On the facet defining inequalities of the mixed-integer bilinear covering set

Abstract

We study the facet defining inequalities of the convex hull of a mixed-integer bilinear covering arising in trim-loss (or cutting stock) problem under the framework of disjunctive cuts. We show that all of them can be derived using a disjunctive procedure. Some of these are split cuts of rank one for a convex mixed-integer relaxation of the covering set, while others have rank at least two. For certain linear objective functions, the rank-one split cuts are shown to be sufficient for finding the optimal value over the convex hull of the covering set. A relaxation of the trim-loss problem has this property, and our computational results show that these rank-one inequalities find the lower bound quickly.

Appendix: Additional proofs

Proposition 10

Let $j, k \in \mathbb {N}$ with $j \ne k$, then the following two relations

$$\begin{aligned}&\pi _0< \sqrt{j(j - 1)}< \pi _0 + 1 \ \text {and}\\&\pi _0< \sqrt{k(k - 1)} < \pi _0 + 1 \end{aligned}$$

cannot hold simultaneously for any non-negative integer $\pi _0$.

Proof

Without loss of generality let $k > j$. Note that, it is equivalent to prove that $\sqrt{k(k - 1)} - \sqrt{j(j - 1)} \ge 1$. This is because, if $\sqrt{k(k - 1)} - \sqrt{j(j - 1)} \ge 1$ holds, then both the values $\sqrt{k(k - 1)}$ and $\sqrt{j(j - 1)}$ cannot lie between two consecutive integers.

Also note that, since the function $f(j) = \sqrt{j(j - 1)}$ is strictly increasing for $j \in \mathbb {N}$, it is sufficient to prove the result when $k = j + 1$, i.e., we show that $\sqrt{j(j + 1)} - \sqrt{j(j - 1)} \ge 1$. Since we are dealing with positive numbers only, in our following steps of proof, we consider only the positive roots. For any $j \in \mathbb {N}$,

$$\begin{aligned}&4 j^2 - 4 j + 1> 4 j^2 - 4 j \\&\quad \Rightarrow \, (2 j - 1)^2> 4 j (j - 1) \\&\quad \Rightarrow \, 2 j - 1> 2 \sqrt{j (j - 1)} \\&\quad \Rightarrow \, j^2 + j> 1 + 2 \sqrt{j (j - 1)} + j^2 - j \\&\quad \Rightarrow \, j (j + 1)> 1 + 2 \sqrt{j (j - 1)} + j (j - 1) \\&\quad \Rightarrow \, j (j + 1)> \left( 1 + \sqrt{j (j - 1)} \right) ^2 \\&\quad \Rightarrow \, \sqrt{j(j + 1)} - \sqrt{j(j - 1)} > 1. \end{aligned}$$

This completes the proof. $\square$

Proposition 11

Let $k \in \mathbb {N}$ with $k \ge 2$. Consider the positive integers $\mu _0, \mu$ with $\mu \ge 1$. If $\mu _0< \mu \sqrt{k(k - 1)} < \mu _0 + 1$ then $k -1 < \frac{\mu _0 +1}{\mu } \le k$ and $k -1 \le \frac{\mu _0}{\mu } < k$.

Proof

Since $\mu _0< \mu \sqrt{k(k - 1)} < \mu _0 + 1$, then $\frac{\mu _0 + 1}{\mu \sqrt{k(k - 1)}} > 1$. Again since $\sqrt{\frac{k - 1}{k}} < 1$, we have

$$\begin{aligned}&\frac{\mu _0 + 1}{\mu \sqrt{k(k - 1)}}> \sqrt{\frac{k - 1}{k}} \\&\quad \Rightarrow \frac{\mu _0 + 1}{\mu } > k - 1 \end{aligned}$$

We show the other side of the desired inequality next. From the given condition we have,

$$\begin{aligned}&\mu \sqrt{k(k - 1)}> \mu _0 \\&\quad \Rightarrow \mu \left\lceil \sqrt{k(k - 1)} \right\rceil > \mu _0 \\&\quad \Rightarrow \mu \left\lceil \sqrt{k(k - 1)} \right\rceil \ge \mu _0 + 1 \ \text {(since the left hand side is integral)} \\&\quad \Rightarrow \mu k \ge \mu _0 + 1 \ \left( \text {since } k \ge \left\lceil \sqrt{k(k - 1)} \right\rceil \right) \\&\quad \Rightarrow k \ge \frac{\mu _0 + 1}{\mu } \end{aligned}$$

Therefore, we have $k -1 < \frac{\mu _0 +1}{\mu } \le k$.

Since $\frac{\mu _0 +1}{\mu } \le k$, we have $\frac{\mu _0}{\mu } < k$. It remains to show $k - 1 \le \frac{\mu _0}{\mu }$. From the given relation we have

$$\begin{aligned}&\mu _0 + 1> \mu \sqrt{k (k - 1)} \\&\quad \Rightarrow \mu _0 + 1> \mu (k - 1) \ \left( \text {since } \sqrt{k(k - 1)} > k - 1 \right) \\&\quad \Rightarrow \mu _0 \ge \mu (k - 1) \ \left( \text {since both sides are integral}\right) \\&\quad \Rightarrow \frac{\mu _0}{\mu } \ge k - 1 \end{aligned}$$

Therefore, we get $k - 1 \le \frac{\mu _0}{\mu } < k$. $\square$