An Improved Distributed Gradient-Push Algorithm for Bandwidth Resource Allocation over Wireless Local Area Network

Abstract

Bandwidth allocation problems over wireless local area network have attracted extensive research recently due to the rapid growth in the number of users and bandwidth-intensive applications. In this paper, a bandwidth allocation problem over wireless local area network with directed topologies is investigated and the global objective function of the problem consists of local downloading and uploading cost with both constraints of feasible allocation region and network resources. An improved high-efficiency gradient-push algorithm is proposed for the bandwidth allocation problem which not only guarantees successful data transmission but also minimizes the global objective function. Compared with the existing distributed algorithms, firstly, we use weighted running average bandwidth to replace the current state variables which can ensure the solution converge to the optimal value asymptotically with probability one. Next, noisy gradient samples are used in the proposed algorithm instead of accurate gradient information which enhances the robustness and expands the scope of application. Theoretical analysis shows the convergence rate of the time-averaged value to the optimal solution. Finally, numerical examples are presented to validate the proposed algorithm.

Appendices

Appendix A: Summary of Notations

See Table 2.

Table 2 Summary of notations

Appendix B: Proof of Theorem 3.1

For any user $i \in v$, it can be obtained by Lemma 3.2:

$$\begin{aligned} {\left\| {{\mathbf{x}_i}(t + 1) - \mathbf{x}} \right\| ^2} \le {\left\| {{\mathbf{x}_i}(t) - \mathbf{x}} \right\| ^2} + {\left\| {c(t) \cdot {{\mathbf{s}}_i}(t + 1)} \right\| ^2} - 2c(t) \cdot {{\mathbf{s}}_i}{(t + 1)^T}\left( {{\mathbf{x}_i}(t) - \mathbf{x}} \right) \end{aligned}$$

(16)

The last term $- 2c(t) \cdot {s_i}{(t + 1)^T}\left( {{\mathbf{x}_i}(t) - \mathbf{x}} \right)$ can be transformed as:

$$\begin{aligned} \begin{aligned}&- 2c(t) \cdot {{\mathbf{s}}_i}{(t + 1)^T}\left( {{\mathbf{x}_i}(t) - \mathbf{x}} \right) \le - 2c(t) \cdot \left[ {{L_i}\left( {{\mathbf{x}_i}(t),{{\mathbf{z}}_i}(t + 1)} \right) - {L_i}\left( {\mathbf{x}},{{\mathbf{z}}_i}(t + 1) \right) } \right] \\ {}&- 2c(t) \cdot {N_i}{\left( {{\mathbf{x}_i}(t)} \right) ^T} \cdot \left( {{\mathbf{x}_i}(t) - \mathbf{x}} \right) - 2c(t) \cdot {N_i}{\left( {{\mathbf{x}_i}(t)} \right) ^T} \cdot \left( {{\mathbf{x}_i}(t) - \mathbf{x}} \right) \end{aligned} \end{aligned}$$

(17)

From (17), take conditional expectation with respect to ${F_t}$ on both sides of (16), we obtain:

$$\begin{aligned} \begin{aligned}&\quad E\left[ {{{\left\| {{\mathbf{x}_i}(t + 1) - \mathbf{x}} \right\| }^2}|{F_t}} \right] \le {\left\| {{\mathbf{x}_i}(t) - \mathbf{x}} \right\| ^2} + E\left[ {{{\left\| {c(t) \cdot {{\mathbf{s}}_i}(t + 1)} \right\| }^2}|{F_t}} \right] \\&- 2c(t) \cdot \left[ {{L_i}\left( {{\mathbf{x}_i}(t),\bar{\varvec{\mu }} (t)} \right) - {L_i}\left( {{\mathbf{x}},\bar{\varvec{\mu }} (t)} \right) + {{\left( {{{\mathbf{z}}_i}(t + 1) - \bar{\varvec{\mu }} (t)} \right) }^T}\left( {{g_i}\left( {{\mathbf{x}_i}(t)} \right) - {g_i}\left( \mathbf{x} \right) } \right) } \right] \\ {}&- 2c(t) \cdot E\left[ {{N_i}{{\left( {{\mathbf{x}_i}(t)} \right) }^T} \cdot \left( {{\mathbf{x}_i}(t) - \mathbf{x}} \right) |{F_t}} \right] \end{aligned} \end{aligned}$$

(18)

We denote ${B(t)} = \max \left\{ {\dfrac{{{B_y}}}{{{\beta (t)}{r^2}}},\dfrac{{{B_y}}}{{{r^3}}}} \right\}$, $B(t) \le \dfrac{{{B_y}}}{{\beta (t) \cdot {r^2}}}$. From Lemma 3, $\left\| {{{\mathbf{z}}_i}(t)} \right\| \le B(t)$ and ${\left\| {{{\mathbf{s}}_i}(t + 1)} \right\| ^2} \le {\left( {{C_f} + {C_g}B(t) + {C_n}} \right) ^2}$. Since $\left\| {{N_i}(\mathbf{x}_i(t))} \right\| \le {C_n}$ and $E\left[ {{{N_i}(\mathbf{x}_i(t))}|{F_t}} \right] = 0$, from the law of total expectation, following inequality holds with probability one.

$$\begin{aligned} \begin{aligned}&\quad c(t) \cdot \sum \nolimits _{i = 1}^m {E\left[ {{L_i}\left( {{{\mathbf{x}}_i}(t),\bar{\varvec{\mu }} (t)} \right) - {L_i}\left( {{\mathbf{x}},\bar{\varvec{\mu }} (t)} \right) } \right] } \\ {}&\le \sum \nolimits _{i = 1}^m {\frac{1}{2}\left\{ {E\left[ {{{\left\| {{{\mathbf{x}}_i}(t) - \mathbf{x}} \right\| }^2}} \right] - E\left[ {{{\left\| {{\mathbf{x}_i}(t + 1) - \mathbf{x}} \right\| }^2}} \right] } \right\} } \\ {}&- 2{B_g} \cdot c(t) \cdot \sum \nolimits _{i = 1}^m {E\left[ {\left\| {{{\mathbf{z}}_i}(t + 1) - \bar{\varvec{\mu }} (t)} \right\| } \right] } + \frac{{{c^2}(t)}}{2}m \cdot {\left( {{C_f} + {C_g}B(t) + {C_n}} \right) ^2} \end{aligned} \end{aligned}$$

(19)

From (10) and Lemma 3.2, the following inequality can be obtained.

$$\begin{aligned}&\quad {\left\| {{{\varvec{\mu }} _i}(t + 1) - {w_i}(t + 1) \cdot {\mathbf{z}}} \right\| ^2} \le {\left\| {{{\hat{\varvec{\mu }} }_i}(t + 1) - {w_i}(t + 1) \cdot {\mathbf{z}}} \right\| ^2} + {c^2}(t)\nonumber \\&\quad \left\| \cdot \frac{{\hat{\mathbf{y}}_i}(t + 1)}{w_i^2(t + 1)} - \beta (t) \cdot \frac{{{\hat{\varvec{\mu }} }_i}(t + 1)}{{w_i}(t + 1)} \right\| ^2 \nonumber \\&+\frac{{2c(t) \cdot {{\hat{\mathbf{y}}}_i}{{(t + 1)}^T}}}{{w_i^2(t + 1)}} \cdot \left( {{{\hat{\varvec{\mu }} }_i}(t + 1) - {w_i}(t + 1) \cdot {\mathbf{z}}} \right) - \frac{{2c(t) \cdot \beta (t) \cdot \hat{\varvec{\mu }} _i^T(t + 1)}}{{{w_i}(t + 1)}}\nonumber \\&\quad \cdot \left( {{{\hat{\varvec{\mu }} }_i}(t + 1) - {w_i}(t + 1) \cdot {\mathbf{z}}} \right) \end{aligned}$$

(20)

From the convexity of norm and ${\hat{\varvec{\mu }} _i}(t + 1) = \sum \nolimits _{j \in N_i^{in}(t)} {{a_{ij}}(t) \cdot {{\varvec{\mu }} _j}(t)}$, we have

$$\begin{aligned} \sum \nolimits _{i = 1}^m {{{\left\| {{{\hat{\varvec{\mu }} }_i}(t + 1) - {w_i}(t + 1) \cdot {\mathbf{z}}} \right\| }^2}} \le {\sum \nolimits _{j = 1}^m {\left\| {{{\varvec{\mu }} _j}(t) - {w_j}(t) \cdot {\mathbf{z}}} \right\| } ^2} \end{aligned}$$

(21)

Since $B(t) \le \dfrac{{{B_y}}}{{\beta (t) \cdot {r^2}}}$, ${\left\| {\dfrac{{{{\hat{\mathbf{y}}}_i}(t + 1)}}{{w_i^2(t + 1)}} - \beta (t) \cdot \dfrac{{{{\hat{\varvec{\mu }} }_i}(t + 1)}}{{{w_i}(t + 1)}}} \right\| ^2} \le \dfrac{{4B_y^2}}{{{r^6}}}$.

We denote ${\bar{\mathbf{y}}}(t) = \dfrac{1}{m}\sum \nolimits _{i = 1}^m {{{\mathbf{y}}_i}(t)}$, the upper bound of the third term of (20) is

$$\begin{aligned} \begin{aligned}&\;\frac{{{{\hat{\mathbf{y}}}_i}{{(t + 1)}^T}}}{{w_i^2(t + 1)}} \cdot \left( {{{\hat{\varvec{\mu }} }_i}(t + 1) - {w_i}(t + 1) \cdot {\mathbf{z}}} \right) \\ {}&\le \left( {{B_t} + \left\| \mathbf{z} \right\| } \right) \left\| {\frac{{{{\hat{\mathbf{y}}}_i}(t + 1)}}{{{w_i}(t + 1)}} - \bar{\mathbf{y}}(t + 1)} \right\| + \left\| {{B_g}} \right\| \left\| {{{\mathbf{z}}_i}(t + 1) - \bar{\varvec{\mu }} (t)} \right\| \\&\quad + \frac{1}{m}\left( {{L_i}\left( {{x_i}(t),\bar{\varvec{\mu }} (t)} \right) - {L_i}\left( {{x_i}(t),{\mathbf{z}}} \right) } \right) \end{aligned} \end{aligned}$$

(22)

For ${\left\| {{{\hat{\varvec{\mu }} }_i}(t)} \right\| ^2} - w_i^2(t + 1){\left\| {\mathbf{z}} \right\| ^2} \le 2\hat{\varvec{\mu }} _i^T(t)\left( {{{\hat{\varvec{\mu }} }_i}(t) - {w_i}(t + 1) \cdot {\mathbf{z}}} \right)$, the last term of (20) satisfies

$$\begin{aligned} - \dfrac{{2c(t) \cdot \beta (t) \cdot \hat{\varvec{\mu }} _i^T(t + 1)}}{{{w_i}(t + 1)}} \cdot \left( {{{\hat{\varvec{\mu }} }_i}(t + 1) - {w_i}(t + 1) \cdot {\mathbf{z}}} \right) \le c(t)\beta (t)m \cdot {\left\| {\mathbf{z}} \right\| ^2} \end{aligned}$$

(23)

For ${{\mathbf{y}}_i}(0) = {g_i}\left( {{x_i}(0)} \right)$, $\left\| {\bar{\mathbf{y}}(t)} \right\| \le {B_g}$, we obtain

$$\begin{aligned}&c(t) \cdot \sum \nolimits _{i = 1}^m {E\left[ {{L_i}({{\mathbf{x}}_i}(t),{\mathbf{z}}) - {L_i}({{\mathbf{x}}_i}(t),\bar{\varvec{\mu }} (t))} \right] } \le \frac{m}{2}\left\{ {\sum \nolimits _{i = 1}^m {E\left[ {\left\| {{{\varvec{\mu }} _i}(t) - {w_i}(t) \cdot {\mathbf{z}}} \right\| } \right] } } \right. \nonumber \\ {}&\left. { - \sum \nolimits _{i = 1}^m {E\left[ {{{\left\| {{{\varvec{\mu }} _i}(t + 1) - {w_i}(t + 1) \cdot {\mathbf{z}}} \right\| }^2}} \right] } + {c^2}(t) \cdot m\frac{{4B_y^2}}{{{r^6}}}} \right\} \nonumber \\&+ m \cdot c(t) \cdot \sum \nolimits _{i = 1}^m {\left( {\left( {{B_t} + \left\| {\mathbf{z}} \right\| } \right) } \right. E\left[ {\left\| {\frac{{{{\hat{\mathbf{y}}}_i}(t + 1)}}{{{w_i}(t + 1)}} - \bar{\mathbf{y}}(t + 1)} \right\| } \right] } \nonumber \\&\left. { + \left\| {{B_g}} \right\| \cdot E\left[ {\left\| {{{\mathbf{z}}_i}(t + 1) - \bar{\varvec{\mu }} (t)} \right\| } \right] } \right) + \frac{{{m^3}}}{2}c(t) \cdot \beta (t){\left\| {\mathbf{z}} \right\| ^2} \end{aligned}$$

(24)

Let ${\mathbf{x}} = {{\mathbf{x}}^*}$, where ${{\mathbf{x}}^*}: = \mathop {\min }\limits _{{\mathbf{x}} \in X} \sum \nolimits _{i = 1}^m {{f_i}({\mathbf{x}})}$, $\sum \nolimits _{i = 1}^m {{g_i}({\mathbf{x}}_i^*)} \le 0$. From the definition of ${L_i}({\mathbf{x}},{\mathbf{z}})$, $\sum \nolimits _{i = 1}^m {E\left[ {{L_i}\left( {{{\mathbf{x}}_i}(t),{\mathbf{z}}} \right) - {L_i}\left( {{{\mathbf{x}}^*},\bar{\varvec{\mu }} (t)} \right) } \right] } \ge \sum \nolimits _{i = 1}^m E\left[ {f_i}\left( {{{\mathbf{x}}_i}(t)} \right) \right. \left. + {{\mathbf{z}}^T} \cdot {g_i}\left( {{{\mathbf{x}}_i}(t)} \right) - {f_i}\left( {{{\mathbf{x}}^*}} \right) \right]$ holds. Thus,

$$\begin{aligned} \begin{aligned}&c(t) \cdot \sum _{i=1}^{m} E\left[ f_{i}\left( \mathbf {x}_{i}(t)\right) -f_{i}\left( \mathbf {x}^{*}\right) +\mathbf {z}^{T} \cdot g_{i}\left( \mathbf {x}_{i}(t)\right) \right] \\ \le&c(t) \cdot \sum _{i=1}^{m} E\left[ L_{i}\left( \mathbf {x}_{i}(t), \mathbf {z}\right) -L_{i}\left( \mathbf {x}^{*}, \overline{\varvec{\mu }}(t)\right) \right] \\=&c(t) \cdot \sum _{i=1}^{m}\left\{ E\left[ L_{i}\left( \mathbf {x}_{i}(t), \mathbf {z}\right) -L_{i}\left( \mathbf {x}_{i}(t), \overline{\varvec{\mu }}(t)\right) \right] +E\left[ L_{i}\left( \mathbf {x}_{i}(t), \overline{\varvec{\mu }}(t)\right) -L_{i}(\mathbf {x}, \overline{\varvec{\mu }}(t))\right] \right\} \end{aligned} \end{aligned}$$

(25)

Let ${\mathbf{z}} = \mathbf{0}$ and sum up $c(t) \cdot \sum \nolimits _{i = 1}^m {E[{f_i}({{\mathbf{x}}_i}(t)) - {f_i}({{\mathbf{x}}^*})]}$ from $t=1$ to $t=T$.

$$\begin{aligned} \begin{aligned}&\quad \sum \nolimits _{t = 1}^T {\sum \nolimits _{i = 1}^m {c(t) \cdot E\left[ {{f_i}\left( {{{\mathbf{x}}_i}(t)} \right) - {f_i}\left( {{{\mathbf{x}}^*}} \right) } \right] } } \\[2.0mm]&\le \sum \nolimits _{t = 1}^T {\frac{m}{2}\left\{ {\sum \nolimits _{i = 1}^m {E\left[ {{{\left\| {{{\varvec{\mu }} _i}(t)} \right\| }^2}} \right] } - \sum \nolimits _{i = 1}^m {E\left[ {{{\left\| {{{\varvec{\mu }} _i}(t + 1)} \right\| }^2}} \right] } + {c^2}(t) \cdot \frac{{4B_y^2}}{{{r^6}}} \cdot m} \right\} } \\[2.0mm]&+ m\sum \nolimits _{t = 1}^T {\sum \nolimits _{i = 1}^m {c(t) \cdot \left( {{B_t} \cdot E\left[ {\left\| {\frac{{{{\hat{\mathbf{y}}}_i}(t + 1)}}{{{w_i}(t + 1)}} - \bar{\mathbf{y}}(t + 1)} \right\| } \right] + \left\| {{B_g}} \right\| \cdot E\left[ {\left\| {{{\mathbf{z}}_i}(t + 1) - \bar{\varvec{\mu }} (t)} \right\| } \right] } \right) } } \\[2.0mm]&+ \sum \nolimits _{t = 1}^T {\sum \nolimits _{i = 1}^m {\frac{1}{2} \cdot \left\{ {E\left[ {{{\left\| {{{\mathbf{x}}_i}(t) - {{\mathbf{x}}^*}} \right\| }^2}} \right] - E\left[ {{{\left\| {{{\mathbf{x}}_i}(t + 1) - {{\mathbf{x}}^*}} \right\| }^2}} \right] } \right\} } } \\[2.0mm]&\mathrm{{ + 2}}{B_g}\sum \nolimits _{t = 1}^T {\sum \nolimits _{i = 1}^m {c(t) \cdot \left\| {{{\mathbf{z}}_i}(t + 1) - \bar{\varvec{\mu }} (t)} \right\| } } + \sum \nolimits _{t = 1}^T {\frac{{{c^2}(t)}}{2} \cdot m \cdot {{\left( {{C_f} + {C_g}B(t) + {C_n}} \right) }^2}} \end{aligned} \end{aligned}$$

(26)

We use ${B_\mu }$ to denote the maximal value of $\left\| {{{\varvec{\mu }} _i}(1)} \right\|$, $\forall i \in v$, i.e. ${B_\mu } = \mathop {\max }\limits _{\forall i \in v} \left\{ {\max \left\{ {\dfrac{{{w_i}(2){B_y}}}{{a \cdot \beta (1) \cdot {r^2}}},\dfrac{{{w_i}(2){B_y}}}{{a \cdot {r^3}}}} \right\} } \right\}$. The first term of (26) satisfies

$$\begin{aligned} \quad \sum \nolimits _{t = 1}^T {\dfrac{m}{2}\left\{ {\sum \nolimits _{i = 1}^m {E\left[ {{{\left\| {{{\varvec{\mu }} _i}(t)} \right\| }^2}} \right] } - \sum \nolimits _{i = 1}^m {E\left[ {{{\left\| {{{\varvec{\mu }} _i}(t + 1)} \right\| }^2}} \right] } } \right\} } \le \dfrac{{{m^2}B_\mu ^2}}{2} \end{aligned}$$

(27)

To determine the upper bound of the third term of (26), let ${{\varvec{\varepsilon }} _{{y_i}}}(t + 1) = {g_i}\left( {{{\tilde{\mathbf{x}}}_i}(t + 1)} \right) - {g_i}\left( {{{\tilde{\mathbf{x}}}_i}(t)} \right)$, from Lemma 3.1, we obtain

$$\begin{aligned} \left\| {\frac{{{{\hat{\mathbf{y}}}_i}(t + 1)}}{{{w_i}(t + 1)}} - \bar{\mathbf{y}}(t + 1)} \right\| \le \frac{8}{\delta }\left( {{\lambda ^t}\sum \nolimits _{j = 1}^m {{{\left\| {{{\mathbf{y}}_j}(0)} \right\| }_1} + \sum \nolimits _{s = 1}^t {{\lambda ^{t - s}}\sum \nolimits _{j = 1}^m {{{\left\| {{{\varvec{\varepsilon }} _{{y_i}}}(s)} \right\| }_1}} } } } \right) \end{aligned}$$

(28)

where the term $\left\| {{{\varvec{\varepsilon }} _{{y_i}}}(t + 1)} \right\|$ satisfies $\left\| {{{\varvec{\varepsilon }}_{{y_i}}}(t + 1)} \right\| = \left\| {{g_i}\left( {{{\mathbf{x}}_i}(t + 1)} \right) - {g_i}\left( {{{\mathbf{x}}_i}(t)} \right) } \right\| \le {C_g} \cdot \left\| {{{\mathbf{x}}_i}(t + 1) - {{\mathbf{x}}_i}(t)} \right\|$. For $\left\| {{{\mathbf{x}}_i}(t + 1) - {{\mathbf{x}}_i}(t)} \right\| \le c(t) \cdot \left\| {{{\mathbf{s}}_i}(t + 1)} \right\| \le c(t) \cdot \left( {{C_f} + {C_g} \cdot {B_t} + {C_n}} \right)$ and $\left\| {{{\varvec{\varepsilon }} _{{y_i}}}(t + 1)} \right\| \le c(t) \cdot {C_g} \cdot \left( {{C_f} + \dfrac{{{C_g} \cdot {B_y}}}{{\beta (t){r^2}}} + {C_n}} \right)$, by Jensen inequality relation, ${\left\| {{{\varvec{\varepsilon }} _{{y_i}}}(t + 1)} \right\| _1} \le \sqrt{2} \cdot c(t) \cdot {C_g} \cdot \left( {{C_f} + \dfrac{{{C_g} \cdot {B_y}}}{{\beta (t){r^3}}} + {C_n}} \right)$ holds. Since $\left\{ {c(t)} \right\}$ and $\left\{ {\dfrac{{c(t)}}{{\beta (t)}}} \right\}$ are non-increasing sequences, $\sum \nolimits _{t = 1}^T {\sum \nolimits _{s = 1}^t {{\lambda ^{t - s}}} } \cdot {c^2}(s) \le \dfrac{1}{{1 - \lambda }}\sum \nolimits _{t = 1}^T {{c^2}(t)}$ and $\sum \nolimits _{t = 1}^T {\sum \nolimits _{s = 1}^t {{\lambda ^{t - s}}} } \cdot \dfrac{{{c^2}(s)}}{{\beta (s)}} \le \dfrac{1}{{1 - \lambda }}\sum \nolimits _{t = 1}^T {\dfrac{{{c^2}(t)}}{{\beta (t)}}}$. Then we obtain

$$\begin{aligned} \begin{aligned}&\;\sum \nolimits _{t = 1}^T {\sum \nolimits _{i = 1}^m {c(t) \cdot E\left[ {\left\| {\frac{{{{\hat{\mathbf{y}}}_i}(t + 1)}}{{{w_i}(t + 1)}} - \bar{\mathbf{y}}(t + 1)} \right\| } \right] } } \\ {}&\le \frac{{8\sqrt{2} }}{{\delta \left( {1 - \lambda } \right) }} \cdot m{C_g}\left( {{C_f} + {C_n}} \right) \sum \nolimits _{t = 1}^T {{c^2}(t)} + \frac{{8\sqrt{2} }}{{\delta \left( {1 - \lambda } \right) }} \cdot m\sum \nolimits _{t = 1}^T {\frac{{{c^2}(t)}}{{\beta (t)}}} \end{aligned} \end{aligned}$$

(29)

The forth term of (26) satisfies $\sum \nolimits _{t = 1}^T {\sum \nolimits _{i = 1}^m {\dfrac{1}{2}\{ {{\left\| {{{\mathbf{x}}_i}(t) - {\mathbf{x}}_i^*} \right\| }^2} - {{\left\| {{{\mathbf{x}}_i}(t + 1) - {\mathbf{x}}_i^*} \right\| }^2}\} } } \le 2B_X^2 \cdot m$.

To determine the upper bound of the fifth term of (26), we let ${{\varvec{\varepsilon }} _{{\mu _i}}}(t + 1) = {\left[ {{{\hat{\varvec{\mu }} }_i}(t + 1) + {c_t} \cdot {g_i}\left( {{{\mathbf{x}}_i}(t + 1)} \right) } \right] _ + } - {\hat{\varvec{\mu }} _i}(t + 1)$, then ${{\varvec{\mu }} _i}(t + 1) = {\hat{\varvec{\mu }} _i}(t + 1) + {{\varvec{\varepsilon }} _{{\mu _i}}}(t + 1)$. From Lemma 3.1, we obtain

$$\begin{aligned} \left\| {{{\mathbf{z}}_i}(t + 1) - \bar{\varvec{\mu }} (t)} \right\| \le \frac{8}{\delta }\left( {{\lambda ^t}\sum \nolimits _{j = 1}^m {{{\left\| {{{\varvec{\mu }} _j}(0)} \right\| }_1} + \sum \nolimits _{s = 1}^t {{\lambda ^{t - s}}\sum \nolimits _{j = 1}^m {{{\left\| {{{\varvec{\varepsilon }} _{{\mu _i}}}(s)} \right\| }_1}} } } } \right) \end{aligned}$$

(30)

Assume that ${{\varvec{\mu }} _i}(0) \in {\mathbb {R}^n_+ }$ holds for any user $i \in v$, and ${\hat{\varvec{\mu }} _i}(t)$ is a convex combination of ${{\varvec{\mu }} _i}(t)$ for any user $i \in v$, ${\hat{\varvec{\mu }} _i}(t) \in {\mathbb {R}^n_+ }$ holds. From Lemma 3.3, we have

$$\begin{aligned} \left\| {{{\varvec{\varepsilon }} _{{\mu _i}}}(t + 1)} \right\| \le \left\| {c(t) \cdot \left( {\dfrac{{{{\hat{\mathbf{y}}}_i}(t + 1)}}{{w_i^2(t + 1)}} - \beta (t)\dfrac{{{{\hat{\varvec{\mu }} }_i}(t + 1)}}{{{w_i}(t + 1)}}} \right) } \right\| \le \dfrac{{2 c(t){B_y}}}{{{r^3}}} \end{aligned}$$

(31)

Then ${\left\| {{{{ \varepsilon }}_{{\mu _i}}}(t + 1)} \right\| _1} \le \dfrac{{2\sqrt{2} c(t){B_y}}}{{{r^3}}}$. Obviously, there exists a scalar $\hat{\mu }$ such that ${\left\| {{{\varvec{\mu }} _i}(0)} \right\| _1} \le \hat{\mu }$ holds for any user $i \in v$. From (30), we obtain $\sum \nolimits _{t = 1}^T {\sum \nolimits _{i = 1}^m {c(t) \cdot \left\| {{{\mathbf{z}}_i}(t + 1) - \bar{\varvec{\mu }} (t)} \right\| } } \le \dfrac{{8m \cdot c(1) \cdot \hat{\mu }}}{{\delta (1 - \lambda )}} + \dfrac{{16\sqrt{2} m{B_y}}}{{\delta (1 - \lambda ){r^3}}}\sum \nolimits _{t = 1}^T {{c^2}(t)}$; thus, the fifth term of inequality (26) is bounded.

$$\begin{aligned} \begin{aligned}&B_{g} \cdot (2+m) \sum _{t=1}^{T} \sum _{i=1}^{m} c(t) \cdot E\left[ \left\| \mathbf {z}_{i}(t+1)-\overline{\varvec{\mu }}(t)\right\| \right] \\ \le&\frac{8 m(2+m) \cdot c(1) \hat{\mu } \cdot B_{g}}{\delta \cdot (1-\lambda )}+\frac{16 \sqrt{2} m(2+m) B_{y} \cdot B_{g}}{\delta \cdot (1-\lambda ) r^{3}} \sum _{t=1}^{T} c^{2}(t) \end{aligned} \end{aligned}$$

(32)

The upper bound of the final term of (26) can be obtained.

$$\begin{aligned} \begin{aligned}&\sum _{t=1}^{T} \frac{c^{2}(t)}{2} \cdot m \cdot \left( C_{f}+C_{g} \cdot B(t)+C_{n}\right) ^{2} \\ \le&m \cdot C_{f}^{2} \cdot \sum _{t=1}^{T} c^{2}(t)+m \cdot \frac{C_{g}^{2} B_{y}^{2}}{r^{4}} \cdot \sum _{t=1}^{T} \frac{c^{2}(t)}{\beta ^{2}(t)}+\sum _{t=1}^{T} m \cdot C_{n}^{2} \cdot c^{2}(t) \end{aligned} \end{aligned}$$

(33)

Then the following inequality relation holds.

$$\begin{aligned} \begin{aligned}&\quad \sum \nolimits _{t = 1}^T {\sum \nolimits _{i = 1}^m {c(t) \cdot E\left[ {{f_i}\left( {{{\mathbf{x}}_i}(t)} \right) - {f_i}\left( {{{\mathbf{x}}^*}} \right) } \right] } } \\ {}&\le \dfrac{{{m^2}{B_\mu }^2}}{2} + \sum \nolimits _{t = 1}^T {\left\{ {\left( {\frac{{4B_y^2 \cdot m}}{{{r^6}}} + m\left( {C_f^2 + C_n^2} \right) } \right) {c^2}(t)} \right\} } \\ {}&+ \frac{{8\sqrt{2} }}{{\delta \left( {1 - \lambda } \right) }} \cdot m{C_g}\left( {{C_f} + {C_n}} \right) \sum \nolimits _{t = 1}^T {{c^2}(t)} + \frac{{8\sqrt{2} }}{{\delta \left( {1 - \lambda } \right) }} \cdot m\sum \nolimits _{t = 1}^T {\frac{{{c^2}(t)}}{{\beta (t)}}} + 2B_T^2 \cdot m \\ {}&+ \frac{{8m(2 + m) \cdot c(1) \cdot \hat{\mu }\cdot {B_g}}}{{\delta \cdot \left( {1 - \lambda } \right) }} + \dfrac{{16\sqrt{2} m(2 + m){B_y} \cdot {B_g}}}{{\delta \cdot \left( {1 - \lambda } \right) {r^3}}}\sum \nolimits _{t = 1}^T {{c^2}(t)} + \dfrac{{mC_g^2B_y^2}}{{{r^4}}}\sum \nolimits _{t = 1}^T {\dfrac{{{c^2}(t)}}{{{\beta ^2}(t)}}} \end{aligned} \end{aligned}$$

(34)

Since ${\tilde{\mathbf{x}}_i}(t + 1)$ is a convex combination of past values of ${{\mathbf{x}}_i}(t)$, ${\tilde{\mathbf{x}}_i}(t + 1) \in {X_i}$ holds for any $t > 0$, $\sum \nolimits _{i = 1}^m {{f_i}\left( {{{\tilde{\mathbf{x}}}_i}(t + 1)} \right) } \le \sum \nolimits _{i = 1}^m {\dfrac{{\sum \nolimits _{r = 1}^{t+1} {c(r) \cdot {f_i}\left( {{{\mathbf{x}}_i}(r)} \right) } }}{{\sum \nolimits _{r = 0}^t {c(r)} }}}$. This implies

$$\begin{aligned} \sum \nolimits _{i = 1}^m {\left( {{f_i}\left( {{{\tilde{\mathbf{x}}}_i}(t + 1)} \right) - {f_i}\left( {{{{\mathbf{x}}_i}^*}} \right) } \right) } \le \sum \nolimits _{i = 1}^m {\frac{{\sum \nolimits _{r = 1}^{t+1} {c(r) \cdot \left( {{f_i}\left( {{{\mathbf{x}}_i}(r)} \right) - {f_i}\left( {{{{\mathbf{x}}_i}^*}} \right) } \right) } }}{{\sum \nolimits _{r = 1}^{t+1} {c(r)} }}} \end{aligned}$$

(35)

From (35), we obtain

$$\begin{aligned} \sum \nolimits _{i = 1}^m {E\left( {{f_i}\left( {{{\tilde{\mathbf{x}}}_i}(t + 1)} \right) - {f_i}\left( {{{{\mathbf{x}}_i}^*}} \right) } \right) } \le \sum \nolimits _{i = 1}^m {\frac{{E\left[ {\sum \nolimits _{r = 1}^{t+1} {c(r) \cdot \left( {{f_i}\left( {{{\mathbf{x}}_i}(r)} \right) - {f_i}\left( {{{{\mathbf{x}}_i}^*}} \right) } \right) } } \right] }}{{\sum \nolimits _{r = 1}^{t+1} {c(r)} }}} \end{aligned}$$

(36)

Since the step-size sequences $\left\{ {c(t)} \right\}$, $\left\{ {\beta (t)} \right\}$ satisfies $c(t) > 0$, $\mathop {\lim }\limits _{t \rightarrow 0} \sum \nolimits _{r = 0}^t {c(r)} = \infty$, $\mathop {\lim }\limits _{t \rightarrow \infty } \sum \nolimits _{r = 0}^t {{c^2}(r)} < \infty$, $\sum \nolimits _{t = 1}^\infty {\dfrac{{{c^2}(t)}}{{\beta (t)}}} < \infty$ and $\sum \nolimits _{t = 1}^\infty {\dfrac{{{c^2}(t)}}{{\beta ^2 (t)}}} < \infty$,

$\mathop {\lim }\limits _{T \rightarrow \infty } \sum \nolimits _{i = 1}^m {E\left[ {{f_i}\left( {{{\tilde{\mathbf{x}}}_i}(t + 1)} \right) - {f_i}\left( {{{{\mathbf{x}}_i}^*}} \right) } \right] = 0}$ holds, Theorem 3.1 is proved.