A first-order block-decomposition method for solving two-easy-block structured semidefinite programs

Abstract

In this paper, we consider a first-order block-decomposition method for minimizing the sum of a convex differentiable function with Lipschitz continuous gradient, and two other proper closed convex (possibly, nonsmooth) functions with easily computable resolvents. The method presented contains two important ingredients from a computational point of view, namely: an adaptive choice of stepsize for performing an extragradient step; and the use of a scaling factor to balance the blocks. We then specialize the method to the context of conic semidefinite programming (SDP) problems consisting of two easy blocks of constraints. Without putting them in standard form, we show that four important classes of graph-related conic SDP problems automatically possess the above two-easy-block structure, namely: SDPs for $\theta$-functions and $\theta _{+}$-functions of graph stable set problems, and SDP relaxations of binary integer quadratic and frequency assignment problems. Finally, we present computational results on the aforementioned classes of SDPs showing that our method outperforms the three most competitive codes for large-scale conic semidefinite programs, namely: the boundary point (BP) method introduced by Povh et al., a Newton-CG augmented Lagrangian method, called SDPNAL, by Zhao et al., and a variant of the BP method, called the SPDAD method, by Wen et al.

Appendix: Ergodic convergence results

This appendix derives an ergodic iteration-complexity bound for Algorithm 1.

We start by stating the weak transportation formula for the $\varepsilon$-subdifferential.

Proposition 10.1

(Proposition 1.2.10 in [8]) Suppose that $f:{\mathcal {Z}}\rightrightarrows {[-\infty ,\infty ]}$ is a closed proper convex function. Let $z^{i},v^{i}\in {\mathcal {Z}}$ and $\varepsilon _{i},\alpha _{i}\in {\mathbb {R}}_{+}$, for $i=1,\ldots ,k$, be such that

$$\begin{aligned} v^{i}\in \partial _{\varepsilon _{i}}f(z^{i}),\quad i=1,\ldots ,k,\qquad \sum _{i=1}^{k}\alpha _{i}=1, \end{aligned}$$

and define

$$\begin{aligned}&z_{a}:=\sum _{i=1}^{k}\alpha _{i}z^{i},\quad v_{a}:=\sum _{i=1}^{k}\alpha _{i}v^{i}, \\&\varepsilon _{a}:=\sum _{i=1}^{k}\alpha _{i} [\varepsilon _{i}+\langle z^{i}-z_{a},v^{i}-v_{a}\rangle _{{\mathcal {Z}}}]= \sum _{i=1}^{k}\alpha _{i}[\varepsilon _{i}+\langle z^{i}-z_{a},v^{i}\rangle _{{\mathcal {Z}}}]. \end{aligned}$$

Then, $\varepsilon _{a}\ge 0$ and $v_{a}\in \partial _{\varepsilon _{a}}f(z_{a})$.

Theorem 10.2

Consider the sequences $\{(x^{k},y^{k})\}, \{({\tilde{x}}^{k},{\tilde{y}}^{k})\}, \{(v_{1}^{k},v_{2}^{k})\}$ and $\{\varepsilon _{k}\}$ generated by Algorithm 1, and the sequences $\{c^{k}\}$ and $\{d^{k}\}$ defined in (26). For every $k\in {{\mathbb {N}}}$, define

$$\begin{aligned}&\varLambda _{k}:=\sum _{i=1}^{k}\lambda _{i}, \quad ({\tilde{x}}_{a}^{k},{\tilde{y}}_{a}^{k}):= \varLambda _{k}^{-1}\sum _{i=1}^{k}\lambda _{i} ({\tilde{x}}^{i},{\tilde{y}}^{i}),\\&(v_{1,a}^{k},v_{2,a}^{k}):=\varLambda _{k}^{-1} \sum _{i=1}^{k}\lambda _{i}(v_{1}^{k},v_{2}^{k}), \quad (c_{a}^{k},d_{a}^{k}):=\varLambda _{k}^{-1} \sum _{i=1}^{k}\lambda _{i}(c^{k},d^{k}) \end{aligned}$$

and

$$\begin{aligned}&\varepsilon _{k}^{1,a}:=\varLambda _{k}^{-1}\sum _{i=1}^{k} \lambda _{i}[\varepsilon _{k}+\langle {\theta ^{-1}c^{i}},{{\tilde{x}}^{i}-{\tilde{x}}_{a}^{k}}\rangle ], \quad \varepsilon _{k}^{2,a}:=\varLambda _{k}^{-1} \sum _{i=1}^{k}\lambda _{i}\langle {d^{i}},{{\tilde{y}}^{i} -{\tilde{y}}_{a}^{k}}\rangle , \nonumber \\&\quad \varepsilon _{k}^{a}:=\varepsilon _{k}^{1,a}+ \varepsilon _{k}^{2,a}. \end{aligned}$$

(61)

Then, for every $k\in {{\mathbb {N}}}$,

$$\begin{aligned}&(\theta ^{-1}v_{1,a}^{k},v_{2,a}^{k})\in \left[ \partial _{\varepsilon _{k}^{1,a}} \left( f+h_{1}+\langle {{\tilde{y}}_{a}^{k}},{\cdot }\rangle \right) ({\tilde{x}}_{a}^{k})\right] \times \left[ \partial _{\varepsilon _{k}^{2,a}} \left( h_{2}^{*}-\langle {{\tilde{y}}_{a}^{k}},{\cdot }\rangle \right) ({\tilde{y}}_{a}^{k})\right] \nonumber \\&\quad \subseteq \partial _{\varepsilon _{k}^{a}} [{\mathcal L}(\cdot ,{\tilde{y}}_{a}^{k}) -{\mathcal L}({\tilde{x}}_{a}^{k},\cdot )] ({\tilde{x}}_{a}^{k},{\tilde{y}}_{a}^{k}) \end{aligned}$$

(62)

and

$$\begin{aligned}&\sqrt{\theta ^{-1}\Vert v_{1,a}^{k}\Vert ^{2}+\Vert v_{2,a}^{k} \Vert ^{2}}\le \max \left\{ \frac{1}{\sigma },\frac{\sqrt{\theta }L}{\sigma _{1}^{2}}\right\} \left( \frac{2\sqrt{\theta }}{k}\right) \sqrt{\theta ^{-1}d_{x,0}^{2}+d_{y,0}^{2}}, \end{aligned}$$

(63)

$$\begin{aligned}&\varepsilon _{k}^{a}\le \max \left\{ 1,\frac{\sqrt{\theta }L\sigma }{\sigma _{1}^{2}}\right\} \left[ \frac{8\sqrt{\theta }}{(1-\sigma _{1})k}\right] \left( \theta ^{-1}d_{x,0}^{2}+d_{y,0}^{2}\right) ,\qquad \end{aligned}$$

(64)

where $d_{x,0}$ and $d_{y,0}$ are defined in (31).

Proof

Let $k\in {\mathbb {N}}$ be given. Note that by (35) and the definition of $\langle {\cdot },{\cdot }\rangle _{\theta }$, we have

$$\begin{aligned} \varepsilon _{k}^{1,a}=\varLambda _{k}^{-1}\sum _{i=1}^{k} \lambda _{i}[\varepsilon _{k}+\langle {c^{i}},{{\tilde{x}}^{i} -{\tilde{x}}_{a}^{k}}\rangle _{\theta }],\quad \varepsilon _{k}^{2,a} =\varLambda _{k}^{-1}\sum _{i=1}^{k}\lambda _{i}\langle {d^{i}},{{\tilde{y}}^{i}-{\tilde{y}}_{a}^{k}}\rangle . \end{aligned}$$

Then, in view of Lemma 4.2 and Theorem 2.4 in [12], we have

$$\begin{aligned} \Vert F({\tilde{x}}_{a}^{k},{\tilde{y}}_{a}^{k}) +(c_{a}^{k},d_{a}^{k})\Vert _{\theta ,1}\le 2 \frac{d_{0}^{\theta }}{\varLambda _{k}}, \quad \varepsilon _{k}^{a}=\varepsilon _{k}^{1,a}+ \varepsilon _{k}^{2,a}\le \left( \frac{8\sigma }{1-\sigma _{1}}\right) \frac{(d_{0}^{\theta })^{2}}{\varLambda _{k}}. \end{aligned}$$

Hence, it follows from the above relations, Lemma 4.2(d) and the fact that $\lambda _{k}\ge {\tilde{\lambda }}$, that

$$\begin{aligned} \Vert (v_{1,a}^{k},v_{2,a}^{k})\Vert _{\theta ,1}\!=\! \Vert F({\tilde{x}}_{a}^{k},{\tilde{y}}_{a}^{k}) \!+\!(c_{a}^{k},d_{a}^{k})\Vert _{\theta ,1}\!\le \! 2 \frac{d_{0}^{\theta }}{\varLambda _{k}}\!\le \! 2 \frac{d_{0}^{\theta }}{k{\tilde{\lambda }}}, \quad \varepsilon _{k}^{a}\!\le \!\left( \frac{8\sigma }{1-\sigma _{1}} \right) \frac{(d_{0}^{\theta })^{2}}{k{\tilde{\lambda }}}. \end{aligned}$$

Using the definition of $\Vert (\cdot ,\cdot )\Vert _{\theta ,1}$, (30) and the definition of ${\tilde{\lambda }}$ in (22), we easily see that the above two inequalities imply (63) and (64). Now, (28), (29), (35), (61) and Proposition 10.1 imply that

$$\begin{aligned} \theta ^{-1}v_{1,a}^{k}\in \partial _{\varepsilon _{k}^{1,a}} (f+h_{1})({\tilde{x}}_{a}^{k})+{\tilde{y}}_{a}^{k},\quad v_{2,a}^{k}\in \partial _{\varepsilon _{k}^{2,a}}(h_{2}^{*}) ({\tilde{y}}_{a}^{k})-{\tilde{x}}_{a}^{k}. \end{aligned}$$

and hence that

$$\begin{aligned} \theta ^{-1}v_{1,a}^{k}\in (\partial _{x,\varepsilon _{k}^{1,a}}{\mathcal L})({\tilde{x}}_{a}^{k},{\tilde{y}}_{a}^{k}),\quad v_{2,a}^{k}\in (\partial _{y,\varepsilon _{k}^{2,a}}{\mathcal L})({\tilde{x}}_{a}^{k},{\tilde{y}}_{a}^{k}). \end{aligned}$$

The above four inclusions are easily seen to imply (62). $\square$