A facility engineer is looking at two proposed alternatives to install anti-poll

Question

A facility engineer is looking at two proposed alternatives to install anti-pollution control equipment mandated by the State of California. Assume equal lives of 25 years, no salvage values, and interest at 25%. The first cost of Alternative A is $20,000 with an annual expenditure of $3,000. The first cost of Alternative B is $25,000 with an annual expenditure of $2,500.

Make the comparison on the basis of present worth. Which alternative is best?

Make the comparison on the basis of capitalized cost. Which alternative is best?

Make the comparison on the basis of annual cost. Which alternative is best?

Explanation / Answer

Present Worth:

Alternative A:

PW = -$20,000 - $2,000(P/A,0.25,25)

PW = -$20,000 -$2,000*3.9848

PW = -$20,000 - $7,969.6 = -$27,969.6

Alternative B:

PW = -$25,000 - $2,500*3.9848

PW = -$25,000 - $9,962 = -$34,962

Alternative A has the least value and hence alternative A is preferred.

Capitalized Cost (CC):

Alternative A:

CC = -$20,000 - $2,000 / 0.25

CC = -$20,000 - $8,000 = -$28,000

Alternative B:

CC = $25,000 - $2,500 / 0.25

CC = -$25,000 - $10,000 = $35,000

Alternative A has the least cost and hence alternative A is the best.

Annual Cost (EUAC):

Alternative A:

EUAC = NPV / (1-1/R^t) / R

EUAC = 20,000 / (1 – 1/0.25^25) / 0.25

EUAC = 20,000 / (1 – 1/8.8817) / 0.25

EUAC = 20,000 / (1-0.1125) /0.25

EUAC = 20,000 / 0.8875 / 0.25

EUAC = 20,000 / 3.55 = $5,633.80

Alternative B:

EUAC = 25,000 / (1-0.1125) / 0.25

EUAC = 25,000 / 3.35 = $7,042.25

Annual cost is lower for alternative A, hence alternative A is preferred.

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