hw4 sol - CE 167: Engineering Economy Assignment 4 Solution...

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Unformatted text preview: CE 167: Engineering Economy Assignment 4 Solution Set Spring 2008 1) Alternative 1 has a 10 year life and alternative 2 has a 15 year life. Assume the cash flow to equivalent lives and choose 30 years to be the common multiple of lives. The cash flows will look as follows: Alt 1 NPV Alt 1 = -50,000 -10,000 (P/F, 10%, 5) -50,000 (P/F, 10%, 10) -10,000 (P/F, 10%, 15) -50,000 (P/F, 10%, 20) -10,000 (P/F, 10%, 25) = -50,000 -10,000 (0.6209) -50,000 (0.3855) -10,000 (0.2394) -50,000 (0.1486) -10,000 (0.0923) = $86,231 Alt 2 NPV Alt 2 = -B - B (P/F, 10%, 15) = - B - B (0.2394)= - $ 1.2394 B The two alternatives are equal when B= $ 69,575 Whenever the investment in Machine B is more than $ 69,575 alternative 1 is more economical. When B is less than $ 69,575 alternative 2 becomes more economical 5 10 15 20 25 30 $B $B 5 10 15 20 25 30 $50,000 $50,000 $50,000 $10,000 $10,000 $10,000 2) Sensitivity Analysis Alt 1 1 2 3 4 5 7 8 $20,000 A=$ 6,500 S=$ 4,000 NPV Alt 1 = -20,000 + (6,500)(P/A, 10%, 8) + (4,000)(P/F, 10%, 8)...
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This note was uploaded on 02/27/2011 for the course CE 167 taught by Professor Horvath during the Fall '08 term at University of California, Berkeley.

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hw4 sol - CE 167: Engineering Economy Assignment 4 Solution...

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