ClassHW_4

# ClassHW_4 - Hw-4 Solutions 1(6.3 Cost of heat loss without...

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Hw-4 Solutions 1. (6.3) Cost of heat loss without insulation is \$2,000 per year. For 1-inch thick insulation (Alternative 1) Cost of heat loss = 12% x 2,000 = 240 Cost of insulation = 0.6 x 1,000 = 600 For 2-inch thick insulation (Alternative 2) Cost of heat loss = 8% x 2,000 = 160 Cost of insulation = 1.1 x 1,000 = 1,100 This is a cost alternative, so choose the insulation which minimizes total cost. By the PW method, PW 1 = 600 240(P/A,6%,10) = \$2,366.42 PW 2 = 1,100 160(P/A,6%,10) = \$2,277.62 Select 2-inch thick insulation . 2. (6.7) PW A (20%) = 28,000 + (23,000 - 15,000)(P/A,20%,10) + 6,000(P/F,20%,10) = \$6,509 PW B (20%) = 55,000 + (28,000 - 13,000)(P/A,20%,10) + 8,000(P/F,20%,10) = \$9,180 PW C (20%) = 40,000 + (32,000 - 22,000)(P/A,20%,10) + 10,000(P/F,20%,10) = \$3,540 This is an investment alternative, so select the project that maximizes profit. Select Alternative B to maximize present worth. Note : If you were to pick the alternative with the highest total IRR, you would have incorrectly selected Alternative A.

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3. (6.13) Alternative ER1 is a geometric gradient series. PW ER1 (12%) = 98,600 + 25,800[1 (1.06/1.12) 6 ]/(0.12 0.06) = \$22,373.65 Alternative ER2 is a uniform gradient series. To find PW, we use the usual trick of an annuity and the uniform gradient series starting at EOY 2 with G = \$150.
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