joint user pairing and resource allocation for multiuser sc-fdma

Joint User Pairing and Resource Allocationfor Multiuser SC-FDMA Transmission


Wolfgang H. Gerstacker and Michael A. Ruder

University of Erlangen-NürnbergChair of Mobile Communications

Eurecom September 14, 2011

OutlineOutline

1 Introduction

2 System Model

3 Equalizers for SC-FDMA

4 Criteria for User Grouping and Resource Allocation

5 Algorithms for Time / Frequency Domain Resource Allocation

6 Numerical Results

7 Conclusions

W.H. Gerstacker et al.W.H. Gerstacker et al. – Joint User Pairing and Resource Allocation for Multiuser SC-FDMA– Joint User Pairing and Resource Allocation for Multiuser SC-FDMA11

IntroductionIntroduction

Basic IdeaBasic Idea

System withNuser usersOne transmit antenna per userOne base station with Nu antennas


IntroductionIntroduction

Considered ScenariosConsidered Scenarios

User Pairing/Grouping:

Nu users transmit on the same time and frequency resourceNu = 2 ⇒ user pairingNu > 2 ⇒ user grouping

Pairing/Grouping Criteria:

CapacityBER

Resource Allocation:

TDMA scheme: transmit in subsequent time slots, using entirebandwidthFDMA scheme: transmit in same time slot, on differentfrequency chunks


System ModelSystem Model

SC-FDMA TransmitterSC-FDMA Transmitter

M -DFTW

SubcarrierMapping

Ko

N -IDFTVH

CPinsertion

Pin

ai[k] Ai[µ] Bi[ν] bi[κ] bci[κ]

Figure: SC-FDMA transmitter structure for user i in pair o.

W: M -point DFT matrixKo: subcarrier mapping matrix for group o ∈ {1, . . . , O}V: N -point DFT matrixPin: cyclic prefix insertion matrix



SC-FDMA ReceiverSC-FDMA Receiver

+CP

removalPout

N -DFTV

n1[κ]

...rc1[κ] r1[κ] R1[ν]

KH1

KHO

Y1,1[µ]

Y1,O[µ]

+CP

removalPout

N -DFTV

nNR[κ]

...rcNR

[κ] rNR[κ] RNR

[ν]KH

1

KHO

YNR,1[µ]

YNR,O[µ]

......

Figure: SC-FDMA BS receiver structure.

nl[κ]: spatially and temporally white Gaussian noise at receive antenna lPout: cyclic prefix removal matrixV: N -point DFT matrixKo: subcarrier mapping matrix for group o ∈ {1, . . . , O}



V-MIMO ChannelV-MIMO Channel

Signal at the lth receive antenna:

rcl[κ] =

Ntx∑i=1

qh∑λ=0

hl,i[λ] bci[κ− λ] + nl[κ]

hl,i[λ]: discrete-time subchannel impulse response characterizesthe transmission from user i to the lth receive antennaincluding transmit and receiver input filtering

λ ∈ {0, 1, . . . , qh}qh: channel orderNtx: number of users being transmitted within one subframe

(Nu for TDMA scheme, Nusers for FDMA scheme)



V-MIMO ChannelV-MIMO Channel

Signal at the lth receive antenna in matrix-vector notation:

rcl =

Ntx∑i=1

Hl,i bci + nl,

Hl,i: channel convolution matrix for transmission from user ito receive antenna l


Equalizers for SC-FDMAEqualizers for SC-FDMA

Linear MMSE Equalizer for SC-FDMALinear MMSE Equalizer for SC-FDMA. . .

. . .

r1[κ]

rNR[κ]

n1[κ]

nNR[κ]

FrequencydomainMMSE

LE

R1[ν]

RNR[ν]

N–DFT

N–DFT

V

V

a1[k]

a2[k]remove

remove

CP

CPrc1[κ]

rcNR[κ]

Pout

Pout

Filter matrix Fo for each pair o in frequency domainError variance of the pth user within pair o:

σ2eu(o,p)

=1

M

M−1∑µ=0

Σ2pp[νoffset + νo + µ]

Σ2pp[ν]: error variance on νth subcarrier for user p


Equalizers for SC-FDMAEqualizers for SC-FDMA

SIC Equalizer for SC-FDMASIC Equalizer for SC-FDMA

+ gHo,η

yc

−

aη

HγQfHo,γ

y

zaγaγ

Figure: SC-FDMA SIC equalizer structure for two users.

Filter vectors fo,γ and go,η for each user γ and η of pair oEqualization order depends on SINR after equalization

SINRbiased,p{o} = σ2au(o,p)

/σ2eu(o,p)

with

σ2eu(o,p)

= σ2au(o,p)

− fHo,p ϕyoau(o,p)[0]


Criteria for User Grouping and Resource AllocationCriteria for User Grouping and Resource Allocation

Random Grouping and Resource AllocationRandom Grouping and Resource Allocation

Random grouping

Random allocation of users to resources

⇒ Huge potential is wasted



Capacity GroupingCapacity Grouping

Capacity of group o:

Co =1

M

M−1∑µ=0

ld det(INR

+SNR

NuHo[νoffset + νo + µ]

×HHo [νoffset + νo + µ]

)Aim: find a mapping function u(o, p) that maximizes the sum ofthe capacities of all individual groups:

Csum =

O∑o=1

Co.



Bit Error Rate Grouping IBit Error Rate Grouping I

Approximation for BER of an uncoded transmission:

BER ≈ Nmin

ld(M)Q

(√d2

min

Eb

N0

),

Nmin: average number of nearest neighbors of a signalpoint of the modulation alphabet

M: size of the modulation alphabetQ(·): complementary Gaussian error integrald2

min: normalized minimum squared Euclidean distance= 3 ld(M)/(M− 1) for QAM constellations



Bit Error Rate Grouping IBit Error Rate Grouping I

Approximation for BER of an uncoded transmission:

BER ≈ Nmin

ld(M)Q

(√d2

min

Eb

N0

),

BER for each user p within the group o can be approximatedby

BERu(o,p)≈Nmin

ld(M)Q(√

ζ SINRub,u(o,p) Gc

),

ζ = 3R (M−1)

Gc: gain of the channel code



Bit Error Rate Grouping IIBit Error Rate Grouping II

Average BER for group o

BER{o} = 1/Nu

Nu∑p=1

BERu(o,p),

Average BER for all Nuser users is

BERtotal =Nu

Nuser

O∑o=1

BER{o},


Algorithms for Time / Frequency Domain Resource AllocationAlgorithms for Time / Frequency Domain Resource Allocation

Algorithm for Time Domain Resource AllocationAlgorithm for Time Domain Resource Allocation

Two-dimensional optimization problem

Hungarian Algorithm (HA) finds solution fast

HA needs polynomial time vs. double factorial for full search

Symmetric cost matrix with entries for each pair



FD Resource Allocation – Problem FormulationFD Resource Allocation – Problem Formulation

Combined optimization of resource allocation and user pairing⇒ three-dimensional optimization problem

A three-dimensional assignment problem is in general NP-hard

Task: find that mapping function u(o, p), that eithermaximizes the sum capacity or minimizes the average BER forall users

Each value of o ∈ {1, 2, . . . , O} stands for the number ofthe considered frequency chunk



FD Resource Allocation – Full SearchFD Resource Allocation – Full Search

Compute sum capacity (Csum) or BER for all possible groups(BERtotal) and group/chunk combinations u(o, p)

Choose combination with the maximum sumcapacity/minimum average BER

If the frequency bandwidth is divided into O = Nuser/Nu

disjoint chunks the total number of combinations is

OFS =

(Nuser

Nu

)·(Nuser −Nu

Nu

)· · ·(

2 ·Nu

Nu

)·(Nu

Nu

)=

Nuser!

(Nu!)Nuser/Nu



FD Resource Allocation – Hungarian AlgorithmFD Resource Allocation – Hungarian Algorithm

user 1

user 2

user 3

user 4

. . .

. . .

user Nuser − 1

user Nuser

RBs

Before HA

user Nuser − 1

user Nuser

user 7

user 8

. . .

. . .

user 1

user 2RBs

After HA

Figure: Example for result of HA for the optimization of the resourceallocation of pairs.

HA finds the best assignment of O given user groups to O chunks.



FD Resource Allocation – Binary Switching Algorithm IFD Resource Allocation – Binary Switching Algorithm I

An exchange of users of different groups can be realized by thebinary switching algorithm (BSA)

Based onSimply try to exchange users between different groupsCompute the resulting cost/weight function BERtotal/Csum

Perform the exchange which yields the maximumdecrease/increase of the cost/weight function of all consideredtrials

To limit complexity of the BSA ⇒ confine the number of usersthat can be switched simultaneously to Nsw

Number of total switches for Nsw = 1

OBSA =Nu · (Nuser −Nu) +Nu · (Nuser − 2Nu) + . . .+Nu ·Nu

=1

2(N2

user −Nu ·Nuser).



FD Resource Allocation – Binary Switching Algorithm IIFD Resource Allocation – Binary Switching Algorithm II

user 1

user 2

user 3

user 4

. . .

. . .

user Nuser − 1

user Nuser

RBs

BSA with user 1

user 1 user 3

user 4

user Nuser − 1

user Nuser

user 1

user 2

user 3

user 4

. . .

. . .

user Nuser − 1

user Nuser

RBs

BSA with user 2

user 2

user 3

user 4

user Nuser − 1

user Nuser

user 1

user 2

user 3

user 4

. . .

. . .

user Nuser − 1

user Nuser

RBs

BSA with user 3

user 3 user Nuser − 1

user Nuser

......



FD Resource Allocation – Combined HA+BSA IFD Resource Allocation – Combined HA+BSA I

Initialization: random groups assigned to random chunks

First step: use HA to find optimum allocation of given groupsto chunksBUT HA does not exchange users between chunks

Second step: BSA exchanges Nsw = 1 users between chunks

Iterations: repeat first and second step for Nit iterations

Convergence of the algorithm is guaranteed in principleBUT resulting solution might not be the optimum one



FD Resource Allocation – Combined HA+BSA IIFD Resource Allocation – Combined HA+BSA II

user 1

user 2

user 3

user 4RBs

Starting point

user 3

user 4

user 1

user 2RBs

HA test

user 1

user 4

user 3

user 2RBs

BSA testsuser 3

user 1

user 4

user 2RBs

user 2

user 4

user 1

user 3RBs

user 3

user 2

user 1

user 4RBs

Example for HA+BSA for the pairing and resource allocation ofNuser = 4 users and Nu = 2 users per group.



FD Resource Allocation – Complexity AnalysisFD Resource Allocation – Complexity Analysis

Number of computations of the capacity/BER for the HA+BSAwith Nit iterations

QHA+BSA =Nit ·(

1Nu

2N2user + (N2

user −Nu ·Nuser))

=Nit ·((

1Nu

2 + 1)N2

user −Nu ·Nuser

).



FD Resource Allocation – Complexity AnalysisFD Resource Allocation – Complexity Analysis

HA+BSAFull Search

Q−→

Nuser −→

Nit

8 10 12 14 16 18 20 22 24100

102

104

106

108

1010

1012

1014

1016

Complexity Q of the full search and of the HA+BSA for Nu = 4and Nit ∈ {1, 2, 5, 10, 20}, respectively.


Numerical ResultsNumerical Results

Numerical Results INumerical Results I

BERCapRandomMMSE LESIC equalizer

BLER

−→

10 · log10(Eb/N0) [dB] −→

MMSE LE

SIC

0 2 4 6 8 10 1210−3

10−2

10−1

100

BLER versus Eb/N0 for TD resource allocation, Nuser = 20,Nu = 2, different pairing criteria, 10 MHz bandwidth.



Numerical Results IINumerical Results II

BERCapRandom

BLER

−→

Nuser −→2 4 6 8 10 12 14 16 18 20

10−3

10−2

10−1

100

BLER versus Nuser for TD resource allocation, Eb/N0 = 8 dB,Nu = 2, different pairing criteria, MMSE LE, 10 MHz bandwidth.



Numerical Results IIINumerical Results III

HA+BSA BERFull Search BERHA+BSA CapFull Search CapRandom

BLER

−→

10 · log10(Eb/N0) [dB] −→0 2 4 6 8 10 12

10−3

10−2

10−1

100

BLER versus Eb/N0 for FD res. alloc., Nuser = 10, Nu = 2,Nit = 20, different pairing criteria, MMSE LE, 10 MHz bandwidth.



Numerical Results IVNumerical Results IV

HA+BSA BERFull search BERHA+BSA CapFull search CapRandom

BLER

−→

10 · log10(Eb/N0) [dB] −→−4 −2 0 2 4 6 8

10−3

10−2

10−1

100

BLER versus Eb/N0 for FD resource allocation, Nuser = 8,Nu = 4, Nit = 10, MMSE LE, 20 MHz bandwidth.



Numerical Results VNumerical Results V

HA+BSA BER Nit = 20

HA+BSA BER Nit = 10

HA+BSA Cap Nit = 20

HA+BSA Cap Nit = 10

Random

BLER

−→

10 · log10(Eb/N0) [dB] −→−4 −2 0 2 4 6 8

10−3

10−2

10−1

100

BLER versus Eb/N0 for FD resource allocation, Nuser = 16,Nu = 4, MMSE LE, 20 MHz bandwidth.



Numerical Results VINumerical Results VI

Nit = 50

Nit = 40

Nit = 30

Nit = 20

Nit = 10

Nit = 5

Nit = 2

Nit = 1Random

BLER

−→

10 · log10(Eb/N0) [dB] −→

Nit

0 2 4 6 8 10 1210−3

10−2

10−1

100

BLER versus Eb/N0 for FD resource allocation, Nuser = 20,Nu = 2, BER grouping, MMSE LE, 20 MHz bandwidth.


ConclusionsConclusions

SummarySummary

Random user grouping and resource allocations wastes hugepotential

Hungarian Algorithm finds optimum solution for twodimensional time domain resource allocation and user pairingproblem

Full search finds optimum solution for three dimensionalfrequency domain resource allocation and user groupingproblem with extremely high complexity

Proposed iterative HA+BSA finds close to optimum solutionfor frequency domain resource allocation and user groupingproblem at much lower complexity


ConclusionsConclusions

Future WorkFuture Work

Lower bound for capacity pairing (orthogonality in randomGaussian channels, famous balls-and-bins problem)

Power allocation not yet implemented

Rate adaptation not considered

4× 4 V-MIMO with two transmit antennas per mobile

⇒ Use of network simulator necessary




Wolfgang H. Gerstacker and Michael A. Ruder

University of Erlangen-NürnbergChair of Mobile Communications

Eurecom September 14, 2011

joint user pairing and resource allocation for multiuser sc-fdma

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