In the classic Mixed-Integer Linear Programming (MILP), the variables are fixed to be either integer or real. I am interested in the following MILP variant, where only one thing different from the classic MILP:

Let $n$ be the number of the MILP variables. One variable (any variable) is real and $n-1$ variables are integers. Notice that we can choose the real variable among all the variables.

  • Is this MILP variant NP-hard?
  • Is there a way to know how to choose the real variable to maximize the chances to reach a feasible solution?

One trivial but naive solution to this problem would run $n$ MILPs, such that for each MILP, one different variable is allowed to be real. The output is the best output among the $n$ running. This solution is NP-hard.

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    $\begingroup$ or.stackexchange.com/questions/6851/… $\endgroup$
    – Kuifje
    Oct 8 at 10:19
  • 1
    $\begingroup$ Interesting. Do you have any real-world examples of how this type of model arises? $\endgroup$ Oct 8 at 13:59
  • $\begingroup$ Yes: Dividing $n$ divisible objects among $m$ agents such that only one object can be shared and no agent envies another agent. $\endgroup$ Oct 14 at 15:56

An alternative approach that requires only one solve and no modification of the model is to modify branch and bound to prune by integrality when at most one integer variable takes a fractional value (rather than the usual requirement that all integer variables are integer-valued). You would also need to disable any presolve/cut routines that assume integrality. You would also need to relax the feasibility checker.

  • $\begingroup$ This is a very elegant solution. In practice, do you think this is easier to do with classical solvers on the market (+ open source ones), compared to modifying the model ? $\endgroup$
    – Kuifje
    Oct 8 at 13:01
  • $\begingroup$ Thanks. I would say that it is easier to modify the model, as in the link you provided, but I would expect modifying the algorithm to be more efficient. $\endgroup$
    – RobPratt
    Oct 8 at 13:47

Here's another single-solve solution. Replace each original variable $x_n$ with a sum of two variables, $x_n=y_n + z_n$, where $y_n$ is integer-valued and $z_n\in [0,1]$. Now define $\lbrace z_1,\dots, z_n\rbrace$ to be a type 1 special ordered set (SOS1). Assuming the solver supports SOS1 constraints, you'll end up with a solution in which at least $n-1$ of the $z$ variables are 0, meaning at least $n-1$ of the $x$ variables are integer.


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