Question

3. ACME Corp. is evaluating three opportunities to make money:
Option A: Replace outdated equipment to reduce defective parts. Initial cost = $200,000. Annual
savings from not having to remake parts: $53,000. Salvage value of equipment at the end of the 10-year
study period: $10,000.
Option B: Change packaging to reduce shipping costs. Initial cost = $180,000. Annual savings in
shipping costs: $30,000. Salvage value of equipment at the end of the 10-year study period: $0.
Option C: Manufacture parts in-house rather than purchase them. Initial cost = $250,000. Annual
savings from eliminating the outside vendor = $60,000. Salvage value of equipment at the end of the
10-year study period: $20,000.
If ACME only has $300,000 of capital available to pay the initial cost, it can only choose one project.
Which project should they choose? Assume ACME’s MARR is 14% per year.

Answer 1

i = 14% = 0.14

For option A

Investment = 200000, annual savings = 53000 salvge value = 10000 after 10 years

Present worth of option A= -200000 + 53000 *(P/A, 14%, 10) + 10000*(P/F, 14%,10)

= -200000 + 53000 *5.2161156 + 10000*0.26974381

= 7915.56

For option B

Investment = 180000, annual savings = 30000 salvge value = 0 after 10 years

Present worth of option B= -180000 + 30000 *(P/A, 14%, 10)

= -180000 + 30000 *5.2161156

= -23516.53

For option C

Investment = 250000, annual savings = 60000 salvge value = 20000 after 10 years

Present worth of option C= -250000 + 60000 *(P/A, 14%, 10) + 20000*(P/F, 14%,10)

= -250000 + 60000 *5.2161156 + 20000*0.26974381

= 68361.81

ACME should choose option A as it has highest net present value