Aggregation-based cutting-planes for packing and covering integer programs

Abstract

In this paper, we study the strength of Chvátal–Gomory (CG) cuts and more generally aggregation cuts for packing and covering integer programs (IPs). Aggregation cuts are obtained as follows: given an IP formulation, we first generate a single implied inequality using aggregation of the original constraints, then obtain the integer hull of the set defined by this single inequality with variable bounds, and finally use the inequalities describing the integer hull as cutting-planes. Our first main result is to show that for packing and covering IPs, the CG and aggregation closures can be 2-approximated by simply generating the respective closures for each of the original formulation constraints, without using any aggregations. On the other hand, we use computational experiments to show that aggregation cuts can be arbitrarily stronger than cuts from individual constraints for general IPs. The proof of the above stated results for the case of covering IPs with bounds require the development of some new structural results, which may be of independent interest. Finally, we examine the strength of cuts based on k different aggregation inequalities simultaneously, the so-called multi-row cuts, and show that every packing or covering IP with a large integrality gap also has a large k -aggregation closure rank. In particular, this rank is always at least of the order of the logarithm of the integrality gap.

Appendices

A Polyhedrality of aggregation closure for dense IPs

We prove the result for the case of covering IPs and a similar proof can be given for the packing case.

Proposition 16

Let $Q = \{ x \in {\mathbb {R}}^n_+ \mid Ax \ge b \}$ be a covering polyhedron with $A \in {\mathbb {Z}}_+^{m \times n},~b \in {\mathbb {Z}}_+^{n}$, $A_{ij} \ge 1$ for all $i \in [m],~j\in [n]$, and $b_i \ge 1$ for all $i \in [m]$. Then, ${\mathcal {A}}_k(Q)$ is a polyhedron.

Proof

The intercept of the hyperplane corresponding to the $i{\text {th}}$ constraint, $A^ix \ge b_i$, of the $j{\text {th}}$ coordinate axis is $\frac{b_i}{A_{ij}}$. It is straightforward to verify that the intercept of any aggregated constraint on the $j{\text {th}}$ coordinate axis belongs to the set $\left[ \min _{i \in [m]} \frac{b_i}{A_{ij}},\max _{i \in [m]} \frac{b_i}{A_{ij}}\right]$. Let $M = \max _{i \in [m],~j \in [n]} \frac{b_i}{A_{ij}}$ and let $T = [0,M]^n \cap {\mathbb {Z}}_+^n$.

Based on the above observation, the set of integer points contained in $\{ x \in {\mathbb {R}}^n_+ \mid (\lambda ^\ell )^\top A x \ge (\lambda ^\ell )^\top b,~ \ell \in [k] \}$ is of the form $S \cup ({\mathbb {Z}}_+^n {\setminus } T)$ where $S \subseteq T$. Since T is a finite set, this completes the proof as the number of distinct integer hulls obtained from k-aggregations is finite. $\square$

B Proof of Proposition 1

Given a convex set $C \subseteq {\mathbb {R}}^n$, its support function $\delta ^*(. \mid C)$ is defined by $\delta ^*(c \mid C) = \sup \{ c^T x \mid x \in C\}$.

Consider packing sets $U \supseteq V$. Since U and V are closed, from Corollary 13.1.1 of [24] we have that $U \supseteq \alpha V$ iff

$$\begin{aligned} \sup _{c \in {\mathbb {R}}^n} \left( \delta ^*(c \mid U) - \delta ^*(c \mid \alpha V)\right) \le 0. \end{aligned}$$

(18)

Since U is a packing set we have the following property. Consider a vector $c \in {\mathbb {R}}^n$, let I be the index set of its negative components, and let ${\tilde{c}}$ be obtained by changing the components of c in I to 0. Then $\delta ^*(c \mid U) = \delta ^*({\tilde{c}} \mid U)$: the direction “$\le$” follows from $U \subseteq {\mathbb {R}}^n_+$; the direction “$\ge$” holds because for every point $x \in U$, if we construct ${\tilde{x}}$ by changing the components in I of x to 0 then ${\tilde{x}} \in U$ and $c^T {\tilde{x}} = {\tilde{c}}^T x$. Since the same holds for $\alpha V$, we have that in Eq. (18) we can take the supremum over only non-negative c’s, and hence it holds iff for all $c \in {\mathbb {R}}^n_+$, $\delta ^*(c \mid U) \le \delta ^*(c \mid \alpha V)$. But since $\delta ^*(c \mid \alpha V) = \alpha \,\delta ^*(c \mid V)$ (Corollary 16.1.1 of [24]), this happens iff for all $c \in {\mathbb {R}}^n_+$, $\delta ^*(c \mid U) \le \alpha \,\delta ^*(c \mid V)$. This concludes the proof. $\square$

C Proof of Proposition 2

Let $Q = \{ x \in {\mathbb {R}}_+^n \mid A^ix \le b_i \ \forall i \in I \}$. We assume that for all $j \in [n]$, there exists $i \in I$ with $A_{ij} > 0$. Otherwise, we can project out the $j{\text {th}}$ variable and continue with the argument as the $j{\text {th}}$ variable is allowed to take any value. Therefore, Q is a bounded set and $Q^{{\mathcal {I}}}$ is a polyhedron. Let $Q^{{\mathcal {I}}}= \{ x \in {\mathbb {R}}_+^n \mid Cx \le d \}$. We next argue that C and d are non-negative to complete the proof.

Note that since ${\varvec{0}} \in Q$, $d \ge 0$. The fact that we can take $C \ge 0$ follows from the following claim.

Claim

Let $C^i x \le d_i$ be a facet-defining inequality for $Q^{{\mathcal {I}}}$ and $C_{i j^{*}} < 0$ for some i and $j^{*}$. Define a vector $\hat{c}$ as $\hat{c}_{j^{*}} = 0$ and $\hat{c}_{j} = C_{ij}$ for all other j. Then ${\hat{c}}^ x \le d_i$ is valid for $Q^{{\mathcal {I}}}$.

Proof

Assume by contradiction that there exists $\hat{x} \in Q \cap {\mathbb {Z}}^n$ such that $\sum _{j = 1}^n \hat{c}_{j}\hat{x}_j > d_i$. Since Q is a packing set, we have that ${\tilde{x}} \in Q \cap {\mathbb {Z}}^n$, where ${\tilde{x}}$ is defined as ${\tilde{x}}_j = \hat{x}_j$ for all $j \in [n] {\setminus } \{j^{*}\}$ and ${\tilde{x}}_j^{*} = 0$. Then $d_i < \sum _{j = 1}^n \hat{c}_{j}\hat{x}_j = \sum _{j = 1}^n \hat{c}_{j}{\tilde{x}}_j = \sum _{j = 1}^n {C}_{ij}{\tilde{x}}_j \le d_i$, a contradiction. $\quad \diamond$