Cost and energy aware service provisioning for mobile client in cloud computing environment

Abstract

Currently, mobile devices are becoming the popular instrument for accessing the cloud environment. Mobile devices are resource and energy limited; they can rely on cloud computing resource to perform computationally intensive operations such as data mining and multimedia processing. This paper presents the cost and energy aware service provisioning scheme for mobile client in mobile cloud, which includes two-stage optimization process. In the first-stage optimization process, the mobile cloud user gives the unique optimal payment to the cloud provider under the cost and energy constraint and optimizes its benefit. In the second-stage optimization process, mobile cloud provider runs multiple VMs to execute the jobs for mobile cloud users; the cloud providers also need to maximize the revenue. The cost and energy aware service provisioning algorithm in mobile cloud is proposed. The proposed algorithm is involved with the cloud datacenter provider’s optimization and mobile cloud user’s optimization, which are conducted by two routines. In the simulation, the proposed cost and energy aware mobile cloud service provisioning algorithm is compared with other related algorithm.

Appendix: Lagrangian relaxation

Lagrangian relaxation is a relaxation technique which works by moving hard constraints into the objective so as to exact a penalty on the objective if they are not satisfied.

Mathematical description

Given an LP(linear programming) problem $x\in \;\mathbb {R}^n$ and $A\;\in \;\mathbb {R}^{m,n}$ of the following form:

$$\begin{aligned} \begin{array}{l} \max \;\;C^TX \\ s.t. \\ Ax\le b \\ \end{array} \end{aligned}$$

If we split the constraints in A such that $A_1 \;\in \;\mathbb {R}^{m_1 ,n}$, $A_2 \in \mathbb {R}^{m_2 ,n}$ and m1 + m2 = m, we may write the system:

$$\begin{aligned} \begin{array}{l} \max \;\;C^TX \\ s.t. \\ (1)\; A_1 x\le b_1 \\ (2)\; A_2 x\le b_2 \\ \end{array} \end{aligned}$$

We may introduce the constraint (2) into the objective:

$$\begin{aligned} \begin{array}{l} \max c^Tx+\lambda ^T( {b_2 -A_2 x}) \\ \quad \quad \hbox {s.t.} \\ \quad \quad (1)\; A_1 x\le b_1 \\ \end{array} \end{aligned}$$

If we let $\lambda =( {\lambda _1 ,\lambda _2 \ldots ,\lambda _{m_2}})$ be nonnegative weights, we get penalized if we violate the constraint (2), and we are also rewarded if we satisfy the constraint strictly. The above system is called the Lagrangian Relaxation of our original problem.

Of particular use is the property that for any fixed set of $\tilde{\lambda }$ values, the optimal result to the Lagrangian Relaxation problem will be no smaller than the optimal result to the original problem. Let $\hat{x}$ be the optimal solution to the original problem, and let $\overline{x}$ be the optimal solution to the Lagrangian Relaxation. We can then see that

$$\begin{aligned} c^T\hat{x}\le c^T\hat{x}+\tilde{\lambda }^T( {b_2 -A_2 \hat{x}})\le c^T\bar{x}+\tilde{\lambda }^T( {b_{2}-A_2 \bar{x}}) \end{aligned}$$

The first inequality is true because $\hat{x}$ is feasible in the original problem and the second inequality is true because $\overline{x}$ is the optimal solution to the Lagrangian Relaxation. This in turn allows us to address the original problem by instead exploring the partially dualized problem

$$\begin{aligned} \min \;P( \lambda ) \text{ s.t. }\;\lambda \ge 0 \end{aligned}$$

where we define $P(\lambda )$ as

$$\begin{aligned} \begin{array}{l} \max \;c^Tx+ \lambda ^T( {b_2 -A_2 x}) \\ \hbox {s.t.} \\ ( 1)\; A_1 x\;\le \,b_1 \\ \end{array} \end{aligned}$$

A Lagrangian Relaxation algorithm thus proceeds to explore the range of feasible $\lambda$ values while seeking to minimize the result returned by the inner P problem. Each value returned by P is a candidate upper bound to the problem, the smallest of which is kept as the best upper bound. If we additionally employ a heuristic, probably seeded by the $\overline{x}$ values returned by P, to find feasible solutions to the original problem, then we can iterate until the best upper bound and the cost of the best feasible solution converge to a desired tolerance.