Activity propagation in systems of linear inequalities and its relation to block-coordinate descent in linear programs

Abstract

We study a constraint propagation algorithm to detect infeasibility of a system of linear inequalities over continuous variables, which we call activity propagation. Each iteration of this algorithm chooses a subset of the inequalities and if it infers that some of them are always active (i.e., always hold with equality), it turns them into equalities. We show that this algorithm can be described as chaotic iterations and its fixed points can be characterized by a local consistency, in a similar way to traditional local consistency methods in CSP such as arc consistency. Via complementary slackness, activity propagation can be employed to iteratively improve a dual-feasible solution of large-scale linear programs in a primal-dual loop – a special case of this method is the Virtual Arc Consistency algorithm by Cooper et al. As our second contribution, we show that this method has the same set of fixed points as block-coordinate descent (BCD) applied to the dual linear program. While BCD is popular in large-scale optimization, its fixed points need not be global optima even for convex problems and a succinct characterization of convex problems optimally solvable by BCD remains elusive. Our result may open the way for such a characterization since it allows us to characterize BCD fixed points in terms of local consistencies.

Appendices

Appendix A: Computing an improving feasible direction

As discussed at the beginning of Section 3.2, if () is infeasible, there exists an improving feasible direction $\bar{y}$ satisfying (). We describe one way of obtaining such a direction based on the propagation algorithm (Algorithm 3). We remark that conditions () define a whole convex cone of improving directions and our algorithm finds one of them based on its precise implementation.

Let y be a dual-feasible point such that $p_\mathcal {B}(\tau (y)) = {\perp }$. This implies infeasibility of (), i.e., non-optimality of y. Consider sequences $(B_l)_{l=1}^L$ and $(J_l)_{l = 1}^L$ where $J_1 \supsetneq J_2 \supsetneq \cdots \supsetneq J_L$, $J_1 = \tau (y)$, $J_{l+1} = p_{B_l}(J_l)$ for every $l \in [L-1]$, and $p_{B_L}(J_L) = {\perp }$. To construct $\bar{y}$, we use the primal-dual pair

$$\begin{aligned} \max&\; 0&\min&\; b^T\hat{y}^l{} & {} \end{aligned}$$

(42a)

$$\begin{aligned} A^ix&= b_i&\hat{y}_i^l&\in \mathbb {R}&\forall i&\in B_l \end{aligned}$$

(42b)

$$\begin{aligned} x_j&= 0&\text {--}&\forall j&\in [n]-J_l \end{aligned}$$

(42c)

$$\begin{aligned} x_j&\ge 0&A_j^T\hat{y}^l&\ge 0&\forall j&\in J_l \end{aligned}$$

(42d)

$$\begin{aligned} \text {--}&\hat{y}_i^l&= 0&\forall i&\in [m]-B_l. \end{aligned}$$

(42e)

and proceed as outlined in Algorithm 5. Note that the primal () is a feasibility problem identical to () if $J_l=J$ and $B_l=B$. Even though Algorithm 5 may seem complicated, it is easy to see that in cases when the blocks B are small, problem () is also small (and thus could even be solvable in closed-form). Correctness of Algorithm 5 is given by the following theorem.

Construction of improving direction.

Theorem 19

Let $J_1=\tau (y)$. If $p_\mathcal {B}(\tau (y))={\perp }$, Algorithm 5 returns a vector $\bar{y}^1$ satisfying ().

Proof

We will proceed by induction, i.e., we claim that for each $l\in [L]$, $\bar{y}^l$ satisfies $A_j^T\bar{y}^l \ge 0$ for all $j\in J_{l}$ and $b^T\bar{y}^l = b^T \bar{y}^L < 0$ is maintained during the whole algorithm. Thus, eventually $A_j^T\bar{y}^1 \ge 0$ holds for all $j \in J_{1} =\tau (y)$.

For the base case with $l=L$, primal () is infeasible due to $p_{B_L}(J_L) = {\perp }$ (see Theorem 15) and dual () is therefore unbounded since it is always feasible. Thus, there exists $\hat{y}^L$ feasible to the dual () that satisfies $b^T\hat{y}^L < 0$. By feasibility, $A_j^T\hat{y}_i^L \ge 0$ for all $j \in J_L$.

For the inductive step, let $l \le L-1$. If condition on line 3 is not satisfied, $A_j^T\bar{y}^l \ge 0$ holds for all $j \in J_l$ trivially by setting $\bar{y}^l$ equal to $\bar{y}^{l+1}$ on line 7 due to our inductive hypothesis.

If condition on line 3 is satisfied, let us focus on (). Since $p_{B_l}(J_l) = J_{l+1}\ne {\perp }$, primal () is feasible with optimal value 0, which is attained by all feasible solutions. Let $x^l$ and $\hat{y}^l$ be in the relative interior of the set of optimal solutions of the primal and dual (), respectively. Since $x^l,\hat{y}^l$ are from the relative interior, they satisfy strict complementary slackness (see Theorem 9), i.e., $x_j = 0 \iff A_j^T\hat{y}^l > 0$ for all $i \in J_l$. By the last statement in Theorem 15, $x_j = 0 \wedge A_j^T\hat{y}^l > 0$ holds for all $j \in J_{l}-J_{l+1}$ because $p_{B_l}(J_l) = J_{l+1}$. For completeness, $x_j > 0 \wedge A_j^T\hat{y}^l = 0$ holds for all $j \in J_{l+1}$.

Notice that $\delta _l$ is well-defined because condition on line 3 was satisfied. Moreover, $\delta _l>0$ due to both $-A_j^T\bar{y}^{l+1}$ and $A_j^T\hat{y}^l$ being positive by definition of $\delta _l$.

We consider the following cases to prove that $A_j^T\bar{y}^l \ge 0$ for all $j\in J_l$:

• If $j \in J_{l+1}$, then $A_j^T\bar{y}^{l+1} \ge 0$ by inductive hypothesis and $A_j^T\hat{y}^l = 0$ by strict complementary slackness, hence $A_j^T \bar{y}^l = A_j^T\bar{y}^{l+1} \ge 0$.

• If $j \in J_{l}-J_{l+1}$, then $A_j^T\hat{y}^l > 0$. If $A_j^T\bar{y}^{l+1} \ge 0$, then $A_j^T \bar{y}^{l} = A_j^T\bar{y}^{l+1} + \delta _l A_j^T\hat{y}^l \ge 0$. On the other hand, if $A_j^T\bar{y}^{l+1} < 0$, it holds by definition of $\delta _l$ that $\delta _l \ge -A_j^T\bar{y}^{l+1}/A_j^T\hat{y}^{l}$, which is after a simple reformulation equivalent to $A_j^T \bar{y}^l = A_j^T\bar{y}^{l+1}+\delta _l A_j^T\hat{y}^l \ge 0$.

Finally, it holds by strong duality that $b^T\bar{y}^l = 0$, which yields $b^T\bar{y}^{l} = b^T \bar{y}^{l+1} + \delta b^T\hat{y}^l = b^T\bar{y}^{l+1} < 0$.$\square$

Appendix B: Faces and B-consistent sets

In this section, we explain the geometric meaning of B-consistent sets, as defined in Section 3.2.1. In detail, we will show that the set of B-consistent sets is order-isomorphic to the set of non-empty faces of the polyhedron $X_B([n])$ (defined in (22)). Consequently, the lattice $(\mathcal {J}_B,\sqsubseteq )$ (see (18)) is isomorphic to the face lattice of $X_B([n])$.

The faces of a convex polyhedron are usually defined using valid inequalities (or supporting hyperplanes) [49, Section 2.1]. However, faces can be also equivalently obtained by forcing subsets of inequalities to be active [29, Section 5.6]. We use this latter definition. Moreover, we define faces only for the polyhedron $X_B([n])$ (see (22)) where $B\subseteq [m]$ is fixed.

Definition 6

Let $B\subseteq [m]$. A set $F\subseteq \mathbb {R}^n$ is a face of the polyhedron $X_B([n])$ if $F=\emptyset$ or $F=X_B(J)$ for some $J\subseteq [n]$. The set of all faces of $X_B([n])$ is denoted by

$$\begin{aligned} \mathcal {F}_B=\{F\mid F \text { is a face of } X_B([n])\}. \end{aligned}$$

(43)

It is immediate that the set of all faces of $X_B([n])$ is finite. Moreover, the set of all faces is closed under intersections, as shown by the following corollary.

Corollary 4

([49, Proposition 2.3]) Let $B\subseteq [m]$. If $F, F'\in \mathcal {F}_B$, then $F\cap F'\in \mathcal {F}_B$.

Proof

This is clear if $F=\emptyset$ or $F'=\emptyset$. Otherwise, there are $J,J'\subseteq [n]$ such that $F=X_B(J)$ and $F'=X_B(J')$. Hence, $X_B(J)\cap X_B(J')=X_B(J\cap J')\in \mathcal {F}_B$.$\square$

Thus, the face set of the polyhedron $X_B([n])$ forms a finite meet-semilattice w.r.t. the partial order given by set inclusion where the meet operation is set intersection. Moreover, $X_B([n])$ is the top element of this meet-semilattice, so $(\mathcal {F}_B,\subseteq )$ is a complete lattice by Theorem 1. This is known as the face lattice [27, 49, 50].

We will now describe the connection between B-consistent sets and the faces of polyhedron $X_B([n])$. Firstly, let us point our attention to the fact that we could require the set $J$ to be B-consistent in Definition 6. We formulate a stronger statement in the following theorem.

Theorem 20

Let $B\subseteq [m]$. For any non-empty $F\in \mathcal {F}_B$, there exists a unique B-consistent set $J\subseteq [n]$ such that $F=X_B(J)$. Conversely, for any B-consistent set $J\subseteq [n]$, the face $X_B(J)$ is non-empty.

Proof

For the first part, let us show that at least one such set exists. By definition, since $F\in \mathcal {F}_B$ is non-empty, there exists $J'\subseteq [n]$ such that $F=X_B(J')$. Clearly, we have that $X_B(J')=X_B(p_B(J'))$ and $p_B(J')$ is B-consistent by Theorem 15. To show that this set is unique, let us proceed by contradiction. Let $J_1,J_2\subseteq [n]$ be B-consistent sets such that $X_B(J_1)=X_B(J_2)=F$ and $J_1\ne J_2$. Without loss of generality, assume $J_2-J_1\ne \emptyset$ and let $j^*\in J_2-J_1$ be arbitrary. We have that $x_{j^*}=0$ for all $x\in X_B(J_1)$ due to $j^*\notin J_1$, so $J_2$ is not B-consistent as () for $J_2$ implies $x_{j^*}=0$ and $j^*\in J_2$.

For the other part, it is clear that any B-consistent set defines a face of the polyhedron by Definition 6. Moreover, this face is non-empty due to () being non-empty for any B-consistent set $J$.$\square$

Following Theorem 20, $X_B$ can be interpreted as a bijection between B-consistent sets and the set of non-empty faces of $X_B([n])$.¹⁸

As already noted earlier in Section 3.2.1, $X_B$ is an isotone mapping, i.e., if $J\subseteq J'\subseteq [n]$, then $X_B(J)\subseteq X_B(J')$. The converse relation also holds if we restrict ourselves to B-consistent sets:

Proposition 21

Let $B\subseteq [m]$ and $J,J'\subseteq [n]$ be B-consistent. If $X_B(J)\subseteq X_B(J')$, then $J\subseteq J'$.

Proof

By contradiction: let $X_B(J)\subseteq X_B(J')$ and $J\nsubseteq J'$. The latter implies that there is $j\in [n]$ such that $j\in J$ and $j\notin J'$. By Definition 5, there exists $x^*\in X_B(J)$ satisfying () for J with $x^*_j>0$. However, by definition of $X_B(J')$ (see (22)), we have $x_j=0$ for all $x\in X_B(J')$ (due to $j\notin J'$), so $x^*\notin X_B(J')$, which contradicts $X_B(J)\subseteq X_B(J')$.$\square$

It seems natural to extend the mapping $X_B$ to obtain the isotone bijection $X_B'{:}\,\mathcal {J}_B \rightarrow \mathcal {F}_B$ defined by

$$\begin{aligned} X'_B(J) = {\left\{ \begin{array}{ll} X_B(J) &{} \text {if } J\subseteq [n]\\ \emptyset &{} \text {if } J={\perp } \end{array}\right. }. \end{aligned}$$

(44)

Clearly, we have that $J\sqsubseteq J' \iff X_B'(J)\subseteq X_B'(J')$ for any $J,J'\in \mathcal {J}_B$. The lattices $(\mathcal {J}_B,\sqsubseteq )$ and $(\mathcal {F}_B,\subseteq )$ are therefore order-isomorphic and $X_B'$ is a lattice isomorphism [24].¹⁹

Remark 10

In some formalisms, $\emptyset$ does not belong to the face lattice of some polyhedra. To be precise, if $(S,\subseteq )$ is the face lattice of some polyhedron, then $(S-\{\emptyset \},\subseteq )$ is a lattice if and only if $\bigcap \{F\mid F\in S-\{\emptyset \}\}\ne \emptyset$, i.e., if there is a minimal non-empty face [50, Section 8]. As an example, for the non-negative orthant $X_\emptyset ([n])=\{x\in \mathbb {R}^n\mid x\ge 0\}$, we have $\bigcap \{F\mid F\in \mathcal {F}_\emptyset -\{\emptyset \}\}=\{0\}$ and $(\mathcal {F}_\emptyset -\{\emptyset \},\subseteq )$ is a lattice where the bottom element is $\{0\}$ (i.e., the singleton set containing the origin).

Consequently, if $(\mathcal {F}_B-\{\emptyset \},\subseteq )$ is a lattice, then ${(\mathcal {J}_B-\{\perp \},\sqsubseteq )}$ is a lattice too. These lattices are again order-isomorphic and $X_B$ is the lattice isomorphism.