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C ONVERGENCE OF D EPARTURES IN T ANDEM N ETWORKS OF / GI / Q UEUES 1

B. P RABHAKAR T.S. M OUNTFORD N. BAMBOS

University of California at Los AngelesLos Angeles, CA 90024

Abstract

We consider an innite series of independent and identical /GI/ queues fed by anarbitrary stationary and ergodic arrival process, A 1 . Let A i be the arrival process to thei th node and let i be the law of A i . Denote by T ( ) the input-output map of the / GI / node; that is, i +1 = T ( i ). It is known that the Poisson process is a xed point for T .

In this paper, we are interested in the asymptotic distribution of the departure processfrom the n th node, n +1 = T n ( 1 ), as n . Using couplings for random walks, thislimiting distribution is shown to be either a Poisson process or a stationary -Poissonprocess (dened below) depending on the joint distribution of A 1 and the service process.This generalizes a result of Vere-Jones [11] and is similar in avour to [10] where Poissonconvergence is established for departures from a series of exponential server queues usingcoupling methods.

1 Introduction

Innite server queues which dispense i.i.d. but otherwise arbitrarily distributed service times(symbolically,

/ GI/

queues) are well-known examples of quasi-reversible nodes. Their

behaviour under Poisson process inputs is well-understood; in particular, they are known topossess the Poisson-in-Poisson-out property (see, for example, [8]). In this paper couplingsfor random walks are used to study the asymptotic behaviour of departure processes in largetandem networks of / GI/ queues under non-Poisson process inputs. Vere-Jones [11] usedthe method of probability generating functionals to study this model. He showed that as astationary, ergodic and weakly mixing point process passes through a series of independentand identical / GI/ nodes, it converges to the Poisson process in distribution (we assumethroughout that the service times of the / GI/ nodes are not equal to a constant almostsurely; if this is the case, the / GI/ queue is a pure delay system).

Simple counterexamples show that this result can be false if the weak mixing conditionis dropped, i.e. the successive departure processes can converge to a non-Poisson processlimit. For example, consider the deterministic arrival process with inter-arrival times exactlyequal to one unit passing through a series of independent and identical / GI/ queues withinteger-valued service times. Since the points of all the successive departure processes willbe separated by integer distances, Poisson convergence is impossible. However, as we shallsee, the departure processes do converge in this case also (albeit to a non-Poisson limit). Thecoupling methods used in this paper allow one to obtain a complete characterization of thislimiting process in a natural way. The details of how this is done are given in Section 2.2.

1 Appears in Probability in the Engineering and Informational Sciences , 10 , pp 487-500.

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In addition to [11], other studies of queueing tandems may be found in [2], [3], [6] and[10]. In [2] Bambos and Prabhakar derive the asymptotic completion times of a nite numberof jobs owing through a long series of queues where the service time of each job at different

queues forms a stationary and ergodic sequence. A similar model is considered by Glynn andWhitt [6] with the service times of individual jobs at the various nodes being distributed inan i.i.d. fashion. Bambos and Walrand [3] consider the G/G/1 queue as an operator thatmaps interarrival time sequences into interdeparture time sequences and study the limitingbehaviour of ows in a series of /G/1 nodes. Mountford and Prabhakar [10] show that thelimit from passing a stationary and ergodic process of rate through an innite series of independent /M/1 queues of service rate 1 is a Poisson process. [10] is similar in avour tothe present work in that both establish Poisson convergence results using coupling methods.We conclude the introduction by providing the following brief summary of the rest of thepaper.

Random walk couplings are used to show that when any stationary and ergodic arrivalprocess is passed through a sequence of / GI/ queues, the successive departure processesconverge in distribution either to a Poisson process or to a stationary -Poisson process(dened later) depending on the type of service distribution. In particular, our result includesVere-Jones result as a special case. Our methods also allow easy extensions to the case of departures from a series of non-identically distributed queues.

2 Convergence of Departures in / GI / Queueing Tandems

In this section we study departures from an innite series of independent and identical / GI/ nodes. Suppose customers arrive at the rst node in this series according to a process A 1given by

A 1 =

n =

t an , (1)

where ta 1 ta0 0 < t a1 tan tan +1 pathwise and x denotes the point massat x. The random variable tan species the arrival time of the n th customer of A 1 . Note that weallow non-simple point process (i.e. batch process) inputs. We will suppose throughout thatarrival processes are stationary and ergodic with respect to the transformations { t , t R},where t A 1 =

n = t an t . Suppose that = E (N

A 1 (0, 1]) is the rate of A 1 , where

N A1

(0, 1] =

n =

1 {0

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values of the service times are integer multiples of c. Otherwise the service is said to benon-lattice .

Suppose customers x and y arrive at a series of independent and identical /GI/ queues.If their service times at queue k are xk and yk respectively, then S m = mk =1 xk yk =m

k =1 Z k is the difference in their total service times in the rst m queues. In the sequel wewill often say that the service of a customer is recurrent or transient, when we actually meanto say that the random walk, S m , induced by the service times is recurrent or transient.

Proposition 1, the main result of this section, introduces the key ideas for establishingweak convergence of departures by rst considering tandems of identical / GI/ queues whoseservice is non-lattice and recurrent. We then show how the same ideas generalize to othercases (lattice and/or transient service) with appropriate modications.

Proposition 1: The limit from passing a stationary and ergodic arrival process of rate ,A 1 = n = t an , through an innite series of independent, identical / GI / queues whose service is non-lattice and recurrent is a Poisson process of rate .Preliminary sketch of proof: We will couple A 1 with an independent rate Poissonprocess P 1 = n = t pn by matching up points (or customers) of A

1 and P 1 . Customersbelonging to A 1 are coloured blue and customers belonging to P 1 are coloured red. So longas a red and a blue customer evolve independently of each other, they perform a random walkas regards to their inter-arrival times at each queue in the series. By noting that this randomwalk is non-lattice and recurrent we conclude that at some nite queue these two customerswill be within of each other, where is arbitrarily small. Once they are sufficiently close, wesay they are coupled (at which point they are coloured yellow) and let them evolve togetherforever. Thus, eventually all the points of A n and P n are close to each other and we get therequired weak convergence, since the P n s are all Poisson. In matching up customers of A 1and P 1 for the purpose of coupling them, we cannot just match up the k th customers (whoarrive at times ta

k and t p

k) because as k becomes very large, the difference in their arrival times

(tak t pk ) will become very large - typically of the order of k. To ensure that we are coupling

customers that are not separated by more than a xed distance, say R, we divide time intolarge blocks of length R. The choice of R is dictated by our desire to ensure that there areapproximately the same number of blue and red customers in the interval ( jR, ( j + 1) R], for j Z . The details are as follows.Proof of Proposition 1: The desired weak convergence (see Daley and Vere-Jones [4],Chapter 9 for details on the weak convergence of point processes) follows once we show thatfor each non-negative continuous function f on R with compact support

f dA n D f dP ,where P is a Poisson process of rate . Since for every n , P n is a rate Poisson process, itwill suffice to show that

f dA n f dP n pr 0. (2)Without loss of generality, suppose [0 , N ] contains the support of f .

Fix > 0 arbitrarily small. We will show that the point processes A n and P n can be

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coupled so that for n sufficiently large

E

f dA n

f dP n (3)

is bounded by a multiple of . Since f has compact support it is uniformly continuous. Hencewe can nd < so that > sup( |f (x) f (y)| : |x y| < ).

By the ergodic theorem,

1N

N A1

(0, N ], N P 1

(0, N ] a.s. (, ) as N where N A

1

(0, N ] N P 1

(0, N ] is the number of points of A 1 (P 1) in (0, N ]. Hence there

is a sufficiently large R R such that the chance that both N A1

(0, R] and N P 1

(0, R]are between R ( 2 ) and R ( + 2 ) is greater than 1 . Fix such an R. Let U be arandom variable chosen uniformly on [0 , R], independently of the two point processes A 1and P 1 and of the service times. Divide up the time axis into disjoint intervals of length R ,( jR + U, ( j + 1) R + U ] j Z . Say that an interval ( jR + U, ( j + 1) R + U ] is good if bothA 1 and P 1 have at least R ( 2 ) points in it. By the joint stationarity of A 1 and P 1 , thechance that for any j Z such an interval is good is greater than 1 . Now, we match upcustomers as follows.

1) If interval ( jR + U, ( j + 1) R + U ] is not good, then no customers in the interval(belonging to either A 1 or P 1) are matched.

2) If the interval ( jR + U, ( j + 1) R + U ] is good, then let the customers of A 1 andP 1 in this interval be matched up, on a one-to-one basis from the left end of the intervalto the right end of the interval, possibly with a few unmatched customers of A 1 or P 1 (notboth) remaining at the right end of the interval. That is, if the arrival times of the customersof A 1 and P 1 in this interval are, respectively, xj1 xj2 xj R ( 2 ) , andyj1 y

j2 y

jR ( 2 ) , then match up the customer of A 1 at x

ji with the customer

of P 1 at yji for i R ( 2 ) . Once the customers at xji and y

ji are matched up, we say

they are partners.

(Remark: Note that without the random variable U , the distribution of unmatched cus-tomers over ( jR, ( j + 1) R] will be uneven with more unmatched customers being foundat the end of the interval than at the beginning. The addition of U makes the matchingoperation, and hence the process of unmatched customers, time stationary.)

Now consider two partners that arrive at times x ji and yji for some j Z . 1 = x

ji y

ji

is the difference between their arrival times to the rst queue. If ak and pk are their service

times at the kth

queue, then the random variable

m = 1 +m 1

i =1ai

pi =

1 +m 1

i =1Z i (4)

is the difference in the arrival times of these customers at the m th queue. Because the serviceis recurrent and non-lattice, m is a recurrent, non-lattice random walk on R with startingvalue 1 . Thus | m | < for some m() a.s. where was chosen at the start of the proof.

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By the coupling scheme, | 1| R a.s. and so the distribution of the rst such m() is tightover matched pairs of partners.

Once the two partners of A 1

and P 1

, situated initially at times xj

i and yj

i respectivelyare within of each other, they are said to be coupled and are coloured yellow. We thenlet them receive identical service times at all subsequent queues. This ensures that theywill forever remain within of each other once they are coupled. Similarly all the otherpartners couple at some queue or the other a.s. and then evolve together. Let C nA and C nP be, respectively, the set of points of A n and P n that are coupled.

Now

f dA n = f dC nA + f d (A n C nA ) and f dP n = f dC nP + f d (P n C nP ) .

Consider the random variable

f d (A n

C nA ). Now

E f d (A n C nA ) f max E (# of points of A n C nA in (0, N ]) . (5)The points of A n C nA in [0, N ) are either those that have partners but have not coupledwith (come within of) them yet; or those that did not have partners in the rst place, i.e.the unmatched points of A 1 .

The expected density of unmatched points of A 1 (and hence of A n for each n) is lessthan or equal to minus the expected density of matched customers. This in turn is lessthan or equal to ( /2)(1 ) ( + 1 / 2).

For each matched x A 1 , the probability of not being coupled after the n th queuedepends only on n and the relative displacement of x from its partner at the rst stage.As this displacement is bounded by R, the probability of not being coupled is bounded bysome c(R, n ) which tends to zero as n tends to innity for xed R. The events of matchedpairs becoming coupled are mutually independent, given their initial positions in A 1 and P 1 .Therefore the density of matched but uncoupled points in A n is bounded by c(R, n ).

Putting the two paragraphs together we nd that the expected number of uncoupledpoints of A n in [0, N ) is at most ( + 1/ 2) + c(R, n ). A similar bound holds for theuncoupled points of P n . Therefore, for n sufficiently large, the chance that there is anuncoupled customer in [0 , N ) for either A n or P n is bounded by 3 (1/ 2 + ). Using this in(5) we get that

E f d (A n C nA ) f max 3 ( + 1 / 2).To establish the theorem, it now suffices to bound f dC

nA f dC

nP .

Divide the points of C nA in the interval [0 , N ] into G nA and B nA , where G nA consists of those (good) points of C nA whose partners also fall in the interval [0 , N ]. B nA then consistsof those (bad) points of C nA whose partners fall in the set [ , 0)(N, N + ]. Likewisedene G nP and B nP . With this,

f dC nA f dC nP f d (G nA G nP ) + f dB nA + f dB nP .5

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The last two terms in the right-hand-side of the above expression are bounded by 2 f max 0 is arbitrarilysmall. In order to proceed, we need the following denitions.

Denition 2: Given c > 0, x (mod c) is the element of [0 , c), which taken away from x,leaves an integer multiple of c.

Given a stationary point process A = n = t n , consider

P n () = 1

N A ([0, nc))t i [0, nc )

t i (mod c ) ,

where x is the point mass situated at x. For each n, P n () can be considered as a prob-ability measure on the compact space [0 , c] with the two end points, 0 and c, identied.To obtain P n () we wrap around the points of A () in [0, nc) about the interval [0 , c].The elementary lemma below shows that, for almost all , P n () converges (in the sense of weak convergence of distributions) to a random limit P A () whenever A is a stationary pointprocess of nite intensity.Lemma 1: Let A =

n = t n be a stationary point process of nite intensity and let

P n () = 1

N A ([0, nc))t i [0, nc )

t i (mod c ) .

Then there exists a random probability measure P A () on [0, c] such that a.s. P n () weakly

P A ().Proof: Let f be a continuous function on [0 , c] with f (0) = f (c). Dene F : R byF () = t i [0, c ) f (t i ). Then F ( jc ()) = t i [jc, (j +1) c) f (t i jc ). Hence

f dP n () = 1N A ([0, nc)) t i [0, nc ) f (t i (mod c)) = 1

N A ([0, nc))

n 1

j =0F ( jc ())

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Since A is t -stationary, it is certainly stationary with respect to shifts {nc , n Z}. Itfollows from the Ergodic Theorem and the nite intensity of A that

limn 1n

n 1

j =0F ( jc ())

exists a.s. and is nite. Therefore lim n f i dP n () exists a.s. and is nite for a countable

dense set of functions f i on [0, c] with f i (0) = f i (c).

Since the sequence of probability measures P n () is dened on the compact space [0 , c]with 0 and c identied, they are tight. Hence there exist weak sub-sequential limits for thesequence P n . By the above argument, all the weak sub-sequential limits are identical. Letthis limit be P A . This concludes the proof.Discussion: It should be noted that the limit in the above lemma can be random even whenA is ergodic. This is because a function that is nc -invariant need not be t -invariant. For

example, if A is the ergodic deterministic process of period 1 and c = 1, then P A

() = X ( ) , where X () is uniformly distributed on [0, 1]. However, for ergodic A , P A ( t ()) =t (mod c ) P A (). To see this, simply note that

P A ( t ())= lim

n

1N A ([t, t + nc))

t i [t,t + nc )

t i t (mod c )

= limn

N A ([0, nc))N A ([t, t + nc))

1N A ([0, nc))

t i [t,t + nc )

t (mod c ) t i (mod c )

= limn

1N A ([0, nc))

t (mod c ) t i [kc, (k + n )c)

t i (mod c ) , where t = kc + t (mod c),

= t (mod c ) limn 1N A ([0, nc))t i [kc, (k + n )c )

t i (mod c )

= t (mod c ) P A ().

Of course, since P A ( t ()) is a measure on [0 , c] with the end points being identied,t (mod c ) P A () is understood to be a translate mod c of P A (). By ergodicity of A ,the orbit of {t ()} as t ranges over R is , making the limiting measure P A almost surelyequal to a uniform translation ( mod c) of some measure on [0, c].Denition 3: A point process A =

n = t n is said to be uniform on [0, c] if

1N A ([0, nc))t i [0, nc )

t i (mod c )

converges almost surely to the uniform distribution on [0 , c] as n .We now obtain the following strengthening of Theorem 1.

Theorem 2: Let A 1 be a stationary and ergodic arrival process of rate and let A n be the departure process from the (n 1) th queue when A 1 is fed through an innite series

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of independent, identical /GI/ queues. Then A n converges in distribution to a Poisson process of rate if and only if (a) The service is non-lattice

or (b) The process A 1 is uniform on [0, c], the span of the service times .

Proof : Theorem 1 takes care of the Poisson convergence when (a) is satised.

To prove (b), we use the same argument as in Theorem 1, except that we must changethe rules for matching customers. This time given a continuous function f with compactsupport (support contained in [0 , N ], say), (a measure of how close the matched customersshould be in order to be declared coupled), and c the span of the service time, we dividetime into large blocks of length Rc (R is an integer). We then divide (0 , c] into 1/ intervalsof length c (without loss of generality 1 / is an integer), and choose R so large that outsidea set of probability 1 , the number of points of A 1 and P 1 in (0, Rc] which are in

R 1

k =0[kc + i c, kc+ ( i + 1) c) for i = 0 , 1, , (1/ 1)

are both between Rc( 2 ) and Rc( + 2 ). That is, we take the realizations of A 1 and P 1over the interval (0 , Rc] and wrap them around (0 , c]. If R is big enough, because A 1 andP 1 are both uniform over c, the distribution of points in (0 , c] after wrapping around willbe close to uniform with a high probability and on the average there will be Rc points inthe above union of intervals. Hence a suitably large choice of R gives the desired minimumof Rc( 2 ) points with probability greater than 1 .

As before we choose an independent uniform U on [0, Rc] and we call an interval ( jRc +U, ( j + 1) Rc + U ] good if the number of points of A 1 and P 1 in ( jRc + U, ( j + 1) Rc + U ]which are in

R 1

k =0

[ jRc + U + kc + i c, jRc + U + kc + ( i + 1) c) for i = 0, 1, , (1/ 1)

are both within Rc( 2 ) and Rc( + 2 ). We match up points in A 1 and P 1 as follows:1) If the interval ( jRc + U, ( j + 1) Rc + U ] is not good, then no customers of A 1 and P 1

are matched.

2) If the interval ( jRc + U, ( j +1) Rc + U ] is good, and the arrival times of the customersof A 1 and P 1 in

R 1

k =0

[ jRc + U + kc + i c, jRc + U + kc + ( i + 1) c)

for each i = 0 , 1, , (1/ 1) are, respectively, xij1 < x

ij2 < < x

ijRc ( 2 ) < , and

yij1 < yij2 < < y

ijRc ( 2 ) < , then we match up the customer at x

ijl with the customer

at yijl for l Rc( 2 ) .The rest of the proof is the same as in part (a) with the recurrent and transient cases

being dealt with appropriately.

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It remains to show that if both conditions (a) and (b) fail then A n cannot tend to aPoisson process of rate in distribution. However, should (a) fail, then the service time hasa distribution with span c, say. If (b) fails, then A 1 is non-uniform on [0 , c]. But Lemma

1 tells us that P A

() = lim n 1

N A ([0 , nc )) t i [0, nc ) t i (mod c ) exists a.s. As noted in thediscussion following Lemma 1, P A ( t ) = t (mod c ) P A (). Therefore the event that P Ais uniform is translation invariant. As A 1 is ergodic, this event must have probability 0 or1. By similar reasoning if P A is not a.s. uniform then it must be a.s. equal to some distinctdistribution on [0, c] or a translation ( mod c) of . But in this case Theorem 3 below showsthat A n cannot tend to a Poisson point process in distribution.

Given a probability measure on Rwith support in [0 , c), dene the measures n , n Z with support in [ nc, (n + 1) c) as translates of by nc units of time. That is n (S ) = (S (mod nc )) for Borel sets S in [nc, (n + 1) c). Note that 0 = .

Denition 4: Given c > 0 and a measure on [0, c), the process P is said to be a -Poisson process of rate if the following conditions hold.

i) The number of points of P in [nc, (n + 1) c) is an i.i.d. sequence, as n varies over Z , witheach marginal being distributed as a Poisson, parameter c , random variable.ii) Each point of P in [nc, (n + 1) c) is distributed over the interval [ nc, (n + 1) c) accordingto n , independent of all other points.

Note that according to the above denition P is the usual Poisson process with param-eter if is the uniform distribution over [0 , c), and it is a batch process of i.i.d. Poissonrandom variables if is a point mass at x [0, c). Hence P is, in general, not a timestationary process. However, by shifting the paths of P uniformly over the interval [0, c),we obtain a stationary -Poisson process .

Theorem 3: Let A 1 = n = t n be a stationary and ergodic process of rate and suppose

limn 1

N A 1 ([0, nc))t i [0, nc )

t i (mod c )

converges a.s. to a translate (mod c) of some distribution on [0, c). Let A n be the departure process from the (n 1) th stage of an i.i.d. series of /GI/ queues when A 1 is inputted tothe rst queue. Suppose the service time of each customer is lattice with span c. Then A nconverges in distribution to a stationary -Poisson process of rate .

Proof: Again we couple A 1 with a stationary -Poisson process P 1 . (In Lemma 2 we showthat P 1 is an invariant distribution for a lattice-type / GI/ queue.) First we note that theergodicity of A 1 implies that there exists a unique probability distribution on [0, c) suchthat a.s.

limn

1

N A 1

([0, nc)) t i [0, nc ) t i (mod c )

converges to a translate ( mod c) of . We construct our stationary -Poisson process P 1 fromA 1 (after possibly enlarging our probability space). This is different form previous couplingswhere the two point processes A 1 and P 1 were chosen to be independent. Given A 1 let ()be the limit of the random measures

limn

1N A 1 ([0, nc))

t i [0, nc )

t i (mod c )

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and let P 1 be a -Poisson process, conditionally independent of A 1 given . Then it is clearthat P 1 is a stationary -Poisson process. Fix arbitrarily small but positive. We pick integerR to be sufficiently large that outside of probability 1 , the number of points of A 1 andP

1

in [0, Rc) which are inR 1

k =0

[kc + i c, kc+ ( i + 1) c) for i = 0 , 1, , (1/ 1)

are both between Rc ()([i , (i + 1) )) 2 and Rc ()([i , (i + 1) )) + 2 . This Rcan of course be chosen in a non-random way. The proof now proceeds exactly as in part (b)of Theorem 2.

Lemma 2: Consider a /GI/ queue whose service is lattice with span 1. If is a probability measure on [0, 1], then a -Poisson process of rate , P , is an invariant distribution for this queue.

Proof: The proof is based on the fact that any two / GI/ queues are interchangeable. (Twoqueueing nodes in tandem are said to be interchangeable if, for arbitrary arrival processes, thelaw of the overall departure process from the tandem is invariant with respect to the orderingof the two nodes. It is obvious that any two / GI/ queues are interchangeable.) Let G1 andG2 be two /GI/ queues and let i denote the service time of a typical customer at node i,i = 1 , 2. Suppose the laws of i are given by

P (1 = k) = pk , k = 0 , 1, ; P (2 S ) = (S ),where S is a Borel subset of [0, 1]. We are required to show that G1(A ) = A .

We rst show that a batch arrival process B , of rate , composed of i.i.d. Poisson randomvariables living on the integers is invariant to the queue G1 . To see this, suppose that B isrepresented by the i.i.d. sequence of Poisson random variables

{X n , n

Z} with E (X n ) = .

Split each X n into random variables {X kn }k 0 by sampling X n in an i.i.d. fashion according tothe probabilities pk . The interpretation is that X kn is the number of customers arriving at timen and requiring k units of service. Because X n is Poisson distributed each X kn is a Poissonrandom variable with parameter p k ; and, for distinct k and j X kn and X jn are independent.Therefore the members of the doubly indexed family of random variables {X kn , n Z , k 0}are mutually independent. The departure process from the node D = {Y n } is given byY n =

k =0 X

kn k . The claim now follows since each Y n is a Poisson random variable of rate

(being the sum of independent Poisson random variables of rate pk ), and for distinct kand j , Y k and Y j are independent. Therefore, G1(B ) = B .

Notice that by denition, A = G2(B ). By interchangeability,

G1(G2(B )) = G2(G1(B )) .

But the left-hand side equals G1(A ) and the right-hand side equals G2(B ) = A . HenceG1(A ) = A and the lemma is proved.

Corollary: Given , a probability measure on [0, 1], a stationary -Poisson process is an invariant distribution for a / GI / queue whose service is lattice with span 1.Remark The method we have advanced can also be easily extended to the case when thequeues in series are not i.i.d. For instance, if the service times have the property that the

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service time of every customer converges almost surely to zero, then the series of queueshas the property that all stationary ergodic inputs converge to a Poisson if and only if thesums of symmetrized service times,

N k =1 (

ak

pk ), do not converge. This follows from the

celebrated 3-series Theorem of Kolmogorov. To see how this applies to our case, note that if the symmetrized service sums do not converge, then by elementary martingale arguments itfollows that the sums must have liminf equal to minus innity and limsup equal to innity.This, together with the assumed property that service times tend to zero almost surely impliesthat the coupling of matched customers must be successful.

Conversely if the symmetrized sums converge, then for each customer, x, there is arandom variable X (x) < a.s. and a deterministic sequence of numbers, c(n), not dependingon x, so that for all n |service time of x at the rst n queues c(n)| < X (x). Thus if theinitial arrival process is a sufficiently extreme mixture of long vacant periods alternating withshort dense batches of arrivals, then no Poisson convergence is possible.

References

[1] V. Anantharam (1993): Uniqueness of stationary ergodic xed point for a /M/k node. Annals of Applied Probability , 3 , 1, 154-173.

[2] N. Bambos, B. Prabhakar (1994): On Innite Queueing Tandems, Systems & Control Letters , 23 , pp305-314.

[3] N. Bambos and J. Walrand (1990): An Invariant Distribution for the G/G/1 Queueing Operator. Ad-vances in Applied Probability , 22 , 254-256.

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[5] R. Durrett (1991): Probability: Theory and Examples . Wadsworth Inc., California.

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