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 November 30, 2018

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 November 30, 2018

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 1 magnitude User 2 frequency . . . Base Station - has knowledge of each user’s channel state information thru ideal feedback from the users User K November 30, 2018

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)? November 30, 2018

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 November 30, 2018

Solution to Proportional Resource Allocation Problem [Shen et. al 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 November 30, 2018

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 November 30, 2018

4-Step Approach Determine number of subcarriers Nk for 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(K) O(KNlogN) O(K) O(N) November 30, 2018

Simple Example 1 = 3/4 2 = 1/4 N = 4 K = 2 P = 10 10 8 7 4 9 6 5 3 November 30, 2018

Step 1: # of Subcarriers/User 10 8 7 4 Nk 3 1 1 = 3/4 9 2 = 1/4 N = 4 K = 2 P = 10 6 5 3 1 2 3 4 November 30, 2018

Step 2: Subcarrier Assignment 10 8 4 7 10 10 8 10 8 7 Rk Rtot 8 4 7 4 7 log2(1+2.5*10)=4.70 log2(1+2.5*8)=4.39 4 log2(1+2.5*7)=4.21 13.3 3 6 5 9 9 log2(1+2.5*9)=4.55 4.55 3 6 5  Nk 3/4 3 1/4 1 1 2 3 4 1 2 3 4 November 30, 2018

Step 3: Power per user P1 = 7.66 P2 = 2.34 10 9 8 7 1 2 3 4 November 30, 2018

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  Nk 3/4 3 1/4 1 1 2 3 4 10 8 7 9 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 November 30, 2018

Simulation Parameters Value Number of Subcarriers (N) 64 Channel Model 6-tap, exponentially decaying power profile with Rayleigh fading Number of Users (K) 4-16 Max. Delay spread 5 ms BER constraint 10-3 Doppler Frequency 30 Hz November 30, 2018

Total Capacity Comparison 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] November 30, 2018

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} November 30, 2018

Computational Complexity 22% average improvement Code developed in floating point C and run on the TI TMS320C6701 DSP EVM run at 133 Mhz November 30, 2018

Memory Complexity 2024 1660 1976 2480 4000 4140 8KN+4K O(KN) 4N+12K Memory Type *Proposed Method *Shen’s Method Program Memory Subcarrier Allocation 2024 1660 Power Allocation 1976 2480 Total 4000 4140 Data Memory System Variables 8KN+4K O(KN) 4N+12K O(N+K) 4N+8K 4N+28K 4N+24K * All values are in bytes November 30, 2018

Summary O(KNlogN) O(N+nK), n9 O(N+K) O(NK) Performance Criterion Proposed Method Shen’s Method Subcarrier Allocation Computational Complexity O(KNlogN) Power Allocation Computational Complexity O(N+nK), n9 O(N+K) Memory Complexity O(NK) Achieved Capacity Higher High Adherence to Proportionality Loose Tight Assumptions on Subchannel SNR None November 30, 2018

Backup Slides November 30, 2018

Step 1: Number of subcarriers per user Determine Nk to satisfy This is achieved by Complexity: O(K) November 30, 2018

Step 2: Subcarrier Assignment O(1) O(KNlogN) November 30, 2018

Step 2: Subcarrier Assignment O( (N-K-N*)K ) O(N*K) November 30, 2018

Step 3: Power allocation among users From subcarrier allocation, we have Hence, power allocation problem is reduced into solving November 30, 2018

Step 3: Power allocation among users Whose solution is: (K) November 30, 2018

Step 4: Power allocation across subcarriers per user Waterfilling across subcarriers for each user: O(K) November 30, 2018