research article transmit beamforming optimization design

Research ArticleTransmit Beamforming Optimization Design forBroadband Multigroup Multicast System

Zilong Zhang1 Xiaodong Xu1 and Yanan Wu2

1Key Laboratory of Wireless-Optical Communications Chinese Academy of Sciences School of Information Science and TechnologyUniversity of Science and Technology of China No 96 Jinzhai Road Hefei Anhui 230027 China2Department of Detection Nanjing Forest Police College Nanjing Jiangsu 210023 China

Correspondence should be addressed to Xiaodong Xu xdxuustceducn

Received 11 December 2014 Accepted 15 April 2015

Academic Editor Mustapha Zidi

Copyright copy 2015 Zilong Zhang 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

Spectral efficient transmission techniques are necessary and promising for future broadband wireless communications where thequality of service (QoS) andor max-min fair (MMF) of intended users are often considered simultaneously In this paper boththe QoS problem and the MMF problem are investigated together for transmit beamforming in broadband multigroup multicastchannels with frequency-selective fading characters We first present a basic algorithm by directly using the results in frequency-flat multigroup multicast systems (Karipidis et al 2008) namely the approximation algorithms in this paper for both problemsrespectively Due to high computational consumption nature of the approximation algorithms two reduced-complexity algorithmsfor each of the two problems are proposed separately by introducing the time-frequency correlations In addition parameters inthe new time-frequency formulations such as the number of optimization matrix variables and the taps of the beamformer withfinite impulse response (FIR) structure can be used to make a reasonable tradeoff between computational burden and systemperformance Insights into the relationship between the two problems and some analytical results of the computational complexityof the proposed algorithms are also studied carefully Numerical simulations indicate the efficiency of the proposed algorithms

1 Introduction

Targeting for supporting high throughput and link reliabilitymultiple-antenna transmission techniques have prevailedin the development of terrestrial wireless communicationsystems such as Long Term Evolution Advanced (LTE-A)[1] and future mobile telecommunication networks [2] Notsurprisingly when equipped with multiple antennas at thetransmit side physical-layer multicasting renders its greatadvantage in spectral efficiency over the other communica-tion mechanisms especially for some particular applicationsincluding network video service and online gaming Inthis regard transmit beamforming has received enormousattentions in the literature where perfect channel stateinformation (CSI) is assumed available at both ends and thechannel between each transmit antenna and receive antennaappears frequency-flat fading property

To provide performance assurance to each of the intendedreceivers in multicast systems the quality of service (QoS)

problem [3] and the max-min fair (MMF) problem [4] areusually formulated and investigated in the literature wherethe criteria ofminimizing the total transmission power undereach userrsquos minimum signal-to-interference-plus-noise ratio(SINR) constraint and maximizing the minimum SINRamong all users under the average transmit power constraintare considered respectively Due to the NP-hardness of theoptimization problems several efficient algorithms have beenproposed to guarantee satisfactory performance in single-group multicast scenario What is more an analytical resultof these two problems was extended to the case of multigroupmulticasting in [5] where a solid algorithm based on bothsemidefinite relaxation (SDR) [6] and Gaussian randomiza-tion was proposed to solve the multigroup QoS problemand then an iterative algorithm based on a one-dimensionalbisection search [7] was also adopted to handle the MMFproblem To achieve improved performance for the multi-group QoS problem the authors of [8] proposed an iterativealgorithm which solves an approximate second-order cone

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015 Article ID 563863 13 pageshttpdxdoiorg1011552015563863

2 Mathematical Problems in Engineering

Table 1 List of related works

Problem description Application scenarios Solving approach SourceQoS Broadcast SDR [3]MMF Broadcast Iterative algorithm [4]QoS Multigroup multicast SDR + Gaussian randomization [5]MMF Multigroup multicast SDR + Gaussian randomization + bisection [5]QoS Multigroup multicast Iterative algorithm [8]MMF Multigroup multicast Iterative algorithm [9]QoS Broadcast with PACs Iterative algorithm + duality [10]MMF Broadcast with PACs Iterative algorithm + duality [11]MMF Multigroup multicast with PACs SDR + Gaussian randomization + bisection [12]

programming (SOCP) problem in each iteration And in[9] an iterative algorithm with low complexity and superiorperformance was further investigated to cope with the multi-group MMF problem Recently contrary to the total trans-mission power constraints transmit beamforming under per-antenna power constraints (PACs) was introduced in [10ndash12] As an extending work of [5] the weighted multigroupmulticast MMF problem with PACs was investigated in [12]For the sake of clear expression related works are listed inTable 1

As aforementioned the design of transmit beamformerhas been primarily studied over multiple-antenna multicas-ting and frequency-flat fading channels and to the authorsrsquobest knowledge few works have been dedicated to the case offrequency-selective fading multicasting scenario Motivatedby the potential advantages of multiple-antenna transmis-sion over the frequency-selective fading channels we areconcerned about the QoS and MMF problems in this paperfor multigroup multicasting The main contributions of thispaper are listed in the following

(1) For the QoS problem an approximation algorithmis firstly derived for broadband systems based onthe idea of narrowband multigroup multicasting in[5] And two reduced-complexity algorithms are alsoproposed from the frequency-domain and the time-domain perspectives separately Some parameters incorrelation with the QoS problem are analyzed andcomputational consumption is comparatively com-puted to showmore insights into the tradeoff betweenperformance and complexity

(2) For the MMF problem the corresponding approxi-mation algorithm and its low-complexity modifica-tions are also proposed in a similar way as that ofQoS problem Furthermore the relationship betweenthe QoS problem and the MMF problem is discussedcarefully followed by a complexity analysis

(3) Simulation experiments demonstrate the effective-ness of the frequency-domain and the time-domainalgorithms for bothQoS problem andMMF problemThe main analysis results that the controlled parame-ters in the proposed algorithms could be used tomakea tradeoff between complexity and performance areverified through the numerical examples

The remainder of this paper is structured as follows InSection 2 the system model and the QoS problem are intro-duced briefly The approximation solution to the QoS prob-lem is derived in Section 3 And Section 4 presents two beam-forming algorithms Section 5 formulates the MMF problemand solves it based on the proposed algorithms Computa-tional complexity analysis of proposed algorithms is givenin Section 6 In Section 7 the performance of the proposedalgorithms is evaluated and discussed Finally the conclusionis summarized in Section 8

Notations In the remainder of this paper boldface upper-case letters and math calligraphy uppercase letters denotematrices and boldface lowercase letters denote vectors (sdot)daggerE(sdot) and tr(sdot) are the conjugate transpose the expectationand the trace operator respectively C indicates the set ofcomplex numbers while otimes is Kronecker product lfloorsdotrfloor definesthe floor function A12 is matrix square root function of Aand 120575(sdot) is the impulse function A sim CN(0 I) means that Ais circularly symmetric complex Gaussian process with zeromean and unit variance matrix

2 System Model and Problem Statement

21 System Model Consider a multigroup multicast systemwith one transmitter (base station) and 119870 receivers (users)Assume the transmitter has 119872119905 antenna elements and eachreceiver is equipped with one antenna The users are splitinto 1 le 119871 le 119870 groups 1198921 1198922 119892119871 each containing119870119892119897

user indices We assume that each user listens to a singlemulticast group that is 119892119897 cap 1198921198971015840 = 0 where 119897 = 1198971015840 andcup119897 119892119897 = 1 119870 A frequency-selective fading channel with119881 effective paths is supposed between each transmit antennaand receive antenna and full CSI is available a priori at thetransmit side throughout this paper

With an 119874-tap FIR beamforming filter the transmittedsignal can be written as follows in space-time domain

s119899 =

119871minus1sum119897=0

119874minus1sum119900=0

w119897119900119909119897119899minus119900 (1)

where w119897119900 isin C119872119905times1 and 119909119897119900 denote the beamforming vectorand the discrete information sequence in association with the

Mathematical Problems in Engineering 3

middot middot middot




119852n

g1

g2

gL

Figure 1 The schematic diagram of a multigroup multicast system

119900th tap of FIR beamforming filter for the 119897th group respec-tively 119899 stands for the time index Figure 1 shows theschematic diagram of a multigroup multicast system With-out loss of generality (WLOG) assume the informationsequence is zero mean with unit variance and mutuallyuncorrelated that is E119909119897119900119909119897119900minus120591 = 120575(120591) sdot 120575(119897) Then for the119898th user in the 119897th group the received signal has the form as

119910119897119898119899 =

119881minus1sumV=0

H119897119898Vs119899minusV + 119906119897119898119899

=

119881minus1sumV=0

H119897119898V

119874minus1sum119900=0

w119897119900119909119897119899minusVminus119900⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟desired signal of the 119897th group

+

119881minus1sumV=0

H119897119898Vsum

1198971015840 =119897

119874minus1sum119900=0

w1198971015840 1199001199091198971015840 119899minusVminus119900⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

hybrid interference signals from the other groups

+ 119906119897119898119899

(2)

where H119897119898V isin C1times119872119905 is the channel impulse response of theVth path between the transmitter and the 119898th user in the 119897thgroup and 119906119897119898119899 is an additive Gaussian noise at the 119898th userwith zero mean and unit variance

For brevity purpose (2) can be represented by 119885-transform that is

119910119897119898119899 = H119897119898W119897 (119911)X119897119899 (119911)

+H119897119898 (119911) sum

1198971015840 =119897

W1198971015840 (119911)X1198971015840 119899 (119911) + 119906119897119898119899(3)

where H119897119898(119911) = sum119881minus1V=0 H119897119898V119911

minusV W119897(119911) = sum119874minus1119900=0 w119897119900119911

minus119900 andX119897119899(119911) = sum

119881+119874minus2119901=0 119909119897119899minus119881minus119874+2+119901119911minus119901

In this regard the total transmission power of the multi-group multicast system becomes

119875119905 =

119871minus1sum119897=0

int1

0

1003817100381710038171003817W119897 (119891)100381710038171003817100381722 119889119891 (4)

Similarly the SINR at the119898th receiver in the 119897th group can beformulated as

SINR119897119898

=int10 H119897119898 (119891)W119897 (119891)Wdagger

119897(119891)Hdagger

119897119898(119891) 119889119891

sum1198971015840 =119897

int10 H119897119898 (119891)W1198971015840 (119891)W1015840dagger

119897(119891)Hdagger

119897119898(119891) 119889119891 + 1205902

119897119898

(5)

where 1205902119897119898

= E119906119897119898119899119906dagger

119897119898119899 H119897119898(119891) = H119897119898(119911)|119911=1198901198952120587119891 and

W119897(119891) = W119897(119911)|119911=1198901198952120587119891

22 QoS Problem Statement With the above-mentionedassumptions and definitions the problem of minimizing thetotal transmission power under the SINR constraints of eachuser 120574119897119898 namely the QoS problem can be expressed as

P1 minW119897(119891)

119875119905

st SINR119897119898 ge 120574119897119898 forall119898 forall119897

(6)

forall119898 isin 119892119897 and forall119897 isin 0 119871 minus 1 To solve this problem thefollowing discrete-time form is usually adopted that is

P2 minW119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

1003817100381710038171003817W119897119894

100381710038171003817100381722

st SINR1015840

119897119898 ge 120574119897119898 forall119898 forall119897

(7)

with

SINR1015840

119897119898

=(1119868) sum

119868minus1119894=0 H119897119898119894W119897119894W

dagger

119897119894Hdagger

119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 H119897119898119894W1198971015840 119894W

dagger

1198971015840 119894Hdagger

119897119898119894+ 1205902

119897119898

(8)

where 119868 is a sufficiently large positive integer H119897119898119894 =

H119897119898(119911)|119911=1198901198952120587119894119868 andW119897119894 = W119897(119911)|119911=1198901198952120587119894119868 Note that problem P2 is a discrete approximation of

problem P1 and its approximate accuracy increases as 119868

approaches infinity In fact it is a quadratically constrainedquadratic programming (QCQP) problem with nonconvexconstraints Moreover as a special case of this problemmultigroup multicasting over frequency-flat fading channel(ie 119881 = 1) has been proven to be NP-hard in [5] ThereforeproblemP2 is NP-hardness whichmotivates us to pursue anapproximate solution of it


3 Approximate Solution

Let Q119897119894 = W119897119894Wdagger

119897119894 and we can get the equivalent form of

problemP2

P3 minQ119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894)

st SINR119903

119897119898ge 120574119897119898

Q119897119894 ⪰ 0

rank (Q119897119894) = 1 forall119894 forall119898 forall119897

(9)

where

SINR119903

119897119898=

(1119868) sum119868minus1119894=0 H119897119898119894Q119897119894H

dagger

119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 H119897119898119894Q1198971015840 119894H

dagger

119897119898119894+ 1205902

119897119898

(10)

forall119894 isin 0 119868 minus 1 forall119898 isin 119892119897 and forall119897 isin 0 119871 minus 1 in thissection

Due to the nonconvex nature of problem P3 we dropthe 119871 sdot 119868 rank-one constraints and obtain a semidefinite pro-gramming (SDP) variation

P4 minQ119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894)

st SINR119903

119897119898ge 120574119897119898

Q119897119894 ⪰ 0 forall119894 forall119898 forall119897

(11)

It is noteworthy that problem P4 can be handled byinterior point method (IPM) [13] and the feasible set of thisproblem is actually a superset of that of problem P3 As aconsequence the optimum objective value of problemP4 iscertainly equal or less than that of problemP3

Proposition 1 The optimal solution Q119897119894 of problem P4satisfies

rank (Q119897119894) le 119870119892119897 forall119894 forall119897

119871minus1sum119897=0

119868minus1sum119894=0

rank2 (Q119897119894) le 119870(12)

Proof In order to prove this proposition the dual problem ofproblemP4 is first considered and formulated as

D1 max120579119897119898

119871minus1sum119897=0

sum119898isin119892119897

1205791198971198981205741198971198981205902119897119898

st Y119897119894 ⪰ 0

120579119897119898 ge 0 forall119894 forall119898 forall119897

(13)

where Y119897119894 = I + sum1198971015840 =119897

sum119898isin11989211989710158401205791198971015840 1198981205741198971015840119898H

dagger

1198971015840 119898119894H1198971015840 119898119894 minus

sum119898isin119892119897

120579119897119898Hdagger

119897119898119894H119897119898119894 and 120579119897119898 is the dual variable associated

with SINR constraints in problem P4 Based on the com-plementarity conditions which are part of the Karush-Kuhn-Tucker (KKT) conditions we have

Y119897119894Q119897119894 = (I+ sum

1198971015840 =119897

sum119898isin1198921198971015840

1205791198971015840 1198981205741198971015840 119898Hdagger

1198971015840 119898119894H1198971015840119898119894

minus sum119898isin119892119897

120579119897119898Hdagger

119897119898119894H119897119898119894)Q119897119894 = 0

(14)

and thenwe can calculate the rank ofΘ119897119894Q119897119894 bydefiningΘ119897119894 =

I+sum1198971015840 =119897

sum119898isin11989211989710158401205791198971015840 1198981205741198971015840 119898H

dagger

1198971015840 119898119894H1198971015840 119898119894 ≻ 0 which immediately

leads torank (Θ119897119894Q119897119894) = rank (Q119897119894)

= rank( sum119898isin119892119897

120579119897119898Hdagger

119897119898119894H119897119898119894Q119897119894)

le sum119898isin119892119897

rank (120579119897119898Hdagger

119897119898119894H119897119898119894Q119897119894)

le 119870119892119897

(15)

On the other hand according to Theorem 32 in [14] itis readily to verify that the optimal solution to problem P4satisfies sum

119871minus1119897=0 sum

119868minus1119894=0 rank

2(Q119897119894) le 119870

From Proposition 1 it appears that when each group hasonly one user that is 119870119892119897

= 1 we have rank(Q119897119894) le 1 whichmeans problem P4 is actually equivalent to problem P3 inthis case which will obviously result in a frequency-selectivefading extension of the work in [3] In general if the solutionof problem P4 meets the rank-one constraint that isrank(Q119897119894) = 1 an eigenvalue decomposition (EVD) of Q119897119894 =

U119897119894Λ119897119894Udagger

119897119894may help to generate the frequency-domain beam-

forming vectors where W119897119894 = U119897119894Λ12119897119894

( 1) Otherwise theGaussian randomization technique [5] is used to obtaincandidates of the beamforming vector that is W

119902

119897119894=

U119897119894Λ12119897119894

V119902

119897119894 119902 = 1 2 119876 where 119876 is the maximum num-

ber of the randomizations and V119902

119897119894isin C119872119905times1 sim CN(0 I)

Note that although these processed candidates satisfyEW

119902

119897119894W

119902dagger

119897119894 = Q119897119894 and rank(W

119902

119897119894W

119902dagger

119897119894) = 1 they may still

violate SINR constraints For each candidate a feasible allo-cated power should thus be figured out by solving a multi-group multicast power control (MMPC) problem that is

M1 min119901119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897119894120573119897119894

st(1119868) sum

119868minus1119894=0 119901119897119894120572119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 1199011198971015840 1198941205721198971015840 119898119894 + 1205902

119897119898

ge 120574119897119898


(16)

where 119901119897119894 denotes the power factor for the beamformerW119902

119897119894

120573119897119894 = W119902

11989711989422 120572119897119898119894 = H119897119898119894W

119902

119897119894W

119902dagger

119897119894Hdagger

119897119898119894 and 1205721198971015840 119898119894 =

H119897119898119894W119902

1198971015840 119894W

119902dagger

1198971015840 119894Hdagger

119897119898119894 ProblemM1 is of linear program (LP)


(1) InputH119897119898119894 120590119897119898 120574119897119898 119876(2) solve SDP problemP4 by IPM(3) if rank(Q119897119894) = 1 forall119894 forall119897 then(4) W119897119894 = U119897119894Λ

12119897119894

( 1)(5) break(6) else(7) calculate EVD Q119897119894 = U119897119894Λ119897119894Udagger

119897119894

(8) for 119902 = 1 to 119876 do(9) generate candidatesW119902

119897119894= U119897119894Λ

12119897119894V119902

119897119894

(10) solve MMPC problemM1 to obtain radic119901119897119894W119902

119897119894

(11) end(12) choose W119897119894 = argmin

radic119901119897119894W119902

119897119894sum

119871minus1119897=0 sum

119868minus1119894=0119901119897119894120573119897119894

(13) end(14)Output W119897119894

Algorithm 1 Approximation algorithm

and can be easily solved by basic convex tools if the optimalsolution existsThe resulting frequency-domain beamformercan thus be generated by radic119901119897119894W

119902

119897119894 and the associated objec-

tive value (1119868) sum119871minus1119897=0 sum

119868minus1119894=0 119901119897119894120573119897119894 is recorded

When 119902 reaches119876 the beamformer corresponding to thebest candidate with minimum objective value can be selectedas the optimal oneThe detailed process of the approximationalgorithm is summarized in Algorithm 1

More clearly the approximation algorithm can be treatedas a direct use of the algorithm in [5] on every frequencybin for frequency-selective multigroup multicast systemThe higher approximation accuracy is the larger 119868 may beintroduced For example an 119868 = 128 or larger is often neededfor practical systems which will lead to extremely highcomputational complexity A computing-strong ability is thusrequired at the transmitter otherwise we may not figure outthe exact beamformers by the approximation algorithmwhenthe channel coefficients change with rapid fluctuation

4 Proposed Beamforming Algorithms

To combat the heavy computation burden of the approxi-mation algorithm new beamforming algorithms which canmake tradeoff between performance and complexity areeagerly demanded in these situations Towards this end twobeamforming algorithms are proposed from perspectives offrequency domain and time-domain respectively in thissection

41 Beamforming Design in Frequency Domain In fact thechannel coefficient vector H119897119898119894 has certain correlation withH1198971198981198941015840 when 119894 and 1198941015840 are close to each other It follows that wecan reduce the complexity of solving the SDP problemP4 byreducing the number of optimization variables More clearlythe frequency-domain channel vectors H119897119898119894

119868minus1119894=0 can now be

divided into several groups Assume each group has119865 vectors(here 119868 should be divisible by 119865) and there are 119868119865 groups for

H119897119898119894119868minus1119894=0 In this way we only need to optimize Q119897119894

119868119865minus1119894=0

and the optimization problem can be converted to

P5 minQ119897119894119868119865minus1119894=0

119865

119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)

st SINR1199031119897119898

ge 120574119897119898

Q119897119894 ⪰ 0


(17)

where

SINR1199031119897119898

=(1119868) sum

119868minus1119894=0 H119897119898119894Q119897lfloor119894119865rfloor+1H

dagger

119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 H119897119898119894Q1198971015840 lfloor119894119865rfloor+1H

dagger

119897119898119894+ 1205902

119897119898

(18)

forall119894 isin 0 119868119865 minus 1 forall119898 isin 119892119897 and forall119897 isin 0 119871 minus 1in this subsection Similarly by dropping the 119871119868119865 rank-oneconstraints an SDP problem can then be obtained as

P6 minQ119897119894119868119865minus1119894=0

119865

119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)

st SINR1199031119897119898

ge 120574119897119898


(19)

and the corresponding MMPC problem follows that

M2 min119901119897119894119868119865minus1119894=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897lfloor119894119865rfloor+1120573119897119894

st(1119868) sum

119868minus1119894=0 119901119897lfloor119894119865rfloor+1120572119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 1199011198971015840 lfloor119894119865rfloor+11205721198971015840 119898119894 + 1205902

119897119898

ge 120574119897119898


(20)

where 119901119897119894 denotes the power factor for the beamformerW

119902

119897119894 and 120573119897119894 = W

119902

119897lfloor119894119865rfloor+122 Accordingly we have

120572119897119898119894 = H119897119898119894W119902

119897lfloor119894119865rfloor+1W119902dagger

119897lfloor119894119865rfloor+1Hdagger

119897119898119894 and 1205721198971015840 119898119894 =

H119897119898119894W119902

1198971015840 lfloor119894119865rfloor+1W119902dagger

1198971015840 lfloor119894119865rfloor+1Hdagger

119897119898119894 Thanks to similar random-

ization process as Algorithm 1 the frequency-domain algo-rithm is summarized in Algorithm 2

As a matter of convenience denote P5(120574 119865) as thefrequency-domain problem with parameters 120574 and 119865 Here120574 ≜ [12057400 12057401 1205740119898 120574119897119898] Obviously the proposedbeamforming design in frequency-domain can be treated asa special case of the approximate solution presented in theprevious section In otherwords if the parameter119865 is set to be1 the frequency-domain QoS problemP5(120574 1) is equivalentto the QoS problemP3(120574)


(1) InputH119897119898119894 120590119897119898 120574119897119898 119865 119876(2) solve SDP problemP6 by IPM(3) if rank(Q119897119894) = 1 forall119894 forall119897 then(4) W119897119894 = U119897119894Λ

12119897119894


119897119894


119897119894= U119897119894Λ

12119897119894V119902

119897119894


119897119894

(11) end(12) choose

W119897119894 = argminradic119901119897119894W

119902

119897119894sum

119871minus1119897=0 sum

119868minus1119894=0119901119897lfloor119894119865rfloor+1120573119897119894

(13) end(14)Output W119897119894

Algorithm 2 Frequency-domain algorithm

The following results shed more lights on the influence ofthe controlling parameter 119865

Proposition 2 Assume problem P5(120574 1198651) is feasible with afixed set of channel vectors SINR constraints and noise powersthe sufficient condition for problemP5(120574 1198652) to be also feasibleis that 1198651 can be divisible by 1198652

Proof Assume Q11989711989410158401198681198651minus11198941015840=0 denotes the optimal solution to

problemP5(120574 1198651) For any 1198652 which satisfies that 1198651 is divis-ible by 1198652 Q1198971198941015840

1198681198651minus11198941015840=0 can be expanded as Qexp

1198971198941198681198652minus1119894=0 where

Qexp119897119894

(1198941015840+1)11986511198652minus1

119894=119894101584011986511198652= Q1198971198941015840 It can be verified that Q

exp119897119894

1198681198652minus1119894=0 is a

feasible solution to problemP5(120574 1198652) by substituting it

Proposition 3 Assume P5(120574 1198651) is feasible with a fixed setof channel vectors SINR constraints and noise powers withoptimal value 1198751 If 1198651 is divisible by 1198652 the optimal value ofproblem P5(120574 1198652) defined as 1198752 is less than or equal to 1198751that is 1198752 le 1198751 and the equality holds up if and only if thesolutions of these two problems are the same

Proof Define the optimal solution to problem P5(120574 1198651) tobe Q1198971198941015840

1198681198651minus11198941015840=0 with optimal value 1198751 From the proof of

Proposition 2 Q11989711989410158401198681198651minus11198941015840=0 can be expanded to Qexp

1198971198941198681198652minus1119894=0

which is a feasible solution to problemP5(120574 1198652)with optimalvalue 1198751 Assume the optimal solution to problemP5(120574 1198652)

is Qopt119897119894

1198681198652minus1119894=0 with optimal value 1198752 Due to the optimality

Qopt119897119894

1198681198652minus1119894=0 is at least a good solution as Qexp

1198971198941198681198652minus1119894=0 thus

we have 1198752 le 1198751 And if the Qopt119897119894

1198681198652minus1119894=0 can be obtained by

Q11989711989410158401198681198651minus11198941015840=0 1198752 = 1198751 holds

42 Beamforming Design in Time-Domain Besides reducingthe number of optimization variables in frequency-domainimmediately an alternative way can also benefit the com-plexity reduction by cutting down the number of the FIR

filter taps from time-domain perspective Assume the time-domain FIR filter has 119862 taps where 119862 is far less than 119868 thetransmitted signal in (1) changes to

s119899 =

119871minus1sum119897=0

119862minus1sum119888=0

w119897119888119909119897119899minus119888 (21)

Define

W119897119894 =

119862minus1sum119888=0

w119897119888119890minus2120587119894119888119868

(22)

which can be converted into an equation in Kronecker form

W119897119894 =

[[[[[[[[

[

1

119890minus1198952120587119894119873

119890minus1198952120587119894(119862minus1)119873

]]]]]]]]

]

119879

otimes I119872119905

⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟K119894

sdot

[[[[[[[

[

w1198970

w1198971

w119897119862minus1

]]]]]]]

]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

W119897

(23)

with K119894 isin C119872119905times119862119872119905 andW119897 isin C119862119872119905times1 In the sequel the totaltransmission power is reformulated accordingly as

1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894) =1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (W119897119894W1015840

119897119894)

=1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (K119894W119897Wdagger

119897Kdagger

119894)

(24)

By definingQ119897 = W119897Wdagger

119897 the QoS problem thus becomes

P7 minQ119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (K119894Q119897Kdagger

119894)

st SINR1199032119897119898

ge 120574119897119898

Q119897 ⪰ 0

rank (Q119897) = 1 forall119898 forall119897

(25)

with

SINR1199032119897119898

=(1119868) sum

119868minus1119894=0 tr (H119897119898119894K119894Q119897Kdagger

119894Hdagger

119897119898119894)

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 tr (H119897119898119894K119894Q1198971015840Kdagger

119894Hdagger

119897119898119894) + 1205902

119897119898

(26)

and forall119898 isin 119892119897 forall119897 isin 0 119871 minus 1 in this subsection Similarto previous two algorithms an SDP problem is obtained afterdropping 119871 rank-one constraints

P8 minQ119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)

st SINR1199032119897119898

ge 120574119897119898

Q119897 ⪰ 0 forall119898 forall119897

(27)


(1) InputH119897119898119894 120590119897119898 120574119897119898 119862 119876(2) solve SDP problemP8 by IPM(3) if rank(Q119897) = 1 forall119897 then(4) W119897 = U119897Λ

12119897

( 1)(5) break(6) else(7) calculate EVD Q119897 = U119897Λ119897U

dagger

119897

(8) for 119902 = 1 to 119876 do(9) generate candidatesW

119902

119897= U119897Λ

12119897V119902

119897


119897


radic119901119897W119902

119897sum

119871minus1119897=0 sum

119868minus1119894=0119901119897120573119897119894

(13) end(14)Output W119897

Algorithm 3 Time-domain algorithm

and the corresponding MMPC problem can be representedas

M3 min119901119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897120573119897119894

st(1119868) sum

119868minus1119894=0 119901119897120572119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 11990111989710158401205721198971015840 119898119894 + 1205902

119897119898

ge 120574119897119898

119901119897 ge 0 forall119898 forall119897

(28)

where 119901119897 is the power factor for the beamformer W119902

119897

and 120573119897119894 = tr(K119894Q119902

119897Kdagger

119894) Accordingly we have 120572119897119898119894 =

H119897119898119894K119894Q119902

119897Kdagger

119894Hdagger

119897119898119894 and 1205721198971015840 119898119894 = H119897119898119894K119894Q

119902

1198971015840Kdagger

119894Hdagger

119897119898119894

Again we can summarize the detailed procedure of the time-domain algorithm in Algorithm 3

From Algorithm 3 it is clear that if the parameter 119862 is setto be 119868 the time-domainQoS problemP7(120574 119862) is equivalentto the QoS problem P3(120574) The following propositions statesome further results according to this problem

Proposition 4 If problem P7(120574 1198621) is feasible with a fixedset of channel vectors SINR constraints and noise powers thesufficient condition for problemP7(120574 1198622) to be also feasible isthat 1198622 should be greater than 1198621

Proof Assume Qopt119897

119871minus1119897=0 denotes the optimal solution to

problem P7(120574 1198621) For any 1198622 which satisfies that 1198622 isgreater than 1198621 the solution Qopt

119897119871minus1119897=0 to problemP7(120574 1198621)

can be expanded as Qexp119897

119871minus1119897=0 where

Qexp119897

119871minus1

119897=0

= [

[

Qopt119897

119871minus1

119897=001198621119872119905times(1198622minus1198621)119872119905

0(1198622minus1198621)119872119905times11986211198721199050(1198622minus1198621)119872119905times(1198622minus1198621)119872119905

]

]

(29)

It can be confirmed that Qexp119897

119871minus1119897=0 is also a feasible solution

to problemP7(120574 1198622) by substituting it

Proposition 5 Assume P7(120574 1198621) is feasible with a fixed setof channel vectors SINR constraints and noise powers withoptimal value 1198753 If 1198622 is greater than 1198621 the optimal valueof problemP7(120574 1198622) defined as 1198754 is less than or equal to 1198753that is 1198754 le 1198753 and the equality holds up if and only if thesolutions of these two problems are the same

The basic idea of the proof process is similar to thatof Proposition 3 and thus omitted here We can replaceQ1198971198941015840

1198681198651minus11198941015840=0 and Qexp

1198971198941198681198652minus1119894=0 by Qopt

119897119871minus1119897=0 and Qexp

119897119871minus1119897=0 and

then the result can be reached

5 Max-Min Fair Problem

In addition to the QoS problem another problem alwaysconsidered in a multigroup multicast system is the MMFproblem The original problem of maximizing the minimumSINR of all users under the total transmission power con-straint can be written as

Q1 maxQ119897119894119905isinR

119905

stSINR119903

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894) le 119875 forall119898 forall119897 forall119894

(30)

In fact this problem contains a special case with multicastover frequency-flat fading channel (119881 = 1) which has beenproven to beNP-hard in [5] therefore problemQ1 is alsoNP-hard By virtue of the idea for solving QoS problem it can berelaxed by dropping the rank constraints

Q2 maxQ119897119894119905isinR

119905

stSINR119903

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


(31)

However contrary to the QoS problem P4 problem Q2cannot be transformed into an SDP problem due to theexistence of119870 nonlinear inequality constraintsThe causes ofthese nonlinear inequality constraints is that the SINR target 119905for all users is no longer a constant but a variable in theMMFproblem

Fortunately problemQ2 can be relaxed and its119870 inequal-ity constraints can be changed into linear constraints for


a given 119905 Thus the bisection search method [5 15] can beused to deal with this problem Note that after getting somebeamforming candidates an MMPC problem is consideredhere

M4 max119901119897119894119905isinR

119905

st(1119868) sum

119868minus1119894=0 119901119897119894120572119897119898119894

(120574119897119898119868) sum1198971015840 =119897

sum119868minus1119894=0 1199011198971015840 1198941205721198971015840 119898119894 + 1205741198971198981205902

119897119898

ge 119905

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897119894120573119897119894 = 119875

119905 ge 0 119901119897119894 ge 0 forall119894 forall119898 forall119897

(32)

where all variables have been defined in Section 3 Due tothe variation property of 119905 problemM4 cannot be solved asan equivalent LP Therefore we continue to rely on bisectionsearch method to solve this problem

Alike the QoS problem the reduced-complexity MMFproblem can also be considered both in frequency domainand in time-domain where the frequency-domain versioncan be formulated as

Q3 maxQ119897119894119868119865minus1119894=0 119905isinR

119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)

le 119875 forall119898 forall119897 forall119894

(33)

Drop the rank constraints and we can obtain the relaxedproblem as follows


119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(34)

Similarly the time-domain version of the MMF problemslooks like


119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

rank (Q119897) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)

le 119875 forall119898 forall119897


119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)


(35)

before and after rank relaxation respectively To solve prob-lems Q3 and Q5 the same idea can be found when solvingproblem Q1 To illustrate the procedure of the proposedalgorithms a general solving framework for QoS and MMFproblems is shown in Figure 2

DenoteQ1(120574 119875)Q3(120574 119875 119865) andQ5(120574 119875 119862) as for prob-lem Q1 problem Q3 and problem Q5 respectively withparticular parameters 120574 119875 119865 and 119862 It can be seen thatif the parameter 119865 is set to be 1 the frequency-domainMMF problemQ3(120574 119875 119865) is equivalent to theMMF problemQ1(120574 119875) Also the time-domainMMFproblemQ5(120574 119875 119862) isequivalent to problem Q1(120574 119875) too if 119862 = 119868

The following analytical results demonstrate the relation-ship between the MMF problems with different parameters

Proposition 6 Assume Qopt1119897119894

1198681198651minus1119894=0 is the optimal solution to

the frequency-domainMMFproblemQ3(120574 119875 1198651)with optimalvalue 1199051 and Qopt2

1198971198941198681198652minus1119894=0 is the optimal solution to problem

Q3(120574 119875 1198652) with optimal value 1199052 The sufficient condition for1199052 ge 1199051 is that 1198651 is divisible by 1198652 and the equality holds up ifand only if the solutions of these two problems are the same



the time-domain MMF problem Q5(120574 119875 1198621) with optimalvalue 1199053 and Qopt2


Q5(120574 119875 1198622) with optimal value 1199054 The sufficient condition for1199054 ge 1199053 is that 1198622 is greater than 1198621 and the equality holds upif and only if the solutions of these two problems are the same


Initialize and input

QoS or MMFproblem

QoS problem MMF problem

Frequency-domain algorithmin Section 41 or

time-domain algorithm inSection 42

in Section 41 or

Section 42

Frequency-domain algorithm

time-domain algorithm inOutput

Bisection process inSection 5 starts

Bisection processconverges

Yes

No

Output

Figure 2 Block-diagram of the solving framework

Furthermore from the problem formulation it appearsthat the MMF problems are always feasible while thingscould be different for the QoS problem The relationshipbetween QoS and MMF problems for narrowband multi-group multicast case is discussed in [5] Results therein canbe also extended to the broadband multigroup multicast case(ie 119881 gt 1) For completeness several valuable conclusionsare drawn here

Proposition 8 For a fixed set of channel vectors and noisepowers the QoS problem P3 is parameterized by 120574 where120574 = [12057400 12057401 1205740119898 120574119897119898] Then it can be represented asP3(120574) Likewise the MMF problem Q1 is parameterized by 120574and 119875 that is Q1(120574 119875) The QoS problem P3 and the MMFproblem Q1 have the relationship as

119875 = P3 (Q1 (120574 119875) 120574) (36)

119905 = Q1 (120574P3 (119905120574)) (37)

Proof Define Qopt119897119894

as the optimal solution to problemQ1(120574 119875) and its corresponding optimal value is 119905opt It is easyto verify that Qopt

119897119894 is a feasible solution to problemP3(119905opt120574)

and the corresponding optimal value is 119875 Assume there is afeasible solution Qfea

119897119894 to problem P3(119905opt120574) and 119875fea is the

associated optimal value which satisfies 119875fea lt 119875 and we candistribute the power 119875 minus 119875fea to all Qfea

119897119894 evenly to obtain

a larger optimal value 119905fea than 119905opt under the same powerconstraint It contradicts the optimality of Qopt

119897119894 for problem

Q1(120574 119875) which means that the assumption of Qfea119897119894

is wrongand (36) has been proved

In order to prove (37) a similar process could be appliedDefine Q1015840opt

119897119894 as the optimal solution and 1198751015840opt as the associ-

ated optimal value to problemP3(119905120574) (ifP3(119905120574) is feasible)Note that Q1015840opt

119897119894 is a solution to problem Q1(120574 1198751015840opt) with

optimal value 119905 Assume there is a feasible solution Q1015840fea119897119894

to problem Q1(120574 1198751015840opt) with optimal value 1199051015840fea gt 119905 Thusthere exists a constant 0 lt 120583 lt 1 which can be multipliedby Q1015840fea

119897119894 and the new solution set 120583Q1015840fea

119897119894 also satisfies the

SINR constraints Obviously the solution set 120583Q1015840fea119897119894

has alower transmission power 1205831198751015840opt lt 1198751015840opt which contradicts theoptimality of Q1015840opt

119897119894 for problemP3(119905120574) thus the assumption

is invalid

Proposition 9 The QoS and MMF problem pairs that isproblems P4 and Q2 problems P5 and Q3 problems P6and Q4 problems P7 and Q5 and problems P8 and Q6 allhave the same relationship between problemsP3 andQ1 Alsocorresponding QoS MMPC and MMF MMPC problem pairshave the same relationship too

6 Complexity Analysis

First of all the computational complexities of solving theQoSproblems are discussed in this section For the approxima-tion algorithm derived in Section 3 the SDP problem P4has 119871119868 matrix variables with size 119872119905 times 119872t and 119870 linearinequality constraints Based on the results in [13] it takesO(119871

051198680511987205119905log(1120598)) iterations and each iteration needs

O(119871311986831198726119905+1198701198711198681198722

119905) arithmetic operations 120598 is the accuracy

of the solution here When solving MMPC problem M1 ittakes O(119871

0511986805log(1120598)) iterations and each iteration needsO(11987131198683 + 119870119871119868) arithmetic operations Assuming the param-eter 120598 is same for all algorithms for the sake of simplicity thetotal computational complexity for solving the QOS problemP2 is thus O((119871

351198683511987265119905

+ 119870119871151198681511987225119905

+ 1198761198713511986835 +

1198761198701198711511986815)log(1120598))For the frequency-domain beamforming algorithm

the SDP problem P6 has 119871119868119865 matrix variables with size119872119905 times 119872119905 and 119870 linear inequality constraints Therefore theinterior point method will take O(1198710511986805119865minus0511987205

119905log(1120598))

iterations and each iteration needs O(119871311986831198726119905119865minus3 +

119870119871119868119865minus11198722119905) arithmetic operations After that it takes

O(11987105

11986805

119865minus05log(1120598)) iterations to solve the MMPC prob-

lem M2 and each iteration requires O(11987131198683119865

minus3+ 119870119871119868119865

minus1)

arithmetic operationsThus the total computational complex-ity for solving the QOS problem P5 can be calculated asO((119871

351198683511987265119905

119865minus35 + 119870119871151198681511987225119905

119865minus15 + 1198761198713511986835119865minus35 +

1198761198701198711511986815119865minus15)log(1120598))For the time-domain beamforming algorithm the SDP

problem P8 has 119871 matrix variables with size 119862119872119905 times 119862119872119905

and 119870 linear inequality constraints According to aboveresults it takes O(119871

0511986205

11987205119905log(1120598)) iterations and each

iteration needsO(119871311986261198726119905+11987011987111986221198722

119905) arithmetic operations

When solving MMPC problem M3 it takes O(11987105log(1120598))iterations and each iteration needs O(1198713 + 119870119871) arithmeticoperations Therefore the total computational complexity


for solving the QOS problem P7 is O((119871351198626511987265119905

+

119870119871151198622511987225119905

+ 11987611987135 + 11987611987011987115)log(1120598))Next we will analyze the computational complexities of

solving theMMFproblemswhich are analyzedDefine119861119878 and119861119871 as the number of the bisection iterations for solving therelaxedMMF problems and correspondingMMPC problemsseparately As mentioned in Section 5 the solving process oftheMMFproblem includes solving119861119878 times the SDPproblemand 119876119861119871 times the LP problem Thus the complexities of theproposed algorithms can be easily figured out based on theresult of the QoS problems For the approximation algo-rithm the frequency-domain beamforming algorithm andthe time-domain beamforming algorithm the overallcomputational complexities are O((119861119878119871

351198683511987265119905

119865minus35 +

119861119878119870119871151198681511987225119905

119865minus15 + 1198761198611198711198713511986835119865minus35 +

1198761198611198711198701198711511986815119865minus15)log(1120598)) O((119861119878119871351198683511987265

119905119865minus35 +

11986111987811987011987115

11986815

11987225119905

119865minus15

+ 11987611986111987111987135

11986835

119865minus35

+

1198761198611198711198701198711511986815119865minus15)log(1120598)) and O((119861119878119871351198626511987265

119905+

119861119878119870119871151198622511987225119905

+11987611986111987111987135+11987611986111987111987011987115)log(1120598)) respectively

For bothQoS problem andMMFproblem it is importantto point out that the computational complexities of thecorresponding frequency-domain algorithm and the time-domain algorithm are always much lower than that ofthe approximation one in practical wireless communicationsystems because the parameter 119868 is usually much larger than119868119865 and 119862 For example when the parameters of the systemare set as 119868 = 128 119871 = 2 119870 = 8 and 119876 = 300 while thelog(1120598) is assumed to be 1 the total arithmetic operations forQoS problems are plotted in Figure 3 It can be verified fromthe figure that with the decrease of 119865 the computational com-plexity of solving the frequency-domain algorithm increasesand larger 119862 leads to higher computational complexity forthe time-domain algorithm Meanwhile the two proposedalgorithms reduced the computational complexity efficiently

Since for MMF problems the computational consump-tion for both the frequency-domain algorithm and the time-domain algorithm are linear with those of QoS problems weignore the illustrative comparison in this section

7 Simulation Results

In this section several numerical examples are illustratedto demonstrate the effectiveness of proposed beamformingalgorithms For simplicity the frequency-selective fadingchannel between each receiver and the transmitter is built asa discrete channel model with 3 effective paths that is119881 = 3and each path is assumed to be independent and identicallydistributed (iid) Rayleigh fading channel The SINR con-straints for all users are the same All experimental results areaveraged over 1000 independent Monte Carlo runs and thenumber of randomizations of the SDR-based method is setto be 119876 = 300 for all algorithms used in our simulations AllSDP problems and LP problems are solved by CVX box [16]To measure the performance the approximation algorithmwith 119868 = 128 is chosen to be a comparable goal

71 Feasible Rate of SDP Problems The first step of theproposed algorithms is solving the relaxed SDP problems

4 6 8 10 12 14Number of transmitted antennas

Tota

l arit

hmet

ic o

pera

tions

for Q

OS

prob

lem

s

Approximation algorithm

108

109

1010

1011

1012

1013

1014

1015

1016

Frequency-domain algorithm F = 2



Time-domain algorithm C = 6



Figure 3 The total arithmetic operations for QoS problems

From the relaxation process we can see that the feasible set ofthe relaxed SDP problems is indeed a superset of one of theoriginal QCQP problems which leads to the following con-clusions If the relaxed problems are not feasible the originalones are not feasible either Rather if the relaxed problems arefeasible the original ones may be not feasible Therefore thefeasibility of the SDP problems is a necessary condition forthe validity of proposed algorithms In this subsection thefeasibility of the SDP problems is evaluated under conditionsof different number of multigroups 119871 transmit antennas 119872119905and users119870 All users are equally distributed into the groupswhichmeans that each group has119870119871 users in the simulation

Figure 4 shows the feasibility of SDPproblems in differentcases Because the approximation algorithm works well in allsituations which has all 100 percent feasibility for all SINRconstraints we just use one red line marked ldquoapproximationrdquoto present it By comparing the line pairs some conclusionscan be obtained For example with all other things beingequal we can get that the frequency-domain SDP problemwith less users is feasible with higher probability by compar-ing the ldquo119871 = 2 119872119905 = 6 119870 = 8 119865 = 8rdquo line and ldquo119871 = 2 119872119905 = 6119870 = 12 119865 = 8rdquo line (for time-domain SDP problem we canuse ldquo119871 = 2 119872119905 = 6 119870 = 8 119862 = 4rdquo line and ldquo119871 = 2 119872119905 = 6119870 = 12 119862 = 4rdquo line) The reason lies in that the morethe users are the more the interuser interference exists inmultigroup system By comparing different line pairs resultscan be concluded as follows

(i) The frequency-domain (or time-domain) SDP prob-lem with less users is feasible with higher probability(just mentioned above)


6 8 10 12 14 16 18 20SINR (dB)

0

01

02

03

04

05

06

07

08

09

1

()

ApproximationL = 2 Mt = 6 K = 8 F = 8

L = 2 Mt = 6 K = 12 F = 8

L = 3 Mt = 6 K = 12 F = 8

L = 3 Mt = 4 K = 12 F = 8

L = 2 Mt = 6 K = 8 C = 4

L = 2 Mt = 6 K = 12 C = 4

L = 3 Mt = 6 K = 12 C = 4

L = 3 Mt = 4 K = 12 C = 4

L = 2 Mt = 6 K = 8 C = 5

L = 2 Mt = 6 K = 8 F = 4

Figure 4 The feasible probability of SDP problems

(ii) The frequency-domain (or time-domain) SDP prob-lem with less groups is feasible with higher probabil-ityThat is because the more the groups are the morethe intergroup interference exists

(iii) The frequency-domain (or time-domain) SDP prob-lem with more transmit antennas is feasible withhigher probability since the more the transmit anten-nas are the more the spatial degrees of freedom canbe exploited

(iv) The frequency-domain SDPproblemwith smaller119865 isfeasible with higher probability The reason has beenproved in Proposition 2 and the simulation resultsverify it

(v) The time-domain SDP problem with larger 119862 isfeasible with higher probability The reason has beenproved in Proposition 4 and the simulation resultsverify it

72 Performance for QoS Problems Next the performancesof the proposed algorithms solving QoS problems are com-pared The system parameters are set as 119871 = 2 119872119905 = 6 and119870 = 8 Other cases with different system parameters are alsoevaluated and almost same results are obtained thus they areomitted here Note that only instances with feasible solutionsare counted and averaged Figure 5 displays the transmissionpower performance under various SINR constraints of pro-posed algorithms and the cumulative distribution function(CDF) under 120574119897119898 = 6 dB forall119898 isin 119892119897 forall119897 isin 0 119871 minus 1is given in Figure 6 It can be seen from Figure 5 that thefrequency-domain algorithm with smaller 119865 value has betterperformance and the time-domain algorithm with larger 119862

6 8 10 12 14 160

5

10

15

20

25

30

35

40

Tran

smiss

ion

pow

er

SINR (dB)

Approximation algorithmFrequency-domain algorithm F = 2






Figure 5 The transmission power performance for QoS problems

Transmission power







05 1 15 2 25 3 350

01

02

03

04

05

06

07

08

09

1

Cum

ulat

ive d

istrib

utio

n fu

nctio

n

Figure 6 The cumulative distribution function for QoS problems

value achieves lower transmission power Meanwhile we canget that these results are not only on average but also in eachof the trials fromFigure 6The validities of Propositions 3 and5 are verified When transmission power performance andcomputational complexity are considered together the time-domain algorithm with 119862 = 6 seems to be the best choice


2 4 6 8 10 12 14 16 18 20 22 246

8

10

12

14

16

18

Transmission power

SIN

R (d

B)







Figure 7 The performance for MMF problems

based on Figures 3 and 5 Taking approximation algorithm asa standard the complexity of solving the QoS problem hasbeen reduced to 1100 by the time-domain algorithm with119862 = 6 while the SINR loss is approximately equal to 1 dB

73 Performance for MMF Problems Finally the proposedalgorithms are adopted to solve the MMF problems We set119871 = 2 119872119905 = 6 and 119870 = 12 as the system parametersThe performances are investigated and shown in Figure 7 Asdiscussed in Section 5 because the MMF problem is closelyrelated with the QoS problem the simulation results are verysimilar to the ones of the QoS problems

From Figure 7 we can see that smaller 119865 for the fre-quency-domain algorithm and larger 119862 for the time-domainalgorithm improve the SINR performance respectively Like-wise the proposed algorithms make a great tradeoff betweenthe systemperformance and computational complexity Sameas previous subsection the time-domain algorithm with 119862 =

6 is the best choice when taking both performance andcomplexity into consideration

8 Conclusion

In this paper the downlink beamforming designs for thebroadband multigroup multicast QoS and MMF problemsare investigated By means of the traditional SDR andGaussian randomization methods two algorithms designedin frequency and time-domains are proposed to solve theQoS problem Then we extend these algorithms to handlethe MMF problem through an iterative bisection search pro-cess Several Monte Carlo simulations indicate the proposed

beamforming designs reduce the computational complexityconsiderably with slight performance loss the complexity ofsolving the QoS problem could be reduced to 1100 by thetime-domain algorithm with 119862 = 6 while the SINR loss isapproximately equal to 1 dB and similar conclusion can begiven for the MMF problem

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported in part by the National NaturalScience Foundation of China (no 61271272) the Intercolle-giate Key Project of Nature Science of Anhui Province (noKJ2012A283) and the National High Technology Researchand Development Program of China (863 Program) withGrant no 2012AA01A502 The authors would like to thankProfessor Xuchu Dai for his helpful discussions

References

[1] A Ghosh R Ratasuk B Mondal N Mangalvedhe and TThomas ldquoLTE-advanced next-generation wireless broadbandtechnologyrdquo IEEE Wireless Communications vol 17 no 3 pp10ndash22 2010

[2] S Chen and J Zhao ldquoThe requirements challenges and tech-nologies for 5G of Terrestrial mobile telecommunicationrdquo IEEECommunications Magazine vol 52 no 5 pp 36ndash43 2014

[3] M Bengtsson and B Ottersten ldquoOptimal downlink beam-forming using semidefinite optimizationrdquo in Proceedings of theAnnual Conference on Communication Control and Computingvol 37 pp 987ndash996 Citeseer 1999

[4] M Schubert andH Boche ldquoSolution of themultiuser downlinkbeamforming problem with individual SINR constraintsrdquo IEEETransactions on Vehicular Technology vol 53 no 1 pp 18ndash282004

[5] E Karipidis N D Sidiropoulos and Z-Q Luo ldquoQuality ofservice and max-min fair transmit beamforming to multiplecochannel multicast groupsrdquo IEEE Transactions on Signal Pro-cessing vol 56 no 3 pp 1268ndash1279 2008

[6] L Vandenberghe and S Boyd ldquoSemidefinite programmingrdquoSIAM Review vol 38 no 1 pp 49ndash95 1996

[7] A Schad and M Pesavento ldquoMultiuser bi-directional commu-nications in cooperative relay networksrdquo in Proceedings of the4th IEEE International Workshop on Computational Advancesin Multi-Sensor Adaptive Processing (CAMSAP rsquo11) pp 217ndash220San Juan Puerto Rico December 2011

[8] N Bornhorst and M Pesavento ldquoAn iterative convex approx-imation approach for transmit beamforming in multi-groupmulticastingrdquo in Proceedings of the IEEE 12th InternationalWorkshop on Signal Processing Advances in Wireless Communi-cations (SPAWC rsquo11) pp 426ndash430 IEEE San Francisco CalifUSA June 2011

[9] A Schad andM Pesavento ldquoMax-min fair transmit beamform-ing for multi-group multicastingrdquo in Proceedings of the 16thInternational ITG Workshop on Smart Antennas (WSA rsquo12) pp115ndash118 March 2012


[10] W Yu and T Lan ldquoTransmitter optimization for the multi-antenna downlink with per-antenna power constraintsrdquo IEEETransactions on Signal Processing vol 55 no 6 pp 2646ndash26602007

[11] G Dartmann X GongW Afzal andG Ascheid ldquoOn the dual-ity of the max-min beamforming problem with per-antennaand per-antenna-array power constraintsrdquo IEEE Transactionson Vehicular Technology vol 62 no 2 pp 606ndash619 2013

[12] D Christopoulos S Chatzinotas and B Ottersten ldquoWeightedfair multicast multigroup beamforming under per-antennapower constraintsrdquo IEEE Transactions on Signal Processing vol62 no 19 pp 5132ndash5142 2014

[13] Y Ye Interior Point AlgorithmsTheory and Analysis JohnWileyamp Sons New York NY USA 1998

[14] Y Huang and D P Palomar ldquoRank-constrained separablesemidefinite programming with applications to optimal beam-formingrdquo IEEE Transactions on Signal Processing vol 58 no 2pp 664ndash678 2010

[15] Y Gao and M Schubert ldquoPower-allocation for multi-groupmulticasting with beamformingrdquo in Proceedings of the IEEEITGWorkshop on Smart Antennas (WSA rsquo06) March 2006

[16] M Grant and S Boyd CVX Matlab Software for DisciplinedConvex Programming Version 20 Beta 2013 httpcvxrcomcvx

Submit your manuscripts athttpwwwhindawicom

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Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Stochastic AnalysisInternational Journal of


Table 1 List of related works

Problem description Application scenarios Solving approach SourceQoS Broadcast SDR [3]MMF Broadcast Iterative algorithm [4]QoS Multigroup multicast SDR + Gaussian randomization [5]MMF Multigroup multicast SDR + Gaussian randomization + bisection [5]QoS Multigroup multicast Iterative algorithm [8]MMF Multigroup multicast Iterative algorithm [9]QoS Broadcast with PACs Iterative algorithm + duality [10]MMF Broadcast with PACs Iterative algorithm + duality [11]MMF Multigroup multicast with PACs SDR + Gaussian randomization + bisection [12]

programming (SOCP) problem in each iteration And in[9] an iterative algorithm with low complexity and superiorperformance was further investigated to cope with the multi-group MMF problem Recently contrary to the total trans-mission power constraints transmit beamforming under per-antenna power constraints (PACs) was introduced in [10ndash12] As an extending work of [5] the weighted multigroupmulticast MMF problem with PACs was investigated in [12]For the sake of clear expression related works are listed inTable 1

As aforementioned the design of transmit beamformerhas been primarily studied over multiple-antenna multicas-ting and frequency-flat fading channels and to the authorsrsquobest knowledge few works have been dedicated to the case offrequency-selective fading multicasting scenario Motivatedby the potential advantages of multiple-antenna transmis-sion over the frequency-selective fading channels we areconcerned about the QoS and MMF problems in this paperfor multigroup multicasting The main contributions of thispaper are listed in the following

(1) For the QoS problem an approximation algorithmis firstly derived for broadband systems based onthe idea of narrowband multigroup multicasting in[5] And two reduced-complexity algorithms are alsoproposed from the frequency-domain and the time-domain perspectives separately Some parameters incorrelation with the QoS problem are analyzed andcomputational consumption is comparatively com-puted to showmore insights into the tradeoff betweenperformance and complexity

(2) For the MMF problem the corresponding approxi-mation algorithm and its low-complexity modifica-tions are also proposed in a similar way as that ofQoS problem Furthermore the relationship betweenthe QoS problem and the MMF problem is discussedcarefully followed by a complexity analysis

(3) Simulation experiments demonstrate the effective-ness of the frequency-domain and the time-domainalgorithms for bothQoS problem andMMF problemThe main analysis results that the controlled parame-ters in the proposed algorithms could be used tomakea tradeoff between complexity and performance areverified through the numerical examples

The remainder of this paper is structured as follows InSection 2 the system model and the QoS problem are intro-duced briefly The approximation solution to the QoS prob-lem is derived in Section 3 And Section 4 presents two beam-forming algorithms Section 5 formulates the MMF problemand solves it based on the proposed algorithms Computa-tional complexity analysis of proposed algorithms is givenin Section 6 In Section 7 the performance of the proposedalgorithms is evaluated and discussed Finally the conclusionis summarized in Section 8

Notations In the remainder of this paper boldface upper-case letters and math calligraphy uppercase letters denotematrices and boldface lowercase letters denote vectors (sdot)daggerE(sdot) and tr(sdot) are the conjugate transpose the expectationand the trace operator respectively C indicates the set ofcomplex numbers while otimes is Kronecker product lfloorsdotrfloor definesthe floor function A12 is matrix square root function of Aand 120575(sdot) is the impulse function A sim CN(0 I) means that Ais circularly symmetric complex Gaussian process with zeromean and unit variance matrix

2 System Model and Problem Statement

21 System Model Consider a multigroup multicast systemwith one transmitter (base station) and 119870 receivers (users)Assume the transmitter has 119872119905 antenna elements and eachreceiver is equipped with one antenna The users are splitinto 1 le 119871 le 119870 groups 1198921 1198922 119892119871 each containing119870119892119897

user indices We assume that each user listens to a singlemulticast group that is 119892119897 cap 1198921198971015840 = 0 where 119897 = 1198971015840 andcup119897 119892119897 = 1 119870 A frequency-selective fading channel with119881 effective paths is supposed between each transmit antennaand receive antenna and full CSI is available a priori at thetransmit side throughout this paper

With an 119874-tap FIR beamforming filter the transmittedsignal can be written as follows in space-time domain

s119899 =

119871minus1sum119897=0

119874minus1sum119900=0

w119897119900119909119897119899minus119900 (1)

where w119897119900 isin C119872119905times1 and 119909119897119900 denote the beamforming vectorand the discrete information sequence in association with the






119852n

g1

g2

gL



119910119897119898119899 =

119881minus1sumV=0

H119897119898Vs119899minusV + 119906119897119898119899

=

119881minus1sumV=0

H119897119898V

119874minus1sum119900=0


+

119881minus1sumV=0

H119897119898Vsum

1198971015840 =119897

119874minus1sum119900=0

w1198971015840 1199001199091198971015840 119899minusVminus119900⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟


+ 119906119897119898119899

(2)



119910119897119898119899 = H119897119898W119897 (119911)X119897119899 (119911)

+H119897119898 (119911) sum

1198971015840 =119897

W1198971015840 (119911)X1198971015840 119899 (119911) + 119906119897119898119899(3)



minus119900 andX119897119899(119911) = sum



119875119905 =

119871minus1sum119897=0

int1

0

1003817100381710038171003817W119897 (119891)100381710038171003817100381722 119889119891 (4)


SINR119897119898

=int10 H119897119898 (119891)W119897 (119891)Wdagger

119897(119891)Hdagger

119897119898(119891) 119889119891

sum1198971015840 =119897

int10 H119897119898 (119891)W1198971015840 (119891)W1015840dagger

119897(119891)Hdagger

119897119898(119891) 119889119891 + 1205902

119897119898

(5)

where 1205902119897119898

= E119906119897119898119899119906dagger

119897119898119899 H119897119898(119891) = H119897119898(119911)|119911=1198901198952120587119891 and

W119897(119891) = W119897(119911)|119911=1198901198952120587119891


P1 minW119897(119891)

119875119905


(6)


P2 minW119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

1003817100381710038171003817W119897119894

100381710038171003817100381722

st SINR1015840

119897119898 ge 120574119897119898 forall119898 forall119897

(7)

with

SINR1015840

119897119898

=(1119868) sum

119868minus1119894=0 H119897119898119894W119897119894W

dagger

119897119894Hdagger

119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 H119897119898119894W1198971015840 119894W

dagger

1198971015840 119894Hdagger

119897119898119894+ 1205902

119897119898

(8)







Let Q119897119894 = W119897119894Wdagger


problemP2

P3 minQ119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894)

st SINR119903

119897119898ge 120574119897119898

Q119897119894 ⪰ 0


(9)

where

SINR119903

119897119898=

(1119868) sum119868minus1119894=0 H119897119898119894Q119897119894H

dagger

119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 H119897119898119894Q1198971015840 119894H

dagger

119897119898119894+ 1205902

119897119898

(10)



P4 minQ119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894)

st SINR119903

119897119898ge 120574119897119898


(11)




119871minus1sum119897=0

119868minus1sum119894=0

rank2 (Q119897119894) le 119870(12)


D1 max120579119897119898

119871minus1sum119897=0

sum119898isin119892119897

1205791198971198981205741198971198981205902119897119898

st Y119897119894 ⪰ 0


(13)

where Y119897119894 = I + sum1198971015840 =119897

sum119898isin11989211989710158401205791198971015840 1198981205741198971015840119898H

dagger

1198971015840 119898119894H1198971015840 119898119894 minus

sum119898isin119892119897

120579119897119898Hdagger



Y119897119894Q119897119894 = (I+ sum

1198971015840 =119897

sum119898isin1198921198971015840

1205791198971015840 1198981205741198971015840 119898Hdagger

1198971015840 119898119894H1198971015840119898119894

minus sum119898isin119892119897

120579119897119898Hdagger

119897119898119894H119897119898119894)Q119897119894 = 0

(14)


I+sum1198971015840 =119897

sum119898isin11989211989710158401205791198971015840 1198981205741198971015840 119898H

dagger



= rank( sum119898isin119892119897

120579119897119898Hdagger

119897119898119894H119897119898119894Q119897119894)

le sum119898isin119892119897

rank (120579119897119898Hdagger

119897119898119894H119897119898119894Q119897119894)

le 119870119892119897

(15)


119871minus1119897=0 sum

119868minus1119894=0 rank

2(Q119897119894) le 119870



U119897119894Λ119897119894Udagger




119902

119897119894=

U119897119894Λ12119897119894

V119902





119902

119897119894W

119902dagger

119897119894 = Q119897119894 and rank(W

119902

119897119894W

119902dagger

119897119894) = 1 they may still


M1 min119901119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897119894120573119897119894

st(1119868) sum

119868minus1119894=0 119901119897119894120572119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 1199011198971015840 1198941205721198971015840 119898119894 + 1205902

119897119898

ge 120574119897119898


(16)


119897119894

120573119897119894 = W119902

11989711989422 120572119897119898119894 = H119897119898119894W

119902

119897119894W

119902dagger

119897119894Hdagger

119897119898119894 and 1205721198971015840 119898119894 =

H119897119898119894W119902

1198971015840 119894W

119902dagger

1198971015840 119894Hdagger




12119897119894


119897119894


119897119894= U119897119894Λ

12119897119894V119902

119897119894


119897119894


radic119901119897119894W119902

119897119894sum

119871minus1119897=0 sum

119868minus1119894=0119901119897119894120573119897119894

(13) end(14)Output W119897119894



119902












119868119865minus1119894=0


P5 minQ119897119894119868119865minus1119894=0

119865

119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)

st SINR1199031119897119898

ge 120574119897119898

Q119897119894 ⪰ 0


(17)

where

SINR1199031119897119898

=(1119868) sum


dagger

119897119898119894

(1119868) sum1198971015840 =119897


dagger

119897119898119894+ 1205902

119897119898

(18)


P6 minQ119897119894119868119865minus1119894=0

119865

119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)

st SINR1199031119897119898

ge 120574119897119898


(19)


M2 min119901119897119894119868119865minus1119894=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897lfloor119894119865rfloor+1120573119897119894

st(1119868) sum


(1119868) sum1198971015840 =119897


119897119898

ge 120574119897119898


(20)


119902

119897119894 and 120573119897119894 = W

119902


120572119897119898119894 = H119897119898119894W119902



119897119898119894 and 1205721198971015840 119898119894 =

H119897119898119894W119902








12119897119894


119897119894


119897119894= U119897119894Λ

12119897119894V119902

119897119894


119897119894

(11) end(12) choose

W119897119894 = argminradic119901119897119894W

119902

119897119894sum

119871minus1119897=0 sum


(13) end(14)Output W119897119894







1198971198941198681198652minus1119894=0 where

Qexp119897119894

(1198941015840+1)11986511198652minus1


exp119897119894

1198681198652minus1119894=0 is a






1198971198941198681198652minus1119894=0


is Qopt119897119894


Qopt119897119894


1198971198941198681198652minus1119894=0 thus



Q11989711989410158401198681198651minus11198941015840=0 1198752 = 1198751 holds



s119899 =

119871minus1sum119897=0

119862minus1sum119888=0

w119897119888119909119897119899minus119888 (21)

Define

W119897119894 =

119862minus1sum119888=0

w119897119888119890minus2120587119894119888119868

(22)


W119897119894 =

[[[[[[[[

[

1

119890minus1198952120587119894119873

119890minus1198952120587119894(119862minus1)119873

]]]]]]]]

]

119879

otimes I119872119905

⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟K119894

sdot

[[[[[[[

[

w1198970

w1198971

w119897119862minus1

]]]]]]]

]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

W119897

(23)


1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894) =1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (W119897119894W1015840

119897119894)

=1119868

119871minus1sum119897=0

119868minus1sum119894=0


119897Kdagger

119894)

(24)



P7 minQ119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)

st SINR1199032119897119898

ge 120574119897119898

Q119897 ⪰ 0


(25)

with

SINR1199032119897119898

=(1119868) sum


119894Hdagger

119897119898119894)

(1119868) sum1198971015840 =119897


119894Hdagger

119897119898119894) + 1205902

119897119898

(26)


P8 minQ119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)

st SINR1199032119897119898

ge 120574119897119898

Q119897 ⪰ 0 forall119898 forall119897

(27)



12119897


dagger

119897


119902

119897= U119897Λ

12119897V119902

119897


119897


radic119901119897W119902

119897sum

119871minus1119897=0 sum

119868minus1119894=0119901119897120573119897119894

(13) end(14)Output W119897



M3 min119901119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897120573119897119894

st(1119868) sum

119868minus1119894=0 119901119897120572119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 11990111989710158401205721198971015840 119898119894 + 1205902

119897119898

ge 120574119897119898

119901119897 ge 0 forall119898 forall119897

(28)


119897

and 120573119897119894 = tr(K119894Q119902

119897Kdagger


H119897119898119894K119894Q119902

119897Kdagger

119894Hdagger

119897119898119894 and 1205721198971015840 119898119894 = H119897119898119894K119894Q

119902

1198971015840Kdagger

119894Hdagger

119897119898119894









119871minus1119897=0 where

Qexp119897

119871minus1

119897=0

= [

[

Qopt119897

119871minus1

119897=001198621119872119905times(1198622minus1198621)119872119905


]

]

(29)






1198681198651minus11198941015840=0 and Qexp

1198971198941198681198652minus1119894=0 by Qopt

119897119871minus1119897=0 and Qexp

119897119871minus1119897=0 and




Q1 maxQ119897119894119905isinR

119905

stSINR119903

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0


(30)


Q2 maxQ119897119894119905isinR

119905

stSINR119903

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


(31)





M4 max119901119897119894119905isinR

119905

st(1119868) sum

119868minus1119894=0 119901119897119894120572119897119898119894

(120574119897119898119868) sum1198971015840 =119897

sum119868minus1119894=0 1199011198971015840 1198941205721198971015840 119898119894 + 1205741198971198981205902

119897119898

ge 119905

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897119894120573119897119894 = 119875


(32)




119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(33)



119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(34)



119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

rank (Q119897) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)



119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)


(35)
















QoS or MMFproblem




in Section 41 or

Section 42





Yes

No

Output




119875 = P3 (Q1 (120574 119875) 120574) (36)

119905 = Q1 (120574P3 (119905120574)) (37)























is invalid





O(119871311986831198726119905+1198701198711198681198722




351198683511987265119905

+ 119870119871151198681511987225119905

+ 1198761198713511986835 +



119905log(1120598))



O(11987105

11986805



minus3+ 119870119871119868119865

minus1)


351198683511987265119905

119865minus35 + 119870119871151198681511987225119905

119865minus15 + 1198761198713511986835119865minus35 +




0511986205







+

119870119871151198622511987225119905



351198683511987265119905

119865minus35 +

119861119878119870119871151198681511987225119905

119865minus15 + 1198761198611198711198713511986835119865minus35 +

1198761198611198711198701198711511986815119865minus15)log(1120598)) O((119861119878119871351198683511987265

119905119865minus35 +

11986111987811987011987115

11986815

11987225119905

119865minus15

+ 11987611986111987111987135

11986835

119865minus35

+

1198761198611198711198701198711511986815119865minus15)log(1120598)) and O((119861119878119871351198626511987265

119905+

119861119878119870119871151198622511987225119905








Tota

l arit

hmet

ic o

pera

tions

for Q

OS

prob

lem

s


108

109

1010

1011

1012

1013

1014

1015

1016












6 8 10 12 14 16 18 20SINR (dB)

0

01

02

03

04

05

06

07

08

09

1

()


L = 2 Mt = 6 K = 12 F = 8

L = 3 Mt = 6 K = 12 F = 8

L = 3 Mt = 4 K = 12 F = 8

L = 2 Mt = 6 K = 8 C = 4

L = 2 Mt = 6 K = 12 C = 4

L = 3 Mt = 6 K = 12 C = 4

L = 3 Mt = 4 K = 12 C = 4

L = 2 Mt = 6 K = 8 C = 5

L = 2 Mt = 6 K = 8 F = 4







6 8 10 12 14 160

5

10

15

20

25

30

35

40

Tran

smiss

ion

pow

er

SINR (dB)








Transmission power







05 1 15 2 25 3 350

01

02

03

04

05

06

07

08

09

1

Cum

ulat

ive d

istrib

utio

n fu

nctio

n




2 4 6 8 10 12 14 16 18 20 22 246

8

10

12

14

16

18

Transmission power

SIN

R (d

B)












8 Conclusion





Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra














119852n

g1

g2

gL



119910119897119898119899 =

119881minus1sumV=0

H119897119898Vs119899minusV + 119906119897119898119899

=

119881minus1sumV=0

H119897119898V

119874minus1sum119900=0


+

119881minus1sumV=0

H119897119898Vsum

1198971015840 =119897

119874minus1sum119900=0

w1198971015840 1199001199091198971015840 119899minusVminus119900⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟


+ 119906119897119898119899

(2)



119910119897119898119899 = H119897119898W119897 (119911)X119897119899 (119911)

+H119897119898 (119911) sum

1198971015840 =119897

W1198971015840 (119911)X1198971015840 119899 (119911) + 119906119897119898119899(3)



minus119900 andX119897119899(119911) = sum



119875119905 =

119871minus1sum119897=0

int1

0

1003817100381710038171003817W119897 (119891)100381710038171003817100381722 119889119891 (4)


SINR119897119898

=int10 H119897119898 (119891)W119897 (119891)Wdagger

119897(119891)Hdagger

119897119898(119891) 119889119891

sum1198971015840 =119897

int10 H119897119898 (119891)W1198971015840 (119891)W1015840dagger

119897(119891)Hdagger

119897119898(119891) 119889119891 + 1205902

119897119898

(5)

where 1205902119897119898

= E119906119897119898119899119906dagger

119897119898119899 H119897119898(119891) = H119897119898(119911)|119911=1198901198952120587119891 and

W119897(119891) = W119897(119911)|119911=1198901198952120587119891


P1 minW119897(119891)

119875119905


(6)


P2 minW119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

1003817100381710038171003817W119897119894

100381710038171003817100381722

st SINR1015840

119897119898 ge 120574119897119898 forall119898 forall119897

(7)

with

SINR1015840

119897119898

=(1119868) sum

119868minus1119894=0 H119897119898119894W119897119894W

dagger

119897119894Hdagger

119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 H119897119898119894W1198971015840 119894W

dagger

1198971015840 119894Hdagger

119897119898119894+ 1205902

119897119898

(8)







Let Q119897119894 = W119897119894Wdagger


problemP2

P3 minQ119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894)

st SINR119903

119897119898ge 120574119897119898

Q119897119894 ⪰ 0


(9)

where

SINR119903

119897119898=

(1119868) sum119868minus1119894=0 H119897119898119894Q119897119894H

dagger

119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 H119897119898119894Q1198971015840 119894H

dagger

119897119898119894+ 1205902

119897119898

(10)



P4 minQ119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894)

st SINR119903

119897119898ge 120574119897119898


(11)




119871minus1sum119897=0

119868minus1sum119894=0

rank2 (Q119897119894) le 119870(12)


D1 max120579119897119898

119871minus1sum119897=0

sum119898isin119892119897

1205791198971198981205741198971198981205902119897119898

st Y119897119894 ⪰ 0


(13)

where Y119897119894 = I + sum1198971015840 =119897

sum119898isin11989211989710158401205791198971015840 1198981205741198971015840119898H

dagger

1198971015840 119898119894H1198971015840 119898119894 minus

sum119898isin119892119897

120579119897119898Hdagger



Y119897119894Q119897119894 = (I+ sum

1198971015840 =119897

sum119898isin1198921198971015840

1205791198971015840 1198981205741198971015840 119898Hdagger

1198971015840 119898119894H1198971015840119898119894

minus sum119898isin119892119897

120579119897119898Hdagger

119897119898119894H119897119898119894)Q119897119894 = 0

(14)


I+sum1198971015840 =119897

sum119898isin11989211989710158401205791198971015840 1198981205741198971015840 119898H

dagger



= rank( sum119898isin119892119897

120579119897119898Hdagger

119897119898119894H119897119898119894Q119897119894)

le sum119898isin119892119897

rank (120579119897119898Hdagger

119897119898119894H119897119898119894Q119897119894)

le 119870119892119897

(15)


119871minus1119897=0 sum

119868minus1119894=0 rank

2(Q119897119894) le 119870



U119897119894Λ119897119894Udagger




119902

119897119894=

U119897119894Λ12119897119894

V119902





119902

119897119894W

119902dagger

119897119894 = Q119897119894 and rank(W

119902

119897119894W

119902dagger

119897119894) = 1 they may still


M1 min119901119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897119894120573119897119894

st(1119868) sum

119868minus1119894=0 119901119897119894120572119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 1199011198971015840 1198941205721198971015840 119898119894 + 1205902

119897119898

ge 120574119897119898


(16)


119897119894

120573119897119894 = W119902

11989711989422 120572119897119898119894 = H119897119898119894W

119902

119897119894W

119902dagger

119897119894Hdagger

119897119898119894 and 1205721198971015840 119898119894 =

H119897119898119894W119902

1198971015840 119894W

119902dagger

1198971015840 119894Hdagger




12119897119894


119897119894


119897119894= U119897119894Λ

12119897119894V119902

119897119894


119897119894


radic119901119897119894W119902

119897119894sum

119871minus1119897=0 sum

119868minus1119894=0119901119897119894120573119897119894

(13) end(14)Output W119897119894



119902












119868119865minus1119894=0


P5 minQ119897119894119868119865minus1119894=0

119865

119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)

st SINR1199031119897119898

ge 120574119897119898

Q119897119894 ⪰ 0


(17)

where

SINR1199031119897119898

=(1119868) sum


dagger

119897119898119894

(1119868) sum1198971015840 =119897


dagger

119897119898119894+ 1205902

119897119898

(18)


P6 minQ119897119894119868119865minus1119894=0

119865

119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)

st SINR1199031119897119898

ge 120574119897119898


(19)


M2 min119901119897119894119868119865minus1119894=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897lfloor119894119865rfloor+1120573119897119894

st(1119868) sum


(1119868) sum1198971015840 =119897


119897119898

ge 120574119897119898


(20)


119902

119897119894 and 120573119897119894 = W

119902


120572119897119898119894 = H119897119898119894W119902



119897119898119894 and 1205721198971015840 119898119894 =

H119897119898119894W119902








12119897119894


119897119894


119897119894= U119897119894Λ

12119897119894V119902

119897119894


119897119894

(11) end(12) choose

W119897119894 = argminradic119901119897119894W

119902

119897119894sum

119871minus1119897=0 sum


(13) end(14)Output W119897119894







1198971198941198681198652minus1119894=0 where

Qexp119897119894

(1198941015840+1)11986511198652minus1


exp119897119894

1198681198652minus1119894=0 is a






1198971198941198681198652minus1119894=0


is Qopt119897119894


Qopt119897119894


1198971198941198681198652minus1119894=0 thus



Q11989711989410158401198681198651minus11198941015840=0 1198752 = 1198751 holds



s119899 =

119871minus1sum119897=0

119862minus1sum119888=0

w119897119888119909119897119899minus119888 (21)

Define

W119897119894 =

119862minus1sum119888=0

w119897119888119890minus2120587119894119888119868

(22)


W119897119894 =

[[[[[[[[

[

1

119890minus1198952120587119894119873

119890minus1198952120587119894(119862minus1)119873

]]]]]]]]

]

119879

otimes I119872119905

⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟K119894

sdot

[[[[[[[

[

w1198970

w1198971

w119897119862minus1

]]]]]]]

]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

W119897

(23)


1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894) =1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (W119897119894W1015840

119897119894)

=1119868

119871minus1sum119897=0

119868minus1sum119894=0


119897Kdagger

119894)

(24)



P7 minQ119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)

st SINR1199032119897119898

ge 120574119897119898

Q119897 ⪰ 0


(25)

with

SINR1199032119897119898

=(1119868) sum


119894Hdagger

119897119898119894)

(1119868) sum1198971015840 =119897


119894Hdagger

119897119898119894) + 1205902

119897119898

(26)


P8 minQ119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)

st SINR1199032119897119898

ge 120574119897119898

Q119897 ⪰ 0 forall119898 forall119897

(27)



12119897


dagger

119897


119902

119897= U119897Λ

12119897V119902

119897


119897


radic119901119897W119902

119897sum

119871minus1119897=0 sum

119868minus1119894=0119901119897120573119897119894

(13) end(14)Output W119897



M3 min119901119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897120573119897119894

st(1119868) sum

119868minus1119894=0 119901119897120572119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 11990111989710158401205721198971015840 119898119894 + 1205902

119897119898

ge 120574119897119898

119901119897 ge 0 forall119898 forall119897

(28)


119897

and 120573119897119894 = tr(K119894Q119902

119897Kdagger


H119897119898119894K119894Q119902

119897Kdagger

119894Hdagger

119897119898119894 and 1205721198971015840 119898119894 = H119897119898119894K119894Q

119902

1198971015840Kdagger

119894Hdagger

119897119898119894









119871minus1119897=0 where

Qexp119897

119871minus1

119897=0

= [

[

Qopt119897

119871minus1

119897=001198621119872119905times(1198622minus1198621)119872119905


]

]

(29)






1198681198651minus11198941015840=0 and Qexp

1198971198941198681198652minus1119894=0 by Qopt

119897119871minus1119897=0 and Qexp

119897119871minus1119897=0 and




Q1 maxQ119897119894119905isinR

119905

stSINR119903

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0


(30)


Q2 maxQ119897119894119905isinR

119905

stSINR119903

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


(31)





M4 max119901119897119894119905isinR

119905

st(1119868) sum

119868minus1119894=0 119901119897119894120572119897119898119894

(120574119897119898119868) sum1198971015840 =119897

sum119868minus1119894=0 1199011198971015840 1198941205721198971015840 119898119894 + 1205741198971198981205902

119897119898

ge 119905

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897119894120573119897119894 = 119875


(32)




119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(33)



119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(34)



119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

rank (Q119897) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)



119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)


(35)
















QoS or MMFproblem




in Section 41 or

Section 42





Yes

No

Output




119875 = P3 (Q1 (120574 119875) 120574) (36)

119905 = Q1 (120574P3 (119905120574)) (37)























is invalid





O(119871311986831198726119905+1198701198711198681198722




351198683511987265119905

+ 119870119871151198681511987225119905

+ 1198761198713511986835 +



119905log(1120598))



O(11987105

11986805



minus3+ 119870119871119868119865

minus1)


351198683511987265119905

119865minus35 + 119870119871151198681511987225119905

119865minus15 + 1198761198713511986835119865minus35 +




0511986205







+

119870119871151198622511987225119905



351198683511987265119905

119865minus35 +

119861119878119870119871151198681511987225119905

119865minus15 + 1198761198611198711198713511986835119865minus35 +

1198761198611198711198701198711511986815119865minus15)log(1120598)) O((119861119878119871351198683511987265

119905119865minus35 +

11986111987811987011987115

11986815

11987225119905

119865minus15

+ 11987611986111987111987135

11986835

119865minus35

+

1198761198611198711198701198711511986815119865minus15)log(1120598)) and O((119861119878119871351198626511987265

119905+

119861119878119870119871151198622511987225119905








Tota

l arit

hmet

ic o

pera

tions

for Q

OS

prob

lem

s


108

109

1010

1011

1012

1013

1014

1015

1016












6 8 10 12 14 16 18 20SINR (dB)

0

01

02

03

04

05

06

07

08

09

1

()


L = 2 Mt = 6 K = 12 F = 8

L = 3 Mt = 6 K = 12 F = 8

L = 3 Mt = 4 K = 12 F = 8

L = 2 Mt = 6 K = 8 C = 4

L = 2 Mt = 6 K = 12 C = 4

L = 3 Mt = 6 K = 12 C = 4

L = 3 Mt = 4 K = 12 C = 4

L = 2 Mt = 6 K = 8 C = 5

L = 2 Mt = 6 K = 8 F = 4







6 8 10 12 14 160

5

10

15

20

25

30

35

40

Tran

smiss

ion

pow

er

SINR (dB)








Transmission power







05 1 15 2 25 3 350

01

02

03

04

05

06

07

08

09

1

Cum

ulat

ive d

istrib

utio

n fu

nctio

n




2 4 6 8 10 12 14 16 18 20 22 246

8

10

12

14

16

18

Transmission power

SIN

R (d

B)












8 Conclusion





Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Let Q119897119894 = W119897119894Wdagger


problemP2

P3 minQ119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894)

st SINR119903

119897119898ge 120574119897119898

Q119897119894 ⪰ 0


(9)

where

SINR119903

119897119898=

(1119868) sum119868minus1119894=0 H119897119898119894Q119897119894H

dagger

119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 H119897119898119894Q1198971015840 119894H

dagger

119897119898119894+ 1205902

119897119898

(10)



P4 minQ119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894)

st SINR119903

119897119898ge 120574119897119898


(11)




119871minus1sum119897=0

119868minus1sum119894=0

rank2 (Q119897119894) le 119870(12)


D1 max120579119897119898

119871minus1sum119897=0

sum119898isin119892119897

1205791198971198981205741198971198981205902119897119898

st Y119897119894 ⪰ 0


(13)

where Y119897119894 = I + sum1198971015840 =119897

sum119898isin11989211989710158401205791198971015840 1198981205741198971015840119898H

dagger

1198971015840 119898119894H1198971015840 119898119894 minus

sum119898isin119892119897

120579119897119898Hdagger



Y119897119894Q119897119894 = (I+ sum

1198971015840 =119897

sum119898isin1198921198971015840

1205791198971015840 1198981205741198971015840 119898Hdagger

1198971015840 119898119894H1198971015840119898119894

minus sum119898isin119892119897

120579119897119898Hdagger

119897119898119894H119897119898119894)Q119897119894 = 0

(14)


I+sum1198971015840 =119897

sum119898isin11989211989710158401205791198971015840 1198981205741198971015840 119898H

dagger



= rank( sum119898isin119892119897

120579119897119898Hdagger

119897119898119894H119897119898119894Q119897119894)

le sum119898isin119892119897

rank (120579119897119898Hdagger

119897119898119894H119897119898119894Q119897119894)

le 119870119892119897

(15)


119871minus1119897=0 sum

119868minus1119894=0 rank

2(Q119897119894) le 119870



U119897119894Λ119897119894Udagger




119902

119897119894=

U119897119894Λ12119897119894

V119902





119902

119897119894W

119902dagger

119897119894 = Q119897119894 and rank(W

119902

119897119894W

119902dagger

119897119894) = 1 they may still


M1 min119901119897119894

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897119894120573119897119894

st(1119868) sum

119868minus1119894=0 119901119897119894120572119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 1199011198971015840 1198941205721198971015840 119898119894 + 1205902

119897119898

ge 120574119897119898


(16)


119897119894

120573119897119894 = W119902

11989711989422 120572119897119898119894 = H119897119898119894W

119902

119897119894W

119902dagger

119897119894Hdagger

119897119898119894 and 1205721198971015840 119898119894 =

H119897119898119894W119902

1198971015840 119894W

119902dagger

1198971015840 119894Hdagger




12119897119894


119897119894


119897119894= U119897119894Λ

12119897119894V119902

119897119894


119897119894


radic119901119897119894W119902

119897119894sum

119871minus1119897=0 sum

119868minus1119894=0119901119897119894120573119897119894

(13) end(14)Output W119897119894



119902












119868119865minus1119894=0


P5 minQ119897119894119868119865minus1119894=0

119865

119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)

st SINR1199031119897119898

ge 120574119897119898

Q119897119894 ⪰ 0


(17)

where

SINR1199031119897119898

=(1119868) sum


dagger

119897119898119894

(1119868) sum1198971015840 =119897


dagger

119897119898119894+ 1205902

119897119898

(18)


P6 minQ119897119894119868119865minus1119894=0

119865

119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)

st SINR1199031119897119898

ge 120574119897119898


(19)


M2 min119901119897119894119868119865minus1119894=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897lfloor119894119865rfloor+1120573119897119894

st(1119868) sum


(1119868) sum1198971015840 =119897


119897119898

ge 120574119897119898


(20)


119902

119897119894 and 120573119897119894 = W

119902


120572119897119898119894 = H119897119898119894W119902



119897119898119894 and 1205721198971015840 119898119894 =

H119897119898119894W119902








12119897119894


119897119894


119897119894= U119897119894Λ

12119897119894V119902

119897119894


119897119894

(11) end(12) choose

W119897119894 = argminradic119901119897119894W

119902

119897119894sum

119871minus1119897=0 sum


(13) end(14)Output W119897119894







1198971198941198681198652minus1119894=0 where

Qexp119897119894

(1198941015840+1)11986511198652minus1


exp119897119894

1198681198652minus1119894=0 is a






1198971198941198681198652minus1119894=0


is Qopt119897119894


Qopt119897119894


1198971198941198681198652minus1119894=0 thus



Q11989711989410158401198681198651minus11198941015840=0 1198752 = 1198751 holds



s119899 =

119871minus1sum119897=0

119862minus1sum119888=0

w119897119888119909119897119899minus119888 (21)

Define

W119897119894 =

119862minus1sum119888=0

w119897119888119890minus2120587119894119888119868

(22)


W119897119894 =

[[[[[[[[

[

1

119890minus1198952120587119894119873

119890minus1198952120587119894(119862minus1)119873

]]]]]]]]

]

119879

otimes I119872119905

⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟K119894

sdot

[[[[[[[

[

w1198970

w1198971

w119897119862minus1

]]]]]]]

]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

W119897

(23)


1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894) =1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (W119897119894W1015840

119897119894)

=1119868

119871minus1sum119897=0

119868minus1sum119894=0


119897Kdagger

119894)

(24)



P7 minQ119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)

st SINR1199032119897119898

ge 120574119897119898

Q119897 ⪰ 0


(25)

with

SINR1199032119897119898

=(1119868) sum


119894Hdagger

119897119898119894)

(1119868) sum1198971015840 =119897


119894Hdagger

119897119898119894) + 1205902

119897119898

(26)


P8 minQ119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)

st SINR1199032119897119898

ge 120574119897119898

Q119897 ⪰ 0 forall119898 forall119897

(27)



12119897


dagger

119897


119902

119897= U119897Λ

12119897V119902

119897


119897


radic119901119897W119902

119897sum

119871minus1119897=0 sum

119868minus1119894=0119901119897120573119897119894

(13) end(14)Output W119897



M3 min119901119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897120573119897119894

st(1119868) sum

119868minus1119894=0 119901119897120572119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 11990111989710158401205721198971015840 119898119894 + 1205902

119897119898

ge 120574119897119898

119901119897 ge 0 forall119898 forall119897

(28)


119897

and 120573119897119894 = tr(K119894Q119902

119897Kdagger


H119897119898119894K119894Q119902

119897Kdagger

119894Hdagger

119897119898119894 and 1205721198971015840 119898119894 = H119897119898119894K119894Q

119902

1198971015840Kdagger

119894Hdagger

119897119898119894









119871minus1119897=0 where

Qexp119897

119871minus1

119897=0

= [

[

Qopt119897

119871minus1

119897=001198621119872119905times(1198622minus1198621)119872119905


]

]

(29)






1198681198651minus11198941015840=0 and Qexp

1198971198941198681198652minus1119894=0 by Qopt

119897119871minus1119897=0 and Qexp

119897119871minus1119897=0 and




Q1 maxQ119897119894119905isinR

119905

stSINR119903

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0


(30)


Q2 maxQ119897119894119905isinR

119905

stSINR119903

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


(31)





M4 max119901119897119894119905isinR

119905

st(1119868) sum

119868minus1119894=0 119901119897119894120572119897119898119894

(120574119897119898119868) sum1198971015840 =119897

sum119868minus1119894=0 1199011198971015840 1198941205721198971015840 119898119894 + 1205741198971198981205902

119897119898

ge 119905

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897119894120573119897119894 = 119875


(32)




119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(33)



119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(34)



119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

rank (Q119897) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)



119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)


(35)
















QoS or MMFproblem




in Section 41 or

Section 42





Yes

No

Output




119875 = P3 (Q1 (120574 119875) 120574) (36)

119905 = Q1 (120574P3 (119905120574)) (37)























is invalid





O(119871311986831198726119905+1198701198711198681198722




351198683511987265119905

+ 119870119871151198681511987225119905

+ 1198761198713511986835 +



119905log(1120598))



O(11987105

11986805



minus3+ 119870119871119868119865

minus1)


351198683511987265119905

119865minus35 + 119870119871151198681511987225119905

119865minus15 + 1198761198713511986835119865minus35 +




0511986205







+

119870119871151198622511987225119905



351198683511987265119905

119865minus35 +

119861119878119870119871151198681511987225119905

119865minus15 + 1198761198611198711198713511986835119865minus35 +

1198761198611198711198701198711511986815119865minus15)log(1120598)) O((119861119878119871351198683511987265

119905119865minus35 +

11986111987811987011987115

11986815

11987225119905

119865minus15

+ 11987611986111987111987135

11986835

119865minus35

+

1198761198611198711198701198711511986815119865minus15)log(1120598)) and O((119861119878119871351198626511987265

119905+

119861119878119870119871151198622511987225119905








Tota

l arit

hmet

ic o

pera

tions

for Q

OS

prob

lem

s


108

109

1010

1011

1012

1013

1014

1015

1016












6 8 10 12 14 16 18 20SINR (dB)

0

01

02

03

04

05

06

07

08

09

1

()


L = 2 Mt = 6 K = 12 F = 8

L = 3 Mt = 6 K = 12 F = 8

L = 3 Mt = 4 K = 12 F = 8

L = 2 Mt = 6 K = 8 C = 4

L = 2 Mt = 6 K = 12 C = 4

L = 3 Mt = 6 K = 12 C = 4

L = 3 Mt = 4 K = 12 C = 4

L = 2 Mt = 6 K = 8 C = 5

L = 2 Mt = 6 K = 8 F = 4







6 8 10 12 14 160

5

10

15

20

25

30

35

40

Tran

smiss

ion

pow

er

SINR (dB)








Transmission power







05 1 15 2 25 3 350

01

02

03

04

05

06

07

08

09

1

Cum

ulat

ive d

istrib

utio

n fu

nctio

n




2 4 6 8 10 12 14 16 18 20 22 246

8

10

12

14

16

18

Transmission power

SIN

R (d

B)












8 Conclusion





Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











12119897119894


119897119894


119897119894= U119897119894Λ

12119897119894V119902

119897119894


119897119894


radic119901119897119894W119902

119897119894sum

119871minus1119897=0 sum

119868minus1119894=0119901119897119894120573119897119894

(13) end(14)Output W119897119894



119902












119868119865minus1119894=0


P5 minQ119897119894119868119865minus1119894=0

119865

119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)

st SINR1199031119897119898

ge 120574119897119898

Q119897119894 ⪰ 0


(17)

where

SINR1199031119897119898

=(1119868) sum


dagger

119897119898119894

(1119868) sum1198971015840 =119897


dagger

119897119898119894+ 1205902

119897119898

(18)


P6 minQ119897119894119868119865minus1119894=0

119865

119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)

st SINR1199031119897119898

ge 120574119897119898


(19)


M2 min119901119897119894119868119865minus1119894=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897lfloor119894119865rfloor+1120573119897119894

st(1119868) sum


(1119868) sum1198971015840 =119897


119897119898

ge 120574119897119898


(20)


119902

119897119894 and 120573119897119894 = W

119902


120572119897119898119894 = H119897119898119894W119902



119897119898119894 and 1205721198971015840 119898119894 =

H119897119898119894W119902








12119897119894


119897119894


119897119894= U119897119894Λ

12119897119894V119902

119897119894


119897119894

(11) end(12) choose

W119897119894 = argminradic119901119897119894W

119902

119897119894sum

119871minus1119897=0 sum


(13) end(14)Output W119897119894







1198971198941198681198652minus1119894=0 where

Qexp119897119894

(1198941015840+1)11986511198652minus1


exp119897119894

1198681198652minus1119894=0 is a






1198971198941198681198652minus1119894=0


is Qopt119897119894


Qopt119897119894


1198971198941198681198652minus1119894=0 thus



Q11989711989410158401198681198651minus11198941015840=0 1198752 = 1198751 holds



s119899 =

119871minus1sum119897=0

119862minus1sum119888=0

w119897119888119909119897119899minus119888 (21)

Define

W119897119894 =

119862minus1sum119888=0

w119897119888119890minus2120587119894119888119868

(22)


W119897119894 =

[[[[[[[[

[

1

119890minus1198952120587119894119873

119890minus1198952120587119894(119862minus1)119873

]]]]]]]]

]

119879

otimes I119872119905

⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟K119894

sdot

[[[[[[[

[

w1198970

w1198971

w119897119862minus1

]]]]]]]

]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

W119897

(23)


1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894) =1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (W119897119894W1015840

119897119894)

=1119868

119871minus1sum119897=0

119868minus1sum119894=0


119897Kdagger

119894)

(24)



P7 minQ119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)

st SINR1199032119897119898

ge 120574119897119898

Q119897 ⪰ 0


(25)

with

SINR1199032119897119898

=(1119868) sum


119894Hdagger

119897119898119894)

(1119868) sum1198971015840 =119897


119894Hdagger

119897119898119894) + 1205902

119897119898

(26)


P8 minQ119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)

st SINR1199032119897119898

ge 120574119897119898

Q119897 ⪰ 0 forall119898 forall119897

(27)



12119897


dagger

119897


119902

119897= U119897Λ

12119897V119902

119897


119897


radic119901119897W119902

119897sum

119871minus1119897=0 sum

119868minus1119894=0119901119897120573119897119894

(13) end(14)Output W119897



M3 min119901119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897120573119897119894

st(1119868) sum

119868minus1119894=0 119901119897120572119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 11990111989710158401205721198971015840 119898119894 + 1205902

119897119898

ge 120574119897119898

119901119897 ge 0 forall119898 forall119897

(28)


119897

and 120573119897119894 = tr(K119894Q119902

119897Kdagger


H119897119898119894K119894Q119902

119897Kdagger

119894Hdagger

119897119898119894 and 1205721198971015840 119898119894 = H119897119898119894K119894Q

119902

1198971015840Kdagger

119894Hdagger

119897119898119894









119871minus1119897=0 where

Qexp119897

119871minus1

119897=0

= [

[

Qopt119897

119871minus1

119897=001198621119872119905times(1198622minus1198621)119872119905


]

]

(29)






1198681198651minus11198941015840=0 and Qexp

1198971198941198681198652minus1119894=0 by Qopt

119897119871minus1119897=0 and Qexp

119897119871minus1119897=0 and




Q1 maxQ119897119894119905isinR

119905

stSINR119903

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0


(30)


Q2 maxQ119897119894119905isinR

119905

stSINR119903

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


(31)





M4 max119901119897119894119905isinR

119905

st(1119868) sum

119868minus1119894=0 119901119897119894120572119897119898119894

(120574119897119898119868) sum1198971015840 =119897

sum119868minus1119894=0 1199011198971015840 1198941205721198971015840 119898119894 + 1205741198971198981205902

119897119898

ge 119905

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897119894120573119897119894 = 119875


(32)




119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(33)



119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(34)



119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

rank (Q119897) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)



119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)


(35)
















QoS or MMFproblem




in Section 41 or

Section 42





Yes

No

Output




119875 = P3 (Q1 (120574 119875) 120574) (36)

119905 = Q1 (120574P3 (119905120574)) (37)























is invalid





O(119871311986831198726119905+1198701198711198681198722




351198683511987265119905

+ 119870119871151198681511987225119905

+ 1198761198713511986835 +



119905log(1120598))



O(11987105

11986805



minus3+ 119870119871119868119865

minus1)


351198683511987265119905

119865minus35 + 119870119871151198681511987225119905

119865minus15 + 1198761198713511986835119865minus35 +




0511986205







+

119870119871151198622511987225119905



351198683511987265119905

119865minus35 +

119861119878119870119871151198681511987225119905

119865minus15 + 1198761198611198711198713511986835119865minus35 +

1198761198611198711198701198711511986815119865minus15)log(1120598)) O((119861119878119871351198683511987265

119905119865minus35 +

11986111987811987011987115

11986815

11987225119905

119865minus15

+ 11987611986111987111987135

11986835

119865minus35

+

1198761198611198711198701198711511986815119865minus15)log(1120598)) and O((119861119878119871351198626511987265

119905+

119861119878119870119871151198622511987225119905








Tota

l arit

hmet

ic o

pera

tions

for Q

OS

prob

lem

s


108

109

1010

1011

1012

1013

1014

1015

1016












6 8 10 12 14 16 18 20SINR (dB)

0

01

02

03

04

05

06

07

08

09

1

()


L = 2 Mt = 6 K = 12 F = 8

L = 3 Mt = 6 K = 12 F = 8

L = 3 Mt = 4 K = 12 F = 8

L = 2 Mt = 6 K = 8 C = 4

L = 2 Mt = 6 K = 12 C = 4

L = 3 Mt = 6 K = 12 C = 4

L = 3 Mt = 4 K = 12 C = 4

L = 2 Mt = 6 K = 8 C = 5

L = 2 Mt = 6 K = 8 F = 4







6 8 10 12 14 160

5

10

15

20

25

30

35

40

Tran

smiss

ion

pow

er

SINR (dB)








Transmission power







05 1 15 2 25 3 350

01

02

03

04

05

06

07

08

09

1

Cum

ulat

ive d

istrib

utio

n fu

nctio

n




2 4 6 8 10 12 14 16 18 20 22 246

8

10

12

14

16

18

Transmission power

SIN

R (d

B)












8 Conclusion





Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











12119897119894


119897119894


119897119894= U119897119894Λ

12119897119894V119902

119897119894


119897119894

(11) end(12) choose

W119897119894 = argminradic119901119897119894W

119902

119897119894sum

119871minus1119897=0 sum


(13) end(14)Output W119897119894







1198971198941198681198652minus1119894=0 where

Qexp119897119894

(1198941015840+1)11986511198652minus1


exp119897119894

1198681198652minus1119894=0 is a






1198971198941198681198652minus1119894=0


is Qopt119897119894


Qopt119897119894


1198971198941198681198652minus1119894=0 thus



Q11989711989410158401198681198651minus11198941015840=0 1198752 = 1198751 holds



s119899 =

119871minus1sum119897=0

119862minus1sum119888=0

w119897119888119909119897119899minus119888 (21)

Define

W119897119894 =

119862minus1sum119888=0

w119897119888119890minus2120587119894119888119868

(22)


W119897119894 =

[[[[[[[[

[

1

119890minus1198952120587119894119873

119890minus1198952120587119894(119862minus1)119873

]]]]]]]]

]

119879

otimes I119872119905

⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟K119894

sdot

[[[[[[[

[

w1198970

w1198971

w119897119862minus1

]]]]]]]

]⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

W119897

(23)


1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (Q119897119894) =1119868

119871minus1sum119897=0

119868minus1sum119894=0

tr (W119897119894W1015840

119897119894)

=1119868

119871minus1sum119897=0

119868minus1sum119894=0


119897Kdagger

119894)

(24)



P7 minQ119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)

st SINR1199032119897119898

ge 120574119897119898

Q119897 ⪰ 0


(25)

with

SINR1199032119897119898

=(1119868) sum


119894Hdagger

119897119898119894)

(1119868) sum1198971015840 =119897


119894Hdagger

119897119898119894) + 1205902

119897119898

(26)


P8 minQ119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)

st SINR1199032119897119898

ge 120574119897119898

Q119897 ⪰ 0 forall119898 forall119897

(27)



12119897


dagger

119897


119902

119897= U119897Λ

12119897V119902

119897


119897


radic119901119897W119902

119897sum

119871minus1119897=0 sum

119868minus1119894=0119901119897120573119897119894

(13) end(14)Output W119897



M3 min119901119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897120573119897119894

st(1119868) sum

119868minus1119894=0 119901119897120572119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 11990111989710158401205721198971015840 119898119894 + 1205902

119897119898

ge 120574119897119898

119901119897 ge 0 forall119898 forall119897

(28)


119897

and 120573119897119894 = tr(K119894Q119902

119897Kdagger


H119897119898119894K119894Q119902

119897Kdagger

119894Hdagger

119897119898119894 and 1205721198971015840 119898119894 = H119897119898119894K119894Q

119902

1198971015840Kdagger

119894Hdagger

119897119898119894









119871minus1119897=0 where

Qexp119897

119871minus1

119897=0

= [

[

Qopt119897

119871minus1

119897=001198621119872119905times(1198622minus1198621)119872119905


]

]

(29)






1198681198651minus11198941015840=0 and Qexp

1198971198941198681198652minus1119894=0 by Qopt

119897119871minus1119897=0 and Qexp

119897119871minus1119897=0 and




Q1 maxQ119897119894119905isinR

119905

stSINR119903

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0


(30)


Q2 maxQ119897119894119905isinR

119905

stSINR119903

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


(31)





M4 max119901119897119894119905isinR

119905

st(1119868) sum

119868minus1119894=0 119901119897119894120572119897119898119894

(120574119897119898119868) sum1198971015840 =119897

sum119868minus1119894=0 1199011198971015840 1198941205721198971015840 119898119894 + 1205741198971198981205902

119897119898

ge 119905

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897119894120573119897119894 = 119875


(32)




119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(33)



119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(34)



119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

rank (Q119897) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)



119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)


(35)
















QoS or MMFproblem




in Section 41 or

Section 42





Yes

No

Output




119875 = P3 (Q1 (120574 119875) 120574) (36)

119905 = Q1 (120574P3 (119905120574)) (37)























is invalid





O(119871311986831198726119905+1198701198711198681198722




351198683511987265119905

+ 119870119871151198681511987225119905

+ 1198761198713511986835 +



119905log(1120598))



O(11987105

11986805



minus3+ 119870119871119868119865

minus1)


351198683511987265119905

119865minus35 + 119870119871151198681511987225119905

119865minus15 + 1198761198713511986835119865minus35 +




0511986205







+

119870119871151198622511987225119905



351198683511987265119905

119865minus35 +

119861119878119870119871151198681511987225119905

119865minus15 + 1198761198611198711198713511986835119865minus35 +

1198761198611198711198701198711511986815119865minus15)log(1120598)) O((119861119878119871351198683511987265

119905119865minus35 +

11986111987811987011987115

11986815

11987225119905

119865minus15

+ 11987611986111987111987135

11986835

119865minus35

+

1198761198611198711198701198711511986815119865minus15)log(1120598)) and O((119861119878119871351198626511987265

119905+

119861119878119870119871151198622511987225119905








Tota

l arit

hmet

ic o

pera

tions

for Q

OS

prob

lem

s


108

109

1010

1011

1012

1013

1014

1015

1016












6 8 10 12 14 16 18 20SINR (dB)

0

01

02

03

04

05

06

07

08

09

1

()


L = 2 Mt = 6 K = 12 F = 8

L = 3 Mt = 6 K = 12 F = 8

L = 3 Mt = 4 K = 12 F = 8

L = 2 Mt = 6 K = 8 C = 4

L = 2 Mt = 6 K = 12 C = 4

L = 3 Mt = 6 K = 12 C = 4

L = 3 Mt = 4 K = 12 C = 4

L = 2 Mt = 6 K = 8 C = 5

L = 2 Mt = 6 K = 8 F = 4







6 8 10 12 14 160

5

10

15

20

25

30

35

40

Tran

smiss

ion

pow

er

SINR (dB)








Transmission power







05 1 15 2 25 3 350

01

02

03

04

05

06

07

08

09

1

Cum

ulat

ive d

istrib

utio

n fu

nctio

n




2 4 6 8 10 12 14 16 18 20 22 246

8

10

12

14

16

18

Transmission power

SIN

R (d

B)












8 Conclusion





Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











12119897


dagger

119897


119902

119897= U119897Λ

12119897V119902

119897


119897


radic119901119897W119902

119897sum

119871minus1119897=0 sum

119868minus1119894=0119901119897120573119897119894

(13) end(14)Output W119897



M3 min119901119897119871minus1119897=0

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897120573119897119894

st(1119868) sum

119868minus1119894=0 119901119897120572119897119898119894

(1119868) sum1198971015840 =119897

sum119868minus1119894=0 11990111989710158401205721198971015840 119898119894 + 1205902

119897119898

ge 120574119897119898

119901119897 ge 0 forall119898 forall119897

(28)


119897

and 120573119897119894 = tr(K119894Q119902

119897Kdagger


H119897119898119894K119894Q119902

119897Kdagger

119894Hdagger

119897119898119894 and 1205721198971015840 119898119894 = H119897119898119894K119894Q

119902

1198971015840Kdagger

119894Hdagger

119897119898119894









119871minus1119897=0 where

Qexp119897

119871minus1

119897=0

= [

[

Qopt119897

119871minus1

119897=001198621119872119905times(1198622minus1198621)119872119905


]

]

(29)






1198681198651minus11198941015840=0 and Qexp

1198971198941198681198652minus1119894=0 by Qopt

119897119871minus1119897=0 and Qexp

119897119871minus1119897=0 and




Q1 maxQ119897119894119905isinR

119905

stSINR119903

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0


(30)


Q2 maxQ119897119894119905isinR

119905

stSINR119903

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


(31)





M4 max119901119897119894119905isinR

119905

st(1119868) sum

119868minus1119894=0 119901119897119894120572119897119898119894

(120574119897119898119868) sum1198971015840 =119897

sum119868minus1119894=0 1199011198971015840 1198941205721198971015840 119898119894 + 1205741198971198981205902

119897119898

ge 119905

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897119894120573119897119894 = 119875


(32)




119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(33)



119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(34)



119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

rank (Q119897) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)



119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)


(35)
















QoS or MMFproblem




in Section 41 or

Section 42





Yes

No

Output




119875 = P3 (Q1 (120574 119875) 120574) (36)

119905 = Q1 (120574P3 (119905120574)) (37)























is invalid





O(119871311986831198726119905+1198701198711198681198722




351198683511987265119905

+ 119870119871151198681511987225119905

+ 1198761198713511986835 +



119905log(1120598))



O(11987105

11986805



minus3+ 119870119871119868119865

minus1)


351198683511987265119905

119865minus35 + 119870119871151198681511987225119905

119865minus15 + 1198761198713511986835119865minus35 +




0511986205







+

119870119871151198622511987225119905



351198683511987265119905

119865minus35 +

119861119878119870119871151198681511987225119905

119865minus15 + 1198761198611198711198713511986835119865minus35 +

1198761198611198711198701198711511986815119865minus15)log(1120598)) O((119861119878119871351198683511987265

119905119865minus35 +

11986111987811987011987115

11986815

11987225119905

119865minus15

+ 11987611986111987111987135

11986835

119865minus35

+

1198761198611198711198701198711511986815119865minus15)log(1120598)) and O((119861119878119871351198626511987265

119905+

119861119878119870119871151198622511987225119905








Tota

l arit

hmet

ic o

pera

tions

for Q

OS

prob

lem

s


108

109

1010

1011

1012

1013

1014

1015

1016












6 8 10 12 14 16 18 20SINR (dB)

0

01

02

03

04

05

06

07

08

09

1

()


L = 2 Mt = 6 K = 12 F = 8

L = 3 Mt = 6 K = 12 F = 8

L = 3 Mt = 4 K = 12 F = 8

L = 2 Mt = 6 K = 8 C = 4

L = 2 Mt = 6 K = 12 C = 4

L = 3 Mt = 6 K = 12 C = 4

L = 3 Mt = 4 K = 12 C = 4

L = 2 Mt = 6 K = 8 C = 5

L = 2 Mt = 6 K = 8 F = 4







6 8 10 12 14 160

5

10

15

20

25

30

35

40

Tran

smiss

ion

pow

er

SINR (dB)








Transmission power







05 1 15 2 25 3 350

01

02

03

04

05

06

07

08

09

1

Cum

ulat

ive d

istrib

utio

n fu

nctio

n




2 4 6 8 10 12 14 16 18 20 22 246

8

10

12

14

16

18

Transmission power

SIN

R (d

B)












8 Conclusion





Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











M4 max119901119897119894119905isinR

119905

st(1119868) sum

119868minus1119894=0 119901119897119894120572119897119898119894

(120574119897119898119868) sum1198971015840 =119897

sum119868minus1119894=0 1199011198971015840 1198941205721198971015840 119898119894 + 1205741198971198981205902

119897119898

ge 119905

1119868

119871minus1sum119897=0

119868minus1sum119894=0

119901119897119894120573119897119894 = 119875


(32)




119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

rank (Q119897119894) = 1

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(33)



119905

stSINR1199031

119897119898

120574119897119898ge 119905

Q119897119894 ⪰ 0

1119868

119871minus1sum119897=0

119868119865minus1sum119894=0

tr (Q119897119894)


(34)



119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

rank (Q119897) = 1

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)



119905

stSINR1199032

119897119898

120574119897119898ge 119905

Q119897 ⪰ 0

1119868

119871minus1sum119897=0

119868minus1sum119894=0


119894)


(35)
















QoS or MMFproblem




in Section 41 or

Section 42





Yes

No

Output




119875 = P3 (Q1 (120574 119875) 120574) (36)

119905 = Q1 (120574P3 (119905120574)) (37)























is invalid





O(119871311986831198726119905+1198701198711198681198722




351198683511987265119905

+ 119870119871151198681511987225119905

+ 1198761198713511986835 +



119905log(1120598))



O(11987105

11986805



minus3+ 119870119871119868119865

minus1)


351198683511987265119905

119865minus35 + 119870119871151198681511987225119905

119865minus15 + 1198761198713511986835119865minus35 +




0511986205







+

119870119871151198622511987225119905



351198683511987265119905

119865minus35 +

119861119878119870119871151198681511987225119905

119865minus15 + 1198761198611198711198713511986835119865minus35 +

1198761198611198711198701198711511986815119865minus15)log(1120598)) O((119861119878119871351198683511987265

119905119865minus35 +

11986111987811987011987115

11986815

11987225119905

119865minus15

+ 11987611986111987111987135

11986835

119865minus35

+

1198761198611198711198701198711511986815119865minus15)log(1120598)) and O((119861119878119871351198626511987265

119905+

119861119878119870119871151198622511987225119905








Tota

l arit

hmet

ic o

pera

tions

for Q

OS

prob

lem

s


108

109

1010

1011

1012

1013

1014

1015

1016












6 8 10 12 14 16 18 20SINR (dB)

0

01

02

03

04

05

06

07

08

09

1

()


L = 2 Mt = 6 K = 12 F = 8

L = 3 Mt = 6 K = 12 F = 8

L = 3 Mt = 4 K = 12 F = 8

L = 2 Mt = 6 K = 8 C = 4

L = 2 Mt = 6 K = 12 C = 4

L = 3 Mt = 6 K = 12 C = 4

L = 3 Mt = 4 K = 12 C = 4

L = 2 Mt = 6 K = 8 C = 5

L = 2 Mt = 6 K = 8 F = 4







6 8 10 12 14 160

5

10

15

20

25

30

35

40

Tran

smiss

ion

pow

er

SINR (dB)








Transmission power







05 1 15 2 25 3 350

01

02

03

04

05

06

07

08

09

1

Cum

ulat

ive d

istrib

utio

n fu

nctio

n




2 4 6 8 10 12 14 16 18 20 22 246

8

10

12

14

16

18

Transmission power

SIN

R (d

B)












8 Conclusion





Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











QoS or MMFproblem




in Section 41 or

Section 42





Yes

No

Output




119875 = P3 (Q1 (120574 119875) 120574) (36)

119905 = Q1 (120574P3 (119905120574)) (37)























is invalid





O(119871311986831198726119905+1198701198711198681198722




351198683511987265119905

+ 119870119871151198681511987225119905

+ 1198761198713511986835 +



119905log(1120598))



O(11987105

11986805



minus3+ 119870119871119868119865

minus1)


351198683511987265119905

119865minus35 + 119870119871151198681511987225119905

119865minus15 + 1198761198713511986835119865minus35 +




0511986205







+

119870119871151198622511987225119905



351198683511987265119905

119865minus35 +

119861119878119870119871151198681511987225119905

119865minus15 + 1198761198611198711198713511986835119865minus35 +

1198761198611198711198701198711511986815119865minus15)log(1120598)) O((119861119878119871351198683511987265

119905119865minus35 +

11986111987811987011987115

11986815

11987225119905

119865minus15

+ 11987611986111987111987135

11986835

119865minus35

+

1198761198611198711198701198711511986815119865minus15)log(1120598)) and O((119861119878119871351198626511987265

119905+

119861119878119870119871151198622511987225119905








Tota

l arit

hmet

ic o

pera

tions

for Q

OS

prob

lem

s


108

109

1010

1011

1012

1013

1014

1015

1016












6 8 10 12 14 16 18 20SINR (dB)

0

01

02

03

04

05

06

07

08

09

1

()


L = 2 Mt = 6 K = 12 F = 8

L = 3 Mt = 6 K = 12 F = 8

L = 3 Mt = 4 K = 12 F = 8

L = 2 Mt = 6 K = 8 C = 4

L = 2 Mt = 6 K = 12 C = 4

L = 3 Mt = 6 K = 12 C = 4

L = 3 Mt = 4 K = 12 C = 4

L = 2 Mt = 6 K = 8 C = 5

L = 2 Mt = 6 K = 8 F = 4







6 8 10 12 14 160

5

10

15

20

25

30

35

40

Tran

smiss

ion

pow

er

SINR (dB)








Transmission power







05 1 15 2 25 3 350

01

02

03

04

05

06

07

08

09

1

Cum

ulat

ive d

istrib

utio

n fu

nctio

n




2 4 6 8 10 12 14 16 18 20 22 246

8

10

12

14

16

18

Transmission power

SIN

R (d

B)












8 Conclusion





Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











+

119870119871151198622511987225119905



351198683511987265119905

119865minus35 +

119861119878119870119871151198681511987225119905

119865minus15 + 1198761198611198711198713511986835119865minus35 +

1198761198611198711198701198711511986815119865minus15)log(1120598)) O((119861119878119871351198683511987265

119905119865minus35 +

11986111987811987011987115

11986815

11987225119905

119865minus15

+ 11987611986111987111987135

11986835

119865minus35

+

1198761198611198711198701198711511986815119865minus15)log(1120598)) and O((119861119878119871351198626511987265

119905+

119861119878119870119871151198622511987225119905








Tota

l arit

hmet

ic o

pera

tions

for Q

OS

prob

lem

s


108

109

1010

1011

1012

1013

1014

1015

1016












6 8 10 12 14 16 18 20SINR (dB)

0

01

02

03

04

05

06

07

08

09

1

()


L = 2 Mt = 6 K = 12 F = 8

L = 3 Mt = 6 K = 12 F = 8

L = 3 Mt = 4 K = 12 F = 8

L = 2 Mt = 6 K = 8 C = 4

L = 2 Mt = 6 K = 12 C = 4

L = 3 Mt = 6 K = 12 C = 4

L = 3 Mt = 4 K = 12 C = 4

L = 2 Mt = 6 K = 8 C = 5

L = 2 Mt = 6 K = 8 F = 4







6 8 10 12 14 160

5

10

15

20

25

30

35

40

Tran

smiss

ion

pow

er

SINR (dB)








Transmission power







05 1 15 2 25 3 350

01

02

03

04

05

06

07

08

09

1

Cum

ulat

ive d

istrib

utio

n fu

nctio

n




2 4 6 8 10 12 14 16 18 20 22 246

8

10

12

14

16

18

Transmission power

SIN

R (d

B)












8 Conclusion





Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










6 8 10 12 14 16 18 20SINR (dB)

0

01

02

03

04

05

06

07

08

09

1

()


L = 2 Mt = 6 K = 12 F = 8

L = 3 Mt = 6 K = 12 F = 8

L = 3 Mt = 4 K = 12 F = 8

L = 2 Mt = 6 K = 8 C = 4

L = 2 Mt = 6 K = 12 C = 4

L = 3 Mt = 6 K = 12 C = 4

L = 3 Mt = 4 K = 12 C = 4

L = 2 Mt = 6 K = 8 C = 5

L = 2 Mt = 6 K = 8 F = 4







6 8 10 12 14 160

5

10

15

20

25

30

35

40

Tran

smiss

ion

pow

er

SINR (dB)








Transmission power







05 1 15 2 25 3 350

01

02

03

04

05

06

07

08

09

1

Cum

ulat

ive d

istrib

utio

n fu

nctio

n




2 4 6 8 10 12 14 16 18 20 22 246

8

10

12

14

16

18

Transmission power

SIN

R (d

B)












8 Conclusion





Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










2 4 6 8 10 12 14 16 18 20 22 246

8

10

12

14

16

18

Transmission power

SIN

R (d

B)












8 Conclusion





Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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