Saturday, August 8, 2015

Optimizing Part of the Objective Function II

Today's post extends yesterday's entry about optimizing a portion of the original objective function. I won't repeat much of the previous post, to save myself some typing.

The actual question motivating these posts, which I originally misread, asked how to minimize, not maximize, the sum of the $K$ largest terms in $\sum_{i=1}^N x_i$ (where $x$ is a vector of decision variables in an optimization model). As I mentioned in the last outing, that (or the symmetric case, maximizing the sum of the $K$ smallest terms), is trickier than what I covered yesterday. Everything from yesterday's post (the introduction of binary variables $z_i$ and auxiliary variables $y_i$ and the added constraints) carries over, with one exception: the objective function is now$$\textrm{minimize} \sum_{i=1}^N y_i.$$On top of all that, we require some additional constraints.


Forget everything that follows and just skip down to the comments section. There's a clever reformulation, posted by Michael Grant in a comment, that gets the job done with a linear number of new variables (all continuous) and a linear number of constraints. I knew another formulation with no integer variables, and in fact only one new variable at all, but with an exponential number of constraints. I was debating whether to post it until I saw the more efficient solution in Michael's comment.

5. Lower bounds on auxiliary variables

If we do not do something to prevent it, the solver will achieve an ideal objective value of 0 by setting $y_i = 0$ regardless of whether $z_i$ is 0 or 1. So we need additional constraints to ensure that $z_i = 1 \implies y_i = x_i$. We can accomplish that with$$y_i \ge x_i - U_i(1 - z_i)\quad \forall i \in \{1,\dots,N\},$$which forces $y_i\ge x_i$ if $z_i = 1$ and is vacuous if $z_i = 0$.

6. Choosing the $K$ largest terms

Choosing the largest of the $x_i$ was automatic when the objective was to maximize the sum. When the objective is minimization, the solver will "want" to choose the smallest of the $x_i$, and we need to constrain it to select the largest values. An easy way to do this is to note that choosing the $K$ largest values is equivalent to saying that every value chosen is at least as big as every value not chosen. Expressed mathematically,$$z_i = 1 \bigwedge z_j = 0 \implies x_i \ge x_j.$$Note that, having assumed $0\le x_k \le U_k\,\forall k$, the most negative that any difference $x_i - x_j$ could be would be $0 - U_j = -U_j$. That leads to the following additional constraints:$$x_i - x_j \ge -U_j(1-z_i+z_j)\quad \forall i,j\in \{1,\dots,N\},\,i\neq j.$$
Coupled with yesterday's formulation, that should get the job done.