# Minimizing a Median

Written by: Paul Rubin

Primary Source: OR in an OB World

$$\def\xorder#1{x_{\left(#1\right)}} \def\xset{\mathbb{X}} \def\xvec{\mathbf{x}}$$A somewhat odd (to me) question was asked on a forum recently. Assume that you have continuous variables $$x_{1},\dots,x_{N}$$ that are subject to some constraints. For simplicity, I’ll just write $$\xvec=(x_{1},\dots,x_{N})\in\xset$$. I’m going to assume that $$\xset$$ is compact, and so in particular the $$x_{i}$$ are bounded. The questioner wanted to know how to model the problem of minimizing the median of the values $$x_{1},\dots,x_{N}$$. I don’t know why that was the goal, but the formulation is mildly interesting and fairly straightforward, with one wrinkle.

The wrinkle has to do with whether $$N$$ is odd or even. Suppose that we sort the components of some solution $$\xvec$$, resulting in what is sometimes called the “order statistic”: $$\xorder 1\le\xorder 2\le\dots\xorder N$$. For odd $$N$$, the median is

$$\displaystyle \xorder{\frac{N+1}{2}}.$$

For even $$N$$, it is usually defined as

$$\displaystyle \left(\xorder{\frac{N}{2}}+\xorder{\frac{N}{2}+1}\right)/2.$$

The odd case is easier, so we’ll start there. Introduce $$N$$ new binary variables $$z_{1},\dots,z_{N}$$ and a new continuous variable $$y$$, which represents the median $$x$$ value. The objective will be to minimize $$y$$. In addition to the constraint $$\xvec\in\xset$$, we use “big-M” constraints to force $$y$$ to be at least as large as half the sample (rounding “half” up). Those constraints are:
\begin{align*}
y & \ge x_{i}-M_{i}z_{i},\ i=1,\dots,N\\
\sum_{i=1}^{N}z_{i} & =\frac{N-1}{2}
\end{align*}with the $$M_{i}$$ sufficiently large positive constants. The last constraint forces $$z_{i}=0$$ for exactly $$\frac{N+1}{2}$$ of the indices $$i$$, which in turn forces $$y\ge x_{i}$$ for $$\frac{N+1}{2}$$ of the $$x_{i}$$. Since the objective minimizes $$y$$, $$z_{i}$$ will be 0 for the $$\frac{N+1}{2}$$ smallest of the $$x_{i}$$ and $$y$$ will be no larger than the smallest of them. In other words, we are guaranteed that in the optimal solution $$y=\xorder{\frac{N+1}{2}}$$, i.e., it
is the median of the optimal $$\xvec$$.

If $$N$$ is even and if we are going to use the standard definition of median, we need twice as many added variables (or at least that’s the formulation that comes to mind). In addition to the $$z_{i}$$, let $$w_{1},\dots,w_{N}$$ also be binary variables, and replace $$y$$ with a pair of continuous variables $$y_{1}$$, $$y_{2}$$. The objective becomes minimization of their average $$(y_{1}+y_{2})/2$$ subject to the constraints
\begin{align*}
y_{1} & \ge x_{i}-M_{i}z_{i}\ \forall i\\
y_{2} & \ge x_{i}-M_{i}w_{i}\ \forall i\\
\sum_{i=1}^{N}z_{i} & =\frac{N}{2}-1\\
\sum_{i=1}^{N}w_{i} & =\frac{N}{2}
\end{align*}where the $$M_{i}$$ are as before. The constraints force $$y_{1}$$ to be at least as large as $$\frac{N}{2}+1$$ of the $$x_{i}$$ and $$y_{2}$$ to be at least as large as $$\frac{N}{2}$$ of them. The minimum objective value will occur when $$y_{1}=\xorder{\frac{N}{2}+1}$$ and $$y_{2}=\xorder{\frac{N}{2}}$$.

The following two tabs change content below. #### Paul Rubin

Professor Emeritus at Michigan State University
I'm an apostate mathematician, retired from a business school after 33 years of teaching mostly (but not exclusively) quantitative methods courses. My academic interests lie in operations research. I also study Tae Kwon Do a bit on the side. 