Homework Answers
i = 10%/ 2 = 5% per semi annual period
t= 4*2 = 8 semi annual periods
incremental initial cost (B-A) = 72000 - 53000 = 19000
incremental operating cost (B-A) = 10000 - 8000 = 2000
incremental income (B-A) = 20000 - 15000 = 5000
incremental income gradient (B-A) = 200 - 200 = 0
incremental salvage (B-A) = 11000 - 9000 = 2000
Let semiannual IRR be i%, then
-19000 + (5000-2000)*(P/A,i%,8) + 2000*(P/F,i%,8) =0
3000*(P/A,i%,8) + 2000*(P/F,i%,8) = 19000
dividing by 1000
3*(P/A,i%,8) + 2*(P/F,i%,8) = 19
using trail and error method
When i = 7%, value of 3*(P/A,i%,8) + 2*(P/F,i%,8) = 3*5.971299 + 2*0.582009 = 19.0779
When i = 8%, value of 3*(P/A,i%,8) + 2*(P/F,i%,8) = 3*5.746639 + 2*0.540269 = 18.3204
using interpolation
i = 7% + (19.0779 - 19) /(19.0779 - 18.3204) *(8%-7%)
i = 7% + 0.102% = 7.1% ~ 7%
option b is correct answer
As semiannual IRR > Semiannual MARR, Machine B should be selected
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