Is it possible to forbid a LP variable from being a specific constant?

ex X <> 0.5

asked 25 Apr '18, 13:21

gtg489p's gravatar image

accept rate: 0%

No. But you can (at the cost of adding a binary variable) enforce that x is either below \(0.5-\varepsilon\) or above \(0.5+\varepsilon\) where \(\varepsilon>0\) is a small tolerance.


answered 25 Apr '18, 13:28

Sune's gravatar image

accept rate: 20%

That's unfortunate. Adding a binary var would defeat the purpose for me, because formulation is itself an LP relaxation of a MIP.

I guess I could still give it a try; the LP relaxation with those few binary variables may solve much faster than the original IP, maybe even as fast as the pure LP relaxation.

(25 Apr '18, 13:44) gtg489p

Instead of solving one LP, you can solve two:

  1. original LP & x <= 0.5 - \epsilon
  2. original LP & x >= 0.5 + \epsilon

For a suitable \epsilon value, as per Sune's suggestion. You can then process the solutions accordingly (e.g. compare the two solutions, to see which one is actually optimal based on the two objective function values). As long as you have one or a few such variables, this approach should work.


answered 26 Apr '18, 15:49

AndyT's gravatar image

accept rate: 7%

edited 26 Apr '18, 17:37

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Asked: 25 Apr '18, 13:21

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Last updated: 26 Apr '18, 17:37

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