Engineering Economics for Civil Engineering-full Lecture Note, By Melese Mengistu

Engineering Economics Ceng 5191
Civil Engineering Department
By: Melese Mengistu (MSc. Construction Engineering and Management)
Lecturer at Dire Dawa University Institute Of Technology- School Of
Civil Engineering & Architecture
E-mail: melesemngst@yahoo.com

Engineering Economics [CENG 5191]
Chapter 1
General Introduction
Lecture # 1
by Melese m.
2

What is Economics
 It is the branch of social science that deals with the production, distribution
and consumption of goods and services and their management.
 It is the study of production and distribution of wealth.
 It is the study of choice and decision-making in the world with limited
resources.
 It is the study of how individuals, businesses & governments use their
limited resources and satisfy unlimited wants.
by Melese m.
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Engineering and Economics
 Engineering activities are means of satisfying human wants and requirements
 Concerns – material, work forces, capital and etc.
 Because of resource constraints, engineering is closely associated with
economics
 Engineering proposals are evaluated in terms of economics [worth & cost]
before it is undertaken.
 Essential pre-requisite of successful engineering application is economic
feasibility
 Engineering economics is a collection of mathematical / analytical techniques that
simplify economic comparison.
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Cont…
 Engineering economics: formulation, estimation and evaluation of the economic
outcomes out of various available alternatives to accomplish a defined purpose.
 Systematic evaluation of the cost and benefit [economic merits] of proposed
technical projects.
 To be economically acceptable [i.e., affordable], solutions to engineering
problems must demonstrate a positive balance of long-term benefits over long-
term costs.
 The Objective of EEs is to balance different types of costs and the performance
[time, safety, reliability, etc.] in the most economical manner.
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Principles of EE
Principle 1: Develop the alternatives
 The alternatives need to be identified and then defined for
subsequent analysis.
 Since the choice [decision] is among alternatives, developing and
defining the comprehensive list of alternatives for detailed
evaluation is important.
 Creativity and innovation are essential.
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Cont…
Principle 2: Focus on the differences
 Only the differences in expected future outcomes among the
alternatives are relevant to their comparison and should be
considered in the decision.
 Outcomes that are common to all alternatives can be disregarded
in the process of comparison and decision.
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Cont…
Principle 3: Use a consistent viewpoint
 The prospective outcomes of the alternatives, selection of the
criteria and other, should be consistently developed from a defined
viewpoint [perspective].
 Usually, the viewpoint of the decision maker would be used.
 For example, the perspective of the employees is used for the
problem of designing the employee benefit package
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Cont…
Principle 4: Use a common unit of measure
 Using a common unit of measurement to enumerate as many of the
prospective outcomes as possible will simplify the analysis of the
alternatives.
 For economic outcomes, a monetary unit such as “birr/dollars” is the
common unit of measure.
 If the outcomes cannot be quantified, describe these consequences
explicitly so that the information is useful to the decision maker in the
comparison of the alternatives. by Melese m.
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Cont…
Principle 5: Consider all relevant criteria
Principle 6: Make uncertainty explicit
 Risk and uncertainty are inherent in estimating the future outcomes
of the alternatives and should be recognized in the analysis and
comparison.
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Cont…
Principle 7: Revisit your decisions
 Improved decision making results from an adaptive process.
 The initial projected outcomes of the selected alternative should
be subsequently compared with the actual results achieved.
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Cont…
Generally:
Engineering economics is an answer to following questions
 Which engineering projects are worthwhile? [project
worthiness]
 Which engineering projects should have a higher priority?
[priority for available alternatives]
 How should the engineering project be designed? [economic
design]
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Cont…
Objective Strategic Economic Decisions
 Develop profitable projects and expansion
 Service quality improvement
 New product and product improvement
 Purchase Equipment and equipment selection
 Equipment Replacement
 Reduction of costs
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Economic Decision Making
 Maximum capital : The out of pocket commitment is the total expense
required for an alternative.
 Pay Back Period: The pay back period for an investment is the number
of years it takes to repay the original invested capital
 Average Annual Rate of Return: The alternatives are evaluated on the
basis of only the average rate of return as expressed in terms of a
percentage (of the original capital).
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Cont…
1. Out of Pocket Commitment/ Maximum capital
Example. A precast concrete factory has to produce 100,000
railway sleepers per year. An economic choice has to be made
between using steel moulds and wooden moulds. The life of steel
mould is estimated to be one year, while that of wooden mould is
one month. The costs of preparing 100,000 steel mould and one set
of wooden mould are 4 Birr. and1 Birr. respectively. It is further
estimated that the labor costs for assembling and removing the steel
and wooden moulds are Birr. 1 Birr & 0.9 per sleeper respectively.
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Cont…
Solution
- The out of pocket commitment for steel mould option =
- The total labor cost incurred for production of 100000 sleepers/yr +
the cost of the steel mould/yr
=Birr. 100,000 x 1 + Birr. 4x100,000 = 500,000 birr/year
- Similarly, the out of pocket commitment for wooden mould option is =
Birr 0.90x100,000 + Birr 100,000x1 = Birr. 190,000./month =12x
190,000= 2,280,000 birr/year
- STEEL is our Engineering Decision.
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Cont…
2. Pay Back Period
Example. A contractor has two brands of excavators A and B to choose
from. Suppose both the brands are available for a down payment of Birr.
400,000. Both brands can be useful for a period of four years.
 Brand A is estimated to give a return of Birr. 50,000 for the first year,
Birr. 150,000 for the second year, and Birr.200, 000 for the third and
fourth year.
 Brand B on the other hand is expected to give a return of Birr.
150,000 for all the four years.
by Melese m.
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Cont…
Solution
- The payback period for Brand A = 3 yrs., as the initial investment
of Birr. 400,000 is recovered in 3 years (50,000 + 150,000 +
200,000= 400,000).
- The method does not consider the returns after the payback
period.
- For Brand B, the return is Birr. 300,000 up to the end of 2nd year
and in 3rd year it equals Birr. 450,000
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Cont…
 Thus the investment amount of Birr. 400,000 is recovered
somewhere between 2nd and 3rd yr, which can be found out by
interpolation.
 Hence the payback period for Brand B = 2 + (3 – 2) *
(400,000-300,000)/ (450,000 – 300,000)) = 2.67 yrs or 2
years and 8 months.
 Here also as in the first case we neglect the return that is
expected beyond the pay back period
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Cont…
3. Average Annual Rate of Return
- We use the previous example
- The average annual return from Brand A=
(50,000+150,000+200,000+200,000)/4 = 600,000/4= 150,000.
- Average annual rate of return for Brand A in %=
(150,000/400,000)*100= 37.5%.
- Here 400,000 is the original invested capital
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Cont…
The average annual return for Brand
B=(150,000+150,000+150,000+150,000)/4 =150,000.
- The average annual rate of return for equipment
B in %= (150,000/400,000)*100= 37.5%.
- Both are equal. Here we might go for B, since having high initial
return
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Time Value of Money
 In most decisions the change in the value of money needs to be
accounted.
 The manifestation of time value of money is called interest.
 Interest ; Money paid by Borrower for the use of funds provided
by the lender
 Interest represents
 Earning power of money
 Risk of non-repayment
 Loss of used of the loaned money
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Time Value of Money
 Therefore time value of money is the relationship between time
and money.
 It is clearly explained in quote;
 “A bird in hand is more than two in bush”
 The reason for the time value of money is inflation, risk and cost
of money
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Interest
 Interest could be simple or compound
 Simple; The interest doesn't attract any interest during the
repayment period
 Compound; The interest amount it self also attracts further interest.
 Consider the following statement by a Bank “ Interest on the deposit
will be payable at the rate of eight percent compound quarterly “
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Cont…
Example: A year for Quarterly compounding of interest:- Four periods
(3 months of each) If the amount was 100 Birr and interest is 8%
yearly.
1st period = 100 + (100*8%/4)
= 102
2nd Period = 102 + (102*8%/4)
= 104.04
3rd Period = 104.04 + (104.04*8%/4)
=106.12
4th Period = 106.12 + (106.12*8%/4)
= 108.24
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End of Chapter 1
General Introduction
Lecture # 1
Thank You!!!
by Melese m.

Chapter 2
Time Value of Money and cash flow
Lecture # 2
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Time Value of Money
Consider the following two options
A. Option 1: Single lump sum payment of 10 million Birr.
B. Option 2: Annual payment of 3 million for 10 years [total of 30 m]
Which one is better from a strictly economic viewpoint?
 Over time money can earn money = interest, therefore the earlier
a sum of money is received, the more it is worth
 Engineering projects are commitments of capital for extended
periods of time, therefore the effect of time on value of money
must always be considered. by Melese m.
3

Time Value of Money
Interest
 Used to indicate a rental for the use of money.
 Same as the rental paid for the use of equipment, building etc.
 Usually expressed as a percentage of the amount owed.
 It is due and payable at the close of each period of time involved
in the agreed transaction [usually every year or month].
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Time Value of Money
Interest Rate [i]
 Rate of capital growth.
 Rate of gain received from an investment over a period of time.
 Usually expressed on an annual basis.
 For the lender, it consists, for convenience, of [1] risk of loss, [2]
administrative expenses, and [3] profit or pure gain.
 For the borrower, it is the cost of using a capital for immediately
meeting his or her needs. by Melese m.
5

Time Value of Money
Time Value of Money [TVM]  Money- Time Relationships
 Means that two equal amount at different points of time do not
have equal value if the interest rate is greater than zero.
 Money has both earning power [it can be put in the bank to earn
interest] and
 purchasing power [Usually decreases over time: inflation]
 Money has a time value because it can earn interest over time.
 One birr today is worth more than one birr tomorrow.
 Failure to pay the bills results in additional charge termed interest.
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Time Value of Money
Interest [ I ] [Simple]
 Total interest is directly proportional to the amount of loan
[principal], the interest rate, and the number of interest periods
I = [P] [n] [i]
 I : total interest
 P : principal
 n : number of interest periods
 i : interest rate per interest period.
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Time Value of Money
Simple Interest [i] :- Example:
 If 1,000.00 birr is borrowed at 14% interest, then interest on the
principal of 1,000.00 birr after one year is 0.14 x 1, 000, or
140.00 birr.
 If the borrower pays back the total amount owed after one year,
she/he will pay 1,140.00 birr.
by Melese m.
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Time Value of Money
Interest [i] [Compounded] Cont’d
 If someone does not pay back any of the amount owed after one
year, then normally the interest owed, but not paid, is considered
now to be additional principal, and thus the interest is
compounded
 After two years he/she will owe 1,140.00 birr + 0.14 X
1,140.00, or 1,299.60.
by Melese m.
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Time Value of Money
Economic Equivalency
 The banker in the previous example normally does not care
whether you pay him 1,140.00 birr after one year or 1,299.60
birr after two years.
 To him, the three values [1,000, 1,140, and 1,299.60 birr] are
equivalent.
 1,000 Birr today is equivalent to 1,140 birr one year from today
and 1,000 Birr today is equivalent to 1,299.60 Birr two years
from today.
 NB: The three values are not equal but equivalent
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Time Value of Money
Economic Equivalency Cont’d
 It is to be noted that:
1. The concept of equivalence involves timing of money, amount of
money receipt/expenses and a specified rate of interest.
 The three preceding values are only equivalent for an interest
rate of 14%, and then only at the specified times.
2. Equivalence means that one sum or series differs from another
only by the accumulated interest at rate i for n periods of time.
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Time Value of Money
Cash Flow Diagram
 It is strongly recommended for situations in which the analyst
needs to visualize what is involved when flows of money occur at
various times.
 Used to visualize the flow money (income & expense ) with respect
to time.
 The usefulness of cash flow diagram for economic analysis
problems is analogous to that of the free body diagrams of
Engineering mechanics problems or control volume in hydraulics.
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Time Value of Money
Cash Flow Diagrams
 P = a present single amount of money
 F = a future single amount of money, after n periods of time
 A = end-of-period cash flows in a uniform series for a specified number
of periods, starting at the end of first period and continuing through the
last period.
 i = the rate of interest per interest period [usually one year]
 n = the number of periods of time [usually years] by Melese m.
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Time Value of Money
Example
Cash Flows Over Time:
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Time Value of Money
In a cash-flow diagram:
 Horizontal line represents time scale,
 Arrows represent cash flows.
 Downward arrows represent expenses [negative cash flows or
cash outflows] and upward arrows represent receipts [positive
cash flows or cash inflows].
 The CFD is dependent on the point of view. In the course, without
explicitly mention, the company’s [investor’s] point of view will be
taken.
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Time Value of Money
Cash-Flow Diagram Cont’d
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Time Value of Money
Example [CFD]
 You are analyzing a project with five-year life. The project
requires a capital investment of $50,000 now, and it will
generate uniform annual revenue of $6,000. Further, the project
will have a salvage value of $4,500 at the end of the fifth year
and it will require $3,000 each year for the operations.
 Develop the cash-flow diagram for this project from the investor’s
viewpoint.
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Time Value of Money
Example [CFD]: Solution
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Time Value of Money
Example
A machine cost $45,000 to purchase. Fuel, oil, grease
[FOG], and minor maintenance are estimated to cost $12.34
per operating hour [those hours when the engine is
operating and the machine is doing work]. A set of tires
cost $3,200 to replace, and their estimated life is 2,800
use hours. A $6,000 major repair will probably be required
after 4,200 hr of use. by Melese m.
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Time Value of Money
Example cont’d
 The machine is expected to last for 8,400 hr, after which it
will be sold at a price [salvage value] equal to 10% of the
original purchase price. it will generate uniform annual revenue of
$ 20, 0000 machine will operate 1,400 hr per year. Draw
cash flow diagram.
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Time Value of Money
Rules for performing arithmetic calculations with cash flows
1. Cash flows cannot be added or subtracted unless they occur at
the same point in time.
2. To move a cash flow forward in time by one time unit, multiply
the magnitude of the cash flow by [1 + i].
3. To move a cash flow backward in time by one time unit, divide
the magnitude of the cash flow by [1 + i].
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Time Value of Money
Financial Engineering Analysis [Single Payment Series]
1. Single Payment Compound-Amount Factor [SPCAF]:
OR
Find F When P is given
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Time Value of Money
Financial Engineering Analysis [Single Payment]
2. Single Payment Present-Worth Factor [SPPWF]: Find P when F is
given;
OR
 Notation: P = F [P/F, i%, N] where the factor in the parentheses
is read "find P given F at i% interest per period for N interest
periods. by Melese m.
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Time Value of Money
Single Payment Analysis
 To calculate the future value F of a single payment P after n
periods at an interest rate i, we make the following calculation:
 At the end of the first period: F1 = P + Pi = P[1+i]
 At the end of the second period: F2 = P + Pi + [P + Pi]i = P[1 + i]2
 At the end of the nth period: Fn = P[1 + i]n
 The future single amount of a present single amount is F = P[1 + i]n
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Time Value of Money
 Note: F is related to P by a factor which depends only on i and n.
 This factor, termed the single payment compound amount factor
[SPCAF], makes F equivalent to P.
 SPCAF may be expressed in a functional form;
 The present single amount of a future single amount is;
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Time Value of Money
 Note: The factor 1/[1+i]n is called the present worth compound
amount factor [PWCAF]
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Time Value of Money
Example 1: [Single Payment Analysis]
 A contractor wishes to set up a revolving line of credit at the bank to
handle his cash flow during the construction of a project.
 He believes that he needs to borrow12,000 Birr with which to set up the
account, and that he can obtain the money at 1.45% per month.
 If he pays back the loan and accumulated interest after 8 months, how
much will he/she have to pay back?
 F = 12,000[1 + 0.0145]8 = 12,000[1.122061]= 13,464.73 =13,465
Birr.
 The amount of interest will be:13,465 - 12,000 = 1,465 Birr.
by Melese m.
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Time Value of Money
E.G 2: [Single Payment Analysis]
 A construction company wants to set aside enough money today in
an interest-bearing account in order to have 100,000 Birr five
years from now for the purchase of a replacement piece of
equipment.
 If the company can receive 8% interest on its investment, how
much should be set aside now to collect the100,000 Birr five
years from now?
by Melese m.
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Time Value of Money
 E.G 2: Solution
 P = 100,000/[I + 0.08]5 =100,000/[1.46933] = 68,058.32 Birr
= 68,060 Birr
 To solve this problem you can also use the interest tables.
 P = 100,000 [P/F, 8%, 5] = 100,000[0.6805832] 68,058.32
Birr= 68,060 Birr.
by Melese m.
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Time Value of Money
Uniform/Equal Payment Series
 Often payments or receipts occur at regular intervals, and such
uniform values can be handled by the use of additional functions.
 Another symbol: A = uniform end-of-period payments or
receipts continuing for a duration of n periods
 If a uniform amount A is invested at the end of each period for n
periods at a rate of interest i per period, then the total equivalent
amount F at the end of the n periods will be:
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Time Value of Money
Uniform Payment Analysis
 By multiplying both sides of above equation by [1+i] and
subtracting from the original equation, the following expression is
obtained:
 Which can be rearrange to give
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Time Value of Money
3. Uniform [Equal payment] Series Compound-Amount Factor
[USCAF]: Find F when A is given;
OR
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Time Value of Money
E.G 3: You plan to deposit $2,000 to your savings account at the
end of every month for the next 15 months starting from the next
month. If the interest rate you can earn is 2% per month how much
money will accumulate immediately after your last deposit at the
end of the 15th month?
 Solution: A = $2,000, i = 2% per month, N = 15 months.
 F =?
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Time Value of Money
4. Uniform [Equal payment] Series Sinking-Fund Factor [USSFF]: Find
A when F is given;
OR
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Time Value of Money
E.G. 4: What uniform monthly amount should you deposit in your
savings account at the end of each month for the following 10
months in order to accumulate $75,000 at the time of the 10th
deposit? Assume that the interest rate you can earn is 4% per
month and the first deposit will be made next month.
 Solution: F = 75; 000, i = 4% per month, N = 10 months.
 A =?
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Time Value of Money
5. Uniform [Equal payment] Series Capital-Recovery Factor [USCRF]:
Find A when P is known;
OR
Note: This is the case of loans [mortgages]
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Time Value of Money
E.G 5: You plan to borrow a loan of $100,000 which you will
repay with equal annual payments for the next 5 years. Suppose
the interest rate you are charged is 8% per year and you will
make the first payment one year after receiving the loan. How
much is your annual payment?
 Solution: P = 100; 000, i = 8% per year, N = 5 years.
 A =?
by Melese m.
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Time Value of Money
6. Uniform [Equal payment] Series Present-Worth Factor [USPWF]:
Find P when A is given;
OR
by Melese m.
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Time Value of Money
Cash Flow Diagram for Single Payment
Cash Flow Diagram for Uniform Payment
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Time Value of Money
E.G 6: How much should you deposit to your savings account now
at an annual interest rate of 10% to provide for 5 end-of-year
withdrawals of $15,000 each?
 Solution: A = 15; 000, i = 10% per year, N = 5 years.
 P =?
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Time Value of Money
Uniform Gradient Payment Series : involve receipts or
disbursements that are projected to increase or decrease by a
uniform amount each period thus contributing an arithmetic series.
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Time Value of Money
 Example: Linear Gradient typical negative, Increasing Gradient:
G = $50.
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Time Value of Money
Arithmetic Gradient Factors
 The “G” amount is the constant arithmetic change from one time
period to the next.
 The “G” amount may be positive or negative!
 The present worth point is always one time period to the left of the
first cash flow in the series or,
 Two periods to the left of the first gradient cash flow!
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Time Value of Money
Derivation: Gradient Component Only: Focus Only on the
gradient Component.
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Time Value of Money
The Present worth point of a linear gradient is always:
 2 periods to the left of the “1G” point or,
 1 period to the left of the very first cash flow in the gradient
series.
DO NOT FORGET THIS!
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Time Value of Money
 Gradient Component
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Time Value of Money
Cont’d
 PW of the Base Annuity is at t = 0
 PW Base Annuity= $100 (P/A,i%,7)
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Time Value of Money
Present Worth [PW]: Linear Gradient
 The present worth of a linear gradient is the present worth of the
two components:
1. The Present Worth of the Gradient Component and,
2. The Present Worth of the Base Annuity flow
 Requires 2 separate calculations!
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Time Value of Money
Cont’d
 The PW of the Base Annuity is simply the Base Annuity: A{P/A, i%,
n} factor
 What is needed is a present worth expression for the gradient
component cash flow.
 We need to derive a closed form expression for the gradient
component.
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Time Value of Money
General CF Diagram – Gradient Part Only
 Begin- Derivation of P/G, i%, n
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Time Value of Money
Cont’d
 Factor out G and re-write as ….. Factoring G out…. P/G factor
 What is inside of the { }’s?
 Replace (P/F’s) with closed-form
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Time Value of Money
 Multiply both sides by (1+i)
 We have 2 equations [1] and [2].
 Next, subtract [1] from [2] and work with the resultant equation.
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Time Value of Money
The P/G factor for i and N
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Time Value of Money
The A/G factor
 Some authors also include the derivation of the A/G factor.
 A/G converts a linear gradient to an equivalent annuity cash flow.
 Remember, at this point one is only working with gradient
component.
 There still remains the annuity component that you must also
handle separately!
by Melese m.
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Time Value of Money
The A/G Factor Cont’d
 Convert G to an equivalent A; How to do it…………
 A/G factor using A/P with P/G
 The results follow…..
 Resultant A/G factor
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Time Value of Money
Example:7 Consider the following cash flow
 Base Annuity: First, The Base Annuity of $100/period
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Time Value of Money
 PW(10%) of the base annuity = $100(P/A,10%,5)
 PW Base = $100(3.7908)= $379.08
 •Not Finished: We need the PW of the gradient component and
then add that value to the $379.08 amount.
 Focus on the Gradient Component
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Time Value of Money
 We desire the PW of the Gradient Component at t = 0
 The Set Up
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Time Value of Money
 Calculating or looking up the P/G,10%,5 factor yields the
following:
 Pt=0 = $100(6.8618) = $686.18 for the gradient PW
 Final Result
 PW(10%) Base Annuity = $379.08
 PW(10%) Gradient Component = $686.18
 Total PW(10%) = $379.08 + $686.18 = $1065.26
by Melese m.
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Time Value of Money
 Note: The two sums occur at t =0 and can be added together: -
concept of equivalence
 Example Summarized; This Cash Flow…
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Time Value of Money
Shifted Gradient
Example 8 : i =10%; Consider the following Cash Flow
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Time Value of Money
 This is a “shifted” negative, decreasing gradient.
 The PW point in time is at t = 3 (not t = o)
 The base annuity is a $600 cash flow for 3 time periods
by Melese m.
62

Time Value of Money
 PW of the Base Annuity: 2 Steps
by Melese m.
63

Time Value of Money
 PW of Gradient Component: G = -$50
by Melese m.
64

by Melese m.
65
 Final Result
 PW(10%) Base Annuity = -$1428.93
 PW(10%) Gradient Component = -$164.46
 Total PW(10%) = -$1428.93- (-$164.46) = -$ 1264.29
Time Value of Money

by Melese m.
66
 Geometric Gradient Series Factors
 It is common for annual revenues and annual costs such as
maintenance, operations, and labor to go up or down by
a constant percentage, for example, +5% or —3% per
year.
 This change occurs every year on top of a starting amount
in the first year of the project
Time Value of Money

by Melese m.
67
 A geometric gradient series is a cash flow series that either increases
or decreases by a constant percentage each period.
 The uniform change is called the rate of change.
 g = constant rate of change, in decimal form, by which cash flow
values increase or decrease from one period to the next.
 The gradient g can be + or —.
 A, = initial cash flow in year 1 of the geometric series
 Pg = present worth of the entire geometric gradient series, including
the initial amount
Time Value of Money

by Melese m.
68
Time Value of Money

by Melese m.
69
Time Value of Money
Multiply both sides by (1+g)/( 1+ i), subtract Equation first
equation from the result/ 2nd equation , factor out Pg, and obtain

by Melese m.
70
Time Value of Money
 When g = i, substitute i for g in the above Equation and
observe that the term 1/(1 + i) appears n times.
The equation for Pg and the (P/A,g,i,n) factor formula are

by Melese m.
71
Time Value of Money
 Example:9
 A coal-fired power plant has upgraded an emission control valve.
The modification costs only $8000 and is expected to last 6 years
with a $200 salvage value. The maintenance cost is expected to
be high at $1700 the first year, increasing by 11% per year
thereafter. Determine the equivalent present worth of the
modification and maintenance cost at interest rate of 8% per
year.

by Melese m.
72
Time Value of Money
 Cash flow diagram

by Melese m.
73
Time Value of Money
 For g ≠i to calculate Pg. Total PT is the sum of three
present worth components.

Time Value of Money
Example 10
A machine cost $45,000 to purchase. Fuel, oil, grease
[FOG], and minor maintenance are estimated to cost $12.34
per operating hour [those hours when the engine is
operating and the machine is doing work]. A set of tires
cost $3,200 to replace, and their estimated life is 2,800
use hours. A $6,000 major repair will probably be required
after 4,200 hr of use. by Melese m.
74

Time Value of Money
E.G. 10 Cont’d
 The machine is expected to last for 8,400 hr, after which it
will be sold at a price [salvage value] equal to 10% of the
original purchase price. A final set of new tires will not be
purchased before the sale. How much should the owner of
the machine charge per hour of use, if it is expected that the
machine will operate 1,400 hr per year? The company's cost
of capital rate is 15%.
by Melese m.
75

Time Value of Money
 First solve for n, the life;
by Melese m.
76

Time Value of Money
Solution
by Melese m.
77

End of Chapter 2
Time Value of Money and cash flow
Lecture # 2
Thank You!!!
by Melese m.

Evaluating Alternative
by Melese m.

Evaluating Alternative/s
Time Value of Money: Applications
 We will learn how to evaluate the profitability and liquidity of a single
problem solution or alternative.
 Minimum Attractive Rate of Return (MARR) is useful for this analysis.
 MARR ["hurdle rate“] is usually organization-specific and determined based
on the following:
1. Cost of money available for investment
2. Number of good projects available for investment
3. Risks involved in investment opportunities
by Melese m.

Central question: Is a proposed project solution economically
profitable? To do so we require the following data.
 Outflow: capital investment and expenditure
 Inflow: revenue, savings, return on capital, salvage value
 Timing of the cash flows
 Minimum Attractive Rate of Return (MARR)
Technique: Converting the cash flows into their equivalent worth
at some point of time using an interest rate called (MARR)
by Melese m.

How to use MARR?
 Use it as an interest rate to convert cash flows into equivalent
worth at some point in time.
 The proposed problem solution [project or alternative] is
profitable if it generates sufficient cash flow to recover the initial
investment and earn an interest rate that is at least as high as
MARR.
by Melese m.

Why not just use the interest rate? Because there may be other
considerations:
 Cost and amount of money available for investment
 Number of good projects available for investment and their
purpose
 Amount of perceived risk associated with an investment
 Estimated administration cost as determined by the planning
horizon
by Melese m.

MARR Cout’d
 One popular approach for establishing MARR involves the
opportunity cost, which arises when there is capital rationing/
Restriction.
 If you do not invest your money on project/Alternative, your money
continues to earn Another Money.
 We assume that MARR is constant throughout the course of the
project and serves as an interest rate in our considerations.
by Melese m.

Quantitative Methods to evaluate profitability of
Alternatives
1) Net Present Worth [NPW]
2) Incremental Net Present Value [INPV]
3) Future Worth [FW]
4) Annual Worth [AW]
5) Rate of Return [ROR]: Internal Rate of Return [IRR] and
External Rate of Return [ERR]
6) Incremental Rate of Return
7) Payback Period with and without interest by Melese m.

1. Net Present Value or Present Worth[NPV]
 It compares alternatives based on their present values at the time
of the initial investment at the MARR.
1) If NPV is positive, the alternative produces a return greater than
the MARR [Accepted].
2) If NPV is zero, the alternative produces a return equal to the
MARR.
3) If NPV is negative, the alternative produces a return less than the
MARR and, if possible, the investment should be rejected.
by Melese m.

Net Present Value Cout’d
 All cash inflows and outflows are discounted to the present time at
the MARR.
 The present worth of a series of cash inflows and outflows at an
interest rate (or MARR) of i% is given by
 where
 i = effective interest rate, or MARR, per period;
 Fk = cash flow at the end of period k; and
 N = number of periods in the planning horizon.
by Melese m.

Net Present Value Cout’d
 The method has two main assumptions:
The future is known with certainty.
Money can be borrowed and lent at the same interest rate.
by Melese m.

E.G. 1: Your company is looking at purchasing a front-end
loader at a cost of $120,000. The loader would have a useful
life of five years with a salvage value of $12,000 at the end of
the fifth year. The loader can be billed out at $95.00 per hour.
It costs $30.00 per hour to operate the frontend loader and
$25.00 per hour for the operator. Using 1,200 billable hours per
year determine the net present value for the purchase of the
loader using a MARR of 20%. Should your company purchase
the loader? by Melese m.

Solution: The hourly profit [HP] on the loader equals the billing
rate less the operation cost and the cost of the operator.
 HP = $95.00 – [$30.00 + $25.00] = $40.00 per hr
 Annual Profit = $40.00/hr x [1,200 hr/yr] = $48,000/yr
Cash Flow
Diagram
by Melese m.

 The present value of the annual profits [PAP] by using USPWF:
PAP = $48,000[(1+0.20)5-1] / [0.20(1+0.20)5] = $143,549
 The PAP is positive because it is a cash receipt.
 The present value of the salvage value [PSV] by using SPPWF
PSV = $12,000/(1+0.20)5 = $4,823
by Melese m.

 The present value purchase price [PPP] of the loader = purchase
price. Because the net present value is measured at the time of
the initial investment.
 The PPP is negative because it is a cash disbursement.
NPV = PAP + PSV + PPP = $143,549 + $4,823 - 120,000 =
$28,372
by Melese m.

 Because the NPV is greater than zero, the purchase of the front-
end loader will produce a return greater than the MARR and your
company should invest in the front-end loader.
 When comparing two alternatives with positive net present values,
the alternative with the largest net present value produces the
most profit in excess of the MARR.
by Melese m.

E.G. Suppose that there are two projects: A and B.
 Project A requires an investment of $10,000 and will return $12,000 in
one year.
 Project B requires an investment of $100,000 investment and will return
$115,000 in one year.
 Suppose that your MARR is 10%, that the projects are mutually
exclusive, and that you can take up either project (i.e., there are no
budget concerns).
 Question: What do you do? First of all, are the projects economically
justified?
by Melese m.

E.G. Cout’d
 By the PW decision rule, we have
 Thus, both projects are economically justified. However, it is not
clear which is better:
by Melese m.

 To compare the alternatives, remember one of the engineering
economic analysis rule: focus on the difference.
 In our current setting, we take it to mean the following:
 The alternative that requires the minimum investment of capital
and produces satisfactory functional results will be chosen,
unless the incremental capital associated with an alternative
having a larger investment can be justified with respect to its
incremental benefits.
by Melese m.

Continuing the previous example,
 Project B requires an additional $90,000 in investment and
produces an additional return of 103,000 in one year over
Project A. (Why?)
 The present worth of the incremental cash flows at MARR = 10%
is
 Conclusion: The use of the additional $90,000 for Project B is
justified, and hence Project B is preferred over Project A.
by Melese m.

 Example: The evaluation of two mutually exclusive alternatives can
be conveniently carried out by cash flow diagrams.
 Consider the following two projects, both of which will run for 4
years:
 They can be represented by the following cash flow diagrams:
by Melese m.

 Our analysis of the alternatives then reduces to applying your
favorite decision rules to each of the diagrams.
 Continuing the above example, if MARR = 10% and we use
the PW decision rule, then
 The last equation shows that Project B is preferred over Project A.
by Melese m.

ASSIGNMENT ONE
Your company needs to purchase a dump truck and has narrowed the selection
down to two alternatives. The 1st alternative is to purchase a new dump truck
for $65,000. At the end of the seventh year the salvage value of the new
dump truck is estimated to be $15,000. The 2nd alternative is to purchase a
used dump truck for $50,000. At the end of the fourth year the salvage value
of the used dump truck is estimated to be $5,000. The annual profits, revenues
less operation costs, are $17,000 per year for either truck. Using a MARR of
18% and a twenty-eight year study period, calculate the net present value for
each of the dump trucks. Which truck should your company purchase?
by Melese m.

2. Incremental Net Present Value [INPV]
 Step 1: Order the alternatives by increasing initial capital
investment.
 Step 2: Find a base alternative [current best alternative]: Cost
alternatives: the first alternative in the ordered list [the one with
the least capital investment].
 Step 3: Evaluate the difference between the next alternative and
the current best alternative.
by Melese m.

Incremental Net Present Value [INPV] Cout’d
 If the incremental cash flow is positive, choose the next alternative
as the current best alternative.
 Otherwise, keep the current best alternative [i.e. negative] and
drop the next alternative from further consideration.
 Step 4: Repeat Step 3 until the last alternative is considered.
Select the current best alternative as the preferred one.
by Melese m.

E.G. 3: Your company is looking at purchasing a new front-end loader
and has narrowed the choice down to four loaders. The purchase price,
annual profit, and salvage value at the end of five years for each of
the loaders is found in figure below. Which front-end loader should
your company purchase based on the incremental net present values
using a MARR of 20% and a useful life of five years?
Cash Flow Loader A [$] Loader B [$] Loader C [$] Loader D [$]
Purchase Price 110,000 127,000 120,000 130,000
Annual Profit 37,000 43,000 40,000 44,000
Salvage Value 10,000 13,000 12,000 13,000
by Melese m.

Solution:
Step 1: Rank the alternative in order of initial cost [purchase price].
Loader A  Loader C  Loader B  Loader D. Because Loader
A has the lowest initial cost [current best alternative].
Step 2: Compare Loader A to Loader C.
 Difference in purchase price is $10,000 [$120,000 - $110,000].
 Difference in annual profit is $3,000 [$40,000 - $37,000].
 Difference in salvage value is $2,000 [$12,000 - $10,000].
by Melese m.

 The difference in the cash flows for these two alternatives is shown
in Figure below.
by Melese m.

 The present value of the difference in annual profits is determined
by using USPWF as follows:
PAP = $3,000[(1 0.20)5 -1]/ [0.20(1+0.20)5] = $8,972
 The present value of the difference in salvage values is
determined by using SPPWF as follows:
PSV = $2,000[1+0.20]5 = $804
 The incremental net present value for the purchase of Loader C in
place of Loader A is calculated as follows:
INPV = $8,972 + $804 + [ -$10,000] = -$224
by Melese m.

 Because the incremental net present value is negative, Loader A
continues to be the current best alternative.
Next, we compare Loader A to Loader B, the loader with the next
lowest initial cost.
 Difference in purchase price is $17,000 [$127,000 - $110,000].
 Difference in annual profit is $6,000 [$43,000 - $37,000].
 Difference in salvage value is $3,000 [$13,000 - $10,000].
by Melese m.

 The difference in the cash flows for these two alternatives is shown
in Figure below.
 PAP = $6,0003 [(1+0.20)5 -1] / 0.20(1+0.20)5 = $17,944
by Melese m.

 The present value of the difference in salvage values is
determined by using SPPWF as follows:
PSV = $3,000[1+ 0.20]5 = $1,206
 The incremental net present value for the purchase of Loader B in
lieu of Loader A is calculated as follows:
INPV = $17,944 + $1,206 + [-$17,000] = $2,150
by Melese m.

 Because the incremental net present value is positive, Loader B
becomes the new current best alternative and Loader A is
eliminated from comparison.
Next, Compare Loader B to Loader D
 Difference in PP is $3,000 [$130,000 - $127,000].
 Difference in AP is $1,000 [$44,000 - $43,000].
 Difference in SV is zero [$13,000 - $13,000].
by Melese m.

PAP = $1,000 [(1 0.20)5 -1] / [0.20(1+0.20)5] = $2,991
 The incremental net present value for the purchase of Loader B in
lieu of Loader D is calculated as follows:
INPV = $2,991 + $0 + [- $3,000] = - $8
 Because the INPV is negative, Loader B continues to be the current
best alternative. Therefore, your company should purchase
Loader B. by Melese m.

3. Future Value [FW]
 It compares alternatives based on their future values at the end of
the study period.
 If the FW is positive, the alternative produces a return greater
than the MARR.
 If the FW is zero, the alternative produces a return equal to the
MARR.
 If the FW is negative, the alternative produces a return less than
the MARR and, if possible, the investment should be rejected.
by Melese m.

 Find the equivalent worth of all cash flows at the end of the study
period by using the MARR as the interest rate.
 Note that FW and PW of a project are equivalent at the interest
rate of i%, i.e., FW = PW [F/P,i%,N].
by Melese m.

E.G. 4: Consider a project that has the following cash flows over a
study period of 5 years:
 Initial investment: $100,000
 Annual revenues: $40,000
 Annual expenses: $5,000
 Salvage value: $20,000
 MARR: 20%.
Solution:
FW(20%)=-100,000(F/P,20%,5)
+(40,000-5,000)(F/A,20%,5)
+20,000 =$31,624.
 Since FW(20%) > 0, the project
is profitable.
by Melese m.

E.G. 5: Your company is looking at purchasing the front-end
loader at cost of $120,000. The loader would have a useful life
of five years with a salvage value of $12,000 at the end of the
fifth year. The annual profit of loader [revenue less operation
cost] is $48,000. Determine the future worth for the purchase of
the loader using a MARR of 20%. Should your company purchase
the loader?
by Melese m.

Solution: The future value of the purchase price is determined by
using SPCAF as follows:
FPP = $120,000 [1+0.20)5 = - $298,598
 The future value of the purchase price is negative because it is a
cash disbursement.
 The FW of the annual profits is determined by using USCAF as
follows:
FAP = $48,0003 [(1+0.20)5 -1] / 0.20 = $357,197
 The FW of the annual profits is positive because it is a cash
receipt.
by Melese m.

 The future value of the salvage value is equal to the salvage
value because the future value is measured at the end of the
study period. The FW of the salvage value is positive because it is
a cash receipt.
 The future worth for purchasing the loader equals the sum of the
future values of the individual cash flows and is calculated as
follows:
 FW = - $298,598 + $357,197 + $12,000 = $70,599 > MARR
 So, it is attractive for the company to purchase
by Melese m.

4. Annual Equivalent [AE]
 It compares alternatives based on their equivalent annual receipts
less the equivalent annual disbursements.
 The AE is calculated by converting the cash receipts and
disbursements into a uniform series of annual cash flows occurring
over the study period using the equations.
 If the AE is positive, the alternative produces a return greater than
the MARR.
by Melese m.

 If the AE is zero, the alternative produces a return equal to the
MARR.
 If the AE is negative, the alternative produces a return less than
the MARR and, if possible, the investment should be rejected.
 Because any PV can be converted to a uniform series by USSFF.
The AE produces the same result as the net present value.
 Similarly, because any FW can be converted to a uniform series
by USCRF, the AE produces the same result as the future value.
by Melese m.

E.G. 6: Consider a project that has the following cash flows over a
study period of 5 years:
 Initial investment: $100,000
 Annual revenues: $40,000
 Annual expenses: $5,000
 Salvage value: $20,000
 MARR: 20%.
Solution:
AE [20%] = $100,000
[A/P,20%,5] + [$40,000 -
$5,000] + $20,000
[A/Sv,20%,5]
by Melese m.

E.G 7: Your company needs to purchase a dump truck and has
narrowed the selection down to two alternatives. The first alternative is
to purchase a new dump truck for $65,000. At the end of the seventh
year the salvage value of the new dump truck is estimated to be
$15,000. The second alternative is to purchase a used dump truck for
$50,000. At the end of the fourth year the salvage value of the used
dump truck is estimated to be $5,000. The annual profits, revenues less
operation costs, are $17,000 per year for either truck. Using a MARR
of 18% calculate the annual worth for each of the dump trucks. Which
truck should your company purchase?
by Melese m.

Solution: Alternative 1[New]  The useful life of the new truck is
seven years, which is used as the study period for the new truck.
 The purchase price for the new truck is converted to a uniform
series of annual cash flows by USCRF as follows:
APP = - $65,000 [0.18(1+0.18)7] / [(1 0.18)7 -1] = - $17,054
 The salvage value for the new truck is converted to a uniform
series of annual cash flows by USSFF as follows:
ASV = $15,000(0.18) / [ (1+0.18)7 -1] = $1,235
by Melese m.

 The annual profits for the new truck are already a uniform series.
 The annual equivalent for purchasing new loader is;
AE [New] = - $17,054 + $1,235 + $17,000 = $1,181
Alternative 2 [Used]
APP = $50,000[0.18(1+0.18)4] / [(1+0.18)4 1] = - $18,587
ASV = $5,000(0.18) / [(1+0.18)4 -1] = $959
The annual equivalent for purchasing used loader is;
AE = - $18,587 + $959 + $17,000 = - $628
by Melese m.

 The new truck has the highest annual equivalent; therefore, your
company should purchase the new truck.
by Melese m.

5. Rate of Return [ROR]analysis
Knowing the anticipated rate of return of an investment permits
decision maker to have more "perceived" confidence in its decision!
The rate of return of a proposed investment is that interest rate
which makes the discounted present worth of the investment equal
to zero.
To calculate the rate of return, simply set up the equation to be
equal to zero and solve for i.
by Melese m.

Example: A contractor is considering the purchase of either a new
track-type tractor for $73,570, which has a 6-year life with an
estimated net annual income of $26,000, or a used track type
tractor for $24,680, with an estimated life of 3 years and no
salvage value and an estimated net annual income of $12,000. If
the contractor's MARR is 20%, which tractor, if any, should be
chosen?
by Melese m.

Solution:
Approach 1. (comparison on the basis of equal lives)
Old Tractor
New Tractor
by Melese m.

Iterative Solution:
by Melese m.

Iterative Solution
by Melese m.

Decision: If MARR is 20%, then the new tractor is selected
because i = 26.9 % is greater than MARR = 20%
by Melese m.

Example Continued
If we assume the salvage value for the new tractor to be $30,000
after 3 years, the NPW [new] will be:
Before the decision can be reached, YOU MUST KNOW YOUR
MARR. by Melese m.

Decision:
If MARR = 20% and 3 year analysis period, we choose old
tractor.
If MARR = 30%, we choose neither tractor - do nothing
alternative.
If the MARR was 15%, which alternative should we select then?
by Melese m.

Both NPW old and NPW new exceed the MARR = 15%.
But since the old tractor yields a higher MARR, should it not be selected?
To answer this question, determine each alternative's net present worth
at 15%.
NPW old = -24,680 + 12,000(P/A,15,3) = $2,719
NPW new = - 73,570 + 26,000(P/A,15,3) + 30,000(P/F,15,3) =
$5,519
According to the above NPW analysis, the new tractor yields a higher
value for a MARR of 15%?
by Melese m.

Continued
Shouldn't the alternative with the higher rate of return would yield
the higher NPW regardless of the assumed interest rate?
NO IT SHOULD NOT!
The initial investments in the tractor examples we used are not the
same.
by Melese m.

6. Incremental Rate of Return analysis.
When we examined the rate of return of each alternative, we
have ignored their respective differences in initial cash flows.
Therefore, we can obtain misleading results through such an
analysis.
To deal with the problem of unequal initial investments, an
incremental rate of return (IROR) analysis is required.
"For alternatives that have a satisfactory rate of return (ROR), what is
the IROR of the difference in the cash flows of the alternatives?"
by Melese m.

To make this analysis, first arrange the alternatives in ascending
order of initial cash flow.
Then compare alternatives, two by two, alternatively rejecting the
alternative with the lower IROR.
by Melese m.

Example Continued
by Melese m.

Example…
NPW new-old = - 48,890 + 14,000(P/A,i,6) + 24,680(P/F,i,3) +
8,000(P/F,i,6) = 0
i = 30.9%
by Melese m.

While the initial investment of $24,680 for the old tractor will
yield a ROR of 21.5%, the incremental increase in initial
investment of $48,890 (by purchasing the new tractor) will yield
an IROR of 30.9%.
Now that all the rates of return are known, a decision can be
reached which is dependent on the MARR.
For a MARR of 20% the ROR of the new tractor is too low, and
therefore the old tractor is chosen.
by Melese m.

For a MARR of 15% both alternatives exceed it and we have to
examine the IROR.
In this case the IROR is higher than the MARR, so we should choose
the new tractor.
by Melese m.

PAYBACK PERIOD
by Melese m.
 The payback period (PBP) is the time required for an initial
investment to be recovered, neglecting the time value of money.
 Example. Determine the payback period for a proposed investment
as follows:

Cont…
by Melese m.
 The sum of the first three yearly cash inflows, $37 000, is less than
the initial investment, $50 000; but the sum of the first four yearly
cash inflows, $55 000, exceeds the initial investment.
 Hence the payback period will be somewhere between 3 and 4
years. Linear interpolation yields

End of Chapter 3
Evaluating Alternative
Lecture # 3
Thank You!!!
by Melese m.

Depreciation cost
Lecture 4
by Melese m.
2

Depreciation Cost
 Depreciation is the loss/decrease in value of a piece of asset
or property over time,
 generally its caused by wear and tear from use,
deterioration, obsolescence, or reduced need arising from
improvement in design & new technology.
 Depreciation of asset makes them less able to
render/generate the service for which it was originally made.
by Melese m.
3

Depreciation Cost
 Depreciation is a noncash cost.
 Therefore, the depreciation does not a part of actual cash flow
and is only used for tax calculation.
 The object of depreciation of this nature in the engineering
economy study is to distribute the initial cost of the asset for less
salvage value over the period of its life.
by Melese m.
4

Depreciation Cost
 The profitable owner of asset must recover this loss during its
useful life.
 Depreciation accounting is the systematic allocation of the costs
of a capital investment over some specific number of years.
by Melese m.
5

Depreciation Cost
 Depreciation is not allowed for all properties.
 A depreciable asset/property must:
 be used in business or produce income,
 have a determinable useful life,
 lose value over its useful life,
 not be inventory, etc..
 Does land has limited life and lost its value over its
useful life????
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Depreciation Cost
 Classification of properties:
 Tangible (can be seen or touched):- real property (land,
buildings, manufacturing facilities, etc. ) and personal property
(machinery, vehicles, equipment, etc.)
 Intangible (patents, copyrights, trademarks, etc.)
 Almost all tangible and intangible property can be depreciated,
except land (determinable life?, lose in value over time?).
 We will focus on tangible properties.
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Depreciation Cost
 Depreciation Concepts and Terminologies
 Basis or cost basis: The initial cost of acquiring an asset (also
known as unadjusted cost basis). (includes the purchase cost,
delivery and installation fees).
 Adjusted (cost) basis: Adjust the original cost basis by allowable
increases or decreases (e.g., cost of improvement to a capital asset
with useful life greater than a year, casualty, or theft loss).
 This is used for computing depreciation deductions.
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Depreciation Cost
Book value (BV): The value of an asset on the accounting records
of a company after the total amount of depreciation deduction
to date has been subtracted from its adjusted cost basis.
 It represents the amount of capital that remains invested in the
property and must be recovered in the future through the
accounting process.
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Depreciation Cost
Market value (MV): The amount that will be earned if the asset is
sold in an open market.
 MV can be different than BV. For example, IT equipment usually
has a MV much lower than its BV due to rapidly changing
technology.
Useful life: The expected (estimated) period that a property will
be used in a trade or business to produce income.
 It is not how long the property will last but how long the owner
expects to productively use it. by Melese m.
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Depreciation Cost
Salvage value (SV): The estimated value of a property at the
end of its useful life.
Recovery period: The number of years over which the cost basis
of a property is recovered through the accounting process.
 This is usually the useful life of the property.
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Depreciation Cost
 Reasons for calculating the depreciation :
1) To calculated or estimate current market value of the property
[equipment etc], for selling/buying purpose.
2) To know book value of the property at each year of useful life,
3) To know amount of capital at each year of useful life that should
be recovered before the property is out of use.
4) To know the tax amount of property [equipment etc], because the
new property [equipment and old property [equipment does not
render/generate equal profit.
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Depreciation Cost
 Types of depreciation:-
1. Physical depreciation
2. Function depreciation
3. Contingent depreciation
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Depreciation Cost
1. Physical Depreciation:-
 Depreciation resulting in physical impairment/damage of an
asset is known as physical depreciation.
 The result in lowering the ability of the asset to render it intended
service.
 The primary cause of physical depreciation is wear and tear
because of its constant use such as abrasion, shocks, vibration, and
impact etc. and the deterioration due to action of elements such
as corrosion of pipe, chemical decomposition.
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Depreciation Cost
2. Function Depreciation:-
 Functional depreciation often called OBSOLESCENCE is defined
as the loss in the value of the property
Caused by:-
 due to change in fashion, design or structure
 due to inadequacy to meet the growing demand,
 necessity of replacement due to new invention be more
economical and more efficient etc.. by Melese m.
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Depreciation Cost
3. Contingent Depreciation:-
 Accident (due to negligence of risk)
 Economical crisis, or other national problems.
 Reduction of supply (natural gas, electricity, water etc.)
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Depreciation Cost
 Information needed for depreciation accounting:
 The purchase price of the piece of equipment, P
 The optimum period of time to keep the equipment or the recovery
period allowed for income tax purposes, N
 The estimated resale value at the close of the optimum period of
time, F [Salvage Value]
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Depreciation Cost
 Depreciation accounting Method [Most Common]
1. Straight-line [SL] Method
2. Sum-of-the-years [SOY] Method
3. Declining-balance [DB] Method
4. Sinking fund method
5. Units-of-Production method
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Depreciation Cost
1. Straight-line [SL] Method: Easiest to calculate and most
widely used in construction.
 The annual amount of depreciation Dm, for any year m, is a
constant value, and thus the book value BV m decreases at a
uniform rate over the useful life of the equipment.
 Depreciation rate: R m = 1/N
 Annual depreciation amount: Dm = R m [P – F] = [P – F]/N
 Book value at year m: BV m = P – [m*Dm]
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Depreciation Cost
 Note: The value [P – F] is often referred to as the depreciable
value of the investment.
E.G. 1: A piece of equipment is available for purchase for
$12,000, has an estimated useful life of 5 years, and has an
estimated salvage value of $2,000. Determine the depreciation
and the book value for each of the 5 years using the SL method.
R m = 1/5 = 0.2
D m= 0.2[12,000 - 2,000] = $2,000 per year
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Depreciation Cost
 BV2 = $12,000 – 2[2,000] = $8,000
 The table of values is:
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Depreciation Cost
 If the equipment is expected to be used about 1,400 hours per
year then its estimated hourly depreciation portion of the
ownership cost is $2,000/1,400 = $1.428 = $1.43 per hour.
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Depreciation Cost
2. Sum-of-the-years [SOY] Method: SOY is an accelerated
depreciation method [fast write-off], which is a term applied to
accounting methods which permit rates of depreciation faster than
straight line.
 The rate of depreciation is a factor R m [depreciation rate] times
the depreciable value [P – F].
 Dm = R m [P-F]
 SOY = N [N+1] / 2
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Depreciation Cost
 R m = [N-m+1]/SOY
 The annual depreciation Dm for mth year [at any age m] is
 D m = {[N-m+1]/SOY}[P-F]
 The book value at the end of year m is
 BV m = P-[P-F] [m(N-m/2+0.5)/SOY]
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Depreciation Cost
E.G. 2: Using the same values as given in Example 1, calculate the
allowable depreciation and the book value for each of the 5 years
using the SOY method.
 SOY = 1+2+3+4+5 = 15 or = N [N+] / 2 = 5[5+1]/2 = 15
 R m = [5-m+1]/15
 D m = R m [12,000-2,000] = [[5-m+1]*10,000]/15
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Depreciation Cost
E.G. Cont’d: Then tabulate the results as follows:
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Depreciation Cost
Declining-balance [DB] Methods:
 This is also accelerated depreciation methods that provide for
even larger portions of the cost of a piece of equipment to be
written off in the early years.
 DB method often more nearly approximates the actual loss in
market value with time.
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Depreciation Cost
 Declining-balance [DB] Cont’d
 Declining methods range from 1.25 times the current book value
divided by the life to 2.00 times the current book value divided
by the life [the latter is termed double declining balance].
 Note: Although the estimated salvage value F is not included in
the calculation,
 The book value cannot go below the salvage value.
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Depreciation Cost
 The following equations are necessary to use the declining-
balance methods.
 The symbol R is used for the depreciation rate for the declining-
balance method of depreciation:
1. For 1.25 declining-balance [1.25DB] method, R = 1.25/N
For 1.50 declining-balance [1.5DB] method, R = 1.50/N
For 1.75 declining-balance [1.75DB] method, R = 1.75/N
For double-declining-balance [DDB] method, R = 2.00/N
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Depreciation Cost
2. The allowable depreciation Dm, for any year m and any
depreciation rate R is
 Dm = R P[1 – R]m-1 or Dm = [BVm-1]*R
3. The book value for any year m is
 BV m = P[1-R]m or BV m = BV m-1- D m provided that BV m > F
 Since [BV] can never go below [F], the declining balance method
must be forced to intersect the value [F] at time [N].
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Depreciation Cost
 E.G. 3: A piece of equipment is available for purchase for
[$12000], has an estimated useful life of [5 years], and an
estimated salvage value of [$2000]. Determine the depreciation
and the book value for each of the 5 years using the DDB
method.
 Solution:
 Calculate R: 2/N = 2/5 = 0.4
 Calculate Dm: Dm = 0.4 [BVm-1]
 Calculate BVm: BVm = BVm-1 - Dm
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Depreciation Cost
 Then tabulate the results as follows:
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Depreciation Cost
 Depreciation Curves for the above Examples
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Depreciation Cost
4. Sinking Fund Method
 Since the calculation of book value at any year may carried out
by using discounting principle i.e. time value of money.
 The amount of annual deposit is so calculated that the
accumulated sum at the end of estimated life of the asset and the
started interest rate will just equal the value of the asset which is
being appreciated.
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Depreciation Cost
This value as before will be either the initial cost of the asset or if
the asset has a salvage value the initial cost will less the salvage
value.
The amount of asset written out any one year is uniform payment
+ the interest charge for that year on the amount already
accumulated in the sinking fund.
uniform series of payments=A = (P-F)[i/(1+i)n- 1]
Depreciation at n year = A(1+i)n-1
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Depreciation Cost
 Assume i is 10% for the above example.
 DM1= 12000- 1638 = 10362
 DM2 = 10362- 1638(1+0.1)
 DM3 = 8560- 1638(1+0.1)2
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Year Depreciation Book value
0 0 12000
1 1638 10362
2 1802 8560
3 1982 6578
4 2180 4398
5 2398 2000

Depreciation Cost
5. Units-of-Production method:
 Depreciation is based on activity (number of units produced)
rather than time.
 Depreciation per unit of production= (B-SVN)/Estimated lifetime
production units.
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Depreciation Cost
 Example: An equipment has a cost basis of $50,000 and a
salvage value of $10,000 after 30,000 hours of use.
A. What is the depreciation rate per hour of use?
B. What is the BV of the equipment after 10,000 hours of
operation?
Solution
A. ($50,000-$10,000)/30,000=$1.33 per hour.
B. $50,000-10,000(1.33)=$36,700. by Melese m.
38

End of Chapter 4
Depreciation Cost
Lecture # 4
Thank You!!!
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Sensitivity Analysis, Break Even Analysis
and benefit cost ratio
Lecture 5
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Sensitivity Analysis
 In all our previous discussions we had arrived at a
particular decision (in the form of acceptance or rejection
of a proposal, and
 Assuming that variables such as initial cost, receipts,
disbursements, interest rate, life of the assets, salvage value
etc were accurate and constant.
 Unfortunately in real life situation it is not the case.
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 Barring few variable, such as initial cost of the asset, rest all the
variables are all our estimates or forecasts which may prove to be
wrong on most of the occasions.
 The life of the asset could be longer or shorter than our estimate;
the interest rate could be higher or lower than the assumed value
and so on.
 The salvage value may be more or less than the assumed value.
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 We may like to know what will happen to the net present worth associated
with a particular investment alternative when some variables like incomings
(receipts) value or outgoings (disbursement) value vary from its expected
value.
 Sensitivity analysis thus is aimed to study the impact of change in the value
of variables on the economic decision in a particular situation.
 In a sense it aims to answer “what if”. For example what will happen if the
annual disbursement value increases by 10% or 20% from the current
value? Will it turn the positive present worth into negative? Will it change
the earlier decision? by Melese m.
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 The changes (increase or decrease in the assumed values) in the
variable values may or may not lead to reversal of our earlier
decision.
 If even a slight change in one variable makes the reversal of
decision from let’s say acceptance of one alternative to the rejection
we say that the variable is highly sensitive.
 Whereas, even if a large change in one variable does not change
the decision we say that the variable is not sensitive or insensitive.
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 The sensitivity analysis is also aimed at identifying the sensitivity of
a particular variable.
 Once the identification has been made of variables in categories
such as highly sensitive, less sensitive or insensitive the management
can focus their attention to the highly sensitive variables.
 That is for such variables they can put more energy and effort in
preparing their estimate.
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by Melese m.
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Sensitivity analysis
Single alternative More than one alternative
Change in one variable at a
time
Change in two variables at
a time
Change in more than two
variables at a time
Change in one variable
at a time
Change in more than two
variables at a time
Change in two variables
at a time
Spider web diagram
Family of curves or
Isoquants
Scenario analysis

 In its simplest form, sensitivity analysis for a single
alternative can be performed and the impact (on decision)
of change in single variable can be studied.
 Then there could be simultaneous changes in two variables
corresponding to a single alternative can be studied.
 Finally for a single alternative we can study the changes in
more than two variables at a time.
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 Sensitivity analysis can be performed for more than one
alternative.
 Here also we can see the impact of variation of one variable
on the decision.
 There can be cases in which we would like to change two
variables at a time or more than two variables at a time and
see their impact on the decision arrived earlier by assuming all
the variables as fixed or constant. by Melese m.
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 Sensitivity analysis can be performed with any method of evaluation
of alternatives for example, present worth analysis, annual cost or
Future worth analysis or internal rate of return method of analysis.
 Also the analysis can be performed at different stages of project
either with the pre tax cash flow or post tax cash flows.
 However, it is preferable to perform sensitivity analysis with post
tax cash flow.
 It is customary to show the results of sensitivity analysis in the form
of sensitivity graphs. by Melese m.
11

Analysis for one alternative
 Example. Consider the following alternatives
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 Net present worth = -500,000 + 100,000(P/A, 12%, 10) – 5,000(P/A,
12%, 10) + 50,000(P/F, 12%, 10) = 52869.4
 Let’s assume that the estimate of incoming value goes wrong and instead
of Birr. 100,000 it is Birr. 90,000.
 Thus the new net present worth keeping all other variables same would be
(-500,000 + 90,000(P/A, 12%, 10) – 5,000(P/A, 12%, 10) +
50,000(P/F, 12%, 10) = -3632.8
 We find that the decision taken on net present worth would get reversed.
 The value of incoming at which net present worth is zero is known as break
even point
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by Melese m.
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-10000
0
10000
20000
30000
40000
50000
60000
90,000 95,000 100,000
Incomings Rs.
Net
Present
Worth
Rs.
Sensitivity graph showing the effect of changes in incomings on net
present worth

 Similar analysis can be performed by changing other variables one
at a time.
 The slope of the sensitivity line indicates the sensitivity of a
particular variable.
 The steeper the slope, the more sensitive the variable is and
 The milder the slope, the less sensitive the variable is.
 Next we take the variables: incomings and service life
together and study their sensitivity.
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Analysis for more than one alternative
 Illustration
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Solution.
 The net present worth of alternative 1 =
-500,000+100,000 (P/A,12%,10)
-5,000(P/A,12%,10)+50,000(P/F,12%,10)
= -500,000+100,000*5.6502-5000*5.6502+50,000*0.3220
=52,869
 The net present worth of alternative 2 =
-500,000+90,000 (P/A,12%,10)
-5,000(P/A,12%,10)+50,000(P/F,12%,10)
= -500,000+90,000*5.6502-5000*5.6502+50,000*0.3220
=8,394
by Melese m.
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 Now let’s change each of these variables one by one. For example,
consider the changes in the variable ‘incoming’.
 In case the ‘incoming’ of alternative 1 changes to 90,000 from the existing
100,000 the new net present worth of alternative 1 changes to - 3663
 We can find that if the incoming value reduces to less than 90,642.9 the
decision is reversed.
 Such analysis addresses the questions such as: at what value of incomings
the alternative 1 is preferred to alternative 2?
 At what service life of the assets, the alternative 1 is preferred to
alternative 2?
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Changes in more than two variables at a time
 Conducting scenario analysis is the best approach when performing
sensitivity analysis involving changes in more than two variables at a time.
 In scenario analysis a number of scenarios such as best scenario,
objective/normal scenario, and worst scenario.
 The objective of such scenario analysis is to get a feel of what happens
under the most favorable or the most adverse configuration of key
variables.
 For example the best, normal and the worst scenario for the previous
example could be as given below:
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
by Melese m.
20
ILLUSTRATION

 Corresponding to each of the above scenarios, the net present worth
can be computed.
 Based on the net present worth values for each of the scenarios the
decision maker would be in a better position to take the decision.
 For example, if the net present worth corresponding to worst scenario is
a large negative value, and the net present worth corresponding to the
objective and best scenarios are low positive value, and moderate
positive value the decision would be not to acquire the asset.
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Benefits of performing sensitivity analysis
 It shows how robust or vulnerable a particular alternative
is to changes in the value of different variables,
 It enables the decision maker to distinguish the sensitive
variables from insensitive variables thus the decision maker
can focus its attention in making the estimate of sensitive
variables.
by Melese m.
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Break Even Analysis
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 Another way of performing sensitivity analysis
 Here we are more concerned about finding the value (called
the break even point) at which the reversal of decision takes
place.
 In the sensitivity analysis not much emphasis was given on
finding this break even value.
 In sensitivity analysis we ask what will happen to the project
if the invoice or billing declines or costs increase or something
else happens.

Break Even Analysis
by Melese m.
24
 We will also be interested in knowing how much should be produced
and sold at a minimum to ensure that the project does not 'lose
money'.
 Such an exercise is called break-even analysis also referred to as cost-
volume-profit analysis.
 The minimum quantity at which loss is avoided is called the break-
even point.
 It addresses the decision of whether:-
 To make or buy decision situation,
 To lease or purchase decision and so on

Break Even Analysis
by Melese m.
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 Making the product: involves two cost elements:
 Fixed costs such as machine renting cost and operation
expenses
 Variable costs such as raw material cost
 Buying the product: involves only one cost element, the selling
price.
 However, the price may either be constant or variable based on
the quantity.

Break Even Analysis
by Melese m.
26
EXAMPLE:
 A ready mix concrete (RMC) manufacturer wants to find out
the minimum production of concrete which will just be able to
recover its total cost incurred in a particular month.
 The total cost (TC) incurred in a month is the sum total of its
indirect cost (IC) and direct cost (DC).
 The indirect costs in this example are those costs which are
incurred irrespective of concrete production taking place or
not.
 However, the direct costs are proportional to the volume or
quantity of production.
 By definition the total cost TC = IC + DC

Break Even Analysis
by Melese m.
27
 Let the sales price fixed by the RMC supplier is P per unit concrete sold.
 If the quantity of concrete sold is n units,
 Revenue (R) = n x P.
 TC = IC + n x DC
 Gross profit Z for the period would be defined as:
 Z = R – TC = n x P – IC – n x DC
 = n x (P-DC) – IC.
 The net profit after taking taxes into account is given by:-
 Z’= Z x (1-t) where t is the tax rate

Break Even Analysis
by Melese m.
28
 In order to determine the concrete quantity n at which the RMC seller just
recovers its total cost we equate total cost to revenue.
 Thus at break even point/ point of intersection of the total cost and
revenue lines.
 TC = R and Profit Z = 0
 n x DC + IC = n x P
 The quantity produced at break even point is denoted with B, thus n = B at
break even.)
 B x DC + IC = B x P
B = IC/ (P-DC)

Break Even Analysis
by Melese m.
29
 The denominator (P-DC) in the expression is also known as
‘contribution’.
 For n less than B, the RMC seller is making losses,
 while for n greater than B, the RMC seller is making profits.
 The ordinate corresponding to this intersection point gives the
break even quantity of concrete to be produced in order to just
recover the total cost incurred by the RMC seller

Linear Break Even Analysis
by Melese m.
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Example.
 A Ready Mix Concrete company sells RMC for a price of Birr.
3500/ cum. If the fixed cost of the company for the production are
Birr. 40,000/month and the variable cost associated with per cum
production of RMC is Birr. 1500/ cum, determine the breakeven
quantity

by Melese m.
31

by Melese m.
32
 The companies try to lower the break even point by resorting to
different means such as
(1) increasing the sales price,
(2) reducing the total cost of production,
(3) increasing the quantity of production.
 These measures are adopted either in isolation or in combination

by Melese m.
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Assumptions of linear break even analysis
by Melese m.
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 In the previous example it was assumed that the ready mix
concrete manufacturing company is into concrete production only.
 Whatever quantity is produced is sold out.
 The per unit direct cost, indirect cost, and sales price associated
with the production are constant over the study period.
 These are also constant over the quantity produced.

Cont…
by Melese m.
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Example.
 DC = $7/ UNIT
 Sales price P= $12/unit
 Indirect cost = $400
 Break even point = $400/ (12- 7) = 80
 The break even point can be lowered to 40 by
 Reducing the indirect cost to $200
 Reducing the direct cost to $2
 Increasing the sales price to $17

How To Maximize Profit
by Melese m.
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 Using the concept of marginal cost, marginal revenue and the
principle of calculus, one can determine the production level at
which the firm would be able to maximize its profit.
Dumping
 In order to utilize the full capacity some products are sold at price P
while the remaining products are sold at lesser price P’.
 Gross Profit Z = n(P-DC) + n’ (P’-DC) –IC
 The Dumping may lead to some problems to the company practicing it.
 Countries prevent such practices by levying heavy duty on imports.

Cont…
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Marginal cost
 Marginal cost is the additional cost incurred by the company to
produce one extra unit of product.
 For example, suppose that the cost to produce 100 cum of
concrete is Birr. 200,000 and the cost to produce 101 cum of
concrete is Birr. 201,000,
 The marginal cost would be equal to (201,000- 200,000)/ (101-
100)= Birr. 1000.

Cont…
by Melese m.
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Marginal revenue
 Marginal revenue is the additional money realized by selling one
extra unit of product.
 For example if the revenue raised by selling 100 cum of concrete is
Birr. 250,000 and by selling 101 cum of concrete the revenue is
252,000,
 The marginal revenue is (252,000-250,000)/(101-100)= Birr.
2,000.

Benefit-Cost Analysis
 It is a systematic process for calculating and comparing
benefits and cost of a project, decision or policy.
 Present and future benefits (income) and costs need to be
estimated to determine the attractiveness (worthiness) of a
new product investment alternative
 It has two purposes:
◦ To determine if it is a sound investment/decision
◦ To provide basis for comparing projects
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Benefit –Cost Analysis
Benefit Cost Analysis
 The B/C ratio is calculated using one of these relations:
B/C=PW of benefits/PW of costs
B/C=AW of benefits/AW of costs
B/C=FW of benefits/FW of benefits
 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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Conventional B/C Ratio
 It is the most widely use B/C ratio in public and private sector.
Where B = benefits of proposed project
I = initial investment in the proposed project
MV = market value at the end of useful life/Salvage Value
O&M = operating and maintenance costs of the
proposed project by Melese m.
41

Modified B/C ratio
 In this method the maintenance and operation (M&O)is in the
numerator and treated as dis benefits
 Salvage value is included in the denominator as a negative
cost by Melese m.
42

Example:
 Analysis of several mutually exclusive road way alignments yield the
following information. (i=7%)
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Incremental analysis procedure based on the BC ratio
method
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Step one. Rank the alternatives in increasing order of total equivalent worth.
The do{nothing alternative is the initial baseline alternative.
Step two. Compute the deference in benefits (ΔB) and costs (ΔC) between
the next least equivalent worth alternative and the baseline alternative, and
calculate the incremental BC ratio ΔB/ΔC.
Step three. If the ratio is at least 1, then the higher equivalent worth
alternative becomes the new baseline. Otherwise, the last baseline
alternative is maintained.
Step four Repeat Step three until all the alternatives have been considered.

Cont…
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Example: Consider the following three projects, market value is at the
end of useful life.

Cont…
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 The present worth of the cost of the projects are.

Cont…
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 The present worth of the benefit of the projects are.

Cont…
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 Project X requires the least capital investment, and its B-
C ratio is
 which is acceptable.
 The incremental project Δ(Y- X) yields a BC ratio of
 Un acceptable.

Cont…
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 The incremental project Δ(Z - X) yields a BC ratio of
Therefore, Project Z is selected as the final result.

End of Chapter 5
Sensitivity Analysis, break-even Analysis
and benefit cost ratio
Lecture #5
Thank You!!!
by Melese m.
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Inflation and Taxation
Lecture 6
by Melese m.
2

Inflation
 Inflation: is a general increase in price level or it a loss in the
purchasing power of money over time.
 Deflation: a general decrease in price level over time or it is an
increase in purchasing power of money over time.
 The prices are likely to change over the lives of most engineering
projects.
 These changes will affect cash flows associated with the projects
 The inflation rate:- is the rate of increase in average prices of
goods and services over a specified time period, usually a year
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What cause Inflation
 When the demand for goods and services surpasses the economy’s
capacity to produce them (ie excess demand over supply) then demand
will pull up prices.
 Excess demand over supply can arise at times of rising general income
levels.
 Increases in the cost of factor inputs eg raw materials, wages etc, when
passed on to the consumer will push up prices.
 A growth in the money supply eg through excessive lending by the
financial sector will reduce the value of the currency circulating round the
economy.
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Measuring the Inflation Rate
 Consumer Price Index (CPI) :- is a measure based on a typical
market basket of goods and services required by the average
consumer: such as food, housing, clothing, transportation, & etc..
 The CPI compares the cost of the typical market basket of goods
and services in a current month with its cost at a previous time, such
as 1 month ago, 1 year ago, or 10 years ago , This reference
period is called as the base period
 Note that ; the CPI does not take into account the over all sort of
consumer behavior. by Melese m.
5

Cont…
Producer Price Index (PPI):- is a measure of general price of producer of
good.
CPI is a good measure of the general price increase of consumer
products. However, it is not a good measure of industrial price
increases.
When performing engineering economic analysis, the appropriate
price indices must be selected to accurately estimate the price increases
of raw materials, finished products, and operating costs.
by Melese m.
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Inflation
Average inflation rate (f):
 To account for the effect of varying yearly inflation rates over a
period of several years, we have to compute a single rate that
represents an average inflation rate.
 Since each year's inflation rate is based on the previous year's
rate, these rates have a compounding effect.
Example: suppose we want to calculate the average inflation rate for a
two-year period. The first year's inflation rate is 4%, and the second
year's rate is 8%, with a base price of $100, Then calculate the average
inflation rate for the two years ?
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Inflation
Step 1: To find the price at the end of the second year, we use the
process of compounding:
price at the end of the second year =100(1+ 0.04) (1+ 0.08) = 112.32
Step 2: To find the average inflation rate f, we
establish the following equivalence equation: 100(1+ f)2
= 112.32
Solving for f yields f = 5.98%
 Thus we can say that the price increases in the last two years are
equivalent to an average annual percentage rate of 5.98% per
year.
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General Vs specific inflation rate
 General inflation rate : is an average inflation rate that is
calculated based on the CPI for all items in the market basket.
 When we use the CPI as a base to determine the average
inflation rate, we obtain the general inflation rate.
 Specific inflation rate(fj): This rate is based on an index (or the
CPI) specific to segment j of the economy.
 The price changes of specific, individual commodities do not always
reflect the general inflation rate.
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Actual (current) money Vs Constant (real) currency
 Actual (current) dollar (An ): is the amount of dollars that will be
paid or received irrespective of how much these dollars are worth.
 It is a Dollar amounts that reflect the inflation or deflation rate.
 Actual dollars are estimates of future cash flows for year n that
take into account any anticipated changes in amount caused by
inflationary or deflationary effects.
 Usually, these amounts are determined by applying an inflation rate
to base-year dollar estimates.
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Cont…
 Constant (Real) dollars: are a measure of worth not an indicator
of the number of dollars paid or received.
 It is a Dollar amounts that reflect the purchasing power of dollars.
 It reflect constant purchasing power independent of the passage of
time.
 In situations where inflationary effects were assumed when cash
flows were estimated, these estimates can be converted to constant (or
real) dollars (base-year dollars) by adjustment.
by Melese m.
11

Inflation
 Thus, in calculating any cash flow equivalence, we need to identify the
nature of the cash flows.
 All cash flow elements are estimated either in constant dollars or in actual
dollars or Some of the cash flow elements are estimated in constant
dollars while others are estimated in actual dollars.
 In the later case, we simply convert all cash flow elements into one type-
either constant or actual dollars.
 Then we proceed with either constant-dollar analysis or actual-dollar
analysis. by Melese m.
12

Inflation
Thus, if the present cost of a commodity is PC, its future cost, FC will be
 where f is annual inflation rate; n is number of years.
Example : An economy is experiencing inflation at the rate of 6% per
year. An item presently costs birr 100. If the 6% inflation rate
continues, what will be the price(cost) of this item in five years?
 Using the formula, FC = 100(1+ 0.06)5
= 133.82.
by Melese m.
13
n
)
l
(
PC
C
F f



Inflation
 Buying power of money decreases as costs increase.
 It shows Buying/purchasing power of money after n years.
Example : An economy is experiencing inflation at an annual rate of
6%. If this continues, what will birr 100 worth five years from now, in
terms of today's birr ?
= 74.73
 100 birr of today, after 5 year buy a product cost 74.73 today.
by Melese m.
14
n
n
)
l
(
F
)
l
(
1
FC
PC
f
P
or
f 



n
)
l
(
F
f
P

 5
)
06
.
0
l
(
100



Inflation
 If interest is being compounded at the same time that inflation is
occurring, then,
 future worth can be determined by combining equations of
compound formula & inflation factor.
 Or defining the composite interest rate,
by Melese m.
15
 
 
n
n
n
f
i
P
f
i
P














l
l
1
l
F
 n
P
F
have
we
f
f

 



 1
l
i

Inflation
Example:
 An engineer has received birr 10,000 from his employer for a
patent disclosure.
 He has decided to invest the money in a 15-year savings certificate
that pays 8% per year, compounded annually.
 What will be the final value of his investment, in terms of today's
birr, if inflation continues at the rate of 6% per year?
by Melese m.
16

Inflation
Solution: A composite interest rate can be determined from the above
equation as:
Substituting this value ,we obtain
by Melese m.
17
0189
.
0
.06
0
l
0.06
0.08





    61
.
13242
0189
.
0
1
000
,
10
1
15





n
P
F 

Strategies For Tackling Inflation
 Dampen down the level of demand by making borrowing and
credit more expensive.
 Encourage increased capacity by encouraging business to expand
and new business to set up eg Enterprise Allowances, Grants etc.
 Ensure wage rises are accompanied by productivity agreements
 Keep tight control of the money supply through reduced lending,
increasing reserve asset ratio and higher interest rates.
by Melese m.
18

Taxation
 A tax is a compulsory charge or payment imposed by government on
individuals or corporations, implemented by law and failure to pay taxes
is punishable by law or.
 Tax is a financial, mandatory, non-equivalent, non-specific charge or other
levy imposed on a taxpayer by a state and the rate of taxes may vary.
 The persons who are taxed have to pay the taxes irrespective of any
corresponding return from the goods or services by the government.
 The taxes can be paid regularly or occasionally based on certain
conditions stipulated by the tax legislation.
by Melese m.
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Taxation
Benefit of tax
 Taxes are a means to redistribute wealth
 Shift $$$ from wealthier to poorer communities
 Promote equity of educational opportunities
 Spread financial burden over as large a group as possible
 Prevent poverty, & civil unrest
 Raising Revenue
 To cover any government expenses for different public projects.
by Melese m.
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Taxation
Benefit of tax cont.…
 Regulation of Consumption and Production
 Encouraging Domestic Industries
 Stimulating Investment
 Reducing Income Inequalities
 Promoting Economic Growth
 Development of Backward Regions
 Ensuring Price Stability
by Melese m.
21

Taxation
Types of Taxes:-
 Corporate Income Tax
 Personal Tax
 Property Tax
 Value –Added Tax
 Custom Duties Tax
 Payroll Tax
 Excise Tax
 With Holding Tax
 Turn-over Tax
 Export Taxes
 Stamp duty tax and etc… by Melese m.
22

Engineering Economics for Civil Engineering-full Lecture Note, By Melese Mengistu

Engineering Economics for Civil Engineering-full Lecture Note, By Melese Mengistu

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