Engineering Economy
Introduction & Syllabus
Dr. Ilham KISSANI
EGR2302-Engineering Economics
EGR2302Al Akhawayn University
Introduction & Syllabus
TOPICS
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About the course
Required textbook
Intended learning outcomes
Student’s efforts
Grading
Course Outline
Students conduct
Honors Supplement
EGR2302-Engineering
Economics
Al Akhawayn University
1-2
About the course
• Many investment opportunities (projects) involve an investment in
return for future income.
• This course deals with the application of economic analysis for making
selection decisions between investment opportunities (alternatives).
• Engineering economics provides the analytical means to consider cash
flows that occur at different times, to evaluate and compare
investment opportunities. EE groups techniques that simplify
alternatives comparison on an economic basis.
•
An engineer who can clearly and correctly communicate the financial
impacts of his ideas and designs, has more power in the decisionmaking process.
EGR2302-Engineering
Economics
Al Akhawayn University
1-3
About the course
• In today’s highly competitive global economy, products and systems
that cannot economically compete in the global market will never be
built.
• Although economics is important, many important factors are difficult
to be translated into dollars.
• Noneconomic criteria: environmental , cultural, social, legal, political.
• For this reason, economic analysis should not be the only criteria in
accepting or rejecting a design or an investment option.
EGR2302-Engineering
Economics
Al Akhawayn University
1-4
About the course
• If you have money now, you can make more money by investing it in a
company that will make a profit. Thus, $100 today is worth more than
$100 tomorrow; money has a time-value component.
• When we plan to make an investment we should analyse its
profitability.
• An investment is characterised by a rate of return, the number of
periods determining the planning horizon. If the investment is
economically justified we make it, otherwise we keep the current
situation.
• Interest tables are used to compute the value of money over time.
One of the strengths of using the Interest tables is that they make it
easy to obtain results.
EGR2302-Engineering
Economics
Al Akhawayn University
1-5
Required textbook
• Engineering Economy 7th edition, by Leland
Blank and Anthony Tarquin, 2011
• Bringing the textbook is essential to discuss
the examples covered in class
EGR2302-Engineering
Economics
Al Akhawayn University
1-6
Intended learning outcomes
1. Have acquaintance with the subject of economic analysis, i.e. be able
to formulate the problem, analyze it, search for alternatives, select
the preferred alternative .
2. Have a mastery of the notion of time value of money including the
concepts of present worth, future worth, annuities, gradient series,
geometric series.
3. Be able to tackle real life issues and practical applications such as
changing rates, inflation, payment operations and bond problems.
5. Be able to compare alternatives using the methods of measuring
investment worth.
6. Have an acquaintance with the economic analysis used in the public
sector.
7. Be able to include depreciation in economic analysis.
EGR2302-Engineering
Economics
Al Akhawayn University
1-7
Student’s efforts
• Participate actively in the class
• Take notes
• Solve selected problems in class
• Any effort will be considered
• You are welcome and highly encouraged to
see me during my office hours if you need
any help with this class.
EGR2302-Engineering
Economics
Al Akhawayn University
1-8
Grading
The course will consist of :
• two mid-term exams and a comprehensive final
exam (60%),
• homework assignments (20%), and
• 20% for 5 quizzes, class attendance and
participation
Final Grade:
The grade points assigned to each letter grade and
their corresponding percentage ranges are as
follows:
EGR2302-Engineering
Economics
Al Akhawayn University
1-9
Course Outline
Week 1
Topics: Foundations of Engineering Economy Ch 1
Week 2
Practice
Topics: Factors: How Time and Interest Affect Money Ch 2
Week 3
Topics: Factors: How Time and Interest Affect Money Ch 2
Practice
Week 4
Quiz1: Chp 1+2
Combining Factors Ch 3
Practice
Nominal and Effective Interest Rates Ch 4
Nominal and Effective Interest Rates Ch 4
Practice
Break
Quiz2: chp 3+4
Mid-term Exam I (1+2+3+4)
Present Worth Analysis Ch 5
Annual Worth Analysis Ch 6
Quiz3(Chp 5+6)+ Excel Application
Rate of Return Analysis Ch 7
Rate of Return Analysis Ch 8
Quiz4 (Chp 7+8) + Excel Application
EXAM II (5+6+7+8)
Benefit/Cost Analysis Ch 9
Excel Application Ch9
Effects of Inflation Ch 14
Quiz5: ch 9+14
Depreciation Methods Ch 16
1 - 10
Quiz6: Ch 16
Review
Week 5
Week 6
Week 7
Week 8
Week 9
Week 10
Week 11
Week 12
Week 13
Week 14
Students conduct
Refer to:
• http://www.aui.ma/DSA/manual_and_for
ms/Student%20Code%20of%20Conduct.p
df
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Economics
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Course Rules
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The class notes are available in the portal.
The homework assignments are announced in class at the end of each chapter and posted
in the portal.
The homework assignments are due one week after their announcement. They are
corrected right after their collection. Late submission is not accepted. The homework
solutions are not posted: you should take notes!
Exams/quizzes respect a structured template to help you structure your thoughts and
answers, I hope you‘ll write your answers in a structured way: write your answers in the
spaces provided in the exam sheet.
Quiz problems are corrected right after their collection and solutions are not posted: you
should take notes!
Exams and their solutions are discussed. The exam grades are given before the following
exam. You can review your grade within the first week following posting and grading.
No make-ups allowed for exams/quizzes without a doctor certificate or other official excuse.
EGR2302-Engineering
Economics
Al Akhawayn University
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Course Rules (continued)
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We will drop the lowest homework/quiz score from grading. This means you can miss 1
homework/quiz and not be penalized for it.
Exams and quizzes will be closed book. You will be allowed 1 page of notes (front and back)
but it must be handwritten (no photocopying). If someone brings the book to the quiz or
the exam session, his/her grade will be reduced by 1 or 2 respectively.
You bring only: a calculator, 1 page of notes, 1 scratch paper.
Calculators can be used but no laptops or cell phones allowed.
Exchanging any material in the exams in not allowed.
Switching to another section should be granted by the instructor, either for classes, quizzes
or exams .
No grades given by email.
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Economics
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Honors Supplement
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Risk analysis
Introduction to computer computations
EE Software
Literature Review
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Economics
Al Akhawayn University
1 - 14
Your Health First!
Engineering Economy
Chapter 1
Foundations of Engineering
Economy
Session 11-3
Dr. Ilham KISSANI
EGR2302-Engineering Economics
EGR2302Al Akhawayn University
Chapter 1 - Foundations of Engineering
Economy
PURPOSE
Introduce the fundamental concepts of engineering
economy
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Economics
Al Akhawayn University
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Chapter 1 - Foundations of
Engineering Economy
TOPICS
1.1. Engineering Economy : Description and Role in decision making
1.2. Performing an Engineering Economy Study
1.3. Professional Ethics and Economic Decisions
1.4. Interest rate and rate of return
1.5. Terminology
1.6. Cash Flows: Estimation and Diagramming
1.7. Economic Equivalence
1.8. Type of interest: Simple and Compound Interest
1.9. MARR
1.10. Computation and spreadsheets
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Sec1.1. Engineering Economy : Description
and Role in decision making
Professionals need tools to:
• analyze different situations of cash flows:
– before-tax/After-tax, with/without inflation, …
– buy or lease a car,
– shall I upgrade my production capacity now or
next year. By machine 1 or machine 2..
– opt for manual work or automated work
• make wise decisions
• Implement the best solution
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Sec1.1. Engineering Economy : Description
and Role in decision making
• Remember: People make decisions – not
“tools”
• Engineering Economy is a set of tools that
assist in decision making – but will not make
the decision for you
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Sec1.1. Engineering Economy : Description
and Role in decision making
Problem Solving Approach
1. Understand the Problem/ Project
2. Identify the feasible alternatives for the decision making
process and define the criteria /Collect all relevant
data/information
3. Make realistic cash flows estimates
4. Identify an economic measure of worth (PW,AW, FW…)
5. Evaluate each alternative
6. Select the “best” alternative
7. Implement the solution and monitor the results
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Sec1.1. Engineering Economy : Description
and Role in decision making
Sec 1.2 Performing a Study
Fundamental terminology:
– Alternative -- stand-alone representation of a specific situation
– Cash flow -- the flow or movement of money at some specific
time
• Estimated inflows (revenues or receipts or benefits or incomes) and
• Estimated outflows (expenses or disbursements or losses or costs) for
an alternative
– Evaluation criteria -- factor used to select the ‘best’ alternative
(usually money=cost or revenue)
– Time value of money -- Change in amount of money over time
(Most important concept in Eng. Econ.)
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Sec 1.2 Performing a Study
• To financially analyze engineering projects, we
need to model the projects in terms of cash flows
• Different plans/options could raise from the
project.
• The alternatives represent the different plans to
model the project.
• Goal: Analyze alternatives in terms of cash flows
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Sec 1.2 Performing a Study
• To resolve the decision-making
problem(project), one must have alternatives
(two or more ways to represent the problem:
Do-Nothing (Status-quo), equal periodic
amounts, etc.)
• Alternative must first be identified adequately
• Estimate the cash flows for the defined
alternatives
• Analyze the cash flows for each alternative
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Sec 1.2 Performing a Study
• Inflows: Estimate flows of money coming into
the firm – revenues, receipts, benefits ,
incomes =positive cash flows
• Outflows: Estimates flows of money leaving an
account- expenses, investment costs,
operating costs, taxes paid = negative cash
flows
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Sec 1.2 Performing a Study
•Taxes represent a significant negative cash flow to the
for-profit firm.
•A realistic economic analysis must assess the impact
of taxes called an AFTER-TAX cash flow analysis
•Not considering taxes is called a BEFORE-TAX Cash
Flow analysis
• For the firm activities (selling finished product, buying raw
materials, storage, transportation, ect.), one should consider
either before-tax Net Cash flows (revenue-cost) for the aftertax Net Cash flows for whole activities.
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Sec 1.3 Professional Ethics and
Economic Decision
• Ethical using practices can be evaluated by using a code of ethics or code of
morals that forms the standards to guide decisions of individuals and
organizations.
• Types of morals and ethics:
– Universal or common morals: These are fundamental moral beliefs held by
virtually all people. Most people agree that to steal, murder, lie, or physically
harm someone is wrong.
– Individual or personal morals: These are the moral beliefs that a person has
and maintains over time. These usually parallel the common morals in that
stealing, lying, murdering, etc. are immoral acts.
– Professional or engineering ethics: Professionals in a specific discipline
(engineers, doctors…) are guided in their decision making and performance
of work activities by a formal standard or code. The code states the
commonly accepted standards of honesty and integrity that each individual
is expected to demonstrate in her or his practice.
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Sec 1.4 - Interest Rate vs ROR
• Interest is a manifestation of time value of money calculated as
difference between an ending amount and a beginning amount of
money
Interest = end amount – original amount
• Interest rate is interest over specified time period
based on original amount
Interest rate (%) =
interest accrued per time unit
x 100%
original amount
• Interest rate and rate of return (ROR) have same numeric
value, but different interpretations
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Sec 1.4 - Interest Rate and ROR Interpretations
Borrower’s perspective
•Take loan of $5,000 for
one year; repay $5,200
Investor’s perspective
(lender)
•Invest (or lend) $5,000 for
one year; receive $ $5,200
•Interest paid = $200
•Interest earned = $ 200
•Interest rate = 200/5,000
= 4%
INTEREST RATE
•Rate of return= 200/5,000
= 4%
RATE OF RETURN (ROR)
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Examples: 1-3 to 1-5
1 - 30
Sec 1.5 - Terminology and Symbols
• t = time unit: start or end of periods; years,
months, etc.
• P = present value of money at time t = 0; $
• F = value of money at a future time t; $
• A = series of equal, end-of-period cash flows;
currency per period, e.g. per year
• n = total number of periods;
• i = compound interest rate or rate of return;
% per t
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Sec 1.5 - Terminology and Symbols
Examples :
Borrow $10,000 today and repay after 5 years
the principal plus the accrued interest at 8%
per year.
Identify all symbols…
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Sec 1.5 - Terminology and Symbols
Examples:
Borrow $2,000 today and repay annually for
10 years starting next year at 7% per year
compounded.
Identify all symbols.
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Sec 1.5 - Terminology and Symbols
•Example : Borrow $2,000 today and repay
annually for 10 years starting next year at 7%
per year compounded. Identify all symbols.
• Given:
P = $2,000
• Find: A = ? per year for 10 years
•
i = 7% per year
•
n = 10 years
•
t = year 1, 2, …, 10
Examples 1-6 to 1-8
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Sec 1.6 – Cash Flow Estimates
• Cash inflow – receipt, revenue, income, saving
• Cash outflows – cost, expense, disbursement, loss
• Net cash flow (NCF) = inflow – outflow
• End-of-period convention: all cash flows and
NCF occur at the end of an interest period
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Sec 1.6 – Cash Flow Diagrams (CFD)
Typical time CFD scale or 5 years
Year 1
Year 5
Time t
0
1
2
3
4
5
Find P in year 0, given 2 cash flows
+ Inflow
P=?
0
1
2
3
4
- Outflow
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5
Sec 1.6 – Cash Flow Diagrams
• Example: Find an amount to deposit 2 years from
now so that a withdraw of $4,000 per year can be
made for 5 years starting 3 years from now. Assume i
= 15.5% per year
See 1-9 , 1-11
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Sec 1.7 - Equivalence
Different sums of money at different times may be equal in economic value
Interest rate = 6% per year
-1
0
$94.34 last year
1
$100 now
$106 one year
from now
Interpretation:
$94.34 last year(=100/1.06),
$100 now,
$106 (=100*1.06) one year from now
are equivalent only at an interest rate of 6% per year
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Sec 1.8 – Simple and Compound Interest
Simple interest is always based on the
original amount, which is also called the
principal
•
Interest per period = (principal)(interest rate)
• Total interest = (principal)(n periods)(interest rate)
Example s: Invest $1000 in a bond at 5% per year simple
Interest each year = 1000(0.05) = $50
Interest over 3 years = 1000(3)(0.05) = $150
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Sec 1.8 – Simple and Compound Interest
Compound interest is based on the principal
plus all accrued interest
• Interest per period = (principal + accrued interest)(interest rate)
n periods
• Total interest = (principal)(1+interest rate)
- principal
Examples: Invest $1000 at 5% per year compounded
Interest, year 1 = 1000(0.05) = $50
Interest, year 2 = 1050(0.05) = $52.50
Interest, year 3 = 1102.5(0.05) = $55.13
3
Interest over 3 years = 1000(1.05) – 1000 = $157.63
• Examples 1-14, 1-15
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Sec 1.9 - ROR and MARR
• Cost of capital (COC) – interest
rate paid for funds to finance
projects
ROR
• MARR – Minimum ROR needed
for an alternative to be justified
and economically acceptable.
MARR > COC.
• If COC = 5% and 7% must be
realized MARR = 12%
• Verify this inequality for
acceptable projects
•
ROR ≥ MARR > COC
1 - 41
EGR2302-Engineering Economics
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Sec 1.10 – Introduction to Solution
by Computer
All spreadsheets and computation with Excel will
be covered at the end of each chapter
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Sec 1.10 – Introduction to Spreadsheet
Functions
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Excel Function
To display
Present value, P
= PV(i%,n,A,F)
Future value, F
= FV(i%,n,A,P)
Annual amount, A
= PMT(i%,n,P,F)
# of periods, n
= NPER(i%,A,P,F)
Compound rate, I, A fixed = RATE(n,A,P,F)
i for input series , A variable
= IRR(first_cell:last_cell)
P for any series
= NPV(i%,second_cell:
last_cell)+first_cell
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Cash flow Diagram
& Examples
• To financially analyze engineering projects, we need
to model the projects in terms of cash flows
• Simple Interest • Compound Interest EGR2302-Engineering
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Cash flow Diagram
& Examples
• Cash flow movements can be visually
displayed through the use of a cash flow
diagram (CFD)
• Cash Flow Diagram describes type, magnitude
and timing of cash flows over project horizon
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Cash flow Diagram
& Examples
Typical cash flow time scale for 5 years
Year 5
Year 1
0
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1
2 Time, t 3
1 - 46
4
5
Cash flow Diagram
& Examples
• The start of the diagram represents the
beginning of the project
• In a cash flow diagram (CFD) the end of
period(t) = the beginning of period (t+1)
• When t = 0, this is the present
• When t = 1, this is the end of the first year (or
beginning of the second year)
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Cash flow Diagram
& Examples
• A cash flow diagram is a picture of a financial project that
illustrates all cash inflows and outflows plotted along a
horizontal time line
• upward arrows = positive flows (inflows)
• downward arrows = negative flows (outflows)
• A specific arrow denotes the Net Cash Flows for a specific
time :
Net Cash Flows(t) = Sum (inflows(t)-outflows(t))
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Cash flow Diagram
& Examples
0
1
P
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2
n
Receive revenues and pay costs over
time.
1 - 49
Cash flow Diagram
& Examples
Net Cash Flows(t) = Sum (inflows(t)-outflows(t))
0
1
P
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2
n
Specify the NET cash flow in each period.
1 - 50
Cash flow Diagram
& Examples
Example 1:
A man borrowed $1,000 at 8% interest.
Two end-of-year payments:
1- At the end of the first year, he will repay half of the $1000
principal plus the interest that is due.
2- At the end of the second year, he will repay the remaining
half plus the interest for the second year.
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Cash flow Diagram
& Examples
Cash flow for this problem:
End of Prd Interest/period Cash flow
Total Owed after pmt
(+Earned; -/paid
0
1
2
$80
$40
-$580 (-$500-$80)
-$540 (-$500 - $40)
$1000
$500 (1080-500-80)
$0 (540-540)
-$1120
CFD of Ex1:
Year 1
$1,000
0
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1
$580
1 - 52
2
$540
(new principal=500$)
Cash flow Diagram
& Examples
Example2
• A rental company spent $2,500 on a new compressor 7 years
ago
• The annual rental income from the compressor has been $750
• Additionally, the $100 spent on maintenance during the first
year has increased each year by $25
• The company plans to sell the compressor at the end of next
year for $150
Construct the cash flow diagram from the company’s perspective
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Cash flow Diagram
& Examples
Cash flow for this problem:
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Cash flow Diagram
& Examples
CFD of Ex2:
See example 1.16
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Cash flow Diagram
& Examples
Example 3:
• You have deposited $1,000 with an interest rate of 3% every 6
months where the interest is computed every 6 months
• How much you will have after 5 years?
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Cash flow Diagram
& Examples
• Cash flow for this problem:
End of Prd
Interest/period Cash flow
Total Owed
(+Earned; -/paid
0
10
$343.92
+$ 1343.92
• CFD Ex3:
1 - 57
$ 1000
$0
Cash flow Diagram
& Examples
Example 4:
• You invested a $1,000 by a deposit in your account that gives
a daily interest of 0.003 where interest is paid monthly.
• If you deposit another $2,000 on the 10th day and
withdraw $500 on the 25th day, what is your cash flow
at the end of the 30th day?
Assume simple interest
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Cash flow Diagram
& Examples
Find the Cash flow for this problem:
End of Prd Interest/period Cash flow
Total Owed after Pmt
(+Earned; -/paid)
0
10
25
30
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30
135
37.5
-1000
-2000
+500
2702,02
1 - 59
1000
3000
2500
0
Cash flow Diagram
& Examples
CFD Example 4:
F= 1000x(1+30x0,003)+2000x(1+20x0,003)500x(1+5x0,003)
= 1000x(1+0,09)+2000x(1+0,06)-500x(1+0,015)
= 2,702.03
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Rule of 72 ( and 100)
• Compound interest case (Table1-4)
• Simple interest case
-Estimates number of years (n) or
interest rate (i) required for an
amount to double in size at a
stated compound interest rate
n ~ 72 / i
- Doubling time is exact,
using rule of 100
-At compound i = 10%, $1,000 doubles to
$2,000 in ~7.2 years
Approximate i ~ 72 / n
-$1,000 doubles to $2,000 in ~7.2 years
for a compound i = 10%
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n = 100/i
At simple i = 10%, $1,000
doubles to $2,000 in
exactly 10 years
i = 100/n
$1,000 doubles in 10
years at 10% simple
interest
Engineering Economy
Chapter 2
How Time and Interest Affect
Money
Sessions 44-6
Dr. Ilham KISSANI
EGR2302-Engineering Economics
EGR2302Al Akhawayn University
Chapter 2 - How Time Affect & Interest
Money
PURPOSE
Analyze the time value of money related to real life
situations using cash flow factors
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Chapter 2 - How Time & Interest
Affect Money
TOPICS
2.1. Single Payment Factors
2.2. Uniform Series: Present Worth Factor, P/A, and Capital Recovery Factor,
A/P
2.3. Sinking Fund Factor, A/F, and Uniform Series Compound Amount Factor,
F/A
2.4. Interpolation in interest table
2.5. Arithmetic Gradient Factors, P/G, A/G
2.6. Geometric Gradient Series Factor
2.7. Determination of Unknown Interest Rate & Unknown Number of Years
2.8. Spreadsheet Application
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Sec2.1. Single Payment Factors
Single-payment compound amount factor
(SPCAF)
This factor allows for obtaining the amount of
money F of a future payment accumulated
after n periods from a single present worth, P,
for a given compounded interest per period
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Sec2.1. Single Payment Factors
Single-payment compound amount factor
(SPCAF)
For n =1
F1 = P + interest
F1 = P + Pi
F1 = P(1 + i)1
For n =2
F2 = F1 + interest for F1
F2 = F1 + F1 i
F2 = (P + Pi) + (P + Pi)i
F2 = P + Pi + Pi + Pi2
F2 = P(1 + 2i + i2)
F2 = P (1 + i)2
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For n=3
F3 = F2 + interest for F2
F3 = P (1 + i)2 + P (1 + i)2i
F3 = P (1 + i)3
For n = n
Fn = P(1+i)n
1 - 66
Sec2.1. Single Payment Factors
Single-payment compound amount factor
(SPCAF)
This factor is referred to as the F/P Factor and
computed using the formula: (1+i)n
(1+i)n is called the single-payment compound
amount factor, SPCAF or the F/P factor.
NOTATION: To find F given P
( F/P , i% , n )
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Sec2.1. Single Payment Factors
• Single-payment present worth factor
(SPPWF)
This factor allows for obtaining the amount of
money, P, of a present payment given the
equivalent future amount
P = F/(1+i)n
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Sec2.1. Single Payment Factors
• Single-payment present worth factor
(SPPWF)
This factor is referred to as the P/F Factor and
computed using the formula: 1/(1+i)n
1/(1+i)n is called the single payment present
worth factor, SPPWF, or the P/F factor.
NOTATION: Find P given F
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( P/F,i% , n )
1 - 69
Sec2.1. Single Payment Factors
Standard notation represented by the general form :
(X/Y, i, n)
Referes to:
X = what is sought
Y = what is given
i = interest rate, %
n = number of periods
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Sec2.1. Single Payment Factors
• One can either use formulas for calculating P or F, or
Interest Tables to make calculations easier
• Interest tables are given at the back of the book
• For a given interest rate and a given number of interest
periods we obtain:
P/F (Present Worth), or F/P (Compound Amount) factors
• When a given interest rate is not given at the back of the
book, we can use :
1) Formulas (exact)
2)Interpolation (approximation)
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Sec2.1. Single Payment Factors
Example : Given: (F/P, 4%, 5), Find: F/P then check the table value
F the future worth is what is sought
P the present worth is what is known
4% is the interest rate
5 is the number of time periods
(F/P, 4%, 5) = (1+i)n = (1+0.04)5 = 1.045
(F/P, 4%, 5) = 1.21673
From Table 9, p. 589 : (F/P, 4%, 5) = 1.2167
Example 2.1, 2.2
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Sec2.2.Uniform Series: Present Worth Factor, P/A,
and Capital Recovery Factor, A/P
• Uniform series present worth factor(USPWF)
is used to calculate P given the number of
payments A over a certain period
Example : to buy a car we can pay its price P
over n periods with equal payments A and
interest rate i% per month
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Sec2.2.Uniform Series: Present Worth Factor, P/A,
and Capital Recovery Factor, A/P
• Uniform series present worth factor (USPWF)
P = A[1 \ (1+i)1] + A[1 \ (1+i)2] …. +A[1 \ (1+i)n)]
P = A{ [1 \ (1+i)1] + [1 \ (1+i)2] …. +[1 \ (1+i)n] }
P/(1+i) = A[1 \ (1+i)2] + A[1 \ (1+i)3] …. +A[1 \
(1+i)n+1)]
-P = -A{ [1 \ (1+i)1] + [1 \ (1+i)2] …. +[1 \ (1+i)n]}
P = A [(1+i)n -1\ i(1+i)n] i≠
≠0
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Sec2.2.Uniform Series: Present Worth Factor, P/A,
and Capital Recovery Factor, A/P
• Uniform series present worth factor(USPWF)
This factor is referred to as the P/A Factor and
computed using the formula:
[(1+i)n -1\ i(1+i)n]
This is called the single payment present
worth factor, USPWF, or the P/A factor.
NOTATION: Find P given A
( P/A , i% , n )
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Sec2.2.Uniform Series: Present Worth Factor, P/A,
and Capital Recovery Factor, A/P
• Uniform Series Capital Recovery Factor(USCR)
is used to calculate the number of payments A
over a certain period given the principal P
Example : how many equal payments A can
we get over n periods with an interest rate i%
if we invest P as principal amount
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Sec2.2.Uniform Series: Present Worth Factor, P/A,
and Capital Recovery Factor, A/P
Uniform Series Capital Recovery Factor(USCR)
A = P [ i(1+i)n / (1+i)n -1] i≠
≠0
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Sec2.2.Uniform Series: Present Worth
Factor, P/A, and Capital Recovery Factor, A/P
• Uniform series capital recovery (USCRF)
This factor is referred to as the A/P Factor and
computed using the formula:
[ i(1+i)n / (1+i)n -1]
This is called the uniform series capital recovery
factor, (USCR), or the A/P factor.
NOTATION: Find A given P
( A/P ,i% , n )
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Sec2.2.Uniform Series: Present Worth Factor, P/A,
and Capital Recovery Factor, A/P
Uniform series capital recovery (USCRF)
Example:
Given: 12%, n=10
Find: Calculate A/P, calculate P/A
A uniform series end of period payment
12% is the interest rate
10 is the number of time periods
(A/P, 12%, 10) = 0.17699 using the formula
(A/P, 12%, 10) = 0.17698 from T17 p.597
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Sec2.2.Uniform Series: Present Worth Factor, P/A,
and Capital Recovery Factor, A/P
Example:
Given: (A/P, 12%, 10)
Find: Calculate A/P, calculate P/A
P/A = 1/(A/P) = 1/.17699
P/A = 5.6500
From T17 p. 597
(P/A, 12%, 10) = 5.6502
--------------------
Example 2.4
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Sec2.3 Sinking Fund Factor, A/F, and Uniform
Series Compound Amount Factor, F/A
Sinking Fund Factor (SFF)
is used to calculate the number of payments A
over a certain period given F
Example : how many equal payments A can
we get over n periods with an interest rate i%
if we expect to pay F as final amount
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Sec2.3 Sinking Fund Factor, A/F, and Uniform Series
Compound Amount Factor, F/A
Sinking Fund Factor (SFF)
A = F [i/(1+i)n -1] i≠
≠0
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Sec2.3 Sinking Fund Factor, A/F, and Uniform Series
Compound Amount Factor, F/A
Sinking Fund Factor (SFF )
This factor is referred to as the A/F Factor and
computed using the formula:
[i/(1+i)n -1]
is called the sinking fund factor (SFF), or the
A/F factor.
NOTATION: Find A given F
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( A/F , i% , n )
1 - 83
Sec2.3 Sinking Fund Factor, A/F, and Uniform
Series Compound Amount Factor, F/A
Uniform Series Compound Amount Factor
(USCAF)
is used to calculate the future amount F given
the number of payments A over a certain
period
Example : how much do we expect to pay for F
as final amount given equal payments A over
n periods with an interest rate i% if
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Sec2.3 Sinking Fund Factor, A/F, and Uniform Series
Compound Amount Factor, F/A
Uniform Series Compound Amount Factor
(USCAF)
F = A [ (1+i)n -1/ i] i≠
≠0
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Sec2.3 Sinking Fund Factor, A/F, and Uniform Series
Compound Amount Factor, F/A
Uniform Series Compound Amount Factor
(USCAF)
This factor is referred to as the F/A Factor and
computed using the formula: [(1+i)n-1/i]
is called the uniform series compound amount
factor, (USCAF), or the F/A factor.
NOTATION: Find F given A
( F/A ,i% , n )
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Sec2.3 Sinking Fund Factor, A/F, and Uniform Series
Compound Amount Factor, F/A
Example:
Given: 12%, n=10
Find: calculate A/F, calculate F/A
A uniform series end of period payment
12% is the interest rate
10 is the number of time periods
(A/F, 12%, 10) = 0.056984
From T17 p. 597: (A/F, 12%, 10) = 0.05698
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Sec2.3 Sinking Fund Factor, A/F, and Uniform Series
Compound Amount Factor, F/A
Example:
Given: (A/F, 12%, 10)
Find: calculate A/F, calculate F/A
F/A = 1/(A/F) = 1/.05698
F/A = 17.5487
From T17 p. 597 : (F/A, 12%, 10) = 17.5487
-----------------Example 2.5, 2.6
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Sec 2.4. Interpolation in interest
table
Use the following basic relationships to obtain the
unlisted value related to a specific factor for a desired i
or n which are not available in the interest tables:
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Sec 2.4. Interpolation in interest
table
Example :
Find the A/P factor for n=10, i = 7.3% from the following
relationships (see table 12 and 13)
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X represents the unlisted
value of A/P
2.7
1 - 90
Sec 2.5. Arithmetic Gradient
Factors, P/G, A/G
• An arithmetic gradient is a cash flow series
that changes (either increases or decreases)
by a fixed amount, from one time to another:
CF(j)= Base amount+(j-1) x G
for any period j
comprised of TWO main components:
The base annuity component + The Gradient component
Note : G is a constant arithmetic change in the CF, it may be
positive or negative. Exp2.8
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Sec 2.5. Arithmetic Gradient
Factors, P/G, A/G
Examples: Cash flows that increase or decrease by a
constant amount G considered as arithmetic
gradient cash flows. The amount of increase is 25
(or decrease is 500) is the gradient.
$2000
$1500
$175
$150
$125
$100
0 1
2
3
$1000
$500
4
0 1
Base = $100
3
4
G = -$500
G = $25
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Base = $2000
Sec 2.5. Arithmetic Gradient Factors,
P/G, A/G
A positive, increasing arithmetic gradient
(borrower perspective)
P = ?
i = g iv e n
B ase A m o u n t
0
1
2
1 G
3
2 G
n -1
n
(n -2 )G
(n -1 )G
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Sec 2.5. Arithmetic Gradient Factors,
P/G, A/G
A1+(n-1)G
A positive, increasing arithmetic gradient
A +(n-2)G
(investor perspective)
1
A1+2G
A1+G
A1
P
0
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2
3
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n-1
n
Sec 2.5. Arithmetic Gradient Factors,
P/G, A/G
Gradient Component only
(n-1)G
(n-2)G
3G
2G
1G
0G
0
1
2
3
4
………..
P at time t = 0 ( 2 periods to the left of 1G)
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n-1
n
Sec 2.5. Arithmetic Gradient Factors,
P/G, A/G
How to derive the P/G factor:
1- we need to decompose PT using 2
formulas:
PT= PA+PG
PT= PA-PG
for positive gradient
for negative gradient
Where:
PT= the total Present Worth
PG= obtained from the Present Worth of the Arithmetic Gradient
PA= obtained from the Present Worth of the Unifom Series
Note: For simplification we refer to PG as P
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Sec 2.5. Arithmetic Gradient
Factors, P/G, A/G
Decomposition of cash flows to find P for increasing
arithmetic Gradient:
$160
$140
$120
$100
0 1
2
G = $20
3
$60
$40
$20
$100
=>
4
+
0 1
2
3
4
0 1
2
3
Base = $100
PT = 100(P/A,i,4) + 20 (P/G,i,4)
Convention : the gradient starts in year 2.
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4
Sec 2.5. Arithmetic Gradient Factors,
P/G, A/G
Decomposition of cash flows to find P for decreasing
arithmetic gradient cash flow:
$2000
$1500
$1000
$500
0 1
2
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3
4
$2000
=>
-
0 1
2
3
4
$1500
$1000
$500
0 1
2
3
4
PT = 2000(P/A,i,4) - 500(P/G,i,4)
1 - 98
Sec 2.5. Arithmetic Gradient Factors,
P/G, A/G
P = G(P/ F,i%,2) + 2G(P/F,i%,3) + …
+ [(n-2)G](P/F,i%,n-1)+[(n-1)G](P/F,i%,n)
P = G{(P / F,i%,2) + 2(P/F,i%,3) + …
+ [(n-2)](P/F,i%,n-1)+[(n-1)](P/F,i%,n)}
 1
2
n-2
n-1 
P=G 
+
+ ... +
+
2
3
n-1
n 
(1+i)
(1+i) 
 (1+i) (1+i)
Multiply both sides by (1+i)
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Sec 2.5. Arithmetic Gradient Factors,
P/G, A/G
Subtracting [1] from [2]:
2
 1
2
n-2
n-1 
P(1+i) =G 
+
+ ... +
+
1
2
n-2
n-1 
(1+i)
(1+i)
(1+i)
(1+i)


1
-
1
 1
2
n-2
n-1 
P=G 
+
+ ... +
+
2
3
n-1
n 
(1+i)
(1+i) 
 (1+i) (1+i)
G  (1 + i ) − 1
n 
−
P= 
n
n
i  i (1 + i )
(1 + i ) 
n
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Sec 2.5. Arithmetic Gradient
Factors, P/G, A/G
To find P for a gradient cash flow that starts
at the end of year 2 and ends at year n we
n

G
(
1
+
i
)
−1
n 
apply:
P=
−

i  i (1 + i ) n

(1 + i ) 
n
or P = G(P/G,i,n)
1  (1 + i ) n − 1
n 
−
Where (P/G,i,n) = 

n
i  i (1 + i )
(1 + i ) n 
( P / G, i %, N ) factor
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Sec 2.5. Arithmetic Gradient
Factors, P/G, A/G
Convert G to an equivalent A
A = G ( P / G, i, n)( A / P, i, n)
1 (1+ i) −1
n  i(1 + i)
−

n
n
n
i  i(1+ i)
(1+ i)  (1 + i) −1
n
n
AG= G
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Sec 2.5. Arithmetic Gradient
Factors, P/G, A/G
To find the uniform annual series, A, for an
arithmetic gradient cash flow G:
1

n
AG = G  −

n
i
(
1
+
i
)
−
1


Where (A/G,i,n) =
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1

n
 i − (1 + i ) n − 1


1 - 103
Sec 2.5. Arithmetic Gradient
Factors, P/G, A/G
To find the uniform annual series, A, for an
arithmetic gradient cash flow G:
1

n
AG = G  −

n
i
(
1
+
i
)
−
1


The equivalent total annual series:
AT = AA+AG = Base amount + AG
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Sec 2.5. Arithmetic Gradient
Factors, P/G, A/G
$700
$600
Example:
$500
$400
$300
$200
$100
0
1
2
3
4
5
6
P = 100 (P/A, 10%,7) + 100 (P/G, 10%,7)
Page 595: (P/A, 10%,7) = 4.8684
and
(P/G, 10%,7) = 12.763
Total PT = 486.84 + 1,276.3 = 1,763.14
Total AT = PT (A/P, 10%, 7) = 100+100 (A/G, 10%, 7) =362,16
See Exp2.9
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7
Sec 2.6. Geometric Gradient Series
Factor
As opposed to the previous analysis in which the gradient
increased or decreased by a constant amount, a geometric
gradient is one in which the increase or decrease is a constant
percentage :
A(j)= A1 x (1+g)j-1 for any period j
P = ?
0
g = g iv e n
i = g iv e n
1
A
1
2
A 1(1 + g )
3
n-1
n
A 1(1 + g )2
A 1 ( 1 + g ) n -1
Here g is positive, it may also be negative
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Sec 2.6. Geometric Gradient Series
Factor
Typical Geometric Increasing Gradient
A1 = the first cash flow in the series
0
1
A1
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2
3
4
……..
n-1
n
A1(1+g)
A1(1+g)2
A1(1+g)3
A1(1+g)n-1
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107
Sec 2.6. Geometric Gradient Series
Factor
Typical Geometric decreasing Gradient
Let A1 = the first cash flow in the series
0
1
2
3
4
……..
n-1
n
A1(1-g)n-1
A1
A1(1-g)3
A1(1-g)2
A1(1-g)
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108
Sec 2.6. Geometric Gradient Series
Factor
Here is a Geometric gradient when the periodic
payment is increasing by a constant percentage:
$133
A1 = $100,
A2 = $100(1+g)
A3 = $100(1+g)2
An = $100(1+g)n-1
$121
$110
$100
0 1
We get :
g=(A2 - A1 )/ A1= 10%
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2
3
4
Sec 2.6. Geometric Gradient Series
Factor
A1
A1 (1 + g ) A1 (1 + g )2
A1 (1 + g )n−1
Pg =
+
+
+ ... +
1
2
3
(1 + i)
(1 + i)
(1 + i)
(1 + i)n
Factor out the A1 value
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Sec 2.6. Geometric Gradient Series
Factor
 1
(1 + g )1 (1 + g ) 2
(1 + g ) n −1 
Pg = A1 
+
+
+ ... +
2
3
n 
(1 + i )
(1 + i ) 
 (1 + i ) (1 + i )
(1)
(1+g)
(1+g)  1
(1 + g )1 (1 + g ) 2
(1 + g ) nn−−1  (2)
= A1
+
+
+ ... +
Pg


2
3
(1+i)
(1+i)  (1 + i ) (1 + i )
(1 + i )
(1 + i ) n 
Subtract (1) from (2) and the result is…..
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Sec 2.6. Geometric Gradient Series
Factor
n

1+g
(1
+
g
)
1 


Pg 
− 1 = A1 
−

n +1
1+ i 
 1+i 
 (1 + i )
n

1+ g  
1 − 
 
1+ i  


Pg = A1
g ≠i


i−g




n A1
Pg =
(1 + i )
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Sec 2.6. Geometric Gradient Series
Factor
Example 2.11:
•Assume maintenance cost will be $1700 the first year.
•Assume an annual increase of 11% per year over a 6-year
time period.
•If the interest rate is 8% per year, determine the present
worth of the future expenses.
•Draw a cash flow diagram to represent the model.
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Sec 2.6. Geometric Gradient Series
Factor
g= 11%;
A1 = $1700;
i = 8%;
n=6
0
1
$1700
2
3
4
5
6
$1700(1.11)1
$1700(1.11)2
$1700(1.11)3
P(8%) = ??
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$1700(1.11)5
1 - 114
Sec 2.6. Geometric Gradient Series
Factor
g= 11% ;
A1 = $1700;
i = 8%;
n=6
0
1
$1700
2
3
4
5
6
$1700(1.11)1
$1700(1.11)2
$1700(1.11)3
P = 1700 x 5.9559
$1700(1.11)5
=10125.11
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Sec 2.7. Unknown Interest Rate
•In some cases, the cash flows and the related number of
period are known, and the interest rate is unknown.
•When single amount, uniform series are involved the
unknown interest rate can be found by direct solution of the
time value of money equation.
•As opposed to solving directly for a rate, it often easier to
solve for the value of the factor and then use the tables to find
the actual rate.
•When non-uniform series are involved, more complex
problems can be solved by trial and error or a numerical
method (chapter 7).
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Sec 2.7. Unknown Interest Rate
One can use either direct calculation or tables
Example 1:
Given a present worth of $100, what interest rate is required to
obtain $1000 in 14 years.
1- Using direct calculation
P = F(1/(1+i))n
or F = P (1+i)n
(1+i)14=1000/100 = 10
1+i = 1.1788
i= .1788 = 17.88%
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Sec 2.7. Unknown Interest Rate
2- Using Tables
P = F(P/F, i, 14)
100 = 1000 (P/F, i, 14)
(P/F, i, 14) = 100/1000 = 0.10
We don’t have an exact value for i related to this factor from the
interest tables
We need to apply interpolation
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Sec 2.7. Unknown Interest Rate
2- Using Tables
Check the P/F factors at n=14 around 0.100
For i=16%
P/F =0.1253
For i=18%
P/F = 0.0985
Using the interpolation :
The required i= 17.82%
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Sec 2.7. Unknown Interest Rate
Example 2:
A laboratory can pay now or pay latter for leasing space for its
equipment . It can pay $72,000 now for 3 years or can pay $30,000
at the end of each of three years.
What is the related interest rate for these cash flows?
72,000 = 30,000(P/A, i, 3)
(P/A, i, 3) = 2.4000
Use interpolation from the tables
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Sec 2.7. Unknown Interest Rate
Example 2:
Using Tables
Examine the n=3 row and the P/A column looking for 2.4000:
12% yields 2.4018 which is too high and
14% yields 2.3216 which is too low, therefore,
the correct answer for i% corresponding to target value
(P/A=2.400) lies between 12% and 14%
Interpolate:
Difference between target P/A value and (P/A,12 %, 3) :
2.4018 – 2.4000 = 0.0018
Total difference between the two available values i.e. (P/A, 12%, 3)
and (P/A, 14%, 3): 2.4018 – 2.3216 = 0.0802
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Sec 2.7. Unknown Interest Rate
Example 2:
The fraction= 44.55
i-12 = 14-12/ 44.55
Interpolated value = 12 +2/44.55
Interpolated value i= 12. 04%
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Sec 2.7. Unknown Interest Rate
The Excel function
IRR(first cell: last cell)
can also be used.
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Sec 2.7. Unknown Number of
Periods
•Techniques similar to those of the previous section
are used to find the number of periods.
•Direct solutions and interpolation of the tables are
possible as with i.
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Sec 2.7. Unknown Number of
Periods
Example 3: Given $100 for P, 3% interest rate. How many years
required until it reaches $150?
1- Using Direct Calculation
P = F(1/(1+i))n
or F = P (1+i)n
(1+.03)n = 150/100 = 1.5
n log (1.03) = log 1.5
n = .1761/.01283
n =13.72 periods
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Sec 2.7. Unknown Number of
Periods
2- Using Tables
P = F(P/F, i, n)
100 = 150(P/F, i, n)
(P/F, i, n) = 100/150 = 0.6667
Check the P/F factors Table 8, p.734 at i=3% for 0.6667
For n=13 , P/F= 0.6810
For n=14, P/F = 0.6611
Use the Interpolation:
(n-13) = 0.72
n = 13 + 0.72
n = 13.72
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Sec 2.7. Unknown Number of
Periods
The Excel function
NPER(%, A,P,F)
can also be used.
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Class Problems
Pb 1: Single Payment Factors
0
1
2
$300
3
4
5
6
7
8
9
10
$400
$600
Find the present and the future worth of these cash flows at n=10 at a
rate of 5% per year.
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Class Problems
Pb 1: Single Payment Factors
0
1
2
$300
3
4
5
6
7
8
9
10
$400
$600
Find the future worth of these cash flows series at n=10 at an interest
rate of 5% per year.
F = - 600 (F/P,5%,10) – 300 (F/P,5%,8) – 400 (F/P,5%,5) = -1931.08
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Class Problems
$700
Pb 2: Single Payment Factors
13
14 15
27
$300
Find the equivalent single payment of these cash flows at n=15 at a
rate of 5% per year.
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Class Problems
$700
Pb 2: Single Payment Factors
13
14 15
27
$300
Find the equivalent single payment of these cash flows at n=15 at an
interest rate of 5% per year.
F = -300(F/P,5%,2) +700(P/F,5%,12) = 59.04
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Class Problems
Pb 3: Uniform Series Factors
P=?
1
A
24
23
2
A
A
A
Find the present worth and future worth of the following yearly
investment :
$50 in a saving account at the end of each year over 24 years for 6%
rate per year, compounded yearly
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Class Problems
Pb 3: Uniform Series Factors
P=?
1
23
2
24
A
A
A
A
Find the present worth and future worth of the following yearly
investment :
$50 in a saving account at the end of each year for 24 years for 6%
interest rate per year, compounded yearly
P= -50 (P/A, 6%, 24) = -50*12.5504=$-627.520
F=-50 (F/A,6%,24) = -$50 * 50.815 = $- 2,540.779
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Class Problems
Pb 4: Non-uniform Series Factors
Find the expressions related to
P, F, A for this series
0
1
2
3
4
5
9
10
10
20
30
40
50
90
100
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Class Problems
Pb 4: Non-uniform Series Factors
0
1
2
3
4
5
9
10
10
20
30
40
Find the expressions related to 50
P, F, A for this series
90
P = -10(P/A, i ,10) -10(P/G, i ,10),
A = -10 -10(A/G, i ,10)
F = -10 (F/A, i ,10) -10(P/G, i ,10)(F/P, i ,10)
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100
Class Problems
Pb 5: Non-uniform Series Factors
Find the expressions related to P, F, A for this series
0
1
2
3
4
5
20
40
60
100
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80
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Class Problems
Pb 5: Non-uniform Series Factors
Find the expressions related to P, F, A for this series
0
1
2
3
4
5
20
40
60
100
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80
P = -100(P/A, i ,5) +20(P/G, i ,5)
A = -100 +20(A/G, i ,5)
F = -100(F/A, i ,5) +20(P/G, i ,5)(F/P, i ,5)
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Engineering Economy
Chapter 3
Combining Factors
Session 66-7
Dr. Ilham KISSANI
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Chapter 3 - Combining Factors
PURPOSE
Analyze the time value of money related to combined
CFs
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Chapter 3 - Combining Factors
TOPICS
3.0. Review of the key concepts of the previous chapter
3.1. Shifted Uniform Series
3.2. Shifted Uniform Series and Randomly Placed Single Amounts
3.3. Shifted Gradients
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Sec 3.0. Review of the notions of
previous chapter
• The present worth is always located one period prior to the
first uniform-series amount when using the P/A factor.
• The future worth is always located in the same period as the
last uniform-series amount when using the F/A factor.
• The present worth of an arithmetic gradient will always be
located two periods before the gradient starts or one period
before the annuity starts , when using the P/G factor.
• The present worth is located one period prior to the first
geometric gradient amount when using the P/G factor.
• In real life situations these assumptions may not be respected
need new techniques to be able to combine the factors to
solve special problems.
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Sec 3.1. Shifted Uniform Series
• A shifted series begins at a time other than the
end of period 1.
• A shifted series is one whose present worth point
is NOT t = 0. (Shifted either to the left or to the
right of t = 0).
• No matter where the series falls on the time line
• Look at figures 3.1, 3.2, 3.3
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Sec 3.1. Shifted Uniform Series
• Dealing with shifted uniform series:
– Use (P/F,i, j) factor to find the PW of each payment aj in
year 0 and then add them P=sum aj(P/F,i, j)
– Use (Fj/P,i, j) factor to find the FW of each payment aj in
year 13 and then add them F= sum aj (F/P,i, 13-j), then
find
P=F(P/F, i%,13)
– Use F/A factor to find the FW of the whole series in year
13 using F= A(F/A, i%, 10), then find P=F(P/F, i%,13)
– Use P3/A factor to find the PW of the whole series in year
3 using P3= A(P/A, i%, 10), then find P=F(P/F, i%,3) where
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Sec 3.1. Shifted Uniform Series
Specific steps to avoid errors:
• Draw the cash flow diagram that defines the
problem and correctly identify inflows and
outflows
• Locate the present worth and future worth for
each series on the CFD
• Determine n for each series
• Draw another CFD and write the desired
equivalent CF
• Substitute the correct factor values and solve
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Sec 3.1. Shifted Uniform Series
Example
0
..
3
4
5
6
..
A = $50/year
P0
P3
P3 of this series is at t = 3 (F is at t=12)
P3 = $50(P/A,i%,9)
P0 = P3(P/F,i%,3)
Example 3.1, 3.2 with Excel
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12
Sec 3.1. Shifted Uniform Series
Example
0
..
3
4
5
6
A = $50/year
P0
P3
F of this series is at t=12
F12 = A(F/A,i%,9)
P0= F12 (P/F, i%, 12)
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..
13
Sec 3.2 Uniform Series and
Randomly Placed Single Amounts
• In this case cash flows are combinations of series
and other single cash flow.
• The method adopted consists of applying the
previous procedures for uniform series , in
addition of using single payment factors for the
randomly placed single amounts
• Examples 3.3, 3.4
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Sec 3.2 Uniform Series and
Randomly Placed Single Amounts
Dealing with uniform series combined with single
amounts:
• Find the PW value of the series then move to t = 0
• find the PW value of the single cash flow at t=0
• Add the equivalent PWs at t = 0
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Sec 3.2 Uniform Series and
Randomly Placed Single Amounts
F4 = $300
Example
A = $500
0
1
2
3
4
5
6
7
i = 10%
F5 = -$400
•Find the PW at t = 0 for this cash flow:
•t = 1 is the PW point for the $500 annuity;
• we use the factor (P/A, i, 3)
•t = 0 is the PW point for the two other single cash flows
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8
Sec 3.2 Uniform Series and
Randomly Placed Single Amounts
• Write the Equivalence Statement
P = $500(P/A,10%,3)(P/F,10%,1)
+
$300(P/F,10%,4)
400(P/F,10%,5)
•Substitute the factor values into the
equivalence equation and solve
• P= $1086.96
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Sec 3.3 Shifted Gradients
• The Present Worth of an arithmetic
gradient (linear gradient) is always located:
– One period to the left of the first cash
flow in the series or,
– Two periods to the left of the “1rst
Gradient” cash flow
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Sec 3.3 Shifted Gradient
• A Conventional Gradient is one whose present
worth point is t = 0.
• A Shifted Gradient is one whose present value
point is removed from time t = 0.
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Sec 3.3 Shifted Gradient
A1+(n-1)G
Conventional Gradient
A1+(n-2)G
A1+2G
A1+G
A1
P
0
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3
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n-1
n
Sec 3.3 Shifted Gradient
A+(j-1)G
Shifted Gradient
A+(j-2)G
A+2G
A+G
A
P
0
1
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3
….
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…..
n
Sec 3.3 Shifted Gradient
G = +$100
Base Annuity = $100
0
1
2
3
4 :::..
:::..
• CF start at t = 3
• G= +100
• A= 100/year from year 3 to year 10;
•i = 10%; Find the PW at t = 0
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9
10
Sec 3.3 Shifted Gradient
• PW of the Base Annuity
Base Annuity = $100
0
1
2
3
4 :::..
:::..
9
10
P2A = $100( P/A,10%,8 ) = $100( 5.3349 ) = $533.49
P0A = $533.49( P/F,10%,2 ) = $533.49( 0.8264 )
= $440.88
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Sec 3.3 Shifted Gradient
• For the gradient component
G = +$100
0
1
2
3
4 :::..
:::..
9
10
•PW of gradient is at t = 2:
•P2G = $100( P/G,10%,8 ) = $100( 16.0287 ) =
$1,602.87
•P0G = $1,602.87( P/F,10%,2 ) = $1,602.87( 0.8264 )
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= $1,324.61
1 - 157
Sec 3.3 Shifted Gradient
• PW for the Base Annuity
– P0A = $440.88
• PW for the Linear Gradient
– P0G = $1,324.61
• The total Present Worth:
$440.88 + $1,324.61 = $1,765.49
Example 3.5- 3.6
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Sec 3.3 Shifted Gradient
• Conventional Geometric Gradient
A1
0
1
2
3
:
:
:
n
For a conventional geometric gradient, PW occurs
at t = 0
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Sec 3.3 Shifted Gradient
• Shifted Geometric Gradient
A1
0
1
2
3
:
:
:
Present worth point is at t = 3 for this example
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n
Sec 3.3 Shifted Gradient
Example : i = 10%/year
0
1
2
3
4
5
6
7
A = $700
P0= PW for the annuity
A1 = $400 at t = 5
g=12% Increase/year
Pg= PW for the gradient
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8
Sec 3.3 Shifted Gradient
•Present Worth of the Gradient at t = 4
Pg = $400 (P/A1,12%,10%,4) = 400x3.73674
= 1,494.70
•Present Worth of the Gradient at t = 0
P0g = $1,494.70( P/F,10%,4) = $1,494.70( 0. 6830 )
P0g = $1,020,88
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Sec 3.3 Shifted Gradient
• PW of the Annuity
i = 10%/year
0
1
2
3
4
A = $700
P0A = 700(P/A,10%,4)
= 700( 3.1699 ) = 2,218.94
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5
6
7
8
Sec 3.3 Shifted Gradient
• PW for Geometric Gradient at t = 0
P0g = $1,020,88
• PW for Annuity
P0A = $2,218.94
• The total Present Worth:
P= $1,020.88 + $2,218.94
= $3,239.82
Example 3.6, 3.7
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Sec 3.3 Shifted Decreasing
Arithmetic Gradients
• Given the following shifted, decreasing gradient:
F3 = $1,000; G=-$100
i = 10%/year
0
1
2
3
4
5
6
P2 =PW at t = 2
P0
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Find the PW at t = 0
1 - 165
7
8
Sec 3.3 Shifted Decreasing
Arithmetic Gradients
F3 = $1,000; G=-$100
$1,000
i = 10%/year
P2
P0
0
1
2
3
4
5
6
7
8
P2 = $1,000( P/A,10%,5 ) – 100( P/G,10%.5 )
P2= $1,000( 3.7908 ) - $100( 6.8618 ) = $3,104.62
P = $3,104.62( P/F,10%,2 ) = $3104.62( 0 .8264 ) = $2,565.65
0
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More examples- CH2
•An engineer plans for his retirement and starts a
saving program.
•He saves $5000 for the first year and in the
following years he increases his savings by 4%
each year.
•How much will he have in his account after 15
years if the interest rate is 12% per year?
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More examples- CH2
A1 = 5000,
i = 0.12 g = 0.04
PW at t =0:
Pg = 5000.[1 – {(1+0.04)15/(1+0.12)15}]/(0.12 - 0.04)
= 41936
The future amount at year 15:
F = 41936.(F/P,12%,15) = $229541.
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More examples- CH2
• A company that manufactures auto parts has budgeted
$30000 per year to pay for tooling over the next five years.
• If the company expects to spend $12000 in year 1, how
much of a uniform (constant) increase each year is the
company expecting in the cost of the tooling if the interest
rate is 10% per year?
(i.e. what is the equivalent arithmetic gradient of the uniformseries amounts of $30000?)
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More examples- CH2
AT = 30000,
AA = 12000,
G=?
AT = AA + AG
30000 = 12000 + G.(A/G,10%,5) = 12000 + G.(1.8101)
G = $9944 per year.
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More examples- Combining Factors
• A taxi owner plans to purchase a car for $10000.
• He also expects to pay $1000 per year for its maintenance
when the warranty on the car expires after 2 years, i.e.
maintenance starts in the 3rd year.
• If he plans to keep the car for 6 years after the warranty
expires, how much money should he allocate for these
expenses:
what will be the present worth of the payments, if the
interest rate is 8% per year?
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More examples- Combining Factors
PT = ?
P2 = ?
i = 8% per year
A = 1000
10000
P2 = 1000.(P/A,8%,6)
Its present worth :
P = P2.(P/F,8%,2) = 1000.(P/A,8%,6).(P/F,8%,2) = 3963.21
Total present worth, PT = 10000 + 3963.21 = 13963.21
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More examples- Combining Factors
• $2000 is invested and pays a rate return of 7% per year.
• If 10 equal annual withdrawals are made from the account, with the first
withdrawal occurring three years after the deposit,
• Using 2 methods, how much can be withdrawn each year as Uniform
Series?
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More examples- Combining Factors
1
2
3
12
2000
•Present worth of the uniform-series amounts
P0= A.(P/A,7%,10).(P/F,7%,2)
2000 = A.(P/A,7%,10).(P/F,7%,2) = A.(7.0236).(0.8734)
A = 326.029
•Alternatively, we can find the future worth of 2000 at year 2
and then we find the equivalent uniform-series amounts A:
A =2000.(F/P,7%,2).(A/P,7%,10)
A = 326.022
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.
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i=7%
More examples- Combining Factors
300
300
4
5
300
12
300
13
Using 2 methods , find :
1.
2.
PW for the cash flow series above, and
The equivalent regular uniform series over t=1..13
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More examples- Combining Factors
(a) PW= 300(P/A,i,10)(P/F,i,3)
(b) PW= 300(F/A,i,10)(P/F,i,13)
(c) A1-13 = 300(P/A,i,10)(P/F,i,3)(A/P,i,13) or,
(d) A1-13 = 300(F/A,i,10)(A/F,i,13)
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More examples- Combining Factors
1
2
1600 1600
3
1600
4
800
5
6
1600 1600
7
8
1600 1600
9
10
1600 1600
Using 2 methods, find equivalent uniform series and
PW over t=1..10, for the cash flow series above.
177
More examples- Combining Factors
1
2
1600 1600
3
1600
4
800
5
6
1600 1600
7
8
1600 1600
9
10
1600 1600
(a) PW = [-1600(P/A,i,3) -800(P/F,i,4) 1600(P/A,i,6)(P/F,i,4)] or,
(b) PW = -1600 (P/A,i,10)+800(P/F,i,4)
(c) A = [-1600(P/A,i,3) -800(P/F,i,4) 1600(P/A,i,6)(P/F,i,4)] * (A/P,i,10) or,
(d) A = -1600 +800(P/F,i,4)(A/P,i,10)
178
More examples- Combining Factors
1
200
2
200
3
4
5
6
700 700
7
700
8
9
10
200 200
Find the PW and the equivalent uniform series over
t=1..10, for the cash flow series above.
179
More examples- Combining Factors
1
200
2
200
3
4
5
6
700 700
7
700
8
9
10
200 200
PW = -200(P/A,i,2) -700(P/A,i,3)(P/F,i,4)200(F/A,i,2)(P/F,i,10)
A = -200(P/A,i,2)(A/P,i,10) 700(P/A,i,3)(P/F,i,4)(A/P,i,10) -200(F/A,i,2)(A/F,i,10)
180
Engineering Economy
Chapter 4
Nominal and Effective Interest
Rates
Session 88-9
Dr. Ilham KISSANI
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Chapter 4
Nominal and Effective Interest Rates
PURPOSE
Make calculations for interest rates and cash flows that occur on
a time basis other than yearly
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Chapter 4- Nominal & Effective
Interest Rate
TOPICS
4.0 Review of the key concepts of the previous chapter
4.1 Nominal & Effective Interest Rate
4.2 Effective Annual Interest rate
4.3 Effective Interest rate for any Time Period
4.4 Payment Period (PP) and Compounding Period (CP)
4.5 Single Amounts with PP ≥ CP
4.6 Series with PP ≥ CP
4.7 Single and Series with PP < CP
4.8 Continuous Compounding
4.9 Varying rates
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Sec 4.0 Review of the notions of
previous chapter
Types of CF
1-REGULAR
• Single payments,
• Uniform series,
• Non uniform series: Arithmetic, Geometric
2-IRREGULAR
• Shifted Series
• Shifted Series and Single Amounts
• Shifted Gradients
• Shifted Decreasing Gradients
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Sec 4.1 Nominal and Effective Interest Rate
Nominal rates
Effective rates
• Definition: Interest rate per
• Definition : Interest rate is
time period does not include
compounded more frequently
any consideration of
than once per year
compounding
• Format: “r% per time period t,
• Format: “r% per time period t”
compounded m-ly” m is the
compounding frequency
• Some nominal statements:
Considering 2% per month, all the following • Some statements related to
are same:
effective rate:
2% per month x 12 months = 24% per year
2% per month x 24 months = 48% per 2
years
2% per month x 6 months = 12%
semiannually
2% per month x 3 months = 6% quarterly
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2% per year, compounded monthly
2% per year, compounded quarterly
13% per quarter, compounded quarterly
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Effective Interest Rate
• Effective Interest Rate
One other type of interest rate that investors and borrowers
should know is called the effective rate, which takes the power of
compounding into account.
For example,
• If a bond pays 6% on an annual basis and compounds
semiannually, then an investor who invests $1,000 in this bond will
receive $30 of interest after the first 6 months ($1,000 x .03), and
$30.90 of interest after the next 6 months ($1,030 x .03).
The investor received a total of $60.90 for the year, which means
that while the nominal rate was 6%, the effective rate was 6.09%.
• Mathematically speaking, the difference between the nominal and
effective rates increases with the number of compounding periods
within a specific time period.
Sec 4.1 Nominal and Effective Interest Rate
•
•
There are two time units associated with
an interest rate statement:
Time period – the basic time unit of the
interest rate. It is the t in the statement
of r % per time period t. The time unit of
1 year is assumed unless otherwise
stated.
Compounding period (CP) – it is defined
by the compounding term in the interest
rate statement.
The compounding frequency (m), which
is the number of times that
compounding occurs within t, the time
period
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Effective rates
• Effective interest rate is the actual
rate that applies for a stated period of
time.
• An effective rate has the
compounding frequency attached to
the nominal rate statement.
• If the compounding period is NOT
stated, it assumed to coincide with
the same stated time period as r
meaning that the nominal and
effective rates are the same.
Effective rate per CP=r/m
Example: 4-1, 4-2
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Sec 4.1 Nominal and Effective Interest Rate
There are 3 ways to express interest rates T4-1:
• 8% per year compounded quarterly: 8% is nominal
and find the effective rate
• Effective 8.243% per year compounded quarterly:
8.243% is the effective rate and may be used directly.
• 8% per year, no compounding period is stated. The
rate is effective only over the time period of one
year; the effective rate for any other time period
must be calculated.
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Sec 4.2. Effective Annual Interest Rates
r = nominal interest rate per year
m = number of compounding periods per year
i = effective interest rate per compounding period CP = r/m
ia = effective interest rate per year
F = P + Pia = P(1+ia)
CP must be compounded through all m periods to obtain the total.
F = P(1+i)m
Consider the F value for a present worth P of $1 and substituting
$1 for P in the two expressions :
1+ia = (1+i)m
ia = (1+i)m –1
r % per year = (i% per CP)(number of CPs per year) = (i)(m)
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University4.3, 4.4
•Al Akhawayn
Example
Sec 4.3 Effective Interest Rates for Any Time
Period PP
• The payment period, PP, is the frequency of payments or
receipts.
• To evaluate cash flows that occur more frequently than annually,
PP<1 year, the effective interest rate over the PP must be used.
• Substituting r/m for the period interest rate in eq. 4.5 yields
r m
Effective i = (1+ ) − 1
m
• i = effective rate per payment period (PP), e.g., quarterly, annually
• r = nominal rate per payment period (PP)
• m = frequency of compounding per payment period ( CP per PP)
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Sec 4.3 Effective Interest Rates for Any Time
Period
Example: Find i per year compounded quarterly ,
for quarterly compounding m = 4 , and
r = 12% per year
r m
Effective i = (1+ ) − 1
m
Stated period for i is YEAR
i = (1 + 0.12/4)4 - 1 = 12.55%
•Stated period for i is QUARTER
i = 3%
Examples 4.5
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Sec 4.3 Effective Interest Rates for Any Time
Period
Nominal
Effective
r = rate/period × periods
r m
Effective i = (1+ ) − 1
m
Example: Rate is 1.5% per month.
Determine nominal rate per
quarter, year, and over 2 years
Example: 1.5% per month compounded
monthly. Determine effective rate per quarter
and per year
Qtr: r = 1.5 × 3 mth = 4.5%
Year: r = 1.5 ×12 mth = 18%
= 4.5 × 4 qtr = 18%
2 yrs: r =1.5 × 24 mth = 36%
= 18 × 2 yrs = 36%
Period is quarter:
r = 1.5 × 3 mth = 4.5% m = 3
i = (1 + 0.045/3)3 – 1 = 4.57% per quarter
Period is year:
r = 18%
m = 12
i = (1 + 0.18/12)12 - 1) = 19.6% per year
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Sec 4.4 Payment Periods (PP)
and Compounding Periods (CP) :PP vs CP
• In a large percentage of equivalency computations, the
frequency of cash flow does not equal the frequency of
interest compounding.
• PP – how often cash flows occur
• CP – how often interest is compounded
• Cash flows may occur monthly whereas compounding may
occur annually or quarterly.
• To correctly perform any equivalence computation, it is
essential that the compounding period and the payment
period be placed on the same time basis, and that the interest
rate be adjusted accordingly.
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Sec 4.4 PP vs CP: Payment Periods (PP)
and Compounding Periods (CP)
• Examples where effective i, PP, CP are involved:
•Semi-annual payment, monthly compounding (PP > CP)
•Consider monthly deposits to a savings account that
compounds interest on a quarterly basis: CP is a quarter; PP is
a month: Monthly deposit, quarterly compounding (PP < CP)
UnusualM
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Sec 4.4 PP vs CP: Payment Periods (PP)
and Compounding Periods (CP)
Key elements to observe about cash flows:
1. Compare length of PP with CP
PP = CP
PP > CP
PP < CP
2. Determine types of cash flows
–
–
Single amounts (P and F)
Series (A, G, g)
3. Determine correct effective i and n (be represented
by the same time unit on both)
Example 4.5, 4.6
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Sec 4.5 Single Amounts with PP≥CP
•
•
Method 1: Determine the effective interest rate over the compounding period CP,
and set n equal to the number of compounding periods between P and F.
– P = F(P/F, effective i% per CP, total number of CP periods n)
– F = P(F/P, effective i% per CP, total number of CP periods n)
Example 4.7
Assume a nominal 12% rate per year compounded semiannually. Find P or F over a
10 years, using an effective per 6 month.
CP =6 months.
The effective rate is 12/2= 6% per semiannual period (6 months) and the total
number of CPs is 10x2=20.
the effective =6% per 6 months and n=20 semiannual periods are used in the P/F
and F/P factors. F = P(F/P, 6% per CP, 20)
Method 2: Determine the effective interest rate for the time period of the nominal
rate and set n equal to the total number of the same time period.
The effective i% per year is 12.36 % and n=10 are used in the P/F and F/P factors.
F = P(F/P, 12.3% per year, 10)
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Sec 4.6 Series with PP≥CP
When cash flows involve a series (A, G, g) and the
payment period equals or exceeds the
compounding period:
– Find the effective i per payment period=PP.
– Determine n as the total number of payment
periods.
Work table 4-6
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Sec 4.6 Series with PP≥CP
• Count number of payments= n
• Determine effective i over same time period
as n
• Use these i and n values in factors
Example:
$700 per month for 3 years at 12% per year
compounded monthly
PP = CP = month
n = 36 months
effective i = 1% per month
F = A(F/A,1%,36)
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Sec 4.6 Series with PP≥CP
• Count number of payments. This is n
• Determine effective i over same time period as n
• Use these i and n values in factors
Example 4.7: Payment of $500 per 6 months for 7 years at 20%
per year compounded quarterly
PP = 6 month and CP = quarter → PP > CP
n = 14 PPs
i = 10% per 6 month
m = 2 CP per PP
effective i per PP= (1 + 0.10/2)2 – 1 = 10.25%
F = A(F/A, 10.25%, 14)
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Sec 4.6 Series with PP≥CP
0
P = $3M
• First step: Find P for n = 10 annual payments
• Period is 1 year
• CP = 6 months; PP = year; PP > CP
• Effective i per year = (1 + 0.08/2)2 – 1 = 8.16%
PT = 3M + 200,000(P/A,8.16%,10) = $4,332,400
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Sec 4.6 Series with PP≥CP
0
P = $3M
• Second step: Find A for n = 20 semi-annual amounts
• Period is six months
• CP = 6 months; PP = 6 months; PP = CP
• Effective i per 6 months = 8%/2 = 4%
Relation: A = 4,332,400(A/P,4%,20) = $318,778
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Sec 4.7 Single Amounts & Series with PP<CP
There are two policies: interperiod cash flows earn no interest or they
earn compound interest:
1-
If no interperiod interest considered , deposits are regarded as
deposited at the end of the compounding period and withdrawals
are regarded as withdrawn at the beginning.
For instance assume monthly compounding, if you make a payment
(deposit)on the 1st that is due on the 30th, you do not get interest
credit.
If you withdraw money from a monthly compounded savings
account 1 day before the end of the month, you get no interest for
that month.
2- If interperiod compounding interest is earned, m<1 because there is
only a fractional part of the CP within one PP.
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Sec 4.7 Single Amounts & Series with PP<CP
1. Approach taken if interest is not paid on interperiod
deposits or withdrawals
• For equivalence computations: Cash flows are ‘moved’ to
match CP time period
• Move cash flows to the beginning or the end of the
compounding period:
– Deposits ( negative cash flows) - to end of current CP
– Withdrawals (positive cash flows) - to beginning of current
CP period
Example (4.11): move monthly deposits to match quarterly compounding.
Now, PP = CP = quarter Find P, F or A using effective i per quarter
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Sec 4.7 Single Amounts & Series with PP<CP
Qtr 1
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Qtr 3
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Qtr4
Sec 4.7 Single Amounts & Series with PP<CP
2. Approach taken if interperiod compounding is
earned, m<1
There is only a fractional part of the CP within one PP
•
•
Relations between CF are determined in the same way as
the previous 2 sections for PP > CP or PP = CP
Example ( p 114): weekly CF and quarterly compounding
m = 1/13 of a quarter , if r=12% compounded quarterly
The effective interest rate per PP:
The effective weekly i% = (1.03)1/13 - 1 =0.228%
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Sec 4.8 Effective Interest Rate for Continuous
Compounding
Continuous compounding practically occurs in
businesses that have a very large number of cash
flows every day.
r m
Effective i = (1+ ) − 1
m
As m → a, conbnuous compounding is approached
Effective i = (℮r – 1)
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Sec 4.8 Effective Interest Rate for
Continuous Compounding
Example 4-12
r = 18% per year compounded continuously
i% per year = (℮ 0.18 - 1) = 19.72% per year
i% per month = (℮ 0.015 - 1) = 1.511% per month
See example 4-13
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Sec 4.9 Varying rates
To find P for future CF (Ft) at different i% values
(it%) of each year t, we assume annual
compounding:
it = annual effective interest rate for year t
(t=1..n)
P = F1 (P/F, i1 , 1) + F2 (P/F, i1 , 1) (P/F, i2 , 1)
+….. +Fn (P/F, i1 , 1) (P/F, i2 , 1) …(P/F, in , 1)
Example 4.14
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Review
1. Using 13% per year compounded monthly, find the
future amount after 5 years of an investment of 2500
now.
• Find: effective interest rate per year
• ia = (1+i)m –1
• i=r/m=13/12=1.08333%=.01083333
• ia = (1+.0108333)12 –1
• ia = 13.80% per year
• Then F = 2500 (F/P, 13.80%,5)
Review
2.Given: Nominal and effective rates of 16 and
16.986 %
• Find: What is the compounding period
• Effective i = (1+r/m)m –1
• 16.986 = (1+.16/m)m – 1
• Trial and error, m=4 yields 16.986, therefore,
• Quarterly
Review
3. Given: An interest rate of 1% is per month, find the
equivalent uniform series of 2-months payments
over 1 year FOR an investment of 2500 now.
• Find: the equivalent effective rate per 2 months
• r/2months = 1% x 2months
• r/2months = 2%
• Effective i = (1+r/m)m –1 = (1+.02/2)2 –1
• Effective i = 2.01% for 2 months
• Then A = 2500 (A/P, 2.01%,6)
Review
4. Given: An effective rate of 6.707 per semi-annual
period, compounded weekly
• Find: The equivalent weekly effective interest rate
• m=26 which is semi-annual, 26 weeks in ½ year.
• Effective i = (1+r/m)m –1
.06707 = (1+r/26)26 – 1
• r/6months = 6.5%
• i/week = 6.5/26
• i/week = .25% per week which is an effective rate
Review
5. A rate of 18% per year, compounded monthly, fill in
the blanks to have equality between these 2 factors:
(P/F, , 3Y) = (P/F,
, 36M)
• the time period t is 1 year. Effective i% per year =
(1+18%)12 –1 = 1.196 – 1 = .196 = 19.6%, n = 3 years
• the time period t is 1 month . Effective i% per
month=1.5%, n = 36
• The P/F factor is the same by both methods: (P/F,
1.5%, 36) = .5851
• (P/F, 19.6%, 3) = .5787 for 20%
Engineering Economy
Chapter 5
Present Worth Analysis
Session 11
Dr. Ilham KISSANI
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Chapter 5
Present Worth Analysis
PURPOSE
Identify types of alternatives; and compare alternatives using
a present worth basis
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Chapter 5- Present Worth Analysis
TOPICS
5.0. Present Worth Analysis
5.1. Mutually Exclusive Alternatives
5.2. Present Worth Analysis of Equal Life Alternatives
5.3. Present Worth Analysis of Different-Life Alternatives
5.4. Future Worth Analysis
5.5. Capitalized Cost Calculation and Analysis, CC
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5.0. Present Worth Analysis
• The PW is always less than the actual cash flow for any interest rate
greater than zero because all P/F factors have a value <1, therefore,
PW values are often referred to as discounted cash flows, DCF.
• The interest rate is referred to as the discount rate.
• Extensions of PW are also covered here:
– Future worth
– Capitalized cost
– Payback period
– Life-cycle costing and
– Bond analysis
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5.1. Mutually Exclusive Alternatives
Types of alternatives
• Mutually exclusive (ME) - only one viable project can be
selected by the economic analysis. Each viable project is a
stand-alone, alternative. These alternatives compete
with one another.
• Independent - more than one project can be selected by the
economic analysis. DN is one of the projects. These
alternatives do NOT compete with one another.
• Do-nothing – DN alternative is selected if none are justified
economically. Maintain status quo/current situation.
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5.1. Mutually Exclusive Alternatives
Nature of alternatives
Cash flows determine whether the alternative is
revenue-based or service-based
All the alternatives evaluated in one particular
engineering economy study must be of the same
type:
– Revenue. Each alternative generates cost and revenue cash
flow estimates. Revenues are dependent upon which
alternative is selected.
– Cost. Each alternative has only cost cash flow estimates.
Revenues are NOT dependent upon the alternative
selected, so these cash flows are assumed to be equal.
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Sec 5.2 – Equal-life ME Alternatives
1- Single alternative
• Calculate PW at stated MARR
• If PW ≥ 0 project is economically justified
Example: MARR = 10%
First cost, P = $-2500
Annual revenue, R = $2000
Annual cost, AOC = $-900
Salvage value, S = $200
Life, n = 5 years
PW = P +S(P/F,10%,5) + (R-AOC)(P/A,10%,5) = -2500 + 200(P/F,10%,5)
+ (2000-900)(P/A,10%,5) = $1794
PW > 0; project is economically justified or viable
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Sec 5.2 – Equal-life ME Alternatives
2- Two or more alternatives
• Calculate PW of each alternative at MARR
• Select alternative with most favorable PW value, that is,
numerically largest PW value
PW1
PW2
$-1,500
-2,500
2,500
$-500
500
1,500
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Selected
Alternative
2
2
1
Sec 5.2 Equal-life ME Alternatives
Example 5.1: 3 ME service machines alternatives.
Revenues are equal. MARR is 10% per year. Select
the most favorable one.
Estimate
Electric-powered
Gas-powered
Solar-powered
First Cost, $
-2,500
-3,500
-6,000
AOC, $
-900
-700
-50
SV, $
200
350
100
life, years
5
5
5
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Sec 5.2 – Equal-life ME Alternatives
Determine PWE , PWG and PWS; select larger PW
PWE = -2500-900(P/A,10%,5)+200(P/F,10%,5)
= $-5788
PWG = -3500-700(P/A,10%,5)+350(P/F,10%,5)
= $-5936
PWS = -6000-50(P/A,10%,5)+100(P/F,10%,5)
= $-6127
PWE is the largest select electric-powered
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Sec 5.3 – Different-life Alternatives
• PW evaluation always requires equal-service between
all alternatives
• Two methods available:
– Study period (same period for all alternatives)
– Least common multiple (LCM) of lives for
alternatives
• Study period method is recommended when LCM is
unrealistic
• Evaluation approach: Determine each PW at stated
MARR; select alternative with numerically largest PW
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Sec 5.3 – Different-life Alternatives
1- Study period Method
Assumptions
– Set a time horizon over which the economic analysis is
conducted
– Only CF occurring during this time period are considered
as relevant, the others occurring beyond this time horizon
are neglected
– An estimated market value at the end of the study period
must be made.
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Sec 5.3 – Different-life Alternatives
2- LCM Method
Assumptions (may be unrealistic at times)
– Same service needed for LCM years (e.g., LCM of 6 and
9 is 18
years)
– Alternatives are repeated for each life cycle
– CF estimates are the same over all life cycles
Evaluation approach:
Obtain LCM, repeat CFs over life cycle for LCM years;
calculate PW over LCM; select alternative with most
favorable PW
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Sec 5.3 – Different-life Alternatives
Example 5.3
Use PW to select lower-cost alternative:
• For 5-year study period
• Using LCM
Assume MARR = 15% per year
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Sec 5.3 – Different-life Alternatives
Study period of 5 years
F = 1,000
PWA = ?
Location A
P = -15,000
A = -3,500
F = 2,000
P = -18,000
A = -3,100
PWB = ?
Location B
For 5 years at i = 15%: PWA = $-26,236 and PWB = $-27,397
Select Location A having the largest PW
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Sec 5.3 – Different-life Alternatives
LCM evaluation
• LCM is 18 years
• The fist cost is repeated in year 0 of each new
cycle :
• A twice (years 6 and 12);
• B once (year 9)
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Sec 5.3 – Different-life Alternatives
For 18 years at MARR = 15%:
PWA = $-45,036; PWB = $-41,384
Select location B having the largest PW
Note: Selection changed from 5-year study period
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Sec 5.4 – Future Worth Analysis
• FW evaluation of alternatives is especially applicable for
LARGE capital investment situations when maximizing the
future worth of a corporation is important (e.g., equipments,
buildings, corporation)
• The future worth, FW, may be determined directly by
calculating the future worth value or by multiplying the PW by
F/P at the MARR. Therefore, FW is an extension of PW
analysis.
• Evaluation approach: Determine FW value from cash flows or
from PW using the F/P factor , n is related to the study period,
or to the LCM of alternatives’ lives.
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Sec 5.4 – Future Worth Analysis
• For one alternative: FW≥0 means the MARR is met or
exceeded
• For two or more mutually exclusive alternatives,
select the one with the numerically largest FW value.
• Example 5.5, p.137
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Sec 5.6 – Capitalized Cost (CC)
• CC= PW of an alternative that will last
‘forever’
• Especially applicable to public project
evaluation (bridges, irrigation, hospitals, etc.)
• CC relation is derived using the limit as
n → a for the P/A factor
PW = A(P/A,i%,n) =
PW = A[1/i ]
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Sec 5.5 – Capitalized Cost (CC)
• Refer to PW as CC when n is large (can be
considered infinite). Then
CC = AW/i
and
AW = CC × i
Example: If $10,000 earns 10% per year, $1,000 is interest
earned annually for eternity.
• Cash flows for CC computations are of two types :
recurring (periodic) and nonrecurring(one time)
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Sec 5.5 – Capitalized Cost (CC)
Procedure to find CC
1. Draw CFD for all nonrecurring CFs and at least 2
cycles of all recurring CFs
2. Calculate PW (CC) for all nonrecurring amounts
3. Find AW for 1 life cycle of all recurring amounts;
then add these to all A series applicable for all
years 1 to a (or long life)
4. Find CC for amount above using CC = AW/i
5. Add all CC values (steps 2 and 4)
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Sec 5.5 – Capitalized Cost (CC)
Example 5.6: Using the 5 step, find CC and A values at i
= 5% with cash flows below. Cycle time is 13 years.
Nonrecurring costs: first $150,000; one-time of $50,000 in year 10
Recurring costs: annual maintenance of $5000 (years 1-4) and $8000
thereafter; upgrade costs $15,000 each 13 years
Step 1
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Sec 5.5 – Capitalized Cost (CC)
2.
CC of nonrecurring costs:
CC1 = -150,000 – 50,000(P/F,5%,10) = $-180,695
3.
AW of recurring $15,000 upgrade:
AW = -15,000(A/F,5%,13) = $-847 per year
AW of recurring maintenance costs years 1 to a:
AW = $-5000 per year forever
4.
CC of extra $3000 maintenance for years 5 to a:
CC2 = -3000/0.05 (P/F,5%,4) = $-49,362
CC for recurring upgrade and maintenance costs:
CC3 = (-847-5000)/0.05 = $-116,940
5.
Total CC obtained by adding all three CC components
CCT = -180,695 – 49,362 – 116,940 = $-346,997
The AW value is the annual cost forever:
AW = CC × i = -346,997(0.05) = $-17,350
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Sec 5.5 – Capitalized Cost (CC)
• For two long-life or infinite-life alternatives:
SELECT ALTERNATIVE HAVING LOWER CC OF COSTS
• For one infinite life and one finite life:
Determine CC for finite life alternative using
AW of 1 life cycle and relation CC = AW/i
SELECT ALTERNATIVE HAVING LOWER CC OF COSTS
• Example 5.6
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EXTRA - Payback Period Analysis
• Payback analysis, also payout analysis, is another extension of
the PW method.
• Payback can take two forms: i>0% and i=0%
• The payback period np is the estimated time it will take for
the estimated revenues to recover the initial investment.
• The np should never be used as the primary measure of
worth to select an alternative. Rather it should be
determined in order to provide initial screening in
combination with an analysis performed using PW or another
method.
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EXTRA - Payback Period Analysis
• To find the discounted payback period at a stated rate i>0%,
calculate the years np that make the following expression
np
correct.
0 = -P + Σ NCFt(P/F,i,t)
• P is the initial investment, NCF is the estimated net cash flow
for each year t. NCF = receipts – disbursements.
• If NCF are equal each year, P/A may be used :
0 = -P + NCF(P/A,i,np)
• After np years, the cash flows will recover the investment and
a return of i%.
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EXTRA - Payback Period Analysis
• It is very important to realize that in payback analysis all net flows
occurring after np years are neglected, unlike other analyzes.
• This np assumption may be quite unfair : that is why this method should
only be used as primary measure to select an alternative.
• No return or simple payback determines np at i=0%. For i=0%, the
preceding equations become:
0 = -P + ΣNCFt and np = P/NCF
• It is incorrect to use the no-return payback period to make final
alternative selections because it:
– Neglects any required return since the time value of money is
omitted.
– Neglects all net cash flows after time np including positive cash flows
that may contribute to the return of the investment.
• Examples 5.7, 5.8 p. 187
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Engineering Economy
Chapter 6
Annual Worth Analysis
Session 12
Dr. Ilham KISSANI
EGR2302-Engineering Economics
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Chapter 6
Annual Worth Analysis
PURPOSE
Make annual worth calculations and compare
alternatives using the annual worth method
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Chapter 6- Annual Worth Analysis
TOPICS
6.0 Review of the key concepts of the previous chapter
6.1. Advantages and Uses of AW Analysis
6.2. Calculation of Capital Recovery and AW Values
6.3. Evaluating Alternatives by AW Analysis
6.4. AW of a Permanent Investment
6.5. LCC analysis
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6.0 Review of the previous chapter
Present Worth Analysis
Extensions of PW covered :
– Future worth
– Capitalized cost
– Payback period
– Life-cycle costing and
– Bond analysis
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Sec 6.1. Advantages and Uses of
AW Analysis
•
•
•
•
•
Annual worth, AW, is easiest to understand and best to use, preferable to PW, FW
and rate of return.
AW is equivalent to the PW and FW values at the MARR for n years. All three can be
determined from each other by the relation: AW = PW(A/P, i, n) = FW(A/F, i, n)
The AW value has to be calculated for only one life cycle. Therefore, it is not
necessary to use the LCM of lives, as it is for the PW or FW analysis.
When alternatives being compared have different lives, the AW method makes the
assumptions that:
– The services provided are needed for at least the LCM of the lives of the
alternatives.
– The selected alternative will be repeated for succeeding life cycles in exactly the
same manner as for the first life cycle.
– All cash flows will have the same estimated values in every life cycle.(These
estimates change with inflation)
If this assumption is not correct: use new CF estimates for each life cycle and use a
study period .
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Sec 6.1. Advantages and Uses of
AW Analysis
Example 6.1
AW of the first life cycle
AW = -15,000(A/P,15%,6) +1000(A/F,15%,6) - 3500
= $-7349 per year
AW over 3 life cycles
AW = -45,036 (A/P,15%,18)
= $-7349 per year
Demonstrates that AW will be the same ove any
number of life cycles
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Sec 6.2. Calculation of Capital
Recovery and AW Values
• Any alternative should have the following cash flow estimates:
– Initial Investment P (total first cost).
– Salvage value S (terminal estimated value of the assets at the end of
their useful life.
– Annual amount A (often the AOC, annual operating cost).
• AW is comprised of two components:
– Capital recovery, CR, for the initial investment P
– The equivalent annual amount, A of annual operating costs (A of AOC)
AW = CR + A
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Sec 6.2. Calculation of Capital
Recovery and AW Values
Or AW = – CR – A
The minus signs represent costs.
– Aw is determined from recurring costs and non-recurring
amounts
– CR is the equivalent annual cost of owning the asset plus
the return on the initial investment.
– If there is a positive salvage value, S, its equivalent annual
value is removed using the A/F factor. According, CR is
CR = -[P(A/P,i,n) – S(A/F,i,n)]
• Example 6.2
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Sec 6.2. Calculation of Capital
Recovery and AW Values
S = $0.5M
0
1
2
6
7
8
AOC = $0.9M
$5M
P = $8M
Capital recovery is the equivalent annual amount to recover $13M at
12% per year if the salvage is $5M after 8 years
CR = -(8+5(P/F,12%,1))(A/P,12%,8) + 0.5(A/F,12%,8)
= $-2,47M per year
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Sec 6.2. Calculation of Capital
Recovery and AW Values
General formula for CR
CR = -P(A/P,i,n) + S(A/F,i,n)
Using (A/F,i,n) = (A/P,i,n) - i
Alternative formula to calculate CR
CR = -(P-S)(A/P,i,n) - Si
The use of the second formula results in the same value:
CR = -(12,46 – 0.5)(A/P,12%,8) - 0.5(0.12)
= $-2,47 per year
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Sec 6.2. Calculation of Capital
Recovery and AW Values
Example:
Add A to CR
AOC =A= $0.9M each year:
AW = -2.47M – 0.9M = $-3.37M per year
Meaning:
The equivalent total revenue must be at least $3.37M per
year to recover the initial PW, A and the required return
12% per year
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Sec 6.3. Evaluating Alternatives by
AW Analysis
• AW is typically the easiest evaluation to perform when the MARR is
specified. The alternative selected has the lowest equivalent annual
cost for service alternatives or the highest equivalent income for
revenues.
• Single project analysis
Calculate AW at stated MARR
Acceptance criterion: If AW ≥ 0, MARR is met or exceeded and the
project is economically justified
• Multiple alternatives
Calculate AW of each alternative at MARR over respective life or
study period
Selection criterion: select alternative with most favorable AW value,
that is, numerically largest AW value. Examples 6.3 - 6.4
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6.3. Evaluating Alternatives by AW
Analysis
Example 6.3:
•Each system costs $4600, has a 5-year useful life, and may be
salvaged for an estimated $300.
•Total operating cost for all systems is $1000 for the first year,
increasing by $100 per year thereafter.
•The MARR is 10%.
•Perform an annual worth evaluation.
•Expected income of 6000 for all 5 systems.
•What annual income is necessary to recover the investment at
the MARR of 10% per year?
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6.3. Evaluating Alternatives by AW
Analysis
CR = - 5(4600)(A/P,10%, 5) + 5(300)(A/F,10%, 5) = -$5822
The annual operating costs and incomes form an
arithmetic gradient series with a base of $550 ($1200 –
$650) in year 1, increasing by $50 per year for 5 years. The
AW relation is:
A = - capital recovery + equivalent net income
= -5822 + (6000-1000) - 100(A/G,10%,5) = -$1003
This is the equivalent 5-year net amount needed to return
the investment and recover the estimated operating costs
at a 10% per year return.
This shows, that the alternative is not financially viable at
MARR = 10%.
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6.3. Evaluating Alternatives by AW
Analysis
• When a study period is specified the AW is
determined for that period regardless of the life
cycles of alternatives
• Study period = 6 years
• Select the alternative having the lowest cost
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6.4. AW of a Permanent Investment
• AW of alternative that will last perpetually
• The alternatives having long lives such as dams, bridges that
the lives are considered infinite in analysis terms.
• The annual worth of the initial investment is the perpetual
annual interest earned on the initial investment, A = Pi.
• This is the equivalent annual worth of capitalized cost (CC)
AW = PW x i = CC x i / Example 6.5 and 6.6
Procedure:
Regular interval cash flows
– find AW over one cycle (add all the A values to the CR)
Non-regular intervals
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6.4. AW of a Permanent Investment
Example 6.5 :
Comparison of short-lived and long-lived alternatives at i = 5%
Alt
P and S
AOC
Life
A
P = $650,000
S = $17,000
A = $170,000
10
P = $4 million
A = $5,000
$30,000 every
5 years
P = $6 million
A = $3,000
B
C
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Method
CR over 10; add
AOC
Perman CR over ∞; add
ent
AOC; add periodic
repair over 5 years
50
CR over 50; add
AOC
6.4. AW of a Permanent Investment
AWA = - 650,000(A/P,5%,10) + 17,000(A/F,5%,10) - 170,000
= $-252,824
AWB = - 4,000,000(0.05) - 5,000 - 30,000(A/F,5%,5)
= $-210,429
AWC = - 6,000,000(A/P,5%,50) - 3,000
= $-331,680
Select proposal B
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Sec 6.5 – Life Cycle Cost Analysis (LCC)
• Another application of PW analysis
• Useful when entire life cycle of a system is under evaluation
(e.g., new product lines, new car model or aircraft model;
introducing new technology)
• PW evaluation must include cost estimates for all stages of
the product or service:
• LCC may be categorized into major phases of acquisition and
operations:
– Acquisition phase: activities prior to the delivery of
products and services(Definition of system requirements,
preliminary design, detailed design)
– Operations phase : Construction, Usage, Phase out
– Example 6.7
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Engineering Economy
Chapter 7
Rate of Return Analysis:
Single Alternative
Session 1313-14
Dr. Ilham KISSANI
EGR2302-Engineering Economics
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Chapter 7
Rate of Return Analysis: Single Alternative
PURPOSE
Understand the meaning of rate ROR and perform ROR
calculations for one alternative
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Chapter 7
Rate of Return Analysis: Single Alternative
TOPICS
7.0 Review of the key concepts of the previous chapter
7.1 Interpretation of a Rate of Return, ROR, Value
7.2 ROR Calculation Using a PW or AW Equation
7.3 Cautions When Using the ROR Method
7.4 Multiple ROR Values
7.5 Techniques for Removing Multiple ROR values
7.6 Rate of Return of a Bond Investment
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Sec 7.0 Review of the key concepts of
the previous chapter
• The AW value has to be calculated for only
one life cycle. Therefore, it is not necessary to
use the LCM of lives, as it is for the PW or FW
analysis
• Compute AW using Capital Recovry
• Compare alternatives in AW–basis
• AW of alternative that will last perpetually
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Sec 7.1 Interpretation of a Rate of
Return, ROR, Value
• ROR is the rate earned on the unrecovered balance of an
investment so that the final payment or receipt brings the
balance to exactly zero with interest,
• or the rate paid on the unpaid balance of borrowed money.
• The ROR is expressed as a percent per period, i=6% per year.
• –100%<i<∞. –100% means that the entire amount is lost.
• The ROR definition does not state that the rate of return is on
the initial amount of the investment, rather the unrecovered
balance which changes each time period.
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Sec 7.1 Interpretation of a Rate of
Return, ROR, Value
Example 7.1
• An investor took a $1000 at i = 10% per year for 4 years to invest it in
home office equipment. From the lender’s perspective, the investment is
expected to produce an equivalent net cash flow of $315.47 for the next 4
years.
A = $1000(A/P,10%,4) = $315.47
• This represents a 10% per year rate of return on the lender’s unrecovered
balance.
• Compute the amount of the unrecovered investment for each of the 4
years using:
– (a) the ROR on the unrecovered balance (the correct basis) and
– (b) the return on the initial $1000 investment.
– (c) Explain the difference between the 2 situations
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Sec 7.1 Interpretation of a Rate of
Return, ROR, Value
(a) Table 7–1 shows the unrecovered balance at the end of each year in
column 6 using the 10% rate on the unrecovered balance at the beginning
of the year.
After 4 years, the total $1000 is recovered, and the balance in column 6 is
exactly zero.
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Sec 7.1 Interpretation of a Rate of
Return, ROR, Value
• (b) Table 7–2 shows the unrecovered balance if the 10% return is always
figured on the initial $1000.
• Column 6 in year 4 shows a remaining unrecovered amount of $138.12,
because only $861.88 is recovered in the 4 years (column 5).
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Sec 7.1 Interpretation of a Rate of
Return, ROR, Value
• (c) $400 of interest is earned if the 10% return each year is based on the
initial amount of $1000.
• Only $261.88 in interest is earned if a 10% return on the unrecovered
balance is used.
• The total investment is totally recovered when the rate is applied to the
unrecovered balance as in part (a).
• Figure 7–1 illustrates the correct interpretation of rate of return.
• Each year the receipt of $315.47 is comprised of 2 parts: 10% interest on
the unrecovered balance in column 2 + the recovered amount in column
5.
• Because rate of return is the interest rate on the unrecovered balance, the
computations in Table 7–1 for part (a) present a correct interpretation of a
10% rate of return.
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Sec 7.1 Interpretation of a Rate of
Return, ROR, Value
Installment financing:
• The situation (b) is sometimes referred to as the installment financing
problem.
• If the purchase is not paid in full by the time the promotion is over usually
6 months or 1 year
• Finance charges are assessed from the original date of purchase.
• The common theme is more interest paid over more time by the
consumer.
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Sec 7.2 ROR Calculation Using a PW or
AW Equation
•
•
•
•
•
•
To determine the ROR, set up the equation using either PW or AW. PWD
represents disbursements and PWR represents receipts.
Using PW : PWD = PWR 0 = - PWD + PWR
or using AW : AWD = AWR 0 = - AWD + AWR
The i value that makes these equations numerically correct is called i*. It is the
root of the ROR relationship. To determine if the cash flow series is viable,
compare i* with the established MARR.
– If i* ≥ MARR, accept the alternative as economically viable
– If i* < MARR, the alternative is not economically viable
In ROR calculations, the objective is to find the interest rate i* at which the cash
flows are equivalent.
There are 2 ways to determine i* once the PW relation has been established:
– Trial and Error
– Excel
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Sec 7.2 ROR Calculation Using a PW or
AW Equation
•
•
By hand:
– Convert all disbursements to into single (P or F) or uniform amounts
– Convert all receipts to either single (P or F) or uniform amounts
– Approximate an interest rate using the tables. This is an estimate.
By Excel: The computer uses a trial and error algorithm.
– Rate(n,A,P,F)
– When cash flows vary from year to year: IRR(first_cell:last_cell, guess). The
guess is i* value at which the computer starts searching.
– Procedure:
• Draw the cash flow diagram
• Set up the ROR
• Enter the cash flows onto the spreadsheet
• Develop the IRR to display i*
• Use the NPV function to develop a chart PW vs. i values. i* is found at
PW= 0 is desired
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Sec 7.2 ROR Calculation Using a PW or
AW Equation
Example
• An engineer for a company constructing one of the world’s tallest
buildings (Shanghai Financial Center in the Peoples’ Republic of China) has
requested that $500,000 be spent now during construction on software
and hardware to improve the efficiency of the environmental control
systems. This is expected to save $10,000 per year for 10 years in energy
costs and $700,000 at the end of 10 years in equipment refurbishment
costs.
• Find the ROR.
• We are trying to find an unknown interest rate (i*) that satisfies the
following:
PW(+ inflows) – PW( - outflows) = 0
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Sec 7.2 ROR Calculation Using a PW or
AW Equation
• Use the trial-and-error procedure based on a PW equation.
0 = - 500,000 + 10,000 (P/A, i*,10) + 700,000 (P/F, i*,10)
• Use the estimation procedure to determine i for the first trial.
• All income will be regarded as a single F in year 10 so that the
P/F factor can be used.
• The P/F factor is selected because most of the cash flow
($700,000) already fits this factor and errors created by
neglecting the time value of the remaining money will be
minimized.
• Only for the first estimate of i, define
P = $500,000, n = 10, and F = 10(10,000) + 700,000 = $800,000.
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Sec 7.2 ROR Calculation Using a PW or
AW Equation
•
•
•
•
•
500,000 = 800,000(P/F, i,10); (P/F, i,10) = 0.625
The roughly estimated i is close to 5%. Use 5% as the first trial because
this approximate rate for the P/F factor (0.6139) is lower than the true
value when the time value of money is considered.
At i = 5%, the ROR equation is:
0 = -500,000 + 10,000(P/A, 5%,10) + 700,000(P/F, 5%,10);
0 = -500,000 + 10,000 (7.7217) + 700,000 (0.6139) ;
when 0 < $6946
The result is positive, indicating that the return is more than 5%.
Try i = 6%. 0 = - 500,000 + 10,000(P/A, 6%,10) + 700,000 (P/F, 6%,10);
where 0 > - $35,519
Since the interest rate of 6% is too high, linearly interpolate between 5%
and 6% .
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Sec 7.2 ROR Calculation Using a PW or
AW Equation
Amount
Trial i
PW value
0
-$500,000 4.00%
$54,004
1
$10,000 4.20%
$44,204
2
$10,000 4.40%
$34,603
3
$10,000 4.60%
$25,198
4
$10,000 4.80%
$15,984
5
$10,000 5.00%
$6,946
6
$10,000 5.20%
-$1,888
7
$10,000 5.40%
-$10,555
8
$10,000 5.60%
-$19,047
9
$10,000 5.80%
-$27,368
10
$710,000 6.00%
-$35,519
ROR =
5.16%
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PW value
Year
$100,000
$50,000
$0
-$50,000
3.8% 4.2% 4.6% 5.0% 5.4% 5.8%
Interest rate i
Sec 7.2 ROR Calculation Using a PW or
AW Equation
Example
• Use AW computations to find the ROR for the cash flows in Example 7.2.
• The AW relations for disbursements (AWD) and receipts (AWR) are
formulated using Equation [7.2].
AWD = 500,000 (A/P, i,10)
AWR = 10,000 + 700,000 (A/F, i,10)
0 = - 500,000 (A/P, i*,10) + 10,000 + 700,000 (A/F, i*,10)
• Trial-and-error solution yields these results:
•
•
At i = 5%, 0 < $900 (=- 500,000 (0.1295) + 10,000 + 700,000 (0.0795) )
At i = 6%, 0 > - $4826 (=- 500,000 (0.13587) + 10,000 + 700,000 (0.07587)
By interpolation, i* = 5.16%, as before.
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Sec 7.3 Cautions When Using the ROR
Method
• When used correctly, the ROR technique will always result in
a good decision i.e. the same one as with PW, AW or FW.
• There are some assumptions and difficulties with ROR
analysis that must be considered when calculating i*.
• From an engineering economic study perspective, the AW or
PW method at a stated MARR should be used in lieu of ROR.
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Sec 7.3 Cautions When Using the ROR
Method
Some difficulties related to ROR analysis
• Multiple i* values - Depending upon the sequence of net cash flow
disbursements and receipts, there may be more than one real-number
root to the ROR equation, resulting in more than one i* value.
• Reinvestment at i*. PW and AW assume reinvestment at the MARR, ROR
assumes reinvestment at i*rate. when i* is not close to the MARR, i* is
not a good basis for decision making.
• Computational difficulty vs. understanding. Spreadsheet solutions are
easier than solutions by hand, but they don’t offer the same level of
understanding as that provided by hand solutions.
• Special procedure for multiple alternatives. To correctly use ROR method
to compare MEA it requires a different procedure introduced in chapter 8.
• When working with two or more alternatives, and when it is important to
know the exact value of i*, a good approach is to determine PW or AW at
the MARR, then follow up with the specific i* for the selected alternative.
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Sec 7.4 Multiple ROR Values
•
•
•
So far we have dealt with conventional (or simple) cash flow series. The algebraic
signs on the net cash flows changed only once, usually from minus in year 0 to plus
at some time during the series.
However, for many series the net cash flows switch between positive and negative
causing more than one sign change. Such a series is called nonconventional (no
simple).
When there is more than one sign change in the net cash flows, it is possible that
there will be multiple i* values.
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7.4 Multiple ROR Values
There are two tests to perform in sequence on the
nonconventional series to determine if there is one
unique or multiple i* values involved in the project.
• The first test is the (Descartes’) rule of signs
• The second is Norstrom’s criterion.
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7.4 Multiple ROR Values
• The first test is the (Descartes’) rule of signs states that the
total number of real-number roots is always less than or
equal to the number of sign changes in the series.
• The second and more discriminating test determines if there
is one, real number, positive i* value. This is the cumulative
cash flow sign test, also known as Norstrom’s criterion. It
states that only one sign change in the series of cumulative
cash flows which starts negatively, indicates that there is one
positive root to the polynomial relation. To perform this test,
determine the series: St = cumulative cash flows through
period t
• Observe the sign of S0 and count the sign changes in the
series S0, S1, . . . , Sn. Only if S0 < 0 and signs change one time
in the series is there a single, real number, positive i*.
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Sec 7.4 Multiple ROR Values
Example 7.4
• (a) Table 7–4 shows the annual cash flows and cumulative
cash flows. Since there are two sign changes in the cash flow
sequence, the rule of signs indicates a maximum of two realnumber i* values. The cumulative cash flow sequence starts
with a positive number S0 = +2000.
• This indicates there is not just one positive root.
• The conclusion is that as many as two i* values can be found.
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Sec 7.4 Multiple ROR Values
• (b) The PW relation is:
PW = 2000 – 500 (P/F,i,1) – 8100 (P/F, i, 2) + 6800(P/F, i, 3)
• Select values of i to find the two i* values, and plot PW vs. i.
The PW values are plotted in Figure 7–5 for i values of 0, 5,
10, 20, 30, 40, and 50%.
• PW is crossing the i axis at approximately i1* = 8 and
i2 * = 41%.
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7.4 Multiple ROR Values
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7.4 Multiple ROR Values
• Example 7.4 (by computer)
$250.00
$200.00
PW value
$150.00
$100.00
$50.00
$0.00
-$50.00
-$100.00
-$150.00
0%
10% 20% 30% 40% 50% 60%
i value
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7.5 Techniques for Removing Multiple
ROR values
• The interest rate obtained from the previous calculations is
known as the internal rate of return (IROR).
• The IROR as stated is the rate of return on the unrecovered
balance of an investment, as defined earlier. The funds that
remain unrecovered are still inside the investment.
• The concept of unrecovered balance becomes important
when positive net cash flows are generated before the end of
a project. A positive net cash flow, once generated, becomes
released as external funds to the project and is not
considered further in an internal rate of return calculation.
• These positive net cash flows may cause a nonconventional
cash flow series and multiple i* values to develop.
• The external rate of return (EROR) methods consider explicitly
these funds
• The dilemma of multiple i* roots is eliminated.
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7.5 Techniques for Removing Multiple
ROR values
1. Modified ROR Approach (EROR): the concept behind is that
positive/neg cash flows from an investment do not necessarily
earn/pay interest at the same rate as the project.
To compensate for this, 2 interest rates are used, ii, as the
investment rate at which positive cash flows are invested (earn
interest), ib, as the borrowing rate at which negative cash flows
are borrowed(pay interest). Then i’ satisfies:
FWn= PW0(F/P, i’%, n)
FWn = future worth of positive CFs at year n at ii
PW0 = present worth of negative CFs at year 0 at ib
2. Return On Invested Capital (ROIC)
The correct approach used to find the ROIC is called the netinvestment procedure . This procedure involves finding the
future worth of the net investment amount, one period in the
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7.5 Techniques for Removing Multiple
ROR values
Net-investment procedure (ROIC):
Mathematically, for each year t set up the relation
Ft = Ft-1 (1+k) + NCFt
Where t= 1,2,…,n and NCFt = net CF in year t
k = ii if Ft-1 > 0
k = i’’ if Ft-1 < 0
• Set net-investment relation for year n equal to zero (Fn= 0)
and solve for i’’.
• i’ value is unique for a stated reinvestment rate
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7.5 Techniques for Removing Multiple
ROR values
The procedure to find i’:
1.Draw a cash flow diagram of the original net cash flows.
2.Develop the series of net investments using the
equation: [7.10]. The result is the last net investment
equation, Fn, expressed in terms of i’’.
3.Set Fn= 0 and find the i’’ value to balance the equation.
Examples 7.5, 7.6
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7.5 Techniques for Removing Multiple
ROR values
Example 7.6: Compute the ROIC of below cash flow
Year
0
1
2
3
Cash Flow
$2,000
-$500
-$8,100
$6,800
Use procedure to determine i’’ for MARR=9%, ii= 12%
• F0=2000>0 invest at 12%
• F1=2000(1.12)-500=1740>0 use ii =12%
• F2= 1740(1.12)-8100=-6151 <0 use ii =12%
• F3= -6151(1+i’’)+6800= 0 then solve i’’= 10.55% which is greater
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7.6 Rate of Return of a Bond Investment
• A bond is a long-term note, issued by a corporation or government who
then becomes the borrower to finance major projects.
• The borrower receives money NOW in return for a promise to pay the
face value V of the bond on a stated maturity date. Bonds are usually
issued in face value amounts of $100, $1000, $5000 or $10,000.
• Bond interest, I, also called bond dividend, is paid periodically between
the time the money is borrowed and the time the face value v is repaid.
• The bond interest is paid c times per year usually quarterly or semiannually. The amount of interest is determined using the stated interest
rate, called the bond coupon rate, b.
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7.6 Rate of Return of a Bond Investment
• Bonds are issued to finance projects of some institutions.
• The cash flow series for a bond investment is conventional
and has one unique i* which is best determined by solving a
PW-based ROR equation.
• Example 7.7, 7.8
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7.6 Rate of Return of a Bond Investment
Example
Gerry is an entry-level engineer at Boeing Aerospace in
California. He took a financial risk and bought a bond from a
different corporation that had defaulted on its interest
payments. The bond bought at 4240 , is an 8% $10,000 bond
with interest payable quarterly. The bond paid no interest for
the first 3 years after Gerry bought it. If interest was paid for
the next 7 years, and then Gerry was able to resell the bond
for $11,000, what rate of return did he make on the
investment? Assume the bond is scheduled to mature 18
years after he bought it.
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7.6 Rate of Return of a Bond Investment
• The bond interest received in years 4 through 10 was:
• The effective rate of return per quarter can be determined by solving
the PW equation developed on a per quarter basis, since this basis
makes PP = CP.
• 0 = - 4240 + 200(P/A, i* per quarter, 28) (P/F, i* per
quarter,12) + 11,000(P/F, i* per quarter, 40)
• The equation is correct for i* = 4.1% per quarter, which is a nominal
16.4% per year, compounded quarterly.
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Assignments
•Find NPV, the PW of the followings non-uniform
cash flows,
•then plot PW vs i to derive the ROR values
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Engineering Economy
Chapter 8
ROR Analysis:
Analysis: Multiple
Alternatives
Session 1515-16
Dr. Ilham KISSANI
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Chapter 8
PURPOSE
Select the best MEA on the basis of ROR analysis of
incremental CFs
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Chapter 8
TOPICS
8.0 Review of key concepts of the previous chapter
8.1 Why Incremental Analysis is Necessary
8.2 Calculation of Incremental CF for ROR Analysis
8.3 Interpretation of Rate of Return On the Extra Investment
8.4 ROR evaluation Using PW: Incremental and Breakeven
8.5 ROR evaluation Using AW
8.6 Incremental ROR Analysis of Multiple, MEA
8.7 Spreadsheet Application
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Sec 8.1 Why Incremental Analysis is
Necessary
• Extension of the previous chapter: 2 or more
MEA are evaluated using ROR comparison.
• The ROR evaluation correctly performed will
result in the same selection as the PW, AW
and FW analyses.
• A different procedure is considered for MEA
evaluation based on ROR comparison.
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Sec 8.1 Why Incremental Analysis is
Necessary
Assume that a company:
• uses a MARR of 16% per year,
• has $90,000 available for investment,
• two alternatives (A and B) are being evaluated.
1-Alternative A requires an investment of $50,000 and has an
internal rate of return i* of 35% per year.
2-Alternative B requires $85,000 and has an i* of 29% per year.
Intuitively we may conclude that the better alternative is the one
that has the larger return, A in this case.
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Sec 8.1 Why Incremental Analysis is
Necessary
• However, this is not necessarily so. While A has the higher
projected return, it requires an initial investment that is much
less than the total money available ($90,000).
• What happens to the investment capital that is left over? It is
generally assumed that excess funds will be invested at the
company’s MARR as learned in the previous chapter.
• Using this assumption, determine the Overall ROR of A and B
to compare the alternatives investments.
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Sec 8.1 Why Incremental Analysis is
Necessary
• If alternative A is selected, $50,000 will return 35% per year. The
remaining $40,000 will be invested at the MARR of 16% per year. The
remaining $ 5000 from B will be invested at the MARR of 16% per year.
• The rate of return on the total capital available, then, will be the weighted
average.
• OVERALL ROR(A)= (50000*0.35 + 40000*0.16)/90000 = 26.6%
• OVERALL ROR(B)= (85000*0.29 + 5000*0.16)/90000 = 28.3%
• Even though the i* for alternative A is higher, alternative B presents the
better overall ROR for the 90000.
• If either a PW or AW comparaison is conducted using the MARR of 16%
per year, alternative B will be chosen.
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Sec 8.2 Calculation of Incremental CF for
ROR Analysis
•Prepare an incremental CF tabulation between 2 alternatives:
•The year column will go from 0 to LCM of lives
•For simplification use the convention:
The alternative with the larger initial investment is regarded as
alternative B.
Incremental cash flow : Cash flow B - Cash flow A
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Sec 8.2 Calculation of Incremental CF for
ROR Analysis
Example 8.1 (equal lives)
• A tool and die company in Pittsburgh is considering the
purchase of a drill press with fuzzy-logic software to improve
accuracy and reduce tool wear. The company has the
opportunity to buy a slightly used machine for $15,000 or a
new one for $21,000.
• Because the new machine is a more sophisticated model, its
operating cost is expected to be $7000 per year, while the
used machine is expected to require $8200 per year.
• Each machine is expected to have a 25-year life with a 5%
salvage value.
• Tabulate the incremental cash flow.
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Sec 8.2 Calculation of Incremental CF for
ROR Analysis
As demonstrated below:
• the subtraction performed is (new - used) since the new
machine has a larger initial cost.
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Sec 8.2 Calculation of Incremental CF for
ROR Analysis
Example 8.2 (different lives)
• Sandersen Meat Processors has asked its lead process
engineer to evaluate two different types of conveyors for the
bacon curing line.
• Type A has an initial cost of $70,000 and a life of 8 years.
• Type B has an initial cost of $95,000 and a life expectancy of
12 years.
• The annual operating cost (AOC) for type A is expected to be
$9000, while the AOC for type B is expected to be $7000.
• The salvage values are $5000 and $10,000 for type A and type
B, respectively.
• Tabulate the incremental cash flow using their LCM.
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Sec 8.2 Calculation of Incremental CF for
ROR Analysis
• The LCM of 8 and 12 is 24 years leading to the following
table.
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Sec 8.2 Calculation of Incremental CF for
ROR Analysis
• The incremental cash flows in year 0 reflect the extra investment or cost
required if the alternative with the larger first cost is selected.
• If the incremental cash flows of the larger investment don’t justify it, we
must select the cheaper one.
• The decision to buy the used or new machine can be made on the basis of
the profitability of investing the extra $6000 in the new machine.
• If the equivalent worth of the savings is greater than the equivalent worth
of the extra investment at the MARR, the extra investment should be
made (i.e., the larger first-cost proposal should be accepted).
• On the other hand, if the extra investment is not justified by the savings,
select the lower-investment proposal.
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Sec 8.3 Interpretation of Rate of Return on
the Extra Investment
•If the ROR available through the incremental cash flow equals or
exceeds the MARR, the alternative assosicated with the extra
investment should be selected.
•For multiple revenue alternatives, calculate the internal rate of
return i* for each alternative, and eliminate all alternatives that
have an i* less than MARR.
•Compare the remaining alternatives incrementally.
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Sec 8.4 ROR evaluation using PW:
Incremental and Breakeven
The Procedure for an incremental ROR analysis for 2 alternatives is as follows:
1- Order the alternatives by initial investment or cost, starting with the
smaller one, called A. the one with the larger initial investment is in the
column labeled B.
2-Develop the CF and incremental CF series using the LCM of years, assuming
reinvestment in alternatives.
3-Draw an incremental CFD, if needed.
4-Count the number of sign changes in the incremental CF series to
determine if multiple RORs may be present. If necessary, use Norstrom’s
criterion on the cumulative incremental CF series to determine if single
positive root existe.
5-Set up the PW equation for the incremental CF series and determine Δi* (BA)using trial and error.
6-Select the economically better alternativeas as follows:
Δi* (B-A)< MARR select A
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Sec 8.4 ROR evaluation using PW:
Incremental and Breakeven
Example 8.3
In 2000, Bell Atlantic and GTE merged to form a giant telecommunications
corporation named Verizon Communications. As expected, some equipment
incompatibilities had to be rectified, especially for long distance and
international wireless and video services. One item had two suppliers - a U.S.
firm (A) and an Asian firm (B). Approximately 3000 units of this equipment
were needed. Estimates for vendors A and B are given for each unit.
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Sec 8.4 ROR evaluation using PW:
Incremental and Breakeven
Determine which vendor should be selected if the MARR is 15% per year.
These are service alternatives, since all cash flows are costs.
Alternatives A and B are correctly ordered with the higher first-cost alternative in
column (2).
There are 3 sign changes in the incremental cash flow series, indicating as many as
three roots. There are also three sign changes in the cumulative incremental series
that starts negatively at S0 = - $5000 and continues to S10 = +$5000, indicating that
more than one positive root may exist.
The ROR equation based on the PW of incremental cash flows is
0 = - 5000 +1900(P/A, i,10) - 11,000(P/F, i, 5) + 2000 (P/F, i, 10)
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Sec 8.4 ROR evaluation using PW:
Incremental and Breakeven
• Assume that the reinvestment rate is equal to the resulting i*.
• Solution of PW-equation for the first root finds results for i*
between 12 and 15%.
• By interpolation i* = 12.65%.
• Since the rate of return of 12.65% on the extra investment is
less than the 15% MARR, the lower-cost vendor A is selected.
• The extra investment of $5000 is not economically justified by
the lower annual cost and higher salvage estimates.
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=IRR(D4:D14)
=NPV($B$1,D5:D14)+D4
316
Sec 8.4 ROR evaluation using PW:
Incremental and Breakeven
Example 8.4
•
•
•
•
•
Bank of America uses a MARR of 30% .
Two alternative software systems and the marketing/delivery plans have been
jointly developed by software engineers and the marketing department. They are
for new online services to passenger cruise ships and military vessels at sea
internationally.
For each system, start-up, annual net income, and salvage value estimates are
summarized below.
(a) Perform the incremental ROR analysis by computer.
(b) Develop the PW vs. i graphs for each alternative and the increment. Which
alternative, if either, should be selected.?
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Sec 8.5 ROR evaluation Using AW
For AW based technique, there are 2 equivalent ways to perform
the evaluation:
• 1- Using the incremental CF over the LCM of alternative lives,
just as for the PW-based relation and derive Δi*.
OR
• 2- Finding the AW for each alternative CF and setting the
difference of 2 AWs equal to zero to find the i* value.
AW(A)=AW(B)
• Example 8.5
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Sec 8.6 Incremental ROR Analysis of
Multiple, MEA
For ROR analysis of multiple MEA, the following criteria are used. Select the
alternative that :
• 1- Requires the largest investment, and
• 2-Indicates that extra investment over another acceptable alternative is
justified.
The incremental ROR evaluation procedure for multiple equal lives alternative
is:
• 1- Order alternatives from smallest to largest initial investment. Record
the annual CF estimates for each equal life alternative
• 2- Calculate i* for the first alternative(Revenue only). In effect, this makes
DN the defender and the first alternative the challenger. If i* < MARR
eliminate the alternative and go to the next one. Repeat this until i* ≥
MARR and define that alternative as the defender. The next alternative is
now the challenger.
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Sec 8.6 Incremental ROR Analysis of
Multiple, MEA
• 3-Determine the incremental CF between the challenger and
defender, using the relation ICF = CCF - DCF set up the ROR
relation.
• 4- Calculate Δ i* for the incremental CF series using a PW, AW,
of FW based equation.
• 5- If Δi* ≥ MARR, the challenger becomes the defender and
the previous defender is eliminated. Conversely, Δi* < MARR,
the challenger is removed, and the defender remains against
the next challenger.
• 6-Repeat steps 3 to 5until only one alternative remains. It is
the selected one.
• Examples 8.6 and 8.7
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Sec 8.7 Spreadsheet Application
PW, AW, and ROR analyses all in one.
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Engineering Economy
Chapter 9
Benefit/Cost Analysis and
Benefit/
Public Sector Economics
Session 1818-19
Dr. Ilham KISSANI
EGR2302-Engineering Economics
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Chapter 9
PURPOSE
Understand public sector economics; evaluate a project
and compare alternatives using the benefit/cost ratio
method
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Chapter 9
TOPICS
9.0 Review of key concepts of the previous chapter
9.1 Public Sector Projects
9.2 Benefit/Cost Analysis of a Single Project
9.3 Alternative Selection Using Incremental B/C Analysis
9.4 Incremental B/C Analysis of MEA
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Sec 9.1 Public Sector Projects
• Public sector projects are owned, used and financed
by the citizenry of any government level. Unlike
private sector, projects are owned by corporations,
partnerships, and individuals.
• Public sector projects have a primary purpose to
provide services for the public good at no profit.
• Partnerships of public entities and private enterprise
are more common as funding for large public
projects becomes more difficult.
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Sec 9.1 Public Sector Projects
Some public sector examples:
• Hospitals and clinics
• Parks and recreation
• Utilities: water, electricity,
gas, sewer, sanitation
• Schools: primary, secondary,
community colleges,
universities
• Economic development
• Convention centers
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• Sports arenas
• Transportation: highways,
bridges, waterways, airports
• Police and fire protection
• Courts and prisons
• Food stamp and rent relief
programs
• Job training
• Emergency relief
• Public housing
• Codes and standards
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Sec 9.1 Public Sector Projects
Difference s in the characteristics of private and public sector
Characteristic
Public Sector
Private Sector
Size of Investment
Larger
Some Large; medium to
small
Life Estimates
Longer:
30 – 50 years
Shorter:
2-25 years
Annual Cash Flow
estimates
No Profit; costs,
Revenues contribute to
benefits and disbenefits profits– costs are
are estimated
estimated
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Sec 9.1 Public Sector Projects
To perform an economic analysis of public alternatives, the costs, the benefits
and the disbenefits must be estimated as accurately as possible in
monetary units.
- Costs. Estimated expenditures for construction, operation and
maintenance of the project less salvage value. (Bridge construction cost;
Investing in new technologies; salaries)
- Benefits. Advantages to the owners, the public (Reduced property taxes;
Lower transportation costs due to less gas used)
- Disbenefits. Expected undesirable or negative consequences to the
owners if the alternative is implemented. usually these are economic
disadvantages estimable in monetary units.(tax rebates not realized)
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Sec 9.1 Public Sector Projects
Characteristic
Public Sector
Private Sector
Funding
Public projects use taxes, fees, bonds ;
taxes and fees are collected from users of
project services;
Benefits, partnerships
Interest rate
Called discount rate, it is considerably
lower than for private projects since no
profit is considered and governments are
exempt from taxes; typical rates range
between 4 to 8% per year
higher
Alternative
selection
Politics and special interest groups make
selection more complex for public
projects; B/C method developed to put
more objectivity into the analysis process
Primarily based on ROR
Environement
of the
evaluation
Politically inclined
Primarily Economic
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Sec 9.1 Public Sector Projects
• Determine viewpoint (perspective) before costs,
benefits, and disbenefits are estimated and before
the evaluation is performed. choose one and
maintain it throughout estimation and analysis.
Several viewpoints exist for any situation which may
alter the classification of CF estimates.
– Citizen
– City Tax base
– Creation/retention of jobs
– Economic development potential
– Particular industry interest
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Sec 9.2 B/C Analysis of a Single Project
Example 1
• The (CIP) Committee has recommended a $5 million bond issue for the
purchase of greenbelt/floodplain land.
• The proposal is referred to as the Greenway Acquisition Initiative.
• Developers immediately opposed the proposal due to the reduction of
available land for commercial development.
• The city engineer and economic development director have made the
following preliminary estimates for some obvious areas, considering the
Initiative’s consequences in maintenance, parks, commercial
development, and flooding over a projected 15-year planning horizon.
• The estimates are not yet classified as costs, benefits or disbenefits.
• If the Greenway Acquisition Initiative is implemented, the estimates are as
follows.
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Sec 9.2 B/C Analysis of a Single Project
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Sec 9.2 B/C Analysis of a Single Project
There are many perspectives to take. 3 are addressed here. The viewpoints
and goals are identified and each estimate is classified as a cost, benefit or
disbenefit.
• Viewpoint 1: Citizens of the city
Goal: Maximize the quality and wellness of citizens with family and
neighborhood as prime concerns.
Costs: 1, 2, 3 - Disbenefits: 4, 5
- Benefits: 6, 7, 8
• Viewpoint 2: City budget.
Goal: Ensure the budget is balanced and sufficient to fund rapidly growing
city services.
Costs: 1, 2, 3, 5
- Disbenefits: 4
- Benefits: 6, 7, 8
• Viewpoint 3: Economic development.
Goal: Promote new commercial and industrial economic development for
creation and retention of jobs.
Costs: 1, 2, 3, 4, 5
- Disbenefits: none
- Benefits: 6, 7, 8
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Sec 9.2 B/C Analysis of a Single Project
• All cost and benefit estimates must be converted to a common equivalent
monetary unit (PW, AW or FW) at the discount rate (interest rate). The B/C
ratio is then calculated using one of these relations:
If B/C >1.0, accept the project as economically acceptable for
the estimates and discount rate applied.
If B/C < 1.0, the project is not economically acceptable.
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Sec 9.2 B/C Analysis of a Single Project
Conventions
– AW and PW are more used than FW
– Revenues have (+) signs
– Costs have (+) signs
– Salvage values are subtracted from costs
– Disbenefit values are subtracted from benefits or
– Disbenefit values are added to costs
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Sec 9.2 B/C Analysis of a Single Project
- Conventional B/C ratio
-The modified B/C ratio includes maintenance and operation (M&O)
costs in the numerator and treats them in a manner similar to
disbenefits. The denominator includes only the initial
investment cost
-Once all amounts are expressed in PW, AW or FW terms, the
modified B/C ratio is calculated as
The modified procedure can change the magnitude of the ratio but not
the decision to accept or reject the project. It makes no difference which
approach is used; the ratio values will differ, but, the (accept/reject)
decision will be the same
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Sec 9.2 B/C Analysis of a Single Project
• The benefit and cost difference measure of worth does not involve a ratio,
and is based on the difference between the PW, AW or FW of benefits and
costs (B – C).
• If (B - C) > 0, the project is acceptable. This method has the advantage of
eliminating the divergence noted when disbenefits are regarded as costs,
because B represents net benefits. For the numbers 10, 8 and 8 the same
result is obtained regardless of how disbenefits are treated.
• Subtracting disbenefits from benefits: B - C = (10 - 8) - 8 = -6
• Adding disbenefits to costs: B - C = 10 - (8 + 8) = -6
• If the numbers 10, 8 and 8 are used to represent the PW of benefits,
disbenefits and costs, respectively, the correct procedure results in B/C =
(10 - 8)/8 = 0.25.
• The incorrect placement of disbenefits in the denominator results in B/C
=10/(8 + 8) =0.625: the method affects the magnitude of the B/C ratio.
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Sec 9.2 B/C Analysis of a Single Project
Example 2
• The Ford Foundation expects to award $15 million in grants to public high
schools to develop new ways to teach the fundamentals of engineering
that prepare students for university-level material.
• The grants will extend over a 10-year period and will create an estimated
savings of $1.5 million per year in faculty salaries and student-related
expenses.
• The Foundation uses a rate of return of 6% per year on all grant awards.
This grant program will share Foundation funding with ongoing activities,
so an estimated $200,000 per year will be removed from other program
funding. To make this program successful, a $500,000 per year operating
cost will be incurred from the regular M&O budget.
• Use the B/C method to determine if the grant program is economically
justified.
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Sec 9.2 B/C Analysis of a Single Project
The project is also not justified by the modified B/C method, as expected. For the (B C) model, B is the net benefit, and the annual M&O cost is included with costs.
B -C = (1,500,000 - 200,000) - (2,038,050 + 500,000) = - $1.24 million
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Since (B - C) < 0, the program is not1 justified.
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Sec 9.3 Alternative Selection Using
Incremental B/C Analysis
Steps to correctly perform a conventional B/C ratio analysis of two
alternatives. Equivalent values can be expressed in PW, AW or FW terms.
1. Determine the total equivalent costs for both alternatives.
2. Order the alternatives by total equivalent cost; smaller first, then larger.
Calculate the incremental cost (∆C) for the larger-cost alternative.
This is the denominator in B/C.
3. Calculate the total equivalent benefits and any disbenefits estimated for
both alternatives. Calculate the incremental benefits (∆ B) for the larger
cost alternative. (This is ∆(B - D) if disbenefits are considered.)
4. Calculate the incremental B/C ratio using Equation [9.2], (B - D)/C.
5. Use the selection guideline to select the higher-cost alternative if
B/C >1.
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Sec 9.3 Alternative Selection Using
Incremental B/C Analysis
Note:
Public projects usually have long lives, long
enough to use capitalized cost
• If lives are short, use LCM method
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Sec 9.3 Alternative Selection Using
Incremental B/C Analysis
Example 9.4
Comparison of 2 designs A and B for a new patient room wing to
the municipal hospital.
Step 1. No disbenefits; use equivalent AW of costs
AWA = 10 M(A/P,5%,30) + 35,000 = $685,500
AWB = 15 M(A/P,5%,30) + 55,000 = $1,030,750
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Sec 9.3 Alternative Selection Using
Incremental B/C Analysis
Step 2.
∆C = 1,030,750 – 685,500 = $345,250
Step 3.
∆B = 450,000 – 200,000 = $250,000
Step 4.
B/C = 250,000/345,250 = 0.72
Step 5.
0.72 < 1, eliminate B; A is selected for the
construction bid.
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Sec 9.3 Alternative Selection Using
Incremental B/C Analysis
Example 9.4 (b)
• When the design A will reduce its income by an estimated
$500,000 per year; some of the day-surgery features of design
A duplicate its services.
• The design B could reduce its annual revenue by an estimated
$400,000; it will eliminate an entire parking lot used by their
patrons for short-term parking.
• The city financial manager stated that these concerns would
be entered into the evaluation as disbenefits of the respective
designs.
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Sec 9.3 Alternative Selection Using
Incremental B/C Analysis
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Sec 9.4 Incremental B/C Analysis of
Multiple Alternatives
• The selection guideline is:
Choose the largest cost alternative that is justified with an
incremental B/C≥1.0 when this selected alternative has been
compared with another justified alternative
• There are two types of benefits estimates:
– Implied benefits based on usage cost estimates :
comparison of alternatives is against each other only
– Direct benefits: Comparison of alternatives is against DN
first, then each other (Like revenue alternatives in ROR
analysis)
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Sec 9.4 Incremental B/C Analysis of
Multiple Alternatives
The procedure for incremental B/C analysis of MEA is:
1. Determine the total equivalent costs for all the alternatives.
2. Order the alternatives, smallest first.
3. Determine the total equivalent benefits.
4. Direct benefits estimation only. Calculate B/C for the first ordered defender
which makes DN the defender and the first alternative the challenger. If
B/C<1, eliminate the challenger. Repeat until B/C≥1. The defender is
eliminated and the next alternative becomes the challenger.
5. Calculate incremental costs and benefits. ∆C = challenger cost - defender
cost; ∆B = challenger benefits – defender benefits. If usage costs are used ∆B
= defender usage costs – challenger user costs.
6. Calculate the incremental B/C for the first challenger compared to the
defender. B/C = ∆B/∆C. If incremental B/C≥1, the challenger becomes the
defender and the previous defender is eliminated. If B/C<1, remove the
challenger and the defender remains against the next challenger. Repeat the
above until only one alternative remains, it is the best one.
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Sec 9.4 Incremental B/C Analysis of
Multiple Alternatives
Example
Step 1. Total equivalent cost is sum of two incentives.
For proposal 1:
AW1 = 250,000(A/P,7%,8) + 25,000 = $66,867
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Sec 9.4 Incremental B/C Analysis of
Multiple Alternatives
• The discount rate used by the EDC is 7% per year.
• Can the current incentive guidelines be used to accept the
winning proposal?
• The viewpoint taken for the economic analysis is that of a
county resident.
• The first-year cash incentives and annual tax reduction
incentives are real costs to the residents.
• Benefits are derived from two components: the decreased
entrance fee estimates and the increased sales tax receipts.
• These will benefit each citizen indirectly through the increase
in money available to those who use the park and through the
city and county budgets where sales tax receipts are
deposited. Since these benefits must be calculated indirectly
from these two components, the initial proposal B/C values
cannot be calculated to initially eliminate any proposals.
• A B/C analysis incrementally comparing two alternatives at a
time must be conducted.
Sec 9.4 Incremental B/C Analysis of
Multiple Alternatives
Step 2. Order alternatives by increasing AW of total costs
Step 3. Compare 2-to-1 over 8 years; use ∆usage costs for ∆B
∆B = entrance fee decrease + sales tax receipt increase
= 50,000 + 10,000 = $60,000
∆C = 93,614 – 66,867 = $26,747
Step 4. ∆B/C = 60,000/26,747 = 2.24
Step 5. 2.24 > 1; eliminate 1; accept 2
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Sec 9.4 Incremental B/C Analysis of
Multiple Alternatives
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Sec 9.4 Incremental B/C Analysis of
Multiple Alternatives
• Compare 3-to-2: ∆B/C = 25,000/40,120 = 0.62
Proposal 2 is accepted
• Compare 4-to-2: ∆B/C = 220,000/120,360 = 1.83
Proposal 2 is eliminated; accept4
Proposal 4 is the best proposal
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Engineering Economy
Chapter 14
Effects of Inflation
Session 20
20--21
Dr. Ilham KISSANI
EGR2302
EGR
2302--Engineering Economics
Al Akhawayn University
Chapter 14
PURPOSE
Consider inflation in an engineering economics analysis
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Chapter 14
TOPICS
14.0 Review of key concepts of the previous chapter
14.1 Understanding the impact of inflation
14.2 PW calculations adjusted for inflation
14.3 FW calculations adjusted for inflation
14.4 Capital Recovery calculations adjusted for inflation
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Sec 14.1 Understanding the impact of
inflation
Definition:
Inflation is an increase in the amount of money necessary to
obtain the same amount of product or service.
• The value of money has decreased : it takes more dollars for
the same amount of goods or services.
This is a sign of inflation.
• To make comparisons between monetary amounts that occur
in different time periods, the future-value dollars must first be
converted to constant-value dollars to represent the same
purchasing power over time.
• Money in one period of time t1 can be compared with money
in another period of time t2 by using the equation:
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Sec 14.1 Understanding the impact of
inflation
Some statements
60 years ago you could buy :
• a bread for 0.05dh,
• a new car for less than 10,000dh,
• and an average house for around 50,000 dh.
What causes inflation?
• Inflation is mainly related to the equilibrium between the increase of
money supply and the growth of production capacity for goods and services.
• Inflation is also related to the equilibrium between the demand and supply
of goods, If demand is growing faster than supply prices will increase.
• Inflation is also related to a rise in production inputs prices (gas, oil…). If
companies' costs go up (taxes, raw materials), they need to increase prices
to maintain their profit margins.
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Sec 14.1 Understanding the impact of
inflation
• McDonald’s Big Mac cost $2.23 in August 2004
• If inflation =4% during the last year, the constant-value is 2.23/(1.04) = $2.14
in August 2003
• The price is $2.23(1.04) = $2.32 in August 2005
• If inflation=4% per year over the next 10 years, the price in 2014 is
$2.23(1.04)10 = $3.30 August 2014
• If inflation =6% per year, the Big Mac cost in 10 years will be $3.99, an
increase of 79%.
• In some areas of the world, hyperinflation may average 50% per year. The Big
Mac in 10 years in this case rises from $2.23 to $128.59.
• This is why countries experiencing hyperinflation must devalue the currency
by factors of 100 and 1000.
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Sec 14. 1 Understanding the impact of
inflation
• Real or inflation-free interest rate i - rate at which interest is
earned when inflation has been removed; presents an actual
gain in purchasing power.
• Inflation-adjusted interest rate if - The market interest rate is
an inflation-adjusted rate (quoted daily). This rate is a
combination of the real interest rate i and the inflation rate f,
and, therefore, it changes as the inflation rate changes. It is also
known as the inflated interest rate.
• Inflation rate f - measure of the rate of change in the value of
the currency.
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Sec 14. 1 Understanding the impact of
inflation
• Deflation is the opposite of inflation in that when deflation is present, the
purchasing power of the monetary unit is greater in the future than at
present.
• Temporary price deflation may occur in specific sectors of the economy due
to the introduction of improved products, cheaper technology, or imported
materials or products that force current prices down.
• However, if deflation occurs at a national level, there may be a lack of money
for new capital. Another result is that individuals and families have less
money to spend due to fewer jobs, less credit, and fewer loans available.
• For example, if deflation is estimated to be 2% per year, an asset that costs
$10,000 today would have a first cost 5 years from now:
10,000(1 - f )n = 10,000(0.98)5 = 10,000(0.9039) = $9039
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Sec 14.2 PW calculations adjusted for
inflation
Conclusions
• At f = 4%, $5000 today inflates to $5849 in 4 years.
• $5000 four years from now has a PW of only $3415 constant-value dollars at
a real interest rate of 10% per year.
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Sec 14.2 PW calculations adjusted for
inflation
•
•
•
The constant-value amount of $5000,
the future-dollar costs at 4% inflation, and
the present worth at 10% real interest with inflation considered. The effect of
compounded inflation and interest rates is large (shaded area).
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Sec 14.2 PW calculations adjusted for
inflation
• To account for inflation in a PW analysis it involves adjusting
the interest formulas. Consider the P/F formula, where i is the
real interest rate.
• F is a future-dollar amount with inflation built in. F can be
converted into today’s dollars using.
• If the term i + f + if is defined as if, the equation becomes:
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Sec 14.2 PW calculations adjusted for
inflation
• The inflation-adjusted interest rate if is defined as
where i = real interest rate; f = inflation rate
• Requires no conversion to CV amounts to obtain PW values
• For i = 10% per year and f = 4% per year, the previous equation yields an
inflated interest rate of 14.4%.
if = 0.10 + 0.04 + 0.10(0.04) = 0.144
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Sec 14.2 PW calculations adjusted for
inflation
Example 1
Given: the donation earns a real 10% per year and assume the inflation rate is
3% per year.
Three options are available:
• Plan A. $60,000 now.
• Plan B. $15,000 per year for 8 years beginning 1 year from now.
• Plan C. $50,000 three years from now and another $80,000, five years from
now.
• Find: Select the plan that maximizes the buying power of the dollars
received.
• which plan should be selected?
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Sec 14.2 PW calculations adjusted for
inflation
Example 1
• For plans B and C, the easiest way to obtain the PW is through
the use of the inflated interest rate.
• PWC is the largest in today’s dollars, select plan C.
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Sec 14.2 PW calculations adjusted for
inflation
Example 2
• A 15-year $50,000 bond that has a dividend rate of 10% per
year, payable semiannually, is currently for sale.
• If the expected rate of return of the purchaser is 8% per year,
compounded semiannually, and if the inflation rate is 2.5% each
6-month period, what is the bond worth now ?
(a) without an adjustment for inflation.
The semiannual dividend : I = [(50,000)(0.10)] / 2 = $2500.
At a nominal 4% per 6 months for 30 periods:
PW = 2500(P/A,4%,30) + 50,000(P/F,4%,30) = $58,645
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Sec 14.2 PW calculations adjusted for
inflation
(b) With inflation: Use the inflated rate if .
if = 0.04 + 0.025 + (0.04)(0.025) = 0.066 per semiannual period
• PW = 2500(P/A, 6.6%,30) + 50,000(P/F, 6.6%,30)
= 2500(12.9244) + 50,000(0.1470)
= $39,660 vs $58,645
• The $18,985 difference in PW values illustrates the negative
impact made by only 2.5% inflation each 6 months (5.06% per
year).
• Purchasing the $50,000 bond means receiving $75,000 in
dividends over 15 years and the $50,000 principal in year 15.
This is worth only $39,660 in constant-value (today’s) dollars.
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Sec 14.2 PW calculations adjusted for
inflation
Example 14.3: using i = 15% and g = 12% for the geometric series.
(a) The PW without an adjustment for inflation
PW = -35,000 - 7,000(P/A,15%,4)
- 7,000(P/A,12%,15%,9)(P/F,15%,4)
= $-83,232
(b) With inflation considered
if = 0.15 + 0.11 + (0.15)(0.11) = 0.2765
PW = -35,000 - 7,000(P/A,27.65%,4) 7,000(P/A,12%,27.65%,9)(P/F,27.65%,4)
= -35,000 - 7000(2.2545) - 30,945(0.3766)= - $62,436
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Sec 14.2 PW calculations adjusted for
inflation
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Sec 14.3 FW calculations adjusted for
inflation
In future worth calculations, a future amount F can have any one of 4 different
interpretations:
• Case 1 - The actual amount of money that will be accumulated at time n.
• Case 2 - The purchasing power of the actual amount accumulated at
time n, but stated in today’s (constant-value) dollars.
• Case 3 - The number of future dollars required at time n to maintain
the same purchasing power as a dollar today; that is, inflation is considered,
but interest is not.
• Case 4 - The number of dollars required at time n to maintain
purchasing power and earn a stated real interest rate. This is the case
applied when a MARR is established.
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Sec 14.3 FW calculations adjusted for
inflation
Illustrations of the 4 different interpretations: P = $1,000; n = 7;
Market return if = 10%; Inflation rate f = 4%
Case 1 - The actual amount of money that will be accumulated at time n.
Actual future amount accumulated at market rate if (both purchasing power
and return included)
F = P(F/P,if,n) = 1,000(F/P,10%,7) = $1,948
Case 2: Purchasing power maintained, but no inflation considered. Use real
interest rate i
• Usually the market (inflation-adjusted) rate if and inflation rate f are
estimated. To get I use:
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Sec 14.3 FW calculations adjusted for
inflation
Therefore, real return i = (0.10 – 0.04)/(1.04) = 5.77%
F = 1,000(F/P,5.77%,7) = $1,481
Case 3: Future amount with no return. Use only the inflation rate f
F = P(F/P,f%,n) = 1,000(F/P,4%,7) = $1,316
Case 4 – The MARR is determined using both inflation and a return to cover
capital increase and expected return
Inflation-adjusted MARR is: MARRf = i + f + if
MARRf = 0.13 + 0.04 + (0.13)(0.04) = 17.52%
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Sec 14.3 FW calculations adjusted for
inflation
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Sec 14.3 FW calculations adjusted for
inflation
Example 14.4
• If the company selects plan A, the equipment will be purchased now for
$200,000.
• If the company selects plan I, the purchase will be deferred for 3 years when
the cost is expected to rise rapidly to $340,000.
• Abbott expects a real MARR of 12% per year. The inflation rate in the country
has averaged 6.75% per year.
A- Inflation not considered
• The real rate (MARR) is i = 12% per year. The cost of plan I is $340,000 three
years from now.
• FWA = -200,000(F/P,12%,3) = -$280,986
• FWI = - $340,000
• Select plan A (purchase now).
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Sec 14.3 FW calculations adjusted for
inflation
• The real rate is 12% and inflation rate is 6.75%.
• The inflation-adjusted MARR:
if = 0.12 + 0.0675 + 0.12(0.0675) = 0.1956
•
•
•
•
Use if to convert PW to a FW value for plan A in future dollars.
FWA = -200,000(F/P,19.56%,3) = - $341,812
FWI = -$340,000
Purchase later (plan I ) is now selected, because it requires
fewer equivalent future dollars. The inflation rate of 6.75% per
year has raised the equivalent future worth of costs by 21.6% to
$341,812.
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Sec 14.4 Capital Recovery calculations
adjusted for inflation
• Capital must be recovered with future inflated dollars
• Less buying power in future means more money needed to
recover present investments plus a return
• Use the inflated interest rate in the A/P formula.
• For example, if $1000 is invested today at a real interest rate of
10% per year when the inflation rate is 8% per year, the
equivalent amount that must be recovered each year for 5 years
in future dollars is:
if = 0.10 + 0.08 + 0.10(0.08) = 0.1880
A = 1000(A/P,18.8%,5) = $325.59
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Sec 14.4 Capital Recovery calculations
adjusted for inflation
Example 14.5
What is the equivalent AW required for 5 years to accumulate
an amount of money with the same purchasing power as
$680.58 today, if the market inflated interest rate is 10% per
year and inflation is 8% per year?
• First, find the inflated (actual amount) dollars required 5 years
from now (case 3).
F = (present buying power)(1 + f )5 = 680.58(1.08)5 = $1000
• The actual amount of AW is calculated using the market
(inflated) interest rate of 10%.
A =1000(A/F,10%,5) = $163.80
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Assignments
Assume i= 6%, and f= 4%, redo with Excel
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Engineering Economy
Chapter 16
Methods of Depreciation
Session 2323-24
Dr. Ilham KISSANI
EGR2302-Engineering Economics
EGR2302Al Akhawayn University
Chapter 16
PURPOSE
Use classical and government-approved methods to
reduce the value of the capital investment in an asset or
natural resource
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Chapter 16
TOPICS
16.0 Review of key concepts of the previous chapter
16.1 Depreciation Terminology
16.2 Straight Line (SL) Depreciation
16.3 Declining Balance and Double Declining Balance Depreciation
16.4 Modified Accelerated Cost Recovery System (MACRS)
16.5 Determining The MACRS Recovery Period
16.6 Depletion Methods
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Sec 16.1 Depreciation Terminology
• Depreciation is the reduction in value of an asset.
• The method used to depreciate an asset is a way to account
for the decreasing value of the asset and to represent the
diminishing value of the capital funds invested in it.
• The annual depreciation amount Dt does not represent an
actual cash flow, nor does it necessarily reflect the actual
usage pattern of the asset during ownership.
• Depreciation may be performed for two reasons:
1. Use by a corporation or business for internal financial
accounting (book depreciation). Equivalent to an expense.
2. Use in tax calculations per government regulations (tax
depreciation). Equivalent to tax savings.
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Sec 16.1 Depreciation Terminology
• Book depreciation indicates the reduced investment in an
asset based upon the usage pattern and expected useful life
of the asset. Several depreciation methods are used to
determine book depreciation: straight line, declining balance,
and the infrequently used sum-of-year digits method.
• Tax depreciation is tax deductible; it can be subtracted from
income when calculating the amount of taxes due each year.
However, the tax depreciation amount must be calculated
using a government approved method.
• Tax depreciation must be calculated using MACRS; book
depreciation may be calculated using any classical method or
MACRS.
• MACRS has the DB and SL methods, in slightly different forms,
embedded in it, but these two methods cannot be used
directly if the annual depreciation is to be tax deductible.
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Sec 16.1 Depreciation Terminology
• First Cost or Unadjusted Basis (B): initial purchase price + all costs incurred in
placing the asset in service
• Book value : the remaining capital investment (undepreciated) on the books after
the total amount of depreciation has been subtracted from the basis. The book
value (BV) is usually determined at the end of each year.
• Recovery Period (n): depreciable life of the asset in years– often set by law
• Market Value (MV) : the amount realized from the asset sale on the open market.
• Salvage Value (S) is the estimated trade-in value or market value at the end the
asset’s useful life.
• Depreciation Rate (dt) : the fraction of the first cost removed by depreciation each
year
• Personal Property : all property except real estate used to conduct business
activities to generate profit or gain(vehicles, equipment, etc.)
• Real Property: real estate , buildings and certain structures /Land is Real Property, but
by law is NOT depreciable for tax
• Half-year convention: assumes that assets are placed in service or disposed of in
midyear, regardless of when these events actually occur during the year.
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Sec 16.1 Depreciation Terminology
Models of depreciation
Classical methods:
• The straight line (SL) model is used for tax and book depreciation.
Accelerated models:
• The declining balance (DB) model, decrease the book value to
zero (or to the salvage value) more rapidly than the straight line
method.
• Sum-of-year digits (SYD) is applied less frequently.
• Modified Accelerated Cost Recovery System (MACRS):Method for
tax depreciation in the U.S.
To determine annual depreciation, Excel functions are available for
the classical methods, straight line, declining balance, and sumof-year digits (SYD).
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Sec 16.2 Straight Line (SL) Depreciation
Straight line (SL): most frequent, BV decreases linearly
over time.
• Depreciation is the same every period
• It writes off capital investment linearly over n years.
• The estimated salvage value is always considered.
• This is the classical, no accelerated depreciation model.
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Sec 16.2 Straight Line (SL) Depreciation
Straight line: Depreciation is the same every period
The fixed annual depreciation:
The depreciation rate:
The book value
after t years of service:
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Sec 16.2 Straight Line (SL) Depreciation
Example 16.1:
• An asset has a first cost of $50,000 with a $10,000 estimated
salvage value after 5 years.
• Find the annual depreciation.
• To obtain the same result using the Excel function: in a single
cell operation use SLN (B,S,n)
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Sec 16.3 Declining Balance and Double
Declining Balance Depreciation
Declining Balance (DB) : Also known as fixed percentage or
uniform percentage method
• The model accelerates depreciation compared to straight line
because the book value is reduced each year by a fixed
percentage.
• Does not include salvage value into DB or DDB for
depreciation calculation.
• The most used rate is twice the SL rate; called double
declining balance (DDB).
• DDB has an implied salvage that may be lower than the
estimated salvage.
• It is not an approved tax depreciation method in the United
States. It is frequently used for book depreciation purposes.
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Sec 16.3 Declining Balance and Double
Declining Balance Depreciation
Max. depr. rate by law:
d max
2
=
n
Book Value amounts (two
methods)
BVt = B (1 − d )t
BVt = BVt −1 − Dt
Actual depreciation rate for year t:
d t = d (1 − d )t −1
Implied Salvage Value
Depreciation for year t:
Implied d for S >0
impS = BVn = B (1 − d ) n
Dt = ( d ) BVt −1
1/ n
S
1−  
B
If BVt-1 not known, apply:
Dt = dB (1 − d ) t −1
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Sec 16.3 Declining Balance and Double
Declining Balance Depreciation
Example 16.2= DDB
A fiber optics testing device is to be DDB depreciated. It has a first cost of
$25,000 and an estimated salvage of $2500 after 12 years.
(a) Calculate the depreciation and book value for years 1 and 4.
(b) Calculate the implied salvage value after 12 years.
------------------The DDB fixed depreciation rate is d = 2/n = 2/12 = 0.1667 per year.
Year 1: D1 = (0.1667)(25,000)(1 - 0.1667)1-1 = $4167
BV1 = 25,000(1 - 0.1667)1 = $20,833
Year 4: D4 = (0.1667)(25,000)(1 - 0.1667)4-1 = $2411
BV4 = 25,000(1 - 0.1667)4 = $12,054
Implied S = 25,000(1 - 0.1667)12 = $2803
Since the estimated S = $2500 is less than $2803, the asset is not fully
depreciated when it reaches its 12-year expected life.
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Sec 16.3 Declining Balance and Double
Declining Balance Depreciation
Example 16.3=DB
• Freeport-McMoRan Mining Company has purchased a computercontrolled gold ore grading unit for $80,000. The unit has an anticipated
life of 10 years and a salvage value of $10,000.
• Use the DB and DDB methods to compare the schedule of depreciation
and book values for each year. ---------------------------------------------------------• An implied DB depreciation rate : d = 1 – (10,000/80,000) 1/10 = 0.1877
• 0.1877 < 2/n = 0.2, so this DB model does not exceed twice the straight
line rate.
• For example: D2 = d(BV1) = 0.1877(64,984) = $12,197
• BV2 = 64,984 - 12,197 = $52,787
• Because we round off to even dollars, $2312 is calculated for depreciation
in year 10, but $2318 is deducted to make BV10 = S = $10,000 exactly.
• Similar calculations for DDB with d = 0.2 are conducted.
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Sec 16.3 Declining Balance and Double
Declining Balance Depreciation
Example 16.3
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Sec 16.4 Modified Accelerated Cost
Recovery System (MACRS)
Modified Accelerated Cost Recovery System (MACRS)
•It is the only approved tax depreciation system in the United
States.
•It automatically switches from DDB or DB to SL depreciation.
•It always depreciates to zero it assumes S = 0. (B is
completely depreciated)
•Recovery periods are specified by property classes.
•Depreciation rates are tabulated and not computed.
•The actual recovery period is 1 year longer due to the imposed
half-year convention.
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Sec 16.4 Modified Accelerated Cost
Recovery System (MACRS)
• MACRS was derived from the 1981 ACRS system and went into
effect in 1986.
• Defines statutory recovery (depreciation) percentages.
• Incorporates the half-year convention.
Dt = dtB
BVt = BVt-1 – Dt
BVt = first cost – sum of accumulated depreciation
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Sec 16.4 Modified Accelerated Cost
Recovery System (MACRS)
dt: depreciation rate over years
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Sec 16.4 Modified Accelerated Cost
Recovery System (MACRS)
Example 16.4
MACRS: D1 + D2 = $133,320 + 177,800 = $311,120
DDB: D1 + D2 = $266,667 + 88,889 = $355,556
The DDB depreciation is larger.
n=3; d(DDB)=0.66
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Sec 16.5 Determining The MACRS Recovery
Period
• The expected useful life of property is estimated in years and used as the n
value in alternative evaluation and in depreciation computations. For book
depreciation, the n value should be the expected useful life. However, when
the depreciation will be claimed as tax deductible, the n value should be
lower.
• The advantage of a recovery period shorter than the anticipated useful life is
leveraged by the accelerated depreciation models that write off more of the
basis B in the initial years.
• The U.S. government requires that all depreciable property be classified into
a property class that identifies its MACRS-allowed recovery period.
• Table 16–4, a summary of material from IRS Publication 946, gives examples
of assets and the MACRS n values.
• Virtually any property considered in an economic analysis has a MACRS n
value of 3, 5, 7, 10, 15, or 20 years.
• The General Depreciation System (GDS) value used in problems and
examples.
• The alternative depreciation system (ADS) recovery period range.
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400
Sec16.6 Depletion Methods
• Depletion applies only to natural resources which cannot be
replaced or repurchased in the same manner as can a
machine, computer, or structure. Like: Timber, Mineral
deposits, Oil and gas, etc.
• There are two methods of depletion - cost depletion and
percentage depletion.
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Sec16.6 Depletion Methods
Cost depletion
• Sometimes referred to as factor depletion, is based on the
level of activity or usage, not time, as in depreciation.
• The cost depletion factor for year t, denoted by pt, is the ratio
of the first cost of the resource to the estimated number of
units recoverable. (page 546)
• The annual depletion charge is pt times the year’s usage or
volume. (page 547)
• The total cost depletion cannot exceed the first cost of the
resource.
• If the capacity of the property is re-estimated some year in
the future, a new cost depletion factor is determined based
upon the undepleted amount and the new capacity estimate.
• Example 16.5
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Sec16.6 Depletion Methods
Percentage depletion
• A constant, stated percentage of the resource’s gross income
may be depleted each year provided it does not exceed 50%
of the company’s taxable income.
• For oil and gas property, the limit is 100% of taxable income.
• The annual depletion amount is calculated as using
percentage depletion, total depletion charges may exceed
first cost with no limitation.
• However, the law also requires that the cost depletion
amount be chosen if the percentage depletion is smaller in
any year.
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Sec16.6 Depletion Methods
Example 16.6
• A gold mine was purchased for $10 million. It has an
anticipated gross income of $5.0 million per year for years 1
to 5 and $3.0 million per year after year 5.
• Assume that depletion charges do not exceed 50% of taxable
income. Compute annual depletion amounts for the mine.
How long will it take to recover the initial investment at i =
0%?
A 15% depletion applies to gold. Depletion amounts are
Years 1 to 5:
0.15(5.0 million) = $750,000
Years thereafter: 0.15(3.0 million) = $450,000
A total of $3.75 million is written off in 5 years, and the
remaining $6.25 million is written off at $450,000 per year.
The total number of years is: 5+6.25M/450000=18.9
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