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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