# How to model If $A \le B$ then $Y = 1$, otherwise $Y = 0$

Somehow I don't get it right.

I would like to model the following conditional:

If $$A\le B$$ then $$Y=1$$ otherwise $$Y=0$$
where $$A, B$$ are reals and $$Y$$ is binary.

I can model as follows:

$$Y \cdot A \le B$$ and linearise this, but then I get into trouble when $$A = 0$$;

In this case $$Y$$ can be anything but I want it to be $$1$$.

If $$A\in[\underline{A},\overline{A}]$$ and $$B\in[\underline{B},\overline{B}]$$, the following big-M constraints enforce $$Y=1\implies A \le B$$ and $$Y=0\implies B \le A$$, respectively: \begin{align} A - B &\le (\overline{A}-\underline{B})(1-Y)\\ B - A &\le (\overline{B}-\underline{A}) Y\\ \end{align} To disambiguate the $$A=B$$ case, you could introduce $$\epsilon$$ in the second constraint, as follows: $$B - A + \epsilon \le (\overline{B}-\underline{A}+\epsilon) Y$$

• But the opposite is not necessarily true. I mean, A≤B does not imply Y=1. I had a look here: yalmip.github.io/tutorial/logicprogramming they propose a solution to the problem IF f(x) <= 0 THEN a else not a, a is binary which matches my problem when setting f(x) to A-B and a to Y – Clement Mar 1 at 17:07
• My suggestion about $\epsilon$ takes care of that case. – RobPratt Mar 1 at 18:31
• Hello Rob Thanks for waisting your time with me. Could you please have a look in the link, I did provide? Somehow I don't get it right there: They speak about a function g(x) but they mean f(x) I suppose: Further they write Z1= 1 -> f(x) <= 0, a = 1 Z2=1 -> f(x) >= 0, a = 0 Is this correct? In my application I get an infeasibility, but it surely is my mistake. Does the case f(x) = 0 generate any problems? By the way my function is x1 - x2, and both are positive reals, the are supposed to model the start time of jobs. – Clement Mar 1 at 18:52
• That link has lots of errors. The place you referenced does not disambiguate $f(x)=0$ and is unnecessarily complicated. You need only one binary variable. Look instead at the section "If a then f(x)<0," which is equivalent to its contrapositive "If f(x) >= 0 then not a." It uses $\epsilon$, just like I suggested. – RobPratt Mar 1 at 18:56
• Can you give me a hint for the magnitude of epsilon? Should it be very close to zero e.g. 1E-8 or significantly distinct from Zero e.g. 0.001? – Clement Mar 2 at 14:53

If $$A$$ is continuous, this logical constraint cannot be represented by a finite set of linear inequalities (see old work by Bob Jeroslow). What you can do is to relax a little.

Saying $$A\le B \implies Y=1$$ is the same as imposing $$Y=0 \implies A > B$$. If you can allow this to become $$A \ge B+\epsilon$$ then the constraint becomes $$A \ge M\cdot Y + (B+\epsilon) (1-Y)$$ where $$M$$ is a lower bound of $$A$$ ($$A$$ is always $$\ge$$).

Of course if $$A$$ is known, e.g., to be integer, than you can choose $$\epsilon = 1$$ without loss of generality.

• $B$ is a variable here, so your proposed constraint is nonlinear. – RobPratt Mar 11 at 17:57