Power Allocation Algorithm for Heterogeneous Cellular Networks Based on Energy Harvesting

Abstract

Cellular network can use renewable energy through energy harvesting technology in green communication. In this paper, power allocation for heterogeneous cellular networks (HetNets) with energy harvesting is proposed to maximize the system energy efficiency. Considering the minimal transmit rate of the users and the battery capacity of the system, a low complexity power allocation algorithm based on fractional programming is proposed to maximize the energy efficiency of the system. Simulation results demonstrate the effectiveness of the proposed algorithm.

This work was partially supported by the National Natural Science Foundation of China (No. 11502039), Scientific and Technological Research Program of Chongqing Municipal Education Commission (No. KJ1600424), PhD research startup foundation of Chongqing University of Posts and Telecommunications (No. A2015-41), and the Science Research Project of Chongqing University of Posts and Telecommunications for Young Scholars (No. A2015-62).

Appendix A

Energy use and storage conditions C1 and C2 can be derived as follows. Let $S_{i,n}={\overline{\mathop L\nolimits _E } \left( {\mathop p\nolimits _{i,n}^R + \mathop p\nolimits _{i,n}^O } \right) }$ be the energy used for femtocell n in time slot i. Then, for time slot 1 to K, we have following inequality constraints.

$$\begin{array}{l} i = 1: \mathrm{{ }}\overline{\mathop L\nolimits _E } \left( {\mathop p\nolimits _{1,n}^R + \mathop p\nolimits _{1,n}^O } \right) \le \mathop {\mathop E\nolimits _{0,n} + E}\nolimits _{1,n} ,\forall n \\ i = 2: \mathrm{{ }}\overline{\mathop L\nolimits _E } \left( {\mathop p\nolimits _{2,n}^R + \mathop p\nolimits _{2,n}^O } \right) \le \mathop E\nolimits _{2,n} + \left( {\mathop E\nolimits _{0,n} + \mathop E\nolimits _{1,n} } \right) - \mathop S\nolimits _{1,n} ,\forall n \\ \mathrm{{ = }}\mathop {\mathop E\nolimits _{0,n} + \mathop E\nolimits _{1,n} + E}\nolimits _{2,n} - \mathop S\nolimits _{1,n} \\ \vdots \\ i = K: \mathrm{{ }}\overline{\mathop L\nolimits _E } \left( {\mathop p\nolimits _{K,n}^R + \mathop p\nolimits _{K,n}^O } \right) \\ \le \mathop E\nolimits _{K,n} + \left( {\mathop E\nolimits _{0,n} + \mathop E\nolimits _{1,n} + \cdots + \mathop E\nolimits _{K - 1,n} } \right) - \left( {\mathop S\nolimits _{K-1,n} + \cdots + \mathop S\nolimits _{1,n} } \right) \\ \mathrm{{ = }}\sum \nolimits _{i = 0}^K {\mathop E\nolimits _{i,n} } - \sum \nolimits _{i = 1}^{K - 1} {\mathop S\nolimits _{i,n} } \\ \end{array}$$

therefore $\mathrm{{ }}\mathop S\nolimits _{j,n} \mathrm{{ \le }}\sum \nolimits _{i = 0}^j {\mathop E\nolimits _{i,n} } - \sum \nolimits _{i = 1}^{j - 1} {\mathop S\nolimits _{i,n} }$ is hold for $j=1,\cdots , K$, move the item $\sum \nolimits _{i = 1}^{j- 1} {\mathop S\nolimits _{i,n} }$ from the right side to the left side, we have

$$\begin{aligned} \mathrm{{ }}\sum \nolimits _{i = {1,n}}^j {\mathop S\nolimits _{i,n} } \mathrm{{ \le }}\sum \nolimits _{i = 0}^j {\mathop E\nolimits _{i,n} } \end{aligned}$$

(11)

substitute $S_{i,n}={\overline{\mathop L\nolimits _E } \left( {\mathop p\nolimits _{i,n}^R + \mathop p\nolimits _{i,n}^O } \right) }$ into (11), we have

$$\sum \nolimits _{i = 1}^j {\overline{\mathop L\nolimits _E } \left( {\mathop p\nolimits _{i,n}^R + \mathop p\nolimits _{i,n}^O } \right) } \le \sum \nolimits _{i = 0}^j {\mathop E\nolimits _{i,n} ,\forall n} ;$$

Therefore C1 is hold. Meanwhile, because capacity limit of the battery, the remaining battery energy in each time slot for femtocell n cannot exceed the battery capacity, more than part of the energy will be discarded:

$$\sum \nolimits _{i = 0}^j {\mathop E\nolimits _{i,n} } - \sum \nolimits _{i = 1}^j {\mathop S\nolimits _i } \le \mathop E\nolimits _{\max ,n} $$

for each time slot $j=1,\cdots ,K$,then we have

$$ \sum \nolimits _{i = 0}^j {\mathop E\nolimits _{i,n} } - \sum \nolimits _{i = 1}^j {\overline{\mathop L\nolimits _E } \left( {\mathop p\nolimits _{i,n}^R + \mathop p\nolimits _{i,n}^O } \right) } \le \mathop E\nolimits _{\max ,n} $$

Therefore C2 is hold.