and each DMG reuses different CUs channels. In a D2D multicast group, the D2D transmitter (DT) multicasts messages to a group of D2D receivers (DRs). Let DGj denote the j-th DMG, and the number of the DRs in DGj is Mj. 
Figure 1 System model of D2D multicast communication
Assume that the c-th channel is reused by the j-th DMG, and then the rate of the c-th CU is defined as:
(1)
where Wc is the bandwidth for CUc; pc is the transmit power of the c-th CU;
is the channel gain from CUc to the BS; pj,c is the power of DT in DGj on the c-th channel;
is the channel gain from DT in DGj to the BS on the c-th channel;
is the variance of the independent zero-mean additive white Gaussian noise (AWGN).
Let 
denote the m-th DR in the j-th DMG. Introduce a new parameter
to evaluate the quality of channel link, and thencan be defined as:
(2)
where
is the channel gain of DT to in DGj on the c-th channel; 
is the channel gain of the c-th CU toin DGj.
Since the capacity of DMG is restricted by the DR with the worst channel condition, let
denote the weakest link in the j-th DMG, and then is expressed as:
(3)
Therefore, the aggregate rate for the j-th DMG on the c-th channel is:
(4)
Note in the above discussion, the loss of CUs’ rates caused by the DMGs is not considered. By introducing the loss of CUs’ rate, we define:
(5)
where the second term in the right hand size denotes the loss of the c-th CU’s rate; ωj (0≤ωj≤1) is the weight factor for the loss of the CUs’ rate in the j-th DMG. Observe that when ωj=0,
. Therefore, the definition in Eq. (5) is more general.
Let
indicate whether or not the c-th channel is reused by the j-th DMG. Accordingly, the EE of the j-th DMG is defined as:
(6)
where κj is the inverse of the power amplifier efficiency and
is the fixed circuit power consumption.
To maximize the EE of the worst DMG link, the jointly channel and power allocation optimization problem can be formulated as:
(7)
where Rreq is the required transmit rate; Pth is the power budget; C1 is the minimum transmit rate requirement constraint; C2 and C3 are the transmission power constraints; C4 and C5 indicate that each channel resource of CUs cannot be shared by multiple DMGs.
For convenience, the key notes used in this paper are listed in Table 1.
Table 1 Summary of key notation
3 Proposed resource allocation scheme
Obviously, Eq. (7) is a mixed integer non- linear fractional programming problem. It is in general difficult to solve. In this section, a two- stage approach, which contains two individual procedures, is proposed to solve the above problem.
3.1 Channel assignment algorithm
In this section, heuristic channel assignment algorithm is proposed to maximize the minimum individual EE, under the constraints C1-C5.
According to the constraints C1-C3, there is 
Set that the maximum transmit power for DT in the j-th DMG on the c-th channel is 
Let
be the worst DR in the j-th DMG. Observe that channel should be allocated to DMG with low 
since it will introduce less interference to the CU. Therefore, define a new statistic
Allocating channels according to 
considers not only the D2D link channel condition but also the interference channel condition.
Let Ωj denote the channel set for the j-th DMG, and the proposed heuristic channel assignment algorithm is tabulated as Algorithm 1. In the first step, each DMG is allocated to a channel with the highest value 
In the second step, the DMG with the lowest energy efficiency has a priority to choose the channel with the highest value 
. If the energy efficiency can be improved by the new channel, then assign it to the DMG. This step is repeated until all channels are consumed. Although the proposed heuristic channel assignment algorithm is suboptimal, it has a lower complexity than the optimal channel allocation algorithm [17]. The complexity of the proposed channel assignment algorithm is approximately linear to the number of the channels, i.e. o(C) [18].
3.2 Power allocation algorithm
For a given channel assignment, the optimization problem Eq. (7) is transformed into the following form:
s.t. C1-C3 (8)
However, the above optimization problem is still non-convex. Fractional programming [19, 20]
Algorithm 1 Channel assignment
1:
Set
2:
For each
3:
Find
according to
4:
Calculate
5:
Update
6:
End for
7:
While
8:
Find 
according to
9:
Find
according to
10:
If 
11:
Update 
and 
12:
Calculate
13:
End if
14:
End while
is usually used to solve the above problem. Let 
and p denote the feasible region and variable vector of Eq. (8). Introduce a new variable λ and define a new function F(λ):
(9)
where
, and F(λ) is a function of variable λ. The objective function in Eq. (9) is a subtractive form of the objective function in Eq. (8).
Let p* and λ* be the optimal solution and optimal objective of Eq. (8), respectively:
(10)
Theorem 1: Eqs. (8) and (9) have the same set of optimal solutions, if and only if:
(11)
Proof: See Appendix A. Similar proof can be found in Refs. [21, 22].
By Theorem 1, it is clear that solving problem in Eq. (8) can be achieved by finding a solution of the equation F(λ*)=0. However, λ* is unknown. To determine the optimal λ*, the generic Dinkelbach algorithm [23] is adopted. Combining with constraints C1-C3, Plb and Pub, which denote the minimum and the maximum power consumption respectively, are given as follows:
(12)
(13)
where 
is the number of elements in Ωj.
Introduce two new variables αmin and αmax, where αmin≥αmax. Define an interval [αmin, αmax] which contains λ*. Then at the t-th iteration, αmin and αmax can be updated via:
(14)
(15)
The proposed iterative power allocation algorithm is described as Algorithm 2.
Algorithm 2 Iterative power allocation algorithm
1:
Calculate Plb and Pub
2:
Set t=1 and 
3:
Choose
and set termination precise ε>0
4:
Repeat
5:
For the given λt,
6:
Solve problem in Eq. (9) to obtain pt
7:
Calculate 
8:
Update the interval 
with Eqs. (14) and (15)
9:
Update
and choose 
10:
If 
11:
Let t=t+1
12:
End if
13:
End for
14
Until
Similar convergence proof of Algorithm 2 can be found in Ref. [23]. In step 2, problem in Eq. (9) is needed to be solved. In the following, how to solve this problem is discussed.
3.2.1 Without loss of CUs’ rate (wj=0)
Since wj=0,
is concave on p, where 
and the objective function of Eq. (9) is concave. Introduce a new variable η, and transform the nonsmooth problem in Eq. (9) to a smooth one:
(16)
Problem in Eq. (16) is convex optimization problem which can be solved efficiently by interior-point algorithm [24]. The computational complexity to solve Eq.(16) is o((JC)3). Since the proposed algorithm without loss of CUs’ rate includes two loops: outer loop (i.e., Algorithm 2) and interior-point (i.e., interior-point algorithm to solve Eq.(16)), the total computational complexity of the algorithm without loss of CUs’ rate is o(tmax(JC)3), where tmax is the maximum iteration number of outer loops (i.e., Algorithm 2).
3.2.2 With loss of CUs’ rate (0j≤1)
In this case, 
is defined as Eq. (5), and Eq. (9) is non-convex and NP-hard, which is hard to find the global optimal solution. However, there is a special structure in Eq. (9). First, define two new functions φj(p) and fj(p) respectively:
(17)
(18)
where φj(p) and fj(p) are concave for p.
Then, Eq. (9) can be expressed by:
(19)
Since φj(p) and fj(p) are concave for p, φj(p)- fj(p) is a D.C (difference of two concave functions) function [25], which can be effectively solved by sequential convex programming [26]. Let pτ be the power allocation at step τ, and pτ+1 is updated by solving the following convex problem:
(20)
As in Eq. (16), further transform Eq. (20) into a smooth and standard convex optimization problem:
(21)
where 
is the gradient of fj(p) at p and the element of is easily given by:
(22)
where n is the index of the ’s element, 1≤n≤J*C.
Eq. (21) is convex optimization problem, which can be solved efficiently by interior-point algorithm [24]. The detailed process for solving Eq. (9) with the loss of CUs’rate is given as Algorithm 3.
Similar convergence analysis of Algorithm 3 can be find in Refs. [26, 27]. Now, we have presented the power allocation algorithm to solve Eq. (8), when the loss of CUs’rate is considered. The algorithm with loss of CUs’ rate includes two loops: outer loop (i.e., Algorithm 2) and inner loop (i.e., Algorithm 3 and interior-point algorithm to solve Eq. (21)).
Algorithm 3 Algorithm for solving problem Eq. (9)
1:
Choose p0 and calculate 
set τ=0
2:
Repeat
3:
Solve problem Eq. (21) to obtain the
solution 
4:
Set τ=τ+1 and 
5:
Calculate
6:
Until 
Let χ denote the total cost of computing fj(p), and its first and second derivatives. The computational complexity of Algorithm 3 is 
[27], where τmax is the maximum iteration number of Algorithm 3. Therefore, the total computational complexity of the algorithm with loss of CUs’rate is 
.
4 Performance analysis
In this section, some numerical results are presented to evaluate the performance of the proposed schemes. Considering that there are 12 cellular users and 4 D2D multicast groups located randomly in a single cell. The BS is located at the center of the cell. The channel gains are assumed to be Rayleigh-fading with an average power gain of 0 dB. Without loss of generality, set Mj=5,Wc=1 MHz, pc=15 dBm, κj=1, 
and ωj=ω.
For convenience, we use WoL algorithm and WL algorithm to denote resource allocation algorithms without and with loss of CUs’ rate, respectively. Figure 2 illustrates the evolution of WoL algorithm and WL algorithm. In Figure 2, set Pth=12 dBm and Rreq=1 Mbps. Note that the algorithm with loss of CUs’ rate consists of two loops, and the effect of outer loop iterations t is only considered. It is observed that both of the algorithms converge to the optimal λ fast. We also find that WL algorithm converges faster than WoL algorithm. According to Algoirthm 2, the optimal λ* must be in the interval 
Equations (14) and (15) can be respectively rewritten as:
(23)
(24)
The size of interval 
is 
Therefore, the size of interval is determined by F(λt) and 
Based on the definition of Plb and Pub, i.e. Eqs. (12) and (13), the value of 
for WoL algorithm is the same as the value for WL algorithm. According to Eqs. (5) and (9), the value of F(λt) for WoL algorithm is larger than the value of F(λt) for WL algorithm. Therefore, the size of interval for WoL algorithm is larger than the size of interval for WL algorithm. More iterative times are need to find the optimal value λ* in larger size of interval. As a result, WL algorithm converges faster than WoL algorithm.
Figure 2 Convergence evolution of proposed algorithms
Figure 3 depicts the EE of the worst case DMG versus weight factor ω for different power budget with Rreq=1 Mbps. Weight factor ω reflects the loss of CUs’ rates caused by the DMGs. In Figure 3, when ω=0, the algorithm becomes WoL algorithm; When
the algorithm becomes WL algorithm. According to Eqs. (5) and (6), the EE of DMG is a monotony decrease function of weight factor ω. Therefore, the EE of the worst case DMG decreases with the growth of weight factor ω.
Figure 3 EE of the worst case DMG vs weight factor
To emphasize the advantages of the proposed schemes, the proposed algorithms with the algorithm in Ref. [28] are compared. For the sake of fairness, the channel allocation and power allocation of D2D users in Ref. [28] are only considered, and set Rreq=1 Mbps. Let the number of D2D multicast groups J=1. Figure 4 depicts the EE of the two algorithms versus the power budget. As can be seen in Figure 4, the EE performance of the proposed algorithms has similar performance, and both of the algorithms can improve the EE compared with the algorithm in Ref. [28]. Because the weight factor ω=0.5 is very small, the algorithm with CUs’rate has similar performance with the algorithm without CUs’rate, and both of the proposed algorithms can always select the optimal power that achieves the highest EE, while the algorithm in Ref. [28] is to maximize the throughput of system.
Figure 4 EE comparison of proposed algorithms and algorithm in Ref. [28]
The EE as a function of the minimal rate requirement in Figure 5 is investigated, where Pth=12 mW. As shown in Figure 5, the EE decreases with the growth of the rate requirements. This is because when the rate requirement is high, in order to meet the rate requirement, much power is allocated to the D2D users (the growth of rate requirements will result in exponentially increase of power consumption). At the same time the growth of the rate requirements also results in more frequently system outage. In addition, the proposed algorithms have higher EE than the algorithm in Ref. [28]. The reason is the same as the reason analyzed in Figure 4.
Figure 6 shows the system throughput of the proposed algorithms and algorithm in Ref. [28] versus power budget. As can be seen, the algorithm in Ref. [28] outperforms the proposed algorithms; at the high power budget region, the throughput of the proposed algorithms become saturated. This is because the algorithm in Ref. [28] is throughput maximization algorithm, which uses all the available power to maximize its throughput, and thus the energy efficiency is impaired. While, the proposed algorithms tend to reduce the transmit power to maximize the system energy efficiency.
Figure 5 EE versus minimal rate requirement
Figure 6 Throughput comparison of proposed algorithms and algorithm in Ref. [28]
The EE of the worst case DMG versus the power budget is depicted in Figure 7 where Rreq=1 Mbps. As can be seen in Figure 7, the EEs of the two algorithms increase with the increasing of the power budget at the beginning. When the power budget is sufficient enough, the EE remains almost unchanged. This is because the lower the power budget is, the more the D2D multicast communications network suffers outage.
The fairness of the proposed algorithm in resource allocation is also investigated. Usually,the fairness index is used to measure the fairness of the algorithm [29], and is defined as: 
Figure 8 shows the fairness index with respect to the number of DMG with Pth=12 dBm and Rreq=1 Mbps. Obviously, fairness index performance of the proposed algorithms is better than that of the algorithm in Ref. [28]. This is because the proposed algorithms use the max-min fairness criterion as the optimization criterion, which can guarantee the fairness of resource allocation. However, the algorithm in Ref. [28] aims to maximize the total EE of the system, and more resource is allocated to the users with better link quality indicator.
Figure 7 EE of the worst case DMG vs power budget
Figure 8 Fairness index vs number of DMGs
The EEs of each DMG under the resource allocation algorithms with and without loss of CUs’ rate are compared in Figure 9, where Pth=12 dBm and Rreq=1 Mbps. As shown in Figure 9, the EE fairness among DMGs is guaranteed in both of the proposed algorithms.
Figure 9 EE of each DMG under proposed algorithms
5 Conclusions
In this paper, a max-min energy efficiency- optimal problem for D2D multicast communications is studied by integrating both channel assignment and power allocation. In addition, both with and without loss of cellular users’ rate are considered in formulated optimization problem. To solve above nonsmooth and mixed integer fractional programming problem, two-step resource allocation algorithms are proposed. Numerical results show that our proposed algorithms can guarantee energy efficiency fairness among multicast groups and improve the energy efficiency of D2D multicast groups. For future work, we will consider the impact of imperfect channel state information on the energy efficiency and how to obtain the optimal solution of the formulated energy efficient problem.
Appendix A
Before proving Theorem 1, the following lemma is given:
Lemma 1：If and only if λ<λ*, there is F(λ)>0. Therefore, F(λ*)≤0.
Proof：Assuming F(λ)≥0, there is a solution p0 (
), which makes the following inequality true:
(A-1)
Eq. (A-1) is equivalent to
Since
there is:
(A-2)
Therefore, when F(λ)≥0, there is λ<λ*.
Assuming λ<λ*, there is a solution p1 
, which makes the following inequality true:
(A-3)
Equation (A-3) is equivalent to the following inequality:
(A-4)
Therefore, when λ<λ*, there is F(λ)>0. In summary, if and only if λ<λ*, F(λ)>0. In addition, since
F(λ) is a monotone decreasing function on λ. Thus, there is F(λ*)≤0.
In the following, the proof of Theorem 1 combining with the above Lemma 1 is presented.
1) Assume that p* is the optimal solutions of Eq. (8), since
which can be rewritten as following form:
(A-5)
According to F(λ*)≤0 in Lemma 1, there is F(λ*)=0. Therefore, the optimal solution of Eq. (8) makes F(λ*)=0.
2) Assuming F(λ*)=0, p* is the optimal solutions of Eq. (9), then:
(A-6)
Equation (A-6) can be rewritten as following form:
(A-7)
Therefore, p* is also the optimal solution of Eq.(8).
Combing with the analysis of (17) and (27), Theorem 1 is proved.
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(Edited by ZHENG Yu-tong)
中文导读
D2D组播通信中的高能效资源分配研究
摘要：研究了D2D组播通信中的资源分配问题，为了获得公平的能量效率，采用最大最小公平准则作为优化准则，以D2D组播的能量效率作为优化目标函数，在考虑最大发射功率约束和最小速率约束的情况下，通过最大化最小D2D组播的能量效率来实现公平的资源分配，并将上述优化问题建模为非凸的、混合整数规划问题。提出了一种次优的资源分配算法来求解上述问题。首先，采用提出的信道分配算法将信道资源分配给不同的D2D组播用户。然后，对于给定的信道分配，提出了两种迭代的功率分配算法。仿真实验结果证明了提出算法的收敛性。另外，与传统的吞吐量最大化算法进行了比较，实验结果表明提出的算法能够提高系统的能量效率，并且保证不同D2D组播之间的公平性。
关键词：D2D；能量效率；资源分配；最大最小公平规则；分式规划
Foundation item: Projects(61801237, 61701255) supported by the National Natural Science Foundation of China; Project(SBH17024) supported by the Postdoctoral Science Foundation of Jiangsu Province, China; Project(15KJB510026) supported by the Natural Science Foundation of the Jiangsu Higher Education Institutions, China; Project(BK20150866) supported by the Natural Science Foundation of Jiangsu Province, China; Projects(NY215046, NY217056) supported by the Introduction of Talent Fund of Nanjing University of Posts and Telecommunications, China
Received date: 2018-08-28; Accepted date: 2019-03-07
Corresponding author: ZUO Jia-kuo, PhD, Lecturer; Tel: +86-15380410922; E-mail: zuojiakuo@njupt.edu.cn; ORCID: 0000-0001- 5940-7559
Central South University Press and Springer-Verlag GmbH Germany, part of Springer Nature 2019