Solved problem on present worth method - 4
Solved problem on present worth method - 4
An
engineer has two alternatives for an elevator installed in a new building. The
details of the alternative for the elevators are as follows:
Engineer’s estimates
|
|||
Initial cost (Rs.)
|
Annual operation & maintenance cost
(Rs.)
|
Service life (Years)
|
|
Alternative
1
|
5,50,000
|
27,500
|
12
|
Alternative
2
|
6,20,000
|
30,000
|
12
|
Determine
which alternative should be accepted, based on the present worth method of
comparison assuming 20% interest rate, compounded annually.
Given data
Method
= Present worth method - cost dominated cash flow
i
= 20%
Alternative 1
P
= Rs. 5,50,000
C1
= C2 = … = C10 = A = Rs. 27,500
n
= 12 years
S
= Rs.0
Alternative 2
P
= Rs. 6,20,000
C1
= C2 = … = C10 = A = Rs. 30,000
n
= 12 years
S
= Rs.0
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
Alternative 1
PW(20%)1 = 5,50,000 + 27,500 (P/A, 20%, 12) - 0
(P/F, 20%, 12)
= 5,50,000 + 27,500 (4.4392) - 0
= 5,50,000 + 1,22,078
PW(20%)1 = Rs. 6,72,078
Alternative 2
PW(20%)2 = 6,20,000 + 30,000 (P/A, 20%, 12) - 0
(P/F, 20%, 12)
= 6,20,000 + 30,000 (4.4392) - 0
= 6,20,000 + 1,33,176
PW(20%)2 = Rs. 7,53,176
Result
In
present worth method cost dominated cash flow, the alternative which has minimum
present worth amount is the best alternative. Therefore the elevator is
purchased from alternative 1 and installed.
