Solved problem on annual equivalent method – 3
Solved problem on annual equivalent method – 3
A
firm desires an economic analysis to determine which of the two machines is
attractive in a given interval of time. The minimum attractive rate of return
for the firm is 20%. The following data are to be used in the analysis:
Details
|
Machine X
|
Machine Y
|
First
cost (₹)
|
₹.
1,50,000
|
₹.
2,40,000
|
Annual
maintenance cost (₹)
|
₹.
0
|
₹.
4,500
|
Estimated
life (years)
|
12
|
12
|
Salvage
value (₹)
|
₹.
0
|
₹.
6,000
|
Which
machine would you choose? Base your answer on annual equivalent cost.
Given data
Method
= Annual equivalent method - Cost dominated cash flow
i
= 20%
Machine X
P
= ₹. 1,50,000
A1
= A2 = … = A10 = A = ₹.
0
n
= 12
S
= ₹. 0
Machine Y
P
= ₹. 2,40,000
A1
= A2 = … = A10 = A = ₹.
4,500
n
= 12
S
= ₹. 6,000
Formula used
AE(i) = P (A/P, i, n) + A - S (A/F, i, n)
Solution
Machine X
AE(20%)1 = 1,50,000 (A/P, 20%, 12) + 0 - 0 (A/F, 20%,
12)
= 1,50,000 (0.2253) + 0 - 0
= 33,795 + 0 - 0
AE(20%)1 = ₹. 33,795
Machine Y
AE(20%)2 = 2,40,000 (A/P, 20%, 12) + 4,500 – 6,000
(A/F, 20%, 12)
= 2,40,000 (0.2253) + 4,500 – 6,000
(0.0253)
= 54,072 + 4,500 – 151.8
AE(20%)2 = ₹. 58,420.20
Result
In annual equivalent
cost method, the alternative which has minimum annual equivalent cost is the
best alternative. Therefore machine
X is the best alternative.
