Galvanized Products is considering purchasing 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 has a salvage value of $5,000 at that time. Over the S-year period, Galvanized Products expects to pay a technician 525,000 per year to maintain the system but will save $35.000 per year through increased efficiencies. Galvanized Products uses a MARR of 18 year to evaluate investments.
a. What is the present worth of this investment?
b. What is the decision rule for judging the attractiveness of investments based on present worth?
c. Should the new computer system be purchased?
(a)
Working notes:
(1) Loan amount = 100,000 x (1/4) = 25,000
(2) First cost = 100,000 x (3/4) = 75,000
(3) Annual loan repayment, years 1-3 = 25,000 / P/A(15%, 3) = 25,000 / 2.2832 = 10,949.54
(4) Annual savings, year 5 = 55,000 + 5,000 salvage value = 60,000
(5) Annual net benefit (NAB) = Annual saving - Annual loan repayment - Annual cost
(6) PV Factor in year N = (1.18)-N.
PW of investment as follows.
Year | Cost | Loan Repayment | Savings | NAB | PV factor@18% | Discounted NAB |
0 | 75,000 | -75,000 | 1.0000 | -75,000.00 | ||
1 | 25,000 | 10,949.54 | 55,000 | 19,050 | 0.8475 | 16,144.46 |
2 | 25,000 | 10,949.54 | 55,000 | 19,050 | 0.7182 | 13,681.74 |
3 | 25,000 | 10,949.54 | 55,000 | 19,050 | 0.6086 | 11,594.70 |
4 | 25,000 | 0 | 55,000 | 30,000 | 0.5158 | 15,473.67 |
5 | 25,000 | 0 | 60,000 | 35,000 | 0.4371 | 15,298.82 |
PW of NAB = | -2,806.61 |
(b)
Decision rule is:
Accept project if PW > 0
Reject project if PW < 0
(c)
Since PW < 0, system should not be purchased.
Galvanized Products is considering purchasing a new computer system for its enterprise data management system.
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