Engineering Economy

Chapter 1: Introduction to
Engineering Economy
 Engineering economy…

• Involves the systematic evaluation of the economic

merits of proposed solutions to engineering
problems.
• Solutions to engineering problems must
demonstrate a positive balance of long-term
benefits over long-term cost.
 There are seven fundamental principles of
engineering economy.

• Develop the alternatives

more alternatives = Quality Decision

• Focus on the differences

• Use a consistent viewpoint
• Use a common unit of measure ($ )
• Consider all relevant criteria
• Make risk and uncertainty explicit
• Revisit your decisions
 Engineering economic analysis
procedure
• Problem definition
• Development of alternatives
• Development of prospective outcomes
• Selection of a decision criterion
• Analysis and comparison of alternatives.
• Selection of the preferred alternative.
• Performance monitoring and post-evaluation of
results.
Example:

A person bought a building for 100,000$.

Spent 10,000 of his own money and mortgage 90,000 from a bank.
Mortgage payment 10,500$ per year.
The expected maintenance 15,000 $ per year.
There are 4 apartments in the building,
Each apartment can be rented for 360$ per month

A. Does this person have a problem? If so, what is it?

Spend: 10,500 + 15,000 = 25,500$ per year

Received: 360 ×12 ×4 = 17,280$ per year
Loss: 25,500 – 17,280 = 8,220$ per year
B. Alternatives:
1. Raise the rent:
minimum increase = 8,220/ (12×4) =171.25$ per apartment
means raise the rent from 360$ to 531.25$ !!!!!!!!!!!!!!

2. Lower the annual maintenance by 8,220$

from 15,000$ to 6780$

3. Sell the building

 Chapter 2: The Time Value of
Money

The objective of Chapter 2 is to explain time value of

money calculations and to illustrate economic
equivalence.
 Interest and Interest Rate
• Interest – the manifestation of the time value of
money
• Fee that one pays to use someone else’s
money
• Difference between an ending amount of
money and a beginning amount of money

➢ Interest = amount owed now – principal

• Interest rate – Interest paid over a time period

expressed as a percentage of principal
 Rate of Return
• Interest earned over a period of time is expressed
as a percentage of the original amount (principal)

interest accrued per time unit

Rate of return (%) = x 100%
original amount

❖ Borrower’s perspective – interest rate paid

❖ Lender’s or investor’s perspective – rate of return
earned
 Commonly used Symbols
t = time, usually in periods such as years or months
P = value or amount of money at a time t
designated as present or time 0
F = value or amount of money at some future
time, such as at t = n periods in the future
A = series of consecutive, equal, end-of-period amounts
of money
n = number of interest periods; years, months
i = interest rate or rate of return per time period;
percent per year or month
 Cash Flows: Terms
• Cash Inflows – Revenues (R), receipts,
incomes, savings generated by projects and
activities that flow in. Plus sign used
• Cash Outflows – Disbursements (D), costs,
expenses, taxes caused by projects and
activities that flow out. Minus sign used
• Net Cash Flow (NCF) for each time period:
NCF = cash inflows – cash outflows = R – D
• End-of-period assumption:
Funds flow at the end of a given interest period
 Cash Flow Diagrams
What a typical cash flow diagram might look
Draw a timeline
like
Always assume end-of-period cash flows

Time
0 1 2 … … … n-1 n
One time
period
F = $100
Show the cash flows (to approximate scale)

0 1 2 … … … n-1 n
Cash flows are shown as directed arrows: + (up) for inflow
P = $-80
- (down) for outflow
 Cash Flow Diagram Example
Plot observed cash flows over last 8 years and estimated sale next
year for $150. Show present worth (P) arrow at present time, t = 0
 Economic Equivalence

Definition: Combination of interest rate (rate of

return) and time value of money to determine
different amounts of money at different points
in time that are economically equivalent

How it works: Use rate i and time t in

upcoming relations to move money (values of
P, F and A) between time points t = 0, 1, …, n
to make them equivalent (not equal) at the
rate i
 Example of Equivalence
Different sums of money at different
times may be equal in economic value
at a given rate $110

Year

0 1
Rate of return = 10% per year

$100 now

$100 now is economically equivalent to $110 one year from

now, if the $100 is invested at a rate of 10% per year.
 Simple Interest: infrequently used

With simple interest, the interest is due along with the

principal (Capital), i.e. the amount borrowed, at the
end of the loan period. For example, borrow $100 for
1 year @ 10% simple interest per year:
• Repay $100 + $10 at the end of one year
• Borrow $100 for 2 years @ 10% simple interest per
year:
• Repay $100 + $20 at the end of two years
 Computation of simple interest
The total interest, I, earned or paid may be computed
using the formula below.

P = principal amount lent or borrowed

N = number of interest periods (e.g., years)
i = interest rate per interest period

The total amount repaid at the end of N interest

periods is P + I.
Example:

If $5,000 were loaned for five years at a simple interest

rate of 7% per year, the interest earned would be

So, the total amount repaid at the end of five years

would be the original amount ($5,000) plus the interest
($1,750), or $6,750.
 Compound interest: frequently used

Compound interest reflects both the remaining principal

and any accumulated interest. For $1,000 at 10%…

(1) (3)=(1)+(2)
Amount owed at (2)=(1)x10% Amount owed
beginning of Interest amount at end of
Period period for period period
1 $1,000 $100 $1,100

2 $1,100 $110 $1,210

3 $1,210 $121 $1,331

Notation used in formulas for compound interest
calculations.

i = effective interest rate per interest period

N = number of compounding (interest) periods
P = present sum of money; equivalent value of one or more
cash flows at a reference point in time; the present
F = future sum of money; equivalent value of one or more cash
flows at a reference point in time; the future
A = end-of-period cash flows in a uniform series continuing
for a certain number of periods, starting at the end of the
first period and continuing through the last
 Finding F when given P

F = P (1+ i) N
F = P ( F/P, i % , N )

Single payment compound amount factor = (1 + i ) N = ( F /P, i % , N )

Example:

F = P (1+ i) N
F = 8000 (1 + 0.1) 4 = 8000 (1.1) 4
F = 8000 1.4641 =11,712.8$

F = P ( F/P, i % , N )
F = 8000 ( F /P,10 % , 4)
F = 8000 (1.4641) =11,712.8$
 Finding P when given F

From F = P (1 + i) N

P = F (1+ i) − N

P = F ( P/F , i % , N )

−N
Single payment present worth factor = (1 + i ) = ( P/F , i % , N )
Example:

P = F (1+ i) − N
P = 10000 (1 + 0.08) −6 = 10000 (1.08) −6 = 10000[1 /(1.08) 6 ]
P = 10000  0.6302 = 6,302$

P = F ( P/F , i % , N )
P =10000 ( P /F , 8 % , 6)
P =10000 (0.6302) = 6,302$
Finding Interest rate (i) when given P, F, and N

F = P (1+ i) N  (1+ i) = ( F /P)1/N

i = (F / P ) −1
1/ N

Finding N when given P, F, and i

F = P (1 + i ) N  (1 + i ) N = ( F /P)
 N log (1 + i) = log( F /P)
log( F /P)
N=
log (1 + i)
Example:

i = (F / P ) −1
1/ N

i = (2.31/ 1.07) −1
1/12

i =1.0662 −1 = 0.0662
i = 6.62% per year
Example:

log( F /P)
N=
log (1 + i)

log( 5 /2.31) log( 2.1645)

N= =
log (1+ 0.0662) log (1.0662)
N = 12.05 years
Interpolation Example:
suppose you wish to have 100,000$ after 27 years, what amount
should be deposited now to provide for it at 7% interest rate per
year.

P = F ( P/F , i % , N )  P =100,000 ( P/F , 7 % , 27)

( P/F , 7 % , 25) = 0.1842

( P/F , 7 % , 30) = 0.1314

25 0.1842 x − 0.1314 0.1842 − 0.1314

27 x =
30 0.1314 27 − 30 25 − 30
x = 0.1631 = ( P/F , 7 % , 27)
P =100,000 0.1631 = 16310$
Example:
suppose you wish to have 100,000$ after 27 years, what amount
should be deposited now to provide for it at 7% interest rate per
year.

P = F (1+ i) − N

P = 100,000 (1+ 0.07) −27 = 100,000  0.1609

P =16093$
 Interpolation Example:
suppose you wish to have 100,000$ after 20 years, what amount
should be deposited now to provide for it at 5.5% interest rate per
year.
P = F ( P/F , i % , N )  P =100,000 ( P/F , 5.5 % , 20)
( P/F , 5 % , 20) = 0.3769
( P/F , 6 % , 20) = 0.3118
5 0.3769 x − 0.3118 0.3769 − 0.3118
5.5 x =
5.5 − 6 5−6
6 0.3118
x = 0.3444 = ( P/F , 5.5 % , 20)
P =100,000 0.3444 = 34,440$

P = F (1+ i)−N = 100,000 (1.055)−20 = 100,000  0.3427 = 34,270$

Relating a Uniform Series (Annuity)(A) to Its
Present (P)and Future (F) Equivalent Values
 Finding F when given A

 (1 + i ) N −1
F = A 
 i 
F = A ( F/ A, i % , N )
(1 + i ) N −1
Uniform series compound amount factor = = ( F / A, i % , N )
i
F = A ( F / A, i % , N )
F = 23000( F / A, 6 % , 40)
F = 23000(154.762) = 3,559,526$
 Finding P when given A

 (1 + i) N −1
P = A N 
 i (1 + i ) 

P = A ( P/ A, i % , N )
(1 + i ) N −1
Uniform series present worth factor = = ( P/ A, i % , N )
i (1 + i ) N
P = A ( P/ A, i % , N )
P =30 ( P/ A, 2 % , 20)
P = 30(16.3514) = 490.54$
 Finding A when given F

 i 
A= F  
 (1 + i ) N
− 1 

A = F ( A/F , i % , N )

i
Uniform series sinking fund factor = = ( A/F , i % , N )
(1 + i) − 1
N
Example:
What uniform annual amount should be deposited each year in
order to accumulate 2143.6$ at the time of the eighth annual
deposit given i = 10%

A = F ( A/F , i % , N )
A = 2143.6 ( A/F ,10 % , 8)
A = 2143.6(0.0874) = 187.45$
 Finding A when given P

 i(1 + i ) N 
A= P  
 (1 + i ) N
− 1 

A = P ( A/P, i % , N )
i (1 + i ) N
Uniform series capital recovery factor = = ( A/P, i % , N )
(1 + i ) − 1
N
A = P( A/P, i % , N )
A =15000 ( A/P, 0.25 % , 36)
A =15000(0.0291) = 436.5$
 Finding The number of cash flows in an annuity (N)

1. Given A, P, and i
 A 
log 
 A − Pi 
N=
Log (1 + i )
P
( P/ A, i % , N ) =
A
2. Given A, F, and i
 Fi + A 
log 
N=  A 
Log (1 + i )
F
( F / A, i % , N ) =
A
 P = A ( P/ A, i % , N )  100,000 =10,000 ( P/ A, 8 % , N )

( P/ A, 8 % , N ) =10

( P/ A, 8 % , 20) = 9.8181  P = 9.818110,000 = 98181$

( P/ A, 8 % , 21) =10.0168  P =10.016810,000 =100168$

10,000$ for 20 years P = 98181$

21th payment equivalents to 1819$ (100,000-98,181) in present
F = 1819 ( F/P, 8 % , 21) =1819  5.0338 = 9,156.5$
  A 
log 
N=  A − Pi 
Log (1 + i )

 10,000 
log 
 10,000 − 100,000  0.08  log( 5)
N= = = 20.91 years
Log (1.08) log(1.08)
Finding The interest rate (i) given A, N, and F or P

 (1 + i) N −1
P = A N 
Can’t be solved for i
 i (1 + i) 

 (1 + i ) N −1
F = A  Can’t be solved for i
 i 
Finding The interest rate (i) given A, N, and F or P

F = A ( F/ A, i % , N )  60,000 = 6,000 ( F/ A, i % , 8)

( F / A, i % , 8) =10
( F/ A, 6 % , 8) = 9.8975
( F/ A, 7 % , 8) =10.2598
Linear Interpolation

i ( F / A, i % , 8)
6% 9.8975
i 10
7% 10.2598

i−7 6−7
=
10 − 10.2598 9.8975 − 10.2598

i = 6.28%