Question

A farm equipment manufacturer is considering two types of controllers for one part of its production facility. Alternative 1 will have an equipment cost of $80,000, an installation cost of $25,000, annual M&O costs of $50,000, and a $5000 salvage value after 4 years. Alternative 2 will have an initial cost of $170,000 (including installation), an annual operating cost of $35,000, and $10,000 salvage value after its 8 year life. At an interest rate of 12% per year, the present worths of the two alternatives are closest to

a)PW1=-$362,500, PW2 = -$341,700

b)PW1=-$253,700, PW2 = -$339,800

c)PW1 = -$414,900, PW2 = -$339,800

d)PW1 = -$339,800, PW2 = -$253,700

Answer 1

Find the PW of both methods

PW(1) = -105000 – 105000(P/F, 12%, 4) – 50000(P/A, 12%, 8) + 5000(P/F, 12%, 4) + 5000(P/F, 12%, 8)

= -105000 – 105000*0.63552 –50000*4.9676 + 5000*0.63552 + 5000*0.40388

= - 414912

PW (2) = -170000 – 35000(P/A, 12%, 8) + 10000(P/F, 12%, 8)

= -170000 – 35000*4.9676 + 10000*0.40388

= -339827

Select PW1 = -$414,900, PW2 = -$339,800