--- title: "Optimize capacity by marginal cost-benefit" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:marginal-cost-benefit-installed-capacity" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/marginal-cost-benefit-installed-capacity site: BCT Engineering --- # Optimize capacity by marginal cost-benefit Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2080 Baishakh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/marginal-cost-benefit-installed-capacity - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question A RoR hydropower plant is proposed in a river. Using the marginal cost and benefit method, optimize the installed capacity with the following data: interest rate = 12%; energy price = \$0.08/kWh; fixed cost = \$60 × 10⁶; variable cost (electro-mechanical) = \$650/kW; annual O&M = 3% of variable cost; project life = 35 years. | % Time | 8.33 | 16.67 | 25.0 | 33.33 | 41.67 | 50.0 | 58.33 | 66.67 | 75.0 | 83.33 | 91.67 | 100 | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Power (kW) | 805 | 604 | 564 | 407 | 300 | 167 | 130 | 115 | 96 | 89 | 74 | 65 | ## Short answer Add capacity as long as marginal benefit exceeds marginal cost. Annual cost per kW is about $99, and the step from 604 to 805 kW gives MB < MC, so optimum installed capacity is 604 kW. ## Answer Capacity should be increased as long as the **marginal benefit (MB) of the added capacity is at least its marginal cost (MC)**. The fixed cost (\$60 × 10⁶) does not depend on the capacity, so it does not enter the marginal comparison. #### Data - Interest $i=12\%$, life $n=35$ years: $CRF=\dfrac{i(1+i)^n}{(1+i)^n-1}=0.12232$ - Annual cost of 1 kW of electro-mechanical equipment $c = 650\,(CRF+0.03) = 650\times(0.12232+0.03) = 99.01$ USD per kW per year - Energy price = \$0.08/kWh #### Method The power duration table gives the percentage of time each power level is equalled or exceeded. Raising the capacity from the next lower power level to the level $P_k$ adds $\Delta P$ kW, which can be used only for the percentage of time $t_k$ for which that power is available: $$ \Delta E = \Delta P \times \frac{t_k}{100} \times 8760,\quad MB = 0.08\,\Delta E,\quad MC = c\,\Delta P $$ | Capacity step (kW) | ΔP (kW) | Time used (%) | ΔE (kWh) | MB (\$) | MC (\$) | MB/MC | |---|---|---|---|---|---|---| | 604 → 805 | 201 | 8.33 | 146,671 | 11,734 | 19,900 | 0.59 | | 564 → 604 | 40 | 16.67 | 58,412 | 4,673 | 3,960 | 1.18 | | 407 → 564 | 157 | 25.0 | 343,830 | 27,506 | 15,544 | 1.77 | | 300 → 407 | 107 | 33.33 | 312,409 | 24,993 | 10,594 | 2.36 | | 167 → 300 | 133 | 41.67 | 485,489 | 38,839 | 13,168 | 2.95 | | 130 → 167 | 37 | 50.0 | 162,060 | 12,965 | 3,663 | 3.54 | | 115 → 130 | 15 | 58.33 | 76,646 | 6,132 | 1,485 | 4.13 | | 96 → 115 | 19 | 66.67 | 110,966 | 8,877 | 1,881 | 4.72 | | 89 → 96 | 7 | 75.0 | 45,990 | 3,679 | 693 | 5.31 | | 74 → 89 | 15 | 83.33 | 109,496 | 8,760 | 1,485 | 5.90 | | 65 → 74 | 9 | 91.67 | 72,273 | 5,782 | 891 | 6.49 | | 0 → 65 | 65 | 100 | 569,400 | 45,552 | 6,435 | 7.08 | #### Decision The ratio MB/MC is below 1 for the step 604 → 805 kW (MB = \$11,734 < MC = \$19,900), so this extra capacity is not worth building. The step 564 → 604 kW has MB/MC = 1.18 (more than 1), so it is justified, and all the lower steps are also justified. **Answer: the optimum installed capacity is 604 kW** (the capacity exceeded 16.67% of the time). (If the curve is interpolated linearly between the 8.33% and 16.67% points, break-even occurs where $0.08\times8760\times t = 99.01$, i.e. at about 14.1% of the time, which corresponds to about 665 kW; with the tabulated power levels, 604 kW is the answer.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the fixed cost of $60 million left out of the marginal comparison? A. It is paid by the government, not the project B. It is already included in the energy price C. It does not change with installed capacity, so it does not affect MB versus MC D. It is smaller than the O&M cost Answer: C. Marginal analysis compares only the costs and benefits that change when capacity is added. The fixed cost is the same for any capacity. (Key point: Fixed cost in marginal analysis) 2. How is the annual cost of 1 kW of electro-mechanical equipment obtained? A. 650 x CRF only B. 650 x (CRF + 0.03), i.e. annualised capital plus 3% O&M C. 650 x 0.03 only D. 650 / 35 years Answer: B. The capital is annualised with the capital recovery factor and the annual O&M of 3% is added, giving about 99 USD per kW per year. (Key point: Annual cost per kW) 3. How is the extra energy of a capacity step from the lower power level to P_k calculated? A. Delta P x 8760 x 100% B. P_k x 8760 x (t_k/100) C. Delta P x (t_k/100) x 8760, since the extra power is available only for the percentage of time t_k D. Delta P x t_k only Answer: C. The added power can be used only for the fraction of time that flow at that level is available, so ΔE = ΔP x t_k/100 x 8760. (Key point: Energy of an added capacity step) 4. Which capacity is optimal and why? A. 805 kW, the largest available power B. 604 kW, because the step 564 to 604 still has MB/MC > 1 while the next step 604 to 805 has MB < MC C. 564 kW, because MB/MC is highest there D. 65 kW, because MB/MC is highest at 100% time Answer: B. Capacity is increased while MB is at least MC. The 604 to 805 step has MB 11,734 < MC 19,900, so it is rejected, and 604 kW is the optimum. 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