Robust Power Control for 5G Small Cell Networks with Sleep Strategy

Abstract

In this paper, the resource scheduling and power control strategy is studied for a two-tier 5G network system, which is comprised of a central macrocell and several small cells. Because cellular deployment is ultra dense in the 5G network, a sleep strategy for small cells is proposed to reduce energy consumption caused by the cellular overlapping coverage. Specifically, the imperfect channel state information (CSI) is considered due to the complex and dynamic communication environment in practice. This work attempts to maximize the resource utilization and enhance the robustness of the power control strategy with imperfect CSI. An optimization problem with probability constraints is formulated, and the high rate requirements of 5G network users are included. A novel method is provided to convert probability constraints into deterministic ones. An iterative algorithm of power control to enhance the network efficiency is proposed further. Finally, numerical results are given to illustrate the effectiveness of the joint resource scheduling scheme.

Appendices

Appendix 1

In this part, the uncertain constraint is transformed into the certainty expression. Consider the transmission rate constraint,

$$\begin{aligned} \text {Pr}\{R_{i,j} \le R_0 \} \le \varepsilon _{i,j}. \end{aligned}$$

(18)

Namely,

$$\begin{aligned} \text {Pr}\{W \log _2\left( 1+\frac{p_{i,j}G_{i,j}{{\overline{g}}}_{i,j}}{I_{i,j}}\right) \le R_0 \} \le \varepsilon _{i,j}. \end{aligned}$$

(19)

where ${{\overline{g}}}_{i,j}$ is the average gain. $I_{i,j}$ is the total interference. The value of $I_{i,j}$ is

$$\begin{aligned} I_{i,j}=\sum _{(m,n)\ne (i,j)}p_{m,n}{g}_{m,n}+N_0 \end{aligned}$$

(20)

$G_{i,j}$ is assumed to be exponential distribution with unit mean.

According to (6), (19) can be transformed into

$$\begin{aligned} \begin{array}{lll} &{} \text {Pr} \{ G_{i,j}<\frac{(2^{\frac{R_0}{W}}-1)(I_{i,j}+N_0)}{ {{\overline{g}}}_{i,j}p_{i,j}}\}, \\ &{}\quad =\int _{0}^{\frac{(2^{\frac{R_0}{W}}-1)(I_{i,j}+N_0)}{ {{\overline{g}}}_{i,j}p_{i,j}}}e^{-x}dx,\\ &{}\quad =1-e^{-\frac{(2^{\frac{R_0}{W}}-1)(I_{i,j}+N_0)}{ {{\overline{g}}}_{i,j}p_{i,j}}}. \end{array} \end{aligned}$$

(21)

The probabilistic constraint of optimal problem can be rewritten as follows

$$\begin{aligned} \begin{array}{lll} &{}\text {Pr}\{R_{i,j} \le R_0 \} \le \varepsilon _{i,j},\\ &{}\quad \Rightarrow 1-e^{-\frac{\left( 2^{\frac{R_0}{W}}-1\right) (I_{i,j}+N_0)}{{{\overline{g}}}_{i,j}p_{i,j}}}\le \varepsilon _{i,j},\\ &{}\quad \Rightarrow \ln (1-\varepsilon _{i,j})\le -\frac{\left( 2^{\frac{R_0}{W}}-1\right) (I_{i,j}+N_0)}{ {{\overline{g}}}_{i,j}p_{i,j}},\\ &{}\quad \Rightarrow {{\overline{g}}}_{i,j}p_{i,j} \ln (1-\varepsilon _{i,j})+(2^{\frac{R_0}{W}}-1)\left( \sum _{(m,n)\ne (i,j)}p_{m,n}{g}_{m,n}+N_0\right) \le 0 \end{array} \end{aligned}$$

(22)

The proof is completed.

Appendix 2

In this section, the detailed process of obtaining optimal power is given.

$$\begin{aligned} L(p,\lambda )\!= & {} \!\sum \limits _{i,j=1}^{n} (R_{i,j}\!-\!c*p_{i,j})\!\!-\!\!\sum \limits _{k=i,j}\lambda _k \left( {{\overline{g}}}_{i,j}p_{i,j} \ln (1\!-\!\varepsilon _{i,j})\!\!\nonumber \right. \\&\left. +\left( 2^{\frac{R_0}{W}}\!-\!1\right) \left( \sum _{(m,n)\ne (i,j)}p_{m,n}{g}_{m,n}\!+\!N_0\right) \right) . \end{aligned}$$

(23)

Namely,

$$\begin{aligned} \begin{array}{cl} \ L(p,\lambda )= &{}\sum \limits _{i,j=1}^{n} \left( \alpha W \log _2\left( \frac{{g}_{i,j}e^{{{\widetilde{p}}}_{i,j}}}{\sum _{(m,n)\ne (i,j)}{p}_{m,n}{g}_{m,n}^j+N_0}\right) +\beta -c*e^{{{\widetilde{p}}}_{i,j}}\right) \\ &{}-\sum \limits _{k=i,j}\lambda _k ({{\overline{g}}}_{i,j}e^{{{\widetilde{p}}}_{i,j}}) \ln (1-\varepsilon _{i,j})+(2^{\frac{R_0}{W}}-1)\left( \sum _{(m,n)\ne (i,j)}p_{m,n}{g}_{m,n}+N_0\right) ).\\ \end{array} \end{aligned}$$

(24)

In order to make it convenient to obtain the derivative, we separate the (i, j)th term including ${{\widetilde{p}}}_{i,j}$ and other terms. The Lagrangian function is rewritten as follows.

$$\begin{aligned} \begin{array}{cl} \ L(p,\lambda )=&{}\alpha W\log _2\left( 1+\frac{{g}_{i,j}e^{{{\widetilde{p}}}_{i,j}}}{\sum _{(m,n)\ne (i,j)}p_{m,n}{g}_{m,n}^j+N_0}\right) +\beta -c*e^{{{\widetilde{p}}}_{i,j}}\\ &{} +\sum \limits _{m,n} \left( \alpha W \log _2\left( \frac{{g}_{m,n}{p}_{m,n}}{e^{{{\widetilde{p}}}_{i,j}}{g}_{i,j}+\sum _{(m,n)\ne (i,j)}{p}_{a,b}{g}_{a,b}+N_0}\right) +\beta -c*{p}_{m,n}\right) \\ &{} -\lambda _k ({{\overline{g}}}_{i,j}e^{{{\widetilde{p}}}_{i,j}} \ln (1-\varepsilon _{i,j})+(2^{\frac{R_0}{W}}-1)({I}_{i,j}+N_0))\\ &{}-\sum \limits _{k\ne i,j}\lambda _k ({{\overline{g}}}_{m,n}p_{m,n} \ln (1-\varepsilon _{i,j})+(2^{\frac{R_0}{W}}-1)({I}_{m,n}+e^{{{\widetilde{p}}}_{i,j}}{g}_{i,j}+N_0)) \end{array} \end{aligned}$$

(25)

After simplification, the derivative of (25) about ${{{\widetilde{p}}}_{ij}}$ is given as follows

$$\begin{aligned} L'(p,\lambda )\!= & {} \!\frac{\alpha }{\ln 2}\!-\!e^{{{\widetilde{p}}}_{i,j}}\left( c\!+\!\sum \limits _{m,n}(\frac{\alpha }{\ln 2} \frac{{g}_{i,j}}{{p}_{m,n}{g}_{m,n}} SIR(e^{{{\widetilde{p}}}_{i,j}}))\!\!\nonumber \right. \\&\left. +\lambda _k {{\overline{g}}}_{i,j}\ln (1\!-\!\varepsilon _{i,j})\!\!+\!\!\sum \limits _{k\ne i,j}\lambda _k({g}_{i,j}(2^{\frac{R_0}{W}}\!\!-\!\!1))\right) \end{aligned}$$

(26)

where

$$\begin{aligned} \ SIR(e^{{{\widetilde{p}}}_{i,j}})=\frac{{g}_{m,n}{p}_{m,n}}{({e^{{{\widetilde{p}}}_{i,j}}{g}_{i,j} +\sum _{(m,n)\ne (i,j)}{p}_{a,b}{g}_{a,b}+N_0})^2}. \end{aligned}$$

(27)

Let (26) be zero, we have

$$\begin{aligned}&\frac{\alpha }{\ln 2}\!-\!e^{{{\widetilde{p}}}_{i,j}}\left( c\!\!+\!\!\sum \limits _{m,n}(\frac{\alpha }{\ln 2} \frac{{g}_{i,j}}{{p}_{m,n}{g}_{m,n}} SIR(e^{{{\widetilde{p}}}_{i,j}})\right) \!\nonumber \\&\quad +\!\lambda _k {{\overline{g}}}_{i,j}\ln (1\!\!-\!\!\varepsilon _{i,j})\!+\!\sum \limits _{k\ne i,j}\lambda _k({g}_{i,j}(2^{\frac{R_0}{W}}\!\!-\!\!1)))\!=\!0. \end{aligned}$$

(28)

We can get the optimal power and the iteration of Lagrangian multiplier as follows

$$\begin{aligned} {p}_{i,j}^*= & {} e^{{{\widetilde{p}}}_{i,j}}\!=\!\frac{\frac{\alpha }{\ln 2}}{c\!\!+\!\!\sum \limits _{m,n}\left( \frac{\alpha }{\ln 2} \frac{{g}_{i,j}}{{p}_{m,n}{g}_{m,n}} SIR(e^{{{\widetilde{p}}}_{i,j}})\right) \!\!+\!\!\lambda _k {{\overline{g}}}_{i,j}\ln (1\!-\!\varepsilon _{i,j})\!\!+\!\!\sum \limits _{k\ne i,j}\lambda _k\left( {g}_{i,j}(2^{\frac{R_0}{W}}\!\!-\!\!1)\right) },\end{aligned}$$

(29)

$$\begin{aligned} \ \mathop {\lambda _k} \limits _{k=i,j}= & {} {{\overline{g}}}_{i,j}{p}_{i,j}\ln (1-\varepsilon _{i,j})+(2^{\frac{R_0}{W}}-1))({I}_{i,j}+N_0). \end{aligned}$$

(30)

The solution is achieved.