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I had a linear programming problem where I am optimizing some function

$$f(x) = \sum_{j}x_jq_jp_j - \sum_{i}\left(\sum_{j}x_jq_jC_{ij} \right) c_i$$

Where $q, p, C, c$ are known.

I now want to extend this problem to something where my $p$ vector depends on the $x$ vector.

That is, I want to change the value of $p_j$ depending on the $x_j$ and original $p_j$. In particular, I want to do it in the following way. Let's take the example where:

$x = [1,0,1]^T$ and $p=[30,20,50]^T$

In this case, I would like my vector $p$ to be altered using the following rule: redistribute the % lost in $p$ following the multiplication with $x$ over all other $p$. So in this case, because the second element of $x$ is 0, I want to redistribute the 20/100 lost over the other $p$, such that $p_{new} = [36,0,60]$ (the other elements became 20% higher)

I am not sure how to incorporate this logic into my linear programming problem.

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  • $\begingroup$ Hey, your x variables are binary and you want p = 0 whenever x = 0, and the summation of p values should be the same right? Should it be in linear format? because you can easily model it using multiplications of the binary variable(which is also possible to linearize later)?? $\endgroup$
    – Oguz Toragay
    May 12, 2020 at 14:36
  • $\begingroup$ I don't think I fully follow.. It is indeed true that p=0 when x=0, but summation of x is not the same. Like in the example, initially it was 30+20+50=100, and after it was 36+0+60=96. $\endgroup$
    – BarkingCat
    May 12, 2020 at 14:43
  • $\begingroup$ Never mind, I didn't fully read the question... $\endgroup$
    – Oguz Toragay
    May 12, 2020 at 14:49
  • $\begingroup$ Are you only redistributing when $x_j=0$ (as opposed to when $x_j$ is "small"? In your example, what would $M$ be? You wrote it as if it would be a multiplicand, but it sounds more like a function of the first summation in the objective. $\endgroup$
    – prubin ♦
    May 12, 2020 at 15:58
  • $\begingroup$ Ah yeah you are right. Multiplying by M makes no sense here. So the solution would probably look different.. Indeed we are only redistributing when $x_j=0$ $\endgroup$
    – BarkingCat
    May 12, 2020 at 18:33

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