Ch9Solutions

# Ch9Solutions - Chapter 9 9.1 In a replacement study the...

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Unformatted text preview: Chapter 9 9.1 In a replacement study, the in-place asset is referred to as the defender 9.2 A consultant’s viewpoint assumes that neither alternative is currently owned. 9.3 The P value for the defender is its current market value . 9.4 (a) The assets annual worth’s over their own life cycle can be used when the study period is unlimited, as long as assets similar to the ones under comparison (including the used defender ) are likely to be available in the future, (b) When the study period is not an even multiple of the asset’s lives, the AW over one life cycle cannot be used because one or more if the assets would not end exactly when the study period ends, rendering the cost estimates wrong, (c) When the study period is an even multiple of the asset’s lives, the AW over one life cycle can be used, as long as assets similar to the ones under comparison (including the used defender ) are likely to be available through the end of the study period. 9.5 AW 1 = -10,000(A/P,10%,1) – 1000 + 7000(A/F,10%,1) = \$-5000 AW 2 = -10,000(A/P,10%,2) – 1000(P/F,10%,1)(A/P,10%,2) + (5000 – 1200)(A/F,10%,2) = \$-4476 AW 3 = -10,000(A/P,10%,3) – [1000(P/F,10%,1) +1200(P/F,10% 2)](A/P,10%,3) + (4200 – 1500)(A/F,10%,3) = \$-3970 AW 4 = -10,000(A/P,10%,4) – [1000(P/F,10%,1) +1200(P/F,10% 2) + 1500(P/F,10%,3)](A/P,10%,4) + (3000 – 2000)(A/F,10%,4) = \$-3894 AW 5 = -10,000(A/P,10%,5) – [1000(P/F,10%,1) + 1200(P/F,10% 2) + 1500(P/F,10%,3) + 2000(P/F,10%,4)](A/P,10%,5) + (2000 – 3000)(A/F,10%,5) = \$-3961 Therefore, ESL is 4 years with AW = \$-3894 9.6 AW 1 = -345,000(A/P,10%,1) – 148,000 + 140,000(A/F,10%,1) = \$-387,500 AW 2 = -345,000(A/P,10%,2) – 148,000 + 140,000(A/F,10%,2) = \$-280,119 AW 3 = -345,000(A/P,10%,3) – 148,000 + 140,000(A/F,10%,3) = \$-244,434 AW 4 = -345,000(A/P,10%,4) – 148,000(P/A,10%,3)(A/P,10%,4) -210,000(P/F,10%,4)(A/P,10%4) = \$-270,197 AW 5 = -345,000(A/P,10%,5) – 148,000(P/A,10%,3)(A/P,10%,5) -210,000(P/A,10%,2)(P/F,10%,3)(A/P,10%,5) = \$-260,337 AW 6 = -345,000(A/P,10%,5) – 148,000(P/A,10%,3)(A/P,10%,6) -210,000(P/A,10%,3)(P/F,10%,3)(A/P,10%,6) = \$-253,813 Therefore, ESL is 3 years with AW = \$-244,434 9-1 9.7 AW 1 = -65,000(A/P,10%,1) – 50,000 + 30,000(A/F,10%,1) = \$-91,500...
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Ch9Solutions - Chapter 9 9.1 In a replacement study the...

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