CoTask: Correlation-aware task offloading in edge computing

Abstract

In this paper, we first study the problem of Correlation-aware Task computation offloading (CoTask) in mobile edge computing. Specifically, considering the correlation among multiple computation tasks, we study how to determine a joint task offloading decision and resource allocation strategy under the constraints of feasible offloading decisions and edge servers’ computation capacity, such that the overall task latency is minimized. Before addressing the challenging CoTask problem, we first investigate the case without considering task correlation, namely, NonCoTask. We prove that, NonCoTask with two combinatorial integer-continuous optimization variables (i.e., integral offloading decision variable and continuous resource allocation variable) can be equivalently solved by solving a 0-1 integer programming problem with respect to the offloading decision variable only, while the closed-form optimal resource allocation can be efficiently obtained under any given offloading decision. The 0-1 integer programming problem further falls into the realm of minimizing a supermodular set function with a matroid base constraint. We then propose a performance-guaranteed algorithm for NonCoTask. Next, we rigorously analyze the performance gap between CoTask and NonCoTask, and develop a correlation-aware offloading decision and resource allocation algorithm with theoretic performance guarantee for CoTask, via a correlation-aware exchange search based on the solution to NonCoTask. Extensive evaluation results show that our proposed algorithm outperforms several state-of-the-art algorithms as well as their enhanced counterparts with correlation in terms of overall task latency.

Appendix

1.1 Appendix A: Proof of Lemma 2

Proof

Recall that $\varOmega (\mathbf {Z})$ is the objective function in NonCoTask-$\mathbf {Z}$. Let $\mathbf {Z}^*:=\{z^*_{ij}\}$ be an optimal solution of NonCoTask-$\mathbf {Z}$ and $\tilde{\mathbf {F}}:=\{\tilde{f}_{ij}\}$ is obtained by (4). The feasibility of $(\mathbf {Z}^*, \tilde{\mathbf {F}})$ is not difficult to verify. In the following, we prove that for any feasible solution of NonCoTask, $(\mathbf {Z}, \mathbf {F})$, $\varPhi (\mathbf {Z}^*, \tilde{\mathbf {F}}) \le \varPhi (\mathbf {Z}, \mathbf {F})$ holds. According to (5) and the definition of $\varOmega (\mathbf {Z})$, we have $\varPhi (\mathbf {Z}, \tilde{\mathbf {F}})=\varOmega (\mathbf {Z})$ and $\varPhi (\mathbf {Z}^*, \tilde{\mathbf {F}})=\varOmega (\mathbf {Z}^*)$. By definition, $\varOmega (\mathbf {Z}^*) \le \varOmega (\mathbf {Z})$ holds for any feasible solution of NonCoTask-$\mathbf {Z}$ (i.e., $\mathbf {Z}$). Then, we obtain $\varPhi (\mathbf {Z}^*, \tilde{\mathbf {F}})=\varOmega (\mathbf {Z}^*)\le \varOmega (\mathbf {Z})=\varPhi (\mathbf {Z}, \tilde{\mathbf {F}})\le \varPhi (\mathbf {Z}, \mathbf {F}),$ where the last inequality works due to the optimality of $\tilde{\mathbf {F}}$. Accordingly, $\varPhi (\mathbf {Z}^*, \tilde{\mathbf {F}})\le \varPhi (\mathbf {Z}, \mathbf {F})$ is established for any $(\mathbf {Z}, \mathbf {F})$, which validates the optimality of $(\mathbf {Z}^*, \tilde{\mathbf {F}})$. Therefore, any optimal solution of NonCoTask-$\mathbf {Z}$ combined with $\mathbf {F}$ derived from (4) is an optimal solution of NonCoTask.

Next, we prove that for any optimal solution of NonCoTask, e.g., $(\mathbf {Z}^*, \mathbf {F}^*)$, $\mathbf {Z}^*$ is an optimal solution of NonCoTask-$\mathbf {Z}$ and $\mathbf {F}^*$ satisfies (4). The feasibility of $\mathbf {Z}^*$ for NonCoTask-$\mathbf {Z}$ is not hard to verify. For any feasible solution $\mathbf {Z}$ of NonCoTask-$\mathbf {Z}$, it naturally satisfies constraint (1) in NonCoTask, which implies that $(\mathbf {Z}, \mathbf {F}^*)$ is feasible for NonCoTask. Be definition, $\varPhi (\mathbf {Z}^*, \mathbf {F}^*) \le \varPhi (\mathbf {Z}, \mathbf {F}^*)$ holds. Since $\varPhi (\mathbf {Z}^*, \mathbf {F}^*)=\varOmega (\mathbf {Z}^*)$ and $\varPhi (\mathbf {Z}, \mathbf {F}^*)=\varOmega (\mathbf {Z})$ hold, we have $\varOmega (\mathbf {Z}^*)=\varPhi (\mathbf {Z}^*, \mathbf {F}^*)\le \varPhi (\mathbf {Z}, \mathbf {F}^*)=\varOmega (\mathbf {Z}),$ which leads to $\varOmega (\mathbf {Z}^*)\le \varOmega (\mathbf {Z})$. The optimality of $\mathbf {Z}^*$ is thus established.

Moreover, since (4) means that $\mathbf {F}^*$ is an optimal solution of NonCoTask under a given $\mathbf {Z}$, proving that $\mathbf {F}^*$ satisfies (4) is equivalent to proving that $\mathbf {F}^*$ is an optimal solution of NonCoTask under given $\mathbf {Z}^*$. For any $\mathbf {F}$ satisfying (2) and (3), $(\mathbf {Z}^*, \mathbf {F})$ is feasible for NonCoTask and $\varPhi (\mathbf {Z}^*, \mathbf {F}^*)\le \varPhi (\mathbf {Z}^*, \mathbf {F})$ thus holds. Then, $\mathbf {F}^*$ is an optimal solution of NonCoTask under given $\mathbf {Z}^*$. Lastly, based on the above proofs, it is easy to validate the equivalence of the optimal values of NonCoTask and NonCoTask-$\mathbf {Z}$. This lemma thus holds.

1.2 Appendix B: Proof of Lemma 4

Proof

First, the nonempty property of $\mathcal {I}$ is obvious due to $N\ge 2$ and $M\ge 2$. Second, if $A\subseteq B\in \mathcal {I}$, then $A\in \mathcal {I}$. If not, there exists at least two different elements $u_1, u_2\in A$ are adjacent. Since $A\subseteq B$ holds, $u_1, u_2\in B$, which contradicts $B\in \mathcal {I}$. Thus, the monotone property is proven. Last, suppose $A, B\in \mathcal {I}$ and $|A|<|B|$. If there does not exist an $u\in B$ such that $A\cup \{u\}\in \mathcal {I}$, then for any element $u\in B$, $A\cup \{u\}\notin \mathcal {I}$. Since $A\in \mathcal {I}$, every element in B is adjacent with some element in A. Due to $|A|<|B|$, there exist at least two elements $u_1, u_2\in B$ adjacent with the same element $w\in A$, which means that $u_1$ and $u_2$ are also adjacent with each other. This obviously contradicts with $B\in \mathcal {I}$. Therefore, $\mathfrak {M}:=\{U, \mathcal {I}\}$ is a matroid.

According to the construction of $\mathfrak {M}$, it is not hard to obtain that the size of the matroid $\mathfrak {M}$ is N, since there are N tasks. Denote $\mathcal {B}(\mathfrak {M})$ be the set of bases of $\mathfrak {M}$. Next, we prove that the constraint of NonCoTask-$\mathbf {Z}$ is equivalent to finding a set $S\in U$ such that $S\in \mathcal {B}(\mathfrak {M})$. In light of the relationship among U, $\mathcal {G}$, and $\mathcal {I}$, $\sum _{j=1}^M z_{ij}=1$ with $z_{ij}\in \{0,1\}$ means finding a vertex from the adjacent vertices associated with task i in $\mathcal {G}$. Since any two vertices associated with different tasks are nonadjacent in $\mathcal {G}$, the constraint of NonCoTask-$\mathbf {Z}$ actually requires N nonadjacent vertices in $\mathcal {G}$, which constitutes a base of $\mathfrak {M}$. The lemma is thus proved.

1.3 Appendix C: Proof of Lemma 5

Proof

We first prove that $\hat{\varPhi }(\mathbf {Z},\mathbf {F}) \le \varPhi (\mathbf {Z},\mathbf {F})$ holds by construction. Recall that in NonCoTask, the optimization of both offloading decision $\mathbf {Z}$ and computation resource allocation $\mathbf {F}$ for each task is unaware of the task correlation and all tasks are executed independently. We divide it into two cases, according to the task correlation under the offloading decision $\mathbf {Z}$. One case is that no correlation precisely exists for all tasks offloaded to any server under $\mathbf {Z}$. In this case, CoTask is equivalent to NonCoTask, which results in $\hat{\varPhi }(\mathbf {Z}, \mathbf {F}) = \varPhi (\mathbf {Z}, \mathbf {F})$.

The other case is that there exist at least two correlated tasks offloaded to one of the edge servers under $\mathbf {Z}$. Without loss of generality, suppose tasks $i_1$, $i_2\in \mathcal {N}$, whose workloads are $w_{i_1}$ and $w_{i_2}$, respectively, are correlated with a common workload $\varDelta w_k$ (group index $k=g(i_1)=g(i_2)$) and offloaded to server $j\in \mathcal {M}$. Assume that they are allocated $f_{i_1j}$ and $f_{i_2j}$ fractions of the processing power $C_j$ in NonCoTask, respectively. Then, the overall latency of these two tasks in NonCoTask is $t_{NC}(i_1, i_2) = \frac{w_{i_1}}{f_{i_1j}C_j} + \frac{w_{i_2}}{f_{i_2j}C_j}.$ We now construct a feasible computation resource allocation $\hat{F}'$ to CoTask under $\mathbf {Z}$, with a smaller overall latency than $t_{NC}(i_1, i_2)$. We consider the computation resource allocation between task $i_1$ and $i_2$ only, while the setting for other tasks remains unchanged. Hence, we only compare the overall latency of task $i_1$ and $i_2$ with and without task correlation. The overall latency of task $i_1$ and $i_2$ in CoTask is $t_{C}(i_1, i_2) = \frac{2\varDelta w_k}{(f_{i_1j}+f_{i_2j})C_j} + \frac{w_{i_1}-\varDelta w_k}{f_{i_1j}C_j} + \frac{w_{i_2}-\varDelta w_k}{f_{i_2j}C_j}.$ Since $\frac{\varDelta w_k}{(f_{i_1j}+f_{i_2j})C_j} < \frac{\varDelta w_k}{f_{i_1j}}$ and $\frac{\varDelta w_k}{(f_{i_1j}+f_{i_2j})C_j} < \frac{\varDelta w_k}{f_{i_2j}}$, we then have $t_{C}(i_1, i_2) < t_{NC}(i_1, i_2)$ holds, which results in $\hat{\varPhi }(\mathbf {Z}, \mathbf {F}) < \varPhi (\mathbf {Z}, \mathbf {F})$.

Next, we prove that $(1-\lambda )\varPhi (\mathbf {Z},\mathbf {F}) \le \hat{\varPhi }(\mathbf {Z},\mathbf {F})$ holds by

$$\begin{aligned} \hat{\varPhi }(\mathbf {Z},\mathbf {F})= & {} \sum \limits _{i=1}^N\sum \limits _{j=1}^M \left[ z_{ij} \frac{d_i}{B_j\log (1+\eta _{ij})}\sum \limits _{\hat{i}=1}^N z_{\hat{i}j} +\right. \nonumber \\&\quad \quad \quad \left. z_{ij} \left( \frac{\varDelta w_{g(i)}}{\sum _{i^{\prime }=1|g(i^{\prime })=g(i)}^N f_{i^{\prime }j} C_j} + \frac{w_{i} - \varDelta w_{g(i)}}{f_{ij} C_j}\right) \right] \nonumber \\\ge & {} \sum \limits _{i=1}^N\sum \limits _{j=1}^M \left[ z_{ij} \frac{d_i}{B_j\log (1+\eta _{ij})}\sum \limits _{\hat{i}=1}^N z_{\hat{i}j} + z_{ij} \frac{w_{i} - \varDelta w_{g(i)}}{f_{ij} C_j}\right] \end{aligned}$$

(13)

$$\begin{aligned}\ge & {} \sum \limits _{i=1}^N\sum \limits _{j=1}^M \left[ z_{ij} \frac{d_i}{B_j\log (1+\eta _{ij})}\sum \limits _{\hat{i}=1}^N z_{\hat{i}j} + z_{ij} \frac{(1-\lambda )w_i}{f_{ij} C_j}\right] \\= & {} (1-\lambda )\varPhi (\mathbf {Z},\mathbf {F}) + \lambda \sum \limits _{i=1}^N\sum \limits _{j=1}^M z_{ij}\frac{d_i}{B_j\log (1+\eta _{ij})}\sum \limits _{\hat{i}=1}^N z_{\hat{i}j}\nonumber \\\ge & {} (1-\lambda )\varPhi (\mathbf {Z},\mathbf {F}), \end{aligned}$$

(14)

where (12), (13), and (14) are due to $\varDelta w_{g(i)}\ge 0$, $\varDelta w_{g(i)} \le \lambda w_i$, and $\lambda z_{ij}\frac{d_i}{B_j\log (1+\eta _{ij})}\sum _{\hat{i}=1}^N z_{\hat{i}j} \ge 0$, respectively.

1.4 Appendix D: Proof of Theorem 3

Proof

Let $\mathbf {Z}^0$ and $\mathbf {F}^0$ be the offloading decision and computation resource allocation derived in step 1 and step 2 in Algorithm 2, respectively. To avoid repetition, the feasibility of $(\mathbf {Z}^0, \mathbf {F}^0)$, $(\hat{\mathbf {Z}}, \hat{\mathbf {F}})$ for both CoTask and NonCoTask is not described again. Based on the previous analysis, we have

$$\begin{aligned} \hat{\varPhi }(\mathbf {Z}^0, \mathbf {F}^0)&\le \varPhi (\mathbf {Z}^0, \mathbf {F}^0)\end{aligned}$$

(15)

$$\begin{aligned}&\le \left( \frac{1}{6}-\varepsilon \right) \varPhi (\mathbf {Z}^*, \mathbf {F}^*) + \left( \frac{5}{6}+\varepsilon \right) H_{\max }\end{aligned}$$

(16)

$$\begin{aligned}&\le \left( \frac{1}{6}-\varepsilon \right) \frac{1}{1-\lambda }\hat{\varPhi }(\hat{\mathbf {Z}}^*, \hat{\mathbf {F}}^*) + \left( \frac{5}{6}+\varepsilon \right) H_{\max }, \end{aligned}$$

(17)

where (15), (16), and (17) are due to Lemma 5, Theorem 2, and Lemma 6, respectively. Moreover, since in Algorithm 2, we always search a better solution of CoTask than $(\mathbf {Z}^0, \mathbf {F}^0)$, we have $\hat{\varPhi }(\hat{\mathbf {Z}}, \hat{\mathbf {F}})\le \hat{\varPhi }(\mathbf {Z}^0, \mathbf {F}^0)$, which results in the performance bound as in the theorem. As to the time complexity, compared with Algorithm 1, steps 4-19 need extra $O(N^2M^2)$ operations. According to Theorem 2, the time complexity of Algorithm 2 is $O\left( \frac{1}{\varepsilon }N^4M^4\log (NM-N)+NM+N^2M^2\right) =O\left( \frac{1}{\varepsilon }N^4M^4\log (NM-N)+N^2M^2\right)$.