Basics of Engineering Economy
Chapter 1
Foundations of Engineering
Economy
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Chapter 1 - Foundations
PURPOSE
TOPICS
 Definition and study
Understand the
fundamental concepts of
engineering economy
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approach
 Interest rate, ROR, and
MARR
 Equivalence
 Interest – simple and
compound
 Cash flow diagrams
 Rules of 72 and 100
 Spreadsheet
introduction
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Sec 1.1 – Definition of Engineering Economy
Sec 1.2 – Elements of a Study
 DEFINITION: Techniques that simplify comparison of
alternatives on an economic basis
 Most project decisions consider additional factors – safety,
environmental, political, public acceptance, etc.
 Fundamental terminology:
 Alternative -- stand-alone solution
 Cash flows -- estimated inflows (revenues) and outflows (costs) for
an alternative
 Evaluation criteria -- Basis used to select ‘best’ alternative;
usually money (currency of the country)
 Time value of money -- Change in amount of money over time
(Most important concept in Eng. Econ.)
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Sec 1.3 - Interest Rate, ROR, MARR
 Interest is a manifestation of time value of money
 Calculated as difference between an ending amount and a beginning
amount of money

Interest = end amount – original amount
 Interest rate is interest over specified time period
based on original amount
 Interest rate (%) =
interest accrued per time unit
x 100%
original amount
 Interest rate and rate of return (ROR) have same
numeric value, but different interpretations
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Sec 1.3 - Interest Rate and ROR Interpretations
Borrower’s perspective
Investor’s perspective
Take loan of $5,000 for one
year; repay $5,350
Invest (or lend) $5,000 for
one year; receive $5,350
Interest paid = $350
Interest earned = $350
Interest rate = 350/5,000
= 7%
INTEREST RATE
Rate of return = 350/5,000
= 7%
RATE OF RETURN
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Sec 1.3 - ROR and MARR
Cost of capital (COC) – interest
rate paid for funds to finance
projects
ROR
MARR – Minimum ROR needed
for an alternative to be
justified and economically
acceptable. MARR ≥ COC.
If COC = 5% and 6% must be
realized, MARR = 11%
Always, for acceptable projects
ROR ≥ MARR > COC
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Sec 1.4 - Equivalence
Different sums of money at different times may be equal
in economic value
Interest rate = 6% per year
-1
$94.34 last year
0
1
$100 now
$106 one
year from
now
Interpretation: $94.34 last year, $100 now, and $106 one year
from now are equivalent only at an interest rate of 6% per year
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Sec 1.5 – Simple and Compound Interest
Simple interest is always based on the original
amount, which is also called the principal
Interest per period = (principal)(interest rate)
Total interest = (principal)(n periods)(interest rate)
Example: Invest $250,000 in a bond at 5% per year simple
Interest each year = 250,000(0.05) = $12,500
Interest over 3 years = 250,000(3)(0.05) = $37,500
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Sec 1.5 – Simple and Compound Interest
Compound interest is based on the principal
plus all accrued interest
Interest per period = (principal + accrued interest)(interest rate)
Total interest = (principal)(1+interest rate)n periods - principal
Example: Invest $250,000 at 5% per year compounded
Interest, year 1 = 250,000(0.05) = $12,500
Interest, year 2 = 262,500(0.05) = $13,125
Interest, year 3 = 275,625(0.05) = $13,781
Interest over 3 years = 250,000(1.05)3 – 250,000 = $39,406
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Sct 1.6 - Terminology and Symbols
 t = time index in periods; years, months, etc.
 P = present sum of money at time t = 0; $
 F = sum of money at a future time t; $
 A = series of equal, end-of-period cash flows;
currency per period, $ per year
 n = total number of periods; years, months
 i = compound interest rate or rate of return;
% per year
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Sct 1.6 - Terminology and Symbols
Example: Borrow $5,000 today and repay
annually for 10 years starting next year at 5% per
year compounded. Identify all symbols.
Given: P = $5,000
i = 5% per year
n = 10 years
Find: A = ? per year
t = year 1, 2, …, 10
(F not used here)
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Sec 1.7 – Cash Flow Estimates
Cash inflow – receipt, revenue, income, saving
Cash outflows – cost, expense, disbursement, loss
Net cash flow (NCF) = inflow – outflow
End-of-period convention: all cash flows and
NCF occur at the end of an interest period
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Sec 1.7 – Cash Flow Diagrams
Year 1
0
Year 5
1
2 Time, t
3
4
5
Typical time
scale or 5
years
+ Cash flow
P=?
0
1
2
3
4
5
Find P in
year 0,
given 3
cash flows
- Cash flow
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Sec 1.7 – Cash Flow Diagrams
Example: Find an amount to deposit 2 years from now
so that $4,000 per year can be available for 5 years
starting 3 years from now. Assume i = 15.5% per year
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Sec 1.8 – Rule of 72 ( and 100)
Approximate n = 72 / i
Estimates # of years (n) for an
amount to double (2X) at a stated
compound interest rate
e.g., at i = 10%, $1,000 doubles to $2,000
in ~7.2 years
 For simple
interest, doubling
time is exact,
using rule of 100
n = 100/i
or
Solution for i estimates compound
rate to double in n years
i = 100/n
Approximate i = 72 / n
 $1,000 doubles in
10 years at 10%
simple interest
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Sec 1.9 – Introduction to Spreadsheet
Functions
To display
Present value, P
Future value, F
Annual amount, A
# of periods, n
Compound rate, i
i for input series
P for input series
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Excel Function
= PV(i%,n,A,F)
= FV(i%,n,A,P)
= PMT(i%,n,P,F)
= NPER(i%,A,P,F)
= RATE(n,A,P,F)
= IRR(first_cell:last_cell)
= NPV(i%,second_cell:
last_cell)+first_cell
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