Exact spectral-like gradient method for distributed optimization

Abstract

Since the initial proposal in the late 80s, spectral gradient methods continue to receive significant attention, especially due to their excellent numerical performance on various large scale applications. However, to date, they have not been sufficiently explored in the context of distributed optimization. In this paper, we consider unconstrained distributed optimization problems where n nodes constitute an arbitrary connected network and collaboratively minimize the sum of their local convex cost functions. In this setting, building from existing exact distributed gradient methods, we propose a novel exact distributed gradient method wherein nodes’ step-sizes are designed according to the novel rules akin to those in spectral gradient methods. We refer to the proposed method as Distributed Spectral Gradient method. The method exhibits R-linear convergence under standard assumptions for the nodes’ local costs and safeguarding on the algorithm step-sizes. We illustrate the method’s performance through simulation examples.

Appendix

Proof of Lemma 4.2

Consider algorithm (8)–(9). For the special case considered here, the update rule (7)–(8) becomes:

$$\begin{aligned} x^{k+1}= & {} W x^{k} - \alpha \, z^k, \end{aligned}$$

(57)

$$\begin{aligned} z^{k+1}= & {} W z^k + x^{k+1}- x^k. \end{aligned}$$

(58)

Denote by $\xi ^k$ the $(2n) \times 1$ vector defined by $\xi ^k = \left( e^k\,,z^k\right)$, where $e^k = x^k - x^\star$. Then, it is easy to show that $\xi ^k$ obeys the following recursion:

$$\begin{aligned} \xi ^{k+1} = E\,\xi ^k, \end{aligned}$$

where E is the $(2n) \times (2n)$ matrix with the following $n \times n$ blocks:¹

$$\begin{aligned} E_{11} = W-J,\,\,\,E_{12} =-\alpha \,I,\,\,\,E_{21} =W-I, \,\,\,E_{22} =W-\alpha \,I. \end{aligned}$$

Consider the eigenvalue decomposition of matrix $W = Q\Lambda Q^T$, where Q is the matrix of orthonormal eigenvectors, and $\Lambda$ is the matrix of eigenvalues ordered in a descending order.

We have that $\lambda _1=1$, and $\lambda _i \in (0,1)$, $i \ne 1$. Note that the matrix E can now be decomposed as follows:

$$\begin{aligned} E = {\widehat{Q}}\,{\widehat{P}}\,{\widehat{\Lambda }}\,{\widehat{P}}^T\,{\widehat{Q}}^T. \end{aligned}$$

Here, ${\widehat{Q}}$ is the $(2n) \times (2n)$ orthonormal matrix with the $n \times n$ blocks at positions (1,1) and (2,2) equal to Q, and zero-off diagonal $n \times n$ blocks; and ${\widehat{P}}$ is an appropriate permutation matrix. Furthermore, ${\widehat{\Lambda }}$ is the $(2n)\times (2n)$ block-diagonal matrix with the $2 \times 2$ diagonal blocks $D_1,\ldots ,D_n$, as follows:

$$\begin{aligned} D_1 = \begin{bmatrix} 0&\quad -\alpha \\ 0&\quad 1-\alpha \end{bmatrix},\,\,\, D_i = \begin{bmatrix} \lambda _i&\quad -\alpha \\ \lambda _i-1&\quad \lambda _i-\alpha \end{bmatrix},\,\,\,i \ne 1. \end{aligned}$$

It is then clear that the matrix E has the same eigenvalues as the matrix ${\widehat{\Lambda }}$, and hence the two matrices have the same spectral radius. Next, by evaluating the eigenvalues of the $2 \times 2$ matrices $D_i,$$i=1,\ldots ,n$, it is straightforward to verify sufficient conditions on $\alpha$ such that the spectral radius $\rho ({\widehat{\Lambda }})$ is strictly less than one, and such that $\rho ({\widehat{\Lambda }})$ is strictly greater than one. Namely, for $i \ne 1$, it is easy to show that $\rho (D_i)<1$ if and only if $\alpha \in (0,{\overline{\alpha }}_i)$, where ${\overline{\alpha }}_i = \frac{(1+\lambda _i)^2}{2}$. On the other hand, for $i=1$, we have that $\rho (D_1) = 1-\alpha < 1$. In view of the fact that $\lambda _i>0$, $i=2,\ldots ,n$, the latter implies that, when $\alpha \le 1/2$, we have that $\rho ({\widehat{\Lambda }})<1$; also, $\rho ({\widehat{\Lambda }})>1$ whenever $\alpha >2$. This in particular implies that $x^k$ converges R-linearly to $x^\star$ if $\alpha \le 1/2$, and that $x^k$ diverges, when $\alpha >2$. The proof is complete. $\square$