In: Economics
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
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.