太赫兹科学与电子信息学报  2020, Vol. 18 Issue (6): 1110-1116     DOI: 10.11805/TKYDA2019114
云无线接入网络的下行链路能效的优化    [PDF全文]
吴波1, 庄莉2, 骆建华2, 范照健2     
1. 国网江苏省电力有限公司 淮安市洪泽区供电分公司,江苏 淮安市 223000;
2. 福建亿榕信息技术有限公司,福建 福州 350003
摘要: 在下行云接入网络(C-RANs)中,核心网络内基站处理单元(BBU)通过前向回传线路向远端无线电头端(RRH)传输数据,然后RRHs通过无线接入链路转发数据。目前,C-RANs的下行链路的能效是C-RANs的研究热点。为此,对C-RANs下行链路能效进行研究,其目的在于最大下行链路的能效。通过对联合用户分配(UA)、RRH激活、数据率分配和信号预编码约束,建立能效最大化表述式,再利用基于逐次凸二次规划的迭代算法求解混合整数非线性问题。仿真结果表明,提出的联合优化设计提高了C-RANs的能效。
关键词: 联合用户分配    逐次凸二次规划    前向回传线路    能效    云接入网络    
Energy efficient based design for downlink cloud radio access networks
WU BO1, ZHUANG li2, LUO Jianhua2, FAN Zhaojian2     
1. Power Supply Branch of Hongze District in Huaian City, State Grid Jiangsu Electric Power Co., Ltd, Huaian Jiangsu 401331, China;
2. Fujian Yirong Information Technology Co.Ltd, Fuzhou Fujian 350003, China
Abstract: In downlink Cloud Radio Access Network(C-RANs), the data is transferred from a Base Band Unit(BBU) in the core network to several remote radio heads via a set of edge routers over capacity-limited fronthaul links. The remote radio heads then send the received signals to their users via radio access links. Now, energy efficiency of C-RANs is a topic issue. Therefore, the general downlink of C-RANs is studied in order to maximize the network energy efficiency by jointly optimizing User Association(UA), Radio Remote Head(RRH) activation, data rate allocation and signal precoding. By employing the successive convex quadratic programming framework, an iterative algorithm is proposed to solve the challenging mixed-integer nonlinear problem formulation. Simulation results show that the proposed joint optimization approach significantly improves the energy efficiency.
Keywords: User Association    successive convex quadratic programming    fronthaul links    energy efficiency    Cloud Radio Access Network    

云无线接入网络(C-RANs)[1]是5G组网的核心技术之一[2]。C-RANs用低成本、低功耗的远端无线电头端(RRH)替代了传统的高成本、高功耗的基站,降低了功耗[3]。基站处理单元(BBU)是C-RANs的核心组件。BBU通过有线前向回传线路(前传)连接RRHs。而RRHs通过无线接入链路连通用户。C-RANs主要由RRH, BBU和高速前向回传(简称前传)组成[4]。BBU池由虚拟机组成,并处理用户传输到核心网的数据流。而RRH接收用户设备信息。C-RANs最大的优势在于:BBU池能够集中处理、分配无线资源和计算资源,提高了能量效率[5]。然而,由于采用联合处理,前传链路需要传输大量的数据,前传链路的有限容量成为C-RANs主要瓶颈之一[6-7]

为了提高C-RANs下行链路的能量效率,可采用以下3个方法:1)提高数据率;2)降低RRH传输功率;3)关掉RRHs。文献[8]将能量效率最大化问题转化成功率最小化问题,但其并没有考虑前传容量约束问题。而文献[9]运用随机矩阵理论,并通过启发式用户分配策略,最大化能量效率。然而,上述的研究工作只集中于单跳前传网络。而实际上,BBU是通过多跳前传网络[10]的多个路由连通RRHs。

为此,本文以多跳C-RANs为对象,并基于前传容量受限的事实,提出基于逐次凸二次规划(Successive Convex Quadratic Programming,SCQP)的迭代算法(SCQP-Iterative,SCQP-I)。SCQP-I算法通过联合用户分配(UA)、RRH激活、数据率分配和信号预编码,最大化网络能效。由于最大化网络能效涉及到多个变量,并且这些变量相互关联,难以用单一表达式表述最大化网络能耗问题。

据此,SCQP-I算法建立新的最大化能效问题表达式,并且其受路由限制、前传容量和最大传输功率的约束。SCQP-I算法引用逐次凸二次规划解决最大化能效问题,进而降低能耗。主要工作可归纳于2点:1)依据UA, RRH激活、数据率分配和信号预编码建立了能效最大化表达式;2)引用逐次凸二次规划求解。

1 系统模型及问题表述 1.1 系统模型

图 1显示了通用C-RANs的下行链路模型。BBU通过前传网络的M个路由器和N个无噪声链路[10]连通${K_{\rm{R}}}$个RRHs。将KR个RRHs构成一个矢量${\kappa _{\rm{R}}}{\rm{ = }}\left\{ {{\rm{1, 2, }} \cdot \cdot \cdot {\rm{, }}{K_{\rm{R}}}} \right\}$

Fig.1 Generic C-RANs downlink model 图 1 通用C-RANs的下行链路模型

对于多跳C-RANs,一定满足$M > 0$$N > {K_{\rm{R}}}$;对于单跳C-RAN,$M = 0$$N = {K_{\rm{R}}}$。令$\pi {\rm{ = }}\left\{ {{\rm{1, 2, }} \cdot \cdot \cdot , M} \right\}$$ \psi \rm{=}\left\{\rm{1, 2},\cdot \cdot \cdot \rm{, }N\right\}$。假定任意前传链路$n \in \psi $的容量为${C_n}$,且${C_n} > 0$,其单位为bit/s。

RRHs通过无线接入链路服务了${\kappa _{\rm{U}}}{\rm{ = }}\left\{ {{\rm{1, 2, }} \cdot \cdot \cdot {\rm{, }}{K_{\rm{U}}}} \right\}$个用户[11],并且允许每个用户连通多个RRHs。每个用户$k \in {\kappa _{\rm{U}}}$安装了${N_{\rm{u}}}$个天线,而每个RRH $i \in {\kappa _{\rm{R}}}$安装了${N_{\rm{r}}}$个天线。

1.2 问题表述

${M_k}$表示BBU传输给每个用户K的消息,并且${M_k}$分布于$\left\{ {1, 2, \cdot \cdot \cdot , {2^{u{R_k}}}} \right\}$中,其中$u$为块长度,${R_k}$为传输消息${M_k}$的数据率[12]。再将消息${M_k}$编码成符号${\chi _k} \in {C^{d \times 1}}$,其中${\chi _k}$来自高斯信道码本$C_k^{\rm{CH}}$。因此,${\chi _k} \sim CN$(0,I)。而$d$表示数据流数,其定义如式(1)所示:

$d = \min \left( {{N_u}, {N_r}} \right)$ (1)

因此,网络的总体吞吐量可依式(2)表述:

${R_{\rm{sum}}} = \sum\limits_{k \in {\kappa _U}} {{R_k}} $ (2)

此外,每个用户的已编码的消息符号通过前传链路路由至RRHs。不同的RRH为用户服务。为此,引用布尔变量表示RRH-用户的分配关系、RRH的激活状态,如式(3)和(4)所示:

$\alpha_{k, i}=\left\{\begin{array}{l}1, \text { if } \mathrm{RRH}\;i \text { serves user } k \\ 0, \text { otherwise }\end{array}\right.$ (3)
$b_{i}=\left\{\begin{array}{l}0, \text { if } \mathrm{RRH}\;i \text { serves no user } \\ 1, \text { otherwise }\end{array}\right.$ (4)

假定服务用户k的RRHs集为${D_k}$,其定义如式(5)所示:

${D_k}{\rm{ = }}\left\{ {i|{a_{k, i}} = 1, i \in {K_{R}}} \right\}$ (5)

然后,BBU通过多跳的前传网络,并以速率${R_k}$${D_k}$传输信息${\chi _k}$

此外,对于${K_{\rm{U}}}$个用户的消息,存在${K_{\rm{U}}}$个组播业务。以${r_{k, n}}$为路由变量,其决定链路$n$上的流量率,其中k表示第k个组播业务。如果${r_{k, n}}$=0,表示在链路$n$上没有传输组播业务k

依据文献[13]的网络编码理论,如果${D_k}$内每个目的节点所获取的速率${R_k}$相互独立,那整条组播业务也能获取速率${R_k}$。因此,将连通RRH $i \in {D_k}$的链路$n$上的组播流量看成独立的流量:

${f_{k, i, n}} \leqslant {r_{k, n}}, \forall k \in {\kappa _{\rm{U}}}, i \in {\kappa _{\rm{R}}}, m \in \pi $ (6)

最终,对前传网络的路由约束可表示为[8]

${f_{k, i, n}} \leqslant {r_{k, n}}, \forall k \in {\kappa _{\rm{U}}}, i \in {\kappa _{\rm{R}}}, n \in \psi $ (7)
${r_{k, n}} \leqslant {a_{k, i}}{C_n}, \forall k \in {\kappa _{\rm{U}}}, n \in I_i^{{\kappa _{\rm{R}}}}$ (8)
$a_{k, i} R_{k} \leqslant \sum\limits_{n \in I_{i}^{\kappa_{R}}} f_{k, i, n}, \forall k \in \kappa_{\mathrm{U}}, i \in \kappa_{\mathrm{R}}, m \in \pi $ (9)
$\sum\limits_{n \in O_{m}^{\pi}} f_{k, i, n}=\sum\limits_{n \in I_{m}^{\pi}} f_{k, i, n}, \forall k \in \kappa_{\mathrm{U}}, i \in \kappa_{\mathrm{R}}, m \in \pi $ (10)
$\sum\limits_{k \in \kappa_{U}} r_{k, n} \leqslant C_{n}, \forall n \in \psi $ (11)
$R_{k} \geqslant R_{\mathrm{QoS}}, \quad r_{k, n} \geqslant 0, f_{k, i, n} \geqslant 0, \quad \forall k \in \kappa_{\mathrm{U}}, i \in \kappa_{\mathrm{R}}, n \in \pi $ (12)

式中:$I_i^{{\kappa _R}}$表示流入第$i$个RRH的链路集;$I_i^\pi $, $O_m^\pi $分别表示路由m上流入和流出链路集。

接下来,先描述每个RRH传播的基带信号以及它们所消耗的功率。最后,通过这些功耗建立网络能效。

1.3 信号模型

一旦通过多跳前传网络,收到来自K个组播业务所传输的消息符号,第i个RRH就能产生已传输的基带信号${\mathit{\boldsymbol{x}}_i} \in {C^{{N_r} \times 1}}$

${\mathit{\boldsymbol{x}}_i}{\rm{ = }}\sum\limits_{k \in {\kappa _U}} {{\mathit{\boldsymbol{F}}_{k, i}}} {\mathit{\boldsymbol{\chi}} _k}$ (13)

式中${\mathit{\boldsymbol{F}}_{k, i}} \in {C^{{N_r} \times d}}$表示${\mathit{\boldsymbol{\chi}}_k}$的预编码矩阵。

假定每个RRH受平均传输功率约束,如式(14)所示:

$E\left\{ {{{\left\| {{\mathit{\boldsymbol{x}}_i}} \right\|}^{\rm{2}}}} \right\} \leqslant {P_i}$ (14)

${\mathit{\boldsymbol{H}}_{k, i}} \in {C^{{N_u} \times {N_r}}}$表示从RRH $i$连通用户k的平稳–衰落信道矩阵。相应地,${\mathit{\boldsymbol{H}}_k}{\rm{ = }}\left[ {{\mathit{\boldsymbol{H}}_{k, 1}}, {\mathit{\boldsymbol{H}}_{k, 2}}, \cdot \cdot \cdot , {H_{k, {\kappa _{\rm{R}}}}}} \right] \in {C^{{N_u} \times {N_R}}}$表示用户k连通所有RRHs的信道矩阵,其中${N_{\rm{R}}} = {\kappa _{\rm{R}}}{N_r}$

假定在传输间隔内,信道状态${\mathit{\boldsymbol{H}}_{k, i}}$仍保持不变,且$k \in {\kappa _{\rm{U}}}$, $i \in {\kappa _{\rm{R}}}$。引入变量${\bar {\mathit{\boldsymbol{F}}}_k}$

$\overline{\boldsymbol{F}}_{k}=\left[\left(\boldsymbol{F}_{k, 1}\right)^{\mathrm{H}},\left(\boldsymbol{F}_{k, 2}\right)^{\mathrm{H}}, \ldots,\left(\boldsymbol{F}_{k, \kappa_{R}}\right)^{\mathrm{H}}\right]^{\mathrm{H}} \in C^{N_{\mathrm{R}} \times d} $ (15)

那用户k所接收的${\mathit{\boldsymbol{y}}_k} \in {C^{{N_u} \times 1}}$可表示为:

$\boldsymbol{y}_{k}=\boldsymbol{H}_{k} \overline{\boldsymbol{F}}_{k} \boldsymbol{\chi}_{k}+\sum\limits_{l \in \kappa_{U} \backslash\{k\}} \boldsymbol{H}_{l} \overline{\boldsymbol{F}}_{l} \boldsymbol{\chi}_{l}+\boldsymbol{n}_{k} $ (16)

式中:$\sum\limits_{l \in {\kappa _U}\backslash \left\{ k \right\}} {{\mathit{\boldsymbol{H}}_l}{{\bar {\mathit{\boldsymbol{F}}}}_l}{\mathit{\boldsymbol{\chi}} _l}} $为干扰项;${\mathit{\boldsymbol{n}}_k} \in {C^{{N_u} \times 1}}$为加性噪声项,其服从$CN\left( {0, \sum\nolimits_k {} } \right)$。将此干扰项看成加性高斯噪声进行处理,可用香农形式表述消息符号${\mathit{\boldsymbol{\chi}} _k}$的传输速率${R_k}$

$R_{k} \leqslant g_{k}(\overline{\boldsymbol{F}})=W \log _{2}\left|\boldsymbol{I}_{N_{u}}+\Pi_{k} \Pi_{k}^{\mathrm{H}} \Xi_{k}^{-1}\right| $ (17)

式中:$W$为总体可用带宽;$\bar {\mathit{\boldsymbol{F}}}{\rm{ = }}{\left\{ {{\mathit{\boldsymbol{F}}_k}} \right\}_{k \in {\kappa _U}}}$, ${\prod _k} = {\mathit{\boldsymbol{H}}_k}{\bar {\mathit{\boldsymbol{F}}}_k}$。而$\Xi _k^{}$的定义如式(18)所示:

$\Xi_{k}=\sum\limits_{l \in \kappa_{\mathrm{U}} \backslash\{k\}} \boldsymbol{H}_{k} \overline{\boldsymbol{F}}_{l} \overline{\boldsymbol{F}}_{l}^{\mathrm{H}} \boldsymbol{H}_{k}^{\mathrm{H}}+\Sigma_{k} $ (18)

$\mathit{\boldsymbol{a}} = {\left\{ {{\alpha _{k, i}}} \right\}_{k \in {\kappa _U}, i \in {\kappa _R}}}$, $\mathit{\boldsymbol{b}} = {\left\{ {{b_i}} \right\}_{i \in {\kappa _R}}}$。用式(19)表示矢量a, bF间的相互关系:

$\alpha_{k, i}=\left\{\begin{array}{c}0, \text { if }\left\langle\bar{\mathit{\boldsymbol{E}}}_{i}^{\mathrm{H}} \bar{\mathit{\boldsymbol{F}}}_{k} \bar{\mathit{\boldsymbol{F}}}_{k}^{\mathrm{H}} \bar{\mathit{\boldsymbol{E}}}_{i}\right\rangle=0, _{\forall k \in \kappa_{\mathrm{U}}, i \in \kappa_{\mathrm{R}}} \\ 1, \text { otherwise }\end{array}\right. $ (19)
${b_i}{\rm{ = }}\left\{ {\begin{array}{*{20}{c}} {{\rm{0, }}{\rm{if}}{\kern 1pt} {\kern 1pt} {a_{k, i}} = 0, \forall k \in {\kappa _U}} \\ {{\rm{1, otherwise}}} \end{array}} \right.\forall i \in {\kappa _{\rm{R}}}$ (20)

依据式(19)和式(20)可知,矢量a, b满足式(21):

${\alpha _{k, i}} \leqslant {b_i} \leqslant \sum\limits_{k \in {\kappa _{\rm{U}}}} {{\alpha _{k, i}}, \forall k \in {\kappa _{\rm{U}}}, i \in } {\kappa _{\rm{R}}}$ (21)

通过式(21)的约束,确保没有用户安排给未激活的RHH。

1.4 功率模型

引用文献[8]的功率消耗模型,在给定传输间隔内,RRH $i \in {\kappa _{\rm{R}}}$所消耗的功率为:

$ {P}_{i}^{\rm{RRH}}=\{\begin{array}{c}{\beta }_{i}{P}_{i}^{\rm{Tx}}+{P}_{i, a},{\rm{if}}\;0<{P}_{i}^{\rm{Tx}}\le {P}_{i}\\ {P}_{i, \chi },{\rm{if}}\;{P}_{i}^{\rm{Tx}}=0\end{array}$ (22)

式中:${\beta _i} > 0$$i \in {\kappa _{\rm{R}}}$表示功率放大效率;$P_i^{\rm{Tx}}$为传输功率;$P_{i, a}^{}$表示支持RRH $i$处于活动状态所需要的功率。$P_{i, \chi }^{}$表示处于休眠状态时功率消耗,且$P_{i, \chi }^{} < P_{i, a}^{}$$P_i^{}$表示RRH $i$的最大传输功率。因此,RRH $i$的传输功率$P_i^{\rm{Tx}}$可表示为:

$P_{i}^{\mathrm{Tx}}=b_{i} \sum\limits_{k \in \kappa_{\mathrm{U}}} \alpha_{k, i}\left\langle\overline{\boldsymbol{E}}_{i}^{\mathrm{H}} \overline{\boldsymbol{F}}_{k} \overline{\boldsymbol{F}}_{k}^{\mathrm{H}} \overline{\boldsymbol{E}}_{i}\right\rangle=b_{i} \sum\limits_{k \in \kappa_{\mathrm{U}}}\left\langle\overline{\boldsymbol{E}}_{i}^{\mathrm{H}} \overline{\boldsymbol{F}}_{k} \overline{\boldsymbol{F}}_{k}^{\mathrm{H}} \overline{\boldsymbol{E}}_{i}\right\rangle=\sum\limits_{k \in \kappa_{\mathrm{U}}}\left\langle\overline{\boldsymbol{E}}_{i}^{\mathrm{H}} \overline{\boldsymbol{F}}_{k} \overline{\boldsymbol{F}}_{k}^{\mathrm{H}} \overline{\boldsymbol{E}}_{i}\right\rangle$ (23)

换而言之,将前传链路$n \in \psi $看成通信信道集,其总容量为${C_n}$,总的功耗为$P_{n, \max }^{\rm{FH}}$。依据文献[8],总的功率消耗可表示为$P_n^{\rm{FH}}$

$P_n^{\rm{FH}}{\rm{ = }}\frac{{\sum\nolimits_{k \in {\kappa _U}} {{r_{k, n}}} }}{{{C_n}}}P_{n, {\rm{\max}} }^{\rm{FH}} = {\alpha _n}\sum\nolimits_{k \in {\kappa _{\rm{U}}}} {{r_{k, n}}} $ (24)

式中${\alpha _n}{\rm{ = }}P_{n, \max }^{\rm{FH}}{\rm{/}}{C_n}$。而${r_{k, n}}$表示链路k上的实际流量率。

从式(22)至式(24)可知,总体网络所消耗的功率可表示为:

$P_{\text {total }}(\overline{\boldsymbol{F}}, \boldsymbol{r}, \boldsymbol{b})=\sum_{i \in \kappa_{R}} P_{i}^{\mathrm{RRH}}+\sum_{n \in \psi} P_{n}^{\mathrm{FH}}=\sum_{i \in \kappa_{R}}\left(\beta_{i} P_{i}^{\mathrm{Tx}}+b_{i} P_{i, \Delta}\right)+\sum_{i \in \psi} \alpha_{n} \sum_{k \in \kappa_{U}} r_{k, n}+P_{\chi}$ (25)

式中:$\bar {\mathit{\boldsymbol{F}}}{\rm{ = }}{\left\{ {{{\bar {\mathit{\boldsymbol{F}}}}_{k, i}}} \right\}_{k \in {\kappa _U}, i \in {\kappa _R}}}$${P_{i, \Delta }} = {P_{i, a}} - {P_{i, \chi }}$${P_\chi } = \sum\nolimits_{i \in {\kappa _R}} {{P_{i, \chi }}} $$\mathit{\boldsymbol{r}} = {\left\{ {{r_{k, n}}} \right\}_{k \in {\kappa _U}, n \in \psi }}$

简之,用功率消耗表示网络的能效[14]。因此,多跳C-RANs的优化问题可表示如下:

$\max\limits_{\boldsymbol{\alpha}, \boldsymbol{b}, \boldsymbol{R}, \overline{\boldsymbol{F}}, \boldsymbol{f}, \boldsymbol{r}} \quad P_{1}=\frac{R_{\mathrm{sum}}}{P_{\text {total }}}$ (26)
${\rm{s}}{\rm{.t}}.{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} (7)\sim(12), (17), (21), (22)$ (27)
$\sum\nolimits_{k \in \kappa_{\mathrm{U}}}\left\langle\overline{\boldsymbol{E}}_{i}^{\mathrm{H}} \overline{\boldsymbol{F}}_{k} \overline{\boldsymbol{F}}_{k}^{\mathrm{H}} \overline{\boldsymbol{E}}_{i}\right\rangle \leqslant P_{i}, \forall i \in \kappa_{\mathrm{R}} $ (28)
$\sum\nolimits_{i \in {\kappa _{_{\rm{R}}}}} {{\alpha _{k, i}} \geqslant 1, {\kern 1pt} {\kern 1pt} {\kern 1pt} \forall k \in {\kappa _{\rm{U}}}} $ (29)

式中:$\mathit{\boldsymbol{R}} = {\left\{ {{R_k}} \right\}_{k \in {\kappa _{\rm{U}}}}}$$\mathit{\boldsymbol{f}} = {\left\{ {{f_{k, i, n}}} \right\}_{k \in {\kappa _{\rm{U}}}, i \in {\kappa _{\rm{R}}}, n \in \psi }}$。式(26)旨在最大化网络吞吐量与总的功率消耗的比值,即最大化网络能效。而式(27)对前传网络的路由进行了约束。式(28)对传输功率进行了限制。式(29)确保没有用户安排给未激活的RHH。

1.5 SCQP-I算法

首先,对上述问题进行转换,如下所述:

$\mathop {\max }\limits_{t, \mathit{\boldsymbol{p}}, \mathit{\boldsymbol{a}}, \mathit{\boldsymbol{f}}, \mathit{\boldsymbol{r}}} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} t$ (30)
${\rm{s}}{\rm{.t}}.{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} (7)\sim(12), (17), (2{\rm{1}}), ({\rm{29}})$ (31)
$tz \leqslant \sum\limits_{k \in {\kappa _{\rm{U}}}} {{R_k}} $ (32)
$z \geqslant \sum\limits_{i \in \kappa_{R}}\left(\beta_{i} \sum\limits_{k \in \kappa_{U}}\left\langle\overline{\boldsymbol{E}}_{i}^{\mathrm{H}} \overline{\boldsymbol{F}}_{k} \overline{\boldsymbol{F}}_{k}^{\mathrm{H}} \overline{\boldsymbol{E}}_{i}\right\rangle+b_{i} P_{i, \Delta}\right)+\sum\limits_{n \in \psi} \alpha_{n} \sum\limits_{k \in \kappa_{U}} r_{k, n}+P_{S}$ (33)
$\left\langle\overline{\boldsymbol{E}}_{i}^{\mathrm{H}} \overline{\boldsymbol{F}}_{k} \overline{\boldsymbol{F}}_{k}^{\mathrm{H}} \overline{\boldsymbol{E}}_{i}\right\rangle \leqslant u_{k, i}, \forall k \in \kappa_{U}, i \in \kappa_{R}$ (34)
${u_{k, i}} \leqslant {\alpha _{k, i}}{P_i}, \forall k \in {\kappa _U}, i \in {\kappa _R}$ (35)
${\alpha _{k, i}} \in \left\{ {{\rm{0, 1}}} \right\}, {b_i} \in \left\{ {0, 1} \right\}, \forall k \in {\kappa _U}, i \in {\kappa _R}$ (36)
$\sum\limits_{k \in {\kappa _U}} {{u_{k, i}}} \leqslant {P_i}, \forall i \in {\kappa _R}$ (37)

式中$\mathit{\boldsymbol{p}} = \left( {\mathit{\boldsymbol{R}}, \mathit{\boldsymbol{F}}, z, \mathit{\boldsymbol{u}}} \right)$$\mathit{\boldsymbol{u}}{\rm{ = }}{\left\{ {{u_{k, i}}} \right\}_{k \in {\kappa _{\rm{U}}}, i \in {\kappa _{\rm{R}}}}}$

由于处理式(36)二值特性,对于$\forall {\alpha _{k, i}}, {b_i} \in \left\{ {{\rm{0, 1}}} \right\}$,存在:

$\sum\limits_{i \in {\kappa _{\rm{R}}}} {\sum\limits_{k \in {\kappa _{\rm{U}}}} {\left( {{\alpha _{k, i}} - \alpha _{k, i}^2} \right)} } + \sum\limits_{i \in {\kappa _{\rm{R}}}} {\left( {{b_i} - b_i^2} \right)} = 0$ (38)

相反,对于所有的${\alpha _{k, i}}, {b_i} \in \left\{ {{\rm{0, 1}}} \right\}$,则存在:

$\sum\limits_{i \in {\kappa _{\rm{R}}}} {\sum\limits_{k \in {\kappa _{\rm{U}}}} {\left( {{\alpha _{k, i}} - \alpha _{k, i}^2} \right)} } + \sum\limits_{i \in {\kappa _{\rm{R}}}} {\left( {{b_i} - b_i^2} \right)} \geqslant 0$ (39)

因此,可以对式(36)进行改写:

$\sum\limits_{i \in {\kappa _{\rm{R}}}} {\sum\limits_{k \in {\kappa _{\rm{U}}}} {\left( {{\alpha _{k, i}} - \alpha _{k, i}^2} \right)} } + \sum\limits_{i \in {\kappa _{\rm{R}}}} {\left( {{b_i} - b_i^2} \right)} \geqslant 0$ (40)
${\rm{0}} \leqslant {\alpha _{k, i}} \leqslant {\rm{1}}, {\rm{0}} \leqslant {b_i} \leqslant {\rm{1, }}\forall k \in {\kappa _{\rm{U}}}, i \in {\kappa _{\rm{R}}}$ (41)

结合式(40)和式(41)可知,式(36)为一个凸约束问题[15]。可将式(30)~式(37)所表述的问题转化为以下形式:

$\mathop {\min }\limits_{\left( {t, \mathit{\boldsymbol{p}}, \mathit{\boldsymbol{a}}, \mathit{\boldsymbol{f}}, \mathit{\boldsymbol{r}}, \mathit{\boldsymbol{b}}} \right) \in \Re } {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} - t$ (42)

式中$\Re {\rm{ = }}\left\{ {\left( {t, \mathit{\boldsymbol{p}}, \mathit{\boldsymbol{a}}, \mathit{\boldsymbol{f}}, \mathit{\boldsymbol{r}}, \mathit{\boldsymbol{b}}} \right){\rm{|(7)}} - {\rm{(12), (17), (21), (29)}}} \right.{\rm{, }}\left. {(3{\rm{3}}) - (3{\rm{5}}), (3{\rm{7}}), (4{\rm{0}}), (4{\rm{1}})} \right\}{\kern 1pt} $

为了求解式(42)问题,将式(42)转换成拉格朗日形式:

$\Im \left( {t, \mathit{\boldsymbol{b}}, \mathit{\boldsymbol{a}}, \lambda } \right){\rm{ = }} - t{\rm{ + }}\lambda \left( {\sum\limits_{i \in {\kappa _R}} {\sum\limits_{k \in {\kappa _U}} {\left( {{\alpha _{k, i}} - \alpha _{k, i}^2} \right) + \sum\limits_{i \in {\kappa _R}} {\left( {{b_i} - b_i^2} \right)} } } } \right)$ (43)

式中$\lambda \geqslant {\rm{0}}$为拉格朗日乘子。

因此,式(42)所表述的问题能够表示为:

$\mathop {\min }\limits_{\left( {t, \mathit{\boldsymbol{p}}, \mathit{\boldsymbol{f}}, \mathit{\boldsymbol{r}}, \mathit{\boldsymbol{a}}, \mathit{\boldsymbol{b}}} \right) \in \Re } \Im \left( {t, \mathit{\boldsymbol{b}}, \mathit{\boldsymbol{a}}, \lambda } \right){\rm{ = }} - t{\rm{ + }}\lambda \left( {\sum\limits_{i \in {\kappa _R}} {\sum\limits_{k \in {\kappa _U}} {\left( {{\alpha _{k, i}} - \alpha _{k, i}^2} \right) + \sum\limits_{i \in {\kappa _R}} {\left( {{b_i} - b_i^2} \right)} } } } \right)$ (44)

通过选择合适$\lambda $值,求解式(44),进而得到式(42)问题的解。为了处理式(9)和式(33),对它们进行改写:

${\left( {{R_k} + {\alpha _{k, i}}} \right)^2} - {\left( {{R_k} - {\alpha _{k, i}}} \right)^2} - 4\sum\limits_{n \in I_i^{{\kappa _{\rm{R}}}}} {{f_{k, i, n}}} \leqslant 0$ (45)
${\left( {t + z} \right)^2} - {\left( {t - z} \right)^2} - 4\sum\limits_{k \in {\kappa _{\rm{U}}}} {{R_k}} \leqslant 0$ (46)

函数${f_1}\left( {x, y} \right) = {\left( {x - y} \right)^2}$为联合凸优函数。在给定点$\left( {{x^{\left( \kappa \right)}}, {y^{\left( \kappa \right)}}} \right)$进行一阶泰勒级数展开,其凸下界表示为:

${\rm{2}}\left( {{x^{\left( \kappa \right)}} - {y^{\left( \kappa \right)}}} \right)\left( {x - y} \right) - {\left( {{x^{\left( \kappa \right)}} - {y^{\left( \kappa \right)}}} \right)^2} \leqslant {\left( {x - y} \right)^2}$ (47)

因此,在给定点$\left( {{t^{\left( \kappa \right)}}, {\mathit{\boldsymbol{p}}^{\left( \kappa \right)}}, {\mathit{\boldsymbol{\alpha}} ^{\left( \kappa \right)}}} \right)$对式(45)和式(46)进行近似:

$\begin{gathered} \quad \;\, {\left( {{R_k} + {\alpha _{k, i}}} \right)^2} - {\rm{2}}\left( {R_k^{\left( \kappa \right)} - \alpha _{k, i}^{\left( \kappa \right)}} \right)\left( {{R_k} - {\alpha _{k, i}}} \right) \\ {\left( {R_k^{\left( \kappa \right)} - \alpha _{k, i}^{\left( \kappa \right)}} \right)^2} - 4\sum\limits_{n \in I_i^{{\kappa _R}}} {{f_{k, i, n}}} \leqslant 0, \forall k \in {\kappa _U}, i \in {\kappa _R} \\ \end{gathered} $ (48)
${\left( {t + z} \right)^2} - {\rm{2}}\left( {{t^{\left( \kappa \right)}} - {z^{\left( \kappa \right)}}} \right)\left( {t - z} \right) + {\left( {{t^{\left( \kappa \right)}} - {z^{\left( \kappa \right)}}} \right)^2} - 4\sum\limits_{k \in {\kappa _{\rm{U}}}} {{R_k}} \leqslant 0$ (49)

算法1表述了求解式(42)的过程。首先,产生迭代的初始点,然后再利用式(50)求解最优解,更新值,再重复Step3, Step4, Step5,直到算法收敛。

$\mathop {\min }\limits_{\left( {t, \mathit{\boldsymbol{p}}, \mathit{\boldsymbol{f}}, \mathit{\boldsymbol{r}}, \mathit{\boldsymbol{a}}, \mathit{\boldsymbol{b}}} \right) \in {\Re ^{\left( \kappa \right)}}} \Im \left( {t, \mathit{\boldsymbol{b}}, \mathit{\boldsymbol{a}}, \lambda } \right){\rm{ = }} - t{\rm{ + }}\lambda \left( {\sum\limits_{i \in {\kappa _R}} {\sum\limits_{k \in {\kappa _U}} {\left( {\left( {1 - 2\alpha _{k, i}^{\left( \kappa \right)}} \right){\alpha _{k, i}} + {{\left( {\alpha _{k, i}^{\left( \kappa \right)}} \right)}^2}} \right) + \left. {\sum\limits_{i \in {\kappa _R}} {\left( {\left( {1 - 2b_i^{\left( \kappa \right)}} \right){b_i} + {{\left( {b_i^{\left( \kappa \right)}} \right)}^2}} \right)} } \right)} } } \right.$ (50)

此外,考虑算法1的收敛性能问题,设置迭代次数阈值N。依据仿真实验数据(见图 5),算法1迭代至10次时,性能趋于平稳。因此,迭代阈值N=20。

2 性能分析 2.1 实验环境

为了更好地分析SCQP-I算法的性能,利用MATLAB软件建立仿真平台。C-RANs的无线接入部分如图 2所示,其中RRHs数为${\kappa _{\rm{R}}} = 7$个,且其位置固定,而用户数${\kappa _{\rm{U}}} = 5$。它们随机地分布于RRHs的覆盖区域。

Fig.2 Simulation scene of wireless network(κR=7, κU=5) 图 2 无线网络仿真场景(κR=7, κU=5)

实验中用到的LTE参数如表 1所示。每个RRH具有2根天线(${N_{\rm{r}}} = 2$),而每个用户具有1根天线(${N_{\rm{u}}} = 1$)。每个RRH在活动状态和休眠状态时所消耗的能量分别为84 W和56 W。

表 1 LTE仿真参数 Table 1 Parameters of LTE

所有用户$i \in {\kappa _{\rm{R}}}$的传输功率参数:${\beta _i} = \beta = 2.8$, ${\alpha _i} = \alpha = {\rm{5}}$。而对于所有的RRH $k \in {\kappa _{\rm{U}}}$,其相关参数:$d = 1$, ${P_i} = P$, $\sum\nolimits_k { = {\sigma ^2}} \mathit{\boldsymbol{I}}$$\lambda = 100$,每类实验独立重复100次,取平均值作为最终的实验数据。

同时,考虑多跳前传和单跳前传网络,如图 3(a)3(b)所示。在多跳前传网络中有10个路由器,25条前传链路。

Fig.3 Simulation platform of multi-hop and single-hop fronthaul network 图 3 多跳和单跳前传网络仿真平台

选择的启发方案HUA[8],其应用于单跳C-RAN网络,将其标记为HUA-SH。在HUA-SH算法中,每个用户分配给${N_{\rm{c}}}$个RRHs,并且具有最大的信道增益。当其应用于多跳C-RAN网络,其标记为HUA-MH。

2.2 数据分析

首先,分析SCQP-I算法的收敛性。图 4显示了SCQP-I算法在单跳网络和多跳网络中的性能,分别标记为Alg.1-SH和SCQP-I-MH。

Fig.4 Convergence of the algorithm 图 4 算法的收敛性

图 4可知,SCQP-I算法能够有效地收敛。当迭代至5次时,$\Im $值趋于零。当迭代至15次后,SCQP-I-SH和SCQP-I-MH都得到很好的收敛。

图 5分析了HUA算法与SCQP-I算法的能效性能。从图 5可知,提出的SCQP-I算法在单跳、多跳C-RANs的能效比HUA分别提高了15%和23%。这归功于SCQP-I算法引用联合优化算法,并利用泰勒级数展开算法进行近似求解,提高了能效。

Fig.5 Average energy efficiency 图 5 能效
3 结论

针对C-RAN网络的下行链路能效问题,提出基于SCQP-I算法优化下行链路能效。基于前传链路的容量限制、最小数据率以及最大RRH的传输功率约束,建立混合整数优化问题,再利用SCQP-I算法求解优化问题。仿真结果表明,提出的SCQP-I算法能够有效地降低能效。

本文仅考虑了简单的C-RANs网络(7个RRHs,5个用户数),并只通过实验仿真分析了算法的性能。后期,将加大C-RANs网络规模,并用实测数据分析算法的性能,这将是后期的工作方向。

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