Question

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A) White Oaks Properties builds strip shopping centers and small malls. The company plans to replace its refrigeration, cooking, and HVAC equipment with newer models in one entire center built 12 years ago. 12 years ago, the original purchase price of the equipment was $800,000 and the operating cost has averaged $290,000 per year. Determine the equivalent annual cost of the equipment if the company can now sell it for $224,000. The company’s MARR is 20% per year.

B) You have two machines under consideration for an improved automated wrapping process for Snickers Fun Size candy bars as detailed below. Compare them on the basis of annual worth at i = 14%.

Machine	C	D
First Cost	$-35,000	$-55,000
Annual Cost per Year	$-10,000	$-15,000
Salvage Value	$10,000	$10,000
Life	3 years	6 years

Answer 1

Answer to Question – A

Purchase price (initial cost) = 800,000

Annual Operating Cost = 290,000

Salvage Value =224,000

MARR = 20%

Life = 12 years

Calculate the Equivalent Annual Cost

AW = -800,000 (A/P, 20%, 12) – 290,000 + 224,000 (A/F, 20%, 12)

AW = -800,000 (0.2253) – 290,000 + 224,000 (0.0253) = -464,572.8

If we ignore the negative sign (taking the absolute value) it will be 464,572.5

Alternatively, we can first calculate the PW

PW = -800,000 – 290,000 (P/A, 20%, 12) + 224,000 (P/F, 20%, 12)

PW = -800,000 – 290,000 (4.4392) + 224,000 (0.1122) = -2,062,235.2

AW = PW (A/P, 20%, 12)

AW = -2,062,235.2 (0.2253) = -464,622

Answer to Question – B

Compare them on the basis of annual worth at i = 14%.

Machine	C	D
First Cost	$-35,000	$-55,000
Annual Cost per Year	$-10,000	$-15,000
Salvage Value	$10,000	$10,000
Life	3 years	6 years

The life of the alternatives is unequal. For the evaluation purpose first convert the unequal life into equal life using the common multiple method. The LCM of 3 years and 6 years is 6 years. Therefore, Machine C is repeated two times, However, Machine D need not to be repeated.

For the calculation of AW of both the machines we need to calculate the PW

PW of Machine C

PW = -35,000 – 35,000 (P/F, 14%, 3) – 10,000 (P/A, 14%, 6) + 10,000 (P/F, 14%, 3) + 10,000 (P/F, 14%, 6)

PW = -35,000 – 35,000 (0.6750) – 10,000 (3.8887) + 10,000 (0.6750) + 10,000 (0.4556)

PW = -86,206

AW of Machine C

Annual Worth = PW (A/P, 14%, 6)

Annual Worth = -86,206 (0.2572) = -22,172

PW of Machine D

PW = – 55,000 – 15,000 (P/A, 14%, 6) +10,000 (P/F, 14%, 6)

PW = – 55,000 – 15,000 (3.8887) +10,000 (0.4556) = -108,774.5

AW of Machine C

Annual Worth = PW (A/P, 14%, 6)

Annual Worth = -108,774.5 (0.2572) = -27,976.8

On the basis of the above AW comparison, we should select the Machine C because it has less cost.