Question

A firm is considering the purchase of a new machine. You have been asked to evaluate the following data. MARR = 10%. What is the incremental rate of return ()? (Use a starting value of 18%.) Enter your answer as a percent, rounded to two decimal places. For example, you would enter 2.34 for 2.34%.

	Machine 1	Machine 2
Initial Cost	$65,000	$55,000
Salvage Value	$5,000	$2,000
Net Annual Benefit	$8,000	$6,000
Useful life (years)	30	30

Answer 1

Incremental initial cost (1 - 2) = 65000 - 55000 = 10000

Incremental salvage (1 - 2) = 5000 - 2000 = 3000

Incremental annual benefit (1 - 2) = 8000 - 6000 = 2000

Let incremental IRR be i%, then

2000*(P/A,i%,30) + 3000*(P/F,i%,30) = 10000

Dividing by 1000

2*(P/A,i%,30) + 3*(P/F,i%,30) = 10

using trail and error method

When i = 18%, value of 2*(P/A,i%,30) + 3*(P/F,i%,30) = 2*5.516806 + 3*0.006975 = 11.054537

When i = 19%, value of 2*(P/A,i%,30) + 3*(P/F,i%,30) = 2*5.234658 + 3*0.005415 = 10.485561

When i = 20%, value of 2*(P/A,i%,30) + 3*(P/F,i%,30) = 2*4.978936 + 3*0.004213 = 9.970511

using interpolation

i = 19% + (10.485561-10) /(10.485561-9.970511)*(20%-19%)

i = 19% + 0.94% = 19.94%