Question

4. A slip sheet manufacturer is considering two machines. An engineer is asked to perform analyses to select the best machine. She prepares the following information for the evaluation. All machines have a useful life of 5 years. The company's MARR is 4% per year. Please use the IRR method to determine which machine should be selected. (25 points) Machine Initial investment Annual expenses Annual revenue Salvage value IRR (%) X $52,000 $5750 $15,250 $23,400 9.10 Y $37,000 $8050 $15,750 $5550 5.62

Answer 1

MARR = 4%

t = 5 yrs

We need to find incremental cash flow between (X-Y)

Machine	X	Y	X-Y
Initial investment	52000	37000	= 52000 - 37000 = 15000
Annual expenses	5750	8050	= 5750 - 8050 = - 2300 (Saving)
Annual Revenue	15250	15750	= 15250 - 15750 = -500 (Loss)
Salvage Value	23400	5550	= 23400 - 5550 = 17850
IRR(%)	9.10	5.62

Now let incremental IRR be i%

PW of incremental cash flow = -15000 + 2300 * (P/A, i%,5) - 500 * (P/A, i%,5) + 17850 * (P/F, i%,5) = 0

-15000 + 1800 * (P/A, i%,5) + 17850 * (P/F, i%,5) = 0

1800 * (P/A, i%,5) + 17850 * (P/F, i%,5) = 15000

Using trail and error method

When i = 10%, value of 1800 * (P/A, i%,5) + 17850 * (P/F, i%,5) = 17906.86

When i = 12%, value of 1800 * (P/A, i%,5) + 17850 * (P/F, i%,5) = 16617.166

When i = 14%, value of 1800 * (P/A, i%,5) + 17850 * (P/F, i%,5) = 15450.276

When i = 15%, value of 1800 * (P/A, i%,5) + 17850 * (P/F, i%,5) = 14908.484

using interpolation

i = 14% + (15450.276 - 15000) / (15450.276 - 14908.484) *(15% - 14%)

i = 14% + 0.83%

i = 14.83%

As incremental IRR is greater than MARR, X is selected and Y is rejected