--- title: "Both types of BCR, gradient income and cost" subject: "Engineering Economics" chapter: 3 key: "semester_6/engineering_economics:bcr-both-growing-12years" source: https://www.bctengineer.com/q/semester_6/engineering_economics/bcr-both-growing-12years site: BCT Engineering --- # Both types of BCR, gradient income and cost Engineering Economics · Chapter 3: Basic Methodologies of Engineering Economic Analysis - Asked in: 2072 Asoj, 6 marks - Video: [Engineering Economics: Benefit - Cost Ratio (Part 1/2)](https://www.youtube.com/watch?v=KJa45xBT1PQ) — MEL Designs - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/engineering_economics/bcr-both-growing-12years - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-3 ## Question Calculate both types of BCR of a project with the following details. MARR = 12%. | Initial Investment | Annual Income | Annual Cost | Useful Life | Salvage Value | |---|---|---|---|---| | Rs. 100000 | Rs. 20000 at the end of first year and increase by 5% per year | Rs. 3000 at the end of first year and increase by Rs. 500 per year | 12 years | 25000 | ## Short answer Compute PW of benefits (growing at 5%) = Rs 154,014, PW of O&M (Rs 500 gradient) = Rs 31,559, salvage PW = Rs 6,417. Conventional B/C = 1.23, modified B/C = 1.31 (both >1, acceptable). ## Answer Assumption: PW formulation is used. Definitions: - Conventional B/C = PW(B) / [I + PW(O&M) − PW(S)] - Modified B/C = [PW(B) − PW(O&M)] / [I − PW(S)] ```text Benefit : 20,000 grows 5%/yr -> 20,000, 21,000, 22,050 ... Cost : 3,000 + 500/yr -> 3,000, 3,500, 4,000 ... Year 12 : salvage 25,000; Year 0: investment 1,00,000 ``` Factors at i = 12%, n = 12: (P/A) = 6.1944, (P/G) = 25.9523, (P/F) = 0.2567 **PW of benefits (geometric, g = 5%)** PW(B) = 20,000 × [1 − (1.05/1.12)^12] / (0.12 − 0.05) = 20,000 × 7.7007 = Rs 154,014 **PW of O&M cost (gradient)** PW(M) = 3,000 (P/A, 12%, 12) + 500 (P/G, 12%, 12) = 3,000 × 6.1944 + 500 × 25.9523 = 18,583 + 12,976 = Rs 31,559 **PW of salvage** PW(S) = 25,000 × 0.2567 = Rs 6,417 **Conventional B/C** = 154,014 / (1,00,000 + 31,559 − 6,417) = 154,014 / 125,142 = **1.231** **Modified B/C** = (154,014 − 31,559) / (1,00,000 − 6,417) = 122,455 / 93,583 = **1.309** **Answer: conventional B/C = 1.23, modified B/C = 1.31. Both are greater than 1, so the project is acceptable at MARR = 12%.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which formula gives PW of a benefit growing 5% per year at i = 12%? A. B1 x (P/A, 12%, 12) B. B1 x (P/G, 12%, 12) C. B1[1 - (1.05/1.12)^12]/(0.12 - 0.05) D. B1 x (P/F, 12%, 12) Answer: C. Geometric series: PW = B1[1 - ((1+g)/(1+i))^n]/(i - g) = 154,014. (Key point: PW of benefits) 2. How is the cost 3,000 rising by 500 per year valued? A. 3,000(P/A) + 500(P/G) B. 3,000(P/A) only C. 3,000(P/G) + 500(P/A) D. 3,500(P/A) Answer: A. A base annuity plus a gradient: 18,583 + 12,976 = 31,559. (Key point: PW of O&M) 3. Which ratio is higher here and why? A. Conventional, since salvage is added B. Modified (1.31), since O&M is netted from benefits C. They are equal D. Conventional, since O&M is smaller Answer: B. When B/C > 1, moving O&M from the denominator to the numerator raises the ratio. (Key point: Conventional vs modified) 4. How is the Rs 25,000 salvage treated? A. Added to benefits at full value B. Ignored C. Divided by 12 D. Discounted from year 12 (x 0.2567) and subtracted from investment side Answer: D. PW(S) = 6,417 is subtracted in the denominator. 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