O堀䘀O刀䐀
Fourth Canadian Edition
E GINEERIN
ECONOMIC
ANA � S
NEWNAN
JONES
WHITTAKER
ESCHENBACH
LAVELLE
Fourth Canadian Edition
ENGINEERING
ECONOMIC
ANALYSIS
NEWNAN
JONES
WHIT吀䄀KER
ESCHENBACH
LAVELLE
O堀䘀O刀䐀
UNIVERSITY PRESS
O堀䘀O刀䐀
UNIVERSITY PRESS
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Library and Archives Canada Cataloguing in Publication
Newnan, Donald G., author
Engineering economic analysis / Donald G. Newnan, John Jones, John
Whittaker, Ted G. Eschenbach, Jerome P. Lavelle. - Fourth Canadian edition.
Includes bibliographical references and index.
Issued in print and electronic 昀漀rmats.
ISBN 978-0-19-902511-4 (hardcover).-ISBN 978-0-19-902522-0 (PDF)
1. Engineering economy-Textbooks. 2. Textbooks. I. Jones, John (John
Dewey), author II. Whittaker, J. D., author III. Eschenbach, Ted, author
IV. Lavelle, Jerome P., author V. Title.
TA177.4.N486 2018
658.15
C2017-905532-1
C2017-905533-X
Cover image: © Ravil Say昀甀llin/123RF
Cover and interior design: Sherill Chapman
Ox昀漀rd University Press is committed to our environment.
Wherever possible, our books are printed on paper which comes 昀爀om
responsible sources.
Printed and bound in Canada
1 2 3 4 - 21 20 19 18
Contents
Preface and Acknowledgements
ix
1 Economic Decisions, Engineering Costs,
and Cost Estimating 2
A Sea of Problems
4
Simple Problems
4
Intermediate Problems
Complex Problems
4
4
The Role of Engineering Economic Analysis
5
Examples of Engineering Economic Analysis
5
The Decision-Making Process
Rational Decision Making
Ethics
6
6
12
Ethical Dimensions in Engineering Decision Making
12
Importance of Ethics in Engineering and Engineering Economy
Engineering Decision Making for Current Costs
Engineering Costs
17
Fixed, Variable, Marginal, and Average Costs
Sunk Costs
14
14
17
19
Opportunity Costs
20
Recurring and Non-Recurring Costs
Incremental Costs
21
Cash Costs versus Book Costs
Life-Cycle Costs
Cost Estimating
21
22
22
23
Types of Estimate
23
Dif昀椀culties in Estimation
Estimating Models
Per-Unit Model
24
25
25
Segmenting Model 25
Cost Indexes
25
Power-Sizing Model 27
Triangulation
28
Improvement and the Learning Curve
Estimating Bene昀椀ts
Cash Flow Diagrams
31
31
Categories of Cash Flow
32
Drawing a Cash Flow Diagram
Summary
32
Problems
34
28
32
2 Accounting and Engineering Economy 50
The Role of Accounting
52
Accounting for Business Transactions
The Balance Sheet
Assets
53
Liabilities
54
53
52
iv
Contents
Equity
54
Financial Ratios Derived from Balance Sheet Data
The Income Statement
55
55
57
Financial Ratios Derived from Income Statement Data
Linking the Balance Sheet, Income Statement, and Capital Transactions
Traditional Cost Accounting
59
Direct and Indirect Costs
59
Indirect Cost Allocation
59
Problems with Traditional Cost Accounting
Other Problems to Watch For
Summary
61
Problems
62
61
3 Interest and Equivalence
Computing Cash Flows
67
70
Compound Interest
Repaying a Debt
Equivalence
66
69
Time Value of Money
Simple Interest
59
70
71
73
Single- Payment Compound Interest Formulas
Nominal and E昀昀ective Interest
Continuous Compounding
75
79
83
Single Payment Interest Factors: Continuous Compounding
Equivalence and Sustainability
Summary
86
Problems
87
86
4 Equivalence for Repeated Cash Flows
Uniform Series Compound Interest Formulas
96
Cash Flows That Do Not Match Basic Patterns
102
Relationships between Compound Interest Factors
The Interest Rate Viewed as Fog
Arithmetic Gradient
10 6
108
108
Reality and the Assumed Uniformity of a, G, and g
114
115
When Compounding Period and Payment Period Di昀昀er
118
Uniform Payment Series: Continuous Compounding at
Normal Rate r per Period
Summary
122
Problems
125
120
5 Present Worth Analysis
140
Assumptions in Solving Economic Analysis Problems
End-of-Year Convention 142
Viewpoint of Economic Analysis Studies
Borrowed Money Viewpoint
Income Taxes
143
94
107
Derivation of Arithmetic Gradient Factors
Geometric Gradient
83
143
142
142
57
Contents
Effect of Inflation and Deflation
Economic Criteria
143
143
Present Worth Techniques
143
1. Useful Lives Equal to the Analysis Period
144
2. Useful Lives Different from the Analysis Period
3. In昀椀nite Analysis Period: Capitalized Cost
145
148
Multiple Alternatives 151
Bond Pricing
153
Future Worth Analysis
Summary
154
156
6 Annual Cash Flow Analysis 176
Problems
156
Annual Cash Flow Calculations
178
Converting a Present Cost to an Annual Cost
Treatment of Salvage Value
Annual Cash Flow Analysis
Analysis Period
178
179
182
182
Analysis Period for a Common Multiple of Alternative Lives
Analysis Period for a Continuing Requirement
In昀椀nite Analysis Period
185
186
187
Building an Amortization Schedule
Types of Mortgage Available
Equity
183
184
Some Other Analysis Period
Mortgages in Canada
183
188
188
Mortgage Interest Rate
Summary
189
Problems
189
188
7 Rate of Return Analysis 200
202
Minimum Attractive Rate of Return
Internal Rate of Return
203
Calculating Rate of Return
204
Plot of NPW versus Interest Rate i
207
Interest Rates When There Are Fees or Discounts
Incremental Analysis
Analysis Period
216
Sensitivity Analysis
217
A Second Pitfall
Summary
223
Problems
224
209
212
220
8 Bene昀椀t-Cost Ratio and Other Analysis Techniques 242
Public Sector Investment Objectives
Viewpoint for Analysis
244
245
Selecting an Interest Rate
247
No Time-Value-of-Money Concept
Cost-of-Capital Concept
247
247
v
vi
Contents
Opportunity-Cost Concept 247
Recommended Concept 248
The Bene昀椀t-Cost Ratio 248
Incremental Benefit-Cost Analysis 249
Elements of the Incremental Bene昀椀t-Cost Ratio Method 249
Other E昀昀ects of Public Projects 252
Project Financing 252
Duration of Project 253
Quantifying and Valuing Benefits and Costs 254
Project Politics 255
Payback Period 257
Summary 262
Problems 262
9 Selection of a Minimum Attractive
Rate of Return 272
Sources of Capital 275
Money Generated from the Firm's Operations 275
External Sources of Money 275
Choice of Sources of Funds 275
Cost of Funds 276
Cost of Borrowed Money 276
Cost of Capital 276
Investment Opportunities 277
Opportunity Cost 277
Choosing a Minimum Attractive Rate of Return 280
Adjusting MARR to Account for Risk and Uncertainty 281
Representative Values of MARR Used in Industry 281
Capital Budgeting, or Selecting the Best Projects 282
Summary 284
Problems 285
10 Uncertainty in Future Events 292
Break-Even Analysis 295
Optimistic and Pessimistic Estimates
Probability 298
Joint Probability Distributions 301
Expected Value 303
Economic Decision Trees 305
Risk 311
Risk versus Return 313
Simulation 315
Summary 318
Problems 319
296
11 Income, Depreciation, and Cash Flow 330
Taxation 332
Basic Aspects of Depreciation 333
Deterioration and Obsolescence 333
Contents
Depreciation and Expenses 334
Types of Property 335
Cost Basis 336
Calculating Depreciation 336
Depreciation Methods 337
Straight-Line Depreciation 337
Sum-of-Years'-Digits Depreciation 339
Declining-Balance Depreciation 341
Unit-of-Production Depreciation 342
Depreciation for Tax Purposes-Capital Cost Allowance 343
The 50% Rule, or Half-Year Convention 344
Accelerated CCA-50% Straight-Line 344
Calculating the CCA-Schedule 8 347
Calculating the CCA and the UCC (without Using Schedule 8)
Depreciation and Asset Disposal 351
Natural Resource Allowances 353
Percentage Depletion 353
Cost Depletion 355
Summary 355
Problems 356
12 After-Tax Cash Flows 366
A Partner in the Business 369
Calculation of Taxable Income 369
Taxable Income of Individuals 369
Corporate Income Taxes 373
Using CCA to Calculate Net Pro昀椀t 375
Accounting and Engineering Economy 376
Calculating Net Cash Flow 379
Acquiring and Disposing of Assets 381
Net Salvage of Land-Capital Gain 381
Net Salvage-Books-Closed Assumption 382
Capital Tax Factors and Books Open 383
Calculating After-Tax Present Worth 385
Working Capital Requirements 388
Initial Working Capital 389
Increasing Working Capital Requirement 389
Loan Financing 390
Loan with Equal Principal Repayment 391
Loan with Equal Annual Repayment 392
Estimating the After-Tax Rate of Return 393
Summary 394
Problems
395
13 Replacement Analysis 408
The Replacement Problem 410
Replacement Analysis Flow Chart 411
Minimum-Cost Life of a New Asset-The Challenger
Defender's Marginal Cost Data 414
412
351
vii
viii
Contents
Replacement Analysis Technique 1: Defender Marginal Costs Can Be Computed and Are
Increasing 416
Replacement Repeatability Assumptions 417
Replacement Analysis Technique 2: Defender Marginal Costs Can Be Computed and Are
Not Increasing 418
Replacement Analysis Technique 3: When Defender Marginal Cost Data Are Not
Available 422
Complications in Replacement Analysis 422
De昀椀ning Defender and Challenger First Costs 422
Repeatability Assumptions Not Acceptable 424
A Closer Look at Future Challengers 425
After-Tax Replacement Analysis 426
Marginal Costs on an After-Tax Basis 426
Minimum-Cost Life Problems 428
Spreadsheets and Replacement Analysis 430
Summary 432
Problems 433
14 Inflation and Price Change 444
Meaning and Effect of Inflation 446
How Does Inflation Happen? 446
Hyperinflation 447
De昀椀nitions for Considering Inflation in Engineering Economy 447
Analysis in Real Dollars versus Actual Dollars 454
Price Change with Indexes 456
What Is a Price Index? 456
Composite versus Commodity Indexes 458
How to Use Price Indexes in Engineering Economic Analysis 459
Cash Flows That Inflate at Different Rates 461
Different Inflation Rates per Period 463
Effect of Inflation on After-Tax Calculations 464
Inflation and the Buy/Lease Decision 467
Inflation and the Cost of Borrowed Money 470
Summary 471
Problems 472
15 Introduction to Project Management 482
Scientific Management 483
The Gantt Chart 484
Critical Path Method (CPM) 487
CPM and PERT 491
The Methods of Operations Research
498
Linear Programming and the Simplex Method 498
Queuing Theory 499
Summary 499
Problems 500
Glossary 504
References 508
Index 510
Preface and Acknowledgements
Engineering is the application of knowledge to develop practical and economical solutions
to the problems that con昀爀ont our civilization. Canada, with an area of 9,984,670 km2 ,
is the second-largest country in the world. But with a population of about 35,300,000,
Canada has less than 0.5% of the world's people. In population density we rank about
230th, with 3.9 people per square kilometre of land area. Because much of the country is
a cold and un昀漀rgiving place, Canadians h愀瘀e needed all the engineering skills they could
muster in order to exist in this 昀漀rmidable environment and, in doing so, to consistently
achieve a position in the top 昀椀ve on the United Nations list of the best countries to live in.
吀栀e history of this country is a history of engineering innovation and adaptability. 吀栀e
ingenious design of the birchbark canoe enabled First Nations people to travel thousands
of kilometres and develop trading routes that spanned the country. In the 1800s, engin­
eers 昀爀om Britain taught us how to construct canals to circumvent rapids; we built on that
knowledge, and with the opening of the St Lawrence Seaway in 1959, ocean-going ships of
commerce were able to bypass even so great an obstacle as Niagara Falls and penetrate the
heart of the continent.
吀栀e promise of a transcontinental railway was what brought British Columbia into the
Canadian Confederation, and again the engineers responded, completing the project in
1885 and opening the western provinces to settlement and development.
Canoes work well in the summer, but winter travel poses special challenges. Joseph­
Armand Bombardier attached a Ford engine to a sleigh, which, with continuing engineer­
ing re昀椀nements, evolved into their snowmobile, the Ski-doo, and its inventor's company
became the aircra昀琀 and light-rail-transit manu昀愀cturer Bombardier, Inc.
吀栀e need to communicate with northern settlements led Canadian engineers to de­
velop the Anik communications satellite. And our need to communicate, wherever we
happen to be, led a University of 圀愀terloo engineering student, Mike Lazaridis, to develop
the BlackBerry and create the company Research In Motion (RIM).
Engineering has always been a mixture of need, practicality, and economics. 吀栀e con­
cepts of ethics, sustainability, and environmental stewardship are now being added to that
mix. 吀栀is book introduces the economic decisions that accompany engineering design,
but it also includes questions that explore those other aspects. Each chapter begins with
a vignette describing a Canadian engineering challenge and inviting the reader to think
about the broader implications of engineering decisions.
Changes to the Fourth Edition
吀栀is edition has signi昀椀cant improvements in organization and coverage. Be昀漀re going
chapter by chapter, we' d like to point out a few global changes made to keep this text the
most current and most use昀甀l in today's courses:
•
•
•
•
•
A new discussion on the environmental impacts of engineering projects has been
added.
Additional Canadian examples are included throughout.
All units are now in SL
We have expanded coverage of money, inflation, and the Bank of Canada.
吀栀e majority of in-chapter Excel explanations have been removed and placed
online.
x
Preface and Acknowledgements
Numerous improvements to the content and wording h愀瘀e been made throughout the
text. 吀栀e text has been condensed and some print material, such as the compound interest
tables, has been placed online.
Changes 昀爀om the third edition include the 昀漀llowing:
•
•
•
•
•
•
䌀䐀
Chapters 1 and 2 have been combined and condensed into one chapter.
Chapter 8 has been eliminated, and some material has been distributed in other
chapters (in昀漀rmation on incremental analysis to Chapter 7, 昀甀ture worth content
to Chapter 5, bene昀椀t-cost ratio to Chapter 8, sensitivity and break-even content
to Chapter 10).
Chapter 17 has been moved up to become Chapter 2.
Chapter 16 has been moved up to become Chapter 8.
A new chapter on project management has been added.
In Chapter 12, the explanation 昀漀r calculation of taxes 昀漀r individuals has been
revised 昀漀r clarity, and the calculation of a昀琀er-tax cash 昀氀ows has been revised.
Icons
吀栀ese marginal icons appear throughout the text to highlight important
key concepts.
⠀　the answer to the problem can be 昀漀und on the companion website: www.oupcanada
吀栀ese icons appear in the Problems section at the end of each chapter to signi昀礀 that
.com/Newnan4Ce
A Special Word on Spreadsheets
In this edition, the authors have deliberately omitted any discussion of how to use any 昀漀rm
of spreadsheet so昀琀ware. Since there are many spreadsheet programs available, and since
their details change 昀爀equently, these details are better obtained 昀爀om online documen­
tation. Students will notice rather quickly that the use of a suitable spreadsheet program
can greatly reduce the labour involved in solving economic problems, and they should be
encouraged to do this.
Acknowledgements
吀栀is textbook is the result of the hard work of a lot of people 昀漀r over 35 years. I, along
with Ox昀漀rd University Press Canada, would like to grate昀甀lly acknowledge the 昀漀llowing
reviewers, as well as those who wished to remain anonymous:
Soha Eid Moussa, University of Guelph
Ti昀昀any Bayley, University of Waterloo
Scott Flemming, Dalhousie University
Michael Bennett, UOIT
Juan Pernia, Lakehead University
Michael H. Wang, University of Windsor
Nadine Ibrahim, University of Toronto
Preface and Acknowledgements
吀栀eir thought昀甀l comments and suggestions have helped to shape the 昀漀urth edition
of this text.
For the 昀漀urth Canadian edition, I would like to thank all the people at OUP
Canada-including Meg Patterson, Lauren Wing, Suzanne Clark, and Liana Spada-and
Doug Linzey of Agricola Communications. Finally, I want to thank my wife, Joan, and my
daughter and editorial critic Ali.
John Jones
Simon Fraser University, 2017
1
Economic Decisions,
Engineering Costs,
and Cost Estimating
Alternative Futures
There is no better place to start an engineering economics text than with a discussion of the future
of the car, for that question embodies many of the technology and economic factors involved.
Hybrids, plug-in electrics, turbocharged diesels, and advanced gasoline engines all compete
for the buyer's (and driver's) attention. In 2016 the lineup included the following sedans:
Model
Technolo最礀
Nissan Leaf
Chevrolet Volt
Toyota Prius
Honda Civic
BMW XS xDrive35d
Chevrolet Cruz
Chevrolet Cruze Eco
Toyota Yaris
Electric
Electric
Hybrid
Hybrid
Diesel
Diesel
Gasoline
Gasoline
Litres per 100 km
MSRP
2.4
3.8
4.7
5.8
6.2
5.1
5.6
4.6
$32,700
$40,000
$26,000
$16,000
$87,000
$24,670
$16,175
$16,365
sjo/iStock Photo
T
SL n
Alternative Futures
Choosing the right vehicle is a complex problem, and evaluating the choices requires much
more information than the price and fuel mileage. Will you be driving in the country or city, or a
mix? How far do you drive in a year? What will the future price of energy (gas, diesel, electricity)
be? How long will you keep the car before you sell it? What will the resale value be? What financing
terms do the sellers offer?
Electric vehicle technology has been slowly improving. Since 2012, the University of Waterloo
has been challenging teams of high-school students to design and build all-electric vehicles to
compete in a 12-volt and a 24-volt endurance race. In the 昀椀rst year of the competition, Bluevale
Secondary School won the 12-volt and 24-volt races, with 41 and 61 laps, respectively. In 2013,
they were narrowly edged out of first place in the 12-volt race by Resurrection Catholic Secondary
School, who completed 51.9 laps, but retained first place in the 24-volt race with an impressive
95 laps.
But the gasoline and diesel engineers have not been standing still. Future drivers will continue
to be o昀昀ered an array of technologies and will have to use engineering economic analysis to find
the correct choices.
QUESTIONS TO CONSIDER
1.
What marketplace dynamics stimulate or suppress the development of alternative-fuel vehicles? What
role, if any, does government have in these dynamics? What additional responsibilities should government
have?
2.
Develop a list of concerns and questions that consumers might have about the conversion to alternative-fuel
vehicles. Which are economic factors, and which are non-economic?
3.
Some environmentalists have proposed producing bio-fuels from crops as an alternative to fossil fuels.
What ethical questions might this involve?
4.
Power companies have a lot of extra capacity at night, and so there is a possibility of their offering
off-peak power at a reduced cost. This could be used by people with plug-in hybrids. How much in­
vestment would a power company have to make in smart circuits for a time-variable pricing policy to
be possible?
LEARNING OBJECTIVES
This chapter will help you
•
describe the economic decision-making process
•
choose appropriate economic criteria for different problems
•
describe common ethical issues in engineering economic decisions
•
de昀椀ne various cost concepts
•
provide speci昀椀c examples of how and why these engineering cost concepts are important
•
de昀椀ne engineering cost estimating
•
explain the three types of engineering estimates, as well as common difficulties encountered
in making them
•
use several common mathematical estimating models in cost estimating
•
discuss the e昀昀ect of the lea爀渀ing curve on cost estimates
•
draw cash flow diagrams to show project costs and benefits
3
4
Cha pter 1 Eco n o m i c Decisions, E n g i neeri ng Costs, a n d Cost Esti mati ng
KEY TERMS
average cost
book cost
cash cost
cash flow d iagram
esti mating by analogy
昀椀xed cost
forgone opportu n ity
incremental cost
learning curve
learn ing cu rve percentage
lea rn ing cu rve rate
life-cycle costing
marg i nal cost
model build i ng
non - recu rri ng cost
opportunity cost
out-of- pocket cost
overhead
per unit model
power-sizi ng model
recu rri ng cost
seg menting model
shadow price
sunk cost
tria ngulation
variable cost
A Sea of Problems
吀栀is book is about making decisions. It develops the tools 昀漀r analyzing and solving eco­
nomic problems commonly 昀愀ced by engineers.
Our emphasis is on solving problems that con昀爀ont 昀椀rms in the marketplace, but
many examples are problems students 昀愀ce in daily life. Let us start by looking at some of
these problems.
Sim ple Problems
Many problems are simple, and good solutions do not require much time or e昀昀ort.
•
•
•
•
Should I pay by cash or credit card?
Do I buy a semester parking pass or use the parking meters?
Should we replace a burned-out motor?
If we use three crates of an item a week, how many crates should we buy at a time?
I ntermed iate Problems
At the middle level of complexity, we 昀椀nd problems that are primarily economic.
•
•
•
•
•
Should I buy or lease my next car?
Which pieces of equipment should be chosen 昀漀r a new assembly line?
Should our o昀케ce upgrade to the newest version of our project management so昀琀ware?
Should I buy a one- or two-semester parking pass?
Should we buy a low-cost press requiring three operators, or a more expensive one
requiring only two operators?
Complex Problems
Complex problems represent a mixture of economic, political, and social and ethical elements.
•
吀栀e question of building pipelines to export Albertan oil overseas is an example
of a complex problem. Pipelines such as the proposed Northern Gateway 昀爀om
Alberta to Kitimat, BC, will bene昀椀t the Albertan economy and create jobs in the
Albertan oil industry. On the other hand, some environmentalists believe that the
risk of oil spills will endanger First Nations lands and the BC coastline. Globally,
The Role of Engineering Economic Analysis
•
•
burning Alberta's oil will increase global warming, with possible adverse conse­
quences worldwide.
吀栀e choice of a girl昀爀iend or a boy昀爀iend (who may later become a spouse) is ob­
viously complex. Economic analysis can be of little or no help.
吀栀e annual budget of a corporation is an allocation of resources, and all pro­
jects are evaluated economically. 吀栀e budget process is also heavily in昀氀uenced by
non-economic 昀漀rces such as power struggles, geographical balancing, and e昀昀ects
on individuals, programs, and pro昀椀ts.
The Role of Engineering Economic Analysis
Engineering economic analysis is most suitable 昀漀r intermediate problems and the eco­
nomic aspects of complex problems. Such problems have the 昀漀llowing characteristics:
1.
2.
3.
吀栀e problem is important eno最� to justify serious thought and e昀昀ort.
吀栀e problem can't be worked out in one's head-that is, a care昀甀l analysis requires
that we organize the problem and all the various consequences.
吀栀e problem has economic a猀瀀ects that are important in reaching a decision.
Vast numbers of problems having these three characteristics are encountered in the
business world and in people's personal lives, so engineering economic analysis is o昀琀en
required.
Examples of Engineering Economic Analysis
Engineering economic analysis 昀漀cuses on costs, revenues, and bene昀椀ts that occur at d昀케er­
ent times. For example, when a nuclear engineer designs a reactor, the construction costs
occur in the near 昀甀ture; bene昀椀ts to the users begin only when construction is 昀椀nished, but
continue 昀漀r a long time.
Engineering economic analysis is used to answer many di昀昀erent questions.
•
•
•
Which engineering projects are worthwhile? Has the mining or petroleum engin­
eer shown that the mineral or oil deposit is worth developing?
Which engineeri渀最 projects should have a h最턀er priori琀礀? Has the electrical engin­
eer shown that replacing the plant's step-down trans昀漀rmer is more urgent than
upgrading the power supply to the machine shop?
How should the engineering project be designed? Has the mechanical engineer
chosen the most economical size of motor? Has the so昀琀ware engineer calculated
the development time 昀漀r writing the next release of the so昀琀ware? Has the aero­
nautical engineer made the best trade-o昀昀s between (1) lighter materials that are
expensive to buy but result in a plane that is cheaper to 昀氀y and (2) he愀瘀ier materi­
als that are cheap to buy but result in a plane that is more expensive to fly?
.
.
Engineering economic analysis can also be used to answer questions that are person­
ally important.
•
How to achieve long-term 昀椀nancial goals: How much should you save each month
to buy a house, retire, or 昀甀nd a trip around the world? Is going to graduate school
a good investment?
0
5
6
Cha pter 1 Eco n o m i c Decisions, E n g i neeri ng Costs, a n d Cost Esti mati ng
•
•
How to compare d昀케erent ways to 昀椀nance purchases: Is it better to 昀椀nance your
car purchase by using the dealer's low-interest loan or by taking the rebate and
borrowing money 昀爀om your bank or credit union?
How to make short- and long-term investment decisions: Should you buy a one- or
two-semester parking pass? Is a higher salary better than stock options?
The Decision - Maki ng Process
Decision making requires that there be at least two alternatives. If only one course of action
is 愀瘀ailable, there is no decision to make. Does decision making, then, consist of choosing
昀爀om among alternative courses of action? Consider the 昀漀llowing situation:
At a race track, a bettor was uncertain which horse to bet on in the next race. He
closed his eyes and pointed his 昀椀nger at the list of horses printed in the racing
program. Upon opening his eyes, he saw that he was pointing to horse number 4.
He hurried o昀昀 to place his bet on that horse.
Does the racehorse selection represent a process of decision making? Clearly, it was a
process of choosing among alternatives. But this process seems inadequate and irrational.
We want to deal with 爀愀tional decision making.
Rational Decision Making
Rational decision making can be a complex process. One possible systematic approach
to this process is shown in Figure 1-1. Although the steps are shown sequentially, it is
common 昀漀r a decision maker to repeat steps, take them out of order, and do some steps
simultaneously. For example, when a new alternative is identi昀椀ed, more data will be re­
quired. Or when the outcomes are summarized, it may become clear that the problem
needs to be rede昀椀ned or new goals established. 吀栀e 昀漀llowing sections describe the ele­
ments listed in Figure 1-1.
1. Recognize 琀栀e Problem
吀栀e starting point in rational decision making is recognizing that a problem exists.
Some years ago, 昀漀r example, it was discovered that several species of ocean 昀椀sh con­
tained substantial concentrations of mercury. 吀栀e decision-making process began with
this discovery, and the rush was on to determine what should be done. Research revealed
that 昀椀sh taken 昀爀om the ocean decades be昀漀re and preserved in laboratories contained
similar concentrations of mercury. 吀栀us the problem had existed 昀漀r a long time but had
not been recognized.
In many situations, recognition is obvious and immediate. A car accident, a cheque
that bounces, a burned-out motor, an exhausted supply of parts all produce the recogni­
tion of a problem. Once we are aware of the problem, we solve it as best we can.
2. De昀椀ne 琀栀e Goal or Objective
吀栀e goal can be a grand, overall goal of a person or a 昀椀rm. For example, a personal goal could
be to lead a pleasant and meaning昀甀l life, whereas a 昀椀rm's goal is usually to operate pro昀椀tably.
But an objective need not be the grand, overall goal of a business or an individual. It
may be narrow and speci昀椀c: "I want to pay o昀昀 the loan on my car by May" and "吀栀e plant
must produce 300 golf carts in the next two weeks" are more limited objectives.
The Decision-Making Process
1 . Recognize the problem
2. De昀椀ne the goal or objective
3. Assemble relevant data
4. Identi昀礀 feasible alternatives
5. Select the criteria to determine the best alternative
6. Construct a model
7. Predict the outcomes or consequences 昀漀r each alternative
8. Choose the best alternative
9. Audit the results
FIGURE 1-1 One possible flow chart of the decision - making process.
3. Assemble Relevant Data
To make a good decision, you must assemble good in昀漀rmation. In addition to all the
published in昀漀rmation, a vast quantity of in昀漀rmation is not written down anywhere but
is stored as individuals' knowledge and experience. 吀栀ere is also in昀漀rmation that re­
mains ungathered. A question like "How many people in your town would be interested in
buying a pair of le昀琀-handed scissors?" cannot be answered by examining published data
or by asking any one person. Market research or other data gathering would be required to
obtain the desired in昀漀rmation.
One part of the data that must be assembled is the time horizon of the problem. How
long will the building or equipment last? How long will it be needed? Will it be scrapped,
sold, or shi昀琀ed to another use? In some cases, such as a road or a tunnel, the life may be
centuries with regular maintenance and occasional rebuilding. However, a shorter period,
such as 50 years, may be chosen as the time horizon so that decisions can be based on more
reliable data.
In engineering decisions, an important source of data is a firm's own account­
ing system. T hese data must be examined carefully. T hat is because accounting data
昀漀cus on past in昀漀rmation, whereas engineering judgment must often be applied to
estimate current and 昀甀ture values. For example, accounting records can show the
past cost of buying computers, but engineering judgment is required to estimate their
future cost.
Financial and cost accounting are designed to show the 昀氀ow of money-speci昀椀cally
costs and bene昀椀ts-in a company's operations. When costs are directly related to speci昀椀c
operations, there is no di昀케culty; but there are other costs unrelated to speci昀椀c operations.
吀栀ese indirect costs, or overhead, are usually allocated to a company's operations and
7
8
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
products by some semi-arbitrary method. 吀栀e results are generally satis昀愀ctory 昀漀r cost­
accounting purposes but may be unreliable 昀漀r use in economic analysis.
To create a meaning昀甀l economic analysis, we must determine the true di昀昀erences be­
tween alternatives, and that might require some adjustment of cost-accounting data. 吀栀e
昀漀llowing example illustrates this situation.
EXAM PLE 1-1
吀栀e cost-accounting records of a large company show the average monthly costs 昀漀r the three-person
printing department. 吀栀e wages and bene昀椀ts of the department make up the 昀椀rst category of direct
labour. 吀栀e company's indirect or overhead costs-such as heat, electricity, and employee insurance­
must be distributed to its various departments in some manner and, like many other 昀椀rms, this one
uses 昀氀oor space as the basis 昀漀r its allocations.
Direct labour (including employee bene昀椀ts)
Materials and supplies consumed
Allocated overhead costs:
200 m2 of 昀氀oor area at $25/m2
$6,000
7,000
5,000
$18,000
吀栀e printing department charges the other departments 昀漀r its services to recover its $18,000 monthly
cost. For example, the charge to run 1,000 copies of an announcement is
Direct labour
Materials and supplies
Overhead costs
Cost to other departments
$7.60
9.80
9.05
$26.45
吀栀e shipping department checks with a commercial printer that would print the same 1,000 copies 昀漀r
$22.95. Although the shipping department needs only about 30,000 copies printed a month, its 昀漀reman
decides to stop using the printing department and h愀瘀e the work done by the outside printer. 吀栀e in­
house printing department objects to this. As a result, the general manager has asked you to study the
situation and recommend what should be done.
SO LUTI O N
Much of the printing department's output reveals the company's costs, prices, and other 昀ꌀnancial in­
昀漀rmation. 吀栀e company president considers the printing department necessary in order to prevent
disclosing such in昀漀rmation to people outside the company.
A review of the cost-accounting charges reveals nothing unusual. 吀栀e charges made by the printing
department cover direct labour, materials and supplies, and overhead. 吀栀e allocation of indirect costs
is a customary procedure in cost-accounting systems, but it may be misleading 昀漀r decision making, as
the 昀漀llowing discussion indicates.
Direct labour
Materials and supplies
Overhead costs
Printing Department
Outside Printer
1,000
Copies
30,000
Copies
1,000
Copies
30,000
Copies
$7.60
9.80
9.05
$26.45
$228.00
294.00
271.50
$793.50
$22.95
$688.50
$22.95
$688.50
The Decision-Making Process
吀栀e shipping department would reduce its cost 昀爀om $793 to $688 by using the outside printer. In that
case, by how much would the printing department's costs decline? We will examine each of the cost
components:
1.
Direct Labou爀⸀ If the printing department had been working overtime, then the overtime could
be reduced or eliminated. But, assuming there was no overtime, how much would the saving
be? It seems unlikely that a printer could be 昀椀red or even put on less than a 40-hour work
week. 吀栀us, although there might be a $228 saving, it is much more likely that there will be no
reduction in direct labour.
2. Materials and S瀀lies. 吀栀ere would be a $294 s愀瘀ing in materials and supplies.
3. Allocated Overhead Costs. 吀栀ere will be no reduction in the printing department's monthly
$5,000 overhead, because there will be no reduction in department 昀氀oor space. (吀栀ough there
may be a slight reduction in the 昀椀rm's power costs if the printing department does less work.)
Furthermore, the 昀椀rm will incur additional expenses in purchasing and accounting to deal
with the outside 昀椀rm.
吀栀e 昀椀rm will save $294 in materials and supplies and may or may not save $228 in direct labour if
the printing department no longer does the shipping department's work. 吀栀e maximum saving would
be $294 + 228 = $522. But if the shipping department obtains its printing 昀爀om the outside printer,
the 昀椀rm must pay $688.50 a month. 吀栀e saving 昀爀om not doing the shipping department work in the
printing department would not exceed $522, and it would probably be only $294. 吀栀e result would be a
net increase in cost to the 昀椀rm. For this reason, the shipping department should be discouraged 昀爀om
sending its printing to the outside printer.
Gathering cost data presents other di昀케culties. One way to look at the 昀椀nancial
consequences-costs and bene昀椀ts-of various alternatives is as 昀漀llows:
•
•
•
Market Consequences. 吀栀ese consequences have an established price in the
marketplace. We can quickly determine raw material prices, machinery costs,
labour costs, and so 昀漀rth.
Extra-Market Consequences. 吀栀ere are other items that are not directly priced in
the marketplace. But by indirect means, a price may be assigned to these items.
(Economists call these prices shadow prices.) Examples might be the cost of an
employee's injury or the value to employees of going 昀爀om a 昀椀ve-day to a 昀漀ur-day
40-hour week.
Intangible Consequences. Numerical economic analysis never 昀甀lly describes the
di昀昀erences between alternatives. We are tempted to leave out consequences that
do not have a signi昀椀cant e昀昀ect on the analysis itself or on the conversion of the
昀椀nal decision into its cash equivalent. How does one evaluate the potential loss of
workers' jobs because of automation? What is the value of landscaping around a
昀愀ctory? 吀栀ese and similar 昀愀ctors may be le昀琀 out of the numerical calculations,
but they should be considered in conjunction with the numerical results in reach­
ing a decision.
9
10
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
4. 䤀搀ent椀昀y Feasi b le Alternatives
It should be kept in mind that unless the best alternative is considered, the result will
always be less than ideal. 吀眀o types of alternatives are sometimes ignored. First, in many
situations a "do-nothing" alternative is feasible. Second, there are o昀琀en feasible (but un­
glamorous) alternatives, such as "Patch it up and keep it running 昀漀r another year be昀漀re
replacing it."
吀栀ere is no 昀漀olproof way to ensure that the best alternative is among the ones being
considered. You should ensure that all conventional alternatives have been listed and then
strive to suggest innovative solutions. Sometimes it can be use昀甀l 昀漀r a group to consider
alternatives in an innovative atmosphere, that is, by b爀愀instorming. Even impractical al­
ternatives may lead to a better possibility.
5. Select 琀栀e Criteria for Determining 琀栀e Best Alternative
To choose the best alternative, we must de昀椀ne what we mean by best. 吀栀ere must be at least
one criterion, or a set of criteria, to evaluate which alternative is best-or, in some un昀漀rtu­
nate cases, the least bad. Here are several possible criteria:
•
•
•
•
•
•
•
Create minimal disturbance to the environment.
Improve the distribution of wealth among people.
Minimize the expenditure of money.
Ensure that the bene昀椀ts to those who gain 昀爀om the decision are greater than the
losses of those who are harmed by the decision. 1
Minimize the time needed to accomplish the goal or objective.
Minimize unemployment.
Maximize pro昀椀t.
Selecting the criteria by which to choose the best alternative will not be easy if di昀昀er­
ent groups support di昀昀erent criteria and hence desire di昀昀erent alternatives. Or the criteria
may conflict. For example, minimizing unemployment may increase the expenditure of
money. Minimizing environmental disturbance may increase the time to complete the
project. Disagreements between management and labour may re昀氀ect disagreements over
the criteria 昀漀r choosing the best alternative.
吀栀e last criterion-maximize pro昀椀t-is the one normally selected in engineering de­
cision making.
6. Construct a Model
At some point in the decision making, the various elements must be brought together.
吀栀e objective, relevant data, feasible alternatives, and selection standards must be
merged.
Constructing the relationships between the decision-making elements is 昀爀equently
called model building. To an engineer, a model may be a physical representation of the
real thing or a set of equations describing the desired relationships. In economic decision
making, the model is usually mathematical.
It is help昀甀l to model only that part of the system important to the problem at hand.
吀栀us the mathematical model of the student capacity of a classroom might be
1
吀栀is is the Kaldor criterion.
lw
. =­
Capac1ty
k
The Decision-Making Process
where
l = length of classroom, in metres
w = width of classroom, in metres
k = classroom arrangement 昀愀ctor
吀栀e equation 昀漀r student capacity of a classroom is a very simple model, yet it may be ad­
equate 昀漀r the problem being solved.
㜀⸀ Predict 琀栀e Outcomes for Each Alternative
A model and the data are used to predict the outcomes 昀漀r each feasible alternative. As
suggested earlier, each alternative might produce a variety of outcomes. Choosing a motor­
cycle rather than a bicycle, 昀漀r example, may make the 昀甀el supplier happy, the neighbours
unhappy, the environment more polluted, and the owner's savings account smaller. But, to
愀瘀oid unnecessary complications, we assume that decisions are based on a single criterion
昀漀r measuring the relative attractiveness of the various alternatives. If necessary, one could
devise a single composite criterion that is the weighted average of several di昀昀erent criteria.
To choose the best alternative, the outcomes 昀漀r each alternative must be stated in a
compa爀愀ble way. Usually the consequences of each alternative are stated in terms of money.
吀栀e consequences can be categorized as 昀漀llows:
•
•
•
Market consequences-where established market prices are 愀瘀ailable
Extra-market consequences-no direct market prices, so priced indirectly
Intangible consequences-valued by judgment, not monetary prices
In the 昀椀rst problems that we will examine, the costs and bene昀椀ts occur over a short
period and can be considered to occur at the same time. More commonly, the various costs
and bene昀椀ts take place over a longer period. We will represent these as a cash 昀氀ow diag爀愀m
to show the timing of the various costs and bene昀椀ts.
For these longer-term problems, a common mistake is to assume that the current
situation will be unchanged if the do-nothing alternative is chosen. For example, current
pro昀椀ts may shrink or vanish as a result of the actions of competitors and the expectations
of customers. As another example, tra昀케c jams normally increase over the years as the
number of vehicles increases.
8. Choose the Best Alternative
Ea爀氀ier we said that choosing the best alternative m愀礀 be simply a matter of determining which
best meets the selection criterion. But the solutions to most problems in economics h瘀였 market
consequences, extra-market consequences, and intangible consequences. Since the intangible
consequences of possible alternatives are le昀琀 out of the numerical calculations, they should be
introduced into the decision making at this point. 吀栀e right choice is the one that best meets
the criteria a昀琀er we have considered both the numerical and intangible consequences.
During the decision making, certain feasible alternatives are eliminated because they
are dominated by other, better alternatives. For example, shopping 昀漀r a computer online
may allow you to buy a custom-con昀椀gured computer 昀漀r less money than a stock computer
in a local store. Buying at the local store is feasible, but dominated.
H愀瘀ing examined the structure of the decision-making process, we can ask: When
is the decision made, and who makes it? If one person per昀漀rms all the steps in decision
making, then she is the decision maker. When she makes the decision is less clear. 吀栀e
selection of the feasible alternatives may be the crucial step, with the rest of the analysis a
methodical process leading to the inevitable decision.
11
0
12
Chapter 1 Economic Decisions, Engineering Costs, and Cost Estimating
9. Audit 琀栀e Results
An audit of the results is a comparison of what happened against the predictions. Do the
results of a decision analysis reasonably agree with its projections? If a new machine tool
was purchased to save labour and improve quality, did it? If the savings are not being
obtained, what was overlooked? 吀栀e audit may help ensure that the projected operating
advantages are ultimately obtained. On the other hand, the economic analysis projections
may have been unduly optimistic. We want to know this, too, so that the mistakes that
led to the inaccurate projection are not repeated. An e昀昀ective way to promote realistic
economic analysis calculations is 昀漀r all people involved to know that there will be an audit
of the results.
Ethics
You must be mind昀甀l of the ethical dimensions of engineering economic analysis and of
your engineering and personal decisions. 吀栀is text can only introduce the topic, which we
hope you will explore in greater depth.
Ethics can be described as distinguishing right and wrong when making decisions.
吀栀is may include establishing systems of beliefs and moral obligations, de昀椀ning values
and 昀愀irness, and determining duties and guidelines 昀漀r conduct. Ethical decision making
requires an understanding of the context of the problem, the possible choices, and the
outcomes of each choice.
Ethical Dimensions in Engineering Decision Making
Ethical issues can arise at every stage of the decision-making process. Ethics is such an
important part of professional and business decisions that ethical codes or standards of
conduct exist 昀漀r professional engineering societies and small and large organizations, to
guide individual and corporate beh愀瘀iour. Written professional codes are common in the
engineering profession, where they serve as a reference 昀漀r new engineers and a basis 昀漀r
legal action against engineers who violate the code.
In Canada, provincial and territorial associations of professional engineers
are responsible 昀漀r the regulation of the practice of engineering. The federation of
these organizations is called Engineers Canada. This is the Engineers Canada code
of ethics:
Registrants shall conduct themselves with integrity and in an honourable and
ethical manner. Registrants shall uphold the values of truth, honesty, and trust­
worthiness and safeguard human life and wel昀愀re and the environment. In keep­
ing with these basic tenets, registrants shall
1.
hold paramount the safety, health, and wel昀愀re of the public and the protection of
the environment, and promote health and safety within the workplace;
2. o昀昀er services, advise on or undertake engineering assignments only in areas of
their competence, and practise in a care昀甀l and diligent manner and in compli­
ance with applicable legislation;
3. act as 昀愀ith昀甀l agents of their clients or employers, maintain con昀椀dentiality,
and avoid con昀氀icts of interest, but, where such con昀氀ict arises, 昀甀lly disclose the
circumstances without delay to the employer or client;
4. keep themselves in昀漀rmed in order to maintain their competence, and advance
their knowledge in the 昀椀eld within which they practise;
Ethics
conduct themselves with integrity, equity, 昀愀irness, courtesy, and good 昀愀ith
toward clients, colleagues, and others, give credit where it is due, and accept, as
well as give, honest and 昀愀ir professional criticism;
6. present clearly to employers and clients the possible consequences if engineering
decisions or judgments are overruled or disregarded;
7. report to their association or other appropriate agencies any illegal or unethical
engineering decisions or practices by engineers or others;
8. be aware of and ensure that clients and employers are made aware of societal and
environmental consequences of actions or projects, and endeavour to interpret
engineering issues to the public in an objective and truth昀甀l manner;
9. treat equitably and promote the equitable treatment of people and in accordance
with human rights legislation;
10. uphold and enhance the honour and dignity of the profession. (CCPE 2012)
5.
In addition, each provincial and territorial association has a code of ethics 昀漀r its
members. Most engineering organizations have similar written standards. For all engin­
eers, di昀케culties arise when their actions are contrary to these written or internal codes.
吀栀e Environment We Live In
吀栀e decision maker must ask who incurs the costs 昀漀r the project and who receives the
bene昀椀ts. Ethical issues can be particularly di昀케cult because there are o昀琀en stakeholders
with opposing viewpoints, and some of the data may be uncertain and hard to quantify.
吀栀e 昀漀llowing are examples of di昀케cult choices:
•
•
•
Protecting the habitat of an endangered species versus 昀氀ood-control projects that
protect people, animals, and structures
Meeting the need 昀漀r electrical power when every means of generation causes
some environmental damage:
• Hydroelectric-loss of land and habitat to reservoir
• Coal-danger to workers 昀爀om underground mining, damage to habitat 昀爀om
open-pit mining, air pollution 昀爀om burning the coal, and an increased rate
of global warming because of the release of carbon dioxide to the atmosphere.
• Nuclear-disposal of radioactive waste
• Fuel oil-air pollution and economic dependence on 昀漀reign sources
• Wind-visual pollution of wind 昀愀rms, killing of birds and bats by whirling
blades
Determining standards 昀漀r pollutants: Is 1 part per million acceptable, or is 1 part
per billion necessary?
Safe琀礀 and Cost
Some of the most common and most di昀케cult ethical dilemmas involve trade-o昀昀s between
safety and cost. If a product is "too safe," it will be too expensive and it will not be used.
And sometimes the cost is borne by one party and the risk by another.
•
•
•
Should the oil plat昀漀rm be designed 昀漀r the 100-year, 500-year, or 1,000-year
hurricane?
Should the auto manu昀愀cturer add run-昀氀at tires, stability control, side-cushion
airbags, and rear-seat airbags to every car?
Are stainless steel valves required, or is it economically better to use less-corrosion­
resistant valves and replace them more o昀琀en?
13
14
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
Emerging 䤀猀sues and "Solutions"
Breaches of the law by the corporate leaders of Enron, Tyco, and other 昀椀rms led to attempts
by governments to prevent, limit, and expose 昀椀nancial wrongdoing within corporations.
One part of the American response has been the Sarbanes-Oxley Act of 2002, which im­
posed requirements on executives and auditing 昀椀rms, and penalties 昀漀r violations.
Globalization is another area of increasing importance 昀漀r ethical considerations. One
reason is that di昀昀erent countries and regions have di昀昀erent ethical expectations. A second
reason is that jobs may be moved to another country because of di昀昀erences in cost, pro­
ductivity, environmental standards, and so on. What m愀礀 昀爀om a Canadian perspective
be a sweatshop may 昀爀om the perspective of a less developed nation be an opportunity to
support many 昀愀milies.
I m porta nce of Eth ics I n Engi neeri ng and Engi neeri ng Economy
Frequently, though engineers and 昀椀rms try to act ethically, mistakes are made-the data
were wrong, the design was changed, or the operating environment was di昀昀erent than
expected. But in other cases, such as the Challe最팀r launch decision, a choice was made to
place expediency above ethics. Under such circumstances, it is the engineers' duty to speak
out, within and if necessary beyond their company, to protect the safety of the public.
O昀琀en, recent engineering graduates are asked, "What is the most important thing you
want 昀爀om your supervisor?" 吀栀e most common response is mentoring and opportunities
to learn and progress. When employees with 5, 15, 25, or more years of experience are
asked the same question, the most common response is integrity. 吀栀is is what your sub­
ordinates, peers, and superiors will expect and value the most 昀爀om you. Integrity is the
昀漀undation 昀漀r long-term career success.
Engineering Decision Making for Current Costs
Some of the easiest kinds of engineering decisions deal with problems of alternative designs,
methods, or materials. If the results of the decision appear in a very short time, one can
quickly add up the costs and bene昀椀ts 昀漀r each alternative. 吀栀en, by using suitable economic
standards, one can identify the best alternative. 吀栀ree examples illustrate these situations.
EXAM PLE 1-2
A concrete aggregate mix must contain at least 31% sand by volume 昀漀r proper batching. One source of
material, which has 25% sand and 75% coarse aggregate, sells 昀漀r $3.00 per cubic metre (m3). Another
source, which has 40% sand and 60% coarse aggregate, sells 昀漀r $4.40/m3. Determine the least cost per
cubic metre of blended aggregates.
SO LUTI O N
吀栀e least cost of blended aggregates will result 昀爀om maximum use of the lower-cost material. 吀栀e higher­
cost material will be used to increase the proportion of sand up to the minimum level (31%) speci昀椀ed.
Let x = Portion of blended aggregates 昀爀om $3.00/m3 source
1 - x = Portion of blended aggregates 昀爀om $4.40/m3 source
Engineering Decision Making for Current Costs
Sand Balance
X (0.25)
15
+ (1 + x)(0.40) = 0.31
0.25x + 0.40 - 0.40x = 0.31
0.3 1 - 0.40
x = ----0.25 - 0.40
吀栀us the blended aggregates will contain
= 0.60
-0.09
-0. 15
60% of $3.00/m3 material
40% of $4.40/m3 material
吀栀e least cost per cubic metre of blended aggregates is
0.60($3.00) + 0.40($4.40) = 1 .80 + 1.76
= $3.56/m3
EXAM PLE 1-3
A machine part is manu昀愀ctured at a unit cost of 40¢ 昀漀r material and 15¢ 昀漀r direct labour. An invest­
ment of $500,000 in tooling is required. 吀栀e order calls 昀漀r 3 million pieces. Halfw愀礀 through the order,
a new method of manu昀愀cture can be put into e昀昀ect that will reduce the unit costs to 34¢ 昀漀r material
and 10¢ 昀漀r direct labour-but it will require $100,000 昀漀r additional tooling. 吀栀is tooling will not be
use昀甀l 昀漀r 昀甀ture orders. Other costs are allocated at 2.5 times the direct labour cost. What, if anything,
should be done?
SO LUTI O N
Since there is only one w 愀礀 to handle the 昀椀rst 1.5 million pieces, our problem concerns only the second
half of the order.
Alternative A: Continue with Present Method
Material cost
1,500,000 pieces x 0.40
1,500,000 pieces x 0.15
Direct labour cost
Other costs
2.50 x direct labour cost =
Cost 昀漀r remaining 1,500,000 pieces
$600,000
225,000
562,500
$1,387,500
continued
16
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
Alternative B: Change the Manu昀愀cturing Method
Additional tooling cost
Material cost
1,500,000 pieces x 0.34 =
1,500,000 pieces x 0.10 =
Direct labour cost
Other costs
2.50 x direct labour cost =
Cost 昀漀r remaining 1,500,000 pieces
$100,000
510,000
150,000
375,000
$1,135,000
Be昀漀re making a 昀椀nal decision, you should closely examine the other costs to see that they do, in 昀愀ct,
vary as the direct labour cost. If they do, the decision would be to change the manu昀愀cturing method.
EXAM PLE 1-4
In the design of a cold-storage warehouse, the speci昀椀cations call 昀漀r a maximum heat transfer through
the warehouse walls of 8 watts per square metre of wall when there is a 30 ° C temperature di昀昀erence
between the inside sur昀愀ce and the outside sur昀愀ce of the insulation. 吀栀e two insulation materials being
considered are listed below:
Insulation Material
Cost per Cubic Metre
Conductivity
(W/m · ° C)
$12.50
14.00
0.039
0.03
Rock wool
Foamed insulation
吀栀e basic equation 昀漀r heat conduction through a wall is
Q = K(�T)
L
where Q = heat transfer, in W/m2 of wall
K = conductivity in W/m- ° C
�T = di昀昀erence in temperature between the two sur昀愀ces, in ° C
L = thickness of insulating material, in metres
Which insulation material should be selected?
SO LUTI O N
吀眀o steps are needed to solve the problem. First, the required thickness of each of the alternative ma­
terials must be calculated. 吀栀en, since the problem is one of providing a 昀椀xed output (heat transfer
through the wall is limited to a 昀椀xed maximum amount), the criterion is to minimize the input (cost).
Required Insulation 吀栀ickness
Rock wool
Foamed insulation
0.039(30)
L
0.03(30)
8=
L
8=
L = 0.IS m
L = 0.1 1 m
Engineering Decision Making for Current Costs
17
Cost of Insulation per Square Metre of Wall
Rock wool
Foamed insulation
Unit cost= Cost/m3 x Insulation thickness, in metres
Unit cost= $12.50 x 0.15 m= $1.87/m2
Unit cost= $14.00 x 0.11 m= $1.54/m2
吀栀e 昀漀amed insulation is the cheaper alternative. However, there are several intangible constraints that
must be considered. O昀琀en, insulation material comes in a small number of standard thicknesses. We
note that 0.15 m is very close to six inches, so we can probably obtain rock wool in that thickness or a
multiple of it. How thick is the 愀瘀ailable wall space? What is the cost of more insulation versus the cost
of cooling the warehouse over its life?
Engineering Costs
Evaluating a set of alternatives requires that many costs be analyzed. In this section we
describe several concepts 昀漀r classi昀礀ing and understanding these costs.
Fixed, Variable, Marginal, and Average Costs
䌀䐀
Fixed costs are constant or unchanging regardless of the level of output or activity. In con­
trast, variable costs depend on the level of output or activity. A marginal cost is the variable
cost 昀漀r one more unit, while the 瘀였rage cost is the total cost divided by the number of units.
For example, many universities charge 昀甀ll-time students a 昀椀xed cost 昀漀r 12 to 18 credit
hours and a cost per credit hour 昀漀r each credit hour over 18. 吀栀us, 昀漀r 昀甀ll-time students
who are taking an overload (more than 18 hours), there is a variable cost that depends on
the level of activity.
吀栀is example can also be used to distinguish between marginal and average costs. A
marginal cost is the cost of one more unit. 吀栀is will depend on how many credit hours the
student is taking. If a student is currently enrolled 昀漀r 12 to 17 hours, one additional hour is
昀爀ee. 吀栀e marginal cost of an additional credit hour is $0. However, if the student is taking
18 or more hours, the marginal cost is the variable cost of one more hour.
To illustrate 瘀였rage costs, the 昀椀xed and variable costs need to be speci昀椀ed. Suppose the
cost of 12 to 18 hours is $1,800 per term and 瘀퐀rload credits are $120 an hour. If a student takes
12 hours, the ave爀愀ge cost is $1,800/12 = $150 per credit hour. If the student were to take 18
hours, the ave�였 cost would decrease to $1,800/18 = $100 per credit hour. If the student takes
21 hours, the ave�였 cost is $102.86 per credit hour [$1,800 + (3 x $120)/21]. Average cost is
thus calculated by dividing the total cost 昀漀r all units by the total number of units. Decision
makers use average cost to attain an overall cost picture of the i瘀팀stment on a per-unit basis.
EXAMPLE 1-5
吀栀e Federation of Student Societies of Engineering (FeSSE) wants to o昀昀er a one-day training course to
help students 昀椀nd jobs and to raise 昀甀nds 昀漀r the Federation. 吀栀e organizing committee is sure that they
can 昀椀nd alumni, local business people, and 昀愀culty to provide the training at no charge. 吀栀us the main
costs will be 昀漀r space, meals, handouts, and advertising.
continued
18
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
吀栀e organizers have classi昀椀ed the costs 昀漀r room rental, room set-up, and advertising as 昀椀xed costs.
吀栀ey also have included the meals 昀漀r the speakers as a 昀椀xed cost. 吀栀eir total of $225 is pegged to a room
that will hold 40 people. If demand is greater than 40, the 昀椀xed costs will also increase.
吀栀e variable costs 昀漀r 昀漀od and bound handouts will be $20 per student. 吀栀e organizing committee
believes that $35 is about the right price to match value to students with their budgets. Since FeSSE has
not o昀昀ered training courses be昀漀re, they are unsure how many students will reserve seats.
Develop equations 昀漀r FeSSE's total cost and total revenue, and determine the number of registra­
tions needed 昀漀r revenue to equal cost.
SO LUTI O N
Let x be the number of students who sign up. 吀栀en,
Total cost = $225 + $20x
Total revenue = $35x
To 昀椀nd the number of student registrants needed, we equate these quantities and solve.
Total cost = Total revenue
$225 + $20x = $35x
$225 = ($35 - $20)x
x = 225/15 = 15 students
1 ,200
1 ,000
Total Revenue ,,; '
y = 35x , , '
800
,
� ,,,, '
,
6 00
400
200
0
Break- Even Point
0
5
15
Customers
20
25
30
FIGURE 1-2 Pro昀椀t- loss break-even chart for Exa mple 1-5.
From Example 1-5 we see how it is possible to calculate total 昀椀xed and total vari­
able costs. Furthermore, these values can be combined into a single total-cost equation as
昀漀llows:
Total cost = Total 昀椀xed cost + Total variable cost
(1-1)
Example 1-5 developed total cost and total revenue equations to describe the training
course proposal. 吀栀ese equations can be used to create what is called a pro昀需t-loss break­
even chart (see Figure 1-2). Both the costs and revenues associated with various levels of
E n g i neeri n g Costs
output (activity) are placed on the same set of x-y axes. 吀栀is makes it possible to illustrate a
break-even point and regions of pro昀需t and loss 昀漀r some business activity. 吀栀ese terms can
be de昀椀ned as 昀漀llows:
Break-even point: 吀栀e level of activity at which the total cost of providing the product,
good, or service is equal to the revenue (or savings) generated.
Pro昀需t region: Values of the variable x greater than the break-even point, where total
revenue is greater than total costs.
Loss region: Values of the variable x less than the break-even point, where total cost is
greater than total revenue.
Notice in Figure 1-2 that the break-even point 昀漀r the number of people in the training
course is 15 people. For more than 15 people, FeSSE will make a pro昀椀t. If fewer than 15
sign up, there will be a loss.
吀栀e 昀椀xed costs of our simple model are only 昀椀xed over a certain range of values 昀漀r x.
In Example 1-5, that range was 1 to 40 students. If zero students sign up, the course could
be cancelled and many of the 昀椀xed costs would not be incurred. Some costs, such as ad­
vertising, might already have been incurred, and there might be cancellation fees. If more
than 40 students sign up, there would be greater costs 昀漀r larger rooms or multiple sessions.
When modelling a speci昀椀c situation, we o昀琀en use linear variable costs and revenues.
However, sometimes the relationship m愀礀 be non-linear. For example, employees are o昀琀en
paid at 150% of their hourly rate 昀漀r overtime hours, so that production levels requiring
overtime have higher variable costs. Total cost in Figure 1-3 is a 昀椀xed cost of $3,000 plus a
variable cost of $200 per unit 昀漀r straight-time production of up to 10 units and $300 per
unit 昀漀r overtime production of up to 昀椀ve more units.
Figure 1-3 can also be used to illustrate marginal and average costs. At a volume of 昀椀ve
units the marginal cost is $200 per unit, whereas at a volume of 12 units the marginal cost
is $300 per unit. At 昀椀ve units the 愀瘀erage cost is $800 per unit, or (3,000 + 200 x 5)/5. At
12 units the average cost is $467 per unit, or (3,000 + 200 x 10 + 300 x 2)/12.
8,000
Total
Cost
6,000
⸀尀 4,000
�
Variable Cost
2,000
Fixed Cost
o �----�----�-�--�
0
5
15
Volume
FIGURE 1-3 Non-linea r va riable costs.
䌀䐀
Sunk Costs
A sunk cost is money already spent as a result of a past decision. Sunk costs should
be disregarded in our engineering economic analysis because current decisions
19
20
Chapter 1 Economic Decisions, Engineering Costs, and Cost Estimating
cannot change the past. For example, dollars spent last year to buy new production
machinery is money that is sunk: there is nothing that can be done now to change
that action.
It is di昀케cult not to be in昀氀uenced by sunk costs. Consider 100 shares of stock in XYZ, Inc.,
bought 昀漀r $15 a share last 礀攀ar. 吀栀e share price has steadily declined over the past 12 months
to a price of $10 a share today. Current decisions must 昀漀cus on the $10 a share that could be
obtained tod愀礀 (as well as 昀甀ture price potential), not the $15 a share that was paid last 礀攀ar.
吀栀e $15 a share paid last year is a sunk cost and has no in昀氀uence on present opportunities.
When we get to Chapters 11 and 12 on depreciation and income taxes, we will 昀椀nd an
exception to the rule that sunk costs should be ignored. When an asset is sold or disposed
of, the cost that was paid 昀漀r it is important in 昀椀guring out how much is owed in taxes. 吀栀is
exception only applies to the a昀琀er-tax analysis of capital assets.
Opportunity Costs
䌀䐀
Every time we use a business resource (equipment, dollars, manpower, etc.) in one activity,
we give up the opportunity to use the same resource in some other activity.
A 昀椀rm that chooses to use the resource in one w愀礀 is losing the bene昀椀ts that would be
derived 昀爀om using it in other ways. 吀栀is loss is the opportunity cost of using the resource
in the chosen activity. Opportunity cost may also be considered a 昀漀rgone opportunity
cost because we are 昀漀rgoing the bene昀椀t that could have been realized.
EXAMPLE 1-6
A distributor of electric pumps must decide what to do with a "lot" of old electric pumps bought three
years ago. Soon a昀琀er the distributor purchased the lot, technological advances made the old pumps less
desirable to customers. 吀栀e pumps are becoming obsolete as they sit in inventory. 吀栀e pricing manager
has the 昀漀llowing in昀漀rmation:
Distributor's purchase price three years ago
Distributor's storage costs to date
Distributor's list price three years ago
Current list price of the same number of new pumps
Amount o昀昀ered 昀漀r the old pumps 昀爀om a buyer two years ago
Current price the lot of old pumps would bring
$7,000
1,000
9,500
12,000
5,000
3,000
Looking at the data, the pricing manager has concluded that the price should be set at $8,000. 吀栀is is
the money that the 昀椀rm has tied up in the lot of old pumps ($7,000 purchase and $1,000 storage), and it was
reasoned that the company should at least recover this cost. Furthermore, the pricing manager has argued
that an $8,000 price would be $1,500 less than the list price 昀爀om three years ago, and it would be $4,000
less than what a lot of new pumps would cost ($12,000 - $8,000). What would be your advice on price?
SOLUTION
Let's look more closely at each of the data items.
Distributor's purchase price three years 最퀀: 吀栀is is a sunk cost that should not be considered in
setting the price today.
E n g i neeri n g Costs
21
Distributor's sto�였 cos琀猀 to date: 吀栀e storage costs 昀漀r keeping the pumps in inventory are sunk
costs; that is, they h愀瘀e been paid. Hence they should not in昀氀uence the pricing decision.
Distributor's list price three years 最퀀: If there have been no willing buyers in the past three years
at this price, it is unlikely that a buyer will emerge in the 昀甀ture. 吀栀is past list price should have no
in昀氀uence on the current pricing decision.
Current list price of newer pumps: Newer pumps now include technology and features that h愀瘀e
made the older pumps less valuable. It is misleading to compare the older pumps directly to those
with new technology. However, the price of the new pumps and the value of the new features help
determine the market value of the old pumps.
Amount o昀昀ered by a buyer two years ago: 吀栀is is a 昀漀rgone opportunity. At the time of the o昀昀er, the
company chose to keep the lot, and thus the $5,000 o昀昀ered became an opportunity cost 昀漀r keeping
the pumps. 吀栀is amount should not in昀氀uence the current pricing decision.
Current price the lot could bring: 吀栀e price a willing buyer in the marketplace o昀昀ers is called the
asset's market value. 吀栀e lot of old pumps in question is believed to h愀瘀e a current market value of
$3,000.
From this analysis, it is easy to see the flaw in the pricing manager's reasoning. In an engineering
economic analysis we deal only with tod愀礀 's and prospective future opportunities. It is impos­
sible to go back in time and change decisions that have been made. T hus the pricing manager
should recommend to the distributor that the price be set at the value that a buyer assigns to the
item: $3,000.
䌀䐀
䌀䐀
Recu rri ng and Non- Recu rri ng Costs
Recurring costs are any expenses that are known and anticipated and that occur at
regular intervals. Non-recurring costs are one-of-a-kind expenses that occur at irregu­
lar intervals and thus are sometimes di昀케cult to plan 昀漀r or anticipate 昀爀om a budgeting
perspective.
In engineering economic analyses, recurring costs are modelled as cash 昀氀ows that
occur at regular intervals (such as every year or every 昀椀ve years). 吀栀eir magnitude can
be estimated, and they can be included in the overall analysis. Non-recurri渀最 costs can be
handled easily in our analysis if we are able to anticipate their timing and size. However,
this is not always easy.
I ncremental Costs
One of the 昀甀ndamental principles in engineering economic analysis is that in the choice
between competing alternatives, the emphasis is on the d昀케erences between those al­
ternatives. 吀栀is is the concept of incremental costs. For instance, you may be interested
in comparing two options 昀漀r leasing a car 昀漀r personal use. 吀栀e two lease options may
have several speci昀椀cs 昀漀r which the costs are the same. However, there may be incremen­
tal costs associated with one option but not with the other. In a comparison of the two
leases, you need look only at the di昀昀erences between the alternatives, not at the common
costs.
22
Chapter 1 Economic Decisions, Engineering Costs, and Cost Estimating
Cash Costs versus Book Costs
A cash cost requires the movement of dollars out of one person's pocket into the pocket of
someone else. When you buy dinner 昀漀r your 昀爀iends or make your monthly car payment,
you are incurring a cash cost.
Book costs do not require the movement of dollars 昀爀om one pocket to another. Rather,
they are cost e昀昀ects of past decisions that are recorded in the accounting books of a 昀椀rm. In
one common book cost, asset depreciation (which we discuss in Chapter 11), the expense
paid 昀漀r a particular business asset is "written o昀昀" on a company's accounting system over
a number of periods. Book costs do not ordinarily represent cash 昀氀ows and thus are not in­
cluded in engineering economic analysis. One exception is the e昀昀ect of asset depreciation
on tax payments-which are cash flows and are included in a昀琀er-tax analyses.
Life-Cycle Costs
䌀䐀
Life-cycle costing refers to designing products, goods, and services with a 昀甀ll and explicit
recognition of the associated costs over their life cycles. 吀眀o key concepts in life-cycle
costing are that the later a design change is made, the higher the cost, and that decisions
made early in the life cycle tend to "lock in" costs that will be incurred later. Figure 1-4
illustrates how costs are committed early in the product life cycle-nearly 70%-90% of all
costs are set during the design phases. At the same time, only 10%-30% of these costs are
incurred be昀漀re the end of the production phase.
Figure 1-5 illustrates that downstream product changes are more expensive and that
upstream changes are easier (and less expensive) to make. When planners try to s愀瘀e
money at an early design stage, the result is o昀琀en a poor design that results in changes
during construction and prototype development. 吀栀ese changes, in turn, cost more than
working out a better design would have cost.
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FIGURE 1-4 Cumulative life-cycle costs committed and dollars spent.
Engineering Costs
High
--------⸀尀 ⸀尀
Ease of
尀言
⸀尀 , ,
Changing Design ' , ,
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Project Phase
FIGURE 1-5
Life-cycle desig n change costs and ease of change.
From Figures 1-4 and 1-5 we see that the time to consider all life-cycle e昀昀ects and
make design changes is during the needs de昀椀nition and conceptual design phases­
be昀漀re a lot of money is committed. Some of the life-cycle e昀昀ects that engineers should
consider at design time are product costs 昀漀r liability, production, material, testing and
quality assurance, and maintenance and warranty. Other life-cycle e昀昀ects are product
features based on customers' input and the e昀昀ects of product disposal on the environ­
ment. Engineers who design products and the systems that produce them should consider
all life-cycle costs.
Cost Estimating
Estimating is the 昀漀undation of economic analysis. As in any analysis procedure, the out­
come is only as good as the numbers used to reach the decision.
Types of Estimate
We can de昀椀ne three general types of estimate whose purposes, accuracy, and underlying
methods are quite di昀昀erent.
Ro最� estimates are order-of-magnitude estimates used 昀漀r high-level planning, 昀漀r de­
termining feasibility, and in the initial planning and evaluation phases of a project.
Rough estimates tend to use back-of-the-envelope numbers. 吀栀ese estimates can be
made quickly and easily; they tend to range 昀爀om 30% higher than the 昀椀nal cost to
60% too low (accuracy of -30% to +60%).
Notice the non-symmetry in the estimating error. 吀栀is is because decision make爀猀
tend to underestimate the magnitude of costs (n 最였tive economic e昀昀ec猀鴀. Also, as
Murphy's Law predicts, there seem to be more ways for results to be worse than ex­
pected than for the resul琀猀 to be better than expected.
23
24
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
Bu最준t estimates are used 昀漀r budgeting purposes at the conceptual or prelimin­
ary design stages of a project. 吀栀ese estimates are more detailed, and they re­
quire additional time and resources. Greater sophistication is used in developing
budget estimates than the rough-order type, and their accuracy is generally - 15%
to +20%.
Detailed estimates are used during the detailed design and contract bidding phases of
a project. 吀栀ese estimates are made 昀爀om detailed quantitative models, blueprints,
product speci昀椀cation sheets, and vendor quotes. Detailed estimates take the most
time and resources to develop and are thus much more accurate than rough or
budget estimates. 吀栀e accuracy of these estimates is generally -3% to +5%.
吀栀e 瀀er limits of +60% for ro最� orde爀Ⰰ ⬀㈀0% for budget, and +5% for detailed
estimates are based on construction data for plants and inf爀愀structure. In the case of
so昀琀ware, research and development, and new military weapons, the percentages are
o昀琀en much higher.
Rough estimates are used 昀漀r general feasibility activities, budget estimates support
budgeting and preliminary design decisions, and detailed estimates are used 昀漀r establish­
ing design details and contracts.
However, this increased accuracy requires added time and resources. Figure 1-6 illus­
trates the trade-o昀昀 between accuracy and cost. In engineering economic analysis, the re­
sources spent must be justi昀椀ed by the need 昀漀r detail in the estimate. For example, during
the project feasibility stages we would not want to use our people, time, and money to
develop detailed estimates 昀漀r unfeasible alternatives that will be quickly eliminated 昀爀om
昀甀rther consideration.
Di昀케culties i n Esti mation
With few exceptions (such as in legal contracts), it is di昀케cult to 昀漀resee 昀甀ture economic
consequences exactly. In this section we discuss several aspects of estimating that make it
a di昀케cult task.
High
Low 开贀________⸀븀________븀
Low
Medium
High
Accuracy of Estimate
FIGURE 1-6 Accu racy versus cost trade-o昀昀 i n esti mati ng.
Cost Estimating
One-o昀ⴀa-䬀椀nd Estimates
Estimated parameters can be 昀漀r one-of-a-kind or 昀椀rst-run projects. 吀栀e 昀椀rst time some­
thing is done, it is di昀케cult to estimate the cost of designing, producing, and maintaining
a product over its life cycle.
吀栀e good news is that there are very few one-of-a-kind estimates to be made in engin­
eering design and analysis. Nearly all new technologies, products, and processes have close
cousins that led to their development. 吀栀e concept of estimation by analogy allows us to
use knowledge of well-understood activities to anticipate costs 昀漀r new activities.
吀椀me and E昀昀ort Availa b le
Our ability to develop engineering estimates is constrained by time and the availability of
person-power. Constraints on time and person-power can make the overall estimating task
di昀케cult. If the estimate does not need much detail (such as when a rough estimate is the
goal), then time and personnel constraints may not be a 昀愀ctor. When detail is critical (such
as in legal contracts), however, requirements must be anticipated and resource use planned.
Estimating Models
吀栀is section develops several estimating models that can be used at the rough, budget, or
detailed design levels. 吀栀e level of detail will depend on the accuracy of the data used.
Per- U n it Model
吀栀e per-unit model uses a per-unit 昀愀ctor, such as cost per square metre, to develop the esti­
mate. 吀栀is is a very simple yet use昀甀l technique, especially 昀漀r developing order-of-magnitude
estimates. 吀栀e per-unit model is commonly used in the construction industry. For example,
you m愀礀 be interested in a new home that is constructed with a certain type of material
and has a speci昀椀c style. Using this in昀漀rmation, a contractor m愀礀 quote a cost of $1,400 per
square metre 昀漀r your home. 吀栀us, if you are interested in a 200-m2 house, your cost would
be 1,400 x 200 = $280,000. Per-unit 昀愀ctors are also used 昀漀r the 昀漀llowing estimates:
•
•
•
•
service cost per customer
safety cost per employee
gasoline cost per kilometre
cost of defects per batch
It is important to note that the per-unit model does not make allowances 昀漀r econ­
omies of scale (the 昀愀ct that higher quantities usually cost less on a per-unit basis).
Seg menting Model
吀栀e segmenting model can be described as "divide and conquer." An estimate is decom­
posed into its individual components, estimates are made at those lower levels, and then
the estimates are added back together. It is much easier to estimate at the lower levels be­
cause they are more readily understood.
Cost I ndexes
Cost indexes are numerical values that record historical change in engineering (and other)
costs. 吀栀e cost index numbers are dimensionless, and they re昀氀ect relative price change
25
26
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
in either individual cost items (labour, material, utilities) or groups of costs (consumer
prices, producer prices). Indexes can be used to update historical costs with the basic ratio
relationship given in Equation 1-2.
Cost at Time A
Cost at Time B
Index v愀氀ue at Time A
Index value at Time B
(1-2)
Equation 1-2 states that the ratio of the cost index numbers at two points in time (A
and B) is equivalent to the dollar cost ratio of the item at the same times (see Example 1-7).
EXAM PLE 1-7
Miriam is interested in estimating the annual labour and material costs 昀漀r a new production 昀愀cility.
She was able to obtain the 昀漀llowing cost data:
Labour Costs
• Labour cost index value was at 124 ten years ago and is 188 today.
• Annual labour costs 昀漀r a similar 昀愀cility were $575,500 ten years ago.
Material Costs
• Material cost index value was at 544 three years ago and is 7 15 today.
• Annual material costs 昀漀r a similar 昀愀cility were $2,455,000 three years ago.
SO LUTI O N
Miriam will use Equation 1-2 to develop her cost estimates 昀漀r annual labour and material costs.
Labour
Annu愀氀 cost today
Annu愀氀 cost 10 years ago
Index v愀氀ue tod愀礀
Index value 10 years ago
188
Annu愀氀 cost tod愀礀 = - X $575,500 = $872,532
124
Materials
Annu愀氀 cost tod愀礀
Annual cost 3 years ago
Index v愀氀ue tod愀礀
Index v愀氀ue 3 years ago
715
Annu愀氀 cost tod愀礀 = - X $2,455,000 = $3,226,700
544
Cost index data are collected and published by several private companies and govern­
ment agencies in Canada and the United States. Canadian data are 愀瘀ailable 昀爀om Statis­
tics Canada. 吀栀e US government publishes data through the Bureau of Labor Statistics of
the Department of Commerce. 吀栀e Statistical Abstract of the United States publishes cost
Estimating Models
indexes 昀漀r labour, construction, and materials. Another use昀甀l source 昀漀r engineering
cost index data is the McGraw Hill publication Engineering News-Record.
Power-Sizing Model
吀栀e power-sizing model is used to estimate the costs of industrial plants and equipment.
吀栀e model "scales up" or "scales down" known costs, thereby accounting 昀漀r economies
of scale that are common in industrial plant and equipment costs. Consider the cost of
building a re昀椀nery. Would it cost twice as much to build the same 昀愀cility with double the
capacity? It is unlikely. 吀栀e power-sizing model uses the exponent (x), called the power­
sizing exponent, to represent economy of scale in size or capacity:
[
l
Cost of equipment A
Size (capaci琀礀) of equipment A
Cost of equipment B - Size (capacity) of equipment B x
(1-3)
where x is the power-sizing exponent, costs of A and B are at the same point in time (have
the same dollar basis), and size or capacity is in the same physical units 昀漀r both A and B.
吀栀e power-sizing exponent (x) can be 1.0 (indicating a linear cost-versus-size or -cap­
acity relationship) or greater than 1.0 (indicating diseconomy of scale), but it is usually
less than 1.0 (indicating economy of scale). Generally the ratio should be less than 2, and
it should never exceed 5. 吀栀is model works best in a middle range-when the plants and
equipment are not very small or very large.
Exponent values 昀漀r plants and equipment of many types may be 昀漀und in several
sources, including industry reference books, research reports, and technical journals.
Such exponent values may also be 昀漀und in Perry's Chemical Engineers' Handbook, Plant
Design and Economics for Chemical Enginee爀猀, and Preliminary Chemical Engineering
Plant Design. Table 1-1 gives power-sizing exponent values 昀漀r several types of industrial
昀愀cilities and equipment. 吀栀e exponent given applies only to equipment within the size
range speci昀椀ed.
Table 1-1 Examples of Power-Sizing Exponent Values
Equipment or Facility
Size Range
Power-Sizing E砀瀀onent
Blower, centri昀甀gal
Compressor
Crystallizer, vacuum batch
Dryer, drum, single atmospheric
Fan, centri昀甀gal
Filter, vacuum rotary drum
Lagoon, aerated
Motor
Reactor, 2 MPa
Tank, atmospheric, horizontal
5-50 m /s
150-1,500 kW
55-780 m2
1-10 m2
10-35 m3/s
1-150 m2
0.002-1 m3/s
4-15 kW
0.4-4 m3
0.4-160 m3
0.59
0.32
0.37
0.40
1.17
0.48
1.13
0.69
0.56
0.57
3
In Equation 1-3, equipment costs 昀漀r both A and B occur at the same point in time. 吀栀is
equation is use昀甀l 昀漀r scaling equipment costs but not 昀漀r updating those costs. When the time
of the desired cost estimate is di昀昀erent 昀爀om the time in which the scaling occurs (as in Equa­
tion 1-3), cost indexes are used 昀漀r the time updating. 吀栀us, in cases like Example 1-8 involving
both scaling and updating, we use the power-sizing model together with cost indexes.
27
28
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
EXAM PLE 1-8
Because of her work in Example 1-7, Miriam has been asked to estimate the cost tod 愀礀 of a 250 m2 heat
exchange system 昀漀r the new plant being analyzed. She has the 昀漀llowing data:
• Her company paid $50,000 昀漀r a 1 00 m2 heat exchanger 昀椀ve years ago.
• Heat exchangers within this range of capacity have a power-sizing exponent (x) of 0.55.
• Five years ago the Heat Exchanger Cost Index (HECI) was 1 ,306; today it is 1 ,487.
SOLUTION
Miriam will 昀椀rst use Equation 1-3 and the 0.55 power-sizing exponent to scale up the cost of the 100 m 2
exchanger to one that is 250 m2 •
Cost of 250 m 2 equipment
250 o.ss
l
=
[
Cost of 100 m 2 equipment
100
250 0.55
Cost of 250 m 2 equipment = [ - ) X 50,000 = $82,763
100
(Note that our data are good to only two or three signi昀椀cant 昀椀gures, so it would be misleading to give
more than three signi昀椀cant 昀椀gures in the calculated cost.) Miriam knows that the $82,800 re昀氀ects only
the scaling up of the cost of the 100 m2 model to a 250 m 2 model. Now she will use Equation 1-2 and the
HECI data to estimate the cost of a 250 m2 exchanger today. Miriam's cost estimate would be
Equipment cost today
Equipment cost 昀椀ve years ago
Index v愀氀ue today
Index v愀氀ue 昀椀ve years ago
1,487
.
cost today = -- x $82,800 = $94,300
Eqwpment
1,306
Tria ngulation
Triangulation is used in engineering surveying. A geographical area is divided into
triangles 昀爀om which the surveyor is able to map points within that region by using three
昀椀xed points and horizontal angular distances to locate 昀椀xed points of interest (e.g., prop­
erty line reference points). Since any point can be located with two lines, the third line
represents a check. Triangulation in cost estimating might involve using di昀昀erent sources
of data or di昀昀erent quantitative models to con昀椀rm the value initially calculated. As deci­
sion makers we should always seek out varied perspectives.
I m provement and the Lea rn ing Cu rve
One common phenomenon, regardless of the task being per昀漀rmed, is that as the number
of repetitions increases, per昀漀rmance becomes 昀愀ster and more accurate. From our own
experience we all know that our 昀椀昀琀ieth repetition of a task takes much less time than we
needed initially.
䌀䐀
Estimating Models
29
吀栀e learning curve captures the relationship between task per昀漀rmance and task
repetition. In general, as output doubles, the unit production time will be reduced to some
昀椀xed percentage: the learning curve percentage or learning curve rate. For example,
it may take 300 minutes to produce the third unit in a production run involving a task
with a 95% learning time curve. In this case the sixth (2 x 3) unit will take 300(0.95)
= 285 minutes to produce. Sometimes the learning curve is known as the progress curve,
improvement curve, experience curve, or manufacturing progress function.
Equation 1-4 gives an expression that can be used 昀漀r estimating time in repetitive
tasks.
(1-4)
where
= time required 昀漀r the 一琀h unit of production
T. ·t · 1 = time required 昀漀r the 昀椀rst (initial) unit of production
N
= number of completed units (cumulative production)
b
= learning curve exponent (slope of the learning curve on a log-log plot)
TN
mi ia
A learning curve is o昀琀en referred to by its percentage learning slope. 吀栀us, a curve with
b = -0.0㜀㐀 is a 95% learning curve because 2-0• 074 = 0.95. 吀栀is equation uses 2 because the
learning curve percentage applies 昀漀r doubling cumulative production. 吀栀e learning curve
exponent is calculated 昀爀om Equation 1-5.
b=
log (learning curve e砀瀀ressed as a decim愀氀)
log 2.0
(1-5)
EXAM PLE 1-9
Calculate the time required to produce the 100th unit of a production run if the 昀椀rst unit took 32.0
minutes to produce and the learning curve rate 昀漀r production is 80%.
SO LUTI O N
T100 = TI X 1o otog0. 80 / log2 .0
TIOO
= 32.0 X 100 -0 , 32 1 9
T100 = 7.27 minutes
It is particularly important to account 昀漀r the learning curve e昀昀ect if the production
run involves a small number of units instead of a large number. When thousands or even
millions of units are being produced, early ine昀케ciencies tend to be averaged out because
of the larger batch sizes. However, in the short run, ine昀케ciencies of the same magnitude
can lead to rather poor estimates of production time requirements, and thus production
cost estimates may be understated. Consider Example 1-10 and the results that might be
observed if the learning curve e昀昀ect is ignored. Notice in this example that a "steady-state"
time is given. A昀琀er reaching steady state, the physical constraints of per昀漀rming the task
prevent any more learning or improvement.
30
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
EXAM PLE 1-10
Estimate the overall labour cost portion 昀漀r a task that has a learning curve rate of 85% and reaches a
steady-state value a昀琀er 16 units of 5 minutes per unit. Labour and bene昀椀ts are $22 per hour, and the task
requires two skilled workers. 吀栀e overall production run is 20 units.
SO LUTI O N
Because we know the time required 昀漀 r the 16th unit, we can use Equation 1-4 to calculate the time
required to produce the 昀椀rst unit.
T1 6 = T1 X 16log0.8 5/log2 .0
5.0 = T1 X 16-0 • 2345
T1 = 5.0 X 16 0. 2345
T1 = 9.6 minutes
Now we use Equation 1-4 to calculate the time requirements 昀漀r each unit in the production run as well
as the total production time required.
TN = 9.6 X N -0. 2345
Unit
Number, N
Time (min.) to Produce
Nth Unit
Cumulative Time
昀爀om 1 to N
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
9.6
8.2
7.4
6.9
6.6
6.3
6.1
5.9
5.7
5.6
5.5
5.4
5.3
5.2
5.1
5.0
5.0
5.0
5.0
5.0
9.6
17.8
25.2
32.1
38.7
45.0
51.1
57.0
62.7
68.3
73.8
79.2
84.5
89.7
94.8
99.8
104.8
109.8
114.8
119.8
吀栀e total cumulative time of the production run is 1 19.8 minutes (2 hours). 吀栀us the total labour cost
estimate would be
2 hours X $22/hour per worker X 2 workers = $88
Estimating Models
31
If we ignore the learning curve e昀昀ect and calculate the labour cost portion 昀爀om only the steady-state
labour rate, the estimate would be
0.083 hours/unit X 20 units X $22/hour per worker X 2 workers = $73.04
吀栀is estimate understates the true cost by about 20%.
Estimating Bene昀椀ts
吀栀is chapter has 昀漀cused on cost terms and cost estimating. However, engineering econo­
mists must o昀琀en also estimate bene昀椀ts. Bene昀椀ts would include sales of products, revenues
昀爀om bridge tolls and electric power sales, cost reductions 昀爀om reduced material or labour
costs, less time spent in tra昀케c jams, and reduced risk of 昀氀ooding.
吀栀e uncertainty in bene昀椀t estimates is also usually asymmetric, with a broader limit
昀漀r negative outcomes. Bene昀椀ts are more likely to be overestimated than underestimated,
so an example set of limits might be (-50%, +20%). One di昀昀erence between cost and
bene昀椀t estimation is that many costs of engineering projects occur in the near 昀甀ture (昀漀r
design and construction), but because bene昀椀ts are o昀琀en 昀甀rther in the 昀甀ture, they are
more di昀케cult to estimate accurately.
䌀䐀
Cash Flow Diagrams
吀栀e costs and bene昀椀ts of engineering projects over time can be summarized on a cash 昀氀ow
diagram (CFD) . A CFD illustrates the size, sign, and timing of individual cash 昀氀ows.
A CFD is created by drawing a segmented horizontal line, divided into time periods.
吀栀e time units on the CFD can be years, months, quarters, or any other consistent time
unit. At each time at which a cash 昀氀ow occurs, a vertical arrow is added-pointing down
昀漀r costs and up 昀漀r revenues or bene昀椀ts. 吀栀ese cash 昀氀ows are drawn approximately to scale.
吀栀e cash 昀氀ows are assumed to occur at the beginning or end of a period. Consider
Figure 1-7, the CFD 昀漀r a speci昀椀c investment opportunity whose cash 昀氀ows are described
as 昀漀llows:
Timing of Cash Flow
Size of Cash Flow
time 0 (now or today)
1 time period 昀爀om today
2 time periods 昀爀om today
3 time periods 昀爀om today
4 time periods 昀爀om today
5 time periods 昀爀om tod愀礀
positive cash 昀氀ow of $100
negative cash 昀氀ow of $100
positive cash 昀氀ow of $100
negative cash 昀氀ow of $150
negative cash 昀氀ow of $150
positive cash 昀氀ow of $50
t
$ 100
+/ t
(+)
(-)
t
$ 100
End of Period 2
is also Beginning
of Period 3
$50
Pe爀℀od 1
,
1 -- 2 -- 3 --- 4 --- 5
0
Time 0
("today")
$ 1 00
l
$ 1 50
FIGURE 1-7 An exa mple of a cash flow d iagram (CFD).
l
$ 150
t
32
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
Categories of Cash Flows
吀栀e expenses and receipts 昀爀om engineering projects usually 昀愀ll into one of the 昀漀llowing
categories:
First cost
expense of building or of buying and installing
Revenues
annual receipts 昀爀om sale of products or services
Operations and maintenance (O&M) annual expense, such as electricity, labour,
and minor repairs
Salvage value receipt at project termination 昀漀r sale or transfer of the equipment
(can be a salvage cost)
Overhaul
major capital expenditure that occurs during the li昀攀 of the asset
Drawi ng a Cash Flow Diagram
吀栀e cash flow diagram shows when all cash 昀氀ows occur. Look at Figure 1-7 and the $100
positive cash 昀氀ow at the end of Period 2. From the timeline one can see that this cash
昀氀ow can also be described as occurring at the beginning of Period 3. In a CFD, the end of
Period N is always the same time as the beginning of Period N + 1. Beginning-of-period
cash 昀氀ows (such as rent, lease, and insurance payments) are thus easy to handle: just draw
your CFD and put them in where they occur. O&M, salvages, revenues, and overhauls are
assumed to be end-of-period cash flows.
吀栀e choice of Time O is arbitrary. For example, it can be when a project is analyzed,
when 昀甀nding is approved, or when construction begins. When construction periods are
short, 昀椀rst costs are assumed to occur at Time 0, and the 昀椀rst annual revenues and costs
start at the end of the 昀椀rst period. When construction periods are long (several years, 昀漀r
example), Time O is usually the date of commissioning-when the 昀愀cility comes on stream.
Point of view is also important when one is drawing a CFD. Consider the simple trans­
action of paying $5,000 昀漀r some equipment. To the 昀椀rm buying the equipment, the cash
flow is a cost and hence negative. To the 昀椀rm selling the equipment, the cash flow is a rev­
enue and hence positive.
O昀琀en two or more cash 昀氀ows occur in the same year, such as an overhaul and an O&M
expense or the salvage value and the last year's O&M expense. Although combining these
into one total cash 昀氀ow per year would simpli昀礀 the cash 昀氀ow diagram, it is better to show
each individually to ensure a clear connection 昀爀om the problem statement to each cash
flow in the diagram.
S U M MARY
Classifyi ng Problems
Many problems are simple and thus easy to solve. Others are of intermediate di昀케culty and
need considerable thought or calculation to evaluate properly. 吀栀ese intermediate prob­
lems tend to have a substantial economic component and hence are good candidates 昀漀r
economic analysis. Complex problems o昀琀en contain human elements along with political
and economic components. Economic analysis is still very important, but the best alterna­
tive must be selected with all criteria in mind.
S u m m a ry
The Decision -Making Process
Engineering decision making refers to solving substantial engineering problems in which
economic aspects dominate. Decision making uses a logical method of analysis to choose
the best alternative 昀爀om among the feasible alternatives.
Some of the unusual aspects of engineering decision making are as 昀漀llows:
Cost-accounting systems, while an important source of cost data, contain alloca­
tions of indirect costs that may be unsuitable 昀漀r use in economic analysis.
2. 吀栀e various consequences-costs and bene昀椀ts-of an alternative may be of three
types:
• Market consequences-there are established market prices.
• Extra-market consequences-there are no direct market prices, but prices
can be assigned by indirect means.
• Intangible consequences-value is determined by judgment, not by
monetary prices.
3. To choose among the alternatives, we organize the market consequences and
extra-market consequences into a cash flow diagram. We will see in Chapter 3
how engineering economic calculations can be used to compare di昀昀erent cash
flows. 吀栀ese outcomes are compared against the selection criteria. From this com­
parison plus the consequences not included in the monetary analysis, the best
alternative is chosen.
4. An essential part of engineering decision making is the audit of results. 吀栀is step
helps to ensure that the projected bene昀椀ts are obtained and to encourage realistic
estimates in analyses.
1.
I m porta nce of Eth ics i n Engi neeri ng a nd Engi neeri ng Economy
One of the gravest responsibilities of an engineer is to protect the safety of the public,
clients, and employees. In addition, the engineer can be responsible 昀漀r the economic per­
昀漀rmance of projects and products on which bonuses and jobs depend. Not surprisingly, in
this environment one of the most valued personal characteristics is integrity.
Estimating Costs
Fixed costs are constant and unchanging as volumes change, whereas variable costs change
as output changes. Fixed and variable costs are used to 昀椀nd a break-even value between
costs and revenues. A marginal cost is 昀漀r one more unit, whereas the average cost is the
total cost divided by the number of units.
Sunk costs result 昀爀om past decisions and should not in昀氀uence our attitude toward
current and 昀甀ture opportunities. Opportunity costs involve the bene昀椀t that is 昀漀rgone
when we choose to use a resource in one activity instead of another. Recurring costs can
be planned and anticipated expenses; non-recurring costs are one-of-a-kind costs that are
o昀琀en more di昀케cult to anticipate.
Incremental costs are economic consequences associated with the di昀昀erences between
two choices of action. Cash costs are also known as out-of-pocket costs. Book costs do not
result in the exchange of money, but rather are costs listed in a 昀椀rm's accounting books.
Life-cycle costs are all costs that are incurred over the life of a product, process, or service.
吀栀us engineering designers must consider life-cycle costs when choosing materials and
33
34
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
components, tolerances, processes, testing, safety, service and warranty, and method of
disposal.
Cost estimating is the process of developing the numbers 昀漀r engineering economic
analysis. Rough estimates give us order-of-magnitude numbers and are use昀甀l 昀漀r high­
level and initial planning as well as 昀漀r judging the feasibility of alternatives. Budget
estimates are more accurate than rough-order estimates, thus requiring more resources to
develop. 吀栀ese estimates are used in preliminary design and budgeting. Detailed estimates
generally have an accuracy of ±3%-5%. 吀栀ey are used during the detailed design and con­
tract bidding phases of a project.
One-of-a-kind estimates will have no basis in earlier work, but this disadvantage can
be addressed through estimation by analogy.
Several general models and techniques 昀漀r developing cost estimates were discussed.
吀栀e per-unit and segmenting models use di昀昀erent levels of detail and costs per square
metre or other unit. Cost index data are use昀甀l 昀漀r updating historical costs to 昀漀rmulate
current estimates. 吀栀e power-sizing model is use昀甀l 昀漀r scaling up or down a known cost
quantity to account 昀漀r economies of scale, with di昀昀erent power-sizing exponents 昀漀r in­
dustrial plants and equipment of di昀昀erent types. Triangulation suggests that to provide a
check one should use multiple perspectives when developing cost estimates. As the number
of task repetitions increases, e昀케ciency improves because of learning or improvement. 吀栀is
is summarized in the learning curve percentage, where doubling the cumulative produc­
tion reduces the time to complete the task.
Cash 昀氀ow estimation must include project bene昀椀ts. 吀栀ese include labour cost savings,
quality costs 愀瘀oided, direct revenue 昀爀om sales, reduced catastrophic risks, improved
tra昀케c 昀氀ow, and cheaper power supplies. Cash flow diagrams are used to model the positive
and negative cash flows of potential investment opportunities. 吀栀ese diagrams provide a
consistent view of the problem (and the alternatives) to support economic analysis.
PROBLEMS
In each chapter of this text, we o昀昀er problems grouped
under particular topics, 昀漀llowed by a section, "Un­
classi昀椀ed," in which the student must decide which of
the methods covered in the chapter is most appropri­
ate 昀漀r the particular problem.
Decision Making
1-1
Some of the 昀漀llowing problems would be suit­
able 昀漀r solution by engineering economic an­
alysis. Which ones?
(a) Would it be better to buy a car with a diesel
engine or a gasoline engine?
(b) Should an automatic machine be pur­
chased to replace three workers now doing
a task by hand?
(c) Would it be wise to enrol 昀漀r an
early-morning class so you could avoid
travelling during the morning rush hour?
(d) Would you be better o昀昀 if you changed
your m愀樀or?
(e) One of the people 礀漀u might marry has a job
that p愀礀s 瘀攀ry little money, while another one
has a professional job with an excellent salary.
Which one should you marry?
1-2
⠀　Which one of the 昀漀llowing problems is most
suitable 昀漀r analysis by engineering economic
analysis?
(a) Some 45¢ chocolate bars are on sale at 12
bars 昀漀r $3. Sandy, who eats a couple of
chocolate bars a week, must decide whether
to buy a dozen at the lower price.
(b) A woman has $150,000 in a bank chequing
account that p愀礀s no interest. She can either
invest it immediately at a desirable interest
rate or wait a week and obtain an interest
rate that is 0.15% higher.
Proble ms
挀儀
1-3
1-4
1-5
Joe backed his car into a tree , damaging the
fender . He has car insurance that wil l pay
for the fender repair . But ifhe files a claim
for payment , they may change his "good
driver " rating downward and charge him
more for car insu爀愀nce in
the future.
1-9
Toward the end of the twent ieth century,
the US government wanted to save money
by closing a small portion of all its m氀툀tary
insta氀氀ations th爀漀ughout the US . Though
many people agreed that saving money was a
desirable goal, the parts of the country that
might be a昀昀ected by a closing soon reacted
negatively. Congress fina氀氀y set up a panel of
people whose task was to develop a 氀椀st of in­
sta氀氀ations to close, with the legislation spe挀椀­
f ying that Congress could not a氀琀er the 氀椀st.
Since the goal was to save money, why was
this p爀漀blem so hard to solve?
1- 6
Suppose you have just two ho urs to ans wer
the fo氀氀owing question. How many people in
your home town wo uld be interes ted in buying
a pair o昀氀eft- handed s挀椀ssors ? Give a step -bystep ou琀氀ine of how you would seek to answer
this question within two hours .
1-7
A university student ca氀挀ulates that he will
have only $5 0 0 per month ava椀氀able for his
housing for the coming year. He is determined
to continue in the un iversity, so he has de挀椀ded
to list all 昀攀asible a氀琀ernat ives for his housing.
吀漀 help him, list five 昀攀asible a氀琀ernatives.
1-8
Describe a situation where a poor altern at ive
was chosen bec ause there was a poor search 昀漀r
better alternatives.
The th ree econom椀挀 c rite ria 昀漀r choosing the
best alte rnative a re minimize input , max i­
mize output , and maximize the diffe rence
between output and input . For each of the 昀漀 l­
lowing situations , what is the right econom椀挀
c riterion?
愀儀 A ma nu昀愀cturer of plast椀挀 d rafting t r i­
angles can sell a氀氀 the triangles he can
p爀漀duce at a fixed price . As he inc reases
p爀漀duction, his unit costs increase as
a result of overtime pay and so 昀漀rth.
The manu昀愀cturer's c爀椀terion should
be- -�
(戀⤀ An architectural and enginee爀椀ng firm
has been 愀眀arded the contract to design
a wharf for a pet爀漀leum company for a
fixed sum of money. The engineering
firm 's criterion should be- - 挀儀 A book pub氀椀sher is about to set the 氀椀st
pr椀挀e (retail pr椀挀e) on a textbook. The
cho椀挀e of a low 氀椀st pr椀挀e would mean
less advertising than would be used 昀漀r
a higher 氀椀st pr椀挀e. The amount of adver­
tising will affect the number of copies
sold. The pub氀椀sher 's criterion should
The uni瘀攀rsity bookstore has put pads of engin­
eering computation paper on sale at half pr椀挀e.
What is the minimum and maximum number
ofpads you might buy during the sale? Explain.
A car manufacturer is considering locating
a car assembly plant in your region. List two
simple, two intermediate, and two complex
problems asso挀椀ated with this proposal.
35
搀儀
1-10
be- ·At an auction of antiques, a bidder 昀漀r
a part椀挀ular porcelain statue would be
tr ying to_
_
_
As in Problem 1 -9, state the right economic cri­
terion 昀漀r each of the 昀漀氀氀owing situations.
愀儀 The engineering school held a ra昀툀e of
a car with t椀挀ke ts se氀픀ng 昀漀r 50¢ each or
three 昀漀r $ 1 . When the s tudents were se氀氀­
ing t椀挀kets, they noted that many people
had tr o uble deciding whether to buy one
or three t椀挀kets . This ind椀挀ates that the
buyers ' criterion was_ _
_ .
(戀⤀ A student organization bo ught a soft­
d爀椀nk machine 昀漀r use in a st udent area .
There was consider able disc ussion o ver
whether they should s et the machine to
charge 50¢, 75¢, or $ 1 per d爀椀nk. The or­
ganiz ation recognized that the number
of so昀琀 dr inks sold wo uld depend on the
pr ice charged. Eventually the decision
was made to charge 75¢ . 吀栀eir cr iter ion
was _ _
븀餀
_
36
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
move into the residence as his replacement.
吀栀e cost of residence was $3,200 昀漀r the two
terms, which Jovis had already paid.
A month a昀琀er he moved into residence, he
decided he would prefer to live in an apart­
ment. 吀栀at week, a昀琀er some searching 昀漀r
a replacement to buy his contract, Jovis had
two o昀昀ers. One student o昀昀ered to move in im­
mediately and to pay Jovis $300 per month
昀漀r the seven remaining months of the school
year. A second student o昀昀ered to move in in
the second term and pay Jovis $1,600 when he
moves in.
Jovis estimates his 昀漀od costs $500 a month
if he lives in residence and $350 if he lives in an
apartment with three other students. His share
of the apartment rent and utilities will be $400
a month. Assume each term is 昀漀ur months
long. Disregard the small di昀昀erences in the
timing of the disbursements or receipts.
(a) What are the three alternatives available
to Jovis?
(b) Evaluate the cost of each of the
alternatives.
(c) What do you recommend that Jovis do?
(c) In many cities, grocery stores 昀椀nd that
their sales are much greater on days
when they have advertised their special
bargains. However, the advertised special
prices do not appear to increase the total
physical volume of groceries sold by a
store. 吀栀is leads us to conclude that many
shoppers' criterion is ___븀謀
搀儀 A recently graduated engineer has de­
cided to return to school in the evenings
to obtain a master's degree. He feels it
should be accomplished in a manner that
will allow him the maximum amount of
time 昀漀r his regular day job plus time 昀漀r
recreation. In working 昀漀r the degree, he
will----·
1-11
Seven criteria are given in the chapter 昀漀r judg­
ing which alternative is best. A昀琀er reviewing
the list, devise three additional criteria that
might be used.
1-12
Suppose you are assigned the task of deter­
mining the route of a new highway through an
older section of town. 吀栀e highway will require
that many older homes be either moved or torn
down. Two possible criteria might be used in
deciding exactly where to locate the highway:
愀儀 Ensure that there are bene昀椀ts 昀爀om the
decision and that no one is harmed by the
decision.
(b) Ensure that the bene昀椀ts to those who
gain 昀爀om the decision are greater than
the losses of those who are harmed by the
decision.
Which criterion will you choose to use in de­
termining the route of the highway? Explain.
1-13
For the project in Problem 1-12, name the m愀樀or
costs and bene昀椀ts. Which are market conse­
quences, which are extra-market consequences,
and which are intangible consequences?
1-14
In the 昀愀ll, Jovis 吀栀ompson decided to live in a
college residence. He signed a contract under
which he was obligated to pay the room rent
昀漀r the 昀甀ll college year. One clause stated that
if he moved out during the year, he could sell
his contract to another student, who would
1-15
An electric motor on a conveyor burned out.
吀栀e 昀漀reman told the plant manager that the
motor had to be replaced. 吀栀e 昀漀reman said
there were no alternatives and asked 昀漀r au­
thorization to order the replacement. In this
situation, is any decision making taking place?
If so, who is making the decision(s)?
1-16
State the alternatives, outcomes, criteria, and
process 昀漀r the selection of your university
program. Did you make the best choice?
1-17
Apply the steps of the decision-making process
to your situation as you decide what to do a昀琀er
graduation.
1-18
One strategy 昀漀r solving a complex problem is
to break it into a group of less complex prob­
lems and then 昀椀nd solutions to the smaller
problems. 吀栀e result is the solution of the com­
plex problem. Give an example in which this
strategy will work. 吀栀en give another example
in which this strategy will not work.
Problems
Eth ics
1 - 19
W hen yo u make pr ofessional decisions involv­
ing i n ve stments in engineer ing p r ojects, what
cr iter ia will yo u use?
1 - 20
Suppose yo u are an engineer working in a
pr ivate engineering 昀椀rm and yo u are asked
to sign doc uments veri昀礀ing in昀漀rmation that
yo u belie ve is not tr ue . Y o u 氀椀ke yo ur work and
yo ur colleag ues in the 昀椀rm, and your 昀愀m椀氀y
depends on your income . What crite爀椀a can
yo u use to g uide your de cision regarding
this iss ue ?
1 -21
1 -22
Use a personal example or a p ub氀椀shed so urce
to analyze what went w爀漀ng or right with re­
spect to eth椀挀s at each of the 昀漀氀氀owing stages of
the de挀椀sion making.
愀儀 Recognize the p爀漀blem.
(戀⤀ Define the goal or objec tive.
挀儀 Assemble rele vant data.
搀儀 Identify 昀攀asible a氀琀ernatives.
攀儀 Select the criteria 昀漀r determining the best
alternative.
樀儀 Construct a model.
最儀 Pred椀挀t the outcomes or consequences 昀漀r
each a氀琀ernative.
(栀⤀ Choose the best a氀琀ernative.
椀儀 Audit the resu氀琀s.
One of the important responsib氀툀ties of muni­
挀椀pal counc椀氀s and school boards is the main­
tenance of pub氀椀c in昀爀astructure, such as 爀漀ads
and schools. 吀栀e decisions made in that area
can be imp爀漀ved by the skills, knowledge, and
perspectives of engineers. Often, this pub氀椀c
爀漀le is a part -time one, since the engineers that
昀甀lfill it will also h愀瘀e 昀甀氀氀 -time jobs as employ­
ees or owners of enginee爀椀ng firms.
愀儀 What are some of the eth椀挀al issues
that can arise 昀爀om confl椀挀ts between
the pub氀椀c and private sector 爀漀les of
eng i neers?
(戀⤀ What guideli nes 昀漀r p爀漀cess or outcomes
are the re i n your hometown?
挀儀 Su mmarize and analyze an example of
such a situation 昀爀om a newspaper or
news magaz i ne .
37
1-23
吀栀e problems of an increasing population and
tra昀케c j ams are o昀琀en addressed through road
improvement projects. 吀栀ese may pit the in­
terests of homeowners and business owners in
the p r oj ect area a gainst the interests of people
tr avel氀椀ng throu gh the improvement proj ect.
愀儀 What ethical issues can arise?
(戀⤀ Summarize and analyze an example of
such a situation from a news paper or
news magazine .
挀儀 What g uide氀椀nes for pro cess or outcomes
can h攀氀p with the eth椀挀al issues of your
exam ple?
1- 24
Economic de velopment and redevelopment
often require significant acreage that is assem­
bled by acquiring sma氀氀er parcels . Sometimes
this is done th爀漀ugh simple purchase, but the
p爀漀per ty of an "unw氀픀ing se氀氀er" can be ac­
quired by the federal government under the
p爀漀visions of the E砀瀀r漀瀀riation Act.
愀儀 What are some of the eth椀挀al issues that
can arise?
(戀⤀ Summarize and analyze an example of
such a situation from a newspaper or
news magazine.
挀儀 What guide氀椀nes for p爀漀cess or out挀漀mes
can help with the eth椀挀al issues in your
挀椀ty or p爀漀vince?
1-25
P爀漀vin挀椀al governments use a variety of ad­
visory and regulatory bodies with such re­
sponsib氀툀ties as the 氀椀censing of p爀漀fessional
engineers and the pri挀椀ng and ope爀愀tion of
regulated ut氀툀ties. Engineers b爀椀ng sk氀픀s,
knowledge, and perspectives that are useful
in this context. Often the pub氀椀c 爀漀le is a part­
time one, and engineers that fulfill it will also
h愀瘀e fu氀氀 -time jobs as employees or owners of
engineering firms.
愀儀 What are some of the eth椀挀al issues
that can arise from confl椀挀ts between
the public and private sector 爀漀les of
eng ineers?
(戀⤀ What guidelines for p爀漀cess or outcomes
?
are there in your p爀漀v ince
(挀⤀ Summarize and analyze an example of
such a situation from a newspaper or
news magazine .
38
C h a p t e r 1 Eco n o m i c D e c i s i o n s, E n g i n ee r i n g Costs, and Cost Est i m a t i n g
1- 2 6
Many engineer s work in pr ov incial go ver nments, and some ar e in high - pr o昀椀le positions
as legislator s, dep ut y minister s, and so on.
Many of these people mo v e between working
in the pr ivate and public sector s .
愀儀 W hat are some of the ethical issues that
can ar ise as individuals shi昀琀 between
pub氀椀c and pr ivate sector positions?
(戀⤀ W hat g uidelines 昀漀r pr ocess or outcomes
昀漀r such sit uations are there in your
pr o vince?
挀儀 Summarize and analyze an example of
such a sit uation 昀爀om a ne wspaper or
ne ws magazine .
1- 27
P愀礀ment 昀漀r o ver time ho urs is reg ulated by
the pr o vin挀椀al go vernment. Because most
engineering work is done as par t of a pr o ject, it is common 昀漀r engineers to be asked
or required to work o ver time as projec ts near
their dead氀椀nes. Sometimes the over time is
paid at time and a half, some times s traight
time, and sometimes the engineer 's salary remains constant e ven when there is overtime.
In a part椀挀ular firm, engineering interns,
engineers, and partners may be treated the
same way or differently.
愀儀 What are some of the legal and eth椀挀al
issues that can arise?
(戀⤀ What is the law or regulation about
overtime in your p爀漀vince?
挀儀 Summarize and analyze the po氀椀挀椀es
of a particular firm or the data on how
common the different p爀愀ct椀挀es are.
1 -28
At both p爀漀vin挀椀al and 昀攀deral levels, legisla­
tors can be involved in "pork barr攀氀" 昀甀nding
of capital pr o j ects. These p爀漀jects may even
bypass the economic evaluation using engin­
eering econo my that normal p爀漀 j ects are sub­
ject to.
愀儀 What are some of the ethical issues that
ca n arise?
(戀⤀ Su mmarize and analyze an example of
such a situation from a newspaper or
news magazine .
What
guidelines 昀漀r p rocess or outcomes
挀儀
a re the re 昀漀r the agency or legislative body
?
involved in your examp le
1-29
At the international level, a common ethics
iss ue that is important to engineering and
p r oject j usti昀椀cation is that of environmental
reg ulation. O昀琀en di ffe rent nations have dif­
ferent envi爀漀nmental standards, and a p roject
or pro duct might be built in any one of several
di ffe rent countries.
愀儀 W hat are some of the ethical issues th at
can ar ise?
(戀⤀ Summa爀椀ze and analyze an example of
such a situation from a newspaper or
news magazine .
挀儀 What guide氀椀ne s are there for pro cess or
outcomes that address the eth椀挀al issues
of your example?
1-30
At the international le vel, a common ethics
issue that is impor tant to engineering and pro ­
ject jus tification is that of worker hea氀琀h and
safe ty. Often different nations h愀瘀e di昀昀erent
standards, and a p爀漀jec t or produc t might be
built in any one of several different countries.
(a) What are some of the eth椀挀al issues that
can arise?
(戀⤀ Summa爀椀ze and analyze an example of
such a situation from a newspaper or
news magazine.
挀儀 What guide氀椀nes for p爀漀cess or outcomes
address the eth椀挀al issues of your example?
1-31
At the international level, engineering de挀椀­
sions are cru挀椀al in ma琀琀ers of "sustainable de­
velopment, " a common eth椀挀s issue.
(a) What are some of the eth椀挀al issues that
can arise?
(戀⤀ Summarize and analyze an example of
such a situation from a newspaper or
news magazine .
挀儀 What guide氀椀nes for p爀漀cess or outcomes
address the eth椀挀al issues of your example?
1-32
At the international level, questions arise about
whether Canada's ban (by the Cor爀甀ption of
䘀漀 reign Public O昀케cials Ac t) on the use ofbr ib­
e ry by Canadians working ab爀漀ad is practical
or app爀漀priate . In some countries government
workers are very poorly paid , and they can sup­
r
port thei families only by accepting money to
"facilitate " a p爀漀cess.
Problems
(a) What are some of the ethical issues that
can arise?
(b) Summarize and analyze an example of
such a situation 昀爀om a newspaper or news
magazine.
(c) What guidelines 昀漀r process or outcomes
address the ethical issues of your example?
1-33
吀栀e decision to launch the Challenger shuttle
has been extensively analyzed. Brie昀氀y sum­
marize the main institutional groups, how the
decision was made, and the ethical principles
that may h愀瘀e been compromised.
1-34
One of the elements in the 昀氀ooding of New
Orleans during Hurricane Katrina was the
昀愀ilure of some of the levees that protected the
city. Outline the role that ethical 昀愀ilures by
engineers may have played in this situation.
How could society structure decision making
to minimize such 昀愀ilures?
1-35
1-36 If you rent a car you can (1) return it with a 昀甀ll
gas tank, (2) return it without 昀椀lling it and pay
$1.15/litre, or (3) accept a 昀椀xed price of $40 昀漀r
gas. 吀栀e local price is $0.97/litre 昀漀r gasoline,
and you expect this car to get 8.5 L/100 km. 吀栀e
car has a 75-litre tank. What choice should you
make if you expect to drive
(a) 250 km?
(b) 400 km?
(c) 800 km?
(d) How do your answers change if stopping
at the 昀椀lling station takes 15 minutes and
your time is worth $30/hour?
1-37
(b) if gas costs $1.00 per litre and your time is
worth $12/hour?
(c) if gas costs $1.25 per litre and your time is
worth $9/hour?
1-38
Toyota has a long history of engineering-driven
high-quality production processes. In 2014 a
series of problems with braking raised ques­
tions about quality and ethics. Summarize
what seems to have caused the problems, how
they have been dealt with, and how success昀甀l
Toyota has been in restoring its reputation.
Cu rrent Costs
Your car gets 9.5 L/100 km at 90 km per hour
(kph) and 1 1.5 L/100 km at l lO kph. At what
speed should you make an 800 km trip
(a) if gas costs $0.90 per litre and your time is
worth $18/hour?
39
Maria, a university student, is getting ready
昀漀r three 昀椀nal examinations. Between now and
the start of exams, she has 15 hours of study
time available. She would like to get the high­
est possible overall grade 愀瘀erage based on her
grades in her math, physics, and engineering
economy classes. She feels she must study at
least two hours 昀漀r each course and, if neces­
sary, will settle 昀漀r the low grade that the lim­
ited study would yield. How much time should
Maria devote to each class if she estimates her
grade in each subject as 昀漀llows:
Mathematics
P栀礀sics
Engineering
Economy
Study
Hours Grade
Study
Hours Grade
Study
Hours Grade
2
3
4
5
6
7
8
1-39
25
35
44
52
59
65
70
2
3
4
5
6
7
8
35
41
49
59
68
77
85
2
3
4
5
6
7
8
50
61
71
79
86
92
96
A city needs to increase its rubbish-disposal 昀愀­
cilities. 吀栀ere is a choice of two disposal areas,
as 昀漀llows.
Area A: A gravel pit with a capacity of 16 million
cubic metres. Owing to the possibility of
high groundwater, however, the provincial
environment ministry has restricted the
lower 2 million cubic metres of 昀椀ll to inert
material only (earth, concrete, asphalt,
p愀瘀ing, brick, etc.). 吀栀e inert material, prin­
cipally clean earth, must be purchased and
hauled to this area 昀漀r the bottom 昀椀ll.
Area B: Capacity is 14 million cubic metres.
吀栀e entire capacity may be used 昀漀r general
rubbish disposal. 吀栀is area will require an
average increase in a round-trip haul of
8 km 昀漀r 60% of the city and a decreased
haul of 3 km 昀漀r 20% of the city. For the
remaining 20% of the city, the haul is the
same distance as 昀漀r Area A.
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
40
organize its assembly operation. Skilled work­
ers, at $33 per hour each, could be assigned to
assemble the test equipment individually. Each
worker would do all the assembly steps, and
it would take 2.6 hours to complete one unit.
An alternative approach would be to set up
teams of 昀漀ur less-skilled workers (at $19 per
hour each) and organize the assembly tasks so
that each worker does part of the assembly. 吀栀e
昀漀ur-person team would be able to assemble a
unit in one hour. Which approach would result
in more-economical assembly?
Assume the 昀漀llowing conditions:
•
•
•
•
•
Cost of inert material placed in Area A
will be $9.40/m3 •
䄀瘀erage speed of trucks 昀爀om last pickup
to disposal site is 25 km/hr.
吀栀e rubbish truck and a two-man crew
will cost $140/hr.
Truck capacity is 4½ tonnes per load, or
20 m3 •
Su昀케cient cover material is available at
all areas; however, inert material 昀漀r the
bottom 昀椀ll in Area A must be hauled in.
Which of the sites do you recommend?
1-40 An oil company is considering adding an addi­
tional grade of 昀甀el at its service stations. To do
this, an additional 12,000-litre tank must be
buried at each station. Discussions with tank
昀愀bricators indicate that the least expensive
tank would be cylindrical with minimum sur­
昀愀ce area. What size of tank should be ordered?
1-41
A 昀椀rm is planning to manu昀愀cture a new prod­
uct. 吀栀e sales department estimates that the
quantity that can be sold depends on the sell­
ing price. As the selling price is increased, the
quantity that can be sold decreases. Numeric­
ally they estimate
P = $35.00 - 0.02 Q
where P = selling price per unit
Q = quantity sold per year
On the other hand, the management estimates
that the 愀瘀erage cost of manu昀愀cturing and
selling the product will decrease as the quan­
tity sold increases. 吀栀ey estimate
C = $4.00 Q + $8,000
where C = cost to produce and sell Q per year
吀栀e 昀椀rm's management wishes to produce and
sell the product at the rate that will maximize
pro昀椀t; that is, where income minus cost will be
a maximum. What quantity should the deci­
sion makers plan to produce and sell each year?
1-42
A manu昀愀cturing 昀椀rm has received a contract
to assemble 1,000 units of test equipment in
the next year. 吀栀e 昀椀rm must decide how to
1-43
A grower estimates that if she picks her apple
crop now, she will obtain 1,000 boxes of apples,
which she can sell at $3 a box. However, she
thinks her crop will increase by 120 boxes of
apples 昀漀r each week she delays picking but that
the price will drop at a rate of 15¢ a box per
week; in addition, she estimates approximately
20 boxes a week will spoil 昀漀r each week she
del愀礀s picking. When should she pick her crop
to obtain the largest total cash return? How
much will she receive 昀漀r her crop at that time?
1-44 On her 昀椀rst engineering job, Joy Hayes was
given the responsibility of determining the
production rate 昀漀r a new product. She has as­
sembled data as indicated on two graphs:
$2,000
$ 1,900
$ 1,800
$ 1,700
$ 1,600
$ 1,500
$ 1,400
$ 1,300
㨀夀 $ 1,200
� $ 1, 100
� $ 1,000
∀츀 $900
$800
$700
$600
$500
$400
$300
$200
$ 100
0
50
100 1 50 200
Output (units/hr)
250
Problems
(a) 75
(b) 125
(c) 250
$16
$15
·a
$ 14
� $13
樀鈀
1- 47
Venus Computer can produce 23,000 per­
sonal computers a year on its daytime shi昀琀.
吀栀e 昀椀xed manu昀愀cturing costs per year
are $2 million, and the total labour cost is
$9, 109,000. To increase its production to
46,000 computers per year, Venus is consid­
ering adding a second shi昀琀. 吀栀e unit labour
cost 昀漀r the second shi昀琀 would be 25% higher
than 昀漀r the day shi昀琀, but the total 昀椀xed
manu昀愀cturing costs would increase only to
$2.4 million 昀爀om $2 million.
(a) Compute the unit manu昀愀cturing cost 昀漀r
the daytime shi昀琀.
(b) Would adding a second shi昀琀 increase or
decrease the unit manu昀愀cturing cost at
the plant?
1-48
Two automatic systems 昀漀r dispensing maps
are being compared by a provincial highway
department. 吀栀e accompanying break-even
chart of the comparison of these systems
(System I vs System II) shows total yearly costs
昀漀r the number of maps dispensed per year
昀漀r both alternatives. Answer the 昀漀llowing
questions.
(a) What is the 昀椀xed cost 昀漀r System I?
(b) What is the 昀椀xed cost 昀漀r System II?
(c) What is the variable cost per map
dispensed 昀漀r System I?
(d) What is the variable cost per map
dispensed 昀漀r System II?
(e) What is the break-even point in terms of
maps dispensed at which the two systems
have equal annual costs?
昀儀 For what range of annual number
of maps dispensed is System I
recommended?
最儀 For what range of annual number
of maps dispensed is System II
recommended?
栀儀 At 3,000 maps a year, 眀栀at are the
marginal and 瘀였rage map costs 昀漀r each
system?
$ 12
·㨀甀 $ 1 1
开开
.s $ 1 0
樀鈀
$9
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
$8
- - - � - - - - J_ _ _ _ J_ _ _ _
$7
- - - � - - - - J_ _ _ _ J_ _ _ _ J_ _ _ _
0
50
�
I
I
I
I
100 150 200
Output (units/hr)
250
(a) Select a suitable economic criterion and estimate the production rate
based on it.
(b) Joy's boss told her: "I want you to maxi­
mize output with minimum input." Joy
wonders if it is possible to meet her boss's
criterion. She asks your advice. What
would you tell her?
Fixed, Va riable, Average,
and Marg i na l Costs
1-45
1-46
41
One of your 昀椀rm's suppliers discounts prices
昀漀r larger quantities. 吀栀e 昀椀rst 1,000 parts are
$13 each. 吀栀e next 2,000 are $12 each. All parts
in excess of 3,000 cost $11 each. What are the
average and marginal costs per part 昀漀r each of
the 昀漀llowing quantities?
(a) 500
(b) 1,500
挀儀 2,500
(d) 3,500
A new machine comes with 100 昀爀ee service
hours over the 昀椀rst year. Additional time costs
$75 per hour. What are the average and mar­
ginal costs per hour 昀漀r the 昀漀llowing quanti­
ties of service hours?
42
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
•
Total Cost, System I: y = 0. 9 0x + 1 . 0
8
-------- - -Cost,
- - - -Total
System II:
y = O.lOx + 5.0
•
1-51
35
o 开贀_______
30
蠁‫_______܀‬开䨀
0
⠀　5
� 25
0
Maps Dispensed per Year (thousands)
A privately owned summer camp 昀漀r youngsters
has the 昀漀llowing data 昀漀r a 12-week session:
Charge per camper
Fixed costs
Variable cost per camper
Capacity
1-50
Consider the accompanying break-even graph
昀漀r an investment, and answer the 昀漀llowing
questions as they pertain to the graph.
40
2
1-49
the break-even point 昀漀r the two rides
in terms of number of visitors
the ranges of visitors per year where
each alternative is preferred
$480 a week
$92,000 a session
$320 a week
200 campers
0
500
750
1 ,000 1 ,250 1 ,500 1 ,750
Output (units/year)
(a) Give the equation 昀漀r total revenue 昀漀r
x units per year.
(b) Give the equation 昀漀r total costs 昀漀r
x units per year.
(c) What is the break-even level of x?
(d) If you sell 1,500 units this year, will you
have a pro昀椀t or loss? How much?
(e) At I,500 units, what are your marginal
and average costs?
(a) Develop the mathematical relationships
昀漀r total cost and total revenue.
(b) What is the total number of campers that
will allow the camp to just break even?
(挀⤀ Wh愀琀 is the pro椀� or loss 昀漀r the 12-week
session if the camp operates at 80% c愀瀀acity?
(d) What are the marginal and average costs
per camper at 80% capacity?
吀眀o new rides are being compared by a local
amusement park in terms of their annual
operating costs. 吀栀e two rides can generate
the same revenue (hence the 昀漀cus on costs).
吀栀e 吀甀mmy Tugger has 昀椀xed costs of $10,000 a
year and variable costs of $2.50 per visitor. 吀栀e
Head Buzzer has 昀椀xed costs of $4,000 per year
and variable costs of $4 per visitor. Answer the
昀漀llowing questions so the amusement park
can make the needed comparison.
(a) Determine mathematically the break­
even number of visitors per year 昀漀r the
two rides to have equal annual costs.
(b) Develop a graph that illustrates the 昀漀llow­
ing (note: put visitors per year on the hori­
zontal axis and costs on the vertical axis):
• accurate total cost lines 昀漀r the two
alternatives (show line, slopes, and
equations)
250
1-52
吀栀ree alternative designs have been created
by Snakisco engineers 昀漀r a new machine
that spreads cheese between the crackers in a
Snakisco snack. 吀栀e costs 昀漀r the three designs
(where x is the annual production rate of boxes
of cheese crackers) are as 昀漀llows:
Design
Fixed Cost
Variable Cost
($/x)
A
$100,000
350,000
600,000
20.Sx
10.Sx
8.0x
B
C
(a) Determine mathematically which of the
machine designs would be recommended
昀漀r di昀昀erent levels of annual production
of boxes of snack crackers. Management
is interested in the production interval
Problems
of 0-150,000 boxes of crackers per year.
Over what production volume would each
design (A or B or C) be chosen?
(b) Depict your solution 昀爀om part (a) graph­
ically, so that management can easily see
the 昀漀llowing:
i. accurate total cost lines 昀漀r each al­
ternative (show line, slopes, and line
equations)
ii. any relevant break-even or crossover
points
iii. ranges of annual production where
each alternative is preferred
iv. clear titles and labels 昀漀r the graph
1-53
A painting operation is per昀漀rmed by a produc­
tion worker at a labour cost of $1.40 per unit. A
robot spray-painting machine, costing $15,000,
would reduce the labour cost to $0.20 per unit.
If the device would be valueless at the end of
three years, what is the minimum number of
units that would have to be painted each year
to justify the purchase of the robot machine?
1-54
Company A has 昀椀xed expenses of $15,000 per
year, and each unit of product has a $0.20 vari­
able cost. Company B has 昀椀xed expenses of
$5,000 per year and can produce the same prod­
uct at a $0.50 variable cost. At what number of
units of annual production will Company A
h愀瘀e the same overall cost as Company B?
1-55
0
1-56
1-57
Heinrich is a manu昀愀cturing engineer with the
Miller Company. He has determined the costs
of producing a new product to be as 昀漀llows:
•
•
•
•
Equipment cost: $288,000/year
Equipment salvage value at EOY5: $41,000
Variable cost per unit of production: $14.55
Overhead cost per year: $48,300
If the Miller Company uses a 昀椀ve-year plan­
ning horizon and the product can be sold 昀漀r
a unit price of $39.75, how many units must be
produced and sold each year to break even?
C = $3,000,000 - $18,000Q + $75 Q 2
A newly hired employee, who previously
worked 昀漀r Bendix, tells Mr Spade that Bendix
is now producing 110 units a year. If the sell­
ing price remains unchanged, Mr Spade won­
ders if Bendix is likely to increase the number
of units produced per year in the near 昀甀ture.
He asks you to look at the in昀漀rmation and tell
him what you are able to deduce 昀爀om it.
A 昀椀rm believes that the sales volume (S ) of its
product depends on its unit selling price (P),
and that S can be determined 昀爀om the equa­
tion S = 100 - 倀⸀ 吀栀e cost (C) of producing the
product is $1,000 + 10S.
(a) Draw a graph with S 昀爀om O to 100 on the
x axis, and total cost and total income
昀爀om O to 2,500 on the y axis. On the graph
draw the line C = $1,000 + 10S. 吀栀en plot
the curve of total income, which is sales
volume (S) x unit selling price ($100 - S).
Mark the break-even points on the graph.
(b) Determine the break-even point (lowest
sales volume where total sales income just
equals total production cost). (䠀椀nt: 吀栀is
may be done by trial and error or by using
the quadratic equation to locate the point
at which pro昀椀t is zero.)
(c) Determine the sales volume (S) at which
the 昀椀rm's pro昀椀t is a maximum. (Hint:
Write an equation 昀漀r pro昀椀t and solve it
by trial and error, or as a minima-maxima
calculus problem.)
Sunk and Other Costs
Mr Sam Spade, the president of 䄀樀ax, recently
read in a report that a competitor named
Bendix has the 昀漀llowing relationship between
cost and production quantity:
where C = total manu昀愀cturing cost per year
and Q = number of units produced per year.
43
1-58
An assembly line can produce 90 units an hour.
吀栀e line's hou爀氀y cost is $4,500 on straight time
⠀琀he 昀椀rst eight hours). Workers are guaranteed a
minimum of six hours a day. 吀栀ere is a 50% pre­
mium 昀漀r overtime, and product椀瘀ity 昀漀r overtime
drops by 5%. What are the 瘀였rage and marginal
costs per unit 昀漀r the 昀漀llowing daily quantities?
(a) 450
(b) 600
(c) 720
44
1-59
1-60
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
A pump has 昀愀iled in a 昀愀cility that will be
completely replaced in three years. A brass
pump costing $6,000 installed will last three
years. However, a used stainless steel pump
that should last three more years has been sit­
ting in the maintenance shop 昀漀r a year. 吀栀e
pump cost $13,000 when it was new. 吀栀e ac­
countants say the pump is now worth $7,000.
吀栀e maintenance supervisor says that it will
cost an extra $500 to recon昀椀gure the pump 昀漀r
the new use and that he could sell it used (as is)
昀漀r $4,000.
(a) What is the book cost of the stainless steel
pump?
(b) What is the opportunity cost of the stain­
less steel pump?
(c) How much cheaper or more expensive
would it be to use the stainless steel pump
rather than a new brass pump?
Bob Johnson decided to buy a new house. A昀琀er
looking at some new housing developments,
he decided that a custom-built house was pref­
erable. He hired an architect to prepare the
drawings. In due time, the architect completed
the drawings and submitted them. Bob liked
the plans; he was less pleased that he had to
pay the architect a fee of $7,000 to design the
house. Bob asked a building contractor 昀漀r a
bid to construct the home on a lot Bob already
owned. While the contractor was working to
assemble the bid, Bob came across a book of
standard house plans. In the book was a house
that he and his wife liked better than the one
designed 昀漀r them by the architect. Bob paid
$200 and obtained a complete set of plans 昀漀r
this other house. Bob then asked the contractor
昀漀r a bid to construct this "stock-plan" house.
In this way Bob felt he could compare the costs
and make a decision. 吀栀e building contractor
submitted the 昀漀llowing bids:
Custom-designed house
Stock-plan house
$258,000
261,000
Bob was willing to p愀礀 the extra $3,000 昀漀r the
stock-plan house. Bob's wife, however, felt they
should go ahead with the custom-designed house,
昀漀r, as she put it, "We can't a昀昀ord to throw 愀眀ay a
set of plans that cost $7,000." Bob agreed, but he
disliked the thought of building a less-desirable
house. 吀栀en he asked your advice. Which house
would you advise him to build? Explain.
1-61
You are re-evaluating the pump choice that
was made last year by your boss. 吀栀e ex­
pected cost savings have not materialized be­
cause the pump is too small. Choice A, which
costs $90,000, is to replace the pump with one
that is the right size and to sell the old one.
Choice B, which costs $100,000, is to buy a
small pump to use in tandem with the exist­
ing pump. 吀栀e two-pump solution has slightly
higher maintenance costs, but it is more ver­
satile and reliable. What criteria should you
use? Which choice would you recommend
and why?
1-62
Consider the situation of owning rental prop­
erties that local university students rent 昀爀om
you 昀漀r the academic year. Develop a set of
costs that could be classi昀椀ed as recurring
and others that could be classi昀椀ed as non­
recurring.
1-63
De昀椀ne the di昀昀erence between cash cost and
book cost. Is engineering economic analysis
concerned with both types of cost? Give an
example of each, and explain the context in
which it is important.
1-64 In your own words, state what the authors
mean by "life-cycle costs." Is it important 昀漀r
a 昀椀rm to be aware of life-cycle costs? Explain
why or why not.
1-65
Most engineering students own a computer.
What costs have you incurred at each stage of
your computer's life cycle? Estimate the total
cost of ownership. Estimate the bene昀椀ts of
ownership. Has it been worth it?
Estimating
1-66 It is o昀琀en more di昀케cult to estimate bene昀椀ts
than costs. Use the example of car ownership to
describe the complicating 昀愀ctors in estimating
the costs and bene昀椀ts.
Problems
1-67
1-68
In looking at Figures 1-4 and 1-5, restate in
your own words what the authors are trying
to get across with these 昀椀gures. Do you agree
that this is an important e昀昀ect 昀漀r companies?
Explain.
Develop an estimate 昀漀r each of the 昀漀llow­
ing situations. Ensure that your answer is not
given to excessive precision.
(a) 吀栀e cost of an 800 km trip by car if gaso­
line costs $1.00 per litre, wear and tear on
the car is $0.35 per kilometre, and our car
uses 11 L/100 km.
(b) 吀栀e total number of hours in the average
human life, if the average life is 75 years.
挀儀 吀栀e number of days it takes to travel
around the equator in a hot air balloon, if
the balloon averages 150 km a day and the
diameter of the earth is ~ 12,800 km.
1-70
Northern 吀甀ndra Telephone ( NT T ) has re­
ceived a contract to install emergency phones
along a new 100 km section of the Snow­
Moose Highway. Fi昀琀y emergency phone
systems will be installed about 2 km apart.
吀栀e material cost of a unit is $125. NT T will
need to run underground communication
lines that cost NT T $7,500 per kilometre (in­
cluding labour) to install. 吀栀ere will also be
a one-time cost of $10,000 to network these
phones into NT T 's current communication
system.
(a) Develop a cost estimate of the project
昀爀om NT T's perspective.
(b) If NT T adds a pro昀椀t margin of 35% to its
costs, how much will it cost the province
to 昀甀nd the project?
You want a cottage built 昀漀r weekend trips, vaca­
tions, and perhaps eventually to retire to. A昀琀er
discussing the project with a local contractor,
you receive an estimate that the total construc­
tion cost of 礀漀ur 200 m2 cottage will be $250,000.
吀栀e percentage of costs is broken down as
昀漀llows:
Percentage of
Total Costs
Cost Items
Construction permits, legal, and title fees
Roadway, site clearing, preparation
Foundation, concrete, masonry
Wallboard, 昀氀ooring, carpentry
Heating, ventilation, air conditioning
Electric, plumbing, communications
Roo昀椀ng, 昀氀ooring
Painting, 昀椀nishing
Total
8
15
13
12
13
10
12
17
100
(a) What is the cost per square metre of the
200 m2 cottage?
(b) If you are also considering a 400 m2
layout option, estimate your construction
costs if
i. all cost items (in the table) change pro­
portionately to the size increase.
ii. the 昀椀rst two cost items do not change
at all; all others are proportionate.
Seg menting Models
1-69
45
1-71
SungSam, Inc. is currently designing a new
digital camcorder that is projected to h愀瘀e the
昀漀llowing per-unit costs to manu昀愀cture:
Cost Categories
Unit Costs
Materials costs
Labour costs
Overhead costs
Total unit cost
$ 63
24
110
$197
SungSam adds 30% to its manu昀愀cturing cost
昀漀r corporate pro昀椀t.
(a) What unit pro昀椀t would SungSam realize
on each camcorder?
(b) What is the overall cost to produce a
batch of 10,000 camcorders?
(c) What would SungSam's pro昀椀t be on the
batch of 10,000 if historical data show
that 1% of product will be scrapped in
manu昀愀cturing, 3% of 昀椀nished product
will go unsold, and 2% of sold product
will be returned 昀漀r re昀甀nd?
(d) How much can SungSam a昀昀ord to pay 昀漀r
a contract that would lock in a 50% reduc­
tion in the unit material cost previously
46
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
trade the old equipment in 昀漀r 15% of its ori­
ginal cost)? Assume also that there has been no
in昀氀ation in equipment prices.
given? If SungSam does sign the contract,
the sales price will not change.
I ndexi ng and Sizing Models
1-72
Estimate the cost of expanding a planned new
clinic by 2,000 m2 • 吀栀e appropriate capacity
exponent is 0.66 and the budget estimate 昀漀r
20,000 m2 was $45 million.
1-73
Your boss is the director of reporting 昀漀r the
Athens County Construction Agency (ACCA).
It has been his job to track the cost of construc­
tion in Athens County. Twenty-昀椀ve years ago
he created the ACCA Cost Index to track these
costs. Costs during the 昀椀rst year of the index
were $120 per square metre of constructed
space (the index value was set at 100 昀漀r that
昀椀rst year). 吀栀is past year a survey of contract­
ors revealed that costs were $720 per square
metre. What index number will your boss
publish in his report 昀漀r this year? If the index
value was 525 last year, what would the cost per
square metre h愀瘀e been last year?
1-74
A new aerated sewage lagoon is required in a
small town. Earlier this year one was built on a
similar site in an a搀樀acent city 昀漀r $2.3 million.
吀栀e new lagoon will be 60% larger. Use the
data in Table 1-1 to estimate the cost of the new
lagoon.
1-75
A re昀椀nisher of antiques named Constance has
been so success昀甀l with her small business that
she is planning to expand her shop and buy
all new equipment. She is going to start en­
larging her shop by purchasing the 昀漀llowing
equipment.
Equipment
Varnish bath
Power scraper
Paint booth
PowerCost of
Capacity
Original Original
Sizing
of New
Capacity Equipment Exponent Equipment
0.2 m3
0.5 kW
0.1 m3
$3,500
250
3,000
0.80
0.22
0.6
0.3 m3
1-76
Original
Equipment
Cost Index
Cost
When Originally Index
Today
Purchased
Varnish bath
Power scraper
Paint booth
1-77
154
780
49
171
900
76
Five years ago, when the relevant cost index
was 120, a nuclear centri昀甀ge cost $40,000. 吀栀e
centri昀甀ge had a capacity of separating 1.5 litres
of ionized solution per second. Today, a centri­
昀甀ge with a capacity of 4.5 litres per second is
needed, but the cost index now is 300. Assum­
ing a power-sizing exponent to re昀氀ect econ­
omies of scale, x, of 0.75, use the power-sizing
model to determine the approximate cost (ex­
pressed in today's dollars) of the new reactor.
Lea rn ing Cu rve Models
1-78
Determine the time required to produce the
2,000th item if the 昀椀rst item requires 180 min­
utes to produce and the learning curve per­
centage is 92%.
1-79
If 200 worker hours were required to produce
the 昀椀rst unit in a production run and 60 worker
hours were required to produce the 7th unit, what
was the learning curve 爀愀te during production?
1-80
Sally is implementing a system of statistical
process control (SPC) charts in her 昀愀ctory in
l kW
0.4 m3
What would be the net cost to Constance to
obtain this equipment (assume that she can
Refer to Problem 1-75 and now assume the
prices 昀漀r the equipment that Constance wants
to replace have not been constant. Use the
cost index data 昀漀r each piece of equipment to
update the costs to the price that would be paid
today. Develop the overall cost 昀漀r Constance,
again assuming the 15% trade-in allowance 昀漀r
the old equipment. Trade-in is based on ori­
ginal cost.
⠀　Problems
an e昀昀ort to reduce the overall cost of scrapped
product. 吀栀e current cost of scrap is $x per
month. If an 80% learning curve is expected
in the use of the SPC charts to reduce the cost
of scrap, what would the percentage reduction
in monthly scrap cost be a昀琀er the charts have
been used 昀漀r 12 months? (Hint: Model each
month as a unit of production.)
1-81
Randy has been asked to develop an estimate of
the per-unit selling price ⠀琀he price that each unit
will be sold 昀漀r) of a new line of handcra昀琀ed book­
lets that o昀昀er excuses 昀漀r missed appointments.
His assistant Doc has collected in昀漀rmation that
Randy will need in developing his estimate:
Cost of direct labour
Cost of materials
Cost of overhead items
Desired pro昀椀t
$20 per hour
$43.75 per batch of
25 booklets
50% of direct labour
cost
20% of total
manu昀愀cturing cost
Doc also 昀椀nds out that (1) they should use a
75% learning curve 昀漀r estimating the cost of
direct labour, (2) the time to complete the 昀椀rst
booklet is estimated at 0.60 hour, and (3) the
estimated time to complete the 25th booklet
should be used as their standard time 昀漀r the
purpose of determining the unit selling price.
What would Randy and Doc's estimate be?
Cash Flow Diagrams
1-83
On 1 December, Al Smith bought a car 昀漀r
$18,500. He paid $5,000 immediately and
agreed to make three additional payments of
$6,000 each (which include principal and in­
terest) at the end of one, two, and three years.
Maintenance 昀漀r the car is projected at $1,000
at the end of the 昀椀rst year and $2,000 at the end
of each subsequent year. Al expects to sell the
car at the end of the 昀漀urth year (a昀琀er paying
昀漀r the maintenance work) 昀漀r $7,000. Using
these 昀愀cts, prepare a table of cash 昀氀ows.
1-84
Bonka Toys is considering a robot that will cost
$20,000. A昀琀er seven years its salvage value will
be $2,000. An overhaul costing $5,000 will be
needed in year 4. O&M costs will be $2,500 per
year. Draw the cash 昀氀ow diagram.
1-85
Pine Village needs some additional recreation
昀椀elds. Construction will cost $225,000, and
annual O&M expenses are $85,000. 吀栀e city
council estim愀琀es th愀琀 the value of the additional
youth leagues is about $190,000 annually. In year
6 another $75,000 will be needed to re昀甀rbish the
昀椀elds. 吀栀e salvage value is estimated to be $100,000
a昀琀er 10 礀攀ars. Draw the cash 昀氀ow diagram.
1-86
Identi昀礀 your m愀樀or cash 昀氀ows 昀漀r the current
university term as 昀椀rst costs, O&M expenses,
salvage values, revenues, overhauls, and so on.
Using a week as the time period, draw the cash
昀氀ow diagram.
Bene昀椀ts Esti mation
1-82
Large projects such as a new bridge at
Windsor-Detroit, new helicopters 昀漀r the Armed
Forces, and major pipelines 昀爀om Alberta to 昀甀el
the Asian and American markets o昀琀en take 5 to
15 years 昀爀om concept to completion.
愀儀 Should bene昀椀t and cost estimates be ad­
justed 昀漀r the greater e昀昀ects of inflation,
government regulatory changes, and
changing local economic environments?
Why or why not?
(b) How does the public budget-making pro­
cess interact with the goal of accurate
bene昀椀t and cost estimating 昀漀r these large
projects?
47
U nclassi昀椀ed
1-87
Quatro Hermanas, Inc., is investigating the
use of some new production machinery as part
of its operations. 吀栀ree alternatives have been
identi昀椀ed, and they have the 昀漀llowing 昀椀xed
and variable costs:
Alternative
A
B
C
Annual
Annual
Variable Costs
per Unit
Fixed Costs
$100,000
200,000
150,000
$20.00
5.00
7.50
48
Cha pter 1 Economic Decisions, Engineering Costs, and Cost Estimating
Determine the ranges of production (units
produced per year) over which each alternative
would be recommended. Be exact. (Note: Con­
sider the range of production to be 0-30,000 units
per year.)
1-88
Electricity is sold 昀漀r $0.12 per kilowatt hour
(kWh) 昀漀r the 昀椀rst 10,000 units each month
and $0.09/欀圀h 昀漀r all remaining units. If a
昀椀rm uses 14,000 欀圀h/month, what are its
average and marginal costs?
1-89
Two manu昀愀cturing companies, located in
cities 150 kilometres apart, h愀瘀e discovered
that they each send their trucks 昀漀ur times a
week to the other city 昀甀ll of cargo and return
empty. Each company pays its driver $185 a
d愀礀 (the round trip takes all d礀言 and has truck
operating costs (excluding the driver) of 37.5¢
a kilometre. How much could each company
save each week if they shared the task, with
each sending its truck twice a week and hauling
the other company's cargo on the return trip?
1-90
1-91
Willie tr愀瘀els 昀爀om city to city to conduct his
business. Every other year he buys a used car
昀漀r about $15,000. 吀栀e car dealer allows about
$8,000 as a trade-in allowance, with the result
that Willie spends $7,000 every other year 昀漀r a
car. Willie keeps accurate records, which show
that all other expenses on his car amount to
30¢ 昀漀r each kilometre he drives. Willie's em­
ployer has two plans by which salesmen are re­
imbursed 昀漀r their car expenses:
愀儀 Willie will receive all his operating
expenses, and in addition will receive
$3,500 each year 昀漀r the decline in value
of the car.
(b) Willie will receive 50¢ per kilometre but
no operating expenses and no deprecia­
tion allowance.
If Willie travels 29,000 km a year, which
method of computation gives him the larger re­
imbursement? At what annual distance do the
two methods give the same reimbursement?
Last year, to help with your New Year's reso­
lutions, you purchased a $500 piece of 昀椀tness
equipment. However, you use it only once a
week on average. It is now December, and you
can sell the equipment 昀漀r $200 (to someone
with a New Year's resolution) and rely on the
university gym until you graduate in May. If
you don't sell until May, you will get only $100.
If you keep the heavy piece, you will have to
pay $25 to move it to the city of your new job.
吀栀ere is no convenient gym at the new loca­
tion. What costs and intangible consequences
are relevant to your decision? What should
you do?
1-92
A 昀愀rmer must decide what combination of
seed, water, fertilizer, and pest control will be
most pro昀椀table 昀漀r the coming year. 吀栀e local
agricultural college did a study of this 昀愀rmer's
situation and prepared the 昀漀llowing table.
Plan
Cost/Hectare
Income/Hectare
A
B
C
D
$1,500
3,750
4,500
5,250
$2,000
4,750
5,625
6,250
吀栀e last page of the college's study was torn o昀昀,
and hence the 昀愀rmer is not sure which plan
the agricultural college recommends. Which
plan should the 昀愀rmer adopt? Explain.
1-93
You must 昀氀y to another city 昀漀r a Friday meet­
ing. If you stay until Sunday morning, your
ticket will cost $200 rather than $800. Hotel
costs are $100 per night. Compare the eco­
nomics with reasonable assumptions 昀漀r meal
expenses. What intangible consequences may
dominate the decision?
1-94
Rose is a project manager at the civil engineer­
ing consulting 昀椀rm of Sands, Gravel, Concrete,
and Waters, Inc. She has been collecting data
on a project in which concrete pillars were
being constructed; however, not all the data
are available. She has been able to 昀椀nd out that
the 10th pillar required 260 person-hours to
construct, and that a 75% learning curve ap­
plied. She is interested in calculating the time
required to construct the 1st and 20th pillars.
Compute the values 昀漀r her.
Problems
1-95
A small machine shop, with 22 kW of connected load, purchases electricity at the 昀漀llowing monthly rates (assume any demand charge
is included in this schedule):
• First 50 欀圀h per kW of connected load at
12.6¢ per 欀圀h
• Next 50 kWh per 欀圀 of connected load at
10.6¢ per kWh
• Next 150 欀圀h per kW of connected load at
6.0¢ per 欀圀h
• All electricity over 250 欀圀h per 欀圀 of
connected load at 5.7¢ per 欀圀h
• 吀栀e shop uses 2,800 欀圀h per month.
(a) Calculate the monthly bill 昀漀r this shop.
What are the marginal and average costs
per kilowatt hour?
(b) Suppose Jennifer, the proprietor of the
shop, has the chance to secure additional
business that will require her to operate
her existing equipment more hours per
day. 吀栀is will use an extra 1,200 kWh per
month. What is the lowest 昀椀gure that she
might reasonably consider to be the "cost"
of this additional energy? What is this per
kilowatt hour?
挀儀 Jennifer contemplates installing certain
new machines that will reduce the labour
time required on certain operations.
吀栀ese will increase the connected load by
7.5 欀圀 but, since they will operate only
on certain special jobs, will add only
100 欀圀h a month. In a study to determine the economy of installing these new
machines, what should be considered as
the "cost" of this energy? What is this per
kilowatt hour?
1-96
49
A small company manu昀愀ctures a certain prod­
uct. Variable costs are $20 per unit and 昀椀xed
costs are $10,875. 吀栀e price-demand relationship
昀漀r this product is P = -0.25D + 250, where P
is the unit sales price of the product and D is the
annual demand. Use the data (and help昀甀l hints)
that 昀漀llow to work out answers to parts (a)-(e).
Total cost = Fixed cost + Variable cost
Revenue = Demand X Price
Pro昀椀t = Revenue - Total cost
Set up your graph with dollars on the y axis
(between $0 and $70,000) and, on the x axis,
demand (units produced or sold, between 0
and 1,000 units).
(a) Develop the equations 昀漀r total cost and
total revenue.
(b) Find the break-even quantity (in terms of
pro昀椀t and loss) 昀漀r the product.
(c) What pro昀椀t would the company obtain by
maximizing its total revenue?
(d) What is the company's maximum
possible pro昀椀t?
(e) Graph neatly the solutions to parts (a), (b),
(c), and (d).
1-97
A labour-intensive process has a 昀椀xed cost of
$338,000 and a variable cost of $143 per unit.
A capital-intensive (automated) process 昀漀r the
same product has a 昀椀xed cost of $1,244,000
and a variable cost of $92.50 per unit. How
many units must be produced and sold at $197
each 昀漀r the automated process to be preferred
to the labour-intensive process?
Accounting and
Engineering Economy
The 吀眀o Faces of AB B
ABB Ltd, a Swedish-Swiss multinational corporation headquartered in Z urich, Switzerland, is a
world leader in the power and automation technology areas. In the late 1990s, ABB was flying high.
This European conglomerate was an engineering giant with a global network of operations and
forward-looking management.
Unlike many of its competitors, ABB was also committed to modernizing its accounting
system. ABB had previously followed the traditional practice of assigning overhead costs to its
divisions on a roughly equal basis. But that practice tended to obscure the fact that some activities
incurred far more costs than others.
So the company spent substantial time and resources switching to an activity-based costing
(ABC) system, which assigns costs to the activities that actually incur them. As a result ABB was able
to zero in on areas where it could cut costs most effectively.
But in other respects ABB's accounting practices were less effective.
One of its most serious problems arose at Combustion Engineering, an American subsidi­
ary that was exposed to numerous asbestos liability claims. For years, ABB had downplayed this
Alto Co. Ltd./Alamy Stock Photo
The Two Faces of AB B
51
potential liability despite warnings from outside analysts. Finally, in late 2002, ABB admitted that its
asbestos liability exceeded the subsidiary's total asset value.
But there was still more bad news. The company also issued a very poor third-quarter earn­
ings report after previously assuring investors that ABB was on target to improve its earnings and
decrease its debt. When incredulous investors asked how that could have happened, ABB manage­
ment blamed "poor internal reporting. "
By now, the price of ABB shares had nose-dived, and credit rating agencies viewed the com­
pany's bonds as little better than junk. In 2003, Combustion Engineering went through bankruptcy
and reached a settlement for the asbestos claims.
ABB subsequently recovered. As of December 2016 it was one of the largest engineering
conglomerates in the world. It has operations in more than 100 countries, with some 132,000
employees and reported global revenue of $ 34 billion for 2016.
QU ESTI ONS TO CONSI D E R
1.
ABB's adoption of activity-based costing received widespread publicity and boosted the company's reputa­
tion for innovative management. How may that have affected investors' assumptions about the company's
other accounting practices?
2.
Outside analysts estimated that ABB lost more than $690 million in 2001. Yet many investors were still
stunned by its poor showing in late 2002. What does that say about the relationship between financial ac­
counting and investor confidence?
3.
ABB was not alone in its financial misery, of course. How did ABB's accounting problems compare with
those of well-known companies such as Nortel, Enron, and WorldCom?
LEARN I N G OBJ ECTIVES
This chapter will help you
•
describe the links between engineering economy and accounting
•
describe the objectives of general accounting, explain what financial transactions are, and
show how they are important
•
use a firm's balance sheet and associated financial ratios to evaluate the firm's health
•
use a firm's income statement and associated financial ratios to evaluate the firm's performance
•
use traditional absorption costing to calculate product costs
•
understand the greater accuracy in product costs available with ABC
KEY TE RMS
a bsorption cost i n g
cu rrent ratio
net pro昀椀t ratio
acid - test ratio
equ ity
pro昀椀t a n d loss state ment
asset
i ncome state ment
q u ick asset
ba la nce sheet
liabi lity
q u ick ratio
52
Cha pter 2 Acco u n t i n g a n d E n g i neeri n g Eco n o my
Engineering economy 昀漀cuses on the 昀椀nancial aspects of projects, whereas accounting
昀漀cuses on the 昀椀nancial aspects of 昀椀rms. 吀栀us the application of engineering economy
is much easier if one has some understanding of accounting principles. One important
accounting topic, depreciation, will be covered in Chapter 1 1.
The Role of Accounting
Accounting data are used to value capital equipment, to decide whether to make or buy
a part, to determine costs and set prices, to set indirect cost rates, and to make decisions
about product mix. Accounting is used in private sector 昀椀rms and public sector agencies,
but 昀漀r simplicity this chapter uses "the 昀椀rm" to designate both. Accountants track the
costs of projects and products, which are the basis 昀漀r estimating 昀甀ture costs and revenues.
吀栀e engineering economics, accounting, and managerial 昀甀nctions are interdepend­
ent. Whether carried out by a single person in a small 昀椀rm or by distinct divisions in a
large 昀椀rm, all are needed.
•
•
•
Engineering economics analyzes the economic impact of design alternatives and
projects over their life cycles.
Accounting determines the dollar impact of past decisions, reports on the eco­
nomic viability of a unit or 昀椀rm, and evaluates potential 昀甀nding sources.
Management allocates available investment 昀甀nds to projects, evaluates unit and
昀椀rm per昀漀rmance, allocates resources, and selects and directs personnel.
Accounting for Business Tra nsactions
A business transaction involves two parties and the exchange of dollars (or the promise
of dollars) 昀漀r a product or service. Every day, millions of transactions take place between
昀椀rms and their customers, suppliers, vendors, and employees. 吀栀e accounting 昀甀nction
records, analyzes, and reports these exchanges.
Transactions can be as simple as payment 昀漀r a water bill, or as complex as the inter­
national transfer of billions of dollars worth of buildings, land, equipment, inventory, and
other assets. Moreover, with transactions, one business event may lead to another-all of
which need to be accounted 昀漀r. Consider, 昀漀r example, the process of selling a robot or
a bulldozer. 吀栀at simple act involves several related transactions: (1) releasing equipment
昀爀om inventory, (2) shipping equipment to the purchaser, (3) invoicing the purchaser, and
昀椀nally (4) collecting 昀爀om the purchaser.
Transaction accounting i瘀�lves more than just reporting: it includes 昀椀nding, synthesiz­
ing, summarizing, and analyzing data. For the engineering economist, historical data housed
in the accounting 昀甀nction are the 昀漀undation 昀漀r estimates of 昀甀ture costs and re瘀攀nues.
Most accounting is done in actual dollars. We will see in Chapter 14 that these are
dollars whose purchasing power may change with inflation. Management has a natural
desire to maximize a 昀椀rm's apparent value. To keep this desire in check, accountants will
value most assets at their acquisition cost, adjusted 昀漀r depreciation and improvements.
If market value is lower than that a搀樀usted cost, the lower value is used. When an asset is
sold, or when a 昀椀rm must be liquidated, current market values must be estimated.
吀栀e accounting 昀甀nction provides data 昀漀r general accounting and cost accountin最⸀ 吀栀is
chapter's presentation begins with the balance sheet and income statement, which are the
two key summaries of 昀椀nancial transactions 昀漀r general accounting. 吀栀at discussion in­
cludes some of the basic 昀椀nancial ratios used 昀漀r short-and long-term evaluations. 吀栀e chap­
ter concludes with a central topic in cost accounting-the allocating of indirect expenses.
The Ba� nce She�
䌀䐀
The Balance Sheet
吀栀e primary accounting statements are the balance sheet and income statement. 吀栀e bal­
ance sheet describes the 昀椀rm's 昀椀nancial condition at a speci昀椀c time, while the income
statement describes the 昀椀rm's per昀漀rmance over a period of time-usually a year.
吀栀e balance sheet lists the 昀椀rm's assets, liabilities, and equity on a speci昀椀ed date. It is a pic­
ture of the organization's 昀椀nancial hea氀琀h. Usually, balance sheets are taken at the end of each
quarter and the 昀椀scal 礀攀ar. 吀栀e balance sheet is based on thefundamental accounting equation:
Assets = Liabilities + Equity
(2-1)
Figure 2-1 illustrates the basic 昀漀rmat of the balance sheet. Notice that in the balance
sheet, as in Equation 2-1, assets are listed on the le昀琀-hand side and liabilities and equity
are on the right-hand side. 吀栀e 昀愀ct that the 昀椀rm's resources are balanced by the sources of
昀甀nds gives the balance sheet its name.
䌀䐀
Assets
In Equation 2-1 and Figure 2-1, assets are owned by the 昀椀rm and have monetary value.
Liabilities are the dollar claims against the 昀椀rm. Equity represents 昀甀nding 昀爀om the 昀椀rm
and its owners (the shareholders). In Equation 2-1, the value 昀漀r retained earnings is always
set so that equity equals assets minus liabilities.
On a balance sheet, assets are listed in order of decreasing liquidity, that is, accord­
ing to how quickly each can be converted to cash. 吀栀us, current assets are listed 昀椀rst, and
Balance Sheet 昀漀r Engineered Industries, December 3 1 , 20 16 (all amounts in $ 1,000s)
Assets
Current assets
Cash
Accounts receivable
Securities
Inventories
(minus)Bad debt provision
Total current assets
Fixed assets
Land
Plant and equipment
(minus)Accumulated depr.
Total 昀椀xed assets
Liabilities
1,940
950
4,100
1,860
- 80
-8,770
335
6,500
-2,350
4,485
Other assets
Prepays/deferred charges
Intangibles
Total other assets
Total assets
140
420
560
13,815
FIGURE 2-1 Sam ple balance sheet.
Current liabilities
Accounts payable
Notes payable
Accrued expense
Total current liabilities
1,150
80
950
2,180
Long-term liabilities
1,200
Total liabilities
3,380
Equity
Pre昀攀rred shares
Common shares
Capital surplus
Retained earnings
Total equity
110
650
930
8,745
10,435
Total liabilities and equity
13,815
53
54
Cha pter 2 Acco u n t i n g a n d E n g i neeri n g Eco n o my
within that category in order of decreasing liquidity are listed cash, receivables, securities,
and inventories. Fixed assets-or property, plant, and equipment-are used to produce and
deliver goods and services, and they are not intended 昀漀r sale. Finally, items such as pre­
payments and intangibles such as patents are listed.
吀栀e term "receivables" comes 昀爀om the w愀礀 billing and p愀礀ment are handled 昀漀r most
business sales. Rather than requesting immediate p愀礀ment 昀漀r every transaction, most busi­
nesses record each transaction and then once a month bill 昀漀r all transactions. 吀栀e total that
has been billed less payments already received is called accounts receivable, or receivables.
䌀䐀
Lia bilities
On the balance sheet, liabilities are divided into two m愀樀or classi昀椀cations-short-term and
long-term. Short-term, or current, liabilities are expenses, notes, and other payable accounts
that are due within one year 昀爀om the balance sheet date. Long-term liabilities include
mortgages, bonds, and loans with later due dates. For Engineered Industries in Figure 2-1,
total current and long-term liabilities are $2,180,000 and $1,200,000, respectively. O昀琀en in
per昀漀rming engineering economic analyses, the working capital 昀漀r a project must be esti­
mated. 吀栀e total amount of working capital available may be calculated with Equation 2-2
as the di昀昀erence between current assets and current liabilities.
Working capital = Current assets - Current liabilities
(2-2)
For Engineered Industries, there would be $8,770,000 - $2,180,000 = $6,590,000
available in working capital.
Eq u ity
Equity is also called owner's equity, or net worth. It includes the value of the owners' share­
holdings and the capital surplus, which are the excess dollars brought in over the nominal
value of the shares when they were issued. 吀栀e capital surplus can also be called additional
paid-in c愀瀀ital, or APIC. Retained earnings are dollars a 昀椀rm chooses to retain rather than
pay out as dividends to shareholders. Retained earnings within the equity component are
the adjustable dollar amount used to bring the balance sheet, and thus the 昀甀ndamental
accounting equation, into balance. For Engineered Industries, total equity value is listed at
$10,435,000. From Equation 2-1 and the assets, liabilities, and equity values in Figure 2-1,
we can write the balance as 昀漀llows:
Assets = Liabilities + Equity
Assets (current, 昀椀xed, other) = Liabilities (current and long-term) + Equity
(4,670,000 + 4,485,000 + 560,000) = (2,180,000 + 1,200,000) + 10,435,000
$13,815,000 = $13,815,000
An example of owner's equity is ownership of a home. Most homes are purchased
with a mortgage loan that is paid o昀昀 at a certain interest rate over 15 to 30 years. At any
point in time, the di昀昀erence between what is owed to the bank (the remaining balance
on the mortgage) and what the house is worth (its appraised market value) is the owner's
equity. In this case, the loan balance is the liability, and the home's value is the asset­
with equi琀礀 being the di昀昀erence. Over time, as the house loan is paid o昀昀, the owner's
equity should increase. However, there are un昀漀rtunate exceptions to this expectation:
The Ba� nce She�
the 2007 sub-prime mortgage crisis in the United States le昀琀 many homeowners with
liabilities greater than the suddenly shrunken market value of their homes, a condition
referred to as an "underwater" mortgage. Even in 2015, about 14% of mortgages in the US
were still underwater.
吀栀e balance sheet is a very use昀甀l tool that shows one view of the 昀椀rm's 昀椀nancial con­
dition at a particular point in time.
䌀䐀
Financial Ratios Derived from Bala nce Sheet Data
One common way to evaluate the 昀椀rm's health is through ratios of quantities on the bal­
ance sheet. Firms in a particular industry will usually have similar values, and exceptions
will o昀琀en indicate 昀椀rms with better or worse per昀漀rmance. Two common ratios used to
analyze the 昀椀rm's current position are the current ratio and the acid-test ratio.
A 昀椀rm's current ratio is the ratio of current assets to current liabilities, as in Equation 2-3.
Current ratio = Current assets/Current liabilities
(2-3)
吀栀is ratio provides insight into the 昀椀rm's solvency over the short term by indicating its
ability to cover current liabilities. Historically, 昀椀rms h愀瘀e aimed to be at or above a ratio of
2.0; however, that depends heavily on the industry as well as the individual 昀椀rm's manage­
ment practices and philosophy. 吀栀e current ratio 昀漀r Engineered Industries in Figure 2-1 is
above 2 (8,770,000/2,180,000 = 4.02).
Both working capital and the current ratio indicate the 昀椀rm's ability to meet currently
maturing obligations. However, neither describes the type of assets owned. 吀栀e acid-test
ratio, or quick ratio, becomes important when considering the 昀椀rm's ability to p愀礀 debt
quickly. 吀栀e acid-test ratio is computed by dividing a 昀椀rm's quick assets (cash, receivables,
and market securities) by total current liabilities, as in Equation 2-4.
Acid-test ratio = Quick assets/Total current liabilities
(2-4)
Current inventories are excluded 昀爀om quick assets because of the time required to
sell these inventories, collect the receivables, and subsequently have the cash on hand to
reduce debt. For Engineered Industries in Figure 2-1, the calculated acid-test ratio is well
above 1: [(1,940,000 + 950,000 + 4,100,000)/2,180,000 = 3.21].
Working capital, current ratio, and acid-test ratio are all indications of the 昀椀rm's 昀椀­
nancial health (status). A thorough 昀椀nancial evaluation would consider all three, including
comparisons with values 昀爀om previous periods and with broad-based industry standards.
䌀䐀
The Income Statement
吀栀e income statement, or pro昀椀t and loss statement, summarizes the 昀椀rm's revenues and
expenses over a month, quarter, or year. Rather than being a snapshot like the balance sheet,
the income statement covers a period of business activity. 吀栀e income statement is used to
evaluate revenue and expenses that occur in the interval between consecutive balance sheet
statements. 吀栀e income statement reports the 昀椀rm's net income (pro昀椀琀⤀ or loss by subtract­
ing expenses 昀爀om revenues. If revenues minus expenses is positive in Equation 2-5, there
has been a pro昀椀t; if it is negative, a loss has occurred.
Revenues - Expenses = Net pro昀椀t (loss)
(2-5)
55
56
Cha pter 2 Acco u n t i n g a n d E n g i neeri n g Eco n o my
To aid in analyzing per昀漀rmance, the income statement in Figure 2-2 separates operat­
ing and non-operating activities and shows revenues and expenses 昀漀r each. Operating rev­
enues are made up of sales revenues (minus returns and allowances), while non-operating
revenues come 昀爀om rents and interest receipts.
Operating expenses produce the products and services that generate the 昀椀rm's rev­
enue stream of cash 昀氀ows. 吀礀pical operating expenses include cost of goods sold, selling
and promotion costs, depreciation, general and administrative costs, and lease payments.
Cost of最漀ods sold (COGS) includes the labour, materials, and indirect costs of production.
Engineers design production systems, speci昀礀 materials, and analyze make/buy
decisions. All these items a昀昀ect a 昀椀rm's COGS. Good engineering design 昀漀cuses not only
on technical 昀甀nctionality but also on cost-e昀昀ectiveness because the design integ爀愀tes the
entire production system. Also of interest to the engineering economist is depreciation
(see Chapter 11), which is the systematic "writing o昀昀" of a capital expense over a period of
years. 吀栀is non-cash expense is important because it represents a decrease in value in the
昀椀rm's capital assets.
吀栀e operating revenues and expenses are shown 昀椀rst so that the 昀椀rm's operating income
昀爀om its products and services can be calculated. Also shown on the income statement are
non-operating expenses, such as interest payments on debt in the 昀漀rm of loans or bonds.
From the data in Figure 2-2, Engineered Industries has total expenses (operating =
$17,1 10,000 and non-operating = $120,000) of $17,230,000. Total revenues are $18,350,000
Income Statement 昀漀r Engineered Industries 昀漀r End of Year 2016 (all amounts in $1,000)
Operating revenues and expenses
Operating revenues
Sales
(minus) Returns and allowances
Total operating revenues
18,900
- 870
18,030
Operating expenses
Cost of goods and services sold
Labour
Materials
Indirect cost
Selling and promotion
Depreciation
General and administrative
Lease p愀礀ments
Total operating expense
6,140
4,640
2,280
930
450
2,160
510
17, 1 1 0
Total operating income
920
Non-operating re瘀攀nues and expenses
Rents
Interest receipts
Interest p愀礀ments
Total non-operating income
20
300
- 120
200
Net income be昀漀re taxes
Income taxes
Net pro昀椀t (loss) 昀漀r Year 2016
1,120
390
730
FIGURE 2-2 Sa mple income statement.
The Income Statement
($18,030,000 + $20,000 + $300,000). 吀栀e net a昀琀er-tax pro昀椀t 昀漀r Year 2016 shown in
Figure 2-2 is $730,000, but it can also be calculated using Equation 2-5:
吀栀ere昀漀re,
Net pro昀椀ts (Loss) = Revenues - Expenses [be昀漀re taxes]
$1,120,000 = 18,350,000 - 17,230,000 [be昀漀re taxes] and with
$390,000 taxes paid
$730,000 = 1,120,000 - 390,000 [a昀琀er taxes]
Financial Ratios Derived from I ncome Statement Data
吀栀e net pro昀椀t ratio (Equation 2-6) equals net pro昀椀ts divided by net sales revenue. Net
sales revenue equals sales minus returns and allowances.
Net pro昀椀t ratio = Net pro昀椀t/Net sales revenue
(2-6)
吀栀is ratio provides insight into the cost e昀케ciency of operations as well as a 昀椀rm's abil­
ity to convert sales into pro昀椀ts. For Engineered Industries in Figure 2-2, the net pro昀椀t ratio
is 730,000/18,030,000 = 0.040 = 4.0%. Like other 昀椀nancial measures, the net pro昀椀t ratio
is best evaluated by comparisons with other time periods and industry benchmarks, and
trends may be more signi昀椀cant than individual values.
Interest coverage, as given in Equation 2-7, is calculated as the ratio of total income to
interest payments-where total income is total revenues minus all expenses except interest
payments.
Interest coverage = Total income/Interest p愀礀ments
(2-7)
吀栀e interest coverage ratio (which 昀漀r industrial 昀椀rms should be at least 3.0) shows
how much revenue must drop to a昀昀ect the 昀椀rm's ability to 昀椀nance its debt. With an
interest coverage ratio of 3.0, a 昀椀rm's revenue would have to decrease by two-thirds
(unlikely) be昀漀re it became impossible to pay the interest on the debt. 吀栀e larger the
interest coverage ratio the better. Engineered Industries in Figure 2-2 has an interest
coverage ratio of
(18,350,000 - 17, 1 10,000)/120,000 = 10.3
Linking the Balance Sheet, I ncome Statement, and Capital Transactions
吀栀e balance sheet and the income statement are separate but linked documents. Under­
standing how the two are linked helps clari昀礀 each. Accounting describes these links as the
articulation between these reports.
吀栀e balance sheet shows a 昀椀rm's assets, liabilities, and equity at a particular point in
time, whereas the income statement summarizes revenues and expenses over an interval.
吀栀ese tabulations can be visualized as a snapshot (a balance sheet) at the beginning of the
period, a video summary over the period (the income statement), and a snapshot at the end
of the period (another balance sheet). 吀栀e income statement and changes in the balance
sheets summarize the business transactions that h愀瘀e occurred during that period.
57
58
Cha pter 2 Acco u n t i n g a n d E n g i neeri n g Eco n o my
吀栀ere are ma渀礀 links between these statements and the cash flows that make up business
transactions, but 昀漀r engineering economic analysis the 昀漀llowing are the most important:
Overall pro昀椀t or loss (income statement) and the starting and ending equity
(balance sheets)
2. Acquisition of capital assets
3. Depreciation of capital assets
1.
吀栀e overall pro昀椀t or loss during the year (shown on the income statement) is reflected
in the change in retained earnings between the balance sheets at the beginning and end
of the year. To 昀椀nd the change in retained earnings (RE), one must also subtract any divi­
dends distributed to the owners and add the value of any new capital stock sold:
REh
q
+ Net income/Loss + New stock - Dividends = RE d
昀케
When capital equipment is purchased, the balance sheet changes, but the income
statement does not. If cash is paid, the cash asset account decrease equals the increase in
the capital equipment account-there is no change in total assets. If a loan is used, then the
capital equipment account increases, and so does the liability item 昀漀r loans. In both cases
the equity accounts and the income statement are unchanged.
吀栀e depreciation of capital equipment is shown as a line on the income statement.
吀栀e depreciation 昀漀r that year equals the change in accumulated depreciation between the
beginning and the end of the year-a昀琀er subtraction of the accumulated depreciation 昀漀r
any asset that is sold or disposed of during that year.
Example 2-1 applies these relationships to the data in Figures 2-1 and 2-2.
EXAM PLE 2-1
For simplicity, assume that Engineered Industries will not pay dividends in 2015 and did not sell any
capital equipment. It did purchase $4,000,000 in capital equipment. Use the linkages just described to
decide what can be said about the values on the balance sheet at the end of 2016.
SO LUTI O N
First, the net pro昀椀t of $730,000 will be added to the retained earnings 昀爀om the end of 2015 to 昀椀nd the
new retained earnings at the end of 2016:
RE I2/ 31 / 2 0 1 6
= $730,000 + $8,㜀㐀5,000 = $9,475,000
Second, the 昀椀xed assets shown at the end of 2015 would increase 昀爀om $6,500,000 to $6,900,000. (Note:
吀栀at is a major investment of the retained earnings in the 昀椀rm's physical assets.)
吀栀ird, the accumulated depreciation would increase by the $450,000 in depreciation shown in the
2016 income statement 昀爀om the $2,350,000 shown in the 2015 balance sheet. 吀栀e new accumulated
depreciation on the 2016 balance sheet would be $2,800,000. Combined with the change in the amount
of capital equipment, the new 昀椀xed asset total 昀漀r 2016 would equal
$335,000 + $6,900,000 - $2,800,000 = $4,435,000
Traditional Cost Accounting
Traditional Cost Accounting
A 昀椀rm's cost-accounting system collects, analyzes, and reports operational per昀漀rmance
data (costs, utilization rates, etc.). Cost-accounting data are used to develop product costs,
to determine the mix of labour, materials, and other costs in a production setting, and to
evaluate outsourcing and subcontracting possibilities.
Di rect a nd I nd i rect Costs
Costs incurred to produce a product or service are traditionally classi昀椀ed as either direct
or indirect (overhead). Direct costs come 昀爀om activities directly associated with the 昀椀nal
product or service produced. Examples include material costs and labour costs 昀漀r engin­
eering design, component assembly, painting, and drilling.
Some organizational activities are di昀케cult to link to speci昀椀c projects, products, or
services. For example, the receiving and shipping areas of a manu昀愀cturing plant are used
by all incoming materials and all outgoing products. Materials and products di昀昀er in their
weight, size, 昀爀agility, value, number of units, packaging, and so on, and the receiving and
shipping costs depend on all those 昀愀ctors. Also, di昀昀erent materials arrive together and
di昀昀erent products are shipped together, so these costs are intermingled and o昀琀en cannot
be tied directly to each product or material.
Other costs, such as the organization's management, sales, and administrative ex­
penses, are di昀케cult to link directly to individual products or services. 吀栀ose indirect, or
overhead, expenses also include machine depreciation, engineering and technical support,
and customer warranties.
䌀䐀
I n d i rect Cost Allocation
To allocate indirect costs to di昀昀erent departments, products, and services, account­
ants use quantities such as direct-labour hours, direct-labour costs, material costs, and
total direct cost. One of those is chosen to be the "burden vehicle." 吀栀e total of all
indirect or overhead costs is divided by the total 昀漀r the burden vehicle. For example, if
direct-labour hours is the burden vehicle, then overhead will be allocated on the basis
of overhead dollar per direct-labour hour. 吀栀en each product, project, or department
will absorb (or be allocated) overhead costs, which will be based on the number of
direct-labour hours each has.
吀栀at is the basis 昀漀r calling traditional costing systems absorption costing. For
decision making, the problem is that the absorbed costs represent average, not incremen­
tal, per昀漀rmance.
Four common ways of allocating overhead are direct-labour hours, direct-labour cost,
direct-materials cost, and total direct cost. 吀栀e 昀椀rst two di昀昀er signi昀椀cantly only if the
cost per hour of labour di昀昀ers 昀漀r di昀昀erent products. Example 2-2 uses direct-labour and
direct-materials cost to illustrate the di昀昀erent choices of burden vehicle.
Problems with Trad itional Cost Accounting
Allocation of indirect costs can distort product costs and the decisions based on those
costs. To be accurate, the analyst must determine which indirect or overhead expenses will
be changed because of an engineering project. In other words, what are the incremental
cash 昀氀ows? For example, vacation and sick leave accrual may be part of overhead, but will
59
60
Cha pter 2 Acco u n t i n g a n d E n g i neeri n g Eco n o my
EXAM PLE 2-2
Industrial Robots does not manu昀愀cture its own motors or computer chips. Its premium prod­
uct di昀昀ers 昀爀om its standard product in having heavier-duty motors and more computer chips 昀漀r
greater 昀氀exibility.
As a result, Industrial Robots manu昀愀ctures a higher 昀爀action of the value of the standard product
itself, and it buys a higher 昀爀action of the premium product's value. Use the 昀漀llowing data to allocate
$850,000 in overhead on the basis of labour cost and materials cost.
Standard
Premium
750
$400
$550
400
$500
$900
Number of units per year
Labour cost (each)
Materials cost (each)
SO LUTI O N
First, the labour and material costs 昀漀r the standard product, the premium product, and in total
are calculated.
Number of units per year
Labour cost (each)
Materials cost (each)
Labour cost
Materials cost
Standard
Premium
Total
750
$400
$550
$300,000
$4 1 2,500
400
$500
$900
$200,000
$360,000
$500,000
$772,500
吀栀en the allocated cost per labour dollar, $1.70, is 昀漀und by dividing the $850,000 in overhead by the
$500,000 in total labour cost. 吀栀e allocated cost per material dollar, $1.100324, is 昀漀und by dividing the
$850,000 in overhead by the $772,500 in materials cost. Now, the $850,000 in allocated overhead is split
between the two products on the basis of labour costs and material costs.
Labour cost
Overhead/labour
Allocation by labour
Material cost
Overhead/material
Allocation by material
Standard
Premium
Total
$300,000
1 .70
5 1 0,000
4 1 2,500
1 . 100324
453,884
$200,000
1 .70
340,000
360,000
1 . 1 00324
396, 1 1 7
$500,000
850,000
0
850,000
If labour cost is the burden vehicle, 60% of the $850,000 in overhead is allocated to the standard prod­
uct. If material cost is the burden vehicle, 53.4% is allocated to the standard product. In both cases,
the $850,000 has been split between the two products. Using total direct costs would produce another
overhead allocation between these two values. However, 昀漀r decisions about product mix and product
prices, incremental overhead costs must be analyzed. All the allocation or burden vehicles are based on
an average cost of overhead per unit of burden vehicle.
Traditional Cost Accounting
they change if the labour content is changed? 吀栀e changes in costs incurred must be esti­
mated. Loadings, or allocations, of overhead expenses cannot be used.
吀栀is issue has become very important because in some 昀椀rms, automation has reduced
direct-labour content to less than 5% of the product's cost. Yet in some of these 昀椀rms, the
basis 昀漀r allocating overhead is still direct-labour hours or cost.
Other 昀椀rms are shi昀琀ing to activity-based costing (ABC), where each activity is linked
to speci昀椀c cost drivers, and the number of dollars allocated as overhead is minimized
(Ligett et al. 1992). Figure 2-3 illustrates the di昀昀erence between ABC and traditional over­
head allocations (Tippet and Hoekstra 1993).
Direct
Labour
Direct
Material
Overhead
Setup
Engineering
Product Cost
FIGURE 2-3 Activity- based costing versus traditional overhead allocation.
Other Problems to Watch For
Centralized accounting systems have o昀琀en been accused by project managers of being too
slow or being "untimely." Because engineering economy is not concerned with the prob­
lem of daily project control, that is a less-critical issue. However, if an organization estab­
lishes numerous 昀椀les and systems so that project managers (and others) have the timely
data they need, the level of accuracy in one or all systems m愀礀 be low. As a result, analysts
making cost estimates will have to consider other internal data sources.
吀栀ere are several cases in which data on equipment or inventory values may be ques­
tionable. When inventory is valued on a "last in, 昀椀rst out" basis, the remaining inventory
may be valued too low. Similarly, land valued at its acquisition cost is likely to be signi昀椀­
cantly undervalued. Finally, capital equipment may be valued at either a low or a high
value, depending on allowable depreciation techniques and company policy.
S U M MARY
Accounting comprises a set of techniques that allow us to track the past 昀椀nancial per昀漀rm­
ance of a company and assess its current 昀椀scal health. Two of the most important tools 昀漀r
these purposes are the balance sheet, which provides a snapshot of the company's 昀椀nances
at a given moment, and the income statement, which tracks cash 昀氀ows into and out of the
company over a 昀椀xed period, typically a year. From the balance sheet we can calculate sta­
tistics, such as the current ratio and acid-test ratio, which measure the company's ability to
pay its debts. From the income statement we can calculate the net pro昀椀t ratio and interest
coverage ratio, which measure the company's pro昀椀tability and ability to meet the interest
due on borrowed money.
61
62
Cha pter 2 Acco u n t i n g a n d E n g i neeri n g Eco n o my
PROBLEMS
Cost accounting is used to keep track of the direct and indirect costs associated with
each of the company's activities. Traditionally, accountants have selected an easy-to-measure
direct cost, such as direct-labour hours, and assumed that the hard-to-measure indirect costs
can be divided between departments in the same ratio as the selected direct cost. 吀栀is method
is known as absorption costing. More recently, some companies have turned to activity-based
costing, which tracks the indirect costs associated with each distinct activity in greater detail.
Accounting
2-1
2-2
2-3
2-4
Explain the accounting 昀甀nction within a 昀椀rm.
What does this 昀甀nction do, and why is it im­
portant? What types of data does it provide?
Manipulation of 昀椀nancial data by the Enron
Corporation was revealed in October 2001.
Firm executives were sentenced to prison.
Arthur Anderson, which had been one of the
"Big 5" accounting 昀椀rms and Enron's auditor,
surrendered its licences to practice as certi昀椀ed
public accountants in August 2002. 吀栀at and
other scandals led the American government
to pass the federal Sarbanes-Oxley (SOX) legislation in July 2002. What are the main parts of
that law? How does it a昀昀ect Canadians?
Insolvency or cash 昀氀ow problems in the US
banking industry started the 昀椀nancial crisis of
2007. Have signi昀椀cant changes in accounting
standards and practices been required by new
legislation? If so, what are the changes?
Accounting and 昀椀nance are required topics 昀漀r
most management degrees, and many engin­
eers do become managers during their careers.
A昀琀er you graduate with a degree in engineer­
ing, what courses in accounting and 昀椀nance
are available to you 昀爀om your or another
nearby university?
Balance Sheet
2-5
2-6
Write short de昀椀nitions 昀漀r the 昀漀llowing terms:
balance sheet, income statement, and 昀甀nda­
mental accounting equation.
Explain the di昀昀erence between short-term and
long-term liabilities.
2-7
Calculate the equity of the Gravel Construc­
tion Company if it has $1 million worth of
assets. Gravel has $127,000 in current liabilities
and $210,000 in long-term liabilities.
2-8
CalcTech has $930,000 in current assets and
$320,000 in 昀椀xed assets less $108,000 in ac­
cumulated depreciation. 吀栀e 昀椀rm's current
liabilities total $350,000, and the long-term lia­
bilities $185,000.
(a) What is the 昀椀rm's equity?
(b) If the 昀椀rm's stock and capital surplus total
$402,000, what is the value of retained
earnings?
2-9
Mama 䰀✀s Baby Monitor Company has current
assets of $5 million and current liabilities of
$2 million. Give the company's working capital
and current ratio.
2-10
From the 昀漀llowing data, taken 昀爀om the bal­
ance sheet of Petey's Widget Factory, deter­
mine the working capital, current ratio, and
acid-test ratio.
Cash
Net accounts and notes receivable
Retailers' inventories
Prepaid expenses
Accounts and notes payable (short-term)
Accrued expenses
2-11
$90,000
1 75,000
2 1 0,000
6,000
322,000
87,000
(a) For Evergreen Environmental Engineer­
ing (EEE), determine the working capital,
current ratio, and acid-test ratio. Evaluate
the company's economic situation with re­
spect to its ability to pay o昀昀 debt.
EEE Balance Sheet Data (thousands)
Cash
Securities
$ 1 10,000
40,000
Problems
Accounts receivable
Retailers' inventories
Prepaid expenses
Accounts p愀礀able
Accrued expenses
160,000
250,000
3,000
351,000
89,000
(b) 吀栀e entries to complete EEE's balance
sheet include the 昀漀llowing:
More EEE Balance Sheet Data (thousands)
Long-term liabilities
Land
Plant and equipment
Accumulated depreciation
Stock
Capital surplus
Retained earnings
$220,000
25,000
510,000
210,000
81,000
15,000
Value not given
Construct EEE's balance sheet.
挀儀 What are EEE's values 昀漀r total assets,
total liabilities, and retained earnings?
2-12
(a) For J&W Graphics Supply, compute the
current ratio. Is this a 昀椀nancially healthy
company? Explain.
J&W Graphics Supply Balance Sheet Data (thousands)
Assets
Cash
Accounts receivable
Inventories
Bad debt provision
Liabilities
Accounts payable
Notes payable
Accrued expenses
$1,740
2,500
900
-75
1,050
500
125
(b) 吀栀e entries on J&W's complete balance
sheet include the 昀漀llowing:
2-13
$950
475
3,100
1,060
680
45
Value not given
Construct J&W's balance sheet.
挀儀 What are J&W's values 昀漀r total assets,
total liabilities, and total earnings?
For Sutton Manu昀愀cturing, determine the
current ratio and the acid-test ratio. Are these
values acceptable? Why or why not?
Su琀琀on Manufacturing Balance Sheet D愀琀a (thousands)
Liabilities
Assets
Current assets
Cash
Accounts
receivable
Inventory
Prepaid expenses
Total current assets
Net 昀椀xed assets
Land
Plant and
equipment
(less accumulated
depreciation)
Other assets
Notes receivable
Intangibles
Total assets
$870
450
Current liabilities
Notes payable
Accounts payable
Accruals
Taxes payable
Current portion
long-term debt
Total current
liabilities
2,670
Long-term debt
O昀케cer debt
(subordinated)
Total liabilities
1,200
60
$500
600
200
30
100
1,430
2,000
200
3,630
1,200 Equi琀礀
3,800
Common shares
1,000
Capital surplus
Retained earnings
Total equity
400
1,200
3,270
200
120 Total liabilities and
equity
6,900
6,900
1670
2-14
If a 昀椀rm has a current ratio less than 2.0 and
an acid-test ratio less than 1.0, will the com­
pany eventually become bankrupt and go out
of business? Explain your answer.
2-15
What is the advantage of comparing 昀椀nancial
statements across periods or against industry
benchmarks rather than looking at statements
associated with a single date or period?
More J&W Balance Sheet Data (thousands)
Long-term liabilities
Land
Plant and equipment
Accumulated depreciation
Stock
Capital surplus
Retained earnings
63
I ncome Statement
2-16
List the two primary general accounting state­
ments. What is each used 昀漀r and how do they
di昀昀er? Which is most important?
2-17
Scarmack's Paint Company has annual sales of
$500,000. If there is a pro昀椀t of $1,000 a day, and
the company operates six days a week, what is
64
Cha pter 2 Accounting a n d Engi neering Economy
the total yearly business expense? All calcula­
tions are on a be昀漀re-tax basis.
2-18
Laila's Surveying Inc. had revenues of $100,000
in 2016. Expenses totalled $60,000. What was
her net pro昀椀t (or loss)?
2-19
For Magdalen Industries, compute the net
income be昀漀re taxes and net pro昀椀t (or loss).
Taxes 昀漀r the year were $1 million.
(a) Calculate net pro昀椀t 昀漀r the year.
(b) Construct the income statement.
(c) Calculate the interest coverage and net
pro昀椀t ratio. Is the interest coverage ac­
ceptable? Explain why or why not.
Non-operating expenses
Interest payments
Total non-operating expenses
Total expenses, E
Net income be昀漀re taxes, R - E
Income taxes
Net pro昀椀t (loss) 昀漀r Year 2016
Linking Balance Sheet and I ncome Statement
2-21
0
2-20
$81
5
70
7
For Andrew's Electronic Instruments, calcu­
late the interest coverage and net pro昀椀t ratio. Is
Andrew's business healthy?
0
2-22
Income Statement 昀漀r Andrew's Electronics
昀漀r End of Year 2016 (thousands)
$395
- 15
380
Magdalen Industries (see Problem 2-19) had
the 昀漀llowing entries in its balance sheet at the
end of last year:
$ 1 5 million
8 million
60 million
In addition to the income statement data 昀漀r
this year in Problem 2-19, we also know that
the 昀椀rm purchased $3 million of equipment
with cash and that depreciation expenses were
$2 million of the $70 million in operating ex­
penses listed in Problem 2-19. 吀栀e 昀椀rm paid no
dividends this year. What are the entries in the
balance sheet at the end of this year 昀漀r
愀儀 plant and equipment?
(b) accumulated depreciation?
(c) retained earnings?
50
25
75
455
200
34
68
20
30
10
10
372
$420,000
480,000
In addition, we also know the 昀椀rm purchased
$800,000 worth of equipment with cash. 吀栀e
昀椀rm paid $200,000 in dividends this year.
What are the entries in the balance sheet at the
end of this year 昀漀r
(a) plant and equipment?
(b) accumulated depreciation?
(c) retained earnings?
Plant and equipment
(less accumulated depreciation)
Retained earnings
Revenues
Operating revenues
Sales
(minus) Returns
Total operating revenues
Non-operating revenues
Interest receipts
Stock revenues
Total non-operating revenues
Total revenues, R
Expenses
Operating expenses
Cost of goods and services sold
Labour
Materials
Indirect cost
Selling and promotion
Depreciation
General and administrative
Lease payments
Total operating expenses
Sutton Manu昀愀cturing (see balance sheet at the
end of last year in Problem 2-13) had the 昀漀l­
lowing entries in this year's income statement:
Depreciation
Pro昀椀t
Magdalen Industries Income Statement
Data (millions)
Revenues
Total operating revenue
(including sales of $48 million)
Total non-operating revenue
Expenses
Total operating expenses
Total non-operating expenses
(interest payments)
22
22
394
61
30
31
Allocating Costs
2-23
Categorize each of the 昀漀llowing costs as direct
or indirect. Assume that a traditional costing
system is in e昀昀ect.
Problems
Machine run costs
Machine depreciation
Material handling costs
Cost of materials
Overtime expenses
Machine operator wages
Utility costs
Support (administrative)
sta昀昀 salaries
2-24
Cost of marketing the product
Cost of storage
Insurance costs
Cost of product sales 昀漀rce
Engineering drawings
Machine labour
Cost of tooling and 昀椀xtures
Model S
Model M
Model G
Direct-material costs
Direct-labour costs
Direct-labour hours
$3,800,000 $1,530,000
380,000
600,000
20,000
64,000
$2,105,000
420,000
32,000
2-25
Par Golf Equipment Company produces two
types of golf bag: the standard and deluxe
models. 吀栀e total indirect cost to be allocated
to the two bags is $35,000. Calculate the net
revenue that Par Golf can expect 昀爀om the sale
of each bag.
(a) Use direct-labour cost to allocate
indirect costs.
(b) Use direct-material cost to allocate in­
direct costs.
Data Item
Direct -labour cost
Direct-material cost
Selling price
Units produced
2-27
吀栀e general ledger of the Fly-Buy-Nite (FBN)
Engineering Company contained the 昀漀llow­
ing account balances. Construct an income
statement. What is the net income be昀漀re taxes
and the net pro昀椀t (or loss) a昀琀er taxes? FBN has
a tax rate of 27%.
Amount (thousands)
Administrative expenses
Subcontracted services
Development expenses
Interest expense
Sales revenue
Selling expenses
$ 2,750
18,000
900
200
30,000
4,500
2-28
R䰀圀-11 Enterprises estimated that indirect
manu昀愀cturing costs 昀漀r the year would be
$60 million and that 12,000 machine-hours
would be used.
(a) Compute the predetermined indirect
cost application rate using machine
hours as the burden vehicle.
(b) Determine the total cost of production
昀漀r a product with direct material
costs of $1 million, direct-labour
costs of $600,000, and 200 machine­
hours.
2-29
For Gee-Whiz Devices calculate the 昀漀llowing:
working capital, current ratio, and acid-test
ratio.
Standard
Deluxe
Gee-Whiz Devices Balance Sheet Data
$50,000
$35,000
$60
1,800
$65,000
$47,500
$95
1,400
Cash
Market securities
Net accounts and notes receivable
Retailers' inventories
Prepaid expenses
Accounts and notes payable (short-term)
Accrued expenses
U nclassi昀椀ed
2-26
(b) If the 昀椀rm's stock and capital surplus
total $305,000, what is the value of re­
tained earnings?
LeGaroutte Industries makes industrial pipe
manu昀愀cturing equipment. Use direct-labour
hours as the burden vehicle, and compute the
total cost per unit 昀漀r each model given in
the table. Total manu昀愀cturing indirect costs
are $15,892,000, and there are 100,000 units
manu昀愀ctured per year 昀漀r Model S, 50,000 昀漀r
Model M, and 82,250 昀漀r Model G.
Item
Matbach Industries has $870,000 in current
assets and $430,000 in 昀椀xed assets less $180,000
in accumulated depreciation. 吀栀e 昀椀rm's cur­
rent liabilities total $330,000, the long-term lia­
bilities $115,000.
(a) What is the 昀椀rm's equity?
65
2-30
$100,000
45,000
150,000
200,000
8,000
315,000
90,000
Turbo Start has current assets totalling
$1.5 million (that includes $500,000 in cur­
rent inventor礀⤀ and current liabilities totalling
$50,000. Find the current ratio and acid-test
ratio. Are the ratios at desirable levels? Explain.
Interest and Eq uivalence
Nuclear Waste Storage
Nuclear power offers a reliable supply of energy, with no greenhouse gas emissions. But spent
fuel from reactors is highly radioactive and must be stored securely for long periods-hundreds of
thousands of years-until the radioactivity has decayed to safe levels. Although skilled engineers
have been working on this problem for many decades, humanity as a whole has no experience in
building anything that can last for such a long period.
The Canadian Nuclear Safety Commission oversees the Nuclear Waste Management Organ­
ization (NWMO), which is responsible for the long-term management of Canadian used nuclear
fuel. In October 2015 the NWMO released a report, " Implementing Adaptive Phased Management,
2016 - 2020, " describing its plan to locate a suitable depository, deep underground in a suitable
rock formation, where the fuel can be placed.
At the end of this chapter, we will look at the implications of engineering economic analysis
for long-term projects of this kind.
Satakorn/iStock Photo
N u clea r Waste Sto ra ge
67
QU ESTI ONS TO CONSI D E R
1.
What strateg ies have other cou ntries co nsidered to h a n d le thei r n u clea r waste storage problems?
2.
U n l i ke m a ny cou ntries, Ca nada has vast a reas of a lmost e m pty territory, pa rti c u la rly i n the fa r N o rt h . Does
this m a ke our n u clea r storage problem easier?
3.
4.
What a lternatives a re there t o b u rying n u clea r waste deep u nd e rg ro u n d ?
B u r n i n g foss i l fuels leads t o the lon g-te rm problem o f g lobal wa r m i ng, wh i le the carbon-free a lternative of
n u clea r powe r lea ds to the very lo ng-term problem of n u clea r waste. What issues wo u ld need to be co n ­
s i d e red to determ i n e which o f these two problems p resents a g reater th reat to futu re generations?
LEARN I N G OBJ ECTIVES
This chapter will help you
•
d e昀椀 n e a n d p rovi d e exa m p les of the time 瘀愀lue of money
•
d isti n g u ish between simple a n d compound interest, a n d use com po u n d i nterest i n e n g i n ee ri n g eco n o m i c a n a lysi s
•
expla i n equi瘀愀lence of cash flows
•
solve problems by u s i n g the s i n g le-payment co m p o u n d i nterest form u la s
•
d isti n g u ish a n d a p ply nominal a n d e昀昀ective interest rates
•
use continuously compounded interest with s i n g le-payment i nterest factors
KEY TE RMS
com pound interest
continuous com pou nding
disbursement
e昀昀ective interest rate
equ ivalence
interest
nominal interest rate
receipt
simple interest
time value of money
In the 昀椀rst chapter we discussed situations where the economic consequences of an
alternative were immediate or took place in a very short time, as in Example 1-2 (the
decision about the design of a concrete aggregate mix) or Example 1-3 (the change of
manu昀愀cturing method). We totalled the various positive and negative aspects, compared
our results, and could quickly reach a decision. However, we cannot do this if the eco­
nomic consequences occur over a considerable period of time.
吀栀e reason we cannot is that money has value over time. We would rather receive $1,000
tod愀礀 than receive $1,000 ten years 昀爀om today. Our preference can be expressed as an interest
rate. In this chapter, 眀攀 describe two concepts involving the time value of money: interest and
cash 昀氀ow e焀甀ivalence. Later in the chapter, nominal interest and e昀昀ecti瘀攀 interest are discussed.
Finally, equations are deri瘀攀d 昀漀r situations where interest is compounded continuously.
䌀䐀
Computing Cash Flows
Installing an expensive piece of machinery in a plant has economic consequences that occur
over an extended period of time. It was bought because it per昀漀rms a use昀甀l 昀甀nction, which
has a cash value. If the machinery was bought on credit, p愀礀ing 昀漀r it may take several
years. We describe the bene昀椀ts and costs as receipts (cash 昀氀owing in) and disbursements
(cash 昀氀owing out) at di昀昀erent points in time. 吀栀is is illustrated by Examples 3-1 and 3-2.
68
Cha pter 3 I nterest a n d Eq u iva le nce
EXAM PLE 3 - 1
吀栀e manager has decided to buy a new $30,000 mixing machine. 吀栀e machine m愀礀 be paid 昀漀r in one
of two ways:
Pay the 昀甀ll price now minus a 3 % discount.
Pay $5,000 now; at the end of one year, p愀礀 $8,000; at the end of each of the next 昀漀ur years,
pay $6,000.
1.
2.
List the alternatives in the 昀漀rm of a table of cash flows.
SO LUTI O N
吀栀 e 昀椀rst plan represents a lump sum o f $29,100 now; the second calls 昀漀 r payments continuing until
the end of the 昀椀昀琀h year. 吀栀ose p愀礀ments total $37,000, but the $37,000 value has no clear meaning,
since it is the sum of several cash 昀氀ows occurring at di昀昀erent times. 吀栀e second alternative can only be
described as a set of cash 昀氀ows.
Both alternatives are shown in the cash 昀氀ow table, with disbursements given negative signs.
End of Year
Pay in Full Now
Pay over 5 Years
0 (now)
- $29, 100
- $5,000
1
0
- 8,000
2
0
-6,000
3
0
- 6,000
4
0
- 6,000
5
0
-6,000
吀栀is can also be presented as a cash 昀氀ow diagram:
!
0 -- 1 -- 2 -- 3 -- 4 -- s
$29, 100
! ,.L ,⸀䰀 ,.L
0 -- 1 --2 -- 3 -- 4 -- s
1st
$8,000
Joo
EXAM PLE 3 - 2
A man borrowed $1,000 昀爀om a bank at 8% interest. He agreed to repay the loan in two end-of-year
payments. At the end of each year, he will repay half of the $1,000 principal amount plus the interest
that is due. Compute the borrower's cash 昀氀ows.
SO LUTI O N
I n engineering economic analysis, we normally refer to the beginning of the 昀椀rst year a s "time 0." At
this point the man receives $1,000 昀爀om the bank. (A positive sign represents a receipt of money, and a
negative sign, a disbursement.) His Time 0 cash 昀氀ow is +$1,000.
At the end of the 昀椀rst year, the man pays 8% interest 昀漀r the use of $1,000 昀漀r one year. 吀栀e interest
is 0.08 x $1,000 = $80. In addition, he repays half the $1,000 loan, or $500. 吀栀ere昀漀re, his end-o뼀鈀ar-I
cash 昀氀ow is -$580.
Computing Cash Flows
69
At the end of the second year, the p愀礀ment is 8% 昀漀r the use of the balance of the principal ($500)
昀漀r the one-year period, or 0.08 x $500 = $40. 吀栀e $500 principal is also repaid 昀漀r a total end-of-Year-2
cash 昀氀ow of -$540. 吀栀e cash 昀氀ow is
End of Year
0 (now)
1
2
Cash Flow
+$1,000
-580
-540
t
$ 1 ,000
0 -- 1 -- 2
i t
$580
$540
Techniques 昀漀r comparing the value of money at di昀昀erent dates are the 昀漀undation
of engineering economic analysis. We must be able to compare, 昀漀r example, a low-cost
motor with a higher-cost motor. If there were no other consequences, we would obviously
prefer the low-cost one. But if the higher-cost motor were more e昀케cient and thereby re­
duced the annual electric power cost, we would want to consider whether to spend more
money now on the motor to reduce power costs in the 昀甀ture. 吀栀is chapter will provide
methods 昀漀r comparing the alternatives to determine which is preferable.
䌀䐀
Time Value of Money
We o昀琀en 昀椀nd that the monetary consequences of any alternative occur over a substantial
period of time-say, a year or more. When monetary consequences occur in a short period
of time, we simply add up the various sums of money. But can we treat money this way
when the time span is greater?
Which would you prefer, $100 cash today or the assurance of receiving $100 a year
昀爀om now? You might decide you would prefer the $100 now because that is one way to be
certain of receiving it. But suppose you were convinced that you would receive the $100
one year hence. Now what would be your answer? A little thought should persuade you
that it would still be more desirable to receive the $100 now. If you had the money now,
rather than a year hence, you would have the use of it 昀漀r an extra year. And if you had no
current use 昀漀r $100, you could let someone else use it.
Money is a valuable asset-so valuable that people are willing to pay to have it avail­
able to use. Money can be rented in roughly the same way that an apartment is rented;
with money the charge is called interest instead of rent. 吀栀e importance of interest is de­
monstrated by the 昀愀ct that banks and savings institutions are continually o昀昀ering to pay
昀漀r the use of people's money, that is, to pay interest.
Our preference 昀漀r having money now rather than money in the 昀甀ture di昀昀ers 昀爀om
person to person. For example, suppose you have just come up with an idea 昀漀r a new cell
phone. If you can get it to market ahead of the competition, you can expect to dominate the
market 昀漀r years to come. So you might be prepared to p愀礀 a high interest rate on the money
that you borrow to get the device market-ready. You might, 昀漀r example, be prepared to pay
an interest rate of 9% or more. 吀栀at is, in exchange 昀漀r having the use of $100 during the
coming year, you are willing to pay $109 or more a year 昀爀om now. If the bank agrees to loan
you the $100 at 9% interest, then the bank manager is implicitly saying that he has a di昀昀er­
ent preference-he would rather have your promise of $109 in a year's time than the $100
right now. So in this case, you assign a higher time value to money than the banker does.
吀栀e bank expresses the time value it puts on money by publishing its interest rate. If you
put a higher time value on money than the bank does, you may 昀椀nd it attractive to borrow
70
Cha pter 3 Interest and Equivalence
money 昀爀om the bank at its published interest rate and pay it back later. If, on the other hand,
you put a lower time value on money than the bank does, you will 昀椀nd it attractive to lend
money to the bank in return 昀漀r their promise to p愀礀 it back, with interest, in the 昀甀ture.
To 昀漀restall a possible con昀甀sion, we should make it clear that our preference 昀漀r h愀瘀ing
money now rather than later has nothing to do with in昀氀ation. In昀氀ation occurs when the
same pizza that cost $5.00 last year now costs $5.50. Until we get to Chapter 14, we assume
that there is no in昀氀ation-the pizza always costs $5.00, but we would nevertheless prefer
having pizza now to the promise of a pizza in a year's time.
Sim ple I nterest
Simple interest is interest that is computed only on the original sum and not on accrued
interest. 吀栀us, if you were to loan a present sum of money P to someone at a simple annual
interest rate i (stated as a decimal) 昀漀r a period of n years, the amount of interest you would
receive 昀爀om the loan would be
Total interest earned = P X i X n = Pin
(3-1)
At the end of n years the amount of money due you, F, would equal the amount of the loan P
plus the total interest earned. 吀栀at is, the amount of money due at the end of the loan would be
F = P + Pin
or F = P (1 + in)
(3-2)
EXAM PLE 3-3
You have agreed to lend a 昀爀iend $5,000 昀漀r 昀椀ve years at a simple-interest rate of 8% a year. How much
interest will you receive 昀爀om the loan? How much will your 昀爀iend pay you at the end of 昀椀ve years?
SO LUTI O N
Total interest earned = Pin = ($5,000)(0.08)(5 yr) = $2,000
Amount due at end of loan = P + Pin = 5,000 + 2,000 = $7,000
In Example 3-3 the interest earned at the end of the 昀椀rst year is (5,000)(0.08)(1) = $400,
but this money is not paid to you until the end of the 昀椀昀琀h year. As a result, the borrower
has the use of your $400 昀漀r 昀漀ur years without paying any interest on it. 吀栀is is how simple
interest works, and it is easy to see why lenders seldom agree to make simple-interest loans.
䌀䐀
Com pound I nterest
With simple interest, the amount earned (昀漀r invested mone礀⤀ or due (昀漀r borro眀攀d mone礀⤀ in
one period does not a昀昀ect the principal 昀漀r interest calculations in later periods. However, this
is not how interest is normally calculated. In practice, interest is computed by the compound
interest method. For a loan, any interest owed but not paid at the end of the year is added to
the balance due. 吀栀us the next 礀攀ar's interest is calculated on the unpaid balance due, which
includes the unpaid interest 昀爀om the preceding period. In this way, compound interest can
be thought of as interest on top of interest. 吀栀is distinguishes compound interest 昀爀om simple
interest. In this section and in the rest of the book, 礀漀u should assume that the rate is a com­
pound interest rate. 吀栀e few exceptions will clea爀氀y state that you should use simple interest.
Time Value of Money
71
EXAM PLE 3-4
To highlight the di昀昀erence between simple and compound interest, rework Example 3-3 using an inter­
est rate of 8% a year compound interest. How will this change a昀昀ect the amount that your 昀爀iend pays
you at the end of 昀椀ve years?
SO LUTI O N
Original loan amount (original principal) = $5,000
Loan term = 昀椀ve years
Interest rate charged = 8% a year compound interest
In the 昀漀llowing table we calculate on a year-to-year basis the total dollar amount due at the end of each
year. Notice that this amount becomes the principal upon which interest is calculated in the next year
(this is the compounding e昀昀ect).
Year
Total Principal (P )
on Which Interest Is
Calculated in Year n
1
2
3
4
5
$5,000
5,400
5,832
6,299
6,803
Interest (I ) Owed at the
End of Year n 昀爀om Year n's
Unpaid Total Principal
$5,000 X 0.08 = 400
5,400 X 0.08 = 432
5,832 X 0.08 = 467
6,299 X 0.08 = 504
6,803 X 0.08 = 544
Total Amount Due at the
End of Year n, New Total
Principal 昀漀r Year n + 1
$5,000 + 400 = 5,400
5,400 + 432 = 5,832
5,832 + 467 = 6,299
6,299 + 504 = 6,803
6,803 + 544 = 7,347
吀栀e total amount due at the end of the 昀椀昀琀h year, $7,347, is the amount that your 昀爀iend will give you to
repay the original loan. Notice that this amount is $347 more than the amount you received 昀漀r lending
the same amount 昀漀r the same period at simple interest. 吀栀is, of course, is because of the e昀昀ect of inter­
est being earned (by you) on top of interest.
Repayi ng a Debt
To understand better the mechanics of interest, let us say you owe $5,000 and must rep愀礀
it in 昀椀ve years, together with 8% annual interest. 吀栀ere are many ways in which debts are
repaid; 昀漀r simplicity, we have selected three speci昀椀c ways 昀漀r our example. Table 3-1 tabu­
lates the three plans.
Plan 1 (Constant Principal), like Example 3-2, repays 1/nth of the principal each year.
So in Plan 1, $1,000 will be paid at the end of each year plus the interest due at the end of
the year 昀漀r the use of money to that point. 吀栀us, at the end of Year I, we will have had
the use of $5,000. 吀栀e interest owed is 8% x $5,000 = $400. 吀栀e end-of-year p愀礀ment is
$1,000 principal plus $400 interest, 昀漀r a total payment of $1,400. At the end of Year 2, an­
other $1,000 principal plus interest will be repaid on the money owed during the year. 吀栀is
time the amount owed has declined 昀爀om $5,000 to $4,000 because of the $1,000 principal
repayment at the end of Year 1. 吀栀e interest payment is 8% x $4,000 = $320, making the
end-of-year payment a total of $1,320. As shown in Table 3-1, the series of payments con­
tinues each year until the loan is 昀甀lly repaid at the end of Year 5.
72
Cha pter 3 Interest and Equivalence
In Plan 2 ⠀䤀nterest On礀销, only the interest due is paid each year, with no principal
payment. Instead, the $5,000 owed is repaid in a lump sum at the end of the 昀椀昀琀h year.
吀栀e end-of-year payment in each of the 昀椀rst 昀漀ur years of Plan 2 is 8% x $5,000 = $400.
吀栀e 昀椀昀琀h year, the payment is $400 interest plus the $5,000 principal, 昀漀r a total of $5,400.
Plan 3 (Constant Payments) calls 昀漀r 昀椀ve equal end-of-year payments of $1,252 each.
In Example 4-3 we will show how the 昀椀gure of $1,252 is computed. By 昀漀llowing the com­
putations in Table 3-1, we see that a series of 昀椀ve payments of $1,252 repays a $5,000 debt
in 昀椀ve years with interest at 8%.
吀栀e cash flow diagrams corresponding to the three plans are shown in Figure 3-1, 眀栀ich
is drawn 昀爀om the bank's point of view-the money lent is shown as a negative quantity, below
the line, while the money repaid is shown as a series of positive quantities above the line.
Table 3-1 Th ree Pla ns for Repayment of $5,000 i n Five Yea rs with
I nterest at 8%
(a)
(b)
Year
Amount Owed
at Beginning
of Year
(c)
(d)
Interest Owed Total Owed at
昀漀r 吀栀at Year, End of Year,
8% X (b)
(b) + (c)
(e)
(f)
Principal
P愀礀ment
Total End-ofYear Payment
Plan 1 : Constant $ 1,000 principal payment plus interest due.
1
$5,000
$ 400
$5,400
$1,000
$1,400
2
4,000
320
4,320
1,000
1,320
3
3,000
240
3,240
1,000
1,240
4
2,000
160
2,160
1,000
1,160
5
1,000
80
1,080
1,000
--
1,080
--
$1,200
$5,000
$6,200
Plan 2: Annual interest payment due and principal payment at end of 昀椀ve years.
1
$5,000
$ 400
$5,400
$0
$ 400
2
5,000
400
5,400
0
400
3
5,000
400
5,400
0
400
4
5,000
400
5,400
0
400
5
5,000
400
5,400
5,000
--
5,400
--
$5,000
$2,000
$7,000
Plan 3: Constant annual payments.
1
$5,000
$ 400
$5,400
$ 852
$1,252*
2
4,148
331
4,479
921
1,252
3
3,227
258
3,485
994
1,252
4
2,233
178
2,411
1,074
1,252
5
1,159
93
1,252
1,159
--
1,252
--
$1,260
*吀栀e exact value is $ 1 ,252.28, which has been rounded to an even dollar amount.
$5,000
$6,260
Time Value of Money
l I t t t
$ 1 ,400
$ 1 , 3 20
$ 1 , 240
$ 1 , 1 60
$ 1 ,0 8 0
0 -- 1 -- 2 -- 3 -- 4 -- 5
l
$5, 000
$5,400
r
'400
--T----r---T---l-
0 -- 1 -- 2 -- 3 -- 4 -- 5
l
$5,000
s
r - - - - - l - - - - - -r- - - - - - l - - - - - -r - - - - - .,., ,
0 -- 1 -- 2 -- 3 -- 4 -- 5
l
$5, 000
FIGURE 3-1 Th ree repayment sched ules for a $5,000 loa n.
With Table 3-1 we have illustrated three di昀昀erent ways of accomplishing the same
task; that is, repaying a debt of $5,000 in 昀椀ve years with interest at 8%. Having described
the alternatives, we will now use them to present the important concept of equivalence.
䌀䐀
Equivalence
When we are indi昀昀erent as to whether we have a quantity of money now or the assurance
of some other 昀甀ture sum or sums of money, we say that the present sum of money is
equivalent to the 昀甀ture sum or series of 昀甀ture sums.
Since the bank in Table 3-1 believes that 8% is a reasonable interest rate, it has no par­
ticular preference about whether it receives $5,000 now or is repaid by Plan 1 of Table 3-1.
吀栀us $5,000 tod愀礀 is equivalent to the series of 昀椀ve end-of-year payments. In 昀愀ct, all three
repayment plans must be equivalent to each other and to $5, 000 now at 8% interest.
Equivalence is an essential 昀愀ctor in engineering economic analysis. In Chapter 1,
we saw how an alternative could be represented by a cash 昀氀ow table. How might two al­
ternatives with di昀昀erent cash 昀氀ows be compared? For example, consider the cash flows 昀漀r
Plans 1 and 2 shown in Figure 3-1.
If you were given your choice between the two alternatives, which one would you choose?
Obviously the two plans have cash 昀氀ows that are di昀昀erent. Plan 1 requires that there be larger
payments in the 昀椀rst 昀漀ur years, but the total p愀礀ments are smaller than the sum of the Plan 2
payments. To make a decision, the cash 昀氀ows must all be moved to the same moment in time
so that they can be compared. 吀栀e technique of equivalence is the way we do that.
Using a selected interest rate, we can determine an equivalent value at some point in
time 昀漀r Plan 1 and a comparable equivalent value 昀漀r Plan 2. 吀栀en we can judge the rela­
tive attractiveness of the two alternatives. Since Plan 1, like Plan 2, repays a present sum
of $5,000 with interest at 8%, the plans are each equivalent to $5,000 now at an interest
rate of 8%.
73
74
Cha pter 3 Interest and Equivalence
Now let us look at the three loan schemes 昀爀om our own point of view rather than the
bank's. We want to borrow money now to develop our invention and expect to realize a
substantial pro昀椀t in the 昀甀ture. So we put a higher value on having money now than the
bank does. Let's suppose that the time value we put on money corresponds to an interest
rate of 9%. Now 昀爀om our point of view, the three schemes may not be equivalent. Which
one is the most attractive?
吀栀e method we use to answer this is the same method that we will be using 昀漀r many
of the problems throughout this text. It consists of three steps. First, we summarize all the
cash 昀氀ows involved in each plan, using a cash 昀氀ow diagram. Figure 3-1 shows all the cash
昀氀ows 昀爀om the bank's point of view. From our own point of view, each of these cash 昀氀ows
is reversed: our debt repayments are income 昀漀r the bank but expenditures 昀漀r us. Next, we
bring all of the cash flows shown in the diagram to a single moment in time. In this case,
let's use the present time, shown as Year 0 on the cash 昀氀ow diagram.
吀栀e beginning of this process 昀漀r Plan 1 is shown in Figure 3-2(i): we take the cash
昀氀ow of $1,400 at Year 1, and replace it by its equivalent value at Year 0, shown by a dotted
arrow. To 昀椀nd the amount that is equivalent to the $1,400, we multiply the latter 昀椀gure by
a conversion 昀愀ctor, (PIP, 0.09, 1).
For the moment, we can think of this 昀愀ctor as a call to a subroutine. 吀栀e way we write
the 昀愀ctor tells us three things:
"PIP" says that we are seeking the present quantity equivalent to a 昀甀ture amount.
"0.09" says that this equivalence is 昀漀r an interest rate of 9%.
"1" says that the 昀甀ture amount is one time period in the 昀甀ture.
$5, 000
$5, 000
1
1 ! t t t
0 -- 1 -- 2 -- 3 -- 4 -- 5
4
1
$ , 00 :
(PIP, 0.09, 1 ) �
I
$ 1 ,400 $ 1 ,320
V
$ 1 ,240
$ 1 , 160
(PIP, 0.09, 1 )
$5. 000
$ 1 ,080
i
1
! i t i
0 -- 1 -- 2 -- 3 -- 4 -- 5
$ 1 ,400 (P/F, 0 . 0 9, I )
$ l , 320 (PIP, o . o9, 2) :
l
$ 1 , 320
I
T
�
(PIP, 0.09, 2 )
i
0 -- 1 -- 2 -- 3 -- 4 -- 5
$ 1,400 (PIP, 0.09, 1 ) :
T
$ 1 ,oso
$ 1, 320 (PIP, 0.09, 2) :
T
T
+
$ 1, 160 (PIP, 0.09, 3 ) :
$ 1, 240 (PIP, 0.09, 4)
$ 1, 0 B 0 (PIP, 0.09, 5) y
FIGURE 3-2 Movi ng a series of futu re cash flows to the present.
(PIP, 0.09, 5)
$ 1 ,240
$ 1 , 160
$ 1,080
Eq u ivalence
Once we know these three things, the numerical value of the conversion 昀愀ctor can be
calculated. 吀栀ere is a simple 昀漀rmula 昀漀r doing this calculation, which we will encounter in
a few pages; alternatively, we can look it up on the OUP website associated with this text,
or obtain it 昀爀om a 昀椀nancial calculator or a spreadsheet program.
Having moved the Year 1 amount back to Year 0, we next move the Year 2 cash 昀氀ow,
this time using the conversion 昀愀ctor (PIP, 0.09, 2) (See Figure 3-2(ii)). You should now be
able to write down the conversion 昀愀ctors 昀漀r the remaining amounts; the 昀椀nal stage is
shown in Figure 3-2(iii).
All the cash 昀氀ows are now at the same moment in time, so we can add them up to get
the present worth (PW), to us, of Plan 1:
PW(Plan 1) = $5,000 - 1,400(PIP, 0.09, 1) - 1,320(PIP, 0.09, 2)
- 1,240(PIP, 0.09, 3) - 1,160(PIP, 0.09, 4)) - 1,0B0(PIP, 0.09, 5)
Similarly, we can write down the present worth, to us, of Plans 2 and 3:
PW(Plan 2) = $5,000 - 400(PIP, 0.09, 1) - 400(PIP, 0.09, 2)
- 400(PIP, 0.09, 3) - 400(PIP, 0.09, 4) - 5,400(PIP, 0.09, 5)
and
PW(Plan 3) = $5,000 - 1,252(PIP, 0.09, 1) - 1,252 (PIP, 0.09, 2)
- 1,252 (PIP, 0.09, 3) - 1,252 (PIP, 0.09, 4) - 1,252 (PIP, 0.09, 5)
(In Chapter 4, we will see that there are shortcuts available 昀漀r simple cash 昀氀ow pat­
terns like those in Plans 2 and 3.)
吀栀e hard work has now been done, and all that remains is the numerical evaluation of
these expressions. We will learn a simple 昀漀rmula 昀漀r this in the next section, but 昀漀r the
present, we can look up the 昀椀ve needed values on the OUP website, remembering that to
昀椀nd the conversion 昀愀ctor (PIP, i, n), we turn to page i, 昀椀nd the column labelled "PIP," and
go down to the row labelled "n":
Factor
Numerical Value
(PIP, 0.09, 1)
0.9174
(PIP, 0.09, 2)
0.8417
(PIP, 0.09, 3)
0.7722
(PIP, 0.09, 4)
0.7084
(PIP, 0.09, 5)
0.6499
Substituting these values into the equations above, we 昀椀nd that the present worth of
Plan 1 is $123.37, while the present worths of Plans 2 and 3 are $194.48 and $130.16, re­
spectively. So 昀爀om our point of view, each of the plans is pro昀椀table, and Plan 2, which
allows us to defer the repayment of the principal 昀漀r the longest time, is the most pro昀椀t­
able of the three.
Single- Payment Com pound Interest Formulas
We will now derive the interest 昀漀rmulas used to calculate the table above. To simpli昀礀 the
presentation, we'll use the 昀漀llowing notation:
75
76
Cha pter 3 Interest and Equivalence
i = interest 爀愀te per interest period. In the equations the interest rate is stated as a
decimal (that is, 9% interest is 0.09).
n = number of interest periods.
P = a present sum of money.
F = a future sum of money at the end of the nth interest period, which is equivalent to
P at the interest rate i.
Suppose a present sum of money P is invested 昀漀r one year1 at interest rate i. At the end of
the year, we should receive back our initial investment P, together with interest equal to iP,
or a total amount P + iP. Factoring P, the sum at the end of one year is P(l + i).
Let us assume that, instead of removing our investment at the end of one year, we
agree to let it remain 昀漀r another year. How much would our investment be worth at the
end of Year 2? 吀栀e end-of-昀椀rst-year sum P(l + i) will draw interest in the second year of
iP(l + i). 吀栀is means that at the end of Year 2 the total investment will become
P(l + i) + iP(l + i)
吀栀is may be rearranged by 昀愀ctoring out P(l + i), which gives
P(l + i)(l + i)
P(l + i) 2
or
If the process is continued 昀漀r a third year, the total amount at the end of the third year will
be P(l + i) 3 ; at the end of n years, we will have P(l + i) 0 • 吀栀e progression looks like this:
Year
1
2
3
n
Amount at Beginning of
Interest Period
p
P( l + i)
P( l + i) 2
P( l + i)n-i
+ Interest 昀漀r Period
= Amount at End
of Interest Period
= P( l + i)
= P( l + i) 2
+ iP
+ iP( l + i)
+ iP( l + i) 2
+ iP( l + i)n-i
= P( l + i) 3
= P( l + i)n
In other words, a present sum P increases in n periods to P(l +
We there昀漀re have a
relationship between a present sum P and its equivalent 昀甀ture sum 䘀⸀
Future sum = (Present sum)(l + i) n
F = P(l + i) n
漀渀.
(3-3)
吀栀is is the si最�e-payment compound amountformula and is written in 昀甀nctional notation
as
F = P(FIP, i, n)
(3-4)
吀栀e notation in parentheses (PIP, i, n) can be read as 昀漀llows:
A more general statement is to speci昀礀 "one interest period" rather than "one year:' Since it is easier to
visualize one year, the derivation uses one year as the interest period.
1
Single-Payment Compound Interest Formulas
77
Find a 昀甀ture sum F, given a present sum P at an interest rate i per interest period,
and n interest periods hence;
or
昀椀nd F, given P, at i, over n.
Functional notation is designed so that the compound interest 昀愀ctors may be written in
an equation in an algebraically correct 昀漀rm. In Equation 3-4, 昀漀r example, the 昀甀nctional
notation is interpreted as
which is algebraically correct. Without proceeding 昀甀rther, we can see that when we derive
a compound interest 昀愀ctor to 昀椀nd a present sum P, given a 昀甀ture sum F, the 昀愀ctor will be
(PIP, i, n); so, the resulting equation would be
P = F(PIF, i, n)
which again is algebraically correct.
EXAM PLE 3-5
If $500 were deposited in a bank savings account, how much would be in the account three years 昀爀om
now if the bank paid 6% interest compounded annually?
We can draw a diagram of the problem.
SO LUTI O N
From the viewpoint of the person depositing the $500, the diagram 昀漀r "today" (Time = 0) through
Year 3 is as 昀漀llows:
F= ?
Receipts ( + $)
0 -- 1 -- 2 -- 3
Disbursements ( - $)
P = 500
n=3
i = 0.06
We need to identify the various elements of the equation. 吀栀e present sum P is $500. 吀栀e interest rate
per interest period is 6%, and in three years there are three interest periods. 吀栀e 昀甀ture sum F can be
computed 昀爀om Equation 3-4:
F = P(I + i) n = 500(1 + 0.06) 3 = $595.50
where P = $500, i = 0.06, n = 3, and F is unknown.
吀栀us if we deposit $500 now at 6% interest, there will be $595.50 in the account in three years.
78
Cha pter 3 Interest and Equivalence
If we take F = P(l + W and solve 昀漀r P, then
P=F
(1 + i) n
l
= F(l + i)- n
吀栀is is the single-payment present worth 昀漀rmula. 吀栀e equation
P = F(l + i)- n
(3-5)
in our notation becomes
P = F(P/F, i, n)
(3-6)
EXAM PLE 3 · 6
If you wish to have $800 in a savings account at the end of 昀漀ur years, and 5% interest will be paid an­
nually, how much should you put into the savings account now?
FORM U LA SOLUTION
P = unknown
n=4
i = 0.05
F = $800
P = F( l + i) -n = 800(1 + 0.05) -4 = 800(0.8227) = $658. 16
吀栀us to h愀瘀e $800 in the s愀瘀ings account at the end of 昀漀ur years, we must deposit $658.16 now.
TABLE SOLUTION
P = F (PIP, i , n) = $800(PIF, 5 % , 4)
From the compound interest tables,
(P/F, 5%, 4) = 0.8227
P = $800(0.8227) = $658.16
Here the problem has an exact answer. In many situations, however, the answer is
rounded o昀昀 since it can be only as accurate as the input in昀漀rmation on which it is based.
EXAM PLE 3 - 7
Suppose the bank changed its interest policy in Example 3-5 to "6% interest, compounded quarterly."
For this situation, how much money would be in the account at the end of three years, assuming a $500
deposit now?
Single-Payment Compound Interest Formulas
79
SO LUTI O N
First, we must understand the meaning of 6% interest, compounded quarterly. 吀栀ere are two elements:
6% interest: Unless otherwise described, it is customary to assume that the stated interest is 昀漀r a oneyear period. 䤀昀 the stated interest is for any other period, the time f爀愀me must be clearly stated.
Compounded quarterly: 吀栀is indicates there are 昀漀ur interest periods per year; that is, an interest period
is three months long.
We know that the 6% interest is an annual rate because if it were 昀漀r a di昀昀erent period, it would have
been stated. Since we are dealing with 昀漀ur interest periods per year, it 昀漀llows that the interest rate per
interest period is 1.5%. For the 昀甀ll three years duration, there are 12 interest periods. 吀栀us
i = 0.015
n = (4 X 3) = 12
P = $500
n
F = P( l + i) = P(FIP, i, n)
= $500(1 + 0.015)12 = $500(F/P, 1.5%, 12)
= $500(1. 196) = $598
F = unknown
A $500 deposit now would yield $598 in three years.
Nominal and E昀昀ective Interest
You may have noticed that when specifying an interest rate we implicitly mention two time
periods: we pay an interest rate of i per time_periodl, compounded every time_period2.
Usually, both these time periods are the same, but it is possible 昀漀r them to be di昀昀erent.
Where they are both the same, the interest rate we have speci昀椀ed is an e昀昀ective interest
rate. Equation 3-3, the tables in the OUP website, and all the 昀漀rmulas we will encounter in
this and the next chapter are based on the assumption that we are dealing with an e昀昀ective
interest rate.
If the two time periods don't match, what we have is a nominal interest rate. Be昀漀re
we can use any of our 昀漀rmulas or look anything up in the OUP website, we must 昀椀rst
convert the nominal rate to an e昀昀ective rate, as illustrated in Example 3-8.
EXAM PLE 3 -8
Consider a person depositing $100 in a bank that pays 5% interest, compounded semi-annually. How
much would be in the savings account at the end of one year?
SO LUTI O N
We can assume that the given rate of 5% interest is an annual rate. But the compounding period is
not annual but semi-annual. So the two time periods don't match, and the 5% is there昀漀re a nominal
rate, which can't use昀甀lly be looked up in tables or substituted into a 昀漀rmula. To convert it into an
continued
80
Cha pter 3 Interest and Equivalence
e昀昀ective rate, we change the 昀椀rst time period to match the second. A 5% annual rate is the same as a
2.5% semi-annual rate. So the e昀昀ective semi-annual interest rate is 2.5% semi-annually, compounded
semi-annually. So we can now apply Equation 3-3 to obtain
F = P(FIP, 0.025, 2) = P( l + i)2 = 100(1 + 0.025)2
= 105.06
We might also want to know what e昀昀ective annual rate is equivalent to the e昀昀ective semi-annual rate of
2.5%. Call the e昀昀ective annual rate ia . If the two rates are equivalent, this means that they will give us
the same sum of money at the end of the same time period.
So F = P(F/P, i, 2) = P(F/P, j, 1)
So 100(1 + i)2 = 105.06 = 100(1 + ia)
So ia = 5.06%
䌀䐀
You will see in advertising-especially relating to mortgages, time payment, and short­
term loans-the terms APR (annual percentage rate) and EIR (e昀昀ective annual interest
rate). 吀栀is is because, in Canada and the United States, lenders are legally required to tell
the borrower the true cost of borrowing (T COB). Canada uses the EIR, which is also called
the APR.
吀栀e American de昀椀nition of annual percentage rate, or APR, is "Charges i洀瀀osed on a
borrower to obtain a mor츁㨀, expressed on an annualized basis as an interest rate." 吀栀e
APR includes the annual interest rate (AIR), loan fees, and points. 吀栀e intent of American
APR is to disclose the total cost of borrowing; however, since it just multiplies the per­
period rate by the number of periods per year, it is actually what in engineering economy
terminology is de昀椀ned as the nominal rate, and it is incomplete unless the number of
compounding periods per year is also speci昀椀ed.
In Canada, the APR as de昀椀ned in the federal Interest Act as what is generally under­
stood to mean e昀昀ective annual interest rate (EIR, or ia), which considers in its derivation
the number of compounding periods per year.
To avoid con昀甀sion, this book will use only the terms nominal and e昀昀ective, but stu­
dents are cautioned that when reading advertisements or contracts, they should check the
context to see which side of the border the ad is 昀爀om.
In Example 3-8 we saw that $100 le昀琀 in the savings account 昀漀r one year increased
to $105.06, so the interest paid was $5.06. 吀栀e e昀昀ective interest rate per year, ia, is $5.06/
$100.00 = 0.0506 = 5.06%.
Using the method presented in Example 3-8, we can derive a general equation 昀漀r the
e昀昀ective interest rate. If a $1 deposit were made to an account that compounded interest m
times a year and paid a nominal interest rate per year, r, the interest 爀愀te per compounding
sub-period would be rim, and the total in the account at the end of one year would be
Nominal and 昀꬀ective Interest
81
If we deduct the $1 principal sum, the expression would be
吀栀ere昀漀re,
E昀昀ective interest rate per year ia = [ 1 + :
r-
1
(3-7)
where r = nominal interest rate per year
m = number of compounding sub-periods per year
Or, substituting the e昀昀ective interest rate per compounding sub-period, i = (爀椀m),
E昀昀ective interest rate per year ia = ( 1 + i t - 1
(3-8)
where i = e昀昀ective interest rate per compounding sub-period
m = number of compounding sub-periods per year
Either Equation 3-7 or 3-8 may be used to compute an e昀昀ective interest rate per year.
EXAM PLE 3 -9
If a savings bank pays 1.5% interest every three months, what are the e昀昀ective and nominal interest
rates per year?
SO LUTI O N
We are going to assume that the rate of 1.5% is the quarterly rate (since it's rather low to be an annual
rate). So the e昀昀ective quarterly interest rate is 1.5%. Now we can use Equation 3-8 to calculate the
e昀昀ective annual interest rate:
E昀昀ective interest rate per year ia = [ 1 + :
r-
= [ I + o;
r1
I = 0.061
= 6. 1%
To 昀椀nd the nominal interest rate per year, we just calculate r = 4 x 1.5% = 6%
Table 3-2 tabulates the e昀昀ective interest rate 昀漀r a range of compounding 昀爀equencies
and nominal interest rates. Note that when a nominal annual interest rate is compounded
annually, the nominal interest rate equals the e昀昀ective interest rate. Note also that increas­
ing the 昀爀equency of compounding (昀漀r example, 昀爀om monthly to continuously) has only
a small e昀昀ect on the e昀昀ective interest rate. But if the amount of money is large, even small
di昀昀erences in the e昀昀ective interest rate can be signi昀椀cant.
82
Cha pter 3 Interest and Equivalence
Ta ble 3-2 N om i na l and E昀昀ective I nterest
Nominal Interest
Rate per Year
E昀昀ective Interest Rate per Year, ia,
When Nominal Rate Is Compounded {%}
r (%)
Yearly
Semi-Annually
Monthly
Daily
Continuously
1
1.0000
1.0025
1.0046
1.0050
1.0050
2
2.0000
2.0100
2.0184
2.0201
2.0201
3
3.0000
3.0225
3.0416
3.0453
3.0455
4
4.0000
4.0400
4.0742
4.0809
4.0811
5
5.0000
5.0625
5.1162
5.1268
5.1271
6
6.0000
6.0900
6.1678
6.1831
6.1837
8
8.0000
8.1600
8.3000
8.3278
8.3287
10
10.0000
10.2500
10.4713
10.5156
10.5171
15
15.0000
15.5625
16.0755
16.1798
16.1834
25
25.0000
26.5625
28.0732
28.3916
28.4025
EXAM PLE 3-10
A loan shark lends money on the 昀漀llowing terms: "If I give you $50 on Monday, you owe me $60 on the
昀漀llowing Monday."
(a) What nominal interest rate per year (r) is the loan shark charging?
(b) What e昀昀ective interest rate per year (i) is he charging?
(c) If the loan shark started with $50 and was able to keep it, as well as all the money he received,
out in loans at all times, how much money would he h愀瘀e at the end of one year?
SO LUTI O N TO (a)
Let i be the nominal weekly interest rate and n be the number of weeks. 吀栀en
吀栀ere昀漀re, i = 20% per week.
F = P(FIP, i, n)
6 0 = S0(FIP, i, 1)
(ꀀ甀 i, 1) = 1.2
Nominal interest rate per year = 52 weeks X 0 .2 0 = 10 .4 0 = 1 , 0 4 0%
SO LUTI O N TO 戀儀
E昀昀ecti瘀攀 interest rate per year ia = [ 1 + :
rr-
= [ 1 + 1 :;0
1
1
= 1 3, 1 05 - 1
= 1 3, 104 = 1 ,3 10 ,400%
Nominal and 昀꬀ective Interest
SOLUTION TO 挀儀
83
F = P(l + it = 50(1 + 0.20) 52
= $655,200
With a nominal interest rate of 1,040% per year and an e昀昀ective interest rate of 1,310,400% per year, if
he started with $50, the loan shark would have $655,200 at the end of one year.
Continuous Compounding
吀眀o variables we have introduced are
r = nominal interest rate per interest period
m = number of compounding sub-periods per time period
Since the interest period is normally one year, the de昀椀nitions become
r = nominal interest rate per year
m = number of compounding sub-periods per year
℀贀 = interest rate per interest period
mn = number of compounding sub-periods in n years
Single- Payment I nterest Factors: Conti nuous Compounding
吀栀e single-payment compound amount 昀漀rmula (Equation 3-3)
F = P(l + it
m愀礀 be rewritten as
If we increase m, the number of compounding sub-periods per year, without limit, m
becomes very large and approaches in昀椀nity, and rim becomes very small and approaches
zero.
吀栀is is the condition of continuous compounding, that is, where the duration of the
interest period decreases 昀爀om some 昀椀nite duration ℀㄀t to an in昀椀nitely small duration dt,
and the number of interest periods per year becomes in昀椀nite. In this situation of continu­
ous compounding,
℀贀 mn
(3-9)
F = P lim + )
mⴀ⬀漀漀
[1
m
84
Cha pter 3 Interest and Equivalence
An important limit in calculus is
1 x
fun ( 1 + x ) 1
x ⴀ堀 0
= 2.71828 = e
(3-10)
If we set x = rim, then mn may be written as (I!x)(rn). As m becomes in昀椀nite, x
becomes 0. Equation 3-9 becomes
x 1'x ]
F = P [ 洀渀 ( l + )
爀洀
xⴀ堀 0
Equation 3-10 tells us the quantity inside the brackets equals e. So returning to
Equations 3-3 and 3-5, we 昀椀nd that
and
F = P(l + i) n
becomes
F = Pern
(3-1 1)
P = F(l + i)-n
becomes
P = Fe-rn
(3-12)
We see that 昀漀r continuous compounding,
or, as shown earlier,
(1 + i) = er
E昀昀ective interest rate per year ia = er - 1
(3-13)
To 昀椀nd compound amount and present worth 昀漀r continuous compounding and a
single payment, we write
Compound amount F = P(ern) = P[F/P, r, n]
Present worth P = F(e-rn) = F[PIF, r, n]
(3-14)
(3-15)
Square brackets around the 昀愀ctors denote continuous compounding.
EXAM PLE 3 - 11
If you were to deposit $2,000 in a bank that pays 5% nominal interest, compounded continuously, how
much would be in the account at the end of two years?
SO LUTI O N
吀栀e equation 昀漀r continuous compounding 昀漀r a single-payment compound amount is
where r = nominal interest rate = 0.05
n = number of years = 2
F = 2,000e <0 • 0 5> <2> = 2,000(1.1052) = $2,210.40
吀栀ere would be $2,210.40 in the account at the end of two years.
Conti n u o u s Com pou n d i n g
85
EXAM PLE 3-12
A bank o昀昀ers to sell savings certi昀椀cates that will pay the purchaser $5,000 at the end of 10 years but will
pay nothing to the purchaser in the meantime. If interest is computed at 6%, compounded continu­
ously, at what price is the bank selling the certi昀椀cates?
SO LUTI O N
n = 10 years
r = 0.06
F = $5,000
P = Fe-rn
P = 5,000e - 0.06 X 10 = 5,000(0.5488) = $2,744
吀栀ere昀漀re, the bank is selling the $5,000 certi昀椀cates 昀漀r $2,744.
EXAM PLE 3-13
How long will it take 昀漀r money to double at 10% nominal interest, compounded continuously?
2 = l eO . l O n
eO . l O n = 2
F = Pe
or
rn
0.10n = ln2 = 0.693
n = 6.93 years
It will take 6.93 years 昀漀r money to double at 10% nominal interest, compounded continuously.
EXAM PLE 3-14
If the s愀瘀ings bank in Example 3-9 changes its interest policy to 6% interest, compounded continuously,
what are the nominal and the e昀昀ective interest rates?
SO LUTI O N
吀栀e nominal interest rate remains at 6% per year.
E昀昀ective interest rate = er - 1
= e0• 06 - 1 = 0.0618
= 6. 18%
86
Cha pter 3 Interest and Equivalence
Equivalence and Sustainability
Recall that at the beginning of this chapter, we described the Canadian government's plan
昀漀r the storage of used nuclear 昀甀el 昀漀r periods of up to 1,000,000 years.
Consider the possibility of radiation leaking 昀爀om the used-昀甀el repository 5,000 years
in the 昀甀ture, and doing a billion dollars worth of damage. Using the 昀漀rmulas we have
derived in this chapter, what is it worth spending now to avoid this damage in the 昀甀ture?
Suppose that the Canadian government can borrow money at 1 % interest.
We apply Equation 3 -5: it is worth spending a present amount P equivalent to the
worth of the damage that we hope to avert, F = $1,000,000,000, 昀椀ve thousand years in
the 昀甀ture.
So P = $1,000,000,000(1 .01)-5,000
吀栀e result of this calculation is an in昀椀nitesimal 昀爀action of a cent, implying that it's not
worth spending anything now to avoid the repository leaking in the 昀甀ture.
吀栀is answer is mathematically correct, but we have a feeling it cannot be right. In gen­
eral, the 昀漀rmulas and methods we cover in this text give reasonable results when applied
to time spans of less than a century, but can be misleading when applied to longer time
periods. 吀栀is is important, since many of the problems humanity currently 昀愀ces require
us to plan on a time scale of centuries or longer.
S U M MARY
吀栀is chapter describes cash flow tables, the time value of money, and equivalence. 吀栀e
single-payment compound interest 昀漀rmulas were derived. It is essential that these con­
cepts and the use of the interest 昀漀rmulas be 昀甀lly understood, since the rest of this book
and the practice of engineering economy are based on them.
吀椀me value of money 吀栀e continuing o昀昀er of banks to pay interest 昀漀r the temporary use
of other people's money is ample proof that there is a time value of money. 吀栀us we
would always choose to receive $100 today rather than the promise of $100 to be paid
at a 昀甀ture date.
Equivalence What sum would you be willing to accept a year hence instead of $100 today?
Your answer will depend on your personal circumstances. If you have an invention
that you believe will generate a large return on an initial investment, you may be pre­
pared to promise a large 昀甀ture sum in exchange 昀漀r $100 that you can invest right
now. If on the other hand you expect to spend the next year in a maximum-security
prison, there is nothing you can do with $100 today except leave it in the bank to
collect interest that you can spend when you get out. So in the 昀漀rmer case, you set a
high value on having 昀甀nds right now and would be prepared to pay a high interest
rate to get them. In the latter case, you are prepared to accept whatever interest rate
the bank will o昀昀er, however low. Once you have determined the interest rate you'd
be prepared to pay, you can use Equation 3 -3 to convert a cash flow at any moment
in time to a cash 昀氀ow of equivalent value to you at any other moment in time.
S u m m a ry
87
Single- Payment Form u las
Compound amount F = P( l + i)n = P(阀甀 i, n)
Present worth P = F( l + i)-n = F(PIF, i, n)
where i = interest rate per interest period (stated as a decimal)
n = number of interest periods
P = a present sum of money
F = a 昀甀ture sum of money at the end of the nth interest period that is equivalent to
P with interest rate i
吀栀is chapter also de昀椀ned simple interest, where interest does not carry over and become
part of the principal in subsequent periods. Unless otherwise speci昀椀ed, all interest rates in
this text are compound rates.
Single-愀준ment Formulas: Continuous Compounding at Nominal Rate r per Period
Compound amount:
Present worth:
F = P(e rn) = P[FIP, r, n]
P = F(e -rn) = F[PIF, r, n]
Note that the square brackets around the 昀愀ctors are used to distinguish continuous
compounding.
PROBLEMS
Eq u ivalence
3-1
Explain the di昀昀erence between simple and
compound interest. Which is more common?
3-2
A woman borrowed $2,000 and agreed to
repay it at the end of three years, together with
10% simple interest a year. How much will she
pay three years hence?
⠀　3-3
A $5,000 loan was to be repaid with 8% simple
annual interest. A total of $5,350 was paid.
How long had the loan been outstanding?
3-4
At an interest rate of 10% a year, $100,000
today is equivalent to how much a year 昀爀om
now?
3-5
Ten years ago, a company invested $450,000 in
a new technology that is now worth $1,000,000.
What rate of interest did the company earn on
a simple-interest basis?
3-6
Imagine a country in which every citizen has
a million dollars in the bank. 吀栀e bank pays
interest at 5%, so every citizen has an annual
income of $50,000. It is possible to live quite
com昀漀rtably on $50,000 a year without doing
any work. But if no one's doing any work,
where does all the money come 昀爀om?
⠀　3 -7
In your own words explain the time value of
money. From your own life (either now or in a
situation that might occur in the 昀甀ture), give
an example in which the time value of money
would be important.
88
Cha pter 3 Interest and Equivalence
3-8
Which is more valuable, $20,000 received now
or $5,000 a year 昀漀r 昀漀ur years? Why?
3-9
Magdalen, Miriam, and Mary were asked to
consider two di昀昀erent cash 昀氀ows: $500 that
they could receive today or $1,000 that they
would receive three years 昀爀om today. Magdalen wanted the $500 today, Miriam chose to
collect $1,000 in three years, and Mary was in­
di昀昀erent. Can you o昀昀er an explanation 昀漀r the
choice made by each woman?
3 -10
How long will it take 昀漀r an investment to
double at 4% a year compounding annually?
3 -11
Assume that you save 1 cent a day 昀漀r 50 years,
that you deposit it in the bank at the end of
each month, and that there are 30.5 days per
month (i.e., you save 30.5 cents each month).
How much do you have a昀琀er 50 years if
(a) the bank does not pay any interest?
(b) the bank pays 2% per month interest?
can he spend on the computer now and still
have an amount that will grow to be $12,000
when he graduates?
3-16
V
䐀⸀
0
3-17
How long would it take to double your money
if you invested it at 4% simple interest? At 4%
compound interest?
3-18
A savings account earns 8% interest. If $1,000
is invested in the account, how many years is
it until each of the 昀漀llowing amounts is on
deposit?
愀儀 $1,360
(b) $2,720
(c) $4,316
(d) $6,848
3-19
Muhammad can get a Certi昀椀cate of Deposit
(CD) at his bank that will pay 3.7% annually
昀漀r 10 years. If he places $5,530 in this CD, how
much will it be worth when it matures?
Use the 昀漀rmula.
Use the interest tables and interpolation.
3-20
Ace Manu昀愀cturing is building a new 昀愀cility
that will cost $44M. Ace will borrow $40M
昀爀om First National Bank and pay the remainder immediately as a down payment.
Ace will pay 7% interest but will make no
payments 昀漀r 昀漀ur years, at which time the
entire amount will be due. How large will
Ace's payment be?
Single- Payment Factor
3 -12
Solve the diagram below 昀漀r the unknown Q,
assuming a 10% interest rate.
t
t
200
Q
0 = 0 -- 1 -- 2 -- 3 -- 4
3 -13
0
3 -14
3 -15
n=4
A man borrowed $750 昀爀om a bank. He agreed
to repay the sum at the end of three years,
together with the interest at 8% per annum.
How much will he owe the bank at the end of
three years?
We know that a certain piece of equipment will
cost $150,000 in 昀椀ve years. How much must be
deposited today to pay 昀漀r it if the interest rate
is 10%?
Alvin's Aunt Agatha gave him $16,000 to start
college. Alvin wants to have $12,000 available
when he graduates in 昀漀ur years to buy a used
car. He wants to buy a new computer now, and
he earns 3% in his savings account. How much
An inheritance will be $20,000. 吀栀e interest
rate 昀漀r the time value of money is 7%. How
much is the inheritance worth now if it will be
received
(a) in 5 years?
(b) in 10 years?
(c) in 20 years?
搀儀 in 50 years?
0
3-21
How much must you invest now at 7.9% interest to accumulate $175,000 in 63 years?
3-22
(a) If $100 at Time 0 will be worth $110 a year
later and was worth $90 a year ago, compute
the interest rate 昀漀r the past year and the interest rate next year.
Problems
3-23
(b) Assume that $90 invested a year ago will
return $110 a year 昀爀om now. What is the
annual interest rate in this situation?
3-29
T he tabulated 昀愀ctors stop at n = 100. How
can they be used to calculate (PIP, i, 150)?
(PIP, i, 200)?
吀栀e 昀漀llowing series of payments will repay a
present sum of $5,000 at an 8% interest rate.
Using single-payment 昀愀ctors, what present
sum is equivalent to this series of payments at
a 10% interest rate?
3-30
In 1990 Mrs John Hay Whitney sold her
painting by Renoir, Au Moulin de la Galette,
depicting an open-air Parisian dance hall, 昀漀r
$7 1 million. 吀栀e buyer also had to pay the auc­
tion house a commission of 10%, making a
total of $78.1 million. 吀栀e Whitney 昀愀mily had
bought the painting in 1929 昀漀r $165,000.
(a) What rate of return did Mrs Whitney re­
ceive on the investment?
(b) Was the rate of return really as high as
you computed in (a)? Explain.
3 -31
A sum of money Q will be received six years
昀爀om now. At 5% annual interest, the present
worth of Q is $60. At the same interest rate,
what will the value of Q be 10 years 昀爀om now?
Year
End-of-夀攀ar Payment
1
2
3
4
5
$1,400
1,320
1,240
1,160
1,080
3-24
What sum of money now is equivalent to
$8,250 two years later if interest is 4% per six­
month period?
3-25
In approximately how many years will a sum
of money double if it is invested at 2% per sixmonth period (semi-annually)?
儀ⴀ
�
3-26
⠀　3-27
3-28
⠀　89
One thousand dollars is borrowed 昀漀r one
year at an interest rate of 1% a month. If the
same sum of money could be borrowed 昀漀r
the same period at an interest rate of 12% per
annum, how much could be saved in interest
charges?
Sally is buying a car that costs $12,000. She
will pay $2,000 immediately and the remaining $10,000 in 昀漀ur annual end-of-year principal payments of $2,500 each. In addition to
the $2,500, she must pay 15% interest on the
unpaid balance of the loan each year. Draw a
cash 昀氀ow table to represent this situation.
吀栀e local bank o昀昀ers to pay 5% interest on savings deposits. In a nearby town, the bank pays
1.25% per three-month period (quarterl礀⤀. A
man who has $3,000 to put in a savings account
wonders whether the higher interest paid in
the nearby town justi昀椀es making the deposit
there. Assuming he will leave all money in the
account 昀漀r two years, how much additional
interest would he obtain 昀爀om the out-of-town
bank over the local bank?
Nominal and E昀昀ective I nterest Rates
3-32
One thousand dollars is invested 昀漀r seven
months at an interest rate of 1% per month,
compounded monthly. What is the nominal
annual interest rate? What is the e昀昀ective
annual interest rate?
3-33
A 昀椀rm charges its credit customers interest
1.75% a month, compounded monthly. What
is the e昀昀ective annual interest rate?
3-34
If the nominal annual interest rate is 12% compounded quarterly, what is the e昀昀ective annual
interest rate?
3-35
A local store, 昀漀r its charge accounts, charges 1.5% each month on the unpaid balance,
compounded monthly. What nominal annual
interest rate is being charged? What is the effective annual interest rate?
3-36
What annual interest rate, compounded quarterly, is equivalent to a 9.31% e昀昀ective annual
interest rate?
3-37
A bank advertises that it pays 7% annual interest, compounded daily, on savings accounts,
⠀　⠀　⠀　90
Cha pter 3 Interest and Equivalence
provided the money is le昀琀 in the account 昀漀r
昀漀ur years. What e昀昀ective annual interest rate
do they pay?
3 -38
3 -39
At the Central Furniture Company, custom­
ers who buy on credit pay an e昀昀ective annual
interest rate of 16.1%, based on monthly com­
pounding. What is the nominal annual inter­
est rate that they pay?
3 - 44
A student bought a $75 used guitar and agreed
to pay 昀漀r it with a single $85 p愀礀ment at the
end of six months. Assuming semi-annual
(every six months) compounding, what is the
nominal annual interest rate? What is the ef­
fective annual interest rate?
Tra昀케c at a certain intersection is currently
2,000 cars a day. A consultant has told the
city that tra昀케c is expected to grow at a con­
tinuous rate of 5% a year 昀漀r the next 昀漀ur
years. How much tra昀케c will be expected at
the end of two years?
3 - 45
A bank pays 10% nominal annual interest
on special three-year certi昀椀cates. What is
the e昀昀ective annual interest rate if interest is
compounded
(a) every three months?
(b) daily?
(c) continuously?
3 - 46
A department store charges 1.75% interest
per month, compounded continuously, on its
customer's charge accounts. What is the nom­
inal annual interest rate? What is the e昀昀ective
annual interest rate?
3 - 47
If you want a 12% annual rate of return, con­
tinuously compounded, on a project that will
yield $6,000 at the end of two and a half years,
how much must you be willing to invest now?
3 - 48
Bank North advertises, "We pay 6.50%, com­
pounded daily." Bank South says, "We pay
6.50%, compounded continuously." If you de­
posit $10,000 with Bank South 昀漀r one year,
how much additional interest will you receive?
3 - 49
Lisa wishes to tour the country with her 昀爀iends.
To do this, she is saving money 昀漀r a bus.
(a) How much money must Lisa deposit in
a s愀瘀ings account paying 8% nominal
annual interest, compounded continu­
ously, in order to have $8,000 in 昀漀ur and
a half years?
3-40
A bank is o昀昀ering to sell six-month certi昀椀cates
of deposit 昀漀r $9,500. At the end of six months,
the bank will pay $10,000 to the certi昀椀cate
owner. Using a six-month interest period,
compute the nominal annual interest rate and
the e昀昀ective annual interest rate.
3 - 41
吀栀e treasurer of a 昀椀rm noted that many in­
voices were received with the 昀漀llowing terms
of payment: "2%-10 days, net 30 days." 吀栀us,
if he were to pay the bill within 10 days of its
date, he could deduct 2%. On the other hand,
if he did not pay the bill promptly, the 昀甀ll
amount would be due 30 days 昀爀om the date of
the invoice. Assuming a 20-day compound­
ing period, what e昀昀ective annual interest
rate is the 2% deduction 昀漀r prompt payment
equivalent to?
3 -42
Jane made an investment of $10,000 in a s愀瘀­
ings account 10 years ago. 吀栀is account paid
interest of 5.5% 昀漀r the 昀椀rst 昀漀ur years and 6.5%
interest 昀漀r the remaining six years. 吀栀e inter­
est charges were compounded quarterly. How
much is this investment worth now?
Conti nuous Com pou nding
3 - 43
(a) 11.98% interest rate compounded
continuously
(b) 12.00% interest rate compounded daily
(c) 12.01% interest rate compounded
monthly
搀儀 12.02% interest rate compounded
quarterly
(e) 12.03% interest rate compounded yearly
Choose the best of the 昀漀llowing 昀椀ve alterna­
tives 昀漀r an investment. Assume the invest­
ment is 昀漀r 昀漀ur years and P = $10,000.
Problems
(b) A 昀爀iend o昀昀ers to pay Lisa $8,000 in 昀漀ur
and a half years if Lisa gives him $5,000
now. Assuming continuous compound­
ing, what is the nominal annual interest
rate of this o昀昀er?
3-50
⠀　3-51
吀栀e I've Been Moved Corporation (IBM) re­
ceives a constant 昀氀ow of 昀甀nds 昀爀om its world­
wide operations. 吀栀is money (in the 昀漀rm of
cheques) is deposited continuously in many
banks with the goal of earning as much inter­
est as possible 昀漀r IBM. One billion dollars is
deposited each month, and the money earns an
愀瘀erage of 0.5% interest a month, compounded
continuously. Assume all the money remains
in the accounts until the end of the month.
(a) How much interest does IBM earn each
month?
(b) How much interest would IBM earn each
month if it held the cheques and made de­
posits to its bank accounts just 昀漀ur times
a month?
A 昀漀rkli昀琀 truck costs $29,000. A company
agrees to purchase such a truck with the
understanding that it will make a single pay­
ment 昀漀r the balance due in three years. 吀栀e
vendor agrees to the deal and o昀昀ers two dif­
ferent interest schedules. 吀栀e 昀椀rst schedule
uses an annual e昀昀ective interest rate of 13%.
吀栀e second schedule uses 12.75% compounded
continuously.
愀儀 Which schedule should the company
accept?
(b) What would be the size of the single
payment?
3-52
How long will it take 昀漀r $10,000, invested at
5% a year, compounded continuously, to triple
in value?
3-53
A man was le昀琀 $50,000 by his uncle. He has
decided to put it into a savings account 昀漀r
the next year or so. He 昀椀nds there are vary­
ing interest rates at savings institutions:
4.375% compounded annually, 4.250% com­
pounded quarterly, and 4.125% compounded
continuously. He wants to choose the savings
⠀　91
institution that will give him the highest return
on his money. What interest rate should he
choose?
3-54
Ace Zenovia Bank and Trust deposits
$2,567,223 of excess capital in the Bank of
Canada. If the Bank of Canada pays 4% in­
terest compounded daily, how much interest
will Zenovia earn by le愀瘀ing the money on
deposit 昀漀r two years? By how much does the
answer change if continuous compounding is
assumed?
U nclassi昀椀ed
3-55
Mr Sansome withdrew $1,000 昀爀om a savings
account and invested it in common shares. At
the end of 昀椀ve years, he sold the shares and re­
ceived a cheque 昀漀r $1,307. If Mr Sansome had
le昀琀 his $1,000 in the savings account, he would
have received an annual interest rate of 5%,
compounded quarterly. Mr Sansome would
like to compute a comparable interest rate on
his common shares investment. If compound­
ing is quarterly, what nominal annual interest
rate did Mr Sansome receive on his investment
in shares? What e昀昀ective annual interest rate
did he receive?
3-56
吀栀e 昀漀llowing cash 昀氀ows are equivalent in
value if the interest rate is i. Which one is more
valuable if the interest rate is 2i?
⠀　椀儀
⠀椀椀⤀
3-57
⠀　t
0 -- 1 -- 2
t
0 -- 1 -- 2 -- 3
In 1995 an anonymous private collector bought
a painting by Picasso, entitled A最팀l Fernandez
de Soto, 昀漀r $29,152,000. 吀栀e picture depicts
Picasso's 昀爀iend de Soto seated in a Barcelona
cafe drinking absinthe. 吀栀e painting was done
in 1903 and valued then at $600. If the painting
was owned by the same 昀愀mily until its sale in
92
Cha pter 3 Interest and Equivalence
1995, what rate of return did they receive on
the $600 investment?
3-58 Rita borrows $5,000 昀爀om her parents. She
repays them $6,000. What is the interest rate if
she pays the $6,000 at the end of
愀儀 Year 2?
(b) Year 3?
(c) Year 5?
搀儀 Year 10?
3-59
A 昀椀rm has borrowed $5,000,000 昀漀r 昀椀ve years
at 10% a year compound interest. 吀栀e 昀椀rm will
make no payments until the loan is due, when
it will pay o昀昀 the interest and principal in one
lump sum. What is the total payment?
3-60 A manu昀愀cturing company made an invest­
ment 10 years ago that is now worth $1,500,000.
How much was the initial investment
愀儀 at a simple-interest rate of 10% a year?
(b) at an interest rate of 10% a year compounding annually?
3-61
Suppose that $2,000 is deposited in an account
that earns 6% interest, compounded annually.
How much is in the account
愀儀 a昀琀er 5 years?
(b) a昀琀er 10 years?
(c) a昀琀er 20 years?
搀儀 a昀琀er 50 years?
(e) a昀琀er 100 years?
3-62
First Bank is sending university alumni invi­
tations to obtain a credit card, with the name
of their university written on it, 昀漀r a nominal
9.9% interest a year a昀琀er six months of 0%
interest. 吀栀ese interest rates apply to the out­
standing debt if it is not paid by a speci昀椀ed
date each month, and interest is compounded
monthly. If you 昀愀il to make the minimum
payment in any month, your interest rate could
increase (without notice) to a nominal 19.99%
a year. Calculate the e昀昀ective annual interest
rates that First Bank is charging in each case.
3-63
Jack deposited $500,000 in a bank 昀漀r six
months. At the end of that time, he with­
drew the money and received $520,000. If
⠀　the bank paid interest based on continuous
compounding,
(a) What was the e昀昀ective annual
interest rate?
(b) What was the nominal annual
interest rate?
M i n i - Cases
3-64 吀栀e United States recently purchased $1 bil­
lion in 30-year zero-coupon bonds 昀爀om a
struggling 昀漀reign nation. 吀栀e bonds yield
4.5% interest per annum. 吀栀e zero-coupon
bonds pay no interest during their 30-year
life. Instead, at the end of 30 years, the US
government is to receive back its $1 billion
together with interest at 4.5% per annum.
A US senator objected to the purchase, argu­
ing that the correct interest rate 昀漀r bonds
like this is 5.25%. 吀栀e result, he said, was a
multi-million dollar gi昀琀 to the 昀漀reign coun­
try without the approval of Congress. Assum­
ing the senator's math is correct, how much
will the 昀漀reign country have saved in interest
when it repays the bonds at 4.5% instead of
5.25% at the end of 30 years?
3-65
吀栀e Apex Company sold a water so昀琀ener to
Mary Smith. 吀栀e price of the unit was $350.
Mary asked 昀漀r a deferred-payment plan, and a
contract was drawn up. Under the contract, the
buyer could delay paying 昀漀r the water so昀琀ener
if she bought the coarse salt 昀漀r recharging the
so昀琀ener 昀爀om Apex. At the end of two years,
the buyer was to pay 昀漀r the unit in a lump
sum, with interest at an e昀昀ective rate of 1.5%
per quarter-year. According to the contract, if
the customer stopped buying salt 昀爀om Apex
at any time be昀漀re two years, the 昀甀ll payment
due at the end of two years would automatically
become due.
Six months later, Mary decided to buy
salt elsewhere and stopped buying 昀爀om Apex,
whereupon Apex asked 昀漀r the 昀甀ll payment
that was to have been due 18 months hence.
Mary was unhappy about this, so Apex of­
fered as an alternative to accept the $350 with
interest at 10% per semi-annual period 昀漀r
the six months that she had been buying salt
Problems
昀爀om Apex. Which of these alternatives should
Mary accept? Explain.
3-66 吀栀e local garbage company charges $6 a
month 昀漀r garbage collection. It had been
the company's practice to send bills to its
100,000 customers at the end of each two­
month period. 吀栀us, at the end of February
it would send a bill to each customer 昀漀r
$12 昀漀r garbage collection during January
and February.
Recently, the 昀椀rm changed its billing date:
it now sends out the two-month bills a昀琀er one
93
month's service has been per昀漀rmed. Bills 昀漀r
January and February, 昀漀r example, are sent at
the end of January. 吀栀e local newspaper points
out that the 昀椀rm is receiving half its money
be昀漀re the garbage collection. 吀栀is unearned
money, the newspaper says, could be invested
temporarily 昀漀r one month at 1% per month
interest by the garbage company to earn extra
income.
Compute how much extra income the
garbage company could earn each year if
it invested the money as described by the
newspaper.
Eq uivalence for Repeated
Cash Flows
Mrs Dashwood's Annuity
At the beginning of Jane Austen's 1811 novel, Sense and Sensibili琀礀, Mr John Dashwood and his
avaricious wife, Fanny, are debating how he can most cheaply discharge his obligations toward his
half-sister, the recently widowed Mrs Henry Dashwood, and her three daughters, Elinor, Marianne,
and Margaret. He first considers giving her a lump sum of £1,500; then, since £1,500 seems a lot to
part with in one lump, he considers paying her an annual sum of £100-an annui琀礀- for as long as
she lives. But, objects Fanny, although Mrs Dashwood is old, such an arrangement will encourage
her to cling to life for an unreasonable time; and, should she survive for more than 15 years, her
half-brother will lose money by the arrangement. (Mrs Dashwood is actually only 40, but Jane
Austen evidently considers this to be near senility; Mrs Dashwood herself says at one point in the
novel that she can scarcely expect to survive another 15 years.)
It is clear from this discussion that Fanny Dashwood cannot have studied engineering
economics; for in fact, Mr Dashwood would save money by the latter alternative, even if Mrs
Henry Dashwood were to live considerably longer than 15 years. Fanny Dashwood's error lies in
temm uzcan/iStock Photo
M rs Das hwood 's Annu ity
95
co m p a r i n g fut u re and p resent s u ms of money. In this cha pter, we will i ntrod u ce factors for fi n d i n g
the p rese nt a n d fut u re equ iva le nts t o a n annuity; t h a t i s , t o a series o f eq u a l payments occ u r r i n g
at reg u la r i nte rva ls.
To resolve Mr Dashwood 's p roblem, note that if he d e posits £1,500 in an acco u n t that pays 10%
i nterest, then in a yea r's t i m e, he ca n with d raw the £150 of accrued i nterest to g ive to h i s ha lf-s ister,
leavi n g £1,500 i n the accou nt. This p rocess ca n be re peated a n n u a lly, eve n if h i s ha lf-s ister we re to
su rvive i nto the twe nty-fi rst centu ry.
Engi neers often encou nter scena rios analogous to Mr Dashwood's, that is, scena rios in which we
expect either to face a u n iform series of expenses or to receive a u n iform series of payments. Some­
ti mes we will know the n u m ber of cash flows i n the series; sometimes, as i n M r Dashwood's case, the
series is of indetermi nate length. The techniq ues we will lea rn i n this cha pter will a llow us to analyze
each of these cases, either by movi ng the cash flows to their eq u iva lent a mount on a pa rticular present
or futu re date, or by converting all other present and future cash flows to equ iva lent u n iform series.
QUESTIONS TO CONSIDER
1.
The solution to M r Das hwood's problem, sugg ested a bove, i m plies that a n i n itial i nvestment of £1,500
ca n generate a n in昀椀nite series of a n n u a l payments of £100. Such a conclusion wou ld seem very stra nge
i n other fi elds of e n g i n eeri n g - for exa m p le, if a n i nventor cla i med to have d eveloped a battery that wo u ld
s u p ply cu rrent fo reve r afte r a n i n itial c h a rg i n g, we wou ld suspect a violati o n of the Fi rst or Seco nd Laws of
Thermodyn a m i cs . Why do these laws not a p p ly to e n g i neeri ng eco n o m i cs?
2.
The p resent va lue of an i nfi n ite series of cash flows is
a) i nfi n ite, u n less the i n terest rate is zero
bl i n fi n ite, if the i nterest rate is zero
c) i nfi n ite, if the i nterest rate is i nfi n ite
d) i nfi n ite, u n less the i nterest rate i s i nfi n ite
3.
Ass u m e that Mr Das hwood sta rted payi n g h i s ha lf-s ister her £100/yea r a n n u ity in 1816, a n d ass u m e that, by
fru g a l livi ng, she d e posits £50/yea r in her own ba n k acco u nt at 5% i nterest. H ow m u c h does she have in her
acco u n t i n 2016, assu m i n g she is still a live?
LEARNING OBJECTIVES
This chapter will help you
•
co nve rt u n iform series of cash flows to thei r equ iva lent p resent or fut u re va lues
•
use a rith metic a n d geometric g ra d i e nts to solve correctly mode lled problems
•
use conti n u o u sly co m po u nded i n terest with uniform payment series
KEY TERMS
annu ity
a rithmetic g radient series
geometric grad ient series
u n iform payment series
96
Cha pter 4 Equivalence for Repeated Cash Flows
Chapter 3 presented the 昀甀ndamental components of engineering economic analysis,
including 昀漀rmulas 昀漀r computing equivalent single sums of money at di昀昀erent points in
time. Most problems we will encounter are much more complex. 吀栀is chapter develops
昀漀rmulas 昀漀r cash flows that are uni昀漀rm series or that increase on an arithmetic or geo­
metric gradient.
U niform Series Compou nd I nterest Formulas
Car loans, house payments, and many other loans are based on a uni昀漀rm p愀礀ment series.
It will o昀琀en be convenient to convert such series to a single payment of equivalent value.
In what 昀漀llows, we will use the symbol A to represent an end-of-period cash receipt or
disbursement in a uni昀漀rm series, continuing 昀漀r n periods.
It is customary to de昀椀ne A as an end-of-period event rather than a beginning-of-period
or middle-of-period event. 吀栀e derivations that 昀漀llow are based on this end-of-period
assumption. One could, of course, derive other equations based on beginning-of-period or
mid-period assumptions, but it is rarely done.
吀栀e horizontal line in Figure 4-1 represents 昀漀ur interest periods. Uni昀漀rm p愀礀ments
A occur at the end of each interest period, so there are as many A's as there are interest
periods n. Figure 4-1 uses 1 January through 31 December, but other one-year periods
could be used.
In the section in Chapter 3 on single-payment 昀漀rmulas, we saw that, given an interest
rate i, a sum P at one point in time is equivalent to a sum F n periods later, according to
the equation
F = P(FIP, i, n) = P (I + i) n
We will now develop a similar 昀愀ctor, (䤀蜀, i, n), that will give us the 昀甀ture amount F
equivalent to a series of n cash flows of size A, and its inverse, (AIF, i, n). We will also 昀椀nd
the corresponding 昀愀ctors, (PIA, i, n) and (AIP, i, n), 昀漀r converting a uni昀漀rm series to its
equivalent present value.
Looking at Figure 4-1, we see that if an amount A is invested at the end of each year
昀漀r 昀漀ur years, the total amount F at the end of 昀漀ur years will be the sum of the compound
amounts of the individual investments.
F
A
t
A
A
t
t
t
夀攀ar o -- 1 -- 2 -- 3 -- 4
-
g
o⸀쨀 _
_
∀䤀
⸀ꔀ
n=4
FIGURE 4·1A The general relationship between A and F.
U n iform Series Co m po u n d I nterest Form u las
97
t
t
+ + + +
+
+
+ t
+
O - l -2-3-4 = 0 - 1 -2-3-4 + 0 - l -2-3-4 + 0 - l -2-3-4 + 0 - l -2-3-4
A
A
A
A
A
A ( l + i )3
+
A
A
A
A ( l + i )2
+
A(l + i )
+
A
0 - 1 -2-3-4
FIGURE 4-1B The general relationsh ip between A and F.
In the general case 昀漀r n years,
Multiplying Equation 4-1 by (1 + i), we have
(1 + i)F = A (1 + ir + • • • + A (1 + i)4
+ A ( l + i)3 + A ( l + i)2 + A (l + i)
(4-2)
Factoring out A and subtracting Equation 4-1 gives
( 1 + i)F = A( l + i) n + • • • + ( 1 + i) 4 + (1 + i) 3 + ( 1 + i) 2 + ( 1 + i)]
-F = A( l + i) n -i + • • • + ( 1 + i) 3 + ( 1 + i) 2 + (1 + i) + 1]
iF = A[(l +
i) n
- 1]
(4-3)
Solving Equation 4-3 昀漀r F gives
(4-4)
吀栀us we have an equation 昀漀r F when A is known. 吀栀e term inside the brackets
( l + ir - 1
i
is called the un椀昀orm series co洀瀀ound amountfactor and is represented by (䤀蜀, i, n).
EXAM PLE 4-1
You deposit $500 in a credit union at the end of each year 昀漀r 昀椀ve years. 吀栀e credit union pays 5%
interest, compounded annually. How much do you have in your account immediately a昀琀er the 昀椀昀琀h
deposit?
continued
98
Cha pter 4 Equivalence for Repeated Cash Flows
SO LUTI O N
吀栀e diagram on the le昀琀 shows the situation 昀爀om your point of view; the one on the right, 昀爀om the
credit union's point of view. Either way, the diagram of the 昀椀ve deposits and the desired computation of
the 昀甀ture sum F matches the situation 昀漀r the uni昀漀rm series compound amount 昀漀rmula:
F
t
t
t
t
t
0 -- 1 -- 2 -- 3 -- 4 -- 5
A
A
A
A
A
• • • • •
A
A
A
A
A
0 -- 1 -- 2 -- 3 -- 4 -- s
F
where A = $500, n = 5, i = 0.05, F = unknown. Filling in the known variables gives
F = $500(FIA, 5%, 5) = $500(5.526) = $2,763
吀栀ere will be $2,763 in the account a昀琀er the 昀椀昀琀h deposit.
If Equation 4-4 is solved 昀漀r A, we h愀瘀e
A=F
i
( I + ir - 1
= F(AIF, i, n)
l
(4-5)
where
is called the un椀昀orm series sinki渀最fund1 昀愀ctor and is written as (AIF, i, n).
A sinkingfund is a separate 昀甀nd into which one makes a uni昀漀rm series of money deposits (A) with the
goal of accumulating some desired 昀甀ture sum (F) by the end of period n.
1
U n iform Series Co m po u n d I nterest Form u las
99
EXAM PLE 4 - 2
Jim wants to buy some electronic equipment 昀漀r $1,000. He decided to save a uni昀漀rm amount at the end
of each month so that he would have the required $1,000 at the end of one year. 吀栀e local credit union
pays 6% interest, compounded monthly. How much would Jim have to deposit each month?
SO LUTI O N
F = $1,000
n = 12
i = 0.5%
A = unknown
A = 1,000(A/F, 0.5%, 12) = 1,000(0.081 1) = $81 . 10
Jim would have to deposit $81.10 each month.
If we use the sinking 昀甀nd 昀漀rmula (Equation 4-5) and substitute 昀漀r F the single-payment
compound amount 昀漀rmula (Equation 3-3), we obtain
A = FI
A=P
i
( 1 + i )" - 1
l
= P(l + i) "
i
( 1 + i )" - 1
i ( l + i) "
] = P(AIP, i, n)
{ l + i )" - 1
(4-6)
We now have an equation 昀漀r determining the value of a series of end-of-period payments,
or disbursements, A, when the present sum P is known.
吀栀e portion inside the brackets
i ( l + i )"
{ l + i )" - 1
is called the un椀昀orm series capital recovery factor and is represented by (A!P, i, n).
吀栀e capital recovery factor is used to 昀椀nd the size of the annual return, A, needed
昀漀r the investor to recover the capital, P, invested at Time 0. 吀栀is is illustrated in
Example 4-3.
EXAM PLE 4- 3
An energy-e昀케cient machine costs $5,000 and has a life of 昀椀ve years. If the interest rate is 8%, how much
will it h愀瘀e to save every year in order 昀漀r the initial capital amount to be recovered?
continued
100
Cha pter 4 Equivalence for Repeated Cash Flows
SO LUTI O N
A
A
A
A
A
+ +
+
+
+
0 -- 1 -- 2 -- 3 -- 4 -- 5
n=5
A = unknown
i = 8%
P = $5,000
A = P(AIP, 8%, 5) = 5,000(0.2505) = $1,252
吀栀e required annual saving to recover the capital is $1,252.
In Example 4-3, with interest at 8%, a present sum of $5,000 is equivalent to 昀椀ve
equal end-of-period disbursements of $1,252. 吀栀is is another way of stating Plan 3 of
Table 3-1.
If the capital recovery 昀漀rmula (Equation 4-6) is solved 昀漀r the present sum P, we
obtain the un椀昀orm series present worth formula
i ( l + i )" - 1
P = A ---- = A(PIA, i, n)
( I + i )"
and
(PIA, i, n) =
(4-7)
i ( I + i) " - I
i ( I + i )"
EXAM PLE 4-4
An investor holds a time-payment purchase contract on some machine tools. 吀栀e contract calls 昀漀r the
p愀礀ment of $140 at the end of each month 昀漀r a 昀椀ve-year period. 吀栀e 昀椀rst payment is due in one month.
He o昀昀ers to sell you the contract 昀漀r $6,800 cash today. If you can otherwise make 1% per month on
your money, would you accept or reject the investor's o昀昀er?
U n iform Series Co m po u n d I nterest Form u las
101
SO LUTI O N
Summarizing the data in a cash 昀氀ow diagram, we h愀瘀e
p
Use the uni昀漀rm series present worth 昀漀rmula to compute the present worth.
P = A(P/A, i, n) = 140(P/A, 1%, 60)
= 140(44.955)
= $6,29 3.70
If you are paying more than $6,293.70, you will receive less than the required 1% per month interest.
Reject the investor's o昀昀er.
EXAM PLE 4-S
Suppose you decided to p愀礀 the $6,800 昀漀r the time-payment purchase contract in Example 4-4. What
monthly rate of return would you obtain on your investment?
SO LUTI O N
In this situation, we know P, A, and n , but we do not know i. 吀栀e problem may be solved by using either
the uni昀漀rm series present worth 昀漀rmula
P = A(℀鄀, i, n)
or the uni昀漀rm series capital recovery 昀漀rmula
A = P(A!P, i, n)
Either way, we have one equation with one unknown.
P = $6,800
i = unknown
A = $140
n = 60
P = A(℀鄀, i, n)
$6,800 = $140(P/A, i, 60)
6, SOO
( PIA, i, 60 ) =
= 48.571
140
continued
102
Cha pter 4 Equivalence for Repeated Cash Flows
We could solve Equation 4-7 algebraically 昀漀r the interest rate i, but it is more convenient to look through
several compound interest tables to 昀椀nd the values of (䤀鄀, i, 60) closest to 48.57 1 and then compute i by
interpolation. Entering values 昀爀om the tables, we 昀椀nd
Interest Rate
O %
b f� ,:
0.75 %
�1
(PIA, i, 60)
5 1 .726
48.57 1 贀꤀
48. 1 74
c
d
吀栀e rate of return, which is between 0.5% and 0.75%, may be computed by a linear interpolation. But the
interest 昀漀rmulas are not linear, so a linear interpolation will not give an exact solution. To minimize
the error, the interpolation should be computed with interest rates as close to the correct answer as
possible. Since alb = c!d, a = b (c!d),
Rate of return i = 0.5% + a
= 0.5% + b(挀℀d)
= 0.50% + 0.25% [ 5 1 .726 - 48.571
l
5 1 .726 - 48. 174
= 0.50% + 0.25% [ 3 • 155 ) = 0.50% + 0.22%
3.552
= 0.72% per mon琀栀
吀栀e monthly rate of return on your investment would be 0.72% a month.
Cash Flows That Do Not Match Basic Patterns
EXAM PLE 4 -6
Using a 15% interest rate, compute the value of F in the 昀漀llowing cash 昀氀ow:
Year
Cash Flow
1
+ 1 00
+ 1 00
+ 1 00
0
3
4
5
-F
100
100
1 00
t t t
!
0-1-2-3-4-5
F
SO LUTI O N
We see that the cash 昀氀ow diagram does not match the sinking 昀甀nd 昀愀ctor diagram. F occurs two per­
iods later, rather than at the same time as the last A. Since the diagrams do not match, the problem is
more di昀케cult than those we've discussed so 昀愀r. 吀栀ink of this as a game in which the objective is to
U n iform Series Co m po u n d I nterest Form u las
103
move the amounts A to the position F, using only the conversion 昀愀ctors we have introduced so 昀愀r. 吀栀e
most elegant solution is the one that uses the fewest conversions.
One rather inelegant solution is to use three (PIP) 昀愀ctors to move each of the individual
$100 payments to the position of 䘀Ⰰ then add them up.
(PIP, 0. 15, 4)
�
�
l l I
$ 100
$ 100
$ 100
!
0 -- 1 -- 2 -- 3 -- 4 -- 5
F
F = F1 + F2 + F3 = lO0(F/P, 15%, 4) + lO0(F/P, 15%, 3) + lO0(F/P, 15%, 2)
= 100(1.749) + 100(1 .521) + 100(1 .322)
= $459.20
吀栀e value of F in the illustrated cash flow is $459.20.
ALTERN䄀吀IVE SOLUTION
(䤀蜀, 0. 1 5, 3 )
�
l l l
!
$ 100
$ 100
$ 100
!
0 -- 1-- 2 -- 3 -- 4 -- 5
(PI倀Ⰰ 0. 15, 2 )
continued
104
Cha pter 4 Equivalence for Repeated Cash Flows
A more elegant solution is to use the (䤀蜀) 昀愀ctor introduced in this chapter. 吀栀is takes the series of
p愀礀ments A to the equivalent sum F1 , and we can then move F1 to F using a single (PIP) 昀愀ctor. We 昀椀rst
solve 昀漀r F1 .
FI = IO0(FIA, 15%, 3) = 100(3.472) = $347.20
And then
F = F1 (阀甀 15%, 2)
= 347.20(1.322)
= $459.00
吀栀e slightly di昀昀erent value 昀爀om the preceding computation is due to rounding in the compound
interest tables.
EXAM PLE 4-7
Consider the 昀漀llowing situation, where P is deposited in a savings account and three later withdrawals
are made, leaving a 昀椀nal balance of zero:
l l l
$20
$20
$20
0 -- 1 -- 2 -- 3 -- 4
吀栀e diagram is not in a standard 昀漀rm, so we again need a multiple-step solution. 吀栀ere are many dif­
ferent ways of computing the answer; we present three of them.
SO LUTI O N 1
We move each withdrawal separately to the present and equate their sum to the initial deposit P:
P = 20(PIF, 15%, 2) + 20(阀甀 15%, 3) + 20(阀甀 15%, 4)
= 20(0.7561) + 20(0.6575) + 20(0.5718)
= $39.71
U n iform Series Co m po u n d I nterest Form u las
105
(PIP, 0. 15, 4)
�
�
I
$20
I
$20
1
$20
0 -- 1 -- 2 -- 3 -- 4
p
SO LUTI O N 2
吀栀e second approach moves the two withdrawals in Years 2 and 3 to their equivalent values in Year 4,
and adds them to the $20 withdrawal already there, giving us a total 昀甀ture amount F at the end of
Period 4:
I
$20
(PIP, 0. 15, 1)
I
$20
1
$20
0 -- 1 -- 2 -- 3 -- 4
p
(䤀ꀀ, 0. 1 5, 4 )
F = 20(FIP, 15%, 2) + 20(FIP, 15%, 1) + 20
吀栀e present equivalent of this amount is P = F(阀甀 15%, 4). We combine these two equations to get
P = [20(阀甀 15%, 2) + 20(阀甀 15%, 1) + 20] (P/F, 15%, 4)
= 20 [(阀甀 15%, 2) + (ꀀ甀 15%, 1) + l ] (P/F, 15%, 4)
= 20 [(1.322) + (1.150) + 11 (0.5718)
= $39.71
continued
106
Cha pter 4 Equivalence for Repeated Cash Flows
SO LUTI O N 3
吀栀e third approach takes advantage of one of the new 昀愀ctors introduced in this chapter to convert the
series of withdrawals to an equivalent sum F at Time 4:
(䤀蜀, 0. 15, 3 )
l l I
$20
$20
$20
0 -- 1 -- 2 -- 3 -- 4
(PIP, 0. 15, 4 )
F = 20(℀蜀, 15%, 3)
We then move F to its present equivalent P using the (ꀀ甀 0.15, 4) 昀愀ctor:
P = 20(F!A, 15%, 3)(P/F, 15%, 4)
= 20(3.472)(0.5718)
= $39.71
Using the (䤀蜀) 昀愀ctor has saved us some calculation; you can imagine how much more it would save if
we had to evaluate a longer series of uni昀漀rm cash flows.
Relationships between Compound Interest Factors
吀栀e reader will no doubt h愀瘀e noticed that there are simple algebraic relationships between
the 昀愀ctors we have developed so 昀愀r. For example,
(FIP, i, n) =
(AIP, i, n) =
(FIA, i, n) =
1
( PIP, i, n)
l
(PIA, i, n)
1
(AIF, i, n)
(4-8)
(4-9)
(4-10)
Relationships between Compound Interest Factors
A less obvious relationship is that the uni昀漀rm series capital recovery 昀愀ctor equals the
uni昀漀rm series sinking 昀甀nd 昀愀ctor plus i:
(A!P, i, n) = (A!F, i, n) + i
(4-1 1)
The Interest Rate Viewed as Fog
㸀Ⰰ
⸀挀
愀⸀
伀氀
0
⸀挀
氀焀
嘀氀
∀伀
稀鐀
�0
䤀攀
䤀攀
䌀⤀
We saw in Chapter 3 that the present value of a 昀甀ture cash 昀氀ow decreases as the interest
rate increases. By examining the value of the (PIA, i, n) 昀愀ctor as i changes, you should
convince yourself that the same is true of a regular series of 昀甀ture cash 昀氀ows. 吀栀is sug­
gests the metaphor of the interest rate as 昀漀g. 吀栀ink of the series of cash flows extending
into the 昀甀ture as an avenue of trees, and think of the interest rate as the thickness of
the 昀漀g. An interest rate of zero corresponds to a perfectly clear day: the trees 昀愀r down
the road appear just as distinct as those close to us. Under these conditions, the gain or
loss of a dollar in the distant 昀甀ture is exactly as important to us as the gain or loss of
a dollar right now. 吀栀e value of (ꀀ甀 0, n) is 1 昀漀r any value of n, and the value of (PIA,
0, n) is n.
As the interest rate increases, the 昀漀g thickens: the trees 昀愀r down the road can now
be seen only indistinctly. Under these circumstances, the importance of 昀甀ture cash 昀氀ows
is reduced; cash 昀氀ows occurring in the distant 昀甀ture will have little e昀昀ect on present
decisions.
As the interest rate approaches in昀椀nity, the 昀漀g becomes a pea-souper: we can see only
the tree immediately adjacent to us, and the rest of the avenue is invisible. (PI倀Ⰰ 漀漀, n) and
(䤀鄀, 漀漀, n) are both zero, whatever the value of n.
What kind of social and economic circumstances might give us an interest rate close
to zero? Under what circumstances might the interest rate become very high?
107
108
Cha pter 4 Equivalence for Repeated Cash Flows
Arithmetic Gradient
It 昀爀equently happens that the cash 昀氀ow series is not a constant amount A. Instead, there is
a uni昀漀rmly increasing series:
A + 4G
A
t
0 -- 1 -- 2 -- 3 -- 4-- s
i
p
Cash 昀氀ows of this 昀漀rm may be resolved into two components:
r
A + 3G
A
l
t
A+G
t
A + 2G
1
A + 4G
t I
3G
A
!
A
A
t t
A
A
t
t
t
l
x
o
2G
I
4G
O --l -- 2 -- 3 -- 4-- S = O --l -- 2 -- 3 -- 4-- S + 0 -- 1 -- 2 -- 3 -- 4-- 5
p
Note that if the problem is solved in this manner, the 昀椀rst cash 昀氀ow in the arithmetic
gradient series becomes zero. 吀栀is is done so that G is the change 昀爀om period to period,
and because the gradient ( G) series is normally used along with a uni昀漀rm series (A).
We already have an equation 昀漀r P ', and we need to derive an equation 昀漀r P ". In this way,
we will be able to write
P = P ' + P "= A (䤀鄀, i, n) + G(䤀需, i, n)
Pause here and look at the tables 昀漀r n = 1. What is the value of (PIG, i, 1) 昀漀r any i ?
Derivation of Arithmetic G radient Factors
吀栀e arithmetic gradient is a series of increasing cash flows as 昀漀llows:
(n - l ) G
I
(n - 2)G
/
t
2G
G
0
⸀尀
t
o -- 1 -- 2 -- 3 꼀똀 n - 1 -- n
Note: Time = 0 here
p
Arithmetic Gradient
吀栀e arithmetic gradient series may be thought of as a series of individual cash flows:
(n - 2) G
2G
(n - l ) G
G
�
0 - 1 - 2 - 3 吀帀 n - l- n + 0 - 1 - 2 - 3 吀帀 n - I - n + • · · + O - I - 2 - 3吀帀 n - I - n + O - I - 2 - 3吀帀 n - I - n
!.
l
!.
吀栀e value of F 昀漀r the sum of the cash 昀氀ows = F1 + F2 + · · · + Fn -i + Fn, or
F = G ( l + i ) n -2 + 2G ( l + i ) n -3 + · · · + (n - 2)(G)( l + i ) l + (n - l )G
Multiply Equation 4-12 by (1 + i) and 昀愀ctor out G, or
(4-12)
(1 + i )F = G [( l + n -l + 2(1 + n -2 + . . . + (n - 2)(1 + i )2 + (n - 1) (1 + i ) 1 ] (4-13)
n
n
Rewrite Equation 4-12 to show other terms in the series:
F = G [( l + ; r -2 + . . . + (n - 3)(1 + i )2 + (n - 2)(1 + i )1 + n - 1]
(4-14)
Subtracting Equation 4-14 昀爀om Equation 4-13, we obtain
F + iF - F = G [( l + i ) n -1 + ( l + i ) n -2 + • · · + (1 + i )2 + (1 + i ) 1 + 1] - nG (4-15)
In the derivation of Equation 4-4, the terms inside the brackets of Equation 4-15 were
shown to equal the uni昀漀rm series compound amount 昀愀ctor:
吀栀us Equation 4-15 becomes
iF = G [
(1 + i - 1
t
Rearranging and solving 昀漀r F, we write
F = �[
I-
nG
(1 + i - 1
-n
t
I
Multiplying Equation 4-16 by the single-payment present worth 昀愀ctor gives
(4-16)
109
110
Cha pter 4 Equivalence for Repeated Cash Flows
(P/G, i, n) = I
(I + ir - in - I
i
i 2 (1 + i) n
(4-17)
Equation 4-17 is the arithmetic gradientpresent worthfactor. Multiplying Equation 4-17
by the sinking 昀甀nd 昀愀ctor, we have
p=
GI
(I + i) n - I _
i
i
(A/G, i, n) = I
n]
I
I
( I + i) n - I
l GI
=
[! -
(I + i) n - in - 1
1=
i
i( I + i) n - i
(I + i) n - in - 1
1
i( I + i) n - i
n
I
(I + i) n - i
(4-18)
Equation 4-18 is the arithmetic g爀愀dient un椀昀orm series 昀愀ctor.
EXAM PLE 4-8
A woman has bought a new car. She wishes to set aside enough money in a bank account to pay the
maintenance on the car 昀漀r the 昀椀rst 昀椀ve years. It has been estimated that the maintenance cost of a car
is as 昀漀llows:
Year
Maintenance Cost
1
2
$120
150
180
210
240
4
5
Assume the maintenance costs occur at the end of each year and that the bank pays 5% interest. How
much should the car owner deposit in the bank now?
SO LUTI O N
210
240
0 -- 1 -- 2 -- 3 -- 4 -- s
Arithmetic Gradient
111
吀栀e cash 昀氀ow may be broken into its two components:
240
2 10
ll
180
150
IIIII
120
120
120
120
1 20
r1
120
60
30
0
!
0 - 1- 2 - 3- 4- 5 = 0 - 1- 2 - 3- 4- 5 + 0 - 1- 2 - 3- 4- 5
j
l
p
A = 1 20
l
G = 30
Both components represent cash 昀氀ows 昀漀r which compound interest 昀愀ctors have been derived. 吀栀e 昀椀rst
is a uni昀漀rm series present worth, and the second is an arithmetic gradient series present worth:
P = A(P!A, 5%, 5) + G(P/G, 5%, 5)
Note that the value of n in the gradient 昀愀ctor is 5, not 4. When we derive the gradient 昀愀ctor, the cash
昀氀ow in the 昀椀rst period is zero 昀漀llowed by (n - 1) terms containing G. Here there are 昀漀ur terms con­
taining G, and it is a 昀椀ve-period gradient.
吀栀us (n - 1) = 4, so n = 5.
P = 120(䤀鄀, 5%, 5) + 30(P/G, 5%, 5)
= 120(4.329) + 30(8.237)
= 519 + 247
= $766
She should deposit $766 in the bank now.
EXAM PLE 4-9
On a certain piece of machinery, it is estimated that the maintenance expense will be as 昀漀llows:
Year
1
2
3
4
Maintenance Cost
$ 1 00
200
300
400
continued
112
Cha pter 4 Equivalence for Repeated Cash Flows
What is the equivalent uni昀漀rm annual maintenance cost 昀漀r the machinery if 6% interest is used?
SO LUTI O N
rl
400
t
200
1 00
0 -- 1 -- 2 -- 3 -- 4
吀栀e 昀椀rst cash flow in the arithmetic gradient series is zero; hence the diagram is not in the proper
昀漀rm 昀漀r the arithmetic gradient equation. As in Example 4-8, the cash 昀氀ow must be resolved into two
components:
l l
400
t
200
100
t
300
0 --1-- 2 -- 3 -- 4
t
300
200
1 00
t
1 00
t
100
t
100
t+
0 --1-- 2 --3-- 4
1 00
0
0 --1-- 2 -- 3 --4
A = 100 + lO0(A/G, 6%, 4) = 100 + 100(1 .427) = $242.70
吀栀e equivalent uni昀漀rm annual maintenance cost is $242.70.
EXAM PLE 4-10
A textile mill in India installed a number of new looms. It is expected that initial maintenance costs and
expenses 昀漀r repairs will be high but will then decline 昀漀r several years:
Year
1
2
3
4
Maintenance and
Repair Costs (rupees)
24,000
1 8,000
1 2,000
6000
What is the projected equivalent annual maintenance and repair cost if interest is 10%?
Arithmetic Gradient
SO LUTI O N
24,000
I
'T
A'
A'
A'
I I I I
12,000
6000
0 -- 1 -- 2 -- 3 -- 4
A'
113
0 -- 1 -- 2 -- 3 -- 4
吀栀e projected cash 昀氀ow is still a cash 昀氀ow (24,000) that de昀椀nes the uni昀漀rm series. However, now the
gradient or change each year is -6,000.
I
24,000
IIII
A = 24,000
18, OOO
12,000
6000
l
0 --1 --2 -- 3 -- 4
= 0 --1 --2 -- 3 -- 4 + 0 -- 1 -- 2 -- 3 -- 4
0
t
6000
! l
12,000
18,000
A = 24,000 - 6,000(A/G, 10%, 4)
= 24,000 - 6,000(1.381)
= 15,714 rupees
吀栀e projected equivalent uni昀漀rm maintenance and repair cost is 15,7 14 rupees per year.
EXAM PLE 4-11
A car's warranty is three years. Upon expiration, annual maintenance starts at $150 and then climbs by
$25 a year until the car is sold at the end of Year 7. Use a 10% interest rate and 昀椀nd the present worth of
these expenses.
11
o -- 1 -- 2 -- 3 -- 4 -- 5 -- 6 -- 7
1 75
200
225
continued
114
Cha pter 4 Equivalence for Repeated Cash Flows
SO LUTI O N
With the uni昀漀rm series and arithmetic gradient series present worth 昀愀ctor, we can compute a present
sum P'.
l
l
0 - - - - - + - - - - - 2 - - - - - - 3 -- 4 -- 5 -- 6 -- 7
p
P'
J
o
l
1 75
200
225
It is important that you examine the location of P ' closely. Because of the way the 昀愀ctor was derived,
there will be one zero value in the gradient series to the right of P '. (If this seems strange or incorrect,
review the beginning of this section on arithmetic gradients.)
P' = A (䤀鄀, i, n) + G (PIG, i, n)
= l 50(PIA, 10% , 4) + 25(PIG, 10% , 4)
= 150(3.170) + 25(4.378) = 475.50 + 109.45 = 584.95
吀栀en
P = P'(PIF, 10% , 3) = 584.95(0.7513) = $439.47
Reality and the Assumed Uniformity of a, G, and g
吀栀e reality of engineering projects is that the annual revenues 昀爀om selling a new product
or the annual bene昀椀ts 昀爀om using a new highway change each year as demand and tra昀케c
levels change. Most annual cash 昀氀ows are not really uni昀漀rm. However, 昀漀recasting mar­
kets and use is di昀케cult, sometimes costly and, in any case, beyond the scope of this text.
And, given limited in昀漀rmation, the most common 昀漀recasts made are that things will stay
the same, change arithmetically, or change geometrically.
吀栀us we de昀椀ne and start with an A that is a uni昀漀rm annual cost, a G that is a uni昀漀rm
annual gradient, and a g (see next section) that is a uni昀漀rm annual rate of increase 昀漀r the
昀漀llowing three reasons:
It is easier to start with simpler models. When necessary we use spreadsheets and
more complex models.
2. 吀栀ese model cash 昀氀ows are convenient 昀漀r bounding the problems o昀琀en encoun­
tered in engineering economic analysis. When dealing with the 昀甀ture, it is wise
to be very explicit about the assumptions used in the analysis. 吀栀en the decision
maker can either agree or change them.
1.
Arithmetic Gradient
3. O昀琀en in the real world, engineering economy is applied in a feasibility or prelim­
inary analysis. At this stage annual cash 昀氀ows 昀漀r costs and revenues are usually
estimated by using A, G, and/or g. Not enough is known about the problem 昀漀r
more detailed estimates.
Geometric Gradient
We saw that the arithmetic gradient is applicable where the period-by-period change
in a cash receipt or payment is a constant amount. 吀栀ere are other situations where the
period-by-period change is a uni昀漀rm rate, g. O昀琀en geometric gradient series can be
traced to population levels or other levels of activity where changes over time are best
modelled as a percentage of the previous year. For example, if the maintenance costs 昀漀r a
car are $100 the 昀椀rst year and they increase at a uni昀漀rm rate, g, of 10% per year, the cash
昀氀ow 昀漀r the 昀椀rst 昀椀ve years would be as 昀漀llows:
Cash Flow
夀攀ar
1
2
3
4
5
100.00
100.00 + 1 0%( 1 00.00) = 1 00( 1 + 0.10) 1
1 1 0.00 + 1 0%( 1 10.00) = 1 00( 1 + 0.10) 2
1 2 1 .00 + 1 0%( 1 2 1 .00) = 1 00( 1 + 0.10) 3
1 3 3 . 1 0 + 1 0%( 1 3 3 . 1 0) = 1 00( 1 + 0.10) 4
$ 1 00.00
1 1 0.00
1 2 1 .00
1 33. 10
146.4 1
0 -- 1 -- 2 -- 3 -- 4 -- 5
100.00 - - 1 10.00- - 1 2 1 .00 - - - 1 33. io ⴀ娀 ⴀ娀 尀尀
146.4 1'
From the table, we can see that the maintenance cost in any year is
$100(1 + 最⤀ t-l
Stated in a more general 昀漀rm,
(4-19)
where
g = uni昀漀rm 爀愀te of cash flow increase or decrease 昀爀om period to period, that is, the
geometric gradient
A 1 = value of cash 昀氀ow at Year 1 ($100 in the example)
A t = value of cash 昀氀ow at any 夀攀ar t
䌀䐀
115
116
Cha pter 4 Equivalence for Repeated Cash Flows
Since the present worth Pt of any cash 昀氀ow A t at interest rate i is
(4-20)
we can substitute Equation 4-19 into Equation 4-20 to get
吀栀is may be rewritten as
⸀縀 = A ( 1 + i ) - 1 1 + g ) -l
t
[
I
1+i
t
(4-21)
吀栀e present worth of the entire gradient series of cash 昀氀ows may be obtained by
expanding Equation 4-21:
(4-22)
In the general case, where i 㜀㴀 g, Equation 4-22 may be written out as 昀漀llows:
p = A (1 + i) n - l + A ( I + i) - 1 1 + g + A ( I + i) - 1 1 + g
[
[
)
)
I
I
I
l+i
l+i
n
1 + g -l
1
+
A
i)•
•
•
(1
+ l
+
[
)
1+i
Let a = A 1 (1 + i) -1 and b = (1 + 最⤀/(1 + i). Equation 4-23 becomes
P = a + a b + a b 2 + • • • + a b n -1
2
(4-23)
(4-24)
Multiply Equation 4-24 by b:
(4-25)
Subtract Equation 4-25 昀爀om Equation 4-24:
P - bP = a - a bn
P⠀氀 - b) = a⠀氀 - bn)
p = a(I - b n )
I-b
Geometric Gradient
117
Replacing the original values 昀漀r a and b, we obtain
n
封甀
1
[
l
l+g n
l+i
=
A
P = AI ( l + i)- I [ l - [
l
I
I+
I+i
(1 + i) - [
(1 + i)
1+i
= AI [
g)
l
+ l
1 - ( 1 + g ) n (1 + i) - n
l+i-1-g
p = A 1 - ( 1 + g) n ( l
_
I[
t-g
i)- n
(4-26)
where i ⬀㴀 g.
吀栀e expression in the brackets of Equation 4-26 is the geometric series present worth
factor where i ⬀㴀 g.
(PIA, g , i, n) = [
l - ( l +. g ) n ( l + i)- n
t-g
]
where i � g.
(4-27)
In the special case of i = g, Equation 4-26 becomes
P = Ai n( l + i)- I
(P 䤀䄀 , g , i, n) = [n( l + i)- I ]
(4-28)
where i = g.
EXAM PLE 4-12
吀栀e 昀椀rst-year maintenance cost 昀漀r a new car is estimated to be $100, and it increases at a uni昀漀rm rate
of 10% a year. Using an 8% interest rate, calculate the present worth of cost of the 昀椀rst 昀椀ve years of
maintenance.
STEP- BY-STEP SOLUTION
Year n
1
2
3
4
5
Maintenance Cost
100.00
100.00 + 10%(100.00)
110.00 + 10%(110.00)
121.00 + 10%(121.00)
133.10 + 10%(133.10)
= 100.00
= 110.00
= 121.00
= 133.10
= 146.41
X
X
X
X
X
(P/F, 8%, n)
0.9259 =
0.8573 =
0.7938 =
0.7350 =
0.6806 =
PW of
Maintenance
$ 92.59
94.30
96.05
97.83
99.65
$480.42
continued
118
Cha pter 4 Equivalence for Repeated Cash Flows
SOLUTION USI NG G EOM ETRIC SERI ES PRESENT WORTH FACTOR
P = A1 1
I
1 - (1 + g) n ( l + i)- n
t. - g
where t ⬀㴀 g
l
- (1. l 0) 5 (1 .08)- 5
= $480.42
-0.02
吀栀e present worth of cost of maintenance 昀漀r the 昀椀rst 昀椀ve years is $480.42.
= 100.00 [ l
When Compounding Period
and Payment Period Di昀昀er
When the various time periods in a problem match, we can generally solve the problem by
simple calculations. 吀栀us in Example 4-3, where we had $5,000 in an account paying 8%
interest, compounded annually, the 昀椀ve equal end-of-year withdrawals are simply com­
puted as 昀漀llows:
A = P (AIP, 8%, 5) = 5,000(0.2505) = $1,252
Consider how this simple problem becomes more di昀케cult if the compounding period
is changed so that it no longer matches the annual withdrawals.
EXAM PLE 4- 13
On 1 January, a woman deposits $5,000 in a credit union that pays 8% nominal annual interest,
compounded quarterly. She wishes to withdraw all the money in 昀椀ve equal yearly sums, beginning
31 December of the 昀椀rst year. How much should she withdraw each year?
SOLUTION
Since the 8 % nominal annual interest rate r i s compounded quarterly, we know that the e昀昀ective interest
rate per interest period, i, is 2%; and there are a total of 4 x 5 = 20 interest periods in 昀椀ve years. For the
equation A = P(A!P, i, n) to be used, there must be as many periodic withdrawals as there are interest
periods, n. In this example we have 昀椀ve withdrawals and 20 interest periods.
I
w
I
w
I
w
I
w
l
w
O - l - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9-10 -l l - 12 - 1 3- 14 - 15 - 16 - 17 - 18 - 19 - 20
j
$5,000
i = 2% per quarter
n = 20 quarters
When Compounding Period and Payment Period Di昀昀er
119
To solve the problem, we must adjust it so that it is in one of the standard 昀漀rms 昀漀r which we have
compound interest 昀愀ctors. 吀栀is means we must 昀椀rst compute either an equivalent A 昀漀r each three­
month interest period or an e昀昀ective i 昀漀r each time period between withdrawals. Let's solve the prob­
lem both ways.
SO LUTI O N 1
Compute an equivalent A 昀漀r each three -month time period.
If we had been required to compute the amount that could be withdrawn quarterly, the diagram would
have been as 昀漀llows:
IllllIIllIllIlIllIll
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
0 - l - 2 - 3 - 4 - S - 6 - 7 - 8 - 9 - 10 - l l - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20
j
i = 2% per quarter
n = 20 quarters
$5,000
A = P (A!P, i, n) = 5,000(A!P, 2%, 20) = 5,000 (0.0612) = $306
Now, since we know A, we can construct the diagram that relates it to our desired equivalent annual
withdrawal 圀⸀
A = $306
=
llIIlIlllIllIIlIIlll
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
0 - l-2 - 3 - 4 - S - 6 - 7 - 8 - 9 - 10 -l l - 12 - 1 3-14-15 - 16 - 17 - 18 - 19 - 20
w
1
w
1
w
1
w
1
w
1
0 - l-2 -3-4- S-6 - 7 - 8 -9 - 10 -l l-12-13-14 - 15 - 16 - 17 - 18 - 19 - 20
Looking at a one-year period,
A = $ 306
l l l I
0 -- 1 -- 2 -- 3 -- 4 = 0 -- 1 -- 2 -- 3 -- 4
i = 2% per quarter
n = 4 quarters
W = A(䤀蜀, i, n) = 306 (䤀蜀, 2%, 4) = 306 (4.122)
= $1,260
continued
120
Cha pter 4 Equivalence for Repeated Cash Flows
SO LUTI O N 2
Compute an e昀昀ective i 昀漀r the time periods between withdrawals.
Between annual withdrawals, 圀Ⰰ there are 昀漀ur interest periods, hence m = 4 compounding sub-periods
per year. Since the nominal interest rate per year, r, is 8%, we can proceed to compute the e昀昀ective
annual interest rate.
E昀昀ective annu愀氀 interest rate i. = [ 1 + :
r-
1 = [ I + o;
s
r-
1
= 0.0824 = 8.24% per year
Now the problem may be redr愀眀n as 昀漀llows:
w
w
w
w
w
l l l l l
0 -- 1 -- 2 -- 3 -- 4 -- 5
i = 8.24% per year
n = 5 years
$5,000
吀栀is diagram may be directly solved to determine the annual withdrawal W with the capital recovery
昀愀ctor:
W = P(AIP, i, n) = 5,000 (AIP, 8.24%, 5)
= 5,000(0.2520} = $1,260
吀栀e depositor should withdraw $1,260 a year.
U n iform Payment Series: Conti nuous Com pou nding at Nominal Rate r per
Period
Let us now substitute the equation i = er - 1 into the equations 昀漀r end-of-period
compounding.
Continuous Compoundi渀最 Sinking Fund
er - 1
[ AIF, r, n ] = -em - 1
Continuous Compoundi渀最 Capital Recovery
[ AIP, r, n ] =
e m (e r - 1)
em - 1
(4-29)
{4-30}
Continuous Co洀瀀oundi渀最 Series Co洀瀀ound Amount
em - 1
[ FIA, r, n ] = -er - 1
{4-31 }
When Compounding Period and Payment Period Di昀昀er
121
Continuous Compounding Series Present Wor琀栀
(4-32)
EXAM PLE 4-14
In Example 4-1, $500 per year was deposited in a credit union that paid 5% interest, compounded annu­
ally. At the end of 昀椀ve years, $2,763 was in the credit union account. How much would there have been
if the institution paid 5% nominal interest, compounded continuously?
SO LUTI O N
A = $500
r = 0.05
l
-
F = A [ FIA, r, n ] = [ -er
1
= $2,769.84
e 爀渀 _ 1 )
n = 5 years
= 500
[
e o.os(s) - 1
e 0 .05 - 1
l
EXAM PLE 4-15
In Example 4-2, Jim wished to save a uni昀漀rm amount each month so he would have $1,000 at the end
of a year. Using 6% nominal interest, compounded monthly, he calculated that he had to deposit $81.10
per month. How much would he have to deposit if his credit union paid 6% nominal interest, com­
pounded continuously?
SO LUTI O N
吀栀e deposits are made monthly; hence, there are 1 2 compounding sub-periods in the one-year time
period.
0.06
• 愀氀 mterest
•
• d = -r = nomm
rate1·mterest perm
= o .005
12
F = $ 1,000
n = 12 computing sub-periods in the one-year period of the problem
[ _ l=
er
1
A = F [ AIF, r, n ] = F -爀渀
e
_ 1
= 1,000 [ 0 • 005013 ] = $81 .07
1,000
[
e 0 -005 - 1
e
o.oos(12)
_ l
l
0.061837
He would h愀瘀e to deposit $81.07 per month. Note that the di昀昀erence between monthly and continuous
compounding is just 3¢ a month.
122
Cha pter 4 Equivalence for Repeated Cash Flows
SUMMARY
吀栀e compound interest 昀漀rmulas described in this chapter, along with those in Chapter 3,
will be referred to throughout the rest of the book. It is very important 昀漀r the reader to
understand the concepts presented and how these 昀漀rmulas are used. 吀栀e 昀漀llowing nota­
tion is used consistently:
i = e昀昀ective interest rate per interest period2 (stated as a decimal)
n = number of interest periods
P = a present sum of money
F = a 昀甀ture sum of money at the end of the nth interest period that is
equivalent to P with interest rate i
A = an end-of-period cash receipt or disbursement in a uni昀漀rm series con­
tinuing 昀漀r n periods; the entire series equivalent to P or F at interest rate i
G = uni昀漀rm period-by-period increase or decrease in cash receipts or dis­
bursements; the arithmetic gradient
g = uni昀漀rm rate of cash 昀氀ow increase or decrease 昀爀om period to period; the
geometric gradient
r = nominal interest rate per interest period
ia = e昀昀ective interest rate per year (annum)
Single- Payment Form u las (Derived in Cha pter 3)
Compound amount:
F = P( l + it = F(FIP, i, n)
Present worth:
P = F( l + it = F(PIF, i, n)
U n iform Series Form u las
Compound amount:
F=
Sinking 昀甀nd:
A [ ( l + 椀琀 - l ] = A(FIA, i, n)
A = Fl
i
[ = F(AIF, i, n)
(1 + i) " - 1
Normally the interest period is one year, but it could be some other period ( e.g., a quarter, month, or
half year).
2
S u m m a ry
Capital recovery:
A = PI
i(I + i) n
] = P(A!P, i, n)
(1 + i) n - 1
P = Al
(I + i) n - I
i = A(P℀䄀, i, n)
i(I + i) n
Present worth:
Arithmetic Gradient Formulas
Arithmetic gradient present worth:
n
p = G (I + i) - in - I = G(P!G, i, n)
]
l
i 2 (1 + i) n
Arithmetic gradient uni昀漀rm series:
A = Gl
l
n
(I + i) n - in - I
= G(AIG, i, n)
] = G I! n
i(I + i) - i
i
(I + i) n - 1
Geometric G rad ient Formulas
Geometric series present worth, where i 尀 g:
n
n
p = A 1 - (1 + g) (I + i)- I = A1 (P/A, g, i, n)
l
I-g
Geometric series present worth, where i = g:
Single- Payment Formulas: Continuous Compounding at Nominal Rate r
per Period
Compound amount:
F = P(e 爀渀 ) = P[FIP, r, n]
Present worth:
P = F(e - 爀渀 ) = F[PIF, r, n]
Note that square brackets around the 昀愀ctors are used to distinguish continuous
compounding.
123
124
Cha pter 4 Equivalence for Repeated Cash Flows
U n iform Payment Series: Conti nuous Com pou nding at Nominal Rate r per
Period
l
Continuous compounding sinking 昀甀nd:
= F [ AIF, r, n ]
A = F [ -e 爀渀 - 1
er - 1
Continuous compounding capital recovery:
[ -l
Continuous compounding series compound amount:
e 爀渀 1
= A [ FIA, r, n ]
F = A -r
e -1
Continuous compounding series present worth:
Nominal I nterest Rate per Yea r, r
吀栀e annual interest rate without considering the e昀昀ect of any compounding.
E昀昀ective I nterest Rate per Yea r, ia
吀栀e annual interest rate taking into account the e昀昀ect of any compounding during the
year.
E昀昀ective interest rate per year (periodic compounding):
or
E昀昀ective interest rate per year (continuous compounding):
Problems
125
PROBLEMS
U niform Annual Cash Flow
4 -1
4-4
For diagrams (a) to (c), compute the unknown
values-B, C, 嘀Ⰰ respectively-using the
minimum number of compound interest
昀愀ctors.
⠀　I
j
100
1 00
200
l
(a)
+ + + +
10
t t t t
10
10
n=?
200
(c)
4-2
4-5
⠀　How much must be deposited now at 5.25%
interest to produce $300 at the end of every
year 昀漀r 10 years?
( b)
4-6
A car may be purchased with a $3,000 down
payment now and 60 monthly payments of
$480. If the interest rate is 12% compounded
monthly, what is the price of the car?
4 -7
A company deposits $2,000 in a bank at the
end of every year 昀漀r 10 years. 吀栀e company
makes no deposits during the subsequent 昀椀ve
years. If the bank pays 8% interest, how much
would be in the account at the end of 15 years?
4-8
A city engineer knows that she will need
$25 million in three years to replace toll booths
on a toll road in the city. Tra昀케c on the road is
estimated to be 20 million vehicles a year. How
much per vehicle should the toll be to cover
the cost of replacing the toll booths? Interest is
10%. (Simplify your analysis by assuming that
the toll receipts are received at the end of each
year in a lump sum.)
4-9
Using linear interpolation, determine the
value of (PIA, 6.5%, 10) 昀爀om the compound
interest tables. Compute this same value using
the equation or a T VM calculator. Why do the
values di昀昀er?
4-10
How many months will it take to pay o昀昀 a $525
debt, with monthly payments of $15 at the end
of each month, if the interest rate is 18%, com­
pounded monthly?
!
V
Compute the value of J 昀漀r the diagram,
given a 10% interest rate and assuming that
the net present value of the six cash 昀氀ows is
zero.
I I I
100
1 00
100
⠀　! ! !
0 -1-2 - 3- 4- 5 -6 - 7- 8
J
J
J
What is the value of n, 昀漀r this diagram, if the
interest rate is 3.5% and the net present value of
the cash 昀氀ows is zero?
4-3
⠀　A = 50
l
t t t t t t t t t
0 - l - 2 - 3 - 4- 5 - 6 - 7 - 8 - 9 - . . . - n
1 ,000
n=?
!
F = 35.95
0 - 1 -2-3- 4 - 5
i = 10%
t t t
0-- 1 -- 2 -- 3 - 4 - • • • - n - 2 - n - I - n
i = 10%
C
10
200
0 -1-2 - 3- 4
0 -1-2 - 3-4- 5
i = 1 0%
n=5
B
A=I
I I I I
I
I
100
200
吀栀e cash 昀氀ows h瘀였 a present value of zero.
Compute the value of n, given a 10% interest rate.
⠀　126
Chapte r 4 Equ iva lence fo r Repeated Ca sh Flows
4-11
Tori is planning to buy a car . The maximum
payment she can make is $3 ,400 a year , and she
can get a car loan at her credit union for 7.3%
interest. Assume her payments w椀氀l be made at
i
the end of each year from 1 to 4 . I f 吀漀r 's old
car can be traded in for $3,325, which is her
down payment, what is the most expensive car
?
she can buy
4-12
A manufacturing firm spends $500 ,000
annually for a mandatory saf ety inspection of
its p爀漀duction lines. A new monitoring tech­
nology would a氀氀ow the company to e氀椀minate
the need for such inspection. If the interest
rate is ㄀　% a year, how much can the company
a昀昀ord to spend on this new technology? The
company wants to recover its investment in
15 years.
4- 13
A student is buying a new car. The car 's
pr椀挀e is $19 ,500, the GST is 5%, and the ti琀氀e,
氀椀cence, and registration 昀攀e are $650 to be
paid in cash. Instead of buying the car now,
the student has de挀椀ded to save money in
equal monthly amounts for 48 months and
then pay cash. If the student earns 0 .75%
per month interest on the money she saves,
how much money is the s tudent saving each
month?
4- 14
4 - 15
4 -16
Rose recently graduated in engineering. Her
employer will give her a raise of $3,500 a year
if she passes the FE exam (Fundamentals of
Engineering). Over a career of 40 years, what
is the present worth of the raise if the interest
rate is 7%?
i
Jo se graduated in engineering f ve years ago .
H is employe r will give him a 爀愀ise of $ ㄀　, 0 0 0
a ye ar when he complete s the requ irements
and become s registere d as a profes sional
engineer. Over a career of 3 5 ye ars , what is
the present worth of the raise if the interest
rate is 8%?
If the university's School of Engineering can
earn 4% on its i nvestments, how much should
be in its s avings account to 昀甀nd one $5,000
scholarship each year 昀漀r 10 years?
4-17
Elias makes seven annua l end -of -year deposits
of $350 into an account ea rning 2 . 5% inte r­
est . How much is in the account at the end of
夀攀ar 11 if he makes no additional deposits and
makes no withd r愀眀als?
4-18
Liam dreams of sta rting his own business 昀漀r
importing co nsumer elect爀漀n 椀挀 p爀漀ducts to
his home country . He estimates that he ca n
earn 5% on his investments and w椀氀l need to
h愀瘀e $300 ,000 at the end of the ㄀　th year if he
wants to give his business a good, solid 昀漀un­
dation. He now has $28 ,850 in his account,
and he be氀椀eves he can save $12 ,000 昀爀om his
income each year, beginning now. He plans to
marry at about the end of the sixth year and
will skip the investment contribution that
year. How 昀愀r below or above his $300, 000
goal will he be?
4-19
Mary p愀礀s rent of $500 a month 昀漀r the nine­
month academic year. She is going to t爀愀vel
the wo爀氀d this summer and won 't be work­
ing. How much must she set aside in her sav­
ings account 昀漀r the three -month summer
ho氀椀days to cover her rent 昀漀r next year? 吀栀e
savings account earns 3% with monthly
compounding.
4-20
Laquita deposits $3, 500 in her re tirement
account e ver y year. I fher account pays an aver­
age of 6% interest and she makes 38 de posits
be昀漀re she retires, how much mone y can she
withd爀愀w in 20 equal payments, beginning one
year after her las t deposit?
4 -21
Determine the break- e ven resale pr椀挀e 1 0 years
昀爀om now of an apar tment ho use that can
be bo ught to day 昀漀r $449, 0 00. 䤀琀s annual net
income is $54, 0 0 0 . 吀栀e owner wants a ㄀　%
annual return on her in vestment .
4 -22
Kelse y Constr uction has purchased a cr ane
th at comes with a three - year war ranty.
Repair costs are expected to average $3, 5 0 0 a
year beginning in Year 4 when the war ranty
expires. Deter mine the present wor th of the
repair costs for the cr ane over its 1 5 - year life .
吀栀e interest r ate is 10%.
Problems
0
A young engineer wishes to become a million­
aire by the time she is 60 years old. She believes
that by care昀甀l investment she can obtain a
15% rate of return. She plans to add a uni昀漀rm
sum of money to her investment program each
year, beginning on her 20th birthday and con­
tinuing through her 59th birthday. How much
money must the engineer set aside 昀漀r this pro­
ject each year?
4 -28
4 -24
What amount will be necessary in order to
purchase, on an engineer's 40th birthday, an
annuity to provide him with 30 equal semi­
annual payments of $1,000 each, the 昀椀rst to be
received on his 50th birthday, if nominal inter­
est is 4% compounded semi-annually?
4 -29
4-25
吀栀e 昀椀rst of a series of equal semi-annual cash
昀氀ows occurs on 1 July 2011, and the last occurs
on 1 January 2024. Each cash 昀氀ow is equal to
$128,000. 吀栀e nominal interest rate is 12%
compounded semi-annually. What single
amount on 1 July 2015 is equivalent to this
cash 昀氀ow system?
4-23
0
4-26
0
4 -27
On 1 January, Frank bought a used car 昀漀r
$7,200 and agreed to pay 昀漀r it as 昀漀llows: one­
third down payment; the balance to be paid in
36 equal monthly payments; the 昀椀rst payment
due 1 February; an annual interest rate of 9%,
compounded monthly.
(a) What is the amount of Frank's monthly
p愀礀ment?
(b) During the summer, Frank made enough
money that he decided to pay o昀昀 the
entire balance due on the car as of
1 October. How much did Frank owe on
1 October?
If i = 12%, 昀漀r what value of B is the present
value 0?
t I I I
800
800
800
B
! ! t
0 - 1 - 2 - 3- 4
I .SB
Compute E 昀漀r the diagram.
r
3 00
T + T tT
t t
1
°
1
127
r
3 00
0 - 1 - 2 - 3-4- 5 - 6 - 7 - 8
i = l0%
E
0
E
Julie bought a house 昀漀r $500,000 and made
a $100,000 down payment. She obtained
a 30-year loan 昀漀r the remaining amount.
Payments were made monthly. 吀栀e nominal
annual interest rate was 9%. A昀琀er 10 years
(120 p愀礀ments) she decided to pay the remain­
ing balance on the loan.
(a) What was her monthly loan p愀礀ment?
(b) What must she have paid (in addition to
her regular 120th monthly payment) to
pay the remaining balance of her loan?
4 -30
A man wants to help pay 昀漀r a university edu­
cation 昀漀r his young daughter. He can a昀昀ord
to invest $6,000 a year 昀漀r the next 昀漀ur years,
beginning on the girl's 昀漀urth birthday. He
wishes to give his daughter $40,000 on her 18th,
19th, 20th, and 21st birthd愀礀s, 昀漀r a total of
$160,000. Assuming 5% interest, what uni昀漀rm
annual investment will he have to make on the
girl's 8th to 17th birthdays?
4 -31
In Table 3-1 in the text, three plans were pre­
sented 昀漀r the repayment of $5,000 in 昀椀ve years
with interest at 8%. Still another way to repay
the $5,000 would be to make 昀漀ur annual end­
of-year payments of $1,000 each, 昀漀llowed by a
昀椀nal payment at the end of the 昀椀昀琀h year. How
much would the 昀椀nal payment be?
0
4 -32
A $1,500 bicycle was purchased on 1 December
with a $150 down payment. 吀栀e balance
is to be paid at the rate of $100 at the end of
each month, with the 昀椀rst payment due on
31 December. 吀栀e last payment may be some
amount less than $100. If interest on the
unpaid balance is computed at 1.5% a month,
how many payments will there be, and what is
the amount of the 昀椀nal payment?
128
Cha pter 4 Equivalence for Repeated Cash Flows
4-33
A company buys a machine 昀漀r $12,000, which
it agrees to pay in 昀椀ve equal annual p愀礀ments,
beginning one year a昀琀er the date of purchase, at
an interest rate of 4% per annum. Immediately
a昀琀er the second payment, the terms of the
agreement are changed to allow the balance
due to be paid o昀昀 in a single payment the next
year. What is the 昀椀nal single payment?
4-34 An engineering student bought a car at a local
used-car lot. Including tax and insurance, the
total price was $6,000. She is to pay 昀漀r the
car in 12 equal monthly p愀礀ments, beginning
with the 昀椀rst payment immediately (in other
words, the 昀椀rst payment was the down pay­
ment). Nominal interest on the loan is 12%,
compounded monthly. A昀琀er six payments (the
down payment plus 昀椀ve additional payments),
she decides to sell the car. A buyer agrees to
p愀礀 a cash amount to pay o昀昀 the loan in 昀甀ll
at the time the next payment is due and also
to p愀礀 the engineering student $2,000. If there
are no penalty charges 昀漀r this early payment
of the loan, how much will the car cost the new
buyer?
4-35
4-36
⠀　A realtor sold a house on 31 August 2016 昀漀r
$150,000 with a 20% down p愀礀ment. 吀栀e buyer
took a 15-year loan on the property with an
e昀昀ective interest rate of 8% per annum. 吀栀e
buyer intends to p愀礀 o昀昀 the loan owed in yearly
payments starting on 31 August 2017.
愀儀 How much of the loan will still be owed
a昀琀er the payment due on 31 August 2019
has been made?
(b) Solve the same problem by separating the
interest and the principal amounts.
To pay 昀漀r a university education 昀漀r her son,
a woman opened an escrow account in which
equal deposits were made. 吀栀e 昀椀rst deposit was
made on 1 January 2000, and the last deposit
was made on 1 January 2017. 吀栀e yearly univer­
sity expenses, including tuition, were estimated
to be $8,000 昀漀r each of 昀漀ur years. Assuming
the interest rate to be 5.75%, how much did the
mother have to deposit each year in the escrow
account 昀漀r her son to draw $8,000 a year 昀漀r
昀漀ur years beginning 1 January 2017?
4-37
An engineer borrowed $3,000 昀爀om the bank,
payable in six equal end-of-year payments at
8%. 吀栀e bank agreed to reduce the interest on
the loan if interest rates declined in Canada
be昀漀re the loan was 昀甀lly repaid. At the end of
three years, at the time of the third payment,
the bank agreed to reduce the interest rate
昀爀om 8% to 7% on the remaining debt. What
was the amount of the equal annual end-of­
year payments 昀漀r each of the 昀椀rst three years?
What was the amount of the equal annual
end-of-year payments 昀漀r each of the last three
years?
4-38
A local 昀椀nance company will lend $10,000 to
a homeowner. It is to be repaid in 24 monthly
payments of $499 each. 吀栀e 昀椀rst payment is
due 30 days a昀琀er the $10,000 is received. What
interest rate per month is the 昀椀nance company
charging?
⠀　⠀　4-39 A woman made 10 annual end-of-year pur­
chases of $1,000 worth of common shares.
吀栀e shares paid no dividends. 吀栀en 昀漀r 昀漀ur
years she held the shares. At the end of the
昀漀ur years she sold all the shares 昀漀r $28,000.
What interest rate did she obtain on her
investment?
Relationsh i p between Factors
4-40 For some interest rate i and some number of
interest periods n, the uni昀漀rm series capital
recovery 昀愀ctor is 0.1728 and the sinking 昀甀nd
昀愀ctor is 0.0378. What is the interest rate? What
is n?
⠀　4-41 If $200 is deposited in a savings account at the
beginning of each of 15 years, and the account
draws interest at 7% per year, how much will be
in the account at the end of 15 years?
⠀　4-42 How can the tables be used to compute (䤀鄀,
5%, 150)? (PIA, 7%, 200)?
Problems
Arith metic G radients
4-43
3 00
I
Assume a 10% interest rate and 昀椀nd S, 吀Ⰰ and x.
In (c), assume that the net present value of the
cash 昀氀ows is zero.
tTI
150
t I ⠀　• t
120
60
30
4 -46
0 -1- 2 - 3- 4
0 - 1- 2 - 3- 4- 5
(a) S = value at t = 0
( b) T = uni昀漀rm value 昀漀r 5 periods
+
!
l
500
t
100
0- 1- 2-3
0-1-2-3
i = 10%
i = 10%
(c)
4-47
I
f 1
100
+ ✀昀
0 - 1 - 2 - 3- 4
0 - 1 - 2 - 3- 4
i = 10%
i = 10%
25
7
(b) F = value at t = 4
(a) C = value at t = 0
Compute the present value of the cash 昀氀ows.
+
TT l
tI
(c) A = uni昀漀rm value 昀爀om t = 1 to 4
For diagrams (a) to (d), compute the present
values of the cash 昀氀ows.
t f I
7
1
2
100
4 -48
⠀　Compute the value of C 昀漀r the diagram,
assuming a 10% interest rate and a net present
value of zero.
✀琀 T
3 00
I
t t t
C
4-49
0 - 1- 2 - 3- 4
0 - 1- 2 - 3- 4- 5
i = 10%
i = 10%
(b)
C
C
If i = 12% and the net present value is zero,
compute G in the diagram.
6G
5G
r t 'f
2G
50
(a)
Use a 15% interest rate to compute the present
value of the cash flows.
0-1- 2- 3- 4- 5- 6- 7- 8
0 - 1 - 2 - 3- 4
i = 10%
4-45
250
200
0-1-2 -3- 4- 5
120
40
l
(d)
i = 10%
Compute the unknown values.
75
50
0 - 1- 2 - 3
(c)
4-44
100
100
0-1- 2-3- 4
i = 1 0%
x=?
I t I
200
4x
T ✀昀
X
129
4G
I
0-1- 2- 3- 4- 5- 6
i i
500
500
130
Cha pter 4 Equivalence for Repeated Cash Flows
4-50 Compute the value of D in the diagram, if the
net present value is zero.
㈀ꐀ
4-53
Find the value of P 昀漀r the 昀漀llowing cash 昀氀ow
diagram, if the net present value is zero.
3 , 000
0 - 1 - 2 - 3 - 4- 5 - 6 - 7- 8
t t t t
D
4-51
D
D
1 ,25 0
i = IO%
D
I TT T
Using a 10% interest rate, compute B in the
diagram, if the net present value is zero.
4 00
l
i
l
l
3 00
P=?
2 00
1 00
4-54 Use a 10% interest rate to compute the present
value of the cash 昀氀ows.
㈀ꐀ
l
0 -1-2 -3-4
TT I TT
5 00
3B
0
4-52
4B
0 -1-2 - 3- 4- 5
吀栀e 昀漀llowing two cash 昀氀ow transactions are
said to be equivalent in terms of economic
desirability at an interest rate of 12% com­
pounded annually. Determine the unknown
value A.
600
A
A
A
tttt
750
= 0 - 1 - 2 - 3 - 4- 5
Cash Flow 2
Consider the 昀漀llowing cash 昀氀ow:
Year
Cash Flow
0
1
-$ 1 00
+so
+60
+70
+80
+ 1 40
2
3
4
5
A
0 -l- 2 - 3 - 4
Cash Flow 1
4-55
Which one of the 昀漀llowing is correct 昀漀r this
cash 昀氀ow, if the net present value is zero?
(1) 100 = 50 + lO(A/G, i, 5) + 50(PIF, i, 5)
( l)
50(P IA, i, 5) + l O(P /G, i, 5) + 50(P/F, i, 5)
=l
100
(3) IO0(AIP, i, 5) = 50 + IO(A/G, i, 5)
(4) None of the equations are correct.
Problems
0
4-56
Consider the 昀漀llowing cash 昀氀ow:
Year
0
1
2
3
4
5
6
7
8
Cash Flow
- $P
+1,000
+850
+700
+550
+400
+400
+400
+400
Alice was asked to compute the value of P 昀漀r
the cash 昀氀ow at 8% interest, assuming the net
present value of the cash 昀氀ows is zero. She
wrote three equations:
(1) P = 1,000(䤀鄀, 8%, 8) - ISO(P/G, 8%, 8)
+ IS0(P/G, 8%, 4)(P/F, 8%, 4)
(2) P = 400(℀鄀, 8%, 8) + 600(℀鄀, 8%, 5)
- IS0(P/G, 8%, 4)
(3) P = IS0(P!G, 8%, 4) + 850(℀鄀, 8%, 4)
+ 400(P!A, 8%, 4)(P/F, 8%, 4)
0
4 -57
4-58
that the 昀漀llowing are reasonable estimates of
the end-of-year maintenance costs:
Year
Maintenance
Cost
Year
Maintenance
Cost
1
2
3
4
5
$500
100
125
150
175
6
7
8
9
10
$200
225
250
275
300
Assuming interest at 6% a year, how much
should the smog control district pay to the
company now to provide 昀漀r the 昀椀rst cost of the
equipment and its maintenance 昀漀r 10 years?
4 -59
Which of the equations is correct?
吀栀e council members of a small town have
decided that the breakwater that protects the
town 昀爀om a nearby river should be rebuilt
and strengthened. 吀栀e town engineer esti­
mates that the cost of the work at the end of
the 昀椀rst year will be $85,000. She estimates
that in subsequent years the annual repair
costs will decline by $10,000, making the
second-year cost $75,000, the third-year
$65,000, and so 昀漀rth. 吀栀e council members
want to know what the equivalent present
cost is 昀漀r the 昀椀rst 昀椀ve years of repair work if
interest is 4%.
A company expects to install smog control
equipment on the exhaust of a gasoline engine.
吀栀e local smog control district has agreed to
pay to the 昀椀rm a lump sum of money to pro­
vide 昀漀r the 昀椀rst cost of the equipment and
maintenance during its IO-year use昀甀l life.
At the end of 10 years the equipment, which
initially cost $10,000, is valueless. 吀栀e com­
pany and smog control district have agreed
131
A debt of $5,000 can be 昀甀lly repaid, with inter­
est at 8%, by the 昀漀llowing payments:
Year
Payment
1
2
3
4
5
$500
1,000
1,500
2,000
X
吀栀e p愀礀ment at the end of the 昀椀昀琀h year is X.
How much is X?
4 - 60
A man is buying a small garden tractor. 吀栀ere
will be no maintenance cost during the 昀椀rst
two years because the tractor comes with two
years' 昀爀ee maintenance. For the third year,
the maintenance is estimated at $80. In subse­
quent years the maintenance cost will increase
by $40 a year. How much would have to be set
aside now at 8% interest to pay the mainten­
ance costs on the tractor 昀漀r the 昀椀rst six years
of ownership?
4 - 61
A sports star can sign a six-year contract
that starts at $1.2 million, with increases of
$0.3 million each year 昀漀r his expected pl愀礀ing
career of six years. It is also possible to sign a
contract that starts at $0.8 million 昀漀r the 昀椀rst
year and increases by $0.2 million each year
昀漀r 10 years (note that some income is deferred
until a昀琀er he retires). If his interest rate 昀漀r the
time value of money is 8%, what is the value of
each choice?
132
Cha pter 4 Equivalence for Repeated Cash Flows
4-62 A college student is buying a new car, which
costs $16,500 plus 8% sales tax. 吀栀e title,
licence, and registration fee is $650. 吀栀e dealer
o昀昀ers her a 昀椀nancing program that starts
with a small monthly payment; the payments
will gradually increase. 吀栀e dealer o昀昀ered to
昀椀nance 80% of the car's price 昀漀r 48 months at
a nominal interest rate of 9% per annum, com­
pounded monthly. 吀栀e 昀椀rst p愀礀ment is $300,
and each successive p愀礀ment will increase by a
constant dollar amount x.
愀儀 How much is the constant amount x?
(b) How much is the 48th p愀礀ment?
4-63 A 昀椀rm has $500,000 a year to pay 昀漀r
replacing machinery over the next 昀椀ve years.
What is the expected cost in Year 1 if the 昀椀rm
has projected that the machinery cost will
increase by $15,000 a year? 吀栀e interest rate
is 10% a year.
4-64 Helen can earn 3% interest in her savings
account. Her daughter Roberta is 1 1 years old
today. Suppose Helen deposits $4,000 today,
and one year 昀爀om today she deposits another
$500. Each year she increases her deposit
by $500 until she makes her last deposit on
Roberta's 18th birthday. What is the annual
equivalent of her deposits, and how much is on
deposit a昀琀er the 18th birthday?
4 - 65 Francisco believes in planning ahead. So he
decides to find out how much he can draw
from a retirement account each year 昀漀r
30 years if he invests exactly $ 1,000 each
year in the account. Suppose that he can
earn 5% on this long-term account, that
he will make 40 deposits, and that he will
make the first withdrawal one year a昀琀er the
last deposit. How big can Francisco's with­
drawals be?
4-66 Perry is a 昀爀eshman. He estimates that tuition,
books, room and board, transportation, and
other incidentals will cost him $13,000 this
year. He expects these costs to rise about
$1,500 each year while he is in college. If it will
take him 昀椀ve years to earn his BS, what is the
present cost of his degree at an interest rate
of 6%? If he earns an extra $10,000 annually
昀漀r 40 years, what is the present worth of his
degree?
4-67 吀栀e 昀漀llowing cash 昀氀ows are equivalent in
value if the interest rate is i. Which one is more
� valuable if the interest rate is 2i?
?
(a)
tt tt
0 - 1 - 2 - 3- 4
(b)
tt t
0 - 1 - 2 - 3- 4
4-68 吀栀e 昀漀llowing cash 昀氀ows are equivalent in
value if the interest rate is i. Which one is more
valuable if the interest rate is 2i?
(a)
tt tt
0 - 1 - 2 - 3- 4
(b)
t
ttI
0 - 1 - 2 - 3- 4
Geometric G radients
4-69 A set of cash flows begins at $20,000 the 昀椀rst
year and increases each year until n = 10 years.
If the interest rate is 8%, what is the present
value when
(a) the annual increase is $2,000?
(b) the annual increase is 10%?
⠀　4-70
A set of cash 昀氀ows begins at $50,000 the 昀椀rst
year, with an increase each year until n = 15
years. If the interest rate is 7%, what is the
present value when
(a) the annual increase is $5,000?
(b) the annual increase is 10%?
4-71
A set of cash flows begins at $20,000 the 昀椀rst
year, with a decrease each year until n = 10
years. If the interest rate is 8%, what is the
present value when
愀儀 the annual decrease is $2,000?
(b) the annual decrease is 10%?
4-72
吀栀e market 昀漀r a product is expected to
increase at an annual rate of 8%. First-year
sales are estimated at $60,000, the horizon is
15 years, and the interest rate is 10%. What is
the present value?
Problems
4-73
Fred is evaluating whether a more e昀케cient
motor with a life of 昀椀ve years should be
installed on an assembly line. Energy savings
are estimated at $400 昀漀r the 昀椀rst year, increasing by 6% annually. If the interest rate is 10%,
what is the present value of the energy savings?
4-74
In Problem 4-73, what is the present value of
the energy savings if they are increasing by
15% annually?
4-75
Mark saves a 昀椀xed percentage of his salary at
the end of each year. 吀栀is year he saved $1,500.
For the next 昀椀ve years, he expects his salary
to rise at an 8% annual rate, and he plans to
increase his savings at the same 8% annual rate.
He invests his money in the stock market. 吀栀us
there will be six end-of-year investments (the
initial $1,500 plus 昀椀ve more). Solve the problem by using the geometric gradient 昀愀ctor.
(a) How much will the investments be worth
at the end of six years if they increase on
the stock market at a 10% annual rate?
(b) How much will Mark have at the end of
six years if his stock market investments
increase only at 8% annually?
⠀　⠀　4-76
4-77
⠀　吀栀e Macintosh Company has an employee savings plan that allows every employee to invest
up to 5% of his or her annual salary. 吀栀e money
is invested in company common stock with the
company guaranteeing that the annual return
will never be less than 8%. Jill was hired at an
annual salary of $52,000. She immediately
joined the savings plan, investing the 昀甀ll 5%
of her salary each year. If Jill's salary increases
at an 8% uni昀漀rm rate, and she continues
to invest 5% of it each year, what amount of
money is she guaranteed to have at the end of
20 years?
吀栀e 昀漀otball coach at a university was given
a 昀椀ve-year employment contract that paid
$225,000 the 昀椀rst year and increased at an
8% uni昀漀rm rate in each subsequent year. At
the end of the 昀椀rst year's 昀漀otball season, the
alumni demanded that the coach be 昀椀red. 吀栀e
alumni agreed to buy his remaining years on
the contract by paying him the equivalent
133
present sum, computed with a 12% interest
rate. How much will the coach receive?
4-78
A 25-year-old engineer is opening an individual retirement account (RRSP) at a bank. Her
goal is to accumulate $1 million in the account
by the time she retires 昀爀om work in 40 years.
吀栀e bank manager estimates she may expect
to receive 8% nominal annual interest, compounded quarterly, throughout the 40 years.
吀栀e engineer believes her income will increase
at a 7% annual rate during her career. She
wishes to start with as low a deposit as possible
to her RRSP now and increase it at a 7% rate
each year. Assuming end-of-year deposits, how
much should she deposit the 昀椀rst year?
4-79
A contractor estimates maintenance costs 昀漀r
a new backhoe to be $275 昀漀r the 昀椀rst month
with a monthly increase of 0.5%. 吀栀e contractor can buy a 昀漀ur-year maintenance contract 昀漀r $18,500 at any point. If the contract
is purchased at the same time as the backhoe, the dealer has o昀昀ered a 10% discount.
Use i = 0.75% per month. What should the
contractor do?
4-80
Zachary has opened a retirement account
that will pay 5% interest each year. He plans
to deposit 10% of his annual salary into the
account 昀漀r 39 years be昀漀re he retires. His 昀椀rst
year's salary is $52,000, and he expects the
salary to grow 4% each year. How much will be
in his account a昀琀er he makes the last deposit?
What uni昀漀rm amount can he withdraw 昀爀om
the account 昀漀r 25 years beginning one year
a昀琀er his last deposit?
4-81
Eddie is a production engineer 昀漀r a major supplier of parts 昀漀r cars. He has determined that
a robot can be installed on the production line
to replace one employee. 吀栀e employee earns
$20 an hour and receives bene昀椀ts worth $8
an hour 昀漀r a total annual cost of $58,240 this
year. Eddie estimates this cost will increase 6%
each year. 吀栀e robot will cost $16,500 to operate 昀漀r the 昀椀rst year with costs increasing by
$1,500 each year. 吀栀e 昀椀rm uses an interest rate
of 15% and a 10-year planning horizon. 吀栀e
134
Cha pter 4 Equivalence for Repeated Cash Flows
robot costs $ 75,000 installed and will have a
salvage value of $ 5 ,000 a昀琀er 10 years. Should
Eddie recommend that purchase of the robot?
Nominal and E昀昀ective I nterest Rates
4-82
4 - 83
4 - 84
Quentin has been using his credit card too much.
His plan is to use only cash until the balance of
$8, 5 㜀㐀 is paid o昀昀. 吀栀e credit card company char­
ges 18% interest, compounded monthly. What
is the e昀昀ective annual interest rate? How much
interest will he owe in the 昀椀rst month's pay­
ment? If he makes monthly p愀礀ments of $225,
how long will it be until it is paid o昀昀?
A bank is o昀昀ering a loan of $2 5 ,000 with a
nominal interest rate of 18% compounded
monthly, payable in 60 months. (Hint: 吀栀e loan
origination fee of 2% will be taken out of the
loan amount.)
(a) What is the monthly payment?
(b) If a loan origination fee of 2% is charged
at the time of the loan, what is the e昀昀ect­
ive interest rate?
0
4 - 87
You are taking out a $2,000 loan. You will pay
it back in 昀漀ur equal amounts, paid every six
months starting three years 昀爀om now. 吀栀e
interest rate is 6% compounded semi-annually.
Calculate the 昀漀llowing:
(a) 吀栀e e昀昀ective interest rate, based on
both semi-annual and continuous
compounding
(b) 吀栀e amount of each semi-annual payment
(c) 吀栀e total interest paid
4-88
In 1 535, King Henry VIII borrowed money
昀爀om his bankers on the condition that he pay
5% of the loan at each 昀愀ir (there were 昀漀ur 昀愀irs
a year) until he had made 40 payments. At that
time the loan would be considered repaid.
What e昀昀ective annual interest did King Henry
pay?
4 - 89
A local bank will lend a customer $10,000 on a
two-year car loan as 昀漀llows:
One of the largest car dealers in the city adver­
tises a three-year-old car 昀漀r sale as 昀漀llows:
Cash price $1 3 , 755, or a down payment of
$1, 375 with 45 monthly payments of $ 361.2 3 .
Susan bought the car and made a down pay­
ment of $2,000. 吀栀e dealer charged her the
same interest rate used in his advertised o昀昀er.
What is the monthly interest rate? How much
will Susan pay each month 昀漀r 45 months?
What e昀昀ective annual interest rate is being
charged?
4-85
0
4-86
Pete borrows $10,000 to buy a car. He must
repay the loan in 48 equal end-of-period
monthly payments. Interest is calculated at
1.2 5% a month. Determine the 昀漀llowing:
愀儀 吀栀e nominal annual interest rate
(b) 吀栀e e昀昀ective annual interest rate
(c) 吀栀e amount of the monthly payment
吀栀e Bawl Street Journal costs $ 580, p愀礀able now,
昀漀r a two-year subscription. 吀栀e newspaper
is published 252 d愀礀s a year (昀椀ve d愀礀s a week,
except holid愀礀s). If a 10% nominal annual inter­
est rate, compounded quarterly, is used,
(a) What is the e昀昀ective annual interest rate
in this problem?
(b) Compute the equivalent interest rate per
1/2 52 of a year.
(c) What is a subscriber's cost per copy of the
newspaper, taking interest into account?
Money to pay 昀漀r car
吀眀o years' interest at 7%:
2 X 0.07 X 10,000
$ 10,000
1,400
$ 1 1,400
1 1,400
24 monthly p愀礀ments = -- = $475.00
24
0
4 -90
吀栀e 昀椀rst payment must be made in 3 0 days.
What is the nominal annual interest rate the
bank is receiving?
A local lending institution advertises the
" 5 1- 50 Club," A person may borrow $2,000 and
repay $ 5 1 昀漀r the next 50 months, beginning
3 0 days a昀琀er receiving the money. Compute
the nominal annual interest rate 昀漀r this loan.
What is the e昀昀ective interest rate?
Problems
0
4 -91
吀栀e Rule of 78s is a commonly used method
of computing the amount of interest when the
balance of a loan is repaid in advance.
Adding the numbers representing
12 months gives
1 + 2 + 3 + 4 + 5 + • • • + 11 + 12 = 78
If a 12-month loan is repaid at the end
of one month, 昀漀r example, the inter­
est the borrower would be charged is
12/78 of the year's interest. If the loan
is repaid at the end of two months,
the total interest charged would be (12
+ 1 1)/78, or 23/78 of the year's interest.
A昀琀er 11 months the interest charge
would there昀漀re be 77/78 of the total
year's interest.
Shannon borrowed $10,000 on 1 January at
9% annual interest, compounded monthly.
吀栀e loan was to be repaid in 12 equal end-of­
period payments. Shannon made the 昀椀rst two
payments and then decided to repay the bal­
ance of the loan when she pays the third pay­
ment. 吀栀us she will pay the third payment plus
an additional sum.
Calculate the amount of this additional sum
(a) using the rule of 78s.
(b) using exact economic analysis methods.
4 -92
Upon the birth of his 昀椀rst child, Dick decided
to establish a savings account to partly pay 昀漀r
his son's education. He plans to deposit $20
a month in the account, beginning when the
boy is 13 months old. 吀栀e savings and loan
association has a current interest policy of 6%
per annum, compounded monthly, paid quar­
terly. Assuming no change in the interest rate,
how much will be in the savings account when
Dick's son turns 16?
4-93
What is the present worth of a series of equal
quarterly payments of $3,000 that extends over
a period of eight years if the interest rate is 10%
compounded monthly?
0
4 -94
What single amount on 1 April 2012 is equivalent to a series of equal, semi-annual cash
135
昀氀ows of $1,000 that starts with a cash 昀氀ow on
1 January 2010 and ends with a cash flow on
1 January 2019? 吀栀e interest rate is 14%, and
compounding is quarterly.
Com pou nding and Payment Periods Di昀昀er
4 -95
A contractor wishes to set up a special 昀甀nd by
making uni昀漀rm semi-annual end-of-period
deposits 昀漀r 20 years. 吀栀e 昀甀nd is to provide
$10,000 at the end of each of the last 昀椀ve years
of the 20-year period. If interest is 8%, com­
pounded semi-annually, what semi-annual
deposit is required?
4 -96
Paco's saving account earns 13% compounded
weekly and receives quarterly deposits of
$38,000. His 昀椀rst deposit was made on
1 October 2006, and the last deposit is sched­
uled 昀漀r 1 April 2022. Tisha's account earns 13%
compounded weekly. Semi-annual deposits of
$18,000 are made into her account, with the
昀椀rst one being made on 1 July 2016 and the last
one on 1 January 2025. What single amount on
1 January 2017 is equivalent to the sum of both
cash flow series?
4 -97
A series of monthly cash 昀氀ows is deposited into
an account that earns 12% nominal interest
compounded monthly. Each monthly deposit
is equal to $2,100. 吀栀e 昀椀rst monthly deposit
was made on 1 June 2016, and the last monthly
deposit will be on 1 January 2023. 吀栀e account
(the series of monthly deposits, 12% nominal
interest, and monthly compounding) also has
equivalent quarterly withdrawals 昀爀om it. 吀栀e
昀椀rst quarterly withdrawal is equal to $5,000
and was made on 1 October 2016. 吀栀e last
$5,000 withdrawal will be made on 1 January
2023. How much remains in the account a昀琀er
the last withdrawal?
Continuous Com pou nding
4 -98
PARC Company has money to invest in an
employee bene昀椀t plan, and you have been
chosen as the plan's trustee. As an employee
yourself, you want to maximize the interest
earned on this investment and have 昀漀und an
136
Cha pter 4 Equivalence for Repeated Cash Flows
account that pays 14% compounded continu­
ously. PARC is providing you with $1,200 a
month to put into the account 昀漀r seven years.
What will be the balance in this account at the
end of the seven-year period?
4-99 Barry, a recent engineering graduate, never
took engineering economics. When he gradu­
ated, he was hired by a prominent architec­
tural 昀椀rm. 吀栀e earnings 昀爀om this job allowed
him to deposit $750 each quarter into a s愀瘀­
ings account. 吀栀ere were two banks o昀昀ering
a s愀瘀ings account in his town (a small town!).
吀栀e 昀椀rst bank was o昀昀ering 4.5% interest
compounded continuously. 吀栀e second bank
o昀昀ered 4.6% compounded monthly. Barry
decided to deposit in the 昀椀rst bank since it
o昀昀ered continuous compounding. Did he
make the right decision?
Spreadsheets for Economic Ana lysis
4-100 Develop a complete amortization table 昀漀r a
loan of $4,500, to be paid back in 24 uni昀漀rm
monthly instalments, based on an interest rate
of 6%. 吀栀e amortization table must include the
昀漀llowing column headings:
Payment Number, Principal Owed
(beginning of period), Interest Owed
in Each Period, Total Owed (end of
each period), Principal Paid in Each
Payment, Uni昀漀rm Monthly Payment
Amount
You must also show the equations used to
calculate each column of the table. You are
encouraged to use spreadsheets. 吀栀e entire
table must be shown.
4-101 吀栀e 昀漀llowing beginning-of-month (BOM)
and end of month (EOM) amounts are to be
deposited in a savings account that pays inter­
est at 9%, compounded monthly:
Today (BOM 1)
EOM 2
EOM 6
EOM 7
BOM lO
$400
270
100
180
200
Set up a spreadsheet to calculate the account
balance at the end of the 昀椀rst year (EOM 12).
吀栀e spreadsheet must include the 昀漀llowing
column headings: Month Number, Deposit
BOM, Account Balance at BOM, Interest
Earned in Each Month, Deposit EOM, Ac­
count Balance at EOM. Use the compound
interest tables to draw a cash flow diagram of
this problem, and solve 昀漀r the account bal­
ance at the EOM 12.
4-102 What is the present worth of cash 昀氀ows that
begin at $20,000 and increase at 7% a year
昀漀r 10 years? 吀栀e interest rate is 9%. Use a
spreadsheet.
4-103 What is the present worth of cash 昀氀ows that
begin at $50,000 and decrease at 12% a year
昀漀r 10 years? 吀栀e interest rate is 8%. Use a
spreadsheet.
4-104 Net revenues at an older manu昀愀cturing plant
will be $2 million 昀漀r this year. 吀栀e net rev­
enue will decrease by 15% a year 昀漀r 昀椀ve years,
when the assembly plant will be closed (at the
end of Year 6). If the 昀椀rm's interest rate is 10%,
calculate the PW of the revenue stream. Use a
spreadsheet.
4-105 What is the present worth of cash 昀氀ows that
begin at $10,000 and increase at 8% a year
昀漀r 昀漀ur years? 吀栀e interest rate is 6%. Use a
spreadsheet.
4-106 What is the present worth of cash 昀氀ows that
begin at $30,000 and decrease at 15% a year
昀漀r six years? 吀栀e interest rate is 10%. Use a
spreadsheet.
4-107 Five annual payments at an interest rate of
9% are made to repay a loan of $6,000. Build
the table that shows the balance due, princi­
pal payment, and interest payment 昀漀r each
payment. What is the annual payment? (Use
a spreadsheet 昀甀nction, not the tables.) What
interest is paid in the last year?
4-108 A newly graduated engineer bought 昀甀rniture
昀漀r $900 昀爀om a local store. Monthly payments
Problems
昀漀r one year will be made. Interest is computed
at a nominal rate of 6%. Build the table that
shows the balance due, principal payment,
and interest payment 昀漀r each payment. What
is the monthly payment? (Use a spreadsheet
昀甀nction, not the tables.) What interest is paid
in the last month?
4-109 Using a spreadsheet, calculate and print out
the balance due, principal payment, and
interest payment 昀漀r each period of a used­
car loan. 吀栀e nominal interest is 12% a year,
compounded monthly. Payments are made
monthly 昀漀r three years. 吀栀e original loan is
昀漀r $1 1,000.
4-110 Using a spreadsheet, calculate and print out
the balance due, principal payment, and inter­
est payment 昀漀r each period of a new car loan.
吀栀e nominal interest is 9% a year, compounded
monthly. Payments are made monthly 昀漀r 昀椀ve
years. 吀栀e original loan is 昀漀r $17,000.
4-111 Your beginning salary is $50,000. You deposit
10% at the end of each year in a savings
account that earns 6% interest. Your salary
increases by 5% a year. What value does
your savings book show a昀琀er 40 years? Use a
spreadsheet.
137
U nclassi昀椀ed
4-114 Ann deposited $100 in her bank savings
account at the end of each month. 吀栀e bank
paid 6% nominal interest, compounded and
paid quarterly. No interest was paid on money
not in the account 昀漀r the 昀甀ll three-month
period. How much was in Ann's account at the
end of three years?
4-115 Penelope borrows $1,000. To repay the amount
she makes 12 equal monthly payments of
$90.30. Determine the 昀漀llowing:
(a) 吀栀e e昀昀ective monthly interest rate
(b) 吀栀e nominal annual interest rate
(c) 吀栀e e昀昀ective annual interest rate
⠀　4-116 What single amount on 1 October 2012 is equal
to a series of $1,000 quarterly deposits made
into an account? 吀栀e 昀椀rst deposit is made on
1 October 2012, and the last deposit occurs on
1 January 2026. 吀栀e account earns 13% com­
pounded continuously.
4-117 A 30-year mortgage 昀漀r $120,000 has been
issued. 吀栀e interest rate is 10%, and payments
are made monthly. Print out the balance due,
principal payment, and interest payment 昀漀r
each period. Use a spreadsheet.
4-112 吀栀e market volume 昀漀r widgets is increas­
ing by 15% a year 昀爀om current pro昀椀ts of
$200,000. Investing in a design change will
allow the pro昀椀t per widget to stay steady;
otherwise they will drop 3% a year. What is
the present worth of the savings over the next
昀椀ve years? Ten years? 吀栀e interest rate is 10%.
Use a spreadsheet.
4-118 A homeowner may upgrade a 昀甀rnace that runs
on 昀甀el oil to a natural gas unit. 吀栀e invest­
ment will be $2,500 installed. 吀栀e cost of the
natural gas will average $60 a month over the
year, instead of the $145 a month that the 昀甀el
oil costs. If the interest rate is 9% per year, how
long will it take to recover the initial invest­
ment? Use a spreadsheet.
4-113 Develop a general-purpose spreadsheet to cal­
culate the balance due, principal payment, and
interest payment 昀漀r each period of a loan. 吀栀e
user's inputs to the spreadsheet will be the loan
amount, the number of payments per year, the
number of years payments are made, and the
nominal interest rate. Submit printouts of your
analysis of a loan in the amount of $15,000
at 8.9% nominal rate 昀漀r 36 months and 昀漀r
60 months of payments.
4-119 A construction 昀椀rm can achieve a $15,000
cost savings in year 1 and increase it by $2,000
each year 昀漀r the next 昀椀ve years by upgrading
some equipment. At an interest rate of 15%,
what is the equivalent annual worth of the
savings?
4-120 A man borrowed $500 昀爀om a bank on
15 October. He must repay the loan in 16 equal
monthly payments, due on the 15th of each
⠀　138
Cha pter 4 Equivalence for Repeated Cash Flows
month, beginning 15 November. If interest is
computed at 1% a month, how much must he
pay each month?
4-121 A bank recently announced an "instant cash"
plan 昀漀r holders of its bank credit cards. A card­
holder m愀礀 receive cash 昀爀om the bank up to
a preset limit (about $500). 吀栀ere is a special
charge of 4% made at the time the "instant cash"
is sent to a cardholder. 吀栀e debt m愀礀 be repaid
in monthly instalments. Each month the bank
charges 1.5% on the unpaid balance. 吀栀e monthly
payment, including interest, m愀礀 be as little as
$10. 吀栀us, 昀漀r $150 of "instant cash," an initial
charge of $6 is made and added to the balance
due. Assume the cardholder makes a monthly
payment of $10 (this includes both principal and
interest). How many months are needed to repay
the debt? If your answer includes a 昀爀action of a
month, round up to the next month.
4-122 A woman makes an investment every three
months at a nominal annual interest rate of
28%, compounded quarterly. Her 昀椀rst invest­
ment was $100, 昀漀llowed by investments
increasing by $20 each three months. 吀栀us the
second investment was $120, the third invest­
ment $140, and so on. If she continues to make
this series of investments 昀漀r a total of 20 years,
what will be the value of the investments at the
end of that time?
4-123 A set of cash 昀氀ows begins at $20,000 the 昀椀rst
year, with an increase each year until n = 10
years. If the interest rate is 10%, what is the
present value when
愀儀 the annual increase is $2,000?
(b) the annual increase is 10%?
4-124 Suzanne is a recent chemical engineering
graduate who has been o昀昀ered a 昀椀ve-year
contract at a remote location. She has been
o昀昀ered two choices. 吀栀e 昀椀rst is a 昀椀xed salary of
$75,000 a year. 吀栀e second has a starting salary
of $65,000 with annual raises of 5% starting in
Year 2. (For simplicity, assume that her salary
is paid at the end of the year, just be昀漀re her
annual vacation.) If her interest rate is 9%,
which option should she take?
4-125 An engineer will deposit 15% of her salary
each year into a retirement 昀甀nd. If her current
annual salary is $80,000 and she expects that
it will increase by 5% each year, what will be
the present worth of the 昀甀nd a昀琀er 35 years if it
earns 5% a year?
4-126 A man buys a car 昀漀r $18,000 with no money
down. He pays 昀漀r the car in 30 equal monthly
payments with interest at 12% per annum,
compounded monthly. What is his monthly
loan p愀礀ment?
4-127 A student wants to have $30,000 when he
graduates 昀漀ur years 昀爀om now to buy a new
car. His grand昀愀ther gave him $10,000 as a
high school graduation present. How much
must the student save each year if he deposits
the $10,000 today and can earn 12% on both
the $10,000 and on his earnings in a mutual
昀甀nd his grand昀愀ther recommends?
4-128 A student is buying a new car. 吀栀e car's price
is $16,500, the sales tax is 8%, and the title,
licence, and registration fee are $450 to be paid
in cash. 吀栀e dealer o昀昀ered to 昀椀nance 90% of the
car's price 昀漀r 48 months at a nominal interest
rate of 9% a year, compounded monthly.
(a) How much cash is paid when the car is
purchased?
(b) How much is the monthly payment?
4-129 Brad will graduate next year. When he begins
working, he plans to deposit $3,000 at the end
of each year into an RRSP retirement account.
If the account pays 4% interest, how much will
be in his account a昀琀er 40 deposits?
4-130 How much money should Timothy and
Ti昀昀any deposit annually 昀漀r 20 years in order
to provide an income of $30,000 a year 昀漀r the
next 10 years? Assume the interest rate is a
constant 4%.
4-131 For some interest rate i and some number of
interest periods n, the uni昀漀rm series capital
recovery 昀愀ctor is 0.1408 and the sinking 昀甀nd
昀愀ctor is 0.0408. What is the interest rate? What
is n?
Problems
4-132 It is estimated that the maintenance cost on a new
car will be $400 the 昀椀rst year. Each subsequent
year, this cost is expected to increase by $100.
How much would 礀漀u need to set aside when 礀漀u
bought a new car to p愀礀 all 昀甀ture maintenance
costs if 礀漀u planned to keep the car 昀漀r seven
years? Assume interest is 5% per annum.
4-133 吀栀e 昀椀rst of a series of equal, monthly cash
flows of $2,000 occurred on 1 April 2016 and
the last of the monthly cash 昀氀ows was made on
1 February 2018. 吀栀is series of monthly cash
flows is equivalent to a series of semi-annual
cash 昀氀ows. 吀栀e 昀椀rst semi-annual cash 昀氀ow was
made on 1 July 2019, and the last semi-annual
cash 昀氀ow will be made on 1 January 2028.
What is the amount of each semi-annual cash
flow? Use a nominal interest rate of 12% with
monthly compounding on all accounts.
M i n i - Cases
4-134 Assume that you plan to retire 40 years 昀爀om
now and that you need $2M to support the life­
style that you want.
139
(a) If the interest rate is 10%, is the 昀漀l­
lowing statement approximately true?
"Waiting 昀椀ve years to start saving
doubles what you must deposit each
year.,,
(b) If the interest rate is 12%, is the required
multiplier higher or lower than 昀漀r the 5%
rate in (a)?
(c) At what interest rate is the 昀漀llowing
statement exactly true? "Waiting 昀椀ve
years to start saving doubles what you
must deposit each year."
4-135 For winners of the Cali昀漀rnia SuperLotto
Plus, the choice is between a lump sum and
annual payments that increase 昀爀om 2.5%
昀漀r the 昀椀rst year, to 2.7% 昀漀r the second year,
and then increase by 0.1% a year to 5.1% 昀漀r
the 26th payment. 吀栀e lump sum is equal
to the net proceeds of bonds purchased to
昀甀nd the 26 payments. 吀栀is is estimated at
45% to 55% of the lump sum amount. At what
interest rate is the present worth of the two
payment plans equivalent if the lump sum is
45%? If it is 55%?
Present Worth Analysis
The Clock of the Long Now
The Clock of the Long Now project is an attempt to build a clock that will run for 10,000 years. It
was first conceived by Daniel Hillis, inventor of the Connection Machine; Stewart Brand, creator of
the Whole Earth Catalogue; and Brian Eno, the musician. The clock will be built in a remote desert
area, reachable only after an arduous hike. It will run with minimal human intervention and will
be built of nearly valueless materials. Its power source cannot be fossil fuels or batteries, since a
10,000-year stockpile of either would be a valuable resource, likely to attract thieves in the same
way as the Pyramids of Egypt attracted looters. Instead, it will get its power from the temperature
difference between day and night. This will allow it to keep track of the time; visitors will also be
able to wind the clock to power its chimes.
The intent of this project is to foster thinking and planning on very long time scales. The eco­
nomic tools covered in this textbook do not appear to work well when extended to time periods
The I n terva l, The Long N ow Foundation, Because We Can Desi g n - B u i ld Studio
The Clock of the Long N ow
141
beyo nd a cent u ry. As we saw i n a p revious cha pter, the p resent worth of bi llions of dolla rs worth
of d a m a g e occ u r r i n g as little as a m i lle n n i u m in the futu re, as calcu lated by the form u la s we have
d evelo ped, i s neg l i g i bly sma ll. I f we wa nt to a c h i eve lo ng -term susta i na b i lity, q u ite d i ffe rent ways
of th i n ki n g will have to be d eveloped . The Long N ow Fou ndation has beg u n to explore such ways
in a series of lectu res, Semina爀猀 about Long-Term 吀栀inking.
A story is told conce r n i n g N ew College at Oxfo rd U n ive rs ity that i llustrates the k i n d of th i n k ­
i n g req u i red . N ew College, desp ite its na me, was fou n ded more t h a n five centu ries a g o . W h e n i t
was fi rst b u i lt, its d i n i ng h a ll was roofed with massive oa k bea ms. B y t h e twentieth centu ry, th ese
bea ms had beco me i nfested with beetles a n d were in need of re placement. The Fellows of the
College wo ndered where they m i g ht fi nd oaks of s u ita ble d i mensions to yield new beams, most of
E n g la n d 's forests havi ng been felled d u ri n g the i nterve n i n g centuries. So they sent fo r the college
fo reste r, a man c h a rged with m a i nta i n i n g the la nds belo n g i n g to the college, who told them that he
a n d h i s p redecessors had been a ntici pati ng this q u estion s i n ce 1379, and had a sta nd of 60 0-year­
old oa ks, pla n ted i n 1379 a n d now ready fo r h a rvesting, that wou ld provi d e a re place ment set of
bea ms. ( I t is not recorded whether the college then pla nted a fresh set of acorns i n a ntici pation of
the re place ment that will come due i n 2579.)
QU ESTIONS TO CONSI DER
1.
Ass u m i n g that the va lue of a new college ce i li n g i n 1979 i s £1 m i llion, and ass u m i ng that N ew College ca n
borrow money at 5% i nterest, what is the most it wou ld have been worth spend i n g i n 1379 to e n s u re that a
re place ment ce i l i n g wou ld be ava i la b le six h u n d red yea rs later?
2.
Ass u m i n g that the va lue of a new college ce i l i n g in 2579 will be £1 m i llion, what is the most it wou ld have
been worth spend i n g in 1979 to e n s u re that a re place ment ceiling wou ld be ready in 2579?
LEARN I NG OBJ ECTIVES
This chapter will help you
•
defi n e the present worth a n d future worth criteria
•
use these criteria to choose between a lternatives
•
a p p ly the criteria i n cases with eq u a l, u n eq u a l, a n d i nfi n ite p roj ect lives
KEY TERMS
analysis period
ca pitalized cost
昀椀 nancing
futu re worth analysis
investment
plan n ing horizon
present worth analysis
142
Cha pter 5 Present Worth Analysis
In Chapters 3 and 4 we presented the concept of equivalence. 吀栀at's important because
we can compare cash 昀氀ows only if we can resolve them into equivalent values. Second, we
derived a series of compound interest 昀愀ctors to 昀椀nd those equivalent values.
Assumptions in Solving Economic Analysis Problems
One of the di昀케culties of solving problems is that most problems tend to be very compli­
cated. It becomes apparent that some simplifying assumptions are needed to make such
problems manageable. 吀栀e trick, of course, is to solve the simpli昀椀ed problem and still be
satis昀椀ed that the solution is applicable to the real problem. In the subsections that 昀漀llow, we
will consider 昀椀ve di昀昀erent situations and explain the customary assumptions that are made.
吀栀ese assumptions apply to all problems and examples unless other assumptions are given.
End-of-Yea r Convention
As we said in Chapter 4, economic analysis textbooks and practice 昀漀llow the end-of­
period convention. 吀栀is makes '✀䄀" a series of end-of-period receipts or disbursements.
(We generally assume in problems that all series of receipts or disbursements occur at the
end of the interest period. 吀栀is allows us to use values 昀爀om our compound interest tables
without any adjustments.)
A cash flow diagram of P, A, and F 昀漀r the end-of-period convention is as 昀漀llows:
Year 0
1 Jan.
End of Year 2
3 1 Dec.
End of Year 1
3 1 Dec. 1 Jan.
!
i
lA
A
p
t
F
If one were to adopt a middle-of-period convention, the diagram would be
Year 0
1 Jan.
!
p
Middle of
Year 1
30 Jun.
i
3 1 Dec. 1 Jan.
I
A
Middle of
Year 2
30 Jun.
i
A
End of Year 2
3 1 Dec.
l
F
As the diagrams illustrate, only A shi昀琀s; P remains at the beginning of period and F at the
end of period, regardless of the convention. Compound interest tables are based on the
end-of-period convention.
Viewpoint of Economic Analysis Studies
When we make economic analysis calculations, we must proceed 昀爀om a point of refer­
ence. Generally, we will want to take the point of view of a total 昀椀rm when doing industrial
economic analyses. Example 1-1 illustrates the problem vividly: a 昀椀rm's shipping depart­
ment decided it could save money by having its printing work done outside rather than by
the in-house printing department. An analysis 昀爀om the viewpoint of the shipping depart­
ment supported this, since it could get 昀漀r $688.50 the same print job that it was paying
$793.50 昀漀r in-house. Further analysis showed, however, that the printing department's
costs would decline less than the amount the 昀椀rm would save by using the commercial
printer. From the viewpoint of the 昀椀rm, the net result would be an increase in total cost.
Assumptions in Solving Economic Analysis Problems
From Example 1-1 we see it is important that the viewpoint of the study be care昀甀lly
considered. Selecting a narrow viewpoint, like that of the shipping department, may result
in a suboptimal decision 昀爀om the 昀椀rm's viewpoint. It is the viewpoint of the total 昀椀rm
that is used in industrial economic analyses. For public sector problems the combined
viewpoint of the government and the citizens is chosen, because 昀漀r many public projects
the bene昀椀ts of 昀愀ster commuting, newer schools, and so on, are received by individuals and
the costs are paid by the government.
䌀䐀
Borrowed Money Viewpoint
In most economic analyses, the proposed alternatives require money to be spent, and so it is
natural to ask about the source of that money. 吀栀us each problem has two monetary aspects:
one is the 昀椀nancing-the obtaining of money; the other is the investment-the spending
of money. 吀栀ese two things should be distinguished 昀爀om each other. When separated, the
problems of obtaining money and of spending it are both logical and straight昀漀rward. Fail­
ure to separate them sometimes produces con昀甀sing results and poor decisions.
吀栀e conventional assumption in economic analysis is that the money required to
昀椀nance projects is obtained at interest rate i.
I ncome Taxes
Income taxes, like in昀氀ation and deflation, must be considered in order to 昀椀nd the real
payo昀昀 of a project. However, taxes will o昀琀en a昀昀ect alternatives in the same way, allowing
us to compare our choices without considering income taxes. So we will put o昀昀 our intro­
duction of income taxes until Chapter 12.
E昀昀ect of I nflation and Deflation
Until we get to Chapter 14, we will assume that prices are stable. 吀栀is means that a ma­
chine that costs $5,000 today can be expected to cost the same several years hence.
Economic Criteria
We have established that, to compare cash 昀氀ows, we must 昀椀rst move them to the same
moment in time. We can choose to move them all to the present, giving us present worth
analysis; to some 昀甀ture date, giving us 昀甀ture worth analysis; or to an equivalent series
of equal cash 昀氀ows at regular intervals, giving us annual cash 昀氀ow analysis (Chapter 6).
Each of these methods, and the rate of return analysis we will encounter in Chapter 7,
will always yield the same recommendation 昀漀r choosing the best alternative(s) among a
set. Some problems, however, may be solved more naturally by one method than another.
We now 昀漀cus on the kinds of problems that are most readily solved by present worth
analysis.
Present Worth Techniques
One of the easiest ways to compare mutually exclusive alternatives is to resolve their con­
sequences to the present time. Care昀甀l consideration must be given to the time period cov­
ered by the analysis. Usually the task to be accomplished has a time period associated with
it. 吀栀e consequences of each alternative must be considered 昀漀r this period of time, which
is usually called the analysis period, planning horizon, or project life.
143
144
Cha pter 5 Present Worth Analysis
吀栀e analysis period 昀漀r an economic study should be determined 昀爀om the situation.
In some industries with rapidly changing technologies, a rather short analysis period or
planning horizon might be in order. Industries with more-stable technologies (like steel­
making) might use a longer period (say, 10 or 20 years), whereas government agencies
o昀琀en use analysis periods extending to 50 years or more. Ensuring long-term sustainabil­
ity requires a planning horizon of centuries or millenniums; humanity, collectively, has
had very little experience planning over such long periods.
吀栀ree di昀昀erent analysis-period situations are encountered in economic analysis prob­
lems with more than one alternative:
1. 吀栀e use昀甀l life of each alternative equals the analysis period.
2. 吀栀e alternatives have use昀甀l lives di昀昀erent 昀爀om the analysis period.
3. 吀栀ere is an in昀椀nite analysis period, n = 漀漀.
1. Usefu l Lives Eq ual to the Ana lysis Period
-- - - -- --
Since di昀昀erent lives and an in昀椀nite analysis period present some complications, we will
begin with two examples in which the use昀甀l life of each alternative equals the analysis
period.
EXAM PLE S-1
A city plans to build an aqueduct to bring water in 昀爀om the mountains in the west. 吀栀e aqueduct can be
built now at a reduced size 昀漀r $300 million and enlarged 25 years hence 昀漀r an additional $350 million.
An alternative is to construct the 昀甀ll-sized aqueduct now 昀漀r $400 million.
Both alternatives would provide the needed capacity 昀漀r the SO-year analysis period. Maintenance
costs are small and may be ignored. At 6% interest, which alternative should be selected?
SO LUTI O N
吀栀is problem illustrates staged construction. 吀栀e aqueduct may be built in a single stage or in a smaller
昀椀rst stage 昀漀llowed many years later by a second stage to provide the additional capacity when needed.
For the Two-Stage Construction
PW of cost = $300 million + $350 million(P/F, 6%, 25)
= $300 million + $81.6 million
= $381 .6 million
For the Single-Stage Construction
PW of cost = $400 million
吀栀e two-stage construction has a smaller present worth of cost and is there昀漀re the preferred construc­
tion plan.
Present Worth Techniques
145
EXAM PLE S - 2
A 昀椀rm is considering which of two mechanical devices to install to reduce costs. Both devices have
use昀甀l lives of 昀椀ve years and no salvage value. Device A costs $1,000 and can be expected to result in
$300 savings annually. Device B costs $1,350 and will provide cost savings of $300 the 昀椀rst year, but
the savings will increase by $50 annually, making the second-year savings $350, the third-year savings
$400, and so 昀漀rth. With interest at 7%, which device should the 昀椀rm purchase?
SO LUTI O N
吀栀e analysis period can b e conveniently chosen a s the use昀甀l life of the devices, which i s 昀椀ve years. 吀栀e
appropriate decision criterion is to maximize the net present worth of bene昀椀ts minus costs.
!
t
Device B
Device A
t t t t
A = 3 00
0 -- 1 -- 2 -- 3 -- 4 -- 5
P = 1 ,000
n = 5 years
t
3 00
!
t
3 50
t T T
400
0 -- 1 -- 2 -- 3 -- 4 -- 5
n = S years
P = 1 , 350
PWA = -1,000 + 300(䤀鄀, 7%, 5) = -1,000 + 300(4.100) = $230
PW8 = -1,350 + 300(PIA, 7%, 5) + 50(PIG, 7%, 5)
= -1,350 + 300(4.100) + 50(7.647) = $262.4
Device B has the larger present worth and is the preferred alternative.
Whenever we are comparing alternatives, we can neglect any costs or bene昀椀ts common
to all alternatives. In Example 5-1, both alternatives provided the same bene昀椀t, so we based
our decision on the comparative present worth of their cost. In Example 5-2, we chose
the alternative in which present worth of bene昀椀ts minus present worth of cost was a max­
imum. 吀栀is criterion is called the net present worth criterion and is written simply as NPW:
Net present worth = Present worth of bene昀椀ts - Present worth of cost
NPW = PW of bene昀椀ts - PW of cost
(5-1)
吀栀e 昀椀eld of engineering economy and this text use present worth (PW), present value
(PV), net present worth (NPW), and net present value (NPV) as synonyms. Sometimes, as in
the de昀椀nition above, net is included to emphasize that both costs and bene昀椀ts are included.
2 . Usefu l Lives Di昀昀erent from the Ana lysis Period
In present worth analysis, the analysis period must always be speci昀椀ed. It 昀漀llows, then,
that each alternative must be considered 昀漀r the entire period. In Examples 5-1 and 5-2, the
use昀甀l life of each alternative was equal to the analysis period. Although that is o昀琀en true,
146
Cha pter 5 Present Worth Analysis
in many situations at least one alternative will have a use昀甀l life di昀昀erent 昀爀om the analysis
period. 吀栀is section describes one w愀礀 to evaluate alternatives with lives di昀昀erent 昀爀om the
study period. Consider the 昀漀llowing example:
EXAM PLE S-3
A purchasing agent is considering the purchase of some new equipment 昀漀r the mailroom. Two di昀昀er­
ent manu昀愀cturers have provided quotations. An analysis of the quotations shows the 昀漀llowing:
Manu昀愀cturer
Cost
Use昀甀l Life
(years)
Speedy
Allied
$1,500
1,600
5
10
End-of-Use昀甀l-Life
Salvage Value
$200
325
吀栀e equipment of both manu昀愀cturers is expected to per昀漀rm at the desired level of (昀椀xed) output.
Which manu昀愀cturer's equipment should be selected? Assume 7% interest and equal maintenance costs.
SO LUTI O N
吀栀e Speedy and the Allied equipment will both provide the same bene昀椀t, but the Allied equipment will
provide it 昀漀r 10 years, while the Speedy will only provide it 昀漀r 昀椀ve. One way of making a 昀愀ir com­
parison would be to suppose that, if we buy the Speedy, we can replace the equipment a昀琀er 昀椀ve years at
the same cost. 吀栀at is, we select an analysis period that is the least common multiple of the lives of the
two alternatives. Our two alternatives now both o昀昀er the same bene昀椀t over a 10-year period, so we can
simply choose the one that does this more cheaply.
Speedy
Allied
PW of cost = $1,500 + (1,500 - 200)(P/F, 7%, 5) - 200(PIF, 7%, 10)
= 1,500 + 1,300(0.7130) - 200(0.5083)
= 1,500 + 927 - 102 = $2,325
PW of cost = 1,600 - 325(PIF, 7%, 10) = 1,600 - 325(0.5083)
= 1,600 - 165 = $1,435
Since it is only the d昀케erences between alternatives that are relevant, maintenance costs may be le昀琀 out
of the economic analysis. So we should buy the Allied equipment.
We have seen that setting the analysis period equal to the least common multiple of the
lives of the two alternatives seems reasonable in Example 5-3. However, what if the alterna­
tives had use昀甀l li瘀攀s of seven and 13 礀攀ars? Here the least common multiple of lives is 91 礀攀ars.
An analysis period of 91 years hardly seems realistic. Instead, a suitable analysis period should
be based on how long the equipment is likely to be needed. 吀栀is may require that terminal
values be estimated 昀漀r the alternatives at some point be昀漀re the end of their use昀甀l lives.
Present Worth Techniques
Alternative 1
Salvage
Value
t
l
l
147
Terminal Value at
End of Year 10
t
!
0 -- 1 -- 2 -- 3 -- 4 -- S-- 6 -- 7 -- 8 -- 9 -- 10 -- 1 1 -- 12 -- 13 -- 14
Replacement
Initial
1
Cost
Cost
椀ⴀ------ 7 - Year Life -------娀阀------- 7 - Year Life ------ⴀ椀
Alternative 2
l
Terminal Value at
End of Year 10
t
!
!
0 -- 1 -- 2 -- 3 -- 4 -- S-- 6 -- 7 -- 8 -- 9 -- 10 -- 1 1 -- 12 -- 13
Initial
Cost
I:
13-Year Life
1
I
椀ⴀ-------- 10-Year Analysis Period --------蜀蠀
FIGURE 5-1 Su peri mposing a 10 -yea r analysis period on 7- and 13 -year alternatives.
As Figure 5-1 shows, it is not necessary 昀漀r the analysis period to equal the use昀甀l life
of an alternative or some multiple of the use昀甀l life. To re昀氀ect the situation properly at the
end of the analysis period, an estimate is required of the market value of the equipment at
that time.
EXAM PLE S-4
A diesel manu昀愀cturer is considering the two alternative production machines depicted in Figure 5-1.
Speci昀椀c data are as 昀漀llows:
Alt. 1
Initial cost
Estimated salvage value at end of useful li昀攀
Use昀甀l life of equipment, in years
Alt. 2
$50,000 $75,000
$10,000 $12,000
7
13
吀栀e manu昀愀cturer uses an interest rate of 8% and wants to use the PW method to compare these al­
ternatives over an analysis period of 10 years.
Alt. I
Alt. 2
Estimated market value, end of 10-year analysis period $20,000 $15,000
continued
148
Cha pter 5 Present Worth Analysis
SO LUTI O N
In this case, the decision maker is setting the analysis period at 10 years rather than accepting a common
multiple of the lives of the alternatives. 吀栀is is a legitimate approach-perhaps the diesel manu昀愀cturer
will be phasing out this model at the end of the 10-year period. In any event, we need to compare the
alternatives over the 10 years.
As illustrated in Figure 5-1, we may assume that Alternative 1 will be replaced by an identical
machine a昀琀er its seven-year use昀甀l life. Alternative 2 has a 13-year use昀甀l life. 吀栀e diesel manu昀愀cturer
has provided an estimated market value of the equipment at the time of the analysis period. We can
compare the two choices over 10 years as 昀漀llows:
PW (Alt 1 .) = -50,000 + (10,000 - 50,000)(PIF, 8%, 7) + 20,000(PIF, 8%, IO)
= -50,000 - 40,000(0.5835) + 20,000(0.4632)
= -$64,076
PW (Alt 2.) = -75,000 + 15,000(P/F, 8%, 10)
= -75,000 + 15,000(0.4632)
= -$68,052
To minimize PW of costs, the diesel manu昀愀cturer should choose Alternative 1.
䌀䐀
3. I n昀椀 n ite Ana lysis Period : Ca pitalized Cost
Another di昀케culty in present worth analysis arises when we encounter an in昀椀nite analysis
period (n = 漀漀). In governmental analyses, a service or condition must sometimes be main­
tained 昀漀r an inde昀椀nitely long period. 吀栀e need 昀漀r roads, dams, pipelines, and so on, is
sometimes considered to be permanent. In these situations a present worth of cost analysis
would have an in昀椀nite analysis period. We call this particular analysis capitalized cost.
In昀椀nite lives are rare in the private sector, but a similar assumption of "inde昀椀nitely
long" horizons is sometimes made. 吀栀is assumes, 昀漀r example, that the 昀愀cility will need
electric motors, mechanical HVAC equipment, and 昀漀rkli昀琀s as long as it operates and that
the life of the 昀愀cility is 昀愀r longer than this equipment. So the equipment can be analyzed
as though the problem horizon were in昀椀nite or inde昀椀nitely long.
Capitalized cost is the present sum of money that would need to be set aside now, at
some interest rate, to yield the 昀甀nds required to provide the service (or whatever) inde昀椀n­
itely. To accomplish this, the money set aside 昀漀r 昀甀ture expenditures must not decline. 吀栀e
interest received on the money set aside may be spent, but the principal may not. When we
stop to think about an in昀椀nite analysis period (as opposed to something relatively short,
like a hundred years), we see that an undiminished principal sum is essential; otherwise
one will run out of money be昀漀re eternity.
In Chapter 3 we saw that
Principal sum + Interest 昀漀r the period = Amount at end of period, or
W
P
P+W
+
Present Worth Techniques
149
If we spend iP, in the next interest period the principal sum P will again increase to
P + iP. 吀栀us, we can again spend iP.
吀栀is concept may be illustrated by John Dashwood's problem 昀爀om the beginning
of Chapter 4. Suppose Mr Dashwood deposits £1,500 in an account that pays 10%. How
much money can he withdraw each year without reducing the balance in the account
below the initial £1,500? At the end of the 昀椀rst year, the £1,500 would have earned
10%(£1,500) = £150 interest. If Mr Dashwood withdraws this interest and pays it to
his half-sister, the £1,500 will remain in his account. At the end of the second year, the
£1,500 balance would again earn 10%(£1,500) = £150. 吀栀is £150 could also be with­
drawn and the account would still have £1,500. 吀栀is procedure could be continued in­
de昀椀nitely, and the bank account would always contain £1,500 at the beginning of each
year. If more or less than £150 is withdrawn, the account will either increase toward 漀漀
or decrease to 0.
吀栀us, 昀漀r any initial present sum P, there can be an end-of-period withdrawal of A
equal to iP each period, and these withdrawals can continue 昀漀rever without diminishing
the initial sum P. 吀栀is gives us the basic relationship
For n = 漀漀,
A = Pi
吀栀is relationship is the key to capitalized cost calculations. Earlier we de昀椀ned capital­
ized cost as the present sum of money that would need to be set aside at some interest rate
to yield the 昀甀nds to provide the desired task or service 昀漀rever. Capitalized cost is there­
昀漀re the P in the equation A = iP. It 昀漀llows that
(5-2)
Capit氀쨀zed cost P = �
t
If we can resolve the desired task or service into an equivalent A, the capitalized cost
can be computed. 吀栀e 昀漀llowing examples illustrate such computations.
EXAM PLE S - 5
How much should you set aside to pay $50 a year 昀漀r maintenance on a gravesite if interest is assumed
to be 4%? For perpetual maintenance, the principal sum must remain undiminished a昀琀er the annual
disbursement is made.
SO LUTI O N
Annu愀氀
disbursement A
. 愀氀ized cost p = -------Cap1t
Interest rate i
50
P = - = $ 1,250
0.04
You should set aside $1,250.
150
Cha pter 5 Present Worth Analysis
EXAM PLE S - &
A city plans a pipeline to transport water 昀爀om a distant area to the city. 吀栀e pipeline will cost $8 million
and will have an expected life of 70 years. 吀栀e city expects it will need to keep the water line in service
inde昀椀nitely. Compute the capitalized cost, assuming 7% interest.
SO LUTI O N
吀栀 e capitalized cost equation
P= A
i
is simple to apply when there are end-of-period disbursements A. Here we have renewals of the pipeline
every 70 years. To compute the capitalized cost, it is 昀椀rst necessary to compute an end-of-period dis­
bursement A that is equivalent to $8 million every 70 years.
!
70 years
140 years
n = 漀漀
$8 million
$8 million
$8 million
!
0 -- = 0
$8 million
!
!
Capitalized Cost
吀栀e $8 million disbursement at the end of 70 years may be resolved into an equivalent A.
o-r �
洀洀nmpm洀洀
n = 70
$8 million
A = F(AIF, i, n) = $8,000,000(AIF, 7%, 70)
= $8,000,000(0.00062) = $4,960
Each 70-year period is identical to this one, and the in昀椀nite series is shown in Figure 5-2.
Capit愀氀ized cost P = $8, 000 , 000 +
= $8, 071, 000
A
= $8, 000 , 000 + 4, 960
i
0. 07
r-. �1mmmmn�nmnmmmn • • mnmn爀渀
Cap1tal1zed
Cost
$8 million
A - 4 , 9 60
FIGURE 5-2 Using the sinking fu nd factor to compute an in昀椀nite series.
ALTERNATIVE SO LUTI O N
Another way o f solving the problem i s to assume the interest period i s 70 years long. Compute an
equivalent interest rate 昀漀r the 70-year period. 吀栀en the capitalized cost may be computed by using
Equation 3-8 昀漀r m = 70
Present Worth Techniques
151
i70 礀爀 = (1 + i1 礀爀 ) 7° - 1 = (1 + 0 . 0 7) 7° - 1 = 1 12.989
$8,000,000
. ⼀d cost = $ 8,000,000 + ---Cap1t
= $ 8,0 71,000
1 12.989
Multiple Alternatives
So 昀愀r the discussion has been based on examples with only two alternatives. But multiple­
alternative problems m愀礀 be solved by exactly the same methods. (吀栀e only reason 昀漀r
愀瘀oiding multiple alternatives was to simplify the examples.) Example 5-7 has multiple
alternatives.
EXAM PLE 5-7
A contractor has been awarded the contract to construct a 6 km tunnel in the mountains. During the
昀椀ve-year construction period, the contractor will need water 昀爀om a nearby stream. She will construct
a pipeline to carry the water to the main construction yard. An analysis of costs 昀漀r various pipe sizes
is as 昀漀llows:
Pipe Sizes (cm)
7
5
Installed cost of pipeline and pump
Cost per hour 昀漀r pumping
9
11
$22,000 $23,000 $25,000 $30,000
$0.65
$0.50
$0.40
$1.20
吀栀e pipe and pump will have a salvage value at the end of 昀椀ve years equal to the cost of removing them.
吀栀e pump will operate 2,000 hours a year. 吀栀e lowest interest rate at which the contractor is willing to
invest money is 7%. Select the alternative with the least present worth of cost.
SO LUTI O N
We can compute the present worth of cost 昀漀r each alternative. For each size of pipe, the present worth
of cost is equal to the installed cost of the pipeline and pump plus the present worth of 昀椀ve years of
pumping costs.
Pipe Size (cm)
5
Installed cost of pipeline and pump
1.20 x 2,000 hr x (PIA, 7%, 5)
0.65 x 2,000 hr x 4.100
0.50 x 2,000 hr x 4.100
0.40 x 2,000 hr x 4.100
Present worth of cost
$22,000
9,840
$31,840
7
9
11
$23,000 $25,000 $30,000
5,330
4,100
3,280
$28,330 $29,100 $33,280
吀栀us, to minimize the present worth of cost, choose the 7 cm pipe.
152
Cha pter 5 Present Worth Analysis
EXAM PLE S - a
吀眀o pieces of construction equipment are being analyzed:
Year
Alt. A
Alt. B
0
1
2
3
4
5
6
7
8
- $2,000
+ 1 ,000
+850
+700
+550
+400
+400
+400
+400
- $ 1 ,500
+700
+300
+300
+300
+300
+400
+500
+600
At an 8% interest rate, which alternative should be chosen?
SO LUTI O N
---t--- -f- -=
Alternative A
t-
l , O O�- - - - 嬀娀 - - - - -g 50I
- - -700
- - 550
r
- l - _ !r - - -,L _ ,,
0 -- l -- 2 -- 3 -- 4 -- 5 -- 6 -- 7 -- 8
2,000
PW of bene昀椀ts = 400(P/A, 8%, 8) + 600(䤀鄀, 8%, 4) - 150(P/G, 8%, 4)
= 400(5.㜀㐀7) + 600(3.312) - 150(4.650) = 3,588.50
PW of cost = 2,000
Net present worth = 3,588.50 - 2,000 = +$1,588.50
700
- -1
---1----f---i----r--t---f---Alternative B
3 00
3 00
3 00
500
4
3 00 - - 00- - - - -
6 00
}
o -- 1 -- 2 -- 3 -- 4 -- 5 -- 6 -- 7 -- 8
1 ,500
Present Worth Techniques
153
PW of bene昀椀ts = 300(P/A, 8%, 8) + (700 - 300)(P/F, 8%, 1)
+ lO0(P/G, 8%, 4)(P/F, 8%, 4)
= 300(5.㜀㐀7) + 400(0.9259) + 100(4.650)(0.7350)
= 2,436.24
PW of cost = 1,500
Net present worth = 2,436.24 - 1,500
= $936.24
To maximize NPW, choose Alternative A.
Bond Pricing
吀栀e calculation in Example 5-9 is done routinely when bonds are bought and sold during
their lives. Bonds are issued at a 昀愀ce, or par, value (usually $1,000), which is received when
the bonds mature. 吀栀ere is a coupon interest rate, which is set when the bond is origin­
ally issued or sold. 吀栀e term co甀瀀on interest dates 昀爀om the time when bonds were paper
rather than electronic, and a paper coupon was detached 昀爀om the bond to be redeemed in
cash. 吀栀e annual interest paid equals the coupon rate times the 昀愀ce value. 吀栀is interest is
usually paid in two semi-annual payments, although some bonds have other compound­
ing periods. A cash flow diagram 昀漀r the remaining interest payments and the 昀椀nal 昀愀ce
value is used with a current interest rate to calculate a price.
EXAM PLE S - 9
A 15-year municipal bond was issued 昀椀ve years ago. Its coupon interest rate is 8%, interest payments
are made semi-annually, and its 昀愀ce value is $1,000. If the current market interest rate is 12.36%, what
should the price of the bond be? Note: 吀栀e issuer of the bond (a government or company) makes interest
payments to the bondholder (at the coupon rate), as well as a 昀椀nal value payment.
SO LUTI O N
吀栀e 昀椀rst 昀椀ve years are past, and there are 2 0 more semi-annual payments. 吀栀e coupon interest rate is the
nominal annual rate or APR. Half of that, 4% of$1,000 = $40, is paid at the end of each six-month period.
吀栀e price of the bond is the PW of the cash 昀氀ows that will be received if the bond is purchased. 吀栀e
cash 昀氀ows are $40 at the end of each of the 20 semi-annual periods and the 昀愀ce value of $1,000 at the
end of period 20.
$ 40 + $ 1 ,000
t t t t t t t t
$40
t
t
0 -- l -- 2 -- 3 -- 4 -- S -- 6 Ĉࠇ 1 7 -- 1 8 -- 1 9 -- 20
continued
154
Cha pter 5 Present Worth Analysis
Since the $40 in interest is received semi-annually, the market or e昀昀ective annual interest rate (i) must
be converted to a semi-annual rate. Using Equation 3-8, we obtain
(1 + i) 2 = 1 + ia = 1 .1236
(1 + i) = 1 .06
i = 6% e昀昀ective semi-annual interest rate
PW = 40(P!A, 6%, 20) + 1,000(P/F, 6%, 20)
= 40(1 1 .470) + 1,000(0.31 18) = $770.60
吀栀e $770.60 is the discounted price, that is, the PW at 12.36% of the cash 昀氀ows 昀爀om the $1,000 bond.
吀栀e $229.40 discount raises the rate of return on the investment 昀爀om a nominal 8% 昀漀r the 昀愀ce value
to 12.36% on an investment of $770.60.
吀栀is example also illustrates why it is better to state cash 昀氀ows separately. At the end of period 20
昀爀om now, there are two cash 昀氀ows, $40 and $1,000. 吀栀e $40 is part of the 20-period uni昀漀rm series, and
the $1,000 is a single cash flow. All of these numbers come directly 昀爀om the problem statement. If the
two 昀椀nal cash 昀氀ows are combined into $1,040, then the $40 uni昀漀rm series has only 19 periods-and it
is easy to err and 昀漀rget that change.
Futu re Worth Analysis
In present worth analysis, alternatives are compared in terms of their present equivalent
value. We can also choose between alternatives by 昀椀nding their equivalent value at some
昀甀ture date. 吀栀is is called future worth analysis. It will immediately be apparent to the
reader that, if we are comparing two alternatives A and B and have just 昀漀und that PW(A) >
PW(B), then their 昀甀ture values a昀琀er n years, at an interest rate i, will be FW(A) = PW(A)
(PIP, i, n) and FW(B) = PW(B)(F/P, i, n). Since we are multiplying each present worth by
the same 昀愀ctor, it must be the case that FW(A) > FW(B); that is, 昀甀ture worth analysis will
always lead us to pick the same alternative as present worth analysis.
吀栀e reader may reasonably raise an objection at this point: if 昀甀ture worth analysis
is always going to give the same answer as the method we've just learned, why bother
learning it? 吀栀e answer is that it is a more natural way of thinking about certain types of
problem. For example, if we are setting aside money to provide ourselves with a pension
on retirement, we're interested in how much money we'll have when we retire. We don't
care what that amount of money would be worth now, since we're not going to be spending
it now.
EXAM PLE 5-10
Ron, a 20-year-old university student, smokes about a carton of cigarettes a week. He wonders how
much money he could accumulate by age 65 if he quit smoking now and put his cigarette money into a
savings account. Cigarettes cost $85 a carton. Ron expects that a s愀瘀ings account would earn 5% inter­
est, compounded semi-annually. Compute the 昀甀ture worth of Ron's savings at age 65.
Future Worth Analys is
SO LUTI O N
155
Semi-annual saving = ($85/carton)(26 weeks) = $2,210
Future worth (FW) = A(䤀蜀, 2.5%, 90) = 2,210(329.2) = $727,532
EXAM PLE 5-11
An East Coast 昀椀rm has decided to establish a second plant in the West. 吀栀ere is a 昀愀ctory 昀漀r sale 昀漀r
$850,000 that, with extensive remodelling, could be used. As an alternative, the company could buy
vacant land 昀漀r $85,000 and have a new plant constructed there. Either way, it will be three years be昀漀re
the company will be able to get a plant into production. 吀栀e timing and costs 昀漀r the 昀愀ctory are as 昀漀llows:
Construct New Plant
Year Component
0
1
2
3
Remodel Available Factory
Component
Cost
Buy land
Design
Construction costs
Production equipment
$85,000
200,000
1 ,200,000
200,000
Cost
Purchase 昀愀ctory
Design
Remodelling
Production equipment
$850,000
250,000
250,000
250,000
If interest is 8%, which alternative has the lower equivalent cost when the 昀椀rm begins production at the
end of Year 3?
SO LUTI O N
New Plant
!
0 -- 1 -- 2 -- 3
0 -- 1 -- 2 -- 3
s ,L
s
, ••
-----
t}
2 00 , 000
.
00 000 r
FW
FW of cost = 85,000(F/P, 8%, 3) + 200,000(F/A, 8%, 3) + 1,000,000(F/P, 8%, 1) = $1,836,000
Remodel Available Factory
! i
0 -- 1 -- 2 -- 3
850,000
2 5 0:000
i
- o ------- 3
FW
continued
156
Cha pter 5 Present Worth Analysis
FW of cost = 85,000(F/P, 8%, 3) + 250,000(F/A, 8%, 3)
= $1,882,000
吀栀e total cost of remodelling the available 昀愀ctory ($1,600,000) is smaller than the total cost of a new
plant ($1,685,000). However, the timing of the expenditures is better with the new plant. 吀栀e new plant
is projected to have the smaller 昀甀ture worth of cost and thus is the preferred alternative.
SUMMARY
Present worth and 昀甀ture worth analyses are suitable 昀漀r many economic analysis problems.
To make valid comparisons, we need to analyze each alternative in a problem over the same
analysis period or planning horizon. If the alternatives do not h瘀였 equal lives, some tech­
nique must be used to achieve a common analysis period. One method is to select an analysis
period equal to the least common multiple of the alternative lives. Another method is to select
an analysis period and estimate end-of-analysis-period salvage values 昀漀r the alternati瘀攀s.
Capitalized cost is the present worth of cost 昀漀r an in昀椀nite analysis period (n = 漀漀).
When n = 漀漀, the 昀甀ndamental relationship is A = iP. Some 昀漀rm of this equation is used
whenever there is a problem with an in昀椀nite analysis period.
吀栀e 昀漀llowing assumptions are routinely made in solving economic analysis problems:
1.
2.
3.
4.
5.
6.
Present sums P are beginning of period, and all series receipts or disbursements A
and 昀甀ture sums F occur at the end of the interest period. 吀栀e compound interest
tables were derived on this basis.
I n industrial economic analyses, the point o f reference 昀爀om which to compute
the consequences of alternatives is the total 昀椀rm. Taking a narrower view of the
consequences can result in suboptimal solutions.
Only di昀昀erences between alternatives are relevant. Past costs are sunk costs and
generally do not a昀昀ect present or 昀甀ture costs. For this reason they are ignored.
吀栀e investment problem is isolated 昀爀om the 昀椀nancing problem. We generally
assume that all money required is borrowed at interest rate i.
For now, 眀攀 assume no taxation and stable prices. Taxation is deferred to Chapter 12.
Similarly, the in昀氀ation-de昀氀ation problem is deferred to Chapter 14.
O昀琀en, uni昀漀rm cash 昀氀ows or arithmetic gradients are reasonable assumptions. How­
ever, spreadsheets simpli昀礀 the 昀椀nding of PW and FW in more complicated problems.
PROBLEMS
Present Value of One Alternative
5-1
Compute the present value, P, 昀漀r the 昀漀llowing
cash 昀氀ows.
I
120
t t t t t
60
60
60
60
60
0 -- 1 -- 2 -- 3 -- 4 -- 5
i = 10%
Problems
5-2
Compute the present value, P, 昀漀r the 昀漀llowing
cash 昀氀ows.
5-7
3 00
A stonecutter assigned to carve the headstone
昀漀r a well-known engineering economist began
with the 昀漀llowing design.
200
100
0
0 -- 1 -- 2 -- 3 -- 4
5-3
4G
3G
5G
2G
G
i = 12%
157
0 -- 1 -- 2 -- 3 -- 4 -- 5
Compute the present value, P, 昀漀r the 昀漀llowing
cash 昀氀ows.
rI
200
100
50
p
He then started the equation as 昀漀llows:
P = G(PIG, i, 6)
0 -- 1 -- 2 -- 3 -- 4
i = 10%
5-4
Compute the present value, P, 昀漀r the 昀漀llowing
cash 昀氀ows.
He realized he had made a mistake. 吀栀e equa­
tion should have been
I t t I
30
30
20
P = G(PIG, i, 5) + G(䤀鄀, i, 5)
20
吀栀e stonecutter does not want to discard the
stone and start over. He asks you to help him
with his problem. What one compound inter­
est 昀愀ctor can be added to make the equation
correct?
0 -- 1 -- 2 -- 3
i = 15%
5-5
Find the value of Q that makes the present
value = 0.
50
t
50
t
50
50
t
t
50
t
50
P = G(P/G, i, 6)(
t
0 -- 1 -- 2 -- 3 -- 4 -- 5
i = 1 2%
5-6
266.20
r ,r"T
242 Ⰰ堀
Ⰰ堀
∀✀
0 -- 1 -- 2 -- 3 -- 4
i = 15%
Write the complete equation.
5-8
Use a geometric gradient 昀漀rmula to compute
the present value, P, 昀漀r the 昀漀llowing cash 昀氀ows.
, i, )
Consider the 昀漀llowing cash flow. At a 6% in­
terest rate, what value of P at the end of Year 1
is equivalent to the bene昀椀ts at the end of
Years 2 to 7?
Year Cash Flow
1
2
3
4
5
6
7
- $P
1 00
200
300
400
500
600
158
5-9
Cha pter 5 Present Worth Analysis
A manu昀愀cturer is considering purchasing
equipment that will have the 昀漀llowing 昀椀nan­
cial e昀昀ects:
annual maintenance cost to $500 a year 昀漀r the
昀椀rst 昀椀ve years and to $1,000 a year 昀漀r the next
昀椀ve years. A昀琀er 10 years the annual mainten­
ance would again be $2,000. If maintenance
costs are the only saving, what investment can
be justi昀椀ed 昀漀r the new sur昀愀ce? Assume inter­
est at 4%.
Year Disbursements Receipts
0
1
2
3
$
0
880
1,980
2,420
1,760
$4,400
660
660
440
220
5-13
A road-building contractor has received a
m愀樀or highway construction contract that will
require 50,000 m3 of crushed stone each year
昀漀r 昀椀ve years. 吀栀e stone can be obtained 昀爀om a
quarry 昀漀r $5.80/m3 . As an alternative, the con­
tractor has decided to try to buy the quarry. He
believes that if he owned the quarry, the stone
would cost him only $4.30/m3 • He thinks he
could resell the quarry at the end of 昀椀ve years
昀漀r $200,000. If the contractor uses a 10% in­
terest rate, how much would he be willing to
pay 昀漀r the quarry?
5-14
IBP Inc. is considering establishing a new ma­
chine to automate a meat-packing process. 吀栀e
machine will save $50,000 in labour annually.
吀栀e machine can be bought 昀漀r $200,000 today
and will be used 昀漀r 10 years. It has a salvage
value of $10,000 at the end of its use昀甀l life. 吀栀e
new machine will require an annual main­
tenance cost of $9,000. 吀栀e corporation has a
minimum rate of return of 10%. Do you rec­
ommend automating the process?
5-15
A local o昀케ce supply 昀椀rm will agree to sell the
equipment to the 昀椀rm now and accept payment
according to the treasurer's schedule. If inter­
est is charged at 3% every two months, with
compounding once every two months, how
much o昀케ce equipment can the Miro Company
buy now?
A 昀椀rm has installed a manu昀愀cturing line 昀漀r
packaging materials. 吀栀e 昀椀rm plans to produce
50 tonnes of packing peanuts at $5,000 a tonne
annually 昀漀r 昀椀ve years, and then 80 tonnes of
packing peanuts a year at $5,500 a tonne 昀漀r
the next 昀椀ve years. What is the present worth
of the expected income? 吀栀e 昀椀rm's minimum
attractive rate of return is 18% per annum.
5-16
Annual maintenance costs 昀漀r a particular
section of highway pavement are $2,000. 吀栀e
placement of a new sur昀愀ce would reduce the
A wholesale company has signed a contract
with a supplier 昀漀r purchases worth $2,000,000
annually. 吀栀e 昀椀rst purchase will be made now
and will be 昀漀llowed by 10 more. Determine the
contract's present worth at a 7% interest rate.
5-17
A new o昀케ce building was constructed 昀椀ve
years ago by a consulting engineering 昀椀rm.
4
If money is worth 6%, should he invest in the
equipment?
5-10
5-11
5-12
In a present worth analysis of certain equip­
ment, one alternative has a net present worth
of +$420, based on a six-year analysis period
that equals the use昀甀l life of the alternative. A
10% interest rate was used in the computations.
吀栀e alternative device is to be replaced at
the end of the six years by an identical item
with the same cost, bene昀椀ts, and use昀甀l life.
Using a 10% interest rate, compute the net
present worth of the alternative equipment 昀漀r
the 12-year analysis period.
On 1 February, the Miro Company needs to
buy some o昀케ce equipment. 吀栀e company is
short of cash and expects to be short 昀漀r several
months. 吀栀e company treasurer has said that
he could pay 昀漀r the equipment as 昀漀llows:
Date
P愀礀ment
1 Apr.
1 June
1 Aug.
1 Oct.
1 Dec.
$150
300
450
600
750
159
Problems
At that time the 昀椀rm obtained a bank loan 昀漀r
$100,000 with a 12% annual interest rate, com­
pounded quarterly. 吀栀e loan is to be repaid in
10 years by equal quarterly payments. 吀栀e loan
can also be repaid at any time without penalty.
As a result of internal changes in the
昀椀rm, it is now proposed to re昀椀nance the loan
through an insurance company. 吀栀e new loan
would be 昀漀r a 20-year term with an interest
rate of 8% a year, compounded quarterly. 吀栀e
new equal quarterly payments would rep愀礀 the
loan in the 20-year period. 吀栀e insurance com­
pany requires the payment of a 5% loan initi­
ation charge (o昀琀en described as a "昀椀ve-point
loan fee"), which will be added to the new loan.
(a) What is the balance due on the original
mortgage if 20 payments have been made
in the last 昀椀ve years?
(b) What is the di昀昀erence between the
equal quarterly payments on the present
bank loan and the proposed insurance
company loan?
5-18
Argentina is considering constructing a bridge
across the Rio de la Plata to connect its northern
coast to the southern coast of Uruguay. If this
bridge is constructed, it will reduce the travel
time 昀爀om Buenos Aires, Argentina, to Sao Paulo,
Brazil, by over 10 hours and has the potential to
impro瘀攀 signi昀椀cantly the 昀氀ow of manu昀愀ctured
goods between the two countries. 吀栀e cost of the
new bridge, which will be the longest bridge in
the world, spanning over 80 kilometres, will be
$700 million. 吀栀e bridge will require an annual
maintenance of $10 million 昀漀r repairs and up­
grades and is estimated to last 80 years. It is es­
timated that 550,000 vehicles will use the bridge
during the 昀椀rst year of operation, and an addi­
tional 50,000 vehicles a year until the 10th year.
吀栀ese data are based on a toll charge of $90 per
vehicle. 吀栀e annual tra昀케c 昀漀r the rest of the life
of the bridge will be one million vehicles a year.
吀栀e Argentine government requires a minimum
rate of return of 9% to proceed with the project.
(a) Does this project provide su昀케cient revenues to o昀昀set its costs?
(b) What considerations are there besides
economic 昀愀ctors in deciding whether to
construct the bridge?
5-19
A student has a job that leaves her with $500
a month in disposable income. She decides
that she will use the money to buy a car. Be昀漀re
looking 昀漀r a car, she arranges a 100% loan
whose terms are $250 a month 昀漀r 36 months
at 18% annual interest. What is the maximum
purchase price that she can a昀昀ord with her
loan?
5-20
吀栀e student in Problem 5-19 昀椀nds a car she
likes, and the dealer o昀昀ers to arrange 昀椀nan­
cing. His terms are 12% interest 昀漀r 60 months
and no down p愀礀ment. 吀栀e car's sticker price is
$24,000. Can she a昀昀ord to buy this car with her
$500 monthly disposable income?
0
5-21
0
5-22
吀栀e student in Problem 5-20 really wants this
particular car. She decides to try to negotiate
a di昀昀erent interest rate. What is the highest
interest rate that she can accept, given a 60month term and payments of $500 per month?
Compute the present value, P, 昀漀r the 昀漀llowing
cash flows.
120
120
120
t l t l t l
50
50
50
0 -- l -- 2 -- 3 -- 4 -- 5 -- 6 -- 7 -- 8 -- 9
i = 10%
5-23
If i = 10%, compute the present value, P, 昀漀r the
昀漀llowing cash 昀氀ows.
t t t t
B
B
B
B
o -- 1 -- 2 -- 3 -- 4 -- 5 -- 6 -- 7 -- . . .
n = 漀漀
5-24
Compute the present value, P, 昀漀r the 昀漀llowing
cash flows.
1,000
1,000
1,000
1,000
I T't" + I TT + I I
2
2
0 - l - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 -l l- 12 - 13 �　i = 8%
5-25
By installing some elaborate inspection equip­
ment on its assembly line, the Robot Corp. can
160
Cha pter 5 Present Worth Analysis
瘀퀀id hiring an extra worker who would have
earned $36,000 a year in wages and an addi­
tional $9,500 a year in employee bene昀椀ts. 吀栀e
inspection equipment has a six-year use昀甀l life
and no salvage value. Use a nominal 18% interest
rate in your calculations. How much can Robot
a昀昀ord to p愀礀 昀漀r the equipment if the wages and
worker bene昀椀ts would have been paid
愀儀 At the end of each year?
(b) Monthly?
(c) Explain why the answer in (b) is larger
than the answer in (a).
5-26
Julie, a young industrial engineer, prepared an
economic analysis 昀漀r some equipment to replace
one production worker. 吀栀e analysis showed
that the present worth of bene昀椀ts (of employing
one less production worker) just equalled the
present worth of the equipment costs, assuming
a 10-year use昀甀l life 昀漀r the equipment. It was de­
cided not to buy the equipment.
A short time later, the production work­
ers won a new three-year union contract that
granted them an immediate 40¢-an-hour wage
increase, plus an additional 25¢-an-hour wage
increase in each of the two subsequent years.
Assume that in each and every 昀甀ture year, a
25¢-an-hour wage increase will be granted.
Julie has been asked to revise her earlier
economic analysis. 吀栀e present worth of bene­
昀椀ts of replacing one production employee will
now increase. Assuming an interest rate of 8%,
by how much will the justi昀椀able cost of the
automation equipment (with a 10-year use昀甀l
life) increase? Assume the plant operates a
single eight-hour shi昀琀, 250 days a year.
If the interest rate is 6%, which alternative
should be selected?
5-28
If produced by Method A, the initial capital
cost of a product will be $100,000, its operat­
ing cost will be $20,000 a year, and its salvage
value a昀琀er three years will be $20,000. With
Method B there is a 昀椀rst cost of $150,000, an
operating cost of $10,000 a year, and a $50,000
salvage value a昀琀er its three-year life. Use a
present worth analysis at a 15% interest rate to
decide which method should be used.
5-29
Quinton's re昀爀igerator has just died. He can get
a basic re昀爀igerator or a more e昀케cient re昀爀iger­
ator with an Energy Star designation. Quinton
earns 4% compounded annually on his invest­
ments, he wants to consider a 10-year planning
horizon, and he will use present worth analysis
to determine the best alternative. What is your
recommendation?
Basic Ener最礀
Unit Star Unit
Initial cost
Delivery & installation
Professional servicing (Year 5)
Annual energy costs
Salvage value (Year 10)
0
吀眀o di昀昀erent companies are o昀昀ering a punch
press 昀漀r sale. Company A charges $250,000 to
deliver and install the device. Company A has
estimated that the machine will have main­
tenance and operating costs of $4,000 a year
and will provide an annual bene昀椀t of $89,000.
Company B charges $205,000 to deliver and
install the device. Company B has estimated
maintenance and operating costs of the press
at $4,300 a year, with an annual bene昀椀t of
$86,000. Both machines will last 昀椀ve years and
can be sold 昀漀r $15,000 昀漀r the scrap metal. Use
an interest rate of 12%. Which machine should
your company buy?
5-31
A battery-manu昀愀cturing plant has been or­
dered to cease discharging acidic waste liquids
containing mercury into the city sewers. As a
result, the 昀椀rm must now adjust the pH and
Two alternative courses of action have the 昀漀llowing schedules of disbursements:
Year
0
1
2
3
4
5
A
B
-$1,300
0 - $100
0 -200
0 -300
0
400
0
500
$1,300 $1,500
800
60
100
55
175
5-30
Lives Match
5-27
700
60
100
120
150
Problems
remove the mercury 昀爀om its waste liquids.
吀栀ree 昀椀rms have provided quotes on the ne­
cessary equipment. An analysis of the quotes
provided the 昀漀llowing table of costs.
Bidder
Fo砀栀氀픀
Instr甀洀e渀琀
Quicksilver
Almaden
Annual
Income
昀爀om
Annual
Inst氀촀ed Operating Mercury Salvage
Recovery V愀氀ue
Cost
Cost
$35,000
$8,000
$2,000
$20,000
40,000
100,000
7,000
2,000
2,200
3,500
0
0
If the installation can be expected to last
20 years and money is worth 7%, which equip­
ment should be purchased?
5-32
A new tennis court complex is planned, and
two alternative construction plans have been
created. Both alternatives will last 18 years, and
the interest rate is 7%. Use present worth an­
alysis to determine which should be selected.
Construction Annual Cost
A
B
5-33
5-34
$500,000
640,000
In order to improve evacuation routes out of
New Orleans in the event of another m愀樀or
disaster such as Hurricane Katrina, the Lou­
isiana Department of Transportation (L-DoT )
is planning to construct an additional bridge
across the Mississippi River. L-DoT is consid­
ering two alternatives: a suspension bridge and
a cantilever bridge. 吀栀e department uses an
interest rate of 8% and plans a SO-year life 昀漀r
either bridge. Which design has the better PW ?
Suspension
Bridge
Cantilever
Bridge
Initial construction costs
Initial land acquisition costs
Annual O&M costs
Annual growth in O&M
$585,000,000
120,000,000
1,500,000
Growing 4%
M愀樀or maintenance (Year 25)
Salvage cost
185,000,000
30,000,000
$470,000,000
95,000,000
2,000,000
Growing
$300,000
210,000,000
27,000,000
Teri is an IE at Smith Manu昀愀cturing in Sara­
sota. She has been studying process line G
to determine if an automated system would
be preferred to the existing labour-intensive
system. If Teri wants to earn at least 15% and
uses a 15-year planning horizon, which al­
ternative is preferable?
Labour
Intensive Automated
Initial cost
Installation cost
First-year maintenance costs
Annual increase in
maintenance costs
First-year labour costs
Annual increase in labour
costs
Salvage value (EOY 15)
5-35
⠀　O&M
$25,000
10,000
161
$0
0
1,500
200
$55,000
15,500
4,800
600
116,000
4%
41,000
4%
5,000
19,000
Tele昀漀no Mexico is expanding its 昀愀cilities to
serve a new manu昀愀cturing plant. 吀栀e new
plant will require 2,000 telephone lines this
year, and another 2,000 lines a昀琀er expansion
in 10 years. 吀栀e plant will operate 昀漀r 30 years.
Option l
Provide one cable now with capacity to serve
4,000 lines. 吀栀e cable cost will be $200,000, and
annual maintenance costs will be $15,000.
Option 2
Provide a cable with capacity to serve 2,000 lines
now and a second cable to serve the other 2,000
lines in 10 years. 吀栀e cost of each cable will be
$150,000, and each cable will have an annual
maintenance of $10,000.
吀栀e telephone cables will last at least 30 years,
and the cost of removing the cables is o昀昀set by
their salvage value. Use a spreadsheet.
(a) Which alternative should be chosen as­
suming a 10% interest rate?
(b) Will your answer to (a) change if the
demand 昀漀r additional lines occurs in 昀椀ve
years instead of 10 years?
5-36
A consulting engineer has been hired to advise
a town how best to proceed with the construc­
tion of a 200,000 m3 water supply reservoir.
Since only 120,000 m3 of storage will be re­
quired 昀漀r the next 25 years, an alternative to
building the 昀甀ll capacity now is to build the
reservoir in two stages. Initially, the reservoir
162
Cha pter 5 Present Worth Analysis
could be built with 120,000 m3 of capacity, and
then, 25 years hence, the additional 80,000 m3
of capacity could be added by increasing the
height of the reservoir.
5-39
Annual
Construction Maintenance
Cost
Cost
Build in two stages
First st最였: 120,000 m3
reservoir
Second stage: Add
80,000 m3 of capacity,
additional construction
and maintenance costs
Build 昀甀ll capaci琀礀 now
200,000 m reservoir
3
$14,200,000
$75,000
$12,600,000
$25,000
$22,400,000
$100,000
If interest is computed at 4%, which construc­
tion plan is preferred?
Lives Di昀昀er
5-37
5-38
A man had to have the mu昀昀ler replaced on
his two-year-old car. 吀栀e repairman o昀昀ered
two alternatives. For $300 he would install a
mu昀툀er guaranteed 昀漀r two years. For $400 he
would install a mu昀昀ler guaranteed "昀漀r as long
as you own the car." Assuming the present
owner expects to keep the car 昀漀r about three
more years, which mu昀툀er would you advise
him to have installed if you thought 20% was
a suitable interest rate and the less expensive
mu昀툀er would last only two years?
5-40 A new alloy can be produced by process A,
which costs $200,000. 吀栀e operating cost will
be $10,000 per quarter, with a salvage value of
$25,000 a昀琀er its two-year life. Process B will
have a 昀椀rst cost of $250,000, an operating cost
of $15,000 per quarter, and a $40,000 salvage
value a昀琀er its 昀漀ur-year life. 吀栀e interest rate is
8% a year compounded quarterly. Use present
value analysis to decide which process should
be selected.
5-41
Bids: installed cost
Engineer's estimates
Service life (years)
Annual oper愀琀ing cost,
including repairs
Salvage value at end of
service life
Itis
$45,000
10
$54,000
15
$2,700
$2,850
$3,000
$4,500
Which process line should be built 昀漀r a new
chemical? 吀栀e market 昀漀r the chemical is ex­
pected to last 昀漀r 20 years. A 9% rate is used
to evaluate new process 昀愀cilities, which are
compared with PW. How much does the better
choice save?
Life
First Cost O☀䴀 Cost/礀攀ar Salvage
A
B
$18M
25M
$4M
6M
$SM
3M
10 years
20 years
5-42 Which equipment is preferred if the 昀椀rm's in­
terest rate is 9%? In PW terms, how great is the
di昀昀erence?
Alternative
An engineer has received two bids 昀漀r an eleva­
tor to be installed in a new building.
Given a 10% interest rate, which bid should
be accepted?
Westinghome
A weekly business magazine o昀昀ers a one-year
subscription 昀漀r $58 and a three-year subscrip­
tion 昀漀r $116. If you thought you would read
the magazine 昀漀r at least the next three years,
and you considered 20% to be a minimum rate
of return, which way would you buy the maga­
zine: with three one-year subscriptions or a
single three-year subscription?
First cost
Annual O&M
Salvage value
Overhaul (Year 6)
Li昀攀
5-43
A
B
$35,000
$40,000
$2,900
$4,200
0
$5,000
$10,000 Not required
10
5
吀栀e Crockett Land Winery must replace its
grape-pressing equipment. 吀栀e two alterna­
tives are the Quik-Skwish and the Stomp­
Master. 吀栀e annual operating costs increase by
12% each year as the machines age. If the inter­
est rate is 9%, which press should be chosen?
Problems
$150,000, and 昀甀ture replacement plants will
also cost $150,000 each. If the trust 昀甀nd earns
8% interest, what is the builder's capitalized
cost to construct the plant and 昀甀ture replace­
ments, and to pay the operating costs?
Quik-S欀眀ish Stomp-Master
First cost
Annual operating costs
Salvage value
Use昀甀l life (years)
$350,000
$28,000
$35,000
5
$500,000
$22,500
$50,000
10
Perpetual Life
5-50
A man who likes cherry blossoms very much
would like to have an urn 昀甀ll of them put on
his grave once each year 昀漀rever a昀琀er he dies.
He intends, in his will, to leave a certain sum
of money in the trust of a local bank to pay the
昀氀orist's annual bill. How much money should
be le昀琀 昀漀r this purpose? Make whatever as­
sumptions you feel are justi昀椀ed by the 昀愀cts
presented. State your assumptions, and com­
pute a solution.
5-51
An elderly lady decided to donate most of her
considerable wealth to charity and to keep 昀漀r
herself only enough money to live on. She feels
that $2,000 a month will amply provide 昀漀r her
needs. She will establish a trust 昀甀nd at a bank
that p愀礀s 6% interest, compounded monthly.
At the end of each month she will withdraw
$2,000. She has arranged that, upon her death,
the balance in the account is to be paid to her
niece, Susan. If she opens the trust 昀甀nd and
deposits enough money to pay herself $2,000
a month in interest as long as she lives, how
much will Susan receive when her aunt dies?
5-52
A trust 昀甀nd is to be established 昀漀r three pur­
poses: (1) to provide $750,000 昀漀r the construc­
tion and $250,000 昀漀r the initial equipment of
a small engineering laboratory; (2) to pay the
$150,000 a year that it costs to operate the lab­
oratory; and (3) to pay 昀漀r $100,000 of replace­
ment equipment every 昀漀ur years, beginning
昀漀ur years 昀爀om now.
At 6% interest, how much money is re­
quired in the trust 昀甀nd to provide 昀漀r the
laboratory and equipment and its perpetual
operation and equipment replacement?
5-53
We want to donate a marble bird bath to the
city park as a memorial to cats. We also want
to set up a perpetual care 昀甀nd to cover 昀甀ture
expenses 昀漀rever. 吀栀e initial cost of the bath
is $5,000. Routine annual operating costs are
5-44 A small dam was built 昀漀r $2 million. 吀栀e
annual maintenance cost is $15,000. If interest
is 5%, compute the capitalized cost of the dam,
including maintenance.
5-45 A depositor puts $25,000 in a savings account
that pays 5% interest, compounded semi­
annually. Equal annual withdrawals are to be
made 昀爀om the account, beginning one year
昀爀om now and continuing 昀漀rever. What is the
maximum annual withdrawal?
5-46 What amount of money deposited 50 years ago
at 8% interest would provide a perpetual pay­
ment of $10,000 a year beginning this year?
5-47
吀栀e president of the E.L. Echo Corporation de­
cided that her 昀椀rm would endow a chair in the
Department of Mechanical Engineering of the
local university. If the professor holding that
chair receives $268,000 a year and the interest
received on the endowment 昀甀nd is expected to
be 8%, what lump sum of money will the Echo
Corporation need to provide to establish the
endowment 昀甀nd?
5-48 吀栀e local botanical society wants to ensure
that the gardens in the town park are prop­
erly cared 昀漀r. 吀栀ey recently spent $100,000 to
plant the gardens. 吀栀ey would like to set up a
perpetual 昀甀nd to provide $100,000 昀漀r 昀甀ture
replantings of the gardens every 10 years. If
interest is 5%, how much money would be
needed to pay the cost of replanting 昀漀rever?
5-49 A home builder must construct a sewage treat­
ment plant and deposit enough money in a
perpetual trust 昀甀nd to pay the $5,000-per­
year operating cost and to replace the treat­
ment plant every 40 years. 吀栀e plant will cost
163
164
5-54
⠀　Cha pter 5 Present Worth Analysis
$200 a year, but every 昀椀昀琀h year the cost will be
$500 to cover major cleaning and maintenance
as well as operation.
愀儀 What is the capitalized cost of this project
if the interest rate is 8%?
(b) How much is the present worth of this pro­
ject if it is to be demolished a昀琀er 75 礀攀ars?
吀栀e 昀椀nal $500 payment in the 75th year
will cover the year's operating costs and
the site reclamation.
5-56
A local symphony association o昀昀ers member­
ships as 昀漀llows:
⠀　Continuing membership, per year: $15
Patron lifetime membership: $375
吀栀e patron membership has been based
on the symphony association's belief that it can
obtain a 4% rate of return on its investment.
If you believed 4% to be an appropriate rate of
return, would you be willing to buy the patron
membership?
Explain why or why not.
A
Initial cost
Annual maintenance
Periodic resur昀愀cing
Resur昀愀cing interval (years)
5-57
Concrete
Steel
5_58
⠀　A city has developed a plan to provide 昀漀r 昀甀ture
municipal water needs. 吀栀e plan proposes an
aqueduct that passes through 150 metres of
tunnel in a nearby mountain. Two alternatives
are being considered. 吀栀e 昀椀rst proposes to
build a 昀甀ll-capacity tunnel now 昀漀r $556,000.
吀栀e second proposes to build a half-capacity
tunnel now (cost = $402,000), which should
be adequate 昀漀r 20 years, and then to build
a second parallel half-capacity tunnel. 吀栀e
maintenance cost of the tunnel lining 昀漀r the
昀甀ll-capacity tunnel is $40,000 every 10 years,
and 昀漀r each half-capacity tunnel it is $32,000
every 10 years.
吀栀e 昀爀iction losses in the half-capacity
tunnel will be greater than in the 昀甀ll-capacity
tunnel. 吀栀e estimated additional pumping
costs in the single half-capacity tunnel will
be $2,000 a year, and 昀漀r the two half-capacity
tunnels it will be $4,000 a year. On the basis of
capitalized cost and a 7% interest rate, which
alternative should be selected?
B
$500,000 $700,000
$35,000 $25,000
$350,000 $450,000
15
10
A new bridge project is being evaluated at
i = 5%. Recommend an alternative based on
the capitalized cost 昀漀r each.
Construction
Multiple Alternatives
5-55
Using capitalized cost, determine which type
of road sur昀愀ce is preferred on a particular sec­
tion of highway. Use a 12% interest rate.
Cost
Annual
O&M
Life
(years)
$50,000,000
40,000,000
$250,000
1,000,000
70
50
A new stadium is being evaluated at i = 6%.
Recommend an alternative 昀漀r the main struc­
tural material based on the capitalized cost
昀漀r each.
Construction Cost
Concrete
Steel
Annual
O&M
$25,000,000 $200,000
21,000,000 1,000,000
Life
(years)
80
60
5-59
A rather wealthy man decided he would like
to arrange 昀漀r his descendants to be well
educated. He would like each child to have
$120,000 昀漀r his or her education. He plans
to set up a perpetual trust fund so that six
children will receive this assistance in each
generation. He estimates that there will be
昀漀ur generations per century, spaced 25 years
apart. He expects the trust to be able to
obtain a 4% rate of return, and the 昀椀rst re­
cipients to receive the money 10 years hence.
How much money should he now set aside in
the trust?
5-60
Kansas Public Service Company wishes to de­
termine the capitalized worth at an interest
rate of 9% of a new windmill with the 昀漀llow­
ing costs.
Problems
Initial costs
Installation costs
Annual O&M costs
M愀樀or overhaul (Year 25)
Expected li昀攀
Salvage value
5-61
5-62
5-63
An open-pit mine must 昀甀nd an account
now to pay 昀漀r maintenance of a tailing pond
in perpetuity (a昀琀er the mine shuts down in
30 years). 吀栀e costs until shutdown are part of
the mine's operating costs. 吀栀e maintenance
costs begin in 30 years at $300,000 annually.
How much must be deposited now if the 昀甀nd
will earn 5% interest?
5-65
A 昀椀rm is considering three mutually exclusive
alternatives as part of a production improve­
ment program. 吀栀e alternatives are
B
C
$10,000 $15,000 $20,000
$1,625 $1,530 $1,890
20
20
10
吀栀e salvage value at the end of the use昀甀l
life of each alternative is zero. At the end of 10
years, Alternative A could be replaced with
another A with identical cost and bene昀椀ts.
吀栀e maximum attractive rate of return is 6%.
Which alternative should be chosen?
5-64 吀栀e General Hospital is evaluating new o昀케ce
equipment o昀昀ered by three companies.
First cost
Maintenance and
operating costs
Annual bene昀椀t
Salvage value
Company
A
Company Company
B
C
$15,000
1,600
$25,000
400
$20,000
900
8,000
3,000
13,000
6,000
11,000
4,500
吀栀e 昀漀llowing costs are associated with three
tomato-peeling machines being considered 昀漀r
use in a canning plant.
If the canning company uses an inter­
est rate of 12%, which is the best alternative?
Use NPW to make your decision. (Note: Con­
sider the least common multiple as the study
period.)
Machine Machine Machine
B
C
A
A 昀椀rm wants to sponsor a new engineering lab
at a local university. 吀栀is requires $2.5 million
to construct the lab, $1.2 million to equip it,
and $600,000 every 昀椀ve years 昀漀r new equip­
ment. If the university endowment earns 6%
interest, what is the required endowment?
Installed cost
Uni昀漀rm annual bene昀椀t
Useful life (years)
0
In each case the interest rate is 15% and the
use昀甀l life of the equipment is 昀漀ur years. Use
NPW analysis to determine the company 昀爀om
which you should purchase the equipment.
$725,000
$143,000
$12,000
$260,000
40 years
$32,000
A
165
First cost
Maintenance and
operating costs
Annual bene昀椀t
Salvage value
Use昀甀l life (years)
$52,000
$15,000
$63,000
$9,000
$67,000
$12,000
$38,000
$13,000
4
$31,000
$19,000
6
$37,000
$22,000
12
5-66 A railway branch line to a land昀椀ll site is to
be constructed. It is expected that the rail­
way line will be used 昀漀r 15 years, a昀琀er which
the land will be returned to agricultural use.
吀栀e railway track and ties will be removed at
that time.
In building the railway line, either
treated or untreated wood ties may be used.
Treated ties have an installed cost of $6 and
a 10-year life; untreated ties cost $4.50 with
a six-year life. If at the end of 15 years the
ties then in place have a remaining use昀甀l
life of 昀漀ur years or more, they will be used
by the railway elsewhere, and they have an
estimated salvage value of $3 each. Any ties
that are removed at the end of their service
life or are too close to the end of their ser­
vice life to be used elsewhere can be sold 昀漀r
$0.50 each.
Determine the most economical plan 昀漀r
the initial railway ties and their replacement
昀漀r the 15-year period. Make a present worth
analysis assuming 8% interest.
5-67
A building contractor obtained bids 昀漀r some
asphalt p愀瘀ing, based on a speci昀椀cation.
166
Cha pter 5 Present Worth Analysis
cc do-nothing" alternatives. Estimates of the
cost and bene昀椀ts are as 昀漀llows:
吀栀ree paving subcontractors quoted the 昀漀llowing prices and terms of payment:
Paving Co.
Price
Quick
$85,000 50% payable immediately
25% payable in 6 months
25% payable at the end of
one year
$82,000 Payable immediately
$84,000 25% payable immediately
75% payable in 6 months
Tartan
Faultless
P愀礀ment Schedule
Alternatives
1
2
3
4
吀栀e building contractor uses a 12% nominal
interest rate, compounded monthly, in this type
of bid analysis. Which p愀瘀ing subcontractor
should be awarded the paving job?
5-68
?
�
Initial cost
Annual bene昀椀t
Salvage value
Use昀甀l life (years)
B
C
$10,000 $15,000 $12,000
$6,000 $10,000 $5,000
$1,000 -$2,000 $3,000
2
4
3
Use an analysis period of 12 years and 10%
interest.
5-69
Consider the 昀漀llowing 昀漀ur alternatives.
吀栀ree are cc do something" and one is "do
nothing."
Cost
Net annual bene昀椀t
Use昀甀l li昀攀 (years)
A
B
C
D
$0
$0
$50
$12
5
$30
$4.5
10
$40
$6
10
At the end of the 昀椀ve-year use昀甀l life of B, a
replacement is not made. If a 10-year analy­
sis period and a 10% interest rate are selected,
which is the preferred alternative?
5-70
A cost analysis is to be made to determine
what, if anything, should be done in a situa­
tion o昀昀ering three cc do-something" and one
$500
600
700
0
$135
100
100
0
5
5
10
0
$0
250
180
0
Use a 10-year analysis period 昀漀r the 昀漀ur
mutually exclusive alternatives. At the end
of 昀椀ve years, Alternatives 1 and 2 may be
replaced with identical alternatives (with
the same cost, bene昀椀ts, salvage value, and
use昀甀l life).
(a) If an 8% interest rate is used, which al­
ternative should be chosen?
(b) If a 12% interest rate is used, which al­
ternative should be chosen?
Given the 昀漀llowing data, use present
worth analysis to 昀椀nd the best alternative, A,
B, or C.
A
End-ofUni昀漀rm Use昀甀l-Life Use昀甀l
Annual
Life
Salvage
Cost Bene昀椀t
Value
(years)
5-71
Consider A to E, 昀椀ve mutually exclusive
alternatives:
A
B
Initial cost
$600 $600
Uni昀漀rm annual bene昀椀ts
100 100
For 昀椀rst 5 years
For last 5 years
50 100
C
D
E
$600 $600 $600
100
110
150
0
150
50
吀栀e interest rate is 10%. If all the alternatives
have a 10-year use昀甀l life, and no salvage value,
which alternative should be chosen?
5-72
An investor has care昀甀lly studied a number
of companies and their common shares.
From his analysis, he has decided that the
shares of six firms are the best of the many
he has examined. T hey represent about the
same amount of risk, and so he would like to
choose one single share in which to invest.
He plans to keep the shares 昀漀r 昀漀ur years,
and he requires a 10% minimum attractive
rate of return.
Which share 昀爀om the table, if any, should
the investor consider buying?
Problems
Annual
Estim愀琀ed
Price 愀琀
Price per End-of-Year
Common Di瘀椀dend
End of
per Share Four Years
Share
Company
Western House
Fine Foods
Mobile Motors
Spartan Products
US Tire
Wine Products
5-73
$23.75
45.00
30.65
12.00
33.38
52.50
$1.25
4.50
0.00
0.00
2.00
3.00
$32
45
42
20
40
60
Six mutually exclusive alternatives, A to F, are
being examined. For an 8% interest rate, which
alternative should be selected? Each alternative
has a six-year use昀甀l life.
Uni昀漀rm
Initial Cost Annual Bene昀椀t
A
B
C
D
E
F
5-74
$ 20.00
35.00
55.00
60.00
80.00
100.00
Initial Cost
Plan (thousands)
1
2
3
4
5
6
7
8
$265
220
180
100
305
130
245
165
Bonds
5-75
A corporate bond has a 昀愀ce value of $1,000,
with its maturity date 20 years 昀爀om today. 吀栀e
bond pays interest semi-annually at a rate of
8% a year, based on the 昀愀ce value. 吀栀e interest
rate paid on similar corporate bonds has de­
creased to a current rate of 6%. Determine the
market value of the bond.
5-76
Calculate the present worth of a 4.5%, $5,000
bond with interest paid semi-annually. 吀栀e
bond matures in 10 years, and the investor
wants to make 8% a year compounded quar­
terly on the investment.
5-77
You bought a $1,000 corporate bond 昀漀r $900
three years ago. It is paying $30 in interest at
the end of every six months, and it matures in
昀漀ur more years.
(a) Compute its coupon rate.
(b) Compute its current value assuming
the market interest rate 昀漀r such
investments is 5% per year, compounded
semi-annually.
5-78
A 6% coupon rate bond has a 昀愀ce value of
$1,000, pays interest semi-annually, and will
mature in 10 years. If the current market rate is
8% interest compounded semi-annually, what
is the bond's price?
5-79
A treasury bond with a 昀愀ce value of $5,000 and
a coupon rate of 6% payable semi-annually was
bought by Kirsten when the market's nominal
rate was 8%. 吀栀e bond matures 20 years 昀爀om
now. What did Kirsten pay 昀漀r the bond?
5-80
A zero-coupon bond with a 昀愀ce value of
$10,000 and maturity date in 昀椀ve years is being
considered 昀漀r purchase by Pam. 吀栀e current
market interest rate is a nominal 10%, com­
pounded quarterly. How much should she pay
昀漀r the bond?
5-81
A provincial government wants to raise
$3 million by issuing bonds. 吀栀e bond's
coupon interest rate was set at 8% per annum,
with semi-annual payments. However, market
$ 6.00
9.25
13.38
13.78
24.32
24.32
吀栀e management of an electronics manu昀愀c­
turing 昀椀rm believes it is desirable to automate
its production 昀愀cility. 吀栀e automated equip­
ment would have a IO-year life with no salvage
value at the end of 10 years. 吀栀e plant engin­
eering department has surveyed the plant and
suggested there are eight mutually exclusive
alternatives available.
Net Annual
Bene昀椀t
(thousands)
$51
39
26
15
57
23
47
33
If the 昀椀rm expects a 10% rate of return, which
plan, if any, should it adopt?
167
168
Cha pter 5 Present Worth Analysis
interest rates have risen to a nominal 9%
interest rate. How much will the province raise
昀爀om issuing $3M in bonds if the bonds mature
in 20 years?
5-82
Kai Tech, a manu昀愀cturing company, needs
to raise $2 million to 昀椀nance an expansion
project. 吀栀e bonds will have a coupon interest
rate of 12%, payable quarterly, and will mature
in 20 years. What will the 昀愀ce value of the
bonds have to be if the bonds have an inter­
est rate of 12% per annum, payable quarterly,
and mature in 20 years? 吀栀e current market
interest rate is a nominal 16% compounded
semi-annually.
5-86
0
t
0 -- 1 -- 2 -- 3 -- 4 -- 5
Compute the 昀甀ture worth 昀漀r the 昀漀llowing
cash 昀氀ows.
•t
3 00
25 0
'f r
o
0 -- 1 -- 2 -- 3 -- 4 -- 5
F
0
5-87
5-84 For the 昀漀llowing cash 昀氀ows compute the
昀甀ture worth.
I l t
4x
3x
2x
0 -- 1 -- 2 -- 3 -- 4
i = 1 5%
For the 昀漀llowing cash 昀氀ows compute the
昀甀ture worth.
⸀⸀ t
5
10
Compute F so that the 昀漀llowing cash 昀氀ows
h愀瘀e a 昀甀ture worth of 0.
t
1 00
25
lfl
30
30
30
30
30
30
0 - 1 - 2 - 3 - 4- 5 - 6 - 7 - 8 - 9 - 10 - 1 1
i = 1 0%
i = 10%
5-88
0
5-89
i = 12%
0
.l l
t
100
+
t
100
+
+
0 -- 1 -- 2 -- 3 -- 4 -- 5 -- 6
2 00
5-85
I
2 00
100
Futu re Worth
5-83
For a 12% interest rate, compute the value of F
so that the 昀漀llowing cash 昀氀ows have a 昀甀ture
worth of 0.
100
1 00
F
Calculate the present worth and the 昀甀ture
worth of a series of 10 annual cash 昀氀ows with
the 昀椀rst cash 昀氀ow equal to $15,000 and each
successive cash 昀氀ow increasing by $1,200. 吀栀e
interest rate is 12%.
Sally deposited $100 a month in her savings
account 昀漀r 24 months. For the next 昀椀ve years
she made no deposits. What is the 昀甀ture worth
in Sally's savings account at the end of the
seven years if the account earned 6% annual
interest, compounded monthly?
5-90 A 20-year-old student decided to set aside $100
on her 21st birthday 昀漀r investment. Each sub­
sequent year through her 55th birthday, she
plans to increase the investment on a $100
arithmetic gradient. She will not set aside
additional money a昀琀er her 55th birthday. If the
student can achieve a 12% rate of return, what
is the 昀甀ture worth of the investments on her
65th birthday?
0
5-91
You can buy a piece of vacant land 昀漀r $30,000
cash. You plan to hold it 昀漀r 15 years and then
sell it at a pro昀椀t. During this period, you would
h愀瘀e to pay annual property taxes of $600.
You would h愀瘀e no income 昀爀om the property.
Problems
Assuming that you want a 10% rate of return,
at what net price would you have to sell it
15 years hence?
5 -92
5 -93
⠀　A man's salary is now $32,000 a year and he
expects to retire in 30 years. If his salary is increased by $600 each year and he deposits 10%
of his yearly salary into a 昀甀nd that earns 7%
interest, what is the 昀甀ture worth of the 昀甀nd
when he retires?
Stamp collecting has become an increasingly
popular-and expensive-hobby. One 昀愀vourite method is to save plate blocks (usually 昀漀ur
stamps with the printing plate number in the
margin) of each new stamp as it is issued. With
rising postage rates and increased numbers of
new stamps being issued, this collecting plan
costs more each year.
Stamps have been a good place to invest
money over the last 10 years because the
demand 昀漀r stamps previously issued has
caused resale prices to increase 18% each
year. Suppose a collector bought $100 worth
of stamps 10 years ago and increased his purchases by $50 a year in each subsequent year.
A昀琀er 10 years of stamp collecting, what is the
昀甀ture worth of the stamp collection?
5 -94
吀栀e interest rate is 16% per annum, and there
are 48 compounding periods a year. 吀栀e principal is $50,000. What is the 昀甀ture worth in
昀椀ve years?
5 -95
In the early 1980s, UK planners were examining alternative sites 昀漀r a new London airport.
吀栀e economic analysis included the value of
the structures that would need to be removed
昀爀om various airport sites. At one airport site,
the 12th-century Norman church of St Michael, in the village of Stewkley, would have to
be demolished. 吀栀e planners used the value of
the 昀椀re insurance policy on the church-a few
thousand pounds sterling-as the value of the
church.
An outraged antiquarian wrote to the
London 吀椀mes that an equally plausible computation would be to assume that the original
cost of the church (estimated at 100 pounds
⠀　169
sterling) should be increased at the rate of
10% a year 昀漀r 800 years. According to that
calculation, what would the 昀甀ture worth of
St Michael's be? (Note: 吀栀ere was great public
opposition to the church being torn down, and
it was spared.)
5 -96
Bill made a budget and planned to deposit
$150 a month in a savings account, beginning
1 September. He did this, but on the 昀漀llowing
1 January, he reduced the monthly deposits to
$100. He made 18 deposits, 昀漀ur at $150 and 14
at $100. If the savings account paid 6% interest, compounded monthly, what was the 昀甀ture
worth of his savings immediately a昀琀er the last
deposit?
5 -97
A company deposits $1,000 in a bank at the beginning of each year 昀漀r six years. 吀栀e account
earns 8% interest, compounded every six
months. What is in the account at the end of
six years?
⠀　5-98
Don Ball is a 55-year-old engineer. According
to mortality tables, a man at age 55 has an average li昀攀 expectancy of 21 more years. Don has
accumulated $48,500 toward his retirement. He
is now adding $5,000 a year to his retirement
昀甀nd. 吀栀e 昀甀nd earns 12% interest. Don will
retire when he can obtain an annual income of
$20,000 昀爀om his retirement 昀甀nd, assuming he
lives to age 76. He will make no provision 昀漀r
a retirement income a昀琀er age 76. What is the
youngest age at which Don can retire?
5 -99
Jean invests $100 in Year 1 and doubles the
amount each year a昀琀er that (so the investment is $100, $200, $400, $800, . . .). If she does
this 昀漀r 10 years, and the investment pays 10%
annual interest, what is the 昀甀ture worth of her
investment?
⠀　5 -100 If you invest $2,500 in a 24-month term de-
posit paying 8.65%, compounded monthly,
what is the 昀甀ture worth when it matures?
5 -101
⠀　A昀琀er receiving an inheritance of $25,000 on
her 21st birthday, Ann deposited all the money
in a savings account with an e昀昀ective annual
170
Cha pter 5 Present Worth Analysis
interest rate of 6%. She decided to make regular deposits beginning with $1,000 on her 22nd
birthday and increasing by $200 each year (i.e.,
$1,200 on her 23rd birthday, $1,400 on her 24th
birthday, etc.). What was the 昀甀ture worth of
Ann's deposits on her 56th birthday?
5 -102
5 -103
⠀　吀栀e Association of General Contractors (AGC)
is endowing a 昀甀nd of $1 million 昀漀r the Construction Engineering Technology Program at
Grambling State University. In doing so, the
AGC established an escrow account in which 10
equal end-of-year deposits that earn 7% compound interest were to be made. A昀琀er seven deposits, the Louisiana legislature revised the laws
relating to the licensing fees that the AGC can
charge its members, and there was no deposit
at the end of Year 8. What must the amount of
the remaining equal end-of-year deposits be to
ensure that the $1 million is 愀瘀ailable on schedule to Grambling State 昀漀r the Construction
Engineering Technology Program?
A new engineer is considering investing in a
registered retirement savings plan (RRSP).
A昀琀er some research, she 昀椀nds an eligible 昀甀nd
with an average annual return of 10%. What is
the 昀甀ture worth of her RRSP at age 65 if she
makes annual deposits of $2,000 beginning on
her 25th birthday? Assume that the 昀甀nd continues to earn an annual return of 10%.
$55,000 at the end of the 昀椀昀琀h year of service.
Assume that all these costs occur at the end of
the relevant period. What is the 昀甀ture value
of all the costs of owning and operating this
machine if the nominal interest rate is 7.2%?
5 -106 A 昀愀mily starts an education 昀甀nd 昀漀r their son
Patrick when he is eight years old, investing
$150 on his eighth birthday and increasing the
yearly investment by $150 per year until Patrick is 18 years old. 吀栀e 昀甀nd pays 9% annual
interest. What is the 昀甀ture worth of the 昀甀nd
when Patrick is 18?
5 -107
⠀　5-108
Let's assume that a late-t眀攀ntieth-century university graduate got a good job and began a s愀瘀ings account. He authorized the bank to transfer
$75 each month 昀爀om his chequing account to
his s愀瘀ings account. 吀栀e bank made the 昀椀rst
withdr愀眀al on 3 July 2012 and is instructed to
make the last withdrawal on 3 July 2030. 吀栀e
bank pays a nominal interest rate of 4.5% and
compounds twice a month. What is the 昀甀ture
worth of the account on 3 July 2030?
5-109
Bob, an engineer, decided to start a university
昀甀nd 昀漀r his son. Bob will deposit a series of
equal, semi-annual cash 昀氀ows, with each deposit
equal to $1,500. Bob made the 昀椀rst deposit on
3 July 2011 and will make the last deposit on 3 July
2031. Joe, a 昀爀iend of Bob's, recei瘀攀d an inheritance on 3 April 2016 and has decided to begin
a uni瘀攀rsity 昀甀nd 昀漀r his daughter. Joe wants to
send his daughter to the same university as Bob's
son. 吀栀ere昀漀re, Joe needs to accumulate the same
amount of money on 3 July 2031 as Bob will h瘀였
accumulated 昀爀om his semi-annual deposits. Joe
never took engineering economics and has no
5 -104 IPS Corp. will upgrade its package-labelling
machinery. It will cost $150,000 to buy the machinery and have it installed. Operation and
maintenance costs are $1,500 per year 昀漀r the
昀椀rst three years, and they increase by $500 per
year 昀漀r the remaining years of the machine's
10-year life. 吀栀e machinery has a salvage value
of 5% of its initial cost. Interest is 10%. What is
the 昀甀ture worth of the cost of the machinery?
5 -105
⠀　A company is considering buying a new bottie-capping machine. 吀栀e initial cost of the
machine is $325,000, and it has a 10-year life.
Monthly maintenance costs are expected to be
$1,200 a month 昀漀r the 昀椀rst seven years and
$2,000 a month 昀漀r the remaining years. 吀栀e
machine requires a major overhaul costing
A bank account pays 19.2% interest with
monthly compounding. A series of deposits
started with a deposit of $5,000 on 1 January 2007. Deposits in the series were to occur
each six months. Each deposit in the series is
昀漀r $150 less than the one be昀漀re it. 吀栀e last
deposit in the series will be due on 1 January
2022. What is the 昀甀ture worth of the account
on 1 July 2024 if the balance was zero be昀漀re
the 昀椀rst deposit and no withdrawals are made?
⠀　Problems
idea how to 昀椀gure out the amount that should
be deposited. He decides to deposit $40,000 on
3 July 2016. Will Joe's deposit be enough? If not,
how much should he put in? Use a nominal in­
terest of 7% with semi-annual compounding on
all accounts.
5-110 A business executive is o昀昀ered a management
job at Generous Electric Company, which o昀昀ers
her a 昀椀ve-year contract that calls 昀漀r a salary
of $62,000 a year, plus 600 shares of GE stock
at the end of the 昀椀ve years. 吀栀is executive is
currently employed by Fearless Bus Company,
which also has o昀昀ered her a 昀椀ve-year contract.
It calls 昀漀r a salary of $65,000, plus 100 shares
of Fearless stock each year. 吀栀e Fearless stock
is currently worth $60 per share and pays
an annual dividend of $2 per share. Assume
end-of-year payments of salary and stock.
Dividends begin to be paid one year a昀琀er the
shares are received. 吀栀e executive believes that
the value of the shares and the dividends will
remain constant. If the executive considers 9%
a suitable rate of return in this situation, what
must the Generous Electric stock be worth per
share to make the two o昀昀ers equally attractive?
Use the 昀甀ture worth analysis method in your
comparison.
5-111 Pick a discretionary expense that you incur
regularly, such as buying clothes monthly,
buying sports tickets monthly, or going to a
movie weekly. Assume that you instead place
the money in an investment account that earns
9% annually. A昀琀er 40 years, how much will be
in the account?
a year. What initial principal or PW will this
repay?
5-114 Assume annual car p愀礀ments of $6,000 昀漀r
昀漀ur years and an interest rate of 6.168% a year.
What initial principal or PW will this repay?
5-115 Assume mortgage payments of $1,000 per
month 昀漀r 30 years and an interest rate of 0.5%
per month. What initial principal or PW will
this repay?
5-116 Assume annual mortgage payments of $12,000
昀漀r 30 years and an interest rate of 6% per year.
What initial principal or PW will this rep愀礀?
5-117 Assume annual mortgage payments of $12,000
昀漀r 30 years and an interest rate of 6.168% per
year. What initial principal or PW will this
repay?
5-118 A construction project has the 昀漀llowing end­
of-month costs. Calculate the PW at a nominal
interest rate of 18%.
5-113 Assume annual car payments of $6,000
昀漀r 昀漀ur years and an interest rate of 6%
Year
O
1
2
Net cash ($)
o
-120,000
-60,000
$30,000
50,000
110,000
430,000
January
February
March
April
May
$520,000
460,000
June
275,000
July
August
95,000
5-119 A 昀愀ctory has had the 昀漀llowing average
monthly heating and cooling costs over the
last 昀椀ve years. Calculate the PW at a nominal
interest rate of 12%.
January
February
March
April
May
June
Spreadsheets
5-112 Assume monthly car payments of $500 a
month 昀漀r 昀漀ur years and an interest rate of
0.5% a month. What initial principal or PW
will this repay?
171
$25,000
19,000
15,000
9,000
12,000
18,000
$29,000
July
33,000
August
September 19,000
8,000
October
16,000
November
28,000
December
5-120 Ding Bell Imports requires a return of 15%
on all projects. If Ding is planning an over­
seas development project with the cash 昀氀ows
shown below, what is the project's net present
value?
3
4
5
6
7
20,000
40,000
80,000
100,000
60,000
172
Cha pter 5 Present Worth Analysis
costs shows that the installed cost, with all con­
trols, would be least 昀漀r natural gas at $30,000;
昀漀r 昀甀el oil it would be $55,000; and 昀漀r coal it
would be $180,000. If natural gas is used rather
than 昀甀el oil, the annual 昀甀el cost will increase
by $7,500. If coal is used rather than 昀甀el oil,
the annual 昀甀el cost will be $15,000 per year
less. Assuming 8% interest, a 20-year analysis
period, and no salvage value, which is the most
economical installation?
5-121 Maverick Enterprises is planning a new prod­
uct. Annual sales, unit costs, and unit revenues
are as tabulated; the 昀椀rst cost of R&D and set­
ting up the assembly line is $42,000. If i is 10%,
what is the PW ?
Year Annual Sales Cost/Unit
I
2
3
4
5
6
$ 5,000
6,000
9,000
1 0,000
8,000
4,000
$3.50
3.25
3.00
2.75
2.50
2.25
Price/Unit
$6.00
5.75
5.50
5.25
4.50
3.00
5-122 Northern Engineering is analyzing a mining
project. Annual production, unit costs, and
unit revenues are in the table. 吀栀e 昀椀rst cost of
the mine set-up is $8 million. If i is 15%, what
is the PW ?
Year
Annual
Production (tonnes)
Cost per
Tonne
Price per
Tonne
I
2
3
4
5
6
7
70,000
90,000
120,000
100,000
80,000
60,000
40,000
$25
20
22
24
26
28
30
$35
34
33
34
35
36
37
U nclassi昀椀ed
5-123 An investor is considering buying a 20-year
corporate bond. 吀栀e bond has a 昀愀ce value
of $1,000 and pays 6% interest a year in two
semi-annual payments. 吀栀us the purchaser
of the bond will receive $30 every six months
in addition to $1,000 at the end of 20 years,
along with the last $30 interest payment.
If the investor wants to receive 8% inter­
est, compounded semi-annually, how much
would he or she be willing to pay 昀漀r the
bond?
5-124 A steam boiler is needed as part of the design
of a new plant. 吀栀e boiler can be 昀椀red by nat­
ural gas, 昀甀el oil, or coal. A decision must be
made on which 昀甀el to use. An analysis of the
5-125 Dr Fog E. Professor is retiring and wants to
endow a chair of engineering economics at
his university. It is expected that he will need
to cover an annual cost of $200,000 昀漀rever.
What lump sum must he donate to the uni­
versity today if the endowment will earn 10%
interest?
5-126 Walt Wallace Construction Enterprises is in­
vestigating the purchase of a new dump truck.
Interest is 9%. 吀栀e cash flows 昀漀r two likely
models are as 昀漀llows:
Model
A
B
First
Cost
$50,000
80,000
Annual
Operating Annu愀氀 Salvage
Income V愀氀ue
Cost
Life
$9,000
1 2,000
IO yr
IO yr
$2,000
1 ,000
$ 1 0,000
30,000
(a) Using present worth analysis, decide which
truck the 昀椀rm should buy, and explain why.
(b) Be昀漀re the construction company can
close the deal, the dealer sells out of
Model B and cannot get any more. What
should the 昀椀rm do now, and why?
5-127 We know that a car costs 60 monthly payments
of $399. 吀栀e car dealer has set us a nominal in­
terest rate of 4.5% compounded daily. What is
the purchase price of the car?
5-128 吀栀e annual income 昀爀om a rented house is
$24,000. 吀栀e annual expenses are $6,000. If
the house can be sold 昀漀r $245,000 at the end
of 10 years, how much could you a昀昀ord to pay
昀漀r it now, if you considered 9% to be a suitable
interest rate?
Problems
5-129 How much would the owner of a building be
justi昀椀ed in paying 昀漀r a sprinkler system that
will save $750 a year in insurance premiums
if the system has to be replaced every 20 years
and has a salvage value equal to 10% of its in­
itial cost? Assume money is worth 7%.
5-130 A machine costs $980,000 and will provide
$200,000 a year in bene昀椀ts. 吀栀e company plans
to use the machine 昀漀r 13 years and will then
sell it 昀漀r scrap, receiving $20,000. 吀栀e com­
pany interest rate is 12%. Should the machine
be purchased?
5-131 Use an eight-year analysis period and a 10%
interest rate to determine which alternative
should be selected:
A
First cost
Uni昀漀rm annual bene昀椀t
Use昀甀l life (years)
B
$5,300 $10,700
$1,800 $2,100
4
8
5-132 North City must choose between two new
snow-removal machines. 吀栀e SuperBlower
has a $70,000 昀椀rst cost, a 20-year life, and an
$8,000 salvage value. At the end of nine years,
it needs a m愀樀or overhaul costing $19,000.
Annual maintenance and operating costs are
$9,000. 吀栀e Sno-Mover will cost $50,000, has
an expected life of 10 years, and has no salvage
value. 吀栀e annual maintenance and operating
costs are expected to be $12,000. Using a 12%
interest rate, which machine should be chosen?
M i n i - Cases
5-133 Bayview's growth is constrained by moun­
tains on one side and the bay on the other. A
bridge across the bay is planned, but which
plan is best? It can be built with a single deck
to meet the needs of the next 20 years, or it
can be built with two decks to meet the needs
of the next 50 years. 吀栀e piers can also be
built to support two decks, but with only one
deck being built now.
Building it all now will cost $160M, and
leaving the top deck 昀漀r later will save $40M.
173
Building that top deck later will cost $70M,
including the cost of tra昀케c disruption. A
single-deck bridge will cost $100M now and
$USM in 20 years. Deck maintenance is $1.4M
per year per deck. Pier maintenance is $1.2M
per year per bridge. If the interest rate is 5%,
which design should be built?
If the two-deck bridge is built immedi­
ately, dedicated lanes 昀漀r buses, carpools, and
bicycles can be added. To evaluate this use eco­
nomically, estimate the cost of the underused
capacity 昀漀r the bridge.
5-134 Florida Power and Light has committed to
building a solar power plant. JoAnne, an
industrial engineer working 昀漀r FPL, has been
charged with evaluating the three current
designs. FPL uses an interest rate of 10% and a
20-year horizon.
Design 1: Flat Solar Panels
A 昀椀eld of "昀氀at" solar panels angled to best catch
the sun will yield 2.6 MW of power and will
cost $87 million initially, with 昀椀rst-year oper­
ating costs at $2 million, growing $250,000
annually. It will produce electricity worth
$6.9 million the 昀椀rst year and increasing by 8%
each year therea昀琀er.
Design 2: Mechanized Solar Panels
A 昀椀eld of mechanized solar panels rotates
昀爀om side to side so that they are always
positioned at right angles to the sun's rays,
maximizing the production of electricity.
吀栀is design will yield 3.1 MW of power and
will cost $101 million initially, with 昀椀rst-year
operating costs at $2.3 million, growing by
$300,000 annually. It will produce electricity
worth $8.8 million the 昀椀rst year and increas­
ing 8% each year therea昀琀er.
Design 3: Solar Collector Field
吀栀is design uses a 昀椀eld of mirrors to 昀漀cus the
sun's rays onto a boiler mounted in a tower.
吀栀e boiler then produces steam and generates
electricity the same way as a coal-昀椀red plant.
吀栀is system will yield 3.3 MW of power and
will cost $91 million initially, with 昀椀rst-year
operating costs at $3 million, growing by
174
Cha pter 5 Present Worth Analysis
$350,000 annually. It will produce electricity
worth $9.7 million the 昀椀rst year and increasing
8% each year therea昀琀er.
5-135 Your grandparents are asking you 昀漀r advice
on when they should start collecting their
pension payments. If they wait until 66, they
will collect $1,150 a month; but if they start
collecting at age 62, they will collect $832 a
month. Assume they live to be 85, and simpli昀礀
by assuming annual payments.
(a) When do the higher payments catch up in
total dollars received with the lower pay­
ment that starts earlier?
(b) If their interest rate is 6%, which plan has
a higher PW ?
Annual Cash Flow Analysis
N INJA Loans
This cha pter is concerned with reg u la r, period i c cash flows. I n you r perso n a l life, the most i m po rt ­
a n t period i c cash i nflow you w i l l encou nter i s , p roba bly, you r paycheq u e . You r most i m porta nt
period ic cash outflow will p robably be a series of mortgage payments.
About a decade ago, mortgages i n North America beca me much easier to obta i n . Bankers dis­
covered that, rather tha n lend ing money to house buyers and wa iting to be repa id, they cou ld i nstead
sell the debt to someone else as a "mortgage-backed secu rity. " Si nce getting repa id by the borrower
was no longer the ba n k's problem, ba n ks relaxed their lend i ng sta ndards so that even people with no
reg u la r job or i ncome cou ld quali昀礀 for a loa n . These loa ns beca me known as "No I ncome, No Job, and
no Assets," or N I NJA, loa ns. The increased n u m ber of people who cou ld q uali昀礀 for a mortgage led to
an increase in the demand for housing, pushing up house prices. This, it was thoug ht, made the whole
enterprise safe - even if the borrower cou ld not pay back the loa n, the ba n k or the purchaser of the
mortgage-backed secu rity wou ld sti ll have a cla i m on their house, which was s u re to go up i n va lue.
Feverpitched/iStock Photo
N I NJA Loans
177
Unfortunately, though perhaps not surprisingly, it turned out that many of the N INJA borrow­
ers were not able to keep up their mortgage payments and defaulted on their loans. It became
clear that the acronym N INJA was particularly appropriate, since, like a ninja, the borrowers were
stealthily vanishing. A large number of repossessed houses now appeared on the market. As a result
of this oversupply, house prices dropped dramatically. Many of the borrowers now found that the
amount they still owed on their house was more than the house's market value, an undesirable
condition known as being "underwater. " The holders of mortgage-backed securities discovered
that the backing for the security was now worthless. This crisis then spread to the rest of the econ­
omy, creating the crash of 2008.
As an engineer, you may encounter mortgages in both your personal and professional life.
In today's housing market, most young people find it necessary to borrow money from the bank
before they can buy their first house, which they will then gradually repay over the first few dec­
ades of their working life. Another growing expense many young people face is the repayment of
their student loans. In some parts of North America, this can overshadow the cost of a first home.
In your professional life, you may have to advise your company on the advisability of borrowing
money to purchase a building or piece of equipment. Lastly, it may happen that, after years of
working for a larger company, you and some of your fellow employees decide to create a start-up
company of your own. Second mortgages on your individual homes are one, high-risk way to
generate the initial funding you will need.
QU ESTIONS TO CONSI DER
1.
What differences were there between the US and the Canadian banking systems before the 2008 crash, and
how did these differences affect the seriousness of its consequences in the two countries?
2.
What has been done to prevent a recurrence of the 2008 crash?
3.
Estimate how much you expect to earn each month in your first job. After deducting necessary living ex­
penses, how much can you afford to put into mortgage payments each month? How much will it cost you
to buy a home to your liking, and how long will it take you to pay for it?
4.
You have two possible sources of funding for a house purchase. One requires regular monthly payments
of $A for 10 years, the other requires regular monthly payments of $8 for 20 years. You calculate that
120A > 2408, and decide that the second alternative is therefore cheaper overall. Is this correct?
LEARN I NG OBJ ECTIVES
This chapter will help you
•
define equivalent uniform annual cost (E唀䄀C)
•
resolve any series of cash flows into its annual cash flow equivalent
•
use annual costs to compare alternatives with equal, common multiple, or continuous lives,
or over some fixed study period
KEY TERMS
amortization
amortization sched u le
equity
in昀椀nite analysis period
mortgage
mortgage docu ment
salvage value
178
Cha pter 6 Annual Cash Flow Analysis
吀栀e 昀甀ndamental idea throughout this text is that, in order to compare di昀昀erent cash flows,
they must 昀椀rst be moved to their equivalent value at the same point in time. We have al­
ready covered present worth and 昀甀ture worth analysis. In this chapter, we consider com­
parisons based on converting cash 昀氀ows to a regular series of equal payments.
Suppose we h瘀였 calculated the present worth of the cash 昀氀ows associated with two al­
ternative strategies, A and B, and 昀漀und that PW(A) > PW(B). We can con瘀攀rt these present
worths to equivalent series of equal payments spread over n years by multiplying each by the
same 昀愀ctor, ⠀䄀 I倀Ⰰ i, n). From this it immediately 昀漀llows that E唀䄀C(A) > E唀䄀C(B); that is,
annual worth analysis will always lead us to choose the same strategy as present worth analysis.
吀栀e reader may reasonably protest at this point that the authors are wasting his or her
time. We already have two methods of reaching the same answer. Why are we introducing
yet a third method if it is only going to con昀椀rm what we already know?
We can o昀昀er two justi昀椀cations 昀漀r this. Firstly, the EUAC method is in widespread use,
so if you 昀椀nd yourself working 昀漀r a company that uses it, you should be 昀愀miliar with it.
Secondly, some situations are most naturally described by a weekly or monthly series of
cash 昀氀ows; 昀漀r example, you may want to know how much spending money you will have
each week a昀琀er you commit to a particular mortgage schedule and an expensive car loan.
A minor note: some texts distinguish between equivalent annual costs, or E唀䄀C, and
equivalent annual worth or bene昀椀t, EU䄀圀 or E唀䄀B. 吀栀e only di昀昀erence between these
terms is a change in sign. To simplify matters, we will use EUAC throughout and leave it to
the reader to distinguish between cash 昀氀owing in and cash 昀氀owing out.
Annual Cash Flow Calculations
Converting a Present Cost to a n An nual Cost
In annual cash 昀氀ow analysis, the goal is to convert money to an equivalent uni昀漀rm annual
cost or bene昀椀t. 吀栀e simplest case is to convert a present sum P to a series of equivalent
uni昀漀rm end-of-period cash 昀氀ows. 吀栀is is illustrated in Example 6-1.
EXAM PLE 6- 1
A student bought $1,000 worth of home 昀甀rniture. If it is expected to last 10 years, what will the equiva­
lent uni昀漀rm annual cost be if interest is 7%? (吀栀e student might, 昀漀r example, need this in昀漀rmation in
order to compare the annual lease costs of a 昀甀rnished versus an un昀甀rnished apartment.)
! x ! ! x ! ! ! ! !
0 -- = O -- l -- 2 -- 3 -- 4 -- 5 --6 -- 7 -- 8 -- 9 -- 10
l
P = 1 ,000
n = 10 years
i = 7%
SO LUTI O N
Equivalent uni昀漀rm annual cost = P(AIP, i, n)
= 1,000(AIP, 7%, 10)
= $142.40
Annual Cash Flow Calculations
179
Treatment of Sa lvage Value
When there is a salvage value at the end of the use昀甀l life of an asset, this lowers the
equivalent uni昀漀rm annual cost.
EXAM PLE 6-2
吀栀e student in Example 6-1 now believes the 昀甀rniture can be sold at the end of 10 years 昀漀r $200. Under
these circumstances, what is the equivalent uni昀漀rm annual cost?
Resale Value
S = 200
l
1 1 1 1 1 1 1 1 1 1
0 -- 1 글 10 = 0 -- 1 -- 2 -- 3 -- 4 -- S--6 -- 7 -- 8 -- 9 -- 10
P = 1,000
SO LUTI ON
For this situation, the problem may be solved by any of three di昀昀erent calculations.
SO LUTI ON 1
E唀䄀C = P(AIP, i, n) - S (AIF, i, n)
= 1,000(A/P, 7%, 10) - 200(A/F, 7%, 10)
= 1,000(0. 1424) - 200(0.0724)
= 142.40 - 14.48 = $127.92
(6-1)
EUAC = PMT(i, n, P, F)
= PMT(7%, 10, -1000, 200)
= $127.90
吀栀is method reflects the annual cost of the cash disbursement minus the annual bene昀椀t of the 昀甀ture
resale value.
SO LUTI ON 2
Equation 6-1 describes a relationship that may be modi昀椀ed by an identity presented in Chapter 4:
(AIP, i, n) = (AIF, i, n) + i
(6-2)
Substituting this into Equation 6-1 gives
EUAC = P(AIF, i, n) + Pi - S (AIF, i, n)
= (P - S)(A/F, i, n) + Pi
= (1,000 - 200)(A/F, 7%, 10) + 1,000(0.07)
= 800(0.0724) + 70 = 57.92 + 70
= $127.92
(6-3)
continued
180
Cha pter 6 Annual Cash Flow Analysis
吀栀is method computes the equivalent annual cost 昀爀om the unrecovered $800 when the 昀甀rniture is
sold, and it adds annual interest on the $1,000 investment.
SO LUTI O N 3
If the value 昀漀r (AIF, i, n) 昀爀om Equation 6-2 is substituted into Equation 6-1, we obtain
EUAC = P(AIP, i, n) - S(AIP, i, n) + Si
= (P - S)(AIP, i, n) + Si
= (1,000 - 200)(A/P, 7%, 10) + 200(0.07)
= 800(0.1424) + 14 = 1 1 3.92 + 14 = $127.92
(6-4)
吀栀is method computes the annual cost of the $800 decline in value during the 10 years,
plus interest on the $200 tied up in the 昀甀rniture as the salvage value.
When there is an initial disbursement P 昀漀llowed by a salvage value S, the annual cost
may be computed in any of the three di昀昀erent ways introduced in Example 6-2:
EUAC = P(AIP, i, n) S(AIF, i, n)
EUAC = (P - S)(AIF, i, n) + Pi
EUAC = (P - S)(AIP, i, n) + Si
(6-1)
(6-3)
(6-4)
Each of the three calculations gives the same results. In practice, the 昀椀rst method is the
most commonly used. 吀栀e EUAC is also known as the c愀瀀ital recovery cost of a project.
EXAM PLE 6-3
Betty owned a car 昀漀r 昀椀ve years. One day she wondered what her uni昀漀rm annual cost 昀漀r maintenance
and repairs had been. She assembled the 昀漀llowing data:
Year
1
2
3
4
5
Maintenance and Repair
Cost 昀漀r 夀攀ar
$ 45
90
180
135
225
Compute the equivalent uni昀漀rm annual cost (EUAC), assuming 7% interest and end-of-礀攀ar disbursements.
SO LUTI O N
吀栀e EUAC may be computed 昀漀r this irregular series of payments in two steps:
1.
2.
Use (PIP) 昀愀ctors to compute the present worth of cost 昀漀r the 昀椀ve years.
With the present cost known, use the ⠀䄀IP) 昀愀ctor to compute EUAC.
Annual Cash Flow Calculations
181
PW of cost = 45(P/F, 7%, 1) + 90(P/F, 7%, 2) + 180(P/F, 7%, 3)
+ 135(PIF, 7%, 4) + 225(PIF, 7%, 5)
= 45(0.9346) + 90(0.8734) + 180(0.8163) + 135(0.7629) + 225(0.7130)
= $531
EUAC = 531(A/P, 7%, 5) = 531(0.2439) = $130
EXAM PLE 6-4
Betty re-examined her calculations and 昀漀und that in her table she had reversed the maintenance and
repair costs 昀漀r Years 3 and 4. 吀栀is is the correct table.
Year
1
2
3
4
5
Maintenance and Repair
Cost 昀漀r Year
$ 45
90
180
135
225
Recompute the E唀䄀C.
SO LUTI O N
吀栀is time the schedule of disbursements is an arithmetic gradient series plus a uni昀漀rm annual cost, as
昀漀llows:
i l
E唀䄀C = 45 + 45(A/G, 7%, 5)
= 45 + 45(1.865)
= $129
• 1 l
0 - 1-2 - 3- 4- 5 = 0 - 1- 2 - 3- 4- 5 + 0 - 1- 2 - 3- 4- 5
11 1 1 1
90
90
135
135
180
180
225
Since the timing of the expenditures is di昀昀erent in Examples 6-3 and 6-4, we would not expect to obtain
the same EUAC.
Note that when there are irregular cash disbursements over the analysis period, a con­
venient method of solution is 昀椀rst to determine the PW of cost; and then use the (AIP)
昀愀ctor to calculate the E唀䄀C.
182
Cha pter 6 Annual Cash Flow Analysis
Annual Cash Flow Analysis
As in the previous chapter, when comparing alternatives, we can always ignore any cash
flows common to each alternative. So if two strategies both yield the same bene昀椀ts, we can
simply choose the one with the lower equivalent annual costs. Conversely, if two strategies
have the same costs, we will choose the one with the greater equivalent annual bene昀椀ts.
Analysis Period
In Chapter 5, we saw that the analysis period is an important consideration in computing
present worth comparisons. A common analysis period must be used 昀漀r all the alterna­
tives. Example 6-5 will show that in annual cash flow analysis, we can sometimes relax this
requirement.
ⴀⴀ-----·�
=
=
=
=
=㴀ⴀ픁㸀----툁㸀-----------贀턀
•-----�
EXAM PLE 6- 5
吀眀o pumps are being considered 昀漀r purchase. If interest is 7%, which pump should be bought?
Pump A
Pump B
Initial cost
$7,000
$5,000
End-of-use昀甀l-li昀攀 salvage value
$1,500
$1,000
12
6
Useful life, in years
SO LUTI O N
吀栀e annual cost 昀漀r 12 years of Pump A can be 昀漀und by using Equation 6-6:
EUAC = (P - S)(AIP, i, n) + Si
= (7,000 - 1,500)(A/P, 7%, 12) + 1,500(0.07)
= 5,500(0. 1259) + 105 = $797
Now compute the annual cost 昀漀r six years of Pump B:
EUAC = (5,000 - 1,000)(A/P, 7%, 6) + 1,000(0.07)
= 4,000(0.2098) + 70 = $909
For a common analysis period of 12 years, we need to replace Pump B at the end of its six-year use昀甀l
life. If we assume that another pump, B', can be obtained, having the same $5,000 initial cost, $1,000
salvage value, and six-year life, the cash flow will be as 昀漀llows:
t
1 ,000
l
l
t
1 ,000
O -- l -- 2 -- 3 -- 4 -- S -- 6 -- 7 -- 8 -- 9 -- 1 0 -- 1 1 -- 12
5,000
5,000
椀ⴀ----- Pump B __________ Replacement Pump B ' ____
6 years
6 years
Analysis Period
183
For the 12-year analysis period, the annual cost 昀漀r Pump B is
EUAC = [5,000 - 1,000(P/F, 7%, 6) + 5,000(P/F, 7%, 6) - 1,000(P/F, 7%, 12)] (A/P, 7%, 12)
= [5,000 - 1,000(0 .6663) + 5,000(0.6663) - 1,000(0.4440)] (0.1259)
= (5,000 - 666 + 3,331 - 444)(0.1259)
= (7,21 1)(0.1259) = $909
吀栀e annual cost of Pump B 昀漀r the six-year analysis period is the same as the annual cost 昀漀r the
12-year analysis period. 吀栀is is not a surprising conclusion when one recognizes that the annual
cost of the 昀椀rst six-year period is repeated in the second six-year period. 吀栀us the lengthy calcu­
lation of EUAC 昀漀r 12 years of Pumps B and B' was not needed. By assuming that the shorter-life
equipment is replaced by equipment with identical economic consequences, we have avoided a lot
of calculations. Choose Pump A.
Analysis Period for a Com mon M u ltiple of Alternative Lives
When the analysis period is a common multiple of the alternative lives (昀漀r example, in
Example 6-5, the analysis period was 12 years, with six- and 12-year alternative lives), a
"replacement with an identical item with the same costs, per昀漀rmance, and so 昀漀rth" is
o昀琀en assumed. 吀栀is means that when an alternative has reached the end of its use昀甀l life,
we assume that it will be replaced with something identical. As shown in Example 6-5, the
result is that the EUAC 昀漀r Pump B with a six-year use昀甀l life is equal to the E唀䄀C 昀漀r the
entire analysis period because we assume Pump B is replaced by Pump B'.
Under these circumstances of identical replacement, we can compare the annual cash
flows computed 昀漀r alternatives on the basis of their own service lives. In Example 6-5, the
annual cost 昀漀r Pump A, based on its 12-year service life, was compared with the annual
cost 昀漀r Pump B, based on its six-year service life.
Analysis Period for a Conti n u i ng Req u i rement
Many times, an economic analysis is undertaken to determine how to provide 昀漀r a more
or less continuing requirement. For example, it might be a continuing requirement to
pump water 昀爀om a well. 吀栀ere is no distinct analysis period. In this situation, the analysis
period is assumed to be long but unde昀椀ned.
If, 昀漀r example, we had a continuing requirement to pump water and if alternative
Pumps A and B had use昀甀l lives of seven and 1 1 years, respectively, what should we do? 吀栀e
customary assumption is that Pump 氀椀s annual cash 昀氀ow (based on a seven-year life) may
be compared to Pump B's annual cash flow (based on an 11-year life). 吀栀is is done without
much concern that the least common multiple of the seven- and 11-year lives is 77 years.
吀栀is comparison of "di昀昀erent-life" alternatives assumes identical replacement (with iden­
tical costs, per昀漀rmance, etc.) when an alternative reaches the end of its use昀甀l life.
吀栀e continuing requirement, which can also be described as an inde昀椀nitely lo渀最 hori­
zon, is illustrated in Example 6-6. Since that is longer than the lives of the alternatives, we
can make the best decision possible given current in昀漀rmation by minimizing E唀䄀C. At a
later time we will make another replacement and there will be more in昀漀rmation on costs
at that time.
184
Cha pter 6 Annual Cash Flow Analysis
EXAM PLE 6 - 6
Pump B in Example 6-5 is now believed to have a nine-year use昀甀l life. Assuming the same initial cost
and salvage value, compare it with Pump A, using the same 7% interest rate.
SO LUTI O N
I f we assume that the need 昀漀 r A or B will exist 昀漀 r some continuing period, the comparison o f costs per
year 昀漀r the unequal lives is an acceptable technique. For 12 years of Pump A:
EUAC = (7,000 - 1,S00)(A/P, 7%, 12) + 1,500(0.07) = $797
For nine years of Pump B:
EUAC = (5,000 - 1,000)(A/P, 7%, 9) + 1,000(0.07) = $684
For minimum EUAC, choose Pump B.
䌀䐀
In昀椀nite Ana lysis Period
At times we have an alternative with a limited (昀椀nite) use昀甀l life in an in昀椀nite analysis
period situation. 吀栀e equivalent uni昀漀rm annual cost may be computed 昀漀r the limited
life. 吀栀e assumption of identical replacement (replacements have identical costs, per昀漀rm­
ance, etc.) is o昀琀en correct. 吀栀us the same EUAC occurs 昀漀r each replacement of the lim­
ited-life alternative. 吀栀e EUAC 昀漀r the in昀椀nite analysis period is there昀漀re equal to the
EUAC computed 昀漀r the limited life. With identical replacement,
EUAC.m昀椀m. te anaIys,. s per10. d = EUAC,cor 11. m1.ted i•cue n
A di昀昀erent situation occurs when there is an alternative with an in昀椀nite life in a problem
with an in昀椀nite analysis period:
EUAC.m昀椀m·te ana1ys,. s per10. d = P(A/P, i, 漀漀) + Any other annual costs
When n = 漀漀, we have A = Pi and, hence, (AIP, i, 漀漀) equals i.
EUACin昀椀nite anal sis pe iod = Pi + Any other annual costs
r
y
EXAM PLE 6-7
In the construction of an aqueduct to expand the water supply of a city, there are two alternatives 昀漀r a
particular portion of the aqueduct. Either a tunnel can be constructed through a mountain, or a pipe­
line can be laid to go around the mountain. If there is a permanent need 昀漀r the aqueduct, should the
tunnel or the pipeline be chosen 昀漀r this particular portion of the aqueduct? Assume a 6% interest rate.
Analysis Period
185
SO LUTI O N
Initial cost
Maintenance
Use昀甀l li昀攀
Salvage value
Tunnel through Mountain
Pipeline around Mountain
$5.5 million
$5 million
0
50 years
0
0
Permanent
0
Tunnel
For the tunnel, with its permanent life, we want (AIP, 6%, 漀漀). For an in昀椀nite life, the capital recovery is
simply the interest on the invested capital. So (AIP, 6%, 漀漀) = i, and we write
EUAC = Pi = $5.5 million(0.06)
= $330,000
Pipeline
EUAC = $5 million(A/P, 6%, 50)
= $5 million(0.0634) = $317,000
For 昀椀xed output, minimize EUAC. Choose the pipeline.
吀栀e di昀昀erence in annual cost between a long life and an in昀椀nite life is small unless an
unusually low interest rate is used. In Example 6-7 the tunnel is assumed to be permanent. For
purposes of comparison, compute the annual cost if an 85-year life is assumed 昀漀r the tunnel.
EUAC = $5,500,000(A/P, 6%, 85)
= $5,500,000(0.0604)
= $332,000
吀栀e di昀昀erence in time between 85 years and in昀椀nity is great indeed; yet the di昀昀erence in
annual costs in Example 6-7 is very small.
Some Other Ana lysis Period
吀栀e analysis period in a particular problem m愀礀 be something other than one of the 昀漀ur
we have described so 昀愀r. It m愀礀 be equal to the life of the shorter-life alternative, the
longer-life alternative, or something entirely di昀昀erent. One must care昀甀lly examine the
consequences of each alternative throughout the analysis period and, in addition, see what
di昀昀erences there might be in salvage values, and so 昀漀rth, at the end of the analysis period.
EXAM PLE 6 - 8
Suppose that Alternative 1 has a seven-year life and a salvage value at the end of that time. 吀栀e replace­
ment cost at the end of seven years may be more or less than the original cost. If the replacement is
retired in less than seven years, it may have a terminal value that exceeds the end-of-life salvage value.
continued
186
Cha pter 6 Annual Cash Flow Analysis
Alternative 2 has a 13-year life and a terminal value whenever it is retired. If the situation indicates that
10 years is the proper analysis period, set up the equations to compute the EUAC 昀漀r each alternative.
Use results 昀爀om Example 5-4 to compute the results.
SO LUTI O N
Alternative 1
EUAC1 = [Initial cost + (Replacement cost - Salvage value)(P/F, i, 7)
- (Terminal value)(P/F, i, IO)] (AIP, i, 10)
= 64,076(AIP, 8%, 10)
= 64,076(0.1490) = $9,547
Alternative 1
Salvage
Value
t
l
l
Terminal Value at
End of 10th Year
t
0 -- 1 -- 2 -- 3 -- 4 -- 5 -- 6 -- 7 -- 8 -- 9 -- 10 -l l - 12 - 13 - 14
Replacement
Initial
!
I
Cost
Cost
椀ⴀ----- 7 - Year Life -----ⴀ㄀帀堀----- 7 - Year Life -----ⴀ椀
I
Alternative 2
l
Terminal Value at
End of 10th Year
t
O -- I -- 2 -- 3 -- 4 -- 5 -- 6 -- 7 -- 8 -- 9 -- 10 -l l - 12 - 13
I:
fu�
Cost
13-year Life
椀ⴀ------- 10 - Year Analysis Period _______言蠀
Alternative 2
I
EUAC2 = [Initial cost - (Terminal value)(P/F, i, IO)] (AIP, i, IO)
= 68,052(AIP, 8%, 10)
= 68,052(0. 1490) = $10,140
Choose Alternative 1.
Mortgages i n Ca nada
Although technically a mortgage is a legal document, most people use the word to mean a
long-term amortized loan that is used 昀漀r buying real property such as a house or land. If
the mortgage p愀礀ments are not made, the lender can take the property and sell it to recover
the outstanding debt.
A mortgage document outlines the terms and conditions 昀漀r repaying the money bor­
rowed: the amount borrowed, the interest rate, the 昀椀rst and last payment dates, the amor­
tization period, and the date the balance is due (the renewal date or term). Prepayment
options and penalties may also be included.
Mortgages in Canada
187
Amortization is the length of time it would take to pay o昀昀 the mortgage-assuming
that the interest rate never changed, all payments were made on time, and no additional
payments were made. In Canada the shortest amortization is usually 昀椀ve years, and the
longest is 40 years, although the norm is 20 to 25 years. 吀栀e amortization of a mortgage is
made up of smaller periods called "terms." A term can be anywhere 昀爀om three months to
25 years. 吀栀e term is the period of time during which the interest rate is 昀椀xed. At the end
of the term, the mortgage can be renewed 昀漀r a new term at the prevailing rates of interest.
Build i ng a n Amortization Sched u le
An amortization schedule lists the 昀漀llowing 昀漀r each p愀礀ment period: loan p愀礀ment, inter­
est paid, principal paid, and remaining balance. For each period the interest paid equals the
interest rate times the balance remaining 昀爀om the period be昀漀re. 吀栀en the principal p愀礀­
ment equals the p愀礀ment minus the interest paid. Finally, this principal payment is applied
to the balance remaining 昀爀om the preceding period to calculate the new remaining balance.
吀栀is kind of calculation will be important when we get to Chapter 12, since interest
payments and repayments of principal are treated di昀昀erently by Revenue Canada.
Figure 6-1 shows this calculation 昀漀r Example 6-9.
EXAM PLE 6-9
An engineer wanted to celebrate graduating and getting a job by buying $2,400 worth of new 昀甀rniture.
Luckily the store was o昀昀ering six-month 昀椀nancing at the low interest rate of 6% per year nominal (really
0.5% per month). Calculate the amortization schedule.
SO LUTI O N
A
B
1
2,400 Initial balance
2
3
0.50%
4
C
D
E
i
6 N
$ 407.0 3 Payment
= - PMT(A2,A3 ,A l )
5
6
7
Month
Principal
Ending
Interest
Payment
Balance
8
0
2,400.00
=Al
9
1
12.00
3 95.0 3
2,004 .9 7
= D8 - C9
10
2
10.02
397 .00
1 ,607 .9 7
1 ,208 .98
11
3
8.04
12
4
6.04
3 98.99
400.98
13
5
4 .04
402.99
14
6
2.03
405.00
尀开
15
16
17
18
尀开
807.99
405.00
0.00
= $A$4 - B l4
= Payment- Interest
= $A$2*D l 3
= rate*previous balance
FIGURE 6-1 Amortization sched ule for fu rniture loa n .
continued
188
Cha pter 6 Annual Cash Flow Analysis
吀栀e 昀椀rst step is to calculate the monthly payment:
A = 2,400(A/P, 0.5%, 6) = 2,400(0. 1696) = $407.03
With this in昀漀rmation the engineer can use the spreadsheet in Figure 6-1 to obtain the amortization
schedule.
吀礀pes of Mortgage 䄀瘀ailable
Be昀漀re the advent of computers and imaginative interest calculations, most mortgages in
Canada were either "conventional mortgages" or "high-ratio mortgages." Conventional
mortgages are 昀漀r 80% (or less) of the appraised value of the property, and consequently
they require the purchaser to make a down payment of at least 20%. High-ratio mortgages
are those where the lender is providing an amount greater than 80% of the appraised value
of the property. Generally, lenders will do this only if an outside agency, usually Canada
Mortgage and Housing Corporation (CMHC), will insure the mortgage. CMHC charges
the borrower a fee 昀漀r this insurance.
Today the 昀漀llowing types of mortgages are also common: open mortgage, variable
rate mortgage, ARM (a搀樀ustable rate mortgage), capped rate mortgage, closed mortgage,
convertible rate mortgage, second mortgage, reverse mortgage, and CHIP mortgage (short
昀漀r Canadian Home Income Plan).
Equity
Equity is the value remaining in a property a昀琀er all mortgages and loans registered against
the title are subtracted 昀爀om the appraised value.
For example:
Appraised value $210,000
minus mortgage $150,000
minus second mortgage $25,000
equals equity $35,000
Mortgage I nterest Rate
In Canada the amount of interest charged to a borrower 昀漀r residential 昀椀rst mortgages is
stated as a percentage, 昀漀r example, 8.25% "compounded semi-annually, not in advance."
"Compounded semi-annually" is an old English term, 昀爀om the days when ledgers were
kept by hand. 吀栀e interest was calculated twice a year and added on to the principal owing,
and then payments were subtracted. Even though we now use computers, the same method
exists. 吀栀us 昀漀r a conventional Canadian mortgage the interest is stated as nominal, calcu­
lated semi-annually with a monthly payment period.
吀栀is means that a 6% mortgage is actually 3% semi-annually. To determine the actual
e昀昀ective monthly rate, one has to solve 昀漀r i in the 昀漀llowing equation:
(1 + i) 6 = 1 .03
i = 0.49% per month
S u m m a ry
SUMMARY
Annual cash 昀氀ow analysis is one more method 昀漀r choosing between alternatives by putting
their associated cash 昀氀ows into a comparable 昀漀rm. When an alternative has an initial cost P
and salvage value S, there are three ways of computing the equivalent uni昀漀rm annual cost:
• EUAC = P(AIP, i, n) - S(AIF, i, n)
• EUAC = (P - S)(AIF, i, n) + Pi
• EUAC = (P - S)(AIP, i, n) + Si
(6-1)
(6-3)
(6-4)
All three equations give the same answer. 吀栀is quantity is also known as the c愀瀀ital recov­
ery cost of the project.
吀栀e relationship between the present worth of cost and the equivalent uni昀漀rm annual
cost is:
• EUAC = (PW of cost)(AIP, i, n)
In present worth analysis there must be a common analysis period. Annual cash flow
analysis, however, allows some 昀氀exibility as long as the necessary assumptions are suitable
in the situation being studied. 吀栀e analysis period may be di昀昀erent 昀爀om the lives of the
alternatives, and if the 昀漀llowing conditions are met, a valid cash 昀氀ow analysis m愀礀 be made.
1.
It is assumed that when an alternative has reached the end of its use昀甀l life, it will
be replaced by an identical replacement (with the same costs, per昀漀rmance, etc.).
2. 吀栀e analysis period is a common multiple of the use昀甀l lives of the alternatives, or
there is a continuing or perpetual requirement 昀漀r the alternative chosen.
If neither condition applies, it is necessary to make a detailed study of the consequences
of the various alternatives over the entire analysis period, with particular attention to the
di昀昀erence between the alternatives at the end of the analysis period.
吀栀ere is very little numerical di昀昀erence between a long-life alternative and a perpetual
alternative. As the value of n increases, the capital recovery 昀愀ctor approaches i. At the
limit, (AIP, i, 漀漀) = i.
PROBLEMS
t
Annual Calculations
6-1
Using a 10% interest rate, compute the EUAC
昀漀r these cash 昀氀ows.
•
15
60
t l l
30
45
0 -- 1 -- 2 -- 3 -- 4
6-2
t
0 -- 1 -- 2 -- 3 -- 4 -- 5
Compute the EUAC 昀漀r these cash 昀氀ows:
i = 15%
100
6-3
100
Compute the EUAC 昀漀r these cash 昀氀ows:
I
60
l t +
45
30
15
0 -- 1 -- 2 -- 3 -- 4
i = 12%
189
190
6-4
Cha pter 6 Annual Cash Flow Analysis
昀漀r the device. 吀栀e purchasing agent was
very 昀漀rtunate to sell the inspection device
昀漀r $60,000, the original price. Compute
the equivalent uni昀漀rm annual cost during
the 昀漀ur years the device was used. Assume
interest at 10% per year.
If i = 6%, compute the EUAC that is equivalent
to the two receipts shown.
I
2 00
t
1 00
0 -- 1 -- 2 -- 3 -- 4 -- S-- 6
6-5
⠀　6-6
6-7
⠀　6-8
When she started work on her 22nd birth­
day, D.B. Cooper decided to invest money
each month with the objective of becoming a
millionaire by the time she reaches age 65. If
she expects her investments to yield 18% per
annum, compounded monthly, how much
should she invest each month?
吀栀e average age of engineering students at
graduation is a little over 23 years. 吀栀is means
that the working career of most engineers
is almost exactly 500 months. How much
would an engineer need to save each month to
become a millionaire by the end of her work­
ing career? Assume a 15% annual interest rate,
compounded monthly.
6-11
A 昀椀rm purchased some equipment at a very
昀愀vourable price of $30,000. 吀栀e equipment
resulted in an annual net saving of $1,000 a
year during the eight years it was used. At
the end of eight years, the equipment was
sold 昀漀r $40,000. Assuming interest is 8%,
did the equipment purchase prove to be
desirable?
6-12
A couple is saving 昀漀r their newborn daughter's
university education. She will need $25,000 a
year 昀漀r a 昀漀ur-year university program, which
she will start when she is 17. What uni昀漀rm
deposits starting three years 昀爀om now and
continuing through Year 16 are needed, if the
account earns 6% interest?
6-13
How much should a new graduate pay in 10
equal annual payments, starting two years
昀爀om now in order to repay a $30,000 loan
he has received today? 吀栀e interest rate is 6%
per year.
6-14
A 昀椀rm is buying an a搀樀acent 1,000-hectare
parcel of land 昀漀r a 昀甀ture plant expansion.
吀栀e price has been set at $30,000 a hectare.
吀栀e payment plan is 25% down, and the bal­
ance two years 昀爀om now. If the transaction
interest rate is 12% a year, what are the two
payments?
⠀　For the diagram, compute the value of D that
results in a net equivalent uni昀漀rm annual cost
(EUAC) of 0.
t t I t t
D
L SD
D
D
0 --I-- 2 -- 3 -- 4 -- 5
l
i = 1 2%
500
⠀　A 昀椀rm that is about to begin pilot operation
could add an optional heat exchanger unit.
A unit is now available 昀漀r $30,000, and it is
estimated that the heat exchanger unit will be
worth $35,000 at the end of eight years 昀漀r use
in other company operations. 吀栀is seemingly
high salvage value is because the $30,000 pur­
chase price is really a rare bargain. If the 昀椀rm
believes 15% is an acceptable rate of return,
what annual bene昀椀t is needed to justi昀礀 the
heat exchanger unit?
A loan of $1,000 is to be repaid in three equal
semi-annual payments. If the annual interest
rate is 7% compounded semi-annually, how
much is each payment?
D
6-9
6-10
An electronics 昀椀rm invested $60,000 in
a precision inspection device. 吀栀e device
cost $4,000 to operate and maintain in the
昀椀rst year and $3,000 in later years. At the
end of 昀漀ur years, the 昀椀rm changed its in­
spection procedure, eliminating the need
Problems
6-15
6-16
吀栀e manager of a small cleaning company
applies 昀漀r a $25,000 loan at an interest rate
of 10% per year. She will repay the loan over
six years with annual payments. 吀栀e third
through sixth payments are $1,500 greater
than the 昀椀rst two. Determine the size of the
payments.
Amanda and Blake have 昀漀und a house, which
because of a depressed real estate market costs
only $201,500. 吀栀ey will put $22,000 down and
昀椀nance the remainder with a 30-year mort­
gage loan 昀爀om the Central Imperial Bank of
Canada at 4.65% interest.
(a) How much is their monthly loan
payment?
(b) How much interest will they p愀礀 in the
second payment?
挀儀 吀栀ey will also have the 昀漀llowing ex­
penses: property taxes of $2,100, home­
owners' insurance of $1,625, and $290
mortgage insurance (in case one of them
dies be昀漀re the loan is repaid, a require­
ment of the bank). 吀栀ese annual amounts
are paid in 12 instalments and added to
the loan payment. What will Amanda and
Blake's 昀甀ll monthly cost be?
(d) If they can a昀昀ord $1,200 a month, can
Amanda and Blake a昀昀ord this house?
6-17
Helen buys a new Ford Focus. She negotiates
a price of $18,400, trades in her 2003 Contour
昀漀r $1,700, puts down an additional $1,000,
and borrows the remainder 昀漀r three years
at 6% interest. How large will her monthly
payments be?
6-18
To reduce her personal carbon 昀漀otprint, Zooey
is buying a new Ford Escape Hybrid. She has
negotiated a price of $21,900 and will trade in
her old Ford Contour 昀漀r $2,350. She will put
another $850 with it and borrow the remain­
der at 7% interest compounded monthly 昀漀r
昀漀ur years. Prepare a payment schedule 昀漀r the
昀椀rst three months of payments.
6-19
For the diagram, compute the value of C that
results in a net equivalent uni昀漀rm annual cost
(EUAC) of 0.
I l t
40
30
20
10
⸀尀
⸀尀
C
C
191
i i i i
0 -- 1 -- 2 -- 3 -- 4 -- 5
C
C
i = 10%
6-20 If interest is 10%, what is the EUAC?
I l t 1 t 'f I
40
30
40
20
o -- 1 -- 2 -- 3 -- 4 -- 5 -- 6 -- 7
6-21
吀栀e maintenance 昀漀reman of a plant, in
reviewing his records, 昀漀und that a large
press had the 昀漀llowing maintenance cost
record:
5 years ago
4 years ago
3 years ago
2 years ago
Last year
$ 600
700
800
900
1 ,000
A昀琀er consulting with a lubrication special­
ist, he changed the preventive maintenance
schedule. He believes that this year mainten­
ance will be $900 and will decrease by $100 a
year in each of the 昀漀llowing 昀漀ur years. If his
estimate of the 昀甀ture is correct, what will be
the equivalent uni昀漀rm annual maintenance
cost 昀漀r the IO-year period? Assume inter­
est at 8%.
6-22
A motorcycle is 昀漀r sale 昀漀r $26,000. 吀栀e motor­
cycle dealer is willing to sell it on the 昀漀llowing
terms:
No down payment; pay $440 at the end
of each of the 昀椀rst 昀漀ur months; pay
$840 at the end of each month a昀琀er that
until the loan has been paid in 昀甀ll
At a 12% annual interest rate compounded
monthly, how many $840 payments will be
required?
192
6-23
Cha pter 6 Annual Cash Flow Analysis
Anna, an engineer, has made a considerable
昀漀rtune. She wishes to start a perpetual schol­
arship 昀漀r engineering students at her alma
mater. 吀栀e scholarship will provide a student
with an annual stipend of $2,500 昀漀r each of
昀漀ur years, plus an additional $5,000 during
the 昀漀urth year to cover job search expenses.
Assume that students graduate in 昀漀ur years,
a new award is given every 昀漀ur years, and the
money is paid at the beginning of each year
with the 昀椀rst award at the beginning of Year 1.
吀栀e interest rate is 8%.
愀儀 Determine the equivalent uni昀漀rm annual
cost (EUAC) of providing the scholarship.
(b) How much money must Anna donate to
the university?
and 31 December). How much money should
Steve put into the account when he opens it on
1 September? What uni昀漀rm monthly deposit
should he make 昀爀om that time on?
6-27
Your company must make a $500,000 balloon
payment on a lease two years and nine months
昀爀om today. You have been directed to deposit
an amount of money quarterly, beginning
today, to provide 昀漀r the $500,000 payment.
吀栀e account pays 4% a year, compounded
quarterly. What quarterly deposit is required?
Note: Lease payments are due at the beginning
of the quarter.
6-28
Linda deposited $30,000 in a savings account
as a perpetual trust. She believes the account
will earn 7% annual interest during the 昀椀rst
10 years and 5% interest therea昀琀er. 吀栀e trust is
to provide a uni昀漀rm end-of-year scholarship
at the university. What uni昀漀rm amount could
be used 昀漀r the student scholarship each year,
beginning at the end of the 昀椀rst year and con­
tinuing 昀漀rever?
6-29
An engineer has a 昀氀uctuating 昀甀ture budget
昀漀r the maintenance of a particular machine.
During each of the 昀椀rst 昀椀ve years, $1,000 per
year will be budgeted. During the second
昀椀ve years, the annual budget will be $1,500 a
year. In addition, $3,500 will be budgeted 昀漀r
an overhaul of the machine at the end of the
昀漀urth year, and another $3,500 昀漀r an over­
haul at the end of the eighth year.
吀栀e engineer asks you to compute the
uni昀漀rm annual expenditure that would be
equivalent to these fluctuating amounts,
assuming interest is 6% per year.
6-24 A machine costs $20,000 and has a 昀椀ve-year
use昀甀l life. At the end of the 昀椀ve years, it can be
sold 昀漀r $4,000. If annual interest is 8%, com­
pounded semi-annually, what is the equivalent
uni昀漀rm annual cost of the machine?
6-25
6-26
Ms Wiggley wants to buy a new house. It will
cost $178,000. 吀栀e lending company will lend
90% of the purchase price at a nominal inter­
est rate of 10.75% compounded weekly, and
Ms Wiggley will make monthly payments.
What is the amount of the monthly payments
if she intends to pay the house o昀昀 in 25 years?
Steve must pay his property taxes in two equal
instalments on 1 December and 1 April. 吀栀e
two payments are 昀漀r taxes 昀漀r the 昀椀scal year
that begins on 1 July and ends the 昀漀llowing
30 June. Steve bought a home on 1 September.
He estimates the annual property taxes will
be $850 a year. Assuming the annual property
taxes remain at $850 a year 昀漀r the next several
years, Steve plans to open a savings account
and to make uni昀漀rm monthly deposits on the
昀椀rst of each month. 吀栀e account is to be used
to pay the taxes when they are due.
To begin the account, Steve deposits a
lump sum equivalent to the monthly pay­
ments that will not have been made 昀漀r the
昀椀rst year's taxes. 吀栀e savings account pays 9%
interest, compounded monthly and payable
quarterly (31 March, 30 June, 30 September,
6-30 A machine has a 昀椀rst cost of $150,000, an
annual operation and maintenance cost of
$2,500, a life of 10 years, and a salvage value
of $30,000. At the end of Years 4 and 8, it
requires major servicing, which costs $20,000
and $10,000, respectively. At the end of Year 5,
it will need to be overhauled at a cost of
$45,000. What is the equivalent uni昀漀rm
annual cost of owning and operating this par­
ticular machine?
Problems
6-31
⠀　吀栀ere is an annual receipt of money that varies
昀爀om $100 to $300 in a 昀椀xed pattern that is re­
peated 昀漀rever. If interest is 10%, compute the
EUAC, also continuing 昀漀rever, that is equiva­
lent to the 昀氀uctuating disbursements.
3 00
1
3 00
6-34
Jenny is an engineer 昀漀r a municipal power
plant. 吀栀e plant uses natural gas, which is cur­
rently obtained 昀爀om an existing pipeline at an
annual cost of $10,000 per year. Jenny is con­
sidering a project to construct a new pipeline.
吀栀e initial cost of the new pipeline would be
$35,000, but it would reduce the annual cost to
$5,000 per year. Assume an analysis period of
20 years and no salvage value 昀漀r either the ex­
isting or new pipeline. 吀栀e interest rate is 6%.
(a) Determine the equivalent uni昀漀rm annual
cost (EUAC) 昀漀r the new pipeline.
(b) Should the new pipeline be built?
6-35
Claude, a salesman, needs a new car 昀漀r use in
his business. He expects to be promoted to a
supervisory job at the end of three years, and
so he will no longer be on the road. 吀栀e com­
pany reimburses salespeople each month at
the rate of 55¢ per kilometre driven. Claude
has decided to drive a low-priced car. He 昀椀nds,
however, that there are three di昀昀erent ways of
obtaining the car:
(1) Purchase 昀漀r cash; the price is $26,000.
(2) Lease the car; the monthly charge is $700
on a 36-month lease, payable at the end of
each month; at the end of the three-year
period, the car is returned to the leasing
company.
(3) Lease the car with an option to buy it at
the end of the lease; pay $720 a month
昀漀r 36 months; at the end of that time,
Claude could buy the car, if he chooses,
昀漀r $7,000.
Claude believes he should use a 12% inter­
est rate. If the car could be sold 昀漀r $7,500 at
the end of three years, which method should
he use to obtain it?
6-36
When he bought his home, Al borrowed
$280,000 at 10% interest to be repaid in 25
equal annual end-of-year payments. A昀琀er
making 10 payments, Al has 昀漀und he can re­
昀椀nance the balance due on his loan at 9% inter­
est 昀漀r the remaining 15 years.
To re昀椀nance the loan, Al must pay the
original lender the balance due on the loan,
plus a penalty charge of 2% of the balance
due; to the new lender he also must pay a
3 00
f T r T ㄀昀 T r T T T r T
0 - l - 2 - 3 - 4 - S - 6 - 7 - 8 - 9 - 10 - 1 1 - 12 - · • • n = i = 10%
6-32
If the owner of a car earns 5% interest on her
investments, determine the equivalent annual
cost of owning a car with the 昀漀llowing costs:
Initial down payment = $2,200
Annual p愀礀ments = $5,500, EOYl - EO夀㐀
Pre-paid insurance = $1,500, growing 8%
annually
Gas, oil, and minor maintenance = $2,000,
growing 10% annually
Replacement tires = $650 at EOY4 & $800 at
EOY8
Major maintenance = $2,400 at EOY5
Salvage value = $3,750 at EOY9
⠀　Annual Comparisons
6-33
⠀　吀栀e Johnson Company pays $2,000 a month
to a trucker to haul waste paper and card­
board to the city dump. 吀栀e material could be
recycled if the company were to buy a $60,000
hydraulic press baler and spend $30,000 a
year 昀漀r labour to operate the baler. 吀栀e baler
has an estimated use昀甀l life of 30 years and
no salvage value. Strapping material would
cost $2,000 a year 昀漀r the estimated 500 bales
a year that would be produced. A waste paper
company will pick up the bales at the plant
and pay Johnson $23 per bale 昀漀r them. Use
an annual cash flow analysis in working this
problem.
(a) If interest is 8%, is it economical to install
and operate the baler?
(b) Would you recommend that the baler be
installed?
193
194
0
6-37
6-38
Cha pter 6 Annual Cash Flow Analysis
$1,000 service charge to obtain the loan. 吀栀e
new loan would be made equal to the bal­
ance due on the old loan, plus the 2% penalty
charge, and the $1,000 service charge. Should
Al re昀椀nance the loan, assuming that he will
keep the house 昀漀r the next 15 years? Use an
annual cash 昀氀ow analysis in working this
problem.
A company must decide whether to provide its
salespeople with company-owned cars or pay
them a mileage allowance and have them drive
their own cars. New cars would cost about
$28,000 each and could be resold 昀漀ur years
later 昀漀r about $1 1,000 each. Annual operating
costs would be $1,200 a year plus 12¢ per kilo­
metre. If the salespeople drove their own cars,
the company would probably pay them 50¢
per kilometre. Calculate the number of kilo­
metres each salesperson would have to drive
each year 昀漀r it to be economically practical
昀漀r the company to provide the cars. Assume a
10% annual interest rate. Use an annual cash
昀氀ow analysis.
吀栀e town of Oak Hills needs more water 昀爀om
Pine Creek. 吀栀e town engineer has selected
two plans 昀漀r comparison: a g爀愀vi琀礀 plan (divert
water at a point 10 km up Pine Creek and pipe
it by gravity to the town) and a pumping plan
(divert water at a point closer to the town
and pump it to the town). 吀栀e pumping plant
would be built in two stages, with half-capacity
installed initially and the other half installed
10 years later.
吀栀e analysis will assume a 40-year life,
10% interest, and no salvage value. Use an
annual cash 昀氀ow analysis to 昀椀nd which plan is
more economical.
Initial investment
Additional investment in
10th year
Operation and
maintenance
Power cost
䄀瘀erage 昀椀rst 10 years
䄀瘀erage next 30 years
Gravity
Pumping
$2,800,000
$1,400,000
None
200,000
10,000/yr
25,000/yr
None
None
50,000/yr
100,000/yr
6-39
A manu昀愀cturer is considering replacing a
production machine tool. 吀栀e new machine,
costing $37,000, would have a life of 昀漀ur years
and no salvage value, but it would save the
昀椀rm $5,000 a year in direct labour costs and
$2,000 a year in indirect labour costs. 吀栀e ex­
isting machine tool was purchased 昀漀ur years
ago at a cost of $40,000. It will last 昀漀ur more
years and will have no salvage value at the end
of that time. It could be sold now 昀漀r $10,000
cash. Assume that money is worth 8% and
that the di昀昀erence in taxes, insurance, and
so 昀漀rth 昀漀r the two alternatives is negligible.
Use an annual cash 昀氀ow analysis to determine
whether the new machine should be bought.
6-40 吀眀o possible routes 昀漀r a power line are under
study. Data on the routes are as 昀漀llows:
Length
First cost
Maintenance
Useful li昀攀, in years
Salvage value
Yearly power loss
Annual property
taxes
0
6-41
Around
the Lake
Under
the Lake
15 km
$5,000/km
$200/km/yr
15
$3,000/km
$500/km
2% of 昀椀rst
cost
5 km
$25,000/km
$400/km/yr
15
$5,000/km
$500/km
2% of 昀椀rst
cost
If 7% interest is used, should the power line be
routed around the lake or under the lake?
An oil re昀椀nery must begin sending its waste
liquids through a costly treatment process
be昀漀re discharging them into a nearby stream.
吀栀e engineering department estimates costs at
$300,000 昀漀r the 昀椀rst year. It is estimated that
with process and plant alterations, the waste
treatment cost will decline by $30,000 each
year. As an alternative, a specialized 昀椀rm, Hy­
dro-Clean, has o昀昀ered a contract to process the
waste liquids 昀漀r 10 years 昀漀r a 昀椀xed price of
$150,000 a year, p愀礀able at the end of each year.
Either way, there should be no need 昀漀r waste
treatment a昀琀er 10 years. 吀栀e re昀椀nery manager
considers 8% to be a suitable interest rate. Use an
annual cash 昀氀ow analysis to determine whether
he should accept the Hydro-Clean o昀昀er.
195
Problems
6-42
Betty buys a car every two years as 昀漀llows. She
makes a down p愀礀ment of $6,000 on a $15,000
car. She pays the balance in 24 equal monthly
p愀礀ments with annual interest at 12%. When
she has made the last p愀礀ment on the loan, she
trades in the two-year-old car 昀漀r $6,000 on a
new $15,000 car, and the cycle begins over again.
Doug decided on a di昀昀erent purchase
plan. He thought he would be better o昀昀 if he
paid $15,000 cash 昀漀r a new car. 吀栀en he would
make a monthly deposit in a savings account
so that, at the end of two years, he would
h愀瘀e $9,000 in the account. 吀栀e $9,000 plus
the $6,000 trade-in value of the car will allow
Doug to replace his two-year-old car by paying
$9,000 昀漀r a new one. 吀栀e bank pays 6% inter­
est, compounded quarterly.
(a) What is Betty's monthly payment to pay
o昀昀 the loan on the car?
(b) A昀琀er he has bought the new car 昀漀r cash,
how much per month should Doug deposit
in his s愀瘀ings account to h愀瘀e enough
money 昀漀r the next car two years hence?
挀儀 Why is Doug's monthly s愀瘀ings account
deposit smaller than Betty's payment?
6-43 吀眀o mutually exclusive alternatives are being
considered.
⠀　Year
0
1
2
3
4
5
A
B
-$3,000 -$5,000
+845 +1,400
+845 +1,400
+845 +1,400
+845 +1,400
+845 +1,400
One of the alternatives must be chosen. Using
a 15% nominal interest rate, compounded
continuously, determine which one. Solve by
annual cash 昀氀ow analysis.
6-44 North Plains Bio昀甀els (NPB) has negotiated a
contract with an oil 昀椀rm to sell 150,000 barrels
of ethanol per year, beginning at EOY4. 吀栀e oil
昀椀rm will pay NPB $10M annually, 昀爀om EOY0
to EOY3 and then $110 per barrel 昀爀om EO夀㐀
through EOY13. If NPB uses an interest rate of
15%, which method should be used to produce
the bio昀甀els?
Algae
Corn
Purchase of land (EOY 0)
Facility construction
(at EOY l)
Annual O&M increasing
6% yearly 昀爀om EOY l
through EOY 13
Raw materials ( corn or
algae) annual increasing
8% yearly 昀爀om EOY 4
through EOY 13
Salvage value (EOY 13)
$1,900,000 $3,800,000
$5,300,000 $7,100,000
$2,450,000 $2,800,000
$1,500,000
$250,000
$3,000,000 $3,600,000
6-45 Which car has a lower EUAC if the owner can
earn 5% in his best investment?
Toyota Corolla Toyota Prius
Initial cost
Annual
maintenance
Annual gas &
oil (increasing
15% yearly)
Salvage value
(EOY 8)
$19,200
1,000
$25,500
1,500
2,500
1,200
8,000
10,000
Di昀昀erent Lives
6-46 A 昀椀rm is choosing between machines that
per昀漀rm the same task in the same amount of
time. Assume the minimum attractive return
is 8%.
First cost
Estimated life, in
years
Salvage value
Annual maintenance
cost
Machine X
Machine Y
$5,000
5
$8,000
12
$0
$0
$2,000
$150
Which machine would you choose?
6-47 A company must decide whether to buy Ma­
chine A or Machine B:
⠀　Initial cost
Use昀甀l life, in years
End-of-useful-li昀攀
salvage value
Annual maintenance
Machine A
Machine B
$10,000
4
$10,000
$20,000
10
$10,000
$1,000
$0
196
Cha pter 6 Annual Cash Flow Analysis
At a 10% interest rate, which machine should
be installed? Use an annual cash flow analysis
in working this problem.
6-51
6-48 Consider the 昀漀llowing two mutually exclusive
alternatives:
Cost
Uni昀漀rm annual bene昀椀t
Use昀甀l life, in years
A
B
$10,000
$1,600
$15,000
$2,400
20
Alternative B may be replaced with an iden­
tical item every 20 years at the same $15,000
cost and will have the same $2,400 uni昀漀rm
annual bene昀椀t. Using a 10% interest rate and
an annual cash 昀氀ow analysis, decide which al­
ternative should be chosen.
First cost
Maintenance and
operating costs
Annual bene昀椀t
Salvage value
Useful li昀攀, in years
6-52
First cost
Annual operating cost
Annual maintenance
Salvage value
Life, in years
$2,400
$1,200
$300
$300
5
$6,000
$750
$50
$600
10
6-50 A suburban taxi company is considering
buying taxis with diesel engines instead of
gasoline engines. 吀栀e cars average 50,000 km
a year.
Vehicle cost
Use昀甀l life, in years
Fuel cost per litre
Mileage, in km/L
Annual repairs
Annual insurance premium
End-of-use昀甀l-life resale
value
Diesel
Gasoline
$24,000
5
$1.10
9
$900
$1,000
$4,000
$19,000
$15,000
$1,600
$25,000
$400
$8,000
$3,000
$13,000
$6,000
10
7
A
B
C
$10,000
$1,000
$150,000
$1,762
$20,000
$5,548
20
5
Assuming that Alternatives B and C are re­
placed with identical replacements at the end
of their use昀甀l lives, and an 8% interest rate,
which alternative should be selected? Use
an annual cash 昀氀ow analysis in working this
problem.
6-53
⠀　Carp, Inc. wants to evaluate two methods of
packaging its products. Use an interest rate of
15% and annual cash 昀氀ow analysis to decide
which is the most desirable alternative.
First cost
Maintenance and
operating costs
+ cost gradient
(begin Year 1)
Annual bene昀椀t
Salvage value
Use昀甀l life, in years
4
$1.20
7
$700
$1,000
$6,000
Use an annual cash 昀氀ow analysis to determine
the more economical choice if interest is 6%.
Machine B
Consider the 昀漀llowing three mutually exclu­
sive alternatives:
Cost
Uni昀漀rm annual
bene昀椀t
Useful life, in
years
⠀　Electric
Machine A
Assume an interest rate of 12%. Use annual
cash 昀氀ow analysis to determine which ma­
chine should be chosen.
6-49 A pump is needed 昀漀r 10 years at a remote
location. 吀栀e pump can be driven by an elec­
tric motor if a power line is extended to the
site. Otherwise, a gasoline engine will be
used. Use an annual cash 昀氀ow analysis and
a 10% interest rate. How should the pump be
powered?
Gasoline
吀栀e manager in a canned 昀漀od processing
plant is trying to decide between two labelling
machines.
A
B
$700,000
$18,000
$1,700,000
$29,000
+$900/yr
+$750/yr
$154,000
$142,000
10
$303,000
$210,000
20
6-54 A university student has been looking 昀漀r new
tires 昀漀r his car and has 昀漀und the 昀漀llowing
alternatives:
⠀　Problems
Tire Warran琀礀 (months)
Price per Tire
12
24
36
48
$39.95
59.95
69.95
90.00
Consider the 昀漀llowing alternatives:
Cost
Uni昀漀rm annual bene昀椀t
Use昀甀l life, in years
A
B
$5,000
$1,500
10
$18,000
$6,000
5
吀栀e analysis period is 10 years, but there will
be no replacement 昀漀r Alternative B a昀琀er 昀椀ve
years. Assuming a 15% interest rate, decide
which alternative should be chosen. Use an
annual cash 昀氀ow analysis.
6-56
A
B
$1,000
$550
3
$1,500
$610
4
At any time a昀琀er the equipment is in­
stalled, it has no salvage value. Assume that
Alternatives A and B will be replaced at the
end of their use昀甀l lives by identical equipment
with the same costs and bene昀椀ts. For a sev­
en-year analysis period and a 10% interest rate,
use an annual cash 昀氀ow analysis to determine
which alternative should be selected.
6-57
$2,500
$200
$100
4
$10,000
$100
$1,000
20
6-58
Dick Dickerson Construction Inc. has asked
you to help them choose a new backhoe. You
have a choice between a wheel-mounted ver­
sion, which costs $60,000 and has an expected
life of 昀椀ve years and a salvage value of $2,000,
and a track-mounted one, which costs $80,000
and has a seven-year life and an expected sal­
vage value of $10,000. Both machines will
achieve the same productivity. Interest is
8%. Which one will you recommend? Use an
annual cost analysis.
6-59
A small manu昀愀cturing company is evaluating
trucks 昀漀r delivering their products. Truck A has
a 昀椀rst cost of $22,000, its operating cost will be
$5,500 per year, and its salvage a昀琀er three years
will be $7,000. Truck B has a 昀椀rst cost of $27,000,
an operating cost of $5,200, and a resale value
of $12,000 a昀琀er 昀漀ur years. At an interest rate of
15% a year, which model should be chosen if an
annual worth analysis is per昀漀rmed?
Some equipment will be installed in a ware­
house that a 昀椀rm has leased 昀漀r seven years.
吀栀ere are two alternatives:
Cost
Uni昀漀rm annual bene昀椀t
Use昀甀l li昀攀, in years
Brick
First cost
Annual maintenance
Salvage value
Service life, in years
吀栀e student feels that the warranty period is
a good estimate of the tire life and that a 10%
interest rate is reasonable. Using an annual
cash 昀氀ow analysis, determine which tires he
should buy.
6-55
Masonite
Uncle Elmo needs to replace the 昀愀mily privy.
吀栀e local sanitary engineering 昀椀rm has submit­
ted two alternative structural proposals with cost
estimates as shown. Which construction should
Uncle Elmo choose if his minimum attractive
rate of return is 6%? Use both a present worth
and an annual cost approach in your comparison.
197
Spreadsheets and Loa ns
6-60 A student loan totals $18,000 at graduation.
吀栀e interest rate is 6%, and there will be 60
payments beginning one month a昀琀er gradua­
tion. What is the monthly payment? What is
owed a昀琀er the 昀椀rst two years of payments? Use
a spreadsheet.
6-61
A year a昀琀er buying her car, Annick has been
o昀昀ered a job in Europe. Her car loan is 昀漀r
$15,000 at a 9% nominal interest rate 昀漀r
60 months. If she can sell the car 昀漀r $12,000,
how much will she be able to keep a昀琀er paying
o昀昀 the loan? Use a spreadsheet.
6-62
(a) You are paying o昀昀 a debt at a nominal 8%
compounded quarterly per year by paying
$400 at the end of each quarter 昀漀r the
198
6-63
Cha pter 6 Annual Cash Flow Analysis
next year. Find the interest paid in the last
$400 payment.
(b) If this debt were to be paid o昀昀 in two
equal p愀礀ments of $1,650 at the end of
this year and at the end of the next year,
昀椀nd the interest paid in the 昀椀rst $1,650
payment. Again the loan rate is a nominal
8% per year compounded quarterly.
of each month over the next 10 years at a nom­
inal annual interest rate of 10%.
(a) What is the constant beginning-of-month
payment?
(b) Of the 昀椀rst payment, what are the interest
and the principal paid?
(c) Of the last payment, what are the interest
and the principal paid?
Sam can a昀昀ord to spend $500 a month on a car.
He 昀椀gures he needs half of that 昀漀r gas, park­
ing, and insurance. He has been to the bank,
and they will loan him 100% of the car's pur­
chase price. (Note: If he had a down payment
saved, he could borrow at a lower rate.)
愀儀 If his loan is at a nominal 12% annual rate
over 36 months, what is the most expen­
sive car he can a昀昀ord?
(b) 吀栀e car he likes costs $14,000, and the
dealer will 昀椀nance it over 60 months at
12%. Can he a昀昀ord it? If not, 昀漀r how
many months will he need to s愀瘀e his
$500 a month?
(c) What is the highest interest rate he can
pay over 60 months and st愀礀 within his
budget if he buys the $14,000 car now?
6-66 A $92,000 conventional Canadian mortgage
has a 30-year term and a 9% nominal interest
rate. Use a spreadsheet.
(a) What is the monthly payment?
(b) A昀琀er the 昀椀rst year of payments, what
昀爀action of the loan has been repaid?
(c) A昀琀er the 昀椀rst 10 years of payments, what
is the outstanding balance?
(d) How much interest is paid in Month 25?
How much principal?
6-64 Energ礀䴀ax Engineering constructed a small
o昀케ce building 昀椀ve years ago. 吀栀e company 昀椀­
nanced it with a bank loan 昀漀r $450,000 over
15 years at 12% interest, with quarterly pay­
ments and compounding. 吀栀e loan can be
repaid at any time without penalty. 吀栀e loan
can be re昀椀nanced through an insurance 昀椀rm
昀漀r 8% over 20 years-still with quarterly com­
pounding and payments. 吀栀e new loan has a
5% loan initiation fee, which will be added to
the new loan.
愀儀 What is the balance due on the original
bank loan (20 payments have been made
in the last 昀椀ve years)?
(b) How much will the payments drop with
the new loan?
(c) How much longer will the proposed loan
run?
6-65
Suppose you graduate with a debt of $42,000
that you must repay. One option is to pay o昀昀
the debt in constant amounts at the beginning
⠀　6-67 A 30-year conventional Canadian mortgage
昀漀r $95,000 is issued at a 9% nominal interest
rate. Use a spreadsheet.
(a) What is the monthly payment?
(b) How long does it take to pay o昀昀 the mort­
gage if $1,000 per month is paid?
(c) How long does it take to p愀礀 o昀昀 the mortgage if double payments are made?
6-68 A 30-year conventional Canadian mortgage
昀漀r $145,000 is issued at a 6% nominal interest
� rate. Use a spreadsheet.
(a) What is the monthly payment?
(b) How long does it take to pay o昀昀 the mort­
gage if $1,000 per month is paid?
(c) How long does it take to pay o昀昀 the mort­
gage if 20% extra is paid each month?
?
U nclassi昀椀ed
6-69 A job can be done with machine A, which costs
$12,500, has annual end-of-year mainten­
ance costs of $5,000, and has a salvage value
of $2,000 a昀琀er three years. Or the job can be
done with Machine B, which costs $15,000, has
end-of-year maintenance costs of $4,000, and
has a salvage value of $1,500 at the end of 昀漀ur
years. 吀栀ese investments can be repeated in the
昀甀ture, and your work is expected to continue
Problems
inde昀椀nitely. Compare the machines with
present worth, annual worth, and capitalized
cost. 吀栀e interest rate is 5% a year.
6-70
An engineer wishes to have $5 million by the
time he retires in 40 years. Assuming a nom­
inal interest rate of 15%, compounded con­
tinuously, what annual sum must he set aside?
6-71
A new car is purchased 昀漀r $12,000 with a 0%
down, 9% loan. 吀栀e loan is 昀漀r 昀漀ur years. A昀琀er
making 30 payments, the owner wants to pay
o昀昀 the balance of the loan. How much is owed?
Use a spreadsheet.
6-72
Zwango Plus Manu昀愀cturing expects that the
昀椀xed costs of keeping its Zephyr Hills Plant
operating will be $1.4M this year. If the 昀椀xed
costs increase by $100,000 each year, what is
the EUAC 昀漀r a IO-year period? Assume the
interest rate is 12%.
6-73
A construction 昀椀rm needs a new small loader.
It can be leased 昀爀om the dealer 昀漀r three years
昀漀r $5,500 per year including all maintenance,
or it can be bought 昀漀r $20,000. 吀栀e 昀椀rm expects the loader to have a salvage value of
$7,000 a昀琀er seven years. 吀栀e maintenance will
be $500 the 昀椀rst year, and then it will increase
by $300 each year. 吀栀e 昀椀rm's interest rate is
12% per year. Compare the EUACS 昀漀r leasing
and buying the loader.
6-74
A $78,000 conventional Canadian mortgage
has a 30-year term and a 9% nominal interest
rate. Use a spreadsheet.
(a) What is the monthly payment?
(b) A昀琀er the 昀椀rst year of payments, what is
the outstanding balance?
挀儀 How much interest is paid in Month 13?
How much principal?
199
M i n i - Cases
6-75
A certain o昀케ce building should last 60 years,
but the owner will sell it at 20 years 昀漀r 40% of
its construction cost. For the 昀椀rst 20 years it
can be leased as Class A space, which is all this
owner operates. When the building is sold, the
cost of the land will be recovered in 昀甀ll.
Land
$2.2M
Building
$4. lM
Annual operating and maintenance
$640,000
Annual property taxes and insurance 4%
(% of initial investment)
If the owner wants a 12% rate of return, what is
the required monthly leasing cost?
(a) Assuming that the building is vacant 5% of
the time, what must the monthly rent be?
(b) Give an example of monthly per-square­
昀漀ot rent 昀漀r Class A space in your
community.
6-76
A 30-unit apartment building should last
35 years, when it will need either to be replaced
or to undergo m愀樀or renovation. Assume the
value of the building at 35 years will be 10% of
its construction cost. Assume it will be sold and
that the cost of the land will be recovered in 昀甀ll.
Land
Building
Annual operating and maintenance
Annual property taxes and insurance
(% of initial investment)
Vacancy rate
$3.2M
$4.SM
$850,000
6%
12%
愀儀 If the owner wants a 15% rate of return,
what does the monthly rent 昀漀r each unit
have to be?
(b) If turning two units into an exercise 昀愀­
cility would decrease the vacancy rate by
昀椀ve percentage points, would that be a
good decision?
Rate of Return Analysis
Pay Now or Pay Later-The Story of the Giant Mine
The G i a nt M i ne was one of the fi rst la rg e-sca le gold m i nes in the N o rthwest Te rritories, a n d it
helped esta blish Ye llowknife as a frontier com m u n ity in the 1940s. Sta rt i n g fro m a n u m ber of pros ­
pect i n g cla i ms i n 1937, Frobisher Explorations sa n k the fi rst shaft i n 1945 a n d pou red the fi rst gold
brick i n 1948. From its sta rt - u p i n 1945, the m i ne went t h rou g h a series of owners ove r the yea rs
u nt i l 1999, when the then owner, Roya l Oak M i nes, went ba n kru pt. Ove r its h a lf ce ntu ry of oper­
ation the mine prod uced more tha n 7.6 m i llion ou nces of gold . (At a n ave ra ge price of $260 an
ou nce, this a mou nts to a tota l output va lue of $1,976,000,000.)
The refracto ry gold at the G i a nt M i ne was " roasted" out of the ore. Th is p rocess co nve rted nat­
u ra l ly occurring a rsenic i nto a rsenic trioxi d e powd e r. By the t i m e t h i s p ractice was d i sco nti n u ed, the
J udy Waytiu k/Ala my Stock Photo
Pay N ow or Pay Later-The Story of the G iant M ine
201
mine had produced some 237,000 tonnes of poisonous arsenic trioxide dust. Originally dumped
into tailings ponds or released to go up the stack, this product, in the early 1950s, started to be
disposed of in underground, mined-out chambers, where it remains to this day.
Putting the arsenic trioxide into these chambers proved an effective short-term measure, but
it was a stop-gap at best. Today, it is the responsibility of the federal government to deal with the
environmental legacy of the mine; none of the previous owners have any remaining responsibil­
ities. The current government proposal, which was passed in August 2014, is to freeze the arsenic
permanently in place with an active thermosiphon system around the chambers, convert to a pas­
sive thermosiphon system after a period of time, and monitor the arsenic into the distant future.
Other parties, including local Aboriginal and civil-society groups, would like either to see the
arsenic removed for disposal elsewhere or to have guarantees that the system will be monitored
in perpetuity and that if technologies emerge that allow for a more permanent solution (e.g., safer
removal of the arsenic), they would be considered. These groups have two particular concerns:
•
Can monitoring and treatment continue in perpetuity, as is required in this case? (How
much faith do we have that the government will still be monitoring and managing the
site in 100, 250, or 1,000 years and that it will have the capacity and desire to fund the
continued upkeep of the system?)
•
Will climate change, seismic shifts, geological instability, or a mixture of these factors
render the proposed freezing system inoperable, leading to the release of arsenic triox­
ide into local waterways from underground?
Estimates of cost for the Frozen Block method are in the range $900 million- $1,000 million,
with an ongoing $2 million per year to monitor and maintain the site.
QUESTIONS TO CONSIDER
1.
The mine has been a profitable venture because the costs to the environment were, in the jargon of econo­
mists, "externalized. " That is, they were passed on to someone else, like the community, the health system,
and future generations. If the original developers had taken responsibility for the long-term environmental
costs, would the mine have been an economically viable undertaking?
2.
Should developers have to pay for the entire proposed reclamation costs in advance of development? If
so, should this amount be based on a lowest-cost estimate, a most-likely cost scenario, or a worst-case
scenario with the option of a refund with interest if this scenario does not come true?
3.
Is economic feasibility a relevant factor when options for the control and disposal of hazardous wastes are
being considered? How heavily should economic feasibility be weighted, and against which other factors
(e.g., engineering and geotechnical feasibility, length of time until probable failure, long-term maintenance
requirements, magnitude of human and ecosystem health risks)?
LEARNING OBJECTIVES
This chapter will help you
•
•
evaluate project cash flows with the inte爀渀al rate of retu爀渀 (IRR) measure
use an incremental rate of return analysis to evaluate competing alternatives
202
Cha pter 7 Rate of Return Analysis
•
conduct a sensitivity analysis by plotting the present worth (PW) of a project against the
interest rate
•
recognize when to calculate the modi昀椀ed inte爀渀al rate of retu爀渀 (M IRR)
KEY TERMS
incremental analysis
incremental investment
internal rate of return
min imum attractive rate
of retu rn
mod ified internal rate
of retu rn
N PW plot
rate of retu rn analysis
In this chapter we will examine one more analysis method, 爀愀te of return. First, the mean­
ing of rate of return is explained and methods of calculating it are illustrated; then, the
method of incremental analysis is presented. Lastly, we describe di昀케culties sometimes en­
countered when computing an interest rate 昀漀r cash 昀氀ow series with multiple sign changes.
Rate of return analysis is the most 昀爀equently used measure in industry. Problems in
computing the rate of return sometimes occur, but its m愀樀or advantage is that it is a single
昀椀gure of merit that is readily understood.
Consider these statements:
•
•
•
儀⤀
吀栀e net present worth on the project is $32,000.
吀栀e equivalent uni昀漀rm annual net bene昀椀t is $2,800.
吀栀e project will produce a 23% annual rate of return.
While none of these statements tells the complete story, the third one measures the
desirability of the project in terms that are widely and easily understood: 昀漀r every dollar
we invest in the project, we expect to have $1.23 a year later.
吀栀ere is another advantage to rate of return analysis. In present worth, 昀甀ture worth,
and annual cash 昀氀ow calculations, you have to select an interest rate-and the exact value
may be di昀케cult and controversial. In rate of return analysis, no interest rate is introduced
into the calculations (except in the case of the modi昀椀ed internal rate of return). Instead,
we compute a rate of return (more accurately called internal rate of return) 昀爀om the
cash 昀氀ow. 吀栀e calculated rate of return is then compared with a pre-selected minimum
attractive rate of return, or simply MARR.
Minimum Attractive Rate of Return
吀栀e purpose of rate of return analysis, like all the methods we study in this text, is to allow
us to decide between di昀昀erent possible projects. Once we have calculated that the rate of
return on a particular project is, 昀漀r example, 23%, what do we do next?
We need some 昀漀rm of benchmark against which we can compare this number. What
could we do with our money if we didn't invest in the project? We could le愀瘀e the money
in our pocket, where it would earn a rate of return of zero. Or we could put it in the bank,
where it would earn whatever interest rate the bank pays on deposits. Putting money in the
bank is simpler than running a business, so unless our project o昀昀ers a rate of return at least
as high as the bank, we may as well leave the money in the bank.
Minimum Attractive Rate of Return
O昀琀en the money we invest in a project is someone else's money. We m愀礀 borrow the
money 昀爀om the bank-at a rate that is typically a little higher than the bank o昀昀ers us
on deposits-or we can get the money 昀爀om shareholders, who will expect some 昀甀ture
recompense in the 昀漀rm of dividends or increased equity. 吀栀e average rate at which we
have to recompense our creditors and investors sets a lower bound on the rate at which we
want our project to earn money.
Lastly, suppose we are already in business and have an ongoing project that earns an
attractive rate of return on any money invested in it. Any new project that we consider
should o昀昀er at least as high a rate; otherwise we may as well put any additional 昀甀nds into
the existing project.
Each of these considerations sets a lower bound on the rate of return at which a pro­
posed project becomes attractive; the highest of these lower bounds is the minimum
attractive rate of return, or MARR. It is discussed in greater detail in Chapter 9.
Internal Rate of Return
Consider a very simple case. You go to the roulette wheel at the casino and bet a sum of
money on red. Red comes up, and you get double your money back. Your rate of return on
your bet is there昀漀re 100%. You earned this 100% rate of return over the relatively short
time it took to place and collect your bet, so the corresponding annual rate of return would
be remarkably high, if you could go on winning.
Instead of gambling, we might buy a painting 昀漀r $5,000 and, a year later, sell it 昀漀r
$10,000. Clearly, our annual rate of return on this investment is 100%. We can also write
this as:
or
5,000 = 10,000(P/F, 100%, 1)
{7-1)
PW = 5,000 - 10,000(P/F, 100%, 1) = 0
{7-2)
吀栀at is, the internal rate of return (IRR) is that value of i at which the equivalent pres­
ent value of all the cash 昀氀ows associated with the project is zero. 吀栀is de昀椀nition o昀琀en seems
counterintuitive when we 昀椀rst encounter it-surely, we think, we would like the present
value to be a large positive number, not zero? But you should be able to convince yourself
that Equation 7-2 is just a restatement of Equation 7-1, and that Equation 7-1 meets our
commonsense de昀椀nition 昀漀r a rate of return.
Consider a third example. We invest $5,000 in a machine tool with a 昀椀ve-year use昀甀l
life and an equivalent uni昀漀rm annual bene昀椀t of $1,252. What rate of return do these cash
昀氀ows correspond to?
t t t t t
$ 1 ,252
0 -- 1 -- 2 -- 3 -- 4 --s
i
$5,000
We write the equation
5,000 = 1,252(P/A, i, 5)
203
204
Cha pter 7 Rate of Return Analysis
or
PW = 5,000 - 1,252{PIA, i, 5) = 0
and solve 昀漀r i. We recognize this case as identical to Plan 3 in Table 3-1, 昀漀r which
i = 8%, and conclude that the internal rate of return 昀漀r the machine-tool purchase
must be 8% .
Calcu lati ng Rate of Retu rn
To calculate a rate of return on an investment, we must convert the various consequences
of the investment into a cash 昀氀ow series. 吀栀en we will solve the cash 昀氀ow series 昀漀r the
unknown value of the IRR, using the equation
Net present worth = 0
{7-3)
It should be immediately obvious that, since equivalent annual cost and 昀甀ture cost
can both be obtained by multiplying PW by the appropriate 昀椀xed 昀愀ctor, the equations
EUAC = 0 or FW = 0 can be substituted 昀漀r Equation 7-3 when convenient.
EXAM PLE 7-1
An engineer invests $5,000 at the end of every year during a 40-year career. If she wants $ 1 million in
savings at retirement, what interest rate must the investment earn?
SO LUTI O N
Using Equation 7-3, we write
Rewriting, we see that
Net PW = 0 = -$5,000(FIA, i, 40) + $1,000,000
(䤀蜀, i, 40) = $1,000,0001$5,000 = 200
We then look at the compound interest tables 昀漀r the value of i where (䤀蜀, i, 40) = 200. If no tabulated
value of i gives this value, then we will interpolate, using the two closest values, or solve exactly with a
calculator or spreadsheet. In this case, (䤀蜀, 0.07, 40) = 199.636, which to three signi昀椀cant digits equals
200. 吀栀us the required rate of return 昀漀r the investment is 7%.
Although this example has been de昀椀ned in terms of an engineer's personal 昀椀nances, we could
just as easily h愀瘀e said, '✀䄀 mining 昀椀rm makes annual deposits of $50,000 into a reclamation 昀甀nd 昀漀r
40 years. If the 昀椀rm must h愀瘀e $10 million when the mine is closed, what interest rate must the invest­
ment earn?" Since the annual deposit and required 昀椀nal amount are 10 times larger in each case, the
FIA 昀愀ctor and the answer are obviously the same.
Calculating Rate of Return
205
EXAM PLE 7-2
An investment resulted in the 昀漀llowing cash 昀氀ow. Compute the rate of return.
✀椀 1 1 l
1 75
2
325
0-- I-- 2 -- 3 --4
- 700
SO LUTI O N
EUAC = 0 = -100 - 75(A/G, i, 4) + 700(A/P, i, 4)
Here the interest rate appears twice in the equation, so we will solve by trial and error. 吀栀e EUAC value
is a 昀甀nction of i. Try i = 5% 昀椀rst:
EUAC (5%) = -100 - 75(AIG, 5%, 4) + 700(AIP, 5%, 4)
= -100 - 75(1.439) + 700(0.2820)
= -208 + 197 = -11
吀栀e EUAC is too low. If the interest rate is increased, EUAC will increase. Try i = 8%:
EUAC(8%) = -100 - 75(A!G, 8%, 4) + 700(A/P, 8%, 4)
= -100 - 75(1 .404) + 700(0.3019)
= -205 + 211 = 6
吀栀is time the E唀䄀C is too large. We see that the true rate of return is between 5% and 8%. Try i = 7%:
EUAC(7%) = -100 - 75(AIG, 7%, 4) + 700(AIP, 7%, 4)
= -100 - 75(1.416) + 700(0.2952)
= 206 - 206 = 0
吀栀e IRR is 7%.
EXAM PLE 7-3
A local 昀椀rm sponsors a student loan program 昀漀r the children of employees. No interest is charged until
graduation, and then the interest rate is 5%. Maria borrows $9,000 every year, and she graduates a昀琀er
continued
206
Cha pter 7 Rate of Return Analysis
昀漀ur years. Since tuition must be paid ahead of time, assume that she borrows the money at the start of
each year.
If Maria makes 昀椀ve equal annual p愀礀ments, how much is each payment? Use the cash 昀氀ow 昀爀om
when she started borrowing the money to when it is all paid back, and then calculate the internal rate
of return 昀漀r Maria's loan. Is this arrangement attractive to Maria?
SO LUTI O N
Maria owes $36,000 at graduation. 吀栀e 昀椀rst step is to calculate the 昀椀ve equal annual payments to repay
this loan at 5%.
Loan p愀礀ment = $36,000(AIP, 5%, 5) = 36,000(0.2310) = $8,316
Maria receives $9,000 by borrowing at the start of each year. She graduates at the end of Year 4. At the
end of Year 4, which is also the beginning of Year 5, interest starts to accrue. She makes her 昀椀rst pay­
ment at the end of Year 5, which is one year a昀琀er graduation.
t t t t
9,000
O --l-- 2 -- 3 -- 4 -- 5 -- 6 -- 7 -- 8 -- 9
i
i
i
- 8, 3 16
i
i
吀栀e next step is to write the present worth equation in 昀愀ctor 昀漀rm, so that we can apply Equation 7-3
and set it equal to 0. 吀栀is equation has three 昀愀ctors, so we will have to solve the problem by picking
interest rates and substituting values. 吀栀e present worth value is a 昀甀nction of i.
PW(i) = 9,000 [1 + (䤀鄀, i, 3)] - 8,316(PIA, i, 5)(PIF, i, 4}
吀栀e 昀椀rst two interest rates we try are 0%, because it is easy; and 3%, because the subsidized rate will be
below the 5% that is charged a昀琀er graduation.
At 0%, any 䤀鄀 昀愀ctor equals n, and any PIP 昀愀ctor equals 1.
PW(0%} = 9,000(4) - 8,316(5) = -5,580
PW(3%} = 9,000(1 + 2.829} - 8,316(4.580)(0.8885} = 620.5
Since PW(i) has opposite signs 昀漀r 0 and 3, there is a value of i between 0% and 3% which is the IRR.
Because the value 昀漀r 3% is closer to 0, the IRR will be closer to 3%. Try 2% next.
PW = (2%} = 9,000(1 + 2.884} - 8,316(4.713)(0.9238) = -1,251
As shown in Figure 7-1, interpolating between 2% and 3% leads to
IRR = 2% + (3% - 2%} [1,251/(1,251 + 620.8)] = 2.67%.
Calculating Rate of Return
吀栀is rate is quite low, and it makes the loan look like a good choice.
1 ,000
( 3 %, 620.5)
e
500
j
0
V
∀✀
V
⸀尀
鸀尀
- 500
z - 1 ,000 ------------(2%, - 1 ,25 1 )
- 1 ,500 ⬀娀--鼀蜀--嬀蜀--蜀딀--娀夀
4
2
3
1
0
Interest Rate (%)
FIGURE 7-1 Plot of PW versus i nterest rate i.
If in Figure 7-1 net present worth (NPW) had been computed 昀漀r a broader range of
values of i, Figure 7-2 would have been obtained. From this 昀椀gure it is apparent that the
error resulting 昀爀om linear interpolation increases as the interpolation width increases.
䌀䐀
Plot of N PW versus I nterest Rate i
吀栀e plot of NPW versus interest rate i is an important source of in昀漀rmation. For a cash
flow where borrowed money is repaid, the NPW plot would appear as in Figure 7-2. 吀栀e
borrowed money is received early in the time period, with a later repayment of an equal
sum plus payment of interest on the borrowed money. In all cases in which interest is
charged and the amount borrowed is 昀甀lly repaid, the NPW at 0% will be negative.
For a cash 昀氀ow representing an investment 昀漀llowed by bene昀椀ts 昀爀om the investment,
the plot of NPW versus i (we will call it an NPW plot 昀漀r convenience) would have the
昀漀rm of Figure 7-3. As the interest rate increases, 昀甀ture bene昀椀ts are discounted more heav­
ily and the NPW decreases.
-
e 12,000
V
V
∀✀
⸀尀
6,000
鸀尀
V
z
0
0
5
10
Interest Rate (%)
15
20
FIGURE 7-2 Replot of N PW versus interest rate i over a la rger ra nge of values.
207
208
Cha pter 7 Rate of Return Analysis
Year
0
1
Cash Flow
+P
-A
-A
-A
-A
2
3
4
+
z O ⴀꄀ-_렀_____
FIGURE 7-3 N PW plot for a typical i nvestment.
Interest is a charge 昀漀r the use of someone else's money or a receipt 昀漀r letting others
use our money. 吀栀e interest rate is almost always positive, but negative interest rates do
occur. A loan with a 昀漀rgiveness provision (meaning not all principal is repaid) can have a
negative rate. Some investments per昀漀rm poorly and have negative rates.
EXAM PLE 7-4
A new corporate bond was initially sold by a stockbroker to an investor 昀漀r $1,000. 吀栀e issuing corpora­
tion promised to pay the bondholder $40 interest on the $1,000 昀愀ce value of the bond every six months,
and to repay the $1,000 at the end of 10 years. A昀琀er one year the bond was sold by the original buyer
昀漀r $950.
(a) What rate of return did the original buyer receive on his investment?
(b) What rate of return can the new buyer (paying $950) expect to receive if he keeps the bond 昀漀r
its remaining nine-year life?
SO LUTI O N TO PART (a)
吀栀e original bondholder sold the bond 昀漀r less ($950) than she paid 昀漀r it ($1,000), so the semi-annual
rate of return is less than the 4% semi-annual rate of return on the bond.
J
950
40
t t
0 -- 1 -- 2
l
1 ,000
Since $40 is received each six months, we will use a six-month interest period to solve the problem. Let
PW of cost = PW of bene昀椀ts, and write
1,000 = 40(P!A, i, 2) + 950(PIF, i, 2)
Calculating Rate of Return
Try i = 1.5%:
209
1,000 = 40(1.956) + 950(0.9707) = 78.24 + 922.17 = 1,000.41
吀栀e interest rate per six months, IRR6 mon' is very close to 1.5%. 吀栀is means the nominal (annual) inter­
est rate is 2 x 1.5% = 3%. 吀栀e e昀昀ective (annual) interest rate = IRR = (1 + 0.015)2 - 1 = 3.02%.
SO LUTI ON TO 倀䄀RT (b}
吀栀e new buyer will redeem the bond 昀漀r more ($1,000) than he paid 昀漀r it ($950). So the semi-annual
rate of return is more than the 4% semi-annual interest rate on the bond.
1 ,000
t t t t t t t t t t t t t t t t t t
A = 40
0 - l - 2 - 3 - 4 - S - 6- 7 - 8 - 9-10 - l l - 12 - 1 3 - 14 - 1 5 - 16 - 1 7 - 1 8
n = 18
950
950 = 40(PIA, i, 18) + 1,000(P/F, i, 18)
Given the same $40 semi-annual interest payments, 昀漀r six-month interest periods we write
Try i = 5%:
950 ± 40(1 1 .690) + 1,000(0.4155) = 467.60 + 415.50 = 883.10
950 ± 40(12.659) + 1,000(0.4936) = 50.36 + 493.60 = 999.96
吀栀e PW of bene昀椀ts is too low. Try a lower interest rate, say, i = 4%:
950 ± 40(12.659) + 1, 000(0.4936) = 50.36 + 493.60
± 999.96
吀栀e value of the six-month rate i is between 4% and 5%. By interpolation,
i = 4% + ( 1%) [
999.96 - 950.00 )
= 4.43%
999.96 - 883. 10
吀栀e nominal annual interest rate is 2 x 4.43% = 8.86%. 吀栀e e昀昀ective annual interest rate or IRR is
(1 + 0.0443)2 - 1 = 9.05%.
I nterest Rates When There Are Fees or Discou nts
O昀琀en when 昀椀rms and individuals borrow money, there are fees charged in addition to the
interest. 吀栀is can be as simple as the underwriting fee that a 昀椀rm is charged when it sells a
bond. In Example 7-5 we add that underwriting fee to the bond in Example 7-4 and look at
the bond 昀爀om the 昀椀rm's perspective rather than the investor's.
210
Cha pter 7 Rate of Return Analysis
EXAM PLE 7-5
吀栀e corporate bond in Example 7-4 was part of a much larger o昀昀ering that the 昀椀rm arranged with the
underwriter. Each of the bonds had a 昀愀ce value of $1,000 and a life of 10 years. Since $40 at 4% of the
昀愀ce value was paid in interest every six months, the bond had a nominal interest rate of 8% a year. If
the 昀椀rm paid the underwriter a 1 % fee to sell the bond, what is the e昀昀ective annual interest rate that the
昀椀rm is paying on the bond?
SO LUTI O N
From the 昀椀rm's perspective, it receives $1,000 minus the fee at Time 0 , then it pays interest every six
months 昀漀r 10 years, and then it pays $1,000 to redeem the bond. 吀栀e 1 % fee reduces to $990 what the
昀椀rm receives when the bond is sold. 吀栀e interest payments are $40 every six months. 吀栀is is easiest to
model with 20 six-month periods.
PW(i) = 990 - 40(䤀鄀, i, 20) - 1,000(P/F, i, 20)
Since the nominal interest rate is 4% every six months, we know that the fee will raise this somewhat.
So let us use the next higher table of 4.5%.
PW(4.5%) = 990 - 40(PIA, 4.5%, 20) - 1,000(P/F, 4.5%, 20)
= 990 - 40(13.008) - 1,000(0.4146) = $55.08
We know that the PW of the interest and 昀椀nal bond payo昀昀 is $1,000 at 4%.
PW(4%) = 990 - 1,000 = -10
Now we interpolate to 昀椀nd the interest rate 昀漀r each six-month period.
i = 4% + (4.5% - 4%)(10)/(10 + 55.08) = 4.077%
吀栀e e昀昀ective annual rate is
i = 1 .040772 - 1 = 0.0832 = 8.32%
Examples 7-4 and 7-5 were about borrowing money through bonds and loans, but
many applications 昀漀r the rate of return are stated in other ways. Example 7-6 is a common
problem on university campuses-buying parking permits 昀漀r an academic year or 昀漀r one
term at a time.
Buying a year's parking permit is investing more money now to avoid paying 昀漀r an­
other shorter permit later. Choosing to buy a shorter permit can be modelled as a loan,
where the money saved by not buying the annual permit is borrowed to be repaid with the
cost of the second-term permit.
Interest Rates When There Are Fees or Discounts
211
EXAM PLE 7-6
An engineering student is deciding whether to buy two one-term parking permits or an annual permit.
吀栀e annual parking permit costs $180, due on 15 August: the term permits are $130, due 15 August and
15 January. What is the rate of return 昀漀r buying the annual permit?
SO LUTI O N
Be昀漀re we solve this mathematically, let u s describe it i n words. We are comparing the $50 cost di昀昀er­
ence now between the two permits with the $130 cost of buying another term permit in 昀椀ve months.
Since the $130 is 2.6 times the $50, it is clear that we will get a high interest rate.
吀栀is is most easily solved by using monthly periods; the payment 昀漀r the second term is 昀椀ve months
later. 吀栀e cash flow table adds a column 昀漀r the incremental di昀昀erence to the in昀漀rmation in the cash
昀氀ow diagram.
1 5 Aug.
t
1 5 Jan.
�
180
=
15 Aug.
15 Jan.
130
130
i i
Time
Annual
Pass
Term
Pass
Incremental
Di昀昀erence
15 Aug.
-$180
-$130
-$50
-130
130
15 Jan.
Setting the two PWs equal to each other, we have
-$180 = -$130 [1+(PIP, imon , 5)]
(PIP, imon , 5) = -$501-$130 = 5113 = 0.3846
Rather than interpolating, we can use the 昀漀rmula 昀漀r the PIP 昀愀ctor.
1/(1 + imon) 5 = 0.3846
(1 + imon) 5 = 2.600
1 + imon = 1.2106
imon = 21 .06%, which is an extremely high rate per month
On an annual basis, the e昀昀ective interest rate is (1.2106 12 - 1) = 891 %. Unless the student is planning to
graduate in January, it is clearly better to buy the permit a year at a time.
212
Cha pter 7 Rate of Return Analysis
䌀䐀
Incremental Analysis
You 昀椀nd yourself obliged to spend a year in a small town in northern Canada. You have
$20,000 in cash. 吀栀e only investment alternatives in town are as 昀漀llows:
Year
Alt. I
Alt. 2
0
1
- $ 1 0,000
+ 1 5,000
- $20,000
+28,000
Any money you have le昀琀 over can be deposited in the town bank at 6% interest.
You 昀椀rst calculate the rates of return on the two alternatives. Alternative 1 has a rate
of return of 50%, while Alternative 2 only returns 40%. So, dimly remembering what you
learned in engineering economics, you decide to go with Alternative 1, since it has the
higher rate of return.
A year later, you evaluate your net worth. You have $15,000 昀爀om Alternative 1, and
$10,600 昀爀om depositing your remaining money in the bank, giving you a total of $25,600.
It occurs to you that if you'd put all your money in Alternative 2 instead, you'd now h愀瘀e
$28,000. What went wrong?
This draws our attention to a hidden pit昀愀ll in the rate of return method: at the
end of the day, we want to maximize our net wealth, not our rate of return. Fortun­
ately we can avoid this pit昀愀ll by using the 昀漀llowing algorithm to compare multiple
alternatives:
1.
2.
3.
4.
5.
6.
Calculate the rate of return 昀漀r each alternative, and discard any 昀漀r which
IRR < MARR.
Arrange the remaining alternatives in ascending order of 昀椀rst cost.
吀栀e alternative with lowest 昀椀rst cost is your current champion.
Calculate the incremental IRR of upgrading 昀爀om the current champion to the
alternative with next-lowest 昀椀rst cost.
If the incremental IRR > MARR, upgrade; otherwise, stick with the current
champion.
Repeat steps 4 and 5 until you run out of alternatives.
To apply this algorithm, you need to know your MARR. In the next section, "Sensi­
tivity Analysis," we will see how to deal with multiple alternatives when you don't know
your MARR.
We apply this algorithm to the two alternatives just considered:
We will select the lesser-cost alternative (Alternative 1), unless we 昀椀nd that the addi­
tional cost of Alternative 2 produces enough additional bene昀椀ts to make it preferable.
We can compute the rate of return on the di昀昀erences between the alternatives.
Year
Alt. I
0
1
- $ 1 0,000
+ 1 5,000
0
Alt. 2
Alt. 2 - Alt. I
- $20,000 - $20,000 - ( - $ 1 0,000) = - $ 1 0,000
+28,000 +28,000 - (+ 1 5,000) = + 1 3,000
= PWA!t. 2-Alt. l = -1 0, 000 + 13, 000 (P/F, i, 1)
We can see that if $10,000 increases to $13,000 in one year, the interest rate must
be 30%. 吀栀e compound interest tables con昀椀rm this conclusion. 吀栀e 30% rate of return
I n cremental Analysi s
213
on the di昀昀erence between the alternatives is 昀愀r higher than the 6% MARR. 吀栀e addi­
tional $10,000 investment to obtain Alternative 2 is superior to depositing the $10,000
in the bank at 6%. To obtain this desirable incremental rate of return, Alternative 2 is
selected.
EXAM PLE 7-7
If the computations above do not convince you, and you still think Alternative 1 would be preferable,
try this problem.
You have $20 in your wallet and two di昀昀erent ways of lending Bill some money.
(a) Lend Bill $10 with his promise of a 50% return. 吀栀at is, he will pay you back $15 at the agreed
time.
(b) Lend Bill $20 with his promise of a 40% return. He will pay you back $28 at the same agreed
time.
You can choose whether to lend Bill $10 or $20. 吀栀is is a one-time situation, and any money not lent to
Bill will remain in your wallet. Which alternative do you choose?
SO LUTI O N
A 50% return on the smaller sum is less rewarding to you than 40% on the larger sum. Since you would
prefer to have $28 than $25 ($15 昀爀om Bill plus $10 remaining in your wallet) a昀琀er the loan is paid, lend
Bill $20.
Incremental rate of return analysis is illustrated by Examples 7-8 and 7-9.
Example 7-8 illustrates a very important point that was not part of our earlier exam­
ples of parking permits, insurance, and equipment. In those cases the decision that we
needed the insurance, permit, or equipment had already been made. We needed only
to decide the best way to obtain it. In Example 7-8 there is a do-nothing alternative that
must be considered.
EXAM PLE 7-8
If an electromagnet is installed on the input conveyor of a coal-processing plant, it will pick up scrap
metal in the coal. Removing this scrap will save an estimated $1,200 a year in costs associated with
damage to the machinery. 吀栀e electromagnetic equipment has an estimated use昀甀l life of 昀椀ve years
and no salvage value. 吀眀o suppliers have been contacted: Leaseco will provide the equipment in
return 昀漀r three beginning-of-year annual payments of $1,000 each; Saleco will provide the equip­
ment 昀漀r $2,783. If the MARR is 10%, should the project be done, and which supplier should be
chosen?
continued
214
Cha pter 7 Rate of Return Analysis
SO LUTI O N
Be昀漀re we analyze which supplier should be chosen, we must decide whether the do-nothing alterna­
tive would be a better choice. Since the 昀椀rst cost of Leaseco is lower than the 昀椀rst cost with Saleco, let
us compare Leaseco with doing nothing. 吀栀is is the same as 昀椀nding the present worth or IRR of the
Leaseco investment.
吀栀e cash 昀氀ow at Time 0 昀漀r Leaseco is -$1,000. At the end of Years 1 and 2, the 昀椀rm spends $1,000
on the lease and saves $1,200 in machinery damage 昀漀r a net of $200. 吀栀en 昀漀r three years the 昀椀rm saves
$1,200 annually. Since the 昀椀rm is investing $1,000 to save $4,000 spread over 昀椀ve years, the arrange­
ment is clearly worthwhile. In 昀愀ct, at the 10% MARR, the PW is $1,813.
To compare Saleco and Leaseco, we 昀椀rst recognize that both will provide equipment with the same
use昀甀l life and bene昀椀ts. In rate of return analysis, the method of solution is to examine the di昀昀erences
between the alternatives. We 昀椀nd the incremental cash 昀氀ows each year to evaluate the additional in­
vestment required by Saleco.
Saleco
Di昀昀erence between
Alternatives: Saleco - Leaseco
-$2,783
-$1,783
1
+1,200
+1,000
2
+1,200
+1,000
Year
0
Leaseco
-$1,000
3
+1,200
+1,200
0
4
+1,200
+1,200
+1,200
+1,200
0
5
0
Compute the NPW at various interest rates on the incremental investment represented by the di昀昀er­
ence between the alternatives.
PW*
Year
n
Cash Flow:
Saleco - Leaseco
At 0%
At 8%
At 20%
At 漀漀%
0
-$1,783
-$1,783
-$1,783
-$1,783
-$1,783
1
+1,000
+1,000
+926
+833
0
2
+1,000
+1,000
+857
+694
0
3
0
0
0
0
0
4
0
0
0
0
0
5
0
0
0
0
0
+217
0
-256
-1,783
NPW =
'Each year the cash flow is multiplied by (PIP, i, n).
At 0%: (PIP, 0%, n) = 1 昀漀r all values of n
䄀琀 %: (PIP, %, 0) = 1
(PlP, %, n) = 0 昀漀r all other values of n
From the plot of these data in Figure 7-4, we see that NPW = 0 at i = 8%.
I n cremental Analysi s
215
Interest Rate (%)
FIGURE 7-4 N PW plot for Example 7-8.
吀栀us, the incremental rate of return-�IRR-of choosing Saleco rather than Leaseco is 8%. 吀栀is is less
than the 10% MARR. 吀栀ere昀漀re choose Leaseco.
EXAM PLE 7-9
A 昀椀rm is considering which of two devices to install to reduce costs. Both devices have use昀甀l lives of
昀椀ve years and no salvage value. Device A costs $1,000 and can be expected to result in $300 in savings
annually. Device B costs $1,300 and will provide cost savings of $300 the 昀椀rst year, but savings will
increase $50 annually, making the second-year savings $350, the third-year savings $400, and so 昀漀rth.
For a 7% MARR, which device should the 昀椀rm buy?
SO LUTI O N
吀栀is problem has been solved by present worth analysis (Example 5-1). 吀栀is time we will use rate of
return analysis, which must be applied to the incremental investment.
Year
Device A
Device B
Di昀昀erence between Alternatives:
Device B - Device A
0
1
2
3
4
5
-$1,000
+300
+300
+300
+300
+300
-$1,350
+300
+350
+400
+450
+soo
-$350
0
+so
+100
+150
+200
For the di昀昀erence between the alternatives, write a single equation with IRR as the only unknown.
PW(i) = 0 = -350 + 50(PIG, i, 5)
We know (PIG, i, 5) = 7, so i must be between 9% (since (PIG, 9%, 5) = 7.111) and 10% (since (PIG, 10%, 5)
= 6.862)
Interpolating, i = [9% + (10% - 9%)(7.111 - 7)]1(7.111 - 6.862) = 9.45%
吀栀e 9.45% IRR is greater than the 7% MARR; there昀漀re, the increment is desirable. Reject Device A
and choose Device B.
216
Cha pter 7 Rate of Return Analysis
Ana lysis Period
In discussions of present worth analysis and annual cash 昀氀ow analysis, an import­
ant consideration is the analysis period. 吀栀is is also true in rate of return analysis.
吀栀e method 昀漀r solving 昀漀r two alternatives is to examine the di昀昀erences between the
alternatives.
In Example 7-10 the analysis period is a common multiple of the alternative service
lives, and identical replacement is assumed. 吀栀is problem illustrates an analysis of the
di昀昀erences between the alternatives over the analysis period.
EXAM PLE 7-10
吀眀o machines are being considered 昀漀r purchase. If the MARR is 10%, which machine should be
bought? Use an IRR analysis comparison.
Machine X
Machine Y
$200
$700
Uni昀漀rm annual bene昀椀t
95
120
End-of-useful-life salvage
value
50
150
Use昀甀l li昀攀, in years
6
12
Initial cost
SO LUTI O N
吀栀 e solution i s based o n a 12-year analysis period and a replacement Machine X that i s identical to the
present Machine X. 吀栀e cash 昀氀ow 昀漀r the di昀昀erences between the alternatives is as 昀漀llows:
Year
Machine X
Machine Y
Di昀昀erence between Alternatives
Machine Y - Machine X
0
1
2
3
-$200
+95
+95
+95
+95
+95
-$700
+120
+120
+120
+120
+120
-$500
+25
+25
+25
+25
+25
+120
+25
+150
+120
+120
+120
+ 120
+120
+25
+25
+25
+25
+25
+120
+150
+25
+100
4
5
6
7
8
9
10
11
12
+95
+95
+95
+95
+95
I n cremental Analysi s
217
PW of cost (di昀昀erences) = PW of bene昀椀ts (di昀昀erences)
500 = 25(䤀鄀, i, 12) + 150(P/F, i, 6) + lO0(P/F, i, 12)
吀栀e sum of the bene昀椀ts over the 12 years is $550, which is only a little greater than the $500 additional
cost. 吀栀is suggests that the rate of return is quite low. Try i = I%.
25(P/A , 1% , 12) + 150(P/F, 1% , 6) + lO0(P/F, 1% , 12)
= 25(1 1.255) + 150(0.942) + 100(0.887) = 5 1 1
吀栀e interest rate i s too low. Try i = 1.5%:
25(䤀鄀, 1 .5% , 12) + 150(P/F, 1 . 5% , 6) + lO0(P/F, 1 .5% , 12)
= 25(10.908) + 150(0.914) + 100(0.836) = 494
吀栀e internal rate of return on the Y - X increment is about 1.3%, 昀愀r below the 10% minimum attractive
rate of return. 吀栀e additional investment to obtain Machine Y yields an unsatis昀愀ctory rate of return;
there昀漀re Machine X is the preferred alternative.
Sensitivity Analysis
It can o昀琀en happen in engineering problems that we need to make tentative conclusions
about a project while one or more important parameters remain undetermined. Under
these circumstances, it is o昀琀en possible to use sensitivity analysis: we construct a graph
showing what per昀漀rmance we would expect 昀漀r a range of plausible values of the undeter­
mined parameter(s).
We saw in the previous section that, in order to apply our algorithm 昀漀r selecting
projects based on incremental rate of return, we needed to know the MARR. If we don't
know the MARR, we can still obtain a basis 昀漀r decision by plotting the project's net
present worth as a 昀甀nction of the unknown MARR, as illustrated in Examples 7-1 1
and 7-12.
吀栀e general method involves the 昀漀llowing steps:
1.
2.
3.
Write down an expression 昀漀r the PW or E唀䄀C of each alternative, with the
unknown MARR as a 昀爀ee parameter.
Plot PW (or EUAC) versus MARR 昀漀r each alternative on the same graph.
Note which alternative gives the maximum PW (or minimum E唀䄀C) at each
value of MARR, and the crossover points between alternatives.
吀栀is method involves a bit more work than the case where we know the MARR but
can be made quite straight昀漀rward and painless if you use a spreadsheet program to con­
struct the graph. Note that you can always get two points on the graph, 昀漀r MARR = 0 and
MARR = 漀漀, with relatively little calculation.
218
Cha pter 7 Rate of Return Analysis
EXAM PLE 7-11
A pressure vessel can be made out of brass, stainless steel, or titanium. 吀栀e 昀椀rst cost and expected life
昀漀r each material are as 昀漀llows:
Brass
Cost
Life, in years
Stainless Steel Titanium
$175,000
10
$100,000
4
$300,000
25
吀栀e pressure vessel will be in the non-radioactive portion of a nuclear power plant that is expected to
have a life of 50 to 75 years. 吀栀e public utility commission and the power company have not yet agreed
on the interest rate to be used 昀漀r making decisions and setting rates. Build a choice table to determine
the best alternative at each interest rate.
SO LUTI O N
吀栀 e pressure vessel will b e replaced repeatedly during the life o f the 昀愀cility, and each material has a dif­
ferent life. 吀栀us, the best way to compare the materials is by using EUAC (see Chapter 6). 吀栀is assumes
identical replacements.
Figure 7-5 graphs the EUAC 昀漀r each alternative. In this case the best alternative at each interest
rate is the material with the lowest EUAC.
吀栀e 昀愀ctor equation is
EUAC = 昀椀rst cost(AIP, i, life)
50,000 �----------✓.�-/
IRR1itanium - stainless = 6 . 3 %
/
40,000 ⬀娀---ⴀ堀----/·-蜀딀-销쀀렀쐀
/
/
Stainless
Brass
stainless - brass
o �--�---�---�--�
1 0,000 ⬀娀---ⴀ堀------ ⬀娀---
Titanium
0
5
IRR
10
Interest Rate (%)
15
= 15.3%
20
-- Brass - - - - Stainless - - Titanium
FIGURE 7-5 EUAC compa rison of alternatives.
吀栀e choice table 昀漀r each material is
Interest Rate
Best Choice
0% � i � 6.3%
6.3% � i � 15.3%
15.3% � i
Titanium
Stainless steel
Brass
Sensitivity Analys is
219
EXAM PLE 7-12
吀栀e 昀漀llowing in昀漀rmation is 昀漀r 昀椀ve mutually exclusive alternatives that have 20-year use昀甀l lives. 吀栀e
decision maker may choose any one of the options or reject them all. Prepare a choice table.
Alternatives
Cost
Uni昀漀rm annual bene昀椀t
A
B
C
D
E
$4,000
639
$2,000
410
$6,000
761
$1,000
117
$9,000
785
SO LUTI O N
Figure 7-6 i s a n NPW graph o f the alternatives constructed by means of a spreadsheet.
1 0,000
--A
Do
8 ,000
-B
-c
6,000
---D
---E
4,000
Do
2,000
Do nothing
0
- 2,000
- 4,000
- 6,000
0
FIGURE 7-6
5
10
Interest Rate (%)
15
20
N PW g ra p h .
吀栀e graph clearly shows that Alternatives D and E are never part of the solution. 吀栀ey are dominated by
the other three. 吀栀e crossover points can either be read 昀爀om the graph (if you have plotted it at a large
enough scale) or 昀漀und by calculating the �IRR of the intersecting graphs.
Calculate the incremental interest rates:
�IRR (C - A)
$6,000 - $4,000 = ($761 - $639)(P/A, i, 20),
�IRR (A - B)
$4,000 - $2,000 = ($639 - $410)(P/A, i, 20),
i = 2%
i = 9.6%
Find where the NPW of B crosses the 0 axis:
�IRR (B)
$2,000 = $410 (䤀鄀, i, 20),
i = 20%
continued
220
Cha pter 7 Rate of Return Analysis
Put these numbers into a choice table:
If
If
If
If
20%
9.6%
2%
MARR �
� MARR �
� MARR �
� MARR
20%
9.6%
2%
do nothing
select B
select A
select C
A 昀椀nal point to note on this example is that if we view the IRRs of the 昀椀ve alternatives, the only
in昀漀rmation we can glean 昀爀om them is that if the do-nothing alternative is available, it will be
chosen if MARR is greater than the largest IRR. 吀栀ere is nothing in the numbers to tell us which
alternatives will be in the 昀椀nal solution set; only the NPW graph will show us where the change
points will be. 1
A
IRR
C
D
E
15% 20% 11%
10%
6%
B
吀栀ere are analytical techniques 昀漀r determining which incremental investments enter the solution set, but they are
beyond the scope of this book and, in any case, are redundant in this era of spreadsheets. Interested readers can consult
Economic Analysis for Engineers and Managers by Sprague and Whittaker (1986).
1
A Second Pitfall
We now discover a second potential pit昀愀ll in the rate of return method.
EXAM PLE 7-13
I have a small research company. A government agency o昀昀ers me a contract: I get $20,000 in start-up
昀甀nds now. If all goes well, I put in $180,000 of my own money in three years' time to develop a
prototype. If the prototype works, I get $200,000 in 昀椀ve years' time. What is my rate of return if all
goes well?
SO LUTI O N
I write down an expression 昀漀r the present worth of the project:
PW (in thousands of dollars) = 20 - 180(P/F, i, 3) + 200(P/F, i, 5) = 0
If i = 0, the PW is 20,000 -180,000 + 200,000 = $40,000
And if i = 漀漀, the PW of the project is $20,000
吀栀is is mildly worrying: we will be looking 昀漀r values of i that make PW = 0, but these values suggest
that the PW may always be positive.
A Second Pitfall
221
Present Worth
50 ⴀ嬀--------------------------娀夀
- 20 �--------------------------�
Rate of Return
FIGURE 7-7 N PW g raph.
Figure 7-7 shows that the curve does cut the horizontal axis, not once but twice. What does this mean?
How can there be two solutions 昀漀r the rate of return?
Let us look at the solution IRR = 80%. 吀栀is is an attractively high rate of return, but can it be
believed?
T he 80% rate of return corresponds to the situation where, as soon as I get the $20,000 昀爀om
the government, I invest it at 80% interest. T hree years later, I have 20,000 (PIF, 80%, 3) = $1 16,000.
So, when called upon to invest $180,000 of my own money, I can cover most of it 昀爀om the return
on my initial investment, leaving me with just $64,000 to come up with out of my own pocket.
At the end of the 昀椀nal two years of the project, that $64,000 brings in $200,000, which is again a
return of 80%.
吀栀e downside to this happy scenario is that I can't actually 昀椀nd anywhere to invest the initial
$20,000 that would give me an 80% rate of return. So my conclusion is based on a 昀椀ction.
We ran into trouble here because the sequence of cash 昀氀ows in this project is unusual. In the
typical situation, we make an investment (or place a bet), and subsequently we receive income (or
collect our winnings). But in this situation, we start o昀昀 with money coming in and are later obliged
to spend our own money. To stay in touch with reality, we can't assume that our initial income can
be invested at an arbitrarily high rate. Instead, we should limit our expectations to what we might
get 昀爀om a bank.
吀栀is insight is the 昀漀undation of the modi昀椀ed internal rate of return method, or
MIRR.
吀眀o external rates of return can be used to ensure that the resulting equation is solv­
able 昀漀r a unique internal rate of return-the MIRR. 吀栀e MIRR is a measure of the attract­
iveness of the cash 昀氀ows, but it is also a 昀甀nction of the two external rates of return.
吀栀e rates that are external to the project's cash 昀氀ows are (1) the rate at which the organ­
ization normally invests and (2) the rate at which it normally borrows. 吀栀ese are external
rates 昀漀r investing, ein瘀✀ and 昀漀r 昀椀nancing, e n · Because pro昀椀table 昀椀rms invest at higher
昀椀
222
Cha pter 7 Rate of Return Analysis
rates than they borrow at, the rate 昀漀r investing is generally higher than the rate 昀漀r 昀椀nan­
cing. Sometimes a single external rate is used 昀漀r both, but this requires the questionable
assumption that investing and 昀椀nancing happen at the same rate.
吀栀e approach is as 昀漀llows:
Combine cash 昀氀ows in each period (t) into a single net receipt, R t, or net
expense, Bf
2. Find the present worth of the expenses with the 昀椀nancing rate.
3. Find the 昀甀ture worth of the receipts with the investing rate.
4. Move the present worth of the expenses into the 昀甀ture using the MIRR, and set
this equivalent to the 昀甀ture worth calculated in Step 3.
1.
吀栀e result is Equation 7-4. 吀栀is equation will have a unique root, since it has a single
negative present worth and a single positive 昀甀ture worth. 吀栀ere is only one sign change in
the resulting series.
(7-4)
吀栀ere are other external rates of return, but the MIRR has historically been the
most clearly de昀椀ned. Since all of the external rates of return are a昀昀ected by the assumed
values 昀漀r the investing and 昀椀nancing rates, none is a true rate of return on the project's
cash 昀氀ow.
Since the IRR sometimes takes on multiple, unrealistic values, while the MIRR
only ever has a single value, you might wonder why economists don't always use the
MIRR. 吀栀e reason is that cash 昀氀ow patterns yielding multiple solutions 昀漀r the IRR
are relatively rare, and the MIRR algorithm requires slightly more work than the IRR
algorithm.
We should mention that some texts use the expression "external rate of return,"
or "ERR," to refer to the concept that we have de昀椀ned as MIRR. 吀栀e same texts use
the expression "auxiliary rate of return" to refer to the concept that we have referred
. .
to as e mv
$ 3 . SM
$2. S M
t
External Investing Rate
$ 1 . SM
+
$0. S M
o -- 1 -- 2 -- 3-- 4 -- 5 -- 6 -- 7
- $0. S M
- $ 1 . SM
- $2. S M
- $ 4M
External �-----------------�
Financing
Rate
P( l + MIRRr = F
FIGURE 7-8 M I RR for the oil well.
A Second Pitfall
223
EXAM PLE 7-14
Adding an oil well to an existing 昀椀eld had the cash flows summarized in Figure 7-8. If the 昀椀rm normally
borrows money at 8% and invests at 15%, 昀椀nd the modi昀椀ed internal rate of return (MIRR).
SO LUTI O N
We apply the algorithm described above:
1.
2.
Each period's cash flow i s already a single net receipt o r expenditure.
Find the present worth of the expenses with the 昀椀nancing rate.
3.
Find the 昀甀ture worth of the receipts with the investing rate.
4.
Use the MIRR to move the present worth 昀漀und in Step 2 into the 昀甀ture, and equate it with the
昀甀ture worth 昀漀und in Step 3.
PW = -4M - 0.5M(PIF, 8%, 5) - I . 5M(PIF, 8%, 6) - 2.5M(PIF, 8%, 7)
= -4M - 0.5M(0.6806) - 1 .5M(0.6302) - 2.5M(0.5835) = -6.㜀㐀4M
FW = 3.5M(䘀䤀P, 15%, 6) + 2.5M(FIP, 15%, 5) + I .5M(FIP, 15%, 4) + 0.5M(䘀䤀P, 15%, 3)
= 3.5M (2.313) + 2.5M(2.01 1) + 1 . 5M(l .㜀㐀9) + 0.5M(l .521) = 16.507M
0 = (1 + MIRR) " (PW) + FW
0 = (1 + MIRR) 7 (-6.744M) + 16.507M
(1 + MIRR) 7 = 16.507M/6.㜀㐀4M = 2.448
(1 + MIRR) = 2.448 1 17 = 1 . 1 364
MIRR = 13.64%
S U M MARY
Internal rate of return is the interest rate i at which the net present worth of the cash flows
associated with a project is zero.
To choose between multiple projects, use incremental rate of return analysis, as out­
lined in this algorithm:
1.
2.
3.
4.
5.
6.
Calculate the rate of return 昀漀r each alternative, and discard any 昀漀r which IRR <
MARR.
Arrange the remaining alternatives in ascending order of 昀椀rst cost.
吀栀e alternative with lowest 昀椀rst cost is your current champion.
Calculate the incremental IRR of upgrading 昀爀om the current champion to the
alternative with next-lowest least cost.
Ifthe incremental IRR > MARR, upgrade; otherwise, stick with the current champion.
Repeat steps 4 and 5 until you run out of alternatives.
224
Cha pter 7 Rate of Return Analysis
When the MARR is unknown, you can instead use sensitivity analysis. 吀栀e general
method involves the 昀漀llowing steps:
Write down an expression 昀漀r the NPW or E唀䄀C of each alternative, with the
unknown MARR as a 昀爀ee parameter.
2. Plot NPW (or EUAC) versus MARR 昀漀r each alternative on the same graph.
3. Note which alternative gives the maximum NPW (or minimum EUAC) at each
value of MARR, and the crossover points between alternatives.
1.
Some patterns of cash 昀氀ows may yield multiple solutions 昀漀r the IRR. In these cases,
you can use the modi昀椀ed internal rate of return, or MIRR:
Combine cash 昀氀ows in each period (t) into a single net receipt, R t, or net
expense, Ef
2. Find the present worth of the expenses with the 昀椀nancing rate.
3. Find the 昀甀ture worth of the receipts with the investing rate.
4. Move the present worth of the expenses into the 昀甀ture using the MIRR, and set
this equivalent to the 昀甀ture worth calculated in Step 3.
1.
吀栀e result is Equation 7-4.
(7-4)
PROBLEMS
Rate of Retu rn
7-1
7-2
7-3
⠀　a month 昀漀r 12 months beginning 30 days
hence. If you choose the time payment plan,
what nominal annual interest rate will you be
charged?
Compute the rate of return 昀漀r the 昀漀llowing
cash flow to within 0.5%.
Year
Cash Flow
0
1 - 10
-$ 100
+27
吀栀e Diagonal Stamp Company, which sells used
postage stamps to collectors, advertises that its
average price has increased 昀爀om $1 to $5 in the
last 昀椀ve years. 吀栀us, management states, in­
vestors who bought stamps 昀爀om Diagonal 昀椀ve
years ago would have received a 100% rate of
return each year.
(a) To check their calculations, compute the
annual rate of return.
(b) Why is your computed rate of return less
than 100%?
A table saw costs $175 at a local store. You may
either pay cash 昀漀r it or pay $35 now and $12.64
7-4
An investment of $5,000 in Biotech common
stock proved to be very pro昀椀table. At the end of
three years the stock was sold 昀漀r $25,000. What
was the rate of return on the investment?
7-5
Helen is buying a $12,375 car with a $3,000
down payment, 昀漀llowed by 36 monthly p愀礀­
ments of $325 each. 吀栀e down payment is paid
immediately, and the monthly payments are
due at the end of each month. What nominal
annual interest rate is Helen p愀礀ing? What ef­
昀攀ctive interest rate?
7- 6
Peter Minuit bought an island 昀爀om the Man­
hattoes Indians in 1626 昀漀r $24 worth of glass
beads and trinkets. 吀栀e 1991 estimate of the
value of land on this island was $12 billion.
What rate of return would the Manhattoes have
Problems
received if they had retained title to the island
rather than selling it 昀漀r $24?
7-7
⠀　An engineer invests $5,800 at the end of every
year of a 35-year career. If he wants $1 million
in savings at retirement, what interest rate must
the investment earn?
7-8
A mining 昀椀rm makes annual deposits of
$250,000 into a reclamation 昀甀nd 昀漀r 20 years.
If the 昀椀rm must have $10 million when the
mine is closed, what interest rate must the
investment earn?
7-9
You spend $1,000 and in return receive two
payments of $1,094.60-one at the end of three
years and the other at the end of six years. Cal­
culate the resulting rate of return.
⠀　7-10 Your cat just won the local feline lottery to
the tune of 3,000 cans of 9-Lives cat 昀漀od (as­
sorted 昀氀愀瘀ours). A local grocer o昀昀ers to take the
3,000 cans and, in return, supply 30 cans a
month 昀漀r the next 10 years. What rate of return,
in terms of nominal annual rate, will you realize
on this deal? (Compute to the nearest 0.01%.)
7-11 A woman went to the Bene昀椀cial Loan Com­
pany and borrowed $3,000. She must pay
$1 19.67 at the end of each month 昀漀r the next
30 months.
(a) Calculate the nominal annual interest rate
she is paying to within ±0.15%.
(b) What e昀昀ective annual interest rate is she
paying?
What rate of return per year will the investor
make over a 30-year period if the salvage value
is ignored? If the property can be sold 昀漀r
$200,000, what is the rate of return?
7-14 For the 昀漀llowing diagram, compute the IRR to
within 0.5%.
50
60
7-15 For the 昀漀llowing diagram, compute the rate of
return.
7-13 An investor has invested $250,000 in a new
rental property. His estimated annual costs
are $6,000, and annual revenues are $20,000.
• t I
15
10
5
25
f I
0 -- 1 -- 2 -- 3 -- 4 -- 5
l
⠀　7-12 Your cousin Jeremy has asked you to bankroll
his proposed business painting houses in the
summer. He plans to operate the business 昀漀r 昀椀ve
years to p愀礀 his w愀礀 through college. He needs
$5,000 to buy an old pickup truck, some ladders,
a paint spr礀였r, and some other equipment. He
is promising to pay you $1,500 at the end of each
summer (昀漀r 昀椀ve 礀攀ars) in return 昀漀r this invest­
ment. Calculate your annual rate of return.
225
42.55
7-16 Consider the 昀漀llowing cash 昀氀ow:
⠀　Year Cash Flow
0
1
2
3
4
- $500
+200
+ 1 50
+ 1 00
+so
Compute the rate of return represented by the
cash 昀氀ow.
7-17 Switching to powder coating technology will
reduce the emission of volatile organic com­
pounds (VOCs) 昀漀r a 昀椀rm's production process.
吀栀e initial cost is $200,000, with annual costs of
$50,000 and revenues of $90,000 in the 昀椀rst year.
Revenues are projected to increase by $2,000
226
Cha pter 7 Rate of Return Analysis
annually a昀琀er Year 1. 吀栀e salvage value 10 years
昀爀om now is projected to be $30,000. What rate of
return will the 昀椀rm make on this investment?
7-18 For the 昀漀llowing diagram, compute the inter­
D est rate at which the costs are equivalent to the
� bene昀椀ts.
t t t t t t
80
80
80
80
80
80
0 -- 1 -- 2 -- 3 -- 4 -- 5 -- 6
l
l
2 00
l
2 00
2 00
7-19 For the 昀漀llowing diagram, compute the rate of
return on the $3,810 investment.
l t l t l t l
500
25 0
500
25 0
500
25 0
500
0 - 1 - 2 - 3 - 4- 5 - 6 - 7 - • • • n = l
3 ,8 10
0
7-20 For the 昀漀llowing diagram, compute the rate of
return.
A = 1 ,000
t
0 I I I I I I I I
412
l lOt t t t t t t t
l
• • • n = 漀漀
5,000
0
7-21 You have just been elected into the Society of
Honourable Engineers. First-year dues are
waived in honour of your election. 吀栀us, your
昀椀rst payment of $200 is due at the end of the
year, and annual dues are expected to increase
3% annually. A昀琀er 40 years you become a life
member and no longer owe any dues. Instead
of paying annual dues, however, you can pay a
one-time $2,000 life membership fee.
愀儀 Show the equation 昀漀r determining the rate
of return 昀漀r buying a life membership.
(b) What is the rate of return?
Net Present Worth versus i a nd Bonds
7-22 For Problem 7-1, graph the PW versus the in­
terest rate 昀漀r values 昀爀om 0% to 50%. Is this the
typical PW graph 昀漀r an investment?
7-23 A man buys a corporate bond 昀爀om a bond
brokerage house 昀漀r $925. 吀栀e bond has a 昀愀ce
value of $1,000 and pays 4% of its 昀愀ce value each
year. If the bond is paid o昀昀 at the end of 10 years,
what rate of return will the man receive?
0
7-24 A well-known industrial 昀椀rm has issued $1,000
bonds that carry a 4% nominal annual interest
rate, paid semi-annually. 吀栀e bonds mature
20 years 昀爀om now, at which time the industrial
昀椀rm will redeem them 昀漀r $1,000 plus the 昀椀nal
semi-annual interest payment. From the 昀椀nan­
cial pages of your newspaper you learn that the
bonds may be purchased 昀漀r $7 15 each ($7 10
昀漀r the bond plus a $5 sales commission). What
nominal annual rate of return would you re­
ceive if you bought the bond now and held it to
maturity 20 years 昀爀om now?
7-25 On 2 April 2004, an engineer bought a $1,000
bond of a Canadian airline 昀漀r $875. 吀栀e bond
paid 6% on its principal amount of $1,000, half
in each of its 1 April and 1 October semi-annual
payments; it repaid the $1,000 principal sum on
1 October 2017. What nominal rate of return
did the engineer receive 昀爀om the bond if he
held it to its maturity (on 1 October 2017)?
7-26 Danielle can purchase a municipal bond with
a par value of $5,000 that will mature in three
and a half years. 吀栀e bond p愀礀s 8% interest
compounded quarterly. If she can buy this bond
昀漀r $4,800, what rate of return will she earn?
7-27 Mike buys a corporate bond with a 昀愀ce value of
$1,000 昀漀r $900. 吀栀e bond matures in 10 years
and pays a coupon interest rate of 6%. Interest is
paid every quarter.
(a) Determine the e昀昀ective rate of return if
Mike holds the bond to maturity.
(b) What e昀昀ective interest rate will Mike get
if he keeps the bond 昀漀r only 昀椀ve years and
sells it 昀漀r $950?
Problems
7-28
A 12%, $50,000 bond is o昀昀ered 昀漀r sale at
$45,000. If the bond interest is p愀礀able monthly
and the bond matures in 20 years, what nom­
inal and e昀昀ective rates of return per year would
the purchaser make on the investment?
7-29
An investor bought a 5%, $5,000 bond 昀漀r
$4,000. 吀栀e interest was payable quarterly, and
the bond's maturity was 20 years. 吀栀e bond was
kept 昀漀r only nine years and sold 昀漀r $4,200 im­
mediately a昀琀er the 36th interest payment was
received. What nominal and e昀昀ective rates of
return per year were made on this investment?
0
7-30
7-31
ABC Corporation has recently issued bonds
paying interest semi-annually and maturing in
10 years. 吀栀e 昀愀ce value of each bond is $1,000,
and the nominal interest rate is 6.8%.
(a) What is the e昀昀ective interest rate an in­
vestor receives?
(b) If a 0.75% fee is deducted by the brokerage
昀椀rm 昀爀om the initial $1,000, what is the
e昀昀ective annual interest rate paid by ABC
Corporation?
ABC Corporation is issuing some zero coupon
bonds, which pay no interest. At maturity in
20 years they pay a 昀愀ce value of $10,000. 吀栀e
bonds are expected to sell 昀漀r $3,118 when issued.
(a) What is the e昀昀ective interest rate an investor receives?
(b) A 1% fee (based on the 昀愀ce value) is
deducted by the brokerage 昀椀rm 昀爀om
the initial sales revenue. What is the ef­
fective annual interest rate paid by ABC
Corporation?
0
0
7-33
吀栀e cash price of a machine tool is $3,500. 吀栀e
dealer is willing to accept a $1,200 down pay­
ment and 24 end-of-month monthly payments
of $110 each. At what e昀昀ective interest rate are
these terms equivalent?
A local bank makes car loans. It charges 4% a
year in the 昀漀llowing manner: if $3,600 is bor­
rowed to be repaid over three years, the bank
interest charge is ($3,600)(0.04)(3 years) = $432.
吀栀e bank deducts the $432 of interest 昀爀om the
$3,600 loan and gives the customer $3,168 in
cash. 吀栀e customer must rep愀礀 the loan by p愀礀ing
$100 ($3,600 + 36), at the end of each month 昀漀r
36 months. What nominal annual interest rate is
the bank actually charging 昀漀r this loan?
7-34
Jan bought 100 shares of Peach Computer stock
昀漀r $18 per share, plus a $45 brokerage commis­
sion. Every six months she received a dividend
昀爀om Peach of 50¢ a share. At the end of two
years, just a昀琀er receiving the 昀漀urth dividend,
she sold the stock 昀漀r $23 a share and paid a
$58 brokerage commission 昀爀om the proceeds.
What annual rate of return did she receive on
her investment?
7-35
A used-car dealer advertises 昀椀nancing at 0% in­
terest over three years with monthly payments.
You must p愀礀 a processing fee of $250 at sign­
ing. 吀栀e car you like costs $6,000.
(a) What is your e昀昀ective annual interest rate?
(b) You believe that the dealer would accept
$5,200 if you paid cash. What e昀昀ective
annual interest rate would you be p愀礀ing if
you 昀椀nanced the car with the dealer?
0
7-36
Discou nts and Fees
7-32
227
0
7-37
A new-car dealer advertises 昀椀nancing at 0% in­
terest over 昀漀ur years with monthly payments or
a $3,000 rebate if you pay cash.
(a) 吀栀e car you like costs $12,000. What ef­
fective annual interest rate would you be
paying if you 昀椀nanced the car with the
dealer?
(b) 吀栀e car you like costs $18,000. What ef­
fective annual interest rate would you be
paying if you 昀椀nanced the purchase with
the dealer?
(c) 吀栀e car you like costs $24,000. What ef­
fective annual interest rate would you be
paying if you 昀椀nanced the purchase with
the dealer?
Some laboratory equipment sells 昀漀r $75,000.
吀栀e manu昀愀cturer o昀昀ers 昀椀nancing at 8% with
annual payments 昀漀r 昀漀ur years 昀漀r up to
$50,000 of the cost. 吀栀e salesman is willing to
cut the price by 10% if you pay cash. What is the
interest rate you would pay by 昀椀nancing?
228
7-38
7-39
Q
V
Cha pter 7 Rate of Return Analysis
A home mortgage with monthly p愀礀ments 昀漀r
30 years is available at 6% interest. 吀栀e home
you are buying costs $120,000, and you have
saved $12,000 to meet the requirement 昀漀r a 10%
down payment. 吀栀e lender charges "points" of
2% of the loan value as a loan origination and
processing fee. 吀栀is fee is added to the initial
balance of the loan.
愀儀 What is your monthly payment?
(b) If you keep the mortgage until it is paid o昀昀
in 30 years, what is your e昀昀ective annual
interest rate?
(c) If you move to a larger house in 10 years
and p愀礀 o昀昀 the loan, what is your e昀昀ective
annual interest rate?
A 昀椀nance company is using the "Money by
Mail" o昀昀er shown in Table P7-39. Calculate the
yearly nominal IRR received by the company if
a customer chooses the loan of $2,000 and ac­
cepts the credit insurance (life and disabilit礀⤀.
7-40 A 昀椀nance company is using the "Money by
Mail" o昀昀er shown in Table P7-39. Calculate the
yearly nominal IRR received by the company if
a customer chooses the $3,000 loan but declines
the credit insurance.
I nvestments and Loa ns
7-41 An investor bought a one-hectare lot on the
outskirts of a city 昀漀r $9,000 cash. Each year she
paid $80 in property taxes. At the end of 昀漀ur
years, she sold the lot 昀漀r a net value of $15,000.
What rate of return did she receive on her
investment?
7-42 An apartment building in your neighbourhood
is 昀漀r sale 昀漀r $960,000. 吀栀e building has 昀漀ur
units, which are rented at $1,100 a month each.
吀栀e tenants have long-term leases that expire
in 昀椀ve years. Maintenance and other expenses
昀漀r care and upkeep are $4,000 annually. A new
Non-negotiable
For the
Amount of $3,000 or $2,000 or $1,000
Pay to the Order of I Feel Rich
Limited Time O昀昀er
For the Amount of $3,000 Dollars
Pay to the
I Feel Rich
Order of
Total of Payments $4,280.40
Number of Monthly Payments
36 Months
Amount Financed $3,246.25
For the Amount of $2,000 Dollars
Pay to the
I Feel Rich
Order of
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cash advance is total amount 昀椀nanced.
吀䄀BLE P7·39.
To borrow $3,000,
$2,000, or $ 1,000
$ 1,000 loan terms
Problems
university is being built in the vicinity, and it
is expected that the building could be sold 昀漀r
$1,200,000 a昀琀er 昀椀ve years.
(a) What is the internal rate of return 昀漀r this
investment?
(b) Should this investment be accepted if your
other options have a rate of return of 7%?
7-43
7-44
An insurance company is o昀昀ering to sell an
annuity 昀漀r $20,000 cash. In return the 昀椀rm
will guarantee to pay the purchaser 20 annual
end-of-year payments, with the 昀椀rst payment
amounting to $1,100. Subsequent p愀礀ments will
increase at a uni昀漀rm 10% rate each year (second
payment is $1,210; third payment is $1,331, etc.).
What rate of return will the purchaser receive if
he buys the annuity?
Fi昀琀een 昀愀milies live in the village of Stony Hill.
Although several water wells have been drilled,
none has produced water. 吀栀e residents take
turns driving a water truck to a 昀椀re hydrant in a
nearby town. 吀栀ey 昀椀ll the truck with water and
then haul it to a storage tank in Stony Hill. Last
year, truck 昀甀el and maintenance cost $3,180.
吀栀is year the residents are seriously considering
spending $100,000 to install a pipeline 昀爀om the
nearby town to their storage tank. What rate of
return would the Stony Hill residents receive on
their new water supply pipeline if the pipeline is
considered to last
(a) 昀漀rever?
(b) 100 years?
(c) 50 years?
Would you recommend that the pipeline be in­
stalled? Explain.
7-45
7-46
An investor bought 100 common shares of
Omega 昀漀r $9,000. She held the stock 昀漀r nine
years. For the 昀椀rst 昀漀ur years she received
annual end-of-year dividends of $800. For the
next 昀漀ur years she received annual dividends
of $400. She received no dividend 昀漀r the ninth
year. At the end of the ninth year she sold her
stock 昀漀r $6,000. What rate of return did she re­
ceive on her investment?
One aspect of obtaining 昀甀rther education is
the prospect of improved 昀甀ture earnings 昀漀r
0
7-47
229
university graduates in comparison to non­
graduates. Sharon estimates that a university
education has a $28,000 equivalent cost at gradu­
ation. She believes she will reap the bene昀椀ts of
her education throughout 40 years of employ­
ment. She thinks that during her 昀椀rst 10 years
out of university, her income will be higher than
that of a non-graduate by $3,000 a year. During
the next 10 years, she expects an annual income
that is $6,000 a year higher. During the last
20 years of employment, she expects an annual
salary that is $12,000 above the level of the
non-graduate. If her estimates are correct, what
rate of return will she receive as a result of her
investment in a university education?
吀栀e 昀漀llowing advertisement appeared in the
Wall Street Journal on 吀栀ursday, 9 February 1995:
"吀栀ere's nothing quite like the Seville
SmartLease. Seville SLS $0 down, $599
a month/36 months.
"First month's lease p愀礀ment of $599 plus $625 re­
昀甀ndable security deposit and a consumer down
p愀礀ment of $0 昀漀r a total of $1,224 due at lease
signing. Monthly p愀礀ment is based on a net cap­
italized cost of $39,264 昀漀r total monthly payments
of $21,564. Payment examples based on a 1995
Seville SLS: $43,658 MSRP including destination
charge. Tax, licence, title fees, and insurance extra.
Option to purchase 愀琀 lease end 昀漀r $27,854. Mile­
age charge of $0.15 per mile over 36,000 miles."
(a) Set up the cash 昀氀ows.
(b) Determine the interest rate (nominal and
e昀昀ective) 昀漀r the lease.
7-48
An engineering student is deciding whether to
buy two one-term parking permits or an annual
permit. 吀栀e annual parking permit costs $100
due 15 August, and the term permits are $65
due 15 August and 15 January. What is the rate
of return 昀漀r buying the annual permit?
7-49
An engineering 昀椀rm can pay 昀漀r its liability in­
surance on an annual or a quarte爀氀y basis. If paid
quarte爀氀y, the insurance costs $18,000. If paid an­
nually, the insurance costs $65,000. What are the
焀甀arte爀氀y rate of return and the nominal and ef­
fecti瘀攀 interest rates 昀漀r paying on an annual basis?
230
Cha pter 7 Rate of Retu rn Analysis
7-50 An engineering 昀椀rm can pay 昀漀r its liability
insurance on an annual or a quarterly basis.
If paid quarterly, the insurance costs $18,000.
If paid annually, the insurance costs $65,000.
Use a spreadsheet to calculate the exact quar­
terly interest rate 昀漀r paying on an annual
basis.
Initial cost
Uni昀漀rm annual bene昀椀t
7-52 For your auto or home insurance, 昀椀nd out the
cost of paying annually or on a shorter term.
What is the rate of return 昀漀r buying the insur­
ance on an annual basis?
I ncremental Analysis
7-53 吀眀o alternatives are as 昀漀llows:
Year
A
B
0
1
2
3
-$2,000
+800
+800
+800
-$2,800
+ 1 , 1 00
+ 1 , 1 00
+ 1 , 1 00
If 5% is considered the minimum attractive rate
of return, which alternative should be selected?
7-54 Consider two mutually exclusive alternatives:
Year
X
y
0
1
2
3
4
- $ 1 00
+35
+35
+35
+35
-$50.0
+ 1 6.5
+ 1 6.5
+ 1 6.5
+ 1 6.5
If the minimum attractive rate of return is 10%,
which alternative should be chosen?
7-55 Two
mutually exclusive alternatives are
being considered. Both have a 10-year use昀甀l
life. If the MARR is 8%, which alternative is
preferable?
$ 1 00.00 $50.00
1 1 .93
1 9.93
7-56 Consider two mutually exclusive alternatives:
7-51 An engineering student must decide whether
to pay 昀漀r auto insurance on a monthly or an
annual basis. If paid annually, the cost is $1,650.
If paid monthly, the cost is $150 at the start
of each month. What is the rate of return 昀漀r
buying the insurance on an annual basis?
B
A
Year
X
y
0
1
2
3
4
-$5,000
-3,000
+4,000
+4,000
+4,000
-$5,000
+2,000
+2,000
+2,000
+2,000
If the MARR is 8%, which alternative should be
selected?
7-57 吀眀o mutually exclusive alternatives, A and B,
are being considered. Both h愀瘀e lives of 昀椀ve
years. A has a 昀椀rst cost of $2,500 and annual
bene昀椀ts of $㜀㐀6. B costs $6,000 and has annual
bene昀椀ts of $1,664.
If the minimum attractive rate of return is
8%, which alternative should be chosen? Solve
the problem by
(a) Present worth analysis
(b) Annual cash 昀氀ow analysis
(c) Rate of return analysis
7-58 A contractor is considering whether to buy or
lease a new machine 昀漀r her layout site work.
Buying a new machine will cost $12,000,
and the machine will have a salvage value
of $1,200 at the end of its use昀甀l life of eight
years. On the other hand, leasing requires an
annual lease payment of $3,000. On the basis
of a 15% MARR and an internal rate of return
analysis, which alternative should the con­
tractor be advised to accept? 吀栀e cash flows
are as 昀漀llows:
Year (n) Alt. A (purchase) Alt. B (lease)
0
1
2
3
4
5
6
7
8
- $ 1 2,000
+ 1 ,200
-$3,000
-3,000
-3,000
-3,000
-3,000
-3,000
-3,000
-3,000
0
Problems
7-59
employer contributed (that is, she diverted 昀爀om
your salar礀⤀ $1,500 each year to an account 昀漀r
your retirement (a 昀爀inge bene昀椀t), and you con­
tributed a matching amount each year. 吀栀e whole
昀甀nd was invested at 5% during that time, and
the value of the account now stands at $30,000.
You are now 昀愀ced with two alternatives. (1) You
m愀礀 le愀瘀e both contributions in the 昀甀nd until
you retire in 35 years, during which time you
will get the 昀甀ture value of this amount at 5%
interest per year. (2) You m愀礀 take out the total
value of "your" contributions, which is $15,000
(one-half of the total $30,000). You can do as you
wish with the money you take out, but the other
half will be lost as 昀愀r as you are concerned. In
other words, you can give up $15,000 tod愀礀 昀漀r
the sake of getting the other $15,000 now. Other­
wise, you must wait 35 years to get the accumu­
lated value of the entire 昀甀nd. Which alternative
is more attractive? Explain your choice.
Two hazardous-environment 昀愀cilities are being
evaluated, with the projected life of each 昀愀cility
being 10 years. 吀栀e cash flows are as 昀漀llows:
Alt. A
Alt. B
$615,000 $300,000
First cost
25,000
10,000
Maintenance and operating cost
92,000
158,000
Annual bene昀椀ts
65,000
-5,000
Salvage value
吀栀e company uses a MARR of 15%. Using
rate of return analysis, which alternative should
be selected?
7-60
7-61
7- 62
A grocery distribution centre is considering
whether to i瘀팀st in RFID or barcode technol­
ogy to track its i瘀팀ntory within the warehouse
and truck loading operations. 吀栀e use昀甀l life of
the RFID and barcode devices is projected to be
昀椀ve years with minimal or zero sa氀瘀age value. 吀栀e
barcode investment cost is $100,000 and can be
expected to s瘀였 at least $50,000 in lost and stolen
products annually. 吀栀e RFID system is estimated
to cost $200,000 and will s瘀였 $30,000 the 昀椀rst
year, with an increase of $15,000 annually a昀琀er the
昀椀rst year. For a 5% MARR, should the manager
invest in the RFID system or the barcode system?
Analyze incrementally using rate of return.
A provincial department of transportation
(DOT ) is considering whether to buy or lease
an RFID tracking system 昀漀r asphalt, concrete,
and gravel trucks to be used in road paving.
Purchasing the RFID system will cost $5,000
per truck, with a salvage value of $1,500 a昀琀er
the RFID system's use昀甀l life of 昀椀ve years. How­
ever, the DOT considering this purchase is also
looking at leasing this same RFID system 昀漀r
an annual payment of $3,500, which includes
a 昀甀ll replacement warranty. Assuming that
the MARR is 11% and using an internal rate
of return analysis, which alternative would
you advise the DOT to consider? Analyze in­
crementally using rate of return. 吀栀e number
of trucks used in a season varies 昀爀om 5,000 to
7,500. Does this matter?
A昀琀er 15 years of working 昀漀r one employer, you
transfer to a new job. During these years your
231
7- 63
㈀쐀
In his will, Francesca's uncle has given her the
choice between two alternatives:
Alternative 1 $2,000 cash
Alternative 2 $150 cash now plus $100 per month 昀漀r
20 months beginning the 昀椀rst day of
next month
(a) At what rate of return are the two alterna­
tives equivalent?
(b) If Francesca thinks the rate of return in
(a) is too low, which alternative should she
select?
7-64
A diesel generator 昀漀r electrical power can be
purchased by a remote community 昀漀r $480,000
and used 昀漀r 10 years, when its salvage value is
$50,000. Or it can be leased 昀漀r $70,000 a year.
(Remember that lease payments occur at the
start of the year.) 吀栀e community's interest rate
is 8%.
(a) What is the interest rate 昀漀r buying versus
leasing? Which is the better choice?
(b) 吀栀e community will spend $80,000 less
each year 昀漀r 昀甀el and maintenance than it
currently spends on buying power. Should
it obtain the generator? What is the rate
of return 昀漀r the generator using the best
昀椀nancing plan?
232
Cha pter 7 Rate of Return Analysis
Ana lysis Period
Spreadsheets
7- 65 吀眀o alternatives are being considered:
7-69 吀栀e Southern Guru Copper Company operates
A
First cost
Uni昀漀rm annual bene昀椀t
Use昀甀l life, in years
B
$9,200 $5,000
$1,850 $1,750
4
8
If the minimum attractive rate of return is 7%,
which alternative should be selected?
7- 66 Jean has decided it is time to buy a new battery
㈀ꔀ
昀漀r her car. Her choices are the 昀漀llowing:
Zappo Kicko
First cost
Guarantee period, in months
$90
$56
12
24
Jean believes the batteries can be expected to
last only 昀漀r the guarantee period. She does not
want to invest extra money in a battery unless
she can expect a 50% rate of return. If she plans
to keep her present car another two years, which
battery should she buy?
7- 67 吀眀o alternatives are being considered:
A
Initial cost
Uni昀漀rm annual bene昀椀t
Use昀甀l life, in years
0
$9,200 $5,000
$1,850 $1,750
4
Base your computations on a MARR of 7% and
an eight-year analysis period. If identical re­
placement is assumed, which alternative should
be chosen?
7- 68 吀眀o investment opportunities are as 昀漀llows:
A
Year
Cash Flow
0
1
2
3
4
5
6
7
$0.5 million investment
3.5 million pro昀椀t
0.9 million pro昀椀t
3.9 million pro昀椀t
8.6 million pro昀椀t
4.3 million pro昀椀t
3.1 million pro昀椀t
6.1 million pro昀椀t
吀栀e legislator divided the $30.4 million total
pro昀椀t by the $0.5 million investment. 吀栀is pro­
duced, he said, a 6,080% rate of return on the
investment. Southern Guru claims the actual
rate of return is much lower. 吀栀ey ask you to
compute their rate of return. Use a spreadsheet.
annual raises of 2%. He will deposit
0 He10%expects
of the annual salary at the end of each year
7-70 A young engineer's starting salary is $55,000.
B
8
a large mine in a South American country. A
legislator said in the National Assembly that
most of the capital 昀漀r the mining operation
was provided by loans 昀爀om the World Bank;
in 昀愀ct, Southern Guru has only $500,000 of its
own money actually invested in the property.
吀栀e cash 昀氀ow 昀漀r the mine is as 昀漀llows:
B
$100
First cost
$150
Uni昀漀rm annual bene昀椀t
$25 $22.25
$0
End-of-useful-life salvage value $20
10
Use昀甀l life, in years
15
At the end of 10 years, Alternative B is not re­
placed. 吀栀us, the comparison is 15 years of A
versus 10 years of B. If the MARR is 10%, which
alternative should be selected?
in a savings account that earns 5%. How much
will the engineer have saved 昀漀r his retirement
a昀琀er 40 years? Use a spreadsheet.
7-71 Find the average starting engineer's salary 昀漀r
your discipline. Find and reference a source 昀漀r
the average annual raise you can expect. If you
deposit 10% of your annual salary at the end of
each year in a savings account that earns 4%,
how much will you have saved 昀漀r retirement
a昀琀er 40 years? Use a spreadsheet.
Sensitivity Ana lysis
吀栀ese problems are organized such that the (a) parts
require graphical analysis and there昀漀re are much
more easily done with spreadsheets, and the (b) parts
require numerical incremental analysis. Some prob­
lems include only one approach.
Problems
7-72 Given the 昀漀llowing, construct a choice table 昀漀r
the Atlas or the Zippy. Construct a choice table
昀漀r interest rates 昀爀om 0% to 100%.
interest rates 昀爀om 0% to 100%.
Year
0
1
7-73
X
Y
Atlas
- $10 - $20
+15 +28
Initial cost
Annual operation and
maintenance cost
Annual bene昀椀t
Salvage value
Useful life, in years
Consider two mutually exclusive alternatives
and the do-nothing alternative:
Year
0
1
2
3
4
Buy X
Buy Y
7-74 Consider three alternatives: A, B, and do noth­
ing. Construct a choice table 昀漀r interest rates
昀爀om 0% to 100%.
B
0
1
2
3
4
5
-$100
+30
+30
+30
+30
+30
-$150
+43
+43
+43
+43
+43
7-75 Your cat's summer kitty-cottage needs a new
roof. You are considering the 昀漀llowing two pro­
posals and you think a 15-year analysis period
is in line with your cat's remaining lives. (吀栀ere
is no salvage value 昀漀r old roofs.)
吀栀atch Slate
First cost
Annual upkeep
Service life, in years
$20
$5
3
$6,700 $16,900
$1,500 $1,200
$4,000
$1,000
3
$4,500
$3,500
6
of $170,000. 吀栀e owner is considering open­
ing a second bookstore on the north side of the
campus. He can rent an existing building 昀漀r 昀椀ve
years with an option to continue the lease 昀漀r a
second 昀椀ve-礀攀ar period. If he opens the second
bookstore, he expects the existing store will lose
some business, which will be gained by 吀栀e North
End, the new bookstore. It will take $500,000 of
store 昀椀xtures and inventory to open 吀栀e North
End. He believes that the two stores will h瘀였 a
combined pro昀椀t of $260,000 a year a昀琀er all the
expenses of both stores h愀瘀e been paid.
吀栀e owner's economic analysis is based on
a 昀椀ve-year period. He will be able to recover his
$500,000 investment at the end of 昀椀ve years by
selling the store 昀椀xtures and inventory.
(a) Construct a choice table 昀漀r interest rates
昀爀om 0% to 100%.
(b) 吀栀e owner will not open 吀栀e North End
unless he can expect a 15% rate of return.
What should he do? Show computations to
justify your decision.
Construct a choice table 昀漀r interest rates 昀爀om
0% to 100%.
A
Zippy
7-77 吀栀e South End bookstore has an annual pro昀椀t
-$100.0 -$50.0
+31.5 +16.5
+31.5 +16.5
+31.5 +16.5
+31.5 +16.5
Year
233
7-78 A stockbroker has proposed two investments
in low-rated corporate bonds paying high in­
terest rates and selling below their stated value
(in other words, junk bonds). 吀栀e two bonds are
rated as equally risky.
$40
$2
5
(a) Construct a choice table 昀漀r interest rates
昀爀om 0% to 100%.
(b) Which roof should you choose if your
MARR is 12%? What is the actual value of
the IRR on the incremental cost?
7-76 Don is a landscaper. He is considering the pur-
chase of a new commercial lawn mower, either
Bond
Current
Annual Market
Bond
Stated Interest Price, 眀椀th
Value Payment Commission Maturitr
Gen Dev
RJR
$1,000
1,000
$94
140
$480
630
15 years
15 years
*At maturity the bondholder receives the last interest payment plus the
stated value of the bond.
234
Cha pter 7 Rate of Return Analysis
(a) Construct a choice table 昀漀r interest rates
昀爀om 0% to 100%.
(b) Which, if any, of the bonds should you buy
if your MARR is 25%?
7-79 George is going to replace his car in three years
when he graduates, but now he needs a radi­
ator repair. 吀栀e local shop has a used radiator
that it will guarantee 昀漀r two years, or they can
install a new one "guaranteed 昀漀r as long as
you own the car." 吀栀e used radiator costs $250,
and the new one $450. If he assumes the used
radiator will last three years but will need to
be replaced so he can sell the car, which should
he buy?
(a) Develop a choice table 昀漀r interest rates
昀爀om 0% to 50%.
(b) George's interest rate on his credit card is
20%. What should he do?
0
7-82 Consider 昀漀ur mutually exclusive alternatives,
each having an eight-year use昀甀l life:
B
7-83 Consider the three alternatives:
Initial cost
Annual bene昀椀t in each of
昀椀rst 5 years
Annual bene昀椀t in each of
subsequent 5 years
First cost
Uni昀漀rm annual
bene昀椀t
Use昀甀l life, in years
End-of-use昀甀l-Hfe
salvage value
Computed rate of
return
A
B
C
$200
$59.7
$300
$77.1
$600
$165.2
5
$0
5
$0
5
$0
15%
9%
11.7%
For what range of values of MARR is Alternative C
the preferable alternative? Put your answer in
the 昀漀llowing 昀漀rm: "Alternative C is preferred
when _% < MARR � _%."
D
A
B
C
$1,500
250
$1,000
250
$2,035
650
450
250
145
Each alternative has a 10-year use昀甀l life and no
salvage value. Construct a choice table 昀漀r inter­
est rates 昀爀om 0% to 100%.
Initial cost
$300 $600 $200
41
35
Uni昀漀rm annual bene昀椀ts
98
7-81 Consider the 昀漀llowing three mutually exclusive
alternatives:
C
(a) Construct a choice table 昀漀r interest rates
昀爀om 0% to 100%.
(b) If the minimum attractive rate of return
is 8%, which alternative should be
selected?
C
Each alternative has a 10-year use昀甀l life and no
salvage value.
(a) Construct a choice table 昀漀r interest rates
昀爀om 0% to 100%.
(b) If the MARR is 8%, which alternative
should be selected?
B
First cost
$1,000 $800 $600 $500
97 122
Uni昀漀rm annual bene昀椀t
122 120
750 500 500
0
Salvage value
7-80 Consider the 昀漀llowing alternatives:
A
A
7-84 A business magazine is 愀瘀ailable 昀漀r $58 昀漀r one
year, $108 昀漀r two years, $153 昀漀r three years, or
$230 昀漀r 昀椀ve years. Assume you will read the
magazine 昀漀r at least the next 昀椀ve years. For
what interest rates do you prefer each payment
plan?
0
7-85 A new 10,000-square-metre warehouse next
door to the Tyre Corporation is 昀漀r sale 昀漀r
$450,000. 吀栀e terms o昀昀ered are $100,000 down
with the balance being paid in 60 equal monthly
payments based on 15% interest. It is estimated
that the warehouse would have a resale value of
$600,000 at the end of 昀椀ve years.
吀礀re has the necessary cash available and
could buy the warehouse but does not need all
the warehouse space at this time. 吀栀e John­
son Company has o昀昀ered to lease half the new
warehouse 昀漀r $2,500 a month.
Tyre presently rents and uses 7,000 square
metres of warehouse space 昀漀r $2,700 a month.
It has the option of reducing the rented space to
2,000 square metres, in which case the monthly
Problems
rent would be $1,000 a month. Or Tyre could
cease renting warehouse space entirely. Tom
Clay, the Tyre Corporation plant engineer, is
considering three alternatives:
1. Buy the new warehouse and rent half the
space to the Johnson Company. In turn,
the space that 吀礀re rents would be reduced
to 2,000 square metres.
2. Buy the new warehouse and cease renting
any warehouse space.
3. Continue as is, with 7,000 square metres
of rented warehouse space.
(a) Construct a choice table 昀漀r interest
rates 昀爀om 0% to 100%.
(b) On the basis of a 20% minimum
attractive rate of return, which
alternative should be chosen?
7-86 A 昀椀rm is considering the 昀漀llowing 昀漀ur alterna­
tives, as well as a 昀椀昀琀h choice to do nothing. Each
alternative has a 昀椀ve-year use昀甀l life.
⠀　2
1
Initial cost
Uniform
annual net
income ($000)
Computed rate
of return
4
3
$100,000 $130,000 $200,000 $330,000
$38.78
$47.48
$91.55
$26.38
10%
15%
12%
6%
(a) Construct a choice table 昀漀r interest rates
昀爀om 0% to 100%.
(b) 吀栀e 昀椀rm's minimum attractive rate of
return is 8%. Which alternative should be
chosen?
7-87 Construct a choice table 昀漀r interest rates 昀爀om
0% to 100%.
Alternatives
A
B
C
D
Initial cost
$2,000 $5,000 $4,000 $3,000
Annual bene昀椀t
$800
$500
$400 $1,300
Salvage value
$2,000 $1,500 $1,400 $3,000
4
5
7
Life, in years
6
7-88 In a particular situation, 昀漀ur mutually exclu­
sive alternatives are being considered. Each of
the alternatives costs $1,300 and has no salvage
value a昀琀er a 10-year life.
⠀　C愀氀cul愀琀ed
R愀琀e of
Re
Annual Bene昀椀t
Alte爀渀ati瘀攀
$100 at end of 昀椀rst year,
increasing by $30 per year
therea昀琀er
$10 at end of 昀椀rst year,
increasing by $50 per year
therea昀琀er
Annual end-of-year bene昀椀t = $260
$450 at end of 昀椀rst year,
declining by $50 per year
therea昀琀er
A
B
C
D
235
10.0%
8.8%
15.0%
18.1 %
(a) Construct a choice table 昀漀r interest rates
昀爀om 0% to 100%.
(b) If MARR is 8%, which alternative should
be selected?
7-89 吀栀e owner of a downtown parking lot has em­
ployed a civil engineering consulting 昀椀rm to
advise him on the economic feasibility of con­
structing an o昀케ce building on the site. Marie,
a newly hired civil engineer, has been assigned
to make the analysis. She has assembled the 昀漀l­
lowing data:
⠀　Alternative
Sell parking lot
Keep parking lot
Build I -storey
building
Build 2-storey
building
Build 3-storey
building
Build 4-storey
building
Build 5-storey
building
Total Net
Total
Annual Revenue
Investment* 昀爀om Property
$0
200,000
400,000
$0
22,000
60,000
555,000
72,000
750,000
100,000
875,000
105,000
1,000,000
120,000
*Includes the value of the land.
吀栀e analysis period is to be 15 years. For all
the alternatives, the property has an estimated
resale (salvage) value at the end of 15 years equal
to the present total investment.
(a) Construct a choice table 昀漀r interest rates
昀爀om 0% to 100%.
(b) If the MARR is 10%, what recommenda­
tion should Marie make?
236
Cha pter 7 Rate of Return Analysis
7-90 吀栀e Pure White Soap Company is considering
adding some processing equipment to the plant
to aid in the removal of impurities 昀爀om certain
raw materials. By adding the processing equip­
ment, the 昀椀rm can purchase lower-grade raw
material at reduced cost and upgrade it 昀漀r use
in its products.
Four di昀昀erent pieces of processing equipment
are being considered:
A
Initial investment
Annual saving in
materials costs
Annual operating
cost
B
C
D
$10,000 $18,000 $25,000 $30,000
4,000
6,000
7,500
9,000
2,000
3,000
3,000
4,000
吀栀e company can obtain a 15% annual return
on its investment in other projects and is will­
ing to invest money in the processing equip­
ment only as long as it can obtain 15% annual
return on each increment of money invested.
Which one, if any, of the alternatives should be
chosen? Use a challenger-defender rate of return
analysis.
7-91 Frequently we read in the newspaper that you
should lease a car rather than buy it. For a typ­
ical 24-month lease on a car costing $9,400,
the monthly lease charge is about $267. At the
end of the 24 months, the car is returned to the
lease company (which owns the car). As an al­
ternative, the same car could be bought with
no down payment and 24 equal monthly pay­
ments, with interest at a nominal annual rate of
12%. At the end of 24 months the car is 昀甀lly
paid 昀漀r. 吀栀e car would then be worth about half
its original cost.
(a) Over what range of nominal be昀漀re-tax
interest rates is leasing the preferred
alternative?
(b) What are some of the 昀愀ctors that would
make leasing more desirable than is indi­
cated in (a)?
7-92 "吀栀e Financial Adviser" is a weekly column in
the local newspaper. Assume you must answer
the 昀漀llowing question: "I need a new car that
I will keep 昀漀r 昀椀ve years. I have three options.
I can (A) pay $19,999 now, (B) make monthly
payments 昀漀r a 6% 昀椀ve-year loan with 0% down,
or (C) make lease p愀礀ments of $299.00 a month
昀漀r the next 昀椀ve years. 吀栀e lease option also
requires an up-昀爀ont payment of $1,000. What
should I do?"
Assume that the number of miles driven
matches the assumptions 昀漀r the lease and that
the value of the car a昀琀er 昀椀ve years is $6,500. Re­
member that lease p愀礀ments are made at the be­
ginning of the month and that the salvage value
is received only if you own the car.
(a) Construct a choice table 昀漀r nominal in­
terest rates 昀爀om 0% to 50%. (You do not
know what the reader's interest rate is.)
(b) If i = 9%, use an incremental rate of return
analysis to recommend which option
should be chosen.
Modi昀椀ed I nternal Rate of Return
Unless the problem asks a di昀昀erent question or pro­
vides di昀昀erent data, graph the PW versus the interest
rate to see whether multiple roots occur. If there is a
unique IRR, it is the project's rate of return. If there
are multiple roots, use an external investing 爀愀te of
12% and an external borrowing 爀愀te of 6% to compute
the MIRR.
7-93 Find the rate of return 昀漀r the 昀漀llowing
cash 昀氀ow:
Year Cash Flow
0
1
2
3
4
-$15,000
+10,000
-8,000
+11,000
+13,000
7-94 A group of businessmen 昀漀rmed a partnership
to buy and race an Indianapolis-type racing
car. 吀栀ey agreed to pay $50,000 昀漀r the car and
associated equipment. 吀栀e payment was to be
in a lump sum at the end of the year. In what
must have been beginner's luck, the group won
a major race the 昀椀rst week and $80,000. 吀栀e rest
of the 昀椀rst year, however, was not as good: at
the end of the 昀椀rst year, the group had to pay
out $35,000 昀漀r expenses plus the $50,000 昀漀r
Problems
the car and equipment. 吀栀e second year was
a poor one: the group had to pay $70,000 just
to clear up the racing debts at the end of the
year. During the third and 昀漀urth years, racing
income just equalled costs. When the group was
approached by a prospective buyer 昀漀r the car,
they readily accepted $80,000 cash, which was
paid at the end of the 昀漀urth year. What rate of
return did the businessmen obtain 昀爀om their
racing venture?
7-95 A student organization, at the beginning of
the 昀愀ll quarter, purchased and operated a so昀琀­
drink vending machine as a means of helping
昀椀nance its activities. 吀栀e vending machine cost
$75 and was installed at a gasoline station near
the university. 吀栀e student organization pays
$75 every three months to the station owner
昀漀r the right to keep the vending machine at
the station. During the year that the student
organization owned the machine, they received
the 昀漀llowing quarterly income 昀爀om it, be昀漀re
making the $75 quarterly payment to the sta­
tion owner:
Quarter Income
Fall
Winter
Spring
Summer
$150
25
125
150
At the end of one year, the student group resold
the machine 昀漀r $50. Determine the quarterly
cash 昀氀ow. 吀栀en determine a quarterly rate of
return, a nominal annual rate, and an e昀昀ective
annual rate.
7-96 Compute the rate of return on the investment
characterized by the 昀漀llowing cash 昀氀ow.
Year Cash Flow
0
1
2
3
4
5
-$l l0
-500
+300
-100
+400
+500
7-97 A 昀椀rm invested $15,000 in a project that appeared
to have excellent potential. Un昀漀rtunately, a
237
lengthy labour dispute in Year 3 resulted in
costs that exceeded bene昀椀ts by $8,000. 吀栀e cash
flow 昀漀r the project is as 昀漀llows:
Year Cash Flow
0
1
2
3
4
5
6
-$15,000
+10,000
+6,000
-8,000
+4,000
+4,000
+4,000
Compute the rate of return 昀漀r the project.
7-98 吀栀e 昀漀llowing cash 昀氀ow has no positive interest
rate. 吀栀e project, which had a projected life of
昀椀ve years, was terminated early.
Year Cash Flow
0
1
2
-$50
+20
+20
吀栀ere is, however, a negative interest rate. Com­
pute its value.
7-99 Consider the 昀漀llowing cash 昀氀ow.
Year Cash Flow
0
1
2
-$100
+240
-143
If the minimum attractive rate of return is 12%,
should the project be undertaken?
7-100 Determine the rate of return on the investment
昀漀r the 昀漀llowing cash flow.
Year Cash Flow
0
1-3
4
5-8
-$3,570
+1,000
-3,170
+1,500
7-101 Bill bought a vacation lot he saw advertised
on television 昀漀r an $800 down payment and
monthly payments of $55. When he visited the
lot, he 昀漀und it was not something he wanted to
own. A昀琀er 40 months he was 昀椀nally able to sell
238
Cha pter 7 Rate of Return Analysis
$60 a year in annual payments on 31 December.
Under the convertible feature of the bond, it
could be converted into 20 common shares by
tendering the bond plus $400 cash. 吀栀e day
a昀琀er the investor received the 31 December
2005 interest payment, she submitted the bond
together with $400 to the XLA Corporation.
In return, she received the 20 common shares.
吀栀e common shares paid no dividends. On 31
December 2007 the investor sold the shares 昀漀r
$1,㜀㐀0, terminating her 昀椀ve-year investment in
XLA Corporation. What rate of return did she
receive?
the lot. 吀栀e new purchaser assumed the balance
of the loan on the lot and paid Bill $2,500. What
rate of return did Bill receive on his investment?
7-102 Compute the rate of return on an investment
having the 昀漀llowing cash 昀氀ow.
Year Cash Flow
0
1
2-9
10
-$850
+600
+200
-1,800
7-103 Compute the rate of return 昀漀r the 昀漀llowing
cash 昀氀ow.
Year Cash Flow
0
1
2
3
4
5
6
7
8
-$200.0
+100.0
+100.0
+100.0
-300.0
+100.0
+200.0
+200.0
-124.5
Year Pump 1 {$000s) Pump2 ($000s}
7-104 An investor is considering two mutually exclu­
sive projects. She can obtain a 6% be昀漀re-tax
rate of return on external investments, but she
requires a minimum attractive rate of return
of 7% 昀漀r these projects. Use a 10-year analy­
sis period to compute the incremental rate of
return 昀爀om investing in Project A rather than
Project B.
Project A:
Project B:
Build Drive-Up Buy Land in
H愀眀aii
Photo Shop
Initial capital
investment
Net uni昀漀rm annual
income
Salvage value 10 years
hence
Computed rate of
return
7-106 A problem o昀琀en discussed in the engineering
economy literature is the "oil-well pump prob­
lem." 1 Pump 1 is a small pump; Pump 2 is a larger
pump that costs more, will produce slightly
more oil, and will produce it more rapidly. If the
MARR is 20%, which pump should be selected?
Assume that any temporary external invest­
ment of money earns 10% per year and that any
temporary 昀椀nancing is done at 6%.
$58,500
$48,500
$6,648
$0
$30,000
$138,000
8%
11%
7-105 In January 2003, an investor bought a convert­
ible debenture bond issued by the XLA Corpor­
ation. 吀栀e bond cost $1,000 and paid interest of
0
1
2
- $100
+70
+70
- $110
+ l l5
+30
U nclassi昀椀ed
7-107 A mine is 昀漀r sale 昀漀r $240,000. It is believed the
mine will produce a pro昀椀t of $65,000 the 昀椀rst
year, but the pro昀椀t will decline by $5,000 a year
a昀琀er that, eventually reaching $0, whereupon
the mine will be worthless. What rate of return
would this $240,000 investment produce 昀漀r the
purchaser of the mine?
7-108 An engineering student is deciding whether to
buy two one-term parking permits or an annual
One of the more interesting exchanges of opinion about this
problem is in Professor Martin Wohl's (1979) "Common Mis­
understandings about the Internal Rate of Return and Net
Present Value Economic Analysis Methods" and the associated
discussion by Pro昀攀ssors Win昀爀ey, Leavenworth, Steiner, and
Bergmann, published in Evaluating Transportation Proposals,
吀爀ansportation Research Record 731, 吀爀ansportation Research
Board, Washington, DC.
1
Problems
permit. Using the dates and costs 昀漀r your uni­
versity, 昀椀nd the rate of return 昀漀r the incremen­
tal cost of the annual permit.
7-109 A bulldozer can be bought 昀漀r $380,000 and used
昀漀r six years, when its salvage value is 15% of the
昀椀rst cost. Or it can be leased 昀漀r $60,000 a year.
(Remember that lease p愀礀ments occur at the start
of the year.) 吀栀e 昀椀rm's interest rate is 12%.
(a) What is the interest rate 昀漀r buying versus
leasing? Which is the better choice?
(b) If the 昀椀rm will receive $65,000 more each
year than it spends on operating and
maintenance costs, should the 昀椀rm obtain
the bulldozer? What is the rate of return
昀漀r the bulldozer, using the best 昀椀nancing
plan?
7-110 At what coupon interest rate will a $20,000
bond yield a nominal 12% interest compounded
quarterly if the purchaser pays $18,000 and the
bond becomes due in 20 years? Assume the
bond interest is payable quarterly.
7-111 A 9%, $10,000 bond that has interest payable
semi-annually sells 昀漀r $8,500. Calculate what
the maturity date should be so that the pur­
chaser may enjoy a 12% nominal rate of return
on this investment.
7-112 A popular magazine o昀昀ers a lifetime sub­
scription 昀漀r $200. Such a subscription may
be given as a gi昀琀 to an in昀愀nt at birth (the
parents can read it in those early years) or
taken out by an individual 昀漀r himself. Nor­
mally, the magazine costs $12.90 a year.
Knowledgeable people say it will probably
continue inde昀椀nitely at this $12.90 rate. What
rate of return would the parents obtain if they
bought a life subscription rather than paying
$12.90 per year beginning immediately? You
may make any reasonable assumptions, but
the compound interest 昀愀ctors must be used
correctly.
7-113 Installing an automated production system
costing $278,000 initially is expected to save
Zia Corporation $52,000 in expenses annually.
If the system needs expenditures of $5,000 on
239
operations and maintenance each year and has
a salvage value of $25,000 at EOYlO, what is the
IRR 昀漀r 10 years' use of this system? If the com­
pany wants to earn at least 12% on all invest­
ments, should it buy this system?
7-114 A used-car dealer advertises 昀椀nancing at 4% in­
terest over three years with monthly payments.
吀栀e buyer must pay a processing fee of $250 at
signing. 吀栀e car you like costs $6,000.
(a) What is your e昀昀ective annual interest
rate?
(b) You believe that the dealer would accept
$5,200 if you paid cash. What e昀昀ective annual interest rate would you be
paying if you 昀椀nanced the car with the
dealer?
(c) Compare these answers with those 昀漀r
Problem 7-35. What can you say about
what matters the most 昀漀r determining the
e昀昀ective interest rate?
7-115 A new machine can be purchased today 昀漀r
$300,000. 吀栀e annual revenue 昀爀om the ma­
chine is calculated to be $67,000, and the equip­
ment will last 10 years. Expect the maintenance
and operating costs to be $3,000 a year and to
increase $600 per year. 吀栀e salvage value of the
machine will be $20,000. What is the rate of
return 昀漀r this machine?
7-116 Consider the 昀漀llowing cash 昀氀ow:
Year Cash Flow
0
1
2
3
4
s
-$400
0
+200
+150
+100
+so
Write one equation 昀漀r the cash 昀氀ow, with i
as the only unknown. In the equation you are
not to use more than two single-payment com­
pound interest 昀愀ctors. (You may use as many
other 昀愀ctors as you wish.) 吀栀en solve your
equation 昀漀r i.
0 will yield a desirable return on his investment.
7-117 吀栀e owner of a corner lot wants to 昀椀nd a use that
240
Cha pter 7 Rate of Return Analysis
A昀琀er much study and calculation, he decides
that the two best alternatives are the 昀漀llowing:
Build Gas Build So昀琀 Ice
Station
Cream Stand
First cost
Annual property taxes
Annual income
Li昀攀 of building, in years
Salvage value
$80,000
$3,000
$11,000
20
$0
$120,000
$5,000
$16,000
20
$0
If the owner wants a minimum attractive rate
of return on his investment of 6%, which of the
two alternatives would you recommend?
M i n i - Cases
7-118 (a) A 昀椀ve-year auto loan 昀漀r $18,000 has
monthly payments at a 9% nominal annual
rate. If the borrower must pay a loan
origination fee of 2%, what is the true ef­
fective cost of the loan?
(b) If the car is sold a昀琀er two years and the
loan is paid o昀昀, what is the e昀昀ective inter­
est rate?
(c) Graph the e昀昀ective interest rate as the time
to sell the car and pay o昀昀 the loan varies
昀爀om one to 昀椀ve years.
7-119 愀儀 A 30-year mortgage 昀漀r $220,000 has
monthly p愀礀ments at a 6% nominal annual
rate. If a borrower's loan origination fee is
3% and it is added to the initial balance,
what is the true e昀昀ective cost of the loan?
(b) If the house is sold a昀琀er six years and the
loan is paid o昀昀, what is the e昀昀ective inter­
est rate?
(c) Graph the e昀昀ective interest rate as the
time to sell the house and pay o昀昀 the loan
varies 昀爀om one to 15 years.
Bene昀椀t- Cost Ratio and
Other Analysis Techniq ues
The Avro Arrow
In 1939, when Canada went to war, the country's aircraft industry consisted of a number of machine
and assembly shops that produced about 40 airplanes in a year. But 1940 brought the invasion
of France, the bombing of England, and a need for factories safe from enemy bombers to build
planes. The government established the Department of Munitions and Supply to build a wartime
production capability.
The National Steel & Car Corporation was taken over by the government, renamed Victory
Aircraft Limited, and set to building British-designed Lancaster bombers and Anson light transports.
By the end of the war, Victory's Malton factory outside Toronto had built a total of 3,634 Avro air­
craft comprising 3,197 Ansons, 430 Mk X Lancasters, six Lancastrian transports, one Mk 15 Lincoln
heavy bomber, and one York transport.
After the war ended, the government sold Victory Aircraft to the British company Hawker Sid­
deley, which used it for its subsidiary AV. Roe Canada Ltd (known as Avro Canada).
Deisgn Pies lnc/Ala my Stock Photo
The Avro Arrow
243
After the war the RCAF needed a fighter that would be suitable for defence against potential
attack by long-range, high-altitude bombers armed with nuclear bombs. In the 1950s Canada
produced two such planes. Avro built the twin-engine, two-seat CF100 fighter, the Canuck; and its
competitor, the Vickers subsidiary Canadair, manufactured the F-86 Sabre jet.
In the early 1950s the RCAF was looking for a supersonic missile-armed replacement for the
Canuck, and in response Avro engineers developed the delta-winged twin-engine CF105, known
as the Avro Arrow.
With a speed of Mach 2 and a cruising altitude of 50,000 feet, the Arrow was considered to be
the most advanced fighter of its day and an example of the excellence of Canadian engineering.
However, its rollout date, 4 October 1957, was the day the world heard the beeping of the first sat­
ellite, the Soviet Sputnik. This changed the nature of the military threat-a nation that could launch
Sputnik could also deliver nuclear warheads using missiles, against which fighter aircraft would be
helpless. The military shifted its attention from planes to missiles and space.
Faced with rapidly escalating costs and a shift in military priorities, on 20 February 1959, the
government cancelled the Arrow program. With the cancellation came a brain drain as the now
unemployed engineers and designers headed south to work in the American aerospace industry.
With them went 20 years of experience and development in high-speed, high-altitude flight.
QU ESTIONS TO CONSI DER
1.
What is the role of government in high-technology industries? The Americans support theirs through space
programs and military contracts. The British and French governments subsidize their aircraft industries in
the cause of national security. The results of the British-French collaboration were the Concorde and Airbus
industries.
2.
Is Canada too small? At the time of cancellation it was argued that no other countries would buy the Arrow
3.
Are Israel and Sweden too small to support aircraft industries?
4.
Can a government do anything about highly trained professionals leaving to work in another country?
and that it was too expensive for Canada to support alone.
Should it?
LEARN I NG OBJ ECTIVES
This chapter will help you
•
recognize the unique objective and viewpoint of public decisions
•
explain methods for determining the interest rates for evaluating public projects
•
use the benefit-cost ratio to analyze projects
•
use an incremental benefit-cost ratio to evaluate a set of mutually exclusive projects
•
discuss the effect of financing, duration, and politics in public-investment analysis
•
describe the payback period method of analysis
KEY TERMS
bene昀椀t-cost analysis
government opportunity cost
i ncremental B/C ratio
opportunity cost
payback period
taxpayer opportun ity cost
244
Cha pter 8 Benefit-Cost Ratio and Other Analysis Techniques
In this chapter, we will deal with two remaining methods 昀漀r choosing between projects:
cost-bene昀椀t analysis, which is usually used in the public sector, and payback period analy­
sis, which is usually used in the private sector.
Public organizations, such as federal, provincial or territorial, and municipal govern­
ments, port authorities, and school districts make investment decisions. For these deci­
sion-making bodies, economic analysis is complicated by several 昀愀ctors that do not a昀昀ect
companies in the private sector. 吀栀ose 昀愀ctors include the overall purpose of investment,
the viewpoint 昀漀r analysis, and the way the interest rate is chosen. 吀栀e overall mission in
the public sector is the same as that in the private sector-to make prudent investment
decisions that promote the organization's overall goals.
吀栀e primary economic measure used in the public sector is the bene昀椀t-cost (B!C) ratio.
吀栀is measure is calculated as a ratio of the equivalent worth (EW) of the project's bene昀椀ts
to the equivalent worth of the project's costs. We have de昀椀ned various kinds of equivalent
worth in the text so 昀愀r: present worth, 昀甀ture worth, and equivalent annual worth. As long
as we calculate EW on the same basis 昀漀r both costs and bene昀椀ts, each of these methods
will give exactly the same value 昀漀r the B/C ratio. If the B/C ratio is greater than LO, the
project under evaluation is accepted; if not, it is rejected. 吀栀e B/C ratio is used to evaluate
both single investments and sets of mutually exclusive projects. 吀栀e uncertainties of quan­
ti昀礀ing cash 昀氀ows, long project lives, and low interest rates all tend to lessen the reliability of
the public sector engineering economic analysis. Nevertheless, if applied correctly, the B/C
ratio technique will alw愀礀s lead to the same decision as each of the other methods we have
studied so 昀愀r. 吀栀e reader may again protest the waste昀甀lness of learning one more method
that will merely con昀椀rm the conclusions of the methods we already know. We acknow­
ledge this protest, but point out that the B/C ratio is a widely used and accepted measure in
government economic analysis and decision making, and if we 昀椀nd ourselves employed in
these sectors, we must be prepared to use the same language as everybody else.
吀栀e 昀椀nal method we will introduce 昀漀r evaluating projects is payback period. 吀栀is, 昀漀r
the 昀椀rst time in this text, will sometimes lead to recommendations that di昀昀er 昀爀om those
of the methods we have previously studied. When we come to this section, we will see what
circumstances might prompt us to use this method.
Public Sector Investment Objectives
Organizations exist to promote the overall goals of those they serve. In private sector 昀椀rms,
investment decisions are based on increasing the 昀椀rm's wealth and economic stability.
Bene昀椀ciaries of investments are generally identi昀椀ed clearly as the owners or shareholders
of the company.
In the public sector, the purpose of investment decisions is sometimes ambiguous.
吀栀eoretically, the purpose of bene昀椀t-cost analysis is to measure (or quanti昀礀) a project so
that one can determine whether it causes a net increase in economic and social wel昀愀re.
But applying this guideline to practical problems may be di昀케cult.
Consider the case of a dam construction project to provide water, electricity, 昀氀ood
control, and recreational 昀愀cilities. Such a project might seem to be advantageous 昀漀r the
entire population of the region. But on closer inspection, decision makers must consider
that the dam will require the loss of land upstream because of backed-up water. Farmers
will lose pasture or crop land, 昀椀sh will not be able to reach their spawning grounds, and
nature lovers will lose woodlands. Perhaps the land to be lost is a breeding ground 昀漀r pro­
tected species, in which case environmentalists may oppose the project. 吀栀e project m愀礀
also have a detrimental e昀昀ect on towns, cities, and regions downstream. How will it a昀昀ect
their water supply?
P u b l i c Sector I nvestment O bjectives
Investment decisions are more di昀케cult in the public sector than in the private sector
owing to the many people, organizations, and political units that may be a昀昀ected. Oppos­
ition to a proposal is more likely in public-investment decisions than in those made by
private sector companies because 昀漀r every group that bene昀椀ts 昀爀om a particular project,
there is usually an opposing group. Many con昀氀icts arise when the project involves the use
of public lands, including industrial parks, housing developments, business districts, road­
ways, sewage plants, power plants, and land昀椀lls. Opposition may be based on the belief
that development of any kind is bad or that the proposed development should not be near
our homes, schools, or businesses.
Consider the decision that a small town might 昀愀ce when considering whether to es­
tablish a municipal rose garden. Although it might seem a bene昀椀cial public investment
with no adverse consequences, analysis of the project must consider all e昀昀ects of the pro­
ject, including potential un昀漀reseen outcomes. Where will visitors park their cars? Will
increased travel around the park necessitate new tra昀케c lights and signs? Will tra昀케c and
visitors to the park increase the noise 昀漀r adjacent homes? Will the garden's special var­
ieties of roses create a disease hazard 昀漀r local gardens? Will the garden require high levels
of fertilizers and insecticides, and where will these substances wind up a昀琀er they have
been applied? Clearly, many issues must be addressed.
Our simple rose garden illustrates how e昀昀ects on all parties involved must be ident昀툀ed,
even 昀漀r projects that seem very use昀甀l. Public decision makers must reach a comprom­
ise between the bene昀椀ts to be enjoyed by some groups and the negative e昀昀ects on other
groups. 吀栀e overall objective is to make prudent decisions that promote the gene爀愀l we氀昀are.
In the public sector, the decision process is not so straight昀漀rward as in the private sector,
since the scope of "general wel昀愀re" is much larger.
A m愀樀or document in the development of bene昀椀t-cost analysis was the United States
Flood Control Act of 1936, which speci昀椀ed that waterway improvements 昀漀r 昀氀ood control
could be made as long as "the bene昀椀ts to whomsoever they accrue [italics added] are in
excess of the estimated costs." Perhaps the overall general objective of investment decision
analysis in government should be a dual one: to promote the general wel昀愀re and to ensure
that the value to those who can potentially bene昀椀t exceeds the overall costs to those who
do not bene昀椀t.
But even this objective is controversial. Do we expect the provincial government of
Alberta to take into account the potential 昀氀ooding of Paci昀椀c Islands as a consequence of
carbon released by exploiting the Albertan oil sands? Do we consider bene昀椀ts accruing to
our remote descendants to be exactly as desirable as bene昀椀ts accruing to those of us alive
today? We are not going to resolve these questions within the 昀爀amework of engineering
economics; we have to recognize that we h愀瘀e come to the limits of what economics can
tell us.
Viewpoint for Analysis
When governmental bodies do economic analysis, an important concern is the viewpoint
of the analysis. Economic analysis, both governmental and industrial, must be based on
a viewpoint. In the case of industry the viewpoint is obvious-a company in the private
sector pays the costs and counts its bene昀椀ts. 吀栀us, both the costs and bene昀椀ts are measured
昀爀om the perspective of the 昀椀rm.
Costs and bene昀椀ts that occur outside the 昀椀rm are referred to as external consequences
(see Figure 8-1). In years past, private sector companies generally ignored the external
consequences of their actions. Ask anyone who has lived near a paper plant, a slaugh­
terhouse, or a steel mill about external consequences! More recently, governments have
245
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Cha pter 8 Benefit-Cost Ratio and Other Analysis Techniques
Plant
Boundary
D
Intern al
Consequences
�--ⴀ吀
Smoke
Stack
⸀Ⰰ
I
◄
• -•• -►
• -•-◄•-◄
• -•
►
圀
Extern al
Conse quences
FIGURE 8-1 I nternal and external conseq uences of an ind ustrial plant.
昀漀rced industry to reduce pollution and other undesirable external consequences, with the
result that today many companies are evaluating the consequences of their action 昀爀om a
broader, or community-oriented, viewpoint.
吀栀e council members of a small town that levies taxes can be expected to take into
account the "viewpoint of the town" in making decisions: unless it can be shown that the
money 昀爀om taxes can be used e昀昀ectively, the town council is unlikely to spend it. But what
happens when the money is contributed to the town by the provincial government, as in
"revenue sharing" or by means of some other provincial grant? A provincial government
may p愀礀 a share of project costs varying 昀爀om 10% to 90%. Example 8-1 illustrates the
viewpoint problem that is created.
EXAM PLE S-1
A municipal project will cost $ 1 million. 吀栀e provincial government will pay 50% of the cost if the
project is undertaken. Although the original economic analysis showed that the PW of bene昀椀ts was
$1.5 million, a subsequent detailed analysis by the city engineer 昀椀nds that a more realistic estimate is
$750,000. 吀栀e city council must decide whether to proceed with the project. What would you advise?
SO LUTI O N
From the viewpoint o f the city, the project i s still a good one. I f the city puts up half the cost ($500,000)
it will receive all the bene昀椀ts ($750,000). On the other hand, 昀爀om an overall viewpoint, the revised es­
timate of $750,000 of bene昀椀ts does not justi昀礀 the $1 million expenditure. 吀栀at illustrates the dilemma
caused by varying viewpoints.
Possible viewpoints that may be taken include those of an individual, a business 昀椀rm or
corporation, a regional municipality, a city, a province, a nation, or a group of nations. To
avoid sub-optimizing, the proper approach is to take a viewpoint at least as broad as those
who pay the costs and those who receive the bene昀椀ts. When the costs and bene昀椀ts are totally
con昀椀ned to a town, 昀漀r example, the town's viewpoint seems to be a reasonable basis 昀漀r
the analysis. But when the costs or the bene昀椀ts are spread beyond the proposed viewpoint,
then the viewpoint should be enlarged to this broader population.
Other than investments in defence and social programs, most of the bene昀椀ts pro­
vided by government projects are realized at a regional or local level. Projects such as dams
Viewpo i n t fo r Analys is
昀漀r electricity, flood control, and recreation, and transportation 昀愀cilities such as roads,
bridges, and harbours all bene昀椀t most those in the region in which they are constructed.
Even smaller-scale projects, such as the municipal rose garden, although 昀甀nded by public
monies at a local or provincial level, provide most bene昀椀t to those nearby. As in the case of
private decision making, it is important to adopt an appropriate and consistent viewpoint,
or perspective, and to designate all the costs and bene昀椀ts that accrue to the prospective
investment 昀爀om that perspective. To shi昀琀 perspective when quantifying costs and bene昀椀ts
could greatly skew the results of the analysis and subsequent decision.
Selecting an Interest Rate
Several 昀愀ctors, not present 昀漀r private sector 昀椀rms, influence the selection of an interest
rate 昀漀r economic analysis in the government sector. Recall that 昀漀r private sector 昀椀rms the
overall goal is wealth maximization, and the rate to use in evaluating projects is chosen
to be consistent with that goal. Most 昀椀rms use cost of capital or opportunity cost when set­
ting an interest rate. 吀栀e goal of public investment, on the other hand, involves the use of
public resources to promote the gene爀愀l we氀昀are and to secure the bene昀椀ts of a given project
to whomsoever they m愀礀 accrue, as long as those bene昀椀ts outweigh the costs. How to set an
interest rate is less clear-cut in this case. 吀栀e possibilities include no interest rate, cost-of­
capital concepts, and opportunity-cost concepts.
No Time-Value-of-Money Concept
In government, monies are obtained through taxation and spent about as quickly as they
are obtained. O昀琀en, there is little delay between collecting money 昀爀om taxpayers and
spending it. (Remember that the federal government collects taxes on every paycheque,
in the 昀漀rm of withholding tax.) 吀栀e collection of taxes, like their disbursement, although
based on an annual budget, is actually a continuous process. Using this line of reasoning,
some would argue that there is little or no time lag between collecting and spending tax
dollars. 吀栀us they would advocate the use of a 0% interest rate 昀漀r economic analysis of
public projects.
Cost-of-Ca pital Concept
Another approach in determining interest rates in public investments is that most levels
of government (federal, provincial, and local) borrow money 昀漀r capital expenditures in
addition to collecting taxes. When money is borrowed 昀漀r a speci昀椀c project, one line of
reasoning is to use an interest rate equal to the cost of borrowed money.
Opportu nity-Cost Concept
Opportunity cost, which is related to the interest rate on the best opportunity 昀漀rgone, m愀礀
take two 昀漀rms in governmental economic analysis: government opportunity cost and tax­
payer opportunity cost. In public decision making, if the interest rate is based on the oppor­
tunity cost to a government agency or other governing body, that interest rate is known as
�퐀rnment opportunity cost. In this case the interest rate is set at that of the best pro­
spective project 昀漀r which 昀甀nding is not 愀瘀ailable. One disadvantage of the government
opportunity-cost concept is that di昀昀erent agencies and subdivisions of government will h瘀였
di昀昀erent opportunities. 吀栀ere昀漀re, political units could use di昀昀erent interest rates, resulting
247
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Cha pter 8 Benefit-Cost Ratio and Other Analysis Techniques
in a project that could be rejected in one branch and accepted in another. Di昀昀ering interest
rates lead to inconsistent evaluation and decisions across government.
Dollars used 昀漀r public investments are generally gathered by taxing the cit­
izens. The concept of taxpayer opportunity cost suggests that a correct interest rate
昀漀r evaluating public investments is that which the taxpayer could have received if
the government had not collected those dollars through taxation. T his philosophy
holds that taxation takes away the taxpayers' opportunity to use the same dollars 昀漀r
investment. The interest rate that the government requires should not be less than
what the taxpayer would have received. It is not economically desirable to take money
昀爀om a taxpayer with a 12% opportunity cost, 昀漀r example, and invest it in a govern­
ment project yielding 4%.
Recom mended Concept
吀栀e general rule of thumb in setting an interest rate 昀漀r government investments has been
to select the largest of the cost of capital-the government opportunity cost, or the tax­
payer opportunity-cost interest rates. However, as is the case in the private sector, there is
no hard and 昀愀st rule universally applied in all circumstances. 吀栀e setting of an interest
rate 昀漀r use in economic analysis is at the discretion of the government entity per昀漀rming
the analysis.
The Bene昀椀t-Cost Ratio
吀栀e bene昀椀t-cost ratio is used almost exclusively in public-investment analysis, primarily
because it is easy to understand. As in present worth analysis, we calculate the equivalent
present value of the bene昀椀ts and the costs expected 昀爀om a project. 吀栀en, rather than sub­
tracting one 昀爀om the other and seeing if the result is positive, we take the ratio of one to
the other and see if the result is greater than unity.
It will be immediately apparent that we would get the same result if, instead of looking
at the ratio of the present worths of bene昀椀ts and costs, we looked at the ratio of their 昀甀ture
worths or equivalent annual worths, since each of these can be obtained by multiplying
numerator and denominator by the same 昀愀ctor.
An economic analysis is per昀漀rmed to help with the goal of making a decision. When
using the B/C ratio, the decision rule is straight昀漀rward:
If the B/C ratio is > 1, then invest; otherwise, don't.
As usual, if the ratio is very close to 1, we may need to do sensitivity analysis and 昀甀r­
ther research.
Despite the apparent simplicity of this method, there is a potential pit昀愀ll hidden in the
de昀椀nition. Recall that we de昀椀ne the ratio as:
EW of net bene昀椀ts to whomsoever they may accrue
B/c rati. o = ------------------EW of costs to the sponsors of the project
Now, consider a proposal to build a toll bridge. 吀栀ere is a certain construction cost,
a昀琀er which there will be an annual income, 昀爀om tolls, and an annual expenditure on
maintenance. We could add the present equivalent of the maintenance costs to the
denominator-this is sometimes referred to as the conventional B/C method. Or we could
The Bene昀椀t-Cost Ratio
regard the di昀昀erence between annual income and the annual maintenance costs as a net
annual bene昀椀t, and add its present worth to the denominator-this is sometimes referred
to as the mod昀툀ed BIC method. 吀栀is shouldn't change our decision-it's the same project
either way-but it will change the numerical value of the B/C ratio.
Fortunately, the invest/don't invest decision will not change; whichever way we calcu­
late it, the two values will both be either less than unity or greater than unity.
It does mean, however, that if we have multiple exclusive projects, we can't sensibly
choose between them just by comparing their B/C ratios. Instead, we must use the incre­
mental cost-bene昀椀t technique described in the next section.
Some texts use the unnecessary word "dis bene昀椀ts." 吀栀ese are the negative e昀昀ects of
government projects felt by some individuals or groups. For example, consider Canada's
national parks. Development projects by the skiing or lumber industries might provide
enormous bene昀椀ts to the recreation or construction sectors while creating simultaneous
disbene昀椀ts 昀漀r environmental groups. You can safely replace the word "disbene昀椀ts" by
"costs" whenever you encounter it. As long as the methods introduced in the next section
are applied consistently, this will always yield correct conclusions.
Incremental Bene昀椀t-Cost Analysis
吀栀e incremental B/C ratio should be used, like the incremental rate of return, to compare
mutually exclusive alternatives. Incremental B/C ratio analysis is consistent with maximiz­
ing the present worth of the alternatives. As with the incremental IRR method, it is not
proper to simply calculate the B/C ratio 昀漀r each alternative and choose the one with the
highest value. Rather, an incremental approach is called 昀漀r.
Elements of the I ncrementa l Bene昀椀t- Cost Ratio Method
1.
Ident椀昀y all relevant alternatives. Decision rules or models can recommend a best
course of action only 昀爀om the set of identi昀椀ed alternatives. If a better alterna­
tive exists but is not considered, it will never be chosen, and the solution will be
sub-optimal. For B/C ratio problems, when the do-nothing option is available, it
is alw愀礀s the "base case" 昀爀om which the incremental method proceeds.
2. (optional) Calculate the B䤀䌀 爀愀tio of each alternative. Once the individual B/C
ratios have been calculated, the alternatives with a ratio less than 1 are eliminated
昀爀om 昀甀rther consideration. 吀栀is step gets the poor per昀漀rmers out of the w愀礀
be昀漀re the incremental procedure is started. 吀栀is step may be omitted, however,
because the incremental analysis method will eliminate the subpar alternatives
in due time.
Note: 吀栀ere is a case where this step must be skipped. If doing nothing is not
an alternative and if all the alternatives h愀瘀e a B/C ratio less than 1, then incre­
mental analysis will select the least-bad of the alternatives.
3. Rank-order the projects. 吀栀e alternatives must be ordered 昀爀om the one with
the lowest costs to the one with the largest. 吀栀e do-nothing alternative always
becomes the 昀椀rst on the ordered list. 吀栀e 昀椀rst alternative on the list starts out as
current champion.
4. Ident椀昀y the increment under conside爀愀tion. 吀栀e 昀椀rst increment considered is
always that of going 昀爀om the lowest-cost alternative (when available, this is the
do-nothing option) to the next-higher-cost alternative. As the analysis proceeds,
any increment identi昀椀ed is always in reference to the current champion.
䌀䐀
249
250
Cha pter 8 Benefit-Cost Ratio and Other Analysis Techniques
5.
Calculate the B℀䌀 爀愀tio for the incremental cash 昀氀ows. First calculate the incremen­
tal bene昀椀ts and the incremental cos琀猀. 吀栀at is done by 昀椀nding the cash 昀氀ows that
represent the di昀昀erence (䐀⸀) between the two alternatives under consideration.
吀栀e incremental B/C 爀愀tio is 䐀⸀B/䐀⸀C.
6. Use the incremental B/C ratio to decide which alternative is better. If the incremen­
tal B/C ratio (䐀⸀B/䐀⸀C) calculated in Step 5 is greater than 1, the increment is jus­
ti昀椀ed and the alternative associated with that increment, the challenger, becomes
the new current champion. If the ratio is less than 1, the increment is not justi昀椀ed,
the challenger is rejected, and the defender continues as current champion.
7. Ite爀愀te to Step 4 until all increments (projects) have been considered. 吀栀e incremen­
tal method continues until all alternatives have been evaluated.
8. Choose the best alternative昀爀om the set of mutually exclusive competi渀最 projects. At
the end of the process, whichever project is the current champion will be the best
of all the alternatives considered.
In Example 8-2 the incremental B/C ratio is used to evaluate a set of mutually exclusive
alternatives.
EXAM PLE S-2
A city may construct and operate two gas-burning power plants and a distribution network to provide
electricity to several city-owned properties. 吀栀e 昀漀llowing costs and bene昀椀ts have been identi昀椀ed:
Primary costs: Construction of the power plant 昀愀cilities; cost of installing the power distribution
network; life-cycle maintenance and operating costs.
Primary bene昀椀ts: Elimination of payments to the current electricity provider; creation of jobs 昀漀r
construction, operation, and maintenance of the 昀愀cilities and distribution network; revenue
昀爀om selling excess power to utility companies; increased employment 昀漀r city residents.
吀栀ere are 昀漀ur possible designs 昀漀r the power plants. Each has a life of 45 years. Use the B/C ratio with
an interest rate of 8% to recommend a course of action.
Values (x $104) 昀漀r Competing Design Alternatives
($ 昀椀gures in tens of thousands)
Project costs
Plant construction
Annual operating and maintenance
Project bene昀椀ts
Annual savings 昀爀om utility pa礀洀ents
Revenue 昀爀om overcapacity
Annual e昀昀ect of jobs created
Other data
Project life, in years
Discounting rate (MARR)
I
II
III
IV
$ 12,500
$ 1 20
$ 1 1 ,000
$480
$ 1 2,500
$325
$ 16,800
$ 1 45
45
8%
45
8%
45
8%
45
8%
$580
$700
$400
$700
$550
$750
$950
$200
$ 1 50
$ 1 ,300
$250
$500
Incremental Bene昀椀t-Cost Analysis
251
SO LUTI O N
Alternatives I to IV and the do-nothing alternative are mutually exclusive choices because one and only
one of them will be chosen. 吀栀ere昀漀re, an incremental B/C ratio method is used to obtain the solution.
Step 1 䤀搀ent椀昀y the alternatives. 吀栀e alternatives are do nothing, and designs I, II, III, and IV.
Step 2
Calculate 琀栀e ℀蔀 爀愀tio for each alternative. In this optional step we calculate the B/C ratio 昀漀r each
alternative based on individual cash 昀氀ows. We will use the ratio of the PW of bene昀椀ts to costs.
B/C ratio (I) = (580 + 700 + 400)(P/A, 8%, 45)/[12,500 + 120(P/A, 8%, 45)] = 1 .46
B/C ratio (II) = (700 + 550 + 750)(P/A, 8%, 45)/[1 1,000 + 480(P/A, 8%, 45)] = 1 .44
B/C ratio (III) = (200 + 950 + l S0)(P/A, 8%, 45)/[12,500 + 325(P/A, 8%, 45)] = 0.96
B/C ratio (IV) = (1,300 + 250 + S00)(P/A, 8%, 45)/[16,800 + 145(P/A, 8%, 45)] = 1.30
Alternatives I, II, and IV all have B/C ratios greater than 1 and thus merit 昀甀rther consider­
ation. Alternative III does not meet the acceptability criterion and could be eliminated 昀爀om
昀甀rther consideration. However, to illustrate that Step 2 is optional, all 昀漀ur design alternatives
will be analyzed incrementally.
Step 3 Rank-order the projects. Here we calculate the PW of costs 昀漀r each alternative. 吀栀e
denominator of the B/C ratio includes 昀椀rst cost and annual O&M costs, so the PW of costs
昀漀r the alternatives are
PW costs (I) = 12,500 + 120(P/A, 8%, 45) = $13,953
PW costs (III) = 12,500 + 325(P/A, 8%, 45) = $16,435
PW costs (II) = 1 1,000 + 480(P/A, 8%, 45) = $16,812
PW costs (IV) = 16,800 + 145(P/A, 8%, 45) = $18,556
吀栀e rank order 昀爀om low to high value of the B/C ratio denominator is as 昀漀llows: do nothing,
I, III, II, IV. So we start o昀昀 with "do nothing" as the champion.
Step 4 䤀搀ent椀昀y the increment under conside爀愀tion.
Step 5 Calculate the B/C ratio.
Step 6 Which alternative is better?
1st Iteration
Step 4 Increment under Consideration (Do Nothing 娀堀 I)
Step s
Step 6
㘀尀Plant construction cost
䐀⸀Annual O&M cost
PW of 㘀尀Costs
䐀⸀Annual utility payment savings
䐀⸀Annual overcapacity revenue
䐀⸀Annual bene昀椀ts of new jobs
PW of 䰀㄀Bene昀椀ts
䐀⸀B/C ratio (PW 㘀尀B)/(PW 䐀⸀C)
Is increment justi昀椀ed?
$12,500
120
13,953
580
700
400
20,342
1.46
Yes
2nd Iteration 3rd Iteration 4th Iteration
(I 娀堀 III)
(I ⴀ⬀ II)
(II ⴀ⬀ IV)
$0
205
2,482
370
500
-250
-4,601
-1.15
No
-$1,500
360
2,859
120
-150
350
3,875
1.36
Yes
$5,800
-335
1,744
600
-300
-250
605
0.35
No
continued
252
Cha pter 8 Benefit-Cost Ratio and Other Analysis Techniques
As an example of these calculations, consider the third increment (I 娀堀 II).
戀⸀Plant construction cost
戀⸀Annual O&M cost
PW of 戀⸀Costs
戀⸀Annual utility payment s愀瘀ings
戀⸀Annual overcapacity revenue
戀⸀Annual bene昀椀ts of new jobs
PW of 戀⸀Bene昀椀ts
戀⸀B/C ratio (PW 戀⸀B)/(PW 戀⸀C)
= 1 1,000 - 12,500 = -$1,500
= 480 - 120 = $360
= -1,500 + 360(P/A, 8%, 45) = $2,859
or = 16,812 - 13,953 = $2,859
= 700 - 580 = $120
= 550 - 700 = -$150
= 750 - 400 = $350
= (120 - 150 + 350)(PIA, 8%, 45)
= $3,875
= 3,875/2,850 = 1 . 36
The analysis in the table proceeded as 昀漀llows: do nothing to Alternative I was justified
(戀⸀B/戀⸀C ratio = 1 .46), so Alternative I became the new champion; Alternative I to Alterna­
tive III was not justi昀椀ed (戀⸀B/戀⸀C ratio = - 1 . 1 5), so Alternative I remained the champion;
Alternative I to Alternative II was justi昀椀ed (戀⸀B/戀⸀C ratio = 1 . 36) , so Alternative II became
the champion; Alternative II to Alternative IV was not justi昀椀ed (戀⸀B/戀⸀C ratio = 0. 35), so
Alternative II remained as champion.
Step 8 Select best alternative. Alternative II is the champion at the end of the process, so we select it as
best alternative.
Other E昀昀ects of Public Projects
Four areas remain that merit discussion in describing the di昀昀erences between government
and non-government economic analysis:
•
•
•
•
the 昀椀nancing of government versus non-government projects
the typical length of government versus non-government project lives
the quantifying and valuing of bene昀椀ts and costs
the general e昀昀ects of politics on economic analysis
Project Financing
Governmental and market-driven 昀椀rms di昀昀er in the way investments in equipment, 昀愀­
cilities, and other projects are 昀椀nanced. In general, 昀椀rms rely on monies 昀爀om individual
investors (through shares and bonds), private lenders, and retained earnings 昀爀om oper­
ations. 吀栀ese sources serve as the pool 昀爀om which investment dollars 昀漀r projects come.
Management's job in the market-driven 昀椀rm is to match 昀椀nancial resources with projects
in a way that keeps the 昀椀rm growing, produces an e昀케cient and productive environment,
and continues to attract investors and 昀甀ture lenders of capital.
On the other hand, the government sector o昀琀en uses taxation and bonds as the source
of investment capital. In government, taxation and revenue 昀爀om operations is adequate to
昀椀nance only modest projects. However, public projects tend to be large in scale (roadw愀礀s,
bridges, and so on), which means that 昀漀r many public projects 100% of the investment
costs must be borrowed-unlike those in the private sector.
Other Effects of Public Projects
253
In Canada, because of the division of powers between di昀昀erent levels of government,
large public projects o昀琀en have shared 昀甀nding 昀爀om federal, provincial, and municipal
sources. Generally the projects are 昀椀nanced 昀爀om current revenues, but 昀漀r large projects
governments have the right to borrow money through bond issues.
Another recent innovation is called P3 昀甀nding, which stands 昀漀r "public-private part­
nership." In these arrangements, a need is identi昀椀ed by the government, but the capital, en­
gineering, construction, and operation of the resulting 昀愀cility are undertaken by a private
corporation. 吀栀e corporation is repaid either by user fees or by payments 昀爀om the govern­
ment. 吀栀e Confederation Bridge that links Prince Edward Island to the mainland is a P3
arrangement; the operating company collects and keeps the tolls according to a speci昀椀ed
arrangement. 吀栀e bypass 昀爀eew愀礀 around Edmonton is another P3, but in this instance,
there are to be no tolls and the provincial government pays the P3 partner an annual fee.
Limitations on the use and sources of borrowed monies make the 昀甀nding of projects
in the public sector much di昀昀erent 昀爀om this process in the private sector. Private sector
昀椀rms are seldom able to borrow 100% of the 昀甀nds required 昀漀r projects, as can be done
in the public sector, but at the same time, private entities do not 昀愀ce restrictions on debt
retirement or the uncertainty of voter approval.
Duration of Project
Government projects o昀琀en have longer lives than those in the private sector. In the pri­
vate sector, projects most o昀琀en have a projected or intended life ranging between 昀椀ve and
15 years. Some markets and technologies change more rapidly and some more slowly, but
a majority of projects 昀愀ll in that interval. Complex advanced manu昀愀cturing technologies,
like computer-aided manu昀愀cturing or flexible automated manu昀愀cturing cells, tend to
have project lives at the longer end of that range.
Government projects usually have lives in the range of 20 to 50 or more years. 吀礀pical
projects are highways, city water and sewer in昀爀astructure, county waste 昀愀cilities, and
provincial museums. Government projects, because they tend to be long-range and large­
scale, usually require substantial 昀甀nding in the early stages. Highway, water and sewer,
and library projects can cost millions of dollars 昀漀r design, surveying, and construction.
吀栀ere昀漀re, it is in the best interest of the advocates of such projects to spread that 昀椀rst cost
over as many years as possible to reduce the annual cost of capital recovery. Using longer
project lives to downplay the e昀昀ects of a large 昀椀rst cost increases the desirability of the
project, as measured by the B/C ratio. Another aspect closely associated with managing
the size of the capital recovery cost in a B/C ratio analysis is the interest rate used 昀漀r dis­
counting. Lower interest rates reduce the capital recovery cost of having money tied up in
a project. Example 8-3 illustrates the e昀昀ects that project life and interest rate can have on
the analysis and acceptability of a project.
EXAM PLE 8-3
Consider a project to build a new high school, needed because of recent (and projected) population
growth. Analyze the project with interest rates of 3%, 10%, and 15% and with horizons of 15, 30, and
60 years.
continued
254
Cha pter 8 Benefit-Cost Ratio and Other Analysis Techniques
Building 昀椀rst costs (design, planning, and construction)
Initial cost 昀漀r roadway and parking 昀愀cilities
First cost to equip and 昀甀rnish 昀愀cility
Annual operating and maintenance costs
Annual savings 昀爀om rented space
Annual bene昀椀ts to community
$10,000,000
5,500,000
500,000
350,000
400,000
1,600,000
SOLUTION
With this project we examine the e昀昀ect that varying project lives and interest rates have on the eco­
nomic value of a public project.
In each case the 昀漀rmula is
1,600,000 + 400,000
B/c rati. o = -------------------( 10,000 + 5,500,000 + 500,000) (AIP, i, n) + 350,000
吀栀e B/C ratio 昀漀r each combination of project life and interest rate is tabulated as 昀漀llows:
Bene昀椀t-Cost Ratio 昀漀r Various
Combinations of Project Life and Interest Rate
Interest
Project Life (years)
3%
10%
15%
15
30
60
1.24
1.79
2.24
0.86
1.03
1.08
0.69
0.76
0.77
From these numbers one can see the e昀昀ect of project life and interest on the analysis and recommen­
dation. At the lower interest rate, the project has B/C ratios above 1 in all cases of project life, while at
the highest rate the ratios are all less than 1. At an interest rate of 10%, the recommendation to invest
changes 昀爀om no at a life of 15 years to yes at 30 and 60 years. A higher interest rate discounts the ben­
e昀椀ts in later years more heavily, so that they may not matter. In this case, the bene昀椀ts 昀爀om Years 31 to
60 add only 0.01 to the B/C ratio at 15% and 0.05 at 10%. At 3% those bene昀椀ts add 0.45 to the ratio. By
manipulating these two parameters (project life and interest rate), it is possible to reach entirely di昀昀er­
ent conclusions about the desirability of the project.
Example 8-3 demonstrates that we must ensure that a long life and a low interest rate
昀漀r a public project are truly appropriate and not chosen solely to make a marginal project
look better.
Quantifying and Valuing Bene昀椀ts and Costs
吀栀e high school in Example 8-3 provided annual bene昀椀ts of $1.6 million to the commun­
ity. If you were evaluating the high school, how would you estimate this? Many public
sector projects like the school and the examples in Table 8-1 have consequences di昀케cult to
state in monetary terms.
Other Effects of Public Projects
255
First, the number of people a昀昀ected by the project-now and through the project's
horizon-have to be counted. 吀栀en a dollar value 昀漀r each person is required. For the high
school it m愀礀 be easy to estimate the number of students. But how much better will the
educational outcomes be with the new school, and how valuable is that improvement? Will
the school be available 昀漀r a昀琀er-hours community uses?
On the other hand, consider the rebuilding of levees (dikes) around New Orleans in
the a昀琀ermath of Katrina. Economic evaluation of the di昀昀erent alternatives requires esti­
mating the number of residents that will be protected by improved levees, a calculation
that is extremely di昀케cult. It requires estimation not only of how the rebuilding of the city
will progress but also of the size and 昀爀equency of 昀甀ture storm surges 昀爀om hurricanes,
whose 昀爀equency and intensity may be changing. Once the number of people and homes
has been estimated over the next 30 to 100 years, it will be necessary to put a value on
property, disrupted lives, and human lives.
Ta ble 8-1
Exa mple Bene昀椀ts and Costs for Public I nvestments
Public Project
Primary Bene昀椀ts
Primary Costs
New city airport just outside city
More 昀氀ights, new business
Increased travel time to airport, more
tra昀케c in outer suburbs, more noise 昀漀r
those living under the 昀氀ight path
Highway b礀瀀 ass around town
Shorter commuting times, reduced con­
gestion on roads
Lost sales to businesses on roads, loss
of agricultural lands
New subway
Faster commuting times, less pollution
Loss of jobs because of bus line clos­
ing, less access to service (fewer stops)
Creation of a county waste disposal 昀愀­
cility versus sending waste to another
region or province
Less costly, 昀愀ster, and more responsive to
customers
Objectionable sights and smells, loss
of market value 昀漀r homeowners, loss
of relatively pristine 昀漀rest land
Construction of a nuclear power plant
Lower energy costs, new industry in area
Environmental risk
Although many people 昀椀nd it di昀케cult to put a dollar value on a human life, there
are many public projects whose main intent is to reduce the number of deaths caused by
昀氀oods, cancer, car accidents, and other causes. 吀栀ose projects are o昀琀en justi昀椀ed by the
value of preventing deaths. 吀栀us, valuing human lives is an inescapable part of public
sector engineering economy.
Because the bene昀椀ts and costs of public projects are o昀琀en di昀케cult to quantify, the es­
timated values will have more uncertainty than is typical 昀漀r private sector projects. 吀栀us
those who 昀愀vour a project and those who oppose it will o昀琀en identi昀礀 di昀昀erent values and
try to have them applied in support of their respective positions.
Project Politics
Political in昀氀uences are felt in nearly every decision made in any organization. Predictably,
some individual or group will support its own particular interests over competing views.
吀栀at situation exists in both 昀椀rms and government. In government the e昀昀ects of politics
are felt continuously at all levels because of the large-scale and multi-purpose nature of
projects, because government decision making involves the use of the citizens' common
pool of money, and because individuals and groups have di昀昀erent values and views. For
example, how important is economic development relative to environmental problems?
256
Cha pter 8 Benefit-Cost Ratio and Other Analysis Techniques
吀栀e guideline 昀漀r public decision making is to produce a net gain in wel昀愀re. However,
it is impossible to please everyone all the time. 吀栀ere昀漀re bene昀椀t-cost analysis measures
the variables in order to determine whether a particular policy or program has resulted in
a net bene昀椀t.
Since government projects tend to be large in scale, the time required to plan,
design, 昀甀nd, and construct them is usually several years. However, the political pro­
cess tends to produce government leaders who support short-term decision making ( be­
cause many government terms of o昀케ce, either elected or appointed, are relatively short).
吀栀erein lies another di昀昀erence between 昀椀rms and government-short-term decision
making, long-term projects.
Because government decision makers are in the public eye more than those in the
private sector, governmental decisions are generally more a昀昀ected by politics. 吀栀us
the decisions that public o昀케cials make may not always be the best 昀爀om an ove爀愀ll per­
spective. If a particular situation exposes a public o昀케cial to ridicule, he or she may
choose an expedient action to eliminate negative exposure (whereas a more care昀甀l
analysis might have been better). Or such a decision maker may placate a small, but
vocal, political group over the interest of the m愀樀ority. Or a public decision maker may
avoid controversy by declining to make any decision on an important, but politically
charged, issue.
EXAM PLE S-4
Consider again Example 8-2, where we evaluated power plant designs. Remember that government
projects are o昀琀en opposed and supported by di昀昀erent groups of citizens. For the decision about the
electric power plant, several political considerations m愀礀 a昀昀ect the evaluation of this project.
•
•
•
•
•
吀栀e mayor has been a strong advocate of workers' rights and has received abundant cam­
paign support 昀爀om organized labour (which is especially important in an industrial city). By
championing this project, the m愀礀or will be seen as pro-labour, thereby bene昀椀ting her bid 昀漀r
re-election, even if the project is not 昀甀nded.
吀栀e regulated electric utilities in the province are strongly against this project, claiming that it
would compete directly with them and take away some of their biggest customers. 吀栀e provid­
ers have a strong lobby and contacts in the provincial cabinet. 吀栀e mayor of a neighbouring city
has already protested that this project is the 昀椀rst step toward "rampant socialism."
Business leaders in the municipalities where the two 昀愀cilities would be constructed 昀愀vour
the project because it would create more jobs and increase the tax base. 吀栀ese leaders promote
the project as a win-win opportunity 昀漀r government and industry, 昀漀r the city can bene昀椀t by
reducing costs and the electric utilities can improve their service by 昀漀cusing more e昀昀ectively
on residential customers and their needs.
吀栀e Chamber of Commerce is promoting this project, proclaiming that it is an excellent example
of "initiating proactive and creative solutions to the problems that this city 昀愀ces."
Federal and provincial regulatory agencies are watching this project closely with respect to
environmental legislation. Speculation is that the plans are to use a high-sulphur grade of local
coal exclusively. 吀栀us "stack scrubbers" would be required, or the high-sulphur coal would
have to be mixed with imported lower-sulphur coal to bring the overall air emissions in line
Other Effects of Public Projects
•
•
257
with federal standards. 吀栀e mayor is using this opportunity to make the point that "the people
of this city don't need regulators to tell us if we can use our own coal!"
吀栀e coal operators and mining unions strongly support this project. 吀栀ey are very pleased
with the increased demands 昀漀r coal and the mayor's pro-labour advocacy. 吀栀ey plan to lobby
strongly in 昀愀vour of the project.
Land-preservation and environmental groups are strongly opposing the proposed proj­
ect. T hey have studied the potentially damaging e昀昀ects of this project on the land and on
water and air quality, as well as on the ecosystem and wildlife, in the areas where the two
昀愀cilities would be constructed. Environmentalists have started a public-awareness cam­
paign urging the premier to act as the "chief steward" of the natural beauty and resources
of the province.
Will the project be 昀甀nded? We can only guess. Clearly, however, we can see the competing in昀氀uences
that can be, and o昀琀en are, part of decision making in the public sector.
Payback Period
We now come to the 昀椀nal method of comparing projects: payback period. 吀栀is method
is o昀琀en used in the private sector, despite certain disadvantages that might at 昀椀rst
appear serious.
Payback period is the period of time required 昀漀r the pro昀椀t or other bene昀椀ts of a
project to equal the cost.
吀栀e rule in all situations is to minimize the payback period. 吀栀e computation of p愀礀­
back period is illustrated in Examples 8-5 and 8-6.
EXAM PLE 8-5
吀栀e cash 昀氀ows 昀漀r two alternatives are as 昀漀llows:
Year
0
1
2
3
4
5
A
B
- $1,000
+200
+200
+1,200
+1,200
+1,200
- $2,783
+1,200
+1,200
+1,200
+1,200
+1,200
You may assume the bene昀椀ts occur throughout the year rather than just at the end of the year. On the
basis of payback period, which alternative should be selected?
continued
258
Cha pter 8 Benefit- Cost Ratio a n d Other Analysis Tec h n i q ues
SO LUTI O N
Alternative A
Payback period is how long it takes 昀漀r the pro昀椀t or other bene昀椀ts to equal the cost of the investment.
In the 昀椀rst two years, only $400 of the $1,000 cost is recovered. 吀栀e remaining $600 cost is recovered in
the 昀椀rst half of Year 3. 吀栀us the payback period 昀漀r Alternative A is two and a half years.
Alternative B
Since the annual bene昀椀ts are uni昀漀rm, the payback period is simply
$2,783/$1,200 per year = 2.3 years
To minimize the payback period, choose Alternative B.
ⴀⴀ-----·=
=
=
=
=
Ęᡓ------眀嬀
=
------ⴀⴀ-----------�
EXAM PLE S - 6
A 昀椀rm is trying to decide which of two scales it should install to check a package-昀椀lling opera­
tion in the plant. If both scales have a six-year life, which one should be selected? Assume an 8%
interest rate.
Alternative
Cost
Atlas scale
Tom 吀栀umb scale
$2,000
3,000
Uni昀漀rm Annual
Bene昀椀t
End- of- Use昀甀l- Life
Salvage Value
$450
600
SO LUTI O N
Atlas scale
Cost
. = ------P礀쌀ack penod
Uni昀漀rm annu愀氀 bene昀椀t
2,000
= -= 4.4 years
450
Tom 吀栀umb scale
Cost
. = -------Payback penod
Uni昀漀rm annu愀氀 bene昀椀t
3,000
= -- = S years
600
$ 1 00
700
Payback Period
-u
§
3,000
-u
2,000
0
1 ,000
.∀✀
㐀⸀氀匀
§
氀匀�
0
�氀匀
䨀攀
1 ,000
㰀漀
�
Cost = $3,000
氀匀
2,000
0
�w �,:㬀阀�
�
椀猀f
� ⴀ㜀
䨀攀
3,000
氀匀�
� ��
�䤀退 錀娀
0儀愀 ⼀⼀�
�
�
∀儀
氀匀
氀匀
㬀椀
0
1
2
3
Year
(a)
4 1 5
4.4
0
2
3
4
5
Year
( b)
FIGURE 8-2 Payback period plots for Example 8 -6: (a) Atlas scale and (b) Tom Th umb scale.
Figure 8-2 illustrates the situation. To minimize payback period, choose the Atlas scale.
吀栀ere are 昀漀ur important points to be understood about payback period calculations:
1. 吀栀is is an approximate, rather than exact, economic analysis calculation.
2. All costs and all pro昀椀ts, or savings of the investment, be昀漀re payback are included
without considering di昀昀erences in their timing.
3. All the economic consequences beyond the payback period are ignored.
4. Being an approximate calculation, the payback period method may or may not
choose the correct alternative.
吀栀is last point-that the payback period method m愀礀 result in the selection of the wrong
alternative-was illustrated by Example 8-6. When payback period is used, the Atlas scale
appears to be the more attractive alternative. Yet if the same problem were solved by the
present worth method, the alternative chosen would be the Tom 吀栀umb scale. 吀栀e reason
昀漀r the di昀昀erence is the $700 salvage value at the end of six years. 吀栀e salvage value occurs
a昀琀er the payback period, so it was ignored in the payback calculation. However, it was
considered in the present worth analysis, which showed correctly that the Tom 吀栀umb
scale was in 昀愀ct more desirable.
But if payback period calculations are approximate and may cause the wrong alter­
native to be chosen, why are they used? First, the calculations can be made readily by
people un昀愀miliar with economic analysis. Second, the concept of payback period is easily
understood. 吀栀ird, minimizing the payback period minimizes the distance we have to
look into the 昀甀ture and hence reduces the uncertainty in our predictions of 昀甀ture costs
and bene昀椀ts.
Payback period measures how long it will take 昀漀r the cost of the investment to be
recovered 昀爀om the bene昀椀ts. Firms are o昀琀en very interested in this time period: a rapid
return of invested capital means that it can be reused sooner 昀漀r other purposes. But one
must not con昀甀se the speed of the return of the investment, as measured by the p愀礀back
259
260
Cha pter 8 Benefit-Cost Ratio and Other Analysis Techniques
period, with economic e昀케ciency. 吀栀ey are two distinctly separate concepts. 吀栀e 昀漀rmer
emphasizes the quickness with which invested 昀甀nds return to a 昀椀rm; the latter considers
the overall pro昀椀tability of the investment.
Example 8-7 illustrates how using the payback period method may result in an unwise
decision.
EXAM PLE S - 7
A 昀椀rm is buying production equipment 昀漀r a new plant. Two di昀昀erent machines are being considered
昀漀r a particular operation.
Installed cost
Net annual bene昀椀t a昀琀er all
annual expenses h愀瘀e been
deducted
Use昀甀l li昀攀, in years
Tempo Machine
Dura Machine
$30,000
$12,000 the 昀椀rst year, declining
$3,000 per year therea昀琀er
$35,000
$1,000 the 昀椀rst year,
increasing $3,000 per
year therea昀琀er
4
8
Neither machine has any salvage value. Compute the payback period 昀漀r each machine.
SOLUTION BASED ON 䄀꤀BACK PERIOD
-
35,000
30,000
25,000
�
∀✀0 20,000
u 15,000
-
8
栀먀
.尀尀㬀娀⼀挀
�
�
30,000
25,000
u 15,000
꜀✀∀꜀
5,000
10,000
5,000
1
2
Year
(a)
FIGURE 8-3
Cost = $35,000
∀✀0 20,000
�
�
10,000
0
35,000
3
4
5
0
1
2
Year
3
4
5
( b)
Payback period plots for Example 8-7: (a) Tem po mach ine and (b) Dura mach ine.
吀栀e Tempo machine has a declining annual bene昀椀t, while the Dura has an increasing annual bene昀椀t.
Figure 8-3 shows that the Tempo has a 昀漀ur-year payback period and the Dura has a 昀椀ve-year payback
period. To minimize the payback period, the Tempo is selected.
Now, as a check on the payback period analysis, compute the rate of return 昀漀r each alternative.
Assume the minimum attractive rate of return is 10%.
Payback Period
261
SOLUTION BASED ON R䄀吀E OF RETU RN
吀栀e cash 昀氀ows 昀漀r the two alternatives are as 昀漀llows:
Year
Tempo Machine
Dura Machine
0
1
2
3
4
5
6
7
8
- $30,000
+ 1 2,000
+9,000
+6,000
+3,000
0
0
0
0
0
-$35,000
+ 1 ,000
+4,000
+7,000
+ 1 0,000
+ 1 3,000
+ 1 6,000
+ 1 9,000
+22,000
+57,000
Tempo Machine
Since the sum of the cash 昀氀ows 昀漀r the Tempo machine is zero, we see immediately that the $30,000
investment just equals the subsequent bene昀椀ts. 吀栀e resulting rate of return is 0%.
Dura Machine
PW(i) = -35,000 + 1,000(P/A, i, 8) + 3,000(P/G, i, 8)
PW(20%) = -35,000 + 1,000(3.837) + 3,000(9.883) = -1,514
吀栀e 20% interest rate has discounted the 昀甀ture bene昀椀ts too much; it is too high. Try i = 15%:
PW(15%) = -35,000 + 1,000(4.487) + 3,000(12.481) = 6,930
= 4,487 + 37,443 = 41,930
吀栀is time, the interest rate is too low. Linear interpolation shows that the rate of return is approximately 19%.
If we use the rate of return method, it is clear that the Dura machine is 昀愀r superior to the Tempo.
On the other hand, the shorter payback period 昀漀r the Tempo does measure the speed of the return of
the investment. 吀栀e conclusion to be drawn is that liquidity and pro昀椀tability may be two quite di昀昀erent
criteria.
From the discussion and the examples, we see that the payback period method
can measure the speed of the return of the investment. 吀栀is might be quite important,
昀漀r example, 昀漀r a company that is short of working capital or 昀漀r a 昀椀rm in an industry
experiencing rapid changes in technology. Calculation of payback period alone, however,
must not be con昀甀sed with a care昀甀l economic analysis. Ignoring the cash flows a昀琀er the
payback period is seldom wise. We have shown that a short payback period does not always
mean that the associated investment is desirable. 吀栀us, payback period is not a suitable
replacement 昀漀r accurate economic analysis.
262
Cha pter 8 Benefit-Cost Ratio and Other Analysis Techniques
SUMMARY
Bene昀椀t-Cost Analysis: Economic analysis and decision making in government are
notably di昀昀erent 昀爀om these processes in the private sector because the objectives
of the public and private sectors are di昀昀erent. Government investments in proj­
ects seek to maximize the ratio of bene昀椀琀猀 to citizens to the costs to citizens and
to the government. Private 昀椀rms, on the other hand, are interested primarily in
maximizing shareholder wealth.
Several 昀愀ctors that do not a昀昀ect private 昀椀rms enter into decision making in government.
吀栀e source of capital 昀漀r public projects is limited primarily to taxes and bonds. Govern­
ments issuing bonds 昀漀r project construction are subject to legislative restrictions on debt
that do not apply to private 昀椀rms. Also, raising tax and bond monies involves sometimes
long and politically charged processes not 昀漀und in the private sector. In addition, govern­
ment projects tend to be larger than those of private 昀椀rms, and government projects a昀昀ect
many more people and groups in the population. 吀栀e bene昀椀ts and costs to the many people
a昀昀ected are di昀케cult to quanti昀礀 and value, unlike the situation in the private sector, where
products and services are sold and the revenue to the 昀椀rm is clearly de昀椀ned. All these 昀愀c­
tors slow down the decision process and make investment decision analysis more di昀케cult
昀漀r government decision makers than 昀漀r those in the private sector.
Another di昀昀erence between the public and private sectors lies in how the interest rate
(MARR) is set 昀漀r economic studies. In the private sector, considerations 昀漀r setting the
rate include the cost of capital and opportunity costs. In government, establishing the in­
terest rate is complicated by uncertainty in specifying the cost of capital and by the issue
of assigning opportunity costs to taxpayers or to the government.
吀栀e bene昀椀t-cost ratio is widely used to evaluate and justi昀礀 government-昀甀nded proj­
ects. 吀栀is measure of merit is the ratio of the equivalent worth of bene昀椀ts to the equivalent
worth of costs. A B/C ratio greater than 1.0 indicates that a project should be invested in
if 昀甀nding sources are available. For considering mutually exclusive alternatives, an in­
cremental analysis is required. 吀栀is method results in the recommendation of the project
with the highest investment cost that can be justi昀椀ed incrementally.
PROBLEMS
Payback Period: We de昀椀ne p愀礀back as the period of time required 昀漀r the pro昀椀t
or other bene昀椀ts of an investment to equal the cost of the investment. Payback is
simple to use and to understand, but it is a poor analysis technique 昀漀r ranking
alternatives. Although it provides a measure of the speed of the return of the in­
vestment, it is not an accurate measure of the pro昀椀tability of an investment.
Objective a nd Viewpoint
8-1
List the potential costs and bene昀椀ts that should
be considered in evaluating a potential nuclear
power plant. What stakeholder viewpoints will
need to be considered?
8-2
A highway bypass will completely circle the
city. Name at least three bene昀椀ts and three
costs. What stakeholder viewpoints will need
to be considered?
8-3
Improvements at a congested intersection re­
quire that the government acquire the prop­
erties on the 昀漀ur corners: two gas stations,
a church, and a bank. Construction will
take a year; 70% of the costs will be paid by
the province and 30% by the city. 吀栀e traf­
昀椀c through the intersection consists mainly
of commuters, local residents, and deliver­
ies to and by local businesses. 吀栀ere is some
through tra昀케c 昀爀om other parts of the met­
ropolitan area.
Problems
What are the bene昀椀ts and costs that must
be considered in evaluating the project? Which
of them must be included 昀爀om the city's view­
point? From the province's viewpoint? What
viewpoint should be used to evaluate the
project?
8-4
⠀　吀栀e province may eliminate a railw愀礀 level
crossing by building an overpass. 吀栀e new
structure, together with the land needed,
would cost $1.8 million. 吀栀e analysis period is
assumed to be 30 years because either the rail­
w愀礀 or the highway above it will be relocated
by then. 吀栀e salvage value of the bridge (actu­
ally, the net value of the land on either side of
the railway tracks) 30 years hence is estimated
to be $100,000. A 6% interest rate is to be used.
Every day about 1,000 vehicles are de­
layed by trains at the level crossing. Trucks
account 昀漀r 40%; 60% are other vehicles. Time
昀漀r truck drivers is valued at $18 an hour and
昀漀r other drivers $5 an hour. Average time
saving per vehicle will be two minutes if the
overpass is built. No time saving occurs 昀漀r
the railway.
吀栀e railway spends $48,000 annually 昀漀r
crossing guards. During the preceding 10year period, the railway has paid out $600,000
in settling lawsuits and accident cases related
to the level crossing. 吀栀e proposed project
will entirely eliminate both these expenses.
吀栀e province estimates that the new overpass
will save it about $6,000 per year in expenses
caused directly by the accidents. 吀栀e overpass,
if built, will belong to the province.
Should the overpass be built? If the
overpass is built, how much should the rail­
way be asked to contribute as its share of the
$1,800,000 construction cost?
Selecting a Rate
8 -5
8-6
Discuss the di昀昀erent concepts that can be used
when setting the discounting rate 昀漀r economic
analysis in the public sector. What is the 昀椀nal
recommendation of this chapter 昀漀r setting
that rate?
吀栀e city's waste disposal department has a
capital budget of $600,000. What is the city's
263
opportunity cost of capital if it has the 昀漀llow­
ing projects to consider? What does that say
about which projects should be done?
Project
First Cost
Rate of
Return {%}
A
B
C
D
E
$100,000
200,000
300,000
200,000
100,000
23
22
17
19
18
8-7
B/C Ratio
at 7%
1.30
1.40
1.50
1.35
1.56
吀栀e province's 昀椀sh and game department has
a capital budget of $9 million. What is the
government's opportunity cost of capital if it
has the 昀漀llowing projects to consider? What
does it tell us about which projects should
be done?
Project
First Cost
Rate of
Return {%}
A
B
C
D
E
F
G
H
$2,000,000
1,000,000
2,000,000
3,000,000
2,000,000
3,000,000
3,000,000
1,000,000
9
14
10
16
13
15
12
11
8-8
B/C Ratio
at 7%
1.23
1.42
1.17
1.45
1.56
1.35
1.32
1.26
A provincial bond has a 昀愀ce value of $10,000.
Interest of $400 is paid every six months. 吀栀e
bond has a life of 20 years. What is the e昀昀ec­
tive rate of interest on the bond? Is that rate
adjusted 昀漀r in昀氀ation? Estimate the govern­
ment's cost of capital 昀漀r this bond.
B/C and Modi昀椀ed Ratio
8-9
瘀∀
⠀　Consider the 昀漀llowing investment opportunity:
Initial cost
Additional cost at end of Year 1
Bene昀椀t at end of Year 1
Annual bene昀椀t per year at end
of Years 2-10
$100,000
150,000
0
20,000
With interest at 7%, what is the bene昀椀t-cost
ratio 昀漀r the project?
264
8-10
Cha pter 8 Benefit-Cost Ratio and Other Analysis Techniques
Calculate the bene昀椀t-cost ratio 昀漀r the 昀漀llow­
ing project.
Required 昀椀rst costs
Annual bene昀椀ts to users
Annual costs to users
Annual cost to government
Project li昀攀
Interest rate
8-11
⠀　$1,200,000
$500,000
$25,000
$125,000
35 years
10%
⠀　A government agency has estimated that a
昀氀ood control project has costs and bene昀椀ts that
are parabolic, according to the equation
( PW of bene昀椀ts) 2 - 22( PW of cost) + 44 = 0
where both bene昀椀ts and costs are stated in mil­
lions of dollars. What is the present worth of
cost 昀漀r the optimal-sized project?
I ncremental Analysis
8-12
8-13
吀栀e Highridge region needs an additional
supply of water 昀爀om Steep Creek. 吀栀e engi­
neer has selected two plans 昀漀r comparison:
G爀愀vi琀礀 plan: Divert water at a point 10 km
up Steep Creek and carry it through a pipeline
by gravity to the district.
Pumping plan: Divert water at a point near
the district and pump it through 2 km of pipe­
line to the district. 吀栀e pumping plant can be
built in two stages, with half-capacity installed
initially and the other half 10 years later.
Gr愀瘀i琀礀
Pumping
$2,800,000
Initial investment
0
Investment in 10th year
Operation, maintenance,
10,000
replacements, per year
Average annual power cost,
0
昀椀rst 10 years
0
next 30 years
$1,400,000
200,000
25,000
50,000
100,000
Use a 40-year analysis period and 8% interest. Salvage values can be ignored. During
the 昀椀rst 10 years, the average use of water will
be less than during the remaining 30 years.
Use the bene昀椀t-cost ratio method to select the
more economical plan.
吀眀o di昀昀erent routes are being considered 昀漀r
a mountain highway construction project. 吀栀e
h最턀 road will require the building of several
bridges, and it navigates around the highest
mountain points, thus requiring more road­
way. 吀栀e low road constructs several tunnels
昀漀r a more direct route through the mountains.
Projected travel volume 昀漀r this new section of
road is 2,500 cars per day. Use the B/C ratio to
determine which alternative should be recom­
mended. Assume that project life is 45 years
and i = 6%.
High Road Low Road
Construction cost per kilometre
Number of kilometres required
Annual bene昀椀t per
car-kilometre
Annual O&M costs per
kilometre
$200,000
35
$0.015
$450,000
10
$0.045
$2,000
$10,000
8-14
吀栀e city engineer has prepared two plans 昀漀r
roads in the city park. Both plans meet an­
ticipated requirements 昀漀r the next 40 years.
吀栀e minimum attractive rate of return 昀漀r the
city is 7%.
Plan A is a three-stage development pro­
gram: $300,000 is to be spent now, 昀漀llowed by
$250,000 at the end of 15 years and $300,000 at
the end of 30 礀攀ars. Annual maintenance will
be $75,000 昀漀r the 昀椀rst 15 礀攀ars, $125,000 昀漀r the
next 15 礀攀ars, and $250,000 昀漀r the 昀椀nal 10 years.
Plan B is a two-stage program: $450,000
is required now, 昀漀llowed by $50,000 at the
end of 15 years. Annual maintenance will be
$100,000 昀漀r the 昀椀rst 15 years and $125,000 昀漀r
subsequent years. At the end of 40 years, the
plan has a salvage value of $150,000.
Use bene昀椀t-cost ratio analysis to deter­
mine which plan should be chosen.
8-15
Evaluate these mutually exclusive alternatives
with a horizon of 15 years and a MARR of 12%.
Initial investment
Annual savings
Annual costs
Salvage value
A
B
C
$9,500
3,200
1,000
6,000
$18,500
5,000
2,750
4,200
$22,000
9,800
6,400
14,000
Problems
Use the 昀漀llowing:
(a) Conventional BIC ratio
(b) Mod昀툀ed BIC ratio
挀儀 Present worth analysis
(d) Internal rate of return analysis
(e) Payback period
8-16
8-17
V
䐀⸀
A SO-metre tunnel must be constructed 昀漀r
a new sewer system 昀漀r a city. One alternative is to build a 昀甀ll-capacity tunnel now 昀漀r
$500,000. 吀栀e other alternative is to build
a half-capacity tunnel now 昀漀r $300,000 and
then to build a second parallel half-capacity
tunnel 20 years hence 昀漀r $400,000. 吀栀e cost
to repair the tunnel lining every 10 years
is $20,000 昀漀r the 昀甀ll-capacity tunnel and
$16,000 昀漀r each half-capacity tunnel.
Determine whether the 昀甀ll-capacity
tunnel or the half-capacity tunnel should be
constructed now. Solve the problem by ben­
e昀椀t-cost ratio analysis, using a 5% interest
rate and a SO-year analysis period. 吀栀ere will
be no tunnel lining repair at the end of the
SO years.
Ta ble PS-18
265
Six mutually exclus瘀츀 i瘀팀stments h瘀였 been iden­
ti昀椀ed 昀漀r evalu愀琀ion by means of the bene昀椀t-cost
ratio method. Assume a MARR of 10% and an
equal project life of 25 years 昀漀r all altern愀琀瘀츀s.
(a) Use annual worth and the B/C ratio to
identify the better alternative.
(b) If this were a set of independent alternatives,
how would you conduct a comparison?
Mutually Exclusive Alternatives
Annualized
I
2
3
4
5
6
Net costs to
sponsor ($M)
Net bene昀椀ts to
users ($M)
15.5
13.7
16.8
10.2
17.0
23.3
20.0
16.0
15.0
13.7
22.0
25.0
8-18
A section of a highw愀礀 needs repair, but the
volume of tra昀케c is so low that few motorists
would bene昀椀t 昀爀om the work. However, tra昀케c
is expected to increase. 吀栀e repair work will
produce bene昀椀ts 昀漀r 10 years a昀琀er it is com­
pleted. See Table PS-18 昀漀r the data.
Should the road be repaired and, if so,
when? Use a 15% MARR.
Data (Costs i n $000s)
Repair in
2 Years
Repair in
4 Years
Repair in
5 Years
Year
Repair
Now
0
-$150
1
5
2
10
-$150
3
20
20
4
30
30
-$150
5
40
40
40
-$150
6
50
50
50
50
7
50
50
50
50
8
50
50
50
50
9
50
50
50
50
10
50
50
50
50
11
0
50
50
50
12
0
50
50
50
13
0
0
50
50
14
15
0
0
50
50
0
0
0
50
266
8-19
Cha pter 8 Benefit-Cost Ratio and Other Analysis Techniques
吀栀e provincial highway department is analyz­
ing the reconstruction of a mountain road. 吀栀e
vehicle tra昀케c increases each year; hence the
bene昀椀ts to the motoring public also increase.
Based on a tra昀케c count, the bene昀椀ts are pro­
jected as 昀漀llows:
Year
End-of-Year Bene昀椀t
2016
2017
2018
2019
2020
2021
$10,000
12,000
14,000
16,000
18,000
20,000
and so on, increasing
$2,000 per year
吀栀e reconstructed pavement will cost
$275,000 when it is installed and will have a
15-year use昀甀l life. 吀栀e construction period
is short; hence a beginning-of-year recon­
struction will result in the listed end-of-year
bene昀椀ts. Assume a 6% interest rate. 吀栀e re­
construction, if done at all, must be done no
later than 2021. Should it be done, and if so, in
what year?
Cha llenges for Public Sector
8-20 Big City Carl, a local politician, is advancing a
project 昀漀r the construction of a new dock and
pier system on the river to attract new com­
merce to the city. A committee appointed by
the mayor (an opponent of Carl's) has devel­
oped the 昀漀llowing estimates 昀漀r the e昀昀ects of
the project.
吀愀ble PS-21
$750,000
Cost of wrecking and removing current
昀愀cilities
Material, labour, and overhead 昀漀r new
construction
Annual operating and maintenance
expenses
Annual bene昀椀ts 昀爀om new commerce
Annual costs to sportsmen in area
Project life
Interest rate
$2,750,000
$185,000
$550,000
$35,000
20 years
8%
(a) Using the B/C ratio, determine whether
the project should be 昀甀nded.
(b) A昀琀er studying the numbers given by the
committee, Big City Carl argued that the
project life should be at least 25 years and
more likely closer to 30 years. Why is he
making that statement, and how did he
arrive at his estimate?
8-21
吀栀e provincial government proposes to con­
struct a multi-purpose water project to pro­
vide water 昀漀r irrigation and municipal use.
In addition there are 昀氀ood control and rec­
reation bene昀椀ts. 吀栀e bene昀椀ts are given in
Table P8-21. 吀栀e annual bene昀椀ts are one­
tenth of the decade bene昀椀ts. 吀栀e operation
and maintenance cost is $15,000 per year.
Assume a SO-year analysis period with no net
project salvage value.
(a) If an interest rate of 5% is used, and a
bene昀椀t-cost ratio of unity, what capital
expenditure can be justi昀椀ed to build the
water project now?
(b) If the interest rate is changed to 8%,
how does it change the justi昀椀ed capital
expenditure?
Data (i n $000s)
Decades
Purpose
First
Second
吀栀ird
Fourth
Fi昀琀h
Municipal
Irrigation
Flood control
Recreation
Totals
$40
350
150
60
-$600
$50
370
150
70
-$640
$60
370
150
80
$660
$70
360
150
80
$660
$110
350
150
90
$700
267
Problems
8-22
8-23
Brie昀氀y describe your sources and methods 昀漀r
estimating the value of
(a) a saved hour of commuting time
(b) the conversion of 15 kilometres of unused
railway tracks near a city of 300,000 into
a new bike path
挀儀 a reduction in annual 昀氀ood risks 昀爀om the
Mississippi River 昀漀r St Louis by 5%
(d) a human life
months each year that the plant is in operation.
Maintenance costs of the case-sealing machine
are expected to be negligible. 吀栀e case-sealing
machine is expected to be use昀甀l 昀漀r 昀椀ve annual
canning seasons and will then have no salvage
value. What is the p愀礀back period? What is the
nominal annual rate of return?
8-28
For the highway bypass in Problem 8-2, dis­
cuss potential data sources and methods 昀漀r
estimating each of the bene昀椀ts and costs.
8-24
Discuss potential data sources and methods 昀漀r
estimating each of the costs and bene昀椀ts identi­
昀椀ed in Problem 8-3 昀漀r the congested intersection.
8-25
吀栀ink about a major government construction
project under w愀礀 in your province or city. Are
the decision makers who originally analyzed
and initiated the project currently in o昀케ce?
How can politicians use political posturing
about government projects?
Year
0
1
2
3-10
8-27
⠀　Able Plastics, an injection-moulding 昀椀rm, has
negotiated a contract with a national chain of
department stores. Plastic pencil boxes are to
be produced 昀漀r a two-year period. Able Plas­
tics has never produced the item be昀漀re and re­
quires all new dies. If the 昀椀rm invests $67,000
昀漀r special removal equipment to unload the
completed pencil boxes 昀爀om the moulding
machine, one machine operator can be elimi­
nated. 吀栀is would s愀瘀e $26,000 a year. 吀栀e re­
moval equipment has no salvage value and is
not expected to be used a昀琀er the two-year pro­
duction contract is completed. 吀栀e equipment,
although useless, would be serviceable 昀漀r about
15 years. What is the payback period? Should
Able Plastics buy the removal equipment?
A cannery is considering installing an auto­
matic case-sealing machine to replace current
hand methods. If they purchase the machine 昀漀r
$3,800 in June, at the beginning of the canning
season, they will save $400 a month 昀漀r the 昀漀ur
Bene昀椀ts ( $)
Costs ($)
1,400
500
300
400
300 in each year
8-29
A car dealer is leasing a small computer with
so昀琀ware 昀漀r $5,000 a year. As an alternative
she could buy the computer 昀漀r $7,000 and
lease the so昀琀ware 昀漀r $3,500 a year. Any time
she decided to switch to some other computer
system, she could cancel the so昀琀ware lease and
sell the computer 昀漀r $500. She decides to buy
the computer and lease the so昀琀ware.
(a) What is the payback period?
(b) If she kept the computer and so昀琀ware
昀漀r six years, what would the bene昀椀t-cost
ratio be if the interest rate is 10%?
8-30
Two alternatives with identical bene昀椀ts are
being considered:
⠀　Payback Period
8-26
A project has the 昀漀llowing costs and bene昀椀ts.
What is the payback period?
⠀　Initial cost
Uni昀漀rm annual cost
Use昀甀l life, in years
A
B
$500
$200
$800
$150
8
8
(a) Compute the payback period if Alterna­
tive B rather than Alternative A is bought.
(b) Use a MARR of 12% and bene昀椀t-cost
ratio analysis to identi昀礀 the alternative
that should be selected.
U nclassi昀椀ed
8-31
吀栀e provincial department of highways may
build a new highway between Edmonton
and Fort Elsewhere, currently a distance of
Cha pter 8 Benefit-Cost Ratio and Other Analysis Techniques
268
444 kilometres. Design 1 is a 昀漀ur-lane highw愀礀
built entirely on the existing route. Design 2 in­
cludes a signi昀椀cant re-routing that would reduce
the mileage to 407 kilometres. Design 3 is a 昀甀lly
access-controlled interprovincial-quality high­
w愀礀 with more re-routing that would reduce the
total mileage to 390 kilometres. 吀栀e bene昀椀ts 昀漀r
this project depend on mileage saved times the
number of vehicles, plus the estimated value 昀漀r
the larger number of trips that will occur with
the shorter and 昀愀ster routes. 吀栀e estimated
bene昀椀ts and costs of the three potential designs
are shown in the table. Doing nothing yields no
costs and no bene昀椀ts. Use incremental analysis
昀漀r the B/C ratio, a planning horizon of 75 years,
and a MARR of 9% to decide which of the de­
signs you would recommend.
Design 1
Design 2
Design 3
8-32
8-33
Initial Cost
($M)
Annual
Maintenance
Cost ($M)
Annual
Bene昀椀t
($M)
684
1,215
2,328
26
42
87
160
297
430
Chungyang Dam is being constructed across
the Hungshui River in southern China. 吀栀e
dam will produce electricity to serve over
500,000 people in the region. 吀栀e initial cost
is 3.7 billion yuan, and annual operating costs
are 39.2 million yuan. A m愀樀or overhaul of
the electricity generation 昀愀cilities estimated
to cost 650 million yuan will take place at the
end of Year 25. 吀栀e dam and generating plant
have no salvage value but must be torn down
and removed at end of Year 50 昀漀r a cost of
175 million yuan. Ishan Electric has a MARR
of 10%. What annual bene昀椀t in yuan is needed
昀漀r a B/C ratio of 1? Using the Internet and
current exchange rates, calculate the annual
bene昀椀t in Canadian dollars.
A light-rail system will connect the airport, the
city centre, and a cluster of high-density hous­
ing on the other side of the river. Name at least
three bene昀椀ts and three costs. What stake­
holder viewpoints will need to be considered?
8-34
A proposed bridge will cost $4 million to
build and $180,000 per year in maintenance.
吀栀e bridge should last 40 years. Bene昀椀ts to the
driving public are estimated to be $900,000 per
year. Damage (not paid) to adjacent property
owners due to noise is estimated to be $250,000
per year. Because of the uncertainty about
what interest rate should be used to evaluate
the project, calculate the break-even annual
interest rate that results in a B/C ratio of 1.
8-35 Research and report on how an agency of your
municipality evaluates public project propos­
als. What interest rate and what economic
measures are used?
8-36
A large project requires an i瘀팀stment of
$200 million. 吀栀e construction will take three
years: $30 million will be spent during the 昀椀rst
year, $100 million during the second year, and
$70 million during the third year of construc­
tion. 吀眀o project operation periods are being
considered: 10 years with the expected net pro昀椀t
of $40 million a year and 20 years with the
expected net pro椀� of $32.5 million a 礀攀ar. For
simplicity of calculations it is assumed that all cash
flows occur at the end of the year. 吀栀e compa渀礀's
minimum required return on investment is 10%.
Calculate the 昀漀llowing 昀漀r each alternative:
(a) 吀栀e payback periods
(b) 吀栀e total equivalent investment cost at the
end of the construction period
(c) 吀栀e equivalent uni昀漀rm annual worth of
the project (use the operation period of
each alternative)
Which operation should be chosen?
M i n i - Cases
Note: 吀栀e data 昀漀r the three problems that 昀漀llow were
contributed by William R. Truran, Stevens Institute
of Technology, and Peter A. Cerenzio, Cerenzio &
Panaro Consulting Engineers.
From Waste to Power and Money
吀栀e world's leading climate scientists h瘀였 con­
cluded that global warming is unequivocal and
Problems
that human activity is the main driver. One such
activity is dumping organic material into land­
昀椀lls, where it gradually decomposes, producing
a signi昀椀cant amount of land昀椀ll gas (LFG). About
50% of LFG is methane, which is a greenhouse
gas that contributes to global warming. Methane
is of particular concern because it traps heat in
the atmosphere with an e昀昀ect 20 times as great
as that of carbon dioxide, which is also a large
contributor to global climate change.
Land昀椀lls can also release volatile organic
compounds that create smog in the 昀漀rm of
ground-level ozone and cause respiratory
problems. Land昀椀lls also generate hazardous
air pollutants, such as benzene and toluene,
and odorous compounds, such as hydrogen
sulphide, that h愀瘀e a detrimental e昀昀ect on the
quality of life 昀漀r people who live nearby.
But humans must produce waste to live.
吀栀at's bad news all around, but using engi­
neering principles, bringing together various
constituencies, and evaluating alternatives
with engineering economy can bring bene昀椀ts
out of almost every one of these issues.
If extracted and used properly, land昀椀ll gas
can be an asset. It has a heating value of between
12 and 20 MJ/m3 , about half the heating value of
natural gas. As a result, more and more owners
of municipal solid waste land昀椀lls extract the gas
and use it to generate electricity. Such electricity
is used directly 昀漀r the power needs of the land­
昀椀ll or is sold to the general power grid.
吀栀e environment bene昀椀ts 昀爀om the gas ex­
traction and combustion because the release to
the environment of methane, volatile organics,
hazardous air pollutants, and odorous gases is
greatly reduced. 吀栀e generation of electrical
power 昀爀om a previously wasted source (i.e.,
methane 昀爀om land昀椀ll gas) reduces the use of
昀漀ssil 昀甀el and the associated emissions. 吀栀e
income 昀爀om selling power can help pay 昀漀r
land昀椀ll operating expenses.
Extracting and using a land昀椀ll gas to gen­
erate electricity is a "6-win" situation.
1. It bene昀椀ts the environment by reducing
unwanted gas emission.
2. It adds electrical power to the grid.
3. It produces cash 昀氀ow 昀漀r the land昀椀ll owner.
269
4. It lowers the use of non-renewable 昀漀ssil
昀甀els.
5. It reduces 昀椀nancial costs to the local
population.
6. It reduces the hazardous, noxious, and
odorous gases 昀漀r those nearby and
downwind.
Data 昀漀r a speci昀椀c application of the technol­
ogy 昀漀llow:
Daily and intermediate cover material is
20% of the land昀椀ll's usable space. Density of
solid waste is 900 kg/m3.
Recovery rates 昀漀r land昀椀ll gas:
• 250 m3/t 昀漀r municipal solid waste
• 125 m3/t 昀漀r construction and demolition
waste
Assume waste composition is 80% municipal
solid waste and 20% construction and demo­
lition waste.
Methane content in the land昀椀ll gas is 50%
昀漀r municipal solid waste and 20% 昀漀r con­
struction and demolition waste.
Heating value:
• Of methane: 37 MJ/m3
• Of 昀甀el oil: 0.39 MJ/L
Assume:
• Furnace e昀케ciency is 88% 昀漀r methane and
82% 昀漀r 昀甀el oil.
• Cost of 昀甀el oil is $0.66/L.
• Heating load 昀漀r residential dwelling is
29,300 kWh per year. Value = $0.12/kWh.
8-37
An economic analysis is needed 昀漀r a new
municipal solid waste land昀椀ll. 吀栀at includes
determining the potential economic bene昀椀t
昀爀om the gas that is generated when solid waste
decomposes. 吀栀e land昀椀ll is proposed at 5.6 ha
with a design capacity of 1,000,000 m3 of ca­
pacity. 吀栀e 昀椀nal capping system will require a
1 m layer over the 5.6 ha, and the waste flow
rate is 120,000 t/yr. Calculate the 昀漀llowing:
(a) 吀栀e life of the land昀椀ll
(b) 吀栀e average annual methane production
(assume all methane production has ceased
by 15 礀攀ars a昀琀er the land昀椀ll is closed)
(c) 吀栀e dollar value of annual methane con­
verted to electricity (ignore collection and
energy production costs)
270
Cha pter 8 Benefit-Cost Ratio and Other Analysis Techniques
8-38
A developer has proposed a 650-unit residen­
tial development beside the land昀椀ll. 吀栀e pro­
posal includes using the gas generated 昀爀om
the land昀椀ll to heat the homes (and mitigate
odours). To determine the economic feasibility
of the proposal, the value of the gas 昀漀r heat­
ing purposes must be determined. Determine
whether the quantity is su昀케cient 昀漀r heating
the development. Is that economically more at­
tractive than using 昀甀el oil? Is it operationally
feasible?
8-39
A 2 ha land昀椀ll must be evaluated 昀漀r economic
viability. As part of the bene昀椀t-cost analysis, the
cost of extracting and treating the land昀椀ll gas
must be determined. Using the 昀漀llowing data,
design a land昀椀ll gas-extraction system and es­
timate the cost to implement such a system:
• Land昀椀ll dimensions are 300 m by 60 m.
• Area of in昀氀uence of a land昀椀ll gas-ex­
traction well is within a 15 m radius.
• Cost of well construction is $3,000 each.
• Cost of wellhead (necessary 昀漀r each well)
is $2,500 each.
• Cost of high-density polyethylene (HDPE)
collection header piping is $115/m.
• Cost of condensate knockouts (located at
low points) is $5,000 each.
• Cost of blower and/or 昀氀are station is
$500,000.
Selection of a Minimum
Attractive Rate of Return
Who Owns the Diamonds, Gold, Oil, and Gas?
The rights to work on the surface of the land are called surface rights. The rights to explore for
the resources below the surface and extract them are called mineral rights. The royalty is the
mineral-rights owner's share of production or revenues. That owner can be a government or the
private holder of freehold mineral rights.
Until the early 1900s, surface rights and mineral rights came with the purchase of land.
Since then, however, mineral rights have belonged to the government and cannot be bought by
individuals or companies, but only leased. Consequently, the mineral rights on more than 90% of
Canada's land are owned either by the provincial governments or, in the case of the territories, by
the federal government, except in the case of Nunavut, where land claims have made about 10% of
the mineral rights of the area the property of the Inuit community. Devolution from the Dominion
government varied from jurisdiction to jurisdiction. In the case of Alberta, for example, the mineral
rights were transferred to the provincial government by the Natural Resources Transfer Act in 1930.
lmagi neGolf/iStock Photo
Who Owns the D i a m o n ds, G o ld, O i l, and Gas?
I n 2007 there was p u blic p ressu re in Albe rta fo r the government to ra ise the roya lty rates it
c h a rged the o i l a n d gas d evelo pers. As a res u lt the government esta blished a review panel cha i red
by the former C EO of a resou rce compa ny; mem bers consisted of two eco n o m i cs p rofessors,
a private secto r economist, an e n g i neer a n d fo rmer p resident of an o i l sa nds compa ny, a n d a n
accou nta nt who was C E O o f a g ro u p o f technology co m pa n ies. On 1 8 Septe m be r 2007, t h e review
pa nel pu blished its re po rt, entitled Our Fair Share, which a rg u ed that Alberta ns were not receivi n g
e n o u g h eco n o m i c re nt fo r t h e i r resou rces a n d recom mended that the roya lty rates be ra ised . I n
expla i n i n g its methods, the panel wrote :
The gove r n m e nt/resou rce owner s h a re target identified by the Pa nel as com petitive
a n d re prese nting a fa i r s h a re fo r Alberta ns a n d Ca n a d i a n s ("sha re" i n cludes Fed e ra l Cor­
porate I ncome Tax) i s based o n a co m p rehensive i nter-j u risd i ctional com petitiveness
a n a lysis. The Pa nel equa lly considered the pers pective of the energy co m p a n ies by
modelling the va rious i nvestment decision - ma ki n g criteria e m p loyed by i n d u stry. I n d u s ­
try decision criteria i nclude rate o f retu rn, n e t present va lue ( N PV), a n d p rofita b i lity ratio
a n a lysis. ( p . 22)
The genera l reco m m e ndation was to ra ise the roya lty rate to a leve l that was a bout in the
m i d d le of what othe r cou ntries c h a rge.
The res ponse was a sava ge atta c k by i n d u stry on the panel. O i l co m p a ny C EOs th reate ned to
ta ke t h e i r i nvestment do lla rs out of Alberta, a n d perso n a l atta cks were made on the com pete nce
and cred i b i lity of the mem bers of the pa nel. One oil i n d ustry C EO eve n paid his e m p loyees to stage
a demonstration on the leg islatu re g ro u n d s . The Ca n a d i a n Association of Petrole u m P rod ucers
fou n d the panel's report to be based u po n "flawed data," " i n correct costs, " a n d opti m istic assu m p ­
tions. The association stated fu rther that Ca nada's i nvestment cli mate was n ot ve ry attractive, a n d
it used t h e g ra p h i n Fig u re 9-1 t o s u p port t h e state ment.
50%
40%
3 9%
3 5%
3 0%
20%
10%
3 0%
26%
22% 22% 21 %
20㤀琀0 20㤀琀0 1 9%
1 8% 1 8%
l S % 14% 14%
1 2% 1 2% 1 2%
FIGURE 9-1 Globa l oi l/gas investment retu rns-昀椀ve-yea r retu rn on cumulative capital costs.
Sou rce : 2 0 0 7 G loba l U pstrea m Performa nce Revi ew, J . S . H e rold I nc . Reprod u ced in CAP P's
Tec h n i ca l Revi ew of the Albe rta Roya l Pa n e l Repo rt, Octo b e r 2 0 0 7, p . 21. With p e r m i ss i o n by
H I S H e rold .
273
274
Cha pter 9 Selection of a M i n i m u m Attractive Rate of Return
O n 25 Octo ber 2007, the Albe rta government a n nou nced a revision to its roya lty structu re ­
one that they expected to i n crease the p rovince's o i l reve n u e rece i pts to a bout $2 b i llion a yea r.
The rate was not as low as the o i l com pa n i es wa nted, but not as h i g h as the pa nel had recom ­
mended . H oweve r, there was a flaw i n the reg u lations that the Alberta g overnment i ntrod uced
in 2009 : a lthou g h oil reve n u es did g o up from $1.5 bi llion to $2 b i llion, and oil sands reve n u es
we nt u p from $1.9 b i llion to $3.9 b i llion, reve n u es from natu ra l gas dropped fro m $6.3 b i llion to
$1.1 b i llion, so the g overnment actua lly took i n a bout $13 . 5 bi llion less ove r the 2 0 0 9 - 2014 period
than they wou ld have done u nd e r the reg u lations p reviously in fo rce .
QUESTIONS TO CONSIDER
1.
T h e o i l co m p a n ies cla i med the government was cha n g i n g the ru les afte r the g a m e had sta rted . T h e govern ­
ment re plied that the c i rcu msta nces had changed a n d that a new rate system was needed . Do you t h i n k that
it is reaso n a b le to adj ust the roya lty reg i me, which is usua lly based on an a mo u n t per u n it, when the ma rket
price of the resou rce tri ples? What a bout when it fa lls?
2.
The panel was com posed of experts fro m i n d ustry a n d from the academy. Why then d i d the i n d u stry fi nd its
conclu sions so d i f昀椀 c u lt to acce pt?
3.
Go t o the W e b a n d 昀椀 n d the fo llow-u p pa nel report p u b lished i n 2016 : htt p : //www.e n e rgy.a lberta . ca/Org/
pdfs/ Roya ltyRepo rtJ a n 2 016. pdf. Sca n the Web fo r com peti ng a rg u me nts.
LEARNING OBJECTIVES
吀栀is chapter will help you
•
defi n e va rious sou rces of ca pita l a n d the costs of those funds to the fi rm
•
se lect a 昀椀 rm's MARR by u s i n g the opportu n ity cost method of a n a lyzi n g i nvestme nts
•
adj u st the fi rm's MARR to account fo r risk a n d u n certa i nty
•
use s p readsheets to d evelop c u m u lative i nvestments a n d the opportu n i ty cost of ca pita l
KEY TERMS
cost of capital
prime rate
treasu ry stock
uncertai nty
WACC
吀栀e preceding chapters have said very little about what interest rate or minimum attract­
ive rate of return is suitable 昀漀r use in a particular situation. A discussion of what interest
rate to use must begin by examining the sources of capital, 昀漀llowed by looking at the
prospective investment opportunities and risk. Only in this way can an interest rate or
minimum attractive rate of return (MARR) be chosen intelligently.
Sources of Capital
Sources of Capital
In broad terms there are three sources of capital available to a 昀椀rm: money generated 昀爀om
the 昀椀rm's operation, borrowed money, and money 昀爀om selling stock.
Money Generated from the Firm's Operations
A major source of capital investment money is retained pro昀椀ts 昀爀om the 昀椀rm's operations.
Overall, industrial 昀椀rms retain about half of their pro昀椀ts and pay out the rest to sharehold­
ers. In addition to pro昀椀t, 昀椀rms also generate money equal to the annual depreciation of their
capital assets. Even a 昀椀rm that earns zero pro昀椀t will still generate money 昀爀om operations
equal to its depreciation charges. (A 昀椀rm with a loss, of course, will h愀瘀e still fewer 昀甀nds.)
External Sou rces of Money
When a 昀椀rm requires money 昀漀r a few weeks or months, it usually borrows 昀爀om banks.
Longer-term unsecured loans (of, say, one to 昀漀ur years) may also be arranged through
banks. Although banks undoubtedly 昀椀nance a lot of capital expenditures, regular bank
loans cannot be considered a source of permanent 昀椀nancing.
Longer-term 昀椀nancing is done by selling bonds to banks, insurance 昀椀rms, pension
昀甀nds, and the public. A wide variety of bonds exist, but most are interest-only loans,
where interest is paid every six months or once a year and the principal is due at the bond's
maturity. Usually, interest rates are stated explicitly; Chapter 7 includes examples of how to
calculate the interest rates. Common maturities are 10 to 30 years, although some extend
to 100 years and a few even longer-in 1883, the Toronto, Grey and Bruce Railw愀礀 issued
bonds 昀漀r £719,000, due to be redeemed 1,000 years later, in 2883.
For a 昀椀rm to be able to raise money by selling bonds, the bond buyers need to be
convinced that the 昀椀rm will still be in business when the bond comes due. So this method
of raising money is not available to small start-up companies, since it is well-known that
most start-up companies 昀愀il within their 昀椀rst 昀椀ve years.
A 昀椀rm can also raise 昀甀nds by issuing new stock (shares of ownership in the 昀椀rm).
Many 昀椀rms have also bought back their own stock at some time in the past; that is called
treasury stock. Another way 昀椀rms can raise 昀甀nds is to sell the treasury stock.
One of the 昀椀nance questions every 昀椀rm must answer is how to maintain a proper bal­
ance between debt (loans and bonds) and equity (stock and retained earnings). 吀栀e debt has a
maturity date, and there are legal obligations to rep愀礀 it unless the 昀椀rm declares bankruptcy.
On the other hand, shareholders expect a higher rate of return to compensate them 昀漀r the
risks of ownership. 吀栀ose who are interested in the models used to calculate the cost of equity
capital are referred to 吀栀e Economic Analysis of Industrial Projec琀猀 (Eschenbach et al. 2012).
Choice of Sou rces of Funds
Choosing the source of 昀甀nds 昀漀r capital expenditures is a decision 昀漀r the 昀椀rm's top execu­
tives and may require approval of the board of directors. When internal operations gener­
ate adequate 昀甀nds 昀漀r the desired capital expenditures, external sources of money are not
likely to be used. But when the internal sources are inadequate, external sources must be
used, or the capital expenditures will have to be deferred or cancelled.
275
276
Cha pter 9 Selection of a Minimum Attractive Rate of Return
Cost of Fu nds
Cost of Borrowed Money
A 昀椀rst step in deciding on a minimum attractive rate of return might be to determine
the interest rate at which money can be borrowed. Longer-term loans or bonds may be
obtained 昀爀om banks, insurance companies, or the variety of places in which substantial
amounts of money accumulate (昀漀r example, the wealth of oil-producing nations).
A large, pro昀椀table corporation might be able to borrow money at the prime rate; that
is, the interest rate that banks charge their best and most sought-a昀琀er customers. All other
昀椀rms are charged an interest rate that is higher (by 0.5% to several percentage points). In
addition to the 昀椀rm's 昀椀nancial strength and ability to repay the loan, the interest rate will
depend on the duration and whether the debt has collateral or is unsecured.
Cost of Capital
Another relevant interest rate i s the cost o f capital, which i s also called the we最턀ted aver­
age cost of capital (圀䄀CC). 吀栀is is the rate 昀爀om all sources of 昀甀nds in the 昀椀rm's overall
capitalization. 吀栀e mechanics of computing the cost of capital are given in Example 9-1.
ⴀⴀ------·= = =
=
=------ⴀⴀ� =
琀欀ⴀⴀ----�------------�㨀䨀
EXAM PLE 9-1
For a particular 昀椀rm, the purchasers of common shares require an 1 1% rate of return, bonds are sold
at a 7% interest rate, and bank loans are available at 9%. Compute the cost of capital, or WACC, 昀漀r the
昀漀llowing capital structure:
Rate of Return
$ 20 million
20 million
60 million
$100 million
Bank loan
Bonds
Common shares and retained earnings
9%
7
11
SOLUTION
吀栀e weighted cost of capital weights the return on each source of capital by the 昀爀action of the total cap­
ital it represents. In this case 20% of the total capital is 昀爀om the bank loan, 20% of the capital is 昀爀om
bonds, and 60% of the capital is 昀爀om equity sources-that is, common stock and retained earnings.
WACCbe昀漀re-taxes = {0 .2){9%) + {0 .2) {7%) + {0 .6) { 1 1%)
= 1.8% + 1 .4% + 6.6% = 9.8%
Note that since that is an ave爀愀ge, the result must be between the lowest rate of return {7%) and the
highest { 1 1%). Since it is a we最턀ted average, the return with the largest weight {60%) has the most e昀昀ect
on the 昀椀nal average.
吀栀e a昀琀er-tax cost of capital is computed a昀琀er considering that interest payments on debt, such as
bank loans and mortgage bonds, are tax-deductible business expenses. 吀栀us,
A昀琀er-tax interest cost = (Be昀漀re-tax interest cost) x (1 - Tax rate)
Cost of Fu nds
277
If we assume that the 昀椀rm pays 40% income tax, the computations become
Bank loan
Bonds
A昀琀er-tax interest cost = 9%(1 - 0.40) = 5.4%
A昀琀er-tax interest cost = 7%(1 - 0.40) = 4.2%
Dividends paid on the ownership in the 昀椀rm (common shares + retained earnings) are not tax­
deductible. 吀栀e cost of capital can also be computed by dividing the total amount of interest by the total
amount of capital. Combining the three components, the a昀琀er-tax interest cost 昀漀r the $100 million of
capital is
$20 million (5.4%) + $20 million (4.2%) + $60 million (1 1%) = $8.52 million
_ $8.52 m픀픀on _
WACC 昀픀er-愀es - 8.52 0븀Ⰰ
• •
$100 m椀氀hon
In practical situations, the cost of capital is o昀琀en di昀케cult to compute. 吀栀e 昀氀uctuation
in the price of common shares, 昀漀r example, makes it di昀케cult to pick a cost, and because of
the fluctuating prospects of the 昀椀rm, it is even more di昀케cult to estimate the 昀甀ture bene昀椀ts
that purchasers of the shares might expect to receive. Given the 昀氀uctuating costs and pros­
pects of 昀甀ture bene昀椀ts, what rate of return do shareholders require?
Investment Opportunities
An industrial 昀椀rm can invest its money in many more places than are available to an indi­
vidual. A 昀椀rm has larger amounts of money, allowing it to make investments that are un­
available to individual investors, with their more limited investment 昀甀nds. 吀栀e Canadian
government, 昀漀r example, borrows money 昀漀r short terms of 90 or 180 days by issuing
certi昀椀cates called treasury bills that o昀琀en pay a higher interest rate than savings accounts.
吀栀e customary minimum purchase is $25,000.
More important, however, is the 昀愀ct that a 昀椀rm conducts a business, which itself o昀昀ers
many investment opportunities. In general, opportunities 昀漀r investing money within
the 昀椀rm tend to be superior to the investment opportunities outside the 昀椀rm. Consider
the available investment opportunities 昀漀r a particular 昀椀rm as outlined in Table 9-1. 吀栀e
cumulative investment required 昀漀r all projects at or above a given rate of return is given
in Figure 9-2.
Figure 9-2 illustrates that a 昀椀rm may have a broad range of investment opportunities
愀瘀ailable at varying rates of return and with varying lives and uncertainties. It may take
some study and searching to 昀椀nd the best investment projects available to a 昀椀rm. Usually,
the good projects will require more money than the 昀椀rm budgets 昀漀r capital investment
projects.
Opportu n ity Cost
We see that there are two independent aspects of investin最⸀ One a猀瀀ect is the source and
quantity of money available for capital investment projects. 吀栀e other is the昀椀rm's investment
opportunities.
278
Cha pter 9 Selection of a Minimum Attractive Rate of Return
Table 9-1 A Fi rm's Ava ila ble I nvestment Opportun ities
Project
Number
Project
Estimated Rate
of Return (%)
Cost {$000)
I渀瘀estment Related to Current Operations
1
2
3
4
New Operations
5
Manu昀愀cture parts that previously had been
purchased
Further processing of products previously sold
in semi-昀椀nished 昀漀rm
Further processing of other products
7
Investment in a di昀昀erent industry
Other investment in a di昀昀erent industry
Overseas investment
Purchase of treasury bills
11
12
45 _
40 -
35 -
l 30
㨀ꌀ
-
⸀⸀
-
■
25 -
-
20 -
--
5-
0
FIGURE 9 - 2
return.
35
250
25
300
300
400
Unlimited
20
10
15
8
18
-
椀稀 1 5 -
10 -
200
200
E砀琀ernal I渀瘀estments
9
40
28
Relocate production to new plants
10
100
100
New Production Facilities
8
30
45
38
1 50
50
50
New equipment to reduce labour costs
Other new equipment to reduce labour costs
Overhaul particular machine to reduce
material costs
New test equipment to reduce defective
products produced
Project Number
2 4 3
5
200 400
1
6
I
600
9
8
I
I
7
11
10
I
ⴀꘀ
12
I
I
I
800 1 ,000 1 ,200 1 ,400 1 ,600 1 ,800 2,000 2,200 2,400
Cumulative Investment ($ X 10 3 )
Cu m u lative i nvestment req u i red for a ll projects at or a bove a g iven rate of
吀栀ese two situations are usually out of balance, with investment opportunities ex­
ceeding the available money supply. 吀栀us some investment opportunities can be chosen
and many must be rejected. Obviously, we want to ensure that all the projects selected are
better than the best project rejected. To do this, we must know something about the rate of
Investment Opportunities
279
return on the best rejected project. 吀栀e best rejected project is the best opportunity 昀漀r­
gone, and this in turn is called the opportunity cost.
Opportunity cost = Cost of the best opportunity 昀漀rgone
= Rate of return on the best rejected project
If the opportunity cost 昀漀r some 昀甀ture period (like the next 12 months) could be
predicted, this rate of return could be one w愀礀 to judge whether to accept or reject any
proposed capital expenditure. Examples 9-2 and 9-3 illustrate this.
EXAM PLE 9-2
Consider the situation represented by Table 9-1 and Figure 9-2. For a capital expenditure budget of
$1.2 million ($1.2 x 106), what is the opportunity cost?
SO LUTI O N
From Figure 9-2 we see that the eight projects with a rate of return of 20% or more require a cumulative
investment of $1.2(x 106). We would take on these projects and reject the other 昀漀ur (Projects 7, 11, 10,
and 12) with rates of return of 18% or less. 吀栀e best rejected project is Project 7, and it has an 18% rate
of return. 吀栀us the opportunity cost is 18%.
EXAM PLE 9-3
Nine independent projects are being considered. Figure 9-3 is based on the 昀漀llowing data.
Project
Cost (thousands)
Uni昀漀rm Annual
Bene昀椀t (thousands)
1
2
3
4
5
6
7
8
9
$100
200
50
100
100
100
300
300
50
$23.85
39.85
34.72
20.00
20.00
18.00
94.64
47.40
7.00
Use昀甀l Li昀攀
(years)
Salvage Value
(thousands)
Computed Rate
of Return
10
10
2
6
10
10
4
10
10
$ 0
0
0
100
100
100
0
100
50
20%
15
25
20
20
18
10
12
14
If a capital budget of $650,000 is available, what is the opportunity cost of capital? With this model,
which projects should be selected?
SO LUTI O N
Looking at the nine projects, we see that some are expected to produce a larger rate of return than
others. It is natural that if we are to choose 昀爀om among them, we will pick those with a higher rate of
continued
280
Cha pter 9 Selection of a Minimum Attractive Rate of Return
return. When the projects are arrayed by rate of return, as in Figure 9-3, Project 2 is the last one 昀甀nded.
吀栀us the opportunity cost of capital is 14% 昀爀om Project 9, the highest ranked un昀甀nded project. Pro­
jects 3, 1, 4, 5, 6, and 2 are the best options.
-
30 똀
� 25
-
言言 20 -
-
-
3
0
FIGURE 9-3
50
1
4
1 50
5
250
6
2
9
8
350 450
650 700
1 ,000
Cumulative Cost of Projects (thousands of $)
7
1,300
Cumulative cost of projects versus rate of return.
Choosing a Minimum Attractive Rate of Return
䌀䐀
If we consider the three concepts related to the cost of money (the cost of borrowed money,
the cost of capital, and opportunity cost), which, if any, should be used as the minimum
attractive rate of return (MARR) in economic analyses?
We know that unless the bene昀椀ts of a project exceed its cost, it cannot add to the prof­
itability of the 昀椀rm. A lower boundary 昀漀r the minimum attractive rate of return must be
the cost of the money invested in the project. It would be unwise, 昀漀r example, to borrow
money at 8% and invest it in a project yielding a 6% rate of return.
Furthermore, we know that no 昀椀rm has an unlimited ability to borrow money.
Bankers-and others who evaluate the limits of a 昀椀rm's ability to borrow money-look
at both the pro昀椀tability of the 昀椀rm and the relationship between the components in the
昀椀rm's capital structure. 吀栀is means that continued borrowing of money will require that
additional shares be sold to maintain an acceptable ratio between owne爀猀hip (equity) and
debt. In other words, money borrowed 昀漀r a particular investment project is only a block of
money 昀爀om the overall capital structure of the 昀椀rm. 吀栀at suggests that the MARR should
not be less than the cost of capital. Finally, we know that the MARR should not be less than
the rate of return on the best opportunity 昀漀rgone. Stated simply,
吀栀e minimum att爀愀ctive rate of return should be equal to the la最휀st of the cost of
borrowed money, the cost of capital, and the opportunity cost.
Adj usti ng MARR to Acco u nt fo r Risk a n d U n certa i nty
Adjusting MARR to Account for Risk and Uncertainty
What actually happens in the 昀甀ture is o昀琀en di昀昀erent 昀爀om our estimates. When we are
昀漀rtunate enough to be able to assign probabilities to a set of possible 昀甀ture outcomes, we
call this a risk situation. We will see in Chapter 10 that techniques like expected value and
simulation may be used when the probabilities are known.
Uncertainty is the term used to describe the condition when the probabilities are not
known. 吀栀us, if the probabilities of 昀甀ture outcomes are known, we have risk, and if the
probabilities are unknown, we have uncertainty. With uncertainty, adjustments 昀漀r risk
are more subjective.
In projects accompanied by normal business risk and uncertainty, the MARR is
used without adjustment. For projects with greater than 瘀였rage risk or uncertainty, most
昀椀rms increase the MARR. As reported in Block (2005), the percentage of 昀椀rms using risk­
adjusted rates varies 昀爀om 66% in retail to 82% in health care. Some of the percentages 昀漀r
other industries are 70% 昀漀r manu昀愀cturing, 73% 昀漀r energy, and 78% 昀漀r technology 昀椀rms.
Table 9-2 shows an example of risk-a搀樀usted MARRs in manu昀愀cturing. Some 昀椀rms use the
same rates 昀漀r all divisions and groups. Other 昀椀rms vary the rates by division 昀漀r strategic rea­
sons. 吀栀ere are even cases when a project-speci昀椀c rate based on that project's 昀椀nancing may
be justi昀椀ed. For example, a 昀椀rm or joint venture m愀礀 be 昀漀unded to develop a speci昀椀c mine,
pipeline, or other resource project. However, risk-a搀樀usted rates do not alw愀礀s work well. It is
preferable to deal explicitly with the probabilities by using the techniques 昀爀om Chapter 10.
Ta ble 9-2 Exa mple of Risk-Adj usted I nterest
MARR Values in Manufacturing
Rate (%)
6
8
10
12
16
20
Applies to
Equipment replacement
New equipment
New product in normal market
New product in related market
New product in new market
New product in 昀漀reign market
Representative Values of MAR R Used in Industry
We argued that the minimum attractive rate of return should be established at the highest
of the cost of borrowed money, the cost of capital, or the opportunity cost.
吀栀e cost of borrowed money will vary 昀爀om enterprise to enterprise, with the lowest
rate being the prime interest rate. 吀栀e prime rate is set by the Bank of Canada and may
change several times in a year; it is widely reported in newspapers and business publica­
tions. As we pointed out, the interest rate 昀漀r 昀椀rms that do not qualify 昀漀r the prime inter­
est rate may be 0.5% to several percentage points higher.
吀栀e cost of the capital of a 昀椀rm is elusive. 吀栀ere is no widely accepted way to compute
it; we know that as a composite value 昀漀r the capital structure of the 昀椀rm, it is convention­
ally higher than the cost of borrowed money. 吀栀e cost of capital is also a 昀甀nction of the
market valuation of the shares (common shares, etc.) of the 昀椀rm, which may 昀氀uctuate
widely, depending on the 昀甀ture earnings prospects of the 昀椀rm. We cannot generalize on
representative costs of capital.
䌀䐀
281
282
Cha pter 9 Selection of a Minimum Attractive Rate of Return
Somewhat related to cost of capital is the computation of the return on total capital
(long-term debt, capital stock, and retained earnings) actually achieved by 昀椀rms. Fortune,
among other magazines, does an annual analysis of the rate of return on total capital.
吀栀e a昀琀er-tax rate of return on total capital 昀漀r individual 昀椀rms ranges 昀爀om 0% to about
40% and 愀瘀erages 8%. Business Week does a periodic survey of corporate per昀漀rmance; it
reports an a昀琀er-tax rate of return on common shares and retained earnings. We would
expect the values to be higher than the rate of return on total capital, and that is the case.
吀栀e a昀琀er-tax return on common shares and retained earnings ranges 昀爀om 0% to about
65% with an average of 14%.
Higher values 昀漀r the MARR are used by 昀椀rms such as high-technology start-ups that
are short of capital. 吀栀ey are also used in industries like petroleum and mining, where
volatile prices increase the risk of poor returns 昀漀r projects. Rates of 25% to 30% are rela­
tively common, and even higher rates are sometimes used. For companies with more­
normal levels of risk, rates of 12% to 15% are more typical.
Note that the values of MARR given earlier are approximations. But the values quoted
are opportunity costs rather than the cost of borrowed money or capital. Firms seldom
seek money to 昀甀nd projects whose expected rates of return are only slightly above the
cost of borrowed money or the cost of capital. While one could make a case that this leads
to good projects being rejected needlessly, one reason that 昀椀rms operate as they do is that
they can 昀漀cus limited resources of people, management, and time on a smaller number
of good projects.
It should be noted that the MARR used by enterprises is typically much higher than
can be obtained by individuals. (Where can you get a 30% a昀琀er-tax rate of return without
excessive risk?) 吀栀e reason appears to be that businesses have much more capital at their
disposal and hence are not obliged to compete with the thousands of individuals in any
region seeking a pro昀椀table investment 昀漀r a few thousand dollars. Instead, they have to
compete only with the relatively small number of entities able to invest $500,000 or more.
吀栀is diminished competition, combined with a higher risk, explains at least some of the
di昀昀erence.
Ca pita l Budgeting, or Selecting the Best Projects
吀栀e opportunity-cost-of-capital approach of ranking projects by their rate of return intro­
duces a new problem. Up to this point we h愀瘀e been analyzing mutually exclusive alterna­
tives, where only one can be chosen. Engineering design problems are this type of problem,
where younger engineers use engineering economy to choose the best alternative design.
At higher levels in the organization, engineering economy is applied to solve a di昀昀er­
ent problem. For example, suppose that 30 projects have passed initial screening and are
being proposed 昀漀r 昀甀nding. Every one of the 30 meets the MARR. 吀栀e 昀椀rm can a昀昀ord
to invest in only some of them. Which ones should be chosen? 吀栀is is called the c愀瀀ital
budgeting problem.
Examples 9-2 and 9-3 applied the opportunity-cost-of-capital approach to the capital
budgeting problem. Firms o昀琀en use this approach as a starting point to rank the projects
昀爀om best to worst. In some cases the ranking by rate of return is used to make the decision.
More o昀琀en, managers then meet and decide which projects will be 昀甀nded by obtaining
a consensus, or a decision by the highest-ranking manager, which will modify the rate­
of-return ranking. At this meeting, business units argue 昀漀r a larger share of the capital
budget, as do plants in the same business, groups at the same plants, and individuals
within the groups. Some considerations, such as strategy, necessity, and the availability
Representative Values of MAR R Used in Industry
283
and capability of particular resources and people, are di昀케cult to represent in the project's
numbers, which are the subject of economic analysis.
Anyone who has ever bought 昀椀recrackers probably used the practical ranking criterion
of "biggest bang 昀漀r the buck" in making a selection. A similar criterion is used by some
昀椀rms to rank independent projects:
Rank independent projects according to their value of net present worth divided by
the present worth of cost. 吀栀e appropriate interest 爀愀te is 䴀䄀RR (as a reasonable
estimate of the cut-o昀昀 爀愀te of return).
EXAM PLE 9-4
Rank the 昀漀llowing nine independent projects in their order of desirability, based on a 14.5% min­
imum attractive rate of return. (To 昀愀cilitate matters, the necessary computations are included in the
tabulation.)
Project
Cost
(thousands)
Uni昀漀rm
Annual
Bene昀椀t
(thousands)
Use昀甀l
Life
(years)
Salvage
Value
(thousands)
1
2
3
4
5
6
7
8
9
$100
200
50
100
100
100
300
300
50
$23.85
39.85
34.72
20.00
20.00
18.00
94.64
47.40
7.00
10
10
2
6
10
10
4
10
10
$ 0
0
0
100
100
100
0
100
50
SO LUTI O N
Ranked by NPW/PW of cost, the projects are listed as 昀漀llows:
Computed
Rate of
Return
Computed
NPW at
14.5%
(thousands)
Computed
NPW/PW
of Cost
20%
15
25
20
20
18
10
12
14
$22.01
3.87
6.81
21.10
28.14
17.91
-27.05
-31.69
-1.28
0.2201
0.0194
0.1362
0.2110
0.2814
0.1791
-0.0902
-0.1056
-0.0256
Project
NPW /PW of Cost
Rate of Return
5
1
4
6
3
2
9
7
8
0.2814
0.2201
0.2110
0.1791
0.1362
0.0194
-0.0256
-0.0902
-0.1056
20%
20
20
18
25
15
14
10
12
With a 14.5% MARR, Projects 1 to 6 are recommended 昀漀r 昀甀nding and 7 to 9 are not. However, they
are ranked in a di昀昀erent order by the present worth index and by the rate-of-return approaches. For
example, Project 3 has the highest ranking 昀漀r the rate of return and is 昀椀昀琀h by the present worth index.
284
Cha pter 9 Selection of a Minimum Attractive Rate of Return
Some consider the present worth index to be a better measure, but that is true only if PW
is applied at the correct interest rate. It is more common 昀漀r 昀椀rms to use rate-of-return
ranking.
If independent projects can be ranked in their order of desirability, the selection of
projects to be included in a capital budget is a simple task. One proceeds down the list of
ranked projects until the capital budget has been exhausted. 吀栀e only di昀케culty with this
scheme is that occasionally the capital budget is more than enough 昀漀r n projects but too
little 昀漀r n + I projects.
In Example 9-4, the capital budget is $550,000. 吀栀is is more than enough 昀漀r the top
昀椀ve projects (sum = $450,000) but not enough 昀漀r the top six projects (sum = $650,000).
When we have this situation it may not be possible to say with certainty that the best use
of a capital budget of $550,000 is to 昀甀nd the top 昀椀ve projects. 吀栀ere may be some other
set of projects that makes better use of the available $550,000. Although some trial-and­
error computations may indicate the proper set of projects, more-elaborate techniques are
needed to prove optimality.
As a practical matter, a capital budget probably has some 昀氀exibility. If in Example 9-4
the tentative capital budget is $550,000, a care昀甀l examination of Project 2 will dictate
whether to expand the capital budget to $650,000 (to be able to include Project 2) or to
drop back to $450,000 (and le愀瘀e Project 2 out of the capital budget). Or perhaps Project 2
can be started in this budget year and 昀椀nished next year.
SUMMARY
吀栀ere are three general sources of capital available to a 昀椀rm. 吀栀e most important one is
money generated 昀爀om the 昀椀rm's operation. 吀栀is has two components: there is the portion
of pro昀椀t that is retained in the business, and there are the 昀甀nds equal to the 昀椀rm's depreci­
ation charges that are available 昀漀r reinvestment.
吀栀e two other sources of capital are 昀爀om outside the operations:
Debt: Borrowed as loans 昀爀om banks, insurance companies, and so 昀漀rth.
Longer-term borrowing 昀爀om selling bonds.
Equity: Sale of equity securities such as common or preferred shares.
Retained pro昀椀ts and cash equal to depreciation charges are the primary sources of
investment capital 昀漀r most 昀椀rms, and the only sources 昀漀r many enterprises.
In selecting a MARR, three values are 昀爀equently considered:
1.
2.
3.
Cost of borrowed money.
Cost of capital. 吀栀is is the composite cost of the components of the 昀椀rm's overall
capitalization.
Opportunity cost. 吀栀is is the rate of return on the best investment project that is
rejected.
吀栀e MARR should be equal to the highest one of these three values.
Be昀漀re this chapter, we had assumed that all worthwhile projects are approved and im­
plemented. But industrial 昀椀rms, like individuals and governments, are usually 昀愀ced with
S u m m a ry
285
more good projects than can be 昀甀nded with the money available. 吀栀e task is to choose the
best projects and reject, or at least delay, the rest.
Capital may be rationed among competing investment opportunities by either rate-of­
return or present worth methods. In many practical situations the results may not always
be the same 昀漀r these two methods.
If projects are ranked by rate of return, a proper procedure is to go down the list until
the capital budget has been exhausted. 吀栀e rate of return at this point is the cut-o昀昀 rate of
return. 吀栀is procedure gives the best group of projects but does not necessarily have them
in the proper priority order.
It has been shown in earlier chapters that the usual business objective is to maximize
NPW, and that is not necessarily the same as maximizing rate of return. One suitable pro­
cedure is to use the ratio (NPW/PW of cost) to rank the projects, letting the MARR equal
the cut-o昀昀 rate of return (which is the opportunity cost of capital). 吀栀is present worth
ranking method will order the projects so that, 昀漀r a limited capital budget, NPW will be
maximized. 吀栀e MARR must equal the cut-o昀昀 rate of return 昀漀r the rate of return and
present worth methods to yield comparable results.
PROBLEMS
Cost of Funds
9-1
Examine the 昀椀nancial pages of your news­
paper (昀漀r example, the Globe and Mail or the
Financial Post), determine the current interest
rate on the debenture bonds of two di昀昀erent
industrial 昀椀rms, and explain why the interest
rates are di昀昀erent 昀漀r these di昀昀erent bonds.
9-2
A small engineering 昀椀rm has borrowed
$125,000 at 8%. 吀栀e partners have invested
another $75,000. If the partners require a 12%
rate of return, what is the 昀椀rm's cost of capital?
(a) Be昀漀re taxes
(b) A昀琀er taxes with a tax rate of 30%
9-3
An engineering 昀椀rm has borrowed $725,000
at 7%. 吀栀e shareholders have invested another
$600,000. 吀栀e 昀椀rm's retained earnings total
$1.2M. 吀栀e return on equity is estimated to be
11%. What is the 昀椀rm's cost of capital?
(a) Be昀漀re taxes
(b) A昀琀er taxes with a tax rate of 30%
9-4
A 昀椀rm's shareholders expect an 18% rate of
return, and there is $22M in common stock
and retained earnings. 吀栀e 昀椀rm has $9M in
loans at an average rate of 8%. 吀栀e 昀椀rm has
raised $14M by selling bonds at an average rate
of 4%. What is the 昀椀rm's cost of capital?
(a) Be昀漀re taxes
(b) A昀琀er taxes with a tax rate of 30%
9-5
A 昀椀rm has 40,000 shares whose current price
is $80.75. Shareholders expect a return of 15%.
吀栀e 昀椀rm has a two-year loan of $900,000 at
6.4%. It has issued 12,500 bonds with a 昀愀ce
value of $1,000, 15 years le昀琀 to maturity,
semi-annual compounding, a coupon interest
rate of 6%, and a current price of $1,090. Using
market values 昀漀r its debt and equity, calculate
the 昀椀rm's cost of capital
(a) Be昀漀re taxes
(b) A昀琀er taxes with a tax rate of 40%
9-6
A university wants to apply the concept of
the 圀䄀CC to set its interest rate 昀漀r analyz­
ing capital projects. It has an endowment of
$68 million, which is earning 6.3% interest.
It is paying 4.5% interest on $29 million in
bonds. It believes that $94 million in general
昀甀nds 昀爀om the taxpayers should be assigned
an interest rate of 13%. What is the university's
cost of capital? Note that only the interest on
the endowment is available to 昀甀nd capital
projects.
286
9-7
9-8
Cha pter 9 Selection of a Minimum Attractive Rate of Return
Assume you h瘀였 $2,000 愀瘀ailable 昀漀r invest­
ment 昀漀r a 昀椀ve-year period. You wish to invest
the money-not just spend it on things that are
昀甀n. 吀栀ere are obviously many alternatives 愀瘀ail­
able. You should be willing to assume a modest
amount of risk of loss of some or all of the money
if necessary, but not a great amount of risk (no
investments in poker games or at horse races).
How would you invest the money? What is your
minimum attractive rate of return? Explain.
吀栀ere are many venture capital syndicates that
consist of a few (say, eight or ten) wealthy people
who combine to make investments in small and
(hope昀甀lly) growing businesses. Usually, the in­
vestors hire a young investment manager (o昀琀en
an engineer with an MBA) to seek and analyze
investment opportunities 昀漀r the group. Would
you estimate that the MARR sought by such a
group is more or less than 12%? Explain.
9-11
9-12
A 昀愀ctory has a $100,000 capital budget. Determine which project(s) should be 昀甀nded
and the opportunity cost of capital. Use a
spreadsheet.
Project
A
B
C
D
9-10
First
Cost
Annual
Bene昀椀ts
Life
(years)
Salvage
Value
$50,000
50,000
50,000
50,000
$ 1 3,500
9,000
1 3,250
9,575
5
10
5
8
$5,000
0
1 ,000
6,000
Chips USA is considering the 昀漀llowing pro­
jects to improve its production process. Chips
has a short life, and so a three-year horizon
is used in evaluation. Which projects should
be done if the budget is $70,000? What is the
opportunity cost of capital? Use a spreadsheet.
Project
First Cost
Bene昀椀t
1
2
3
4
5
6
7
$20,000
30,000
10,000
5,000
25,000
1 5,000
40,000
$ 1 1 ,000
14,000
6,000
2,400
1 3,000
7,000
2 1 ,000
Project
First Cost
Annual
Bene昀椀t
1
2
3
4
5
6
7
$200,000
300,000
1 00,000
50,000
250,000
1 50,000
400,000
$ 50,000
70,000
40,000
12,500
75,000
32,000
125,000
Li昀攀
(years)
15
10
5
10
5
20
5
Which projects should be done if the budget is
$100,000? What is the opportunity cost of capital? Use a spreadsheet.
Li昀攀
Annual Salvage
Project (years) First Cost Bene昀椀t Value
Opportu n ity Cost of Capital
9-9
National Motors's Rock Creek plant is considering the 昀漀llowing projects to improve the
company's production process. Which projects should be done if the budget is $500,000?
What is the opportunity cost of capital? Use a
spreadsheet.
1
2
3
4
5
6
7
20
20
30
15
25
10
15
$20,000
20,000
20,000
20,000
20,000
20,000
20,000
$4,000
3,200
3,300
4,500
4,500
5,800
4,000
$ 20,000
1 0,000
-20,000
1 0,000
Risk-Adj usted MARR
9-13
Use the examples of risk-adjusted interest
rates 昀漀r manu昀愀cturing projects in Table 9-2.
Assume Project B in Problem 9-9 is a new prod­
uct in a new market. What is the interest rate
昀漀r evaluating this project? Should it be done?
9-14
Use the examples of risk-adjusted interest
rates 昀漀r manu昀愀cturing projects in Table 9-2.
Assume Project 1 in Problem 9-12 is a new
product in a 昀漀reign market. What is the inter­
est rate 昀漀r evaluating this project? Should it be
done?
Capital Budgeti ng
9-15
Each of the 昀漀llowing 10 independent projects
has a IO-year life and no salvage value.
Problems
Uni昀漀rm
Annual
Computed
Bene昀椀ts
Cost
Rate of
Return
Project (thousands) (thousands)
1
2
3
4
5
6
7
8
9
10
9-16
⠀　$ 5
15
10
30
5
20
5
20
5
10
$1.03
3.22
1.77
4.88
1.19
3.83
1.00
3.69
1.15
2.23
9-17
Al is planning his Christmas shopping 昀漀r
seven people. To quanti昀礀 how much his vari­
ous relatives would e渀樀oy receiving presents
昀爀om a list of possibilities, Al has assigned
appropriateness units (called "ohs") 昀漀r each
present 昀漀r each person (Table P9-17). A rating
of 昀椀ve ohs represents a present that the re­
cipient would really like. A rating of 昀漀ur ohs
indicates the recipient would like it 昀漀ur-昀椀昀琀hs
as much, three ohs three-昀椀昀琀hs as much, and
so 昀漀rth. A zero rating indicates an unsuitable
present that cannot be given to that person.
Everyone must get a present.
吀栀e objective is to maximize total ohs that
can be obtained with the selected budget.
(a) How much will it cost to buy the seven
presents that the people would like best
if there is ample money 昀漀r Christmas
shopping?
(b) If the Christmas shopping budget is set
at $112, which presents should be bought,
and what is their total appropriateness
rating in ohs?
(c) If the Christmas shopping budget must
be cut to $90, which presents should be
bought, and what is their total appropri­
ateness rating in ohs?
9 -18
A 昀椀nancier has a sta昀昀 of three people whose
job it is to examine possible business ven­
tures 昀漀r her. Periodically they present their
昀椀ndings to her. On a particular occasion,
they presented the 昀漀llowing investment
opportunities:
⠀　16%
17
12
10
20
14
15
13
19
18
吀栀e projects have been proposed by the sta昀昀 of
the Ace Card Company. 吀栀e MARR of Ace has
been 12% 昀漀r several years.
(a) If there is ample money available, what
projects should Ace approve?
(b) List all the acceptable projects in their
order of desirability.
If
only $55,000 is available, which projects
挀儀
should be approved?
Ten capital spending proposals have been
made to the budget committee as the members
prepare the annual budget 昀漀r their 昀椀rm. Each
independent project has a 昀椀ve-year life and no
salvage value.
Uni昀漀rm
Annual
Computed
Bene昀椀t
Initial Cost
Rate of
Project (thousands) (thousands)
Return
A
B
C
D
E
F
G
H
I
J
$10
15
5
20
15
30
25
10
5
10
$2.98
5.58
1.53
5.55
4.37
9.81
7.81
3.49
1.67
3.20
15%
25
16
12
14
19
17
22
20
18
(a) On the basis of a MARR of 14%, which
projects should be approved?
(b) Rank-order all the projects in order of
desirability.
挀儀 If only $85,000 is available, which projects
should be approved?
287
Project A: 吀栀is is a project 昀漀r the use
of the commercial land the 昀椀nancier
already owns. 吀栀ere are three mutually
exclusive alternatives:
• Al. Sell the land 昀漀r $500,000.
• A2. Lease the property 昀漀r a car-washing
business. An annual income, a昀琀er all costs
(property taxes, etc.) of $98,700 would
be received at the end of each year 昀漀r
20 years. It is believed that at the end of
the 20 years, the property could be sold 昀漀r
$750,000.
• A3. Construct an o昀케ce building on the
land. 吀栀e building will cost $4.5 million
to construct and will not produce any net
288
Cha pter 9 Selection of a Minimum Attractive Rate of Return
Ta ble P9-17 Data
"Oh" Rating of Gi昀琀 If Given to Various Family Members
Prospective Gi昀琀
1. $20 box of candy
2. $12 box of cigars
3. $16 necktie
4. $20 shirt or blouse
5. $24 sweater
6. $30 camera
7. $6 calendar
8. $168 magazine subscription
9. $18 book
10. $16 game
Father
Mother
Sister
Brother
Aunt
Uncle
Cousin
4
3
2
5
3
1
0
4
3
2
4
0
0
3
4
5
0
3
4
2
2
0
0
4
5
2
1
4
2
3
1
5
0
0
4
3
1
2
3
2
2
4
2
0
income 昀漀r the 昀椀rst two years. 吀栀e prob­
abilities of various levels of rental income,
a昀琀er all expenses, 昀漀r the next 18 years are
as 昀漀llows:
Annual Rental Income
Probability
$1,000,000
1,100,000
1,200,000
1,900,000
0.1
0.3
0.4
0.2
吀栀e property (building and land) probably can
be sold 昀漀r $3 million at the end of 20 years.
Project B: An insurance company is
seeking to borrow money 昀漀r 90 days
at 13.75% per annum, compounded
continuously.
Project C: A 昀椀nancier owns a manu昀愀ctur­
ing company. 吀栀e 昀椀rm wants additional
working capital to allow it to increase its
inventories of raw materials and 昀椀nished
products. An investment of $2 million
will allow the company to obtain sales
that in the past it had to 昀漀rgo. 吀栀e addi­
tional capital will increase company
pro昀椀ts by $500,000 a year. 吀栀e 昀椀nancier
can recover this additional investment
by ordering the company to reduce its
inventories and to return the $2 mil­
lion. For planning purposes, assume the
1
3
4
4
5
0
4
3
2
1
3
4
2
1
3
1
4
2
0
1
0
1
1
3
3
2
additional investment will be returned at
the end of 10 years.
Project D: 吀栀e owners of Sunrise maga­
zine are seeking a loan of $500,000 昀漀r
10 years at a 16% interest rate.
Project E: 吀栀e Galveston Bank has indi­
cated a willingness to accept a deposit
of any sum of money over $100,000, 昀漀r
any desired duration, at a 14.06% inter­
est rate, compounded monthly. It seems
likely that this interest rate will be avail­
able 昀爀om Galveston, or some other bank,
昀漀r the next several years.
Project F: A car rental company is seek­
ing a loan of $2 million to expand its
昀氀eet of cars. 吀栀e company o昀昀ers to repay
the loan by paying $1 million at the end
of Year 1 and $1,604,800 at the end of
Year 2 .
(a) I f there i s $ 4 million available 昀漀 r invest­
ment now (or $4.5 million if the Project
A land is sold), which projects should
be selected? What is the MARR in this
situation?
(b) If there is $9 million available 昀漀r invest­
ment now (or $9.5 million if the Project
A land is sold), which projects should be
selected?
Problems
9-19
吀栀e Raleigh Soap Company has been o昀昀ered
a 昀椀ve-year contract to manu昀愀cture and pack­
age a leading brand of soap 昀漀r Taker Bros. It
is understood that the contract will not be ex­
tended past the 昀椀ve years because Taker Bros
plans to build its own plant nearby. 吀栀e con­
tract calls 昀漀r 10,000 tonnes (1 tonne equals
1,000 kg) of soap a year. Raleigh normally
produces 12,000 tonnes of soap a year, so pro­
duction 昀漀r the 昀椀ve-year period would be in­
creased to 22,000 tonnes. Raleigh must decide
what changes, if any, to make to accommodate
the increased production. Five projects are
under consideration.
Project 1: Increase liquid storage capacity.
Raleigh has been 昀漀rced to buy caustic
soda in tank-truck quantities owing to
inadequate storage capacity. If another
liquid caustic soda tank is installed to
hold 1,000 cubic metres, the caustic soda
may be purchased in railway tank car
quantities at a more 昀愀vourable price. 吀栀e
result would be a saving of 0.1¢ per kilo­
gram of soap. 吀栀e tank, which would cost
$83,400, has no net salvage value.
Project 2: Acquire another sulphonation
unit. 吀栀e present capacity of the plant
is limited by the sulphonation unit. 吀栀e
additional 12,000 tonnes of soap cannot
be produced without an additional sul­
phonation unit. Another unit can be in­
stalled 昀漀r $320,000.
Project 3: Expand the packaging depart­
ment. With the new contract, the pack­
aging department must either work two
eight-hour shi昀琀s or have another pack­
aging line installed. If the two-shi昀琀 oper­
ation is used, a 20% wage premium must
be paid 昀漀r the second shi昀琀. 吀栀e premium
would amount to $35,000 a year. 吀栀e
second packaging line could be installed
昀漀r $150,000. It would have a $42,000 sal­
vage value at the end of 昀椀ve years.
Project 4: Build a new warehouse. 吀栀e
existing warehouse will be inadequate
289
昀漀r the greater production. It is estimated
that 400 square metres of additional
warehouse is needed. A new warehouse
can be built on a lot beside the existing
warehouse 昀漀r $225,000, including the
land. 吀栀e annual taxes, insurance, and
other ownership costs would be $5,000 a
year. It is believed the warehouse could be
sold at the end of 昀椀ve years 昀漀r $200,000.
Project 5: Lease a warehouse. An alterna­
tive to building an additional warehouse
would be to lease warehouse space. A
suitable warehouse one kilometre away
could be leased 昀漀r $15,000 a year. 吀栀e
$15,000 includes taxes, insurance, and so
昀漀rth. 吀栀e annual cost of moving materi­
als to this more remote warehouse would
be $34,000 a year.
吀栀e contract o昀昀ered by Taker Bros is a
昀愀vourable one, which Raleigh Soap plans to
accept. Raleigh management has set a 15%
be昀漀re-tax minimum attractive rate of return
as the criterion 昀漀r any of the projects. Which
projects should be undertaken?
9-20 Mike's microbrewery is considering produc­
tion of a new ale called Mike's Honey Harvest
Brew. To produce his new o昀昀ering, he is con­
sidering two independent projects. Each of the
projects has two mutually exclusive alterna­
tives, and each alternative has a use昀甀l life of
10 years and no salvage value. Mike's MARR
is 8%. In昀漀rmation about the projects and al­
ternatives is given in the 昀漀llowing table:
Project or Alternative
Project 1. Buy new
fermenting tanks
Alt. A: 20 m3 tank
Alt. B: 60 m3 tank
Project 2. Buy bottle 昀椀ller
and capper
Alt. A: 2,500-bottle/hour
machine
Alt. B: 5,000-bottle/hour
machine
Cost
Annual
Bene昀椀t
$ 5,000
10,000
$1,192
1,992
15,000
3,337
25,000
4,425
290
Cha pter 9 Selection of a Minimum Attractive Rate of Return
Use incremental rate of return analysis to
complete the 昀漀llowing worksheet.
by present worth methods? (Limit your
answer to integral values.)
Project/
Alternative
Annual
Cost, P Bene昀椀t, A
IA
lB-lA
2A
2B-2A
$ 5,000
5,000
15,000
10,000
Uni昀漀rm
Annual Computed
Project
Bene昀椀ts
Cost
Rate of
Proposal (thousands) (thousands) Return
$1,192
800
3,337
A/P,
i, 10
IRR
0.2385 20%
0.1601
Project 1
Alt. A
Alt. B
Alt. C
Project 2
Alt. A
Alt. B
Project 3
Alt. A
Alt. B
Project 4
Use this in昀漀rmation to determine
愀儀 which projects should be 昀甀nded if only
$15,000 is available
(b) the cut-o昀昀 rate of return if only $15,000
is available
(c) which projects should be 昀甀nded if
$25,000 is available
U nclassi昀椀ed
9-21
Project
First Cost
A
B
C
D
$15,000
20,000
30,000
25,000
40,000
50,000
35,000
60,000
75,000
$ 4,429
6,173
9,878
6,261
11,933
11,550
6,794
12,692
14,058
E
F
G
H
I
9-22
Li昀攀
(years)
4
4
4
5
5
5
8
8
8
At Red Deer Products, 昀漀ur project proposals
(three with mutually exclusive alternatives) are
being considered. All the alternatives have a
10-year use昀甀l life and no salvage value.
愀儀 Use rate of return methods to determine
which set of projects should be under­
taken if the capital budget is limited to
about $100,000.
(b) For a budget of $100,000, what interest
rate should be used in rationing capital
$4.61
9.96
2.39
13%
15
20
20
35
4.14
6.71
16
14
25
10
10
5.56
2.15
1.70
18
17
11
(c) Using the interest rate determined in
part (b), rank-order the eight di昀昀erent
investment opportunities by means of the
present worth method.
搀儀 For a budget of about $100,000 and the
ranking in part (c), which of the invest­
ment opportunities should be selected?
吀栀e WhatZit Company has decided to 昀甀nd six
of nine project proposals 昀漀r the coming budget
year. Determine the next capital budget 昀漀r
WhatZit. What is the MARR? Use a spreadsheet.
Annual
Bene昀椀ts
$25
50
10
9-23
A 昀椀rm's shareholders expect a 15% rate of
return, and there is $12M in common stock
and retained earnings. 吀栀e 昀椀rm has $SM in
loans at an average rate of 7%. 吀栀e 昀椀rm has
raised $SM by selling bonds at an average rate
of 6%. What is the 昀椀rm's cost of capital?
(a) Be昀漀re taxes
(b) A昀琀er taxes with a tax rate of 30%
9-24 A 昀椀rm has 60,000 shares whose current price
is $45.90. 吀栀ose shareholders expect a return
of 14%. 吀栀e 昀椀rm has a three-year loan of
$1,900,000 at 7.3%. It has issued 22,000 bonds
with a 昀愀ce value of $1,000, 20 years le昀琀 to ma­
turity, semi-annual compounding, a coupon
interest rate of 7%, and a current price of $925.
Using market values 昀漀r its debt and equity,
what is the 昀椀rm's cost of capital?
(a) Be昀漀re taxes
(b) A昀琀er taxes with a tax rate of 34%
Uncertainty in
Future Events
Technological Uncertainty and the Consequence
of Failure: Waste Disposal and Wild Game
The Swan Hills Waste Treatment Centre is an important N orth American centre for the treatment
and disposal of a number of particularly toxic and environmentally harmful substances. It is the
only licensed full-spectrum PC B treatment and disposal facility in Canada, the only incinera­
tion facility in Canada permitted to dispose of ozone-depleting substances (O DS), and the only
facility in N orth America licensed to treat and destroy certain dioxin- and furan-contaminated
materials.
But the path to the treatment centre's design, development, and construction was a rocky
one. In the 1970s, increased industrial activity in western Canada contributed to an increase
in the volume of hazardous waste being produced. After an exhaustive study, a detailed
regional assessment, an information campaign, and local referendums, the province of Alberta
Rya n Jackson/Edmonton Journal. Material reproduced with the express permission of: Edmonton Jou rnal, a d ivision of Postmedia N etwork I nc.
Techno logical Uncertainty and the Consequence of Fai lure: Waste Disposal and W i ld Game
commissioned the construction of the largest and most advanced incinerator in Canada, to be
built in Swan Hills northwest of Edmonton. A noteworthy element in this process was that it
faced the N IM BY (not in my back yard) issue squarely by inviting the communities that might be
affected to openly debate and decide on the trade-off of employment Uobs in the plant) versus
environmental risk.
The Swan Hills Special Waste Treatment Centre opened in 1987 with the capacity to pro­
cess more than 20,000 tonnes of waste per year. It could incinerate organic liquids and solids,
treat inorganic liquids and solids, and stabilize and decontaminate hazardous wastes and landfill­
contaminated bulk solids. It was the most comprehensive and integrated treatment facility in North
America. Unfortunately, the technology was not as clean or as reliable as its designers had thought
it would be.
In 1995, the plant was taken over by Bovar Inc., which increased the capacity of the inciner­
ator to 35,000 tonnes a year. Then, in October of 1996, there was a mechanical failure of a high­
temperature expansion joint in the flue gas duct of the furnace that treats electrical transformers.
The result of the failure was that toxic gases were released into the atmosphere. As a consequence,
the government of Alberta issued a public health warning to local First Nations communities not to
eat wild game from the area around the plant. The following is an excerpt from the Alberta Health
Public H ealth Advisory issued in 1996 :
Avoid eating wild game taken from a 30 km radius of the Swan Hills Treatment Centre.
This 30 km radius includes a safety factor of approximately 10 times that of the potential
range of game in the area. If wild game from the area has already been eaten, simply avoid
eating any more of the meat. Again no health risk has been identified and the consump­
tion of potentially contaminated meat would have to occur over a number of years before
it could lead to any adverse health effects.
The technological failure, combined with the warning from Alberta Health, caused some
First Nations members to become suspicious of their water, plants, medicines, and wild food,
and many curtailed their traditional land use practices. The impact of limiting their traditional
hunting and fishing included becoming more sedentary and more reliant on store-bought foods,
which are high in fats and sugar. Currently, these First Nations communities have high rates of
diabetes and various social problems. Although these complex social and health issues cannot
be attributed directly to the release of toxic gases, it is extremely difficult to rebuild confidence
in the safety of traditional foods. The full cost of the environmental impact, clean-up, and mon ­
itoring has not been assessed, but Alberta taxpayers have certainly lost millions of dollars. Not
only did the curtailment of traditional land use practices by First N ations have a harmful effect on
the health of the people, but the transmission of traditional knowledge about those activities has
been severed as well.
Now, years after the release, the risk is still being monitored by analyzing wild game for traces
of dioxins, furans, and PCBs.
Growth and change bring many new technologies, ranging from waste disposal and recyc­
ling to clean-coal technologies. Developing these kinds of technology in a way that makes them
effective and financially viable, especially in light of the need for sustainability and the issue of
climate change, is one of the most difficult challenges facing the world. An understanding of how
new technologies evolve and penetrate the marketplace is central to commercializing these tech­
nologies successfully. Even with a successful innovation pathway, significant resources can be
expended in moving a new technology from basic research through the research-and-innovation
chain to commercial acceptance.
293
294
Cha pte r 10 Uncertainty in Future Events
QU ESTI ONS TO CON S I D E R
1.
How can a company foresee and manage the risk of a future event such as the discovery of downwind
pollution or accidental spills?
2.
What are some ways in which a company might estimate the cost of such an event?
3.
Before the 1970s it was common (and legal) for manufacturers to dispose of chemicals and other poten­
tially hazardous materials in landfills. How might companies foresee and prepare for future laws that might
penalize activities that are legal today?
4.
One risk-mitigation practice has been to put potentially dangerous facilities, such as nuclear power plants,
incinerators, and chemical plants, in remote locations where few people would be affected by them. Is this
practice still viable?
LEARN I N G OBJ ECTIVES
吀栀is chapter will help you
•
use a range of estimated variables to evaluate a project
•
describe possible outcomes with probability distributions
•
combine probability distributions for individual variables into joint probability distributions
•
use expected values for economic decision making
•
use economic decision trees to describe and solve more complex problems
•
measure and consider risk when making economic decisions
•
understand how simulation can be used to evaluate economic decisions
KEY TE RMS
domi nated
economic simulation
e昀케cient
expected va lue (EV)
Monte Ca rlo method
most li kely esti mate
optim istic esti mate
pessimistic estimate
In practically every chapter of this book, there are cash 昀氀ow tables and diagrams that de­
scribe precisely the costs and bene昀椀ts 昀漀r 昀甀ture years. How do we know so precisely what
will happen many years in the 昀甀ture? In the case of this book, we, the authors, simply
make the numbers up. In real life, making reliable predictions is more di昀케cult, so we need
to take into account the resultant uncertainty.
吀栀e 昀椀rst thing we can do is to determine a break-even point 昀漀r each variable whose
value is uncertain; that is, the threshold value at which our best course of action will
change. 吀栀is is illustrated in Example 10-2. Secondly, we can supplement our estimate of
the most likely outcome with optimistic and pessimistic limits (Example 10-3). Finally, in
those cases where we have numerical estimates of the probabilities of various outcomes, we
can use the tools of probability theory to calculate the expected value and variance of the
pro昀椀tability of each alternative open to us.
Break- Even Analysis
295
Break- Even Analysis
Example 10-1 illustrates the sensitivity of a conclusion to uncertainty in one of the input
parameters.
EXAM PLE 10 - 1
吀眀o alternatives are being considered. 吀栀e best estimates 昀漀r the various consequences are as 昀漀llows:
Cost
Net annual bene昀椀t
Use昀甀l life, in years
End-of-use昀甀l-life salvage value
A
B
$1,000
$150
10
$100
$2,000
$250
10
$400
If interest is 3.5%, which alternative has higher net present worth (NPW)?
SOLUTION
Alternative A
Alternative B
NPW = -1,000 + 150(P/A, 3.5%, 10) + lO0(P/F, 3.5%, 10)
= -1,000 + 150(8.317) + 100(0.7089)
= -1,000 + 1,248 + 71
= $319
NPW = -2,000 + 250(P/A, 3.5%, 10) + 400(P/F, 3.5%, 10)
= -2,000 + 250(8.317) + 400(0.7089)
= -2,000 + 2,079 + 284
= $363
Alternative B, with its higher NPW, would be chosen.
Modi昀椀cation of Example 10 -1
Suppose that at the end of 10 years, the actual salvage value 昀漀r B were $300 instead of the $400 best
estimate. If all the other estimates were correct, is B still the preferred alternative?
SOLUTION
Revised B
NPW = -2,000 + 250(P/A, 3.5%, 10) + 300(P/F, 3.5%, 10)
= -2,000 + 250(8.317) + 300(0.7089)
= -2,000 + 2,079 + 213
= $292
A is now the preferred alternative.
296
Cha pter 10 U n certa i nty in Futu re Eve nts
Example 10-1 shows that the change in the salvage value of Alternative B results in a
change of preferred alternative. 吀栀us, a more thorough analysis of Example 10-1 would con­
sider (1) which values are uncertain, (2) whether the uncertainty is ±5% or -50% to +80%,
and (3) which uncertain values lead to di昀昀erent decisions. Step 3 of this process leads us to
break-even analysis. 吀栀is identi昀椀es the threshold value of each variable at which our deci­
sion would change, and is illustrated in Example 10-2.
EXAM PLE 10 - 2
Use the data 昀爀om Example 10-1 to compute the sensitivity of the decision to the Alternative B salvage
value by computing the break-even value.
For break-even between the alternatives,
NPWA = NPWB
319 = -2,000 + 250(PIA, 3.5%, 10) + Salvage value8 (P/F, 3.5%, 10)
= -2,000 + 250(8.317) + Salvage value/0.7089)
At the break-even point,
S愀氀vage v愀氀ue 8
319 + 2,000 - 2,079
0.7089
240
= $339
0.7089
When the Alternative B salvage value > $339, B is preferred; when < $339, A is preferred.
Break-even analysis helps by answering the question, How much variability can a
parameter have be昀漀re the decision will be a昀昀ected? 吀栀e preferred decision depends on
whether the salvage value is above or below the break-even value, and the economic dif­
ference between the alternatives is small when the salvage value is close to break-even. But
break-even analysis does not solve the problem of how to take the variability of parameters
into account in an economic analysis. 吀栀at will be considered next.
儀⤀
Optimistic and Pessimistic Estimates
When creating estimates, it is usually more realistic to describe parameters with a range
of possible values than a single value. A range could include an optimistic estimate, a
most likely estimate, and a pessimistic estimate. 吀栀en economic analysis can determine
whether the decision changes within the range of projected values.
EXAM PLE 10 -3
A 昀椀rm is considering an investment. 吀栀e most likely data values were 昀漀und during the feasibility study.
Analyzing past data 昀爀om similar projects shows that optimistic values 昀漀r the 昀椀rst cost and the annual
bene昀椀t are 5% better than most likely values. Pessimistic values are 15% worse. 吀栀e 昀椀rm's most experi­
enced project analyst has estimated the values 昀漀r the use昀甀l life and salvage value.
Optimistic and Pessimistic Estimates
Optimistic
Cost
Net annual bene昀椀t
Use昀甀l life, in years
Salvage value
$950
$2 10
12
$ 1 00
Most Likely
$ 1 ,000
$200
10
$0
297
Pessimistic
$ 1 , 1 50
$ 1 70
8
$0
Compute the rate of return 昀漀r each estimate. If a 10% be昀漀re-tax minimum attractive rate of return
is required, is the investment justi昀椀ed under all three estimates? If it is only justi昀椀ed under some esti­
mates, how can these results be used?
SO LUTI O N
Optimistic Estimate
PW = 0 = -$950 + 210(䤀鄀, IRRopt' 12) + lO0(PIF, IRRopt ' 12)
IRRopt = i (n, A, P, F) = i (12, 210, -950, 100)
= 19.8%
Most Likely Estimate
PW = 0 = -$1,000 + 200(䤀鄀, IRRmost likely' 10}
(䤀鄀, IRRmost l1"keI·"1 10) = 1,0001200 = 5 ➔ IRRmost l1"ke1y = 15.1%
or
Pessimistic Estimate
or
IRRmost J1· ke1y = i (n, A, P, F) = i ( l O, 200, -1,000, 0) = 15.1%
PW = 0 = -$1,150 + 170(䤀鄀, IRRpess , 8)
(PIA, IRRpess , 8) = 1,1501170 = 6.76 ➔ IRRpess = 3.9%
IRRpess = i (n, A, P, F) = i (8, 170, -1,150, 0) = 3.9%
From the calculations we conclude that the rate of return 昀漀r this investment is most likely to be
15.1% but might range 昀爀om 3.9% to 19.8%. 吀栀e investment meets the 10% MARR criterion 昀漀r two of
the estimates. 吀栀ese estimates can be considered to be scenarios of what may happen with this project.
Since one scenario suggests that the project is not attractive, we need to have a method of weighting the
scenarios or considering how likely each is.
Example 10-3 made separate calculations 昀漀r the sets of optimistic, most likely,
and pessimistic values. However, if there are more than a few uncertain variables, it
is unlikely that all will simultaneously take their most optimistic values ( best case), or
that all will simultaneously take their most pessimistic values (worst case). It is more
likely that many parameters are the most likely values, some are optimistic, and some
are pessimistic.
298
Cha pter 10 U n certa i nty in Futu re Eve nts
吀栀is can be addressed by using Equation 10-1 to calculate average or mean values 昀漀r
each parameter. Equation 10-1 puts 昀漀ur times as much weight on the most likely value as
on the other two. 吀栀is equation has a long history of use in project management to esti­
mate activity completion times. It is an approximation to the beta distribution.
Optimistic v愀氀ue + 4( Most l椀欀ely v愀氀ue) + Pessimistic v愀氀ue
Mean v愀氀ue = �----------�----------
(10-1)
吀栀is approach is illustrated in Example 10-4.
EXAM PLE 10 -4
Solve Example 10-3 by using Equation 10-1. Compute the resulting mean rate of return.
SO LUTI O N
Compute the mean 昀漀r each parameter:
Mean cost = [950 + 4(1,000) + 1,150]/6 = 1,016.7
Mean net annual bene昀椀t = [210 + 4(200) + 170]/6 = 196.7
Mean use昀甀l life = [12 + 4(10) + 8]/6 = 10.0
Mean salvage life = 100/6 = 16.7
Compute the mean rate of return:
PW of cost = PW of bene昀椀t
$1,016.7 = 196.7(P/A, IRRbeta> 10) + 16.7(P/F, IRRbeta> 10)
IRRb eta = i (n, A, P, F) = i (10, 196.7, -1,016.7, 16.7)
= 14. 3%
Example 10-3 gave a most likely rate of return (15.1%), which di昀昀ers 昀爀om the mean
rate of return (14.3%) computed in Example 10-4. 吀栀ese values are di昀昀erent because the
昀漀rmer is based exclusively on the most likely values and the latter takes into account the
variability of the parameters.
In examining the data, we see that the pessimistic values are 昀甀rther away 昀爀om
the most likely values than are the optimistic values. 吀栀is is a common occurrence. For
example, a savings of 10% to 20% m愀礀 be the maximum possible, but a cost overrun can
be 50%, 100%, or even more. 吀栀is causes the resulting weighted mean values to be less
昀愀vourable than the most likely values. As a result, the mean rate of return, in this example,
is less than the rate of return based on the most likely values.
Probability
What is the probability of getting a head when 昀氀ipping a coin? If the coin is 昀愀ir, both
the heads and tails come up with a probability of 50%. We could reach this conclusion
experimentally, by counting the number of times heads comes up over many trials, or by
Probabi lity
reasoning 昀爀om the observed symmetry of the coin. 吀栀ese two methods correspond to the
frequentist and intuitionist approaches to probability theory, respectively.
In most situations of economic interest, we do not have experimental data on which
to base 昀爀equentist estimates of probability. Instead, the probabilities we start 昀爀om will be
based on past data, expert judgment, or a combination of both. Past data on weather and
climate, on project completion times and costs, and on highway tra昀케c are combined with
expert judgment to 昀漀recast 昀甀ture events. Once we have these initial estimates, the stan­
dard tools of probability theory can be applied.
An example based on long-run relative 昀爀equencies is the PW of a 昀氀ood protection
dam that depends on the probabilities of di昀昀erent-sized 昀氀oods. 吀栀is might be based on
data 昀爀om past floods acquired over many years of observation. A challenge 昀愀ced by insur­
ance companies at the beginning of the twenty-昀椀rst century is that global climate change
may alter the likelihood of extreme weather events, so the 昀氀ood probabilities calculated
昀爀om historical records may no longer apply in the 昀甀ture.
All the data in an engineering economy problem may have some uncertainty. How­
ever, small uncertainties may be ignored so that more analysis can be done with the large
uncertainties. For example, the price of an o昀昀-the-shelf piece of equipment may vary by
only ±5%. 吀栀is price could there昀漀re be treated as a known or deterministic value. On the
other hand, demand over the next 20 years will have more uncertainty. Demand should
be analyzed as a random variable. We should establish probabilities 昀漀r di昀昀erent values
of demand.
吀栀ere are also logical or mathematical rules 昀漀r probabilities. If an outcome can never
happen, its probability is 0. If an outcome will certainly happen, its probability is 1. Prob­
abilities cannot be negative or greater than 1; they must lie within the interval [O, l], as
indicated below in Equation 10-2.
Probabilities are de昀椀ned so that the sum of probabilities 昀漀r all possible outcomes
is 1 (Equation 10-3). An exploration well drilled in a potential oil 昀椀eld will have three
possible outcomes (dry hole, non-commercial quantities, or commercial quantities) whose
probabilities will sum to 1.
Equations 10-2 and 10-3 can be used to check that probabilities are valid. If the prob­
abilities 昀漀r all but one outcome are known, the equations can be used to 昀椀nd the unknown
probability 昀漀r that outcome (see Example 10-5).
䤀㨀 P( outcomei ) = 1,
0 < Probability < 1
j=l to K
where there are K outcomes
(10-2)
(10-3)
In a probability course many probability distributions, such as the normal, uni昀漀rm,
and beta, are presented. 吀栀ese continuous distributions describe a large population of
data. However, 昀漀r engineering economy it is more common to use two to 昀椀ve outcomes
with discrete probabilities-even though the two to 昀椀ve outcomes only represent or
approximate the range of possibilities.
吀栀is is done 昀漀r two reasons. First, using seven to 10 outcomes would be 昀愀lse accur­
acy; we get the likelihood of each outcome 昀爀om our experts, but we can't really expect
our experts to provide that level of detail. Second, each outcome requires more analysis.
In most cases the two to 昀椀ve outcomes represent the best trade-o昀昀 between representing
the range of possibilities and the amount of calculation required. Example 10-5 illustrates
these calculations.
299
--- - - -300
Cha pter 10 U n certa i nty in Futu re Eve nts
EXAM PLE 10 -5
What are the probability distributions 昀漀r the annual bene昀椀t and life 昀漀r the 昀漀llowing project?
吀栀e most likely value of the annual bene昀椀t is $8,000 with a probability of 60%. 吀栀ere is a 30% prob­
ability that it will be $5,000, and the highest value that is likely is $10,000. A life of six years is twice as
likely as a life of nine years.
SO LUTI O N
Probabilities are given 昀漀r only two of the possible outcomes 昀漀r the annual bene昀椀t. 吀栀e third value is
昀漀und 昀爀om the 昀愀ct that the probabilities 昀漀r the three outcomes must sum to 1 (Equation 10-3).
1 = P(bene昀椀t is $5,000) + P(bene昀椀t is $8,000) + P(bene昀椀t is $10,000)
P(bene昀椀t is $10,000) = 1 - 0.6 - 0.3 = 0.1
吀栀e probability distribution can then be summarized in a table. Figure 10-1 shows the histogram, or
relative 昀爀equency diagram.
Annual Bene昀椀t
Probability
$5,000
0.3
$8,000
0.6
$ 10,000
0.1
0. 7
0.6
a攀嬀 0.4
0.5
e
o .3
0.2
0. 1
0 ���-�����������
8 ,000
1 0,000
0
5,000
Annual Bene昀椀t ($)
FIGURE 10 -1
Probability d istri bution for annual bene昀椀t.
For the probability distribution of project life, the problem statement tells us
P(life is 6 years) = 2P(life is 9 years)
If we assume that these are the only two possible lifespans, Equation 10-3 can be used to write a second
equation 昀漀r their likelihoods:
Combining these, we write
P(6) + P(9) = 1
2P(9) + P(9) = 1
P(9) = 1/3
P(6) = 2/3
吀栀e probability distribution 昀漀r the life is P(6) = 66.7% and P(9) = 33.3%.
J o i n t Pro b a b i lity D i stri butions
301
Joint Probability Distributions
Example 10-5 constructed probability distributions 昀漀r the annual bene昀椀t and life of a pro­
ject. We would like to construct a similar probability distribution 昀漀r the present worth of
the project. 吀栀at present worth depends on both input probability distributions, so we need
to construct the joint probability distribution 昀漀r the di昀昀erent combinations of their values.
We assume that two random variables, such as the annual bene昀椀t and life, are unrelated
or statistically independent. 吀栀at means that the joint probability of a combined event
(Event A de昀椀ned on the 昀椀rst variable and Event B on the second variable) is the product of
the probabilities 昀漀r the two events. 吀栀at is Equation 10-4:
If A and B are independent, then P(A and B) = P(A) X P(B)
(10-4)
For example, 昀氀ipping a coin and rolling a die are statistically independent. 吀栀us, the
probability of {昀氀ipping a head and rolling a 4} equals the probability of a {heads} = 1/2 times
the probability of a {4} = 1/6, 昀漀r a joint probability = 1/12.
吀栀e number of outcomes in the joint distribution is the product of the number of out­
comes in the distribution of each variable. 吀栀us, 昀漀r the coin and the die, there are 2 times
6 combinations. Each of the two outcomes 昀漀r the coin is combined with each of the six
outcomes 昀漀r the die.
Some variables are not statistically independent, and the calculation of their joint
probability distribution is more complex. For example, a project with low revenues may
be terminated early and one with high revenues may be kept operating as long as possible.
In these cases annual cash 昀氀ow and project life are not independent. Although this type
of relationship can sometimes be modelled with economic decision trees (covered later
in this chapter), we will limit our coverage in this text to the simpler case of independent
variables.
Example 10-6 uses the three values and probabilities 昀漀r the annual bene昀椀t and the
two values and probabilities 昀漀r the life to construct the six possible combinations. 吀栀en
the values and probabilities are constructed 昀漀r the PW of the project.
EXAM PLE 10 - 6
吀栀e project described in Example 10-5 has a 昀椀rst cost of $25,000. 吀栀e 昀椀rm uses an interest rate of 10%.
Assume that the probability distributions 昀漀r annual bene昀椀t and life are unrelated or statistically in­
dependent. Calculate the probability distribution 昀漀r the PW.
SO LUTI O N
Since there are three outcomes 昀漀 r the annual bene昀椀t and two outcomes 昀漀 r the life, there are six com­
binations. 吀栀e 昀椀rst 昀漀ur columns of the 昀漀llowing table show the six combinations of life and annual
bene昀椀t. 吀栀e probabilities in Columns 2 and 4 are multiplied to calculate the joint probabilities in
Column 5. For example, the probability of a low annual bene昀椀t and a short life is 0.3 x 2/3, which
equals 0.2, or 20%.
吀栀e PW values include the $25,000 昀椀rst cost and the results of each pair of annual bene昀椀t and life.
For example, the PW 昀漀r the combination of high bene昀椀t and long life is
PW$10,000,9 = -25,000 + 10,000(P/A, 10%, 9) = -25,000 + 10,000(5.759) = $32,590
continued
302
Cha pter 10 U n certa i nty in Futu re Eve nts
Annual Bene昀椀t
Probability
Life (years)
Probability
Joint Probabili琀礀
PW
$ 5,000
8,000
10,000
5,000
8,000
10,000
30%
60
10
30
60
10
6
6
6
9
9
9
66.7%
66.7
66.7
33.3
33.3
33.3
20.0%
40.0
6.7
10.0
20.0
3.3
100.0%
-$ 3,224
9,842
18,553
3,795
21,072
32,590
Figure 10-2 shows the probabilities 昀漀r the PW. 吀栀is is called the histogram, relative 昀爀equency
distribution, or probability distribution 昀甀nction.
40
30
o ���-��-��-������-����
- 3,224 0
FIGURE 10 -2
3,795
9,842
18,553 2 1,072
Present Worth ($)
32,590
Proba bi lity d istri bution fu nction for PW.
吀栀is probability distribution 昀甀nction shows that there is a 20% chance of having a negative PW. It
also shows that there is a small, 3.3%, chance of the PW being $32,590. 吀栀e three values used to describe
possible annual bene昀椀ts 昀漀r the project and the two values 昀漀r life have been combined to describe the
uncertainty in the PW of the project.
Creating a distribution, as in Example 10-6, gives us a much better understanding of
the possible PW values along with their probabilities. 吀栀e three possibilities 昀漀r the annual
bene昀椀t and the two 昀漀r the life are representative of the much broader set of possibilities
that really exist. Labelling a value as optimistic, most likely, or pessimistic is a good way to
represent the uncertainty about the variable.
Similarly, the six values 昀漀r the PW represent a much broader set of possibilities. 吀栀e 20%
probability of a negati瘀攀 PW is one measure of risk that we will talk about later in the chapter.
Some problems, such as Examples 10-3 and 10-4, have so many variables or di昀昀erent
outcomes that it is arithmetically burdensome to construct the joint probability distribution.
If the values in Equation 10-1 are treated as a discrete probability distribution 昀甀nction, the
probabilities are 1/6, 2/3, and 1/6. With an optimistic, most likely, and pessimistic outcome
昀漀r each of 昀漀ur variables, there are 34 = 81 combinations. In Examples 10-3 and 10-4, the
salvage value has only two distinct values, so there are still 3 x 3 x 3 x 2 = 54 combinations.
J o i n t Pro b a b i lity D i stri butions
303
When the problem is important enough, the e昀昀ort to construct the joint probability
distribution is worthwhile. It gives the analyst and the decision maker a better under­
standing of what may happen. It is also needed to calculate measures of the risk of a pro­
ject. While spreadsheets can automate the arithmetic, simulation (described at the end
of the chapter) can be a better choice when there are a large number of variables and
combinations.
䌀䐀
Expected Value
For any probability distribution we can compute the expected value (EV ), or weighted
arithmetic average (mean). To calculate the EV, each outcome is weighted by its probabil­
ity, and the results are summed.
吀栀e EV is a weighted average, like a student's grade point average (GPA). For a GPA,
the student's grade in each class is weighted by the number of credits. For the EV of a prob­
ability distribution, the weights are the probabilities.
吀栀is is described in Equation 10-5. We saw in Example 10-4 that these expected values
can be used to compute a rate of return. 吀栀ey can also be used to calculate a present worth,
as in Example 10-7.
Expected value = OutcomeA X P(A) + OutcomeB X P(B) + . . .
(10-5)
EXAM PLE 10 -7
吀栀e 昀椀rst cost of the project in Example 10-5 is $25,000. Use the expected values 昀漀r annual bene昀椀ts and
life to estimate the present worth. Use an interest rate of 10%.
SO LUTI O N
EVbene昀椀t = 5,000(0.3) + 8,000(0.6) + 10,000(0.1) = $7,300
EVlife = 6(2/3) + 9(1/3) = 7 years
Using these values, the PW is
PW(EV) = -25,000 + 7,300(P/A, 10%, 7) = -25,000 + 7,300(4.868) = $10,536
(Note: 吀栀is is not the expected value of the present worth, EV(PW); rather, it is the present worth of the
expected values, PW(EV). 吀栀is is an easy value to calculate and one that approximates the EV(PW),
which will be computed 昀爀om the joint probability distribution 昀漀und in Example 10-8.)
Example 10-7 is a simple way to approximate the project's expected PW. But the true
EV(PW ) is somewhat di昀昀erent. To 昀椀nd it, we must use the joint probability distribution
昀漀r bene昀椀t and life and the resulting probability distribution 昀甀nction 昀漀r PW that was
derived in Example 10-6. Example 10-8 shows the expected value of the PW, or the
EV(PW).
304
Cha pter 10 U n certai nty in Futu re Events
EXAM PLE 10 -8
Use the probability distribution 昀甀nction of the PW that was derived in Example 10-6 to calculate the
EV(PW). Does this indicate an attractive project?
SOLUTION
吀栀e table 昀爀om Example 10-6 can be reused with one additional column 昀漀r the weighted values of the
PW (= PW x probability). 吀栀en, the EV(PW ) is calculated by summing the column of present worth
values that have been weighted by their probabilities.
Annual
Bene昀椀t
$ 5,000
8,000
10,000
5,000
8,000
10,000
Probability
Life
(years)
Probability
30%
60
10
30
60
10
6
6
6
9
9
9
66.7%
66.7
66.7
33.3
33.3
33.3
Joint
Probability
PW
PW x Joint
Probability
20.0%
40.0
6.7
10.0
20.0
3.3
100.0%
-$ 3,224
9,842
18,553
3,795
21,072
32,590
EV(PW) =
-$ 645
3,937
1,237
380
4,214
1,086
$10,209
With an expected PW of $10,209, this is an attractive project. Although there is a 20% chance of a nega­
tive PW, the possible positive outcomes are larger and more likely.
吀栀e $10,209 value is more accurate than the approximate value calculated in Example 10-7. 吀栀e
values di昀昀er because PW is a non-linear 昀甀nction of the life. 吀栀e more-accurate value of $10,209 is lower
because the annual bene昀椀t values 昀漀r the longer life are discounted by 1/(1 + i ) 昀漀r more years.
--- - - -·
In Examples 10-7 and 10-8, the question was whether the project had a positive PW.
With two or more alternatives, our goal would be to maximize the expected PW. Obviously,
if we were to choose between alternatives on the basis of cost rather than worth, our goal
would be to minimize the expected cost. Example 10-9 uses the rule of minimizing the EV
of the E唀䄀C to choose the best height 昀漀r a dam.
EXAM PLE 10 -9
A dam is being considered to reduce river 昀氀ooding. But if a dam is built, how high should it be? Increas­
ing the dam's height will (1) reduce the probability of a 昀氀ood, (2) reduce the damage when 昀氀oods occur,
and (3) cost more. Which dam height minimizes the expected total annual cost? 吀栀e province uses an
interest rate of 5% 昀漀r 昀氀ood protection projects, and all the dams should last 50 years.
Dam
Height (m)
No da爀渀
7
10
13
First
Cost
Annual P(昀氀ood)
> Height
Damage If
Flood Occurs
$
0
700,000
800,000
900,000
0.25
0.05
0.01
0.002
$800,000
500,000
300,000
200,000
Expected Value
305
SO LUTI O N
吀栀e easiest way to solve this problem is to choose the dam height with the lowest equivalent uni昀漀rm
annual cost (E唀䄀C). Calculating the EUAC of the 昀椀rst cost requires us to multiply the 昀椀rst cost by
(AIP, 5%, SO). For example, 昀漀r a dam 7 metres high, this is 700,000(AIP, 5%, SO) = $38,344.
Calculating the annual expected cost of flood damage 昀漀r each alternative is simpli昀椀ed because
the term 昀漀r P(no flood) is zero, since the cost of the damage 昀漀r no flood is $0. 吀栀us we need to cal­
culate only the term 昀漀r flooding. 吀栀is is done by multiplying the P(昀氀ood) by the damage if a 昀氀ood
happens. For example, the expected annual cost of flood damage with no dam is 0.25 x $800,000,
or $200,000.
吀栀en the E唀䄀C of the 昀椀rst cost and the expected annual 昀氀ood damage are added together to 昀椀nd
the total EUAC 昀漀r each height. 吀栀e IO-metre dam is somewhat cheaper than the 13-metre dam.
Height of
Dam (m)
EUAC of Annual P(昀氀ood)
> Height
First Cost
No dam
7
10
13
$
0
38,344
43,821
49,299
0.25
0.05
0.01
0.002
X
X
X
X
Damage If Flood
Occurs
Expected Annual
Flood Damage
Total Expected
E唀䄀C
$800,000
500,000
300,000
200,000
$200,000
25,000
3,000
400
$200,000
63,344
46,821
49,699
Economic Decision Trees
Some engineering projects are more complex, and evaluating them properly is correspond­
ingly more complex. For example, consider a new product with potential sales volumes
ranging 昀爀om low to high. If the sales volume is low, the product may be discontinued
early in its potential life. On the other hand, if sales volume is high, additional capacity
may be added to the assembly line and new product variations may be added. 吀栀is can be
modelled with a decision tree.
吀栀e 昀漀llowing symbols are used to model decisions with decision trees:
D1
Decision node D- Dz Decision maker chooses one of the available paths.
Dx
C1
Chance node
X
Represents a probabilistic event. Each possible outcome (C C
o2c
2
�:
Py�
Cy
Outcome node 䌀眀
Pruned branch �
has a probability (P1 , P2 , • • • , P) associated with it.
1,
2 , • • • , Cy )
Shows result 昀漀r a particular path through the decision tree.
吀栀e double hatch mark indicates that a branch has been pruned because
another branch has been chosen. 吀栀is can happen only at decision nodes,
not at chance nodes. 吀栀e term "pruned" refers to the gardener's practice of
trimming, or pruning, branches to make a tree or bush healthier.
306
Cha pter 10 U n certa i nty in Futu re Eve nts
Figure 10-3 illustrates how decision nodes, chance nodes, and outcome nodes can be
used to describe the structure of the problem. Details such as the probabilities and costs
can be added on the branches that link the nodes. With the branches 昀爀om decision and
chance nodes, the model becomes a decision tree.
7: Net Revenue
= $0
4 : Net Revenue
Year 1 = $ 100K
8: Revenue
$ 100K/year
2: Volume 昀漀r
New Product
Medium
Volume
P = 0.6
5: Net Revenue
Year 1 = $Z OO K
i
Year 2 • • • n
9: Net Revenue
= $600K/year
�-----� � First Cost = $S OO K
1: Build New
Product
6: Net Revenue
Year 1 = $ 400K
10: Net Revenue
= $400K/year
t=0
t= I
t = 2, . . . , FIGURE 10-3 Economic decision tree for new prod uct.
Figure 10-3 illustrates the construction of a decision tree by starting at the decision
that must be made and then adding chance and decision nodes in the proper logical
order.
To make the decision once the tree has been constructed, calculations begin with the
昀椀nal (昀甀rthest to the right) nodes in the tree. Since they are the 昀椀nal nodes, enough in昀漀r­
mation is available to evaluate them. We then move le昀琀ward through the tree. At decision
nodes the rule is to maximize expected worth (or minimize expected cost). At chance
nodes an expected value 昀漀r worth or cost is calculated.
Once all nodes that branch out 昀爀om a node h愀瘀e been evaluated, the originating node
can be evaluated. If the originating node is a decision node, choose the branch with the
highest worth and put that value in the node. If the originating node is a chance node,
Eco n o m i c Decision Trees
307
calculate the expected value and put that value in the node. 吀栀is process "rolls back"
values 昀爀om the terminal nodes in the tree to the initial decision. Example 10-10 illustrates
this process.
EXAM PLE 10 -10
What decision should be made about the new product summarized in Figure 10-3? What is the expected
value of the product's PW ? 吀栀e 昀椀rm uses an interest rate of 10% to evaluate projects. If the product is
terminated a昀琀er one year, the capital equipment has a salvage value of $550,000 昀漀r use with other new
products. If the equipment is used 昀漀r eight years, the salvage value is $0.
SO LUTI O N
Decision trees are evaluated by starting with the end outcome nodes and the decisions that lead to
them. In this case the decisions are whether to terminate a昀琀er one year if the sales volume is low and
whether to expand a昀琀er one year if the sales volume is high.
吀栀e decision to terminate the product depends on which is more valuable, the $550,000 salvage
value of the equipment or the revenue of $100,000 per year 昀漀r seven more years. 吀栀e worth (PW 1 ) of
the revenue stream at the end of Year 1 shown in node 8 is
PW1 昀漀r node 8 = 100,000(P/A, 10%, 7)
= 100,000(4.868) = $486,800
吀栀us, terminating the product and using the equipment 昀漀r other products is better. We enter the two
"present worth" values at the end of Year 1 in nodes 7 and 8. We make the arc to node 7 bold to indicate
that it is our preferred choice at node 4. We use a double hatch mark to show that we are pruning the arc
to node 8 since it has been rejected as an inferior choice at node 4.
吀栀e decision to expand at node 6 could be based on whether the $800,000 昀椀rst cost 昀漀r expansion
can be justi昀椀ed by an annual increase in revenues of $200,000 昀漀r seven years. However, this is di昀케cult
to show on the tree. It is easier to calculate the present worth values at the end of Year 1 昀漀r each of the
two choices. 吀栀e worth (PW 1 ) of node 9 (e砀瀀and) is
PW1 昀漀r node 9 = -800,000 + 600,000(P/A, 10%, 7)
= -800,000 + 600,000(4.868)
= $2,120,800
吀栀e value of node 10 (continue without e砀瀀andi最需 is
PW 1 昀漀r node 10 = 400,000(P/A, 10%, 7)
= 400,000(4.868)
= $1,947,200
Since this is $173,600 less than the expansion node, the expansion should happen if volume is high.
Figure 10-4 summarizes what we know at this stage of the process.
continued
308
Cha pter 10 U n certa i nty in Futu re Eve nts
Ⰰ☀
./
�
�----� �
7: Net Revenue
= $0
PW 1 = $55 0, 000
4 : Net Revenue
Year 1 = $ 100K
8: Revenue
$ 100K/year
2: Volume 昀漀r
New Product
Medium
Volume
P = 0. 3
5: Net Revenue
Year 1 = $Z OO K
I
✀戀
Year 2 . . . n
9: Net Revenue
= $600K/year
PW1 = $2,120,8 00
--- _
∀䤀 � First Cost = $S OO K
t Reven_u_
e
6 :_N
_e_
嬀娀
Year 1 = $ 400K
1 : Build New
Product
10: Net Revenue
= $ 400K/year
t=0
FIGURE 10-4
t= I
t = 2, . . . , Partia lly solved decision tree for new prod uct.
吀栀e next step is to calculate the PW (Time 0) at nodes 4, 5, and 6.
PW at node 4 = (100,000 + 550,000)(P/F, 10%, 1) = 650,000(0.9091) = $590,915
PW at node 5 = (200,000)(P/A, 10%, 8) = 200,000(5.335) = $1,067,000
PW at node 6 = [400,000 - 800,000 + 600,000(P/A, 10%, 7)] (P/F, 10%, 1)
= [-400,000 + 600,000(4.868)] (0.9091) = $2,291,660
Now the expected value at node 2 can be calculated
EV at node 2 = 0.3(590,915) + 0.6(1,067,000) + 0.1(2,291,660) = $1,046,640
Since the cost of choosing node 2 is $1,000,000, the expected PW of proceeding with the product is
$46,640. 吀栀is is greater than the $0 昀漀r not building the project. So the decision is to build. Figure 10-5
is the decision tree at the 昀椀nal stage.
Eco n o m i c Decision Trees
�
./
0,9 1 5
PW = $5_9_
�__ ___� �$
4 : Net Revenue
Year 1 = $ 1 00K
309
7: Net Revenue
= $0
PW 1 = $550,000
8: Revenue
$ 1 00K/year
EV = $1,046,640
2: Volume 昀漀r
New Product
PW = $46,640
Medium
Volume
P = 0.6
PW = $ 1 ,067,000
5: Net Revenue
Year I = $200K
Year 2 • • • n
PW = $2,2 9 1 ,660
1 : Build New
Product
6: Net Revenue
Year 1 = $ 4 00K
10: Net Revenue
= $ 4 00K/year
t=0
t= I
t = 2, . . . , FIGURE 10-5 Solved decision tree for new prod uct.
Example 10-10 is representative of many problems in engineering economy. 吀栀e main
criterion is to maximize expected worth. However, as shown in Example 10-1 1, other cri­
teria, such as risk, are used in addition to expected worth.
EXAM PLE 10 -11
Consider the economic evaluation of collision and comprehensive insurance (against 昀椀re, the昀琀, etc.) 昀漀r
a car. 吀栀is insurance is usually required by lenders, but once the car has been paid 昀漀r, this insurance is
not required. (Liability insurance is a legal requirement.)
continued
310
Cha pter 10 U n certa i nty in Futu re Eve nts
Figure 10-6 begins with a decision node with two alternatives 昀漀r the next year. Insurance will cost
$800 a year with a $500 deductible if there is a loss. 吀栀e other option is to self-insure, which means to go
without buying collision and comprehensive insurance. 吀栀en if there is a loss, the owner must replace
the car with money 昀爀om savings or a loan, or do without a vehicle until he or she can a昀昀ord to replace it.
$0
$500
$0
$ 1 3,000
FIGURE 10 - 6 Decision tree for buyi ng auto collision insurance.
吀栀ree accident severities are used to represent the range of possibilities: a 90% chance of no acci­
dent, a 7% chance of a small accident (at a cost of $300, which is less than the deductible), and a 3%
chance of totalling the $13,000 car. Since our driving habits are likely to be the same with and without
insurance, the accident probabilities are the same 昀漀r both chance nodes.
Even though this is a text on engineering economy, we have simpli昀椀ed the problem and ignored
the di昀昀erence in timing of the cash 昀氀ows. Insurance payments are made at the beginning of the cov­
ered period, and accident costs occur during the covered period. Since car insurance is usually paid
semi-annually, the results of the economic analysis are not changed signi昀椀cantly by the simpli昀椀cation.
We 昀漀cus on the new concepts of expected value, economic decision trees, and risk.
What are the expected values 昀漀r each alternative, and what decision is recommended?
SO LUTI O N
吀栀e expected values are computed 昀爀om Equation 10-5. If the car is insured, the maximum cost equals
the deductible of $500. If the car is self-insured, the cost is the cost of the accident.
EVaccident w/ins . = (0 .9)(0) + (0 .07)(300) + (0 .03)(500) = $36
EVaccident w/o ins . = (0.9)(0) + (0 .07)(300) + (0 .03)(13,000) = $411
Eco n o m i c Decision Trees
311
吀栀us buying insurance lowers the expected cost of an accident by $375. To evaluate whether we
should buy insurance, we must also account 昀漀r the cost of the insurance. 吀栀ere昀漀re, these expected
costs are combined with the $0 昀漀r self-insuring (total $41 1) and the $800 昀漀r insuring (total $836).
吀栀us self-insuring has an EV cost that is $425 less per year (= $836 - $41 1). 吀栀at is not surprising,
since the premiums collected must cover both the cost of operating the insurance company and the EV
of the payouts.
吀栀is is also an example of expected values alone not determining the decision. Buying insurance has
an expected cost that is $425 a year higher, but the insurance limits the maximum loss to $500 rather
than $13,000. 吀栀e $425 may be worth spending to avoid that risk.
Risk
Risk can be thought of as the chance of getting an outcome other than the expected
value-with an emphasis on something negative. One common measure of risk is the
probability of a loss (see Example 10-6). 吀栀e other common measure is the standard
deviation (a), which measures the dispersion of outcomes about the expected value. For
example, many students have used the normal distribution in other classes. 吀栀e normal
distribution has 68% of its outcomes within ± 1 standard deviation of the mean and 95%
within ±2 standard deviations of the mean.
Mathematically, the standard deviation is de昀椀ned as the square root of the variance.
And variance is de昀椀ned as the weighted average of the squared di昀昀erence between the
outcomes of the random variable X and its mean. 吀栀us, the larger the di昀昀erence between
the mean and the values, the larger the variance and the standard deviation. 吀栀is is
Equation 10-6:
Standard de瘀椀ation (u) = Ⰰ䨀[EV(X - mean) 2 ]
(10-6)
Squaring the di昀昀erences between individual outcomes and the EV ensures that posi­
tive and negative deviations both receive positive weights. Consequently, negative values
昀漀r the standard deviation are impossible. 吀栀e standard deviation is O if only one outcome
is possible. Otherwise, the standard deviation is positive.
In practice it is easier to use Equation 10-7, which is equivalent to Equation 10-6:
Standard deviation (u) = Ⰰ䨀{EV(X 2 ) - [EV(X)] 2 }
= ✓{Outcome1 X P(A) + Outcome� X P(B) + • • • - e砀瀀ected v愀氀ue 2 }
(10-7)
(10-7')
吀栀is equation de昀椀nes standard deviation as the square root of the di昀昀erence between
the average of the squares and the square of the average. Standard deviation is used instead
of variance because the standard deviation is measured in the same units as the EV. 吀栀e
variance is measured in "squared dollars," a unit that is dimensionally correct but conveys
little intuitive meaning.
吀栀e standard deviation is use昀甀l when the standard deviation of each among several
alternatives is calculated and compared. But 昀椀rst, here are some examples of calculating
the standard deviation.
312
Cha pter 10 U n certa i nty in Futu re Eve nts
EXAM PLE 10 -12
Consider the economic evaluation of collision and comprehensive (昀椀re, the昀琀, etc.) insurance 昀漀r a car.
One example was described in Figure 10-6. 吀栀e probabilities and outcomes are summarized in the cal­
culation of the expected values, which was done with Equation 10-5.
= (0.9)(0) + (0.07)(300) + (0.03)(500) = $36
EVaccident /
w o ins. = (0.9)(0) + (0.07) (300 ) + (0.03)(13,0 00) = $ 411
EVaccident w/ins.
Calculate the standard deviations 昀漀r insuring and not insuring.
SO LUTI O N
吀栀 e 昀椀rst step i s to calculate the EV(outcome2) 昀漀 r each.
= (0.9)(02) + (0.07)(3002) + (0.03) (5002) = $13,800
2
2
2
2
EV accident /
w o ins. = (0.9)(0 ) + (0.07) (300 ) + (0.03)( 13,000 ) = $5,0 76,300
2
EV accident w/ins.
吀栀en the standard deviations can be calculated.
= ⸀䨀EV㬀Ⰰ/ins. - (EVw/ins. )
Uw/ins.
2
= ⸀樀(13,800 - 36 2 ) = Ⰰ䨀12,504 = $ 1 1 2
Uw/o ins. = ⸀樀(5,076,300 - 41 1 2 ) =
✓4,907,379 = $2,215
As described in Example 10-11, the expected cost of insuring is $836 (= $36 + $800) and the
expected cost of self-insuring is $41 1. 吀栀us the expected cost of not insuring is about half the cost of
insuring. But the standard deviation of self-insuring is 20 times as large. It is clearly riskier. As stated
be昀漀re, this is an example of the 昀愀ct that expected values alone do not determine the decision. Buying
insurance has an expected cost that is $425 per year higher, but the insurance limits the maximum loss
to $500 rather than $13,000. 吀栀e $425 may be worth spending to avoid that risk.
Which choice is preferred depends on how much risk one is com昀漀rtable with. In general, people
prefer to avoid risk. On the other hand, gambling resorts such as Las Vegas and Atlantic City exist only
because gamblers actively seek out opportunities to put their money at risk.
--- - - -·
EXAM PLE 10 -13
Using the probability distribution 昀漀r the PW 昀爀om Example 10-6, calculate the standard deviation of
the PW.
Risk
313
SO LUTI O N
吀栀e 昀漀llowing table adds a column 昀漀r (PW 2) (probability) to calculate the EV(PW 2).
Annual
Bene昀椀t
Probability
$ 5,000
8,000
10,000
5,000
8,000
10,000
30%
60
10
30
60
10
Life
(years) Probability
6
6
6
9
9
9
66.7%
66.7
66.7
33.3
33.3
33.3
Joint
Probability
20.0%
40.0
6.7
10.0
20.0
3.3
PW
- $ 3,224
9,842
18,553
3,795
21,072
32,590
EV
PW2 x
PW x
Probability Probability
- $ 645
3,937
1,237
380
4,214
1,086
$10,209
$ 2,079,480
38,747,954
22,950,061
1,442,100
88,797,408
35,392,740
$189,409,745
Standard deviation (u) = ⸀樀{EV(X 2 ) - [EV(X)] 2 }
u = ⸀樀{ l s9,40S,74S - [10,209] 2 } = ⸀䨀ss,1s2,064 = $9,229
Risk versus Return
A graph of risk versus return is one way to consider these items together. Figure 10-7 in
Example 10-14 illustrates the most common 昀漀rmat. Risk measured by standard deviation
is placed on the x axis, and return measured by expected value is placed on the y axis. 吀栀is
is usually done with internal rates of return of alternatives or projects.
EXAM PLE 10 - 14
A large 昀椀rm is discontinuing an older product, and thus some 昀愀cilities are becoming available 昀漀r other
uses. 吀栀e 昀漀llowing table summarizes eight new projects that would use the 昀愀cilities. Considering ex­
pected return and risk, which projects are good candidates? 吀栀e 昀椀rm believes it can earn 4% on a risk­
昀爀ee investment in government securities (labelled as Project F).
Project
IRR
Standard Deviation
1
2
3
4
5
6
7
8
F
13.1%
12.0
7.5
6.5
9.4
16.3
15.1
15.3
4.0
6.5%
3.9
1.5
3.5
8.0
10.0
7.0
9.4
0.0
continued
314
Cha pter 10 U n certa i nty i n Futu re Eve nts
SO LUTI O N
It is 昀愀r easier to answer the question if we use Figure 10-7. Since a larger expected return is better, we
want to choose projects that are as high up on the graph as possible. Since a lower risk is better, we want
to choose projects that are as 昀愀r le昀琀 as possible. 吀栀e graph lets us examine the trade-o昀昀 of accepting
more risk 昀漀r a higher return.
20
•8
6
•5
0
FIGURE 10-7
2
4
6
Standard Deviation of IRR (%)
8
Risk-versus- retu rn g raph.
First, we can eliminate Projects 4 and 5. 吀栀ey are dominated projects. Project 4 is dominated by
Project 3, which has a higher expected return and a lower risk. Project 5 is dominated by Projects 1, 2,
and 7. All three have a higher expected return and a lower risk.
Second, we look at the e昀케cient 昀爀ontier. 吀栀is is the line in Figure 10-7 that connects Projects F, 3, 2,
7, and 6. Depending on the trade-o昀昀 that we want to make between risk and return, any of these could
be the best choice.
Project 1 appears to be inferior to Projects 2 and 7. Project 8 appears to be inferior to Projects 7 and 6.
Projects 1 and 8 are inside and not on the e昀케cient 昀爀ontier.
吀栀ere are models of risk and return that can allow us to choose between Projects F, 3, 2, 7, and 6;
but those models are beyond what is covered here.
吀栀e 昀漀llowing is a simple rule of thumb 昀漀r comparing the risk and return of a
project:
If the expected present worth is at least double the standard deviation of the
present worth, the project is relatively safe.
For comparison, remember that in a normal distribution, about 2.5% of the values are less
than two standard deviations below the mean.
S i m u lation
315
Simulation
䌀䐀
Simulation is a more advanced approach to considering risk in engineering economy
problems. As the 昀漀llowing examples show, spreadsheet 昀甀nctions and add-in packages
make simulation easier to use 昀漀r economic analysis.
吀栀e 昀漀rm of economic simulation known as the Monte Carlo method uses random
sampling 昀爀om the probability distributions of one or more variables to analyze an eco­
nomic model 昀漀r many iterations. For each iteration, all variables with a probability distri­
bution are randomly sampled. 吀栀ese values are used to calculate the worth of the project.
吀栀en the results of all iterations are combined to create a probability distribution 昀漀r the
worth.
Simulation can be done by hand by means of a table of random numbers-if there are
only a few random variables and iterations. However, the results are more reliable as the
number of iterations increases, so in practice the calculation is usually computerized. 吀栀is
can be done in Excel, using the RAND() 昀甀nction to generate random numbers, as shown
in Example 10-15.
吀栀e probability distributions presented earlier in this chapter (and in some end­
of-chapter problems) used two or three discrete outcomes. 吀栀is limited the number of
combinations that we needed to consider. Simulation makes it easy to use continuous
probability distributions like the uni昀漀rm, normal, exponential, log normal, binom­
ial, and triangular. Examples 10-15 and 10-16 use the normal and the discrete uni昀漀rm
distributions.
EXAM PLE 10 - 15
ShipM4U is considering installing a new, more-accurate scale, which will reduce the error in calcu­
lating postage charges and save $250 a year. 吀栀e use昀甀l life of the scale is believed to be distributed
uni昀漀rmly over 12, 13, 14, 15, and 16 years. 吀栀e initial cost of the scale is estimated to be distributed
normally with a mean of $1,500 and a standard deviation of $150.
Use Excel to simulate 25 random samples of the problem, and compute the rate of return 昀漀r each
sample. Construct a graph of rate of return versus 昀爀equency of occurrence.
SO LUTI O N
吀栀is problem i s simple enough that a table with the values o f the life and the 昀椀rst cost o f each iteration
can be constructed. From these values and the annual s愀瘀ings of $250, the IRR 昀漀r each iteration can be
calculated with the RATE 昀甀nction. 吀栀ese are shown in Figure 10-8. 吀栀e IRR values are summarized in
a relative 昀爀equency diagram in Figure 10-9.
continued
316
Cha pter 10 U n certa i nty in Futu re Eve nts
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
Min
Max
Iteration
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
250
B
C
Annual Savings
Life
First Cost
12
16
1,500
150
Mean
Std dev
12
15
12
16
14
12
12
14
15
12
16
12
1,277
1,546
16.4%
13.9%
1, 193
1,728
1,500
1,415
1,610
19.2%
1 1 .7%
12.7%
16.0%
1 1 .2%
IRR
15
12
14
13
14
1,523
1,628
1,40 1
1,341
1,683
23
24
25
15
14
15
14
12
16
13
16
1,434
1,335
1,468
1,469
1,409
1,484
1,594
1,342
1,309
1,541
1,564
1,590
1,3 1 1
Mean
Std dev
14
2
1,468
135
18
19
20
21
22
FIGURE 10-8
D
12.4%
13.3%
15.5%
1 5.2%
10.2%
15.4%
15.4%
14.5%
13.9%
15.3%
14.7%
12.8%
16.8%
17.0%
12. 1 %
14.0%
12.2%
17.7%
14.4%
2.2%
Excel spreadsheet for simulation (N = 25) .
(Note: Each time Excel recalculates the spreadsheet, di昀昀erent values 昀漀r all the random numbers
are generated. 吀栀us the results depend on the set of random numbers, and your results will be di昀昀erent
if you create this spreadsheet.)
S i m u lation
317
6
5
4
最Ⰰ 3
2
1
0
9
FIGURE 10-9
10
11
12
13
14
IRR (%)
15
16
17
18
19
Graph of I RR va lues.
Stand-alone simulation programs and commercial spreadsheet add-in packages,
such as @Risk and Crystal Ball, have probability distribution 昀甀nctions to use 昀漀r each
input variable. In Example 10-16 the 昀甀nctions RiskUni昀漀rm and RiskNormal are used.
吀栀e packages also collect values 昀漀r the output variables, such as the IRR 昀漀r Example
10-16. From these distributions the simulation package can calculate the expected return,
P (loss), and the standard deviation of the return.
Example 10-16 uses @Risk to simulate 1,000 iterations of PW 昀漀r the data in Example 10-15.
A simulation package makes it easy to do more iterations. More important, since it is much
easier to use di昀昀erent probability distributions and parameters, more-accurate models can
be built. Because the models are easier to build, they are less likely to contain errors.
EXAM PLE 10 -16
Consider the scale described in Example 10-15. Generate 1,000 iterations and construct a 昀爀equency
distribution 昀漀r the scale's rate of return.
SO LUTI O N
吀栀e 昀椀rst IRR (cell AS) of 14.01% that is computed in Figure 10-10 is based on the average life and the
average 昀椀rst cost. 吀栀e second IRR (cell Al l) of 14.01% is computed by @Risk, using the average of each
distribution. 吀栀e cell content is the RAT E 昀漀rmula with its RiskUni昀漀rm and RiskNormal 昀甀nction;
however, spreadsheets with @Risk 昀甀nctions display by de昀愀ult the results of using average values.
吀栀e RAT E 昀甀nction contains two @Risk 昀甀nctions: Ris欀唀ni昀漀rm and RiskNormal. 吀栀e uni昀漀rm
distribution has the minimum and maximum values as parameters. 吀栀e normal distribution has the
average and standard deviation as parameters.
continued
318
Cha pter 10 U n certa i nty in Futu re Eve nts
I
1
A
- 1 ,500
2
3
150
12
minimum value of life
4
16
maximum value of life
5
250
annual bene昀椀t
8
9
E
F
standard deviation of 昀椀rst cost
I
6
7
D
C
B
average 昀椀rst cost
IRR computed using averages
14.01 % = RATE {(A3 + A4 )/2, A5, A l )
I
10
IRR 昀漀r each simulation iteration
11
14.01 %
= RATE {Risk Uniform(A3 , A4), AS , Risk Normal (A l , A2))
13
14. 1 5%
14
2. 1 7%
= average value of Al 1 (IRR) 昀爀om 1 ,000 iterations
= standard deviation of A l I (IRR) 昀爀om 1 ,000 iterations
I
12
I
15
16
17
16
18
14 尀尀
19
12 尀尀
20
21
22
23
24
25
26
27
-
l 10 尀尀
-
.€ 8 尀尀
]
尀尀
0
6 尀尀
2 尀尀
7
28
--
-
8.5
,n
10
-
-
�
--
-
4 尀尀
0
Distribution 昀漀r IRR (cell All)
-
-
-
1 1 .5
13
14.5 16
IRR (%)
1 7.5
,n,n,n,言됀,_ 19 20.5
-
FIGURE 10-10 Simulation spreadsheet for Exa m ples 10 -15 and 10 - 16.
吀栀e third IRR (cell A13) is the average 昀漀r 1,000 iterations. It will change each time the simulation is done.
吀栀e graph in Figure 10-10 with 1,000 iterations is much smoother than the graph 昀爀om Example 10-15,
where 25 iterations were done.
S U M MARY
For economic analysis, the 昀甀ture has to be estimated, and there are several ways to do
this. Precise estimates will not ordinarily be correct, but they are considered to be the best
single values to represent what we think will happen. We can tell which parameters are
most important to estimate correctly by 昀椀nding their break-even points.
S u m m a ry
319
A simple way to represent uncertainty is through a range of estimates 昀漀r each vari­
able, such as optimistic, most likely, and pessimistic. A wide range of prospective results
may be examined by using the optimistic values to solve the problem and then using the
pessimistic values. Solving the problem with the most likely values yields a good single
value estimate. Extremes with all optimistic values or all pessimistic values are less
likely-it is more probable that a mix of optimistic, most likely, and pessimistic values
will occur.
One approach taken 昀爀om project management uses the 昀漀llowing weighted values
instead of a range of estimates.
Estimate
Relative Weight
Optimistic
Most likely
Pessimistic
1
4
1
吀栀e most commonly used approach 昀漀r decision making relies on expected values. Here,
known or estimated probabilities 昀漀r 昀甀ture events are used as weights 昀漀r the correspond­
ing outcomes.
Expected value = Outcome A x ProbabilityA + OutcomeB X ProbabilityB + • • •
Expected value is the most use昀甀l and the most 昀爀equently used technique 昀漀r estimat­
ing the attractiveness of a project.
Risk, as measured by standard deviation, and the probability of a loss is also important
in evaluating projects. Since projects with higher expected returns also 昀爀equently have
higher risk, it is use昀甀l in decision making to evaluate the trade-o昀昀s between risk and
return. More complicated problems can be summarized and analyzed by using decision
trees, which allow problems to be logically evaluated with sequential chance, decision, and
outcome nodes.
Where the elements of an economic analysis are stated in terms of probability distri­
butions, a repetitive analysis of a random sample is o昀琀en done. 吀栀e Monte Carlo approach
relies on the 昀愀ct that a random sampling of increasing size becomes a better and better
estimate of the possible outcomes. 吀栀e large number of computations means that simula­
tion is usually computerized.
PROBLEMS
Ra nge of Estimates
0
10 -1
Telephone poles are an example of things that
have varying use昀甀l lives. Telephone poles,
once installed in a location, remain in use昀甀l
service until one of a variety of events occurs.
(a) Name three reasons that a telephone pole
might be removed 昀爀om use昀甀l service at
a particular location.
(b) You are to estimate the total use昀甀l life
of telephone poles. If the pole is removed
昀爀om an original location while it is still
serviceable, it will be installed elsewhere.
Estimate the optimistic life, most likely
life, and pessimistic life 昀漀r telephone poles.
What percentage of all telephone poles
would you expect to h愀瘀e a total use昀甀l life
greater than your estimated optimistic life?
320
10 - 2
10 -3
0
C h a p t e r 10 U n ce r t a i nty in F u tu r e E v e n t s
吀栀e p ur chase of a used pick- up tr uck for $9, 0 0 0
is being consider ed . Recor ds for other vehi­
cles sho w that costs for oil, tir es, and r epair s
r o ughly equal the cost for fuel . F uel costs ar e
$99 0 a year if the tr uck is dr iven 16, 0 0 0 kilo ­
i
metres . 吀栀e salvage value after f ve year s of use
d rops by about $ 0 . 05 per kilometre. Find the
equivalent uni昀漀rm annual cost if the interest
r ate is 8% . Ho w much does this change if the
annual m椀氀eage is 24, 0 0 0 km? 8 , 0 0 0 km?
A he at exchanger is being insta氀氀ed as p ar t
of a pl ant mo dern ization . It co sts $ 8 0 , 0 0 0 ,
in挀氀uding in stallation, and is exp ec te d to
reduce the o verall plant 昀甀el cost by $2 0 , 0 0 0
a ye ar. Es timate s of the use昀甀l life of the heat
exchanger r ange 昀爀om an op timis tic 1 2 years
to a pessimis tic 昀漀ur ye ars. 吀栀e mos t likely
value is five years. Assume the heat exchanger
has no salvage value at the end of its use昀甀l
氀椀昀攀 .
愀儀 Determine the pessimis t椀挀, most 氀椀kely,
and optimistic rates of return.
(戀⤀ Use the r ange of estimates to compute the
mean 氀椀fe and determine the estimated
be昀漀re -tax rate of return.
10 - 4
For the data in P爀漀blem ㄀　-2 assume that the
8, 000 -, 16, 000 -, and 24, 000 -k椀氀ometre values
are, respectively, pessimist椀挀, most 氀椀kely, and
optimistic estimates. Use a weighted estimate
to ca氀挀ulate the equivalent annual cost.
10 -5
A new two -lane r oad is needed in a part of
town that is g爀漀wing. At some point the road
will need to h愀瘀e 昀漀ur lanes to handle the
anti挀椀pated tra昀케c. If the 挀椀ty 's optimistic esti­
mate of g爀漀wth is used, the expansion will be
needed in 昀漀ur years . For the most like氀礀 and
pessimistic estimates, the expansion will be
needed in eight and 15 years, respectively. 吀栀e
expansion w氀픀 cost $4. 2 million. Use an inter­
est 爀愀te of 8% .
愀儀 What is the PW 昀漀r each scenario, and
what is the range of values?
(戀⤀ U sing Equation 10-1, calculate the mean
value of the expansion's P W.
Proba bi lities
10 - 6
W hen a pair of dice are tossed, the results m 愀礀
be a ny whole number from 2 to 1 2 . In the game
of craps one can win by tossing either a 7 or an
1 1 on the 昀椀rst 爀漀ll . What is the prob ab氀툀ty of
doing this? (Hint: There are 36 ways that a pair
of six- sided d椀挀e can be tossed . What portion
of them result in either a 7 or an 1 1 ?)
10 -7
Over the last 10 years, the hurdle , or discount,
rate for proj ects from the firm 's research and
de velopment division has been 10% tw椀挀e, 1 5%
three times, and 20% the rest of the time. There
is no recognizable pa琀琀ern. Ca氀挀ulate the prob­
ab氀툀ty dis tribution for next year 's discount
r ate.
10-8
Sales and profits for a new product are uncer­
琀愀in. The marketing department has pred椀挀ted
that sales might be as high as ㄀　, 000 units a
year with a probab氀툀ty of ㄀　%. The most 氀椀kely
value is 7, 000 units a year. The pessimistic
value is estimated to be 5, 000 units a year with
a p爀漀bab氀툀ty of 20%. Manufacturing and mar­
keting together h愀瘀e estimated the most 氀椀kely
unit p爀漀fit to be $32. The pessimistic value of
$24 has a p爀漀bab氀툀ty of 0.3, and the optimistic
value of $38 has a p爀漀bab氀툀ty of 0.2 . Construct
the p爀漀bab氀툀ty distributions for sales and unit
p爀漀fits.
10-9
A road between 夀攀氀氀owknife, NW T , and Uran­
ium City, Saskatchewan, wi ll have a most likely
construction cost of $4 m氀픀ion per km. Doub­
ling this cost is considered to h愀瘀e a p爀漀bab氀툀ty
of 30%, and cu琀琀ing it by 25% to h愀瘀e a p爀漀b­
ab椀氀ity of ㄀　%. The governme nt uses an interest
爀愀te of 8% for these types of p爀漀jects, and the
0
爀漀ad should last 40 years. What is the p爀漀babil­
ity distr ibution of the equivalent annual con­
struction cost per km?
321
Problems
10 -10 You recently had a car accident that was your
⠀　昀愀ult. If you have another accident or receive
another moving violation within the next
three years, you will become part of the "as­
signed risk" pool, and you will pay an extra
$600 a year 昀漀r insurance. If the probability of
an accident or moving violation is 20% a year,
what is the probability distribution of your
"extra" insurance payments over the next 昀漀ur
years? Assume that insurance is purchased an­
nually and that violations register at the end of
the year-just in time to a昀昀ect next year's in­
surance premium.
Shi昀琀s/ Savings/
D愀礀
Year Probability
1
2
3
probability distribution 昀漀r the annual pro昀椀t.
Assume that the sales and unit pro昀椀ts are sta­
tistically independent.
value. 吀栀e 昀椀rm uses an interest rate of 12% to
evaluate engineering projects. 吀栀e project has
an uncertain 昀椀rst cost and net revenue.
First Cost
p
Net Revenue
p
$300,000
400,000
600,000
0.2
0.5
0.3
$70,000
90,000
1 00,000
0.3
0.5
0.2
(a) What is the joint probability distribution
昀漀r 昀椀rst cost and net revenue?
(b) De昀椀ne optimistic, most likely, and pes­
simistic scenarios by using both optimis­
tic, both most likely, and both pessimistic
estimates. What is the present worth 昀漀r
each scenario?
10 -13 Modifying an assembly line has a 昀椀rst cost of
$80,000, and its salvage value is $0. 吀栀e 昀椀rm's
interest rate is 9%. 吀栀e savings shown in the
table depend on whether the assembly line
runs one, two, or three shi昀琀s, and on whether
the product is made 昀漀r three or 昀椀ve years.
3
5
0.6
0.4
Expected Value
10 -14 Annual savings due to an energy e昀케ciency
project have a most likely value of $30,000.
吀栀e high estimate of $40,000 has a probabil­
ity of 0.2, and the low estimate of $20,000 has
a probability of 0.3. What is the EV 昀漀r the
annual savings?
10 -12 A project has a life of 10 years and no salvage
⠀　0.3
0.5
0.2
(a) What is the joint probability distribution
昀漀r savings per year and use昀甀l life?
(b) De昀椀ne optimistic, most likely, and pes­
simistic scenarios by using both optimis­
tic, both most likely, and both pessimistic
estimates. Use a life of 昀漀ur years as the
most likely value. What is the present
worth 昀漀r each scenario?
Joint Proba bi lities
10 -11 For the data in Problem 10-8 construct the
$ 1 5,000
30,000
45,000
Use昀甀l
Life
(years) Probability
10-15 Two instructors announced that they "grade
⠀　on the curve"; that is, give a 昀椀xed percentage of
each of the various letter grades to each of their
classes. 吀栀eir curves are as 昀漀llows:
Grade
Instructor A
Instructor B
A
1 0%
15
45
15
15
15%
15
30
20
20
B
C
D
F
If a random student came to you and said that
his object was to enrol in the class in which he
could expect the higher grade point average,
which instructor would you recommend?
10 -16 For the data in Problem 10-7, compute the EV
昀漀r the next year's discount rate.
322
Cha pter 10 U n certa i nty in Futu re Eve nts
casino pays 35 times the amount bet if the
ball drops into the bettor's numbered slot in
the roulette wheel. In addition, the bettor re­
ceives back the original $5 bet. Estimate how
much money the woman is expected to win
or lose.
10-17 For the data in Problem 10-9 calculate the EV
of the equivalent annual construction cost per
km.
10-18 A university 昀漀otball team has 10 games sched­
uled 昀漀r next season. 吀栀e business manager
wishes to estimate how much money the team
can be expected to h愀瘀e le昀琀 over a昀琀er paying
the season's expenses, including any expenses
昀漀r a post-season exhibition game. From re­
cords 昀漀r the past season and estimates by
in昀漀rmed people, the business manager has as­
sembled the 昀漀llowing data:
⠀　Situation
Regular season
Win 3 games
Win 4 games
Win 5 games
Win 6 games
Win 7 games
Win 8 games
Win 9 games
Win 10 games
Post-season
exhibition
game
Probability
0.10
0.15
0.20
0.15
0.15
0.10
0.07
0.03
0.10
Situation
Regular season
Win 5 or
fewer
games
Win 6 to 8
games
Win 9 or 10
games
Post-season
exhibition
最愀me
Net Income
$250,000
400,000
600,000
Additional
income of
$100,000
What is the expected net income 昀漀r the team
next season?
10-19 In some casinos, craps is a popular gambling
game. One of the many bets available is the
"Hard-way 8." A $1 bet in this 昀愀shion will win
the player $4 if in the game the pair of dice
come up 4 and 4 be昀漀re one of the other ways
of totalling 8. For a $1 bet, what is the player's
expected return?
10 -20 A woman went to a casino with $500 and
placed 100 bets of $5 each, one a昀琀er another,
on the same number on the roulette wheel.
吀栀ere are 38 numbers on the wheel, and the
⠀　10-21 For the data in Problems 10-2 and 10-4, assume
that the optimistic probability is 20%, the most
likely is 50%, and the pessimistic is 30%.
(a) What is the expected value of the equiva­
lent uni昀漀rm annual cost?
(b) Compute the expected value 昀漀r the
number of kilometers and the corres­
ponding equivalent uni昀漀rm annual cost.
(c) Do the answers to (a) and (b) match? Why
or why not?
10-22 For the data in Problem 10-3, assume that the
optimistic probability is 15%, the most likely is
80%, and the pessimistic is 5%.
(a) What is the expected value of the rate of
return?
(b) Compute the expected value 昀漀r the life
and the corresponding rate of return.
(c) Do the answers to (a) and (b) match? Why
or why not?
⠀　10-23 From the data in Problem 10-8, calculate the
EV of sales and unit pro昀椀ts. From the data
in Problem 10-1 1 calculate the EV of annual
pro昀椀t. Are these results consistent?
10-24 Assume that the pessimistic and optimistic
estimates in Problem 10-5 have 40% and 20%
probabilities respectively.
愀儀 What is the expected PW of the expan­
sion costs?
(b) What is the expected number of years
until the expansion?
(c) What is the PW of the expansion cost
if you use the expected number of years
until the expansion?
Problems
(d) Do your answers to (a) and (c) match?
If not, why not?
10-25 An industrial park is being planned 昀漀r a tract
of land near the river. To prevent flood damage
to the industrial buildings that will be built on
this low-lying land, an earthen embankment
can be constructed. 吀栀e height of the embank­
ment will be determined by an economic an­
alysis of the costs and bene昀椀ts. 吀栀e 昀漀llowing
data have been gathered.
Embankment Height
above Roadw愀礀 (m)
Initial Cost
2.0
2.5
3.0
3.5
4.0
$100,000
165,000
300,000
400,000
550,000
Flood Level above Average Frequency that Flood
Roadw愀礀 (m)
Level Will Exceed Height in Col. 1
2.0
2.5
3.0
3.5
4.0
Once in 3 years
Once in 8 years
Once in 25 years
Once in 50 years
Once in 100 years
吀栀e embankment can be expected to last 50 years
and will require no maintenance. Whenever the
flood water 昀氀ows over the embankment, there
is $300,000 worth of damage. Should the em­
bankment be built? If so, to which of the 昀椀ve
heights above the roadway? A 12% rate of return
is required.
10-26 If your interest rate is 8%, what is the expected
value of the present worth of the "extra" insur­
ance payments in Problem 10-10?
⠀　10-27 Should the project in Problem 10-12 be under­
taken if the 昀椀rm uses expected value of present
worth to evaluate engineering projects?
323
(a) Compute the PW 昀漀r each combination
of 昀椀rst cost and revenue and the corres­
ponding expected worth.
(b) What are the expected 昀椀rst cost, expected
net revenue, and corresponding present
worth of the expected values?
(c) Do the answers 昀漀r (a) and (b) match?
Why or why not?
10-28 An energy e昀케ciency project has a 昀椀rst cost
of $300,000, a life of 10 years, and no salvage
value. Assume that the interest rate is 9%.
吀栀e most likely value 昀漀r annual savings is
$60,000. 吀栀e optimistic value 昀漀r annual sav­
ings is $80,000 with a probability of 0.2. 吀栀e
pessimistic value is $40,000 with a probability
of 0.3.
(a) What are the expected annual savings
and the expected PW ?
(b) Compute the PW 昀漀r the pessimistic,
most likely, and optimistic estimates of
the annual savings. What is the expected
PW ?
(c) Do the answers 昀漀r the expected PW
match? Why or why not?
10-29 For this question, use the data in Problem
10-13. Use a spreadsheet.
(a) What are the expected savings per year
and life, and the corresponding present
worth 昀漀r the expected values?
(b) Compute the present worth 昀漀r each
combination of savings per year and life.
What is the expected present worth?
(c) Do the answers 昀漀r (a) and (b) match?
Why or why not?
Decision Trees
10-30 吀栀e tree in Figure Pl0-30 has probabilities
a昀琀er each chance node and PW values 昀漀r each
terminal node. What decision should be made?
What is the EV?
324
Cha pter 10 U n certa i nty in Futu re Eve nts
0.4 --娀椀 12,000
0.6 --娀帀
0.4 --娀帀 D2
8,100
0.6 --娀帀 4,000
0.4 --娀帀 9,000
0.6 --娀帀
5,000
FIGURE P10·30
10-31 吀栀e tree in Figure Pl0-31 has probabilities
a昀琀er each chance node and PW values 昀漀r each
terminal node. What decision should be made?
What is the expected value?
0.2 --娀帀 15,000
12,000
0.3 --娀帀 9,000
0.4 --娀帀 16,000
0.6 --� 8,000
o .3 --娀帀1 - 10,000 1
0.7 --娀쌀 D2
0.2 --娀帀 10,000
o.s --� 7,000
0.9 --ⴀ㰀 8,000
0. l --娀帀 4,000
FIGURE P10·31
10-32 A decision has been made to make certain re­
pairs to the outlet works of a small dam. For
a particular 100 cm gate valve, there are three
available alternatives:
愀儀 Le愀瘀e the valve as it is.
(b) Repair the valve.
(c) Replace the valve.
If the valve is le昀琀 as it is, the probability,
over the life of the project, of a 昀愀ilure of the
valve seats is 60%, of a 昀愀ilure of the valve
stem is 50%, and of a 昀愀ilure of the valve body
is 40%.
If the valve is repaired, the probability of a
昀愀ilure of the seats over the life of the project is
Problems
40%, of a 昀愀ilure of the stem is 30%, and of a 昀愀il­
ure of the body is 20%. If the valve is replaced,
the probability, over the life of the project, of
a 昀愀ilure of the seats is 30%, of a 昀愀ilure of the
stem is 20%, and of a 昀愀ilure of the body is 10%.
吀栀e present worth of cost of 昀甀ture repairs
and service disruption of a 昀愀ilure of the seats
is $10,000, the present worth of cost of a 昀愀ilure
of the stem is $20,000, and the present worth of
cost of a 昀愀ilure of the body is $30,000. 吀栀e cost
of repairing the valve now is $10,000, and the
cost of replacing it is $20,000. If the criterion is
to minimize expected costs, which alternative
is best?
10-33 A 昀愀ctory building is located in an area subject
to occasional 昀氀ooding by a nearby river. You
have been brought in as a consultant to deter­
mine whether flood-proo昀椀ng of the building is
economically justi昀椀ed. 吀栀e alternatives are as
昀漀llows:
(1) Do nothing. Damage in a moderate 昀氀ood
is $10,000 and in a severe flood, $2S,000.
(2) Alter the 昀愀ctory building at a cost of
$1S,000 to withstand moderate flooding
without damage and to withstand severe
昀氀ooding with $10,000 worth of damage.
(3) Alter the 昀愀ctory building at a cost of
$20,000 to withstand a severe 昀氀ood with­
out damage.
In any year the probability of flooding is as 昀漀l­
lows: 0.7, no 昀氀ooding of the river; 0.2, moderate
昀氀ooding; and 0.1, severe 昀氀ooding. If interest is
1S% and a lS-year analysis period is used, what
do you recommend?
0
10-34 Five years ago a dam was constructed to im­
pound irrigation water and to provide 昀氀ood
protection 昀漀r the area below the dam. Last
winter a 100-year flood caused extensive
damage both to the dam and to the surround­
ing area. 吀栀is was not surprising, since the
dam was designed 昀漀r a SO-year 昀氀ood.
吀栀e cost of repairing the dam now will be
$2S0,000. Damage in the valley below amounts
to $7S0,000. If the spillw愀礀 is redesigned at a cost
of $2S0,000 and the dam is repaired 昀漀r another
$2S0,000, the dam m愀礀 be expected to with­
stand a 100-year flood without being damaged.
325
However, the storage capacity of the dam will
not be increased and the probability of damage
to the surrounding area below the dam will be
unchanged. A second dam can be constructed
up the river 昀爀om the existing dam 昀漀r $1 million.
吀栀e capacity of the second dam would be more
than adequate to provide the desired 昀氀ood pro­
tection. If the second dam is built, a redesign of
the existing dam spillw愀礀 will not be necessary,
but the $2S0,000 worth of repairs must be done.
吀栀e development in the area below the
dam is expected to be complete in 10 years.
A new 100-year 昀氀ood in the meantime will
cause a $1 million loss. A昀琀er 10 years the loss
would be $2 million. In addition, there would
be $2S0,000 worth of spillway damage if the
spillway is not redesigned. A SO-year flood is
also likely to cause about $200,000 worth of
damage, but the spillway would be adequate.
Similarly, a 2S-year flood would cause about
$S0,000 worth of damage.
吀栀ere are three alternatives: (1) repair the ex­
isting dam 昀漀r $2S0,000 but make no other alter­
ations, (2) repair the existing dam ($2S0,000)
and redesign the spillway to withstand a 100year flood ($2S0,000), and (3) repair the exist­
ing dam ($2S0,000) and build the second dam
($1 million). On the basis of an expected annual
cash flow analysis and a 7% interest rate, which
alternative should be selected? Dr愀眀 a decision
tree to describe the problem clearly.
10-35 In Problems 10-13 and 10-29, how much is it
worth to the 昀椀rm to be able to extend the life
of the product life by three years, at a cost of
$S0,000, at the end of the initial use昀甀l life of
the product? Use a spreadsheet.
Risk
10-36 For the data in Problems 10-7 and 10-16 com­
pute the standard deviation of the interest rate.
10-37 For the data in Problems 10-9 and 10-17 com­
pute the standard deviation of the equivalent
annual cost per km.
10-38 For the data in Problem 10-28 compute the
standard deviation of the present worth.
326
Cha pter 10 U n certa i nty in Futu re Eve nts
10 -39 吀栀e Graham Telephone Company may invest
in new switching equipment. 吀栀ere are three
possible outcomes, having net present worth
of $6,570, $8,590, and $9,730 respectively.
吀栀e probability of those outcomes is 0.3,
0.5, and 0.2 respectively. Calculate the ex­
pected return and risk associated with this
proposal.
10-45 A 昀椀rm is choosing a new product. 吀栀e 昀漀llow­
ing table summarizes six new possible prod­
ucts. Considering expected return and risk,
which products are good candidates? 吀栀e 昀椀rm
believes it can earn 4% on a risk-昀爀ee invest­
ment in government securities (labelled as
Product F).
10-40 A new machine will cost $25,000. 吀栀e machine
is expected to last 昀漀ur years and have no salvage value. If the interest rate is 12%, determine the return and the risk associated with
the purchase.
p
Annual Sa瘀椀ngs
0.3
$7,000
0.4
$8,500
Product
IRR
Standard Deviation
1
2
3
4
5
6
F
10.4%
9.8
6.0
12. 1
12.2
13.8
4.0
3.2%
2.3
1.6
3.6
8.0
6.5
0.0
0.3
$9,500
10-41 Wh愀琀 is your risk associ愀琀ed with Problem 10-26?
10-42 Measure the risk 昀漀r Problems 10-12 and 10-27
using the P (loss), range of PW values, and
standard deviation of the PWs.
10-43 (a) In Problems 10-13 and 10-29, using the P
(loss) and standard deviation of the PWs,
describe the risk.
(b) How much do the answers change if the
possible life extension in Problem 10-34 is
allowed?
Risk versus Retu rn
10-44 A 昀椀rm wants to select one new research and
development project. 吀栀e 昀漀llowing table sum­
marizes six possibilities. Considering expected
return and risk, which projects are good can­
didates? 吀栀e 昀椀rm believes it can earn 5% on a
risk-昀爀ee investment in government securities
(labelled as Project F).
儀⤀
Project
IRR
Standard Deviation
1
2
3
4
5
6
F
15.8%
12.0
10.4
12.1
14.2
18.5
5.0
6.5%
4.1
6.3
5.1
8.0
10.0
0.0
Simulation
10-46 吀栀e 昀椀rst cost of a project is $25,000, and it has
no salvage value. 吀栀e interest rate 昀漀r evalua­
tion is 7%. 吀栀e project's life is 昀爀om a discrete
uni昀漀rm distribution that takes on the values
7, 8, 9, and 10. 吀栀e annual bene昀椀t is distributed
normally with a mean of $4,400 and a stan­
dard deviation of $1,000. Using Excel's RAND
昀甀nction, simulate 25 iterations. What are the
expected value and standard deviation of the
present worth?
儀⤀
10-47 A 昀愀ctory's power bill is $55,000 a year. 吀栀e
昀椀rst cost of a small geothermal power plant is
distributed normally with a mean of $150,000
and a standard deviation of $50,000. 吀栀e power
plant has no salvage value. 吀栀e interest rate 昀漀r
evaluation is 8%. 吀栀e life of the project is 昀爀om
a discrete uni昀漀rm distribution that takes on
the values 3, 4, 5, 6, and 7. (吀栀e life is relatively
short because of corrosion.) 吀栀e annual oper­
ating cost is expected to be about $10,000 a
year. Using Excel's RAND 昀甀nction, simulate
25 iterations. What are the expected value and
standard deviation of the present worth?
U nclassi昀椀ed
10-48 A new engineer is evaluating whether to use
a larger-diameter pipe 昀漀r a water line. It will
cost $350,000 more initially, but it will reduce
Problems
pumping costs. 吀栀e optimistic, most likely, and
pessimistic projections 昀漀r annual savings are
$30,000, $20,000, and $5,000, with respective
probabilities of 20%, 50%, and 30%. 吀栀e inter­
est rate is 6%-8%, and the water line should
h愀瘀e a life of 40 years.
(a) What is the PW 昀漀r each estimated value?
What is the expected PW ?
(b) Compute the expected annual savings
and expected PW.
挀儀 Do the answers 昀漀r the expected PW
match? Why or why not?
(c) Does the answer to (b) match the present
worth 昀漀r the most likely value? Why or
why not?
of the cost of 昀椀re insurance 昀漀r his $200,000
0 home.
From a 昀椀re rating bureau he 昀漀und the
10 -52 An engineer decided to make a care昀甀l analysis
昀漀llowing risk of 昀椀re loss in any year.
Outcome
10-50 吀栀e chief uncertainty about a new product
is its annual net revenue. So 昀愀r, $35,000 has
been spent on development, but an additional
$30,000 is needed to 昀椀nish development. 吀栀e
昀椀rm's interest rate is 10%. Use a spreadsheet.
(a) What is the expected PW 昀漀r deciding
whether to proceed?
(b) Find the P (loss) and the standard devia­
tion 昀漀r proceeding.
Probability
Net revenue
Life, in years
Bad
State
OK
Great
0.3
-$15,000
5
0.5
$15,000
5
0.2
$20,000
10
transmission line. It will cost
0 higher-voltage
$250,000 more initially, but it will reduce trans­
10-51 A new engineer is deciding whether to use a
mission losses. 吀栀e optimistic, most likely, and
pessimistic projections 昀漀r annual savings are
$20,000, $15,000, and $8,000 respectively. 吀栀e
interest rate is 6%, and the transmission line
should have a life of 30 years.
(a) What is the present worth 昀漀r each esti­
mated value?
(b) Use the range of estimates to compute the
mean annual savings, and then determine
the present worth.
Probability
0.986
0.010
0.003
0.001
No 昀椀re loss
$ 10,000 昀椀re loss
40,000 昀椀re loss
200,000 昀椀re loss
10-49 Alice took a mid-term examin愀琀ion in physics
and rece瘀츀d a mark of 65. 吀栀e mean was 60, and
the standard deviation was 20. Bill received a
mark of 14 in mathematics, 眀栀ere the exam mean
was 12 and the standard deviation was 4. Which
student ranked higher in his class? Explain.
327
(a) Compute his expected 昀椀re loss in any
year.
( b) He 昀椀nds that the expected 昀椀re loss in any
year is less than the $550 annual cost of
昀椀re insurance. In 昀愀ct, an insurance agent
explains that this is always true. Never­
theless, the engineer buys 昀椀re insurance.
Explain why this is or is not a logical
decision.
10 -53 A robot has just been installed at a cost of
$81,000. It will have no salvage value at the end
of its use昀甀l life. Use a spreadsheet.
Savings
per Year
$ 18,000
20,000
22,000
Probability
Use昀甀l Life
(years)
Probability
0.2
0.7
0.1
12
5
4
1/6
2/3
1/6
(a) What is the joint probability distribution
昀漀r savings per year and use昀甀l life?
(b) De昀椀ne optimistic, most likely, and pes­
simistic scenarios by using both optimis­
tic, both most likely, and both pessimistic
estimates. What is the rate of return 昀漀r
each scenario?
10-14 has a 昀椀rst cost of $150,000, a life
0 ofProblem
10 years, and no salvage value. Assume that
10 -54 吀栀e energy e昀케ciency project described in
the interest rate is 8%.
(a) What is the equivalent uni昀漀rm annual
worth 昀漀r the expected annual savings?
328
Cha pter 10 U n certa i nty in Futu re Eve nts
(b) Compute the equivalent uni昀漀rm annual
worth 昀漀r the pessimistic, most likely,
and optimistic estimates of the annual
s愀瘀ings. What is the expected value of the
equivalent uni昀漀rm annual worth?
(c) Do the answers to (a) and (b) match? Why
or why not?
10-55 A man wants to decide whether to invest $1,000
in a 昀爀iend's speculative venture. He will do so
if he thinks he can get his money back in one
year. He believes the probabilities of the various
outcomes at the end of one year are as 昀漀llows:
Result
$2,000 (double his money)
1,500
1,000
500
0 (lose ever礀琀hing)
Probability
0.3
0.1
0.2
0.3
0.1
What would be his expected outcome if he in­
vests the $1,000?
10-56 吀栀e construction time 昀漀r a bridge depends
on the weather. 吀栀e project is expected to take
250 days if the weather is dry and hot. If the
weather is damp and cool, the project is ex­
pected to take 350 days. Otherwise, it is ex­
pected to take 300 days. Historical data suggest
that the probability of cool, damp weather is
30% and that of hot, dry weather is 20%. Find
the probability distribution and expected com­
pletion time 昀漀r the project.
Income, Depreciation,
and Cash Flow
Taxes and Incentives
When governments decided that smoking was bad for the people and, more to the point, was
costing governments a lot of money for health care, they started to use advertising and public
information campaigns to help people make informed choices-and they raised taxes to make the
smoking habit almost prohibitively expensive. Those are two of the ways that governments can
influence what people do. But there are others.
To encourage development in a certain region of the country, governments may offer grants
and tax incentives and, in some cases, may even subsidize the salaries of the workers. In the 1970s,
Nova Scotia, which was a place of high unemployment and limited prospects, invited the Michelin
Tire Company to establish a plant in Bridgewater, and it offered grants and favourable tax and
labour legislation to encourage the company. The initiative was a great success; by 2011 there were
three Michelin plants in Nova Scotia employing about 3,500 people. The photo below, from 1971,
heralds the production of Michelin's first tire in Canada.
Michelin N orth America (Canada) I nc.
Taxes and I n centives
331
Another exa m p le of a su ccessfu l government i nce ntive occu rred in the 1990s, when o i l was
selli ng fo r $20 a ba rrel a n d the cost of b u i ld i ng an o i l sa nds pla n t was a bout $5 b i llion. The then
Albe rta g overnment offe red the o i l com pa n ies a n extraord i na r i ly low roya lty rate of o n ly 1%, which
was to stay i n effect u nt i l p rod uction had p rod u ced e n o u g h p rofit to pay off the construction costs .
Then the roya lty rate wou ld rise to 20% of net p rofits. The o i l co m p a n ies res ponded by sta rt i n g to
b u i ld the pla nts a n d u p g ra d e rs necessa ry to d evelop the resou rce. By 2007, with o i l ove r $100 a
ba rrel a n d five p la n ts u n d e r construction, the Albe rta government revised the terms a n d chan ged
the oil roya lty structu re to match world sta n d a rds more closely.
O n e t i me-tested a n d relia ble way of i n d ucing desi red behavi o u r is with a ta rgeted tax i n ce n ­
tive. Fi rms u nd e rsta nd t h e t i m e va lue o f money, a n d t h e government, b y cha n g i n g t h e de p reciation
ru les, ca n change the rate at which fi rms recove r thei r capital i nvestments a n d ca n there by encour­
age ce rta i n kinds of i nvestment. For exa m p le, i n the 2007 budget, the federa l government changed
the ru les to permit accelerated recove ry of money i nvested i n clean-energy-generation eq u i pment,
such as certa i n wave- a n d tidal-e nergy eq u i pment, active-sola r eq u i pment, s m a ll photovolta ic a n d
fixed-location fuel-cell syste ms, biogas-prod uction eq u i pme nt, a n d ce rta i n other types o f eq u i p ­
ment related t o t h e envi ron mentally frie n d ly generation o f energy.
QU ESTIONS TO CONSI DER
1.
I t is not u n u s u a l fo r gove r n m e nts to change the ru les afte r a p roj ect has sta rted by cha n g i n g the tax reg u la ­
tio ns, i nc rea s i n g roya lti es, o r eve n dema n d i n g some form o f pa rti c i pation i n t h e project. H ow ca n o n e a l low
fo r t h i s poss i b i lity in the preparation of an e n g i neeri ng eco nomy stu dy?
2.
T h e Ca nad i a n Taxpaye rs Fed e ration, i n a re port o n g overnment fi n a n c i a l assistance entitled O n the Dole,
wrote, "Corporate we lfa re a n d fi n a n c i a l assista nce p rog ra ms serve neither b u s i n esses nor taxpaye rs a n d
s h o u ld be scra p ped . Government med d l i n g i n the economy is expens ive, u neq u a l, u nfa i r, a n d u n necessary. "
D iscuss t h i s state ment i n light of N ova Scotia's expe rie nce with M i cheli n .
LEARN I NG OBJ ECTIVES
This chapter will help you
•
describe d e p reciation, deterioration, a n d obsolescence
•
d isti n g u ish between va rious types of d e p recia ble property a n d d i ffere ntiate between depreci ­
ation expenses a n d other b u s i n ess expenses
•
use historical d e preciation methods to calcu late the annual depreciation charge a n d book
瘀愀lue ove r the asset's life
•
expla i n the d i fferences between h istorica l d e p reciation methods a n d the ca pita l cost a llow­
a nce system (CCA)
•
use CCA to calcu late a llowa ble annual depreciation charge a n d book value ove r the asset's
life fo r va rious asset classes
•
acco u n t fo r capital gains and losses, loss on disposal of 昀椀xed assets, a n d recaptured CCA from
the d i s posa l of a depreciated b u s i n ess asset
•
use the unit-of-production a n d depletion d e p reciation methods as needed i n e n g i neeri ng
economic a n a lysi s problems
332
Cha pter 11 Income, Depreciation, and Cash Flow
KEY TERMS
a n n u a l depreciation
book value
ca pital gain
decli n i ng - ba la nce
depreciation
depletion
deprecia ble life
dou ble-decli ning balance
expensed item
i ntangi ble property
loss on d isposal
personal property
rea l property
reca ptured CCA
reca ptu red depreciation
straight- line depreciation
sum-of-yea rs' -dig its (SOYD)
depreciation
tangi ble property
u n it-of- prod uction (UOP)
depreciation
Taxation
We have so 昀愀r dealt with a variety of economic analysis problems and many techniques
昀漀r their solution. In the process we have avoided discussing income taxes, which are an
important element in most economic analyses. Now we can move to more realistic-and
more complex-situations.
堀夀Z Compa渀礀
Income Statement
For the year ending 25 April 2014
Revenue
Costs
Sales ofp爀漀duct
Cha最휀s for services
$
Total Revenue
$
Cost of Goods Sold
Labour wages
Materials
Utilities
Machines (a portion of the cost)
Factory buildings (a portion of the cost)
Selling Costs
Advertising
Sale commissions
Administration Costs
Administ爀愀tive salaries
O昀케ce rental
Financing Costs
Interest paid on debt
Total Costs
Net Income be昀漀re 甀찀es
FIGURE 11-1 The i ncome statement.
$
----
Taxation
Our government taxes individuals and businesses to support its processes-law­
making, education, public health, transport, domestic and 昀漀reign economic policy­
making, even the making and issuing of money. 吀栀e omnipresence of taxes requires that
they be included in economic analyses, and that means we must understand the way taxes
are imposed. 吀栀e measure of a business's success is its annual pro昀椀t (or loss). 吀栀is measure
is arrived at by taking all the revenue that was earned in the year and subtracting 昀爀om
that the expenses incurred in earning that income. 吀栀is calculation is shown in the income
statement 昀漀r the business (Figure 11-1). For expenses incurred on a weekly, monthly, or
hourly basis, the calculation is reasonably straight昀漀rward. However, when you consider
equipment that you p愀礀 昀漀r now but continues to per昀漀rm 昀漀r several years (the items
"machines" and "buildings" in Figure 11-1), the question arises as to what portion of the
initial cost of these items should be allocated to which year's expenses.
Depreciation is the mechanism 昀漀r allocating the cost of a long-lived property over
a number of years 昀漀r the purpose of calculating annual income. Chapter 11 examines
depreciation, and Chapter 12 illustrates how depreciation is used in income tax computa­
tions. 吀栀e goal is to make decisions about engineering projects, not 昀椀nal tax calculations.
Basic Aspects of Depreciation
吀栀e word depreciation is de昀椀ned as a "decrease in value." 吀栀is is ambiguous because value
has several meanings. In economic analysis, value m愀礀 refer either to market value-that
is, the monetary value others place on property-or value to the owner.
Deterioration and Obsolescence
A machine m愀礀 depreciate because it is deterio爀愀ti渀最 or wearing out and no longer per­
昀漀rming its 昀甀nction as well as when it was new. Many kinds of machinery need increased
maintenance as they age and there is a slow but continuing 昀愀ilure of individual parts. In
other types of equipment, the quality of output may decline because of wear on compon­
ents and resulting poorer mating of parts. Anyone who has worked to maintain a car has
observed deterioration due to 昀愀ilure of individual parts such as 昀愀n belts, mu昀툀ers, and bat­
teries, and the wear on components such as bearings, piston rings, and alternator brushes.
Depreciation is also caused by obsolescence. A machine is described as obsolete when
it is no longer needed or use昀甀l. A machine may be in excellent working condition, yet may
still be obsolete. In the 1970s, mechanical business calculators with hundreds of gears and
levers became obsolete. 吀栀e advance of integrated circuits resulted in a completely di昀昀er­
ent and 昀愀r superior approach to calculator design. 吀栀us, mechanical calculators rapidly
declined or depreciated in value.
If your car depreciated in the last year, that means it has declined in market value.
It has less value to potential buyers. On the other hand, a manager who says a piece of
machinery has depreciated may be describing a machine that has deteriorated because
of use or because it has become obsolete compared to newer machinery. Both situations
indicate that the machine has declined in value to the owner.
吀栀e accounting profession de昀椀nes depreciation in yet another way, as allocating the
cost of an asset over its depreciable life. 吀栀us, we now have three distinct de昀椀nitions of
depreciation:
1. Decline in market value of an asset
2. Decline in value of an asset to its owner
3. Systematic allocation of the cost of an asset over its depreciable life
333
334
Cha pter 11 Income, Depreciation, and Cash Flow
Depreciation a nd Expenses
It is this third de昀椀nition that is used to compute depreciation 昀漀r business assets. Business
costs are generally either expensed or depreciated. Expensed items, such as labour, util­
ities, materials, and insurance, are part of regular business operations and are "consumed"
over short periods of time (sometimes recurring). 吀栀ese costs do not lose value gradually.
For tax purposes they are subtracted 昀爀om business revenues when they occur. Expensed
costs reduce income taxes because businesses are able to write o昀昀 their 昀甀ll amount when
they occur.
In contrast, business costs 昀爀om capital assets (buildings, 昀漀rkli昀琀s, computers, etc.)
are not 昀甀lly written o昀昀 when they occur. A capital asset loses value gradually and must be
written o昀昀 or depreciated over an extended period. For instance, consider a plastic-mould
injection machine used to produce the beverage cups 昀漀und at sporting events. 吀栀e plastic
pellets melted into the cup shape lose their value as raw material directly a昀琀er manu昀愀c­
turing. 吀栀e cost of the raw material is there昀漀re expensed immediately. On the other hand,
the plastic-mould injection machine itself will lose value over time, and thus its costs (pur­
chase price and installation expenses) are written o昀昀 over its depreciable life or recovery
period. 吀栀is is o昀琀en di昀昀erent 昀爀om the asset's use昀甀l or most-economic life. Depreciable
life is determined by the depreciation method used to spread out the cost-many types of
depreciated assets operate well beyond their depreciable life.
Depreciation is a non-cash cost that requires no exchange of dollars 昀爀om one hand to
another. Companies do not write cheques to pay their depreciation expenses. Rather, it is
a business expense that is allowed by the government to o昀昀set the loss in value of business
assets. Remember, the company has already paid 昀漀r the asset up 昀爀ont, and depreciation is
simply a way to claim this "business expense" over time. Depreciation deductions reduce
the taxable income of businesses and thus reduce the amount of tax paid. Since taxes are
cash 昀氀ows, depreciation must be considered in a昀琀er-tax economic analyses.
In general, business assets can be depreciated only if they meet the 昀漀llowing basic
requirements:
•
•
•
吀栀e property must be used 昀漀r business purposes to produce income.
吀栀e property must have a use昀甀l life that can be determined, and this life must be
longer than one year.
吀栀e property must be an asset that decays, gets used up, wears out, becomes obso­
lete, or loses value to the owner over time
EXAM PLE 11-1
Consider the costs that are incurred by a local pizza business. Identify each cost as either expensed or
depreciated and explain why.
•
•
•
•
Cost of pizza dough and toppings
Cost of new delivery van
Cost of wages 昀漀r janitor
Cost of 昀甀rnishings in dining room
Basic Aspects of Depreciation
•
•
Cost of a new baking oven
Utility costs 昀漀r so昀琀 drink re昀爀igerator
SO LUTI O N
Cost Item
T礀瀀e of Cost
圀栀y
Pizza dough and toppings
New delivery van
Wages 昀漀r janitor
Furnishings in dining room
New baking oven
唀琀ilities 昀漀r so昀琀 drink re昀爀igerator
Expensed
Depreciated
Expensed
Depreciated
Depreciated
Expensed
Life < 1 year, loses value immediately
Meets 3 requirements 昀漀r depreciation
Life < 1 year, loses value immediately
Meets 3 requirements 昀漀r depreciation
Meets 3 requirements 昀漀r depreciation
Life < 1 year, loses value immediately
Types of Property
吀栀e rules 昀漀r depreciation are linked to the classi昀椀cation of business property as either
tangible or intangible. Tangible property is 昀甀rther classi昀椀ed as either real or personal.
Tangible property can be seen, touched, and felt.
Real property includes land, buildings, and all things growing on, built upon,
constructed on, or attached to the land.
Personal property includes equipment, 昀甀rnishings, vehicles, o昀케ce machinery,
and anything else that is tangible, excluding those assets de昀椀ned as real
proper琀礀.
Intangible property is all property that has value to the owner but cannot be
directly seen or touched. Examples include patents, copyrights, trademarks, trade
names, and 昀爀anchises.
Many di昀昀erent types of properties that wear out, decay, or lose value can be depreci­
ated as business assets. 吀栀is wide range includes photocopiers, helicopters, buildings,
interior 昀甀rnishings, production equipment, and computer networks. Almost all tangible
property can be depreciated.
One important and notable exception is land, which is never depreciated. Land does
not wear out, lose value, or have a determinable use昀甀l life and thus does not qual­
ify as depreciable property. Rather than decreasing in value, most land becomes more
valuable as time passes. In addition to the land itself, expenses 昀漀r clearing, grading,
preparing, planting, and landscaping are not generally depreciated because they have
no 昀椀xed use昀甀l life. Other tangible property that cannot be depreciated includes 昀愀ctory
inventory, containers considered as inventory, and leased property. 吀栀e leased property
exception highlights the 昀愀ct that only the owner of property may claim depreciation
expenses.
Tangible properties used in both business and personal activities, such as a car used in
a consulting engineering 昀椀rm that is also used to take the owner's children to school, can
be depreciated but only in proportion to the use 昀漀r business purposes. Accurate records
indicating the portion of use 昀漀r business and personal activities are required.
335
336
Cha pter 11 Income, Depreciation, and Cash Flow
Cost Basis
吀栀e cost basis of an asset represents the total cost of acquiring it and getting it into working
order. It is not just the price. It includes such necessary expenses as engineering, account­
ing and legal fees, 昀爀eight, site preparation, installation costs, and commissioning. It is this
total cost that must be charged as an expense over the life of the asset.
Calculating Depreciation
Figure 1 1-2 illustrates the method of allocating the total depreciation charges over the
asset's depreciable life. 吀栀e vertical axis is labelled book value (BV ), and the curve of
asset cost minus depreciation charges made starts at the cost basis and declines to the
salvage value:
Book value = Cost basis - Depreciation charges made to date
Looked at another way, BV is the remaining unallocated cost of the asset.
In Figure 1 1-2, book value goes 昀爀om a value of B at Time O in the recovery period to a
value of S at Time 5. 吀栀us, BV changes over an asset's recovery period. 吀栀e equation used
to calculate the BV of an asset over time is
,, ------------------------ T
B
Cost
Basis
',
'
',
''
Total Depreciation
Charges
' ',
Curve Values Depend on
' ' , Depreciation Method
'''
''
,,
',,
',
- - - - - - - - - - - - - - - - - - - - - - -, �s
Salvage
Value
I
0
2
4
3
5
堀謀
- Depreciable Life (years) 椀ⴀ
FIGU RE 11-2 General depreciation.
t
BVt = Cost basis - � d;
i=l
where
(1 1-1)
BVt = book value of the depreciated asset at the end of time t
Cost basis = B = dollar amount that is being depreciated. 吀栀is includes the purchase
price of the asset, as well as any other costs necessary to make the asset
"ready 昀漀r use."
� di = the sum of depreciation deduction taken 昀爀om Time O to Time t, where d; is
i=1
the depreciation deduction in Year i.
t
Basic Aspects of Depreciation
Equation 11-1 shows that year-to-year depreciation charges reduce an asset's BV over
its life. 吀栀e 昀漀llowing section describes methods that are or have been allowed under the
Income Tax Act 昀漀r quanti昀礀ing these yearly depreciation deductions.
Depreciation Methods
Depreciation accounting, the allocation of the cost 昀漀r capital expenditures over time, is
part of how accountants represent the 昀椀nancial per昀漀rmance of a business. 吀栀e particular
depreciation method chosen is the one that best represents the actual decline in value
of the property. In this w愀礀 it is possible to calculate an annual pro昀椀t or loss. However,
昀漀r taxation purposes, governments speci昀礀 exactly how depreciation is to be calculated.
In Canada the method of depreciation 昀漀r income tax purposes is known as capital cost
allowance; in the United States it is called the Modi昀椀ed Accelerated Cost Recovery System.
Accounting depreciation methods can be categorized as 昀漀llows:
General Depreciation Methods
吀栀ese methods include the st爀愀最턀t-line, sum-o昀ⴀyears'-digits, declining-balance, and
unit-鰀혀roduction methods. Each method requires estimates of an asset's use昀甀l life and
salvage value. Firms elect which method to use 昀漀r assets, and there is little uni昀漀rmity in
how depreciation expenses are reported.
Tax Reporting Depreciation Me琀栀ods
Canada. 吀栀e capital cost allowance ( CCA) is the portion of the capital cost of certain depre­
ciable property that a corporation m愀礀 deduct 昀爀om income earned during the year. 吀栀e CCA
method has the 昀漀llowing features: (1) classes into which property is grouped, (2) declining­
balance depreciation of the grouped property at a government-speci昀椀ed percentage, and (3)
reduction of the amount eligible 昀漀r deduction by 50% in the year of acquisition.
USA. 吀栀e Modi昀椀ed Accelerated Cost Recovery 匀礀stem (MACRS) has been in e昀昀ect since the Tax
Reform Act of 1986 (TRA-86). 吀栀e MACRS has the 昀漀llowing features: (1) A small number of cat­
egories are cre愀琀ed, and all depreci愀琀ed assets are assigned to one particular c愀琀egory. (2) 吀栀e need
to estim愀琀e salvage values is elimin愀琀ed because all assets areful氀礀 depreci愀琀ed 瘀퐀r their rec瘀퐀ry
period. (3) 吀栀e annual depreci愀琀ion percentages are modi昀椀ed to include a half-year co瘀팀ntion 昀漀r
the 昀椀rst and 昀椀nal years. (4) 吀栀e rec瘀퐀ry periods used to calcul愀琀e annual depreciation accelera琀攀
the write-o昀昀 of capital costs more quickly than did the historical methods-hence the name.
In this chapter our primary purpose is to describe the CCA depreciation method.
However, it is use昀甀l to 昀椀rst describe the 昀漀ur general depreciation methods. 吀栀ey are still
used in some countries, and both the CCA and the MACRS are variants of them.
Straight- Li ne Depreciation
吀栀e simplest and best-known depreciation method is straight-line depreciation. To cal­
culate the constant annual depreciation charge, the total amount to be depreciated, B - S,
is divided by the depreciable life, in years, N:
Annual depreciation charge = dt =
(B - S)
N
(11-2)
N is used 昀漀r the depreciation period because it may be shorter than n, the horizon or
project life.
337
338
Cha pter 11 Income, Depreciation, and Cash Flow
EXAM PLE 11- 2
Consider the 昀漀llowing ($ in thousands):
Cost of the asset, B
Depreciable life, in years, N
Salvage value, S
$900
5
$70
Compute the straight-line depreciation schedule.
SO LUTI O N
900 - 70
B-S
Annual depreciation charge = dt = -- = --- = $166
N
5
Depreciation 昀漀r
Year,
Year t ($000),
t
dt
Sum of Depreciation
Charges Up to Year t
t
($000),
dj
䤀㨀
i= l
$166
166
166
166
166
2
3
4
5
$166
332
498
664
830
Book Value at the
End of Year ($000),
BVt = B - 䤀㨀dj
j= l
900 - 166 = 734
900 - 332 = 568
900 - 498 = 402
900 - 664 = 236
900 - 830 = 70 = S
吀栀is situation is illustrated in Figure 1 1-3. Notice that dt is constant at $166,000 each year 昀漀r 昀椀ve years,
and that the asset has been depreciated down to a BV of $70,000, which was the estimated salvage value.
0
0
紀贀
900
166
Straight Line Annual
Depreciation
Charge
-
734
568
402
236
70
0
s
4
3
2
1
Time, Depreciable Life = N (years)
FIGURE 11-3 Straight-line depreciation.
5
1 66
D e p reciati o n M ethods
339
吀栀e straight-line (SL) method is o昀琀en used 昀漀r intangible property. For example,
Veronica's 昀椀rm bought a patent in April that was not acquired as part of acquiring a busi­
ness. She paid $6,800 昀漀r this patent and will use the straight-line method to depreciate it
over 17 years with no salvage value. 吀栀e annual depreciation is $400 (= $6,800/17). Since
the patent was purchased in April, the deduction must be pro-rated over the nine months
of ownership. 吀栀is year the deduction is $300 (= $400 x 9/12), and next year she can begin
taking the 昀甀ll $400 per year.
S u m -of-Yea rs'- Dig its Depreciation
Another method 昀漀r allocating an asset's cost minus salvage value 瘀퐀r its depreciable life is
,
called sum-o�鈀ars -digits (SOYD) depreci愀琀ion. 吀栀is method results in larger-than-straight­
line depreciation charges during an asset's ea爀氀y years and smaller charges as the asset nears the
end of its depreciable life. Each year, the depreciation charge is a 昀爀action of the total amount
to be depreciated (B - S). 吀栀e denominator of the 昀爀action is the sum of the 礀攀ars' digits. For
example, if the depreciable life is 昀椀ve 礀攀ars, 1 + 2 + 3 + 4 + 5 = 15 = SOYD. 吀栀en 5/15, 4/15,
3/15, 2/15, and 1/15 are the 昀爀actions 昀爀om Year 1 to Year 5. Each 礀攀ar the depreciation charge
shrinks by 1/15 of B - S. Because this change is the same every year, SOYD depreciation can
be modelled as an arithmetic gradient, G. 吀栀e equations can also be written as
I
Sum-of-years ' -digits
depreciation charge
any year
=
where
Remaining depreciable life
at beginning of year
(Tot愀氀 amount depreciated)
Sum-of-years ' -digits
昀漀r tot愀氀 depreciable life
dt =
N-t+1
(B - S)
SOYD
(1 1-3)
dt = depreciation charge in any year t
N = number of years in depreciable life
SOYD = sum-of-years'-digits, calculated as N(N + 1)/2 = SOYD
B = cost of the asset made ready 昀漀r use
S = estimated salvage value a昀琀er depreciable life
EXAM PLE 11-3
Compute the SOYD depreciation schedule 昀漀r the situation in Example 11-2 ($ in thousands).
Cost of the asset, B
Depreciable life, in years, N
Salvage value, S
$900
5
$70
continued
340
Cha pter 11 I n come, Depreciation, a n d Cash Flow
SO LUTI O N
SOYD =
吀栀us,
dl =
= 15
5X6
2
5-l+l
(900 - 70) = 277
15
d2 =
5-2+1
(900 - 70) = 221
15
d4 =
5-4+1
(900 - 70) = 1 1 1
15
d3 =
d5 =
Year,
t
Depreciation
昀漀r Year t
($000), dt
1
2
3
4
5
$277
221
166
111
55
5-3+1
(900 - 70) = 166
15
5-5+1
(900 - 70) = 55
15
Sum of Depreciation
Charges Up to Year
t ($000),
Book Value at
End of Year t ($000),
䤀㨀 d
t
j= l
t
B嘀琀 = B - E d
j
j= l
$277
498
664
775
830
900 - 277 = 623
900 - 498 = 402
900 - 664 = 236
900 - 775 = 125
900 - 830 = 70 = S
吀栀ese data are plotted in Figure 11-4.
SOYD Depreciation
B
Charges
277
0 623
22 1
40 2
0
0
㨀먀
236
125
70
0
s
2
3
4
Time, Depreciable Life = N (years)
FIGURE 11-4
j
5
S u m -of-yea rs' - d i g its depreciation.
D e p reciati o n M ethods
341
Decli n i n g - Balance Depreciation
Declining-balance depreciation applies a constant depreciation 爀愀te (D) to the declining
book value of the property. 吀栀e 昀愀ctor D is used to determine the depreciation charge d1 昀漀r
a given year t, as 昀漀llows:
D = the 昀爀action of the year's initial book value that is lost during the year
dl = D X B
d2 = D X (B - d1 ) = D X B(l - D)
d3 = D X (B - d1 - d2) = D X B(l - D) 2
吀栀us 昀漀r any year,
dn = DB(l - D) n -1 = D X BVn�
(11-4a)
and the book value, BVn , at the end of n years will be
(1 1-4b)
Historically, be昀漀re the days of CCA and MACRS, companies depreciated their assets over
their depreciable lives either at a straight-line rate of 1/N or a declining-balance rate of
2/N. 吀栀us the value of an asset with a 10-year depreciable life would be reduced at 10%
of the cost basis each year or at 20% of the BV each year. Because the declining-balance
percentage was twice the straight-line, the method came to be known as double-declining
balance, or DDB, with general equations:
Double-declining b愀氀ance d1 = ㌀谀 (Book v愀氀ue 1 _1 )
N
(1 1-4c)
EXAM PLE 11-4
Compute the declining-balance depreciation schedule 昀漀r the situations in Examples 1 1-2 and 1 1-3
($ in thousands):
Cost of the asset, B
Declining-balance rate
Salvage value, S
SO LUTI O N
Year,
t
Depreciation 昀漀r Year t 昀爀om
Equation 1 1 -4a ($000), dt
1
2
3
4
5
40% X 900 = 360
40% X 540 = 216
40% X 324 = 130
40% X 194 = 078
40% X 116 = 046
Figure 1 1-5 illustrates the situation.
$900
40%
$70
Sum of Depreciation Charges
Up to Year t ($000),
$360
576
706
784
830
䤀㨀 d
t
i=l
i
($000), BVt = B - Edi
Book Value at End of Year t
t
i=l
900 - 360 = 540
900 - 576 = 324
900 - 706 = 194
900 - 784 = 116
900 - 830 = 70
continued
342
Cha pter 11 Income, Depreciation, and Cash Flow
Declining- Balance
Depreciation Charges
B
360
5 40
�
0
0
㨀먀
216
334
194
1 16
70
0
s
1
2
4
3
Time, Depreciable Life = N (years)
5
FIGURE 11-5 Declin ing - balance depreciation.
吀栀e 昀椀nal salvage value of $70 昀漀r Examples 11-2, 11-3, and 11-4 was chosen to match the ending
value 昀漀r the DDB method. 吀栀is does not normally happen.
Unit-of- Production Depreciation
At times, there may be situations where the recovery of depreciation on a particular asset
is more closely related to use than to time. In these few situations (and they are rare), the
unit-of-production (UOP) depreciation in any year is
Production 昀漀r year
. i. on m
. any year = ------------ (B - S)
Uop deprec1at
Tot愀氀 lifetime production 昀漀r asset
(1 1-5)
吀栀is method might be use昀甀l 昀漀r machinery that processes natural resources if the resour­
ces will be exhausted be昀漀re the machinery wears out. It is not an acceptable method 昀漀r
general use in depreciating industrial equipment.
EXAM PLE 11- 5
For purposes of comparison with previous examples, assume that equipment costing $900,000 has
been purchased 昀漀r use in a sand and gravel pit. 吀栀e pit will operate 昀漀r 昀椀ve years, while a nearby air­
port is being rebuilt and paved. 吀栀en the pit will be shut down and the equipment removed and sold
U n it- of- Prod uction Depreciation
343
昀漀r $70,000. Compute the UOP depreciation schedule if the airport reconstruction schedule calls 昀漀r
40,000 cubic metres of sand and gravel as 昀漀llows:
Year Sand and Gr愀瘀el Required (m3}
4,000
8,000
16,000
8,000
4,000
1
2
3
4
5
SO LUTI O N
吀栀e cost basis, B, is $900,000. 吀栀e salvage value, S, is $70,000. 吀栀e total lifetime production 昀漀r the asset
is 40,000 cubic metres of sand and gravel. From the airport reconstruction schedule, the 昀椀rst-year UOP
depreciation would be
First-year UOP depreciation = (4,000 m 3/40,000 m3) x ($900,000 - $70,000) = $83,000
Similar calculations 昀漀r the subsequent 昀漀ur years give the complete depreciation schedule:
Year UOP Depreciation ($000)
1
2
3
4
5
$83
166
332
166
83
$830
Note that the unit-of-production depreciation charge in any year is based on the actual production 昀漀r
the year rather than on the scheduled production.
Depreciation for Tax Purposes-Capital
Cost Allowance
吀栀e Canada Revenue Agency (CRA) is the agency responsible 昀漀r collecting income taxes.
吀栀e Canadian government is continually reviewing and revising the tax laws, and inter­
pretations by courts and CRA a昀昀ect how they are applied. Minor changes are made in
these laws every year, and major revisions are made periodically. 吀栀e 昀漀llowing discussion
is intended to give the reader an understanding of how the tax laws a昀昀ect investment deci­
sions. However, Canadian tax law 昀椀lls many volumes the size of this book. 吀栀ere昀漀re, if
you have to do a real-world economic study, check with an accountant or a tax specialist to
ensure that you have included any special tax considerations.
Canadian income tax law permits corporations to depreciate most physical capital
assets by the declining-balance method at a rate speci昀椀ed in the tax legislation. With
very few exceptions, the taxpayer is given no option as to the depreciation method or the
depreciation rate to be used. Since the depreciation allowance 昀漀r income tax purposes
344
Cha pter 11 Income, Depreciation, and Cash Flow
represents an allowable deduction in computing taxable income, it is referred to as capital
cost allowance (CCA).
Canadian legislation is also very speci昀椀c about how the capital cost allowance is to be
calculated. 吀栀e prescribed method is asset-class accounting. 吀栀at is, all assets of a single
class are grouped together into a single ledger account. When additional assets of that class
are acquired, their cost is added to the account; when assets are disposed of, the proceeds
are deducted 昀爀om the account. 吀栀e capital cost allowance 昀漀r any year is the account total
at the end of the year times the capital cost allowance rate 昀漀r the class. In the 昀椀rst year
of ownership of a depreciable asset, only one-half of the maximum rate can be applied.
Table 1 1-1 lists examples of asset classes and rates.
吀栀e Income Tax Act uses special terms when referring to CCA. 吀栀ese are the most
common ones:
Book Depreciation Term
Tax Depreciation Term
asset
depreciation
cost base
book value
salvage value
property
capital cost allowance
capital cost
undepreciated capital cost
proceeds 昀爀om disposition
The 50% Ru le, or Half-Year Convention
Generally, property acquired that is available 昀漀r use during the taxation year is eligible
昀漀r only 50% of the normal maximum CCA 昀漀r the year. You can claim 昀甀ll CCA 昀漀r that
property in subsequent taxation years. However, 昀漀r most classes of assets, the income
昀爀om disposal of assets in that class during the year is 昀椀rst subtracted 昀爀om the cost of
acquisitions made in the same year. Consequently, the e昀昀ect of the half-year rule is miti­
gated when there are major disposals of 昀椀xed assets.
Certain properties-Classes 12, 13, 14, 15, 23, 24, 27, 29, and 34, and those acquired
through non-arm's-length transfers-are exempt 昀爀om the 50% rule. See Table 11-1.
Accelerated CCA-50% Straight- Line
As was mentioned in the Michelin case at the beginning of this chapter, the government
will o昀琀en use taxes as a means of encouraging speci昀椀c behaviour. For example, the way
CCA is calculated can be changed. A more rapid charging of depreciation expense trans­
fers the taxes payable to later years and thus 昀爀ees up money 昀漀r today's operation. 吀栀at
is important, and sometimes crucial, when a business is starting a new line or product.
Assets placed in Class 29 are eligible 昀漀r a 50% straight-line rate calculated as 昀漀llows: claim
up to 25% in the 昀椀rst year, 50% in the second year, and the remaining 25% in the third year.
Since the government is interested in encouraging manu昀愀cturing investment and
clean energy, it periodically makes changes in the depreciation rules. 吀栀e 昀漀llowing are
described as "some of the more signi昀椀cant examples" in the Tax Planning Guide 2016-2017
http://www.taxplanningguide.ca/tax-planning-guide/section-I-businesses/calculating­
depreciation/):
•
Manu昀愀cturing and processing (M&P) machinery and equipment that would
otherwise qualify 昀漀r a 30% CCA (Class 43) will, 昀漀r a limited time, quali昀礀 昀漀r
Depreciation for Tax Purposes-Capital Cost Allowance
345
Ta ble 11-1 CCA Asset Classes
CCA classes of commonly used business assets
Class
Rate (%) Description
1
4
Most buildings you bought a昀琀er 1987 and the cost of certain additions or alterations made a昀琀er 1987.
吀栀e rate 昀漀r eligible non-residential buildings acquired a昀琀er March 18, 2007, and used in Canada to
manu昀愀cture and process goods 昀漀r sale or lease includes an additional allowance of 6% ( total 10%).
For all other eligible non-residential buildings in this class, the rate includes an additional allowance of
2% (total 6%). To be eligible 昀漀r the additional allowances, elections h愀瘀e to be 昀椀led. For more in昀漀rma­
tion, see Class 1 {4%).
3
5
Most buildings acquired be昀漀re 1988 (or 1990, under certain conditions). Also includes the cost of
additions or alterations made a昀琀er 1987. For more in昀漀rmation, see Class 3 {5%).
6
10
Frame, log, stucco on 昀爀ame, galvanized iron, or corrugated metal buildings that meet certain condi­
tions. Class 6 also includes certain fences and greenhouses. For more in昀漀rmation, see Class 6 {10%).
8
20
Property that you use in your business that is not included in another class. Also included are data
network in昀爀astructure equipment and systems so昀琀ware 昀漀r that equipment acquired be昀漀re March 23,
2004. For more in昀漀rmation, see Class 8 {20%) and Class 46 {30%).
10
30
General-purpose electronic data-processing equipment (commonly called computer hardware) and
systems so昀琀ware 昀漀r that equipment acquired be昀漀re March 23, 2004, or a昀琀er March 22, 2004, and
be昀漀re 2005 if you made an election.
Motor vehicles and some passenger vehicles. For more in昀漀rmation, see Class 10 {30%) and
Class 10.1 {30%).
10.1
30
A passenger vehicle not included in Class 10. For more in昀漀rmation, see Class 10.1 {30%).
12
100
吀栀e cost limit 昀漀r access to Class 12 (100%) treatment is $500 昀漀r tools acquired on or a昀琀er May 2, 2006,
and medical and dental instruments and kitchen utensils acquired on or a昀琀er May 2, 2006. For more
in昀漀rmation, see Class 12 {100%).
13
Varies
Leasehold interest. You can claim CCA on a leasehold interest, but the maximum rate depends on the
type of leasehold interest and the terms of the lease.
14
Varies
Patents, 昀爀anchises, concessions, or licences 昀漀r a limited period. Your CCA is the lesser of the total of
the capital cost of each property spread out over the life of the property, or the undepreciated capital
cost to the taxpayer as of the end of the tax year of property of that class.
16
40
Taxis, vehicles you use in a daily car-rental business, coin-operated video games or pinball machines
acquired a昀琀er February 15, 1984, and 昀爀eight trucks acquired a昀琀er December 6, 1991, that are rated
higher than 11,788 kilograms.
17
8
Roads, parking lots, sidewalks, airplane runw愀礀s, storage areas, or similar sur昀愀ce construction.
29
Varies
Eligible machinery and equipment used in Canada to manu昀愀cture and process goods 昀漀r sale or lease,
acquired a昀琀er March 18, 2007, and be昀漀re 2016, that would otherwise be included in Class 43.
38
30
Most power-operated, movable equipment you bought a昀琀er 1987 that was used 昀漀r exc愀瘀ating, moving,
placing, or compacting earth, rock, concrete, or asphalt.尀騀
43
30
Eligible machinery and equipment, used in Canada to manu昀愀cture and process goods 昀漀r sale or lease,
that are not included in Class 29. For more in昀漀rmation, see Class 43 {30%).
46
30
Data network in昀爀astructure equipment and systems so昀琀ware 昀漀r that equipment acquired a昀琀er
March 22, 2004. If acquired be昀漀re March 23, 2004, include them in Class 8 {20%). For more in昀漀rma­
tion, see Class 46 {30%).
50
55
General-purpose electronic data-processing equipment (commonly called computer hardware) and
systems so昀琀ware 昀漀r that equipment, including ancillary data-processing equipment acquired a昀琀er
March 18, 2007, and not included in Class 29. For more in昀漀rmation, see Class 50 {55%).
Source: Canada Revenue Agency. Reproduced with permission of the Minister of Public Works and Government Services Canada, 2016.
l♦I
Canada Revenue
Agency
Agence d u revenu
du Canada
Capital Cost A氀氀owance (CCA)
(2006 and later t愀砀 years)
挀묀ed甀氀e s
Code0603
Protected B
when completed
Corporation's name
Business nwnber
洀渀 □
⠀⤀
㨀㬀
儀⤀
∀伀
尀言
尀言
For more information, see the section called "Capital Cost Allowance" in the 吀㈀ Co爀瀀oration Income Tax Guide.
Is the corporation electing under Regu!ation ll0I(Sq)?
1
Class
number
洀愀
2
Undepreciated
capital cost
at the beginning
of 琀栀e year
(amount 昀爀om
column 13
oflast year's
schedule 8)
洀洀
3
4
Cost of
acquisitions
during the year
(new proper琀礀
must be av椀촀able
昀漀r use)
Adjustments 愀渀d
transfers (show
愀洀ounts that w氀픀
reduce the
undepreciated
capital cost in
brackets)
(see note 1
below)
m
(see note 2
below)
m
Yes
□
No
%
10
Recapture
of capital cost
氀촀owance
(see note 4
below)
(see note 5
below)
5
6
7
8
9
Proceeds of
Undepredated
capital cost
(column 2 plus
column 3 plus or
minus colwnn 4
minus column 5)
50% rule
(1/2 of the
愀洀ount, if any,
by which the
net cost of
acquisitions
exceeds
column 5)
Reduced
undepreciated
capit愀氀 cost
(col眀渀n 6 m椀渀us
column 7)
CCA
dispositions
during the year
(愀洀ount not to
exceed the
capital cost)
m
(see note 3
below)
洀氀
rate
愀愀
昀洀
11
Terminal loss
琀洀
12
Capital cost
氀촀owance
(昀漀r declin椀渀g
balance method,
colwnn 8
multiplied by
column 9, or a
lower amount
(see note 6
below)
13
Undepreciated
capital cost at
the end of the
year (col眀渀n 6
minus colwnn
12)
m
昀洀
㨀萀
⠀⤀
0
3
℀䐀
0
氀氀
∀伀
爀漀⠀⤀
�
;5
?
儀⤀
㨀萀
愀⸀
∀✀㨀㬀
ⰀⰀ
⠀⤀
儀⤀
[
To愀s
Note 1.
Include any property acquired in previous years that has now become available 昀漀r use. 吀栀is property would have been previously excluded
昀爀om column 3. List separately any acquisitions that are not subject to the 50% rule, see Re最甀lation 1 100(2) and (2.2).
Note 2. Enter in column 4, "Adjustments and transfers", amounts that increase or reduce the undepreciated capital cost (column 6). Items that
inĻ㬸 the undepreciated capital cost include amounts transferred under section 85, or transferred on amalgamation or 眀椀nding-up of a
subsidiary. Items that reduce the undepreciated capital cost (show amounts that reduce the undepreciated capital cost in brackets) include
government assistance received or entitled to be received in the year, or a reduction of capital cost a昀琀er the application of section 80. See the
吀㈀ Co爀瀀oration Income Tax Guide 昀漀r other examples of adjustments and trans昀攀rs to include in column 4.
Note 3. The net cost of acquisitions is the cost of acquisitions (column 3) plus or minus certain adjustments and trans昀攀rs 昀爀om column 4. For
exceptions to the 50% rule, see Interpretation Bulletin IT-285, Capital Cost Allowance - General Comments.
Enter the total of column 10 on line 107 of Schedule 1.
Enter the total of column 1 1 on line 404 of Schedule 1.
Enter the total of column 12 on line 403 of Schedule 1.
Note 4. Enter a rate only if you are using the declining balance method. For any other method (昀漀r example the straight-line method, where
calculations are always based on the cost of acquisitions), enter NIA. Then enter the amount you are claiming in column 12.
Note 5.
For every entry in column 10, the nRecapture of capital cost allowance" there must be a corresponding entry in column 5, "Proceeds of
dispositions during the yearn . The recapture and terminal loss rules do not apply to passenger vehicles in Class 10.l.
Note 6.
If the tax year is shorter than 365 days, prorate the CCA claim. Some classes of property do not have to be prorated. See the 吀㈀ Corporation
Income Tax Guide 昀漀r more in昀漀rmation.
T2 SCH 8 ( 17)
( Ce 昀漀rmulaire est disponible en 昀爀an㄀爀ais. )
FIGURE 11-6 CRA T2 Schedule 8 Form: Capital Cost Allowance (CCA) (2006 and later yea rs) .
Source: Canada Revenue Agency. Reproduced with permission of the Minister of Public Works and Government Services Canada, 2016.
C愀渀搀숀
Depreciation for Tax Purposes-Capital Cost Allowance
•
•
347
an accelerated write-o昀昀. Eligible purchases acquired be昀漀re 2016 will qualify 昀漀r
a 50% straight-line accelerated CCA rate and will be placed in Class 29. Eligible
purchases acquired a昀琀er 2015 and be昀漀re 2026 will qualify 昀漀r a 50% declin­
ing-balance accelerated CCA rate (new Class 53).
Buildings acquired a昀琀er March 18, 2007 and used 昀漀r M&P in Canada qualify
昀漀r a CCA rate of 10% as opposed to 4%. To claim the 10% rate, at least 90% of
the 昀氀oor space must be used in manu昀愀cturing or processing in Canada. Other
non-residential buildings acquired a昀琀er March 18, 2007 that are not used 90%
昀漀r M&P may qualify 昀漀r a CCA rate of 6% as opposed to 4%. To claim the higher
CCA rate, the taxpayer must elect to include the building in a separate prescribed
class (Class 1). 吀栀is election is made by attaching a note in the taxpayer's income
tax return 昀漀r the tax year in which the building is acquired.
吀栀e CCA rate 昀漀r computer hardware and systems so昀琀ware has gradually
increased 昀爀om 30% to 55%, except 昀漀r a temporary period 昀爀om January 28, 2009
until January 31, 201 1 when such property was eligible 昀漀r a 100% write-o昀昀. Such
purchases are now placed in Class 50, which is eligible 昀漀r a 55% CCA rate.
Calcu lating the CCA-Sched u le 8
Capital cost allowance is claimed by 昀椀lling in the CRA 昀漀rm called Schedule 8, which is
reproduced in Figure 1 1-6. 吀栀e use of this 昀漀rm 昀漀r calculating capital cost allowance is
outlined in Example 11-6 below.
Table 11-1 is a partial list and description adapted 昀爀om the Income Tax Guide 2016 of
the most common capital cost allowance (CCA) classes. A complete list can be 昀漀und in
Schedule II of the Income Tax Regulations.
EXAM PLE 11-6
A 昀椀rm has six vehicles; the make, age, and current BV of each are as 昀漀llows:
Age in Years Description
5
1
3
22
7
Chev Van
H礀甀ndai Sedan
Honda Accord
Ford Explorer
Dodge 1/2 ton pick-up
Total book value
Book Value (undepreciated capital cost)
$22,465
31,620
18,732
2,419
11,563
$86,799
What depreciation deduction (CCA) is permitted at the end of the current year (Year 1)?
continued
348
Cha pter 11 Income, Depreciation, and Cash Flow
SO LUTI O N
According to tax legislation (see Table 11-1), automotive equipment is a Class 10 asset and can be
depreciated at a CCA rate of 30%. 吀栀ere昀漀re, all the vehicles would be grouped into a single CCA schedule as 昀漀llows:
吀栀e Schedule 8 昀漀r Year 1
2
1
UndepreciClass
number ated capital
cost at the
beginning of
the year ( un depreciated
capital cost at
the end of the
year 昀爀om last
year's CCA
schedule)
3
Cost of acquisitions
during the
year (new
property
must be
愀瘀ailable
昀漀r use)
12
5
6
7
8
9
Proceeds Undepreci- 50% rule Reduced
CCA Capital
ated capof dis(1/2 of the undepreci- rate % cost alpositions ital cost
amount, ated caplowance
(Column 8
during
(Column 2 if any, by ital cost
multithe year
which the (Column 6
plus
minus
(amount Column 3 net cost
plied by
Column 9;
of acqui- Column 7)
not to
plus or
or a lower
minus
exceed
sitions
the capital Column 4 exceeds
amount)
Column 5)
minus
cost)
Column 5)
13
Undepreciated capital
cost at
the end of
the year
(Column 6
minus
Column 12)
10
$-
$-
$60,759
$86,799
$86,799
$-
$86,799
30%
$26,040
吀栀e Year 1 CCA = $26,040.
Action. In Year 2 the company sells the Honda Accord 昀漀r $20,000. what CCA is permitted at the end
of Year 2?
Calculation. When the company disposes of an asset, the instructions in Schedule 8 state that the
proceeds 昀爀om the disposal are subtracted 昀爀om the current total BV of the asset class. 吀栀e sale of the
Honda is treated as 昀漀llows. 吀栀e proceeds are entered in Column 5 and subtracted 昀爀om the capital cost
in Column 2 to produce a reduced total 昀漀r the class.
吀栀e Schedule 8 昀漀r Year 2
1
2
Class
Undepreciated
number capital cost at
the beginning
of the year
(undepreciated
capital cost at
the end of the
year 昀爀om last
year's CCA
schedule)
3
Cost of acquisitions
during the
year (new
property
must be
愀瘀ailable
昀漀r use)
5
6
Proceeds Undepreciof disated cappositions ital cost
during
(Column 2
the year plus
(amount Column 3
plus or
not to
minus
exceed
the capital Column 4
minus
cost)
Column 5)
10
$-
$20,000
$60,759
吀栀e Year 2 CCA = $12,228.
$40,759
7
9
12
13
8
Undepreci50% rule Reduced
CCA Capital
ated capital
( 1/2 of the undepreci- rate % cost
amount, ated capallowance cost at
(Column 8 the end of
if any, by ital cost
multithe year
which the (Column 6
minus
net cost
(Column 6
plied by
Column 9; minus
of acqui- Column 7)
or a lower Column 12)
sitions
amount)
exceeds
Column
5)
$-
$40,759
30%
$12,228
$28,532
Notice that the current BV of the Honda = $18,732 x (1 - 30%) = $13,1 12. So the sale resulted in a
recapture of ⠀␀20, 000 - 13, 1 12 = $6,888) capital cost allowance. However, recaptures and losses are not
explicitly calculated 昀漀r a particular asset: only the class total is adjusted. 吀栀e result of the Schedule 8
worksheet is that the recaptures or losses are added to or subtracted back into income at the same CCA
Depreciation for Tax Purposes-Capital Cost Allowance
349
rate at which they were taken out. 吀栀is is covered in more detail in Chapter 12 under Acquiring and
Disposing of Assets.
Action. In Year 3 the company buys a Toyota Land Cruiser 昀漀r $26,000. What CCA is permitted at the
end of Year 3?
Calculation. In the year that a company acquires an asset, the 50% rule permits only one-half the normal
CCA. 吀栀e 昀甀ll capital cost of acquisitions is added in Column 3, but then one-half of the net amount 昀漀r
the year (acquisitions minus proceeds) is taken in Column 7, and the account total is reduced by that
amount to arrive at a reduced undepreciated capital cost (Column 8) 昀爀om which to calculate the CCA.
吀栀e Schedule 8 昀漀r Year 3
1
2
Class
Undeprecinumber ated capital
cost at the
beginning of
the year (undepreciated
capital cost at
the end of the
year 昀爀om last
year's CCA
schedule)
10
$28,532
5
3
Proceeds
Cost of
acquisitions of disduring the positions
year (new during
property
the year
must be
(amount
available 昀漀r not to
use)
exceed
the capital
cost)
$26,000
吀栀e Year 3 CCA = $12,459.
$0
6
7
8
9
Undepreci- 50% rule Reduced
CCA
ated cap(1/2 of the undepreci- rate
amount, ated cap- %
ital cost
(Column 2 if any, by ital cost
which the (Column 6
plus
minus
Column 3 net cost
plus or
of acqui- Column 7)
sitions
minus
Column 4 exceeds
Column 5)
minus
Column 5)
$54,532
$13,000
$41,532
30%
12
13
Capital
Undepreciated capital
cost allowance
cost at
(Column 8 the end of
multithe year
(Column 6
plied by
Column 9; minus
or a lower Column 12)
amount)
$12,459
$42,072
吀栀e e昀昀ect of the calculation method is that one-half the capital cost, ($26,000/2) = $13,000, is
added to the previous year's total of $28,532 to arrive at the CCA 昀漀r second and subsequent years. No
昀甀rther adjustment is necessary. 吀栀is is because of the half-year rule.
Action. In Year 4 no capital assets were acquired or disposed of. What CCA is permitted at the end of
the year?
Calculation. With no additions and no subtractions, the Year 4 CCA is just a multiplication of the
beginning-of-year undepreciated amount by the CCA rate.
吀栀e Schedule 8 昀漀r Year 4
2
5
6
1
3
Proceeds UndepreciUndepreciated Cost of
Class
ated capnumber capital cost at acquisitions of disthe beginning during the positions ital cost
of the year
year (new during
(Column 2
the year plus Column
(undepreciated property
capital cost at must be
(amount 3 plus or
minus
the end of the available
not to
year 昀爀om last 昀漀r use)
exceed
Column
year's CCA
the capital 4 minus
Column 5)
schedule)
cost)
7
8
9
50% rule Reduced CCA
(1/2 of the undepreci- rate %
amount, ated capif any, by ital cost
which the (Column
net cost
6 minus
of acqui- Column 7)
sitions
exceeds
Column 5)
10
$-
$42,072
$0
吀栀e Year 4 CCA = $12,622.
$0
$42,072
$42,072
30%
12
13
UndepreciCapital
cost alated capital
cost at
lowance
(Column the end of
8 multi- the year
(Column
plied by
Column 9; 6 minus
or a lower Column
amount) 12)
$12,622
$29,450
continued
350
Cha pter 11 Income, Depreciation, and Cash Flow
Action. In Year 5 the company sells the Ford Explorer 昀漀r $850 and buys process control equipment 昀漀r
$25,000. What CCA is permitted at the end of Year 5?
Calculation. 吀栀e Income Tax Act lists process control equipment in Class 10, and so the capital cost
is lumped in with the automobiles. In this year there is both an acquisition (Column 3) and a disposal
(Column 5), and it is the net amount that is subject to the 50% rule.
吀栀e Schedule 8 昀漀r Year 5
2
1
3
Class
UndepreciCost of acnumber ated capital
quisitions
during the
cost at the
beginning of year (new
the year (un- property
must be
depreciated
capital cost at available
the end of the 昀漀r use)
year 昀爀om last
year's CCA
schedule)
10
$29,450
$25,000
5
7
8
9
6
Proceeds Undepreci- 50% rule Reduced CCA
of disated cap- (1/2 of the undepreci- rate %
amount, ated cappositions ital cost
during
(Column 2 if any, by ital cost
the year
which the (Column 6
plus
minus
(amount Column 3 net cost
not to
of acqui- Column 7)
plus or
minus
sitions
exceed
the capital Column 4 exceeds
Column 5)
minus
cost)
Column s)
$850
$53,600
$12,075
$41,525
30%
12
13
Capital
Undepreciated capital
cost
allowance cost at
(Column 8 the end of
multithe year
(Column 6
plied by
Column 9; minus
or a lower Column 12)
amount)
$12,458
$41,143
吀栀e Year 5 CCA = $12,458.
Action. In Year 6 the company sells all the remaining vehicles 昀漀r $9,000. What CCA is permitted at
the end of Year 6?
Calculation. 吀栀e proceeds 昀爀om the sale are entered in Column 5 and serve to reduce the total undepreciated capital cost of the Class 10 assets owned to $32,143. From this amount the annual CCA is calculated.
吀栀e Schedule 8 昀漀r Year 6
2
5
7
1
3
8
9
6
Proceeds Undepreci- 50% rule Reduced
CCA
Class
UndepreciCost of
number ated capital acquisitions of disated cap(1/2 of the undepreci- rate %
during the positions ital cost
amount, ated capcost at the
beginning of year (new during
(Column 2 if any, by ital cost
the year (un- property
which the (Column 6
the year plus
minus
(amount Column 3 net cost
depreciated must be
of acqui- Column 7)
plus or
capital cost at available 昀漀r not to
minus
sitions
exceed
the end of the use)
the capital Column 4 exceeds
year 昀爀om last
year's CCA
minus
Column 5)
cost)
schedule)
Column s)
10
$41,143
$0
$9,000
$32,143
$32,143
$-
30%
12
13
UndepreciCapital
ated capital
cost
allowance cost at
(Column 8 the end of
multithe year
(Column 6
plied by
Column 9; minus
or a lower Column 12)
amount)
$9,643
$22,500
吀栀e Year 6 CCA = $9,643.
Observe that at this time the only Class 10 asset that the company has is one-year-old process con­
trol equipment that, if we were to consider it alone, has an undepreciated capital cost or book value of
l
Book value = Original capital cost minus Accumulated depreciation
= $25,000 - [ $25;
000
= $21,250
X 30%
Depreciation for Tax Purposes-Capital Cost Allowance
351
However, the amount on which the CCA is calculated is $32,143. 吀栀is illustrates the di昀昀erence
between the asset class account totals and actual book values. 吀栀e account totals re昀氀ect the transactions
昀漀r the asset class and are a result of assets that are no longer possessed. Generally, as long as the com­
pany possesses at least a single asset in any one class, the account remains open and the CCA is calcu­
lated annually. If the last asset in a class is sold, then the account books 昀漀r that asset class are closed
and the remaining balance (loss or recapture) is added to that year's income.
Calculating the CCA and the U CC (without using
Schedule 8)
Since, in engineering economics, we deal with individual assets and not with asset classes,
it is necessary to calculate the CCA amounts 昀漀r the particular asset under consideration.
吀栀e capital cost allowance 昀漀r year n (CCA), and the undepreciated capital cost at the end
of year n (UCC) can be calculated 昀漀r the declining-balance case by the 昀漀llowing 昀漀rmulas:
CCA n = Capital cost allowance 昀漀r year n
P = Asset cost basis
d = CCA rate
CCA 1 = P [ : J 昀漀r n = 1
(11-6)
CCA n = P [ 1 - : 1 (1 - d) n - 2 Jor n > 2
(1 1 -7)
UCC n = Undepreciated capital cost at the end of year n
UCC n = P [ 1 - : ) (1 - d) n - 1
(11-8)
In the case of Class 29 Accelerated CCA, the 昀漀rmulas are
CCA l = 0.25P
CCA 2 = 0.S0P
CCA 3 = 0.25P
吀栀e undepreciated capital cost is also referred to as the book value of the asset. 吀栀is is
because it is the value shown in the company's books of account.
Depreciation and Asset Disposal
In the normal conduct of business, assets are bought and sold. For income tax purposes
in Canada, the cost of acquisition and proceeds 昀爀om disposal are totalled and added to
the appropriate column in Schedule 8, and the calculation of annual permissible CCA is
automatic. However, in engineering economy studies, it is o昀琀en necessary to know about
a particular asset-how much depreciation is taken and what the eventual salvage value
(SV) is; that is, the a昀琀er-tax value at the time of disposal. To calculate this, it is necessary
to compare the asset's market value, MV (what a willing buyer pays), with the asset's book
value, BV, to determine if the depreciation deductions taken match the actual decrease in
value. When they do not, an a搀樀ustment is made to the amount of taxes paid.
352
Cha pter 11 Income, Depreciation, and Cash Flow
吀栀e precise a搀樀ustment can be a complex calculation, depending, as it does, on the
time of the disposal, the time the money is realized, and the time period 昀漀r which the tax
return is 昀椀led. Here we will consider only the straight昀漀rward case, assuming that the asset
is disposed of at the end of the year, immediately a昀琀er the depreciation 昀漀r that year has
been taken, and that the adjustment is included in the calculations 昀漀r that year. For more
complex cases, it is recommended that a tax specialist be consulted.
When an asset is disposed of, the important question is which is larger: (1) the BV
(what we show in our accounting records a昀琀er applying the rules set by the government) or
(2) the asset's MV (what a willing buyer pays). If the BV is lower than the market value, too
much CCA has been deducted 昀爀om taxable income. On the other hand, if the BV is higher
than the market value, not enough CCA has been deducted. In either case, the amount of
tax owed changes.
Recaptured CCA (see Figure 11-7) occurs when an asset is sold 昀漀r more than its cur­
rent BV. If more than the original cost basis is received, only the amount up to the original
cost basis is recaptured depreciation. Recaptured depreciation represents the over-expense
in depreciation that has been claimed. In other words we have taken too much expense 昀漀r
the asset's loss in value.
Loss on disposal (see Figure 1 1-8) occurs when the market value is less than the book
value. In the accounting records, we've exchanged an asset worth its book value 昀漀r some­
thing less-which is a loss. In this case a company has not claimed enough depreciation
expense.
Capital gains (see Figure 11-9) occur when the asset is sold 昀漀r more than its original
cost. 吀栀e excess over the original cost basis is the capital gain. As described in Chapter 12,
the tax rate on such gains is sometimes lower than 昀漀r ordinary income, but there are
complicated rules that consider how long the investment has been held. In most engin­
eering economic analyses, capital gains occur only on land because business and produc­
tion equipment and 昀愀cilities almost always lose value over time. Capital gains are much
more likely on non-depreciated assets such as stocks, bonds, real estate, jewellery, art, and
collectibles.
Cost basis
$ 10,000
Market value $6,500
$5,000
Book value
Cost basis
$ 10,000
Market value $2,250
$5,000
Book value
$ 12,000 -
$ 12,000
10,000 -
-I -
10,000
8,000 -
6,000 -
- - - - - - - -- - - -- -- �Recaptured CCA
4,000 -
2,000 -
, 开贀_븀贀开⸀,
0 ⴀ吀娀脀吀娀吀蜀'-贀븀謀븀
Cost
Basis
Market
Value
Book
Value
8,000
6,000
Loss on
Disp?�!l
4,000
2,000
0
Cost
Basis
Market
Value
Book
Value
If Cost basis > Market value > Book value, there is Recaptured CCA.
Recaptured CCA = Market value minus Book value = $ 1,500
If Book value > Market value, there is a loss on disposal.
Loss on disposal = Book value minus Market value = $2,750
FIGURE 11-7 Reca ptu red CCA.
FIGURE 11-8 Loss on disposal.
Depreciation and Asset Disposal
Cost basis
$ 10,000
Market value $ 1 3 ,000
$5,000
Book value
$ 14,000
12,000
Capital Gain
10,000
8,000
Recaptured CCA
6,000
4 ,000
2,000
0 开贀븀贀贀븀謁ᴀ_謀븀贀謀븀ĝᷬ_븀贀开Ⰰ
Cost
Market
Book
Basis
Value
Value
If Market value > Cost basis, there is a capital 最愀in plus recaptured depreciation.
Capital gain = Market value minus Cost basis = $ 3 ,000
Recaptured depreciation = Cost basis minus Book value = $5,000
FIGURE 11-9 Ca pital gain.
Natural Resource Allowances
Depletion refers to the consumption of exhaustible natural resources as a result of their
removal. Since depletion covers such things as mineral properties, oil and gas wells, and
standing timber, removal may take the 昀漀rm of digging up minerals, producing petroleum
or natural gas 昀爀om wells, or cutting down trees.
In Canada, the federal and provincial governments collect income tax, but each prov­
ince owns its natural resources and so the tax and royalty regimes surrounding these
activities vary across the country. In the United States, depletion is recognized 昀漀r income
taxes 昀漀r the same reason depreciation is-capital investment is being consumed or used
up. 吀栀us a portion of the gross income should be considered a return of the capital invest­
ment. For mining companies in Canada, the use of an allowance (the lesser of 25% of
resource pro昀椀ts or the amount of earned depletion) was discontinued in federal legislation
in 1990, although existing mines were grand昀愀thered in.
But Canadian mining companies don't just mine in Canada. In 昀愀ct, in 2012, Canadian
companies in the extract瘀츀 sector had interests in more than 8,000 properties in 100 countries­
representing approximate氀礀 12% of Canada's direct investment abroad. 吀栀ose companies
account 昀漀r almost half the mining activities in the wo爀氀d. As such, they are subject to the rules
of many di昀昀erent jurisdictions.
Although not used in Canada, depletion is a standard method in the US and in many
other countries. 吀栀e 昀漀llowing sections illustrate the two American methods of calculating
depletion: percent最였 depletion and cost depletion.
Percentage Depletion
Percentage depletion is sometimes used 昀漀r mineral property and some oil or gas wells.
吀栀e allowance is a certain percentage of the property's gross income during the year. 吀栀is
is an entirely di昀昀erent concept than depreciation. Unlike depreciation, which allocates
353
354
Cha pter 11 Income, Depreciation, and Cash Flow
-- - - -- -·
cost over use昀甀l life, the percentage depletion allowance is based on the property's gross
income.
Since percentage depletion is computed on the income rather than the cost of the prop­
erty, the total depletion m愀礀 exceed the cost of the property. In computing the allowable
percentage depletion on a property in any year, United States regulations specify that the
percentage depletion allowance cannot exceed 50% of the property's taxable income com­
puted without the depletion deduction. 吀栀e percentage depletion calculations are illus­
trated by Example 11-7.
EXAM PLE 11-7
A coal mine has a gross income of $250,000 昀漀r the year. Mining expenses equal $210,000. Compute the
allowable percentage depletion deduction.
SO LUTI O N
From Table 11-2, coal has a 10% depletion allowance. 吀栀e percentage depletion deduction is computed
昀爀om gross mining income. 吀栀en the taxable income must be computed. 吀栀e allowable percentage
depletion deduction is limited to the computed percentage depletion or 50% of taxable income, which­
ever is smaller.
Ta ble 11-2 Percentage Depletion Allowa nce for Selected Items
T礀瀀e of Deposit
Percentage
22
15
15
10
5
14
Lead, zinc, nickel, sulphur, uranium
Oil and gas (small producers only)
Gold, silver, copper, iron ore
Coal and sodium chloride
Sand, gravel, stone, clam and oyster shells, brick, and tile clay
Most other minerals and metallic ores
Computed Percentage Depletion
Gross income 昀爀om mine
Depletion percentage
Computed percentage depletion
$250,000
X 10%
$25,000
Taxable Income Limitation
Gross income 昀爀om mine
Less: expenses other than depletion
Taxable income 昀爀om mine
Deduction limitation
Taxable income limitation
$250,000
- 210,000
40,000
X 50%
$20,000
Since the taxable income limitation ($20,000) is less than the computed percentage depletion ($25,000),
the allowable percentage depletion deduction is $20,000.
Natural Resource Allowances
355
Cost Depletion
吀栀e calculation of depreciation relied on the cost of an asset, its depreciable life, and its
salvage value to apportion the cost minus salvage value over the depreciable life. In some
cases where the asset is used at 昀氀uctuating rates, we might use the unit-of-production
(UOP) method of depreciation. For mines, oil wells, and standing timber, 昀氀uctuating pro­
duction rates are the usual situation. 吀栀us, cost depletion is computed like UOP deprecia­
tion; the calculation uses the 昀漀llowing values:
1.
2.
3.
Property cost
Estimated number of recoverable units (tonnes of ore, cubic metres of gravel, bar­
rels of oil, million cubic metres of natural gas, thousand board feet of timber, etc.)
Salvage value, if any, of the property
EXAM PLE 11-7
Assume that an oil reservoir is estimated to contain 150,000 barrels (bbl.) and the investment cost to
develop the reserve is $1,250,000. 吀栀en the unit depletion rate would be calculated as
(Cost basis of reserve/Number of recoverable units) = l,250,000/150,000 = $8.33 per bbl.
吀栀e depletion allowance 昀漀r a year when 5,000 bbl. of oil were produced would be
Number of units produced X Unit depletion rate = 5,000 X $8.33 = $41,667
As previously stated, in the case of mineral property and some oil and gas wells, the depletion deduc­
tion can be based on either cost or percentage depletion, depending upon the rules of the jurisdiction.
SUMMARY
Depreciation is part of computing income taxes in economic analysis. 吀栀ere are three dis­
tinct de昀椀nitions of depreciation:
1.
2.
3.
Decline in asset's market value
Decline in asset's value to its owner
Systematic allocating of the asset's cost over its depreciable life
While the 昀椀rst two de昀椀nitions are used in everyday discussions, it is the third, or
accountant's, de昀椀nition that is used in tax computations and in this chapter.
Book value is the remaining unallocated cost of an asset, or
Book value = Asset cost - Depreciation charges made to date
吀栀is chapter describes how depreciable assets are written o昀昀 (or claimed as a business
expense) over a period of years instead of expensed in a single period (like wages, material
356
Cha pter 11 Income, Depreciation, and Cash Flow
costs, etc.). 吀栀e depreciation methods described include the historical methods: st爀愀ight­
line, sum-of-the-years'-digits, and declining-balance. 吀栀ese methods required estimating
the asset's salvage value and depreciable life.
Current Canadian tax law speci昀椀es that CCA be used to allocate the asset's cost over
time. 吀栀us, when one is using CCA, it is o昀琀en necessary to consider recaptured deprecia­
tion. 吀栀is is the excess of SV over BV, and it is taxed as ordinary income. Similarly, losses
on sale or disposal are taxed as ordinary expenses.
UOP depreciation relies on usage to quanti昀礀 the loss in value. UOP is appropriate 昀漀r
assets that lose value as based on the number of units produced; 昀漀r example, the number
of stampings a die has made or the tons of gravel ground up by the crusher. However, this
method is not considered to be acceptable 昀漀r most business assets.
Natural resources are a provincial jurisdiction in Canada, and so the rules governing
royalties and write-o昀昀s vary across the country. Cost depletion is based on the 昀爀action of
the resource that is removed or sold. In the USA, 昀漀r minerals and some oil and gas wells,
an alternative calculation called percentage depletion is allowed. Percentage depletion is
based on income, so the total allowable depletion deductions may exceed the invested cost.
PROBLEMS
(c) Double-declining-balance depreciation
搀儀 CCA depreciation as a Class 43 asset
(e) CCA depreciation as a Class 29 asset
Depreciation Sched u les
11-1
⠀　A depreciable asset costs $10,000 and has an
estimated SV of $1,600 at the end of its six-year
depreciable life. Compute the depreciation
schedule 昀漀r this asset by both SOYD deprecia­
tion and DDB depreciation.
11-2
A million-dollar oil drilling rig has a six-year
depreciable life and a $75,000 salvage value at
the end of that time. Determine which one of
the 昀漀llowing methods provides the preferred
depreciation schedule: DDB or SOYD. Show the
depreciation schedule 昀漀r the preferred method.
11-3
A new machine tool is being purchased 昀漀r
$16,000 and is expected to have a zero sal­
vage value at the end of its 昀椀ve-year use昀甀l life.
Compute the DDB depreciation schedule 昀漀r
this capital asset. Assume that any remaining
depreciation is claimed in the last year.
11-4
Some special handling devices can be obtained
昀漀r $12,000. At the end of 昀漀ur years, they can be
sold 昀漀r $3,500. Compute the depreciation sched­
ule 昀漀r the devices by the 昀漀llowing methods:
愀儀 Straight-line depreciation
(b) Sum-of-years'-digits depreciation
⠀　11-5
A company treasurer must determine the best
depreciation methods the 昀椀rm should use 昀漀r
o昀케ce 昀甀rniture that costs $50,000 and has
zero SV at the end of a IO-year depreciable life.
Compute the depreciation schedule using the
methods listed:
(a) Straight-line
(b) Double-declining-balance
(c) Sum-of-years'-digits
搀儀 Capital cost allowance
11- 6
吀栀e Acme Chemical Company bought $45,000
worth of research equipment, which it believes
will have zero SV at the end of its 昀椀ve-year life.
Compute the depreciation schedule 昀漀r the
equipment by each of the 昀漀llowing methods:
(a) Straight-line
(b) Sum-of-years'-digits
(c) Double-declining-balance
搀儀 CCA as Class 8 assets
11-7
Consider a $6,500 piece of machinery, with
a 昀椀ve-year depreciable life and an estimated
$1,200 salvage value. 吀栀e projected use of the
⠀　⠀　Problems
吀栀ey are based on the same initial cost, use昀甀l
life, and salvage value. Identify each schedule
as one of the 昀漀llowing:
• Schedule 43
• Sum-of-years' -digits depreciation
• 150% declining-balance depreciation
• Double-declining-balance depreciation
• Unit-of-production depreciation
machinery when it was purchased and its actual
production to date are shown below.
Actual Production
Projected
Year Production (tonnes)
(tonnes)
1
2
3
4
5
3,000
5,000
[Not
yet
known]
3,500
4,000
4,500
5,000
5,500
Compute the depreciation schedule using
愀儀 straight-line
(b) sum-of-years'-digits
挀儀 double-declining-balance
(d) unit-of-production (昀漀r 昀椀rst two years
onl礀⤀
(e) CCA-30% rate
11-8
⠀　11-9
A large pro昀椀table corporation purchased a
small jet plane 昀漀r use by the 昀椀rm's executives
in January. 吀栀e plane cost $1.5 million and
will be kept 昀漀r 昀椀ve years. Compute the CCA
depreciation schedule 昀漀r 昀椀ve years.
For an asset that 昀椀ts into the CCA "Property
that you use in your business that is not in­
cluded in any other class . . ." designation,
show in a table the depreciation and book
value over a 10-year life of use. 吀栀e cost basis
of the asset is $10,000.
11-10 A company that manu昀愀ctures 昀漀od and bev­
erages in the vending industry has purchased
some handling equipment that cost $75,000
and will be depreciated as a Class 43 asset.
Show in a table the yearly depreciation amount
and book value of the asset over 10 years of
depreciation life.
⠀　11-11 Consider 昀椀ve depreciation schedules:
Year
A
B
1
2
3
4
5
6
$45.00
36.00
27.00
18.00
9.00
$35.00
20.00
30.00
30.00
20.00
C
$21.75
36.98
25.88
18.12
12.68
8.36
D
E
$58.00
34.80
20.88
12.53
7.52
$43.50
30.45
21.32
14.92
10.44
357
11-12 吀栀e depreciation schedule 昀漀r an asset with
a SV of $90 at the end of the recovery period
has been computed by several methods. Iden­
ti昀礀 the depreciation method used 昀漀r each
schedule.
⠀　Year
A
B
1
2
3
4
5
6
$323.3
258.7
194.0
129.3
64.7
$212.0
339.2
203.5
122.1
122.1
61.1
C
970.0 1,060.0
E
D
$424.0 $194.0 $212.00
254.4 194.0 339.20
152.6 194.0 203.52
91.6 194.0 122.11
47.4 194.0
73.27
970.0
970.0
950.10
11-13 吀栀e depreciation schedule 昀漀r a computer has
been calculated by several methods. 吀栀e esti­
mated SV of the equipment at the end of its six­
year use昀甀l life is $600. Identi昀礀 the resulting
depreciation schedules.
⠀　Year
A
B
C
D
1
2
3
4
5
6
$2,114
1,762
1,410
1,057
705
352
$2,000
1,500
1,125
844
633
475
$1,600.00
2,560.00
1,536.00
921.60
552.96
331.78
$1,233
1,233
1,233
1,233
1,233
1,233
11-14 Hillsborough Architecture and Engineering,
Inc. has purchased a blueprint printing machine
昀漀r $18,800. 吀栀is printer 昀愀lls under CCA Class
43. Prepare a depreciation table 昀漀r this printer.
11-15 If Hillsborough Architecture and Engineer­
ing (in Problem 11-14) purchases a new o昀케ce
building in May 昀漀r $5.7M, determine the
allowable depreciation 昀漀r each year.
11-16 Al Ja昀愀r Jewel Co. purchased, 昀漀r $50,000, a crys­
tal extraction machine that has an estimated
358
Cha pter 11 Income, Depreciation, and Cash Flow
SV of $10,000 at the end of its eight-year
use昀甀l life. Compute the depreciation sched­
ule using
(a) CCA Class 43
(b) straight-line depreciation
(c) sum-of-the-years'-digits depreciation
(d) double-declining-balance depreciation
11-17 Gamma Cruise, Inc. bought a new utility truck
昀漀r $35,000. Its SV is $7,500 a昀琀er its use昀甀l life
of 昀椀ve years. Calculate the depreciation sched­
ule using
(a) CCA Class 10
(b) the SOYD method
11-18 A pro昀椀table company making earthmoving
equipment is considering an investment of
$100,000 in manu昀愀cturing equipment that
will have a 昀椀ve-year use昀甀l life and a $20,000
salvage value. Use a spreadsheet 昀甀nction to
compute the CCA Class 43 depreciation sched­
ule. Show the total depreciation taken as well as
the PW of the depreciation charges discounted
at the MARR of 10%.
0
11-19 O昀케ce equipment whose initial cost is $100,000
has an estimated actual life of six years, with
an estimated salvage value of $10,000. Prepare
tables listing the annual costs of depreciation
and the BV at the end of each six years, based
on straight-line, sum-of-years'-digits, and
CCA depreciation. Use a spreadsheet to calcu­
late the depreciation amounts.
11-20 You are equipping an o昀케ce. 吀栀e total o昀케ce
equipment will have a 昀椀rst cost of $1,750,000.
You expect the equipment will last 10 years.
Use a spreadsheet to compute the CCA depre­
ciation schedule.
0
11-21 Unit-of-production depreciation is being used
昀漀r a machine that, on the basis of usage, has
an allowable depreciation charge of $6,500
the 昀椀rst year, increasing by $1,000 each year
until complete depreciation. If the cost basis
of the machine is $1 10,000, set up a deprecia­
tion schedule that shows depreciation charge
and BV over the 10-year use昀甀l life of the
machine.
11-22 A custom-built production machine is being
depreciated by the unit-of-production method.
吀栀e machine costs $65,000 and is expected to
produce 1.5 million units, a昀琀er which it will h瘀였
a $5,000 salvage value. In the 昀椀rst two 礀攀ars of
operation the machine was used to produce
140,000 units each 礀攀ar. In the third and 昀漀urth
years, production went up to 400,000 units a
year. A昀琀er that, annual production returned to
135,000 units. Use a spreadsheet to de瘀攀lop a de­
preciation schedule showing the machine's depre­
ciation allowance and BV over its depreciable life.
Com pa ring Depreciation Methods
0
11-23 T ELCO Corp. has leased some industrial land
near its plant. It is building a small warehouse
on the site at a cost of $250,000. 吀栀e building
will be ready 昀漀r use on 1 January. 吀栀e lease
will expire 15 years a昀琀er the building is occu­
pied. 吀栀e warehouse will then belong to the
landowner, with the result that there will be
no salvage value to T ELCO. 吀栀e warehouse is
to be depreciated by either CCA at 10% or SL
depreciation. If the interest rate is 15%, which
depreciation method should be selected?
0
11-24 吀栀e White Swan Talc Company bought
$120,000 of mining equipment 昀漀r a small talc
mine. 吀栀e mining engineer's report states that
the mine contains 40,000 cubic metres of com­
mercial-quality talc. 吀栀e company plans to
mine all the talc in the next 昀椀ve years as 昀漀llows:
Year
Production (m3 )
I
2
3
4
5
1 5,000
1 1 ,000
4,000
6,000
4,000
At the end of 昀椀ve years, the mine will be
exhausted and the mining equipment will be
worthless. 吀栀e company accountant must now
decide whether to use sum-of-years'-digits de­
preciation or unit-of-production depreciation.
吀栀e company considers 8% to be a reasonable
time value of money. Compute the depreci­
ation schedule 昀漀r each of the two methods.
Problems
Which method would you recommend that
the company adopt? Show the computations to
justi昀礀 your decision.
11-25 A small used delivery van can be purchased
昀漀r $18,000. At the end of its use昀甀l life (eight
years), the van can be sold 昀漀r $2,500. Deter­
mine the PW of the depreciation schedule
based on 10% interest using
(a) straight-line depreciation
(b) sum-of-the-years'-digits depreciation
挀儀 CCA Class 10
(d) double-declining-balance depreciation
11-26 吀栀e XYZ Block Company bought a new o昀케ce
computer and other depreciable computer
hardware 昀漀r $4,800. During the third year, the
computer is declared obsolete and is donated
to the local community college. Using an in­
terest rate of 15%, calculate the PW of the de­
preciation deductions. Assume that no SV was
initially declared and that the machine was ex­
pected to last 昀椀ve years.
(a) Straight-line depreciation
(b) Sum-of-the-years'-digits depreciation
挀儀 CCA Class 52
搀儀 Double-declining-balance depreciation
11-27 Some equipment costs $1,000, has a 昀椀ve-year
depreciable life, and will have an estimated
$50 salvage value at the end of that time. You
have been assigned the problem of determin­
ing whether to use straight-line or SOYD de­
preciation. At a 10% interest rate, which is the
preferable depreciation method 昀漀r this prof­
itable corporation? Use a spreadsheet to show
your computations of the di昀昀erence in present
worths.
0
11-28 吀栀e FOURX Corp. has purchased $12,000
worth of experimental equipment. 吀栀e an­
ticipated SV is $400 at the end of its 昀椀ve-year
depreciable life. 吀栀is pro昀椀table corporation is
considering two methods of depreciation: sum­
of-years'-digits and double-declining-balance.
If it uses 7% interest in its comparison, which
method do you recommend? Show computa­
tions to support your recommendation. Use a
spreadsheet to develop your solution.
359
Depreciation and Book Value
0
11-29 Use Canadian tax depreciation 昀漀r each of the
assets, 1-3, to calculate the items (a), (b), and (c).
1. A light general-purpose truck used by a de­
livery business, cost = $17,000.
2. Production equipment used by a Detroit automaker to produce vehicles, cost = $30,000.
3. Cement buildings used by a construction
昀椀rm, cost $130,000.
(a) 吀栀e asset class
(b) 吀栀e depreciation deduction 昀漀r Year 3
(c) 吀栀e book value (UCC) of the asset a昀琀er
six years
11-30 On 1 July, Sarah paid $600,000 昀漀r a commer­
cial building and an additional $150,000 昀漀r the
land on which it stands. Four years later, also
on 1 July, she sold the property 昀漀r $850,000.
Compute the CCA depreciation 昀漀r each of the
昀椀ve calendar years during which she had the
property and the gain or loss on disposal.
11-31 A group of investors has 昀漀rmed New Corpo­
ration to buy a small hotel. 吀栀e asking price
is $150,000 昀漀r the land and $850,000 昀漀r the
hotel building. If the purchase takes place in
June, compute the CCA depreciation 昀漀r the
昀椀rst three calendar years. 吀栀en assume the
hotel is sold in June of the 昀漀urth year, and
compute the CCA 昀漀r that year.
0
11-32 A company is considering buying a new piece
of machinery. A 10% interest rate will be used
in the computations. 吀眀o models of the ma­
chine are available.
Initial cost
End-of-useful-Hfe
salvage value, S
Annual operating
cost
Use昀甀l life, in years
Machine I
Machine II
$20,000
$25,000
$15,000 昀椀rst 10
years, $20,000
therea昀琀er
25
$80,000
$18,000
20
$100,000
(a) Using equivalent uni昀漀rm annual cost,
determine which machine should be
purchased.
(b) What is the capitalized cost of Machine I?
360
Cha pter 11 Income, Depreciation, and Cash Flow
(c) Machine I is purchased and a 昀甀nd is
set up to replace Machine I at the end of
20 years. Compute the required uni昀漀rm
annual deposit.
搀儀 Machine I will produce an annual saving
of material of $28,000. What is the rate of
return if Machine I is installed?
(e) What will the BV of Machine I be a昀琀er
two years if sum-of-years'-digits deprecia­
tion is used?
(f) What will the BV of Machine II be a昀琀er
three years if double-declining-balance
depreciation is used?
(g) Assuming that Machine II is a Class 43
asset, what would the CCA depreciation
be in the third year?
11-33 Sarah recently purchased an asset that she
intends to use 昀漀r business purposes in her
small tourism business. 吀栀e asset is CCA
Class 45 and has a life of 昀椀ve years. Sarah
purchased the asset 昀漀r $85,000 and uses
a salvage value 昀漀r tax purposes of $15,000
(when applicable). Answer the 昀漀llowing
questions 昀漀r Sarah.
(a) Using CCA depreciation, what is the BV
a昀琀er 昀漀ur years?
(b) Using CCA depreciation, what is the de­
preciation 昀漀r the sixth year?
(c) Using CCA depreciation, what is the BV
a昀琀er eight years?
搀儀 Using CCA depreciation, what is the BV
a昀琀er two years ?
(e) Using CCA depreciation, what is the sum
of the depreciation charges through the
昀椀昀琀h year?
(f) Using SL depreciation (with no half-year
convention), what is the BV a昀琀er the third
year?
(g) Using SL depreciation (with no half-year
convention), what is the BV a昀琀er the
eighth year?
11-34 Mary, Medhi, Marcos, and Marguerite have
bought a small warehouse in St Pete Beach.
If they paid $745,000 昀漀r the unit in Septem­
ber, how much will their depreciation be in
the 15th year? (Note: 吀栀is asset is a Class 6
property.)
11-35 Mud搀礀 Meadows Earthmoving can purchase a
bulldozer 昀漀r $80,000. A昀琀er seven years of use,
the bulldozer should h愀瘀e a SV of $15,000. What
depreciation is allowed 昀漀r this asset in Year 5 昀漀r
(a) earthmoving equipment, which is
Class 38 (rate 30%)?
(b) straight-line depreciation?
(c) sum-of-the-years'-digits depreciation?
搀儀 150% declining-balance depreciation?
11-36 An asset costs $150,000 and has a SV of $15,000
a昀琀er 10 years. What is the depreciation charge
昀漀r the 昀漀urth year, and what is the book value
at the end of the eighth year with
(a) CCA Class 8?
(b) straight-line depreciation?
(c) sum-of-the-years'-digits depreciation?
搀儀 double-declining-balance depreciation?
11-37 A 昀椀ve-axis milling machine costs $180,000
and will be scrapped a昀琀er 10 years. Compute
the book value and depreciation 昀漀r the 昀椀rst
two years using
(a) CCA Class 43
(b) straight-line depreciation
(c) sum-of-the-years'-digits depreciation
搀儀 double-declining-balance depreciation
(e) 150% declining-balance depreciation
11-38 To meet increased delivery demands, Mary
Moo Dairy just bought 15 new delivery trucks.
Each truck cost $30,000 and has an expected
life of 昀漀ur years. 吀栀e trucks can each be sold
昀漀r $8,000 a昀琀er 昀漀ur years. Using CCA Class
10 depreciation, determine (a) the depreciation
allowance 昀漀r Year 2 昀漀r the 昀氀eet and (b) the BV
of the 昀氀eet at the end of Year 3.
11-39 A minicomputer purchased in 2010 costing
$12,000 has no sa氀瘀age value a昀琀er 昀漀ur years.
What is the depreciation allowance 昀漀r Year 2 and
BV at the end of th愀琀 礀攀ar with CCA Class 52?
11-40 Blank Lobes, Inc. just purchased a new psy­
chograph machine 昀漀r $100,000. 吀栀e expected
resale value a昀琀er 昀漀ur years is $15,000. Deter­
mine the BV a昀琀er two years using
(a) CCA Class 29 depreciation
(b) straight-line depreciation
Problems
挀儀 sum-of-the-years' digits depreciation
(d) 150% declining-balance depreciation
11-41 A used drill press costs $55,000, and delivery
and installation charges add $5,000. 吀栀e SV
a昀琀er eight years is $15,000. Compute the accu­
mulated depreciation through Year 4 using
(a) CCA Class 43 depreciation
(b) straight-line depreciation
挀儀 sum-of-the-years'-digits depreciation
(d) double-declining-balance depreciation
11-42 A pump in an ethylene production plant costs
$15,000. A昀琀er nine years, the SV is declared
at $0.
(a) Determine depreciation charge and book
value 昀漀r Year 9 using straight-line, sum­
of-the-years'-digits, and CCA Class 43.
(b) Find the PW of each depreciation schedule if the interest rate is 5%.
Gai n/Loss on Disposa l
11-43 An asset with a 30% CCA rate costs $50,000 and
was purchased on 1 January 2001. Calculate any
depreciation recapture, ordinary losses, or cap­
ital gains associated with selling the equipment
on 31 December 2003 昀漀r $15,000, $25,000, and
$60,000. Consider two cases of depreciation 昀漀r
the problem: if the CCA method is used, and if
straight-line depreciation over an eight-year life
is used with a $10,000 salvage value.
0
11-44 A $150,000 asset has been depreciated with the
straight-line method over an eight-year life.
吀栀e estimated SV was $30,000. At the end of
the 昀椀昀琀h year the asset was sold 昀漀r $90,000.
From a tax perspective, what is happening at
the time of disposal, and what is the dollar
amount?
11-45 O'Leary Engineering Corp. has been depre­
ciating a $50,000 machine 昀漀r the last three
years. 吀栀e asset was just sold 昀漀r 60% of its 昀椀rst
cost. What is the size of the recaptured depre­
ciation or loss at disposal if the 昀漀llowing de­
preciation methods are used?
(a) Sum-of-years'-digits depreciation with
N = 8 and S = 2,000
361
(b) Straight-line depreciation with N = 8 and
S = 2,000
(c) CCA depreciation, classi昀椀ed as a Class
43 property
11-46 A numerically controlled milling machine
was bought 昀漀r $95,000 昀漀ur years ago. 吀栀e es­
timated SV was $15,000 a昀琀er 15 years. What
is the machine's book value a昀琀er 昀椀ve years of
depreciation? If the machine is sold 昀漀r $20,000
early in Year 7, how much gain on sale or re­
captured depreciation is there? Assume
(a) CCA Class 43
(b) straight-line depreciation
(c) sum-of-the-years'-digits depreciation
(d) 150% declining-balance depreciation
11-47 A computer costs $9,500, and its SV in 昀椀ve
years is negligible. What is the BV a昀琀er three
years? If the machine is sold 昀漀r $1,500 in
Year 5, how much gain or recaptured deprecia­
tion is there? Assume
(a) CCA Class 29
(b) straight-line depreciation
(c) sum-of-the-years'-digits depreciation
(d) double-declining-balance depreciation
11-48 A belt conveyor purchased 昀漀r $140,000 has
shipping and installation costs of $20,000. It
was expected to last six years, a昀琀er which it
would be dismantled at a cost of $5,000 and
sold 昀漀r $25,000. Instead, it lasted 昀漀ur years,
and several workers were permitted to take it
apart on their own time 昀漀r reassembly at a pri­
vate technical school. How much gain, loss, or
recaptured depreciation is there? Assume
(a) CCA Class 43
(b) straight-line depreciation
(c) sum-of-the-years'-digits depreciation
(d) 150% declining-balance depreciation
Depletion
0
11-49 When a major highway was to be constructed
nearby, a 昀愀rmer realized that a dry stream­
bed running through his property might be a
valuable source of sand and gr愀瘀el. He shipped
samples to a testing laboratory and learned that
the material met the requirements 昀漀r certain
362
Cha pter 11 Income, Depreciation, and Cash Flow
low-grade 昀椀ll. 吀栀e 昀愀rmer contacted the high­
way construction contractor, who o昀昀ered 65¢
per cubic metre 昀漀r 45,000 cubic metres of sand
and gravel. 吀栀e contractor would build a haul
road and would use his own equipment. All ac­
tivity would take place during a single summer.
吀栀e 昀愀rmer hired an engineering student
昀漀r $2,500 to count the truckloads of material
hauled away. 吀栀e 昀愀rmer estimated that two
hectares of streambed had been stripped of the
sand and gravel. 吀栀e 640-hectare 昀愀rm had cost
him $300 a hectare, and the 昀愀rmer felt the prop­
erty had not changed in value. He knew that
there had been no use 昀漀r the sand and gravel
be昀漀re the construction of the highway, and he
could 昀漀resee no 昀甀ture use 昀漀r any of the re­
maining 50,000 cubic metres of sand and gravel.
Determine the 昀愀rmer's depletion allowance.
11-50 Mr H. Salt purchased a one-eighth interest in
a producing oil well 昀漀r $45,000. Recoverable
oil reserves 昀漀r the well were estimated at that
time at 15,000 barrels, one-eighth of which
represented Mr Salt's share of the reserves.
During the subsequent year, Mr Salt received
$12,000 as his one-eighth share of the gross
income 昀爀om the sale of 1,000 barrels of oil.
From this amount, he had to pay $3,000 as
his share of the expense of producing the oil.
Compute the depletion allowance that US tax
rules would allow Mr Salt 昀漀r the year.
11-51 A piece of machinery has a cost basis of
$45,000. Its SV will be $5,000 a昀琀er 10,000 hours
of operation. With units-of-production
depreciation, what is the allowable deprecia­
tion rate per hour? What is the book value a昀琀er
4,000 hours of operation?
11-52 Mining recently began on a new deposit of 10
million tonnes of ore (2% nickel and 4% copper).
Annual production of 350,000 tonnes begins
this year. XYZ Mining Company expects to
reco瘀攀r 90% of the metal in the ore. 吀栀e net
smelter return (amount the company will re­
ceive) is $7.30/kg 昀漀r nickel and $4.75/kg 昀漀r
copper. Mining operation costs are expected
to be $1.00/kg of metal recovered. XYZ paid
$600 million 昀漀r the deposits. What is the maxi­
mum depletion allowance each year 昀漀r the mine?
⠀　11-53 During the construction of a highway bypass,
earthmoving equipment costing $40,000 was
purchased 昀漀r use in transporting 昀椀ll 昀爀om the
borrow pit. At the end of the 昀漀ur-year proj­
ect, the equipment will be sold 昀漀r $20,000.
吀栀e schedule 昀漀r moving 昀椀ll calls 昀漀r a total of
30,000 cubic metres during the project. In the
昀椀rst year, 40% of the total 昀椀ll is required; in the
second year, 30%; in the third year, 25%; and
in the 昀椀nal year, the remaining 5%. Determine
the units-of-production depreciation schedule
昀漀r the equipment.
11-54 Eastern Slopes Coal Co. expects to produce
125,000 tonnes of coal annually 昀漀r 15 years.
吀栀e deposit cost $3M to acquire; the annual
gross revenues are expected to be $9.50/t, and
the net revenues are expected to be $4.25/t.
(a) Compute the annual depletion on a cost
basis.
(b) Compute the annual depletion on a percentage basis.
11-55 A 2,500-hectare tract of timber is purchased by
the Houser Paper Company 昀漀r $1,200,000. 吀栀e
acquisitions department at Houser estimates
the land will be worth $275 a hectare once the
timber is cleared. 吀栀e materials department es­
timates that a total of 5 million board feet of
timber is available 昀爀om the tract. 吀栀e harvest
schedule calls 昀漀r equal amounts of the timber
to be harvested each year 昀漀r 昀椀ve years. Deter­
mine the depletion allowance 昀漀r each year.
11-56 吀栀e Piney Copper Company bought an
ore-bearing tract of land 昀漀r $7,500,000. 吀栀e
geologist 昀漀r Piney estimated the recoverable
copper reserves to be 450,000 tonnes. During
the 昀椀rst year, 50,000 tonnes were mined and
40,000 tonnes were sold 昀漀r $4,000,000. Ex­
penses (not including depletion allowances)
were $2,500,000. What are the percentage de­
pletion and the cost depletion allowances?
11-57 吀栀e Red River oil 昀椀eld will become less produc­
tive each year. Rojas Brothers is a small com­
pany that owns Red River, which is eligible 昀漀r
percentage depletion. Red River cost $2.5M to
acquire, and oil will be produced over 15 years.
Initial production costs are $4 per barrel, and
Problems
363
the wellhead value is $10 per barrel. 吀栀e 昀椀rst
year's production is 90,000 barrels, which will
decrease by 6,000 barrels a year.
(a) Compute the annual depletion (each year
may be cost based or percentage based).
(b) What is the PW at i = 12% of the deple­
tion schedule?
salvage value is $20,000, and expected life is
10 years. Compute the PW of the depreciation
deductions assuming
(a) straight-line depreciation
(b) sum-of-the-years'-digits depreciation
(c) CCA Class 50
(d) double-declining-balance depreciation
11-58 An automated assembly line is purchased 昀漀r
$1,250,000. 吀栀e company has decided to use
units-of-production depreciation. At the end
of eight years, the line will be scrapped 昀漀r an
estimated $250,000. Using the 昀漀llowing in昀漀r­
mation, determine the depreciation schedule
昀漀r the assembly line.
11-62 吀栀e RX Drug Company has just purchased a
capsulating machine 昀漀r $76,000. 吀栀e plant en­
gineer estimates the machine has a use昀甀l life
of 昀椀ve years and no salvage value. Compute the
depreciation schedule 昀漀r the machine by
(a) straight-line depreciation
(b) sum-of-years'-digits depreciation
(c) CCA at 30% rate
(d) CCA as a Class 29 asset
Year
Production Level
1
2
3
4
5
6
7
8
5,000 units
10,000 units
15,000 units
15,000 units
20,000 units
20,000 units
10,000 units
5,000 units
U nclassi昀椀ed
11-59 Metal Stampings, Inc., can purchase a new
昀漀rging machine 昀漀r $100,000. A昀琀er 20 years
of use the 昀漀rge should h愀瘀e a salvage value of
$15,000. What depreciation is allowed 昀漀r this
asset in Year 3 昀漀r
(a) CCA Class 43 depreciation?
(b) straight-line depreciation?
挀儀 double-declining-balance depreciation?
11-60 A pro昀椀table company making earthmoving
equipment is considering an investment of
$100,000 in equipment that will have a 昀椀ve­
year use昀甀l life and a $20,000 salvage value. If
money is worth 10%, which one of the 昀漀llow­
ing three methods of depreciation would be
preferable?
(a) Straight-line method
(b) Double-declining-balance method
挀儀 CCA at 30% rate
⠀　11-61 Loretta Livermore Labs purchased R&D equip­
ment costing $200,000. 吀栀e interest rate is 5%,
11-63 A heavy-construction 昀椀rm has been awarded
a contract to build a large concrete dam. It is
expected that eight years will be needed 昀漀r the
work. 吀栀e 昀椀rm will buy $600,000 worth of spe­
cial equipment 昀漀r the job. During the prepa­
ration of the job-cost estimate, the 昀漀llowing
utilization schedule was computed 昀漀r the spe­
cial equipment:
⠀　Utilization
Year
1
2
3
4
(hr/yr)
6,000
4,000
4,000
1,600
Utilization
Year
5
6
7
8
(hr/yr)
800
800
2,200
2,200
It is estimated that at the end of the job, the
equipment can be sold at auction 昀漀r $60,000.
(a) Compute the sum-of-years' -digits depreciation schedule.
(b) Compute the unit-of-production depreci­
ation schedule.
11-64 A machine cost $245,000, with delivery and
installation charges amounting to $15,000. 吀栀e
declared SV was $20,000. Early in Year 4, the
company changed its product mix and 昀漀und
that it no longer needed the machine. One of
its competitors agreed to buy the machine 昀漀r
$1 10,000. Find the loss, gain, or recapture de­
preciation on the sale. 吀栀e machine is a CCA
Class 43 asset.
364
Cha pter 11 Income, Depreciation, and Cash Flow
11-65 Prairie Gravel expects to produce 60,000 tonnes
of gravel annually 昀漀r 昀椀ve years. 吀栀e deposit
cost $150K to acquire; the annual gross reve­
nues are expected to be $9 per tonne, and the
net revenues are expected to be $4 per tonne.
(a) Compute the annual depletion on a cost
basis.
(b) Compute the annual depletion on a
percentage basis.
11-66 For its 昀愀bricated metal products, the Able
Corp. is buying $10,000 worth of special tools
that have a 昀漀ur-year use昀甀l life and no sal­
vage value. Compute the depreciation charge
昀漀r the second year by each of the 昀漀llowing
methods:
(a) DDB
(b) SOYD
(c) CCA 昀漀r Class 12 assets
11-67 Equipment costing $20,000 that is a CCA 30%
asset is disposed of during the second year 昀漀r
$14,000. Calculate any depreciation recapture,
losses, or capital gains associated with disposal
of the equipment.
⠀　11-68 American Pulp Corp. (APC) has entered
into a contract to harvest timber in the state
of Georgia 昀漀r $450,000. 吀栀e total estimated
available harvest is 150 million board feet.
(a) What is the depletion allowance 昀漀r Years
1 to 3 if 42 million, 45 million, and 35
million board feet are harvested by APC
in those years?
(b) A昀琀er three years, the total available
harvest 昀漀r the original tract was
re-estimated at 180 million board feet.
Compute the depletion allowances 昀漀r
Years 4 and beyond.
⠀　After-Tax Cash Flows
On with the Wind
Global warming and the erratic price of oil have convinced many that we must seriously investigate
non-carbon sources of energy. One solution is to rely more on renewable sources of energy, such as
solar power and wind. The technology for such energy sources has been around for many years, and
if good intentions were all it took, we would be getting much of our electricity from wind turbines.
Canada is a land of ice, snow, and natural resources, and, as anyone who has ventured into the
Crowsnest Pass in southern Alberta knows, it is a land of wind. A chinook wind sweeps down the
eastern slopes of the Rocky Mountains and emerges near the town of Cowley, Alberta. It was there,
on Cowley Ridge, that Canada's first commercial wind plant was built in 1993. After subsequent
phases in 1994 and additions in 2001, the Cowley plant had 57 Kenetech turbines, with a total rated
capacity of 21.4 MW, and 15 Nordex N60 turbines, each rated at 1.3 MW. The Nordex turbines were
mounted on 46-metre towers, had a diameter of 60 metres, and operated in wind speeds of up
to 100 kilometres an hour. The plant was decommissioned in 2016; this did not represent a shift
away from wind energy but was a consequence of a replacement analysis of the kind we will be
considering in Chapter 13.
M ichael Wheatley/Alamy Stock Photo
On with the W i n d
The tra nsition to g reater use of wind powe r will req u i re s i g n ificant i nvestments in i nfra ­
structu re, especia lly costly wi nd t u r b i n es, a n d few i nvestors a re willi ng to com m it thei r money
without a solid expectation of a com petitive retu r n . I t was not tec h n o logy, but rather cost facto rs,
that p revi ously kept wi nd e n e rgy fro m beco m i n g an attractive i nvestment. As rece ntly as the late
198 0s, wi nd-g e n e rated power cost rou g h ly twice as m u c h to prod uce as energy from conve n ­
tional sou rces.
In the past few decad es, howeve r, the price of wind energy has decreased d ra matica lly. The
America n W i n d Energy Association (AW EA) re ports that m a ny modern wind pla n ts ca n now p ro ­
d u ce power a t prices com petitive with conventional sou rces. N ot s u rpris i n g ly, i nvestment i n wi nd
powe r has a lso i ncreased su bsta ntia lly.
H ow d i d t h i s h a p pen? I n pa rt, it was d u e to a dva nces in wind t u r b i n e technology. But chan ges
i n tax policy ca n a lso a lter i nvestors' behaviou r. For i n sta nce, i n the U n ited States the d evelopment
of wi nd power was helped by the Ene最 Policy Act of 1992, which a l lowed electricity s u p p liers
a " p roduction tax c red it" of 1.5¢ per ki lowatt h o u r (later ra ised to 1.9¢ to acco u n t for i nflation) .
I n Ca nada, all levels of g overnment too k steps to promote i nvestment i n wi nd-energy p roj ects.
One exa m p le i s Natu ra l Resou rces Canada's W i n d Powe r Prod uction I n centive (W P P I ) , which is
desig ned to su bsid ize a bout h a lf the cost prem i u m of prod u c i n g wind energy ove r conventional
energy sou rces. The Ca n a d i a n Income Tax Act a lso p rovi des a n u m ber of i n centives. For exa m p le,
ce rta i n wi n d-powe r p rod uction eq u i pment may be eli g i ble fo r accelerated a n d e n h a nced d e p reci ­
ation expense cla i ms . I n the 2005 fede ra l b u d g et, the federal government a n nou nced that cer­
ta i n re newa b le - e n e rgy- generation eq u i pment (inclu d i ng wind t u r b i nes) acq u i red afte r 22 Febru a ry
2005 wou ld be eli g i b le fo r d e p reciation (fo r tax pu rposes) at a rate of 5 0 % .
W i n d powe r has conti nued t o expa n d . Data from the Ca nad i a n W i n d Energy Association
(Ca nW EA) shows that i nsta lled capac ity rose fro m 256 megawatts i n 2 0 0 2 to 11, 8 9 8 i n 2016.
The g rowth is expected to conti n u e, a n d Ca n W EA believes that by 2025 wind energy will p ro ­
vide 2 0 % o f Ca nada's electrica l energy.
12,000
1 1 ,000
蠀묀
∀✀
I
Ģ⊤ Annual ⬀娀 Cumulative
I
10,000
9,000
8,000
.£
7,000
6,000
5,000
∀✀
4,000
3,000
2,000
1 ,000
0
1998 1 999 2000 200 1 2002 2003 2004 2005 2006 2007 2008 2009 20 10 20 1 1 20 12 20 1 3 2014 20 15 20 16
Year
I nstalled Wind Power Capacity in Ca nada (in megawatts)
Sou rce : N RCAN . Data fro m 199 8 - 2014 fou n d at http: //www. n rca n .gc. ca/sites/w眀眀. n rca n.gc.ca/
昀椀 les/energy/i mages/ l n stalled%20Wi nd%20 Power%20Capacity%20i n%20Canada. png, data from
2015 - 16 fou n d at https : //ca nwea .ca/wp - content/uploads/2017/03/Canada - Cu rrent - l nsta lled ­
Ca pacity_e. pdf
367
368
Cha pter 12 Afte r-Tax Cash Flows
QU ESTI ONS TO CON S I D E R
1.
Canada's tota l GHG em issions i n 2014 were 732 megato n nes (Mt) of carbon d ioxide equ iva le nt. I f wi nd
turbi nes we re generating 20% of Ca n a d i a n electricity today, they wou ld be p reventing the release of a p ­
p roxi mately 2 4 Mt o f g reenhouse gases eve ry yea r. Does t h i s benefit m a ke t h e governme nt's encourage­
ment to the wi nd i n d u stry sensi ble?
2.
What the government g ives, the government can ta ke away. In the U n ited States the prod uction tax cred it ex­
pi red at the end of 2001. As a resu lt, an esti mated $3 bi llion worth of wind projects were suspended, and h u n ­
d reds o f workers were la id off. Fortu nately for t h e i nd u stry, t h e cred it was su bseq ue ntly re-i ntrod uced, most
rece ntly thro u g h 2013. H ow cou ld you d ea l with this type of u ncerta i nty in an engi neeri ng economy stu dy?
3.
Propone nts o f the wi nd p rod uction t a x cred it i nsist t h a t the e n e rgy m a r ket has neve r b e e n g e n u i nely "free."
They point out that other types of energy p rod uction (nota bly coa l, n u clea r powe r, a n d o i l) have obta i ned
la rg e government s u bs i d ies ove r the yea rs . In fact, these p ropone nts a rg u e, ca rbo n-based sou rces enjoy a
tremendous h i d d e n s u bsidy beca use the envi ron menta l a n d health effects of these more-polluting energy
sou rces a re not paid fo r by coa l a n d oil p rod u ce rs but a re borne by soci ety at la rg e . Is this a va lid a rg u me nt?
4.
Deve lo p i n g a wi nd power p roj ect ta kes m a ny yea rs and req u i res the co m m itment of la rge s u ms of i nvest­
ment ca pita l befo re the p roject beg i n s to return a p rofit. What i s the effect on i nvestment when a tax cred it
is a l lowed to expi re o r is exte nded for o n ly a few yea rs?
LEARN I N G OBJ ECTIVES
吀栀is chapter will help you
•
calcu late taxes due, or taxes owed, fo r both i n d ivi d u a ls a n d corporations
•
u nd e rsta nd the i ncremental natu re of the i n d ivi d u a l a n d corporate tax rates used for calculat­
ing taxes on i ncome
•
calcu late a co m b i ned i ncome tax rate fo r p rovi ncial a n d fed eral i ncome taxes a n d select a n
a p p rop riate tax rate fo r e n g i neeri ng eco n o m i c a n a lyses
•
use a n after-tax table to fi nd the afte r-tax cash flows fo r a p ros pective i nvestment p roj ect
•
calcu late measu res of merit-such as p resent wo rth, a n n u a l worth, payba ck period, i ntern a l
rate of retu rn, a n d ben efit- cost ratio- from d eveloped afte r-tax c a s h flows
•
eva luate i nvestment a lternatives on a n afte r-tax basis, i nclud i n g asset d isposa l
KEY TE RMS
average tax rate
basic federal tax
books-closed assu m ption
books-open assu m ption
d ividend
i ncome statement
marg i nal tax rate
worki ng capita l
Be渀樀amin Franklin said that the only inevitable things in life are death and taxes. In this chap­
ter we will examine the structure of income taxes in Canada. 吀栀ere is, of course, a wide var­
iety of taxes, ranging 昀爀om sales taxes (GST, PST, and HST ) to gasoline taxes, property taxes,
provincial and federal income taxes, and so 昀漀rth. Here we will concentrate on income taxes
because they o昀琀en h瘀였 a direct e昀昀ect on the economic viability of an engineering project.
A Partner in the Business
First, we must understand the way in which taxes are imposed. Since the previous
chapter, concerning depreciation, is an integral part of this analysis, it is essential to under­
stand the principles covered there. Having understood the mechanism of depreciation, we
will see how income taxes a昀昀ect our economic analysis.
A Partner in the Business
Probably the most straight昀漀rward way to understand income taxes is to consider the gov­
ernment as a partner in every business activity. As a partner, the government shares in the
pro昀椀ts 昀爀om every success昀甀l venture. In a somewhat more complex way, the government
shares in the losses of unpro昀椀table ventures too. 吀栀e tax laws are complex, and it is not our
purpose to explain them 昀甀lly. 1 Instead, we will examine the 昀甀ndamental concepts of the
income tax laws-but the reader must realize that there are exceptions and variations to
almost every statement we make!
Calculation of Taxable Income
At the mention of income tax, one can visualize dozens of elaborate and complex calcu­
lations. Yet income taxes are just another type of disbursement. Our economic analysis
calculations in earlier chapters have dealt with all sorts of disbursements: operating costs,
maintenance, labour and materials, and so 昀漀rth. Now we simply add one more prospective
disbursement to the list-income tax.
Taxa ble I ncome of I nd ividuals
吀栀e amount of federal income tax to be paid depends on taxable income and the income tax
rates. To begin, you, as an individual, must calculate your taxable income. You do this by
adding up your income 昀爀om all sources, and then subtracting the permitted deductions.
Gross income = Wages and salary
+ Interest income
+ Dividend income
+ Rent and royalty income
+ Capital gains
+ Other income
Deductions = Retirement plan contribution
+ Union and professional dues
+ Child care expenses
+ Attendant care expenses
+ Business investment losses
+ Moving expenses
+ Spousal support payments
+ Interest on money borrowed 昀漀r investment and others
Gross income - Deductions = Taxable income
吀栀ere are many government and private sources that describe income taxation in detail. O昀케cial, up-to­
date in昀漀rmation on rates and rules can be 昀漀und on the website of the Canada Revenue Agency ( CRA) at
http://眀.cra-arc.gc.ca.
1
369
370
Cha pter 12 After-Tax Cash Flows
吀䄀BLE 12 - 1 I n d ividual Federal I ncome Tax Structu re (2016 rates)
T愀砀able Income
䌀䐀
Tax Rate
First
$45,282
15%
Amount between
$45,282 and $90,563
20.5%
Amount between
$90,563 and $140,388
26%
Amount between
$140,388 and $ 200,000
29%
Amount above
$200,000
33%
To calculate the tax owed, we multiply each part of the taxable income by the appro­
priate tax rate. Federal tax rates are called "progressive" because the larger the taxable
income, the larger the percentage that is required in taxes. Table 12-1 shows the federal tax
rates 昀漀r 2016.
From time to time, Parliament amends the Income Tax Act and changes the rates. 吀栀e
break points-that is, the amounts of income at which you move 昀爀om one rate to the next
higher rate-are changed every year because they are tied to the cost of living. (See also
Chapter 14, on in昀氀ation.)
From the federal income tax, individuals are allowed to subtract non-re昀甀ndable tax
credits. 吀栀ere is a personal exemption amount, shown in Table 12-2, and credits are also
given 昀漀r items such as post-secondary tuition and Canada Pension Plan contributions.
吀栀e credit is calculated by multiplying the total amount to be credited by the lowest tax
rate. 吀栀is will be illustrated in Examples 12-1 and 12-2. Note that these exemptions are
non-refundable; that is, even if applying the exemption shows that the government owes
you a tax re昀甀nd, you won't get one.
吀栀e total amount of federal income tax is calculated by applying the rates in Table 12-1
to the taxable income and then subtracting any tax credits. 吀栀e result is called the basic
federal tax.
Provincial tax rates vary 昀爀om province to province, and are also progressive. Each
province also allows a personal exemption; two examples are shown in Table 12-2.
Two concepts that we will use extensively in this and subsequent chapters are average
tax rate and marginal tax rate. 吀栀e average tax rate is the ratio of total taxes p愀礀able to
taxable income. Total taxes payable include provincial and federal taxes as well as all sur­
taxes. 吀栀e marginal rate refers to the tax bracket, or step, that you are in, and it is the rate
that will be charged on the next dollar made.
Tot愀氀 t愀砀es payable
Average t愀砀 rate = ----樀娀脀蜀T砀숀ble income
Marginal tax rate = 吀栀e tax rate that applies to the next taxable dollar
Examples 12-1 and 12-2 look at the di昀昀erent average and marginal rates in Manitoba
and Ontario.
Ca lcu lati o n of Taxa b le I n co me
371
吀栀ese two provinces were chosen because they illustrate two di昀昀erent ways of taxing
income. Residents of each province must 昀椀rst calculate the basic federal tax; that calcula­
tion and rate are the same 昀漀r every province. 吀栀e provincial taxes, like the federal one, are
progressive taxes, the rate increasing as the taxable income increases; and Ontario adds a
tax on the tax, called a surtax.
A consequence of these di昀昀erent schemes is a very di昀昀erent marginal tax 昀漀r the dif­
ferent income levels. A person with a taxable income of $30,000 would have a marginal tax
rate of 25.8% in Manitoba, 21.05% in Ontario. If a person has a taxable income of $150,000,
the corresponding rates are 39% and 46.4%. So, 昀爀om a tax point of view, it is better to live
in Ontario if you h愀瘀e a low taxable income, and Manitoba if you have a high one. 吀栀e tax
calculations are illustrated in Examples 12-1 and 12-2.
As well as the rates, the exemptions also vary 昀爀om province to province.
Table 12-2 Basic Personal Exemption (2016)
Federal
$11,474
Manitoba
9,134
Ontario
10,011
EXAM PLE 12 -1
I nd ividual Tax i n Man itoba (2016)
A Manitoban has a taxable income of $107,000. Find the 昀漀llowing:
(a) Total taxes p愀礀able
(b) 䄀瘀erage tax rate
(c) Marginal tax rate
(d) A昀琀er-tax income
SO LUTI O N
I
Data
Taxable income
$107,000
First, we calculate the federal tax. We allot the taxable income to the various tax bands shown in
Table 12-1. 吀栀e 昀椀rst $45,282 is taxed at the lowest rate, 15%. 吀栀is leaves us with $61,7 18 to allocate to
the higher bands. 吀栀e next $45,282 of income 昀愀lls into the band between $45,282 and $90,563, taxed
at 20.5%. 吀栀e 昀椀nal $16,437 of income 昀愀lls into the 26% band, so our marginal tax rate at the federal
level is 26%. We calculate the amount of tax owed on each 昀爀action of income and sum the results to
get $20,348.52.
continued
372
Cha pter 12 After-Tax Cash Flows
Next, we subtract the federal personal exemption shown in Table 12-2. 吀栀e table tells us that we needn't
p愀礀 tax on the 昀椀rst $1 1,474. 吀栀is $1 1,4㜀㐀 would have been taxed at the lowest rate, 15%, so we multiply
it by 15% and subtract it 昀爀om total taxes owed.
We then go through a similar calculation at the provincial level.
Tax per Level
2016 Federal Individual Rates
On the 昀椀rst
$45,282.00
15%
$6,792,03
昀爀om
$45,282.00
to
$90,563.00
20.5%
$9,282.60
昀爀om
$90,563.00
to
$107,000.00
26%
$4,723.62
$20,348.52
$11,474
Less non-re昀甀ndable tax credit
15%
Basic 昀攀deral tax =
$1,721.10
$ 18,627.42
2016 Manitoba Provincial Tax
On the 昀椀rst
$31,000.00
昀爀om
$31,000.00
above
$67,000.00
$67,000
10.8%
$3,348.00
12.8%
$4,608.00
17.4%
$6,960.00
$14,916.00
$9,134.00
Less non-re昀甀ndable tax credit
10.8%
Total federal & provincial taxes p愀礀able =
Average Tax Rate = Total Federal & Provincial Taxes P愀礀able/Taxable Income =
Marginal Tax Rate = Tax P愀礀able on Next Dollar
= 26% + 17.4% =
A昀琀er-tax income = $74,443.05
EXAM PLE 12 -2
I n d ividual Tax i n Onta rio (2016)
An Ontarian has a taxable income of $107,000. Find the 昀漀llowing:
(a) Total taxes payable
(b) 䄀瘀erage tax rate
(c) Marginal tax rate
SO LUTI O N
I
Data
Taxable income
$107,000
$986.47.47
$32,556.95
30.4%
43.4%
Ca lcu lati o n of Taxa b le I n co me
Tax per Level
2016 Federal Individual Rates
On the 昀椀rst
$45,282
15%
$6,792.30
昀爀om
$45,282
to
$90,563.00
20.8%
$9,282.60
昀爀om
$90,563.00
to
$107,000/00
26%
$4,273.62
$20,348.52
$11,474.00
Less non-re昀甀ndable tax credit
15%
Basic federal tax =
$1,721.10
$ 19,790.39
2016 Ontario Provincial Tax
On the 昀椀rst
昀爀om
$41,536.00
above
$83,075.00
to
$41,536.00
5.05%
$2,097.57
$83,075.00
9.15%
$3,800.82
11.16%
$2,670.03
Regular Ontario tax =
$ 8,568.42
Ontario Surtax
On the amount the tax exceeds $4,484 =
$4,084.42
@ 20%
$816.88
On the amount the tax exceeds $5,739 =
$2,829.42
@ 36%
$1,018.59
Ontario surtax =
$1,835.47
Less non-re昀甀ndable tax credit
$10,011.00
5.05%
Total federal & provincial taxes payable =
䄀瘀erage tax rate = Total federal & provincial taxes p愀礀able/taxable income =
Marginal tax rate = Tax p愀礀able on next dollar
A昀琀er-tax income = 78,474.24
= 26% + 1 1 . 16% X [ 1 + (20% + 36%)] =
$505.55
$28,525.76
26.66%
43.4%
Corporate Income Taxes
Engineering economy is usually practised in a corporate environment, and it is corporate
income taxes, as opposed to individual income taxes, that are our main 昀漀cus.
Just as individuals pay personal tax on the income they h愀瘀e earned, corporations pay
tax on the money that they have earned-that is, their net income or pro昀椀t. Although the
calculation is a bit more complex because of credit terms and time lags and di昀케culties in
collection (the cash 昀氀ow is not always at the same time as the transaction), the corporate
accountants apply what are called generally accepted accounting principles (GAAP) to a
corporation's account to try to represent what has actually happened. Depreciation is done
according to the rules and 昀漀rms of capital cost allowance provided by the Canada Rev­
enue Agency.
In an e昀昀ort to be precise, the Income Tax Act de昀椀nes certain speci昀椀c accounting con­
cepts. For example, depreciation of a machine is called capital cost allowance because it
is not really the machine wearing out but rather the amount of the original capital cost
373
374
Cha pter 12 After-Tax Cash Flows
that the government will allow you to deduct 昀爀om this year's income to arrive at a tax­
able income. Special terms are used by CRA to describe property and amounts. 吀栀e more
important ones were listed in Chapter 11 and are repeated here.
Book Depreciation Term
Tax Depreciation Term
asset
depreciation
cost base
book value
salvage value
property
capital cost allowance
capital cost
undepreciated capital cost
proceeds 昀爀om disposition
Unlike personal taxes, corporate taxes are not progressive and do not increase as the
amount of income increases. 吀栀e percentage rate does vary, depending on the type of
business and the province, and there is a Small Business Deduction (SBD) that results in a
lesser tax being charged on the 昀椀rst $400,000 of taxable income. However, since many en­
gineering economy projects are undertaken by large enterprises, we can be com昀漀rtable in
the assumption that a single marginal rate, t, is applied to all the taxable income. Start-up
昀椀rms, special research enterprises, and small businesses are sometimes eligible 昀漀r a var­
iety of credits and incentives, but those are beyond the scope of this course and can best be
explained by a tax expert.
吀栀e terms income and pro昀椀t are o昀琀en used interchangeably to describe the amount
of money a corporation earns. To 愀瘀oid con昀甀sion, we use the word income to describe
amounts be昀漀re the application of income taxes (e.g., net income be昀漀re tax, taxable
income, income be昀漀re interest and taxes), and pro昀椀t to mean what is le昀琀 a昀琀er income
taxes have been subtracted.
Pro昀椀t is calculated on an accounting statement called the income statement. A sim­
pli昀椀ed example is given in Figure 12-1. In the right-hand column are 昀漀rmulas that can be
used to directly calculate net pro昀椀t and be昀漀re-tax cash 昀氀ow.
Table 12 -3 2015 Marg i na l Tax Rate on Active Busi ness I ncome over
$500,000
Province or Territory
British Columbia
Alberta
Saskatchewan
Manitoba
Ontario
Quebec
New Brunswick
Nova Scotia
Prince Edward Island
Newfoundland and Labrador
Yukon
Northwest Territories
Nuna瘀甀t
Provincial Rate
Federal Rate
Combined Rate
11.0%
10/12%
12.0%
12.0%
11.5%
11.9%
12.0%
16.0%
16.0%
14.0%
15.0%
11.5%
12.0%
15%
15%
15%
15%
15%
15%
15%
15%
15%
15%
15%
15%
15%
26.0%
25/27%
27.0%
27.0%
26.5%
26.9%
27.0%
31.0%
31.0%
29.0%
30.0%
26.5%
27.0%
Corporate Income Taxes
375
INCOME S吀䄀TEMENT
For ABC Corporation
For the year ending 7 January 2017
Operating revenue
Operating costs
CCA
OR
- oc
Be昀漀re-tax cash 昀氀ow
Debt interest
BTCF
- CCA
-I
甀찀able income
Less income tax (at rate t)
OR - OC - CCA - I
- t( OR - OC - CCA - I )
( OR - OC - CCA - I ) ( 1 - t)
Net pro昀椀t
FIGURE 12-1 Sim plified income statement formula.
Using CCA to Calculate Net Pro昀椀t
EXAM PLE 12 -3
A construction company (t = 27%) purchased a new bulldozer 昀漀r $220,000 (CCA rate = 30%). 吀栀e
expected annual revenues and costs that will be created by the machine are
Operating revenue (OR) = $324,000
Operating cost (OC) = $96,000
Find the net pro昀椀t 昀漀r Years 1 and 2.
SO LUTI O N
Unless speci昀椀cally stated otherwise, the tax rate t given in a question is assumed to be the marginal tax
rate, and the income tax payable is the taxable income multiplied by t.
Year 1 CCA = $220,000 X 0.5 X 30%
Year 2 CCA = ($220,000 - 33,000) x 30%
= $33,000 (half-year rule applies)
= $56,100
Income Statements 昀漀r
Year 1
Year 2
OR
$324,000
96,000
33,000
$324,000
96,000
56,100
$195,000
52,650
$171,900
46,413
$142,350
$125,487
oc
CCA
Taxable income
Less income tax (27%)
Net pro昀椀t
376
Cha pter 12 After-Tax Cash Flows
Accounting and Engineering Economy
Engineering econo洀礀 studies use accounting practices and 昀漀rmulas to determine what will
happen, but there are fundamental di昀昀erences bet眀攀en the t眀漀 昀椀elds. First, accountants are
trying to measure 眀栀at has happened, 眀栀ereas engineering economists look 昀漀rward and try
to predict what may happen. Second, accountants try to allocate funds in time to attach them to
units or products, whereas engineering economy is concerned with when 琀栀e cash 昀氀ows occu爀⸀
吀栀e principal accounting documents that deal with cash 昀氀ows are
1.
2.
the income statement: tells what happened over the past year
the cash 昀氀ow statement: lists the sources and uses of cash
We are concerned with flows, so a useful comparison is with a pipeline. Figure 12-2 shows
money flowing into the 昀椀rm and branching o昀昀 to pay for the di昀昀erent items.
From this diagram we can develop the formulas for calculating the net funds from
operations. Beginning at the top, the operating revenue stream is divided into two streams,
operating cost and the before-tax cash flow. 吀栀us,
Operating revenue (OR) = Operating cost (OC) + Be昀漀re-tax cash 昀氀ow (BTCF)
$$ From Sales $$
Operating Revenue (OR)
Operating Cost
(OC)
Be昀漀re - Tax Cash Flow (BTCF)
Debt Interest
Taxable Income
CCA
Paid to Employees,
Suppliers, Utilities
Net Pro昀椀t
昀딀er-Tax Cash Flow (䄀吀CF)
Interest Payments
Retained
Earnings
Net Funds 昀爀om
Operations
FIGURE 12-2 The operations cash flow pipeline.
Dividends
Income
Tax
Acco u nt i n g a n d E n g i neeri n g Eco n o my
To calculate the a昀琀er-tax cash 昀氀ow (AT CF), it is 昀椀rst necessary to calculate the taxable
income. In the diagram, the BT CF stream is partitioned into interest payments on debt (I),
CCA, and taxable income. 吀栀us,
BTCF = Debt interest (I) + CCA + Taxable income
rearranging,
Taxable income = BTCF - Debt interest (I) - CCA
吀栀en, we need to determine the net pro昀椀t. From the diagram,
Taxable income = Net pro昀椀t + Income tax
Net pro昀椀t = Taxable income - Income tax
where
Income tax = Taxable income X Tax rate (t)
So,
Net pro昀椀t = Taxable income x (1 - t)
= (OR - OC - CCA - I) x (1 - 琀⤀
(12-1)
吀栀is is the same 昀漀rmula that was developed in the income statement of Figure 12-1,
and it is a 昀甀ndamental relationship that we use later when converting cash flow diagrams
昀爀om be昀漀re-tax to a昀琀er-tax.
When the corporation borrows money (called debt capital), it commits itself to
making interest payments. 吀栀e money is received as a loan, and the lender does not own
any portion of the corporation. Interest payments are not a昀昀ected by the pro昀椀tability
(or lack thereof) of the corporation, but the pro昀椀tability can be very much a昀昀ected by
the interest payments. Borrowing and repayment a昀昀ect the cash 昀氀ow in two ways: (1) by
the actual transfer of 昀甀nds and (2) by changing the amount of taxes paid. Since the
interest is a cost of doing business, it is deducted 昀爀om revenue when taxable income is
calculated.
Recall the di昀昀erent loan rep愀礀ment schedules that we considered in Chapter 3. Even
though we may repay the same amount of money to the bank every month, as time goes
on, a decreasing 昀爀action of that amount will go to interest and an increasing 昀爀action will
go to repayment of debt. 吀栀ese two components of our loan p愀礀ments 昀漀llow di昀昀erent
routes through Figure 12-2: repayment of debt comes out of the net 昀甀nds emerging at the
bottom of the diagram. 吀栀is is 昀甀rther discussed under "Loan Financing," toward the end
of this chapter.
In the pipeline diagram, Figure 12-2,
A昀琀er-tax cash 昀氀ow (ATCF) = Net pro昀椀t + CCA + Debt interest (I)
377
378
Cha pter 12 After-Tax Cash Flows
Substituting, we get
ATCF = Taxable income x (1 - 琀⤀ + CCA + I
ATCF = [BTCF - I - CCA] x (1 - 琀⤀ + CCA + I
ATCF = BTCF( l - t) - I X (1 - 琀⤀ - CCA X (1 - t) + CCA + I
ATCF = [ OR - OC] (l - t) + I X t + CCA X t
(12-2)
ATCF = OR (1 - 琀⤀ - OC(l - t) + I X t + CCA x t
䌀䐀
Corporations are owned by their shareholders. 吀栀e corporation sells shares, and the
shares entitle the buyer to a percentage of ownership of the corporation. 吀栀e money that
the corporation receives as a result of this sale is called equity. Dividends are amounts of
money, a portion of the pro昀椀t, that are paid to the shareholders.
To 昀椀nd the net cash 昀爀om operations,
Net cash 昀爀om operations = OR(l - 琀⤀ - OC (l - t) -1(1 -琀⤀
+ CCA x t - Dividends
(12-3)
Net cash 昀爀om operations = (1 - t) [OR - OC - I] + CCA X t - Dividends
= Net pro昀椀t + CCA - Dividends
(12-4)
But operations are not the only source or use of cash 昀漀r a corporation. Sometimes the
corporation will buy or sell its production equipment, or borrow money, or repay debt, or
issue or repurchase shares. A company's 昀漀unders may sell shares to raise money, at the
cost of diluting their ownership of the company. A昀琀er the company has started to prosper,
they m愀礀 use excess cash to buy those shares back, increasing the 昀爀action of equity that
they own.
In Figure 12-3, a large tub is shown below the pipeline to catch all the cash 昀氀ows and
to store the excess cash. Some companies have very large tubs; 昀漀r example, Apple had
$203 billion in cash reserves in 2015.
吀栀e pipes coming into the tub are sources, and the drainpipes are uses. 吀栀e pipes at
the top show that the net 昀甀nds 昀爀om operations, new equity, new debt, and proceeds 昀爀om
disposal of assets all add to the pool of 昀甀nds 愀瘀ailable to the corporation.
吀栀e drains on 昀甀nds are the repurchase of equity, the repayment of debt, and the pur­
chase of assets. 吀栀us, a 昀椀rm's net cash 昀氀ow is
Net cash 昀氀ow = Net cash 昀爀om operations
+ New equity
+ New debt
+ Proceeds 昀爀om asset disposal
Acco u nt i n g a n d E n g i neeri n g Eco n o my
- Repurchase of equity
- Repayment of debt (principal)
- Purchase of assets
Net Funds 昀爀om
Operations
New Equity
379
(12-5)
Proceeds 昀爀om Disposal of Assets
New Debt
Repurchase
of Equity
Repayment
of Debt
Purchase
of Assets
FIGURE 12-3 Sou rces and uses of cash .
Calcu lating Net Cash Flow
EXAM PLE 12 -4
A construction company (t = 27%) bought a new bulldozer 昀漀r $220,000 (CCA rate
吀栀e expected annual revenues and costs that will be created by the machine are
30%).
Operating cost (OR) = $324,000
Operating cost (OC) = $96,000
Find the net cash 昀氀ow 昀漀r Years 1 and 2.
SO LUTI O N
吀栀ese are the same data as in Example 12-3, and that solution, net pro昀椀t calculation, is shown in italics.
Below the net pro昀椀t line, we continue the table to show the actual cash 昀氀ow. We do this by providing
lines 昀漀r each of the items in Equations 12-4 and 12-5.
吀栀ere is a negative cash 昀氀ow at the start of Year 1 that occurred when we bought the bulldozer.
continued
380
Cha pter 12 After-Tax Cash Flows
Cash Flow at the
Beginning
of Year 1
(Time O)
End of
Year 1
End of
Year 2
CCA
$324,000
96,000
33,000
$324,000
96,000
56, 1 00
Taxable income
Less income tax (27%)
$1 95,000
52,650
$1 71,900
46,413
Net pro昀椀t
$142,350
$125,487
$142,350
33,000
$125,487
56, 100
$175,350
$181,587
oc
ORA
Calculation of Net Cash Flow
Net pro昀椀t
+ CCA
- Dividends
+ New equity
+ Proceeds 昀爀om asset disposal
- Repurchase of equity
- Rep愀礀ment of debt
- Purchase of assets
$220,000
Net Cash Flow
-$220,000
Note: Net pro昀椀t and net cash 昀氀ow can be very di昀昀erent numbers. Because the result of the CCA de­
duction is to reduce the taxable income but not the cash flow, the actual e昀昀ect of a CCA amount d is to
increase the tax flow by an amount t x CCA. 吀栀is is illustrated in the derivation of Equation 12-4, which
shows that the CCA is added to the net pro昀椀t to get the net 昀甀nds.
We can now return to the 昀愀miliar world of cash flow diagrams. In this situation we started with
the 昀漀llowing before-tax cash 昀氀ows:
0
i
$220,000
t
t
$324,000
$324,000
1
2
i
$96,000
i
$96,000
which, a昀琀er taking into account CCA and taxes, produced the 昀漀llowing a昀琀er-tax cash 昀氀ows:
t
$ 175,350
t
$ 181 ,587
0 --- 1 --- 2 ---
i
$220,000
Acco u nt i n g a n d E n g i neeri n g Eco n o my
381
Acq u i ring and Disposing of Assets
䌀䐀
In an ongoing business, c愀瀀ital assets are acquired and disposed of regula爀氀y, and the account­
ing system deals with this by adding them to and subtracting them 昀爀om the asset pool. For an
economic study of a particular project or item of equipment, when an asset is acquired, the cost
basis is paid and becomes part of the initial i瘀팀stment that must meet the MARR criterion.
When an asset is sold or otherwise disposed o昀Ⰰ there must be some reconciliation with the cash
昀氀ow in that 礀攀ar of the project. To do this it is necessary to calculate a net salvage value.
Under Canadian tax rules any loss on disposal or rec愀瀀tured CCA is allocated continu­
ously by the declining-balance mechanism as long as the account exists (theoretically on to
in昀椀nit礀⤀. 吀栀is was illustrated in Example 1 1-6, where we saw the CCA continuing even a昀琀er
the asset had been sold. In the next main section, Capital Tax Factors and Books Open, 昀漀r­
mulas are derived to accommodate this. 吀栀ere昀漀re, the usual assumption made about asset
disposal is that any di昀昀erence between the book value and the disposal price will continue
to be allocated at the regular CCA rate. 吀栀is is referred to as the books-open assumption.
When we are working with spreadsheets, the continuing depreciation is not a conven­
ient assumption because it stretches the calculations 昀愀r into the 昀甀ture. For such cases, it
is a reasonable procedure to calculate the recaptured CCA or loss and apply it to the 昀椀nal
year's income statement, especially when we are dealing with a small adjustment to a dis­
tant estimate. 吀栀is is tantamount to closing the account book on that asset class, and thus
it is known as the books-closed assumption.
吀栀e exception to both of the above assumptions is the situation where there is a c愀瀀ital
gain. A capital gain occurs when an asset is sold 昀漀r more than its cost basis. 吀栀e important
thing about a capital gain is that it is taxed at a di昀昀erent rate 昀爀om regular income. In Canada
the tax on capital gains is only one-half the marginal tax rate. Capital gains tax must be paid
at the time the asset is sold and the gains are realized. (Review Figures 11-7 to 1 1-9.)
Net Sa lvage of La nd-Ca pita l Gain
Land, 昀漀r which CCA is not allowed and which usually increases in value, provides the
least complicated example 昀漀r calculating net salvage values, so we will consider it 昀椀rst.
EXAM PLE 12 - 5
Five years ago, anticipating expansion, the XYZ Company bought the lot next to its current 昀愀ctory 昀漀r
$2,300,400. Over the ensuing period it modi昀椀ed its production system and began to make extensive use
of outsourcing. 吀栀us, despite increasing sales, it 昀漀und it used less space, not more. Consequently it sold
the lot and, a昀琀er paying 昀漀r advertising, legal fees, and commissions, realized a sum of $3,427,958.25. If
the company's marginal tax rate is 27%, what is the net salvage value of the land?
SO LUTI O N
Capital gain = Realized value - Cost basis
= $3,427,958.25 - $2,300,400 = $1,127,558.25
Capital gains tax = t x 0.5 x capital gain
= 27% X 0.5 X $1,127,558.25 = $152,220.36
Net salvage value = Realized value - Capital gains tax
= $3,427,958.25 - $152,220.36 = $3,275.737.89
382
Cha pter 12 After-Tax Cash Flows
Net Sa lvage- Books- Closed Assu m ption
For depreciable assets the situation is more complicated because we must reconcile the CCA
taken with the actual loss in value that occurred. Usually when a depreciable asset is sold,
the money received is less than the original cost basis; there昀漀re, there is no capital gain. In
that case the books-closed situation is one where the di猀瀀osal tax e昀昀ect is considered at the
time of sale. If there is a recapture, there is a tax to be paid; if there is a loss, there is a tax
credit to be received. Both situations can be accommodated in a single 昀漀rmula.
Net salvage value = S + DTE
NSV = S ( l - 琀⤀ + Bi
(12-6)
where
t = marginal tax rate
S = salvage value (net proceeds 昀爀om disposal of the asset)
Ba = book value (UCC) at disposal
DTE = disposal tax e昀昀ect
DTE = t X (Ba - S)
NS V = Net salvage value = a昀琀er-tax net proceeds 昀爀om disposal of an asset
EXAM PLE 12 - 6
As a result of the outsourcing, the XYZ Company auctioned o昀昀 its production equipment (Class 43CCA rate = 30%) 昀漀r $320,000; and its 昀氀eet of trucks (Class 10-CCA rate = 30%) 昀漀r $176,000. 吀栀e cost
basis of the equipment was $1,500,000, and the current UCC is $415,283. 吀栀e cost basis of the trucks
was $480,000, and the current UCC is $98,000. If the company's marginal tax rate is 31%, what is the
net salv最였 value of the equipment and vehicles?
SO LUTI O N
Equipment DTE = t X (Ba - S)
= 31% X ($415,283 - $320,000)
= $29,524
NSV = S + DTE
= $320,000 + $29,524
= $349,524
With the equipment there was a loss on disposal (asset sold 昀漀r less than UCC), which resulted in a tax
credit of $38,1 13, so the actual net salv最였 value is greater than the selling price.
Trucks DTE = t X (Ba - S)
= 31% X ($98,000 - $176,000)
= -$24,180
NSV = S + DTE
= $176,000 + (-$24,180)
= $151,820
Since the trucks sold 昀漀r more than their book value, there was recaptured CCA and consequent tax
liability. 吀栀us the actual net salvage value is less than the selling price.
Acco u nt i n g a n d E n g i neeri n g Eco n o my
383
Occasionally we 昀椀nd situations where there are both capital gains and recaptured
CCA. In these cases, as was outlined in Chapter 1 1, you can only recapture up to the
amount of the original cost basis. Money received above that is capital gain. Example 12-7
considers such a situation.
EXAM PLE 12 -7
The XYZ Company also had a 1973 Aston Martin sedan in mint condition that it had been storing
in a shed on the land. T he car had been bought in 1973 by the company's 昀漀under, who, un昀漀rtu­
nately, was stricken with gout and could not drive it. T hus it had remained parked 昀漀r the last
39 years and was depreciated on the books 昀爀om its original purchase price of $32,000 at a CCA
rate of 30%.
Since the 昀漀under was dead, the current board felt they could safely sell the car and did so at a public
auction. It was bought by an eccentric car collector 昀漀r $225,000. If the company's marginal tax rate is
26.5%, what is the net salv最였 value (NSV) of the Aston Martin?
SO LUTI O N
Use equation 1 1-8 to calculate the current book value (UCC) of the car:
B39 = ucc39 = $32,000(1 - o .3/2)(1 - o .3) 39 -1
= $0.04 � $0
Tax on recapture = t X (B 39 - Cost basis)
= 26.5% X ($0.0 - $32,000) = - $8,480
Capital gains tax = t X 0.5 X (Realized value - Cost basis)
= 26.5% X 0.5 X ($225,000 - $32,000) = $25,572
DTE = 吀愀x on recapture + Capital gains tax
= -$8,480 + -$25,572 = - $34,052
NSV = S + DTE
= $225,000 - $34,052 = $190,948
All the variants can be easily accommodated in an Excel worksheet.
Capital Tax Factors and Books Open
In the normal course of events, as long as a corporation stays in business the books
stay open and the annual CCA calculation has traces of all the assets that have come
and gone. If we consider just one single asset, the basic cash 昀氀ow pattern 昀漀r an asset
depreciated according to the CCA method, including the half-year rule, is illustrated
in Figure 12-4.
吀栀e purchase of an asset 昀漀r an amount P generates an in昀椀nite series of depreciation
deductions that result in positive cash 昀氀ows of tax credits.
384
Cha pter 12 After-Tax Cash Flows
p
FIGURE 12-4 Cash flow pattern of tax cred its from CCA.
吀栀us the tax credits 昀漀rm an in昀椀nite series. 吀栀e present worth of the tax credits also
昀漀rms an in昀椀nite series, and we can calculate the sum as 昀漀llows.
吀栀e present worth of the tax credits resulting 昀爀om an asset of cost basis P is
and calculating the sum of the series, this reduces to
[ꄀ�l[l + ½j
PW = p
i+d
I+i
and so the present worth of the a昀琀er-tax cost of an asset is
PW = P
1 - [ ꄀ� )[ 1 + ½j
i+d
(12-7)
I+i
吀栀e value in brackets is called the capital tax factor and is abbreviated CT F.
吀栀e same logic applies when we dispose of an asset. 吀栀at is, the tax liabilities resulting
昀爀om the sale of an asset 昀漀r an amount S produce the pattern shown in Figure 12-5.
s
S( l - d) 2 dt
S dt
S( l - d) dt
FIGURE 12-5 Salvage value tax e昀昀ect.
S( 1 - d) dt
3
S( l - d)4 dt
S( l - d) s dt
Capita l Tax Facto rs a n d Books Open
385
吀栀e present worth of these cash 昀氀ows can likewise be calculated as 昀漀llows:
Std
S(l - d)td S(l - d) 2 td S(l - d) 3 td
PW = -- + --+ ---+ ---+•··
( 1 - i) 3
(1 + i)
(1 + i) 2
( 1 + i) 4
吀栀ere昀漀re the present worth of the a昀琀er-tax cost of salvage at the end of period n is
PW =
s[1 - �]
i+d
(12-8)
which is called the capital salvage factor and is abbreviated CSP.
Calculating After-Tax Present Worth
We can now use the taxation in昀漀rmation to calculate engineering economy measures­
present worths, annual worths, and rates of return-of a昀琀er-tax cash flows. Present worth
calculations are illustrated in Example 12-8.
EXAM PLE 12 -8
A capital expenditure of $246,000 is made 昀漀r equipment (CCA Class 8, 20% rate). 吀栀e investment is
expected to generate the 昀漀llowing be昀漀re-tax cash 昀氀ows over the next 10 years.
Year
Be昀漀re-Tax Cash
Flow Amount
1
2
3
4
5
6
7
8
9
10
$ 30,000
40,000
50,000
60,000
70,000
80,000
90,000
1 00,000
1 1 0,000
120,000
At the end of 10 礀攀ars the equipment is sold 昀漀r $20,000. 吀栀e marginal tax rate is 35%, and the MARR = 12%.
Find the present 眀漀rth of the investment.
SO LUTI O N
吀栀is problem can be solved in two ways: 昀椀rst, by using the capital tax 昀愀ctors and assuming that the
account books stay open; second, by using a spreadsheet program. 吀栀e advantages and shortcomings of
each method will be discussed a昀琀er the example.
continued
386
Cha pter 12 After-Tax Cash Flows
Solution Using CTFs
吀栀e be昀漀re-tax cash 昀氀ow diagram (values in thousands) is
SV = $20K
+
$120K
$246K
To convert 昀爀om a be昀漀re-tax situation to an a昀琀er-tax situation, Formula 12-2 and the capital tax 昀愀ctor
derivation show us that it is necessary only to
•
•
•
multiply the cost and revenues (be昀漀re-tax cash 昀氀ows) by (1 - t).
multiply the depreciable capital investment amounts by CT F.
multiply the proceeds 昀爀om disposal of capital assets (cash salvage values) by CSF.
䨀开 1 1 l
吀栀ere昀漀re, the corresponding a昀琀er-tax cash flow diagram is
$ 3 0K( l - t)
1
t
$40K ( l - t)
$ 50K ( l - t)
$ 60K( l - t)
$ 70K ( l - t)
开䰀
2 븀 3
4 开䰀 5 䨀开 6
$246K(crF)
And the calculations are as 昀漀llows:
t = 35%
d = 20%
i = 12%
$ 90K( l - t) $ 1 00K( l
$80K ( l - t)
7
8
- t) $ 1 1 0K(
9
SV = $2 0K (csF)
+
$120K( l - t)
l - t)
10
Calculating After-Tax Present Worth
CTF = 1 - [
l+
½l
i+d [ l+i
ꄁ㸀l
l[
= 1 1 - [ 0.35 X 0.20 1 + 0. 12 I 2
0. 12 + 0.20
1 + 0. 12
= 0.7930
387
lI
l
+
= 1 1 - [ 0.35 X 0.20 I
0. 12
= 0.7813
0.20
PW = -$246K (CTF) + $35K (1 - t)(䤀鄀, i, n) + $10K (1 - t)
(PIG, i, n) + $20K (CSF) (PIP, i, n)
= -246 X 0.7930 + 35 X (1 - 0.35) X (䤀鄀, 12%, 10) + 10 X (10.35)
X (PIG, 12%, 10) + 20 X 0.7813 X (PIP, 12%, 10)
= -195,070 + 128,538 + 131,651 + 5,032
= +$70,151 뤀蘀 $70,000
Solution in a Spreadsheet
To analyze this in a spreadsheet it is convenient to assume that the account books are closed and that a
net (a昀琀er-tax) salvage value can be used. 吀栀is can be calculated explicitly, as 昀漀llows:
Data
n=
MARR = i =
A=
10
12%
$ 35,000
DTE =
Equipment P =
S=
d=
t=
$ 10,000
$246,000
$ 20,000
20%
35%
Calculation of Net Salvage
UCC at Year 10 =
Proceeds S =
Loss on di猀瀀osal =
Tax e昀昀ect DTE =
Net salvage = DTE + S =
$29,716
20,000
9,716
3,401
$23,401
continued
388
Cha pter 12 After-Tax Cash Flows
吀栀e spreadsheet tabular 昀漀rmat takes the relationships 昀爀om the corporate cash 昀氀ow pipeline of Figure 12-2.
End of Year
0
BT CF
10,400
3,640
- Income tax
= Net pro昀椀t
= Net 䄀吀CF
2
3
4
5
6
7
8
9
10
$35,000 $45,000 $55,000 $65,000 $75,000 $85,000 $95,000 $105,000 $115,000 $125,000
7,429
24,600 44,280 35,424 28,339 22,671 18,137 14,510 11,608
9,286
- CCA
= Taxable income
+ CCA
= ATCF 昀爀om
operations
Cap investment
+ Net salvage
1
720
252
19,576
6,852
36,661
12,831
52,329
18,315
66,863
23,402
80,490
28,172
93,392 105,714 117,571
32,687 37,000 41,150
6,760
468 12,724 23,830 34,014 43,461 52,319 60,705 68,714 76,421
24,600 44,280 35,424 28,339 22,671 18,137 14,510 11,608
7,429
9,286
$246,000
(246,000)
31,360
44,748
48,148
52,169
56,685
61,598
66,828
72,313
78,000
83,850
23,401
31,360 44,748
48,148
52,169
56,685
61,598
66,828
72,313
78,000 107,251
Using the Excel 一倀Vfunction:
Net AT CF present worth = $70,565
吀栀e di昀昀erence between the two answers ($70,565 - 70,151 = $414) is due to the di昀昀erent assumptions­
books open or books closed. For most engineering economy studies that involve long time periods and
small salvage values, the di昀昀erence is not signi昀椀cant.
Both the spreadsheet tabular calculation and the tax 昀愀ctor method h瘀였 advantages. 吀栀e tax 昀愀ctors are
use昀甀l when a quick feasibility check is desired and the estimates are based on either arithmetic or geometric
series. But if there are discontinuous cash 昀氀ows, such as a m愀樀or revenue or cost item in a particular year, or
a situation when it is necessary to monitor cash and working capital requirements care昀甀lly throughout the
project, then the spreadsheet provides a more complete picture. 吀栀e spreadsheet method is also extremely
use昀甀l when one is doing a "What-if' analysis or experimenting with di昀昀erent methods of 昀椀nancing.
Suppose you 昀椀nd a product that you can make 昀漀r 10¢ and sell 昀漀r 25¢. You borrow money
䌀䐀 and
buy equipment, you 昀椀nd 昀愀ctory space, you enter into contracts with reputable ma­
Working Capital Requirements
terial suppliers, and you hire a motivated and trained work昀漀rce-but still there is more.
You need money to operate.
吀栀e material suppliers want to be paid when they deliver, or at least within the month. For
the workers, p礀씀ay is Friday. 吀栀e landlord expects the rent in advance! And even 昀漀r products
that are stamped out in the thousands, there is still the need to inspect them, package them, in­
瘀攀ntory them, ship them to a retailer, and wait until they are sold to the 昀椀nal consumer and the
money comes back to you. 吀栀ere is a time lag bet眀攀en when money is spent in production and
when money returns 昀爀om sales. To co瘀攀r this time lag it is necessary, 愀琀 the start of an oper­
ation, to i渀樀ect a sum of money into the operation. 吀栀is money is referred to as working capital.
O昀琀en it is necessary to i渀樀ect it only at the start-in the 昀椀rst few months. 吀栀en the
returns 昀爀om sales start coming in and the expenses of today are covered by the cash re­
ceipts resulting 昀爀om the sales of products manu昀愀ctured months earlier. 吀栀is situation of
cash balancing could, barring seasonal 昀氀uctuations, continue 昀漀r the life of the product.
吀栀en, when the product is discontinued and manu昀愀cturing ceases, the money continues
to come in 昀漀r several months as the product in the supply chain is used up.
Working Capital Requirements
389
Examples 12-9 and 12-10 illustrate how to deal with working capital requirements.
I n itial Worki ng Capital
EXAM PLE 12 -9
A company was adding a new product line that needed $80,000 of Class 43 equipment ( CCA rate = 30%)
and initial working capital of $55,000. 吀栀e product would h愀瘀e production costs of $79,000 a year and
annual revenues of $167,000. 吀栀e product would be manu昀愀ctured 昀漀r 昀椀ve years and then discontinued,
and then the working capital would be recovered and the equipment sold 昀漀r $5,000. Find the equiva­
lent uni昀漀rm annual worth with MARR = 10% and t = 29%.
SO LUTI O N
吀栀e a昀琀er-tax diagram shows the working capital going in at Time 0, and coming back at the end of Year 5.
Equipment = $5,000 X CSF
Working Capital = $55,000
r
5 @ $ 167,000 ( 1 - t)
- - - 1 ⴀ⬀ 2 ±:oo± · ⴀ⬀ , ⴀꠀ
Equipment = $80,000 X CTF
Working Capital = $55,000
Now we use the annual worth 昀漀rmula on the a昀琀er-tax cash 昀氀ows.
CTF = 1 - [(0.29 X 0.30)/(0. 1 + 0.30)] (1 .05/1 . 1) = 0.7924
CSF = 1 - [(0.29 X 0.30)/(0.1 + 0.30)] = 0.7825
E䄀가 = - ($80,000 X CTF + $55,000)(A/P, 10%, 5) + ($167,000 - $79,000)(1 - 0.29)
+ ($5,000 X CSF + $55,000)(A/F, 10%, 5)
= -$31,251 + $62,480 + $9,650
= $40,898 뤀蠀 $41,000
I ncreasing Worki ng Ca pita l Req u i rement
EXAM PLE 12 -10
A company was adding a new product line that required $80,000 worth of Class 43 equipment (CCA
rate = 30%) and initial working capital of $55,000. 吀栀e working c愀瀀ital requirement would increase
at 18% a year. 吀栀e product would have production costs of $79,000 a year and annual revenues of
$167,000. 吀栀e product would be manu昀愀ctured 昀漀r 昀椀ve years and then discontinued; then the work­
ing capital would be recovered and the equipment sold 昀漀r $5,000. Find the E䄀가 with MARR = 10%
and t = 29%.
continued
390
Cha pter 12 After-Tax Cash Flows
SO LUTI O N
吀栀e a昀琀er-tax diagram shows the initial working capital going in at Time 0. 吀栀en there are the annual
additions to ensure that there is enough working capital to meet the increasing requirements. Since the
initial working capital remains throughout the project, it is necessary only to add an amount annually
to cover the percentage increase.
Working Capital Requirement Increasing at 18%
Year
Working Capital at
Beginning of Year
1
2
3
4
5
$ 55,000
$ 64,900
$ 76,582
$ 90,376
$106,633
Amount Added at End of Year
$55,000 X 18% = $9,900
($55,000 X 18%)(1 + 18%) = $11,682
($55,000 X 18%)(1 + 18%) 2 = $13,785
($55,000 X 18%)(1 + 18%) 3 = $16,266
Total Available 昀漀r
Ne砀琀 Year
$ 64,900
$ 76,582
$ 90,376
$106,633
$106,633
吀栀e increasing working capital thus 昀漀rms an n - 1 long geometric series as shown on the cash 昀氀ow diagram.
Equipment = $5,000 X CSF
Working Capital = $ 106,6 33
, ꄀ蜀,堀娀 ,堀娀• 堀娀
5 @ $ 167,000 ( 1 - t)
t
$9,900
s ⴀꠀ
5 @ $79,000 ( 1 - t)
t
$ 1 1 ,682
t
$ 1 3 , 785
$ 16,266
Equipment = $80,000 X CTF
Working Capital = $55,000
吀栀e annual equivalent 昀漀rmula on the a昀琀er-tax cash 昀氀ows is the same as in Example 12-9 but with the
addition of a geometric series term 昀漀r 昀漀ur periods and an increased amount of working capital recovered
at the end of Period 5.
CTF = 1 - [(0.29 X 0.30)/(0.1 + 0.30)] (1 .05/1.1) = 0.7924
CSF = 1 - [(0.29 X 0.30)/(0.1 + 0.30)] = 0.7825
EUAW = - [$80,000 X CTF + $55,000 + ($55,000 X 0.18)(P/A, 18%, 10%, 4)] (A/P, 10%, 5%)
+ ($167,000 - $79,000)(1 - 0.40) + ($5,000 X CSF + $106,033)(A/F, 10%, 5)
= -$41,815 + $62,480 + $18,107
= $38,772 $38,800
✀뀀
Loan Financing
Interest, which is money paid 昀漀r the use of money, is an expense of doing business.
吀栀us interest is deducted 昀爀om the be昀漀re-tax income. 吀栀e repayment of the loan princi­
pal, however, is just the returning of money borrowed and so must come 昀爀om the a昀琀er-tax
cash 昀氀ow. Incorporating interest repayment into a capital tax 昀愀ctor 昀漀rmulation is beyond
the scope of this text; here we will deal with interest only in the spreadsheet 昀漀rmulation.
Loa n F i n a n c i n g
391
Loa n with Eq ual Pri nci pa l Repayment
EXAM PLE 12 -11
A company was adding a new product line that required $80,000 of Class 43 equipment (CCA
rate = 30%) and initial working capital of $55,000. 吀栀e product would have production costs of $79,000
a year and annual revenues of $167,000. 吀栀e product would be manu昀愀ctured 昀漀r 昀椀ve years and then
discontinued, the working capital would be recovered, and the equipment would be sold 昀漀r $5,000. To
assist in 昀椀nancing the project, the company is borrowing $100,000 at 12% interest. 吀栀e loan interest is
to be paid yearly, and the principal is to be repaid in 昀椀ve equal annual payments. Find the equivalent
uni昀漀rm annual worth if MARR = 10% and t = 29%.
SO LUTI O N
吀栀e data are the same a s i n Example 12-10, s o we can use that spreadsheet, with the addition of a line in
the income statement 昀漀r the loan interest and a line in the "sources and uses" portion 昀漀r the loans and
repayment amounts. 吀栀ese changes are shown below in bold昀愀ce type.
Data
n=
MARR = i =
Loan interest =
R=
C=
WORKING CAPI吀䄀L = WC =
Equipment P =
S=
d=
t=
0
5
10%
12%
$167,000
$ 79,000
$ 55,000
$ 80,000
$ 5,000
30%
27%
1
2
3
4
5
Revenue
- Costs
- CCA
Loan interest
$167,000
79,000
12,000
12,000
$167,000
79,000
20,400
9,600
$167,000
79,000
14,280
7,200
$167,000
79,000
9,996
4,800
$167,000
79,000
6,997
2,400
= Taxable income
- Income tax
76,000
22,040
67,600
19,604
73,720
21,379
78,004
22,621
81,003
23,491
= Net pro昀椀t
+ CCA
53,960
12,000
47,996
20,400
52,341
14,280
55,383
9,996
57,512
6,997
65,960
68,396
66,621
65,379
64,509
(20,000)
(20,000)
(20,000)
(20,000)
(20,000)
8,285
55,000
45,960
43,396
46,621
45,379
107,794
End of Year
= ATCF 昀爀om operations
- Cap investment
Loan (repayment)
+ Net salvage
WC =
= Net 䄀吀CF
$(80,000)
100,000
(55,000)
(35,000)
continued
392
Cha pter 12 After-Tax Cash Flows
0
1
2
3
4
5
$ 80,000
100,000
3,284.77
8,284.77
$68,000
80,000
$47,600
60,000
$33,320
40,000
$23,324
20,000
$16,327
End of Year
UCC of equipment
Outstanding principal
Disposal tax e昀昀ect =
Net salvage =
Using the Excel NPV and PMT昀甀nctions on the Net ACTF row:
Present worth = $ 1 79,732
EUAW = $47,41 3
Loa n with Eq ual An nual Repayment
EXAM PLE 12 -12
A company was adding a new product line that required $80,000 of Class 43 equipment ( CCA rate = 30%)
and initial working capital of $55,000. 吀栀e product would have production costs of $79,000 a year and
annual revenues of $167,000. 吀栀e product would be manu昀愀ctured 昀漀r 昀椀ve years and then discontinued,
the working capital would be recovered, and the equipment would be sold 昀漀r $5,000. To assist in 昀椀nancing
the project, the company is borrowing $100,000 at 12% interest. 吀栀e loan is to be repaid in 昀椀ve equal annual
payments. Find the E䄀가 if MARR = 10% and t = 29%.
SO LUTI O N
吀栀 e capital recovery 昀愀ctor i s used to calculate the payment amount.
Equal annual payment = $100,000(A/P, 12%, 5)
= $27,741
Data
n=
MARR = i =
Loan in琀攀rest =
R=
C=
WC =
Equipment P =
S=
d=
t=
5
10%
12%
$167,000
$ 79,000
$ 55,000
$ 80,000
$ 5,000
30%
29%
A repayment schedule is then used to calculate the annual principal and interest payments:
Year
Amount Owing
at Start of Year
1
2
3
4
5
$100,000
84,259
66,629
46,884
24,769
Interest
Principal
Rep愀礀ment
Amount Owing
at End of Year
$12,000
10,111
7,995
5,626
2,972
$15,741
17,630
19,745
22,115
24,769
$84,259
66,629
46,884
24,769
0
Loa n F i n a n c i n g
1
2
3
4
5
Revenue
- Costs
- CCA
Loan interest
= Taxable income
- Income tax
$167,000
79,000
12,000
12,000
76,000
22,040
$167,000
79,000
20,400
10,1 1 1
67,600
19,604
$167,000
79,000
14,280
7,995
73,720
21,379
$167,000
79,000
9,996
5,626
78,004
22,621
$167,000
79,000
6,997
2,972
81,003
23,491
+ CCA
53,960
12,000
47,996
20,400
52,341
14,280
55,383
9,966
57,512
6,997
65,960
68,396
66,621
65,379
64,509
- 15,741
- 1 7,630
- 19,745
-22, 1 1 5
-24,769
8,285
55,000
50,219
50,766
46,876
43,264
103,025
0
End of Year
= Net pro昀椀t
䄀吀CF 昀爀om operations
- Cap investment
Loan (rep愀礀ment)
+ Net salvage
Working capital = WC =
= Net AT CF
-$80,000.00
100,000.00
55,000.00
(35,000.00)
Using the Excel 一倀V and PMTfunctions on the Net ATCF row:
Present worth = $ 181,348
EUAW = $47,839
Estimating the After-Tax Rate of Return
Example 12-12 is the same problem as Example 12-9 except that it uses borro眀攀d money to
partially 昀椀nance the project. 吀栀e e昀昀ect of 昀椀nancing was to change the EUAW 昀爀om $41,033 to
$47,839. 吀栀is increase in value results even though the money is borrowed at a rate greater than
the MARR value (12% 瘀攀rsus 10%)! 吀栀at is because the interest 眀退ich is the cost of borrowing
the mone礀⤀ is deductible 昀爀om revenue be昀漀re the taxes are calcul愀琀ed, so the actual rate of
borrowing the money is reduced by (1 - 琀⤀. In this instance, 12%(1 - 29%) = 8.5% is the cost
of borrowing the money. 吀栀e equation relating the be昀漀re- and a昀琀er-tax cost of debt capital is
(12-9)
where
= be昀漀re-tax cost t of debt capital
t = marginal tax rate
idt = a昀琀er-tax cost t of debt capital
id
吀栀ere is no shortcut to computing the a昀琀er-tax rate of return 昀爀om the be昀漀re-tax rate
of return. One possible exception is when non-depreciable assets and 昀椀nancing are repaid
in a lump sum at the end of the project. In this special case, we h愀瘀e
A昀琀er-tax rate of return = (1 - Marginal tax rate) X (Be昀漀re-tax rate of return)
吀栀is relationship may be help昀甀l 昀漀r selecting a trial a昀琀er-tax rate of return when the
be昀漀re-tax rate of return is known. It must be emphasized, however, that this relationship
is almost alw愀礀s only a rough approximation.
393
394
Cha pter 12 After-Tax Cash Flows
SUMMARY
Since income taxes are part of most situations, no realistic economic analysis can ignore
their consequences. Income taxes make the government a partner in many business ven­
tures. 吀栀us the government bene昀椀ts 昀爀om all pro昀椀table ventures and shares in the losses of
unpro昀椀table ventures.
Individuals pay income taxes to both the federal and provincial governments. 吀栀e
personal federal tax is based on a progressive scale in which the higher the taxable income,
the higher the marginal tax rate.
For corporations, taxable income equals gross income minus all ordinary and ne­
cessary expenditures (except capital expenditures) and depreciation and depletion char­
ges. 吀栀e income tax computation (whether 昀漀r an individual or a corporation) is relatively
straight昀漀rward; it consists of multiplying the taxable income by the tax rate. 吀栀e proper
rate to use in an economic analysis is the marginal tax rate applicable to the increment of
taxable income being considered.
To introduce the e昀昀ect of income taxes into an economic analysis, the starting point is
a be昀漀re-tax cash flow. 吀栀en the depreciation schedule and interest p愀礀ments are deducted
昀爀om appropriate parts of the be昀漀re-tax cash 昀氀ow to obtain taxable income. Income taxes
are obtained by multiplying taxable income by the proper tax rate. Be昀漀re-tax cash 昀氀ow
minus income taxes equals the a昀琀er-tax cash 昀氀ow.
Accounting statements show pro昀椀t and loss, not the timing of cash flows. 吀栀us it is
necessary 昀漀r an engineering economist to understand the accounting statements, but the
in昀漀rmation contained in them must be interpreted. 吀栀e economy studies need to consider
timing; to convert between systems, use the 昀漀llowing equation:
Net cash 昀氀ow = Net cash 昀爀om operations
+ New equity
+ New debt
+ Proceeds 昀爀om asset disposal
- Repurchase of equity
- Repayment of debt (principal)
- Purchase of assets
Working capital also a昀昀ects the cash flow and cash requirements and must be
included.
吀栀e process of CCA calculations is such that assets depreciate 昀漀rever. To make the prob­
lems tractable, we must make some assumptions 昀漀r the tax implications of asset disposal.
吀栀e books-open assumption is easiest 昀漀r manual calculation; the books-closed is easiest
昀漀r spreadsheets. 吀栀e error introduced by using one instead of the other is usually minimal.
When dealing with non-depreciable assets, there is a nominal relationship between
be昀漀re-tax and a昀琀er-tax rate of return. It is
A昀琀er-tax rate of return = (1 - Tax rate)(Be昀漀re-tax rate of return)
吀栀ere is no simple relationship between be昀漀re-tax and a昀琀er-tax rate of return in the
more usual case of investments involving depreciable assets.
Problems
395
PROBLEMS
and necessary" expenses and depreciation have
been deducted. At present the business is oper­
ated as a proprietorship; that is, Jane pays per­
sonal federal income tax on the entire $90,000.
For tax purposes, it is as if she had a job that
paid her a salary of $90,000 a year.
As an alternative, Jane is considering in­
corporating the business. If she does, she will
pay herself a salary of $30,000 a year 昀爀om the
corporation. 吀栀e corporation will then pay
taxes on the remaining $60,000 at the small
business rate (federal is 1 1% and Ontario is
4.5%) and retain the balance of the money as
a corporate asset. 吀栀us Jane's two alternatives
are to operate the business as a proprietor­
ship or as a corporation. Jane is single and has
non-re昀甀ndable tax credits: $12,000 federal
and $9,500 provincial. Which alternative will
result in a smaller total payment of taxes to the
government?
吀栀ese problems can be solved by hand, but most can
be solved much more easily with a spreadsheet.
Individual Taxes
12-1
⠀　12-2
12-3
⠀　An unmarried taxpayer with no dependents
expects a taxable income of $62,000 in a given
year. His non-re昀甀ndable tax credits are ex­
pected to be $12,000.
(a) What will his federal income tax be?
(b) He is considering an additional activity
expected to increase his taxable income. If
this increase should be $16,000 and there
should be no change in deductions or
non-re昀甀ndable tax credits, what will the
increase be in his federal income tax?
Mary has a $65,000 adjusted gross income
and $12,000 federal and $18,000 provincial
non-re昀甀ndable tax credits. Compute the total
tax she would pay as a resident of
(a) Manitoba
(b) Ontario
Bill, a Manitoban, worked during school and
during the 昀椀rst two months of his summer
vacation. A昀琀er 昀愀ctoring in his deductions,
he 昀漀und that he had a total taxable income of
$7,800 and non-re昀甀ndable tax credits ($10,500
federal and $8,600 provincial). Bill's employer
wants him to work another month during the
summer, but Bill had planned to spend the
month hiking. If an additional month's work
would increase Bill's taxable income by $2,600,
how much more money would he have a昀琀er
paying the income tax?
12-4
Do Problem 12-3 as if Bill were a resident of
Ontario.
12-5
Jane lives in Ontario and operates a manage­
ment consulting business. 吀栀e business has
been success昀甀l and now produces a taxable
income of $90,000 a year a昀琀er all "ordinary
⠀　12-6
Do Problem 12-5 as if Jane were a resident
of Ontario, where the small business tax rate
is 0%.
12-7
An unmarried woman in British Columbia
with a taxable income of about $100,000 has a
federal incremental tax rate of 26% and a prov­
incial incremental tax rate of 12.29%. What is
her combined incremental tax rate?
⠀　12-8
A $10,000 commercial bond that has a 10%
bond rate and matures in 昀椀ve years can be
bought 昀漀r $8,000. Interest is paid at the end
of each year 昀漀r the next 昀椀ve years. Find the
annual a昀琀er-tax rate of return of this invest­
ment. Assume a 29% tax rate applies.
12-9
A 30-year-old buys, 昀漀r $8,000, an $8,000
Registered Retirement Savings Plan (RRSP)
bond that will pay 5% ($400 per annum) into
the plan 昀漀r the next 40 years and then be re­
deemed 昀漀r $8,000 plus accumulated interest.
吀栀e payments go into the RRSP and also earn
396
Chapte r 12 Afte r - Tax Cash Flows
5% a year. Because it is an RRSP , the purchaser
receives an immediate tax rebate of $8 ,000
times the marginal tax rate , and p愀礀s no taxes
on the annual interest earned . At age 70 , when
the RRSP is redeemed , taxes must be paid on
both the original $8,000 and the accumulated
interest. I f your marginal tax rate is 36% when
you are 30, and 32% when 礀漀u are 70, what is
the rate of return on th is investment?
Corporate Taxes
12- 10 A company wants to set up a new o昀케ce in a
country where the corpo爀愀te tax 爀愀te is as
fo氀氀ows: 15% of first $50 ,000 p爀漀fits, 25% of
next $25,000, 34% of next $25,000, and 39%
of everything over $㄀　0 ,000. Executives es­
timate that they will h愀瘀e g爀漀ss revenues of
$500,000, total costs of $300 ,000, $30 ,000 in
a氀氀owable tax deductions, and a one -time busi­
ness start-up credit of $8 ,000. What is taxable
income for the first year, and how much should
the company expect to pay in taxes?
0
12- 11 Wo爀氀dWide Oil Company purchased two large
compressors for $125 ,000 each. One com­
pressor was insta氀氀ed in the firm 's Alberta
refinery and is being depre挀椀ated by the CCA
method. The other compressor was placed in
the Oklahoma refinery, where it is being de­
pre挀椀ated by sum-of-years ' -digits depre挀椀ation
with zero salvage value. Ass ume the company
pays federal income taxes each year and the
tax rate is constant. The corpo爀愀te acco unting
department noted that the two compressors
are being depre挀椀ated differen琀氀y and won­
ders whether the corporation will wind up
paying more income taxes over the life of the
equipment as a re sult of th is . What do you tell
them?
0
12-12 Nova Scotia has a corp orate tax rate of 16% of
taxable income. The federal rate is 1 5% . If a
corporation has a taxable income of $ 1 50,000,
what is the total provincial and federal income
tax it must p ay? Comp ute its combined incre ­
mental provincial and 昀攀deral tax rates.
12-13
To inc rease its ma rket sha re , Sole B r other Inc .
de挀椀ded to bo r r ow $50 ,000 from its banke r 昀漀r
the pu rchase of newspaper adve rtising 昀漀r its
shoe reta椀氀 line . The loan is to be paid in 昀漀ur
equal annua p愀礀ments with 15% inte rest . 吀栀e
l
loan is discounted six points . 吀栀e six "points "
a re an additio nal inte rest charge of 6% of the
loan, deducted immediately . 吀栀is additional
interest (6% [50 ,000
$3 ,000) means the
]
actual amount received 昀爀om the $50 ,000 loan
is $47, 000 . The $3 ,000 additional interest may
be deducted as 昀漀ur $750 additional annual in­
terest p愀礀ments. What is the after -tax interest
爀愀te on this loan?
=
12-14 The L ynch Bull investment company suggests
that Steve, a wea氀琀栀礀 investor (his incremental
income tax 爀愀te is 40%) , consider the 昀漀氀氀owing
investment.
Buy corpo爀愀te bonds on the New 夀漀rk
Stock Exchange with a 昀愀ce value (par value)
of $㄀　0 ,000 and a 5% coupon rate (the bonds
pay 5% of $㄀　0,000, which equals $5, 000 inter­
est per year) . These bonds can be purchased at
their present market value of $75, 000. At the
end of each year, Steve will receive the $5, 000
interest, and at the end of five years, when the
bonds mature, he will receive $㄀　0, 000 plus the
last $5, 000 of interest.
Steve will pay for the bonds by borr o w­
ing $50, 000 at ㄀　% interes t for five years. 吀栀e
$5, 000 interes t paid on the loan each year
will equal the $5, 000 of interes t income from
the bonds. As a res ult Ste ve will ha ve no ne t
taxable income during the f ive years because
of this bond p urchase and borr o wing mone y
scheme . At the end of five years, Ste ve will re ­
ceive $ 1 0 0, 0 0 0 plus $ 5 , 0 0 0 interes t 昀爀om the
b onds and will repay the $ 5 0, 0 0 0 lo an and
pay the last $ 5 , 0 0 0 intere s t . 吀栀e net re s ult is
that he will have a $ 2 5 , 0 0 0 capital ga i n ; that
i s , he will re c eive $ 1 0 0 , 0 0 0 昀爀om a $ 75 , 0 0 0 in­
ve stment. (No te : This sit uation repres ents a n
actual recommendation of a br oker a ge 昀椀rm .)
愀儀 Comp ute Ste ve's after- tax r ate of return
on this dual bond- plus -loan in vestment
packa ge.
Problems
(b) What would Steve's a昀琀er-tax rate of return
be if he purchased the bonds 昀漀r $75,000
cash and did not borrow the $50,000?
H istorical Depreciation
12-15 Albert decided to buy an old duplex as an in­
vestment. A昀琀er looking 昀漀r several months, he
昀漀und a desirable duplex that could be bought
昀漀r $300,000 cash. He decided that he would
rent both sides of the duplex and determined
that the total expected income would be
$1,500 a month. 吀栀e total monthly expenses
昀漀r property taxes, repairs, gardening, and so
昀漀rth are estimated at $750. For tax purposes,
Al plans to depreciate the building by the cap­
ital cost allowance method at a 10% rate and
assumes that the building has a 20-year re­
maining li昀攀 and no salvage value. Of the total
$300,000 cost of the property, $250,000 repre­
sents the value of the building and $50,000 is
the value of the lot. Assume that Al is in the
38% incremental income tax bracket (com­
bined provincial and federal taxes) through­
out the 20 years.
In this analysis Al estimates that the
income and expenses will remain constant at
their present levels. If he buys and holds the
property 昀漀r 20 years, what a昀琀er-tax rate of
return can he expect to receive on his invest­
ment, using the 昀漀llowing assumptions?
(a) Al believes the building and the lot can be
sold at the end of 20 years 昀漀r the $50,000
estimated value of the lot.
(b) A more optimistic estimate of the 昀甀ture
value of the lot is that the property can be
sold 昀漀r $380,000 at the end of 20 years.
12-16 Zeon, a large, pro昀椀table corporation, is con­
sidering adding some automatic equipment
to its production 昀愀cilities. An investment of
$120,000 will produce an initial annual bene­
昀椀t of $29,000, but the bene昀椀ts are expected to
decline by $3,000 a year, making second-year
bene昀椀ts $26,000, third-year bene昀椀ts $23,000,
and so 昀漀rth. If the 昀椀rm uses sum-of-years' digits depreciation, an eight-year use昀甀l life,
397
and $12,000 salvage value, will it obtain the
desired 6% a昀琀er-tax rate of return? Assume
that the equipment can be sold 昀漀r its $12,000
salvage value at the end of the eight years.
Also assume a 46% income tax rate 昀漀r prov­
incial and federal taxes combined.
12-17 A group of business people 昀漀rmed a corpor­
ation to lease a piece of land 昀漀r 昀椀ve years at
the intersection of two busy streets. 吀栀e cor­
poration has invested $50,000 in car-washing
equipment. 吀栀ey will depreciate the equipment
by sum-of-years' -digits depreciation, assuming
a $5,000 salvage value at the end of the 昀椀ve­
year use昀甀l life. 吀栀e corporation is expected to
have a be昀漀re-tax cash 昀氀ow, a昀琀er meeting all
expenses of operation (except depreciation),
of $20,000 the 昀椀rst year, declining $3,000 per
year in 昀甀ture years (second year = $17,000,
third year = $14,000, etc.). 吀栀e corporation has
other income and is taxed at a combined cor­
porate tax rate of 20%. If the projected income
is correct, and the equipment can be sold 昀漀r
$5,000 at the end of 昀椀ve years, what a昀琀er-tax
rate of return would the corporation receive
昀爀om this venture?
12-18 A 昀椀rm is considering the 昀漀llowing investment
project:
Year
0
1
2
3
4
5
Be昀漀re-Tax Cash
Flow (thousands)
-$1,000
+soo
+340
+244
+100
+100
+ 125 salvage value
吀栀e project has a 昀椀ve-year use昀甀l life with
a $125,000 salvage value, as shown. Double­
declining-balance depreciation will be used,
assuming the $125,000 salvage value. 吀栀e
income tax rate is 34%. If the 昀椀rm requires a
10% a昀琀er-tax rate of return, should the project
be undertaken?
398
Cha pter 12 After-Tax Cash Flows
12-19 吀栀e Shellout Corp. has a 34% tax rate and
owns a piece of petroleum-drilling equipment
that costs $100,000 and will be depreciated at a
CCA rate of 30%. Shellout will lease the equip­
ment to others and each year receive $30,000
in rent. At the end of 昀椀ve years, the 昀椀rm will
sell the equipment 昀漀r $35,000. What is the af­
ter-tax rate of return Shellout will receive 昀爀om
this equipment investment?
will remain unchanged. 吀栀e loan will be repaid
by 昀漀ur equal end-of-year p愀礀ments of $25,240.
Prepare an expanded cash flow table that takes
into account both the special tools and the loan.
(a) Compute the a昀琀er-tax rate of return 昀漀r the
tools, taking into account the $80,000 loan.
(b) Explain why the rate of return obtained
in part (a) is di昀昀erent 昀爀om that obtained
in Problem 12-21.
⠀　Hints:
1. Interest on the loan is 10%, $25,240 = 80,000
(A!P, 10%, 4). Each payment is made up of both
interest and principal. 吀栀e interest portion 昀漀r
any year is 10% of the balance due at the begin­
ning of the year.
2. Interest payments are tax-deductible (i.e.,
they reduce taxable income and thus taxes
paid). Principal p愀礀ments are not. Separate
each $25,240 payment into interest and princi­
pal portions.
3. 吀栀e Year O cash flow is $20,000 (100,000
- 80,000).
4. A昀琀er-tax cash 昀氀ow will be be昀漀re-tax cash
flow - interest payment - principal payment
- taxes.
12-20 A mining corporation purchased $120,000
worth of production machinery and depreci­
ated it by SOYD depreciation, a 昀椀ve-year de­
preciable life, and zero salvage value. 吀栀e
corporation is a pro昀椀table one that has a 34%
incremental tax rate. At the end of 昀椀ve years
the mining company changed its method of
operation and sold the production machinery
昀漀r $40,000. During the 昀椀ve years the machin­
ery was used, it reduced mine operating costs
by $32,000 a year, be昀漀re taxes. If the company
MARR is 12% a昀琀er taxes, was the investment
in the machinery a satis昀愀ctory one?
12-21 An automobile manu昀愀cturer is buying some
special tools 昀漀r $100,000. 吀栀e tools are being
depreciated by using double-declining-balance
depreciation, a 昀漀ur-year depreciable life, and
a $6,250 salvage value. It is expected that the
tools will actually be kept in service 昀漀r six
years and then sold 昀漀r $6,250. 吀栀e be昀漀re-tax
bene昀椀t of owning the tools is as 昀漀llows:
⠀　Year
Be昀漀re-Tax Cash Flow
1
$30,000
30,000
35,000
40,000
10,000
10,000
$6,250 selling price
2
3
4
5
6
Compute the a昀琀er-tax rate of return 昀漀r this
investment situation, assuming a 46% incre­
mental tax rate.
12-22 吀栀is is a continuation of Problem 12-21. Instead
of paying $100,000 cash 昀漀r the tools, the cor­
poration will p愀礀 $20,000 now and borrow the
remaining $80,000. 吀栀e depreciation schedule
12-23 A project will require the i瘀팀stme渀琀 of $108,000 in
equipment (sum-o�鈀ars'-digits depreciation with
a depreciable life of eig栀琀 years and zero sa瘀촀ge
value) and $25,000 in r愀眀 m愀琀erials (not deprecia­
ble). 吀栀e annual project income a昀琀er all e砀瀀enses
e砀挀ept depreci愀琀ion h瘀였 been paid is projected to
be $24,000. 䄀琀 the end of eight years the project
will be discontinued and the $25,000 i瘀팀stment
in r愀眀 materials will be rec瘀퐀red. Assume a 34%
income tax rate 昀漀r this corporation. 吀栀e corpor­
ation wants a 15% a昀琀er-tax rate of return on its
i瘀팀stments. Determine by present 眀漀rth analysis
眀栀ether this project should be undertaken.
⠀　12-24 A salad oil bottling plant can either buy caps 昀漀r
the glass bottles at 5¢ each or install $500,000
worth of plastic moulding equipment and
manu昀愀cture the caps at the plant. 吀栀e manu­
昀愀cturing engineer estimates that the material,
labour, and other costs would be 3¢ per cap.
(a) If 12 million caps a year are needed and
the moulding equipment is installed,
what is the payback period?
⠀　Problems
(b) 吀栀e plastic moulding equipment would be
depreciated by straight-line depreciation
with a 昀椀ve-year use昀甀l life and no salvage
value. Assuming a 40% income tax rate,
what is the a昀琀er-tax payback period, and
what is the a昀琀er-tax rate of return?
12-25 A 昀椀rm manu昀愀ctures padded shipping bags.
A cardboard carton should contain 100 bags,
but since the machine operators 昀椀ll the card­
board cartons by eye, the cardboard carton
m愀礀 contain anywhere 昀爀om 98 to 123 bags
(average = 105.5 bags).
Management realizes that they are giving
away 5.5% of their output by over昀椀lling the
cartons. 吀栀e solution would be to weigh each
昀椀lled shipping carton. Underweight cartons
would have additional shipping bags added,
and overweight cartons would have some ship­
ping bags removed. By weighing, management
believes that the average number of bags per
carton could be reduced to 102, and almost no
cartons would contain fewer than 100 bags. 吀栀e
weighing equipment would cost $18,600. 吀栀e
equipment would be depreciated by straight­
line depreciation, with a 10-year depreciable
life and a $3,600 salvage value at the end of
10 years. 吀栀e equipment quali昀椀es 昀漀r a 10% in­
vestment tax credit. One person, hired at a cost
of $16,000 a year, would be needed to operate
the weighing equipment and to add or remove
padded bags 昀爀om the cardboard cartons. Each
year 200,000 cartons would be checked on the
weighing equipment, with an average removal
of 3.5 padded bags per carton, with a manu­
昀愀cturing cost of 3¢ a bag. 吀栀is large pro昀椀table
corporation has a 50% combined incremental
tax rate. Assume a 10-year study period 昀漀r
the analysis and an a昀琀er-tax MARR of 20%.
Compute
(a) the a昀琀er-tax present worth
(b) the a昀琀er-tax internal rate of return
挀儀 the a昀琀er-tax simple payback period
⠀　12-26 ACDC Company is considering the installa­
tion of a new machine that costs $150,000. 吀栀e
machine is expected to lead to net income of
$44,000 per year 昀漀r the next 昀椀ve years. Using
straight-line depreciation, no salvage value,
and an e昀昀ective income tax rate of 29%, 昀椀nd
399
the a昀琀er-tax rate of return 昀漀r this investment.
If the company's a昀琀er-tax MARR rate is 12%,
would this be a good investment or not?
12-27 Fast Construction Equipment Rentals (FCER)
purchases a new 5-tonne-rated crane 昀漀r rental
to its customers. 吀栀is crane costs $1,125,000
and is expected to last 昀漀r 25 years, at which
time it will have an expected salvage value of
$147,000. FCER earns $195,000 be昀漀re-tax cash
昀氀ow each year in rental income 昀爀om this crane;
its total taxable income each year is between
$10M and $ISM, and the marginal tax rate is
31%. If FCER uses straight-line depreciation
and a MARR of 15%, what is the present worth
of the a昀琀er-tax cash 昀氀ow 昀漀r this equipment?
Should the company invest in this crane?
12-28 A computer-controlled milling machine will
cost Ajax Manu昀愀cturing $65,000 to purchase
plus $4,700 to install.
(a) If the machine will have a salvage value
of $6,600 at EOY 20, how much can
Ajax charge annually to depreciation of
this equipment? Ajax uses straight-line
depreciation.
( b) What is the book value of the machine at
EOY 3?
12-29 A 昀愀rmer bought a new harvester 昀漀r $120,000.
吀栀e operating expenses 愀瘀eraged $10,000 a
year, but the harvester saved the 昀愀rmer $40,000
a year in labour costs. It was depreciated over
a life of 昀椀ve years by the SOYD method, as­
suming a salvage value of $30,000. 吀栀e 昀愀rmer
sold the harvester 昀漀r only $10,000 at the end of
the 昀椀昀琀h year. Given an income tax rate of 31%
and a MARR of 5% per year, determine the af­
ter-tax net present worth of this investment.
Ca pital Cost Allowance
12-30 Sam, a success昀甀l Alberta businessman, is
considering erecting a small building on a
commercial lot. A local 昀甀rniture company
is willing to lease the building 昀漀r $9,000 per
year, paid at the end of each year. It is a net
lease, which means the 昀甀rniture company
must also pay the property taxes, 昀椀re insur­
ance, and all other annual costs. 吀栀e 昀甀rniture
400
0
Cha pter 12 After-Tax Cash Flows
company will require a 昀椀ve-year lease with an
option to buy the building and land on which it
stands 昀漀r $125,000 a昀琀er 昀椀ve years. Sam could
have the building constructed 昀漀r $82,000. He
could sell the commercial lot now 昀漀r $30,000,
the same price he paid 昀漀r it. Sam currently has
an annual taxable income 昀爀om other sources
of $123,000. He would depreciate the commer­
cial building at a CCA rate of 4%. He believes
that at the end of the 昀椀ve-year lease he could
easily sell the property 昀漀r $125,000 ($30,000
昀漀r the lot and $90,000 昀漀r the building). What
is the a昀琀er-tax present worth of this 昀椀ve-year
venture if Sam uses a 10% a昀琀er-tax MARR?
12-31 One January, Geraldine bought a small house
and lot 昀漀r $99,700. She estimated that $9,700 of
this amount represented the value of the land.
She rented the house 昀漀r $6,500 a year during
the 昀漀ur years she owned it. Expenses 昀漀r prop­
erty taxes, maintenance, and so 昀漀rth were
$500 a year. For tax purposes the house was
depreciated at a CCA rate of 10%. At the end of
昀漀ur years the property was sold 昀漀r $105,000
($90,000 昀漀r the house and $15,000 昀漀r the
land). Geraldine is married and works as an
engineer. She estimates that her incremental
combined tax rate is 36%. What a昀琀er-tax rate
of return did she obtain on her investment?
12-32 A corporation with a 34% income tax
rate is considering the 昀漀llowing investment
in research equipment, and has projected the
bene昀椀ts as 昀漀llows:
Year
0
1
2
3
4
5
6
Be昀漀re-T愀砀 Cash Flow
- $5,000,000
+200,000
+800,000
+ 1 ,760,000
+ 1 ,376,000
+576,000
+288,000
Prepare an a昀琀er-tax cash flow table assuming a
CCA rate of 30%.
(a) What is the a昀琀er-tax rate of return?
(b) What is the be昀漀re-tax rate of return?
0
12-33 An engineer is working on the layout of a new
research and experimentation 昀愀cility. Two
plant operators will be required. If, however,
an additional $100,000 worth of instrumenta­
tion and remote controls were added, the plant
could be run by a single operator. 吀栀e total be­
昀漀re-tax cost of each plant operator is projected
to be $35,000 per year. 吀栀e instrumentation
and controls will be depreciated at a CCA rate
of 20%. If this corporation (the corporate tax
rate is 34%) invests in the additional instru­
mentation and controls, how long will it take
昀漀r the a昀琀er-tax bene昀椀ts to equal the $100,000
cost? In other words, what is the a昀琀er-tax pay­
back period?
12-34 A special power tool 昀漀r plastic products costs
$400,000 and has a 昀漀ur-year use昀甀l life, no sal­
vage value, and a two-year be昀漀re-tax payback
period. Assume uni昀漀rm annual end-of-year
bene昀椀ts.
(a) Compute the be昀漀re-tax rate of return.
(b) Compute the a昀琀er-tax rate of return,
based on a CCA rate of 100% and a 34%
corporate income tax rate.
12-35 A pro昀椀table wood products corporation is con­
sidering buying a parcel of land 昀漀r $50,000,
building a small 昀愀ctory at a cost of $200,000
( CCA rate = 4%), and equipping it with $150,000
worth of machinery (CCA rate = 30%). Assume
the plant is put in service 1 October. 吀栀e be­
昀漀re-tax net annual bene昀椀t 昀爀om the project is
estimated at $70,000 a year. 吀栀e analysis period
is to be 昀椀ve 礀攀ars, and the planners assume the
total property (land, building, and machiner礀⤀
will be sold at the end of 昀椀ve years, also on 1 Oc­
tober, 昀漀r $328,000 ($50,000 昀漀r land, $166,470
昀漀r building, and $111,530 昀漀r equipment). Com­
pute the a昀琀er-tax cash flow at a 34% income
tax rate. If the corporation's criterion is a 15%
a昀琀er-tax rate of return, should it proceed with
the project?
0
12-36 A small vessel was purchased by a chemical
company 昀漀r $55,000 and was to be depreci­
ated at a CCA rate of 15%. Its requirements
changed suddenly, and the chemical company
Problems
leased the vessel to an oil company 昀漀r six
years at $10,000 a year. 吀栀e lease also pro­
vided that the vessel could be purchased at
the end of six years by the oil company 昀漀r
$35,000. At the end of the six years, the oil
company exercised its option and bought the
vessel. 吀栀e chemical company has a 34% in­
cremental tax rate. Compute its a昀琀er-tax rate
of return on the vessel.
12-37 Xon, a small oil company, bought a new petrol­
eum-drilling rig 昀漀r $1,800,000. Xon will de­
preciate the rig using a 39% CCA rate. 吀栀e rig
has been leased to a drilling company, which
will pay Xon $450,000 a year 昀漀r eight years. At
the end of eight years the rig will belong to the
drilling company. If Xon has a 34% incremen­
tal tax rate and a 10% a昀琀er-tax MARR, does
the investment appear to be satis昀愀ctory?
12-38 吀栀e pro昀椀table Palmer Golf Cart Corp. is con­
sidering investing $300,000 in special tools 昀漀r
some of the plastic golf cart components. Exec­
utives of the company believe the present golf
cart model will continue to be manu昀愀ctured
and sold 昀漀r 昀椀ve years, a昀琀er which a new cart
design will be needed, together with a di昀昀er­
ent set of special tools. 吀栀e saving in manu­
昀愀cturing costs, owing to the special tools, is
estimated to be $150,000 a year 昀漀r 昀椀ve years.
Assume CCA rate of 30% 昀漀r the special tools
and a 39% income tax rate.
⠀　(a) What is the a昀琀er-tax payback period 昀漀r
this investment?
(b) If the company wants a 12% a昀琀er-tax rate
of return, is this a desirable investment?
12-39 Michael is contemplating a $10,000 investment
in a methane gas generator. He estimates his
gross income would be $2,000 the 昀椀rst year and
would increase by $200 each year over the next
10 years. His expenses of $200 the 昀椀rst year
would increase by $200 each year over the next
10 years. He would depreciate the generator by
a CCA rate of 20%. A 10-year-old methane gen­
erator has no market value. 吀栀e income tax rate
is 40%. (Remember that recaptured deprecia­
tion is taxed at the same 40% rate.)
401
(a) Construct the a昀琀er-tax cash flow 昀漀r the
10-year project life.
(b) Determine the a昀琀er-tax rate of return on
this investment. Michael thinks it should
be at least 8%.
(c) If Michael could sell the generator 昀漀r
$7,000 at the end of the 昀椀昀琀h year, would
his rate of return be better than if he kept
it 昀漀r 10 years? You don't have to actually
昀椀nd the rate of return, just do enough cal­
culations to see whether it is higher than
that of part (b).
12-40 Steve has a house and lot 昀漀r sale 昀漀r $210,000.
It is estimated that the land is worth $40,000,
the house $170,000. Anthea is purchasing the
house on 1 January to rent and plans to own
it 昀漀r 昀椀ve years. A昀琀er 昀椀ve years, it is expected
that the house and land can be sold on 31 De­
cember 昀漀r $225,000 ($40,000 昀漀r the land,
$185,000 昀漀r the house). Total annual expenses
(maintenance, property taxes, insurance, etc.)
are expected to be $5,000. 吀栀e house would
be depreciated at a CCA rate of 10%. Anthea
wants a 15% a昀琀er-tax rate of return on her in­
vestment. You may assume that Anthea has an
incremental income tax rate of 27% in each of
the 昀椀ve years. Capital gains are taxed at 13.5%.
Determine the 昀漀llowing:
(a) 吀栀e annual depreciation
(b) 吀栀e capital gain (loss) resulting 昀爀om the
sale of the house
(c) 吀栀e annual rent Anthea must charge to
produce an a昀琀er-tax rate of return of 15%
12-41 Carolyn owns a data processing company.
She plans to buy an additional computer 昀漀r
$20,000, use it 昀漀r three years, and sell it 昀漀r
$10,000. She expects that the use of the com­
puter will produce a net income of $8,000 a
year. 吀栀e combined federal and provincial in­
cremental tax rate is 45%. Using a CCA rate
of 55% and an interest rate of 12%, complete
Table P12-41 to determine the net present
worth of the a昀琀er-tax cash flow.
⠀　12-42 Specialty Machining, Inc. bought a new
multi-turret turning centre 昀漀r $250,000.
402
Cha pter 12 After-Tax Cash Flows
吀愀ble P12-41 Worksheet for Problem 12-41
Year
0
1
2
3
Be昀漀re-Tax
Cash Flow
Capital Cost
Allowance
- $20,000
+8,000
+8,000
+8,000
+ 1 0,000
吀栀e machine generated new revenue of $80,000
a year. Operating costs 昀漀r the machine aver­
aged $10,000 a year. In accordance with CRA
regulations, the machine was depreciated as a
Class 43 Asset (30%). 吀栀e centre was sold 昀漀r
$75,000 a昀琀er 昀椀ve years of service. 吀栀e com­
pany uses an a昀琀er-tax MARR of 12% and is in
the 35% tax bracket. Determine the a昀琀er-tax
net present worth of this asset over the 昀椀ve­
year service period.
12-43 Fleet rental car company purchased 10 new
cars 昀漀r a total cost of $180,000. 吀栀e cars gen­
erated income of $150,000 per year and in­
curred operating expenses of $60,000 per year.
吀栀e company uses Class 16 CCA depreciation
(40%), and its marginal tax rate is 27%. 吀栀e
10 cars were sold at the end of the third year
昀漀r a total of $75,000. Assuming a MARR of
10% and using NPW, determine if this was a
good investment on an a昀琀er-tax basis.
12-44 An investor bought a racehorse 昀漀r $1 million.
吀栀e horse's average winnings were $700,000
per year, and expenses averaged $200,000
per year. 吀栀e horse was retired a昀琀er three
years, at which time it was sold to a breeder
昀漀r $175,000. Assuming accelerated 50%
straight-line CCA (25% -50% -25%) and
an income tax rate of 36%, determine the
investor's a昀琀er-tax rate of return on this
investment.
12-45 Vanguard Solar Systems is building a new
manu昀愀cturing 昀愀cility to be used 昀漀r the
Taxable
Income
Income
Tax {27%)
A昀琀er-Tax
Cash Flow
Present
Worth
( 1 2%)
Net Present Worth =
production of solar panels. Vanguard uses a
MARR of 15% and Class 29 (50% straight-line
depreciation) and is in the 31% tax bracket.
吀栀e building will cost $2.75 million, and
the equipment (Class 43 Property) will cost
$1.55 million plus installation costs of
$135,000. O&M costs are expected to be
$1.3 million the 昀椀rst year, increasing 6% an­
nually. If the 昀愀cility opens in the month of
March and sales are as shown, determine the
present worth of the a昀琀er-tax cash 昀氀ows 昀漀r
the 昀椀rst 昀椀ve years of operation.
Year
1
2
3
4
5
Sales
$2, 1 00,000
3,200,000
3,800,000
4,500,000
5,300,000
12-46 XYZ Electric Company is planning a m愀樀or up­
grade in its computerized demand-management
system. To accommodate this upgrade, a build­
ing will be constructed on land already owned
by the company. 吀栀e building (CCA Class 1) is
estimated to cost $1.8M and will be opened in
August 2015. 吀栀e computer equipment (CCA
Class 50) 昀漀r the building will cost $2.75M, and
all o昀케ce equipment (CCA Class 10) will cost
$225,000. Annual expenses 昀漀r operating this
昀愀cility (labour, materials, insurance, energy,
etc.) are expected to be $325,000 during 2015.
Use of the new demand-management system
is expected to decrease 昀甀el and other costs 昀漀r
the company by $1.8M in the 昀椀rst year (2015).
Problems
this project, how much could he pay to buy the
adjoining land with the old building? Assume
that the analysis period is 10 years and that the
parking lot could always be sold to recover the
costs of buying the property and demolishing
the old building.
If the company expects to earn 9% on its in­
vestments, is in the 35% tax bracket, and uses
a 20-year planning horizon, determine the es­
timated a昀琀er-tax cash 昀氀ow 昀爀om this project in
2015.
12-47 Mid-Canada Shipping is considering pur­
chasing a new barge 昀漀r use on its Great Lakes
routes. 吀栀e barge will cost $13.2 million and is
expected to generate an income of $7.5 million
the 昀椀rst year (growing by $1M each year),
with additional expenses of $2.6 million the
昀椀rst year (growing by $400,000 per year). If
Mid-Canada uses a CCA rate of 30%, is in
the 27% tax bracket, and has a MARR of 12%,
what is the present worth of the 昀椀rst 昀漀ur years
of a昀琀er-tax cash 昀氀ows 昀爀om this barge? Would
you recommend that Mid-Canada purchase
this barge? (Hin琀猀: Remember that all expenses
are deducted 昀爀om income be昀漀re calculating
the taxes. Round to the nearest tenth of a mil­
lion dollars.)
Solving for U n knowns
12-48 A store owner, Justin, believes his business
has su昀昀ered 昀爀om the lack of adequate cus­
tomer parking space. 吀栀us, when he was of­
fered an opportunity to buy an old building
and lot next to his store, he was interested. He
would demolish the old building and make
o昀昀-street parking 昀漀r 20 customers' cars. He
estimates that the new parking would increase
his business and produce an additional be昀漀re­
income-tax pro昀椀t of $7,000 a year. It would
cost $2,500 to demolish the old building. Jus­
tin's accountant advised that the total value of
the land 昀漀r tax purposes would be considered
to be the combined cost of buying the property
and demolishing the old building, and it would
not be depreciable. Justin would spend an addi­
tional $3,000 right away to put a light gr愀瘀el
sur昀愀ce on the lot. 吀栀is expenditure, he be­
lieves, may be charged as an operating expense
immediately and need not be capitalized. To
compute the tax consequences of adding the
parking lot, he estimates that his combined in­
cremental income tax rate will average 40%. If
Justin wants a 15% a昀琀er-tax rate of return 昀爀om
403
12-49 吀栀e management of a private hospital is con­
sidering the installation of an automatic tele­
phone switchboard, which would replace a
manual switchboard and eliminate the attend­
ant operator's position. 吀栀e class of service
provided by the new equipment is estimated to
be at least equal to the present method of oper­
ation. To provide telephone service, 昀椀ve oper­
ators currently work three shi昀琀s a day, 365 days
a year. Each operator earns $25,000 a year.
Company-paid bene昀椀ts and overhead are 25%
of wages. Money costs 8% a昀琀er income taxes.
Combined income taxes are 40%. Annual
property taxes and maintenance are 2.5% and
4% of investment, respectively. Depreciation is
15-year straight-line. Disregarding in昀氀ation,
how large an investment in the new equipment
can be economically justi昀椀ed by savings ob­
tained by eliminating the present equipment
and labour costs? 吀栀e existing equipment has
zero salvage value.
12-50 A contractor has to choose one of the 昀漀llowing alternatives in per昀漀rming earth-moving
contracts:
(a) Purchase a heavy-duty truck 昀漀r $35,000.
Salvage value is expected to be $8,000 at
the end of the vehicle's seven-year depre­
ciable life. Maintenance is $2,500 a year.
Daily operating expenses are $200.
(b) Hire a similar unit 昀漀r $550 a day. Using a
10% a昀琀er-tax rate of return, calculate how
many days a year the truck must be used
to justi昀礀 its purchase. Base your calcula­
tions on straight-line depreciation and a
50% income tax rate.
⠀　12-51 吀栀e Able Corporation is considering the in­
stallation of a small electronic testing device
昀漀r use in conjunction with a government con­
tract the 昀椀rm has just won. 吀栀e device will cost
$20,000 and have an estimated salvage value
404
Cha pter 12 After-Tax Cash Flows
of $5,000 in 昀椀ve years when the contract is
昀椀nished. 吀栀e 昀椀rm will depreciate the instru­
ment by the sum-of-years'-digits method,
using 昀椀ve years as the use昀甀l life and a $5,000
salvage value. Assume that Able pays 50%
corporate income taxes and uses 8% a昀琀er tax
in economic analysis. What minimum equal
annual bene昀椀t must Able obtain before taxes in
each of the 昀椀ve years to justi昀礀 purchasing the
electronic testing device?
12-52 A house and lot are 昀漀r sale 昀漀r $155,000. It is
estimated that $45,000 is the land value and
$110,000 is the value of the house. 吀栀e net rental
income would be $12,000 a year a昀琀er taking
all expenses, except depreciation, into account.
吀栀e house would be depreciated by straight­
line depreciation, with a 27.5-year deprecia­
ble life and zero salvage value. Mary Silva, the
prospective buyer, wants a 10% a昀琀er-tax rate
of return on her investment a昀琀er considering
both annual income taxes and a capital gain
when she sells the house and lot. At what price
would she have to sell the house at the end of
10 years to achieve her objective, given that
the value of the lot is now $75,000? You may
assume that Mary has an incremental income
tax rate of 27% in each of the 10 years.
⠀　12-53 A corporation is considering buying a medium­
sized computer that will eliminate a task that
must be per昀漀rmed by three shi昀琀s a day, seven
days a week, except 昀漀r one eight-hour shi昀琀
every week when the operation is shut down 昀漀r
maintenance. Four people are needed to per­
昀漀rm the d愀礀 and night tasks; thus the computer
will replace 昀漀ur employees. Each employee
costs the company $32,000 a year ($24,000 in
direct wages plus $8,000 a year in other com­
pany employee costs). It will cost $18,000 a year
to maintain and operate the computer. 吀栀e
computer will be depreciated by sum-of-years' digits depreciation with a six-year depreciable
life, at which time it will be assumed to h瘀였
zero salvage value. 吀栀e corporation has a com­
bined incremental tax rate of 50%. If the 昀椀rm
wants a 15% rate of return a昀琀er considering
income taxes, how much can it a昀昀ord to p愀礀 昀漀r
the computer?
12-54 A sales engineer has the 昀漀llowing alternatives
to consider in touring her sales territory.
(a) Buy a two-year old car 昀漀r $14,500. Sal­
vage value is expected to be about $5,000
a昀琀er three more years. Maintenance
and insurance cost is $1,000 in the 昀椀rst
year and increases at the rate of $500 a
year in subsequent years. Daily operating
expenses are $50 a day.
(b) Rent a similar car 昀漀r $80 a day. At a 12%
a昀琀er-tax rate of return, how many days
a year must she use the car to justi昀礀
its purchase? Assume that she is in the
30% incremental tax bracket. Use a CCA
rate of 30%.
⠀　12-55 You recently bought a mini-supercomputer
昀漀r $10,000 to allow 昀漀r tracking and analysis
of real-time changes in stock and bond prices.
Assume you plan to spend half your time tend­
ing to the stock market with this computer and
the other half 昀漀r personal use. Also assume
you can depreciate your computer by a CCA
Class 50. How much tax savings will you have
in the next 昀椀ve years, if any? Use a tax rate of
28%.
Multiple Alternatives
12-56 Use the a昀琀er-tax IRR method to evaluate the
昀漀llowing three alternatives 昀漀r a CCA Class
50 asset and o昀昀er a recommendation. 吀栀e af­
ter-tax MARR is 25%, the project life is 昀椀ve
years, and the 昀椀rm has a combined incremen­
tal tax rate of 45%.
Alt.
A
B
C
First
Costs
Annual
Cost
Salvage
Value
$ 1 4,000
1 8,000
1 ,000
$2,500
1 ,000
5,000
$ 5,000
1 0,000
0
12-57 A small business corporation is considering
whether to replace some equipment in the
plant. An analysis shows there are 昀椀ve alterna­
tives in addition to the do-nothing option,
Alternative A. Each has a 昀椀ve-year use昀甀l life
with no salvage value. Straight-line deprecia­
tion would be used.
Problems
Cost
(thousands)
Alternatives
A
B
C
D
E
F
0
$0
25
10
5
15
30
Be昀漀re-Tax Uni昀漀rm
Annual Bene昀椀ts
(thousands)
$0.0
7.5
3.0
1.7
5.0
8.7
吀栀e corporation has a combined income tax
rate of 20%. Prepare a choice table to guide
the corporation in selecting the most desirable
alternative.
12-58 A corporation with $7 million in annual tax­
able income is considering two alternatives:
Be昀漀re-Tax Cash Flow
Year
0
1-10
11-20
2
3
4
5
Alt. 1
Alt. 2
- $10,000
+4,500
0
- $20,000
+4,500
+4,500
Both alternatives will be depreciated by
straight-line depreciation, assuming a 10-year
depreciable life and no salvage value. Neither
alternative is to be replaced at the end of its
use昀甀l life. If the corporation has a tax rate of
34% and a minimum attractive rate of return
of 10% a昀琀er taxes, which alternative should it
choose? Solve the problem by
(a) present worth analysis
(b) annual cash 昀氀ow analysis
挀儀 rate of return analysis
(d) 昀甀ture worth analysis
(e) bene昀椀t-cost ratio analysis
(f) any other method you choose
12-59 吀眀o mutually exclusive alternatives are being
considered by a pro昀椀table corporation with an
annual taxable income between $5 million and
$10 million.
Be昀漀re-Tax Cash Flow ( $000)
Year
Alt. A
Alt. B
0
1
-$3,000
+$1,000
-$5,000
+$1,000
0
405
+$1,200
+$1,400
+$2,600
+$2,800
+$1,000
+$1,000
+$1,000
+$1,000
Both alternatives have a 昀椀ve-year use昀甀l
anddepreciable life and no salvage value.
Alternative A would be depreciated by sum-of­
years'-digits depreciation, and Alternative B by
straight-line depreciation. If the MARR is 10%
a昀琀er taxes, and the tax rate is 34%, which al­
ternative should be chosen?
12-60 A large, pro昀椀table corporation is considering
two mutually exclusive capital investments:
Initial cost
Uni昀漀rm annual bene昀椀t
End-of-depreciable-Hfe
salvage value
Depreciation method
End-of-use昀甀l-life
salvage value (expected)
Depreciable life, in years
Use昀甀l life, in years
Alt. A
Alt. B
$11,000
$3,000
$33,000
$9,000
$2,000
SL
$3,000
SOY D
$2,000
3
5
$5,000
4
5
If the 昀椀rm's a昀琀er-tax minimum attractive rate
of return is 12% and its incremental income tax
rate is 34%, which project should be selected?
12-61 Padre owns a small business and has a mar­
ginal tax rate of 36%. He is considering
昀漀ur mutually exclusive alternative models of
machinery. Which machine should be selected
on an a昀琀er-tax basis? 吀栀e a昀琀er-tax MARR is
15%. Assume that each machine is CCA Class
43 (30%) and can be sold 昀漀r a market value
that is 25% of the purchase cost, and the pro­
ject life is 10 years.
Model
First cost
Annual costs
I
II
III
IV
$9,000
25
$8,000
200
$7,500
300
$6,200
600
12-62 LoTech Welding can buy a computer package
昀漀r $175,000 and depreciate it as a CCA Class
43 asset (45%). Annual maintenance would be
$9,800, and its salvage value a昀琀er eight years
406
Cha pter 12 After-Tax Cash Flows
is $15,000. 吀栀e package can also be leased 昀漀r
$35,000 a year on an "all costs" inclusive lease
(maintenance costs included). Lease payments
are due at the beginning of each year, and they
are tax-deductible. 吀栀e 昀椀rm's combined tax rate
is 29%. If the 昀椀rm's a昀琀er-tax interest rate is 9%,
which alternative has the lower E唀䄀C, and by
how much?
12-63 For Problem 12-62, assume that Class 12
(100%) depreciation can be considered since
this is primarily computer so昀琀ware. Also
assume that the 昀椀rm is pro昀椀table. How much
lower is the annual cost of purchasing the
machine?
12-64 For Problem 12-62, assume that the purchase
will be 昀椀nanced with a 昀漀ur-year loan whose
interest rate is 12%.
愀儀 Graph the EUAC 昀漀r the purchase 昀漀r
昀椀nancing 昀爀actions ranging 昀爀om 0%
to 100%.
(b) Assume that Accelerated CCA (Problem
12-63) is available 昀漀r the package. Graph
the EUAC 昀漀r the purchase 昀漀r 昀椀nancing
昀爀actions ranging 昀爀om 0% to 100%.
U nclassi昀椀ed
12-65 A plant can be purchased 昀漀r $1,000,000 or it
can be leased 昀漀r $200,000 per year. 吀栀e annual
income is expected to be $800,000, with an
annual operating cost of $200,000. 吀栀e resale
value of the plant is estimated to be $400,000
at the end of its 10-year life. 吀栀e company's
combined federal and provincial income tax
rate is 40%. A straight-line depreciation can be
used over the 10 years with the 昀甀ll 昀椀rst-year
depreciation rate.
愀儀 If the company uses the a昀琀er-tax min­
imum attractive rate of return of 10%,
should it lease or purchase the plant?
(b) What is the break-even rate of return of
purchase versus lease?
12-66 VML Industries needs specialized yarn­
manu昀愀cturing equipment 昀漀r operations over
the next three years. 吀栀e 昀椀rm could buy the
machinery 昀漀r $95,000 and depreciate it at a
CCA rate of 30%. Annual maintenance would
be $7,500, and it would have a sa氀瘀age value of
$25,000 a昀琀er three years. An alternative would
be to lease the same machine 昀漀r $45,000 per year
on an "all costs" inclus瘀츀 lease (maintenance
costs included in lease payment). Lease payments
are due at the beginning of each 礀攀ar. VML uses
an a昀琀er-tax MARR of 18% and a combined tax
rate of 31%. Do an a昀琀er-tax present worth analy­
sis to determine which option is preferable.
12-67 Using a marginal tax rate of 31%, calculate the
a昀琀er-tax rate of return on a $5,000 corporate
bond that costs $4,600, pays $500 interest at
the end of each year 昀漀r 昀椀ve years, and is then
redeemed 昀漀r $5,000.
12-68 A pro昀椀table incorporated business is considering
an investment in equipment h愀瘀ing the 昀漀llow­
ing be昀漀re-tax cash 昀氀ow. 吀栀e equipment will be
depreciated by CCA with a 25% rate.
Year
0
1
2
3
4
5
6
Be昀漀re-Tax
Cash Flow
($000)
- $12,000
+1,727
+2,414
+2,872
+3,177
+3,358
+1,997
+ 1,000 salvage value
If the 昀椀rm wants a 9% a昀琀er-tax rate of return
and its incremental income tax rate is 34%, de­
termine by annual cash 昀氀ow analysis whether
the investment is desirable.
12-69 吀栀e Ogi Corporation, a construction company,
bought a used pickup truck 昀漀r $14,000 and used
a CCA rate of 30% in the income tax return.
During the time the company had the truck,
they estimated that it saved $5,000 a year. At the
end of 昀漀ur years, Ogi sold the truck 昀漀r $3,000.
吀栀e combined income tax rate 昀漀r Ogi is 27%.
Find the a昀琀er-tax rate of return 昀漀r the truck.
12-70 吀栀e e昀昀ective combined tax rate in an
owner-managed corporation is 40%. An outl愀礀
of $2 million 昀漀r certain new assets is under
Problems
consideration. It is estimated that 昀漀r the next
eight years, these assets will be responsible 昀漀r
annual receipts of $600,000 and annual dis­
bursements (other than 昀漀r income taxes) of
$250,000. A昀琀er this time, they will be used only
昀漀r stand-by purposes, and no 昀甀ture excess of
receipts over disbursements is estimated.
(a) What is the prospective rate of return
be昀漀re income taxes?
(b) What is the prospective rate of return
a昀琀er taxes if straight-line depreciation
can be used to write o昀昀 these assets 昀漀r
tax purposes in eight years?
挀儀 What is the prospective rate of return
a昀琀er taxes if it is assumed that these
assets must be written o昀昀 昀漀r tax purposes
over the next 20 years, using straight-line
depreciation?
12-71 Sole Brother Inc. is a shoe outlet of a major
shoe manu昀愀cturing company located in
Montreal. It uses accounts payable as one of its
昀椀nancing sources. Shoes are delivered to Sole
Brother with a 3% discount if payment on the
invoice is received within 10 days of delivery.
By paying a昀琀er the 10-day period, Sole is bor­
rowing money and p愀礀ing (giving up) the 3%
discount. Although Sole is not required to pay
interest on delayed payments, the shoe manu­
昀愀cturers require that payments not be delayed
beyond 45 days a昀琀er the invoice date. To be
sure of paying within 10 days, Sole decides to
pay on the 昀椀昀琀h day. Sole has a marginal cor­
porate income tax of 40% (combined provin­
cial and federal). By paying within the 10-day
period, Sole is avoiding paying a 昀愀irly high
price to retain the money owed to shoe manu­
昀愀cturers. What would have been the e昀昀ective
annual a昀琀er-tax interest rate?
12-72 A 昀椀rm has invested $14,000 in machinery with a
seven-礀攀ar use昀甀l life. 吀栀e machinery will h瘀였 no
sa氀瘀age value, as the cost to rem瘀퐀 it will equal
its scrap value. 吀栀e uni昀漀rm annual bene昀椀ts 昀爀om
the machinery are $3,600. For a 47% income tax
rate, and sum-o�鈀ars'-digits depreci愀琀ion, com­
pute the a昀琀er-tax rate of return.
12-73 A large pro昀椀table company, in the 40% tax
bracket, is considering the purchase of a new
407
piece of equipment. It will yield bene昀椀ts of
$10,000 in Year 1, $15,000 in Year 2, $20,000
in Year 3, and $25,000 in Year 4. 吀栀e CCA rate
is 30%. It is expected that the equipment will
be sold at the end of the 昀漀urth year 昀漀r 20% of
its purchase price. What is the maximum price
the company can pay 昀漀r the equipment if its
a昀琀er-tax MARR is 10%?
12-74 Machine X costs $248,751 and has annual
operating and maintenance costs of $9,980.
Machine Y costs $264,500 and has annual
operating and maintenance costs of $5,120.
Both machines are Class 43, which speci昀椀es
a CCA rate of 30%. 吀栀e company needs the
machines 昀漀r 11 years, and at the end of Year
11 Machine X can be sold 昀漀r $12,257 and Ma­
chine Y can be sold 昀漀r $13,033. 吀栀e company's
MARR is 10%, and its marginal taxation rate
is 27%. Do an a昀琀er-tax analysis to determine
which machine should be chosen.
⠀　12-75 Christopher wants to add a 2-kW solar photo­
voltaic system to his home and has received a
quote 昀爀om an installer 昀漀r $19,750. 吀栀e federal
government will give him a tax credit of 30%
on his combined taxes. His marginal rate is
29%. Provincial law requires the utility com­
pany to buy back all excess power generated
by the system. Christopher's estimated annual
power bill of $2,000 will be eliminated by the
solar system. He expects to receive a cheque 昀漀r
$600 昀爀om the power company every year 昀漀r
his excess production. If the tax credits are re­
ceived at EOY 1 and Christopher saves $2,000
a year as well as receiving income of $600
income at the end of each year, use present
worth to determine if the system is economic
in eight years. Assume that Christopher earns
3% on all his investments.
12-76 Katie's Butter and Egg Business is such that she
pays an e昀昀ective tax rate of 40%. Katie is con­
sidering the purchase of a new 吀甀rbo Churn
昀漀r $25,000. 吀栀is churn has an estimated life
of 昀漀ur years and a salvage value of $5,000. It
is expected to increase net income by $8,000 a
year 昀漀r each of the 昀漀ur years of use. If Katie
works with an a昀琀er-tax MARR of 10% and
CCA of 30%, should she buy the churn?
⠀　Replacement Analysis
The $7 Billion U pgrade
In 2009 Intel Corporation announced it was spending $ 7 billion in 2009 and 2010 to modern­
ize and update its silicon-wafer manufacturing plants in the US. The upgrade will allow Intel
to manufacture 32-nanometre (nm) chips in fabs originally built for 45-nm and 65-nm chips.
Upgrade work will include remodelling the plant's interior and purchasing new wafer fabrication
tools.
Despite the price tag for the project, Intel planned on saving money overall by upgrading the
existing plants instead of designing and building a new one. The upgraded plants use newer pro­
cesses that allow chips to hold smaller and faster transistors, and the new larger wafer accommo­
dates more chips. All these changes lead to lower production costs.
The company also noted that by deciding to remain in its current locations, it was able to
retain its highly skilled workforce.
Gideon Mendel/Corbis/Getty I mages
The $7 Billion U pg rade
409
QU ESTIONS TO CONSI DER
1.
Intel made the decision to upgrade existing plant facilities rather than build new ones. In its analysis it
claimed to have saved money by doing so. Use the Internet to search for articles that discuss what new
manufacturing plants cost in industries of different types. Does Intel's $ 7 billion upgrade decision seem in
line? What issues are you not considering if you think the cost seems very high?
2.
Companies that upgrade existing production assets may or may not scrap their current assets. Give a real­
istic scenario for each case.
3.
The upgrade-or-replace decision often includes economic as well as non-economic factors. List three eco­
nomic and three non-economic factors that may have been at the top of Intel's list in making the decision.
4.
Companies that build new manufacturing plants often site them outside North America. Discuss the ethical
considerations that are or should be part of a firm's decision process.
LEARN I NG OBJ ECTIVES
This chapter will help you
•
recast an equipment reinvestment decision as a challenger-ve爀猀us-defender analysis
•
choose the correct economic analysis technique to apply
•
calculate the minimum-cost life of economic challengers
•
use criteria such as repea tability assumption for replacement ana礀�is and marginal cost data
•
perform replacement problems on an after-tax basis
for the defender to choose the appropriate economic analysis techniques
KEY TERMS
bathtu b cu rve
challenger
defender
marg i nal cost
m i n i m u m -cost life
replacement analysis
replacement repeata bi lity
assum ption
Up to this point we have considered the evaluation and selection of new alternatives. Which
new car or production machine should we buy? Which new material handling system or
ceramic grinder should we install? More o昀琀en, however, economic analysis weighs existing
versus new 昀愀cilities. For most engineers the problem is less likely to involve building a new
plant than how best to keep a present plant operating economically. We are not choosing
between new ways to per昀漀rm the desired task. Instead, we have equipment per昀漀rming
the task, and the question is: should the existing equipment be retained or replaced? 吀栀is
adversarial situation gives rise to the terms defender and challenger. 吀栀e defender is the
existing equipment; the challenger is the best available replacement equipment. 吀栀e eco­
nomic evaluation of the existing defender and its challengers is the domain of replacement
analysis.
410
Cha pter 13 Replacement Analysis
The Replacement Problem
It may be reasonable to replace an existing asset because of obsolescence, depletion, or
deterioration 昀爀om aging. In each of these cases, the ability of an asset to produce a desired
output is compromised and it may be economical to replace the existing asset. We de昀椀ne
each of these situations.
Obsolescence: the technology of an asset is surpassed by newer or di昀昀erent technologies.
For example, today's personal computers (PCs)-with more RAM, 昀愀ster clock speeds,
larger hard drives, and more power昀甀l central processors-have made older PCs
obsolete.
Depletion: the gradual loss of market value of an asset as it is consumed or exhausted. Oil
wells and stands of timber are examples of such assets. In most cases the asset will be
used until it is depleted, at which time a replacement asset will be obtained. Depletion
was treated in Chapter 1 1 .
Deterioration caused by agi渀最: the loss i n value of some asset due to the aging process.
Production machinery and other business assets eventually become aged. As a result
of the loss in 昀甀nctionality caused by aging, there are usually additional operating and
maintenance expenses to maintain the asset at its operating e昀케ciency.
Aging equipment o昀琀en has a greater risk of breakdowns. Planned replacements can be
scheduled to minimize the cost of disruptions. Unplanned breakdowns can be very costly
or even, as with an airplane engine, catastrophic.
吀栀ere are variations of the replacement problem: an existing asset may be abandoned
or retired, augmented by a new asset but kept in service, or overhauled. 吀栀ese variations
are most easily considered as potential new challengers.
Replacement problems are normally analyzed by looking only at the costs of the exist­
ing and replacement assets. Since the assets usually per昀漀rm the same 昀甀nction, the value
of using the vehicle, machine, or other equipment can be ignored. If the new asset has new
features or better per昀漀rmance, these can be included as a cost saving.
Alternatives in a replacement problem almost alw愀礀s h愀瘀e d昀케erent lives, and the prob­
lem includes picking the best one. 吀栀is is because an existing asset will o昀琀en be kept 昀漀r at
most a few years longer, whereas the potential replacements may have lives of any length.
吀栀us replacement problems 昀漀cus on annual marginal costs and on EUAC values. We can
calculate present costs, but only as a step in calculating E唀䄀C values.
Expenditures are normally monitored by means of annual bu最준ts. One important
昀愀cet of a budget is the allocation of money 昀漀r new capital expenditures, either new 昀愀cili­
ties or replacement and upgrading of existing 昀愀cilities.
Replacement analysis may recommend that certain equipment be replaced, with the
cost included in the capital expenditures budget. Even if no recommendation to replace
is made, such a recommendation may be made subsequently. At some point, the existing
equipment will be replaced, either when it is no longer necessary or when better equip­
ment is 愀瘀ailable. 吀栀us, the question is not 椀昀 the defender will be replaced, but when it
will be replaced. 吀栀is leads us to the 昀椀rst question in the comparison of defender and
challenger:
Should we replace the defender now, or should we keep itfor one or more additional
years?
The Replacement Problem
If we do decide to keep the asset 昀漀r another year, we will o昀琀en re-analyze the prob­
lem next year. 吀栀e operating environment and costs may change, or new challengers with
lower costs or better per昀漀rmance may emerge.
Replacement Analysis Flow Chart
Figure 13-1 is a 昀氀ow chart 昀漀r conducting a replacement analysis.
Defender --�
Identi昀礀
Alternatives
尀娀--
䄀瘀ailable
Defender
Marginal
Cost Increasing
An愀氀ysis Technique I
Compare the next-year
marginal cost of the defender
with the EUAC of the
challenger.
Best Challenger
Not Available
Find Lowest
EUAC 昀漀r
Defender
Find
E唀䄀C Over
Given Life
An愀氀ysis Technique 2
Compare the lowest E唀䄀C
of the defender with the
EUAC of the challenger at
its minimum cost life.
An愀氀ysis Technique 3
Lacking defender marginal cost data,
compare the E唀䄀C of defender over
its remaining use昀甀l life with the EUAC
of challenger at its minimum cost life.
No
FIGURE 13-1 Replacement analysis flow chart.
Looking at the chart, we see that three di昀昀erent replacement analysis techniques can
be used under di昀昀erent circumstances. 吀栀e correct technique depends on the data avail­
able 昀漀r the alternatives and how the data behave over time.
We see that the 昀椀rst step is to identify the alternatives.
䤀昀 the defender proves more economical, it will be kept. 䤀昀 the challenger proves more
economical, it will be installed.
In this comparison the defender is being evaluated against a challenger that has been
selected 昀爀om a set of mutually exclusive competing challengers. Figure 13-2 illustrates this
as a drag race between the defender and the challenger. 吀栀e challenger competing against
the defender has emerged 昀爀om an earlier competition among a set of potential challen­
gers. Any of the methods discussed previously in this text 昀漀r evaluating sets of mutually
exclusive alternatives could be used to identi昀礀 the best challenger to race against the de­
fender. However, it is important to note that the comparison of these potential challengers
should be made at each alternative's minimum-cost l椀昀e. 吀栀is concept is discussed next.
411
412
Cha pter 13 Replacement Analysis
⠀䐀 吀栀e minimum-cost life of any new asset is the number of years at which the equivalent
Minimum-Cost Life of a New Asset-The Challenger
uni昀漀rm annual cost (EUAC) of ownership is minimized. 吀栀is minimum-cost life is o昀琀en
shorter than either the physical or use昀甀l life of the asset because of increasing operating
and maintenance costs in the later years of asset ownership. 吀栀e challenger selected to
compete with the defender is the one having the lowest minimum-cost life of all the com­
peting mutually exclusive challengers.
Mutually Exclusive Set of Challengers
"Current Champio渀✀' Asset
(previously implemented)
vs
Challenger- I
Challenger-2
Challenger- 3
Challenger-k
"Winner" of Challengers Comparison
FIGURE 13-2 Defender-cha llenger com pa rison.
To calculate the minimum-cost life of an asset, we determine the EUAC 昀漀r each pos­
sible life less than or equal to the use昀甀l life. As illustrated in Example 13-1, the E唀䄀C
tends to be high if the asset is kept only a few years, then decreases to some minimum
value, and then increases again as the asset ages. By identi昀礀ing the number of years at
which the EUAC is a minimum and then keeping the asset 昀漀r that number of years, we are
minimizing the yearly cost of ownership.
EXAM PLE 13 -1
A piece of machinery costs $7,500 and has no salvage value a昀琀er it is installed. 吀栀e manu昀愀cturer's
warranty will pay the 昀椀rst year's maintenance and repair costs. In the second year, maintenance costs
will be $900, and they will increase on a $900 arithmetic gradient in subsequent years. Also, operating
expenses 昀漀r the machinery will be $500 the 昀椀rst year and will increase on a $400 arithmetic gradient in
the 昀漀llowing years. If interest is 8%, compute the use昀甀l life of the machinery that minimizes the E唀䄀C.
Minimum-Cost Life of a New Asset-The Challenger
413
SO LUTI O N
If Retired at the End of Year n
Year, n
E唀䄀C of Capital
Recovery Costs:
$7,S00(A/P, 8%, n)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
$8,100
4,206
2,910
2,264
1,878
1,622
1,440
1,305
1,200
1,117
1,050
995
948
909
876
EUAC of Maintenance
and Repair Costs:
$900(A/G, 8%, n)
E唀䄀C of Operating
Costs: (A/G, 8%, n)
E唀䄀C Total
$ 500
692
880
1,062
1,238
1,410
1,578
1,740
1,896
2,048
2,196
2,338
2,476
2,609
2,738
$8,600
5,331
4,644
4,589堀娀
4,779
5,081
5,443
5,834
6,239
6,650
7,063
7,470
7,871
8,265
8,648
$
0
433
854
1,264
1,661
2,048
2,425
2,789
3,142
3,484
3,816
4,136
4,446
4,746
5,035
吀栀e total EUAC data are plotted in Figure 13-3. We see that the minimum-cost life of the machinery is
昀漀ur years, with a minimum EUAC of $4,589.
1 1 ,000
1 0,000
9,000
8,000
7,000
§
6,000
u
5,000
4,000
3,000
2,000
1 ,000
0
­
尀尀
尀尀
- -尀尀
尀尀
_ 尀尀 尀言-ⴀⴀ- - - - · - -- - - · - - 尀言 - -• - - ⴀⴀ - - •
娀蠀ꈀ㴀娀-ⴀⴀ - -
E唀䄀C of Maintenance
and Repair
贀尀 - - ⸀尀
⸀⸀-1
尀尀 _ _
开开 开开
尀尀 - -
E唀䄀C of Operating
EUAC of Capital Recovery
-栀嬀娀묀-2
3
4
5
6
7
Year
8
9
10
11
12
13
FIGURE 13-3 Plot of costs for Exa m ple 13 -1.
Looking at Figure 13-3 more closely, we see the e昀昀ects of each of the individual cost
components on total EUAC (capital recovery, maintenance and repair, and operating
expense EUACs) and how they behave over time. 吀栀e total EUAC curve of most assets
tends to 昀漀llow this concave shape-high at the beginning because of capital recovery costs,
and high at the end because of increased maintenance, repair, and operating expenses.
14
15
414
Cha pter 13 Replacement Analysis
吀栀is curve is o昀琀en referred to as a bathtub curve because of its shape. 吀栀e minimum
EUAC occurs somewhere between these high points.
Like many pieces of installed equipment, the machinery considered in Example 13-1
had no salvage value. However, some assets, such as a car, can easily be sold 昀漀r a value
that depends on the car's age and condition. Another possible complication is that repair
costs may be reduced in early years by a warranty. 吀栀e resulting cost curves will look like
Figure 13-3, but the calculations take more work, and a spreadsheet is very help昀甀l.
Defender's Marginal Cost Data
Once the participants in the defender-challenger comparison have been identi昀椀ed, two
speci昀椀c questions regarding marginal costs must be answered: Do we have ma爀最inal cost data
for the defender? and Are the defender's marginal costs increasing on a year-to-year basis? Let
us 昀椀rst de昀椀ne marginal cost and then discuss why these two questions are important.
Marginal costs, as opposed to an EUAC, are the year-by-year costs of keeping an
asset. 吀栀ere昀漀re, the "period" of any yearly marginal cost of ownership is always one year.
In our analysis, marginal cost is compared with EUAC, which is an end-of-year cash flow.
吀栀ere昀漀re, the marginal cost is also calculated as end-of-year cash 昀氀ow.
吀栀e total marginal cost 昀漀r any year can include the capital recovery cost (loss in
market value and lost interest 昀漀r the year), yearly operating and maintenance costs, yearly
taxes and insurance, and any other expenses that occur during that year. To calculate the
yearly marginal cost of ownership of an asset, it is necessary to have estimates of an asset's
market value on a year-to-year basis over its use昀甀l life, as well as ordinary yearly expenses.
Example 13-2 illustrates how total marginal cost can be calculated 昀漀r an asset.
EXAM PLE 13 - 2
A new piece of production machinery has the 昀漀llowing costs.
= $25,000
Investment cost
Annual operating and maintenance cost = $2,000 in Year 1, then increasing by $500 per year
Annual cost 昀漀r risk of breakdown
= $5,000 per year 昀漀r 3 years, then increasing by $1,500 per year
Use昀甀l life
= 7 years
MARR
= 15%
Calculate the marginal cost of keeping this asset over its use昀甀l life.
SO LUTI O N
From the data we can easily 昀椀nd the marginal costs 昀漀r O&M and risk of breakdowns. However, to
calculate the marginal capital recovery cost, we need estimates of each year's market value:
Year
1
2
3
4
5
6
7
Market Value
$18,000
13,000
9,000
6,000
4,000
3,000
2,500
Defender's Marginal Cost Data
415
We can now calculate the production machinery's marginal cost (year-to-year cost of ownership) over
its seven-year use昀甀l life.
Year, n
1
2
3
4
5
6
7
Loss in Market
Value in Year n
$25,000 - 1 8,000 = 7,000
1 8,000 - 13,000 = 5,000
13,000 - 9,000 = 4,000
9,000 - 6,000 = 3,000
6,000 - 4,000 = 2,000
4,000 - 3,000 = 1 ,000
3,000 - 2,500 = 500
Forgone Interest
in Year n
$25,000(0. 1 5) = 3,750
1 8,000(0. 15) = 2,700
13,000(0. 15) = 1 ,950
9,000(0. 1 5) = 1 ,350
6,000(0. 1 5) = 900
4,000(0. 1 5) = 600
3,000(0. 1 5) = 450
O&M
Cost in
Year n
Cost of
Breakdown
Risk
in Year n
Total
Marginal
Cost
in Year n
$2,000
2,500
3,000
3,500
4,000
4,500
5,000
$5,000
5,000
5,000
6,500
8,000
9,500
1 1 ,000
$ 1 7,750
1 5,200
1 3,950
14,350
14,900
1 5,600
1 6,950
Notice that each year's total marginal cost includes loss in market value, 昀漀rgone interest, O&M cost,
and cost 昀漀r risk of breakdowns. For example, the Year 5 marginal cost of $14,900 is calculated as 2,000
+ 900 + 4,000 + 8,000.
Do We Have Marginal Cost Data for 琀栀e Defender?
Figure 13-1 indicates that, in order to decide which replacement technique to use, it is ne­
cessary to know whether marginal cost data are available 昀漀r the defender asset. Usually,
in engineering economic problems, annual savings and expenses are given 昀漀r all alterna­
tives. However, as in Example 13-2, it is also necessary to have year-to-year salvage value
estimates in order to calculate total marginal costs. If the total marginal costs 昀漀r the de­
fender can be calculated, and if they are rising 昀爀om year to year, then replacement analysis
technique 1 should be used to compare the defender to the challenger.
Are 吀栀ese Marginal Costs 䤀渀creasing?
We have seen that it is important to know whether the marginal cost 昀漀r the defender is
increasing 昀爀om year to year. 吀栀is is determined by inspecting the total marginal cost of
owning the defender over its remaining life. In most replacement analyses the defender
is nearing the end of its economic life. 吀栀e question usually is whether we should replace
it now, next year, or perhaps the year a昀琀er. In the early years of an asset's life we rarely
analyze whether it is time 昀漀r replacement. 吀栀us the defender's marginal costs are usually
increasing, as in Example 13-3.
EXAM PLE 13 -3
An asset purchased 昀椀ve years ago 昀漀r $75,000 can be sold tod 愀礀 昀漀r $15,000. Operating expenses will be
$10,000 this year but will increase by $1,500 per year. It is estimated that the asset's market value will
decrease by $1,000 per year over the next 昀椀ve years. If the MARR used by the company is 15%, calculate
the total marginal cost of ownership of this old asset (that is, the defender) 昀漀r each of the next 昀椀ve years.
continued
416
Cha pter 13 Replacement Analysis
SO LUTI O N
We calculate the total marginal cost of maintaining the old asset 昀漀r the next 昀椀ve-year period as 昀漀llows:
Year, n
1
2
3
4
5
Loss in Market
Value in Year n
$ 1 5,000 - 14,000 = 1 ,000
14,000 - 13,000 = 1 ,000
1 3,000 - 12,000 = 1 ,000
12,000 - 1 1 ,000 = 1 ,000
1 1 ,000 - 10,000 = 1 ,000
Interest in Year n
$ 1 5,000(0. 15) = 2,250
14,000(0. 1 5) = 2, 1 00
13,000(0. 15) = 1 ,950
1 2,000(0. 1 5) = 1 ,800
1 1 ,000(0. 15) = 1 ,650
Operating
Cost in Year n
Marginal
Cost in Year n
$ 1 0,000
1 1 ,500
13,000
14,500
1 6,000
$ 1 3,250
14,600
1 5,950
1 7,300
1 8,650
We can see that marginal costs increase in each subsequent year of ownership. When
the condition of increasing marginal costs 昀漀r the defender has been met, then the defender­
challenger comparison should be made with replacement analysis technique 1.
Replacement Analysis Technique 1: Defender Marginal
Costs Can Be Computed and Are Increasing
When our 昀椀rst method of analyzing the defender asset against the best available challenger
is used, the comparison involves the marginal cost of the defender and the minimum-cost
l椀昀e of the challenger.
When the marginal cost of the defender is increasing 昀爀om year to year, we will main­
tain that defender as long as the marginal cost of keeping it one more year is less than the
minimum EUAC of the challenger. 吀栀us, our decision rule:
Maintain the defender as long as the marginal cost of ownership for one more year
is less than the minimum EUAC of the challe最팀r. When the marginal cost of the
defender becomes greater than the minimum EUAC of the challenge爀Ⰰ replace the
d昀쨀nder with the challe最팀r.
吀栀is technique assumes that the current best challenger, with its minimum EUAC,
will be available and unchanged in the 昀甀ture. However, it is easy to update a replacement
analysis when marginal costs 昀漀r the defender change or when there is a change in the cost
or per昀漀rmance of available challengers. Example 13-4 illustrates the use of this technique
昀漀r comparing defender and challenger assets.
EXAM PLE 13 -4
Taking the machinery in Example 13-2 as the challe最팀r and the machinery in Example 13-3 as the
defender, use replacement analysis technique 1 to determine when, if at all, a replacement decision
should be made.
Replacement Analysis Technique 1
417
SO LUTI O N
Replacement analysis technique 1 should be used only in the condition of increasing marginal costs 昀漀r
the defender. Since these marginal costs are increasing 昀漀r the defender (昀爀om Example 13-3), we can com­
pare the marginal costs of the defender against the minimum EUAC of the challenger. In Example 13-2
we calculated only the marginal costs of the challenger; now it is necessary to calculate the challenger's
minimum E唀䄀C. 吀栀e EUAC of keeping this asset 昀漀r each year of its use昀甀l life is worked out as 昀漀llows.
Year, n
Challenger
Total Marginal
Cost in Year n
Present Cost If Kept
through Year n (PC)
1
2
3
4
5
6
7
$17,750
15,200
13,950
14,350
14,900
15,600
16,950
[ I7,750 (PIF, 15%, I)]
PC 1 + l5,200 (PIF, 15%, 2)
PC2 + l3,950 (PIF, 15%, 3)
PC3 + I4,350 (PIF, 15%, 4)
PC4 + l4,900 (PIF, 15%, 5)
PCS + l5,600 (PIF, 15%, 6)
PC6 + I6,950 (PIF, 15%, 7)
E唀䄀C if Kept
through Year n
x (AIP, 15%, 1) = $17,750
x (AIP, 15%, 2) = 16,560
x (AIP, 15%, 3) = 15,810
x (A/P, 15%, 4) = 15,520
x (AIP, 15%, 5) = 15,430
x (AIP, 15%, 6) = 15,450
x (AIP, 15%, 7) = 15,580
A minimum E唀䄀C of $15,430 is attained 昀漀r the challenger at Year 5, which is the challenger's minimum-cost
l椀昀e. We proceed by comparing this value with the ma爀最inal costs of the defender 昀爀om Example 13-3:
Year, n
1
2
3
4
5
Defender Total
Marginal Cost
in Year n
$13,250
14,600
15,950
17,300
18,650
Challenger
Minimum EUAC
Comparison Result
and Recommendation
$15,430
15,430
15,430
Since $13,250 is less than $15,430, keep de昀攀nder.
Since $14,600 is less than $15,430, keep defender.
Since $15,950 is greater than $15,430, replace defender.
On the basis of the data given 昀漀r the challenger and 昀漀r the defender, we would keep the defender 昀漀r
two more years and then replace it with the challenger because at that point the defender's marginal
cost of another year of ownership would be greater than the challenger's minimum EUAC.
䌀䐀 Using the challenger's minimum E唀䄀C implies two assumptions: (1) the best challenger
Replacement Repeatability Assumptions
will be available with the same minimum E唀䄀C in the 昀甀ture, and (2) the period of needed
service is inde昀椀nitely long. In other words, we assume that once the decision has been
made to replace, there will be an inde昀椀nite cycle of replacing the current defender asset
with the current best challenger. 吀栀ese assumptions must be satis昀椀ed 昀漀r our calculations
to be correct into an inde昀椀nite 昀甀ture. However, because the near 昀甀ture is more important
economically than the distant 昀甀ture, and because our analysis is done with the best data
currently available, our results and recommendations are robust or stable 昀漀r reasonable
changes in the estimated data.
418
Cha pter 13 Replacement Analysis
吀栀e repeatability assumptions together are much like the repeatability assumptions
that allowed us to use the annual-cost method to compare competing alternatives with dif­
ferent use昀甀l lives. We call them the replacement repeatability assumptions. 吀栀ey allow
us to simpli昀礀 the comparison of the defender and the challenger. Stated 昀漀rmally these
two assumptions are as 昀漀llows:
1.
吀栀e currently available best challenger will continue to be available in subsequent
years and will be unchanged in its economic costs. When the defender is ultimately
replaced, it will be replaced with this challenger. Any challengers put into service
will also eventually be replaced with the same currently available challenger.
2. 吀栀e service of the asset is needed inde昀椀nitely. 吀栀us the challenger, once put into
service, will continually be replaced in repeating, unchanged cycles.
If these two assumptions are satis昀椀ed completely, our calculations are exact. O昀琀en,
however, 昀甀ture challengers represent 昀甀rther improvements, so that Assumption 1 is not
satis昀椀ed. Although the calculations are no longer exact, the repeatability assumptions
allow us to make the best decision we can with the data we have. If the defender's mar­
ginal cost is increasing, once it rises above the challenger's minimum EUAC, it will con­
tinue to be greater. Under the repeatability assumptions, we would never want to incur a
defender's marginal cost that was greater than the challenger's minimum EUAC. 吀栀us, we
use replacement analysis technique 1 when the defender's marginal costs are increasing.
Replacement Analysis Technique 2 : Defender Marginal
Costs Can Be Computed and Are Not Increasing
If the defender's marginal costs do not increase, we have no guarantee that replacement
analysis technique 1 will produce the alternative of greatest economic advantage. Con­
sider the new asset in Example 13-2, which has marginal costs that begin at $17,750, then
decrease over the next years to a low of $13,950, and then increase therea昀琀er to $16,950
in Year 7. If the new asset is evaluated one year a昀琀er implementation, it will not have in­
creasing marginal costs. Defenders in the early stages do not usually 昀椀t the requirements
of replacement analysis technique 1. In the situation graphed in Figure 13-3, such defender
assets would be in the downward slope of a concave marginal cost curve. Example 13-5
details why replacement analysis technique 1 can only be applied when defenders h愀瘀e
consistently increasing marginal cost curves. If the marginal cost is not increasing, we use
replacement analysis technique 2: calculate the defender's minimum E唀䄀C to see whether
the replacement should occur immediately. If not, the replacement occurs a昀琀er the de­
fender's minimum-cost life, when the marginal costs will be increasing. 吀栀en replacement
analysis technique 1 applies again.
EXAM PLE 13 - 5
Let us look again at the defender and challenger assets in Example 13-4. 吀栀is time let us arbitrarily
change the defender's marginal costs 昀漀r its 昀椀ve-year use昀甀l life. Now when, if at all, should the defender
be replaced with the challenger?
Replacement Analysis Technique 2
Year, n
Defender 吀漀tal Marginal
Cost in Year n
1
2
3
4
5
$16,000
14,000
13,500
15,300
17,500
419
SO LUTI O N
In this case the total marginal costs of the defender are not consistently increasing 昀爀om year to year.
However, if we ignore this 昀愀ct and apply replacement analysis technique 1, the recommendation would
be to replace the defender now, because the defender's marginal cost 昀漀r the 昀椀rst year ($16,000) is greater
than the minimum EUAC of the challenger ($15,430). 吀栀is would be the wrong choice.
Since the defender's marginal cost is greater than the challenger's minimum E唀䄀C in the second
to 昀漀urth years, we must calculate the EUAC of keeping the defender asset each of its remaining 昀椀ve
years when i = 15%.
Year, n
Present Cost If Kept n Years (PC)
EUAC If Kept n Years
1
16,000(PIF, 15%, 1)
PC 1 + 14,000(P/F, 15%, 2)
PC2 + 13,500(PIF, 15%, 3)
PC3 + 15,300(PIF, 15%, 4)
PC4 + 17,500(PIF, 15%, 5)
x (AIP, 15%, 1) = $16,000
x (AIP, 15%, 2) = 15,070
x (AIP, 15%, 3) = 14,618
x (AIP, 15%, 4) = 14,754
x (AIP, 15%, 5) = 15,162
2
3
4
5
吀栀e minimum EUAC of the defender 昀漀r three years is $14,618, which is less than
the challenger's minimum E唀䄀C of $15,430. 吀栀us, under the replacement repeatability
assumptions we will keep the defender 昀漀r at least three years. We must still decide how
much longer. 吀栀e defender's EUAC begins to rise in Year 4 because the marginal costs are
increasing. 吀栀us, we can use replacement analysis technique 1 昀漀r Year 4 and later. 吀栀e
defender's marginal cost in Year 4 is $15,300, which is $130 below the challenger's min­
imum EUAC of $15,430. Since the defender's marginal cost of $17,500 is higher in Year 5,
we replace it with the new challenger at the end of Year 4. Notice that we did not keep the
defender 昀漀r its minimum-cost life of three years; we kept it 昀漀r 昀漀ur years.
If the challenger's minimum EUAC were less than the defender's minimum EUAC
of $14,618, the defender would be immediately replaced. Example 13-5 illustrates several
potentially con昀甀sing points about replacement analysis.
•
•
•
If the defender's marginal cost data are not increasing, the defender's minimum
E唀䄀C must be calculated.
If the defender's minimum EUAC exceeds the challenger's minimum EUAC,
replace immediately. If the defender's minimum EUAC is lower than the chal­
lenger's minimum E唀䄀C, under the replacement repeatability assumptions the
defender will be kept at least the number of years 昀漀r its minimum EUAC.
A昀琀er this number of years, replace when the defender's increasing marginal cost
exceeds the challenger's minimum E唀䄀C.
420
Cha pter 13 Replacement Analysis
吀栀e problem statement 昀漀r Example 13-6 illustrates a second approach to calculating the
defender's marginal costs 昀漀r its capital costs. 吀栀e value at the beginning of the year is multi­
plied by (1 + i), and the salvage value at year's end is subtracted. Each year's total marginal
cost also includes the operations and maintenance costs. 吀栀e solution to Example 13-6
details the calculation of the minimum EUAC when the defender's data are presented as
costs and salvage values in each year rather than as marginal costs. Notice that this is
calculated in the same way as the minimum-cost life was calculated 昀漀r new assets.
EXAM PLE 13 - 6
A 昀椀ve-year-old machine, whose current market value is $5,000, is being analyzed to determine its
minimum EUAC at a 10% interest rate. Salvage value (S) and maintenance estimates and the corres­
ponding marginal costs are given in the 昀漀llowing table.
Calculating Marginal Costs
Data
Year
Salvage Value
0
1
2
3
4
5
6
7
8
9
10
11
$5,000
4,000
3,500
3,000
2,500
2,000
2,000
2,000
2,000
2,000
2,000
2,000
O&M Cost
s,_ 1 (1 + ;)
-St
Marginal Cost
0
1 00
200
300
400
500
600
700
800
900
1 ,000
$5,500
4,400
3,850
3,300
2,750
2,200
2,200
2,200
2,200
2,200
2,200
-$4,000
- 3,500
-3,000
-2,500
-2,000
-2,000
-2,000
-2,000
-2,000
-2,000
-2,000
$ 1 ,500
1 ,000
1 ,050
1 , 1 00
1 , 1 50
700
800
900
1 ,000
1 , 1 00
1 ,200
$
SO LUTI O N
Because the marginal costs h愀瘀e a complex, non-increasing pattern, we must calculate the minimum
EUAC of the defender.
If Retired at End of Year n
Years
Kept, n
0
1
2
3
4
5
6
7
8
9
10
11
Salvage
Value (S) at
End of Year n
P = $5,000
4,000
3,500
3,000
2,500
2,000
2,000
2,000
2,000
2,000
2,000
2,000
Maintenance
Cost 昀漀r Year
$
0
100
200
300
400
500
600
700
800
900
1 ,000
E唀䄀C of Capital Recovery
(P - S) x (AIP, 10%, n) + Si
$ 1 , 100 + 400
864 + 350
804 + 300
789 + 250
79 1 + 200
689 + 200
6 1 6 + 200
562 + 200
52 1 + 200
488 + 200
462 + 200
E唀䄀C of
Maintenance
IOO(A/G, 10%, n)
$
0
48
94
138
181
222
262
300
337
372
406
We observe that the minimum EUAC, $1,058, occurs if the defender is kept another 9 years.
Total
EUAC
$ 1 ,500
1 ,262
1 , 198
1 , 1 77
1 , 1 72
1,1 1 1
1 ,078
1 ,062
1 ,058
1 ,060
1 ,068
Replacement Analysis Technique 2
421
EXAM PLE 13 -7
We must decide whether existing (defender) equipment in an industrial plant should be replaced. A
$4,000 overhaul must be done now if the equipment is to be kept in service. Maintenance is $1,800 in
each of the next two years, a昀琀er which it increases by $1,000 each year. 吀栀e defender has no present
or 昀甀ture salvage value. 吀栀e equipment described in Example 13-1 is the challenger (EUAC = $4,589).
Should the defender be kept or replaced if the interest rate is 8%?
Year
---2
3 -- 4 -- 5
1
0
l
4,000
L .⸀䰀 l l
2,800
3,800
4,800
FIGURE 13-4 Overhaul and maintenance costs for the defender in Exa mple 13 -7.
SO LUTI O N
吀栀e 昀椀rst step i s to determine the defender's lowest EUAC. 吀栀e pattern of overhaul and maintenance
costs (Figure 13-4) suggests that if the overhaul is done, the equipment should be kept 昀漀r several years.
吀栀e computation is as 昀漀llows:
If Retired at End of Year n
Year, n
E唀䄀C of Overhaul
$4,000(A/P, 8%, n)
E唀䄀C of Maintenance $1,800 + $1,000
Gradient 昀爀om Year 3 On
Total E唀䄀C
1
2
3
4
5
$4,320
2,243
1,552
1,208
1,002
$1,800
1,800
1,800 + 308*
1,800 + 683t
1,800 + 1,079
$ 6,120
4,043
3,660ⴀ堀
3,691
3,881
'For 琀栀e 昀椀rst three 礀攀ars, the maintenance is $1,800, $1,800, and $2,800. 吀栀us, E唀䄀C = 1,800 + l ,000(AIF, 8%, 3) = 1,800 + 308.
tE唀䄀C = 1,800 + l ,000(P!G, 8%, 3 ) (P/F, 8%, l ) (A℀倀, 8%, 4) = 1,800 + 683.
吀栀e lowest EUAC of the defender is $3,660. In Example 13-1, the challenger's minimum-cost life
was 昀漀ur years with an EUAC of $4,589. If we assume the equipment is needed 昀漀r at least 昀漀ur years, the
defender's EUAC ($3,660) is less than the challenger's EUAC ($4,589). Keep the defender.
If the defender's and challenger's cost data do not change, we can use replacement analysis tech­
nique 1 to determine when the defender should be replaced. We know 昀爀om the minimum EUAC
calculation that the defender should be kept at least three years. Is this the best life? 吀栀e 昀漀llowing
table computes the marginal cost to answer this question.
Year, n
Overhaul Cost
Maintenance Cost
Marginal Cost to Extend Service
0
1
2
3
4
5
$4,000
0
0
0
0
0
$
0
1,800
1,800
2,800
3,800
4,800
$6,120 = 4,000(1.08) + 1,800
1,800
2,800
3,800
4,800
continued
422
Cha pter 13 Replacement Analysis
吀栀e de昀攀nder's lowest EUAC occurs in Year 3, and Year 5 is the 昀椀rst year a昀琀er Year 3 in which the
$4,800 marginal cost exceeds the challenger's $4,589 minimum EUAC. 吀栀us, the defender should be
kept 昀漀ur more years if costs do not change. (Note that if the defender can be overhauled again a昀琀er
three or 昀漀ur years; that might be an even better choice.)
Replacement Analysis Technique 3 : When Defender
Marginal Cost Data Are Not Available
In this case, we simply compare the defender's E唀䄀C over its stated useful l椀昀e and the
challenger's minimum EUAC.
If the defender's marginal cost data are not known and cannot be estimated, it is im­
possible to apply replacement analysis techniques 1 or 2 to decide when the defender should
be replaced. Instead we must assume that the defender's stated use昀甀l life is the only one
to consider. In the real world the most likely scenario 昀漀r this approach involves a 昀愀­
cility-wide overhaul every three, 昀椀ve, ten, etc., years. Pipelines and many process plants,
such as re昀椀neries, chemical plants, and steel mills, must shut down to do m愀樀or mainten­
ance. All equipment is overhauled or replaced with a new challenger as needed, and the
昀愀cility is expected to operate until the next maintenance shutdown.
吀栀e defender's EUAC over its remaining use昀甀l life is compared with the challenger's
EUAC at its minimum-cost life, and the lower cost is chosen.
Complications in Replacement Analysis
De昀椀 n ing Defender and Challenger Fi rst Costs
Because the defender is already in service, analysts o昀琀en misunderstand what 昀椀rst cost to
assign it. Example 13-8 demonstrates this problem.
EXAM PLE 13 - 8
A model SK-30 was purchased two years ago 昀漀r $1,600; it has been depreciated by straight-line depreci­
ation with a 昀漀ur-year life and zero salvage value. Because of recent innovations, the current price of the
SK-30 is $995. An equipment 昀椀rm has o昀昀ered a trade-in allowance of $350 昀漀r the SK-30 on a new $1,200
model EL-40. Some discussion revealed that without a trade-in, the EL-40 can be bought 昀漀r $1,050.
吀栀us the originally quoted price of the EL-40 was overstated to allow a larger trade-in allowance. 吀栀e
true current market value of the SK-30 is probably only $200. In a replacement analysis, what value
should be assigned to the SK-30?
SO LUTI O N
In the example, 昀椀ve di昀昀erent dollar amounts relating to the SK-30 have been mentioned:
1.
2.
Original cost: It cost $1,600 two years ago.
Current cost: It now sells 昀漀r $995.
Co m p licati ons i n Replacement Analys is
423
3. Book value: 吀栀e original cost less two years of depreciation is 1,600 - 2/4(1,600 - 0) = $800.
4. Trade-in value: 吀栀e o昀昀er was $350.
5. Market value: 吀栀e estimate was $200.
We know that an economic analysis is based on the current situation, not on the past. In Chapter 1
we introduced the expression sunk costs to emphasize that past expenses have no relevance to current
decisions.
We want to use actual cash 昀氀ows 昀漀r each alternative. Here the question is: what value should be
used in an economic analysis 昀漀r the SK-30? 吀栀e relevant cost is the $200 current market value 昀漀r the
equipment. 吀栀e original cost, the present cost, the book value, and the trade-in value are all irrelevant.
At 昀椀rst glance, the trade-in value of an asset would appear to be a suitable present
value 昀漀r the equipment. But o昀琀en the trade-in price is in昀氀ated alo渀最 with the price 昀漀r the
new item. (吀栀is practice is so common in new-car showrooms that the term overt爀愀de is
used to describe the excessive portion of the trade-in allowance. 吀栀e buyer is also quoted
a higher price 昀漀r the new car.) Distorting the present value of the defender or the price of
the challenger can be serious because these distortions do not cancel out in an economic
analysis.
Example 13-8 illustrated that of the several di昀昀erent values that can be assigned to the
defender, the correct one is the current market value. If a trade-in value is obtained, care
should be taken to ensure that it actually represents a 昀愀ir market value.
Determining the value 昀漀r the challenger's installed cost should be less di昀케cult. In
such cases the 昀椀rst cost is usually made up of purchase price, sales tax, installation costs,
and other items that occur initially only once if the challenger is selected. 吀栀ese values are
usually straight昀漀rward to obtain. One aspect to consider in assigning a 昀椀rst cost to the
challenger is the potential disposition (or market or salvage) value. We must not arbitrarily
subtract the disposition cost of the defender 昀爀om the 昀椀rst cost of the challenger asset-this
practice can lead to an incorrect analysis.
As described in Example 13-8, the correct 昀椀rst cost to assign to the defender SK-30 is
its $200 current market value. 吀栀is value represents the present economic bene昀椀t that we
would be fo最ing to keep the defender. 吀栀is can be called our opportunity 昀椀rst cost. If, in­
stead of assuming that this is the defender's opportunity cost, we assume it is a cash bene昀椀t
to the challenger, a potential error arises.
吀栀e error lies in the incorrect use of a cash 昀氀ow perspective when the lives of the
challenger and the defender are not equal, as is u