i2 n a At 11 1 i2 2t
t51
Hence,
n
F 5 a At 1F Z P i%,n 2 t2
(2.15)
t51
Determining the Future Worth of a Series of Cash Flows
Given the series of cash flows in Figure 2.8 from example 2.6, determine
the future worth at the end of the eighth period using an interest rate of
6 percent per interest period.
Given: Cash flow series in Figure 2.8; i 5 6% per interest period.
Find: F
F 5 $3001F Z P 6%,72 2 $3001F Z P 6%,52 1 $2001F Z P 6%,42
1 $4001F Z P 6%,22 1 $200
5 $30011.503632 2 $30011.338232 1 $20011.262482
1 $40011.123602 1 $200
5 $951.56
EXAMPLE
Video Example
KEY DATA
SOLUTION
37
38
Chapter 2
Time Value of Money Calculations
Alternately, since we know the present worth is equal to $597.02,
F 5 $597.021F Z P 6%,82
5 $597.0211.593852
5 $951.56
Obtaining the present worth and future worth equivalents of a cash flow
series by summing the individual present worth and future worth, respectively, can be time consuming if many cash flows are included in the series.
However, the Excel® NPV financial function is well suited for determining the present worth of a series of cash flows. This function computes the
net present value or net present worth of a specified range of cash flows.
Importantly, the value obtained occurs one time period prior to the first
cash flow in the range given. Its parameters, in order, are interest rate (i),
followed by the individual cash flows.
Using the Excel NPV Function
EXAMPLE
Video Example
Recall Example 2.6, depicted in Figure 2.8. The Excel® NPV worksheet
function is well suited for such a problem. However, as shown in Figure 2.9,
no blank cells can be included in the range of cash flows, and every time
period must be accounted for. Finally, since the cash flow at time zero is, in
fact, zero, there is no need to add the value of C3 to the NPV calculation. As
FIGURE 2 . 9
Example 2.6
The Excel® NPV and FV Financial Functions Used to Solve
2-3
Multiple Cash Flows
before, the future worth can be obtained by using the NPV value. Notice, the
value obtained for the present worth, $597.02, is within 2¢ of the value
obtained in Example 2.6 using the interest tables in Appendix A, and the
value obtained for the future worth, $951.56, is identical to the value obtained
in Example 2.7 using the interest tables in Appendix A.
2.3.2
Uniform Series of Cash Flows
LEARNING O BJECTI VE: Perform time value of money calculations for a
uniform series of cash flows with annual compounding.
A uniform series of cash flows exists when all cash flows in a series are
equally sized and spaced. In the case of a uniform series the present worth
equivalent is given by
n
P 5 a A11 1 i2 2t
(2.16)
t51
where A is the magnitude of an individual cash flow in the series.
Letting X 5 (1 1 i)21 and bringing A outside the summation yields
n
P 5 A a Xt
t51
n
5 AX a X t21
t51
Letting h 5 t 2 1 gives the geometric series
n21
P 5 AX a Xh
(2.17)
h50
Since the summation in Equation 2.17 represents the first n terms of a
geometric series, the closed-form value for the summation is given by
n21
h
aX 5
h50
1 2 Xn
12X
(2.18)
Replacing X with (1 1 i)21 yields the following relationship between P and A
11 1 i2 n 2 1
(2.19)
P5A
i11 1 i2 n
more commonly expressed as
P 5 A1P Z A i%,n2
(2.20)
where (P Z A i%,n) is referred to as the uniform series, present worth factor
and is tabulated in Appendix A for various values of i and n.
Video Lesson:
Transformations—
Uniform Series
39
40
Chapter 2
Time Value of Money Calculations
Computing the Present Worth of a Uniform Series of Cash Flows
EXAMPLE
Troy Long wishes to deposit a single sum of money in a savings account
so that five equal annual withdrawals of $2,000 can be made before depleting the fund. If the first withdrawal is to occur 1 year after the deposit and
the fund pays interest at a rate of 5 percent compounded annually, how
much should he deposit?
KEY DATA
SOLUTION
Given: A 5 $2,000; i 5 5%; n 5 5
Find: P
Because of the relationship of P and A, as depicted in Figure 2.10, in which
P occurs one period before the first A, we see that
P 5 $2,0001P Z A 5%,52
5 $2,00014.329482
5 $8,658.96
0
A
A
1
2
End of Period
A
A
n–1
n
P occurs one period before the first A
P
P = A(P|A i%,n)
A = P(A|P i%,n)
FIGURE 2.10
CFD of the Relationship Between P and A in a Loan Transaction
Thus, if $8,658.96 is deposited in a fund paying 5 percent compounded
annually, then five equal annual withdrawals of $2,000 can be made. After
the fifth withdrawal, the fund will be depleted.
The Excel® PV function can be used to determine the present worth
of a uniform series. Recall, when we used the PV function previously, we
included two commas in consecutive order because equal-sized cash flows
per period were not present for the application. Now, with a uniform series
of cash flows being present, an entry is required for the third parameter
(A). Therefore, when the present worth of a uniform series is desired, given
i, n, and A, the following can be entered in any cell in an Excel® spreadsheet: 5PV1i,n,2A2. As before, a negative sign is used for A because the
PV function reverses the sign of A when calculating the value of P.
Using the Excel® PV worksheet function,
P 5PV15%,5,220002 5 8658.95
2-3 Multiple Cash Flows
Computing the Present Worth of a Delayed Uniform Series
of Cash Flows
EXAMPLE
In Example 2.9, suppose the first withdrawal does not occur until 3 years
after the deposit. How much should be deposited?
Given: A 5 $2000; i 5 5%; n 5 5
Find: P at time 0
KEY DATA
As depicted in Figure 2.11, the value of P to be determined occurs at t 5 0,
whereas a straightforward application of the (P Z A 5%,5) factor will yield a
single sum equivalent at t 5 2, one period before the first A. Hence, to
determine the present worth, the value obtained after using the (P Z A 5%,5)
factor must be moved backward in time 2 years. The latter operation is
performed using the (P Z F 5%,2) factor. Therefore,
SOLUTION
P 5 $2,0001P Z A 5%,52 1P Z F 5%,22
5 $2,00014.329482 10.907032
5 $7,853.94
(+)
0
$2,000
1
(–)
2
$2,000
3
4
End of Year
$2,000
$2,000
$2,000
5
6
7
i = 5%
FIGURE 2.11
CFD for Example 2.10
Again, the Excel® PV function can be used to determine the present worth
for this example. The example also provides an opportunity to show how
an Excel® worksheet function can be embedded in another Excel® worksheet function. Specifically, for this example, the following can be entered
in any cell in an Excel® spreadsheet:
P 5PV15%,2,,2PV15%,5,220002 2 5 $7853.93
Notice, two PV function calculations are performed; we will refer to them
as the “outside” PV function and the “inside,” or embedded, PV function.
By entering two commas in the outside PV function, the next entry is a single sum, future worth amount. (Notice, as before, the use of a negative sign.)
41
42
Chapter 2
Time Value of Money Calculations
Instead of entering a dollar amount for the single sum, future worth amount,
we entered the inside PV function to calculate the present worth of the uniform series over time periods 3 through 7. Since the value obtained by the
inside PV function occurred at the end of the second time period, a value of
2 was entered for n in the outside PV function.
The reciprocal relationship between P and A can be expressed as
i11 1 i2 n
A5P
11 1 i2 n 2 1
(2.21)
or as
A 5 P1A Z P i%,n2
(2.22)
Capital Recovery Factor
The amount of annual
savings or recovery funds
required to justify a capital
investment over some
period of time, n.
The expression (A Z P i%,n) is called the capital recovery factor, since it
provides the annual savings or recovery of funds (A) required to justify the
capital investment (P). The (A Z P i%,n) factor is used frequently in both
personal financing and in comparing economic investment alternatives.
EXAMPLE
What Size Uniform Withdrawals Can Occur?
Video Example
KEY DATA
SOLUTION
Suppose Rachel Townsley deposits $10,000 into an account that pays 8 percent interest compounded annually. If she withdraws 10 equal annual
amounts from the account, with the first withdrawal occurring 1 year after
the deposit, how much can she withdraw each year in order to deplete the
fund with the last withdrawal?
Given: P 5 $10,000; i 5 8% compounded annually; n 5 10
Find: A
Since we know that A and P are related by
A 5 P1A Z P i%,n2
then
A 5 $10,0001A Z P 8%,102
5 $10,00010.149032
5 $1490.30
2-3 Multiple Cash Flows
The Excel® PMT worksheet function can be used to determine the uniform series equivalent to a single sum, present amount. (PMT is shorthand
for “payment.”) This function computes the size of a payment when P is
borrowed. The parameters of the PMT function, in order of occurrence,
are interest rate, number of periods, present amount, future amount, and
type. As with the PV and FV functions, type indicates end-of-period or
beginning-of-period. To obtain the uniform series equivalent of a single
sum, present amount, the latter two parameters are omitted.
Using the Excel® PMT worksheet function,
A 5PMT18%,10,2100002 5 1490.29
Determining the Size of Delayed Uniform Withdrawals
EXAMPLE
Suppose, in Example 2.11, the first withdrawal is delayed for 2 years, as
depicted in Figure 2.12. How much can be withdrawn each of the 10 years?
(+)
0
1
(–)
2
A
A
A
A
A
A
A
A
A
A
3
4
5
6
7
End of Year
8
9
10
11
12
i = 8%
$10,000
FIGURE 2.12
CFD for the Deferred Payment Example
Given: P 5 $10,000; i 5 8% compounded annually; n 5 10
Find: Amount in the fund at t 5 2 (V2); then find A
KEY DATA
The amount in the fund at t 5 2 equals
SOLUTION
V2 5 $10,0001F Z P 8%,22
5 $10,00011.166402
5 $11,664
43
44
Chapter 2
Time Value of Money Calculations
Therefore, the size of the equal annual withdrawal will be
A 5 $11,664.001A Z P 8%,102
5 $11,664.0010.149032
5 $1,738.29
The Excel® FV worksheet function can be used to determine the future
worth of a uniform series. To solve the example problem, one Excel®
worksheet function will be embedded in another. Specifically, the following can be entered in any cell in an Excel® spreadsheet:
A 5PMT18%,10,2FV18%,2,,2100002 2 5 $1738.28
Notice, in this case, the FV function was embedded in the PMT function.
The “inside” financial function calculated the future value of $10,000
moved forward 2 years at 8 percent annual compound interest.
Instead of using the embedded approach, separate Excel® calculations could be performed, as shown below:
V2 5FV18%,2,,2100002 5 $11,664.00
A 5PMT18%,10,2116642 5 $1738.28
The future worth of a uniform series is obtained by recalling that
F 5 P11 1 i2 n
(2.23)
Substituting Equation 2.19 into Equation 2.23 for P and reducing yields
11 1 i2 n 2 1
F5A
(2.24)
i
or, equivalently,
F 5 A1F Z A i%,n2
(2.25)
where (F Z A i%,n) is referred to as the uniform series, future worth factor.
EXAMPLE
Determining the Future Worth of a Uniform Series of Cash Flows
If Luis Jimenez makes annual deposits of $1,000 into a savings account for
30 years, how much will be in the fund immediately after his last deposit
if the fund pays 6 percent interest compounded annually?
2-3
Given: A 5 $1000; i 5 6% compounded annually; n 5 30
Find: F
Multiple Cash Flows
KEY DATA
F 5 $1,0001F Z A 6%,302
5 $1,000179.058192
5 $79,058.19
SOLUTION
The Excel® FV worksheet function can be used to determine the future
worth of a uniform series. Using the Excel® FV worksheet function,
F 5FV16%,30,10002 5 $78,058.19
The reciprocal relationship between A and F is easily obtained from Equation 2.24. Specifically, we find that
A5F
i
11 1 i2 n 2 1
(2.26)
or, equivalently,
A 5 F1A Z F i%,n2
(2.27)
The expression (A Z F i%,n) is referred to as the sinking fund factor, since
it is used to determine the size of a deposit to place (sink) into a fund in
order to accumulate a desired future amount. As depicted in Figure 2.13, F
occurs at the same time as the last A. Thus, the last A or deposit earns no
interest.
F occurs at the same time as the last A
F
F = A(F|A i%,n)
A = F(A|F i%,n)
End of Period
0
1
2
n–1
A
A
A
n
A
CFD of the Relationship Between A and F in an
Investment Scenario
FIGURE 2.13
Sinking Fund Factor
The amount of savings
to deposit in order to
accumulate a desired
future amount.
45
46
Chapter 2
Time Value of Money Calculations
Who Wants to Be a Millionaire?
EXAMPLE
Suppose Crystal Wilson wants to accumulate $1,000,000 by the time she
retires in 40 years. If she earns 10 percent on her investments, how much
must she invest each year in order to realize her goal?
KEY DATA
SOLUTION
Given: F 5 $1,000,000; i 5 10%; n 5 40
Find: A
Applying Equation 2.27,
A 5 $1,000,0001A Z F,10%,402
5 $1,000,00010.00225942
5 $2,259.40/year
Using the Excel® PMT worksheet function gives
A 5PMT110%,40,,210000002 5 $2,259.41/year
2.3.3 Gradient Series of Cash Flows
LEARNING OBJECTIVE: Perform time value of money calculations for a
Video Lesson:
Transformations—Gradient
and Geometric Series
Gradient Series A series
characterized by cash flows
that increase by a constant
amount (G) each period.
gradient series of cash flows with annual compounding.
A gradient series of cash flows occurs when the value of a given cash flow
is greater than the value of the previous cash flow by a constant amount, G,
the gradient step. Consider the series of cash flows depicted in Figure
2.14. The series can be represented by the sum of a uniform series and a
gradient series. By convention, the gradient series is defined to have the
first positive cash flow occur at the end of the second time period. The size
of the cash flow in the gradient series occurring at the end of period t is
given by
At 5 1t 2 12G
t 5 1, p , n
(2.28)
As an illustration, if an individual receives an annual bonus and the size of
the bonus increases by $100 each year, then the series is gradient. Also,
operating and maintenance costs tend to increase over time because of
inflation and a gradual deterioration of equipment; such costs often are
approximated by a gradient series.
2-3
Composite series
A1 + (n – 2)G
A1
0
1
A1 + G
2
A1 + (n – 1)G
A1 + 2G
...
3
End of Period
n–1
n
P = P1 + P2
=
0
A1
A1
1
2
Uniform series
A1
...
3
End of Period
A1
n–1
n
+
P1
Gradient series
(n – 2)G
G
0
A1
1
2
(n – 1)G
2G
...
3
End of Period
n–1
n
P2
CFD of a Combination of Uniform and Gradient Series in an
Investment Scenario
FIGURE 2.14
The present worth equivalent of a gradient series is obtained by
recalling
n
P 5 a At 11 1 i2 2t
t51
(2.29)
Multiple Cash Flows
47
48
Chapter 2
Time Value of Money Calculations
Substituting Equation 2.28 into Equation 2.29 gives
n
P 5 G a 1t 2 12 11 1 i2 2t
(2.30)
t51
which reduces to
P 5 Gc
1 2 11 1 ni2 11 1 i2 2n
d
i2
(2.31)
or, equivalently,
P 5 G1P Z G i%,n2
(2.32)
where (P Z G i%,n) is the gradient series, present worth factor and is tabulated in Appendix A.
The uniform series equivalent to the gradient series is obtained by
multiplying the value of the gradient series, present worth factor by the
value of the (A Z P i%,n) factor:
A 5 Gc
n
1
2 1A Z F i%,n2 d
i
i
or, equivalently,
A 5 G1A Z G i%,n2
(2.33)
where the factor (A Z G i%,n) is referred to as the gradient-to-uniform series
conversion factor and is tabulated in Appendix A.
To obtain the future worth equivalent of a gradient series at time n,
multiply the value of the (A Z G i%,n) factor by the value of the (F Z A i%,n)
factor to obtain the (F Z G i%,n) factor:
11 1 i2 n 2 11 1 ni2
d
F 5 Gc
i2
(2.34)
The (F Z G i%,n) gradient series, future worth factor is not tabulated in
Appendix A.)
Often a cash flow series is the sum or difference of a uniform series
and a gradient series. To determine the present and future worth equivalents of such a composite, one can deal with each special type of series
separately.
2-3 Multiple Cash Flows
Determining the Present Worth of a Gradient Series
(Sitting on Top of a Uniform Series)
Maintenance costs for a particular production machine increase by $1,000/
year over the 5-year life of the equipment. The initial maintenance cost is
$3,000. Using an interest rate of 8 percent compounded annually, determine the present worth equivalent for the maintenance costs.
For this problem, it is very helpful to sketch the cash flow diagram. As
shown in Figure 2.15, the original sketch (composite series) can then be
converted into two parts showing the uniform and gradient portions,
respectively, of the series. Visualizing the flows in this way suggests an
efficient approach to determining the solution.
Given: The maintenance costs—a composite series of cash flows as shown in
Figure 2.15
Find: P of the composite series
Composite series
0
1
2
3
4
5
$3,000
$4,000
$5,000
$6,000
$7,000
=
Uniform series
0
1
2
3
4
5
$3,000
$3,000
$3,000
$3,000
$3,000
+
Gradient series
0
1
2
$1,000
3
4
5
$2,000
$3,000
$4,000
FIGURE 2.15
CFD for Example 2.15
EXAMPLE
Video Example
KEY DATA
49
50
Chapter 2
Time Value of Money Calculations
SOLUTION
Note that the composite series may be converted to a uniform series plus a
gradient series (see Figure 2.15).
Converting the gradient series to an equivalent uniform series gives
AG 5 $1,0001A Z G 8%,52
5 $1,00011.846472
5 $1,846.47
(Notice that n equals 5 even though only four positive cash flows are present
in the gradient series.)
Adding the “converted” uniform series to the “base” uniform series
gives A 5 $1,846.47 1 $3,000, or $4,846.47. Therefore, converting the
uniform series to its present worth equivalent,
P 5 $4,846.471P Z A 8%,52
5 $4,846.4713.992712
5 $19,350.55
To determine the future worth equivalent, we can use either the uniform
series or present worth equivalent. Using the uniform series equivalent
gives
F 5 $4,846.471F Z A 8%,52
5 $4,846.4715.866602
5 $28,432.30
Similarly, using the present worth equivalent, the future worth equivalent is
F 5 $19,350.551F Z P 8%,52
5 $19,350.5511.469332
5 $28,432.34
Excel® does not have a special worksheet function for gradient series. However, the NPV worksheet function can be used. Specifically, the following
entry can be made in any cell: 5NPV18%,3000,4000,5000,6000,70002 or
51000*NPV(8%,3,4,5,6,7). The result is a present worth of $19,350.56;
once the present worth is known, the uniform series equivalent and future
worth equivalent can be obtained using the PMT and FV worksheet
functions:
A 5PMT18%,5,219350.562 5 $4846.47
F 5FV18%,5,,219350.562 5 $28,432.32
2-3
2.3.4
Multiple Cash Flows
51
Geometric Series of Cash Flows
LEARNING OBJECTIVE: Perform time value of money calculations for a
geometric series of cash flows with annual compounding.
A geometric series of cash flows as depicted in Figure 2.16, occurs when
the size of a cash flow increases (decreases) by a fixed percent from one
time period to the next. If j denotes the percent change in a cash flow’s size
from one period to the next, the size of the tth cash flow can be given by
At 5 At21 11 1 j2
t 5 2, p , n
or, more conveniently,
At 5 A1 11 1 j2 t21
t 5 1, p , n
(2.35)
The geometric series is used to represent the growth (positive j) or decay
(negative j) of costs and revenues undergoing annual percentage changes.
As an illustration, if labor costs increase by 10 percent a year, then the
resulting series representation of labor costs will be geometric.
A1 (1 + j)(n – 1)
A1 (1 + j)n – 2
A1
0
1
A1 (1 + j)
A1 (1 + j)2
...
2
3
End of Period
n–1
n
P
FIGURE 2.16
CFD for an Increasing Geometric Series
The present worth equivalent of the cash flow series is obtained by
substituting Equation 2.35 into Equation 2.9 to obtain
n
P 5 A1 11 1 j2 21 a 11 1 j2 t 11 1 i2 2t
(2.36)
t51
This can be written as:
1 2 11 1 j2 n 11 1 i2 2n
d
i2j
P5 •
n A1/ 11 1 i2
A1 c
i?j
i5j
(2.37)
Geometric Series A series
characterized by cash flows
that increase or decrease by
a constant percentage (j%)
each period.
52
Chapter 2
Time Value of Money Calculations
or
P 5 A1 1P Z A1 i%, j%,n2
(2.38)
where (P Z A1 i%,j%,n) is the geometric series, present worth factor and is
tabulated in Appendix A for various values of i, j, and n.
Determining the Present Worth of a Geometric Series
EXAMPLE
A company is considering purchasing a new machine tool. In addition to
the initial purchase and installation costs, management is concerned about
the machine’s maintenance costs, which are expected to be $1,000 at the
end of the first year of the machine’s life and increase 8 percent/year thereafter. The machine tool’s expected life is 15 years. Company management
would like to know the present worth equivalent for expected costs. If the
firm’s time value of money is 10 percent/year compounded annually, what
is the present worth equivalent?
KEY DATA
Given: Machine maintenance costs – a geometric series with A1 5 $1,000,
j 5 8%, i 5 10%, n 5 15
Find: P – the present worth equivalent of all maintenance costs
SOLUTION
P 5 $1,0001P Z A1 10%,8%,152
5 $1,000112.030402
5 $12,030.40
Using the NPV worksheet function, the present worth for the increasing geometric series of maintenance costs is $12,030.40, as shown in Figure 2.17. Alternately, the present worth can be obtained by entering the
following in any cell:
10%,1000,1000*1.08,1000*1.08^2,1000*1.08^3,1000*1.08^4,
1000*1.08^5,1000*1.08^6,1000*1.08^7,1000*1.08^8,
5NPV ≥
¥
1000*1.08^9,1000*1.08^10,1000*1.08^11,1000*1.08^12,
1000*1.08^13,1000*1.08^14
or
51000*NPV(10%,1,1.08,1.08^2,1.08^3,1.08^4,1.08^5,1.08^6,1.08^7,
1.08^8,1.08^9,1.08^10,1.08^11,1.08^12,1.08^13,1.08^14)
2-3 Multiple Cash Flows
FIGURE 2.17
Excel® Solution to Example 2.16
The resulting present worth, $12,030.40, is identical to that obtained using
the spreadsheet approach in Figure 2.17. (When more than 10 values are
included in the range of cash flows, the spreadsheet approach is much preferred to entering the individual cash flow values in the NPV worksheet
function.)
A uniform series equivalent to the geometric series is obtained by multiplying the value of the geometric series, present worth factor by the value
of the (A Z P i%,n) factor resulting in
1 2 11 1 j2 n 11 1 i2 2n
i11 1 i2 n
dc
d
i2j
11 1 i2 n 2 1
¥
A5 ≥
i11 1 i2 n21
nA1a
b
11 1 i2 n 2 1
A1 c
i?j
(2.39)
i5j
or
A 5 A1(A Z A1 i%,j%,n)
(2.40)
53
54
Chapter 2
Time Value of Money Calculations
where (A Z A1 i%,j%,n) is the geometric-to-uniform series conversion factor.
This factor is not tabulated in Appendix A.
The future worth equivalent of the geometric series is obtained by
multiplying the value of the geometric series, present worth factor by the
(F Z P i%,n) factor to obtain
11 1 i2 n 2 11 1 j2 n
d
A1 c
i2j
F5 µ
nA1 11 1 i2 n21
i?j
i5j
(2.41)
or
F 5 A1 1F Z A1 i%,j%,n2
where (F Z A1 i%,j%,n) is the geometric series, future worth factor and is
tabulated in Appendix A.
Determining the Future Worth of a Geometric Series
EXAMPLE
Video Example
KEY DATA
SOLUTION
Mattie Bookhout receives an annual bonus and deposits it in a savings
account that pays 8 percent compounded annually. The size of her
bonus increases by 10 percent each year; her initial deposit is $500.
Determine how much will be in the fund immediately after her tenth
deposit.
Given: A1 5 $500, i 5 8%, j 5 10%, and n 5 10
Find: F
The value of F is given by
F 5 $5001F ZA1 8%,10%,102
5 $500121.740872
5 $10,870.44
Excel® does not have a worksheet function for geometric series. However, as with gradient series, the NPV worksheet function can be used to
determine the present worth, uniform series, and future worth equivalents to a geometric series. For the example, the future worth for the
geometric series is $10,870.44, as shown in Figure 2.18.
2-4 Compounding Frequency
FIGURE 2.18
2-4
55
Excel® Solution to Example 2.17
COMPOUNDING FREQUENCY
LEARNING O BJECTI VE: Perform time value of money calculations for mul-
tiple compounding periods per year using the period interest rate and the
effective annual interest rate.
Thus far, when referring to an interest rate, we said that it was x% compounded annually or x% annual compound interest. While it is true that
practically all engineering economic analyses incorporate annual compounding, in personal financing, compounding typically occurs more frequently than once a year. For example, credit cards charge interest of,
say, 1½ percent on the unpaid balance of the account each month. As a
result, if you owe $1,000 and do not pay it by the monthly payment deadline, your balance owed increases to $1,015. If you do not make any payments, the interest owed the next month will be 1½ percent of $1,015.
Hence, monthly compounding is at work. (In addition, with many credit
cards, penalties are added for lack of payment and they, too, draw interest.)
In the case of 1½ percent per month, an alternate way of expressing
the interest rate is 18 percent per annum compounded monthly, or 18 percent per year per month. When expressed in this form, 18 percent is known
as the nominal annual interest rate. We designate the nominal rate3 as r.
3
In the text, we will consider nominal interest rates that are annual, and compounding periods
that are either annual or more frequent than annual.
Nominal Annual Interest
Rate The annual interest
rate without adjustments
for compounding.
56
Chapter 2
Time Value of Money Calculations
Period Interest Rate The
nominal annual interest
rate divided by the number
of interest periods per year.
In the examples presented thus far, cash flows occurred on an annual
basis and money was compounded annually. When cash flow frequency or
compounding frequency (or both) is not annual, one of two approaches
must be employed: the period interest rate approach or the effective
interest rate approach.
2.4.1 Period Interest Rate Approach
To utilize the period interest rate approach, we must define a new term—
the period interest rate:
Period interest rate 5
Nominal annual interest rate
Number of interest periods per year
When the interest period and the compounding period are the same
(monthly), the factors in Appendix A can be applied directly. Note, however, the number of interest periods (n) must be adjusted to match the
new frequency, as in Examples 2.18 and 2.19.
Determining Future Worth with Multiple Compounding
Periods per Year
EXAMPLE
Two thousand dollars is invested in an account. What is the account balance after 3 years when the interest rate is:
a. 12 percent per year compounded monthly?
b. 12 percent per year compounded semiannually?
c.
KEY DATA
SOLUTION
12 percent per year compounded quarterly?
Given: P 5 $2,000; nominal interest rate 5 12%; duration of investment 5
3 years
Find: For each scenario: period interest rate; number of interest
periods; F
a. Nominal annual interest rate 5 12%/year
Number of interest periods/year 5 12 months/year
Period interest rate 5
12%/year
5 1%/month
12 months/year
2-4 Compounding Frequency
Number of interest periods 5 3 years(12 months/year) 5 36 months
F 5 $2,0001F Z P 1%,362
5 $2,00011.430772
5 $2,861.54
b. Nominal annual interest rate 5 12%/year
Number of interest periods/year 5 2 semiannual periods/year
Period interest rate 5
12%/year
5 6%/semiannual period
2 semiannual periods/year
Number of interest periods 5 3 years (2 semiannual periods/year) 5
6 semiannual periods
F 5 $2,0001F Z P 6%,62
5 $2,00011.418522
5 $2,837.04
c.
Nominal annual interest rate 5 12%/year
Number of interest periods/year 5 4 quarters/year
Period interest rate 5
12%/year
5 3%/quarter
4 quarters/year
Number of interest periods 5 3 years (4 quarters/year) 5 12 quarters
F 5 $2,0001F Z P 3%,122
5 $2,00011.425762
5 $2,851.54
Determining Car Payments
Rebecca Carlson purchases a car for $25,000 and finances her purchase
by borrowing the money at 8 percent per year compounded monthly; she
pays off the loan with equal monthly payments for 5 years. What will be
the size of her monthly loan payment?
EXAMPLE
Video Example
57
58
Chapter 2
Time Value of Money Calculations
KEY DATA
SOLUTION
Given: P 5 $25,000; nominal annual interest rate 5 8%/year; duration of
loan 5 5 years; number of interest periods/year 5 12 months/year
Find: Period interest rate; number of interest periods; A
Period interest rate 5
8%/year
5 0.66667%/month
12 months/year
Number of interest periods 5 5 years(12 months/year) 5 60 months
A 5 $25,0001A Z P 0.66667%,602 5 $25,000 £
0.006666711.00666672 60
§
11.00666672 60 2 1
5 $506.91/month
Using the Excel® PMT worksheet function,
A 5PMT10.08/12,60,2250002 5 $506.91
2.4.2 Effective Annual Interest Rate
Video Lesson:
Effective Annual
Interest Rate
Effective Annual Interest
Rate The annual interest
rate that is equivalent to
the period interest rate.
The second approach to solving problems when compounding is not
annual is the effective interest rate approach. The effective annual interest
rate is the annual interest rate that is equivalent to the period interest rate as
previously calculated.
For example, if the interest rate is 12 percent per year compounded
quarterly, then the nominal annual interest rate is 12 percent, and there are
four interest periods per year. Thus, the period interest rate is 3 percent per
quarter. Hence, $1 invested for 1 year at 3 percent per quarter has a future
worth of
F 5 $1(F Z P 3%,4) 5 $1(1.12551) 5 $1.12551
To obtain the same value in 1 year requires an annual compound interest
rate of 12.551 percent. This value is called the effective annual interest
rate and is given by (1.03)4 2 1 5 0.12551, or 12.551 percent.
The Excel® EFFECT worksheet function can be used to determine
the effective annual interest rate. This function has the following parameters: r (nominal rate) and m (number of compounding periods in a
year).
2-4 Compounding Frequency
The general equation for the effective annual interest rate, ieff, is
ieff 5 11 1 r/m2 m 2 1 5 EFFECT1r,m2
(2.43)
where r is the nominal annual interest rate and m is the number of interest
periods per year.
Calculating the Effective Annual Interest Rate
EXAMPLE
Calculate the effective annual interest rate for each of the following cases:
(a) 12 percent per year compounded quarterly; (b) 12 percent per year
compounded monthly; (c) and 12 percent per year compounded every
minute.
a. Twelve percent per year compounded quarterly: r 5 12% ; m 5 4
From Equation 2.43,
ieff 5 (1 1 0.12/4)4 2 1
5 (1.03)4 2 1
5 0.12551 5 12.551%
Using the Excel® EFFECT worksheet function gives the same result:
ieff 5EFFECT112%,42 5 12.551%
b. Twelve percent per year compounded monthly: r 5 12%; m 5 12
From Equation 2.43,
ieff 5 (1 1 0.12/12)12 2 1
5 (1.01)12 2 1
5 0.12683 5 12.683%
Using the Excel® EFFECT worksheet function gives the same result:
ieff 5EFFECT112%,122 5 12.683%
c.
Twelve percent per year compounded every minute: r 5 12%; m 5
525,600
From Equation 2.43, to eight decimal places,
ieff 5 11 1 0.12/525,6002 525,600 2 1
5 0.12749684 5 12.749684%
Using the Excel® EFFECT worksheet function gives
ieff 5EFFECT112%,5256002 5 12.749684%
SOLUTION
59
60
Chapter 2
Time Value of Money Calculations
Making Monthly House Payments
EXAMPLE
Greg Wilhelm borrowed $100,000 to purchase a house. He agreed to
repay the loan with equal monthly payments over a 30-year period at a
nominal annual interest rate of 6 percent compounded monthly. The
closing fee on the loan is $2,000.
a. What is Greg’s monthly payment on the loan?
b. What is Greg’s monthly payment if he chooses to finance the closing
fee along with the loan?
c. What is the effective annual interest rate on the loan?
KEY DATA
Given: Loan amount 5 $100,000; closing cost 5 $2,000; nominal annual
r 5 6% (compounded monthly); n 5 360
Find: A, ieff
SOLUTION
a. In order to calculate Greg’s monthly payment we first need to deter-
mine the period interest rate:
period interest rate 5
6%/year
5 0.5%/month
12 months/year
Greg’s monthly payment on the 30-year loan is calculated as
follows:
A 5 $100,0001A Z P 0.5%,3602
5 $100,00010.00599552
5 $599.55
or, using the Excel® PMT worksheet function,
A 5PMT10.06/12,360,21000002 5 $599.55
b. If Greg chooses to finance the closing costs also, then his monthly pay-
ment will be
A 5 $102,0001A Z P 0.5%,3602
5 $102,00010.00599552
5 $611.54
or
A 5PMT10.06/12,360,21020002 5 $611.54
2-4 Compounding Frequency
c.
If Greg made 360 payments of $611.54/month for $102,000, then the
effective annual interest rate on the loan would be
ieff 5EFFECT(12*RATE(360,611.54,2102000),12) 5 6.1678%
2.4.3 When Compounding and Cash Flow Frequencies Differ
In the previous example, the frequency of compounding coincided with
the frequency of cash flows—for example, monthly compounding and
monthly cash flows. What if they are not the same?
The approach used in this book assumes that money deposited during
a compounding period earns interest regardless of when it is deposited.
This approach is consistent with other DCF assumptions made throughout
the text.
Let r denote the nominal annual interest rate for money and m denote
the number of compounding periods in a year; let k denote the number of
cash flows in a year, and let i denote the effective interest rate per cash flow
period. The value of i is obtained as follows:
i 5 11 1 r/m2 m/k 2 1
Video Lesson:
Effective Rate per
Payment Period
(2.44)
Equation 2.44 results from setting the effective annual interest rate for the
stated compounding frequency of money equal to that for the cash flow
frequency:
11 1 i2 k 2 1 5 11 1 r/m2 m 2 1
(2.45)
and solving for i.
When Cash Flow Frequency Does Not Match
Compounding Frequency
EXAMPLE
What size monthly payments should occur when $10,000 is borrowed at
8 percent per year compounded quarterly and the loan is repaid with
36 equal monthly payments?
Given: P 5 $10,000; r 5 8% (compounded quarterly); n 5 36
Find: A
KEY DATA
61
62
Chapter 2
Time Value of Money Calculations
SOLUTION
From Equation 2.44, r 5 0.08, k 5 12, and m 5 4. Therefore,
i 5 11 1 0.08/42 4/12 2 1
5 0.006623 or 0.6623%/month
Knowing the effective monthly interest rate, the monthly payment can be
determined:
A 5 $10,0001A Z P 0.6623%,362
5 $10,0003 10.0066232 11.0066232 36 y11.0066232 36 2 1 4
5 $313.12
Using the Excel® PMT worksheet function,
A 5PMT11.02^11/3221,36,2100002 5 $313.12.
SUMMARY
KEY CONCEPTS
1. Learning Objective: Construct a cash flow diagram (CFD) depicting the
cash inflows and outflows for an investment alternative. (Section 2.1)
The CFD is a powerful visual tool used to depict an investment alternative.
The graphic depicts the timing, magnitude and direction of the cash flow.
Cash flows can represent present values, future values, uniform series, gradient series or geometric series. Cash flows are positive or negative depending upon the perspective (such as borrower versus lender) from which they
are drawn.
$5,000
$5,000
$5,000
(+)
0
(–)
1
2
3
4
Time
$2,000
$3,000
$4,000
FIGURE 2 . 19
An Example Cash Flow Diagram
5
Summary 63
2. Learning Objective: Perform time value of money calculations for single
cash flows with annual compounding. (Section 2.2)
In engineering economic analysis, it often is useful to transform the
present sum of a single cash flow into a future sum or vice versa. The equations and factor notation for doing so are summarized in the table that
follows.
Single Cash Flows
Present Worth
of a Future
Payment
Future Worth
of a Present
Payment
Formula
P 5 F 11 1 i 2
Factor Notation
2n
F 5 P 11 1 i 2 n
Cash Flow Diagram
P 5 F 1P Z F i%,n2
F
F 5 P 1F Z P i%,n2
0
1
2
n–1
End of Period
n
P
3. Learning Objective: Distinguish between uniform, irregular, gradient, and
geometric series of cash flows. (Section 2.3)
Series of cash flows may exhibit a pattern or be irregular. Recognizing patterns when they occur can facilitate analysis of a particular investment
alternative. Three common patterns of cash flows are uniform, gradient,
and geometric:
■
■
■
Uniform cash flows involve sums of equal (or uniform) value for each
period. An example is the uniform payments of a monthly car loan or
home mortgage payment.
Gradient series cash flows increase by a constant value in each period. For
example, maintenance costs for a piece of equipment may increase by
a constant value each year.
Geometric series cash flows increase or decrease by a constant percentage each period. For example, labor costs may increase by a constant percentage each year.
The use of a CFD is helpful in illustrating the conversion of these positive
and negative cash flows into a single value.
4. Learning Objective: Perform time value of money calculations for multiple
cash flows with annual compounding. (Section 2.3)
It is useful—and common—to transform series of cash flows for engineering economic analysis into some single equivalent cash flow, such as a present or future worth value. The equations and factor notation for making such
transformations for irregular, gradient, and geometric series are summarized
in the table that follows.
64
A5Pc
F 5 Ac
A5Fc
• Uniform Series
from Present Value
• Future Worth of
Uniform Series
• Uniform Series
from Future Value
P 5 Gc
A 5 Gc
F 5 Gc
• Uniform Series
Equivalent of a
Gradient Series
• Future Worth of a
Gradient Series
d
d
d
i2
11 1 i 2 n 2 11 1 ni 2
n
1
2 1A Z F i%,n2 d
i
i
i2
d
1 2 11 1 ni 211 1 i 2 2n
i
d
11 1 i 2 n 2 1
i
11 1 i 2 n 2 1
11 1 i 2 n 2 1
i 11 1 i 2 n
i 11 1 i 2
n
11 1 i 2 2 1
• Present Worth of a
Gradient Series
Gradient Series
Cash Flows
P 5 Ac
• Present Worth of
Uniform Series
n
d
P 5 A1 11 1 i 2 21 1 A 2 11 1 i 2 22 1 A 3 11 1 i 2 23 1 p
1 An21 11 1 i 2 21n212 1 An 11 1 i 2 2n
Irregular Series
Cash Flows
Uniform Series
Cash Flows
Formula
Series Cash Flows
F 5 G 1F Z G i%,n2
A 5 G 1A Z G i%,n2
P 5 G 1P Z G i%,n2
A 5 F1A Z F i%,n2
F 5 A1F Z A i%,n2
A 5 P1A Z P i%,n2
P 5 A1P Z A i%,n2
t51
P 5 a At 1P Z F i%,t 2
n
Factor Notation
P
0
0
P
0
P
0
1
A
1
1
A
1
A1
3
2
G
A
n–1
A
...
A
n –1
3
End of Period
2G
End of Period
2
...
End of Period
2
End of Period
A
A2
2
A3
n
A
F
n
A
n–1
An – 1
n–1
(n – 2)G
Cash Flow Diagram
n
(n – 1)G
An
n
• Future Worth of a
Geometric Series
• Uniform Series
Equivalent of a
Geometric Series
• Present Worth of a
Geometric Series
Geometric Series
Cash Flows
nA1 y11 1 i 2
i2j
1 2 11 1 j 2 n 11 1 i 2 2n
d
i5j
iZj
F5 µ
i2j
11 1 i 2 n 2 11 1 j 2 n
nA1 11 1 i 2 n21
A1 c
A1 c
d
i5j
i?j
i 11 1 i 2 n
1 2 11 1 j 2 n 11 1 i 2 2n 2
d c
d
i2j
11 1 i 2 n 2 1
A5 µ
i 11 1 i 2 n21
nA1 a
b
11 1 i 2 n 2 1
P5 µ
A1 c
i5j
iZj
F 5 A1 1F Z A1 i%,j%,n2
A 5 A1(A Z A1 i%,j%,n)
P 5 A1 1P Z A1 i%,j%,n2
P
0
1
A1
2
...
3
End of Period
A1 (1 + j)
A1 (1 + j)2
n–1
A1 (1 +
n
A1 (1 + j)(n – 1)
j)n – 2
65
66
Chapter 2
Time Value of Money Calculations
5. Learning Objective: Perform time value of money calculations for multiple compounding periods per year by utilizing the period interest rate and
the effective annual interest rate. (Section 2.4)
Although many engineering economic analyses assume annual compounding, in personal financing, compounding typically occurs more frequently. This compounding can be significant and should not be ignored
for economic decision making. When cash flow frequency or compounding frequency (or both) is not annual, one of two approaches must be
employed: the period interest rate approach or the effective interest rate
approach.
a.
The interest rate using the period interest rate approach will match the
interest rate to the compounding frequency, for example the interest
rate per month or quarter (where the monthly or quarterly designation
represents the compounding frequency). The equation for the period
interest rate is:
Period interest rate 5
b.
Nominal annual interest rate
Number of interest periods per year
The effective annual interest rate represents the annual interest rate
that is equivalent to the period interest rate, thus annualizing the period interest rate. The formula for the effective annual interest rate is:
ieff 5 11 1 rym2 m 2 1 5 EFFECT1r,m2
(2.43)
When cash flow and compounding frequencies differ, the effective
interest rate per cash-flow period i is obtained by the equation
i 5 11 1 rym2 m/k 2 1
(2.44)
where r 5 the nominal annual interest rate for money, m 5 the number of
compounding periods in a year, and k 5 the number of cash flows in a year.
KEY TERMS
Capital Recovery Factor, p. 42
Cash Flow Diagram (CFD), p. 24
Compounding, p. 26
Effective Annual Interest Rate, p. 58
Future Value, p. 28
Geometric Series, p. 51
Gradient Series, p. 46
Interest Rate, p. 26
Nominal Annual Interest Rate, p. 55
Period Interest Rate, p. 56
Present Value, p. 27
Sinking Fund Factor, p. 45
Uniform Series, p. 26
Summary 67
Problem available in WileyPLUS
GO Tutorial Tutoring Problem available in WileyPLUS
Video Solution Video Solution available in WileyPLUS
FE-LIKE PROBLEMS
1.
If you want to triple your money at an interest rate of 6% per year compounded annually, for how many years would you have to leave the money in
the account?
a. 12 years
c. 32 years
b. 19 years
d. Cannot be determined without knowing the
amount invested.
2.
Let F be the accumulated sum, P the principal invested, i the annual
compound interest rate, and n the number of years. Which of the following correctly relates these quantities?
a. F 5 P (1 1 in)
c. F 5 P (1 1 n)i
n
b. F 5 P (1 1 i)
d. F 5 P (1 1 ni)n 2 1
3.
Consider the following cash flow diagram. What is the value of X if
the present worth of the diagram is $400 and the interest rate is 15% compounded annually?
200
X
0
a. $246
b. $165
1
X
2
3
c. $200
d. $146
4.
The plan was to leave $5,000 on deposit in a savings account for 15 years
at 6.5% interest compounded annually. It became necessary to withdraw
$1,500 at the end of the 5th year. How much will be on deposit at the end of
the 15-year period?
a. $11,359
c. $12,043
b. $9,359
d. $10,043
5.
A deposit of $3,000 is made in a savings account that pays 7.5% interest
compounded annually. How much money will be available to the depositor at
the end of 16 years?
a. $8,877
c. $9,542
b. $10,258
d. $943
68
Chapter 2
Time Value of Money Calculations
6.
If you invest $5,000 three years from now, how much will be in the account 15 years from now if i 5 10% compounded annually?
a. $8,053
c. $20,886
b. $15,692
d. $27,800
7.
The maintenance costs of a car increase by $200 each year. This cash
flow pattern is best described by which of the following?
a. gradient series
c. infinite series
b. geometric series
d. uniform series
8.
Your company seeks to take over Good Deal Company. Your company’s
offer for Good Deal is $3,000,000 in cash upon signing the agreement followed by 10 annual payments of $300,000 starting one year later. The time
value of money is 10%. What is the present worth of your company’s offer?
a. $3,000,000
c. $4,843,380
b. $2,281,830
d. $5,281,830
9.
What is the effective annual interest rate if the nominal annual interest rate
is 24% per year compounded monthly?
a. 2.00%
c. 26.82%
b. 24.00%
d. 27.12%
10.
A young engineer calculated that monthly payments of $A are required
to pay off a $5,000 loan for n years at i% interest, compounded annually. If
the engineer decides to borrow $10,000 instead with the same n and i%, her
monthly payments will be $2A.
a. True
b. False
c. Cannot be determined without knowing the value of n and i
d. Cannot be determined without knowing the value of n or i
11.
The president of a growing engineering firm wishes to give each of
20 employees a holiday bonus. How much needs to be deposited each month
for a year at a 12% nominal rate, compounded monthly, so that each employee will receive a $2,500 bonus?
a. $2,070
c. $3,940
b. $3,840
d. $4,170
12.
Under what circumstances are the effective annual interest rate and the
period interest rate equal?
a. Never
b. If the number of compounding periods per year is one
c. If the number of compounding periods per year is infinite
d. Always
13.
A child receives $100,000 as a gift which is deposited in a 6% bank account compounded semiannually. If $5,000 is withdrawn at the end of each
half year, how long will the money last?
a. 21.0 years
c. 25.0 years
b. 15.5 years
d. 18.0 years
Summary 69
PROBLEMS
Introduction
1.
You are offered $200 now plus $100 a year from now for your used
computer. Since the sum of those two amounts is $300, the buyer suggests
simply waiting and giving you $300 a year from now. You know and trust the
buyer, and you typically earn 5.0% per year on your money. So, is the offer
fair and equitable?
2. You are offered $500 now plus $500 one year from now. You can earn
6% per year on your money.
a. It is suggested that a single fair amount be paid now. What do you consider fair?
b. It is suggested that a single fair amount be paid one year from now. What
do you consider fair?
3. What words comprise the abbreviation “DCF”? Tell/describe/define what it
means in 10 words or less.
4. State the four DCF rules.
Section 2.1 Cash-Flow Diagrams
5. Pooi Phan needs $2,000 to pay off her bills. She borrows this amount from a
6.
7.
8.
9.
bank with plans to pay it back over the next four years at $X per year. Draw
a cash flow diagram from the bank’s perspective.
A laser cutting machine is purchased today for $23,000. There are no maintenance costs for the next two years. Maintenance at the end of year 3 is expected to be $2,000, with each subsequent year’s maintenance costs exceeding
the previous year’s by $1,000. An increase in revenues of $14,000 per year is
expected. The planning horizon is 6 years. Draw the cash flow diagram.
Rodeo Jeans are stonewashed under a contract with independent USA Denim
Company. USA Denim purchased two semiautomatic machines that cost
$19,000 each at (t 5 0). Annual operating and maintenance costs are $15,000
per machine. Two years after purchasing the machines, USA Denim made
them fully automatic at a cost of $12,000 per machine. In the fully automatic
mode, the operating and maintenance costs are $6,000 the first year, increasing by $1,000 each year thereafter. The contract with Rodeo Company is for
8 years. Draw the cash flow diagram for all of USA Denim’s investment and
other costs assuming the contract will not be extended beyond 8 years.
Video Solution You rent an apartment for $550 per month, payable at
the beginning of the month. An initial deposit of $450 is required. Utilities
are an additional $150 per month payable at the end of the month. The deposit is refundable at the time you move out, assuming a clean apartment in
good condition. Draw a monthly cash flow diagram, assuming you keep the
apartment for 12 full months.
GO Tutorial Kaelyn borrows $30,000 from her grandfather today to cover
her college expenses. She agrees to repay the loan, with the first payment due
5 years from today in the amount of $2,000. No payment is made at the end of
70
Chapter 2
Time Value of Money Calculations
year 6. Starting 7 years from today, a series of 5 annual end-of-year payments
is made, with the first in the amount of $X. Each subsequent payment is $1,500
greater than the previous payment. Draw the cash flow diagram of this transaction from the grandfather’s perspective.
10.
Video Solution David is borrowing $150,000 from Hartford Bank to open
Road and Off-Road Bicycle Shop. David expects it to take a few years before
the shop earns a sizeable profit, so he has arranged for no payments on the
loan until the end of the fourth year. The first and second payments are due 4
and 5 years, respectively, from today in the amounts of $20,000 each. Starting
at the end of year 6, a series of 4 annual end-of-year payments will be made.
The first of these is $X. Each subsequent payment is $8,000 greater than the
previous payment. Draw the cash flow diagram from David’s perspective.
11. Draw a cash flow diagram depicting the net cash flows associated with the
purchase, operation, and disposition of a synthetic rubber blending machine.
The cash flow components are shown below. Your CFD should have only one
arrow at any given time period, reflecting the net of that period’s cash flows.
At t 5 0 (now), purchase blender for $62,000.
At t 5 0, install at cost of $8,000.
At t 5 1, savings generated by blender is $10,000.
At t 5 1, maintenance costs of $800.
At t 5 2, savings generated by blender are $12,000.
At t 5 2, maintenance costs of $1,200.
At t 5 3, savings generated by blender is $18,000.
At t 5 3, maintenance costs of $1,600.
At t 5 4, 5, 6, 7, 8, 9, 10, savings generated are $24,000 and maintenance
is $4,000.
At t 5 10, the blender is sold for $8,000.
At t 5 10, blender removal costs are $1,600.
12.
Video Solution Today you borrow $10,000 to pay for your expected college costs over the next four years, including a master’s degree. Two years
from now, you determine that you need an additional $4,000 so you borrow
this additional amount. Starting four years from the original loan (two years
from the second loan), you begin to repay your combined debt by making annual payments of $2,880. You will make these payments for 10 years. Draw
a cash flow diagram of this situation from your perspective.
Section 2.2.1
Single Cash Flows—Future Worth Calculations
If you deposit $5,000 4 years from today, how much will you be able to withdraw 10 years from today if interest is 8.5% per year compounded annually?
14.
Video Solution How much will a $25,000 investment today be worth in
10 years if it earns 7% annual compound interest?
15. Use the six approaches from Example 2.4 to determine to the nearest year
how long it takes for an investment to double if the interest is compounded
annually at the following rates.
a. 5%
d. 15%
b. 7%
e. 20%
c. 10%
13.
Summary 71
16. On August 1, 1958, first-class postage for a 1-ounce envelope was 4¢. On
August 1, 2007, a first-class stamp for the same envelope cost 41¢. What
annual compound increase in the cost of first-class postage was experienced
during the 49-year period?
17.
What is the smallest integer-valued annual compound interest that will
result in an investment tripling in value in less than or equal to 10 years?
18.
Video Solution You purchase a quarter section (160 acres) of land for
$176,000 today and sell it in exactly 9 years for $525,000 at auction. At what
annual compound rate did the value of your land grow?
19.
With interest at 9% compounded annually, what is the fewest number of
years (integer-valued) required for money to double in magnitude?
20. At what interest rate will money:
a. Double itself in 10 years?
b. Triple itself in 10 years?
c. Quadruple itself in 10 years?
21.
How much money can be withdrawn at the end of the investment
period if:
a. $1,000 is invested at 8%/year compounded annually for 10 years?
b. $5,000 is invested at 11%/year compounded annually for 4 years?
c. $13,000 is invested at 9%/year compounded annually for 7 years?
d. $25,000 is invested at 10%/year compounded annually for 3 years?
22. If you invest $1,500 today and withdraw $2,500 in 3 years, what interest rate
was earned?
23.
What will be the amount accumulated by each of the following present
investments?
a. $3,000 invested for 7 years at 14% compounded annually.
b. $1,600 invested for 17 years at 12% compounded annually.
c. $20,000 invested for 38 years at 16% compounded annually.
d. $3,500 invested for 71 years at 8% compounded annually.
e. $5,000 invested for 34 years at 11.5% compounded annually.
24. How long, to the nearest year, does it take an investment at 6% compounded
annually to (approximately):
a. Double itself?
b. Triple itself?
c. Quadruple itself?
25.
What rate of interest compounded annually is involved if:
a. An investment of $10,000 made now will, 10 years from now, result in a
receipt of $23,674?
b. An investment of $2,000 made 18 years ago has increased in value to
$15,380?
c. An investment of $2,500 made now will, 5 years from now, result in a
receipt of $4,212?
72
Chapter 2
Time Value of Money Calculations
Section 2.2.2
Single Cash Flows—Present Worth Calculations
26. Develop a single spreadsheet that allows you to calculate F Z P and P Z F fac-
tors. For each cell where the calculation is performed, place above it a small
“control panel” that allows you to enter whatever numbers you need to perform the calculation. For example, for the F Z P factor, your control panel
needs to include i and n. Test each of your factors against the tables in Appendix A to ensure they work right. Keep this handy. It will be useful to you.
27.
Charlotte wishes to accumulate $100,000 in a savings account in 10 years.
If she wishes to make a single deposit today and the bank pays 4% compounded annually on deposits of this size, how much should Charlotte deposit
in the account?
28. How much money would have to be deposited today to accumulate:
a.
b.
c.
d.
$10,000 after 6 years if the investment earns 5%/year compounded annually?
$6,500 after 4 years if the investment earns 8%/year compounded annually?
$3,400 after 12 years if the investment earns 6%/year compounded annually?
$13,500 after 5 years if the investment earns 10%/year compounded
annually?
29.
GO Tutorial What deposit today is required for it to be worth $150,000 in
25 years if the deposit earns 5% annual compound interest?
30.
Video Solution What present amount of money must be deposited at
11% interest compounded annually to grow to $15,000 in 9 years? Give
your answer to the nearest penny.
a. Use the tables provided in Appendix A.
b. Use the formula, either in a spreadsheet or using a calculator.
c. Use the PV function in Excel®.
31.
For your 21st birthday, your grandfather offers you a gift of $1,000
today. However, you have the choice of waiting 3 years and receiving $1,500
or waiting 5 years and receiving $3,000. If your money grows at a rate of
8% compounded annually, which alternative should you choose?
32. What is the present value of the following future receipts?
a.
b.
c.
d.
e.
$19,000 5 years from now at 9% compounded annually
$8,300 12 years from now at 15% compounded annually
$6,200 53 years from now at 12% compounded annually
$13,000 18 years from now at 19.2% compounded annually
$5,000 10 years from now at 8% compounded annually
33. Calculate using the interest formula the factor (PZF11.5%,37). Compare that
to the result obtained using Excel®’s PV function.
34. Jason takes out a loan at 10% compounded annually for 7 years. At the end
of this period, he pays off the loan at a value of $23,384.61. What amount
did he borrow?
35. How much money today is equivalent to $10,000 in 12 years, with interest
at 10% compounded annually?
Summary 73
36. You want to withdraw a single sum amount of $6,000 from an account at
the end of 7 years. This withdrawal will zero out the account. What single
sum of money deposited today is required if the account earns 12% per year
compounded annually?
a. Use the tables provided in Appendix A.
b. Use the P Z F formula directly in Excel® or your calculator.
c. Use an appropriate Excel® function.
37. If a fund pays 12 percent compounded annually, what single deposit now will
accumulate $12,000 at the end of the 10th year? If the fund pays 6 percent
compounded annually, what single deposit now is required in order to accumulate $6,000 at the end of the tenth year?
Section 2.3.1
Irregular Series Cash Flows
38. The cash flow profile for an investment is given below and the interest rate
is 6.5% compounded annually.
End of
Year
Net Cash
Flow
End of
Year
Net Cash
Flow
0
$0
4
1
2$500
5
$500
2
$200
6
3
$400
7
2$200
$100
2$300
a. Find the future worth of this cash flow series using the actual cash flows.
b. Find the present worth of this series using the actual cash flows.
c. Find the present worth using the future worth.
39.
If you invest $2,000 today, withdraw $1,000 in 3 years, deposit $3,000 in
5 years, deposit $1,500 in 8 years, and withdraw the entire sum three years
after the final deposit, how much will you withdraw? Interest is 7%.
Video Solution An investment has the following cash flow series where
40.
interest is 8%:
End of
Year
Cash
Flow
End of
Year
Cash
Flow
0
1
2
3
$300
$300
$600
$0
$800
$700
2$500
5
6
7
8
4
2$300
$600
a. Determine the present worth of the series.
b. Determine the future worth of the series at the end of the 8th year.
c. Find the worth of the series at the end of year 2.
41.
Ben deposits $5,000 now into an account that earns 7.5% interest compounded annually. He then deposits $1,000 per year at the end of the 1st and
2nd years. How much will the account contain 10 years after the initial deposit?
74
Chapter 2
Time Value of Money Calculations
42. Renaldo borrows $8,000 from his aunt today to help pay for college ex-
penses. He agrees to repay the loan according to the following schedule, at
a rate of 6%/year compounded annually.
End of
Year
Cash
Flow
End of
Year
Cash
Flow
0
$8,000
5
2$X
1
$0
6
2($X 1 100)
2
$0
7
2($X 1 400)
3
$0
8
2$2X
4
$0
a. Draw the cash flow diagram from Renaldo’s perspective.
b. Find the value of X such that the loan is fully repaid with the last payment.
c. What is the dollar amount of each of the 4 payments.
43.
The cash flow profile for an investment is given below and the interest
rate is 8% compounded annually.
End of
Year
0
1
2
3
Net Cash
Flow
$0
$500
2$200
$600
End of
Year
Net Cash
Flow
4
5
2$100
$300
6
$200
Find the present worth of this series using the actual cash flows.
Find the future worth of this cash flow series using the actual cash flows.
Find the future worth using the present worth.
Find the worth of the series at EOY 4 using the individual cash flows.
Find the present worth using the worth at EOY 4.
44. The manager at a Sherwin-Williams store has decided to purchase a new
$30,000 paint mixing machine with high-tech instrumentation for matching
color and other components. The machine may be paid for in one of two ways:
(1) pay the full price now, less a 3% discount, or (2) pay $5,000 now, $8,000
one year from now, and $6,000 at the end of each of the next 4 years. If
interest is 12% compounded annually, determine which way is best for the
manager to make the purchase.
45.
Ken loans his grandson Rex $20,000 at 5.5% per year to help pay for
executive chef schooling in Florida. Rex requires 3 years of schooling before
beginning to earn a salary. He agrees to pay Ken back the loan following the
schedule below:
a.
b.
c.
d.
e.
End of
Year
Cash
Flow
End of
Year
Cash
Flow
0
$20,000
5
2$2X
1
$0
6
2$3X
2
$0
7
2$4X
3
$0
8
2$5X
4
2$X
Summary 75
a.
b.
c.
d.
Draw the cash flow diagram from Ken’s perspective.
Find the value of X such that the loan is fully repaid with the last payment.
What is the dollar amount of each of the 5 payments.
Quite by surprise, following successful on-time completion of all payments, Ken gives back to Rex all interest paid. For how much does Ken
write the check?
46. Maria deposits $1,200, $500, and $2,000 at t 5 1, 2, and 3, respectively. If
the fund pays 8 percent compounded per period, what sum will be accumulated in the fund at (a) t 5 3, and (b) t 5 6?
47. Juan deposits $1,000 in a savings account that pays 8 percent compounded
annually. Exactly 2 years later he deposits $3,000; 2 years later he deposits
$4,000; and 4 years later he withdraws all of the interest earned to date
and transfers it to a fund that pays 10% compounded annually. How much
money will be in each fund 4 years after the transfer?
Section 2.3.2
Uniform Series Cash Flows
48. A debt of $1,000 is incurred at t 5 0. What is the amount of four equal
payments at t 5 1, 2, 3, and 4 that will repay the debt if money is worth
10 percent compounded per period?
49. Develop a single spreadsheet that allows you to calculate any of the P Z A, A Z P
F Z A, or A Z F, factors. For each cell where the calculation is performed, place
above it a small “control panel” that allows you to enter whatever numbers
you need to perform the calculation. For example, for the F Z A factor, your
control panel needs to include i and n. Test each of your factors against the
tables in Appendix A to ensure they work right. Keep this handy. It will be
useful to you.
50. Jason has been making equal annual payments of $7,500 to repay a college
loan. He wishes to pay off the loan immediately after having made an annual payment. He has eight payments remaining. With an annual compound
interest rate of 6%, how much should Jason pay?
51.
Each and every year $7,500 is invested at 4% annual compound interest.
a. What is the value of the investment portfolio after 20 years? After 25 years?
After 30 years?
b. Repeat part (a) if the investment is at 5% annual compound interest.
c. Based upon your answers to (a) and (b), what conclusions can be drawn
regarding the impact of the interest earned versus the duration of the
investment?
52. Five deposits of $500 each are made a t 5 1, 2, 3, 4, and 5 into a fund paying
6 percent compounded per period. How much will be accumulated in the
fund at (a) t 5 5, and (b) t 5 10?
53.
GO Tutorial Using a 5% annual compound interest rate, what investment
today is needed in order to withdraw $5,000 annually:
a. For 10 years?
b. For 10 years if the first withdrawal does not occur for 3 years?
76
Chapter 2
Time Value of Money Calculations
54. If money is worth more than 0% to you, would you rather receive $10,000/
year for 5 years or receive $5,000/year for 10 years? What is your preference
if you must pay these amounts, rather than receive them?
55.
Determine the present worth of 5 equal annual deposits of $1,200 at the
end of years 1 through 5, followed by 4 equal annual withdrawals of $700 at
the end of years 4 through 7. Note that both years 4 and 5 will have a deposit
and a withdrawal. Interest is 5%.
56. Juan borrows $25,000 at 7% compounded annually. If the loan is repaid in
five equal annual payments, what will be the size of Juan’s payments if the
first payment is made one year after borrowing the money?
57. What equal annual deposits must be made at t 5 2, 3, 4, 5, and 6 in order
to accumulate $25,000 at t 5 8 if money is worth 10 percent compounded
annually?
58. Eight equal deposits of $1,000 are made at the end of each year into a fund
paying 8%.
a. What is the present worth, 1 year before the first deposit?
b. What is the future worth, immediately after the last deposit?
c. What is the future worth, 3 years after the last deposit?
59.
Adriana wishes to accumulate $2,000,000 in 35 years. If 35 end-of-year
deposits are made into an account that pays interest at a rate of 7% compounded annually, what size deposit is required each year to meet Adriana’s
stated objective?
60. If annual deposits of $1,000 are made into a fund paying 12 percent interest
compounded annually, how much money will be in the fund immediately
after the 5th deposit?
61. An amount equal to $50,000 is borrowed at 7% annual compound interest.
a. What size equal annual payment is required if the first of 5 payments is
made one year after receiving the $50,000?
b. What size payment is required if the first payment is not made until 4 years
after receipt?
62. You purchase a house for $250,000 directly from the buyer who owns the
home outright. You pay a 20% down payment. You sign a first mortgage and
the buyer agrees to finance the remaining $200,000 at 7% annual compound
interest with annual end-of-year payments over 12 years. How much is a
single yearly payment?
63.
You deposit $1,000 in a fund at the end of each year for a 10-year period.
The fund pays 5% compounded annually. How much money is available to
withdraw immediately after your last deposit?
64. You take out a loan to buy a new audio system. Your equal annual payments
are 20% of the amount you borrowed. The interest rate on the loan is 7%
compounded annually.
a. Determine the number of years you will be required to make payments.
(This number may be a nonwhole one such as 4.791, for example.)
Summary 77
b. If you make the same payment for an integer number of years, rounding
up from your answer in part (a), what interest rate will you be paying?
65.
You take out a loan to build a swimming pool in the back yard of your
new home. Your equal annual payments are 1/6th of the amount you borrowed. If it will take you 7 years to fully repay the loan, what is the interest
rate on the loan?
66. What uniform series over the interval [11,20] will be equivalent to a uniform
series of $10,000 cash flows over the interval [1,10] based on:
a. A 6% interest rate?
b. A 10% interest rate?
67.
What uniform series over the interval [1,8] will be equivalent to a uniform series of $10,000 cash flows over the interval [3,10] based on:
a. A 6% interest rate?
b. A 10% interest rate?
68. Janie deposits $10,000 in the bank today. Starting 3 years from now, she makes
equal withdrawals of $1,000 for 5 years and then withdraws the remaining
amount 10 years from now. How much will she be able to withdraw 10 years
from now, assuming the bank pays 6% compounded annually?
69. You plan to open a retirement account. Your employer will match 50% of
your deposits up to a limit on the match of $2,500 per year. You believe the
fund will earn 12% over the next 30 years, and you will make 30 deposits of
$5,000, plus 50% employer matching, totaling $7,500 per year.
a. How much money will be in the account immediately after the last
deposit?
b. How much total money will you put into the fund?
c. How much total money will your employer put into the fund?
d. How much will the total investment earnings be?
e. If you want the account to last for 30 years (30 withdrawals), starting
1 year after the last deposit, what amount will you be able to withdraw
each year?
f. If you want the account to last forever, what amount will you be able to
withdraw each year?
70. How much money can be withdrawn at the end of 15 years if:
a. $2,000 is deposited at the end of each year and earns 5%/year com-
pounded annually?
b. $3,000 is deposited at the end of each year for 10 years and no deposits are
made thereafter, where the fund earns 8%?
c. $2,000 is deposited at the end of years 1 through 5, $4,000 is deposited at
the end of years 6 through 10, and $6,000 is deposited at the end of years
11 through 15, with all deposits earning 8%?
71.
How much money can be withdrawn at the end of the investment period if:
a. $4,000 is invested at the end of each of 3 years at 5%/year compounded
annually, with the lump sum then shifted into an investment paying 8%/
year for 5 additional years?
78
Chapter 2
Time Value of Money Calculations
b. $12,000 is invested at the end of each of 10 years at 10%/year com-
pounded annually, with the lump sum then shifted into an investment
paying 5%/year for 3 additional years?
c. $18,000 is invested at the end of each of 5 years at 9%/year compounded
annually, with the lump sum then shifted into an investment paying 7%/
year for 8 additional years?
72.
Video Solution You decide to open an IRS approved retirement account
at your local brokerage firm. Your best estimate is that it will earn 9%. At the
end of each year for the next 25 years, you will deposit $4,000 per year into
the account (25 total deposits). Three years after the last deposit, you will
begin making annual withdrawals.
a. How much money is in the account one year before the first withdrawal?
b. If you want to make 30 annual withdrawals, what amount will you be
able to withdraw each year?
c. If you want the account to last forever, what amount will you be able to
withdraw each year?
73.
In planning for your retirement, you have decided that you would like
to be able to withdraw $60,000 per year for a 10 year period. The first withdrawal will occur 20 years from today.
a. What amount must you invest today if your return is 10% per year?
b. What amount must you invest today if your return is 15% per year?
74. Determine the equivalent annual cash flow of this series at 10% interest:
End of
Year
75.
Cash
Flow
End of
Year
Cash
Flow
0
2$2,500
5
$0
1
$3,000
6
2
3
4
$4,500
$0
7
8
2$1,000
$7,000
$3,000
2$5,000
Fishing Designs has arranged to borrow $15,000 today at 12% interest.
The loan is to be repaid with end-of-year payments of $3,000 at the end of
years 1 through 4. At the end of year 5, the remainder will be paid. What is
the year 5 payment?
76. You deposit $X in an account on your 25th, 30th, and 35th birthdays. The
account pays 9%. You intend to withdraw your savings in 10 equal annual
withdrawals on your 41st, 42nd, . . . , 50th birthdays, just depleting your
account. Just after making the withdrawal on your 45th birthday, you have
$32,801.60 left in the account. What is $X?
77.
You have $20,000 that you put on deposit on your 30th birthday at
5% compounded annually. On your 40th birthday, the account begins earning
6%. Then, on your 50th birthday it begins earning 7%. You plan to withdraw
equal annual amounts on each of your 61st, 62nd, . . . , 70th birthdays.
Summary 79
a. How much will be your annual withdrawal?
b. On your way to the bank on your 65th birthday, you decide to withdraw
the entire amount remaining. How much do you withdraw?
78. Develop a mathematical relationship for finding the accumulated amount
F at the end of n years that will result from a series of n beginning-of-year
payments each equal to B if these payments are placed in an account for
which the interest rate is i%/year.
a. Express the relationship between F and B in terms of the factors listed in
the tables of Appendix A.
b. Express the relationship between F and B in terms of i and n.
c. Demonstrate that your answers to (a) and (b) are equivalent by calculating the value of F using B 5 $1,000, n 5 5, and i 5 10% for each
approach.
Section 2.3.3
79.
Gradient Series Cash Flows
On Juan’s 26th birthday, he deposited $7,500 in a retirement account.
Each year thereafter he deposited $1,000 more than the previous year. Using
a gradient series factor, determine how much was in the account immediately after his 35th deposit if:
a. The account earned annual compound interest of 5%.
b. The account earned annual compound interest of 6%.
80. Develop a single spreadsheet that allows you to calculate the P Z G, F Z G, and
A Z G factors. For each cell where the calculation is performed, place above it
a small “control panel” that allows you to enter whatever numbers you need
to perform the calculation. For example, for the P Z G factor, your control
panel needs to include i and n. Test the P Z G and A Z G factors against the tables
in Appendix A to ensure they work right. Keep this handy. It will be useful
to you.
81. John borrows $10,000 at 18 percent compounded annually. He pays off the
loan over a 5-year period with annual payments. Each successive payment is
$700 greater than the previous payment. How much was the first payment?
82. Solve problem 81 for the case in which each successive payment is $700 less
than the previous payment.
83.
A small company wishes to set up a fund that can be used for technology
purchases over the next 6 years. Their forecast is for $12,000 to be needed at
the end of year 1, decreasing by $2,000 each year thereafter. The fund earns
8% per year. How much money must be deposited to the fund at the end of
year 0 to just deplete the fund after the last withdrawal?
84.
Video Solution Deposits are made at the end of years 1 through 7 into an
account paying 6% per year interest. The deposits start at $5,000 and increase
by $1,000 each year. How much will be in the account immediately after the
last deposit?
80
Chapter 2
Time Value of Money Calculations
85.
You want to be able to withdraw from a savings account $800 at the end
of year 1, $900 at the end of year 2, $1,000 at the end of year 3, and so on
over a total of 5 years. How much money must be on deposit right now, at
the end of year 0, to just deplete the account after the 5 withdrawals if interest is 5% compounded annually?
86. Consider the following cash flow profile:
EOY
0
1
2
3
4
Cash Flow
EOY
2$75,000
$3,000
$6,000
$9,000
$12,000
Cash Flow
5
$15,000
6
7
8
$18,000
$21,000
$24,000
Using a gradient series factor, determine the present worth equivalent for the
cash flow series using an annual compound interest rate of:
a. 6%
b. 7%
87.
A person you trust foresees the need for a loan and suggests that you loan
them $2,000 at the end of year 1, $1,000 at the end of year 2, nothing in year
3, and then they will pay you $1,000 in year 4, $2,000 in year 5, and $3,000
in year 6. They note that you will pay out a total of $3,000 to them, and then
they will pay back $6,000 to you, allowing you to “double your money.” If
you are able to make 12% per year on your investments, determine the present worth of this series of cash flows.
88. A $90,000 investment is made. Over a 5-year period, a return of $30,000
occurs at the end of the first year. Each successive year yields a return that
is $3,000 less than the previous year’s return. If money is worth 5%, use
a gradient series factor to determine the equivalent present worth for the
investment.
89.
A series of 10 end-of-year deposits is made that begins with $7,000 at the
end of year 1 and decreases at the rate of $300 per year with 10% interest.
a. What amount could be withdrawn at t 5 10
b. What uniform annual series of deposits (n 5 10) would result in the same
accumulated balance at the end of year 10.
90. Consider the following cash flow profile:
EOY
Cash Flow
EOY
0
2$45,000
$12,000
$11,000
$10,000
$9,000
5
$8,000
6
7
8
$7,000
$6,000
$5,000
1
2
3
4
Cash Flow
With a compounded annual interest rate of 6%, what single sum of money at
the end of the sixth year will be equivalent to the cash flow series?
Summary 81
91. In Problem 90, what uniform annual series over [4,7] will be equivalent to
the cash flow profile if money is worth 6% compounded annually?
92. In Problem 90, suppose the positive-valued cash flows are replaced by a
positive gradient series. If the cash flow at end-of-year 8 is $10,000, what
gradient step is required for the cash flow profiles to be equivalent?
Consider a loan of $10,000 and the following pattern of cash flows.
93.
End of
Year
0
Cash
Flow
2$10,000
$3,000
$4,000
1
2
End of
Year
Cash
Flow
3
$5,000
4
$6,000
a. What is the interest rate that makes the present worth equal to $0.00?
b. Using the interest rate determined in part (a), and leaving the 2$10,000
at year 0 in place, determine the equal annual incomes that are equivalent
to the gradient series in years 1, 2, 3, and 4?
94. Consider the following cash flow profile:
EOY
Cash Flow
EOY
0
2$50,000
$13,000
$12,000
$11,000
$10,000
5
1
2
3
4
6
7
8
Cash Flow
$9,000
$8,000
$7,000
$6,000
What is the present worth equivalent for the cash flow series with an interest
rate of 12%?
95. In Problem 94, using an interest rate of 10%, what uniform series over the
closed interval [1,8] is equivalent to the cash flow profile shown?
96. In Problem 94, using an interest rate of 8%, what single sum of money oc-
curring at the end-of-year 8 is equivalent to the cash flow profile shown?
97. In Problem 94, with an interest rate of 6%, what increasing gradient series is
equivalent to the cash flow profile shown if the gradient series sought has a
value of X at EOY 5 1 and a value of 8X at EOY 5 8?
98. A series of 25 end-of-year deposits is made that begins with $1,000 at the
end of year 1 and increases at the rate of $200 per year with a 12% interest
rate compounded annually.
a. What amount can be withdrawn at t 5 25?
b. What uniform annual series of deposits (n 5 25) would result in the same
accumulated balance at t 5 25?
82
Chapter 2
Time Value of Money Calculations
99. Piyush has recently inherited 20 million INR (Indian rupees) from his late
Uncle Scrooge. To keep Piyush from spending his money immediately,
Scrooge made arrangements for the inheritance to be deposited at the time
of his death into an account paying 5%. Further arrangements instructed the
bank to pay Piyush 2 million INR at the end of the 1st year, 2 million 1 X
INR at the end of the 2nd year, 2 million 1 2X INR at the end of the 3rd year,
2 million 1 3X INR at the end of the 4th year, and so on for a period of
10 years, just depleting the fund after the 10th payment.
a. What is the value of X?
b. How much is in the fund immediately after the 5th withdrawal?
100. Miller Machining needs to purchase a piece of machinery to be able to com-
pete on a new contract with a first-tier automotive supplier. The machinery
will cost $140,000 and the owner arranges to borrow the entire amount at
8% interest. The initial payment 1 year after purchase is $11,000 with successive payments increasing each year by $X. The last payment is to be
made 6 years after the purchase.
a. By how much ($X) does the payment increase each year?
b. What is the amount of the final payment?
c. Suppose that, at the last minute, the company decides to purchase the
same machinery at the same rate (8%), with payments decreasing by
$7,500 each year. How much is the first payment?
101.
Below is an equation to compute the present value of a cash flow series.
Determine the cash flow profile that is implied by the equation.
P 5 27,000 1 [1,850 1 200(A Z G 8%,6)](P Z A 8%,6)(P Z F 8%,4)
102. A cash flow profile starts with $2,000 and increases by $1,000 each year
up to $21,000 at time 20. Then, it starts again with $21,000 at time 21 and
decreases by $1,000 each year to $2,000 at year 40. You desire to convert it
to an equivalent gradient series beginning at year 1 with $X and continuing
through year 40 with $500 increases each year (ending at $X 1 19,500 at
time 40). Interest is 8% compounded annually. What is X?
103. Your friend claims that the following series of payments is absolutely worth-
less since they add up to $0:
End of
Year
Cash
Flow
0
$100
$80
$60
$40
$20
$0
1
2
3
4
5
End of
Year
6
7
8
9
10
Cash
Flow
2$20
2$40
2$60
2$80
2$100
The time value of money is 18%. Determine the present value of these cash
flows.
Summary 83
104. Below is an equation to compute the present value of a cash flow series.
Determine the cash flow profile that is implied by the equation.
P 5 800 1 950(P Z A i%,4) 2 450(P Z G i%,4) 2 600(P Z A i%,3)(P Z F i%,4)
105. Land is purchased for $75,000. It is agreed for the land to be paid for over
a 5-year period with annual payments and using a 12 percent annual compound interest rate. Each payment is to be $3,000 more than the previous
payment. Determine the size of the last payment.
106. Solve Problem 105 for the case in which each successive payment is to be
$3,000 less than the previous payment.
107. What single sum of money at the end of the 3rd year is equivalent to a
payment series of $10,000 the 1st year, $9,000 the 2nd year, . . . , down to
$6,000 the 5th year? Assume that money has a time value of 10%/year compounded annually.
108. Develop a mathematical relationship for finding the accumulated amount F
at the end of n years of a geometric series where the interest is i%. Put differently, you already have access to a (P Z G i%,n) factor. Develop an (F Z G i%,n)
factor.
a. Express the (F Z G i%,n) factor in terms of the existing factors listed in the
tables of Appendix A.
b. Express the (F Z G i%,n) factor in terms of i and n.
c. Demonstrate that your answers to (a) and (b) are equivalent by calculating the value of F using a first payment of $0, increasing by $1,000 each
year with n 5 5 and i 5 10% for each approach.
109. Kim deposits $1,000 in a savings account. Four years after the deposit, half
the account balance is withdrawn. Then, $2,000 is deposited annually for an
8-year period, with the first deposit occurring 2 years after the withdrawal.
The total balance is withdrawn 15 years after the initial deposit. If the account earned interest of 8 percent compounded annually over the 15-year
period, how much was withdrawn at each withdrawal point?
110. An easy payment plan offered by a local electronics store for your new au-
dio system calls for end-of-year payments of $2,000, $2,500, $3,000, and
$3,500 at the ends of years 1 through 4 respectively. Your money is well
invested and earns a consistent 10% per year.
a. What is the present worth of these payments?
b. If you prefer to make equal annual payments having the same present
worth, how much would they be?
Section 2.3.4
Geometric Series Cash Flows
111. Solve Problem 81 for the case in which each successive payment is to be
10% greater than the previous payment.
112. Solve problem 81 for the case in which each successive payment is to be
10% less than the previous payment.
84
Chapter 2
Time Value of Money Calculations
113. Solve Problem 105 for the case in which each successive payment is to be
12% greater than the previous payment.
114. Solve problem 105 for the case in which each successive payment is to be
20% less than the previous payment.
115. Consider the following cash flow profile:
EOY
Cash Flow
EOY
0
2$45,000
$12,000
$11,000
$10,000
$9,000
5
$8,000
6
7
8
$7,000
$6,000
$5,000
1
2
3
4
Cash Flow
Suppose the positive-valued cash flows are now replaced by a geometric
series. If the cash flow at end-of-year 1 is $10,000, what geometric rate is required for the cash flow profiles to be equivalent? Interest is at a compounded
annual rate of 6%.
116.
GO Tutorial Suppose you make 30 annual investments in a fund that pays
5% compounded annually. If your first deposit is $7,500 and each successive
deposit is 5% greater than the preceding deposit, how much will be in the
fund immediately after the 30th deposit?
117. In Problem 116, how much will be in the fund immediately after the 30th de-
posit if the fund pays 6% compounded annually and each successive deposit
is 6% greater than the preceding deposit?
118. Develop a single spreadsheet that allows you to calculate the (P Z A1 i%, j%,n),
(F Z A1, i%, j%,n), and (A Z A1 i%, j%,n) factors. Note that the (A Z A1 i%,j%,n)
factor is not in Appendix A. For each cell where the calculation is performed, place above it a small “control panel” that allows you to enter
whatever numbers you need to perform the calculation. For example, for the
(P Z A1 i%,j%,n) factor, your control panel needs to include i, j, and n. Test
the first two factors using i 5 7%, j 5 5% and n 5 10 against the tables
in Appendix A to ensure they work right. Then, using A1 5 $1,000, i 5
7%, and j 5 5%, calculate A using the third factor and check it against
(P Z A1 i%,j%,n)*(A Z P i%,n). Keep this handy. It will be useful to you.
119.
A famous high-volume calculus text generates royalties beginning with
$60,000 in the first year and declining each year by 40% of the previous year
due to used sales and competition. The author is on a 4-year cycle of revision. Determine the present worth of one complete cycle of royalties if the
author’s time value of money is 7%.
120. An easy payment plan offered by a local electronics store for your new
audio system calls for end-of-year payments of $2,000 at the end of year 1,
increasing by 15% each year thereafter through year 4. Your money is well
invested and earns a consistent 10% per year.
Summary 85
a. What is the present worth of these payments?
b. If you prefer to make equal annual payments having the same present
worth, how much would they be?
121.
Video Solution A $90,000 investment is made. Over a 5-year period, a
return of $30,000 occurs at the end of the first year. Each successive year
yields a return that is 10% less than the previous year’s return. If money is
worth 5%, what is the equivalent present worth for the investment?
122. You are preparing the business plan for a new company. A net revenue
analysis covering the first 6 years is required for obtaining financing. Net
revenue in year 1 is expected to be $50,000 and increase by 15% each
year, thereafter. If i 5 12% and the net revenue is assumed to be an endof-year cash flow, what is the present value of the cash flow series over the
6 years?
123. On your child’s first birthday, you open an account to fund her college edu-
cation. You deposit $2,000 to open the account. Each year, on her birthday,
you make another deposit, with each being 10% larger than the previous
deposit. The account pays interest at 5% per year compounded annually.
How much money is in the account immediately after the deposit on her
18th birthday?
124. A cash flow series is increasing geometrically at the rate of 6% per year. The
initial cash flow at t 5 1 is $1,000. The increasing payments end at t 5 20.
The interest rate in effect is 15% compounded annually. Find the present
amount at t 5 0 that is equivalent to this cash flow series.
125. A small company wishes to set up a fund that can be used for technology
purchases over the next 6 years. Their forecast is for $9,000 to be needed at
the end of year 1, increasing by 5% each year thereafter. The fund earns 10%
per year. How much money must be deposited to the fund at the end of year
0 to just deplete the fund after the last withdrawal?
126. A boat is purchased by financing $50,000. The loan is to be paid for over
a 5 year period with annual payments based on a 15% compounding rate
per year. Each successive payment is scheduled to be 10% greater than the
previous one.
a. Determine the size of the smallest payment.
b. Determine the size of the largest payment.
127. You want to be able to withdraw $1,000 from a savings account at the end
of year 1, with withdrawals increasing by 10% each year thereafter over a
total of 5 years. How much money must be on deposit right now, at the end
of year 0, to just deplete the account after the 5 withdrawals if interest is 5%
compounded annually?
128. Deposits are made at the end of years 1 through 7 into an account paying 5%
per year interest. The deposits start at $4,000 and increase by 15% each year.
How much will be in the account immediately after the last deposit?
86
Chapter 2
Time Value of Money Calculations
129.
GO Tutorial Susan gets a job upon completion of her MSME degree with
a mechanisms design firm. Her starting salary is $70,000; each successive
year she gets a 5% raise. Assuming she deposits 10% of her salary each year
into a fund earning 8% interest, how much money will she have in 10 years
for donation to her university?
130. In a new, highly automated factory, labor costs are expected to decrease at an
annual rate of 5%; material costs will increase at an annual rate of 4%; overhead costs will increase at 8%. The labor, material, and overhead costs at the
end of the first year are $2 million, $3 million, and $1.6 million, respectively.
The time value of money rate is 11% and the time horizon is 7 years.
a. Determine the dollar value for each cost category (labor, material, overhead)
for each year and the total cost for each year. (Hint: Use a spreadsheet)
b. Determine the present worth of each cost category and the total cost.
c. Determine the annual worth over 7 years that is equivalent to the present
worth of the total cost.
131. In a new, highly automated factory, labor costs are expected to decrease at an
annual compound rate of 5 percent; material costs are expected to increase
at an annual compound rate of 6 percent; and energy costs are expected to
increase at an annual compound rate of 3 percent. The labor, material, and
energy costs the first year are $3 million, $2 million, and $1,500,000.
a. What will be the value of each cost during each of the first 5 years?
b. Using an interest rate of 10 percent compounded annually, what uniform
annual costs over a 5-year period would be equivalent to the cumulative
labor, material, and energy costs?
132.
On Juan’s 26th birthday, he invested $7,500 in a retirement account.
Each year thereafter he deposited 8% more than the previous deposit. The
account paid annual compound interest of 5%. How much was in the account
immediately after his 35th deposit?
133.
In Problem 132, if Juan decided to wait 10 years before investing for
retirement, how much would he have to invest on his 36th birthday to have
the same account balance on his 60th birthday?
134.
In Problem 132, what uniform annual investment is required to achieve
the same account balance?
135. Carlson Photography receives royalties based on the use of their photo-
graphs with a major client. Every year, the client makes deposits in Carlson’s
bank account that earns 6% compounded annually. The client increases the
amount they deposit into the Carlson account by 4%. If the client initially
gives Carlson $15,000, how much will the account have in 5 years?
Section 2.4.1
Period Interest Rate
136. How many monthly payments are required to repay a loan of $12,000 with
an interest rate of ¾% per month and end-of-month payments of $400?
Summary 87
137.
A total of $50,000 is borrowed and repaid with 60 monthly payments,
with the first payment occurring one month after receipt of the $50,000.
The stated interest rate is 6% compounded monthly. What monthly payment
should be made?
138. A refrigerator sold for $500. The store financed the refrigerator by charging
0.5% monthly interest on the unpaid balance. If the refrigerator is paid for
with 30 equal end-of-month payments:
a. What will be the size of the monthly payments?
b. If the first payment is not made until one year after the purchase, what will
be the size of the monthly payments?
139.
You decide to open a retirement account at your local bank that pays 8%/
year/month (8% per year compounded monthly). For the next 20 years you
will deposit $400 per month into the account, with all deposits and withdrawals occurring at the end of the month. On the day of the last deposit, you will
retire. Your expenses during the first year of retirement will be covered by your
company’s retirement plan. As such, your first withdrawal from your retirement account will occur on the day exactly 12 months after the last deposit.
a. What monthly withdrawal can you make if you want the account to last
15 years?
b. What monthly withdrawal can you make if you want the account to last
forever (with infinite withdrawals)?
140. Wei Min opens a retirement account that pays 8%/year/month. For the next
30 years he deposits $300 per month into it, with all deposits occurring at the
end of the month. On the day of the last deposit, Wei Min retires. As a benefit
to retirees, the bank increases the interest rate to 12%/year/quarter from that
time on. His first withdrawal will occur exactly 2 years after his last deposit.
He then plans to make equal quarterly withdrawals from the account.
a. What is the balance of the account immediately after the last monthly
deposit?
b. What is the balance of the account one quarter before the first quarterly
withdrawal?
c. What quarterly amounts can be withdrawn to last for 15 years?
141.
Video Solution Your boss, who never took an engineering economy
course is buying a new house and needs your help in answering some questions. The loan amount will be in the “jumbo loan” category of $600,000
at (1) 7.0% per year compounded monthly over 30 years, or (2) 6.625%
compounded monthly over 15 years. There are no loan initiation fees, points
paid, or other charges. Prepayment, if desired, can be done without penalty.
a. What is the monthly payment for plan (1)?
b. What is the monthly payment for plan (2)?
c. What is the effective annual interest rate for plan (1)?
d. What is the effective annual interest rate for plan (2)?
e. What is the total interest paid over the life of loan (1)?
f. What is the total interest paid over the life of loan (2)?
88
Chapter 2
Time Value of Money Calculations
142. Yavuz wishes to make a single deposit P at time t 5 0 into a fund paying
15 percent compounded quarterly such that $1,000 payments are received at
t 5 1, 2, 3, and 4 (periods are 3-month intervals), and a single payment of
$7,500 is received at t 5 12. What single deposit is required?
143.
You invest $10,000 in a fund that pays 7% per year for 5 years. How much
is in the fund at the end of 5 years if (forgetting leap years and making
“convenient” assumptions):
a. Compounding is annual?
b. Compounding is quarterly?
c. Compounding is monthly?
d. Compounding is daily?
144.
Video Solution How much money must be deposited now in order to be
able to withdraw $10,000 in 4 years if interest is 5% compounded quarterly?
145.
If you deposit $4,000 into an account paying 6% per year compounded
semiannually, how much will you have in the account after 10 years?
146. Lynn borrows $5,000 at 15 percent per year compounded monthly. She
wishes to repay the loan with 12 end-of-month payments. She wishes to
make her first payment 1 month after receiving the $5,000. She also wishes
that, after the first payment, the size of each payment be 10 percent greater than
the previous payment. What is the size of her 6th payment?
147. Solve problem 146 for the case in which the size of each payment is $60
greater than the previous payment.
148. Mary Lib purchases a house for $450,000. She makes a down payment of
$40,000 at the time of purchase, and the balance is financed at 6.0 percent
compounded monthly, with monthly payments made over a 10-year period.
a. What is the size of the monthly payments?
b. If the loan period had been 20 years, what would have been the size of the
monthly payments?
149.
GO Tutorial What equal monthly investment is required over a period of
40 years to achieve a balance of $2,000,000 in an investment account that
pays monthly interest of ¾%?
Section 2.4.2
Effective Annual Interest Rate
150. Develop a single spreadsheet that allows you to (1) calculate the effective
annual interest rate given the nominal annual interest rate and the number of
compounding periods per year, and (2) calculate the nominal annual interest
rate given the effective annual interest rate and the number of compounding
periods per year. Use the “control panel” approach that allows you to enter
whatever numbers you need to perform the calculation. Keep this handy. It
will be useful to you.
151.
What is the effective annual interest rate for 10% compounded (a) semiannually, (b) every four months, (c) quarterly, (d) every other month,
(e) monthly?
Summary 89
152.
Video Solution What is the effective annual interest rate for 5% compounded (a) semi-annually, (b) every four months, (c) quarterly, (d) every
other month, (e) monthly?
153.
You borrow $2,000 from Gougo’s, a well-known loan consolidation
outfit. The loan is an “unbelievably low” 2.5% per month compounded
monthly. You have 2 years to pay back the loan.
a. What is the nominal interest rate?
b. What is the effective interest rate?
c. If you wait until the end of year 2 to pay it off in one lump sum, how
much must you pay? Use the “period interest rate” approach.
d. If you wait until the end of year 2 to pay it off in one lump sum, how
much must you pay? Use the “effective interest rate” approach.
e. Of your payment in parts (c) or (d), how much is interest?
f. Suppose you make equal end-of-month payments. How much is the
monthly amount?
154. How much money must be invested in an account that pays 6% per year in-
terest to be worth $20,000 at the end of 8 years if (forgetting leap years and
making “convenient” assumptions):
a. Interest is compounded annually?
b. Interest is compounded semi-annually?
c. Interest is compounded quarterly?
d. Interest is compounded monthly?
e. Interest is compounded weekly?
f. Interest is compounded daily?
g. Interest is compounded hourly?
h. Interest is compounded minutely?
i. Interest is compounded secondly?
155.
You have your eyes on a new automobile costing $25,000. If you had the
$25,000 and wrote a check for that amount, you could drive off in your new
car. You don’t have it, and must finance $20,000 through the dealership at
15%/year/month over a 5-year period. The dealer then proceeds to add on
a 1.25% loan initiation fee of $250. Also, they have a prepaid loan closeout
fee of another $250. Then there is the paperwork filing and storage fee of
another $100 and another prepaid loan maintenance fee of only $8/month
or $480. At this point, they are speaking very fast and assure you that these
little “required” amounts are routine and can be rolled into your loan. They
figure your monthly payment for the $20,000 loan as A 5 $21,080 (A Z P
1.25,60) 5 $501.49.
a. What is the monthly rate of “interest” you are really paying for the
$20,000 loan?
b. What is the nominal annual “interest” rate you are really paying for the
$20,000 loan?
c. What is the effective annual “interest” rate you are really paying for the
$20,000 loan?
90
Chapter 2
Time Value of Money Calculations
156. Naihui and Haiyan deposit $250 into their joint account at the end of each
month. They want to have a total of $12,000 in their account after 40 months:
a. What monthly rate of interest must they earn?
b. What nominal annual rate of interest must they earn?
c. What effective annual rate of interest must they earn?
157. Barbara makes four consecutive annual deposits of $2,000 in a savings ac-
count that pays interest at a rate of 10 percent compounded semiannually.
How much money will be in the account 2 years after the last deposit?
158.
Video Solution You are down on your luck and need a loan, quick!
You locate Mr. Loa N. Shark who advertises weekly loans for “an almost
imperceptibly small rate” of only 3%, prepaid at the time of the loan. You
sign over your federal tax refund for $1,000 to Mr. Shark, with proof that it
is correct and will be forthcoming from the IRS in one week.
a.
b.
c.
d.
How much money does Mr. Shark hand you?
How much weekly interest are you really paying?
What is the nominal annual interest rate?
What is the effective annual interest rate?
Section 2.4.3
159.
Differing Frequencies of Compounding and Cash Flows
Daniel deposits $20,000 into an account earning interest at 6% per year
compounded quarterly. He wishes to withdraw $400 at the end of each
month. For how many months can he make these withdrawals?
160. Daniel deposits $20,000 into an account earning interest at 6% per year
compounded monthly. He wishes to withdraw $1,200 at the end of each
quarter. For how many quarters can he make these withdrawals?
161.
GO Tutorial Mario and Claudia deposit $100 into their joint account at
the end of each month. If their account earns 7%/year/quarter (7% per year
compounded quarterly), how long will it take them to have a total of $15,000 in
their savings account?
162.
Video Solution A total of $50,000 is borrowed and repaid with 60 monthly
payments, with the first payment occurring one month after receipt of the
$50,000. The stated interest rate is 6% compounded quarterly. What monthly
payment is required?
3
ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E —
S AM U E L WAS HI N G TO N TAK E S A LOAN
Fifteen years after graduating in electrical engineering and accepting employment with Texas Instruments, Samuel Washington decides to establish a
consulting business. Although he has invested wisely for the past 15 years,
the value of his investments is only $325,000. After developing a business
plan, he realizes he will need $250,000 on hand initially, plus $150,000 each
successive year, to cover the expenses of an office and an assistant. He is
unsure about how much of his own money he should use and how much to
borrow. In talking to the loan officer of a local bank, he learns that the bank
will charge him annual compound interest of 6 percent for a 5-year loan
period or 5.5 percent for a 10-year loan period. Over the past 10 years, Samuel
earned an average of 5.25 percent annually on his investments; he believes
he will continue to earn at least that amount on his investment portfolio. If
he borrows money, he can repay the loan in several ways: pay accumulated
interest monthly, plus pay the principal at the end of the loan period; make
equal monthly payments; make monthly payments that increase like a gradient series; make monthly payments that increase like a geometric series; or
make a lump sum payment at the end of the loan period. Because this is a
business investment, any interest paid can be deducted from his taxable
income.
DISCUSSION QUESTIONS:
1. Discuss the quantitative (economic) tradeoffs that Samuel should consider when he decides how much money to use from his personal
savings versus borrowing money from the bank.
92
EQUIVALENCE, LOANS,
AND BONDS
2. What qualitative (noneconomic) factors should Samuel consider when
he decides how much money to use from his personal savings versus
borrowing money from the bank?
3. What types of assumptions is Samuel making when he determines his
loan needs? What happens if these assumptions do not hold?
4. How might Samuel set out to secure funding for this proposed business venture?
LEARNING OBJECTIVES
When you have finished studying this chapter, you should be able to:
1. Compare the equivalence between two or more cash flow profiles
(Section 3.1).
2. Analyze immediate payment and deferred payment loans, including
payment amount, remaining balance, and interest and principal per
payment (Section 3.2).
3. Analyze investments in bonds and determine the purchase price, selling price, and return on such investments (Section 3.3).
4. Calculate the worth of a cash flow profile with variable interest rates
(Section 3.4).
5. Explain the Annual Percentage Rate (APR) commonly calculated for a
home mortgage (Section 3.5).
INTRODUCTION
This chapter builds on the foundation established in Chapter 2. With one
exception, in this chapter you will learn how to analyze the financial needs
Samuel Washington faces. How to incorporate income tax effects in his
analysis is reserved for Chapter 9. In addition to learning how to perform
93
94
Chapter 3
Equivalence, Loans, and Bonds
the analysis Samuel performed, you will learn how to analyze various
bond transactions, and how to determine the amount of each loan payment that is an interest versus a principal payment. Throughout the chapter, you will see how to perform the analyses using tabulated values of
compound interest factors and using Excel®’s financial functions.
3-1
EQUIVALENCE
LEARN I N G O B JEC T I V E : Compare the equivalence between two or more
cash flow profiles.
Equivalence The state
of being equal in value,
such as with two cash flow
profiles.
In engineering economic analyses, equivalence means “the state of being
equal in value.” The concept is primarily applied in the comparison of two
or more cash flow profiles. A commonly used approach to determine
equivalence is to compare the present worth of the cash flow profiles. If
they are equal, then the cash flow profiles are equivalent. Equivalence
problems, however, generally involve a parameter whose value is to be
determined in order for the two cash flow profiles to be equivalent. Typically, the parameter is the interest rate or a cash flow. For example, are the
following two cash flow profiles equivalent at 15 percent compounded
annually?
Cash Flow Profile 1: Receive $1,322.50 two years from today
Cash Flow Profile 2: Receive $1,000 today.
Computing the present worths for the two cash flow profiles yields
PW112 5 $1,322.501P Z F 15%,22
5 $1,322.5011.152 22
5 $1,322.5010.756142 5 $1,000
5PV115%,2,,21322.52 5 $1,000
PW122 5 $1,000
Because the present worths are equal, the two cash flow profiles are
equivalent.
Notice, we did not obtain the value of the (P Z F 15%,2) factor from
Appendix A. Instead, we calculated the value of the factor to five decimal
places in order to demonstrate that equivalency existed. (That is why the
Excel® results were identical to the calculated results.)
Based on the foundation established in Chapter 2, it should be obvious
that the worths of the two cash flow profiles will be the same at any particular point in time, e.g., t 5 2 and t 5 6.
3-1 Equivalence
A Uniform Series Equivalency of a Decreasing
Gradient Series
EXAMPLE
Using an 8 percent discount rate, what uniform series over five periods,
[1, 5], is equivalent to the cash flow profile given in Figure 3.1?
Video Example
i = 8%
$500
$400
$300
$200
$100
0
1
2
3
4
5
6
7
The Decreasing Gradient CFD for Example 3.1
FIGURE 3.1
The cash flow profile in Figure 3.1 consists of the difference in a uniform
series of $500 and a gradient series, with G 5 $100. A uniform series equivalent of the cash flow profile can be obtained over the interval [2, 6] as follows:
SOLUTION
A32, 64 5 $500 2 $1001A Z G 8%,52
5 $500 2 $10011.846472
5 $315.35
$315.35
$315.35
$315.35
$400
$315.35
i = 8%
$500
$315.35
Figure 3.2 shows the CFD for the uniform series that is equivalent to the
original decreasing gradient series.
2
3
4
5
6
$300
$200
$100
0
1
FIGURE 3.2
2
3
4
5
6
7
=
0
1
Equivalent Gradient and Uniform CFDs for Example 3.1
But we still need to convert the uniform series from interval [2, 6] to [1, 5].
To do so, the entire series must be shifted backward in time one time unit.
From DCF Rule #4, we move money backward in time one time unit by
dividing by 1 plus the interest rate. Hence, as shown in Figure 3.3, the
equivalent worth over the interval [1, 5] is
A31, 54 5 $315.351P Z F 8%,12
5 $315.3510.925932
5 $291.99
7
95
$300
$291.99
$400
$291.99
i = 8%
$500
$291.99
Equivalence, Loans, and Bonds
$291.99
Chapter 3
$291.99
96
1
2
3
4
5
$200
$100
0
1
FIGURE 3.3
2
3
4
5
6
=
7
0
6
7
Equivalent Gradient CFD and Converted Uniform CFD for Example 3.1
Consequently, a uniform series of $291.99 over the interval [1, 5] is equivalent to the cash flow profile given in Figure 3.1. (If you have doubts concerning the equivalence, compare their present worths using an 8 percent
interest rate.)
Determining an Equivalent Gradient Step
EXAMPLE
Consider Figure 3.4. Determine the value of X that makes the two cash
flow profiles equivalent using a TVOM of 15 percent.
i = 15%
0
1
2
3
4
=
$200
0
1
2
3
4
$200
$300
$300
$200 + X
$400
$200 + 2X
$200 + 3X
FIGURE 3 . 4
SOLUTION
CFDs for Example 3.2
Equating the future worths of the two cash flow profiles at t 5 4 gives
$2001F Z A 15%,42 1 $1001F Z A 15%,32 1 $100 5
3$200 1 X1A Z G 15%,42 4 1F Z A 15%,42
3-1 Equivalence
Eliminating the common term of $2001F Z A 15%,42 yields
$10013.472502 1 $100 5 X11.326262 14.993382
Solving for X gives a value of $67.53.
This example offers an opportunity to use either Excel®’s SOLVER
tool or Excel®’s GOAL SEEK tool. To use SOLVER, the data are entered
in a spreadsheet as shown in Figure 3.5. The present worth of the cash
flows to the left of the equality in Figure 3.5 is set equal to the present
worth of the cash flows to the right of the equality.
FIGURE 3.5
Using the Excel® SOLVER Tool to Solve Example 3.2
Notice, the unknown is given by cell E10. The target cell, E9, is to be
made equal to $826.71, which is the present worth of the equality’s lefthand side. The cash flows on the equality’s right-hand side appropriately
include the value of cell E10. When SOLVER is applied, the value in cell
E10 is changed to $67.53 with the result that the present worth in cell B9
equals the present worth in cell E9.
97
98
Chapter 3
Equivalence, Loans, and Bonds
Determining an Equivalent Interest Rate
EXAMPLE
For what interest (discount) rate are the two cash flow profiles shown in
Figure 3.6 equivalent?
i=?
$3,500
$3,000
$2,500
$2,000
$1,500
$1,500 $1,500 $1,500 $1,500 $1,500
0
1
2
3
4
5
=
0
1
2
3
4
5
$4,000
$7,000
FIGURE 3 . 6
SOLUTION
CFDs for Example 3.3
Converting each cash flow profile to a uniform series over the interval
[1, 5] gives
2$4,0001A Z P i%,52 1 1,500 5 27,0001A Z P i%,52 1 1,500 1
$5001A Z G i%,52
or
$3,0001A Z P i%,52 5 $5001A Z G i%,52
which reduces to
1A Z G i%,52 5 61A Z P i%,52
Searching through the interest tables at n 5 5, the value of the
(A Z G i%,5) factor is six times the value of the (A Z P i%,5) factor for an
interest rate between 12 percent and 15 percent. Specifically, with a 12
percent interest rate,
1A Z G 12%,52 2 61A Z P 12%,52 5 1.77459 2 610.277412 5 0.11013
and, using a 15 percent interest rate,
1A Z G 15%,52 2 61A Z P 15%,52 5 1.72281 2 610.298322 5 20.06711
3-1 Equivalence
Interpolating for i gives
i 5 0.12 1 10.15 2 0.122 10.110132y10.11013 1 0.067112
or
i 5 0.13864
Therefore, using a discount rate of approximately 13.864 percent will
establish an equivalence relationship between the cash flow profiles given
in Figure 3.6.
Excel® can be used to determine the equivalency. As shown in Figure 3.7, we set the present worth of the equality’s left-hand side equal to
its right-hand side. Excel®’s NPV function is used to compute the present
worth of each cash flow stream. In cell E11 is entered the difference in the
present worth of the left-hand side (B10) and the present worth of the
right-hand side (E10). Each present worth is calculated using a value for
the interest rate given in cell E12.
FIGURE 3.7
Using the Excel® SOLVER Tool to Solve Example 3.3
99
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Chapter 3
Equivalence, Loans, and Bonds
The Excel® SOLVER tool is used to set the target cell to E11.
(Because we want the two present worths to be equal, E11 must equal
zero.) To make E11 equal 0, cell E12’s value, the interest rate, must be
changed. We initialize the underlying search performed by SOLVER by
setting E12 equal to 10 percent. After applying SOLVER, as shown in
Figure 3.8, a value of 13.8677 percent is obtained for the interest rate,
which is quite close to the value obtained by interpolating values obtained
from tables in Appendix A. (The Excel® GOAL SEEK tool also can be
used to solve the example.)
FIGURE 3 . 8
3-2
Solution to Example 3.3
INTEREST PAYMENTS
AND PRINCIPAL PAYMENTS
LEARN I N G O B JEC T I V E : Analyze immediate payment and deferred payment
Video Lesson:
Loans—Interest and
Principal
Principal Payment The
principal on a loan refers
to the amount borrowed.
Thus, the principal
payment (or equity
payment) is the portion
of the loan payment that
reduces the unpaid balance.
loans, including payment amount, remaining balance, and interest and principal per payment.
In Chapter 9, we examine the after-tax effects of borrowing money. For
businesses, interest charges can be deducted from taxable income. Similarly, for primary residences, interest payments are tax deductible. For
these reasons, it is useful to know how to compute the amount of a loan
payment that is interest, with the balance of the payment being a principal
payment.
What do we mean by a principal payment? The principal on a loan is
the amount borrowed. The unpaid balance on a loan is the sum of the
unpaid interest and the unpaid principal. As loan payments are made, over
3-2
Interest Payments and Principal Payments
time the unpaid balance on a loan declines until the final payment reduces
the unpaid balance to zero.
In determining the amount of a loan payment that is interest, it is helpful to remember the following: The first thing paid in repaying a loan is
interest. If the accumulated unpaid interest on a loan exceeds the magnitude of the loan payment then all of the payment is an interest payment.
Loan payments do not include a principal payment if the unpaid interest
equals or exceeds the magnitude of the loan payment.
Consider a loan in the amount of $10,000 at an interest rate of 1 percent/
month. If the first loan payment is made one month after receiving the
$10,000, then the amount of unpaid interest at the time of the first payment
is $10,000(0.01) or $100. Hence, the amount of the first payment that will
be a principal payment is equal to the magnitude of the loan payment less
the $100 interest payment. (Principal payments are often referred to as
equity payments. We use the terms interchangeably.)
Suppose you borrow $P and repay the loan with n payments at times t
through t 1 n 2 1, where t $ 1. The payments to be made are denoted by
At, At11, . . . , At1n21 .
Immediately after making the kth payment, you decide to pay off the
loan. The amount you repay is given by the present worth of the n 2 k payments remaining to be paid. Likewise, if you decide to pay off the loan
immediately after making the k 1 1st payment, the amount owed is equal
to the present worth of the n 2 k 2 1 payments remaining to be paid. The
difference in the two present worth values is the amount of principal in the
k 1 1st payment.
The segregation of a loan payment into interest and principal components can be performed for any set of loan payments. However, because the
most commonly used payment schedule is one in which the set of payments
is a uniform series, we will develop results for such a payment schedule. Two
cases are considered: immediate payment loans and deferred payment loans.
3.2.1
Immediate Payment Loans
Consider a loan in which the first payment is made one interest period after
receipt of the loan principal. If $P is borrowed at t 5 0 with a period interest
rate of i%/period for n periods and is repaid with n equal end-of-period
payments (starting at t 5 1), then the magnitude of each payment, $A, is
given by $A 5 $P(A Z P i%,n). Such a loan is called an immediate payment
loan. From the preceding discussion, the amount of principal remaining to
be repaid immediately after making a payment at time t, Ut, is given by
Ut 5 A1P Z A i%,n 2 t2
where A, n, and i are defined as noted above.
(3.1)
Immediate Payment
Loans When the loan
payment begins one
interest period after
receipt of the principal.
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Chapter 3
Equivalence, Loans, and Bonds
A related quantity is the payoff quantity. Payofft is the total amount
required to pay off the loan at time t, including both the current payment
and the unpaid balance.
(3.2)
Payofft 5 A 1 Ut 5 A 1 A1P Z A i%,n 2 t2
5 A31 1 1P Z A i%,n 2 t2 4
(3.3)
The interest accrued during any payment period is given by the product
of the unpaid balance at the beginning of the period and the period interest
rate. Letting It denote the portion of payment t that goes to pay interest, we
have
(3.4)
It 5 iUt21
Substituting the results of Equation 3.1 into Equation 3.4, the following relationship can be derived.
(3.5)
It 5 A 31 2 1P Z F i%,n 2 t 1 12 4
The portion of the payment that does not pay accrued interest goes to
principal reduction. Letting Pt denote the portion of payment t that is a
principal payment (goes to reduce principal), we have
or
Pt 5 A 2 It
(3.6)
Pt 5 A1P Z F i%,n 2 t 1 12
(3.7)
Excel® has two financial functions that apply directly to this material:
IPMT and PPMT. The IPMT worksheet function determines the amount
of a periodic payment that is interest and has the following parameters:
interest rate, period for which the value is sought, number of periodic payments made, present amount, future amount, and type. The PPMT worksheet function determines the amount of a periodic payment that reduces
the unpaid principal on a loan; it has the same parameters as IPMT. For
conventional loans, the future amount and type parameters are not needed.
EXAMPLE
Purchasing a Car
Sara Beth wants to purchase a used car in excellent condition. She has decided
on a car with low mileage that will cost $20,000. After considering several
alternatives, she identified a local lending source that will charge her an
interest rate of 6 percent per annum compounded monthly for a 48-month
loan: (a) What will be the size of her monthly payments? (b) What will be the
remaining balance on her loan immediately after making her 24th payment?
(c) If she chooses to pay off the loan at the time of her 36th payment, how
much must she pay? (d) What portion of her 12th payment is interest?
(e) What portion of her 12th payment is an equity payment?
3-2
Interest Payments and Principal Payments
Given: P 5 $20,000; i 5 6%; compounded monthly; n 5 48
Find: A, U24, Payoff36, I12, P12
a.
A 5 P1A Z P i%,n2
Period interest rate 5 6 percent/12 months 5 0.5 percent/month
A 5 $20,0001A Z P 0.5%,482
5 $20,00010.023492
5 $469.80
5 PMT10.5%,48,2200002
5 $469.70
b.
Ut 5 A1P Z A i%,n 2 t2 n 5 48 t 5 24 n 2 t 5 48 2 245 24
U24 5 A1P Z A 0.5%,242
5 $469.80122.562872
5 $10,600.04
5 PV10.5%,24,2PMT10.5%,48,2200002 2
5 $10,597.79
c.
d.
Payofft 5 A 1 A1P Z A i%,n 2 t2
n 5 48 t 5 36 n 2 t 5 48 2 36 5 12
Payoff36 5 A 1 A1P Z A 0.5%,122 5 $469.80 1 $469.80111.618932
5 $5,928.37
5 PV10.5%,12,2PMT10.5%,48,2200002 2
1 PMT10.5%,48,2200002
5 $5,927.12
It 5 iUt21
n 5 48 t 5 12 t 2 1 5 12 2 1 5 11
U11 5 A1P Z A 0.5%,372 5 $469.80133.702502 5 $15,833.44
I12 5 0.0051$15,833.442 5 $79.17
5 IPMT10.5%,12,48,2200002 5 $79.15
e.
Pt 5 A 2 It
t 5 12
P12 5 A 2 I12
5 $469.80 2 $79.17 5 $390.63
5 PPMT10.5%,12,48,2200002 5 $390.55
KEY DATA
SOLUTION
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Chapter 3
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3.2.2 Deferred Payment Loans
Deferred Payment
Loans When the loan
payment begins more than
one interest period after
receipt of the principal.
Repayment of loans does not always begin one interest period after receipt
of the principal. For any number of reasons, some individuals and organizations arrange to delay the beginning of payment for one or more interest
periods. Such loans are called deferred payment loans and are the subject
of this section.
Determining the amount of a deferred payment that is interest versus
equity is not as straightforward as it is for immediate payment loans. However, a tabular approach can be used in this situation, as illustrated in
Example 3.5. (Importantly, until the accumulated interest resulting from
deferring payments is paid, no principal payments occur.)
To formalize the computation of interest payments and principal payments when deferred payments are made on a loan, we introduce the following notation. Let
Princ
IR
UB
UIB
5 loan principal
5 interest rate on the loan
5 unpaid balance at the beginning of the interest period
5 unpaid accumulated interest immediately before making a
payment
UIA 5 unpaid accumulated interest immediately after making a
payment
AO 5 amount owed just before making a payment
Int 5 interest earned during the period 5 UB 3 IR
5 size of the deferred payment
Ad
IPmt 5 amount of the payment that is an interest payment
PPmt 5 amount of the payment that is a principal payment 5 Ad 2 IPmt
The amount of a payment that is interest is either the accumulated
interest immediately before making a payment or the entire payment,
whichever is the lesser, as expressed in Equation (3.8):
IPmt 5 min1UIB; Ad 2
(3.8)
The difference in the payment and the interest payment is the principal
payment. Example 3.5 illustrates the use of this procedure with a deferred
payment loan.
EXAMPLE
Interest and Equity in Deferred Payments
The owner of a small business borrows $10,000 at 15 percent annual compound interest. Five equal annual payments will be made to repay the loan,
but the first will not occur until 4 years after receipt of the principal amount.
How much of each payment will be interest and principal?
3-2
Interest Payments and Principal Payments
Given: P 5 $10,000; IR 5 15%; n 5 5
Find: Deferred payment amount (Ad); principal and interest amounts for
each payment
KEY DATA
The size of the new, deferred payments will be
SOLUTION
105
Ad 5 $10,0001F Z P 15%,32 1A Z P 15%,52
5 $10,00011.520882 10.298322 5 $4,537.09
5 PMT115%,5,2FV115%,3,,2100002 5 $4,537.01
Table 3.1 summarizes the calculations to determine the amount of each
payment that will be interest and equity. Because no payments are made
until 4 years after receiving the $10,000, the amount owed immediately
before making the first deferred payment equals $10,000(F Z P 15%, 4), or
$17,490.10 (based on the compound interest values given in Appendix A)
or $17,490.06 (using Excel®’s FV function). Note that the last column of
Table 3.1, unpaid balance after payment (UBA), contains the value carried
into the following payment period as unpaid balance before payment (UB).
The first deduction from a payment is the interest on the unpaid balance. Because the amount owed ($17,490.06) includes $7,490.06 in accumulated interest, which is greater than a full payment ($4,537.01), all of
the first payment is an interest payment.
Likewise, the amount owed before making the second payment
($14,896.01) includes accumulated interest of $4,896.01, which is also
greater than a full payment ($4,537.09). Therefore, all of the second payment is an interest payment.
TABLE 3.1
Year
Interest and Equity Payments in a Deferred Payment Loan
Unpaid
Balance
Before
Payment
(UB)
Interest
During
Year (Int)
Unpaid
Interest
Before
Payment
(UIB)
Amount
Owed (AO)
Loan
Payment
(Ad )
Interest
Payment
(IPmt)
Principal
Payment
(PPmt)
Unpaid
Interest
After
Payment
(UIA)
Unpaid
Balance
After
Payment
(UBA)
1
$10,000.00
$1,500.00
$1,500.00
$11,500.00
$0.00
$0.00
$0.00
$1,500.00
$11,500.00
2
$11,500.00
$1,725.00
$3,225.00
$13,225.00
$0.00
$0.00
$0.00
$3,225.00
$13,225.00
3
$13,225.00
$1,983.75
$5,208.75
$15,208.75
$0.00
$0.00
$0.00
$5,208.75
$15,208.75
4
$15,208.75
$2,281.31
$7,490.06
$17,490.06
$4,537.01
$4,537.01
$0.00
$2,953.06
$12,953.06
5
$12,953.06
$1,942.96
$4,896.01
$14,896.01
$4,537.01
$4,537.01
$0.00
$359.01
$10,359.01
6
$10,359.01
$1,553.85
$1,912.86
$11,912.86
$4,537.01
$1,912.86
$2,624.15
$0.00
$7,375.85
7
$7,375.85
$1,106.38
$1,106.38
$8,482.23
$4,537.01
$1,106.38
$3,430.63
$0.00
$3,945.22
8
$3,945.22
$591.78
$591.78
$4,537.01
$4,537.01
$591.78
$3,945.22
$0.00
UBt 5 UBAt–1
Intt 5
IRt 3 UBt
UIBt 5
Intt 1 UIAt21
AOt 5
UBt 1 Intt
Adt
IPmt 5
min(UIBt; Adt)
PPmtt 5
Adt 2 IPmtt
UIAt 5
UIBt 2 IPmtt
$0.00
UBAt 5
AOt 2 Adt
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Chapter 3
Equivalence, Loans, and Bonds
The amount owed at the time of the third payment ($11,912.86)
includes accumulated unpaid interest totaling $1,912.86, which is less
than a full payment ($4,537.01). Therefore, the $1,912.86 in accumulated
interest is an interest payment; the $2,624.15 remaining in the $4,537.01
payment reduces principal.
Just before making the fourth payment, $8,482.23 is owed, $1,106.38
of which is unpaid interest. Therefore, the $1,106.38 in interest earned
during the year is deducted from the payment, and the $3,430.63 balance
reduces principal.
Finally, the amount owed just before the fifth and final payment is
made totals $4,537.01, the size of a payment. The interest charged during
the year equals $591.78. Therefore, the balance of $3,945.22 is a principal
payment and reduces the unpaid balance to zero.
3-3
Bond A long-term note
issued by a borrower to
a lender, typically for the
purpose of financing a
large project.
Face or Par Value The
stated value (or face value)
on an individual bond.
Redeem To pay a bond
holder a value as specified
by the terms and conditions
of the bond. A unit
issuing a bond is obligated
to redeem it.
Maturity A specified
period of time at which a
bond reaches its par value.
Bond Rate The amount
that an issuing unit is
obligated to pay on the par
value of the bond during
the interim between the
date of issuance and the
date of redemption.
BOND INVESTMENTS
LEARN I N G O B JEC T I V E : Analyze investments in bonds and determine the
purchase price, selling price, and return on such investments.
A bond is a long-term note issued by the borrower (normally a corporation or governmental agency) to the lender, typically for the purpose of
financing a large project. This note specifies terms of repayment and other
conditions.
Bond investments are attractive vehicles for applying the DCF models
developed in Chapter 2. Also, they prepare you for material to be covered
in subsequent chapters.
Individual bonds are normally issued in denominations such as $1,000.
The stated value on the individual bond is termed the face or par value.
The par value is to be repaid by the issuing organization at the end of a
specified period of time, say 5, 10, 15, 20, or even 50 years. Thus, the issuing unit is obligated to redeem the bond at par value at maturity.
Furthermore, the issuing unit is obligated to pay a stipulated bond rate
on the face value during the interim between date of issuance and date of
redemption. This might be 10 percent per year payable quarterly, 8 percent
per year payable semiannually, 9 percent per year payable annually, and so
forth. For the purposes of the following examples, it is emphasized that the
bond rate applies to the par value of the bond.
We now employ the following notation:
P 5 the purchase price of a bond
F 5 the sales price (or redemption value) of a bond
3-3
Bond Investments
V 5 the par or face value of a bond
r 5 the bond rate (coupon rate) per interest period
i 5 the yield rate (return on investment or rate of return) per interest
period
n 5 the number of interest (coupon) payments received by the bondholder
A 5 Vr 5 the interest or coupon payment received per interest period
The general expression relating these terms is
P 5 Vr1P Z A i%,n2 1 F1P Z F i%,n2
(3.9)
Three types of bond problems are considered:
Given P, r, n, V, and a desired i, find the sales price F.
2. Given F, r, n, V, and a desired i, find the purchase price P.
3. Given P, F, r, n, and V, find the yield rate i that has been realized.
1.
Each of these cases is illustrated in the following examples.
Determining the Selling Price for a Bond
On January 1, 2011, Austin plans to pay $1,050 for a $1,000, 12 percent
semiannual bond. He will keep the bond for 3 years, receive six coupon
payments, and then sell it. How much should he sell the bond for in order
to receive a yield of 10 percent compounded semiannually?
EXAMPLE
Video Example
Given: P 5 $1,050; V 5 $1,000; n 5 6; r 5 6%; i 5 5%
Find: F
KEY DATA
The present worth for the bond investment is given by
SOLUTION
P 5 Vr1P Z A 5%,62 1 F1P Z F 5%,62
or
$1,050 5 1$1,0002 10.062 15.075692 1 F10.746222
Solving for F yields a value of $998.98. As long as the selling price at the
end of 3 years is at least $998.98, the bond will yield a return of at least
10 percent compounded semiannually.
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Chapter 3
Equivalence, Loans, and Bonds
The Excel® FV worksheet function is well suited for this example. Recall that
the syntax for the FV worksheet function is FV(rate,nper,pmt,pv,type).
Therefore, entering
5FV15%,6,60,210502
in any cell in an Excel® spreadsheet will produce the following result:
$998.99. (The 1¢ difference in results is due to round-off errors in the
interest tables.)
Determining the Purchase Price for a Bond
EXAMPLE
Emma plans to purchase a $1,000, 12 percent semiannual bond, hold it for
3 years, receive six coupon payments, and redeem it at par value. What is
the maximum amount she should pay for the bond if she wants to earn at
least 14 percent compounded semiannually on her investment?
KEY DATA
SOLUTION
Given: F 5 $1,000; V 5 $1,000; n 5 6; r 5 6%; i 5 7%
Find: P
The present worth for the bond investment is given by
P 5 Vr1P Z A 7%,62 1 F1P Z F 7%,62
or
P 5 1$1,0002 10.062 14.766542 1 $1,00010.666342
Solving for P yields a value of $952.33. As long as the purchase price
is no greater than $952.33, Emma’s rate of return on her investment will be
at least 14 percent compounded semiannually.
The Excel® PV worksheet function matches perfectly with this bond
calculation. Recall the PV syntax is PV(rate,nper,pmt,fv,type). Therefore, entering
5PV17%,6,260,210002
in any cell in an Excel® spreadsheet will give the answer sought: $952.33.
3-3
Determining the Rate of Return for a Bond Investment
Bond Investments
EXAMPLE
Charlotte purchased a $1,000, 12 percent quarterly bond for $1,020, kept
it for 3 years, received twelve coupon payments, and sold it for $950. What
was her quarterly yield on her bond investment? What was her effective
annual rate of return?
Given: P 5 $1,020; V 5 $1,000; F 5 $950; r 5 3%
Find: i, ieff
KEY DATA
Setting the present worth of her investment equal to 0 gives
SOLUTION
2$1,020 1 1$1,0002 10.032 1P Z A i%,122 1 $9501P Z F i%,122 5 $0
Solving by trial and error, we first let i 5 2 percent/quarter (8 percent
compounded quarterly):
2$1,020 1 $30110.575342 1 $95010.788492 . $0 2 $46.32570 . $0
Next, letting i 5 2½ percent/quarter (10 percent compounded quarterly):
2$1,020 1 $30110.257782 1 $95010.743562 , $0 2 $5.88520 , $0
Linear interpolation gives
0.02 1 0.0051$46.325702/ 1$46.32570 1 5.885202 5 0.02444
or 2.444 percent/quarter (9.776 percent compounded quarterly).
The effective annual return is
ieff 5 3 11 1 0.024442 4 2 14100% 5 10.140%
The Excel® RATE and EFFECT worksheet functions are ideally
suited for this example. Note, there will be some minor differences in values
due to rounding. Entering the following in any cell of an Excel® spreadsheet
will yield the quarterly rate:
iqtr 5 RATE112,30,21020,9502 5 2.442%
Embedding the RATE function in the EFFECT function allows us to calculate the effective annual return on the bond transaction:
ieff 5 EFFECT14*RATE112,30,21020,9502,42
5 10.132%
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Chapter 3
Equivalence, Loans, and Bonds
3-4
VARIABLE INTEREST RATES
LEARN I N G O B JEC T I V E : Calculate the worth of a cash flow profile with vari-
able interest rates.
Thus far we have considered interest rates to be fixed over the duration of
the financial transaction. Recent experience indicates that such a situation
is not likely if the time period of interest extends over several years.
Considering a single sum of money and discrete compounding, if it
denotes the interest rate appropriate during time period t, the future worth
equivalent for a single sum of money can be expressed as
F 5 P11 1 i1 2 11 1 i2 2 . . . 11 1 in21 2 11 1 in 2
Extending the consideration of variable interest rates to cash flow
series, the present worth of a series of cash flows can be represented as
P 5 A1 11 1 i1 2 21 1 A2 11 1 i1 2 21 11 1 i2 2 21 1 . . .
1 An11 1 i1 2 21 11 1 i2 2 21 . . . 11 1 in 2 21 (3.14)
The future worth of a series of cash flows with variable interest rates
can be given as
F 5 An 1 An21 11 1 in 2 1 An22 11 1 in21 2 11 1 in 2 1 . . .
1 A1 11 1 i2 2 11 1 i3 2 . . . 11 1 in21 2 11 1 in 2
EXAMPLE
(3.15)
A Cash Flow Series with Variable Interest Rates
Consider the CFD given in Figure 3.9 with the appropriate interest rates
indicated. Determine the present worth, future worth, and uniform series
equivalents for the cash flow series.
$300
$200
(+)
10%
0
$200
10%
1
8%
2
8%
3
12%
4
5
(–)
$200
FIGURE 3 . 9
CFD for Variable Interest Rate Example 3.9
3-5 Annual Percentage Rate
Computing the present worth gives
SOLUTION
P 5 $2001P Z F 10%,12 2 $2001P Z F 10%,12 1P Z F 10%,12
1 $3001P Z F 8%,12 1P Z F 10%,12 1P Z F 10%,12
1 $2001P Z F 12%,12 1P Z F 8%,12 1P Z F 8%,12 1P Z F 10%,12 1P Z F 10%,12
5 $2001P Z F 10%,12 2 $2001P Z F 10%,22 1 $3001P Z F 8%,12 1P Z F 10%,22
1 $2001P Z F 12%,12 1P Z F 8%,22 1P Z F 10%,22 5 $372.63
The future worth is given by
F 5 $200 1 $3001F Z P 8%,12 1F Z P 12%,12 2 $2001F Z P 8%,22 1F Z P 12%,12
1 $2001F Z P 10%,12 1F Z P 8%,22 1F Z P 12%,12
5 $200 1 $30011.080002 11.120002 2 $20011.166402 11.120002
1 $20011.100002 11.166402 11.120002 5 $589.01
The uniform series equivalent is obtained as follows:
P 5 A1P Z F 10%,12 1 A1P Z F 10%,22 1 A1P Z F 8%,12 1P Z F 10%,22
1 A1P Z F 8%,22 1P Z F 10%,22 1 A1P Z F 12%,12 1P Z F 8%,22
1P Z F 10%,22
$372.63 5 A3 10.909092 1 10.826452 1 10.925932 10.826452
1 10.857342 10.826452 1 10.892862 10.857342 10.826452 4
5 3.841958A
A 5 $96.99
Thus, $96.99/time period for five time periods is equivalent to the
original cash flow series.
This example provides an opportunity to use an Excel® worksheet
function not previously used in this text: FVSCHEDULE. Its syntax is
FVSCHEDULE(principal,schedule), and it is applied to single cash flows.
For the example,
F 5 FVSCHEDULE1200,50.1,0.08,0.08,0.1262
2FVSCHEDULE1200,50.08,0.08,0.1262
1 FVSCHEDULE1300,50.08,0.1262 1 200
5 $589.01
3-5
ANNUAL PERCENTAGE RATE
LEARNING O BJECTI VE: Explain the Annual Percentage Rate (APR) commonly calculated for a home mortgage.
In Chapter 2, we defined the effective annual interest rate as the annual compounding rate that was equivalent to the stated interest rate. Specifically, we
111
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Chapter 3
Equivalence, Loans, and Bonds
Annual Percentage Rate
(APR) A term used by
lenders to disclose the
annual interest rate to
borrowers. This term is not
calculated in a consistent
way from lender to lender,
and may reflect a nominal
annual interest rate, an
effective annual interest
rate, or something else.
showed that 12 percent per year compounded quarterly had an effective
annual interest rate of 12.551 percent.
Unfortunately, in loan transactions lenders are not required to provide
the effective annual interest rate to borrowers. Instead, they are required to
provide the annual percentage rate for the financial transaction. Denoted
APR, the annual percentage rate specified might be a nominal annual
interest rate, an effective annual interest rate, or something quite different
from either the nominal or effective annual rate.
An APR is commonly calculated for home mortgages. Many variations exist in how APR values are calculated. The differences in results
may well be due to differences in various assumptions, including the
following:
■
■
■
■
■
when the first payment is to be made (immediately, in 30 days, or in
45 days)
whether interest is prepaid (at the beginning of the month) or postpaid
(at the end of the month)
whether points are applied to only the loan principal or to the sum of
the loan principal and ‘‘other’’ closing costs
the number of days considered in a year (365 versus 360—the product
of 30 days per month and 12 months per year)
the number of decimal points included in the calculations
Two basic approaches are used to compute the APR for a home mortgage, although there are many variations on each. The first approach,
which we call additive, assumes that the buyer finances the closing costs;
hence, the payments are based on the sum of the closing costs and the
amount needed to purchase the house. Because the net cash flow for the
buyer at the time of closing is the purchase price for the house, the APR is
the nominal annual rate that equates the monthly payment to the lender
with the amount needed to purchase the house.
The second approach, which we call subtractive, assumes that the
buyer pays all closing costs at the time of closing and makes mortgage
payments based only on the price of the house purchased. Because the
net cash flow for the buyer at the time of closing is the difference in
what is obtained to purchase the house and the closing costs paid to the
lender, the APR is the nominal annual rate that equates monthly payments made to the lender with the buyer’s net cash flow at the time of
closing.
The additive approach is more commonly used than the subtractive
approach. Among the APRadd calculators available is eFunda’s (www.
efunda.com) for fixed rate loans. A Web-based APR calculator for
adjustable rate mortgages is available at http://www.dinkytown.net.
Summary 113
KEY CONCEPTS
1. Learning Objective: Compare the equivalence between two or more cash
flow profiles. (Section 3.1)
If two or more cash flow profiles have equal values, they are said to be
equivalent. Determining whether cash flow profiles are equivalent—or
not—enables us to decide whether we should prefer one over another. A
common problem in engineering economic analysis is to determine the
value of a parameter, often the interest rate, which makes two or more cash
flow profiles equivalent.
2. Learning Objective: Analyze immediate payment and deferred payment
loans, including payment amount, remaining balance, and interest and principal per payment. (Section 3.2)
A loan payment consists of two parts, the principal and the interest. Recall
that the first thing paid in repaying a loan is the interest. The portion of the
payment that exceeds the interest charge will go toward paying the loan
principal. Immediate payment loans begin paying the loan off immediately
after the first interest period. With deferred loans, the loan payment begins
more than one interest period after receipt of the principal. Interest charges
can be deducted from taxable income; we will revisit this topic in Chapter 9,
“Income Taxes.”
3. Learning Objective: Analyze investments in bonds and determine the
purchase price, selling price, and return on such investments. (Section 3.3)
Bonds are important financial instruments for raising capital to finance
projects in both the public and private sectors. Analyzing bond investments
includes establishing the purchase price or selling price for a bond and
determining the rate of return on a bond investment given the purchase and
sales prices.
The general expression relating the terms associated with a bond is
P 5 Vr1P Z A i%,n2 1 F1P Z F i%,n2
where
P 5 the purchase price of a bond
F 5 the sales price (or redemption value) of a bond
V 5 the par or face value of a bond
r 5 the bond rate (coupon rate) per interest period
i 5 the yield rate (return on investment or rate of return) per interest period
n 5 the number of interest (coupon) payments received by the bondholder
A 5 Vr 5 the interest or coupon payment received per interest period
SUMMARY
114
Chapter 3
Equivalence, Loans, and Bonds
4. Learning Objective: Calculate the worth of a cash flow profile with variable interest rates. (Section 3.4)
A simplifying assumption made for an engineering economic analysis
often holds interest rates fixed over time. This is not necessarily a valid
assumption, however, especially in cases where the time period is long.
When interest rates vary over time, one must adjust the analysis to consider the varying interest rates.
5. Learning Objective: Explain the Annual Percentage Rate (APR) commonly
calculated for a home mortgage. (Section 3.5)
There are many variations in calculating the APR including the additive
and subtractive approach discussed in this chapter. In addition, the APR
calculation is dependent on various assumptions including when the first
payment is to be made, whether interest is prepaid or postpaid, how points
are applied, the number of days in a year, and the precision of the calculations. Caution should be taken when considering an APR given these variations. A Web-based additive calculator can be used to illustrate calculating
the APR under various assumptions.
KEY TERMS
Annual Percentage Rate (APR), p. 112
Bond, p. 106
Bond Rate, p. 106
Deferred Payment Loans, p. 104
Equivalence, p. 94
Immediate Payment Loans, p. 101
Maturity, p. 106
Par Value, p. 106
Principal Payment, p. 100
Redeem, p. 106
Problem available in WileyPLUS
GO Tutorial Tutoring Problem available in WileyPLUS
Video Solution Video Solution available in WileyPLUS
FE-LIKE PROBLEMS
1.
What single sum of money at t 5 4 is equivalent to receiving $5,000 at t 5 1,
$6,000 at t 5 2, $7,000 at t 5 3, and $8,000 at t 5 4 if money is compounded
at a rate of 8% per time period?
a. $28,857
c. $30,892
b. $26,000
d. $33,363
Summary 115
2.
If the interest rate is 10% per year, what series of equal annual payments
is equivalent to the following series of decreasing payments: $5,000, $4,000,
$3,000, $2,000, $1,000?
a. $3,000
b. $3,000(1 1 0.1) 5 $3,300
c. [$5,000(F Z P 10%,4) 1 $4,000(F Z P 10%,3) 1 $3,000(F Z P 10%,2) 1
$2,000(F Z P 10%,1) 1 1,000]/5
d. $5,000 2 $1,000(A Z G 10%,5)
3.
You borrow $5,000 at 10% per year and will pay off the loan in 3 equal
annual payments starting one year after the loan is made. The end-of-year
payments are $2,010.57. Which of the following is true for your payment at
the end of year 2?
a. Interest is $500.00 and principal is $1,510.57.
b. Interest is $450.00 and principal is $1,560.57.
c. Interest is $348.94 and principal is $1,661.63.
d. Interest is $182.78 and principal is $1,827.79.
4.
You borrow $10,000 at 15% per year and will pay off the loan in three
equal annual payments with the first occurring at the end of the fourth year
after the loan is made. The three equal annual payments will be $6,661.08.
Which of the following is true for your first payment at the end of year 4?
a. Interest 5 $6,661.08; principal 5 $0.00
b. Interest 5 $2,281.31; principal 5 $4,379.77
c. Interest 5 $1,500.00; principal 5 $5,161.08
d. Interest 5 $0.00; principal 5 $6,661.08
5.
You purchase a $10,000 bond with a bond rate of 6% per year payable semiannually for 2 years. You pay $9,600 for the bond. Which statement is correct?
a. Semiannual cash flows will be 2$9,600, $300, $300, $300, and $9,900,
and the bond will earn more than 10%
b. Semiannual cash flows will be 2$9,600, $300, $300, $300, and $9,900,
and the bond will earn less than 10%
c. Semiannual cash flows will be 2$9,600, $300, $300, $300, and $10,300,
and the bond will earn more than 10%
d. Semiannual cash flows will be 2$9,600, $300, $300, $300, and $10,300,
and the bond will earn less than 10%
6.
Consider a cash flow and interest profile as shown:
Cash Flow at
End of Year
Interest Rate
During Year
Year 0
Year 1
Year 2
Year 3
2$1,000
$3,000
$2,000
$1,000
NA
6%
8%
10%
The worth at the end of Year 3 of these cash flows is:
a. $5,000.00
c. $5,994.56
b. $5,504.72
d. $5,440.00
116
Chapter 3
Equivalence, Loans, and Bonds
7.
A $200,000 bond having a bond rate of 8% payable annually is purchased
for $190,500 and kept for 6 years, at which time it is sold. How much should
it sell for in order to yield a 7% effective annual return on the investment?
a. $168,000
c. $174,000
b. $171,000
d. $177,000
8.
A house is to be purchased for $180,000 with a 10% down payment,
thereby financing $162,000 with a home loan and mortgage. There are no
“points” or other closing charges associated with the loan. A conventional
30-year loan is used at 7.5%, resulting in monthly payments of $1,132.73.
The interest portion of the first monthly payment will be what?
a. $1,012.50
c. $120.23
b. $682.73
d. The answer cannot be determined without
more information.
9.
$150,000 is deposited in a fund that pays 5% annual compound interest for
2 years, 3% annual compound interest for 2 years, and 4% annual compound
interest for 2 years. If uniform annual withdrawals occur over the 6-year
period, what will be the magnitude of the annual withdrawals?
a. $27,689.63
c. $28,804.50
b. $28,614.29
d. $29,552.62
PROBLEMS
Section 3.1
Equivalence
1. If $7,000 is borrowed and repaid with four quarterly payments of $600 dur-
ing the first year and four quarterly payments of $1,500 during the second
year after receiving the $7,000 loan, what is the effective annual interest rate
for the loan?
2. Quarterly deposits of $1,000 are made at t 5 1, 2, 3, 4, 5, 6, and 7. Then,
withdrawals of size A are made at t 5 12, 13, 14, and 15. If the fund pays
interest at a quarterly compounding rate of 4%, what value of A will deplete
the fund with the fourth withdrawal?
3. What uniform annual series of cash flows over a 12-year period is equivalent
to an investment of $5,000 at t 5 0, followed by receipts of $600 per year for
11 years and a final receipt of $1,600 at t 5 12 if the investor’s time value of
money is 6% per time period?
4. An automobile is “priced” at $7,000. A buyer may purchase the car for
$6,500 now or, alternatively, the buyer can make a down payment of $1,000
now and pay the remaining $6,000 in 8 equal quarterly payments (over two
years) at 8% compounded quarterly.
a. If the buyer’s TVOM is 10% per year compounded quarterly, would the
buyer prefer to pay the $6,500 outright, or make the down payment and
the quarterly payments?
Summary 117
b. What is the effective annual interest rate at which these two payment
options are equivalent?
5.
Video Solution Consider the following two cash flow series of payments:
Series A is a geometric series increasing at a rate of 8% per year. The initial
cash payment at the end of year 1 is $1,000. The payments occur annually
for 5 years. Series B is a uniform series with payments of value X occurring
annually at the end of years 1 through 5. You must make the payments in
either Series A or Series B.
a. Determine the value of X for which these two series are equivalent if your
TVOM is i 5 6.5%.
b. If your TVOM is 8%, would you be indifferent between these two series
of payments? If not, which do you prefer?
c. If your TVOM is 5%, would you be indifferent between these two series
of payments? If not, which do you prefer?
6. Kinnunen Company wishes to give its customers three options on payments
for office equipment when the initial purchase price is over a certain amount.
For example, the following three payment plans are options on a typical purchase, and Kinnunen wants to be sure they are equivalent at a their TVOM
of 14%. Determine the values of Q and R.
End of Year
0
1
2
3
4
5
Option 1
Option 2
Option 3
$0
1,800
1,800
1,800
1,800
1,800
$0
Q
2Q
3Q
4Q
5Q
$0
R
(1.1)R
(1.1)2R
(1.1)3R
(1.1)4R
Consider the following three cash flow series:
7.
End of
Year
0
1
2
3
4
5
Cash Flow
Series A
Cash Flow
Series B
Cash Flow
Series C
2$1,000
X
1.5X
2.0X
2.5X
3.0X
2$2,500
3,000
2,500
2,000
1,500
1,000
Y
Y
Y
2Y
2Y
2Y
Determine the values of X and Y so that all three cash flows are equivalent at
an interest rate of 15% per year compounded yearly.
118
Chapter 3
Equivalence, Loans, and Bonds
8. Consider the following three cash flow series:
End of
Year
0
1
2
3
4
5
Cash Flow
Series A
Cash Flow
Series B
Cash Flow
Series C
$3.0X
2.5X
2.0X
1.5X
1.0X
$1,000
1,500
2,000
2,500
3,000
21,000
22,500
2Y
2Y
2Y
Y
Y
Y
Determine the values of X and Y so that you are indifferent between all three
cash flows if your TVOM is 11% per year compounded yearly.
9.
Zetterberg Builders is given two options for making payments on a brush
hog. Find the value of X such that they would be indifferent between the two
cash flow profiles if their TVOM is 12% per year compounded yearly.
End of Year
0
1
2
3
4
5
Section 3.2
Series 1
Series 2
$150
$200
$250
$300
$0
$0
$0
$0
$35X
$25X
$15X
$5X
Interest Payments and Principal Payments
10.
Video Solution In order to buy a car, you borrow $25,000 from a friend at
12%/year compounded monthly for 4 years. You plan to repay the loan with
48 equal monthly payments.
a. How much are the monthly payments?
b. How much interest is in the 23rd payment?
c. What is the remaining balance after the 37th payment?
d. Three and one-half years after borrowing the money, you decide to pay
off the loan. You have not yet made the payment due at that time. What is
the payoff amount for the loan?
11.
CTL (Concrete Testing Lab) borrowed $80,000 for new equipment at
8% per year, compounded quarterly. It is to be paid back over three years in
equal quarterly payments. For each part below, use both the interest tables
and the Excel® financial functions. Compare answers between the two.
a. How much interest is in the 6th payment?
b. How much principal is in the 6th payment?
c. What principal is owed immediately following the 6th payment?
12. Med Diagnostics, Inc. borrowed $200,000 from a lender for a new blood
analyzer module to improve the accuracy and consistency of its tests. The
Summary 119
rate was 6.0%, 2 percent above the prime rate. The loan was to be paid back
in equal monthly amounts over 7 years.
a. How much is the monthly payment?
b. Two years (24 months) of payments have been made. What is the principal
remaining after 2 years?
c. The prime rate has now risen, and the bank can make loans to other customers,
if it has the available capital, for 8.5% payable in equal monthly payments.
The bank contacts Med Diagnostics and offers to let them pay off the loan
immediately for the amount owed at the end of year 2, less $X. What is the
range of $X that would be beneficial to both the bank and Med Diagnostics?
13.
GO Tutorial Aerotron Electronics has just bought a used delivery truck
for $15,000. The small business paid $1,000 down and financed the rest,
with the agreement to pay nothing for the entire first year and then to pay
$536.83 at the end of each month over years 2, 3, and 4.
a.
b.
c.
d.
e.
What nominal interest rate is Aerotron paying on the loan?
What effective interest rate are they paying?
How much of the 14th month’s payment is interest? How much is principal?
How much of the 18th month’s payment is interest? How much is principal?
How much of the 22nd month’s payment is interest? How much is principal?
14. BioElectroMechanical Systems (BEMS) is a startup company with high po-
tential and little available cash. They obtain $500,000 for necessary technology
from a venture capitalist, who charges them 24% compounded monthly. The
agreement calls for no payment until the end of the first month of the 4th year,
with equal monthly payments thereafter for 3 complete years (36 payments).
a. How much are the monthly payments?
b. What is the total interest paid to the lender?
c. What is the total principal paid to the lender?
d. If BEMS is doing incredibly well and would like to pay off the debt immediately after making the 24th payment in month 60, how much must they pay?
15. $100,000 is borrowed at 6% compound annual interest, with the loan to be
repaid with 10 equal annual payments.
a. If the first payment is made one year after receiving the $100,000, how
much of the third payment will be an interest payment?
b. If the first payment is made four years after receiving the $100,000, how
much of the first payment will be an interest payment?
c. If the first payment is made four years after receiving the $100,000, how
much of the last payment will be an interest payment?
16. $25,000 is borrowed at an annual compound rate of 8%. The loan is repaid
with 5 annual payments, each of which is $500 greater than the previous
payment. How much of the 2nd payment will be a principal payment?
17. $25,000 is borrowed at an annual compound rate of 8%. The loan is repaid
with 5 annual payments, each of which is 10% greater than the previous
payment. How much of the 2nd payment will be a principal payment?
120
Chapter 3
Equivalence, Loans, and Bonds
Section 3.3
Bond Investments
18. Five 15-year bonds each having a face value of $1,000 and a coupon rate of
6% per 6 months payable semiannually were purchased for $7,000 8 years
ago, and the 16th coupon payment was just made. What can they be sold for
now to a buyer if that buyer’s desired return is 4% per 6 months?
19.
You have just purchased ten municipal bonds, each with a $1,000 par
value, for $9,500. You purchased them immediately after the previous owner
received semiannual coupon payments. The bond rate is 6.6% per year payable semiannually. You plan to hold the bonds for 5 years, selling them immediately after you receive the coupon payment. If your desired nominal
yield is 12% per year compounded semiannually, what will be your minimum selling price for the bonds?
20. You wish to purchase a $1,000 bond from a friend who needs the money.
There are 7 years remaining until the bond matures, and interest payments
are quarterly. You decide to offer $750.08 for the bond because you want to
earn exactly 16% per year compounded quarterly on the investment. What is
the annual bond rate of interest?
21.
Video Solution Leann just sold ten $1,000 par value bonds for $9,800. The
bond coupon rate was 6% per year payable quarterly. Leann owned the bonds
for 3 years. The first coupon payment she received was 3 months after she
bought the bonds. She sold the bonds immediately after receiving her 12th
coupon payment. Leann’s yield on the bond was 12% per year compounded
quarterly. Determine how much Leann paid when she purchased the bonds.
22. Eight bonds were purchased for $8,628.16. They were kept for 5 years and
coupon payments were received at the end of each of the 5 years. Immediately following receipt of the 5th coupon payment, the owner sells each
bond for $62.50 more than its par value. The bond coupon rate is 8%, and
the owner’s money yields a 10% annual return.
a. Draw a clear, completely labeled cash flow diagram of the entire bond
transaction using dollar amounts where they are known and using $X to
represent the face value of the bond.
b. Determine the face value of each bond.
23. Ten bonds are purchased for $9,855.57. They are kept for 5 years and coupon
payments are received at the end of each of the 5 years. Immediately following
the owner’s receipt of the 5th coupon payment, the owner sells each bond
for $50 less than its par value. The bond coupon rate is 8%, and the owner’s
money yields a 10% annual return.
a. Draw a clear, completely labeled cash flow diagram of the entire bond
transaction using dollar amounts where they are known and using $X to
represent the face value of the bond.
b. Determine the face value of each bond.
24. Twenty $1,000 municipal bonds are offered for sale at $18,000. The bond
coupon rate is 6% per year payable semiannually. The bonds will mature and
be redeemed at face value 5 years from now. If you purchase the bonds, the
Summary 121
first premiums will be received 6 months from today. You have decided that
you will invest $18,000 in the bonds if your effective annual yield is at least
8.16%. Will you buy these bonds? Why or why not?
25.
Shannon purchases a bond for $952.00. The bond matures in 3 years,
and Shannon will redeem it at its face value of $1,000. Coupon payments are
paid annually. If Shannon will earn a yield of 12%/year compounded yearly,
what is the bond coupon rate?
26. One hundred $1,000 bonds having bond coupon rates of 8% per year pay-
able annually are purchased for $95,250 and kept for 6 years, at which time
they are sold. Determine the selling price that will yield a 7% effective
annual return on the investment.
27. One hundred $1,000 bonds having bond rates of 8% per year payable annually
are available for purchase. If you purchase them and keep them until they mature in 4 years, what is the maximum amount you should pay for the bonds if
you wish to earn no less than a 7% effective annual return on your investment?
28. One hundred $1,000 bonds having a bond rate of 8% per year payable
quarterly are purchased for $97,500, kept for 4 years, and sold for $95,000.
Determine the effective annual return on the bond investment.
29. Five $1,000 bonds having a bond rate of 8% per year payable quarterly are pur-
chased for $4,940 and kept for 6 years, at which time they are sold. Determine
the selling price that yields a 6% effective annual return on the investment.
30. Twenty-five $1,000 bonds having a bond rate of 8% per year payable quar-
terly are purchased for $22,250 and kept for 5 years. Assume that the bonds
are sold immediately after receiving the coupon payments at the end of the
5th year. What must they be sold for in order to earn a 6% effective annual
return on the investment?
Section 3.4 Variable Interest Rates
31. True or False: A single investment of $10,000 is to be made. Two investment
alternatives exist. Alternative A pays interest rates of 1%, 2%, 3%, 4%, 5%,
6%, 7%, 8%, 9%, and 10%, each of the next 10 years. The interest rates paid
with Alternative B are the reverse order of those available for Alternative A.
To maximize the value of the investment portfolio at the end of the 10-year
period, you should not choose investment B.
32. True or False: Two investments are available; they require the same equal
annual investments over a 10-year period. Investment A will pay interest rates
of 1%, 2%, 3%, 4%, 5%, 6%, 7%, 8%, 9%, and 10%, each of the next 10 years;
the annual interest rates paid with Investment B are the reverse of those paid
by Investment A. To maximize the value of the investment portfolio at the end
of the 10-year investment period, you should choose investment A.
33. Based on the interest rates and cash flows shown in the cash flow diagram,
determine the value of $X.
122
Chapter 3
Equivalence, Loans, and Bonds
$25,000
$X
0
1
4%
2
4%
3
4%
4
3%
5
3%
6
3%
7
5%
8
5%
9
5%
$50,000
34. Based on the interest rates and cash flows shown in the cash flow diagram,
determine the value of $X.
$10,000
$X
0
1
4%
2
4%
3
3%
4
5%
$150,000
35. Based on the interest rates and cash flows shown in the cash flow diagram,
determine the value of $X.
$X
$10,000
0
1
3%
2
3%
3
3%
4
5%
5
5%
6
5%
7
5%
8
4%
9
4%
10
4%
$150,000
36. Maria deposits $2,000 in a savings account that pays interest at an annual
compound rate of 3%. Two years after the deposit, the interest rate increases
to 4% compounded annually. A second deposit of $3,000 is made immediately
after the interest rate changes to 4%. How much will be in the fund 7 years
after the second deposit?
37. In Problem 36, how much will be in the fund 7 years after the second deposit
if the rates of interest are switched (i.e., 4% for 2 years and 3% for 7 years)?
38. You deposit $10,000 in a fund that pays compound annual interest equal to
3%. One year later the interest rate changes to 4%, 2 years later the interest
Summary 123
rate changes to 5%, and 3 years later the interest rate changes to 6%. How
much will you have in the fund after 4 years if you withdrew $2,500 at the
end of each of the previous 3 years?
39.
Charlie has $10,000 to invest for a period of 5 years. The following three
alternatives are available to Charlie:
■
■
■
Account I pays 4% for the 1st year, 6% for year 2, 8% for year 3, 10% for
year 4, and 12% for year 5, all with annual compounding.
Account II pays 12% for the 1st year, 10% for year 2, 8% for year 3, 6% for
year 4, and 4% for year 5, all with annual compounding.
Account III pays interest at the rate of 7.96294% per year for all 5 years.
Based on the available balance at the end of year 5, which alternative is
Charlie’s best choice?
40.
You have $2,000 that you want to invest at the beginning of each of
5 years. The following alternatives are available to you:
■
■
■
An investment that pays 7% for year 1, 6% for year 2, 5% for year 3, 4% for
year 4, and 3% for year 5.
An account that pays 3% for year 1, 4% for year 2, 5% for year 3, 6% for
year 4, and 7% for year 5.
An account that pays 5% per year each year.
On the basis of available balance at the end of the 5th year, which alternative
is the best choice?
41.
GO Tutorial Jimmy deposits $4,000 now, $2,500 3 years from now, and
$5,000 6 years from now. Interest is 5% for the first 3 years and 7% for the
last 3 years.
a. How much money will be in the fund at the end of 6 years?
b. What is the present worth of the fund?
c. What is the uniform series equivalent of the fund (uniform cash flow at
end of years 1 through 6)?
Consider the following cash flows and interest rates:
42.
End of
Year
0
1
2
3
Interest Rate
During Period
10%
8%
12%
Cash Flow at
End of Period
$0
$2,000
2$3,000
$4,000
a. Determine the future worth of this series of cash flows.
b. Determine the present worth of this series of cash flows.
c. Determine a 3-year uniform annual series that is equivalent to the original
series.
ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E :
C ONO C O P HI L L I P S
In 2011 ConocoPhillips was one of the largest integrated energy companies
and refiners in the United States and abroad, based on market capitalization
and oil and natural gas proved reserves and production.The company is known
globally for its technological expertise in exploration and production, reservoir
management and exploitation, 3-D seismic technology, high-grade petroleum
coke upgrading, and sulfur removal.
Headquartered in Houston, Texas, ConocoPhillips operates in more than
30 countries. As of December 31, 2011, the company had 29,800 employees
worldwide and assets of $153.2 billion. Its capital expenditures and investments in 2011 totaled $13.3 billion; total revenues were $251.2 billion. Its debtto-capital ratio at the end of 2011 was 26 percent, compared to 25 percent at
the end of 2010.
The company’s core business activities include petroleum exploration and
production; petroleum refining, marketing, supply, and transportation; and
natural gas gathering, processing, and marketing. Also, the company invests in
emerging businesses—power generation, carbon-to-liquids, technology solutions, and emerging technologies such as renewable fuels and alternative energy
sources—that provide current and potential future growth opportunities. In
2010, the company initiated a three-year strategic plan to reposition itself by
implementing a spinoff of its downstream businesses into a new company,
Phillips 66. In this move, ConocoPhillips will become an independent exploration
and production company. These changes are expected to be complete in 2012.
In its Form 10-K for 2011, the company provided the following comments
regarding “standardized measure of discounted future net cash flows relating
to proved oil and gas reserve quantities”:
“In accordance with SEC and FASB requirements, amounts were computed using 12-month average prices and end-of-year costs (adjusted
only for existing contractual changes), appropriate statutory tax rates
and a prescribed 10 percent discount factor. Twelve-month average
124
PRESENT WORTH
prices are calculated on the unweighted arithmetic average of the firstday-of-the-month price for each month within the 12-month period
prior to the end of the reporting period. . . . For all years, continuation
of year-end economic conditions was assumed. The calculations were
based on estimates of proved reserves, which are revised over time as
new data become available.”
ConocoPhillips and numerous other organizations rely on present worth
analysis in making investment decisions.To list the other firms that use present
worth analysis would surely require listing every organization that uses discounted cash flow methods.
DISCUSSION QUESTIONS:
1. Do you think that small organizations rely on present worth analysis
as well, or is this type of analysis important only for extremely large
firms such as ConocoPhillips?
2. The statement about the calculations being based on estimates of proved
reserves suggests to stakeholders that this is a somewhat volatile business. What measures can an engineering economist take to ensure that a
rigorous analysis is performed to instill investor confidence?
3. Are you surprised to see the detailed explanation of how the numbers
were calculated?
4. In what other parts of their business might ConocoPhillips rely on
present worth analysis?
LEARNING OBJECTIVES
When you have finished studying this chapter, you should be able to:
1. List eight DCF methods for comparing economic alternatives
(Section 4.1.1).
125
126
Chapter 4
Present Worth
2. Differentiate between ranking methods and incremental methods of
determining economic worth. (Section 4.1.2)
3. Calculate a present worth (PW) converting all cash flows to a single
sum equivalent at time zero for a given interest rate. (Section 4.2)
4. Perform an economic analysis of public investments utilizing the benefitcost analysis for a given interest rate. (Section 4.3)
5. Calculate the discounted payback period (DPBP) for a given interest
rate to determine how long it takes for the cumulative present worth
(PW) to be positive. (Section 4.4)
6. Calculate the capitalized worth (CW) of an investment for a given interest
rate when the planning horizon is infinitely long. (Section 4.5)
INTRODUCTION
Recall the fifth step in the 7-step SEAT process described in Chapter 1 for
performing an engineering economic analysis: Compare the alternatives.
Implicit in this step is a decision regarding the method(s) to be used in
making the comparison. In this chapter, you will learn how results obtained from present worth analysis compare with other methods of
measuring economic worth. Present worth analysis is the most popular
DCF method of comparing investments. You will also learn about three
methods that are special cases of the present worth method: benefit-cost
ratio, discounted payback period, and capitalized worth.
Systematic Economic Analysis Technique
1.
2.
3.
4.
5.
6.
7.
4-1
Identify the investment alternatives.
Define the planning horizon.
Specify the discount rate.
Estimate the cash flows.
Compare the alternatives.
Perform supplementary analyses.
Select the preferred investment.
COMPARING ALTERNATIVES
LEARN I N G O B JEC T I V E : List eight DCF methods for comparing economic
Video Lesson:
Project Analysis Methods
alternatives.
This section summarizes the eight DCF methods that are covered in detail
in Chapters 4–6. It also touches on several important considerations when
4-1
comparing alternatives, regardless of method, including whether the analysis is before-tax or after-tax, whether the lives of the alternatives are equal
or unequal, and the concept of the do-nothing alternative.
4.1.1
Methods of Comparing Economic Worth
The eight DCF methods that can be used in comparing investment alternatives are present worth, benefit-cost ratio, discounted payback period, capitalized worth, annual worth, future worth, internal rate of return, and
external rate of return.
These methods may be described briefly as follows:
1.
2.
3.
4.
5.
6.
7.
8.
The present worth (PW) method converts all cash flows to a single
sum equivalent at time zero using i 5 MARR.
The benefit-cost ratio (B/C) method determines the ratio of the present
worth of benefits (savings or positive-valued cash flows) to the negative of the present worth of the investment(s) (or negative-valued cash
flows) using i 5 MARR.
The discounted payback period (DPBP) method determines how
long it takes for the cumulative present worth to be positive using
i 5 MARR.
The capitalized worth (CW) method determines the present worth
(using i 5 MARR) when the planning horizon is infinitely long.
The annual worth (AW) method converts all cash flows to an equivalent uniform annual series of cash flows over the planning horizon
using i 5 MARR.
The future worth (FW) method converts all cash flows to a single sum
equivalent at the end of the planning horizon using i 5 MARR.
The internal rate of return (IRR) method determines the interest
rate that yields a future worth (or present worth or annual worth) of
zero.
The external rate of return (ERR) method determines the interest
rate that equates the future worth of the invested capital to the future
worth of recovered capital (when the latter is computed using the
MARR.)
4.1.2
Ranking and Incremental Methods of Economic Worth
LEARNING O BJECTI VE: Differentiate between ranking methods and incre-
mental methods of determining economic worth.
The eight DCF methods can be divided into two groups: ranking and
incremental methods.
Comparing Alternatives
127
128
Chapter 4
Present Worth
Ranking Methods
Present worth, future worth, annual worth, and capitalized worth are ranking methods; as such, the alternative having the greatest PW, FW, AW, or
CW over the planning horizon is the economic choice and would be recommended, absent nonmonetary criteria.1 DPBP also is a ranking method,
but the goal is to identify the investment alternative with the shortest payback period.
Incremental Methods
Internal rate of return, external rate of return, and benefit-cost ratio are
incremental methods; as such, the preferred alternative is the one that satisfies Principle #6: Money should continue to be invested as long as each
additional increment of investment yields a return that is greater than the
investor’s TVOM.
Incremental solutions that are equivalent to ranking solutions can be
obtained using PW, FW, AW, and CW. However, there seems to be little
reason to employ the more cumbersome incremental solution, since it is
much simpler to compute the value of the PW, FW, AW, or CW and recommend the one having the greatest value over the planning horizon.
4.1.3
Equivalence of Methods
Six of the eight analysis methods are equivalent: PW, FW, AW, IRR, ERR,
and B/C. That is, when applied correctly, all six of these methods will
yield the same recommendation regarding investment alternatives. The
CW and DPBP methods are not guaranteed to result in a recommendation
identical to that obtained using any of the six equivalent economic worth
methods. Examples in this and later chapters will illustrate why this is so.
4.1.4
Before-Tax versus After-Tax Analysis
In using a measure of economic worth to compare investment alternatives,
you can employ either before-tax or after-tax cash flows. Be consistent,
however! It is either/or but not both in the same analysis. If the comparison is
based on before-tax cash flows, then use a before-tax MARR; likewise, if the
comparison is based on after-tax cash flows, then use an after-tax MARR.
Although we believe it is usually best to perform after-tax economic
justifications, we will not cover tax issues until after discussing a variety
of methods used to compare economic alternatives.2 Consequently, in this
1
Rather than make the statement ‘‘absent nonmonetary criteria’’ every time to indicate which
investment is recommended, henceforth it will be assumed that the single criterion is monetary
and the objective is to maximize economic worth.
2
Income taxes and after-tax analysis are presented in Chapter 9.
4-1
and the following two chapters, you may consider the analyses to be either
before-tax (with before-tax cash flows and a before tax MARR) or aftertax (with after-tax cash flows and an after-tax MARR).
4.1.5
Equal versus Unequal Lives
When comparing investment alternatives, they must be compared over a
common time period, called the planning horizon. (Recall Principle #8:
Compare investment alternatives over a common period of time.) If the
duration of the planning horizon differs from the useful lives of the alternatives, then (at the end of the planning horizon) cash flow estimates must
be provided for the terminal or salvage values for alternatives with lives
greater than the planning horizon; for alternatives having useful lives less
than the planning horizon, replacement decisions must be made and cash
flow estimates must be provided for the replacements.
4.1.6 A Single Alternative
‘‘A single alternative’’ is an oxymoron. If there is no choice, then there is
no alternative. When we consider ‘‘a single alternative,’’ however, we are
considering doing something versus doing nothing. Assuming that the donothing alternative is feasible, you have two options when faced with the
question “Should I invest in an opportunity or not?”: Invest, or don’t invest.
In evaluating the ‘‘invest’’ option, we assume that the cash flow estimates
reflect the differences in doing nothing versus doing something. On that
basis,
■
■
■
■
when using PW, FW, and AW, we choose to ‘‘do something’’ if the
measure of economic worth has a value greater than zero;3
when using the B/C method, we choose to ‘‘do something’’ if the B/C
ratio is greater than 1; and
when using the IRR and ERR methods, we choose to ‘‘do something’’
if the measure of economic worth has a value greater than the MARR.
For DPBP, the decision to invest or not depends on the prescribed
acceptable value.
Implicit in the criteria above is an opportunity cost assumption.
Namely, we assume we are currently earning a return on our money equal
to the MARR. Therefore, the decision to invest in a particular opportunity
reduces to the following: Will we make more money by investing here or
by leaving our money in an investment pool that earns a return equal to the
3
As discussed in a later section, CW . 0 is not generally a viable test, because the do-nothing
alternative is not feasible for most of its applications.
Comparing Alternatives
129
130
Chapter 4
Present Worth
MARR? If we will not make more money by investing in the opportunity
in question, then we should leave our money in the investment pool. We
also assume the investment in question has risks comparable to those of
the investment pool.
4-2
Video Lesson:
Present Worth
Present Worth (PW) The
value of all cash flows
converted to a single sum
equivalent at time zero
using i 5 MARR.
PRESENT WORTH CALCULATIONS
LEARN I N G O B JEC T I V E : Calculate a present worth (PW) converting all cash
flows to a single sum equivalent at time zero for a given interest rate.
In this section we focus on what is variously referred to as present worth,
net present worth, present value, and net present value. Present worth
analysis uses the MARR to express the economic worth of a set of cash
flows, occurring over the planning horizon, as a single equivalent value at
the current time, often referred to as “time now” or “time zero.”
4.2.1 Present Worth of a Single Alternative
When using the present worth method to evaluate whether an investment
should be made, the decision depends on whether the present worth is
positive. If so, then the investment is recommended. Recalling our work in
Chapter 2, the present worth of an investment can be expressed mathematically as follows:
n
PW1i%2 5 a At 11 1 i2 2t
(4.1)
i50
Present Worth of a Single Investment Opportunity
EXAMPLE
To automatically insert electronic components in printed circuit boards
for a cell phone production line, a $500,000 surface mount placement
(SMP) machine is being evaluated by a manufacturing engineer. Over
the 10-year planning horizon, it is estimated that the SMP machine will
produce annual after-tax cost savings of $92,500. The engineer estimates
the machine will be worth $50,000 at the end of the 10-year period.
Based on the firm’s 10 percent after-tax MARR, should the investment
be made?
KEY DATA
Given: The cash flows outlined in Figure 4.1; MARR 5 10%; planning
horizon 5 10 years
Find: PW of the investment
4-2
Present Worth Calculations
$50,000
(+)
$92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500
0
1
2
3
4
5
6
7
8
9
10
(–)
$500,000
FIGURE 4.1
CFD for Example 4.1
From our work in Chapter 2, we compute the present worth for the investment:
SOLUTION
PW 5 2$500,000 1 $92,5001P Z A 10%,102 1 $50,0001P Z F 10%,102
5 2$500,000 1 $92,50016.144572 1 $50,00010.385542
5 $87,649.73
or, using Excel®,
PW 5 PV110%,10,292500,2500002 2 500000
5 87,649.62
Since PW . $0, the investment is recommended.
With a present worth of $87,649.62, there can be little doubt that the
investment is a good one for the company. However, it is useful to examine
how the cumulative present worth behaves over the 10-year period.
Figure 4.2 shows that the cumulative present worth begins at
2$500,000 with the purchase of the new machine and increases over the
planning horizon to a final value of $87,649.62. Notice that the investment
“loses money” (has a negative present worth) for the first 8 years. It is only
in the ninth year that the cumulative present worth becomes positive.
Therefore, if something unforeseen occurred, causing the company to
abandon the investment before the ninth year, would the investment still be
EXPLORING THE
SOLUTION
131
132
Chapter 4
Present Worth
FIGURE 4 . 2
Spreadsheet of Cumulative Present Worth Over the Planning Horizon
profitable? The answer to the question depends on the SMP machine’s
salvage value at the time it is abandoned.
We will examine cumulative present worth in more detail in Section 4.4,
which discusses the discounted payback period.
4.2.2 Present Worth of Multiple Alternatives
As a ranking method, PW is easily applied when choosing the preferred
alternative from among several mutually exclusive alternatives: Choose
the one with the greatest PW over the planning horizon. Mathematically,
the objective is
n
Maximize PWj 5 a Ajt 11 1 MARR2 2t
;j
(4.2)
t50
Present Worth of Two Alternatives
EXAMPLE
Video Example
Entertainment Engineers, Inc., is an Ohio-based design engineering firm
that designs rides for amusement and theme parks all over the world. Two
alternative designs are under consideration for a new ride called the Scream
Machine at a theme park located in Florida. The two candidate designs
differ in complexity, cost, and predicted revenue. The first alternative
4-2
Present Worth Calculations
design (A) will require an investment of $300,000 and is estimated to
produce after-tax revenue of $55,000 annually over a 10-year planning
horizon. The second alternative design (B) will require an investment of
$450,000 and is expected to generate annual after-tax revenue of $80,000.
A negligible salvage value is assumed for both designs. Theme park management could decide to “do nothing”; if so, the present worth of doing
nothing will be zero. An after-tax MARR of 10 percent is used. Which
alternative design, if either, should the theme park select?
Given: The cash flows outlined in Figure 4.3; MARR 5 10%; planning
horizon 5 10 years
Find: PW of each alternative
KEY DATA
$55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000
(+)
0
1
2
3
4
5
6
7
8
9
10
(–)
Alternative A
$300,000
$80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000
(+)
0
1
2
3
4
5
(–)
Alternative B
$450,000
FIGURE 4.3
CFDs for Example 4.2
6
7
8
9
10
133
134
Chapter 4
Present Worth
SOLUTION
Letting A denote the alternative design for the $300,000 initial investment and B denote the other alternative, the present worth of each is as
follows.
Alternative A
PWA 5 2$300,000 1 $55,0001P Z A 10%,102
5 2$300,000 1 $55,00016.144572
5 $37,951.35 . $0.00 1therefore, A is better than doing nothing2
or, using Excel®,
PWA 5 PV110%,10,2550002 2 30000
5 37,951.19 . $0.00 1therefore, A is better than doing nothing2
Alternative B
PWB 5 2$450,000 1 $80,0001P Z A 10%,102
5 2$450,000 1 $80,00016.144572
5 $41,565.60 . $37,951.35 1therefore, B is better than A2
or, using Excel®,
PWB 5 PV110%,10,2800002 2 450000
5 41,565.37 + 37,951.19 1therefore, B is better than A2
Based on the present worth analysis, the more expensive alternative (B) is
recommended. It has the greater present worth.
EXPLORING THE
SOLUTION
The theme park’s manager is interested in learning how the present worth
for the two design alternatives is affected by the MARR value used. As
shown in Figure 4.4, Alternative B has the greatest present worth for all
values of MARR less than approximately 10.5 percent, but, thereafter,
Alternative A has the greatest present worth. The manager also noticed
that for MARR values greater than 13 percent, neither alternative has a
positive present worth.
4-2 Present Worth Calculations
135
$400,000
$350,000
$300,000
Present Worth
$250,000
$200,000
$150,000
$100,000
$50,000
$0
0%
2%
4%
6%
8%
10%
12%
14%
16%
18%
20%
–$50,000
–$100,000
–$150,000
MARR
Alternative A
FIGURE 4.4
Alternative B
Plot of Present Worths for Example 4.2
4.2.3 Present Worth of One-Shot Investments
Occasionally, investments are available only once. Such cases are called
one-shot investments. When one-shot investments are being considered, the
planning horizon is defined to be equal to the longest life among the investment alternatives. Then, the present worth is computed for each alternative.
Selecting from Among Multiple One-Shot Investments
Consider the two cash flow diagrams given in Figure 4.5. Both alternatives
are one-shot investments. As such, we cannot predict what investment
alternatives might be available in the future. However, the minimum attractive rate of return, 15 percent, reflects the opportunity to reinvest recovered
capital. Which alternative is preferred?
One-Shot Investment An
investment that is available
only once.
EXAMPLE
136
Chapter 4
Present Worth
$6000
$5000
$4500
$4000
$3500 $3500 $3500
$3000
$2000
$1000
(+)
0
1
2
3
4
0
1
2
3
4
5
6
(–)
Alternative 2
Alternative 1
$4000
MARR = 15%
$5000
FIGURE 4.5
KEY DATA
CFDs for Example 4.3
Note that the one-shot investments in question have different durations.
Therefore, we will use a 6-year planning horizon, with zero cash flows
occurring in years 5 and 6 for Alternative 1.
Given: The cash flows outlined in Figure 4.5; MARR 5 15%; planning
horizon 5 6 years
Find: PW of each alternative
SOLUTION
Using present worth analysis, the following results are obtained:
PW1 115%2 5 2$4,000 1 $3,5001P Z A 15%,42 1 $1,0001P Z F 15%,42
5 2$4,000 1 $3,50012.854982 1 $1,00010.571752
5 $6,564.18
5 NPV115%,3500,3500,3500,45002 2 4000
5 6564.18
PW2 115%2 5 2$5,000 1 $1,0001P Z A 15%,62 1 $1,0001P Z G 15%,62
5 2$5,000 1 $1,00013.784482 1 $1,00017.936782
5 $6,721.26
5 NPV115%,1000,2000,3000,4000,5000,60002 2 5000
5 6721.26
Thus, we would recommend Alternative 2.
4-3
4-3
Benefit-Cost Analysis
BENEFIT-COST ANALYSIS
LEARNING O BJECTI VE: Perform an economic analysis of public invest-
ments utilizing the benefit-cost analysis for a given interest rate.
Government units fund projects using money taken, usually in the form of
taxes, from the public. They then provide goods or services to the public
that would be infeasible for individuals to provide on their own. While
they are not in business to make a profit, it is important that they make
wise investment decisions. Projects should provide benefits for the public’s greater good that exceed the costs of providing those benefits. The
most frequently used method in evaluating government (local, state, or
federal) projects is benefit-cost analysis.
Four classes cover the spectrum of projects that government enters:
cultural development, protection, economic services, and natural resources.
■
■
■
■
Cultural development is enhanced through education, recreation, and
historic and similar institutions or preservations.
Protection is achieved through military services, police and fire protection, and the judicial system.
Economic services include transportation, power generation, and
housing loan programs.
Natural resource projects entail wildland management, pollution control, and flood control.
Some projects belong in more than one area. For example, flood control is a form of assistance for some, provides transportation and power
generation for others, and also relates to natural resource benefits.
Many government projects are huge, having first costs of tens or hundreds of millions of dollars. They may have long lives, such as 50 years for
a bridge or a dam. The multiple-use concept is common, as in wildland
management projects, where economic (timber), wildlife preservation
(deer, squirrel), and recreation projects (camping, hiking) are each considered uses of importance. The benefits or enjoyment of some government
projects are often completely out of proportion to the financial support of
specific individuals or groups. Also, there are often multiple government
agencies that have an interest in a project. For example, there is often federal support of state road projects. Finally, some public-sector projects are
not easily evaluated due to difficulty in estimating benefits. Also, it may be
many years before their benefits are realized.
4.3.1
Benefit-Cost Calculations for a Single Alternative
Benefit-cost analysis is recommended for a formal economic analysis of
government programs or projects. Its two most common forms are the
Video Lesson:
Benefit-Cost Analysis
137
138
Chapter 4
Present Worth
Benefit-Cost Ratio (B/C)
The ratio of the present
worth of benefits (savings
or positive-valued cash
flows) to the negative of
the present worth of
the investment(s) (or
negative-valued cash
flows) using i 5 MARR.
benefit-cost ratio (B/C) or, equivalently, a measure of benefits minus
costs (B 2 C). With either of these two forms, both the benefits (B) and
costs (C) are typically expressed as present worth or annual worth monetary figures, using an appropriate discount rate for the time value of
money.
The mathematics of benefit-cost analysis is quite simple, although its
application and quantification of benefits can be challenging. If
Bjt 5 net public benefits associated with alternative j during year t, t 5
1;2; . . . ; n
Cjt 5 net government costs associated with alternative j during year t,
t 5 0;1;2; . . . ; n, and
i 5 appropriate interest or discount rate,
then the B/C criterion may be expressed mathematically, using the present
worth of all net benefits over the present worth of all net costs, as
n
B/Cj 1i2 5
a Bjt 11 1 i2
2t
a Cjt 11 1 i2
2t
t51
n
(4.3)
t50
Although Equation 4.3 is a ratio of present worths of benefits to costs, it
could just as well be a ratio of annual worths of benefits to costs.
The benefits minus costs criterion (B 2 C) is expressed as
n
1B 2 C2 j 1i2 5 a 1Bjt 2 Cjt 2 11 1 i2 2t
(4.4)
t50
which is similar to the present worth method described earlier in this chapter.
It can be used directly to compare alternatives and will always be consistent with a proper B/C ratio analysis.
Benefit-Cost Analysis for a Single Alternative
EXAMPLE
Video Example
Consider a 10-year public investment program that will commit the government to a stream of expenditures appearing in the Cost column of
Figure 4.6. Real benefits appear in the Benefit column. Discounting takes
place at a rate of 7 percent. Is this program desirable?
4-3 Benefit-Cost Analysis
FIGURE 4.6
Costs and Benefits for Public Investment Program, i 5 7%
The present value of benefits is $1,424,102, and the present value of costs
is $1,063,987, so the net present value of benefits minus costs is $360,115.
The program is desirable when considered alone.
It is also common to calculate the ratio of benefits to costs, resulting in
B/C 5 $1,424,102/1,063,987 5 1.33. When the B/C ratio is greater than
1.00, the program is desirable when considered alone.
4.3.2 Benefit-Cost Calculations for Multiple Alternatives
When two or more project alternatives are being compared using a B/C
ratio, the analysis should be done on an incremental basis—that is, let the
alternative with the lower present worth of costs be Alternative 1 and let
the other be Alternative 2. Then, the incremental benefits of the second
alternative over the first, DB2–1(i), are divided by the incremental costs of
the second over the first, DC2–1(i)—that is,
n
2t
a 1B2t 2 B1t 2 11 1 i2
DB221 1i2
t51
DB/C221 1i2 5
5 n
DC221 1i2
2t
a 1C2t 2 C1t 2 11 1 i2
(4.5)
t50
Note that if the first alternative is to do nothing, the incremental B/C
ratio is also the straight B/C ratio for the second alternative. As long as
DB/C2–1(i) exceeds 1.0, Alternative 2 is preferable to Alternative 1; otherwise, Alternative 1 is preferred to Alternative 2. The winner of these is then
compared on an incremental basis with another alternative. These pairwise comparisons continue until all alternatives have been exhausted and
only the best project remains.
SOLUTION
139
140
Chapter 4
Present Worth
With the B 2 C criterion, an incremental basis may be used following
the same rules as for the B/C ratio, preferring Alternative 2 to Alternative
1 as long as the following condition holds:
n
D 1B 2 C2 221 1i2 5 DB221 1i2 2 DC221 1i2 5 a 3B2t 1i2 2 B1t 1i2 4 11 1 i2 2t
n
t51
2 a 3C2t 1i2 2 C1t 1i2 4 11 1 i2 2t $ 0
(4.6)
t50
The incremental basis of Equation 4.6 is not required when using the
B 2 C criterion, as long as benefits and costs are known for each alternative. In this case, Equation 4.4 could be used directly for each alternative
and the maximum value selected.
EXAMPLE
Benefit-Cost Analysis of Three Routes
The state of Washington must decide between three highway alternatives
to replace an old winding road. The length of the current route is 26 miles.
Planners agree that the old road must be replaced or overhauled; they
cannot keep it as it is.
■
■
■
Alternative A is to overhaul and resurface the old road at a cost of
$2 million/mile. Resurfacing will also cost $2 million/mile at the end
of each 10-year period. Annual maintenance will cost $10,000/mile.
Alternative B is to cut a new road following the terrain. It will be only
22 miles long. Its first cost will be $3 million/mile with resurfacing at
10-year intervals costing $2 million/mile with annual maintenance at
$12,000/mile.
Alternative C also involves a new highway to be built along a 20.5-mile
straight line. Its first cost will be $4 million/mile with resurfacing at
10-year intervals costing $2 million/mile and with annual maintenance costing $20,000/mile. This increase over Routes A and B is due
to additional roadside bank retention efforts.
Traffic density along each of the three routes will fluctuate widely from day
to day but will average 4,000 vehicles/day throughout the 365-day year. This
volume is composed of 350 light trucks, 250 heavy trucks, and 80 motorcycles, and the remaining 3,320 are automobiles. The average operation cost for
each of these vehicles is $0.70, $1.10, $0.30, and $0.60 per mile, respectively.
There will be a time savings because of the different distances along
each of the routes, as well as different speeds that each of the routes will
sustain. Route A will allow heavy trucks to average 35 miles/hour, while
the other vehicles can maintain 45 miles/hour. For each of Routes B and C,
these numbers are 40 miles/hour for heavy trucks and 50 miles/hour for
4-3
Benefit-Cost Analysis
other vehicles. The cost of time for all commercial traffic is valued at
$25/hour and for noncommercial vehicles, $10/hour. Twenty-five percent
of the automobiles and all of the trucks are considered commercial.
Finally, there is a significant safety factor to be included. An excessive
number of accidents per year have occurred along the old winding road.
Route A will reduce the number of vehicles involved in accidents to 105,
and Routes B and C are expected to involve only 75 and 50 vehicles in
accidents, respectively. The average cost per vehicle in an accident is estimated to be $18,000, considering actual property damages, lost time and
wages, medical expenses, and other relevant costs.
The Washington planners want to compare the three alternative routes
using benefit-cost criteria, specifically the popular benefit-cost ratio.
The benefit-cost analysis includes the government first costs, resurfacing
costs, and maintenance costs plus the public operational, time, and accident costs that follow.
Annual public operational costs are:
Route A: [(350)(0.70) 1 (250)(1.10) 1 (80)(0.30) 1 (3,320)(0.60)]
(26)(365) 5 $24,066,640/year
Route B: [(350)(0.70) 1 (250)(1.10) 1 (80)(0.30) 1 (3320)(0.60)]
(22)(365) 5 $20,364,080/yr
Route C: [(350)(0.70) 1 (250)(1.10) 1 (80)(0.30) 1 (3320)(0.60)]
(20.5)(365) 5 $18,975,620/yr
Annual public time costs:
Route A: [(350)(1/45)(25) 1 (250)(1/35)(25) 1 (80)(1/45)(10)
1 (3,320)(0.25)(1/45)(25) 1 (3,320)(0.75)(1/45)(10)]
(26)(365) 5 $13,335,710/year
Route B: [(350)(1/50)(25) 1 (250)(1/40)(25) 1 (80)(1/50)(10)
1 (3,320)(0.25)(1/50))(25) 1 (3,320)(0.75)(1/50)(10)]
(22)(365) 5 $10,119,808/year
Route C: [(350)(1/50)(25) 1 (250)(1/40)(25) 1 (80)(1/50)(10)
1 (3,320)(0.25)(1/50))(25) 1 (3,320)(0.75)(1/50)(10)]
(20.5)(365) 5 $9,429,821/year
Annual accident costs per vehicle:
Route A: (105)(18,000) 5 $1,890,000/year
Route B: (75)(18,000) 5 $1,350,000/year
Route C: (50)(18,000) 5 $900,000/year
All relevant government and public costs are summarized in Figure 4.7.
SOLUTION
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Present Worth
FIGURE 4 . 7
Summary of Data for Three Routes of Example 4.5
Since planners have not yet defined the “benefits” per se, public benefits
are taken as the incremental reduction in user costs between each pair of
alternatives evaluated. Then, the incremental benefits are compared against
the respective incremental costs needed to achieve the incremental benefits.
The incremental benefits and costs for Route B as compared to Route A
using a real discount rate of 8 percent are given as follows:
DBB2A 18%2 5 Public costsA 18%2 2 Public costsB 18%2
$39,292,350 2 $31,833,888 5 $7,458,462
DCB2A 18%2 5 Government costsB 18%2 2 Government costsA 18%2
$8,775,501 2 $8,009,533 5 $765,968
That is, for an incremental expenditure of $765,968/year, the government
can provide added benefits of $7,458,462/year for the public. The appropriate benefit-cost ratio is then
DB/CB2A 18%2 5
DBB2A 18%2
$7,458,462
5
5 9.74
DCB2A 18%2
$765,968
4-3 Benefit-Cost Analysis
This clearly indicates that the additional funds for Route B are worthwhile,
and Route B is desired over Route A.
Using a similar analysis, the benefits, costs, and DB/C ratio may
now be calculated to determine whether or not Route C is preferable to
Route B:
DBC2B 18%2 5 Public costsB 18%2 2 Public costsC 18%2
$31,833,888 2 $29,305,441 5 $2,528,447
DCC2B 18%2 5 Government costsC 18%2 2 Government costsB 18%2
$10,162,134 2 $8,775,501 5 $1,386,633
DBC2B 18%2 5
DBC2B 18%2
$2,528,447
5
5 1.82
DCB2A 18%2
$1,386,633
This benefit-cost ratio, being greater than 1.00, shows that the additional
expenditure of $1,386,633/year to build and maintain Route C would provide commensurate benefits in public savings of $2,528,447/year. Of the
three alternatives, Route C is preferred.
Figure 4.8 illustrates the benefit-cost calculations using Excel®.
FIGURE 4.8
Benefit-Cost Calculations for Example 4.5
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Present Worth
4-4
DISCOUNTED PAYBACK PERIOD
LEARN I N G O B JEC T I V E : Calculate the discounted payback period (DPBP) for
Video Lesson:
Payback Period
and Discounted
Payback Period
Discounted Payback Period
(DPBP) The amount of
time required for the
cumulative present worth
to become positive using
i 5 MARR.
a given interest rate to determine how long it takes for the cumulative present worth (PW) to be positive.
The time required for an investment to be fully recovered, including the
time value of money, is the amount of time required for the cumulative
present worth to become positive using i 5 MARR; it is called the
discounted payback period (DPBP) for an investment. Note that the DPBP
may be a fractional or whole-year amount.
In this section, we determine the DPBP for single and multiple investment alternatives.
4.4.1 Discounted Payback Period for a Single Alternative
We begin with a relatively simple example of a DPBP analysis.
Computing the DPBP of a Single Investment Opportunity
EXAMPLE
Video Example
KEY DATA
An initial $400,000 investment in new production equipment will yield
annual positive cash flows of $150,000 in year 1. Annual cash flows will
decrease by $25,000 each year thereafter. The new equipment has a useful life of 7 years. MARR is 10% per year. Determine the DPBP of this
project.
Given: Net cash flows as shown in Table 4.1; MARR 5 10%, planning
horizon 5 7 years
Find: DPBP
Net Cash Flows
for Example 4.6
TABLE 4.1
EOY
Net Cash Flow
0
–$400,000
1
150,000
2
125,000
3
100,000
4
75,000
5
50,000
6
25,000
7
0
4-4 Discounted Payback Period
To determine the DPBP for this investment, we must calculate the time
required for the cumulative present worth to equal 0. As shown in Table 4.2,
the cumulative PW reaches 0 and the investment is fully recovered between
years 5 and 6. Assuming continuous cash flows requires interpolation to
calculate the DPBP of 5 1 (2,927.02/14,111.85) 5 5.21 years. If we
assume an end-of-year cash flow convention, the DBPB is 6 years.
SOLUTION
Present Worth and Cumulative Present Worth of Example 4.6
TABLE 4.2
Cash Flows
EOY
Cumulative
PW of CF
Net CF
(P/F 10%, n)
PW of CF
0
2$400,000
1.00000
2400,000
2400,000
1
150,000
0.90909
136,363.64
2263,636.36
2
125,000
0.82645
103,305.79
2160,330.58
3
100,000
0.75131
75,131.48
285,199.10
4
75,000
0.68301
51,226.01
233,973.09
5
50,000
0.62092
31,046.07
22,927.02
6
25,000
0.56447
14,111.85
11,184.83
7
0
0.51316
0
To illustrate discounted payback period for a single alternative using Excel,
we revisit Example 4.1.
Using the Excel® NPER Tool to Determine DPBP of a Single
Investment Opportunity
EXAMPLE
In Example 4.1, suppose management asked the manufacturing engineer
to determine how long it takes for the new SMP machine to recover fully
its initial cost of $500,000, including the time value of money.
Given: The cash flows shown in Figure 4.1; MARR 5 10%; planning
horizon 5 10 years
Find: DPBP using the Excel NPER function
KEY DATA
The Excel® NPER worksheet function can be used to determine how long
it takes for the $500,000 investment to be recovered, based on an annual
return of $92,500. What is not known, however, is what the SMP machine’s
salvage value will be if its useful life is less than the 10-year planning
horizon. To provide management with more information regarding the
recovery period, the manufacturing engineer did the following analysis.
SOLUTION
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Chapter 4
Present Worth
First, the engineer used the Excel® NPER worksheet function, assuming a negligible salvage value:
yrs to recover the investment
5 NPER 110%, 92500,25000002 5 8.16 years
Next, the engineer estimated the salvage value if the machine was sold at any
time before the end of the planning horizon. Not knowing exactly how the
salvage value would decrease with use, the engineer made two approximations: The salvage value decreases as a geometric series, and the salvage
value decreases as a gradient series. The engineer computed the geometric
rate required to yield a $50,000 salvage value after 10 years and obtained a
value of 20.6 percent; then, he computed the gradient step required and
obtained a value of $45,000. (As an exercise, you should verify the engineer’s calculations.) Using the salvage values shown in Figure 4.9, the
FIGURE 4.9
Plot of Cumulative Present Worths for Example 4.7
4-4 Discounted Payback Period
engineer calculated the present worth, assuming the SMP machine was kept
for 1, 2, . . . , 10 years.
Based on the salvage value decreasing at a geometric rate of 20.6 percent each year, as shown in Figure 4.9, the investment in the new SMP
machine is fully recovered during the seventh year. Based on the salvage
value decreasing at a gradient rate of $45,000 per year, the investment is
fully recovered during the third year.
Management was much more comfortable making the $500,000
investment after learning that the present worth over the planning horizon
was $87,649.62 and that, if things went badly, the investment would be
fully recovered in no more than 7 years and, possibly, as quickly as 3 years,
depending on the salvage value for the SMP machine.
We do not recommend using DPBP to identify the investment that is to
be made from among a set of mutually exclusive investment alternatives.
Instead, we recommend it be used as a supplemental tool, just as it was used
in the SMP example. If DPBP is used as a stand-alone measure of economic
worth, it is difficult to understand how one would decide if the DPBP value
obtained was acceptable or not. Such decisions would have to be arbitrary,
because we know of no rational basis for saying, in the case of the previous
example, that a DPBP value of 5 is acceptable but a value of 6 is not.
As illustrated in the example, if the salvage value for an investment of
$P in an asset is negligible, regardless of how long the asset is used, then
the Excel® NPER worksheet function can be used to determine the DPBP,
when the annual returns are a uniform annual series of $A,
DPBP 5 NPER(MARR,A,2P)
If, on the other hand, salvage value decreases with usage according to
some mathematical relationship, then the Excel® SOLVER and GOAL
SEEK tools can be used to determine the DPBP. In the previous example,
salvage value was assumed to decline either geometrically or ‘‘gradiently.’’
The following example illustrates the calculations involved in determining
the DPBP for the SMP investment.
Using the Excel® SOLVER Tool to Calculate the DPBP
EXAMPLE
Use the Excel® SOLVER tool to determine the DPBP for the SMP investment.
Figure 4.10 shows a spreadsheet with the relevant input parameters.
KEY DATA
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Chapter 4
Present Worth
Excel® SOLVER Set Up to Calculate the DPBP in Example 4.8 with Gradient
Decrease in Salvage Value
FIGURE 4.10
SOLUTION
When salvage value decreases at an annual rate of 20.6 percent, the formula for salvage value after n years of use is $500,000(0.794)n. Likewise,
when it decreases by $45,000 each year, the salvage value after n years is
$500,000 − $45,000n.
To determine the DPBP, the value of n that makes PW equal zero can be
obtained using SOLVER, twice: once for the geometric decrease in salvage value and again for the gradient or linear decrease in salvage value.
After entering a “trial value” in cell B2 for the DPBP when salvage
value decreases as a geometric series, SOLVER is used to determine the
value of B2 that makes present worth (cell B6) equal 0. As shown in Figure
4.11, SOLVER yields a value of 6.95 years when salvage value decreases
geometrically over time.
When salvage value decreases as a gradient series, a “trial value” for
DPBP is entered in cell B8 and SOLVER is used to determine the value of
cell B8 that makes present worth (cell B12) equal 0. As shown in Figure
4.11, SOLVER yielded a value of 2.17 years when salvage value decreases
by $45,000 per year.
4-4 Discounted Payback Period
FIGURE 4.11 Excel® SOLVER Solutions for the DPBP in Example 4.8
with Geometric and Gradient Decreases in Salvage Value
When using DPBP, salvage values should not be ignored. However,
determining salvage values for periods of use ranging from 1 year to n
years tends to be a very inexact process. The need to know salvage values
for all possible periods of use is a limitation of the DPBP method.
4.4.2 Discounted Payback Period for Multiple Alternatives
We have emphasized the use of DPBP as a supplemental tool when comparing investment alternatives. We do not recommend it as the sole basis
for choosing the preferred alternative. To understand why, consider the
following example.
Using DPBP When Choosing Among Three Alternatives
EXAMPLE
Recall Example 4.2 involving a theme park and two design alternatives.
Now we add a third design alternative, which requires an initial investment
of $150,000 and will produce annual after-tax revenue represented by a
decreasing gradient series, with a revenue of $45,000 the first year and
decreasing by $5,000 per year to a final value of 0 in the last year. Which
alternative has the smallest DPBP?
Given: The cash flows shown in Figure 4.12; MARR 5 10%
Find: DPBP for each alternative
KEY DATA
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Chapter 4
Present Worth
$55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000
(+)
0
1
2
3
4
5
6
7
8
9
10
(–)
Alternative A
$300,000
$80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000
(+)
0
1
2
3
4
5
6
7
8
9
10
(–)
Alternative B
$450,000
$45,000
(+)
0
1
$40,000
2
$35,000
3
$30,000
4
$25,000
$20,000
5
$15,000
6
7
$10,000
8
$5,000
9
$0
10
(–)
Alternative C
$150,000
FIGURE 4.12
SOLUTION
CFDs for the Alternatives in Example 4.9
The present worth values for the three alternatives (A, B, and C), for DPBP
years are given by
PWA 5 2$300,000 1 $55,0001P Z A 10%, DPBPA 2
PWB 5 2$450,000 1 $80,0001P Z A 10%, DPBPB 2
PWC 5 2$150,000 1 $45,0001PZ A 10%, DPBPC 2 2
$5,0001P ZG 10%, DPBPC 2
4-4 Discounted Payback Period
The Excel® NPER function can be used to solve for the values of DPBPA
and DPBPB that equates the present worth to 0:
DPBPA 5 NPER110%,255000,3000002 5 8.273 years
DPBPB 5 NPER110%,280000, 4500002 5 8.674 years
The Excel® SOLVER tool is used to obtain the value of DPBPC , as shown
in Figure 4.13.
DPBPC 5 6.273 years
Thus, alternative C has the smallest DPBP.
FIGURE 4.13
Excel® SOLVER Solution for DPBPC in Example 4.9
Based on the discounted payback period, the rank-ordered preference for
the alternatives is C, B, A—exactly the reverse order of their present worth
values. The present worth for C can be shown to be
PWC 5 2$150,000 1 $45,0001P Z A 10%,102 2 $5,0001P Z G 10%,102
5 2$150,000 1 $45,00016.144572 2 $5,000122.891342
5 $12,048.95
or, using Excel®,
PWC 5 1000*NPV110%,45,40,35,30,25,20,15,10,5,02 2 150000
5 12,048.81
whereas, using Excel®, PWA 5 37,951.19 and PWB 5 41,565.37.
Figure 4.14 illustrates how the three present worths change over the
duration of the planning horizon.
EXPLORING THE
SOLUTION
151
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Chapter 4
Present Worth
$50,000
$0
–$50,000
0
1
2
3
4
5
6
7
8
9
10
Present Worth
–$100,000
–$150,000
–$200,000
–$250,000
–$300,000
–$350,000
–$400,000
–$450,000
Investment Year
Alternative A
FIGURE 4. 1 4
Alternative B
Alternative C
Discounted Payback Period Analysis for Example 4.9
As Example 4.9 illustrates, the DPBP can lead to wrong conclusions
regarding the investment alternative that maximizes economic worth.
4-5
Video Lesson:
Capitalized Worth
Capitalized Worth (CW)
The value of all cash flows
converted to a single sum
equivalent at time zero
using i 5 MARR when
the planning horizon is
infinitely long.
CAPITALIZED WORTH
LEARN I NG O B JEC T I V E : Calculate the capitalized worth (CW) of an investment for a given interest rate when the planning horizon is infinitely long.
A special type of cash flow series is a perpetuity, a uniform series that
continues indefinitely. This is a special case, because an infinite series of
cash flows would rarely be encountered in the business world. However,
for such very long-term investment projects as bridges, tunnels, railways,
highways, hydroelectric dams, nuclear power plants, forest harvesting, or
the establishment of endowment funds where the estimated life is 50 years
or more, assuming that an infinite cash flow series exists will generally be
a good approximation.
The present worth equivalent of an infinitely long series of cash flows
is called the capitalized worth (CW). Because, in most applications, the
capitalized worth is being computed for investments that have few if any
positive returns, the capitalized worth is more generally referred to as the
capitalized cost (CC) of an undertaking.
The capitalized worth method is applicable only if there is reason to
believe a series of cash flows will extend indefinitely into the future.
4-5 Capitalized Worth
Because it does not use the same planning horizon as PW, FW, and AW,
there is no reason to assume that the results will be the same as those that
would occur when using a finite planning horizon. If one wants to compute
the present worth of an infinitely long uniform series of cash flows, which
we call capitalized worth, then
CW 5 A1P Z A i%,q2 5
A
i
(4.7)
Therefore, if the project in question involves an indefinite repetition of a
life cycle, then one can convert the life cycle costs to an annual equivalent
and divide the result by the MARR to obtain the CW.
4.5.1 Capitalized Worth for a Single Alternative
As with the other measures of economic worth, capitalized worth can be
used in the absence of alternatives. Because capitalized worth calculations
are generally performed when the do-nothing alternative is not feasible,
however, the requirement that CW . 0 in deciding whether or not to proceed with the investment is not appropriate. If only costs occur, then our
objective is to minimize capitalized cost.
Capitalized Cost of Building Maintenance
EXAMPLE
Every 10 years, the dome of the state capitol building has to be cleaned,
sandblasted, and retouched. It costs $750,000 to complete the work. Using
a 5 percent MARR, what is the capitalized cost for refurbishing the capitol
dome?
Given: Cash flow profile in Figure 4.15
Find: Capitalized Cost (CC)
KEY DATA
0
10
20
30
n=∞
$750,000
$750,000
$750,000
$750,000
$750,000
(+)
(–)
FIGURE 4.15
CFD for Example 4.10
CC 5 $750,0001A Z P 5%,102/0.05
5 $750,00010.129502/0.05
5 $1,942,500
SOLUTION
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Present Worth
Alternatively,
CC 5 $750,000 1 $750,0001A Z F 5%,102/0.05
5 $750,000 1 $750,00010.079502/0.05
5 $1,942,500
The alternative approach yielded the same result because of the following
relationship between the A Z P and A Z F factors: (A Z P i%,n) 5 (A Z F i%,n) 1 i.
EXPLORING THE
SOLUTION
A useful interpretation of the CC is as follows: If $1,942,500 is deposited in a
fund that pays 5 percent annual compound interest, then $750,000 can be paid
out every 10 years, forever, to cover the cost of cleaning the capitol dome.
Capitalized Cost for Highway Construction
EXAMPLE
A new highway is to be constructed, and asphalt paving will be used. The
asphalt will cost $150 per foot, including the material and the paving operation. The asphalt is expected to last 5 years before requiring resurfacing.
It is anticipated that the cost of resurfacing will remain the same per foot.
Concrete drainage ditches will be installed on each side of the highway;
they will each cost $7.75 per foot to install. Ditches will have to be replaced
every 15 years; the cost of replacing them will also be $7.75 per foot. Four
pipe culverts will be installed every mile; each culvert will cost $8,000 and
will last 10 years; replacement culverts will cost $10,000 each, indefinitely. Annual maintenance of the highway will cost $9,000 per mile.
Cleaning each culvert will cost $1,250 per year. Cleaning and maintaining
each ditch will cost $3.75 per foot per year. Determine the capitalized cost
(CC) per mile of highway using a MARR of 5 percent.
KEY DATA
Given: The relevant data is organized in Table 4.3
TABLE 4.3
Data for Highway Construction Costs
Paving
Initial Cost
Duration
Replacement Cost
$150/ft
$150/ft
5 yrs
$9,000/mi
1 yr
Ditches (install)
$7.75/ft
15 yrs
(cleaning/maintenance)
$3.75/ft
1 yr
4 3 $8,000/mi
4 3 $1,250/mi
10 yrs
1 yr
Maintenance
Culverts (install 1 replace)
(cleaning)
$7.75/ft
4 3 $10,000/mi
4 3 $1,250/mi
4-5
Capitalized Worth
Find: CC(highway/mile) 5 CC(paving/mile) 1 CC(maintenance/mile) 1
CC(ditches/mile) 1 CC(culverts/mile)
First, let’s compute the capitalized cost of paving the road and the ditches.
Because paving must occur every 5 years, there are two approaches that
can be used: (1) convert the initial paving of the highway and ditches to
an annual equivalent and then divide by the MARR to obtain the capitalized cost, or (2) treat the initial paving as a present cost and convert the
resurfacing and replacing costs to an annual equivalent and then divide by
the MARR to obtain the capitalized cost of resurfacing. The following
results:
1. CC1paving/mile2 5 5,280 ft/mi3 $150/ft1A Z P 5%,52
1 $7.75/ft1A Z P 5%,152 4 /10.052
5 $3,737,409
5 1PMT115%,5,25280*1502
1 PMT15%,15,25280*7.7522/0.05
5 $3,737,487
2. CC1paving/mile2 5 5,280 ft/mi5$150/ft 3 1 1 1A Z F 5%,52 4
1 $7.75/ft 3 1 1 1A Z F 5%,152 4 6/10.052
5 $3,737,409
5 5280*1150 1 PMT15%,5,,21502/0.05
1 7.75 1 PMT15%,15, ,27.75/0.05
5 $3,737,487
To complete the example, we add the capitalized cost for maintaining
the highway and ditches, plus the installation and maintenance of the
culverts:
CC1maintenance/mile2 5 $9,000/0.05 5 $180,000
CC1ditches/mile2 5 3 122 15,280 ft/mi2 1$3.75/ft2 4/ 10.052 5 $792,000
CC1culverts/mile2 5 142 3$8,000 1 $1,250/0.05
1 $10,0001A Z F 5%,102/0.05 4
5 $195,600
5 4* 18000 1 1250/0.05
1 PMT15%,10,, 2100002/0.052
5 $195,604
SOLUTION
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Chapter 4
Present Worth
Therefore,
CC1highway/mile2 5 $3,737,487 1 $180,000 1 $792,000 1 $195,604
5 $4,905,091
The capitalized cost for paving the highway and ditches is $3,737,487. The
capitalized cost to maintain the highway and ditches is $972,000. The capitalized cost to install culverts every 10 years and maintain them annually
is obtained by using the alternative approach in Example 4.10, since the
replacement costs differ from the initial installation cost; the result is a
capitalized cost of $195,604. Therefore, the overall capitalized cost to
build and maintain the highway is $4,905,091 per highway mile.
(Based on state highway estimates, the initial cost to construct a 4-lane
divided highway comparable to interstates is currently approximately
$25 million per highway mile. The initial cost to construct a 2-lane highway, as above, is approximately $1 million per highway mile, depending
on terrain and a host of other factors. Hence, the cost estimates provided
above are somewhat conservative.)
4.5.2 Capitalized Worth for Multiple Alternatives
When multiple alternatives exist, the alternative having the greatest CW
over the infinitely long planning horizon is recommended. Again, because
capitalized worth alternatives usually involve costs, not revenues, and
costs are designated with positive signs, the alternative with the smallest
CC is recommended.
Capitalized Cost for Water Delivery
EXAMPLE
Video Example
In a developing country, two alternatives are under consideration for delivering water from a mountainous area to an arid area in the country’s southern
region. A coated heavy-gauge plastic pipeline can be installed, with pumps
spaced appropriately along the pipeline. Alternatively, a canal can be built;
however, it will have greater water loss than the pipeline, due to evaporation
and poaching along the canal route. To compensate for the water loss, the
canal will have a greater carrying capacity than the pipeline.
It is estimated it will cost $125 million to install the pipeline. Major
replacements are planned every 15 years at a cost of $10 million. Pumping
and other annual operating and maintenance costs are estimated to be
$5 million.
Summary 157
The canal will cost $200 million to construct; its annual operating and
maintenance costs are anticipated to be $1 million. Major upgrades of the
canal are anticipated every 10 years, at a cost of $5 million.
Based on a 5 percent MARR and an infinitely long planning horizon,
which alternative has the lowest capitalized cost?
Given:
KEY DATA
Initial Cost
Annual O&M
Major
Pipeline
$125 million
$5 million
$10 million every
15 years
Canal
$200 million
$1 million
$5 million every
10 years
Find: Capitalized Cost (CC) of each alternative
Pipeline
CC 5 $125,000,000 1 3$10,000,0001A Z F 5%,152 1 $5,000,0004/0.05
5 $234,268,000.00
5 125000000 1 1PMT15%,15,2100000002 1 50000002/0.05
5 $234,268,457.52
SOLUTION
Canal
CC 5 $200,000,000 1 3$5,000,0001A Z F 5%,102 1 $1,000,0004/0.05
5 $227,950,000.00
5 200000000 1 1PMT15%,10, ,250000002 1 10000002/0.05
5 $227,950,457.50
The canal has the smaller capitalized cost and would be recommended.
KEY CONCEPTS
1. Learning Objective: List eight DCF methods for comparing economic
alternatives. (Section 4.1.1)
The eight DCF methods for comparing economic alternatives are:
1. Present worth (PW)
2. Benefit-cost ratio (B/C)
3. Discounted payback period (DPBP)
SUMMARY
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Chapter 4
Present Worth
4.
5.
6.
7.
8.
Capitalized worth (CW)
Annual worth (AW)
Future worth (FW)
Internal rate of return (IRR)
External rate of return (ERR)
2. Learning Objective: Differentiate between ranking methods and incremental methods of determining economic worth. (Section 4.1.2)
Present worth (PW), future worth (FW), annual worth (AW), and capitalized worth (CW) are all examples of ranking methods to determine economic worth. The alternative having the greatest worth value over the
planning horizon is the most attractive. The discounted payback period
(DPBP) method is another ranking method, but its goal is to identify the
investment with the shortest payback period.
The incremental methods include the internal rate of return (IRR), external rate of return (ERR), and the benefit-cost ratio (B/C). The preferred alternative is that which satisfies Principle #6: Money should continue to be
invested as long as each additional increment of investment yields a return
that is greater than the investor’s time value of money (TVOM).
3. Learning Objective: Calculate a present worth (PW) converting all cash flows
to a single sum equivalent at time zero for a given interest rate. (Section 4.2)
If the PW is positive, then the investment is recommended. When comparing multiple investments, the alternative with the highest PW is preferred.
Mathematically, this is expressed as
n
Maximize PWj 5 a Ajt 11 1 MARR2 2t
;j
(4.2)
t50
4. Learning Objective: Perform an economic analysis of public investments
utilizing benefit cost analysis for a given interest rate. (Section 4.3)
Public investments such as cultural development, public protection, economic services, and protecting the environment usually involve large
investments undertaken with tax dollars. In order to see if a public investment is attractive, we must ensure that the benefits for the public’s greater
good exceed the costs of providing those benefits, or in other words the
B/C ratio is greater than 1. Mathematically the B/C ratio is expressed as
n
B/C j 1i2 5
2t
a Bjt 11 1 i2
t51
n
a Cjt 11 1 i2
t50
(4.3)
2t
Summary 159
5. Learning Objective: Calculate the discounted payback period (DPBP) for a
given interest rate to determine how long it takes for the cumulative present
worth (PW) to be positive. (Section 4.4)
The DPBP shows the time for an investment to be fully recovered, including the time value of money (TVOM). When used as a stand-alone measure of economic worth, it is not always clear if the DPBP value obtained
is acceptable or not. Therefore, DPBP should not be used to identify the
investment that is to be made from a set of mutually exclusive investment
alternatives. Instead, it is best used as a supplemental tool with other economic worth analyses.
6. Learning Objective: Calculate the capitalized worth (CW) of an investment for a given interest rate when the planning horizon is infinitely long.
(Section 4.5)
Although an infinite series of cash flows would rarely be encountered in
the real world, the CW method is a good approximation for very long-term
investment projects such as bridges, tunnels, railways, dams, nuclear
power plants and others. In these cases we assume that cash flows will
extend indefinitely into the future. The alternative having the greatest CW
is preferred. The equation for CW is
CW 5 A1P Z A i%,q2 5 A/i
(4.7)
KEY TERMS
Benefit-cost Ratio (B/C), p. 138
Capitalized Worth (CW), p. 152
Discounted Payback Period
(DPBP), p. 144
One-Shot Investment, p. 135
Present Worth (PW), p. 130
Problem available in WileyPLUS
GO Tutorial Tutoring Problem available in WileyPLUS
Video Solution Video Solution available in WileyPLUS
FE-LIKE PROBLEMS
1.
When using present worth to evaluate the attractiveness of a single investment alternative, what value is the calculated PW compared to?
a. 0.0
c. 1.0
b. MARR
d. WACC
160
Chapter 4
Present Worth
2.
A natural gas well is projected to produce $200,000 in profit during its
first year of operation, $190,000 the second year, $180,000 the third year, and
so on, continuing this pattern. If the well is expected to produce for a total of
10 years, and the effective annual interest rate is 8 percent, which of the
following most closely represents the present worth of the well?
a. $1,770,000
c. $1,253,000
b. $1,508,000
d. $1,082,000
3.
The present worth of a multiyear investment with all positive cash flows
(incomes) other than the initial investment is PW . 0 at MARR 5 i%. If
MARR changes to (i 1 1)%, the present worth will be
a. Less than $10,000.
b. Equal to $10,000.
c. Greater than $10,000.
d. Cannot determine without the cash flow profile and a value for i
4.
Consider the following cash flow diagram. Which of the expressions is
not valid for the present worth?
$250 $250 $250
$200
$150
$100
+0
–
a.
b.
c.
d.
1
2
3
4
5
6
10%/yr
P 5 1001P Z A 10%,62 1 501P Z G 10%,32 1 1501P Z A 10%,32 1P Z F 10%,32
P 5 1001P Z A 10%,32 1 501P Z G 10%,32 1 2501P Z A 10%,32 1P Z F 10%,32
P 5 2501P Z A 10%,62 2 501P Z G 10%,32
P 5 1001P Z A 10%,42 1 501P Z G 10%,42 1 2501P Z A 10%,22 1P Z F 10%,42
5.
Ivan, an industrial engineering student, is working on a homework problem for Engineering Economy. He needs to calculate the PW at 12 percent of
a cash flow series with $1,000 at t 5 3, $1,500 at t 5 4, and $2,000 at t 5 5.
If Ivan uses the equation P 5 1,000(P Z A 12%,3) 1 500(P Z G 12%,3), where
is the P now located?
a. t 5 4
c. t 5 1
b. t 5 2
d. t 5 0
6.
The owner of a cemetery plans to offer a perpetual care service for grave
sites. The owner estimates that it will cost $120 per year to maintain a grave
site. If the interest rate is 8 percent, what one-time fee should the owner
charge for the perpetual care service?
a. $96
c. $1,500
b. $120
d. $12,000
7.
Consider a palletizer at a bottling plant that has a first cost of $150,000,
operating and maintenance costs of $17,500 per year, and an estimated net
salvage value of $25,000 at the end of 30 years. Assume an interest rate of
Summary 161
8 percent. What is the present equivalent cost of the investment if the planning horizon is 30 years?
a. $335,000
c. $360,000
b. $344,500
d. $395,500
8.
The heat loss through the windows of a home is estimated to cost the
homeowner $412 per year in wasted energy. Thermal windows will reduce
heat loss by 93 percent and can be installed for $1,232. The windows will have
no salvage value at the end of their estimated life of 8 years. Determine the net
present equivalent value of the windows if the interest rate is 10 percent.
a. $412
c. $1,044
b. $812
d. $1,834
9.
An inline filter has an estimated life of 9 years. By adding a purifier to
the filter, savings of $300 in annual operating costs can be obtained. Annual
interest on capital is 8 percent. Compute the maximum expenditure justifiable
for the purifier.
a. $24
b. $33
c. $300
d. $1,875
10.
The city council has approved the building of a new bridge over Running
Water Creek. The bridge will cost $17,000 for initial construction and have an
annual maintenance cost of $1,000. The council plans to withdraw money from
the city’s Bridges and Highways account to open a special account to cover
the initial construction and to fund a perpetuity to cover the maintenance costs
forever. How much money must be withdrawn from the Bridges and Highways
account if the city can expect to earn 5 percent on the special account?
a. $1,000
c. $18,000
b. $17,000
d. $37,000
11.
Two projects, A and B, are analyzed using ranking present worth analysis
with MARR at i%. It is found that PW(A) . PW(B). If MARR is changed to
(i 1 1)%, what will be the relationship between PW(A) and PW(B)?
a. PW(A) . PW(B)
c. PW(A) , PW(B)
b. PW(A) 5 PW(B)
d. Cannot be determined without the cash
flow profiles
12.
A library shelving system has a first cost of $20,000 and a useful life of
10 years. The annual maintenance is expected to be $2,500. The annual benefits
to the library staff are expected to be $9,000. If the effective annual interest rate
is 10 percent, what is the benefit-cost ratio of the shelving system?
a. 1.51
c. 1.73
b. 2.24
d. 1.56
13.
When using the benefit-cost ratio measure of worth, what benchmark is
the calculated ratio compared to in determining if an individual investment is
attractive?
a. 0.0
c. 1.0
b. MARR
d. IRR
162
Chapter 4
Present Worth
14.
When using a benefit cost ratio analysis to evaluate multiple alternatives,
which of the following approaches is acceptable?
a. Ranking approach only
b. Incremental approach only
c. Either incremental or ranking
d. Neither incremental nor ranking
15.
Elm City is considering a replacement for its police radio. The benefits
and costs of the replacement are shown below. What is the replacement’s
benefit-cost ratio if the effective annual interest rate is 8 percent?
Purchase Cost: $7,000
Annual Savings: $1,500
Life: 15 years
a. 3.21
c. 1.76
b. 1.83
d. 1.34
PROBLEMS
Section 4.1
1.
Comparing Alternatives
Match the measures of worth in the first column with an appropriate definition from the second column.
Measure of Worth
Definitions
(a)
Annual worth
(1)
(b)
(2)
(c)
Discounted payback
period
Capitalized worth
(d)
External rate of return
(4)
(e)
Future worth
(5)
(f)
Internal rate of return
(6)
(g)
Present worth
(7)
(3)
Converts all cash flows to a single sum
equivalent at t 5 (planning horizon) using
i 5 MARR
Converts all cash flows to a single sum
equivalent at t 5 0 using i 5 MARR
Converts all cash flows to an equivalent
uniform series over the planning horizon
Determines an interest rate that yields a
PW (or FW or AW) of 0
Determines how long it takes for the
cumulative present worth to be positive
at i 5 MARR
Determines the interest rate that equates
the future worth of invested capital to
the future worth of recovered capital
invested at i 5 MARR
Determines the PW when the planning
horizon is infinitely long
2. GeoWorld Systems uses a subset of the following questions during the inter-
view process for new engineers. For each of the following cases, determine if
“the project” or “do nothing” is preferred. The value of MARR in each case
is 14 percent.
a. The present worth of the project is $1,367.
b. The internal rate of return of the project is 12.9 percent.
c. The annual worth of the project is −$632.
Summary 163
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
n.
o.
p.
q.
The benefit cost ratio of the project is 1.08.
The future worth of the project is $3.75.
The external rate of return of the project is 15.3 percent.
The present worth of the project is −$47.
The internal rate of return of the project is 14.7 percent.
The annual worth of the project is $6,775.
The benefit cost ratio of the project is 0.97.
The future worth of the project is −$13,470.
The external rate of return of the project is 3.7 percent.
For the cases (if any) in which “do nothing” was preferred, what assumption is being made about the return generated by the “uninvested” funds?
Is it possible that the values stated in (a) and (b) were correctly calculated
on the same project? If not, why?
Is it possible that the values stated in (c) and (d) were correctly calculated
on the same project? If not, why?
What do you know must be true about the present worth for the project
in (j)?
What do you know must be true about the internal rate of return for the
project in (j)?
3. Match the measures of worth in the first column with an appropriate decision
rule for preferring a project over “do nothing.”
Measure of Worth
Decision Rule
(a) Annual worth
(b) Benefit-cost ratio
(c) External rate of return
(d) Future worth
(e) Internal rate of return
(f) Present worth
Section 4.2.1
4.
(1) Measure of worth is greater than 0
(2) Measure of worth is greater than 1
(3) Measure of worth is greater than MARR
Present Worth of a Single Alternative
Video Solution DuraTech Manufacturing is evaluating a process improvement project. The estimated receipts and disbursements associated with
the project are shown below. MARR is 6 percent/year.
End of Year
0
1
2
3
4
5
Receipts
Disbursements
$0
$0
$2,000
$4,000
$3,000
$1,600
$5,000
$200
$300
$600
$1,000
$1,500
a. What is the present worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on present worth?
c. Should DuraTech implement the proposed process improvement?
164
Chapter 4
Present Worth
5.
Bailey, Inc., is considering buying a new gang punch that would allow them
to produce circuit boards more efficiently. The punch has a first cost of $100,000
and a useful life of 15 years. At the end of its useful life, the punch has no salvage value. Annual labor costs would increase $2,000 using the gang punch, but
annual raw material costs would decrease $12,000. MARR is 5 percent/year.
a. What is the present worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on present worth?
c. Should Bailey buy the gang punch?
6. Repeat Problem 5 assuming that floor support and vibration dampening must be
added for the gang punch. These one-time first costs are estimated to be $35,000.
7. Repeat Problem 5 assuming that the cost of capital has increased due to weak
market conditions and MARR is now 6 percent/year.
8.
Carlisle Company has been cited and must invest in equipment to reduce
stack emissions or face EPA fines of $18,500 per year. An emission reduction
filter will cost $75,000 and will have an expected life of 5 years. Carlisle’s
MARR is 10 percent/year.
a. What is the present worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on present worth?
c. Is the filter economically justified?
d. State at least one noneconomic factor that might influence this decision.
9. Repeat parts (a) through (c) of Problem 8 assuming that Carlisle’s board of
directors has decided that because this action is based on a Federal government citation, no financial gain should be expected and the appropriate value
of MARR is 0.
10. Fabco, Inc., is considering purchasing flow valves that will reduce annual
operating costs by $10,000 per year for the next 12 years. Fabco’s MARR
is 7 percent/year. Using a present worth approach, determine the maximum
amount Fabco should be willing to pay for the valves.
11. Eddie’s Precision Machine Shop is insured for $700,000. The present yearly
insurance premium is $1.00 per $100 of coverage. A sprinkler system with an
estimated life of 20 years and no salvage value can be installed for $20,000.
Annual maintenance costs for the sprinkler system are $400. If the sprinkler
system is installed, the system must be included in the shop’s value for insurance purposes, but the insurance premium will reduce to $0.40 per $100 of
coverage. Eddie uses a MARR of 15 percent/year.
a. What is the present worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on present worth?
c. Is the sprinkler system economically justified?
12. Quilts R Us (QRU) is considering an investment in a new patterning attach-
ment with the cash flow profile shown in the table below. QRU’s MARR is
13.5 percent/year.
Summary 165
EOY
0
Cash Flow
2$1,400
$0
$500
$500
$500
$500
$0
$500
1
2
3
4
5
6
7
EOY
Cash Flow
8
$600
9
10
11
12
13
14
15
$700
$800
$900
−$1,000
−$2,000
−$3,000
$1,400
a. What is the present worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on present worth?
c. Should QRU invest?
13.
GO Tutorial Galvanized Products is considering purchasing a new computer system for their enterprise data management system. The vendor has
quoted a purchase price of $100,000. Galvanized Products is planning to borrow one-fourth of the purchase price from a bank at 15 percent compounded
annually. The loan is to be repaid using equal annual payments over a 3-year
period. The computer system is expected to last 5 years and has a salvage
value of $5,000 at that time. Over the 5-year period, Galvanized Products
expects to pay a technician $25,000 per year to maintain the system but will
save $55,000 per year through increased efficiencies. Galvanized Products
uses a MARR of 18 percent/year to evaluate investments.
a. What is the present worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on present worth?
c. Should the new computer system be purchased?
14. Jupiter is considering investing time and administrative expense on an effort
that promises one large payoff in the future, followed by additional expenses
over a 10-year horizon. The cash flow profile is shown in the table below.
Jupiter’s MARR is 12 percent/year.
EOY
0
Cash Flow (K$)
EOY
Cash Flow (K$)
2$2
6
1
2$10
7
2$10
2
2$12
8
2$12
$200
3
2$14
9
2$14
4
2$16
10
2$100
5
2$18
a. What is the present worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on present worth?
c. Should Jupiter invest?
166
Chapter 4
Present Worth
15.
Imagineering, Inc., is considering an investment in CAD-CAM compatible design software with the cash flow profile shown in the table below.
Imagineering’s MARR is 18 percent/year.
EOY
Cash Flow (M$)
EOY
Cash Flow (M$)
0
2$12
4
$5
1
2$1
$5
$2
5
$5
6
7
$2
$5
2
3
a. What is the present worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on present worth?
c. Should Imagineering invest?
16. Home Innovation is evaluating a new product design. The estimated receipts
and disbursements associated with the new product are shown below. MARR
is 10 percent/year.
End of Year
Receipts
Disbursements
0
1
2
3
4
$0
$600
$600
$700
$700
$1,000
$300
$300
$300
$300
5
$700
$300
a. What is the present worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on present worth?
c. Should Home Innovation pursue this new product?
17.
Video Solution Aerotron Electronics is considering purchasing a water
filtration system to assist in circuit board manufacturing. The system costs
$40,000. It has an expected life of 7 years at which time its salvage value will
be $7,500. Operating and maintenance expenses are estimated to be $2,000 per
year. If the filtration system is not purchased, Aerotron Electronics will have to
pay Bay City $12,000 per year for water purification. If the system is purchased,
no water purification from Bay City will be needed. Aerotron Electronics must
borrow half of the purchase price, but they cannot start repaying the loan for
2 years. The bank has agreed to three equal annual payments, with the first payment due at the end of year 2. The loan interest rate is 8 percent compounded
annually. Aerotron Electronics’ MARR is 10 percent compounded annually.
a. What is the present worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on present worth?
c. Should Aerotron Electronics buy the water filtration system?
18.
Video Solution Nancy’s Notions pays a delivery firm to distribute its products in the metro area. Delivery costs are $30,000 per year. Nancy can buy a
used truck for $10,000 that will be adequate for the next 3 years. Operating and
Summary 167
maintenance costs are estimated to be $25,000 per year. At the end of 3 years,
the used truck will have an estimated salvage value of $3,000. Nancy’s MARR
is 24 percent/year.
a. What is the present worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on present worth?
c. Should Nancy buy the truck?
Section 4.2.2
19.
Present Worth of Multiple Alternatives
The engineering team at Manuel’s Manufacturing, Inc., is planning to purchase an enterprise resource planning (ERP) system. The software and installation from Vendor A costs $380,000 initially and is expected to increase revenue
$125,000 per year every year. The software and installation from Vendor B
costs $280,000 and is expected to increase revenue $95,000 per year. Manuel’s
uses a 4-year planning horizon and a 10 percent per year MARR.
a. What is the present worth of each investment?
b. What is the decision rule for determining the preferred investment based
on present worth ranking?
c. Which ERP system should Manuel purchase?
20.
Parker County Community College (PCCC) is trying to determine
whether to use no insulation or to use insulation that is either 1 inch thick
or 2 inches thick on its steam pipes. The heat loss from the pipes without
insulation is expected to cost $1.50 per year per foot of pipe. A 1-inch thick
insulated covering will eliminate 89 percent of the loss and will cost $0.40
per foot. A 2-inch thick insulated covering will eliminate 92 percent of the
loss and will cost $0.85 per foot. PCCC Physical Plant Services estimates
that there are 250,000 feet of steam pipe on campus. The PCCC Accounting
Office requires a 10 percent/year return to justify capital expenditures. The
insulation has a life expectancy of 10 years. Determine which insulation (if any)
should be purchased using a ranking present worth analysis.
21.
GO Tutorial Nadine Chelesvig has patented her invention. She is offering
a potential manufacturer two contracts for the exclusive right to manufacture
and market her product. Plan A calls for an immediate single lump sum payment to her of $30,000. Plan B calls for an annual payment of $1,000 plus
a royalty of $0.50 per unit sold. The remaining life of the patent is 10 years.
Nadine uses a MARR of 10 percent/year.
a. What must be the uniform annual sales volume of the product for Nadine
to be indifferent between the contracts, based on a present worth analysis?
b. If the sales volume is below the volume determined in (a), which contract
would the manufacturer prefer?
22. Quantum Logistics, Inc., a wholesale distributor, is considering the construc-
tion of a new warehouse to serve the southeastern geographic region near the
Alabama–Georgia border. There are three cities being considered. After site
visits and a budget analysis, the expected income and costs associated with
locating in each of the cities have been determined. The life of the warehouse
is expected to be 12 years and MARR is 15 percent/year.
168
Chapter 4
Present Worth
City
Initial Cost
Net Annual Income
Lagrange
Auburn
Anniston
$1,260,000
$1,000,000
$1,620,000
$480,000
$410,000
$520,000
a. What is the present worth of each site?
b. What is the decision rule for determining the preferred site based on present
worth ranking?
c. Which city should be recommended?
23.
Tempura, Inc., is considering two projects. Project A requires an investment of $50,000. Estimated annual receipts for 20 years are $20,000; estimated
annual costs are $12,500. An alternative project, B, requires an investment of
$75,000, has annual receipts for 20 years of $28,000, and has annual costs of
$18,000. Assume both projects have a zero salvage value and that MARR is
12 percent/year.
a. What is the present worth of each project?
b. Which project should be recommended?
24. DelRay Foods must purchase a new gumdrop machine. Two machines are avail-
able. Machine 7745 has a first cost of $10,000, an estimated life of 10 years,
a salvage value of $1,000, and annual operating costs estimated at $0.01 per
1,000 gumdrops. Machine A37Y has a first cost of $8,000, a life of 10 years,
and no salvage value. Its annual operating costs will be $300 regardless of
the number of gumdrops produced. MARR is 6 percent/year, and 30 million
gumdrops are produced each year.
a. What is the present worth of each machine?
b. What is the decision rule for determining the preferred machine based on
present worth ranking?
c. Which machine should be recommended?
25.
Xanadu Mining is considering three mutually exclusive alternatives, as
shown in the table below. MARR is 10 percent/year.
EOY
A001
B002
C003
0
2$210
$80
$90
$100
$110
2$110
$60
$60
$60
$70
2$160
$80
$80
$80
$80
1
2
3
4
a. What is the present worth of each alternative?
b. Which alternative should be recommended?
26. Two storage structures, given code names Y and Z, are being considered for a
military base located in Sontaga. The military uses a 5 percent/year expected
Summary 169
rate of return and a 24-year life for decisions of this type. The relevant characteristics for each structure are shown below.
First Cost
Estimated Life
Estimated Salvage Value
Annual Maintenance Cost
Structure Y
Structure Z
$4,500
12 years
None
$1,000
$10,000
24 years
$1,800
$720
a. What is the present worth of each machine?
b. What is the decision rule for determining the preferred machine based on
present worth ranking?
c. Which structure should be recommended?
27. Several years ago, a man won $27 million in the state lottery. To pay off the
winner, the state planned to make an initial $1 million payment immediately
followed by equal annual payments of $1.3 million at the end of each year
for the next 20 years. Just before receiving any money, the man offered to sell
the winning ticket back to the state for a one-time immediate payment of
$14.4 million. If the state uses a 6 percent/year MARR, should it accept the
man’s offer? Use a present worth analysis.
28. Final Finishing is considering three mutually exclusive alternatives for a new
polisher. Each alternative has an expected life of 10 years and no salvage
value. Polisher 1 requires an initial investment of $20,000 and provides annual
benefits of $4,465. Polisher 2 requires an initial investment of $10,000 and
provides annual benefits of $1,770. Polisher 3 requires an initial investment of
$15,000 and provides annual benefits of $3,580. MARR is 15 percent/year.
a. What is the present worth of each polisher?
b. Which polisher should be recommended?
29.
An environmental consultant is considering the installation of a water
storage tank for a client. The tank is estimated to have an initial cost of
$426,000, and annual maintenance costs are estimated to be $6,400 per year.
As an alternative, a holding pond can be provided a short distance away at
an initial cost of $180,000 for the pond plus $90,000 for pumps and piping.
Annual operating and maintenance costs for the pumps and holding pond are
estimated to be $17,000. The planning horizon is 20 years, and at that time,
neither alternative has any salvage value. Determine the preferred alternative
based on a present worth analysis with a MARR of 20 percent/year.
30. Dark Skies Observatory is considering several options to purchase a new
deep-space telescope. Revenue would be generated from the telescope by
selling “time and use” slots to various researchers around the world. Four
possible telescopes have been identified in addition to the possibility of
not buying a telescope if none are financially attractive. The table below
details the characteristics of each telescope. A present worth ranking analysis
is to be performed.
170 Chapter 4
Present Worth
Useful Life
First Cost
Salvage Value
Annual Revenue
Annual Expenses
T1
T2
T3
T4
10 years
$600,000
$70,000
$400,000
$130,000
10 years
$800,000
$130,000
$600,000
$270,000
10 years
$470,000
$65,000
$260,000
$70,000
10 years
$540,000
$200,000
$320,000
$120,000
a. Determine the preferred telescope if MARR is 25 percent/year.
b. Determine the preferred telescope if MARR is 42 percent/year.
31.
GO Tutorial Value Lodges owns an economy motel chain and is considering building a new 200-unit motel. The cost to build the motel is estimated
at $8,000,000; Value Lodges estimates furnishings for the motel will cost
an additional $700,000 and will require replacement every 5 years. Annual
operating and maintenance costs for the motel are estimated to be $800,000.
The average rental rate for a unit is anticipated to be $40/day. Value Lodges
expects the motel to have a life of 15 years and a salvage value of $900,000
at the end of 15 years. This estimated salvage value assumes that the furnishings are not new. Furnishings have no salvage value at the end of each
5-year replacement interval. Assuming average daily occupancy percentages
of 50 percent, 60 percent, 70 percent, and 80 percent for years 1 through 4,
respectively, and 90 percent for the fifth through fifteenth years, MARR of
12 percent/year, 365 operating days/year, and ignoring the cost of land,
should the motel be built? Base your decision on a present worth analysis.
32. RealTurf is considering purchasing an automatic sprinkler system for its
sod farm by borrowing the entire $30,000 purchase price. The loan would
be repaid with four equal annual payments at an interest rate of 12 percent/
year. It is anticipated that the sprinkler system would be used for 9 years and
then sold for a salvage value of $2,000. Annual operating and maintenance
expenses for the system over the 9-year life are estimated to be $9,000 per
year. If the new system is purchased, cost savings of $15,000 per year will
be realized over the present manual watering system. RealTurf uses a MARR
of 15 percent/year for economic decision making. Based on a present worth
analysis, is the purchase of the new sprinkler system economically attractive?
Section 4.2.3
33.
Present Worth of One-Shot Investments
Consider the two one-shot investment alternatives shown in the table below.
Neither alternative is expected to be available again in the future. MARR is
11 percent/year. Based on a present worth analysis, which alternative is preferred?
EOY
0
1
2
3
4
5
6
7
Alternative W
Alternative X
2$100,000
$20,000
$20,000
$50,000
$80,000
2$150,000
$40,000
$45,000
$50,000
$55,000
$110,000
$60,000
$65,000
$70,000
Summary 171
34.
Video Solution Two new opportunities are being considered for a venture
capital firm. Both are one-time opportunities with no option for renewal. The
firm uses a 12 percent/year expected rate of return for decisions of this type.
The relevant characteristics for each option are shown below. Based on a
present worth analysis, which option is preferred?
Initial Investment
Estimated Life
Expected Annual Return
35.
Option 1
Option 2
$100,000
12 years
$16,500
$75,000
9 years
$14,300
Technology Innovations is planning to purchase one of two chip insertion
machines. Due to the pace of technological change in this area, it is realistic
to assume that these are one-shot investments. The expected cash flows for
each machine are shown below. MARR is 8 percent/year. Based on a present
worth analysis, which machine is preferred?
Initial Investment
Estimated Life
End of Life Salvage
Annual Income
Annual Expense
Section 4.3.1
E Series
M Series
$40,000
7
$10,000
$19,400
$10,000
$60,000
5
$0
$26,000
$6,000
Benefit-Cost Calculations for a Single Alternative
36.
Video Solution The Oklahoma City Zoo has proposed adding to their
Web site a major segment providing a virtual tour of the grounds and animals,
suitable for both routine enjoyment and educational purposes in classrooms.
Survey data indicate that this will have either a neutral or positive effect upon
actual zoo attendance. The Web site will be professionally done and have
an initial cost of $325,000. Upkeep, refreshing the videos, and developing
videos for scientific research and entertainment will cost another $80,000 per
year. The zoo is expected to be in operation for an indefinite period; however,
a study period of only 10 years for the Web site is to be assumed, with only a
residual (salvage) value of $60,000 for the archival value being anticipated.
Interest is 7 percent. An estimated 100,000 persons will visit the e-zoo in the
first year, increasing by 30,000 each year, and they will receive, on the average, an additional $0.80 of benefit per visit when the new area is complete.
On the basis of B/C analysis, should the Web site be supported for funding?
37.
Ten cavemen with a remaining average life expectancy of 10 years use a
path from their cave to a spring some distance away. The path is not easily
traveled due to 100 large stones that could be removed. The annual benefit to
each individual if the stones were removed is $6. Each stone can be removed
at a cost of $1. The interest rate is 2 percent.
a. Compute the benefit-cost ratio for the individual if he alone removed the
100 stones.
172 Chapter 4
Present Worth
b. Compute the benefit-cost ratio for the individual if the task was under-
taken collectively, with each individual removing 10 stones.
c. What maximum amount may be charged by a manager who organizes the
group effort if the minimum acceptable benefit-cost ratio is 2?
38. The Logan Public Library in Iowa serves long-term residents, “bedroom
community” residents who work in Omaha, and all of Harrison County. A
renovation is planned, especially to include access to more electronic volumes, modernized computer facilities, and quicker check-in and checkout. In
addition, two small meeting rooms with modernized e-access are needed. The
cost of the renovation, including cabling, will be $33,000, the new equipment
will cost $21,000, and e-volume access initiation will be $17,500. Maintenance is expected to run an additional $3,500 per year, plus $4,000 for renewed e-volume access. The interest rate is 8 percent, the planning horizon is
10 years, and the building renovation is expected to have a salvage value of
30 percent, with no salvage value for equipment or e-volume access. It is estimated that an additional 2,500 visits to the library will occur in the first year,
increasing by 500 per year thereafter. It is estimated that the average benefits
due to the new facilities, equipment, and access will be $2.00 per person per
visit.
a. Should the city government vote to approve the plans? Use PW and calcu-
late B 2 C.
b. What is the smallest benefit per person that will make this project desirable?
39.
Lincoln Park Zoo in Chicago is considering a renovation that will improve some physical facilities at a cost of $1,800,000. Addition of new species will cost another $310,000. Additional maintenance, food, and animal
care and replacement will cost $145,000 in the first year, increasing by
3 percent each year thereafter. The zoo has been in operation since 1868 and
is expected to continue indefinitely; however, it is common to use a 20-year
planning horizon on all new investments. Salvage value on facilities after
20 years will be 40 percent of initial cost. Interest is 7 percent. An estimated
1.5 million visits per year are made to the zoo, and the cost remains free yearround. How much additional benefit per visit, on average, must the visitors
perceive to justify the renovation?
40. The Boundary Waters Canoe Area Wilderness (BWCA) located in north-
eastern Minnesota, has 1 million acres of wilderness, 1,000 waterways, and
1,500 miles of canoe routes. While some youth, for example those in the
Boy Scouts, are on high-adventure treks for 10 days at a time, it is common
to have the rest of their family take advantage of the opportunity to also enjoy the BWCA area. A new area for this purpose is to be developed and will
have an equivalent annual cost of $30,000, including initial cost (design,
clearing, potable water, restrooms, showers, road, etc.), operating, upkeep,
and security. Approximately 250 families will camp for 8 days each during
the summer season. Another 2,200 persons will be admitted for single-day
use of the facilities during the summer season. Although there are currently
no fees charged, the average family camping in the area is willing to pay
$12.00/night for the privilege, with some willing to pay more and some
Summary 173
less. The average day user would be willing to pay $4.00/day, again some
more and some less.
a. What is the anticipated B/C ratio of this recreation area?
b. What is the value of B 2 C?
Section 4.3.2
41.
Benefit-Cost Calculations for Multiple Alternatives
Recent development near Eugene, Oregon, has identified a need for improved access to Interstate 5 at one location. Civil engineers and public planners are considering three alternative access plans. Benefits are estimated for
the public in general; disbenefits primarily affect some local proprietors who
will see traffic pattern changes as undesirable. Costs are monetary for construction and upkeep, and savings are a reduction in cost of those operations
today that will not be necessary in the future. All figures are relative to the
present situation, retention of which is still an alternative, and are annualized
over the 20-year planning horizon.
Alternative
A
B
C
Benefits
Disbenefits
Costs
Savings
$200,000
$37,000
$150,000
$15,000
$300,000
$69,000
$234,000
$31,000
$400,000
$102,000
$312,000
$42,000
a. What is the B/C ratio for each of these alternatives?
b. Using incremental B/C ratio analysis, which alternative should be selected?
c. Determine the value of B 2 C for each alternative.
42. A highway is to be built connecting Maud and Bowlegs. Route A follows the
old road and costs $4 million initially and $210,000/year thereafter. A new
route, B, will cost $6 million initially and $180,000/year thereafter. Route C
is an enhanced version of Route B with wider lanes, shoulders, and so on.
Route C will cost $9 million at first, plus $260,000 per year to maintain. Benefits to the users, considering time, operation, and safety, are $500,000 per
year for A, $850,000 per year for B, and $1,000,000 per year for C. Using a
7 percent interest rate, a 15-year study period, and a salvage value of 50 percent
of first cost, determine which road should be constructed.
43.
GO Tutorial The city of Columbus has identified three options for a
public recreation area suitable for informal family activities and major
organized events. As with most alternatives today, there are benefits, disbenefits, costs, and some savings. These have been estimated with the help
of an external planning consultant and are identified in the table below. In
each case, these are annualized over a 10-year planning horizon.
Benefits
Disbenefits
Costs
Savings
Option 1
Option 2
Option 3
$400,000
$78,000
$235,000
$25,000
$550,000
$125,000
$390,000
$65,000
$575,000
$180,000
$480,000
$90,000
174
Chapter 4
Present Worth
a. Determine the B/C ratio for each project. Can you tell from these ratios
which option should be selected?
b. Determine which option should be selected using the incremental B/C ratio.
c. Determine which option should be selected using B 2 C for each option.
d. At what value of Option 2 costs are you indifferent between Option 1 and
Option 2?
44.
Video Solution An improvement to the roadway is desired from Philmont
Scout Ranch to Springer in northeastern New Mexico. Alternative N (for north)
costs $2,400,000 initially and $155,000/year thereafter. Route SA (for south,
Alternative A) will cost $4,200,000 initially, and $88,000/year thereafter.
Route SB is the same as SA with wider lanes and shoulders. It costs $5,200,000
initially with maintenance at $125,000/year. User costs considering time,
operation, and safety are $625,000 for N, $410,000 for SA, and $310,000 for
SB. The salvage values for N, SA, and SB after 20 years are 20 percent of
initial cost, respectively. Using a MARR of 7 percent and a 20-year study
period, which should be constructed?
a. Use an incremental B/C analysis.
b. Use a B 2 C analysis.
c. Which route is preferred if 0 percent interest is used?
45.
Video Solution A relocation of a short stretch of rural highway feeding into Route 390 northwest of Dallas is to be made to accommodate new
growth. The existing road is now unsafe, and improving it is not an alternative. Alternate new route locations are designated as East and West. The
initial investment by government highway agencies will be $3,500,000 for
East and $5,000,000 for West. Annual highway maintenance costs will be
$120,000 for East and $90,000 for the shorter location West. Relevant annual
road user costs, considering vehicle operation, time en route, fuel, safety,
mileage, and so on, are estimated as $880,000 for East and only $660,000 for
West. Assume a 20-year service life and i 5 7 percent.
a. Clearly identify the annual equivalent benefits and costs of route West over
route East.
b. Compute the appropriate B/C ratio(s) and decide whether East or West
should be constructed.
46. Lynchburg has two old four-lane roads that intersect, and traffic is controlled by
a standard green, yellow, red stoplight. From each of the four directions, a left
turn is permitted from the inner lane; however, this impedes the flow of traffic
while a car is waiting to safely turn left. The light operates on a two-minute
cycle with 60 seconds of green-yellow and 60 seconds of red for each direction. Approximately 10 percent of the 12,000 vehicles using the intersection
each day are held up for an extra 2 full minutes and average 3 extra start-stop
operations, solely due to the left-turn bottleneck. These delays are realized
during 300 days per year. A start-stop costs 3 cents per vehicle, and the cost
of the excess waiting is $18/hour for private traffic and $45/hour for commercial traffic. Approximately 3,000 of the vehicles are commercial, with the
remainder being private. The potential benefit to the public is that the cost of
Summary 175
extra waiting and start-stops can be reduced by 90 percent through a project to
widen the intersection to include specific left-turn lanes and use of dedicated
left-turn arrows. If the planning horizon is 10 years and the city uses a 7 percent
interest rate, what is the most that can be invested in the project and maintain a
B/C ratio of 1.0 or greater? There will be no additional maintenance cost.
Section 4.4.1
Discounted Payback Period for a Single Alternative
47.
Reconsider Problem 5 (repeated here). Bailey, Inc., is considering buying
a new gang punch that would allow them to produce circuit boards more efficiently. The punch has a first cost of $100,000 and a useful life of 15 years. At
the end of its useful life, the punch has no salvage value. Labor costs would
increase $2,000 per year using the gang punch, but raw material costs would
decrease $12,000 per year. MARR is 5 percent/year.
a. What is the discounted payback period for this investment?
b. If the maximum attractive DPBP is 3 years, what is the decision rule for
judging the worth of this investment?
c. Should Bailey buy the gang punch based on DPBP?
48.
Video Solution Home Innovations is evaluating a new product design.
The estimated receipts and disbursements associated with the new product
are shown below. MARR is 10 percent/year.
End of Year
0
1
2
3
4
5
Receipts
Disbursements
$0
$600
$600
$700
$700
$700
$1,000
$300
$300
$300
$300
$300
a. What is the discounted payback period for this investment?
b. If the maximum attractive DPBP is 3 years, what is the decision rule for
judging the worth of this investment?
c. Should Home Innovations buy the gang punch based on DPBP?
Section 4.4.2
49.
Discounted Payback Period for Multiple Alternatives
Reconsider Problem 19 (repeated here). The engineering team at Manuel’s
Manufacturing, Inc., is planning to purchase an enterprise resource planning
(ERP) system. The software and installation from Vendor A costs $380,000
initially and is expected to increase revenue $125,000 per year every year.
The software and installation from Vendor B costs $280,000 and is expected
to increase revenue $95,000 per year. Manuel’s uses a 4-year planning horizon and a 10 percent per year MARR.
a. What is the discounted payback period of each investment?
b. Which ERP system should Manuel purchase if his decision rule is to select
the system with the shortest DPBP?
c. Does this decision agree or disagree with the results of the present worth
analysis in Problem 21?
176
Chapter 4
Present Worth
50.
Video Solution Octavia Bakery is planning to purchase one of two ovens.
The expected cash flows for each oven are shown below. MARR is 8 percent/
year.
Model 127B
Model 334A
$50,000
10
$10,000
$19,400
$10,000
$80,000
5
$0
$26,000
$6,000
Initial Investment
Estimated Life
End of Life Salvage
Annual Income
Annual Expense
a. What is the discounted payback period for each investment?
b. Which oven should Octavia Bakery purchase if they wish to minimize the
DPBP?
51.
Reconsider Problem 22 (repeated here). Quantum Logistics, Inc., a
wholesale distributor, is considering the construction of a new warehouse to
serve the southeastern geographic region near the Alabama–Georgia border.
There are three cities being considered. After site visits and a budget analysis,
the expected income and costs associated with locating in each of the cities
have been determined. The life of the warehouse is expected to be 12 years,
and MARR is 15 percent/year.
City
Initial Cost
Net Annual Income
Lagrange
Auburn
Anniston
$1,260,000
$1,000,000
$1,620,000
$480,000
$410,000
$520,000
a. What is the discounted payback period for each location?
b. Which city should Quantum Logistics select if they wish to minimize the
DPBP?
c. Is this recommendation consistent with a present worth analysis recom-
mendation in Problem 22?
Section 4.5
Capitalized Worth
52. A municipality is planning on constructing a water treatment plant at an initial
cost of $10,000,000. Every 5 years, major repairs and cleanup are required
at a cost of $2,000,000. Due to the necessity to remove sludge and make minor repairs, annual costs of operating the treatment plant are estimated to be
$700,000, $775,000, $850,000, $925,000, and $1,000,000 each year leading
up to the 5-year major repair and cleanup. Based on a 4 percent/year TVOM,
what is the capitalized cost for the water treatment plant?
53.
Video Solution Two incinerators are being considered by a waste management company. Design A has an initial cost of $2,500,000, has annual
operating and maintenance costs of $800,000, and requires overhauls every
5 years at a cost of $1,250,000. Design B is more sophisticated, including
computer controls; it has an initial cost of $5,750,000, has annual operating
and maintenance costs of $600,000, and requires overhauls every 10 years
at a cost of $3,000,000. Using a 5 percent/year interest rate, determine the
capitalized cost for each design and recommend which should be chosen.
Summary 177
54.
Video Solution A flood control project at Pleasant Valley dam is projected
to cost $2,000,000 today, have annual maintenance costs of $50,000, and have
major inspection and upkeep after each 5-year interval costing $250,000. If
the interest rate is 10 percent/year, determine the capitalized cost.
55.
GO Tutorial The gaming commission is introducing a new lottery game
called Infinite Progresso. The winner of the Infinite Progresso jackpot will
receive $1,000 at the end of January, $2,000 at the end of February, $3,000
at the end of March, and so on up to $12,000 at the end of December. At the
beginning of the next year, the sequence repeats starting at $1,000 in January and ending at $12,000 in December. This annual sequence of payments
repeats indefinitely. If the gaming commission expects to sell a minimum of
1 million tickets, what is the minimum price they can charge for the tickets
to break even, assuming the commission earns 6 percent/year/month on its
investments and there is exactly one winning ticket.
56. A generous benefactor donates $500,000 to a state university. The donation
is to be used to fund student scholarships. Determine the dollar amount of
scholarships that can be given out each year under each of the following conditions. The state university earns 4 percent per year on its investments.
a. The donation is a quasi-endowment designed to last 20 years.
b. The donation is a quasi-endowment designed to last 30 years.
c. The donation is a quasi-endowment designed to last 50 years.
d. The donation is an endowment designed to last forever.
57. Reconsider the data from Problem 56. Plot a graph of scholarship dollars
versus number of years, where the number of years varies from 1 year to 100
years by 2-year increments. Dollars should be on the y-axis and years on the
x-axis.
58. A prospective venture has the following cash flow profile over a 5-year
horizon. What is the capitalized worth at 6 percent/year assuming the pattern
repeats indefinitely?
End of Year
0
1
2
3
4
5
59.
Receipts
Disbursements
$100
$300
$500
$400
$350
$250
$1,100
$50
$250
$150
$100
$0
You decide to open an individual retirement account (IRA) at your local
bank that pays 8 percent/year compounded annually. At the end of each of the
next 40 years, you will deposit $4,000 into the account. Three years after your
last deposit, you will begin making annual withdrawals. What annual amount
will you be able to withdraw if you want the withdrawals to last
a. 20 years?
b. 30 years?
c. forever?
5
ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E :
J OS H L I U
Josh Liu has been investing in the stock market for several years, beginning
when he was a freshman in high school. His recent investments in Google
and Akamai stock have been quite successful. Based on a stellar academic
record as an undergraduate engineering student, Josh is contemplating
graduate school, with interests in pursuing an MBA. Due to his interests in
investments, Josh has a keen interest in financial engineering and possibly
working on Wall Street following completion of his master’s degree.
With an eye toward the future, Josh is interested in knowing what
annual investments and annual returns on investments are required for him
to amass an investment portfolio valued at $2 million within 30 years. He is
also interested in knowing how long it will take for him to achieve a net
worth in excess of $1 million with this same investment strategy.
Not only do individual investors establish financial goals for the future,
but so do businesses. Although we are not aware of firms using future worth
analysis as the principal tool to measure the economic worth of investment
alternatives, future worth is used as a supplement to present worth in an
engineering economic analysis. Also, some firms use future worth analysis to
estimate terminal values in business acquisition analysis. The most popular
use of future worth analysis is in retirement planning.
DISCUSSION QUESTIONS:
1. Thus far, Josh’s annual investments have been in the stock market.
What other investment options might you suggest to Josh? What factors should Josh consider in his stock investment portfolio?
178
ANNUAL WORTH
AND FUTURE WORTH
2. Josh wants to know how long it will take him to become a millionaire.
(Don’t we all!) What factors will affect this timeline?
3. To reach his $2 million goal within 30 years, Josh could determine the
necessary uniform annual savings. However, one might expect Josh’s
investment capacity to grow as his career advances. How would this
impact an investment plan for Josh?
4. Think of a specific example of a business acquisition or a capital facilities project where a future analysis would be an important consideration. Who might be most interested in this terminal value of the
investment? Why?
LEARNING OBJECTIVES
When you have finished studying this chapter, you should be able to:
1. Calculate an annual worth (AW) converting all cash flows to an equivalent uniform annual series of cash flows over the planning horizon for
a given interest rate. (Section 5.1)
2. Calculate a future worth (FW) converting all cash flows to a single sum
equivalent at the end of the planning horizon for a given interest rate.
(Section 5.2)
INTRODUCTION
In this chapter, we continue the discussion of using measures of economic
worth to compare mutually exclusive investment alternatives. Specifically,
we will look at annual worth analysis and future worth analysis.
179
180
Chapter 5
Annual Worth and Future Worth
Systematic Economic Analysis Technique
1.
2.
3.
4.
5.
6.
7.
Annual Worth (AW) The
value of all cash flows
converted to an equivalent
uniform annual series
of cash flows over the
planning horizon using
i 5 MARR.
Identify the investment alternatives
Define the planning horizon
Specify the discount rate
Estimate the cash flows
Compare the alternatives
Perform supplementary analyses
Select the preferred investment
Future worth analysis uses the MARR to express the economic worth
of a set of cash flows, occurring over the planning horizon, as a single
equivalent value at an ending time called “the future.” Future worth analysis
typically is used as a supplement to present worth analysis, rather than a
principal tool, in engineering economic analysis. It is also of great interest
to individuals in determining what the value of their investment portfolio
will be at some future time, such as for retirement planning.
In engineering economic analysis, annual worth is more frequently
used than future worth. As the name implies, annual worth is used to
express economic equivalency in the form of a uniform annual series over
the planning horizon. We begin this chapter by looking at annual worth
analysis methods.
5-1
ANNUAL WORTH
LEARN I N G O B JEC T I V E : Calculate an annual worth (AW) converting all cash
Video Lesson:
Annual Worth
flows to an equivalent uniform annual series of cash flows over the planning
horizon for a given interest rate.
In this section, we learn how to determine the uniform series equivalent for
the cash flows that occur for an investment alternative during the planning
horizon, and, when multiple mutually exclusive alternatives are available
for investment, how to choose the one that maximizes economic worth.
5.1.1 Annual Worth of a Single Alternative
Recalling our work in Chapter 2, the annual worth of an investment can be
expressed mathematically as follows:
n
AW 5 c a At 11 1 MARR2 n2t d 1A Z F MARR%,n2
t50
(5.1)
5-1 Annual Worth
Of course, the annual worth can also be calculated using Equation 4.1 to
compute the present worth and multiplying the result by (A Z P MARR%,n).
As with present worth analysis, the decision to pursue an investment
opportunity depends on AW . 0, assuming the do-nothing alternative is
feasible and has a negligible net cost. If the annual worth is positive, then
the investment will be recommended.
Annual Worth of a Single Alternative
EXAMPLE
We continue to use the acquisition of the SMP machine, introduced in
Example 4.1, as our example of a single alternative. Recall, it involved a
$500,000 investment, with annual returns of $92,500 and a $50,000 salvage
value at the end of the 10-year planning horizon with a MARR of 10%.
Given: The cash flows outlined in Figure 5.1; MARR 5 10%; planning
horizon 5 10 years
Find: AW of the investment. Is this investment recommended?
KEY DATA
$50,000
(+)
$92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500
0
1
2
3
4
5
6
7
8
9
10
(–)
$500,000
FIGURE 5.1
CFD for Example 5.1
The annual worth for the investment will be
AW 5 2$500,0001A Z P 10%,102 1 $92,500 1 $50,0001A Z F 10%,102
5 2$500,00010.162752 1 $92,500 1 $50,00010.062752 5 $14,262.50
or, using Excel®,
AW 5 PMT110%,10,500000,2500002 1 92500 5 14,264.57
SOLUTION
181
182
Chapter 5
Annual Worth and Future Worth
Since the choice is either do nothing or invest in the SMP machine and
obtain an annual worth of $14,264.57 for the 10-year planning horizon, we
choose to make the investment. Of course, the same choice would occur as
long as AW . $0.
In the previous chapter, we examined the impact of changes in the
MARR on the economic worth of an investment. We did so, however, only
when we were considering multiple alternatives. There was no reason for
not performing a similar analysis for a single alternative. But we chose to
wait until now to do so.
EXAMPLE
Using Excel® to Examine the Impact of Changes in the MARR
Let’s examine the behavior of annual worth for the SMP machine when
the MARR changes values, ranging from 0 percent to 20 percent. The plot
of annual worth for the 10-year planning horizon is provided in Figure 5.2.
FIGURE 5.2
Analyzing the Effect on AW of Changes in MARR for Example 5.2
5-1 Annual Worth
Also shown in the figure is the result of using the Excel® SOLVER tool
to determine the value of the MARR that makes the annual worth equal 0;
the value obtained is 13.80 percent. (In the next chapter, we explore the
internal rate of return, which is the MARR value that makes the annual
worth, present worth, or future worth of an investment equal 0.)
5.1.2 Annual Worth of Multiple Alternatives
When attempting to determine the preferred alternative from among multiple mutually exclusive alternatives and when using annual worth as the
measure of economic worth, choose the alternative that maximizes annual
worth over the planning horizon of n years. Mathematically, the objective
can be stated as
n
Maximize AWj 5 c a At j 11 1 MARR2 n2t d 1A Z F MARR%,n2 (5.2)
;j
t50
Using Annual Worth to Choose Between Two Alternatives
In Chapter 4, we considered the installation of a new ride (the Scream
Machine) at a theme park in Florida. Two alternative designs (A and B)
were considered: Alternative A required a $300,000 investment, produced net annual after-tax revenues of $55,000, and had a negligible
salvage value at the end of the 10-year planning horizon. Alternative B
required a $450,000 investment, generated net annual after-tax revenues of $80,000, and also had a negligible salvage value at the end of
the 10-year planning horizon. The do-nothing (DN) alternative was
feasible and had an economic worth of $0. Based on a 10 percent
MARR and using an annual worth measure, which design, if either,
should be chosen?
Given: The cash flows outlined in Figure 5.3; MARR 5 10%; planning
horizon 5 10 years
Find: AW of each investment alternative. Which investment is recommended?
EXAMPLE
Video Example
KEY DATA
183
184
Chapter 5
Annual Worth and Future Worth
$55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000
(+)
0
1
2
3
4
5
6
7
8
9
10
(–)
Alternative A
$300,000
$80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000
(+)
0
1
2
3
4
5
6
7
8
9
(–)
Alternative B
$450,000
FIGURE 5.3
CFDs for Example 5.3
AWA 5 2$300,0001A Z P 10%,102 1 $55,000
5 2$300,00010.162752 1 $55,000
5 $6,175.00
5 PMT110%,10,3000002 1 55000 5 6176.38
SOLUTION
Likewise,
AWB 5 2$450,0001A Z P 10%,102 1 $80,000
5 2$450,00010.162752 1 $80,000
5 $6,762.50
5 PMT110%,10,4500002 1 80000 5 6764.57
Since AWB . AWA . AWDN, Design B is recommended.
10
5-2
5-2
Future Worth
185
FUTURE WORTH
LEARNING O BJECTI VE: Calculate a future worth (FW) converting all cash
flows to a single sum equivalent at the end of the planning horizon for a
given interest rate.
Video Lesson:
Future Worth
Now we focus on what is variously referred to as future worth, future value,
and terminal value. Future worth analysis uses the MARR to express the economic worth of a set of cash flows, occurring over the planning horizon, as a
single equivalent value at an ending or termination time called ‘‘the future.’’
In the previous chapter, we noted that many investors prefer to express
the economic worth of the set of cash flows as a single monetary sum at a
point in time called ‘‘the present.’’ To obtain the single sum equivalent, we
discount cash flows that occur at various points in time in the future. Hence,
the name discounted cash flow analysis.
As popular as discounting money is, we have found that people have
far more difficulty understanding discounting than compounding. For
many, it is easier to grasp the notion of money growing in value as one
moves forward in time, rather than shrinking in value as one moves backward in time. For them, future worth analysis is more intuitively appealing
than present worth analysis.
Yet another reason for performing future worth analysis is goal setting.
When performing financial planning, many individuals are interested in knowing what the value of their investment portfolio will be at some particular
point in the future. For them, future worth is more relevant than present worth.
In this section, we learn how to make decisions regarding the economic viability of a single investment using future worth analysis. We also
learn how to determine the point in time when an investment begins to
‘‘make money.’’ In addition, we learn how to use future worth analysis to
choose from among multiple investment alternatives the one having the
greatest economic worth. Finally, we learn how to maximize the value of
an investment portfolio by considering both the money invested in a particular alternative and the available capital that is not invested.
Future Worth (FW) The
value of all cash flows
converted to a single sum
equivalent at the end of
the planning horizon using
i 5 MARR.
5.2.1
Future Worth of a Single Alternative
Recalling our work in Chapter 2 and letting i denote the MARR, the future
worth of an investment can be expressed mathematically as follows:
n
FW 5 a At 11 1 MARR2 n2t
(5.3)
t50
As with present worth analysis, the decision to pursue an investment
opportunity is dependent on FW . 0. If the future worth is positive, then
the investment will be recommended.
186
Chapter 5
Annual Worth and Future Worth
Future Worth of a Single Investment
EXAMPLE
Previously, we considered the acquisition of a new surface mount placement (SMP) machine having an initial cost of $500,000. It was anticipated
that the investment would result in annual operating and maintenance
costs being reduced by $92,500 per year, after taxes. The manufacturing
engineer estimated the machine would be worth $50,000 at the end of the
10-year planning horizon. Using a 10 percent after-tax MARR and future
worth analysis, should the investment be made?
KEY DATA
SOLUTION
Given: The cash flows outlined in Figure 5.1; MARR 5 10%; planning
horizon 5 10 years
Find: FW of the investment. Is this investment recommended?
The future worth for the investment will be
FW 5 2$500,0001F Z P 10%,102 1 $92,5001F Z A 10%,102 1 $50,000
5 2$500,00012.593742 1 $92,500115.937422 1 $50,000
5 $227,341.40
or, using Excel®,
FW 5 FV110%,10,292500,5000002 1 50000 5 227,340.55
Since FW . $0, the investment is recommended.
We noted previously that future worth analysis is appropriate when
planning to achieve a particular financial goal at some point in the future.
The following example illustrates the use of future worth analysis in financial planning.
Using Future Worth to Achieve a Financial Goal
EXAMPLE
A recent engineering graduate decided to begin an investment program at
the age of 23, with the hope of achieving an investment goal of $5 million
by age 58. If a gradient series describes the engineer’s investment pattern
over the 35-year period and if the annual return on the engineer’s investments is approximately 6.5 percent, what gradient step is required to
achieve the goal if the first of the 36 investments equals $5,000?
KEY DATA
Given: F 5 $5 million; i 5 6.5%; A1 5 $5,000; n 5 36
Find: G
5-2
Because the investment’s future worth is given and the unknown is the size
of the gradient step (G), the following equation is to be solved for G:
G 5 3$5,000,0001A Z F 6.5%,362 2 $5,000 4/ 1A Z G 6.5%,362
where
Future Worth
187
SOLUTION
1A Z F 6.5%,362 5 0.065/ 3 11.0652 36 2 1 4 5 0.0075133
and
1A Z G 6.5%,362 5 511.0652 36 2 31 1 3610.06524 6/50.065 311.0652 36 2 14 6
5 11.22339
Therefore,
G 5 3$5,000,00010.00751332 2 $5,000 4/11.22339 5 $2,901.66
SOLVER can be used to solve for G. As shown in Figure 5.4, we let
cell C20 contain the value of G. Then we generate the deposits by adding
Set Up to
Use the Excel® SOLVER
Tool to Determine the
Gradient Step Needed to
Achieve a Financial Goal
F I G U RE 5 . 4
188
Chapter 5
Annual Worth and Future Worth
C20 to the preceding deposit. The balance in the investment account
is computed by adding the most recent deposit to the product of the
previous balance and 1.065. After 36 years, when the engineer is 58, the
balance is to be $5 million. (If G 5 $3,000, the final balance is
$5,146,882.19.)
Figure 5.5 contains the solution obtained using SOLVER. Notice,
SOLVER is set up to make F19 equal 5000000 by changing C20. As
shown, if G 5 $2,901.67, then a $5 million balance will occur in the
investment account after 36 years.
FIGURE 5.5
EXPLORING THE
SOLUTION
Excel® SOLVER Solution to Example 5.5
What if the investments do not earn 6.5 percent? To gain an understanding of the impact on the future worth of the investments, Figure 5.6 was
developed, showing the growth in the investment portfolio over time
for various annual returns on the investment, where the formula for
(F/G i%,n) is given in Chapter 2. In anticipation of annual returns being
between 6 and 8 percent, the engineer anticipates the size of the investment portfolio will be between $4.6 million and $6.4 million after the
thirty-sixth deposit.
5-2
FIGURE 5.6
Future Worth
Impact of Annual Returns on an Investment Portfolio
5.2.2 Future Worth of a Multiple Alternatives
When attempting to determine the preferred alternative from among multiple mutually exclusive alternatives and when using future worth as the measure of economic worth, choose the alternative that maximizes FW. Mathematically, letting MARR denote the interest rate used, the objective is to
n
Maximize FWj 5 a Ajt 11 1 MARR2 n2t
;j
(5.4)
t50
Using Future Worth to Choose Between Two Alternatives
Recall previously, we compared economically two design alternatives
(A and B) for a new ride (the Scream Machine) at a theme park. Alternative A required a $300,000 investment, produced after-tax net annual revenue of $55,000, and had a negligible salvage value at the end of the
EXAMPLE
Video Example
189
190
Chapter 5
Annual Worth and Future Worth
10-year planning horizon. Alternative B required a $450,000 investment,
generated after-tax net annual revenue of $80,000, and also had a negligible
salvage value at the end of the 10-year planning horizon. The do-nothing
alternative is feasible. Based on an after-tax MARR of 10 percent and using
a future worth analysis, which alternative, if any, is the economic choice?
(The do-nothing alternative is assumed to have a future worth of $0.)
KEY DATA
Given: The cash flows outlined in Figure 5.3; MARR 5 10%; planning
horizon 5 10 years
Find: FW of each investment alternative. Which investment is recommended?
SOLUTION
FWA 5 2$300,0001F Z P 10%,102 1 $55,0001F Z A 10%,102
5 2$300,00012.593742 1 $55,000115.937422
5 $98,436.10
or, using Excel®,
FWA 5 FV110%,10,255000,3000002 5 $98,435.62
Likewise,
FWB 5 2$450,0001F Z P 10%,102 1 $80,0001F Z A 10%,102
5 2$450,00012.593742 1 $80,000115.937422
5 $107,810.60
or, using Excel®,
FWB 5 FV110%,10,280000,450000 5 $107,809.86
Since FWB . FWA . $0, Design Alternative B is recommended.
EXAMPLE
Using Future Worth to Choose a Retirement Plan
During a meeting with a human resources representative of her new
employer, a recent engineering graduate learns that she needs to choose
between two retirement plans. One plan involves investments that are
matched by the employer; the plan is managed by a committee of employees
in the firm. Another option is to establish an investment plan that she
will manage. In both cases, deposits are tax-deferred; also in both cases,
5-2
Future Worth
withdrawal of funds before age 62 will result in a significant financial penalty. Her current age is 22.
Under the first plan, up to 4 percent of an employee’s annual compensation is matched by the employer. Funds are invested in a mix of securities, including the company’s own stock; however, no more than 20 percent of the investment portfolio is allowed to be invested in the firm’s
stock. The portfolio includes a mix of stocks, bonds, and U.S. Treasury
notes. Over the past 15 years, the investment portfolio has increased in
value at an annual compound rate of 6 percent.
Under the second plan, individuals can manage their own investment
portfolios. Although there are limits on the investments that can be made,
a number of riskier choices are available. Individuals who choose to manage their own portfolio have to pay a management fee of 1.5 percent of
their annual deposit. The company still matches up to 4 percent of the
employee’s annual compensation.
After considering the investments available in the second plan, the
new employee narrows the choices to a set of investments that historically
have earned between 2 and 12 percent annually.
If the employee’s current salary is $55,000, she invests the maximum
allowed, and her annual salary increases at an annual rate of 5 percent,
which retirement plan should she choose?
Given: The employee and plan details summarized below.
First Plan
KEY DATA
Second Plan
4% of annual salary invested 1
4% company match
4% of annual salary invested 1
4% company match
Current annual salary 5 $55,000
Current annual salary 5 $55,000
Annual salary increase 5 5%
Annual salary increase 5 5%
Historical portfolio annual growth 5 6%
Historical portfolio annual growth 5
2% to 12%
No management fees
Annual management fee 5 1.5%
Find: FW of each plan.
Under the first plan, her investment portfolio will have the following value
after 40 years:
FW1 5 210.042 1$55,0002 1F Z A16%,5%,402
5 $4,4003 11.062 40 2 11.052 40 4/ 10.06 2 0.052
5 $1,428,120.90
SOLUTION
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Chapter 5
Annual Worth and Future Worth
Under the second plan, in the pessimistic case (earns 2 percent/year), her
investment portfolio will have the following value after 40 years:
FW2p 5 210.042 1$55,0002 10.9852 1F Z A12%,5%,402
5 $4,334 3 11.022 40 2 11.052 40 4/ 10.02 2 0.052
5 $698,055.57
Under the second plan, in the optimistic case (earns 12 percent/year), her
investment portfolio will have the following value after 40 years:
FW2o 5 210.042 1$55,0002 10.9852 1F Z A112%,5%,402
5 $4,334 3 11.122 40 2 11.052 40 4/ 10.12 2 0.052
5 $5,325,308.50
After giving the matter considerable thought, she decided to choose the
second plan and manage her own investment portfolio. What would you
have done? Given the impact on investment returns of the global economic
collapse of 2008, how reasonable are the annual increases in salary and the
annual returns on her investment over a 40-year period?
5.2.3 Portfolio Analysis
In Chapter 1 we noted the assumption that any money not invested in a
candidate alternative remains in an investment pool and earns a return
equal to the MARR. A portfolio analysis looks at the value of an entire
portfolio of investments, including the money in the investment pool.
Future worth is a convenient means of conducting a portfolio analysis.
EXAMPLE
Future Worth of the Portfolio
In the case of the two design alternatives (A and B) for the Scream
Machine, the most expensive one required a $450,000 investment. For the
design alternative to be feasible, $450,000 must have been available for
investment. If it is not invested in a new ride at the theme park, it could be
invested and earn 10 percent annual compound returns. Compare the
future worth of each alternative over a 10-year planning horizon using an
investment portfolio approach, including residual capital in the investment pool.
5-2
Future Worth
Given: The cash flows outlined in Figure 5.3; MARR 5 10%; planning
horizon 5 10 years
Find: FW of the investment portfolio for the do-nothing alternative, Alternative A, and Alternative B (including the investment and residual capital)
KEY DATA
The future worth for the do-nothing alternative (DN) would be
SOLUTION
FWDN 5 $450,0001F Z P 10%,102 5 $450,00012.593742 5 $1,167,183.00
or, using Excel®,
FWDN 5 FV110%,10,,2450,0002 5 $1,167,184.11
The $450,000 invested in Design Alternative B will result in annual revenue
of $80,000. If the recovered funds are added to the investment pool, in
10 years they will be worth
FWB 5 $80,0001F Z A 10%,102 5 $80,000115.937422 5 $1,274,993.60
or, using Excel®,
FWB 5 FV110%,10,2800002 5 $1,274,993.97
If Design Alternative A is chosen for investment, $55,000 will be recovered annually for 10 years. Placing the $55,000 in the investment pool will
result in a future value of
FWA 5 $55,0001F Z A 10%,102 5 $55,000115.937422 5 $876,558.10
or, using Excel®,
FWA 5 FV110%,10,2550002 5 $876,558.35
In addition, since $450,000 was available and only $300,000 was required
for Alternative A, $150,000 of residual capital (RC) remains in the investment pool and earns an annual return of 10 percent. Therefore, the $150,000
will grow at a rate of 10 percent compounded annually to achieve a value of
FWRC 5 $150,0001F Z P 10%,102 5 $150,00012.593742 5 $389,061
or, using Excel®,
FWRC 5 FV110%,10,,21500002 5 $389,061.37
Hence, if Design Alternative A is chosen, the investment portfolio will total
$876,558.10 1 $389,061.00, or $1,265,619.10. Or, using Excel®, the investment portfolio will total $876,558.35 1 $389,061.37, or $1,265,619.72.
In summary, if neither design is selected, the investment portfolio will
have a value, based on results from Excel®, of $1,167,184.11. If Design B
is installed, the investment portfolio will have a value of $1,274,993.97. If
193
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Chapter 5
Annual Worth and Future Worth
Design A is installed, the total value of the investment portfolio will be
$1,265,619.72. Hence, to maximize the investment portfolio, Design B
should be selected, and the value of the overall investment portfolio will be
$1,274,993.97 2 $1,265,619.72, or $9374.25 greater than if Design A is
purchased.
SUMMARY
KEY CONCEPTS
1. Learning Objective: Calculate an annual worth (AW) converting all cash
flows to an equivalent uniform annual series of cash flows over the planning
horizon for a given interest rate. (Section 5.1)
If the AW is positive, then the investment is recommended. When comparing multiple investments, the alternative with the highest AW is preferred.
Recall that in engineering economic analysis, AW is more frequently used
than FW.
Mathematically, the objective of AW analysis can be stated as follows:
n
Maximize AWj 5 c a At j 11 1 MARR2 n2t d 1A Z F MARR%,n2 (5.2)
;j
t50
2. Learning Objective: Calculate a future worth (FW) converting all cash
flows to a single sum equivalent at the end of the planning horizon for a
given interest rate. (Section 5.2)
The FW analysis is typically used as a supplement to the PW analysis, but
offers a unique forward-looking perspective to the investor to assist with
goal setting. If the FW is positive, then the investment is recommended.
When comparing multiple investments, the alternative with the highest
FW is preferred.
Mathematically, the objective of FW analysis can be stated as follows:
n
Maximize FWj 5 a Ajt 11 1 MARR2 n2t
;j
t50
KEY TERMS
Annual Worth (AW), p. 180
Future Worth (FW), p. 185
(5.4)
Summary 195
Problem available in WileyPLUS
GO Tutorial Tutoring Problem available in WileyPLUS
Video Solution Video Solution available in WileyPLUS
FE-LIKE PROBLEMS
1.
Consider a palletizer at a bottling plant that has a first cost of $150,000,
operating and maintenance costs of $17,500 per year, and an estimated net
salvage value of $25,000 at the end of 30 years. Assume an interest rate of
8 percent. What is the annual equivalent cost of the investment if the planning
horizon is 30 years?
a. $29,760
c. $31,980
b. $30,600
d. $35,130
2.
When using annual worth to evaluate the attractiveness of a single alternative, what value is the calculated AW compared to?
a. PW
c. 0.0
b. FW
d. MARR
3.
The annual worth of an alternative is 0. Which of the following is (are)
also true?
(1) PW 5 0
(2) FW 5 0
a. (1) only
c. Both (1) and (2)
b. (1) only
d. Neither (1) nor (2)
4.
The overhead costs in a highly automated factory are expected to increase
at an annual compound rate of 10 percent for the next 7 years. The overhead
cost at the end of the first year is $200,000. What is the annual worth of
the overhead costs for the 7-year period? The time value of money rate is
8 percent/year.
a. $263,250
c. $200,000
b. $231,520
d. $187,020
5.
The operating and maintenance expenses for a mining machine are expected to be $11,000 in the first year and increase by $800 per year during the
15-year life of the machine. What uniform series of payments would cover
these expenses over the life of the machine? Interest is 10 percent/year compounded annually.
a. $11,000
c. $13,423
b. $4,223
d. $15,223
6.
Scott wants to accumulate $2,500 over a period of 7 years so that a cash
payment can be made for roof maintenance on his summer cottage. To have
this amount when it is needed, he will make annual deposits at the end of each
196
Chapter 5
Annual Worth and Future Worth
year into a savings account that earns 8 percent annual interest per year. How
much must each annual deposit be?
a. $244
c. $280
b. $259
d. $357
7.
On the day your daughter is born, you deposit $1,000 in a college savings
account that earns 8 percent compounded annually. On each of her birthdays
thereafter, up to and including her eighteenth birthday, you deposit an additional $1,000. How much money is in the college account the day after her
eighteenth birthday?
a. $37,450
c. $41,450
b. $38,950
d. $46,800
8.
A deposit of $800 is planned for the end of each year into an account paying
8 percent/year compounded annually. The deposits were not made for the
tenth and eleventh years. All other deposits were made as planned. What
amount will be in the account after the deposit at the end of year 25?
a. $55,397
c. $59,537
b. $55,397
d. $53,597
9.
If you invest $3,000 three years from now, how much will be in the account
15 years from now if i 5 8 percent compounded annually?
a. $3,500
c. $9,415
b. $7,555
d. $9,516
10.
Consider a palletizer at a bottling plant that has a first cost of $150,000,
has operating and maintenance costs of $17,500 per year, and an estimated
net salvage value of $25,000 at the end of 30 years. Assume an interest rate of
8 percent/year. What is the future equivalent cost of the investment if the
planning horizon is 30 years?
a. $3,371,000
c. $3,623,000
b. $3,467,000
d. $3,980,000
PROBLEMS
Note to Instructors and Students
Many of the problems in this chapter are similar to problems in previous chapters.
This similarity is intentional. It is designed to illustrate the use of different measures of merit on the same problem.
Section 5.1.1 Annual Worth of a Single Alternative
1.
Carlisle Company has been cited and must invest in equipment to reduce
stack emissions or face EPA fines of $18,500 per year. An emission reduction
filter will cost $75,000 and have an expected life of 5 years. Carlisle’s MARR
is 10 percent/year.
Summary 197
a. What is the annual worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on annual worth?
c. Is the filter economically justified?
2. DuraTech Manufacturing is evaluating a process improvement project. The
estimated receipts and disbursements associated with the project are shown
below. MARR is 6 percent/year.
End of Year
0
1
2
3
4
5
Receipts
Disbursements
$0
$0
$2,000
$4,000
$3,000
$1,600
$5,000
$200
$300
$600
$1,000
$1,500
a. What is the annual worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on annual worth?
c. Should DuraTech implement the proposed process improvement?
3.
Video Solution Eddie’s Precision Machine Shop is insured for $700,000.
The present yearly insurance premium is $1.00 per $100 of coverage. A
sprinkler system with an estimated life of 20 years and no salvage value can
be installed for $20,000. Annual maintenance costs for the sprinkler system
are $400. If the sprinkler system is installed, the system must be included in
the shop’s value for insurance purposes, but the insurance premium will reduce to $0.40 per $100 of coverage. Eddie uses a MARR of 15 percent/year.
a. What is the annual worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on annual worth?
c. Is the sprinkler system economically justified?
4. Fabco, Inc., is considering the purchase of flow valves that will reduce annual
operating costs by $10,000 per year for the next 12 years. Fabco’s MARR is
7 percent/year. Using an annual worth approach, determine the maximum
amount Fabco should be willing to pay for the valves.
5.
Galvanized Products is considering the purchase of a new computer system for their enterprise data management system. The vendor has quoted a
purchase price of $100,000. Galvanized Products is planning to borrow onefourth of the purchase price from a bank at 15 percent compounded annually.
The loan is to be repaid using equal annual payments over a 3-year period.
The computer system is expected to last 5 years and has a salvage value
of $5,000 at that time. Over the 5-year period, Galvanized Products expects
to pay a technician $25,000 per year to maintain the system but will save
$55,000 per year through increased efficiencies. Galvanized Products uses a
MARR of 18 percent/year to evaluate investments.
198
Chapter 5
Annual Worth and Future Worth
a. What is the annual worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on annual worth?
c. Should the new computer system be purchased?
6. Quilts R Us (QRU) is considering an investment in a new patterning attach-
ment with the cash flow profile shown in the table below. QRU’s MARR is
13.5 percent/year.
EOY
Cash Flow
EOY
8
$600
1
2
3
4
2$1,400
$0
$500
$500
$500
9
10
11
12
$700
$800
$900
2$1,000
5
$500
13
2$2,000
6
$0
14
7
$500
15
2$3,000
$1,400
0
Cash Flow
a. What is the annual worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on annual worth?
c. Should QRU invest?
7.
Imagineering, Inc., is considering an investment in CAD-CAM compatible design software with the cash flow profile shown in the table below.
Imagineering’s MARR is 18 percent/year.
EOY
Cash Flow (M$)
EOY
Cash Flow (M$)
0
2$12
4
$5
1
2$1
$5
$2
5
$5
6
7
$2
$5
2
3
a. What is the annual worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on annual worth?
c. Should Imagineering invest?
8. Jupiter is considering investing time and administrative expense on an effort
that promises one large payoff in the future, followed by additional expenses
over a 10-year horizon. The cash flow profile is shown in the table below.
Jupiter’s MARR is 12 percent/year.
EOY
Cash Flow (K$)
EOY
Cash Flow (K$)
0
2$2
6
$200
1
2$10
7
2$10
2$12
2
2$12
8
3
2$14
9
2$14
4
2$16
10
2$100
5
2$18
Summary 199
a. What is the annual worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on annual worth?
c. Should Jupiter invest?
9.
Video Solution Aerotron Electronics is considering the purchase of a
water filtration system to assist in circuit board manufacturing. The system
costs $40,000. It has an expected life of 7 years at which time its salvage
value will be $7,500. Operating and maintenance expenses are estimated to
be $2,000 per year. If the filtration system is not purchased, Aerotron Electronics will have to pay Bay City $12,000 per year for water purification. If
the system is purchased, no water purification from Bay City will be needed.
Aerotron Electronics must borrow half of the purchase price, but they cannot
start repaying the loan for 2 years. The bank has agreed to three equal annual
payments, with the first payment due at the end of year 2. The loan interest rate is 8 percent compounded annually. Aerotron Electronics’ MARR is
10 percent compounded annually.
a. What is the annual worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on annual worth?
c. Should Aerotron Electronics buy the water filtration system?
10. Home Innovation is evaluating a new product design. The estimated receipts
and disbursements associated with the new product are shown below. MARR
is 10 percent/year.
End of Year
0
1
2
3
4
5
Receipts
Disbursements
$0
$600
$600
$700
$700
$700
$1,000
$300
$300
$300
$300
$300
a. What is the annual worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on annual worth?
c. Should Home Innovations pursue this new product?
11.
Mayberry, Inc., is considering a design change that will cost $6,000 and
will result in an annual savings of $1,000 per year for the 6-year life of the
project. A cost of $2,000 will be avoided at the end of the project as a result
of the change. MARR is 8 percent/year.
a. What is the annual worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on annual worth?
c. Should Mayberry implement the design change?
200 Chapter 5
Annual Worth and Future Worth
12. Nancy’s Notions pays a delivery firm to distribute its products in the metro
area. Delivery costs are $30,000 per year. Nancy can buy a used truck for
$10,000 that will be adequate for the next 3 years. Operating and maintenance costs are estimated to be $25,000 per year. At the end of 3 years, the
used truck will have an estimated salvage value of $3,000. Nancy’s MARR is
24 percent/year.
a. What is the annual worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on annual worth?
c. Should Nancy buy the truck?
Section 5.1.2 Annual Worth of Multiple Alternatives
13.
The engineering team at Manuel’s Manufacturing, Inc., is planning to
purchase an enterprise resource planning (ERP) system. The software and
installation from Vendor A costs $380,000 initially and is expected to increase revenue $125,000 per year every year. The software and installation
from Vendor B costs $280,000 and is expected to increase revenue $95,000
per year. Manuel’s uses a 4-year planning horizon and a 10 percent per year
MARR.
a. What is the annual worth of each investment?
b. What is the decision rule for determining the preferred investment based
on annual worth ranking?
c. Which ERP system should Manuel purchase?
14.
GO Tutorial Video Solution Parker County Community College (PCCC)
is trying to determine whether to use no insulation or to use insulation that
is either 1 inch thick or 2 inches thick on its steam pipes. The heat loss from
the pipes without insulation is expected to cost $1.50 per year per foot of
pipe. A 1-inch thick insulated covering will eliminate 89 percent of the loss
and will cost $0.40 per foot. A 2-inch thick insulated covering will eliminate
92 percent of the loss and will cost $0.85 per foot. PCCC Physical Plant
Services estimates that there are 250,000 feet of steam pipe on campus. The
PCCC Accounting Office requires a 10 percent/year return to justify capital
expenditures. The insulation has a life expectancy of 10 years. Determine
which insulation (if any) should be purchased using annual worth analysis.
15.
Final Finishing is considering three mutually exclusive alternatives for a
new polisher. Each alternative has an expected life of 10 years and no salvage
value. Polisher 1 requires an initial investment of $20,000 and provides annual
benefits of $4,465. Polisher 2 requires an initial investment of $10,000 and
provides annual benefits of $1,770. Polisher 3 requires an initial investment of
$15,000 and provides annual benefits of $3,580. MARR is 15 percent/year.
a. What is the annual worth of each polisher?
b. Which polisher should be recommended?
16.
GO Tutorial Quantum Logistics, Inc., a wholesale distributor, is considering the construction of a new warehouse to serve the southeastern geographic
Summary 201
region near the Alabama–Georgia border. There are three cities being considered. After site visits and a budget analysis, the expected income and costs
associated with locating in each of the cities have been determined. The life
of the warehouse is expected to be 12 years, and MARR is 15 percent/year.
City
Initial Cost
Net Annual Income
Lagrange
Auburn
Anniston
$1,260,000
$1,000,000
$1,620,000
$480,000
$410,000
$520,000
a. What is the annual worth of each site?
b. What is the decision rule for determining the preferred site based on
annual worth ranking?
c. Which city should be recommended?
17.
Nadine Chelesvig has patented her invention. She is offering a potential
manufacturer two contracts for the exclusive right to manufacture and market
her product. Plan A calls for an immediate single lump sum payment to her
of $30,000. Plan B calls for an annual payment of $1, 000 plus a royalty of
$0.50 per unit sold. The remaining life of the patent is 10 years. Nadine uses
a MARR of 10 percent/year. What must be the uniform annual sales volume
of the product for Nadine to be indifferent between the contracts, based on an
annual worth analysis?
18.
GO Tutorial DelRay Foods must purchase a new gumdrop machine. Two
machines are available. Machine 7745 has a first cost of $10,000, an estimated
life of 10 years, a salvage value of $1,000, and annual operating costs estimated
at $0.01 per 1,000 gumdrops. Machine A37Y has a first cost of $8,000, a
life of 10 years, and no salvage value. Its annual operating costs will be $300
regardless of the number of gumdrops produced. MARR is 6 percent/year, and
30 million gumdrops are produced each year.
a. What is the annual worth of each machine?
b. What is the decision rule for determining the preferred machine based on
annual worth ranking?
c. Which machine should be recommended?
19.
Final Finishing is considering three mutually exclusive alternatives for a
new polisher. Each alternative has an expected life of 10 years and no salvage
value. Polisher 1 requires an initial investment of $20,000 and provides annual
benefits of $4,465. Polisher 2 requires an initial investment of $10,000 and
provides annual benefits of $1,770. Polisher 3 requires an initial investment of
$15,000 and provides annual benefits of $3,580. MARR is 15 percent/year.
a. What is the annual worth of each polisher?
b. Which polisher should be recommended?
20. Tempura, Inc., is considering two projects. Project A requires an investment
of $50,000. Estimated annual receipts for 20 years are $20,000; estimated
annual costs are $12,500. An alternative project, B, requires an investment
202
Chapter 5
Annual Worth and Future Worth
of $75,000, has annual receipts for 20 years of $28,000, and has annual costs
of $18,000. Assume both projects have zero salvage value and that MARR is
12 percent/year.
a. What is the annual worth of each project?
b. Which project should be recommended?
21.
Alpha Electronics can purchase a needed service for $90 per unit. The
same service can be provided by equipment that costs $100,000 and that will
have a salvage value of 0 at the end of 10 years. Annual operating costs for
the equipment will be $7,000 per year plus $25 per unit produced. MARR is
12 percent/year.
a. Based on an annual worth analysis, should the equipment be purchased if
the expected production is 200 units/year?
b. Based on an annual worth analysis, should the equipment be purchased if
the expected demand is 500 units/year?
c. Determine the breakeven value for annual production that will return
MARR on the investment in the new equipment.
22. Packaging equipment for Xi Cling Wrap costs $60,000 and is expected to
result in end of year net savings of $23,000 per year for 3 years. The equipment will have a market value of $10,000 after 3 years. The equipment can be
leased for $21,000 per year, payable at the beginning of each year. Xi Cling’s
MARR is 10 percent/year. Based on an annual worth analysis, determine if
the packaging equipment should be purchased or leased.
23.
Orpheum Productions in Nevada is considering three mutually exclusive
alternatives for lighting enhancements to one of its recording studios. Each
enhancement will increase revenues by attracting directors who prefer this
lighting style. The following table shows the cash flow details, in thousands
of dollars, for these enhancements. MARR is 10 percent/year. Based on an
annual worth analysis, which alternative (if any) should be implemented?
End of Year
Light Bar
Sliding Spots
Reflected Beam
0
2$6,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
2$14,000
$3,500
$3,500
$3,500
$3,500
$3,500
$3,500
2$20,000
$0
$2,300
$4,600
$6,900
$9,200
$11,500
1
2
3
4
5
6
24. RealTurf is considering purchasing an automatic sprinkler system for its sod
farm by borrowing the entire $30,000 purchase price. The loan would be repaid
with four equal annual payments at an interest rate of 12 percent/year/year. It
is anticipated that the sprinkler system would be used for 9 years and then sold
for a salvage value of $2,000. Annual operating and maintenance expenses for
the system over the 9-year life are estimated to be $9,000 per year. If the new
system is purchased, cost savings of $15,000 per year will be realized over the
present manual watering system. RealTurf uses a MARR of 15 percent/year for
Summary 203
economic decision making. Based on an annual worth analysis, is the purchase
of the new sprinkler system economically attractive?
Section 5.2.1
25.
Future Worth of a Single Alternative
An investment has the following cash flow profile. MARR is 12 percent/
year. What is the minimum value of X such that the investment is attractive
based on a future worth measure of merit?
End of Year
Cash Flow
0
2$30,000
$6,000
$13,500
$X
$13,500
1
2
3
4
26. A 22-year-old engineering graduate wants to accumulate $2,000,000 to be avail-
able when she retires 40 years from today. She investigates several investment
options and decides to invest in a stock market index fund after discovering that
the long-term average return for the stock market is 10.4 percent per year. Since
this will be a tax-sheltered account, she plans to ignore the impact of taxes.
a. If she plans to make 40 uniform annual deposits starting 1 year from today,
what is the dollar amount of the required deposits?
b. If she makes the first of the 40 deposits starting today rather than 1 year
from today, what is the dollar amount of the required deposits?
c. If she plans to make the first payment 1 year from today and each annual
payment will be $200 greater than the previous year’s payment,
(i) what is the dollar amount of the first deposit?
(ii) what is the dollar amount of the last deposit?
d. If she plans to make the first payment 1 year from today and each annual
payment will be 5 percent greater than the previous year’s payment,
(i) what is the dollar amount of the first deposit?
(ii) what is the dollar amount of the last deposit?
27. Reconsider the situation described in Problem 26. Assume that rather than an-
nual deposits, she makes monthly deposits. The first deposit will be 1 month
from today, and the last deposit will be 40 years from today. Assume that the
stock market return is 10.4 percent per year compounded monthly.
a. If she plans to make uniform monthly deposits, what is the dollar amount
of the monthly deposit?
b. If she earns a 4 percent annual raise each year throughout her career (starting
1 year from today) and adjusts her monthly deposits by the same 4 percent
each year, how much will be in the account immediately after the last deposit?
28. You decide to set up a college fund for your 10-year-old child and plan to
make annual deposits into the account each year on your child’s birthday.
Because “other things” consistently use more of your money than anticipated,
your deposits are actually somewhat erratic. One year even resulted in a withdrawal. The account earns 5 percent per year.
204
Chapter 5
Annual Worth and Future Worth
Birthday
10
11
12
13
14
Deposit
$1,000
$1,000
$2,000
$1,500
Birthday
Deposit
15
16
17
18
$3,000
$2,500
$2,000
$2,000
2$1,500
a. How much is in the account immediately after the deposit on your child’s
eighteenth birthday?
b. How much would you have needed to deposit on each birthday to accu-
mulate the same total if you had started on your child’s tenth birthday and
made equal annual deposits with no withdrawals?
c. How much would you have needed to deposit on each birthday to accumulate the same total if you had started on your child’s first birthday and
made equal annual deposits with no withdrawals?
29.
You decide to open an individual retirement account (IRA) at your local
bank that pays 8 percent/year/year. At the end of each of the next 40 years,
you will deposit $2,000 per year into the account (40 total deposits). Three
years after the last deposit, you will begin making annual withdrawals. If you
want the account to last 30 years (30 withdrawals), what amount will you be
able to withdraw each year?
30. You decide to open an individual retirement account (IRA) at your local
stockbroker that pays 10 percent/year/year for the life of the account. You
deposit $2,000 today to open the account. For the next 41 years, you will
deposit $2,000 per year into the account at the end of each year. There are a
total of forty-two $2,000 deposits. Exactly 1 year after the last deposit, you
will start making withdrawals.
a. What is the balance in the account immediately after the last deposit?
b. What annual withdrawal can you make if you want the withdrawals to last
15 years?
c. What annual withdrawal can you make if you want the withdrawals to last
20 years?
d. What annual withdrawal can you make if you want the withdrawals to last
25 years?
31.
On your child’s first birthday, you open an account to fund his college education. You deposit $300 to open the account. Each year, on his birthday, you make
another deposit. Each subsequent deposit is 8 percent larger than the previous.
The account pays interest at 5 percent/year compounded annually. How much
money is in the account immediately after the deposit on his eighteenth birthday?
32.
Video Solution Financial planners (and engineering economists) unanimously encourage people to start early in planning for retirement. To illustrate
this point, they frequently produce a table similar to the one below. Fill in the
blank cells in this table assuming that your goal is to have $1,000,000 on your
sixty-fifth birthday and that deposits start on the birthday shown and continue
annually in the same amount on each birthday up to and including your sixtyfifth birthday. Assume that interest is compounded annually at 10 percent/year.
Summary 205
Birthday of
First Deposit
Amount of Required
Annual Deposit
25
30
35
40
45
50
55
60
65
33.
$1,000,000
Video Solution Financial planners (and engineering economists) unanimously encourage people to seek out the highest rate of return possible within
their personal level of risk tolerance. To illustrate this point, they frequently
produce a table similar to the one below. Fill in the blank cells in this table
assuming that your goal is to have $1,000,000 on your sixty-fifth birthday and
that deposits start on your twenty-sixth birthday and continue annually in the
same amount on each birthday up to and including your sixty-fifth birthday.
Interest Rate Earned
Amount of Required
Annual Deposit
4 percent/year
5 percent/year
6 percent/year
7 percent/year
8 percent/year
9 percent/year
34.
Video Solution Bailey, Inc., is considering buying a new gang punch that
would allow them to produce circuit boards more efficiently. The punch has a
first cost of $100,000 and a useful life of 15 years. At the end of its useful life,
the punch has no salvage value. Annual labor costs would increase $2,000
using the gang punch, but annual raw material costs would decrease $12,000.
MARR is 5 percent/year.
a. What is the future worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on future worth?
c. Should Bailey buy the gang punch?
35.
Carlisle Company has been cited and must invest in equipment to reduce
stack emissions or face EPA fines of $18,500 per year. An emission reduction
filter will cost $75,000 and have an expected life of 5 years. Carlisle’s MARR
is 10 percent/year.
a. What is the future worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on future worth?
c. Is the filter economically justified?
206
Chapter 5
Annual Worth and Future Worth
36. DuraTech Manufacturing is evaluating a process improvement project. The
estimated receipts and disbursements associated with the project are shown
below. MARR is 6 percent/year.
End of Year
0
1
2
3
4
5
Receipts
Disbursements
$0
$0
$2,000
$4,000
$3,000
$1,600
$5,000
$200
$300
$600
$1,000
$1,500
a. What is the future worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on future worth?
c. Should DuraTech implement the proposed process improvement?
37.
Galvanized Products is considering the purchase of a new computer system for their enterprise data management system. The vendor has quoted a
purchase price of $100,000. Galvanized Products is planning to borrow onefourth of the purchase price from a bank at 15 percent compounded annually.
The loan is to be repaid using equal annual payments over a 3-year period.
The computer system is expected to last 5 years and has a salvage value
of $5,000 at that time. Over the 5-year period, Galvanized Products expects
to pay a technician $25,000 per year to maintain the system but will save
$55,000 per year through increased efficiencies. Galvanized Products uses a
MARR of 18 percent/year to evaluate investments.
a. What is the future worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on future worth?
c. Should the new computer system be purchased?
38. Quilts RUs (QRU) is considering an investment in a new patterning attach-
ment with the cash flow profile shown in the table below. QRU’s MARR is
13.5 percent/year.
EOY
0
1
2
3
4
5
6
7
Cash Flow
2$1,400
$0
$500
$500
$500
$500
$0
$500
EOY
Cash Flow
8
$600
9
10
11
12
13
14
15
$700
$800
$900
−$1,000
−$2,000
−$3,000
$1,400
a. What is the future worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on future worth?
c. Should QRU invest?
Summary 207
39.
Video Solution Imagineering, Inc., is considering an investment in CADCAM compatible design software with the cash flow profile shown in the
table below. Imagineering’s MARR is 18 percent/year.
EOY
Cash Flow (M$)
EOY
Cash Flow (M$)
0
2$12
4
$5
1
2$1
$5
$2
5
$5
6
7
$2
$5
2
3
a. What is the future worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on future worth?
c. Should Imagineering invest?
40. Jupiter is considering investing time and administrative expense on an effort
that promises one large payoff in the future, followed by additional expenses
over a 10-year horizon. The cash flow profile is shown in the table below.
Jupiter’s MARR is 12 percent/year.
EOY
Cash Flow (K$)
EOY
Cash Flow (K$)
0
2$2
6
$200
1
2$10
7
2$10
2$12
2
2$12
8
3
2$14
9
2$14
4
2$16
10
2$100
5
2$18
a. What is the future worth of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on future worth?
c. Should Jupiter invest?
Section 5.2.2
41.
Future Worth of a Multiple Alternatives
The following three investment opportunities are available. The returns
for each investment for each year vary, but the first cost of each is $20,000.
Based on a future worth analysis, which investment is preferred? MARR is
9 percent/year.
End of Year
1
2
3
4
Investment 1
Investment 2
Investment 3
$8,000
$9,000
$10,000
$11,000
$11,000
$10,000
$9,000
$8,000
$9,500
$9,500
$9,500
$9,500
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Chapter 5
Annual Worth and Future Worth
42. The engineering team at Manuel’s Manufacturing, Inc., is planning to pur-
chase an enterprise resource planning (ERP) system. The software and installation from Vendor A costs $380,000 initially and is expected to increase
revenue $125,000 per year every year. The software and installation from Vendor B costs $280,000 and is expected to increase revenue $95,000 per year.
Manuel’s uses a 4-year planning horizon and a 10 percent per year MARR.
a. What is the future worth of each investment?
b. What is the decision rule for determining the preferred investment based
on future worth ranking?
c. Which ERP system should Manuel purchase?
43.
Parker County Community College (PCCC) is trying to determine
whether to use no insulation or to use insulation that is either 1 inch thick
or 2 inches thick on its steam pipes. The heat loss from the pipes without
insulation is expected to cost $1.50 per year per foot of pipe. A 1-inch thick
insulated covering will eliminate 89 percent of the loss and will cost $0.40
per foot. A 2-inch thick insulated covering will eliminate 92 percent of the
loss and will cost $0.85 per foot. PCCC Physical Plant Services estimates
that there are 250,000 feet of steam pipe on campus. The PCCC Accounting Office requires a 10 percent/year return to justify capital expenditures.
The insulation has a life expectancy of 10 years. Determine which insulation
(if any) should be purchased using a ranking future worth analysis.
44. Nadine Chelesvig has patented her invention. She is offering a potential man-
ufacturer two contracts for the exclusive right to manufacture and market
her product. Plan A calls for an immediate single lump sum payment to her
of $30,000. Plan B calls for an annual payment of $1,000 plus a royalty of
$0.50 per unit sold. The remaining life of the patent is 10 years. Nadine uses
a MARR of 10 percent/year. What must be the uniform annual sales volume
of the product for Nadine to be indifferent between the contracts based on a
future worth analysis?
45.
Quantum Logistics, Inc., a wholesale distributor, is considering the construction of a new warehouse to serve the southeastern geographic region
near the Alabama–Georgia border. There are three cities being considered.
After site visits and a budget analysis, the expected income and costs associated with locating in each of the cities have been determined. The life of the
warehouse is expected to be 12 years and MARR is 15 percent/year.
City
Initial Cost
Net Annual Income
Lagrange
Auburn
Anniston
$1,260,000
$1,000,000
$1,620,000
$480,000
$410,000
$520,000
a. What is the future worth of each site?
b. What is the decision rule for determining the preferred site based on future
worth ranking?
c. Which city should be recommended?
Summary 209
46.
Video Solution Final Finishing is considering three mutually exclusive
alternatives for a new polisher. Each alternative has an expected life of 10 years
and no salvage value. Polisher 1 requires an initial investment of $20,000
and provides annual benefits of $4,465. Polisher 2 requires an initial investment of $10,000 and provides annual benefits of $1,770. Polisher 3 requires
an initial investment of $15,000 and provides annual benefits of $3,580.
MARR is 15 percent/year.
a. What is the future worth of each polisher?
b. Which polisher should be recommended?
47.
Xanadu Mining is considering three mutually exclusive alternatives, as
shown in the table below. MARR is 10 percent/year.
EOY
0
1
2
3
4
A001
B002
C003
2$210
$80
$90
$100
$110
2$110
$60
$60
$60
$70
2$160
$80
$80
$80
$80
a. What is the future worth of each alternative?
b. Which alternative should be recommended?
48. Yani has $12,000 for investment purposes. His bank has offered the
following three choices.
a. A special savings certificate that will pay $100 each month for 5 years and
a lump sum payment at the end of 5 years of $13,000
b. Buy a share of a racehorse for $12,000 that will be worth $20,000 in
5 years
c. Put the money in a savings account that will have an interest rate of
12 percent per year compounded monthly.
Use a future worth analysis to make a recommendation to Yani.
49.
Two numerically controlled drill presses are being considered by the production department of Zunni’s Manufacturing; one must be selected. Comparison data is shown in the table below. MARR is 10 percent/year.
Initial Investment
Estimated Life
Estimated Salvage Value
Annual Operating Cost
Annual Maintenance Cost
Drill Press T
Drill Press M
$20,000
10 years
$5,000
$12,000
$2,000
$30,000
10 years
$7,000
$6,000
$4,000
a. What is the future worth of each drill press?
b. Which drill press should be recommended?
210
Chapter 5
Annual Worth and Future Worth
50.
Video Solution Alpha Electronics can purchase a needed service for $90
per unit. The same service can be provided by equipment that costs $100,000
and that will have a salvage value of zero at the end of 10 years. Annual
operating costs for the equipment will be $7,000 per year plus $25 per unit
produced. MARR is 12 percent/year.
a. Based on a future worth analysis, should the equipment be purchased if
the expected production is 200 units/year?
b. Based on a future worth analysis, should the equipment be purchased if
the expected production is 500 units/year?
c. Determine the breakeven value for annual production that will return
MARR on the investment in the new equipment.
51.
The management of Brawn Engineering is considering three alternatives
to satisfy an OSHA requirement for safety gates in the machine shop. Each
gate will completely satisfy the requirement, so no combinations need to be
considered. The first costs, operating costs, and salvage values over a 5-year
planning horizon are shown below. Using a future worth analysis with a
MARR of 20 percent/year, determine the preferred gate.
End of Year
Gate 1
Gate 2
Gate 3
0
2$15,000
2$19,000
2$24,000
1
2$6,500
2$5,600
2$4,000
2
2$6,500
2$5,600
2$4,000
3
2$6,500
2$5,600
2$4,000
4
2$6,500
2$5,600
2$4,000
5
2$6,500 1 $0
2$5,600 1 $2,000
2$4,000 1 $5,000
52. RealTurf is considering purchasing an automatic sprinkler system for its
sod farm by borrowing the entire $30,000 purchase price. The loan would
be repaid with four equal annual payments at an interest rate of 12 percent/
year. It is anticipated that the sprinkler system would be used for 9 years
and then sold for a salvage value of $2,000. Annual operating and maintenance expenses for the system over the 9-year life are estimated to be
$9,000 per year. If the new system is purchased, cost savings of $15,000
per year will be realized over the present manual watering system. RealTurf
uses a MARR of 15 percent/year for economic decision making. Based on a
future worth analysis, is the purchase of the new sprinkler system economically attractive?
53.
Orpheum Productions in Nevada is considering three mutually exclusive alternatives for lighting enhancements to one of its recording studios.
Each enhancement will increase revenues by attracting directors who prefer this lighting style. The cash flow details, in thousands of dollars, for
these enhancements are shown in the chart below. MARR is 10 percent/
year. Based on a future worth analysis, which alternative (if any) should be
implemented?
Summary 211
End of Year
Light Bar
Sliding Spots
Reflected Beam
0
2$6,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
2$14,000
$3,500
$3,500
$3,500
$3,500
$3,500
$3,500
2$20,000
$0
$2,300
$4,600
$6,900
$9,200
$11,500
1
2
3
4
5
6
54. Deep Seas Submarine must implement a new engine in its submarines to
meet the needs of clients who desire quieter operation. Two designs, both
technologically feasible, have been created, and Deep Seas wishes to know
which one to pursue. Design 1 would require an up-front manufacturing cost
of $15,000,000 and will cost $2,500,000 per year for 3 years to swap out the
engines in all its current submarines. Design 2 will cost $20,000,000 up front,
but due to a higher degree of compatibility will only require $1,500,000 per
year to implement. MARR is 10 percent/year. Based on a future worth analysis,
determine which design should be chosen.
Section 5.2.3
55.
Portfolio Analysis
An investor has $100,000 to invest in a business venture, or she can
earn 10 percent/year with a $100,000 certificate of deposit for 4 years.
Three possible business ventures have been identified. Any money not invested in the business venture can be put into a bank account that earns
7 percent/year. Based on a future worth analysis, what should be done with
the $100,000?
End of Year
0
1
2
3
4
BV01
BV02
BV03
2$35,000
$0
$0
$0
$50,000
2$80,000
$10,000
$10,000
$10,000
$90,000
2$60,000
$0
$40,000
$0
$40,000
56. Reconsider data from Problem 53 (Orpheum Productions lighting enhance-
ment). Assume that any money not invested in the lighting enhancements will
be placed in an interest-bearing account earning MARR and will be used for
future studio modernization projects.
a. Use the total portfolio approach to examine the future worth of each
alternative.
b. Compare the future worth results from Problem 53 with your FW results
for part (a). Explain your conclusions from this comparison.
6
ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E :
MOTORO L A S O L U T I O N S
Motorola was founded in Illinois in 1928. From car radios, to televisions, to
two-way radios, to automatic identification technologies, Motorola evolved
into a communications company that today is designing, building, marketing, and selling products, services, and applications globally. On January 4,
2011, Motorola’s mobile devices and home businesses were spun off as a
separate company, Motorola Mobility Holdings, Inc., and subsequently were
sold to Google in 2012. Following the separation of Motorola Mobility, the
original company was renamed Motorola Solutions.
In its 2011 Annual Report, Motorola Solutions Chairman & CEO Greg
Brown stated, “Our strategy is focused on providing mission-critical communication solutions to government and enterprise customers.” The key words
are mission-critical communication solutions. First responders depend on
Motorola Solutions products in “moments that matter”—from military
maneuvers, to emergency response, to rescue missions.
If you see someone at an airport or police personnel using a two-way
radio, it probably is a Motorola Solutions product. If you receive a delivery
from FedEx, information regarding the delivery very likely was communicated using Motorola Solutions products. In addition, there are numerous
behind-the-scenes applications of Motorola’s products, including barcode
readers in distribution centers.
Motorola Solutions, Inc. had annual revenues in 2011 of $8.2 billion. As of
December 31, 2011, it and its subsidiaries had approximately 23,000 employees
in 65 countries. During 2011, it had more than 100,000 customers in 100 countries. Approximately 7,000 engineers and scientists were involved in performing
research and development; R&D expenditures totaled about $1 billion.
A technology company, Motorola Solutions is highly disciplined in managing its capital. Discounted cash flow methods are used throughout the
company to ensure that shareholders receive attractive returns on their
invested capital. The required return on a particular capital investment is
212
RATE OF RETURN
based on weighted average cost of capital calculations, as well as considerations of the type of investment, its duration, and risks involved in the particular investment. The required return on an investment can change from
year to year due to economy dynamics, but the need to perform present
worth and internal rate of return calculations never stops.
DISCUSSION QUESTIONS:
1. What factors would determine the rate of return values that Motorola
Solutions would use for their investment decisions?
2. Motorola Solutions is a high-tech company with significant new product
development. What unique considerations would such a company
have with respect to expected returns on their investments as opposed
to a “lower-tech” company?
3. Motorola Solutions is a global company with operations around the
world. What complexity does this introduce into their capital investment decisions?
4. What financial gains might the parent company Motorola have sought
as it spun off Motorola Mobility (sold to Google in 2012)?
LEARNING OBJECTIVES
When you have finished studying this chapter, you should be able to:
1. Compute the Internal Rate of Return (IRR) for an individual investment
and perform incremental comparisons of mutually exclusive investment alternatives to determine the one that maximizes economic
worth. (Section 6.1)
213
214
Chapter 6
Rate of Return
2. Apply Descartes’ rule of signs and Norstrom’s criterion to test for multiple roots when using the IRR method. (Section 6.1.2)
3. Compute the External Rate of Return (ERR) for an individual investment and perform incremental comparisons of mutually exclusive
investment alternatives to determine the one that maximizes economic worth. (Section 6.2)
INTRODUCTION
Although this chapter does not have worth in its title, rates of return are
also measures of economic worth. Instead of measuring economic worth
in dollars, here we measure it in percentages.
Among the DCF measures of economic worth, rates of return are
probably the second most popular among corporations, ranked just behind present worth. For personal investment decision making, however,
rates of return are used more frequently than present worth. No doubt
the popularity of this analysis method is due to investors’ familiarity with
interest rates and the ease with which investment returns can be compared with costs of capital for investment.
Like the other DCF methods we have studied so far, rates of return
can be used to compare mutually exclusive alternatives and choose the
one having the greatest economic worth. However, they tend to be used
more frequently in industry as supplements to one of the traditional
‘‘worth’’ methods—present, future, or annual worth.
Systematic Economic Analysis Technique
1.
2.
3.
4.
5.
6.
7.
Identify the investment alternatives
Define the planning horizon
Specify the discount rate
Estimate the cash flows
Compare the alternatives
Perform supplementary analyses
Select the preferred investment
Unlike present, future, and annual worths, rates of return are not ranking methods. When used to choose from among mutually exclusive alternatives on the basis of monetary considerations, incremental analysis is
required when using rates of return. When performed correctly, however,
6-1 Internal Rate of Return Calculations
215
the incremental analysis will result in the same investment alternative
being recommended as when using one of the ranking methods.
Although many different rates of return exist, here we consider only
two, both of which are discounted cash flow methods: internal rate of return
and external rate of return.
In this chapter, we learn how to compare mutually exclusive investment alternatives using rate of return methods. In the case of the internal
rate of return method, we learn that multiple solutions can occur, and we
figure out how to deal with them.
6-1
INTERNAL RATE OF RETURN
CALCULATIONS
LEARNING O BJECTI VE: Compute the Internal Rate of Return (IRR) for an
individual investment and perform incremental comparisons of mutually
exclusive investment alternatives to determine the one that maximizes
economic worth.
Video Lesson:
Rate of Return
The internal rate of return is also referred to as the discounted cash flow
rate of return, the cash flow rate of return, the rate of return (ROR), the
return on investment (ROI), and the true rate of return. The more common
name, however, is internal rate of return (IRR). Mathematically, investment j’s internal rate of return, denoted i*j , satisfies the following equality:
Internal Rate of Return
(IRR) The interest rate
that makes the present
worth, the future worth,
and the annual worth equal
to 0. Also referred to as the
discounted cash flow rate
of return, the cash flow
rate of return, the rate of
return (ROR), the return on
investment (ROI), and the
true rate of return.
n
0 5 a Ajt 11 1 i*j 2 n2t
(6.1)
t50
In words, the internal rate of return is the interest rate that makes the
future worth of an investment equal 0. Likewise, it is the interest rate that
equates the present worth and annual worth to 0. If the internal rate of
return is at least equal to the MARR, the investment should be made.
6.1.1 Single Alternative
As with other DCF methods, when applied to a single alternative, IRR is
used to determine whether the investment opportunity is preferred to the
do-nothing alternative.
The Surface-Mount Placement Machine Investment
Recall the manager of an electronics manufacturing plant who was asked
to approve the purchase of a surface mount placement (SMP) machine
having an initial cost of $500,000 in order to reduce annual operating and
maintenance costs by $92,500 per year. At the end of the 10-year planning
EXAMPLE
216
Chapter 6
Rate of Return
horizon, it was estimated that the SMP machine would be worth $50,000.
Using a 10 percent MARR and internal rate of return analysis, should the
investment be made?
Given: The cash flows shown in Figure 6.1; MARR 5 10%; planning
horizon 5 10 years
Find: The IRR of the investment. Is this investment recommended?
KEY DATA
$50,000
(+)
$92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500 $92,500
0
1
2
3
4
5
6
7
8
9
10
(–)
$500,000
FIGURE 6.1
SOLUTION
CFD for Example 6.1
Setting the future worth for the investment equal to 0 gives
0 5 2$500,0001F Z P i*%,102 1 $92,5001F Z A i*%,102 1 $50,000
Recalling the formulas used to compute capital recovery cost, the following annual worth formulation can be solved for the internal rate of return:
0 5 1$500,000 2 $50,0002 1A Z P i*,102 1 $50,000i* 2 $92,500
For i 5 12%,
$450,0001A Z P 12%,102 1 $50,00010.122 2 $92,500 5 2$6,850
For i 5 15%,
$450,0001A Z P 15%,102 1 $50,00010.152 2 $92,500 5 $4,685
Interpolating for i* gives 13.78%.
6-1 Internal Rate of Return Calculations
Alternately, one can use the Excel® RATE worksheet function to solve
for i* in this example:
i* 5 RATE110,92500,2500000,500002
5 13.8%
Because i* . 10 percent, the investment is recommended.
Example 6.1 illustrates one of the attractions of the IRR method. We can
say “Invest $500,000 to obtain a 13.8 percent return on your investment.”
This is often more informative or appealing than a dollar figure such as
present or future worth.
6.1.2 Multiple Roots
LEARNING O BJECTI VE: Apply Descartes’ rule of signs and Norstrom’s
criterion to test for multiple roots when using the IRR method.
Descartes’ rule of signs, as applied to internal rate of return analysis, indicates there will be at most as many positive rates of return as there are sign
changes in the cash flow profile. For example, one change of sign, usually
one or more negative cash flows followed by one or more positive cash
flows will have at most one positive IRR value. Three sign changes, for
example, will have at most three positive IRR values. Most cash flow profiles encountered in practice, however, will have a unique internal rate of
return, despite multiple changes in sign. Example 6.2 illustrates a cash
flow profile with multiple internal rates of return.
Multiple Roots
To illustrate a cash flow profile having multiple roots, consider the data
given in Table 6.1. The future worth of the cash flow series will be 0 using
a 20, 40, or 50 percent interest rate.
FW1 120%2 5 2$4,00011.22 3 1 $16,40011.22 2 2 $22,32011.22 1
$10,080 5 0
3
2
FW2 140%2 5 2$4,00011.42 1 $16,40011.42 2 $22,32011.42 1
$10,080 5 0
3
2
FW3 150%2 5 2$4,00011.52 1 $16,40011.52 2 $22,32011.52 1
$10,080 5 0
EXAMPLE
217
Chapter 6
Rate of Return
TABLE 6.1
Cash Flow Profile
EOY
CF
0
−$4,000
1
$16,400
2
−$22,320
3
$10,080
A plot of the future worth for this example is given in Figure 6.2. It is
simply a plot of the future worth evaluated at values of i from 10% to 60%
in increments of 2%.
$60
$40
Future Worth
218
$20
$0
10%
20%
30%
40%
50%
60%
–$20
–$40
MARR
FIGURE 6 . 2
Plot of Future Worth for Example 6.2
In addition to Descartes’ rule of signs, Norstrom’s criterion can be applied
to determine if there is at most one real positive internal rate of return. If
the cumulative cash flow series begins with a negative value and changes
only once to a positive value, then there exists at most a single positive
internal rate of return. Example 6.3 illustrates a cash flow profile having
multiple sign changes, yet only a single positive IRR value. This example
also illustrates Norstrom’s criterion that if the cumulative cash flow series
begins with a negative number and changes only once to a positive-valued
series, then there exists a unique positive rate of return. In the example,
interestingly, if the additional investment is $200,000 instead of $150,000,
Norstrom’s criterion is not met, but a single positive internal rate of return
still exists. Norstrom’s criterion is a sufficient, not a necessary, condition
for at most a single real positive rate of return to exist.
6-1 Internal Rate of Return Calculations
Rate of Return with Multiple Sign Changes and a Single Root
EXAMPLE
Julian Stewart invested $250,000 in a limited partnership to drill for natural gas. His investment yielded annual returns of $45,000 the first year,
followed by annual increases of $10,000 until the sixth year, at which time
an additional $150,000 had to be invested for deeper drilling. Starting in
the seventh year, following the supplemental investment, the annual returns
decrease by $10,000 annually from $85,000 to $5,000. What is the IRR of
Julian’s investment when future worth is maximized?
Given: The cash flows for Julian’s investment are shown in column B of
the spreadsheet in Figure 6.3.
Find: IRR when FW is at a maximum.
KEY DATA
From the plot of future worth shown in Figure 6.3, it is evident that a
single root exists, at i* 5 19.12%, and using the Excel® SOLVER tool,
FW is maximized when MARR equals 8.5469%. For the investment
IRR 5 19.12%.
SOLUTION
FIGURE 6.3
IRR for a Natural Gas Investment
219
220
Chapter 6
Rate of Return
6.1.3 Multiple Alternatives
In computing the internal rate of return for each of several investments, we
continue to search for the interest rate that equates the economic worth
to 0. Mathematically, from Equation 6.1, the internal rate of return for
alternative j, denoted ij*, satisfies the following equality:
n
0 5 a Ajt 11 1 i*j 2 n2t
t50
Rate of return methods must be used incrementally when comparing
mutually exclusive investment alternatives. That is, let the investment with
the lowest initial cost be Alternative 1 (or the base alternative), let the
investment with the next lowest cost be Alternative 2, and so on. Then,
analyze each alternative in order of increasing initial cost. For alternatives
2 and beyond, rate of return is determined for the additional increment of
investment (above the base alternative), rather than the entire cost. As long
as each successive alternative’s rate of return exceeds the MARR, it is
preferred to the previous (lower cost) investment. There is value in investing the additional incremental capital for a subsequent alternative when its
rate of return is equal to or exceeds the MARR.
The preferred alternative will not necessarily have the greatest internal
rate of return. Whereas the present, future, and annual worths are compared to 0, the comparison with rate of return analyses is with the MARR.
The following example illustrates why incremental analysis is used when
performing rate of return comparisons of investment alternatives.
IRR Analysis with Mutually Exclusive Alternatives
EXAMPLE
Video Example
KEY DATA
Recall the example of the theme park in Florida that is considering two
designs for a new ride called the Scream Machine. The first design alternative (A) requires a $300,000 investment and will produce net annual aftertax revenue of $55,000 over the 10-year planning horizon; the second
alternative (B) requires a $450,000 investment and will produce net annual
after-tax revenue of $80,000 annually. Both alternatives are expected to
have negligible salvage values after the 10-year planning horizon. Based
on a 10 percent MARR and an IRR comparison, which design (if either)
should be chosen?
Given: The cash flows outlined in Figure 6.4; MARR 5 10%; planning
horizon 5 10 years
Find: Recommended investment using incremental IRR analysis.
6-1
Internal Rate of Return Calculations
$55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000
(+)
0
1
2
3
4
5
6
7
8
9
10
(–)
Alternative A
$300,000
$80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000
(+)
0
1
2
3
4
5
6
7
8
9
10
(–)
Alternative B
$450,000
FIGURE 6.4
CFDs for Example 6.4
First, we examine each alternative looking for the smallest initial investment, in this case, Design A. Next, we compute the IRR for Design A.
Recall that the computed IRR will be compared with the MARR.
iA* 5 RATE110,255000,3000002
5 12.87% . MARR 5 10%
Because Design A is justified (because its return exceeds the MARR), the
next step is to compute the internal rate of return on the $150,000 incremental investment in Design B (over and above in initial investment in
Design A):
*
5 RATE110,225000,1500002
iB2A
5 10.56% . MARR 5 10%
SOLUTION
221
222
Chapter 6
Rate of Return
Because the internal rate of return for the incremental investment in Design
B is greater than the MARR (albeit only slightly), this incremental investment needed to acquire Design B is justified. Therefore, the overall internal rate of return for Design B can be computed as follows:
iB* 5 RATE110,280000,4500002
5 12.11%
In summary, Design B is the preferred alternative. The return on the first
$300,000 is 12.87%, and the return on the last $150,000 is 10.56% for an
overall return of 12.11%.
EXPLORING THE
SOLUTION
Note that if the IRR(s) had been calculated for each design independently
and then ranked, the wrong investment decision would have been made:
Design A would have been ranked higher with an IRR of 12.87% as compared to Design B with a lower IRR of 12.11%. An incremental investment analysis is required when comparing mutually exclusive alternatives using IRR.
Example 6.4 illustrates Principle #6: Continue to invest as long as
each additional increment of investment yields a return that is greater than
the investor’s TVOM.
When comparing mutually exclusive investment alternatives, each of
which has well behaved cash flows, it is quite likely that multiple roots will
occur. The reason for this is, except for the initial step, incremental analyses
are performed. When one well-behaved cash flow series is subtracted from
another, there is no guarantee that the difference in the cash flow series
will be well behaved, as the following example illustrates.
IRR Analysis with Regular Cash Flow Series for Alternative
Investments, but Irregular Incremental Analysis Cash Flows
EXAMPLE
Two mutually exclusive investment alternatives are being considered. The
MARR is 12 percent. Alternative 1 requires an initial investment of
$100,000; it returns $33,600 in year 1, $72,320 in year 2, and $39,920 in
year 3. It has a regular cash flow, with a single change of sign and an IRR
of 20.8122%. Alternative 2 requires an initial investment of $104,000 and
has equal annual returns of $50,000 over the three years. It also has a regular cash flow, with a single change of sign. Which alternative is preferred?
KEY DATA
Given: The cash flow profiles for the alternatives are given in Table 6.2.
Find: FW of the incremental investment.
6-1
TABLE 6.2
EOY
Data for Example 6.5
CF(1)
CF(2)
CF(2-1)
0
2$100,000.00
2$104,000.00
2$4,000.00
1
$33,600.00
$50,000.00
$16,400.00
2
$72,320.00
$50,000.00
2$22,320.00
3
$39,920.00
$50,000.00
$10,080.00
IRR 5
Internal Rate of Return Calculations
20.8122%
20.0000%
IRR 5
40.0000%
IRR 5
50.0000%
To determine which of the two alternatives is preferred, they are ordered by
increasing investment. Because Alternative 1 has the smaller initial investment, it is considered first; it is justified, because IRR1 5 20.8122% .
MARR . 12%.
Next considered is Alternative 2. The incremental investment required
to move from Alternative 1 to Alternative 2 is $4,000 initially, followed by
incremental returns of $16,400, 2$22,320, and $10,080. Note that this is
an irregular incremental cash flow, with three changes of sign. In fact, this
incremental cash flow is identical to the cash flow of Example 6.2, analyzed previously, with multiple IRR values of 20%, 40%, and 50%. The
future worth plot is seen in Figure 6.5; it is identical to Figure 6.2.
Using Figure 6.5, we can conclude that if MARR , 20% or if 40% ,
MARR , 50%, the incremental investment yields a positive future worth and
therefore Alternative 2 is preferred to Alternative 1. If 20% , MARR , 40%
or if 50% , MARR, the future worth of the incremental investment is negative and Alternative 1 is preferred. Because MARR is 12%, Alternative 2 is
selected.
$60
Future Worth
$40
$20
$0
10%
20%
30%
40%
50%
60%
–$20
–$40
MARR
FIGURE 6.5 Plot of Future Worth of Incremental Investment CF(2-1) for Example 6.5
SOLUTION
223
224
Chapter 6
Rate of Return
6-2
EXTERNAL RATE OF RETURN
CALCULATIONS
Compute the External Rate of Return (ERR) for an
individual investment and perform incremental comparisons of mutually
exclusive investment alternatives to determine the one that maximizes
economic worth.
LEARN I N G O B JEC T I V E
External Rate of Return
(ERR) The interest rate
that makes the absolute
value of the future worth of
negative-valued cash flows
equal to the future worth of
positive-valued cash flows
that are reinvested at the
MARR.
Because of the possibilities of multiple roots when using the internal rate
of return method, an alternative approach called the external rate of
return (ERR) method was developed.1 The approach is based on the
following equation:
n
n
t50
t50
n2t
5 a Ct 11 1 i¿2 n2t
a Rt 11 1 r2
(6.2)
where Rt denotes the positive-valued cash flows in a cash flow series and
Ct denotes the absolute value of the negative-valued cash flows in a cash
flow series; r is the reinvestment rate, which we call the MARR; and i9 is
the external rate of return.
The reasoning in creating the ERR went like this: Any funds remaining in the investment pool are assumed to earn returns equal to the MARR.
Therefore, money recovered from an investment should also earn returns
equal to the MARR. The interest rate that makes the future worth of negative-valued cash flows equal to the future worth of positive-valued cash
flows that are reinvested at the MARR is the external rate of return. The
word external makes the point that recovered funds are not reinvested at a
rate equal to the internal rate of return, and the return earned on recovered
capital is external to the investment in question.
Two features of the ERR are important: There exists a unique solution
(no multiple rates of return), and the ERR is always between the IRR and the
MARR. Hence, if IRR . MARR, then IRR . ERR . MARR; likewise, if
IRR , MARR, then IRR , ERR , MARR; and if IRR 5 MARR, then
IRR 5 ERR 5 MARR.
6.2.1
Single Alternative
We begin by considering ERR for a single-alternative, the now-familiar
SMP machine.
1
White, J. A., K. E. Case, and M. H. Agee, ‘‘Rate of Return: An Explicit Reinvestment Rate
Approach,’’ Proceedings of the 1976 AIIE Conference, American Institute of Industrial
Engineers, Norcross, GA, 1976.
6-2 External Rate of Return Calculations
Computing the External Rate of Return for a Single Alternative
EXAMPLE
The investment of $500,000 in a surface mount placement machine is
expected to reduce manufacturing costs by $92,500 per year. At the end of
the 10-year planning horizon, it is expected the SMP machine will be
worth $50,000. Using a 10% MARR and an external rate of return analysis, should the investment be made?
Given: The cash flows outlined in Figure 6.1; MARR 5 10%; planning
horizon 5 10 years
Find: The ERR of the investment. Is this investment recommended?
KEY DATA
The only negative-valued cash flow is the initial investment. Therefore,
Equation 6.2 becomes
SOLUTION
$500,0001F Z P i¿,102 5 $92,5001F Z A 10%,102 1 $50,000
$500,00011 1 i¿2 10 5 $92,5001F Z A 10%,102 1 $50,000
11 1 i¿ 2 10 5 3 $92,500115.937422 1 $50,000 4 /$500,000 5 3.048423
i¿ 5 11.79117%
or
i¿ 5 RATE110,,2500000,FV110%,10,29250021500002
5 11.79117%
Since i9 . MARR, the machine is justified economically.
The ERR can also be obtained using the Excel® MIRR worksheet
function, only if the first cash flow is negative and all subsequent cash flows
are positive. The modified internal rate of return [MIRR] as a measure of
economic worth is not covered in this text. The syntax for the Excel® MIRR
worksheet function is 5MIRR(values,finance,rate,reinvest_rate); for
purposes of computing the ERR, the finance rate is ignored. As shown
in Figure 6.6, using the MIRR worksheet function, we obtain the exact
figure for the ERR: 11.79118%. Recall, for this example, IRR 5 13.8%.
The IRR function may also be used to determine the ERR, even if
there are negative cash flows in other than year 0, as follows.
Define two new cash flow series: CF(1) and CF(2). Using the
Excel® IF function, separate the original cash flow roots into two
series: {Rt, t 5 0, . . . ,10} and {Ct, t 5 0, . . . ,10}.
2. Next, develop a new cash flow series {CFt, t 5 0, . . . ,10}, which,
except for t 5 10, is the negative of {Ct}.
1.
225
226
Chapter 6
Rate of Return
3. Then, compute the future worth of the positive-valued cash flows and
add it to the negative cash flow (although in this example none exists)
in the last year of the planning horizon.
4. The resulting series (E3:E13 in Figure 6.6) contains zeros, the negative-valued cash flow from the original cash flow series, and the future
worth of all positive-valued cash flows; the latter (shown in cell E13)
is obtained using the MARR of 10 percent.
5. Recalling the definition of the external rate of return (the interest rate
that makes the absolute value of the future worth of the negativevalued cash flows equal to the future worth of the positive-valued cash
flows, based on the MARR), use the Excel® IRR worksheet function
to solve for the ERR, as indicated in cell E14. ERR 5 11.79118%.
FIGURE 6 . 6
EXAMPLE
ERR Solution to Example 6.6
Using the ERR to Resolve a Multiple Root Problem
Recall the data in Example 6.2 that led to multiple solutions to the IRR
formulation. The following four cash flows occurred at t 5 0, 1, 2, and 3:
2$4,000, $16,400, 2$22,320, and $10,080, respectively. Using a MARR
of 12 percent, find the ERR.
6-2 External Rate of Return Calculations
From Equation 6.2,
$4,0001F Z P i¿,32 1 $22,3201F Z P i¿,12 5 $16,4001F Z P 12%,22 1 $10,080
To determine the value of i9 that satisfies the equality, trial-and-error
methods can be used, followed by interpolation. Due to the multiple negative signs, the Excel® MIRR function cannot be used to obtain the
ERR value. However, the Excel® SOLVER and GOAL SEEK tools can
be used. Figure 6.7 provides the SOLVER setup and solution. Both
SOLVER and GOAL SEEK yield a value of 12.0911% for the ERR
obtained.
Shown in Table 6.3 are ERR values for various values of the MARR.
Notice, for MARR , 20%, ERR . MARR; for 20% , MARR , 40%,
ERR , MARR; for 40% , MARR , 50%, ERR . MARR; and for
MARR . 50%, ERR , MARR.
FIGURE 6.7
Excel® SOLVER Setup and Solution to Example 6.7
SOLUTION
227
228
Chapter 6
Rate of Return
TABLE 6.3
ERR Solutions for Various MARR Values in Example 6.7
MARR
ERR
MARR
ERR
MARR
ERR
0%
0.4654%
20%
20.0000%
40%
40.0000%
2%
2.3768%
22%
21.9900%
42%
42.0030%
4%
4.2999%
24%
23.9837%
44%
44.0049%
6%
6.2338%
26%
25.9805%
46%
46.0052%
8%
8.1775%
28%
27.9799%
48%
48.0037%
10%
10.1302%
30%
29.9812%
50%
50.0000%
12%
12.0911%
32%
31.9840%
52%
51.9939%
14%
14.0592%
34%
33.9877%
54%
53.9850%
16%
16.0399%
36%
35.9919%
56%
55.9732%
18%
18.0144%
38%
37.9962%
58%
57.9581%
6.2.2 Multiple Alternatives
When mutually exclusive investment alternatives exist, the external rate of
return method can be used to select the economically preferred one. However, like the internal rate of return method, it must be applied incrementally. Although there are multiple alternatives and each has its own external
rate of return (ij9), there is a common reinvestment rate (r). Mathematically,
for alternative j, the following equality must hold:
n
n
t50
t50
n2t
5 a Cjt 11 1 ij¿2 n2t
a Rjt 11 1 r2
(6.3)
where Rjt denotes the positive-valued cash flows and Cjt denotes the absolute value of the negative-valued cash flows in the cash flow series for
alternative j.
EXAMPLE
Using the ERR to Compare Two Alternatives for the Scream Machine
Recall the example involving two design alternatives for a new ride called
the Scream Machine, the first of which requires an initial investment of
$300,000 and yields net annual revenue of $55,000 over a 10-year planning period; the second requires an initial investment of $450,000 and
yields a net annual revenue of $80,000 over a 10-year period. Using the
ERR, and based on a 10 percent MARR, which alternative is preferred?
6-2
External Rate of Return Calculations
Given: The cash flows outlined in Figure 6.4; MARR 5 10%; planning
horizon 5 10 years
Find: ERR for each alternative and the incremental investment.
KEY DATA
The following steps are taken:
1. Compute the ERR for the smaller initial investment alternative,
Design A:
SOLUTION
$300,00011 1 i¿A 2 10 5 $55,0001F Z A 10%,102
11 1 i¿A 2 10
5 $55,000115.937422/$300,000
i¿A
5 11.31814% . MARR 5 10%
or
i¿A 5 RATE110,,2300000,FV110%,10,2550002 2
i¿A 5 11.31814% . MARR 5 10%
Design Alternative A is justified economically.
2. Compute the ERR for the $150,000 incremental investment to move
from Design A to Design B:
$150,00011 1 i¿B 2A 2 10 5 $25,0001F Z A 10%,102
11 1 i¿B 2A 2 10
5 $25,000115.93742/$150,000
i¿B 2A
5 10.26219% . MARR 5 10%
or
i¿B 2A 5 RATE110,,2150000,FV110%,10,2250002 2
5 10.26219% . MARR 5 10%
The incremental investment to accept Design B is justified economically.
3. Design Alternative B will have an overall external rate of return of
$450,00011 1 i¿B 2 10 5 $80,0001F Z A 10%,102
11 1 i¿B 2 10
5 $80,000115.93742/$450,000
i¿B
5 10.97611%
or
i¿B 5 RATE110,,2450000,FV110%,10,2800002 2
5 10.97611%
229
230
Chapter 6
Rate of Return
ERR Analysis with Regular Cash Flow Series for Alternative
Investments, but Irregular Incremental Analysis Cash Flows
EXAMPLE
Recall Example 6.5 where two mutually exclusive alternative investments
are considered. The MARR is 12 percent. The cash flow profiles for the
alternatives are given in Table 6.2. Alternative 1 requires an initial investment of $100,000; it returns $33,600 in year 1, $72,320 in year 2, and
$39,920 in year 3. It has a regular cash flow, with a single change of sign
and an ERR of 17.7031%. Alternative 2 requires an initial investment of
$104,000 and has equal annual returns of $50,000 over the three years. It
also has a regular cash flow, with a single change of sign. Using ERR,
which alternative is preferred?
FIGURE 6 . 8
SOLUTION
Data and ERR Analysis for Example 6.9
To determine which of the two alternatives is preferred, they are ordered
by increasing investment. Because Alternative 1 has the smaller initial
investment, it is considered first. It is justified, because, as shown in Figure
6.8, ERR1 5 17.7031% . MARR . 12%.
Next considered is Alternative 2. The incremental investment required
to move from Alternative 1 to Alternative 2 is $4,000 initially, followed by
$16,400, 2$22,320, and $10,080. Note that this is an irregular incremental
cash flow, with three changes of sign. Of course, it is identical to the incremental cash flow of Example 6.5, analyzed previously, which had multiple
IRR values of 20%, 40%, and 50%. Likewise, it is identical to the incremental cash flow of Example 6.7, analyzed previously, which had a unique
ERR of 12.0911%.
Since ERR221 5 12.0911% . MARR 5 12%, Alternative 2 is selected.
Summary 231
SUMMARY
KEY CONCEPTS
1. Learning Objective: Compute the Internal Rate of Return (IRR) for an individual investment and perform incremental comparisons of mutually exclusive investment alternatives to determine the one that maximizes economic
worth. (Section 6.1)
The IRR is a popular method to measure economic worth, especially for
personal investment decision making. Unlike present, future, and annual
worth methods, the IRR is not a ranking method, but rather incremental
analysis must be performed to select from a set of mutually exclusive alternatives. The IRR method can return multiple solutions.
Mathematically, investment j’s internal rate of return, denoted ij*, satisfies the following equality:
n
0 5 a Ajt 11 1 ij* 2 n2t
(6.1)
t50
In words, the internal rate of return is the interest rate that makes the
future worth of an investment equal 0.
2. Learning Objective: Apply Descartes’ rule of signs and Norstrom’s criterion to test for multiple roots when using the IRR method. (Section 6.1.2)
Descartes’ rule of signs and Norstrom’s criterion are common methods
used to test for multiple solutions with the IRR method. Descartes’ rule of
signs states that there will be at most as many positive rates of return as
there are sign changes in the cash flow profile. Norstrom’s criterion is
applied to determine if there is at most one real positive IRR.
3. Learning Objective: Compute the External Rate of Return (ERR) for an
individual investment and perform incremental comparisons of mutually
exclusive investment alternatives to determine the one that maximizes economic worth. (Section 6.2)
Like the IRR, the ERR is a popular method to measure economic worth.
It is not a ranking method, but rather incremental analysis must be performed to select from a set of mutually exclusive alternatives. The ERR is
“external” to make the point that the recovered funds are not reinvested at
a rate equal to the IRR, and the return earned on recovered capital is
external to the investment in question. The ERR method returns a single
solution.
The ERR is based on the following equation:
n
n
t50
t50
n2t
5 a Ct 11 1 i¿2 n2t
a Rt 11 1 r2
(6.2)
232
Chapter 6
Rate of Return
where Rt denotes the positive-valued cash flows in a cash flow series and
Ct denotes the negative-valued cash flows in a cash flow series; r is the
reinvestment rate, which we call the MARR; and i9 is the external rate of
return.
KEY TERMS
External Rate of Return (ERR),
p. 224
Internal Rate of Return (IRR),
p. 215
Problem available in WileyPLUS
GO Tutorial Tutoring Problem available in WileyPLUS
Video Solution Video Solution available in WileyPLUS
FE-LIKE PROBLEMS
1.
Consider the following cash flow diagram. What is the value of X if the
internal rate of return is 15%?
X
0
1
X
2
$400
a. $246
b. $255
2.
c. $281
d. $290
If the internal rate of return (IRR) of a well-behaved investment alternative is equal to MARR, which of the following statements about the other
measures of worth for this alternative must be true?
PW 5 0
AW 5 0
a. I only
c. Neither I nor II
b. II only
d. Both I and II
Summary 233
3.
An investment is guaranteed to have a unique value of IRR if which of the
following is true?
a. alternating positive and negative cash flows
b. an initial negative cash flow followed by all positive cash flows
c. a unique value for ERR
d. a positive PW at MARR
4.
What is the internal rate of return of the following cash flow diagram?
0
1
$30
$31
2
3
$15
$30
a. 20.0%
b. 18.2%
A snow cone machine at an ice cream shop costs $15,000. The machine is
expected to generate profits of $2,500 each year of its 10 year useful life. At
the end of the 10 years the machine will have a salvage value of zero. Within
what interest rate range does the IRR fall?
a. Less than 10%
c. 12% to 14%
b. 10% to 12%
d. Greater than 14%
The next two questions are based on the following “present worth versus
interest rate” graph for a well-behaved investment.
A
Present Worth
5.
c. 17.5%
d. 15.0%
E
C
B
Interest Rate
D
234
Chapter 6
Rate of Return
6.
If the interest rate at B is 20%, then which of the following best describes
the analysis of the investment?
a. the IRR of the investment is less than 20%
b. the IRR of the investment is equal to 20%
c. the IRR of the investment is greater than 20%
d. none of the above are true
7.
The IRR of this investment is located at which point?
a. A
c. D
b. C
d. E
8.
A company is considering two alternatives, one of which must be implemented. Of the two projects, A has the higher maintenance cost, but B has the
higher investment cost. The appropriate (and properly calculated) incremental IRR is 17.6%. Which alternative is preferred if the Minimum Attractive
Rate of Return is 20%.
a. A
b. B
c. the company is indifferent between A and B
d. cannot be determined from the information given
9.
Using an incremental internal rate of return (IRR) analysis the decision to
replace the “current best” by the “challenger” is based on what decision rule?
a. the internal rate of return of the increment is greater than the external rate
of return
b. the internal rate of return of the increment is greater than the internal rate
of return of the previous increment
c. the internal rate of return of the increment is greater than zero
d. the internal rate of return of the increment is greater than MARR
10.
If the IRR of Alternative A is 18%, the IRR of Alternative B is 16%, and
MARR is 12%, which of the following is correct?
a. alternative B is preferred over alternative A
b. alternative A is preferred over alternative B
c. not enough information is given to determine which alternative is preferred
d. neither alternative A nor alternative B is acceptable
11.
When conducting an incremental analysis, what step must always be
taken immediately prior to beginning the pairwise comparisons?
a. order the alternatives from highest to lowest initial investment
b. order the alternatives from lowest to highest present worth
c. order the alternatives from lowest to highest internal rate of return
d. order the alternatives from lowest to highest initial investment
12.
Consider the IRR and ERR measures of worth. If we define a “root” to mean
a value for the measure that results in PW 5 0, then which of the following
statements is true?
a. IRR can have multiple roots and ERR can have multiple roots
b. IRR has only a single root but ERR can have multiple roots
Summary 235
c. ERR has only a single root but IRR can have multiple roots
d. IRR has only a single root and ERR has only a single root
13.
Consider the calculation of an external rate of return (ERR). The positive
cash flows in the cash flow profile are moved forward to t 5 n using what
value of i in the (FZP,i,n–t) factors?
a. 0
b. the unknown value of ERR (i9)
c. MARR
d. IRR
PROBLEMS
Note to Instructors and Students
Many of the problems in this chapter are similar to problems in previous chapters.
This similarity is intentional. It is designed to illustrate the use of different measures of merit on the same problem.
Introduction
1. Match the measures of worth in the first column with one (or more) of the
analysis approaches that is (are) appropriate for that measure.
Measure of Worth
Analysis Approach
(a) Annual Worth
(b) External Rate of Return
(c) Future Worth
(d) Internal Rate of Return
(e) Present Worth
(1) Ranking Approach
(2) Incremental Approach
2. Match the measures of worth in the first column with the appropriate unit of
measure that results from the analysis.
Measure of Worth
Resulting Units of Measure
(a) Annual Worth
(b) External Rate of Return
(c) Future Worth
(d) Internal Rate of Return
(e) Present Worth
Sections 6.1.1 and 6.1.2
(1) dollars
(2) percentage
IRR Calculations—Single Alternative
3. Draw a cash flow diagram of any investment that has both of the following
properties:
(1) the investment has a four-year life,
(2) the investment has a 10%/yr internal rate of return
236
Chapter 6
Rate of Return
4. An investment has the following cash flow profile. For each value of MARR
below, what is the minimum value of X such that the investment is attractive
based on an internal rate of return measure of merit?
End of Year
Cash Flow
0
2$30,000
$6,000
$13,500
$X
$13,500
1
2
3
4
a. MARR is 12%/yr
b. MARR is 15%/yr
c. MARR is 24%/yr
5.
d. MARR is 8%/yr
e. MARR is 0%/yr
Nu Things, Inc. is considering an investment in a business venture with
the following anticipated cash flow results:
EOY
Cash Flow
EOY
Cash Flow
0
2$70,000
20,000
19,000
18,000
17,000
16,000
15,000
7
14,000
14
7,000
8
9
10
11
12
13
13,000
12,000
11,000
10,000
9,000
8,000
15
16
17
18
19
20
6,000
5,000
4,000
3,000
2,000
1,000
1
2
3
4
5
6
EOY
Cash Flow
Assume MARR is 20% per year. Based on an internal rate of return analysis
(1) determine the investment’s worth; (2) state whether or not your results
indicate the investment should be undertaken; and (3) state the decision rule
you used to arrive at this conclusion.
6. Carlisle Company has been cited and must invest in equipment to reduce
stack emissions or face EPA fines of $18,500 per year. An emission reduction
filter will cost $75,000 and have an expected life of 5 years. Carlisle’s MARR
is 10%/yr.
a. What is the internal rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on internal rate of return?
c. Is the filter economically justified?
7.
Fabco, Inc. is considering the purchase of flow valves that will reduce
annual operating costs by $10,000 per year for the next 12 years. Fabco’s
MARR is 7%/yr. Using an internal rate of return approach, determine the
maximum amount Fabco should be willing to pay for the valves.
Summary 237
8. Imagineering, Inc. is considering an investment in CAD-CAM compatible
design software with the cash flow profile shown in the table below. Imagineering’s MARR is 18%/yr.
EOY
Cash Flow (M$)
0
2$12
1
2$1
$5
$2
$5
$5
$2
$5
2
3
4
5
6
7
a. What is the internal rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on internal rate of return?
c. Should Imagineering invest?
9. DuraTech Manufacturing is evaluating a process improvement project. The
estimated receipts and disbursements associated with the project are shown
below. MARR is 6%/yr.
End of Year
0
1
2
3
4
5
Receipts
Disbursements
$0
$0
$2,000
$4,000
$3,000
$1,600
$5,000
$200
$300
$600
$1,000
$1,500
a. What is the internal rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on internal rate of return?
c. Should DuraTech implement the proposed process improvement?
10. A design change being considered by Mayberry, Inc. will cost $6,000 and
will result in an annual savings of $1,000 per year for the 6 year life of the
project. A cost of $2,000 will be avoided at the end of the project as a result
of the change. MARR is 8%/yr.
a. What is the internal rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on internal rate of return?
c. Should Mayberry implement the design change?
238
Chapter 6
Rate of Return
11. Home Innovation is evaluating a new product design. The estimated receipts
and disbursements associated with the new product are shown below. MARR
is 10%/yr.
End of Year
0
1
2
3
4
5
Receipts
Disbursements
$0
$600
$600
$700
$700
$700
$1,000
$300
$300
$300
$300
$300
a. What is the internal rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on internal rate of return?
c. Should Home Innovations pursue this new product?
12. A project has been selected for implementation. The net cash flow (NCF)
profile associated with the project is shown below. MARR is 10%/yr.
EOY
0
1
2
3
4
5
6
NCF
2$70,000
$30,000
$30,000
$30,000
$30,000
$30,000
$30,000 1 $2,000
a. What is the internal rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on internal rate of return?
c. Is the project economically justified?
13.
Bailey, Inc. is considering buying a new gang punch that would allow
them to produce circuit boards more efficiently. The punch has a first cost of
$100,000 and a useful life of 15 years. At the end of its useful life, the punch
has no salvage value. Labor costs would increase $2,000 per year using the
gang punch but raw material costs would decrease $12,000 per year. MARR
is 5%/yr.
a. What is the internal rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on internal rate of return?
c. Should Bailey buy the gang punch?
14. Eddie’s Precision Machine Shop is insured for $700,000. The present yearly
insurance premium is $1.00 per $100 of coverage. A sprinkler system with an
estimated life of 20 years and no salvage value can be installed for $20,000.
Annual maintenance costs for the sprinkler system are $400. If the sprinkler
Summary 239
system is installed, the system must be included in the shop’s value for insurance purposes but the insurance premium will reduce to $0.40 per $100 of
coverage. Eddie uses a MARR of 15%/yr.
a. What is the internal rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on internal rate of return?
c. Is the sprinkler system economically justified?
GO Tutorial Nancy’s Notions pays a delivery firm to distribute its products in
the metro area. Shipping costs are $30,000 per year. Nancy can buy a used truck
for $10,000 that will be adequate for the next 3 years. Operating and maintenance
costs are estimated to be $25,000 per year. At the end of 3 years, the used truck
will have an estimated salvage value of $3,000. Nancy’s MARR is 24%/yr.
a. What is the internal rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on internal rate of return?
c. Should Nancy buy the truck?
16. Brock Associates invested $80,000 in a business venture with the following
results:
15.
EOY
0
1
2
3
4
CF
2$80,000
10,000
16,000
22,000
28,000
EOY
CF
5
$28,000
6
7
8
22,000
16,000
10,000
MARR is 12%. Determine the internal rate of return and whether or not this
is a desirable venture.
17. Shrewd Endeavors, Inc. invested $70,000 in a business venture with the
following cash flow results:
EOY
CF
0
2$70,000
20,000
19,000
18,000
17,000
16,000
15,000
14,000
13,000
12,000
11,000
1
2
3
4
5
6
7
8
9
10
EOY
CF
11
10,000
12
13
14
15
16
17
18
19
20
9,000
8,000
7,000
6,000
5,000
4,000
3,000
2,000
1,000
MARR is 10%. Determine the internal rate of return and whether or not this
is a desirable venture.
240
Chapter 6
Rate of Return
18. An investment of $20,000 for a new condenser is being considered. Esti-
mated salvage value of the condenser is $5,000 at the end of an estimated life
of 6 years. Annual income each year for the 6 years is $8,500. Annual operating expenses are $2,300. Assume money is worth 15% compounded annually.
Determine the internal rate of return and whether or not the condenser should
be purchased.
19. Smith Investors places $50,000 in an investment fund. One year after making
the investment, Smith receives $7,500 and continues to receive $7,500 annually until 10 such amounts are received. Smith receives nothing further until
15 years after the initial investment, at which time $50,000 is received. Over
the 15-year period, determine the internal rate of return and whether or not
this is a desirable investment if MARR 5 10%.
20.
Video Solution What do you know about the mathematical value of the
internal rate of return of a project under each of the following conditions?
a.
b.
c.
d.
e.
f.
g.
h.
i.
21.
the present worth of the project is greater than zero
the present worth of the project is equal to zero
the present worth of the project is less than zero
the future worth of the project is greater than zero
the future worth of the project is equal to zero
the future worth of the project is less than zero
the annual worth of the project is greater than zero
the annual worth of the project is equal to zero
the annual worth of the project is less than zero
Value Lodges is the owner of an economy motel chain. Value Lodges is
considering building a new 200-unit motel. The cost to build the motel is
estimated at $8,000,000; Value Lodges estimates furnishings for the motel
will cost an additional $700,000 and will require replacement every 5 years.
Annual operating and maintenance costs for the motel are estimated to be
$800,000. The average rental rate for a unit is anticipated to be $40/day. Value
Lodges expects the motel to have a life of 15 years and a salvage value of
$900,000 at the end of 15 years. This estimated salvage value assumes that
the furnishings are not new. Furnishings have no salvage value at the end of
each 5-year replacement interval. Assuming average daily occupancy percentages of 50%, 60%, 70%, 80% for years 1 through 4, respectively, and
90% for the fifth through fifteenth years, MARR of 12%/yr, 365 operating
days/year, and ignoring the cost of land, should the motel be built? Base your
decision on an internal rate of return analysis.
22. Baon Chemicals Unlimited purchases a computer-controlled filter for
$100,000. Half of the purchase price is borrowed from a bank at 15%
compounded annually. The loan is to be paid back with equal annual payments over a 5-year period. The filter is expected to last 10 years, at which
time it will have a salvage value of $10,000. Over the 10-year period, the
operating and maintenance costs are expected to equal $20,000 in year 1
and increase by $1,500/yr each year thereafter. By making the investment,
Summary 241
annual fines of $50,000 for pollution will be avoided. Baon expects to earn
12% compounded annually on its investments. Based on an internal rate
of return analysis, determine whether the purchase of the filter is economically justified.
23.
Video Solution RealTurf is considering purchasing an automatic sprinkler system for its sod farm by borrowing the entire $30,000 purchase price.
The loan would be repaid with four equal annual payments at an interest rate
of 12%/yr/yr. It is anticipated that the sprinkler system would be used for
9 years and then sold for a salvage value of $2,000. Annual operating and
maintenance expenses for the system over the 9-year life are estimated to be
$9,000 per year. If the new system is purchased, cost savings of $15,000 per
year will be realized over the present manual watering system. RealTurf uses
a MARR of 15%/yr for economic decision making. Based on an internal rate
of return analysis, is the purchase of the new sprinkler system economically
attractive?
24. Galvanized Products is considering the purchase of a new computer system
for their enterprise data management system. The vendor has quoted a purchase price of $100,000. Galvanized Products is planning to borrow onefourth of the purchase price from a bank at 15% compounded annually. The
loan is to be repaid using equal annual payments over a 3-year period. The
computer system is expected to last five years and has a salvage value of
$5,000 at that time. Over the 5-year period, Galvanized Products expects
to pay a technician $25,000 per year to maintain the system but will save
$55,000 per year through increased efficiencies. Galvanized Products uses a
MARR of 18%/yr to evaluate investments.
a. What is the internal rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on internal rate of return?
c. Should the new computer system be purchased?
25. Aerotron Electronics is considering the purchase of a water filtration sys-
tem to assist in circuit board manufacturing. The system costs $40,000.
It has an expected life of 7 years at which time its salvage value will be
$7,500. Operating and maintenance expenses are estimated to be $2,000
per year. If the filtration system is not purchased, Aerotron Electronics will
have to pay Bay City $12,000 per year for water purification. If the system
is purchased, no water purification from Bay City will be needed. Aerotron
Electronics must borrow half of the purchase price, but they cannot start
repaying the loan for two years. The bank has agreed to three equal annual
payments, with the first payment due at end of year 2. The loan interest rate
is 8% compounded annually. Aerotron Electronics’ MARR is 10% compounded annually.
a. What is the internal rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on internal rate of return?
c. Should Aerotron Electronics buy the water filtration system?
242
Chapter 6
Rate of Return
26. Delta Dawn’s Bakery is considering purchasing a new van to deliver bread.
The van will cost $18,000. Two-thirds ($12,000) of this cost will be borrowed.
The loan is to be repaid with four equal annual payments (first payment at
t 5 1) based on an interest rate of 4%/yr. It is anticipated that the van will be
used for 6 years and then sold for a salvage value of $500. Annual operating
and maintenance expenses for the van over the 6-year life are estimated to
be $700 per year. If the van is purchased, Delta will realize a cost savings of
$3,800 per year. Delta uses a MARR of 6%/yr. Based on an internal rate of
return analysis, is the purchase of the van economically attractive?
Consider the following cash flow profile and assume MARR is 10%/yr.
27.
EOY
0
2$100
$25
$25
$60
1
2
3
a.
b.
c.
d.
NCF
EOY
NCF
4
2$30
$60
$25
5
6
What does Descartes’ rule of signs tell us about the IRR(s) of this project?
What does Norstrom’s criterion tell us about the IRR(s) of this project?
Determine the IRR(s) for this project.
Is this project economically attractive?
Video Solution Consider the following cash flow profile and assume
28.
MARR is 10%/yr.
EOY
0
2$100
$25
$25
$25
1
2
3
a.
b.
c.
d.
NCF
EOY
NCF
4
$25
5
6
$25
$25
What does Descartes’ rule of signs tell us about the IRR(s) of this project?
What does Norstrom’s criterion tell us about the IRR(s) of this project?
Determine the IRR(s) for this project.
Is this project economically attractive?
Consider the following cash flow profile and assume MARR is 10%/yr.
29.
EOY
0
1
2
3
a.
b.
c.
d.
NCF
2$100
$15
$15
$15
EOY
NCF
4
$15
5
6
$15
$15
What does Descartes’ rule of signs tell us about the IRR(s) of this project?
What does Norstrom’s criterion tell us about the IRR(s) of this project?
Determine the IRR(s) for this project.
Is this project economically attractive?
Summary 243
30. Consider the following cash flow profile and assume MARR is 10%/yr.
EOY
0
NCF
EOY
NCF
2$100
$25
4
$250
5
2$200
2
$200
6
2$100
3
2$100
1
a.
b.
c.
d.
What does Descartes’ rule of signs tell us about the IRR(s) of this project?
What does Norstrom’s criterion tell us about the IRR(s) of this project?
Determine the IRR(s) for this project.
Is this project economically attractive?
Consider the following cash flow profile and assume MARR is 10%/yr.
31.
EOY
0
1
2
EOY
NCF
4
2$950
$700
5
6
2$750
$900
3
a.
b.
c.
d.
NCF
2$100
$800
2$800
What does Descartes’ rule of signs tell us about the IRR(s) of this project?
What does Norstrom’s criterion tell us about the IRR(s) of this project?
Determine the IRR(s) for this project.
Is this project economically attractive?
32. Consider the following cash flow profile and assume MARR is 10%/yr.
EOY
0
1
2
3
NCF
EOY
2$101
$411
2$558
$253
NCF
4
$2
5
6
$8
2$14
a. Determine the IRR(s) for this project.
b. Is this project economically attractive?
33. Quilts R Us (QRU) is considering an investment in a new patterning attachment
with the cash flow profile shown in the table below. QRU’s MARR is 13.5%/yr.
EOY
Cash Flow
EOY
2$1,400
$0
$500
$500
$500
8
$600
1
2
3
4
9
10
11
12
$700
$800
$900
2$1,000
5
$500
13
2$2,000
6
$0
14
7
$500
15
2$3,000
$1,400
0
Cash Flow
244
Chapter 6
Rate of Return
a. What is the internal rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on internal rate of return?
c. Should QRU invest?
Section 6.1.3
IRR Calculations—Multiple Alternatives
34. Quantum Logistics, Inc., a wholesale distributor, is considering the construc-
tion of a new warehouse to serve the southeastern geographic region near the
Alabama-Georgia border. There are three cities being considered. After site
visits and a budget analysis, the expected income and costs associated with
locating in each of the cities have been determined. The life of the warehouse
is expected to be 12 years and MARR is 15%/yr. Based on an internal rate of
return analysis, which city should be recommended?
City
Initial Cost
Net Annual Income
Lagrange
Auburn
Anniston
$1,260,000
$1,000,000
$1,620,000
$480,000
$410,000
$520,000
35. Two mutually exclusive proposals, each with a life of 5 years, are under con-
sideration. MARR is 12%. Each proposal has the following cash flow profile:
EOY
0
1
2
3
4
5
NCF(A)
NCF(B)
2$30,000
$9,300
$9,300
$9,300
$9,300
$9,300
2$42,000
$12,625
$12,625
$12,625
$12,625
$12,625
Determine which alternative the decision maker should select using the internal rate of return method.
36. Chingos and Daughters Construction is considering three investment propos-
als: A, B, and C. Proposals A and B are mutually exclusive, and Proposal C is
contingent on proposal B. The cash flow data for the investments over a 10-year
planning horizon are given below. The company has a budget limit of $1 million
for investments of the type being considered currently. MARR 5 15%.
Initial investment
Planning horizon
Salvage values
Annual receipts
Annual disbursements
NCF(A)
NCF(B)
NCF(C)
$600,000
10 years
$70,000
$400,000
$130,000
$800,000
10 years
$130,000
$600,000
$270,000
$470,000
10 years
$65,000
$260,000
$70,000
Determine which alternative should be selected using the internal rate of
return method.
Summary 245
37. ZeeZee’s Construction Company has the opportunity to select one of four
projects (A, B, C, or D) or the null (Do Nothing) alternative. Each project
requires a single initial investment and has an internal rate of return as shown
in the first table below. The second table shows the incremental IRR(s) for
pairwise comparisons between each project and all other projects with a
smaller initial investment.
Investments and IRR(s)
Project
null
A
B
C
D
Initial Investment
IRR
$0
$600,000
$800,000
$470,000
$540,000
0.0%
44.0%
40.0%
39.2%
36.0%
Incremental IRR(s)
Increment
Incremental IRR
B-A
B-D
B-C
A-D
A-C
D-C
28.3%
48.8%
41.4%
116.5%
61.0%
18.4%
For each of the values of MARR below indicate which project is preferred
based on an incremental IRR analysis.
a. MARR 5 50%
b. MARR 5 41%
c. MARR 5 25%
38. A large company has the opportunity to select one of seven projects: A, B, C,
D, E, F, G, or the null (Do Nothing) alternative. Each project requires a single
initial investment as shown in the table below. Information on each alternative was fed into a computer program that calculated the IRR for each project
as well as all the pertinent incremental IRR(s) as shown in the table below.
Project
A
B
C
D
E
F
G
Incremental Rate of Return of “Row”—“Column”
Initial
Investment
Null
A
B
C
D
E
F
$10,000
12,000
13,000
15,000
16,000
18,000
23,000
10%
9
8
7
6
5
7
7%
2
9
5
8
3
0.1%
5
1
2
8
9%
6
5
7
3%
5
4
5%
3
2%
246
Chapter 6
Rate of Return
For example, the IRR for Project A is 10% and the incremental IRR of Project C
minus Project B (C-B) is 0.1%. For each value of MARR below indicate which
project is preferred and the evaluations you made to arrive at this conclusion.
a. MARR 5 12%
d. MARR 5 3.5%
b. MARR 5 9.5%
e. MARR 5 1.5%
c. MARR 5 8%
39.
Dark Skies Observatory is considering several options to purchase a new
deep space telescope. Revenue would be generated from the telescope by selling “time and use” slots to various researchers around the world. Four possible
telescopes have been identified in addition to the possibility of not buying a
telescope if none are financially attractive. The table below details the characteristics of each telescope. An internal rate of return analysis is to be performed.
Useful Life
First Cost
Salvage Value
Annual Revenue
Annual Expenses
T1
T2
T3
T4
10 years
$600,000
$70,000
$400,000
$130,000
10 years
$800,000
$130,000
$600,000
$270,000
10 years
$470,000
$65,000
$260,000
$70,000
10 years
$540,000
$200,000
$320,000
$120,000
a. Determine the preferred telescope if MARR is 25%/yr.
b. Determine the preferred telescope if MARR is 42%/yr.
40. Orpheum Productions in Nevada is considering three mutually exclusive
alternatives for lighting enhancements to one of its recording studios. Each
enhancement will increase revenues by attracting directors who prefer this
lighting style. The cash flow details, in thousands of dollars, for these enhancements are shown in the chart below. MARR is 10%/yr. Based on an internal
rate of return analysis, which alternative (if any) should be implemented?
End of Year
Light Bar
Sliding Spots
Reflected Beam
0
2$6,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
2$14,000
$3,500
$3,500
$3,500
$3,500
$3,500
$3,500
2$20,000
$0
$2,300
$4,600
$6,900
$9,200
$11,500
1
2
3
4
5
6
41.
Yani has $12,000 for investment purposes. His bank has offered the
following three choices.
1) A special savings certificate that will pay $100 each month for 5 years and
a lump sum payment at the end of 5 years of $13,000;
2) Buy a share of a racehorse for $12,000 that will be worth $20,000 in 5 years;
3) Put the money in a savings account that will have an interest rate of 12%
per year compounded monthly.
Use an internal rate of return analysis to make a recommendation to Yani.
Summary 247
42. On-Site Testing Service has received four investment proposals for consid-
eration. Two of the proposals, X1 and X2, are mutually exclusive. The other
two proposals, Y1 and Y2 are also mutually exclusive. Proposal Y1 is contingent on X1 and Y2 is contingent on X2. Other than these restrictions, any
combination of proposals (including null) is feasible. MARR is 10%/yr. The
expected cash flows for the proposals are shown below. An internal rate of
return analysis is to be conducted.
End of Year
0
1 through 8
X1
X2
Y1
Y2
2$10,000
$1,600
2$15,000
$2,600
2$6,000
$2,500
2$9,000
$3,500
a. List all the alternatives to be considered.
b. Determine which (if any) proposals On-Site Testing should accept.
43. Arnold Engineering has available two mutually exclusive investment propos-
als, A and B. Their net cash flows are as shown in the table over a 10-year
planning horizon. MARR is 12%.
EOY
NCF(A)
NCF(B)
0
2$40,000
8,000
8,000
8,000
8,000
8,000
8,000
8,000
8,000
8,000
8,000
2$30,000
9,000
8,500
8,000
7,500
7,000
6,500
6,000
5,500
5,000
4,500
1
2
3
4
5
6
7
8
9
10
Determine which alternative the decision maker should select using an internal rate of return approach.
44. A firm is faced with four investment proposals, A, B, C, and D, having the cash
flow profiles shown below. Proposals A and C are mutually exclusive, and
Proposal D is contingent on Proposal B being chosen. Currently, $750,000 is
available for investment and the firm has stipulated a MARR of 10%.
Initial investment
Planning horizon
Annual receipts
Annual disbursements
Salvage value
NCF(A)
NCF(B)
NCF(C)
NCF(D)
$400,000
10 years
$205,000
$110,000
$50,000
$400,000
10 years
$215,000
$125,000
$50,000
$600,000
10 years
$260,000
$120,000
$100,000
$300,000
10 years
$230,000
$150,000
$50,000
Determine which alternative the decision maker should select. Use the internal rate of return method.
248
Chapter 6
Rate of Return
45.
Three alternatives are being considered by the management of Brawn
Engineering to satisfy an OSHA requirement for safety gates in the machine
shop. Each of the gates will completely satisfy the requirement, so no combinations need to be considered. The first costs, operating costs, and salvage
values over a 5-year planning horizon are shown below. Using an internal rate
of return analysis with a MARR of 20%/yr, determine the preferred gate.
End of Year
Gate 1
Gate 2
Gate 3
0
2$15,000
2$19,000
2$24,000
1
2$6,500
2$5,600
2$4,000
2
2$6,500
2$5,600
2$4,000
3
2$6,500
2$5,600
2$4,000
4
2$6,500
2$5,600
2$4,000
5
2$6,500 1 $0
2$5,600 1 $2,000
2$4,000 1 $5,000
46. Five projects form the mutually exclusive, collectively exhaustive set under
consideration. The cash flow profiles for the five projects are given in the
table below.
Life
Initial Investment
Salvage Value
Annual Revenues
Annual Expenses
null
A
B
C
D
10 years
0
0
0
0
10 years
$600,000
$70,000
$400,000
$130,000
10 years
$800,000
$130,000
$600,000
$270,000
10 years
$470,000
$65,000
$260,000
$70,000
10 years
$540,000
$200,000
$320,000
$120,000
Information on each project was fed into a computer program that calculated
the IRR(s) and incremental IRR(s) as shown in the table below. Unfortunately, when the table was printed, one of the cells was overprinted with X’s
and was unreadable. As the resident expert on incremental IRR analysis, you
have been asked to assist.
Incremental Rate of Return of “Row”—“Column”
Project
Null
C
D
A
C
D
A
B
39%
36%
44%
40%
18%
XXXXX
41%
117%
49%
28%
a. Specify the incremental cash flow profile that must be analyzed to deter-
mine the value in the overprinted incremental IRR cell.
b. Determine the incremental IRR value that belongs in the overprinted cell.
c. If MARR is 37%/yr, which project is preferred?
d. Based on the data in the table, if MARR is 40%, specify whether the present
worth of each project would be positive, negative, or zero when evaluated
at MARR?
Summary 249
47. DelRay Foods must purchase a new gumdrop machine. Two machines are avail-
able. Machine 7745 has a first cost of $10,000, an estimated life of 10 years,
a salvage value of $1,000, and annual operating costs estimated at $0.01 per
1,000 gumdrops. Machine A37Y has a first cost of $8,000, a life of 10 years,
and no salvage value. Its annual operating costs will be $300 regardless of the
number of gumdrops produced. MARR is 6%/yr and 30 million gumdrops
are produced each year. Based on an internal rate of return analysis, which
machine should be recommended?
48.
Video Solution The engineering team at Manuel’s Manufacturing, Inc.
is planning to purchase an Enterprise Resource Planning (ERP) system. The
software and installation from Vendor A costs $380,000 initially and is expected to increase revenue $125,000 per year every year. The software and installation from Vendor B costs $280,000 and is expected to increase revenue
$95,000 per year. Manuel’s uses a 4 year planning horizon and a 10% per
year MARR. Based on an internal rate of return analysis, which ERP system
should Manuel purchase?
49.
Final Finishing is considering three mutually exclusive alternatives for
a new polisher. Each alternative has an expected life of 10 years and no
salvage value. Polisher I requires an initial investment of $20,000 and provides annual benefits of $4,465. Polisher II requires an initial investment
of $10,000 and provides annual benefits of $1,770. Polisher III requires an
initial investment of $15,000 and provides annual benefits of $3,580. MARR
is 15%/yr. Based on an internal rate of return analysis, which polisher should
be recommended?
50. Xanadu Mining is considering three mutually exclusive alternatives as shown
in the table below. MARR is 10%/yr. Based on an internal rate of return
analysis, which alternative should be recommended?
EOY
A001
B002
C003
0
2$210
$80
$90
$100
$110
2$110
$60
$60
$60
$70
2$160
$80
$80
$80
$80
1
2
3
4
51.
Video Solution Parker County Community College (PCCC) is trying
to determine whether to use no insulation or to use insulation 1-inch thick
or 2-inches thick on its steam pipes. The heat loss from the pipes without
insulation is expected to cost $1.50 per year per foot of pipe. A 1-inch thick
insulated covering will eliminate 89% of the loss and will cost $0.40 per
foot. A 2-inch thick insulated covering will eliminate 92% of the loss and
will cost $0.85 per foot. PCCC Physical Plant Services estimates that there
are 250,000 feet of steam pipe on campus. The PCCC Accounting Office
requires a 10%/yr return to justify capital expenditures. The insulation has
250
Chapter 6
Rate of Return
a life expectancy of 10 years. Determine which insulation (if any) should be
purchased using an internal rate of return analysis.
52. Several years ago, a man won $27 million in the State Lottery. To pay off the
winner, the State planned to make an initial $1 million payment today followed by equal annual payments of $1.3 million at the end of each year for
the next 20 years. Just before receiving any money, the man offered to sell
the winning ticket back to the State for a one-time immediate payment of
$14.4 million. If the State uses a 6%/yr MARR, should the State accept the
man’s offer? Use an internal rate of return analysis.
53.
Calisto Launch Services is an independent space corporation and has
been contracted to develop and launch one of two different satellites. Initial
equipment will cost $750,000 for the first satellite and $850,000 for the second.
Development will take five years at an expected cost of $150,000 per year to
the first satellite; $120,000 per year for the second. The same launch vehicle
can be used for either satellite and will cost $275,000 at the time of the launch
five years from now. At the conclusion of the launch, the contracting company
will pay Calisto $2,500,000 for either satellite.
Calisto is also considering whether they should consider launching both
satellites. Because Calisto would have to upgrade its facilities to handle two
concurrent projects, the initial costs would rise by $150,000 in addition to
the first costs of each satellite. Calisto would need to hire additional engineers and workers, raising the yearly costs to a total of $400,000. An additional compartment would be added to the launch vehicle at an additional
cost of $75,000. As an incentive to do both, the contracting company will
pay for both launches plus a bonus of $1,000,000. Using an internal rate
of return analysis with a MARR of 10%/yr, what should Calisto Launch
Services do?
54. Packaging equipment for Xi Cling Wrap costs $60,000 and is expected to
result in end-of-year net savings of $23,000 per year for 3 years. The equipment will have a market value of $10,000 after 3 years. The equipment can be
leased for $21,000 per year, payable at the beginning of each year. Xi Cling’s
MARR is 10%/yr. Based on an internal rate of return analysis, determine if
the packaging equipment should be purchased or leased.
Section 6.2.1
ERR Calculations—Single Alternative
55. Consider the following cash flow profile and assume MARR is 10%/yr.
EOY
0
1
2
3
NCF
2$100
$800
2$750
$900
EOY
4
5
6
NCF
2$950
$700
2$800
Summary 251
a. Determine the ERR for this project.
b. Is this project economically attractive?
Consider the following cash flow profile and assume MARR is 10%/yr.
56.
EOY
0
NCF
2$100
$25
$25
$25
1
2
3
EOY
NCF
4
$25
5
6
$25
$25
a. Determine the ERR for this project.
b. Is this project economically attractive?
57. Consider the following cash flow profile and assume MARR is 10%/yr.
EOY
0
NCF
2$100
$25
$25
$60
1
2
3
EOY
NCF
4
2$30
$60
$25
5
6
a. Determine the ERR for this project.
b. Is this project economically attractive?
58. Consider the following cash flow profile and assume MARR is 10%/yr.
EOY
0
NCF
2$101
$411
1
2
2$558
$253
3
EOY
NCF
4
$2
5
6
2$14
$8
a. Determine the ERR for this project.
b. Is this project economically attractive?
Consider the following cash flow profile and assume MARR is 10%/yr.
59.
EOY
0
1
2
3
NCF
2$100
$15
$15
$15
EOY
NCF
4
$15
5
6
$15
$15
a. Determine the ERR for this project.
b. Is this project economically attractive?
252
Chapter 6
Rate of Return
60. Consider the following cash flow profile and assume MARR is 10%/yr.
EOY
0
NCF
EOY
NCF
2$100
$25
4
$250
5
2$200
2
$200
6
2$100
3
2$100
1
a. Determine the ERR for this project.
b. Is this project economically attractive?
61.
Nu Things, Inc. is considering an investment in a business venture with
the following anticipated cash flow results:
EOY
0
1
2
3
4
5
6
Cash Flow
EOY
Cash Flow
2$70,000
20,000
19,000
18,000
17,000
16,000
15,000
7
14,000
14
EOY
Cash Flow
7,000
8
9
10
11
12
13
13,000
12,000
11,000
10,000
9,000
8,000
15
16
17
18
19
20
6,000
5,000
4,000
3,000
2,000
1,000
Assume MARR is 20% per year. Based on an external rate of return analysis
(1) determine the investment’s worth; (2) state whether or not your results
indicate the investment should be undertaken; and (3) state the decision rule
you used to arrive at this conclusion.
62. Solve problem 16, except use the external rate of return approach.
63. Solve problem 17, except use the external rate of return approach.
64. Solve problem 18, except use the external rate of return approach.
65. Solve problem 19, except use the external rate of return approach.
66. Quilts R Us (QRU) is considering an investment in a new patterning attach-
ment with the cash flow profile shown in the table below. QRU’s MARR is
13.5%/yr.
EOY
0
1
2
3
4
5
6
7
Cash Flow
2$1,400
$0
$500
$500
$500
$500
$0
$500
EOY
Cash Flow
8
$600
9
10
11
12
13
14
15
$700
$800
$900
−$1,000
−$2,000
−$3,000
$1,400
Summary 253
a. What is the external rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on external rate of return?
c. Should QRU invest?
67.
Aerotron Electronics is considering the purchase of a water filtration
system to assist in circuit board manufacturing. The system costs $40,000. It
has an expected life of 7 years at which time its salvage value will be $7,500.
Operating and maintenance expenses are estimated to be $2,000 per year. If
the filtration system is not purchased, Aerotron Electronics will have to pay
Bay City $12,000 per year for water purification. If the system is purchased,
no water purification from Bay City will be needed. Aerotron Electronics
must borrow half of the purchase price, but they cannot start repaying the
loan for two years. The bank has agreed to three equal annual payments,
with the first payment due at end of year 2. The loan interest rate is 8%
compounded annually. Aerotron Electronics’ MARR is 10% compounded
annually.
a. What is the external rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on external rate of return?
c. Should Aerotron Electronics Fit buy the water filtration system?
68. Home Innovations is evaluating a new product design. The estimated receipts
and disbursements associated with the new product are shown below. MARR
is 10%/yr.
End of Year
Receipts
Disbursements
0
1
2
3
4
$0
$600
$600
$700
$700
$1,000
$300
$300
$300
$300
5
$700
$300
a. What is the external rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on external rate of return?
c. Should Home Innovations pursue this new product?
69.
GO Tutorial Galvanized Products is considering the purchase of a new
computer system for their enterprise data management system. The vendor
has quoted a purchase price of $100,000. Galvanized Products is planning
to borrow one-fourth of the purchase price from a bank at 15% compounded
annually. The loan is to be repaid using equal annual payments over a 3-year
period. The computer system is expected to last five years and has a salvage
value of $5,000 at that time. Over the five year period, Galvanized Products
expects to pay a technician $25,000 per year to maintain the system but will
254
Chapter 6
Rate of Return
save $55,000 per year through increased efficiencies. Galvanized Products
uses a MARR of 18%/yr to evaluate investments.
a. What is the external rate of return of this investment?
b. What is the decision rule for judging the attractiveness of investments
based on external rate of return?
c. Should the new computer system be purchased?
Section 6.2.2
ERR Calculations—Multiple Alternatives
70. Three alternatives are being considered by the management of Brawn
Engineering to satisfy an OSHA requirement for safety gates in the machine shop. Each of the gates will completely satisfy the requirement so
no combinations need to be considered. The first costs, operating costs,
and salvage values over a 5-year planning horizon are shown below. Using
an external rate of return analysis with a MARR of 20%/yr, determine the
preferred gate.
End of Year
71.
Gate 1
Gate 2
Gate 3
0
2$15,000
2$19,000
2$24,000
1
2$6,500
2$5,600
2$4,000
2
2$6,500
2$5,600
2$4,000
3
2$6,500
2$5,600
2$4,000
4
2$6,500
2$5,600
2$4,000
5
2$6,500 1 $0
2$5,600 1 $2,000
2$4,000 1 $5,000
Calisto Launch Services is an independent space corporation and has
been contracted to develop and launch one of two different satellites. Initial
equipment will cost $750,000 for the first satellite and $850,000 for the second. Development will take five years at an expected cost of $150,000 per
year to the first satellite; $120,000 per year for the second. The same launch
vehicle can be used for either satellite and will cost $275,000 at the time of
the launch five years from now. At the conclusion of the launch, the contracting company will pay Calisto $2,500,000 for either satellite.
Calisto is also considering whether they should consider launching both
satellites. Because Calisto would have to upgrade its facilities to handle two
concurrent projects, the initial costs would rise by $150,000 in addition to
the first costs of each satellite. Calisto would need to hire additional engineers and workers, raising the yearly costs to a total of $400,000. An additional compartment would be added to the launch vehicle at an additional
cost of $75,000. As an incentive to do both, the contracting company will
pay for both launches plus a bonus of $1,000,000. Using an external rate
of return analysis with a MARR of 10%/yr, what should Calisto Launch
Services do?
Summary 255
72. Tempura, Inc. is considering two projects. Project A requires an investment
of $50,000. Estimated annual receipts for 20 years are $20,000; estimated
annual costs are $12,500. An alternative project, B, requires an investment
of $75,000, has annual receipts for 20 years of $28,000 and annual costs of
$18,000. Assume both projects have a zero salvage value and that MARR is
12%/yr. Based on an external rate of return analysis, which project should be
recommended?
73. Solve problem 35, except use the external rate of return approach.
74. Solve problem 36, except use the external rate of return approach.
75. Solve problem 43, except use the external rate of return approach.
76. Solve problem 44, except use the external rate of return approach.
ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E :
J. B. HU N T T R AN S P O RT S E RVI C E S
J. B. Hunt Transport Services Inc. (JBHT) is one of the largest surface transportation companies in North America. A full truckload transportation firm
that delivers goods all over the United States, its approximately 10,500
drivers make short- and long-haul deliveries. In 2011, JBHT’s consolidated
revenue was $4.5 billion, spread across its four business segments: intermodal (JBI), dedicated contract services (DCS), truck (JBT), and integrated
capacity solutions (ICS). Intermodal is a partnership between JBHT and
most major North American rail carriers to deliver freight throughout the
continental United States, Canada, and Mexico; full containers are delivered by JBI between customers and railroads. In 2011, JBI generated
59 percent of the company’s total revenue. Dedicated contract services
involve partnerships with major corporations such as Home Depot, Wal-Mart,
Procter & Gamble, and Coca-Cola; DCS provides logistics services for its
partners, often driving trucks and trailers showing the customers’ logos
rather than JBHT’s logo; in 2011, DCS accounted for 22 percent of total
revenue. The truck segment, consisting of truckload dry-van shipping, is the
most visible, with its JBHT trucks regularly traveling U.S. interstates and
highways; in 2011, JBT represented 11 percent of total revenue. The newest
segment, integrated capacity solutions, provides transportation solutions
to customers by utilizing third-party carriers, as well as JBHT-owned equipment. In a sense, ICS serves as a freight broker; it purchases transportation
services on behalf of its customers and bills its customers for services provided. ICS services include flatbed, refrigerated, less-than-truckload (LTL),
and a variety of dry-van and intermodal solutions. In 2011, ICS generated
8 percent of total revenue.
With more than 9,100 company-owned tractors and 75,000 trailers and
containers used to ship nearly 3.4 million loads in 2011, JBHT is constantly
faced with decisions regarding the acquisition and replacement of tractors
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REPLACEMENT
ANALYSIS
and trailers. Also, using its own fleet versus contracting with independent
truckers and purchasing versus leasing over-the-road equipment are examples
of decisions the company must make on a regular basis.
In a typical year, JBHT purchases hundreds of new over-the-road tractors, each of which has a retail price of approximately $100,000. In addition, JBHT annually purchases a large quantity of trailers and vans at an
average cost of approximately $25,000. Also, JBHT annually purchases
several hundred intermodal containers for approximately $15,000 each,
plus the chassis to haul the containers (each chassis costs approximately
$13,000).
Tractors tend to be replaced after driving approximately 600,000 miles;
during the life of a tractor, it incurs annual repair and maintenance expenses
in excess of $7,000. Trade-in values on used tractors vary depending on use,
but it is typical for JBHT to receive approximately $25,000 for a used tractor
and $10,000 for a used trailer or van. Clearly, with hundreds of millions of
dollars being spent annually on tractors and trailers, determining the most
economic time to replace its moving equipment is a key to JBHT’s economic
performance.
DISCUSSION QUESTIONS:
1. Do you suppose that JBHT makes replacement decisions for its equipment on an item-by-item level or for a fleet of equipment? What might
be some of the advantages and disadvantages to replacement at the
fleet level?
2. What factors might influence the useful life of the JBHT equipment?
3. Why use miles instead of years (time) as a parameter for replacement?
4. How might safety factor into the replacement analysis decision?
257
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Replacement Analysis
LEARNING OBJECTIVES
When you have finished studying this chapter, you should be able to:
1. Explain the terms defender, challenger, obsolescence, and sunk costs
as they pertain to replacement analysis. (Section 7.1)
2. Perform a replacement analysis using a cash flow approach. (Section 7.2)
3. Perform a replacement analysis using an opportunity cost approach.
(Section 7.2)
4. Determine the optimum replacement interval in cases where an asset
will be used for many years. (Section 7.3)
INTRODUCTION
One of the most common investment alternatives considered by businesses and individuals involves the replacement of an asset. No doubt,
you have made many replacement decisions, such as replacing a car, a
computer, a calculator, a television, a cell phone service provider, and
so forth. Businesses make replacement decisions on an ongoing basis.
Replacement decisions are influenced by economics, capacity, quality of
service provided, changing requirements, prestige, fads, and a host of
other factors.
In Chapters 4, 5, and 6, we examined various ways to compare investment alternatives. We considered two situations: a single alternative
(where the decision was between investing and not investing) and a set
of mutually exclusive investment alternatives (where we recommended
the alternative having the greatest economic worth). In both situations,
the alternatives in question could have been replacement alternatives.
In this chapter, we consider two equivalent approaches that can be
used to perform a replacement analysis: the cash flow approach and the
opportunity cost approach. We conclude the chapter with a consideration
of the optimum replacement frequency for equipment that will be used
year in and year out; such analyses are called optimum replacement
interval analyses.
7-1
FUNDAMENTALS OF
REPLACEMENT ANALYSIS
LEARN I N G O B JEC T I V E : Explain the terms challenger, defender, obsoles-
cence, and sunk costs as they pertain to replacement analysis.
Because decisions to replace versus continuing to use an asset occur so
frequently, a body of literature has evolved on this subject. We use the term
7-1
Fundamentals of Replacement Analysis
replacement analysis when referring to the comparison of investment
alternatives that involve replacing an asset or service provider.
Although replacement decisions occur for a variety of reasons, the
following are typical:
The current asset, which we call the defender, has developed deficiencies, such as high setup cost, excessive maintenance expense, declining productivity, high energy cost, limited capability, or physical
impairment.
2. Potential replacement assets, which we call the challengers, are available and have a number of advantages over the defender, such as new
technology that is quicker to set up and easier to use, lower labor cost,
lower maintenance expense, lower energy cost, higher productivity, or
additional capabilities.
3. A changing external environment, including
a. changing user and customer preferences and expectations;
b. changing requirements;
c. new, alternative ways of obtaining the functionality provided by the
defender, including the availability of leased equipment and thirdparty suppliers; and
d. increased demand that cannot be met with the current equipment—
either supplementary or replacement equipment is required to meet
demand.
1.
Obsolescence is a frequently cited reason for replacing an asset. Various types of obsolescence can occur, including the following:
Functional obsolescence, which can result from physical deterioration of the defender, increased demand that exceeds the defender’s
capacity, or new requirements that the defender cannot meet.
2. Technological obsolescence, which occurs through the introduction
of new technology, such that challengers possess capabilities not present
in the defender.
3. Economic obsolescence, which occurs when the economic worth of
one or more challengers exceeds the defender’s economic worth.
1.
Replacement analyses are basically just another type of alternative
comparison. As such, the same systematic seven-step approach described
in Chapter 1 can be used. Likewise, the consistent measures of economic
worth described previously can be used.
Yet, replacement decisions seem to pose unique difficulties. Why?
Perhaps it is because an “emotional” attachment to present equipment
can occur. Also, perhaps it is due to the presence of sunk costs. Sunk
costs are past costs that have no bearing on current decisions. (A long
259
Replacement Analysis
The comparison of
investment alternatives
that involve replacing an
asset or service provider.
Defender The current
asset being compared in a
replacement analysis.
Challenger The potential
replacement asset being
compared in a replacement
analysis.
Functional Obsolescence
Resulting from physical
deterioration of the
defender, increased
demand that exceeds the
defender’s capacity, or
new requirements that the
defender cannot meet.
Technological Obsolescence
Occurs through the
introduction of new
technology, such that
challengers possess
capabilities not present
in the defender.
Economic Obsolescence
Occurs when the economic
worth of one or more
challengers exceeds the
defender’s economic worth.
Sunk Costs Past costs
that have no bearing on
current decisions.
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Chapter 7
Replacement Analysis
history of maintenance and repair can cause owners to be reluctant to
replace equipment—doing so might appear as an admission that mistakes
were made in the past by keeping the equipment longer than it should
have been.)
Systematic Economic Analysis Technique
1.
2.
3.
4.
5.
6.
7.
Identify the investment alternatives
Define the planning horizon
Specify the discount rate
Estimate the cash flows
Compare the alternatives
Perform supplementary analyses
Select the preferred investment
Based on a DCF comparison, many companies use equipment long
after replacements would be justified economically. Why? Several reasons
come to mind:
1. The firm is currently making a profit, so there is no compelling reason
2.
3.
4.
5.
6.
to invest in new technology.
The current equipment still works and produces a product of acceptable quality—an “if it isn’t broken, don’t replace it” attitude prevails.
There are risks and uncertainties associated with change—replacing
the “tried and true” proven defender with an unproven challenger
is viewed as too risky. (Recall the cartoon character Pogo saying,
“Change is good! You go first!”)
A decision to replace existing technology is a stronger commitment, for a period of time into the future, than continuing to use the
defender—given the rapidly changing world and the growth of outsourcing and offshoring, some managers are reluctant to invest new
capital in domestic operations.
Due to limitations on investment capital, replacements are given secondary consideration over investments that expand operations or add
new capabilities (even though the replacement might have a greater
economic worth).
Uncertainty regarding the future—the defender has a track record
insofar as annual costs are concerned, the challenger is unproven,
estimates of future demand might not materialize, and annual costs
estimates for the challenger could be incorrect.
7-2
Cash Flow and Opportunity Cost Approaches
261
7. The psychological impact of sunk costs—it is very hard for some peo-
ple to ignore what was spent in the past; as a result, they continue to
waste money by supporting equipment they should replace.
8. The technological improvement trap—because technological improvements occur so frequently, some are reluctant to replace the defender with this year’s challenger, because next year’s challenger will
be less expensive and will have greater capability than this year’s.
9. Some companies prefer to be “technology followers” instead of “technology leaders”—they want others to debug the new generation of
technology, and since new technology is being introduced so rapidly,
they can never bring themselves to make the replacement decision.
10. Management is concerned about taking a hit on the quarterly financial
statement by writing off an asset that is not fully depreciated, regardless of the DCF analysis.
7-2
CASH FLOW AND OPPORTUNITY
COST APPROACHES
LEARNING O BJECTI VES: Perform a replacement analysis using a cash flow
approach. Perform a replacement analysis using an opportunity cost approach.
As noted, two equivalent approaches in a replacement analysis are the cash
flow and the opportunity cost approaches. The two approaches differ in
how the market value for the currently owned asset is treated if replacement occurs and if replacement does not occur.
The cash flow approach, also called the insider’s viewpoint approach,
“follows the money.” Specifically, with the cash flow approach, for each
replacement alternative, cash flows are shown for each alternative for
each year in the planning horizon. If replacement does not occur, the
cash flow shown in year zero is often zero. If money must be expended
in order to continue the use of the current asset, however, then a negative-valued cash flow is shown in year zero. On the other hand, if a
replacement occurs, then the amount of money received in trade-in or the
amount of money required to dispose of the current asset is shown in
year zero.
With the opportunity cost approach, the salvage value or market
value of the current asset is treated as its investment cost if it is retained.
Opportunity cost refers to the cost of a foregone alternative (or opportunity) that is incurred in order to pursue another alternative. In the opportunity cost approach to replacement analysis, by deciding to keep the asset,
one gives up the opportunity to receive a monetary amount for it. The opportunity cost approach is also known as the outsider’s viewpoint approach.
Salvage Value The
estimated value of an asset
at the end of its useful
life. Also referred to as the
market value.
Opportunity Cost
The cost of a forgone
alternative (or opportunity)
that is incurred in order to
pursue another alternative.
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Equivalent Uniform Annual
Cost The EUAC is the
equivalent uniform annual
cost to keep or operate
an asset.
An example will illustrate how the cash flow and opportunity cost
approaches are performed. This example introduces a new term, EUAC, or
equivalent uniform annual cost. The EUAC is the annual cost to keep or
operate the asset. It is similar to the annual worth, AW.
EXAMPLE
Cash Flow Approach to Replacement Analysis
Video Example
KEY DATA
A chemical mixer was purchased 8 years ago for $100,000. If retained, it
will require an investment of $50,000 to upgrade it; if upgraded, it will
cost $35,000/year to operate and maintain (O&M) and will have a negligible salvage value after 5 years. A new mixer can be purchased for
$120,000; it will have an annual O&M cost of $15,000 and a salvage
value of $40,000 after 5 years. Alternatively, a mixer can be leased with
5 beginning-of-year lease payments of $20,000; O&M costs will be
$18,000/year. If the mixer is replaced, the old mixer can be sold on the
used equipment market for $15,000. Using an insider’s approach, what
are a) the EUAC of keeping the current mixer, b) the EUAC of replacing
with a new mixer, and c) the EUAC of replacing with a leased mixer? The
MARR is 10%.
Given: MARR 5 10%. Using a cash flow or insider’s approach, the cash
flows are shown below for each replacement alternative.
EOY
CF(Keep)
CF(Replace)
CF(Lease)
0
2$50,000.00
2$105,000.00
2$5,000.00
1–4
2$35,000.00
2$15,000.00
2$38,000.00
5
2$35,000.00
$25,000.00
2$18,000.00
Find: EUAC for each alternative.
SOLUTION
The EUACs for each alternative are as follows:
a) EUAC(Keep) 5 $35,000 1 $50,000(A Z P 10%,5) 5 $35,000 1 $50,000
(0.26380) 5 $48,190.00
5 PMT110%,5,2500002 1 35000 5 48,189.87
b) EUAC(Replace) 5 $15,000 1 $105,000(A Z P 10%,5) – $40,000
(A Z F 10%,5) 5 $15,000 1 $105,000(0.26380) – $40,000(0.16380) 5
$36,147.00
5 PMT110%,5,2105000,400002 1 15000 5 $36,146.84
7-2 Cash Flow and Opportunity Cost Approaches
c) EUAC(Lease) 5 $18,000 1 $20,000(F Z P 10%,1) – $15,000(A Z P 10%,5) 5
$18,000 1 $20,000(1.10) – $15,000(0.26380) 5 $36,043.00
5 PMT110%,5,150002 1 18000 2 FV110%,1,,200002 5 $36,043.04
Notice, the $100,000 purchase price for the mixer is a sunk cost. It does
not affect the cash flows over the planning horizon.
Opportunity Cost Approach to Replacement Analysis
EXAMPLE
The replacement problem in Example 7.1 can also be addressed by the
opportunity cost approach.
Given: MARR 5 10%. Using an opportunity cost or outsider’s viewpoint
approach, the cash flows are shown below for each alternative.
EOY
CF(Keep)
CF(Replace)
CF(Lease)
0
2$65,000.00
2$120,000.00
2$20,000.00
1–4
2$35,000.00
2$15,000.00
2$38,000.00
5
2$35,000.00
$25,000.00
2$18,000.00
KEY DATA
Find: EUAC for each alternative.
The EUACs for each alternative are as follows:
a) EUAC(Keep) 5 $35,000 1 $65,000(A Z P 10%,5) 5 $35,000 1 $65,000
(0.26380) 5 $52,147.00
5 PMT110%,5,2650002 1 35000 5 $52,146.84
b) EUAC(Replace) 5 $15,000 1 $120,000(A Z P 10%,5) – $40,000
(A Z F 10%,5) 5 $15,000 1 $120,000(0.26380) – $40,000(0.16380) 5
$40,104.00
5 PMT110%,5,2120000,400002 1 15000 5 $40,103.80
c) EUAC(Lease) 5 $18,000 1 $20,000(F Z P 10%,1) 5 $18,000 1 $20,000
(1.10) 5 $40,000.00
5 18000 2 FV(10%,1,,20000) 5 $40,000.00
SOLUTION
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Chapter 7
Replacement Analysis
EXPLORING THE
SOLUTION
Notice, the same conclusion is reached with the two approaches:
leasing has the lowest equivalent uniform annual cost. Furthermore,
since the EUAC values obtained using the two approaches differ by
$15,000(A Z P10%,5) 5 $3,957.00 or 5 PMT(10%,5,215000) 5 $3,956.96,
the difference in the EUAC(Keep) and either the EUAC(Replace) or the
EUAC(Lease) will be the same for the cash flow and the opportunity cost
approaches:
EUAC(K–R) 5 $52,147.00 – $40,104.00 5 $12,043.00
(opportunity cost approach)
5 $48,190.00 – $36,147.00 5 $12,043.00
(cash flow approach)
EUAC(K–L) 5 $52,147.00 – $40,000.00 5 $12,147.00
(opportunity cost approach)
5 $48,190.00 – $36,043.00 5 $12,147.00
(cash flow approach).
Although the two approaches are equivalent, we prefer the cash flow
approach. We believe it is more direct, especially when multiple replacement candidates are available and each offers a different trade-in value for
the current asset. Also, when income taxes are being considered, the cash
flow approach, in our view, is simpler to use than the opportunity cost
approach. Consequently, unless we specify otherwise, in solving the problems at the end of the chapter, use the cash flow approach.
7-3
OPTIMUM REPLACEMENT INTERVAL
LEARN I N G O B JEC T I V E : Determine the optimum replacement interval in
cases where an asset will be used for many years.
Optimum Replacement
Interval (ORI) The interval
at which an asset should
be replaced to minimize
cost (or maximize worth).
In this section, we consider situations in which a particular asset will be
needed for an indefinite period of time. When it wears out, it will be replaced
by an identical asset, as many times as necessary. Examples would include
over-the-road trailers, lift trucks, a basic machine tool, and rental cars,
among others. For engineering economic analysis, we would like to know
the optimum replacement interval for such an asset—the interval at
which it should be replaced to minimize cost or maximize worth.
As equipment ages, operating and maintenance (O&M) costs tend to
increase. At the same time, the cost of ownership tends to decrease with
7-3
Optimum Replacement Interval
Dollars/year
Equivalent uniform annual cost
Annual operating
and maintenance cost
Capital recovery cost
Life, years
EUAC Components Used to Determine the Optimum
Replacement Interval
FIGURE 7.1
increased usage. The sum of the ownership and operating costs can often
be represented by Figure 7.1. In this case, the equivalent uniform annual
cost is a convex function of the life of the equipment.
The cost of ownership is often called the capital recovery cost and is
computed as follows:
CR 5 P1A Z P i%,n2 2 F1A Z F i%,n2
(7.1)
where P is the initial investment in the equipment, F is its salvage value
after n years of use, and the time value of money is i%.
By converting the acquisition and disposal cost (salvage value) to a
uniform annual series and by converting the O&M cost to a uniform
annual series, one obtains the EUAC for a given value of n. By enumerating or searching over n, the optimum replacement interval (ORI) is
obtained.
Computing the Optimum Replacement Interval
Five hundred thousand dollars is invested in a surface mount placement
machine. Experience shows that the salvage value for the SMP machine
decreases by 25% per year. Hence, its salvage value equals $500,000(0.75n)
after n years of use. O&M costs increase at a rate of 15% per year; the
O&M cost equals $125,000 the first year of use. Based on a MARR of
10%, what is the ORI and what is the minimum EUAC (EUAC*)?
Video Lesson:
Capital Recovery Cost
Capital Recovery Cost
represents the cost of
ownership, a value that
typically decreases with
increased usage.
EXAMPLE
Video Example
265
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Chapter 7
Replacement Analysis
KEY DATA
SOLUTION
Given: Initial investment (P) 5 $500,000, Salvage value (F ) 5
$500,000(0.75n), O&M 5 $125,000 in year one, increasing 15% per year
Find: ORI and EUAC*
For the example problem, EUAC 5 [$500,000 1 $125,000(P Z A1
10%,15%,n)](A Z P 10%,n) – $500,000(0.75n)(A Z F 10%,n). As shown below,
ORI 5 5 years and EUAC* 5 $276,611.
n
(A|P 10%,n)
(P |A1 10%,15%,n)
(A|F 10%,n)
EUAC
4
0.31547
3.89190
0.21547
$277,119
5
0.26380
4.97789
0.16380
$276,611
6
0.22961
6.11325
0.12691
$278,729
From Figure 7.1, will the ORI increase or decrease if the slope of the
annual O&M cost curve increases? In general, the more expensive it becomes
to operate and maintain equipment, the sooner it needs to be replaced.
Hence, when the slope of the annual O&M cost curve is increased, the ORI
will tend to decrease (due to integer values of n, the ORI might remain
unchanged for some changes in the slope of the annual O&M cost curve).
From Figure 7.1, will the ORI increase or decrease if the initial investment increases? In general, the more expensive an asset (all other parameters
remaining constant), the longer it will be used. Hence, when the initial
investment is increased, the ORI will tend to increase.
By varying other parameters in Example 7.3, you will find that increasing the MARR tends to increase ORI and increasing the rate of decrease in
salvage value tends to increase ORI.
Three observations come to mind regarding ORI computations.
Implicit in the solution procedure we used to obtain the ORI is an
assumption that the planning horizon equals an integer multiple of the
ORI value obtained.
2. Subsequent replacements must have cash flow profiles that are identical to their predecessors.
3. For the ORI to be truly optimal, it must minimize present worth over
the planning horizon.
1.
Suppose the planning horizon is fixed and greater than the ORI. How
might one determine the optimum time to replace the asset? The following
example addresses this scenario.
7-3
Optimum Replacement Interval
Optimum Replacement Timing with a Given Planning Horizon
EXAMPLE
For the SMP machine in the previous example, suppose a planning
horizon of 9 years is to be used. Recall, we obtained an ORI of 5 years.
Since the planning horizon is not an integer multiple of 5, which of the
following replacement sequences should we use: {5,4}; {6,3}; {7,2};
{8,1}; {9,0}?
Given: Initial investment (P) 5 $500,000, Salvage value (F ) 5
$500,000(0.75n), O&M 5 $125,000 in year one, increasing 15% per year,
Planning horizon 5 9 years
Find: The replacement sequence that minimizes PW of costs given a
9-year planning horizon.
KEY DATA
Shown below is the EUAC for values of n ranging from 1 to 9.
SOLUTION
n
(A|P 10%,n)
(P |A1 10%,15%,n)
(A|F 10%,n)
EUAC
1
1.10000
0.90909
1.00000
$300,000
2
3
0.57619
0.40211
1.85950
2.85312
0.47619
0.30211
$288,095
$280,737
4
5
0.31547
0.26380
3.89190
4.97789
0.21547
0.16380
$277,119
$276,611
6
7
0.22961
0.20541
6.11325
7.30022
0.12961
0.10541
$278,729
$283,112
8
9
0.18744
0.17364
8.54113
9.83846
0.08744
0.07364
$289,462
$297,599
For a replacement sequence {x, y}, the present worth over the 9-year planning horizon is given by
PW{x, y} 5 EUAC(x)(P Z A 10%,x) 1 EUARC(y)(P Z A 10%,y)(P Z F 10%,x)
Hence, the present worth for each of the sequences is as shown below.
For a 9-year planning horizon, the first SMP machine should be replaced
after 5 years and the successor should be used for the balance of the
planning horizon.
{x, y}
PW{x, y}
{5,4}
$1,593,998.52
{6,3}
{7,2}
$1,608,014.16
$1,634,857.37
{8,1}
{9,0}
$1,671,520.06
$1,713,885.95
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Chapter 7
Replacement Analysis
Perhaps you wondered why we did not consider the sequences {1,8},
{2,7}, {3,6}, and {4,5}. Based on the time value of money, the replacement interval having the smallest EUAC should be placed first in the
sequence. An examination of the EUAC for various values of n shows that
we always chose {x, y} sequences such that EUAC(x) , EUAC(y).
SUMMARY
KEY CONCEPTS
1. Learning Objective: Explain the terms defender, challenger, obsolescence, and sunk costs as they pertain to replacement analysis. (Section 7.1)
In a replacement analysis, the defender (existing asset) is compared to one
or more challengers (potential replacement assets). Often, replacement
analysis is undertaken because of the obsolescence of the defender—it
lacks needed functional or technological capabilities, or its economic
worth is less than that of the challenger(s). Sunk costs are past costs associated with the defender and have no bearing on current decisions.
2. Learning Objective: Perform a replacement analysis utilizing a cash flow
approach. (Section 7.2)
The cash flow approach, also called the insider’s viewpoint approach,
“follows the money.” By this it is meant that for each replacement alternative, cash flows are shown each year for each alternative throughout the
planning horizon. The cash flow and the opportunity cost approaches for
replacement analysis are equivalent; however, the cash flow approach is
preferred because it is more direct and simpler to evaluate when income
taxes are considered.
3. Learning Objective: Perform a replacement analysis utilizing an opportunity cost approach. (Section 7.2)
In the opportunity cost approach, also called the outsider’s point viewpoint
approach, the salvage value of the current asset is treated as its investment
cost if it is retained; thus, by deciding to keep the asset, one gives up the
opportunity to receive a monetary amount for it.
4. Learning Objective: Determine the optimum replacement interval in
cases where an asset will be used for many years. (Section 7.3)
Often an asset is used for many years, for example in the case of over-theroad trailers and rental cars. In this situation, it is common to assume an
Summary 269
identical asset that will be used over an indefinite period of time. We
assume that the planning horizon equals an integer multiple of the ORI
value obtained, future cash flow profiles are identical to their predecessors,
and the present worth must be minimized over the planning horizon.
KEY TERMS
Capital Recovery Cost, p. 265
Challenger, p. 259
Defender, p. 259
Economic Obsolescence, p. 259
Equivalent Uniform Annual Cost
(EUAC), p. 262
Functional Obsolescence, p. 259
Opportunity Cost, p. 261
Optimum Replacement
Interval (ORI), p. 264
Replacement Analysis, p. 259
Salvage Value, p. 261
Sunk Costs, p. 259
Technological Obsolescence, p. 259
Problem available in WileyPLUS
GO Tutorial Tutoring Problem available in WileyPLUS
Video Solution Video Solution available in WileyPLUS
FE-LIKE PROBLEMS
1.
A company owns a 6-year old gear hobber that has a book value of
$60,000. The present market value of the hobber is $80,000. A new gear hobber can be purchased for $450,000. Using an insider’s point of view, what is
the net first cost of purchasing the new gear hobber?
a. $310,000
c. $390,000
b. $370,000
d. $450,000
2.
A company owns a 6-year old gear hobber that has a book value of $60,000.
The present market value of the hobber is $80,000. A new gear hobber can be
purchased for $450,000. Using an outsider’s point of view, what is the net first
cost of purchasing the new gear hobber?
a. $310,000
c. $390,000
b. $370,000
d. $450,000
3.
In evaluating a piece of equipment for its optimum replacement interval,
the following table of equivalent uniform annual costs is obtained. What is
the optimum replacement interval for the equipment?
270
Chapter 7
Replacement Analysis
n
EUAC(n)
1
2
3
4
1582.00
1550.00
1575.00
1580.00
a. 1 year
b. 2 years
4.
A radiology clinic is considering buying a new $700,000 x-ray machine,
which will have no salvage value after installation since the cost of removal
will be approximately equal to its sales value. Maintenance is estimated at
$24,000 per year as long as the machine is owned. After ten years the x-ray
source will be depleted and the machine must be scrapped. Which of the
following represents the most economic life of this x-ray machine?
a.
b.
c.
d.
5.
c. 3 years
d. 4 years
One year, since it will have no salvage after installation
Five years, since the maintenance costs are constant
Ten years, because maintenance costs don’t increase
Cannot be determined from the information given
Which of the following is not an approach to replacement analysis?
c. outsider viewpoint
d. supply chain approach
a. cash flow approach
b. insider viewpoint
6.
The most common approach to determining the planning horizon for
replacement analysis is which of the following?
a. Shortest life
c. Longest life
b. Median Life
d. Least Common Multiple
7.
A company owns a 5-year old turret lathe that has a book value of $20,000.
The present market value of the lathe is $16,000. A new turret lathe can be
purchased for $45,000. Using a before tax analysis and an outsider’s point of
view, what is the first cost of keeping the old lathe?
a. $29,000
c. $20,000
b. $45,000
d. $16,000
8.
A company owns a 5-year old turret lathe that has a book value of $20,000.
The present market value of the lathe is $16,000. A new turret lathe can be
purchased for $45,000. Using a before tax analysis and an insider’s point of
view, what is the first cost of the new lathe?
a. $29,000
c. $25,000
b. $45,000
d. $16,000
9.
What two cost categories form the trade off that leads to an optimum
replacement interval?
a. Direct costs and indirect costs
b. Insider costs and outsider costs
c. Operating & maintenance costs and capital recovery costs
d. Sunk costs and opportunity costs
Summary 271
10.
Increasing the magnitude of the initial investment tends to ____________
the optimum replacement interval.
a. Decrease
c. Reverse
b. Increase
d. Not affect
PROBLEMS
Section 7.1 Fundamentals of Replacement Analysis
1. Identify something you own, perhaps even something you still use regularly.
a. Give a list of at least 6 potential reasons why you might consider replacing
the identified item.
b. Identify at least two possible “challengers” that you might consider to
replace the item, and tell for each challenger the main reason you would
consider it.
2. Give two examples each of (i) functional obsolescence, (ii) technological
obsolescence, and (iii) economic obsolescence for items that you or your
family own.
3. Ten reasons why companies use equipment long after replacements would
be justified economically are given in the text. In many cases, these reasons
do not apply just to companies; rather, they apply to us as individuals. Give
specific examples of at least 3 of the 10 reasons that apply to things owned by
you or your family.
Section 7.2 Cash Flow and Opportunity Cost Approaches
4. The Container Corporation of America is considering replacing an automatic
painting machine purchased 9 years ago for $700,000. It has a market value
today of $40,000. The unit costs $350,000 annually to operate and maintain.
A new unit can be purchased for $800,000 and will have annual O&M costs
of $120,000. If the old unit is retained, it will have no salvage value at the
end of its remaining life of 10 years. The new unit, if purchased, will have a
salvage value of $100,000 in 10 years. Analyze this using an EUAC measure
and a MARR of 20% to perform a before-tax analysis to see if the automatic
painting machine should be replaced if the old automatic painting machine is
taken in as a trade-in for its market value of $40,000.
a. Use the cash flow approach (insider’s viewpoint approach).
b. Use the opportunity cost approach (outsider’s viewpoint approach).
5.
A specialty concrete mixer used in construction was purchased for
$300,000 7 years ago. Its annual O&M costs are $105,000. At the end of
the 8-year planning horizon, the mixer will have a salvage value of $5,000.
If the mixer is replaced, a new mixer will require an initial investment of
$375,000 and at the end of the 8-year planning horizon, the new mixer will
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have a salvage value of $45,000. Its annual O&M cost will be only $40,000
due to newer technology. Analyze this using an EUAC measure and a MARR
of 15% to see if the concrete mixer should be replaced if the old mixer is sold
for its market value of $65,000.
a. Use the cash flow approach (insider’s viewpoint approach).
b. Use the opportunity cost approach (outsider’s viewpoint approach).
6. Online Educators (OE), a not-for-profit firm exempt from taxes, is consider-
ing replacement of some electronic equipment associated with its distance
learning (DL) facility. They spent $100,000 on the equipment 3 years ago and
have depreciated it to a current book value of $40,000. Its end-of-year O&M
costs are $9000 per year. Today’s newer technology, including high definition
digital TV, has a price tag of $125,000 and, after some negotiation, OE negotiates a price of either (1) $108,000 in cash or (2) $91,000 in cash plus the
current equipment as a trade-in. OE checked around and determined it could
do no better by selling the soon-to-be-obsolete equipment to someone else.
OE will use a 5-year planning horizon. If the newer equipment is purchased,
it will have end-of-year O&M costs of $8,000 and a salvage value of $20,000
at that time. If the old equipment is retained, it will have to be supplemented
in years 3, 4, and 5 by leasing a hi-def add-on unit costing $30,000 per year
payable at the beginning of the year with additional end-of-year O&M costs
of $7,000. The old equipment will also have no salvage value at the end of the
planning horizon. MARR is 10%.
a. What is the market value of the old equipment?
b. Use the cash flow approach (insider’s viewpoint approach) to determine
whether to keep or replace the current equipment.
c. What is the market value of the old equipment?
d. Use the opportunity cost approach (outsider’s viewpoint approach) to
determine whether to keep or replace the current equipment.
7.
A currently-owned shredder for use in a refuse-powered electrical generating plant has a present net realizable value of $200,000 and is expected to
have a market value of $10,000 after 4 years. Operating and maintenance disbursements are $100,000 per year. An equivalent shredder can be leased for
$200 per day plus $80 per hour of actual use as determined by an hour meter,
with both components assumed to be paid at the end of the year. Actual use
is expected to be 1500 hours and 250 days per year. Using a 4 year planning
horizon, a before-tax analysis, and a MARR of 15%, determine the preferred
alternative using the annual cost criterion.
a. Consider only the above information and use the cash flow approach
(insider’s viewpoint approach).
b. Consider the additional information that the annual cost of operating
without a shredder is $190,000.
c. Consider only the above information and use the opportunity cost approach
(outsider’s viewpoint approach).
d. Consider the additional information that the annual cost of operating
without a shredder is $190,000 and use the opportunity cost approach
(outsider’s viewpoint approach).
Summary 273
8. Dell is considering replacing one of its material handling systems. It has an
annual O&M cost of $48,000, a remaining operational life of 8 years, and
an estimated salvage value of $6,000 at that time. A new system can be purchased for $175,000. It will be worth $50,000 in 8 years, and it will have
annual O&M costs of only $17,000 per year due to new technology. If the
new system is purchased, the old system will be traded in for $55,000, even
though the old system can be sold for only $45,000 on the open market. Leasing a new system will cost $31,000 per year, payable at the beginning of the
year, plus operating costs of $15,000 per year payable at the end of the year.
If the new system is leased, the existing material handling system will be sold
for its market value of $45,000. Use a planning horizon of 8 years, an annual
worth analysis, and MARR of 15% to decide which material handling system
to recommend: (i) keep existing, (ii) trade in existing and purchase new, or
(iii) sell existing and lease.
a. Use the cash flow approach (insider’s viewpoint approach).
9.
Allen Construction purchased a crane 6 years ago for $130,000. They
need a crane of this capacity for the next five years. Normal operation costs
$35,000 per year. The current crane will have no salvage value at the end
of 5 more years. Allen can trade in the current crane for its market value of
$40,000 toward the purchase of a new one that costs $150,000. The new crane
will cost only $8000 per year under normal operating conditions and will
have a salvage value of $55,000 after 5 years. If MARR is 20%, determine
which option is preferred.
a. Use the cash flow approach (insider’s viewpoint approach).
b. Use the opportunity cost approach (outsider’s viewpoint approach).
10. A division of Raytheon owns a 5-year old turret lathe that has a non-tax
book value of $24,000. It has a current market value of $18,000. The expected decline in market value is $3,000 per year from this point forward to
a minimum of $3,000. O&M costs are $8,000 per year. Additional capability is needed. If the old lathe is kept, that new capability will be contracted
out for $13,000, assumed payable at the end of each year. A new turret
lathe has the increased capability to fulfill all needs, replacing the existing turret lathe and requiring no outside contracting. It can be purchased
for $65,000 and will have an expected life of 8 years. Its market value is
expected to be $65,000(0.7t ) at the end of year t. Annual O&M costs are
expected to equal $10,000. MARR is 15% and the planning horizon is
8 years.
a. Clearly show the cash flow profile for each alternative using a cash flow
approach (insider’s viewpoint approach).
b. Using an EUAC and a cash flow approach, decide which is the more favorable alternative.
c. Clearly show the cash flow profile for each alternative using an opportunity cost approach (outsider’s viewpoint approach).
d. Using an EUAC comparison and an opportunity cost approach, decide
which is the more favorable alternative.
274 Chapter 7
Replacement Analysis
11.
GO Tutorial Five years ago, a multi-axis NC machine was purchased for
the express purpose of machining large, complex parts used in commercial
and military aircraft worldwide. It cost $350,000, had an estimated life of
15 years, O&M costs of $50,000 per year. It was originally thought to have a
salvage value of $20,000 at the end of 15 years, but is now believed to have
a remaining life of 5 years with no salvage value at that time. With business
booming, the existing machine is no longer sufficient to meet production needs.
It can be kept and supplemented by purchasing a new smaller Machine S for
$210,000 that will cost $37,000 per year for O&M, have a life of 10 years,
and a salvage value of $210,000(0.8t ) after t years. As an alternative, a larger,
faster, and more capable Machine L can be used alone to replace the current
machine. It has a cash price without trade-in of $450,000, O&M costs of
$74,000 per year, a salvage value of $450,000(0.8t ) after t years, and a 15 year
life. The present machine can be sold on the open market for a maximum of
$70,000. MARR is 20% and the planning horizon is 5 years.
a. Clearly show the cash flow profile for each alternative using a cash flow
approach (insider’s viewpoint approach).
b. Using an EUAC and a cash flow approach, decide which is the more favorable alternative.
c. Clearly show the cash flow profile for each alternative using an opportunity cost approach (outsider’s viewpoint approach).
d. Using an EUAC comparison and an opportunity cost approach, decide
which is the more favorable alternative.
12. Five years ago, ARCHON, a regional architecture/contractor firm, purchased
an HVAC unit for $25,000 which was expected to last 15 years. It will have
a salvage value of $0 in 10 more years. The annual operating cost of this
unit started at $2,000 in the first year and has increased steadily at $250 per
year ever since; last year the cost was $3,000. Its book value is now $13,000.
ARCHON has been phenomenally successful due to their reputation for highly
functional, high quality, cost effective designs and construction. They are building a new wing at their regional headquarters to accommodate a much larger
computer design emphasis requiring larger, faster computers, architectural
printers, e-storage for a construction repository of previous designs, and an
increased human heat load. They can buy an additional unit to air condition
the new wing for $18,000. It will have a service life of 15 years, net salvage of
$0 at that time, and a $3,000 market value after 10 years. It will have annual
operating costs of $1,800 in the first year, increasing at $100 per year. As an
alternative, ARCHON can buy a new unit to heat and cool the entire building
for $35,000. It will last for 15 years, have a net salvage of $0 at that time; however, it will have a market value of $8,500 after 10 years. It will have first-year
operating costs of $3,700/year, increasing at $200 per year. The present unit can
be sold now for $7,000. MARR is 20%, and the planning horizon is 10 years.
a. Clearly show the cash flow profile for each alternative using a cash flow
approach (insider’s viewpoint approach).
b. Using a PW analysis and a cash flow approach, decide which is the more
favorable alternative.
Summary 275
c. Clearly show the cash flow profile for each alternative using an opportu-
nity cost approach (outsider’s viewpoint approach).
d. Using a PW analysis and an opportunity cost approach, decide which is
the more favorable alternative.
13.
Video Solution Clear Water Company has a down-hole well auger that
was purchased 3 years ago for $30,000. O&M costs are $13,000 per year.
Alternative A is to keep the existing auger. It has a current market value of
$12,000 and it will have a $0 salvage value after 7 more years. Alternative B
is to buy a new auger that will cost $54,000 and will have a $14,000 salvage
value after 7 years. O&M costs are $6,000 for the new auger. Clear Water
can trade in the existing auger on the new one for $15,000. Alternative C
is to trade in the existing auger on a “treated auger” that requires vastly
less O&M cost at only $3,000 per year. It costs $68,000, and the trade-in
allowance for the existing auger is $17,000. The “treated auger” will have
a $18,000 salvage value after 7 years. Alternative D is to sell the existing
auger on the open market and to contract with a current competitor to use
their equipment and services to perform the drilling that would normally
be done with the existing auger. The competitor requires a beginning-ofyear retainer payment of $10,000. End-of-year O&M cost would be $6,000.
MARR is 15% and the planning horizon is 7 years.
a. Clearly show the cash flow profile for each alternative using a cash flow
approach (insider’s viewpoint approach).
b. Using an EUAC and a cash flow approach, decide which is the more favor-
able alternative.
c. Clearly show the cash flow profile for each alternative using an opportu-
nity cost approach (outsider’s viewpoint approach).
d. Using an EUAC comparison and an opportunity cost approach, decide
which is the more favorable alternative.
14. Apricot Computers is considering replacing its material handling system and
either purchasing or leasing a new system. The old system has an annual operating and maintenance cost of $32,000, a remaining life of 8 years, and an
estimated salvage value of $5,000 at that time.
A new system can be purchased for $250,000; it will be worth $25,000 in
8 years; and it will have annual operating and maintenance costs of $18,000/
year. If the new system is purchased, the old system can be traded in for
$20,000.
Leasing a new system will cost $26,000/year, payable at the beginning of the year, plus operating costs of $9,000/year, payable at the end
of the year. If the new system is leased, the old system will be sold for
$10,000.
MARR is 15%. Compare the annual worths of keeping the old system,
buying a new system, and leasing a new system based upon a planning horizon of 8 years.
a. Use the cash flow approach.
b. Use the opportunity cost approach.
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Chapter 7
Replacement Analysis
15. Fluid Dynamics Company owns a pump that it is contemplating replacing.
The old pump has annual operating and maintenance costs of $8,000/year:
it can be kept for 4 years more and will have a zero salvage value at that
time.
The old pump can be traded in on a new pump. The trade-in value is
$4,000. The new pump will cost $18,000 and have a value of $9,000 in
4 years and will have annual operating and maintenance costs of $4,500/
year.
Using a MARR of 10%, evaluate the investment alternative based upon
the present worth method and a planning horizon of 4 years.
a. Use the cash flow approach.
b. Use the opportunity cost approach.
16. A company owns a 5-year old turret lathe that has a book value of $25,000.
The present market value for the lathe is $16,000. The expected decline in
market value is $2,000/year to a minimum market value of $4000. maintenance plus operating costs for the lathe equal $4,200/year.
A new turret lathe can be purchased for $45,000 and will have an expected
life of 8 years. The market value for the turret lathe is expected to equal
$45,000(0.70)k at the end of year k. Annual maintenance and operating cost is
expected to equal $1,600. Based on a 12% before-tax MARR, should the old
lathe be replaced now? Use an equivalent uniform annual cost comparison, a
planning horizon of 7 years, and the cash flow approach.
17. Esteez Construction Company has an overhead crane that has an estimated
remaining life of 7 years. The crane can be sold for $14,000. If the crane is
kept in service it must be overhauled immediately at a cost of $6,000. Operating and maintenance costs will be $5,000/year after the crane is overhauled.
After overhauling it, the crane will have a zero salvage value at the end of the
7-year period. A new crane will cost $36,000, will last for 7 years, and will
have a $8000 salvage value at that time. Operating and maintenance costs are
$2,500 for the new crane. Esteez uses an interest rate of 15% in evaluating
investment alternatives. Should the company buy the new crane based upon
an annual cost analysis?
a. Use the cash flow approach.
b. Use the opportunity cost approach.
18. A small foundry is considering the replacement of a No. 1 Whiting cupola
furnace that is capable of melting gray iron only with a reverberatory-type
furnace that can melt gray iron and nonferrous metals. Both furnaces have
approximately the same melting rates for gray iron in pounds per hour. The
foundry company plans to use the reverberatory furnace, if purchased, primarily for melting gray iron, and the total quantity melted is estimated to be
about the same with either furnace. Annual raw material costs would therefore be about the same for each furnace. Available information and cost estimates for each furnace are given below.
Cupola furnace. Purchased used and installed 8 year ago for a cost of $20,000.
The present market value is determined to be $8,000. Estimated remaining life
Summary 277
is somewhat uncertain but, with repairs, the furnace should remain functional
for 7 years more. If kept 7 years more, the salvage value is estimated as
$2,000 and average annual expenses expected are:
Fuel
Labor (including maintenance)
Payroll taxes
Taxes and insurance on furnace
Other
$35,000
$40,000
10% of direct labor costs
1% of purchase price
$16,000
Reverberatory furnace. This furnace costs $32,000. Expenses to remove the
cupola and install the reverberatory furnace are about $2,400. The new furnace has an estimated salvage value of $3,000 after 7 years of use and annual
expenses are estimated as:
Fuel
Labor (operating)
Payroll taxes
Taxes and insurance on furnace
Other
$29,000
$30,000
10% of direct labor costs
1% of purchase price
$16,000
In addition, the furnace must be relined every 2 years at a cost of $4,000/
occurrence. If the foundry presently earns an average of 15% on invested
capital before income taxes, should the cupola furnace be replaced by the
reverberatory furnace?
a. Use the cash flow approach.
b. Use the opportunity cost approach.
19. Metallic Peripherals, Inc. has received a production contract for a new
product. The contract lasts for 5 years. To do the necessary machining operations, the firm can use one of its own lathes, which was purchased 3 years
ago at a cost of $16,000. Today the lathe can be sold for $8,000. In 5 years
the lathe will have a zero salvage value. Annual operating and maintenance
costs for the lathe are $4,000/year. If the firm uses its own lathe it must also
purchase an additional lathe at a cost of $12,000, its value in 5 years will
be $3,000. The new lathe will have annual operating and maintenance costs
of $3,500/year.
As an alternative, the presently owned lathe can be traded in for $10,000
and a new lathe of larger capacity purchased for a cost of $24,000; its value
in 5 years is estimated to be $8,000, and its annual operating and maintenance
costs will be $6,000/year.
An additional alternative is to sell the presently owned lathe and subcontract the work to another firm. Company X has agreed to do the work for the
5-year period at an annual cost of $12,000/end-of-year.
Using a 15% interest rate, determine the least-cost alternative for performing the required production operations.
a. Use the cash flow approach.
b. Use the opportunity cost approach
278
Chapter 7
Replacement Analysis
20. A machine was purchased 5 years ago at $12,000. At that time, its estimated
life was 10 years with an estimated end-of-life salvage value of $1,200. The
average annual operating and maintenance costs have been $14,000 and are
expected to continue at this rate for the next 5 years. However, average annual revenues have been and are expected to be $20,000. Now, the firm can
trade in the old machine for a new machine for $5000. The new machine has
a list price of $15,000, an estimated life of 10 years, annual operating plus
maintenance costs of $7,500, annual revenues of $13,000, and salvage values
at the end of the jth year according to
Sj 5 $15,000 2 $1,500j,
for j 5 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Determine whether to replace or not by the annual worth method using
a MARR equal to 15% compounded annually. Use a 5-year planning horizon
and the cash flow approach.
21. A building supplies distributor purchased a gasoline-powered fork-lift truck 4
years ago for $8,000. At that time, the estimated useful life was 8 years with
a salvage value of $800 at the end of this time. The truck can now be sold for
$2,500. For this truck, average annual operating expenses for year j have been
Cj 5 $2,000 1 $4001 j 2 12
Now the distributor is considering the purchase of a smaller battery-powered
truck for $6,500. The estimated life is 10 years, with the salvage value
decreasing by $600 each year. Average annual operating expenses are
expected to be $1,200. If a MARR of 10% is assumed and a 4-year planning
horizon is adopted, based on a cash flow approach should the replacement
be made now?
22. Kwik-Kleen Car Wash has been experiencing difficulties in keeping its
equipment operational. The owner is faced with the alternative of overhauling the present equipment or replacing it with new equipment. The
cost of overhauling the present equipment is $8,500. The present equipment has annual operating and maintenance costs of $7,500. If it is overhauled, the present equipment will last for 5 years more and be scrapped at
zero value. If it is not overhauled, it has a trade-in value of $3,200 toward
the new equipment.
New equipment can be purchased for $28,000. At the end of 5 years the
new equipment will have a resale value of $12,000. Annual operating and
maintenance costs for the new equipment will be $3,000.
Using a MARR of 12%, what is your recommendation to the owner of
the car wash? Base your recommendation on a present worth comparison and
the cash flow approach.
23. A firm is contemplating replacing a computer they purchased 3 years ago
for $400,000. It will have a salvage value of $20,000 in 4 years. Operating
and maintenance costs have been $75,000/year. Currently the computer has
a trade-in value of $100,000 toward a new computer that costs $300,000 and
has a life of 4 years, with a salvage value of $50,000 at that time. The new
computer will have annual operating and maintenance costs of $80,000.
Summary 279
If the current computer is retained, another small computer will have to
be purchased in order to provide the required computing capacity. The smaller
computer will cost $150,000, has a salvage value of $20,000 in 4 years, and
has annual operating and maintenance costs of $30,000.
Using an annual worth comparison before taxes, with a MARR of 15%,
determine the preferred course of action using a cash flow approach.
24. National Chemicals has an automatic chemical mixture that it has been using
for the past 4 years. The mixer originally cost $18,000. Today the mixer can
be sold for $10,000. The mixer can be used for 10 years more and will have
a $2,500 salvage value at that time. The annual operating and maintenance
costs for the mixer equal $6,000/year.
Because of an increase in business, a new mixer must be purchased. If
the old mixer is retained, a new mixer will be purchased at a cost of $25,000
and have a $4,000 salvage value in 10 years. This new mixer will have annual
operating and maintenance costs equal to $5,000/year.
The old mixer can be sold and a new mixer of larger capacity purchased
for $32,000. This mixer will have a $6,000 salvage value in 10 years and will
have annual operating and maintenance costs equal to $8,000/year.
Based on a MARR of 15% and using a cash flow approach, what do you
recommend?
25. The Ajax Specialty Items Corporation has received a 5-year contract to pro-
duce a new product. To do the necessary machining operations, the company
is considering two alternatives.
Alternative A involves continued use of the currently owned lathe. The
lathe was purchased five years ago for $20,000. Today the lathe is worth
$8,000 on the used machinery market. If this lathe is to be used, special
attachments must be purchased at a cost of $3,500. At the end of the 5-year
contract, the lathe (with attachments) can be sold for $2,000. Operating and
maintenance costs will be $7,000/year if the old lathe is used.
Alternative B is to sell the currently owned lathe and buy a new lathe at
a cost of $25,000. At the end of the 5-year contract, the new lathe will have
a salvage value of $13,000. Operating and maintenance costs will be $4,000/
year for the new lathe.
Using an annual worth analysis, should the firm use the currently owned
lathe or buy a new lathe? Base your analysis on a minimum attractive rate of
return of 15% and use a cash flow approach.
26. The Telephone Company of America purchased a numerically controlled
production machine 5 years ago for $300,000. The machine currently has
a trade-in value of $70,000. If the machine is continued in use, another machine, X, must be purchased to supplement the old machine. Machine X costs
$200,000, has annual operating and maintenance costs of $40,000, and will
have a salvage value of $30,000 in 10 years. If the old machine is retained, it
will have annual operating and maintenance costs of $55,000 and will have a
salvage value of $15,000 in 10 years.
As an alternative to retaining the old machine, it can be replaced with
Machine Y. Machine Y costs $400,000, has anticipated annual operating
280
Chapter 7
Replacement Analysis
and maintenance costs of $70,000, and has a salvage value of $140,000 in
10 years.
Using a MARR of 15%, a cash flow approach, and a present worth comparison, determine the preferred economic alternative.
27. A highway construction firm purchased a particular earth-moving machine
3 years ago for $125,000. The salvage value at the end of 8 years was estimated to be 35% of first cost. The firm earns an average annual gross revenue
of $105,000 with the machine and the average annual operating costs have
been and are expected to be $65,000.
The firm now has the opportunity to sell the machine for $70,000 and subcontract the work normally done by the machine over the next 5 years. If the
subcontracting is done, the average annual gross revenue will remain $105,000
but the subcontractor charges $85,000/end-of-year for these services.
If a 15% rate of return before taxes is desired, use a cash flow approach
to determine by the annual worth method whether or not the firm should
subcontract.
28. Bumps Unlimited, a highway contractor, must decide whether to overhaul a
tractor and scraper or replace it. The old equipment was purchased 5 years
ago for $130,000; it had a 12-year projected life. If traded for a new tractor
and scraper, it can be sold for $60,000. Overhauling the equipment will cost
$20,000. If overhauled, O&M cost will be $25,000/year and salvage value
will be negligible in 7 years.
If replaced, a new tractor and scraper can be purchased for $150,000.
O&M costs will be $12,000/year. Salvage value after 7 years will be $35,000.
Using a 15% MARR, should the equipment be replaced?
Section 7.3
29.
Optimum Replacement Interval
You plan to purchase a car for $28,000. Its market value will decrease by
20% per year. You have determined that the IRS-allowed mileage reimbursement rate for business travel is about right for fuel and maintenance at $0.485
per mile in the first year. You anticipate that it will go up at a rate of 10% each
year, with the price of oil rising, influencing gasoline, oils, greases, tires,
and so on. You normally drive 15,000 miles per year. What is the optimum
replacement interval for the car? Your MARR is 9%.
30. Griffin Dewatering purchases a wellpoint pump connected to a skid-mounted
diesel engine for $14,000. Its market value for salvage purposes decrease
by 30% each year. When installed on a construction job, a wellpoint system
operates virtually 24/7, and operating and maintenance costs will be $3,500
the first year, increasing by $600 each year thereafter. What is the optimum
replacement interval if MARR 5 15%?
31. Polaris Industries wishes to purchase a multiple-use in-plant “road test”
simulator that can be used for ATVs, motorcycles, and snow mobiles. It
takes digital data from relatively short drives on a desired surface—from
smooth to exceptionally harsh—and simulates the ride over and over while
Summary 281
the vehicle is mounted to a test stand under load. It can run continuously if
desired, and provides opportunities to redesign in areas of poor reliability.
It costs $128,000 and its market value decreases by 30% each year. Operating costs are modest; however, maintenance costs can be significant due to
the rugged use. O&M in the first year is expected to be $10,000, increasing by 25% each subsequent year. MARR is 15%. What is the optimum
replacement interval?
32. A granary purchases a conveyor used in the manufacture of grain for trans-
porting, filling, or emptying. It is purchased and installed for $70,000 with
a market value for salvage purposes that decreases at a rate of 20% per year
with a minimum of $3,000. Operation and maintenance is expected to cost
$14,000 in the first year, increasing by $1,000 per year thereafter. The granary uses a MARR of 15%. What is the optimum replacement interval for the
conveyor?
33.
GO Tutorial Milliken uses a digitally controlled “dyer” for placing intricate
and integrated patterns on manufactured carpet squares for home and commercial use. It is purchased for $400,000. Its market value will be $310,000 at
the end of the first year and drop by $40,000 per year thereafter to a minimum
of $30,000. Operating costs are $20,000 the first year, increasing by 8% per
year. Maintenance costs are only $8,000 the first year but will increase by
35% each year thereafter. Milliken’s MARR is 20%. Determine the optimum
replacement interval for the dyer.
34. A firm is presently using a machine that has a market value of $11,000 to do
a specialized production job. The requirement for this operation is expected
to last only 6 years more, after which it will no longer be done. The predicted
costs and salvage values for the present machine are:
Year
Operating cost
Salvage value
1
2
3
4
5
6
$1,500
8,000
$1,800
6,000
$2,100
5,000
$2,400
5,000
$2,700
3,000
$3,000
2,000
A new machine has been developed that can be purchased for $17,000 and
has the following predicted cost performance.
Year
Operating cost
Salvage value
1
2
3
4
5
6
$ 1,000
13,000
$ 1,100
11,000
$ 1,200
10,000
$1,300
9,000
$1,400
8,000
$1,500
7,000
If interest is at 0%, when should the new machine be purchased?
35. A particular unit of production equipment has been used by a firm for a
period of time sufficient to establish very accurate estimates of its operating and maintenance costs. Replacements can be expected to have identical cash flow profiles in successive life cycles if constant worth dollar
estimates are used. The appropriate discount rate is 15%. Operating and
282
Chapter 7
Replacement Analysis
maintenance costs for a unit of equipment in its tth year of service,
denoted by Ct, are as follows:
t
Ct
t
1
2
3
4
5
$6,000
7,500
9,150
10,950
12,900
6
7
8
9
10
Ct
$15,000
17,250
19,650
22,200
24,900
Each unit of equipment costs $45,000 initially. Because of its special design, the unit of equipment cannot be disposed of at a positive salvage value
following its purchase; hence, a zero salvage value exists, regardless of the
replacement interval used.
a. Determine the optimum replacement interval assuming an infinite planning horizon. (Maximum feasible interval 5 10 years.)
b. Determine the optimum replacement interval assuming a finite planning horizon of 15 years, with Ct11 5 Ct 1 $1,500 1 $150(t 2 1) for t 5 10, 11, . . .
c. Solve parts (a) and (b) using a discount rate of 0%.
d. Based on the results obtained, what can you conclude concerning the
effect the discount rate has on the optimum replacement interval?
36. Given an infinite planning horizon, identical cash flow profiles for successive
life cycles, and the following functional relationships for Ct, the operating
and maintenance cost for the ith year of service for the unit of equipment in
current use, and Fn, the salvage value at the end of n years of service:
Ct 5 $4,00011.102 t
t 5 1, 2, . . . , 12
Fn 5 $44,00010.502 n
n 5 0, 1, 2, . . . , 12
Determine the optimum replacement interval assuming a MARR of (a) 0%,
(b) 10%. (Maximum life 5 12 years.)
37. $100,000 is invested in equipment having a negligible salvage value regard-
less of when the equipment is replaced. O&M costs equal $25,000 the first
year and increase $5,000 per year. Based on a MARR of 10%, what are the
optimum replacement interval and minimum EUAC?
38. $100,000 is invested in equipment having a negligible salvage value regard-
less of when the equipment is replaced. O&M costs equal $25,000 the first
year and increase 15% per year. Based on a MARR of 10%, what are the
optimum replacement interval and minimum EUAC?
39. $250,000 is invested in equipment having a negligible salvage value regard-
less of the number of years used. O&M costs equal $60,000 the first year and
increase $12,000 per year. Based on a MARR of 10%, what are the optimum
replacement interval and minimum EUAC?
40. $250,000 is invested in equipment having a salvage value equal to
$250,000(0.80n ) after n years of use. O&M costs equal $60,000 the first year
and increase $8,000 per year. Based on a MARR of 10%, what are the optimum replacement interval and minimum EUAC?
Summary 283
41. $225,000 is invested in equipment having a salvage value equal to
$200,000(0.75n ) after n years of use. O&M costs equal $45,000 the first year
and increase $15,000 per year. Based on a MARR of 10%, what is optimum
replacement interval and what is the minimum EUAC?
42. $500,000 is invested in equipment having a negligible salvage value regard-
less of the number of years used. O&M costs equal $125,000 the first year
and increase $25,000 per year. Based on a MARR of 10%, what are the optimum replacement interval and minimum EUAC?
43. $500,000 is invested in equipment having a salvage value equal to
$500,000(0.80n ) after n years of use. O&M costs equal $125,000 the first
year and increase $15,000 per year. Based on a MARR of 10%, what are the
optimum replacement interval and minimum EUAC?
44. $500,000 is invested in equipment having a salvage value equal to
$500,000(0.75n ) after n years of use. O&M costs equal $125,000 the first
year and increase 15% per year. Based on a MARR of 10%, what are the
optimum replacement interval and minimum EUAC?
45. $500,000 is invested in equipment having a salvage value equal to
$500,000(0.65n ) after n years of use. O&M costs equal $125,000 the first
year and increase 10% per year. Based on a MARR of 10%, what are the
optimum replacement interval and minimum EUAC?
46. True or False: Consider an optimum replacement interval problem of the type
considered in this chapter. If O&M costs are a gradient annual series, the optimum replacement interval tends to increase as the magnitudes of the gradient
step decreases.
47. True or False: Consider an optimum replacement interval problem of the type
considered in this chapter. If the annual O&M cost is an increasing gradient series “sitting on top of ” a base uniform series, then decreasing the magnitude of
the base uniform series will tend to increase the optimum replacement interval.
48. True or False: Consider an optimum replacement interval problem of the type
considered in this chapter. If salvage value is negligible regardless of how
long the equipment is used and O&M cost is represented by a uniform annual series, then the equipment in question should not be replaced until it no
longer can function.
49. True or False: Consider an optimum replacement interval problem of the type
considered in this chapter. If O&M costs are a geometric series, then decreasing the magnitude of the O&M cost in the first year will tend to decrease the
optimum replacement interval.
50. True or False: Consider an optimum replacement interval problem of the type
considered in this chapter. If the ORI equals 5 years based on a salvage value
equal to 0.60n, where n is the years used before replacement, then, if the salvage value for a replacement becomes 0.80n the new ORI will not be greater
than 5 years.
8
ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E :
S C HNE I DE R N AT I O N AL
Based on 2011 revenues, Schneider National is a $3.4 billion privately held
company in the transportation and logistics sector. Headquartered in Green
Bay, Wisconsin, its customers include many of the global companies on
Fortune magazine’s list of top 100 companies. Schneider operates within
North America and China, it employs 17,400 associates, and it is structured
into three principal business units—truckload, intermodal, and logistics.
The largest business unit, truckload, includes 9,860 company drivers and
1,775 owner-operators (independent drivers with their own tractors), operating 11,800 tractors and 31,500 trailers. Intermodal, a business that combines
the geographic reach and delivery of trucking with the efficient long-haul
transportation of railroads, is the second largest business unit: It employs
1,350 company drivers, 140 owner-operators, and 13,500 containers to service
customers in the intermodal market. Schneider Logistics has 2,000 associates
worldwide and provides a variety of value-added services to customers, including Transportation Management (Brokerage), Port Logistics (Transloading and
Distribution, Inland Logistics Management, and China Solutions), Supply
Chain Management (Transportation Network Design), Supplier Management,
Integrated Delivery Services (LTL and Cross-Dock), and Schneider Consulting.
Each of Schneider’s businesses has a different level of asset intensity. The
entire portfolio, however, results in $1 billion of invested capital showing on
its balance sheet. As a result, it is critical that each increment of investment
produce positive contributions to after-tax income, ensuring value creation
for shareholders commensurate with its level of risk. Engineering economic
analysis principles and tools are used by Schneider in allocating capital and
human resources decisions.
Three recent examples of economic analyses performed include the design
of a trailer fleet, the use of trailer-tracking devices, and a strategic acquisition.
■
284
Economic analyses showed that increasing the purchase price of duraplate 53-foot Wabash trailers by adding a 22-inch “high rail” aluminum
DEPRECIATION
strip at the bottom of the trailer’s sides reduced a significant amount of
damage from lift-truck forks and mishandling by trailer-pool hostlers. The
analysis included reductions in maintenance expense, plus savings by
reducing the number of trailers due to less repair time required.
■
The incorporation of economic analysis at the project’s beginning significantly influenced the ultimate design and functionality of a trailer-tracking
device installed on all of Schneider’s trailers. The initial results evidenced
that knowing where trailers were at all times added significant value for
Schneider and its customers.
■
In pursuing a strategic acquisition, Schneider required that the investment produce a series of cash flows that meet or exceed its weighted
average cost of capital, adjusted for risk. Because of the inevitability of
future downside considerations, as well as the cyclical nature of the transportation market, the economic justifications were required to include
well-demonstrated and documented scenario and risk-management plans
showing how the returns on the target acquisitions would be impacted.
Because Schneider National is privately owned, it is not required to satisfy
United States Securities and Exchange Commission rules that apply to publicly traded corporations. As a result, less pressure exists to meet or exceed
estimates of quarterly earnings developed by financial analysts. Financial
decisions at Schneider are based on after-tax returns on investment.
In contrast to Schneider, U.S. publicly traded corporations generally
maintain two sets of books, or accounting records—and it is perfectly legal.
Consequently, two different depreciation schedules are maintained: one for
financial reporting and the other for computing income taxes to be paid. We
consider both in this chapter. In the following chapter, we address the tax
consequences of various depreciation allowances.
When a firm’s capital assets are distributed around the world, when
income-tax laws vary from country to country, and when capital equipment
is purchased, replaced, and modified frequently, it is a challenge to maintain
accurate records regarding the value of assets owned by the shareholders.
285
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Depreciation
DISCUSSION QUESTIONS:
1. What impact does depreciation have on capital investment decision
making at Schneider?
2. What will be the impact of these capital investment decisions on aftertax income for Schneider?
3. Although maintaining two sets of books is legal, what concerns, if any,
do you have about this practice?
4. In the previous chapter on replacement analysis, we introduced J. B.
Hunt Transport Services Inc. (JBHT), another transportation and logistics
company. What differences do you see between JBHT and Schneider
with respect to depreciation of their significant volume of assets?
Recall that JBHT is a public company based in North America.
LEARNING OBJECTIVES
When you have finished studying this chapter, you should be able to:
1. Describe the impact that depreciation has on design decisions from an
income-tax perspective. (Section 8.1)
2. Define a depreciable property and distinguish whether this property is
tangible or intangible, and if tangible, whether the property is personal or real. (Section 8.2)
3. Calculate depreciation using the Straight-Line (SLN), Declining Balance
(DB), and Double Declining Balance (DDB) methods. (Section 8.3)
4. Determine at what point it is optimal to switch from a declining balance rate to a straight-line rate. (Section 8.3.3)
5. Calculate depreciation using the Modified Accelerated Cost Recovery
System (MACRS). (Section 8.4).
INTRODUCTION
Depreciation
The concept
that takes
into account
the fact that
most property
decreases in
value with use
and time.
Most property decreases in value with use and time—that is, it depreciates.
Depreciation is the subject of this chapter. We present it to ensure that
you understand the implications of making investments in things that
must (by U.S. tax law) be depreciated versus those that do not. Also,
we want to ensure that you understand how a depreciation method can
impact the profitability of an investment alternative.
In addition to learning the mechanics of calculating depreciation
charges, we want you to gain an understanding of why depreciation and
the accompanying income-tax treatments are important in performing
engineering economic analyses. Engineers make decisions in the
8-1 The Role of Depreciation in Economic Analysis
process of designing products and processes that can have significant
after-tax effects.
So, as you learn the material in the chapter, don’t lose sight of WHY
you are learning about depreciation. Remember, you are learning about
depreciation because your design decisions can affect the way investments and annual operating costs are treated from an income-tax perspective. In the next chapter you will further explore the topic of income
taxes as they relate to engineering economic analyses.
8-1
THE ROLE OF DEPRECIATION
IN ECONOMIC ANALYSIS
LEARNING O BJECTI VE: Describe the impact that depreciation has on design
decisions from an income-tax perspective.
We use a cash flow approach throughout this book. For that reason, we
seldom use the word depreciation, because depreciation is not a cash flow.
Typically, there are two cash flows associated with a capital asset: what
you pay for it and what you get when you sell or trade it in for a replacement. Depreciation is an artifice that reflects the decrease in the asset’s
value over time or with usage.
A principal reason for developing the concept of depreciation is to
allow a reasonably accurate report to the owners of a business regarding its
value at any given point in time. Another reason is to allow reasonable estimates to be made concerning the cost of doing business; prices need to be
set at levels that will recapture investments made in depreciable property—
property that is used to produce the goods and services sold by the business.
As such, when the time comes to replace an asset, funds will be available to
do so (unless prices have increased over the asset’s life).
A report to the owners of the business regarding its profitability, based
strictly on cash flows, can be misleading. For example, suppose you own a
business with annual revenues that exceed annual expenses by, say, $1 million. This year, however, you spend $1.5 million on revenue-producing
equipment that will be used for 5 years. If all you count is cash, then a cash
flow calculation will indicate that you lost $0.5 million this year. On the
other hand, if you spread the $1.5 million investment out over the 5-year
period of its use, then your business will be accurately portrayed as being
profitable.
To spread the investment costs over the useful life of the equipment
purchased, there are various approaches you can use. We consider a number of them in this chapter.
Video Lesson:
Depreciation
287
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Chapter 8
Depreciation
Depreciation Allowances
The periodic amounts that
may be deducted from
taxable income—these
amounts are treated as
expenses even though they
are not cash flows.
As good as the foregoing arguments are for considering depreciation,
they are not the reason for including a treatment of depreciation in this
book. We include it because U.S. income-tax law permits depreciation
allowances to be deducted from taxable income; in other words, depreciation allowances can be treated as expenses even though they are not cash
flows. As a result, depreciation affects income taxes, which are cash flows.
Therefore, when comparing investment alternatives using an after-tax
analysis, an accurate consideration of deprecation becomes important.
For example, whether to design a production process that will be
performed by people versus one that will be performed by robots is a
typical choice made by mechanical and manufacturing engineers. But
the after-tax consequences are quite different, depending on the choice
made.
If people perform the work, labor costs are treated as expenses and are
fully deducted from taxable income in the year in which they occur. If
robots perform the work, however, only the robots’ operating costs are
treated as expenses. The acquisition cost for the robots must be depreciated. As such, the investment will be recovered over multiple years.
8-2
LANGUAGE OF DEPRECIATION
LEARN I N G O B JE C T I V E : Define a depreciable property and distinguish
whether this property is tangible or intangible, and if tangible, whether the
property is personal or real.
Depreciable Property
Property meeting the
specific requirements as
defined by the U.S. Internal
Revenue service whereby
it is used in business or
held for the production of
income; has a life that can
be determined, and that
life must be longer than
one year; and is something
that wears out, decays,
gets used up, becomes
obsolete, or loses value
from natural causes.
Business expenditures are either expensed or depreciated. Expensing an
expenditure is akin to depreciating it fully in the year in which it occurs.
Expenditures for labor, materials, and energy are examples of items that
are fully deducted from taxable income. On the other hand, expenditures
for production equipment, vehicles, and buildings cannot be fully deducted
from taxable income in the year in which they occur; instead, they must be
spread out or distributed over some allowable recovery period.
Technically speaking, U.S. tax law permits deduction from taxable
income of a reasonable allowance for wear and tear, natural decay or
decline, exhaustion, or obsolescence of property used in a trade or business or of property held for producing income. Specifically, the Internal
Revenue Service requires that the following requirements be met for
depreciable property:
It must be used in business or held for the production of income.
2. It must have a life that can be determined, and that life must be longer
than a year.
3. It must be something that wears out, decays, gets used up, becomes
obsolete, or loses value from natural causes.
1.
8-3
Straight-Line and Declining Balance Depreciation Methods
Depreciable property may be tangible or intangible. Tangible property
can be seen or touched and can be categorized as personal or real.
Personal property is goods such as cars, trucks, machinery, furniture,
equipment, and anything that is tangible except real property. Real property
is land and generally anything that is erected on, growing on, or attached
to it. Land, by itself, does not qualify for depreciation, but the buildings,
structures, and equipment on it do qualify; land does not qualify because
it does not pass the third requirement, given above. In contrast to tangible property, intangible property cannot be seen or touched but has
value to the owner; it includes copyrights, brands, software, goodwill,
formulas, designs, patents, trademarks, licenses, information bases, and
franchises.
A term used synonymously with depreciation is amortization. Generally speaking, depreciation is associated with personal property and real
property (other than land), whereas amortization is associated with intangible property. The U.S. income tax code defines amortization as ‘‘the
recovery of certain capital expenditures, that are not ordinarily deductible,
in a manner that is similar to straight-line depreciation.’’
(While intangible property is amortized (depreciated) for financial
reporting purposes, it is expensed for income tax purposes. Business
expenditures for software, for example, can be deducted from taxable
income, although they are typically spread out uniformly over the amortization period for financial reports.)
8-3
STRAIGHT-LINE AND DECLINING
BALANCE DEPRECIATION METHODS
LEARNING O BJECTI VE: Calculate depreciation using the Straight-Line (SLN),
Declining Balance (DB), and Double Declining Balance (DDB) methods.
There are several methods of calculating depreciation. Firms may use any
of the methods discussed in this section (SLN, DB, DDB, or DDB with
switch to SLN) for accounting and financial reporting purposes, though
use of the straight-line (SLN) method is most common. For income tax
purposes, however, firms must use one of the Modified Accelerated Cost
Recovery System (MACRS) methods discussed later in this chapter.
8.3.1
Straight-Line Depreciation (SLN)
Straight-line depreciation is the oldest of the depreciation methods in
use today. As the name implies, annual depreciation charges form a uniform annual series. If P denotes the cost basis, i.e., the amount paid for
acquisition and installation of a depreciable asset, and F denotes the asset’s
289
Tangible Property
Property that can be seen
or touched and can be
categorized as personal
or real.
Personal Property
Goods such as cars, trucks,
machinery, furniture
equipment and anything
that is tangible except real
property.
Real Property Land and
generally anything that is
erected on, growing on, or
attached to it.
Intangible Property
Property that cannot be
seen or touched, but has
value to the owner such
as copyrights, brands,
software, goodwill,
formulas, designs, patents,
trademarks, licenses,
information bases, and
franchises.
Amortization As defined
by the U.S. income tax code
“the recovery of certain
capital expenditures,
that are not ordinarily
deductible, in a manner
that is similar to straightline depreciation.”
Straight-Line Depreciation
(SLN) The depreciation
method that treats annual
depreciation charges as a
uniform annual series.
Cost Basis The amount
paid for acquisition
and installation of a
depreciable asset.
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Chapter 8
Depreciation
salvage value at the end of the useful life, recovery period, or depreciable
life of n years, then the annual SLN depreciation charge dt for year t is
dt 5 1P 2 F2/n
5 SLN1P,F,n2
Book Value The
undepreciated portion of
an asset determined by
subtracting the cumulative
depreciation charge
from the cost basis.
Also referred to as the
unrecovered investment,
the adjusted cost basis,
and the adjusted basis.
(8.1)
where SLN is the Excel® worksheet function for calculating the annual depreciation charge. The syntax for the SLN function, SLN(cost,salvage,life), is
self-explanatory.
The undepreciated portion of an asset is called the book value. Also
referred to as the unrecovered investment, the adjusted cost basis, and the
adjusted basis, the book value at the end of year t (Bt), based on straight-line
depreciation, is the cost basis less the cumulative depreciation charge, or
Bt 5 P 2 td
5 P 2 t 1P 2 F2/n
(8.2)
Hence, book value is a straight line. Further, since the salvage value is
incorporated in the depreciation calculation, book value at the end of the
recovery period is equal to the salvage value.
While corporate income taxes, and consequently engineering project
cash flows, are based on MACRS depreciation methods, we know of companies that require the use of straight-line depreciation in engineering economic justifications. Why? We were given the following reasons:
Management does not want an investment’s economic viability to be
decided on the basis of the depreciation method used. Recognizing that
cash flow estimates are just that, estimates, and they are likely to prove
to be incorrect, management does not feel that it is justified to be so
precise regarding depreciation when errors in the cash flow estimates
can easily offset any gains made by using accelerated depreciation.
2. Management wishes to use a standardized approach and, because of
its simplicity, chooses to use SLN in all economic justifications.
1.
Straight-Line Depreciation Applied to the SMP Machine
EXAMPLE
An SMP machine costs $500,000 and has a $50,000 salvage value after
10 years of use. Using straight-line depreciation, what are the depreciation
charge and book value for the machine after year 5?
KEY DATA
Given: Cost Basis P 5 $500,000; Salvage Value F 5 $50,000; Useful Life
n 5 10 yrs
Find: d5, B5
8-3
Straight-Line and Declining Balance Depreciation Methods
The following values occur for t 5 5, with SLN:
SOLUTION
d5 5 1$500,000 2 $50,0002/10
5 $45,000
5 SLN1500000,50000,102
5 45,000
B5 5 $500,000 2 51$45,0002
5 $275,000
8.3.2
291
Declining Balance and Double Declining Balance
Depreciation (DB and DDB)
Declining balance depreciation is an accelerated depreciation method. As
such, it provides larger depreciation charges in the early years and smaller
depreciation charges in the later years.
Accelerated depreciation methods are desirable because they take into
account the time value of money. Assuming a TVOM greater than 0,
because depreciation charges can be deducted from taxable income, aftertax present worth is maximized when the present worth of depreciation
charges is maximized. Just as one prefers to receive money sooner rather
than later, due to the TVOM, so should one prefer to depreciate an asset
sooner rather than later.
Where SLN produced a uniform annual series of depreciation charges,
DB produces a negative geometric series of depreciation charges. With
DB, the depreciation charge in a given year is a fixed percentage of the
book value at the beginning of the year. Letting p denote the DB percentage or depreciation rate used,
dt 5 pBt21
5 pP11 2 p2 t21
(8.3)
and
Bt 5 P11 2 p2 t
(8.4)
Notice, salvage value is not incorporated in the calculations for DB. As a
result, the book value at the end of the cost recovery period is unlikely to
equal the salvage value obtained for the asset. (We will discuss the incometax implications of book values not equaling salvage values in Chapter 9.)
What depreciation rate or value of p should be used? Among the
most often used values are 125 percent, 150 percent, and 200 percent
(the maximum amount allowed by the IRS) of the straight-line rate, which
Declining Balance (DB)
A type of accelerated
depreciation method
that produces a negative
geometric series of
depreciation charges
providing larger
depreciation charges in the
early years and smaller
depreciation charges in the
later years.
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Depreciation
is 1/n. If it is desired to use a percentage such that the book value equals
the salvage value after n years, then, from Equation 8.4,
F 5 P(1 2 p)n
Solving for p gives
p 5 1 2 1F Z P2 1/n
5 2RATE1n,,2P,F2
Double Declining Balance
(DDB) A type of
accelerated depreciation
method that is twice, or
200 percent, the straightline rate.
(8.5)
Notice the minus sign between the equal sign and the Excel® RATE
worksheet function. Also, when p 5 2/n, which is twice, or 200 percent,
the straight-line rate, the DB plan is called double declining balance
(DDB) depreciation.
Excel® has three declining balance worksheet functions: DB, DDB,
and VDB. The VDB, or variable declining balance function, returns the
depreciation of an asset for any period specified, including partial periods,
using the DDB method or another specified method. The answers obtained
from the Excel® DB function sometimes differ from answers obtained
using Equations 8.3 and 8.4. Therefore, if Excel® functions are to be used,
DDB and VDB are recommended. The syntaxes for DDB and VDB are
DDB(cost,salvage,life,period,factor); and
VDB(cost,salvage,life,start_period,end_period,factor,no_switch)
where
cost 5 the cost basis of the asset
salvage 5 the salvage value of the asset
life 5 the cost recovery period of the asset
period 5 the time period for which the depreciation charge is
desired
factor 5 the depreciation rate (p) to use; in the case of DDB, if
omitted, it is assumed to equal 2, the double-declining rate
start_period 5 the starting period for which depreciation is
calculated; must use the same units as life
end_period 5 the ending period for which depreciation is
calculated; must use the same units as life
no_switch 5 a logical value specifying whether to switch to
straight-line depreciation when the straight-line depreciation is
greater than the declining balance calculation
The Excel® DDB function uses a depreciation rate equal to 2/n. When
using the Excel® VDB function, switching from declining balance to
8-3 Straight-Line and Declining Balance Depreciation Methods
straight-line depreciation is an option; if the declining balance depreciation rate is not specified, it is assumed to equal 2/n.
We address the concept of switching depreciation methods in the next
section. Since we have introduced the syntax for the Excel® VDB function,
note that when no_switch is TRUE, Excel® does not switch to straight-line
depreciation even when the depreciation is greater than the declining balance calculation; if no_switch is FALSE or omitted, Excel® switches to
straight-line depreciation when it is greater than the declining balance depreciation. Finally, all arguments except no_switch must be positive numbers;
therefore, the cost basis is not shown as a negative-valued cash flow.
Declining Balance Depreciation Applied to the SMP Machine
EXAMPLE
For the SMP machine in Example 8.1, what are the depreciation charge
and book value after year 5 using declining balance depreciation?
Given: Cost Basis P 5 $500,000; Salvage Value F 5 $50,000; Useful Life
n 5 10 years
Find: d5, B5
KEY DATA
Before computing d5, B5, we use Equation 8.5 to determine the depreciation rate that equates the book value at the end of the recovery period to the
salvage value.
Recall, a $500,000 investment is made in the SMP machine; it has an
estimated $50,000 salvage value after 10 years of usage. Therefore,
SOLUTION
p 5 1 2 1$50,000/$500,00020.1
5 20.5672%
5 2RATE110,,500000,500002
5 20.5672%
Unfortunately, this rate exceeds 2/n and is not permitted by the IRS.
So, suppose the declining balance rate is to be twice the straight-line
rate. Letting p 5 2/10, from Equations 8.3 and 8.4,
d5 5 0.201$500,0002 10.802 4
5 $40,960.00
B5 5 $500,00010.802 5
5 $163,840.00
Using the Excel® DDB worksheet function gives
d5 5 DDB1500000,50000,10,52
5 $40,960.00
293
294
Chapter 8
Depreciation
When using the Excel® VDB worksheet function, it is necessary to specify
a beginning and ending year for the calculation of the depreciation charge.
To compute the depreciation charge for t 5 5,
d5 5 VDB1500000,50000,10,4,5,22
5 $40,960.00
To obtain the book value using the Excel® DDB, and VDB worksheet
functions, it is necessary to compute the depreciation charges for all previous years and then subtract the cumulative depreciation charges from the
cost basis.
Table 8.1 provides the depreciation charges obtained and the associated book values for the $500,000 SMP investment for t 5 1, . . . , 10 using
the Excel® worksheet functions SLN, DDB, and VBD. Notice that the
final book value with DDB does not equal $50,000. Also, notice that the
depreciation charges are identical for VDB and DDB until year 9, at which
time VDB switches from a declining balance rate to a straight-line rate.
TABLE 8.1
t
Depreciation Allowances and Book Values for the SMP Investment
SLN dt
0
SLN Bt
DDB dt
$500,000.00
1
$45,000.00
2
3
DDB Bt
VDB dt
$500,000.00
VDB Bt
$500,000.00
$455,000.00
$100,000.00
$400,000.00
$100,000.00
$400,000.00
$45,000.00
$410,000.00
$80,000.00
$320,000.00
$80,000.00
$320,000.00
$45,000.00
$365,000.00
$64,000.00
$256,000.00
$64,000.00
$256,000.00
4
$45,000.00
$320,000.00
$51,200.00
$204,800.00
$51,200.00
$204,800.00
5
$45,000.00
$275,000.00
$40,960.00
$163,840.00
$40,960.00
$163,840.00
6
$45,000.00
$230,000.00
$32,768.00
$131,072.00
$32,768.00
$131,072.00
7
$45,000.00
$185,000.00
$26,214.40
$104,857.60
$26,214.40
$104,857.60
8
$45,000.00
$140,000.00
$20,971.52
$83,886.08
$20,971.52
$83,886.08
9
$45,000.00
$95,000.00
$16,777.22
$67,108.86
$16,943.04
$66,943.04
10
$45,000.00
$50,000.00
$13,421.77
$53,687.09
$16,943.04
$50,000.00
8.3.3
Switching from DDB to SLN with the Excel®
VDB Function
LEARN I N G O B JEC T I V E : Determine at what point it is optimal to switch from
a declining balance rate to a straight-line rate.
As noted, the Excel® VDB worksheet function includes an optional feature:
It can switch from using a specified declining balance rate to a straight-line
8-3 Straight-Line and Declining Balance Depreciation Methods
rate when it is optimum to do so. But what criterion is used to define
optimum? Accelerated depreciation methods are preferred to straight-line
depreciation but only up to a point, and VDB determines that point.
The declining balance methods front-end load depreciation charges.
Because declining balance methods compute depreciation charges using a
constant percentage of the book value, however, toward the recovery
period’s latter stages, the amount of depreciation charged drops off precipitously with declining balance methods. And, as was the case with DDB
in the previous example, the book value calculated with declining balance
methods might not reach the salvage value during the recovery period.
The Excel® VDB function computes the depreciation charge that
would result if declining balance continued to be used and compares it
with the depreciation charge that would result if straight-line depreciation
were used for the balance of the recovery period. Hence, to maximize the
present worth of depreciation charges, the optimum time to switch is the
first time the depreciation charge with straight-line depreciation is greater
than would result if declining balance were continued. Hence, a switch to
straight-line depreciation occurs at the first year for which
Bt21 2 F
. pBt21
n 2 1t 2 12
(8.6)
Notice, the estimated salvage value is used in determining the straight-line
depreciation component, even though it is neglected in the DDB calculations. Switching to straight-line depreciation is never desirable if the estimated salvage value, F, exceeds Bn, the declining balance unrecovered
investment for the last year of the recovery period.
Switching from DDB to SLN with the SMP Machine
EXAMPLE
For the SMP machine in previous examples, what is the optimal point for
the depreciation method to switch from DDB to SLN?
Given: P 5 $500,000, F 5 $50,000, n 5 10, and p 5 0.2 for DDB.
Find: The t for which
KEY DATA
Bt21 2 F
. pBt21
n 2 1t 2 12
When t 5 1, the left-hand side (LHS) of Equation 8.6 is ($500,000 2
$50,000)/10 5 $45,000; the right-hand side (RHS) of Equation 8.6 is
0.20 ($500,000) 5 $100,000. Because LHS , RHS, try t 5 2.
When t 5 2, the LHS of Equation 8.6 is ($400,000 2 $50,000)/9 5
$38,888.89; the RHS of Equation 8.6 is 0.20 ($400,000) 5 $80,000.
Because LHS , RHS, try t 5 3.
SOLUTION
295
296
Chapter 8
Depreciation
This iterative approach continues until the LHS of equation 8.6
exceeds the RHS as it does in year t 5 9. That is the point at which depreciation switches from DDB to SLN, and the depreciation charges dt remain
the same from that point until t 5 n.
The calculations performed and the process used to determine the
optimum time to switch from DDB to SLN are shown in Figure 8.1.
FIGURE 8.1
8-4
Depreciation and Book Value Using DDB Switching to SLN
MODIFIED ACCELERATED COST RECOVERY
SYSTEM (MACRS)
LEARN I N G O B JEC T I V E : Calculate depreciation using the Modified Accelerated Cost Recovery System (MACRS).
Modified Accelerated
Cost Recovery System
(MACRS) The only
depreciation method
approved by the U.S. IRS
for computing income-tax
liability.
Although SLN is the most commonly used depreciation method in the
United States for purposes of financial reporting, Modified Accelerated
Cost Recovery System (MACRS) is the only depreciation method
approved by the IRS for computing income-tax liability. In general,
MACRS depreciation should be used for engineering economic analysis
because we are interested in analyzing the actual cash flows (including
taxes) involved in an investment decision.
8-4
Modified Accelerated Cost Recovery System (MACRS) 297
Most depreciable property placed in service after 1986 qualifies for
MACRS. There are two variations of MACRS: the General Depreciation
System (GDS) and the Alternative Depreciation System (ADS). For
most applications of interest to us, GDS is based on double declining balance, switching to straight-line depreciation. ADS is based on straight-line
depreciation, with a longer recovery period than GDS. Both MACRS-GDS
and MACRS-ADS have pre-established recovery periods for most property;
in the case of MACRS-GDS, most property is assigned to eight property
classes, which establish the number of years over which the cost basis is to
be recovered.
Both MACRS-GDS and MACRS-ADS include a feature we have not
used previously: a half-year convention. It is assumed that, on average, a
property is used for half of the first year of service. Similarly, it is assumed
that it is used for half of the last year of service. The letter H is used to
designate the half-year convention. Hence, 200% DBSLH-GDS denotes
double declining balance, switching to straight-line depreciation with the
half-year convention applied to the first and last years of service over the
GDS recovery period. Exceptions to the half-year convention are 27.5-year
and 39-year property.
8.4.1
MACRS-GDS
Depreciable tangible property may be assigned to one of the eight MACRS
(MACRS-GDS unless otherwise specified) classes, shown in Table 8.2.
From the property class descriptions, it is obvious that determining the
class to which a given property belongs is a complex task. For that reason,
if there is any uncertainty regarding the appropriate class to use, we recommend you consult tax professionals or review publications available
from the IRS, Department of the Treasury, U.S. Government.
The Excel® VDB function can be used easily to calculate the depreciation allowance for an asset, given its MACRS property class. For
example, letting P 5 1, F 5 0, n 5 5, start_period 5 0.0, end_period 5
0.5, and factor 5 2, VDB will determine the first year’s depreciation rate
for 5-year property to be 20%. Repeating this for five more periods (0.5,
1.5; 1.5, 2.5; 2.5, 3.5; 3.5, 4.5; and 4.5, 5.0) will determine the remaining
five rates. However, it is much easier to use the values provided in Table 8.3
for 3-, 5-, 7-, 10-, 15-, and 20-year property. For 27.5- and 39-year property, use the percentages given in Table 8.4, which are based on the month
in which the property was placed in service. For 3-, 5-, 7-, and 10-year
property, the annual depreciation charge is based on 200% DBSLH-GDS;
for 15- and 20-year property, the annual depreciation charge is based on
150% DBSLH-GDS; for 27.5- and 39-year property, the annual depreciation charge is based on SLM, or straight-line depreciation, mid-month
convention.
Modified Accelerated Cost
Recovery System–General
Depreciation System
(MACRS-GDS) The
MACRS depreciation
method based on double
declining balance
switching to straight-line
depreciation.
Modified Accelerated
Cost Recovery System–
Alternative Depreciation
System (MACRS-ADS)
The MACRS depreciation
method based on straightline depreciation, with a
longer recovery than GDS.
Half-Year Convention
A feature used in both
MACRS-GDS and
MACRS-ADS assuming
that on average, a property
is used for half of the first
year of service and half of
the last year of service.
298
Chapter 8
TABLE 8.2
Depreciation
MACRS-GDS Property Classes
Property Class
Personal Property
3-Year Property
Qualifying property with a class life of 4 years or less, including: tractor
units for over-the-road use; special tools for manufacturing motor vehicles;
special handling devices for manufacture of food and beverages; and special
tools for manufacturing rubber, finished plastic, glass, and fabricated metal
products
5-Year Property
Qualifying property with a class life of more than 4 years but less than 10 years,
including automobiles and light, general-purpose trucks; certain research and
experimentation equipment; alternative energy and biomass property; computers
and peripheral equipment; satellite space segment property; data-handling
equipment (typewriters, calculators, copiers, printers, facsimile machines, etc.);
heavy, general-purpose trucks; timber cutting assets; offshore oil and gas well
drilling assets; certain construction assets; computer-based telephone central
office switching equipment; and many assets used for the manufacture of knitted
goods, carpets, apparel, medical and dental supplies, chemicals, and electronic
components
7-Year Property
Qualifying property with a class life of 10 or more years but less than 16 years,
property without any class life and not included in the 27.5- and 39-year categories,
including office furniture, fixtures, and equipment; theme and amusement park
assets; assets used in the exploration for and production of petroleum and
natural gas deposits; most assets used for manufacturing such things as food
products, spun yarn, wood products and furniture, pulp and paper, rubber
products, finished plastic products, leather products, glass products, foundry
products, fabricated metal products, motor vehicles, aerospace products, athletic
goods, and jewelry
10-Year Property
Qualifying property with a class life of 16 or more years but less than 20 years,
including vessels, tugs, and similar water transportation equipment; petroleum
refining assets; assets used in the manufacture of grain, sugar, vegetable oil
products, and substitute natural gas-coal gasification
15-Year Property
Qualifying property with a class life of 20 or more years but less than 25 years,
including land improvements, such as sidewalks, roads, drainage facilities, sewers,
bridges, fencing, and landscaping; cement manufacturing assets; some water and
pipeline transportation assets; municipal wastewater treatment plants; telephone
distribution plant assets; certain electric and gas utility property; and some
liquefied natural gas assets
20-Year Property
Qualifying property with a class life of 25 years or more, other than real property
with a class life of 27.5 years or more, plus municipal sewers, including such assets
as farm buildings; some railroad structures; some electric generating equipment,
such as certain transmission lines, pole lines, buried cable, repeaters, and much
other utility property
Property Class
Real Property
27.5-Year Property
Residential rental property, including a rental home or structure for which
80% or more of the gross rental income for the tax year is rental income from
dwelling units. A dwelling unit is a house or an apartment used to provide living
accommodations in a building or structure, but not a unit in a hotel, motel, inn,
or other establishment in which more than one-half of the units are used on a
transient basis
39-Year Property
Nonresidential real property: depreciable property with a class life of 27.5 years or
more and that is not residential rental property
MACRS-GDS percentages for 3-, 5-, 7-, and 10-year property
are 200% DBSLH and 15- and 20-year property are 150% DBSLH
TABLE 8.3
EOY
3-Year
Property
5-Year
Property
7-Year
Property
10-Year
Property
15-Year
Property
20-Year
Property
1
2
33.33
44.45
20.00
32.00
14.29
24.49
10.00
18.00
5.00
9.50
3.750
7.219
3
4
14.81
7.41
19.20
11.52
17.49
12.49
14.40
11.52
8.55
7.70
6.677
6.177
11.52
5.76
8.93
8.92
9.22
7.37
6.93
6.23
5.713
5.285
8.93
4.46
6.55
6.55
5.90
5.90
4.888
4.522
9
10
6.56
6.55
5.91
5.90
4.462
4.461
11
12
3.28
5.91
5.90
4.462
4.461
13
14
5.91
5.90
4.462
4.461
15
16
5.91
2.95
4.462
4.461
5
6
7
8
17
18
4.462
4.461
19
20
4.462
4.461
21
2.231
TABLE 8.4
a. MACRS-GDS percentages for 27.5-year residential rental property using mid-month convention
Month in Tax Year Property Placed in Service
Year
1
2
3
4
5
6
7
1.970% 1.667%
8
9
1.364% 1.061%
10
11
12
1
3.485% 3.182% 2.879% 2.576% 2.273%
2–9
3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636%
0.758% 0.455% 0.152%
10–26*
3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637% 3.637%
11–27** 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636%
28
1.970% 2.273% 2.258% 2.879% 3.182% 3.485% 3.636% 3.636% 3.636% 3.636% 3.636% 3.636%
29
0.152% 0.455% 0.758%
1.061% 1.364% 1.667%
b. MACRS-GDS percentages for 39-year nonresidential real property using mid-month convention
Month in Tax Year Property Placed in Service
Year
1
2
3
4
5
6
7
1.391% 1.177%
8
9
11
2–39
2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564% 2.564%
40
0.107% 0.321% 0.535% 0.749% 0.963%
1.605% 1.819%
0.535% 0.321%
12
2.461% 2.247% 2.033% 1.819% 1.605%
1.177% 1.391%
0.963% 0.749%
10
1
0.107%
2.033% 2.247% 2.461%
*Even-numbered year.
**Odd-numbered year.
299
300
Chapter 8
Depreciation
For a given property class, the sum of the depreciation rates equals
100 percent. Therefore, the asset is fully depreciated over the recovery
period. The depreciation allowance for a given year is the product of
the cost basis and the depreciation rate taken from Table 8.3 or 8.4, as
appropriate.
Letting pt denote the MACRS depreciation rate for year t, the
depreciation charge for year t and the book value at the end of year t are
given by
dt 5 pt P
(8.7)
and
t
t
j51
j51
Bt 5 P 2 a dj 5 P a1 2 a pj b
(8.8)
Applying MACRS with the SMP Machine Investment
EXAMPLE
Video Example
What are the depreciation allowances and book values for the SMP
machine using MACRS depreciation and property classes?
KEY DATA
Use Tables 8.2 and 8.3 to determine the property class and MACRS
percentages for the SMP machine.
Given: P 5 $500,000
Find: dt and Bt
SOLUTION
The IRS allows the $500,000 investment in a surface mount placement
machine to qualify as 5-year property. Using the MACRS percentages for
5-year properties and applying Equations 8.7 and 8.8, the depreciation
charges and the book values for the SMP machine are:
d1
d2
d3
d4
d5
d6
5 0.201$500,0002 5 $100,000
5 0.321$500,0002 5 $160,000
5 0.1921$500,0002 5 $96,000
5 0.11521$500,0002 5 $57,600
5 0.11521$500,0002 5 $57,600
5 0.05761$500,0002 5 $28,800
B1 5 $500,000 2 $100,000 5 $400,000
B2 5 $400,000 2 $160,000 5 $240,000
B3 5 $240,000 2 $96,000 5 $144,000
B4 5 $144,000 2 $57,600 5 $86,400
B5 5 $86,400 2 $57,600 5 $28,800
B6 5 $28,800 2 $28,800 5 $0
Many variations exist in how depreciation charges are computed for the
wide range of depreciable properties that exist. As noted at the chapter’s
beginning, our objective is to ensure that you understand the after-tax
Summary 301
consequences of the design decisions that engineers make. Because
depreciation affects income taxes, engineers must understand how depreciation allowances are determined.
8.4.2 MACRS-ADS
Although MACRS-GDS is far more popular, taxpayers may elect to
claim MACRS-ADS deductions. The MACRS-ADS method is required
for use on some property, including property (1) used predominantly
outside the United States; (2) having any tax-exempt use; (3) financed by
tax-exempt bonds; or (4) imported and covered by executive order of the
U.S. president. The MACRS-ADS method is simply straight-line depreciation with either a half-year (SLH) or a mid-month (SLM) convention,
as appropriate.
Other than public sector applications, it is unlikely that a taxpayer
would choose to use MACRS-ADS depreciation. Hence, we have limited
detailed consideration to MACRS-GDS.
KEY CONCEPTS
1. Learning Objective: Describe the impact that depreciation has on design
decisions from an income-tax perspective. (Section 8.1)
Most property deceases in value with use and time—it depreciates. U.S.
income-tax law permits depreciation allowances to be treated as expenses
and deducted from taxable income. This is an important concept to consider, as it can affect the way investments and annual operating costs are
treated from an after-tax (as opposed to before-tax) perspective. The
depreciation method used can significantly impact the present worth of
depreciation allowances and thereby can significantly impact the income
taxes a business pays.
2. Learning Objective: Define a depreciable property and distinguish whether
this property is tangible or intangible, and if it is tangible, whether the property
is personal or real. (Section 8.2)
The Internal Revenue Service requires the following requirements for
depreciable property:
■
■
■
It must be used in business or held for the production of income.
It must have a life that can be determined, and that life must be longer
than a year.
It must be something that wears out, decays, gets used up, becomes
obsolete, or loses value from natural causes.
SUMMARY
302
Chapter 8
Depreciation
Depreciable property may be tangible or intangible. Tangible property can
be seen or touched and can be categorized as personal or real. Personal
property is goods such as cars, trucks, machinery, furniture, equipment,
and anything that is tangible except real property. Real property is land and
generally anything that is erected on, growing on, or attached to it. Intangible property cannot be seen or touched but has value to the owner; it
includes copyrights, brands, software, goodwill, formulas, designs, patents,
trademarks, licenses, information bases, and franchises.
3. Learning Objective: Calculate depreciation using the Straight-Line
(SLN), Declining Balance (DB), and Double Declining Balance (DDB) methods.
(Section 8.3)
SLN, DB, and DDB represent the first category of depreciation methods
typically used for financial reporting purposes. In SLN, depreciation
charges are uniform across the useful life of the property. DB and DDB are
extensions of the SLN with intent to accelerate the depreciation schedule.
The following table summarizes the key equations for depreciation and
book values using these methods.
Depreciation Method
Depreciation Value
Book Value
Straight Line
dt 5 (P 2 F )/n
Bt 5 P – td
5 P 2 t (P 2 F)/n
Declining Balance
dt 5 pBt21
5 pP(1 2 p)t21
Bt 5 P (1 2 p)t
4. Learning Objective: Determine at what point it is optimal to switch from
a declining balance rate to a straight-line rate. (Section 8.3.3)
It is often desirable to switch from using a specified declining balance rate
to a straight-line rate at the point where the straight-line depreciation
charges exceed those specified using the declining balance rate. The equation that defines this point is:
Bt21 2 F
. pBt21
n 2 1t 2 12
(8.6)
5. Learning Objective: Calculate depreciation using the Modified Accelerated Cost Recovery System (MACRS). (Section 8.4)
The second category of depreciation methods is for income-tax purposes.
In the United States, MACRS is the only depreciation method approved
for use by the IRS. MACRS is an accelerated cost recovery system using
a half-year convention for various types of properties. Accelerated depreciation methods are preferred to nonaccelerated depreciation methods
because of the desire to maximize the present worth of depreciation allowances. The recovery period for depreciable property is specified by the IRS
and is usually shorter than the useful life of the property.
Summary 303
KEY TERMS
Amortization, p. 289
Book Value, p. 290
Cost Basis, p. 289
Declining Balance (DB), p. 291
Depreciable Property, p. 288
Depreciation, p. 286
Depreciation Allowances, p. 288
Double Declining Balance (DDB),
p. 292
Modified Accelerated Cost Recovery
System (MACRS), p. 296
Modified Accelerated Cost Recovery
System–Alternative Depreciation
System (MACRS-ADS), p. 297
Modified Accelerated Cost Recovery
System–General Depreciation
System (MACRS-GDS), p. 297
Personal Property, p. 289
Real Property, p. 289
Half-Year Convention, p. 297
Intangible Property, p. 289
Straight-Line Depreciation, p. 289
Tangible Property, p. 289
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FE-LIKE PROBLEMS
1.
a.
b.
c.
d.
Which of the following is not a requirement for an asset to be depreciable?
It must have a life longer than 1 year
It must have a basis (initial purchase plus installation cost) greater than
$1,000
It must be held with the intent to produce income
It must wear out or get used up
2.
A lumber company purchases and installs a wood chipper for $200,000.
The chipper is classified as MACRS 7-year property. The chipper’s useful
life is 10 years. The estimated salvage value at the end of 10 years is $25,000.
Using Straight Line depreciation, compute the first year depreciation.
a. $28,571.43
c. $17,500.00
b. $20,000.00
d. $25,000.00
3.
Which of the following is not true about depreciation?
Depreciation is not a cash flow
To be depreciable, an asset must have a life longer than one year
A 5-year property will generate a regular MACRS-GDS depreciation
deductions in six fiscal years.
For MACRS-GDS an estimate of the salvage values is required
a.
b.
c.
d.
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4.
The depreciation deduction for year 11 of an asset with a 20-year useful
life is $4,000. If the salvage value of the asset was estimated to be zero and
straight line depreciation was used to calculate the depreciation deduction for
year 11, what was the initial cost of the asset?
a. $42,105
c. $72,682
b. $67,682
d. $80,000
5.
The depreciation deduction for year 11 of a 15-year property with a 20-year
class life is $4,000. If the salvage value of the asset is estimated to be $5,000
and MACRS-GDS is used to calculate the depreciation deduction for year 11,
what was the initial cost of the asset?
a. $42,105
c. $72,682
b. $67,682
d. $80,000
6.
MACRS-GDS deductions are a combination of which other methods of
depreciation?
a. Sum of Years Digits and Straight Line
b. Sum of Years Digits and Declining Balance
c. Double Declining Balance and 150% Declining Balance
d. Double Declining Balance and Straight Line
7.
Production equipment used in the bottling of soft drinks (MACRS-GDS,
10 year property) is purchased and installed for $630,000. What is the depreciation deduction in the 4th year under MACRS-GDS?
a. $90,720
c. $72,576
b. $78,687
d. $48,510
8.
The depreciation allowance for a $100,000 MACRS-GDS asset was
$8,550 after its third year. What was the depreciation allowance after its
second year?
a. $8,550
c. $18,000
b. $9,500
d. Cannot be determined with the
information given
9.
A lumber company purchases and installs a wood chipper for $200,000.
The chipper is classified as MACRS 7-year property. The chipper’s useful
life is 10 years. The estimated salvage value at the end of 10 years is $25,000.
Using MACRS depreciation, compute the first year depreciation.
a. $17,500.00
c. $25,007.50
b. $20,000.00
d. $28,580.00
10.
An x-ray machine at a dental office is MACRS 5-year property. The x-ray
machine costs $6,000 and has an expected useful life of 8 years. The salvage
value at the end of 8 years is expected to be $500. Assuming MACRS depreciation, what is the book value at the end of the third year?
a. $1,584
c. $3,916
b. $1,728
d. $4,272
Summary 305
11.
Which of the following is (are) required to calculate MACRS-GDS depreciation deductions?
I. Property Class
III. First Cost
II. Salvage Value
IV. Annual Maintenance Costs
a. I and III only
c. I, II, and III
b. II and III only
d. I, II, III, and IV
PROBLEMS
Section 8.1 The Role of Depreciation in Economic Analysis
1. Michelin is considering going “lights out” in the mixing area of the business
that operates 24/7. Currently, personnel with a loaded cost of $600,000 per
year are used to manually weigh real rubber, synthetic rubber, carbon black,
oils, and other components prior to manual insertion in a Banbary mixer that
provides a homogeneous blend of rubber for making tires. New technology
is available that has the reliability and consistency desired to equal or exceed the quality of blend now achieved manually. It requires an investment of
$3.75 million, with $80,000 per year operational costs and will replace all of
the manual effort described above.
a. How are the current manual expenditures handled for tax purposes?
b. How would the new technology expenditures be handled for tax purposes?
2. Taxes are paid each year on some measure of financial gain. We typically
think of financial gain as “cash inflows minus cash outflows,” and yet simply
subtracting outflows due to capital investments in plant and equipment that
will be used over multiple years is not allowed.
a. Why are these outflows not allowed to just be subtracted?
b. Where do depreciation allowances fit into the tax picture, especially since
they are not cash flows?
3. SteelTubes had sales of $300 million this year. Expenses were $250 million.
Aside from these figures, the company also invested in new mills for carbon
steel tubing, complete with peripheral loading, straightening, and coiling
equipment plus facility reconfiguration totaling $14 million. SteelTubes believes the usable life of the mill will be only seven years, owing to technological advances. There were no other financial considerations.
a. Looking strictly at cash flows, what will be reported as the financial gain
or loss? Is this a fair representation of financial performance?
b. If, for internal financial reporting, the manufacturer writes off equal
amounts of the capital investment over the usable life, beginning this year,
what will be the reported financial gain or loss?
4. Which of the following statements are TRUE?
a. The reason for including a treatment of depreciation in this book is to
allow you to develop a reasonably accurate report to the owners of a business regarding its value at any given point in time.
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b. Depreciation spreads investment costs over the useful life of equipment
purchased.
c. Depreciation allowances can be treated as expenses because they are cash
flows.
d. Depreciation affects income taxes, which are cash flows.
Section 8.2
5.
Language of Depreciation
For each of the following assets, state whether the asset is tangible/
intangible property, personal/real property, and depreciable/nondepreciable
property.
a. A cell phone tower.
b. A plot of land for your personal use.
c. A computer used in your job.
d. A Mooney viscometer used in a polymers lab.
e. An electric generator purchased by a public utility.
6. For each of the following assets, state whether the asset is tangible/intangible
property, personal/real property, and depreciable/nondepreciable property.
a. A melt-indexer used in a company research lab.
b. A plot of land for the production of income.
c. A restaurant franchise.
d. An amateur radio tower attached to land with multiple antennas on it.
e. Fencing and landscaping around an office complex.
7.
For each of the following assets, state whether the asset is tangible/
intangible property, personal/real property, and depreciable/nondepreciable
property.
a. A tractor (part of tractor-trailer rig - 18-wheeler)
b. A copyright.
c. An all-in-one copier, scanner, fax machine used in your business.
d. A rental home for the purpose of generating rental income.
8. For each of the following assets, state whether the asset is tangible/intangible
property, personal/real property, and depreciable/nondepreciable property.
a. A computer used for personal e-mail, blogging, and hobbies.
b. A file cabinet in your business office.
c. A commercial delivery truck.
d. An office complex for business.
Section 8.3
9.
Straight-Line and Declining Balance Depreciation Methods
Video Solution A high-precision programmable router for shaping furniture components is purchased by Henredon for $190,000. It is expected to
last 12 years and have a salvage value of $5,000. Calculate the depreciation
deduction and book value for each year.
a. Use straight-line depreciation.
b. Use declining balance depreciation using a rate that ensures the book
value equals the salvage value.
Summary 307
c. Use double declining balance depreciation.
d. Use double declining balance switching to straight line depreciation.
10. A land grant university has upgraded its Course Management System
(CMS), integrating the system throughout all of its main campus and
branch campuses around the state. It has purchased a set of 15 servers
and peripherals for needs associated with the CMS. The total cost basis is
$120,000 and expected use will be 5 years after which they will have no
projected value. Calculate the depreciation deduction and book value for
each year.
a. Use straight-line depreciation.
b. Use declining balance depreciation using a rate that ensures the book
value equals the salvage value.
c. Use double declining balance depreciation.
d. Use double declining balance switching to straight line depreciation.
11.
A small truck is purchased for $17,000. The truck is expected to be of use
to the company for 6 years, after which it will be sold for $3,500.
a. Use straight-line depreciation.
b. Use declining balance depreciation using a rate that ensures the book
value equals the salvage value.
c. Use double declining balance depreciation.
d. Use double declining balance switching to straight line depreciation.
12.
Video Solution A surface mount PCB placement/soldering line is to be
installed for $1.6 million. It will have a salvage value of $100,000 after 5 years.
Determine the depreciation deduction and the resulting unrecovered investment during each year of the asset’s life.
a. Use straight-line depreciation.
b. Use declining balance depreciation using a rate that ensures the book
value equals the salvage value.
c. Use double declining balance depreciation.
d. Use double declining balance switching to straight line depreciation.
13.
GO Tutorial A tractor for over-the-road hauling is purchased for $90,000.
It is expected to be of use to the company for 6 years, after which it will be
salvaged for $4,000. Calculate the depreciation deduction and the unrecovered
investment during each year of the tractor’s life.
a. Use straight-line depreciation.
b. Use declining balance depreciation using a rate that ensures the book
value equals the salvage value.
c. Use double declining balance depreciation.
d. Use double declining balance switching to straight line depreciation.
14. WindPower Inc designs and commissions the manufacture of a wind powered
inverter-based constant voltage generator for research and experimentation
with low-rated, highly variable, wind fields as a form of alternative energy.
The unit cost $35,000 plus $3,000 for shipping and installation. After 3 years,
WindPower had no further use for the experimental unit and was able to sell
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it for $2,000, less $500 for removal. WindPower had depreciated the generator cost over the 3 years with an estimated life of 5 years and a terminal book
value of $1,000. All of the following parts relate to financial reporting, not
computing income taxes.
a. What is the total investment cost (basis) for this generator?
b. What is the net market value actually received after three years?
c. By what amount did the book value differ from the net market value at the
end of 3 years if the following depreciation method was used?
i. Straight-line depreciation.
ii. Declining balance depreciation using a rate that ensures the book value
equals the salvage value.
iii. Double declining balance depreciation.
iv. Double declining balance switching to straight line depreciation.
15. A digitally controlled plane for manufacturing furniture is purchased on
April 1 by a calendar-year taxpayer for $66,000. It is expected to last 12 years
and have a salvage value of $5,000. Calculate the depreciation deduction
during years 1, 4, and 8.
a. Use straight-line depreciation.
b. Use declining balance depreciation, with a rate that ensures the book value
equals the salvage value.
c. Use double declining balance depreciation.
d. Use declining balance depreciation, switching to straight line depreciation.
16. A file server and peripherals are purchased in December by a calendar-year
taxpayer for $8,000. The server will be used for 6 years and be worth $200 at
that time. Calculate the depreciation deduction during years 1, 3, and 6.
a. Use straight-line depreciation.
b. Use declining balance depreciation, with a rate that ensures the book value
equals the salvage value.
c. Use double declining balance depreciation.
d. Use declining balance depreciation, switching to straight line depreciation.
Section 8.4
Modified Accelerated Cost Recovery System (MACRS)
17. Which of the following statements are FALSE?
a. MACRS is the only depreciation method approved by the IRS for comput-
ing income-tax liability and it is also the most commonly used method in
the United States for financial reporting.
b. MACRS stands for Modified Annuitized Cost Recovery System.
c. MACRS-GDS is based on double declining balance switching to straight
line depreciation.
d. MACRS-ADS is based on straight line depreciation.
18. Which of the following statements are TRUE?
a. MACRS-GDS uses a half-year convention, whereas MACRS-ADS does not.
b. The half-year convention has the effect of depreciating over n – 1 full
years (2, 3, . . . , n), and two half-years (1 and n 1 1).
Summary 309
c. The investment’s property class establishes the number of years over
which the cost basis is to be recovered (depreciated).
d. In general, MACRS-GDS has a longer recovery period (depreciation
period) than MACRS-ADS.
19. A mold for manufacturing medical supplies (MACRS-GDS 5-year property)
is purchased at the beginning of the fiscal year for $30,000. The estimated
salvage value after 10 years is $3,000. Calculate the depreciation deduction
and the book value during each year of the asset’s life using MACRS-GDS
allowances.
20. A panel truck (MACRS-GDS 5-year property) is purchased for $17,000. The
truck is expected to be of use to the company for 6 years, after which it will
be sold for $2,500. Calculate the depreciation deduction and the book value
during each year of the asset’s life using MACRS-GDS allowances.
21. A digitally controlled plane for manufacturing furniture (MACRS-GDS
7-year property) is purchased on April 1 by a calendar-year taxpayer for
$66,000. It is expected to last 12 years and have a salvage value of $5,000.
Calculate the depreciation deduction during years 1, 4, and 8 using MACRSGDS allowances.
22. Material-handling equipment used in the manufacture of grain products
(MACRS-GDS 10-year property) is purchased and installed for $180,000.
It is placed in service in the middle of the tax year. If it is removed just before the end of the tax year approximately 4.5 years from the date placed in
service, determine the depreciation deduction during each of the tax years
involved using MACRS-GDS allowances.
23. Repeat the previous problem (problem 22) if the material-handling equipment
is removed just after the tax year, again using MACRS-GDS allowances.
24. Electric utility transmission and distribution equipment (MACRS-GDS 20-year
property) is placed in service at a cost of $300,000. It is expected to last 30 years
with a salvage value of $15,000. Determine the depreciation deduction and
the book value during each year of the first 4 tax years using MACRS-GDS
allowances.
25. A business building (MACRS-GDS 39-year property) is placed in service by
a calendar-year taxpayer on January 4 for $300,000. Calculate the depreciation deduction for years 1 and 10, assuming the building is kept longer than
10 years, using MACRS-GDS allowances.
26. A building used for the overhaul of dewatering systems (MACRS-GDS 39-year
property) is placed in service on October 10 by a calendar-year taxpayer for
$140,000. It is sold almost 4 years later on August 15. Determine the depreciation deduction during years 1, 3, and 5 and the book value at the end of
years 1, 3, and 5 using MACRS-GDS allowances.
27. A rental apartment complex (MACRS-GDS 27.5-year property) is placed in
service by a calendar-year taxpayer on January 4 for $200,000. If the apartments are kept for 5 years and 2 months (sold on March 6), determine the
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depreciation deduction during each of the 6 tax years involved using MACRSGDS allowances.
28. A rental house (MACRS-GDS 27.5-year property) is placed in service by a
calendar-year taxpayer during July for $70,000. Determine the depreciation
deduction and resulting book value for each applicable year using MACRSGDS allowances.
29. For each of the following assets, state both the MACRS-GDS property class,
if applicable, and the specific depreciation method to be used (e.g., 15-year
property, 150 percent DBSLH).
a. A melt-indexer used in a company research lab.
b. A plot of land for the production of income.
c. A restaurant franchise.
d. An amateur radio tower attached to land with multiple antennas on it.
e. Fencing and landscaping around an office complex.
30. For each of the following assets, state both the MACRS-GDS property class,
if applicable, and the specific depreciation method to be used (e.g., 15-year
property, 150 percent DBSLH).
a. A cell phone tower.
b. A plot of land for your personal use.
c. A computer used in your job.
d. A Mooney viscometer used in a polymers lab.
e. An electric generator purchased by a public utility.
31. For each of the following assets, state both the MACRS-GDS property class,
if applicable, and the specific depreciation method to be used (e.g., 15-year
property, 150 percent DBSLH).
a. A computer used for personal e-mail, blogging, and hobbies.
b. A file cabinet in your business office.
c. A commercial delivery truck.
d. An office complex for business.
32. For each of the following assets, state both the MACRS-GDS property class,
if applicable, and the specific depreciation method to be used (e.g., 15-year
property, 150 percent DBSLH).
a. A tractor (part of tractor-trailer rig, an 18-wheeler).
b. A copyright.
c. An all-in-one copier, scanner, fax machine used in your business.
d. A rental home for the purpose of generating rental income.
33.
Electric generating and transmission equipment is placed in service at
a cost of $3,000,000. It is expected to last 30 years with a salvage value of
$250,000.
a. What is the MACRS-GDS property class?
b. Determine the depreciation deduction and the unrecovered investment
during each of the first 4 tax years.
Summary 311
34. Bell’s Amusements purchased an expensive ride for their theme and amuse-
ment park situated within a city-owned Expo Center. Bell’s had a multi-year
contract with Expo Center. The ride cost $1.2 million. Bell’s anticipated that
the ride would have a useful life of 12 years, after which the net salvage value
would be $0. After 4 years, the city and Bell’s were unable to come to an
agreement regarding an extended contract. In order to expedite Bell’s departure, Expo Center agreed to purchase the ride and leave it in place. Right at
the end of the 4th fiscal year, Expo Center paid to Bell’s the unrecovered investment (remember the half-year convention for MACRS-GDS). Determine
the amount paid, assuming:
a. Straight line depreciation used for valuation purposes was agreed upon.
b. MACRS-GDS depreciation used for tax purposes (state the property class)
was agreed upon.
35.
A virtual mold apparatus for producing dental crowns permits an infinite
number of shapes to be custom constructed based upon mold imprints taken
by dentists. It costs $28,500 and is purchased at the beginning of the tax year.
It is expected to last 9 years with no salvage value at that time. The dental
supplier depreciates assets using MACRS, and yet values assets of the company using straight line depreciation. Determine the depreciation allowance
and the unrecovered investment for each year:
a. For tax purposes (be sure to identify the MACRS-GDS property class).
b. For company valuation purposes.
36. Milliken uses a digitally controlled “dyer” for placing intricate and integrated
patterns on manufactured carpet squares for home and commercial use. It is
purchased on April 1 for $350,000. It is expected to last 8 years and have a
salvage value of $30,000.
a. What is the MACRS-GDS property class?
b. Determine the depreciation deduction during each year of the asset’s
8 year life.
c. Determine the unrecovered investment at the end of each of the 8 years.
37.
GO Tutorial A high-precision programmable router for shaping furniture
components is purchased by Henredon for $190,000. It is expected to last
12 years. Calculate the depreciation deduction and book value for each year
using MACRS-GDS allowances.
a. What is the MACRS-GDS property class?
b. Assume the complete allowable depreciation schedule is used.
c. Assume the asset is sold during the 5th year of use.
38. Material handling equipment used in the manufacture of sugar is purchased
and installed for $250,000. It is placed in service in the middle of the tax
year and removed from service 5.5 years later. Determine the MACRS-GDS
depreciation deduction during each of the tax years involved assuming:
a. What is the MACRS-GDS property class?
b. The equipment is removed one day before the end of the tax year.
c. The equipment is removed one day after the end of the tax year.
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39.
GO Tutorial Englehard purchases a slurry-based separator for the mining of clay that costs $700,000 and has an estimated useful life of 10 years, a
MACRS-GDS property class of 7 years, and an estimated salvage value after
10 years of $75,000. It was financed using a $200,000 down payment and a
loan of $500,000 over a period of 5 years with interest at 10%. Loan payments
are made in equal annual amounts (principal plus interest) over the 5 years.
a. What is the amount of the MACRS-GDS depreciation taken in the 3rd
year?
b. What is the book value at the end of the 3rd year?
c. Returning to the original situation, what is the amount of the MACRSGDS depreciation taken in the 3rd year if the separator is also sold during
the 3rd year?
40. A portable concrete test instrument used in construction for evaluating and
profiling concrete surfaces (MACRS-GDS 5-year property class) is purchased in December by a calendar-year taxpayer for $22,000. The instrument
will be used for 6 years and be worth $2,000 at that time.
a. Calculate the depreciation deduction for years 1, 3, and 6.
b. If the instrument is sold in year 4, determine the depreciation deduction
for years 1, 3, and 4.
41. A mold for manufacturing powdered metal firearm parts is purchased by
Remington at the beginning of the fiscal year for $120,000. The estimated
salvage value after 8 years is $10,000. Calculate the depreciation deduction
and book value for each year using MACRS-GDS allowances.
a. What is the MACRS-GDS property class?
b. Assume the complete depreciation schedule is used.
c. Assume the asset is sold during the 6th year of use.
42. A hydroprocessing reactor, an asset used in petroleum refining, is placed into
service at a cost of $2.7 million. It is thought to have a useful life, with turnarounds and proper maintenance, of 18 years and will have a salvage value of
nothing at that time.
a. What is the MACRS-GDS property class?
b. Determine the depreciation deduction and the unrecovered investment
over the depreciable life of the reactor.
43. A manufacturing system for fabricated metal products is purchased in the
middle of the fiscal year for $800,000. The estimated salvage value after
10 years is $130,000.
a. What is the MACRS-GDS property class?
b. Determine the depreciation deduction during each year of the asset’s
10 year life.
c. Determine the unrecovered investment at the end of each of the 10 years.
44.
Video Solution A surface mount PCB placement/soldering line for the
manufacture of electronic components is to be installed for $1.6 million with
an expected life of 6 years. Determine the depreciation deduction and the
Summary 313
resulting unrecovered investment during each year of the asset’s life using
MACRS-GDS allowances.
a. What is the MACRS-GDS property class?
b. Assume the line equipment will be sold shortly after the 5th year.
c. Assume the line equipment is sold during the 3rd year of use.
45.
GO Tutorial A gas-powered electric generator is purchased by a public
utility as part of an expansion program. It is expected to be useful, with proper
maintenance, for an estimated 30 years. The cost is $17 million, installed.
The salvage value at the end of 30 years is expected to be 10% of the original
cost, not counting installation.
a. What is the MACRS-GDS property class?
b. Determine the depreciation deduction and the unrecovered investment for
years 1, 5, and the last depreciable year of the generator.
46. A tractor for over-the-road hauling is purchased for $90,000. It is expected
to be of use to the company for 6 years, after which it will be salvaged for
$4,000. Calculate the depreciation deduction and the unrecovered investment
during each year of the tractor’s life using MACRS-GDS allowances.
a. What is the MACRS-GDS property class?
b. Assume the tractor is used for the full 6 years
c. Assume the tractor is sold during the 4th year of use.
d. Assume the tractor is sold during the 3rd year of use.
47. Pretend that you have misplaced your MACRS tables. Develop the tables
for a property class of 3 years assuming 200% DB depreciation switching to
straight line; half-year convention; salvage value equal to $0. Your answers
should match those for MACRS-GDS 3-year property.
48. Pretend that you have misplaced your MACRS tables. Develop the tables
for a property class of 5 years assuming 200% DB depreciation switching to
straight line; half-year convention; salvage value equal to $0. Your answers
should match those for MACRS-GDS 5-year property.
49.
Suppose the IRS has decided to institute a new MACRS-GDS property
class of only 2 years. It will follow the usual depreciation conventions, determined in the same way as 3, 5, 7, and 10-year property. Determine the yearly
MACRS-GDS percentages for each year.
50. Suppose the IRS has decided to institute a new MACRS-GDS property class
of 4 years. It will follow the usual depreciation conventions, determined in
the same way as 3, 5, 7, and 10-year property. Determine the yearly MACRSGDS percentages for each year.
51.
A nonresidential business building is placed in service by a calendar year
taxpayer on March 3 for $300,000.
a. What is the MACRS-GDS property class?
b. Calculate the depreciation deduction for years 1, 4, and 8 if it is kept longer
than 8 years.
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Depreciation
c. Calculate the depreciation deduction for years 1, 4, and 8 if it is sold on
August 12 in the 8th calendar year.
d. Why is it highly unlikely the building will ever be completely depreciated?
52. Now that you are making the big bucks, your spouse has decided to venture
into the rental property business. He purchases a rental house and after making some improvements it has a basis of $85,000. He places it in service as a
calendar-year taxpayer during May and sells it in September, just over 4 years
later.
a. What is the MACRS-GDS property class?
b. Determine the depreciation deduction during each of the years involved.
c. Determine the unrecovered investment during each of the years involved.
53.
A permanent steel building used for the overhaul of dewatering systems
(engines, pumps, and wellpoints) is placed in service on July 10 by a calendar
year taxpayer for $240,000. It is sold almost 5 years later on May 15.
a. What is the MACRS-GDS property class?
b. Determine the depreciation deduction during each of the years involved.
c. Determine the unrecovered investment during each of the years involved.
54. A residential rental apartment complex is placed in service by a calendar year
taxpayer on February 27 for $530,000. The apartments are kept for slightly
more than 6 years and sold on March 6.
a. What is the MACRS-GDS property class?
b. Determine the depreciation deduction during each of the 7 years involved.
c. Determine the unrecovered investment during each of the 7 years involved.
55.
Video Solution Equipment for manufacturing vegetable oil products is
purchased from Alfa. Items such as oil expellers, filter presses, and a steam
generator are purchased for $1.2 million. These devices are expected to be
used for 11 years with no salvage value at that time. Compare MACRS to traditional depreciation methods by calculating yearly depreciation allowances,
present worth of the depreciation allowances, and book value for each year
using each of the following. MARR is 9%.
a. MACRS-GDS as is proper over its property class depreciation life.
b. DDB taking a full deduction in the first year, with the last deduction in
year 10.
c. DDB switching to straight line, taking a full deduction in the 1st year, with
the last deduction in year 10.
d. SLN taking a full deduction in the 1st year, with the last deduction in year 10.
56. A building with business offices, a reception area, and numerous small diag-
nosis rooms is placed in service by a group of three orthopedic surgeons on
January 4 for $650,000.
a. What is the MACRS-GDS property class?
b. Calculate the depreciation deduction for years 1, 4, and 7 if it is kept longer
than 7 years.
Summary 315
c. Calculate the depreciation deduction for years 1, 4, and 7 if it is sold on
July 1 in the 7th calendar year.
d. Why is it highly unlikely the building will ever be completely depreciated?
57. Ultra-clean special handling devices used in the filling process for the manu-
facture of baby food are placed into use at a cost of $850,000. These devices
are expected to be useful for 4 years with a salvage value of nothing at that
time. Compare MACRS to traditional depreciation methods by calculating
yearly depreciation allowances, present worth of the depreciation allowances,
and book value for each year using each of the following. MARR is 11%.
a. MACRS-GDS as is proper over its property class depreciation life.
b. DDB taking a full deduction in the first year, with the last deduction in
year 3.
c. DDB switching to straight line, taking a full deduction in the 1st year, with
the last deduction in year 3.
d. SLN taking a full deduction in the 1st year, with the last deduction in year 3.
ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E : I N T E L
With revenue of $54 billion in 2011, Intel is the world’s largest semiconductor
chip maker. The chances are great that your computer includes one or more
of Intel’s products: chips, printed circuit boards, and other semiconductor
components. More specifically, Intel’s products include microprocessors,
chipsets, motherboards, flash memory, wired and wireless connectivity
products, communications infrastructure components (including network
processors), application and cellular baseband processors, and products for
networked storage. Intel’s products appear in computers, as well as in servers and networking and communications products.
Intel has long been a leader in silicon process technology and manufacturing. As of December 31, 2011, Intel employed 100,100 people worldwide,
with approximately 55 percent of those employees located in the United
States. A substantial number of Intel’s employees are engineers or computer
scientists. Intel invests heavily in R&D and is committed to investing in
world-class technology development, especially in the design and production of integrated circuits. In 2011, R&D expenditures totaled $8.4 billion.
Semiconductor manufacturing is very expensive. Many of Intel’s competitors do not own manufacturing, assembly, and test facilities. Instead,
they contract with third parties to perform those functions. Intel considers
its ownership of fabrication facilities to be one of its most significant strategic advantages. Intel’s plans for 2011 included building and expanding their
22 nm process technology manufacturing capacity.
DISCUSSION QUESTIONS:
1. Intel is a high-tech company with significant R&D. What income tax
considerations would such a company have as compared to a “lowertech” company?
316
INCOME TAXES
2. What tax implications might there be for a global company such as
Intel?
3. Do you suspect that Intel is making large capital investments? If so,
what impact will depreciation have on these investments?
4. What barriers of entry would competitors of Intel have entering into
this market?
LEARNING OBJECTIVES
When you have finished studying this chapter, you should be able to:
1. Calculate corporate income taxes, including the effective, incremental,
and marginal tax rates. (Section 9.1)
2. Analyze the after-tax implications for investment alternatives using
retained earnings. (Section 9.2)
3. Analyze the after-tax implications for investment alternatives using
borrowed capital. (Section 9.3)
4. Illustrate the different tax consequences for the common example of a
lease vs. purchase alternative. (Section 9.4)
INTRODUCTION
From our discussion of depreciation methods, we know that income taxes
can significantly impact the economic viability of a capital investment.
Tax dollars are cash flows, and therefore it is necessary to consider them
explicitly, just like costs of wages, equipment, materials, and energy. One
of the major factors affecting income taxes is depreciation. Although depreciation allowances are not cash flows, their magnitudes and timing
affect income taxes. Therefore, proper knowledge and application of tax
317
318
Chapter 9
Income Taxes
laws can make the economic difference between accepting or rejecting an
investment alternative, as well as between profit or loss on the corporate
bottom line.
The chapter begins with basic concepts of income taxes, focusing
on corporate, not personal, investments. Next, we consider the impact
of depreciation allowances on after-tax cash flows and demonstrate the
after-tax effects of accelerated depreciation methods. Then we expand
our consideration of after-tax consequences to the use of borrowed
capital, because interest paid by businesses is deductible from taxable
income.
The taxes a corporation pays represent a real cost of doing business
and, consequently, affect the cash flow profile of projects. For this reason, it is wise to perform economic analyses on an after-tax basis. Aftertax analysis procedures are identical to before-tax procedures; however,
cash flows are adjusted for taxes paid or saved.
Corporations typically pay a variety of taxes, including ad valorem
(property), sales, excise (a tax on the manufacture, sale, or consumption
of a commodity), and income taxes. Among the various types of taxes
paid, corporate income taxes tend to be the most significant when performing an economic analysis, because most of the other taxes are not
affected by the kinds of investments generally included in an engineering economic analysis. Income taxes are assessed on gross income less
certain allowable deductions and on gains resulting from the disposal of
property.
Federal and state income tax regulations are detailed, intricate, and
subject to change over time. Hence, only general concepts and procedures for calculating after-tax cash flow profiles and performing after-tax
analyses are emphasized here. Furthermore, due to the diversity of state
laws, only federal income taxes are considered.
Systematic Economic Analysis Technique
1.
2.
3.
4.
5.
6.
7.
Identify the investment alternatives
Define the planning horizon
Specify the discount rate
Estimate the cash flows
Compare the alternatives
Perform supplementary analyses
Select the preferred investment
9-1
9-1
Corporate Income Tax Rates
319
CORPORATE INCOME TAX RATES
LEARNING O BJECTI VE: Calculate corporate income taxes, including the
effective, incremental, and marginal tax rates.
Video Lesson:
Corporate Taxes
For income-tax purposes, a corporation includes associations, business
trusts, joint stock companies, insurance firms, and trusts and partnerships
that actually operate as associations or corporations. Corporate income
tax, however, is not limited to traditional business organizations. Engineers, doctors, lawyers, and other professional people may be treated as
corporations if they have formally organized under state professional association acts.
Ordinary federal income-tax rates imposed on corporations are given in
Table 9.1. The tax rates shown are current for 2012; in fact, they have not
changed since 1993. As a matter of interest, the highest corporate tax bracket
has ranged from 1 percent in 1915 to 12 percent in 1918, to 13.5 percent in
1927, to 19 percent in 1939, to 24 percent in 1940, to 40 percent in 1945,
to 52.8 percent in 1969, to 46 percent from 1979 to 1986, to 34 percent
from 1988 to 1992, and, most recently, to 35 percent from 1993 to 2012.
Because income tax laws do change and may differ from those described
in the text, consult the U.S. Internal Revenue Service website (http://www.
irs.gov) for up-to-date information on U.S. tax laws.
TABLE 9.1
Corporate Income Tax Rates for Tax Years Beginning January 1, 1993 and Beyond
Taxable Income (TI), in $
Tax Rate (t)
Income Tax (T)
0 , TI # 50,000
0.15
0.15(TI)
50,000 , TI # 75,000
0.25
7,500 1 0.25(TI 2 50,000)
75,000 , TI # 100,000
0.34
13,750 1 0.34(TI 2 75,000)
100,000 , TI # 335,000
0.39(0.34 1 0.05)
22,250 1 0.39(TI 2 100,000)
335,000 , TI # 10,000,000
0.34
113,900 1 0.34(TI 2 335,000)
10,000,000 , TI # 15,000,000
0.35
3,400,000 1 0.35(TI 2 10,000,000)
15,000,000 , TI # 18,333,333
0.38(0.35 1 0.03)
5,150,000 1 0.38(TI 2 15,000,000)
18,333,333 , TI
0.35
0.35(TI)
Computing Income Taxes for Corporations
A small company is currently forecasting a taxable income of $50,000 for
the year. The company’s owner is considering an investment that will
increase taxable income by $45,000. If the investment is pursued and the
anticipated return occurs, what will be the magnitude of the increase in
income taxes caused by the new investment? What will it be if the company is currently forecasting a taxable income of $400,000 for the year?
EXAMPLE
320
Chapter 9
Income Taxes
KEY DATA
Given: Base taxable income 1 5 $50,000; Base taxable income 2 5
$400,000; Increase in annual income with investment 5 $45,000
Find: Increase in income taxes for each scenario
SOLUTION
From Table 9.1, with a base taxable income of $50,000, the federal income
tax will be 0.15($50,000), or $7,500. The income tax for a taxable income
of $95,000 will be $13,750 1 0.34($95,000 2 $75,000) 5 $13,750 1
0.34($20,000) 5 $20,550. The magnitude increase in income tax caused by
the new investment is therefore $20,550 2 $7,500 5 $13,050.
Again, from Table 9.1, with a base taxable income of $400,000, the
federal income tax will be $113,900 1 0.34($400,000 2 $335,000) 5
$113,900 1 0.34($65,000) 5 $136,000, or 34 percent of $400,000. In this
case, every dollar of the additional $45,000 in taxable income will be taxed
at 34 percent, so the increase in taxable income will be 0.34($45,000) 5
$15,300, for a total tax of $151,300.
Effective Tax Rate The
income tax divided by the
taxable income. Also referred
to as the average tax rate.
Incremental Tax Rate The
incremental income tax
divided by the incremental
investment.
From Example 9.1, we can define three tax rates for the corporation:
1.
The effective tax rate, or average tax rate, which equals the income
tax divided by the taxable income
2. The incremental tax rate, which is the average rate charged to the
incremental investment
Marginal Tax Rate The
tax rate that will apply to
the last dollar included in
taxable income.
3. The marginal tax rate, which is the tax rate that will apply to the last
EXAMPLE
Computing Effective, Incremental, and Marginal Income Tax Rates
dollar included in taxable income
In Example 9.1, what are the effective, incremental, and marginal tax rates
when the base taxable income is $50,000?
KEY DATA
Given: Base taxable income 5 $50,000; Income tax on base taxable
income 5 $7,500; Increase in taxable income with investment 5 $45,000;
Taxable income with investment 5 $95,000; Income tax on taxable
income with investment 5 $20,550; Increase on taxable income with
investment 5 $13,050
Find: Effective, incremental, and marginal tax rates
9-1
When the base taxable income was $50,000, the corporation had an effective tax rate of 15 percent: $7,500/$50,000. With the addition of $45,000
in taxable income, the new effective tax rate became $20,550/$95,000 5
21.63 percent.
The $45,000 increase in taxable income caused taxes to increase from
$7,500 to $20,550. Therefore, $13,050/$45,000, or 29 percent, is the incremental tax rate.
The marginal tax rate was 34 percent, because taxable income of
$95,000 is in the 34 percent tax bracket. Therefore, the last $1 added to
taxable income contributed 34¢ to the company’s income taxes.
Conservatively, when performing economic justifications, an investment alternative’s after-tax economic worth should be computed using the
marginal tax rate.
Taxable income must first be determined before any tax rate can be
applied. Taxable income is gross income less allowable deductions. Gross
income is income in a general sense less any monies specifically exempt
from tax liability. Corporate deductions are subtracted from gross income
and commonly include items such as salaries, wages, repairs, rent, bad
debts, taxes (other than income), charitable contributions, casualty losses,
interest, and depreciation. Interest and depreciation are of particular
importance, because we can control them to some extent through financing
arrangements and accounting procedures. Taxable income is represented
pictorially in Figure 9.1. These components are not all cash flows, because
the depreciation allowance is treated simply as an expense in determining
taxable income.
Gross income
(income less
tax exemptions)
Taxable income
Depreciation
allowance
FIGURE 9.1
Deductions
(other than
interest on
borrowed
money and
depreciation)
Interest on
borrowed money
Pictorial Representation of Taxable Income
Corporate Income Tax Rates
SOLUTION
321
322
Chapter 9
Income Taxes
9-2
AFTER-TAX ANALYSIS USING RETAINED
EARNINGS (NO BORROWING)
LEARN I N G O B JEC T I V E : Analyze the after-tax implications for investment
alternatives using retained earnings.
Before proceeding with after-tax economic analyses, it will be useful to introduce the concept of a before-tax MARR (MARRBT) and an after-tax MARR
(MARRAT). An approximation to account for income taxes when they are
not explicitly considered in our calculations is to set MARRBT greater than
MARRAT to account for the effective income tax rate. Hence, an approximation of MARRAT will include the effect of income taxes as follows:
MARRAT < MARRBT 11 2 effective income tax rate2
After-Tax Cash Flow
(ATCF) The amount
remaining after income
taxes and deductions
(including interest, but
excluding depreciation
allowances) are subtracted
from gross income.
Before-Tax Cash Flow
(BTCF) A term used
when no borrowed
money is involved, equal
to the gross income less
deductions (excluding
depreciation allowances).
Before-Tax-and-Loan Cash
Flow (BT&LCF) A term
used when borrowed
money is involved, equal
to the gross income less
deductions (not including
either depreciation or
principal or interest on
the loan).
The base elements needed to calculate after-tax cash flow (ATCF) for
an alternative are summarized in Figure 9.2, which shows that the ATCF
is the amount remaining after income taxes and deductions, including
interest but excluding depreciation allowances, are subtracted from gross
income. In many of the following tables and spreadsheets, we simplify our
terminology by speaking of before-tax cash flow (BTCF). This term is
used when no borrowed money is involved; it equals gross income less
deductions, not including depreciation. The term before-tax-and-loan
cash flow (BT&LCF) is used when borrowed money is involved; it equals
gross income less deductions, not including either depreciation or principal or interest on the loan.
The following notation will be used for after-tax analysis of an investment alternative’s economic worth, without borrowed capital:
BTCF 5 before-tax cash flow
DWO 5 depreciation write-off or allowance
TI 5 taxable income
itr 5 income tax rate
T 5 income tax
ATCF 5 after-tax cash flow
Gross income
(income less
tax exemptions)
After-tax
cash flow
Income taxes
FIGURE 9 . 2
Deductions
(other than
interest on
borrowed
money and
depreciation allowance)
Interest on
borrowed money
Pictorial Representation of After-Tax Cash Flow (ATCF)
9-2 After-Tax Analysis Using Retained Earnings (No Borrowing)
323
The following equation holds for investments in depreciable property:
ATCF 5 BTCF 2 T
(9.1)
T 5 itr1TI2
(9.2)
TI 5 BTCF 2 DWO
(9.3)
ATCF 5 BTCF11 2 itr2 1 itr1DWO2
(9.4)
where
where
Therefore,
Because state income taxes are generally deductible expenses when computing federal income taxes, the overall income tax rate (itr) should equal
itr 5 str 1 ftr11 2 str2
(9.5)
where
str 5 state tax rate, and ftr 5 federal tax rate.
In the text, a value of 40 percent will be used for the combined income tax
rate in most examples and end-of-chapter problems. The 40 percent tax
rate is obtained using a 7 percent state income-tax rate and a 35 percent
federal income tax rate. From Equation 9.5,
0.07 1 0.3511 2 0.072 5 0.3955
or
39.55%.
Depending on the particular state (and country) in which the investment is
made, a different marginal income tax rate might be more appropriate.
Income Tax Rate (itr) The
income tax rate reflecting
the combined federal and
state income tax rates. This
rate is typically not additive,
as state income taxes
are generally deductible
expenses when computing
federal income tax.
State Tax Rate (str) The
state income-tax rate.
Federal Tax Rate (ftr) The
federal income-tax rate.
9.2.1 Single Alternative
In Examples 9.3 through 9.5 we focus on after-tax analysis, using retained
earnings, for a single investment alternative. Later, we will look at a similar analysis for multiple alternatives.
After-Tax Analysis with SLN Depreciation
EXAMPLE
Recall the $500,000 investment in the surface mount placement (SMP)
machine in an electronics manufacturing plant. It produced $92,500 in net
revenue (after taxes) for 10 years, plus a $50,000 salvage value at the end
of the 10-year period. Now, to determine the after-tax economic worth of
the investment, we will use a 40 percent income-tax rate and will perform
an after-tax analysis using a 10 percent after-tax MARR.
Given: P 5 $500,000, itr 5 40%, MARRAT 5 10%, ATCF (years 1–10) 5
$92,500, S 5 $50,000
Find: PWAT, AWAT, FWAT, IRRAT, ERRAT
KEY DATA
324
Chapter 9
Income Taxes
SOLUTION
TABLE 9.2
In the previous chapters, the $92,500 net revenue was given to be an aftertax figure. Although we did not specify the depreciation method used to
generate the ATCF amount of $92,500, it was based on straight-line depreciation over a 10-year recovery period. As shown in Table 9.2, a BTCF of
$124,166.67 results in an ATCF of $92,500 when straight-line depreciation is used with a $50,000 salvage value.
Before-Tax and After-Tax Analysis of the SMP Investment with SLN Depreciation
EOY
BTCF
0
1
2$500,000.00
$124,166.67
DWO
$45,000.00
2
3
$124,166.67
$124,166.67
$45,000.00
$45,000.00
4
5
$124,166.67
$124,166.67
6
7
8
9
TI
T
ATCF
$79,166.67
$31,666.67
2$500,000.00
$92,500.00
$79,166.67
$79,166.67
$31,666.67
$31,666.67
$92,500.00
$92,500.00
$45,000.00
$45,000.00
$79,166.67
$79,166.67
$31,666.67
$31,666.67
$92,500.00
$92,500.00
$124,166.67
$124,166.67
$45,000.00
$45,000.00
$79,166.67
$79,166.67
$31,666.67
$31,666.67
$92,500.00
$92,500.00
$124,166.67
$124,166.67
$45,000.00
$45,000.00
$79,166.67
$79,166.67
$31,666.67
$31,666.67
$92,500.00
$92,500.00
10
MARRBT 5
$174,166.67
16.667%
$45,000.00
$79,166.67
$31,666.67
MARRAT 5
$142,500.00
10%
PWBT 5
FWBT 5
$96,229.49
$449,547.99
PWAT 5
FWAT 5
$87,649.62
$227,340.55
AWBT 5
IRRBT 5
$20,406.41
21.64%
AWAT 5
IRRAT 5
$14,264.57
13.80%
ERRBT 5
18.74%
ERRAT 5
11.79%
The computation of ATCF is as shown in Table 9.2. In the tenth year,
notice how the salvage value is handled. The $500,000 investment is not
fully recovered through depreciation allowances. Straight-line depreciation allowances of $45,000 over a 10-year period reduce the book value to
$50,000 at the end of the 10-year planning horizon. Because the book
value exactly equals the salvage value, there is no income tax associated
with the sale of the SMP machine for $50,000. Therefore, the taxable
income in year 10 is identical to that in each of the nine previous years.
EXPLORING THE
SOLUTION
Table 9.2 also provides values for the various DCF-based measures of
economic worth treated in previous chapters. As expected, the BTCF
values and the after-tax measures of economic worth shown in Table 9.2
are identical to those obtained in Chapters 4 through 6.
If, in the previous chapters, BTCF was used and a before-tax analysis
was performed, then a before-tax minimum attractive rate of return
(MARRBT) must be used. An estimate of MARRBT is obtained by dividing
9-2 After-Tax Analysis Using Retained Earnings (No Borrowing)
325
the after-tax minimum attractive rate of return (MARRAT) by 1 minus the
income tax rate. Hence, with a 40 percent tax rate and a 10 percent MARRAT,
a MARRBT of 16.667 percent would be used.
After-Tax Analysis with MACRS Depreciation
EXAMPLE
As noted in Chapter 8, the SMP machine purchased for $500,000 qualifies
as 5-year property for MACRS depreciation. What is the after-tax effect on
the measures of economic worth if MACRS is used with a 40% tax rate
and a 10 percent MARRAT?
Given: P 5 $500,000, itr 5 40%, MARRAT 5 10%, ATCF (years 1–10) 5
$92,500, S 5 $50,000
Find: PWAT, AWAT, FWAT, IRRAT, ERRAT
KEY DATA
Table 9.3 provides the results of the after-tax analysis. Notice, because
MACRS fully recovers the investment, the book value at the end of year 10
is 0. If the equipment is sold for $50,000, then the full amount of the salvage
value is taxable income. (When a depreciable asset is sold for more than its
book value, the gain is called depreciation recapture and is taxed as ordinary
income under current tax law.)
SOLUTION
TABLE 9.3
After-Tax Analysis of the SMP Investment with MACRS Depreciation
EOY
BTCF
0
1
2$500,000.00
$124,166.67
DWO
TI
T
$100,000.00
$24,166.67
$9,666.67
2$500,000.00
$114,500.00
ATCF
2
3
$124,166.67
$124,166.67
$160,000.00
$96,000.00
2$35,833.33
$28,166.67
−$14,333.33
$11,266.67
$138,500.00
$112,900.00
4
5
$124,166.67
$124,166.67
$57,600.00
$57,600.00
$66,566.67
$66,566.67
$26,626.67
$26,626.67
$97,540.00
$97,540.00
6
7
$124,166.67
$124,166.67
$28,800.00
$0.00
$95,366.67
$124,166.67
$38,146.67
$49,666.67
$86,020.00
$74,500.00
8
9
$124,166.67
$124,166.67
$0.00
$0.00
$124,166.67
$124,166.67
$49,666.67
$49,666.67
$74,500.00
$74,500.00
10
MARRBT 5
$174,166.67
16.667%
$0.00
$174,166.67
$69,666.67
$104,500.00
PWBT 5
FWBT 5
$96,229.49
$449,547.99
PWAT 5
FWAT 5
$123,988.64
$321,594.61
AWBT 5
IRRBT 5
$20,406.41
21.64%
AWAT 5
IRRAT 5
$20,178.58
16.12%
ERRBT 5
18.74%
ERRAT 5
12.46%
326
Chapter 9
Income Taxes
Examples 9.3 and 9.4 bring out some important observations. Note
that the after-tax PW, FW, AW, IRR, and ERR values at the bottom of
Table 9.3 using MACRS depreciation are all better than the after-tax PW,
FW, AW, IRR, and ERR values at the bottom of Table 9.2 using SLN
depreciation. As pointed out in Chapter 8, because depreciation is deducted
from taxable income, accelerated depreciation yields a greater economic
worth than occurs using straight-line depreciation.
If the salvage value exactly matches the book value at the planning
horizon’s end as in Example 9.3 using SLN depreciation, there are no
tax implications. If the salvage value exceeds the book value, as it does
in Example 9.4 where the SMP machine was sold for $50,000 more
than the book value at the horizon’s end, the excess amount is taxed at
the itr (40%). If the salvage value is less than the book value at the horizon’s end, the difference is termed a book loss and is deducted from
taxable income. In general, we recommend that the following relationship be used to compute taxable income in the year of property disposal.
Specifically, if disposal occurs at the end of year n at a salvage value
of Fn,
TIn 5 BTCFn 3including Fn 4 2 DWOn 2 Bn
(9.6)
In words, taxable income in the year of a depreciable asset’s disposal
equals the before-tax cash flow, including the salvage value, less the depreciation allowance, less the book value at the time of disposal.
Also, if depreciable property is disposed of before the end of the
recovery period, when using MACRS depreciation, a half-year allowance (in the case of personal property) or midmonth allowance (in the
case of real property) is permitted in the year of disposal. For example,
if an asset is sold during the sixth tax year and it is 7-year property, then
the MACRS depreciation allowance percentages are 14.29 percent,
24.49 percent, 17.49 percent, 12.49 percent, 8.93 percent, and 4.46 percent;
normally, the depreciation percentage for the sixth year would be
8.92 percent.
Not all expenditures are used to acquire depreciable property. Consider, for example, an expenditure on a consulting study to identify costsaving opportunities. Another example of alternatives with different tax
consequences is leasing versus purchasing equipment. Likewise, let’s not
forget expenditures on advertising to generate increased revenue. For tax
purposes, these types of expenditures are ‘‘expensed’’ or ‘‘written off’’ in
the year in which they occur. Other expenditures that can be expensed
include software and most R&D expenses. The following example shows
the after-tax effects of expenditures that can be expensed versus those that
must be capitalized and depreciated.
9-2 After-Tax Analysis Using Retained Earnings (No Borrowing)
After-Tax Analysis with Non-Depreciable Expenditures
EXAMPLE
In Example 9.4, $500,000 was invested in depreciable property to obtain
before-tax annual revenues of $124,166.67 each of the following 9 years
and $174,166.67 the tenth year. Now suppose the expenditure is on a consulting study that identifies cost-saving opportunities that will produce the
same BTCF. With a 40 percent income-tax rate and an after-tax MARR of
10 percent, what will be the after-tax measures of economic worth for the
investment?
Given: P 5 $500,000, itr 5 40%, MARRAT 5 10%, BTCF (years 1–9) 5
$124,166.67, BTCF (year 10) 5 $174,166.67
Find: PWAT, AWAT, FWAT, IRRAT, ERRAT
KEY DATA
Table 9.4 provides the results of the after-tax analysis.
SOLUTION
After-Tax Analysis of a $500,000 Investment in a
Consulting Study
TABLE 9.4
EOY
BTCF
TI
T
ATCF
0
2$500,000.00
2$500,000.00
2$200,000.00
2$300,000.00
1
$124,166.67
$124,166.67
$49,666.67
$74,500.00
2
$124,166.67
$124,166.67
$49,666.67
$74,500.00
3
$124,166.67
$124,166.67
$49,666.67
$74,500.00
4
$124,166.67
$124,166.67
$49,666.67
$74,500.00
5
$124,166.67
$124,166.67
$49,666.67
$74,500.00
6
$124,166.67
$124,166.67
$49,666.67
$74,500.00
7
$124,166.67
$124,166.67
$49,666.67
$74,500.00
8
$124,166.67
$124,166.67
$49,666.67
$74,500.00
9
$124,166.67
$124,166.67
$49,666.67
$74,500.00
10
$174,166.67
$174,166.67
$69,666.67
$104,500.00
PWAT 5
$169,336.56
FWAT 5
$439,215.43
AWAT 5
$27,558.75
IRRAT 5
21.64%
ERRAT 5
15.03%
Now, Examples 9.4 and 9.5 show some interesting observations. Note
that the values at the bottom of Table 9.4 when the investment is expensed
327
328
Chapter 9
Income Taxes
are all better than the values at the bottom of Table 9.3 when the investment is capitalized and depreciated using MACRS.
One way to think about this is that expensing is the extreme of
accelerated depreciation. Rather than spreading out the asset’s value
evenly over time (as in SLN), or in a relatively accelerated fashion (with
DDB), we are able to achieve the most tax benefit by expensing the total
amount in year 0. Expensing would be preferred to depreciating, but tax
law dictates when we can and cannot expense, as well as how we must
depreciate.
9.2.2 Multiple Alternatives
An after-tax comparison of investment alternatives follows the procedures
described in previous chapters. The only changes are that we are now
explicitly calculating ATCF and using MARRAT. Depending on the alternatives being considered, different depreciation classes might apply to the
alternatives or some may even be expensed.
After-Tax Comparison of Alternatives with Different Property Classes
EXAMPLE
A manufacturing firm is considering different investments that qualify
for different property classes. Alternative A is for specialized tools that
qualify as 3-year property, require an investment of $300,000, and provide before-tax annual savings of $70,000 per year over the 10-year
planning horizon. Alternative B is for production equipment that qualifies as 7-year property, requires an investment of $450,000, and provides
savings of $102,000 per year over the 10-year horizon. An after-tax
MARR of 10 percent applies, and with a 40 percent income tax rate this
is equivalent to a before-tax MARR of 16.67 percent. Which alternative
is preferred?
KEY DATA
Given: Alternative A: P 5 $300,000; 3-year property; itr 5 40%; MARRAT 5
10%; BTCF (years 1–10) 5 $70,000
Alternative B: P 5 $450,000; 7-year property; itr 5 40%; MARRAT 5
10%; BTCF (years 1–10) 5 $102,000
Find: PWBT and PWAT of Alternatives A and B
SOLUTION
As shown in Table 9.5, Alternative A is the preferred investment, with an
after-tax present worth of $57,940.19 versus $55,908.46 for Alternative B.
On a before-tax basis, however, Alternative B is preferred to Alternative A.
9-2 After-Tax Analysis Using Retained Earnings (No Borrowing)
TABLE 9.5
329
After-Tax Comparison of Investment Alternatives with Different Property Classes
EOY
BTCF(A)
0
2$300,000.00
1
$70,000.00
2
3
DWO(A)
Tl(A)
T(A)
ATCF(A)
$99,990.00
2$29,990.00
2$11,996.00
$81,996.00
$70,000.00
$133,350.00
2$63,350.00
2$25,340.00
$95,340.00
$70,000.00
$44,430.00
$25,570.00
$10,228.00
$59,772.00
4
$70,000.00
$22,230.00
$47,770.00
$19,108.00
$50,892.00
5
$70,000.00
$0.00
$70,000.00
$28,000.00
$42,000.00
6
$70,000.00
$0.00
$70,000.00
$28,000.00
$42,000.00
7
$70,000.00
$0.00
$70,000.00
$28,000.00
$42,000.00
$42,000.00
2$300,000.00
8
$70,000.00
$0.00
$70,000.00
$28,000.00
9
$70,000.00
$0.00
$70,000.00
$28,000.00
$42,000.00
10
$70,000.00
$0.00
$70,000.00
$28,000.00
$42,000.00
BTPW(A) 5
$30,095.51
ATPW(A) 5
$57,940.19
EOY
BTCF(B)
0
2$450,000.00
1
$102,000.00
$64,305.00
$37,695.00
$15,078.00
$86,922.00
2
$102,000.00
$110,205.00
2$8,205.00
2$3,282.00
$105,282.00
3
$102,000.00
$78,705.00
$23,295.00
$9,318.00
$92,682.00
4
$102,000.00
$56,205.00
$45,795.00
$18,318.00
$83,682.00
5
$102,000.00
$40,185.00
$61,815.00
$24,726.00
$77,274.00
6
$102,000.00
$40,140.00
$61,860.00
$24,744.00
$77,256.00
7
$102,000.00
$40,185.00
$61,815.00
$24,726.00
$77,274.00
8
$102,000.00
$20,070.00
$81,930.00
$32,772.00
$69,228.00
9
$102,000.00
$0.00
$102,000.00
$40,800.00
$61,200.00
10
$102,000.00
$0.00
$102,000.00
$40,800.00
$61,200.00
ATPW(B) 5
$55,908.46
BTPW(B) 5
DWO(B)
Tl(B)
T(B)
ATCF(B)
2$450,000.00
$30,996.31
After-Tax Comparison of Manual versus Automated Solutions
In a distribution center, loads to be shipped have been palletized manually.
A proposal has been made to use a robot to perform the palletizing operation. Because cartons to be palletized are not dimensionally uniform, a
vision system coupled with optimization software will be required with
the robot.
Currently, two people perform palletizing. The labor cost for this is
$50,000 per year. A fully equipped robot to perform the task will cost
EXAMPLE
Video Example
330
Chapter 9
Income Taxes
$125,000 and will have annual operating costs of $500. The robot qualifies
as 3-year property. If a tax rate of 40 percent and a MARR of 10 percent
are used, should the robot be purchased? Use a 5-year planning horizon
and perform an after-tax annual worth analysis; assume a salvage value of
$25,000 for the robot after 5 years of use.
TABLE 9.6
EOY
KEY DATA
Given: Manual Alternative: A 5 $50,000/year; itr 5 40%; MARR 5
10%; N 5 5
Robotic Alternative: P 5 $125,000; A 5 $500; 3-year property; itr 5
40%; MARR 5 10%; N 5 5; SV 5 $25,000
Find: EUACAT
SOLUTION
Manual palletizing: Because the labor cost can be expensed in the
year in which it occurs, the after-tax equivalent uniform annual cost
of manually palletizing equals $50,000(0.60), or $30,000. Robotic
palletizing: As shown in Table 9.6, the equivalent uniform annual cost of
robotic palletizing equals $19,840.63. Therefore, the robot is justified
economically.
After-Tax Comparison of Manual versus Robotic Palletizing
BTCF
DWO
TI
T
ATCF
0
−$125,000.00
1
−$500.00
$41,662.50
−$42,162.50
−$16,865.00
$16,365.00
2
−$500.00
$55,562.50
−$56,062.50
−$22,425.00
$21,925.00
3
−$500.00
$18,512.50
−$19,012.50
−$7,605.00
$7,105.00
4
−$500.00
$9,262.50
−$9,762.50
−$3,905.00
$3,405.00
5
$24,500.00
$0.00
$24,500.00
$9,800.00
$14,700.00
EUACBT 5
$29,379.75
EUACAT 5
$19,840.63
9-3
−$125,000.00
AFTER-TAX ANALYSIS USING
BORROWED CAPITAL
LEARN I N G O B JEC T I V E : Analyze the after-tax implications for investment
alternatives using borrowed capital.
The previous after-tax analyses assumed the investments were paid for
using retained earnings. However, depreciable property is frequently
9-3 After-Tax Analysis Using Borrowed Capital
331
purchased using borrowed funds. Because interest is deductible from taxable income, it is useful to consider how an after-tax analysis is performed
when using borrowed funds.
When borrowed funds are used, additional notation is needed. Let
PPMT 5 principal payment
IPMT 5 interest payment
LCF 5 loan cash flow 5 PPMT 1 IPMT
The following formulas apply when borrowed funds are used:
ATCF 5 BT&LCF 2 LCF 2 T
(9.7)
After-tax cash flow equals before-tax-and-loan cash flow, less the loan
payment, less the income tax. As before,
T 5 itr 1TI 2
but, now,
TI 5 BT&LCF 2 IPMT 2 DWO
(9.8)
Taxable income equals before-tax-and-loan cash flow, less the interest
paid, less the depreciation allowance. Therefore,
ATCF 5 BT&LCF11 2 itr2 2 LCF 1 itr 1DWO 1 IPMT2
(9.9)
Finally, in the year of the depreciable property’s disposal,
TIn 5 BT&LCFn 1including Fn 2 2 IPMTn 2 DWOn 2 Bn
(9.10)
Because interest on a loan can be deducted from taxable income, the effective after-tax interest rate paid on borrowed funds is
ieff 5 i 11 2 itr2
(9.11)
MARRBT is therefore given by MARRAT /(1 – itr). Hence, if the incometax rate is 40 percent, and MARRAT is 10 percent, then MARRBT
is 0.10/0.60, or 16.667 percent. When the interest rate, for example i 5
12 percent, is less than MARRBT, one should borrow as much as possible
and delay repaying the loan as long as possible; when the interest rate is
greater than MARRBT, one should borrow as little as possible and repay
the principal as quickly as possible.
After-tax analysis of a single alternative when money is borrowed is
very similar to the analysis of a single alternative financed using retained
earnings. When borrowed money is used, however, the interest payment
on the loan (IPMT) must be subtracted from BT&LCF to determine taxable income (TI). Also, remember that both the principal and interest payments are cash flows. This is demonstrated in Example 9.8.
Loan Cash Flow (LCF)
The cash flow reflecting
the sum of the principal
payment and the interest
payment.
332
Chapter 9
Income Taxes
Using Excel to Analyze After-Tax Effects of Borrowed Funds—
Single Alternative
EXAMPLE
Now, let’s examine four payment plans as they are used to secure $300,000
financing for the $500,000 SMP machine acquisition considered in earlier
examples. The plans are: (a) Plan 1—pay the accumulated interest at the
end of each interest period and pay the principal at the end of the loan
period; (b) Plan 2—make equal principal payments, plus interest on the
unpaid balance at the end of the period; (c) Plan 3—make equal end-ofperiod payments; and (d) Plan 4—make a single payment of principal and
interest at the end of the loan period. The loan interest rate 5 12%.
KEY DATA
Given: Loan amount 5 $300,000; P 5 $500,000; itr 5 40%;
MARRAT 5 10%; BT&LCF (years 1–9) 5 $124,166.67; BT&LCF
(year 10) 5 $174,166.67
Find: PWAT, AWAT, FWAT, IRRAT, ERRAT for Plans 1, 2, 3, and 4
SOLUTION
a) Figure 9.3 depicts the after-tax analysis for Plan 1, in which only the
interest is paid each year; at the end of year 10, the $300,000 principal is paid.
F IGURE 9.3
Using Plan 1
After-Tax Analysis of the SMP Investment with $300,000 of Borrowed Capital Repaid
9-3 After-Tax Analysis Using Borrowed Capital
333
As shown, the after-tax present worth (PWAT), based on a 12 percent interest
rate on the $300,000 loan and a 10 percent MARRAT, is $175,603.01. When
no borrowed funds were used, PWAT was $123,988.64 (see Example 9.4).
Borrowing the $300,000 at 12 percent compound annual interest increased
PWAT, because interest is deducted from taxable income.
Notice, because of the single payment of principal in year 10, ATCF is
negative. Hence, there are two negative values in the ATCF column. Descartes’ rule of signs indicates that either two or zero roots exist; the two
roots occur at –18.62 percent and at 41.54 percent. Because multiple negative values exist, the Excel® MIRR worksheet function cannot be used to
compute ERR. However, by following the process first outlined in Example 6.6, the IRR function can be used to calculate ERRAT. As shown in
Figure 9.3, ERRAT equals 17.16 percent.
b) Table 9.7 provides the after-tax analysis for Plan 2, in which equal annual
principal payments are made, plus interest on the unpaid loan balance. As
shown, PWAT is $156,374.28. Plan 2 is not as attractive as Plan 1. Why?
Because the after-tax interest rate is less than the MARR, it is better to delay
paying principal.
After-Tax Analysis of the SMP Investment with $300,000 of Borrowed Capital Repaid
Using Plan 2
TABLE 9.7
MARRAT 5 10%
Income Tax
Rate 5 40%
EOY
BT&LCF
PPMT
0
2$500,000.00
2$300,000.00
1
$124,166.67
2
3
4
Interest
Rate 5 12%
IPMT
DWO
TI
Tax
ATCF
$30,000.00
$36,000.00
$100,000.00
−$11,833.33
−$4,733.33
$62,900.00
$124,166.67
$30,000.00
$32,400.00
$160,000.00
−$68,233.33
−$27,293.33
$89,060.00
$124,166.67
$30,000.00
$28,800.00
$96,000.00
−$633.33
−$253.33
$65,620.00
$124,166.67
$30,000.00
$25,200.00
$57,600.00
$41,366.67
$16,546.67
$52,420.00
$54,580.00
2$200,000.00
5
$124,166.67
$30,000.00
$21,600.00
$57,600.00
$44,966.67
$17,986.67
6
$124,166.67
$30,000.00
$18,000.00
$28,800.00
$77,366.67
$30,946.67
$45,220.00
7
$124,166.67
$30,000.00
$14,400.00
$0.00
$109,766.67
$43,906.67
$35,860.00
8
$124,166.67
$30,000.00
$10,800.00
$0.00
$113,366.67
$45,346.67
$38,020.00
9
$124,166.67
$30,000.00
$7,200.00
$0.00
$116,966.67
$46,786.67
$40,180.00
10
$174,166.67
$30,000.00
$3,600.00
$0.00
$170,566.67
$68,226.67
$72,340.00
PWAT 5
$156,374.28
FWAT 5
$405,594.61
AWAT 5
$25,449.19
IRRAT 5
28.51%
ERRAT 5
16.54%
334
Chapter 9
Income Taxes
c) Table 9.8 contains the results for Plan 3, in which equal annual payments are made over the 10-year period. For Plan 3, PWAT is $160,734.89.
As expected, because principal payments increase over time, Plan 3 does
not perform as well as Plan 1, where no principal payments are made until
the end of the loan period.
After-Tax Analysis of the SMP Investment with $300,000 of Borrowed Capital Repaid
Using Plan 3
TABLE 9.8
MARRAT 5 10%
Income Tax
Rate 5 40%
EOY
BT&LCF
PPMT
0
−$500,000.00
−$300,000.00
1
$124,166.67
2
Interest
Rate 5 12%
IPMT
DWO
TI
Tax
ATCF
$17,095.25
$36,000.00
$100,000.00
−$11,833.33
−$4,733.33
$75,804.75
$124,166.67
$19,146.68
$33,948.57
$160,000.00
−$69,781.90
−$27,912.76
$98,984.18
3
$124,166.67
$21,444.28
$31,650.97
$96,000.00
−$3,484.30
−$1,393.72
$72,465.14
4
$124,166.67
$24,017.59
$29,077.65
$57,600.00
$37,489.02
$14,995.61
$56,075.81
5
$124,166.67
$26,899.71
$26,195.54
$57,600.00
$40,371.13
$16,148.45
$54,922.97
6
$124,166.67
$30,127.67
$22,967.58
$28,800.00
$72,399.09
$28,959.64
$42,111.78
7
$124,166.67
$33,742.99
$19,352.26
$0.00
$104,814.41
$41,925.76
$29,145.66
8
$124,166.67
$37,792.15
$15,303.10
$0.00
$108,863.57
$43,545.43
$27,525.99
9
$124,166.67
$42,327.21
$10,768.04
$0.00
$113,398.63
$45,359.45
$25,711.97
10
$174,166.67
$47,406.47
$5,688.78
$0.00
$168,477.89
$67,391.16
$53,680.26
−$200,000.00
PWAT 5
$160,734.89
FWAT 5
$416,904.92
AWAT 5
$26,158.86
IRRAT 5
31.65%
ERRAT 5
16.68%
d) Figure 9.4 provides the results for Plan 4, in which no payment is made
until the end of the loan period. Notice, with an after-tax PW worth of
$162,184.44, Plan 4 ranks second to Plan 1 in maximizing PWAT.
With Plan 4, no payment is made until year 10. Hence, the most
negative cash flow occurs at that time. Descartes’ rule of signs indicates
either two or zero roots exist; as shown in Figure 9.4, the two roots
occur at –3.65 percent and at 53.49 percent. Because multiple negative
values exist, the Excel® MIRR worksheet function cannot be used to
compute ERR; however, as shown, ERR can be calculated using the
Excel® IRR worksheet function with the modified ATCF column and
equals 16.73 percent. Notice, PWAT is maximized when MARRAT equals
9-3 After-Tax Analysis Using Borrowed Capital
F IGURE 9.4
After-Tax Analysis of SMP Investment with $300,000 of Borrowed Capital Repaid Using Plan 4
9.328 percent. (We used the Excel® SOLVER tool to obtain the MARRAT
value that maximized PWAT.)
Given the results obtained for the four payment plans, it is anticipated
that after-tax PW will increase as the amount borrowed increases.
Figure 9.5 presents the results for Plan 1 if the investment is entirely
paid for using borrowed funds. As expected, the PWAT of $210,012.58
is greater than the PWAT of $175,603.01 that occurred when only
$300,000 was borrowed.
When 100 percent of investment capital is obtained by borrowing,
no initial investment occurs at the beginning of the ATCF series. As
shown in Figure 9.5, the only negative-valued cash flow occurs at the
end of year 10. As such, increases in MARR will increase PWAT. This is
just the opposite of what we observed previously. Because there are no
negative-valued cash flows before the planning horizon’s end, IRR and
ERR are not defined. (Notice, PWAT is maximized when MARR 5
14.332 percent.)
EXPLORING THE
SOLUTION
335
336
Chapter 9
FIGURE 9.5
Income Taxes
After-Tax Analysis of $500,000 Investment with 100% Borrowed Capital Repaid Using Plan 1
Example 9.8 concerns a single investment alternative. After-tax analysis
of multiple alternatives with borrowed capital is similar. Again, comparison
of alternatives is based on ATCF instead of BTCF, the use and repayment of
borrowed capital, and the use of MARRAT instead of MARRBT. Also, depending on the alternatives being considered, different depreciation allowances
might apply to the alternatives. Examples 9.6, 9.7, and 9.8 present ample
guidance for after-tax analysis of multiple alternatives using borrowed capital.
9-4
LEASING VERSUS PURCHASING EQUIPMENT
LEARN I N G O B JEC T I V E : Illustrate the different tax consequences for the
common example of a lease vs. purchase alternative.
As mentioned previously, another example of alternatives with different
tax consequences is leasing versus purchasing.
9-4 Leasing Versus Purchasing Equipment
After-Tax Comparison of Leasing versus Purchasing
337
EXAMPLE
The Acme Brick Company is considering adding five lift trucks to its fleet.
Both purchasing and leasing were discussed with the fork-truck supplier.
If the lift trucks are purchased, they will have a first cost of $18,000; annual
operating and maintenance costs are estimated to be $3,750 per truck. At
the end of the 5-year planning horizon, the lift trucks are estimated to have
salvage values of $3,000 each. The lift trucks qualify as MACRS 3-year
property.
If the lift trucks are leased, beginning-of-year lease payments will be
$5,900 per truck. The supplier includes maintenance in the lease price.
However, the company must pay annual operating costs of $1,800 per
truck. Lease payments can be expensed for tax purposes.
Using an after-tax MARR of 12 percent and an income-tax rate of 40
percent, should the company lease the lift trucks?
Given Buy: P 5 $18,000; A 5 $3,750 per truck; SV 5 $3,000 per truck
at end of Year 5; itr 5 40%; MARRAT 5 12%
Lease: A 5 $5,900 per truck plus $1,800 per truck for annual operating cost
Find: Should the trucks be leased or bought assuming that the trucks
qualify as a MACRS 3-year property?
KEY DATA
As shown in Table 9.9, an after-tax analysis indicates the lift trucks should
be leased. The present worth cost of purchasing is $96,486.74, whereas the
SOLUTION
TABLE 9.9
EOY
After-Tax Comparison of Purchasing versus Leasing Lift Trucks
BTCF(P)
DWO(P)
Tl(P)
T(P)
ATCF(P)
0
2$90,000.00
1
2$18,750.00
$29,997.00
2$48,747.00
2$19,498.80
$748.80
2
2$18,750.00
$40,005.00
2$58,755.00
2$23,502.00
$4,752.00
3
2$18,750.00
$13,329.00
2$32,079.00
2$12,831.60
2$5,918.40
4
2$18,750.00
$6,669.00
2$25,419.00
2$10,167.60
2$8,582.40
5
2$3,750.00
$0.00
2$3,750.00
2$1,500.00
2$2,250.00
PWBT(P) 5
EOY
2$90,000.00
PWAT(P) 5
2$140,045.81
BTCF(L)
DWO(L)
Tl(L)
2$96,486.74
T(L)
ATCF(L)
0
2$29,500.00
2$29,500.00
2$11,800.00
2$17,700.00
1
2$38,500.00
$0.00
2$38,500.00
2$15,400.00
2$23,100.00
2
2$38,500.00
$0.00
2$38,500.00
2$15,400.00
2$23,100.00
3
2$38,500.00
$0.00
2$38,500.00
2$15,400.00
2$23,100.00
4
2$38,500.00
$0.00
2$38,500.00
2$15,400.00
2$23,100.00
5
$9,000.00
$0.00
2$9,000.00
2$3,600.00
2$5,400.00
PWBT(L) 5
2$132,783.18
PWAT(L) 5
2$90,926.87
338
Chapter 9
Income Taxes
present worth cost of leasing is $90,926.87. The $5,559.87 difference in
present worth costs is probably great enough for the company to lease the
lift trucks.
Interestingly, as shown in Table 9.8, if the decision had been made
using a before-tax MARR of 0.12/(1 2 0.4), or 20 percent, the before-tax
present worth cost of purchasing is $140,045.81, and the present worth
cost of leasing is $132,783.18. Again, the $7,262.63 difference in the present worths is probably great enough for the company to decide to lease the
lift trucks.
SUMMARY
KEY CONCEPTS
1. Learning Objective: Calculate corporate income taxes, including the
effective, incremental, and marginal tax rates. (Section 9.1)
The taxes a corporation pays are a true cost of doing business and may
impact selection of the best investment alternative. Therefore, it is best to
perform economic analyses on an after-tax basis so that the tax effect can
be considered. Income-tax rates and regulations change rapidly, so consulting an expert in income taxes is advisable when income taxes will play
a major role in determining the economic viability of an investment.
2. Learning Objective: Analyze the after-tax implications for investment
alternatives using retained earnings. (Section 9.2)
The viability of or preference for an investment alternative may change
when tax implications are considered. In general, the faster an investment
is depreciated, the greater its after-tax present worth, as illustrated in several chapter examples.
The after-tax cash flow of an investment alternative without borrowed
capital can be expressed as:
ATCF 5 BTCF 11 2 itr2 1 itr 1DWO2
where
BTCF 5 before-tax cash flow
DWO 5 depreciation write-off or allowance
itr 5 income tax rate
ATCF 5 after-tax cash flow
(9.4)
Summary 339
The overall income tax rate can be expressed as
itr 5 str 1 ftr 11 2 str2
(9.5)
where str 5 state income-tax rate, and ftr 5 federal income-tax rate.
3. Learning Objective: Analyze the after-tax implications for investment
alternatives using borrowed capital. (Section 9.3)
For borrowed capital, interest paid by businesses is deductible from
taxable income. This necessitates performing an after-tax analysis in
order to determine the attractiveness of the investment. We learned that
using someone else’s money can possibly make even more money for
ourselves.
The after-tax cash flow of an investment alternative with borrowed
capital can be expressed as:
ATCF 5 BT&LCF11 2 itr2 2 LCF 1 itr 1DWO 1 IPMT2
(9.9)
where
BT&LCF 5 before-tax-and-loan cash flow
LCF 5 loan cash flow
IPMT 5 interest payment
In the year of the depreciable property’s disposal,
TIn 5 BT&LCFn 1including Fn 2 2 IPMTn 2 DWOn 2 Bn
(9.10)
where
TIn 5 taxable income in the year of property’s disposal
Bn 5 book value at time of disposal
Fn 5 salvage value at time of disposal
Because interest on a loan can be deducted from taxable income, the
effective after-tax interest rate paid on borrowed funds is
ieff 5 i 11 2 itr2
(9.11)
4. Learning Objective: Illustrate the different tax consequences for the
common example of a lease vs. purchase alternative. (Section 9.4)
The decision to lease vs. purchase is a common investment alternative that
should be analyzed using an after-tax analysis, as income-tax considerations can change the recommendation regarding the investment to be
made. For example, in the lease versus purchase case, the DWO allowance
can make the purchase alternative quite attractive.
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KEY TERMS
After-Tax Cash Flow (ATCF), p. 322
Before-Tax-and-Loan Cash Flow
(BT&LCF), p. 322
Before-Tax Cash Flow (BTCF), p. 322
Effective Tax Rate, p. 320
Federal Tax Rate (ftr), p. 323
Income Tax Rate (itr), p. 323
Incremental Tax Rate, p. 320
Loan Cash Flow (LCF), p. 331
Marginal Tax Rate, p. 320
State Tax Rate (str), p. 323
Problem available in WileyPLUS
GO Tutorial Tutoring Problem available in WileyPLUS
Video Solution Video Solution available in WileyPLUS
FE-LIKE PROBLEMS
1.
The marginal tax rate on a corporate income of $87,000 is closest to
which of the following? (Table 9.1 or similar is required for this question.)
a. 15.0%
c. 25.0%
b. 20.5%
d. 34.0%
2.
The average tax rate on a corporate income of $87,000 is closest to which
of the following? (Table 9.1 or similar is required for this question.)
a. 15.0%
c. 25.0%
b. 20.5%
d. 34.0%
3.
The correctly calculated taxes due on a corporate taxable income of
$13,000,000 are closest to which of the following? (Table 9.1 or similar is
required for this question.)
a. $3,400,000
c. $4,450,000
b. $4,420,000
d. $4,550,000
4.
When a business calculates taxable income from gross income, which of
the following is true?
a. depreciation, interest, and principal are all subtracted
b. depreciation and interest are subtracted, principal is not
c. depreciation is subtracted, interest and principal are not
d. interest and principal are subtracted, depreciation is not
5.
When considering the use of debt capital to finance a project, the upper
limit for the interest rate on an attractive loan can be determined by which of
the following?
Summary 341
a.
b.
c.
d.
6.
MARR
MARR * (1 1 tax rate)
MARR/(1 2 tax rate)
MARR * (1 2 tax rate)
Consider the following data extracted from an after-tax cash flow calculation.
Before-Tax Cash Flow 5 22,500
Loan Principal Payment 5 7,434
Loan Interest Payment 5 892
MACRS Depreciation Deduction 5 7,405
Taxes Due 5 5,397
Which of the following is closest to the after-tax cash flow?
a. $1,372
c. $8,806
b. $8,777
d. $16,211
7.
Consider the following data extracted from an after-tax cash flow
calculation.
Before Tax Cash Flow 5 $22,500
Loan Principal Payment 5 $5,926
Loan Interest Payment 5 $2,400
MACRS Depreciation Deduction 5 $16,665
Which of the following is closest to the Taxable Income?
a. 2$2,491
c. $3,435
b. 2$91
d. $14,174
8.
Consider the following data for 2012 from an after-tax cash flow analysis.
What is the loan interest payment for 2012?
Before Tax Cash Flow 5 $20,000
Loan Principal Payment 5 $4,018
Depreciation Deduction 5 $8,920
Taxable Income 5 $8,018
Taxes Due 5 $2,726
After Tax Cash Flow 5 $10,194
a. $1,274
c. $7,062
b. $3,062
d. $11,080
9.
Consider the following data for 2011 from an after tax cash flow analysis.
What is the taxable income for 2011?
Before Tax Cash Flow 5 $23,000
Loan Principal Payment 5 $3,203
Loan Interest Payment 5 $3,877
Depreciation Deduction 5 $12,490
Taxes Due 5 $2,255
After Tax Cash Flow 5 $21,530
a. $40,000
c. $6,633
b. $35,540
d. $28,460
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10.
Consider the following data for 2010 from an after tax cash flow analysis.
What is the after tax cash flow for 2010?
Before Tax Cash Flow 5 $23,000
Loan Principal Payment 5 $3,203
Loan Interest Payment 5 $3,877
Depreciation Deduction 5 $12,490
Taxable Income 5 $6,633
Taxes Due 5 $2,255
a. $20,744
c. $3,430
b. $13,665
d. $1,175
PROBLEMS
Introduction
1. What are the various taxes paid in the USA by corporations? Which of those
has the most significant impact upon economic analyses?
2. Income taxes are calculated based on gross income less certain allowable
deductions. They are also assessed on gains resulting from the disposal of
property. What is a 10 word or less definition appropriate for a corporation,
based on Wikipedia, for each of the following factors?
a. Gross income
b. Expenses
c. Depreciation
d. Interest
e. Property (e.g., equipment) disposition
3. For each statement in parts (a) to (d), give a short answer or indicate True or
False.
a. Which of the following is a cash flow: (1) depreciation, (2) loan interest
paid, and/or (3) income tax.
b. We know that depreciation law has changed dramatically since 1950;
however, tax law has not changed. (T or F)
c. Depreciation method affects taxes owed. (T or F)
d. In an alternative evaluation, the inclusion of taxes will change the amount
of the measures of merit (e.g., PW or AW), but will not change which
alternative is selected as most desirable. (T or F)
4. Explain the primary effect of (a) the Economic Recovery Act of 1981,
(b) the Tax Reform Act of 1986, and (c) the Omnibus Reconciliation Act
of 1993.
5. Explain why it is generally preferred to do economic analyses on an after-tax
basis, rather than on a before-tax basis.
Summary 343
Section 9.1 Corporate Income Tax Rates
6. What is the federal income tax for each of the following corporate taxable
incomes?
$25,000
$70,000
$95,000
$200,000
a.
b.
c.
d.
e.
f.
g.
h.
$1,000,000
$12,000,000
$17,000,000
$25,000,000
7.
Calculate the corporate income tax for each of the following corporate
taxable incomes. For each, determine the effective (average) tax rate and also
the marginal tax rate.
a. $15,000
d. $400,000
b. $88,000
e. $16,700,000
c. $180,000
8.
Video Solution Calculate the corporate income tax for each of the following corporate taxable incomes. For each, determine the effective (average)
tax rate and also the marginal tax rate.
a. $12,000
d. $1,000,000
b. $65,000
e. $19,300,000
c. $220,000
9.
Determine the smallest taxable income on which:
a. The very last dollar is taxed at 25% or more.
b. Every dollar is taxed at 25% or more.
c. Every dollar is taxed at 34% or more.
10. Determine the smallest taxable income on which:
a. The very last dollar is taxed at 35% or more.
b. Every dollar is taxed at 34% or more.
c. Every dollar is taxed at 35% or more.
11.
TenTec in Sevierville, TN makes commercial and amateur radio equipment including receivers, transceivers, antenna tuners, linear amplifiers, etc.
Their taxable income last year was $720,000. They have established a new
line of electronic equipment that is on sale this year, and expected to add
another $140,000 to taxable income.
a. Determine the effective (average) tax rate on all of last year’s taxable
income.
b. Determine the effective (average) tax rate on all of this year’s taxable
income.
c. Determine the incremental tax rate.
d. Determine the marginal tax rate this year.
12. An engineer named Don maintains a small incorporated civil engineering
consulting practice. Last year, Don’s firm’s taxable income was $41,000.
During that year, he spent a great deal of non-billable time developing computerized tools for analyzing changes in water runoff due to new highway,
344
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Income Taxes
building, and parking lot construction. Given his expertise and his ability to
sign off as a licensed professional engineer (he took the FE exam in college),
this year his company has been given the opportunity to take on two projects
that will increase taxable income by $175,000.
a. Determine the effective (average) tax rate on all of last year’s taxable
income.
b. Determine the effective (average) tax rate on all of this year’s taxable
income.
c. Determine the incremental tax rate.
d. Determine the marginal tax rate this year.
13.
Noise Sniffers Inc (NSI) is a contractor to public utilities providing electrical service to homes and businesses. Most small- and mid-sized municipal
utilities do not have the expertise or the equipment to go out and “sniff ” noise
caused by foreign objects on power lines, poor grounding on poles, cracked
or carbonized lightning arrestors, etc. Success means reduction in energy loss
to the utility and reduction in static to those using radios nearby. Early on,
ham radio operator K5KC suggested NSI personnel get training from the
American Radio Relay League and purchase the right detection equipment.
NSI now travels throughout the state to help utilities in need. Taxable income
last year was only about $30,000 for this sporadic part-time work. This year,
NSI is adding a larger adjoining state as a customer and will enjoy an increase
of about $40,000 in taxable income.
a. Determine the effective (average) tax rate on all of last year’s taxable
income.
b. Determine the effective (average) tax rate on all of this year’s taxable
income.
c. Determine the incremental tax rate.
d. Determine the marginal tax rate this year.
14. Matrix Service Company is an industrial service contractor with a strong
reputation in refining, power, petrochemicals, gas/LNG, and related areas.
Their base taxable income is $19.2 million on sales of $490 million. They
have landed two large above-ground storage tank projects and expect taxable
income to increase to $25.7 million on sales of $540 million next year.
a. Determine the effective (average) tax rate on all of last year’s taxable
income.
b. Determine the effective (average) tax rate on all of this year’s taxable
income.
c. Determine the incremental tax rate.
d. Determine the marginal tax rate for next year.
Section 9.2.1 After-Tax Analysis Using Retained Earnings—Single
Alternative (No Borrowing)
15.
West Mountain Radio, an electronics firm specializing in equipment interfaces, had been in a marginal tax bracket of 39% and then this year had an
Summary 345
increase in taxable income of $150,000. Their increase in taxes was exactly
$54,750. For each of the following, determine the smallest amount that meets
the conditions given above.
a. What was their taxable income last year?
b. What was their total income tax this year?
16. Pneumatics, a startup firm of four persons, had an increase in taxable income
of $20,000 this year. Their increase in taxes was exactly $6350.
a. What was their taxable income last year?
b. What was their total income tax this year?
17.
Acme Universal is a micro-cap that had a $3,000,000 drop in taxable
income this year. Their drop in taxes was $1,030,000. For each of the
following, determine the largest amount that meets the conditions given
above.
a. What was their taxable income last year?
b. What was their total income tax this year?
18. Westwood Specialties had an increase in taxable income of $5,000,000 this
year. Their increase in taxes was exactly $1,720,000.
a. What was their taxable income last year?
b. What was their total income tax this year?
19. An air purifier for use in manufacturing semiconductors is placed in service
with a first cost of $50,000. It will be used for 8 years, have an annual gross
income less operating expenses of $14,000 and will have no salvage value.
Corporate income taxes are 40 percent.
a. Determine the after-tax cash flows for years 0–8 if depreciation allowances are $10,000, $16,000, $9,600, $5,760, $5,760, $2,880, $0, and $0
during the 8 years (MACRS-GDS 5-year property class).
b. Determine the after-tax cash flows for years 0–8 if depreciation allowances are $10,000 for years 1–5 and $0 in years 6–8 (SLN depreciation
over only the first 5 years).
20. A special handling device for the manufacture of food is placed in service. It
costs $30,000 and has a salvage value of $2,000 after a useful life of 5 years.
The device generates a savings of $14,000 per year. Corporate income taxes
are 40 percent. Find the after-tax cash flow for each year if:
a. Depreciation allowances are $9,999, $13,335, $4,443, $2,223, and $0 for
years 1–5 (MACRS-GDS 3-year property class).
b. Depreciation allowances are $7,500 for each of years 1–4 and $0 in year 5
(SLN depreciation only over the first 4 years).
21. A specially coated mold for manufacturing tires (MACRS-GDS 3-year
property) costs $35,000 and has a salvage value of $1,750 after a useful life
of 5 years. The mold generates a net savings of $14,000 per year. The corporate tax rate is 40 percent. Find the after-tax cash flow for each year using
MACRS-GDS allowances.
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Chapter 9
Income Taxes
22. Raytheon wishes to use an automated environmental chamber in the manu-
facture of electronic components. The chamber is to be used for rigorous
reliability testing and burn-in. It is installed for $1.4 million and will have
a salvage value of $200,000 after 8 years. Its use will create an opportunity
to increase sales by $650,000 per year and will have operating expenses of
$250,000 per year. Corporate income taxes are 40%. Develop tables using a
spreadsheet to determine the ATCF for each year, and the after-tax PW, AW,
IRR and ERR if the chamber is kept for 8 years. After-tax MARR is 10%.
a. Use straight-line depreciation (no half-year convention).
b. Use MACRS-GDS and state the appropriate property class.
c. Use double declining balance depreciation (no half-year convention, no
switching).
23.
GO Tutorial Milliken uses a digitally controlled “dyer” for placing intricate and integrated patterns on manufactured carpet squares for home and
commercial use. It is purchased for $400,000. It is expected to last 8 years
and have a salvage value of $30,000. Increased net income due to this dyer is
$95,000 per year. Milliken’s tax rate is 40% and the after-tax MARR is 12%.
Develop tables using a spreadsheet to determine the ATCF for each year, and
the after-tax PW, AW, IRR and ERR after 8 years.
a. Use straight-line depreciation (no half-year convention).
b. Use MACRS-GDS and state the appropriate property class.
c. Use double declining balance depreciation (no half-year convention, no
switching).
24. Suppose Milliken has an opportunity with similar cash flows to those for a
digitally controlled “dyer,” although there are no depreciable items. They can
invest in a marketing study by a blue-ribbon consultancy costing $400,000.
Expected net returns are $95,000 per year over 7 years and $125,000 during
the 8th year. Milliken’s tax rate is 40% and the after-tax MARR is 12%.
a. Develop tables using a spreadsheet to determine the ATCF for each year,
and the after-tax PW, AW, IRR and ERR after 8 years.
b. Compare the results of part (a) with those of problem 23(b) where
MACRS-GDS is used. Explain the differences.
25. A subsidiary of AEP places in service electric generating and transmission
equipment at a cost of $3,000,000. It is expected to last 30 years with a salvage value of $250,000. The equipment will increase net income by $500,000
in the first year, increasing by 2.4% each year thereafter. The subsidiary’s tax
rate is 40% and the after-tax MARR is 9%. There is some concern that the
need for this equipment will last only 10 years and need to be sold off for
$550,000 at that time. Develop tables using a spreadsheet to determine
the ATCF for each year, and the after-tax PW, AW, IRR and ERR after only
10 years to see if the venture would be worthwhile economically.
a. Use straight-line depreciation (no half-year convention).
b. Use MACRS-GDS and state the appropriate property class.
c. Use double declining balance depreciation (no half-year convention, no
switching).
Summary 347
26. Bell’s Amusements purchased an expensive ride for their theme and amuse-
ment park situated within a city-owned Expo Center. Bell’s had a multiyear
contract with Expo Center. The ride cost $1.35 million, installed. Gross
income from the ride was $420,000 per year, with operating expenses of
$120,000. Bell’s anticipated that the ride would have a useful life of 12 years,
after which the net salvage value would be $0. After 4 years, the city and
Bell’s were unable to come to an agreement regarding an extended contract.
In order to expedite Bell’s departure, Expo Center agreed to purchase the ride
and leave it in place. Right at the end of the 4th fiscal year, Expo Center paid
to Bell’s the $900,000 unrecovered investment based on using straight line
depreciation. Corporate income taxes are 40% and the after-tax MARR is
9%. Develop tables using a spreadsheet to determine the ATCF for each year,
and the after-tax PW, AW, IRR and ERR after 4 years.
a. Use straight-line depreciation (no half-year convention).
b. Use MACRS-GDS and state the appropriate property class.
c. Use double declining balance depreciation (no half-year convention, no
switching).
27.
A high-precision programmable router for shaping furniture components is purchased by Henredon for $190,000. It is expected to last 12 years
and have a salvage value of $5,000. It will produce $45,000 in net revenue
each year during its life. Corporate income taxes are 40% and the after-tax
MARR is 10%. Develop tables using a spreadsheet to determine the ATCF
for each year, and the after-tax PW, AW, IRR and ERR if the router is kept
for 12 years.
a. Use straight-line depreciation (no half-year convention).
b. Use MACRS-GDS and state the appropriate property class.
c. Use double declining balance depreciation (no half-year convention, no
switching).
28. Henredon can spend $190,000 now for a design portfolio with a different fur-
niture look inspired by some of the ultra-modern culture in the metropolitan
areas of Dubai. While some consider this a gamble, it is generally conceded
that such a new line can result in increased net revenues of $45,000 per year
for 11 years plus $50,000 in the 12th year. Henredon’s marginal tax rate is
40% and MARR is 10% on the after-tax cash flows.
a. Develop tables using a spreadsheet to determine the ATCF for each year,
and the after-tax PW, AW, IRR and ERR after 12 years.
b. Compare the results of part (a) with those of problem 27(b) where
MACRS-GDS is used. Explain the differences.
29.
A tractor for over-the-road hauling is to be purchased by AgriGrow for
$90,000. It is expected to be of use to the company for 6 years, after which
it will be salvaged for $4,000. Transportation cost savings are expected to be
$170,000 per year; however, the cost of drivers is expected to be $70,000 per
year and operating expenses are expected to be $63,000 per year, including
fuel, maintenance, insurance and the like. The company’s marginal tax rate is
40% and MARR is 10% on after-tax cash flows. Suppose that, to AgriGrow’s
348
Chapter 9
Income Taxes
surprise, they actually dispose of the tractor at the end of the 4th tax year for
$6,000. Develop tables using a spreadsheet to determine the ATCF for each
year, and the after-tax PW, AW, IRR and ERR after only 4 years.
a. Use straight-line depreciation (no half-year convention).
b. Use MACRS-GDS and state the appropriate property class.
c. Use double declining balance depreciation (no half-year convention, no
switching).
30. AgriGrow can invest in a “100-day” short-term project costing $90,000 to
improve customer service. They believe the return on the project will be a net
increase in sales of $37,000 per year over 3 years and $43,000 in the 4th year.
AgriGrow’s marginal tax rate is 40% and MARR is 10% on the after-tax cash
flows.
a. Develop tables using a spreadsheet to determine the ATCF for each year,
and the after-tax PW, AW, IRR, and ERR after 4 years.
b. Compare the results of part (a) with those of problem 29(b) where
MACRS-GDS is used. Explain the differences.
31. Specialized production equipment is purchased for $125,000. The equip-
ment qualifies as 5-year equipment for MACRS-GDS depreciation. The
BTCF profile for the acquisition, shown below, includes a $30,000 salvage
value at the end of the 5-year planning horizon. A 40 percent tax rate applies. Develop tables using a spreadsheet to determine the ATCF for each
year and the after-tax PW, AW, IRR, and ERR values if the after-tax MARR
is 10 percent.
EOY
0
1
2
3
4
5
BTCF
–$125,000
50,000
60,000
70,000
80,000
120,000
32. A company purchases a machine for $800,000. The equipment qualifies as
5-year property for MACRS-GDS depreciation. Before-tax cash flows are as
shown below, including a $200,000 salvage value after 5 years. Using a 40
percent income tax rate, determine the ATCF for each year and the after-tax
PW, AW, IRR, and ERR using an 8 percent MARRAT.
EOY
0
1
2
3
4
5
BTCF
–$800,000
100,000
200,000
300,000
400,000
700,000
Summary 349
33. An investment of $800,000 is made in equipment that qualifies as 3-year equip-
ment for MACRS-GDS depreciation. The BTCF profile for the investment is
given below, including a $200,000 salvage value at the end of the 5-year planning horizon. A 40 percent tax rate applies and the after-tax MARR is 8 percent.
Determine the ATCF for each year and the after-tax PW, AW, IRR, and ERR.
EOY
0
1
2
3
4
5
BTCF
2$800,000
100,000
200,000
300,000
400,000
700,000
34. In Problem 33, suppose the equipment still qualifies as MACRS-GDS 3-year
property, but is sold for $300,000 after 3 years of use. A 40 percent tax rate
applies and the after-tax MARR is 8 percent. Determine the ATCF for each
year and the after-tax PW, AW, IRR, and ERR.
Section 9.2.2 After-Tax Analysis Using Retained Earnings—Multiple
Alternatives (No Borrowing)
35.
GO Tutorial Two investments involving a virtual mold apparatus for producing dental crowns qualify for different property classes. Investment A has
a cost of $58,500, lasts 9 years with no salvage value, and costs $150,000 per
year in operating expenses. It is in the 3-year property class. Investment B has
a cost of $87,500, lasts 9 years with no salvage value, and costs $125,000 per
year. Investment B, however, is in the 7-year property class. The company
marginal tax rate is 40% and MARR is an after-tax 10%.
a. Based upon the use of MACRS-GDS depreciation, compare the AW of
each alternative to determine which should be selected.
b. What must be Investment B’s cost of operating expenses for these two
investments to be equivalent?
36. A virtual mold apparatus for producing dental crowns permits an infinite
number of shapes to be custom constructed based upon mold imprints taken
by dentists. Two models are available. One costs $58,500 and is expected
to last 9 years with no salvage value at that time. Costs of use are $30 per
crown and 5,000 crowns per year are produced. The other mold apparatus
costs $87,500, lasts 9 years, has no salvage value, and is less costly to use at
$25 per crown. The dental supplier depreciates assets using MACRS, and yet
values assets of the company using straight line depreciation. The marginal
tax rate is 40% and MARR is an after-tax 10%.
a. Based upon the use of MACRS-GDS depreciation (be sure to state the
property class), compare the AW of each alternative to determine which
should be selected.
b. Based upon the use of straight line (no half year) depreciation, compare
the AW of each alternative to determine which should be selected.
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Chapter 9
Income Taxes
37.
Video Solution A portable concrete test instrument used in construction
for evaluating and profiling concrete surfaces (MACRS-GDS 5-year property
class) is under consideration by a construction firm for $22,000. The instrument
will be used for 6 years and be worth $2000 at that time. The annual cost of use
and maintenance will be $9500. Alternatively, a more automated instrument
(same property class) available from the manufacturer costs $29,000, with use
and maintenance costs of only $7500 and salvage value after 6 years of $3000.
The marginal tax rate is 40% and MARR is an after-tax 12%. Determine which
alternative is less costly, based upon comparison of after-tax annual worth.
38. Two investments involving a granary qualify for different property classes.
Investment A costs $70,000 with $3000 salvage value after 16 years and
is depreciated as MACRS-GDS in the 10-year property class. Investment
B costs $110,000 with a $4000 salvage value after 16 years and is in the
MACRS-GDS 5-year property class. Operation and maintenance for each is
expected to be $18,000 and $14,000 per year, respectively. The marginal tax
rate is 40%, and MARR is 9% after taxes.
a. Determine which alternative is less costly, based upon comparison of
after-tax annual worth.
b. What must the cost of the second (more expensive) investment be for there
to be no economic advantage between the two?
39.
A granary has two options for a conveyor used in the manufacture of
grain for transporting, filling, or emptying. One conveyor can be purchased
and installed for $70,000 with $3000 salvage value after 16 years. The other
can be purchased and installed for $110,000 with $4000 salvage value after
16 years. Operation and maintenance for each is expected to be $18,000 and
$14,000 per year, respectively. The granary uses MACRS-GDS depreciation,
has a marginal tax rate of 40%, and a MARR of 9% after taxes.
a. Determine which alternative is less costly, based upon comparison of
after-tax annual worth.
b. What must the cost of the second (more expensive) conveyor be for there
to be no economic advantage between the two?
40. A firm may invest $30,000 in a numerically controlled lathe for use in furniture
manufacturing (MACRS-GDS 7-year property). It would last for 11 years and
have zero salvage value at that time. Alternatively, the firm could invest $X in a
methods improvement study (this is non-depreciable). Each of the investment
alternatives will yield an increase in income of $15,000 for 11 years. If the tax
rate for the firm is 40 percent, for what value of $X will the firm be indifferent
between the two investment alternatives? The after-tax MARR is 15 percent.
(Hint: Excel® SOLVER would be an excellent way to determine the answer.)
41. Two mutually exclusive alternatives, A and B (both MACRS-GDS 5-year
property), are available. Alternative A requires an original investment of
$100,000, has a useful life of 6 years, annual operating costs of $2,500, and
a salvage value at the end of year k given by $100,000(0.70)k. Alternative
B requires an original investment of $150,000, has a life of 8 years, zero
annual operating costs, and a salvage value at the end of year k given by
$150,000(0.80)k. The after-tax MARR is 15 percent, and a 40 percent tax
Summary 351
rate is applicable. Perform an annual worth comparison and recommend the
least-cost alternative.
a. Use a planning horizon of 6 years.
b. Use a planning horizon of 8 years and assume that an identical A will be
purchased for the final two years of use.
Section 9.3 After-Tax Analysis Using Borrowed Capital—Single Alternative
42. What is the difference or distinction being made when we speak of before-tax
cash flows and before-tax and loan cash flows. Be precise in your answer.
43.
Chevron Phillips has put into place new laboratory equipment for the production of chemicals; the cost is $1,800,000 installed. CP borrows 45% of all
capital needed and the borrowing rate is 12.5%. In the first year, 25% of the
principal borrowed will be paid back. The throughput rate for in-process test
samples has increased the capacity of the lab, saving a net of $X per year. In
this first year, depreciation is $360,000 and taxable income is $328,000.
a. What is the “gross income” or annual savings $X?
b. Determine the income tax for the first year assuming a marginal tax rate
of 40%.
c. What is the after-tax cash flow for the first year?
44. Hyundai USA has numerous robotic welders as well as robotic checkers with
vision. One underbody robotic welder was installed and is increasing productivity by 2.5% in one area. The result is a savings of $500,000 per year.
Deductible expenses other than depreciation and interest associated with the
installed robotic welder are only $120,000. Depreciation is $171,480 this year.
Interest on borrowed money is $60,500 and no principal is paid back this year.
a. Determine the taxable income for the year.
b. Determine the income tax for the year assuming a marginal tax rate of 40%.
c. Determine the after tax cash flow for the year.
45.
GO Tutorial Abbott placed into service a flexible manufacturing cell
costing $850,000 early this year for production of their analytical testing
equipment. Gross income due to the cell is expected to be $750,000 with
deductible expenses of $475,000. Depreciation is based on MACRS-GDS
and the cell is in the 7 year property class calling for a depreciation percentage of 14.29% or $121,465 in the first year. Half of the cell cost is financed at
11% with principal paid back in equal amounts over 5 years. The first year’s
interest is therefore $46,750 while the principal payment is $85,000.
a. Determine the taxable income for the first year.
b. Determine the tax paid due to the cell during the first year using a 40%
marginal tax rate.
c. Determine the after-tax cash flow for the first year.
46. A Boeing contractor responsible for producing a portion of the landing
gear for huge airliners experienced a storm-related power glitch during the
multi-axis milling, to tolerances less than 0.001 inch, of a large and complex part. The value already in the part, plus the equipment damage, was
$300,000. Risk analysis indicates that a similar event might occur once
352
Chapter 9
Income Taxes
per year on average if nothing is done. PolyPhaser, a leader in lightning and
surge protection was commissioned to do a turnkey installation to protect
this critical portion of the process. The first cost is $480,000 installed. A
total of $275,000 is borrowed at a rate of 12% per year and no principal is
repaid in the first year. Deductible annual costs are $Y, and depreciation is
MACRS-GDS in the 7-year property class, or 14.29% in the first year. The
taxable income is $15,000.
a. What is the value of the “deductions” or $Y in the first year?
b. What is the income tax paid in the first year assuming a marginal tax rate
of 40%?
c. What is the after-tax cash flow for the first year?
Note: Problems 47–55 use one or more of the following loan plans:
■
Plan 1—pay the accumulated interest at the end of each interest period
and pay the principal at the end of the loan period.
■
■
■
Plan 2—make equal principal payments, plus interest on the unpaid
balance at the end of the period.
Plan 3—make equal principal-plus-interest end-of-period payments.
Plan 4—make a single payment of principal and interest at the end of the
loan period.
47. An investment of $250,000 is made in equipment that qualifies as MACRS-
GDS 7-year property. The before-tax cash flow profile for the investment
is given below, including a $100,000 salvage value at the end of the 5-year
planning horizon. A loan is taken for 80 percent of the investment capital at
an annual compound interest rate of 18 percent and the loan is repaid over a
5-year period. A 40 percent tax rate and an MARRAT of 7 percent apply.
EOY
0
1
2
3
4
5
BTCF
–$250,000
40,000
40,000
40,000
40,000
140,000
Determine the PW of the ATCFs using:
a. Loan payment Plan 1
b. Loan payment Plan 2
c. Select which of the two loan payment plans is preferred, and explain why
it is preferred in terms of the relationship between the loan rate and the
MARRAT.
48. A company purchases a machine for $800,000. The equipment qualifies as
5-year property for MACRS-GDS depreciation. The machine is paid for by
borrowing $500,000, to be repaid over a 5-year period at an annual compound interest rate of 12 percent. Before-tax cash flows are as shown on the
next page, including a $200,000 salvage value after 5 years. The income tax
rate is 40 percent. An MARRAT of 10% applies.
Summary 353
EOY
0
BTCF
2$800,000
100,000
200,000
300,000
400,000
700,000
1
2
3
4
5
Determine the PW of the ATCFs using:
a. Loan payment Plan 2.
b. Loan payment Plan 4.
c. State which of the two loan payment plans is preferred, and explain why it is
preferred in terms of the relationship between the loan rate and the MARRAT.
49. Specialized production equipment is purchased for $125,000. The equip-
ment qualifies as 5-year equipment for MACRS-GDS depreciation. Suppose
40 percent of the investment capital is borrowed at an annual compound rate
of 18 percent and the loan is repaid over a 4-year period. The BTCF profile
for the acquisition, shown below, includes a $30,000 salvage value at the end
of the 5-year planning horizon. A 40 percent tax rate applies. Develop tables
using a spreadsheet to determine the ATCF for each year and the after-tax
PW, AW, IRR, and ERR values if the after-tax MARR is 10 percent.
EOY
0
1
2
3
4
5
BTCF
–$125,000
50,000
60,000
70,000
80,000
120,000
Determine the PW of the ATCFs using:
a. Loan payment Plan 1.
b. Loan payment Plan 3.
c. State which of the two loan payment plans is preferred, and explain why it is
preferred in terms of the relationship between the loan rate and the MARRAT.
50. Hyundai USA has numerous robotic welders as well as robotic checkers with
vision. One underbody robotic welder costing $1,200,000 (7-year property
class) was installed and is increasing productivity by 2.5% in one area. The result is a savings of $500,000 per year. Deductible expenses other than depreciation and interest associated with the installed robotic welder are only $120,000.
Hyundai borrowed $550,000 at 11% for 5 years. They plan to keep the welder
for 8 years. Hyundai’s marginal tax rate is 40% and their MARR is 9% after
taxes. Determine the PW, FW, AW, IRR, and ERR for the investment if:
a.
b.
c.
d.
The loan is paid back using Plan 1.
The loan is paid back using Plan 2.
The loan is paid back using Plan 3.
The loan is paid back using Plan 4.
354
Chapter 9
Income Taxes
51.
GO Tutorial Abbott placed into service a flexible manufacturing cell costing $850,000 early this year. They financed $425,000 of it at 11% per year
over 5 years. Gross income due to the cell is expected to be $750,000 with
deductible expenses of $475,000. Depreciation is based on MACRS-GDS
and the cell is in the 7 year property class. Abbott’s marginal tax rate is 40%,
MARR is 10% after taxes, and they expect to keep the cell for 8 years. Determine the PW, FW, AW, IRR, and ERR for the investment if:
a.
b.
c.
d.
The loan is paid back using Plan 1.
The loan is paid back using Plan 2.
The loan is paid back using Plan 3.
The loan is paid back using Plan 4.
52. A Boeing contractor responsible for producing a portion of the landing gear
for huge airliners experienced a storm-related power glitch during the multiaxis milling, to tolerances less than 0.001 inch, of a large and complex part.
The value already in the part, plus the equipment damage, was $300,000. Risk
analysis indicated that a similar cost might occur once per year on average if
nothing is done. PolyPhaser, a leader in lightning and surge protection was
commissioned to do a turnkey installation to protect the process from similar yearly losses. The first cost is $480,000 installed. A total of $275,000 is
borrowed at a rate of 12% per year for the entire 10 year planning horizon.
Deductible annual operating and maintenance costs are $Y, and depreciation
is MACRS-GDS in the 7-year property class. The marginal tax rate is 40%,
MARR is 10%, and the expected life of the PolyPhaser equipment is 10 years.
There is no salvage value. Use Goal Seek or Solver in Excel® to determine the
value of Y such that MARR is exactly achieved, no more and no less, if:
a.
b.
c.
d.
53.
The loan is paid back using Plan 1.
The loan is paid back using Plan 2.
The loan is paid back using Plan 3.
The loan is paid back using Plan 4.
Video Solution Raytheon wishes to use an automated environmental
chamber in the manufacture of electronic components. The chamber is to be
used for rigorous reliability testing and burn-in. It is installed for $1.4 million,
$600,000 of which is borrowed at 11% for 5 years, and will have a salvage
value of $200,000 after 8 years. Its use will create an opportunity to increase
sales by $650,000 per year and will have operating expenses of $250,000 per
year. Corporate income taxes are 40%. Develop tables using a spreadsheet to
determine the ATCF for each year, and the after-tax PW, AW, IRR and ERR if
the chamber is kept for 8 years. After-tax MARR is 10%. Determine for each
year the ATCF and the PW, FW, AW, IRR, and ERR for the investment if (for
parts a, b, and c, use straight-line depreciation (over 8 years with no half-year
convention), and for parts d, e, and f, use MACRS-GDS depreciation with the
appropriate property class):
a,d. The loan is paid back using Plan 1.
b,e. The loan is paid back using Plan 2.
c,f. The loan is paid back using Plan 3.
Summary 355
54. A subsidiary of AEP places in service electric generating and transmission line
equipment at a cost of $3,000,000 with half of it borrowed at 11% over 8 years.
It is expected to last 30 years with a salvage value of $250,000. The equipment
will increase net income by $500,000 in the first year, increasing by 2.4% each
year thereafter. The subsidiary’s tax rate is 40% and the after-tax MARR is
9%. There is some concern that the need for this equipment will last only
10 years and need to be sold off for $550,000 at that time. Develop tables using
a spreadsheet to determine the ATCF for each year, and the after-tax PW, AW,
IRR and ERR after only 10 years to see if the venture would be worthwhile
economically if (for parts a, b, and c, use straight-line depreciation (no halfyear convention), and for parts d, e, and f, use MACRS-GDS depreciation):
a,d. The loan is paid back using Plan 1.
b,e. The loan is paid back using Plan 2.
c,f. The loan is paid back using Plan 3.
55.
Chevron Phillips has put into place new laboratory equipment for the
production of chemicals; the first cost is $1,800,000 installed. Chevron
Phillips borrows 45% of all capital needed and the borrowing rate is 12.5%
over 4 years. The throughput rate for in-process test samples has increased
the capacity of the lab, with a net savings of $X per year. Depreciation
follows MACRS-GDS, MARR is 11%, and the planning horizon is 6 years
with a salvage value of $250,000 at that time. Use Goal Seek or Solver in
Excel® to determine the value of X such that MARR is exactly achieved, no
more and no less, if:
a. The loan is paid back using Plan 1.
b. The loan is paid back using Plan 2.
c. The loan is paid back using Plan 3.
d. The loan is paid back using Plan 4.
56. An asset is purchased for $90,000 with the intention of keeping it for 10 years,
but is sold at the end of year 3. A total of $30,000 was borrowed money that
was to be repaid over three years in equal annual payments, including principal and interest. The depreciation is correct for the appropriate MACRS
Recovery Period.
LOAN INT
1 RCVD
2 PAID
LOAN PRINC
1 RCVD
2 PAID
BTLCF
1 RCVD
2 PAID
LOAN
1 RCVD
2 PAID
0
2$90,000.00
$30,000.00
1
$35,000.00
2$4,500.00 2$8,639.31
2
$35,000.00
2$3,204.10
3
$40,000.00
EOY
a.
b.
c.
d.
e.
DEPR
BK VALUE
TI
TAX
1 RCVD
2 PAID
$90,000.00
$12,861.00
$7,870.50
ATCF
1 RCVD
2 PAID
PWATCF
2$60,000.00
2$60,000.00
$77,139.00 $17,639.00
2$7,408.38
$14,452.31
$12,903.85
$9,754.90
2$4,097.06
$17,763.63
$14,161.06
$47,227.50
What is the property class of the asset?
What is the value of MARR?
What is the salvage value received at the end of year 3?
What is the loan interest rate?
Determine the value of the entries in the BOLD BORDERED cells.
356
Chapter 9
Income Taxes
57.
A project has a first cost of $180,000, an estimated salvage value of $20,000
after 6 years, and other economic attributes as detailed in the table below. Unfortunately, as the end of year 4 neared, the project had to be abandoned and the
market value of the asset at that time was different from the original estimated
salvage value. There was a loan to help finance the project. It was being paid
back in equal annual installments (the yearly principal plus interest payment
was the same). The remaining principal had to be paid at the end of year 4 when
the project was stopped. Following is the table (Note! The shaded cells would
normally have something in them but have been erased for this exercise.):
BTLCF
1 IN;
2 OUT
LOAN
PRINC
1 IN;
2 OUT
LOAN
PRINC
REMAIN
0
2$180,000.00
$70,000.00
$70,000.00
1
$40,000.00
2$9,785.71
$60,214.29
2$4,900.00 $36,000.00
2
$40,000.00
$49,743.59
2$4,215.00
3
$40,000.00
EOY
4
LOAN INT
1 IN;
2 OUT
DEPR
BK VALUE
TI
$180,000.00
2$2,697.80 $10,368.00
a.
b.
c.
d.
e.
ATCF
1 IN;
2 OUT
PWATCF
2$110,000.00 2$110,000.00
$144,000.00
$25,638.29
$22,891.33
$7,853.40
$1,957.95
2$38,539.93
TAX
1 IN;
2 OUT
2$704.62
$24,609.43
$41,472.00 $15,462.20
Fill in all of the blanks having BOLD BORDERS.
What is the value of MARR?
What is the loan interest rate?
What is the MACRS property class?
What is the tax rate?
Section 9.3 After-Tax Analysis Using Borrowed
Capital—Multiple Alternatives
Note: Problems 58–63 use one or more of the following loan plans:
■
Plan 1—pay the accumulated interest at the end of each interest period
and pay the principal at the end of the loan period.
■
Plan 2—make equal principal payments, plus interest on the unpaid balance at the end of the period.
■
Plan 3—make equal principal-plus-interest end-of-period payments.
■
Plan 4—make a single payment of principal and interest at the end of the
loan period.
58. Recall problem 36. Suppose both models will require a loan of $40,000 at an
interest rate of 17 percent, with payment using Plan 1 over a period of 4 years.
Rework the problem.
59. Recall problem 37. Suppose both alternatives will require a loan of 50 percent of
the investment cost, an annual interest rate of 18 percent, with payment using
Plan 3 over the 6-year planning horizon. Rework the problem.
60. Recall problem 38. Suppose both investments will require a loan of $14,000
at an interest rate of 10 percent, with payment using Plan 4 over a period of
3 years. Rework the problem.
Summary 357
61. Recall problem 39. Suppose both conveyors will require a loan of $14,000
at an interest rate of 10 percent, with payment using Plan 2 over a period of
3 years. Rework the problem.
62. Recall problem 40. Suppose both alternatives will involve a loan of 40 percent
of the investment cost, with payment over 10 years using Plan 4 and a sweet
Federal government loan rate of 2 percent. Rework the problem.
63. Recall problem 41. Suppose both options will involve a loan of 40 percent of
the investment cost, with payment over a) 6 years and b) 8 years using Plan 1
and a sweet Federal government loan rate of 2 percent. Rework the problem.
Section 9.4 Purchasing versus Leasing Equipment
64. Griffin Dewatering is considering three alternatives. The first is the purchase
of a permanent steel building to house their existing equipment for the overhaul of dewatering systems (engines, pumps, and well points). The building
can be put into service for $240,000 in early January of this year. The planning horizon for this is 10 years, at which time the building can be sold for
$120,000 in late December. Maintenance and upkeep of the equipment, plus
labor and materials for overhaul, costs $130,000 per year. A second alternative is to lease a building for $15,000 per year at the beginning of each year
in which case all operating costs are identical except for an additional $4000
per year cost due to the inconvenient location of the lease property. Third,
they could simply contract out the overhaul work for $170,000 per end-ofyear, with an immediate credit through salvage of their present equipment for
$45,000. Marginal taxes are 40% and the after-tax MARR is 12%.
a. Determine the annual worth associated with buying the building. Be sure
to give the appropriate MACRS-GDS property class.
b. Determine the annual worth of leasing.
c. Determine the annual worth of contracting out the work.
d. Determine the annual contract price that makes contracting and leasing
economically equivalent.
65.
Michelin is considering going “lights out” in the mixing area of the business that operates 24/7. Currently, personnel with a loaded cost of $600,000
per year are used to manually weigh real rubber, synthetic rubber, carbon
black, oils, and other components prior to manual insertion in a Banbary
mixer that provides a homogeneous blend of rubber for making tires (rubber
products). New technology is available that has the reliability and consistency desired to equal or exceed the quality of blend now achieved manually.
It requires an investment of $2.5 million, with $110,000 per year operational
costs and will replace all of the manual effort described above. The planning
horizon is 8 years and there will be a $300,000 salvage value at that time for
the new technology. Marginal taxes are 40% and the after-tax MARR is 10%.
a. Determine the annual cost of purchasing the new technology.
b. Determine the annual cost of continuing with the manual mixing.
c. Determine the amount of the investment in new technology that would
make the two alternatives equivalent.
ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E :
C ATER P I L L AR
In 1890, Benjamin Holt and Daniel Best competed to see who could produce
the best steam-powered tractor for use in farming. By 1925, they had merged
their companies, forming the Caterpillar Tractor Company. Headquartered in
Peoria, Illinois, Caterpillar today is the world’s leading manufacturer of construction and mining equipment, diesel and natural gas engines, industrial
gas turbines, and diesel-electric locomotives.
Caterpillar’s chairman and CEO, Douglas R. Oberhelman, stated in the
2011 Annual Report, “Business has changed. The industries we serve have
changed. Competition is tough, customer requirements are demanding, and
the industries we serve are booming, and they are complicated.”
In 2011, the company had sales and revenues totaling $60.1 billion, an
increase of 41 percent from 2010. Caterpillar is a global company with seventy
percent of sales and revenue in 2011 occurring outside the United States. In
2011, Caterpillar’s worldwide employment was 125,099 of which 53,236 were
within the United States and 71,863 were located outside the United States.
Caterpillar’s profits were $7.15 billion, compared with $3.96 billion in 2010, and
capital expenditures totaled $2.6 billion, up from $1.58 billion in 2010. R&D
expenditures in 2011 were $2.3 billion, representing 3.8 percent of sales.
In its 2011 Annual Report, the company acknowledges that its business
is highly sensitive to global economic conditions and economic conditions in
the industries and markets that it serves. Inflation is an important economic
factor, especially when one considers that Caterpillar is making significant
investments to increase their production of construction equipment not only
within the United States, but also in Brazil and Asia, regions that have experienced tremendous inflation over the past decade. Global companies such
as Caterpillar must pay close attention to inflation rates in the countries in
which they operate and serve. Inflation rates can differ markedly from one
economy to another, as do income-tax laws.
358
INFLATION
DISCUSSION QUESTIONS:
1. In 2011, Caterpillar generated 70 percent of its sales and revenue
from outside the United States. Why is this relevant to our study of
inflation?
2. Given high inflation rates in regions in which Caterpillar has
production facilities, what decisions might management consider
making?
3. Inflation rates in the United States have been fairly steady in recent
years. Does Caterpillar need to develop different business strategies
for its ventures within and outside the United States?
4. The inflation rate in different regions of the world is outside Caterpillar’s control. What can the company do to protect itself from the
adverse effects of inflation such as rising commodity or component
prices and rising labor rates?
LEARNING OBJECTIVES
When you have finished studying this chapter, you should be able to:
1. Explain the effects of inflation on purchasing power and how the inflation rate is estimated. (Section 10.1)
2. Conduct a before-tax analysis with inflation by correctly incorporating
inflationary effects in both the cash flows and the discount rate.
(Section 10.2)
3. Conduct an after-tax analysis with inflation by correctly adjusting for
situations in which depreciation allowances are not permitted to
increase with inflation. (Section 10.3)
4. Conduct an after-tax analysis with inflation using borrowed capital.
(Section 10.4)
359
360
Chapter 10
Inflation
INTRODUCTION
Previously, we have said little about inflation and its effect on engineering
economic justification. In Chapter 1, we stated, “Money has time value in
the absence of inflation.” In Chapter 2, we indicated that annual increases
in expenditures represented by a geometric series of cash flows could be
due to inflation.
In this chapter, we examine the before-tax and after-tax effects of inflation on the economic worth of an investment when capital is borrowed
and when it comes from retained earnings. We show that inflation can
significantly impact the economic viability of a capital investment. For
that reason, when comparing investment alternatives in an inflationary
economy, it is important to give explicit consideration to inflation.
Systematic Economic Analysis Technique
1.
Identify the investment alternatives
2. Define the planning horizon
3. Specify the discount rate
4. Estimate the cash flows
5. Compare the alternatives
6. Perform supplementary analyses
7.
10-1
Select the preferred investment
THE MEANING AND MEASURE OF INFLATION
LEARN I N G O B JEC T I V E : Explain the effects of inflation on purchasing power
and how the rate is estimated.
Inflation A decrease in
the purchasing power of
money that is caused by
an increase in general
price levels of goods
and services without an
accompanying increase in
the value of the goods and
services.
Consumers understand the impact of inflation on their ability to purchase
goods and services. As the value of a dollar diminishes over time, the
effects of inflation are manifested. More precisely, we can say that inflation
is a decrease in the purchasing power of money that is caused by an
increase in general price levels of goods and services without an accompanying increase in the value of the goods and services. Inflationary pressure
is created when more dollars are put into an economy without an accompanying increase in goods and services. In other words, printing more
money without an increase in economic output generates inflation.
Inflation can have such a significant impact on an investment’s economic worth that it should be considered when comparing investment
10-1 The Meaning and Measure of Inflation 361
alternatives. Although it is important to consider the impact of inflation
on investments made within one country, it is especially important to do
so in multinational investment situations. The inflation rates in different
countries can be dramatically different. Hence, a firm that is faced with
making decisions concerning investments of capital in various nations
must give strong consideration to the inflationary conditions in the countries in question.
As noted, inflation adversely affects the purchasing power of money.
To illustrate what we mean by the purchasing power of money, suppose a
firm purchases 1 million pounds of a particular material each year, and the
price of the material increases by 3 percent per year. The quantity of material the firm can purchase with a fixed amount of money—that is, the purchasing power of the firm’s money—decreases over time. The only way
the firm can afford to continue purchasing the material is to decrease its
usage rate or to increase its source of funds. In the latter case, the firm
might increase the price of the products it sells; if so, then the purchasing
power of its customers’ money will be decreased. In these situations, the
continuing spiral of price increases does not contribute real increases to
the firm’s profits; instead, it results in an inflated representation of the
firm’s profits. The overall process of price increases without accompanying increases in the quality or value of the goods or services is referred to
as inflation.
How is the inflation rate determined? The approach varies, depending on what rate is desired. For example, the most commonly used way
of estimating the inflation rate for consumers in the United States is the
consumer price index, developed by the U.S. Department of Labor’s
Bureau of Labor Statistics. Published monthly, the consumer price
index (CPI) measures the price changes that occur from one month to
the next for a specified set of products. Also called the retail price index,
it is the index most often referred to by the U.S. media when referring to
inflation rates.
The CPI is a market basket rate, in that it is based on 80,000 prices that
are recorded in 87 urban areas. The so-called market basket is purchased
each month and reflects what consumers buy for their day-to-day living.
The market basket includes more than 200 categories of expenditures,
arranged in eight major groups: food and beverages; housing; apparel;
transportation; medical care; recreation; education and communication;
and other goods and services.
The Bureau of Labor Statistics has collected data since 1913. Major
changes in definitions and methodology have occurred over the years. The
CPI reflects spending patterns, not costs of living, for two population
groups in the United States: all urban consumers, and urban wage earners
and clerical workers.
Consumer Price Index
(CPI) A value developed
by the U.S. Department
of Labor’s Bureau of
Labor Statistics used to
determine inflation rates,
measuring the price
changes that occur from
one month to the next for
a specified set of products.
Also known as the retail
price index.
362
Chapter 10
Inflation
Table 10.1 shows CPI values from 1913 through 2012. The table and
Figure 10.1 also show the year-to-year percentage changes in the CPI.
These percentages are the estimates of inflation used by many businesses
and financial institutions. As shown, double-digit annual inflation has
occurred in the past. From 2007 through 2012, annual inflation rates have
been 2.8%, 3.8%, 20.4%, 1.6%, 3.2%, and 2.1% respectively.
TABLE 10.1
Year
CPI
1913
9.9
1914
10.0
1915
10.1
1916
CPI Data 1913–2012
% Change
Year
CPI
Year
CPI
1946
19.5
% Change
8.3%
1979
72.6
% Change
11.3%
1.0%
1947
22.3
14.4%
1980
82.4
13.5%
1.0%
1948
24.1
8.1%
1981
90.9
10.3%
10.9
7.9%
1949
23.8
21.2%
1982
96.5
6.2%
1917
12.8
17.4%
1950
24.1
1.3%
1983
99.6
3.2%
1918
15.1
18.0%
1951
26.0
7.9%
1984 103.9
4.3%
1919
17.3
14.6%
1952
26.5
1.9%
1985 107.6
3.6%
1920
20.0
15.6%
1953
26.7
0.8%
1986 109.6
1.9%
1921
17.9
210.5%
1954
26.9
0.7%
1987 113.6
3.6%
1922
16.8
26.1%
1955
26.8
20.4%
1988 118.3
4.1%
1923
17.1
1.8%
1956
27.2
1.5%
1989 124.0
4.8%
1924
17.1
0.0%
1957
28.1
3.3%
1990 130.7
5.4%
1925
17.5
2.3%
1958
28.9
2.8%
1991 136.2
4.2%
1926
17.7
1.1%
1959
29.1
0.7%
1992 140.3
3.0%
1927
17.4
21.7%
1960
29.6
1.7%
1993 144.5
3.0%
1928
17.1
21.7%
1961
29.9
1.0%
1994 148.2
2.6%
1929
17.1
0.0%
1962
30.2
1.0%
1995 152.4
2.8%
1930
16.7
22.3%
1963
30.6
1.3%
1996 156.9
3.0%
1931
15.2
29.0%
1964
31.0
1.3%
1997 160.5
2.3%
1932
13.7
29.9%
1965
31.5
1.6%
1998 163.0
1.6%
1933
13.0
25.1%
1966
32.4
2.9%
1999 166.6
2.2%
1934
13.4
3.1%
1967
33.4
3.1%
2000 172.2
3.4%
1935
13.7
2.2%
1968
34.8
4.2%
2001 177.1
2.8%
1936
13.9
1.5%
1969
36.7
5.5%
2002 179.9
1.6%
1937
14.4
3.6%
1970
38.8
5.7%
2003 184.0
2.3%
1938
14.1
22.1%
1971
40.5
4.4%
2004 188.9
2.7%
1939
13.9
21.4%
1972
41.8
3.2%
2005 195.3
3.4%
1940
14.0
0.7%
1973
44.4
6.2%
2006 201.6
3.2%
1941
14.7
5.0%
1974
49.3
11.0%
2007 207.3
2.8%
1942
16.3
10.9%
1975
53.8
9.1%
2008 215.3
3.8%
1943
17.3
6.1%
1976
56.9
5.8%
2009 214.5
20.4%
1944
17.6
1.7%
1977
60.6
6.5%
2010 218.1
1.6%
1945
18.0
2.3%
1978
65.2
7.6%
2011 224.9
3.2%
2012 229.6
2.1%
10-1 The Meaning and Measure of Inflation 363
Percent Annual Change
20.0%
15.0%
10.0%
5.0%
0%
–5.0%
–10.0%
2012
2005
1998
1991
1984
1977
1970
1963
1956
1949
1942
1935
1928
1921
1914
–15.0%
Year
FIGURE 10.1
CPI Annual Percent Changes 1913–2012
Some people contend that the CPI does not measure correctly the
inflation they experience, because they do not purchase everything included
in the CPI market basket calculation. As an example, if they do not intend
to buy or sell a house, they argue that the changes in house prices have no
influence on them.
Further, if they are vegetarians, they argue that the change of prices for
beef, pork, and chicken does not affect them. Similar arguments are made
for a large number of the items in the market basket the Bureau of Labor
Statistics uses to calculate the CPI.
While it is true that few, if any, people purchase all of the items
included in the CPI market basket, it is incorrect to conclude that such
price increases do not impact everyone. Today’s economy is interconnected.
Consider vegetarians. How can their household budgets be impacted by
increases in meat prices? If owners of stores where vegetarians shop eat
meat and increase the prices in their stores because of price increases for
meat, then vegetarians are impacted indirectly by increases in meat prices.
In general, if store owners increase prices in reaction to price increases for
items in the CPI market basket they purchase, then all of their customers are
indirectly affected by the price increases in the CPI market basket. This
phenomenon causes all of us to be impacted by price increases in things we
do not purchase.
The same occurs in business. A soft-drink manufacturer that purchases large quantities of sugar but very little oil is still impacted when
oil prices increase. Oil is consumed in manufacturing the plastic bottles
and aluminum cans the soft-drink manufacturer uses; also, price increases
for oil impact the cost of fuel to heat and air-condition the offices and
364
Chapter 10
Inflation
Producer Price Index
(PPI) A family of over
10,000 indexes published
monthly by the Bureau
of Labor Statistics. PPI
measures price changes
from the seller’s or
producer’s perspective.
Higher Education Price
Index (HEPI) A value
used to determine inflation
rates specifically within the
higher education sector.
production areas. Price increases in one segment of the economy are
generally felt in all segments of the economy.
Because most businesses do not experience inflation as reflected by
the market basket used to measure the CPI, a different index is used for
producers of goods and services. The producer price index (PPI) is a
family of over 10,000 indexes published monthly by the Bureau of
Labor Statistics. It measures price changes from the seller’s or
producer’s perspective. Formerly called the wholesale price index, the
PPI generally provides a better measurement of the inflation effects for
a particular business or for the purchase of particular equipment than
the CPI.
Another index, the Higher Education Price Index (HEPI), has been
developed to measure the year-to-year changes in higher education costs.
Due to rapid increases in expenditures for library holdings, Internet
upgrades, wireless connectivity, computing and laboratory equipment,
health insurance, utilities, and faculty salaries in some key academic disciplines, in general, HEPI increases faster than CPI.1
10-2
BEFORE-TAX ANALYSIS
LEARN I N G O B JEC T I V E : Conduct a before-tax analysis with inflation by correctly incorporating inflationary effects in both the cash flows and the discount rate.
Constant Dollars Dollars
considered as being
free from inflation. Also
known as real dollars,
today’s dollars, inflationfree dollars, constant
purchasing power dollars,
constant-value dollars, and
constant-worth dollars.
Then-Current
Dollars Dollars that
include inflation. Also
known as future dollars,
nominal dollars, actual
dollars, and inflated
dollars.
When incorporating inflation effects in economic analyses, it is essential
for the discount rate and cash flows to be consistent. Specifically, if inflation is to be incorporated explicitly, then inflationary effects must be incorporated in both the cash flows and in the discount rate. Likewise, if inflation is not to be incorporated in the analysis, then both the cash flows and
the discount rate must exclude inflation or be inflation-free. Otherwise, the
analysis will be flawed.
To facilitate a consideration of inflation in economic analyses, it is
useful to define two terms: constant dollars and then-current dollars.
Constant dollars are free of inflation; they are also called real dollars,
today’s dollars, inflation-free dollars, constant purchasing power dollars,
constant-value dollars, and constant-worth dollars. Then-current dollars
include inflation; they are also called future dollars, nominal dollars,
actual dollars, and inflated dollars.
1
For a year-to-year comparison of CPI and HEPI, see White, J. A., K. E. Case, D. B. Pratt,
Principles of Engineering Economic Analysis, 6th edition, John Wiley & Sons, NY, 2012; for
more information on CPI, see the Web site for the Bureau of Labor Statistics (www.bls.gov/
cpi/cpifact2htm); and for more information on HEPI, see the Web site for the Commonfund
Institute (www.commonfund.org/Commonfund).
10-2 Before-Tax Analysis
It is also useful to define three rates: the real interest rate, the inflation rate, and the combined rate. The real interest rate does not include
inflation; it is a pure discount rate—one that expresses the real desired
return on investment. The inflation rate is expressed as a percent and
represents the loss of a dollar’s purchasing power in 1 year. The combined rate, also called the market rate and the inflation-adjusted interest
rate, includes both the real required return on investment and the inflation rate.
Letting ir denote the real interest rate, f denote the inflation rate, and ic
denote the combined rate, the following relationship exists among the
three rates:
11 1 ic 2 5 11 1 ir 2 11 1 f 2
(10.1)
i c 5 ir 1 f 1 ir 1 f 2
(10.2)
which reduces to
The same relationships exist among the real minimum attractive rate
of return (MARRr), the inflation rate, and the combined minimum attractive rate of return (MARRc):
MARRc 5 MARRr 1 f 1 MARRr 1 f 2
Real Interest Rate An
interest rate that does
not include inflation. It
is a pure discount rate,
one that expresses the
real desired return on an
investment.
Inflation Rate A rate
representing the loss of a
dollar’s purchasing power
in 1 year.
Combined Rate An
inflation rate that includes
both the real required
return on investment and
the inflation rate. It is also
known as the market rate
and the inflation-adjusted
interest rate.
(10.3)
When given ic and f, the value of ir can be obtained from Equation
10.2. Specifically,
ir 5 1ic 2 f 2/ 11 1 f 2
(10.4)
MARRr 5 1MARRc 2 f 2/ 11 1 f 2
(10.5)
Likewise,
Calculating the Real Interest Rate
EXAMPLE
If inflation averages 4 percent per year and your return on an investment, based
on then-current dollars, is 10 percent, what is your real return on investment?
Given: f 5 4%, ic 5 10%
Find: ir
365
KEY DATA
366
Chapter 10
Inflation
SOLUTION
From Equation 10.4,
ir 5 1ic 2 f 2/ 11 1 f 2 5 10.10 2 0.042/ 11 1 0.042
5 0.057692 or 5.7692%
Letting $Tk denote the magnitude of a then-current cash flow at the
end of year k and $Ck denote the magnitude of a constant-dollar cash flow
at the end of year k, the following relationship holds:
or
Likewise,
or
$Tk 5 $Ck 11 1 f 2 k
(10.6)
$Tk 5 $Ck 1F Z P f%,k2
(10.7)
$Ck 5 $Tk 11 1 f 2 2k
(10.8)
$Ck 5 $Tk 1P Z F f %,k2
(10.9)
Before-Tax Impact of Inflation on the Present Worth of Raw
Material Cost
EXAMPLE
Video Example
KEY DATA
SOLUTION
A small business spent $125,000 for raw material during 2012. It anticipates the cost of raw material will increase, due to inflation, at a rate of
3 percent per year. It also anticipates that the amount of raw material
required will increase at a rate of 8 percent per year due to increased
demand for the firm’s products. If the business has a 10 percent real
required return on its investments, what is the present worth of the cost of
raw material over the 5-year period (2013–2017)?
Given: Cash flows shown in Table 10.2, f 5 3%, MARRr 5 10%
Find: PW of raw material cost
Table 10.2 shows the yearly cash flows for this example in constant and thencurrent dollars. The constant-dollar cash flows equal $125,000(1.08)k 2 2012
and the then-current cash flows equal $125,000[1.08(1.03)] k 2 2012.
The present worth of $591,724.35 for the constant-dollar cash flows
can be obtained using the Excel® NPV worksheet function with a 10%
MARRr. Similarly, the present worth of $591,724.35 for the then-current
cash flows is obtained using a 10% 1 3% 1 10%(3%) or 13.3% MARRc.
10-3 After-Tax Analysis 367
Alternatively, the present worths can be calculated by hand using calculated or tabulated values, as shown in Table 10.2.
TABLE 10.2
k
Cash Flows and Present Worth for Example 10.2
$Ck
(P |F,
10%,n)
PWk in
Constant $
$Tk
(P |F,
13.3%,n)
PWk in
ThenCurrent $
2013
$135,000.00 0.90909
$122,727.27 $139,050.00 0.88261
$122,727.27
2014
$145,800.00 0.82645 $120,495.87 $154,679.22 0.77900
$120,495.87
2015
$157,464.00 0.75131
$118,305.03 $172,065.16 0.68756
$118,305.03
2016
$170,061.12
0.68301
$116,154.03 $191,405.29 0.60685
$116,154.03
2017
$183,666.01 0.62092
$114.042.14 $212,919.24 0.53561
$114.042.14
PWtotal 5
$591,724.35
PWtotal 5
$591,724.35
10-3
AFTER-TAX ANALYSIS
LEARNING O BJECTI VE: Conduct an after-tax analysis with inflation by cor-
rectly adjusting for situations in which depreciation allowances are not permitted to increase with inflation.
In previous chapters, one of two assumptions was made: Our before-tax
analysis was based on cash flows and interest rates that were inflation-free
or that included inflation effects. When cash flows are expressed in constant dollars and the minimum attractive rate of return is a real required
return, nothing special needs to be done to incorporate inflation when performing before-tax analyses. That is not true for after-tax analyses!
Why the difference? Because, in the United States, depreciation allowances are not permitted to increase with inflation. Depreciation allowances
are expressed in then-current dollars, not constant dollars. Therefore,
either the effects of inflation must be “stripped” out of the depreciation
allowances in order to express them in constant dollars, or all other cash
flows must be expressed in then-current dollars.
Using Excel to Analyze the After-Tax Impact of Inflation on the
SMP Investment
Recall the $500,000 investment in a surface-mount placement machine
that produces net annual savings, before taxes, of $124,166.67 for
10 years, plus a $50,000 salvage value at the end of the 10-year period.
EXAMPLE
368
Chapter 10
Inflation
The estimates were in constant dollars. The SMP machine qualified as
5-year property for MACRS-GDS depreciation. A 40 percent income tax
rate and a MARRr of 10 percent were used to perform an after-tax analysis. To illustrate the effect of inflation in after-tax analyses, suppose annual
inflation equals 4 percent. What are the PW and IRR values of the investment in then-current dollars?
KEY DATA
Given: Cash flows and depreciation as outlined in Figure 10.2, MARRr 5
10%, f 5 4%, income tax 5 40%
Find: PW of the SMP investment with 4% inflation, IRRc, IRRr
FIGURE 10.2
After-Tax, Then-Current Dollar Analysis of the SMP Investment
with 4% Inflation
SOLUTION
As shown in Figure 10.2, then-current, before-tax cash flows ($TBTCF)
are obtained by increasing constant dollar, before-tax cash flows
($CBTCF) at an annual compound rate of 4 percent; depreciation allowances are in then-current dollars. The resulting then-current, after-tax
present worth is $109,201.18, compared to $123,988.64 if inflation is
negligible. Likewise, when inflation is negligible, the after-tax internal
rate of return is 16.12 percent. With 4 percent inflation, the combined
after-tax internal rate of return is 19.99 percent; therefore, the real aftertax internal rate of return is
IRRr 5 10.1999 2 0.042/ 11.042 5 0.1538, or 15.38%.
10-4 After-Tax Analysis with Borrowed Capital
An alternative approach is to convert the depreciation allowances to constant dollars using Equation 10.8 or 10.9. As shown in Figure 10.3, the
constant dollar, after-tax present worth is $109,201.18, which is identical
to the then-current result. The real after-tax internal rate of return is
15.38 percent.
FIGURE 10.3
After-Tax, Constant Dollar Analysis of the SMP Investment with
4% Inflation
10-4
AFTER-TAX ANALYSIS
WITH BORROWED CAPITAL
LEARNING O BJECTI VE: Conduct an after-tax analysis with inflation using
borrowed capital.
When money is borrowed at fixed rates, the loan payments (principal
and interest) are then-current cash flows. Depreciation allowances are
also then-current cash flows. Because taxable income is reduced by
interest payments and depreciation allowances, it is more convenient to
perform after-tax analyses using a then-current approach when both
inflation and borrowed funds are present. If one wishes to use a constant
EXPLORING THE
SOLUTION
369
370
Chapter 10
Inflation
dollar approach, however, then the principal payments, interest payments, and depreciation allowances must be converted to constantdollar cash flows. Equation 10.8 or 10.9 can be used to perform the
conversion.
Using Excel to Analyze the Impact of Inflation on the SMP
Investment When Money Is Borrowed
EXAMPLE
Suppose $300,000 is borrowed at a 12 percent annual compound interest
rate and is repaid over a 10-year period using one of four payment plans.
This is the same scenario we considered in Example 9.8, but we now
include inflation of 4 percent/per year in our analysis. To review, the four
payment plans are: (a) Plan 1—pay the accumulated interest at the end of
each interest period and repay the principal at the end of the loan period;
(b) Plan 2—make equal principal payments, plus interest on the unpaid
balance at the end of the period; (c) Plan 3—make equal end-of-period
payments; and (d) Plan 4—make a single payment of principal and interest
at the end of the loan period.
Which of the payment plans should be used if the MARRr equals
10 percent? (As before, a 40 percent income tax rate is used.)
KEY DATA
Given: i 5 12%, f 5 4%, MARRr 5 10%, tax rate 5 40%, $300,000
borrowed with each of the four payment plans
Find: PWAT for each payment plan
SOLUTION
Figure 10.4 presents the results for Plan 1. Tables 10.3 through 10.5 provide the after-tax results for Plans 2, 3, and 4, respectively.
■
■
■
■
Plan 1 yields an after-tax present worth of $220,132.29;
Plan 2 yields an after-tax present worth of $182,165.68;
Plan 3 yields an after-tax present worth of $191,157.40;
Plan 4 yields an after-tax present worth of $232,335.64.
On this basis, Plan 4 should be used to repay the $300,000 loan over a
10-year period. (Similar conclusions are reached using either FW or AW
then-current comparisons.)
Notice that for Plans 1 and 4 multiple negative values exist for ATCF,
so, the Excel® MIRR worksheet function cannot be used to compute
the ERR. Instead, as was done in Example 9.8, the ERR is calculated
using the Excel® IRR worksheet function with the modified ATCF
column.
10-4 After-Tax Analysis with Borrowed Capital
371
FIGURE 10.4 After-Tax Analysis of the SMP Investment with $300,000 Borrowed, Repaid with Plan 1, and 4% Inflation
After-Tax Analysis of the SMP Investment; $300,000 Borrowed and Repaid
with Plan 2; 4% Inflation
TABLE 10.3
MARRr 5 10%
Income Tax Rate 5 40%
EOY
0
1
2
3
4
5
6
7
8
9
10
$TBTCF
$TPPMT
2$500,000.00
$129,133.34
$134,298.67
$139,670.62
$145,257.44
$151,067.74
$157,110.45
$163,394.87
$169,930.66
$176,727.89
$257,809.22
2$300,000.00
$30,000.00
$30,000.00
$30,000.00
$30,000.00
$30,000.00
$30,000.00
$30,000.00
$30,000.00
$30,000.00
$30,000.00
Interest Rate 5 12%
Inflation Rate 5 4%
$TIPMT
$36,000.00
$32,400.00
$28,800.00
$25,200.00
$21,600.00
$18,000.00
$14,400.00
$10,800.00
$7,200.00
$3,600.00
$TDWO
$TTI
$TTax
$TATCF
$100,000.00
$160,000.00
$96,000.00
$57,600.00
$57,600.00
$28,800.00
$0.00
$0.00
$0.00
$0.00
2$6,866.66
2$58,101.33
$14,870.62
$62,457.44
$71,867.74
$110,310.45
$148,994.87
$159,130.66
$169,527.89
$254,209.22
2$2,746.67
2$23,240.53
$5,948.25
$24,982.98
$28,747.10
$44,124.18
$59,597.95
$63,652.26
$67,811.16
$101,683.69
PW$T 5
FW$T 5
AW$T 5
IRRc 5
ERRc 5
IRRr 5
ERRr 5
2$200,000.00
$65,880.00
$95,139.20
$74,922.37
$65,074.47
$70,720.64
$64,986.27
$59,396.92
$65,478.40
$71,716.73
$122,525.53
$182,165.68
$699,401.91
$35,470.47
35.34%
22.05%
30.13%
17.36%
372
Chapter 10
Inflation
After-Tax Analysis of the SMP Investment; $300,000 Borrowed and Repaid
with Plan 3; 4% Inflation
TABLE 10.4
MARRr 5 10%
Income Tax Rate 5 40%
EOY
0
1
2
3
4
5
6
7
8
9
10
Interest Rate 5 12%
Inflation Rate 5 4%
$TBTCF
$TPPMT
2$500,000.00
$129,133.34
$134,298.67
$139,670.62
$145,257.44
$151,067.74
$157,110.45
$163,394.87
$169,930.66
$176,727.89
$257,809.22
2$300,000.00
$17,095.25
$19,146.68
$21,444.28
$24,017.59
$26,899.71
$30,127.67
$33,742.99
$37,792.15
$42,327.21
$47,406.47
TABLE 10.5
$TIPMT
$36,000.00
$33,948.57
$31,650.97
$29,077.65
$26,195.54
$22,967.58
$19,352.26
$15,303.10
$10,768.04
$5,688.78
$TDWO
$TTI
$100,000.00
$160,000.00
$96,000.00
$57,600.00
$57,600.00
$28,800.00
$0.00
$0.00
$0.00
$0.00
2$6,866.66
2$59,649.90
$12,019.65
$58,579.79
$67,272.20
$105,342.87
$144,042.61
$154,627.56
$165,959.85
$252,120.44
$TTax
$TATCF
2$2,746.67
2$23,859.96
$4,807.86
$23,431.91
$26,908.88
$42,137.15
$57,617.04
$61,851.02
$66,383.94
$100,848.18
PW$T 5
FW$T 5
AW$T 5
IRRc 5
ERRc 5
IRRr 5
ERRr 5
2$200,000.00
$78,784.75
$105,063.38
$81,767.51
$68,730.28
$71,063.61
$61,878.05
$52,682.57
$54,984.39
$57,248.70
$103,865.79
$191,157.40
$733,924.46
$37,221.30
38.58%
22.34%
33.25%
17.63%
After-Tax Analysis of the SMP Investment; $300,000 Borrowed and Repaid with Plan 4; 4% Inflation
MARRr 5 10%
Income Tax Rate 5 40%
EOY
$TBTCF
$TPPMT
0
1
2
3
4
5
6
7
8
9
10
2$500,000.00
$129,133.34
$134,298.67
$139,670.62
$145,257.44
$151,067.74
$157,110.45
$163,394.87
$169,930.66
$176,727.89
$257,809.22
2$300,000.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$300,000.00
Interest Rate 5 12%
Inflation Rate 5 4%
$TIPMT
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$631,754.46
$TDWO
$TTI
$100,000.00
$29,133.34
$160,000.00
2$25,701.33
$96,000.00
$43,670.62
$57,600.00
$87,657.44
$57,600.00
$93,467.74
$28,800.00
$128,310.45
$0.00
$163,394.87
$0.00
$169,930.66
$0.00
$176,727.89
$0.00 2$373,945.24
$TTax
$TATCF
2$200,000.00
$11,653.33
$117,480.00
2$10,280.53
$144,579.20
$17,468.25
$122,202.37
$35,062.98
$110,194.47
$37,387.10
$113,680.64
$51,324.18
$105,786.27
$65,357.95
$98,036.92
$67,972.26
$101,958.40
$70,691.16
$106,036.73
2$149,578.10 2$524,367.15
$232,335.64
PW$T 5
$892,023.06
FW$T 5
$45,239.33
AW$T 5
58.76%
IRRc 5
23.57%
ERRc 5
52.66%
IRRr 5
18.82%
ERRr 5
$TATCF
2$200,000.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$1,659,897.55
Summary 373
In Chapter 9, PWAT was $175,603.01 for Plan 1; $156,374.28 for Plan 2;
$160,734.89 for Plan 3; and $162,184.44 for Plan 4. A comparison of the
results indicates PWAT differences varied significantly among the payment
plans. With borrowed funds, however, inflation produced a greater PWAT
for each plan, as summarized in Table 10.6.
Notice the impact of inflation on the preferred plan. With no inflation,
Plan 1 is preferred; with 4 percent inflation, Plan 4 is preferred.
EXPLORING THE
SOLUTION
Comparison of PWAT for Four Payment Plans, with and
Without Inflation
TABLE 10.6
Plan
PWAT (0% Inflation)
PWAT (4% Inflation)
Difference
1
$175,603.01
$220,132.29
$44,529.28
2
$156,374.28
$182,165,68
$25,791.40
3
$160,734.89
$191,157.40
$30,422.51
4
$162,184.44
$232,335.64
$70,151.20
KEY CONCEPTS
1. Learning Objective: Explain the effects of inflation on purchasing power
and how the rate is estimated. (Section 10.1)
Inflation reduces the purchasing power of money and is an important factor
to consider in any type of engineering economic decision, especially one
where the decision spans multiple years. Inflation rates can be estimated in
many ways, including:
■
■
CPI: Within the United States, the consumer price index (CPI) is often
used to estimate the inflation rate. The CPI measures price changes that
occur from one month to the next for a specified bundle of products.
Even if one purchases products outside this selected set, they will tend to
see similar impacts of inflation, as we live in a global society where the
economy is so interconnected. For example, increases in the price of oil
in the Middle East may increase the price of consumer products in the
United States, which uses the Middle East oil to make gasoline to fuel the
trucks that transport the products to the stores where the consumers shop.
PPI: Because not all businesses experience inflation as reflected by
the market basket used to measure CPI, the producer price index (PPI)
is an alternative method to be used that is comprised of a family of
over 10,000 indexes.
SUMMARY
374 Chapter 10
Inflation
2. Learning Objective: Conduct a before-tax analysis with inflation by correctly incorporating inflationary effects in both the cash flows and the discount rate. (Section 10.2)
If inflation is to be considered in a before-tax analysis, then both the cash
flows and the discount rate must be adjusted to include inflation or exclude
inflation; otherwise, the analysis will be flawed. One must carefully consider whether constant dollars or then-current dollars should be used for
the analysis. When performed correctly, the constant dollars approach and
the then-current dollars approach will yield identical present worths.
The relationship between the real interest rate (ir), inflation rate ( f ),
and combined rate (ic) can be expressed as:
ic 5 ir 1 f 1 ir 1 f 2
(10.2)
The relationships between the real minimum attractive rate of return
(MARRr), the inflation rate, and the combined minimum attractive rate of
return (MARRc) can be expressed as:
MARRc 5 MARRr 1 f 1 MARRr 1 f 2
(10.3)
Two key equations relating constant-dollars ($Ck ) and then-current
dollars ($Tk ) at the end of year k are:
and
$Tk 5 $Ck 11 1 f 2 k
(10.6)
$Ck 5 $Tk 11 1 f 2 2k
(10.8)
3. Learning Objective: Conduct an after-tax analysis with inflation by correctly adjusting for the situation that in the U.S. depreciation allowances are
not permitted to increase with inflation. (Section 10.3)
After-tax analyses can be a bit trickier than before-tax analyses when inflation is considered because in the United States, depreciation allowances
are not permitted to increase with inflation. When performing after-tax
analyses, there are two options:
■
■
Remove the effects of inflation from the depreciation allowances in
order to express them in constant dollars;
Or, express other cash flows in then-current dollars. This is often the
simplest option.
Inflation reduces the after-tax present worth of an investment in capital
equipment unless borrowed capital is used.
4. Learning Objective: Conduct an after-tax analysis with inflation using
borrowed capital. (Section 10.4)
When money is borrowed at fixed rates, the loan payments and depreciation
allowances are then-current cash flows. It is easiest to perform after-tax
Summary 375
analyses using a then-current approach when both inflation and borrowed
funds are present because the taxable income will be reduced by interest
payments and depreciation allowances. Depending on the fraction of investment capital that is borrowed, inflation can increase the after-tax present
worth of an investment in capital equipment when using borrowed capital.
KEY TERMS
Combined Rate, p. 365
Constant Dollars, p. 364
Consumer Price Index (CPI), p. 361
Higher Education Price
Index (HEPI), p. 364
Inflation, p. 360
Inflation Rate, p. 365
Producer Price Index (PPI), p. 364
Real Interest Rate, p. 365
Then-Current Dollars, p. 364
Problem available in WileyPLUS
GO Tutorial Tutoring Problem available in WileyPLUS
Video Solution Video Solution available in WileyPLUS
FE-LIKE PROBLEMS
1.
Logan is conducting an economic evaluation under inflation using the
then-current approach. If the inflation rate is j and the real time value of
money rate is d, which of the following is the interest rate he should use for
discounting the cash flows?
a. j
c. j 1 d
b. d
d. j 1 d 1 dj
2.
Mike’s Veneer Shop owns a vacuum press that requires annual maintenance.
Mike has a contract to cover the maintenance expenses for the next five years.
The contract calls for an annual payment of $600 with adjustment each year
for inflation. Inflation is expected to hold constant at 6%/yr over this period.
The then-current cash flow pattern for this expense is best described by which
of the following?
a. Uniform series
b. Gradient series
c. Geometric series
d. Continuous series
3.
Mike’s Veneer Shop owns a vacuum press that requires annual maintenance. Mike has a contract to cover the maintenance expenses for the next
376
Chapter 10
Inflation
five years. The contract calls for an annual payment of $600 with adjustment
each year for inflation. Inflation is expected to hold constant at 6%/yr over
this period. The constant dollar cash flow pattern for this expense is best
described by which of the following?
a. Uniform series
b. Gradient series
c. Geometric series
d. Continuous series
4.
An economist has predicted that there will be a 7% per year inflation of
prices during the next ten years. If this prediction proves to be correct, an item
that presently sells for $10 would sell for what price in ten years?
a. $5.08
c. $17.00
b. $10.70
d. $19.67
5.
If the real discount rate is 7% and the inflation rate is 10%, which of the
following interest rates will be used to find the present worth of a series of
cash flows that are in then-current dollars?
a. 10.0%
c. 7.0%
b. 17.7%
d. 10.7%
6.
If the real discount rate is 7% and the inflation rate is 10%, which of the
following interest rates will be used to find the present worth of a series of
cash flows that are in constant-worth dollars?
a. 10.0%
c. 7.0%
b. 17.7%
d. 10.7%
7.
When done correctly, what is the relationship between the present worth
of an alternative calculated using a then-current approach and the present
worth of the alternative calculated using a constant-worth approach?
a. They are equal
b. Then-current PW is higher because it uses inflated dollars
c. Constant worth PW is higher because it uses a lower discount rate
d. Cannot be determined without knowing the cash flows and inflation rate
8.
Ten years ago Jennifer bought an investment property for $100,000. Over
the ten year period inflation has held consistently at 3% annually. If Jennifer
expects a 13%/yr real rate of return, what would she sell the property for
today?
a. $116,000
b. $134,400
c. $339,500
d. $456,200
9.
As reported by the Bureau of Labor Statistics, the CPI for 2005 was 585.0
(using a Base Year of 1967 5 100). The CPI for 2006 was 603.9. Based on
this data, what was the inflation rate for 2006?
a. 3.23%
c. 6.04%
b. 5.85%
d. 18.9%
Summary 377
PROBLEMS
Section 10.1 The Meaning and Measure of Inflation
1. What is a good working definition of “inflation” in 10 words or less?
2. Give four examples of goods or services that have exhibited inflation in recent
years.
3. What is the relationship between “inflation” and “deflation”? Give an example
of deflation experienced in your everyday life.
4. Give two examples of goods or services that you have seen inflate dramati-
cally and also deflate dramatically over the past few years.
5. What is meant by a “market basket rate?”
6. What are the differences between the Consumer Price Index and the Producer
Price Index?
7. What is the Higher Education Price Index (HEPI), where does it fit in with the
CPI and PPI, and how is the HEPI related to the CPI?
8. Suppose a friend argues that the CPI does not represent them because they do
not purchase some of the things, including big ticket items, in the market basket.
Can they conclude that the CPI is irrelevant to them? Explain your reasoning.
9.
The CPI-U (U.S. city average, all items) has the following annual averages:
Year
Index
2002
2003
2004
2005
2006
179.9
184.0
188.9
195.3
201.6
a. For each year from 2003 to 2006 determine the annual inflation rate in
percent to two decimal places.
b. Since inflation, like interest, is compounded from period to period (e.g.,
year to year), estimate the overall annual inflation rate per year from 2002
to 2006. Suggestion! Do not simply average the rates of part (a).
10. The CPI-U for Americans 62 years of age and older (some of your professors
and some of your authors are interested in this!) present the following annual
inflation rates in percent:
Year
Rate %
2001
2002
2003
2004
2005
1.6
2.4
1.9
3.3
3.4
378
Chapter 10
Inflation
a. Assuming the index value in year 2000 was 100.0, determine the index for
each year from 2001 to 2005 to one place after the decimal.
b. Since inflation, like interest, is compounded from period to period (e.g.,
year to year), estimate the overall annual inflation rate per year from 2002
to 2006. Suggestion! Do not simply average the rates given above.
11.
The Korean manufacturing output per hour index is given as follows for
years 2001–2005 as follows:
Year
Index
2001
2002
2003
2004
2005
214.8
235.8
252.2
281.2
305.1
a. For each year from 2002 to 2005 determine the rate of increase in Korean
manufacturing output per hour to two decimal places.
b. Since this index, like inflation, is compounded from period to period (e.g.,
year to year), estimate the overall annual rate of increase in Korean manufacturing output per hour from 2002 to 2005. Suggestion! Do not simply
average the rates of part (a).
12. The Korean hourly compensation rates of increase are presented as follows:
Year
Rate %
2001
2002
2003
2004
2005
–14.84
13.75
10.09
18.87
18.79
a. Assuming the index value in year 2000 was 165.9, determine the index for
each year from 2001 to 2005 to one place after the decimal.
b. Since this index, like inflation, is compounded from period to period (e.g.,
year to year), estimate the overall annual rate of increase per year from
2001 to 2005. Suggestion! Do not simply average the rates given above.
Section 10.2
13.
Before-Tax Analysis
You are earning 5.2% on a certificate of deposit. Inflation is running 3.5%.
What is the real rate of return on your investment?
14. You are considering a bond that pays annually at 6.2%. Inflation is projected
to be running 4.0%. What will be your real interest rate?
15.
Array Solutions requires a 14.0% return on their projects. Analysis shows
that even though they have been earning the desired 14.0%, their real return appears to be only 10.0% when they look at what they can buy with their returns.
a. Explain why there is this discrepancy.
b. Determine the inflation rate.
Summary 379
16. Chevron-Phillips requires a real return of 14.2%. If inflation is running 3.8%,
what must be their MARR or “hurdle rate” on capital investments when using
then-current dollars in analyses?
17.
Suppose you want to earn a real interest rate of 5%. For inflation rates of
0.0, 1.0, 2.0, . . ., 9.0, 10.0, 15.0, 20.0, and 50.0 percent, determine the combined rate of interest you must earn.
18. How much must you invest exactly 5 years from now to have $500,000 in
today’s buying power 20 years from now? You can invest your money at 10%
per year and inflation runs 4%.
19.
A software company’s labor requirements currently cost $350,000/year. The
labor hour requirements are expected to increase by 10% per year over the next
five years. If inflation is 4%, determine the labor costs after five years using:
a. Then-current dollars.
b. Constant-worth dollars.
20. You invested $10,000 on January 1, 2004, at 7% interest compounded annu-
ally. You have not touched the investment since that date. You are planning to
take your money and close out the investment on January 1, 2014.
a. If average inflation is 3.7%, what has been your “real” annual interest rate?
b. At the time you originally invested, there was a boat you admired costing
$8,000. Over the years, boats are inflating at a rate of 7%. You were also
interested in a marine navigation system costing $4,000; similar systems
are dropping in price at a 2% rate. If you decide to buy one of each when
you close out the account, how much will the purchases cost you?
c. How much money will you have left over (or be short) after your purchases?
21.
Your department is budgeting miscellaneous expenses for the next 5 years.
Your best guess at the annual inflation rate is 3.9% and the combined MARR
is 15%. Expenses currently run $14,500 per year. Assume that expenses are
end-of-year payments.
a. Determine the then-current dollar amounts for years 1, 2, 3, 4, and 5.
b. Determine the constant dollar amount for years 1, 2, 3, 4, and 5.
c. Determine the PW of the then-current dollar amounts.
d. Determine the PW of the constant dollar amounts.
22. Shea is pricing materials (wood, wire, pipe, etc.) for new home construction
on a “per unit” basis. Inflation on materials has been running at 16.0% for the
past three years and is expected to remain at that rate for the next 10 years. The
actual dollars paid for expenses at the end of year 3 for a “unit” are $55,000.
a. What did the same set of materials cost 3 years ago?
b. If the trend continues, what will they cost 10 years from now?
23.
Video Solution Global steel prices have a year-over-year inflationary rate
increase of 12.4%. Tube Fab purchased $700,000 of a particular carbon steel
during the year just ended right now, and they intend to purchase the same
quantity at the end of each of the next 5 years. Tube Fab earns a real rate of
16.0% on their money.
a. Determine the then-current amounts they will pay for steel at the end of
each of the next 5 years.
380
Chapter 10
Inflation
b. Determine the constant value amounts they will pay for steel at the end of
each of the next 5 years.
c. Determine Tube Fab’s PW of expenditures over the next 5 years using
then-current dollars.
d. Determine Tube Fab’s PW of expenditures over the next 5 years using
constant-value dollars.
24. Global steel prices have a year-over-year inflationary rate increase of 12.4%.
Tube Fab purchased $700,000 of a particular carbon steel during the year
just ended right now. Their business has been increasing and they intend to
purchase 20% more steel each year, over the previous year’s purchase, for the
next 5 years. Tube Fab earns a real rate of 9.0% on their money.
a. Determine the then-current amounts they will pay for steel at the end of
each of the next 5 years.
b. Determine the constant value amounts they will pay for steel at the end of
each of the next 5 years.
c. Determine Tube Fab’s PW of expenditures over the next 5 years using
then-current dollars.
d. Determine Tube Fab’s PW of expenditures over the next 5 years using
constant-value dollars.
25.
Padayappa has now retired after 40 years of employment. He just made
an annual deposit to his investment portfolio and realized he has $2 million
(not counting home, cars, furniture, etc.). His money has been earning 7% per
year and inflation has been running 4% per year over the past 40 years.
a. What equal amount of money did he put into his investment at the end of
each year?
b. What is the buying power of his $2 million in terms of a base 40 years ago?
c. If he could buy a TV 40 years ago for $400, what would a comparable one
cost today if the consumer electronics inflation rate is 23%?
26. A 24-year old December 2012 graduate has decided he wants the equivalent of
$2,500,000 in January 1, 2013 buying power to be available exactly 40 years
later on January 1, 2053. He plans to make his first investment of $Z on January 1, 2014 and every year thereafter, with the last payment of $Z on January
1, 2053. He can earn 8% on his money and expects inflation to run 5%.
a. How many actual dollars will there be in his account immediately after his
last deposit?
b. What is $Z?
27.
An investment of $8,000 is made at time 0 with returns of $3,500 at the
end of each of years 1–4, with all monetary amounts being in real dollars.
Inflation is running 7% per year over that time. Also, the real rate of return is
15% per year. Determine the present worth of the investment using both real
dollars and then-current dollars.
28. The winner of a lottery is given a choice of $1,000,000 cash today or
$2,000,000 paid out as follows: $100,000 cash per year for 20 years with the
first payment today and 19 subsequent annual payments thereafter. The
inflation rate is expected to be constant at 4%/year over the award period and
the winner’s TVOM (real interest rate) is 3.5%/year.
Summary 381
a. Which choice is better for the winner? Neglect the effect of taxes, life
span, and uncertainty.
b. At what value of inflation are the two choices economically equivalent?
c. What would you do if you do NOT neglect the effect of life span and
uncertainty?
29. A family wishes to provide for their child’s college education. Being a bit risk
averse, they plan to invest in stable, yet unspectacular, opportunities yielding
a 6.0% return. Their best guess at inflation is 4.0% for the foreseeable future.
The plan to make investments on the child’s birthday (USA-style—first birthday is one year after date of birth), every year from age 1 through 18. They
envision their child needing $100,000 at the beginning of the first year of
college, with inflated amounts to follow for 3 more years. The first $100,000
will be needed right at the end of the 18th birthday’s investment—right at the
beginning of the 19th year.
a. What equal amount of money must they invest at the end of each year?
b. If the parents decide that their earning power will increase, and each year
they will invest 10% more, how much must they invest in the first year?
c. If a grandparent offers to put up the entire sum of money needed, on the
date of the child’s birth, what sum must they put up?
30. Labor costs over a 4-year period have been forecast in then-current dollars as
follows: $10,000, $12,000, $15,000, and $17,500. The general inflation rate
for the 4 years is forecast to be 5%. Determine the constant dollar labor costs
for each of the 4 years.
31. Yearly labor costs of a highway maintenance group are currently $420,000/
year. If labor costs increase at a 12% rate and general inflation increases at
9%, determine for each of the next 6 years the labor cost in then-current and
constant dollars.
32. If you desire a real return of 8% on your money and inflation is running at
3%, what combined rate of return should you require on your investments?
33. Mellin Transformers, Inc. uses a required return of 15% in all economic jus-
tifications. Inflation is anticipated to be 5% over the foreseeable future. What
real discount rate are they implicitly using?
34. The following material costs are anticipated over a 5-year period: $9,000,
$11,000, $14,000, $18,000, and $23,000. It is estimated that a 4% inflation
rate will apply over the time period in question. The material costs given
above are expressed in then-current dollars. The time value of money, excluding inflation, is estimated to be 7%. Determine the present worth equivalent
for material cost using the following:
a. Then-current costs
b. Constant, worth costs
35. The inflation rates for 4 years are forecast to be 3%, 3%, 4%, and 5%. The
interest rate exclusive of inflation is anticipated to be 6%, 5%, 4%, and 5%
over this same period. If labor is projected to be $1,000, $15,000, $2,000,
and $1,000 in then-current dollars, during those years, determine the present
worth equivalent for labor cost.
382
Chapter 10
Inflation
36. A landfill has a first cost of $270,000. Annual operating and maintenance
costs for the first year will be $40,000. These costs will increase at 11 percent annually. Income for dumping rights at the landfill will be held fixed at
$120,000 per year. The landfill will be operating for 10 years. Inflation will
average 4.5 percent, and a real return of 3.6 percent is desired. What is the
present worth of this project using
a. a then-current analysis?
b. a constant-worth analysis?
37. The unit price of personal computers, measured in constant dollars, is expected
to decrease at an annual rate of 10%. However, the number of microcomputers
purchased by the company is expected to increase over a 5-year period at an annual rate of 30%. The unit price of a personal computer is currently $2,000. This
year the company will purchase 100 microcomputers at the current price. Using
a real interest rate of 8% and an inflation rate of 6%, determine the following:
a. the yearly expenditures for personal computers over the 5-year period,
measured in constant worth dollars.
b. the yearly expenditures for personal computers over the 5-year period,
measured in then-current dollars.
c. the single sum equivalent at the end of the current year of the expenditures
on microcomputers over the 5-year period, measured in current dollars.
d. the future worth equivalent of the expenditures on microcomputers at the
end of the 5-year period, measured in then-current dollars.
38. $90,000 is invested in a program to reduce the material requirements in a
production process. As a result of the investment, the annual material requirement is reduced by 10,000 pounds. The present unit cost of the material is
$2 per pound. The price of a pound of the material is expected to increase
at an annual rate of 8%, due to inflation. Determine the combined interest
rate that equales the present worth of the savings to the present worth of the
investment over a 5-year period; based on the result obtained, determine the
real interest rate that equates the two present worths.
39. A rental car agency rents cars for an average price of $45/day. The num-
ber of cars owned in 2010 totals 150. The number of days rented per car in
2010 equals 200. Hence, the rental revenue in 2010 totals $45(150)(200), or
$1,350,000. If the agency increases the number of cars at an annual rate of
10%, the number of days rented per car each year remains constant, and the
average rental rate increases 5%/year due to inflation, what is the present
worth of rental income in 2015 if the agency’s real MARR is 10%? (Allow
for a fractional valued number of cars.)
40. In 2005, Motorola introduced a new smart phone on the market at a retail
price of $250; the cost of producing the phone was $50/unit. During 2005,
5 million smart phones were sold. The number of smart phones sold increased at an annual rate of 10% through 2009; in 2010, Apple introduced a
new product that caused the demand for the Motorola smart phone to equal
one-half the number sold in 2009. Over the 5-year period (2006, 2007, 2008,
2009, and 2010), the retail price for the Motorola smart phone (measured
Summary 383
in then-current dollars) decreased at an annual rate of 20% and the cost of
producing the smart phone (measured in constant or 2005 dollars) decreased
at an annual rate of 25%. Over the 5-year period, annual inflation was 2%.
Motorola has a real required return on investment of 10%. What was the
cumulative present worth of the profits for the Motorola smart phone for the
years 2005 through 2010?
41. True or False: when the real return on investment is 6% and the combined
minimum attractive rate of return is 10%, then inflation must be less than 4%.
42. True or False: if the combined minimum attractive rate of return is 12% and
inflation is 3%, then an investment that yields a real external rate of return of
9% is not justified economically.
43. True or False: if the real MARR is 10% and inflation is 4%, then an invest-
ment that yields a combined external rate of return of 13.5% is not justified
economically.
Section 10.3 After-Tax Analysis
44. An economic analysis is being performed in real (not actual) dollars. The com-
pany’s combined MARR is 10 percent, and the inflation rate is 4 percent. The
asset has a first cost of $10,000. It will be depreciated as MACRS 3-year property using rates of 33.33 percent, 44.45 percent, 14.81 percent, and 7.41 percent. What depreciation amount will be shown in year 3 of the analysis?
45. Electronic Games is moving very quickly to introduce a new interrelated set
of video games. The initial investment for equipment to produce the necessary electronic components is $9 million. The salvage value after 6 years is
$700,000. Anticipated net contribution to income is $6 million the first year,
decreasing by $1 million each year for 6 years, with all dollar amounts expressed in real dollars. Depreciation follows MACRS 5-year property, taxes
are 40 percent, the real MARR is 18 percent, and inflation is 4 percent.
a. Determine the actual after-tax cash flows for each year.
b. Determine the PW of the after-tax cash flows.
c. Determine the AW of the after-tax cash flows.
d. Determine the FW of the after-tax cash flows.
e. Determine the combined IRR of the after-tax cash flows.
f. Determine the combined ERR of the after-tax cash flows.
g. Determine the real IRR of the after-tax cash flows.
h. Determine the real ERR of the after-tax cash flows.
46. Reconsider Problem 45 exactly as written.
a.
b.
c.
d.
e.
f.
g.
h.
Determine the real after-tax cash flows for each year.
Determine the PW of the after-tax cash flows.
Determine the AW of the after-tax cash flows.
Determine the FW of the after-tax cash flows.
Determine the real IRR of the after-tax cash flows.
Determine the real ERR of the after-tax cash flows.
Determine the combined IRR of the after-tax cash flows.
Determine the combined ERR of the after-tax cash flows.
384
Chapter 10
Inflation
47.
Video Solution Henredon purchases a high-precision programmable
router for shaping furniture components for $190,000. It is expected to last
12 years and have a salvage value of $5,000. It will produce $45,000 in net
revenue each year during its life. All dollar amounts are expressed in real dollars. Depreciation follows MACRS 7-year property, taxes are 40 percent, the
real after-tax MARR is 10 percent, and inflation is 3.9 percent.
a. Determine the actual after-tax cash flows for each year.
b. Determine the PW of the after-tax cash flows.
c. Determine the AW of the after-tax cash flows.
d. Determine the FW of the after-tax cash flows.
e. Determine the combined IRR of the after-tax cash flows.
f. Determine the combined ERR of the after-tax cash flows.
g. Determine the real IRR of the after-tax cash flows.
h. Determine the real ERR of the after-tax cash flows.
48. Reconsider Problem 47 exactly as written.
a.
b.
c.
d.
e.
f.
g.
h.
49.
Determine the real after-tax cash flows for each year.
Determine the PW of the after-tax cash flows.
Determine the AW of the after-tax cash flows.
Determine the FW of the after-tax cash flows.
Determine the real IRR of the after-tax cash flows.
Determine the real ERR of the after-tax cash flows.
Determine the combined IRR of the after-tax cash flows.
Determine the combined ERR of the after-tax cash flows.
GO Tutorial Raytheon wishes to use an automated environmental chamber in the manufacture of electronic components. The chamber is to be used
for rigorous reliability testing and burn-in. It is installed for $1.4 million and
will have a salvage value of $200,000 after 8 years. Its use will create an
opportunity to increase sales by $650,000 per year and will have operating
expenses of $250,000 per year. All dollar amounts are expressed in real dollars. Depreciation follows MACRS 5-year property, taxes are 40 percent, the
real after-tax MARR is 10 percent, and inflation is 4.2 percent.
a. Determine the actual after-tax cash flows for each year.
b. Determine the PW of the after-tax cash flows.
c. Determine the AW of the after-tax cash flows.
d. Determine the FW of the after-tax cash flows.
e. Determine the combined IRR of the after-tax cash flows.
f. Determine the combined ERR of the after-tax cash flows.
g. Determine the real IRR of the after-tax cash flows.
h. Determine the real ERR of the after-tax cash flows.
50. Reconsider Problem 49 exactly as written.
a.
b.
c.
d.
e.
f.
Determine the real after-tax cash flows for each year.
Determine the PW of the after-tax cash flows.
Determine the AW of the after-tax cash flows.
Determine the FW of the after-tax cash flows.
Determine the real IRR of the after-tax cash flows.
Determine the real ERR of the after-tax cash flows.
Summary 385
g. Determine the combined IRR of the after-tax cash flows.
h. Determine the combined ERR of the after-tax cash flows.
51. Henredon purchases a high-precision programmable router for shaping fur-
niture components for $190,000. It is expected to last 12 years and have a
salvage value of $5,000. It will produce $45,000 in net revenue each year during its life. All dollar amounts are expressed in actual dollars. Depreciation
follows MACRS 7-year property, taxes are 40 percent, the actual after-tax
MARR is 14.62 percent, and inflation is 4.2 percent.
a. Determine the real after-tax cash flows for each year.
b. Determine the PW of the after-tax cash flows.
c. Determine the AW of the after-tax cash flows.
d. Determine the FW of the after-tax cash flows.
e. Determine the real IRR of the after-tax cash flows.
f. Determine the real ERR of the after-tax cash flows.
g. Determine the combined IRR of the after-tax cash flows.
h. Determine the combined ERR of the after-tax cash flows.
52. Reconsider Problem 51 exactly as written.
a.
b.
c.
d.
e.
f.
g.
h.
Determine the actual after-tax cash flows for each year.
Determine the PW of the after-tax cash flows.
Determine the AW of the after-tax cash flows.
Determine the FW of the after-tax cash flows.
Determine the combined IRR of the after-tax cash flows.
Determine the combined ERR of the after-tax cash flows.
Determine the real IRR of the after-tax cash flows.
Determine the real ERR of the after-tax cash flows.
53. A surface mount placement machine is purchased for $500,000. The SMP
machine qualifies as 5-year equipment for MACRS-GDS depreciation. The
before-tax cash flows, in constant dollars, include an annual uniform series
of $120,000 plus a $100,000 salvage value at the end of the 4-year planning
horizon. A 40% tax rate applies. Inflation is 3%/yr. The real ATMARR is 8%.
a. Determine the after-tax cash flows, in constant dollars, for each year.
b. Determine the present worth for the investment.
c. Determine the real internal rate of return for the investment.
54. An investment of $600,000 is made in equipment that qualifies as 3-year equip-
ment for MACRS-GDS depreciation. The before-tax cash flows, measured
in constant dollars, for the investment consist of a uniform annual series of
$200,000 plus a $200,000 salvage value at the end of the 5-year planning horizon. A 40% tax rate and 3% inflation rate apply. The real ATMARR is 10%.
a. Determine the after-tax cash flows, in constant dollars, for each year.
b. Determine the present worth for the investment.
c. Determine the real internal rate of return for the investment.
55. Specialized production equipment is purchased for $125,000. The equipment
qualifies as 5-year equipment for MACRS-GDS depreciation. The BTCF
profile for the acquisition, in then-current dollars, is an increasing $10,000
gradient series, beginning with $50,000 the first year. In addition, a $30,000
386
Chapter 10
Inflation
salvage value occurs at the end of the 5-year planning horizon. A 40% tax rate
applies. Inflation is 3%/yr. The real ATMARR is 9%.
a. Determine the after-tax cash flows, in constant dollars, for each year.
b. Determine the present worth for the investment.
c. Determine the real internal rate of return for the investment.
56. A manufacturing company decides to purchase a computer for $800,000. The
equipment qualifies as 5-year equipment for MACRS-GDS depreciation. The
constant-dollar before-tax cash flows can be represented by a $25,000 increasing gradient series; the BTCF the first year is $125,000; and a $100,000
salvage value occurs at the end of the 7-year planning horizon. A 40% tax rate
applies. Inflation is 5%/yr. The real ATMARR is 10%.
a. Determine the after-tax cash flows, in constant dollars, for each year.
b. Determine the present worth for the investment.
c. Determine the real internal rate of return for the investment.
57. A construction company decides to purchase a crane for $250,000. The crane
qualifies as 5-year equipment for MACRS-GDS depreciation. The BTCF profile
for the acquisition, expressed in constant dollars, consists of an annual uniform
series of $50,000, plus a $50,000 salvage value at the end of the 7-year planning
horizon. A 40% tax rate applies. Inflation is 4%/yr. The real ATMARR is 8%.
a. Determine the after-tax cash flows, in constant dollars, for each year.
b. Determine the present worth for the investment.
c. Determine the real internal rate of return for the investment.
58. An investment of $450,000 is made in equipment that qualifies as 5-year
equipment for MACRS-GDS depreciation. The then-current dollar before tax
cash flows are given by a $50,000 increasing gradient series, with the cash
flow the first year equaling $100,000. In addition, a $50,000 then-current salvage value occurs at the end of the 5-year planning horizon. A 40% tax rate
and 4% inflation rate apply. The real ATMARR is 8%.
a. Determine the after-tax cash flows, in constant dollars, for each year.
b. Determine the present worth for the investment.
c. Determine the real internal rate of return for the investment.
Section 10.4 After-Tax Analysis with Borrowed Capital
59. A ham radio operator wishes to borrow $160,000 to construct a world-class
antenna system, transceiver, and amplifier at an electrically quiet location that
can be accessed remotely and controlled via the Internet. Microphone, Morse
code, radio teletype, slow-scan TV, and a host of other modes may be used for
contesting, amassing DX awards, and chatting from anywhere in the world.
She borrows the money at 8.5 percent. Inflation is running 3.8 percent. Her
combined MARR is 9 percent. The loan is to be paid back over 5 years. What
is the amount to be paid at each year-end and the PW (using both then-current
and constant dollar approaches) if repayment follows
a. Plan 1 (pay accumulated interest each year and principal at the end of the
last year)?
b. Plan 2 (make equal annual principal payments each year, plus interest on
the unpaid balance)?
Summary 387
c. Plan 3 (make equal annual payments)?
d. Plan 4 (make a single payment of principal and interest at the end of the
last year)?
60. For the situation stated in Problem 59, let the interest on borrowed money
go from 5 percent to 15 percent in 1 percent increments. For each borrowing
rate, which payment plan is preferred? Do your answers match what is predicted in the text?
61. Steinway R&D is pursuing the development of an attachment that can easily
clean the inside of grand pianos. This innovation will require a loan of
$500,000 for fabrication and testing of several units. Inflation is 3.9 percent,
and the loan is available to them at a rate of 10 percent. Their combined
MARR is 17 percent. The loan is to be paid back over 4 years. What is the
amount to be paid at the end of each year and the PW (using both then-current
and constant-dollar approaches) if payment follows
a. Plan 1 (pay accumulated interest each year and principal at the end of the
last year)?
b. Plan 2 (make equal annual principal payments each year, plus interest on
the unpaid balance)?
c. Plan 3 (make equal annual payments)?
d. Plan 4 (make a single payment of principal and interest at the end of the
last year)?
62. For the situation stated in Problem 61, let the interest on borrowed money
go from 7 percent to 20 percent in 1 percent increments. For each borrowing
rate, which payment plan is preferred? Do your answers match what is predicted in the text?
63. Electronic Games is moving very quickly to introduce a new interrelated
set of video games. The initial investment for equipment to produce the
necessary electronic components is $9 million, with $4 million borrowed at
12 percent over 6 years and paying only the interest each year and the entire
principal in the last year. The salvage value after 6 years is $700,000. Anticipated net contribution to income is $6 million the first year, decreasing by
$1 million each year for 6 years, with all dollar amounts expressed in real
dollars. Depreciation follows MACRS 5-year property, taxes are 40 percent,
the real MARR is 18 percent, and inflation is 4 percent.
a. Determine the actual after-tax cash flows for each year.
b. Determine the PW of the after-tax cash flows.
c. Determine the AW of the after-tax cash flows.
d. Determine the FW of the after-tax cash flows.
e. Determine the combined IRR of the after-tax cash flows.
f. Determine the combined ERR of the after-tax cash flows.
g. Determine the real IRR of the after-tax cash flows.
h. Determine the real ERR of the after-tax cash flows.
64. Reconsider Problem 63 exactly as written.
a. Determine the real after-tax cash flows for each year.
b. Determine the PW of the after-tax cash flows.
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Chapter 10
Inflation
c.
d.
e.
f.
g.
h.
65.
Determine the AW of the after-tax cash flows.
Determine the FW of the after-tax cash flows.
Determine the real IRR of the after-tax cash flows.
Determine the real ERR of the after-tax cash flows.
Determine the combined IRR of the after-tax cash flows.
Determine the combined ERR of the after-tax cash flows.
Video Solution Henredon purchases a high-precision programmable
router for shaping furniture components for $190,000. It is expected to last
12 years and have a salvage value of $5,000. Henredon will borrow $100,000
at 13 percent over 6 years, paying only interest each year and paying all the
principal in the sixth year. It will produce $45,000 in net revenue each year
during its life. All dollar amounts are expressed in real dollars. Depreciation follows MACRS 7-year property, taxes are 40 percent, the real after-tax
MARR is 10 percent, and inflation is 3.9 percent.
a.
b.
c.
d.
e.
f.
g.
h.
Determine the actual after-tax cash flows for each year.
Determine the PW of the after-tax cash flows.
Determine the AW of the after-tax cash flows.
Determine the FW of the after-tax cash flows.
Determine the combined IRR of the after-tax cash flows.
Determine the combined ERR of the after-tax cash flows.
Determine the real IRR of the after-tax cash flows.
Determine the real ERR of the after-tax cash flows.
66. Reconsider Problem 65 exactly as written.
a.
b.
c.
d.
e.
f.
g.
h.
67.
Determine the real after-tax cash flows for each year.
Determine the PW of the after-tax cash flows.
Determine the AW of the after-tax cash flows.
Determine the FW of the after-tax cash flows.
Determine the real IRR of the after-tax cash flows.
Determine the real ERR of the after-tax cash flows.
Determine the combined IRR of the after-tax cash flows.
Determine the combined ERR of the after-tax cash flows.
GO Tutorial Raytheon wishes to use an automated environmental chamber
in the manufacture of electronic components. The chamber is to be used for
rigorous reliability testing and burn-in. It is installed for $1.4 million and will
have a salvage value of $200,000 after 8 years. Raytheon will borrow $800,000 at
12 percent to be paid back over 8 years. The environmental chamber will create
an opportunity to increase sales by $650,000 per year and will have operating
expenses of $250,000 per year. All dollar amounts are expressed in real dollars.
Depreciation follows MACRS 5-year property, taxes are 40 percent, the real
after-tax MARR is 10 percent, and inflation is 4.2 percent. Determine the actual
after-tax cash flows for each year and the PW, FW, AW, IRRc, ERRc, IRRr, and
ERRr for each of the four loan payment plans discussed in the chapter:
a. Plan 1
d. Plan 4
b. Plan 2
e. Which is the preferred plan
c. Plan 3
for Raytheon?
Summary 389
68. Electronic Games is moving very quickly to introduce a new interrelated set
of video games. The initial investment for equipment to produce the necessary electronic components is $9 million, with $4 million borrowed at
12 percent over 6 years. The salvage value after 6 years is $700,000. Anticipated net contribution to income is $6 million the first year, decreasing by
$1 million each year for 6 years, with all dollar amounts expressed in real
dollars. Depreciation follows MACRS 5-year property, taxes are 40 percent,
the real MARR is 18 percent, and inflation is 4 percent. Determine the actual
after-tax cash flows for each year and the PW, FW, AW, IRRc, ERRc, IRRr,
and ERRr for each of the four loan payment plans discussed in the chapter:
a. Plan 1
d. Plan 4
b. Plan 2
e. Which is the preferred plan
c. Plan 3
for Electronic Games?
69. A construction company decides to purchase a crane for $250,000. The crane
qualifies as 5-year equipment for MACRS-GDS depreciation. The BTCF
profile for the acquisition, expressed in constant dollars, consists of an annual uniform series of $50,000, plus a $50,000 salvage value at the end of
the 7-year planning horizon. The crane is paid for by borrowing $100,000.
The loan is to be repaid over a 5-year period at an annual compound interest
rate of 12%. A 40% tax rate applies. Inflation is 4%/yr. The real ATMARR is
8%. Among the four payment plans discussed in the chapter, use the one that
maximizes ATPW for the borrower.
a. Determine the present worth for the investment.
b. Determine the real internal rate of return for the investment.
70. An investment of $250,000 is made in equipment that qualifies as 7-year
equipment for MACRS-GDS depreciation. Measured in constant dollars, the
investment yields annual returns of $40,000, plus a salvage value of $100,000
at the end of the 5-year planning horizon. $200,000 of the investment capital
is obtained by borrowing money at an annual compound interest rate of 18%
and the loan is repaid over a 5-year period. A 40% tax rate and 3% inflation
rate apply. The ATMARR, is 7%. Among the four payment plans discussed
in the chapter, use the one that maximizes the after-tax present worth and
determine the after-tax present worth for the investment.
71. An investment of $600,000 is made in equipment that qualifies as 3-year
equipment for MACRS-GDS depreciation. The before-tax cash flows, measured in constant dollars, for the investment consist of a uniform annual series
of $200,000 plus a $200,000 salvage value at the end of the 5-year planning
horizon. $400,000 of the investment capital is obtained by borrowing at an
annual compound interest rate of 10% and repaying the loan over a 5-year
period. A 40% tax rate and 3% inflation rate apply. The real ATMARR is
10%. Among the four payment plans discussed in the chapter, use the one that
maximizes ATPW.
a. Determine the present worth for the investment.
b. Determine the real internal rate of return for the investment.
ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E : B P
2010 was a difficult year for BP, one of the world’s largest oil and gas companies. The Gulf of Mexico oil spill in April claimed eleven lives, injured multiple
people, affected the incomes of thousands, and damaged the environment
immeasurably. Carl-Henric Svanberg, BP’s Chairman, wrote in the 2010
Annual Report, “In the days after the accident in the Gulf of Mexico the company faced a complex and fast-changing crisis. With oil escaping into the
ocean, uncertainty grew around our ability to seal the well and restore the
areas affected. This was an intense period, with the situation worsening
almost daily. Our meeting with President Obama on 16 June 2010 provided
reassurance to the U.S. government that BP would do the right thing in the
Gulf, and this marked a turning point. Through diligence and invention, our
teams stopped the flow of oil in July and completed relief-well operations in
September.”
Subsequent investigations concluded that what happened in the Gulf of
Mexico was a complex accident involving multiple causes and multiple parties. In financial terms, BP paid out over $26 billion in 2010 and 2011 in pre-tax
dollars to cover oil spill response costs, meet claims and litigation expenses,
support research, promote tourism, and help restore the environment.
2011 was a year of recovery, consolidation, and change for BP. In 2011, the
board outlined three priorities for the company related to putting safety first,
regaining trust and creating value. The company developed a 10-point strategic plan to support these priorities. The first point, as outlined in their 2011
Annual Report, states, “We are determined that BP will deliver world-class
performance in safety, risk management and operational discipline. We will
be a company that systematically applies our global standards as a single
team.”
As successful as BP had been until April of 2010, it cannot accurately predict what the future holds for it and its product portfolio. The uncertainty is
due mainly to a host of exogenous factors, such as political instability in the
Middle East, weather, earthquakes, the overall economy, and environmental
390
BREAK-EVEN,
SENSITIVITY, AND
RISK ANALYSIS
concerns, among others. BP is not alone in this regard. No business or government is immune to the impact of unknown and unknowable factors on
economic outcomes of capital investments. In fact, the Form 10-K filings for
publicly traded companies include extensive lists of risks that can significantly affect the profitability of the companies.
DISCUSSION QUESTIONS:
1. Is the risk experienced by BP limited to financial risk? If not, what other
risks are present?
2. Is the exposure to risk at BP limited to their shareholders and employees?
If not, who else is at risk?
3. What are some specific steps that BP can take to limit their risks?
4. How might sensitivity analysis be used to mitigate the risk faced by BP?
5. Given the complex investment decisions made by a company such as
BP, is an economic break-even analysis sufficient to make decisions? If
not, what other factors might need to be considered and how might
this be done?
LEARNING OBJECTIVES
When you have finished studying this chapter, you should be able to:
1. Calculate the break-even value using a break-even analysis approach.
(Section 11.1)
2. Perform a sensitivity analysis to examine the impact on the measure of
economic worth and the robustness of the economic decision when values of one or more parameters vary over specified ranges. (Section 11.2)
391
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Chapter 11
Break-Even, Sensitivity, and Risk Analysis
3. Perform risk analysis using analytical solutions to derive exact values
or estimates for the expected value and standard deviation of the measure of economic worth. (Section 11.3.1)
4. Perform risk analysis using simulation solutions to derive exact values
or estimates for the expected value and standard deviation of the measure of economic worth. (Section 11.3.2)
INTRODUCTION
To date, we have assumed complete certainty when specifying the values
of all parameters involved in an economic justification. We have assigned
quantities such as the magnitude of the initial investment, the magnitudes of the annual cash flows, the length of the planning horizon, the
value of the discount rate, the terminal salvage value for the investment,
and inflation rates.
However, unexpected things happen, such as Hurricane Sandy,
which devastated parts of the Caribbean and northeastern and midAtlantic portions of the United States in late October 2012; BP’s oil spill in
the Gulf of Mexico in 2010; the global economic crisis in 2009 and 2010;
and the United States terrorist attacks on September 11, 2001, to name
just a recent few. Each example is a major event that impacted many
lives and companies. There are also thousands of smaller events that do
not produce headlines but that affect the financial outcomes of capital
investments. And, you don’t have to read the newspaper to know how
rapidly the price of a barrel of oil changes; all you have to do is fill your
car or truck with gasoline or diesel fuel.
In this chapter, we examine several techniques that are used to increase the confidence of managers making capital investment decisions
in the face of uncertainty. As noted in Chapter 1, this chapter addresses
the sixth step in the systematic economic analysis technique: perform
supplementary analyses.
Systematic Economic Analysis Technique
1.
2.
3.
4.
5.
6.
7.
Identify the investment alternatives
Define the planning horizon
Specify the discount rate
Estimate the cash flows
Compare the alternatives
Perform supplementary analyses
Select the preferred investment
11-1
Break-Even Analysis
393
We begin the chapter by addressing situations where we want to know
what single value for a particular parameter will make someone indifferent
as to whether or not to make an investment. Such analyses are termed
break-even analyses.
The second type of supplementary analysis we consider is sensitivity
analysis. In performing such analyses, we are interested in learning how
sensitive the economic worth for one or more investments is to various
values of one or more parameters.
The third type of supplementary analysis considered is risk analysis.
In contrast to sensitivity analysis, probabilities are assigned to various
values of one or more parameters, and a probabilistic statement is made
regarding the economic worth for one or more investments. Typically,
the probabilistic statement takes the form of “the probability of the
investment having a positive-valued present worth is . . .” or “the probability of the internal rate of return for the investment being greater than
the minimum attractive rate of return is . . .” or “the investment alternative having the greatest probability of having a positive-valued present
worth is . . .”.
11-1 BREAK-EVEN ANALYSIS
LEARNING O BJECTI VE: Calculate the break-even value using a break-even
analysis approach.
Break-even analysis is normally used when an accurate estimate of a
parameter’s value cannot be provided, but intelligent judgments can be
made as to whether or not the value is less than or greater than some breakeven value. The break-even value is the value at which someone is indifferent as to whether or not to make an investment. More precisely, it is the
value of a parameter at which the measure of economic worth being used
(PW, FW, AW) is equal to zero. In performing a break-even analysis, we
want to determine the break-even value.
Although we did not label it as such, in Chapter 3 we performed
several break-even analyses and referred to the process as equivalence.
In particular, we posed a situation in which we determined the value of
a particular parameter in order for two cash flow profiles to be equivalent. Another way of stating the problem could have been, “Determine
the value of X that will yield a break-even situation between two alternatives.” In this case, X denotes the parameter whose value is to be
determined. Additionally, when the cash flow profiles for two alternatives are equivalent, a break-even situation can be said to exist between
two alternatives.
The internal rate of return itself is a break-even value, because it is the
interest rate that equates to 0 the economic worth of an investment. In
Video Lesson:
Break-Even Analysis
Break-Even Analysis A
method used when an
accurate estimate of a
parameter’s value cannot
be provided, but intelligent
judgments can be made as
to whether or not the value
is less than or greater than
some break-even value.
Break-Even Value The
value of a parameter at
which the measure of
economic worth equates
to zero. The term breakeven point is sometimes
used instead of break-even
value; however, within this
text, we will use the latter
term.
394
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
other words, it is the interest rate that equates the present worth of the
positive-valued cash flows to the present worth of the negative-valued cash
flows. Stated another way, the internal rate of return is the break-even
value for the reinvestment rate, because such a reinvestment rate will yield
a future worth of 0 for either an individual alternative or the differences in
cash flows for two alternatives. (Similarly, the external rate of return is a
break-even value, because it is the interest rate that equates the future
worth of negative-valued cash flows to the future worth of positive-valued
cash flows when the positive-valued cash flows are reinvested and earn
interest equal to the minimum attractive rate of return.)
Recall, in Chapter 4 we introduced another example of break-even
analysis. Specifically, we determined how long it would take to recover the
initial investment, which is termed the discounted payback period, when
the TVOM is greater than 0. (Also, recall that the Excel® NPER worksheet function can be used to determine how long uniform annual savings
must occur in order to fully recover an initial investment, given a desired
return on investment.)
In an engineering economic analysis, break-even analysis is often
used to determine the level of annual savings required to justify a particular capital investment. In a replacement study, break-even analysis is often
performed on the purchase price or salvage value of the replacement alternatives. Likewise, break-even analysis might be used to determine how
long annual savings must occur in order to economically justify a particular investment.
EXAMPLE
Determining the Break-Even Value for Units Sold
The Gizmo Manufacturing Company is considering making and selling a
new product. The following data have been provided to management:
Sales price
$17.50/unit
Equipment cost
$250,000
Incremental overhead cost
$50,000/year
Sales and marketing cost
$150,000/year
Operating and maintenance cost
$25/operating hour
Production time/1,000 units
100 hours
Packaging and shipping cost
$0.50/unit
Planning horizon
5 years
Minimum attractive rate of return
15 percent
11-1
Break-Even Analysis
Some managers are reluctant to launch a new product because of the
uncertainty of future sales. To provide management with information that
might make it easier to draw the correct conclusion, a break-even analysis
will be performed for annual sales required to economically justify introducing the new product.
What is the break-even value of units sold annually?
Given: Product cost parameters as defined in the chart above
KEY DATA
Find: Break-even value of units sold annually (X units/year)
Assuming a negligible salvage value for all equipment at the end of the
5-year planning horizon and letting X denote the number of units sold
annually for the product, the annual worth for the investment alternative
can be determined as follows:
SOLUTION
AW115%2 5 2$250,0001A Z P 15%,52 2 $50,000
2 $150,000 2 0.11$252X 2 $0.50X 1 $17.50X
5 2$274,578.89 1 $14.50X
Setting the annual worth equal to 0 and solving for X gives a break-even
value of 18,936.475 units per year.
How should management interpret this break-even value? Suppose they
are confident that annual sales will be between 20,000 and 30,000 units.
Because this projection is greater than the break-even value, the product
should be launched. Clearly, it is not necessary to know precisely what the
annual sales volume will be; instead, we need to know if it will be greater
than or less than the break-even value.
The break-even value will not always fall outside the range of what are
judged to be realistic values of a parameter. For example, if annual sales
will be between 15,000 and 20,000 units per year, it is not obvious what
decision should be made.
A graphical representation of the example is given in Figure 11.1. The
chart is referred to as a break-even chart, because one can determine
graphically the break-even point by observing the value of X when annual
revenue equals annual cost.
EXPLORING THE
SOLUTION
395
396
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
$600,000
$500,000
Profit region
$400,000
Annual cost
$300,000
Loss region
$200,000
Annual revenue
$100,000
Break-Even point
18,937 units
$0
0
FIGURE 11.1
3,000
6,000
9,000
12,000 15,000 18,000
Annual Sales (units)
21,000 24,000 27,000
30,000
Break-Even Analysis for the Annual Sales of a New Product
Break-Even Analysis with Excel for the SMP Investment
EXAMPLE
Recall the $500,000 investment in a surface-mount placement machine,
treated in previous chapters. a) Based on a 10-year planning horizon, a
$50,000 salvage value, and a 10 percent MARR, what annual savings are
required for the investment to break even? b) Assume $92,500 is an accurate estimate of the annual savings that will result from the SMP investment, but it is not clear how long the machine will be used. What is the
break-even value for the investment’s duration?
SOLUTION
a. Letting X denote the break-even value for annual savings, the follow-
ing annual worth relationship must hold:
$500,000 1A Z P 10%,102 5 X 1 $50,000 1A Z F 10%,102
11-1
or
X 5 $500,000 1A Z P 10%,102 2 $50,000 1A Z F 10%,102
Using the Excel® PMT function,
X 5 PMT110%,10,2500000,500002
5 $78,235.43
b. To determine the break-even value for the investment’s duration when
we do not know how long the SMP machine will be used, an assumption is needed regarding its salvage value.
First, suppose salvage value decreases linearly from $500,000 to
$50,000 over a 10-year period. With salvage value decreasing at a rate
of $45,000 per year, setting the EUAC of purchasing the SMP machine
equal to the annual worth of the annual savings and salvage value gives
$500,0001A Z P 10%,n2 5 $92,500 1 1$500,000 2 $45,000n2 1A Z F 10%,n2.
What value of n is required for the equality to hold? As shown in
Figure 11.2, using the Excel® SOLVER tool yields a value of 2.17
years for the break-even value of n.
Using the Excel® SOLVER Tool to Determine the Break-Even
Value for the Planning Horizon
FIGURE 11.2
Break-Even Analysis
397
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Chapter 11
Break-Even, Sensitivity, and Risk Analysis
Second, suppose salvage value decreases geometrically from $500,000
to $50,000 over a 10-year period. Using the Excel® RATE function, the
geometric rate is
j 5 RATE110,,2500000,500002
5 220.5672%
Therefore,
$500,0001A Z P 10%,n2 5 $92,500 1 $500,00010.794328n 2 1A Z F 10%,n2
As shown in Figure 11.2, using the Excel® SOLVER tool yields a value
of 6.95 years for the break-even value of n.
If we assume the salvage value is not a function of n and remains
$50,000 regardless of the investment’s duration, then the Excel® NPER
function can be used to determine the break-even value of n:
n 5 NPER192500,2500000,500002
5 7.57668 years
Likewise, if the salvage value is negligible, regardless of investment duration, the break-even value of n is given by
n 5 NPER192500,25000002
5 8.15972 years
11-2 SENSITIVITY ANALYSIS
LEARN I N G O B JEC T I V E : Perform a sensitivity analysis to examine the impact
Video Lesson:
Sensitivity Analysis
Sensitivity Analysis A
method used to determine
the impact on the measure
of economic worth when
values of one or more
parameters vary over
specified ranges.
on the measure of economic worth and the robustness of the economic decision when values of one or more parameters vary over specified ranges.
Sensitivity analyses are performed to determine the impact on the measure
of economic worth when values of one or more parameters vary over specified ranges. If modest changes of parameter values adversely affect an
investment’s economic worth, then it is said to be sensitive to changes in the
particular parameters. On the other hand, if the economic choice is not
affected by significant changes in the values of one or more parameters, then
the decision is said to be insensitive to changes in the parameter values.
We perform sensitivity analyses in several chapters, although we do
not always refer to them as such. For example:
■
in Chapters 4 and 5, we examine the impact on economic worth of the
minimum attractive rate of return;
11-2
■
■
■
■
Sensitivity Analysis
in Chapter 7, we examine the sensitivity of the optimum replacement
interval to changes in the initial investment, the salvage value, the rate
of increase in annual operating and management costs, and the minimum attractive rate of return;
in Chapter 9, we examine the sensitivity of after-tax present worth to
changes in the depreciation method and to changes in the amount of
money borrowed;
in Chapter 10, we examine the sensitivity of after-tax present worth to
changes in the inflation rate; we also examine the impact of inflation
on the preferred loan repayment method; and
in Chapter 12, we examine the sensitivity of the optimum investment
portfolio to changes in the level of investment capital available and the
minimum attractive rate of return.
A derisive term occasionally used to describe analytical models is
GIGO—“garbage-in, garbage-out.” In other words, what you get out of the
model is no better than what you put into it. While one cannot always trust
the results obtained from models of reality, perfect information is not often
required to produce correct decisions. Sensitivity analysis can be used to
determine if, in fact, less-than-perfect estimates of the parameter values
will result in the best decision being made.
Sensitivity Analysis for a Single Alternative
To illustrate how a sensitivity analysis might be performed, we consider
once more the SMP investment: a $500,000 initial investment, annual savings of $92,500 for a 10-year period, and a salvage value of $50,000. As
before, a 10 percent MARR applies.
Let’s consider how sensitive the annual worth for the investment is to
errors in estimating the initial investment, the annual savings, the salvage
value, the investment’s duration, and the MARR. Specifically, for an error
range of 650 percent for each parameter, what is the impact on AW?
If it is assumed that all estimates are correct except that for annual savings,
the alternative’s annual worth can be given as
AW110%2 5 2$500,0001A Z P 10%,102
1 $50,0001A Z F 10%,102 1 $92,50011 1 X2
where X denotes the percent error (decimal equivalent) in estimating
the value for annual savings. Plotting annual worth as a function of the
EXAMPLE
Video Example
SOLUTION
399
400 Chapter 11
Break-Even, Sensitivity, and Risk Analysis
$70,000
$50,000
Initial
investment
$40,000
Annual Worth
$60,000
Annual
savings
$30,000
$20,000
$10,000
–50%
–40%
Planning
horizon
–30%
–20%
–10%
$0
0%
MARR
10%
20%
Salvage
value
30%
40%
50%
–$10,000
–$20,000
–$30,000
–$40,000
FIGURE 11.3
Sensitivity Analysis for Example 11.3
percent error in estimating the value of annual savings yields the straight
line having positive slope in Figure 11.3. Performing similar analyses for
the initial investment required, the salvage value, the investment’s duration (planning horizon), and the MARR yields the remaining results given
in Figure 11.3.
(Note: For the example, we assumed the salvage value was $50,000,
regardless of the investment’s duration. In reality, salvage value depends
on investment life. Also, we used annual worth instead of present worth
in the sensitivity analysis because of the variability of the planning
horizon.)
As shown in Figure 11.3, the investment’s net annual worth is affected
differently by errors in estimating the values of the various parameters.
Based on the slopes of the sensitivity curves, annual worth is most sensitive to errors in estimating the values of the initial investment, the annual
11-2 Sensitivity Analysis
savings, and the planning horizon. It is insensitive to changes in salvage
value and is moderately sensitive to changes in the MARR.
One may also explore the impact to the break-even value during the
sensitivity analysis. The break-even values for the individual parameters are $587,649.62 for the initial investment, $78,235.43 for the annual
savings, 2$177,340.55 for the salvage value, 7.5767 years for the
investment’s duration, and 13.8 percent for the MARR. These values
are determined by equating the alternative’s annual worth to zero and
solving for the break-even value in question. (As noted previously, the
discounted payback period (DPBP) is the break-even value for the
investment duration, and the internal rate of return (IRR) is the breakeven value for the MARR.)
Sensitivity Analysis for Multiple Alternatives
EXPLORING THE
SOLUTION
EXAMPLE
Recall, in previous chapters we considered two alternative designs for a
new ride called the Scream Machine at a theme park in Florida. Design A
had an initial cost of $300,000 and net annual after-tax revenues of
$55,000; Design B had an initial investment of $450,000 and net annual
after-tax revenues of $80,000. A 10 percent MARR was used over the
10-year planning horizon. Using sensitivity analysis, determine under
what circumstances Design A will be preferred over Design B, and vice
versa.
Given: The cash flows outlined in Figure 11.4; MARR 5 10%; planning
horizon 5 10 years
Find: Sensitivity of design choice to estimation errors in annual revenue
KEY DATA
Letting x denote the error in estimating the annual revenue for Design A,
and letting y denote the error in estimating the annual revenue for Design B,
the present worth for each is as follows:
SOLUTION
PWA 5 2$300,000 1 $55,00011 1 x2 1P Z A 10%,102
and
PWB 5 2$450,000 1 $80,00011 1 y2 1P Z A 10%,102
401
402
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
$55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 $55,000
(+)
0
1
2
3
4
5
6
7
8
9
10
(–)
$300,000
Alternative A
$80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000
(+)
0
1
3
2
4
5
6
7
8
9
10
(–)
$450,000
Alternative B
FIGURE 11.4
CFDs for Example 11.4
For Design A to be preferred over Design B, PW(A) . PW(B). Therefore,
assuming either Design A or Design B must be chosen, Design A will be
preferred so long as
2$300,000 1 $55,000(1 1 x)(P Z A 10%,10) .
2$450,000 1 $80,000(1 1 y)(P Z A 10%,10)
or
$3,614.18 2 $337,951.20x 1 $491,565.00y , $0
Solving for y,
y , 20.007352385 1 0.6875x
As shown in Figure 11.5, Design A is preferred for combinations of estimation errors that fall below the indifference diagonal plotted. Design B is
11-2
Sensitivity Analysis
Estimation errors for Design B
40%
Choose Design B
–50%
–40%
–30%
–20%
–10%
30%
20%
10%
0%
0%
10%
20%
30%
40%
Estimation errors for Design A
–10%
–20%
–30%
Choose Design A
–40%
FIGURE 11.5
Sensitivity Analysis for Example 11.4
preferred for error combinations above the indifference diagonal. If the
estimates are correct for Design B, an increase of 1.07 percent in annual
revenue for Design A will cause it to be the preferred choice. Likewise,
assuming the estimates for Design A are correct, a decrease of 0.735 percent in annual revenue for Design B will make it no longer be the preferred
choice. Therefore, the preferred design is sensitive to errors in estimating
annual revenue.
The sensitivity analyses we’ve considered so far involve just one
parameter at a time. In practice, estimation errors occur for multiple parameters simultaneously. A popular method of performing multiparameter
sensitivity analysis is to provide three estimates for each parameter subject
to estimation error. The estimates, typically, represent optimistic, pessimistic, and most likely values for each parameter. The following example
illustrates this approach.
50%
403
404
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
Multiparameter Sensitivity Analysis
EXAMPLE
For Example 11.4, suppose (for both designs) there is uncertainty concerning the values for the initial investments required and the annual revenue
that will result. Specifically, suppose the estimates shown in Table 11.1 are
available for the four parameters. Under what circumstances is Design A
preferred over Design B, and vice versa?
SOLUTION
From Table 11.1, assuming the pessimistic scenario occurs for each,
Design A is preferred. If, however, the optimistic or the most likely scenario occurs, then Design B is preferred.
TABLE 11.1 Optimistic, Pessimistic, and Most Likely Estimates for Four
Scream Machine Parameters
Parameter
Pessimistic
Most Likely
Initial Investment (A)
Optimistic
$285,000
$310,000
$300,000
Initial Investment (B)
$400,000
$510,000
$450,000
Annual Revenue (A)
$65,000
$40,000
$55,000
Annual Revenue (B)
$85,000
$70,000
$80,000
PWA
$114,396.86
2$64,217.32
$37,951.19
PWB
$122,288.20
2$79,880.30
$41,565.37
It seems unlikely that pessimistic estimates will occur for all four
parameters and for both designs. Likewise, it is unlikely that the optimistic estimates will occur for all four parameters for both designs.
And, unfortunately, neither is it likely that the most likely estimates will
be realized for each of the eight parameters (four for each design).
Instead, combinations of pessimistic, optimistic, and most likely values
generally will occur.
There are 81 possible combinations of the four parameters and three
estimates (34 5 81) for each design. After computing the present worth for
each design and for each possible combination, we show in Figure 11.6 the
preferred design. In 41 cases of the 81 combinations considered, Design A
has the greatest present worth.
Some might conclude that Design A is best because more than 50 percent of the combinations favor it. However, there is no reason to believe
each combination is equally likely to occur. In order to explicitly account
for the likelihood of outcomes, we may assign probabilities to one or more
11-3 Risk Analysis
405
parameters and conduct a risk analysis. This process is demonstrated in
Example 11.7 of the next section.
FIGURE 11.6
Considering 81 Possible Combinations of 3 Estimates and 4 Parameters
11-3 RISK ANALYSIS 1
Risk analysis is performed when probabilities can be assigned to various
values of one or more parameters. We define risk analysis as the process of
incorporating explicitly random variation in one or more parameters. For
example, estimates of probabilities for possible values of annual savings
and estimates of probabilities for possible salvage values might be used to
analyze the economic worth of investing in a new machine tool.
In comparison with sensitivity analyses, risk analyses attempt to
reflect the imprecision inherent in assigning values to parameters in an
engineering economic analysis. The imprecision is represented in the
form of a probability distribution. The parameters are treated as random
variables.
Probability distributions for the random variables in question are often
based on subjective probabilities. Occasionally, there might be historical
1
A basic understanding of probability theory and Monte Carlo simulation is assumed for this
section.
Risk Analysis The
process of incorporating
explicitly random variation
in one or more parameters.
406
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
data on which the probability distributions are based. Typically, the more
distant in the future an event is, the less precise is our estimate of the value
of the event’s outcome. Hence, letting the variance reflect our degree of
precision, we expect the variance of the probability distributions to increase
with time.
Among the probability distributions commonly used in risk analysis
are the normal distribution and the beta distribution. Examples of these are
depicted in Figures 11.7 and 11.8. For discussions of several probability
distributions and their process generators in the context of simulation, see
any number of simulation texts.
Using either analytic or simulation approaches, risk analysis develops
exact values or estimates for the expected value and standard deviation of the
measure of economic worth. In addition, the probability of present worth,
future worth, or annual worth being greater than 0 and the probability of internal rate of return or external rate of return being greater than the MARR are
typically determined in risk analyses, either analytically or with simulation.
The magnitudes of cash flows, the planning horizon’s duration, and the
value of the MARR frequently can be considered to be random variables.
For example, the cash flows occurring in a given year are often functions
0.135%
2.145%
13.59%
34.13%
34.13%
68.26%
95.45%
99.73%
FIGURE 11. 7
Normal Distribution
13.59%
2.145%
11-3
FIGURE 11.8
Risk Analysis
407
Sample Beta Distributions
of several other factors, such as selling prices, market size and share, market growth rate, investment required, inflation rate, tax rates, operating
costs, fixed costs, and salvage values of all assets. The values of a number
of these random variables can be correlated with each other and can be
autocorrelated. Consequently, an analytical development of the probability distribution for the measure of economic worth is not easily achieved
in most real-world situations. Thus, simulation is widely used in performing risk analyses.
11.3.1 Analytical Solutions
LEARNING O BJECTI VE: Perform risk analysis using analytical solutions to
derive exact values or estimates for the expected value and standard deviation of the measure of economic worth.
To illustrate the use of an analytical approach to develop the probability
distribution for present worth, consider the following present worth
formulation:
n
PW 5 a At 11 1 i2 2t
(11.1)
t50
Suppose the cash flows, At , are random variables with expected values
E(At) and variances V(At). Because the expected value of a sum of random
variables is given by the sum of the expected values of the random variables, the expected present worth is given by
n
E1PW2 5 a E 3At 11 1 i2 2t 4
t50
(11.2)
Autocorrelated
Correlated with itself over
time.
408
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
Further, because the expected value of the product of a constant and a
random variable is given by the product of the constant and the expected
value of the random variable, Equation 11.2 reduces to
n
E1PW2 5 a E1At 2 11 1 i2 2t
(11.3)
t50
Hence, the expected present worth of a series of cash flows is found by
summing the present worths of the expected values of the individual cash
flows.
To determine the variance of present worth, we recall that the variance
of the sum of statistically independent random variables is the sum of
the variances of the random variables; also, we recall that the variance of
the product of a constant and a random variable equals the product of the
square of the constant and the variance of the random variable. Hence,
from Equation 11.1, when the random cash flows are statistically independent, the variance of present worth is given by
n
V1PW2 5 a V1At 2 11 1 i2 22t
(11.4)
t50
When the annual cash flows are statistically independent, based on the
Central Limit Theorem, we can expect the distribution of present worth to
approximate a normal distribution. The Excel® NORMSDIST function
can be used to approximate the probability of present worth being greater
than 0. The syntax for the NORMSDIST function provides the probability
of a normally distributed random variable with mean m and standard deviation s being less than or equal to z by entering 5NORMSDIST((z-m)/s)
in any cell. Due to the symmetry of the normal distribution, the probability
of a normally distributed random variable with mean m and standard deviation s being greater than or equal to z can be obtained by entering
5NORMSDIST((m-z)/s) in any cell. Hence, for z 5 0,
Pr(PW . 0) 5NORMSDIST(E(PW)/SD(PW))
(11.5)
where E(PW) and SD(PW) denote the expected value and standard deviation of present worth, respectively.
EXAMPLE
Risk Analysis for the SMP Investment
To illustrate the calculation of the expected value and the variance of
present worth, recall the SMP investment: $500,000 initial investment;
$92,500 annual savings; $50,000 salvage value; 10-year planning horizon;
11-3
Risk Analysis
409
and 10 percent MARR. Suppose the annual savings and the salvage value
are random variables, distributed as shown in Table 11.2.
TABLE 11.2 Means, Variances, Standard Deviations, and Probability Distributions for Annual Savings
and Salvage Value for the SMP Investment
A
p(A)
Ap(A)
A2p(A)
SV
p(SV)
SVp(SV)
SV2p(SV)
$75,000
0.070
$5,250
393,750,000
$40,000
0.10
$4,000
160,000,000
$80,000
0.095
$7,600
608,000,000
$45,000
0.20
$9,000
405,000,000
$85,000
0.131
$11,135
946,475,000
$50,000
0.40
$20,000
1,000,000,000
$90,000
0.178
$16,020
1,441,800,000
$55,000
0.20
$11,000
605,000,000
$95,000
0.183
$17,385
1,651,575,000
$60,000
0.10
$6,000
360,000,000
$100,000
0.181
$18,100
1,809,999,998
Sum
1.00
$50,000
2,530,000,000
$105,000
0.162
$17,010
1,786,050,000
Sum
1.000
$92,500
8,637,649,998
E(A) 5
$92,500
E (SV ) 5
$50,000
V (A) 5
81,400,001.7
V (SV ) 5
30,000,000
SD(A) 5
$9,022.19
S D (SV ) 5
$5,477.23
Computing the expected value for the annual cash flow gives $92,500.
Similarly, the expected value for the salvage value is $50,000.
The variance for annual savings is
V1A2 5 E1A2 2 2 3E1A2 2 4
5 192,5002 2 2 8,637,649,998
5 81,400,001.7
and the variance for salvage value is
V1SA2 5 E1SV2 2 2 3E1SV2 4 2
5 150,0002 2 2 2,530,000,000
5 30,000,000
Table 11.3 provides the results of applying the expected values and
variances for the annual savings and salvage value to obtain the statistical parameters for the SMP investment’s present worth. As noted, the
expected present worth is $87,649.62, and the variance of present
worth is 334,461,261.82. Notice, the expected cash flow in the tenth
year equals the sum of the expected annual savings and the expected
salvage value; likewise, the variance of the cash flow in the tenth year
equals the sum of the annual savings variance and the salvage value
variance.
410
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
TABLE 11.3 Computing the Expected Value and Variance for the Present Worth of the SMP Investment,
Plus the Probability of the Present Worth Being Greater than Zero
E(A) 5
V (A) 5
SD(A) 5
EOY(t)
$92,500
81,400,001.7
$9,022.19
E (CF)
(1.10)⫺t
E (CF)
(1.10)⫺t
E(SV) 5
V (SV) 5
SD(SV) 5
$50,000
30,000,000
$5,477.23
V (CF)
(1.10)⫺2t
V (CF)
(1.10)⫺2t
0
2$500,000
1.0000
2$500,000.00
0.0
1.0000
0.00
1
$92,500
0.9091
$84,090.91
81,400,001.7
0.8264
67,272,728.68
2
$92,500
0.8264
$76,446.28
81,400,001.7
0.6830
55,597,296.43
3
$92,500
0.7513
$69,496.62
81,400,001.7
0.5645
45,948,178.87
4
$92,500
0.6830
$63,178.74
81,400,001.7
0.4665
37,973,701.54
5
$92,500
0.6209
$57,435.22
81,400,001.7
0.3855
31,383,224.41
6
$92,500
0.5645
$52,213.84
81,400,001.7
0.3186
25,936,549.10
7
$92,500
0.5132
$47,467.13
81,400,001.7
0.2633
21,435,164.55
8
$92,500
0.4665
$43,151.93
81,400,001.7
0.2176
17,715,012.02
9
$92,500
0.4241
$39,229.03
81,400,001.7
0.1799
14,640,505.80
10
$142,500
0.3855
$54,939.92
111,400,001.7
0.1486
16,558,900.41
E(PW) 5
$87,649.62
V (PW) 5
SD(PW) 5
Pr(PW + 0)* 5
334,461,261.82
$18,288.28
0.999999178
*Central Limit Theorem approximation
EXAMPLE
Risk Analysis for the Scream Machine Investment
Recall the example involving two design alternatives (A and B) for a
new ride (the Scream Machine) in a Florida theme park. Design A has
an initial cost of $300,000 and net annual after-tax revenue of $55,000;
Design B has an initial investment of $450,000 and net annual after-tax
revenue of $80,000. A 10 percent MARR was used over the 10-year
planning horizon. Now, suppose annual revenue for each machine is a
statistically independent random variable. Further, suppose annual revenue for Design A is normally distributed with a mean of $55,000 and
a standard deviation of $5,000; annual revenue for Design B is normally distributed with a mean of $80,000 and a standard deviation of
$7,500. What is the probability of each design having a positive present
worth? What is the probability of Design B having a greater present
worth than Design A?
11-3
Given: MARR 5 10%; planning horizon 5 10 years
Risk Analysis
411
KEY DATA
Design A: Initial cost 5 $300,000; Annual Revenue is normally
distributed with mean 5 $55,000 & standard deviation 5 $5,000
Design B: Initial cost 5 $450,000; Annual Revenue is normally
distributed with mean 5 $80,000 & standard deviation 5 $7,500
Find: Pr[PW(A) . 0], Pr[PW(B) . 0], Pr[PW(B) . PW(A)]
Because the annual revenues are statistically independent, normally distributed random variables, present worth will be normally distributed. The
expected present worth for each design will be the same as in Chapter 4:
$37,951.19 for Design A and $41,565.37 for Design B. Calculations for
the variance and standard deviation for present worth for each design are
summarized in Table 11.4. Also shown is the variance for the difference in
cash flows, (B 2 A). Notice, to determine which of the two designs is more
profitable, we take advantage of Pr[PW(B) . PW(A)] being the same as
Pr[PW(B 2 A) . 0].
TABLE 11.4
EOY(t)
SOLUTION
Risk Analysis for the Selection of the Scream Machine Design
(1.10)⫺2t
V (At)
(1.10)⫺2t
V (At)
V (Bt)
(1.10)⫺2t
V (Bt)
V (Bt 2 At)
V (Bt 2 At)
(1.10)⫺2t
0
1.0000
0
0.0000
0
0.0000
1
0.8264
25,000,000
20661157.0248
56,250,000
46487603.3058
81,250,000
67148760.3306
2
0.6830
25,000,000
17075336.3841
56,250,000
38419506.8643
81,250,000
55494843.2484
3
0.5645
25,000,000
14111848.2513
56,250,000
31751658.5655
81,250,000
45863506.8169
4
0.4665
25,000,000
11662684.5052
56,250,000
26241040.1368
81,250,000
37903724.6420
5
0.3855
25,000,000
9638582.2357
56,250,000
21686810.0304
81,250,000
31325392.2661
6
0.3186
25,000,000
7965770.4428
56,250,000
17922983.4962
81,250,000
25888753.9390
7
0.2633
25,000,000
6583281.3577
56,250,000
14812383.0547
81,250,000
21395664.4124
8
0.2176
25,000,000
5440728.3948
56,250,000
12241638.8882
81,250,000
17682367.2829
9
0.1799
25,000,000
4496469.7477
56,250,000
10117056.9324
81,250,000
14613526.6801
0.1486
25,000,000
3716090.7006
56,250,000
8361204.0764
81,250,000
10
V [PW(A)] 5
SD[PW(A)] 5
Pr[PW(A) + 0]14 5
101351949.0447
$10,067.37
0.999918
V [PW(B)] 5
SD[PW(B)] 5
Pr[PW(B) + 0]* 5
*Central Limit Theorem approximations
V [PW(B 2 A)] 5
SD[PW(B 2 A)] 5
Pr[PW(B 2 A) + 0]* 5
228041885.3507
$15,101.06
0.997043
12077294.7770
329393834.3954
$18,149.21
0.578922
412
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
Recall, the variance of the difference in two statistically independent
random variables is the sum of the variances of the two random variables.
Therefore, the variance of the difference in annual revenues equals the
sum of (5,000)2 for Design A and (7,500)2 for Design B, or 81,250,000.
Notice, the probability of Design A having a positive present worth is
0.999918, the probability of Design B having a positive present worth is
0.997043, and the probability of Design B having a present worth greater
than the present worth for Design A is 0.578922. Hence, the probability
of Design A being the best choice economically is 0.421078, or there is a
58 percent chance that Design B is best and a 42 percent chance that
Design A is best.
The probabilities were calculated using the Excel® NORMSDIST
function as follows:
Pr3PW1A2 . 0 4 5 NORMSDIST137951.19/10067.372
5 0.999918
Pr3PW1B2 . 04 5 NORMSDIST141565.37/15101.062
5 0.997043
Pr3PW1B 2 A2 . 0 4 5 NORMSDIST13614.18/18149.212
5 0.578922
11.3.2
Simulation Solutions2
LEARN I N G O B JEC T I V E : Perform risk analysis using simulation solutions to
derive exact values or estimates for the expected value and standard deviation of the measure of economic worth.
Some of the major reasons for using simulation in risk analysis include the
following:
1.
2.
3.
4.
5.
2
Except for the simplest problems, analytical solutions are difficult to
obtain.
Simulation is useful in selling a system modification to management.
Simulation can be used as a verification of analytical solutions.
Simulation is very versatile.
Less background in mathematical analysis and probability theory is
generally required.
We assume the reader is familiar with Monte Carlo simulation. A number of other analytical
solutions and a more extensive discussion of simulation solutions are presented in White, J. A.,
K. E. Case, and D. B. Pratt, Principles of Engineering Economic Analysis, 6th edition, John
Wiley & Sons, Inc., NY, 2012.
11-3 Risk Analysis
Some of the major disadvantages of simulation are the following:
1.
2.
3.
4.
5.
Simulation can be quite time-consuming to formulate.
Simulations introduce a source of randomness not present in analytical
solutions (sampling error).
Monte Carlo simulations do not reproduce the input distribution exactly
(especially the tails of the distribution).3
Validation of the simulation model is easily overlooked.
Simulation is so easily applied that it is often used when analytical
solutions can be more easily obtained at considerably less cost.
Computer software is available to support simulation analyses. Among
the vast array of options, we have found Pallisade Corporation’s Excel®based software, @RISK, to be easy to use. In addition, Microsoft’s VBA
(Visual Basic for Applications) software language can be used with Excel®
to perform simulations of engineering economic investments.
@RISK software includes the option of using either Monte Carlo
simulation or Latin Hypercube simulation. Practically every known probability distribution is included in @RISK’s menu of input distributions.
Among the discrete probability mass functions available are the binomial,
discrete, discrete uniform (rectangular), hypergeometric, negative binomial, and Poisson distributions; the continuous probability density functions include, among others, the beta, chi square, Erlang, gamma, geometric, lognormal, normal, Pareto, PERT, triangular, uniform, and Weibull
distributions.
The following examples illustrate the use of @RISK 4.5 simulation
software in performing engineering economic analyses.
Monte Carlo Simulation with Excel and @Risk for the
SMP Investment
Recall, in Example 11.6, we performed an analytical analysis of the SMP
investment, with annual savings and salvage value assumed to be statistically independent random variables. The probability distributions used are
given in Table 11.5. The analytical solution yielded an expected present
worth of $87,649.62, a variance for present worth of 334,461,261.82, and
a 0.99999918 probability of a positive-valued present worth.
3
Latin hypercube simulation, included in Pallisade Corporation’s @RISK software, uses
stratified sampling of the input distributions to force sampling across the entire range of values
of the random variables. It incorporates “sampling without replacement,” because only one
sample is drawn randomly from a stratification or stratum.
EXAMPLE
413
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
TABLE 11.5
Probability Distributions for Example 11.8
Annual Savings
Probability
Salvage Value
Probability
$75,000
0.070
$40,000
0.1
$80,000
0.095
$45,000
0.2
$85,000
0.131
$50,000
0.4
$90,000
0.178
$55,000
0.2
$95,000
0.183
$60,000
0.1
$100,000
0.181
$105,000
0.162
We now perform a Monte Carlo simulation using @RISK; 100,000 simulated investments yielded an average present worth of $87,677.55, a variance of 336,070,498, and an estimate of 0.99999914 (based on the central
limit theorem) for the probability of a positive-valued present worth. The
number of present worths that were greater than 0 totaled 100,000; not
once was the present worth less than 0. This should not be surprising,
because the expected number of instances of present worth being less than
0 is (0.00000082)(100,000), or 0.082. The histogram obtained from the
Monte Carlo simulation is provided in Figure 11.9.4
Distribution for PW(SMP)/B13
2.500
Values in 10^ -5
414
Mean=87677.55
2.000
1.500
1.000
0.500
0.000
0
40
5%
80
120
160
Values in Thousands
90%
5%
56.9239
117.3692
PW Histogram from 100,000 Simulation Trials in Example 11.8.
(The Figure was Generated with the Help of @RISK, a Product of Palisade
Corporation, Ithaca, NY; www.palisade.com.)
FIGURE 11. 9
4
Many of the figures and tables in this section were generated with the help of @RISK 4.5,
a software product of Palisade Corporation, Ithaca, NY: www.palisade.com, or call 800-432RISK (7475).
11-3 Risk Analysis
Because we can only determine the internal rate of return numerically,
we did not consider the distribution of internal rate of return in Example
11.6. However, with @Risk software and the Excel® IRR function, we
can obtain an approximation of the distribution of IRR, as well as an estimate of the probability of IRR being less than the MARR. (Because
Pr(IRR , MARR) 5 Pr(PW , 0), we already know how many times IRR
is less than 10 percent during the 100,000 simulation trials—zero!)
The Monte Carlo simulation yielded an estimate of 13.80126 percent
for the expected internal rate of return for the SMP investment when
annual savings and salvage value were statistically independent random
variables. The sample standard deviation obtained for IRR was 0.7926308
percent. As noted, all simulated investments yielded an IRR value greater
than or equal to the MARR. The histogram obtained from the Monte Carlo
simulation is provided in Figure 11.10.
Distribution for IRR(SMP)/B14
50
45
40
35
30
25
20
15
10
5
0
0.1
Mean=0.1380126
0.1175
5%
0.135
0.1525
5%
90%
.1247
0.17
.1508
FIGURE 11.10 IRR Histogram from 100,000 Simulation Trials in Example 11.8.
(The Figure was Generated with the Help of @RISK, a Product of Palisade
Corporation, Ithaca, NY; www.palisade.com.)
Monte Carlo Simulation with Excel and @Risk for the Scream
Machine Investment
In Example 11.7, we presented analytical results for the two design alternatives being considered for the Scream Machine. Specifically, we estimated the probabilities of Design A being profitable, of Design B being
EXAMPLE
415
416
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
profitable, and of Design B having a greater present worth than Design A.
Here, we duplicate the analysis but with Monte Carlo simulation, and we
assume net annual revenue produced by the new ride is a statistically independent random variable.
Because the revenue in any given year is independent of the revenue in
any other year, as shown in Figure 11.11, the Excel® NPV function is used to
calculate present worth. The histograms for PW(A), PW(B), and PW(B 2 A)
resulting from the 100,000 simulated investments are given in Figure 11.12.
Of the 100,000 simulation trials, all but 12 resulted in a positive-valued
present worth for Design A, all but 280 resulted in a positive-valued present
worth for Design B, and 57,877 trials resulted in Design B having a greater
present worth than Design A. Hence, the probability of Design B being the
most economic is estimated to be equal to 0.57877.
Recall, solving analytically for the mean and variance for PW(A),
PW(B), and PW(B 2 A) yielded probability estimates of positive-valued
present worths equal to 0.999918, 0.997043, and 0.578922, respectively,
compared with the simulated results of 0.99988, 0.99720, and 0.57877,
respectively. Also, the Monte Carlo simulation produced average present
worths of $37,963.06 for Design A, $41,564.11 for Design B, and
$3,601.05 for the difference in Designs B and A. The exact expected values
are $37,951.19 for Design A, $41,565.37 for Design B, and $3,614.18 for
the difference in Designs B and A.
Setup for @RISK Monte Carlo Simulation of Example 11.9. (The Figure
was Generated with the Help of @RISK, a Product of Palisade Corporation, Ithaca, NY;
www.palisade.com.)
FIGURE 11.11
11-3
Risk Analysis
Values in 10^ –5
Distribution for PW(A)/B13
4.000
3.500
3.000
2.500
2.000
1.500
1.000
0.500
0.000
–10
Mean=37963.06
15
40
Values in Thousands
5%
90%
21.3354
65
90
5%
54.4994
PW
Histogram from 100,000
Simulated Investments
in Design A
F I G U RE 11. 1 2 a
Distribution for PW(B)/C13
3.000
Mean=41564.11
Values in 10^ –5
2.500
2.000
1.500
1.000
0.500
0.000
–40
0
40
80
Values in Thousands
5%
90%
16.9294
66.3692
120
5%
PW
Histogram from 100,000
Simulated Investments
in Design B
F I G U RE 11. 1 2 b
Distribution for PW(B – A)/D13
2.500
Mean=3601.05
Values in 10^ –5
2.000
1.500
0.500
0.000
–80
PW
Histogram from 100,000
Simulated Incremental
Investments Between
Designs B and A. (The
Figures were Generated
with the Help of @RISK,
a Product of Palisade
Corporation, Ithaca, NY;
www.palisade.com.)
F I G U RE 11. 1 2 c
1.000
–60
–40
–20
0
Values in Thousands
90%
5%
–26.1902
20
40
33.4125
60
5%
80
100
417
418
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
The Monte Carlo simulation results produced slightly smaller estimates for the probabilities of a positive-valued present worth for Design A
and of Design B being more profitable than Design A, as well as the
expected values for present worth of Design B and the difference in
Designs B and A. However, the differences in the results obtained analytically and with simulation are, for all practical purposes, negligible.
SUMMARY
KEY CONCEPTS
1. Learning Objective: Calculate the break-even value using a break-even
analysis approach. (Section 11.1)
The break-even value is the value at which we are indifferent between two
alternatives, that is, they are equivalent. The internal rate of return itself is
a break-even value because it is the interest rate at which the economic
worth of an investment equals zero. In an engineering economic analysis,
break-even analysis is often used to determine the level of annual savings
required to justify a particular capital investment.
2. Learning Objective: Perform a sensitivity analysis to examine the impact
on the measure of economic worth and the robustness of the economic decision when values of one or more parameters vary over specified ranges.
(Section 11.2)
Sensitivity analysis is performed when we want to gauge the impact on the
economic worth if one or more parameters take on values over some specified range; typically, we are interested in knowing what percent change in
a parameter’s value will result in a different recommendation regarding an
investment. A popular approach of parameter estimation is to consider
three estimates for the value of the parameter typically depicted as an optimistic, pessimistic, and most likely value for a parameter. Compared to
sensitivity analysis, risk analysis attempts to reflect the imprecision inherent in assigning values to parameters in an engineering economic analysis.
This imprecision is represented in the form of a probability distribution,
and the parameters themselves are treated as random variables.
3. Learning Objective: Perform risk analysis using analytical solutions to
derive exact values or estimates for the expected value and standard deviation of the measure of economic worth. (Section 11.3.1)
Risk analysis incorporates probabilistic estimates with the values of the
parameters; typically we want to know the probability of an investment
being profitable or the probability of the measure of economic worth having
Summary 419
at least a particular value. In some cases analytical approaches can be used,
and the probabilities of certain outcomes can be obtained mathematically.
However, analytical solutions can be difficult to obtain except for the simplest problems, thus simulation solutions are often preferred.
4. Learning Objective: Perform risk analysis using simulation solutions to
derive exact values or estimates for the expected value and standard deviation of the measure of economic worth. (Section 11.3.2)
Simulation is a popular approach for performing risk analysis for more
complex situations where an analytical solution would be difficult to
obtain. Numerous computer software packages are available to support
simulation analyses. Simulation approaches have certain advantages and
disadvantages. Advantages are that simulation is useful in selling a system
modification to management; they can be used to verify an analytical solution; they are versatile; and less background in mathematical analysis and
probability theory is generally required. Some of the major disadvantages
of simulation are that simulations can be quite time-consuming; they introduce a source of randomness not present in analytical solutions in the form
of sampling error; Monte Carlo simulations do not reproduce the input
distribution exactly; validation of the simulation model is often overlooked; and it is so easily applied that it is often used when analytical solutions can be easily obtained at considerably less cost.
KEY TERMS
Autocorrelated, p. 407
Break-Even Analysis, p. 393
Break-Even Value, p. 393
Risk Analysis, p. 405
Sensitivity Analysis, p. 398
Problem available in WileyPLUS
GO Tutorial Tutoring Problem available in WileyPLUS
Video Solution Video Solution available in WileyPLUS
FE-LIKE PROBLEMS
1.
If the total cost for producing widgets can be represented by TC 5 8,000 1
0.75*X, where X is the number of widgets produced and total revenue can
be represented by TR 5 4.00*x, what is the break-even value for number of
widgets produced?
a. 1,684
c. 2,462
b. 2,000
d. 3,763
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
The next five questions refer to the following sensitivity graph:
PW
420
–30
–20
–10
3500
3000
2500
2000
1500
1000
500
0
–500 0
–1000
–1500
10
20
30
Percent Change
Annual Revenue
Initial Investment
Salvage Value
a.
b.
c.
d.
The analysis is most sensitive to changes in which component?
Annual Revenue
Initial Investment
Salvage Value
Cannot be determined from the information given
a.
b.
c.
d.
The analysis is least sensitive to changes in which component?
Annual Revenue
Initial Investment
Salvage Value
Cannot be determined from the information given
2.
3.
4.
What is the numeric value of the present worth of the original project (i.e.,
no changes)?
a. 210
b. 20
c. 1,000
d. Cannot be determined from the information given
5.
What percentage change in initial investment would cause the project to
become unattractive?
a.
b.
c.
d.
6.
210
120
11,000
Cannot be determined from the information given
If a line for “annual expenses” was to be added to the graph what slope
would you expect the line to have?
a. Positive slope (line rises as it goes left to right)
b. Negative slope (line falls as it goes left to right)
c. Zero slope (horizontal line)
d. Infinite slope (vertical line)
Summary 421
7.
Which of the following is not a method typically used for supplementary
analysis of engineering economy problems?
a. break-even analysis
c. risk analysis
b. depreciation analysis d. sensitivity analysis
8.
The probability of weather related crop damage during the growing season
in a typical year is given by the following table. If the interest rate is 8%, what
is the expected present worth of crop damage over the next five years?
Value of Crop Damage
Probability
$0
$100,000
$200,000
$300,000
a. $57,000
b. $167,000
60%
25%
13%
2%
c. $228,000
d. $285,000
9.
Gooey Bites sells snack packs for $3 per pack. Variable expenses involved
in producing snack packs are estimated to be $1 per pack and fixed costs for
operating the production line are estimated to be $14,000. How many snack
packs must Gooey Bites sell to break even?
a. 14,000
c. 4,667
b. 3,500
d. 7,000
10.
Reconsider the previous problem. After making changes to the production
line, Gooey Bites made a profit of $36,000 by selling 20,000 snack packs.
Variable costs were modified by the line changes but fixed costs were unaffected. What is the new variable cost per pack?
a. 0.33
c. 1.00
b. 0.50
d. 1.50
PROBLEMS
Introduction
1. Uncertainty can impact many elements of an engineering economic analysis.
Given the list of factors below, rank them from most to least uncertain and
briefly justify why you ranked them in the order you did.
Factor List (alphabetic)
First cost
MARR
Operating and maintenance costs
Planning horizon
Salvage value
422
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
2. Match the terms in the first column with an appropriate definition from the
second column.
Terms
Definitions
(a) break-even analysis
(b) sensitivity analysis
(c) risk analysis
Section 11.1
3.
(1) Determining how the worth of a investment
changes with changes in one or more
parameters
(2) Determining probabilistic statements
about the worth of an investment based on
probabilistic values assigned to one or more
parameters
(3) Determining the indifference value of a
particular parameter
Break-Even Analysis
Cecil’s Manufacturing is considering production of a new product. The
sales price would be $10.25 per unit. The cost of the equipment is $100,000.
Operating and Maintenance costs are expected to be $3,500 annually. Based
on a 7-year planning horizon and a MARR of 12%, determine the number of
units that must be sold annually to achieve break-even.
4. Reconsider Problem 3. Determine whether each of the following statements
is true or false by determining the new break-even for each case. Each case is
independent of the other cases.
a. If the cost of the equipment doubles, the break-even volume will double.
b. If the revenue per unit doubles, the break-even volume will halve.
c. If the O&M costs double, the break-even volume will double.
5.
The Fence Company is setting up a new production line to produce top
rails. The relevant data for two alternatives are shown below.
Installed Cost
Expected Life
Salvage Value
Variable Cost per top rail
Flow Line
Manufacturing Cell
$15,000
5 years
$0
$6.00
$10,000
5 years
$0
$7.00
a. Based on MARR of 8%, determine the annual rate of production for which
the alternatives are equally economical.
b. If it is estimated that production will be 300 top rails per year, which alter-
native is preferred and what will be the total annual cost?
6. A manufacturer offers an inventor the choice of two contracts for the exclu-
sive right to manufacture and market the inventor’s patented design. Plan I
calls for an immediate single payment of $50,000. Plan II calls for an annual
payment of $2,000 plus a royalty of $1.00 for each unit sold. The remaining
life of the patent is 10 years. MARR is 10% per year.
Summary 423
a. What must be the uniform annual sales to make Plan I and Plan II equally
attractive?
b. If fewer than the number in (a) are scheduled for production and sales,
which plan is more attractive?
7.
A pipeline contractor can purchase a needed truck for $40,000. Its estimated life is six years and it has no salvage value. Maintenance is estimated
to be $2,400 per year. Operating expense is $60 per day. The contractor can
hire a similar unit for $150 per day. MARR is 7%.
a. How many days per year must the services of the truck be needed such that
the two alternatives are equally costly?
b. If the truck is needed for 180 days/year, should the contractor buy the
truck or hire the similar unit? Determine the dollar amount of annual savings
generated by using the preferred alternative rather than the non-preferred.
8.
Video Solution A firm has the capacity to produce 650,000 units of product per year. At present, it is operating at 64% of capacity. The firm’s income per unit is $1.00, annual fixed costs are $192,000, and variable costs are
$0.376 per unit of product.
a. What is the firm’s current annual profit or loss?
b. At what volume of product does the firm break even?
c. What would be the profit or loss at 80% of capacity?
9.
Spending $1,500 more today for a hybrid engine rather than a gasoline
engine will result in annual fuel savings of $300. How many years must this
savings continue in order to justify the extra investment if money is worth
10% per year, compounded annually?
10. The Cooper Company is considering investing in a recuperator. The recu-
perator will have an initial cost of $20,000 and a service life of 10 years.
Operating and maintenance costs for the first year are estimated to be $1,500,
increasing by $100 every year thereafter. It is estimated that the salvage value
of the recuperator will be 20% of its initial cost. The recuperator will result
in equal annual fuel savings throughout its service life. Assuming MARR is
12%, what is the minimum value of fuel savings for which the recuperator is
attractive?
11.
Snow Valley Ski Resort has been contracting snow removal from its parking lots at a cost of $400/day. A snow removal machine can be purchased for
$25,000. The machine is estimated to have a useful life of 6 years with a zero
salvage value at that time. Annual costs for operating and maintaining the
equipment are estimated to be $5,000. Determine the break-even value for
the number of days per year that snow removal is required in order to justify
purchasing the snow removal machine. MARR is 12%/yr.
12. A small utility company is considering the purchase of a power driven post
hole digger. The equipment would cost $8,000, and have an estimated life of
8 years and a salvage value of $1,000 at that time. Annual maintenance costs
for the digger are estimated to be 15% of the first cost regardless of its level
424
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
of usage. Operations costs are $40 per day with an output rate of 25 holes per
day. At present, holes are manually dug at a rate of 1.5 per day by a laborer
whose marginal cost is $11.20 per day. Determine the break-even value in
holes per year. MARR is 8%.
13.
GO Tutorial To make a batch of 1,000 units, it is estimated that 120 direct
labor hours are required at a cost of $10 per hour. Direct material costs are
estimated at $1,500 per batch. The overhead costs are calculated based on
an overhead rate of $7.50 per direct labor hour. The item can be readily purchased from a local vendor for $4 per unit.
a. Should the item be manufactured or purchased?
b. What is the break-even value for the overhead rate (dollars per direct labor
hour)? Assume that the material costs, labor hours, and labor costs do not
change.
14. A manufacturer of precision castings has the capacity to produce 1,000,000
castings per year. Each sells for $15. The variable cost per unit to produce the
casting is $9 per casting. Annual fixed costs for the plant are $3,500,000.
a. If the plant is currently operating at 50% of capacity, how much profit
(loss) is being earned?
b. What percent of production capacity is required for break-even?
15.
Video Solution A new manufacturing plant costs $5 million to build.
Operating and maintenance costs are estimated to be $45,000 per year and
a salvage value of 25% of the initial cost is expected. The units the plant
produces are sold for $35 each. Sales and production are designed to run
365 days per year. The planning horizon is 10 years. Find the break-even value
for the number units sold per day for each of the following values of MARR.
a. MARR is 5%
b. MARR is 10%
c. MARR is 15%
16. A subsidiary of a major furniture company manufactures wooden pallets. The
plant has the capacity to produce 300,000 pallets per year. Presently the plant
is operating at 70% of capacity. The selling price of the pallets is $18.25 per
pallet and the variable cost per pallet is $15.75. At zero output, the subsidiary
plant’s annual fixed costs are $550,000. This amount remains constant for any
production rate between zero and plant capacity.
a. With the present 70% of capacity production, what is the expected annual
profit or loss for the subsidiary plant.
b. What annual volume of sales (units) is required in order for the plant to
break even?
c. What would be the annual profit or loss if the plant were operating at 90%
of capacity?
d. If fixed costs could be reduced by 40%, what would be the new break-even
sales volume?
17.
Cowboy Metal Cutting produces a laser cut part based on customer orders.
The number of units requested on a customer’s order for the laser cut part can
Summary 425
vary from 1 unit to 150 units. Cowboy has determined that three different
cutting machines can be used to produce this part. An economic analysis of
production costs has produced the data in the table below.
Cutting
Tool ID
CT 1
CT 2
CT 3
Fixed Cost
per Order
Variable Cost
per Unit
$300
$750
$500
$9.00
$3.00
$5.00
a. For all order sizes between 1 and 150, determine the preferred (most
economical) cutting machine for an order of that size.
b. For an order of size 75, what is the minimum cost production?
18. Two 100-horsepower motors are under consideration by the Mighty Machin-
ery Company. Motor Q costs $5000 and operates at 90% efficiency. Motor R
costs $3,500 and is 88% efficient. Annual operating and maintenance costs
are estimated to be 15% of the initial purchase price. Power costs 3.2¢/
kilowatt-hour. How many hours of full-load operation are necessary each
year in order to justify the purchase of motor Q?
Use a 15-year planning horizon; assume that salvage values will equal
20% of the initial purchase price; and let the MARR be 15%. (Note: 0.746
kilowatts 5 1 horsepower.)
19. An aluminum extrusion plant manufactures a particular product at a variable
cost of $0.04 per unit, including material cost. The fixed costs associated
with manufacturing the product equal $30,000/year. Determine the breakeven value for annual sales if the selling price per unit is (a) $0.40, (b) $0.30,
and (c) $0.20.
20. Owners of a nationwide motel chain are considering locating a new mo-
tel in Snyder, Arkansas. The complete cost of building a 150-unit motel
(excluding furnishings) is $5 million; the firm estimates that the furnishings in the motel must be replaced at a cost of $1,875,000 every 5 years.
Annual operating and maintenance cost for the facility is estimated to
be $125,000. The average rate for a unit is anticipated to be $55/day. A
15-year planning horizon is used by the firm in evaluating new ventures of
this type; a terminal salvage value of 20% of the original building cost is
anticipated; furnishings are estimated to have no salvage value at the end
of each 5-year replacement interval; land cost is not to be included. Determine the break-even value for the daily occupancy percentage based on a
MARR of (a) 0%, (b) 10%, (c) 15%, and (d) 20%. (Assume that the motel
will operate 365 days/year.)
21. A consulting engineer is considering two pumps to meet a demand of
15,000 gallons/minute at 12 feet total dynamic head. The specific gravity
of the liquid being pumped is 1.50. Pump A operates at 70% efficiency
and costs $12,000; pump B operates at 75% efficiency and costs $18,000.
Power costs $0.015/kilowatt-hour. Continuous pumping for 365 days/year
426
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
is required (i.e., 24 hours/day). Using a MARR of 10% and assuming equal
salvage values for both pumps, how many years of service are required for
pump B to be justified economically? (Note: Dynamic head times gallon/
minute times specific gravity divided by 3960 equals horsepower required.
Horsepower times 0.746 equals kilowatts required.)
22. A business firm is contemplating the installation of an improved material-
handling system between the packaging department and the finished goods
warehouse. Two designs are being considered. The first consists of an automated guided vehicle system (AGVS) involving three vehicles on the loop.
The second design consists of a pallet conveyor installed between packaging and the warehouse. The AGVS will have an initial equipment cost of
$280,000 and annual operating and maintenance costs of $50,000. The pallet
conveyor has an initial cost of $360,000 and annual operating and maintenance costs of $35,000. The firm is not sure what planning horizon to use in
the analysis; however, the salvage value estimates given in the following table
have been developed for various planning horizons. Using a MARR of 10%,
determine the break-even value for N, the planning horizon.
Salvage Value Estimates
N
AGVS
Pallet Conveyor
1
2
3
4
5
6
$230,000
185,000
145,000
110,000
80,000
55,000
$300,000
245,000
200,000
160,000
125,000
95,000
23. A manufacturing plant in Michigan has been contracting snow removal at
a cost of $400/day. The past 3 years have produced heavy snowfalls, resulting in the cost of snow removal being of concern to the plant manager. The
plant engineer has found that a snow-removal machine can be purchased for
$25,000; it is estimated to have a useful life of 6 years, and a zero salvage
value at that time. Annual costs for operating and maintaining the equipment
are estimated to be $5,000. Determine the break-even value for the number of
days per year that snow removal is required in order to justify the equipment,
based on a MARR of (a) 0%, (b) 10%, and (c) 15%.
24. The motor on a gas-fired furnace in a small foundry is to be replaced. Three
different 15-horsepower electric motors are being considered. Motor X sells for
$2,500 and has an efficiency rating of 90%; motor Y sells for $1,750 and has
a rating of 85%; motor Z sells for $1,000 and is rated to be 80% efficient. The
cost of electricity is $0.065/kilowatt-hour. An 8-year planning horizon is used,
and zero salvage values are assumed for all three motors. A MARR of 12% is to
be used. Assume that the motor selected will be loaded to capacity. Determine
the range of values for annual usage of the motor (in hours) that will lead to the
preference of each motor. (Note: 0.746 kilowatts 5 1 horsepower.)
Summary 427
25. A machine can be purchased at t 5 0 for $20,000. The estimated life is
5 years, with an estimated salvage value of zero at that time. The average
annual operating and maintenance expenses are expected to be $5,500. If
MARR 5 10%, what must the average annual revenues be in order to be
indifferent between (a) purchasing the machine, or (b) doing nothing?
26. Two condensers are being considered by the Ajax Company. A copper con-
denser can be purchased for $5,000; annual operating and maintenance costs
are estimated to be $500. Alternatively, a ferrous condenser can be purchased
for $3,500; since the Ajax company has not had previous experience with
ferrous condensers, they are not sure what annual operating and maintenance
cost estimate is appropriate. A 5-year planning horizon is to be used, salvage
values are estimated to be 15% of the original purchase price, and a MARR
of 20% is to be used. Determine the break-even value for the annual operating
and maintenance cost for the ferrous condenser.
Section 11.2 Sensitivity Analysis
27.
A pork processing facility is considering the installation of either a storage facility or a holding pond. A biosystems engineer has been hired to evaluate the economic tradeoffs for the two alternatives. The engineer estimates
the cost of the storage facility to be $213,000, with annual costs for maintenance to be $3,200 per year. She estimates the cost of the pond to be $90,000,
plus $45,000 for pumps and piping, and annual operating and maintenance
costs for the holding pond are estimated to be $8,500. The engineer estimates
the life of the storage facility and the pond to be around 20 years, but is concerned about the accuracy of this estimate. She decides to do a sensitivity
analysis.
a. Develop the equation that she would use to determine how sensitive the
economic decision is to changes in life. Use a MARR of 15%.
b. Determine which alternative is preferred for lives ranging from 15 to
25 years.
28. You have been asked to perform a sensitivity analysis on a company’s
plan to modernize its facilities to determine the impact of possible errors
in estimating the net annual savings. The initial investment in the modernization is $30,000. The expected net annual savings are $13,000. The
salvage value is $7,000 after a planning horizon of seven years. MARR is
12% per year.
a. Determine if the modernization is economically attractive based on the
initial estimates and an Annual Worth (AW) analysis.
b. Determine the AW if the net annual savings change by the following percentages from the initial estimate: 280%, 260%, 240%, 220%, 120%, 140%.
c. Determine the percentage change in net annual savings that causes a reversal in the decision regarding the attractiveness of the project.
29. Plot a sensitivity graph for annual worth versus initial cost, annual revenue,
and salvage value for the data in the table on the next page. Vary only one
428
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
parameter at a time, each within the range of 220% to 120%. MARR is
3%/yr. Project life is 4 years.
Initial Cost
Annual Revenue
Salvage Value
$120,000
$25,000
$35,000
Based on your graph, which parameter shows the most sensitivity? the least?
30.
Video Solution A new project will cost $80,000 initially and will last
for seven years, at which time its salvage value will be $2,500. Annual revenues are anticipated to be $15,000 per year. For a MARR of 12%/yr, plot
a sensitivity graph for annual worth versus initial cost, annual revenue, and
salvage value, varying only one parameter at a time, each within the range
of 1/2 50%.
31. In problem 20 suppose the following pessimistic, most likely, and optimistic
estimates are given for building cost, furnishings cost, annual operating and
maintenance costs, and the average rate per occupied unit.
Building cost
Furnishings cost
Annual operating and
maintenance costs
Pessimistic
Most Likely
Optimistic
$7,500,000
$5,000,000
$4,000,000
3,000,000
1,875,000
1,000,000
200,000
125,000
75,000
35/day
55/day
75/day
Average rate
Determine the pessimistic and optimistic limits on the break-even value for
the daily occupancy percentage based on a MARR of 12%. Assume the motel
will operate 365 days/year.
32. Plot a sensitivity graph for annual worth versus initial cost, annual revenue,
and salvage value for the data in the table below. Vary only one parameter at
a time, each within the range of 220% to 120%. MARR is 20%/yr. Project
life is 10 years.
Initial Cost
Annual Revenue
Salvage Value
$800,000
$330,000
$130,000
Based on your graph, which parameter shows the most sensitivity? the least?
33. A warehouse modernization plan requires an investment of $3 million in
equipment. At the end of the 10-year planning horizon, it is anticipated the
equipment will have a salvage value of $600,000. Annual savings in operating and maintenance costs due to the modernization are anticipated to total
$1,500,000/year. A MARR of 10% is used by the firm. Perform a sensitivity
analysis to determine the effects on the economic feasibility of the plan due
to errors in estimating the initial investment required, and the annual savings.
Summary 429
34.
The owners of a discount motel chain are considering building a new
motel. Optimistic, pessimistic, and most likely estimates of several key parameters have been obtained from the local builders and the chamber of commerce. These estimates are shown in the table below. The life of the motel is
estimated to be 15 years and MARR is 20%.
Parameter
Pessimistic
Most Likely
Optimistic
Initial Cost
$10,500,000
$8,875,000
$6,000,000
Annual Operating
Annual Revenue
$350,000
$175,000
$150,000
$1,500,000
$2,500,000
$3,500,000
a. Based only on the pessimistic estimates, is the new motel economically
attractive?
b. Based only on the most likely estimates, is the new motel economically
attractive?
c. Based only on the optimistic estimates, is the new motel economically
attractive?
d. Based on a beta approximation of expected value estimates, is the new
motel economically attractive?
35.
GO Tutorial Initial estimates of the parameters for an investment are given
below.
Parameter
Initial Investment
Net Annual Receipts
Project Life
Salvage Value
MARR
Initial Estimate
Sensitivity
$15,000
none
$2,500
230%, 0%, 130%
10 years
220%, 0%, 120%
$500
250%, 0%, 150%
15% per year
none
You wish to do a multiparameter sensitivity analysis based on the sensitivities
shown. AW is the preferred measure of worth.
a. How many values of AW need to be calculated?
b. Determine the AW values.
36. Reconsider the data in Problem 35. Based on a present worth measure of
worth, complete a multi-parameter sensitivity analysis that examines all possible combinations of the estimates.
Section 11.3 Risk Analysis
37. An investment of $15,000 is to be made into a savings account. The interest
rate to be paid each year is uncertain; however, it is estimated that it is twice
as likely to be 6% as it is to be 4%, and it is equally likely to be either 6%
or 8%. Determine the probability distribution for the amount in the fund after
3 years, assuming the interest rate is not autocorrelated.
430
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
38. Assume an initial investment of $12,000, annual receipts of $4,500, and an
uncertain life for the investment. Use a 15% MARR. Let the probability distribution for the life of the investment be given as follows:
N
p(N )
1
2
3
4
5
6
7
0.10
0.15
0.20
0.25
0.15
0.10
0.05
Analytically determine the probability of the investment being profitable.
39. In problem 38 suppose the MARR is not known with certainty and the
following probability distribution is anticipated to hold:
i
p(N )
0.10
0.12
0.15
0.20
0.60
0.20
Use analytical methods to determine the probability of the investment being
profitable.
40. In problem 38 suppose the magnitude of the annual receipts (R) is subject to
random variation. Assume that each annual receipt will be identical in value
and the annual receipt has the following probability distribution:
R
p(R)
$5,000
$4,500
$4,000
0.20
0.60
0.20
Use analytical methods to determine the probability of the investment’s being
profitable.
41. In problem 38 suppose the minimum attractive rate of return is distributed as
given in problem 39 and suppose the annual receipts are distributed as given
in problem 40. Analytically determine the probability that the investment will
be profitable.
42. In problem 41 suppose the initial investment is equally likely to be either
$9,000 or $11,000. Analytically determine the probability that the investment
will be profitable.
43. Suppose n 5 4, i 5 0%, a $11,000 investment is made, and the receipt in year
j, j 5 1, . . . , 4, is statistically independent and distributed as in problem 40.
Determine the probability distribution for present worth.
Summary 431
44. Consider an investment alternative having a 6-year planning horizon and
expected values and variances for statistically independent cash flows as
given below:
j
E(Cj )
V(Cj )
0
625 3 104
1
2$22,500
4,000
2
5,000
25 3 104
3
6,000
36 3 104
4
7,000
49 3 104
5
8,000
64 3 104
6
9,000
81 3 104
16 3 104
Using a discount rate of 10%, determine the expected values and variances
for both present worth and annual worth. Based on the central limited theorem, compute the probability of a positive present worth; compute the probability of a positive annual worth.
45. Solve problem 44 using a discount rate of (a) 0%, (b) 15%.
46. Two investment alternatives are being considered. Alternative A requires an
initial investment of $15,000 in equipment; annual operating and maintenance costs are anticipated to be normally distributed, with a mean of $5,000
and a standard deviation of $500; the terminal salvage value at the end of the
8-year planning horizon is anticipated to be normally distributed, with a mean
of $2,000 and a standard deviation of $800. Alternative B requires end-ofyear annual expenditures over the planning horizon. The annual expenditure
will be normally distributed, with a mean of $8,000 and a standard deviation
of $750. Using a MARR of 15%, what is the probability that Alternative A is
the most economic alternative?
47. In problem 46, suppose the MARR were 10% with probability 0.25, 12%
with probability 0.50, and 15% with probability 0.25; what is the probability
that Alternative A is the most economic alternative?
48. Company W is considering investing $12,500 in a machine. The machine
will last N years, at which time it will be sold for L. Maintenance costs for
this machine are estimated to increase by 10%/year over its life. The maintenance cost for the first year is estimated to be $1,500. The company has 10%
MARR. Based on the probability distributions given below for N and L, what
is the expected equivalent uniform annual cost for the machine?
N
L
p(N)
6
8
10
$5,000
3,000
1,000
0.2
0.4
0.4
432
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
49.
An initial investment of $22,500 results in independent annual receipts of
$6,250 until the end of the project life. The probability distribution for the life
of the project is shown in the table below. MARR is 15%/yr.
Life (years)
4
5
6
7
8
Probability
0.10
0.25
0.45
0.15
0.05
For the following questions, determine an analytical solution:
a. Determine the probability that the present worth of the project is greater
than zero.
b. Determine the probability that the present worth of the project is greater
than $1,000.
50. Main Electric is deciding whether to invest $10,600,000 in a new plant.
An analyst forecasts that the plant will generate the independent random
cash flows shown in the table below at the end of each year. MARR is
15%/yr.
End of Year
0
1
2
3
4
5
Expected Net
Cash Flow
Standard Deviation
of Net Cash Flow
2$19,700,000
$1,200,000
$3,600,000
$6,000,000
$9,600,000
$11,000,000
$120,000
$240,000
$650,000
$750,000
$1,080,000
$0
For the following questions, determine an analytical solution:
a. Determine the mean and standard deviation of the present worth.
b. If the present worth is normally distributed, what is the probability that
present worth is greater than zero?
Assume each end-of-year cash flow is normally distributed with the mean and
standard deviation shown in the table above.
c. Using a Monte Carlo simulation with 10,000 iterations, estimate the mean
and standard deviation of present worth and the probability of positive
present worth.
51. A new CNC mill is expected to cost $263,000 and have a useful life of
6 years. The net annual savings generated by the mill are independent from
year to year and are estimated to follow a uniform distribution with a lower
bound of $45,000 and an upper bound of $55,000. MARR is 12%/yr.
Summary 433
For the following questions, determine an analytical solution:
a. Determine the probability that the present worth of the CNC mill is
positive.
b. Using a Monte Carlo simulation with 10,000 iterations, estimate the mean
and standard deviation of present worth and the probability of positive
present worth.
52. One of two mutually exclusive alternatives must be selected for implementa-
tion. Alternative A is an equipment purchase; Alternative B is a lease arrangement with annual payments. The characteristics of the two investments are
shown in the table below. Use an 8 year planning horizon and a MARR of
15%/yr.
Alt.
A
A
A
B
Parameter
Initial Cost
Annual Maintenance
Salvage Value
End-of-Year Lease Payment
Mean
Std. Dev.
Distribution
$13,000
$5,000
$2,000
$7,500
None
$500
$800
$750
None
Normal
Normal
Normal
For the following questions, determine an analytical solution:
a. Determine the probability that Alternative A is the preferred alternative.
b. Using a Monte Carlo simulation with 10,000 iterations, estimate the mean
and standard deviation of present worth and the probability of positive
present worth.
53. An investment of $5,000 is expected to generate the probabilistic returns
shown in the table below over its three year life. Assume the annual cash
flows are independent and that the distribution of present worth is normal.
EOY
1
2
3
Annual Return
$2,500 with probability 0.4 or $1,800 with probability 0.6
$3,000 with probability 0.5 or $2,000 with probability 0.5
$3,500 with probability 0.7 or $2,500 with probability 0.3
For the following questions, determine an analytical solution:
a. Determine the probability that the present worth is negative.
b. Using a Monte Carlo simulation with 10,000 iterations, estimate the probability that the present worth is negative.
54. A project under consideration has a 10 year projected life. The initial invest-
ment for the project is estimated to have a mean of $10,000 and a standard
deviation of $1,000. The annual receipts are independent with each year’s
expected return having a mean of $1,800 and a standard deviation of $200.
MARR is 12%.
434
Chapter 11
Break-Even, Sensitivity, and Risk Analysis
For the following questions, determine an analytical solution:
a. Determine the probability that the present worth is negative.
Assume the initial investment and annual receipts are independent and normally distributed.
b. Using a Monte Carlo simulation with 10,000 iterations, estimate the probability that the present worth is negative.
A proposed project has the following cash flow estimates.
55.
End of Year
0
1
2
3
4
5
Mean Net
Cash Flow
Standard Deviation
of Cash Flow
2$32,000
$4,000
$8,000
$12,000
$12,000
$12,000
$1,000
$2,000
$3,000
$5,000
$6,000
$7,000
Assuming independent cash flows, a normally distributed net present value,
and a minimum attractive rate of return of 18%, determine the following.
For the following questions, determine an analytical solution:
a. the mean and standard deviation of net present value
b. the probability that the net present value is positive
c. the probability that the net present value is greater than $5,000
Assume the initial investment and annual receipts are normally distributed.
d. Using a Monte Carlo simulation with 10,000 iterations, estimate the probability that the present worth is positive and estimate the probability that
the present worth is greater than $5,000.
56. A proposed project has the following cash flow estimates.
End of Year
0
1
2
Mean Net
Cash Flow
Standard Deviation
of Cash Flow
2$800,000
$1,000,000
$1,000,000
$250,000
$450,000
$600,000
Assuming independent cash flows, a normally distributed net present value,
and a minimum attractive rate of return of 15%, determine the following.
For the following questions, determine an analytical solution:
a. the mean and standard deviation of net present value
b. the probability that the net present value is negative
c. the probability that the net present value is greater than $1,000,000
Assume the initial investment and annual receipts are normally distributed.
d. Using a Monte Carlo simulation with 10,000 iterations, estimate the probability that the present worth is negative.
Summary 435
57. A $5,000 process improvement project is expected to increase annual ex-
penses for the next 3 years by an average of $20,000 with a standard deviation of $3,000. The annual savings generated over the 3 years will average
$24,000 with a standard deviation of $4,000. MARR is 20%. Assume independent cash flows.
For the following questions, determine an analytical solution:
a. Assuming that present worth is normally distributed, determine the probability that the process improvement will result in a loss.
b. Assuming that present worth is normally distributed, determine the probability that the process improvement will result in a present worth of
$10,000 or greater.
c. Using a Monte Carlo simulation with 10,000 iterations, estimate the probability that the process improvement will result in a loss and the probability that the present worth is $10,000 or greater.
ENGIN E E R I N G E C O N O MI C S I N P R AC T I C E :
AB B OTT L AB O R ATO R I E S
Founded by Dr. Wallace C. Abbott more than a century ago, incorporated in
1900, and headquartered in Abbott Park, Illinois, Abbott Laboratories is a
global, broad-based health care products company with four reportable revenue segments: pharmaceutical, nutritional, diagnostic, and vascular products. Its chairman and CEO, Miles D. White, a mechanical engineering graduate from Stanford, noted, “There’s one fundamental fact that everyone at
Abbott Laboratories understands: the purpose of our company is to improve
lives.”
2011 marked an extraordinary year for the company, in which it announced plans to separate Abbott into two leading healthcare companies
by the end of 2012. The new company is a research-based pharmaceutical
company named AbbVie Inc. that will focus on branded drugs. According to
a report by CNBC on November 29, 2012, “this split is meant to free Abbott
from the risk and obligations of developing pharmaceutical drugs, leaving it
with a more predictable business built around nutritional formula, generic
drugs and heart stents.” In preparation for this split, according to its 2011
annual report, the company has recently expanded its presence in emerging
markets and is working to rebuild its pharmaceutical pipeline. These strategic actions also include significant growth globally, and in 2011 the company reported that its sales outside the United States exceeded domestic
sales.
As of December 31, 2011, Abbott employed approximately 91,922
people around the world. Abbott owns or leases 19 plants in the United
States and Puerto Rico, and has manufacturing facilities in 15 other countries including Argentina, Brazil, Canada, England, France, Germany,
Ireland, Italy, Japan, Mexico, Pakistan, Singapore, Spain, Sweden, and the
Netherlands.
Abbott’s sales in 2011 totaled $38.9 billion, up from $35.2 billion in 2010.
Capital expenditures increased from $1.02 billion in 2010 to $1.5 billion in
436
CAPITAL BUDGETING
2011 in support of upgrading and expanding manufacturing, R&D, and
investments in information technology and administrative support facilities
in all four business segments.
As is true of health care innovators, Abbott invests heavily in R&D:
$4.1 billion in 2011, up from $3.7 billion in 2010, the majority of which was
concentrated in pharmaceutical products.
Health care technology is an R&D-intensive industry, it is highly competitive, and it is very dynamic. Changes occur rapidly. Consequently, firms
in the industry must be highly adaptable. Their very survival might depend
on the speed with which an acquisition or divestiture is made, or how quickly
a new product is brought to market.
Leveraging R&D investments, every effort is made at Abbott to launch
new products faster than the competition. Therefore, Abbott must manage
its R&D portfolio carefully. Many opportunities exist for investment, but wise
choices must be made.
DISCUSSION QUESTIONS:
1. The announced spin-off of the pharmaceutical portion of Abbott’s
business is meant to reduce risk for investors. What other benefits
might be realized?
2. Abbott invests heavily in capital. How might this be affected with the
spin-off and what changes do you predict?
3. In addition to a sharp increase in capital expenditures from 2010 to
2011, we notice a similar sharp increase in R&D. Are these two occurrences random or linked?
4. Despite the fact that a capital investment limit is established for a fiscal
year, capital investment considerations can expand well beyond one
year. What is driving this lengthier period of time?
437
438
Chapter 12
Capital Budgeting
LEARNING OBJECTIVES
When you have finished studying this chapter, you should be able to:
1. Explain the importance of capital budgeting. (Section 12.1)
2. Solve the capital budgeting problem with independent, indivisible
investments as a binary linear programming problem. (Section 12.2)
3. Solve the capital budgeting problem with independent, divisible
investments. (Section 12.3)
4. State some of the practical considerations in capital budgeting.
(Section 12.4)
INTRODUCTION
Thus far in the text we have addressed how to determine if an investment is
fiscally attractive and which one of multiple mutually exclusive investment
alternatives is the most attractive economically. In this chapter we address
a different task: determining the investment portfolio when capital is limited
and many economically viable investments are available that are independent, rather than mutually exclusive. This is the task of capital budgeting.
Our study of capital budgeting in this chapter will focus on comparing
alternatives and selecting the preferred investments (where the portfolio
is developed). Due to the complexity of these analyses, we will use the
Excel® SOLVER tool to help organize the necessary data and automate the
calculations for all of the Example problems.
Systematic Economic Analysis Technique
1.
2.
3.
4.
5.
6.
7.
Identify the investment alternatives
Define the planning horizon
Specify the discount rate
Estimate the cash flows
Compare the alternatives
Perform supplementary analyses
Select the preferred investment
12-1 THE CLASSICAL CAPITAL
BUDGETING PROBLEM
LEARN I N G O B JEC T I V E : Explain the importance of capital budgeting.
Our focus in this chapter is on quantitative approaches to optimize financial
returns when capital is limited. Choosing from among a set of investments
12-1 The Classical Capital Budgeting Problem
those that will be pursued, subject to a limitation on capital available for
investment, is generally referred to as the capital rationing problem or the
capital budgeting problem. We chose the latter because it is more commonly used in the engineering economics literature.
What do you do when you have far more economically attractive
investments than can be funded with the available investment capital? Nice
problem to have, isn’t it? It is certainly better than the reverse—having
more investment capital available than fiscally attractive investments. But,
still, the question must be answered: How do you choose from among a set of
really, really good investments? Capital budgeting provides a methodology
for solving this problem.
The “abundance of riches” scenario occurs far more frequently than
you might imagine. Pharmaceutical, chemical, and semiconductor companies, among others, typically must make choices among investments. They
must forgo making some investments that will generate returns significantly greater than their cost of capital. Indeed, companies that employ
large numbers of engineers are frequently faced with deciding how to
ration scarce investment capital. Otherwise, the engineers are not as effective as they should be.
The amount of money a company budgets for capital expenditures
(often called CAP EX or Cap-X) generally varies from year to year. It
ranges from being significantly higher than annual depreciation to a level
substantially below annual depreciation, depending on market conditions;
and it depends on a combination of recent history and near-term future
expansion plans, the condition of the overall economy, and other similar
factors. However, it is not a positive sign regarding a company’s fiscal
health if its capital expenditures are substantially less than its annual
depreciation over a prolonged period of time.
Why not borrow the money necessary to make investments when the
after-tax present worth will be positive after including the cost of debt
service? For publicly traded companies, the stock market usually reacts
negatively when a firm’s ratio of debt-to-equity capital increases significantly. Likewise, issuing additional stock to obtain equity capital is viewed
negatively by shareholders, because it dilutes the fraction of the firm’s
assets represented by a share of stock. In addition, rating agencies will
downgrade a company when its debt-to-equity ratio increases dramatically, causing the company’s cost of capital to increase and making the
investment community nervous. So, the reality is, a firm will not always be
able to invest in projects that have positive after-tax present worths; choices
will have to be made.
Typically, companies create a hierarchy of approval levels for capital expenditures. Such practices generate what are called size gates. For
example, in a corporation with multibillion-dollar sales, one might
allow capital expenditures requiring less than a million dollars to be
439
Capital Budgeting
Problem The need to
choose from among a set
of investments those that
will be pursued, subject
to a limitation on capital
available for investment.
Also referred to as a capital
rationing problem.
CAP EX The amount of
money a company budgets
for capital expenditures.
Also referred to as Cap-X.
440
Chapter 12
Capital Budgeting
approved by the head of a division, those requiring more than a million
dollars but less than $10 million to be approved by the head of a business
unit within the corporation, those requiring more than $10 million but
less than $30 million to be approved by the corporation’s chief financial
officer, and those requiring more than $30 million to be approved by the
board of directors.
Although the numbers vary from company to company, size gates are
frequently used in large firms. As examples, Eastman Chemical Company
and Motorola Solutions use them.
Where size gates occur, choices must be made, and the selection process does not occur only at the organization’s highest levels. Division
heads, business heads, chief financial officers, and boards of directors frequently must choose from among attractive investment alternatives. How
do they do it? That is the subject of this chapter.
In this chapter, we show
how to formulate a capital budgeting problem with independent, indivisible investments as a binary linear programming problem and how
to solve (using Excel®’s SOLVER tool) reasonably sized problems
by maximizing the investment portfolio’s present worth;
2. how to add mutually exclusive, contingent, “either/or” and other constraints to a formulation of a capital budgeting problem involving
indivisible investments; and how to determine (using SOLVER) the
investment portfolio that maximizes its present worth;
3. how to formulate a capital budgeting problem involving independent,
divisible investments as a linear programming problem and how to
solve (using SOLVER) reasonably sized problems by maximizing
the investment portfolio’s present worth or by “filling the investment
portfolio bucket” with investments ranked in order of their internal
rates of return.
1.
12-2 CAPITAL BUDGETING PROBLEM
WITH INDIVISIBLE INVESTMENTS
LEARN I N G O B JEC T I V E : Solve the capital budgeting problem with independent, indivisible investments as a binary linear programming problem.
Indivisible Investment
An investment that must
be pursued entirely or not
at all.
Investments are called indivisible when they must be pursued entirely or
not at all. Hence, a binary decision variable (xj 5 0 or 1) is used in a mathematical formulation of the capital budgeting problem to denote if investment j is to be included in the investment portfolio. Our objective is to
maximize the present worth of the investment portfolio. Letting cj denote
12-2 Capital Budgeting Problem with Indivisible Investments
the capital investment required for investment j and letting C denote the
total amount of investment capital available, the capital budgeting problem
can be formulated as follows:
Maximize
subject to
PW1x1 1 PW2x2 1 . . . 1 PWn21xn21 1 PWn xn
c1x1 1 c2x2 1 . . . 1 cn21xn21 1 cn xn ⱕ C
xj 5 10, 12
j 5 1, . . . , n
(12.1)
(12.2)
(12.3)
Because xj equals 1 when the investment is included in the portfolio and
equals 0 otherwise, Equation 12.1 is the present worth for the investment
portfolio. The first constraint (Equation 12.2) assures that the total investment required for the portfolio is no greater than the amount of capital
available. The second constraint (Equation 12.3) affirms that the decision
variables are binary.
The mathematical optimization problem is a binary linear programming (BLP) problem. An early heuristic solution procedure, the LorieSavage procedure (named for its developers, J. H. Lorie and L. J. Savage),
was used to obtain, hopefully, good, if not optimal, solutions. For small
sized problems, enumeration can be used. However, we do not recommend
forming all possible 2n combinations of the n investments and, for those
not exceeding the capital limit, choosing the one having the greatest
present worth. For problems with few investments, that might be feasible,
but for a relatively small example with n equal to 10, there are 1,024
possible solutions.
For our purposes, we will solve the BLP formulation of the capital
budgeting problem using the Excel® SOLVER tool. It is well suited
for small-sized problems of this type. However, we must caution that
SOLVER is not guaranteed to yield an optimum solution. The search
algorithm embedded in SOLVER can terminate prematurely and not
produce an optimum solution.
Solving a Capital Budgeting Problem with the Excel®
SOLVER Tool
To illustrate using SOLVER in solving a BLP formulation of the capital
budgeting problem with indivisible investments, suppose you are presented with six different “one shot” investment opportunities with the
parameters given in Figure 12.1. Your MARR is 10% and you have a
5-year planning horizon. Two investments (5 and 6) terminate in fewer
EXAMPLE
Video Example
441
442
Chapter 12
Capital Budgeting
than 5 years. You have a total of $120,000 to invest. The investment
decision for each opportunity is a binary decision. Hence, you can
only invest in an opportunity once; hence, you cannot pursue investment 3 six times for a present worth of $12,149.69. The objective is to
choose investment opportunities from among the six available in order
to maximize present worth and not exceed the $120,000 limitation
on capital.
SOLUTION
FIGURE 12.1
In Figure 12.1, for row k, k 5 2, . . . , 7, the entry in cell Hk is obtained
using the Excel® worksheet function, SUMPRODUCT(Bk:Gk,B9:G9);
for row k, k 5 9, 10, and 11, the entry in cell Hk is obtained using the
Excel® worksheet function, SUM(Bk:Gk); PW in row 8 is obtained using
the Excel® NPV worksheet function; IRR in cell H13 is obtained using
the Excel® IRR worksheet function; and for column j, j 5 B, . . . , G, the
entries in cells j10 and j11 equal the product of the entries in cells j 2 and
j9 and cells j 8 and j 9, respectively.
To use the Excel® SOLVER tool, the parameter values are as shown
in Figure 12.2. Two constraints are needed: one that requires the values of
the decision variables to be binary valued and one that requires that the
total amount invested be less than or equal to the amount of capital available. The solution obtained is shown in Figure 12.3. Specifically, $118,000
is to be invested in opportunities 1, 2, 3, 4, and 6; investment opportunity
5 is not pursued. The resulting PW is $15,930.12 and the resulting IRR for
the investment portfolio is 15.21%.
SOLVER set up for Example 12.1
12-2 Capital Budgeting Problem with Indivisible Investments
FIGURE 12.2
SOLVER parameters for Example 12.1
FIGURE 12.3
SOLVER solution to Example 12.1
443
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Chapter 12
Capital Budgeting
A nice feature of the Excel® SOLVER tool is its ability to handle additional constraints. Consider, for example, a type of constraint that is considered throughout the text—the mutually exclusive or “either, neither, but not
both” constraint. This takes the form of requiring that the sum of the mutually exclusive decision variables be less than or equal to 1. In particular,
suppose investment opportunities g and h are mutually exclusive. In such a
case, the following constraint can be incorporated in the spreadsheet for
solution using SOLVER: xg 1 xh # 1. The constraint ensures that both xg
and xh cannot equal 1; hence, investment in both g and h cannot occur.
Another type of constraint arises when not all investment opportunities
are independent. In particular, suppose investment opportunity k cannot be
pursued unless investment opportunity j is pursued; in other words, suppose
k is contingent on j. In such a case, the following constraint can be added to
the SOLVER parameters: xk # xj. The constraint ensures that xk cannot
equal 1 if xj equals 0. In fact, the only way xk can equal 1 is for xj to equal 1.
A third type of limitation that can be incorporated in the BLP formulation is the “either/or” contingent constraint. To illustrate the concept, suppose opportunity r cannot be pursued unless either opportunity s or opportunity t is pursued. (If both opportunities s and t are pursued, then
opportunity r can also be pursued.) In such a case, the following constraint
can be incorporated in the spreadsheet for solution using the Excel®
SOLVER tool: xr # xs 1 xt. The constraint ensures that xr cannot equal 1
unless at least one of the two opportunities (s and t) is pursued.
Finally, a fourth type of limitation that the Excel® SOLVER tool can
accommodate is the “at least, but not more than” constraint. For instance,
suppose at least u but not more than v investments can be funded. In this
case, the sum of the decision variables must be greater than or equal to u
and less than or equal to v. Thus, two constraints must be added to the set
of SOLVER parameters, both keying on a cell that contains the sum of the
decision variables.
Incorporating Additional Constraints in the Capital
Budgeting Problem with the Excel® SOLVER Tool
EXAMPLE
To illustrate the addition of constraints in the capital budgeting problem,
suppose in the previous example that no more than 4 investments can be
pursued. Also, suppose investment opportunities 2 and 4 are mutually
exclusive and investment opportunity 2 is contingent on either investment
opportunity 1 or investment opportunity 3 being pursued. What is the
impact on PW and IRR of these additional constraints?
SOLUTION
From Figure 12.4, we find the solution obtained using the Excel® SOLVER
tool. Now, investment opportunities 3 and 4 are not pursued, PW is reduced
12-2 Capital Budgeting Problem with Indivisible Investments
to $12,693.07, and IRR is 15.05%. The amount of investment capital
required is $103,000, with the $17,000 balance earning the MARR of 10%.
The Excel® SOLVER tool parameters for the constrained problem
are shown in Figure 12.4. Three additional constraints are added to those
used to solve Example 12.1. Specifically, a constraint is added that the
value of cell D13 has to be less than or equal to 1; this satisfies the mutually exclusive constraint involving investments 2 and 4, because the value
of cell D13 is the sum of the decision variables for investments 2 (C9) and
4 (E9). Another constraint requires that the value of cell D14 is greater
than or equal to the value of cell C9; this satisfies the contingent requirement that investment 2 (C9) cannot be pursued if neither investment 1 (B9)
nor investment 3 (D9) is pursued, because cell D14 contains the sum of the
decision variables for investments 1 and 3. The third new constraint is that
the sum of the decision variables (H9) is less than or equal to 4; this satisfies the constraint that no more than 4 investments be pursued.
FIGURE 12.4
SOLVER solution for Example 12.2, with SOLVER parameters shown
445
446
Chapter 12
Capital Budgeting
Because small-sized BLP problems can be solved using the Excel®
SOLVER tool, it is a relatively simple matter to perform sensitivity analysis
such as described in Chapter 11. For example, one can easily determine how
sensitive the optimum investment portfolio is to changes in, say, the limit on
investment or changes to the MARR. These factors can simply be adjusted in
small increments 1/2 the original value such as 1/2 10%, 1/2 20%, etc.,
so that the decision maker can gain insight into the robustness of the solution.
12-3 CAPITAL BUDGETING PROBLEM
WITH DIVISIBLE INVESTMENTS
LEARN I N G O B JEC T I V E : Solve the capital budgeting problem with independent, divisible investments.
Divisible Investment An
investment that may be
pursed in a fractional
portion.
In the previous section, the investment opportunities were considered to be
indivisible. You either invested in all of an opportunity or you did not
invest in it at all. Investing in a fractional portion was not permitted. In this
section, we examine capital budgeting when, in fact, investment opportunities are divisible.
Certain oil and gas exploration investments are often divisible, because
it is not unusual for several individuals to go together to invest in drilling for
oil or gas. Individual investors do not have to be equal. They share proportionately in the investment and in the returns. Similar divisible investment
opportunities can occur when a block of stock is offered for sale; the same
situation can exist for bonds, land, hotels, shopping centers, and so forth.
With divisible investments, the decision variable is changed from
investing fully (xj 5 1) versus not investing at all (xj 5 0) in investment j
to investing wholly, partially, or not at all (0 # pj # 1) in investment j,
where pj is the decision variable representing the percentage of investment
j to be pursued. The capital budgeting problem with divisible investments
can be formulated mathematically as follows:
Maximize
subject to
PW1 p1 1 PW2 p2 1 . . . 1 PWn21 pn21 1 PWn pn
c1 p1 1 c2 p2 1 . . . 1 cn21 pn21 1 cn pn ⱕ C
0 ⱕ pj ⱕ 1
j 5 1, . . . , n
(12.4)
(12.5)
(12.6)
Optimizing the Investment Portfolio with the Excel® SOLVER Tool
When Investments Are Divisible
EXAMPLE
Video Example
To illustrate using the Excel® SOLVER tool to determine the optimum
investment portfolio when investments are divisible, recall Example
12.1. The setup for a SOLVER solution is provided in Figure 12.5. The
12-3 Capital Budgeting Problem with Divisible Investments
binary decision variables are replaced by fractional valued decision
variables having values ranging from 0% to 100%. The SOLVER
parameters also are shown in the figure. Notice, the objective function
(Equation 12.4) and constraints (Equations 12.5 and 12.6) are incorporated in the SOLVER parameters. Also, notice that the constraint
0 # pj # 1 is implemented by using two constraints: B9:G9 ,5 1 and
B9:G9 .5 0.
The investment portfolio obtained is to fully invest in opportunities 1,
2, 3, 4, and 6 and to partially invest (6.67%) in opportunity 5. The PW for
the portfolio is $16,073.69; the IRR for the portfolio is 15.17%. Not surprisingly, all of the investment capital is used (Why is this not surprising?
Can you think of a situation in which not all of the investment capital will
be used? What if not all investment opportunities have positive-valued
present worths?)
FIGURE 12.5
SOLVER parameters and solution for Example 12.3
447
448
Chapter 12
Capital Budgeting
When partial funding of investments is allowed, salvage value equals
the initial investment, and annual returns are a uniform annual series, it is
quite easy to obtain the optimum investment portfolio. Taking advantage of
the special structure imposed on the investments, the optimum investment
portfolio is obtained by (a) ranking the investment opportunities on their
internal rates of return, and (b) forming the portfolio by “filling the investment bucket,” starting with the opportunity having the greatest internal
rate of return and proceeding sequentially until the “bucket” is full. The
following example illustrates the solution procedure.
Optimizing the Investment Portfolio with the Excel® SOLVER Tool
When Investments Are Divisible and Have a Special Structure
EXAMPLE
To illustrate using the “bucket filling” solution procedure when divisible
investments have the special structure of a) salvage value equaling the
initial investment and b) annual returns being a uniform series, consider
the data for five investments shown in Table 12.1.
SOLUTION
If $150,000 is available for investment, the order in which investments will be
placed in the “investment bucket” is 1, 3, 4, 5, and 2. Because the sum of the
investments is $200,000, not all five investments will “fit” in the “bucket.”
After selecting investment 1, $135,000 remains available for investment;
after adding investment 3 to the investment portfolio, $95,000 remains available; after selecting investment 4, $45,000 remains available. The next investment to be selected is investment 5. However, only $45,000 of the $70,000
investment will be pursued, filling the “bucket.” Hence, the optimum values
of the decision variables are: p1 5 p3 5 p4 5 100%, p5 5 64.29% (45/70),
and p2 5 0%. The IRR for the portfolio is $3,750 1 $9,250 1 $11,250 1
$14,250 (45/70) 5 $33,410.71 divided by $150,000, or 22.27%.
EXPLORING THE
SOLUTION
As an exercise, solve the example using the solution procedure from
Example 12.3 and verify that using an IRR ranking procedure actually
results in maximizing present worth, not IRR. To do so, you will need to
select a planning horizon, such as 5 years or 10 years, and calculate the
present worth for each investment.
TABLE 12.1
Characteristics of Five Investment Opportunities
Investment
Opportunity
Initial Investment
1
2
3
4
5
$70,000
$15,000
$25,000
$40,000
$50,000
Annual Return
$3,750
$5,000
$9,250
$11,250
$14,250
Salvage Value
$15,000
$25,000
$40,000
$50,000
$70,000
Internal Rate
of Return
25.00%
20.00%
23.13%
22.50%
20.36%
12-4
Practical Considerations in Capital Budgeting
With divisible investments, consideration of mutually exclusive, contingency, and other constraints requires thought about what each means.
For example, if investment opportunity 1 is contingent on investment
opportunity 3, does that mean p1 cannot be greater than p3 or that p1 cannot
be greater than zero unless p3 is greater than zero? These (and other) considerations are incorporated in end-of-chapter problems.
12-4 PRACTICAL CONSIDERATIONS
IN CAPITAL BUDGETING
LEARNING O BJECTI VE: State some of the practical considerations in capital
budgeting.
There are several practical considerations we should mention relative to
capital budgeting. For example, even though a capital investment limit is
established for a fiscal year, it is seldom the case that all prospective investments will be known or available for analysis at a particular point in time
during the year.
Instead of determining at the beginning of a fiscal year which of a
known set of investment opportunities will be funded, it is more common
to make capital investment decisions throughout the year. To ensure that
the firm can fund a highly attractive investment that materializes toward
the latter part of the year, some portion of the capital investment funds
might be held in reserve for this purpose. Likewise, because of the uncertainty regarding new investments that might materialize, some investment
decisions are postponed, but with the proviso that investments will be
made if no others materialize that are more attractive financially.
Despite these practicalities, the material presented on capital budgeting is valuable. At various times during a fiscal year, capital investment
decisions are made. Furthermore, when they are made, they are done so in
the face of multiple competing alternative uses for the investment capital.
During a fiscal year, you must often answer multiple times the question, How do you choose from among a set of really, really good investments? Hopefully, you have a better idea of the considerations that should
go into answering this question as well as some tools you can use to determine the optimum investment portfolio.
In closing, we should note that many criticisms have been made
regarding the various mathematical programming formulations of the capital budgeting problem that have been presented in research journals. Some
want to incorporate in the formulation of the capital budgeting problem the
ability to borrow money at stated interest rates, others want to do multiyear
capital budgeting (as though they are prescient regarding future investment opportunities that will arise), still others want to incorporate probabilistic considerations, some argue for optimizing a firm’s annual cash
449
450
Chapter 12
Capital Budgeting
flow, and a significant number of scholars advocate incorporating multiple
criteria. While each of these preferences might be appropriate for specific
applications, such formulations are beyond the scope of an introductory
text. For those who wish to pursue capital budgeting beyond the brief
introduction offered in this chapter, we direct you to the vast capital budgeting and capital rationing literature.
SUMMARY
KEY CONCEPTS
1. Learning Objective: Explain the importance of capital budgeting. (Section 12.1)
Capital budgeting (or capital rationing) is an important concept whereby an
investment portfolio is formed from a set of economically attractive investment opportunities. The capital budgeting problem assumes an abundant
number of independent investment opportunities and limited capital. The
objective is then to maximize the worth of the investment given various constraints. Mathematical programming is an appropriate tool to solve the capital budgeting problem. The use of a heuristic approach to solve the problem
provides the opportunity to obtain optimal or near-optimal solutions. We
saw how the Excel SOLVER tool can be used to solve such problems.
2. Learning Objective: Solve the capital budgeting problem with independent, indivisible investments as a binary linear programming problem.
(Section 12.2)
An indivisible investment must be pursued entirely or not at all. Mathematically, this investment type is considered binary or 0–1, and can be formulated
as a binary linear programming (BLP) problem. The use of an optimization
algorithm to solve this problem can be complicated and is beyond the scope
of this textbook; however, powerful heuristic routines exist to provide optimal or near-optimal solutions. Excel SOLVER utilizes a heuristic approach
and can be used to solve this type of problem, although it should be noted that
SOLVER is not guaranteed to yield an optimum solution. Constraints can be
added to the capital budgeting problem when not all investment opportunities
are independent. Excel is also useful to perform sensitivity analysis, including analyzing the sensitivity of the amount of investment capital available and
the MARR used on the optimum investment portfolio.
3. Learning Objective: Solve the capital budgeting problem with independent, divisible investments. (Section 12.3)
A divisible investment may be pursued in a fractional portion. A common
case of this is for certain oil and gas exploration investments involving
the pooling together of resources by multiple investors. Compared to the
Summary 451
indivisible investment, the divisible investment allows for a partial investment, thus eliminating the binary variable constraint. Additional constraints
can be added when not all investment opportunities are independent and
sensitivity analysis can be performed.
4. Learning Objective: State some of the practical considerations in capital
budgeting. (Section 12.4)
There are several practical considerations relative to capital budgeting. A
common one involves the timing of decision, specifically, that all capital
budgeting investment decisions are not always made at the beginning of
the year. New investment opportunities become available throughout the
year, thus the investment decisions are made on an ongoing basis. Because
of this, a company may wish to hold a portion of its capital in reserve for
future investment opportunities.
KEY TERMS
CAP EX, p. 439
Capital Budgeting Problem, p. 439
Divisible Investment, p. 446
Indivisible Investment, p. 440
Problem available in WileyPLUS
GO Tutorial Tutoring Problem available in WileyPLUS
Video Solution Video Solution available in WileyPLUS
FE-LIKE PROBLEMS
1.
Sarah is considering two investment proposals. Proposal A is to purchase
a new computer. Proposal B is to purchase a new printer. She will not buy the
printer unless she buys the computer. The relationship between Proposals A
and B is best described by which of the following?
a. B and A are mutually exclusive
b. B is contingent on A
c. A is contingent on B
d. Not enough information is given to determine a relationship
2.
Consider a capital budgeting formulation where the binary variables x1
and x2 are used to represent the acceptance (xi 5 1) or rejection (xi 5 0) of
each alternative. A mutual exclusivity constraint between the two alternatives
can be represented by which of the following?
a. x1 1 x2 ,5 1
c. x1 1 x2 .5 1
b. x2 ,5 x1
d. x1 ,5 x2
452
Chapter 12
Capital Budgeting
3.
Consider a capital budgeting formulation where the binary variables x1
and x2 are used to represent the acceptance (xi 5 1) or rejection (xi 5 0) of
each alternative. The requirement that x2 is contingent upon x1 can be represented by which of the following?
a. x1 1 x2 ,5 1
c. x1 1 x2 .5 1
b. x2 ,5 x1
d. x1 ,5 x2
4.
Consider a capital budgeting formulation where the binary variables x1
and x2 are used to represent the acceptance (xi 5 1) or rejection (xi 5 0) of
each alternative. The requirement that the null alternative is not feasible can
be represented by which of the following?
a. x1 1 x2 ,5 1
c. x1 1 x2 .5 1
b. x2 ,5 x1
d. x1 ,5 x2
5.
Sebastian is about to compare a set of mutually exclusive and indivisible
alternatives using a ranking approach. Which of the following is not an appropriate measure of worth?
a. Present worth
c. Annual worth
b. Future worth
d. Internal rate of return
6.
If six investment proposals are under consideration, how many investment
combinations must be evaluated if a complete enumeration approach is being
used?
a. 6
c. 62 5 36
b. 2*6 5 12
d. 26 5 64
7.
Which of the following is not an approach which can be used to perform
a capital budgeting economic analysis?
a. Box-Jenkins algorithm
b. Excel® SOLVER
c. Exhaustive enumeration
d. Lori-Savage formulation
8.
To determine an optimal portfolio of investments when the available
choices are divisible, the investment choices should first be ranked in increasing order based on which of the following?
a. FW
c. IRR
b. Initial investment
d. PW
9.
Consider the following binary linear programming formulation of a capital budgeting problem.
Max
1,200 x1 1 600 x2 1 950 x3 1 1,650 x4
s.t.
15,000 x1 1 20,000 x2 1 25,000 x3 1 30,000 x4 ,5 70,000
x1 1 x2 ,5 1
x4 ,5 x3
x1, x2, x3, x4 5 (0, 1)
The first cost of project x3 is
a. $70,000
c. $950
b. $25,000
d. $x4
Summary 453
10.
Consider the following binary linear programming formulation of a capital budgeting problem.
Max
1,200 x1 1 600 x2 1 950 x3 1 1,650 x4
s.t.
15,000 x1 1 20,000 x2 1 25,000 x3 1 30,000 x4 ,5 70,000
x1 1 x2 ,5 1
x4 ,5 x3
x1, x2, x3, x4 5 (0, 1)
Projects x3 and x4 are
a. Mutually exclusive
b. x3 is contingent on x4
c. x4 is contingent on x3
d. Not related
11.
Consider the following binary linear programming formulation of a capital budgeting problem.
Max
1,200 x1 1 600 x2 1 950 x3 1 1,650 x4
s.t.
15,000 x1 1 20,000 x2 1 25,000 x3 1 30,000 x4 ,5 70,000
x1 1 x2 ,5 1
x4 ,5 x3
x1, x2, x3, x4 5 (0, 1)
The capital budget limit is
a. $90,000
c. $30,000
b. $70,000
d. $4,400
PROBLEMS
Section 12.2
Capital Budgeting Problem with Indivisible Investments
1. True or False: In solving a classical capital budgeting problem using binary
linear programming (BLP), the objective function can be either the sum of
present worths or the sum of annual worths without affecting the optimum
investment portfolio.
2. Aerotron Radio Inc. has $250,000 available and its engineering staff has pro-
posed the following indivisible investments. With each, Aerotron can exit at
the end of its planning horizon of 5 years and have its initial investment returned. In addition, each year Aerotron will receive the annual return shown
below. MARR is 12%.
Investment
1
2
3
4
5
Initial Investment
Annual Return
$75,000
$65,000
$50,000
$80,000
$100,000
$10,800
$12,000
$7,500
$13,750
$15,750
454
Chapter 12
Capital Budgeting
For the original problem:
a. Which investments should Aerotron select for the optimum portfolio?
b. What is the present worth for the optimum investment portfolio?
c. What is the IRR for the optimum investment portfolio?
In addition to the original problem statement, let investments 1 and 4 be mutually exclusive and investment 3 be contingent on investment 2:
d. Now, which alternatives should Aerotron select?
e. What is the present worth for the optimum investment portfolio?
f. What is the IRR for the optimum investment portfolio?
Consider the original problem:
g. Determine the optimum portfolio (state the investments selected and the
portfolio PW) using (1) the current limit on investment capital, (2) plus
20%, and (3) minus 20%.
h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) plus 20%, and (3) minus 20%.
3.
Polaris Industries has $1,250,000 available for additional innovations on
the Victory Vision motorcycle. These include the five indivisible, equal-lived
alternatives, each of which guarantees the investment can be exited after
6 years with the initial investment returned. In addition, each year Polaris will
receive an annual return as noted below. MARR is 15%.
Investment
1
2
3
4
5
Initial Investment
Annual Return
$350,000
$300,000
$250,000
$500,000
$400,000
$90,000
$85,000
$75,000
$130,000
$115,000
For the original problem:
a. Which alternatives should Polaris select for the optimum portfolio?
b. What is the present worth for the optimum investment portfolio?
c. What is the IRR for the portfolio?
In addition to the original problem statement, Polaris has noted that investments 1, 2, and 4 are mutually exclusive, and marketing believes at least
3 investments must be made.
d. Which alternatives should now be selected?
e. What is the present worth for the optimum investment portfolio?
f. What is the IRR for the optimum investment portfolio?
Return to the original problem statement:
g. Determine the optimum portfolio (state the investments selected and the
portfolio PW) using (1) the current limit on investment capital, (2) plus
20%, and (3) minus 20%.
h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) plus 20%, and (3) minus 20%.
Summary 455
4. CustomMetalworks is considering the expansion of their cable fabrication
business for towers, rigging, winches, and many other uses. They have available $250,000 for investment and have identified the following indivisible
alternatives, each of which will provide an exit with full return of the investment at the end of a 5 year planning horizon. Each year, CustomMetalworks
will receive an annual return as noted below. MARR is 12%.
Investment
1
2
3
4
5
Initial Investment
Annual Return
$25,000
$40,000
$85,000
$100,000
$65,000
$7,500
$12,000
$20,000
$22,000
$17,000
For the original problem:
a. Which alternatives should be selected by CustomMetalworks?
b. What is the present worth for the optimum investment portfolio?
c. What is the IRR for the optimum investment portfolio?
In addition to the original opportunity statement, CustomMetalworks has
determined that investments 3 and 4 are mutually exclusive and investment
5 is contingent on either investment 1 or 2 being funded.
d. Now, which alternatives should be selected?
e. What is the present worth for the optimum investment portfolio?
f. What is the IRR for the optimum investment portfolio?
Reconsider the original problem:
g. Determine the optimum portfolio (state the investments selected and the
portfolio PW) using (1) the current limit on investment capital, (2) plus
20%, and (3) minus 20%.
h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) plus 20%, and (3) minus 20%.
5.
Gymnastics4Life is a high-end facility for beginning, intermediate, and
elite gymnasts. The latter are drawn from the nearby region for exclusive
and dedicated training. In order to maintain their edge, G4L trustees wish to
invest up to $350,000 in new methods for critical evaluation and training and
are considering the following independent, indivisible, investments, each of
which guarantees return of the initial investment at the end of a planning horizon of 7 years. In addition, G4L will receive annual returns as noted below.
MARR is 12%.
Investment
1
2
3
4
5
Initial Investment
Annual Return
$150,000
$130,000
$100,000
$160,000
$200,000
$24,000
$22,000
$15,000
$25,000
$30,000
456
Chapter 12
Capital Budgeting
For the original problem:
a. Which alternatives should G4L select to form the optimum portfolio?
b. What is the present worth for the optimum portfolio?
c. What is the IRR for the optimum investment portfolio?
During review, the G4L trustees judge investments 2 and 5 to be considered
as mutually exclusive.
d. Which alternatives should now be selected?
e. What is the present worth for G4L’s new optimum investment portfolio?
f. What is the IRR for the portfolio?
Consider the original problem:
g. Determine the optimum portfolio (state the investments selected and the
portfolio PW) using (1) the current limit on investment capital, (2) plus
20%, and (3) minus 20%.
h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) plus 20%, and (3) minus 20%.
6. Yaesu America wishes to enhance their already fine line of electronic equip-
ment for commercial and individual use. Their engineering staff has proposed
5 independent, indivisible, equal-lived investments, cutting across different
product lines, with each estimated to return the initial investment if it is exited
after a planning horizon of 5 years. In addition, each year, Yaesu is projected
to receive an annual return as noted below. They have available $1,250,000 to
invest and their MARR is 10%.
Investment
1
2
3
4
5
Initial Investment
Annual Return
$400,000
$300,000
$200,000
$600,000
$500,000
$50,000
$36,000
$25,000
$69,000
$55,000
For the original problem:
a. Which alternatives should Yaesu America select as optimal?
b. What is the present worth for the selected portfolio?
c. What is the IRR for the optimum set of investments?
In addition to the original problem statement, Yaesu America has noted that
investment 4 is contingent on investment 2.
d. Now, which alternatives should be selected?
e. What is the present worth for the portfolio?
f. What is the IRR for the portfolio?
Consider the original opportunity statement:
g. Determine the optimum portfolio (state the investments selected and the
portfolio PW) using (1) the current limit on investment capital, (2) plus
20%, and (3) minus 20%.
h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) plus 20%, and (3) minus 20%.
Summary 457
7.
Suppose your own consulting firm has been doing well and you believe
it is time to make a move to add a new, related area of engineering services. To do so, you have identified the following 5 independent, indivisible, equal-lived investments, each of which guarantees you can exit it after
4 years and have your initial investment returned to you. Each year, you
receive an annual return as noted below. Your MARR is 10% and you have
$250,000 to invest.
Investment
1
2
3
4
5
Initial Investment
Annual Return
$45,000
$60,000
$85,000
$100,000
$75,000
$4,000
$7,000
$9,000
$12,000
$11,000
For the original problem:
a. Which alternatives should you select to form the optimum portfolio?
b. What is the present worth of your selected portfolio?
c. What is the IRR for the optimum portfolio?
In addition to the original problem statement, you now believe that investments 4 and 5 should be considered mutually exclusive.
d. Which alternatives should you now select?
e. What is the present worth for this portfolio?
f. What is the IRR now?
Reconsider the original problem:
g. Determine the optimum portfolio (state the investments selected and the
portfolio PW) using (1) the current limit on investment capital, (2) plus
20%, and (3) minus 20%.
h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) plus 20%, and (3) minus 20%.
8. A laboratory within Bayer is considering the five indivisible investment
proposals below to further upgrade their diagnostic capabilities to ensure
continued leadership and state-of-the-art performance. The laboratory uses
a 10-year planning horizon, has a MARR of 10%, and a capital limit of
$1,000,000.
Investment
1
2
3
4
5
Initial
Investment
Annual
Receipts
Annual
Disbursements
Salvage Value
$300,000
$400,000
$450,000
$500,000
$600,000
$205,000
$230,000
$245,000
$260,000
$290,000
$125,000
$130,000
$140,000
$135,000
$150,000
$50,000
$50,000
$60,000
$75,000
$75,000
458
Chapter 12
Capital Budgeting
For the original opportunity statement:
a. Which alternatives should be selected to form the optimum portfolio for
the lab?
b. What is the present worth for the optimum investment portfolio?
c. What is the IRR for the portfolio?
In addition to the original opportunity statement, Bayer declares that investments 2 and 4 are mutually exclusive, investment 5 is contingent on 2 being
funded, and at least two investments must be made.
d. Now, which alternatives should be selected by Bayer?
e. What is the present worth for the resulting investment portfolio?
f. What is the resulting IRR?
Again consider the original opportunity statement:
g. Determine the optimum portfolio (state the investments selected and the
portfolio PW) using (1) the current limit on investment capital, (2) plus
20%, and (3) minus 20%.
h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) plus 20%, and (3) minus 20%.
9. A division of Conoco-Phillips is involved in their periodic capital budgeting
activity and the engineering and operations staffs have identified ten indivisible investments with cash flow parameters shown below. Conoco-Phillips uses
a 10-year planning horizon and a MARR of 10% in evaluating such investments. The division’s capital limit for this budgeting cycle is $2,500,000.
Investment
1
2
3
4
5
6
7
8
9
10
Initial Investment
Annual Return
Salvage Value
$150,000
$200,000
$225,000
$275,000
$350,000
$400,000
$475,000
$500,000
$550,000
$600,000
$35,000
$38,000
$45,000
$60,000
$75,000
$95,000
$110,000
$85,000
$120,000
$125,000
$25,000
$50,000
$22,500
$27,500
$55,000
$75,000
$50,000
$100,000
$75,000
$75,000
For the original problem statement:
a. Which alternatives should Conoco-Phillips select?
b. What is the present worth of the optimum portfolio?
c. What is the IRR for the portfolio?
In addition to the original problem statement, Conoco-Phillips has noted that
investments 1 and 3 are mutually exclusive, investment 4 is contingent on either investment 2 or investment 5 being funded, and at least five investments
must be made.
d. Which alternatives should now be selected?
e. What is the present worth for the new portfolio?
f. What is the IRR for the investment portfolio?
Summary 459
Consider the original problem:
g. Determine the optimum portfolio (state the investments selected and the
portfolio PW) using (1) the current limit on investment capital, (2) plus
20%, and (3) minus 20%.
h. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) plus 20%, and (3) minus 20%.
10. A lending firm is considering 6 independent and indivisible investment alter-
natives which, at any time the firm chooses, can be exited with a full refund
of the initial investment. A total of $200,000 is available for investment, and
the MARR is 10% (Note! There is no planning horizon specified, so the firm
can choose any number of years they wish—the optimum portfolio and the
IRR will remain the same since the initial investment and the salvage value
are the same, and the annual returns are constant each year.):
Alternative
1
2
3
4
5
6
Initial Investment
Annual Return
$25,000
$35,000
$30,000
$40,000
$60,000
$50,000
$2,600
$3,750
$3,050
$4,775
$6,750
$5,850
For the original problem:
a. Which alternatives should the lending firm select as optimal?
b. What is the present worth for the optimum portfolio?
c. What is the IRR for the portfolio?
Several possible constraints have been identified for additional analysis by
the lending firm. Determine (1) the optimum investment portfolio, (2) the
present worth, and (3) the IRR when:
d. Investments 4 and 5 are mutually exclusive.
e. Investment 1 is contingent on investment 2 being pursued.
f. Exactly four investments must be pursued.
g. All of the constraints d, e, and f are considered simultaneously.
Reconsider the original problem:
h. Determine the optimum portfolio (state the investments selected and the
portfolio PW) using (1) the current limit on investment capital, (2) plus
20%, and (3) minus 20%.
i. Determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) plus 20%, and (3) minus 20%.
11.
Rex Electric has decided to move into low-rise (2–8 floors) commercial
building electrical wiring. After great success in upscale residential and small
commercial wiring, they have identified four independent and indivisible
investments, any or all of which will help make the move to the next level.
Rex Electric’s MARR is 10%, and $500,000 is available for investment
immediately, with $175,000 available for follow-up investment the next year.
The cash flows are shown below, in thousands of dollars.
460
Chapter 12
Capital Budgeting
EOY
0
1
2
3
4
5
6
7
8
9
10
CF(1)
CF(2)
CF(3)
CF(4)
2$50
2$100
$50
$50
$50
$50
$50
$50
$50
$50
$75
2$125
2$75
$70
$70
$70
$70
$70
$70
$70
$70
$100
2$200
$50
$50
$50
$50
$75
$75
$75
$75
$75
$75
2$250
$75
$75
$75
$75
$75
$85
$85
$85
$85
$100
a. Which alternatives should be selected by Rex Electric?
b. What is the present worth for the selected investment portfolio?
c. What is the IRR for the optimum portfolio?
Using SOLVER for sensitivity analysis,
d. determine the optimum portfolio (state the investments selected and the
portfolio PW) using (1) the current limit on investment capital, (2) plus
20%, and (3) minus 20%.
e. determine the optimum portfolio (state the investments selected and the portfolio PW) using (1) the current MARR, (2) plus 20%, and (3) minus 20%.
12. Consider the following five indivisible investment alternatives which, at the
end of 5 years, fully refund the initial investment. Given the annual returns
shown below, $150,000 of investment capital available, and a MARR of 10%,
determine the optimum investment portfolio. What are the PW and IRR for
the investment portfolio?
Alternative
Initial Investment
Annual Return
$75,000
$60,000
$80,000
$55,000
$90,000
$8,750
$7,250
$9,750
$6,500
$10,750
A
B
C
D
E
13. Consider the six indivisible investment alternatives shown below. The planning
horizon is 5 years. The MARR is 12%. $50,000 is available for investment.
Alternative
Initial Investment
Annual Return
Salvage Value
$8,000
$15,000
$10,000
$20,000
$19,000
$12,000
$3,200
$4,750
$3,070
$5,950
$5,150
$4,250
$1,000
$1,750
$1,100
$2,000
$2,100
$1,200
A
B
C
D
E
F
a. Which investments should be made in order to maximize present worth?
b. Solve part a) when investments B and D are mutually exclusive and F is
contingent on E.
Summary 461
14. Consider the six indivisible investment alternatives shown below. The planning
horizon is 8 years. The MARR is 15%. $60,000 is available for investment.
Initial Investment
Annual Return
Salvage Value
M
N
O
P
Q
$8,000
$15,000
$10,000
$20,000
$19,000
$3,200
$4,750
$3,070
$5,950
$5,150
$1,000
$1,750
$1,100
$2,000
$2,100
R
$12,000
$4,250
$1,200
Alternative
a. Which investments should be made in order to maximize present worth?
b. Solve part a when investments N and P are mutually exclusive and R is
contingent on Q.
15. The City of Clyde, Ohio is using the binary linear programming (BLP) for-
mulation of a capital budgeting problem shown below. Assume $175,000 is
available for investment and all investments are indivisible; MARR 5 12%;
n 5 10 yrs. Given the incomplete SOLVER parameter box shown, respond to
parts a–f below if it is desired to obtain an optimum investment portfolio.
462
Chapter 12
Capital Budgeting
Specify the contents of the target cell.
Specify the contents for By Changing Cells.
Specify all constraints to be added.
Now, suppose the original problem is modified so that investments 1 and
2 are mutually exclusive. Show how you would incorporate that constraint
in SOLVER.
e. Now, suppose the original problem is modified so investment 2 is contingent on investment 3 being pursued. Show how you would incorporate that
constraint in SOLVER.
f. Now, suppose the original problem is modified to specify that at least three
and no more than four investments can be pursued. Show how you would
incorporate that constraint in SOLVER.
a.
b.
c.
d.
16. The American Radio Relay League is using a binary linear programming
(BLP) formulation to select from among several possible investments. Each
investment continues for 5 years. Given the incomplete SOLVER parameter
box shown, respond to parts a–f below in order to obtain the optimum investment portfolio. Assume $180,000 is available for investment and all investments are indivisible.
Summary 463
Specify the contents of the target cell.
Specify the contents for By Changing Cells.
Specify all constraints to be added.
Now, suppose the original problem is modified so that investments 4 and
5 are mutually exclusive. Show how you would incorporate that constraint
in SOLVER.
e. Now, suppose the original problem is modified so that investment 6 is contingent on investment 5 being pursued. Show how you would incorporate
that constraint in SOLVER.
f. Now, suppose the original problem is modified so that exactly 3 investments must be pursued. Show how you would incorporate that constraint
in SOLVER.
a.
b.
c.
d.
Section 12.3
17.
Capital Budgeting Problem with Divisible Investments
True or False: In solving a capital budgeting problem involving investment opportunities that are divisible (i.e., you can invest in portions of the
opportunities instead of “all or nothing”), you rank the opportunities on the
basis of present worth and add to the portfolio opportunities and fractions of
opportunities, beginning with the largest present worth, until “the investment
bucket is filled.”
18. True or False: In solving a capital budgeting problem involving investment
opportunities that are divisible (i.e., you can invest in portions of the opportunities instead of “all or nothing”), you rank the opportunities on the basis
of internal rate of return and add to the portfolio opportunities and fractions
of opportunities, beginning with the largest internal rate of return, until “the
investment bucket is filled.”
19. Aerotron Radio Inc. has $250,000 available and its engineering staff has pro-
posed the following divisible investments. With each, Aerotron can exit at the
end of its planning horizon of 5 years and have its initial investment returned.
In addition, each year Aerotron will receive the annual return shown below.
MARR is 12%.
Investment
1
2
3
4
5
Initial Investment
Annual Return
$75,000
$65,000
$50,000
$80,000
$100,000
$10,800
$12,000
$7,500
$13,750
$15,750
a. Determine the optimum portfolio, including which investments are fully
or partially (if partial, give percentage) selected. You may use Excel®;
do not use SOLVER.
b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel®
and SOLVER.
464
Chapter 12
Capital Budgeting
c. Determine the optimum portfolio and its PW, specifying which invest-
ments are fully or partially (give percentage) selected using (1) the current
MARR, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER.
d. Determine the optimum investment portfolio and its PW when investments 1, 2, and 3 are divisible and investments 4 and 5 are indivisible. Use
Excel® and SOLVER.
20. Polaris Industries has $1,250,000 available for additional innovations on the
Victory Vision motorcycle. These include the five divisible, equal-lived alternatives, each of which guarantees the investment can be exited after 6 years
with the initial investment returned. In addition, each year Polaris will receive
an annual return as noted below. MARR is 15%.
Investment
1
2
3
4
5
Initial Investment
Annual Return
$350,000
$300,000
$250,000
$500,000
$400,000
$90,000
$85,000
$75,000
$130,000
$115,000
a. Determine the optimum portfolio, including which investments are fully
or partially (if partial, give percentage) selected. You may use Excel®;
do not use SOLVER.
b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel®
and SOLVER.
c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
MARR, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER.
d. Determine the optimum investment portfolio and its PW when investments 1, 2, and 3 are indivisible and investments 4 and 5 are divisible. Use
Excel® and SOLVER.
21.
CustomMetalworks is considering the expansion of their cable fabrication business for towers, rigging, winches, and many other uses. They have
available $250,000 for investment and have identified the following divisible
alternatives, each of which will provide an exit with full return of the investment at the end of a 5 year planning horizon. Each year, CustomMetalworks
will receive an annual return as noted below. MARR is 12%.
Investment
1
2
3
4
5
Initial Investment
Annual Return
$25,000
$40,000
$85,000
$100,000
$65,000
$7,500
$12,000
$20,000
$22,000
$17,000
Summary 465
a. Determine the optimum portfolio, including which investments are fully
or partially (if partial, give percentage) selected. You may use Excel®;
do not use SOLVER.
b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel®
and SOLVER.
c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
MARR, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER.
d. Determine the optimum investment portfolio and its PW when investments 1, 2, and 5 are divisible and investments 3 and 4 are indivisible. Use
Excel® and SOLVER.
22. Gymnastics4Life is a high-end facility for beginning, intermediate, and elite
gymnasts. The latter are drawn from the nearby region for exclusive and dedicated training. In order to maintain their edge, G4L trustees wish to invest up to
$350,000 in new methods for critical evaluation and training and are considering the following independent, divisible, investments, each of which guarantees
return of the initial investment at the end of a planning horizon of 7 years. In
addition, G4L will receive annual returns as noted below. MARR is 12%.
Investment
1
2
3
4
5
Initial Investment
Annual Return
$150,000
$130,000
$100,000
$160,000
$200,000
$24,000
$22,000
$15,000
$25,000
$30,000
a. Determine the optimum portfolio, including which investments are fully
or partially (if partial, give percentage) selected. You may use Excel®;
do not use SOLVER.
b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel®
and SOLVER.
c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
MARR, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER.
d. Determine the optimum investment portfolio and its PW when investments 1 and 2 are divisible and investments 3, 4, and 5 are indivisible. Use
Excel® and SOLVER.
23.
Yaesu America wishes to enhance their already fine line of electronic
equipment for commercial and individual use. Their engineering staff has
proposed 5 independent, divisible, equal-lived investments, cutting across
different product lines, with each estimated to return the initial investment if
it is exited after a planning horizon of 5 years. In addition, each year, Yaesu
466
Chapter 12
Capital Budgeting
is projected to receive an annual return as noted below. They have available
$1,250,000 to invest and their MARR is 10%.
Investment
1
2
3
4
5
Initial Investment
Annual Return
$400,000
$300,000
$200,000
$600,000
$500,000
$50,000
$36,000
$25,000
$69,000
$55,000
a. Determine the optimum portfolio, including which investments are fully
or partially (if partial, give percentage) selected. You may use Excel®;
do not use SOLVER.
b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel®
and SOLVER.
c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
MARR, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER.
d. Determine the optimum investment portfolio and its PW when investments 2 and 4 are divisible and investments 1, 3, and 5 are indivisible. Use
Excel® and SOLVER.
24. Suppose your own consulting firm has been doing well and you believe it is
time to make a move to add a new, related area of engineering services. To
do so, you have identified the following 5 independent, divisible, equal-lived
investments, each of which guarantees you can exit it after 4 years and have
your initial investment returned to you. Each year, you receive an annual
return as noted below. Your MARR is 10% and you have $250,000 to invest.
Investment
1
2
3
4
5
Initial Investment
Annual Return
$45,000
$60,000
$85,000
$100,000
$75,000
$4,000
$7,000
$9,000
$12,000
$11,000
a. Determine the optimum portfolio, including which investments are fully
or partially (if partial, give percentage) selected. You may use Excel®;
do not use SOLVER.
b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel®
and SOLVER.
c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
MARR, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER.
Summary 467
d. Determine the optimum investment portfolio and its PW when invest-
ments 2 and 4 are indivisible and investments 1, 3, and 5 are divisible. Use
Excel® and SOLVER.
25. A laboratory within Bayer is considering the five divisible investment propos-
als below to further upgrade their diagnostic capabilities to ensure continued
leadership and state-of-the-art performance. The laboratory uses a 10-year
planning horizon, has a MARR of 10%, and a capital limit of $1,000,000.
Investment
1
2
3
4
5
Initial
Investment
Annual
Receipts
Annual
Disbursements
Salvage
Value
$300,000
$400,000
$450,000
$500,000
$600,000
$205,000
$230,000
$245,000
$260,000
$290,000
$125,000
$130,000
$140,000
$135,000
$150,000
$50,000
$50,000
$60,000
$75,000
$75,000
a. Determine the optimum portfolio, including which investments are fully
or partially (if partial, give percentage) selected. You may use Excel®;
do not use SOLVER.
b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel®
and SOLVER.
c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
MARR, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER.
d. Determine the optimum investment portfolio and its PW when all investments except investment 3 are divisible and at least 2 investments must
be pursued fully or partially and no more than 3 can be pursued fully or
partially. Use Excel® and SOLVER.
26. A division of Conoco-Phillips is involved in their periodic capital budgeting
activity and the engineering and operations staffs have identified ten divisible
investments with cash flow parameters shown below. Conoco-Phillips uses
a 10-year planning horizon and a MARR of 10% in evaluating such investments. The division’s capital limit for this budgeting cycle is $2,500,000.
Investment
1
2
3
4
5
6
7
8
9
10
Initial Investment
Annual Return
Salvage Value
$150,000
$200,000
$225,000
$275,000
$350,000
$400,000
$475,000
$500,000
$550,000
$600,000
$35,000
$38,000
$45,000
$60,000
$75,000
$95,000
$110,000
$85,000
$120,000
$125,000
$25,000
$50,000
$22,500
$27,500
$55,000
$75,000
$50,000
$100,000
$75,000
$75,000
468
Chapter 12
Capital Budgeting
a. Determine the optimum portfolio, including which investments are fully
b.
c.
d.
e.
f.
27.
or partially (if partial, give percentage) selected. You may use Excel®;
do not use SOLVER.
Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel®
and SOLVER.
Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
MARR, (2) plus 20%, and (3) minus 20%. Use Excel® and SOLVER.
Determine the optimum investment portfolio and its PW when investments 1 through 5 are divisible and investments 6 through 10 are indivisible. Use Excel® and SOLVER.
Determine the optimum investment portfolio and its PW when investments 1 through 5 are indivisible and investments 6 through 10 are divisible. Use Excel® and SOLVER.
Determine the optimum investment portfolio and its PW when: investments 1, 3, and 5 are mutually exclusive; making any investment in 4 is
contingent on either investment 2 or investment 5 being fully or partially
funded; at least five investments must be made, albeit partially; investments 1 through 5 are indivisible and investments 6 through 10 are divisible. Use Excel® and SOLVER.
A lending firm is considering 6 independent and divisible investment alternatives which, at any time the firm chooses, can be exited with a full refund
of the initial investment. A total of $200,000 is available for investment, and
the MARR is 10% (Note! There is no planning horizon specified, so the firm
can choose any number of years they wish—the optimum portfolio and the
IRR will remain the same since the initial investment and the salvage value
are the same, and the annual returns are constant each year.):
Initial Investment
Annual Return
1
2
3
4
5
Investment
$25,000
$35,000
$30,000
$40,000
$60,000
$2,600
$3,750
$3,050
$4,775
$6,750
6
$50,000
$5,850
a. Determine the optimum portfolio, including which investments are fully
or partially (if partial, give percentage) selected. You may use Excel®;
do not use SOLVER.
b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
limit on investment capital, (2) plus 20%, and (3) minus 20%. Use Excel®
and SOLVER.
Summary 469
c. Determine the optimum portfolio and its PW, specifying which invest-
ments are fully or partially (give percentage) selected using (1) the
current MARR, (2) plus 20%, and (3) minus 20%. Use Excel® and
SOLVER.
d. Determine the optimum investment portfolio and its PW when investments
1 through 3 are indivisible and investments 4 through 6 are divisible. Use
Excel® and SOLVER.
e. Determine the optimum investment portfolio when all of the investments
are divisible, but fractional investments are limited to 0%, 25%, 50%,
75%, or 100%. Use Excel® and SOLVER.
28. Rex Electric has decided to move into low-rise (2–8 floors) commercial
building electrical wiring. After great success in upscale residential and small
commercial wiring, they have identified four independent and divisible investments, any or all of which will help make the move to the next level. Rex
Electric’s MARR is 10%, and $500,000 is available for investment immediately, with $175,000 available for follow-up investment the next year. The
cash flows, in thousands of dollars, are shown below.
EOY
CF(1)
CF(2)
CF(3)
CF(4)
0
2$50
2$125
1
2$100
$50
$50
$50
$50
$50
$50
$50
$50
$75
2$75
$70
$70
$70
$70
$70
$70
$70
$70
$100
2$200
$50
2$250
$75
$50
$50
$50
$75
$75
$75
$75
$75
$75
$75
$75
$75
$75
$85
$85
$85
$85
$100
2
3
4
5
6
7
8
9
10
a. Determine the optimum portfolio, including which investments are fully
or partially (if partial, give percentage) selected. You may use Excel®;
do not use SOLVER.
b. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the current
limit on investment capital at the end of year 0, (2) plus 20%, and (3)
minus 20%. Use Excel® and SOLVER.
c. Determine the optimum portfolio and its PW, specifying which investments are fully or partially (give percentage) selected using (1) the
current MARR, (2) plus 20%, and (3) minus 20%. Use Excel® and
SOLVER.
d. Determine the optimum investment portfolio and its PW when investments 2 and 4 are indivisible, investments 1 and 3 are divisible, and only
$125,000 in investment capital will be available for follow-up investment
at the end of year 1. Use Excel® and SOLVER.
470
Chapter 12
Capital Budgeting
29. Consider the six divisible investment alternatives shown below. The planning
horizon is 5 years. The MARR is 12%. $50,000 is available for investment.
Alternative
Initial Investment
Annual Return
Salvage Value
A
B
C
D
E
$8,000
$15,000
$10,000
$20,000
$19,000
$3,200
$4,750
$3,070
$5,950
$5,150
$1,000
$1,750
$1,100
$2,000
$2,100
F
$12,000
$4,250
$1,200
a. What proportion of each investment is to be included in the optimum
investment portfolio?
b. Solve part a when full or partial investment cannot be made in more than
3 of the investments.
c. Solve part a when investments D and F are mutually exclusive.
30. Consider the six divisible investment alternatives shown below. The planning
horizon is 8 years. The MARR is 15%. $60,000 is available for investment.
Initial Investment
Annual Return
Salvage Value
M
N
O
P
Q
Alternative
$8,000
$15,000
$10,000
$20,000
$19,000
$3,200
$4,750
$3,070
$5,950
$5,150
$1,000
$1,750
$1,100
$2,000
$2,100
R
$12,000
$4,250
$1,200
a. What proportion of each investment is to be included in the optimum
investment portfolio?
b. Solve part a when full or partial investment cannot be made in more than
3 of the investments.
c. Solve part a when investments P and R are mutually exclusive.
APPENDIX A
Single Sums
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
80
84
90
96
100
108
120
132
144
180
240
360
480
600
Gradient Series
To Find F
Given P
(F|P i %,n)
To Find P
Given F
(P|F i %,n)
To Find F
Given A
(F|A i %,n)
To Find A
Given F
(A|F i%,n)
To Find P
Given A
(P|A i%,n)
To Find A
Given P
(A|P i%,n)
To Find P
Given G
(P|G i%,n)
To Find A
Given G
(A|G i%,n)
1.00250
1.00501
1.00752
1.01004
1.01256
1.01509
1.01763
1.02018
1.02273
1.02528
1.02785
1.03042
1.03299
1.03557
1.03816
1.04076
1.04336
1.04597
1.04858
1.05121
1.05383
1.05647
1.05911
1.06176
1.06441
1.06707
1.06974
1.07241
1.07510
1.07778
1.09405
1.10503
1.12733
1.13297
1.13864
1.14720
1.16162
1.19695
1.20595
1.22110
1.23335
1.25197
1.27087
1.28362
1.30952
1.34935
1.39040
1.43269
1.56743
1.82075
2.45684
3.31515
4.47331
0.99751
0.99502
0.99254
0.99006
0.98759
0.98513
0.98267
0.98022
0.97778
0.97534
0.97291
0.97048
0.96806
0.96565
0.96324
0.96084
0.95844
0.95605
0.95367
0.95129
0.94892
0.94655
0.94419
0.94184
0.93949
0.93714
0.93481
0.93248
0.93015
0.92783
0.91403
0.90495
0.88705
0.88263
0.87824
0.87168
0.86087
0.83546
0.82922
0.81894
0.81080
0.79874
0.78686
0.77904
0.76364
0.74110
0.71922
0.69799
0.63799
0.54922
0.40703
0.30165
0.22355
1.00000
2.00250
3.00751
4.01503
5.02506
6.03763
7.05272
8.07035
9.09053
10.11325
11.13854
12.16638
13.19680
14.22979
15.26537
16.30353
17.34429
18.38765
19.43362
20.48220
21.53341
22.58724
23.64371
24.70282
25.76457
26.82899
27.89606
28.96580
30.03821
31.11331
37.62056
42.01320
50.93121
53.18868
55.45746
58.88194
64.64671
78.77939
82.37922
88.43918
93.34192
100.78845
108.34739
113.44996
123.80926
139.74142
156.15817
173.07425
226.97269
328.30200
582.73688
926.05950
1389.32309
1.00000
0.49938
0.33250
0.24906
0.19900
0.16563
0.14179
0.12391
0.11000
0.09888
0.08978
0.08219
0.07578
0.07028
0.06551
0.06134
0.05766
0.05438
0.05146
0.04882
0.04644
0.04427
0.04229
0.04048
0.03881
0.03727
0.03585
0.03452
0.03329
0.03214
0.02658
0.02380
0.01963
0.01880
0.01803
0.01698
0.01547
0.01269
0.01214
0.01131
0.01071
9.9218E203
9.2296E203
8.8145E203
8.0769E203
7.1561E203
6.4038E203
5.7779E203
4.4058E203
3.0460E203
1.7160E203
1.0798E203
7.1977E204
0.99751
1.99252
2.98506
3.97512
4.96272
5.94785
6.93052
7.91074
8.88852
9.86386
10.83677
11.80725
12.77532
13.74096
14.70420
15.66504
16.62348
17.57953
18.53320
19.48449
20.43340
21.37995
22.32414
23.26598
24.20547
25.14261
26.07742
27.00989
27.94004
28.86787
34.38647
38.01986
45.17869
46.94617
48.70484
51.32644
55.65236
65.81686
68.31075
72.42595
75.68132
80.50382
85.25460
88.38248
94.54530
103.56175
112.31206
120.80407
144.80547
180.31091
237.18938
279.34176
310.58071
1.00250
0.50188
0.33500
0.25156
0.20150
0.16813
0.14429
0.12641
0.11250
0.10138
0.09228
0.08469
0.07828
0.07278
0.06801
0.06384
0.06016
0.05688
0.05396
0.05132
0.04894
0.04677
0.04479
0.04298
0.04131
0.03977
0.03835
0.03702
0.03579
0.03464
0.02908
0.02630
0.02213
0.02130
0.02053
0.01948
0.01797
0.01519
0.01464
0.01381
0.01321
0.01242
0.01173
0.01131
0.01058
9.6561E203
8.9038E203
8.2779E203
6.9058E203
5.5460E203
4.2160E203
3.5798E203
3.2198E203
0.00000
0.99502
2.98009
5.95028
9.90065
14.82630
20.72235
27.58391
35.40614
44.18420
53.91328
64.58858
76.20532
88.75874
102.24409
116.65666
131.99172
148.24459
165.41059
183.48508
202.46341
222.34096
243.11313
264.77534
287.32301
310.75160
335.05657
360.23340
386.27760
413.18468
592.49878
728.73988
1040.05520
1125.77667
1214.58847
1353.52863
1600.08454
2265.55685
2447.60694
2764.45681
3029.75923
3446.86997
3886.28316
4191.24173
4829.01247
5852.11160
6950.01441
8117.41331
1.1987E104
1.9399E104
3.6264E104
5.3821E104
7.0581E104
0.00000
0.49938
0.99834
1.49688
1.99501
2.49272
2.99001
3.48689
3.98335
4.47940
4.97503
5.47025
5.96504
6.45943
6.95339
7.44694
7.94008
8.43279
8.92510
9.41698
9.90845
10.39951
10.89014
11.38036
11.87017
12.35956
12.84853
13.33709
13.82523
14.31296
17.23058
19.16735
23.02092
23.98016
24.93774
26.37098
28.75142
34.42214
35.83048
38.16942
40.03312
42.81623
45.58444
47.42163
51.07618
56.50843
61.88128
67.19487
82.78122
107.58631
152.89019
192.66991
227.25400
471
Time Value of Money Factors Discrete Compounding 0.25%
n
Uniform Series
0.25%
TABLE A-a-1
0.50%
Time Value of Money Factors Discrete Compounding 0.50%
472 Appendix A
TABLE A-a-2
Single Sums
Uniform Series
Gradient Series
n
To Find F
Given P
(F|P i %,n)
To Find P
Given F
(P|F i %,n)
To Find F
Given A
(F|A i %,n)
To Find A
Given F
(A|F i%,n)
To Find P
Given A
(P|A i%,n)
To Find A
Given P
(A|P i%,n)
To Find P
Given G
(P|G i%,n)
To Find A
Given G
(A|G i%,n)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
80
84
90
96
100
108
120
132
144
180
240
360
480
600
1.00500
1.01003
1.01508
1.02015
1.02525
1.03038
1.03553
1.04071
1.04591
1.05114
1.05640
1.06168
1.06699
1.07232
1.07768
1.08307
1.08849
1.09393
1.09940
1.10490
1.11042
1.11597
1.12155
1.12716
1.13280
1.13846
1.14415
1.14987
1.15562
1.16140
1.19668
1.22079
1.27049
1.28323
1.29609
1.31563
1.34885
1.43204
1.45363
1.49034
1.52037
1.56655
1.61414
1.64667
1.71370
1.81940
1.93161
2.05075
2.45409
3.31020
6.02258
10.95745
19.93596
0.99502
0.99007
0.98515
0.98025
0.97537
0.97052
0.96569
0.96089
0.95610
0.95135
0.94661
0.94191
0.93722
0.93256
0.92792
0.92330
0.91871
0.91414
0.90959
0.90506
0.90056
0.89608
0.89162
0.88719
0.88277
0.87838
0.87401
0.86966
0.86533
0.86103
0.83564
0.81914
0.78710
0.77929
0.77155
0.76009
0.74137
0.69830
0.68793
0.67099
0.65773
0.63834
0.61952
0.60729
0.58353
0.54963
0.51770
0.48763
0.40748
0.30210
0.16604
0.09126
0.05016
1.00000
2.00500
3.01502
4.03010
5.05025
6.07550
7.10588
8.14141
9.18212
10.22803
11.27917
12.33556
13.39724
14.46423
15.53655
16.61423
17.69730
18.78579
19.87972
20.97912
22.08401
23.19443
24.31040
25.43196
26.55912
27.69191
28.83037
29.97452
31.12439
32.28002
39.33610
44.15885
54.09783
56.64516
59.21803
63.12577
69.77003
86.40886
90.72650
98.06771
104.07393
113.31094
122.82854
129.33370
142.73990
163.87935
186.32263
210.15016
290.81871
462.04090
1004.51504
1991.49073
3787.19108
1.00000
0.49875
0.33167
0.24813
0.19801
0.16460
0.14073
0.12283
0.10891
0.09777
0.08866
0.08107
0.07464
0.06914
0.06436
0.06019
0.05651
0.05323
0.05030
0.04767
0.04528
0.04311
0.04113
0.03932
0.03765
0.03611
0.03469
0.03336
0.03213
0.03098
0.02542
0.02265
0.01849
0.01765
0.01689
0.01584
0.01433
0.01157
0.01102
0.01020
9.6086E203
8.8253E203
8.1414E203
7.7319E203
7.0057E203
6.1021E203
5.3670E203
4.7585E203
3.4386E203
2.1643E203
9.9551E204
5.0214E204
2.6405E204
0.99502
1.98510
2.97025
3.95050
4.92587
5.89638
6.86207
7.82296
8.77906
9.73041
10.67703
11.61893
12.55615
13.48871
14.41662
15.33993
16.25863
17.17277
18.08236
18.98742
19.88798
20.78406
21.67568
22.56287
23.44564
24.32402
25.19803
26.06769
26.93302
27.79405
32.87102
36.17223
42.58032
44.14279
45.68975
47.98145
51.72556
60.33951
62.41365
65.80231
68.45304
72.33130
76.09522
78.54264
83.29342
90.07345
96.45960
102.47474
118.50351
139.58077
166.79161
181.74758
189.96787
1.00500
0.50375
0.33667
0.25313
0.20301
0.16960
0.14573
0.12783
0.11391
0.10277
0.09366
0.08607
0.07964
0.07414
0.06936
0.06519
0.06151
0.05823
0.05530
0.05267
0.05028
0.04811
0.04613
0.04432
0.04265
0.04111
0.03969
0.03836
0.03713
0.03598
0.03042
0.02765
0.02349
0.02265
0.02189
0.02084
0.01933
0.01657
0.01602
0.01520
0.01461
0.01383
0.01314
0.01273
0.01201
0.01110
0.01037
9.7585E203
8.4386E203
7.1643E203
5.9955E203
5.5021E203
5.2640E203
0.00000
0.99007
2.96037
5.90111
9.80260
14.65519
20.44933
27.17552
34.82436
43.38649
52.85264
63.21360
74.46023
86.58346
99.57430
113.42380
128.12311
143.66343
160.03602
177.23221
195.24341
214.06109
233.67676
254.08203
275.26856
297.22805
319.95231
343.43317
367.66255
392.63241
557.55983
681.33469
959.91881
1035.69659
1113.81615
1235.26857
1448.64580
2012.34779
2163.75249
2424.64551
2640.66405
2976.07688
3324.18460
3562.79343
4054.37473
4823.50506
5624.58677
6451.31165
9031.33557
1.3416E104
2.1403E104
2.7588E104
3.1974E104
0.00000
0.49875
0.99667
1.49377
1.99003
2.48545
2.98005
3.47382
3.96675
4.45885
4.95013
5.44057
5.93018
6.41896
6.90691
7.39403
7.88031
8.36577
8.85040
9.33419
9.81716
10.29929
10.78060
11.26107
11.74072
12.21953
12.69751
13.17467
13.65099
14.12649
16.96205
18.83585
22.54372
23.46242
24.37781
25.74471
28.00638
33.35041
34.66794
36.84742
38.57628
41.14508
43.68454
45.36126
48.67581
53.55080
58.31029
62.95514
76.21154
96.11309
128.32362
151.79491
168.31425
Appendix A 473
Single Sums
To Find P
Given F
(P|F i %,n)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
80
84
90
96
100
108
120
132
144
180
240
360
480
600
1.00750
1.01506
1.02267
1.03034
1.03807
1.04585
1.05370
1.06160
1.06956
1.07758
1.08566
1.09381
1.10201
1.11028
1.11860
1.12699
1.13544
1.14396
1.15254
1.16118
1.16989
1.17867
1.18751
1.19641
1.20539
1.21443
1.22354
1.23271
1.24196
1.25127
1.30865
1.34835
1.43141
1.45296
1.47483
1.50827
1.56568
1.71255
1.75137
1.81804
1.87320
1.95909
2.04892
2.11108
2.24112
2.45136
2.68131
2.93284
3.83804
6.00915
14.73058
36.10990
88.51826
0.99256
0.98517
0.97783
0.97055
0.96333
0.95616
0.94904
0.94198
0.93496
0.92800
0.92109
0.91424
0.90743
0.90068
0.89397
0.88732
0.88071
0.87416
0.86765
0.86119
0.85478
0.84842
0.84210
0.83583
0.82961
0.82343
0.81730
0.81122
0.80518
0.79919
0.76415
0.74165
0.69861
0.68825
0.67804
0.66301
0.63870
0.58392
0.57098
0.55004
0.53385
0.51044
0.48806
0.47369
0.44620
0.40794
0.37295
0.34097
0.26055
0.16641
0.06789
0.02769
0.01130
To Find F
Given A
(F|A i %,n)
1.00000
2.00750
3.02256
4.04523
5.07556
6.11363
7.15948
8.21318
9.27478
10.34434
11.42192
12.50759
13.60139
14.70340
15.81368
16.93228
18.05927
19.19472
20.33868
21.49122
22.65240
23.82230
25.00096
26.18847
27.38488
28.59027
29.80470
31.02823
32.26094
33.50290
41.15272
46.44648
57.52071
60.39426
63.31107
67.76883
75.42414
95.00703
100.18331
109.07253
116.42693
127.87899
139.85616
148.14451
165.48322
193.51428
224.17484
257.71157
378.40577
667.88687
1830.74348
4681.32027
1.1669E104
Gradient Series
To Find A
Given F
(A|F i%,n)
To Find P
Given A
(P|A i%,n)
To Find A
Given P
(A|P i%,n)
To Find P
Given G
(P|G i%,n)
To Find A
Given G
(A|G i%,n)
1.00000
0.49813
0.33085
0.24721
0.19702
0.16357
0.13967
0.12176
0.10782
0.09667
0.08755
0.07995
0.07352
0.06801
0.06324
0.05906
0.05537
0.05210
0.04917
0.04653
0.04415
0.04198
0.04000
0.03818
0.03652
0.03498
0.03355
0.03223
0.03100
0.02985
0.02430
0.02153
0.01739
0.01656
0.01580
0.01476
0.01326
0.01053
9.9817E203
9.1682E203
8.5891E203
7.8199E203
7.1502E203
6.7502E203
6.0429E203
5.1676E203
4.4608E203
3.8803E203
2.6427E203
1.4973E203
5.4623E204
2.1361E204
8.5696E205
0.99256
1.97772
2.95556
3.92611
4.88944
5.84560
6.79464
7.73661
8.67158
9.59958
10.52067
11.43491
12.34235
13.24302
14.13699
15.02431
15.90502
16.77918
17.64683
18.50802
19.36280
20.21121
21.05331
21.88915
22.71876
23.54219
24.35949
25.17071
25.97589
26.77508
31.44681
34.44694
40.18478
41.56645
42.92762
44.93161
48.17337
55.47685
57.20267
59.99444
62.15396
65.27461
68.25844
70.17462
73.83938
78.94169
83.60642
87.87109
98.59341
111.14495
124.28187
129.64090
131.82705
1.00750
0.50563
0.33835
0.25471
0.20452
0.17107
0.14717
0.12926
0.11532
0.10417
0.09505
0.08745
0.08102
0.07551
0.07074
0.06656
0.06287
0.05960
0.05667
0.05403
0.05165
0.04948
0.04750
0.04568
0.04402
0.04248
0.04105
0.03973
0.03850
0.03735
0.03180
0.02903
0.02489
0.02406
0.02330
0.02226
0.02076
0.01803
0.01748
0.01667
0.01609
0.01532
0.01465
0.01425
0.01354
0.01267
0.01196
0.01138
0.01014
8.9973E203
8.0462E203
7.7136E203
7.5857E203
0.00000
0.98517
2.94083
5.85250
9.70581
14.48660
20.18084
26.77467
34.25438
42.60641
51.81736
61.87398
72.76316
84.47197
96.98758
110.29735
124.38875
139.24940
154.86708
171.22969
188.32527
206.14200
224.66820
243.89233
263.80295
284.38879
305.63869
327.54162
350.08668
373.26310
524.99236
637.46933
886.84045
953.84863
1022.58522
1128.78691
1313.51888
1791.24629
1917.22249
2132.14723
2308.12830
2577.99605
2853.93524
3040.74530
3419.90409
3998.56214
4583.57014
5169.58283
6892.60143
9494.11617
1.3312E104
1.5513E104
1.6673E104
0.00000
0.49813
0.99502
1.49066
1.98506
2.47821
2.97011
3.46077
3.95019
4.43836
4.92529
5.41097
5.89541
6.37860
6.86055
7.34126
7.82072
8.29894
8.77592
9.25165
9.72614
10.19939
10.67139
11.14216
11.61168
12.07996
12.54701
13.01281
13.47737
13.94069
16.69462
18.50583
22.06906
22.94756
23.82115
25.12233
27.26649
32.28818
33.51631
35.53908
37.13566
39.49462
41.81073
43.33112
46.31545
50.65210
54.82318
58.83144
69.90935
85.42103
107.11448
119.66198
126.47762
Time Value of Money Factors Discrete Compounding 0.75%
n
To Find F
Given P
(F|P i %,n)
Uniform Series
0.75%
TABLE A-a-3
1.00%
Time Value of Money Factors Discrete Compounding 1.00%
474 Appendix A
TABLE A-a-4
Single Sums
Uniform Series
Gradient Series
n
To Find F
Given P
(F|P i %,n)
To Find P
Given F
(P|F i%,n)
To Find F
Given A
(F|A i%,n)
To Find A
Given F
(A|F i%,n)
To Find P
Given A
(P|A i %,n)
To Find A
Given P
(A|P i%,n)
To Find P
Given G
(P|G i%,n)
To Find A
Given G
(A|G i%,n)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
80
84
90
96
100
108
120
132
144
180
240
360
480
600
1.01000
1.02010
1.03030
1.04060
1.05101
1.06152
1.07214
1.08286
1.09369
1.10462
1.11567
1.12683
1.13809
1.14947
1.16097
1.17258
1.18430
1.19615
1.20811
1.22019
1.23239
1.24472
1.25716
1.26973
1.28243
1.29526
1.30821
1.32129
1.33450
1.34785
1.43077
1.48886
1.61223
1.64463
1.67769
1.72852
1.81670
2.04710
2.10913
2.21672
2.30672
2.44863
2.59927
2.70481
2.92893
3.30039
3.71896
4.19062
5.99580
10.89255
35.94964
118.64773
391.58340
0.99010
0.98030
0.97059
0.96098
0.95147
0.94205
0.93272
0.92348
0.91434
0.90529
0.89632
0.88745
0.87866
0.86996
0.86135
0.85282
0.84438
0.83602
0.82774
0.81954
0.81143
0.80340
0.79544
0.78757
0.77977
0.77205
0.76440
0.75684
0.74934
0.74192
0.69892
0.67165
0.62026
0.60804
0.59606
0.57853
0.55045
0.48850
0.47413
0.45112
0.43352
0.40839
0.38472
0.36971
0.34142
0.30299
0.26889
0.23863
0.16678
0.09181
0.02782
8.4283E203
2.5537E203
1.00000
2.01000
3.03010
4.06040
5.10101
6.15202
7.21354
8.28567
9.36853
10.46221
11.56683
12.68250
13.80933
14.94742
16.09690
17.25786
18.43044
19.61475
20.81090
22.01900
23.23919
24.47159
25.71630
26.97346
28.24320
29.52563
30.82089
32.12910
33.45039
34.78489
43.07688
48.88637
61.22261
64.46318
67.76889
72.85246
81.66967
104.70993
110.91285
121.67152
130.67227
144.86327
159.92729
170.48138
192.89258
230.03869
271.89586
319.06156
499.58020
989.25537
3494.96413
1.1765E104
3.9058E104
1.00000
0.49751
0.33002
0.24628
0.19604
0.16255
0.13863
0.12069
0.10674
0.09558
0.08645
0.07885
0.07241
0.06690
0.06212
0.05794
0.05426
0.05098
0.04805
0.04542
0.04303
0.04086
0.03889
0.03707
0.03541
0.03387
0.03245
0.03112
0.02990
0.02875
0.02321
0.02046
0.01633
0.01551
0.01476
0.01373
0.01224
9.5502E203
9.0161E203
8.2189E203
7.6527E203
6.9031E203
6.2528E203
5.8657E203
5.1842E203
4.3471E203
3.6779E203
3.1342E203
2.0017E203
1.0109E203
2.8613E204
8.5000E205
2.5603E205
0.99010
1.97040
2.94099
3.90197
4.85343
5.79548
6.72819
7.65168
8.56602
9.47130
10.36763
11.25508
12.13374
13.00370
13.86505
14.71787
15.56225
16.39827
17.22601
18.04555
18.85698
19.66038
20.45582
21.24339
22.02316
22.79520
23.55961
24.31644
25.06579
25.80771
30.10751
32.83469
37.97396
39.19612
40.39419
42.14719
44.95504
51.15039
52.58705
54.88821
56.64845
59.16088
61.52770
63.02888
65.85779
69.70052
73.11075
76.13716
83.32166
90.81942
97.21833
99.15717
99.74463
1.01000
0.50751
0.34002
0.25628
0.20604
0.17255
0.14863
0.13069
0.11674
0.10558
0.09645
0.08885
0.08241
0.07690
0.07212
0.06794
0.06426
0.06098
0.05805
0.05542
0.05303
0.05086
0.04889
0.04707
0.04541
0.04387
0.04245
0.04112
0.03990
0.03875
0.03321
0.03046
0.02633
0.02551
0.02476
0.02373
0.02224
0.01955
0.01902
0.01822
0.01765
0.01690
0.01625
0.01587
0.01518
0.01435
0.01368
0.01313
0.01200
0.01101
0.01029
0.01008
0.01003
0.00000
0.98030
2.92148
5.80442
9.61028
14.32051
19.91681
26.38120
33.69592
41.84350
50.80674
60.56868
71.11263
82.42215
94.48104
107.27336
120.78340
134.99569
149.89501
165.46636
181.69496
198.56628
216.06600
234.18002
252.89446
272.19566
292.07016
312.50472
333.48630
355.00207
494.62069
596.85606
820.14601
879.41763
939.91752
1032.81478
1192.80614
1597.86733
1702.73397
1879.87710
2023.31531
2240.56748
2459.42979
2605.77575
2898.42028
3334.11485
3761.69441
4177.46642
5330.06592
6878.60156
8720.43230
9511.15793
9821.23859
0.00000
0.49751
0.99337
1.48756
1.98010
2.47098
2.96020
3.44777
3.93367
4.41792
4.90052
5.38145
5.86073
6.33836
6.81433
7.28865
7.76131
8.23231
8.70167
9.16937
9.63542
10.09982
10.56257
11.02367
11.48312
11.94092
12.39707
12.85158
13.30444
13.75566
16.42848
18.17761
21.59759
22.43635
23.26863
24.50495
26.53331
31.23861
32.37934
34.24920
35.71704
37.87245
39.97272
41.34257
44.01029
47.83486
51.45200
54.86764
63.96975
75.73933
89.69947
95.92002
98.46384
Appendix A 475
Single Sums
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
80
84
90
96
100
108
120
132
144
180
240
360
480
600
1.01250
1.02516
1.03797
1.05095
1.06408
1.07738
1.09085
1.10449
1.11829
1.13227
1.14642
1.16075
1.17526
1.18995
1.20483
1.21989
1.23514
1.25058
1.26621
1.28204
1.29806
1.31429
1.33072
1.34735
1.36419
1.38125
1.39851
1.41599
1.43369
1.45161
1.56394
1.64362
1.81535
1.86102
1.90784
1.98028
2.10718
2.44592
2.53879
2.70148
2.83911
3.05881
3.29551
3.46340
3.82528
4.44021
5.15400
5.98253
9.35633
19.71549
87.54100
388.70068
1725.91392
Gradient Series
To Find P
Given F
(P|F i%,n)
To Find F
Given A
(F|A i%,n)
To Find A
Given F
(A|F i%,n)
To Find P
Given A
(P|A i %,n)
To Find A
Given P
(A|P i%,n)
To Find P
Given G
(P|G i%,n)
To Find A
Given G
(A|G i%,n)
0.98765
0.97546
0.96342
0.95152
0.93978
0.92817
0.91672
0.90540
0.89422
0.88318
0.87228
0.86151
0.85087
0.84037
0.82999
0.81975
0.80963
0.79963
0.78976
0.78001
0.77038
0.76087
0.75147
0.74220
0.73303
0.72398
0.71505
0.70622
0.69750
0.68889
0.63941
0.60841
0.55086
0.53734
0.52415
0.50498
0.47457
0.40884
0.39389
0.37017
0.35222
0.32692
0.30344
0.28873
0.26142
0.22521
0.19402
0.16715
0.10688
0.05072
0.01142
2.5727E203
5.7940E204
1.00000
2.01250
3.03766
4.07563
5.12657
6.19065
7.26804
8.35889
9.46337
10.58167
11.71394
12.86036
14.02112
15.19638
16.38633
17.59116
18.81105
20.04619
21.29677
22.56298
23.84502
25.14308
26.45737
27.78808
29.13544
30.49963
31.88087
33.27938
34.69538
36.12907
45.11551
51.48956
65.22839
68.88179
72.62710
78.42246
88.57451
115.67362
123.10349
136.11880
147.12904
164.70501
183.64106
197.07234
226.02255
275.21706
332.31981
398.60208
668.50676
1497.23948
6923.27961
3.1016E104
1.3799E105
1.00000
0.49689
0.32920
0.24536
0.19506
0.16153
0.13759
0.11963
0.10567
0.09450
0.08537
0.07776
0.07132
0.06581
0.06103
0.05685
0.05316
0.04988
0.04696
0.04432
0.04194
0.03977
0.03780
0.03599
0.03432
0.03279
0.03137
0.03005
0.02882
0.02768
0.02217
0.01942
0.01533
0.01452
0.01377
0.01275
0.01129
8.6450E203
8.1232E203
7.3465E203
6.7968E203
6.0715E203
5.4454E203
5.0743E203
4.4243E203
3.6335E203
3.0091E203
2.5088E203
1.4959E203
6.6790E204
1.4444E204
3.2241E205
7.2467E206
0.98765
1.96312
2.92653
3.87806
4.81784
5.74601
6.66273
7.56812
8.46234
9.34553
10.21780
11.07931
11.93018
12.77055
13.60055
14.42029
15.22992
16.02955
16.81931
17.59932
18.36969
19.13056
19.88204
20.62423
21.35727
22.08125
22.79630
23.50252
24.20002
24.88891
28.84727
31.32693
35.93148
37.01288
38.06773
39.60169
42.03459
47.29247
48.48897
50.38666
51.82219
53.84606
55.72457
56.90134
59.08651
61.98285
64.47807
66.62772
71.44964
75.94228
79.08614
79.79419
79.95365
1.01250
0.50939
0.34170
0.25786
0.20756
0.17403
0.15009
0.13213
0.11817
0.10700
0.09787
0.09026
0.08382
0.07831
0.07353
0.06935
0.06566
0.06238
0.05946
0.05682
0.05444
0.05227
0.05030
0.04849
0.04682
0.04529
0.04387
0.04255
0.04132
0.04018
0.03467
0.03192
0.02783
0.02702
0.02627
0.02525
0.02379
0.02115
0.02062
0.01985
0.01930
0.01857
0.01795
0.01757
0.01692
0.01613
0.01551
0.01501
0.01400
0.01317
0.01264
0.01253
0.01251
0.00000
0.97546
2.90230
5.75687
9.51598
14.15685
19.65715
25.99494
33.14870
41.09733
49.82011
59.29670
69.50717
80.43196
92.05186
104.34806
117.30207
130.89580
145.11145
159.93161
175.33919
191.31742
207.84986
224.92039
242.51321
260.61282
279.20402
298.27192
317.80191
337.77969
466.28302
559.23198
759.22956
811.67385
864.94093
946.22770
1084.84285
1428.45610
1515.79039
1661.86513
1778.83839
1953.83026
2127.52438
2242.24109
2468.26361
2796.56945
3109.35041
3404.60974
4176.90718
5101.52883
5997.90267
6284.74422
6368.48047
0.00000
0.49689
0.99172
1.48447
1.97516
2.46377
2.95032
3.43479
3.91720
4.39754
4.87581
5.35202
5.82616
6.29824
6.76825
7.23620
7.70208
8.16591
8.62767
9.08738
9.54502
10.00062
10.45415
10.90564
11.35507
11.80245
12.24778
12.69106
13.13230
13.57150
16.16385
17.85148
21.12993
21.92950
22.72110
23.89362
25.80834
30.20472
31.26052
32.98225
34.32581
36.28548
38.17929
39.40577
41.77373
45.11844
48.22338
51.09900
58.45945
67.17640
75.84012
78.76193
79.65216
Time Value of Money Factors Discrete Compounding 1.25%
n
To Find F
Given P
(F|P i %,n)
Uniform Series
1.25%
TABLE A-a-5
1.50%
Time Value of Money Factors Discrete Compounding 1.50%
476 Appendix A
TABLE A-a-6
Single Sums
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
80
84
90
96
100
108
120
132
144
180
240
360
480
600
Uniform Series
Gradient Series
To Find F
Given P
(F|P i %,n)
To Find P
Given F
(P|F i %,n)
To Find F
Given A
(F|A i %,n)
To Find A
Given F
(A|F i%,n)
To Find P
Given A
(P|A i%,n)
To Find A
Given P
(A|P i%,n)
To Find P
Given G
(P|G i%,n)
To Find A
Given G
(A|G i%,n)
1.01500
1.03023
1.04568
1.06136
1.07728
1.09344
1.10984
1.12649
1.14339
1.16054
1.17795
1.19562
1.21355
1.23176
1.25023
1.26899
1.28802
1.30734
1.32695
1.34686
1.36706
1.38756
1.40838
1.42950
1.45095
1.47271
1.49480
1.51722
1.53998
1.56308
1.70914
1.81402
2.04348
2.10524
2.16887
2.26794
2.44322
2.92116
3.05459
3.29066
3.49259
3.81895
4.17580
4.43205
4.99267
5.96932
7.13703
8.53316
14.58437
35.63282
212.70378
1269.69754
7579.23459
0.98522
0.97066
0.95632
0.94218
0.92826
0.91454
0.90103
0.88771
0.87459
0.86167
0.84893
0.83639
0.82403
0.81185
0.79985
0.78803
0.77639
0.76491
0.75361
0.74247
0.73150
0.72069
0.71004
0.69954
0.68921
0.67902
0.66899
0.65910
0.64936
0.63976
0.58509
0.55126
0.48936
0.47500
0.46107
0.44093
0.40930
0.34233
0.32738
0.30389
0.28632
0.26185
0.23947
0.22563
0.20029
0.16752
0.14011
0.11719
0.06857
0.02806
4.7014E203
7.8759E204
1.3194E204
1.00000
2.01500
3.04522
4.09090
5.15227
6.22955
7.32299
8.43284
9.55933
10.70272
11.86326
13.04121
14.23683
15.45038
16.68214
17.93237
19.20136
20.48938
21.79672
23.12367
24.47052
25.83758
27.22514
28.63352
30.06302
31.51397
32.98668
34.48148
35.99870
37.53868
47.27597
54.26789
69.56522
73.68283
77.92489
84.52960
96.21465
128.07720
136.97278
152.71085
166.17264
187.92990
211.72023
228.80304
266.17777
331.28819
409.13539
502.21092
905.62451
2308.85437
1.4114E104
8.4580E104
5.0522E105
1.00000
0.49628
0.32838
0.24444
0.19409
0.16053
0.13656
0.11858
0.10461
0.09343
0.08429
0.07668
0.07024
0.06472
0.05994
0.05577
0.05208
0.04881
0.04588
0.04325
0.04087
0.03870
0.03673
0.03492
0.03326
0.03173
0.03032
0.02900
0.02778
0.02664
0.02115
0.01843
0.01437
0.01357
0.01283
0.01183
0.01039
7.8078E203
7.3007E203
6.5483E203
6.0178E203
5.3211E203
4.7232E203
4.3706E203
3.7569E203
3.0185E203
2.4442E203
1.9912E203
1.1042E203
4.3312E204
7.0854E205
1.1823E205
1.9794E206
0.98522
1.95588
2.91220
3.85438
4.78264
5.69719
6.59821
7.48593
8.36052
9.22218
10.07112
10.90751
11.73153
12.54338
13.34323
14.13126
14.90765
15.67256
16.42617
17.16864
17.90014
18.62082
19.33086
20.03041
20.71961
21.39863
22.06762
22.72672
23.37608
24.01584
27.66068
29.91585
34.04255
34.99969
35.92874
37.27147
39.38027
43.84467
44.84160
46.40732
47.57863
49.20985
50.70168
51.62470
53.31375
55.49845
57.32571
58.85401
62.09556
64.79573
66.35324
66.61416
66.65787
1.01500
0.51128
0.34338
0.25944
0.20909
0.17553
0.15156
0.13358
0.11961
0.10843
0.09929
0.09168
0.08524
0.07972
0.07494
0.07077
0.06708
0.06381
0.06088
0.05825
0.05587
0.05370
0.05173
0.04992
0.04826
0.04673
0.04532
0.04400
0.04278
0.04164
0.03615
0.03343
0.02937
0.02857
0.02783
0.02683
0.02539
0.02281
0.02230
0.02155
0.02102
0.02032
0.01972
0.01937
0.01876
0.01802
0.01744
0.01699
0.01610
0.01543
0.01507
0.01501
0.01500
0.00000
0.97066
2.88330
5.70985
9.42289
13.99560
19.40176
25.61574
32.61248
40.36748
48.85681
58.05708
67.94540
78.49944
89.69736
101.51783
113.93999
126.94349
140.50842
154.61536
169.24532
184.37976
200.00058
216.09009
232.63103
249.60654
267.00017
284.79585
302.97790
321.53101
439.83026
524.35682
703.54615
749.96361
796.87737
868.02846
988.16739
1279.79379
1352.56005
1473.07411
1568.51404
1709.54387
1847.47253
1937.45061
2112.13479
2359.71143
2588.70855
2798.57842
3316.90537
3870.69117
4310.71648
4415.74120
4438.58047
0.00000
0.49628
0.99007
1.48139
1.97023
2.45658
2.94046
3.42185
3.90077
4.37721
4.85118
5.32267
5.79169
6.25824
6.72231
7.18392
7.64306
8.09973
8.55394
9.00569
9.45497
9.90180
10.34618
10.78810
11.22758
11.66460
12.09918
12.53132
12.96102
13.38829
15.90092
17.52773
20.66667
21.42772
22.17938
23.28936
25.09296
29.18927
30.16306
31.74228
32.96677
34.73987
36.43810
37.52953
39.61708
42.51851
45.15789
47.55119
53.41614
59.73682
64.96618
66.28833
66.58749
Appendix A 477
Single Sums
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
80
84
90
96
100
108
120
132
144
180
240
360
480
600
Gradient Series
To Find F
Given P
(F|P i %,n)
To Find P
Given F
(P|F i%,n)
To Find F
Given A
(F|A i %,n)
To Find A
Given F
(A|F i%,n)
To Find P
Given A
(P|A i%,n)
To Find A
Given P
(A|P i%,n)
To Find P
Given G
(P|G i%,n)
To Find A
Given G
(A|G i%,n)
1.01750
1.03531
1.05342
1.07186
1.09062
1.10970
1.12912
1.14888
1.16899
1.18944
1.21026
1.23144
1.25299
1.27492
1.29723
1.31993
1.34303
1.36653
1.39045
1.41478
1.43954
1.46473
1.49036
1.51644
1.54298
1.56998
1.59746
1.62541
1.65386
1.68280
1.86741
2.00160
2.29960
2.38079
2.46485
2.59653
2.83182
3.48721
3.67351
4.00639
4.29429
4.76538
5.28815
5.66816
6.51204
8.01918
9.87514
12.16063
22.70885
64.30730
515.69206
4135.42921
3.3163E104
0.98280
0.96590
0.94929
0.93296
0.91691
0.90114
0.88564
0.87041
0.85544
0.84073
0.82627
0.81206
0.79809
0.78436
0.77087
0.75762
0.74459
0.73178
0.71919
0.70682
0.69467
0.68272
0.67098
0.65944
0.64810
0.63695
0.62599
0.61523
0.60465
0.59425
0.53550
0.49960
0.43486
0.42003
0.40570
0.38513
0.35313
0.28676
0.27222
0.24960
0.23287
0.20985
0.18910
0.17642
0.15356
0.12470
0.10126
0.08223
0.04404
0.01555
1.9391E203
2.4181E204
3.0154E205
1.00000
2.01750
3.05281
4.10623
5.17809
6.26871
7.37841
8.50753
9.65641
10.82540
12.01484
13.22510
14.45654
15.70953
16.98445
18.28168
19.60161
20.94463
22.31117
23.70161
25.11639
26.55593
28.02065
29.51102
31.02746
32.57044
34.14042
35.73788
37.36329
39.01715
49.56613
57.23413
74.26278
78.90222
83.70547
91.23016
104.67522
142.12628
152.77206
171.79382
188.24499
215.16462
245.03739
266.75177
314.97378
401.09620
507.15073
637.75045
1240.50595
3617.56017
2.9411E104
2.3625E105
1.8950E106
1.00000
0.49566
0.32757
0.24353
0.19312
0.15952
0.13553
0.11754
0.10356
0.09238
0.08323
0.07561
0.06917
0.06366
0.05888
0.05470
0.05102
0.04774
0.04482
0.04219
0.03981
0.03766
0.03569
0.03389
0.03223
0.03070
0.02929
0.02798
0.02676
0.02563
0.02018
0.01747
0.01347
0.01267
0.01195
0.01096
9.5534E203
7.0360E203
6.5457E203
5.8209E203
5.3122E203
4.6476E203
4.0810E203
3.7488E203
3.1749E203
2.4932E203
1.9718E203
1.5680E203
8.0612E204
2.7643E204
3.4001E205
4.2327E206
5.2772E207
0.98280
1.94870
2.89798
3.83094
4.74786
5.64900
6.53464
7.40505
8.26049
9.10122
9.92749
10.73955
11.53764
12.32201
13.09288
13.85050
14.59508
15.32686
16.04606
16.75288
17.44755
18.13027
18.80125
19.46069
20.10878
20.74573
21.37173
21.98695
22.59160
23.18585
26.54275
28.59423
32.29380
33.14121
33.95972
35.13545
36.96399
40.75645
41.58748
42.87993
43.83614
45.15161
46.33703
47.06147
48.36790
50.01709
51.35632
52.44385
54.62653
56.25427
57.03205
57.12904
57.14113
1.01750
0.51316
0.34507
0.26103
0.21062
0.17702
0.15303
0.13504
0.12106
0.10988
0.10073
0.09311
0.08667
0.08116
0.07638
0.07220
0.06852
0.06524
0.06232
0.05969
0.05731
0.05516
0.05319
0.05139
0.04973
0.04820
0.04679
0.04548
0.04426
0.04313
0.03768
0.03497
0.03097
0.03017
0.02945
0.02846
0.02705
0.02454
0.02405
0.02332
0.02281
0.02215
0.02158
0.02125
0.02067
0.01999
0.01947
0.01907
0.01831
0.01778
0.01753
0.01750
0.01750
0.00000
0.96590
2.86447
5.66334
9.33099
13.83671
19.15057
25.24345
32.08698
39.65354
47.91623
56.84886
66.42596
76.62270
87.41495
98.77919
110.69257
123.13283
136.07832
149.50799
163.40134
177.73847
192.49999
207.66706
223.22138
239.14512
255.42098
272.03215
288.96226
306.19544
415.12498
492.01087
652.60539
693.70101
735.03220
797.33210
901.49545
1149.11809
1209.77384
1309.24819
1387.15838
1500.87981
1610.47158
1681.08862
1816.18525
2003.02686
2170.82384
2320.13512
2668.57763
3001.26781
3219.08332
3257.88395
3264.17380
0.00000
0.49566
0.98843
1.47832
1.96531
2.44941
2.93062
3.40895
3.88439
4.35695
4.82662
5.29341
5.75733
6.21836
6.67653
7.13182
7.58424
8.03379
8.48048
8.92431
9.36529
9.80341
10.23868
10.67111
11.10069
11.52744
11.95135
12.37243
12.79069
13.20613
15.63986
17.20665
20.20838
20.93167
21.64424
22.69310
24.38848
28.19476
29.08986
30.53289
31.64417
33.24089
34.75560
35.72112
37.54939
40.04685
42.26985
44.24036
48.85131
53.35183
56.44341
57.02676
57.12476
Time Value of Money Factors Discrete Compounding 1.75%
n
Uniform Series
1.75%
TABLE A-a-7
2.00%
Time Value of Money Factors Discrete Compounding 2.00%
478 Appendix A
TABLE A-a-8
Single Sums
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
80
84
90
96
100
108
120
132
144
180
240
360
480
600
Uniform Series
Gradient Series
To Find F
Given P
(F|P i %,n)
To Find P
Given F
(P|F i %,n)
To Find F
Given A
(F|A i %,n)
To Find A
Given F
(A|F i%,n)
To Find P
Given A
(P|A i%,n)
To Find A
Given P
(A|P i%,n)
To Find P
Given G
(P|G i%,n)
To Find A
Given G
(A|G i%,n)
1.02000
1.04040
1.06121
1.08243
1.10408
1.12616
1.14869
1.17166
1.19509
1.21899
1.24337
1.26824
1.29361
1.31948
1.34587
1.37279
1.40024
1.42825
1.45681
1.48595
1.51567
1.54598
1.57690
1.60844
1.64061
1.67342
1.70689
1.74102
1.77584
1.81136
2.03989
2.20804
2.58707
2.69159
2.80033
2.97173
3.28103
4.16114
4.41584
4.87544
5.27733
5.94313
6.69293
7.24465
8.48826
10.76516
13.65283
17.31509
35.32083
115.88874
1247.56113
1.3430E104
1.4458E105
0.98039
0.96117
0.94232
0.92385
0.90573
0.88797
0.87056
0.85349
0.83676
0.82035
0.80426
0.78849
0.77303
0.75788
0.74301
0.72845
0.71416
0.70016
0.68643
0.67297
0.65978
0.64684
0.63416
0.62172
0.60953
0.59758
0.58586
0.57437
0.56311
0.55207
0.49022
0.45289
0.38654
0.37153
0.35710
0.33650
0.30478
0.24032
0.22646
0.20511
0.18949
0.16826
0.14941
0.13803
0.11781
0.09289
0.07324
0.05775
0.02831
8.6290E203
8.0156E204
7.4459E205
6.9167E206
1.00000
2.02000
3.06040
4.12161
5.20404
6.30812
7.43428
8.58297
9.75463
10.94972
12.16872
13.41209
14.68033
15.97394
17.29342
18.63929
20.01207
21.41231
22.84056
24.29737
25.78332
27.29898
28.84496
30.42186
32.03030
33.67091
35.34432
37.05121
38.79223
40.56808
51.99437
60.40198
79.35352
84.57940
90.01641
98.58653
114.05154
158.05702
170.79177
193.77196
213.86661
247.15666
284.64666
312.23231
374.41288
488.25815
632.64148
815.75446
1716.04157
5744.43676
6.2328E104
6.7146E105
7.2289E106
1.00000
0.49505
0.32675
0.24262
0.19216
0.15853
0.13451
0.11651
0.10252
0.09133
0.08218
0.07456
0.06812
0.06260
0.05783
0.05365
0.04997
0.04670
0.04378
0.04116
0.03878
0.03663
0.03467
0.03287
0.03122
0.02970
0.02829
0.02699
0.02578
0.02465
0.01923
0.01656
0.01260
0.01182
0.01111
0.01014
8.7680E203
6.3268E203
5.8551E203
5.1607E203
4.6758E203
4.0460E203
3.5131E203
3.2027E203
2.6708E203
2.0481E203
1.5807E203
1.2259E203
5.8274E204
1.7408E204
1.6044E205
1.4893E206
1.3833E207
0.98039
1.94156
2.88388
3.80773
4.71346
5.60143
6.47199
7.32548
8.16224
8.98259
9.78685
10.57534
11.34837
12.10625
12.84926
13.57771
14.29187
14.99203
15.67846
16.35143
17.01121
17.65805
18.29220
18.91393
19.52346
20.12104
20.70690
21.28127
21.84438
22.39646
25.48884
27.35548
30.67312
31.42361
32.14495
33.17479
34.76089
37.98406
38.67711
39.74451
40.52552
41.58693
42.52943
43.09835
44.10951
45.35539
46.33776
47.11235
48.58440
49.56855
49.95992
49.99628
49.99965
1.02000
0.51505
0.34675
0.26262
0.21216
0.17853
0.15451
0.13651
0.12252
0.11133
0.10218
0.09456
0.08812
0.08260
0.07783
0.07365
0.06997
0.06670
0.06378
0.06116
0.05878
0.05663
0.05467
0.05287
0.05122
0.04970
0.04829
0.04699
0.04578
0.04465
0.03923
0.03656
0.03260
0.03182
0.03111
0.03014
0.02877
0.02633
0.02586
0.02516
0.02468
0.02405
0.02351
0.02320
0.02267
0.02205
0.02158
0.02123
0.02058
0.02017
0.02002
0.02000
0.02000
0.00000
0.96117
2.84581
5.61735
9.24027
13.68013
18.90349
24.87792
31.57197
38.95510
46.99773
55.67116
64.94755
74.79992
85.20213
96.12881
107.55542
119.45813
131.81388
144.60033
157.79585
171.37947
185.33090
199.63049
214.25924
229.19872
244.43113
259.93924
275.70639
291.71644
392.04045
461.99313
605.96572
642.36059
678.78489
733.35269
823.69753
1034.05570
1084.63929
1166.78677
1230.41912
1322.17008
1409.29734
1464.75275
1569.30251
1710.41605
1833.47151
1939.79497
2174.41310
2374.87999
2483.56794
2498.02683
2499.77521
0.00000
0.49505
0.98680
1.47525
1.96040
2.44226
2.92082
3.39608
3.86805
4.33674
4.80213
5.26424
5.72307
6.17862
6.63090
7.07990
7.52564
7.96811
8.40732
8.84328
9.27599
9.70546
10.13169
10.55468
10.97445
11.39100
11.80433
12.21446
12.62138
13.02512
15.38087
16.88850
19.75559
20.44198
21.11638
22.10572
23.69610
27.22341
28.04344
29.35718
30.36159
31.79292
33.13699
33.98628
35.57742
37.71142
39.56755
41.17381
44.75537
47.91102
49.71121
49.96426
49.99585
Appendix A 479
Single Sums
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
80
84
90
96
100
108
120
132
144
180
240
360
480
600
Gradient Series
To Find F
Given P
(F|P i %,n)
To Find P
Given F
(P|F i %,n)
To Find F
Given A
(F|A i %,n)
To Find A
Given F
(A|F i%,n)
To Find P
Given A
(P|A i%,n)
To Find A
Given P
(A|P i%,n)
To Find P
Given G
(P|G i%,n)
To Find A
Given G
(A|G i%,n)
1.03000
1.06090
1.09273
1.12551
1.15927
1.19405
1.22987
1.26677
1.30477
1.34392
1.38423
1.42576
1.46853
1.51259
1.55797
1.60471
1.65285
1.70243
1.75351
1.80611
1.86029
1.91610
1.97359
2.03279
2.09378
2.15659
2.22129
2.28793
2.35657
2.42726
2.89828
3.26204
4.13225
4.38391
4.65089
5.08215
5.89160
8.40002
9.17893
10.64089
11.97642
14.30047
17.07551
19.21863
24.34559
34.71099
49.48957
70.56029
204.50336
1204.85263
4.1822E104
1.4517E106
5.0389E107
0.97087
0.94260
0.91514
0.88849
0.86261
0.83748
0.81309
0.78941
0.76642
0.74409
0.72242
0.70138
0.68095
0.66112
0.64186
0.62317
0.60502
0.58739
0.57029
0.55368
0.53755
0.52189
0.50669
0.49193
0.47761
0.46369
0.45019
0.43708
0.42435
0.41199
0.34503
0.30656
0.24200
0.22811
0.21501
0.19677
0.16973
0.11905
0.10895
0.09398
0.08350
0.06993
0.05856
0.05203
0.04108
0.02881
0.02021
0.01417
4.8899E203
8.2998E204
2.3911E205
6.8886E207
1.9846E208
1.00000
2.03000
3.09090
4.18363
5.30914
6.46841
7.66246
8.89234
10.15911
11.46388
12.80780
14.19203
15.61779
17.08632
18.59891
20.15688
21.76159
23.41444
25.11687
26.87037
28.67649
30.53678
32.45288
34.42647
36.45926
38.55304
40.70963
42.93092
45.21885
47.57542
63.27594
75.40126
104.40840
112.79687
121.69620
136.07162
163.05344
246.66724
272.63086
321.36302
365.88054
443.34890
535.85019
607.28773
778.18627
1123.69957
1616.31893
2318.67634
6783.44532
4.0128E104
1.3940E106
4.8389E107
1.6796E109
1.00000
0.49261
0.32353
0.23903
0.18835
0.15460
0.13051
0.11246
0.09843
0.08723
0.07808
0.07046
0.06403
0.05853
0.05377
0.04961
0.04595
0.04271
0.03981
0.03722
0.03487
0.03275
0.03081
0.02905
0.02743
0.02594
0.02456
0.02329
0.02211
0.02102
0.01580
0.01326
9.5778E203
8.8655E203
8.2172E203
7.3491E203
6.1330E203
4.0540E203
3.6680E203
3.1117E203
2.7331E203
2.2556E203
1.8662E203
1.6467E203
1.2850E203
8.8992E204
6.1869E204
4.3128E204
1.4742E204
2.4920E205
7.1735E207
2.0666E208
5.9537E210
0.97087
1.91347
2.82861
3.71710
4.57971
5.41719
6.23028
7.01969
7.78611
8.53020
9.25262
9.95400
10.63496
11.29607
11.93794
12.56110
13.16612
13.75351
14.32380
14.87747
15.41502
15.93692
16.44361
16.93554
17.41315
17.87684
18.32703
18.76411
19.18845
19.60044
21.83225
23.11477
25.26671
25.72976
26.16624
26.77443
27.67556
29.36509
29.70183
30.20076
30.55009
31.00241
31.38122
31.59891
31.96416
32.37302
32.65979
32.86092
33.17034
33.30567
33.33254
33.33331
33.33333
1.03000
0.52261
0.35353
0.26903
0.21835
0.18460
0.16051
0.14246
0.12843
0.11723
0.10808
0.10046
0.09403
0.08853
0.08377
0.07961
0.07595
0.07271
0.06981
0.06722
0.06487
0.06275
0.06081
0.05905
0.05743
0.05594
0.05456
0.05329
0.05211
0.05102
0.04580
0.04326
0.03958
0.03887
0.03822
0.03735
0.03613
0.03405
0.03367
0.03311
0.03273
0.03226
0.03187
0.03165
0.03129
0.03089
0.03062
0.03043
0.03015
0.03002
0.03000
0.03000
0.03000
0.00000
0.94260
2.77288
5.43834
8.88878
13.07620
17.95475
23.48061
29.61194
36.30879
43.53300
51.24818
59.41960
68.01413
77.00020
86.34770
96.02796
106.01367
116.27882
126.79866
137.54964
148.50939
159.65661
170.97108
182.43362
194.02598
205.73090
217.53197
229.41367
241.36129
313.70284
361.74994
455.02547
477.48033
499.51915
531.74111
583.05261
693.12255
717.69785
756.08652
784.54337
823.63021
858.63770
879.85405
917.60126
963.86347
999.75206
1027.33721
1076.33852
1103.54910
1110.79761
1111.09932
1111.11069
0.00000
0.49261
0.98030
1.46306
1.94090
2.41383
2.88185
3.34496
3.80318
4.25650
4.70494
5.14850
5.58720
6.02104
6.45004
6.87421
7.29357
7.70812
8.11788
8.52286
8.92309
9.31858
9.70934
10.09540
10.47677
10.85348
11.22554
11.59298
11.95582
12.31407
14.36878
15.65016
18.00890
18.55751
19.09021
19.86004
21.06742
23.60363
24.16342
25.03534
25.68056
26.56665
27.36151
27.84445
28.70719
29.77366
30.61110
31.26319
32.44883
33.13397
33.32473
33.33300
33.33332
Time Value of Money Factors Discrete Compounding 3.00%
n
Uniform Series
3.00%
TABLE A-a-9
4.00%
Time Value of Money Factors Discrete Compounding 4.00%
480 Appendix A
TABLE A-a-10
Single Sums
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
80
84
90
96
100
108
120
132
144
180
240
360
480
600
Uniform Series
To Find F
Given P
(F|P i %,n)
To Find P
Given F
(P|F i%,n)
To Find F
Given A
(F|A i%,n)
To Find A
Given F
(A|F i%,n)
1.04000
1.08160
1.12486
1.16986
1.21665
1.26532
1.31593
1.36857
1.42331
1.48024
1.53945
1.60103
1.66507
1.73168
1.80094
1.87298
1.94790
2.02582
2.10685
2.19112
2.27877
2.36992
2.46472
2.56330
2.66584
2.77247
2.88337
2.99870
3.11865
3.24340
4.10393
4.80102
6.57053
7.10668
7.68659
8.64637
10.51963
16.84226
18.94525
23.04980
26.96500
34.11933
43.17184
50.50495
69.11951
110.66256
177.17433
283.66180
1164.12891
1.2246E104
1.3552E106
1.4997E108
1.6596E110
0.96154
0.92456
0.88900
0.85480
0.82193
0.79031
0.75992
0.73069
0.70259
0.67556
0.64958
0.62460
0.60057
0.57748
0.55526
0.53391
0.51337
0.49363
0.47464
0.45639
0.43883
0.42196
0.40573
0.39012
0.37512
0.36069
0.34682
0.33348
0.32065
0.30832
0.24367
0.20829
0.15219
0.14071
0.13010
0.11566
0.09506
0.05937
0.05278
0.04338
0.03709
0.02931
0.02316
0.01980
0.01447
9.0365E203
5.6442E203
3.5253E203
8.5901E204
8.1658E205
7.3790E207
6.6680E209
6.0255E211
1.00000
2.04000
3.12160
4.24646
5.41632
6.63298
7.89829
9.21423
10.58280
12.00611
13.48635
15.02581
16.62684
18.29191
20.02359
21.82453
23.69751
25.64541
27.67123
29.77808
31.96920
34.24797
36.61789
39.08260
41.64591
44.31174
47.08421
49.96758
52.96629
56.08494
77.59831
95.02552
139.26321
152.66708
167.16472
191.15917
237.99069
396.05656
448.63137
551.24498
649.12512
827.98333
1054.29603
1237.62370
1702.98772
2741.56402
4404.35813
7066.54508
2.9078E104
3.0613E105
3.3880E107
3.7492E109
4.1490E111
1.00000
0.49020
0.32035
0.23549
0.18463
0.15076
0.12661
0.10853
0.09449
0.08329
0.07415
0.06655
0.06014
0.05467
0.04994
0.04582
0.04220
0.03899
0.03614
0.03358
0.03128
0.02920
0.02731
0.02559
0.02401
0.02257
0.02124
0.02001
0.01888
0.01783
0.01289
0.01052
7.1806E203
6.5502E203
5.9821E203
5.2312E203
4.2018E203
2.5249E203
2.2290E203
1.8141E203
1.5405E203
1.2078E203
9.4850E204
8.0800E204
5.8720E204
3.6476E204
2.2705E204
1.4151E204
3.4390E205
3.2666E206
2.9516E208
2.6672E210
2.4102E212
To Find P
Given A
(P|A i%,n)
0.96154
1.88609
2.77509
3.62990
4.45182
5.24214
6.00205
6.73274
7.43533
8.11090
8.76048
9.38507
9.98565
10.56312
11.11839
11.65230
12.16567
12.65930
13.13394
13.59033
14.02916
14.45112
14.85684
15.24696
15.62208
15.98277
16.32959
16.66306
16.98371
17.29203
18.90828
19.79277
21.19513
21.48218
21.74758
22.10861
22.62349
23.51564
23.68041
23.91539
24.07287
24.26728
24.42092
24.50500
24.63831
24.77409
24.85890
24.91187
24.97852
24.99796
24.99998
25.00000
25.00000
Gradient Series
To Find A
Given P
(A|P i%,n)
To Find P
Given G
(P|G i%,n)
To Find A
Given G
(A|G i%,n)
1.04000
0.53020
0.36035
0.27549
0.22463
0.19076
0.16661
0.14853
0.13449
0.12329
0.11415
0.10655
0.10014
0.09467
0.08994
0.08582
0.08220
0.07899
0.07614
0.07358
0.07128
0.06920
0.06731
0.06559
0.06401
0.06257
0.06124
0.06001
0.05888
0.05783
0.05289
0.05052
0.04718
0.04655
0.04598
0.04523
0.04420
0.04252
0.04223
0.04181
0.04154
0.04121
0.04095
0.04081
0.04059
0.04036
0.04023
0.04014
0.04003
0.04000
0.04000
0.04000
0.04000
0.00000
0.92456
2.70255
5.26696
8.55467
12.50624
17.06575
22.18058
27.80127
33.88135
40.37716
47.24773
54.45462
61.96179
69.73550
77.74412
85.95809
94.34977
102.89333
111.56469
120.34136
129.20242
138.12840
147.10119
156.10400
165.12123
174.13846
183.14235
192.12059
201.06183
253.40520
286.53030
347.24455
361.16385
374.56381
393.68897
422.99665
481.01697
493.04083
511.11614
523.94309
540.73692
554.93118
563.12487
576.89491
592.24276
602.84668
610.10550
620.59757
624.45902
624.99290
624.99992
625.00000
0.00000
0.49020
0.97386
1.45100
1.92161
2.38571
2.84332
3.29443
3.73908
4.17726
4.60901
5.03435
5.45329
5.86586
6.27209
6.67200
7.06563
7.45300
7.83416
8.20912
8.57794
8.94065
9.29729
9.64790
9.99252
10.33120
10.66399
10.99092
11.31205
11.62743
13.40181
14.47651
16.38322
16.81225
17.22324
17.80704
18.69723
20.45519
20.82062
21.37185
21.76488
22.28255
22.72360
22.98000
23.41455
23.90573
24.25074
24.49056
24.84525
24.98040
24.99973
25.00000
25.00000
Appendix A 481
Single Sums
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
80
84
90
96
100
108
120
132
144
180
240
360
480
600
To Find F
Given P
(F|P i %,n)
To Find P
Given F
(P|F i%,n)
To Find F
Given A
(F|A i %,n)
To Find A
Given F
(A|F i%,n)
1.05000
1.10250
1.15763
1.21551
1.27628
1.34010
1.40710
1.47746
1.55133
1.62889
1.71034
1.79586
1.88565
1.97993
2.07893
2.18287
2.29202
2.40662
2.52695
2.65330
2.78596
2.92526
3.07152
3.22510
3.38635
3.55567
3.73346
3.92013
4.11614
4.32194
5.79182
7.03999
10.40127
11.46740
12.64281
14.63563
18.67919
33.54513
38.83269
49.56144
60.24224
80.73037
108.18641
131.50126
194.28725
348.91199
626.59580
1125.27603
6517.39184
1.2174E105
4.2476E107
1.4821E110
5.1711E112
0.95238
0.90703
0.86384
0.82270
0.78353
0.74622
0.71068
0.67684
0.64461
0.61391
0.58468
0.55684
0.53032
0.50507
0.48102
0.45811
0.43630
0.41552
0.39573
0.37689
0.35894
0.34185
0.32557
0.31007
0.29530
0.28124
0.26785
0.25509
0.24295
0.23138
0.17266
0.14205
0.09614
0.08720
0.07910
0.06833
0.05354
0.02981
0.02575
0.02018
0.01660
0.01239
9.2433E203
7.6045E203
5.1470E203
2.8661E203
1.5959E203
8.8867E204
1.5344E204
8.2143E206
2.3542E208
6.7474E211
1.9338E213
1.00000
2.05000
3.15250
4.31013
5.52563
6.80191
8.14201
9.54911
11.02656
12.57789
14.20679
15.91713
17.71298
19.59863
21.57856
23.65749
25.84037
28.13238
30.53900
33.06595
35.71925
38.50521
41.43048
44.50200
47.72710
51.11345
54.66913
58.40258
62.32271
66.43885
95.83632
120.79977
188.02539
209.34800
232.85617
272.71262
353.58372
650.90268
756.65372
971.22882
1184.84483
1594.60730
2143.72821
2610.02516
3865.74499
6958.23971
1.2512E104
2.2486E104
1.3033E105
2.4348E106
8.4953E108
2.9641E111
1.0342E114
1.00000
0.48780
0.31721
0.23201
0.18097
0.14702
0.12282
0.10472
0.09069
0.07950
0.07039
0.06283
0.05646
0.05102
0.04634
0.04227
0.03870
0.03555
0.03275
0.03024
0.02800
0.02597
0.02414
0.02247
0.02095
0.01956
0.01829
0.01712
0.01605
0.01505
0.01043
8.2782E203
5.3184E203
4.7767E203
4.2945E203
3.6669E203
2.8282E203
1.5363E203
1.3216E203
1.0296E203
8.4399E204
6.2711E204
4.6648E204
3.8314E204
2.5868E204
1.4371E204
7.9924E205
4.4473E205
7.6730E206
4.1072E207
1.1771E209
3.3737E212
9.6692E215
To Find P
Given A
(P|A i%,n)
0.95238
1.85941
2.72325
3.54595
4.32948
5.07569
5.78637
6.46321
7.10782
7.72173
8.30641
8.86325
9.39357
9.89864
10.37966
10.83777
11.27407
11.68959
12.08532
12.46221
12.82115
13.16300
13.48857
13.79864
14.09394
14.37519
14.64303
14.89813
15.14107
15.37245
16.54685
17.15909
18.07716
18.25593
18.41807
18.63347
18.92929
19.40379
19.48497
19.59646
19.66801
19.75226
19.81513
19.84791
19.89706
19.94268
19.96808
19.98223
19.99693
19.99984
20.00000
20.00000
20.00000
Gradient Series
To Find A
Given P
(A|P i%,n)
To Find P
Given G
(P|G i%,n)
To Find A
Given G
(A|G i%,n)
1.05000
0.53780
0.36721
0.28201
0.23097
0.19702
0.17282
0.15472
0.14069
0.12950
0.12039
0.11283
0.10646
0.10102
0.09634
0.09227
0.08870
0.08555
0.08275
0.08024
0.07800
0.07597
0.07414
0.07247
0.07095
0.06956
0.06829
0.06712
0.06605
0.06505
0.06043
0.05828
0.05532
0.05478
0.05429
0.05367
0.05283
0.05154
0.05132
0.05103
0.05084
0.05063
0.05047
0.05038
0.05026
0.05014
0.05008
0.05004
0.05001
0.05000
0.05000
0.05000
0.05000
0.00000
0.90703
2.63470
5.10281
8.23692
11.96799
16.23208
20.96996
26.12683
31.65205
37.49884
43.62405
49.98791
56.55379
63.28803
70.15970
77.14045
84.20430
91.32751
98.48841
105.66726
112.84611
120.00868
127.14024
134.22751
141.25852
148.22258
155.11011
161.91261
168.62255
206.62370
229.54518
269.24673
277.91478
286.10125
297.51040
314.34316
345.14853
351.07215
359.64605
365.47273
372.74879
378.55553
381.74922
386.82363
391.97505
395.14839
397.08516
399.38626
399.95729
399.99982
400.00000
400.00000
0.00000
0.48780
0.96749
1.43905
1.90252
2.35790
2.80523
3.24451
3.67579
4.09909
4.51444
4.92190
5.32150
5.71329
6.09731
6.47363
6.84229
7.20336
7.55690
7.90297
8.24164
8.57298
8.89706
9.21397
9.52377
9.82655
10.12240
10.41138
10.69360
10.96914
12.48719
13.37747
14.89431
15.22326
15.53372
15.96645
16.60618
17.78769
18.01759
18.35260
18.58209
18.87120
19.10436
19.23372
19.44125
19.65509
19.78900
19.87192
19.97238
19.99803
19.99999
20.00000
20.00000
Time Value of Money Factors Discrete Compounding 5.00%
n
Uniform Series
5.00%
TABLE A-a-11
6.00%
Time Value of Money Factors Discrete Compounding 6.00%
482 Appendix A
TABLE A-a-12
Single Sums
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
80
84
90
96
100
108
120
132
144
180
240
360
480
600
Uniform Series
Gradient Series
To Find F
Given P
(F|P i %,n)
To Find P
Given F
(P|F i%,n)
To Find F
Given A
(F|A i%,n)
To Find A
Given F
(A|F i%,n)
To Find P
Given A
(P|A i%,n)
To Find A
Given P
(A|P i%,n)
To Find P
Given G
(P|G i%,n)
To Find A
Given G
(A|G i%,n)
1.06000
1.12360
1.19102
1.26248
1.33823
1.41852
1.50363
1.59385
1.68948
1.79085
1.89830
2.01220
2.13293
2.26090
2.39656
2.54035
2.69277
2.85434
3.02560
3.20714
3.39956
3.60354
3.81975
4.04893
4.29187
4.54938
4.82235
5.11169
5.41839
5.74349
8.14725
10.28572
16.39387
18.42015
20.69689
24.65032
32.98769
66.37772
79.05692
105.79599
133.56500
189.46451
268.75903
339.30208
540.79597
1088.18775
2189.64755
4406.00107
3.5897E104
1.1842E106
1.2886E109
1.4022E112
1.5259E115
0.94340
0.89000
0.83962
0.79209
0.74726
0.70496
0.66506
0.62741
0.59190
0.55839
0.52679
0.49697
0.46884
0.44230
0.41727
0.39365
0.37136
0.35034
0.33051
0.31180
0.29416
0.27751
0.26180
0.24698
0.23300
0.21981
0.20737
0.19563
0.18456
0.17411
0.12274
0.09722
0.06100
0.05429
0.04832
0.04057
0.03031
0.01507
0.01265
9.4522E203
7.4870E203
5.2780E203
3.7208E203
2.9472E203
1.8491E203
9.1896E204
4.5669E204
2.2696E204
2.7858E205
8.4449E207
7.7605E210
7.1316E213
6.5536E216
1.00000
2.06000
3.18360
4.37462
5.63709
6.97532
8.39384
9.89747
11.49132
13.18079
14.97164
16.86994
18.88214
21.01507
23.27597
25.67253
28.21288
30.90565
33.75999
36.78559
39.99273
43.39229
46.99583
50.81558
54.86451
59.15638
63.70577
68.52811
73.63980
79.05819
119.12087
154.76197
256.56453
290.33590
328.28142
394.17203
533.12818
1089.62859
1300.94868
1746.59989
2209.41674
3141.07519
4462.65050
5638.36806
8996.59954
1.8120E104
3.6477E104
7.3417E104
5.9826E105
1.9736E107
2.1476E110
2.3370E113
2.5431E116
1.00000
0.48544
0.31411
0.22859
0.17740
0.14336
0.11914
0.10104
0.08702
0.07587
0.06679
0.05928
0.05296
0.04758
0.04296
0.03895
0.03544
0.03236
0.02962
0.02718
0.02500
0.02305
0.02128
0.01968
0.01823
0.01690
0.01570
0.01459
0.01358
0.01265
8.3948E203
6.4615E203
3.8977E203
3.4443E203
3.0462E203
2.5370E203
1.8757E203
9.1774E204
7.6867E204
5.7254E204
4.5261E204
3.1836E204
2.2408E204
1.7736E204
1.1115E204
5.5188E205
2.7414E205
1.3621E205
1.6715E206
5.0669E208
4.6563E211
4.2789E214
3.9322E217
0.94340
1.83339
2.67301
3.46511
4.21236
4.91732
5.58238
6.20979
6.80169
7.36009
7.88687
8.38384
8.85268
9.29498
9.71225
10.10590
10.47726
10.82760
11.15812
11.46992
11.76408
12.04158
12.30338
12.55036
12.78336
13.00317
13.21053
13.40616
13.59072
13.76483
14.62099
15.04630
15.65003
15.76186
15.86139
15.99054
16.16143
16.41558
16.45585
16.50913
16.54188
16.57870
16.60465
16.61755
16.63585
16.65135
16.65906
16.66288
16.66620
16.66665
16.66667
16.66667
16.66667
1.06000
0.54544
0.37411
0.28859
0.23740
0.20336
0.17914
0.16104
0.14702
0.13587
0.12679
0.11928
0.11296
0.10758
0.10296
0.09895
0.09544
0.09236
0.08962
0.08718
0.08500
0.08305
0.08128
0.07968
0.07823
0.07690
0.07570
0.07459
0.07358
0.07265
0.06839
0.06646
0.06390
0.06344
0.06305
0.06254
0.06188
0.06092
0.06077
0.06057
0.06045
0.06032
0.06022
0.06018
0.06011
0.06006
0.06003
0.06001
0.06000
0.06000
0.06000
0.06000
0.06000
0.00000
0.89000
2.56924
4.94552
7.93455
11.45935
15.44969
19.84158
24.57677
29.60232
34.87020
40.33686
45.96293
51.71284
57.55455
63.45925
69.40108
75.35692
81.30615
87.23044
93.11355
98.94116
104.70070
110.38121
115.97317
121.46842
126.85999
132.14200
137.30959
142.35879
170.03866
185.95682
212.03505
217.45738
222.48229
229.32225
239.04279
255.51462
258.45274
262.54931
265.21627
268.39461
270.79093
272.04706
273.93570
275.68459
276.64619
277.17002
277.68647
277.77417
277.77777
277.77778
277.77778
0.00000
0.48544
0.96118
1.42723
1.88363
2.33040
2.76758
3.19521
3.61333
4.02201
4.42129
4.81126
5.19198
5.56352
5.92598
6.27943
6.62397
6.95970
7.28673
7.60515
7.91508
8.21662
8.50991
8.79506
9.07220
9.34145
9.60294
9.85681
10.10319
10.34221
11.62977
12.35898
13.54854
13.79643
14.02666
14.34112
14.79095
15.56537
15.70583
15.90328
16.03302
16.18912
16.30814
16.37107
16.46659
16.55629
16.60636
16.63398
16.66165
16.66646
16.66667
16.66667
16.66667
Appendix A 483
Single Sums
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
80
84
90
96
100
108
120
132
144
180
240
360
480
600
Gradient Series
To Find F
Given P
(F|P i %,n)
To Find P
Given F
(P|F i %,n)
To Find F
Given A
(F|A i %,n)
To Find A
Given F
(A|F i%,n)
To Find P
Given A
(P|A i%,n)
To Find A
Given P
(A|P i%,n)
To Find P
Given G
(P|G i%,n)
To Find A
Given G
(A|G i%,n)
1.07000
1.14490
1.22504
1.31080
1.40255
1.50073
1.60578
1.71819
1.83846
1.96715
2.10485
2.25219
2.40985
2.57853
2.75903
2.95216
3.15882
3.37993
3.61653
3.86968
4.14056
4.43040
4.74053
5.07237
5.42743
5.80735
6.21387
6.64884
7.11426
7.61226
11.42394
14.97446
25.72891
29.45703
33.72535
41.31500
57.94643
130.50646
159.87602
224.23439
293.92554
441.10298
661.97663
867.71633
1490.89820
3357.78838
7562.38275
1.7032E104
1.9457E105
1.1275E107
3.7858E110
1.2712E114
4.2684E117
0.93458
0.87344
0.81630
0.76290
0.71299
0.66634
0.62275
0.58201
0.54393
0.50835
0.47509
0.44401
0.41496
0.38782
0.36245
0.33873
0.31657
0.29586
0.27651
0.25842
0.24151
0.22571
0.21095
0.19715
0.18425
0.17220
0.16093
0.15040
0.14056
0.13137
0.08754
0.06678
0.03887
0.03395
0.02965
0.02420
0.01726
7.6625E203
6.2548E203
4.4596E203
3.4022E203
2.2670E203
1.5106E203
1.1525E203
6.7074E204
2.9782E204
1.3223E204
5.8713E205
5.1395E206
8.8694E208
2.6414E211
7.8666E215
2.3428E218
1.00000
2.07000
3.21490
4.43994
5.75074
7.15329
8.65402
10.25980
11.97799
13.81645
15.78360
17.88845
20.14064
22.55049
25.12902
27.88805
30.84022
33.99903
37.37896
40.99549
44.86518
49.00574
53.43614
58.17667
63.24904
68.67647
74.48382
80.69769
87.34653
94.46079
148.91346
199.63511
353.27009
406.52893
467.50497
575.92859
813.52038
1850.09222
2269.65742
3189.06268
4184.65058
6287.18543
9442.52329
1.2382E104
2.1284E104
4.7954E104
1.0802E105
2.4330E105
2.7796E106
1.6107E108
5.4083E111
1.8160E115
6.0977E118
1.00000
0.48309
0.31105
0.22523
0.17389
0.13980
0.11555
0.09747
0.08349
0.07238
0.06336
0.05590
0.04965
0.04434
0.03979
0.03586
0.03243
0.02941
0.02675
0.02439
0.02229
0.02041
0.01871
0.01719
0.01581
0.01456
0.01343
0.01239
0.01145
0.01059
6.7153E203
5.0091E203
2.8307E203
2.4598E203
2.1390E203
1.7363E203
1.2292E203
5.4051E204
4.4060E204
3.1357E204
2.3897E204
1.5905E204
1.0590E204
8.0765E205
4.6983E205
2.0853E205
9.2576E206
4.1102E206
3.5977E207
6.2086E209
1.8490E212
5.5066E216
1.6400E219
0.93458
1.80802
2.62432
3.38721
4.10020
4.76654
5.38929
5.97130
6.51523
7.02358
7.49867
7.94269
8.35765
8.74547
9.10791
9.44665
9.76322
10.05909
10.33560
10.59401
10.83553
11.06124
11.27219
11.46933
11.65358
11.82578
11.98671
12.13711
12.27767
12.40904
13.03521
13.33171
13.73047
13.80075
13.86212
13.93994
14.03918
14.17625
14.19636
14.22201
14.23711
14.25333
14.26413
14.26925
14.27613
14.28146
14.28383
14.28488
14.28564
14.28571
14.28571
14.28571
14.28571
1.07000
0.55309
0.38105
0.29523
0.24389
0.20980
0.18555
0.16747
0.15349
0.14238
0.13336
0.12590
0.11965
0.11434
0.10979
0.10586
0.10243
0.09941
0.09675
0.09439
0.09229
0.09041
0.08871
0.08719
0.08581
0.08456
0.08343
0.08239
0.08145
0.08059
0.07672
0.07501
0.07283
0.07246
0.07214
0.07174
0.07123
0.07054
0.07044
0.07031
0.07024
0.07016
0.07011
0.07008
0.07005
0.07002
0.07001
0.07000
0.07000
0.07000
0.07000
0.07000
0.07000
0.00000
0.87344
2.50603
4.79472
7.64666
10.97838
14.71487
18.78894
23.14041
27.71555
32.46648
37.35061
42.33018
47.37181
52.44605
57.52707
62.59226
67.62195
72.59910
77.50906
82.33932
87.07930
91.72013
96.25450
100.67648
104.98137
109.16556
113.22642
117.16218
120.97182
141.19902
152.29277
169.49812
172.90512
176.00368
180.12433
185.76774
194.63648
196.10351
198.07480
199.30463
200.70420
201.70162
202.20008
202.90990
203.51031
203.80529
203.94887
204.06737
204.08131
204.08163
204.08163
204.08163
0.00000
0.48309
0.95493
1.41554
1.86495
2.30322
2.73039
3.14654
3.55174
3.94607
4.32963
4.70252
5.06484
5.41673
5.75829
6.08968
6.41102
6.72247
7.02418
7.31631
7.59901
7.87247
8.13685
8.39234
8.63910
8.87733
9.10722
9.32894
9.54270
9.74868
10.83213
11.42335
12.34467
12.52868
12.69673
12.92146
13.23209
13.72976
13.81365
13.92735
13.99895
14.08122
14.14047
14.17034
14.21323
14.24997
14.26826
14.27726
14.28479
14.28569
14.28571
14.28571
14.28571
Time Value of Money Factors Discrete Compounding 7.00%
n
Uniform Series
7.00%
TABLE A-a-13
8.00%
Time Value of Money Factors Discrete Compounding 8.00%
484 Appendix A
TABLE A-a-14
Single Sums
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
36
40
48
50
52
55
60
72
75
80
84
90
96
100
108
120
132
144
180
240
360
480
600
Uniform Series
Gradient Series
To Find F
Given P
(F|P i %,n)
To Find P
Given F
(P|F i%,n)
To Find F
Given A
(F|A i%,n)
To Find A
Given F
(A|F i%,n)
To Find P
Given A
(P|A i%,n)
To Find A
Given P
(A|P i%,n)
To Find P
Given G
(P|G i%,n)
To Find A
Given G
(A|G i%,n)
1.08000
1.16640
1.25971
1.36049
1.46933
1.58687
1.71382
1.85093
1.99900
2.15892
2.33164
2.51817
2.71962
2.93719
3.17217
3.42594
3.70002
3.99602
4.31570
4.66096
5.03383
5.43654
5.87146
6.34118
6.84848
7.39635
7.98806
8.62711
9.3