# If Then Constraint Linear Programming

I want to write the following constraint: If A=1 and B <= m then C=1 ( where A and C are binary, m is a constant and B is continuous).

My solution requires some kind of hack in the model, a Big-$$M$$ value, which is pretty common, but also a tolerance $$\varepsilon$$, which is not very desirable. I hope that I do not miss something easy here but reckon there is no way around it.
To model a condition of the form $$(X \wedge Y) \to Z$$ for binary $$X, Y, Z$$, you can use $$\tfrac12(X + Y) \leq Z + \tfrac12.$$ We can verify this by inserting: $$X=Y=1$$ yields to $$1$$ on the left hand side, thus, $$Z$$ has to be at least $$\frac12$$, and since it is binary, it must be $$1$$. If either $$X$$ or $$Y$$ is $$0$$ (or both), we have at most $$\frac12$$ on the left hand side and in this case $$Z$$ might be either $$0$$ or $$1$$.
We also need to transform the constraint $$B \leq m$$ as a binary variable in order to use the template above. This can be done by introducing a new binary variable $$B'$$ which must assume the vaue $$1$$ iff $$B \leq m$$. We also need a Big-$$M$$ value which must be greater than any value that $$B$$ can assume (note that for numerical reasons in practice, $$M$$ should be not too large). Moreover, we have $$\varepsilon$$, a small tolerance value to get a strict $$>$$-inequality (in practice you would choose the difference to the next greater number from $$m$$ that can be represented exactly on your machine). We now add the constraints $$B \leq m + (1-B')M\tag{1}$$ $$B \geq m + \varepsilon - B'M\tag{2}$$ $$(1)$$ guarantees that $$B' = 0$$ if $$B > m$$, $$(2)$$ guarantees that $$B' = 1$$ if $$B \leq m$$. We verify this again: If $$B \leq M$$, $$(1)$$ and $$(2)$$ can be satisfied with $$B' = 1$$, but not with $$0$$. If $$B > m$$, we need $$B' = 0$$ to satisfy $$(1)$$ and then $$(2)$$ is satisfied since $$B$$ is at least $$\varepsilon$$ larger than $$m$$.
Putting it all together, we obtain your constraint $$\tfrac12(A + B') \leq C + \tfrac12,\\ B \leq m + (1-B')M,\\ B \geq m + \varepsilon - B'M.$$
• Yes $B'$ is a new binary variable that does not occur anywhere else in your IP. – ttnick Jul 9 at 5:15