Energy-Efficient Resource Allocation in Multi-Cell OFDMA

Introduction System Model Resource Allocation and Scheduling Optimization Simulation Conclusions

Energy-Efficient Resource Allocation in Multi-CellOFDMA Systems with Limited Backhaul Capacity

Derrick Wing Kwan Ng

University of British Columbia, Canada

Supervisor: Prof. Robert SchoberUniversity Erlangen-Nuremberg, Germany

Derrick Wing Kwan Ng FAU 1/27


Outline

1 IntroductionOFDMA and BS cooperationLimited Backhaul and Energy efficiency

2 System ModelMulti-Cell OFDMA network modelAverage Weighted System ThroughputPerformance Measure - Energy Efficiency

3 Resource Allocation and Scheduling OptimizationOptimization Problem FormulationOptimization Solution -Fractional Programming

4 Simulation

5 Conclusions






4 Simulation

5 ConclusionsDerrick Wing Kwan Ng FAU 3/27


Introduction - OFDMA, BS cooperation, limited backhaul,Energy efficiency

Orthogonal frequency division multiple access (OFDMA)

Flexibility in resource allocation. (Per subcarrier adaptivemodulation)Exploits multi-user diversity by selecting the best user for eachsubcarrier

Cooperative Networks

Basic idea: Single antenna terminals in the multiuser systemcan share their antennas and create a virtual MIMOcommunication system.

1 Base stations cooperate with each other via a backbonefixed-line connection.

Advantages: Reducing the co-channel interference effectivelyvia joint resource allocation.

2 Relaying

Advantages: Low cost implementation with promisingperformance, i.e., coverage extension and diversity gains.





Flexibility in resource allocation. (Per subcarrier adaptivemodulation)

Exploits multi-user diversity by selecting the best user for eachsubcarrier





2 Relaying











2 Relaying











2 Relaying











2 Relaying











2 Relaying











2 Relaying











2 Relaying











2 RelayingAdvantages: Low cost implementation with promisingperformance, i.e., coverage extension and diversity gains.






4 Simulation



OFDMA, BS cooperation, limited backhaul, Energyefficiency

Backhaul Capacity [1][2][3] Ideal backhaul ⇒ unlimited control signals,

users channel information, and precoding data can always be exchanged.

In practice: Limited backhaul capacity. Finite information

Energy Efficiency (Green communications) Combining a few technologies:

Free lunch?

Power consumptions in circuitries, RF, and backhaul areignored.

[1] O. Somekh, O. Simeone, Y. Bar-Ness, A. Haimovich, and S. Shamai, “Cooperative Multicell

Zero-Forcing Beamforming in Cellular Downlink Channels,” IEEE Trans. Inform. Theory, vol. 55, pp. 3206–3219, Jul. 2009.

[2] R. Zhang, “Cooperative Multi-Cell Block Diagonalization with Per-Base-Station Power Constraints,”

IEEE J. Select. Areas Commun., vol. 28, pp. 1435 –1445, Dec. 2010.

[3] G. Dartmann, W. Afzal, X. Gong, and G. Ascheid, “Joint Optimization of Beamforming, User

Scheduling, and Multiple Base Station Assignment in a Multicell Network,” in Proc. IEEE WirelessCommun. and Networking Conf., Mar. 2011, pp. 209 –214.








Free lunch?















Free lunch?













4 Simulation



System Model

Mobile

MobileMobile

MobileMobile

Mobile

MobileMobile

MobileBase Station

Backhaul connection topology

Central Unit

= BS= User= Backhaul connections with limited

capacities for data exchange

= Backhaul connections for control signals

Figure: A multi-cell system with M = 3 cells with a fully connected backhaullink topology.

Universal frequency reuse, i.e., Frequency reuse factor = 1.

Each transceiver is equipped with a single antenna.

All base stations are cooperating with each other.

Full connection backhaul topology.



System Model Base Station

Mobile

Relayed signalOverheard

signal Relay

Base StationBase Station

Backhaul connection

MobileJoint

transmission

Joint transmission

Figure: BSs are connected with backhaul with limited capacities.

The transmitted signal from BS m to all selected users on subcarrier i is given by∑k∈S(i)

xkm(i) =

∑k∈S(i)

wkBm

(i)√

PkBm

(i)uk (i). (1)

The received signal from M BSs at user k on subcarrier i is given by

Y k (i) =

(M∑

c=1

HkBc

(i)wkBc

(i)√

PkBc

(i)lkBc

)uk (i) (2)

+M∑

m=1

∑j∈S(i)

j 6=k

√P j

Bm(i)lk

BmHk

Bm(i)w j

Bm(i)uj (i)

︸︷︷︸Multiple Access Interference

+zk (i),



Instantaneous Channel Capacity: BSs→ User kBase Station

Mobile

Relayed signalOverheard

signal Relay

Base StationBase Station

Backhaul connection

MobileJoint

transmission

Joint transmission

Figure: BSs are connected with backhaul with limited capacities.

Given perfect CSI at the receiver, the channel capacity between all thecooperating BSs and user k on subcarrier i with subcarrier bandwidth B

nFis

given by

C k (i) =BnF

log2

(1 + Γk (i)

), (3)

Γk (i) =

∣∣∣∑Mc=1 Hk

Bc(i)w k

Bc(i)√

PkBc

(i)lkBc

∣∣∣2σ2

z + I k (i), (4)

I k (i) =∑

j∈S(i)j 6=k

∣∣∣ M∑m=1

√P j

Bm(i)w j

Bm(i)√

lkBm

HkBm

(i)∣∣∣2, (5)






4 Simulation



Average weighted system throughput

The weighted system capacity is defined as the total number ofbits successfully delivered to the K mobile users and is given by

U(P,W,S) =M∑

m=1

∑k∈Am

αk

nF∑i=1

sk (i)C k (i),

where P, W, and S are the power, precoding coefficient, andsubcarrier allocation policies, respectively. sk (i) ∈ 0, 1 is thebinary subcarrier allocation variable.






4 Simulation



Energy Efficiency

Power dissipation in the system:

UTP (P,W,S) = PC ×M︸︷︷︸Circuits power consumption

+ δ × PBH︸︷︷︸Backhauls power consumption

+M∑

m=1

K∑k=1

nF∑i=1

εPkBm

(i)|wkBm

(i)|2sk (i)

︸︷︷︸Power consumption in the RF PAs

, (6)

The objective function, energy efficiency (bit/Joule), is given by

Ueff (P,W,S) =U(P,W,S)

UTP (P,W,S).






4 Simulation



Problem Formulation

Problem (Optimization Problem Formulation)

maxP,W,S

Ueff (P,W,S)

s.t. C1:K∑

k=1

nF∑i=1

|wkBm

(i)|2PkBm

(i)sk (i) ≤ PTm , ∀m, [Max. transmit power per BS]

C2:M∑

m=1

K∑k∈Am

nF∑i=1

sk (i)C k (i) ≥ Rmin, [Min. data date requirement]

C3:K∑

k∈Am

nF∑i=1

sk (i)C k (i) ≤ Rmaxm = minRBm1,RBm2

, . . . ,RBmNm, ∀m,

[Backhaul capacity limit]

C4:K∑

k=1

sk (i) ≤ M, ∀i , C5: sk (i) = 0, 1,∀i , k, [Subcarrier constraint]

C6: PkBm

(i) ≥ 0, ∀i , k,m, [Power constraint]






4 Simulation



Objective Function Transformation

Without loss of generality, we define the maximum energy efficiency q∗ of theconsidered system as

q∗ =U(P∗,W∗,S∗)

UTP (P∗,W∗,S∗) = maxP,W,S

U(P,W,S)

UTP (P,W,S). (7)

Theorem

The maximum energy efficiency q∗ is achieved if and only if

maxP,W,S

U(P,W,S)− q∗UTP (P,W,S)

= U(P∗,W∗,S∗)− q∗UTP (P∗,W∗,S∗) = 0, (8)

for U(P,W,S) ≥ 0 and UTP (P,W,S) > 0.



Equivalent Objective Function

maxP,R,S

Usec (P,R,S)− q∗UTP(P,R,S)

s.t. C1, C2, C3, C4, C5, C6.

Questions:

How to find the optimal q∗?How to solve the above optimization?



Dinkelbach Method

Table: Iterative Resource Allocation Algorithm.

Algorithm 1 Iterative Resource Allocation Algorithm

1: Initialize the maximum number of iterations Lmax and the maximum tolerance ε2: Set maximum energy efficiency q = 0 and iteration index n = 03: repeat Main Loop4: Solve the inner loop problem for a given q and obtain resource allocation policies

P ′,W ′,S′5: if U(P ′,W ′,S′)− qUTP (P ′,W ′,S′) < ε then6: Convergence = true

7: return P∗,W∗,S∗ = P ′,W ′,S′ and q∗ = U(P′,W′,S′)UTP (P′,W′,S′)

8: else9: Set q = U(P′,W′,S′)

UTP (P′,W′,S′)and n = n + 1

10: Convergence = false11: end if12: until Convergence = true or n = Lmax



Power Allocation Optimization

Based on the Dinkelbach iterative algorithm, we solve the following non-convexoptimization problem for a given parameter q:

maxP,W,S

U(P,W,S)− qUTP (P,W,S)

s.t. C1, C2, C3, C4, C5, C6.

Yet,

Joint optimization of precoding vector, power allocation, and subcarrierallocation ⇒ Non-convex problem ⇒ Computational infeasible for number ofvariables.

Precoding scheme: Zero-forcing beamforming.

Subcarrier allocation: Semi-orthogonal user selection.

Power allocation optimization is still non-convex for a given fixed precodingscheme and subcarrier allocation.





maxP,W,S


s.t. C1, C2, C3, C4, C5, C6.

Yet,









maxP,W,S


s.t. C1, C2, C3, C4, C5, C6.

Yet,









maxP,W,S


s.t. C1, C2, C3, C4, C5, C6.

Yet,







Theorem

Let P and D denote the optimal values of the primal and the dualproblem, respectively. For a given selected user set and ZFBFtransmission, if the number of subcarriers is sufficiently large, thenstrong duality holds and the duality gap is zero, i.e., P = D.



Primal Problem, Perturbation Function, Dual Problem

Without loss of generality, the primal optimization problem can bewritten in general form as

P = maxpk

i ≥0

nF∑i=1

fi 〈pki 〉

s.t.

nF∑i=1

gi 〈pki 〉 ≤ 0. (9)

Then, the perturbation function v〈y〉 and perturbation vector y aredefined as:

v〈y〉 = maxpk

i ≥0

nF∑i=1

fi 〈pki 〉

s.t.

nF∑i=1

gi 〈pki 〉 ≤ y. (10)



Primal Problem, Perturbation Function, Dual Problem

The Lagrangian function of (9) can be expressed as

L〈pki ,u〉 =

nF∑i=1

fi 〈pki 〉 − uT

(gi 〈pk

i 〉)

(11)

where u ∈ RL,u ≥ 0 is a vector of Lagrange multipliers. Thus, thecorresponding dual problem is given by

D = minu≥0

maxpk

i

L〈pki ,u〉. (12)



Geometric Interpretation of Perturbation Function and Dual Problem

Hyperplane with suboptimal u

y

Hyperplane with optimal u

P

Perturbation functionv(y)

Dual optimal = Primal optimal

Figure: Zero duality gap when maximizing a concave problem.



Geometric Interpretation of Duality Gap and Perturbation Function

y

P

Perturbation functionv(y)

Hyperplane with optimal uDual optimal

Primal optimal

Duality gap

Figure: Non-Zero duality gap when maximizing a concave problem.



Zero Duality Gap

Theorem

If the perturbation function v〈y〉 is a concave function of y, thenthe duality gap is zero despite the convexity of the primal problem,i.e., D = P.

We need to prove that v〈y〉 is a concave function of y, i.e.,

v〈ρy + (1− ρ)x〉 ≥ ρv〈y〉+ (1− ρ)v〈x〉 (13)

for 0 ≤ ρ ≤ 1, where x ∈ RL is another perturbation vector suchthat x− y 6= 0.How to apply this result to our problem?

Frequency sharing (Large number of subcarriers)

Frequency sharing in multicarrier systems ⇒ Concave perturbationfunction ⇒ Zero Duality gap ⇒ Dual Optimal = Primal Optimal



Layer 2 Master Problem - minimization of Dual problem w.r.t

Layer 11 st sub-problem

Layer 12 nd sub-problem

Layer 1th sub-problem

Fn

,μ γGradient update on

,μ γ

[ ] [ ]k F k FP sn n

[2] [2]k kP s

[1] [1]k kP s

,μ γ,μ γ

,μ γ

Figure: An illustration of dual decomposition.

Dual Problem → Dual Decomposition

D = minλ,β, γ≥0

maxP

L(λ,β, γ,P).

L(λ,β, γ,P) =M∑

m=1

nF∑i=1

K∑k∈Am∩S⊥(i)

(αk + γ − βm)C k (i)− γRmin +M∑

m=1

βmRmaxm

−M∑

m=1

λm

( nF∑i=1

K∑k∈S⊥(i)

|wkBm

(i)|2PkBm

(i)− PTm

)

− q( M∑

m=1


nF∑i=1

εPkBm

(i)|wkBm

(i)|2 + δPBH + PC × M),



Dual Decomposition Solution in each interaction

Solution of Layer 1 Sub-Problem: The closed-form power allocation for the BSs serving user k in subcarrier i for agiven parameter q is obtained as

PkBm

(i) =

[B/nF (αk + γ − βm)

ln(2)Ωk (i)−

σ2z

|γk (i)|2

]+

and PkBa

(i) = PkBm

(i) ∀a 6= m, where Ωk (i) =( M∑

c=1

(λc + qε)|wkBc

(i)|2).

Solution of Layer 2 Master Problem: The dual function is differentiable and, hence, the gradient method can beused to solve the Layer 2 master problem which leads to

λm(c + 1) =[λm(c)− ξ1(c)×

(PTm −

nF∑i=1

K∑k∈S⊥(i)

|wkBm

(i)|2PkBm

(i))]+, ∀m,

γ(c + 1) =[γ(c)− ξ2(c)×

( M∑m=1

nF∑i=1


C k (i)− Rmin

)]+,

βm(c + 1) =[βm(c)− ξ3(c)×

(Rmaxm −

nF∑i=1


C k (i))]+

, ∀m.



Results – Duality Gap in Non-Convex Optimization Problem

10 20 30 40 50 600

0.5

1

1.5

2

2.5

3x 10

−7

PT (dBm)

Dua

lity

gap

R

maxm

= R1

Rmax

m

= R2

Rmax

m

= R3

Figure: Duality gap versus the maximum transmit power allowance at each BS, PT ,for different backhaul capacities Rmaxm .



Results – Energy Efficiency for Different Limited Backhaul Capacities

10 15 20 25 30 35 40 45 50 550

0.5

1

1.5

2

2.5

3

3.5

4

4.5x 10

5

PT (dBm)

Ene

rgy

effic

ienc

y (b

it/Jo

ule)

Proposed algorithm and baseline

Rmax

m

= 11 Mb/s

Baseline R

maxm

=34 Mb/s,

Rmax

m

= 44Mb/s,

Proposed algorithm, R

maxm

=34 Mb/s,

Rmax

m

= 44Mb/s,

Figure: Energy efficiency (bit-per-Joule) versus the maximum transmit powerallowance at each BS PT . Baseline scheme is a resource allocator which maximizesthe system throughput.



Results – Average Capacity for Different Limited Backhaul Capacities

10 15 20 25 30 35 40 45 50 556

8

10

12

14

16

18

20

22

24

PT (dBm)

Ave

rage

sys

tem

cap

acity

(bi

t/s/H

z/ce

ll)

Baseline

Rmax

m

= R2, R

maxm

= R3

Proposed algorithm and baseline

Rmax

m

= R1

Proposed algorithm, Rmax

m

= R2,

Rmax

m

= R3

Figure: Average outage capacity (bit/s/Hz) versus the maximum transmit powerallowance at each BS, PT , for different resource allocation algorithms and differentbackhaul capacities with K = 45 users. Baseline scheme is a resource allocators whichmaximizes the system throughput.



Conclusions

The resource allocation algorithm design for BS cooperativeOFDMA systems was formulated as a non-convexoptimization problem.

Power allocation was optimized for energy efficiencymaximization for the non-convex optimization problem.

We showed that when the number of subcarriers is sufficientlylarge, the duality gap is practically zero despite thenon-convexity of the primal problem.

An efficient power allocation was obtained by solving the dualproblem.

Simulation results demonstrated the trade-off betweenbackhaul capacity and energy efficiency.



Conclusions








Conclusions








Conclusions








Conclusions








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Energy-Efficient Resource Allocation in Multi-Cell OFDMA