Collaboration among mobile agents for efficient energy allocation in mobile grid

Abstract

The use of mobile devices in grid environments may have two interaction aspects: devices are considered as users of grid resources or as grid resources providers. Due to the limitation constraints on energy and processing capacity of mobile devices, their integration into the Grid is difficult. In this paper, we investigate the cooperation among mobile devices to balance the energy consumption and computation workloads. Mobile devices can have different roles such as buyer devices and seller devices. In the mobile grid, the energies of mobile devices are uneven, energy-poor devices can exploit other devices with spare energy. Our model consists of two actors: A buyer device agent represents the benefits of mobile buyer device that intends to purchase energy from other devices. A seller device agent represents the profits of mobile seller device that is willing to sell spare energy to other devices. The objective of optimal energy allocation in mobile grid is to maximize the utility of the system without exceeding the energy capacity, expense budget and the deadline. A collaboration algorithm among mobile agents for efficient energy allocation is proposed. In the simulation, the performance evaluation of collaboration algorithm among mobile agents is conducted.

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.

1.1 Mathematical description

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

$$ \begin{array}{*{20}{c}} {{ \max }\,{C^T}X} \hfill \\{s.t.} \hfill \\{Ax \leqslant b} \hfill \\\end{array} $$

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:

$$ \begin{array}{*{20}{c}} {{ \max }\,{C^T}X} \hfill \\{s.t.} \hfill \\{\;\;({1})\,{A_1}x \leqslant {b_1}} \hfill \\{\;\;({2})\,{A_2}x \leqslant {b_2}} \hfill \\\end{array} $$

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

$$ \begin{array}{*{20}{c}} {{ \max }\,{c^T}x + {\lambda^T}\left( {{b_{{2}}} - {A_{{2}}}x} \right)} \hfill \\{s.t.} \hfill \\{\,\,({1})\,{A_1}x \leqslant {b_1}} \hfill \\\end{array} $$

If we let $\mathop{\lambda }\nolimits = (\mathop{\lambda }\nolimits_1, \mathop{\lambda }\nolimits_2 \cdots, \mathop{\lambda }\nolimits_{{\mathop{m}\nolimits_2 }} )$ 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 $\mathop{\lambda }\limits^{\sim }$ 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

$$ {c^T}\hat{x} \leqslant {c^T}\hat{x} + {\tilde{\lambda }^T}({b_2} - {A_2}\hat{x}) \leqslant {c^T}\bar{x} + {\tilde{\lambda }^T}({b_2} - {A_2}\bar{x}) $$

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

$$ { \min }\,P(\lambda )\,{\hbox{s}}.{\hbox{t}}.\,\lambda \geqslant 0 $$

where we define P(λ) as

$$ \begin{array}{*{20}{c}} {{ \max }\,{c^T}x + {\lambda^T}\left( {{b_{{2}}} - {A_{{2}}}x} \right)} \hfill \\{s.t.} \hfill \\{({1})\,{A_1}x \leqslant {b_1}} \hfill \\\end{array} $$

A Lagrangian Relaxation algorithm thus proceeds to explore the range of feasible λ 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.