How to use maths to optimize hydro turbines?

Question

I need to prove that for a given total power produced by $n$ turbines, the best power production distribution between the turbines is the even distribution.

All the turbines have the same $\eta = f(P)$ curve. Here is an example of a curve:

This function is known to be concave ($f'$ is decreasing, $f''$ is less than or equal to zero). How can I prove that for a given number of active turbines, the most efficient power production distribution between the turbines is the even distribution?

I tried to prove with the axioms

$f'$ is decreasing

$f'' \leq 0$

$\sum_{i=1}^{n} P_i = P_{total}$

that

$max \sum_{i=1}^{n} f(P_i)$ is obtained for $P_1 = ... = P_n$

but didn't succeed.

Thanks in advance for your help

Answer 1

Assume there are indices $i$ and $j$ such that $P_i\neq P_j.$ Now we create a new configuration $P^{\star}_1,\ldots,P^{\star}_n$ with $P^{\star}_k = P_k$ for $k\not\in\{i,j\}$ and $P^{\star}_i=P^{\star}_j = \frac{1}{2}(P_i+P_j).$ Then $\sum_{k=1}^n P^{\star}_k = \sum_{k=1}^n P_k= P_{total}$ and $$ \sum_{k=1}^n f(P^{\star}_k) - \sum_{k=1}^n f(P_k) = 2\left( f\left(\frac{1}{2}(P_i+P_j)\right) - \frac{1}{2}\left(f(P_i)+f(P_j)\right)\right) > 0 $$ because $f$ is concave. In other words, if the $P_i$ are not all the same, you can obtain a better configuration simply by moving two of them towards their arithmetic mean. This is turn means that all $P_i$ must be the same in the optimal configuration.

Answer 2

Suppose that $f(p) = \max(0,f_0-(p-p_0)^2)$ then the problem can be stated as

$$ \max \sum_{k=1}^nf(p_k)\ \ \text{s. t.}\ \ \sum_{k=1}^n p_k = P $$

and the corresponding lagrangian

$$ L(p_1,\cdots,p_k,\cdots,p_n,\lambda) = \sum_{k=1}^nf(p_k)+\lambda \left(\sum_{k=1}^n p_k - P\right) $$

with stationary points at the solutions of

$$ \nabla L = \cases{2(p_k-p_0)+\lambda = 0, k=1,\cdots, n\\ \sum_{k=1}^n p_k - P=0} $$

or at $p_k = p_0-\frac{\lambda}{2}$, then $n\left(p_0-\frac{\lambda}{2}\right)=P$ or $\lambda = 2\left(p_0-\frac P2\right)$. Now knowing $\lambda$ we follow with

$$ 2(p_k-p_0)+2\left(p_0-\frac Pn\right) = 0\Rightarrow p_k = \frac Pn $$