CoTM-M 5271, Development and Construction Economics Lecture Note

Department of Construction Technology and Management

Name of instructor: Haftom B. Academic year: VI Semester: I Program: Extension

Course title: Development and Construction Economics course Code:

COTM 5271

Pre-requisite: Introduction to Economics Credit hour: 03

Course objective: Students will able to: Ensure the feasibility of infrastructure investment with
regard to the basic planning principle called front End Assessment – of science together with
their economic evaluation.

Course content

1. Development and Construction Economics

1.1 Introduction to construction economics 1.3 basic economics terms
1.2 Why Construction Economics? 1.4 Cash flow diagram
2 Basic Economics Principles
2.1 Time Value of Money 2.3 Equivalence Calculations

2.2 Methods of calculating Interest 2.4 Nominal and Effective Interest Rate

3 Investment Appraisal Methods

3.1 Decision Support Methods and Techniques 3.6 Internal Rate of Return Method
3.2 Payback period 3.7 Cost benefit analysis
3.3 Present worth or Net Present Value Method
3.4 ) Equivalence Future worth
3.5 Annual Cost /Income Method (Equivalent Cost / Income Method)
4 Depreciation
4.1 definition 4.3 Depreciation Methods
4.2 Classification of depreciation
5 Inflation
5.1 Definition 5.2 Cause of Inflation
Attendance Requirements will have Minimum of 90% during lectures
and the Evaluation system will be:
Assignments………………………..….10%
TEST 1…………………………………15%
TEST 2…………………………………15%

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CoTM-M 5271, Development and Construction Economics Lecture Note

Project …………………………………20%
FINAL EXAM…………………………40%

TOTAL……………………………….. 100%

Reference

Economic and financial analysis for engineering and project management

[Chan s. park], fundamentals of engineering economics

[Spon press], construction economics

CHAPTER ONE
INTRODUCTION TO DEVELOPMENT AND CONSTRUCTION ECONOMICS

1.1 Introduction to construction economics

 The Cost, Time, Quality and Safety of constructions should no longer be left to chances and
mere guesses or risks;
 Investments should not be left to chances of successes;
 Economics and Finance are not any longer limited to the disciplinarians;
 Construction time and cost becomes essential constraints; etc
Construction Economics is a subject dealing with the concepts, skills, methods and techniques
of economics in order to support decisions made by investors, implementing agents, etc.;
through indicating the feasibility and priority of their engagements in construction related
businesses.

1.2 Why Construction Economics is important?

Construction Economics important because:
 Investments need to be appraised and shown viable;
 Demands need to be studied and try to much with supplies;
 Alternatives need to be compared to reach into optimal choices;
 Projects need to be prioritized due to scarcity of resources / inputs and increasing and dire
demands;

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 Accountability of using such scarce resources demands proof of utmost ethical

considerations; and
 Decision of investments based on the above requirements called for the importance of
Construction economics.
1.2 Introduction to Economics terms

Principal (P): – Initial amount of money borrowed or invested.

Interest rate (i): – cost or price of money for a given time

Interest period (n):– frequency of interest calculated or it’s number of interest period

Future amount of money (F):- results from the cumulative effects of the interest rate over a
number of interest periods.
Plan for receipts or disbursements (A):- that yields a particular cash flow pattern over a specified
length of time. (For example, we might have a series of equal monthly payments that repay a
loan.)
Payback period: - Number of years to recover the initial investment and a stated rate of return
Supply:-is the amount of products (services, goods and works) that producers / suppliers are
willing or making available for sell in the market.

Demand:-is the amount of products (services, goods and works) that buyers are willing or able to
buy from the market.

1.3 CASH FLOW DIAGRAM

To financial analyze engineering project, we need the project in terms of cash flow. Cash flow
represents the movement of money at some specific time over some period time. Outflows
represent cash that is leaving on account such as withdrawal (expenses or disbursements) and
inflow represent cash that is entering in an account such as a deposit (revenue and receipts).

The following conventions are used in the construction of the cash flow diagram:

* The horizontal axis represents time

* The vertical axis represents inflow and outflow
* Outflow are shown by downward arrows

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* Inflows are shown by upward arrows

Inflow

CFD time

Outflow

CHAPTER TWO

Basic Economics Principles

2.1 Time Value of Money

Money can have different values at different times. This is because money can be used to earn
more money between the different instances of time. Obviously, 10,000 birr now is worth more
than 10,000 birr a year from now even if there is no inflation. This is because it can earn money
during the interval. One could deposit the money in the bank and earn interest on it. This is the
earning power of money over time and is called time value of money.
Note we have to be careful not to confuse the earning power of money, which is related to
interest rate, with the buying power of money, which is related to inflation. Inflation will be
discussed in the last chapter.
2.2 Methods of calculating Interest

Interest:-is what you earn when you let people borrow your money .Some call it the price of
renting your money .Interest can be thought of as the price a lender charges a borrower for the
use of his money .A borrower pays interest charges for the opportunity to do something now that
otherwise would have to be delayed or would never be done.
Interest is the difference between an ending amount of money and the beginning amount
 There are two types of interest:

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Interest paid: - when a person borrows money and repays a larger amount
Interest revenue: - when a person saved, invested, or lent money and obtains a return of a larger
amount
Numerical values are the same for both yet they are different in interpretation
Interest = total amount now – original amount
Where the interest is paid over a specific time unit
•If the interest is expressed as a percentage of the original amount then it is called the interest
rate and expressed as in the following:

•Interest rate (%) = Interest accrued x 100%

Original amount

[Example-1]
•An employee borrows 10,000 birr on May 1 and must repay a total of 10,700 birr exactly 1 year
later. Determine the interest amount and interest rate paid.
Solution
•Interest paid = 10,700 - 10,000 = 700 birr

•Interest rate = 700/10,000 = 7% per year

[Example-2]
•A company plans to borrow 20,000 birr from a bank for 1 year at 9% interest for a new
recording equipment. Compute the interest and the total amount due after 1 year
Solution
Interest = 20,000 x 0.09 = 1,800 Birr
Total amount due = 20,000 + 1,800 = 21,800 Birr

Simple and Compound Interest

In the previous examples, the interest period was 1 year and the interest amount was calculated at
the end of one period also the interest is compounded yearly. When more than one interest period
is involved (e.g. after 3 years), it is necessary to state whether the interest is accrued on a simple
or compound basis from one period to the next.

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•In simple interest: - Interest calculated on the principal only.not recommended for construction
related businesses

If you borrow P birr at yearly interest rate i, at the end of the year, the interest is Pi, and the
total amount you have to pay back to the lender is P + Pi.
There fore
F1= P+Pi =P(1+i)
F2=F1+Pi =P+Pi+Pi =P (1+2i)
F3=F2+Pi=P(1+3i)……
F = p +( ip ) n= P ( 1+ in )

•Compound interest:-interest is calculated on the principal plus the total amount of interest
accumulated in all previous periods.

F = p (1+i )= p+ pi
1
2

F = p (1+i )+ p ( 1+i )i= p (1+i )(1+i )= p (1+i )

2
22

F = p (1+i ) + p( 1+i ) i
3
2 3

= p (1+i ) ( 1+i )= p ( 1+i )

n

F = p ( 1+i )
n

[Example-3]
•If an engineer borrows 1,000 birr at 5% per year for 3 years, (a) How much would you pay at
the end of year three with simple interest? (b) How much would you pay at the end of year three
with compound interest?
Solution

P=1000birr i=5% n=3years

(a) F=P(1+in) (b) F=P(1+i)n

=1000(1+0.05*3) =1000(1+0.05)3
=1,150birr =1,157.63 birr

2.3 Equivalence Calculations

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Equivalence calculations are usually made to compare alternatives. Comparisons between widely
differing series of the cash flows can thus be made at a given interest rate. When sum of money
are described as being equivalent, it is not intended to mean that they are equal, but that taking
time value of money in to account and give risk free condition (that is, that there is a certainty
that nothing will go wrong), there will no reason to prefer one particular sum to different but
equivalent one at other point in time. The importance of the concept of equivalence cannot be
overemphasized in engineering economics, since this is the basis upon which comparisons of
expenditure or receipts over a period of years, or for that matter indefinitely can be compared.
[Example -4]

You are offered either 10,000 birr now or 25,937.425 birr after n yrs. i=10%

a) which offer do you accept if n=10, n=8, n=15

n 10

F= p ( 1+i ) =10 , 000 (1+0 . 1) =25937 . 425

If n=10 years, they are both equivalent

If n = 8, 15 but i remains the same

F = 10,000 (1.1)8 =21435.88

I would take 25937.42 after 8 yrs.

F= 10,000(1.1)15 = 41,772.48, I would take now

b) You have seen that 25937.425 is equivalent of 10000 now after 10 yrs if i=10%
Are these cash flows also equivalent at the end of year 5?

5 5

F = p (1+i) =10 , 000(1. 1) =16105 .1

51 1
−5

F =25937 . 425(1+i) =(1.

52
25937 . 425
1)
=16105. 1 5

This shows that the two cash flows will be equivalent provided that i remain to be 10%.

c) If i=12% are these cash flows still equivalent?

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10

F =10 , 000( 1+0 . 12 ) =31058 . 48>25937 . 425

This shows that the change in i has made the two cash flows no more equivalent.

Generally

Equivalence as a basis for determining time value of money

 Equivalence does not mean money today and money in the future are equal
 Equivalence do not consider risks but can be made to consider if quantifiable and could
be framed / estimated time wise
 Equivalence can make different scenarios comparable at different times
 Non – equivalency indicates an initiative for choice depending on which provides better
value for money
 For construction related business often today’s money is much more important;
specifically in our country context

Types of cash flow

In order to develop interest formulas, we need to categorize cash flows. These categories are:

1) Single Cash Flow

2) Equal (uniform) series
3) Linear Gradient
4) Geometric gradient
5) Irregular Series
1) Single Cash Flow: - deals with only two amounts; a single present amount P with its future
amount.
a) Future Sum –This is the value that certain amount P, now, will be equivalent after n periods
and i.
I, using calculator
F = P (1+i) n

(1+i) n = Single payment compound Amount factor (SPCAF)

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II, Using compound interest table

F=P ( F / P ,i ,n )
III, using computer (Excel)
=FV (i%, n, 0,- PV)

b) Present value (worth) – this is the reverse of (a)

I, using calculator
F
P= n

(1 +i )
1 = SPPWF
(1 + i )
n

II, Using compound interest table

P= F ( P / F , i , n )
III, using computer (Excel)
=PV (i%, n, 0, -FV)

c) To calculate i, knowing the other 3, requires trial & error. But one can calculate it using excel
through

=RATE (n, o,-P, F)

d) To calculate n one uses,

=NPER (i, o,-P, F)

Or using the following formula

F= P (1+i) n

( 1+i ) = FP
n log ( 1+i )=log ( )F
P

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log ( F / P )
n=
log ( 1+i )
2) Uniform Series

a) Using calculator

n−1 n−2 1

F= A (1+i ) + A ( 1+i ) +. . . . . . . . . . .+ A ( 1+i ) + A . . . . . . { 1 }

2 n

F ( 1+i )= A ( 1+i )+ A ( 1+i ) +. . . . . . . . . . .+ A (1+i ) . . . . . . . . . { 2 }

Subtracting 1 from 2

F ( 1+i ) − F = A ( 1+i ) − A
n

F ( 1+i−1)= A {( 1+i ) −1 }
n

{ }
n

( 1+ i ) −1
∴ F= A i
n

( 1+ i ) −1
i = Uniform Series Compound Amount factor
b) Using compound interest table

F=A (F/A, i, n)

C) Using computer (Excel)

=FV (i%, n,-A, 0)

Time Shifts in a uniform Series

If the initial deposit takes place at time zero and the last payment is zero

In this case each payment will be compounded for one more year

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F= A
i {
( 1+ i )n −1
}
( 1+i )
or

Thus,
F= A
i {
( 1+ i )n −1
}
+ A1 { F / p ,i , n }− A n

Using Excel

F=FV (i, n, -A1,, 1)

Example

a) Suppose you want to deposit 1000 birr at the end of each year for 5 yrs. If i=10% what will be
your final amount after 5 yrs.

F= A {
( 1+i )n −1
i }
=1000 {
( 1 .1 )5 −1
0.1 }
=6105 .10 birr

b) If the initial deposit is made at t=0

F 5=1000 ( 1+i )5 +1000 (1+i)4 +1000(1+i)3 +1000 (1+i)2 +1000(1+i)1

=1610. 51+1464 . 1+1331+1210+1100
=6715 .61 birr
or
F 5=(6105 .10 )(1+i)=6715 . 61
or

F 5=1000
1{
( 1+i )n −1
}
+ Aa {( 1+i )5 }− A=6715 .61

c) Using excel
=FV (10%, 5,-1000,,1)

B) Capital Recovery Factor (Annuity)

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F= A {
( 1+ i )n −1
i }
A=F
i
{
( 1+ i )n −1 }
=P (1+i )
n i
{
( 1+i )n −1 }
A=P
{ i (1+i )n
( 1+i )n −1 }
Using Excel: A= PMT (i, n, p, f, type)

C) Present worth factor (P/A, i, n)

P= A
{
( 1+ i ) n−1
i (1+i )n }
Excel: P = PV (I, n, A, F, O)

Sinking Fund Factor

Finding A given F, i, n

A=F
[ i
( 1+i )n−1 ]
=F ( A / F , i , n )

Composite Series (Single & Uniform Series)

Treat the single payment & the uniform series separately.

1) A contractor signed a lump sum contract agreement for 30 million birr on April 1, 2005. The
terms of the contract were 3*106 advance, 2.4*106 per year for the first 5 yrs (first payment after
one year) and 3*106 per year thereafter (first payment at year 6) for 5 years. If i=8% how much is
the contract value now?

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6 6 6

P=3∗10 + 2 . 4∗10 ( P / A , 8 , 5 )+3∗10 ( P / A , 8 , 5 )( P / F , 8 , 5 )

6 6 6

=3∗10 + 2 . 4∗10 ( 3 . 9927 )+3∗10 (3 . 9927 )( 0 . 6806 )

= 20.7*106 birr

3) Arithmetic(Linear) Gradient Series

An arithmetic gradient is a cash flow series that either increases or decreases by a constant
amount. The cash flow, whether income or disbursement, changes by the same arithmetic
amount each period .The amount of the increase or decrease is the gradient (G)

For example, if an engineer predicts that the cost of maintaining a machine will increase by 5000
birr per year until the machine is retired, a gradient series is involved and the amount of the
gradient is 5000birr

G ˃ 0 → increasing gradient

G < 0 → decreasing gradient

Case i)

To find P we can use (P/F, i, n) for each end of year value.

2
( n−1 ) G
∴ p=0+G/(1+i ) + 2( 1+i
G
)
+ 3G
( 1+i )
+. . . . . .+ ( 1+i ) 3 4 n

P=∑ ( n−1)G( 1+i )

−n

n=1

1
G=b and x=
Let ( 1+i )
2 3 4 n

∴ P=0 +bx +2 bx +3 bx +. . . . . .+ ( n−1 ) bx

=bx { 0+ x +2 x +3 x +. . . . .+ x }
2 3 n−1

( n−1 )

The sum of the series

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{0+ x +2 x +3 x +. . . . . . . . .+(n−1) x }
2 3 n−1

{ }
n−1 n

1−nx +(n−1) x
=x ( 1−x )2

Therefore,

P=bx { } ``
2
1−nx n−1 +(n−1 )x n
( 1−x )2
Replacing the original values

{ }
n

( 1+i ) −in−1
P=G i 2 ( 1+i )n

{ }
n

( 1+i ) −1
P= A i ( 1+ i )n

{ }=G {
n

( 1+i ) −1
∴A i ( 1+ i )n
( 1+i )n −in−1
i 2 ( 1+i )n }
{ }{
n

( 1+i ) −in−1
A=G i 2 ( 1+i )n
i ( 1+i )n
(1+i )n −1 }
A=G { }
n

( 1+i ) −in−1
i [ ( 1+i )n −1 ]

Example:

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A contractor has purchased a D9- dozer having a useful life of 5 yrs. The maintenance cost for
the 1st year is estimated to be 10,000 birr and increases by 2500 birr per year. Maintenance is
made at the end of each year. The company wants to deposit a certain lump sum now to cover the
maintenance cost. How much shall it deposit if i=3%.

P=P + P =( P/ A , i , n )+( P/ G , i , n )
A G

{ }+G {
n

( 1+i ) −1
=A i ( 1+i )n
( 1+i )n −in−1
i 2 ( 1+i )n }
=10000( 4 . 5797 )+2500 { 8 . 888 }
=45797+22220
=68017 birr
P = A ( P / A , i , n )+ G ( A / G , i , n )( P / A , i , n
=10 , 000( 4 . 5797 )+ 2500 ( 1 . 9409 )( 4 . 5797 )
=68018 birr

Geometric Gradient Series

Construction costs can involve cash flows that increase or decrease through time, by a constant
percentage (geometric) named as compound growth. Such growth can mostly be due to inflation.

If g represents the % age change, for the first payment

F =A n 1
( 1+g )
n−1
, n=1 ,2 ,3 ....n
Eg. if g=10% and n=5,A1=100

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5−1

F =100( 1+ 0 . 1 ) =100 (1 . 4641)=146 . 41

n

−n

P= F ( 1+i ) = A 1
( 1+g )
n−1
( 1+ i )
−n

n n

P=∑ A ( 1+i ) =∑ (1+

−n n
n−1 A 1 −n
( 1+ g ) ( 1+g ) ( 1+i )
1 g)
1 1

A n

=1+ g 1
∑1
( 1+g )n
(1+i )n

A 1
1+ g is a constant and if we denote it by K and

1+ g = x
1+ i

Subtracting 2 from 1

P ( 1− x )=k ( x− x )
n+1

n+1

x−x
P= K 1− x x≠1
Replacing the original values

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( )( ) [ ]
n+ 1
1+ g ( 1+ g ) 1 ( 1+ g )n
A 1+i
−
( 1+i )n+ 1
−
1+i ( 1+i )n+ 1
P= 1
=A
1+ g
1− (
1+ g
1+i ) 1 (1+i )− ( 1+ g )
(1+i )

[ ]= A {
( 1+ g )n
1−
=A 1
( 1+ i )n
( 1+i )− (1+ g ) 1
1−( 1+ g )n ( 1+i )−n
i− g }
P= A 1
{ 1− (1+ g )n ( 1+i )−n
i− g } …i≠ g
n

1−( 1+ g ) ( 1+i )
−n

lim ( i− g ) = n
1+i

∴ P= A ( n1+i ) ⋯when i= g 1

P= A ( P / F , g , i , n )
1

Example
Engineers at a specific company need to make some modifications to an existing machine. The
modification costs only $8,000 and is expected to last 6 years with a $1,300 salvage value. The
maintenance cost is expected to be high at $1,700 the first year, increasing by 11% per year
thereafter. Determine the equivalent present worth of the modification and maintenance cost. The
interest rate is 8% per year.

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Solution
Draw CFD

The present worth value is comprised of three components

 The present modification cost = $8,000

 The present value of the future salvage value

 The present value of all the maintenance values throughout the 6 years and these are
represented by the geometric gradient series

•PT = –8,000 + 1,300(P/F,8%,6) – Pg

•Pg = A1(P/A,g,i,n) → A1(P/A,11%,8%,6) = 1−[(1+0.11)/(1+0.08)]6
(0.08−0.11)
= 5.9559
•PT = – 8,000 + 819.26 – 1,700×5.9559 = $ –17,305.85
5. Irregular (Mixed) Series
A series of cash flows may be irregular, in that it does not exhibit a regular overall pattern

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2.4 Nominal and Effective Interest Rates

The difference between nominal and effective interest rates is that nominal means once
per year and effective means compounding more than once per year.

If the annual interest rate is 12% compounded monthly, it implies that money is compound 12
times in a year with i= 12/12 =1% per month.

∴ If 1000 birr is deposited

12

F =P ( 1+ i / 12 ) = P ( 1 +0 . 01 ) =1000 ( 1 .1268 )
12
12

=1126 . 8
If compounding was annually

F=1000 ( 1+0 .12 ) =1120 birr

In terms of annual interest compounding monthly gives an interest of

1126 . 8 −1000 ∗100 [ % ]=12 . 68 %

1000

Thus, 1% per month is the monthly effective interest rate and 12.68% is the annual effective
interest rate.

If ia denote the effective annual interest rate M the number of compounding per year (the number
of interest periods) and ir is the nominal or market interest rate (annual rate of interest)

then,

i = F−
P
a
P

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( )
m
ir
P 1+ −P
m
i =Pa

i =( 1 +m
a
ir
) −1

Effective Interest Rates per Payment Period

Let

C = the number interest periods per payment period

K = number of payment periods per year

ir = nominal interest rate

m= number of compounding/ year

⇒ i e
=effective rate per payment period

( )
c
ir
1+
⇒ i e
= c∗k
−1

Example: Suppose that you make quarterly deposits into a savings account that earns 8% interest
compounded monthly. Compute the effective interest rate per quarter.

Given: r = 8%. C = 3 interest periods per quarter, K = 4 quarterly payments per year, and M = 12
interest periods per year.
Find: i.
we compute the effective interest rate per quarter as

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i=(1+0.08/12)3-1
=2.013%
The annual effective interest rate ia is (1 + 0.02013)4-1 = 8.24%.

CHAPTER THREE
Investment Appraisal Methods

3.1 Decision Support Methods and Techniques

In general, the decision-making process involves to evaluate projects feasibility and the
identification of the best alternative among different alternatives. There may initially be many
alternatives, nevertheless only a few will be feasible and actually evaluated. Therefore, each
alternative is a stand-alone option that involves description and best estimates of parameters such
as first cost (purchase prices, transportation, and installation), estimated annual incomes and
expenses, salvage value, interest rate, etc.

To select an alternative among different ones, the measure-of-worth values are compared .This is
simply the result of engineering economy analysis .Once the alternatives are evaluated and
compared, the best alternative is selected and implemented. Keep in mind that alternatives
represent projects are economically and technologically viable.

Project Categories

To help formulate alternatives, projects are categorizing as one of the following:

Mutually exclusive: - Only one of the viable projects can be selected by the economic analysis.
Each viable project is an alternative and competes among each other (when an engineer must
select the one best diesel-powered engine from several competing models)
Independent: - More than one viable project may be selected by the economic analysis. They do
not compete among each other.
The common methods used for financial appraisal of projects are:-
A) Pay back Method
B) Present worth or Net Present Value Method
C) Equivalence Future worth

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D) Annual Cost /Income Method (Equivalent Cost / Income Method)

E) Internal Rate of Return Method

3.1) Pay back Method

This method uses the number of years it takes to pay back the initial investments from profits of
the investment. In computing the payback period, one can either consider time value of money or
disregard it. When one considers time value of money, it is called discounted payback method,
otherwise it is conventional.

The drawback of the payback period is fails to measure profitability; that is, it assumes that no

profit is made during the payback period. Requires an arbitrary cutoff point, ignores cash flows

beyond the cutoff date.

Example: For a dozer purchased at a cost of 3 million birr determine the payback period if the
hourly rental rate is 900 birr/hrs and the cost for fuel, operator and maintenance is 150 birr/hrs.
(Assume the 8hr working hour per day and 6 days per week)
A) calculate the conventional payback period and
B) the discounted payback period if discount rate is 10%
Solution
Yearly profit = (900-150) *8*6*52 = 1,872,000
Conventional payback period discounted payback period

Period Net cash flow Cumulative Cost of funds Cumulative cash flow
cash flow

0 -3,000,000 -3,000,000 0 -3,000,000

1 1,872,000 -1,128,000 0.1*3,000,000= -1,428,000

-300,000

2 1,872,000 744,000 0.1*-1,428,000 301,200

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= -142,800

3 1,872,000

Payback period 1.6years 1.76 ears

3.2) Present worth Analysis

In order to evaluate projects, one needs to use discounted cash flow techniques (DCF). One of
these is the method of net present worth (NPW) or net present value (NPV).

In this method all cash inflows and outflows of a given project (having a given project life) are
brought to time 0. If the difference between the inflows minus the outflows is positive then the
project is acceptable. If it is to compare among various projects, the one having more positive
value is economically the best alternative.

3.2.1 Evaluation of a Single Project

Step 1: Determine the interest rate that the firm wishes to earn on its investments. This interest
rate is often referred to as either a required rate of return or a minimum attractive rate of return
(MARR). Usually this selection is a policy decision made by top management. It is possible for
the MARR to change over the life of a project. But for now, we will use a single rate of interest
when calculating PW.
Step 2: Estimate the service life of the project.
Step 3: Estimate the cash inflow for each period over the service life.
Step 4: Estimate the cash outflow for each period over the service life.
Step 5: Determine the net cash flows for each period (net cash flow =cash inflow –cash outflow).
Step 6: Find the present worth of each net cash flow at the MARR. Add up these present-worth
figures;
their sum is defined as the project's PW. That is,

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A A A
PW ( i ) = ( 1+ i) + ( 1+ i) +. . . .+ ( 1+i )
0
0
1
1
n
n

n
A
=∑ ( 1+i ) n
n
0

Where PW (i) = PW calculated at i,

A, = net cash flow at the end of period n,
i = MARR (or cost of capital), and
n = service life of the project.

Step 7: In this context, a positive PW means that the equivalent worth of the inflows is greater
than the equivalent worth of the outflows, so the project makes a profit. Therefore, if the PW (i)
is positive for a single project, the project should be accepted; if it is negative, the project should
be rejected.
If PW (i) > 0, accept the investment.
If PW(i) = 0, remain indifferent.
If PW(i) < 0, reject the investment.
Example:

A construction company is considering the purchase of a dump truck costing 652,500 birr and
the project cash benefits over a 3 year is estimated as follow

End of year Net cash flow

0 -652,500.

1 212,280.

2 237,858.

3 485,112.

You have been asked by the owner of the company to evaluate the economic merit of the
purchasing of the dump truck. The firm's MARR is known to be 15%.

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PW =−652500+212280 ( P / F ,15 , 1 ) +237858 ( P / F , 15 ,2 )

+485112 ( P / F , 15 ,3 )
=30 , 911 birr
Since the project results in a surplus of 30,911 birr, the project is acceptable. It is returning a
profit greater than 15%.
3.2.2 Guidelines for Selecting a MARR
Return is what you get back in relation to the amount you invested. Return is one way to evaluate
how your investments in financial assets or projects are doing in relation to each other and to the
performance of investments in general. Let us look first at how we may derive rates of return.
Conceptually, the rate of return that we realistically expect to earn on any investment is a
function of three components:
Risk -free real return,
Inflation factor, and
Risk premium(s).
In evaluating projects with this method one will be encountered with different cash flow cases
and accordingly, one has to treat every problem case by case and compute the present worth.

The cases are:

1) Uniform cash flow

2) Uneven cash flow
3) etc
The above cases may be for a definite period or perpetual life (n>40 yrs).

In case of definite and short project life one shall draw and observe the cash flow diagrams and
discount their PW at time zero.

In case of perpetual projects life like irrigation, dams, hydropower plants, etc one uses the
method of capitalized equivalent.

If a uniform series of cash flow exists, then

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{ }
n

( 1+i ) −1
PW = A ( P / A ,i ,n ) = A i ( 1+i )n

For perpetual life n → α

∴ lim { n→ α
( 1+i )n −1
i ( 1+i )n } i
= 1

Therefore,

PW = A ( 1i ) = Ai
Example:

Maintenance money for a new building is to be deposited now. The expected cost estimated is as
follows:

a) 40,000 per year for the first 5 yrs

b) 50,000 “ “ from year 6-10
c) 60,000 “ “ perpetually there after
a) if i=13%, how large shall be deposited
b) What is the perpetual equivalent annual cost?
PW =40 , 000 ( P / A , 13 ,5 ) +50000 ( P / A , 13 ,5 ) ( P / F ,13 , 5 )
60000
+ ( P / F ,13 , 10 )
i
=40000 ( 3 .5172 ) +50000 ( 3 . 5172 ) ( 0 .5428 )
60000
+ ( 0. 2946 )
0 .13
=140688 +95457+135969

Solution =372114

b)
A =PW ∗i=372114∗0 .13=48374 .82

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Comparison of Alternatives

Note if you have two or more alternatives, Calculate the PW of each alternative at the MARR
and select the one with the highest PW, as long as all the alternatives have the same service lives.
When the present worth method is used to compare mutually exclusive alternatives that have
different lives, then the PW of the alternatives must be compared over the same number of years
and end at the same time. A fair comparison can be made only when the PW values
represent costs (and receipts) and associated with equal periods.
Alternatives with equal lives
•The competing alternatives have equal lives. For evaluating the alternatives, the present worth
of both the competing alternatives are found out.

•The alternative with the maximum present worth is the most economical alternative.

•For cost dominated cash flow diagrams the alternative with the lowest present cost is chosen.

•In case of cash flow diagrams involving both costs and revenues the net or difference of present
worth of revenues and costs are found. This is referred to as net present worth or net present
value (NPV).
The method of comparison of NPV is quite popular for evaluation of alternatives.

Alternatives with unequal lives

•the alternatives do not have an equal life period of service (they are not co-terminus).
•decision to choose between two batching plants, which may have different service lives ,say 5
years and 10 years.
•Here in one case, there would be a need to replace the plant at the end of five years, and any
cost likely to be incurred at that point in time, should be appropriately accounted for in the
budgeting at the outset.
•The least common multiple method and the study (analysis) period method are two approaches
to solve this class of problems.
Analysis Period
This is the time span over which the economic effects of an investment will be evaluated
a) Analysis period equals project lives

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b) Project lives longer than analysis period

c) “ “ shorter “ “ “

Analysis Period Equal Project Lives

In this case, calculate the present worth of each alternative and select the one with the highest
NPW.

Example:

Project Net Cash flow

N A B C D

0 -1000 -1000 -1000 -1000

1 0 600 -1200 900

2 0 800 800 900

3 3000 1500 1500 1800

Which alternative is the best if i=10%

PW A =−1000 +3000 ( P/ F ,10 , 3 )

PW B =−1000+600 ( P / F ,10 , 1 ) + 800 ( P/ F , 10 ,2 ) +1500 ( P/ F , 10 , 3 )
PW C =−1000−1200 ( P / F ,10 , 1 ) + 800 ( P/ F , 10 ,2 ) +1500 ( P/ F , 10 , 3 )
PW D=−1000 +900 ( P/ F , 10 , 1 ) + 900 ( P / F , 10 ,2 ) +1800 ( P/ F , 10 , 3 )

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PWA=1253.94 birr

PWB=1333.58 birr

PWC=-301.87 birr

PWD=1914.35 birr

Therefore alternative D is the best

Project Life Longer than Analysis Period

Analyze the project up to the end of the analysis period and any value due to the extra life will be
brought to the end of the analysis period as a salvage value.

Example:

It is required to excavate 400,000 tons of soil; a contractor is given the assignment to finish the
job in 2 years.

The contractor wanted to do the job with excavators. He found two models of excavators with
different service life and cost.

Model A costs 1,290,000 birr and has a life of 6,000 hrs before overhaul.

Operation costs 344,000 birr/yr for 2000 hrs.

At this rate it will serve for 3 yrs with a salvage value of 215,000 birr.

Model B costs 2,064,000 birr & has a life of 12,000 hours before overhaul. It has an operation
cost of 193,500 birr for 2000 hrs of operation per year. It’s has a total life of 6 yrs with a salvage
value of 258,000 birr. To finish in two years, both models require two pieces of equipment.

Which option is the best for the two years operation if I =15%

Solution

1) Amount to be excavated =400,000 tons

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Time given to Finnish = 2 years

Two excavator options exist

Model A Model B

Initial Cost =1,290,000 birr initial cost=2,064,000

Life =6000 hours in 3 yrs life =12000 hrs in6 yrs

O&M = 344000 birr/yr O&M =193500 birr for 2000 hrs operation for
2000 hrs operation

S = 215000 birr s =258000 birr

Two pieces of each model are required.

If I =15%, which model is better

Solution: Analysis period =2 yrs

Model A

6
PW ( A )=−1. 29∗10 −344000 ( P/ A , 15 ,2 ) +215000( P/F , 15 ,1) ( P/ F ,15 , 2 )
6
=−1. 29∗10 −344000 ( 1. 6257 ) +215000 ( 0 . 8696 ) ( 0 .7561 )
=−1 ,707 , 877 .32 birr∗2=−3 , 415 ,754 . 64 birr
Model B

PW ( B)= [ −2. 064∗10 −193500 ( P/ A , 15 ,2 ) +258000 ( P/ F ,15 , 6 ) ] ×2

6

=−4534079 . 1 birr
Project Life Shorter than Analysis Period

In this case, one needs to think of replacement projects to continue at the end until the analysis
period.

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The replacement project can exactly be the same or different based on the decision of keeping
the same technology or otherwise.

When project lives are shorter than the required service period, we must consider how, at the end
of the project lives, we will satisfy the rest of the required service period. Replacement projects
must be implemented when the initial project has reached the limits of the useful life-are needed
in such case.

Sufficient replacement projects must be analyzed to match or exceed the required service period.

Whether we select exactly the same alternative, or a new technology as the replacement project,
we are ultimately likely to have some unused portion of the equipment to consider as salvage
value. Just as in the case when project lives are longer than the analysis period. On the other
hand, we may decide to lease the necessary equipment or subcontract the remaining work for the
duration of the analysis period and not worry about salvage values.

Analysis Period is not specified

When an analysis period is not specified, either explicitly, it is up to the analyst to choose an
appropriate one. When the alternatives have equal lives, this is an easy selection. When the lives
of alternative differ, we must select an analysis period that allows us to compare different-lived
projects on an equal time basis, i.e., a common service period by using the lowest common
multiple of project lives.

Other Common Analysis Periods

In some cases, the lowest common multiple of project lives is a non wise analysis period to
consider. If you were to consider alternatives with lives of 10 and 12 years, a 120-year analysis
period may lead to inaccuracies, since over a long period of time we can be less and less
confident about ability to install identical replacement projects with identical costs and benefits.
In a case like this, it would be reasonable to use the useful life of one of the alternatives by either
factoring in a replacement project or salvaging the remaining useful life as the case may be. The
important rule is to compare both projects on the same time basis.

C) Equivalent future worth analysis

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The future worth of an alternative (FW) can be used to compare alternatives

• Once the FW value is determined, the selection guidelines are the same as with PW analysis
•For one alternative, if FW ≥ 0 means the MARR is met or exceeded
•For two mutually exclusive alternatives, select the one with the numerically larger FW value

D) Annual /Cost or Income /Method: - Equivalent Cost Method

In this case all the cash flow is converted to an equal uniform series of cost or income. Then for
mutual exclusive alternatives, the one with higher annual income or lower annual cost will be
opted.

E.g. two mutually exclusive alternatives are being considered for the water supply of a small
town with a constant water demand of 2*106 m3/yr for 20yrs. Both alternatives are equally
attractive from the technical, social & political point of view.
Alternative A
The water can be supplied by a neighboring town at a price of 0.40 birr /m3
Alternative B
Water can be supplied from wells and will be treated and transported to town. This requires a
total investment of 5.5*106 birr. The investment will have a value of 1*106 birr at the end of
20yrs period. Annual maintenance and operation costs are 260,000 birr and remain constant
during the 20 yrs period. Make an economic comparison by using equivalent cost method if:-
a) i = 4% & b) i = 12%
Soln: -(a)
Alt. A
0.4*2*106 = 800,000 birr /yr
AC = 800,000 birr/yrs
Alt.B
AC = 5.5 * 106 (A/P, 4%, 20) + 260,000 – 1*106 (A/F, 4%, 20)
= 5.5 *106  0.07358 + 260,000 - 1106  0.03358
= 631,000 birr/hrs
Therefore Alternative B is better
Advantage of the annual worth analysis

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 Eliminates the LCM problem

 Only you evaluate one life cycle of a project
 The result is reported in terms of birr/period
 If two or more alternatives possess unequal lives then one need only evaluate the AW for
any given cycle.
Analysis for Different Cycles
 6-year project Find the AW of any 6 – year cycle
 9-year project Find the AW of any 9 – year cycle
 And then compare the AW6/yr to AW9/yr
E) Internal Rate of Return
Rate of return is termed as internal Rate of Return; yield; marginal efficiency of capital.

Defn.

 It is the breakeven interest rate, which equals the present worth of projects’ cash outflows
to the present worth of its cash inflows

PW (i)=PW cash outflows−PW cash inflows=0

A A1 An
⇒ PW (i)=(1+i ) + (1+i ) +.....+ (1+i ) =0
0
0 1 n

Example; Suppose a purchase of equipment is made for 1000 birr and repays 402.1 birr per year
for 3 years. Determine internal rate of return

1000=402.1(P/A, i, 3)

 1000 = 402.1{(1+i)-1 + (1+i)-2 + (1+i)-3}

 Let (1+i)-1 = x
 Therefore, 1000/402.1 = x + x2 + x3
 x + x2 + x3 - 2.49 = 0
 x=1, 0.51=0 not true

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 x=0.9, 0=0

 ⇒ i=10 %

PW (10 %)=−1000+402 .1( P / A ,10 ,3)=0

Checking using project balance:

1 2 3 4 5

Year Unpaid Return on (2) Payment Unpaid

balance at
balance
Received year end

(2+3+4)

0 -1000 0 0 -1000

1 -1000 -100 402.1- -697.9

2 -697.9 -69.8 402.1 -365.6

3 -365.6 -36.5 402.1 0

⇒ The bank can breakeven at 10% rate of interest.

For the previous example, i = 10% is called IRR

Therefore the rate of return, equates the present worth, future worth and annual equivalent worth
of the entire cash flow series to O.

Methods of Finding rate of Return

1) Simple Investment (Conventional)

This is a cash flow with one sign change only

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2) Non-simple Investment

This is a cash flow with multiple sign changes

Computational Methods

A) Direct Solution
Direct mathematical solution can be obtained if:

i) there is only a two flow transaction of cash flow series or

ii) The projects' service life does not exceed 2 yrs.

Example:

Consider the following net cash flow

Project A Project B

0 -8000 -16000

1 0 +10400

2 0 +12000

3 0 0

4 12000 0

Project A

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pw(i )=−8000+12000( Fp , i , 4)=0

¿

=12000 ( P/ F ,i , 4 ) =8000
1
4 =0. 67⇒ ( 1+i ) =1.5
4

( 1+i )
1+i=1 .10668 ⇒i=0 .106681

i=10 . 6681 %

Project B

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Pw ¿¿¿ ¿
¿
B. Trial and Error Methods

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In the trial and error method, an estimated guess of i is assumed and Pw(i*) is equated to
zero.

If Calculated Pw(i*) < 0 then i* is decreased

" " Pw(i*) > 0 " " " increased

And the computation is repeated until Pw (i) = 0

Example

It is proposed to purchase equipment for rental purposes. The initial cost is 10,000 birr
for the 1st year of ownership; 2700 birr is estimated as the excess of receipts over
disbursements for everything except income tax. Rental rates decline with age and up
keep costs increase. The net decrease is 150 birr/yr. The machine has a life time of 15
yrs and will cost 1000 birr as salvage. Income taxes are 1050, 975 and will decrease by
75 birr every year thereafter what is i*?

Solution

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[ ( ) ( )]
15
p (1+i ) −15i−1
Pw (i )= 2700 ,i,15 −150 2 15
A i ( 1+i))

1000
+ 15 −10,000− ( 1050 A ,i,14)−75( G ,i,5)
(1+i )
p P
[ ¿ ]
P
A ( ) ( ) ( )
P
=2700 ,i,15 −150 ,i,15 + 1000 ¿
G
P ,i,15¿ ¿
F
¿¿

¿
¿
PW (9) > 0 increase i

In such a case one can make a linear interpolations and,

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i∗ = 9 % + ( 10 %−9 % )
[ 289. 32−0
289. 32−(−221 . 93 ) ]
=9 %+1 % [ 0 .566 ]
=9 %+0 .566 %=9 .566 %

Computer Solution Method

It is also possible to use computer aided economic analysis software of excel.

Using excel

I* can be calculated by

=IRR ( Range, ( guess) )

Range = Cash flow

guess= any assumed value of i

Accept/Reject Decisions Using R of R

Relationship with Pw analysis

In the present worth analysis, a given alternative is said to be better or acceptable than the other
for a given interest rate. Changing the interest rate may change the decision.

For an individual project also, Pw > 0 for a given range of i.

The present worth of this profile indicates that the project is acceptable for i values between

¿ ¿
0 ∧ i ⇒ 0<i<i
In the above case it was a unique i* and was simple to make accept / reject decision.

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The problem is when we have non simple investments with multiple i*s.

It is difficult to choose which i * to use to make accept/reject decision. Therefore, the i * value
fails to give a clue for decision.

Decision Rule for Simple Investments

The i* value obtained by equating NPW=0 gives the break-even rate of return. But companies are
not interested break-even only. Thus, they establish, as a company policy, what they expect as a
minimum attractive rate of return (MARR) for their investment.

Therefore, for an investment to be accepted or rejected, the following criteria are used.

If IRR > MARR - Accept

IRR = MARR - Indifferent

IRR < MARR - Reject

This is used when we calculate single projects.

Example

1) Consider an investment project with the following cash flow

N Cash flow

0 -5000

1 0

2 4840

3 1331

Is this project acceptable at MARR=10%

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NPW =−5000+0 ( p/ F ,i ,1 ) +4840 ( P/ F ,i ,2 ) +133 ( P/ F ,i ,3 )

i∗¿ 10% IRR=MARR−indifferent
An investor wants to establish a construction company. In order to do that he purchased land at
n=o, for 1.5*106 birr. The building and garage requires 3*10 6 birr, which is completed in one
year. As soon as this is ready, end of first year, equipments worth of 4*10 6 are purchased to start
up. During the 1st operating year revenue of 3.5 *106 birr is obtained. This revenue increases at
5% per year from the previous year for the next 9 yrs. After 10 yrs the revenue remained
constant for 3 yrs and the company phased out.

The salvage value at the end its 13 years service life is

2 * 106 for land

1.4 * 106 for building

0.5 * 106 for equipment

Annual operating and maintenance costs are about 40% of revenue each year. If the investor's
MARR= 15%, is this a good investment?

6 6 6
NPW =−1 .5∗10 −7∗10 ( P /F ,i , 1) +2 .1∗10 ( P /6 ,5 %,i , 10 ) ( P /F , i, 1 )
6
+3 .257∗10 ( P / A ,i , 3 ) ( P/F , i, 10 ) +3 . 9∗10 ( P /F , i, 14 )
6
Assume i=15⇒ NPV =52. 6∗10
i=30 ⇒

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Cost

Year Cash Out Revenue Net Cash Cumulative IRR

flow
Cash Inflow Flow Cash flow

0 -1500000 0 -150000 -1500000

1 -7000000 0 -7000000 -8500000

2 -1400000 3500000 2100000 -6400000 26.47%

3 -1470000 3675000 2205000 -4195000

4 -1543500 3858750 2315250 -1879705

5 -1620675 4051688 2431012,5 551262.5

6 -1701709 4254272 2552563.125 3103826

7 -1786794 4466985 2680191.281 5784017

8 -1876134 4690335 2814200.845 8598218

9 -1969941 4924851 2954910.888 11553129

10 -2068438 5171094 3102656.432 14655785

11 -2068438 5171094 3102656.432 17758442

12 -2068438 5171094 3102656.432 20861098

13 -2068438 9071094 7002656.432 27863754

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Mutually Exclusive Projects

General

When evaluating mutually exclusive alternatives using NPW or AE, the highest worthy or
the least cost projects are selected. But, when one uses the Rate of Return comparison, it
may not be the highest IRR, which shall be selected.

Example

Consider the following cash flow let MARR=10%

n A B

0 - 1000 - 5000

1 2000 7000

IRR 100% 40%

Pw(10%) 818 1364

According to PW criteria, B is more attractive, while IRR of A is greater than IRR of B.

The reason is that IRR is a relative measure while NPW, FW, AE are absolute money (Birr)
values. Thus, IRR does not consider the scale of Investment.

Thus, to decide using Rate of Return analysis one has to use another technique called
Incremental Investment Analysis.

For the previous example

n A B B-A

0 - 1000 - 5000 - 4000

1 2000 7000 + 5000

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5000 5000 ⇒ i=0 .25

PWiB-A = -4000 + 1+i = 0 ⇒ 1+i= 4000

=25% which is greater than MARR (10%)

Criteria

IR >MAR selct xalign¿ ¿IR <MAR ~ xrSub{size8{1} {}#size12{size20{italIR rSub{size8{xrSub{size6{2} -x} rSub{size8{1} ~= selct eithr one.¿
x2− 1 2 ¿ x−
21

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