--- title: "Best of P and Q for 5-year study period" subject: "Engineering Economics" chapter: 4 key: "semester_6/engineering_economics:best-project-pq-study-5yr" source: https://www.bctengineer.com/q/semester_6/engineering_economics/best-project-pq-study-5yr site: BCT Engineering --- # Best of P and Q for 5-year study period Engineering Economics · Chapter 4: Comparative Analysis of Alternatives - Asked in: 2068 Jestha, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/engineering_economics/best-project-pq-study-5yr - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-4 ## Question Select the best project. Required study period is 5 years. | | Project P | Project Q | |---|---|---| | Initial Investment (Rs.) | 5,00,000 | 3,50,000 | | Annual net revenue (Rs.) | 2,00,000 | 1,75,000 | | Salvage value (Rs.) | 50,000 | 35,000 | | Useful life (Years) | 6 | 5 | | MARR | 10% | 10% | ## Short answer Calculate present worth over the 5-year study: PW(P) = Rs. 289,203, PW(Q) = Rs. 335,120. Incremental IRR of P over Q is about -2.1% below 10% MARR. Project Q is best. ## Answer Study period = 5 years, MARR = 10%. #### Assumptions - Project Q has a life equal to the study period (5 years): salvage Rs. 35,000 is received at year 5. - Project P has a life of 6 years but is needed for 5 years only. It is sold at the end of year 5; as no other market value is given, take the stated salvage (Rs. 50,000) as the year-5 market value. #### Cash-flow diagram (Rs., '000) ```text P: 200 200 200 200 200+50 | | | | | 0 --1-----2-----3-----4------5--> year v 500 Q: 175 175 175 175 175+35 | | | | | 0 --1-----2-----3-----4------5--> year v 350 ``` #### Present worth (P/A, 10%, 5) = 3.7908, (P/F, 10%, 5) = 0.6209 PW(P) = -500,000 + 200,000(3.7908) + 50,000(0.6209) = -500,000 + 758,157 + 31,046 = **Rs. 289,203** PW(Q) = -350,000 + 175,000(3.7908) + 35,000(0.6209) = -350,000 + 663,388 + 21,732 = **Rs. 335,120** #### Check by incremental IRR (P over Q) Extra investment 150,000, extra net revenue 25,000 per year, extra salvage 15,000 at year 5. PW of increment at 10% = -150,000 + 25,000(3.7908) + 15,000(0.6209) = -45,917 < 0 Incremental IRR ≈ -2.1% < 10%, so the extra investment in P is not justified. | Project | IRR | PW at 10% | |---|---|---| | P | 30.1% | 289,203 | | Q | 42.1% | 335,120 | **Answer: Select Project Q** (higher PW, Rs. 335,120, and the extra investment in P earns less than MARR). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is P (6-year life) treated for a 5-year study period? A. Used 6 years B. Sold at year 5, stated salvage taken as market value C. Repeated D. Salvage ignored Answer: B. Assumed sold at year 5 for Rs 50,000. (Key point: Study period) 2. Which PWs? A. Both negative B. P 500,000 C. P 289,203; Q 335,120 D. P 335,120; Q 289,203 Answer: C. Q has higher PW. (Key point: PW values) 3. What does incremental analysis of P over Q show? A. IRR 42% B. Extra is justified C. IRR about -2.1% < 10%, extra not justified D. IRR 30% Answer: C. PW of increment is -45,917. (Key point: Incremental) 4. Which is selected? A. P B. Q C. Neither D. Both Answer: B. Higher PW and extra investment in P unjustified. 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