Homework 4 Solution

Homework 4 Solution - HW 5 EE366 McCann Park 4th Edition...

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HW 5 EE366 McCann Park 4 th Edition 5.38 (a): For mutually exclusive alternative with differing lifetimes, we either need a co- terminated study period or the indefinite service life with the repeatability assumption. (b) Using the indefinite service life assumption and the AEW analysis approach. AEW1 = -\$900(A/P, 10%, 3) + 200(A/F, 10%, 3) – 400 AEW1 = -\$361.89 + 60.42 – 400 = -\$701.47 AEW2 = -\$1800(A/P, 10%, 8) + 500(A/F, 10%, 8) – 300 AEW2 = -\$337.32 + 43.70 – 300 = -\$593.62 Choose A2 because it has the lower AEW cost. (c) The AEW of alternative A1 is unchanged. It has a three year life. Set the AEW of A1 equal to the AEW of the modified A2 and use that to compute the required salvage value. -\$701.47 = -\$1800(A/P, 10%, 3) + S(A/F, 10%, 3) – 300 -\$701.47 = -\$723.78 + 0.3021S – 300 S = \$322.31/0.3021 = \$1,067 5.39 (a): AEW1 = -\$18k(A/P, 12%, 5) + 2k(A/F, 12%, 5) – 2k AEW1 = -\$4.993 + 0.315 – 2 = \$6,678 AEW2 = -\$15k(A/P, 12%, 3) + 1k(A/F, 12%, 3) – 2.1k AEW2 = -\$6.245 + 0.296 – 2.1 = -\$8,048 Choose B1 for lower annual cost. (b) For a 10 year study period, repeat B1 twice and repeat B2 three times in full plus a final replacement cycle of 1 year. We can use NPW because the study period is the same for both alternatives.

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