How to formulate (linearize) a maximum function in a constraint? Suppose $C = \max \{c_1, c_2\}$, where both $c_1$ and $c_2$ are variables. If the objective function is minimizing $C$, then it can be simply done by applying $C \geqslant c_1$, and $C \geqslant c_2$. But if the objective function is non-regular, e.g. earliness tardiness, the value of $C$ will be larger than the maximum of $c_1$ and $c_2$. So my question is how to formulate it correctly?

  • $\begingroup$ Do you know an upper bound on $c_1$ and $c_2$? $\endgroup$ – LarrySnyder610 Jun 24 at 3:45
  • $\begingroup$ Actually $c_1$ and $c_2$ are both variables and we have no tight bound on them. $\endgroup$ – Mostafa Jun 24 at 3:48

(I'm going to change $c$ to $x$ in my answer, since $c$ is usually used for cost coefficients, not decision variables.)

We want a set of constraints that enforces $X = \max\{x_1,x_2\}$. Define a new binary decision variable $y$, which will equal 1 if $x_1 > x_2$, will equal 0 if $x_1 < x_2$, and could equal either if $x_1 = x_2$. Let $M$ be a constant such that $x_1,x_2 \le M$ in any "reasonable" solution to the problem.

The following constraints enforce the definition of $y$: $$\begin{align} x_1 - x_2 & \le My \\ x_2 - x_1 & \le M(1-y) \end{align}$$ Then, the following constraints enforce $X = \max\{x_1,x_2\}$: $$\begin{align} X & \ge x_1 \\ X & \ge x_2 \\ X & \le x_1 + M(1-y) \\ X & \le x_2 + My. \end{align}$$ The first two constraints say $X \ge \max\{x_1,x_2\}$, as you suggested in the question. Combined with these constraints, the last two constraints say that $X = x_1$ if $x_1 > x_2$ (so $y=1$) and $X = x_2$ if $x_2 > x_1$ (so $y=0$).


  • $\begingroup$ Is there a particular name for this transformation? I am trying to find some type of academic source that I can reference this from for a paper I am working on. $\endgroup$ – D.Gray yesterday
  • $\begingroup$ @D.Gray Maybe, but I don't know of one. It's just a standard(ish) trick. If you really want to find a standard name, you could post that as a new question. $\endgroup$ – LarrySnyder610 yesterday

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