Optimal Scheme of Resource Allocation for Ultra-reliable and Low-latency in Machine Type Communications Based on Non-orthogonal Multiple Access with Short Block Transmission
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摘要: 针对机器类通信(MTC)应用场景的业务特征和服务质量(QoS)要求，该文考虑基于非正交多址(NOMA)的MTC中短分组/短编码块传输，探讨MTC中基于NOMA的高可靠低迟延无线资源优化问题。首先，上行传输是基于NOMA的MTC通信的瓶颈，考虑无线蜂窝网络中支持NOMA和高可靠低迟延性能要求，该文建立了上行无线资源优化的系统模型；然后，分析上行传输迟延，导出基于距离的链路可靠性函数；进一步，以迟延、可靠性和带宽为约束下条件，提出一种最大化中心用户和速率的无线资源分配算法，并给出算法的收敛性证明和复杂度分析；最后，实验仿真验证了所提算法的性能优势。Abstract: For the service characteristics and Quality of Service (QoS) requirements of Machine Type Communications (MTC), short-packet/short-coded block transmission in MTC based on Non-Orthogonal Multiple Access (NOMA) is considered in this paper, and the resource optimization problem of the Ultra-Reliable and Low-Latency (URLL) in MTC based on NOMA is discussed. Currently, uplink transmission is a bottleneck of MTC based on NOMA. Firstly, considering the performance requirements supporting NOMA and high reliability and low latency in wireless cellular networks, a system model for uplink wireless resource optimization is established. Then, the uplink transmission delay is analyzed and the link reliability function based on distance is derived. Further, with the constraints of delay, reliability and bandwidth, a wireless resource allocation algorithm for maximizing the sum rates of central users is proposed, and also the convergence proof and complexity analysis of the algorithm are given. Finally, the simulation results show the performance advantages of the proposed optimal scheme.
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表 1 算法1的具体流程
算法1 用户2的和速率最大化算法 步骤1 输入${N_k}$, ${r_k}$，当${N_k} \le {N_{\max }}$时，初始值${B_{lb}} = 0$, ${B_{ub}} = {W_{\rm{c}}}$, ${B_0} = \frac{{{B_{lb}} + {B_{ub}}}}{2}$； 步骤2 当${B_{ub}} - {B_{lb}} > {\delta _b}$，通过二分法计算得到使${f_u}\left( {{r_k},{N_k},{B_i},\varepsilon _i^1,\varepsilon _i^2} \right)$最小化的$\varepsilon _i^1$, $\varepsilon _i^2$； 步骤3 若${f_u}\left( {{r_k},{N_k},{B_i},\varepsilon _i^1,\varepsilon _i^2} \right) > {\varepsilon ^{\max }}$，更新初始值${B_{lb}} = {B_i}$, ${B_i} = \frac{{{B_{1b}} + {B_{ub}}}}{2}$，或者${B_{ub}} = {B_i}$, ${B_i} = \frac{{{B_{1b}} + {B_{ub}}}}{2}$；否则转步骤4； 步骤4 若${f_u}\left( {{r_k},{N_k},{B_i},\varepsilon _i^1,\varepsilon _i^2} \right) < {\varepsilon ^{\max }}$, ${B_k}({N_k}) = {B^{\rm{opt}}}$或者${B_k}({N_k}) = {\rm{NaN}}$； 步骤5 求出$N_k^*{\rm{ = }}\mathop {\arg }\limits_{{N_k}} \min {N_k}{B_k}({N_k})$, $B_k^*{\rm{ = }}{B_k}(N_k^*)$； 步骤6 把$N_k^*$, $B_k^*$代入到$R_k^2$中求出最优的$R_k^{2 * }$。 
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