Energy Efficiency Optimization for Massive MIMO Backhaul Networks with Imperfect CSI and Full Duplex Small Cells

Abstract

In this paper, a massive MIMO enabled backhaul model for two-tier heterogeneous networks with in-band full-duplex transmission and imperfect CSI is provided. Because of a massive number of antennas at the macro base station and densification of small cells, circuit power consumption increases in the system. It is requisite to study energy efficiency of such backhaul networks. This paper aims at maximizing EE for optimal user association, spectrum allocation, and power allocation while accounting for QoS and backhaul constraints. A joint optimization problem is formulated as a non-convex mixed integer non-linear problem which provides an un-affordable computational complexity. To crack the problem efficiently, it is progressively divided into sub-problems until each one is convex and is solved separately to obtain the optimal solution. Furthermore, complexity of the proposed algorithm is evaluated and Jain’s fairness index is measured. Simulation results of the proposed distributive algorithm are shown and compared with that of the existing algorithm (Max SINR) to manifest the strength of the proposed algorithm. The impact of self-interference cancellation factor and channel estimation error on EE is clearly reflected by the numerical results. In addition, it is observed that using the proposed algorithm results in better EE than increasing the number of antennas.

Appendices

Steps for Calculating $\Delta p_{j}$

Let $f_{1}$ be the optimization function given in (27). Then, first and second order partial derivatives of $f_{1}$ with respect to $p_{j}$ are derived as below

$$\begin{aligned} \frac{{\partial f_{1}}}{{\partial {p_j}}} =&\sum \limits _{i = 1}^K {\sum \limits _{j = 1}^M {\frac{{{G_{i,j}}}}{{\ln 2\left( {I + \sigma _n^2} \right) .\left( {1 + {\Upsilon _{i,j}}} \right) .{{\log }_2}\left( {1 + {\Upsilon _{i,j}}} \right) }}} \, + \sum \limits _{j = 1}^M {\frac{{{U_j}}}{{{\varepsilon _j}{P_j}}}} } \\ \nonumber +&\sum \limits _{j = 1}^M {\frac{{{\lambda _2}{U_j}\left( {1 - \eta } \right) }}{{\left( {1 + {\Upsilon _j}} \right) }}\left( {\frac{{ - {\zeta _j}{p_m}{G_{j,0}}}}{{{{\left( {{\zeta _j}{p_j} + {I_s}} \right) }^2}}}} \right) } + \,\sum \limits _{i = 1}^K {\sum \limits _{j = 1}^M {\frac{{{\lambda _2}\left( {1 - \eta } \right) {G_{i,j}}}}{{\left( {I + \sigma _n^2} \right) .\left( {1 + {\Upsilon _{i,j}}} \right) }}} } \end{aligned}$$

(35)

and

$$\begin{aligned} \frac{{{\partial ^2}{f_1}}}{{\partial p_{i,j}^2}} =&\sum \limits _{i = 1}^K {\sum \limits _{j = 1}^M {\frac{{ - G_{i,j}^2}}{{{{\left( {I + \sigma _n^2} \right) }^2}{{\left( {1 + {\Upsilon _{i,j}}} \right) }^2}{{\left( {\log \left( {1 + {\Upsilon _{i,j}}} \right) } \right) }^2}}}.\left( {1 + \log \left( {1 + {\Upsilon _{i,j}}} \right) } \right) } \,}\nonumber \\ -&\sum \limits _{j = 1}^M {\frac{{{\lambda _2}{U_j}\left( {1 - \eta } \right) }}{{{{\left( {1 + {\Upsilon _j}} \right) }^2}}}\times {{\left( {\frac{{{\zeta _j}{p_m}{G_{j,0}}}}{{{{\left( {{\zeta _j}{p_j} + {I_s}} \right) }^2}}}} \right) }^2} }\times \left( {1 - \frac{{2\left( {1 + {\Upsilon _j}} \right) \left( {{\zeta _j}{p_j} + {I_s}} \right) }}{{{p_m}{G_{j,0}}}}} \right) \nonumber \\ -&\sum \limits _{i = 1}^K {\sum \limits _{j = 1}^M {\frac{{{\lambda _2}\left( {1 - \eta } \right) G_{i,j}^2}}{{{{\left( {I + \sigma _n^2} \right) }^2}{{\left( {1 + {\Upsilon _{i,j}}} \right) }^2}}}} }- \sum \limits _{j = 1}^M {\frac{{{U_j}}}{{{{\left( {{\varepsilon _j}{P_j}} \right) }^2}}}} \end{aligned}$$

(36)

where $I={{p_m}{G_{i,0}} + \sum \nolimits _{j' = 1/j}^M {{p_{j'}}{G_{i,j'}}} }$ and $I_{s}=\sum \nolimits _{j' = 1/j}^M {{p_{j'}}{G_{i,j'}}} + {\sigma _n^2}$. Then, from the Newton’s method [39], $\Delta {p_{j}}$ can be evaluated as follows

$$\begin{aligned} \Delta {p_{j}} = \left| {{{\partial f_{1}} \over {\partial {p_{j}}}}} \right| \big / \left| {{{{\partial ^2}f_{1}} \over {\partial p_{j}^2}}} \right| . \end{aligned}$$

(37)

Procedure for Calculating the Value of $\eta$

Let’s proceed with $f_{2}$ provided in (31). $f_{2}$ is differentiated with respect to $\eta$ and is equated to zero. Then, the quadratic equation in $\eta$ is attained as follows

$$\begin{aligned} A{\eta ^2} + \left( {\sum \limits _{i = 1}^K {{y_{i,0}}} + \sum \limits _{i = 1}^K {\sum \limits _{j = 1}^M {{y_{i,j}}} - A} } \right) \eta - \sum \limits _{i = 1}^K {{y_{i,0}}} = 0, \end{aligned}$$

(38)

where $A = \sum \nolimits _{i = 1}^K {{\lambda _1}\left( {{y_{i,0}}{a_{i,0}}L - \sum \nolimits _{j = 1}^M {{y_{i,j}}{a_{i,j}}} } \right) }$.

Afterward, the roots of the aforementioned equation can be obtained by using the Quadratic Equation Method [44] as follows

$$\begin{aligned} \begin{array}{l} \eta = \\ {{ - \left( {\sum \nolimits _{i = 1}^K {{y_{i,0}}} + \sum \nolimits _{i = 1}^K {\sum \nolimits _{j = 1}^M {{y_{i,j}}} - A} } \right) \pm \sqrt{{{\left( {\sum \nolimits _{i = 1}^K {{y_{i,0}}} + \sum \nolimits _{i = 1}^K {\sum \nolimits _{j = 1}^M {{y_{i,j}}} - A} } \right) }^2} + 4 \times A \times \sum \nolimits _{i = 1}^K {{y_{i,0}}} } } \over {2 \times A}}. \end{array} \end{aligned}$$

(39)

If $B={\left( {\sum \nolimits _{i = 1}^K {{y_{i,0}}} + \sum \nolimits _{i = 1}^K {\sum \nolimits _{j = 1}^M {{y_{i,j}}} - A} } \right) }$ then, (39) is redrafted as

$$\begin{aligned} \eta = {{ - B \pm \sqrt{{B^2} + 4 \times A \times \sum \nolimits _{i = 1}^K {{y_{i,0}}} } } \over {2 \times A}}. \end{aligned}$$

(40)