I'm trying to maximize this problem using MILP (mixed-integer linear programming).

K represents constant values and decision variables are x1, x2 , x3(binary);

$\max \sum_{i = 1}^{p}(x2_{i}\cdot k - x1_{i}\cdot k) $

subject to:

$x4_{i} = x4_{i-1} + x1_{i} \cdot x3_{i} - (1 - x3_{i})\cdot k\cdot x2_{i} $

Since I have two decision variables multiplying in this restriction, I believe I am unable to solve this optimization as a linear problem. Thus, I though about doing the following transformation that would have the same effect and enable a solution by MILP.

$x4_{i} = x4_{i-1} + x1_{i} - k\cdot x2_{i} $

$x2_{i} \leq x2MAX_{i} \cdot (1-x3_{i}) $

$x2_{i} \geq x2MIN_{i} \cdot (1-x3_{i}) $

$x1_{i} \leq x1MAX_{i} \cdot x3_{i} $

$x1_{i} \geq x1MIN_{i} \cdot x3_{i} $

I believe this would solve the problem as xMIN and xMAX are constant values. Can anyone tell me if this makes sense or does this not apply?

  • $\begingroup$ what do you mean? @RodrigodeAzevedo $\endgroup$ – MiguelL Oct 28 '19 at 22:39
  • 1
    $\begingroup$ See this post for how to linearize a product of binary variables. $\endgroup$ – Rob Pratt Oct 28 '19 at 23:06

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