Solved problem on present worth method - 2
Solved problem on present worth method - 2
The
cash flow of two project proposals is as given below. Each of has an expected
life of 10 years. Select the best project which is based on the present worth
method of comparison assuming 18% interest rate, compounded annually.
|
|
Initial outlay (Rs.)
|
Annual Revenue (Rs.)
|
Salvage value after 10 years (Rs.)
|
|
Project
1
|
-7,50,000
|
2,00,000
|
50,000
|
|
project
2
|
-9,50,000
|
2,25,000
|
1,00,000
|
Given data
Method
= Present worth method - revenue dominated cash flow
i
= 18%
n
= 10 years
Project 1
P
= Rs. 7,50,000
R1
= R2 = … = R10 = A = Rs. 2,00,000
S
= Rs. 50,000
Project 2
P
= Rs. 9,50,000
R1
= R2 = … = R10 = A = Rs. 2,25,000
S
= Rs. 1,00,000
Formula used
Here
the revenue is equal for all the years.
Therefore
PW(i) = -P + A (P/A,
i, n) + S (P/F, i, n)
Solution
Technology 1
PW(18%)1 = -7,50,000 + 2,00,000 (P/A, 18%, 10) + 50,000
(P/F, 18%, 10)
= -7,50,000 + 2,00,000 (4.4941) +
50,000 (0.1911)
= -7,50,000 + 8,98,820 + 9,555
PW(18%)1 = Rs. 1,58,375
Project 2
PW(18%)2 = -9,50,000 + 2,25,000 (P/A, 18%, 10) + 1,00,000
(P/F, 18%, 10)
= -9,50,000 + 2,25,000 (4.4941) +
1,00,000 (0.1911)
= -25,00,000 + 10,11,172.50 +
19,110
PW(18%)2 = Rs. 80,282.50
Result
In present worth
method revenue dominated cash flow, the alternative which has maximum present
worth amount is the best alternative. Therefore alternative 1 i.e project 1 is the
best alternative.
