Solved problem on annual equivalent method – 4
Solved problem on annual equivalent method – 4
Two
possible routes for laying an electrical power line are under study. The
following data are to be used in the analysis:
Around the lake
|
Under the lake
|
|
Length
|
15km
|
5km
|
First
cost
|
₹.
1,50,000 per km
|
₹.
7,50,000 per km
|
Annual
maintenance cost
|
₹.
6,000 per km per year
|
₹.
12,000 per km per year
|
Yearly
power loss
|
₹.
15,000 per km
|
₹.
15,000 per km
|
Estimated
life
|
15 years
|
15 years
|
Salvage
value
|
₹.
90,000 per km
|
₹.
1,50,000 per km
|
If
10% interest used, should the power line be routed around the lake or under the
lake?
Given data
Method
= Annual equivalent method - Cost dominated cash flow
i
= 10%
Around the lake
Initial
investment = P = ₹.
1,50,000 * 15 = ₹. 22,50,000
Maintenance
cost per year = ₹. 6,000 * 15 = ₹. 90,000
Power
loss per year = ₹. 15,000 * 15 = ₹. 2,25,000
Total
cost per year = ₹. 90,000 + ₹. 2,25,000 = ₹. 3,15,000
A1
= A2 = … = A10 = A = ₹.
3,15,000
n
= 15
Salvage
value S = ₹. 90,000 * 15 = ₹. 13,50,000
Under the lake
Initial
investment = P = ₹.
1,50,000 * 5 = ₹. 37,50,000
Maintenance
cost per year = ₹. 12,000 * 5 = ₹. 60,000
Power
loss per year = ₹. 15,000 * 5 = ₹. 75,000
Total
cost per year = ₹. 60,000 + ₹. 75,000 = ₹. 1,35,000
A1
= A2 = … = A10 = A = ₹.
1,35,000
n
= 15
Salvage
value S = ₹. 1,50,000 * 5 = ₹. 7,50,000
Formula used
AE(i) = P (A/P, i, n) + A - S (A/F, i, n)
Solution
Around the lake
AE(10%)1 = 22,50,000 (A/P, 10%, 15) + 3,15,000 – 13,50,000
(A/F, 10%, 15)
= 22,50,000 (0.1315) + 3,15,000 –
13,50,000 (0.0315)
= 2,95,875 + 3,15,000 – 42,525
AE(10%)1 = ₹. 5,68,350
Under the lake
AE(10%)2 = 37,50,000 (A/P, 10%, 15) + 1,35,000 – 7,50,000
(A/F, 10%, 15)
= 37,50,000 (0.1315) + 1,35,000 –
7,50,000 (0.0315)
= 4,93,125 + 1,35,000 – 23,625
AE(10%)2 = ₹. 6,04,500
Result
In annual equivalent
cost method, the alternative which has minimum annual equivalent cost is the
best alternative. Therefore laying
the route around the lake is the best route.
