Polynomial-complexity, Low-delay Scheduling for SCFDMA-based Wireless Uplink Networks (Technical Report)

0.23 MB·English

by  Shreeshankar Bodas

Polynomial-complexity, Low-delay Scheduling for
SCFDMA-based Wireless Uplink Networks
(Technical Report)
Shreeshankar Bodas, Bilal Sadiq
Qualcomm, Inc., Bridgewater, NJ 08807
3 Abstract—Uplink scheduling/resource allocation under the frequencysub-bandsf ,f ,f .Then(x,f ),(x,f ),(y,f )is
1 2 3 1 2 3
1 single-carrier FDMA constraint is investigated, taking into ac- a valid SCFDMA allocation, while (x,f ),(x,f ),(y,f ) is
0 countthequeuingdynamicsatthetransmitters.Underthesingle- 1 3 2
not. We refer to this additionalconstraint as the single-carrier
2 carrierconstraint,theproblemofMaxWeightscheduling,aswell
n as that of determining if a given number of packets can be constraint. The main reason for the choice of SCFDMA for
a served from all theusers, are shown to beNP-complete. Finally, the uplink is that it results in a lower PAPR (peak-to-average
J amatching-basedschedulingalgorithmispresentedthatrequires power ratio) than OFDMA.
only a polynomial number of computations per timeslot, and in In this paper, we show that the single-carrier constraint
7
the case of a system with large bandwidth and user population,
alone is enough to make certain scheduling problems hard
provablyprovidesagood delay(small-queue)performance, even
] (formally,NP-complete).TheclassicMaxWeightscheduler[2]
I under the single-carrier constraint.
N is throughput-optimalfor the uplink network under very mild
In summary, the resultsin first part of the papersupport the
assumptions on the arrival and channel processes (see [3]),
. recent push to remove SCFDMA from the Standards, whereas
s
but selecting a weight-maximizing schedule is NP-complete
c those in the second part present a way of working around the
[ (Theorem2).Anothernatural,myopic,“greedy”schedulerfor
single-carrier constraint if it remains in the Standards.

theschedulingproblemdescribedinSectionIIIoperatesasfol-
1 IndexTerms—Uplinkscheduling,single-carrierFDMA,Batch-
lows: given a queue-lengthvector and a matrix of the rates at
v and-allocate
9 which the frequency sub-bands can serve the individual user-
7 I. INTRODUCTION queues, does there exist an allocation that serves x packets
i
2 In the recent years, we have witnessed an explosion in the fromthe user-queueQ ? Thisschedulerisinterestingbecause
i
1
numbersandcapabilitiesofhand-heldwirelesscommunication by choosing appropriate values of x s in every scheduling
. i
1 devices, and consequently their data consumption. Real-time, period, the per-user queues can be kept small. For example,
0 i.e., delay-constrained data traffic (voice/video/gaming/...) the values of x can be chosen to equalize the queue-lengths
i
3
constitutes a significant fraction of the overall over-the-air afterservice.Forthedownlinkschedulingproblem,inabsence
1
data demand. The demand for high-quality data, and in large of the single-carrier constraint, this scheduler is shown to
:
v
quantities, is ever-growing, but the wireless resources are not have good delay properties [4]; but under the single-carrier
i
X growing nearly as fast. It is therefore important to design constraint, implementing it requires solving an NP-complete
r efficientmethodsofsharingtheresourcesacrossmultipleusers problem (Theorem 1).
a in order to guarantee a good quality of service. In this paper, In the light of these negativeresults, we focuson a simple,
we focus on the problem of resource allocation on the uplink i.i.d.arrivalandchannelmodel,anddesignanalgorithmcalled
(user to base-station) of wireless networks. Batch-and-allocate (BA) scheduler as the main contribution
The 3GPP LTE (Long-Term Evolution) standard has cho- of this paper. This scheduler results in a good delay (small-
sen the single-carrier frequency division multiple access queue)performanceforthesystem,andcanbeimplementedin
(SCFDMA) technology as the uplink multiple access tech- polynomial number of computations per timeslot, even under
nology [1]. The SCFDMA can be thought of as a special the single-carrier constraint.
case of the orthogonal frequency division multiple access Thequalitativemessagesfromthepaperare:(i)Thesingle-
(OFDMA) technology used for the downlink of 3GPP LTE. carrierconstraint,whileattractivefromapoweramplifierpoint
In OFDMA, the available bandwidth at the base-station is of view, severely restricts the class of possible scheduling
partitioned into a number of orthogonalfrequency sub-bands, policies. There has been a recent push to remove it from
and a given user can be allocated any subset of the frequency the standards (e.g., clustered SCFDMA [5], [6]) and this
sub-bandsforhis/herdownlinktrafficundertheconditionthat paper can be seen as an argument in its favor. (ii) Although
a given frequency sub-band can be allocated at most one the uplink scheduling problem is intractable under the single
user. In SCFDMA, there is an additional constraint that a carrier constraint, we can guarantee a good quality of service
given user can be allocated only consecutive frequency sub- for “regular” arrival and channel processes, if the system has
bands.Forexample,considerasystemwith2usersx,y and3 a large number of users and proportionally large bandwidth.
2
M i
II. RELATEDWORK 3) There exists α∈(0,1) such that p =1−α.
i
(cid:24)K(cid:25)
Scheduling and resource allocation for the wireless uplink iP=0
We make the assumption p >0 for all 0≤i≤M only to
network is a well-investigated problem. Researchers have i
avoidtrivialities;ourresultsorprooftechniquesareinnoway
studied this problem from the point of view of maximizing
dependent upon this assumption. We also assume that M >
a system-wide utility function [7], [8], [9], orderwise delay-
K, since otherwise, allocating just one server (with highest
optimal scheduling [10], successive interference cancellation
supported rate K) is enough to serve all the new arrivals to
to allow for simultaneous transmissions from users [11], and
a queue in a given timeslot, and the single-carrier constraint
so on. A majority of the previouswork on the problem either
in the problem can be easily circumvented by the matching-
does not consider the single-carrier constraint, or allows for
based algorithmsfor the downlink, such as those in [12]. Our
fractional server (i.e., frequency sub-band) allocation, thus
objectiveistodefineaservicepolicy,quantifiedbytherandom
circumventing the inherently discrete nature of the allocation
variables Y (t) ∈ {0,1} for i,j ∈ [n] and for all t, where
problem. In wireless uplink systems where frequency sub- ij
Y (t)=1 if the server S serves the queue Q in timeslot t,
bands are grouped together, the fractional server allocation ij j i
and 0 otherwise. The random variables Y (t) are allowed to
is a reasonable assumption. A recurring theme in the prior ij
dependupontheentirepastofthesystem andthearrivalsand
workistoinitiallyignorethesingle-carrierconstraint,comeup
channelrealizationsinthe(current)timeslott,butarerequired
withanallocationofthefrequencysub-bandstotheusersthat
to satisfy the following conditions:
optimizesacertainobjective,andthenuseheuristicstomodify
1) n Y (t)≤1 for all i,j,t.
thatallocationtoincorporatethesingle-carrierconstraint.This i=1 ij
approach usually leads to a loss of performance. In contrast, 2) PIf Yir(t) = Yis(t) = 1 for some 1 ≤ r < s ≤ n, then
Y (t)=1 for all r <j <s, all i∈[n].
inthispaper,westrictly adhereto thesingle-carrierconstraint ij
even in the algorithm design part, and do not perform any The first condition implies that a given server can serve at
fractional server allocations. We present an algorithm that mostonequeuein anytimeslot. Thesecondconditionmodels
is designed with the single-carrier constraint in mind, and the single-carrier constraint. The queues evolve according to
which yieldsa goodsmall-bufferperformanceundera variety n
+
of changes to the basic system model. To the best of our Q (t)= Q (t−1)+A (t)− X (t)Y (t) . (1)
i i i ij ij
knowledge,thisisthefirstcharacterizationofthesmall-queue (cid:16) Xj=1 (cid:17)
performance of the uplink network in the large- system limit. Our objective is to define a scheduling policy that, for every
III. SYSTEMMODEL integer b≥0, results in a strictly positive value of
We consider a discrete-time queuing system with −1
I(b):=liminf logP max Q (t)>b ,
n queues and n servers, as shown in Figure 1. n→∞ n (cid:18)1≤i≤n i (cid:19)
Here the n queues where P(·) refers to the stationary distribution of the queue-
rqeuperueessenatt ththeen puapcliknekt A1(t) X11(t) S1 length process. The function I(·) is called the rate-function
transmitters, and the n Q1 in large deviations theory [13]. In order to guarantee a good
servers represent the A2(t) S2 sthmeal“lo-qvueerfluoewp”erpforormbaabnicliet,y,oiu.er.,trtuheeopbrjoebcatibvielitiys otof tmheiniemveiznet
nfreqouertnhcoygonsaulb-buapnlidnsk. Q2 X2n(t) {max1≤i≤nQi(t) > b}. In real systems with a large number
of users and proportionallylarge bandwidth,the rate-function
The queues can store
maximization is a useful and reasonable surrogate for this
any number of packets An(t) Xnn(t) Sn objective. If I(b) > 0, then the probability of the overflow
until they are served,
so that there are Qn event rapidly diminishes to 0 with the system-size. Hence in
thispaper,wefocusonpoliciesthatresultinastrictlypositive
no dropped packets. Fig.1. SystemModel
Table I summarizes the
notation used throughout this paper. Q Thesetofnqueues{Q1,...,Qn}
Arrivalandchannelprocesses:Weassumethatthearrivals S Thesetofnservers{S1,...,Sn}
Qi(t) ThelengthofQi attheendoftimeslott
to the queues and the channel realizations are i.i.d across Qˆ(t) max{Qi(t):1≤i≤n}
queues, servers, and timeslots. More precisely, Xij(t) ThenumberofpacketsthattheserverSj canpotentiallyserve
1) The number of arrivals to Q at the beginning of fromQi intimeslott
i Ai(t) Thenumberofarrivals toQi atthebeginning oftimeslott
timeslottarei.i.d.acrosstimeslotsandqueues,andobey [n] Theset{i:1≤i≤n}
P(A (t)=m)=p for 0≤m≤M, p >0 for all i, a+ max(a,0)
i m i
and M p =1. |A| Thecardinality ofsetA
i=0 i R+ Theset[0,∞)ofnonnegative realnumbers
2) ThePnumber of packets that the server Sj can serve ∆k Theprobability simplexinRk
fromQ intimeslottarei.i.dacrossqueues,serversand
i TABLEI
timeslots,andobeyP(Xij(t)=k)=qkfor0≤k ≤K, NOTATION
q ≥0 for all i, and K q =1.
i i=0 i
P
3
value of the rate-function. The assumption 3 is a necessary Definition 2 (MaxWeight problem (PM)): Considerasetof
condition for the rate function to be nonzero, even without queues [Q ,...,Q ] with lengths [L ,...,L ], and a set of
1 k 1 k
the single-carrier constraint [14]. Our main contribution is servers {S ,...,S }, where the server S can serve X
1 m j ij
an algorithm that yields a positive value of the rate-function packets from the queue Q . A finite integer W ≥ 0 is given.
i
under this assumption. LetY =1iftheserverS isallocatedtoQ ,and0otherwise.
ij j i
Note: In the rest of the paper, for simplifying notation, Determine if, under the single-carrier allocation constraint,
we make statements like “allocate n/2 servers to a queue.” there exists an allocation of the servers to the queues with
What we actually mean is the integer part (or floor) of k m L X Y ≥W. ⋄.
i=1 j=1 i ij ij
the corresponding fraction. We never make fractional server PIn Pthe (PM) problem, we refer to the quantity
allocations. We are interested in the large deviations results k m L X Y as the weight of the allocation.
i=1 j=1 i ij ij
(n large). In this regime, the rounding has no effect on the PTheoPrem 2: The MaxWeight problem (PM) is NP-
analysis. We do not discuss this issue further in this paper. complete.
Proof: Please see Appendix B.
IV. COMPUTATIONALHARDNESS
V. THE BATCH-AND-ALLOCATEALGORITHM
In this section, we establish that in the presence of the
ThecomputationalhardnessresultsinSectionIVimplythat
single-carrier constraint, certain (otherwise simple and in-
unless P=NP, there does not exist a computationally efficient
teresting) scheduling policies are NP-complete. We use a
scheduling algorithm that guarantees throughput optimality
construction almost identical to the one from [15]. In [15],
under general arrival and channel conditions. On the other
the authors establish the NP-hardness of the single-carrier
hand, the user-experienced quality of service is crucially
scheduling problem in the context of proportionally fair (PF)
dependent upon a good delay performance. Hence we focus
scheduling. Their reduction can be modified to suit in our
on designing a computationally tractable algorithm that gives
case. The reasons that we provide a detailed account here, as
a good delay performance under a restricted class of arrival
opposed to merely citing their result, are: (i) their result is
and channel processes, namely, i.i.d. arrivals and channels
not directly applicable in our case: it is concerned with PF
with a bounded support, as specified in Section III. We call
scheduling,and (ii) their constructionis crypticto the authors
thisalgorithmtheBatch-and-allocate(BA)algorithm.We first
of this paper, with a number of key proof details missing.
define the Selective-allocate (SA) algorithm that is used as a
In the multi-queuemulti-serversetup describedhere, a nat-
“black-box” in the BA algorithm.
ural, myopic way to minimize the probability that the longest
queueexceedsagivenconstantbistoselect,ineverytimeslot, Selective-allocate (SA) algorithm:
that allocation of the servers to the queuesthat minimizes the Input:
maximumqueue-length.Thisrequiresansweringthequestion: 1) An integer k ≥1.
can a queue Qi be allocated at least wi units of service, 2) A bipartite graph G(U ∪ V,E) with |V| ≥ k|U|. Let
i ∈ [n]? A simpler question as defined in Definition 1 is: U ={u ,...,u } and V ={v ,...,v }.
1 x 1 y
can a total of W packets be drained from the queues? Our Steps:
objective is to show that even this simpler problem is NP-
1) Partition the nodesin the set {v ,...,v } into disjoint
complete under the single-carrier constraint. 1 kx
subsetsV ,...,V suchthatV ={v ,...,v }.
Definition 1 (Packet-draining problem (PD)): Consider 1 x i (i−1)k+1 ik
Let V′ :={V ,...,V }.
1 x
a queue-length vector [Q ,...,Q ] and a set of servers
1 k 2) Construct a new graph H(U ∪V′,E′) where an edge
{S ,...,S }, where the server S can serve X packets
1 m j ij (u ,V ) is present in E′ if the node u is connected to
from the queue Q . A finite integer W ≥ 0 is given. i j i
i every node in the set V in the original graph G.
j
Determine if, under the single-carrier allocation constraint,
3) Find a largest cardinality matching M in the graph H,
there exists an allocation of the servers to the queues that
breaking ties arbitrarily.
serves a total of at least W packets. ⋄
Output: The matching M. ⋄
Theorem 1: The packet-draining problem (PD) is NP-
complete. The SA algorithm groups the nodes in the set V into sets
Proof: Please see Appendix A. of size k each, and matches each such group Vi to that node
We now focus on the problem of MaxWeight scheduling uj ∈ U that is connected to each node in the group Vi. One
under the single-carrier constraint. This classic scheduling al- can think of each node in the set U as a queue, each node in
gorithmwasintroducedin [2] andisknownto bethroughput- the set V as a server, and the presence of an edge signifies
optimal (i.e., makes the queue-length Markov chain positive that the server can serve the given queue. An example of the
recurrent if there is any other algorithm that can do so) in a SA algorithm for the case k = 2 is shown in Figure 2. Here
variety of situations, including under the single-carrier con- thesolidedgesinthegraphH representthematchingM.We
straint, even under more general (e.g., correlated) arrival and write M=SA(k,G) for the output of the SA algorithm.
channelprocesses[3].Butasisestablishednext,implementing Batch-and-allocate (BA) algorithm:
it is computationally intractable unless P=NP. Input:
4
v1 Definen′′ =n′−(dc+1+dc+2+···+dm0). Definethe
set of servers T satisfying
u1 v2 u1 V1 c
n′
v |T |=(c+1)n′′+c(d −n′′)+ .
3 c c
2(m −a+1)
0
u2 v4 u2 V2 EnsurethattheserversinT areconsecutivelynumbered
i
v5 for all i∈{a,a+1,...,m0}.
Case a=m +1: Define the set of servers T =S,
u3 v6 u3 V3 the set of all0the servers. m0
v 5) Allocating servers to queues:
7 The constructed graph H
Case a≤m :
The given graph G with the matching M (solid lines) 0
a) For everyi∈{c+1,c+2,...,m }, let G be the
0 i
Fig.2. SAalgorithm -example restriction of the graph G(Q∪S,E) where the set
ofqueuesis restrictedto indicesin D , andthe set
i
of servers to T . Compute M = SA(i+1,G ).
i i i
1) The vector of queue-lengths, Q1(t−1),...,Qn(t−1). For every i ∈ {a,a+1,...,c−1}, let Gi be the
2) The vector of arrivals, A1(t),...,An(t). restriction of the graph G(Q∪S,E) where the set
3) The channel realizations, Xij(t) for i,j ∈[n]. of queues is restricted to indices in Di, and the
Steps: set of servers to Ti. Compute Mi =SA(i,Gi). If
i=0, compute M =SA(1,G ).
1) Calculate Qˆ(t−1):= max Q (t−1). If X (t) < K 0 0
i ij b) Let D′ ⊆ D be any subset satisfying |D′| = n′′,
1≤i≤n c c c
for some pair (i,j), then set Xij(t) = 0 for that pair and Dc′′ = Dc \ Dc′. Let Tc′ ⊆ Tc be a subset
and use this value of Xij(t) throughout the rest of the satisfying|Tc′|=(c+1)n′′+n′/(4(m0−a+1))and
algorithm. T′′ =T \T′.EnsurethattheserversinT′,T′′ are
c c c c c
2) For 1≤r ≤m0, define consecutively numbered.Let G′c (resp. G′c′) be the
restrictionofGwherethesetofqueuesisrestricted
Dr := {i∈[n]:Qˆ(t−1)+(r−1)K+1≤ to indices in D′ (resp. G′′), and the set of servers
c c
Q (t−1)+A (t)≤Qˆ(t−1)+rK} to T′ (resp. T′′). Compute M′ = SA(c+1,G′)
i i c c c c
and M′′ =SA(c,G′′). Let M =M′ ∪M′′.
c c c c c
to be the set of queue-indices i such that the queue Qi For a ≤ i ≤ m0, allocate the servers to the queues as
needstobeallocatedexactlyrserverstoensureQi(t)≤ dictated by Mi : if (Qx,Wy) ∈ Mi for some queue
Qˆ(t−1). Let Q with x∈D and a set of servers W , then allocate
x i y
the servers in W to Q , etc., and accordingly define
D ={i∈[n]:Q (t−1)+A (t)=Qˆ(t−1)} y x
0 i i the allocation random variables Y (t).
ij
Case a = m + 1 : Let G be the restriction of
be the set of queue-indices i such that after arrivals, 0 m0
the graph G(Q ∪ S,E) where the set of queues is
the queue-lengthof Q is the maximumqueue-lengthat
i restricted to indices in D , and the set of servers to
the end of the previous timeslot. We allocate servers to m0
T =S.ComputeM =SA(n/m ,G ).Allocate
only some of the queues in the sets D ,0 ≤ i ≤ m . m0 m0 0 m0
i 0 the servers to the queues as dictated by M .
Let d =|D |. m0
i i 6) Update the queue-lengths to account for service as per
3) Let 0 ≤ a ≤ m + 1 be the smallest integer such
0 Equation (1).
that m0 id ≤ n. Here a = m + 1 implies that
i=a i 0 Output:
the pPrevious summation is vacuous (equal to 0), i.e.,
4) mCa0sdema0≤>mn.0L:eLtent′c:=∈{na−,aP+1mi=,0a..i.d,im. 0}bethelargest 21)) TThhee fialnloalcaqtuioenuse,-lYeinjg(tth)sf,oQrii(,tj).∈[n].
integersuchthatdc+dc+1+···+dm0 ≥n′/2.Foreach Informally, the algorithm tries to reduce the queue-length
i ∈ {c+1,c+2,...,m0} define the set of servers Ti of each of the queues after arrivals, to the maximum queue-
satisfying length before arrivals. In order to limit the number of search
possibilities, the algorithm only considers channels that have
n′
|T |=(i+1)d + . themaximumrate= K.Thealgorithmgroupsthequeuesinto
i i
2(m −a+1)
0 disjoint sets such that the queues in each group require the
same number of servers to attain a queue-length less that
For each i ∈ {a,a+ 1,...,c − 1}, define the set of
or equal to the maximum queue-length at the end of the
servers T satisfying
i
previoustimeslot. It then determinesthe numberof serversto
n′ allocatetothequeuesineachgroup,whichissomewhatmore
|T |=id + .
i i 2(m −a+1) than the bare-minimum required number of servers to reduce
0
5
each queue-length to the desired value. It assigns subsets of 1) In a given timeslot, it increases with probability that is
consecutively-numberedservers to each group of queues. The exponentiallysmallin n, andif itincreases, the amount
SA algorithm is used to make assignment decisions within of increase is no more than M, which is a constant
each set of queues and the respective group of servers. independent of n.
Some features of the algorithm are: (i) This is a real-time 2) Over a constant number of timeslots, it decreases with
algorithm; it does not need to know the statistical system at least a constant (=1/2) probability.
parameters(e.g.,theprobabilities)inordertobeimplemented. Thus, it is reasonableto expectthat the stationary distribution
(ii)Thisalgorithmresultsinastrictlypositivevalueoftherate ofthemaximumqueue-lengthisstronglyconcentratednear0,
function(Theorem3).(iii)Thisalgorithmcanbeimplemented which is formally established next.
in polynomial time (Theorem 4).
Theorem 3: Under the BA algorithm, the stationary distri-
Inordertolimitcomplexity,thealgorithmtreatsthesmaller
bution of the maximum queue-length in the system obeys
channel-rates as 0. In spite of this “wastage,” the algorithm
gives a good small-queue performance (Theorem 3). So the liminf −1logP max Q (t)>b
i
messageis:forgooddelayperformance,evenunderthesingle- n→∞ n (cid:18)1≤i≤n (cid:19)
carrier constraint, it is enough to focus on the highest-rate b+1 α 1
≥ min τ(ǫ), log >0.
channels alone. We first establish an important property of M (cid:18) 4m (m +1) 1−qm0(cid:19)
0 0 K
the SA algorithm.
Lemma 1: Consider a graph G(U ∪V,E) with |V| = r ≥
Proof: Please see Appendix F.
k|U|. Suppose that for any pair of nodes u ∈ U,v ∈ V, the
ThustheproposedBAalgorithmresultsinastrictlypositive
edge (u,v) is present in E with probability q, independently
value of the rate function. Next we analyze its complexity.
of all other random variables. Let M = SA(k,G). Then for
Theorem 4: The BA algorithm can be implemented in
r large enough, P(|M|<|U|)≤3⌊r/k⌋(1−qk)⌊r/k⌋.
O(n2.5) computations per timeslot.
Proof: Please see Appendix C.
Proof: Please see Appendix G.
Note that the RHS of the above expression tends to 0 as
We conclude this section by showing that there is a finite
r →∞forafixedk.Nowourobjectiveistoshowthatunder
upper bound on the rate-function under any algorithm. The
the BA algorithm, in every timeslot, the probability that the
purpose is to establish that in the multi-queue multi-server
maximum queue-lengthin the system increases is “small” for
setup considered in this paper, the probability of the overflow
n large. Define m :=⌈M/K⌉.
0 event decays like e−n at best; not like e−n2 or e−nlogn, etc.
Lemma 2: Fix any ǫ ∈ (0,α/(2Mm )). Define the set B
0 ǫ Theorem 5: Fix θ ∈ (0,M/K − 1). Define C = {x ∈
of probability measures “near” the distribution of the arrival θ
∆ : M ix ≥K(1+θ)} and
process, as M+1 i=0 i
P
M
Bǫ :={[x0,...,xM]∈∆M+1 : |xi−pi|<ǫ∀ 0≤i≤M}. ξ(θ)= inf y logyi.
i
M y y∈∆M+1\CθXi=0 pi
For ǫ∈R , define τ(ǫ):= inf y log i.
+ y∈∆M+1\BǫXi=0 i pi Thenunderanyalgorithmforallocatingserverstothequeues,
Here τ : R+ → R+∪{∞}. Fix any ρ ∈ (0,1). Then under liminf −1logP max Q (t)>b ≤ b+1 ξ(θ).
the BA algorithm, for n large enough, for any timeslot t, n→∞ n (cid:18)1≤i≤n i (cid:19) (cid:24) θ (cid:25)
P Qˆ(t+1)>Qˆ(t)
(cid:16) (cid:17) Proof: Please see Appendix H.
≤ e−nρτ(ǫ)+3m0(cid:22)4m0(nmα0+1)(cid:23)(1−qKm0)j4m0(nmα0+1)k. forTthhuesrtahteerefuinscatitomnousntdaercothnestBanAt-faalcgtoorritghamp.from optimality
Proof: Please see Appendix D. VI. EXTENSIONS
We now show that for n large, the probability that in a
The BA algorithm presented in Section V can be easily
constant number of timeslots, the maximum queue-length in
extended to a variety of cases of interest.
the system decreases is at least 1/2.
(i) Unequal number of queues and servers: This case
Lemma 3: UndertheBAalgorithm,fornlarge,thereexists is of practical importance, because in typical uplink wireless
a constant integer k0 such that systems,thenumberofactiveusersissmallerthanthenumber
of orthogonal frequency sub-bands. The BA algorithm can
1
P Qˆ(t+k )<Qˆ(t)−1 Qˆ(t)>0 ≥ . be easily modified to utilize this “extra” service capacity, as
0
(cid:16) (cid:12) (cid:17) 2 follows. Suppose we have a system with n users and rn
Further, k = 4 is a valid choice(cid:12)(cid:12). frequency sub-bands (servers) for some r ≥ 1. We refer to r
0 α astheover-provisionfactor.Inthestep4oftheBAalgorithm,
Proof: Ple(cid:6)ase(cid:7) see Appendix E. we give r times as many servers to each group of queues D
i
AsaresultofLemmas2and3,themaximumqueue-length compared to the case of n queues and n servers. As a result,
in the system has the following behavior: the rate-function lower bound of Theorem 3 scales up by a
6
factor of r. Formally, under the BA algorithm, the stationary 100 50users,effectiveload=0.62
distributionofthemaximumqueue-lengthinthesystemobeys
10−1
−1
liminf logP max Q (t)>b
n→∞ n (cid:18)1≤i≤n i (cid:19) b)10−2
r(b+1) α 1 ≥
≥ M min(cid:18)τ(ǫ),4m0(m0+1)log1−qKm0(cid:19)>0. Qmax10−3
We omit the proof details. Pr(10−4 BA,q=0.7,r=1.5
BA,q=0.8,r=1.5
(ii) Different priorities to queues: The BA algorithm can BA,q=0.9,r=1.5
be used in the case where the queueshave differentpriorities. 10−5 BBAA,,qq==00..78,,rr==22
BA,q=0.9,r=2
In this set up, we are interested in minimizing the probability 10−60 WithoutSCconstraint 5 10 15
of the event {maxi∈[n]aiQi(t)>b} where 0<amin ≤ai ≤ Maximumqueue-length(b)
1 are given numbers. The BA algorithm instead operates on
the “effective” queue-lengths, namely, a Q (t), to yield rate- Fig.3. Performance oftheBAalgorithm
i i
function results similar to Theorem 3.
VII. SIMULATION RESULTS VIII. CONCLUSIONS
We now analyze the performance of the proposed Batch- We considered the problem of user-scheduling in the wire-
and-allocate (BA) algorithm through simulations. The goals less uplink networks. The distinguishing feature that makes
arethreefold:(i)Therate-functionresultsfortheBAalgorithm this problem harder than the OFDM downlink scheduling
areasymptotic,i.e.,asthenumberofusers(n)andthenumber problem is the presence of the single-carrier constraint. We
ofsub-bandstendtoinfinity.Wewanttounderstandhowlarge showedthatunderthesingle-carrierconstraint,theMaxWeight
nneedstobe,togetagoodsmall-bufferperformance.(ii)We problem and the packet-draining problem are NP-complete.
want to understand the (good) impact of having more fre- We presented the Batch-and-allocate algorithm that has poly-
quencysub-bandsthanthenumberofusers,whichistypically nomial complexity per timeslot, and a good small-queue per-
the case in today’s wireless uplink systems. (iii) We want to formanceforaclassofboundedarrivalandchannelprocesses.
comparetheBAalgorithm’sperformancetoanOFDMA-based The algorithm is robust to changes in the system-model. The
greedy algorithm in [16] that operates in the absence of the results were validated through analysis and simulations.
single-carrier constraint, in order to quantify the performance
ACKNOWLEDGMENTS
lossduetothesingle-carrierconstraint.Inthesimulations,we
run the OFDMA-basedalgorithmwith as manyserversas the Theauthorswouldlike to thankNileshKhudeandSaurabh
users (i.e., over-provisionfactor, r =1). Tavildar for helpful discussions.
Forsimulationpurpose,wearbitrarilyassumeanarrivalpro-
REFERENCES
cessdistributionoftheform(x+1)e−x onaboundedsupport
[1] [Online]. Available: http://www.3gpp.org/lte
{0,1,...,5}, normalized. We assume that the channel-rates
[2] L. Tassiulas and A. Ephremides, “Stability properties of constrained
areeither0or2packetspertimeslot.ThusM =5andK =2 queueing systems and scheduling policies for maximum throughput in
in the paper’snotation.We refer to the quantity M p i multihop radio networks,” IEEE Trans. Automat. Contr., vol. 4, pp.
i=1 i K 1936–1948, December1992.
as the effective load. In our case, the effective Pload is a(cid:6)bou(cid:7)t [3] A.Eryilmaz,R.Srikant,andJ.Perkins,“Stableschedulingpolicies for
62%. We vary the channel ON probability, q, from 0.7 to fadingwirelesschannels,”IEEE/ACMTrans.Network.,vol.13,pp.411–
0.9, and plot the empirical probability of buffer overflow v/s 424,April2005.
[4] S. R. Bodas, “High-performance Scheduling Algorithms for Wireless
buffer-size, averaged over 106 timeslots.
Networks,” Ph.D.dissertation, TheUniversity ofTexas atAustin,Dec.
The results are presented in Figure 3. As we can see, the 2010.
presence of the single-carrierconstraintsignificantlydegrades [5] [Online].Available:http://www.3gpp.org/ftp/Specs/archive/36 series/36.912/
[6] Renesas Mobile Europe Ltd, “LTE Rel-
thesmall-bufferperformance:thebufferoverflowprobabilities
12 and Beyond,” 2012. [Online]. Available:
in the absence of the single carrier constraint are substan- http://www.3gpp.org/ftp/workshop/2012-06-11 12 RAN REL12/Docs/RWS-120022.zip
tially lower than otherwise. We see that the buffer overflow [7] J.Huang,V.G.Subramanian,R.Agrawal,andR.Berry,“JointSchedul-
ingandResource Allocation inUplink OFDMSystemsforBroadband
probabilitydecreaseswithincreasingsystem-size,asexpected:
WirelessAccessNetworks,”IEEEJ.Sel.AreasCommun.,vol.27,no.2,
the overflow probability is exponentially small in the system- pp.226–234, Feb.2009.
size. We also see that changing the over-provisioning factor [8] B. Rengarajan, A. Stolyar, and H. Viswanathan, “Self-organizing Dy-
namicFractionalFrequencyReuseontheUplinkofOFDMASystems,”
from1.5to2providessomeperformanceboost.Thisconfirms
inProc.Conf.onInformationSciencesandSystems(CISS),Mar.2010.
that the BA algorithm can seamlessly utilize more frequency [9] R.MadanandS.Ray, “UplinkResource Allocation forFrequency Se-
sub-bands. Most interestingly, the asymptotic rate-function lectiveChannelsandFractionalPowerControlinLTE,”inInternational
Conference onCommunications (ICC),Jun.2011.
results for the BA algorithm already manifest themselves
[10] M. Neely, “Order Optimal Delay for Opportunistic Scheduling in
to give a good small-buffer performance at n = 50. We Multi-User Wireless Uplinks and Downlinks,” IEEE Transactions on
have seen a comparable performance for the case n = 40. Networking, vol.16,no.5,pp.1188–1199, Oct.2009.
[11] M.MollanooriandM.Ghaderi,“OntheComplexityofWirelessUplink
Thus, the proposed BA algorithm yields a good small-queue
Scheduling with Successive Interference Cancellation,” in Proc. Ann.
performance at realistic system-sizes. AllertonConf.Communication, ControlandComputing, Sep.2011.
7
[12] S.Bodas,S.Shakkottai,L.Ying,andR.Srikant,“SchedulinginMulti- 1) For every x∈[n] and for each server Sj ∈Aℓ,x, define
Channel Wireless Networks: Rate Function Optimality in the Small- X(Q ,S )=1. For every x∈[n] and for each server
ℓ,i j
BufferRegime,”inProc.SIGMETRICS/Performance Conf.,Jun.2009.
S ∈A , define X(Q ,S )=0.
[13] A.DemboandO.Zeitouni, LargeDeviations Techniques andApplica- j r,x ℓ,i j
tions,2nded. Springer-Verlag NewYork,Inc.,1998. 2) Fix x ∈ [2n − 1], x odd. For a server in
[14] S.Bodas,S.Shakkottai,L.Ying,andR.Srikant,“SchedulingforSmall B with index xT + (x − 1)(n + 2) + j, define
x
Delay in Multi-rate Multi-channel Wireless Networks,” in Proc. IEEE
X(Q ,S ) = i + 1 if i = j, and 0
Infocom, Apr.2011. ℓ,i xT+(x−1)(n+2)+j
[15] S.-B. Lee, I. Pefkianakis, A. Meyerson, S. Xu, and S. Lu, otherwise.
“Proportional Fair Frequency-Domain Packet Scheduling for 3GPP 3) Let v ,v ,... be the nodes that have an out-
LTE Uplink,” in Proc. IEEE Infocom, Apr. 2009. [Online]. Available: r,g1 r,g2
going edge to the node v , with g > g >
http://www.cs.ucla.edu/wing/publication/papers/Lee.TR-090001.pdf ℓ,i 1 2
[16] S. Bodas, S. Shakkottai, L. Ying, and R. Srikant, “Low-complexity .... Fix x ∈ [2n − 1], x even, and a server in
SchedulingAlgorithmsforMulti-channelDownlinkWirelessNetworks,” B with index xT + (x − 1)(n + 2) + g . De-
x j
inProc.IEEEInfocom, Mar.2010.
fine X(Q ,S ) = n + 1 − g and
[17] J. Kleinberg and E. Tardos, Algorithm Design. Pearson Education, ℓ,i xT+(x−1)(n+2)+g1 1
2006. X(Qℓ,i,SxT+(x−1)(n+2)+gj)=n+1−(gj−1−gj) for
[18] J. E. Hopcroft and R. M. Karp, “An n5/2 Algorithm for Maximum j > 1. Define X(Q ,S ) = 0 for all
ℓ,i xT+(x−1)(n+2)+j
Matchings in Bipartite Graphs,” SIAM Journal on Computing, vol. 2,
other values of j ∈[n+2].
no.4,pp.225–231,Dec.1973.
Perform the same construction for a queue Q with A
r,i r,x
APPENDIX A replacingA and vice-versain step 1, and the words“even”
ℓ,x
PROOF OF THEOREM1 and “odd” replacing each other in steps 2 and 3. Define the
The problem (PD) clearly belongs to the class NP: a number W =2nT +(2n−1)(n+2).
certificate is an allocation of the servers to the queues that Here the steps 1 and 2 are generic and apply to any graph
serves a total of at least W packets from the queues. In G, while step 3 is dependent upon the graph structure. We
order to show that it is NP-complete, we use a reduction to now establish some basic properties of the possible server
the Hamiltonian path problem, which is NP-complete ([17], allocations for the above construction under the single carrier
Ch. 8). The Hamiltonian path problem asks: given a directed constraint.
graphG, doesitcontaina (directed)path,starting andending
Property 1: If a queue, say Q is not allocated any server
ℓ,i
at any node, that visits every node exactly once?
from the set ∪ A , then the maximum total number of
x∈[n] ℓ,x
Reduction:
packets that can be served from the queues is less than 2nT.
Given a directed graph G(V,E) with |V|=n, we construct a
Proof: A server in a set A can serve at most 1 packet.
ℓ,x
directed bipartite graph G′(V ∪V ,E′) as follows: for every
ℓ r Further,thatpacketmustbe froma queuelabeledQℓ,x′. Thus
node v ∈ V, define two nodes v ∈ V and v ∈ V .
i ℓ,i ℓ r,i r the maximum number of packets served by all the servers in
Connect v to v via a directed edge. If a directed edge
ℓ,i r,i the set A is T. Suppose that for all x∈[n], the servers in
ℓ,x
(v ,v ) exists in E, then introduce a directed edge from v
i j r,i thesetsA serve1packeteach.Sincebyhypothesisatmost
ℓ,x
to v . That is, all the incoming edges to v are connected
ℓ,j i n−1 queues in {Q ,...,Q } can be allocated a server
ℓ,1 ℓ,n
to v and all the outgoing edges are connected to v . One
ℓ,i r,i in A , by the pigeonhole principle, at least one queue Q
ℓ,x ℓ,j
can easily show that the graph G has a Hamiltonian cycle iff
must be served by servers in A and A for some x,
G′ has; we omit the proof. We call the graph G′ the bipartite ℓ,x ℓ,x+r
some r ≥ 1. Consequently, as a result of the single carrier
version of G.
constraint, all the servers in A must be allocated to Q ,
r,x ℓ,j
Define T =2n(n+1)(n+2). Consider an instance of the
eachserving0packetsfromQ .Thusthemaximumnumber
ℓ,j
problem(PD) with 2nqueues,each with 2nT+(2n−1)(n+
ofpacketsthatcanbeservedbytheserversinA is(n−1)T.
r,·
2) packets, and 2nT +(2n−1)(n+2) servers. The servers
The total numberof packetsthat can be served by the servers
are grouped in 4n−1 sets: 2n sets of T servers each, and
in A is at most nT.
ℓ,·
2n−1 sets of n+2 servers each. Let the sets of T servers
For x ∈ [2n−1], the maximum number of packets that
be called A ,...,A ,A ,...,A , and the sets of n+2
ℓ,1 ℓ,n r,1 r,n can be served by a server in B is n+1. The total number
x
servers be called B ,...,B . The servers within a set are
1 2n−1 of servers in any set B is n + 2. Thus the total number
x
consecutively indexed. We use the symbol C < D to denote
of packets that can be served by all the servers in ∪ B is
x x
that the servers in the set C have lower indices than those in
(2n−1)(n+1)(n+2).SinceT =2n(n+1)(n+2),we have
the set D. We order the servers such that
(n−1)T +nT +(2n−1)(n+1)(n+2)<2nT. ♠
By symmetry, the above property also holds for a queue
A <B <A <B <A <B <A
ℓ,1 1 r,1 2 ℓ,2 3 r2 Q . Thus, any allocation that serves W = 2nT + (2n −
<B <A <···<B <A . r,i
4 ℓ,3 2n−1 r,n 1)(n + 2) packets must allocate server(s) from ∪ A
x∈[n] ℓ,x
Let the set of 2n queuesbe Qℓ,1,...,Qℓ,n,Qr,1,...,Qr,n. (resp. ∪x∈[n]Ar,x) to each queue Qℓ,i (resp. Qr,i).
Let X(Q ,S ) denote the number of packets that the server
ℓ,i j Property 2: Exactly one of the following statements is true:
S can serve from the queue Q , and similarly for Q .
j ℓ,i r,i 1) There exists a permutation σ (resp. π) of [n] such that
Fix a queue Q . Note that a server in the set B has the
ℓ,i x all the servers in A (resp. A ) are allocated to Q
index xT +(x−1)(n+2)+j for j ∈[n+2]. ℓ,i r,i ℓ,σi
(resp. Q ).
r,πi
8
2) TheallocationservesatotalofW′ <2nT packetsfrom First suppose there exists an allocation that serves at least W
all the queues. packets. Then it must be normal, and for every B , it serves
x
Proof: If statement 1 holds, then evidently a total of 2nT exactly 2 queues, one from {Qℓ,1,...,Qℓ,n} and the other
or more packets are served, so statement 2 cannot hold. from {Qr,1,...,Qr,n}, and the same queues Qℓ,i and Qr,j
If statement 1 does not hold, then WLG suppose a queue that are served by the adjacent servers in sets Aℓ,· and Ar,·.
Qℓ,i is allocated servers from Aℓ,x and Aℓ,x+r for some x, Thus the queues Qℓ,σ1,Qr,π1,Qℓ,σ2,Qr,π2,...,Qℓ,σn,Qr,πn
some r ≥ 1. As before, the servers in A are allocated to are served in order in consecutive server blocks. Consider the
r,x
Qℓ,i, serving 0 packets. Hence, as established in the proof of path vℓ,σ1 →vr,π1 →vℓ,σ2 →vr,π2 →···→ vℓ,σn →vr,πn.
Property1,theallocationservesatotalofW′ <2nT packets, This is a valid path in the graph G′ (Property 4) and because
thus statement 2 holds. ♠ σ,π are permutations, it visits every node exactly once.
Thus, if an allocationserves2nT or more packets,then for Therefore it is a Hamiltonian path.
every x ∈ [2n−1], the set of servers in B serve at most 2 Next suppose that there is a Hamiltonian path in G′,
x
queues, and the queues (if two) are of the form (Qℓ,i,Qr,j). WLG call it vℓ,σ1 → vr,π1 → vℓ,σ2 → vr,π2 → ··· →
v → v . Then allocating to the queue Q the
Property 3: Let an allocation serve a total of at least 2nT ℓ,σn r,πn ℓ,σi
servers in A , to the queue Q the servers in π , and the
packets from all the queues. If all the servers in a set B are ℓ,i r,j j
x drain-maximizing allocations for each B (which is possible
allocated to the same queue, say Q , then the total number x
ℓ,i because of Property 4), we get an allocation that serves
of packets served by the servers in B is at most n+1.
x exactly W =2nT +(n+2)(2n−1) packets. This completes
Proof:If x is odd, then exactly 1 server from B can serve a
x the reduction. Since T = O(n3), this is a polynomial-time
nonzeronumberofpacketsfromQ ,andthatnumberequals
ℓ,i reduction. Therefore the problem (PD) is NP-complete.
i + 1. If x is even, then the server B can serve at most
x
n+1−1≤n+1 packets from Qℓ,i. ♠ APPENDIX B
An allocation of servers to the queues is said to be normal
PROOF OF THEOREM2
ifthereexistsapermutationσ (resp.π)of[n]suchthatallthe
The problem (PM) clearly belongs to the class NP: a
serversinA (resp.A )areallocatedtoQ (resp.Q ).
ℓ,i r,i ℓ,σi r,πi certificate is an allocation of the servers to the queues that
Property 4: Fix x ∈ [n], x odd. Under a normal allocation, has a weight of at least W. To show that it is NP-complete,
let the servers in Bx serve two queues (Qℓ,i,Qr,j). If there we use the same reduction to the Hamiltonian path problem
exists a directed edge (vℓ,i,vr,j)∈E′, then the servers in Bx as before, we consider each queue to be of length = 1 packet,
serveatotalofatmostn+2packets,else,serveatmostn+1 and ask the question whether W = 2nT +(2n−1)(n+2)
packets. units of total service can be offered, which translates to a
Proof: Suppose the servers in Bx serve two queues schedule-weight of W. We omit the details.
(Q ,Q ) with (v ,v )∈E′. There is exactly one server
ℓ,i r,j ℓ,i r,j
St in Bx that serves Qℓ,i at a nonzero rate of i+1 packets. APPENDIX C
If this server St is not allocated to Qℓ,i, then the number of PROOF OF LEMMA1
packets served from Q is at most n+1 by Property 3: the
rj Let z = ⌊r/k⌋. Adding dummy nodes if necessary to the
number of packets served from Q cannot be more than if
r,j set U, and removing some nodes if necessary from the set V,
all the servers in B are allocated to Q .
x r,j we construct a graph G′(U′ ∪V′,E′) where |V′| = kz and
If S is allocated to Q , then because x is odd and the
t ℓ,i |U′|=z. For a pair of nodes u′ ∈U′ and v′ ∈V′,
allocation is normal, the servers in B with indices less than
x
t are allocated to S . The maximum number of packets that 1) If u′ ∈ U, then for any v′ ∈ V′ ⊆ V, (u′,v′) ∈ E′ if
t
and only if (u′,v′)∈E.
can be served from Q by allocating to it all the servers in
r,j
2) If u′ ∈/ U, then for any v′ ∈ V′, the edge (u′,v′) ∈ E′
B with indices higher than t is n−i+1, implying a total
x
with probability q, independently of all other random
of n+2 packets at most.
Iftheredoesnotexistadirectededge(v ,v )inE′,then, variables.
ℓ,i r,j
even after allocating S to Q and all the serversin B with Group the nodes in the set V′ as described in the SA
t ℓ,i x
indiceshigherthanttoQ ,themaximumnumberofpackets algorithm,to get a bipartite graphG′′(U′∪V′′,E′′) where V′′
r,j
served from Q is at most n−z+1 for z > i, implying a isthesetofgroupsofnodesinV′,andnodesu′ ∈U′,v′′ ∈V′′
r,j
total of at most n+1 packets. ♠ are connectedbyan edgein E′′ if the nodeu′ is connectedto
If the allocationof serversin B to the queues(Q ,Q ) every node in the group V′′. Thus between any pair of nodes
x ℓ,i r,j
serves a total of n+2 packets, we call it a drain-maximizing in U′×V′′, an edge exists with probability qk.
allocation for B . For z large enough, the graph G′′ has a perfect matching
x
A similar statement to Property 4 can be proved for B M′′ withprobabilityatleast1−3z(1−qk)z ([12],Lemma1).
x
for even x, and an edge (Q ,Q ) ∈ E′. We are now in Removing the “dummy” nodes that were added to get the set
r,j ℓ,i
a position to prove that a Hamiltonian path exists in G′ if U′ from U, we get a matching M as the output of the SA
and only if there exists an allocation of servers to the queues algorithm with |U|=|M|. That is, a perfect matching in the
that serves at least W = 2nT + (n + 2)(2n − 1) packets. graph G′′ (deterministically) yields a matching of cardinality
9
|U| as the output of the SA algorithm. Therefore, for r large Fix any i ∈ {1,2,...,m }. We have |T | ≥ i|D | +
0 i i
enough, P(|M|<|U|)≤3⌊r/k⌋(1−qk)⌊r/k⌋. n′/(2(m0−a+1))≥ i|Di|+n′/(2(m0+1)), and |Ti|/i≥
|D |+n′/(2m (m +1))≥|D |+nα/(4m (m +1)). Thus,
i 0 0 i 0 0
APPENDIX D from Lemma 1,
TheproofproceedPsRinOOtwFoOsFteLpEs:MfiMrsAtw2eshowthatforlarge P(|Mi|=|Di|)≥1−3(cid:22)4m (nmα+1)(cid:23)(1−qKm0)j4m0(nmα0+1)k.
0 0
n, with high probability, a=0 holds in the step 3 of the BA
algorithm. In the process, we show that the the number of Hence, by the union bound,
“excess servers” n′ (step 3 of the BA algorithm) is at least
P(|M |=|D | ∀ i∈[m ])
nα/2 with high probability. Next, under the condition a = 0 i i 0
aQˆn(dt)n}′i≥s snmαa/ll2., we show that the probability of {Qˆ(t+1)> ≥ 1−3m0(cid:22)4m0(nmα0+1)(cid:23)(1−qKm0)j4m0(nmα0+1)k.
Step 1: Combiningtheresultsofsteps1and2andonceagainusing
For 0 ≤ i ≤ M, let p′ := |{k ∈ [n] : A (t+1) = i}|/n the union bound,
i k
be the fraction of the n queues that see exactly i arrivals
in the timeslot t + 1. Let p’ = [p′,p′,...,p′ ]. Choose P Qˆ(t+1)>Qˆ(t)
0 1 M (cid:16) (cid:17)
tahneyorǫem∈ (([01,3α],/(T2hMm.m20.)1).,10s)a,yfoǫr=anyα/ρ(4∈M(m0,01)).,BfoyrSnanlaorvg’es ≤ e−nρτ(ǫ)+3m0(cid:22)4m0(nmα0+1)(cid:23)(1−qKm0)j4m0(nmα0+1)k,
enough, P(p’ ∈/ Bǫ) ≤ e−nρτ(ǫ). Since the set ∆M+1\Bǫ is completing the proof.
compact and the function g(y)= M y log(y /p ) is lower
i=0 i i i
semicontinuous([13],Chapter2,EPxercise2.1.22),theinfimum APPENDIX E
in the definition of τ(·) is achieved and is strictly positive PROOF OF LEMMA3
(∵ g(y) = 0 ⇔ y = p,p ∈ B and g(y) ≥ 0 for all y). Thus Supposeattheendoftimeslott,themaximumqueue-length
ǫ
τ(ǫ)>0, implying ismandthenumberofqueuesatlengthmisx.Ourobjective
is to show that at the end of timeslot t+1, with probability
P(|p −p′|<ǫ,∀ i∈{0,1,...,M})≥1−e−nρτ(ǫ).
i i at least 1−e−nφ for some φ>0,
Let Qˆ(t)=m. Define the set Cr :={i∈[n]:(r−1)K + 1) the maximum queue-length is at most m, and
1 ≤ Ai(t +1) ≤ rK}. Since Qi(t) ≤ m for all i, Dr ⊆ 2) the number of queues at the maximum is at most (x−
im=0rCi. Hence, nα/4)+.
S |D | ≤ |C |+|C |+···+|C | Since x ≤ n, the properties 1, 2 and the union bound imply
r r r+1 m0 that with probability at least 1−k e−nφ, at the end of k =
= n(p′ +p′ +···+p′ ) 0 0
(r−1)K+1 (r−1)K+2 M 4 timeslots, the maximum queue-length is at most m−1.
α
implying (cid:6) F(cid:7)irst consider the case x = n, i.e., all the queues in the
system are equal in length. From Lemma 2, for n large, the
m0 r|Dr|= m0 r|Dr| ≤ n m0 i M p′j sporombeabθili>ty0th,astoQˆth(et+pro1p)e≥rtym1iissuspatpisefir-ebdo.uNndexedt,bthyeeB−Anθa1lgfoor-
Xr=1 Xr=0 Xi=1 j=(i−X1)K+1 1
  rithmallocatestothequeuesinthesetsD ,D ,...,D
M c+1 c+2 m0
i one more server than is necessary to bring their length to m,
= n p′
Xi=1 i(cid:24)K(cid:25) and also for n′′ =n′−(dc+1+dc+2+···+dm0) queues in
D . Thus, at the end of timeslot t+1, the number of queues
M c
(a) i at length m is at most (n − n′/2)+, and by the proof of
≤ n (p +ǫ)
Xi=1 i (cid:24)K(cid:25) Lemma 2, the probability of this event is at least 1−e−nθ2
for some θ > 0. Since n′ ≥ nα/2 with probability at least
≤ n(1−α)+nǫMm , 2
0
1−e−nθ3 for some θ3 > 0 (from the proof of Lemma 2), if
wherethestep(a)holdswithprobabilityatleast1−e−nρτ(ǫ). wechooseφ=min(θ ,θ ,θ ),thentheproperty2issatisfied
1 2 3
Since ǫ<α/(2Mm0), we have mr=00r|Dr|≤n−nα/2, or for the case x = n. The case x < n is almost identical; we
a=0 and n′ ≥nα/2 in the stepP3 of the BA algorithm,with omit the details for the sake of brevity.
probability at least 1−e−nρτ(ǫ).
Step 2: APPENDIX F
We assume thata=0 andn′ ≥nα/2 in thestep 3of the BA PROOF OF THEOREM3
algorithm.ConsidertheeventE thateachofthequeuesinthe The proof is almost identical to that of Theorem 5 in [14].
i
setD areallocatedatleastiservers.IftheeventE occursfor Inparticular,Lemma3showsthatthemaximumqueue-length
i i
every i ∈ {1,2,...,m }, then the maximum queue-length at in the system decreases by at least 1 (provided it is nonzero
0
the end of timeslot t+1 is at most m. This event(E ) occurs to begin with) over a constant number of timeslots, with
i
if, in the server allocation step (step 5) of the BA algorithm, probability at least 1/2. Lemma 2 shows that in a given
the matching obeys |M |=|D |. timeslot, it increases by at most M, and the probability of
i i
10
this increase it at most e−nζ for some ζ = ζ(ǫ,α,ρ) > 0,
for n large. Using the same stationary distribution bounding
techniques as those in the proof of Theorem 5 in [14], we
conclude that
−1
liminf logP max Q (t)>b
i
n→∞ n (cid:18)1≤i≤n (cid:19)
b+1
≥ ζ(ǫ,α,ρ)
M
b+1 α 1
= min ρτ(ǫ), log >0,
M (cid:18) 4m (m +1) 1−qm0(cid:19)
0 0 K
implying the desired result because ρ < 1 is arbitrary (for-
mally, taking the limit of both sides as ρ→1).
APPENDIX G
PROOF OF THEOREM4
Thesteps1 and6 oftheBA algorithmcanbe performedin
O(n2)computationseach.Thesteps2and4canbeperformed
in O(n) computations each. The step 3 can be performed in
O(1) computations.
Step 5 requires finding largest cardinality matchings in
bipartite graphs. Given a bipartite graph with O(n) nodes,
the largest cardinality matching can be found in O(n2.5)
computations [18]. In our case, we need to find largest car-
dinality matchingsin bipartite graphs with 2n ,2n ,...,2n
1 2 w
nodes respectively with n +n +···+n = n. Hence the
1 2 w
computationaleffortisO(n2.5+n2.5+···+n2.5)=O(n2.5).
1 2 w
Thus, the BA algorithm can be implemented in O(n2.5)
computations per timeslot.
APPENDIX H
PROOF OF THEOREM5
Consider the following event that leads to overflow: fix
θ ∈ (0,M/K −1), and for t = b+1 timeslots up to and
0 θ
including the timeslot 0, the total(cid:6)num(cid:7)ber of arrivals to all
the queues have an empirical mean ≥nK(1+θ). That is, if
f (t) = 1 n 1{A (t) = i}, then for −t < t ≤ 0, we
i n j=1 j 0
have M Pif (t)≥K(1+θ). Since the system can serve at
i=0 i
mostPnK packets in a given timeslot, this event leads to an
overflow at the end of timeslot 0 under any algorithm.
Analyzing the probability of the event that leads to
overflow: Fix any ρ ∈ (0,1). By Sanov’s theorem ([13],
Thm. 2.1.10), for any timeslot t, the probability of the em-
pirical mean of the arrivals exceeding K(1+ θ) is at least
e−nρξ(θ) for n large.Since θ <M/K−1, the set ∆ \C
M+1 θ
is nonempty: [0,0,...,0,1] ∈ ∆ \ C . Hence, by the
M+1 θ
usual arguments of compactness and lower semicontinuity,
the infimum in the definition of ξ(·) is achieved and is finite
and strictly positive. By the independence of arrivals across
timeslots, the probability of overflow event is thus at least
e−nρt0ξ(θ), implying (because ρ<1 is arbitrary)
−1 b+1
liminf logP max Q (t)>b ≤ ξ(θ).
i
n→∞ n (cid:18)1≤i≤n (cid:19) (cid:24) θ (cid:25)