Two methods can be used to produce expansion an­chors. Method A costs $80,000 initially and will have a $15,000 salvage value after 3 years. The op­erating cost with this method will be $30,000 in year 1, increasing by $4000 each year. Method B will have a first cost of $120,000, an operating cost of $8000 in year 1, increasing by $6500 each year, and a $40,000 salvage value after its 3-year life. At an interest rate of 12% per year, which method should be used on the basis of a present worth analysis?

What will be an ideal response?


PWA = -80,000 – [30,000(P/A,12%,3) + 4000(P/G,12%,3)] + 15,000(P/F,12%,3)
= -80,000 – [30,000(2.4018) + 4000(2.2208)] + 15,000(0.7118)
= $-150,260

PWB = -120,000 – [8,000(P/A,12%,3) + 6500(P/G,12%,3)] + 40,000(P/F,12%,3)
= -120,000 – [8,000(2.4018) + 6500(2.2208)] + 40,000(0.7118)
= $-125,178

Select Method B

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