Question

In: Economics

A pump in a production plant costs $15,000. After 9 years, the salvage value is declared...

A pump in a production plant costs $15,000. After 9 years, the salvage value is declared at $0. Determine the depreciation charge and book value for Year 9 using Straight Line, SOYD, and 7-years MACRS depreciation. Then, find the PW of each depreciation schedule if the interest rate is 15%.

Solutions

Expert Solution

Answer:

(a) (i) Under SL Method:

Annual depreciation ($) = Cost / Useful life = $15,000 / 9 = $1,666.67
Depreciation schedule is as under:

SLM
Year Beginning-of-year Book Value ($) Annual Depreciation ($) End-of-Year Book Value ($)
1 $15,000.00 $1,666.67 $13,333.33
2 $13,333.33 $1,666.67 $11,666.66
3 $11,666.66 $1,666.67 $9,999.99
4 $9,999.99 $1,666.67 $8,333.32
5 $8,333.32 $1,666.67 $6,666.65
6 $6,666.65 $1,666.67 $4,999.98
7 $4,999.98 $1,666.67 $3,333.31
8 $3,333.31 $1,666.67 $1,666.64
9 $1,666.64 $1,666.67 $0.00

(ii) Under SOYD method:
Sum-of-years-digit (SOYD) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
Annual depreciation in year N = Cost x (Number of years remaining at beginning of year N / SOYD)
Depreciation schedule is as under:

SOYD
Year Asset Cost ($) Depreciation Rate Annual Depreciation ($) End-of-Year Book Value ($)
1 $15,000.00 9/45 $3,000.00 $12,000.00
2 $15,000.00 8/45 $2,666.67 $9,333.33
3 $15,000.00 7/45 $2,333.33 $7,000.00
4 $15,000.00 6/45 $2,000.00 $5,000.00
5 $15,000.00 5/45 $1,666.67 $3,333.33
6 $15,000.00 4/45 $1,333.33 $2,000.00
7 $15,000.00 3/45 $1,000.00 $1,000.00
8 $15,000.00 2/45 $666.67 $333.33
9 $15,000.00 1/45 $333.33 $0.00

(iii) Under MACRS:
Depreciation schedule is as under:

MACRS
Year Asset Cost ($) Depreciation Rate (%) Annual Depreciation ($) End-of-Year Book Value ($)
1 $15,000.00 14.29 $2,143.50 $12,856.50
2 $15,000.00 24.49 $3,673.50 $9,183.00
3 $15,000.00 17.49 $2,623.50 $6,559.50
4 $15,000.00 12.49 $1,873.50 $4,686.00
5 $15,000.00 8.93 $1,339.50 $3,346.50
6 $15,000.00 8.92 $1,338.00 $2,008.50
7 $15,000.00 8.93 $1,339.50 $669.00
8 $15,000.00 4.46 $669.00 $0.00

(b) Present Worth (PW) of depreciation are computed as follows.

(i) Present Worth (PW) of depreciation Under SL Method:

Year SLM Depreciation ($) PV Factor @15% Discounted SLM Depreciation ($)
(A) (B) (A) x (B)
1 $1,666.67 0.8696 $1,449.34
2 $1,666.67 0.7561 $1,260.17
3 $1,666.67 0.6575 $1,095.84
4 $1,666.67 0.5718 $953.00
5 $1,666.67 0.4972 $828.67
6 $1,666.67 0.4323 $720.50
7 $1,666.67 0.3759 $626.50
8 $1,666.67 0.3269 $544.83
9 $1,666.67 0.2843 $473.83
PW of Depreciation ($) = $7,952.68

(ii) Present Worth (PW) of depreciation Under SOYD method:

Year SOYD Depreciation ($) PV Factor @15% Discounted SOYD Depreciation ($)
(A) (B) (A) x (B)
1 $3,000.00 0.8696 $2,608.80
2 $2,666.67 0.7561 $2,016.27
3 $2,333.33 0.6575 $1,534.16
4 $2,000.00 0.5718 $1,143.60
5 $1,666.67 0.4972 $828.67
6 $1,333.33 0.4323 $576.40
7 $1,000.00 0.3759 $375.90
8 $666.67 0.3269 $217.93
9 $333.33 0.2843 $94.77
PW of Depreciation ($) = $9,396.50

(iii) Present Worth (PW) of depreciation Under MACRS method :

Year MACRS Depreciation ($) PV Factor @15% Discounted MACRS Depreciation ($)
(A) (B) (A) x (B)
1 $2,143.50 0.8696 $1,863.99
2 $3,673.50 0.7561 $2,777.53
3 $2,623.50 0.6575 $1,724.95
4 $1,873.50 0.5718 $1,071.27
5

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