LOSSES IN M/GI/m/n QUEUES arXiv:math/0506033v5 [math.PR ... · arrival rate, and let µ be the...

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LOSSES IN M/GI/m/n QUEUES

VYACHESLAV M. ABRAMOV

Abstract. The M/GI/m/n queueing system with m homogeneous serversand the finite number n of waiting spaces is studied. Let λ be the customersarrival rate, and let µ be the reciprocal of the expected service time of acustomer, where λ = mµ. The paper proves that under this assumption theexpected number of losses during a busy period remains the same for all n ≥ 1,while in the particular case of the Markovian system M/M/m/n the expected

number of losses during a busy period is mm

m!for all n ≥ 0. Under the addi-

tional assumption that the probability distribution function of a service timebelongs to the class NBU or NWU, the paper establishes simple inequalitiesfor those expected numbers of losses in M/GI/m/n queueing systems.

1. Introduction

Analysis of loss queueing systems is very important from both the theoretical andpractical points of view. In the literature on queueing systems, the informationabout M/GI/m/n queueing systems (n ≥ 1) is very scanty, because explicit resultsfor characteristics of these queueing systems are hard to obtain. (In the presentpaper, for multiserver queueing systems the notation M/GI/m/n is used, where mdenotes the number of servers and n denotes the number of waiting places. Anothernotation which is also acceptable in the literature is M/GI/m/m+ n.) From thepractical point of view, M/GI/m/n queueing systems serve as a model for telephonesystems, where n is the maximally possible number of calls that can wait in the line.The loss probability is one of the most significant performance measures. In thepresent paper, we study the expected number of losses during a busy period underthe assumption that the arrival rate (λ) is equal to the maximum service capacity(mµ), which seems to be the most interesting from the theoretical point of view.Although this characteristic under the assumption of the paper is less meaningfulthan the loss probability, it is also practically useful since it explains the behaviourof losses for different values m and n and can help us to evaluate the loss probabilityunder additional information. In this paper, we prove that under the assumptiongiven in the paper, the expected number of losses in M/GI/m/n queueing systems(n ≥ 1) is the same. This means that the result obtained by simulating the expectednumber of losses in the M/GI/m/1 queueing system enables us to judge about thatexpected number of losses in any M/GI/m/n queueing system with an arbitrary(possibly large) n.

In several recent papers (see Abramov [1], [2], [4], Righter [21], Wolff [30]), ithas been shown that for the M/GI/1/n queueing system, under mutually equalexpectations of interarrival and service time, the expected number of losses during

1991 Mathematics Subject Classification. 60K25.Key words and phrases. Loss systems; M/GI/m/n queueing system; busy period; level-

crossings; stochastic order; coupling.

1

http://arxiv.org/abs/math/0506033v5

2 VYACHESLAV M. ABRAMOV

a busy period is equal to 1 for all n ≥ 0. In the particular case of the M/M/1/nqueueing system, this fundamental property of losses is associated with the propertyof level-crossings in the symmetric random walk (e.g. Szekely [24], Wolff [29], p.411).

Extensions of the aforementioned result to queueing systems with batch arrivalshave been studied in Abramov [5], Wolff [30] and Pekoz, Righter and Xia [20]. Ap-plications of the aforementioned property of losses in M/GI/1/n queueing systemscan be found in [9] for analysis of lost messages in telecommunication systems andin [11] for optimal control of large dams.

Characterization problems associated with losses in the GI/M/1/n queue andproperties of losses in the M/M/m/n and MX/M/m/n queueing systems are stud-ied in Pekoz, Righter and Xia [20]. One of the results of the aforementioned paperof Pekoz, Righter and Xia [20] related to the M/M/m/n queueing system statesthat for λ = mµ, the expected number of losses during a busy period is equal tomm

m! for all n ≥ 0.The results for M/GI/1/n and M/M/m/n queueing systems lead to the fol-

lowing open question: Does the result on losses in M/M/m/n queueing systemsremain true for those M/GI/m/n too?

The answer to this question is negative. Simple examples show that the resultis not longer valid for M/GI/m/n queueing systems when m ≥ 2 and n ≥ 1. Asimple counterexample can be built for M/GI/2/1 queueing systems having theservice time distribution G(x) = 1 − pe−µ1x − qe−µ2x, p + q = 1. The analysis ofthe stationary characteristics for these systems, resulting in an analysis of lossesduring a busy period, can be provided explicitly. Specifically, the structure of the9× 9 Markov chain intensity matrix for the states of the Markov chain associatedwith an M/GI/2/1 queueing system shows a clear difference between the structureof the stationary probabilities in M/GI/2/1 queues and that in M/GI/2/0 queuesgiven by the Erlang-Sevastyanov formulae. So, the parameters p, q, µ1 and µ2 canbe chosen such that the expected number of losses during busy periods in these twoqueueing systems will be different.

Nevertheless, for M/GI/m/n queueing systems with m ≥ 2 there is a propertysimilar to that forM/GI/1/n queueing systems established before. Let Lm,n denotethe number of losses during a busy period of the M/GI/m/n queueing system, letλ, µ be the arrival rate and, respectively, the reciprocal of the expected service time,and let m, n denote the number of servers and, respectively, the number of waitingplaces. We will prove that, under the assumption λ = mµ, the expected numberof losses during a busy period of a M/GI/m/n queueing system, ELm,n, is thesame for all n ≥ 1, which is not generally the same as that for the M/GI/m/0 lossqueueing system. In addition, if the probability distribution function of the servicetime belongs to the class NBU (New Better than Used), then ELm,n = cmm

m! , wherea constant c ≥ 1 is independent of n ≥ 1. In the opposite case of the NWU (New

Worse than Used) service time distribution we correspondingly have ELm,n = cmm

m!with a constant c ≤ 1 independent of n ≥ 1 as well. (The constant c becomesequal to 1 in the case of exponentially distributed service times.) Recall that aprobability distribution function Ξ(x) of a nonnegative random variable is said tobelong to the class NBU if for all x ≥ 0 and y ≥ 0 we have Ξ(x + y) ≤ Ξ(x)Ξ(y),where Ξ(x) = 1− Ξ(x). If the opposite inequality holds, i.e. Ξ(x+ y) ≥ Ξ(x)Ξ(y),then Ξ(x) is said to belong to the class NWU.

LOSSES IN M/GI/m/n QUEUES 3

Another question related to the results of this paper is as follows: What isthe behavior of losses in M/GI/m/n queues when λ 6= mµ ? For the single serverqueuesM/GI/1/n andGI/M/1/n the asymptotic behaviour of losses as n increasesto infinity has been studied in [4] and [8]. Asymptotic analysis of multiserverqueueing system GI/M/m/n as n increases to infinity has been studied in [12].The similar analysis for M/GI/m/n is very difficult, however when m and n areboth large, the heavy-traffic analysis for these and more general systems has beenprovided in [26] and [27].

The proof of the main results of this paper is based on an application of thelevel-crossing approach to the special type stationary processes. The constructionof the level-crossings approach used in this paper is similar to that of the earlierpapers of the author (e.g. [1], [3], [6], [7], [10] and [13]) as well as the paper ofPechinkin [19].

Throughout the paper, it is assumed that m ≥ 2. (This is not the loss ofgenerality since the case m = 1 is known, see [4], [21] and [30].)

The paper is organized as follows. In Section 2, which is the first part of thepaper, M/M/m/n queueing systems are studied. The results obtained for Markov-ian systems are then used in Section 3, which is the second part of the paper, inorder to study M/GI/m/n queueing systems. The study in both of Sections 2 and3 is based on the level-crossing approach. The construction of level-crossings forM/M/m/n queueing systems is then developed for M/GI/m/n queueing systemsas follows. We consider the stationary processes associated with these queueing sys-tems and establish the stochastic relations between the times spent in state m− 1associated with m−1 busy servers during a busy period of M/GI/m/n (n ≥ 1) andM/GI/m−1/0 queueing systems. To prove these stochastic relations, we develop alevel-crossing approach from the paper of Pechinkin [19], which is adapted here forthe problems of the present paper. The inequalities obtained are crucial and thenused to prove the main results of the paper in Section 4. In Section 5 we discuss apossible development of the results for MX/GI/m/n queueing systems with batcharrivals.

2. The M/M/m/n queueing system

In this section we study the Markovian M/M/m/n loss queueing system with theaid of the level-crossings approach in order to establish some relevant propertiesof this queueing system. Those properties are then developed for M/GI/m/nqueueing systems in the following sections.

Let f(j), 1 ≤ j ≤ n+m+ 1, denote the number of customers arriving during abusy period who, upon their arrival, meet j− 1 customers in the system. It is clearthat f(1) = 1 with probability 1. Let tj,1, tj,2,. . . , tj,f(j) be the instants of arrivalof these f(j) customers, and let sj,1, sj,2,. . . , sj,f(j) be the instants of the servicecompletions when there remain only j − 1 customers in the system. Notice, thattn+m+1,k = sn+m+1,k for all k = 1, 2, . . . , f(n+m+ 1).

For 1 ≤ j ≤ n+m let us consider the intervals

(2.1) (tj,1, sj,1], (tj,2, sj,2], . . . , (tj,f(j), sj,f(j)].

Then, by incrementing index j we have the following intervals

(2.2) (tj+1,1, sj+1,1], (tj+1,2, sj+1,2], . . . , (tj+1,f(j+1), sj+1,f(j+1)].


Delete the intervals of (2.2) from those of (2.1) and merge the ends, that is eachpoint tj+1,k with the corresponding point sj+1,k, k = 1, 2, ..., f(j+1). Then f(j+1)has the following properties. According to the property of the lack of memory of ex-ponential distribution, the residual service time, after the procedure of deleting theinterval and merging the ends as it is indicated above, remains exponentially dis-tributed with parameter µmin(j,m). Therefore, the number of points generated bymerging the ends within the given interval (tj,1, sj,1] coincides in distribution withthe number of arrivals of the Poisson process with rate λ during an exponentiallydistributed service time with parameter µmin(j,m). Namely, for 1 ≤ j ≤ m − 1we obtain

E{f(j + 1)|f(j) = 1} =

∞∑

u=1

u

∫∞

0

e−λx (λx)u

u!jµe−jµxdx =

λ

jµ.

Considering now a random number f(j) of intervals (2.1) we have

(2.3) E{f(j + 1)|f(j)} =λ

jµf(j).

Analogously, denoting the load of the system by ρ = λmµ , for m ≤ j ≤ m + n we

have

(2.4) E{f(j + 1)|f(j)} =λ

mµf(j) = ρf(j).

The properties (2.3) and (2.4) mean that the stochastic sequence

(2.5)

{f(j + 1)

(µλ

)jj∏

i=1

min(i,m),Fj+1

}, Fj = σ{f(1), f(2), . . . , f(j)},

forms a martingale.It follows from (2.5) that for 0 ≤ j ≤ m− 1

(2.6) Ef(j + 1) =λj

j!µj,

and for m ≤ j ≤ m+ n

(2.7) Ef(j + 1) =λm

m!µmρj−m.

For example, when ρ = 1 from (2.7) we obtain the particular case of the result of

Pekoz, Righter and Xia [20]: ELm,n=Ef(n + m + 1) = mm

m! for all n ≥ 0, whereLm,n denotes the number of losses during a busy period of the M/M/m/n queueingsystem.

Next, let B(j) be the period of time during a busy period of the M/M/m/nqueueing system when there are exactly j customers in the system. For 0 ≤ j ≤m+ n we have:

(2.8) λEB(j) = Ef(j + 1).

Now, introduce the following notation. Let Tm,n denote a busy period of theM/M/m/n queueing system, let Tm,0 denote a busy period of the M/M/m/0queueing system with the same arrival and service rates as in the initial M/M/m/nqueueing system, and let ζn denote a busy period of M/M/1/n queueing system


with arrival rate λ and service rate µm. From (2.6)-(2.8) for the expectation of abusy period of the M/M/m/n queueing system we have

(2.9) ETm,n =

n+m∑

j=0

EB(j) =

m−1∑

j=1

λj

jj!µj+1+

λm

mm!µm+1

n∑

j=0

ρj .

In turn, for the expectation of a busy period of the M/M/m/0 queueing system wehave

(2.10) ETm,0 =

m−1∑

j=0

EB(j) =

m∑

j=1

λj

jj!µj+1,

where (2.10) is the particular case of (2.9) where n = 0.It is clear that a busy period Tm,n contains one busy period Tm−1,0, where the

subscript m − 1 underlines that there are m − 1 servers, and a random numberof independent busy periods, which will be called orbital busy periods. Denote anorbital busy period by ζn. (It is assumed that an orbital busy period ζn starts atinstant when an arriving customer finds m− 1 servers busy and occupies the mthserver, and it finishes at the instant when after a service completion there at thefirst time remain only m − 1 busy servers.) Therefore, denoting the independent

sequence of identically distributed orbital busy periods by ζ(1)n , ζ

(2)n ,..., we have

(2.11) Tm,n = Tm−1,0 +

κ∑

i=1

ζ(i)n ,

where κ is the random number of the aforementioned orbital busy periods. Itfollows from (2.9), (2.10) and (2.11)

(2.12) E

κ∑

i=1

ζ(i)n =λm

mm!µm+1

n∑

j=0

ρj .

On the other hand, the expectation of an orbital busy period ζn is

Eζn =1

mµ

n∑

j=0

ρj

(this can be easily checked, for example, by the level-crossings method [1], [6] andan application of Wald’s identity [14], p. 384), and we obtain

(2.13) Eκ =λm

m!µm.

Thus, Eκ coincides with the expectation of the number of losses during a busyperiod in the M/M/m/0 queueing system. In the case ρ = 1 we have Eκ = mm

m! .

3. M/GI/m/n queueing systems

In this section, the inequalities between the times spent in the state m− 1 of theM/GI/m/n (n ≥ 1) and M/GI/m/0 queueing systems during their busy periodsare derived.

Consider two queueing systems: M/GI/m/n (n ≥ 1) and M/GI/m/0 bothhaving the same arrival rate λ and probability distribution function of a servicetime G(x), 1

µ =∫∞

0xdG(x) < ∞. Let Tm,n(m − 1) denote the time spent in the

state m − 1 during its busy period (i.e. the total time during a busy period when


m−1 servers are occupied) of the M/GI/m/n queueing system, and let Tm,0(m−1)have the same meaning for the M/GI/m/0 queueing system.

We prove the following lemma.

Lemma 3.1. Under the assumption that the service time distribution G(x) belongsto the class NBU (NWU),

(3.1) Tm,n(m− 1) ≥st (resp. ≤st) Tm,0(m− 1),

Proof. . The proof of the lemma is relatively long. In order to make it transparentand easily readable we strongly indicate the steps of this proof given by severalpropositions (properties). There are also six figures illustrating the constructionsin the proof. Each of these figures contain two graphs. The first (upper) of themindicates the initial (or intermediate) possible path of the process (sometimes two-dimensional), while the second (lower) one indicates the part of the path of oneor two-dimensional process after a time scaling or specific transformation (e.g. inFigure 4). Arc braces in the graphs indicate the intervals that should be deletedand their ends merged.

For the purpose of the present paper we use strictly stationary processes of order1 or strictly 1-stationary processes. Recall the definition of a strictly stationaryprocess of order n (see [18], p.206).

Definition 3.2. The process ξ(t) is said to be strictly stationary of order n orstrictly n-stationary, if for a given positive n < ∞, any h and t1, t2,. . . , tn therandom vectors(

ξ(t1), ξ(t2), . . . , ξ(tn))and

(ξ(t1 + h), ξ(t2 + h), . . . , ξ(tn + h)

)

have identical joint distributions.

If n = 1 then we have strictly 1-stationary processes satisfying the property:

P{ξ(t) ≤ x} = P{ξ(t+ h) ≤ x}.

The probability distribution function P{ξ(t) ≤ x} in this case will be called limitingstationary distribution.

The class of strictly 1-stationary processes is wider than the class of strictlystationary processes, where it is required that for all finite dimensional distributions

P{(ξ(t1), ξ(t2), . . . , ξ(tk)) ∈ Bk} = P{(ξ(t1 + h), ξ(t2 + h), . . . , ξ(tk + h)) ∈ Bk},

for any h and any Borel set Bk ⊂ Rk. The reason of using strictly 1-stationary

processes rather than strictly stationary processes themselves is that, the operationof deleting intervals and merging the ends is algebraically close with respect tostrictly 1-stationary processes, and it is not closed with respect to strictly stationaryprocesses. The last means that if ξ(t) is a strictly 1-stationary process, then forany h > 0 the new process

ξ1(t) =

{ξ(t), if t ≤ t0,

ξ(t+ h), if t > t0

is also strictly 1-stationary and has the same one-dimensional distribution as theoriginal process ξ(t). The similar property is not longer valid for strictly stationaryprocesses. If ξ(t) is a strictly stationary process, then generally ξ1(t) is not strictlystationary.


In the following the prefix ‘strictly’ will be omitted, so strictly stationary andstrictly 1-stationary processes will be correspondingly called stationary and 1-stationary processes.

Let us introduce m independent and identical stationary renewal processes (de-noted below xm(t)) with a renewal period having the probability distribution func-tion G(x).

On the basis of these renewal processes we build the stationary m-dimensionalMarkov process xm(t) = {ξ1(t), ξ2(t), . . . , ξm(t)}, the coordinates ξk(t), k =1,2,. . . ,m of which are the residual times to the next renewal times in time moment t,following in an ascending order.

Let us now consider the two m-dimensional 1-stationary Markov processes (i.e.m independent one-dimensional 1-stationary Markov processes) corresponding tothe M/GI/m/n (n ≥ 1) and M/GI/m/0 queueing systems. These 1-stationaryprocesses ym,n(t) and ym,0(t) are constructed as follows. Let Qm,n(t) denote thestationary queue-length process (the number of customers in the system) of theM/GI/m/n queue, and let Qm,0(t) denote the stationary queue-length processcorresponding to the M/GI/m/0 queueing system. Then we have the Markovprocess

ym,n(t) ={η(m,n)1 (t), η

(m,n)2 (t), . . . , η(m,n)

m (t), Qm,n(t)},

where {η(m,n)m−νm,n(t)+1(t), η

(m,n)m−νm,n(t)+2(t), . . . , η

(m,n)m (t)

}

are the ordered residual service times corresponding to νm,n(t) = min{m,Qm,n(t)}customers in service in time t, and

{η(m,n)1 (t), η

(m,n)2 (t), . . . , η

(m,n)m−νm,n(t)

(t)}= {0, 0, . . . , 0}

all are zeros. Analogously we define the stationary Markov process

ym,0(t) ={η(m,0)1 (t), η

(m,0)2 (t), . . . , η(m,0)

m (t), Qm,0(t)},

only replacing the index n with 0.Let us delete all time intervals of the process ym,n(t) related to the M/GI/m/n

queueing system (n ≥ 1) where there are more than m − 1 and less than m − 1customers and merge the ends. Remove the last component of the obtained processwhich is trivially equal to m−1. Then we get the new (m−1)-dimensional Markovprocess (in the following the prefix ‘Markov’ will be omitted and only used in theplaces where it is meaningful):

ym−1,n(t) ={η(m,n)1 (t), η

(m,n)2 (t), . . . , η

(m,n)m−1 (t)

},

the components of which are now denoted by hat. All of the components of thisvector are 1-stationary, which is a consequence of the existence of the limiting

stationary probabilities of the processes η(m,n)j (t), j = 1, 2, . . . ,m (e.g. Takacs [25])

and consequently those of the processes η(m,n)j (t), j = 1, 2, . . . ,m. The joint limiting

stationary distribution of the process ym−1,n(t) can be obtained by conditioning ofthat of the processes ym,n(t) given Qm,n(t) = m− 1.

The similar operation of deleting intervals and merging the ends, where thereare less than m− 1 customers in the system, for the process ym−1,0(t) is used. We


correspondingly have

ym−1,0(t) ={η(m,0)1 (t), η

(m,0)2 (t), . . . , η

(m,0)m−1 (t)

}.

We establish the following elementary property related to the M/GI/1/n queue-ing systems, n=0,1,. . .

Property 3.3.

(3.2) P{η(1,n)1 (t) ∈ B1 | Q1,n(t) ≥ 1

}= P{x1(t) ∈ B1},

for any Borel set B1 ⊂ R1.

Proof. Delete all of the intervals where the server is idle and merge the correspond-ing ends (see Figure 1). Then in the new time scale, the processes all are structuredas a stationary renewal process with the length of a period having the probabilitydistribution function G(x). Therefore (3.2) follows. �

In order to establish similar properties in the case m = 2 let us first study theproperties of 1-stationary processes.

Recall (see Definition 3.2) that if ξ(t) is a 1-stationary process, then for anyh and t0 the probability distributions of ξ(t0) and ξ(t0 + h) are the same. Theresult remains correct (due to the total probability formula) if h is replaced byrandom variable ϑ with some given probability distribution, which is assumed tobe independent of the process ξ. Namely, we have:

P{ξ(t0 + ϑ) ≤ x} =

∫∞

−∞

P{ξ(t0 + h) ≤ x}dP{ϑ ≤ h}

= P{ξ(t0) ≤ x}

∫∞

−∞

dP{ϑ ≤ h}

= P{ξ(t0) ≤ x}.

That is, ξ(t0) and ξ(t0 + ϑ) have the same distribution.

The above property will be used for the following construction of the sequenceof 1-stationary processes ξ(1)(t), ξ(2)(t), . . . , having identical one-dimensional dis-tributions.

Let ξ(0)(t) = ξ(t) be a 1-stationary process, let t1 be an arbitrary point, andlet ϑ1 be a random variable with some given probability distribution, which isindependent of the process ξ(0)(t). Let us build a new process ξ(1)(t) as follows.Put

ξ(1)(t) =

{ξ(0)(t), for all t < t1,

ξ(0)(t+ ϑ1), for all t ≥ t1.

Since the probability distributions of ξ(t) and ξ(t+ ϑ1) are the same for all t ≥ t1,then the processes ξ(t) and ξ(1)(t) have the same one-dimensional distributions,and ξ(1)(t) is a 1-stationary process as well.

With a new point t2 and a random variable ϑ2, which is assumed to be indepen-dent of the process ξ(0)(t) and random variable ϑ1 (therefore, it is also independent


Residual service times in the original M/GI/1/n queueing system

Residual service times of the scaled process in the M/GI/1/n queueing system

Figure 1. Residual service times for the original and scaled pro-cesses of the M/GI/1/n queueing system.


of the process ξ(1)(t)) by the same manner one can build the new 1-stationaryprocess ξ(2)(t). Specifically,

ξ(2)(t) =

{ξ(1)(t), for all t < t2,

ξ(1)(t+ ϑ2), for all t ≥ t2.

The new process ξ(2)(t) has the same one-dimensional distribution as the processesξ(0)(t) and ξ(1)(t). The procedure can be infinitely continued, and one can ob-tain the infinite family of 1-stationary processes, having the same one-dimensionaldistribution.

The points t1, t2,. . . in the above construction are assumed to be some fixed(non-random) points. However, the construction also remains correct in the case ofrandom points t0, t1,. . . of Poisson process, since according to the PASTA property[28] the limiting stationary distribution of a 1-stationary process in a point of aPoisson arrival coincides with the limiting stationary distribution of the same 1-stationary process in an arbitrary non-random point.

Let us now formulate and prove a property similar to Property 3.3 for m = 2.We have the following.

Property 3.4. For the M/GI/2/0 queueing system we have:

(3.3) P{y1,0(t) ∈ B1} = P{x1(t) ∈ B1},

Proof. In order to simplify the explanation in this case, let us consider two auxiliarystationary one-dimensional processes ζ1,0(t) and ζ2,0(t). The first process describesa residual service time in the first server, and the second one describes a residualservice time in the second server. If the ith server (i = 1, 2) is free in time t,then we set ζi,0(t) = 0. The stationary processes ζ1,0(t) and ζ2,0(t) are assumedto be equivalent in the sense that if at the moment of arrival of a customer bothof the servers are free, then each of these servers can be occupied with the equalprobability 1

2 .Thus we have two alternating stationary renewal processes having service periods

and idle times. These both processes are independent of each other.Let us delete the time intervals where the both servers are simultaneously idle,

and merge the corresponding ends (see Figure 2). The new processes are denoted

by ζ1,0(t) and ζ2,0(t). This two-dimensional 1-stationary process characterizes thesystem where in any time t at least one of two servers is busy. If an arrivingcustomer finds one of the servers busy, he/she occupies another server. Considerthe event of arrival of a customer in the system when only one server (let it be thefirst) is busy, i.e the customer occupies the second server. Let τ∗ be the moment ofthis arrival. Then according to the PASTA property, the probability distribution

of ζ1,0(τ∗) coincides with the limiting stationary distribution of ζ1,0(t) given that

this first server is busy. Let χ be the service time of the customer, who arrivesat moment τ∗. Then, according to the aforementioned property of 1-stationary

processes, the probability distribution of ζ1,0(τ∗ + χ) coincides with the limiting

stationary distribution of ζ1,0(t) given that this first server is busy.Thus, in any point of arrival or service completion of one process the proba-

bility distribution of another one coincides with the limiting stationary stationarydistribution of that process given that it is busy.

Let τ∗∗ denote denote the moment of the first service completion in one of twoservers following after the moment τ∗. Then, at the endpoint τ∗∗ of the interval


[τ∗, τ∗∗) the distribution of a residual service time will be the same as at the momentτ∗.

Deleting the interval [τ∗, τ∗∗) and merging the ends τ∗ and τ∗∗ (see Figure 3)we obtain the following structure of the 1-stationary process y1,0(t).

In the points where idle intervals deleted and the ends are merged we haverenewal points: one of periods is finished and another is started. In the other pointswhere the intervals of type [τ∗, τ∗∗) are deleted and their ends are merged we havethe points of ‘interrupted’ renewal processes, where because τ∗ is the moment ofPoisson arrival, according to the PASTA property the stationary distribution ofthe residual time started from the merged point τ∗ until the next renewal pointcoincides with the limiting stationary distribution of the renewal process, i.e. withthe distribution of x1(t). (Notice, that the intervals of type [τ

∗, τ∗∗) are an analogueof the intervals [s1,k, t1,k) considered in the Markovian case in Section 2.)

By amalgamating the residual service times of the first and second servers givenin the the lower graph in Figure 3, one can built a typical one-dimensional 1-stationary process y1,0(t), the limiting stationary distribution of which coincideswith that of x1(t). (see Figure 4).

Therefore the processes y1,0(t) and x1(t) have the identical one-dimensional dis-tribution, and relation (3.3) is proved. �

Let us develop Property 3.4 to the case m = 3 and then to the case of arbitrarym > 1 for the M/GI/m/0 queueing systems. Namely, we have the following.

Property 3.5. For the M/GI/m/0 queueing system we have:

(3.4) P{ym−1,0(t) ∈ Bm−1} = P{xm−1(t) ∈ Bm−1},

where Bm−1 is an arbitrary Borel set of Rm−1.

Proof. The proof will be concentrated in the casem = 3 for the 1-stationary processy2,0(t), which is associated with the paths of the M/GI/3/0 queueing system whereonly two servers are busy.

Prior studying this case, we first study the case of the M/GI/2/0 queueingsystem with two busy servers. Then using also arguments of the proof of Property3.4 enables us to extend the result related to the M/GI/2/0 queueing systems withtwo busy servers for the 1-stationary process y2,0(t) of the M/GI/3/0 queueingsystem.

As in the proof of Property 3.4 in the case M/GI/2/0 queueing system with twobusy servers, we will consider two stationary one-dimensional processes ζ1,0(t) andζ2,0(t). However the idea of the proof is different. The proof explicitly uses the factthat the class of 1-stationary processes is algebraically closed with respect to theoperations of deleting intervals and merging the ends.

Let us delete the idle intervals of the process ζ1,0(t) and merge the ends. Theninstead of this process we get a stationary renewal process as in the above casem = 1 (Property 3.3).

After deleting the same time intervals of the second stationary process ζ2,0(t),it will be transformed as follows. Let t∗ be a moment of Poisson arrival when acustomer occupies the first server. (Recall that owing to the known properties ofPoisson process, the stream of arrival to each of i servers (i = 1, 2) is Poisson.) Then,according to the PASTA property, ζ2,0(t

∗) = ζ2,0(t) in distribution. Therefore afterdeleting all of the idle intervals of the second server and merging the ends, after the


Residual service times in the original M/GI/2/0 queueing system

Residual service times of the scaled process in the M/GI/2/0 queueing system

after deleting the intervals when both of the servers are idle, and merging the

ends

Figure 2. Residual service times for the original and scaled pro-cesses of the M/GI/2/0 queueing system after deleting intervalswhere both of the servers are idle, and merging the ends.



after deleting the intervals when both of the servers are idle, and merging the

ends


after deleting the intervals when both of the servers are idle and both are busy,

and merging the ends

Figure 3. Residual service times for the original and scaled pro-cesses of the M/GI/2/0 queueing system after deleting intervalswhere both of the servers are idle and both are busy, and mergingthe ends.


*

1

*

2

Figure 4. A typical 1-stationary process of residual service timesis obtained by amalgamating residual service times of the first andsecond servers. τ∗1 and τ∗2 are the points where the intervals of type[τ∗, τ∗∗) are deleted, and the ends are merged


first time scaling (i.e. removing corresponding time intervals, see Figure 5) insteadof the initial 1-stationary process ζ2,0(t) we obtain the new 1-stationary process

with equivalent one-dimensional distribution. This process is denoted by ζ2,0(t).

Notice, that the process ζ2,0(t) is obtained from the process ζ2,0(t) by construct-ing a sequence of 1-stationary processes described above.

Then we have the two-dimensional process the first component of which is x1(t)

and the second one is ζ2,0(t). For our convenience this first component is provided

with upper index, and the two-dimensional vector looks now as{x(1)1 (t), ζ2,0(t)

}.

Let us repeat the above procedure, deleting the remaining idle intervals of thesecond server and merging the ends. We get the 1-stationary process being equiv-alent in the distribution to the stationary renewal process x1(t) and denoted now

x(2)1 (t).

Upon this (final) time scaling the first process x(1)1 (t) is transformed as follows.

Let t∗∗ be a random point of Poisson arrival when the second server is occupied.Applying the PASTA property once again, for the first component of the process we

obtain that x(1)1 (t∗∗) coincides in one-dimensional distribution with x

(1)1 (t). Thus,

after deleting the entire idle intervals and merging the ends, we finally obtain the

two-dimensional process{x(1)1 (t),x

(2)1 (t)

}each component of which has the same

one-dimensional distribution as this of the process x1(t). The dynamic of this timescaling is shown in Figure 6.

The next result that we prove now is independence of the processes x(1)1 (t) and

x(2)1 (t). To prove this independence, we again use the PASTA property and exis-

tence of the limiting stationary distributions independently of the initial conditions(see [25]).

Indeed, in each point of the Poisson arrival where one process changes its value,the other process has the same distribution as x1(t), and then in an arbitrary pointt each of the processes has the same distribution as x1(t). Then numbering the

merged points of the process x(1)1 (t) as t∗∗1 , t∗∗2 , . . . , we have:

(3.5)

P{x(1)1 (t) ≤ x1, x

(2)1 (t) ≤ x2

}= lim

j→∞

P{x(1)1 (t∗∗j ) ≤ x1, x

(2)1 (t) ≤ x2

}

= limj→∞

P{x(1)1 (t∗∗j ) ≤ x1

}P{x(2)1 (t) ≤ x2

}

= P{x1(t) ≤ x1}P {x1(t) ≤ x2} .

The first equality of (3.5) is a consequence of the PASTA property given from theabove construction. The second equality of (3.5) is a consequence of the existence

of the limiting stationary one-dimensional distribution of the processes x(1)1 (t) and

x(2)1 (t). The point t is fixed, while the sequence of points t∗∗j increases unbound-

edly. Since the limiting distribution is independent of point t we have the secondinequality of (3.5). The last equality of (3.5) is, again, a consequence of the PASTAproperty.

Now, using the arguments of the proof of Property 3.4 one can easily extend theresult obtained now for M/GI/2/0 queueing system to the M/GI/3/0 queueingsystem, and thus prove (3.4) for the M/GI/3/0 queueing system.

The idea of proof is as follows. We first follow the proof of Property 3.5 forthe M/GI/2/0 queueing system, i.e. delete all intervals where the first, second and


The processes )(0,1

t and )(0,2

t

The transformed processes )(0,1

t and )(0,2

t after deleting the intervals and

merging the ends

Figure 5. The dynamic of time scaling for a queueing systemwith two servers after deleting the idle intervals in the first serverand merging the ends


The processes x1(t) and )(~

0,2t

The processes x1(1)

(t) and x1(2)

(t) obtained from the processes x1(t) and )(~

0,2t by

the scaling procedure

Figure 6. The dynamic of time scaling for a queueing system withtwo servers after deleting the idle intervals in the second server andmerging the ends


third servers are idle simultaneously and merge corresponding ends. Then we deleteall intervals and merge the ends when the first, second and third servers all togetherhave only a single customer in service. Then we consider intervals [τ∗, τ∗∗) similarto those considered in the proof of Property 3.4. Delete these intervals, merge theends and use the relevant arguments of the proof of Property 3.4.

Then the extension of the result for an arbitrary m ≥ 2 follows by induction. �

Now we will establish a connection between the processes ym−1,n(t), ym−1,0(t)and xm−1(t). We start from the case m = 2.

Property 3.6. Under the assumption that the probability distribution functionG(x) belongs to the class NBU (NWU) we have

(3.6) P{y1,n(t) ≤ x} ≤ (resp. ≥) P{x1(t) ≤ x}.

Proof. Along with the 1-stationary processes y1,n(t), let us introduce another 1-

stationary processes Y2,n(t). This last process is related to the same M/GI/2/nqueueing system as the process y1,n(t), and is obtained by deleting time intervalswhen there are more than two and less than two customers in the system, andmerging the ends.

Using the same arguments as in the proof of Property 3.5 one can prove thatthe components of this process are generated by independent 1-stationary processesand having the same distribution. (However, it will be shown later that the limitingstationary distributions of these one-dimensional processes differ from that of theprocess x1(t).)

Let us go back to the initial process y2,n(t), n ≥ 1, to delete the time intervalswhere the both servers are idle and merge the corresponding ends. We also removethe last component corresponding to the queue-length Q2,n(t). (The exact value ofthe queue-length Q2,n(t) is irrelevant here and is not used in our analysis.) In thenew time scale we obtain the two-dimensional process y2,n(t).

Similarly to the proof of relation (3.3) we have time moments τ∗ and τ∗∗. Thefirst of them is a moment of arrival of a customer to the system with one busyand one free server, and the second one is the following after τ∗ moment of servicecompletion of a customer when there remain one busy server only. The time interval[τ∗, τ∗∗) is an orbital busy period. (The concept of orbital busy period is defined inSection 2 for Markovian systems. For M/GI/m/n queueing systems this conceptis the same.) It can contain queueing customers waiting for their service. Lettbegin be a moment of arrival of a customer during the orbital busy period [τ∗, τ∗∗)who occupies a waiting place, and let tend be the following after tbegin moment oftime when after the service completion the queue space becomes empty again. Aperiod of time [tbegin, tend) is called queueing period. (Note that for the M/GI/m/nqueueing system, the intervals of type [τ∗, τ∗∗) are an analogue of the intervalsof type [sm−1,k, tm−1,k) in the Markovian queueing system M/M/m/n, and theintervals of type [tbegin, tend) are an analogue of the intervals of type [sm,k, tm,k) inthat Markovian queueing system M/M/m/n.)

Since tbegin is an instant of Poisson arrival, the two-dimensional distribution ofy2,n(t

begin) coincides with the corresponding limiting stationary distribution of the

1-stationary process Y2,n(t).However, in the point tend, the probability distribution of y2,n(t

end) is different.More specifically, one of the components, say first, is a random variable having


the probability distribution G(x). Then, the second component, because of theaforementioned properties of 1-stationary processes, coincides in distribution with

a one-dimensional component of the process Y2,n(t). Therefore if we delete thequeueing period [tbegin, tend) and merge the corresponding ends, then before themerged point there is one probability distribution, while in the merged point itselfit is another.

Specifically, in the case where G(x) belongs to the class NBU, the 1-stationaryprocess y2,n(t) satisfies the property y2,n(t

begin) ≤st y2,n(tend). (The stochastic in-

equality between vectors means the stochastic inequality between their correspond-ing components.) Therefore, after deleting all of the queueing periods and mergingthe corresponding ends we obtain the new 1-stationary Markov process, which ac-cording to the theorem of Kalmykov [15] (see also [17]) will be stochastically greaterthan the 1-stationary Markov process y2,0(t). Therefore, after deleting the orbitalbusy periods and merging the corresponding ends we obtain that y1,n(t) ≥st y1,0(t),or using the earlier result, relation (3.3), we obtain y1,n(t) ≥st x1(t). Respectively,the case where G(x) belongs to the class NWU leads to the opposite stochasticinequality y1,n(t) ≤st x1(t).

The case of an arbitrary m ≥ 2 follows by induction. The arguments for thisproof are the same. �

From the above results for the Markov processes the statement of the lemmafollows. The stochastic inequalities between Tm,n(m − 1) and Tm,0(m − 1) followby the coupling arguments. The lemma is completely proved. �

The following study of the properties of Tm,n(m − 1), n ≥ 1, concerns the caseλ = mµ only. We have the following lemma.

Lemma 3.7. Under the assumptions λ = mµ the expectation ETm,n(m− 1) is thesame for all n ≥ 1.

Proof. The proof of this result is based on the renewal reward theorem. ForM/GI/m/n queueing systems with different n=1,2,. . . under the assumption ofthe lemma the expected number of queueing periods during a busy period of thissystem is the same for all n=1,2,. . . , and it coincides with the expected numberof orbital busy periods during a busy period. (Each orbital busy period containsone queueing period in average.) Therefore, during a long-run period the numberof events that the different Markov processes ym−1,n(t) change their values afterdeleting queueing periods and merging the ends (as exactly explained in the proofof Lemma 3.1) is the same in average, and according to the renewal reward theo-rem (e.g. Ross [22], Karlin and Taylor [16]) the stationary characteristics of all ofthese Markov processes ym−1,n(t), n=1,2,. . . , are the same. Hence, the expectationETm,n(m−1) is the same for all n=1,2,. . . as well, and the statement of Lemma 3.7is proved. �

4. Theorems on losses in M/GI/m/n queueing systems

The results obtained in the previous section enable us to establish theorems forthe number of losses in M/GI/m/n queueing systems during their busy periods.Using Lemma 3.7 one can prove the following general theorem on losses.

Theorem 4.1. Under the assumption λ = mµ, the expected number of losses duringa busy period of the M/GI/m/n queueing system is the same for all n ≥ 1.


Proof. To prove this theorem we use the following construction similar to that usedto prove Lemma 3.1. We delete all the intervals where the number of customersin the M/GI/m/n queueing system (n ≥ 1) is less than m and merge the corre-sponding ends. The process obtained is denoted ym,n(t). This is the 1-stationaryprocess of orbital busy periods.

The stationary departure process, together with the arrival Poisson process ofrate λ and the number of waiting places n describes the stationaryM/G/1/n queue-length process (with dependent service times). As soon as a busy period is finished(in our case it is an orbital busy period, see Section 2 for the definition), a systemimmediately starts a new busy period by attaching a new customer into the system.This unusual situation arises because of the construction of the process. There isno idle period, and servers all are continuously busy. Thus, the busy period, whichis considered here, is one of the busy periods attached one after another.

Let T be a large period of time, and during that time there are K(T ) busyperiods of the M/G/1/n queueing system (which does not contain idle times asmentioned). Let L(T ) and ν(T ) denote the number of lost and served customersduring time T . We have the following formula

(4.1) limT→∞

1

EK(T )

(EL(T ) + Eν(T )

)= lim

T→∞

1

EK(T )

(λT + EK(T )

),

which can be motivated by renewal arguments as follows.

There are m independent copies x(1)1 (t), x

(2)1 (t), . . . , x

(m)1 (t) of the stationary

renewal process, which model the process ym,n(t). (In fact, we have m 1-stationaryprocesses, which have the same distributions as m independent renewal processes

x(1)1 (t), x

(2)1 (t), . . . , x

(m)1 (t).) Let 1 ≤ i ≤ m, and let C1, C2,. . .CKi(T ) be such

points of busy period starts associated with the renewal process x(i)1 (t) (one of those

m independent and identically distributed renewal processes), where Ki(T ) denotesthe total number of these regeneration point indexed by i. Since at the moments C1,C2,. . .CKi(T ) of the busy period starts the distribution of the stationary Markovprocess of residual times is the same, then the number of losses during each ofthese busy periods has the same distribution, and one can apply the renewal rewardtheorem.

By renewal reward theorem we have:

(4.2) limT→∞

1

mEKi(T )

(EL(T ) + Eν(T )

)= lim

T→∞

1

mEKi(T )

(λT +mEKi(T )

).

Taking into account that

limT→∞

EK(T )

EKi(T )= m,

we arrive at (4.1). Thus, we bypass the fact that the times between departures aredependent, and thus (4.1) is obtained by application of the renewal reward theoremby a usual manner, like in the case of independent times between departures.

Relationship (4.1) has the following explanation. The left-hand side term EL(T )+Eν(T ) is the expectation of the numbers of lost customers plus the expectation ofthe number of served customers, and the right-hand side term λT + EK(T ) is theexpectation of the number of arrivals during time T plus the expectation of thenumber of attached customers. Let us now write the second equation. We have

(4.3) limT→∞

1

EK(T )Eν(T ) = lim

T→∞

1

EK(T )mµT.


Let us now introduce the following notation. Let ζn denote the length of anorbital busy period, and let Ln and νn correspondingly denote the number of lostand served customers during that orbital busy period. Using the arguments of [5],we prove that ELn = 1 for all n ≥ 1.

Indeed, from (4.1) and (4.3) we have the equations:

(4.4) ELn + Eνn = λEζn + 1,

(4.5) Eνn = mµEζn.

The substitution λ = mµ into the system of equations (4.4) and (4.5) yields ELn =1. Hence, an application of Lemma 3.7 together with Wald’s identity leads tothe desired result. Since the expectation ETm,n(m − 1) and the expectation ofthe number of orbital busy periods during a busy period of M/GI/m/n both areindependent of n, then the result follows. �

In turn, an application of Lemma 3.1 and the arguments of Theorem 4.1 enablesus to prove the following result.

Theorem 4.2. Let λ = mµ. Then, under the assumption that G(x) belongs to theclass NBU, for the number of losses in M/GI/m/n queueing systems, n ≥ 1, wehave

(4.6) ELm,n =cmm

m!,

where the constant c ≥ 1 depends on m and the probability distribution G(x) but isindependent of n.

Under the assumption that G(x) belongs to the class NWU we have (4.6) butwith constant c ≤ 1.

Proof. Notice first, that for the expected number of losses in M/GI/m/0 queueingsystems we have

ELm,0 =mm

m!This result follows immediately from the Erlang-Sevastyanov formula [23], so thatthe expected number of losses during a busy period of the M/GI/m/0 queueingsystem is the same that this of the M/M/m/0 queueing system. The expectednumber of losses during a busy period of the M/M/m/0 queueing system, ELm,0 =mm

m! , has been established in Section 2.In the case where G(x) belongs to the class NBU according to Lemma 3.1 we

have ETm,n(m− 1) ≥ ETm,0(m− 1), and therefore, the expected number of orbitalbusy periods in the M/GI/m/n queueing system (n ≥ 1) is not smaller than thisin the M/GI/m/0 queueing system. Therefore, repeating the proof of Theorem 4.1leads to the inequality ELm,n ≥ ELm,0 and consequently to the desired result. IfG(x) belongs to the class NWU, then we have the opposite inequalities, and finallythe corresponding result that is stated in the formulation of the theorem. �

5. Batch arrivals

The case of batch arrivals is completely analogous to the case of ordinary (non-batch) arrivals. In the case of a Markovian MX/M/m/n queueing system one canalso apply the level-crossing method to obtain equations analogous to (2.6)-(2.13).The same arguments as in Sections 3 and 4 in extended form can be used forMX/GI/m/0 queueing systems.


Acknowledgement

The research was supported by Australian Research Council, Grant #DP0771338.The author thanks all of three anonymous referees for their constructive comments.

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School of Mathematical Sciences, Monash University, Building 28M, Clayton Cam-

pus, Wellington road, VIC-3800, Australia

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