On a Generalization of the Chvátal-Gomory Closure

Abstract

Many practical integer programming problems involve variables with one or two-sided bounds. Dunkel and Schulz (2012) considered a strengthened version of Chvátal-Gomory (CG) inequalities that use 0–1 bounds on variables, and showed that the set of points in a rational polytope that satisfy all these strengthened inequalities is a polytope. Recently, we generalized this result by considering strengthened CG inequalities that use all variable bounds. In this paper, we generalize further by considering not just variable bounds, but general linear constraints on variables. We show that all points in a rational polyhedron that satisfy such strengthened CG inequalities form a rational polyhedron.

A Proof of Lemma 5

First of all, Claims 1 and 2 below are proved inside the proof of Lemma 4.10 in [6].

Claim 1

$\mu \ge \mathbf {0}$ and $\text {supp}(\mu )=\text {supp}(\lambda )$.

Claim 2

$\mu b=\min \left\{ \mu Ax:x\in {P^{\uparrow }}\right\}$ and therefore $(\mu A,\mu b)\in \varPi _{{P^{\uparrow }}}$.

Since $\text {supp}(\mu )=\text {supp}(\lambda )$ by Claim 1 and $Ar^j\ge \mathbf {0}$ for all $j\in N_r$, it follows that $r\text {-supp}(\mu A)=r\text {-supp}(\lambda A)$, and therefore, $t(\mu , A) = t(\lambda , A)$.

The next claim extends Claim 3 of Lemma 4.10 in [6].

Claim 3

Let $Q = \left\{ x \in {{\,\mathrm{cone}\,}}\left\{ e^1,\ldots ,e^{n_1},r^1,\ldots ,r^{n_r}\right\} :\;\mu b \le \mu A x \le \mu b + \varDelta \right\}$. There is no point $x\in Q$ that satisfies

$$\begin{aligned} \sum _{i=1}^{\ell } p_ia_ix \ge 1+ \sum _{i=1}^{\ell } p_ib_i. \end{aligned}$$

(14)

Proof

Suppose for a contradiction that there exists $\tilde{x} \in Q$ satisfying (14). Recall that for the index k defined in (8), the inequality $\mu Ar^k>0$ holds. Let $v = \frac{\mu b}{\mu Ar^k}r^k$. Then $\mu A v =\mu b$ and $v\in Q$. In addition, for the index $\ell$ defined in (10), we have $\sum _{i=1}^\ell p_ia_iv =0$ since $k\not \in \bigcup _{i=1}^{t-1}r\text {-supp}(a_i)$ and $a_ir^k=0$ for $i\le t-1$. As $\tilde{x} \in Q$ satisfies (14) and $v\in Q$ satisfies $\sum _{i=1}^\ell p_ia_iv =0$, we can take a convex combination of these points to get a point $\bar{x} \in Q$ such that

$$\begin{aligned} \sum _{i=1}^\ell p_ia_i \bar{x}=1+\sum _{i=1}^\ell p_ib_i \quad \Rightarrow \quad \sum _{i=1}^\ell p_i(a_i \bar{x}-b_i) =1. \end{aligned}$$

(15)

As $\mu A \bar{x} \le \mu b + \varDelta$, we have

$$\begin{aligned} \sum _{i=1}^\ell \mu _i(a_i\bar{x}-b_i) \le -\sum _{j=\ell +1}^m\mu _j(a_j\bar{x}-b_j) + \varDelta . \end{aligned}$$

(16)

As in [6], we can rewrite (16) as:

$$\begin{aligned} \frac{\lambda _\ell }{p_\ell }\left( 1 + \sum _{i=1}^\ell \varepsilon _i(a_i\bar{x}-b_i)\right)&\le -\sum _{j=\ell +1}^m\mu _j(a_j\bar{x}-b_j) + 2\varDelta \nonumber \\&\le \sum _{j=\ell +1}^m\mu _jb_j + 2\varDelta \le \lambda _{\ell +1}(mB + 2D) =~\frac{1}{2}\lambda _{\ell +1}M_1 \end{aligned}$$

(17)

where the second inequality in (17) follows from the assumption that $A\in \mathbb {Z}^{m\times n}$ and $b\in \mathbb {Z}^m$ satisfy (3), the third inequality follows from the fact that $\mu _i = \lambda _i \le \lambda _{\ell +1}$ for $i=\ell +1, \ldots , m$ by (12) and that $b_j \le B$ by Definition 2, and the last equality simply follows from the definition of $M_1$.

Next, we obtain a lower bound on the first term in (17). As $a_i\bar{x}\ge 0$, $b_i\ge 0$, and $\varepsilon _i \in [ -\varepsilon , \varepsilon ]$, we have

$$\begin{aligned} \sum _{i=1}^\ell \varepsilon _i\left( a_i\bar{x}-b_i\right) = \sum _{i=1}^\ell \varepsilon _i a_i\bar{x}- \sum _{i=1}^\ell \varepsilon _i b_i \ge - \varepsilon \sum _{i=1}^\ell (a_i\bar{x} + b_i). \end{aligned}$$

(18)

Following the same argument in Claim 3 of Lemma 4.10 in [6], we can show that $-\varepsilon \sum _{i=1}^\ell (a_i\bar{x} + b_i) \ge -\frac{1}{2}$. Then it follows from (18) that $\sum _{i=1}^\ell \varepsilon _i(a_i\bar{x}-b_i)\ge -{1}/{2}$. So, the left hand side of (17) is lower bounded by ${\lambda _\ell }/{2p_\ell }$.

Since the first term in (17) is at least ${\lambda _\ell }/{2p_\ell }$, we obtain $\lambda _\ell \le p_\ell \lambda _{\ell +1} M_1$ from (17), implying in turn that $M_\ell < p_\ell M_1$ as we assumed that $\lambda _\ell > M_\ell \lambda _{\ell +1}$ as in (10). However, (11) implies that $M_\ell \ge p_\ell M_1$, a contradiction.   $\square$

Claim 4

$\mu A x\ge \lceil \mu b \rceil _{S,\mu A}$ dominates $\lambda A x\ge \lceil \lambda b \rceil _{S,\lambda A}$.

Proof We will first show that

$$\begin{aligned} \mu b\le \lceil \mu b \rceil _{S,\mu A}\le \mu b+\varDelta \end{aligned}$$

(19)

holds. Set $(\alpha ,\beta )=(\mu A,\mu b)$. By Claim 2, we have that $\beta =\min \{\alpha x:x\in {P^{\uparrow }}\}$. As the extreme points of ${P^{\uparrow }}$ are contained in ${{\,\mathrm{conv}\,}}(S)$, it follows that $\beta \ge \min \{\alpha z:z\in S\}$. If $\beta =\min \{\alpha z:z\in S\}$, then $\beta = \lceil \beta \rceil _{S,\alpha }$. Thus we may assume that $\beta >\min \{\alpha z:z\in S\}$, so there exists $z'\in S$ such that $\beta >\alpha z'$. Remember that by (7), $\varDelta =\min \{\lambda Ar^j:j\in r\text {-supp}(\lambda A)\}$, and let j be such that $\lambda Ar^j=\varDelta$. As $r\text {-supp}(\lambda A)=r\text {-supp}(\mu A)$, we have $\alpha r^j>0$ and $\kappa =({\beta - \alpha z'})/{\alpha r^j}>0$. Therefore $z''=z'+\lceil \kappa \rceil r^j\in S$. Observe that

$$\begin{aligned} \beta = \alpha z' + (\beta -\alpha z') =\alpha \left( z'+\kappa r^j\right) \le \alpha \left( z'+\lceil \kappa \rceil r^j\right) =\beta +\alpha r^j(\lceil \kappa \rceil -\kappa ) \le \beta +\alpha r^j. \end{aligned}$$

As $\lambda \ge \mu$, we have $\varDelta \ge \alpha r^j$ implying $\beta \le \alpha z''\le \beta +\varDelta$ and (19) holds, as desired.

Using (19), we will show that $\mu A x\ge \lceil \mu b \rceil _{S,\mu A}$ dominates $\lambda A x\ge \lceil \lambda b \rceil _{S,\lambda A}$. Let $z\in S$ be such that $\mu A z=\lceil \mu b \rceil _{S,\mu A}$. As z is integral and $\mu b\le \lceil \mu b \rceil _{S,\mu A}\le \mu b+\varDelta$ by (19), Claim 3 implies that $\sum _{i=1}^\ell p_ia_i z < 1 + \sum _{i=1}^\ell p_ib_i$, and therefore, $\sum _{i=1}^\ell p_ia_i z = \sum _{i=1}^\ell p_ib_i - f$ for some integer $f \in [0,\sum _{i=1}^\ell p_ib_i]$. Consider $z+fr^j\in S$ and observe that

$$\begin{aligned} \lambda A\left( z+fr^j\right) =\lambda A z+f\lambda Ar^j&=\left( \mu A+\varDelta \sum _{i=1}^\ell p_ia_i\right) z+\varDelta \sum _{i=1}^\ell p_i(b_i - a_iz)\\ {}&=\lceil \mu b \rceil _{S,\mu A}+\varDelta \sum _{i=1}^\ell p_ib_i. \end{aligned}$$

Since $\lceil \mu b \rceil _{S,\mu A} \ge \mu b$, we must have $\lceil \mu b \rceil _{S,\mu A}+\varDelta \sum _{i=1}^\ell p_ib_i\ge \mu b+\varDelta \sum _{i=1}^\ell p_ib_i=\lambda b.$ Then $\lceil \mu b \rceil _{S,\mu A}+\varDelta \sum _{i=1}^\ell p_ib_i\ge \lceil \lambda b \rceil _{S,\lambda A}$. Then the inequality $\lambda A x\ge \lceil \lambda b \rceil _{S,\lambda A}$ is dominated by $\mu A x\ge \lceil \mu b \rceil _{S,\mu A}$, as the former is implied by the latter and a nonnegative combination of the inequalities in $Ax\ge b$, as required.

   $\square$