--- title: "Best project for 4-year and 8-year service periods" subject: "Engineering Economics" chapter: 4 key: "semester_6/engineering_economics:select-best-project-fw-4yr-irr" source: https://www.bctengineer.com/q/semester_6/engineering_economics/select-best-project-fw-4yr-irr site: BCT Engineering --- # Best project for 4-year and 8-year service periods Engineering Economics · Chapter 4: Comparative Analysis of Alternatives - Asked in: 2069 Bhadra, 4+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/select-best-project-fw-4yr-irr - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-4 ## Question From the following information select the best project. | | Project A | Project B | |---|---|---| | Initial Investment | 35,000 | 50,000 | | Annual Revenue | 16,450 | 25,000 | | Annual costs | 3,000 | 13,830 | | Useful life | 4 years | 8 years | | Salvage value at the end of useful life | 0 | 0 | MARR = 10% When service period required is: i) 4 years by FW method ii) 8 years by IRR method with PW formulation ## Short answer For 4-year service, FW method: Project A has FW Rs. 11,178 while B has FW Rs. -21,365, so choose A. For 8-year service, incremental IRR of B over A is 3.17% below 10% MARR, so A is best. ## Answer Net annual revenue: A = 16,450 - 3,000 = Rs. 13,450; B = 25,000 - 13,830 = Rs. 11,170. MARR = 10%. #### (i) Service period 4 years, FW method Project A completes its 4-year life; B is used for only 4 of its 8 years. Assumption: the only market value given for B is zero at the end of its life, so the market value of B at the end of year 4 is taken as zero (nothing recovered). FW(A) = -35,000(F/P, 10%, 4) + 13,450(F/A, 10%, 4) (F/P, 10%, 4) = 1.4641, (F/A, 10%, 4) = 4.6410 FW(A) = -51,244 + 62,421 = **Rs. 11,178** FW(B) = -50,000(1.4641) + 11,170(4.6410) = -73,205 + 51,840 = **Rs. -21,365** **Answer (i): Select A** (positive FW; B is negative if its assets have no resale value at year 4). #### (ii) Service period 8 years, IRR with PW formulation A is repeated once (second machine bought at year 4 for the same cost); B runs its full 8 years. ```text Year 0 1-3 4 5-8 A -35,000 13,450 -21,550 13,450 (-35,000 + 13,450 at yr 4) B -50,000 11,170 11,170 11,170 B - A -15,000 -2,280 +32,720 -2,280 ``` IRR of each project on its own (8-year flows): - A: PW(i) = -35,000 + 13,450(P/A,i,8) - 35,000(P/F,i,4) = 0 → IRR = 19.73% - B: PW(i) = -50,000 + 11,170(P/A,i,8) = 0 → IRR = 15.08% Both exceed 10%. Incremental (B - A): PW(i) = -15,000 - 2,280(P/A,i,8) + 35,000(P/F,i,4) = 0 At i = 0%: +1,760; at 4%: -433; so IRR(B - A) ≈ **3.17%**, which is less than MARR = 10%. (PW at 10% = -3,258, negative.) **Answer (ii): Select A.** The extra Rs. 15,000 in B earns only 3.2% < 10%. (Also PW(A) = 12,849 > PW(B) = 9,591.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a 4-year service period, how is B treated? A. Repeated B. Used 4 of 8 years, market value taken as zero C. Full life D. Salvage 50,000 Answer: B. With no market value given, nothing is recovered at year 4. (Key point: 4-year FW) 2. Which FWs? A. Both positive B. Both zero C. A +11,178; B -21,365 D. A -21,365; B +11,178 Answer: C. A is selected in part (i). (Key point: FW result) 3. For 8 years, how is A handled? A. Salvage added B. Repeated once, second machine at year 4 C. Used 4 years only D. Ignored Answer: B. A is bought again at year 4. (Key point: 8-year repeat) 4. Why is A selected? A. A has negative PW B. B has higher IRR C. MARR is zero D. Incremental IRR of B-A is 3.17% < 10% Answer: D. The extra Rs 15,000 in B earns less than MARR. 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