A Low Complexity Algorithm for Proportional Resource Allocation in OFDMA Systems

1. A Low Complexity Algorithm for Proportional Resource Allocation in OFDMA Systems Ian C. Wong, Zukang Shen, Jeffrey G. Andrews, and Brian L. Evans The University of Texas at Austin Wireless Networking and Communications Group September 3, 2014

2. Orthogonal Frequency Division Multiplexing (OFDM) • Adapted by current wireless standards • IEEE 802.11a/g, Satellite radio, etc… • Broadband channel is divided into many narrowband subchannels • Multipath resistant • Equalization simpler than single-carrier systems • Uses time or frequency division multiple access channel magnitude carrier subchannel frequency September 3, 2014

3. User 1 User K Orthogonal Frequency Division Multiple Access (OFDMA) • Adapted by IEEE 802.16a/d/e BWA standards • Allows multiple users to transmit simultaneously on different subchannels • Inherits advantages of OFDM • Exploits multi-user diversity User 2 magnitude frequency . . . Base Station - has knowledge of each user’s channel state information thru ideal feedback from the users September 3, 2014

4. Resource Allocation in OFDMA • Given: • N - number of subchannels • K - number of users • P - base station total transmit power • Hk,n -channel gain for user k on subcarrier n • BER - bit error rate (maximum) • f - objective function • How do we allocate the N subchannels and P total power to the K users to optimize the objective function f while satisfying the bit error rate (BER)? September 3, 2014

5. Proportional Resource Allocation in OFDMA Systems • Maximize the overall system throughput while maintaining proportionality among users • Useful for service level differentiation • Very difficult to solve exactly (Nonlinear Mixed-Integer Programming Problem) N - # subchannels K - # users P - BTS Power B - Bandwidth Hk,n - channel gain Objective function Exclusive subcarrier assignment Non-zero power No subcarrier sharing Power constraint Proportionality constraint September 3, 2014

6. Solution to Proportional Resource Allocation Problem [Shen et.al. 2003] • Subchannel allocation step • Greedy algorithm – allow the user with the least allocated capacity/proportionality to choose the best subcarrier O(KNlogN) • Power allocation step • General Case • Solution to a set of K non-linear equations in K unknowns – Newton-Raphson methods O(nK) • High-channel to noise ratio case • Function root-finding O(nK), n=number of iterations, typically 10 for the ZEROIN subroutine September 3, 2014

7. Proposed Low Complexity Solution • Key Ideas • Relax strict proportionality constraint • In practical scenarios, rough proportionality is acceptable • Require a predetermined number of subchannels to be assigned to simplify power allocation • Reduced power allocation to a solution of linear equations O(K) • Achieved higher capacity with lower complexity, while maintaining acceptable proportionality • Does not need a high channel-to-noise ratio assumption September 3, 2014

8. 4-Step Approach O(K) • Determine number of subcarriers Nkfor each user • Assign subcarriers to each user to give rough proportionality • Assign total power Pk for each user to maximize capacity • Assign the powers pk,n for each user’s subcarriers (waterfilling) O(KNlogN) O(K) O(N) September 3, 2014

9. 10 8 7 4 Simple Example N = 4 K = 2 P = 10 1 = 3/4 9 2 = 1/4 6 5 3 September 3, 2014

10. 10 8 7 4 1 2 3 4 Step 1: # of Subcarriers/User 1 = 3/4 9 2 = 1/4 N = 4 K = 2 P = 10 6 5 3 September 3, 2014

11. 9 10 8 7 10 10 10 4 8 7 10 8 4 7 9 4 6 10 4 5 6 8 7 5 3 3 1 1 2 2 3 3 4 4 Step 2: Subcarrier Assignment Rk Rtot log2(1+2.5*10)=4.70 log2(1+2.5*8)=4.39 13.3 log2(1+2.5*7)=4.21 log2(1+2.5*9)=4.55 4.55 September 3, 2014

12. 10 10 8 9 7 1 2 3 4 Step 3: Power per user P1 = 7.66 P2 = 2.34 September 3, 2014

13. 10 10 8 9 7 1 2 3 4 Step 4: Power per subcarrier • Waterfilling across subcarriers for each user P1 = 7.66 P2 = 2.34 p1,1= 2.58 p1,2= 2.55 p1,3= 2.53 p2,1= 2.34 Data Rates: R1 = log2(1 + 2.58*10) + log2(1 + 2.55*8) + log2(1 + 2.53*7) = 13.39008 R2 = log2(1+ 2.34*9) = 4.46336 September 3, 2014

14. Simulation Parameters September 3, 2014

15. Total Capacity Comparison N = 64 SNR = 38dB SNR Gap = 3.3 Based on 10000 channel realizations Proportions assigned randomly from {4,2,1} with probability [0.2, 0.3, 0.5] September 3, 2014

16. Proportionality Comparison Based on the 16-user case, 10000 channel realizations per user Normalized rate proportions for three classes of users using proportions {4, 2, 1} September 3, 2014

17. Computational Complexity 22% average improvement Code developed in floating point C and run on the TI TMS320C6701 DSP EVM run at 133 Mhz September 3, 2014

18. Memory Complexity * All values are in bytes September 3, 2014

19. Summary September 3, 2014

20. Backup Slides September 3, 2014

21. Step 1: Number of subcarriers per user • Determine Nk to satisfy • This is achieved by • Complexity: O(K) September 3, 2014

22. Step 2: Subcarrier Assignment O(1) O(KNlogN) September 3, 2014

23. Step 2: Subcarrier Assignment O( (N-K-N*)K ) O(N*K) September 3, 2014

24. Step 3: Power allocation among users • From subcarrier allocation, we have • Hence, power allocation problem is reduced into solving September 3, 2014

25. Step 3: Power allocation among users • Whose solution is: (K) September 3, 2014

26. Step 4: Power allocation across subcarriers per user • Waterfilling across subcarriers for each user: O(K) September 3, 2014