Chapter 4
More Interest Formulas
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EGR 403 Capital Allocation Theory
Dr. Phillip R. Rosenkrantz
Industrial & Manufacturing Engineering Department
Cal Poly Pomona
EGR 403 - The Big Picture
• Framework: Accounting & Breakeven Analysis
• “Time-value of money” concepts - Ch. 3, 4
• Analysis methods
–
–
–
–
Ch. 5 - Present Worth
Ch. 6 - Annual Worth
Ch. 7, 8 - Rate of Return (incremental analysis)
Ch. 9 - Benefit Cost Ratio & other techniques
• Refining the analysis
– Ch. 10, 11 - Depreciation & Taxes
– Ch. 12 - Replacement Analysis
EGR 403 - Cal Poly Pomona - SA6
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Components of Engineering
Economic Analysis
• Calculation of P and F are fundamental.
• Some problems are more complex and
require an understanding of added
components:
– Uniform series.
– Arithmetic or geometric gradients.
– Nominal and effective interest rates (covered in
presentation #5 on Chapter 3).
– Continuous compounding.
EGR 403 - Cal Poly Pomona - SA6
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Uniform Payment Series
Capital Recovery Factor
• The series of uniform payments that will
recover an initial investment.
A = P(A/P, i, n)
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Uniform Payment Series
Compound Amount Factor F
• The future value of an investment based on
periodic, constant payments and a constant
interest rate.
F = A(F/A, i, n)
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Example 4-1
1000
At 5%/year
Year
Cash in
0
0
1
$500
2
$500
3
$500
4
$500
5
$500
Cash out
0
500
1
500
2
500
3
500
4
500
5
0
1
2
3
4
5
0
-1000
-2000
-2763
-3000
Year Cash In
Cash Out
-$2763
F = $500(F/A, 5%, 5) = $500(5.526) = $2763
EGR 403 - Cal Poly Pomona - SA6
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Uniform Payment Series
Sinking Fund Factor
• The constant periodic amount, at a constant interest
rate that must be deposited to accumulate a future
value.
A = F(A/F, i, n)
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Uniform Payment Series
Present Worth Factor
The present value of a series of uniform
future payments.
P = A(P/A, i, n)
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Example 4-6
F’ = $100(F/A, 15%, 3) = $347.25
F’’ = $347.25(F/P, 15%, 2) = $459.24
Year
Cash flow
1
$100
2
$100
3
$100
4
$0
5
F
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Example 4-7
Finding the Present
Value (P) for each
cash flow is
sometimes the easiest
way to find the
equivalent P.
Year
Cash flow
0
P
1
0
2
$ 20
3
$ 30
4
$ 20
P = $20(P/F, 15%, 2) + $30(P/F, 15%, 3) +
$20(P/F, 15%, 4) = $46.28
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Arithmetic Gradient
• A uniform increasing
amount.
• The first cash flow is
always equal to zero.
• G = the difference
between each cash
amount.
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G = $10
11
Arithmetic Gradient combined
with a Uniform Series
• Decompose the cash flows into a uniform series and a
pure gradient. Then add or subtract the Present Value of
the gradient to the Present Value of the Uniform series
• Example 4-8: Use P/G factor to find present value of
the pure gradient portion of the cash flow
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Arithmetic Gradient
Uniform Series Factor
A pure gradient (uniformly increasing amount) can
also be converted into the equivalent present value
of uniform series:
AG = G(A/G, i, n)
See Example 4-9: Notice that the uniform series
portion of the cash flow was subtracted to separate
the pure gradient.
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Geometric Series
Present Worth Factor
Sometimes cash flows increase at a constant rate
rather than a constant amount. Inflation, for
example, could be reflected in a cash flow
diagram that way. The equivalent present value of
a geometrically increasing amount. g = the rate of
increase (e.g., .05)
P = A(P/A, g, i, n) where (P/A, g, i, n) must be computed
from equation 4-30 or 4-31
• Example 4-12 uses g = .10 and i = .08
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