Question

Galvanized Products is considering the purchase of a new computer system for their 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 annuel payments over a 3-year period. The computer system is expected to last five years and has 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 effeciencies. 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

Purchase Price = 100,000

Borrowing 1/4th from a bank at 15% interest.

Borrowed = 100,000 * ¼ = 25,000

Cash payment made = 75,000

To be repayable in 3 annual equal payments

Annual Payments = 25,000 (A/P, 15%, 3)

Annual Payments = 25,000 (0.43798) = 10,949.5

Life = 5 years

Salvage Value = 5,000

Annual Maintenance Cost = 25,000

Annual Savings = 55,000

Calculate Net Annual Savings

Net Annual Savings = 55,000 – 25,000 = 30,000

MARR = 18%

a. Calculate IRR of the investment.

Calculating IRR using the Trial and Error Method

Let the rate of interest is 16%. Calculate PW of the cash flows at 16%.

PW = -75,000 – 10,949.5 (P/A, 16%, 3) + 30,000 (P/A, 16%, 5) + 5,000 (P/F, 16%, 5)

PW = -75,000 – 10,949.5 (2.245889) + 30,000 (3.274293) + 5,000 (0.476113)

PW = 1,018

The PW is positive. Increase the rate if interest to get the negative PW. Increase the rate f interest to 17%. Calculate PW at 17%.

Let the rate of interest is 17%. Calculate PW of the cash flows at 17%.

PW = -75,000 – 10,949.5 (P/A, 17%, 3) + 30,000 (P/A, 17%, 5) + 5,000 (P/F, 17%, 5)

PW = -75,000 – 10,949.5 (2.209584) + 30,000 (3.199346) + 5,000 (0.456111)

PW = -933

Using interpolation

IRR = 16% + [1,018 – 0 ÷ 1,018 – (-933)] * 1%

IRR = 16.52%

B. What is the decision rule?

If IRR > MARR – Select

If IRR < MARR - Reject

C. Should the new computer system be purchased?

No. The Computer system cannot be purchased because the MARR is 18% and the IRR is 16.52%. As the IRR is less than MARR, reject the purchase.