Question

Kermit is considering purchasing a new computer system. The purchase price is $148645. Kermit will borrow one-fourth of the purchase price from a bank at 10 percent per year compounded annually. The loan is to be repaid using equal annual payments over a 3 year period. The computer system is expected to last 5 years and has a salvage value of $8162 at that time. Over the 5-year period, Kermit expects to pay a technician $20,000 per year to maintain the system but will save $61926 per year through increased efficiencies. Kermit uses a MARR of 12 percent to evaluate investments. What is the net present worth for this new computer system?

Answer 1

Price = $ 148645

Borrowed amount = (1/4)*148645 = $ 37,161.25

Down payment = $ 111,483.75

Equal payment = $ 37,161.25/(P/A,10%,3) = $ 14,943.08

Salvage value = $ 8162

Technician = $ 20,000 (starting from end of 1st year)

Saving = $ 61,926 Every year

MARR = 12%

Useful life = 5 years

PW = -111,483.75 -14,943.08(P/A,12%,3) -20,000(P/A,12%,4) +61,926(P/A,12%,5) + 8162(P/F,12%,5)

PW = - 111,483.75 - 14,943.08*2.402 - 20,000*3.037 + 61,926*3.605 + 8,162*0.4326

PW = - 111,483.75 - 35,890.75 - 60,746.98

+ 232,229.30 + 372.88

Present with = $ 24,480.7

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