Question

A company is considering two investment alternatives. Alternative A is a new machine that costs $50,000 and will last for ten years with no salvage value. It will save the company $5859 per year and the savings will increase by $2080 each year. Alternative B is a is a machine that will cost $75,000 and last 10 years. The salvage value at the end of 10 years is $25,000. It will save $11681 per year. Find the present worth of each alternative if the company as a MARR of 10.8%/year.

Answer 1

Alternative A:

PW = -50,000 + [5859 + 2080(A/G, 10.8%, 10)] * (P/A, 10.8%, 10)

      = -50,000 + [5859 + 2080(3.6685)] * (5.9389)

      = -50,000 + 80,112.67

      = $30,112.67

Alternative B:

PW = -75,000 + 11,681 (P/A, 10.8%, 10) + 25,000(P/F, 10.8%, 10)

      = -75,000 + 11,681 (5.9389) + 25,000(0.3586)

      = -75,000 + 69,372.29 + 8,965

      = $3,337.29