Use the formula below, and explain every step please. Don't use Excell, use the formula on the page.
Thank you!!
Initial Investment = 260,000
Annual O&M Cost = 20,000
Annual Benefits = 60,000
Net Annual Benefits = 60,000 – 20,000 = 40,000
Periodic maintenance at every 4 years = 30,000 (at 4th year and at 8th year)
Salvage Value = 20,000
Life = 10 years
MARR = 5%
Calculate the ROR and suggest whether to buy/reject the purchase of wielding machine.
Calculating ROR using Trial and Error Method.
Let the MARR is 5%
Calculate the PW of the investment at MARR of 5%
PW = -260,000 – 30,000 (P/F, 5%, 4) – 30,000 (P/F, 5%, 8) + 40,000 (P/A, 5%, 10) + 20,000 (P/F, 5%, 10)
PW = -260,000 – 30,000 (0.8227) – 30,000 (0.6768) + 40,000 (7.7217) + 20,000 (0.6139) = 16,161
The PW is positive. So, increase the rate of interest to get negative PW.
Increase the MARR to 6% and calculate the PW.
PW = -260,000 – 30,000 (P/F, 6%, 4) – 30,000 (P/F, 6%, 8) + 40,000 (P/A, 6%, 10) + 20,000 (P/F, 6%, 10)
PW = -260,000 – 30,000 (0.7921) – 30,000 (0.6274) + 40,000 (7.3601) + 20,000 (0.5584) = 2,987
The PW is still positive. So, increase the rate of interest to get negative PW.
Increase the MARR to 7% and calculate the PW.
PW = -260,000 – 30,000 (P/F, 7%, 4) – 30,000 (P/F, 7%, 8) + 40,000 (P/A, 7%, 10) + 20,000 (P/F, 7%, 10)
PW = -260,000 – 30,000 (0.7629) – 30,000 (0.5820) + 40,000 (7.0236) + 20,000 (0.5083) = -9,237
Now, use interpolation and calculate the IRR.
IRR = 6% + [2,987 – 0 ÷ 2,987 – (-9,237)] * 1%
IRR = 6.24%
As the IRR is greater than the given MARR, the decision should be to BUY the wielding machine.
Use the formula below, and explain every step please. Don't use Excell, use the formula on...
The formula is NPW but I'm a little confused on how it is implemented. Write step by step, please. Don't use Excell please Given 2 alternatives: A B First Cost 4,000 6,000 500 1,000 2,200 Annual Cost Annual Bene fit 2,000 Life 5 10 1,000 Salvage 3,000 If i 10%, find the better alternative computing NPW of both alternatives Assume alternative A is replaced at the end of its useful life. Solution: gr pW