Solved problem on future worth method - 5
Solved problem on future worth method - 5
Jothi
micro castings Ltd. is planning to replace its muffle furnace. It has received
tenders from three different original manufactures of muffle furnace. Other comparative details are as follows:
Details
|
Manufacturers
|
||
1
|
2
|
3
|
|
Initial
cost (₹)
|
₹.
70,00,000
|
₹.
80,00,000
|
₹.
90,00,000
|
Annual
operation and maintenance cost (₹)
|
₹.
6,00,000
|
₹.
5,00,000
|
₹.
4,00,000
|
Life
(years)
|
12
|
12
|
12
|
Salvage
value at the end of vehicle life (₹)
|
₹.
5,00,000
|
₹.
4,00,000
|
₹.
6,00,000
|
Select the best alternative based on
future worth method at i = 13%, compounded annually.
Given data
Method
= Future worth method - cost dominated cash flow
i
= 13%
Manufacturer 1
P
= ₹. 70,00,000
C1
= C2 = … = C12 = A = ₹.
6,00,000
n
= 12
S
= ₹. 5,00,000
Manufacturer 2
P
= ₹. 80,00,000
C1
= C2 = … = C12 = A = ₹.
5,00,000
n
= 12
S
= ₹. 4,00,000
Manufacturer 3
P
= ₹. 90,00,000
C1
= C2 = … = C12 = A = ₹.
4,00,000
n
= 12
S
= ₹. 6,00,000
Formula used
FW(i)
= P (F/P, i, n) + A (F/A, i, n) - S
Solution
Manufacturer 1
FW(13%)1 = 70,00,000 (F/P, 13%, 12) + 6,00,000
(F/A, 13%, 12) – 5,00,000
= 70,00,000 (4.335) + 6,00,000 (25.650)
– 5,00,000
=
3,03,45,000+ 1,53,90,000 – 5,00,000
FW(13%)1 = ₹. 4,52,35,000
Manufacturer 2
FW(13%)2 = 80,00,000 (F/P, 13%, 12) + 5,00,000
(F/A, 13%, 12) – 4,00,000
= 80,00,000 (4.335) + 5,00,000 (25.650)
– 4,00,000
= 3,46,80,000 + 1,28,25,000 – 4,00,000
FW(13%)2 = ₹. 4,71,05,000
Manufacturer 3
FW(13%)3 = 90,00,000 (F/P, 13%, 12) + 4,00,000
(F/A, 13%, 12) – 6,00,000
= 90,00,000 (4.335) + 4,00,000 (25.650)
– 6,00,000
= 3,90,15,000 + 1,02,60,000 – 6,00,000
FW(13%)3 = ₹. 4,86,75,000
Result
In future worth
method cost dominated cash flow, the alternative which has minimum future worth
amount is the best alternative. Therefore Jothi micro castings Ltd plan to buy a muffle
furnace from manufacturer 1.
