Cutting planes from extended LP formulations

Abstract

Given a mixed-integer set defined by linear inequalities and integrality requirements on some of the variables, we consider extended formulations of its continuous (LP) relaxation and study the effect of adding cutting planes in the extended space. In terms of optimization, extended LP formulations do not lead to better bounds as their projection onto the original space is precisely the original LP relaxation. However, adding cutting planes in the extended space can lead to stronger bounds. In this paper we show that for every 0–1 mixed-integer set with n integer and k continuous variables, there is an extended LP formulation with $2n+k-1$ variables whose elementary split closure is integral. The proof is constructive but it requires an inner description of the LP relaxation. We then extend this idea to general mixed-integer sets and construct the best extended LP formulation for such sets with respect to lattice-free cuts. We also present computational results on the two-row continuous group relaxation showing the strength of cutting planes derived from extended LP formulations.

Appendix: Proof of Lemma 3.7

Let $n=n_1+n_2$. As $P$ is anti-blocking, it is defined by nonredundant system of inequalities $P=\{x\in {\mathbb {R}}^n:Ax\le b;~x\ge 0\}$ where $A,b\ge 0$. Assume that the cut $c^{T} x\le d$ can be derived using the split set $S(\pi ,{ \gamma })$ and let $c_k<0$ for some index k.

As $c^{ T}x\le d$ is valid for $P^1=\{x\in {\mathbb {R}}^n:Ax\le b, \pi ^{ T} x\le {\gamma }, x\ge 0\}$, there exists multipliers ${ \mu _1,\mu _2\in {\mathbb {R}}_{+}}$ and $\lambda \in {\mathbb {R}}^n_{+}$, such that

$$\begin{aligned} c=\mu _1 a+\mu _2\pi -\lambda ,\quad \text {and}\quad d \ge \mu _1f+\mu _2{\gamma }, \end{aligned}$$

where $a^{T}x\le f$ is a non-negative linear combination of the rows of $Ax\le b$ (and therefore $a\ge 0$). Now consider the split set $S(\pi ^+,{ \gamma })$ where $\pi ^+$ is identical to $\pi$ except $\pi ^+_k=0$. Further let $\lambda ^+$ be identical to $\lambda$ except $\lambda ^+_k=\mu _1 a_k\ge 0$. Now, observe that the inequality ${ {\bar{c}}^T}x\le d$ is valid for $\bar{P}^1=\{x\in {\mathbb {R}}^n:Ax\le b,~{ (\pi ^+)^T}x\le { \gamma }, x\ge 0\}$ where ${ \bar{c}}=\mu _1 a+\mu _2\pi ^+-\lambda ^+$ and note that ${ \bar{c}}$ is identical to c except ${ \bar{c}}_k=0$.

Using the same argument for $P^2=\{x\in {\mathbb {R}}^n:Ax\le b, \pi ^{ T} x\ge {\gamma }+1, x\ge 0\}$ establishes that ${ {\bar{c}}^T}x\le d$ is valid for $P^2$ and therefore is a valid split cut for $P^{\textit{IP}}$.

Repeating this process for all negative components of c completes the proof. $\square$

1.1 Proof of Theorem 3.9

Given a point $(x,y) \in X$, let $T_{ny} = \{j \in N : y_j > 0\}$, i.e., $T_{ny}$ stands for the indices of nonzero components of $y$. Let $y_{max}$ be the maximum value among all components of $y$, and let $T_{max} = \{j \in N : y_j = y_{max}\}$. Let

$$\begin{aligned} T_2= & {} \{ i \in N : (18) \text { for index { i} is tight for } (x,y)\} \text{ and } \\ T_3= & {} \{ i \in N : (19) \text { for index { i} is tight for } (x,y)\}. \end{aligned}$$

If $(x,y)$ is an extreme point of X, it must be the unique solution of a list L of some 2n linearly independent constraints from (17) to (20). From among these 2n constraints, let $L_1, L_2, L_3, L_{y}$ stand for, respectively, the sets of indices of constraints of types (17)–(20) that are contained in L. By definition, we have

$$\begin{aligned}&L_2 \subseteq T_2,\quad L_3 \subseteq T_3, \quad T_{ny} \ne \emptyset \ { \Longrightarrow } \ T_{max} \subseteq T_{ny}, \end{aligned}$$

(42)

$$\begin{aligned}&L_{y} \subseteq N {\setminus } T_{ny}\quad \text{ and }\quad |L_1| + |L_2| + |L_3| + |L_{y}| = 2n. \end{aligned}$$

(43)

Using these definitions, we first show some properties which we need in the proof.

Lemma 6.1

Let $(x,y)$ be an extreme point of X with $y\ne 0$. Then

• (i) $T_2,T_3 \subseteq T_{max}$ and $|L_2\cup L_3|+|L_{y}|\le n.$

• (ii) $|T_{ny}| \ge 3$.

• (iii) $|L_2 \cap L_3| \le 1$ and $|L_1| \ge n-1$.

• (iv) $T_{max} = T_{ny}$.

Proof

• (i) Let $k \in T_{max}$. If the inequality (18) is tight for some index $i \in N {\setminus } T_{max}$ then $y_i < y_k$, and inequality (18) for index k is violated, contradicting the fact that $(x,y) \in X$. Therefore $T_2 \subseteq T_{max}$. The proof of the fact that $T_3 \subseteq T_{max}$ is very similar. To see the last inequality, notice that by inequality (42) we have $L_2,L_3\subset T_{max}\subset T_{ny}$, and therefore $|L_2\cup L_3|+|L_{y}|\le n$.

• (ii) Let $y_{i_1} >0$ for some $i_1 \in N$ and let $i_2 \in N {\setminus }\{i_1\}$. Adding the constraints in (18) corresponding to the indices $i_1$ and $i_2$, we obtain

  $$\begin{aligned} 2\sum _{j \in N} y_j \ge 3(y_{i_1} + y_{i_2}) \Rightarrow 2\sum _{j \in N {\setminus } \{i_1, i_2\}} y_j \ge y_{i_1} + y_{i_2}, \end{aligned}$$

  which implies that at least one of the components of $y$ different from $i_1, i_2$ is nonzero. Repeating the same argument with this new component and $y_{i_1}$ shows that there are at least three such entries.

• (iii) Suppose that $|L_2 \cap L_3| \ge 2$. Then there are two indices $i_1 \ne i_2$ such that the corresponding constraints from (18) and (19) are contained in L. But subtracting $\sum _{j \in N}y_j = 3y_{i_2}$ from $\sum _{j \in N} y_j = 3y_{i_1}$, we get $y_{i_1} - y_{i_2} = 0$. Similarly, subtracting $2 = \sum _{j \in N} (2x_j - y_j) + 2y_{i_2}$ from $2 = \sum _{j \in N} (2x_j - y_j) + 2y_{i_1}$, we also get $y_{i_1} - y_{i_2} = 0$, which means that the above four constraints from L are not linearly independent, a contradiction. Therefore $|L_2 \cap L_3| \le 1$. But this fact, combined with (i) yields $|L_2| + |L_3| \le n-|L_{y}| + 1$. Then Eq. (43) implies that $|L_1| \ge n-1$.

• (iv) Note that $T_{max} \subseteq T_{ny}$ by (42). For contradiction, assume $|T_{max}| \le |T_{ny}| - 1$ and therefore $0 < y_\ell < y_{max}$ for some index $\ell \in N$. If $|L_2| = 0$, then (i) and Eq. (43) imply that

  $$\begin{aligned}&|L_3| \le |T_3| \le |T_{max}| \le |T_{ny}|-1 \Rightarrow |L_2| + |L_3| \le |T_{ny}|-1\\&\quad \qquad \Rightarrow |L_2| + |L_3| + |L_y| \le n-1. \end{aligned}$$

  As $|L_1| \le n$, this means that L cannot have 2n constraints, a contradiction. Therefore, we can assume that $|L_2| \ne 0$. Let $k \in L_2$, i.e., $\sum _{j \in N} y_j = 3y_k$. Then, (18) implies that $y_k \ge y_i$ for all $i \in N$. Thus, $y_k \in T_{max}$ and $y_k = y_{max}$. As $0 < y_\ell < y_k$, there has to be at least another index of $y$ different from $\ell$ such that the corresponding component is positive and strictly less than $y_k$ in order for the constraint $\sum \nolimits _{j \in N} y_j = 3y_k$ to hold. Then we have $|T_{max}|\le |T_{ny}|-2$. This fact, combined with (i) and (iii) and Eq. (42) implies that $|L_2|+|L_3| \le |T_{ny}|-1$. As $|L_1| \le n$, using Eq. (43), we get $|L| \le 2n-1$, a contradiction. Therefore $T_{max} = T_{ny}$.

$\square$

Proof of Theorem 3.9

We will first prove that $\{q^1, \ldots , q^m\} \subseteq X$; as X is convex this will imply that $Q^n \subseteq X$. Let $q^j = (x^j, y^j)$ for some $1 \le j \le m$. If $x^j = \mathbf{0}$ or if $x^j$ is a unit vector in ${\mathbb {R}}^n$, then $y^j = {\varvec{0}}$, and the constraints (17)–(20) are trivially satisfied. Let . The $i^\mathrm{th}$ component of $y^j$ is 1 if and only if the $i^\mathrm{th}$ component of $x^j$ is 1/2. Therefore (17) holds with equality for all $i=1, \ldots , n$; in this case the constraints (19) simply become $y_i \le 1$ for $i \in N$, which are all satisfied by $y^j$. Finally, at least three components of $x^j$ are 1/2, therefore at least three components of $y^j$ are 1. This, combined with the fact that all components of $y^j$ are at most 1, implies that all constraints in (18) are satisfied by $q^j$. Therefore $Q^n \subseteq X$.

For the reverse inclusion, we will show that each extreme point of X belongs to $V^{\prime }$. Note that X is a pointed polyhedron as Eqs. (20) and (17) imply that it is contained in ${\mathbb {R}}^n_+\times {\mathbb {R}}^n_+$. Furthermore, inequalities (19) together with inequalities (17) imply that $2y_i \le 2$ and therefore $y_i \le 1$ for all $i \in N$. In addition, when all $y_i$ are bounded, (19) also implies that all $x_i$ are bounded as well. Therefore, X is in fact a polytope and is equal to the convex hull of its extreme points.

Let $(x,y) \in {\mathbb {R}}^n \times {\mathbb {R}}^n$ be an extreme point of X. If $x_i = 1$ for some index $i \in N$, rewriting (19) as

$$\begin{aligned} 2 \ge \sum _{j \in N {\setminus } \{i\} } (2x_j-y_j) + (2x_i + y_i), \ \forall i \in N, \end{aligned}$$

(44)

it is clear that $x_i = 1 \Rightarrow y_i = 0$ and $2x_j = y_j$, for all $j \in N {\setminus } \{i\}$. Moreover, for any index $k \in N {\setminus } \{i\}$, Eq. (44) with i replaced by k becomes

$$\begin{aligned} 2 \ge \sum _{j \in N {\setminus } \{k\} } (2x_j-y_j) + (2x_k + y_k) = (2x_i - y_i) + (2x_k + y_k). \end{aligned}$$

Together with $x_i = 1$ and $y_i = 0$, the above inequality implies that $2x_k + y_k = 0 \Rightarrow x_k = y_k = 0$ for $k \in N {\setminus } \{i\}$. Therefore $(x,y) = (e_i, \mathbf{0})$ which shows that $(x,y) \in V^{\prime } \subseteq Q^n$.

From now on, we assume that $x_i < 1$, for all $i \in N$. We first assume $y= \mathbf{0}$. By Lemma 6.1(iii), we have $|L_1| \ge n-1$, and therefore at most one of the inequalities of type (17) is not tight for $(x,y)$. Therefore at most one component of x is nonzero. This implies that $(x,y)$ is a convex combination of $(e_k, \mathbf{0})$ and $(\mathbf{0}, \mathbf{0})$ for some $k \in N$, and therefore $(x,y)$ is equal to one of these two points and is contained in $V^{\prime }$.

Henceforth assume that $y\ne \mathbf{0}$. If $L_3 = \emptyset$, then L does not contain any constraints of type (19), and therefore $x = \mathbf{0}, y= \mathbf{0}$ is a solution of the constraints in L. As these constraints have a unique solution, we have $(x,y) = (\mathbf{0}, \mathbf{0})$ which contradicts the assumption that $y\ne \mathbf{0}$. Therefore, $L_3 \ne \emptyset$.

We will now consider two cases.

Case 1 All constraints of type (17) are tight for $(x,y)$. Let $k \in L_3$. As $2x_i - y_i = 0$ for all $i \in N$, the tight inequality (19) for index k implies that $2 = 2y_k \Rightarrow y_k = 1$. But this means $y_{max} = 1$, and, by Lemma 6.1(iv), all nonzero components of $y$ have value 1. Furthermore, Lemma 6.1(ii) implies that at least three components of $y$ have value 1. This combined with the fact that $2x_i = y_i$ for all $i \in N$ means that and therefore $(x,y) \in V^{\prime }$.

Case 2 At least one constraint of type (17) is not tight for $(x,y)$. Therefore, $|L_1| \le n-1$ and by Lemma 6.1(iii) $|L_1| = n-1$. Let $\{\ell \}= N\setminus L_1$, therefore $2x_\ell > y_\ell$. It also follows that $y_{max} < 1$ as $y_k = 1$ for any $k \in N$ would imply that $2x_i - y_i = 0$ for all $i \in N$, a contradiction.

Further, if $|L_2 \cap L_3| = 0$, then $|L_2 \cup L_3|=|L_2| + | L_3|$ and Lemma 6.1(i) would imply that $|L_2| + | L_3|+|L_y|\le n$. This, in turn, implies that $|L_1| = n$, a contradiction. Therefore $|L_2 \cap L_3| \ge 1$, and $L_2\not =\emptyset$. Consider an index $k\in L_2$, then $y_k = y_{max}$ because $L_2\subseteq T_{max}$. As all nonzero components of $y$ have value $y_{max}$ by Lemma 6.1(iv), and inequality (18) is tight for index k, we conclude that $|T_{ny}|=3$.

Case 2a Let $\ell \not \in T_{ny}$. Let $\delta =\min \{y_{max}/3, x_\ell , 1-x_\ell \}>0$. Define $(x^1,y^1)$ by

$$\begin{aligned} x^1_i = \left\{ \begin{array}{ll} x_i + \delta /2, &{}\quad \text{ if } \;i \in T_{ny} \\ x_i - \delta , &{}\quad \text{ if }\; i = \ell \\ x_i, &{}\quad \text{ o.w. }\end{array},\right. \quad y^1_i = \left\{ \begin{array}{ll} y_i + \delta , &{}\quad \text{ if } \;i \in T_{ny} \\ y_i,&{}\quad \text{ o.w. }\end{array}\right. \end{aligned}$$

and $(x^2,y^2) = 2(x,y) - (x^1, y^1)$. Then both $(x^1,y^1)$ and $(x^2,y^2)$ belong to X and $(x,y)$ is a convex combination of these points, contradicting the fact that it is an extreme point of X.

Case 2b Let $\ell \in T_{ny}$. Let $\delta =\min \{1-x_\ell +y_{\ell }{/}2, y_{max}{/}3, (2x_\ell - y_\ell ){/}2\}>0$. Define $(x^1,y^1)$ by

$$\begin{aligned} x^1_i = \left\{ \begin{array}{ll} x_i + \delta {/}2, &{}\quad \text{ if } \;i \in T_{ny}{\setminus }\{\ell \} \\ x_i - \delta {/}2, &{}\quad \text{ if }\; i = \ell \\ x_i &{}\quad \text{ o.w. }\end{array},\right. \quad y^1_i = \left\{ \begin{array}{ll} y_i + \delta , &{}\quad \text{ if }\;i \in T_{ny} \\ y_i, &{}\quad \text{ o.w. }\end{array}\right. \end{aligned}$$

and $(x^2,y^2) = 2(x,y) - (x^1, y^1)$. Then both $(x^1,y^1)$ and $(x^2,y^2)$ belong to X and $(x,y)$ is a convex combination of these points, contradicting the fact that it is an extreme point of X.

As the claim holds for both cases, the proof is complete. $\square$