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EGR 403 Capital Allocation Theory Dr. Phillip R. Rosenkrantz

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### Chapter 4More Interest FormulasClick here for Streaming Audio To Accompany Presentation (optional)

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

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

Uniform Payment SeriesCapital Recovery Factor

- The series of uniform payments that will recover an initial investment.

A = P(A/P, i, n)

EGR 403 - Cal Poly Pomona - SA6

Uniform Payment SeriesCompound 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)

EGR 403 - Cal Poly Pomona - SA6

Year

Cash in

Cash out

0

0

1

$500

2

$500

3

$500

4

$500

5

$500

-$2763

Example 4-1

At 5%/year

F = $500(F/A, 5%, 5) = $500(5.526) = $2763

EGR 403 - Cal Poly Pomona - SA6

Uniform Payment SeriesSinking 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)

EGR 403 - Cal Poly Pomona - SA6

Uniform Payment SeriesPresent Worth Factor

The present value of a series of uniform future payments.

P = A(P/A, i, n)

EGR 403 - Cal Poly Pomona - SA6

Year

Cash flow

1

$100

2

$100

3

$100

4

$0

5

F

Example 4-6F’ = $100(F/A, 15%, 3) = $347.25

F’’ = $347.25(F/P, 15%, 2) = $459.24

EGR 403 - Cal Poly Pomona - SA6

Year

Cash flow

0

P

1

0

2

$ 20

3

$ 30

4

$ 20

Example 4-7Finding the Present Value (P) for each cash flow is sometimes the easiest way to find the equivalent P.P = $20(P/F, 15%, 2) + $30(P/F, 15%, 3) + $20(P/F, 15%, 4) = $46.28

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

EGR 403 - Cal Poly Pomona - SA6

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

EGR 403 - Cal Poly Pomona - SA6

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 403 - Cal Poly Pomona - SA6

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 403 - Cal Poly Pomona - SA6

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