Solved problem on future worth method - 6
Solved problem on future worth method - 6
A
company must decide whether to buy a machine A or machine B
Machine A
|
Machine B
|
|
Initial
cost (₹)
|
₹.
4,00,000
|
₹.
8,00,000
|
Annual
maintenance cost (₹)
|
₹.
40,000
|
₹.
0
|
Life
(years)
|
5
|
5
|
Salvage
value at the end of machine life (₹)
|
₹.
2,00,000
|
₹.
5,50,000
|
Select the best alternative based on
future worth method at i = 15%, compounded annually.
Given data
Method
= Future worth method - cost dominated cash flow
i
= 15%
Machine A
P
= ₹. 4,00,000
C1
= C2 = … = C12 = A = ₹.
40,000
n
= 5
S
= ₹. 2,00,000
Machine 2
P
= ₹. 8,00,000
C1
= C2 = … = C12 = A = ₹.
0
n
= 5
S
= ₹. 5,50,000
Formula used
FW(i)
= P (F/P, i, n) + A (F/A, i, n) - S
Solution
Machine 1
FW(15%)1 = 4,00,000 (F/P, 15%, 5) + 40,000 (F/A,
15%, 5) – 5,00,000
= 4,00,000 (2.011) + 40,000 (6.742)
– 5,00,000
=
8,04,400 + 2,69,680 – 5,00,000
FW(15%)1 = ₹. 5,74,080
Machine 2
FW(15%)2 = 8,00,000 (F/P, 15%, 5) + 0 (F/A, 15%, 5)
– 5,50,000
= 8,00,000 (2.011) + 0 (6.742) – 5,50,000
= 16,08,800 + 0 – 5,50,000
FW(15%)2 = ₹. 10,58,800
Result
In future worth
method cost dominated cash flow, the alternative which has minimum future worth
amount is the best alternative. Therefore machine A is the best alternative.
