Question

The problem below must be solved using rate of return analysis. Type B equipment has an installed cost of $9,000, a uniform annual benefit of $1,600, a salvage value of $3000, and a useful life of 6 years.

Type A equipment has an installed cost of $10,000, a uniform annual benefit of $1,700, a salvage value of $3800, and a useful life of 6 years. If the MARR is 7%, which type of equipment should be selected, A or B?

Justify your answer.

(answer: ∆i* = 7.22%; A since ∆i* > MARR)

Please show all work by hand to arrive at answer.

Answer 1

Increment A over B has PW = -10000 + 9000 + (1700 - 1600)(P/A, i%, 6) + (3800 - 3000)(P/F, i%, 6)

= -1000 + 100(P/A, i%, 6) + 800(P/F, i%, 6)

= -1000 + 100*((1+i)^6-1)/(i(1+i)^6) + 800(1+i)^-6

For finding IRR ∆i*, use interpolation so that PW = 0

When i* = 7%, PW = -1000 +100*((1.07)^6-1)/(0.07(1.07)^6) + 800(1.07)^-6 = 9.727

When i* = 8%, PW = -1000 +100*((1.08)^6-1)/(0.08(1.08)^6) + 800(1.08)^-6 = -33.576

Interpolation gives ∆i* = 7% + (8% - 7%)*(9.727/(9.727 + 33.576)) = 7.224%

Since ∆i* > MARR, A is accepted.