Question

A job can be done with Machine A that costs $12,500 and has annual end-of-year maintenance costs of $5000; its salvage value after 3 years is $2000. Or the job can be done with Machine B, which costs $15,000 and has end-of-year maintenance costs of $4000 and a salvage value of $1500 at the end of 4 years. These investments can be repeated in the future, and your work is expected to continue indefinitely.

Use present worth, annual worth, and capitalized cost to compare the machines. The interest rate is 5%/year.

Answer 1

Answer and Explanation :

As per the data given in the question,

Present value = Amount ÷ (1 + r)^n

Machine A

Year Amount Discount Factor PV

1 $5,000 1.05 $4,761.90

2 $5,000 $4,535.15

3 $5,000 $4,319.19

Total $13,616.24

Now

Present value of salvage value =$2,000 ÷ 1.05^3 = $1,727.68

Present worth of Machine A is

= -$12,500 - $13,616.24 + $1,727.68

= -$24,388.56

Similarly Present worth of Machine B = -$15,000 - $4,000 ÷ 1.05 -$4,000 ÷ (1.05)^2 - $4,000 ÷ 1.05^3 - $4,000 ÷ 1.05^4 + $1,500 ÷ 1.05^4

=-$24,658.94

Based on the comparison between Machine A and Machine B

Machine A is better because it has higher present worth

Annual worth:

For machine A = -$12,500(A/PA,5%,3) -$5,000+$2,000(A/F,5%,3)

=-$12,500 × 0.367 - $5,000 + $2,000 × 0.317

= -$8,953.5

For Machine B:

= -$15,000(A/P,5%,4) - $4,000 + $1,500(A/F,5%,4)

= -$7,882.16

Based on the comparison between Machine A and Machine B

Machine B is better because it has higher annual worth

Capitalized cost:

Machine A :

= -$12,500+$2,000(P/F,5%,3) - $5,000 ÷ 0.05

= -$12,500 + $2,000 × 0.86 - $5,000 ÷ 0.05

= -$110,772

Machine B :

=-$15,000(P/F,5%,4) - $4000 ÷ 0.05

=-$15,000 × 0.82 - $4,000 ÷ 0.05

= -$93,765.9

Based on the comparison between Machine A and Machine B

Machine B is better because it has lower capitalized cost