I am solving an integer program (IP) whose constraint matrix is not totally unimodular (TU). The linear programming (LP) relaxation and the original IP always have the same optimal solution, or the LP relaxation is always optimal — in the hundreds of random instances I chose. Since the constraint matrix is not TU, I do not have a sufficient condition to verify if indeed this should always hold true or not (since TU is only a sufficient condition and not necessary).

Since I am unable to find a counterexample, my questions are:

  • What other tests exist (such as TU) by which I can check if the LP solution should always solve the IP or not?

  • We also have a proof that this problem is NP complete. Does it follow that the LP relaxation cannot solve this IP?

  • $\begingroup$ Is there an integer point in the LP optimal region ? $\endgroup$ – reuns Jun 12 '17 at 17:26
  • $\begingroup$ @user1952009 as I wrote in my post- the LP optimal solution is integer.for the instances I have tested. $\endgroup$ – bissi Jun 12 '17 at 21:07
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    $\begingroup$ The fact you find 'always' integer solution does not mean the continuous relaxation has always optimal integer solutions. It could be luck. You need a proof. $\endgroup$ – AndreaCassioli Jun 13 '17 at 8:34
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    $\begingroup$ @AndreaCassioli exactly my point. That's my question. What are sufficient conditions, other than TU, to check if the LP solution wil always be the IP solution? $\endgroup$ – bissi Jun 13 '17 at 10:43
  • $\begingroup$ I'm sure there are a variety of other conditions. For instance, the so-called RIP property guarantees that certain minimum-cardinality models can be solved by replacing the cardinality function with an $\ell_1$ norm. $\endgroup$ – Michael Grant Jun 21 '17 at 0:10

TU is one of the nice properties that can guarantee LP relaxation solves the integer programs. Generally speaking, if one can find the convex hull description, then the integer program can be solved as linear program. For example, the uncapacitated lot sizing (ULS) problem with $(l,s)$ inequalities, see Barany at al. 1984). The $(l,s)$ inequalities define the facet of the ULS and together with the original description of ULS give the convex hull, hence the formulation still give integer solutions without integer restriction. Another example is the convex hull description for uncapacitated lot sizing problem with backlogging (see Kucukyavuz and Pochet 2009).

So I don't know what your problem is, but it is possible all the constraints are facet-defining therefore even if it is not TU, it still solves the problem as LP.

Finding convex hull is problem specific and it is very challenging since many complicated problems might have exponential number of facet-defining inequalities in their convex hull description. A more practical way is to find several strong facet-defining inequalities and develop efficient separation algorithm inside the branch-and-cut algorithm to achieve $0\%$ optimality gap within reasonable time limit, e.g., $1$ hour. Another way to find convex hull is to construct extended formulation through projection since extended formulation very often is more compact (but not necessary).

  • $\begingroup$ While your comments are correct in general, they do not answer my two specific questions. $\endgroup$ – bissi Jan 9 '18 at 15:09
  • $\begingroup$ TU is a property not a test, my answer is mainly describing the theory. You can google polyhedral combinatorics. $\endgroup$ – user515371 Jan 26 '18 at 17:58

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