# Mixed integer linear programs - variables taking value in a set

 1 How does one deal with Mixed integer linear programs where one bunch of variables takes values from a set. Example $$x \in \{1,2,3,4\}$$, $$y \in \{0.5,0.6,0.8\}$$ and other variables are integer or continuous. Thanks in advance, asked 17 Jun '14, 01:17 pondy 43●1●4 accept rate: 0% fbahr ♦ 4.6k●7●16

 4 There might be more efficient ways, but a simple one to implement is to define binary variables for each variable and have the binary variables multiplied by their corresponding values. You should also add a constraint that enforce selecting at most one of the binary variables (i.e., values). For example, for $$x$$ you would have binary variables $$w_1$$, $$w_2$$, $$w_3$$, and $$w_4$$. To represent $$x$$ in the objective function or a constraint, you would write $$w_1+2w_2+3w_3+4w_4$$. Finally, you should add the constraint $$w_1+w_2+w_3+w_4 \le 1$$. Please note that this does not work well with large problems that have many discrete variables as you have described. EDIT (Thanks to @Sune): The inequality in the constraint $$w_1+w_2+w_3+w_4 \le 1$$ assumes that zero is also part of the acceptable values. If that's not the case, then you should use the equality version. answered 17 Jun '14, 02:30 Ehsan ♦ 4.8k●3●11●22 accept rate: 16% 2 Minor comment: maybe you even need an equality instead of the "less than one" if the variable has to take a value in the set. Otherwise x might take a value of zero if it is optimal to have it at its lower bound. (17 Jun '14, 04:03) Sune @Sune: Good point. If the $$x$$ does not take zero as its value, the formulation could be improved by removing the first the binary variable. (17 Jun '14, 04:31) Ehsan ♦ Perhaps we are just thinking of different ways of formulating it, but I think equality is required if x wants to be at its lower bound. Take for example $$\min\{ x:x\in \{ 1,2,3,5,8\}\}$$. The optimal value is clearly 1. If we formulate that as $$\min w_1+2w_2+3w_3+5w_4+8w_5 \text{ such that: } w_1+w_2+w_3+w_4+w_5\leq 1 \text{ all binary}$$ the optimal value will be zero. However, if we use an equality, we find the right solution, namely $$x=1$$. (17 Jun '14, 07:26) Sune @Sune: I think we're on the same page. If $$0$$ is not part of the acceptable values for $$x$$, then equality is necessary for the constraint. My point in the previous comment was that if the minimum acceptable value for $$x$$ is $$1$$ and not $$0$$, then you could write $$x$$ as $$1+w_2+2w_3+4w_4+7w_5$$ in constraints and just add the constraint $$w_2+w_3+w_4+w_5 = 1$$. (17 Jun '14, 07:43) Ehsan ♦ @Ehsan: Then we are indeed on the same page! (17 Jun '14, 08:59) Sune
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