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: $$B - A + \epsilon \le (\overline{B}-\underline{A}+\epsilon) Y$$

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  • $\begingroup$ 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 $\endgroup$ – Clement Mar 1 at 17:07
  • $\begingroup$ My suggestion about $\epsilon$ takes care of that case. $\endgroup$ – RobPratt Mar 1 at 18:31
  • $\begingroup$ 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. $\endgroup$ – Clement Mar 1 at 18:52
  • $\begingroup$ 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. $\endgroup$ – RobPratt Mar 1 at 18:56
  • $\begingroup$ 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? $\endgroup$ – 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.

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  • 1
    $\begingroup$ $B$ is a variable here, so your proposed constraint is nonlinear. $\endgroup$ – RobPratt Mar 11 at 17:57

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