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We have two constraints: $$ \left[p_1+ q_1e^{\gamma s_1 (1-\alpha_{11}^{*})}\right] \left[p_1+ q_1e^{-\gamma s_1 \alpha_{11}(n_1-1)}\right] \left[p_2+ q_2e^{-\gamma n_2s_2 \alpha_{12}}\right] = 1 $$ and $$ n_2\alpha_{12}+(n_1-1)\alpha_{11} + \alpha_{11}^{*} = 1, $$ subject to $$0 \leq \alpha_{i,j},\alpha_{11}^{*} \leq 1.$$ We are maximizing the function $$ f(\mathbf{\alpha}) = \alpha_{12}. $$ Here, $p_i + q_i \equiv 1$ for all $i$ (i.e. these are probabilities), and the counts $n_i \in \mathbb{N},$ whereas all the other parameters ($s_i,\gamma$) are positive (and $s_i$ are quite "large", if that is relevant). When solving an optimization problem with some values of the parameters using ampl, we get that $\alpha_{11}^{*} = 1$ and $\alpha_{ij} = 0.$ How to prove this formally?

EDIT: The first constraint is crucial to the problem studied.

EDIT': How to prove the above trivial solution is the only solution possible?

Ilonpilaaja
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  • If you remove your first constraint, you have a linear programming problem. Have you checked the convexity of the first constraint? – JMJ Dec 21 '20 at 16:42
  • If you require $p_j+q_j=1$, then why the exponential terms? The brackets in the first constraint will not be equal to $1$ unless the values of $\alpha$ are precisely as you said, so that all the exponential factors are equal to unity – Yuriy S Dec 21 '20 at 16:52
  • What I mean is, $\alpha^*{11}=1$ and $\alpha{i, j}=0$ is an obvious solution to all the constraints. I am not at all certain that any other solution exists – Yuriy S Dec 21 '20 at 17:02

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