AAIT

School of Civil and Environmental

Engineering

Engineering Economics (CEng 4232)

Chapter 2:Cost of Money

February 2025
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 Content
• Cost of Money
– Interest
– Time value of money
– Economic equivalence

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 Cost of Money
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• Interest
• Money borrowed form financial institution is expected to be repaid over time by an
amount greater than the amount borrowed.
• It is evident that financial institutions lend money expecting repayment (including
interest) greater than the borrowed amount over time.
“Interest is the cost of having money available for use.”
• In a financial world, money it self is a commodity, and like other goods that are
bought and sold, money costs money. Interest rate used to measure the cost of
money.
• Interest is the fee paid or a fee earned for the use of money. It is return on capital
(capital is the invested money and resource).
• Interest= Ending amount - Beginning amount = Amount owed now- Principal
• Interest rate is a percentage added to an amount of money over a specified
length of time.
• Interest rate (%)
Interest added per time unit
*100%
Principal
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Interest
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• For example if a firm owns the capital and invests it in a project, then the
project should return that capital plus interest as cost saving or added revenues.
• The interest rate that is appropriate depends on many factors including risks,
economic conditions, and time frame.

Interest paid Interest earned

Interest rate Rate of return

Borrowed money - Lent money
- Saved “
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- Invested
Interest
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Definition of Terms
• P=Value or amount of money at a time designated as the present time t=0. Initial
deposit, Investment made at t=0.
• F= Value or amount of money at some future time.
• A= Series of equal consecutive end of period amounts of money
• n= Number of interest periods (year, month, day).
• i= interest period per time period(percent per year, percent per month).
• t= stated time period (years, months, days).

Types of Interest
1. Simple Interest
2. Compound Interest

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 Interest
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Types of Interest
1. Simple interest: is computed only on original sum (principal), not on prior
interest earned and left in the account.
• Interest paid (earned) on only the original amount, or principal, borrowed (lent).
• A bank account, for example, may have its simple interest every year: in this case, an
account with $1000 initial principal and 20% interest per year would have a balance
of $1200 at the end of the first year, $1400 at the end of the second year, and so
on.
F=P(1+ ni)
End of 1st year: F =1000(1+(1*0.2))=1200
End of 2nd year: F=1000(1+(2*0.2))=1400
• Total interest earned (charged) is linearly proportional to:
– the initial amount of principal (loan)
– interest rate
– number of time periods of commitment
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 Interest
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2. Compound Interest: is interest paid (earned) on any previous interest
earned, as well as on the principal borrowed (lent). It arises when interest is
added to the principal of a deposit or loan, so that, from that moment on, the
interest that has been added also earns interest. This addition of interest to the
principal is called compounding.
• A bank account, for example, may have its interest compounded every year: in this
case, an account with $1000 initial principal and 20% interest per year would have
a balance of $1200 at the end of the first year, $1440 at the end of the second year
and soon.
F  P1  i  • Interest earned (charged) for a period
n

is based on
- Development of the equation: o Remaining principal plus
o Accumulated (unpaid) interest at the
Period Beginning of End of period
beginning of the period
period value value

1 P P(1+i) Period Beginning of End of period value

period value
•

2 P(1+i) P(1+i) (1+i)

3 P(1+i)2 P(1+i)2(1+i) 1 1000 =1000(1+0.2)=1200
… … … 2 P(1+i)=1200 =1000(1.22)=1440
N P(1+i)N-1 P(1+i)N
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 Interest
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Example: i=20%
F  P1  i 
n
• Simple Interest: F=P(1+ni) • Compound Interest:
End of
t Pi Beginning of Pi End of Year
Year t
year (1) (2) =(1)+(2)
0 1000
1 200 1200 0 1,000
2 200 1400
1 1,000 200 1,200

2 1,200 240 1,440

• “Interest is compounded”: means it is computed and then added to the total

owed or deposited.

• Define:
– Nominal interest rates
– Effective Interest rates
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 Interest
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• Cash flow diagram
– Describes inflow and outflow of money overtime.

(+) UP

(-) Down

• The horizontal line is time scale. Moving from left to right with progression of
time. Beginning of first year is traditionally defined as “Time 0”.
• Arrow placed at the end of the period signify cash flows.
– Downward arrow represent expense, disbursement= Negative cash flow,
outflow “–”
– Upward arrow represent Receipts= Positive cash flow, Inflow “+”
• Cash flow diagram is dependent on the point of view.
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 Interest
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• Cash flow diagram
• Cash flow diagram is dependent on the point of view.
Example:
Cash flow diagram for banks and depositor: depending on which viewpoint is taken
the diagram can simply be reversed. The depositor and the bank have opposite
perspectives on cash in and cash out for the initial deposit and the final withdrawal.

Depositor

Bank

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 Interest
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• Example
1. An earth moving company is considering
purchase of a piece of heavy equipment. The cash
flow diagram for the following anticipated cash
flows:
First cost=$120,000
Operating & maintenance cost=$30,000 per year
Overhaul cost=$35,000 in year 3
Salvage value=$40,000 after 5 years

2. An amount P is deposited now so that an equal annual amount of A1=2000 per year for
the first 5 years, starting 1year after the deposit. And a different annual withdrawal of
A2= 3000 per year for the following 3 years. Draw the cash flow diagram if i=8.5% per
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year. [Exercise]
 Interest
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• Cash flow diagram
Multiple Payment Series :
Single Payment Series
Uniform (Equal) or Unequal
F

0 1 2 N
P

Gradient Payment Series: A=R1=R2=….Rn

Linear and Geometric

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 Interest
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1. Single Payment Series

P F

F  P1  i 
n

1 i n : Single payment compound

amount factor
F=Unknown

FP
1
PF
P=Known
1  i n
1
n : Single payment present
1  i  worth factor
F=Unknown

P=Unknown 13
Time Value of Money
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2. Multiple Payment Series: Uniform/ Even

- To convert from a present worth (P) - To convert from a future value (F) to
to a uniform series or annuity (A). a uniform series or annuity (A).

 i 
 i1  i n  AF 
A  P   (1  i) n
 1 
 (1  i) n
 1   i  Sinking fund
Capital  i1  i n   (1  i) n  1 factor
recovery    
   FA / F , i %, n
n
factor  (1 i) 1 
 PA / P, i %, n

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 Time Value of Money
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2. Multiple Payment Series: Uniform/ Even
- To convert from a uniform series or - To convert from a uniform series or
annuity (A) to a present worth (P) annuity (A) to a future value (F).

 (1  i) n  1
FA 
 1  i n - 1  i 
P  A n 
 i(1  i)   (1  i) n  1 Equal payment
Equal payment  1  i n - 1   series compound
 i 
series present  n 
amount factor
worth factor  i(1  i) 
 AF / A, i %, n
 AP / A, i %, n
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 Time Value of Money
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2. Multiple Payment Series: Uniform/ Even
• Example: If $10,000 is borrowed and payments of $2000 are made each year for 9
years, what is the interest rate?

 1  i n - 1  i1  i n 
P  A A  P 
n   
n
i(1  i)  (1 i) 1 
 
 AP / A, i %, n  PA / P, i %, n

 1  i 9 - 1  i1  i n 
10,000  2,000  9  A  P 
 i(1  i)   (1  i) n
 1 
0.2   A / P, i %,9

• Either solve the equation or use tables for interest factors and find that the interest
rate is between 13% and 14%. These capital recovery factors,
(A/P,.13,9) = 0.1949 and (A/P,.14,9) = 0.2022 include the value of .2.
• We interpolate for the value of i.
i = 0.13 + (0.2 – 0.1949)(0.14 – 0.13)/(0.2022 – 0.1949) = 13.7%
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 Time Value of Money
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2. Multiple Payment Series: Uneven
x

y z
0
1 2 3 4 Uneven payment series P

Uneven payment series  F

Y Z
0 0 0
+ +
1 2 3 4 1 2 3 4 1 2 3 4
P2
P3
P1
P=P1+P2+P3
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 Time Value of Money
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2. Multiple Payment Series: Uneven
Exercise: Determine the present value for the given uneven payment series. Take i=8%
2500

5000 P1= 2500 (P/F, 8%, 1)= 2,314.81

3000
0 P2=3000 (P/F, 8%, 2)= 2,572.02
1 2 3 4 P3= 5000 (P/F, 8%, 4)=3,675.15

P=P1+P2+P3=8,562

P 2500

3000 5000
0 0 0
+ +
1 2 3 4 1 2 3 4 1 2 3 4

P2
P3
P1

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 Time Value of Money
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3. Gradient Payment Series: Linear and Geometric
• A gradient series of cash flows occurs when the value of a give cash flow is
greater than the previous cash flow by a constant amount, G, gradient step.

• Linear Gradient Series

 1  i n - in - 1  1  i n - in - 1
F  G  P  G 
 1  i  i 
2
 i 
n 2

 GF / G, i %, n  GP / G, i %, n 19
 Time Value of Money
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3. Gradient Payment Series: Linear and Geometric
• Linear Gradient Series
– Gradient as composite

(a)Increasing gradient series

(b)Decreasing gradient series

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 Time Value of Money
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3. Gradient Payment Series: Linear and Geometric
• Geometric Gradient Series
• The geometric cash flow series occurs
when the size of a cash flow increases  1  (1  g ) N (1  i )  N 
or decreases by a fixed percent from  A1   if i  g
  i  g 
one point to the next. P
• Percent change in a cash flow’s size  A  N  if i  g
from one point to the next is denoted  1
 1 i 
by g or j.

A1(1-g)

A1(1+g) A1(1-g) (N-1)

Increasing geometric series

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Decreasing geometric series
 Find Given Symbol Factor Name
E P F (P/F, i ,n) (1+i)-n Single payment present worth factor
c F P (F/P, i ,n) (1+i)n Single payment compound amount
o factor
n P A (P/A, i ,n) (1  i ) n  1 Uniform series present worth factor
i 1  i 
n
o
A P (A/P, i ,n) i (1  i ) n Uniform series capital recovery
m
1  i n  1 factor
i (1  i ) n  1
F A (F/A, i ,n) Uniform series compound amount
c i factor
A F (A/F, i ,n) i Uniform series sinking fund factor
(1  i ) n  1
F P G (P/G, i ,n) [1  (1  ni )(1  i )  n ] Gradient series present worth
i2 factor
a (1  i ) n  (1  ni ) Gradient series to uniform series
A G (A/G, i ,n)
c i[(1  i ) n  1] conversion factor
t P A1,j
n n
(P/A1,i, j, n) 1  (1  j )  (1  i) , i≠j Geometric series present worth
o i j factor

r F A1,j (F/A1,i, j, n) (1  i ) n  (1  j ) n , i≠j Geometric series future worth

i j factor
s 22
 Time Value of Money
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Multiple Payment Series: Uniform/ Even
• Example: Consider two investment choices that both require an initial
out flow of 20, 000 birr and an expected revenue as shown by respective
cash flow diagrams. Which one should be chosen?
10 k 10 k 10 k 10 k

0 0
1 2 3 4 5 1 2 3 4 5
Choice 2
Choice 1
1 PV 1 2 PV 2
• PV of expected future revenue
0 -20,000 -20,000 -20, 000 -20,000
1 1
PF
1  i n 2 10,000 8,264.46
3 10,000 7,513.15

4 10,000 6,830.13
5 10,000 6,209.21
-4,905 -6,278 23
 Time Value of Money
-- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Example: A consulting engineer is considering two investment alternatives (A and
B) having cash flow alternative shown below.
Assume an equivalent i=10%
Alternative A: is an investment in a land Alternative B: is for computer and the
development venture. Several other limited software required to provide specialized
partners are considering purchasing land, computer-design capabilities for clients. The
subdividing it, and selling land parcels over a engineer anticipate that competition will
5 yr period (an increase in land value is develop quickly if his plan proves successful, a
anticipated). 50 k
declining revenue profile is anticipated.
40 k 50 k
30 k 40 k
20 k 30 k
10 k 20 k 10 k
0
100,000
1 2 3 4 5
100,000
1 2 3 4 5
A PV A B PV B
0 (100,000) (100,000) (100,000) (100,000)
1 10,000 9,090.91 50,000 45,454.55
2 20,000 16,528.93 40,000 33,057.85
3 30,000 22,539.44 30,000 22,539.44
4 40,000 27,320.54 20,000 13,660.27
5 50,000 31,046.07 10,000 6,209.21
6,526 20,921 24
 Time Value of Money
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Gradient Payment Series: Linear
• Example: Maintenance cost for a particular production machine increase by
$1000/yr over the 5 year life of the equipment. The initial maintenance cost is
$3000. Using an interest rate of 8 % compounded annually, determine the present
worth equivalent for the maintenance cost.

0 1 2 3 4 5
= 0 1 2 3 4 5 + 0 1 2 3 4 5

$3000 $1000 $2000

$4000
$5000 A=$3000 $3000
$4000
$6000
$7,000
 1  i n - 1  1  i n - in - 1
P  A n 
P  G 
   1  i  i 
n 2
 i 1  i 
 AP / A, i %, n  GP / G, i %, n
 3000[ P / A,8%,5]  1000[ P / G,8%,5]
 3000 * 3.9927  1000 * 7.37243
Pu  $11,978.13 PG  $7,372.43
P=PU+PG=19,350.56
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 Time Value of Money
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Gradient Payment Series: Geometric
• Example: Assume you receive an annual bonus and deposit it in a saving account
that pays 8 percent compounded annually.Your initial bonus is 500 birr and the
size of your bonus increases by 10% each year. Determine how much will be in the
fund immediately after your 10th deposit.
A1=500, i=8%, g=10%, and n=10 years

End of Cash  1  (1  g ) n (1  i )  n 
Year(n) flow  A1 
  ig
 if i  g  F  P1  i n
0 0 P 
1 500  A  n  if i  g
2 550  1  1  i 
3 605
4 666  (1  i ) n  (1  g ) n 
5 732 F  A1  
6 805  i  g 
7 886 = 500(F/A1,8%,10%,10)
8 974
9 1072 = 500* 21.74
10 1179 F= 10,870.44
P= 10,870.44/(1.0810)=5,035.12 26
 Economic Equivalence
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• Equivalence in engineering economics analysis means “ the state of being
equal in value.”
• Two cash flows series are equivalent at some specified interest rate i%, if the
present worth are equal using an interest rate of i%.
• Two or more cash flow profiles are equivalent if their time value of money worth
at a common point in time are equal.
• There are certain rules that one should follow to make these calculations.

 They need to have a common time basis;

 Equivalence is dependent on interest rate; and
 Equivalence is maintained regardless of anything.

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 Economic Equivalence
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• Example : What single sum of money at t=0 is equivalent to the cash table below.

PA2 P2

A1=100
PA1 P1 A2 =100
0
1 2 3 4 5 6 7 8

P? P=P400+PA1+PA2
400 P400=-400(P/F,10%,1)
PA1=[100(P/A,10%,3)(P/F,10%,1)]
PA2=[100(P/A,10%,3)(P/F,10%,5)]
P=16.85
• Exercise : What single sum of money at t=6 is equivalent to the cash table below.

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 Assignment-1 To be submitted next class
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1. Using an 10% discount rate, what uniform series over five periods, [1,5], is
equivalent to the cash flow given in Figure 1.
600
500
400
250 100 Figure 1
0
1 2 3 4 5 6 7 8
2. For what interest rate are the two cash flows shown in Figure 2 equivalent ?
$3,500 A=$1500
$3,000
$2,500
$1,500 0
$1,000 1 2 3 4 5
=
0
1 2 3 4 5 $5,000 Figure 2

$7,000

3. Define and clarify with examples

– Nominal interest rates and
– Effective Interest rates 29
Thank You

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