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 
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% + (10.48556110) /(10.4855619.970511)*(20%19%)
i = 19% + 0.94% = 19.94%
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