Let's assume we have event $i=1,2,\cdots,k$, denoted as $\text{event}_i$. We know for a fact that $\text{event}_i$ is smaller then $\text{event}_{i+1}$ i.e., $\text{event}_i \leq \text{event}_{i+1}$. Now we have given some events $a,b,c,d \leq k$. How do I formulate the constraint that: if $\text{event}_a \leq \text{event}_b$ then $\text{event}_c \leq \text{event}_d$?

My try was:

\begin{align}\text{event}_a + Mz &> \text{event}_b\\\text{event}_c + M(1-z) &\leq \text{event}_d\end{align} where $z \in\{0,1\}$, $M$ large.

  • $\begingroup$ Given that you have weak inequalities, do you mean that each event is no larger than the subsequent event? $\endgroup$
    – TheSimpliFire
    Commented Dec 4, 2022 at 15:36
  • $\begingroup$ I must have that if event a is before event b then event c must be before event d $\endgroup$ Commented Dec 4, 2022 at 15:38
  • $\begingroup$ Your formulation seems to be ok. Just try with a solver or in excel to confirm. $\endgroup$ Commented Dec 4, 2022 at 15:39
  • $\begingroup$ How would you formulate a constraint which has to be like: Either event a is before event b or event a is before event c, or both. $\endgroup$ Commented Dec 4, 2022 at 15:47
  • $\begingroup$ "Either a is before b or a is before c" is the same as "if b is before a then a is before c". $\endgroup$
    – prubin
    Commented Dec 4, 2022 at 17:04

1 Answer 1


You want to enforce the logical implication $$\text{event}_a \leq \text{event}_b \implies \text{event}_c \leq \text{event}_d.$$ Introduce a binary variable $z$ and enforce \begin{align} \text{event}_a \leq \text{event}_b \implies z = 1 \tag1\label1\\ z = 1 \implies \text{event}_c \leq \text{event}_d \tag2\label2 \end{align} Equivalently, by contraposition of \eqref{1}, \begin{align} z = 0 \implies \text{event}_a > \text{event}_b \tag3\label3\\ z = 1 \implies \text{event}_c \leq \text{event}_d \tag4\label4 \end{align} Now big-M modeling yields \begin{align} \text{event}_b - \text{event}_a &< M_1 z \tag5\label5\\ \text{event}_c - \text{event}_d &\le M_2(1-z) \tag6\label6 \end{align} This is very similar to what you tried, except that your second $M$ needs the opposite sign. Because MILP disallows strict inequalities, you need to introduce a positive tolerance $\epsilon$ and replace \eqref{5} with $$\text{event}_b - \text{event}_a +\epsilon \le M_1 z.$$ The effect is that you are enforcing $$\text{event}_a - \epsilon < \text{event}_b \implies \text{event}_c \leq \text{event}_d.$$


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