1. Chapter 4 More Interest Formulas Click here for Streaming Audio To Accompany Presentation (optional)
EGR 403 Capital Allocation Theory
Dr. Phillip R. Rosenkrantz
Industrial & Manufacturing Engineering Department
Cal Poly Pomona

2. EGR 403 - Cal Poly Pomona - SA6
EGR 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 Replacement Analysis
EGR Cal Poly Pomona - SA6

3. 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.
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4. 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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5. 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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6. EGR 403 - Cal Poly Pomona - SA6
Example 4-1
At 5%/year
$500
Cash in
-$2763 5 4 3 2 1
Cash out
Year
F = $500(F/A, 5%, 5) = $500(5.526) = $2763
EGR Cal Poly Pomona - SA6

7. 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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8. 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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9. EGR 403 - Cal Poly Pomona - SA6
Example 4-6
F’ = $100(F/A, 15%, 3) = $347.25
F’’ = $347.25(F/P, 15%, 2) = $459.24 F 5 $0 4
$100 3 2 1
Cash flow
Year
EGR Cal Poly Pomona - SA6

10. EGR 403 - Cal Poly Pomona - SA6
Example 4-7 Finding the Present Value (P) for each cash flow is sometimes the easiest way to find the equivalent P.
$ 20 4
$ 30 3 2 1 P
Cash flow
Year
P = $20(P/F, 15%, 2) + $30(P/F, 15%, 3) + $20(P/F, 15%, 4) = $46.28
EGR Cal Poly Pomona - SA6

11. EGR 403 - Cal Poly Pomona - SA6
Arithmetic Gradient
A uniform increasing amount.
The first cash flow is always equal to zero.
G = the difference between each cash amount.
G = $10
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12. 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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13. 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.
EGR Cal Poly Pomona - SA6

14. 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
EGR Cal Poly Pomona - SA6