Question

Question 3 (20 points) Compare the two mutually exclusive alternatives on the basis of their capitalized costs at i= 10% per year. First cost, $ Annual cost, $/year Salvage value, $ Life, years -110,000 –54,000 9,000 - 700,000 -15.000 2,000,000

Answer 1

Given,

MARR = 10% per year

Alternative A:

First cost = -$110,000

Annual Cost = -$54,000

Salvage value = $9,000

Useful life = 5 years

From the compound interest table, we obtain

(A/P, 10%, 5) = 0.2638

(A/F, 10%, 5) = 0.1638

Annual Worth of alternative A = -$110,000(A/P, 10%, 5) -$54,000 + $9,000(A/F, 10%, 5)

= -$110,000*0.2638 -$54,000 + $9,000*0.1638

= -$81,543.8

Capitalized cost of alternative A = -$81,543.8(P/A, 10%, ∞) = -$81,543.8 * 1/0.10 = -$815,438

Alternative B:

First cost = -$700,000

Annual Cost = -$15,000

Salvage value = $2,000,000

Useful life = ∞

Annual Worth of alternative B = -$700,000(A/P, 10%, ∞) -$15,000 + $2,000,000(A/F, 10%, ∞)

= -$700,000* 0.10 -$15,000 + $2,000,000 * 0

= -$85,000

Capitalized cost of alternative B = -$85,000(P/A, 10%, ∞) = -$85,000*1/0.10 = -$850,000

As the capitalized cost of alternative A is higher than that of alterative B (please note that the negative sign is included while making the comparison), alternative A is selected.