# Chunk-Based Energy Efficient Resource Allocation in OFDMA Systems

Authors: Yong Li; Zhengguang Zheng; Mingyue Zhao; Shan Jin; Zhangqin Huang
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

Energy efficiency (EE) capacity analysis of the chunk-based resource allocation is presented
by considering the minimum spectrum efficiency (SE) constraint in downlink multiuser orthogonal frequency
division multiplexing (OFDM) systems. Considering the minimum SE requirement, an optimization problem
to maximize EE with limited transmit power is formulated over frequency selective channels. Based on this
model, a low-complexity energy efficient resource allocation is proposed. The effects of system parameters,
such as the average channel gain-to-noise ratio (CNR) and the number of subcarriers per chunk, are evaluated.
Numerical results demonstrate the effectiveness of the proposed scheme for balancing the EE and SE.


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## Body

## 1. Introduction

The unprecedented expansion of high data rate wireless networks has triggered steadily increase in energy consumption and left a significant environmental footprint. It is reported that the energy usage for radio access can account for up to more than 70 percent of the total energy bill for a mobile operator [1, 2]. Thus, making information and communication technologies (ICT) applications greener can not only reduce the greenhouse gas emissions but also help operators attain long-term profitability.Orthogonal frequency division multiplexing access (OFDMA) has become a promising technique due to its improved immunity to fast fading and flexibility in resource allocation [3]. Most of the research on resource allocation in OFDMA systems assumed single subcarrier allocation, where water-filling power distribution algorithm can be used [3–5]. However, the single-subcarrier-based resource allocation schemes need comparative more overhead and complicated implementation when employed with large number of subcarriers [6]. In order to reduce the overhead and complexity, the correlation between adjacent subcarriers should be considered. A set of contiguous subcarriers are grouped into one chunk and a chunk is regarded as the minimum unit for bandwidth allocation to users.In previous works, chunk-based resource allocation schemes for OFDMA systems mainly intend to increase the system capacity or spectrum efficiency (SE) [6–8]. Reference [6] gives the optimal chunk allocation that maximizes a utility function of average user rates for a wireless OFDMA system under different power control policies. Reference [7] presents the performance analysis of the chunk-based allocation for downlink multiuser uncoded OFDMA systems with adaptive modulation. Reference [8] studies the joint chunk and power and bit allocation schemes to maximize the system capacity under a total transmit power constraint.Besides the capacity improvement, the energy efficiency (EE) has been becoming equally or even more important than SE for green radio. Energy efficient design has received much attention from both industry and academia [1]. The EE optimization for single-subcarrier-based OFDMA systems has been investigated in [9–11]. Reference [9] addresses the energy efficient link adaptation to achieve the maximum EE for frequency-selective fading channels. Reference [10] studies the EE maximization subject to the minimum rate constraint. Reference [11] considers the EE optimization of OFDMA systems while ensuring the minimum rate requirement for each user. However, the single-subcarrier-based resource allocation is assumed in all of them, which incurs additional energy consumption due to its high implementation complexity.The above discussions motivate us to investigate the EE optimization for chunk-based OFDMA systems. In this paper, we formulate an optimization problem to balance the tradeoff between SE and EE. The objective function is the EE which is measured by instantaneous “bits-per-Joule.” The SE is assured by imposing a required minimum data rate into the optimization problem. This paper proposes a low-complexity algorithm to obtain high EE for each channel realization. Furthermore, for a fixed chunk assignment, an optimal power allocation with closed-form expression is also derived.The rest of this paper is organized as follows. In Section2, the system model and optimization objective are described. In Section 3, the chunk-based resource allocation scheme consisting of power allocation and chunk scheduling are derived. Section 4 provides numerical results. Finally, conclusions are drawn in Section 5.


## 2. System Model

Consider a single cellular downlink OFDMA system consisting ofK active users served by one base station. The system bandwidth B is divided into M chunks, each with Q orthogonal narrowband subcarriers. Let 𝒦={1,2,…,K} and ℳ={1,2,…,M} denote the sets of active users and chunks, respectively. In order to avoid interference among different users, one chunk is exclusively assigned to at most one user. Assume that each subcarrier is a Rayleigh fading channel which introduces a zero-mean circularly symmetric complex Gaussian noise.The broadband channel is frequency selective and the normalized correlation coefficient betweenn1th subcarrier and n2th subcarrier of any one user can be written as [6]
(1)ρn1,n2=E{hk,n1*hk,n2}=11+((n1-n2)W/fc)2,
where (·)* denotes complex conjugate, fc is the channel coherence bandwidth, and W is the bandwidth of each subcarrier. In this paper, perfect instantaneous channel state information is assumed to be available at the base station. Let hk,m,q denote the frequency response for the kth user at the qth subcarrier of chunk m.Letℳk be the index set of chunks assigned to user k. Denote pk,m,q as the transmit power of user k on subcarrier q of chunk m; then, the achievable data rate and total transmit power of user k are
(2)Rk=∑m∈ℳk‍∑q=1Q‍Wf(pk,m,q,gk,m,q),Pk=∑m∈ℳk‍∑q=1Q‍pk,m,q,
for k∈𝒦, where gk,m,q=|hk,m,q|2/σ2 is the channel gain-to-noise, ratio (CNR), σ2 is the power of additive Gaussian noise and f(pk,m,q,gk,m,q)=log2(1+pk,m,qgk,m,q).To guarantee the quality-of-service (QoS), the overall data rateR should be restricted by
(3)R=∑k=1K‍Rk≥Rˇ,
where Rˇ is the minimum rate requirement.According to [11], EE is defined as the number of transmitted bits per unit energy consumption; thus, the EE can be given as
(4)U=RαP+Pc,
where α is the transmit power factor and the circuit power Pc is
(5)Pc=Ps+βR,
where Ps is the static circuit power and β is the dynamic circuit power per unit data rate.For a practical system, the total transmit powerP is limited, and we have
(6)P=∑k=1K‍Pk≤P^,
where P^ is the maximum allowable total transmit power.Our objective is to optimize the chunk and power allocation in order to obtain the maximum EE under the required minimum rate constraint. Hence, the optimization problem can be formulated as(7)maxℳk∈ℳ,pk,m,qU=RαPt+Pc(8)s.t.ℳk⋂ℳk′=∅,∀k≠k′,(9)⋃k=1K‍ℳk⊆ℳ,(10)P≤P^,(11)R≥Rˇ.


## 3. Resource Allocation

The optimization problem in ((7), (8), (9), (10), and (11)) is a mixed integer programming problem with nonlinear constraints. It is generally very hard to solve since the feasible set is not convex. Ideally, chunk and transmit power should be allocated jointly to achieve the optimal solution. There are KM possible chunk allocations with K users and M chunks. The maximum EE over all these chunk allocation schemes is the global maximum, and the corresponding chunk and power allocation is the optimal resource allocation scheme. However, the computational complexity is too high. In order to reduce complexity, we can divide the solving process into two steps: chunk allocation and power distribution. In this two-step method, we first consider chunk allocation then the power distribution. Section 3.1 describes a chunk allocation and the latter subsection presents the optimal power distribution for a fixed chunk allocation.


### 3.1. Chunk Allocation

For any given maximum total transmit powerP- without the required minimum rate constraint, the equivalent problem is a sum-rate maximization problem. Let us write the Lagrangian function
(12)L(pk,m,q,λ)=∑k=1K‍∑m∈ℳk‍∑q∈m‍f(pk,m,q,gk,m,q)+λ(P--∑k=1K‍∑m∈ℳk‍∑q∈m‍pk,m,q),
where λ is the Lagrange multiplier.The dual problem is defined as(13)θ=minλ≥0L(λ),
where the dual objective is given by
(14)θ(λ)=maxpk,m,qL(pk,m,q,λ).The optimal power distribution should maximizeθ(λ), for all k, m, q, for all λ. More specifically speaking, the kmth user which maximize θ(λ) will be selected as the winner on chunk m in this iteration
(15)km=argmaxk∈𝒦∑q∈m‍(f(pk,m,q,gk,m,q)-λpk,m,q).The optimal transmit power for userk on subcarrier q in chunk m is
(16)pk,m,q={max(μ-1gk,m,q,0),ifk=km,0,otherwise,
where the power level is μ=1/λln2.The core principle of the chunk allocation scheme is to assign the chunk with high capacity as much as possible for each user. Since both the throughput and the EE are nondecreasing with the number of subcarriers. In order to achieve high EE, we can setP-=P^ and obtain the optimal chunk scheduling by iteratively exchanging information and updating the Lagrange multiplier. The goal of maximizing EE while ensuring a required minimum data rate is achieved by the power distribution in the next subsection.


### 3.2. Power Distribution

According to the above discussion, the data rate is maximized when the power is distributed among the subcarriers using water-filling algorithm. For a given chunk allocation, the optimal power distribution and system throughput continuously change as the power level varies. Then, the optimization problem can be reformulated as(17)maxpk,m,qU=RαPt+Pc(18)s.t.P≤P^,(19)R≥Rˇ.It can be seen that the system data rateR is a strictly increasing function of power level μ for μ≥0. Let {gn}n=1N, (N=MQ) be the sequence of {gkm,m,q}, for all m, q, in descending order and let pn be the corresponding power.Lemma 1.
The minimum power levelμˇ that satisfies R=Rˇ is
(20)μˇ=min1≤I≤N2Rˇ/IW(∏n=1I‍gn)-(1/I).Proof.
Please see AppendixA.Lemma 2.
The maximum power levelμ^ to maximize system rate under the total transmit power P^ constraint is
(21)μ^=min1≤I≤N1I(P^+∑n=1I‍1gn).Proof.
Please see AppendixB.Since we assume that the required minimum data rate can be achieved under the constraint (6), then μˇ≤μ^ and the problem ((17), (18), and (19)) can be rewritten as
(22)maxμU=RαPt+Pc(23)s.t.μˇ≤μ≤μ^.The primal variable of problem ((22), (23)) is one dimension, and the direct way to solve this problem is the brute search method over the region [μˇ,μ^]. However, brute search has high complexity to achieve exact solution which poses a prohibitive computational burden at the base station. In order to solve this problem efficiently, we can further analyze the relationship between EE and power level.Lemma 3.
To a certain determined chunk allocation, the EE is a quasiconcave function of power levelμ.Proof.
Please see AppendixC.According to Lemma3, EE U is a quasiconcave function of power level μ; hence, the optimal unconstrained power level always exists. Differentiating (22) with respect to μ and setting the derivative to zero, we can obtain
(24)∂U∂μ=F(1+PsαN~μ-G1μ-ln(μG2))=0
or
(25)PsG2αN~e-G1G2e=μG2eln(μG2e),
where e is the base of the natural logarithm, F is a positive variable, N~ is the number of subcarriers with transmit power pn>0, G1, and G2 are defined as
(26)G1=1N~∑n=1N~‍1gn,G2=(∏n=1N~‍gn)1/N~,
respectively. Let X=ln(μG2/e) and let Y=(PsG2/αN~e)-(G1G2/e); then, (25) can be expressed as
(27)XeX=Y.
Its solution is X=W0(Y), where W0(·) denotes the real branch of the Lambert function [12]. Substituting this solution into (27), the optimal unconstrained power level μ~ without any constraint is
(28)μ~=eG2exp[W0(PsG2αN~e-G1G2e)].Figure1 illustrates three possible curves of the relation between EE and power level that satisfy all constraints. So the optimal power level of the problem ((17), (18), and (19)) is
(29)μ*=min(max(μ~,μˇ),μ^).Figure 1
EE-power level relation in downlink OFDMA.


## 3.1. Chunk Allocation

For any given maximum total transmit powerP- without the required minimum rate constraint, the equivalent problem is a sum-rate maximization problem. Let us write the Lagrangian function
(12)L(pk,m,q,λ)=∑k=1K‍∑m∈ℳk‍∑q∈m‍f(pk,m,q,gk,m,q)+λ(P--∑k=1K‍∑m∈ℳk‍∑q∈m‍pk,m,q),
where λ is the Lagrange multiplier.The dual problem is defined as(13)θ=minλ≥0L(λ),
where the dual objective is given by
(14)θ(λ)=maxpk,m,qL(pk,m,q,λ).The optimal power distribution should maximizeθ(λ), for all k, m, q, for all λ. More specifically speaking, the kmth user which maximize θ(λ) will be selected as the winner on chunk m in this iteration
(15)km=argmaxk∈𝒦∑q∈m‍(f(pk,m,q,gk,m,q)-λpk,m,q).The optimal transmit power for userk on subcarrier q in chunk m is
(16)pk,m,q={max(μ-1gk,m,q,0),ifk=km,0,otherwise,
where the power level is μ=1/λln2.The core principle of the chunk allocation scheme is to assign the chunk with high capacity as much as possible for each user. Since both the throughput and the EE are nondecreasing with the number of subcarriers. In order to achieve high EE, we can setP-=P^ and obtain the optimal chunk scheduling by iteratively exchanging information and updating the Lagrange multiplier. The goal of maximizing EE while ensuring a required minimum data rate is achieved by the power distribution in the next subsection.


## 3.2. Power Distribution

According to the above discussion, the data rate is maximized when the power is distributed among the subcarriers using water-filling algorithm. For a given chunk allocation, the optimal power distribution and system throughput continuously change as the power level varies. Then, the optimization problem can be reformulated as(17)maxpk,m,qU=RαPt+Pc(18)s.t.P≤P^,(19)R≥Rˇ.It can be seen that the system data rateR is a strictly increasing function of power level μ for μ≥0. Let {gn}n=1N, (N=MQ) be the sequence of {gkm,m,q}, for all m, q, in descending order and let pn be the corresponding power.Lemma 1.
The minimum power levelμˇ that satisfies R=Rˇ is
(20)μˇ=min1≤I≤N2Rˇ/IW(∏n=1I‍gn)-(1/I).Proof.
Please see AppendixA.Lemma 2.
The maximum power levelμ^ to maximize system rate under the total transmit power P^ constraint is
(21)μ^=min1≤I≤N1I(P^+∑n=1I‍1gn).Proof.
Please see AppendixB.Since we assume that the required minimum data rate can be achieved under the constraint (6), then μˇ≤μ^ and the problem ((17), (18), and (19)) can be rewritten as
(22)maxμU=RαPt+Pc(23)s.t.μˇ≤μ≤μ^.The primal variable of problem ((22), (23)) is one dimension, and the direct way to solve this problem is the brute search method over the region [μˇ,μ^]. However, brute search has high complexity to achieve exact solution which poses a prohibitive computational burden at the base station. In order to solve this problem efficiently, we can further analyze the relationship between EE and power level.Lemma 3.
To a certain determined chunk allocation, the EE is a quasiconcave function of power levelμ.Proof.
Please see AppendixC.According to Lemma3, EE U is a quasiconcave function of power level μ; hence, the optimal unconstrained power level always exists. Differentiating (22) with respect to μ and setting the derivative to zero, we can obtain
(24)∂U∂μ=F(1+PsαN~μ-G1μ-ln(μG2))=0
or
(25)PsG2αN~e-G1G2e=μG2eln(μG2e),
where e is the base of the natural logarithm, F is a positive variable, N~ is the number of subcarriers with transmit power pn>0, G1, and G2 are defined as
(26)G1=1N~∑n=1N~‍1gn,G2=(∏n=1N~‍gn)1/N~,
respectively. Let X=ln(μG2/e) and let Y=(PsG2/αN~e)-(G1G2/e); then, (25) can be expressed as
(27)XeX=Y.
Its solution is X=W0(Y), where W0(·) denotes the real branch of the Lambert function [12]. Substituting this solution into (27), the optimal unconstrained power level μ~ without any constraint is
(28)μ~=eG2exp[W0(PsG2αN~e-G1G2e)].Figure1 illustrates three possible curves of the relation between EE and power level that satisfy all constraints. So the optimal power level of the problem ((17), (18), and (19)) is
(29)μ*=min(max(μ~,μˇ),μ^).Figure 1
EE-power level relation in downlink OFDMA.


## 4. Numerical Results

In this section, we provide some simulation results to verify the effectiveness of the algorithm proposed in the previous sections. The system parameters are assumed as follows. The number of active users is 6 and the system bandwidth is divided into 256 subcarriers, each with bandwidthW=40 kHz. The channel is frequency selective and independent for all users. Assume that the correlation bandwidth between any two subcarriers of one user is the same for all users and fc=200 kHz. The Rayleigh fading channel can be generated based on the method in [6, 13]. For sake of simplicity, all users have an identical average CNR and the transmit power factor, α, is assumed to be 3.Figure2 shows the performance of EE among the different sizes of each chunk. It can be seen, from Figures 2 and 3, that both EE and SE increase with the average CNR. This is because that energy efficient design tends to use less transmit power when CNR increases. The single-subcarrier-based resource allocation yields the highest system EE and SE. The gap in terms of EE or SE between Q=16 and Q=8 is significant. The gap between Q=8 and Q=4 becomes smaller, and the gap between Q=1 and Q=2 is even negligible. The chunk-based resource allocation has almost the same performance if the size of the chunk is smaller than the coherence bandwidth. The reason is that when the size of chunk decreases, the autocorrelation value among the channels of subcarriers within a chunk increases.Figure 2
System EE versus average CNR in the case thatPs=15 w and β=0.Figure 3
System SE versus average CNR in the case thatPs=15 w and β=0.Figure4 compares and indicates the influence of minimum SE requirement on the relation between EE and SE with different static circuit power and dynamic circuit power factors. Figure 5 shows the SE performance corresponding to the curves in Figure 4. From there, the EE decreases with the circuit power while the SE increases with the static circuit power and is independent of dynamic circuit power factor. From the curves of EE-versus-Rˇ, when the minimum rate requirement is smaller or larger than a certain threshold, energy efficient design leads to same performance. The reason is that in the lower Rˇ regime, the energy-efficient transmission strategy is to operate exactly at the optimal unconstrained point which means μ*=μ~. When Rˇ goes beyond the threshold, EE will decrease approximately linearly with Rˇ because the transmit power dominates the total power consumption and data rate is logarithmic in transmit power. However, when SE is large enough, the EE is flat again. This means that the system has to work with maximum transmit power to reach the required minimum data rate and μ*=μ^.Figure 4
System EE versus average CNR in the case thatCNR=15 dB.Figure 5
System SE versus average CNR in the case thatCNR=15 dB.


## 5. Conclusion

In this paper, the EE-SE relation in a single cell downlink OFDMA system is investigated and the performance analysis of the chunk-based resource allocation is presented with energy-efficient design under the constraints on the minimum SE requirement and total available power. When the number of subcarriers per chunk is given, both EE and SE increase when the coherence bandwidth increase from a very small value. The performance of chunk-based resource allocation is very close to that of single-subcarrier-based allocation scheme when the bandwidth of each chunk is smaller than coherence bandwidth.Simulation results verify the effectiveness of the proposed scheme in practical and demonstrate that it is much more suitable to green communications. Furthermore, the impact of the number of users and minimum SE requirement to the EE are also analyzed.



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