03 the attractiveness

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B. PRABHAKARLindley'srecursionrelates the inter-departureimes to the inter-arrivalimes andservicetimes via waitingtimes as follows:

WA(n + 1, 1)= [WA(n, 1)+ S(n, 1) - A(n, 1)]+,(2) A(n,2) = [A(n,1)- S(n, 1)- WA(n,1)]++ S(n+ 1, 1).The process A2 is inputto the second queue from which we obtainA3 as thedeparture rocess,andso on. Ingeneral,Ak= {A(n, k)}nEz is the arrivalsprocessat the kth queue with {WA(n,k)}neZas the corresponding et of waitingtimes.Thus,A(n, k) is the inter-arrivalimebetween the nth and the (n + 1)stcustomersat the kth queue and WA(n,k) is the waitingtime of the nth customerat the kthqueue.Using the resultof Loynes [9] inductively,one obtainsthatAkis stationaryandergodicfor each k, with IE(A(1,k)) = r. Let T denote thequeueing operatorandrepresent he queueingtandemas Ak+1 = 7k (A1), k > 1.DEFINITION. A stationary and ergodic arrivals process I = {I(n)}nz withE(I (1)) = T > 1 is saidto be a fixedpointor an invariantdistribution t rate 1/rfor a ./GII/ queue whose service times aredistributedas or,if T(I) equals I indistribution.Thefollowingtheorem s the main result of thepaper.THEOREM. Supposethat a ./GI/1 queuewithservice distribution admitsa rate Ir fixedpoint I. LetA1 be a rate 1 ergodicstationaryarrivalprocess toan infinitetandemof independentcopies of the ./GI/1 queue. ThenTk(Al) -- Iin distribution as k -> oo.We shall prove the theoremby coupling A' with an independentprocess I1,which has the same distributionas I. Similarmethods,but differentcouplings,were used in [12-14] to establish the distributional onvergenceof departuresn atandemof queueswith differentassumptionson theservice distribution.Chang[5]used the couplingsof this paperto show thata queue which offers i.i.d. servicesof unbounded upport an have at most one fixedpointof a givenrate. The relatedliteratures surveyed n more detail at theend of the paper.2. The coupling. The processA' is coupledwith a processII, distributed sthe fixedpointI. The assumptionsare thatA1 andI1 aremutually ndependent ndalso independent of the service processes {S(n, k)}nEZ,kEZ+ Let Ik = {I (n, k))}nEbe the inputat node k whenI1 is inputat node 1, andlet WI(n, k) be the waitingtimeof the nth customerof Ik. Then,we havethe followingrecursions:

Wl(n + 1,k) = [WI(n, k) + S(n, k) - I(n, k)]+,(3)

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I(n, k + 1) = [I(n, k)- S(n, k) - Wl(n, k)]+ + S(n + 1,k).


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FIXED POINTS OF A */GI/1 QUEUEA similarrecursion or the processAk is given by

WA(n + 1, k) = [WA(n, k) + S(n, k) - A(n, k)]+,(4) A(n, k + 1) = [A(n, k) - S(n, k) - WA(n, k)]+ + S(n + 1, k).From(3) we obtain

(5) I(n, k + 1)- WI(n +1, k) = I(n, k)-S(n, k)-Wl(n, k)+ S(n +1, k),andfrom(4) we obtain(6) A(n,k+ 1)- WA(n+ 1,k) = A(n,k)-S(n,k)- WA(n,k)+S(n+ 1,k).Subtracting5) from(6) we get

[A(n,k+ 1)- I(n,k + 1)]- [WA(n+ 1,k)- WI(n + 1,k)](7) = [A(n, k) - I(n, k)] - [WA(n, k)- WI(n, k)].A littlealgebranow yields

[WA(n +J,k)- WI(n + J,k)]J-1= j[A(n +j,k+1) -I(n+ j,k+l)]

(8) i=O-[A(n+ j, k)-I(n+ j, k)])

+ [WA(n, k)- WI(n, k)].Equation 8) will form the basisof thecouplingargument.Throughout he rest of the paper,it is helpful to imagine that there are twoqueues at each node k, one for the A customers and one for the I customers.This makes explicit the notion that customers of one process do not influencethe waiting of the customers of the other process. The coupling between thetwo processes merely consists of providingcustomersnumberedn with identicalservice times, S(n, k), distributed .i.d. over n andk. We will refer to each of thetwo queuesas the A-queueand the I-queue, respectively.

2.1. A coloring scheme. The next step is to introducea coloring scheme forour processes. Since A1 and I1 are both of rate 1/r and are not identical (elseTheorem 1 is triviallytrue), there must exist pointsof crossing. Thatis, thereexist disjointrandomsets of integersA1 = {n EZ:A(n, 1)> I(n, 1)},I1 = {n E Z: I(n, 1) > A(n, 1)},

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2240~~~B.RABHAKARwhere A' dominates11or the otherway around.Call these the sets of domination.The stationarityandergodicityof (A', I') implies that of (Al, 11); therefore, hedensityof Al,

d (A)lim# of pointsof Al in [-N, N]N--oo 2N+1Iis well defined andalmost surelyequalto a positiveconstant.Similarly,d(I 1 isalmostsurelya positiveconstant.Let r = sup{m r:M Eilr(I1, 1) = inf{m > b(l, 1):mMEAl).

For n > 2, recursivelydefiner(n, 1) and b(n, 1) as follows:b(n, 1) =inf{m > r(n - 1,9 ): MEI1,r(n, 1) = inf{m > b(n, 1) :m E A').

Let P 0, 1) = sup{m< b(1, 1) :m E A 1) andb (0, ) = suptm < r?(0,1) :m EForn


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FIXED OINTS FA./GI/1QUEUESince d(A1) > 0 andd(l1) > 0 a.s., r(n, 1) andb(n, 1) are almostsurelyfinitefor everyn, and the infima n the above definitionsareminima. Define

,R(n, 1) ={k EZ: r(n, 1) k < b(n + 1, 1)},(11) S(n, 1) = {kEZ: b(n, 1) < k < r(n, 1)}to be nonoverlappingntervalsof integerswhichalmostsurelypartitionZ. Finally,let

R(n, 1)= A(k,1)-I (k, 1) andkeR(n,1)(12) B(n, 1) = I (k, 1)-A(k, 1).keS(n, 1)

We are now ready to introduce the coloring scheme. One thinks of the setsof domination A1 and j1 as the supportof red and blue bubbles, respectively.That is, r(n, 1) is the point at which the nth red bubblebegins, R(n, 1) is theinterval over which it is supported,and R(n, 1) is its volume. Withreference toFigure 1, R(1, 1) is the sum of the lengths of the vertical red lines in eR(1,1) =[r(l, 1), b(2, 1)). A similar nterpretationmaybe madefor the blue bubbles.It is crucial thatfor m =:n the shades of the mth and the nth red bubblesaredistinct.Thus,we think of the nth redbubbleas being colored with an nth shadeof red. Similarly,one is able to distinguishbetween the variousshades of blue.See Figure1.

2.2. Sketchof theproof. By the ergodicityof (A1, I'), the densities of theredandblue bubbles atthefirststageareexactly equal.This follows from the fact thatlimnoo((El=-n A(j, 1) - I(j, 1))/(2n + 1)) = 0, which implies

lim J=-n(A(, 1)- I(j, 1)){jEAi}n oo 22n 1E -n l(j 1) - A(j, l))l{jEllim l=-nn-*oo 2n + 1

thatis,volume of redin [-n, n] volume of blue in [-n, n] Alim lim = d(l),n->oo 2n + n1oo 2n+l1

whered(l) is the averagevolume of red(orblue) perarrivalat stage 1.At any stagek, the arrivalprocesses (Ak, Ik) arejointly ergodic.Hence,by theergodictheorem,d E(A(1, k) -(1, k)|)2

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B. PRABHAKARis almost surely a constant and equals the averagevolume of red (or blue) perarrivalat stagek. Chang[5] has shown

E(IA(1,k)- I(1, k)) > E(IA(l, k+ 1) - I(1, k+ 1)1)for each k. Hence the d(k) are monotonically nonincreasing.Given that I1 is afixedpointforthe queue,the desired weakconvergencewill follow fromshowingthatd(k) convergesto zero almostsurelyas k-- oo.We shall do this by observing the evolution of each individualred and bluebubbleas thetwo processes passthrough he series of queues.Thus,for eachn, thequantitiesr(n, k), b(n, k), R(n, k), B(n, k), R(n, k) andB(n, k) are derived romthe correspondingquantitiesat stagek - 1 and the serviceprocessat stagek - 1.Foreach k, the nth red(resp.,blue)bubble s imaginedto be coloredwith the nthshadeof red(resp.,blue). Wethen show that thered and blue bubblescanceleachother out andtherefore hat R(n, k) and B(n, k) decreasemonotonicallyto zeroas k--oo.The main difficulty in establishing the monotone decrease of R(n, k) andB(n, k) to zerois that t is possibleforthe red bubbles o accumulate arawayfromthe bluebubblesand not cancelthem out.We address hisproblemby showingthat(i) the orderingbetweenred andblue bubblesis always maintained, hatis, theycannot overtakeeach other,and (ii) since the services are independent he twotypes of bubbleswill be forced to interactand must thereforecancel each otherout.The detailsfollow.

3. Preliminary lemmas. We simplifythe notationas follows: Letd(n, k) = WA(n, k)- WI(n, k) Vn e Z, k e Z+,(13) da(n, k)= A(n, k)- I(n, k) Vn e Z, k e Z+.

Mnemonically,dW, .) is the difference n waitingtimes betweencustomersofthe two processes.In this notation(8) reads:J-1

(14) dw(n+J,k)= o( +l)-+j,k+l)-da(n+j,k))+dw(n,k).

LEMMA . The ollowing holdfor any n and k:(i) A(n,k) < S(n,k) ~ A(n,k+ 1)= S(n+ 1,k) andda(n,k+ 1) < 0;(ii) I(n,k) < S(n,k) = I(n,k + 1) = S(n + 1,k) andda(n,k + 1) > 0.PROOF. Given that A(n,k) < S(n,k), it follows from (4) and the non-

negativity of WA(n,k) that A(n,k + 1) = S(n + 1,k). From (3) we getI(n, k + 1) > S(n + 1, k). Therefore,da(n, k + 1) < 0, and similarlywith (ii).D

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FIXEDPOINTSOF A /GI/1 QUEUELEMMA2. For any n and k:(i) dw(n,k) > da(n, k) =~da(n, k ? 1) < Oanddw(n + 1,k) > 0;

(ii) dw(n, k) < da(n, k) ??da(n, k + 1) > 0 and dw(n + 1, k) < 0.Additionally,(iii) if under(i), S(n, k) > I (n, k), then da(n, k + 1) = 0;(iv) if under(ii), S(n, k) > A(n, k), thenda(n, k ? 1) = 0.

Hence, under (iii) and (iv), (14) implies dw(n + 1, k) = dw(n, k) - da (n, k).PROOF. Equations (3) and (4) imply (i). Consider (iii). From part (i) we havethatda(n, k + 1) < 0. Frompart(ii) of Lemma 1 we have thatda(n, k + 1) > 0.Therefore, da (n, k + 1) = 0. The proofs of (ii) and (iv) are similar. LILEMMA 3. Thefollowing holdfor any n and k:(i) dw(n, k) > 0, da(n, k) > 0 =~ da(n, k ? 1) < da(n, k), dw(n + 1, k) da(n, k), d w(n + 1, k) >dW(n, k).

PROOF. We prove (i). From (3) and (4) we get thatda(n, k + 1) = [A n, k) - WA(n, k) - S(n, k)]+ - [I (n, k) - WI n, k) - S(n, k)]+.If [A(n, k) - WA(n, k) - S(n, k)]+ = 0, then da(n, k + 1) < 0 < da(n, k).

On the other hand, if A(n, k) - WA(n, k) - S(n, k) > 0, thenda(n, k + 1) = A(n, k) - WA(n, k) - S(n, k) - [I(n, k) - WI1(n,k) - S(n, k)]+

0 - da(n, k + 1) < 0;(ii) da(n, k + 1) < 0=dw(n + 1, k) > 0;(iii) dw(n?+ 1,k) < 0=da (n, k + 1) > 0;(iv) da(n, k + 1) > 0=?dw(n + 1, k) < 0.

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B. PRABHAKARPROOF. Weverify(i) and(ii). From(3) and(4) we get

dW(n+ 1,k) > 0 e A(n, k) - WA(n, k) < I (n, k) - WI (n, k)'> A(n,k + 1) I(n,k + l)= da(n,k+ 1) Wl(n + 1, k)= dw(n+l1,k)>O. O

LEMMA5. Thefollowing holdfor any n and k:(i) dW(n+ 1, k) > 0, da(n, k) > O dW(n, k) > dW(n+ 1, k);

(ii) dw(n + 1, k) < 0, da(n, k) < 0 =dw (n, k) < dw (n + 1, k).PROOF. Underhypothesis i), Lemma4 impliesda(n, k+ 1) < 0. Usingthisindw(n, k) = dw(n + , k) - da(n, k+ l)+da(n, k) we getdW(n, k) > dw(n+ 1, k).The proofof (ii) is similar. [ILEMMA6. Thefollowing holdfor any n and k:(i) dw(n, k) > 0, da(n, k) < 0 dW(n + 1, k) > 0, da(n, k + 1) < 0;

(ii) dW(n, k) < 0, da(n, k) > 0 = dw(n + 1, k) < O, da(n, k + 1) > 0.Theprooffollows fromLemma2.4. The evolution of the bubbles. From the processes (A1,I) and (A2,12)and the serviceprocessat node 1, we deduce the statusof the red andbluebubblesat node 2. Iteases theexpositionto do thisgradually,o firstconsider he evolutionof bubbles n certainsimplesituations.One can isolatethreebasicpossibilitiesforbubbleevolutionandallotherpossibilitiescanbe described n termsof these three.Possibility a: bubbles can only move to the right. Consider only the first redbubbleand ignore all others,that is, from the processes A1 andI1 construct he

process A' = {A(n, l)}neZ as follows: A(n, 1) = A(n, 1) for all n E R(1, 1) andA(n, 1) = I(n, 1) for all n ? ((1, 1). We then modify the definition of ,R(1, 1)andlet it equalthe set {/: r(l, 1) < 1 < q}, whereq = min{n > r(l, 1): A(k, 1) =I(k, 1), Vk > n}. Since A(n, 1) > I(n, 1) for each n E Z, WA(n, 1) < WI(n, 1)and A(n,2) > I(n,2). For k = 1,2, let da(n,k) = A(n,k) - I(n,k) anddW(n, k) = WA(n, k) - WI (n, k).

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B. PRABHAKAR

Since d(b, 2) < , by Lemma 4, dw(b + 1,1) > 0 and En=r( )da(n, 1) > O,bydefinitionof r(l, 1). Therefore, t follows thatB (1, 2) < d (r(1, 1), 1) and thatB(1, 2) < B(1, 1).We shall now prove that B(1, 1) - B(1, 2) = R(1, 1) - R(1,2). This willestablishbothpart iii), and n conjunctionwith B(1, 2) < B(1, 1) it will also showthatR(1,2) < R(1, 1).Let i = min{n > q: W1(n, 1) = 0}. By the stability of the I-queue, it againfollows that i < oo. It also follows (as before) thatdw(i, 1) = 0. We use this andthe fact thatdw(b(1, 1), 0) = 0 as follows:

i-iO= dW(i, 1)= da(n, 2)-da(n, 1)+ dW(b(l, 1), 1)

n=b(l,1)= R(1, 2)- B(1,2) - R(, 1) + B(1, 1)

orR(1, 1)- R(1, 2) = B(1, 1)- B(1, 2).

This concludes theproofof the lemma. DREMARK. Again, the lemma establishes that bubbles only move to theright, that their volumes do not increase, and that they do not overtake oneanother.Volumecancellations are equal andhappen,in this case, when the bluebubble moves into the red one. This movement is manifestedby the conditiondW(r(1, 1), 1) > 0. For,this is the only conditionunderwhich B(1, 2) < B(1, 1).To conclude Possibility ,B,we need to consider the case that at least one ofb andr is not finite.Accordingly,we state the followingdefinitions:(a) If b =-oo andr < oo, definer(1,2) = r, b(1,2) = r(l,2), S(1,2) = 0and R(1, 2) = [r(l, 2), q'], whereq' is as definedearlier.(b) If b > -oo and r = oo, define b(1,2) = b, r(l,2) = oo, B(1,2)

[b(1,2),q ], where q = min{n > b(1,2):A(k, 2) = I(k,2) Vk > n} andJ(1,2)= 0.(c) If lbl= Irl= oo, defineb(1, 2) = r(l, 2) = oo ando(1, 2) =


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FIXED POINTS OF A ./GI/1 QUEUELet i = min{n:da(k, O)= 0 = dW(k,0) Vk > n}. Note that the stabilityof bothqueuesensures thati < oo. From

i-10 = dW(i, )= E da(n, 2) - da(n, 1) + d(b(, 1), 1),n=b(l, 1)= R(1,2)- B(1, 2)- R(1, 1)+ B(1, 1)

orR(1, 1)- R(1, 2) = B(1, 1)- B(1, 2),

we obtain a proofof the lemma. DThe last of the threepossibilitiesconcernsthe mixing of two neighboringblue(red)bubbles,afterthe red(resp.blue)bubblebetween them has been cancelled.Possibilityy: the mixing of bubblesof the same color. This time consider thefirst two blue bubbles and the first redbubbleonly. Thatis, set A(n, 1) = A(n, 1)for n E (S(1, 1) U3(1, 1) U S(2, 1)) and A(n, 1) = I(n, 1) otherwise.Again,we modify the definition of S(2, 1), settingit equal to {b(2, 1) < n < q}, whereq = minn > b(2, 1): A(k, 1) = I(k, 1) Vk > n}.Given theprecedingdiscussion of possibilitiesa and,, thedynamicsof bubbleevolutionundery areeasy to understand.Define bl = inf{n: A(n, 2) < I(n, 2)}, b2 = sup{n:A(n, 2) < I(n, 2)},r1 =inf{n:A(n, 2) > I(n, 2)}, and r2 = sup{n:A(n, 2) > I(n, 2)}. Following earlier

conventions, he infimum(supremum)of the emptyset equalsoo (resp. -oo).LEMMA11. Supposebl, b2, rF and r2 are all finite. Thenexactlyone of the

following must be true:(i) bl _ Ofor all n E [rl, r2],and(iii) rl r(l, 1).CASE 1 [rF> b(2, 1)]. This implies F2> b2 > b(2, 1). Since da(b2,2) < 0,by Lemma 4 we get that dw b2 + 1, 1) > 0. This togetherwith the two facts:b2 b(2, 1), andda(n, 1) < 0 for alln > b(2, 1) implies,recursivelyvia Lemma6,thatda(n, 2) < 0 for all n > b(2, 1). This contradicts 2 > b2.

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B. PRABHAKARCASE2 [r(l, 1) < rl < b(2, 1)]. Lemma4 implies dW(rl+ 1, 1) < 0. Since

da(n, 1) > 0 for n e [r(l, 1), b(2, 1) - I], Lemma 6 implies da(n,2) > 0 forsuch n. This further mplies b2 > b(2, 1). Now a similarargument o the one inCase 1makes t impossiblefor there o be ann > b2such thatda(n, 2) > 0. Again,this contradicts r2 > b2.

Therefore, if bl < rl, either bl < b2 < rl < r2 [this proves (i)] or bl < r1 r2.Finally, suppose that rl < bl. For contradictionsuppose that b1 < r2. Thepreceding argumentsmake it clear thatr\ > r(l, 1), hence b1 > r(l, 1). We firstclaim that bl > b(2, 1). If not r(l, 1) < rl < bl < b(2, 1). But, da(rl, 1) > 0

implies dw(ri + 1, 1) < 0 (by Lemma4); and, in conjunctionwith da(n, 1) > 0for n E [r(l, 1), b(2, 1) - 1] this further mplies (via Lemma6) thatda(n, 2) > 0forn E [r(l, 1), b(2, 1) - 1].This contradictsbl < b(2, 1).Thus, our assumption hat b1 < r2 leads to the conclusion b(2, 1) < bl < r2.Since da(b1,2) < 0, Lemma 4 implies dw(bl + 1, 1) > 0. This and the factda(n, 1) < 0 for n > b(2, 1) imply(via Lemma6) thatda(n, 2) < 0 for all suchn.This contradictsr2 > b1. Therefore, f rl < bl, it mustbe that r < r2'< bl < b2andthe lemmais proved. O

REMARK. The essence of the lemma is that bubbles do not overtake orinterspersebetweenone another.Thatis, there areuninterruptedunsof blue andred volumes whenever these are not zero. This intuitivestatement s mademoreprecisein the next few lemmas.Suppose that rl, r2,b1 and b2 are all finite and bl < rl < r2 < b2. Defineb(1,2) = , r(1,2) = rl andb(2, 2) = minn : 2 < n < b2, d(n, 2) < 0}. Alsodefine B(, (1,2) - ,2) da(n, 2) R(1,2) (,2)= n, 2) and B2 2)deineB 1) - ---.,n=b(1,2) '- b(2.2)da(n, 2). The previouslemma implies B(1, 2) > 0, R(1, 2) > 0 andB(2, 2) > 0.

LEMMA 2. Supposethatrl, r2, bl and b2 are all finite and bl < rl < r2 b(1, 1), r(l, 2) > r(1, 1) and b(2, 2) > b(2, 1),(ii) B(l, 2) < B(1, 1), R(1, 2) < R(1, 1) and B(2, 1) < B(2, 2), and(iii) R(1, 1) - R(1, 2) = B(1, 1) + B(2, 1) - B(1, 2) - B(2, 2).

PROOF. Statement(i) follows from Possibilities a and fBand the proof ofLemma11.

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FIXED OINTS FA ./GI/1QUEUEIt againfollows from Possibility B that B(1, 2) < B(1, 1). We shall provethelemma by showing that B(2, 2) < B(2, 1) and R(1, 1) - R(1,2) = B(1, 1) +

B(2, 1) - B(1, 2) - B(2, 2). Weproceedby establishingsomepreliminary laims.CLAIM1. Forb = max{n E [b(1, 2), r(l, 2) - 1] da(n, 2) < 0}, it holds thatb < b(2, 1).Supposeto thecontraryb > b(2, 1).Then,sinceda b, 2) < 0, Lemma4 implies

dW(b + 1, 1) > 0. And since da(n, 1) < 0 for all n > b(2, 1), this further impliesvia Lemma6 thatda(n, 2) < 0 for all n > b. This contradicts he finitenessof r2and establishesthe claim.CLAIM . dW(b(2, 1), 1) < 0.Suppose dW(b(2,1), 1) > 0. Since da(n, 2) < 0 for all n > b(2, 1), Lemma 6tells us that da(n, 2) < 0 for all n > b(2, 1). Thereforer(1, 2) < b(2, 1). Now,

given thatda(r(1, 2), 2) > 0, Lemma4 implies dW(r(1,2) + 1, 1) r(l, 1), this further mplies (via Lemma 6) that dw(n, 1) < 0 for n E[r(1, 2) + 1,b(2, 1)]. This contradictsourassumption hatdw(b(2, 1), 1) > 0 andprovesthe claim.Now, given that b < b(2, 1), it follows from the definition of b(2, 2) thatda(n, 2) > 0 for all n E [b(2, 1), b(2, 2) - 1]. Therefore 14) gives(17) d(i +1,1)= E da(n,2)-da(n, l)+ d(b(2, 1),1),

n=b(2, 1)where i = min{n> b(2, 1): dW(k,1) = 0 Vk > n}. Rewriting 17) we get

b(2,2)-10= L da(n,2)-B(2,2)+B(2, l)+ w(b(2,1), 1)n=b(2, 1)

orb(2,2)-1B(2, 1)- B(2, 2) = -dw(b(2, 1), 1)- d(n, 2).n=b(2, 1)

If da(n, 2) = 0 for all n E [b(2, 1), b(2, 2) - 1], the aboveequation mmediatelygives B(2, 1) - B(2, 2) = -d(b(2, 1), 1) > 0.Else, r2> b(2, 1) forr2 as definedearlier. n this case note that ,=b(2 1d) (n,2) = r=(2 ,1da(n, 2). Therefore it suffices to show -d(b(2, 1), 1) -Er=2 ) da(n, 2) > 0 in order to conclude B(2, 1) > B(2, 2).n=b(2, ) .

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B. PRABHAKARBut, this follows immediately from considering

r~2 i2dw(b(2 1), 1) =dw(2 + 1, 1)- d1)(n,2)+ da(n, 1n=b(2, 1) n=b(2, 1)

r2edw(b (2, 1), 1)+ ): da(n, 2)n=b(2, 1)

rdw(w2+ 1, 1 + >C da n, 1) < 0n=b(2,1)since da (q2, 2) > 0 implies dw(z2 ? 1, 1) r(1,1) and b(2, 2) > b(2,l1),(ii) B(1,2) < B(l, 1) and R(l,2) < R(1,l ),

(iii) R(1, 1)- R(1, 2) B(1, 1) ? B(2, 1)- B(1, 2)- B(2, 2).PROOF. Part (i) follows from Possibilities a and 8, and clearly b(2, 2) = oc >b(2, 1).It follows as in the proofof Lemma 12 thatdw(b(2, 1), 1) < 0. (In words,thismeans no blue volume from the first blue bubble enters the second blue bubble.)As a consequence,it is straightforwardo inferfromPossibilityf thatB(1, 2) b(2,1):dw(k,l)=OVk>n}. D

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FIXEDPOINTSOFA */GI/1 QUEUENext suppose bL,b2, F, and F2 are all finite and Fl < F2 < LI < b2. Thendefine b(1, 2) = r(1, 2) = Fl and b(2, 2) = b1. Set B(1, 2) = 0, define R(1, 2)~-r2 baan )2n=r(1l2)da (n, 2) and B(2, 2) = - >n=b(2,2)a (1, 2).LEMMA14. If r1,r2,b1 and b2 are allfinite and Fl r 1, 1) and b(2, 2) > b(2, 1),(ii) R(1, 2) < R(1, 1) and B(2, 2) < B(2, 1),(iii) R(l, 1) - R(l, 2) = B(1, 1) + B(2, 1) - B(1, 2) - B(2, 2).Theprooffollows from Case 8, similarlyas in theproofof Lemma 13.Possibility y': only the redbubblesurvives. SupposethatL1andb2are not fi-nite. In this case, there s no blue at the second stage.Defineb(l, 2) = r(l, 2) = Fjand set b(2,2) = co. Also set B(1,2) = B(2, 2) = 0 and define R(1,2) =

Lnr(,2) da(n, 2).It is clear that b(1, 2) > b(1, 1), r(1, 2) > r(l, 1) and b(2, 2) > b(2, 1).LEMMA 15. Suppose that bl and b2 are notfinite. Then R(l, 2) = R(l, 1) -B(1,1) - B(2,l). In particular, B(1,2) < B(1,1), R(1,2) < R(1,1) andB(2, 2) b(2, 1): dw(k, 1) = 0 Vk > n}. Given that dw(b(l, 1), 1) = 0,it follows immediately hatR(1,2) = R(l, l) - B(1, 1) - B(2, 1). D

Possibility y *: the red bubble is completely cancelled. Suppose that rFI nd F2are not finite. In this case, there is no red at the second stage. Define b(l, 2)=r(l, 2) = b(2,2)=b1 andEnb(1,2) da(n, 2)

B(l,l1)+B(2,1~)B(2, 2)=B(2, 1) -n=b(1,2)

B(l, 1) ? B(2, 1)R(l,2)=0.Note that it is possible for r(1, 2) < r(l, 1) and b(2, 2) < b(2, 1). This is acrucialdeviation from all previouscases, necessitatedby reasons detailed in theremarkafter the lemma.

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B. PRABHAKARLEMMA16. Suppose that rl and r2 are notfinite. Then:

(i) b(1, 2) < r(1, 2) < b(2, 2),(ii) B(i, 2) < B(i, )for i = 1,2 and R(1,2) < R(1, 1) and(iii) B(1, 1) + B(2, 1)- B(1, 2)- B(2,2)) = R(, 1).

PROOF. Statement (i) follows by definition, while (ii) and (iii) follow fromiO= dW(i+ 1, 1) = jda(n, 2) - da(n, 1) + dW(b(l, 1), 1),

n=b(l, 1)where i = min{n > b(2, 1) dw(k, 1) = 0 V1k n}. Observing that dw(b(l, 1),1) = 0 completes the proof of the lemma. D

REMARK. The definitions of quantities at the second stage are so as topreserve orderings between bubble start points and to ensure that bubble volumesdo not grow. The above lemma shows that when the red bubble is fully cancelled,it nullifies anequalamountof bluefrom the blue bubblesput together.The amountof volume taken out of each blue bubble is proportional to its original size. Thus,we do not keep an account of whether or not a specific blue bubble contributedto the cancelling of the red. This is both unnecessary and can lead to needlesscomplication, as seen below.First, without elaboration, here are the ways (and concomitant conditions) inwhich the red bubble can be cancelled: (i) the first blue bubble cancels all ofthe red bubble [dW(r(l, 1), 1) > R(1, 1) and dW(b(2, 1), 1) > 0], (ii) each of theblue bubbles contributes to the cancellation of the red bubble [dw(r(l, 1), 1) > 0,dw(b(2, 1), 1) = dW(r(l, 1), 1) - R(1, 1) < 0, and da(n, 2) < 0 for n > b(2, 1)]and (iii) only the second bubble cancels the red bubble [dW(r(l, 1), 1) = 0,-dW(b(2, 1), 1) = R(1, 1) and da(n, 2) > 0 for n > b(2, 1)].Note that in situations (ii) and (iii) above dW(b(2, 1), 1) < 0; or, in words, noblue volume enters the second blue bubble. This ensures that the two shades of bluedo not mix. But, if dW(b(2, 1), 1) > 0, as can happen in (i), the two shades of bluedo mix. This can make it impossible to decide b(2, 2) so as to satisfy the followingconditions simultaneously: (a) B(1, 2) < B(1, 1) and (b) B(2, 2) < B(2, 1).For example, suppose that B(1, 1) = B(2, 1) = 100, R(1, 1) = 50, dW(r(1, 1),1) = 100 and dw(b(2, 1), 1) = 25. Also suppose that da(l, 2) = 25 for some1E [r(l, 1), b(2, 1)- 1] and that da(m, 2) = 100, da(n, 2) = 25 for some m < n[b(2, 1), oo). Observe that no choice of b(2, 2) can satisfy conditions (a) and (b)above.

A simple way out is to set r(l, 2) = b(2, 2) = b(1, 2), and divide the volume ofblue on the output side proportionately among the two blue bubbles. Althoughthis choice can cause r(1, 2) < r(1, 1), it must be seen as a consequence ofconvenience. It will be clear that this causes no problems in the rest of theargument.

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FIXEDPOINTSOFA ./GI/1 QUEUE4.1. The equilibrium evolution. Using the ideas of the previous section,we will now describe the evolution of the bubbles in equilibrium. Consider

the processes A1, I1 and {S(n, l)}nZ. The quantities r(n, 1), b(n, 1), R(n, 1),B(n, 1), R(n, 1) and ?(n, 1) are as defined in (9), (11) and (12).We describe the procedure for determining r(n, 2) and b(n, 2) for each n E Z.From these one can deducethe quantitiesR(n, 2), B(n, 2), R (n, 2) and S(n, 2).As in the previous section the sequence {dW(n, l)}ne, plays a key role inthe determination of r(n,2) and b(n,2). For what follows, it is helpful tomake a connection between the sign of dw(n, 1) and what it means for bubblemovements at n. Accordingly, depending on whether dW(n, 1) = 0, dw(n, 1) < 0or dw (n, 1) > 0, there is a movement from n - 1 to n, respectively, of nothing, redor blue of volume Id (n, 1)1.Consider 8 = {n: da(n, 2) > 0 infinitely often}. By the joint ergodicity ofA2 and I2, P (8) = 0 or 1. If P (8) = Othen da (n, 2) < 0 for every n a.s. [ergodicityclearly rules out that da(n, 2) < 0 for finitely many n with positive probability].But this last fact togetherwith E(A(n, 2)) = E(I(n, 2)) implies thatA2 = 12 a.s.Thus,if P () = 0 the proofof Theorem 1 is complete.Therefore, suppose P(8) = 1. Note that this implies the co-existence ofinfinitely many blue and red bubbles at the second stage. We will now give aprocedure or determiningb(1, 2) and r(1, 2), and hence for b(n, 2) and r(n, 2)for everyn.First consider the processes A2 and I2. These arejointly ergodic, and hence it ispossible to apply the procedure of Section 2.1 and obtain bubbles. Let b(n, 2)and r(n, 2) be the start points of the bubbles, and let B(n, 2) and R(n, 2) bethe corresponding bubble volumes. Note that r(n - 1, 2) < b(n, 2) < r(n, 2) forevery n. We need variables e(n, 2) and f(n, 2) which mark the end points of thebubbles in order to proceed. Thus, let

e(n, 2) = max{k e [b(n, 2), r(n, 2) - 1]: da(n, 2) < 0},f(n, 2) = max{k E [r(n, 2), b(n + 1, 2) - 1]: da(n, 2) > 0}.

Note that b(n,2) < e(n,2) < r(n, 2) < f(n,2) < b(n + 1,2) for every n.With these definitions, the following procedure relates r(-,2) and b(.,2) to(-, 2) and b(, 2).Determining b(1, 2). Clearly b(1, 1) E [b(k, 2), b(k + 1, 2) - 1] for some k.(a) If b(1, 1) E [b(k, 2), e(k, 2)], set b(1, 2) = b(k, 2).(b) If b(l, 1) E [e(k, 2) + 1, (k, 2)] and r(1, 1) < f(k, 2), set b(1,2)=r(k, 2).

(b') If b(1, 1) E [e(k, 2) + 1, r(k, 2)] and r(l, 1) > f (k, 2), set b(l, 2) = b(k +1,2).(c) If b(1, 1) E [r(k, 2) + 1, f(k, 2)] and r(1, 1) E [b(1, 1) + 1, f (k, 2)], setb(1, 2) = r(k, 2).

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B. PRABHAKAR(c') If b(1, 1) E [J(k,2) + 1, f(k, 2)] and r(1, 1) 4 [b(1, 1) + 1, f(k, 2)], set

b(l, 2) = b(k + 1, 2).(d) If b(l, 1) E [f(k, 2) + 1, b(k + 1, 2) - 1], set b(l, 2) = b(k ? 1, 2).Determining r(1, 2). Clearly r(1, 1) e [?(k, 2), i (k + 1,2) - 1] for some k.(e) If r(l, 1) E [iz(k,2), f(k, 2)], set r(1, 2) = i (k, 2).(f ) If r (1, 1) E [f (k, 2) + 1, b(k + 1, 2)] and b(2, 1) < e(k + 1, 2), set r (1, 2)b(k + 1,2).(f') Ifr(I,,I)EE[f(k, 2) +I, b(k +1, 2)3and b(2, 1)>> (k +1, 2), set r(1,2)==i~(k+ 1,2).(g) If r(l, 2) E [b(k + 1, 2) + 1, (k + 1, 2)] and b(2, 1) E [r(l, 1) + 1, i(k +1,2)], set r(1, 2) = ,(k + 1,2).

(g') If r(l, 1) E [b(k + 1, 2) + 1, i(k + 19 2)] and b(2, 1) B [r(l, 1) + 19e(k +192)], set r(l, 2) = r (k+ 1, 2).(h) If r(1, 1) E [j(k, 2) + 11T(k + 1,2) - i], set r(l, 2) = i(k + 1,2).

LEMMA17. Ifb(1,1) < b(p, 2) then b(1,2) < b(p, 2).PROoF. If p = k, for k as defined in the above procedure, then from case (a),

b(Ig 1) =b(p, 2) =b(1,2). If p > k + 1, then note that i(k, 2) i(k + 1,2) > r(n, 1), it follows from cases (b),(b'), (c) and (c') that b(n + 1, 2) > F k ? 1,2) = r(n, 2).Finally, under (h), we again see that b(n + 1, 1) > r(n, 1) > i(k + 1,2). Againfrom cases (b), (b'), (c) and (c') it follows that b(n + 1, 2) > i (k + 1, 2) = r(n, 2).This concludes the proof of the lemma. D

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FIXED POINTS OF A ./GI/1 QUEUELEMMA 9. At the start of a red bubble at the second stage, there is alwaysexactly one more r(., 2) than there are b(., 2)'s. Similarly,at the start of a blue

bubble at the second stage, there is always exactly one more b(., 2) than thereare r(., 2)'s.PROOF. For concreteness,considerr(1, 2). By the order-preservationstab-lished in Lemma 18, it suffices to show that r(1, 2) = r(k, 2) = b(k + 1, 2) =.. = r(k + m, 2) for some k and m > 0.Consider e(1,2). If max{p:b(p, 1) < e(1,2)} > max{q:r(q, 1) < e(1,2)},then we claim thatthereexists anr(j, 1) E [e(1, 2) + 1,r(l, 2)]. Supposenot.Thismeans da(n, 1) < 0 for all n E [e(1,2) + 1,r(1, 2)]. By Lemma4, dw(e(1, 2) +1, 1) > 0 andfrom the fact thatda(n, 1) 0. This is becausedw(e(1, 2) + 1, 1) > 0, by Lemma 4. Since da(n, 1) < 0 for n e [e(1,2) +1,r(J, 1) - 1], by a recursiveuse of Lemma6, we get thatdw(r(J, 1), 1) > 0).]On the otherhand,supposethatmax{p: b(p, 1) < e(1, 2)} < max{q:r(q, 1) 0 for all n E [f(0,2) + 1, (l, 2)] (because b(Q, 1) < r(Q, 1) e(1,2)) it follows recursivelyfrom Lemma 6that da(n,2) > 0 for all n E [f(0, 2) + 1, (1, 2)]. This contradictsb(1,2) E[f(0, 2) + 1, (1,2)].By ourprocedure cases(f) and(f)], r(Q, 2) = r(l, 2). Andfromtheprocedurefor b(Q, 1), we get that b(Q, 2) < r(1,2). This also identifies r(Q, 1) as thesmallest bubble startpointon the inputside thatgets mapped o r(l, 2).[Weagainnote thatdw(r(Q, 1), 1) > 0. Supposenot. Then,since da(n, 1) > 0for all n E [r(Q, 1), e(1, 2)], a repeateduse of Lemma 6 implies thatda(n, 2) > 0for all suchn. This contradictsda(e(l, 2), 2) < 0.]Thus, in both cases, we see thatthere exists a k such thatr(k, 2) = r(1, 2) >b(k, 2).Now, we shall identify r(k + m, 2). This is easy to see from our procedure:k + m = max{l:r(l, 1) > r(k, 1) andr(l, 1) < f(1, 2)}. Note that it follows fromourprocedure hatb(k + m + 1,2) > r(l, 2).This concludes theproofof the lemma. DNote from the aboveproofthatr(k, 1) < r(1, 2). Wealso claimthatdw(r(k, 1),1)> 0.

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B. PRABHAKARCOROLLARY. For r(k, 2) as definedin theproof of Lemma19, r(k, 1) 0.Both statements have been established during the proof of Lemma 19.REMARK. While Lemma 18 demonstrates that our procedure for determiningr(n, 2) and b(n, 2) preserves order, Lemma 19 is the more important. It records

precisely the identity of the bubbles in the input process that contribute to thevolume of a bubble at the output. That is, consider r(1, 2) = r(k, 2) = - =r(k +m, 2). This implies (as in case y *) that the only possible shades in the volumeR(, 2) are the red shades k to k + m. It also implies that all the intermediate blueshades have been completely cancelled. This influences the following definition ofbubble volumes at the output stage.

Determining B(1, 2). Consider b(l, 2). If b(l, 2) = r(n, 2) for some n, then setB(1, 2) = 0. Else, b(l, 2) = b(n, 2) for some n. Suppose b(n, 2) = b(k, 2) = =b(l,2) = = b(k + m, 2). Then set

B(n, 2)(1,2) B(i, 1)

As in Possibility y*, this credits each of the blue bubbles proportionately orvanquishing the intermediate red bubbles.

Determining R(1, 2). Similarly as above.

LEMMA 0. Foreveryn, R(n, 2) < R(n, 1) and B(n, 2) < B(n, 1).PROOF. We establish R(1,2) < R(1, 1). There is nothing to prove if

R(1, 2) = 0. Else, there is a p such that r(p, 2) = r(k, 2) = . r(l,2) = =r(k + m, 2) for some k and m. And

R(p,2)R(1,2) - R(1, 1)( p, 2)km^R(i, 1)Therefore, t sufficesto provethat

k+m(18) R(p, 2) < E R(i, 1).i=kFirst note that r(k, 1) < r?(p, 2) and from Corollary 1 dW(r(k, 1), 1) > 0. Alsonote that dw(f(p, 2) + 1, 1) < 0 and dw(e(p, 2) + 1, 1) > 0, using Lemma 4. Weconsider two cases.

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FIXED POINTS OF A ./GI/1 QUEUEexists a.s., andmust be treatedas a randomquantity.But

yn [Xk(n)] +_,n Xk(/)]+d = limd(k) = lim lim =-n limlim =k k n 2n+ 1 n k 2n+ 1where the limit interchanges due to thefollowing.First,observethat

d(k) > limsup nn 2n+ 1foreveryk, due to themonotonicityof [Xk 1)]+.By Fatou's emma,for each n weget

imin [X1 (l)]+ - [X(1)]+ -X k(l)]+iminf < liminfk 2n + 1 k 2n + lI-n =-nwhich implies liminfn((E_l=_n[X(l)]+)/(2n + 1)) > limsupkd(k). Therefored = x, makingx an almostsureconstant.Thus,the assumptiond(k) -/ 0 leads to the co-existence,withprobability1, ofeverblue and everredbubbles in A1 andI1 of strictly positive volume per arrivalequal to d. Since the shades of all redand blue bubbles aredistinct,the everredsand everblues have distinct shades. By Lemma 21, the orderingof the red andblue bubbles,and hence of the everredandeverbluebubbles,is preservedateachstagek.Consider the subprocess {X(l)l\X(l)>l>,1 E Z} and its support set E(1)={l: IX(I)I > E}. Call this the processof chosen everredand everbluesegmentsand observe that it is stationarysince {X(1), E Z} is stationary.Now 6(1)can be written as the disjoint union of two sets: Gr(1) = {l:X(l) > e} andgb(l) = {I:X(l) < -6}, which support chosen everred and everblue segmentsrespectively.Giventhatd > 0, foranye E (0, d) we get thatthedensityof pointsinb 1) and 8r(1) for this choice of Eis strictlypositivea.s. (butpossibly random).Fix one such E and observe that the process of chosen segments appearsas analternating equenceof everredand everbluesegments.Consider heleft endpointsof a run of chosen everbluesegments,and et ?Cb(1) C gb(1) be the set of integerswhich supportthese segments.The process of left chosen everbluesegments isalso stationary since chosen everbluesegmentsare stationary)andtherefore heset ?db(l) has a possiblyrandomdensitywhichmustbe strictlypositivea.s. [elsegb(1) cannothave strictlypositivedensity].A crucialconsequenceof ourconstruction s thatany two chosen left everbluesegmentsmust be shadedwithdifferentcolors of blue, since theyareseparatedbychosen everredsegments.Now considerthe startingpoint of the blue bubbles towhich the chosen left everbluesegmentsbelong, and write tb(1) for the integerswhich form these starting points. Note that the set of points in Sb(1) form astationarysequence. Since there is a one-to-one correspondencebetween points

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B. PRABHAKARin C b(1) and 8b(1), their densities are almost surely equal. In particular, hedensity of -8b(1),denoted by the random variable C, is almost surely strictlypositive.Thus, we have obtained at stage 1 the existence of blue bubbles with thefollowing properties:

(a) their volumes are at least ?;(b) betweenanytwo of them there is a redbubble with volume at least e; and(c) their startpointshavedensityC > 0 a.s.By definition, for each n the nth everredand everblue bubbles retain theirvolume atevery stagek. Andby Lemma21 theoriginalorderingbetweeneverredsandeverblues is preserved hroughout.This allows us to use the same argumentas above at each stage k and obtain, under our assumption that d(k) -- 0, theexistence of blue bubblessatisfying properties a), (b) and(c) listed above ateverystagek.Proceeding, choose 8 > 0 such that P(C > 8) > 8. Write C = Cl + Cg, whereCl is thedensityof the startpoints,1,of blue bubblessatisfyingproperty a) abovewith the additionalproperty hat there is anotherblue with volume at least e andstartpoint 1' such that l' < I + 2/8 and there is a red bubble with volume at leastE between them. Let Cg be the density of the startpoints of the remainingbluebubblessatisfying properties a) and (b). By definition,Cg < 8/2 a.s. ThereforeCl > 8/2 wheneverC > 8. Choose 8 so that2/8 is an integer.Therefore,our above arguments mply thatfor any k there exist blue bubbles

satisfyingthe following properties:(1) theirvolumes are at least s;(2) there s a red bubblewithvolume at leastEcontained nthe interval 1,1+ 2/8],where I is the startpointof the bluebubble;and(3) the densityof theirstartpointsis at least 8/2 withprobabilityat least 8.

Foranyk consider the event E:{thedensityof the startpointsof bluebubblessatisfying1 and2 > 8/2}.

Thisevent is shift-invariantndcontained n thejointlyergodicprocesses(Ak,Ik).By property3, P(E) > 8. ThereforeP(E) = 1. We record this in the followinglemma.

LEMMA 2. If d > O,thenthere exist strictlypositive E and 8 not dependingon k such that thereexist blue bubblesin (Ak,Ik) satisfyingproperties 1 and 2above. Further, he startpoints of these blue bubbleshave, withprobability1, adensityat least 8/2.

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FIXEDPOINTSOFA ./GI/1 QUEUE5.1. Unbounded ervice times. We specializeto service times whose supportis unbounded and deal with bounded service times in the next section. By the

independenceof Ik from the service process, the i.i.d. nature and unboundedsupportof the service times {S(n, k), n E Z}, we obtain for the nth arrivalof theprocessIk(19) P S(i,k)>I(i,k)forallie n,n+l+- Ik)>0 a.s.Given that the above conditionalprobability s strictlypositive for each arrivalof Ik, a smallenoughchoice of y gives the followinglemma.

LEMMA23. Ifd > 0, then there exist strictly positive E,8 and y not dependingon k such that there exist blue bubbles in (Ak, Ik) with thefollowing properties:(A) their volumes are at least e;(B) there is a red bubble with volume at least 8 contained in the interval

[1,1+ 2/3], where I is the start point of the blue bubble; and(C) P(S(i,k) > I(i,k)for all i E [1,1 + +2/]Ik) > y and whose start pointshave density at least 8/3 a.s.

We shallproveTheorem1 afterstatingthe followinglemma.LEMMA24. Let I = b(n, k) be the start point of a blue bubble whose volumeB(n, k) > E, and let r(m, k) > b(n, k), m > n, start a red bubble whose volumeR(m, k) > E. Further suppose that both bubbles are contained in the interval[1, + L] (i.e., R(m, k) C [, 1 + L]). If S(i, k) > I(i, k)for all i E [1, + L], theneither B(n, k + 1) = 0 or R(m, k + 1) = 0.PROOF. From the proofof Lemma 18 we know thatb(n, k + 1) equalseitherthe startof a red bubbleor thestartof a bluebubbleatstagek+ 1.Firstsuppose hatb(n, k+ 1) equalsthe startof a redbubble.Then,by theprocedure ordeterminingB(n, k + 1), it follows thatB(n, k + 1) = 0.Next suppose that b(n, k + 1) equals the start of a blue bubble. We claimr(m, k+ 1) = b(n, k+ 1);hencer(m, k+ 1) equalsthestartof abluebubblewhichimplies R(m, k + 1) = 0. To establish the claim, first observe that da (n, k + 1) > 0for all n E [1,1+ L] by (ii) of Lemma 1. Therefore,one of the following mustbetrue:(1) b(n, k + 1) < 1 or (2) b(n, k + 1) > I + L.Undercase (1), sinceb(n, k+ 1) < b(n, k) [i.e., the startpointof thebluebubblemoved left to the startpoint, say b(p, k + 1), of a blue bubble at stage k + 1]we are in Case (a) of the procedurefor determiningb(n, k + 1). Accordingly,b(n, k) E [b(p, k+ 1), e(p, k+ 1)], wheree(p, k+ 1) is as defined n theprocedurefor deciding b(n, k + 1). Since da(n, k + 1) > 0 for all n E [1,1+ L], it followsfrom the definition of e(p, k + 1) that e(p, k + 1) > 1 + L and in fact that

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FIXED POINTSOFA ./GI/1 QUEUEREMARK. In words, the above lemma states that under the event Gk if thereis anymovementof red volume to therightat locationL in the interval[1,1+ 2/8]

(dw(L + 1,k) < 0), therecannotbe anyblue volume left in [1, L] (da(i, k+ 1) > 0for all i e [1,L]). That is, if any red volume moves to the right under Gk wemay infer that the blue bubble of volume at least E has been fully cancelled. Theevent Gk, therefore, ensures that red volume stays in [1,1 + 2/8] so long as thereis any blue volume in this interval to the left of (or, ahead of) the red bubble.

PROOFOF LEMMA25. If dW(L + 1, k) < 0 then WI(L + 1,k) > WA(L +1, k) > 0. But, if W (L + 1, k) > 0 then since I(i, k) > S(i, k) for all i E [ + 1, L],it follows inductively from the recursion

WI i + 1,k) = [WI i, k) + S(i, k) - I(i, k)]+that WI (i, k) > 0 for all i E [1 + 1, L]. Or, equally, that I(i, k) - S(i, k) -WI(i, k) < 0 for all i E [1, L]. Now, from the equation

I(i,k + 1)= [I(i,k) - S(i,k) - WI(i,k)]+ + S(i,k + 1)we deduce that I(i,k + 1) = S(i + 1,k) for all i e [1,L]. Since it is alwaystrue that A(i, k + 1) > S(i + 1,k) [see equation(4)], we get thatda(i, k + 1) =A(i, k+ ) - I(i, k+ 1) >O for alli e [, L].

COROLLARY . Let G = n'k+K Gp. If, under G, dW(L + 1, p) < Ofor someL E [1,1+ 2/8] and p E[k, k + K], thenda(i, p + 1) > Oforall i [1,L].The event G ensures that the red bubble does not move out of the interval[1,1 + 2/8] so long as there is blue volume in it during stages k through k + Kto the left of the red bubble.We now consider the second part:ensuring the blue bubble is forced to the rightand cancels the red volume. Toward this end consider the event

H = {S(l, m) > b for all m E [k, k + K]}.On the event G n H the service time of customer 1 is greater than the servicetimes of all the subsequent 1 + 2/8 customers by at least b - a during stages kthrough k + K. Given the first-come-first-served nature of the service discipline,this implies customer I will be slowed down during stages k through k + Kallowing customers 1 + 1 through / + I + 2/8 to catch up. Now the event Fbounds the separation between customers 1 through / + 1 + 2/8. Therefore, we areguaranteed that under F n G n H customers 1 through 1+ 1+ 2/8 will be served inone busy cycle at stage k + K. That is, the interdeparturetimes from stage k + Kfor the I-process will all equal service times: I (i, k + K + 1) = S(i + 1, k + K)for all i E [1,1+ 2/8]. We establish the above formally in the following lemma.

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B. PRABHAKARLEMMA 6. Assumethe event F n G n H holds. Thenthere existsa K' < Ksuch thatda(i, k + K' + 1) > Oforall i E [1,1+ 2/3].PROOF. Recursively rom(5) we obtain

1+2/8 I(i, k+ 1)- W (l + +2/3, k)i=l

1+2/8= I(i,k)-S(l,k) + S(l + +2/, k)-W(1, k).i=l

SettingCj = E1/= I (i, j), theabovebecomesCk+l = Ck- S(l, k) + S(1+ 1 + 2/S, k) - W(l, k) + W (1+ 1+ 2/8, k).

Supposethat WI(1+ 1 + 2/8, j) = 0 for all j E [k,k + K]. Then, applyingtheprevious equationrecursively,we get that Ck+K< Ck- K(b - a). This impliesCk+K< 0 on the event F n H, which is a contradiction.Therefore, t mustbe thatthereis a K' < K suchthatWI(1+ 1+ 2/, K') > 0.Now from Wl(n + 1,j) = [W'(n,j) + S(n, j) - I(n,j)+ we get thatWI(i, K') > 0 for all i E [1+ 1, 1+ 1+ 2/8], since S(i, K') < I (i, K') for all suchi(under he eventG). Thisimpliesvia (3) thatI (i, k+ K' + 1) = S(i + 1,k+ K') forall i [1,1 + 2/8]. Since A(n, m + 1) > S(n + 1,m) foreveryn andm [from(4)],it follows thatda(i, k + K' + 1) > 0 for all i E [1,1+ 2/8].Therefore,underthe event F n G n H, we have ensured thatduringstages k

throughk+ K' (i) no red volume leaves [1,1+ 2/8] (event G), and(ii) andonlyredvolumeremains n theinterval 1,1+ 2/5] atstagek+ K' (undereventF n G n H).This implies one of the following must have occurredby stage k + K': (i) theblue bubbleof volumeEwipedout theredbubbleand moved outsidethe interval,(ii) the blue bubblemoved into andgot cancelledby the red bubbleor (iii) somered volume enteredthe interval from the left and cancelled the blue bubble. Inall cases it follows that eithera red or a blue volume of E was cancelledbetweenstagesk andk+ K'. (Weomit a tediousargument, imilar o theone in theproofofLemma24, that dentifies he startpointsof thered andblue bubblesovermultiplestagesandinfers the abovevolumeloss.)Continuing, since Ik is a fixed point, it must satisfy all propertiesof anydepartureprocess. In particular,departureprocesses stochasticallydominatetheservice process: D(n,k) > S(n + 1,k - 1) for all n and k. Hence Ik muststochasticallydominatethe services. From this andthe independenceof servicesfromarrivals t follows that for each customer1the conditionalprobabilityof theevent G giventhe processIk is strictlypositive.Therefore,a small enoughchoiceof V2 nsures hatthedensityof customers1 EIkforwhich P(F n G n H IIk)> v2is at least 1/r - v2.

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B. PRABHAKARCOROLLARY . If a stationary and ergodicfixed point exists at mean T, thenit is unique.COROLLARY5. Suppose I1 and 12 are two stationary and ergodic fixed

points for a */GI/1 queue at means rl and T2, respectively. If Tl < T2, then 12stochasticallydominatesII; thatis, there exists a joint distributiony of I1 and12such that undery, I (n) < I2(n)for everyn a.s.PROOF. Let F1 = {F(n, 1),n E Z} be distributedas 12 and define A1 ={A(n, 1),n E Z}, where A(n, 1) = (rt/r2)F(n, 1), to be anotherergodic arrivalprocessof meantr. PassF1 and A1 througha tandemof ./GI/I queues givingthenth customers of both processes the same service time, S(n, k), at each stage k.Since A(n, 1) < F(n, 1) for all n, we get [from(3) and(4)] thatA(n, k) < F(n, k)for all n and k. Let Ykbe thejointdistribution f (Ak,Fk).By Theorem1, (Ak,Fk)convergesto (A?, F??)= (1, 12) in distribution.Let y,o be thejoint distributionof (A?, F??)and note that12 stochasticallydominatesI1 undery,o. Tofinish,set

Y = yoo.6. Conclusions and related work. Assumingthe existence of ergodicfixedpoints for a first-come-first-served/GII/ queue, we have used coupling argu-ments to show thattheyare attractors.As a consequencewe havealso seen thatanergodicfixedpointat mean r > 1 is unique.We note thatwhile the argumentsdonotplace anyrestrictionson the service time distributionotherthan thatit haveafinite meanand be nonconstant), hey rely cruciallyon the first-come-first-served

natureof the servicediscipline.Earlierwork on the uniquenessand attractivenessof fixed points for variousqueuescanbe foundin [1, 5], Section 9.4 of [6], [12-15]. While this list of papersis not exhaustive,combined with the referencescontainedin them, they give afullerpictureof earlierwork. The references[5] and[12] areparticularly elevantfor thepresentpaper.The existence of ergodic fixed points at some rates 1/r < 1 has beenestablishedby Mairesseand Prabhakar10] assuming hatthe service times satisfyf P(S(O,O) > u)1/2 du < oo. We referthe reader o [10] forprecisedetails andfora statementof furtherwork. The result of [10] relies on the work of Baccelli,Borovkovand Mairesse[2], GlynnandWhitt [7] and Martin[11] concerning helimiting behaviorof waiting times in a tandem of ./GI/1 queues with arbitraryinputprocesses.

Acknowledgments. I thank Jean Mairesse and Tom Mountfordfor manydiscussionson fixedpointtheorems orqueues.I thankChandraNairforcarefullyreading hepaperandhelping clarifythepresentation.

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FIXED POINTSOFA ./GI/1 QUEUEREFERENCES

[1] ANANTHARAM,. (1993). Uniquenessof stationaryergodic fixed point for a ./M/k node.Ann.Appl.Probab.3 154-173.[2] BACCELLI, ., BOROVKOV,. and MAIRESSE, . (2000). Asymptotic results on infinitetandemqueueingnetworks.Probab.TheoryRelated Fields 118 365-405.[3] BACCELLI,. and BREMAUD, . (1987). PalmProbabilitiesand StationaryQueues.LectureNotes in Statist.41. Springer,New York.[4] BURKE,P. J. (1956). The outputof a queueingsystem.Oper.Res. 4 699-704.[5] CHANG,C. S. (1994). On the input-outputmap of a G/G/1 queue. J. Appl. Probab. 311128-1133.

[6] DALEY,D. J. andVERE-JONES, . (1988). AnIntroductiono the Theoryof Point Processes.Springer,New York.[7] GLYNN,P.andWHITT,W. (1991). Departuresrommanyqueuesin series.Ann.Appl.Probab.1 546-572.[8] GRAY,R. M. (1988). Probability,RandomProcesses and ErgodicProperties.Springer,NewYork.[9] LOYNES,R. M. (1962). The stabilityof a queuewithnon-independentnter-arrival nd servicetimes. Proc. CambridgePhilos. Soc. 58 497-520.[10] MAIRESSE, . and PRABHAKAR, . (2003). The existence of fixed points for the ./GI/1queue.Ann. Probab.31 2216-2236.[11] MARTIN, . (2002). Large andemqueueingnetworkswithblocking.QueueingSystemsTheoryAppl.41 45-72.[12] MOUNTFORD,. S. and PRABHAKAR,. (1995). On the weak convergenceof departuresfroman infinitesequenceof ./M/1 queues.Ann.Appl.Probab.5 121-127.[13] MOUNTFORD,. S. andPRABHAKAR,. (1996). TheCesaro imitof departuresromcertain*/GI 1 queueingtandems.In StochasticNetworks:TheoryandApplications(F.P.Kelly,S. ZacharyandI. Ziedins,eds.) 309-322. ClarendonPress,Oxford.[14] PRABHAKAR,., MOUNTFORD,. S. and BAMBOS,N. (1996). Convergenceof departuresin tandemnetworksof ./GI/oo queues.Probab.Engrg.Inform.Sci. 10 487-500.

[15] VERE-JONES, . (1968). Some applicationsof probabilitygenerating unctionals o the studyof input-output treams.J. Roy.Statist. Soc. Ser B 30 321-333.DEPARTMENTSOF ELECTRICALENGINEERING

AND COMPUTERSCIENCESTANFORD UNIVERSITYSTANFORD,ALIFORNIA4305-9510E-MAIL: [email protected]

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