Optimal Power Scheduling for SIC-Based Uplink Wireless Networks with Guaranteed Real-Time Performance

Abstract

The k-SIC technology can support at most k parallel transmissions, thus it has the prominent capability of providing fast media access, which is vital for real-time industrial wireless networks. However, it suffers from high power consumption because high interference has to be overcome. In this paper, given the real-time performance requirement of an uplink network supporting k-SIC, we study how to minimize aggregate power consumption of users by power scheduling. We prove that the problem is solvable in polynomial time. A universal algorithm with complexity of O(n ³ ) is proposed for k-SIC, where n is the number of transmitters. For the special case of k = 2, another algorithm with complexity of O(L ⁴ ) is presented, where L is the frame length. Simulation results reveal that both the aggregate power consumption and the maximal transmit power will be exponentially declined with further relaxation of the real-time performance.

Appendix

Lemma A.1

Assume $a_{1} \ge a_{2} \ge \ldots \ge a_{n}$, $b_{1} \le b_{2} \le \cdots \le b_{n}$, then the optimal solution to the problem

$$ \begin{aligned} & \mathop {\hbox{min} }\limits_{{X_{ij} }} \mathop \sum \limits_{i,j = 1..n} X_{ij} a_{i} b_{j} \\ {\text{s}} . {\text{t}} .\,\sum\nolimits_{i = 1..n} {X_{ij} = 1} \,\,\forall j \in \left[ {1..n} \right] & ;\,\,\,\,\sum\nolimits_{j = 1..n} {X_{ij} = 1} \,\,i \in \left[ {1..n} \right]; X_{ij} \in \left\{ {0,1} \right\}; \\ \end{aligned} $$

is $X_{ij} = \left\{ {\begin{array}{*{20}c} {1\,\,\, \, {\text{for }}\,{\text{all}}\, i = j} \\ {0 \,\,\,\, {\text{for all}} \,i \ne j} \\ \end{array} } \right.$.

Proof.

We prove it using mathematical induction.

It is easy to prove the case n = 2 since $a_{1} b_{1} + a_{2} b_{2} \ge a_{1} b_{2} + a_{2} b_{1}$.

Assume that the lemma is true when n = k − 1.

For n = k, order that $\beta_{i} = b_{i} - b_{1}$ for $\forall i \in [2..k]$. We now only need to prove that $a_{1} b_{1} + a_{2} \left( {b_{1} + \beta_{2} } \right) + \ldots + a_{k} \left( {b_{1} + \beta_{k} } \right) \le b_{1} a_{i1} + \left( {b_{1} + \beta_{2} } \right)a_{i2} + \ldots + \left( {b_{1} + \beta_{k} } \right)a_{ik}$, where $(a_{i1} ,a_{i2} , \ldots ,a_{ik} )$ is any permutation of $\{ a_{1} ,a_{2} , \ldots ,a_{k} \}$.

• Case 1. $a_{i1} = a_{1}$. The lemma can be proved by the induction assumption of n = k − 1.

• Case 2. $a_{i1} \ne a_{1}$. Since $\left( {a_{i1} ,a_{i2} , \ldots ,a_{ik} } \right)$ is any permutation of $\{ a_{1} ,a_{2} , \ldots ,a_{k} \}$, therefore, $a_{i1} + a_{i2} + \ldots + a_{ik} = a_{1} + a_{2} + \ldots + a_{k}$, and there exist a $j \in \{ 2, \ldots ,k\}$, $a_{ij} = a_{1}$. Thus, $b_{1} a_{i1} + \left( {b_{1} + \beta_{2} } \right)a_{i2} + \ldots + \left( {b_{1} + \beta_{k} } \right)a_{ik}$ = $b_{1} (a_{1} + a_{2} + \ldots + a_{k} ) + \beta_{2} a_{i2} + \beta_{3} a_{i3} + \ldots + \beta_{k} a_{ik}$ = $b_{1} (a_{1} + a_{2} + \ldots + a_{k} ) + \beta_{2} a_{i2} + \beta_{3} a_{i3} + \ldots + \beta_{j - 1} a_{{i\left( {j - 1} \right)}} + \beta_{j} a_{1} + \beta_{j + 1} a_{{i\left( {j + 1} \right)}} + \ldots + \beta_{k} a_{ik} \ge b_{1} (a_{1} + a_{2} + \ldots + a_{k} ) + \beta_{2} a_{i2} + \beta_{3} a_{i3} + \ldots + \beta_{j - 1} a_{{i\left( {j - 1} \right)}} + \beta_{j} a_{i1} + \beta_{j + 1} a_{{i\left( {j + 1} \right)}} + \ldots + \beta_{k} a_{ik}$.

Thus, the case can be proved if $\beta_{2} a_{2} + \beta_{3} a_{3} + \ldots + \beta_{k} a_{k} \le \beta_{2} a_{i2} + \beta_{3} a_{i3} + \ldots + \beta_{j - 1} a_{{i\left( {j - 1} \right)}} + \beta_{j} a_{i1} + \beta_{j + 1} a_{{i\left( {j + 1} \right)}} + \ldots + \beta_{k} a_{ik}$.

The above inequality can be proved by the induction assumption for $\{ a_{1} ,a_{2} , \ldots ,a_{k} \}$ and $\{ \beta_{1} ,\beta_{2} , \ldots ,\beta_{k} \}$.

In conclusion, the lemma is proved. □