A tandem queueing model with coupled processors

A tandem queueing model with coupled processors

Citation for published version (APA):

Resing, J. A. C., & Örmeci, E. L. (2002). A tandem queueing model with coupled processors. (SPOR-Report :

reports in statistics, probability and operations research; Vol. 200208). Technische Universiteit Eindhoven.

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Published: 01/01/2002

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Ja ques Resing

Department of Mathemati s and Computing S ien e

Eindhoven University of Te hnology

P.O. Box 513 5600 MB Eindhoven The Netherlands Lerzan  Orme i 

Department of Industrial Engineering

Ko University

Sariyer-Istanbul

Turkey

Abstra t

We onsideratandemqueueingmodel onsistingoftwostations. Spe ialfeatureof themodelis thatthetotalservi e apa ityofthestationstogetheris onstant. When bothstationsarenonempty,agivenproportionofthis apa ityisallo atedto therst station andthe remainingpart to these ond station. However,if one of thestations be omesempty,thetotal apa ityofthetwostationstogetherisallo atedtotheother station.

Themodelismotivatedbyasituationen ounteredinmulti-a ess ommuni ationin ableTVnetworks. Beforeusersarea tuallyallowedtotransmitdataovera ommuni- ation hannel,theyrsthavetoobtainakindofgrantinordertoavoid ollisions.The total apa ity of the ommuni ation hannel is divided overthe two dierent stages: allo ationofthegrantsononehandandtransmissionofa tualdataontheotherhand. Westudy thetwo-dimensionalMarkovpro ess representingthenumbersofjobs in the twostations. A fun tional equation for the generating fun tion of the stationary distributionofthisMarkovpro essisderivedandthesolutionofthefun tionalequation is obtained. In the analysis we use the theory of Riemann-Hilbert boundary value problems.

In thispaperwe onsidera tandemqueueing model onsistingof two stations. Jobs arrive

at therst station a ording to a Poisson pro ess. After re eiving servi e at thisstation,

theymoveto these ond station,andupon ompletionofservi eatthese ondstation they

leave thesystem. The amount of work that a job requires at a station is an exponentially

distributed random variable. The total servi e apa ity of the two stations together is

onstant. Whenbothstationsarenonempty,agivenproportionofthe apa ityisallo ated

to station 1,and theremainingproportionisallo ated to station2. However, ifone ofthe

stationsisempty,thetotal servi e apa ityofthestationsisallo atedtotheotherstation.

The model we onsider is motivated by the following situation en ountered in able TV

networks.

Cable TV networks are urrently being upgraded to enable bidire tional

ommuni a-tions between the network terminations (NTs) at the ustomer premises and a entrally

lo ated head end (HE). In order to oordinate upstream transmission (i.e., from NTs to

HE)a mediuma essproto olis needed. This proto ol anbe arequest-grant me hanism

onsisting of two stages. At the rst stage, an NT whi h has data to transmit sends a

request to the HE in a dedi ated time slot to spe ify the number of data slots it needs.

If only one NT sends a request in a ertain time slot, then the HE re eives the request

su essfully. If more NTssend a request simultaneouslyin a ertain time slot, a ollision

o urs, upon whi h a ollision resolution algorithm (CRA) is started for these NTs. The

NTsinvolvedinthe ollisionhavetoretransmittheirrequest. Hen e,forarequesttorea h

the HE su essfully, a random number of time slots is needed, depending on the number

of NTs involved in the ollision and the CRA employed by the system. Upon re eiving

a request su essfully, the HE starts the se ond stage of the me hanism, the a tual data

transmission,bysendingagrant to the orrespondingNTto transmitits datainspe ied

data slots. Note that also the a tual transmission of data from the NTs to the HE needs

a random numberof timeslots sin e ea h NT hasa dierent amount of data to transmit.

Furthermore, the apa ity ofthe upstream hannelis dividedbetweenthese two stages by

theappropriateuseoftimeslots. Someofthetimeslotsarededi atedtodatatransmission

of NTsalready havingagrant, andtherest is dedi atedto requestsof NTsnotyet having

a grant. In ourmodel,servi eat station 1 represents thepro ess of re eivingthe requests,

whereas servi e at station 2 represents the transmissionof the a tual data orresponding

to thesu essfully re eived requests. Hen e, the total server apa ity represents thetotal

upstreambandwidth,andits allo ationtothetwostations orrespondstothetime-sharing

of theupstream hannelbythetwo stages des ribed above.

What isa leverwayto dividethetotal servi e apa ityoverthetwo individualservi e

stations? In[6 ℄, Klimov onsiderstheminimization ofthe average holding ostsina

time-sharingqueueingsystemwitha numberofstationsinseriesattendedbyasingleserver. In

the aseoftwostationsinserieswiththeobje tive ofminimizingtheaveragesojourntime,

Klimov's resultsimplythat the optimalpoli y would be to allo ate the whole apa ityto

the se ond station whenever this station is not empty. However, in the above mentioned

appli ation, the poli y to rst allo ate time slots fordata transmissionto the NTs whi h

requestshavealreadybeenre eived,andallo ateonlytheremainingtimeslotsforre eiving

newrequests, turnsoutto benotverysensible. Thereasonforthisisthatthere isaround

trip delay (RTD) on the ollision feedba k. Upon a ollision, the HE announ es that a

ollision o urred at a ertain slot and all the NTs whi h tried to send a request in that

a queueing model,but also may notbe ignored ompletely dueto its substantialee t on

thewhole pro ess. Sala et al. [11 ℄ show, throughsimulations, that wheneverthe feedba k

delayislong,meansojourntimesatea hstage an beshortenedbyallo atingrequestslots

ona more regularbasis. That iswhywestudythemodelinwhi haxed partofthetotal

servi e apa ity isalways allo atedto therst station,and onlytheremaining partto the

se ondstationwhenbothstationshaveat leastone job. Onlywhenoneofthetwostations

isempty, thetotal servi e apa ity isallo ated to theother station.

Systems in whi h the servi e rates of stations hange at the moments that one of the

stationsbe omesempty, areknownintheliteratureassystemswith oupledpro essors. In

a pioneeringpaper, Fayolle and Iasnogorodski [4 ℄ were therst to onsider su h a system.

They analyzed two oupledservers in parallel with exponential servi e times and derived

a solutionforthe generatingfun tionof the stationary distributionof the Markovpro ess

des ribingthenumberofjobsinbothqueues,usingthetheoryofRiemann-Hilbertboundary

value problems. Konheim, Meilijsonand Melkman [7 ℄ determined the generatingfun tion

of the joint queue length distribution in the ompletely symmetri ase (identi al arrival

and servi e rate at both servers) using a uniformization method. In Cohen and Boxma

[3 ℄, the ordinary oupled pro essor model is analyzed forthe ase of generally distributed

servi etimes. Ourmodel anbeviewedasthetandem versionofthemodelin[3,4 ,7℄. Like

theordinary oupled pro essormodel,ourmodel willalso be analyzed usingthetheory of

boundaryvalue problems.

Another way to dividethe total servi e apa ity over theindividual servi estations is

to ompletelyallo atethetotalservi e apa itytoone ofthetwo stationsinanalternating

order. This would lead to a polling system with two stations in tandem attended by a

singleserver. Forsome tandempollingsystemswithdierenttypesofswit hingrules,su h

as gated and exhaustive servi e, Katayama [5 ℄ has given expli itexpressions forthe mean

sojourn timeof jobsinthesystem.

Therest ofthepaperisorganized inthefollowingway. Inthenext se tion,wedes ribe

in detail the model under onsideration. In se tion 3, we derive a fun tional equation for

the generating fun tion of the stationary joint distribution of the numberof jobs in both

queues. This fun tional equation is analysed in se tion 4 for the two extreme ases in

whi h the total apa ity is allo ated to one of the two stations, even if both stations are

nonempty. Fortheintermediate ases,inwhi hthestationsreallyshare the apa itywhen

bothstationsarenonempty, thefun tionalequationisstudiedinse tion5. First,thekernel

ofthefun tionalequationisanalyzedandafterthataboundaryvalueproblemisformulated

andits solutionispresented. Inse tion6,webrie ydis ussaslightlymore generalmodel.

Asaspe ial aseofthismoregeneralmodel,weprovethewell-knownprodu tformsolution

for thestationary distribution of the ordinarytandem queue with exponential interarrival

and servi e times using the theory of boundary value problems. We on lude this paper

We onsider a tandemqueueing model onsisting oftwo stations. Jobsarrive at station 1

a ording to a Poisson pro ess with rate , and they demand servi e from both stations

beforeleavingthesystem. Ea hjobrequiresanexponentialamountofworkwithparameter

 j

at station j, j = 1;2. The total servi e apa ity of the two servi e stations together is

xed. Without lossof generalitywe assumethat thistotal servi e apa ityequalsone unit

ofworkpertimeunit. Wheneverbothstationsarenonempty,aproportionpofthe apa ity

isallo atedtostation1,andtheremainingpart(1 p)isallo atedtostation2. Thus,when

thereisatleastonejobatea hstation,thedeparturerateofjobsatstation1is 1

pandthe

departurerate ofjobsat station2is  2

(1 p). However, whenone of thestationsbe omes

empty,thetotalservi e apa ityisallo atedtotheotherstation. Hen e,thedeparturerate

at that station,say station j, is temporarilyin reased to  j

. In thesequel we will denote

with j

== j

theaverage amountof work pertimeunit requiredat stationj, j=1;2.

Clearly,thetwo-dimensionalpro essX(t)=(X 1 (t);X 2 (t)),whereX j (t),j=1;2,isthe

numberofjobs at station j at timet, isa Markov pro ess. Thetransition rate diagram of

thispro ess isgiven inFigure 1.

Figure1: The transition ratediagram ofthe system

Under theergodi ity ondition

 1

+ 2

<1; (1)

the pro ess X(t) has a unique stationary distribution. In the sequel we are interested in

Letusdenotewith(n;k)thestationaryprobabilityofhavingn ustomersinstation1and

k ustomersinstation2. Fromthetransitionrate diagramofthemodel,we anderivethe

set ofbalan eequations

(0;0) =  2 (0;1); ( + 1 )(n;0) = (n 1;0)+(1 p) 2 (n;1); n1; ( + 2 )(0;1) =  1 (1;0)+ 2 (0;2); (+p 1 +(1 p) 2 )(n;1) = (n 1;1)+ 1 (n+1;0)+(1 p) 2 (n;2); n1; (+ 2 )(0;k) = p 1 (1;k 1)+ 2 (0;k+1); k 2; (+p 1 +(1 p) 2 )(n;k) = (n 1;k)+p 1 (n+1;k 1)+(1 p) 2 (n;k+1); n1; k2:

Now we dene, forjxj1;jyj1,thejoint probabilitygeneratingfun tion

P(x;y):= X n0 X k0 (n;k)x n y k :

Fromthebalan eequationsitfollowsthatP(x;y)satisesthefollowingfun tionalequation

 (+p 1 +(1 p) 2 )xy x 2 y p 1 y 2 (1 p) 2 x  P(x;y) =  (1 p)[ 1 y(y x)+ 2 x(y 1)℄  P(x;0) +  p[ 2 x(1 y)+ 1 y(x y)℄  P(0;y) +  p 2 x(y 1)+(1 p) 1 y(x y)  P(0;0): (2)

The onstantP(0;0) an bedeterminedbysubstitutingx=( 1 y 2 )=( 1 y  2 (y 1))in(2).

Forthis hoi e ofx,boththefa tor infrontofP(x;0) andthefa torinfrontofP(0;y)are

equalto zero, and hen e equation(2) redu esto

P(  1 y 2  1 y  2 (y 1) ;y)=  2 (y 1)  2 (y 1)+y(1  1 y 2  1 y  2 (y 1) ) P(0;0): (3)

Now, letting y " 1 in (3), we obtain P(0;0) = 1  1

 2

. This result an, of ourse,

be explainedby thefa t that, independent of p, the two stations together always work at

apa ity 1 (if there is work in thesystem) and thefa t that  1

+ 2

equals the amount of

work brought into thesystempertimeunit.

How anwendthesolutionP(x;y)ofthefun tionalequation(2)? Inthenextse tion,

wewillgivetheexpli itsolutionforP(x;y) inthespe ial ases p=0andp=1. Afterthat

we show, inse tion 5,howforthe ase 0<p<1thesolutionof (2) an beobtained using

In the ase p=0, resp. p=1, the model that we onsider an be alternativelyviewed as

a tandem queueing model with one single server for both stations together, in whi h the

server gives preemptive priority to station 2, resp. station 1. It turns out that for these

ases, thefun tionalequation(2) an be solved relativelyeasily. Thisismainlydueto the

fa tthateither thefa tor infront ofP(0;y) inequation (2), in asep=0,orthe fa torin

frontofP(x;0),in asep=1,isequaltozero. Infa t,in asep=0,themodelwe onsider

is well-known. However, as faraswe know, themodelis notstudied before in ase p=1.

Therefore, we willparti ularlypayattention to thelatter aseinthisse tion.

4.1 The ase p=0

If p = 0, every time the server has ompleted a servi e of a job at station 1, he will

immediately ontinue theservi eof thissame job at station 2,dueto thefa t thatservi e

atstation2haspriority. Hen e,theanalysisofthemodelessentiallyredu estotheanalysis

ofasingleM=C 2

=1queue,inwhi htheservi etime onsistsoftwoexponentialphaseswith

parameters 1

and  2

respe tively. Thismodel an,forexample,also beanalysedusingthe

spe tral expansion method (see [8 ℄) orthe matrix-geometri method (see [10 ℄). Equation

(2)redu es inthis aseto

 (+ 2 )xy x 2 y  2 x  P(x;y)=   1 y(y x)+ 2 x(y 1)  P(x;0)+   1 y(x y)  P(0;0): (4)

Now, be ause fory= 2

=(+ 2

x) thefa tor infront ofP(x;y) in(4)iszero, alsothe

righthandsideof (4)shouldbe equalto zero. Hen e,

P(x;0) = 1  1  2 1  1 x(1+ 2  2 x) 1 2x : (5)

Substituting(5)in(4), we obtainafter straightforwardbutlengthy al ulations

P(x;y)= (1  1  2 )(1+ 2 (y x)) 1 ( 1 + 2 + 1  2 )x+ 1  2 x 2 : (6) 4.2 The ase p=1

If p=1, themodel isa tandemqueue with a singleserverand preemptive priorityforthe

rst queue. Equation (2)redu esinthis ase to

 (+ 1 )xy x 2 y  1 y 2  P(x;y)=   2 x(1 y)+ 1 y(x y)  P(0;y)+   2 x(y 1)  P(0;0): (7)

Now,forx=(y),theuniquerootintheunit ir leoftheequationx 2 (+ 1 )x+ 1 y=0,

therighthandside of(7) shouldagainbeequal to zero. Hen e,we obtain

P(0;0) =  1  2 y (1 (y)) 1 y  P(0;y); (8)

P(0;y)= 1  1  2 1  2 y (1 (y)) 1 y : (9)

Furthermore, substitutionof(8) in(7)gives

P(x;y)=  1 x(1 (y))+x y ( 1 +1)x  1 x 2 y P(0;y): (10)

A ni e probabilisti explanationfor theresults inequations (9)and (10) an be given.

Firstremarkthattheroot(y) anbeinterpretedasthegeneratingfun tionofthenumber

of jobs served in a busy period of an M=M=1 queue with arrival rate  and servi e rate

 1

(see e.g., Cohen[2℄, page 190). Now, if we look at our model only duringperiodsthat

the rst queue is empty (i.e., we glue together idle periodsof the rst queue), the se ond

queue behaves as an M X

=M=1 queue with arrival rate , bat h size generating fun tion

(y) and servi e rate  2

. Hen e, thefun tion P(0;y)=P(0;1), i.e. the generatingfun tion

of the onditional distribution of the number of jobs in the se ond queue given that the

rst queue is empty, is the same as the generating fun tion of the number of jobs in the

above mentioned M X

=M=1 queue. The latter one is well known (see e.g., [2℄, page 387)

and immediatelygivesequation (9).

Toexplainequation(10)weintrodu etherandomve tor(Y 1

;Y 2

),denotingthe

station-arynumberof ustomersinthesystem(Y 1

)andthestationarynumberof ustomersalready

served inthe urrent busy period (Y 2

),at apointintimeinwhi htheserver isbusy inan

M=M=1queuewitharrivalintensityand servi eintensity 1

. Furthermore,letQ(x;y)be

thegeneratingfun tionof(Y 1

;Y 2

). Thefun tionQ(x;y) anbestraightforwardlyobtained

by studying the two-dimensional Markov pro ess orresponding to (Y 1

;Y 2

). This pro ess

hasalmostthesametransition ratesastheonesinFigure1withp=1,onlytheratesnear

theverti alboundarydier. Itturnsoutthat Q(x;y) is equalto

Q(x;y)= (1  1 )x(x (y)) ( 1 +1)x  1 x 2 y : (11) Now, ifwe denote by(X 1 ;X 2

) thestationary numberof jobsatthe twostationsinour

model at arbitrary pointsintime, and by(0;X (i) 2

) thesame quantitiesduringidleperiods

of therst station,thenwe have

(X 1 ;X 2 ) d = ( (0;X (i) 2 ); with probability1  1 ; (0;X (i) 2 )+(Y 1 ;Y 2 ); with probability 1 : (12)

Here, the random ve tors (0;X (i) 2 ) and (Y 1 ;Y 2

) are furthermore independent. Hen e,

be- ause therandomve tor (0;X (i) 2

)has generatingfun tionP(0;y)=P(0;1), we have

P(x;y)= P(0;y) P(0;1) (1  1 + 1 Q(x;y)): (13) UsingP(0;1) =1  1

, and ombination of(11) and (13), dire tlygives(10).

Remark: De ompositionresult (12)also holdsforgenerally distributedservi e times and

In thisse tion we will derive the solutionof fun tionalequation (2) for 0<p <1. A key

role isplayed bythe kernel

K(x;y):=(+p 1 +(1 p) 2 )xy x 2 y p 1 y 2 (1 p) 2 x:

5.1 Zeros of the kernel

Be ause the kernel K(x;y) is, for ea h x, a polynomial of degree 2 in y, we have that

for every value of x there are two possible values of y, say y 1 (x) and y 2 (x), su h that K(x;y 1 (x))=K(x;y 2 (x))=0.

Lemma 1 Thealgebrai fun tiony(x)denedbyK(x;y(x))=0hasfourrealbran hpoints

0=x 1 <x 2 1<x 3 <x 4 .

Proof: Bran h points are zeros of the dis riminant,D(x), of the equation K(x;y) = 0 as

fun tionof y, i.e., D(x)=  x 2 (+p 1 +(1 p) 2 )x  2 4p(1 p) 1  2 x:

Clearly, D(0)=0,D(x)<0 forsmallpositivex,D(1)0,D((+p 1 +(1 p) 2 )=)<0 and lim x!1

D(x)=1. Hen e,thelemma follows. 2

Lemma 2 For ea h x 2 [x 1

;x 2

℄, the two roots y 1

(x) and y 2

(x) are omplex onjugate.

Hen e, the interval [x 1

;x 2

℄ is mapped by x 7! y(x) onto a losed ontour L, whi h is

symmetri withrespe t to the real line.

Proof: Follows dire tly from the fa t that the dis riminant D(x) is zero for x = x 1

and

x=x 2

and negativeforx2(x 1 ;x 2 ). 2

L

x

y (x)

y (x)

x - plane

y - plane

1

1

2

2

x

x

0

Figure 2: The ontour L

In Figure 2,the result of Lemma 2 isillustrated. In thesequel we willdenote the interior

ofthe ontourLbyL +

. Finally, noti ethatforapointy(x)onthe ontourLwehavethat

y(x)y(x)= (1 p) 2 x p 1 ; (14)

Next,we willformulate a boundaryvalue problemforthe fun tionP(0;y).

Lemma 3 The fun tion P(0;y) is regular in the domain L +

and satises for y 2 L the

ondition ImP(0;y)=Im p 2  2 y(y 1)+(1 p)y[p 1 y (1 p) 2 ℄ p[y((1 p) 2 p 1 y)+p 2 y(y 1)℄ P(0;0) ! : (15)

Proof: Forzeropairs (x;y) of thekernelforwhi hP(x;y) is nite,we have

 (1 p)[ 1 y(y x)+ 2 x(y 1)℄  P(x;0) +  p[ 2 x(1 y)+ 1 y(x y)℄  P(0;y) +  p 2 x(y 1)+(1 p) 1 y(x y)  P(0;0) =0: (16)

We an rewrite thisequation,bysubstituting(1 p) 2 x=p 1 yy (see (14)), in P(0;y)= p 2  2 y(y 1)+(1 p)y[p 1 y (1 p) 2 ℄ p[ y((1 p) 2 p 1 y)+p 2 y(y 1)℄ P(0;0)+ 1 p p P(x;0): (17) Now, if (1 p) 2 p1 x 2

1,then L lies entirely withinthe unit ir le (y(x 2

) is the point on

L with largest absolute value). Hen e, P(0;y) is regular inL +

. Finally, (15) follows from

(17) bytaking x2[x 1

;x 2

℄ and usingthat P(x;0) isreal forthose x.

If (1 p) 2 p 1 x 2

>1,thenP(0;y(x)) an be ontinuedanalyti allyoverthe interval[x 1

;x 2

℄

via equation(16), be auseP(x;0) isregularon thisinterval. Hen e,theanalyti

ontinua-tionof P(0;y)is niteat y=y(x 2

). Be auseP(0;y) hasapowerseriesexpansionaty =0

with positive oeÆ ients, this impliesthat P(0;y) is regular for jyj <y(x 2

) and hen e in

L +

. 2

Lemma 3 shows that the determination of P(0;y) redu es to the determination of the

solutionofthe followingRiemann-Hilbertboundaryvalue problemon the ontour L:

Determine a fun tionP(0;y) su h that

1. P(0;y) isregular fory2L +

and ontinuous fory2L +

[L.

2. Re [iP(0;y)℄= (y),fory2L,

where (y)= Im p 2  2 y(y 1)+(1 p)y[p 1 y (1 p) 2 ℄ p[y((1 p) 2 p 1 y)+p 2 y(y 1)℄ P(0;0) ! :

Thestandardwaytosolvethistypeofboundaryvalueproblem(see,e.g.,Muskhelishvili

[9 ℄) is to transform the boundary ondition (15), by using onformal mappings, to a

on-dition on the unit ir le. Let z = f(y) be the onformal map of L +

onto the unit ir le

C +

=fz : jzj <1g and denote by y =f 0

(z) the inverse mapping,i.e., the onformalmap

Determine a fun tionH(z) su hthat

1. H(z) is regularforz2C +

and ontinuousforz2C +

[C.

2. Re [iH(z)℄= (z),~ forz2C, where~ (z)= (f 0

(z)),

then P(0;y) = H(f(y)) is the solution of the original problem. The solution of problem

(P), a so- alledDiri hletproblem on the ir le,is well-known (see [9℄) and given by

H(z)= 1 2 Z C ~ (w) w+z w z dw w +K ;

whereK issome onstant.

In thisway, P(0;y) hasbeenformally determined. Substitution,rst in (16)to obtain

P(x;0)andafterthatin(2),thenyieldsP(x;y),sothatthegeneratingfun tionofthejoint

stationary distribution of the queue lengths in thetandem queue has been obtained. In a

futurestudythedetailsofthisanalysiswillbeprovided. Forexample,thedeterminationof

the onformalmapgenerallyposesaninterestingproblemintheanalysisoftheseboundary

valueproblems.

Remark: Inmostproblems,forthedeterminationof the onformalmapf andtheinverse

onformalmap f 0

, a numeri al te hnique (e.g., Theodorsen'spro edure, see [3 ℄)has to be

applied. However,forthisspe i problem,anexpli itexpressionforthe onformalmapping

f(y) an be found (see the paperof Blan [1℄, in whi h the time-dependent behaviour of

theordinarytandemqueue without oupledpro essors is studied).

6 Generalization

The model that we onsidered sofar inthe paperis a spe ial ase of the followingmodel.

Thesystemhastwostationsintandem,ea hstationhavingitsownserver. Customersarrive

at station 1 a ording to a Poissonpro ess withrate , and they require an exponentially

distributedservi e time from both stationsbefore leaving thesystem. The servi erate at

station j is equal to rate  j

whenever both stations have at leastone ustomer. If one of

thestationsbe omesempty, theservi erate at theother station hanges from  j

to   j .

Forthismodelthefun tionalequation be omes

 (+ 1 + 2 )xy x 2 y  1 y 2  2 x  P(x;y) =  (  1  1 )y(y x)+ 2 x(y 1)  P(x;0) +  (  2  2 )x(1 y)+ 1 y(x y)  P(0;y) +  (  2  2 )x(y 1)+(  1  1 )y(x y)  P(0;0): (18)

After al ulations,similartothosedoneinthepreviousse tions,wegetforyonthe ontour

L, Im  (  2  2 ) 1 y(1 y)+ 1 y( 1 y  2 ) (  1  1 )y(  1 y  2 )+ 1  2 y(1 y) P(0;y)  = Im  (  2  2 ) 1 y(1 y) (  1  1 )y( 1 y  2 ) (  1  1 )y(  1 y  2 )+ 1  2 y(1 y) P(0;0)  :

tion P(0;y) onthe ontour L ofthe followingform:

Determine a fun tionP(0;y) su h that

1. P(0;y) isregular fory2L +

and ontinuous fory2L +

[L.

2. Re [g(y)P(0;y)℄= (y),fory 2L.

ThestudyofthismoregeneralRiemann-Hilbertboundaryvalueproblemwillbeatopi

forfurther resear h. Inthe remaining part we restri t ourattention to thesolutionof the

problem forthespe ial ase ofan ordinarytandemqueue, i.e.,   j

= j

, j=1;2.

6.1 The ordinary tandem queue

Inthe ase  j

= j

,j=1;2,itisof oursewell-knownthatthestationaryjointdistribution

ofthenumberofjobsat thetwostationshasaprodu tform. Wenowshowhowthisresult

followsfrom(18). Forzeropairs(x;y)ofthekernelforwhi hP(x;y)isnite,wehave,from

(18),  1 y(x y)P(0;y)= 2 x(1 y)P(x;0): (19)

Multiplyingbothsidesby((1 y))=( 1  2 ),weobtain  2

y(x y)(1 y)P(0;y)= 1 x(1 y)(1 y)P(x;0); (20) wherenow j == j

. Clearly, forreal x,therighthandsideof(20) isreal,andfurthermore

fory on the ontour L,we have  2

yy= 1

x. Usingthese two fa ts,we on ludethat

Im((x y)( 2

y 

1

x)P(0;y))=0:

Finally, usingagainthat (x;y) is azeropair of thekernel, thisredu esto

Im((1  2

y)P(0;y))=0:

The solutionofthisboundaryvalueproblem isgiven by

P(0;y)= K 1  2 y ;

whereK isa onstant. Substitutingthisin(19) gives (again using 2 yy= 1 x) P(x;0)= K 1  1 x :

Finally, substitutingtheformulas forP(0;y) and P(x;0) in(18) gives

P(x;y)= K (1  1 x)(1  2 y) :

Inthisway, wendtheprodu tformsolutionforthetandemqueueingsystemdire tlyfrom

Inthispaperwe analyseda tandemqueueingmodel onsistingoftwostationsinwhi hthe

total servi e apa ity of the two stationstogether is onstant. The servi e apa ity of the

individualstationsdepends onwhetherornotone ofthestationsisempty. The stationary

joint distributionof thenumberof jobsinthetwo stationsisanalysed, usingthe theoryof

boundaryvalue problems.

The numeri al evaluation of the solution is a topi for further resear h. Furthermore,

the analysis of the more general tandem queueing model with oupled pro essors, brie y

des ribedinse tion 6,willalso bepart ofa future study.

A knowledgement The authors like to thank Onno Boxma (Eindhoven University of

Te hnology) and DeeDenteneer (PhilipsResear h,Eindhoven)forstimulating dis ussions.

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