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

As per the question, the alternatives have unequal lives. To calculate the NPW, use the common multiple method and converting the unequal life into equal life. The alternative A and B has the life of 2 years and 3 years respectively.

The LCM of 2 years and 3 years is 6 years. So, the common study period will be 6 years.

Alternative A is to be repeated 3 times and Alternative B is to be repeated 2 times.

Alternative A	Alternative B
Initial Cost = 1,200 Annual revenues (reduction in maintenance cost) = 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	Initial Cost = 1,650 Annual revenues (reduction in maintenance cost) = 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 it can be seen that the NPW of Alternative B is highest. So, select Alternative B.