- 2066 Magh · 8 marks
P vs Q: 10-year study and capitalized worth
Short answer
Annual worth: AW(P) = Rs 74.79, AW(Q) = Rs 120.91. For 10-year study, PW = AW × 5.6502: P = Rs 422.57, Q = Rs 683.18; select Q. For infinite life, capitalized worth = AW / 0.12: P = Rs 623.24, Q = Rs 1007.60; select Q.
Select the best project (i) if study period is 10 years (ii) using capitalized worth method.
| Project | Initial Investment (NRs.) | Annual Revenue (NRs.) | Annual Expenses (NRs.) | Salvage Value | N | MARR |
|---|---|---|---|---|---|---|
| P | 5000 | 3000 | 2000 | 1000 | 8 year | 12% |
| Q | 3500 | 2000 | 800 | 350 | 4 year | 12% |
Net annual cash flow = revenue − expenses: P = 3000 − 2000 = 1000; Q = 2000 − 800 = 1200. The lives are unequal (8 and 4 years), so the projects are compared on annual worth (AW), which is valid under the repeatability assumption.
Assumption: the study period of 10 years is not a multiple of either life, so each project is assumed to be repeated with the same cash flows, and the AW of one cycle applies over the whole study period.
AW of each project (MARR = 12%)
Project P (N = 8):
AW(P) = −5000(A/P,12%,8) + 1000(A/F,12%,8) + 1000
= −5000(0.20130) + 1000(0.08130) + 1000
= −1006.51 + 81.30 + 1000 = 74.79
Project Q (N = 4):
AW(Q) = −3500(A/P,12%,4) + 350(A/F,12%,4) + 1200
= −3500(0.32923) + 350(0.20923) + 1200
= −1152.32 + 73.23 + 1200 = 120.91
(i) Study period = 10 years
PW over 10 years = AW × (P/A, 12%, 10), with (P/A,12%,10) = 5.6502.
| Project | AW | PW (10 yr) |
|---|---|---|
| P | 74.79 | 422.57 |
| Q | 120.91 | 683.18 |
Q has the larger PW, so Q is the better project for a 10-year study period.
(ii) Capitalized worth (infinite period)
CW = AW / i (the projects are repeated for ever).
- CW(P) = 74.79 / 0.12 = 623.24
- CW(Q) = 120.91 / 0.12 = 1007.60
Answer: Q is selected in both cases (PW₁₀ = NRs 683; CW = NRs 1008), since it has the higher AW and hence the higher PW and CW.
✓ Answer checked by Claude Opus 5.5 (high effort)
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