Distributed algorithm for computation offloading in mobile edge computing considering user mobility and task randomness

Abstract

Recent years have witnessed substantial research efforts on computation offloading for mobile edge computing (MEC) systems. User mobility is an intrinsic trait of many MEC applications, which has posed significant challenges for realizing reliable computing. However, existing works studying this problem mainly focus on the movements of users while another high-dynamic behavior due to the randomness of computation task is largely ignored. To fill this gap, in this paper, we formulate the computation offloading decision problem in MEC system as a combinatorial optimization problem, and then we use Log-Sum-Exp function to approximate the optimal objective. Thereafter, we construct a Markov chain with steady-state distribution specifying to our problem in a distributed manner, such that the user mobility problem is transformed into the state transition problem. Moreover, this Markov chain is further extended to consider a dynamic scenario where the number of active users in the MEC system changes due to the random arrivals of new computation task or completions of old tasks. Numerical results show that our proposed computation offloading distributed algorithm can converge very fast to the optimal solution, and has a provable performance with a guaranteed loss bound.

Appendix

Appendix A: Proof of Theorem 2

Proof

First, it is not hard to see that the designed Markov chain is irreducible because all the states $f\in \mathscr {F}$ are able to reach each other within a finite number of transitions. Next, we will validate that the stationary distribution $p_f^{*}(\varvec{y})$ of this Markov chain is exactly given by Eq. (20).

We use $f\rightarrow f'$ to represent the event that the MEC system will enter into state $f'$ after leaving state f when the timer counts down to 0. Let $p_{f\rightarrow f'}$ denote the corresponding probability of this event. When leaving state f, the system will enter into any state $f' \in \varLambda _f$ with equal probability, thus $p_{f\rightarrow f'}$ can be obtained as

$$\begin{aligned} p_{f\rightarrow f'}=\frac{1}{|\varLambda _f|},\forall f\in \mathscr {F}, f'\in \varLambda _f \end{aligned}$$

(46)

where $\varLambda _f$ is the set of targeting states for f, and $|\varLambda _f|$ is given by

$$\begin{aligned} |\varLambda _f|=\sum _{u_i\in \mathscr {U}}(|\mathscr {M}_{u_i}|-1) \end{aligned}$$

(47)

The counting down rate for each user is given by

$$\begin{aligned} \iota _{u_i}=\frac{|\mathscr {M}_{u_i}|-1}{\exp (\alpha +\frac{1}{2}\beta (E_{i}^{f}-E_{i}^{f'}))} \end{aligned}$$

(48)

Combining Eqs. (46), (47) and (48), we can obtain the transmission rate from state f to state $f'$ as

$$\begin{aligned} q_{f,f'}&=\sum _{u_i\in \mathscr {U}}\iota _{u_i}\times p_{f\rightarrow f'}\nonumber \\&=\frac{\sum _{u_i\in \mathscr {U}}(|\mathscr {M}_{u_i}|-1)}{\exp (\alpha +\frac{1}{2}\beta (E_{i}^{f}-E_{i}^{f'}))}\times \frac{1}{\sum _{u_i\in \mathscr {U}}(|\mathscr {M}_{u_i}|-1)}\nonumber \\&=\frac{1}{\exp (\alpha )}\exp (-\frac{1}{2}\beta (E_{i}^{f}-E_{i}^{f'}))\nonumber \\&=\frac{1}{\exp (\alpha )}\exp (-\frac{1}{2}\beta (y_{f'}-y_{f})) \end{aligned}$$

(49)

We can show that the balance equation function $p_f^{*}(\varvec{r})q_{f,f'}=p_{f'}^{*}(\varvec{y})q_{f',f}$, $\forall f, f' \in \mathscr {F}$ holds. And according to Theorem 1.3 and Theorem 1.4 in Ref. [33], the designed Markov chain based on Algorithm 1 is time-reversible with stationary distribution given by Eq. (20).

This completes the proof. $\square$

Appendix B. Proof of Theorem 3

Proof

To attain the lower bound and upper bound for the mixing time $\tau _\mathrm{{mix}}(\varepsilon )$, we leverage the idea proposed in Ref. [27] using the spectral analysis method in Refs. [29, 34].

Since $y_\mathrm{{min}}\le y_{f}\le y_\mathrm{{max}}$ and $|\mathscr {F}|\le \prod _{i=1}^n|\mathscr {M}_{u_i}|$, we have

$$\begin{aligned} \sum _{f\in \mathscr {F}}\exp (\beta y_f)\le |\mathscr {F}|\exp (\beta y_\mathrm{{max}})\le \zeta \exp (\beta y_\mathrm{{max}}) \end{aligned}$$

(50)

Therefore, the minimum of stationary distribution can be obtained as

$$\begin{aligned} p_\mathrm{{min}}\triangleq \min _{f\in \mathscr {F}}p_f^{*}= & {} \min _{f\in \mathscr {F}}\frac{\exp (\beta y_f)}{\sum _{f\in \mathscr {F}}\exp (\beta y_f')}\nonumber \\\ge & {} \frac{\exp (\beta y_\mathrm{{min}})}{\zeta \exp (\beta y_\mathrm{{max}})}\nonumber \\= & {} \frac{1}{\zeta }\exp (-\beta (y_\mathrm{{max}}-y_\mathrm{{min}})) \end{aligned}$$

(51)

Let $Q=\{q_{f,f'}, \forall f,f' \in \mathscr {F}\}$ denote the transition rate matrix of the designed Markov chain, then we can further construct a discrete-time Markov chain $Z(\nu )$ with transition rate matrix as $\varPsi =I+\frac{Q}{\kappa }$, where I is the identity matrix, and $\vartheta$ is the uniform parameter which will be determined later.

For each feasible configuration $f\in \mathscr {F}$, it has at most $\delta =\sum _{u_i\in \mathscr {U}}(|\mathscr {M}_{u_i}|-1)$ targeting states, and the transition rate is bounded by

$$\begin{aligned} q_{f,f'}= & {} \frac{1}{\exp (\alpha )}\exp (-\frac{1}{2}\beta (y_{f'}-y_{f}))\nonumber \\\le & {} \exp (\frac{1}{2}\beta (y_\mathrm{{max}}-y_\mathrm{{min}})-\alpha ) \end{aligned}$$

(52)

Then we can get the value for the uniform parameter $\kappa$, that is

$$\begin{aligned} \sum _{f'\ne f}q_{f,f'}\le \delta \exp (\frac{1}{2}\beta (y_\mathrm{{max}}-y_\mathrm{{min}})-\alpha )=\kappa \end{aligned}$$

(53)

Let $\rho$ represent the second-largest eigenvalue of the transition rate matrix $\varPsi$, using the uniformization theorem and spectral gap inequality in Refs. [29, 34], we can get

$$\begin{aligned} \frac{\exp (-\kappa \tau (1-\rho ))}{2}\le \max _{f\in \mathscr {F}}\Vert H_t(f)-\varvec{P}^{*}\Vert _{TV} \le \frac{\exp (-\kappa \tau (1-\rho ))}{2\sqrt{p_\mathrm{{min}}}}. \end{aligned}$$

(54)

Therefore, we have

$$\begin{aligned} \frac{1}{\kappa (1-\rho )}\ln {\frac{1}{2\varepsilon }}\le \tau _\mathrm{{mix}}(\varepsilon ) \le \frac{1}{\kappa (1-\rho )}\left[ \frac{1}{2}\ln {\frac{1}{p_\mathrm{{min}}}}+\ln {\frac{1}{2\varepsilon }}\right] \end{aligned}$$

(55)

Let $\varPhi$ denote the Conductance for a time-reversible, ergodic Markov chain, which is defined as

$$\begin{aligned} \varPhi \triangleq \min _{B\in \mathscr {F},\varUpsilon _{B}\in (0,1/2]}\frac{J(B,\bar{B})}{\varUpsilon _B}, \end{aligned}$$

(56)

where B is any subset of $\mathscr {F}$, and

$$\begin{aligned} \varUpsilon _B=\sum _{f\in B}p_f^{*}, \end{aligned}$$

(57)

$$\begin{aligned} J(B,\bar{B})=\sum _{f\in B,f'\in \bar{B}}p_f^{*}\varPsi (f,f'). \end{aligned}$$

(58)

Then, according to the Cheeger’s inequality in Refs. [29, 34], we have

$$\begin{aligned} 1-2\varPhi \le \rho \le 1-\frac{1}{2}\varPhi ^2. \end{aligned}$$

(59)

Combining (55) and (59) results in

$$\begin{aligned} \frac{1}{2\kappa \varPhi }\ln {\frac{1}{2\varepsilon }}\le \tau _\mathrm{{mix}}(\varepsilon ) \le \frac{2}{\kappa \varPhi ^2}\left[ \frac{1}{2}\ln {\frac{1}{p_\mathrm{{min}}}}+\ln {\frac{1}{2\varepsilon }}\right] . \end{aligned}$$

(60)

To derive the lower bound of the mixing time $\tau _\mathrm{{mix}}(\varepsilon )$, we will first obtain the upper bound of $\varPhi$. For $\forall B\in \mathscr {F},\varUpsilon _{B}\in (0,1/2]$,

$$\begin{aligned} \varPhi= & {} \min _{B\in \mathscr {F},\varUpsilon _{B}\in (0,1/2]}\frac{J(B,\bar{B})}{\varUpsilon _{B}} \nonumber \\\le & {} \frac{1}{\varUpsilon _{B'}}\sum _{f\in B',f'\in \bar{B'}}p_f^{*}\varPsi (f,f')\nonumber \\= & {} \frac{1}{\varUpsilon _{B'}}\sum _{f\in B'}p_f^{*}\sum _{f'\in \bar{B'}}\varPsi (f,f')\nonumber \\\le & {} \frac{1}{\varUpsilon _{B'}}\sum _{f\in B'}p_f^{*}\nonumber \\= & {} 1. \end{aligned}$$

(61)

By combining Eqs. (53), (60) and (61), the lower bound of mixing time $\tau _\mathrm{{mix}}(\varepsilon )$ can be obtained as

$$\begin{aligned} \tau _\mathrm{{mix}}(\varepsilon )\ge \ln {\frac{1}{2\varepsilon }}\frac{1}{2\kappa } = \ln {\frac{1}{2\varepsilon }}\cdot \frac{\exp (\alpha -\frac{\beta }{2}(y_\mathrm{{max}}-y_\mathrm{{min}}))}{2\delta }. \end{aligned}$$

(62)

Next we calculate the upper bound of the mixing time $\tau _\mathrm{{mix}}(\varepsilon )$. For $\forall f, f'\in \mathscr {F}$, and $q_{f,f'}\ne 0$, we have

$$\begin{aligned} q_{f,f'}=\exp \left( \frac{1}{2}\beta (y_{f'}-y_{f})-\alpha \right) \ge \exp \left( \frac{1}{2}\beta (y_\mathrm{{min}}-y_\mathrm{{max}})-\alpha \right) . \end{aligned}$$

(63)

Combining Eqs. (56) and (63), the lower bound of $\varPhi$ can be derived as

$$\begin{aligned} \varPhi\ge & {} \min _{B\in \mathscr {F},\varUpsilon _{B}\in (0,1/2]}J(B,\bar{B})\nonumber \\\ge & {} \min _{f\ne f', \varPsi (f,f')>0}J(f,f')\nonumber \\= & {} \min _{f\ne f', \varPsi (f,f')>0}p_f^{*}\varPsi (f,f')\nonumber \\= & {} \min _{f\ne f', \varPsi (f,f')>0}p_f^{*}\frac{q_(f,f')}{\kappa }\nonumber \\\ge & {} \frac{p_\mathrm{{min}}}{\kappa }\exp \left( \frac{1}{2}\beta (y_\mathrm{{min}}-y_\mathrm{{max}})-\alpha \right) . \end{aligned}$$

(64)

Finally, the combination of Eqs. (51), (53), (59) and (64) would achieve

$$\begin{aligned} \tau _\mathrm{{mix}}(\varepsilon )\le & {} \frac{2}{\kappa \varPhi ^2}\left[ \ln {\frac{1}{2\varepsilon }}+\frac{1}{2}\ln {\frac{1}{p_\mathrm{{min}}}}\right] \nonumber \\\le & {} \frac{2\kappa \exp (2\alpha -\beta (y_\mathrm{{max}}-y_\mathrm{{min}}))}{p_\mathrm{{min}}^2}\left[ \ln {\frac{1}{2\varepsilon }}+\frac{1}{2}\ln {\frac{1}{p_\mathrm{{min}}}}\right] \nonumber \\\le & {} 2\delta \zeta ^2\exp \left[ \alpha +\frac{3}{2}\beta (y_\mathrm{{max}}-y_\mathrm{{min}})\right] \cdot \nonumber \\&\left[ \frac{1}{2}\ln {\zeta }+\ln {\frac{1}{2\varepsilon }}+\frac{\beta }{2}(y_\mathrm{{max}}-y_\mathrm{{min}})\right] . \end{aligned}$$

(65)

This completes the proof. $\square$

Appendix C. Proof of Theorem 5

Proof

We use $f_n\rightarrow f_{n+1}$ to denote the event that the system will enter state $f_{n+1}$ after leaving state $f_n$ when one user is activated due to the arrival of new computation task. Let $p_{f_n\rightarrow f_{n+1}}$ denote the probability of this event, which can be calculated as

$$\begin{aligned} p_{f_n\rightarrow f_{n+1}}=\frac{\exp (-\beta y_{f_{n+1}}^{n+1})}{\sum _{l'\in \mathscr {L},f'_{n+1}=f_n\cup {l'}}\exp \left( -\beta y_{f'_{n+1}}^{n+1}\right) }. \end{aligned}$$

(66)

As the new computation tasks arrive according to a Poisson process with parameter $\lambda$, the users will be activated with a rate equal to $\lambda$. Thus the system will leave state $f_n$ and enter into state $f_{n+1}$ at a rate of $\lambda$. In this case, the transition rate from state $f_n$ to $f_{n+1}$ can be obtained as

$$\begin{aligned} q_{f_n\rightarrow f_{n+1}}=\frac{\lambda \exp (-\beta y_{f_{n+1}}^{n+1})}{\sum _{l'\in \mathscr {L},f'_{n+1}=f_n\cup {l'}}\exp \left( -\beta y_{f'_{n+1}}^{n+1}\right) }. \end{aligned}$$

(67)

When an old computation task is completed, one of the active users will go into the rest, and the system will enter into state $f_n$ with probability

$$\begin{aligned} p_{f_{n+1}\rightarrow f_{n}}= & {} \frac{\exp (-\beta y_{f_{n}}^{n})}{\sum _{l'\in \mathscr {L},f'_{n+1}=f_n\cup {l'}}\exp (-\beta y_{f'_{n+1}}^{n+1})}\cdot \nonumber \\&\frac{\sum _{f_{n+1}\in \mathscr {F}_{n+1}}\exp \left( -\beta y_{f_{n+1}}^{n+1}\right) }{\sum _{f_n\in \mathscr {F}_n}\exp \left( -\beta y_{f_n}^n\right) }. \end{aligned}$$

(68)

There are totally $n+1$ active users in the MEC system, each user enters into the resting state when its task is completed with parameter $\mu$, thus the system leaves state $f_{n+1}$ with a rate equal to $(n+1)\mu$. Then the transition rate from state $f_{n+1}$ to $f_{n}$ can be calculated as

$$\begin{aligned} q_{ f_{n+1}\rightarrow f_{n}}= & {} \frac{(n+1)\mu \exp \left( -\beta y_{f_{n}}^{n}\right) }{\sum _{l'\in \mathscr {L},f'_{n+1}=f_n\cup {l'}}\exp \left( -\beta y_{f'_{n+1}}^{n+1}\right) }\cdot \nonumber \\&\frac{\sum _{f_{n+1}\in \mathscr {F}_{n+1}}\exp \left( -\beta y_{f_{n+1}}^{n+1}\right) }{\sum _{f_n\in \mathscr {F}_n}\exp \left( -\beta y_{f_n}^n\right) }. \end{aligned}$$

(69)

Consequently, the proposed transition rule realizes a time-reversible Markov chain, because the transition rate satisfies:

• First, we can verify that the constructed Markov chain is irreducible because all the configurations can reach each other within a finite number of transitions.

• Second, it is no hard to observe that the constructed Markov chain is time-reversible since the balance equation holds between any two adjacent states $f_n$ and $f_{n+1}$.

• Finally, according to the result in Ref. [27], our constructed Markov chain will converge and has a stationary distribution given by Eq. (43).

This completes the proof. $\square$