Question

The machines shown below are under consideration for an improvement to an automated candy bar wrapping process.

	Machine C	Machine D
First cost, $	–40,000	–75,000
Annual cost, $/year	–15,000	–10,000
Salvage value, $	12,000	25,000
Life, years	3	6

(Source: Blank and Tarquin)

Question 1 (10 points)

Based on the data provided and using an interest rate of 5% per year, the correct equation to calculate the Capital Recovery “CR” of Machine C is:

1. CR_C = –40,000(P/A, 5%, 3) + 12,000(F/A, 5%, 3)
2. CR_C = –40,000(A/P, 5%, 3) + 12,000(A/F, 5%, 3) –15,000(A/P, 5%,3)
3. CR_C = –40,000(A/P, 5%, 3) + 12,000(A/F, 5%, 3) –15,000
4. CR_C = –40,000(A/P, 5%, 3) + 12,000(A/F, 5%, 3)

Question 2 (10 points)

Based on the data provided and using an interest rate of 5% per year, the Capital Recovery “CR” of Machine C is closest to:

(All the alternatives presented below were calculated using compound interest factor tables including all decimal places)

1. CR_C = –$14,688
2. CR_C = –$18,494
3. CR_C = –$6,117
4. CR_C = –$10,882

Question 3 (10 points)

Based on the data provided and using an interest rate of 5% per year, the correct equation to calculate the Annual Worth “AW” of Machine C is:

1. AW_C= –40,000(P/A, 5%, 3) + 12,000(F/A, 5%, 3) –15,000
2. AW_C = –40,000(A/P, 5%, 3) + 12,000(A/F, 5%, 3) –15,000(A/P, 5%, 3)
3. AW_C = –40,000(A/P, 5%, 3) + 12,000(A/F, 5%, 3) –15,000
4. AW_C = –40,000(A/P, 5%, 3) + 25,000(A/F, 5%, 3) –10,000

Question 4 (10 points)

Based on the data provided and using an interest rate of 5% per year, the Annual Worth “AW” of Machine C is closest to:

(All the alternatives presented below were calculated using compound interest factor tables including all decimal places)

1. AW_C = –$25,882
2. AW_C = –$86,098
3. AW_C = –$21,117
4. AW_C = –$16,390

Answer 1

1.

Capital recovery = First cost (A/P, i, n) + Salvage value (A/F, i, n)

Capital recovery equation for machine C is:

CRC = - 40,000(A/P, 5%, 3) + $ 12,000 (A/F, 5%, 3)

Hence option “d” is correct answer.

2.

CRC = - $ 40,000(A/P, 5%, 3) + $ 12,000 (A/F, 5%, 3)

       = - $ 40,000 x 0.3672 + $ 12,000 x 0.3172

       = - $ 14,688 + $ 3,806.40 = - $ 10,881.60 or -$ 10,882

Hence option “d” is correct answer.

3.

Annual Worth = First cost (A/P, i, n) + Salvage value (A/F, i, n) – Annual cost

Annual Worth equation for machine C is:

AWC = - 40,000(A/P, 5%, 3) + $ 12,000 (A/F, 5%, 3) - $ 15,000

Hence option “c” is correct answer.

4.

AWC = - 40,000(A/P, 5%, 3) + $ 12,000 (A/F, 5%, 3) - $ 15,000

         = - $ 40,000 x 0.3672 + $ 12,000 x 0.3172 - $ 15,000

         = - $ 40,000 x 0.3672 + $ 12,000 x 0.3172 - $ 15,000

         = - $ 14,688 + $ 3,806.40 - $ 15,000

         = - $ 25,881.60 or -$ 25,882

Hence option “a” is correct answer.