--- title: "P vs Q: 10-year study and capitalized worth" subject: "Engineering Economics" chapter: 4 key: "semester_6/engineering_economics:pq-study-10-years-capitalized-worth" source: https://www.bctengineer.com/q/semester_6/engineering_economics/pq-study-10-years-capitalized-worth site: BCT Engineering --- # P vs Q: 10-year study and capitalized worth Engineering Economics · Chapter 4: Comparative Analysis of Alternatives - Asked in: 2066 Magh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/engineering_economics/pq-study-10-years-capitalized-worth - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-4 ## Question 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% | ## 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. ## Answer 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%) ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are P and Q compared on annual worth? A. Their lives (8 and 4 years) are unequal and repeatability is assumed B. They have equal lives C. AW ignores the time value of money D. The salvage values are equal Answer: A. With unequal lives, AW of one cycle can be compared directly under the repeatability assumption. (Key point: Why AW is used) 2. Which expression gives AW(P)? A. -5000(A/F,12%,8) + 1000(A/P,12%,8) + 1000 B. -5000(A/P,12%,8) + 1000(A/F,12%,8) + 1000 C. -5000(A/P,12%,8) - 1000(A/F,12%,8) + 1000 D. -5000(P/A,12%,8) + 1000(P/F,12%,8) + 1000 Answer: B. Capital recovery uses (A/P), salvage is converted with (A/F), and net annual cash flow (3000-2000) is added. (Key point: AW formula with salvage) 3. How is capitalized worth obtained from AW? A. CW = AW x i B. CW = AW x (P/A,12%,10) C. CW = AW / i D. CW = AW / N Answer: C. For an infinite repeated horizon, CW = AW / i, e.g. 120.91/0.12 = 1007.60 for Q. (Key point: Capitalized worth) 4. Which project is selected and why? A. P, because it has the larger investment B. P, because its life is longer C. Neither, since AW values are small D. Q, because its AW (120.91) is higher than P's (74.79), giving higher PW and CW Answer: D. Q has higher AW and so higher PW over 10 years (683) and higher CW (1008). 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