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 |
 Rahul Sunny answered 8 months agoRelated SolutionsA company purchases a piece of equipment for $15,000. After nine years, the salvage value is...A company purchases a piece of
equipment for $15,000. After nine years, the salvage value is $900.
The annual insurance cost is 5% of the purchase price, the
electricity cost is $600/yr, and the maintenance and replacement
parts cost is $120/yr. The nominal annual interest rate is 9.6%,
compounded monthly. Neglecting taxes, what is most nearly the
present worth of the equipment if it is expected to earn the
company $4500 per year?
$2300
$2800
$3200
$3500
69. Machine costs $150,000, and lasts 10 years with salvage value of $15,000. Annual operating costs...69. Machine costs $150,000, and lasts 10 years
with salvage value of $15,000. Annual operating costs are $50,000.
If you want to make a 20% return on investment, what is the minimum
annual revenue required from this machine? Show all work for full
credit.
A copier costs X and will have a salvage value of Y after n years. i)...A copier costs X and will have a salvage value of Y after n
years.
i) using the straight line method, the annual depreciation expense
is $1000.
ii) using the sum of the years digits method, the depreciation
expense in year 3 is $800.
iii) using the declining balance method, the depreciation expense
is 33.125% of the book value at the beginning of the year.
Calculate X (provide excel spreadsheet in your answer).
Pump A Initial Cost: 7000 Salvage Value:1200 Useful Life:12 Pump B Initial Cost: 5000 Salvage Value:...Pump A
Initial Cost: 7000
Salvage Value:1200
Useful Life:12
Pump B
Initial Cost: 5000
Salvage Value: 1000
Useful Life: 6
Two pumps are being considered for purchase. If interest is 9%,
which pump should be bought?
A project ends after 4 years. The after tax salvage value is 12,860 and the released...A project ends after 4 years. The after tax salvage value is
12,860 and the released working capital equals 40,000 at the end of
the project. Assume a 9.82% WACC and compute the present value of
the terminal cash flows.
Which of the following is a correct calculation for after-tax salvage value if salvage value and...Which of the following is a correct calculation for after-tax
salvage value if salvage value and book value differ?
A) Salvage – T(book value – salvage)
B) (Initial cost - Salvage)T – (book value – salvage)
C) Salvage – T(salvage – (initial cost – accumulated
depreciation))
D) (Salvage – accumulated depreciation)T – (salvage
– book value)
New equipment costs $675,000 and is expected to last for five years with no salvage value....New equipment costs $675,000 and is expected to last for five
years with no salvage value. During this time, the company will use
a 30% CCA rate. The new equipment will save $120,000 annually
before taxes. If the company's required return is 12% and the tax
rate is 35%, what is the present value of the equipment's CCA tax
shield?Please show all the calculations by which you came up with
the final answer.
$189,710
$159,710
$169,710
$199,710
$179,710
An asset costs $10,000 and has a depreciable life of 10 years and a salvage value...An asset costs $10,000 and has a depreciable life of 10 years
and a salvage value of $3,000. Determine the book (asset) value at
the end of the 9th year using each of the following methods of
depreciation (a) double-declining-balance method (b)
textbook-declining-balance method (Matheson formula), and (c)
sum-of-years’ digits method.
A company ,purchases a new machine with an initial cost of $15,000 and a salvage value...A company ,purchases a new machine with an initial cost of
$15,000 and a salvage value of $1,800. Net revenues in year 1 are $
8,000,$8,150 in year 2, and increase by $150 each year the
following eight years. Use a MARR of 12% .
a. What is the annual worth ?
b. Determine the IRR of this machine
please show detailed and clear solution with or without using
EXCEL. thank you
a brand new machine is estimated to have a salvage value of P10000 after 10 years...a brand new machine is estimated to have a salvage value of
P10000 after 10 years and book value of P30,000 after 5 years, what
is the initial cost of the machine? (assume straight line
depreciation)
ADVERTISEMENT
ADVERTISEMENT
Latest Questions
ADVERTISEMENT
|