Need help with this problem... An electrical utility is experiencing a sharp pow
ID: 1170184 • Letter: N
Question
Need help with this problem...
An electrical utility is experiencing a sharp power demand that continues to grow at a high rate in a certain local area Two alternatives are under consideration. Each is designed to provide enough capacity during the next 25 years, and both will consume the same amount of fuel, so fuel cost is not considered in the analysis Alternative A. Increase the generating capacity now so that the ultimate demand can be met without additional expenditures later. An investment of $29 million would be required, and it is estimated that this plant facility would be in service for 25 years and have a salvage value of S0.7 million. The annual operating and maintenance costs (including income taxes) would be $0.6 million Alternative B. Spend $17 million now and follow this expenditure with future additions during the 10th year and the 15th year. These additions would cost 520 million and $13 million, respectively. The facility would be sold 25 years from now with a salvage value of $1.1 million. The annual operating and maintenance costs (including income taxes) will be $250,000 initially and will increase to 50.35 milion after the second addition (from the 11th year to the 15th year) and to $0.45 million during the final 10 years. (Assume that these costs begin one year subsequent to the actual addition.) On the basis of the present-worth criterion, if the firm uses 19% as a MARR, which alternative should be undertaken? Note: Adopt incremental cost approach Click the icon to view the interest factors for discrete compounding when 19% per year The present worth of Alternative A is S million. (Round to one decimal place.)Explanation / Answer
Solution:
Increamental Cost Approach (Amount in Million $)
(Note:Negative figure suggests saving while positive suggests cost)
If we choose Alternative B :
= 17 - 29
= -12
2. Increamental Recurring Operational Costs :
= 0.25 - 0.6
= -0.35
= 20 + 0.25 - 0.6
= 19.65
= 0.35 - 0.6
= -0.25
= 13 + 0.35 - 0.6
= 12.75
= 0.45 - 0.6
= -0.15
3. Salvage Value at the end of 25th year
= 1.1 - 0.7
= 0.4
Present Worth of the alternative
Since the net present worth is negative (Saving), Alternative B should be undertaken.
Alternative B will give an extra saving of $ 9.2371 Million
Amount in Million $
Note : Positive figures suggests cost while negative suggests saving
1. Initial Investment = 29
2. Recurring Operating Cost for 25 years = 0.6
3. Salvage Value at the end of 25th year = 0.7
Present Value
Net Present Worth of Alternate A is $32.1 Million
Year Cost/Saving Present Value Factor Present Value 0 -12 1 -12 1 to 9 -0.35 4.16333 -1.4571 10 19.65 0.1756 3.4505 11 to 14 -0.25 0.46335 -0.1158 15 12.75 0.0736 0.9384 16 to 25 -0.15 0.31929 -0.0479 Salvage Value -0.4 0.01292 -0.0052 NET PRESENT VALUE -9.2371