In order to replace an old process for producing a polymer that reduces friction loss in engines, two current-technology machines have been identified. Process K will have a first cost of $160,000, an operating cost of $7000 per month, and a salvage value of $50,000 after 1 year and $40,000 after its maximum 2-year life.

Process L will have a first cost of $210,000, an operating cost of $5000 per month, and salvage values of $100,000 after 1 year, $70,000 after 2 years, $45,000 after 3 years, and $26,000 after its maximum 4-year life. You have been asked to de­termine which process is better using a study pe­riod of (a) 1 year, (b) 2 years, and (c) 3 years. The company’s MARR is 12% per year compounded monthly. (Note: Problem 11.46 continues this analysis via spreadsheet with the added option of upgrading the in-place process.)


All evaluations are performed at an effective i = 1% per month

(a) For 1-year study period

AWK,1 = -160,000(A/P,1%,12) – 7000 + 50,000(A/F,1%,12)
= -160,000(0.08885) –7000 + 50,000(0.07885)
= $-17,274

AWL,1 = -210,000(A/P,1%,12) – 5000 + 100,000(A/F,1%,12)
= -210,000(0.08885) –5000 + 100,000(0.07885)
= $-15,774

Process L is better
(b) For 2-year study period

AWK,2 = -160,000(A/P,1%,24) – 7000 + 40,000(A/F,1%,24)
= -160,000(0.04707) –7000 + 40,000(0.03707)
= $-13,048

AWL,2 = -210,000(A/P,1%,24) – 5000 + 70,000(A/F,1%,24)
= -210,000(0.04707) –5000 + 70,000(0.03707)
= $-12,290

Process L is better

(c) For 3-year study period, repurchase process K machine for 1 year after 24 months

AWK,3 = -160,000(A/P,1%,36) – 7000
+ (-160,000 + 40,000)(P/F,1%,24)(A/P,1%,36) + 50,000(A/F,1%,36)
= -160,000(0.03321) – 7000 – 120,000(0.7876)(0.03321)
+ 50,000(0.02321)
= $-14,292

AWL,3 = -210,000(A/P,1%,36) – 5000 + 45,000(A/F,1%,36)
= -210,000(0.03321) –5000 + 70,000(0.02321)
= $-10,349

Process L is still better

Trades & Technology

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A delivery car had a first cost of $30,000, an annual operating cost of $12,000, and an estimated $4000 salvage value after its 6-year life. Due to an economic slowdown, the car will be retained for only 2 years and must be sold now as a used vehicle. (a) At an interest rate of 10% per year, what must the market value of the 2-year-old vehicle be in order for its AW value to be the same as the AW for a full 6-year life cycle? (b) Compare your answer in (a) with the first cost and expected salvage after 6 years. Is the required market value a reasonable one, in your opinion?

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