Engineering Economics
John Ayers
September 17, 2004
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Engineering Economics
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•
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•
•
•
•
•
•
Why is it important?
Value and Interest
Cash Flow Diagrams and Patterns
Equivalence of Cash Flow Patterns
Evaluating Alternatives
Break-Even Analysis
Income Tax and Depreciation
Inflation
Conclusion
Why do we care about
Engineering Economics?
• Engineering designs are intended to produce good
results.
• They are accompanied by undesirables (costs).
• If outcomes are evaluated in dollars, and “good” is
defined as profit, then decisions will be guided by
engineering economics.
• This process maximizes goodness only if all
outcomes are anticipated and can be monetized.
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Value and Interest
• The “value” of money depends on the
amount and when it is received or spent.
Example: What amount must be paid to settle a current debt of $1000 in
two years at an interest rate of 8% ?
Solution: $1000 (1 + 0.08) (1 + 0.08) = $1166
$1000
1
2
$1166
Cash Flow Diagrams
P-Pattern
F-Pattern
A-Pattern
G-Pattern
“present”
1
2
3
n
“future”
1
2
3
n
“annual”
1
2
3
n
“gradient”
1
2
3
n
Equivalence of Cash Flow
Patterns
To Find Given Multiply By
F
P
( F / P )in
P
F
( P / F )in
A
P
( A / P )in
A
G
( A / G )in
Formula
(1  i) n
1
(1  i) n
i (1  i ) n
(1  i ) n  1
1
n

i (1  i ) n  1
Example: A new circuit board component insertion tool will
save $50,000 in production costs each year and will have a
life of seven years. What is the highest price that can be
justified for the tool using a 12% interest rate?
50k 50k 50k 50k
50k 50k 50k
Solution:
1
P
2
3
4
5
6
7
n
(
1

i
)
1
%
P  ( P / A)12
A

A
7
n
i (1  i )
(1  0.12) 7  1

$50,000
7
0.12(1  0.12)
 4.56  $50,000  $228k
Evaluating Alternatives
•
•
•
•
•
Annual Equivalent Cost Comparisons
Present Equivalent Cost Comparisons
Incremental Approach
Rate of Return Comparisons
Benefit/Cost Comparisons
Minimum Attractive Rate of Return (MARR): The lowest rate of return
that the organization will accept.
Annual Equivalent Cost
Comparison
• Incomes are converted to an A-pattern.
• Costs are converted to an A-pattern.
• The costs are subtracted from the incomes to
determine the ANEV.
• Mutually Exclusive Alternatives – choose the one
with highest ANEV
• Independent Alternatives – choose all with
positive ANEV
ANEV: Annual Net Equivalent Value
Example: A new circuit board component insertion tool is
needed. Which should you buy?
Model
Price
Annual Maintenance Salvage Value Life
JACO
$220k
$20k
$30k
10 years
Cheepo
$100k
$35k
0
5 years
Solution: The ANEV is calculated for each:
JACO:
%
10%
ANEV  ( A / P )10
220
k

20
k

(
A
/
F
)
10
10 30k
 35.8k  20k  1.9k  53.9k
Cheepo:
%
ANEV  ( A / P )10
5 100k  35k
 $61.4k
JACO
Present Equivalent Cost
Comparison
• Incomes and costs are converted to P-patterns.
• The costs are subtracted from the incomes to
determine the PNEV.
• Mutually Exclusive Alternatives – choose the one
with highest PNEV
• Independent Alternatives – choose all with
positive PNEV
PNEV: Present Net Equivalent Value, also called “life cycle cost,”
“present worth,” “capital cost,” and “venture worth.”
Incremental Approach
• For a set of mutually exclusive alternatives, only
the differences in amounts need to be considered.
Model
Price
Annual Maintenance Salvage Value Life
JACO
$220k
$20k
$30k
10 years
Cheepo
$100k
$35k
0
5 years
JACO- Cheepo:
%
10%
10%
PNEV  120k  ( P / A)10
15
k

(
P
/
F
)
100
k

(
P
/
F
)
10
5
10 30k
 120k  92.2k  62.1k  11.6k  $45.9k
JACO
Rate of Return Method
• ANEV or PNEV is formulated
• From this, we solve for the interest rate that will
give zero ANEV or PNEV
• This interest rate is the ROR of the alternative
• For mutually exclusive alternatives, the one with
the highest ROR is chosen
• For independent alternatives, all with a ROR
greater than MARR are accepted
ROR: Rate of Return on Investment
Benefit/Cost Comparisons
• The benefit/cost ratio is determined from
B
uniform net annual benefits

C annual equivalent of initial cost
• For mutually exclusive alternatives, the one
with the highest B/C is chosen.
• For independent alternatives, all with B/C >
1 are accepted.
The MARR is used to determine the numerator (benefits).
Break-Even Analysis
• Break-even point: the value of an independent
variable such that two alternatives are equally
attractive.
• For values above the break-even point, one
alternative is preferred.
• For values below the break-even point, the other is
preferred.
• Break-even analysis is useful when dealing with a
changing variable (such as MARR).
Income Tax and Depreciation
• Businesses pay the IRS a tax:
 gross revenue - operating costs 
TAX  R  

 - interest paid - depreciation 
• Depreciation: method of charging the initial cost
of an asset against more than one year.
• An asset is depreciable if :
– It is used to produce income,
– Has a life greater than one year, but
– Decays, wears out, becomes obsolete, or gets used up.
ACRS: Accelerated Cost Recovery System, used by IRS since 1980.
Inflation
• The buying power of money changes with time.
• Inflation, if anticipated, can be put to good use by
fixing costs and allowing income to rise by
– Entering long-term contracts for materials or wages
– Purchasing materials long before they are needed
– Stockpiling product for sale later.
Conclusion
• For-profit enterprises exist to make money.
• Non-profit entities also make decisions to
maximize the goodness of outcomes by
assigning dollar values.
• Your engineering decisions will be shaped
by economics.
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