Question

Two alternatives for purchasing a new printing machine (from providers A and B) are being considered for a production upgrade of a printing facility. Alternative A has a life of 2 years, first cost of $1200, annual reduction in maintenance cost (can be treated as revenue in this cash flow) of $650, and salvage value after 2 years of $250. Alternative B has a life of 3 years, first cost of $1650, annual reduction in maintenance cost of $790, and salvage value after 3 years of $250. MARR = 8%. Alternatives are replicable in the future. Calculate the NPW of each alternative. Show calculation steps leading to this choice and provide explanations whenever possible.

Answer 1

In the given question, the life of the alternatives is not equal. So, for the evaluation, use the common multiple method and convert the unequal life into equal life. The Alternative A has 2 years of life and Alternative B has 3 years of life. The LCM of 2 and 3 is 6. Hence, the common time period will be 6 years. Alternative A is to repeated 3 times and Alternative B is to be repeated 2 times.

NPW of Alternative A

Initial Cost = 1,200

Annual revenues = 650 per year

Salvage Value = 250

MARR = 8%

NPW = -1,200 – 1,200 (P/F, 8%, 2) – 1,200 (P/F, 8%, 4) + 650 (P/A, 8%, 6) + 250 (P/F, 8%, 2) + 250 (P/F, 8%, 4) + 250 (P/F, 8%, 6)

NPW = -1,200 – 1,200 (0.85734) – 1,200 (0.73503) + 650 (4.62288) + 250 (0.85734) + 250 (0.73503) + 250 (0.63017)

NPW = 449.66

NPW of Alternative B

Initial Cost = 1,650

Annual revenues = 790 per year

Salvage Value = 250

MARR = 8%

NPW = -1,650 – 1,650 (P/F, 8%, 3) + 790 (P/A, 8%, 6) + 250 (P/F, 8%, 3) + 250 (P/F, 8%, 6)

NPW = -1,650 – 1,650 (0.79383) + 790 (4.62288) + 250 (0.79383) + 250 (0.63017)

NPW = 1,048.25

From the above calculations, select the Alternative B.