A fallacious argument in integer programming

A commonly used argument in integer programming goes as follows. Suppose we want to show that inequality ax \le b does not induce a facet of a polyhedron. If we know valid inequalities (1) and (2) that add up to ax \le b, then we have shown that ax\le b cannot induce a facet. Right?

Not quite.

One problem occurs when inequalities (1) and (2) are multiples of ax\le b. For example, suppose that each inequality is (ax)/2 \le b/2.

What if the inequalities are not multiples of ax \le b?

Still no cigar.

Another problem occurs when the polyhedron is not full-dimensional. Consider the set of all (x,y)\in [0,1]^2 such that x\le 0 (meaning that we have an implicit equality constraint x=0). We want to argue that the inequality x+y\le 1 does NOT induce a facet. This seems to follow from the valid inequalities x\le 0 and y \le 1 which sum to x+y\le 1. However, the inequality x+y\le 1 DOES induce a facet; the dimension of our feasible region is one, the inequality x+y\le 1 is valid, and the face where x+y=1 has dimension zero.


One thought on “A fallacious argument in integer programming

  1. Stefano Coniglio

    That’s why you should *never* work with not full-dimensional polytopes (or, at least, bring them down to some “canonical” full-dimensional representation)! 😉


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