1) A new machine tool costs $500,000, and has a useful life of 10 years. Its est
ID: 2563610 • Letter: 1
Question
1) A new machine tool costs $500,000, and has a useful life of 10 years. Its estimated salvage value at the end of year 10 is $50,000.
(a) Determine the depreciation for years 1-10 using the straight-line method (with the assumed salvage value), and also using double declining balance.
(b) Compute the present worth of the two series of depreciation amounts, assuming that you have a minimum acceptable rate of return of 12% per year.
(c) What does your answer to part (b) above tell you about which depreciation method would be preferable for you? Explain.
Explanation / Answer
1)(a)
Straight line depreciation = (Cost – Residual value)/Useful life
= ($ 500,000 - $ 50,000)/10
=$ 450,000/10 = $ 45,000 per year
Double decline depreciation = 2 × Straight-line depreciation rate × Book value at the beginning of the year = 2 x 1/10 x $ 500,000 = $ 2 x 0.1 x $ 500,000 = $ 100,000
Year
Beginning book value
Double decline depreciation
Ending book value
1
$ 500,000.00
$ 100,000.00
$ 400,000.00
2
$ 400,000.00
$ 80,000.00
$ 320,000.00
3
$ 320,000.00
$ 64,000.00
$ 256,000.00
4
$ 256,000.00
$ 51,200.00
$ 204,800.00
5
$ 204,800.00
$ 40,960.00
$ 163,840.00
6
$ 163,840.00
$ 32,768.00
$ 131,072.00
7
$ 131,072.00
$ 26,214.40
$ 104,857.60
8
$ 104,857.60
$ 20,971.52
$ 83,886.08
9
$ 83,886.08
$ 16,777.22
$ 67,108.86
10
$ 67,108.86
$ 17,108.86
$ 50,000.00
b.
Straight line depreciation
Year(n)
Depreciation(D)
PV factor F= 1/(1+i)n
Present Worth =F x D
1
$ 45,000
1/(1+0.12)1
0.892857143
$ 40,178.57
2
$ 45,000
1/(1+0.12)2
0.797193878
$ 35,873.72
3
$ 45,000
1/(1+0.12)3
0.797193878
$ 35,873.72
4
$ 45,000
1/(1+0.12)4
0.711780248
$ 32,030.11
5
$ 45,000
1/(1+0.12)5
0.635518078
$ 28,598.31
6
$ 45,000
1/(1+0.12)6
0.567426856
$ 25,534.21
7
$ 45,000
1/(1+0.12)7
0.506631121
$ 22,798.40
8
$ 45,000
1/(1+0.12)8
0.452349215
$ 20,355.71
9
$ 45,000
1/(1+0.12)9
0.403883228
$ 18,174.75
10
$ 45,000
1/(1+0.12)10
0.360610025
$ 16,227.45
Total PW
$ 275,644.97
Double decline depreciation
Year(n)
Depreciation(D)
PV factor F= 1/(1+i)n
Present Worth =F x D
1
$ 100,000.00
1/(1+0.12)1
0.892857143
$ 89,285.71
2
$ 80,000.00
1/(1+0.12)2
0.797193878
$ 63,775.51
3
$ 64,000.00
1/(1+0.12)3
0.797193878
$ 51,020.41
4
$ 51,200.00
1/(1+0.12)4
0.711780248
$ 36,443.15
5
$ 40,960.00
1/(1+0.12)5
0.635518078
$ 26,030.82
6
$ 32,768.00
1/(1+0.12)6
0.567426856
$ 18,593.44
7
$ 26,214.40
1/(1+0.12)7
0.506631121
$ 13,281.03
8
$ 20,971.52
1/(1+0.12)8
0.452349215
$ 9,486.45
9
$ 16,777.22
1/(1+0.12)9
0.403883228
$ 6,776.04
10
$ 17,108.86
1/(1+0.12)10
0.360610025
$ 6,169.63
Total PW
$ 320,862.19
(c) Straight line depreciation method is preferable as the present worth of this method is less than double decline method.
Year
Beginning book value
Double decline depreciation
Ending book value
1
$ 500,000.00
$ 100,000.00
$ 400,000.00
2
$ 400,000.00
$ 80,000.00
$ 320,000.00
3
$ 320,000.00
$ 64,000.00
$ 256,000.00
4
$ 256,000.00
$ 51,200.00
$ 204,800.00
5
$ 204,800.00
$ 40,960.00
$ 163,840.00
6
$ 163,840.00
$ 32,768.00
$ 131,072.00
7
$ 131,072.00
$ 26,214.40
$ 104,857.60
8
$ 104,857.60
$ 20,971.52
$ 83,886.08
9
$ 83,886.08
$ 16,777.22
$ 67,108.86
10
$ 67,108.86
$ 17,108.86
$ 50,000.00
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