Solved problem on future worth method - 4
Solved problem on future worth method - 4
OLA
taxi company is considering to buy taxis with diesel engines
instead of petrol engines. The cars average travel is 50,000km per year, with a useful life
of three years for the taxi with the petrol engine and diesel engine. Other
comparative information are as follows:
Information
|
Diesel
|
Petrol
|
Vehicle
cost
|
₹.
5,00,000
|
₹.
4,00,000
|
Fuel
cost per litre
|
₹.
9
|
₹.
24
|
Mileage
in km/litre
|
30
|
20
|
Annual
insurance premium
|
₹.
500
|
₹.
500
|
Salvage
value at the end of vehicle life
|
₹.
70,000
|
₹.
1,00,000
|
Determine the more economical choice
based on future worth method at i = 15%, compounded annually.
Given data
Method
= Future worth method - cost dominated cash flow
i
= 15%
n
= 3
Diesel Car
P
= ₹. 5,00,000
Total
fuel cost per year = (50,000/30) * 9 = ₹.
15,000
Annual
cost = Fuel cost per year + Insurance premium per year = 15,000 + 500 = ₹. 15,500
Therefore,
C1 = C2 = C3 = A = ₹. 15,500
S
= ₹. 70,000
Petrol Car
P
= ₹. 4,00,000
Total
fuel cost per year = (50,000/20) * 24 = ₹.
60,000
Annual
cost = Fuel cost per year + Insurance premium per year = 60,000 + 500 = ₹. 60,500
Therefore,
C1 = C2 = C3 = A = ₹.
60,500
S
= ₹. 1,00,000
Formula used
FW(i)
= P (F/P, i, n) + A (F/A, i, n) - S
Solution
Diesel Car
FW(15%)1 = 5,00,000 (F/P, 15%, 3) + 15,500 (F/A,
15%, 3) – 70,000
= 5,00,000 (1.521) + 15,500 (3.472)
– 70,000
= 7,60,500 + 53,816 – 70,000
FW(15%)1 = ₹. 7,44,316
Petrol Car
FW(15%)2 = 4,00,000 (F/P, 15%, 3) + 60,500 (F/A,
15%, 3) – 1,00,000
= 4,00,000 (1.521) + 60,500
(3.472) – 1,00,000
= 6,08,400 + 2,10,056 – 1,00,000
FW(15%)2 = ₹. 7,18,456
Result
In future worth
method cost dominated cash flow, the alternative which has minimum future worth
amount is the best alternative. Therefore petrol car is the best alternative.
