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I have a problem that I managed to write as a binary integer linear program. As a natural first step, I relaxed the integrality constraint to solve a regular LP. To my surprise, the solutions where all integral (either $0$ or $1$).

I then ran some simulations on random instances of my problem and, to my surprise, all were integral.

I started looking into the constraint matrix to try to prove it is totally unimodular, but managed to find a counterexample [An instance whose constraint matrix is not totally unimodular].

So I am left puzzled. I've been running simulations for the past two days on random instances of the problem in hope to find a counterexample [Where the solution of the LP are non integral] to no avail.

What is the next step after total unimodularity that you can explore to prove that the solution of an LP formulation is integral?

Rodrigo de Azevedo
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AspiringMat
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  • To my knowledge, the literature states that the linear program $\max{c^Tx:Ax\leqslant b, x\geqslant0}$ has an integral optimal solution for all integer vectors $b$ for which it has a finite optimal value if and only if $A$ is totally unimodular. Have you tried varying the RHS? – Math1000 Jan 22 '20 at 05:22
  • @Math1000 My RHS ($b$) is actually constant for all instances [and integral either $0,1,$ or $-1$]. So it is mainly the matrix A that I've been changing based on the different instances. – AspiringMat Jan 22 '20 at 05:44
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    http://or.stackexchange.com may be a better home for this question. – Rodrigo de Azevedo Jan 22 '20 at 11:35
  • Just to be clear, the instance that was not TUM still coughed up an integral solution? – prubin Jan 23 '20 at 20:24
  • @prubin Yes, exactly – AspiringMat Jan 24 '20 at 00:03
  • Are you maximizing a linear function over a constraint set whose extreme points all have integer coordinates? – passerby51 May 22 '22 at 04:48

1 Answers1

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When TUM fails, you try a weaker notion called Total Dual Integrality (https://en.wikipedia.org/wiki/Total_dual_integrality)

Severus Tux
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