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Example problem on sinking fund method of depreciation

Example problem on sinking fund method of depreciation -1


A company has purchased an equipment whose first cost is Rs. 1,00,000 with an estimated life of eight years. The estimated salvage value of the equipment at the end of its lifetime is Rs. 20,000. Determine the depreciation charge and book value at the end of various years using the sinking fund method of depreciation with an interest rate of 12%, compounded annually.

Given data
Purchase price, P = . 1,00,000
Salvage value, F = . 20,000
Life of an asset, n = 8 years
i = 12%

Formula used
A = (P-F) * (A/F,i,n)
Dt = (P-F) * (A/F,i,n) * (F/P,i,t-1)
Bt = Bt–1 – Dt

Solution

End of Year
Annual fixed depreciation
A = (P-F) * (A/F,i,n)
in
Depreciation [Dt = (P-F) * (A/F,i,n) * (F/P,i,t-1)]
in
Book Value
(Bt   = Bt–1  – Dt)
in
0
6,504

1,00,000
1
6,504
(1,00,000-20,000) * (A/F,12%,8) * (F/P,12%,0) =
(80,000) * (0.0813) * (1) = 6,504
93,496.00
2
6,504
(1,00,000-20,000) * (A/F,12%,8) * (F/P,12%,1) = (80,000) * (0.0813) * (1.120) = 7,284.48
86,211.52
3
6,504
(1,00,000-20,000) * (A/F,12%,8) * (F/P,12%,2) = (80,000) * (0.0813) * (1.254) = 8,158.62
78,052.90
4
6,504
9,137.65
68,915.25
5
6,504
10,234.17
58,681.08
6
6,504
11,462.27
47,218.81
7
6,504
12,837.74
34,381.07
8
6,504
14,378.27
20,002.80



Example problem on sinking fund method of depreciation -2


A company has purchased an equipment whose first cost is Rs. 1,00,000 with an estimated life of eight years. The estimated salvage value of the equipment at the end of its lifetime is Rs. 20,000. Determine the depreciation charge at the end of 5th and book value at the end of 7th year using the sinking fund method of depreciation with an interest rate of 12% compounded annually.

Given data
Purchase price, P = . 1,00,000
Salvage value, F = . 20,000
Life of an asset, n = 8 years
t1 = 5
t2 = 7

Formula used
Dt = (P-F) * (A/F,i,n) * (F/P,i,t-1)
Bt = P-[(P-F) * (A/F,i,n) * (F/A,i,t)]

Solution
D5       = (P-F) * (A/F,i,n) * (F/P,i,t-1)
            = (1,00,000-20,000) * (A/F,12%,8) * (F/P,12%,5-1)
            = (1,00,000-20,000) * (A/F,12%,8) * (F/P,12%,4)
            = (80,000) * (0.0813) * (1.574)
D5       =. 10,237.30

B7        = P-[(P-F) * (A/F,i,n) * (F/A,i,t)]
            = 1,00,000-[(1,00,000-20,000) * (A/F,12%,7) * (F/A,12%,7)]
            = 1,00,000-[(80,000) * (0.0813) * (10.089)]
B7        =. 34,381.10

Result
At the end of 5th year,
Depreciation charge = . 10,237.30
At the end of 7th year,
         Book value = . 34,381.10