Resource Allocation in Heterogenous Full-duplex OFDMA Networks: Design and Analysis

# Resource Allocation in Heterogenous Full-duplex OFDMA Networks: Design and Analysis

Peyman Tehrani1, Farshad Lahouti2, Michele Zorzi3

1 Computer Science Department, University of California, Irvine, USA 2Electrical Engineering Department, California Institute of Technology, USA 3Department of Information Engineering, University of Padova, Italy
Emails: peymant@uci.edu, lahouti@caltech.edu, zorzi@dei.unipd.it
###### Abstract

Recent studies indicate the feasibility of full-duplex (FD) bidirectional wireless communications. Due to its potential to increase the capacity, analyzing the performance of a cellular network that contains full-duplex devices is crucial. In this paper, we consider maximizing the weighted sum-rate of downlink and uplink of an FD heterogeneous OFDMA network where each cell consists of an imperfect FD base-station (BS) and a mixture of half-duplex and imperfect full-duplex mobile users. To this end, first, the joint problem of sub-channel assignment and power allocation for a single cell network is investigated. Then, the proposed algorithms are extended for solving the optimization problem for an FD heterogeneous network in which intra-cell and inter-cell interferences are taken into account. Simulation results demonstrate that in a single cell network, when all the users and the BSs are perfect FD nodes, the network throughput could be doubled. Otherwise, the performance improvement is limited by the inter-cell interference, inter-node interference, and self-interference. We also investigate the effect of the percentage of FD users on the network performance in both indoor and outdoor scenarios, and analyze the effect of the self-interference cancellation capability of the FD nodes on the network performance.

Index Terms: Full-duplex, self-interference, resource allocation, OFDMA, femto cell, heterogeneous.

## I Introduction

This paper has been presented in part at the IEEE International Conference on Communication (ICC), Kuala Lumpur, Malaysia, May 2016.

In wireless communications, separation of transmission and reception in time or frequency has been the standard practice so far. However, through simultaneous transmission and reception in the same frequency band, wireless full-duplex has the potential to double the spectral efficiency. Due to this substantial gain, full-duplex technology has recently attracted noticeable interest in both academic and industrial worlds. The main challenge in full-duplex (FD) bidirectional communication is self-interference (SI) cancellation. In recent years, many attempts have been made to cancel the self-interference signal [1, 2, 3, 4]. In [5], it is shown that dB SI cancellation is achievable, and by jointly exploiting analog and digital techniques, SI may be reduced to the noise floor.

A full-duplex physical layer in cellular communications calls for a re-design of higher layers of the protocol stack, including scheduling and resource allocation algorithms. In [6], the performance of an FD-based cellular system is investigated and an analytic model to derive the average uplink and downlink channel rate is provided. A resource allocation problem for an FD heterogeneous orthogonal frequency-division multiple access (OFDMA) network is considered in [7], in which the macro base station (BS) and small cell access points operate in either FD or half-duplex (HD) MIMO mode, and all mobile nodes operate in HD single antenna mode. In [8], using matching theory, a sub-channel allocation algorithm for an FD OFDMA network is proposed. In both [7] and [8] only a single sub-channel is assigned to each of the uplink users in which they transmit with constant power. Resource allocation solutions are proposed in [9] and [10] for FD OFDMA networks with perfect FD nodes (SI is canceled perfectly).

Recent research reports investigate resource allocation in multi-cell FD networks. In [11], a sub-optimal resource management algorithm is presented for the sum rate maximization of a small multi-cell system, including FD base stations and HD mobile users. In [12], the problem of maximizing a network-wide rate-based utility function subject to uplink (UL) and downlink (DL) power constraints is studied in a flexible duplex system, in which UL/DL channels are allowed to have partial overlap via fine-tuned bandwidth allocation. For simplicity, it is assumed that the number of sub-channels and the users are exactly the same. In [13], the problem of decoupled UL-DL user association, which allows users to associate with different BSs for UL and DL transmissions, is investigated in a multi-tier FD Network. In [14], weighted sum rate maximization in a FD multi-user multi-cell MIMO network is studied. A user scheduling and power allocation method for ultra-dense FD small-cell networks is presented in [15]. In [13] [14] and [15], the sub-channel allocation problem is not investigated since a single channel network is assumed. The most related work to the current research is [16], in which, a radio resource management solution for an OFDMA FD heterogeneous cellular network is presented. The algorithm jointly assigns the transmission mode, and the user(s) and their transmit power levels for each frequency resource block to optimize the sum of the downlink and uplink rates. The users are assumed to use a single class of service. A sub-optimal resource allocation algorithm is then proposed which takes into account both intra-cell and inter-cell interferences. The sub-optimal power adjustment algorithm is designed under the assumption of high SINR, where the rate of an FD-FD or FD-HD link is independent of power variations.

In this paper, we consider a general resource allocation problem in a heterogeneous OFDMA-based network consisting of imperfect FD macro BS and femto BSs and both HD and imperfect FD users. We aim to maximize the downlink and uplink weighted sum-rate of femto users while protecting the macro users rates. The weights allow for users to utilize differentiated classes of service, accommodate both frequency or time division duplex for HD users, and prioritize uplink or downlink transmissions. To be more realistic, imperfect SI cancellation in FD devices is assumed and FD nodes suffer from their SI. A contribution of the current work is to consider the presence of a mixture of FD and HD users, which enables us to quantify the percentage of FD users needed to capture the full potential of FD technology in wireless OFDMA networks. We also analyze the effect of the SI cancellation level on the network performance, which to our knowledge has not been studied in prior works. We will show that when the SI cancellation capability is worse than a specified threshold, then the throughput of an all FD user network would not be larger than the throughput of an all HD user network. Moreover, we will analyze this threshold theoretically and compare its outcome with simulation results.

The remainder of this paper is organized as follows. In Section II, the basic system model of a single cell FD network is given and the optimization problem is formulated. In Section III, a sub-channel allocation algorithm for selecting the best pair in each sub-channel is presented. Power allocation is considered in Section IV. A theoretical approach for deriving the SI cancellation coefficient threshold is proposed in Section V. In Section VI, the optimization problem for an FD heterogeneous network is presented. Numerical results for the proposed methods are shown in Section VII. Finally, the paper is concluded in Section VIII.

## Ii System Model And Problem Statement

We consider a single cell network that consists of a full-duplex base-station (BS) and a total of half-duplex and full-duplex users. For communications between the nodes and the BS, we assume that an OFDMA system with sub-channels is used. All sub-carriers are assumed to be perfectly synchronized, and so there is no interference between different sub-channels. Since the base-station operates in full-duplex mode, it can transmit and receive simultaneously in each sub-channel. In each timeslot the base-station is to properly allocate the sub-channels to the downlink or uplink of appropriate users and also determine the associated transmission power in an optimized manner. We assume that the base-station and the FD users are imperfect full-duplex nodes that suffer from self-interference. We define a self-interference cancellation coefficient to take this into account in our model and denote it by , where indicates that SI is canceled perfectly and means no SI cancellation. For simplicity, we assume the same self-interference cancellation coefficient for BS and FD users, but consideration of different coefficients would be possible. In this paper, the goal is to maximize the weighted sum-rate of downlink and uplink users with a total power constraint at the base-station and a transmission power constraint for each user.

We define the downlink weighted sum-rate as

 Rd=K∑k=1∑n∈Sk,dwklog(1+gk(n)pk,d(n)Nk+Ik,j(n)pj,u(n))\vspace−0mm (1)

And the uplink weighted sum-rate as

 Ru=K∑j=1∑n∈Sj,uvjlog(1+gj(n)pj,u(n)N0+βpk,d(n)) (2)

The variables used in the above equations are introduced in Table I. We assume here that the channel is reciprocal, i.e., uplink and downlink channel gains are the same. We further assume that the receiver noise powers in different sub-channels are the same. The term in (1) denotes the interference: When user is a FD device and both downlink and uplink of sub-channel are allocated to it , , else is the channel gain between uplink user and downlink user . We assume that the base-station knows all the channel gains, the noise powers, and the SI cancellation coefficient and weights assigned to the downlink and uplink of all users.

Let and denote the maximum available transmit power for the base-station and for user , respectively. Then the proposed design optimization problem, denoted by P1, can be formulated as follows

 {P}1:maximizepk,d,pj,u,Sj,u,Sk,d,∀k,jRd+Ru (3)
 subject to K∑k=1∑n∈Sk,dpk,d(n)≤P0 (4)
 ∑n∈Sj,upj,u(n)≤Pj∀j (5)
 pj,u(n),pk,d(n)≥0∀j,k,n (6)
 Si,d∩Sj,d=ϕ,Si,u∩Sj,u=ϕ∀i≠j (7)
 ∪Kj=1 Sj,u⊆{1,2,...,N},∪Kk=1 Sk,d⊆{1,2,...,N} (8)
 Sk,u∩Sk,d=ϕif user k is HD (9)

where (4) and (5) indicate the power constraint on the BS and the users, respectively. Constraint (6) shows the non-negativity feature of powers; (7) come from the fact that a sub-channel cannot be allocated to two distinct users simultaneously; (8) indicate that we have no more than sub-channels, and the last constraint accounts for the half-duplex nature of the HD users.

The general resource allocation problem presented is combinatorial in nature because of the channel allocation issue and addressing it together with power allocation in an optimal manner is challenging, especially as the number of users and sub-channels grow. Moreover, the non-convexity of the rate function makes the power allocation problem itself challenging even for a fixed sub-channel assignment. Here, we invoke a two step approximate solution. First, we determine the allocation of downlink and uplink sub-channels to users and then determine the transmit power of the users and the base-station on their allocated sub-channels. In other words, we first specify the sets and and then determine the variables , . In the next Section, we introduce our sub-channel allocation algorithm.

## Iii Sub-channel Allocation

The sub-channel allocation problem, denoted by P2, can be formulated as follows

 {P}2: maximizeSj,u,Sk,d Rd+Ru subject to (7)-(11)

To solve the problem P2, we should first solve the following power allocation problem, denoted by P3, to maximize the weighted sum-rate in a single sub-channel and for a fixed pair of uplink and downlink users. Since a single sub-channel is being considered in P3, we have dropped the variable in the notation.

 {P}3:maxpk,d,pj,uL(pk,d,pj,u)= wklog(1+gkpk,dNk+Ik,jpj,u)+vjlog(1+gjpj,uN0+βpk,d)
 0≤pk,d≤Pmax1 (10) 0≤pj,u≤Pmax2 (11)

Here, and are the maximum allowable transmit powers.

###### Proposition 1.

For a fixed downlink user and uplink user , the optimal pair of powers that optimizes P3 belongs to the following set.

 S={(0,Pmax2),(Pmax1,0),(Pmax1,Pmax2), (pak,d,Pmax2),(Pmax1,paj,u)}

where

 pak,d=−B−√B2−4AC2A,paj,u=−E−√E2−4DF2D (12)

and

 A =wkgkβ2,B=2wkN0gkβ+(wk−vj)βgkgjpj,u (13) C =wkgkN20+wkgkgjN0pj,u−vjNkgjpj,uβ−vjgjβIk,jp2j,u (14) D =vjgjI2k,j,E=2vjNkgjIk,j+(vj−wk)Ik,jgkgjpk,d (15) F =vjgjN2k+vjgkgjNkpk,d−wkN0gkpk,dIk,j−wkgkβIk,jp2k,d (16)
###### Proof.

Computing the derivative with respect to and setting it to zero we have:

 ∂L∂pk,d=0⟹Ap2k,d+Bpk,d+C=0

where , and are defined above. It is evident that , and if then . When the above quadratic equation either has no zeros in or has only one zero where the function changes sign from to indicating a local minimum for . Therefore, in both cases the maximum is attained at a boundary point or . But when , could be negative, and the smaller root of the quadratic equation could be positive. In this case, the maximum is attained at or . By similar analysis for one sees that if then the maximum is attained at a boundary point or and when the maximum is attained at or . As a result, when the optimal transmission powers belong to the following set,

 Popt1={(0,Pmax2),(Pmax1,0),(Pmax1,Pmax2),(Pmax1,paj,u)}.

Otherwise, if , they belong to the set below

 Popt2={(0,Pmax2),(Pmax1,0),(Pmax1,Pmax2),(pak,d,Pmax2)}.

The cases and cannot be the optimal solutions of P3 , because they are dominated by and which give a larger . Therefore, optimal powers could be found by checking the members of the set S and picking the one that corresponds to the largest . ∎

Based on the above Proposition one can find the best uplink-downlink pair in each sub-channel by choosing the one with the largest value of . This involves only operations. Now we can present our sub-channel allocation algorithm to solve Problem P2, in which we employ a sub-optimum power allocation scheme. First, for each sub-channel , we find the best channel gain among all users and denote it by . Then, we sort the sub-channels based on the value of . In other words. we find a sub-channel permutation such that . Then, starting from sub-channel , we seek and that maximize . At the first iteration, we set , and for iteration set and where and indicate the number of sub-channels to be allocated to the BS’s downlink transmission and to user ’s uplink transmission, respectively, in the th iteration. The proposed sub-channel allocation algorithm is summarized below.

The complexity of finding the best user in each sub-channel is and for sub-channels is . Similarly, the complexity of finding the best pair in each sub-channel is and doing so for sub-channels requires operations. Since the complexity of sorting values is , then the overall computational complexity of the proposed sub-channel allocation algorithm is .

## Iv Power Allocation

The power allocation problem, denoted by P4, can be formulated as follows

 {P}4: maximizepk,d,pj,u Rd+Ru subject to (4)-(6)

Due to the interference terms, the power allocation problem is non-convex. Here, we use the “difference of two concave functions/sets” (DC) programming technique [17] to convexify this problem. In this procedure, the non-concave objective function is expressed as the difference of two concave functions, and the discounted term is approximated by its first order Taylor series. Hence, the objective becomes concave and can be maximized by known convex optimization methods. This procedure runs iteratively, and after each iteration the optimal solution serves as an initial point for the next iteration until the improvement diminishes in iterations. In [18], the DC approach is used to formulate optimized power allocation in a multiuser interference channel, and in [19], the DC optimization method is used to optimize the energy efficiency of an OFDMA device to device network. Here, we rewrite the objective function of P4 in DC form as follows

 maxpf(p)−h(p)
 f(p) =K∑k=1∑n∈Sk,dwklog(Nk+Ik,j(n)pj,u(n)+gk(n)pk,d(n)) +K∑j=1∑n∈Sj,uvjlog(N0+βpk,d(n)+gj(n)pj,u(n))
 h(p) =K∑k=1∑n∈Sk,dwklog(Nk+Ik,j(n)pj,u(n)) +K∑j=1∑n∈Sj,uvjlog(N0+βpk,d(n))

where

 p=[pk1,d(1),...,pkN,d(N),pj1,u(1),,...,pjN,u(N)]T

is the downlink and uplink transmitted power vector, and and denote the uplink and downlink users that are selected for the th sub-channel after the sub-channel allocation phase. Now, the objective is a DC function. To write the Taylor series of the discounted function ), we need its gradient, that can be easily derived as follows.

 ∇h(p)=[uj1βln(2)1N0+βpk1,d(1)),...,ujNβln(2)NN0+βpkN,d(N)), wk1Ik1,j1(1)ln(2)1Nk1+Ik1,j1(1)pj1,u(1)),..., wkNIkN,jN(N)ln(2)1NkN+IkN,jN(N)pjN,u(N))]T

To make the problem convex, is approximated with its first order approximation at point . We start from a feasible at the first iteration, and at the th iteration is generated as the optimal solution of the following convex program

 p(t+1)=argmaxp f(p)−h(p(t))−∇hT(p(t))(p−p(t)) subject to  (4)−(6)

Since is a concave function, its gradient is also its super gradient so we have

 h(p)≤h(p(t))+∇hT(p(t))(p−p(t)),∀p

and we can deduce

 h(p(t+1))≤h(p(t))+∇hT(p(t))(p(t+1)−p(t)).

Then it can be proved that in each iteration the solution of problem P4 is improved as follows

 f(p(t+1))−h(p(t+1))≥ f(p(t+1))−h(p(t))−∇hT(p(t))(p(t+1)−p(t)) =maxpf(p)−h(p(t))−∇hT(p(t))(p−p(t)) =f(p(t))−h(p(t)).

According to the above equations, the objective value after each iteration is either unchanged or improved and since the constraint set is compact it can be concluded that the above DC approach converges to a local maximum.

## V Analyzing Self-Interference Cancellation Coefficient Threshold

In [20], through simulations it has been observed that in a network that contains an imperfect FD BS and some imperfect FD and HD users, when the self-interference cancellation coefficient is larger than a specified threshold, there is no difference between the throughput of an all HD user network and an all FD user network. Here we wish to analyze this threshold.
Recall that in FD networks there are four possible types of connections in a given sub-channel:

1. HD downlink

2. HD uplink

3. Joint downlink and uplink for two distinct users over an FD BS

4. A full-duplex bidirectional connection between an FD user and an FD BS

Employing an FD user in a cell with an FD capable BS can increase the throughput when the 4th case is more appealing than the other cases in at least one sub-channel. Considering sub-channel , we assume that user is the best downlink user, user is the best uplink user, users and are the best downlink and uplink pair, and user is the best FD node for FD communication with the BS. Here the best user, is the user who gives the highest weighted rate with the same power than the rest. The rate of the four previous cases are presented below (we drop the sub-channel index for simplicity)

 Rd=wdlog(1+gdpBdNd) (17)
 Ru=vulog(1+gupuBN0) (18)
 Rdu= (19)
 Rf=wflog(1+gfpBfNf+βpfB)+vflog(1+gfpfBN0+βpBf) (20)

where, and are the transmission powers form BS to user and from user to the BS, respectively. The other variables were introduced in Table I. In short, using an FD user in the network could be beneficial when these conditions hold in at least one sub-channel:

 Rf>Rd (21)
 Rf>Ru (22)
 Rf>Rdu. (23)

Since we wish to focus on parameter , and to avoid dealing with other parameters, we introduce some simplifications. First, we assume the sum rate case where, . Second, we assume that the noise powers at the BS and at the users are the same. Third, we assume that the transmission power from the BS to all users is the same and is equal to the average BS power, i.e., . Fourth, we assume that the transmission powers from different users to the BS are the same and are equal to the average user power, i.e., . Fifth, the channel gains , , , , , are random variables and in the sum rate case the best downlink user, the best uplink user and the best FD user are all the same , because the channel is reciprocal and the user with maximum channel gain is selected for all of these three cases. If we assume that the number of users in the network is , then random variable can be defined as: , where is the random channel gain between the BS and the user , that itself is a multiplication of an exponential random variable, , with unit power and a path loss random variable that depends on the path loss model and the distance between the BS and the th user whose pdf is shown by (for ). We assume is a random channel gain between two users and which are distributed uniformly in a circle with radius .We also assume that and are the maximum and the second maximum channel gain between users.

Due to the randomness of the channel gains, itself is a random variable and here we wish to derive its distribution. According to conditions (21) - (23), we have:

1. The FD rate should be bigger than the HD downlink rate, so we have:

 log(1+gfPBSN0+βPuser)+log(1+gfPuserN0+βPBS)>log(1+gdPBSN0)

After some manipulations, this is reduced to the inequality , where:

 a1 =gdP2BSPuser (24) b1 =N0gfPuser(PBS−Puser) (25) c1 =N20gfPuser+N0g2fPBSPuser (26)

which holds for . Therefore, due to condition (21) .

2. By writing the condition (22) and doing the same procedure as the previous part we arrive at inequality where:

 a2 =guP2userPBS (27) b2 =N0guPBS(Puser−PBS) (28) c2 =N20gfPBS+N0g2fPBSPuser (29)

which holds for . Therefore, according to (22) .

3. By writing the condition (23) and doing the same procedure as the previous parts we arrive at the inequality , where:

 a3=P3BSPuserga (30) b3=P2userPBSgbN0+P3userPBSgbgab+P2userP2BSgbga+ P2BSPusergaN0+P3BSgaN0−P2userPBSgfN0− P3BSgfN0−gfP3BSPusergab−P3userPBSgfgab (31) c3=P2userPBSgbN0ga+P3usergabN0gb+P2userN20gb+ PuserPBSN20ga+P2BSPusergbN0ga+P2userPBSgbN0gab+ N20PBSgbPuser+2N20P2BSga−P3usergfgabN0− P2BSgfgabN0Puser−P2BSg2fN0Puser−N20gfPBSPuser− 2P2BSN20gf−P2userN20gf−P2BSPusergfgabN0− PBSgabP2userN0gf−g2fP2BSP2usergab (32) d3=PBSN30ga+PuserN30gb+P2userN20gbgab+ PuserPBSN20gbga−P2userN0g2fgabPBS−P2usergfN20gab− PusergfN20)gabPBS−Puserg2fN20PBS−PusergfN30− PBSgfN30 (33)

The above cubic function has the following three roots:

 x1= A+B−b3 (34) x2= −12(A+B)+i√32(A−B)−b3 (35) x3= −12(A+B)−i√32(A−B)−b3 (36)

where

 A= 3 ⎷−q2+√q24+p327 (37) B= 3 ⎷−q2−√q24+p327 (38) p= −b23+c (39) q= 2b327−bc3+d (40)
 b=b3a3,c=c3a3,d=d3a3 (41)

It is evident that . Also, it can be shown that the value of is always negative, therefore, it is deduced that the cubic function has at least one positive real root. Therefore, due to the condition (23), is the smallest positive real root of this cubic function, where is given by:

 R(x)={xif x is real and positive∞Otherwise.

Finally, we arrive at the following proposition:

###### Proposition 2.

For a wireless cell with an FD BS and users with imperfect SI cancellation factor , FD operation is advantageous from the perspective of the network throughput performance if .

In Section VII, we will compare the outcome of this analysis with simulation results.

## Vi Two-Tier Heterogeneous Full Duplex Network

In this section, we consider a two-tier heterogeneous full-duplex OFDMA network. This system includes a macrocell FD BS and multiple femto cell FD BSs along with their associated HD and FD users. Our goal is to maximize the uplink and downlik weighted sum rate of femto cell users while provisioning for the macrocell user’s uplink and downlink data rate. Assume that the numbers of femto cells and available sub-channels are and , respectively, and the number of users related to the th BS is . We denote the set of all BSs as , where the macro BS is indexed by 0. The variables used in the following equations are summarized in Table II.

The downlink rate in cell is given by:

 gk,m(n)pk,d,m(n)Nk,m+∑Kmj=1Ik,j,m(n)pj,u,m(n)+DICm,k(n)+UICm,k(n)) (42)

where and are the downlink and uplink inter-cell interference on sub-channel in the th cell for user , i.e.:

 DICm,k(n)=∑m′∈Ω\textbackslash{m}gk,m,m′(n)pd,m′(n) (43)
 UICm,k(n)=∑m′∈Ω\textbackslash{m}Km′∑j=1gk,m,j,m′(n)pj,u,m′(n) (44)

Similarly the uplink rate in cell is given by:

 Ru,m=Km∑j=1∑n∈Sj,u,mvjlog(1+ gj,m(n)pj,u,m(n)Nm+β∑Kmk=1pk,d,m(n)+DICm(n)+UICm(n)) (45)

where and are the downlink and uplink inter-cell interference on sub-channel at the th BS, i.e.:

 DICm(n)=∑m′∈Ω\textbackslash{m}gm,m′(n)pd,m′(n) (46)
 UICm(n)=∑m′∈Ω\textbackslash{m}Km′∑j=1gj,m′,m(n)pj,u,m′(n) (47)