DTIC ADA534847: Coordinated Beamforming for MISO Interference Channel: Complexity Analysis and Efficient Algorithms

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1. REPORT DATE (DD-MM-YYYY) 2. REPORT TYPE 3. DATES COVERED (From - To)
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4. TITLE AND SUBTITLE 5a. CONTRACT NUMBER
Coordinated Beamforming for MISO Interference Channel: W911NF-09-1-0279
Complexity Analysis and Efficient Algorithms
5b. GRANT NUMBER
5c. PROGRAM ELEMENT NUMBER
611102
6. AUTHORS 5d. PROJECT NUMBER
Ya-Feng Liu, Yu-Hong Dai and Zhi-Quan Luo
5e. TASK NUMBER
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University of Minnesota - Minneapolis 906.00
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14. ABSTRACT
In a cellular wireless system, users located at cell edges often suffer significant out-of-cell interference. Assuming
each base station is equipped with multiple antennas, we can model this scenario as a multiple-input single-output
(MISO) interference channel. In this paper we consider a coordinated beamforming approach whereby multiple
base stations jointly optimize their downlink beamforming vectors in order to simultaneously improve the data rates
of a given group of cell edge users. Assuming perfect channel knowledge, we formulate this problem as the
15. SUBJECT TERMS
MISO interference channel, coordinated beamforming, complexity, cyclic coordinate descent algorithm, global convergence.
16. SECURITY CLASSIFICATION OF: 17. LIMITATION OF 15. NUMBER 19a. NAME OF RESPONSIBLE PERSON
a. REPORT b. ABSTRACT c. THIS PAGE ABSTRACT OF PAGES Zhi-Quan (Tom) Luo
UU UU UU UU 19b. TELEPHONE NUMBER
612-625-0242
Standard Form 298 (Rev 8/98)
Prescribed by ANSI Std. Z39.18
Report Title
Coordinated Beamforming for MISO Interference Channel: Complexity Analysis and Efficient Algorithms
ABSTRACT
In a cellular wireless system, users located at cell edges often suffer significant out-of-cell interference. Assuming each
base station is equipped with multiple antennas, we can model this scenario as a multiple-input single-output (MISO)
interference channel. In this paper we consider a coordinated beamforming approach whereby multiple base stations
jointly optimize their downlink beamforming vectors in order to simultaneously improve the data rates of a given group
of cell edge users. Assuming perfect channel knowledge, we formulate this problem as the maximization of a system
utility (which balances user fairness and average user rates), subject to individual power constraints at each base
station. We show that, for the single carrier case and when the number of antennas at each base station is at least two,
the optimal coordinated beamforming problem is NP-hard for both the harmonic mean utility and the proportional
fairness utility. For general utilities, we propose a cyclic coordinate descent algorithm, which enables each transmitter
to update its beamformer locally with limited information exchange, and establish its global convergence to a stationary
point. We illustrate its effectiveness in computer simulations by using the space matched beamformer as a benchmark.
In a cellular wireless system, users located at cell edges often suffer significant out-of-cell interference. Assuming each
base station is equipped with multiple antennas, we can model this scenario as a multiple-input single-output (MISO)
interference channel. In this paper we consider a coordinated beamforming approach whereby multiple base stations
jointly optimize their downlink beamforming vectors in order to simultaneously improve the data rates of a given group
of cell edge users. Assuming perfect channel knowledge, we formulate this problem as the maximization of a system
utility (which balances user fairness and average user rates), subject to individual power constraints at each base
station. We show that, for the single carrier case and when the number of antennas at each base station is at least two,
the optimal coordinated beamforming problem is NP-hard for both the harmonic mean utility and the proportional
fairness utility. For general utilities, we propose a cyclic coordinate descent algorithm, which enables each transmitter
to update its beamformer locally with limited information exchange, and establish its global convergence to a stationary
point. We illustrate its effectiveness in computer simulations by using the space matched beamformer as a benchmark.
1
Coordinated Beamforming for MISO Interference
Channel: Complexity Analysis and Efficient
∗
Algorithms
Ya-Feng Liu†, Yu-Hong Dai†, and Zhi-Quan Luo‡
Abstract—In a cellular wireless system, users located at cell channel.Weconsiderjointoptimalbeamformingacrossmulti-
edges often suffer significant out-of-cell interference. Assuming plebasestationstosimultaneouslyimprovethedataratesofa
each base station is equipped with multiple antennas, we can
givengroupofcelledgeusers.Assumingthatthechannelstate
model this scenario as a multiple-input single-output (MISO)
information (CSI) is known, we formulate this problem as the
interference channel. In this paper we consider a coordinated
beamforming approach whereby multiple base stations jointly maximizationofasystemutility(whichbalancesuserfairness
optimize their downlink beamforming vectors in order to simul- andaverageuserrates),subjecttoindividualpowerconstraints
taneously improve the data rates of a given group of cell edge at each base station. We show that, for the single carrier case
users. Assuming perfect channel knowledge, we formulate this
andwhenthenumberofantennasateachbasestationisatleast
problem as the maximization of a system utility (which balances
two, the optimal coordinated beamforming problem is NP-
userfairnessandaverageuserrates),subjecttoindividualpower
constraints at each base station. We show that, for the single hard for both the harmonic mean utility and the proportional
carrier case and when the number of antennas at each base fairness utility. This NP-hardness result is in contrast to the
station is at least two, the optimal coordinated beamforming single antenna case for which the same optimization problem
problem is NP-hard for both the harmonic mean utility and the
isconvexforboththeharmonicmeanandproportionalfairness
proportional fairness utility. For general utilities, we propose a
utility functions [1]. For the min-rate utility, this problem is
cyclic coordinate descent algorithm, which enables each trans-
mittertoupdateitsbeamformerlocallywithlimitedinformation known to be also solvable in polynomial time [1], [2].
exchange, and establish its global convergence to a stationary In addition to the complexity analysis, we propose a prac-
point. We illustrate its effectiveness in computer simulations by tical iterative cyclic coordinate descent algorithm for the
using the space matched beamformer as a benchmark.
multi-cell coordinated beamforming problem by exploiting
the separability of power constraints. We prove the global
IndexTerms—MISOinterferencechannel,coordinatedbeam- convergence of this cyclic coordinate descent algorithm (to
forming, complexity, cyclic coordinate descent algorithm, global
a stationary point). Numerical experiments are also presented
convergence.
to illustrate the effectiveness of the proposed algorithm.
I. INTRODUCTION A. Related Work
In a conventional wireless cellular system, base stations Downlink beamforming has been studied extensively in the
from different cells communicate with their respective remote single cell setup [3], [4]. For the multi-cell interference chan-
terminalsindependently.Signalprocessingisperformedonan nel,thereference[5]consideredcoordinatedbeamformingfor
intra-cell basis, while the out-of-cell interference is treated as the minimization of total weighted transmitted power across
background noise. This architecture often causes undesirable the base stations subject to individual signal-to-interference-
serviceoutagestouserssituatednearcelledgeswheretheout- plus-noise-ratio(SINR)constraintsattheremoteusers.Itturns
of-cellinterferencecanbesevere.Sincetheconventionalintra- out this problem can be transformed into a convex second
cell signal processing can not effectively mitigate the impact orderconicprogramming(SOCP)andefficientlysolved.How-
of inter-cell interference, we are led to consider coordinated ever, the maximization of weighted sum rates for a multi-cell
base station beamforming across multiple cells in order to interferencechannelunderindividualpowerconstraintsisNP-
improve the services to edge users. In this paper, we focus on hardevenforthesingleantennaandthesinglecarriercase[1].
the downlink scenario where the base stations are equipped In fact, more is known about the single antenna interference
with multiple antennas and model it as a MISO interference channelcase.Forinstance,ifthesystemutilityischangedinto
either the geometric mean rates (i.e., proportional fairness),
∗ Ya-Feng Liu and Yu-Hong Dai are supported by the National Natural the harmonic mean rates, or the min-rate, the corresponding
Science Foundation of China, Grant number 10831006, and the CAS Grant utility maximization problem (for the single tone case) can
kjcx-yw-s7-03. Zhi-Quan Luo is supported by the Army Research Office,
be converted to a convex optimization problem and solved
GrantnumberW911NF-09-1-0279,andbytheNationalScienceFoundation,
GrantnumberCMMI-0726336. efficiently to global optimality [1], [6]. However, when the
†StateKeyLab.ofScientificandEngineeringComputing,Beijing,100190, number of tones is more than two, then all of the aforemen-
China.(e-mail:{yafliu,dyh}@lsec.cc.ac.cn)
tioned power control problems are NP-hard. The focus of this
‡ Department of Electrical Computer Science Engineering, University of
Minnesota,Minneapolis,55455,USA.(e-mail:[email protected]) paper is to study the multi-antenna case (MISO interference
2
channel), analyze the complexity of the corresponding utility base station k, while s is a complex scalar denoting the
k
maximization problems, and propose a practical algorithm to informationsignalforuserk withE|s |2 =1.Thetransmitter
k
solve them. vectorofthej-thbasestationisv s .Thenthesignalreceived
j j
Inadditiontotheaforementionedutilitybasedformulations, by user k can be described as
various base station cooperation techniques have been pro-
(cid:88)K
posedtomitigateinter-cellinterferences,includingmulti-point y = h† (v s )+z , 1≤k ≤K, (1)
k jk j j k
coordinated transmission, or network multi-input multi-output
j=1
(MIMO) transmission [7–16]. For example, the distributed
where z is the additive white Gaussian noise (AWGN) with
or decentralized approaches are proposed for coordinated k
variance σ2/2 per real dimension. Treating interference as
transmitterbeamformingvectorsinMISOinterferencechannel k
noise, we can write the SINR of each user as
in [7], [10], [13], [14], [16], [17], some of which are based
on dual uplink channels. In particular, a distributed pricing |h† v |2
algorithm for power control and beamformer design in the SINRk = σ2+(cid:80)kk k|h† v |2. (2)
k j(cid:54)=k jk j
MISO interference networks is proposed in [17]. At each
iteration, transmitters try to maximize their utility minus the Adopting an utility, we can formulate the optimal coordinated
total interference cost, i.e., the summation of the interference downlink beamforming problem as
pricetimesthereceivedinterference,andtheinterferenceprice max H(r ,r ,...,r )
1 2 K
indicates the marginal decrease in the corresponding user’s (cid:195) (cid:33)
uwtiitlhitythdeupearttoicualamrarregcieniavleri.ncrease in interference associated s.t. rk =log 1+ σ2+(cid:80)|h†kkvk|h|2† v |2 , (3)
k j(cid:54)=k jk j
The references [11], [12] show that coordination enables
(cid:107)v (cid:107)2 ≤P , 1≤k ≤K,
the cellular network to enjoy a greater spectral efficiency. k k
Mostofthesecooperativetechniquesrequireeachbasestation whereP denotesthepowerbudgetofbasestationk,andH(·)
k
to have not only full/partial CSI but also the knowledge of denotes the system utility which may be any of the following
actual independent data streams to all remote terminals. With 1 (cid:88)K
the complete sharing of data streams and CSI, the multi-cell • Weighted sum-rate utility: H1 = K wkrk, with
scenario is effectively reduced to a single cell interference k=1
weight w ≥0.
management problem with either total [18] or per-group-of- k
antenna power constraints [19], [20]. Among the major draw-
(cid:195) (cid:33)
1/K
backs of these techniques (in comparison to the utility based (cid:89)K
approaches)aretheirstringentrequirementonbasestationco- • Proportional fairness utility: H2 = rk ⇔
ordination,thelargedemandonthecommunicationbandwidth k=1
of backhaul links, as well as the heavy computational load 1 (cid:88)K
logr .
associated with the increasing number of cells [21], [22]. The K k
k=1
references [8], [15], [23], [24] provided characterizations of (cid:195) (cid:33)
(cid:88)K
tNhaesahcheiqeuvialbibleriuramtewrehgiicohnaisndinperfofivceidenthteinextihsteensceensoefathuantiqtuhee • Harmonic mean utility: H3 =K/ rk−1 .
k=1
achievable rates are bounded by a constant, regardless of the • Min-rate utility: H4 = min rk.
available transmit power. See [23], [25] for the recent results 1≤k≤K
of the MISO channel.
According to [23], [24], problem (3) can be written in a more
Notation: The notation for this paper is as follows: Lower
general form
caseboldfaceisusedforcolumnvectors.(·)T and(·)† denote
transpose and Hermitian transpose. (x,y) denotes a two- max H(r ,r ,...,r )
1 2 K
dimensional row vector. (cid:107)·(cid:107) denotes the Euclidean norm. As- (cid:195) (cid:33)
h† V h
dsuemnoitnegifts(xgr)aidsieanmt aunltdi-vHaersiasbialne,fruenscpteiocnti,ve∇lyf.(x) and ∇2f(x) s.t. rk =log 1+ σk2+(cid:80)kjk(cid:54)=kkh†jkkkVjhjk , (4)
Trace(V )≤P , V (cid:60)0, 1≤k ≤K,
k k k
II. PROBLEMFORMULATION
where V is the transmit covariance matrix at transmitter k.
k
Consider a cellular system in which there are K base The results in [23], [24] state that problem (4) has a rank-one
stations each equipped with L transmit antennas. The K base optimal solution for each V . This implies that problem (3)
k
stations wish to transmit respectively to K mobile receivers and problem (4) are equivalent. We focus on formulation (3)
eachhavingonlyasingleantenna.Eachbasestationcandirect in this paper.
abeamtoitsintendedreceiverinsuchawaythattheresulting The above beamforming problem (3) can be nonconvex
interference to the other mobile units is small. Consider the in general due to the nonlinear equality constraints. Given
single carrier case, and let h ∈ CL denote L-dimensional nonnegative weights (w ,w ,...,w ), the optimal tuple of
jk 1 2 K
complex channel vector between base station j and receiver rates (r ,r ,...,r ) of (3) should lie on the boundary of
1 2 K
k. Let v ∈ CL denote the beamforming vector used by the achievable rate region. See [8], [9], [15], [23], [24] for
k
3
various effort to characterize the achievable rate region of the assuming the availability of a feasible solution that meets
interference channel. the threshold requirement. More formally, we say a nondeter-
Inpractice,thechoiceofutilitiesdependsonasuitablecom- ministic algorithm solves a decision version of optimization
promise between system performance (total rates achievable) problem if we can verify each “true” instance of the problem
and user fairness. The sum-rate utility H focuses entirely using a sequence of nondeterministic steps (i.e., involving
1
on system performance, while the min-rate utility H places random guesses). If the number of nondeterministic steps is
4
the highest emphasis on user fairness. The other two choices polynomial, then the algorithm is said to have a nondetermin-
H and H represent an appropriate tradeoff between the two isticpolynomialrunningtime.Forexample,foranysymmetric
2 3
extremes. matrixQ∈Rn×n andathresholdvalueL,considertheprob-
lem of deciding if there exists a binary vector x ∈ {−1,1}n
III. COMPLEXITYANALYSIS such that xTQx ≤ L. A nondeterministic algorithm to solve
this problem is to guess a binary vector x and then check
In this section, we investigate the complexity status of the
if xTQx ≤ L indeed holds. Such a binary vector x exists
optimalcoordinateddownlinkbeamformingproblem(3)under
for all “true” instances of the problem, and the verification
various choices of system utilities. We provide a complete
processrequirespolynomiallymanystepsalthoughsomesteps
analysis on when the problem is NP-hard and also identify
may involve a random guess (of a component of x). In this
subclassesoftheproblemthataresolvableinpolynomialtime.
case, the problem is solvable in nondeterministic polynomial
time. The class NP contains precisely those decision version
A. Computational complexity theory: a brief background
ofoptimizationproblemsthataresolvableinnondeterministic
Generically, an optimization problem can be described by polynomial time. Clearly, P is contained in NP. It is widely
the minimization of an objective function over a feasible conjecturedthatP(cid:54)=NP,orequivalently,thereareproblemsin
region. A decision version of the minimization problem is NP which are not solvablein (deterministic) polynomial time.
to decide if the feasible region contains a vector at which A subset of problems in NP are called NP-complete. NP-
the objective function value is below a given threshold. The complete problems are those in NP which are most difficult
answer to the decision problem is binary, true or false, and to solve, in the sense that if any one of them is in P, so is
there is no need to identify what the solution is. The decision every other problem in NP. There are many well known NP-
version is typically easier to solve than the original optimiza- complete problems such as the traveling salesman problem,
tionproblemwhichrequiresthedeterminationofan(globally) the 3-colorability problem, and so on. The latter problem is
optimalsolution.Thesizeofanoptimizationprobleminstance to decide if the nodes of a given graph can be colored in
isdefinedastheminimumlengthofabinarystringrequiredto three colors so that each adjacent pair of nodes are colored
describetheobjectivefunctionandthefeasibleregion.Wesay differently. The 3-colorability problem is clearly in NP since
an algorithm solves the decision version of an optimization we can easily check if a guessed coloring scheme meets the
problem if for each instance of the problem, the algorithm requirement. There is no known polynomial time algorithm
correctly gives “true” or “false” answer. We can define the to solve the 3-colorability problem. In fact, if this problem is
runningtimeofanalgorithmasthemaximumnumberofbasic solvable in polynomial time (i.e., in P), then every problem in
computational steps (e.g., number of arithmetic operations) NP is solvable in polynomial time, or equivalently P=NP. A
required to solve the decision version of an optimization problemP issaidtobeNP-hardifitisatleastashardasthose
problem of a given size. Typically, the algorithm’s running NP-completeproblems,whichmeansthatthepolynomialtime
time is a function of the problem size. solvability of P would imply every NP-complete problem is
Inthecomputationalcomplexitytheory[26],[30],thereare in P. A NP-hard problem may not be in NP. For example, the
twoimportantclassesofoptimizationproblems,PandNP.The binary quadratic minimization problem min xTQx
x∈{−1,1}n
class P contains optimization problems which are solvable (or isNP-hard,sinceitisnotknowntobeinNP,andisatleastas
decidable)byanalgorithmwhoserunningtimegrowsatmost hard as the NP-complete problem of deciding if there exists
asapolynomialfunctionoftheinputsize.Itturnsoutthatthe a binary vector x such that xTQx ≤ L, where the threshold
class P is rather robust to the actual definition of input size. value L is given. A problem is called strongly NP-hard if
For example, P is invariant if we alternatively define the input it cannot be solved by a pseudo polynomial time algorithm
size of an optimization problem as the sum of the problem unless P=NP.
dimension, the number of constraints and the binary length To prove a problem P is NP-complete, we need to show
of the input data and threshold value. We say an algorithm two things. First, we verify the problem is in NP. This step is
is a pseudo polynomial time algorithm if its running time usuallyeasy.Second,weneedtoshowP isatleastasdifficult
is a polynomial when we bound the size of numbers in the as a known NP-complete problem. This can be accomplished
inputbyaconstant.Forexample,manydynamicprogramming byastandardtechniquecalledpolynomialtimetransformation.
algorithms (e.g., Viterbi algorithm) are pseudo polynomial In a polynomial time transformation, we pick a known NP-
time since their state space are typically exponential if the complete problem and show that it is equivalent to a special
numbers in the input are not bounded. case of P. More precisely, we take an arbitrary instance of
TheclassNP,whichstandsforNondeterministicPolynomial a known NP-complete problem, construct a special instance
time, consists of decision version of optimization problems (withpolynomialsize)ofP,andthenestablishtheequivalence
whose “true” instances can be verified in polynomial time, ofthetwoinstances.ToshowaproblemisNP-hard,wesimply
4
ignorestep1,asthereisnoneedtoshowP isinNP.Toshow it is the True or the False value). Otherwise it is said to be
P is strongly NP-hard, we need to additionally ensure the satisfied in the NAE (Not-All-Equal) sense.
special instance of P we construct involves only numbers of Noticethatadisjunctiveclausemustbesatisfiedifitistohave
constant size, i.e., numbers whose combined bit length does the NAE property. We now define some decision problems
not increase with the problem dimension or the number of over Boolean variables.
constraints. Definition 3.2: MAX-UNANIMITY problem: given a pos-
itive integer M and m disjunctive clauses defined over n
B. Maximization of the Weighted Sum-Rate Utility Boolean variables, we ask whether there exists a truth assign-
1 (cid:88)K mentsuchthatthenumberofunanimousdisjunctiveclausesis
ConsiderthesystemutilityH1 = K wkrk.Inthesingle atleastM.Whenthenumberofliteralsineachclauseistwo,
k=1 we denote the corresponding problem as MAX-2UNANIMITY
antenna case (L = 1), the original system optimization prob-
problem.Wheneachclausecontainsthreeliterals,theproblem
lem(3)becomesthefollowing(5),wherex =(cid:107)v (cid:107)2, α =
k k jk of determining whether there exists a truth assignment under
(cid:107)h (cid:107)2/(cid:107)h (cid:107)2 and γ =σ2/(cid:107)h (cid:107)2. Problem (5) is known
jk kk k k kk which at least M clauses are satisfied in the NAE sense is
tobeNP-hard[1]evenwhentheweightsw areallequal,and
k called the NAE-SAT problem.
the proof is based on a polynomial time transformation from
The NAE-SAT problem is known to be NP-complete [30].
the maximum independent set problem (which is known to
Our next lemma says that the MAX-2UNANIMITY problem
be NP-complete). Thus, the general case of L ≥ 1 is also
is also NP-complete.
NP-hard. For various special MISO channels, the sum-rate
Lemma 3.1: The MAX-2UNANIMITY problem is NP-
maximization problem can still be solved in polynomial time,
complete.
see [25], [27–29].
Proof: We construct a polynomial time transformation
fromtheNAE-SATproblem.ItcanbecheckedthattheMAX-
C. Maximization of the Harmonic Mean Utility
2UNANIMITYproblemisintheclassNP.Givenadisjunctive
We now study the complexity status of the optimal coor-
clause with three literals, c = x∨y∨z, let us construct the
dinated downlink beamforming problem (3) defined by the
following six clauses, each involving only two literals:
harmonic mean rate utility.
Theorem 3.1 (Harmo(cid:195)nic Mean(cid:33)Utility): For the harmonic R(c): x∨y¯, x∨z¯, y∨x¯, y∨z¯, z∨x¯, z∨y¯. (6)
(cid:88)K
mean utility H3 = K/ rk−1 , the optimal coordinated It can be checked that R(c) has the following properties:
k=1 1) The number of unanimous clauses (i.e., all literals hav-
downlink beamforming problem can be transformed into a
ing the same value) in R(c) is at most four.
convex optimization problem when L = 1, but is NP-hard
2) The clause c is satisfied in the NAE sense if and only
when L≥2.
if the number of unanimous clauses in R(c) is four.
When there is only one transmit antenna (L = 1), the
reference[1]showsthattheharmonicmeanratemaximization Now given any instance φ of NAE-SAT problem, we con-
can be transformed into an equivalent convex problem. We structacorrespondinginstanceR(φ)ofMAX-2UNANIMITY
thus focus on the case L≥2. Notice that the harmonic mean problem as follows: for each clause c = α∨β∨γ of φ, we
utility maximization problem is a continuous optimization add to R(φ) the six clauses in (6), with x, y, z replaced
problem. To show its NP-hardness, we need to transform with the literals α, β, γ respectively. In this way, if φ has
a known NP-hard discrete problem to the harmonic mean m clauses, then R(φ) will have 6m clauses. Let M = 4m.
maximization problem. To facilitate this transformation, it is Then properties 1 and 2 imply that all clauses in φ are
necessary to induce certain discrete structure to the optimal simultaneously satisfiable in the NAE sense if and only if at
solutions to the harmonic mean maximization problem. This least M = 4m clauses in R(φ) can be made unanimous. In
is accomplished by using the concavity of the harmonic particular, suppose that 4m clauses in R(φ) are unanimous
mean utility with respect to each beamforming vector v . In under a given truth assignment. Since, by property 1, each
k
particular, Lemma 3.2 (proved in Appendix I) shows that we group R(c) of six clauses can have at most four unanimous
can constrain the optimal beamforming vectors to be taken clauses,itfollowsthatexactlyfourclausesmustbeunanimous
from two orthogonal vectors h or h . in each group. By property 2, this further implies that each
a b
The NP-hardness proof of Theorem 3.1 is based on a clause in φ is satisfied in NAE sense. Conversely, any truth
transformation from a variant of the 3-SAT [30] problem. assignmentthatsatisfiesaclausecintheNAEsensewillgive
To describe this variant, we need to define the UNANIMITY risetofourunanimousclauses.Thus,ifallmclausesinφare
property and the NAE (stands for “not-all-equal”) property of satisfied in the NAE sense, then there will be 4m unanimous
a disjunctive clause1. clauses in R(φ). Finally, this transformation is in polynomial
Definition 3.1: For a given truth assignment to a set of time.
Boolean variables, a disjunctive clause is said to be UNANI- An immediate corollary of Lemma 3.1 is that the MAX-
MOUSifallliteralsintheclausehavethesamevalue(whether UNANIMITY problem is NP-complete. However, if we ask
whether all of the m clauses (i.e., M = m) can be made
1Recall that for a given set of Boolean variables, a literal is defined as
unanimous,thenthecorrespondingMAX-UNANIMITYprob-
eitheraBooleanvariableoritsnegation,whileadisjunctiveclauserefersto
alogicalexpressionconsistingofthelogical“OR”ofliterals. lem, simply called UNANIMITY problem, can be solved in
5
(cid:195) (cid:33)
(cid:88)K x
max w r , subject to r =log 1+ (cid:80) k , 0≤x ≤P , 1≤k ≤K, (5)
k k k γ + α x k k
k=1 k j(cid:54)=k jk j
√
polynomial time (using a tree search technique). Also, notice channel vectors are h=( N,0)T and their noise powers are
that a disjunctive clause is unanimous under a given truth 1. Let v ,i = 1,2,...,n, denote the transmit beamforming
4i
assignment if and only if c is not satisfied in NAE sense. This vector of the “variable user” 4i. For example, the clause
implies that MIN-3UNANIMITY problem is NP-complete. c =x ∨x¯ corresponds to j =1 and is associated with two
1 2 3
Since these results are not needed in the subsequent analysis, “clause users” which are denoted by 4n+1, 4n+2. Since
we state them without proof. π(1) = 2 and τ(1) = −3, the two users 4n+1 and 4n+2
Proposition 3.1: The following is true: experience interferences from “variable users” 4|π(1)| = 8
1) MAX-UNANIMITY problem is NP-complete. and 4|τ(1)| = 12. The corresponding interference terms for
2) MIN-3UNANIMITY problem is NP-complete. thesetwo“clauseusers”4n+1,4n+2are|h†av8|2+|h†bv12|2
3) UNANIMITY problem is solvable in polynomial time. and |h†bv8|2+|h†av12|2, respectively.
We are now ready to prove Theorem 3.1. Proof: Let
the utility in (3) be given by the harmonic mean function H . The correspondence between MAX-2UNANIMITY prob-
3
ConsideraninstanceofMAX-2UNANIMITYproblemφwith lemandtheoptimalcoordinateddownlinkbeamformingprob-
clauses c1,c2,...,cm defined over the variables x1,x2,...,xn lem(7)islistedinTableI.Noticethatr4n+2j canbeobtained
and an√integer M. Let ha = (1,0)T, hb = (0,1)T and from clause cj according to Table I and r4n+2j−1 can be
h = ( N,0)T, where N is a large positive number (to be obtained from r4n+2j by swapping ha with hb.
specifiedlater).Wewriteeachclausec =α ∨β ,withα , β
j j j j j
TABLEI
taken from {x ,x ,...,x ,x¯ ,x¯ ,...,x¯ }. Let us define two
1 2 n 1 2 n
mappings VARIABLECORRESPONDENCE
π, τ :{1,2,...,m}(cid:55)→{±1,±2,...,±n} MAX-2UNANIMITY Problem(7)
variablexi beamformingvectorv4i
such that clausecj ratesr4n+2j andr4n+2j−1
π(j)=(cid:189) −ii,, iiff ααj ==xx¯i,, and τ(j)=(cid:189) −ii,, iiff ββj ==xx¯i., lliitteerraallxx¯ii iinntteerrffeerreenncceetteerrmm||hh†a†bvv44ii||22
j i j i
Forinstance,ifc =x ∨x¯ ,thenwehaveα =x ,β =x¯ ,
4 3 5 4 3 4 5 We first fix some easy variables of (7) to simplify the
with π(4) = 3 and τ(4) = −5. For i = ±1,±2,...,±n, we
problem. Since each of the beamforming vectors v ,
4n+2j
define (cid:189)
v , j = 1,2,...,m, and v , i = 1,2,...,n, l =
h , if i>0, 4n+2j−1 4i−l
hi = ha, if i<0. 1,2,3, appears exactly once in the objective function, it
b
follows that, by optimality, these beamforming vectors must
Given an instance of MAX-2UNANIMITY problem, we con- be matched to the corresponding channel vectors. That is,
struct the following (7) as an instance of (3) (the inverse v∗ = v∗ = (1,0)T, and v∗ = v∗ =
4n+2j 4n+2j−1 4i−1 4i−2
of harmonic mean utility minimization is equivalent to the v∗ = (1,0)T. Substituting these optimal beamforming
4i−3
harmonic mean rate utility maximization) with a total of vectors into (7) yields (8). It only remains to determine the
K =4n+2m users. Herein, each Boolean variable xi corre- optimal beamforming vectors v4∗i, i = 1,2,...,n. For this
spondstofourusers,includinguser4i(called“variableuser”) purpose, we need the following key lemma whose proof is
and user 4i−1,4i−2 and 4i−3 (called “auxiliary variable relegated to Appendix I.
users”); while each clause cj corresponds to a pair of users, Lemma 3.2: If N ≥2(e40m−1), then the optimal beam-
i.e., user 4n+2j and 4n+2j−1 (called “clause users”). forming vectors {v∗ } for the optimization problem (8) must
4i
In our construction (7), each (variable, auxiliary variable or be either h =(1,0)T or h =(0,1)T.
a b
clause) user k is associated with a transmitter beamforming It follows from Lemma 3.2 that the optimal beamforming
vector vk,k =1,2,...,4n+2m. vector v4∗i of (8) is either ha or hb. In either case, the sum of
In (7), the n “variable users” c√ommu√nicate interference inverse rates
free. Their channel vectors are ( 0.9, 0.9)T and their
noise power are 1. The 3n “auxiliary variable users”
1 1 1 1 1 1 1
4i−1,4i−2,4i−3,i = 1,2,...,n, do suffer from crosstalk + + + = + +
r r r r log1.9 log101 log2
interference from the “variable user” 4i. That is, the 4i 4i−1 4i−2 4i−3
(cid:44) C
interference channel vectors from the “variable user” 4i are
(1,1)T,(1,0)T,(0,1)T, respectively; the direct link channel is a constant C. Thus, regardless of whether v∗ = h or
4i a
vectors are (1,0)T,(10,0)T and (10,0)T; the self noise h , the first sum in the objective function of (8) remains
b
power at each “auxiliary variable user” is zero. For the unchanged and equals nC. Thus, we only need to consider
“clauseusers”4n+2j and4n+2j−1, j =1,2,...,m,their the second sum in the objective function of (8). Notice that
6
(cid:181) (cid:182) (cid:181) (cid:182)
(cid:88)n 1 1 1 1 (cid:88)m 1 1
min + + + + +
r r r r r r
4i 4i−1 4i−2 4i−3 4n+2j 4n+2j−1
i=1 (cid:179) √ √ (cid:180) j=1
s.t. r =log 1+|( 0.9, 0.9)v |2 ,
4i 4i
(cid:181) (cid:182)
|(1,0)v |2
r =log 1+ 4i−1 ,
4i−1 |(1,1)v |2
(cid:181) 4i (cid:182)
|(10,0)v |2
r =log 1+ 4i−2 ,
4i−2 |(1,0)v |2
(cid:181) 4i (cid:182) (7)
|(10,0)v |2
r =log 1+ 4i−3 ,
4i−3 |(0,1)v |2
(cid:195) 4i (cid:33)
|h†v |2
r =log 1+ 4n+2j ,
4n+2j 1+|h† v |2+|h† v |2
(cid:195) π(j) 4|π(j)| τ(j) 4|τ(j)| (cid:33)
|h†v |2
r =log 1+ 4n+2j−1 ,
4n+2j−1 1+|h† v |2+|h† v |2
−π(j) 4|π(j)| −τ(j) 4|τ(j)|
(cid:107)v (cid:107)2 ≤1, 1≤i≤n, 1≤j ≤m, 1≤k ≤4n+2m.
k
(cid:181) (cid:182) (cid:181) (cid:182)
(cid:88)n 1 1 1 1 (cid:88)m 1 1
min + + + + +
r r r r r r
4i 4i−1 4i−2 4i−3 4n+2j 4n+2j−1
i=1 (cid:161) (cid:162) j=1
s.t. r =log 1+0.9|(1,1)v |2 ,
4i (cid:161) 4i (cid:162)
r =log 1+1/|(1,1)v |2 ,
4i−1 (cid:161) 4i (cid:162)
r4i−2 =log(cid:161)1+100/|(1,0)v4i|2(cid:162), (8)
r =log 1+100/|(0,1)v |2 ,
4i−3 (cid:179) (cid:179) 4i (cid:180)(cid:180)
r =log 1+N/ 1+|h† v |2+|h† v |2 ,
4n+2j (cid:179) (cid:179) π(j) 4|π(j)| τ(j) 4|τ(j)| (cid:180)(cid:180)
r =log 1+N/ 1+|h† v |2+|h† v |2 ,
4n+2j−1 −π(j) 4|π(j)| −τ(j) 4|τ(j)|
(cid:107)v (cid:107)2 ≤1, 1≤i≤n, 1≤j ≤m.
4i

1 1
 + , if c is unanimous,
1 1 log(1+N/3) log(1+N/1) j
+ =
r4n+2j r4n+2j−1  1 + 1 , if c is not unanimous.
log(1+N/2) log(1+N/2) j
the value of each term in the second sum only depends on Finally, given an instance of the MAX-2UNANIMITY
whether clause c is unanimous (see the second equation in problem, we can construct the harmonic mean rate maxi-
j
the next page). Since mization problem (7) in polynomial time. Since the MAX-
2UNANIMITY problem is NP-complete (Lemma 3.1), it fol-
1 1 2
+ < lows that the optimal coordinated beamforming problem (3)
log(1+N/3) log(1+N) log(1+N/2)
with harmonic mean utility is NP-hard.
from Claim 1 in Appendix I, it follows that second sum
of (8) will be smaller if more clauses are satisfied unani- A few remarks are in order. First, it follows from the proof
mously. Therefore, the minimum of (8) is only related to the of Theorem 3.1 that even if the optimal transmit power levels
maximum number of unanimous clauses in the given MAX- are known (i.e., (cid:107)vk(cid:107)2 ≤ Pk is replaced with (cid:107)vk(cid:107)2 = Pk),
2UNANIMITY problem. Specifically, the minimum of (8) is the problem of finding the optimal beamforming directions of
no more than harmonic mean rate maximization problem is still NP-hard.
Second, we have set the noise powers of users 4i−1,4i−
M M 2(m−M)
nC+ + + (9) 2,4i−3,i=1,2,...,n, to zero in (7). These settings simplify
log(1+N/3) log(1+N) log(1+N/2)
the proof and do not reduce any generality. We could have
ifandonlyifthereexistsanappropriatetruthassignmentsuch used small noise power values in the proof (even though
that at least M clauses are made unanimous for the given some extra argument is needed), since there is a positive
MAX-2UNANIMITYproblem.Thus,wehavetransformedthe gap between the global optimal value and the local optimal
problem of MAX-2UNANIMITY problem to the problem of values of (7). Finally , our proof actually implies that there
checking if problem (7) will have an optimal value below the is a positive probability (measure) that a randomly generated
above threshold (9). MISO coordinated beamforming problem under the harmonic
7
meanutilityisNP-hardtosolve.Inparticular,bycontinuity,all (Lemma3.1),itfollowsthattheoptimalcoordinateddownlink
slightlyperturbedversionsoftheconstructedinstance(7)(i.e., beamforming problem with utility H is also NP-hard.
2
channel vectors, noise/transmit powers are slightly changed)
will be equivalent to the MAX-2UNANIMITY problem. This
E. Maximization of Min-Rate Utility
isbecausethereisapositive(andconstant)jumpintheglobal
Let the system utility function be given by H = H . In
optimal value of the constructed example when the optimal 4
this case, the problem can be solved in polynomial time for
value of the corresponding MAX-2UNANIMITY problem
arbitrary L and K [1], [2]. Specifically, letting
increases by one. When channel conditions change slightly,
this one-to-one correspondence between the optimal values of r = min {r },
k
the two problems and the property of the discrete jump in the 1≤k≤K
optimal value of the constructed MISO problem remain valid. the min-rate utility maximization problem becomes
max r
(cid:195) (cid:33)
D. Maximization of Proportional Fairness Utility |h† v |2
Like the harmonic mean utility, we have the following s.t. r ≤log 1+ σ2+(cid:80)kk k|h† v |2 , (12)
hardness result. k j(cid:54)=k jk j
Theorem 3.2 (Proportional Fairness Utility): For the pro-
(cid:195) (cid:33)1/K (cid:107)vk(cid:107)2 ≤Pk, 1≤k ≤K.
(cid:89)K
portional fairness utility H = r , the optimal co- Given a r ≥ 0, we can efficiently check if there exists
2 k
k=1 vk,k = 1,2,...,K, such that the constraints in (12) are
ordinateddownlinkbeamformingproblemcanbetransformed
satisfied. This feasibility problem is a second order cone
into a convex optimization problem when L = 1, but is NP-
programming, which can be solved efficiently using interior-
hard when L≥2.
point methods. The following theorem is a generalization of
Proof: ThefirstpartofTheorem3.2isprovedin[1].For
the result of [2], which deals with the single-cell case.
thesecondpart,theargumentissimilartothatofTheorem3.1.
Theorem 3.3 (Min-Rate Utility): For the min-rate utility,
We only give a proof outline below.
the optimal coordinated downlink beamforming problem can
First,wehavethefollowinglemmawhoseproofisprovided
be solved in polynomial time with arbitrary K and L.
in Appendix II. (cid:181) (cid:182) Proof: We give the following bisection algorithm for
1
Lemma 3.3: The function f(x) = loglog 1+ solving (12).
σ2+x
is strictly convex in x≥0 for any σ.
Second, given any MAX-2UNANIMITY problem, an in- A Polynomial Time Algorithm for Min-Rate
stance(10)oftheoptimalcoordinateddownlinkbeamforming Utility Maximization
problem (3) with utility logH2 (equivalent to proportional Step 1. Initialization: Choose r(cid:96) and ru such that the
fairness utility maximization) and 3n + 2m users is con- optimal ropt lies in [r(cid:96),ru] and a tolerance (cid:178).
structed as follows. Notice that each global optimal solution Step 2. If ru−r(cid:96) ≤(cid:178), stop, else go to Step 3.
of (10) must have v3∗i−1 = v3∗i−2 = (1,0)T, i = 1,2,...,n, Step 3. Let rmid = (r(cid:96)+ru)/2 and solve an SOCP
andv∗ =v∗ =(1,0)T, j =1,2,...,m.Moreover, problem to check the feasibility problem of (12)
3n+2j 3n+2j−1
we consider the following parametric optimization problem with r = rmid. If feasible, let r(cid:96) = rmid, else set
(similar to (19) in the harmonic mean case): ru =rmid and go to Step 2.
max logr +logr +logr According to standard analysis of path-following interior-
3 2 1
(cid:179) √ √ (cid:180) point methods, Step 3 can be finished in O(K3.5L3.5) time.
s.t. r =log 1+|( 0.1, 0.1)v |2 , With regards to initial choices for r and r , we can let v¯ =
3 3 √ (cid:96) u k
(cid:161) (cid:162) (11)
r =log 1+1/(σ2+|(1,0)v |2) , hkk Pk/(cid:107)hkk(cid:107) (space matched beamformer) and
2 (cid:161) 3 (cid:162) (cid:195) (cid:33)
r1 =log 1+1/(σ2+|(0,1)v3|2) , |h† v¯ |2
(cid:107)v3(cid:107)=t, r(cid:96) =mkinlog 1+ σ2+(cid:80)kk k|h† v¯ |2 ,
k j(cid:54)=k jk j
where σ > 0 is a constant and t is a parameter. The global (cid:195) (cid:33)
maxima of (11) should be (t,0)T and (0,t)T when σ is |h† v¯ |2
r =minlog 1+ kk k .
small.Furthermore,theoptimumvalueof(11)isanincreasing u k σ2
k
function with respect to t∈[0,1]. Using an argument similar
to that of the harmonic mean case for (8), each globally It takes log2((ru−r(cid:96))/(cid:178)) iterations to reach tolerance
optimal beamforming solution v3∗i of (10) should be either (cid:178). Thus, a total of O(K3.5L3.5log2((ru−r(cid:96))/(cid:178))) arithmetic
h or h when N ≥3(e6m−1). When restricted to solutions operations are needed in the worst case. Since
a b
of the form v∗ = h or h , the maximum of (10) is only
3i a b r −r ≤log(1+P (cid:107)h (cid:107)2/σ2)≤P (cid:107)h (cid:107)2/σ2,
linearlyrelatedtothemaximumnumberofunanimousclauses u (cid:96) k kk k k kk k
inthegivenMAX-2UNANIMITYproblem.Thus,maximizing the second equation in the next page holds true, which is a
the number of unanimous clauses is the same as solving polynomial in the length of input data (all the channel vectors
(10). Since MAX-2UNANIMITY problem is NP-complete h , noise power σ2 and power upper bound P ). Thus, the
kj k k