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I have a problem where I have a bunch of inequalities in the form:

  • $a_{1,1}b_{1,1} + a_{1,2}b_{1,2} + ... + a_{1,n}b_{1,n} > a_{2,1}b_{2,1} + a_{2,2}b_{2,2} + ... + a_{2,n}b_{2,n}$

  • $a_{2,1}b_{2,1} + a_{2,2}b_{2,2} + ... + a_{2,n}b_{2,n} > a_{3,1}b_{3,1} + a_{3,2}b_{3,2} + ... + a_{3,n}b_{3,n}$

  • $a_{1,1}b_{1,1} + a_{1,2}b_{1,2} + ... + a_{1,n}b_{1,n} < a_{3,1}b_{3,1} + a_{3,2}b_{3,2} + ... + a_{3,n}b_{3,n}$

  • etc.

I've randomly distributed the < and > signs in the inequalities for the example; they can be any of ≤, ≥, <, or >. All the b variables are dummy variables (either 0 or 1).

I want to find values of all the a variables, assuming possible given the data. This seems like a linear programming problem, but I'm struggling to convert it to one.

One further problem is that I may come to a point where there is not solution by classic linear programming (ex. the example above). In this case, I would want to find the point that is "closest" to being a valid point (think OLS error reduction). I know how to do this for linear programming (another reason I jumped to lp), but am unsure on how to do it for the problem statement above. As a side note, I plan to do this mainly computationally.

What method would I use to solve the above?

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    $\begingroup$ What is a "dummy variable"? Do you mean they are constants or actual (binary) optimization variables in your model? $\endgroup$
    – Michal Adamaszek
    Nov 30, 2020 at 8:10
  • $\begingroup$ They're binary data variables, so for any equation we will end up with something of the form $a_{1,1} + a_{1,3} + \dots + a_{1,n} > a_{2,3} + a_{2,7} + \dots + a_{2,n}$. The b variables are there for expository ease $\endgroup$
    – user760900
    Nov 30, 2020 at 8:13
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    $\begingroup$ So the $b$ come as part of input? Well then you already have an LP: every constraint is a linear inequality in the $a_{i,j}$s. You can maybe move all terms to one side to get a more standard form if necessary. $\endgroup$
    – Michal Adamaszek
    Nov 30, 2020 at 8:15
  • $\begingroup$ ah that's embarrassing. if you would like to make that an answer, i can accept it $\endgroup$
    – user760900
    Nov 30, 2020 at 8:17

1 Answer 1

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Accepting Micheal Adamaszek's answer that he commented. It's already in LP form, just move the terms to one side to get to standard form.

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