A nonautonomous-differential-inclusion neurodynamic approach for nonsmooth distributed optimization on multi-agent systems

Abstract

This paper considers a category of nonsmooth distributed optimization on multi-agent systems, where agents own privacies and collectively minimize a sum of local cost functions. Taking the restrictions on communication among agents into consideration, a nonautonomous-differential-inclusion neurodynamic approach is proposed over a weighed topology graph. The convergence of neural network is analyzed and its state exponentially converges to an optimal solution of distributed optimization under certain conditions. Since no additional conditions are required to guarantee the convergence, the neural network is superior to distributed algorithms based on penalty method, which need to estimate penalty parameters. Compared with some existed approaches, the neural network has the advantage of possessing fewer state variables. Finally, illustrative examples and an application in distributed quantile regression are delineated to testify the effectiveness of the presented neural network.

Appendix

Proposition 4

There exists a constant $\sigma _0>0$, such that for any $\sigma >\sigma _0$, the minimum points of $D_{\sigma }(\mathbf{x} )$ are also the optimal solutions of distributed optimization (2) and vice versa.

Proof

Let $\varOmega _\sigma$ be the set of minimum points of $D_{\sigma }(\mathbf{x} )$ and ${{\bar{x}}}=\frac{1}{n}\sum \nolimits _{i=1}^n x_i$. Then it has $\sum \nolimits _{k=1}^n\parallel x_k-{\bar{x}}\parallel \le \frac{1}{n}\sum \nolimits _{k=1}^n\sum \nolimits _{l=1}^n\parallel x_k-x_l\parallel .$ According to Assumption 2, for any k, $l\in \{1,2,...,n\}$, there must exist a path $P_{kl}\in {\mathcal {E}}$ that connects the two vertexes. Noticing that weighed graph ${\mathcal {G}}$ is undirected and connected, we have

$$\begin{aligned} \begin{array}{ll} a_{\rm min}\cdot \sum \limits _{k=1}^n \parallel x_k-{\bar{x}}\parallel &{}\le \frac{1}{2n}\sum \limits _{k=1}^n\sum \limits _{l=1}^n\sum \limits _{(i,j)\in P_{kl}} a_{ij}\parallel x_i-x_j\parallel \\ &{}\le \frac{n}{2}\sum \limits _{i=1}^n \sum \limits _{j\in {\mathcal {N}}_i} a_{ij} \parallel x_i-x_j\parallel , \end{array} \end{aligned}$$

(34)

where $a_{\rm min}$ is an edge of least weight in the weighed graph ${\mathcal {G}}$. Due to Assumption 1, it can be immediately obtained that

$$\begin{aligned} \sum _{i=1}^n f_i(x_i)-\sum _{i=1}^n f_i({\bar{x}})\ge - {\tilde{M}} \sum _{i=1}^n\parallel x_i -{\bar{x}}\parallel , \end{aligned}$$

(35)

where ${\tilde{M}}=\max \nolimits _{i\in {\mathcal {V}}}\{M_i\}$. Then, in the light of (35), let $\sigma _0=n{\tilde{M}}/2a_{\rm min}$, then

$$\begin{aligned} D_{\sigma }(\mathbf{x} )= & {} \sum \limits _{i=1}^n f_i(x_i)+\sigma \sum \limits _{i=1}^n\sum \limits _{j\in {\mathcal {N}}_i} {{a_{ij} }} \parallel x_i-x_j\parallel \nonumber \\= & {} \sum \limits _{i=1}^n f_i({\bar{x}})+\sum \limits _{i=1}^n f_i(x_i)-\sum \limits _{i=1}^n f_i({{\bar{x}}})\nonumber \\&+\sigma \sum \limits _{i=1}^n\sum \limits _{j\in {\mathcal {N}}_i} {{a_{ij} }} \parallel x_i-x_j\parallel \nonumber \\\ge & {} \sum \limits _{i=1}^n f_i({\bar{x}})+(\sigma -\sigma _0)\sum \limits _{i=1}^n\sum \limits _{j\in {\mathcal {N}}_i} {{a_{ij} }}\parallel x_i-x_j\parallel . \end{aligned}$$

(36)

$\square$

In the following part, we will prove the equivalence of minimum points of $f(\mathbf{x} )$ and $D_{\sigma }(\mathbf{x} )$ with the help of (36).

Suppose $\mathbf{x}^*=\mathrm {col}\{x^*,x^*,...,x^*\}$ is an optimal solution to distributed optimization problem (2). By the definition of $D_{\sigma }(\mathbf{x} )$, it can be immediately obtained that $\mathbf{x}^*$ also minimizes $D_{\sigma }(\mathbf{x} )$ on ${\mathbb {R}}^{mn}$. That is to say,

$$\begin{aligned} \varOmega ^*\subseteq \varOmega _\sigma . \end{aligned}$$

Let $f_{\rm min}$ be the minimum value of $D_{\sigma }(\mathbf{x} )$. In the above analysis, $f_{\rm min}$ is also minimum value of $D_{\sigma }(\mathbf{x} )$ on $\varOmega =\{\mathbf{x }\in {\mathbb {R}}^{mn}|x_i=x_j\}$.

Suppose that $\mathbf{y} ^*=\mathrm {col}\{y_1^*,y_2^*,...,y_n^*\}$ minimizes $D_{\sigma }(\mathbf{x} )$, then for any $\mathbf{y} =\mathrm {col}\{y,y,...,y\}\in \varOmega$, it is apparent that

$$\begin{aligned} D_{\sigma }(\mathbf{y} )=\sum _{i=1}^nf_i(y)+\sigma \sum \limits _{i=1}^n \sum \limits _{j\in {\mathcal {N}}_i} {{a_{ij}}} \parallel y-y\parallel =\sum _{i=1}^nf_i(y). \end{aligned}$$

(37)

By the fact that $\mathbf{y}^*$ minimizes $D_{\sigma }(\mathbf{x} )$ and (36), we have

$$\begin{aligned}&D_{\sigma }(\mathbf{y} ^*)=\sum \limits _{i=1}^nf_i(y_i^*)+\sigma \sum \limits _{i=1}^n \sum \limits _{j\in {\mathcal {N}}_i} {{a_{ij} }} \parallel y_i^*-y_j^*\parallel \nonumber \\&\quad \ge \sum \limits _{i=1}^n f_i\left( \frac{1}{n}\sum \limits _{i=1}^ny_i^*\right) +(\sigma -\sigma _0)\sum \limits _{i=1}^n\sum \limits _{j\in {\mathcal {N}}_i}{{a_{ij} }} \parallel y_i^*-y_j^*\parallel . \end{aligned}$$

(38)

Since the vector $\mathrm {col}\{\frac{1}{n}\sum \limits _{i=1}^ny_i^*,\frac{1}{n}\sum \limits _{i=1}^ny_i^*,...,\frac{1}{n}\sum \limits _{i=1}^ny_i^*\}\in \varOmega,$ then

$$\begin{aligned} \sum \limits _{i=1}^n f_i\left( \frac{1}{n}\sum \limits _{i=1}^ny_i^*\right) \ge f_{\rm min}. \end{aligned}$$

(39)

Combining (38) with (39), it has

$$\begin{aligned} D_{\sigma }(\mathbf{y} ^*)\ge f_{\rm min}+(\sigma -\sigma _0)\sum \limits _{i=1}^n\sum \limits _{j\in {\mathcal {N}}_i} {{a_{ij} }}\parallel y_i^*-y_j^*\parallel . \end{aligned}$$

(40)

If $\mathbf{y}^*\notin \varOmega$, there must exist $k,l\in \{1,2,...,n\}$ such that $\Vert y_k^*-y_l^*\Vert > 0$. Hence, the condition $\sigma >\sigma _0$ and (40) determine that

$$\begin{aligned} D(\mathbf{y} ^*)> f_{\rm min}, \end{aligned}$$

which contradicts with the fact that $D(\mathbf{y} ^*)=\min \nolimits _\mathbf{x \in {\mathbb {R}}^{mn}}{D_{\sigma }(\mathbf{x} )}\le f_{\rm min}$. Thus, we have $\mathbf{y}^*\in \varOmega$, that is $y_i^*=y_j^*$ for $i,j\in \{1,2,...,n\}.$

Note $\mathbf{y} ^*=\mathrm {col}\{{\tilde{y}}^*,{\tilde{y}}^*,...,{\tilde{y}}^*\}$. Due to (37) and (38), it follows that

$$\begin{aligned} f(\mathbf{y} )= & {} \sum \limits _{i=1}^nf_i(y)\ge \sum \limits _{i=1}^n f_i({\tilde{y}}^*)\\&+(\sigma -\sigma _0)\sum \limits _{i=1}^n\sum \limits _{j\in {\mathcal {N}}_i} {{a_{ij}}} \parallel {\tilde{y}}^*-{\tilde{y}}^*\parallel . \end{aligned}$$

That is to say,

$$\begin{aligned}\begin{array}{ll} f(\mathbf{y} )\ge f(\mathbf{y} ^*), \end{array} \end{aligned}$$

for any $\mathbf{y} \in \varOmega$. Therefore, $\mathbf{y} ^*$ is also an optimal solution to distributed optimization (2) and $\varOmega _\sigma \subseteq \varOmega ^*$, which completes the proof.