An asset has the estimated salvage values for various lives, shown in the table

Question

An asset has the estimated salvage values for various lives, shown in the table below. For each possible life from 1 to 6 by 1, determine the capital recovery cost for MARR of 8%/year.

0

1

2

3

4

5

6

Carry all interim calculations to 5 decimal places and then round your final answer to the nearest dollar. The tolerance is ±50.

EOY

0

1

2

3

4

5

6

NCF -$125,000 $30,000 $30,000 $30,000 $30,000 $30,000 $30,000 Estimated SV $125,000 $105,000 $85,000 $75,000 $65,000 $60,000 $57,000 Capital Recovery Cost $______

$_____

$_____

$_____

$_____

$_____

Explanation / Answer

Capital recovery cost, CR = I (A/P, i, N) - S (A/F, i, N)

Where,

I = Initial investment = $ 125,000

i = Interest rate = 8 %

N = No. of Period

S = Salvage value received

Year 1:

CR 1= $ 125,000 x (A/P, 8 %, 1) - S (A/F, 8 %, 1)

      = $ 125,000 x 1.0800 - $ 105,000

      = $ 135,000 - $105,000 = $ 30,000

Year 2:

CR 2= $ 125,000 x (A/P, 8 %, 2) - S (A/F, 8 %, 2)

      = $ 125,000 x 0.56077 - $ 85,000 x 0.48077

      = $ 70,096.25 - $ 40,865.45 = $ 29,230.80 or $ 29,231

Year 3:

CR 3= $ 125,000 x (A/P, 8 %, 3) - S (A/F, 8 %, 3)

      = $ 125,000 x 0.38803 - $ 75,000 x 0.30803

      = $ 48,503.75 - $ 23,102.25 = $ 25,401.50 or $ 25,402

Year 4:

CR 4= $ 125,000 x (A/P, 8 %, 4) - S (A/F, 8 %, 4)

      = $ 125,000 x 0.30192 - $ 65,000 x 0.22192

      = $ 37,740 - $ 14,424.80 = $ 23,315.20 or $ 23,315

Year 5:

CR 5= $ 125,000 x (A/P, 8 %, 5) - S (A/F, 8 %, 5)

      = $ 125,000 x 0.25046 - $ 60,000 x 0.17046

      = $ 31,307.50 - $10,227.60 = $ 21,079.90 or $ 21,080

Year 6:

CR 6= $ 125,000 x (A/P, 8 %, 6) - S (A/F, 8 %, 6)

      = $ 125,000 x 0.21632 - $ 57,000 x 0.13632

      = $ 27,040 - $7,770.24 = $ 19,269.7600 or $ 19,270

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.