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 1/4th 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 5-year period, Galvanized Products expects to pay a technician $25,000 per year to maintain the system but will save $45,000 per year through increased efficiencies. Galvanized Products uses a MARR of 21%/year to evaluate investments.

What is the future worth of this investment?

Answer 1

(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.