April 21, 2023

A Low Complexity Algorithm for Proportional Resource Allocation

in OFDMA SystemsIan C. Wong, Zukang Shen,

Jeffrey G. Andrews, and Brian L. Evans

The University of Texas at AustinWireless Networking and Communications Group

April 21, 2023

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

subchannel

frequency

mag

nitu

de

carrier

channel

April 21, 2023

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

frequency

mag

nitu

de

Base Station - has knowledge of each user’s channel state information thru ideal feedback from the users

User 2

User 1

. . .

User K

April 21, 2023

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)?

April 21, 2023

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)

Objective function

Exclusive subcarrier assignment

Non-zero power

No subcarrier sharing

Power constraint

Proportionality constraint

N - # subchannelsK - # usersP - BTS PowerB - BandwidthHk,n - channel gain

April 21, 2023

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

April 21, 2023

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

April 21, 2023

4-Step Approach

1. Determine number of subcarriers Nk for each user

2. Assign subcarriers to each user to give rough proportionality

3. Assign total power Pk for each user to maximize capacity

4. Assign the powers pk,n for each user’s subcarriers (waterfilling)

O(K)

O(N)

O(KNlogN)

O(K)

April 21, 2023

Simple Example

2 = 1/4

1 = 3/4

3

6 5

9

N = 4K = 2P = 10

108 7

4

April 21, 2023

Step 1: # of Subcarriers/User

108 7

4

3

6 5

9

Nk

3

1

1 2 3 4

2 = 1/4

1 = 3/4

N = 4K = 2P = 10

April 21, 2023

Step 2: Subcarrier Assignment

Nk

3/4 3

1/4 11 2 3 4

3

6 5

9

1 2 3 4

Rk

log2(1+2.5*10)=4.70

log2(1+2.5*7)=4.21

108

478

47

47

1010log2(1+2.5*8)=4.39

Rtot

13.3

9log2(1+2.5*9)=4.55 4.55

4

108

10108 7

3

6 5

April 21, 2023

Step 3: Power per user

1 2 3 4

10108 7

9

P1 = 7.66 P2 = 2.34

April 21, 2023

Step 4: Power per subcarrier

Nk

3/4 3

1/4 1

p1,1= 2.58 p1,2= 2.55p1,3= 2.53 p2,1= 2.34 Data Rates:R1 = log2(1 + 2.58*10) + log2(1 + 2.55*8)

+ log2(1 + 2.53*7) = 13.39008R2 = log2(1+ 2.34*9) = 4.46336

P1 = 7.66 P2 = 2.34

1 2 3 4

10108 7

9

• Waterfilling across subcarriers for each user

April 21, 2023

Simulation ParametersParameter Value Parameter 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 s

BER constraint 10-3 Doppler Frequency

30 Hz

April 21, 2023

Total Capacity Comparison

4 6 8 10 12 14 164.45

4.5

4.55

4.6

4.65

4.7

4.75

number of users

cap

acity

(bit/

s/H

z)

Proposed Method

Shen's Method

N = 64SNR = 38dBSNR Gap = 3.3

Based on 10000 channelrealizations

Proportions assigned randomly from {4,2,1} with probability [0.2, 0.3, 0.5]

April 21, 2023

Proportionality Comparison

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 160

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

User Number (k)

No

rmal

ized

Ra

te P

rop

ort

ion

s

Proportions

Proposed Method

Shen's Method

Based on the 16-user case,10000 channelrealizations per user

Normalized rate proportions for three classes of users using proportions {4, 2, 1}

April 21, 2023

Computational Complexity

2 4 6 8 10 12 14 160

1

2

3

4

5

6

7

8

9

10x 10

5 DSP Imlementation Clock cycle count

Number of users

Clo

ck c

ycle

s

Channel Allocation-shenPower Allocation-shen

Channel Allocation-wong

Power Allocation-wong

Total-shenTotal-wong Code developed

in floating point C and run on the TI TMS320C6701 DSP EVM run at 133 Mhz

22% average improvement

April 21, 2023

Memory ComplexityMemory Type *Proposed Method *Shen’s Method

Program

Memory

Subcarrier Allocation

2024 1660

Power Allocation

1976 2480

Total 4000 4140Data Memory

System Variables

8KN+4K

O(KN)

8KN+4K

O(KN)Subcarrier Allocation

4N+12K

O(N+K)

4N+8K

O(N+K)Power Allocation

4N+28K

O(N+K)

4N+24K

O(N+K)

* All values are in bytes

April 21, 2023

SummaryPerformance Criterion Proposed Method Shen’s Method

Subcarrier Allocation Computational Complexity

O(KNlogN) O(KNlogN)

Power Allocation Computational Complexity

O(N+nK), n9 O(N+K)

Memory Complexity O(NK) O(NK)Achieved Capacity Higher High

Adherence to Proportionality

Loose Tight

Assumptions on Subchannel SNR

None High

April 21, 2023

Backup Slides

April 21, 2023

Step 1: Number of subcarriers per user

Determine Nk to satisfy

This is achieved by

Complexity: O(K)

April 21, 2023

Step 2: Subcarrier Assignment

O(1)

O(KNlogN)

April 21, 2023

Step 2: Subcarrier Assignment

O( (N-K-N*)K )

O(N*K)

April 21, 2023

Step 3: Power allocation among users

From subcarrier allocation, we have

Hence, power allocation problem is reduced into solving

April 21, 2023

Step 3: Power allocation among users

Whose solution is:

(K)

April 21, 2023

Step 4: Power allocation across subcarriers per user

Waterfilling across subcarriers for each user:

O(K)