Solved problem on determination of economic life of an asset - 2
Solved problem on economic life of an asset - 2
The following table gives the operation cost, maintenance cost and salvage value at the end of every year of a machine whose purchase value is Rs. 20,000. Find the economic life of the machine assuming that the interest rate is 12%, compounded annually.
End
of year
|
Operation
cost at the end of year (₹)
|
Maintenance
cost at the end of year (₹)
|
Salvage
value at the end of year (₹)
|
1
|
3000
|
300
|
9000
|
2
|
4000
|
400
|
8000
|
3
|
5000
|
500
|
7000
|
4
|
6000
|
600
|
6000
|
5
|
7000
|
700
|
5000
|
6
|
8000
|
800
|
4000
|
7
|
9000
|
900
|
3000
|
8
|
10000
|
1000
|
2000
|
9
|
11000
|
1100
|
1000
|
10
|
12000
|
1200
|
0
|
Given data
First
cost = ₹. 20,000
i
= 12%
Solution
End
of year
|
Operation
cost at the end of year (₹)
|
Maintenance
cost at the end of year (₹)
|
Total
cost at the end of year (₹)
|
1
|
3000
|
300
|
3300
|
2
|
4000
|
400
|
4400
|
3
|
5000
|
500
|
5500
|
4
|
6000
|
600
|
6600
|
5
|
7000
|
700
|
7700
|
6
|
8000
|
800
|
8800
|
7
|
9000
|
900
|
9900
|
8
|
10000
|
1000
|
11000
|
9
|
11000
|
1100
|
12100
|
10
|
12000
|
1200
|
13200
|
Result
Machine should be replaced at the end of 5th year, when i = 12%. (i.e.) Economic life of the machine is 5 years.
