Power control and group proportional fairness for frequency domain resource allocation in L-SC-FDMA based LTE uplink

Abstract

Resource allocation in Long Term Evaluation (LTE) uplink is challenging because of two reasons: (1) it requires contiguous resource block allocation and (2) high complexity due to per user based power allocation among resource blocks. This paper presents a low complexity, optimal power resource allocation scheme for LTE uplink. We propose Virtual Cluster-based Proportional Fairness scheme that exploits the link adaptation information available at MAC layer to form virtual clusters. The distributed proportional fair scheduler ensures a minimum throughput for all users in the coverage area by assigning contiguous resource blocks, proportional to the throughput and the number of users in a particular cluster or group. Then, it allocates power to each resource block through water-filling algorithm which gives 20.2 % increase in average user throughout compared to equal power allocation. Simulations have been performed using practical scenarios of uniformly distributed users in Rician fading environment. Mathematical framework has been devised for network planning to get the best possible fairness with promising level of quality of service.

Appendices

Appendix 1: Number of RB in a group

In order to ensure the fairness with minimum guaranteed throughput, the number of RBs in a group should be proportional to the group’s modulation scheme and the number of users in that group, i.e.,

$$\begin{aligned} N_{K_{i}}=f(M_{K_{i}}).g(K_{i}).N \quad i=1,2,3 \end{aligned}$$

(22)

where $f(M_{K_{i}})$ is modulation dependent function and $g(K_{i})$ is number of users dependent component. LTE-A supports QPSK, 16, and 64 QAM in the uplink. Our proposed scheme divide the users in three groups $K_{1}$, $K_{2}$, and $K_{3}$ corresponding to QPSK, 16, and 64 QAM, respectively. eNB assigns the QPSK modulation to the UEs with low SNR and 64 QAM to the UEs with high SNR. To get the comparable throughput for UEs in group $K_{1}$ and $K_{2}$ we need more RBs. For example, one subcarrier allocated to UE in group $K_{3}$ gives baud rate of 6 bits per symbol where as it gives 4 bits per symbol and 2 bits per symbol in case of 16 QAM and QPSK, respectively. In order to get the same throughput we need to allocate RBs in the following proportion

$$\begin{aligned} f(M_{K_{1}})=\frac{log_{2}(M_{max})/log_{2}(M_{K_{1}})}{log_{2}\prod _{i=1}^{\upnu }(M_{max}-M_{K_{i}})} \end{aligned}$$

(23)

This expression assigns the fair proportion of RBs to users in different groups as shown in Table 1.

Table 1 RB allocation

Furthermore, the number of RBs for a particular group should be proportional to the number of users in that group. To realize this fact we introduce the following function

$$\begin{aligned} g(K_{1})=\frac{K_{1}}{f(M_{K_{1}})K_{1}+f(M_{K_{2}})K_{2}+f(M_{K_{3}})K_{3}} \end{aligned}$$

(24)

Substituting (23) and (24) in (22), we get the expression (11). Example:- $N=25$, $K_{1}=7$, $K_{2}=2$, and $K_{3}=1$, then using Eq. (11), we get

• $N_{K_{1}}=21$

• $N_{K_{2}}=3$

• $N_{K_{3}}=1$

which is the fair distribution of RBs amongst the three groups.

Appendix 2: Complexity order

The proposed algorithm first determines the number of RCs and the number of RBs per RC for each group, then within the group it performs a linear search on the users and RCs in order to find the user-RC pair that maximizes the utility function. For proposed algorithm, the result of [9, (12) modifies to the following:

$$\begin{aligned}&Complexity\approx {O}\left( \sum _{i=1}^{\upnu }N_{RC,K_{i}}^{2}K_{i}\right) \nonumber \\&\approx {O}\left( \sum _{i=1}^{\upnu }\left( \frac{N_{K_{i}}-r_{i}}{q_{i}}\right) ^{2}K_{i}\right) \end{aligned}$$

(25)

where $q_{i}$ and $r_{i}$ are quotient and remainder from $N_{K_{i}}/K_{i}$, respectively, and $N_{RC,K_{i}}$ is the number of RCs allocated to $K_{i}$ group. From the elementary calculus,

$$\begin{aligned} \{a^{2}+b^{2}+c^{2}+...\}<\{a+b+c+...\}^{2} \end{aligned}$$

(26)

therefore, our proposed algorithm provides a gain of at least $2\left( \frac{N_{K_{1}}}{q_{1}}\frac{N_{K_{2}}}{q_{2}}+\frac{N_{K_{2}}}{q_{2}}\frac{N_{K_{3}}}{q_{3}}+\frac{N_{K_{1}}}{q_{1}}\frac{N_{K_{3}}}{q_{3}}\right)$ (when $r_{i}=0$).