Optimize capacity by marginal cost-benefit

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:

$$\Delta E = \Delta P \times \frac{t_k}{100} \times 8760,\quad MB = 0.08\,\Delta E,\quad MC = c\,\Delta P$$

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.)