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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)