Example problem on replacement of an asset - 2
Example problem on replacement of an asset - 2
Three
years back, a municipality purchased a 10hp motor for pumping drinking water.
Its useful life was estimated to be 10years. Its annual operating and
maintenance cost is Rs. 1,500. Due to rapid development of that locality, the
municipality is unable to meet the current demand for water with the existing
motor. The municipality can cope with the situation either by augmenting an
additional 5hp motor or replacing the existing 10hp motor with the new 15hp
motor. The details of these motors are given in the following table.
Old 10hp motor
|
New 5hp motor
|
New 15hp motor
|
|
Purchase cost (Rs.)
|
18,000
|
7,500
|
25,000
|
Life (years)
|
10
|
7
|
7
|
Salvage value (Rs.)
|
1,300
|
1,000
|
3,500
|
Annual operating and maintenance cost
(Rs.)
|
1,500
|
900
|
500
|
The current market value of the 10hp motor is
Rs.12,000. Using an interest rate of 15%, find the best alternative.
Given data
Old 10hp motor
Purchase
price = ₹. 18,000
Present
value (P) = ₹. 12,000
Salvage
value (F) = ₹. 1,300
Annual
maintenance cost (A) = ₹. 1,500
Remaining
life = 7 years
Interest
rate = 15%
New 5hp motor
Purchase
price = ₹. 7,500
Salvage
value (F) = ₹. 1,000
Annual
maintenance cost (A) = ₹. 900
Useful
life = 7 years
Interest
rate = 15%
New 15hp motor
Purchase
price = ₹. 25,000
Salvage
value (F) = ₹. 3,500
Annual
maintenance cost (A) = ₹. 500
Useful
life = 7 years
Interest
rate = 15%
Formula used
AE(i)
= [(P – F )*(A/P, i, n)] + (F*i) + A
Solution
Alternative 1: Old 10hp
motor + New 5hp motor
AE(15%)Old
10hp = [(P – F)*(A/P, 15%,
7)] + (F*i) + A
= [(12,000 – 1,300)*(0.2404)]
+ (1,300*0.12) + 1,500
= [10,700*0.2404]+ (1,300*0.15)
+ 1,500
= 2,578.28 + 195 + 1,500
AE(15%)Old
10hp = ₹. 4,273.28
New 5hp motor
AE(15%)New
5hp = [(P – F)*(A/P,
15%, 7)] + (F*i) + A
= [(7,500 – 1,000)*(0.2404)]
+ (1,000*0.12) + 900
= [6,500*0.2404]+ (1,000*0.15)
+ 900
= 2,578.28 + 150 + 900
AE(15%)New
5hp = ₹. 2,612.60
Annual
cost of alternative 1 = AE(15%)Old 10hp + AE(15%)New 5hp
= 4,273.28 + 2,612.60
Annual
cost of alternative 1 = ₹. 6,885.88
Alternative 2: New 15hp
motor
AE(15%)
New 15hp = [(P – F)*(A/P,
15%, 7)] + (F*i) + A
= [(25,000 – 3,500)*(0.2404)]
+ (3,500*0.12) + 500
= [21,500*0.2404]+ (3,500*0.15)
+ 500
= 5,168.60 + 525 + 500
AE(15%)
New 15hp = ₹. 6,193.60
