End-to-end energy-efficient resource allocation in device-to-device communication underlaying cellular networks

The Journal of China Universities of Posts and Telecommunications April 2015, 22(2): 24–30 www.sciencedirect.com/science/journal/10058885

http://jcupt.xsw.bupt.cn

End-to-end energy-efficient resource allocation in device-to-device communication underlaying cellular networks Xu Quansheng, Ji Hong, Li Xi (

), Xiong Danni

Key Laboratory of Universal Wireless Communications, Beijing University of Posts and Telecommunications, Beijing 100876, China

Abstract A proposed resource allocation (RA) scheme is given to device-to-device (D2D) communication underlaying cellular networks from an end-to-end energy-efficient perspective, in which, the end-to-end energy consumptions were taken into account. Furthermore, to match the practical situations and maximize the energy-efficiency (EE), the resource units (RUs) were used in a complete-shared pattern. Then the energy-efficient RA problem was formulated as a mixed integer and non-convex optimization problem, extremely difficult to be solved. To obtain a desirable solution with a reasonable computation cost, this problem was dealt with two steps. Step 1, the RU allocation policy was obtained via a greedy search method. Step 2, after obtaining the RU allocation, the power allocation strategy was developed through quantum-behaved particle swarm optimization (QPSO). Finally, simulation was presented to validate the effectiveness of the proposed RA scheme. Keywords

energy-efficiency, resource allocation, D2D communication, mixed integer and non-convex optimization problem

1 Introduction D2D communication underlaying cellular network is a promising network architecture, which has attracted much attention by its advantages [1–5]. Since cellular user equipments (CUEs) and D2D-pair user equipments (DUEs) share the same frequency resource, the interference between CUEs and DUEs is a serious problem. Proper RA is very crucial to control the interference [1–5]. Furthermore, due to limited energy supply and need of environmental-friendly transmission [6], the energyefficient communication is drawing increasing attentions. As a result, developing an efficient RA algorithm that meets the interference constraint and improves the EE of D2D networks becomes significant [5]. Although much work has been done for efficient RA in D2D communication underlaying cellular networks [1–5], it cannot achieve desirable EE performance. First, the feature that D2D communication is a short-distance communication was not fully considered. In Received date: 11-07-2014 Corresponding author: Li Xi, E-mail: [email protected] DOI: 10.1016/S1005-8885(15)60635-5

D2D networks, the communication distance is short, and the transmission power is relatively small [7]. Therefore, the circuit power consumption of receiver plays an importance role in total energy expenditure. EE enhancement can be achieved only if energy consumptions of the entire communication chains are considered [6], thus when designing energy-efficient RA for D2D networks, it is necessary to consider end-to-end energy consumption, including the circuit energy consumption of the transmitter, transmission energy consumption, and the circuit energy consumption of the receiver. Second, to simplify the problem formulation, most of existing RA schemes have a lot of restricts, including the restricts that each CUE can only occupy one sub-channel, and each sub-channel can only be reused by one DUE, etc [1–5]. However, these restricts are impractical and some restricts even limit the achievable EE performance. Thus, to provide a more practical and efficient RA scheme, these restricts should be tackled carefully. Third, most of existing RA algorithms of D2D communication mainly focuses on how to reuse the resources of CUEs [2–4]. The method of first allocating the resources to the CUEs, and then considering reusing

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Xu Quansheng, et al. / End-to-end energy-efficient resource allocation in device-to-device communication…

CUEs’ resources for the DUEs is not a optimal solution, because the RA for CUEs and DUEs are independent. To obtain better EE performance, system resources should be allocated to CUEs and DUEs simultaneously. Although limited works have been done on this aspect [1], there are some impractical restricts just be stated in the above paragraph. The authors investigated RA in D2D communication underlaying cellular networks from an end-to-end energy-efficient perspective. The resource unit (RU) and power allocation are considered together to improve the EE performance. The distinct features of this article are summarized as follows: 1) Considering the intrinsic features of D2D communication, a RA scheme is proposed from an end-to-end energy-efficient perspective, which is different from most of existing works [1–5] and can ensure the proposed scheme to achieve better EE performance. 2) To provide more practical and efficient RA algorithm, the RUs in this paper are used in a complete-shared pattern, where both CUEs and DUEs can occupy multi-RUs and one RU can be simultaneously reused by multi-DUEs. 3) Unlike the most of existing RA schemes that focus on how to reuse the resources of CUEs [2–4]. In this paper, resources are allocated to the CUEs and DUEs simultaneously, which is benefit to EE improvement.

2 System model and problem formulation A single cell uplink resource sharing was investigated. It was assumed that there are K C CUEs and K D DUEs. The sets of these two user classes are denoted as ϕ C and

ϕD

respectively.

Orthogonal

frequency

division

multiplexing access (OFDMA) was employed to support multiple accesses, and there are totally N sub-channels. One sub-channel in one time slot was denoted as a RU. The bandwidth of one RU is W, the time duration of one RU is T. One scheduling period contains M time slots. The channel gain between CUE i and the BS on RU (n, m) is

g ni ,B,m , the channel gain of DUE j is g nj,,Dm . Similarly, the channel gains of the interference links, from the transmitter of DUE j to the BS, and from CUE i to the receiver of DUE j are given as g nj,,Bm and g ni ,, mj , respectively. The classical metric ‘bit/J’ [6] was adopted to evaluate the EE performance. The amount of transmitted data

25

during one scheduling was given as KC

KD

Rtot = ∑ R + ∑ R Dj C i

i =1

(1)

j =1

where, RiC and R Dj are the amount of transmitted data of CUE i and DUE j, respectively. N M  a i | g i ,B |2 pni , m  RiC = ∑∑ TW lb  1 + n , m D-C n , m  I i + N 0W  n =1 m =1  N M  bnj, m | g nj,,Dm |2 qnj, m  R Dj = ∑∑ TW lb 1 + DC-D   + N 0W  Ij n =1 m =1 

( 2)

(3)

KC

where I iD-C = ∑ bnj, m | g nj,,Bm |2 qnj, m is the interference from j =1

DUEs to the CUE i that use the same RU (n, m) . I DC-D = j KD

∑

k =1, k ≠ j

KC

bnk, m | g nk,,mj |2 qnk, m + ∑ ani , m | g ni ,, mj |2 pni , m is the interi =1

ference from the other DUEs and CUE. ani , m and bnj, m are Boolean variables. ani , m = 1 ( bnj, m = 1 ) denotes that RU (n, m) is allocated to CUE i (DUE j); otherwise,

ani , m = 0 ( bnj, m = 0 ). 0 ≤ pni , m ≤ p peak

and 0 ≤ qnj, m ≤

q peak denote the transmission power of CUE i and DUE j on RU (n, m) , respectively. p peak and q peak are peak power thresholds for CUE and DUE, respectively. N 0 is noise power spectral density. The total energy consumption is calculated as following. The total number of time slots where there are data for M  N  CUE i and DUE j are formulated as M iC = ∑ f  ∑ ani , m  m =1  n =1  M  N  and M Dj = ∑ f  ∑ anj, m  , respectively, where f ( x) is m =1  n =1  an integer step function. f ( x) = 0 when x = 0 ,

otherwise,

f ( x) = 1 . Similarly, the total number of

time

slots where there are data for BS is C M  N K  M C = ∑ f  ∑∑ ani , m  . Assume that the work mode of m =1  n =1 i =1  CUEs only has two modes, i.e., transmitting mode and sleep mode. The circuit power of UEs and BS at sleep mode is 0 . The circuit power of CUE i at transmitting mode is pit ,C . The circuit power of BS at receiving mode is p r ,B . The work mode of DUEs includes transmitting mode, receiving mode, and sleep mode. The circuit power of DUE j at transmitting mode, and receiving mode are q tj,D and q rj ,D , respectively. Then q Dj = q tj,D + q rj ,D is the

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The Journal of China Universities of Posts and Telecommunications

circuit power of DUE j, and the total energy consumption can be expressed as C KD  N M K Ptot = T  ∑∑ (∑ pni , m + ∑ qnj, m ) + M C p r ,B + j =1  n =1 m =1 i =1 KC KD  (4) M iC pit ,C + ∑ M Dj q Dj  ∑ i =1 j =1  We assumed that both CUE i and DUE j have their quality of service (QoS) demands in terms of minimum . Then the transmission rates RiCmin and R Dmin j energy-efficient RA can be formulated as Eq. (5), where the constraint 1 (C1) means that each RU can only be are allocated to one CUE at most. Pi Cmax and Q Dmax j maximum transmission power thresholds of CUE i and DUE j, respectively. Rtot  max  i j i j an ,m , bn ,m , pn ,m , qn ,m P tot   s.t.  KC  i C1: ∑ an , m ≤ 1; ∀n, m  i =1  N (5) Cmax i C2 : ∑ pn , m ≤ Pi ; ∀n, m, i   n =1  N  ∀ C3 : ∑ qnj, m ≤ Q Dmax ; n , m , j j  n =1  C4 : RiC ≥ RiCmin ; ∀i ∈ ϕ C   D Dmin D C5 : R j ≥ R j ; ∀j ∈ ϕ  The RA problem in Eq. (5) is a mixed integer and non-convex optimization problem, which is prohibitively difficult to solve. To make the problem tractable, a two-stage approach was considered. Especially, the RA was separated into two individual procedures, RU allocation and power distribution as well.

3 RU allocation via greedy search method Greedy search method was adopted to design the RU allocation. Since there are QoS demands, the RU allocation is separated into two steps. First, the RU allocation should assign some RUs to each UE to guarantee its QoS demand. Second, the remaining resources are allocated properly to achieve the best EE performance. The detail steps of the proposed RU allocation will be described in Algorithm 1, where N t denotes the set of RUs, N rC is the residual RUs for CUEs.

ψ iC and ψ Dj are the sets of RUs that be allocated to CUE

2015

i and DUE j, respectively. φnC,m is the index of CUE that

be allocated with RU (n, m) . φnD,m is the set of DUEs that be allocated with RU (n, m) . pmres,i and qmres,j are residual power for CUE i and DUE j on time slot m, respectively. In the initialization, R C = [ R1C , R2C ,..., RKCC ] =

R D = [ R1D , R2D ,..., RKDD ] = 0 ,

0 ,

ani , m = 0 ,

bnj, m = 0 ,

N rC = N t , ψ iC = ∅ , ψ Dj = ∅ , φnC,m = 0 , φnD,m = ∅ ,

pmres,i = Pi Cmax , and qmres, j = Q Dmax . j At the first round, the UE whose rate is the farthest away from its rate demand has higher priority to get a new RU, i.e., i* or j * in Algorithm 1. For selected UE, the RU with the highest achievable EE will be chosen, i.e., RU (n* , m* ) in Eqs. (6) and (7)

ηiC (n* , m* ) ≥ ηiC (n′, m′); ∀(n′, m′) ∈ N rC

(6)

η (n , m ) ≥ η (n′, m′); ∀(n′, m′) ∈ N

(7)

*

D j*

*

*

*

D j*

t

where, ηiC* (n* , m* ) and η Dj* (n* , m* ) are the EE of the CUE i* and DUE

j * after allocating a new RU

(n* , m* ) , respectively. To simplify the analysis, the transmission power of all CUEs and DUEs are temporarily set as p peak and q peak , respectively. Then transmission power can be omitted when calculating ηiC* (n* , m* ) and

η Dj (n* , m* ) . Thus, *

  

ηiC (n* , m* ) = ( RiC + ri C (n* , m* ) ) T   *

*

*

M

∑

*    m =1, m ≠ m

 N *  f  ∑ ani , m  +  n =1 

C     M  N K   1 pit ,C +  ∑ f  ∑∑ ani , m  + 1 pir ,B        m =1, m ≠ m*  n =1 i =1   

−1

(8) where, ri* (n , m ) is the achievable transmission data of C

*

*

CUE i* on RU (n* , m* ) , which is given as

 | g ni ,B |2 p peak * , m* ri (n , m ) = TW lb  1 + D-C  I * new (n* , m* ) + N 0W i  C

*

*

   

(9)

new where, I iD-C (n* , m* ) is the interference from DUEs, *

new ( n* , m * ) = I iD-C *

∑

j∈φ D* * n ,m

|g nj*,B, m* |2 q peak

(10)

Note that if the RU (n* , m* ) is allocated to the new CUE i* , CUE i* will generate interference to all the DUEs that use RU (n* , m* ) . This kind of interference can be calculated as

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Xu Quansheng, et al. / End-to-end energy-efficient resource allocation in device-to-device communication…

∑

new − D (n* , m* ) = I DC j

k ∈φ D*

n ,m*

*

,k ≠ j

|g nk*, ,jm* |2 q peak + | g ni *,, mj * |2 p peak

(11) From Eqs. (10) and (11), the interference is very complex. To handle the interference, based on Ref. [8] two interference constraints I C and I D are introduced for CUEs and DUEs, respectively. Then, we have following inequations new (n* , m* ) ≤ I C (12) I iD-C * new − D I DC (n* , m* ) ≤ I D ; ∀j ∈ φnD* , m* j

(13)

27

The second round was proposed to improve the EE. Since DUEs can reuse the RU resources of CUEs, then the RU can be regarded as ‘soft-resource’ which can change with the network situations, such as interference level, network loads, etc. However, the power resources are limited and fixed. Therefore, to achieve better EE, the limited power resources should be allocated efficiently. That means, for each UE with residual power, the most proper RU, i.e., the RU that can achieve the highest EE associated with this UE, will be allocated. Algorithm 1 RU allocation via greedy search method

Similarly, the η Dj* (n* , m* ) can be obtained as

R Dj* + rjD* (n* , m* )

η (n , m ) = D j*

*

*

(14)

 M  N *   T  ∑ f  ∑ bnj, m  + 1 q Dj    m =1, m ≠ m*  n =1 D * * where, rj (n , m ) is given as

I

∑

(n , m ) = *

*

|g

k ∈φ D*

k , j* 2 n* , m*

| q

peak

φ C*

+ | gn ,m

n , m* * *

Initialization

2: 3:

First round for guaranteeing QoS while N rC ≠ ∅ & (min( RiC − RiCmin ) < 0 | min( R jD − R Dmin ) < 0) do j

4:

*

   

5:

(15)

, j* 2

| p

peak

7:

(16)

and all the DUEs that use RU (n* , m* ) ,

I

(n , m ) = *

∑

|g

j∈φ D*

j ,B 2 n* , m*

| q

peak

+|g

| q

peak

D

*

Dmin

≤ Rj / Rj *

*

*

For i , find (n , m* ) satisfies Eqs. (6), (12)–(13); i* ,B

peak

( I C + N 0W )) , pmres,i = pmres,i −

i* n* ,m*

= 1 , φn , m = i , ψ = ψ ∪

2

Ri = Ri + TW lb(1 + | g n* ,m* | p C

*

*

peak

*

, N r = N r / (n , m ) , a C

C

*

*

C *

*

*

8: 9:

*

11:

(

i* ,D

2

R j = R j + TW lb 1 + | g n* ,m* | q D

D

*

*

peak

peak

)

( I D + N 0W ) , qmres, j = qmres, j −

*

*

*

*

end if

belonging to φnD* , m* , and the CUE φnC* , m* as

13: Second round for improving EE performance 14: For each pmres,i > 0 and qmres,k > 0

∑

|g

φ C*

|g nk*, ,jm* |2 q peak +

15: while pmres,i > 0

j

16:

,j 2 n , m* n* , m*

*

, bnj* ,m* = 1 , φnD ,m = φnD ,m ∪ { j * } , ψ Dj = ψ Dj ∪ {( n* , m* )} ;

the interference from the new DUE j , the other DUEs

k ∈φ D* * , k ≠ n ,m

*

*

12: end while

new -D I DCD (n* , m* ) = j*

*

C i

else For j * , find (n* , m* ) satisfies Eqs. (7), (19)–(21);

q

n ,m*

*

C i

{( n , m )} ;

(17)

And all the DUEs that use the RU (n* , m* ) will suffer

*

D

*

, ∀j ∈ ϕ , respectively;

Dmin

≤ Rj / Rj

Cmin

C

*

10: j * ,B 2 n* , m*

D

*

p

new -C the interference I DD (n* , m* ) from the new DUE j * j*

*

C

if Ri / Ri

6:

The CUE i that also uses the RU (n* , m* ) will suffer

*

*

Dmin

and R j / R j

n ,m*

DDnew -C j*

i* and j * satisfy RiC / RiCmin ≤ RiC / RiCmin , ∀i ∈ ϕ C ,

Find D

 | g nj*,D, m* |2 q peak rjD* (n* , m* ) = TW lb 1 + DC-Dnew * *  I* (n , m ) + N 0W j  DC-D new j*

1:

*

| p peak + | g nj* ,,mj * |2 q peak

(18)

17:

(20)

new -D I DCD (n* , m* ) ≤ I D ; ∀j ∈ φnD* , m* j*

(21)

Therefore, RU (n* , m* ) satisfies Eqs. (6), (12)–(13) will be allocated to CUE i* . And RU (n* , m* ) satisfies

i ,B

2

Ri = Ri + TW lb(1 + | g n* ,m* | p C

C

peak

( I C + N 0W )) , pmres,i = pmres,i −

p peak , ani * ,m* = 1 , N rC = N rC / (n* , m* ) , φnC* ,m* = i , ψ iC = ψ iC ∪

Thus, the following inequations should be satisfied, new (n* , m* ) ≤ I D (19) I DC-D * j new -C I DD (n* , m* ) ≤ I C ; φnC, m ≠ 0 j*

Find (n* , m* ) satisfies Eqs. (6), (12)–(13);

{( n* , m* )} ;

18: end while 19: while qmres,k > 0 20:

Find (n* , m* ) satisfies Eqs. (7), (19)–(21); *

21: RkD = RkD + TW lb(1 + | g ni *,D,m* |2 q peak ( I D + N 0W )) , qmres,k = qmres,k − q peak , bnk* ,m* = 1 , φnD ,m = φnD,m ∪ {k } , ψ Dj = ψ Dj ∪ {( n* , m* )} ; *

*

*

*

*

Eqs. (7), (19)–(21) will be allocated to DUE j . The procedure terminates until all RUs are exhausted, or the rate requirements of all CUEs and DUEs are satisfied.

22: end while 23: Output the allocation policies ani ,m and bnj,m .

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The Journal of China Universities of Posts and Telecommunications

4 Power allocation for given RU allocation After RU allocation, BS only needs to perform the power allocation. Therefore, the RA problem in Eq. (5) can be reduced to: Rtot  max  i j pn ,m , qn ,m P  tot (22)  s.t.  C2,C3,C4,C5 Unfortunately, the problem is still difficult to solve. In this section, a method based on QPSO [9] was proposed to tackle this problem. Generally, QPSO has three important parts, i.e., particle position, fitness function, and evolution equation. The position of each particle represents the possible solution. In this situation, the solution is the possible power allocation. Power resources of CUE i (DUE j) should not be allocated to the RU that not allocated to CUE i (DUE j). Therefore, the particle location D is constituted by pni , m , ∀(n, m) ∈ψ iC ; and qnj, m ,

∀(n, m) ∈ψ Dj .

where, α > 0 is the penalty factor, and Fp is the penalty function which can be given as

all particles, where Blb (t ) is the best position of the l-th particle. Furthermore, Ll (t ) = θ Blb (t ) + (1 − θ )G b (t ) , where

θ ∈ (0,1) is a random variable, and G b (t ) denotes the global best position of all particles. Based on the QPSO, the power allocation algorithm was developed, which was given in Algorithm 2. According to Ref. [6], the computational complexity of the proposed C KD  K  algorithm is O  LT iter  ∑ ||Ψ i C || + ∑ ||Ψ jD ||   , where   j =1  i =1   || ⋅ || denotes the cardinality of a set. Therefore, if the proposed algorithm has good convergence property, the complexity is acceptable. The convergence of the proposed algorithm is evaluated in Sect. 5. Algorithm 2 Energy-efficient power allocation 1: Initialization: set the PopSize L, maximum iteration times T iter ;

Fp = ∑ (max(0, RiCmin − RiC )) 2 +

Dmin j

for t = 1, 2,..., T iter

3:

Calculates M b (t ) and Ll (t ) according to Eqs. (26) and (27), respectively;

4:

for l = 1, 2,..., L

5:

Update the Dl (t + 1) according to Eq. (25);

6:

if U [ Dl (t + 1)] > U [ Blb (t )] , then BS sets Blb (t + 1) = Dl (t + 1) ; else sets Blb (t + 1) = Blb (t ) ; end if if U [ Blb (t + 1)] > U [G b (t )] , then BS sets G b (t + 1) = Blb (t + 1) ;

l = l +1 ;

8: 2

  N i Cmax    max  0, ∑ pn , m − Pi ∑ ∑  + i =1 m =1   n =1  KD

2:

else sets G b (t + 1) = G b (t ) ; end if

i =1

M

choose a best position from Blb (1) as G b (1) ;

7:

KC

∑ (max(0, R

coefficient, u and r are two random variables between 0 1 1 and 1 . M b (t ) = ∑ Blb (t ) is the mean best position of L L

Initialize the power allocation Dl (1) . Set Blb (1) = Dl (1) , and

The fitness function is constructed by the original optimization problem, which is used to evaluate the quality of the obtained solution. Using the method of penalty function, the fitness function is given as: R (2 3) U = tot − α Fp Ptot

KC

2015

9:

end for

10:

t = t +1

11: end for

− R Dj )) 2 +

12: Obtain the optimized power allocation pni ,m and qnj,m .

j =1

2

  N j Dmax    max  0, ∑ qn , m − Q j   ∑∑ j =1 m =1   n =1  where max(⋅, ⋅) will return the larger value. KD M

5 Performance evaluation and discussion (24)

Assume there are L particles, and then the evolution equation of particle l [9] can be given as: 1  Dl (t + 1) = Ll (t ) + β | M b (t ) − Dl (t ) | ln ; r ≥ 0.5   (25) u  1 Dl (t + 1) = Ll (t ) − β | M b (t ) − Dl (t ) | ln ; r < 0.5   u where t is iteration times, β is contraction-expansion

Consider a hexagonal cell with a radius of 500 m, where the BS is centred, and potential DUEs and CUEs are uniformly distributed. Unless specifically noted, N = 25 , K C = 20 , K D = 10 . The scheduling period is 10 ms, M = 20 , T = 0.5 ms. Referring to the parameter settings in Refs. [3,5,7], we assume RiCmin ( MT ) = 300 kbit/s,

Pi Cmax = 24 R Dmin j

dBm,

( MT ) = 500

pit ,C = 10

dBm ,

∀i ∈ ϕ C ; and

kbit/s, Q Dmax = 21 dBm, q Dj = 13 dBm, j

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Xu Quansheng, et al. / End-to-end energy-efficient resource allocation in device-to-device communication…

∀j ∈ ϕ D . Furthermore, N 0W = −114 dBm, W = 180 kHz, p r ,B = 12 dBm, p peak = 19 dBm, q peak = 16 dBm. Fig. 1 illustrates the effect of I C and I D on the EE. It can be observed that the EE first increases with the increase of I C and I D , and then decreases. When I C and I D are small enough, the effect of interference on the transmission rate is not so serious. Then, the increase of I C and I D results in more efficient RU reuse and leads to higher EE. However, when I C and I D are large enough, the effect becomes more serious. Therefore, increasing I C and I D will result in lower EE. In the rest simulations, to obtain desirable EE performance, I C = −70 dBm and I D = −50 dBm are chosen

Fig. 1

29

To validate the effectiveness of the proposed RA scheme, we compare it with two traditional RA schemes. Comparison Scheme 1 named as “RA only considers transmitter energy consumption” which only considers the transmitter energy consumption as shown in Refs. [5,8]. Comparison Scheme 2 named as “RA only considers receiver energy consumption” which only considers the receiver energy consumption as in Ref. [6]. Fig. 3 shows the EE of different RA schemes versus the number of sub-channels. The EE increases gradually as the number of sub-channels increases. The reason is that as the number of sub-channels increase, more bandwidth resources are available, and then better EE can be obtained. In addition, we can also find that the proposed scheme can achieve the best EE performance. This is because that all energy consumptions are considered in the proposed RA scheme.

EE vs. maximum interference temperature

Fig. 2 illustrates the convergence of the proposed power allocation.

Fig. 3

EE vs. the number of sub-channels

Fig. 4 shows the EE of different RA schemes versus the number of UEs ( K C / K D = 2) .

Fig. 2

Convergence of the proposed power allocation

It can be seen that if K C + K D ≤ 30 ( K C / K D = 2) the proposed algorithm can always converges to 95% of the upper bound performance within T iter = 200 iterations. For our considered system bandwidth, 30 users are considerable big. Therefore, the proposed power allocation algorithm is of good convergence performance. In the simulations, T iter = 250 is chosen, that can ensure the convergence of power allocation.

Fig. 4

EE vs. the number of UEs

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The Journal of China Universities of Posts and Telecommunications

The EE decreases gradually as the number of UEs increase. This can be explained as follows. First, when the number of UEs is little, the RUs are relatively sufficient, then better EE is easy to obtain. Second, the receiver circuit energy consumption is considered in our formulation, and it increases with the number of UEs, which result in low EE. Furthermore, Fig. 4 also proves that the proposed RA scheme can achieve best EE performance.

6 Conclusions In this article, the problem of energy-efficient RA in D2D communication underlaying cellular networks is investigated from an end-to-end energy-efficient perspective. The time-frequency RU and power allocation policies are obtained through the proposed two-stage RA scheme with a reasonable computational complexity. Simulation results show that the proposed RA scheme can effectively improve the EE performance comparing with existing work. Acknowledgements This work was supported by the National Science Foundation for

2015

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Young Scientists of China (61302080).

(Editor: Wang Xuying)