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Homework answers / question archive / A power plant is being considered in the dead sea location
A power plant is being considered in the dead sea location. For an initial investment
of $130 million, annual net revenues are estimated to be $15 million in years 1–5 and $20 million
in years 6–20. Assume no residual market value for the plant.
a. What is the simple payback period for the plant?
b. What is the discounted payback period when the MARR is 9% per year?
c. Using an equivalency technique (FW, PW, or AW), MARR is 9% per year, would you
recommend investing in this project?
(a)
Simple Payback period (PBP) is the time by when cumulative cash flow is equal to zero.
| Year | Cash flow ($M) | Cumulative Cash Flow ($M) |
| 0 | -130 | -130 |
| 1 | 15 | -115 |
| 2 | 15 | -100 |
| 3 | 15 | -85 |
| 4 | 15 | -70 |
| 5 | 15 | -55 |
| 6 | 20 | -35 |
| 7 | 20 | -15 |
| 8 | 20 | 5 |
PBP lies between years 7 & 8.
PBP = 7 + (Absolute value of cumulative discounted cash flow, year 7 / Discounted cash flow, year 8)
= 7 + (15 / 20)
= 7 + 0.75
= 7.75 years
(b)
Discounted payback period (DPBP) is the time by when cumulative discounted cash flows becomes zero.
| Year | Cash flow ($M) | PV Factor @9% | Discounted Cash Flow ($M) | Cumulative Discounted Cash Flow ($M) |
| 0 | -130 | 1.0000 | -130.00 | -130.00 |
| 1 | 15 | 0.9174 | 13.76 | -116.24 |
| 2 | 15 | 0.8417 | 12.63 | -103.61 |
| 3 | 15 | 0.7722 | 11.58 | -92.03 |
| 4 | 15 | 0.7084 | 10.63 | -81.40 |
| 5 | 15 | 0.6499 | 9.75 | -71.66 |
| 6 | 20 | 0.5963 | 11.93 | -59.73 |
| 7 | 20 | 0.5470 | 10.94 | -48.79 |
| 8 | 20 | 0.5019 | 10.04 | -38.75 |
| 9 | 20 | 0.4604 | 9.21 | -29.54 |
| 10 | 20 | 0.4224 | 8.45 | -21.10 |
| 11 | 20 | 0.3875 | 7.75 | -13.34 |
| 12 | 20 | 0.3555 | 7.11 | -6.23 |
| 13 | 20 | 0.3262 | 6.52 | 0.29 |
DPBP lies between years 12 & 13.
DPBP = 12 + (Absolute value of cumulative discounted cash flow, year 12 / Discounted cash flow, year 13)
= 12 + (6.23 / 6.52)
= 12 + 0.96
= 12.96 years
(c)
We use PW method.
PW ($M) = - 130 + 15 x P/A(9%, 5) + 20 x P/A(9%, 15) x P/F(9%, 5)
= - 130 + 15 x 3.8897 + 20 x 8.0607 x 0.6499
= - 130 + 58.35 + 104.77
= 33.12
Since PW > 0, project is acceptable.
