The single server queue with synchronized services

The single server queue with synchronized services Antonis Economou1 , Stella Kapodistria2 and Jacques Resing3 [email protected], [email protected] and [email protected] 1

Department of Mathematics University of Athens Panepistemioupolis, 15784 Athens, Greece 2

Department of Statistics and Actuarial-Financial Mathematics University of Aegean 83200 Karlovassi, Samos, Greece 3

Department of Mathematics and Computer Science Eindhoven University of Technology P.O. Box 513, 5600 MB Eindhoven, The Netherlands July 28, 2010

Abstract We consider a single server queueing system with generally distributed synchronized services. More specifically, customers arrive according to a Poisson process and there is a single server that provides service, if there is at least one customer present in the system. Upon the initialization of a service, all present customers start to get service simultaneously. We consider the gated version of the model, that is, customers who arrive during a service time do not receive service immediately but wait for the beginning of the next service time. At service completion epochs, all served customers decide simultaneously and independently whether they will leave the system or stay for another service. The probability that a served customer gets another service is the same for all customers. We study the model and derive its main performance measures that include the equilibrium distribution of the number of customers at service completion epochs and in continuous time, the busy period and the sojourn time distributions. Moreover, we prove some limiting results regarding the behavior of the system in the extreme cases of the synchronization level. Several variants and extensions of the model are also discussed. Keywords: queueing system; synchronized services; repeated services; embedded distribution; equilibrium distribution; busy period; sojourn time; conditional residual service time.

1

Introduction

The majority of studies in the queueing literature concerns systems where the customers are served one by one or in batches of fixed size. Moreover, there exist many studies of systems with batch services, where the successive batch sizes are independent and identically distributed 1

random variables. These systems represent satisfactorily production or telecommunication processes where the clients are served sequentially. However, in some modern applications, it is reasonable to assume that all present customers are served concurrently. This is the case for example in webseminars or other educational activities, where a system can process all present customers simultaneously. Then, after the completion of a service cycle, all present customers pass a testing procedure and some of them leave the system, while the others have to repeat their service. This gives rise to queueing systems with synchronized repeated services. There are also other kinds of synchronization arising in queueing systems, for example in production networks, where some of the stations are control or assembly stations. In the present paper, we consider a basic single server queueing system with generally distributed synchronized services. In this model, the server performs service cycles with durations that follow a general service time distribution. We suppose that all customers that are present at the beginning of a service constitute a service batch and so they are served concurrently as a whole. At a service completion epoch, every served customer remains in the system with probability q or departs with probability p = 1 − q, independently of the others. The particular feature of this model is the presence of binomial type jumps at the service completion epochs. A Markovian queue with setup times that encompasses this feature has been recently studied by Economou and Kapodistria (2009). We aim to extend such a study in the more general framework of non-Markovian queueing models. Models with this type of binomial jumps have been studied by Brockwell et al. (1982), Neuts (1994) and more recently by Economou (2004), Artalejo et al. (2007) and Economou and Fakinos (2008) in population processes subject to mild catastrophic events. Indeed, in these studies, the evolution of a population is influenced by disasters that occur according to some point process (usually renewal and especially Poisson process). Once a catastrophe occurs, every individual of the population survives with probability q or dies with probability p = 1 − q. Therefore, the population size is reduced at the catastrophe epochs according to some binomial distribution and we observe again this type of binomial jumps. Recently, Adan et al. (2009) and Economou and Kapodistria (2010) studied two vacation queueing models, where the customers become impatient when the server is on vacation and they perform synchronized abandonments. These models are motivated by remote systems, where customers have to wait for a certain transport facility to abandon the system. Then, whenever the facility visits the system, the present customers decide whether to leave the system or not. More specifically, every present customer remains in the system with probability q or abandons the system with probability p = 1 − q, independently of the others. The analysis of these models extends the analysis of Altman and Yechiali (2006) and Yechiali (2007), in the framework of synchronized abandonments. Again, the singular feature of these models is the presence of binomial type jumps at the abandonment epochs. The paper is organized as follows. In Section 2, we describe the model and introduce the appropriate notation. In Section 3, we carry out the transient and the equilibrium analysis of the number of customers in the system at service completion epochs and we derive several formulas and iterative algorithmic schemes. In Section 4, we deduce the equilibrium distribution of the number of customers in continuous time, by appealing to a regenerative approach originated in Hordijk and Tijms (1976). Subsequently, in Section 5, we treat the busy period of the model, while in Section 6 we study the sojourn time of a customer in the system. Moreover, we use a 2

technique inspired by Kerner (2008) to compute the conditional residual service time distribution given the number of customers in the system. This enables us to derive also the conditional sojourn time distribution of a customer, given that he finds a certain state upon his arrival to the system. In Section 7, we deal with some limiting regimes, concerning the behavior of the system in extreme cases of the synchronization feature. Finally, the study concludes in Section 8, where we present several numerical results and discuss several variants of the model, possible extensions and open problems.

2

Model description and notation

We consider the following queueing model. Individuals arrive to the system according to a Poisson process at rate λ. There is one server and the service times are independent identically distributed random variables with a general (absolute) continuous distribution B(t), probability ˜ density function b(t), Laplace–Stieltjes transform (LST) B(s) = E[e−sB ] and finite moments n E[B ], where the random variable B represents a generic service time. The excess, or residual, service time is denoted by Be . Its distribution Be (t), the probability density function be (t), the ˜e (s) and the moments E[B n ] are given respectively by LST B e Rt n+1 ] ˜ (1 − B(u))du (1 − B(t)) ˜e (s) = 1 − B(s) , E[Ben ] = E[B Be (t) = 0 , be (t) = , B . E[B] E[B] sE[B] (n + 1)E[B] (2.1) Upon the initialization of a service time, all present customers start to get service concurrently. Customers arriving during a service time do not begin to receive service immediately, but they have to wait for the beginning of the next service, i.e. they do not join the current service batch. In a sense, we can say that this is a gated queueing system. At a service completion epoch, every served customer either departs from the system with probability p ∈ (0, 1) or remains to repeat his service with probability q = 1 − p, independently of the other individuals. We will refer to this system as the single server queue with synchronized repeated services. We are interested primarily in studying the continuous-time process {Z(t) : t ≥ 0} that records the total number of customers in the system. We will also consider the continuous-time process {(X(t), Y (t)) : t ≥ 0} which provides a more detailed description of the state of the system. More specifically, we define X(t) to be the number of customers in service at time t and Y (t) to be the number of customers waiting for service at time t, t ≥ 0. The state process {(X(t), Y (t))} is not in general Markovian (except in the case where B(t) is exponential) nor is {Z(t)}. To overcome this difficulty, we define and study the embedded process at service completion epochs, which is easily seen to be a discrete-time Markov chain. Then, we exploit the semi-regenerative nature of {Z(t)} to obtain its equilibrium distribution.

3

The state distribution of the embedded Markov chain

Let Bn denote the duration of the n-th service time and τn the epoch of the n-th service time completion, just after it has been decided which of the served customers will stay and the waiting customers have been added to those that are to repeat their service. We define Xn = X(τn ), n ≥ 0, to be the number of customers in service at the n-th service time completion which coincides with the number of customers in system (since Y (τn ) = 0). We 3

can easily see that {Xn : n ≥ 0} is a Markov chain of which the dynamics is described by the basic relation   In,1 + Hn , if Xn−1 = 0, n−1 (3.1) Xn = XP  In,i + Hn , if Xn−1 ≥ 1,  i=1

where In,i are independent and identically distributed binary random variables with Pr[In,i = 1] = q, (In,i = 1 if and only if the i-th customer of the n-th service batch stays in the system at time τn ) and Hn is independent of In,i and denotes the number of events in the Poisson arrival process with rate λ during the n-th service time Bn . The Markov chain {Xn } can be easily shown to be positive recurrent using Pakes’ (1969) Theorem 2: an irreducible and aperiodic Markov chain {Xn } is ergodic provided that for all but finitely many values of k, the drift at state k, E[Xn − Xn−1 |Xn−1 = k], is bounded above by a negative constant. For our model we observe that {Xn } is irreducible, aperiodic and satisfies the condition of stability since E[Xn − Xn−1 |Xn−1 = k] = E[

k X

In,i + Hn − k|Xn−1 = k]

i=1

= λE[B] − kp < 0 , k >

λE[B] . p

The process {Xn } is a kind of a branching process with immigration and a particular immigration mechanism at state 0. Its analysis is done by conditioning. We easily obtain that ˜ − z)), E[z Xn |Xn−1 = 0] = (1 − q + qz)B(λ(1 j ˜ Xn E[z |Xn−1 = j] = (1 − q + qz) B(λ(1 − z)), j ≥ 1.

(3.2) (3.3)

Let PXn (z) denote the probability generating function (PGF) of Xn . Then, by using (3.2) and (3.3) we have ˜ ˜ PXn (z) = PXn−1 (0)(1 − q + qz)B(λ(1 − z)) + (PXn−1 (1 − q + qz) − PXn−1 (0))B(λ(1 − z)) ˜ ˜ = PX (1 − q + qz)B(λ(1 − z)) − PX (0)B(λ(1 − z))q(1 − z). (3.4) n−1

n−1

We are now in position to find an expression for the PGF PXn (z) in closed form. We have the following theorem. Theorem 3.1 Suppose that the system is initially empty. Then the PGF of the transient distribution of the embedded Markov chain {Xn } of the number of customers in the system at service completion epochs is given from PXn (z) =

n−1 Y

j ˜ B(λq (1 − z)) − (1 − z)

j=0

n X

PXn−k (0)q k

k−1 Y

j ˜ B(λq (1 − z)) ,

(3.5)

j=0

k=1

where the transient probabilities PXn (0), n ≥ 0, of an empty system are calculated through the recursive scheme PX0 (0) = 1, PXn (0) =

n−1 Y j=0

(3.6) j ˜ B(λq )−

n X

PXn−k (0)q k

k=1

4

k−1 Y j=0

j ˜ B(λq ), n ≥ 1.

(3.7)

Proof. Equation (3.4) gives immediately ˜ ˜ − z))q(1 − z) − z)) − PXn−2 (0)B(λ(1 PXn−1 (z) = PXn−2 (1 − q + qz)B(λ(1

(3.8)

and substituting 1 − q + qz for z we obtain ˜ ˜ − z))q 2 (1 − z). (3.9) − z)) − PXn−2 (0)B(λq(1 PXn−1 (1 − q + qz) = PXn−2 (1 − q 2 + q 2 z)B(λq(1 Plugging equation (3.9) into (3.4) yields ˜ ˜ − z))B(λ(1 − z)) PXn (z) = PXn−2 (1 − q 2 + q 2 z)B(λq(1 2 ˜ ˜ − z))B(λ(1 − z))q (1 − z) −PXn−2 (0)B(λq(1 ˜ − z))q(1 − z). −PX (0)B(λ(1 n−1

(3.10)

We can further continue by substituting PXn−2 (1−q 2 +q 2 z) using (3.4) and so on. This iterating procedure yields PXn (z) = PX0 (1 − q n + q n z)

n−1 Y

j ˜ B(λq (1 − z))

j=0

−(1 − z)

n X

PXn−k (0)q k

k−1 Y

j ˜ B(λq (1 − z)).

(3.11)

j=0

k=1

But PX0 (z) = 1 since the system is initially empty so we obtain (3.5). Setting z = 0 in (3.5), we can compute PXn (0) recursively as described in the scheme (3.6)-(3.7).  Remark 3.1 By multiplying equations (3.6)- (3.7) by z n and summing for all n ≥ 0, we obtain after a bit of algebraic manipulation the generating function of (PXn (0) : n ≥ 0) as P∞ n Qn−1 ˜ ∞ j X j=0 B(λq ) n=0 z n PXn (0)z = P∞ . (3.12) Q j) ˜ (qz)n n−1 B(λq n=0

Here, the empty product

−1 Q

j=0

n=0

is by definition equal to 1. If we are interested in the first passage

j=0

time probabilities to state 0 starting from 0, i.e. fn = Pr[Xn = 0, Xk 6= 0, for 1 ≤ k ≤ n − 1|X0 = 0], then it is well known (see e.g. Karlin and Taylor (1975)Pp.63, Equation (2.5.8) n or Norris (1997) generating function F (z) = ∞ n=1 fn z satisfies the P∞ Exercise n1.5.3) that the−1 relationship n=0 PXn (0)z = (1 − F (z)) . Hence, we obtain that the generating function of the first passage time probabilities to 0 starting from 0 is given as P∞ n Qn−1 ˜ n j n=0 z (1 − q ) j=0 B(λq ) F (z) = . (3.13) P∞ n Qn−1 ˜ z B(λq j ) n=0

j=0

We now proceed with of the equilibrium distribution (pk : k ≥ 0) of {Xn }. To P∞ the study k this end, let PX (z) = k=0 pk z be the equilibrium PGF of the number of customers at service completion epochs. Then, similarly to the derivation of (3.4) we obtain the relation ˜ ˜ PX (z) = PX (1 − q + qz)B(λ(1 − z)) − p0 B(λ(1 − z))q(1 − z). Using (3.14) we obtain the following analogue of Theorem 3.1. 5

(3.14)

Theorem 3.2 The PGF of the equilibrium distribution of the embedded Markov chain {Xn } of the number of customers in the system at service completion epochs is given from PX (z) =

∞ Y

j ˜ B(λq (1 − z)) − (1 − z)p0

j=0

∞ X

qk

k=1

k−1 Y

j ˜ B(λq (1 − z)) ,

(3.15)

j=0

where the probability of an empty system at service completion epochs is given as Q∞ ˜ j j=0 B(λq ) . p0 = P∞ Q j) ˜ q k k−1 B(λq

(3.16)

j=0

k=0

Proof. Iterating equation (3.14) in a similar manner as in the proof of Theorem 3.1 and taking the limit as the number of iterations goes to infinity yields equation (3.15), while p0 is obtained by setting z = 0 in equation (3.15) and solving for the unknown probability. Note that it is easy to check the absolute convergence of the involved series in the unit disk, by using the ratio criterion. The convergence of the infinite products is also easily established (see e.g. Ahlfors (1979), Theorem 6 in Section 5.2.2).  Remark 3.2 Using Theorems 3.1 and 3.2 we can calculate the moments of the transient and the equilibrium distributions of the number of customers in the system at service completion epochs. For example, by taking derivatives in (3.4) and evaluating at z = 1 we obtain the recursive relationship E[Xn ] = qE[Xn−1 ] + λE[B] + qPXn−1 (0), n ≥ 1. By iterating and using E[X0 ] = 0 we obtain the expression n

E[Xn ] = λ

X 1 − qn q k PXn−k (0) n ≥ 0. E[B] + 1−q

(3.17)

k=1

Similarly, by computing the derivatives of (3.4) of higher orders, using Leibnitz’s rule and evaluating at z = 1, we can obtain in principle all the factorial moments of the transient distribution. For brevity, we will report here the expressions for the factorial moments only for the equilibrium distribution. (n) Let E[X n ] = E[X(X − 1) · · · (X − n + 1)] = PX (1) be the factorial moment of order n of the equilibrium distribution of the number of customers in the system at service completion epochs. Then, differentiating (3.14) n times and setting z = 1 yields (n) PX (1)

n   X n (k) ˜ (n−k) (0)λn−k + np0 qλn−1 (−1)n−1 B ˜ (n−1) (0), n ≥ 1. = P (1)q k (−1)n−k B k X k=0

which after some algebraic manipulations can be simplified to 1 (λE[B] + p0 q) 1−q n−1   1 X n k n−k q λ E[B n−k ](E[X k ] + p0 δk,1 ), n ≥ 2, 1 − qn k

E[X 1 ] = E[X] =

(3.18)

E[X n ] =

(3.19)

k=0

6

where δk,i denotes the Kronecker’s symbol defined as ( 1 , if k = i δk,i = 0 , if k 6= i . Equation (3.19) constitutes a stable recursive scheme with initial condition for E[X 1 ] = E[X] given by (3.18). The recursion given by (3.19) has coefficients that depend on E[B n ], so a closed-form solution seems impossible. Nevertheless, a closed-form expression for E[X n ] can be obtained for certain cases, such as the exponential (by applying results from the theory of q-series – see e.g. Economou and Kapodistria (2009, 2010)) or the deterministic case. Remark 3.3 Unfortunately, we were not able to give a probabilistic interpretation for the expressions (3.5) and (3.15) for PXn (z) and PX (z) respectively, in the sense that we cannot describe the random variables Xn and X in terms of underlying random variables of the model. However, if we consider the excess distribution of Xn and use (3.5) we obtain that 1 − PXn (z) E[Xn ](1 − z)

=

1−

Qn−1 ˜ j j=0 B(λq (1 − z))

n X PXn−k (0)q k

Qk−1 ˜ j j=0 B(λq (1 − z))

+ E[Xn ](1 − z) E[Xn ] k=1 Qn−1 ˜ n k−1 j X Y j=0 B(λq (1 − z)) (n) 1 − (n) j ˜ = g0 + g B(λq (1 − z)) Pn−1 j k (λE[B] j=0 q )(1 − z) k=1 j=0

=

n X

(n)

gk GX (n) (z),

(3.20)

k

k=0

where (n) g0

=

(n)

=

gk

GX (n) (z) = 0

GX (n) (z) = k

λE[B]

Pn−1 j=0

qj

, E[Xn ] q k PXn−k (0) , 1 ≤ k ≤ n, E[Xn ] Qn−1 ˜ B(λq j (1 − z)) 1 − j=0 , Pn−1 j (λE[B] j=0 q )(1 − z) k−1 Y

j ˜ B(λq (1 − z)), 1 ≤ k ≤ n.

(3.21) (3.22) (3.23)

(3.24)

j=0

Note, now, that GX (n) (z) can be easily seen to be the PGF of the number of Poisson arrivals 0 P j at rate λ in a time interval distributed according to the excess distribution of n−1 j=0 q Bj , where B0 , B1 , . . . , Bn−1 are independent identically distributed random variables having the service time distribution. Similarly, GX (n) (z), k = 1, 2, . . . , n, are the PGFs of the number of Poisson arrivals k P j at rate λ during time intervals distributed as the random variables k−1 j=0 q Bj . Pn (n) Note also that k=0 gk = 1 because of (3.21)-(3.22) and (3.17). Therefore, the expression (3.20) shows that a random variable that corresponds to the excess distribution of Xn can be (n) thought of as the finite mixture of the non-negative integer random variables Xk whose PGFs (n) are given by (3.23)-(3.24) with mixing probabilities gk given by (3.21)-(3.22). This expression may also have some intuitive justification. We do not elaborate further on this point now, 7

but we stress that such considerations have been proved fruitful for deriving the limiting results reported in Section 7. Moreover, the representation (3.20) can be used to study approximation and stochastic ordering issues for {Xn }. 1−PX (z) A formula analogous to (3.20) can be easily derived for the PGF E[X](1−z) . Then, the corresponding random variable can be thought of as the infinite mixture of known non-negative integer random variables with known mixing probabilities. We suppress the details as they are almost identical to the above reasoning.

4

The equilibrium state distribution in continuous time

We are interested in studying the equilibrium distribution (πn : n ≥ 0) of the process {Z(t)} that records the number of customers in system in continuous time. We follow a regenerative approach which originated in Hordijk and Tijms (1976) (see also Tijms (1994)). Theorem 4.1 The equilibrium distribution of the number of customers in the system at arbitrary instants, (πn ), can be computed in terms of the equilibrium embedded distribution (pk ) of {Xn } using the relations π0 = πn =

p0 , p0 + λE[B] R∞ R∞ n−1 Pn −λx (1 − B(x))dx + λ λp0 0 (λx) k=1 pk 0 (n−1)! e

(4.1) (λx)n−k −λx (1 (n−k)! e

p0 + λE[B] n ≥ 1.

The equilibrium PGF ΠX (z) = ΠX (z) =

− B(x))dx , (4.2)

P∞

n=0 πn z

n

is given from

˜e (λ(1 − z)) p0 + λE[B](PX (z) − p0 (1 − z))B , p0 + λE[B]

(4.3)

where PX (z) is the PGF of the equilibrium distribution of the number of customers in the system ˜e (s) is given in (2.1). at service completion epochs given in equation (3.15) and B Proof. We define the generic random variables T : the length of a regenerative cycle, i.e. the time elapsed between two consecutive service completion epochs after which the system is empty, Tn : the amount of time during the cycle that there are n customers in the system, n ≥ 0. Then, by the renewal–reward theorem, we have that πn =

E[Tn ] , n ≥ 0. E[T ]

(4.4)

The expected total number of service completions in a cycle equals to 1/p0 . It now immediately follows, from Wald’s theorem, that E[T ] =

1 1 + E[B] . λ p0

(4.5)

Now, we divide the cycle into a number of disjoint intervals separated by the service completion epochs and calculate E[Tn ] in terms of these intervals. We define the random variables 8

Mk : the number of service completion epochs in a cycle that leave k customers in the system, k ≥ 0, Ak,j : the amount of time that j customers are present in the system during a service time that started with k customers. Then

pk , k ≥ 0, p0

E[Mk ] = and Z

B

(4.6) 

1[N (x) = j − k]dx , j ≥ k ≥ 0,

E[Ak,j ] = E 0

where 1[A] denotes the indicator random variable of an event A, N (x) is the number of arrivals according to a Poisson process at rate λ at time x and B is a random variable representing a generic service time, which is independent of the Poisson process. Then, we have that Z ∞Z y E[Ak,j ] = Pr[N (x) = j − k]dxdB(y) Z0 ∞ 0 Z ∞ = Pr[N (x) = j − k] dB(y)dx x Z0 ∞ = Pr[N (x) = j − k](1 − B(x))dx 0 Z ∞ (λx)j−k −λx e (1 − B(x))dx, j ≥ k ≥ 0. (4.7) = (j − k)! 0 We can now see that E[T0 ] =

1 , λ

(4.8)

E[Tn ] = E[A1,n ] +

n X pk k=1

p0

E[Ak,n ], n ≥ 1.

(4.9)

Substituting (4.8),(4.9) and (4.5) in (4.4), taking into account (4.7), we immediately obtain equations (4.1) and (4.2). The relation (4.3) follows after straightforward calculations, using the relations reported in (2.1). 

5

The busy period distributions

We will now study the busy period of the model, i.e. the time from the arrival of a customer at an empty system till the next epoch that the system is empty again. The duration of a busy period L can be thought of as a first passage time starting from just entering at state 1 to reach ˜ 0. To obtain the LST L(s) = E[e−sL ] of L, we use an indirect approach and more specifically we first study the process {K(t) : t ≥ 0} that records the age of the current busy period. Since the process {K(t)} is not Markovian, we define the embedded discrete-time Markov process {Kn : n ≥ 0}, where Kn = K(τn ) is the age of the current busy period just after the n-th service completion epoch. Note that similarly to (3.1) we have now the relation   (In,1 + Hn , Bn ) , if Xn−1 = 0, XP n−1 (Xn , Kn ) = (5.1)  In,i + Hn , Kn−1 + Bn ) , if Xn−1 ≥ 1, ( i=1

9

where Xn , In,i , Hn and Bn have been defined in Section 3. We are interested in studying the transient and equilibrium distributions of the process {(Xn , Kn ) : n ≥ 0}. To this end we define the mixed (PGF-LST) transforms ΦXn ,Kn (z, s) = E[z Xn e−sKn ], n ≥ 0. Observe that Bn and Hn are dependent for any fixed n with PGF-LST transform ˜ + λ(1 − z)), E[z Hn e−sBn ] = B(s so (5.1) yields ˜ + λ(1 − z)) E[z Xn e−sKn |Xn−1 = 0, Kn−1 = u] = (1 − q + qz)B(s ˜ + λ(1 − z)), j ≥ 1. E[z Xn e−sKn |Xn−1 = j, Kn−1 = u] = (1 − q + qz)j e−su B(s

(5.2) (5.3)

Now, we can combine (5.1), (5.2) and (5.3) and we obtain Z ∞ E[z Xn e−sKn |Xn−1 = 0, Kn−1 = u]d Pr[Xn−1 = 0, Kn−1 ≤ u] ΦXn ,Kn (z, s) = 0

+

∞ Z X j=1 ∞

Z

∞

E[z Xn e−sKn |Xn−1 = j, Kn−1 = u]d Pr[Xn−1 = j, Kn−1 ≤ u]

0

˜ + λ(1 − z))d Pr[Xn−1 = 0, Kn−1 ≤ u] (1 − q + qz)B(s

= 0

+

∞ Z X j=1

∞

˜ + λ(1 − z))d Pr[Xn−1 = j, Kn−1 ≤ u] (1 − q + qz)j e−su B(s

0

˜ + λ(1 − z))ΦX ,K (0, 0) = (1 − q + qz)B(s n−1 n−1 ˜ +B(s + λ(1 − z))[ΦX ,K (1 − q + qz, s) − ΦX n−1

n−1 ,Kn−1

n−1

(0, s)].

(5.4)

Let ΦX,K (z, s) denote the equilibrium PGF-LST for the distribution of {Xn , Kn }. Then, similarly to (5.4) we obtain the relation ˜ + λ(1 − z))ΦX,K (1 − q + qz, s) ΦX,K (z, s) = B(s ˜ + λ(1 − z))[(1 − q + qz)ΦX,K (0, 0) − ΦX,K (0, s)]. +B(s

(5.5)

Relation (5.5) is the key to obtain ΦX,K (z, s) using a procedure similar to the proof of Theorem ˜ 3.2 starting from (3.14) and then we can easily derive the LST L(s) of the duration of the busy period L. Theorem 5.1 The LST of the duration of the busy period is given by P∞ Q ˜ + λq j ) (1 − q k+1 ) kj=0 B(s k=0 ˜ L(s) = , Re(s) > 0. P∞ Qk−1 ˜ j k=0 j=0 B(s + λq )

(5.6)

Proof. Equation (5.5) has the form f (z) = A(z)f (1 − q + qz) + C(z).

(5.7)

and can be solved iteratively, as in the proofs of Theorems 3.1 and 3.2. Indeed, iterating equation (5.7) with respect to z and taking the limit as the number of iterations goes to infinity, provided that the infinite series and the infinite product converge, yields f (z) =

∞ X k=0

C(1 − q k + q k z)

k−1 Y

A(1 − q j + q j z) + f (1)

j=0

∞ Y j=0

10

A(1 − q j + q j z).

(5.8)

We can easily check that the convergence in the case of equation (5.5) holds and therefore (5.8) is applicable and yields that ∞ k X Y k+1 k+1 ˜ + λq j (1 − z)) ΦX,K (z, s) = [(1 − q + q z)ΦX,K (0, 0) − ΦX,K (0, s)] B(s j=0

k=0

+ΦX,K (1, s)

∞ Y

˜ + λq j (1 − z)), |z| ≤ 1, Re(s) ≥ 0. B(s

(5.9)

j=0

For z = 0, s = 0 equation (5.9) yields ΦX,K (0, 0) = −

∞ X

q k+1 ΦX,K (0, 0)

k Y

j ˜ B(λq ) + ΦX,K (1, 0)

j=0

k=0

∞ Y

j ˜ B(λq ).

Since ΦX,K (1, 0) = 1, we can solve for ΦX,K (0, 0) and we obtain Q∞ ˜ j j=0 B(λq ) , ΦX,K (0, 0) = P∞ Q j) ˜ q k k−1 B(λq k=0

(5.10)

j=0

(5.11)

j=0

which nicely agrees with (3.16). For z = 0, Re(s) > 0 equation (5.9) yields ΦX,K (0, s) =

∞ X

[(1 − q

k+1

)ΦX,K (0, 0) − ΦX,K (0, s)]

Q∞

˜ + λq j ), Re(s) > 0, B(s

(5.12)

j=0

k=0

since

k Y

˜

+ λq j ) = 0 for Re(s) > 0. Solving (5.12) for ΦX,K (0, s) we obtain Qk ˜ P∞ j k+1 ) j=0 B(s + λq ) k=0 (1 − q ΦX,K (0, s) = ΦX,K (0, 0), Re(s) > 0. P∞ Qk−1 ˜ B(s + λq j )

j=0 B(s

k=0

Note now that

(5.13)

j=0

ΦX,K (0, s) ˜ L(s) = E[e−sL ] = E[e−sK |X = 0] = . ΦX,K (0, 0)

Equations (5.13) and (5.14) imply readily (5.6).

(5.14) 

Remark 5.1 A similar analysis can be carried out for the number N of service completions in a P busy period, which constitutes a discrete analogue of the duration L. Note that although L= N i=1 Bi , where B1 , B2 , . . . are the successive service times in the busy period, we cannot ˜ obtain the PGF PN (z) of N using the LST L(s) of L, since B1 , B2 , . . . are not independent of ˜ ˜ N and hence the relation L(s) = PN (B(s)) is not valid. However, we can obtain PN (z) using an indirect approach. We can consider the sequence of random variables {Mn : n ≥ 0}, where Mn is defined to be the number of service completions in the current busy period at time τn , i.e. ( 1, if Xn−1 = 0, Mn = (5.15) 1 + Mn−1 , if Xn−1 ≥ 1. Then {(Xn , Mn ) : n ≥ 0} is a discrete-time Markov chain that can be studied using the line of argument that we used for {(Xn , Kn )}. Let PX,M (z, u) denote the joint PGF for the equilibrium distribution of {(Xn , Mn )}. Then we can obtain the relation ˜ PX,M (z, u) = uB(λ(1 − z))PX,M (1 − q + qz, u) ˜ +uB(λ(1 − z))[(1 − q + qz)PX,M (0, 1) − PX,M (0, u)], 11

(5.16)

which is the analogue of (5.5). Using (5.16) and proceeding as in the proof of Theorem 5.1, we derive that P∞ Qk ˜ k+1 )uk+1 j k=0 (1 − q j=0 B(λq ) PX,M (0, 1), |u| < 1. (5.17) PX,M (0, u) = P∞ k Qk−1 ˜ j j=0 B(λq ) k=0 u Note that for the PGF PN (z) of the number of service completion in a busy period we have that PX,M (0, z) PX,M (0, 1) P∞ Qk ˜ k+1 k+1 j )z k=0 (1 − q j=0 B(λq ) , P∞ k Qk−1 ˜ j) z B(λq k=0 j=0

PN (z) = E[z N ] = E[z M |X = 0] = =

(5.18)

which coincides with F (z) given in (3.13) as expected.

6

The sojourn time distributions

We now consider a tagged customer and let W denote its sojourn time in the system, i.e. the time since his arrival till his departure from the system. The following Theorem 6.1 provides the LST of W . Theorem 6.1 The LST of the sojourn time W of a customer in the system is given as " # ˜ ˜ 1 − B(s) (1 − q) B(s) ˜ (s) = π0 + (1 − π0 ) , W ˜ E[B]s 1 − q B(s)

(6.1)

where π0 is the probability of an empty system given in equation (4.1). Proof. Since customers arrive to the system according to a Poisson process, by the PASTA property, the tagged customer finds the system empty upon arrival with probability π0 given by (4.1), in which case he starts to be served immediately. Otherwise, that is if he finds the server busy with probability 1 − π0 , he has to wait for the current service time to finish. The distribution of this remaining service time, coincides with the excess distribution Be (t) of the service time, a fact that becomes clear by considering the system exclusively when the server is busy. PJ The overall service time of a customer is equal to j=1 Bj , where B1 , B2 , . . . are independent identically distributed random variables representing the successive service times till the customer leaves the system and J is independent of them and represents the number of services that will be required. We can easily see that J is geometrically distributed with P (J = j) = (1 − q)q j−1 , j = 1, 2, . . .. Let Be be a random variable, independent of B1 , B2 , . . . and J, having the excess service time distribution Be (t). By the above reasoning we have that the sojourn time W has the representation (P J with probability π0 j=1 Bj , W = PJ Be + j=1 Bj , with probability 1 − π0 . 12

Then the LST of W assumes the form ˜ ˜ ˜ (s) = π0 (1 − q)B(s) + (1 − π0 )B ˜e (s) (1 − q)B(s) . W ˜ ˜ 1 − q B(s) 1 − q B(s) ˜e (s) yields (6.1). Using (2.1) for B



We are now interested to study the conditional distribution of the sojourn time of a customer given that he finds upon arrival the system at state (X(t), Y (t)) = (n, m) for given numbers n and m. To obtain a Markovian description of the system we consider the augmented process {(X(t), Y (t), R(t)) : t ≥ 0}, where R(t) is the residual service time at time t (defined to be 0 when X(t) = 0). Since the overall service time of a customer is a geometric sum of independent service times, ˜ B(s) we have seen that its LST is (1−q) . Therefore, what we have to determine is the LST of ˜ 1−q B(s) the equilibrium remaining service time of a customer, given that he finds the system at state (X(t), Y (t)) = (n, m). Note also that, because of the PASTA property, the equilibrium probability that an arriving customer finds the system in a given state coincides with the corresponding equilibrium probability at arbitrary instants. We will now present a recursive scheme for the computation of the conditional distributions of the remaining service time, given the state (X(t), Y (t)). To this end we introduce the following notation: πn,m (r): the equilibrium probability density of having n customers in service, m customers in queue and a remaining service time r, n ≥ 1, m ≥ 0 and r ≥ 0, R∞ πn,m = 0 πn,m (r)dr: the marginal equilibrium probability of having n customers in service and m customers in queue, n ≥ 1, m ≥ 0, π

(r)

bn,m (r) = n,m πn,m : the equilibrium conditional probability density of the residual service time given there are n customers in service and m customers in queue, n ≥ 1, m ≥ 0 and r ≥ 0, R ˜n,m (s) = ∞ e−sr bn,m (r)dr: the LST of the conditional residual service time given there are n B 0 customers in service and m customers in queue, n ≥ 1 and m ≥ 0, ˜ n,m (s): the conditional LST of the sojourn time of a customer given he finds upon arrival n W customers in service and m customers in queue, n ≥ 1 and m ≥ 0. Below, we state a recursive scheme for the computation of bn,m (r). Lemma 6.1 The equilibrium conditional probability densities of the residual service time given there are n customers in service and m customers in queue are computed using the relation R∞ λeλr r e−λu bn,m−1 (u)du R∞ bn,m (r) = , n ≥ 1, m ≥ 1, r ≥ 0 , (6.2) 1 − 0 e−λu bn,m−1 (u)du with initial condition bn,0 (r) =

λeλr ˜ 1 − B(λ)

Z

∞

b(u)e−λu du, n ≥ 1, r ≥ 0 .

r

13

(6.3)

Proof. By considering the evolution of the process {(X(t), Y (t), R(t))} in a time interval [t, t + δt] and taking δt → 0+ we obtain in equilibrium the following Chapman-Kolmogorov differential equations ∞   X k 0 π1,0 (r) = λπ1,0 (r) − λb(r)π0 − q(1 − q)k−1 b(r)πk,0 (0) 1 k=1 ∞   X k − (1 − q)k b(r)πk,1 (0), r ≥ 0, (6.4) 0 k=1   n−1 ∞ X X k 0 πn,0 (r) = λπn,0 (r) − q n−m (1 − q)k−n+m b(r)πk,m (0) n−m m=0 k=n−m ∞   X k − (1 − q)k b(r)πk,n (0), n ≥ 2, r ≥ 0, (6.5) 0 k=1

0 πn,m (r) = λπn,m (r) − λπn,m−1 (r), n ≥ 1, m ≥ 1, r ≥ 0 , dπ

(6.6)

(r)

0 where πn,m (r) denotes the derivative n,m . dr Integrating both sides of equations (6.4)–(6.6) we obtain ∞   X k π1,0 (0) = −λπ1,0 + λπ0 + q(1 − q)k−1 πk,0 (0) 1 k=1 ∞   X k + (1 − q)k πk,1 (0), 0 k=1   n−1 ∞ X X k πn,0 (0) = −λπn,0 + q n−m (1 − q)k−n+m πk,m (0) n−m m=0 k=n−m ∞   X k (1 − q)k πk,n (0), n ≥ 2, + 0

(6.7)

(6.8)

k=1

πn,m (0) = −λπn,m + λπn,m−1 , n ≥ 1, m ≥ 1 .

(6.9)

Substituting equations (6.7)–(6.8) into (6.4)–(6.5) yields 0 πn,0 (r) = λπn,0 (r) − b(r)πn,0 (0) − λb(r)πn,0 , n ≥ 1, r ≥ 0.

Solving the differential equations (6.10) and (6.6) we get Z ∞ λr πn,0 (r) = (λπn,0 + πn,0 (0))e b(u)e−λu du, n ≥ 1, r ≥ 0, r Z ∞ λr −λu πn,m (r) = λe e πn,m−1 (u)du, n ≥ 1, m ≥ 1, r ≥ 0 .

(6.10)

(6.11) (6.12)

r

Setting r = 0 in equation (6.11) we obtain πn,0 (0) = λπn,0

˜ B(λ) , n ≥ 1. ˜ 1 − B(λ)

Combining (6.13) and (6.11) and dividing by πn,0 we obtain (6.3), while (6.12) yields Z πn,m−1 λr ∞ −λu bn,m (r) = λ e e bn,m−1 (u)du, n ≥ 1, m ≥ 1, r ≥ 0 . πn,m r 14

(6.13)

(6.14)

Taking r = 0 in equation (6.12) we obtain that Z ∞ e−λu πn,m−1 (u)du n ≥ 1, m ≥ 1 . πn,m (0) = λ

(6.15)

0

Combining equations (6.15) and (6.9) yields Z ∞ e−λu πn,m−1 (u)du = −πn,m + πn,m−1 , n ≥ 1, m ≥ 1 , 0

which gives πn,m−1 πn,m

= =

1 1−

R∞ 0

e−λu bn,m−1 (u)du

1 , n ≥ 1, m ≥ 1 . ˜ 1 − Bn,m−1 (λ)

Substituting (6.16) in (6.14) yields (6.2).

(6.16) (6.17) 

Theorem 6.2 The LST of the residual service time given that the number of customers in service is n and the number of waiting customers is m can be computed by the recursive scheme ˜n,m (s) = B

˜n,m−1 (λ) − B ˜n,m−1 (s) B λ , n ≥ 1, m ≥ 1 , ˜n,m−1 (λ) − 1 λ−s B

(6.18)

˜ ˜ λ B(λ) − B(s) , n ≥ 1. ˜ λ−s B(λ) −1

(6.19)

with initial condition ˜n,0 (s) = B

The LST of the sojourn time of a customer given that he finds upon arrival n customers in service and m waiting customers is then ˜ ˜ n,m (s) = B ˜n,m (s) (1 − q)B(s) , n ≥ 1 , m ≥ 1. W ˜ 1 − q B(s)

(6.20)

Proof. Multiplying equation (6.3) by e−sr and integrating with respect to r we obtain Z ∞ Z ∞ 1 λr ˜ Bn,0 (s) = λe e−λu b(u)due−sr dr, n ≥ 1 . (6.21) ˜ 1 − B(λ) 0 r Changing the order of integration and proceed with straightforward calculations yields easily (6.19). Similarly, multiplying (6.2) by e−sr and integrating with respect to r we obtain Z ∞ Z ∞ 1 λr ˜n,m (s) = R∞ λe bn,m−1 (u)e−λu due−sr dr, n ≥ 1, m ≥ 1. B 1 − 0 bn,m−1 (u)e−λu du 0 r (6.22) Again, by changing the order of integration and a bit of algebraic manipulation we derive easily (6.18). Consider, now, a customer that finds upon arrival n customers in service and m waiting customers. Then the equilibrium probability density of the remaining current service time coincides with πn,m (r), because of the PASTA property. The overall sojourn time of the customer will be the sum of the remaining service time and the service time of the customer which are independent random variables. Moreover, the total service time of the customer has 15

LST

˜ (1−q)B(s) ˜ 1−q B(s)

and we obtain immediately (6.20).



˜n,m (s) and It is interesting to observe that equations (6.18)-(6.20) show that the LSTs B ˜ Wn,m (s) do not depend on n but only on m. This shows that given the number of waiting customers m that a tagged customer finds upon arrival, his sojourn time and the number of customers in service that he finds upon arrival are independent random variables.

7

Limiting regimes

We now turn our attention to the behavior of the model under some extreme limiting regimes of the synchronization feature, when q → 1− and q → 0+ . To this end, we consider a fixed arrival ˜ rate λ and a service time distribution B(t) with LST B(s) and we let B represent a random variable with this distribution. We consider a family of models, one for each q ∈ (0, 1), that operate according to the dynamics described in Section 2. The model parameterized by q has t ˜ (q) (s) = B((1 ˜ − q)s) arrival rate λ(q) = λ, service time distribution B (q) (t) = B( 1−q ) with LST B (i.e. its service time is distributed as B (q) = (1 − q)B), and the probability that a customer repeats his service is q itself. As we have seen, the number of services till a customer leaves the system, J (q) , is geometrically distributed with Pr[J (q) = j] = (1 − q)q j−1 , j = 1, 2, . . .. Therefore the mean total service 1 time of a customer in the model with parameter q is E[J (q) ]E[B (q) ] = 1−q × (1 − q)E[B] = E[B], fixed for all q ∈ (0, 1). So we can think that all models in the family have the ‘same’ basic input characteristics (Poisson arrival process at rate λ, ‘shape’ of service times according to B and mean total service time per customer E[B]) and they differ in the level of synchronization. Indeed, the case q → 1− corresponds to no synchronization since all customers in service depart almost singly. This happens because in the case q → 1− , the service times are very short but the customers repeat their service with very high probability so they tend to behave almost independently and leave the system one by one. On the other extreme, in the case q → 0+ , all present customers tend to leave the system as a whole after a service completion. The case q → 0+ corresponds to full synchronization. Let X (q) denote a random variable having the equilibrium distribution that corresponds to the number of customers in the system at service completion epochs for the system with parameter q. Then according to Theorem 3.2 we have PX (q) (z) = =

∞ Y j=0 ∞ Y

˜ B((1 − q)q j λ(1 − z)) − (1 − z)PX (q) (0)

∞ X

q

k=1

k

k−1 Y

˜ B((1 − q)q j λ(1 − z))

j=0

˜ B((1 − q)q j λ(1 − z))

j=0

−(1 − z)

∞ k−1 Y PX (q) (0) X ˜ (1 − q)q k B((1 − q)q j λ(1 − z)) 1−q

(7.1)

j=0

k=1

and PX (q) (0) 1−q

Q∞ =

P∞

k=0 (1

j ˜ j=0 B((1 − q)q λ) Q ˜ − q)q k k−1 j=0 B((1 −

16

q)q j λ)

.

(7.2)

Therefore to obtain the limits of PX (q) (z) for q → 1− and q → 0+ we need to study the behavior of the infinite products and the series that appear in (7.1) and (7.2). To this end, we first provide some probabilistic interpretations in the following Lemma 7.1. Lemma 7.1 Let B0 , B1 , B2 , . . . be independent identically distributed random variables with dis˜ tribution B(t), LST B(s) and finite moments E[B n ], n = 1, 2, . . .. We also consider a random ˜e (s) variable Be having the excess distribution of B(t), i.e. with distribution Be (t) and LST B given from (2.1). Let also I be a geometrically distributed random variable with Pr[I = i] = (1 − q)q i , i = 0, 1, . . .. We assume that the sequence of random variables B0 , B1 , B2 , . . ., the random variable Be and the random variable I are mutually independent. Define the random variables ∞ X U (q) = (1 − q)q j Bj , (7.3) j=0 I−1 X

Ue(q) =

(1 − q)q j Bj + (1 − q)q I Be ,

(7.4)

(1 − q)q j Bj .

(7.5)

j=0 I−1 X

V (q) =

j=0 (q)

Then the LSTs of U (q) , Ue

and V (q) are given respectively from ∞ Y

˜ (q) (s) = U

˜ B((1 − q)q j s),

(7.6)

j=0

U˜e

(q)

1−

(s) =

V˜ (q) (s) =

Q∞

˜

j=0 B((1

− q)q j s)

,

(7.7)

˜ B((1 − q)q j s).

(7.8)

E[B]s ∞ X

(1 − q)q k

k−1 Y j=0

k=0

Proof. Using the definition of the infinite product and the LST, the independence of B0 , B1 , . . . and the Lebesgue’s dominated convergence theorem we have that ∞ Y

˜ B((1 − q)q j s) =

j=0

=

lim

n→∞

n Y

E[e−s(1−q)q

=

=

]

j=0

lim E[e−s

Pn

j=0 (1−q)q

n→∞ P j −s ∞ j=0 (1−q)q Bj

= E[e which proves (7.6). We also have Q j ˜ 1− ∞ j=0 B((1 − q)q s) = E[B]s

jB j

jB

j

]

],

(7.9)

  n   Y 1 ˜ lim 1 − B((1 − q)q j s)  E[B]s n→∞  j=0

n

k−1

k=0

j=0

X Y 1 ˜ ˜ lim (1 − B((1 − q)q k s)) B((1 − q)q j s) E[B]s n→∞ ∞ k−1 X ˜ 1 − B((1 − q)q k s) Y ˜ (1 − q)q k B((1 − q)q j s). (1 − q)q k E[B]s j=0

k=0

17

(7.10)

k s) ˜ 1−B((1−q)q (1−q)q k E[B]s

is the LST of (1 − q)q k Be so the LST in the last expression in (7.10) P j is seen to correspond to the geometric mixture of the random variables k−1 j=0 (1 − q)q Bj + (1 − q)q k Be , for k = 0, 1, . . . with mixing probabilities (1 − q)q k . This proves (7.7). Finally, (7.8) is obtained immediately by evaluating the LST of V (q) given in (7.5) by conditioning on I. 

Note now that

We now compute the limits of the infinite products and the series that appear in (7.1) and (7.2) as q → 1− , using the probabilistic interpretations that we report in Lemma 7.1. ˜ (q) (s) and V˜ (q) (s) reported in (7.6) and (7.8) as q → 1− Lemma 7.2 The limits of the LSTs U are given respectively from ∞ Y

lim

q→1−

lim

q→1−

∞ X

(1 − q)q

1 − e−sE[B] . E[B]s

˜ B((1 − q)q j s) =

j=0

k=0

(7.11)

j=0 k−1 Y

k

˜ B((1 − q)q j s) = e−sE[B] ,

(7.12)

Proof. Using (7.3) we have E[U (q) ] = V ar[U (q) ] =

∞ X (1 − q)q j E[Bj ] = E[B], j=0 ∞ X

(1 − q)2 q 2j V ar[Bj ] =

j=0

(1 − q)V ar[B] . 1+q

(7.13)

(7.14)

Consider an arbitrary  > 0. Then, using Chebyshev’s inequality, we have that Pr[|U (q) − E[B]| ≥ ] ≤

V ar[U (q) ] (1 − q)V ar[B] = → 0, q → 1− . 2  (1 + q)2

(7.15)

˜ (q) (s) Therefore, we obtain that U (q) converges in probability to E[B] as q → 1− ; hence its LST U −sE[B] tends to e . Using (7.6) we obtain (7.11). (q) To proceed to the proof of (7.12), we construct a coupling of Ue and V (q) on the same probability space using the same random variables B0 , B1 , B2 , . . ., Be and I of Lemma 7.1. Consider now an arbitrary  > 0. Using Markov’s inequality yields E[(1 − q)q I Be ]  (1 − q)E[q I ]E[Be ] (1 − q)E[B 2 ] = → 0, q → 1− .  2(1 + q)E[B]

Pr[|Ue(q) − V (q) | ≥ ] = Pr[(1 − q)q I Be ≥ ] ≤ = (q)

(7.16)

So we have that Ue − V (q) converges in probability to 0 as q → 1− . On the other hand we have (q) that Ue is a random variable that corresponds to the excess distribution of U (q) (because of (7.6) and (7.7)). But we have seen that U (q) converges in probability to E[B] as q → 1− . There(q) fore Ue converges in distribution to the excess distribution of the constant random variable (q) E[B], as q → 1− , i.e. Ue converges in distribution to the uniform distribution on [0, E[B]]. (q) Then, by Slutsky’s theorem, we obtain that V (q) which is written as the difference of Ue and 18

(q)

Ue − V (q) converges in distribution to the uniform distribution on [0, E[B]]. Since the LST of −sE[B] the uniform distribution on [0, E[B]] is 1−e we obtain readily (7.12).  E[B]s We can now deduce the limiting behavior of the model as q → 1− . Theorem 7.1 Given an arrival rate λ > 0 and a service time distribution B(t), we consider a family of models of the single server queue with synchronized services with the same arrival rate λ and the same mean total service time per customer E[B], parameterized by q ∈ (0, 1). For t any q ∈ (0, 1), the q-model has arrival rate λ(q) = λ, service time distribution B (q) (t) = B( 1−q ) and service repeat probability q. Let PX (q) (z) and ΠX (q) (z) be the PGFs of the equilibrium distributions of the number of customers in system at service completion epochs and at arbitrary instants respectively. Then in the limiting regime of no synchronization, q → 1− , we have lim PX (q) (z) =

q→1−

e−λE[B](1−z) − e−λE[B] 1 − e−λE[B]

(7.17)

and lim ΠX (q) (z) = e−λE[B](1−z) ,

q→1−

(7.18)

i.e. the number of customers in the system at arbitrary instants follow a Poisson distibution with rate λE[B]. Proof. We take the limit in (7.1) and (7.2) as q → 1− , using (7.11) and (7.12). We then obtain lim PX (q) (z) = e−λE[B](1−z) − (1 − z) lim

q→1−

lim

q→1−

q→1−

PX (q) (0) 1−q

=

PX (q) (0) 1 − e−λE[B](1−z) · , 1−q λE[B](1 − z)

λE[B]e−λE[B] , 1 − e−λE[B]

(7.19) (7.20)

that imply immediately (7.17). Equation (7.18) follows now easily, using (4.3) and (7.17).  To study the limiting regime of full synchronization, we need the analogue of Lemma 7.2 as q → 0+ . This is stated below. ˜ (q) (s) and V˜ (q) (s) reported in (7.6) and (7.8) as q → 0+ Lemma 7.3 The limits of the LSTs U are given respectively from lim

q→0+

lim

q→0+

∞ X

(1 − q)q

k=0

k

∞ Y

˜ ˜ B((1 − q)q j s) = B(s),

(7.21)

˜ B((1 − q)q j s) = 1.

(7.22)

j=0 k−1 Y j=0

Proof. Using (7.3) we have E[U (q) − (1 − q)B0 ] =

∞ X

(1 − q)q j E[Bj ] = qE[B] → 0, q → 0+ .

(7.23)

j=1

Since U (q) − (1 − q)B0 is a non-negative random variable for any q ∈ (0, 1), we obtain that U (q) − (1 − q)B0 converges in probability to 0 as q → 0+ . Moreover, we have obviously that 19

(1−q)B0 converges in distribution to B(t), as q → 0+ . By Slutsky’s theorem we deduce that U (q) converges in distribution to B(t) and therefore we obtain (7.21). Relation (7.22) follows using (7.5), since it is easily seen that for q → 0+ , the random variable V (q) converges in probability to 0.  We can now deduce the limiting behavior of the model as q → 0+ . We state the results in the following Theorem 7.2, whose proof is omitted as it is very similar to the proof of Theorem 7.1. Theorem 7.2 Given an arrival rate λ > 0 and a service time distribution B(t), we consider a family of models of the single server queue with synchronized services with the same arrival rate λ and the same mean total service time per customer E[B], parameterized by q ∈ (0, 1). For t any q ∈ (0, 1), the q-model has arrival rate λ(q) = λ, service time distribution B (q) (t) = B( 1−q ) and service repeat probability q. Let PX (q) (z) and ΠX (q) (z) be the PGFs of the equilibrium distributions of the number of customers in system at service completion epochs and at arbitrary instants respectively. Then in the limiting regime of full synchronization, q → 0+ , we have ˜ − z)) lim PX (q) (z) = B(λ(1

q→0+

(7.24)

and lim ΠX (q) (z) =

q→0+

˜ ˜ ˜ ˜e (λ(1 − z)) B(λ) + λE[B](B(λ(1 − z)) − B(λ)(1 − z))B . ˜ B(λ) + λE[B]

(7.25)

Remark 7.1 Equation (7.25) after rearranging some terms can be written as lim ΠX (q) (z) =

q→0+

˜ B(λ) λE[B] ˜ ˜e (λ(1 − z)) + B(λ)z B ˜ ˜ B(λ) + λE[B] B(λ) + λE[B] ˜ ˜ λE[B] (B(λ(1 − z)) − B(λ)) ˜ ˜e (λ(1 − z)) . + (1 − B(λ)) B ˜ ˜ B(λ) + λE[B] 1 − B(λ)

This expression can be interpreted probabilistically as follows. Observing the system at an arbitrary instant, there exist three possibilities for the observation epoch: either it lies in an idle period of the system, or it lies in the first service time of a busy period or it lies in any but the first ˜ B(λ) service time of a busy period. The corresponding probabilities are then respectively B(λ)+λE[B] , ˜ ˜ λE[B]B(λ) ˜ B(λ)+λE[B]

and

˜ λE[B](1−B(λ)) . ˜ B(λ)+λE[B]

In the first case, the conditional PGF of the number of customers

in system is identically 1 (no customers in system). In the second case, the conditional PGF ˜e (λ(1 − z)) corresponding to the customer that initiated the first service time plus all cusis z B tomers that have arrived at the system during the current service time. In the third case, the ˜ ˜ B(λ)) ˜e (λ(1 − z)) corresponding to the number of customers that conditional PGF is (B(λ(1−z))− B ˜ 1−B(λ) accumulated during the previous service time, given that it is different from 0, plus all customers that have arrived at the system during the current service time. Remark 7.2 Equations (7.1) and (7.2) become indeterminate forms as q → 1− . In this sense the main result of Theorem 7.1, i.e. equation (7.17), depends strongly on the particular assumption B (q) = (1 − q)B for the scaling of the service times in the sequence of the q-models. Other possible scalings of the service time distribution forcing the service times to tend to 0, as q → 1− , would result to different limiting regimes (for example by considering B (q) = α(1 − q)m B, for 20

some α > 0 and m ≥ 1). Thus we have to stress the non-uniqueness of the limit in (7.17) as q → 1− . It is the special scaling B (q) = (1 − q)B that yields the results of Theorem 7.1. On the other hand, equations (7.1) and (7.2) are not indeterminate forms for q → 0+ . The uniqueness of (7.24) is clear by setting q = 0 in (7.1).

8

Numerical results, conclusions and extensions

In the previous sections we performed a thorough analysis of the single server queueing model with synchronized repeated services, by seeking for transform expressions (PGFs and LSTs) for the main probabilistic descriptors of the system, including the number of customers at service completion epochs and in continuous time, the busy period and the sojourn times. We also studied the limiting regimes of the model, when the level of synchronization assumes its extreme values. From a theoretical ‘seeking for transforms’ point of view, the study of the model seems complete. In the present section we proceed in the computational - algorithmic implementation of the theoretical results by studying several numerical scenarios. The numerical findings shed further light on the effect of the various model parameters and distributions on the system performance. More concretely, in the course of implementing the results of the paper, we have performed several numerical experiments by keeping all but one parameter fixed and study the mean number of customers in the system as a function of the varying parameter. The effect of λ and E[B] on the mean number of customers in the system at service completion epochs, E[X], appears to be what is normally expected, i.e., E[X] is increasing in λ and E[B]. Much more interesting is the effect of the abandonment probability p on the probability of an empty system, p0 , and on the mean number of customers E[X]. In the numerical scenarios presented below we consider constant, Erlang, exponential and hyper-exponential service time distributions.

Figure 1: p0 vs p

Figure 2: p0 vs truncation level

In Figure 1 we present the graphs of the probability p0 , as a function of the abandonment probability p, while keeping λ and E[B] fixed. As we can see, p0 is an increasing convex function of p. Moreover, in this figure we plot four curves, each corresponding to a different service time distribution (constant, Erlang, exponential and hyper-exponential of order 2 with balanced 21

means - see Tijms (1994)). We observe that p0 increases as the coefficient of variation of the service time, cB , increases, while keeping E[B] fixed. The corresponding mean number of customers E[X] which is given by (3.18) is also an increasing function of cB , when E[B] is kept fixed. This agrees with the usual observation that congestion increases with variability. The computations of the values of p0 in Figure 1 have been done by numerically evaluating formula (3.16). Of course formula (3.16) involves an infinite product in the numerator and an infinite series in the denominator which cannot be further simplified. Therefore, the numerical implementation requires truncation. In general, it is quite obvious from the terms in formula (3.16) that the higher the probability q, the higher the truncation level that we have to use for an accurate approximation of p0 . However, our numerical experiments have shown that in general an appropriate truncation level is achieved quickly as long as q is not very close to 1. To demonstrate this fact, we have plotted in Figure 2 a graph for the four cases of different service time distributions (constant, Erlang, exponential and hyper-exponential) that depicts the values of formula (3.16) with respect to increasing truncation levels. We can see that the convergence of the truncated formula (3.16) is attained quickly. In the scenario presented in Figure 2, we see that with any truncation level higher that 25, we obtain an accurate approximation for p0 . We have also to state here that even in the cases of exponentially distributed or deterministic service times, it is not possible to arrive at closed-form results. It seems that the problem of the determination of the equilibrium distributions of the various probabilistic descriptors is inherently difficult and one has to turn to numerical methods, using truncation, to implement the results. In the case of exponentially distributed service times with rate µ, the PGF PX (z) assumes the form ∞  p0 qµ  Y p0 qµ µ PX (z) = 1 + − . (8.1) j j λ λq + µ − λq z λ j=0 P∞ The infinite product in (8.1) corresponds to the sum j=0 Gj of infinitely many independent geometric random variables G with probability mass functions gj (n) = Pr[Gj = n] = j   µ µ+λq j

λq j µ+λq j

n

, n ≥ 0. Even in this case an exact expression for the equilibrium probabilities is not possible. However, effective approximations at any desired accuracy level can be obtained (see e.g. Economou (2004)). In general, in the case of service times with rational LSTs, the PGFs PX (z) can be written in the form of basic hypergeometric series, also known as q-hypergeometric series. This has been also observed in the context of other Markovian models with binomial transitions (see e.g. Economou and Kapodistria (2009, 2010) and Adan, Economou and Kapodistria (2009)). However, one has again to appeal to numerical methods for performing the computations. In Table 1 we present E[X] for various values of p, while keeping all other parameters fixed. We can observe that E[X] is decreasing with respect to the abandonment probability p. This nicely agrees with our intuition that ‘the greater the abandonment probability, the less the congestion of the system’. By inspecting the entries of Table 1 in each row, we observe that E[X] is almost insensitive to the distribution of the service times. What counts is the arrival rate and the mean service time. This can be justified by equation (3.18). Indeed, the distribution of the service times affects only p0 in (3.18). For models where the work entering per time unit, λE[B], is big enough, we have that the dominant term of (3.18) is λE[B] 1−q and this justifies the 22

λ = 1, E[B] = 1 p

Constant

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

10.0003 5.0180 3.3914 2.5908 2.1057 1.7709 1.5183 1.3158 1.1463

Erlang Exponential Hyper-Exponential cB = 0.7 cB = 1.3 10.0007 10.0014 10.0023 5.0284 5.0393 5.0494 3.4124 3.4314 3.4471 2.6156 2.6368 2.6530 2.1303 2.1503 2.1648 1.7928 1.8101 1.8221 1.5363 1.5501 1.5592 1.3287 1.3385 1.3447 1.1532 1.1585 1.1617

Table 1: E[X] versus p for the different distributions of the service time almost insensitivity. The same is true for models with small repeat probability q and/or heavy traffic (small p0 ). The dominant term is again λE[B] 1−q and the system is almost insensitive. In the variant of the model where the service is provided even when there are no customers in the system (i.e. in the case of a Poisson arrival process subject to renewal generated binomial catastrophes) we have that E[X] = λE[B] 1−q and so the system is insensitive (see Economou (2004)). On an intuitive level the original system and this variant exhibit very similar behavior in case we have long busy periods. So we expect intuitively the almost insensitivity of the original model to hold, based on the exact insensitivity of the variant. Finally, to illustrate the usefulness of the LST formula (5.6) that concerns the length of the busy period of the model, we present the probability mass/density functions of the busy period for several numerical scenarios. These functions have been computed by numerically inverting ˜ the LSTs L(s) of the busy period, provided in formula (5.6) for various service time distributions (constant, Erlang, exponential and hyper-exponential). For the case of constant service time we plot the probability mass function, while for the other three cases, i.e. when the service time distribution is Erlang, exponential and hyper-exponential we plot the probability density function of the busy period distribution. In all graphs we assume that λ = 1 and E[B] = 1, and we consider three cases for p: 0.5, 0.65, and 0.8. Again, the numerical findings in Figure 3 confirm the intuitive expectation that the mean number of customers decreases as the abandonment probability increases. Approximations based on the results reported here as well as on the probabilistic interpretations of the various quantities that we have provided should be further explored. Stochastic ordering results that will provide bounds for the performance measures of our model using simpler models with known performance measures are also of interest. Another direction for future research concerns the study of other semi-Markovian models with this type of binomial transitions, for example models with synchronized abandonments, binomial catastrophes etc. Up to now the main contributions in the literature deal with Markovian models. The ideas and the techniques of the present study can be further used for the study of other general semi-Markovian models with binomial transitions.

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Figure 3: The probability mass/density functions of the busy period distribution when B is constant, Erlang, exponential and hyperexponential

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