Question

I just need help solving for alternative B, I think maybe I am just doing the formula wrong to solve it, so any and all assistance would be appreciated.

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 $25 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 $1.0 million. The annual operating and maintenance costs (including income taxes) would be $0.4 milliorn. 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 $21 million and $14 million, respectively. The facility would be sold 25 years from now with a salvage value of $1.25 million. The annual operating and maintenance costs (including income taxes) will be $450,000 initially and will increase to $0.55 million after the second addition (from the 11th year to the 15th year) and to S0.65 million during the final 10 years (Assume that these costs begin one year subsequent to the actual addition.) On the basis of the present-worth criterion, if the firm uses 20% as a MARR, which alternative should be undertaken? Note: Adopt incremental cost approach Click the icon to view the interest factors for discrete compounding when i 20% per year The present worth of Alternative A is$ 27 million. (Round to one decimal place.) The present worth of Alternative B is million. (Round to one decimal place.)

Answer 1

Answer:

MARR=20%

Expenditure at year 0=$17 million

PV of expenditure at year 0 =$17/(1+20%)^0=17 million

Expenditure at year 10th=$21 million

PV of expenditure at year 10th =$21/(1+20%)^10=3.39 million

Expenditure at year 15 th=$14 million

PV of expenditure at year 15th =$14/(1+20%)^15=0.91 million

PV of expenditure=17+3.39+0.91=$21.3 millions

Salvage Value=$1.25 million

PV of Salvage Value=1.25/(1+20%)^25=$0.013 millions
Annual Operating cost for 1-10 years=$0.45 millions

PV of Annual Operating cost for 1-10 years=$0.45*(1-(1+20%)^-10)/20%=1.89 millions

Annual Operating cost for 11-15 years=$0.55 millions

PV of Annual Operating cost for 11-15 years=$0.55*(1-(1+20%)^-5)/(20%*(1+20%)^10)=0.27 millions

Annual Operating cost for 15-25 years=$0.65 millions

PV of Annual Operating cost for 15-25 years=$0.65*(1-(1+20%)^-10)/(20%*(1+20%)^15)=0.18 millions

PV of Annual Operating cost=1.89+0.27+0.18=$2.34 millions

PW of Alternate B=-PV of expenditure+PV of Salvage Value-PV of Annual Operating cost=-21.3+0.013-2.34

PW of Alternate B=-$23.627