Research Article A Novel Energy-Saving Resource Allocation...

Research ArticleA Novel Energy-Saving Resource Allocation Scheme inLTE-A Relay Networks

Jen-Jee Chen Chi-Wen Luo and Zeng-Yu Chen

Department of Electrical Engineering National University of Tainan Tainan 70005 Taiwan

Correspondence should be addressed to Jen-Jee Chen jjchenmailnutnedutw

Received 21 December 2015 Revised 25 June 2016 Accepted 3 July 2016

Academic Editor Gabriel-Miro Muntean

Copyright copy 2016 Jen-Jee Chen et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The relay node (RN) in Long-Term Evolution-Advanced (LTE-A) networks is used to enhance the coverage of high data rate andsolve the coverage hole problem Considering the limited energy nature of User Equipment (UE) connecting to the RN insteadof Evolved Node B (eNB) is a better choice for cell-edge UE items In this paper on the premise of compatibility to the LTE-Aresource allocation specification we discuss an uplink radio resource uplink path modulation and coding scheme (MCS) andtransmit power allocation problem for energy conservation in LTE-A relay networks The objective is to minimize the total energyconsumption of UE items while guaranteeing the constraints of UE itemsrsquo quality of service (QoS) bit-error-rate (BER) totalsystem resource and maximum transmit power Since the problem is NP-complete and the scheduling period in LTE-A is short(the subframe length is only 1ms) we propose an efficient method to solve the problem The complexity analysis shows the timecomplexity of the proposed heuristics is119874(1198992) Simulation results demonstrate that our algorithm can effectively reduce the energyconsumption of UE items and guarantee usersrsquo service quality

1 Introduction

In recent years the third-generation partnership project(3GPP) has proposed the Long-Term Evolution (LTE) [1] andLTE-Advanced (LTE-A) [2] to support mobile and broad-band wireless access in cellular systems In LTELTE-A theOrthogonal Frequency Division Multiple Access (OFDMA)is selected as the downlink access technology which provideshigh spectrum efficiency while in the uplink the Single-Carrier Frequency Division Multiple Access (SC-FDMA)technique is employed to reduce the Peak-to-Average PowerRatio (PAPR) Relay is one of the key features in LTE-A [3]where relays can enhance the coverage of high data ratesincrease the throughput of cell-edge users solve the coveragehole problem and raise bandwidth utilization by spatialreuse Two types of relays are introduced in the LTE-A TypeI relays act like eNBs to the attached UE items and havetheir own physical identities On the contrary Type II relaysare transparent to the UE items and do not have physicalidentities Like most wireless networks energy saving isalways an important issue for UE items due to the battery

capacity restriction Deploying relays cell-edge UE items areable to save more power by connecting to the eNB via relays

In this paper we study the fundamental energy con-servation problem in LTE-A uplink with Type I relays Weconsider an uplink resource pathMCS and power allocationproblem The objective is to minimize the total energy con-sumption of UE items while guaranteeing their constraintsof relay network frame structure maximum transmit powerBER and QoS Low power consumption is particularlyimportant for UE itemsrsquo batteries which can extend theirlifetime Todayrsquos wireless networks are characterized by afixed spectrum assignment policy Reference [4] shows thatthe average around 60 of the spectrum remains unutilizedThis motivates us to exploit the idle spectrum to decrease thepower consumption of UE items and thus increase UE itemsrsquobattery lifetime and the spectrum utilization

In the literature much work has been done for the uplinkresource allocation in LTELTE-A networks Reference [5]proposes the optimal SC-FDMA resource allocation algo-rithm based on a pure binary-integer program to maxi-mize the total user-weighted system capacity Reference [6]

Hindawi Publishing CorporationMobile Information SystemsVolume 2016 Article ID 3696789 14 pageshttpdxdoiorg10115520163696789

2 Mobile Information Systems

presents a set of resource allocation schemes for LTEuplink toachieve the proportional fairness of users while maintaininggood system throughput However the above studies [5 6] donot take relays into consideration For Type I relay networks[7 8] show how to achieve a good trade-off between systemthroughput and global proportional fairness over in-bandand out-band relay networks respectively But both ofthem focus on the downlink resource allocation and energyconservation is not the concern In IEEE 80216 [9] defines aresource allocation problem which aims at the minimizationof energy consumption of UE items The authors discuss therelationship between the MCSs and the energy consumptionof UE The result shows that the UE can decrease (respincrease) its power consumption by choosing a lower (resphigher) level of MCS but spend more (resp less) physicalresources Reference [10] continues and extends the energy-conserved resource allocation problem in IEEE 80216jHowever both studies [9 10] are not valid for LTELTE-AReference [11] examines the effect of Physical Resource Block(PRB) allocation on LTEUErsquos uplink transmission power andenergy consumption Simulation results show that for eachsubframe to allocate asmany PRBs as possible to a single useris more energy efficient than sharing PRBs among severalusers In [12] to improve the energy efficiency user terminalscooperate with each other in transmitting their data packetsto the base station (BS) by exploiting multitypes of wirelessinterfaces To be specific when two UE items are close toeach other they first exchange their data with short rangecommunication interfaces Once the negotiation is done theyshare the antennas to transmit their data to BS by employingdistributed space-time coding Reference [13] proposes twopower-efficient resource schedulers for LTE uplink systemssubject to rate delay contiguous allocation and maximumtransmit power constraints Reference [14] proposes a greenopportunistic and efficient Resource Block (RB) allocationalgorithm for LTE uplink networks which maximizes thesystem throughput in an energy efficient way subject to usersrsquoQoS requirements and SC-FDMA constraints Reference [15]proposes an energy efficient Medium Access Control (MAC)scheme for multiuser LTE downlink transmission whichutilizes the multiuser gain of the MIMO channel and themultiplexing gain of themultibeamopportunistic beamform-ing technique Reference [16] discusses the buffer-overflowand buffer-underflow problems in the LTE-A relay networkand presents adynamic flow control method to minimizethe buffer-overflow and buffer-underflow probabilities Ref-erence [17] discusses the relay selection power allocationand subcarrier assignment problem and proposes a two-level dual decomposition and subgradient method and twolow-complexity suboptimal schemes to maximize the systemthroughput Reference [18] examines the weighted powerminimization problem and jointly optimizes the bandwidthand power usage under constraints on required rate band-width and transmit power So far there is no existing workstudying the LTE-A relay network uplink energy conser-vation issue by considering the uplink path determinationradio resource scheduling and MCS and transmit powerallocation at the same time

BS

RN RN

MUERUE RUE

Backhaul link

Backhaul link

Access linkAccess linkDirect link

Figure 1 The architecture of the LTE-A relay network

In this paper we propose a novel energy-saving resourceand power allocation scheme in LTE-A relay networksOur contributions can be summarized as follows Firstly tothe best of our knowledge this is the first work to studythe energy-conserved uplink resource allocation problem inLTE-A relay networks The proposed method schedules andallocates radio resource uplink path MCS and transmitpower at the same time Multiple realistic factors are consid-ered in the paper such as usersrsquo required data rate and BERLTE-A relay network frame structure and system capacityand the maximum transmit power constraint Secondly weprove the problem to be NP-complete This means that it isimpossible to conduct the optimal solution for the problemin limited time Thirdly a theoretical analysis is done toshow that the complexity of the proposed heuristics is119874(1198992)Finally we conduct a series of simulations to evaluate theperformance of the proposed scheme The simulation resultsconfirm our motivation and show that the proposed methodcan significantly reduce the overall energy consumptionof UE items compared to other schemes guarantee usersrsquothroughput and increase only few extra delays (less than10ms)

The rest of the paper is organized as follows Section 2gives the preliminaries Section 3 presents our energy-conserved uplink resource allocation heuristics Theoreticalcomplexity analysis is given in Section 4 Simulation resultsare shown in Section 5 Section 6 concludes the paper

2 Preliminaries

In this section we first illustrate and define the systemmodelof LTE-A relay networks Then the energy cost model usedin this paper is described Finally we define the energy-conserved uplink resource allocation problem in LTE-A relaynetworks and prove it to be NP-complete [19] To facilitatethe readability Notations shown at the end of the papersummarizes the notations frequently used throughout thepaper

21 System Model In an LTE-A relay network there is oneeNB with 119872 fixed relay nodes (RNs) and 119873 UE items asshown in Figure 1 RNs are deployed to help relay databetween cell-edge UE items and eNB to improve the signal

Mobile Information Systems 3

7 SC-FDMA symbols

12su

bcar

riers

(180

kHz)

1 subframe (1ms)

One uplink slot (05ms)

Figure 2 One TTI is composed of 2 consecutive RBs where eachRB is a 12 (subcarriers) times 7 (symbols) two-dimensional array

quality There is no direct communication between UE itemsor RNs All UE items roam in the eNBrsquos coverage Wecall the UE items transmitting data by eNB ldquoMUErdquo andthe UE items transmitting data by RN ldquoRUErdquo Backhaullinks access links and direct links are the links between theeNB and RNs RUEs and RNs and the eNB and MUEsrespectively In the relay network the resource allocationunit is 2 consecutive Resource Blocks (RBs) in time domaincalled oneTransmission Time Interval (TTI) One RB is a two-dimensional array (12 subcarriers times 7 symbols) One TTIwith two consecutive RBs is as shown in Figure 2 There aretwo types of radio frame structures Time Division Duplex(TDD) mode and Frequency Division Duplex (FDD) mode[20] In TDD the radio resource is divided into frames eachis of 10ms One frame is composed of 10 subframes of 1mseach (as shown in Figure 3) and each subframe is dividedinto two slots The LTE-A allows the resource managementto schedule the resource on a subframe basis In otherwords the shortest scheduling period in LTE-A is 1ms LTE-A supports seven different uplink-downlink configurationsfor the TDD mode as shown in Table 1 Table 2 [21] showsthe subframe configurations for eNB-RN (backhaul link)uplink and downlink in LTE-A relay networks We call thesubframes configured for eNB-RN communication ldquoback-haul subframesrdquo in which both the eNB-RN and eNB-MUEcommunications are allowed On the contrary the subframeswhich are left blank are called ldquononbackhaul subframesrdquo in

Table 1 TDD frame uplink-downlink configuration

Uplink-downlinkconfiguration

Subframe number0 1 2 3 4 5 6 7 8 9

0 D S U U U D S U U U1 D S U U D D S U U D2 D S U D D D S U D D3 D S U U U D D D D D4 D S U U D D D D D D5 D S U D D D D D D D6 D S U U U D S U U D

which the RN-RUE and eNB-MUE communications areallowed Note that in this paper we skip the FDD mode andfocus on the TDD mode Actually our method can apply onboth LTE-A TDD and FDD modes

Figure 4 shows an example which demonstrates how sub-frames are configured in LTE-A relay networks when TDDeNB-RN transmission subframe configuration 1 in Table 2 isused Since configuration 1 adopts uplink-downlink configu-ration 1 in Table 1 subframes 2 3 7 and 8 are for the uplinkand subframes 0 4 5 and 9 are for the downlink In the above8 subframes subframes 3 and 9 are for uplink and downlinkbackhaul subframes respectively in configuration 1 Soin relay networks the other 6 subframes that is subframes0 2 4 5 7 and 8 are nonbackhaul subframes In relay net-works the RN-RUE transmission (access link) is only allowedto use the nonbackhaul subframes while the eNB-RN trans-mission (backhaul link) can only allocate the resource inthe backhaul subframes The eNB-MUE transmission (directlink) is able to use both kinds of subframes

22 EnergyModel Theenergy cost of eachUE119894 119894 = 1 119873

is 119864119894= 119875119894times119879119894 where 119875

119894is the transmit power (in mW) of UE

119894

and 119879119894is the amount of allocated resources (in TTI or symbol

time) to UE119894 In each schedule the required physical resource

of UE119894depends on its MCS MCS

119894 and the data request 120575

119894

(in bits) 119879119894can be derived by 119879

119894= lceil120575119894rate(MCS

119894)rceil In fact

LTE-A uses Channel Quality Indicators (CQIs) to report thecurrent channel condition and each CQI = 119896 119896 = 1 15has its corresponding MCS (denoted by MCS(CQI = 119896))and rate (denoted by rate(CQI = 119896) the unit is bitsTTI)[22] Furthermore for different CQI and different BER (120585)it requires different Signal-to-Interference-plus-Noise Ratio(SINR) Figure 5 shows the required SINR over different 120585 fordifferent CQIs [23] With Figure 5 we can get each UE

119894rsquos

required SINR SINR(CQI119894 120585119894) accordingly For the commu-

nication pair (119894 119895) the perceived SINR (in dB) of receiver 119895can be written as

SINR119894119895

= 10 times log10

119875119894119895

119861 times 1198730+ 119868119894119895

(1)

where 119875119894119895is the received power at receiver 119895 119861 is the effective

bandwidth (in Hz) 1198730is the thermal noise level and 119868

119894119895is

the interference from transmitters other than 119894 which can


One radio frame T = 10ms

Subframe 0 Subframe 2 Subframe 3 Subframe 4 Subframe 5 Subframe 7 Subframe 8

OnesubframeT = 1ms

Oneslot

Subframe 9

DwPTS GP UpPTS DwPTS GP UpPTS

Figure 3 Frame structure of LTE-A TDDmode

1 2 3 4 5 6 8 970Subframenumber

Subframeconfiguration

nB-UD nonbackhaul uplinkdownlink subframe

B-UD backhaul uplinkdownlink subframe

S nB-U B-U nB-D nB-D S nB-U B-DnB-UnB-D

Figure 4 A subframe configuration example for LTE-A relay networks with TDD eNB-RN transmission subframe configuration 1

Table 2 Supported configurations for TDD eNB-RN transmission

Subframe configuration eNB-RN uplink-downlinkconfiguration


0

1

D U1 U D2 D U D3 U D D4 U D U D5

2

U D6 D U7 U D D8 D U D9 U D D D10 D U D D11 3 U D D12 U D D D13

4

U D14 U D D15 U D D16 U D D D17 U D D D D18 5 U D


10minus3

10minus2

10minus1

100

BER

0 5 10 15 20 25minus5

SINR (dB)

Figure 5 Error ratio for different CQIs (the 99 confidenceintervals are depicted in red)

be evaluated by 119868119894119895

= sum119894 =119895

119875119894119895 Ignoring shadow and fading

effect 119875119894119895can be derived by

119875119894119895

=

119866119894times 119866119895times 119875119894

119871119894119895

(2)

where 119866119894and 119866

119895are the antenna gains at UE

119894and RN

119895

respectively and 119871119894119895is the path loss from 119894 (UE

119894) to 119895 (RN

119895or

the eNB) To save UE119894rsquos energy we can minimize its transmit

power subject to the required minimum SINR that is usingMCS(CQI

119894= 119896) UE

119894rsquos data can be correctly decoded by

receiver 119895 with a guaranteed BER 120585119894only when

SINR119894119895

ge SINR (CQI119894= 119896 120585119894) (3)

Consequently by integrating (1) (2) and (3) the requiredtransmit power 119875

119894of UE

119894subject to the applied MCS(CQI

119894)

and requested 120585119894for the communication pair (119894 119895) is

119875119894ge

10SINR(CQI119894 120585119894)10 times (119861 times 119873

0+ 119868119894119895) times 119871119894119895

119866119894times 119866119895

(4)

23 Problem Definition The uplink energy conservationproblem is defined as below We assume that in the LTE-A relay network there is one eNB with 119872 fixed RNs and119873 UE items For each UE

119894 119894 = 1 119873 it has an average

uplink traffic demand 120575119894bitsframe granted by the resource

management of the eNB UE items can uplink data to theeNB either directly or indirectly through RNs Suppose thatthe relative distances between eNBRNs and UE items canbe estimated through existing techniques The objective ofthe problem is to minimize the total energy consumptionof UE items while guaranteeing their required 120585

119894and traffic

demands being all delivered to the eNB subject to the totalamount of physical resources and the maximum transmitpower constraints Without loss of generality we assumethat the total amounts of physical resources for backhaul

and nonbackhaul subframes are 119865B and 119865nB TTIs per framerespectively To solve the problem we have to determine theuplink path resource allocation uplink transmit power 119875

119894

and the used CQI119894of each UE

119894

Theorem 1 The energy conservation problem is NP-complete

Proof To simplify the proof we consider the case of nospatial reuse in the UE-RN transmissions and each UE hasalready selected an appropriate RN according to the channelcondition So each UE can select an MCS to deliver datato RN and each MCS costs different energy consumptionThus the energy cost of one UE item using a specific MCS isuniquely determinedThen we formulate the uplink resourceallocation problem as a decision problem energy-conserveduplink resource allocation decision (EURAD) problem asbelow Given the network topology119866 and the demand of eachUE item we ask whether or not there exists oneMCS set 119878MCSsuch that with the corresponding selectedMCSs all UE itemscan conserve the total amount of energy119876 and satisfy each oftheir demands and the total amount of required RBs is notgreater than the frame size 119880 Then we will show EURADproblem to be NP-complete

We first show that the EURAD problem belongs to NPGiven a problem instance and a solution containing the MCSset it definitely can be verified whether or not the solution isvalid in polynomial time Thus this part is proved

We then reduce the multiple-choice knapsack (MCK)problem [24] which is known to be NP-complete to theEURAD problem When the reduction is done the EURADproblem is proved to be NP-complete

Before the reduction let us briefly introduce the MCKproblem first The MCK problem is a problem in combi-natorial optimization Given a set of 119899 disjointed classes ofobjects where each class 119894 contains119873

119894objects for each object

119883119894119895 119894 = 1 119899 119895 = 1 119873

119894 it has a weight 119906

119894119895and a

profit 119902119894119895 For each class 119894 one and only one object must be

selected that is sum119873119894forall119895=1

119868119894119895

= 1 119894 = 1 119899 where 119868119894119895

= 1

when object119883119894119895is picked and chosen otherwise 119868

119894119895= 0The

problem is to determine which 119899 objects shall be included ina knapsack to maximize the total object profit and the totalweight has to be less than or equal to a given limit119880 and119880 isalso called the capacity constraint So the MCK problem canbe formally formulated as below

max119899

sum

forall119894=1

119873119894

sum

forall119895=1

119902119894119895119868119894119895

subject to119899

sum

forall119894=1

119873119894

sum

forall119895=1

119906119894119895119868119894119895

le 119880

119873119894

sum

forall119895=1

119868119894119895

= 1 119894 = 1 119899

119868119894119895

= 0 1 119894 = 1 119899 119895 = 1 119873119894

(5)

To reduce the MCK problem to the EURAD probleman instance of the MCK problem is constructed as below


Consider that there are 119899 disjointed classes of objects whereeach class 119894 contains 119873

119894objects In each class 119894 every object

119883119894119895

has a profit 119902119894119895

and a weight 119906119894119895 Besides there is a

knapsack with capacity of 119880 The MCK problem is no largerthan 119880 and the total object profit is 119876

An instance of the EURAD problem is also constructedas follows Let 119899 be the number of UE items Each UE

119894has

119873119894MCSs to its connected eNBRN When UE

119894selects one

MCS 119909119894119895 119895 = 1 119873

119894 it will conserve energy of 119902

119894119895(which

is compared to the energy consumption when UE119894uses its

best level of MCS) and the system should allocate RB(s) ofa total size of 119906

119894119895to transmit UE

119894rsquos data to the connected

eNBRN The total frame space is 119880 Our goal is to let all UEitems conserve energy of 119876 and satisfy their demands In thefollowing we will show that theMCK problem has a solutionif and only if the EURAD problem has a solution

Suppose that we have a solution to the EURAD problemwhich is one MCS set 119878MCS with UE itemsrsquo conserved energyand RB allocations Each UE item chooses exact one MCSwhich is able to satisfy its demand The total size of requiredRBs cannot exceed 119880 and the conserved energy of all UEitems is119876 By viewing the availableMCSs of one UE item as aclass of objects and the total number of RBs119880 as the capacityof the knapsack theMCSs in 119878MCS constitute a solution to theMCK problem This proves the only if part

Conversely let 11990911205721

11990921205722

119909119899120572119899

be a solution to theMCKproblemThen for eachUE

119894 119894 = 1 119899 we select one

MCS such that UE119894conserves energy of 119902

119894120572119894and the number

of allocated RB(s) to transmit UE119894rsquos data to its connected

eNBRN is 119906119894120572119894 In this way the conserved energy of all UE

items will be 119876 and the overall RB is no larger than 119880 Thisconstitutes a solution to the EURAD problem thus provingthe only if part

3 Proposed Method

This section illustrates our proposed heuristics The methodis composed of two phases In the first phase each UEselects an uplink path according to the channel condition andadopts the lowest level of MCS that is MCS(CQI = 1) forpower saving If the amount of required radio resources ofUE items exceeds the system capacity the second phase isthen executed The second phase exploits spatial reuse (orconcurrent transmission) and high level of MCS to increasethe radio resource usage efficiency LTE-A relay networksallow multiple UE items to utilize the same radio resourceand transmit concurrently to each of their serving RNs innonbackhaul subframes called spatial reuse Both spatialreuse and high levelMCSs help the reduction of total requiredTTIs of the system In the end the total amounts of requiredTTIsmustmeet the systemcapacity119865B and119865nB andUE itemsrsquorequirements have to be guaranteed

31 Phase I Initialization and Uplink Path Selection Thereare 119872 + 1 candidate uplink paths for UE items that is RN

119895

119895 = 0 119872 Note that RN0is used to represent the central

eNB Initially set 119878119877119895= 0 for eachRN

119895Then for eachUE

119894 119894 =

1 119873 select the RN119895lowast where 119895

lowast= argmax

forall119895SINR

119894119895

as the uplink path and set 119878119877119895lowast = 119878

119877

119895lowast + UE

119894 To minimize

119864total each UE119894applies CQI

119894= 1 This leads to eNBRNs

must allocate more RBs to UE items But in phase I we omitthe total radio resource constraint temporarily The requiredamount of TTIs for UE

119894to deliver data to its connecting RN

119895

can be derived by

119879UE RN119894

= lceil120575119894

rate (CQI119894= 1)

rceil (6)

subsequently RN119895requires radio resource119879RN BS

119894in backhaul

subframes to forward the received data to the eNB119879RN BS119894

canbe conducted by

119879RN BS119894

= sum

119895=1119872

119909119894119895times lceil

120575119894

rate (CQI = 15)rceil (7)

where 119909119894119895

= 1 when RN119895is UE

119894rsquos uplink path otherwise

119909119894119895

= 0 Then check whether sumforall1198941199091198940 =1

119879UE RN119894

le 119865nB andsum119873

119894=1(119879

RN BS119894

+119879UE RN119894

) le 119865B +119865nB or not If yes terminate thealgorithm and return each UE

119894rsquos resource allocation (119879UE RN

119894

and 119879RN BS119894

) uplink path MCS and uplink transmit power119875119894= (10

SINR(CQI119894 120585119894)10 times119861 times1198730times 119871119894119895)(119866119894times119866119895) (refer to (4))

Otherwise go to phase II for further execution

32 Phase II Energy-Saving Resource Allocation Phase II isto satisfy UE itemsrsquo requests with the least additional energyconsumption To reduce the total amount of required RBswe first exploit the concurrent transmission In a concurrenttransmission group 119892

119896 member UE items connect to dif-

ferent eNBRNs and use the same RBs to deliver data Thisreduces the demand of UE items in 119892

119896from sum

forall119894isin119892119896119879UE RN119894

to max119879UE RN119894

| forall119894 isin 119892119896 However the UE items in the

same group will interfere with each other such that the UEitems have to spend extra transmit power to guarantee 120585

119894 To

minimize the additional power consumption we have to findinterference-free UE items to form groups Hence a weightfunction (119882

119894) is defined to evaluate UE items in the network

119882119894of UE

119894 119894 = 1 119873 can be expressed by

119882119894

= 120572 times

(119889119894119895)minus119908

(minℓ=1119873

119889ℓ119895

| 119909ℓ119895

= 0)minus119908

+ 120573

times120575119894

maxℓ=1119873

120575ℓ| 119909ℓ119895

= 0

+ (minus120574)

times (1 + Δ times 119905119894)

times sum

forall120592120592 =119895(sum119873

ℓ=1119909ℓ120592) =0

(119889119894120592)minus119908

(minℓ=1119873

119889ℓ120592

| 119909ℓ120592

= 0)minus119908

(8)

where120572120573 and 120574 are normalized coefficients and120572+120573minus120574 = 1119908 is the spreading factor 119905

119894denotes the number of times

that UE119894has been excluded from concurrent transmission


groups and Δ is the normalized coefficient The values of thethree coefficients 120572 120573 and 120574 control the relative importanceof three factors path loss data quantity and interferencerespectively To form 119892

119896 for each RN

119895 119895 = 0 119872 we

choose one ungroupedUE itemwith themaximumweight inallUE items connecting toRN

119895 that is 119894lowast = argmax

forall119894isin119878119877

119895

119882119894

Then calculate the required transmission power 119894of each

UE119894in 119892119896 where

119894must be able to guarantee 120585

119894 To prevent

119892119896from selecting the UE items which seriously interfere with

others or are interfered with we will check whether 119864119896

=

sumforall119894119894isin119892119896

(119894times 119879

UE RN119894

) is greater than the energy threshold119864th119896or not If yes it means that some communication pairs

suffer great interference from other UE items in 119892119896 The

threshold119864th119896is set to the summation of the required transmit

energy of all UE items in 119892119896as concurrent transmission is

not applied and the same amount of TTIs is consumed as thecase of concurrent transmission If serious interference existsin 119892119896 the exclusion algorithm will be triggered to remove

someUE items from 119892119896The detail of the exclusion algorithm

will be described later After all UE items are assignedconcurrent transmission groups if UE itemsrsquo requests are stillnot satisfied we consider increasing the MCS level of UEitems

For each 119892119896 119896 = 1 119870 (assume there are totally

119870 concurrent transmission groups and 119870 le 119873) we firstcalculate the energy consumption and required number ofRBs of all feasible CQI settingsWe define the penalty function119875119891(119896 119909 119910) to evaluate 119892

119896rsquos penalty when changing its CQI

setting from a low level 119909 to a high level 119910 where 119909 and 119910

are vectors The penalty function is defined as

119875119891(119896 119909 119910) =

Δ119864119896

119909119910

Δ119860119896

119909119910

=

119864119896

119910minus 119864119896

119909

119860119896

119909minus 119860119896

119910

(9)

where 119864119896

119910and 119864

119896

119909are the amount of energy consumption

of 119892119896using MCS(CQI

119892119896= 119910) and MCS(CQI

119892119896= 119909)

respectively and 119860119896

119909and 119860

119896

119910are the number of required RBs

of 119892119896by adopting MCS(CQI

119892119896= 119909) and MCS(CQI

119892119896= 119910)

respectively The group with the least penalty is preferred toupgrade its CQIs Note that uplink resource arrangement hasto follow the resource constraints of backhaul and nonback-haul subframes The algorithm of phase II is as below

(1) For each UE119894 119894 = 1 119873 calculate119882

119894

(2) Set 1198781198771015840

119895= 119878119877

119895for 119895 = 0 119872 119878 = UE

119894 119894 =

1 119873 119896 = 1 119879accessall = sum

forall1198941199091198940 =1119879UE RN119894

and119879all = sum

119873

119894=1(119879

RN BS119894

+ 119879UE RN119894

)

(3) For each 1198781198771015840

119895 choose the UE

119894lowast isin 119878

1198771015840

119895 where 119894

lowast=

argmaxforallUE119894isin119878119877

1015840

119895

119882119894 and set 119892

119896= 119892119896+ UE119894lowast

(4) Calculate 119894for each UE

119894isin 119892119896(refer to (4)) If

119864119896le 119864

th119896 go to the next step otherwise execute the

exclusion algorithm to remove themost infeasible UEfrom 119892

119896(assume it is UE

ℓ) Then set 119892

119896= 119892119896minus UE

ℓ

and update 119905ℓ= 119905ℓ+ 1 and119882

ℓ Repeat step (4)

(5) If |119892119896| gt 1 update 119879

accessall = 119879

accessall minus

sumforall119894isin1198921198961199091198940 =1

119879UE RN119894

+ max119879UE RN119894

| forall119894 isin 119892119896 and

119879all = 119879all minus sumforall119894isin119892119896

119879UE RN119894

+ max119879UE RN119894

| forall119894 isin 119892119896

Set 1198781198771015840

119895= 1198781198771015840

119895minus 119892119896for 119895 = 0 119872 and 119878 = 119878 minus 119892

119896

If 119879accessall le 119865nB and 119879all le 119865B + 119865nB terminate the

algorithm and return the result of resource allocationgrouping uplink path MCS configuration anduplink transmit power If 119878 = 0 go back to step (3)otherwise go to the next step

(6) For each group 119892119896 119896 = 1 119870 form the MCS con-

figuration pattern matrix 119860119896= [119909119896

1 119909

119896

I119896] where

119909119896

weierp= [119909119896

weierp1 119909

119896

weierp|119892119896|]119879 and 119909

119896

weierpis one of feasible MCS

configuration patterns for 119892119896 Then calculate the

energy consumption 119864119896

weierpand the number of required

RBs 119879UE RN119896weierp

for each 119909119896

weierp Note that without loss

of generality we assume that 1198641198961

le sdot sdot sdot le 119864119896

I119896and

119879UE RN1198961

ge sdot sdot sdot ge 119879UE RN119896I119896

(how to efficiently formthe I

119896feasible MCS configuration patterns for 119892

119896is

discussed in Section 34)(7) For each 119892

119896 calculate the penalties from 119909

119896

1to all

possible MCS configuration 119909119896

weierp weierp = 2 I

119896

(8) First consider the set of groups 119860 which can onlybe assigned resource in 119865nB that is 119860 = 119892

119896|

exist119894 isin 119892119896 1199091198940

= 0 For all groups in 119860 select theminimum 119875

119891(119896lowast 119909lowast 119910lowast) and then change 119892

119896lowast rsquos MCS

configuration from 119909lowast to 119910

lowast update 119892119896lowast rsquos required

physical resource and transmit power and recalculateits penalties from 119910

lowast to 119909119896

weierp weierp = (119910

lowast+ 1) I

119896

Check whether new 119879accessall le 119865nB or not If yes go

to the next step otherwise repeat step (8)(9) In this step we consider satisfying the 119865B + 119865nB

constraint The operation is the same as the previousstep but we set 119860 = 119892

119896| forall119896 Each time after

changing a grouprsquos MCS configuration (assume it isgroup 119892

119896lowast) check whether new 119879all le 119865B + 119865nB or

not If yes stop the algorithm and return each UE119894rsquos

119894 = 1 119873 resource allocation grouping resultuplink path MCS and transmit power otherwiserepeat step (9)

33 Exclusion Algorithm When 119864119896gt 119864

th119896 it represents that

some UE items in 119892119896cause severe interference with other

concurrent transmission pairs in the group We use Figure 6to explain this Assume that UE

0 UE1 UE2 and UE

3are in

a concurrent transmission group and RN0(ie eNB) RN

1

RN2 and RN

3are their serving base stations respectively

Take UE1and its serving base station RN

1 for example

Figures 6(a) and 6(b) show the received interference andtransmit interference respectively As shown in Figure 6(a)for UE

1and RN

1 the received interference 119868119903

11= 11987501

+11987521

+

11987531 On the other hand the transmit interference generated

by the transmission pair (UE1RN1) can be calculated by

119868119905

11= 11987510

+11987512

+11987513 Sum up 119868119903

11and 11986811990511 we then derive the

total interference 119868sum11

of the transmission pair (UE1RN1)


RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3

(a) Received interference for (UE1RN1)

RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3

(b) Transmit interference from UE1

Figure 6 An example of the total interference of a transmission pair (UE1RN1)

When 119864119896gt 119864

th119896occurs we must exclude the UE which

causes severe interference from 119892119896to increase the energy

efficiency The detail is as follows

(1) Without loss of generality for the UE items in 119892119896 we

reindex them asUE119898 119898 = 1 |119892

119896| and denote the

set of their uplink eNBRNs by 120598119896 Next for each UE

119898

and its corresponding RN119899 calculate the received

interference 119868119903119898119899

by

119868119903

119898119899= sum

forallUE120572isin119892119896120572 =119898119875120572119899 (10)

Then for each UE119898 calculate the transmit interfer-

ence 119868119905119898119899

as follows

119868119905

119898119899= sum

forallRN120573isin120598119896120573 =119899119875119898120573

(11)

(2) For eachUE119898 119898 = 1 |119892

119896| calculate 119868sum

119898119899= 119868119903

119898119899+

119868119905

119898119899

(3) From all derived 119868sum119898119899

in the previous step select themaximum one 119868

sum119898lowast119899lowast and exclude the pair (119898lowast 119899lowast)

from 119892119896

34 Listing All I119896Feasible MCS Configuration Patterns for

119892119896 For each 119892

119896 the number of possible MCS configurations

is 15|119892119896| Listing and trying all the configurations will havea tremendous cost Actually for a group 119892

119896 only 15 times |119892

119896|

combinations out of 15|119892119896| (even less) need to be consideredLet us discuss this Consider a group 119892

119896= UE

1 UE

|119892119896|

and one of its MCS configurations 119909119896weierp= [119909119896

weierp1 119909

119896

weierp|119892119896|]119879

assume that applying 119909119896weierpwould consume resource 119879UE RN119896

weierp=

max119879UE RN119894

(119909119896

weierp119894) | forall119894 = 119879

UE RN1

(119909119896

weierp1) that is UE

1requires

the largest number of RBs in 119892119896as 119909119896weierpis used In this case

enhancing any UErsquos MCS other than UE1in 119892119896

doesnot reduce the amount of required radio resources butonly increases the energy consumption of 119892

119896 This means

that MCS configurations [119909119896

weierp1 (119909119896

weierp2+ 1) sdot sdot sdot 15 (119909

119896

weierp3+

1) sdot sdot sdot 15 (119909119896

weierp|119892119896|+ 1) sdot sdot sdot 15]

119879 do not have to be taken intoaccount In other words each time only the UE with the

largest amount of required RBs has to be considered In thisway we can greatly reduce the computing complexity Thedetailed procedure of listing all feasible MCS configurationpatterns for a concurrent transmission group 119892

119896is stated as

below

(1) For a group 119892119896 initialize all member UE itemsrsquo MCS

level to MCS(CQI = 1) Calculate each of theirrequired amounts of RBs and the total amount ofenergy consumption Set weierp = 1 and 119909

119896

weierp= [119909119896

weierp1=

MCS(CQI = 1) 119909119896

weierp|119892119896|= MCS(CQI = 1)]

119879

(2) Select the UE with the largest amount of requiredRBs in 119892

119896 If there is a tie randomly select one If

the selected UErsquos MCS level is MCS(CQI = 15) orthe required amount of TTIs is one then go to step(3) if not increase its CQI by one set weierp = weierp + 1calculate 119892

119896rsquos new total amount of required RBs and

total energy consumption and record this candidateMCS configuration pattern 119909

119896

weierp Then repeat step (2)

(3) Check the recorded MCS configuration patterns insteps (1) and (2) If there is more than 1 patternrequiring the same amount of RBs only reserve theone with the least total energy consumption

By the above listing method for each group 119892119896 the total

number of feasible MCS configuration patterns I119896 would

be less than 15 times |119892119896| and even less which is a significant

improvement compared to 15|119892119896|

Theorem 2 For each concurrent transmission group 119892119896 the

amount of feasible MCS configuration patternsI119896le 15times |119892

119896|

4 Complexity Analysis

In this section we analyze the complexity of the proposedmethod Assume there are 119872 RNs and 119873 UE items and theworst case analysis will be illustrated The whole methodcan be divided into two parts The first part includes theuplink path selection and grouping algorithm while thesecond part deals with MCS level reselection The two parts


will be analyzed separately first In the end we sum up thecomplexities of the two parts

Part I Analysis For each UE item calculate 119872 + 1 channelconditions for 119872 RNs and the eNB and then select the bestone from119872+ 1 candidate base stations which will cost

119874 (2 times 119873 (119872 + 1)) sim 119874 (119873119872) (12)

For the spatial reuse group formulation we first calculate theweight of each UE item and this costs 119874(119873) Then selectone UE item with the maximum weight from each RN

119895 119895 =

0 119872 Assume that for each RN119895 119895 = 0 119872 there are

119873119895UE items connecting to it and 119873

0+ sdot sdot sdot + 119873

119872= 119873 So

selecting UE items to form group costs

119874 (1198731) + sdot sdot sdot + 119874 (119873

119872+1) sim 119874 (119873) (13)

Calculate the transmit powers of UE items in a group cost atmost

119874((119872 + 1)2) sim 119874 (119872

2) (14)

Calculate 119864th119896and determine whether a group shall exclude

UE items or not which needs

119874 (119872 + 1) sim 119874 (119872) (15)

If the result is to exclude someUE (UE items) from the groupexecute the exclusion algorithm In the exclusion algorithmwe first find out the UE which has to be excluded Calculatethe transmit interference and received interference of a UEcost 119874(119872 + 119872) Then for a group of UE items the totalcomplexity is

119874 ((119872 + 1) times (119872 +119872)) sim 119874 (1198722) (16)

To find out the UEwith themaximum total interference costs

119874 (119872 + 1) sim 119874 (119872) (17)

After exclusion we have to update the transmit powers of UEitems in the group and check whether the exclusion is neededor not Consider the worst case that the exclusion will berepeatedly executed until there is only oneUE item remainingin the group Then the complexity for finding a spatial reusegroup is

119874 (119872) times (119874 (1198722) + 119874 (119872) + 119874 (119872

2) + 119874 (119872))

sim 119874 (1198723)

(18)

where (119874(1198722)+119874(119872)+119874(1198722)+119874(119872)) is the summation of

(14) (15) (16) and (17) In a worst case we will form at most119873 single member groups and the complexity is

(119874 (119873) + 119874 (119873) + 119874 (1198723)) times 119874 (119873)

sim 119874 (1198732) + 119874 (119873119872

3)

(19)

The first 119874(119873) is the complexity of updating weights aftereach time grouping a groupThe second119874(119873) is the complex-ity of selecting119872 + 1 UE items out of119873 UE items to form agroup The third 119874(119872

3) is the complexity of (18)

Therefore the complexity of Part I is

119874 (119873119872) + 119874 (1198732) + 119874 (119873119872

3) (20)

by summing (12) and (19) up

Part II Analysis For each group 119892119896 119896 = 1 119870 at most

15 times |119892119896| CQI combinations have to be listed For each group

this costs 119874(15|119892119896|) Because |119892

1| + |119892

2| + sdot sdot sdot + |119892

119870| = 119873

the total complexity of listing all CQI combinations can beexpressed as

119874 (15119873) sim 119874 (119873) (21)

Then calculate the penalty table for each groupThis involvesthe transmit power and consumed energy calculation So thecomplexity of calculating the penalty table for a group 119892

119896is

119874 (151003816100381610038161003816119892119896

1003816100381610038161003816) times 119874 (151003816100381610038161003816119892119896

1003816100381610038161003816

2

) sim 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) (22)

The upper bound of (22) is119874(1198723)when the group size |119892119896| =

119872+1 For119870 groups the total complexity is119874(119870) times119874(|119892119896|3)

Selecting the minimum penalty costs 119874(119873) For the selectedgroup we enhance the CQI and then update the penaltytable of the selected group The updating cost is 119874(15|119892

119896|) sim

119874(|119892119896|)

Above MCS level reselection will be repeated until thetotal number of required resources of UE items is less than orequal to the total systembandwidth For theworst case all UEitems have to be upgraded to the highest level of CQI to meetthe requirement In this case the preceding steps must beexecuted 15119873 times An alternative way to evaluate theexecution time is as below Assume that the total number ofrequired resources is sum

forall119894119877119894 where 119877

119894is the largest amount

of required TTIs of group 119894 when CQI = 1 is used Foreach time we upgrade the CQI of a group at least 1 TTI canbe reduced from the number of total required resources SoMCS reselectionmust be executed atmost (sum

forall119894119877119894minus(119865B+119865nB))

times Therefore the execution time of MCS reselection canbe expressed as

119871 = min119874 (15119873) (sum

forall119894

119877119894minus (119865B + 119865nB)) (23)

So the total complexity of Part II is

119874 (119873) + 119874 (119870) times 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) + 119871 times (119874 (119873) + 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816))

le 119874 (119873) + 119874 (1198731198722) + 119871 times (119874 (119873))

le 119874 (119873) + 119874 (1198731198722) + 119874 (15119873) times (119874 (119873))

sim 119874 (1198732) + 119874 (119873119872

2)

(24)

Combining Part I (20) and Part II (24) the total complex-ity is

119874(1198732) + 119874 (119873119872

3) (25)


Table 3 The parameters in our simulation

Parameter ValueChannel bandwidth 10MHzIntersite distance (ISD) 500m (Case 1)

Channel model

119871(119877) = 119875119871LOS(119877) times Prob(119877) + (1 minus Prob(119877)) times 119875119871119873LOS(119877)

119877 distance in kilometerseNB-UE119875119871LOS(119877) = 1034 + 242 log 10(119877)119875119871119873LOS(119877) = 1311 + 428 log 10(119877)

Prob(119877) = min(0018119877 1) times (1 minus exp(minus1198770063)) + exp(minus1198770063)RN-UE119875119871LOS(119877) = 1038 + 209 log 10(119877)119875119871119873LOS(119877) = 1454 + 375 log 10(119877)

Prob(119877) = 05 minusmin(05 5 exp(minus0156119877)) +min(05 5 exp(minus119877003))eNB maximum transmit power 30 dBmeNB maximum antenna gain 14 dBiRN maximum transmit power 30 dBmRNmaximum antenna gain 5 dBiUE maximum transmit power 23 dBmUE maximum antenna gain 0 dBiThermal noise minus174 dBm

Traffic

Case 1Audio 4ndash25 kbitssVideo 32ndash384 kbitssData 60ndash384 kbitssCase 2Audio 4ndash25 kbitss

Consider that119872 is usually a finite constant so the complexityof the proposed method is 119874(1198732)

5 Simulation Results

We develop a simulator in MATLAB to verify the effec-tiveness of our heuristics The system parameters in thesimulation are listed in Table 3 [3] We consider three typesof traffic audio video and data [25] Two traffic cases areapplied in the simulation TrafficCase 1 ismixed trafficwhereeachUE item executes an audio video or data flowwith equalprobability On the other hand Traffic Case 2 only containsaudio traffic The network contains one eNB and six RNs(119872 = 6) RNs are uniformly deployed inside the 23 coveragerange of the eNB to get the best performance gain In defaultwe set the factors 120572 120573 and 120574 to 1 to get the best performanceand adopt TDDmode uplink-downlink configuration 1 thatis there are 4 uplink subframes per frame The ratio ofuplink backhaul subframe and uplink nonbackhaul subframeis 1 3 We compare the performances of four methods (1)OEA (Opportunistic and Efficient RB Allocation) [14] (2)EPAR (Equal Power Allocation with Refinement) [17] (3) ourproposed scheme without relay nodes and (4) our proposedscheme

Figures 7(a) and 7(b) evaluate the total energy con-sumption of UE items under different number of UE items

(119873) when Traffic Cases 1 and 2 are applied respectivelyBoth figures show that as 119873 increases the total amount ofenergy consumption of UE items increases for all methodsOEA consumes the most energy because UE items alwaysconnect to the eNB and select the most efficient MCS fortransmission EPAR performs better than OEA because cell-edge UE items can choose to connect with RNs instead ofthe eNB and this reduces the energy consumption Withour energy-saving resource allocation method the proposedscheme (wo relay) performs the second Results show thatour proposed scheme performs the best in all methods Thismeans that spatial reuse and RNs do help the reductionof total energy consumption of UE items In Figure 7(b)our heuristics still performs the best compared to the other3 methods Obviously the spatial reuse and energy-savingresource allocation do help to conserve UE itemsrsquo energyOne interesting thing is that when 119873 is large EPAR andthe proposed scheme (wo relay) consume almost the sameenergy This is because relay improves the SINR of cell-edgeusers thus reducing the energy consumption of edge users

Figures 8(a) and 8(b) evaluate the bandwidth utilizationunder different number of UE items for Traffic Cases 1 and 2respectively OEA and EPAR always pursue the most efficientMCSWhen the traffic load is light the bandwidth utilizationhurts and results inmuch idle bandwidth On the other handthe proposed scheme and proposed scheme wo relay get the


OEAEPAR

Proposed scheme (wo relay)Proposed scheme

15 20 25 30 35 40 45 50 5510N

Ener

gy co

nsum

ptio

n(W

lowastsu

bfra

me-

time)

00005

0010015

0020025

0030035

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

Ener

gy co

nsum

ptio

n

000002000040000600008

000100012000140001600018

(Wlowast

subf

ram

e-tim

e)

(b) Traffic Case 2

Figure 7 The impact of119873 on the total energy consumption (119872 = 6)

OEAEPAR


10 20 30 40 50 601N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 14020N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(b) Traffic Case 2

Figure 8 The impact of119873 on the bandwidth utilization (119872 = 6)

best bandwidth utilization in all four methods The resultsshow that our proposedmethods can improve the bandwidthutilization and save more energy for UE items

Figures 9(a) and 9(b) show the impact of 119873 on thethroughput for Traffic Cases 1 and 2 respectively As shownin the figures as 119873 increases the throughput of all schemesincreasesWe can see that the proposedmethods can guaran-tee all the traffic demand being served like OEA and EPARThis means that when the network load is light our schemescan well utilize the idle bandwidth to reduce UE itemsrsquo uplinktransmit power On the contrary when the network load isheavy our schemes will select efficient MCS for UE itemsto reduce each of their required physical radio resourcessuch that the admitted data rates of UE items can still besatisfied So our proposed schemes can not only providesimilar throughput like OEA and EPAR but also save UEitemsrsquo energy

Figure 10 shows the average extra data transmission delayof the proposed schemes and EPAR against OEA Comparedto OEA EPAR causes a longer delay because RUEs haveto deliver their data to the eNB via RNs But in OEA UEitems directly transmit their data to the eNB The proposed

schemes have a longer delay compared to both OEA andEPAR because they utilize more physical resources to deliverdata thus resulting in more extra data packet buffering delayAs119873 increases the result shows that the extra delay does notalways increase (when119873 le 20) but decreases after119873 is morethan 20This is becauseOEAneedsmore time to deliver usersrsquodata when traffic load is heavy but the proposed schemesconsume the same time and upgrade UE itemsrsquo MCS levelinstead Our proposed methods slightly increase the delay ofdata transmission but the average extra delay is nomore than5ms as shown in Figure 10 It should be acceptable

In Figure 11 we discuss the effect of subframe configu-ration on the total energy consumption of UE items In theTDD mode LTE-A relay network it supports four kinds ofuplink nonbackhaul and backhaul subframe configurations(1) 1 uplink nonbackhaul subframe and 1 uplink backhaulsubframe per frame (1a 1b) (2) 2 uplink nonbackhaul sub-frames and 1uplink backhaul subframeper frame (2a 1b) (3)2 uplink nonbackhaul subframes and 2 uplink backhaul sub-frames per frame (2a 2b) and (4) 3 uplink nonbackhaul sub-frames and 1 uplink backhaul subframe per frame (3a 1b) Asshown in Figure 11 no matter which subframe configurations


OEAEPAR


15 20 25 30 35 40 45 50 5510N

0100020003000400050006000700080009000

Thro

ughp

ut (k

bps)

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

0

1000

1500

500

2000

2500

3000

Thro

ughp

ut (k

bps)

(b) Traffic Case 2

Figure 9 The impact of119873 on the throughput (119872 = 6)

EPAROEAProposed scheme (wo relay)OEAProposed schemeOEA

10 20 30 40 50 601N

0

2

4

6

8

10

Extr

a del

ay (m

s)

Figure 10 The average extra data transmission delay of all schemescompared to OEA (119872 = 6 Traffic Case 1)

1a 1b2a 1b 3a 1b

2a 2b

Ener

gy co

nsum

ptio

n

(wo relay) schemeEPAROEA ProposedProposed scheme

Method

times10minus3

0

5

10

15

20

25

(Wlowast

subf

ram

e-tim

e)

Figure 11 The impact of subframe configurations on the totalenergy consumption (119873 = 35 and119872 = 6 Traffic Case 1)

are used our method always gets the best power saving in allschemes For OEA and EPAR the performances are almostthe same for all four kinds of subframe configurations Thisis because they always use the most efficient MCS no matterwhether the uplink radio resources are many or few Theproposed schemes reduce the energy consumption of UEitems by well utilizing the idle radio resource Thereforethe result shows that the total energy consumption of UEitems decreases in the proposed methods as the number ofuplink subframe increases (number of uplink subframes perframe is increased from 2 (1a 1b) to 4 (2a 2b or 3a 1b))When the network has more radio resources UE items canchoose to use lower level of MCS to transmit data andsave energy Comparing subframe configurations 2a 2b and3a 1b Figure 11 shows that the latter can conserve moreenergyThe higher number of nonbackhaul subframesmeansthere aremore resources that can be used byMUEs andRUEsbut the backhaul subframe can only be utilized by MUEsObviously the former provides more flexibility This is whysubframe configuration 3a 1b conducts better energy savingthan that of 2a 2b

In Figure 12 Traffic Case 2 is applied to evaluate the effectof subframe configuration on the total energy consumptionof UE items The proposed scheme performs the best in all 4schemes Compared to the previous experiment as shown inFigure 11 Figure 12 shows that the performance differencesamong all four schemes become smaller This is because inTraffic Case 2 the data size is small compared to the numberof radio resources provided in one single TTI then in ourimplementation OEA and EPAR will automatically apply alow level MCS to fill up the whole space of assigned radioresource This is why we see a closer performance among thefour schemes

Then Figure 13 evaluates the total energy consumption ofUE items over different ratio of 120573120572 Figure 13 presents that as120573120572 increases the total energy consumption decreases when120573120572 le 1This means that factor 1 (path loss factor) and factor2 (data size factor) of (8) have equal importance forweight119882

119894

When choosing the reuse group the distance between a UEitem and the connected RN and the size of the data requestare both significant factors for energy conservation



Method

1a 1b2a 1b 3a 1b

2a 2b

0

times10minus3

Ener

gy co

nsum

ptio

n

010203040506070809

(Wlowast

subf

ram

e-tim

e)


0 04 06 08 1 1202120573120572

096

097

098

099

1

Nor

mal

ized

ener

gy co

nsum

ptio

n

Figure 13The impact of 120573120572 on the total energy consumption (119873 =

40 and119872 = 3)

Figure 14 shows the total energy consumption overdifferent 120574 wherewe set120572 = 120573 = 1 It can be seen that the totalenergy consumption performs the worst when 120574 = 0 Thismeans that 120574 does help the selection of spatial reuse groupsWith a nonzero 120574 we can filter out unsuitable UE items whenforming reuse groups

6 Conclusion

In this paper we investigate the energy conservation issueof the uplink path uplink radio resource MCS and mobiledevice transmit power allocation in LTE-A relay networksWe have proposed heuristics to conserve UE itemsrsquo energyby exploiting RNs MCS BER transmit power and spatialreuse To save energy the key factors are how to determinethe most energy-saving MCS of UE items and how toselect interference-free spatial reuse groups To find the bestsettings we have defined the weight and penalty functionsfor evaluation Simulation results show that our scheme can

02 04 08060 1120574

0

02

04

06

08

1

Nor

mal

ized

ener

gy co

nsum

ptio

n

Figure 14 The impact of 120574 on the total energy consumption where120573 = 120572 = 1 (119873 = 50 and119872 = 3)

significantly reduce the total energy consumption of UEitems compared to other schemes and has good bandwidthutilization Compared with OEA and EPAR schemes ourproposed energy-saving resource allocation method willslightly increase the delay of data but the extra delay is lessthan one frame (no more than 10ms) Usersrsquo required QoSBER and throughput can all be guaranteed

Notations

119873 Number of UE items119872 Number of RNs119865B The total amount of TTIs for uplink

backhaul subframes per frame119865nB The total amount of TTIs for uplink

nonbackhaul subframes per frame119875119894 The transmit power of UE

119894

119864119894 The energy cost of UE

119894

120575119894 The uplink traffic demand of UE

119894per

frame119879UE RN119894

The amount of required TTIs for UE119894to

deliver data to its connected RN119879RN BS119894

The amount of required TTIs for UE119894rsquos

connected RN to deliver data to the eNB119882119894 The weight of UE

119894

119892119896 The concurrent transmission group 119896

119864th119896 Energy threshold of 119892

119896

119864119896

119909 Total amount of energy consumption of

119892119896when using CQI 119909

119860119896

119909 Total amount of required uplink TTIs

for 119892119896when using CQI 119909

119868119905

119898119899 Transmit interference for the

transmission pair (UE119898RN119899)

119868119903

119898119899 Received interference for the


119889119894119895 The distance between UE

119894and RN

119895

119905119894 Number of exclusion times of UE

119894

rate(CQI = 119896) The code rate when using CQI 119896 (inbitsTTI)


MCS(CQI = 119896) The corresponding MCS when usingCQI 119896

119861 Effective bandwidth (in Hz)1198730 Thermal noise

119866119894 Antenna gain of node 119894

119875119894119895 The received power from transmitter 119894

to receiver 119895119868119894119895 The interference to receiver 119895 from

transmitters other than 119894

119871119894119895 The path loss from transmitter 119894 to

receiver 119895

Competing Interests

The authors declare that they have no competing interests

Acknowledgments

This research is sponsored by MOST 104-2221-E-024-005

References

[1] D Astely E Dahlman A Furuskar Y Jading M Lindstromand S Parkvall ldquoLTE the evolution ofmobile broadbandrdquo IEEECommunications Magazine vol 47 no 4 pp 44ndash51 2009

[2] 3GPP TR 36913 v1200 ldquoRequirements for further advance-ments for E-UTRA (LTE-Advanced)rdquo September 2014

[3] 3GPP ldquoFurther advancements for E-UTRA physical layeraspectsrdquo 3GPP TR 36814 v900 2010

[4] P K Dalela A Nayak V Tyagi and K Sridhara ldquoAnalysis ofspectrumutilization for existing cellular technologies in contextto cognitive radiordquo in Proceedings of the 2nd International Con-ference on Computer and Communication Technology (ICCCTrsquo11) pp 585ndash588 Allahabad India September 2011

[5] I C Wong O Oteri and W McCoy ldquoOptimal resourceallocation in uplink SC-FDMA systemsrdquo IEEE Transactions onWireless Communications vol 8 no 5 pp 2161ndash2165 2009

[6] S-B Lee I Pefkianakis A Meyerson S Xu and S LuldquoProportional fair frequency-domain packet scheduling for3GPP LTE uplinkrdquo in Proceedings of the 28th Conference onComputer Communications (IEEE INFOCOM rsquo09) pp 2611ndash2615 IEEE Rio de Janeiro Brazil April 2009

[7] Z Ma W Xiang H Long and W Wang ldquoProportional fairresource partition for LTE-advanced networks with type I relaynodesrdquo in Proceedings of the IEEE International Conference onCommunications (ICC rsquo11) pp 1ndash5 Kyoto Japan June 2011

[8] G Liebl TM deMoraes A Soysal and E Seidel ldquoFair resourceallocation for the relay backhaul link in LTE-Advancedrdquo in Pro-ceedings of the EEE Wireless Communications and NetworkingConference (WCNC rsquo12) pp 1196ndash1201 Shanghai China April2012

[9] J-P Yoon W-J Kim J-Y Baek and Y-J Suh ldquoEfficient uplinkresource allocation for power saving in IEEE 80216 OFDMAsystemsrdquo in Proceedings of the IEEE 67th Vehicular TechnologyConference (VTC Spring rsquo08) pp 2167ndash2171 Singapore May2008

[10] J-M Liang Y-C Wang J-J Chen J-H Liu and Y-C TsengldquoEnergy-efficient uplink resource allocation for IEEE 80216jtransparent-relay networksrdquoComputer Networks vol 55 no 16pp 3705ndash3720 2011

[11] M Lauridsen A R Jensen and P Mogensen ldquoReducingLTE uplink transmission energy by allocating resourcesrdquo inProceedings of the IEEE 74th Vehicular Technology Conference(VTC Fall rsquo11) pp 1ndash5 September 2011

[12] Y Zou J Zhu and B Y Zheng ldquoEnergy efficiency of networkcooperation for cellular uplink transmissionsrdquo in Proceedings ofthe IEEE International Conference onCommunications (ICC rsquo13)pp 4394ndash4398 IEEE Budapest Hungary June 2013

[13] M Kalil A Shami and A Al-Dweik ldquoQoS-aware power-efficient scheduler for LTE uplinkrdquo IEEE Transactions onMobileComputing vol 14 no 8 pp 1672ndash1685 2015

[14] F Z Kaddour E Vivier LMrouehM Pischella and PMartinsldquoGreen opportunistic and efficient resource block allocationalgorithm for LTE uplink networksrdquo IEEE Transactions onVehicular Technology vol 64 no 10 pp 4537ndash4550 2015

[15] R Imran M Shukair N Zorba O Kubbar and C VerikoukisldquoA novel energy saving MIMO mechanism in LTE systemsrdquo inProceedings of the IEEE International Conference on Communi-cations (ICC rsquo13) pp 2449ndash2453 Budapest Hungary June 2013

[16] P-C Lin R-G Cheng and Y-J Chang ldquoA dynamic flowcontrol algorithm for LTE-advanced relay networksrdquo IEEETransactions onVehicular Technology vol 63 no 1 pp 334ndash3432014

[17] M S Alam J W Mark and X S Shen ldquoRelay selectionand resource allocation for multi-user cooperative OFDMAnetworksrdquo IEEE Transactions on Wireless Communications vol12 no 5 pp 2193ndash2205 2013

[18] N Krishnan R D Yates N B Mandayam and J S PanchalldquoBandwidth sharing for relaying in cellular systemsrdquo IEEETransactions on Wireless Communications vol 11 no 1 pp 117ndash129 2012

[19] T H Cormen C E Leiserson R L Rivest and C SteinIntroduction to Algorithms MIT Press Cambridge Mass USA3rd edition 2009

[20] 3GPP ldquoPhysical channels and modulationrdquo 3GPP TS 36211v1300 2015

[21] 3GPP TS 36216 v1300 ldquoPhysical layer for relaying operationrdquoDecember 2015

[22] 3GPP ldquoE-UTRA physical layer proceduresrdquo 3GPP TS 36213v1200 2013

[23] J Blumenstein J Ikuno J C Prokopec andM Rupp ldquoSimulat-ing the long term evolution uplink physical layerrdquo inProceedingsof the ELMAR pp 141ndash144 IEEE Zadar Croatia September2011

[24] H Kellerer U Pferschy and D Pisinger Knapsack ProblemsSpringer Berlin Germany 2004

[25] 3GPP ldquoServices and service capabilitiesrdquo 3GPP TS 22105v1000 2011

Submit your manuscripts athttpwwwhindawicom

Computer Games Technology

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014


Distributed Sensor Networks


Advances in

FuzzySystems

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014


ReconfigurableComputing

Hindawi Publishing Corporation httpwwwhindawicom Volume 2014


Applied Computational Intelligence and Soft Computing

thinspAdvancesthinspinthinsp

Artificial Intelligence

HindawithinspPublishingthinspCorporationhttpwwwhindawicom Volumethinsp2014

Advances inSoftware EngineeringHindawi Publishing Corporationhttpwwwhindawicom Volume 2014


Electrical and Computer Engineering

Journal of

Journal of

Computer Networks and Communications


Hindawi Publishing Corporation

httpwwwhindawicom Volume 2014

Advances in

Multimedia


Biomedical Imaging


ArtificialNeural Systems

Advances in


RoboticsJournal of



Computational Intelligence and Neuroscience

Industrial EngineeringJournal of


Modelling amp Simulation in EngineeringHindawi Publishing Corporation httpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Human-ComputerInteraction

Advances in

Computer EngineeringAdvances in



presents a set of resource allocation schemes for LTEuplink toachieve the proportional fairness of users while maintaininggood system throughput However the above studies [5 6] donot take relays into consideration For Type I relay networks[7 8] show how to achieve a good trade-off between systemthroughput and global proportional fairness over in-bandand out-band relay networks respectively But both ofthem focus on the downlink resource allocation and energyconservation is not the concern In IEEE 80216 [9] defines aresource allocation problem which aims at the minimizationof energy consumption of UE items The authors discuss therelationship between the MCSs and the energy consumptionof UE The result shows that the UE can decrease (respincrease) its power consumption by choosing a lower (resphigher) level of MCS but spend more (resp less) physicalresources Reference [10] continues and extends the energy-conserved resource allocation problem in IEEE 80216jHowever both studies [9 10] are not valid for LTELTE-AReference [11] examines the effect of Physical Resource Block(PRB) allocation on LTEUErsquos uplink transmission power andenergy consumption Simulation results show that for eachsubframe to allocate asmany PRBs as possible to a single useris more energy efficient than sharing PRBs among severalusers In [12] to improve the energy efficiency user terminalscooperate with each other in transmitting their data packetsto the base station (BS) by exploiting multitypes of wirelessinterfaces To be specific when two UE items are close toeach other they first exchange their data with short rangecommunication interfaces Once the negotiation is done theyshare the antennas to transmit their data to BS by employingdistributed space-time coding Reference [13] proposes twopower-efficient resource schedulers for LTE uplink systemssubject to rate delay contiguous allocation and maximumtransmit power constraints Reference [14] proposes a greenopportunistic and efficient Resource Block (RB) allocationalgorithm for LTE uplink networks which maximizes thesystem throughput in an energy efficient way subject to usersrsquoQoS requirements and SC-FDMA constraints Reference [15]proposes an energy efficient Medium Access Control (MAC)scheme for multiuser LTE downlink transmission whichutilizes the multiuser gain of the MIMO channel and themultiplexing gain of themultibeamopportunistic beamform-ing technique Reference [16] discusses the buffer-overflowand buffer-underflow problems in the LTE-A relay networkand presents adynamic flow control method to minimizethe buffer-overflow and buffer-underflow probabilities Ref-erence [17] discusses the relay selection power allocationand subcarrier assignment problem and proposes a two-level dual decomposition and subgradient method and twolow-complexity suboptimal schemes to maximize the systemthroughput Reference [18] examines the weighted powerminimization problem and jointly optimizes the bandwidthand power usage under constraints on required rate band-width and transmit power So far there is no existing workstudying the LTE-A relay network uplink energy conser-vation issue by considering the uplink path determinationradio resource scheduling and MCS and transmit powerallocation at the same time

BS

RN RN

MUERUE RUE

Backhaul link

Backhaul link

Access linkAccess linkDirect link

Figure 1 The architecture of the LTE-A relay network

In this paper we propose a novel energy-saving resourceand power allocation scheme in LTE-A relay networksOur contributions can be summarized as follows Firstly tothe best of our knowledge this is the first work to studythe energy-conserved uplink resource allocation problem inLTE-A relay networks The proposed method schedules andallocates radio resource uplink path MCS and transmitpower at the same time Multiple realistic factors are consid-ered in the paper such as usersrsquo required data rate and BERLTE-A relay network frame structure and system capacityand the maximum transmit power constraint Secondly weprove the problem to be NP-complete This means that it isimpossible to conduct the optimal solution for the problemin limited time Thirdly a theoretical analysis is done toshow that the complexity of the proposed heuristics is119874(1198992)Finally we conduct a series of simulations to evaluate theperformance of the proposed scheme The simulation resultsconfirm our motivation and show that the proposed methodcan significantly reduce the overall energy consumptionof UE items compared to other schemes guarantee usersrsquothroughput and increase only few extra delays (less than10ms)

The rest of the paper is organized as follows Section 2gives the preliminaries Section 3 presents our energy-conserved uplink resource allocation heuristics Theoreticalcomplexity analysis is given in Section 4 Simulation resultsare shown in Section 5 Section 6 concludes the paper

2 Preliminaries

In this section we first illustrate and define the systemmodelof LTE-A relay networks Then the energy cost model usedin this paper is described Finally we define the energy-conserved uplink resource allocation problem in LTE-A relaynetworks and prove it to be NP-complete [19] To facilitatethe readability Notations shown at the end of the papersummarizes the notations frequently used throughout thepaper

21 System Model In an LTE-A relay network there is oneeNB with 119872 fixed relay nodes (RNs) and 119873 UE items asshown in Figure 1 RNs are deployed to help relay databetween cell-edge UE items and eNB to improve the signal


7 SC-FDMA symbols

12su

bcar

riers

(180

kHz)

1 subframe (1ms)











is 119864119894= 119875119894times119879119894 where 119875


119894





119894


119894= lceil120575119894rate(MCS



119894rsquos



SINR119894119895

= 10 times log10

119875119894119895

119861 times 1198730+ 119868119894119895

(1)



119894119895is





OnesubframeT = 1ms

Oneslot

Subframe 9












0

1


2


4



10minus3

10minus2

10minus1

100

BER

0 5 10 15 20 25minus5

SINR (dB)


be evaluated by 119868119894119895

= sum119894 =119895



119875119894119895

=

119866119894times 119866119895times 119875119894

119871119894119895

(2)

where 119866119894and 119866


119894and RN

119895


119894) to 119895 (RN

119895or



119894= 119896) UE



SINR119894119895

ge SINR (CQI119894= 119896 120585119894) (3)


119894of UE


119894)


119875119894ge


0+ 119868119894119895) times 119871119894119895

119866119894times 119866119895

(4)


119894 119894 = 1 119873 it has an average



119894and traffic



119894


119894







119883119894119895 119894 = 1 119899 119895 = 1 119873


119894119895and a



119868119894119895

= 1 119894 = 1 119899 where 119868119894119895

= 1


119894119895= 0The


max119899

sum

forall119894=1

119873119894

sum

forall119895=1

119902119894119895119868119894119895

subject to119899

sum

forall119894=1

119873119894

sum

forall119895=1

119906119894119895119868119894119895

le 119880

119873119894

sum

forall119895=1

119868119894119895

= 1 119894 = 1 119899

119868119894119895

= 0 1 119894 = 1 119899 119895 = 1 119873119894

(5)





119883119894119895

has a profit 119902119894119895




119894has


119894selects one

MCS 119909119894119895 119895 = 1 119873


119894119895(which








11990921205722

119909119899120572119899


119894 119894 = 1 119899 we select one


119894120572119894and the number




3 Proposed Method



119895




119894 119894 =


lowast= argmax

forall119895SINR

119894119895


119877

119895lowast + UE

119894 To minimize





119895

can be derived by

119879UE RN119894

= lceil120575119894

rate (CQI119894= 1)

rceil (6)


119894in backhaul


canbe conducted by

119879RN BS119894

= sum

119895=1119872

119909119894119895times lceil

120575119894


where 119909119894119895



119909119894119895


119879UE RN119894


119894=1(119879

RN BS119894

+119879UE RN119894



119894

and 119879RN BS119894







119896from sum

forall119894isin119892119896119879UE RN119894

to max119879UE RN119894



119894 To



119882119894of UE


119882119894

= 120572 times

(119889119894119895)minus119908

(minℓ=1119873

119889ℓ119895

| 119909ℓ119895

= 0)minus119908

+ 120573

times120575119894

maxℓ=1119873

120575ℓ| 119909ℓ119895

= 0

+ (minus120574)

times (1 + Δ times 119905119894)

times sum

forall120592120592 =119895(sum119873

ℓ=1119909ℓ120592) =0

(119889119894120592)minus119908

(minℓ=1119873

119889ℓ120592

| 119909ℓ120592

= 0)minus119908

(8)






119896 for each RN

119895 119895 = 0 119872 we



forall119894isin119878119877

119895

119882119894


UE119894in 119892119896 where


119894 To prevent



=

sumforall119894119894isin119892119896

(119894times 119879

UE RN119894













119875119891(119896 119909 119910) =

Δ119864119896

119909119910

Δ119860119896

119909119910

=

119864119896

119910minus 119864119896

119909

119860119896

119909minus 119860119896

119910

(9)

where 119864119896

119910and 119864

119896



119892119896= 119910) and MCS(CQI

119892119896= 119909)


119909and 119860

119896



119892119896= 119909) and MCS(CQI

119892119896= 119910)



119894

(2) Set 1198781198771015840

119895= 119878119877

119895for 119895 = 0 119872 119878 = UE

119894 119894 =

1 119873 119896 = 1 119879accessall = sum

forall1198941199091198940 =1119879UE RN119894

and119879all = sum

119873

119894=1(119879

RN BS119894

+ 119879UE RN119894

)

(3) For each 1198781198771015840



1198771015840

119895 where 119894

lowast=


1015840

119895

119882119894 and set 119892

119896= 119892119896+ UE119894lowast


119894isin 119892119896(refer to (4)) If

119864119896le 119864





119896= 119892119896minus UE

ℓ

and update 119905ℓ= 119905ℓ+ 1 and119882

ℓ Repeat step (4)

(5) If |119892119896| gt 1 update 119879

accessall = 119879

accessall minus

sumforall119894isin1198921198961199091198940 =1

119879UE RN119894

+ max119879UE RN119894

| forall119894 isin 119892119896 and


119879UE RN119894

+ max119879UE RN119894

| forall119894 isin 119892119896

Set 1198781198771015840

119895= 1198781198771015840


119896





1 119909

119896

I119896] where

119909119896

weierp= [119909119896

weierp1 119909

119896

weierp|119892119896|]119879 and 119909

119896






for each 119909119896




I119896and

119879UE RN1198961




119896is



119896

1to all


weierp weierp = 2 I

119896


119896|

exist119894 isin 119892119896 1199091198940







lowast to 119909119896


lowast+ 1) I

119896













0 UE1 UE2 and UE

3are in


1

RN2 and RN



1 for example


1and RN


11= 11987501

+11987521

+



119868119905

11= 11987510

+11987512

+11987513 Sum up 119868119903





RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3


RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3



When 119864119896gt 119864








119898


interference 119868119903119898119899

by

119868119903

119898119899= sum

forallUE120572isin119892119896120572 =119898119875120572119899 (10)


ence 119868119905119898119899

as follows

119868119905

119898119899= sum

forallRN120573isin120598119896120573 =119899119875119898120573

(11)

(2) For eachUE119898 119898 = 1 |119892


119898119899= 119868119903

119898119899+

119868119905

119898119899




from 119892119896


119892119896 For each 119892



119896 only 15 times |119892

119896|


119896= UE

1 UE

|119892119896|


weierp1 119909

119896

weierp|119892119896|]119879


weierp=

max119879UE RN119894

(119909119896

weierp119894) | forall119894 = 119879

UE RN1

(119909119896

weierp1) that is UE

1requires




119896 This means


weierp1 (119909119896


119896

weierp3+

1) sdot sdot sdot 15 (119909119896




119896is stated as

below



119896

weierp= [119909119896

weierp1=

MCS(CQI = 1) 119909119896

weierp|119892119896|= MCS(CQI = 1)]

119879






119896









119896|






119874 (2 times 119873 (119872 + 1)) sim 119874 (119873119872) (12)


119895 119895 =




119872= 119873 So


119874 (1198731) + sdot sdot sdot + 119874 (119873

119872+1) sim 119874 (119873) (13)


119874((119872 + 1)2) sim 119874 (119872

2) (14)



119874 (119872 + 1) sim 119874 (119872) (15)


119874 ((119872 + 1) times (119872 +119872)) sim 119874 (1198722) (16)


119874 (119872 + 1) sim 119874 (119872) (17)


119874 (119872) times (119874 (1198722) + 119874 (119872) + 119874 (119872

2) + 119874 (119872))

sim 119874 (1198723)

(18)



(119874 (119873) + 119874 (119873) + 119874 (1198723)) times 119874 (119873)

sim 119874 (1198732) + 119874 (119873119872

3)

(19)




119874 (119873119872) + 119874 (1198732) + 119874 (119873119872

3) (20)





1| + |119892


119870| = 119873


119874 (15119873) sim 119874 (119873) (21)


119896is

119874 (151003816100381610038161003816119892119896

1003816100381610038161003816) times 119874 (151003816100381610038161003816119892119896

1003816100381610038161003816

2

) sim 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) (22)




119896|) sim

119874(|119892119896|)


forall119894119877119894 where 119877



forall119894119877119894minus(119865B+119865nB))


119871 = min119874 (15119873) (sum

forall119894

119877119894minus (119865B + 119865nB)) (23)


119874 (119873) + 119874 (119870) times 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) + 119871 times (119874 (119873) + 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816))

le 119874 (119873) + 119874 (1198731198722) + 119871 times (119874 (119873))

le 119874 (119873) + 119874 (1198731198722) + 119874 (15119873) times (119874 (119873))

sim 119874 (1198732) + 119874 (119873119872

2)

(24)


119874(1198732) + 119874 (119873119872

3) (25)




Channel model





Traffic









OEAEPAR


15 20 25 30 35 40 45 50 5510N

Ener

gy co

nsum

ptio

n(W

lowastsu

bfra

me-

time)

00005

0010015

0020025

0030035

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

Ener

gy co

nsum

ptio

n

000002000040000600008

000100012000140001600018

(Wlowast

subf

ram

e-tim

e)

(b) Traffic Case 2


OEAEPAR


10 20 30 40 50 601N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 14020N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(b) Traffic Case 2








OEAEPAR


15 20 25 30 35 40 45 50 5510N

0100020003000400050006000700080009000

Thro

ughp

ut (k

bps)

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

0

1000

1500

500

2000

2500

3000

Thro

ughp

ut (k

bps)

(b) Traffic Case 2



10 20 30 40 50 601N

0

2

4

6

8

10

Extr

a del

ay (m

s)


1a 1b2a 1b 3a 1b

2a 2b

Ener

gy co

nsum

ptio

n


Method

times10minus3

0

5

10

15

20

25

(Wlowast

subf

ram

e-tim

e)





119894




Method

1a 1b2a 1b 3a 1b

2a 2b

0

times10minus3

Ener

gy co

nsum

ptio

n

010203040506070809

(Wlowast

subf

ram

e-tim

e)


0 04 06 08 1 1202120573120572

096

097

098

099

1

Nor

mal

ized

ener

gy co

nsum

ptio

n


40 and119872 = 3)


6 Conclusion


02 04 08060 1120574

0

02

04

06

08

1

Nor

mal

ized

ener

gy co

nsum

ptio

n



Notations




119894


119894


119894per






119894



119896

119864119896


119892119896when using CQI 119909

119860119896



119868119905



119868119903




119894and RN

119895


119894










receiver 119895

Competing Interests


Acknowledgments


References

































Advances in

FuzzySystems


Volume 2014












Journal of

Journal of





Advances in

Multimedia


Biomedical Imaging



Advances in


RoboticsJournal of










Advances in




7 SC-FDMA symbols

12su

bcar

riers

(180

kHz)

1 subframe (1ms)











is 119864119894= 119875119894times119879119894 where 119875


119894





119894


119894= lceil120575119894rate(MCS



119894rsquos



SINR119894119895

= 10 times log10

119875119894119895

119861 times 1198730+ 119868119894119895

(1)



119894119895is





OnesubframeT = 1ms

Oneslot

Subframe 9












0

1


2


4



10minus3

10minus2

10minus1

100

BER

0 5 10 15 20 25minus5

SINR (dB)


be evaluated by 119868119894119895

= sum119894 =119895



119875119894119895

=

119866119894times 119866119895times 119875119894

119871119894119895

(2)

where 119866119894and 119866


119894and RN

119895


119894) to 119895 (RN

119895or



119894= 119896) UE



SINR119894119895

ge SINR (CQI119894= 119896 120585119894) (3)


119894of UE


119894)


119875119894ge


0+ 119868119894119895) times 119871119894119895

119866119894times 119866119895

(4)


119894 119894 = 1 119873 it has an average



119894and traffic



119894


119894







119883119894119895 119894 = 1 119899 119895 = 1 119873


119894119895and a



119868119894119895

= 1 119894 = 1 119899 where 119868119894119895

= 1


119894119895= 0The


max119899

sum

forall119894=1

119873119894

sum

forall119895=1

119902119894119895119868119894119895

subject to119899

sum

forall119894=1

119873119894

sum

forall119895=1

119906119894119895119868119894119895

le 119880

119873119894

sum

forall119895=1

119868119894119895

= 1 119894 = 1 119899

119868119894119895

= 0 1 119894 = 1 119899 119895 = 1 119873119894

(5)





119883119894119895

has a profit 119902119894119895




119894has


119894selects one

MCS 119909119894119895 119895 = 1 119873


119894119895(which








11990921205722

119909119899120572119899


119894 119894 = 1 119899 we select one


119894120572119894and the number




3 Proposed Method



119895




119894 119894 =


lowast= argmax

forall119895SINR

119894119895


119877

119895lowast + UE

119894 To minimize





119895

can be derived by

119879UE RN119894

= lceil120575119894

rate (CQI119894= 1)

rceil (6)


119894in backhaul


canbe conducted by

119879RN BS119894

= sum

119895=1119872

119909119894119895times lceil

120575119894


where 119909119894119895



119909119894119895


119879UE RN119894


119894=1(119879

RN BS119894

+119879UE RN119894



119894

and 119879RN BS119894







119896from sum

forall119894isin119892119896119879UE RN119894

to max119879UE RN119894



119894 To



119882119894of UE


119882119894

= 120572 times

(119889119894119895)minus119908

(minℓ=1119873

119889ℓ119895

| 119909ℓ119895

= 0)minus119908

+ 120573

times120575119894

maxℓ=1119873

120575ℓ| 119909ℓ119895

= 0

+ (minus120574)

times (1 + Δ times 119905119894)

times sum

forall120592120592 =119895(sum119873

ℓ=1119909ℓ120592) =0

(119889119894120592)minus119908

(minℓ=1119873

119889ℓ120592

| 119909ℓ120592

= 0)minus119908

(8)






119896 for each RN

119895 119895 = 0 119872 we



forall119894isin119878119877

119895

119882119894


UE119894in 119892119896 where


119894 To prevent



=

sumforall119894119894isin119892119896

(119894times 119879

UE RN119894













119875119891(119896 119909 119910) =

Δ119864119896

119909119910

Δ119860119896

119909119910

=

119864119896

119910minus 119864119896

119909

119860119896

119909minus 119860119896

119910

(9)

where 119864119896

119910and 119864

119896



119892119896= 119910) and MCS(CQI

119892119896= 119909)


119909and 119860

119896



119892119896= 119909) and MCS(CQI

119892119896= 119910)



119894

(2) Set 1198781198771015840

119895= 119878119877

119895for 119895 = 0 119872 119878 = UE

119894 119894 =

1 119873 119896 = 1 119879accessall = sum

forall1198941199091198940 =1119879UE RN119894

and119879all = sum

119873

119894=1(119879

RN BS119894

+ 119879UE RN119894

)

(3) For each 1198781198771015840



1198771015840

119895 where 119894

lowast=


1015840

119895

119882119894 and set 119892

119896= 119892119896+ UE119894lowast


119894isin 119892119896(refer to (4)) If

119864119896le 119864





119896= 119892119896minus UE

ℓ

and update 119905ℓ= 119905ℓ+ 1 and119882

ℓ Repeat step (4)

(5) If |119892119896| gt 1 update 119879

accessall = 119879

accessall minus

sumforall119894isin1198921198961199091198940 =1

119879UE RN119894

+ max119879UE RN119894

| forall119894 isin 119892119896 and


119879UE RN119894

+ max119879UE RN119894

| forall119894 isin 119892119896

Set 1198781198771015840

119895= 1198781198771015840


119896





1 119909

119896

I119896] where

119909119896

weierp= [119909119896

weierp1 119909

119896

weierp|119892119896|]119879 and 119909

119896






for each 119909119896




I119896and

119879UE RN1198961




119896is



119896

1to all


weierp weierp = 2 I

119896


119896|

exist119894 isin 119892119896 1199091198940







lowast to 119909119896


lowast+ 1) I

119896













0 UE1 UE2 and UE

3are in


1

RN2 and RN



1 for example


1and RN


11= 11987501

+11987521

+



119868119905

11= 11987510

+11987512

+11987513 Sum up 119868119903





RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3


RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3



When 119864119896gt 119864








119898


interference 119868119903119898119899

by

119868119903

119898119899= sum

forallUE120572isin119892119896120572 =119898119875120572119899 (10)


ence 119868119905119898119899

as follows

119868119905

119898119899= sum

forallRN120573isin120598119896120573 =119899119875119898120573

(11)

(2) For eachUE119898 119898 = 1 |119892


119898119899= 119868119903

119898119899+

119868119905

119898119899




from 119892119896


119892119896 For each 119892



119896 only 15 times |119892

119896|


119896= UE

1 UE

|119892119896|


weierp1 119909

119896

weierp|119892119896|]119879


weierp=

max119879UE RN119894

(119909119896

weierp119894) | forall119894 = 119879

UE RN1

(119909119896

weierp1) that is UE

1requires




119896 This means


weierp1 (119909119896


119896

weierp3+

1) sdot sdot sdot 15 (119909119896




119896is stated as

below



119896

weierp= [119909119896

weierp1=

MCS(CQI = 1) 119909119896

weierp|119892119896|= MCS(CQI = 1)]

119879






119896









119896|






119874 (2 times 119873 (119872 + 1)) sim 119874 (119873119872) (12)


119895 119895 =




119872= 119873 So


119874 (1198731) + sdot sdot sdot + 119874 (119873

119872+1) sim 119874 (119873) (13)


119874((119872 + 1)2) sim 119874 (119872

2) (14)



119874 (119872 + 1) sim 119874 (119872) (15)


119874 ((119872 + 1) times (119872 +119872)) sim 119874 (1198722) (16)


119874 (119872 + 1) sim 119874 (119872) (17)


119874 (119872) times (119874 (1198722) + 119874 (119872) + 119874 (119872

2) + 119874 (119872))

sim 119874 (1198723)

(18)



(119874 (119873) + 119874 (119873) + 119874 (1198723)) times 119874 (119873)

sim 119874 (1198732) + 119874 (119873119872

3)

(19)




119874 (119873119872) + 119874 (1198732) + 119874 (119873119872

3) (20)





1| + |119892


119870| = 119873


119874 (15119873) sim 119874 (119873) (21)


119896is

119874 (151003816100381610038161003816119892119896

1003816100381610038161003816) times 119874 (151003816100381610038161003816119892119896

1003816100381610038161003816

2

) sim 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) (22)




119896|) sim

119874(|119892119896|)


forall119894119877119894 where 119877



forall119894119877119894minus(119865B+119865nB))


119871 = min119874 (15119873) (sum

forall119894

119877119894minus (119865B + 119865nB)) (23)


119874 (119873) + 119874 (119870) times 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) + 119871 times (119874 (119873) + 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816))

le 119874 (119873) + 119874 (1198731198722) + 119871 times (119874 (119873))

le 119874 (119873) + 119874 (1198731198722) + 119874 (15119873) times (119874 (119873))

sim 119874 (1198732) + 119874 (119873119872

2)

(24)


119874(1198732) + 119874 (119873119872

3) (25)




Channel model





Traffic









OEAEPAR


15 20 25 30 35 40 45 50 5510N

Ener

gy co

nsum

ptio

n(W

lowastsu

bfra

me-

time)

00005

0010015

0020025

0030035

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

Ener

gy co

nsum

ptio

n

000002000040000600008

000100012000140001600018

(Wlowast

subf

ram

e-tim

e)

(b) Traffic Case 2


OEAEPAR


10 20 30 40 50 601N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 14020N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(b) Traffic Case 2








OEAEPAR


15 20 25 30 35 40 45 50 5510N

0100020003000400050006000700080009000

Thro

ughp

ut (k

bps)

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

0

1000

1500

500

2000

2500

3000

Thro

ughp

ut (k

bps)

(b) Traffic Case 2



10 20 30 40 50 601N

0

2

4

6

8

10

Extr

a del

ay (m

s)


1a 1b2a 1b 3a 1b

2a 2b

Ener

gy co

nsum

ptio

n


Method

times10minus3

0

5

10

15

20

25

(Wlowast

subf

ram

e-tim

e)





119894




Method

1a 1b2a 1b 3a 1b

2a 2b

0

times10minus3

Ener

gy co

nsum

ptio

n

010203040506070809

(Wlowast

subf

ram

e-tim

e)


0 04 06 08 1 1202120573120572

096

097

098

099

1

Nor

mal

ized

ener

gy co

nsum

ptio

n


40 and119872 = 3)


6 Conclusion


02 04 08060 1120574

0

02

04

06

08

1

Nor

mal

ized

ener

gy co

nsum

ptio

n



Notations




119894


119894


119894per






119894



119896

119864119896


119892119896when using CQI 119909

119860119896



119868119905



119868119903




119894and RN

119895


119894










receiver 119895

Competing Interests


Acknowledgments


References

































Advances in

FuzzySystems


Volume 2014












Journal of

Journal of





Advances in

Multimedia


Biomedical Imaging



Advances in


RoboticsJournal of










Advances in






OnesubframeT = 1ms

Oneslot

Subframe 9












0

1


2


4



10minus3

10minus2

10minus1

100

BER

0 5 10 15 20 25minus5

SINR (dB)


be evaluated by 119868119894119895

= sum119894 =119895



119875119894119895

=

119866119894times 119866119895times 119875119894

119871119894119895

(2)

where 119866119894and 119866


119894and RN

119895


119894) to 119895 (RN

119895or



119894= 119896) UE



SINR119894119895

ge SINR (CQI119894= 119896 120585119894) (3)


119894of UE


119894)


119875119894ge


0+ 119868119894119895) times 119871119894119895

119866119894times 119866119895

(4)


119894 119894 = 1 119873 it has an average



119894and traffic



119894


119894







119883119894119895 119894 = 1 119899 119895 = 1 119873


119894119895and a



119868119894119895

= 1 119894 = 1 119899 where 119868119894119895

= 1


119894119895= 0The


max119899

sum

forall119894=1

119873119894

sum

forall119895=1

119902119894119895119868119894119895

subject to119899

sum

forall119894=1

119873119894

sum

forall119895=1

119906119894119895119868119894119895

le 119880

119873119894

sum

forall119895=1

119868119894119895

= 1 119894 = 1 119899

119868119894119895

= 0 1 119894 = 1 119899 119895 = 1 119873119894

(5)





119883119894119895

has a profit 119902119894119895




119894has


119894selects one

MCS 119909119894119895 119895 = 1 119873


119894119895(which








11990921205722

119909119899120572119899


119894 119894 = 1 119899 we select one


119894120572119894and the number




3 Proposed Method



119895




119894 119894 =


lowast= argmax

forall119895SINR

119894119895


119877

119895lowast + UE

119894 To minimize





119895

can be derived by

119879UE RN119894

= lceil120575119894

rate (CQI119894= 1)

rceil (6)


119894in backhaul


canbe conducted by

119879RN BS119894

= sum

119895=1119872

119909119894119895times lceil

120575119894


where 119909119894119895



119909119894119895


119879UE RN119894


119894=1(119879

RN BS119894

+119879UE RN119894



119894

and 119879RN BS119894







119896from sum

forall119894isin119892119896119879UE RN119894

to max119879UE RN119894



119894 To



119882119894of UE


119882119894

= 120572 times

(119889119894119895)minus119908

(minℓ=1119873

119889ℓ119895

| 119909ℓ119895

= 0)minus119908

+ 120573

times120575119894

maxℓ=1119873

120575ℓ| 119909ℓ119895

= 0

+ (minus120574)

times (1 + Δ times 119905119894)

times sum

forall120592120592 =119895(sum119873

ℓ=1119909ℓ120592) =0

(119889119894120592)minus119908

(minℓ=1119873

119889ℓ120592

| 119909ℓ120592

= 0)minus119908

(8)






119896 for each RN

119895 119895 = 0 119872 we



forall119894isin119878119877

119895

119882119894


UE119894in 119892119896 where


119894 To prevent



=

sumforall119894119894isin119892119896

(119894times 119879

UE RN119894













119875119891(119896 119909 119910) =

Δ119864119896

119909119910

Δ119860119896

119909119910

=

119864119896

119910minus 119864119896

119909

119860119896

119909minus 119860119896

119910

(9)

where 119864119896

119910and 119864

119896



119892119896= 119910) and MCS(CQI

119892119896= 119909)


119909and 119860

119896



119892119896= 119909) and MCS(CQI

119892119896= 119910)



119894

(2) Set 1198781198771015840

119895= 119878119877

119895for 119895 = 0 119872 119878 = UE

119894 119894 =

1 119873 119896 = 1 119879accessall = sum

forall1198941199091198940 =1119879UE RN119894

and119879all = sum

119873

119894=1(119879

RN BS119894

+ 119879UE RN119894

)

(3) For each 1198781198771015840



1198771015840

119895 where 119894

lowast=


1015840

119895

119882119894 and set 119892

119896= 119892119896+ UE119894lowast


119894isin 119892119896(refer to (4)) If

119864119896le 119864





119896= 119892119896minus UE

ℓ

and update 119905ℓ= 119905ℓ+ 1 and119882

ℓ Repeat step (4)

(5) If |119892119896| gt 1 update 119879

accessall = 119879

accessall minus

sumforall119894isin1198921198961199091198940 =1

119879UE RN119894

+ max119879UE RN119894

| forall119894 isin 119892119896 and


119879UE RN119894

+ max119879UE RN119894

| forall119894 isin 119892119896

Set 1198781198771015840

119895= 1198781198771015840


119896





1 119909

119896

I119896] where

119909119896

weierp= [119909119896

weierp1 119909

119896

weierp|119892119896|]119879 and 119909

119896






for each 119909119896




I119896and

119879UE RN1198961




119896is



119896

1to all


weierp weierp = 2 I

119896


119896|

exist119894 isin 119892119896 1199091198940







lowast to 119909119896


lowast+ 1) I

119896













0 UE1 UE2 and UE

3are in


1

RN2 and RN



1 for example


1and RN


11= 11987501

+11987521

+



119868119905

11= 11987510

+11987512

+11987513 Sum up 119868119903





RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3


RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3



When 119864119896gt 119864








119898


interference 119868119903119898119899

by

119868119903

119898119899= sum

forallUE120572isin119892119896120572 =119898119875120572119899 (10)


ence 119868119905119898119899

as follows

119868119905

119898119899= sum

forallRN120573isin120598119896120573 =119899119875119898120573

(11)

(2) For eachUE119898 119898 = 1 |119892


119898119899= 119868119903

119898119899+

119868119905

119898119899




from 119892119896


119892119896 For each 119892



119896 only 15 times |119892

119896|


119896= UE

1 UE

|119892119896|


weierp1 119909

119896

weierp|119892119896|]119879


weierp=

max119879UE RN119894

(119909119896

weierp119894) | forall119894 = 119879

UE RN1

(119909119896

weierp1) that is UE

1requires




119896 This means


weierp1 (119909119896


119896

weierp3+

1) sdot sdot sdot 15 (119909119896




119896is stated as

below



119896

weierp= [119909119896

weierp1=

MCS(CQI = 1) 119909119896

weierp|119892119896|= MCS(CQI = 1)]

119879






119896









119896|






119874 (2 times 119873 (119872 + 1)) sim 119874 (119873119872) (12)


119895 119895 =




119872= 119873 So


119874 (1198731) + sdot sdot sdot + 119874 (119873

119872+1) sim 119874 (119873) (13)


119874((119872 + 1)2) sim 119874 (119872

2) (14)



119874 (119872 + 1) sim 119874 (119872) (15)


119874 ((119872 + 1) times (119872 +119872)) sim 119874 (1198722) (16)


119874 (119872 + 1) sim 119874 (119872) (17)


119874 (119872) times (119874 (1198722) + 119874 (119872) + 119874 (119872

2) + 119874 (119872))

sim 119874 (1198723)

(18)



(119874 (119873) + 119874 (119873) + 119874 (1198723)) times 119874 (119873)

sim 119874 (1198732) + 119874 (119873119872

3)

(19)




119874 (119873119872) + 119874 (1198732) + 119874 (119873119872

3) (20)





1| + |119892


119870| = 119873


119874 (15119873) sim 119874 (119873) (21)


119896is

119874 (151003816100381610038161003816119892119896

1003816100381610038161003816) times 119874 (151003816100381610038161003816119892119896

1003816100381610038161003816

2

) sim 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) (22)




119896|) sim

119874(|119892119896|)


forall119894119877119894 where 119877



forall119894119877119894minus(119865B+119865nB))


119871 = min119874 (15119873) (sum

forall119894

119877119894minus (119865B + 119865nB)) (23)


119874 (119873) + 119874 (119870) times 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) + 119871 times (119874 (119873) + 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816))

le 119874 (119873) + 119874 (1198731198722) + 119871 times (119874 (119873))

le 119874 (119873) + 119874 (1198731198722) + 119874 (15119873) times (119874 (119873))

sim 119874 (1198732) + 119874 (119873119872

2)

(24)


119874(1198732) + 119874 (119873119872

3) (25)




Channel model





Traffic









OEAEPAR


15 20 25 30 35 40 45 50 5510N

Ener

gy co

nsum

ptio

n(W

lowastsu

bfra

me-

time)

00005

0010015

0020025

0030035

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

Ener

gy co

nsum

ptio

n

000002000040000600008

000100012000140001600018

(Wlowast

subf

ram

e-tim

e)

(b) Traffic Case 2


OEAEPAR


10 20 30 40 50 601N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 14020N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(b) Traffic Case 2








OEAEPAR


15 20 25 30 35 40 45 50 5510N

0100020003000400050006000700080009000

Thro

ughp

ut (k

bps)

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

0

1000

1500

500

2000

2500

3000

Thro

ughp

ut (k

bps)

(b) Traffic Case 2



10 20 30 40 50 601N

0

2

4

6

8

10

Extr

a del

ay (m

s)


1a 1b2a 1b 3a 1b

2a 2b

Ener

gy co

nsum

ptio

n


Method

times10minus3

0

5

10

15

20

25

(Wlowast

subf

ram

e-tim

e)





119894




Method

1a 1b2a 1b 3a 1b

2a 2b

0

times10minus3

Ener

gy co

nsum

ptio

n

010203040506070809

(Wlowast

subf

ram

e-tim

e)


0 04 06 08 1 1202120573120572

096

097

098

099

1

Nor

mal

ized

ener

gy co

nsum

ptio

n


40 and119872 = 3)


6 Conclusion


02 04 08060 1120574

0

02

04

06

08

1

Nor

mal

ized

ener

gy co

nsum

ptio

n



Notations




119894


119894


119894per






119894



119896

119864119896


119892119896when using CQI 119909

119860119896



119868119905



119868119903




119894and RN

119895


119894










receiver 119895

Competing Interests


Acknowledgments


References

































Advances in

FuzzySystems


Volume 2014












Journal of

Journal of





Advances in

Multimedia


Biomedical Imaging



Advances in


RoboticsJournal of










Advances in




10minus3

10minus2

10minus1

100

BER

0 5 10 15 20 25minus5

SINR (dB)


be evaluated by 119868119894119895

= sum119894 =119895



119875119894119895

=

119866119894times 119866119895times 119875119894

119871119894119895

(2)

where 119866119894and 119866


119894and RN

119895


119894) to 119895 (RN

119895or



119894= 119896) UE



SINR119894119895

ge SINR (CQI119894= 119896 120585119894) (3)


119894of UE


119894)


119875119894ge


0+ 119868119894119895) times 119871119894119895

119866119894times 119866119895

(4)


119894 119894 = 1 119873 it has an average



119894and traffic



119894


119894







119883119894119895 119894 = 1 119899 119895 = 1 119873


119894119895and a



119868119894119895

= 1 119894 = 1 119899 where 119868119894119895

= 1


119894119895= 0The


max119899

sum

forall119894=1

119873119894

sum

forall119895=1

119902119894119895119868119894119895

subject to119899

sum

forall119894=1

119873119894

sum

forall119895=1

119906119894119895119868119894119895

le 119880

119873119894

sum

forall119895=1

119868119894119895

= 1 119894 = 1 119899

119868119894119895

= 0 1 119894 = 1 119899 119895 = 1 119873119894

(5)





119883119894119895

has a profit 119902119894119895




119894has


119894selects one

MCS 119909119894119895 119895 = 1 119873


119894119895(which








11990921205722

119909119899120572119899


119894 119894 = 1 119899 we select one


119894120572119894and the number




3 Proposed Method



119895




119894 119894 =


lowast= argmax

forall119895SINR

119894119895


119877

119895lowast + UE

119894 To minimize





119895

can be derived by

119879UE RN119894

= lceil120575119894

rate (CQI119894= 1)

rceil (6)


119894in backhaul


canbe conducted by

119879RN BS119894

= sum

119895=1119872

119909119894119895times lceil

120575119894


where 119909119894119895



119909119894119895


119879UE RN119894


119894=1(119879

RN BS119894

+119879UE RN119894



119894

and 119879RN BS119894







119896from sum

forall119894isin119892119896119879UE RN119894

to max119879UE RN119894



119894 To



119882119894of UE


119882119894

= 120572 times

(119889119894119895)minus119908

(minℓ=1119873

119889ℓ119895

| 119909ℓ119895

= 0)minus119908

+ 120573

times120575119894

maxℓ=1119873

120575ℓ| 119909ℓ119895

= 0

+ (minus120574)

times (1 + Δ times 119905119894)

times sum

forall120592120592 =119895(sum119873

ℓ=1119909ℓ120592) =0

(119889119894120592)minus119908

(minℓ=1119873

119889ℓ120592

| 119909ℓ120592

= 0)minus119908

(8)






119896 for each RN

119895 119895 = 0 119872 we



forall119894isin119878119877

119895

119882119894


UE119894in 119892119896 where


119894 To prevent



=

sumforall119894119894isin119892119896

(119894times 119879

UE RN119894













119875119891(119896 119909 119910) =

Δ119864119896

119909119910

Δ119860119896

119909119910

=

119864119896

119910minus 119864119896

119909

119860119896

119909minus 119860119896

119910

(9)

where 119864119896

119910and 119864

119896



119892119896= 119910) and MCS(CQI

119892119896= 119909)


119909and 119860

119896



119892119896= 119909) and MCS(CQI

119892119896= 119910)



119894

(2) Set 1198781198771015840

119895= 119878119877

119895for 119895 = 0 119872 119878 = UE

119894 119894 =

1 119873 119896 = 1 119879accessall = sum

forall1198941199091198940 =1119879UE RN119894

and119879all = sum

119873

119894=1(119879

RN BS119894

+ 119879UE RN119894

)

(3) For each 1198781198771015840



1198771015840

119895 where 119894

lowast=


1015840

119895

119882119894 and set 119892

119896= 119892119896+ UE119894lowast


119894isin 119892119896(refer to (4)) If

119864119896le 119864





119896= 119892119896minus UE

ℓ

and update 119905ℓ= 119905ℓ+ 1 and119882

ℓ Repeat step (4)

(5) If |119892119896| gt 1 update 119879

accessall = 119879

accessall minus

sumforall119894isin1198921198961199091198940 =1

119879UE RN119894

+ max119879UE RN119894

| forall119894 isin 119892119896 and


119879UE RN119894

+ max119879UE RN119894

| forall119894 isin 119892119896

Set 1198781198771015840

119895= 1198781198771015840


119896





1 119909

119896

I119896] where

119909119896

weierp= [119909119896

weierp1 119909

119896

weierp|119892119896|]119879 and 119909

119896






for each 119909119896




I119896and

119879UE RN1198961




119896is



119896

1to all


weierp weierp = 2 I

119896


119896|

exist119894 isin 119892119896 1199091198940







lowast to 119909119896


lowast+ 1) I

119896













0 UE1 UE2 and UE

3are in


1

RN2 and RN



1 for example


1and RN


11= 11987501

+11987521

+



119868119905

11= 11987510

+11987512

+11987513 Sum up 119868119903





RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3


RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3



When 119864119896gt 119864








119898


interference 119868119903119898119899

by

119868119903

119898119899= sum

forallUE120572isin119892119896120572 =119898119875120572119899 (10)


ence 119868119905119898119899

as follows

119868119905

119898119899= sum

forallRN120573isin120598119896120573 =119899119875119898120573

(11)

(2) For eachUE119898 119898 = 1 |119892


119898119899= 119868119903

119898119899+

119868119905

119898119899




from 119892119896


119892119896 For each 119892



119896 only 15 times |119892

119896|


119896= UE

1 UE

|119892119896|


weierp1 119909

119896

weierp|119892119896|]119879


weierp=

max119879UE RN119894

(119909119896

weierp119894) | forall119894 = 119879

UE RN1

(119909119896

weierp1) that is UE

1requires




119896 This means


weierp1 (119909119896


119896

weierp3+

1) sdot sdot sdot 15 (119909119896




119896is stated as

below



119896

weierp= [119909119896

weierp1=

MCS(CQI = 1) 119909119896

weierp|119892119896|= MCS(CQI = 1)]

119879






119896









119896|






119874 (2 times 119873 (119872 + 1)) sim 119874 (119873119872) (12)


119895 119895 =




119872= 119873 So


119874 (1198731) + sdot sdot sdot + 119874 (119873

119872+1) sim 119874 (119873) (13)


119874((119872 + 1)2) sim 119874 (119872

2) (14)



119874 (119872 + 1) sim 119874 (119872) (15)


119874 ((119872 + 1) times (119872 +119872)) sim 119874 (1198722) (16)


119874 (119872 + 1) sim 119874 (119872) (17)


119874 (119872) times (119874 (1198722) + 119874 (119872) + 119874 (119872

2) + 119874 (119872))

sim 119874 (1198723)

(18)



(119874 (119873) + 119874 (119873) + 119874 (1198723)) times 119874 (119873)

sim 119874 (1198732) + 119874 (119873119872

3)

(19)




119874 (119873119872) + 119874 (1198732) + 119874 (119873119872

3) (20)





1| + |119892


119870| = 119873


119874 (15119873) sim 119874 (119873) (21)


119896is

119874 (151003816100381610038161003816119892119896

1003816100381610038161003816) times 119874 (151003816100381610038161003816119892119896

1003816100381610038161003816

2

) sim 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) (22)




119896|) sim

119874(|119892119896|)


forall119894119877119894 where 119877



forall119894119877119894minus(119865B+119865nB))


119871 = min119874 (15119873) (sum

forall119894

119877119894minus (119865B + 119865nB)) (23)


119874 (119873) + 119874 (119870) times 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) + 119871 times (119874 (119873) + 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816))

le 119874 (119873) + 119874 (1198731198722) + 119871 times (119874 (119873))

le 119874 (119873) + 119874 (1198731198722) + 119874 (15119873) times (119874 (119873))

sim 119874 (1198732) + 119874 (119873119872

2)

(24)


119874(1198732) + 119874 (119873119872

3) (25)




Channel model





Traffic









OEAEPAR


15 20 25 30 35 40 45 50 5510N

Ener

gy co

nsum

ptio

n(W

lowastsu

bfra

me-

time)

00005

0010015

0020025

0030035

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

Ener

gy co

nsum

ptio

n

000002000040000600008

000100012000140001600018

(Wlowast

subf

ram

e-tim

e)

(b) Traffic Case 2


OEAEPAR


10 20 30 40 50 601N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 14020N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(b) Traffic Case 2








OEAEPAR


15 20 25 30 35 40 45 50 5510N

0100020003000400050006000700080009000

Thro

ughp

ut (k

bps)

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

0

1000

1500

500

2000

2500

3000

Thro

ughp

ut (k

bps)

(b) Traffic Case 2



10 20 30 40 50 601N

0

2

4

6

8

10

Extr

a del

ay (m

s)


1a 1b2a 1b 3a 1b

2a 2b

Ener

gy co

nsum

ptio

n


Method

times10minus3

0

5

10

15

20

25

(Wlowast

subf

ram

e-tim

e)





119894




Method

1a 1b2a 1b 3a 1b

2a 2b

0

times10minus3

Ener

gy co

nsum

ptio

n

010203040506070809

(Wlowast

subf

ram

e-tim

e)


0 04 06 08 1 1202120573120572

096

097

098

099

1

Nor

mal

ized

ener

gy co

nsum

ptio

n


40 and119872 = 3)


6 Conclusion


02 04 08060 1120574

0

02

04

06

08

1

Nor

mal

ized

ener

gy co

nsum

ptio

n



Notations




119894


119894


119894per






119894



119896

119864119896


119892119896when using CQI 119909

119860119896



119868119905



119868119903




119894and RN

119895


119894










receiver 119895

Competing Interests


Acknowledgments


References

































Advances in

FuzzySystems


Volume 2014












Journal of

Journal of





Advances in

Multimedia


Biomedical Imaging



Advances in


RoboticsJournal of










Advances in






119883119894119895

has a profit 119902119894119895




119894has


119894selects one

MCS 119909119894119895 119895 = 1 119873


119894119895(which








11990921205722

119909119899120572119899


119894 119894 = 1 119899 we select one


119894120572119894and the number




3 Proposed Method



119895




119894 119894 =


lowast= argmax

forall119895SINR

119894119895


119877

119895lowast + UE

119894 To minimize





119895

can be derived by

119879UE RN119894

= lceil120575119894

rate (CQI119894= 1)

rceil (6)


119894in backhaul


canbe conducted by

119879RN BS119894

= sum

119895=1119872

119909119894119895times lceil

120575119894


where 119909119894119895



119909119894119895


119879UE RN119894


119894=1(119879

RN BS119894

+119879UE RN119894



119894

and 119879RN BS119894







119896from sum

forall119894isin119892119896119879UE RN119894

to max119879UE RN119894



119894 To



119882119894of UE


119882119894

= 120572 times

(119889119894119895)minus119908

(minℓ=1119873

119889ℓ119895

| 119909ℓ119895

= 0)minus119908

+ 120573

times120575119894

maxℓ=1119873

120575ℓ| 119909ℓ119895

= 0

+ (minus120574)

times (1 + Δ times 119905119894)

times sum

forall120592120592 =119895(sum119873

ℓ=1119909ℓ120592) =0

(119889119894120592)minus119908

(minℓ=1119873

119889ℓ120592

| 119909ℓ120592

= 0)minus119908

(8)






119896 for each RN

119895 119895 = 0 119872 we



forall119894isin119878119877

119895

119882119894


UE119894in 119892119896 where


119894 To prevent



=

sumforall119894119894isin119892119896

(119894times 119879

UE RN119894













119875119891(119896 119909 119910) =

Δ119864119896

119909119910

Δ119860119896

119909119910

=

119864119896

119910minus 119864119896

119909

119860119896

119909minus 119860119896

119910

(9)

where 119864119896

119910and 119864

119896



119892119896= 119910) and MCS(CQI

119892119896= 119909)


119909and 119860

119896



119892119896= 119909) and MCS(CQI

119892119896= 119910)



119894

(2) Set 1198781198771015840

119895= 119878119877

119895for 119895 = 0 119872 119878 = UE

119894 119894 =

1 119873 119896 = 1 119879accessall = sum

forall1198941199091198940 =1119879UE RN119894

and119879all = sum

119873

119894=1(119879

RN BS119894

+ 119879UE RN119894

)

(3) For each 1198781198771015840



1198771015840

119895 where 119894

lowast=


1015840

119895

119882119894 and set 119892

119896= 119892119896+ UE119894lowast


119894isin 119892119896(refer to (4)) If

119864119896le 119864





119896= 119892119896minus UE

ℓ

and update 119905ℓ= 119905ℓ+ 1 and119882

ℓ Repeat step (4)

(5) If |119892119896| gt 1 update 119879

accessall = 119879

accessall minus

sumforall119894isin1198921198961199091198940 =1

119879UE RN119894

+ max119879UE RN119894

| forall119894 isin 119892119896 and


119879UE RN119894

+ max119879UE RN119894

| forall119894 isin 119892119896

Set 1198781198771015840

119895= 1198781198771015840


119896





1 119909

119896

I119896] where

119909119896

weierp= [119909119896

weierp1 119909

119896

weierp|119892119896|]119879 and 119909

119896






for each 119909119896




I119896and

119879UE RN1198961




119896is



119896

1to all


weierp weierp = 2 I

119896


119896|

exist119894 isin 119892119896 1199091198940







lowast to 119909119896


lowast+ 1) I

119896













0 UE1 UE2 and UE

3are in


1

RN2 and RN



1 for example


1and RN


11= 11987501

+11987521

+



119868119905

11= 11987510

+11987512

+11987513 Sum up 119868119903





RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3


RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3



When 119864119896gt 119864








119898


interference 119868119903119898119899

by

119868119903

119898119899= sum

forallUE120572isin119892119896120572 =119898119875120572119899 (10)


ence 119868119905119898119899

as follows

119868119905

119898119899= sum

forallRN120573isin120598119896120573 =119899119875119898120573

(11)

(2) For eachUE119898 119898 = 1 |119892


119898119899= 119868119903

119898119899+

119868119905

119898119899




from 119892119896


119892119896 For each 119892



119896 only 15 times |119892

119896|


119896= UE

1 UE

|119892119896|


weierp1 119909

119896

weierp|119892119896|]119879


weierp=

max119879UE RN119894

(119909119896

weierp119894) | forall119894 = 119879

UE RN1

(119909119896

weierp1) that is UE

1requires




119896 This means


weierp1 (119909119896


119896

weierp3+

1) sdot sdot sdot 15 (119909119896




119896is stated as

below



119896

weierp= [119909119896

weierp1=

MCS(CQI = 1) 119909119896

weierp|119892119896|= MCS(CQI = 1)]

119879






119896









119896|






119874 (2 times 119873 (119872 + 1)) sim 119874 (119873119872) (12)


119895 119895 =




119872= 119873 So


119874 (1198731) + sdot sdot sdot + 119874 (119873

119872+1) sim 119874 (119873) (13)


119874((119872 + 1)2) sim 119874 (119872

2) (14)



119874 (119872 + 1) sim 119874 (119872) (15)


119874 ((119872 + 1) times (119872 +119872)) sim 119874 (1198722) (16)


119874 (119872 + 1) sim 119874 (119872) (17)


119874 (119872) times (119874 (1198722) + 119874 (119872) + 119874 (119872

2) + 119874 (119872))

sim 119874 (1198723)

(18)



(119874 (119873) + 119874 (119873) + 119874 (1198723)) times 119874 (119873)

sim 119874 (1198732) + 119874 (119873119872

3)

(19)




119874 (119873119872) + 119874 (1198732) + 119874 (119873119872

3) (20)





1| + |119892


119870| = 119873


119874 (15119873) sim 119874 (119873) (21)


119896is

119874 (151003816100381610038161003816119892119896

1003816100381610038161003816) times 119874 (151003816100381610038161003816119892119896

1003816100381610038161003816

2

) sim 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) (22)




119896|) sim

119874(|119892119896|)


forall119894119877119894 where 119877



forall119894119877119894minus(119865B+119865nB))


119871 = min119874 (15119873) (sum

forall119894

119877119894minus (119865B + 119865nB)) (23)


119874 (119873) + 119874 (119870) times 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) + 119871 times (119874 (119873) + 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816))

le 119874 (119873) + 119874 (1198731198722) + 119871 times (119874 (119873))

le 119874 (119873) + 119874 (1198731198722) + 119874 (15119873) times (119874 (119873))

sim 119874 (1198732) + 119874 (119873119872

2)

(24)


119874(1198732) + 119874 (119873119872

3) (25)




Channel model





Traffic









OEAEPAR


15 20 25 30 35 40 45 50 5510N

Ener

gy co

nsum

ptio

n(W

lowastsu

bfra

me-

time)

00005

0010015

0020025

0030035

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

Ener

gy co

nsum

ptio

n

000002000040000600008

000100012000140001600018

(Wlowast

subf

ram

e-tim

e)

(b) Traffic Case 2


OEAEPAR


10 20 30 40 50 601N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 14020N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(b) Traffic Case 2








OEAEPAR


15 20 25 30 35 40 45 50 5510N

0100020003000400050006000700080009000

Thro

ughp

ut (k

bps)

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

0

1000

1500

500

2000

2500

3000

Thro

ughp

ut (k

bps)

(b) Traffic Case 2



10 20 30 40 50 601N

0

2

4

6

8

10

Extr

a del

ay (m

s)


1a 1b2a 1b 3a 1b

2a 2b

Ener

gy co

nsum

ptio

n


Method

times10minus3

0

5

10

15

20

25

(Wlowast

subf

ram

e-tim

e)





119894




Method

1a 1b2a 1b 3a 1b

2a 2b

0

times10minus3

Ener

gy co

nsum

ptio

n

010203040506070809

(Wlowast

subf

ram

e-tim

e)


0 04 06 08 1 1202120573120572

096

097

098

099

1

Nor

mal

ized

ener

gy co

nsum

ptio

n


40 and119872 = 3)


6 Conclusion


02 04 08060 1120574

0

02

04

06

08

1

Nor

mal

ized

ener

gy co

nsum

ptio

n



Notations




119894


119894


119894per






119894



119896

119864119896


119892119896when using CQI 119909

119860119896



119868119905



119868119903




119894and RN

119895


119894










receiver 119895

Competing Interests


Acknowledgments


References

































Advances in

FuzzySystems


Volume 2014












Journal of

Journal of





Advances in

Multimedia


Biomedical Imaging



Advances in


RoboticsJournal of










Advances in





119896 for each RN

119895 119895 = 0 119872 we



forall119894isin119878119877

119895

119882119894


UE119894in 119892119896 where


119894 To prevent



=

sumforall119894119894isin119892119896

(119894times 119879

UE RN119894













119875119891(119896 119909 119910) =

Δ119864119896

119909119910

Δ119860119896

119909119910

=

119864119896

119910minus 119864119896

119909

119860119896

119909minus 119860119896

119910

(9)

where 119864119896

119910and 119864

119896



119892119896= 119910) and MCS(CQI

119892119896= 119909)


119909and 119860

119896



119892119896= 119909) and MCS(CQI

119892119896= 119910)



119894

(2) Set 1198781198771015840

119895= 119878119877

119895for 119895 = 0 119872 119878 = UE

119894 119894 =

1 119873 119896 = 1 119879accessall = sum

forall1198941199091198940 =1119879UE RN119894

and119879all = sum

119873

119894=1(119879

RN BS119894

+ 119879UE RN119894

)

(3) For each 1198781198771015840



1198771015840

119895 where 119894

lowast=


1015840

119895

119882119894 and set 119892

119896= 119892119896+ UE119894lowast


119894isin 119892119896(refer to (4)) If

119864119896le 119864





119896= 119892119896minus UE

ℓ

and update 119905ℓ= 119905ℓ+ 1 and119882

ℓ Repeat step (4)

(5) If |119892119896| gt 1 update 119879

accessall = 119879

accessall minus

sumforall119894isin1198921198961199091198940 =1

119879UE RN119894

+ max119879UE RN119894

| forall119894 isin 119892119896 and


119879UE RN119894

+ max119879UE RN119894

| forall119894 isin 119892119896

Set 1198781198771015840

119895= 1198781198771015840


119896





1 119909

119896

I119896] where

119909119896

weierp= [119909119896

weierp1 119909

119896

weierp|119892119896|]119879 and 119909

119896






for each 119909119896




I119896and

119879UE RN1198961




119896is



119896

1to all


weierp weierp = 2 I

119896


119896|

exist119894 isin 119892119896 1199091198940







lowast to 119909119896


lowast+ 1) I

119896













0 UE1 UE2 and UE

3are in


1

RN2 and RN



1 for example


1and RN


11= 11987501

+11987521

+



119868119905

11= 11987510

+11987512

+11987513 Sum up 119868119903





RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3


RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3



When 119864119896gt 119864








119898


interference 119868119903119898119899

by

119868119903

119898119899= sum

forallUE120572isin119892119896120572 =119898119875120572119899 (10)


ence 119868119905119898119899

as follows

119868119905

119898119899= sum

forallRN120573isin120598119896120573 =119899119875119898120573

(11)

(2) For eachUE119898 119898 = 1 |119892


119898119899= 119868119903

119898119899+

119868119905

119898119899




from 119892119896


119892119896 For each 119892



119896 only 15 times |119892

119896|


119896= UE

1 UE

|119892119896|


weierp1 119909

119896

weierp|119892119896|]119879


weierp=

max119879UE RN119894

(119909119896

weierp119894) | forall119894 = 119879

UE RN1

(119909119896

weierp1) that is UE

1requires




119896 This means


weierp1 (119909119896


119896

weierp3+

1) sdot sdot sdot 15 (119909119896




119896is stated as

below



119896

weierp= [119909119896

weierp1=

MCS(CQI = 1) 119909119896

weierp|119892119896|= MCS(CQI = 1)]

119879






119896









119896|






119874 (2 times 119873 (119872 + 1)) sim 119874 (119873119872) (12)


119895 119895 =




119872= 119873 So


119874 (1198731) + sdot sdot sdot + 119874 (119873

119872+1) sim 119874 (119873) (13)


119874((119872 + 1)2) sim 119874 (119872

2) (14)



119874 (119872 + 1) sim 119874 (119872) (15)


119874 ((119872 + 1) times (119872 +119872)) sim 119874 (1198722) (16)


119874 (119872 + 1) sim 119874 (119872) (17)


119874 (119872) times (119874 (1198722) + 119874 (119872) + 119874 (119872

2) + 119874 (119872))

sim 119874 (1198723)

(18)



(119874 (119873) + 119874 (119873) + 119874 (1198723)) times 119874 (119873)

sim 119874 (1198732) + 119874 (119873119872

3)

(19)




119874 (119873119872) + 119874 (1198732) + 119874 (119873119872

3) (20)





1| + |119892


119870| = 119873


119874 (15119873) sim 119874 (119873) (21)


119896is

119874 (151003816100381610038161003816119892119896

1003816100381610038161003816) times 119874 (151003816100381610038161003816119892119896

1003816100381610038161003816

2

) sim 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) (22)




119896|) sim

119874(|119892119896|)


forall119894119877119894 where 119877



forall119894119877119894minus(119865B+119865nB))


119871 = min119874 (15119873) (sum

forall119894

119877119894minus (119865B + 119865nB)) (23)


119874 (119873) + 119874 (119870) times 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) + 119871 times (119874 (119873) + 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816))

le 119874 (119873) + 119874 (1198731198722) + 119871 times (119874 (119873))

le 119874 (119873) + 119874 (1198731198722) + 119874 (15119873) times (119874 (119873))

sim 119874 (1198732) + 119874 (119873119872

2)

(24)


119874(1198732) + 119874 (119873119872

3) (25)




Channel model





Traffic









OEAEPAR


15 20 25 30 35 40 45 50 5510N

Ener

gy co

nsum

ptio

n(W

lowastsu

bfra

me-

time)

00005

0010015

0020025

0030035

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

Ener

gy co

nsum

ptio

n

000002000040000600008

000100012000140001600018

(Wlowast

subf

ram

e-tim

e)

(b) Traffic Case 2


OEAEPAR


10 20 30 40 50 601N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 14020N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(b) Traffic Case 2








OEAEPAR


15 20 25 30 35 40 45 50 5510N

0100020003000400050006000700080009000

Thro

ughp

ut (k

bps)

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

0

1000

1500

500

2000

2500

3000

Thro

ughp

ut (k

bps)

(b) Traffic Case 2



10 20 30 40 50 601N

0

2

4

6

8

10

Extr

a del

ay (m

s)


1a 1b2a 1b 3a 1b

2a 2b

Ener

gy co

nsum

ptio

n


Method

times10minus3

0

5

10

15

20

25

(Wlowast

subf

ram

e-tim

e)





119894




Method

1a 1b2a 1b 3a 1b

2a 2b

0

times10minus3

Ener

gy co

nsum

ptio

n

010203040506070809

(Wlowast

subf

ram

e-tim

e)


0 04 06 08 1 1202120573120572

096

097

098

099

1

Nor

mal

ized

ener

gy co

nsum

ptio

n


40 and119872 = 3)


6 Conclusion


02 04 08060 1120574

0

02

04

06

08

1

Nor

mal

ized

ener

gy co

nsum

ptio

n



Notations




119894


119894


119894per






119894



119896

119864119896


119892119896when using CQI 119909

119860119896



119868119905



119868119903




119894and RN

119895


119894










receiver 119895

Competing Interests


Acknowledgments


References

































Advances in

FuzzySystems


Volume 2014












Journal of

Journal of





Advances in

Multimedia


Biomedical Imaging



Advances in


RoboticsJournal of










Advances in




RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3


RN0 (BS)

UE1

UE2

UE3UE0

RN1

RN2

RN3



When 119864119896gt 119864








119898


interference 119868119903119898119899

by

119868119903

119898119899= sum

forallUE120572isin119892119896120572 =119898119875120572119899 (10)


ence 119868119905119898119899

as follows

119868119905

119898119899= sum

forallRN120573isin120598119896120573 =119899119875119898120573

(11)

(2) For eachUE119898 119898 = 1 |119892


119898119899= 119868119903

119898119899+

119868119905

119898119899




from 119892119896


119892119896 For each 119892



119896 only 15 times |119892

119896|


119896= UE

1 UE

|119892119896|


weierp1 119909

119896

weierp|119892119896|]119879


weierp=

max119879UE RN119894

(119909119896

weierp119894) | forall119894 = 119879

UE RN1

(119909119896

weierp1) that is UE

1requires




119896 This means


weierp1 (119909119896


119896

weierp3+

1) sdot sdot sdot 15 (119909119896




119896is stated as

below



119896

weierp= [119909119896

weierp1=

MCS(CQI = 1) 119909119896

weierp|119892119896|= MCS(CQI = 1)]

119879






119896









119896|






119874 (2 times 119873 (119872 + 1)) sim 119874 (119873119872) (12)


119895 119895 =




119872= 119873 So


119874 (1198731) + sdot sdot sdot + 119874 (119873

119872+1) sim 119874 (119873) (13)


119874((119872 + 1)2) sim 119874 (119872

2) (14)



119874 (119872 + 1) sim 119874 (119872) (15)


119874 ((119872 + 1) times (119872 +119872)) sim 119874 (1198722) (16)


119874 (119872 + 1) sim 119874 (119872) (17)


119874 (119872) times (119874 (1198722) + 119874 (119872) + 119874 (119872

2) + 119874 (119872))

sim 119874 (1198723)

(18)



(119874 (119873) + 119874 (119873) + 119874 (1198723)) times 119874 (119873)

sim 119874 (1198732) + 119874 (119873119872

3)

(19)




119874 (119873119872) + 119874 (1198732) + 119874 (119873119872

3) (20)





1| + |119892


119870| = 119873


119874 (15119873) sim 119874 (119873) (21)


119896is

119874 (151003816100381610038161003816119892119896

1003816100381610038161003816) times 119874 (151003816100381610038161003816119892119896

1003816100381610038161003816

2

) sim 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) (22)




119896|) sim

119874(|119892119896|)


forall119894119877119894 where 119877



forall119894119877119894minus(119865B+119865nB))


119871 = min119874 (15119873) (sum

forall119894

119877119894minus (119865B + 119865nB)) (23)


119874 (119873) + 119874 (119870) times 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) + 119871 times (119874 (119873) + 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816))

le 119874 (119873) + 119874 (1198731198722) + 119871 times (119874 (119873))

le 119874 (119873) + 119874 (1198731198722) + 119874 (15119873) times (119874 (119873))

sim 119874 (1198732) + 119874 (119873119872

2)

(24)


119874(1198732) + 119874 (119873119872

3) (25)




Channel model





Traffic









OEAEPAR


15 20 25 30 35 40 45 50 5510N

Ener

gy co

nsum

ptio

n(W

lowastsu

bfra

me-

time)

00005

0010015

0020025

0030035

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

Ener

gy co

nsum

ptio

n

000002000040000600008

000100012000140001600018

(Wlowast

subf

ram

e-tim

e)

(b) Traffic Case 2


OEAEPAR


10 20 30 40 50 601N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 14020N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(b) Traffic Case 2








OEAEPAR


15 20 25 30 35 40 45 50 5510N

0100020003000400050006000700080009000

Thro

ughp

ut (k

bps)

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

0

1000

1500

500

2000

2500

3000

Thro

ughp

ut (k

bps)

(b) Traffic Case 2



10 20 30 40 50 601N

0

2

4

6

8

10

Extr

a del

ay (m

s)


1a 1b2a 1b 3a 1b

2a 2b

Ener

gy co

nsum

ptio

n


Method

times10minus3

0

5

10

15

20

25

(Wlowast

subf

ram

e-tim

e)





119894




Method

1a 1b2a 1b 3a 1b

2a 2b

0

times10minus3

Ener

gy co

nsum

ptio

n

010203040506070809

(Wlowast

subf

ram

e-tim

e)


0 04 06 08 1 1202120573120572

096

097

098

099

1

Nor

mal

ized

ener

gy co

nsum

ptio

n


40 and119872 = 3)


6 Conclusion


02 04 08060 1120574

0

02

04

06

08

1

Nor

mal

ized

ener

gy co

nsum

ptio

n



Notations




119894


119894


119894per






119894



119896

119864119896


119892119896when using CQI 119909

119860119896



119868119905



119868119903




119894and RN

119895


119894










receiver 119895

Competing Interests


Acknowledgments


References

































Advances in

FuzzySystems


Volume 2014












Journal of

Journal of





Advances in

Multimedia


Biomedical Imaging



Advances in


RoboticsJournal of










Advances in






119874 (2 times 119873 (119872 + 1)) sim 119874 (119873119872) (12)


119895 119895 =




119872= 119873 So


119874 (1198731) + sdot sdot sdot + 119874 (119873

119872+1) sim 119874 (119873) (13)


119874((119872 + 1)2) sim 119874 (119872

2) (14)



119874 (119872 + 1) sim 119874 (119872) (15)


119874 ((119872 + 1) times (119872 +119872)) sim 119874 (1198722) (16)


119874 (119872 + 1) sim 119874 (119872) (17)


119874 (119872) times (119874 (1198722) + 119874 (119872) + 119874 (119872

2) + 119874 (119872))

sim 119874 (1198723)

(18)



(119874 (119873) + 119874 (119873) + 119874 (1198723)) times 119874 (119873)

sim 119874 (1198732) + 119874 (119873119872

3)

(19)




119874 (119873119872) + 119874 (1198732) + 119874 (119873119872

3) (20)





1| + |119892


119870| = 119873


119874 (15119873) sim 119874 (119873) (21)


119896is

119874 (151003816100381610038161003816119892119896

1003816100381610038161003816) times 119874 (151003816100381610038161003816119892119896

1003816100381610038161003816

2

) sim 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) (22)




119896|) sim

119874(|119892119896|)


forall119894119877119894 where 119877



forall119894119877119894minus(119865B+119865nB))


119871 = min119874 (15119873) (sum

forall119894

119877119894minus (119865B + 119865nB)) (23)


119874 (119873) + 119874 (119870) times 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816

3

) + 119871 times (119874 (119873) + 119874 (1003816100381610038161003816119892119896

1003816100381610038161003816))

le 119874 (119873) + 119874 (1198731198722) + 119871 times (119874 (119873))

le 119874 (119873) + 119874 (1198731198722) + 119874 (15119873) times (119874 (119873))

sim 119874 (1198732) + 119874 (119873119872

2)

(24)


119874(1198732) + 119874 (119873119872

3) (25)




Channel model





Traffic









OEAEPAR


15 20 25 30 35 40 45 50 5510N

Ener

gy co

nsum

ptio

n(W

lowastsu

bfra

me-

time)

00005

0010015

0020025

0030035

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

Ener

gy co

nsum

ptio

n

000002000040000600008

000100012000140001600018

(Wlowast

subf

ram

e-tim

e)

(b) Traffic Case 2


OEAEPAR


10 20 30 40 50 601N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 14020N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(b) Traffic Case 2








OEAEPAR


15 20 25 30 35 40 45 50 5510N

0100020003000400050006000700080009000

Thro

ughp

ut (k

bps)

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

0

1000

1500

500

2000

2500

3000

Thro

ughp

ut (k

bps)

(b) Traffic Case 2



10 20 30 40 50 601N

0

2

4

6

8

10

Extr

a del

ay (m

s)


1a 1b2a 1b 3a 1b

2a 2b

Ener

gy co

nsum

ptio

n


Method

times10minus3

0

5

10

15

20

25

(Wlowast

subf

ram

e-tim

e)





119894




Method

1a 1b2a 1b 3a 1b

2a 2b

0

times10minus3

Ener

gy co

nsum

ptio

n

010203040506070809

(Wlowast

subf

ram

e-tim

e)


0 04 06 08 1 1202120573120572

096

097

098

099

1

Nor

mal

ized

ener

gy co

nsum

ptio

n


40 and119872 = 3)


6 Conclusion


02 04 08060 1120574

0

02

04

06

08

1

Nor

mal

ized

ener

gy co

nsum

ptio

n



Notations




119894


119894


119894per






119894



119896

119864119896


119892119896when using CQI 119909

119860119896



119868119905



119868119903




119894and RN

119895


119894










receiver 119895

Competing Interests


Acknowledgments


References

































Advances in

FuzzySystems


Volume 2014












Journal of

Journal of





Advances in

Multimedia


Biomedical Imaging



Advances in


RoboticsJournal of










Advances in






Channel model





Traffic









OEAEPAR


15 20 25 30 35 40 45 50 5510N

Ener

gy co

nsum

ptio

n(W

lowastsu

bfra

me-

time)

00005

0010015

0020025

0030035

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

Ener

gy co

nsum

ptio

n

000002000040000600008

000100012000140001600018

(Wlowast

subf

ram

e-tim

e)

(b) Traffic Case 2


OEAEPAR


10 20 30 40 50 601N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 14020N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(b) Traffic Case 2








OEAEPAR


15 20 25 30 35 40 45 50 5510N

0100020003000400050006000700080009000

Thro

ughp

ut (k

bps)

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

0

1000

1500

500

2000

2500

3000

Thro

ughp

ut (k

bps)

(b) Traffic Case 2



10 20 30 40 50 601N

0

2

4

6

8

10

Extr

a del

ay (m

s)


1a 1b2a 1b 3a 1b

2a 2b

Ener

gy co

nsum

ptio

n


Method

times10minus3

0

5

10

15

20

25

(Wlowast

subf

ram

e-tim

e)





119894




Method

1a 1b2a 1b 3a 1b

2a 2b

0

times10minus3

Ener

gy co

nsum

ptio

n

010203040506070809

(Wlowast

subf

ram

e-tim

e)


0 04 06 08 1 1202120573120572

096

097

098

099

1

Nor

mal

ized

ener

gy co

nsum

ptio

n


40 and119872 = 3)


6 Conclusion


02 04 08060 1120574

0

02

04

06

08

1

Nor

mal

ized

ener

gy co

nsum

ptio

n



Notations




119894


119894


119894per






119894



119896

119864119896


119892119896when using CQI 119909

119860119896



119868119905



119868119903




119894and RN

119895


119894










receiver 119895

Competing Interests


Acknowledgments


References

































Advances in

FuzzySystems


Volume 2014












Journal of

Journal of





Advances in

Multimedia


Biomedical Imaging



Advances in


RoboticsJournal of










Advances in




OEAEPAR


15 20 25 30 35 40 45 50 5510N

Ener

gy co

nsum

ptio

n(W

lowastsu

bfra

me-

time)

00005

0010015

0020025

0030035

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

Ener

gy co

nsum

ptio

n

000002000040000600008

000100012000140001600018

(Wlowast

subf

ram

e-tim

e)

(b) Traffic Case 2


OEAEPAR


10 20 30 40 50 601N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 14020N

0

02

04

06

08

1

Band

wid

th u

tiliz

atio

n

(b) Traffic Case 2








OEAEPAR


15 20 25 30 35 40 45 50 5510N

0100020003000400050006000700080009000

Thro

ughp

ut (k

bps)

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

0

1000

1500

500

2000

2500

3000

Thro

ughp

ut (k

bps)

(b) Traffic Case 2



10 20 30 40 50 601N

0

2

4

6

8

10

Extr

a del

ay (m

s)


1a 1b2a 1b 3a 1b

2a 2b

Ener

gy co

nsum

ptio

n


Method

times10minus3

0

5

10

15

20

25

(Wlowast

subf

ram

e-tim

e)





119894




Method

1a 1b2a 1b 3a 1b

2a 2b

0

times10minus3

Ener

gy co

nsum

ptio

n

010203040506070809

(Wlowast

subf

ram

e-tim

e)


0 04 06 08 1 1202120573120572

096

097

098

099

1

Nor

mal

ized

ener

gy co

nsum

ptio

n


40 and119872 = 3)


6 Conclusion


02 04 08060 1120574

0

02

04

06

08

1

Nor

mal

ized

ener

gy co

nsum

ptio

n



Notations




119894


119894


119894per






119894



119896

119864119896


119892119896when using CQI 119909

119860119896



119868119905



119868119903




119894and RN

119895


119894










receiver 119895

Competing Interests


Acknowledgments


References

































Advances in

FuzzySystems


Volume 2014












Journal of

Journal of





Advances in

Multimedia


Biomedical Imaging



Advances in


RoboticsJournal of










Advances in




OEAEPAR


15 20 25 30 35 40 45 50 5510N

0100020003000400050006000700080009000

Thro

ughp

ut (k

bps)

(a) Traffic Case 1

OEAEPAR


40 60 80 100 120 140 160 18020N

0

1000

1500

500

2000

2500

3000

Thro

ughp

ut (k

bps)

(b) Traffic Case 2



10 20 30 40 50 601N

0

2

4

6

8

10

Extr

a del

ay (m

s)


1a 1b2a 1b 3a 1b

2a 2b

Ener

gy co

nsum

ptio

n


Method

times10minus3

0

5

10

15

20

25

(Wlowast

subf

ram

e-tim

e)





119894




Method

1a 1b2a 1b 3a 1b

2a 2b

0

times10minus3

Ener

gy co

nsum

ptio

n

010203040506070809

(Wlowast

subf

ram

e-tim

e)


0 04 06 08 1 1202120573120572

096

097

098

099

1

Nor

mal

ized

ener

gy co

nsum

ptio

n


40 and119872 = 3)


6 Conclusion


02 04 08060 1120574

0

02

04

06

08

1

Nor

mal

ized

ener

gy co

nsum

ptio

n



Notations




119894


119894


119894per






119894



119896

119864119896


119892119896when using CQI 119909

119860119896



119868119905



119868119903




119894and RN

119895


119894










receiver 119895

Competing Interests


Acknowledgments


References

































Advances in

FuzzySystems


Volume 2014












Journal of

Journal of





Advances in

Multimedia


Biomedical Imaging



Advances in


RoboticsJournal of










Advances in





Method

1a 1b2a 1b 3a 1b

2a 2b

0

times10minus3

Ener

gy co

nsum

ptio

n

010203040506070809

(Wlowast

subf

ram

e-tim

e)


0 04 06 08 1 1202120573120572

096

097

098

099

1

Nor

mal

ized

ener

gy co

nsum

ptio

n


40 and119872 = 3)


6 Conclusion


02 04 08060 1120574

0

02

04

06

08

1

Nor

mal

ized

ener

gy co

nsum

ptio

n



Notations




119894


119894


119894per






119894



119896

119864119896


119892119896when using CQI 119909

119860119896



119868119905



119868119903




119894and RN

119895


119894










receiver 119895

Competing Interests


Acknowledgments


References

































Advances in

FuzzySystems


Volume 2014












Journal of

Journal of





Advances in

Multimedia


Biomedical Imaging



Advances in


RoboticsJournal of










Advances in











receiver 119895

Competing Interests


Acknowledgments


References

































Advances in

FuzzySystems


Volume 2014












Journal of

Journal of





Advances in

Multimedia


Biomedical Imaging



Advances in


RoboticsJournal of










Advances in










Advances in

FuzzySystems


Volume 2014












Journal of

Journal of





Advances in

Multimedia


Biomedical Imaging



Advances in


RoboticsJournal of










Advances in



Research Article A Novel Energy-Saving Resource Allocation...

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Transcript of Research Article A Novel Energy-Saving Resource Allocation...