Question

Two products being considered for purchase.

Product 1 has an initial cost of $25,000, an estimated life of 8 years, annual operating cost of $3000, and an estimate salvage value of $4000.

Product 2 has an initial cost of $35,000, an estimated life of 12 years, annual operating cost of $3200, an estimated salvage value of $5500. Plus, product 2 will require a major overhaul costing $ 5000 at the end of the sixth year.

Use an interest rate of 12%,

Determine which is the better optioan based on using the present worth method.

Answer 1

PW of costs, Product 1 ($) = 25,000 + 3,000 x P/A(12%, 8) - 4,000 x P/F(12%, 8)

= 25,000 + 3,000 x 4.9676** - 4,000 x 0.4039**

= 25,000 + 14,903 - 1,616

= 38,287

PW of costs, Product 2 ($) = 35,000 + 3,200 x P/A(12%, 12) + 5,000 x P/F(12%, 6) - 5,500 x P/F(12%, 12)

= 35,000 + 3,200 x 6.1944** + 5,000 x 0.5066** - 5,500 x 0.2567**

= 35,000 + 19,822 + 2,533 - 1,412

= 55,943

Since Product 1 has lower PW of costs, Product 1 is a better option.

**From P/A and P/F factor tables

[Since we're computing PW of costs, Present Value of salvage value, a cash inflow, is deducted]