Engineering Economics
November 3, 2004
1
Engineering Economy
• It deals with the concepts and techniques
of analysis useful in evaluating the worth
of systems, products, and services in
relation to their costs
2
Engineering Economy
• It is used to answer many different
questions
– Which engineering projects are worthwhile?
• Has the mining or petroleum engineer shown that
the mineral or oil deposits is worth developing?
– Which engineering projects should have a
higher priority?
• Has the industrial engineer shown which factory
improvement projects should be funded with the
available dollars?
– How should the engineering project be
designed?
• Has civil or mechanical engineer chosen the best
thickness for insulation?
3
Basic Concepts
• Cash flow
• Interest Rate and Time value of money
• Equivalence technique
4
Cash Flow
• Engineering projects generally have economic
consequences that occur over an extended
period of time
– For example, if an expensive piece of machinery is
installed in a plant were brought on credit, the simple
process of paying for it may take several years
– The resulting favorable consequences may last as
long as the equipment performs its useful function
• Each project is described as cash receipts or
disbursements (expenses) at different points in
time
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Categories of Cash Flows
• The expenses and receipts due to
engineering projects usually fall into one of
the following categories:
– First cost: expense to build or to buy and install
– Operations and maintenance (O&M): annual
expense, such as electricity, labor, and minor
repairs
– Salvage value: receipt at project termination for
sale or transfer of the equipment (can be a
salvage cost)
– Revenues: annual receipts due to sale of products
or services
– Overhaul: major capital expenditure that occurs
during the asset’s life
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Cash Flow diagrams
•
•
•
•
•
•
Shows money flow as a function of time
x-axis is time and y-axis is magnitude of
money.
both positive and negative payments are
shown as arrows with upward or
downward directions .
7
An Example of Cash Flow Diagram
• A man borrowed $1,000 from a bank at 8%
interest. Two end-of-year payments: at the
end of the first year, he will repay half of the
$1000 principal plus the interest that is due.
At the end of the second year, he will repay
the remaining half plus the interest for the
second year.
• Cash flow for this problem is:
End of year
0
1
2
Cash flow
+$1000
-$580 (-$500 - $80)
-$540 (-$500 - $40)
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Cash Flow Diagram
$1,000
1
2
0
$580
$540
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• Example
• A mechanical device will cost $20,000
• when purchased. Maintenance will cost
$1000 per year. The device will generate
revenues of $5000 per year for 5 years.
• The salvage value is $7000.
10
11
Interest
• Interest is the fee paid for borrowed money.
• Simple interest is the interest that is
computed on the original principal only.
• If I denotes the interest on a principal P
(in dollars) at an interest rate of r per year
for t years, then we have
I = Prt
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• The accumulated amount A, the sum of the
principal and interest after t years is given
by
A = P + I = P + Prt
= P(1 + rt)
and is a linear function of t
13
Example
• A bank pays simple interest at the rate of
8% per year for certain deposits.
• If a customer deposits $1000 and makes no
withdrawals for 3 years, what is the total
amount on deposit at the end of three
years?
• What is the interest earned in that period?
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Solution
• Using the accumulated amount formula
with P = 1000,
r = 0.08, and t = 3, we
see that the total amount on deposit at the
end of 3 years is given by
• A = P(1+rt)
• A = 1000(1+0.08*3)
or $1240.
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Example
• Wayne earns 4.7% simple interest for 5
years on $4500. How much interest does he
earn?
• Put the numbers in the formula I = Prt.
• Change the percent to a decimal.
• Multiply.
• I = 4500(4.7%)5
• = 4500(0.047)5
• = 1057.50 Wayne would earn $1057.50
interest.
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Compound Interest
• COMPOUND INTEREST is interest paid on
both the original principal and the interest
earned previously.
• The formula for compound interest is
• A = P(1 + r)t ,
• where A = total amount including previous
interest earned, P = principal, r = interest
rate, and t = time
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Time Value of Money
• Money has value
– Money can be leased or rented
– The payment is called interest
– If you put $100 in a bank at 9% interest for one time
period you will receive back your original $100 plus
$9
Original amount to be returned = $100
Interest to be returned = $100 x .09 = $9
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Compound Interest
• Interest that is computed on the original
unpaid debt and the unpaid interest
• Compound interest is most commonly
used in practice
• Total interest earned = In = P (1+i)n - P
– Where,
• P – present sum of money
• i – interest rate
• n – number of periods (years)
I2 = $100 x (1+.09)2 - $100 = $18.81
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Future Value of a Loan With
Compound Interest
• Amount of money due at the end of a loan
– F = P(1+i)1(1+i)2…..(1+i)n or F = P (1 + i)n
– Where,
• F = future value and P = present value
• Referring to slide #10, i = 9%, P = $100 and say
n= 2. Determine the value of F.
F = $100 (1 + .09)2 = $118.81
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Notation for
Calculating a Future Value
• Formula:
F=P(1+i)n is the
single payment compound amount factor.
• Functional notation:
F=P(F/P,i,n) F=5000(F/P,6%,10)
• F =P(F/P) which is dimensionally
correct.
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Notation for
Calculating a Present Value
• P=F(1/(1+i))n=F(1+i)-n is the
single payment present worth factor.
• Functional notation:
P=F(P/F,i,n) P=5000(P/F,6%,10)
Interpretation of (P/F, i, n): a present sum P,
given a future sum, F, n interest periods
hence at an interest rate i per interest
period
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Spreadsheet Function
P = PV(i,N,A,F,Type)
F = FV(i,N,A,P,Type)
i = RATE(N,A,P,F,Type,guess)
Where, i = interest rate, N = number of interest
periods, A = uniform amount, P = present sum
of money, F = future sum of money, Type = 0
means end-of-period cash payments, Type =
1 means beginning-of-period payments,
guess is a guess value of the interest rate
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Equivalence
• Relative attractiveness of different
alternatives can be judged by using the
technique of equivalence
• We use comparable equivalent values of
alternatives to judge the relative
attractiveness of the given alternatives
• Equivalence is dependent on interest rate
• Compound Interest formulas can be
used to facilitate equivalence
computations
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Technique of Equivalence
• Determine a single equivalent value at a
point in time for plan 1.
• Determine a single equivalent value at a
point in time for plan 2.
Both at the same interest rate and at the same time point.
•Judge the relative attractiveness of the
two alternatives from the comparable
equivalent values.
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Engineering Economic Analysis
Calculation
• Generally involves compound interest
formulas (factors)
• Compound interest formulas (factors) can
be evaluated by using one of the three
methods
– Interest factor tables
– Calculator
– Spreadsheet
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Given the choice of these two plans
which would you choose?
Year
0
1
2
3
4
5
Total
Plan 1
$1,000
$1,000
$1,000
$1,000
$1,000
$5,000
Plan 2
$5,000
$5,000
To make a choice the cash flows must be altered
so a comparison may be made.
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Resolving Cash Flows to Equivalent Present
Values
• P = $1,000(PA,10%,5)
• P = $1,000(3.791) =
$3,791
• P = $5,000
• Alternative 2 is better
than alternative 1 since
alternative 2 has a
greater present value
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An Example of Future Value
• Example: If $500 were deposited in a
bank savings account, how much would
be in the account three years hence if
the bank paid 6% interest compounded
annually?
• Given P = 500, i = 6%, n = 3, use F =
FV(6%,3,,500,0) = -595.91
• Note that the spreadsheet gives a
negative number to find equivalent of P.
If we find P using F = -$595.91, we get
P = 500.
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An Example of Present Value
• Example 3-5: If you wished to have
$800 in a savings account at the end of
four years, and 5% interest we paid
annually, how much should you put into
the savings account?
• n = 4, F = $800, i = 5%, P = ?
• P = PV(5%,4,,800,0) = -$658.16
• You should use P = $658.16
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Economic Analysis Methods
• Three commonly used economic analysis
methods are
• Present Worth Analysis
• Annual Worth Analysis
• Rate of Return Analysis
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Present Worth Analysis
• Steps to do present worth analysis for a
single alternative (investment)
– Select a desired value of the return on
investment (i)
– Using the compound interest formulas bring
all benefits and costs to present worth
– Select the alternative if its net present worth
(Present worth of benefits – Present worth of
costs) ≥ 0
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Present Worth Analysis
• Steps to do present worth analysis for
selecting a single alternative (investment)
from among multiple alternatives
– Step 1: Select a desired value of the return on
investment (i)
– Step 2: Using the compound interest formulas
bring all benefits and costs to present worth
for each alternative
– Step 3: Select the alternative with the largest
net present worth (Present worth of benefits –
Present worth of costs)
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Present Worth Analysis
• A construction enterprise is investigating the
purchase of a new dump truck. Interest rate is
9%. The cash flow for the dump truck are as
follows:
• First cost = $50,000, annual operating cost =
$2000, annual income = $9,000, salvage value
is $10,000, life = 10 years. Is this investment
worth undertaking?
• P = $50,000, A = annual net income = $9,000 $2,000 = $7,000, S = 10,000, n = 10.
• Evaluate net present worth = present worth of
benefits – present worth of costs
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Present Worth Analysis
• Present worth of benefits = $9,000(PA,9%,10) =
$9,000(6.418) = $57,762
• Present worth of costs = $50,000 +
$2,000(PA,9%,10) - $10,000(PF,9%,10)=
$50,000 + $2,000(6..418) - $10,000(.4224) =
$58,612
• Net present worth = $57,762 - $58,612 < 0  do
not invest
• What should be the minimum annual benefit for
making it a worthy of investment at 9% rate of
return?
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Present Worth Analysis
• Present worth of benefits = A(PA,9%,10)
= A(6.418)
• Present worth of costs = $50,000 +
$2,000(PA,9%,10) - $10,000(PF,9%,10)=
$50,000 + $2,000(6..418) - $10,000(.4224)
= $58,612
• A(6.418) = $58,612  A = $58,612/6.418
= $9,312.44
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Cost and Benefit Estimates
• Present and future benefits (income) and
costs need to be estimated to determine
the attractiveness (worthiness) of a new
product investment alternative
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Annual costs and Income for a Product
• Annual product total cost is the sum of
annual material, labor, and overhead
(salaries, taxes, marketing expenses,
office costs, and related costs), annual
operating costs (power, maintenance,
repairs, space costs, and related
expenses), and annual first cost minus the
annual salvage value.
• Annual income generated through the
sales of a product = number of units sold
annuallyxunit price
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Rate of Return Analysis
• Single alternative case
• In this method all revenues and costs of
the alternative are reduced to a single
percentage number
• This percentage number can be compared
to other investment returns and interest
rates inside and outside the organization
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Rate of Return Analysis
• Steps to determine rate of return for a
single stand-alone investment
– Step 1: Take the dollar amounts to the same
point in time using the compound interest
formulas
– Step 2: Equate the sum of the revenues to the
sum of the costs at that point in time and
solve for i
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Rate of Return Analysis
• An initial investment of $500 is being
considered. The revenues from this
investment are $300 at the end of the first
year, $300 at the end of the second, and
$200 at the end of the third. If the desired
return on investment is 15%, is the project
acceptable?
• In this example we will take benefits and
costs to the present time and their present
values are then equated
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Rate of Return Analysis
• $500 = $300(PF, i, n=1) + 300(PF, i, n=2) +
$200(PF, i, n=3)
• Now solve for i using trial and error method
• Try 10%: $500 = ? $272 + $247 + $156 = $669
(not equal)
• Try 20%: $500 = ? $250 + $208 + $116 = $574
(not equal)
• Try 30%: $500 = ? $231 + $178 + $91 = $500
(equal)  i = 30%
• The desired return on investment is 15%, the
project returns 30%, so it should be
implemented
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