Question

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You have been asked to evaluate two alternatives, X and Y, that may increase plant capacity for manufacturing high-pressure hydraulic hoses. The parameters associated with each alternative have been estimated. Which one should be selected on the basis of a present worth comparison at an interest rate of 13% per year? Why is yours the correct choice?

Alternative	X	Y
First Cost	$-45,000	$-70,000
Maintenance cost, per Year	$-15000	$-6000
Salvage Value	$2,000	$8,000
Life	5 years	5 years

Answer 1

ANSWER:

I = 13%

N = 5 YEARS

a) alternative x:

pw = first cost + miantenance cost(p/a,i,n) + salvage value(p/f,i,n)

pw = -45,000 - 15,000(p/a,13%,5) + 2,000(p/f,13%,5)

pw = -45,000 - 15,000 * 3.517 + 2,000 * 0.5428

pw = -45,000 - 52,755 + 1,085.6

pw = -96,669.4

b) alternative y:

pw = first cost + miantenance cost(p/a,i,n) + salvage value(p/f,i,n)

pw = -70,000 - 6,000(p/a,13%,5) + 8,000(p/f,13%,5)

pw = -70,000 - 6,000 * 3.517 + 8,000 * 0.5428

pw = -70,000 - 21,102 + 4,342.4

pw = -86,759.6

since the present worth of alternative y is more then that of alternative x , therefore we will choose alternative y.