Analyzing the Convergence of Federated Learning with Biased Client Participation

Abstract

Federated Learning (FL) is a promising decentralized machine learning framework that enables a massive number of clients (e.g., smartphones) to collaboratively train a global model over the Internet without sacrificing their privacy. Though FL’s efficacy in non-convex problems is proven, its convergence amidst biased client participation lacks theoretical study. In this paper, we analyze the convergence of FedAvg on non-convex problems, which is the most renowned FL algorithm. We assume even data distribution but non-IID among clients, and elucidate the convergence rate of FedAvg in situations characterized by biased client participation. Our analysis reveals that biased client participation can significantly reduce the precision of the FL model. We validate this through trace-driven experiments, demonstrating that unbiased client participation results in 11% to 50% higher test accuracy compared to extremely biased client participation.

Appendiex

Within this section, we will provide proofs for Lemma 1 and Lemma 2.

1.1 Proof of Lemma 1

For any $t \ge 0$, there exits a $t - t_0 \le E$, and $\omega _k^{t_0} = \omega ^{t_0}$ for all $k=1,2,...,N$. Similar to previous work [17], we have

$$\begin{aligned} \, \begin{aligned} \mathbb {E} {\left\| \omega _k^t-\omega ^t \right\| }^2 &= \mathbb {E} {\left\| (\omega _k^t - \omega ^{t_0}) -(\omega ^t - \omega ^{t_0}) \right\| }^2\\ &\le \mathbb {E} {\left\| \omega _k^t - \omega ^{t_0} \right\| }^2\\ &\le \eta _t^2 E \sum _{i=t_0}^t \mathbb {E} \left\| \nabla F_k(\omega _k^i,\xi _k^i)\right\| ^2\\ &\le \eta _t^2 E^2 G^2. \end{aligned} \end{aligned}$$

1.2 Proof of Lemma 2

Since $n_k = \frac{n}{N}$ for all $n_k$, $\sum _{k=1}^{N} M_k^t = mt$, we can derive that

$$\begin{aligned} \, \begin{aligned} \sum _{k\in S_t} p_k^2 = \sum _{k\in S_t} \left( \frac{\frac{n}{N} M_k^t}{ \sum _{k=1}^{N} \frac{n}{N} M_k^t} \right) ^2 = \sum _{k\in S_t} {\left( \frac{M_k^t}{mt} \right) }^2. \end{aligned} \end{aligned}$$

(22)

Utilizing the $\rho$-smoothness property of $F(\omega )$, the subsequent inequality can be derived:

$$\begin{aligned} \, \begin{aligned} \mathbb {E} F(\omega ^{t+1}) \le \mathbb {E} F(\omega ^t) + \mathbb {E} \left\langle \nabla F(\omega ^t), \omega ^{t+1}-\omega ^t \right\rangle +\frac{\rho }{2} \mathbb {E} {\left\| \omega ^{t+1}-\omega ^t \right\| }^2. \end{aligned} \end{aligned}$$

(23)

By applying the fact: $\mathbb {E} \left\| x \right\| ^2 = \mathbb {E} \left[ \left\| x - \mathbb {E} x \right\| ^2 \right] + \left\| \mathbb {E} x \right\| ^2$, we can obtain

$$\begin{aligned} \, \begin{aligned} \mathbb {E}& {\left\| \omega ^{t+1}-\omega ^t \right\| }^2 = \eta _t^2 \mathbb {E} {\left\| \frac{N}{m} \sum _{k\in S_t} p_k \nabla F_k(\omega _k^t,\xi _k^t) \right\| }^2\\ &= \eta _t^2 \mathbb {E} {\left\| \frac{N}{m} \sum _{k\in S_t} p_k \left[ \nabla F_k(\omega _k^t,\xi _k^t) - \nabla F_k(\omega _k^t) \right] \right\| }^2 + \eta _t^2 \mathbb {E} {\left\| \frac{N}{m} \sum _{k\in S_t} p_k \nabla F_k(\omega _k^t) \right\| }^2. \end{aligned} \end{aligned}$$

(24)

Since each client works in parallel and independently and according to Assumption 3, we have

$$\begin{aligned} \, \begin{aligned} \mathbb {E} {\left\| \omega ^{t+1}-\omega ^t \right\| }^2 &= \frac{\eta _t^2 N^2}{m^2} \sum _{k\in S_t} p_k^2 \mathbb {E} {\left\| \nabla F_k(\omega _k^t,\xi _k^t) - \nabla F_k(\omega _k^t) \right\| }^2\\ &\quad + \eta _t^2 \mathbb {E} {\left\| \frac{N}{m} \sum _{k\in S_t} p_k \nabla F_k(\omega _k^t) \right\| }^2\\ &\le \frac{\eta _t^2 N^2 \delta ^2}{m^2} \sum _{k\in S_t} p_k^2 + \eta _t^2 \mathbb {E} {\left\| \frac{N}{m} \sum _{k\in S_t} p_k \nabla F_k(\omega _k^t) \right\| }^2. \end{aligned} \end{aligned}$$

(25)

We further note that

$$\begin{aligned} \, \begin{aligned} \mathbb {E}& \left\langle \nabla F(\omega ^t), \omega ^{t+1}-\omega ^t \right\rangle = - \eta _t \mathbb {E} \left\langle \nabla F(\omega ^t), \frac{N}{m} \sum _{k\in S_t} p_k \nabla F_k(\omega ^t,\xi _k^t) \right\rangle \\ & = - \eta _t \mathbb {E} \left[ \mathbb {E} \left[ \left\langle \nabla F(\omega ^t), \frac{N}{m} \sum _{k\in S_t} p_k \nabla F_k(\omega ^t,\xi _k^t) \right\rangle \Bigg | \xi ^t \right] \right] \\ & = - \eta _t \mathbb {E} \left\langle \nabla F(\omega ^t), \mathbb {E} \left[ \frac{N}{m} \sum _{k\in S_t} p_k \nabla F_k(\omega ^t,\xi _k^t) \Big | \xi ^t \right] \right\rangle \\ & = \begin{array}{c} \underbrace{ - \eta _t \mathbb {E} \left\langle \nabla F(\omega ^t), \frac{N}{m} \sum _{k\in S_t} p_k \nabla F_k(\omega ^t) \right\rangle }\\ A1 \end{array}. \end{aligned} \end{aligned}$$

(26)

Firstly, for bound A1, we can obtain

(27)

Secondly, $\omega ^{t+1} = \frac{N}{m} \sum _{k \in s_t} p_k \omega _{k}^{t+1}$ according to Eq. (4), therefore we can obtain $\nabla F(\omega ^{t+1}) = \frac{N}{m} \sum _{k \in s_{t+1}} p_k \nabla F_k(\omega ^{t+1})$�[24]. For bound A2, we can obtain

$$\begin{aligned} \, \begin{aligned} A2 &= \frac{\eta _t}{2} \mathbb {E} \left\| \frac{N}{m} \sum _{k\in S_t} p_k \nabla F_k(\omega _k^t) - \frac{N}{m} \sum _{k\in S_t} p_k \nabla F_k(\omega ^t) \right\| ^2\\ &= \frac{\eta _t}{2} \mathbb {E} \left\| \frac{N}{m} \sum _{k\in S_t} p_k \left[ \nabla F_k(\omega _k^t) - \nabla F_k(\omega ^t) \right] \right\| ^2. \end{aligned} \end{aligned}$$

(28)

According to the Cauchy-Buniakowsky-Schwarz inequality, we have

$$\begin{aligned} \, \begin{aligned} A2 &\le \frac{\eta _t N^2}{2m^2} \sum _{k\in S_t} p_k^2 \sum _{k\in S_t} \mathbb {E} \left\| \nabla F_k(\omega _k^t) - \nabla F_k(\omega ^t) \right\| ^2. \end{aligned} \end{aligned}$$

(29)

By using Assumption 1, we can obtain

$$\begin{aligned} \, \begin{aligned} A2 & \le \frac{\eta _t \rho ^2 N^2}{2m^2} \sum _{k\in S_t} p_k^2 \sum _{k\in S_t} \mathbb {E} \left\| \omega _k^t - \omega ^t \right\| ^2. \end{aligned} \end{aligned}$$

(30)

By using Lemma 1, we can derive the bound of A2 as

$$\begin{aligned} \, \begin{aligned} A2 & \le \frac{\eta _t^3 \rho ^2 N^2 E^2 G^2}{2m} \sum _{k\in S_t} p_k^2. \end{aligned} \end{aligned}$$

(31)

Upon substituting Eq. (31) into Eq. (27), we arrive at the upper bound for A1 as follows:

$$\begin{aligned} \, \begin{aligned} A1 \le - \frac{\eta _t}{2} \mathbb {E} {\left\| \nabla F(\omega ^t) \right\| }^2 &- \frac{\eta _t}{2} \mathbb {E} \left\| \frac{N}{m} \sum _{k\in S_t} p_k \nabla F_k(\omega _k^t) \right\| ^2\\ &\quad + \frac{\eta _t^3 \rho ^2 N^2 E^2 G^2}{2m} \sum _{k\in S_t} p_k^2. \end{aligned} \end{aligned}$$

(32)

By combining the results of Eq. (25), Eq. (26) and Eq. (32), we can obtain

$$\begin{aligned} \, \begin{aligned} F(\omega ^{t+1}) &\le F(\omega ^t) - \frac{\eta _t}{2} \mathbb {E} {\left\| \nabla F(\omega ^t) \right\| }^2\\ &\quad - \frac{\eta _t - \eta _t^2 \rho }{2} \mathbb {E} \left\| \frac{N}{m} \sum _{k\in S_t} p_k \nabla F_k(\omega _k^t) \right\| ^2\\ &\quad + \frac{\eta _t^3 \rho ^2N^2 E^2 G^2}{2m} \sum _{k\in S_t} p_k^2 + \frac{\eta _t^2 \rho N^2 \delta ^2}{2m^2} \sum _{k\in S_t} p_k^2. \end{aligned} \end{aligned}$$

(33)

The conclusion that $0 \le \eta _t \le \frac{1}{\rho }$ can be obtained from the setting $\eta _t = \frac{1}{\rho } \sqrt{\frac{1}{T}}$, we can obtain

(34)

By dividing both the left side and the right side by $\frac{\eta _t}{2}$, we have

(35)

According to Eq. (13), we have $\sum _{k\in S_t} p_k^2 = \sum _{k\in S_t} {\left( \frac{M_k^t}{mt} \right) }^2$. As $\eta _t = \frac{1}{\rho } \sqrt{\frac{1}{T}}$, we can sum Eq. (35) from $t=0$ to $T-1$ and obtain

where $\omega ^*$ is the optimal solution.