Low-Complexity Beam Allocation for Switched-Beam Based Multiuser Massive MIMO Systems Reading Notes 1

I. INTRODUCTION

This paper aims to develop a low-complexity beam allocation algorithm to maximize the total data rate of a switched -beam based multi-user massive MIMO system using the Butler method with NNA uniform linear array of$N$$K$ user services form a large number of$N$ fixed beams, that is:$N\gg K.$ _ The beam allocation problem is formulated as a combinatorial optimizationproblem with two constraints, that is, each user can be served by at most one beam, and each beam can serve at most one user. The complexity of obtaining the optimal solution for brute force search is$O\left(N^K\right)$, in the number of beamsNNProblems that are difficult to solve when$N\; is\; large.$

This paper proposes a suboptimal low-complexity beam allocation ( LBA ) algorithm based on submodular optimization theory , which is a powerful tool for solving combinatorial optimization problems [37]. Specifically, the original optimization problem is first reconstructed into a non-monotone submodular maximization problem under two partition matroid constraints. Due to the non - monotone nature of the objective function, the The problem still has high computational complexity. In order to reduce the complexity, the non-monotone sub-modulus maximization problem is further decoupled into two sub-problems, including a beam -user association sub-problem and a beam allocation sub- problem. -problem ), this sub-problem is a monotone submodular maximization problem subject to a single partition matroid constraint , which can be efficiently solved by a greedy algorithm. Then the LBA algorithm is proposed by combining the solutions of these two sub-problems. Simulation results show that compared with optimal brute force search, our LBA algorithm can achieve almost the same sum data rate, but the complexity is only$O(K\log N)$。

Please note that in order to maximize data rates, some users may not be served, causing delays for unserved users. Therefore, it is very important to study how many users the system can serve simultaneously. This performance is expressed in the paper by the service ratio , which is defined as the ratio of the number of users served to the total number of users. An explicit expression for the average service ratio (i.e., the average service ratio at the user location) is obtained and shown to be the number of beams $N$ and the number of usersKKMonotonically growing function of the ratio of$K.$Simulation results verify that the parsing results can approximate the average service ratio well, thus providing important insights into service latency.

The remainder of this article is organized as follows. Section 2 introduces the system model and problem formulation. A low-complexity beam allocation algorithm is proposed in Section III, followed by simulation results and discussions in Section IV, and conclusions are summarized in Section V.

In this article, $\mathbb{E}[\cdot ]$ represents the expectation operator. $x\sim \mathcal{C}\mathcal{N}\left(u,p^2\right)$means that the mean is$u$ , the variance is$p^2$ complex Gaussian random variables. $|X|representsthe$ setXXThe base of$X.$$2^X$ represents the setXXThe power set of$X.$$X\cap Y$ 和 $X\cup Y$ respectively represents the set$X$ and setYYThe intersection and union of$Y.$$X\backslash Y$ represents the set$Y$ is in collectionXXRelative complement in$X$$\varnothing$ represents the empty set. $\left(\begin{array}{l}n\\ k\end{array}\right)$ represents the binomial coefficient, that is, from$From\; n$ different sets of elements, select$The\; number\; of\; ways\; of\; k$ elements. $\left\{\begin{array}{l}n\\ k\end{array}\right\}$ represents the second kind of Stirling number, that is,$The\; set\; of\; n$ different elements is divided into$The\; number\; of\; methods\; for\; k$ non-empty subsets.

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II. SYSTEM MODEL AND PROBLEM FORMULATION

As shown in Figure 1(a), consider having KKA multi-user switched-beam basedsystem$with\; K$$BS\; of\; N$ linear arrays of equally spaced identical isotropic antenna units, forming$Downlink\; transmission\; of\; N$ fixed beams. Assume$K$ users are evenly distributed inside the circular cell with unit radius, and each user is equipped with an antenna. The BSis located in the center of the cell, and all BS antenna units are at a distance of$d=$placed at equal intervals at$0.5$$\lambda$$\lambda$ is the propagation wavelength. $\left(r_k,i_k\right)$ represents user$k$ position.

N = 2 i N =2^i is formed by applying Butler's method$N=2^{Beam\; of\; i}$ (where$i\ge 1$ is an integer), any beam$n$， $n=1,2，\cdot \cdot \cdot ,$ an angle of departure (AoD) of the$N$ signal θ θThe normalized array factor ofθ$is$ given by [24]

$$A_n\left(i\right)=\frac{\sin \left(0.5N\pi \cos i-b_n\right)}{N\sin \left(0.5p\cos i-\frac{1}{N}{b}_n\right)}\text{\hspace{1em}(1)}$$

Figure 1(b) illustrates $N=$The array pattern ( array patterngenerated according to (1) and (2) at$16$ o'clock , in which the beam index increases from 1 to$N$。

Add user $k$ as a reference user. Different from multipath channels in traditional cellular systems [38]-[40], assuming the existence ofLOSchannels at millimeter wave frequencies, userkkThe AoDof the received signal at$k$ is$i_k$, the corresponding received power can be written as [41]

$$P_k=\sum _{n=1}^N{c}_{k,n}\cdot p_n\cdot D_n\left(i_k\right)\cdot r_k^{-}\text{\hspace{1em}(3)}$$

2: Directivity is a measure of the directivity of a single antenna relative to an isotropic antenna radiating the same total power. In other words, directivity is the ratio of the power density of an anisotropic antenna relative to an isotropic antenna radiating the same total power [42].

在(3)中， c k , n ∈ { 0 , 1 } c_{k, n} \in\{0,1\} ck,n​∈{ 0,1 } represents the beam allocation indicator. If beam$n$ is assigned to user$k$，$c_{k,n}=1$。

max ⁡ { c k , n } ∀ k , ∀ n ∑ k = 1 K R k (10a) \max _{\left\{c_{k, n}\right\}_{\forall k, \forall n}} \sum_{k=1}^{K} R_{k}\tag{10a} { ck,n​}∀k,∀n​max​k=1∑K​Rk​(10a)

 s.t.  ∑ n = 1 N c k , n ≤ 1 , ∀ k ∈ { 1 , 2 , ⋯   , K } ,  (10b) \text { s.t. } \sum_{n=1}^{N} c_{k, n} \leq 1, \quad \forall k \in\{1,2, \cdots, K\} \text {, }\tag{10b}  s.t. n=1∑N​ck,n​≤1,∀k∈{ 1,2,⋯,K}, (10b)

$$\sum _{k=1}^K{c}_{k,n}\le 1,\forall n\in \{1,2,\cdots ,N\}\text{\hspace{1em}(10c)}$$

c k , n ∈ { 0 , 1 } , ∀ k ∈ { 1 , 2 , ⋯   , K } , ∀ n ∈ { 1 , 2 , ⋯   , N } (10d) \begin{array}{l} c_{k, n} \in\{0,1\}, \quad \forall k \in\{1,2, \cdots, K\}, \\ \forall n \in\{1,2, \cdots, N\} \end{array}\tag{10d} ck,n​∈{ 0,1},∀k∈{ 1,2,⋯,K},∀n∈{ 1,2,⋯,N}​( 10d )

(10b) and (10c) respectively follow the constraint that each user can select at most one beam for transmission, and each beam can be used by at most one user to avoid severe intra-beam interference.

For the combinatorial optimization problem given by (10a)-(10d), in all $(N+1{)}^{Brute\; force\; search\; among\; K}$ possible beam assignments will result in$N$ 's massive MIMO system has unbearably high complexity.

In the literature [43]-[47], a widely used method to solve the combinatorial optimization problem is to combine the indicators $c_{k,n}$Relax to a continuous variable between 0 ~ 1 and convert the objective function into a convex function. It can then be solved efficiently by a convex optimization algorithm. However, in our example, there are $N\times The\; K$ index needs to be optimized, even if relaxation is performed, it will still lead to large$The\; computational\; complexity\; of\; N$ is prohibitively high. Therefore, in this paper, we turn to submodular optimization,which has proven to be a powerful tool for solving combinatorial optimization problems [37]. In the next section, the beam assignment problem will be reformulated as a submodularmaximization problemmatroid constraints

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III. BEAM ALLOCATION DESIGN BASED ON SUBMODULAR OPTIMIZATION

Before reformulating our beam assignment problem as a submodular optimization problem, let us first give the definitions of submodular functions and matroids given in [37] as follows.

A. Basic definition

Definition 1: Let $U$ is a finite base set,$2^U$为$The\; power\; set\; of\; U$ (that is, the set of all subsets of U, including the empty set and$U$ itself). a set function$f\left(S\right)$ , the input is$S\subseteq U$ (即$S\in 2^U$ ), the output is real value, denoted as$f:2^U\to R$ , can be called submodular if

$$f(S)+f(T)\ge f(S\cap T)+f(S\cup T)\text{\hspace{1em}(11)}$$

For any $S,T\subseteq U.$ _ An equivalent definition of the submodular function is

f ( S ∪ { e } ) − f ( S ) ≥ f ( T ∪ { e } ) − f ( T ) (12) f(S \cup\{e\})-f(S) \geq f(T \cup\{e\})-f(T)\tag{12} f(S∪{ e})−f(S)≥f(T∪{ e})−f(T)(12)

For any $S\subseteq T\subseteq U$ 和 $e\in U\backslash T$ , that is, the marginal gain of adding an additional element to the set decreases with the size of the set. Intuitively, if a set function is submodular, then its marginal gain is diminishing as the set size is increased by adding more elements to it.

In particular, a set function $f\left(S\right)$ is monotonic if

$$f(S)\le f(T)\text{\hspace{1em}(13)}$$

For any $S\subseteq T\subseteq U$.

B. Problem Reformulation

U = { u 1 , 1 , u 1 , 2 , ⋯   , u 1 , N ; u 2 , 1 , u 2 , 2 , ⋯   , u 2 , N ; ⋯   ; u K , 1 , u K , 2 , ⋯   , u K , N } , (15) \begin{aligned} U=&\left\{ u_{1,1}, u_{1,2}, \cdots, u_{1, N} ; u_{2,1}, u_{2,2}, \cdots, u_{2, N} ; \cdots ;\right. \\ &\left. u_{K, 1}, u_{K, 2}, \cdots, u_{K, N}\right\}, \end{aligned}\tag{15} U=​{ u1,1​,u1,2​,⋯,u1,N​;u2,1​,u2,2​,⋯,u2,N​;⋯;uK,1​,uK,2​,⋯,uK,N​},​(15)

And the beam allocation set $S$为UUA subset of$U$$u_{k,n}\in SIf$ beam n is assigned to user k, that is, ck,n= 1, ∀k, n; otherwise, uk,n∈/ S for any beam allocation set S⊆U, the objective function of (10a) can be written as

$$R_S(S)=\sum _{u_{k,n}\in S}{\log }_2\left(1+\frac{\frac{P_t}{|S|}{D}_n\left(i_k\right)r_k^{-}}{p_0^2+{\sum }_{u_{j,}\in S,j\ne k}\frac{P_t}{|S|}{D}_l\left(i_k\right)r_k^{-}}\right),\text{\hspace{1em}(16)}$$

According to (3) and formulas (6)-(9).

The constraint can be written as the intersection of two partitioning matrices on the ground set U. Specifically, let us divide the ground set U into K disjoint subsets, U1U, U2U,···,UKU, where UkU={ uk,1, uk,2, ···, uk,N} is the set containing all possible beam assignments for user K, and the superscript indicates the partitioning of the ground set according to the user index. Since the beam allocation index ck,n = 1 if uk,n belongs to the beam allocation set S, that is, uk,n∈S, then the constraints given in (10b) can be written as S∈IU, where

$\frac{1}{p+2+\frac{1}{p}+ϵ}$is a given parameter with a small value, $p$ ismatroid constraints. It is also shown in [48] that this algorithm requires at most$O\left(\frac{1}{ϵ}p{q}^4\log q\right)$ local operations, where q is the size of the ground set. In our case, the size of the ground set q = |U| = KN, and the number of matrix constraints ? p = 2, the number of local operations required is O ( 1 ϵ ( KN ) 4 log ⁡ ( KN )$O\left(\frac{1}{ϵ}(KN{)}^{4}log(KN)\right)$ , when the number of beams N is large, the complexity is still very high. In the next section, the beam allocation problem given by (20a)-(20c) is decoupled into two sub-problems, and based on this, a low-complexity beam allocation algorithm is proposed.