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Machines that have the following cost are under consideration for a new manufacturing process. Which is the best alternative

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Answer #1

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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