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 $55,000 per year through increased efficiencies. Galvanized Products uses a MARR of 18%/year to evaluate investments.

Part a

What is the annual worth of this investment?

Answer 1

Purchase Price (initial cost) = -100,000

Borrowed = 1/4^th at 15% interest repayable using equal annual payments over a 3-year period.

Therefore, borrowed = -100,000 * ¼ = -25,000

Cash Payment = -100,000 – (-25,000) = -75,000

Annual Repayment of borrowed money

A = P (A/P, 15%, 3)

A = -25,000 (0.43798) = -10,949.5

Life of the investment = 5 years

Salvage Value = 5,000

Annual Maintenance Cost = 25,000

Annual Savings = 55,000

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

MARR = 18%

Calculate the Annual Worth

Step 1 – Calculate PW

PW = -75,000 – 10,949.5 (P/A, 18%, 3) + 30,000 (P/A, 18%, 5) + 5,000 (P/F, 18%, 5)

PW = -75,000 – 10,949.5 (2.174272) + 30,000 (3.127171) + 5,000 (0.437109)

PW = -2,806.52

Step 2 – Calculate Annual Worth

AW = PW (A/P, 18%, 5)

AW = -2,806.52 (0.319777)

AW = -897.46

Annual worth of the investment is -$897.46

Alternatively

AW = -75,000 (A/P, 18%, 5) – [-10,949.5 (P/A, 18%, 3)] (A/P, 18%, 5) + 30,000 + 5,000 (A/F, 18%, 5)

AW = -75,000 (0.319777) – [-10,949.5 (2.174272)] (0.319777) + 30,000 + 5,000 (0.139777)

AW = -897.38