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I have the following optimization problem: $$ \mbox{maximize } j^{*} \mbox{ subject to:} \sum_{j^{*}\leq j\leq J} \min({\bf A}_j,{\bf B}_j) \geq \lambda, \lambda \in \mathbb{R} \mbox{ and } {\bf A}_j,{\bf B}_j > 0 \forall j $$ where the values of $\bf A \in \mathbb{R}^n$ and $\bf B \in \mathbb{R}^n$ (constants) are dependent on their index $j$ in an arbitrary relationship (i.e. their values are arbitrarily predefined). $\lambda$ here is a constant and independent of $j$ or $j^*$. I am looking for an approach to convert this into a series of linear constraints if it is possible.

I have came across this question and this question which deals with the conversion of constraints containing minimum or maximum functions. However I am not sure if a similar method is possible when the summation function is wrapped over the minimum function, or whether the lack of knowledge on the nature of the entries of $\bf A$ and $\bf B$ mean that any attempt would not be possible.

(One point that might be worth nothing is that the constraint might be relaxed because it can be inferred that: $$ \min\left\{\sum_{j^{*}\leq j\leq J} {\bf A}_j, \sum_{j^{*}\leq j\leq J} {\bf B}_j\right\} \geq \sum_{j^{*}\leq j\leq J} \min({\bf A}_j,{\bf B}_j) $$ from which a conversion to linear constraints is indeed possible, from which feasible but suboptimal solutions can still be found. However in the context of my problem this constraint is best not to be relaxed.)

jackson95
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  • Is the min function being used here to mean coordinate-wise minimization?
  • – prubin Jun 09 '20 at 17:40
  • @RobPratt $A_j$ and $B_j$ are constants – jackson95 Jun 10 '20 at 01:40
  • @prubin It means that $\lambda$ is not a function of $j$ or $J^*$. Since $A_j$ and $B_j$ are scalars, the min function could be treated as coordinate-wise minimization. – jackson95 Jun 10 '20 at 01:41
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    Because everything is constant except $j^$, you can just check the inequality for each $j^$ from $J$ down to $1$, stopping as soon as it is satisfied. No solver required. – RobPratt Jun 10 '20 at 01:58
  • @RobPratt Yes actually you are right, any linear or bisection search would work. Thanks! – jackson95 Jun 10 '20 at 14:33
  • @jackson95, you have to be careful with other searches if $A_j$ or $B_j$ can be negative. – RobPratt Jun 10 '20 at 14:35
  • @RobPratt Thanks for pointing out. I've amended the post to state that $A_j, B_j > 0 \forall j$. – jackson95 Jun 10 '20 at 14:37