Question

A small aerospace company is evaluating two alternatives: the purchase of an automatic-feed machine and a manual-feed machine for a product’s finishing process. The auto-feed machine has an initial cost of $23,000, an estimated salvage value of $4,400 and a predicted life of 10 years. One person will operate the machine at a cost of $12 an hour. The expected output is 8 tons per hour. Annual maintenance and operating cost is expected to be $3,500.

The manual-feed machine has a first cost of $8,000, no expected salvage value, a 5-year life and an output of 6 tons per hour. However, three workers will be required at $8 an hour each. The machine will have an annual maintenance and operation cost of $1,500. All invested capital is expected to generate a market return of 10% per year before taxes.

How many tons per year must be finished in order to justify the higher purchase cost of the auto-feed machine? If a requirement to finish 2,000 tons per year is anticipated, which machine should be purchased?

Answer 1

Let breakeven tons be n, then

Cost per ton for automatic option = 12 / 8 = 1.5

Cost per ton for manual option = 8*3 / 6 = 4

EUAC of automatic option = 23000*(A/P, 10%,10) - 4400*(A/F,10%,10) + 3500 + 1.5*n

= 23000*0.162745 - 4400*0.062745 + 3500 + 1.5*n

= 6967.057 + 1.5*n

EUAC of manual option = 8000*(A/P, 10%,5) + 1500 + 4*n

= 8000*0.263797 + 1500 + 4*n

= 3610.376 + 4*n

As per given condition

6967.057 + 1.5*n = 3610.376 + 4*n

n = (6967.057 - 3610.376) / 2.5 = 1342.67 ~ 1343 (Nearest Whole number)

As 2000 units are required, which is greater than 1343, automatic option should be selected