ENGR 112
Economic Analysis
Engineering Economic Analysis



Evaluates the monetary aspects of the
products, projects, and processes that
engineers design
Aids in the decision making process when
alternatives are compared
Especially useful when resources are limited


Replace vs. keep
Build vs. buy
Time Value of Money

$1 today is more valuable than $1 a year
later
0
5
10
Years

Engineering economy adjusts for the time
value of money to balance current and future
revenues and cost
Capital vs. Interest

Capital


Invested money and resources
Whoever owns it should expect a return
from whoever uses it

Bank lends you money



Buy a car
Go to college
Firm invests in a project


Buy equipment
Buy “knowledge”
Capital vs. Interest

Interest



Return on capital
Typically expressed as an interest rate for a
year
Interest $ amount
Interest rate =
Capital $ amount
Interest rates are needed to evaluate
engineering projects!!
“Interest”-ing Example
The engineering group of Baker Designs
must decide whether to spend $90K (K for
thousands) on a new project. This project
will cost $5K per year for operations, and it
will increase revenues by $20K annually.
Both the costs and the revenues will
continue for 10 years. Should the project be
done?
Eschenbach, T.G., Engineering Economy: Applying Theory to Practice, Irwin, 1995, p.22
Types of Interest

Nomenclature




P = Initial deposit (or principal)
N = Number of years
i = Interest rate per year
F = Future value after N years
Types of Interest

Simple Interest

The interest calculated for N periods is
based on the initial deposit
F0 = P
F1 = P + P*i = P*(1 + i)
F2 = P + P*i + P*i = P* (1 + i + i) = P* (1 + 2i)
…
…
Fn = P + P*i + P*i + … = P* (1 + i + i + …) = P* (1 + Ni)
Types of Interest

Compound Interest



Interest type used in engineering economic
evaluations
Interest is computed on the current balance that
has not yet been paid
Includes accrued interest
F0 = P
F1 = P + P*i = P *(1 + i)
F2 = P1 + P1 * i = P1 *(1+i) = P *(1+i) *(1+i) = P *(1+i)2
…
…
Fn = Pn-1 + Pn-1 * i = Pn-1 *(1+i) = P *(1+i)n-1 *(1+i) = P
*(1+i)n
Simple vs. Compound Interest
If $100 is deposited in a savings account, how much is in the
account at the end of each year for 20 years, if interest is
deposited in the account and no withdrawals are made? Assume a
5% interest per year.
Year
0
1
2
3
4
5
6
7
8
9
10
F (Simple) F (Compound)
100
100.00
105
105.00
110
110.25
115
115.76
120
121.55
125
127.63
130
134.01
135
140.71
140
147.75
145
155.13
150
162.89
Year
11
12
13
14
15
16
17
18
19
20
F (Simple) F (Compound)
155
171.03
160
179.59
165
188.56
170
197.99
175
207.89
180
218.29
185
229.20
190
240.66
195
252.70
200
265.33
Simple vs. Compound Interest
Total in Account at year End
300
Compound
250
200
150
Simple
100
0
5
10
End of year
15
20
In Class Problem #1
If $500 is deposited in a savings
account, would a 5% simple
interest rate be better than a 3%
compound interest rate if you
were planning to keep the money
for 3 years? For 35 years?
Assume that all earned interest is
deposited in the account and no
withdrawals are made.
Example #1
Sam Bostro borrows $4000 from his parents for his
final year of college. He agrees to repay it 3 years
later in one payment to which a 7% compound
interest rate will be applied.
a) How much does he repay?
F = (4000)(1 + 0.07)3 = $4,900.17
b) How much of this is interest and how much is
principal?
F = P + I  P = $4,000; I = $900.17
Example #2
Susan Cardinal deposited $500 in her
savings account and six years later the
account has $600 in it. What compound rate
of interest has Susan earned on her capital?
F = P(1+i)n  i = [F/P]1/n – 1
i = [600/500] 1/6 – 1 = .031
i = 3.1%
In Class Problem #2
Your friend is willing to loan you
$1500 to buy a new computer,
but you must agree to pay him
back $2500 when you graduate
in 4 years? What is the
compound interest rate you will
be paying your friend?
Cash Flow Diagrams


Pictorial description of when and how
much money is spent or received
Summarizes the economic aspects of an
engineering project
0
1
2
3
4
5
Cash Flow Categories

First Cost


Operation and maintenance (O&M)


Expense to build or to buy and install
Annual expenses (electricity, labor, minor repairs,
etc.)
Salvage value


Receipt at project termination for sale or transfer
of equipment
There can also be a salvage COST
Cash Flow Categories

Revenues


Overhauls


Annual receipts due to sale of products or services
Major capital expenditures that occurs part way
through the life of the asset
Prepaid expenses

Annual expenses that must be paid in advance
(e.g., leases, insurance)
Examples
a)
b)
Ernie’s Earthmoving is considering the purchase of a
piece of heavy equipment. What is the cash flow
diagram if the following cash flows are anticipated?
First cost
$120K
O&M cost
$30k per year
Overhaul cost
$35K in year 3
Salvage value
$40K after 5 years
How would the cash flow diagram change if Ernie
decides to lease the equipment (at $25K per year)
instead of purchasing it?
Solution
40K
a)
0
1
2
30K
30K
0
4
5
30K
30K
4
5
30K +
35K
120K
b)
3
1
2
3
25K
25 +
30K
25 +
30K
25K +
30K +
35K
25K +
30K
30K
Frequency of Compounding


Interest rates are typically specified on
an annual basis
However, interest is often compounded
more often




Semiannually
Quarterly
Monthly
Daily
Frequency of Compounding

How does that change our formula?


m = # of compounding periods per year
n = # of compounding periods


n=mxN
Quarterly? 
m=4
n = 4N  F = P(1+i/m)m*N = P(1+i/4)4N

Monthly? 
m=12
n = 12N  F = P(1+i/12)12N
Example
Suppose you borrow $3000 at the beginning of your senior
year to meet college expenses. If you make no payments
for 10 years and then repay the entire amount of the loan,
including accumulated interest, how much money will you
owe? Assume interest is 6% per year, compounded
a) Annually
m=1
P=3,000
b) Quarterly
m=4
N=10 years
c) Monthly
m=12
i=6% per year
d) Daily
m=365
How significant is the frequency of compounding?
Solution
FA = P(1+i/1)1N = 3000(1+0.06/1)10 =
$5,372.54
FQ = P(1+i/4)4N = 3000(1+0.06/4)40 =
$5,442.05
FM = P(1+i/12)12N = 3000(1+0.06/12)120 =
$5,458.19
FD = P(1+i/365)1N = 3000(1+0.06/365)3650 =
$5,466.08
Solution
Frequency of Compunding
College Loan Payment Options
Daily
$3,000.00
$2,466.09
Monthly
$3,000.00
$2,458.19
Quarterly
$3,000.00
$2,442.06
Annually
$3,000.00
$2,372.54
$0
$1,000
$2,000
$3,000
$4,000
Amount owed
$5,000
$6,000
Principal
Interest
In Class Problem #3
You need to borrow $500 to pay
for the text books you will need
for your sophomore level courses.
Which is the better loan
assuming that you will pay the
entire loan amount plus interest
when you graduate in 3 years?


3.0% compounded annually
2.5% compounded quarterly