An engineering company has two alternatives for improving a production line. Alternative A has first costs...

An engineering company has two alternatives for improving a production line. Alternative A has first

costs of $3,000, a salvage value of $500, and a service life of 3 years. Alternative B has equipment

costs of $4,500, a salvage value of $1,000, and a service life of 6 years. Use a present worth analysis

to determine which alternative is economically superior with i=10%. Your solution will be closest to:

a. Alternative A @ - $3,300

b. Alternative B @ - $3,300

c. Alternative A @ -$3,950

d. Alternative B @ -$3,950

Solutions

Expert Solution

Ans: Alternative B @ -$3,950

Explanation:

Since the life of both the alternative is different, so, the LCM of 3 & 6 will be taken for calculation of PW. The LCM is 6. In 6 years, alternative A can be installed 2 times.

PW of alternative A = -3,000 - [(3,000 - 500) (P/F, 10%, 3)] + 500(P/F, 10%, 6)

= -3,000 - 2,500(0.7513) + 500(0.5645)

= -3,000 - 1878.25 + 282.25

= -4,596

PW of alternative B = -4,500 + 1,000(P/F, 10%, 6)

= -4,500 + 1,000(0.5645)

= -4,500 + 564.5

= -3,935.5 [it is closest to $3,950]

So, alternative B will be selected.

Rahul Sunny answered 1 year ago

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