Distributed algorithms for convex problems with linear coupling constraints

Abstract

Distributed and parallel algorithms have been frequently investigated in the recent years, in particular in applications like machine learning. Nonetheless, only a small subclass of the optimization algorithms in the literature can be easily distributed, for the presence, e.g., of coupling constraints that make all the variables dependent from each other with respect to the feasible set. Augmented Lagrangian methods are among the most used techniques to get rid of the coupling constraints issue, namely by moving such constraints to the objective function in a structured, well-studied manner. Unfortunately, standard augmented Lagrangian methods need the solution of a nested problem by needing to (at least inexactly) solve a subproblem at each iteration, therefore leading to potential inefficiency of the algorithm. To fill this gap, we propose an augmented Lagrangian method to solve convex problems with linear coupling constraints that can be distributed and requires a single gradient projection step at every iteration. We give a formal convergence proof to at least $\varepsilon$-approximate solutions of the problem and a detailed analysis of how the parameters of the algorithm influence the value of the approximating parameter $\varepsilon$. Furthermore, we introduce a distributed version of the algorithm allowing to partition the data and perform the distribution of the computation in a parallel fashion.

The classical gradient projection algorithm defined in (2) is not distributable

Consider the general case in which S is not separable due to the presence of the constraints h that couple the different blocks of variables $x_{(\nu )}$, i.e. $m > 0$. To solve Problem (1), one could think to employ the following naive parallel version of the classical gradient projection algorithm (whose original generic iteration is defined in (2)):

$$\begin{aligned} x^{k+1}_{(\nu )} = {\mathscr {P}}_{S_\nu (x^{k}_{(\nu )})} \left[ x^{k}_{(\nu )} - \alpha _k \nabla f(x^k)_{(\nu )} \right] , \qquad \nu = 1, \ldots , N, \end{aligned}$$

where $x^k \in S$ and the decomposed subsets $S_\nu$ are defined in the following way

$$\begin{aligned} S_\nu (x_{(\nu )}^k) \triangleq \left\{ x_{(\nu )} \in X_\nu \, : \, A_{*(\nu )} x_{(\nu )} = A_{*(\nu )} x^k_{(\nu )} \right\} \subseteq \mathfrak {R}^{n_\nu }, \qquad \nu = 1, \ldots , N. \end{aligned}$$

Let $\{x^k\}$ be the sequence produced by this algorithm. The sets $S_\nu$ are fixed during the iterations and depend only on the starting point $x^0$:

$$\begin{aligned} S_\nu (x^{\nu ,k}) = S_\nu (x_{(\nu )}^0) = S_\nu ^0, \qquad \forall \, k \ge 0, \qquad \nu = 1, \ldots , N. \end{aligned}$$

A fixed point ${\overline{x}}$ for $\{x^k\}$ is therefore a solution of the following variational inequality problem (see e.g. [1, 15, 16, 25])

$$\begin{aligned} {\overline{x}} \in X, \;\; h({\overline{x}}) = 0, \qquad \nabla f({\overline{x}})^T (x - {\overline{x}}) \ge 0, \qquad \forall \, x \in \prod _{\nu =1}^N S_\nu ^0. \end{aligned}$$

(10)

On the other hand, computing a solution of Problem (1), being a fixed point for the iterations defined in (2), turns out to be a solution $x^*$ of this different variational inequality

$$\begin{aligned} x^* \in X, \;\; h(x^*) = 0, \qquad \nabla f(x^*)^T (x - x^*) \ge 0, \qquad \forall \, x \in S. \end{aligned}$$

(11)

Notice that the point ${\overline{x}}$, solution of the variational inequality (10), could not be a solution of the variational inequality (11), and therefore of Problem (1). This is due to the fact that, for any feasible starting guess $x^0 \in S$, we obtain only

$$\begin{aligned} \displaystyle \prod _{\nu = 1}^N S_\nu ^{0} \subseteq S, \end{aligned}$$

but not the other inclusion in general. Actually the fixed point ${\overline{x}}$ of $\{x^k\}$ is only an equilibrium of the (potential) generalized Nash equilibrium problem (see e.g. [14, 17, 36]) whose generic player $\nu \in \{1,\ldots , N\}$ solves the following optimization problem that is parametric with respect to all the blocks of variables $x_{(\mu )}$ of the other players $\mu \ne \nu$:

$$\begin{aligned} \begin{array}{rl} \underset{x_{(\nu )}}{\text{ minimize }} \;\; &{} f(x)\\ \text {s.t. } \; &{} h(x) = 0, \;\; x_{(\nu )} \in X_\nu . \end{array} \end{aligned}$$