Question

Galvanized Products is considering the purchase of a new computer system for its enterprise data management system. The vendor has quoted a purchase price of $100,000. Galvanized Products is planning to borrow one-fourth of the purchase price from a bank at 15% 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 have a salvage value of $5,000 at that time. Over the 5-year period, Galvanized Products expects to pay a technician $25,000 per year to maintain the system but will save $55,000 per year through increased efficiencies. Galvanized Products uses a MARR of 18%/yr to evaluate investments. 

a. What is the internal rate of return of this investment? 

b. What is the decision rule for judging the attractiveness of investments based on internal rate of return? 

c. Should the new computer system be purchased? 

Answer 1

Price of computer system = $ 100,000

1/4 th is borrowed at a rate of 15% for 3 years

Borrowed amount = (1/4)*100000 = $ 25,000

Equated annual payment = 25,000(A/P,15%,3)

0.15 Equated Amount = 25,000 x ; 1-1.15-3

Equated Amount = 25,000 X 4.3797

Equated Amount = $ 10.949.42

Saving = $ 55,000 per year

Payment to technician = $ 25,000 per year

Salvage value = $ 5,000

Useful life = 5 years

Refer the attached picture for the Net Cash flow

Net CF Year Cash Outflow Installment Saving sv -75000 -25000 / -10949.42 55000|| -25000 -10949.42 55000|| -25000 -10949.42 55000 -25000 55000 -25000 55000|5000 Internal Rate of Return -75000 19050.58 19050.58 19050.58 30000 35000 16.52%

As we can see using excel function the IRR comes out to be 16.52%.

I will calculate the IRR manually too refer the calculation below. We will be using the trial and error method and as we know if we discount the Net Cash Flow at IRR the NPW will be equal to zero.

NPW = -75,000+19,050.58(P/A,1%, 3)+30,000(P/F,1%, 4)+ 35,000(P/F, 1%,5)

Assume the IRR = 16%

NPW = -75,000+19,050.58x = 1 - 1.16-3 - +30,000x1.16-4+ 0.16 35,000 x 1.16-5

NPW = -75,000+ 19,050.58 x 2.24588+30,000 x 0.55229+ 35,000 X 0.47611

On solving the above equation we get

NPW = $ 1,018.17774

Now assume rate = 17%

NPW = -75,000+19,050.58x = 1 - 1.17-3 - +30,000x 1.17-4+ 35,000 x 1.17-5 0.17

NPW = -75,000+ 19,050.58 x 2.20958+30,000 x 0.53365+ 35.000 x 0.45611

On sol ing the above equation we get

NPW = -$ 932.742073

Now applying linear interpolation to determine the IRR

NPW16% IRR= 16+(17 – 16) * N PW6% – NPW17%

1,018.17774 IRR = 16+1% 1018.17774 - (-932.74207)

IRR = 16+0.52189

IRR 16.52%

A. Internal rate of return = 16.52%

B. When the IRR is greater than MARR the project must be selected. But if the IRR is less than MARR than reject the project.

C. Here, MARR = 18% and IRR = 16.52%

That is IRR is less than MARR. Hence, the computer system must not be purchased.

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