Geom/G 1,G 2 /1/1 REPAIRABLE ERLANG LOSS SYSTEM WITH CATASTROPHE AND SECOND OPTIONAL SERVICE

Transcription

1 J Syst Sci Complex (2011) 24: Geom/G 1,G 2 /1/1 REPAIRABLE ERLANG LOSS SYSTEM WITH CATASTROPHE AND SECOND OPTIONAL SERVICE Yinghui TANG Miaomiao YU Cailiang LI DOI: /s Received: 26 September 2008 / Revised: 23 July 2009 c The Editorial Office of JSSC & Springer-Verlag Berlin Heidelberg 2011 Abstract This paper studies a single server discrete-time Erlang loss system with Bernoulli arrival process and no waiting space. The server in the system is assumed to provide two different types of services, namely essential and optional services, to the customer. During the operation of the system, the arrival of the catastrophe will break the system down and simultaneously induce customer to leave the system immediately. Using a new type discrete supplementary variable technique, the authors obtain some performance characteristics of the queueing system, including the steady-state availability and failure frequency of the system, the steady-state probabilities for the server being idle, busy, breakdown and the loss probability of the system etc. Finally, by the numerical examples, the authors study the influence of the system parameters on several performance measures. Key words Catastrophe, discrete supplementary variable technique, Erlang loss system, repairable queueing system, second optional service. 1 Introduction In recent years, discrete-time queues have been widely used in the performance analysis of communication systems. Much interest has been created due to their applicability in broadband integrated service digital network (BISDN), asynchronous transfer model (ATM), and related technologies. But in most discrete-time queueing models analyzed, it is always assumed that there is infinite or finite waiting space [1 2]. However, in some practical situations, queueing system has no waiting room for customers, so when a new customer finds server busy, he will leave the system immediately, that is, block customers cleared (BCC). An application of the Yinghui TANG School of Mathematics & Software Science, Sichuan Normal University, Chengdu , China. tangyh@uestc.edu.cn. Miaomiao YU School of Mathematics & Software Science, Sichuan Normal University, Chengdu , China; Schoolof Science, Sichuan University of Science and Engineering, Zigong , China. mmyu75@163.com. Cailiang LI Department of Information and Computer Science, Chengdu Electro-Mechanical College, Chengdu , China. This research is supported by the National Natural Science Foundation of China under Grant No , Specialized Research Fund for the Doctoral Program of Higher Education of China under Grant No , and the Scientific Research Fund of Sichuan Provincial Education Department under Grant No. 10ZA136. This paper was recommended for publication by Editor Shouyang WANG.

2 Geom/G 1,G 2/1/1 REPAIRABLE ERLANG LOSS SYSTEM 555 BCC assumption arises in telephone traffic engineering, where calls that find all trunks busy are given a busy signal and will be rejected by the system. Actually, the study of queues with no waiting space has been started with Erlang loss model [3] and has attracted the interest of some researchers. Madan [4] dealt with an M/G/1/1 system which has no waiting capacity. He considered the case where the server offers two types of services having general service times with different probability densities. Later, Sapna [5] generalized Madan s model to include the case of the server unreliability. The term unreliable server implies that the server is subject to unpredictable breakdowns. Various operating characteristics were obtained by Sapna and in the special cases of exponentially distributed service times, a closed form solution was also obtained. Although many continuous-time queueing models with no waiting space have been studied extensively in the past years, their discrete-time counterparts received very little attention in literature. In fact, to the best of our knowledge, the only work about this topic in discrete-time can be found in [6]. The purpose of this paper is to study the discrete-time repairable Erlang loss system with catastrophe and second optional service. In our queueing model, the arrival of catastrophe will make the customer be in service lost and also make the server breakdown. By using a new analytic technique, some important reliability indices such as the system availability, rate of occurrence of failure are all derived. Furthermore, the explicit expressions of the steady-state and loss probabilities of the queueing system are also obtained. In addition, it should be noted that there are crucial differences in the analytic technique details. The rest of this paper is organized as follows. Section 2 gives the mathematical description of the considered queueing model. In Section 3, we establish the difference equations of our queueing system and get the Z transforms of the state probabilities. In Section 4, using Z transform technique, we get the system performance measures including the steady-state availability, failure frequency and loss probability of the queueing system. In Section 5, some numerical illustrations are provided, followed by the conclusions in Section 6. 2 Model Description and Some Notations The catastrophe arrivals affect the queue behavior in a variety of ways. But most of papers have appeared about this topic only considered a catastrophe removes all the work present in the system and neglected the effect of catastrophe on the server. In this paper, we study catastrophes and unreliable server simultaneously. By considering transient analysis of this queueing system, some important reliability indices are given. The system under consideration is a Geom/G 1,G 2 /1/1 discrete-time repairable Erlang loss system with catastrophe and second optional service. We have precisely the following assumptions: 1) The time axis is divided into equal intervals called slots and all queueing activities occur at the slot boundaries. Let the time axis be marked by 0, 1,,k,. Consider the epoch k and suppose that the departures and completion epochs of the repairs take place in (k,k), and the customer arrivals, the catastrophe arrivals and the beginning epochs of the repairs occur in (k, k ), that is, we consider early arrival system (EAS) policy in the present paper. The model under consideration can be viewed through a self-explanatory figure (see Figure 1). 2) Customer and catastrophe arrival occur according to a Bernoulli process with parameter λ(0 <λ<1,λ =1 λ) andα(0 <α<1,α =1 α), respectively, where λ and α are the probabilities that a customer and catastrophe arrive the system in a slot, respectively. Further, we assume that catastrophe will make the customer being in service lost and also make the server breakdown.

3 556 YINGHUI TANG MIAOMIAO YU CAILIANG LI 3) The system has no waiting space (i.e., Erlang loss system) and so an arriving customer enters the system for service when the server is idle and no catastrophe arrive the system. If the server is busy or undergoing repair, the arriving customer will leave the system immediately. k k k (k1) k1 (k1) Potential customer departure epoch End of repair epoch Potential customer arrival epoch Potential catastrophe arrival epoch Beginning of repair epoch Outside observer s epoch Figure 1 Various time epoch in our queueing model 4) The server offers two types of services, namely essential and optional. The time of the essential service η 1 follows a general distribution with probability mass function (p.m.f.) gk 1 =Pr{η 1 = k},k =1, 2, and probability generating function (p.g.f.) ψ 1 (z). 5) Define G 1 m 1 = i=m 11 g1 i, m 1 =0, 1, 2, as the tail probability of the essential service time. Let s 1 m 1 be the conditional probability of completion of essential service in a slot, given that the elapsed essential service time is m 1.So,s 1 m 1 =Pr{η 1 = m 1 1 η 1 >m 1 } = gm 1 11/G 1 m 1, m 1 =0, 1, 2,. 6) On completion of the essential service, a customer desires to have the second optional service with probability p and leaves the system with probability q, wherep q =1. 7) The time of the optional service η 2 also follows a general distribution with p.m.f. gk 2 = Pr {η 2 = k},k =1, 2, and p.g.f. ψ 2 (z). Similarly, we define G 2 m 2 = i=m 21 g2 i, m 2 = 0, 1, 2, as the tail probability of the optional service time and let s 2 m 2 be the conditional probability of completion of the optional service in a slot, given that the elapsed optional service time is m 2. So, we also have s 2 m 2 =Pr{η 2 = m 2 1 η 2 >m 2 } = gm 2 21 /G2 m 2, m 2 =0, 1, 2,. 8) When the server fails, it is repaired immediately. The time required to repair is a positive random variable τ with general distribution h k =Pr{τ = k}, k =1, 2, and p.g.f. φ(z). It is assumed that after repair the server is as good as a new one. We also define H n = i=n1 h i, n =0, 1, 2, as the tail probability of the repair time. Let r n be the conditional probability of completion of repair in a slot, given that the elapsed repair time is n. Clearly, r n =Pr{τ = n 1 τ = n} = h n1 /H n, n =0, 1, 2,. Furthermore, we assume that if the server is already down, the catastrophe has no effect on the server. 9) Finally, various stochastic processes involved in the system are independent of each other. 3 Mathematical Formulations and Analysis In this section, we first develop the difference equations for the queueing system by treating the elapsed essential service, optional service and repair times as discrete supplementary variables, respectively. Then we solve these difference equations and get Z transforms of the state probabilities.

4 Geom/G 1,G 2/1/1 REPAIRABLE ERLANG LOSS SYSTEM 557 For establishment the difference equations to govern the system, we introduce the following state probabilities: P 0 (k): the server is idle at epoch k ; P 1 (k, m 1 ): the server is providing essential service to the customer at epoch k and the elapsed essential service time is m 1, m 1 =0, 1, 2, ; P 2 (k, m 2 ): the server is providing optional service to the customer at epoch k and the elapsed optional service time is m 2, m 2 =0, 1, 2, ; F (k, n): the server is under repair at epoch k and the elapsed repair time is n, n = 0, 1, 2,. Similar to the arguments of Cox [7], we can obtain the difference equations for the system as follows: P 0 (k 1)=α λ P 0 (k) α λ qp 1 (k, m 1 )s 1 m 1 α λ P 2 (k, m 2 )s 2 m 2 α λ F (k, n)r n, (1) P 1 (k 1,m 1 1)=α P 1 (k, m 1 )(1 s 1 m 1 ), (2) P 2 (k 1,m 2 1)=α P 2 (k, m 2 )(1 s 2 m 2 ), (3) F (k 1,n1)=F (k, n)(1 r n ). (4) These sets of equations are to be solved under the following boundary conditions P 1 (k 1, 0) = α λp 0 (k) P 2 (k 1, 0) = F (k 1, 0) = αp 0 (k) α λqp 1 (k, m 1 )s 1 m 1 α λp 2 (k, m 2 )s 2 m 2 α λf (k, n)r n, (5) pα P 1 (k, m 1 )s 1 m 1, (6) αp 1 (k, m 1 ) αp 2 (k, m 2 ) αf (k, n)r n. (7) Further, we assume that the initial condition is P 1 (0, 0) = 1, the others are all equal to zero. In the following sequel, let f(z) = f(k)zk be the Z transform of any arbitrary function f(k). On taking Z transform with respect to k in Equation (1), we get P 0 (k 1)z k = α λ P 0 (k)z k α λ s 2 m 2 α λ qs 1 m 1 P 2 (k, m 2 )z k P 1 (k, m 1 )z k α λ r n F (k, n)z k.

5 558 YINGHUI TANG MIAOMIAO YU CAILIANG LI Using initial condition, we have P 0 (k 1)z k = 1 [ P0 (z) P 0 (0) ] = 1 z z P 0 (z), So the Z transform of Equation (1) is given by P 0 (z) =zα λ P0 (z) zα λ q P 1 (z,m 1 )s 1 m 1 zα λ P2 (z,m 2 )s 2 m 2 zα λ F (z,n)rn. (8) Similarly, taking Z transform with respect to k in Equations (2) (7) and also using the initial condition, we get P 1 (z,m 1 1)=α z P 1 (z,m 1 )(1 s 1 m 1 ), (9) P 2 (z,m 2 1)=α z P 2 (z,m 2 )(1 s 2 m 2 ), (10) F (z,n 1)=z F (z,n)(1 r n ), (11) 1[ P1 (z,0) 1 ] = λα P0 (z) z P 2 (z,0) = F (z,0) = zα P 0 (z) α λq P 1 (z,m 1 )s 1 m 1 α λ P 2 (z,m 2 )s 2 m 2 α λ F (z,n)r n, (12) pα z P 1 (z,m 1 )s 1 m 1, (13) From Equations (9) and (10), we can get αz P 1 (z,m 1 ) αz P 2 (z,m 2 ) αz F (z,n)r n. (14) m P 1 (z,m 1 )=(α z) m1 P (z,0) (1 s 1 i )=(α z) m1 G 1 m 1 P1 (z,0), m 1 =0, 1, 2,, (15) i=0 m P 2 (z,m 2 )=(α z) m2 P (z,0) (1 s 2 i )=(α z) m2 G 2 m 2 P2 (z,0), m 2 =0, 1, 2,. (16) i=0 Substituting (15) into Equation (13) gives P 2 (z,0) = Inserting (17) in (16), we get p(α z) m11 P1 (z,0)g 1 m 11 = p P 1 (z,0)ψ 1 (α z). (17) P 2 (z,m 2 )=(α z) m2 G 2 m 2 pψ 1 (α z) P 1 (z,0). (18)

6 Geom/G 1,G 2/1/1 REPAIRABLE ERLANG LOSS SYSTEM 559 Solving Equation (11), we have n 1 F (z,n)=z n F (z,0) (1 r i )=z n H n F (z,0). (19) Substituting (15), (18), and (19) into Equation (14), we obtain F (z,0)= αz P 0 (z)αz P 1 (z,0) α F (z,0) i=0 (α z) m1 G 1 m 1 αzp P 1 (z,0)ψ 1 (α z) z n1 H n r n = αz P 0 (z)αz P 1 (z,0) 1 ψ 1(α z) 1 α z Further, from (20) we can get (α z) m2 G 2 m 2 αzp P 1 (z,0)ψ 1 (α z) 1 ψ 2(α z) 1 α z α F (z,0)φ(z). F (z,0) = αz[ P0 (z)(1 α z) P 1 (z,0)(1 ψ 1 (α z)) p P 1 (z,0)ψ 1 (α z)(1 ψ 2 (α z)) ] (1 α z)(1 αφ(z)) Substituting (15), (18), (19), and (21) into Equation (8), we get on simplification P 0 (z) =. (20) λ ω(z)(1 α z)(1 αφ(z)) α λ αzφ(z)(1 ω(z)) (1 λ α z)(1 α z)(1 αφ(z)) α λ αzφ(z)(1 α z) P 1 (z,0), (21) where ω(z) =qψ 1 (α z)pψ 1 (α z)ψ 2 (α z). By Substituting (15), (18), (19), (21), and (22) into Equation (12), we have P 1 (z,0) = [(1 α z)(1 αφ(z))(1 αφ(z) α λ z)] {(1 αφ(z) α λ z)[(1 α z)(1 αφ(z))(1 λω(z)) α αλzφ(z)(1 ω(z))] α λz [λ ω(z)(1 α z)(1 αφ(z)) α λ αzφ(z)(1 ω(z))]} 1. (22) By now, for all the state probabilities of the queueing system, we give the explicit expressions of their Z transforms. 4 Steady-State Performance Measures With these explicit expressions obtained in Section3,wearereadytoderivesomeimportant steady-state performance measures of the system. First, we give two important lemmas, which will be frequently used in the sequel. Lemma 1 [8] If z < 1, a n > 0 and Ã(z) = a nz n, then lim (1 z)ã(z) =α< lim z 1 n { 1 n } n a k = α<. k=1 Lemma 2 [8] If lim b { n = b<, then lim 1 n n n n k=1 b } k = b.

7 560 YINGHUI TANG MIAOMIAO YU CAILIANG LI 4.1 System Availability By the definition of transient availability, the system availability at time k (k =0, 1, 2, ) is given by A(k) =P 0 (k) The Z transform of A(k) isgivenby Ã(z) = P 0 (z) P 1 (z,m 1 ) P 1 (k, m 1 ) P 2 (z,m 2 ) P 2 (k, m 2 ). [ λ ω(z)(1 α z)(1 αφ(z)) α λ αzφ(z)(1 ω(z)) = (1 λ α z)(1 α z)(1 αφ(z)) α λ αzφ(z)(1 α z) 1 ψ 1(α z) 1 α pψ 1 (α z) 1 ψ 2(α ] z) z 1 α z [(1 α z)(1 αφ(z))(1 αφ(z) α λ z)] {(1 αφ(z) α λ z)[(1 α z)(1 αφ(z))(1 λω(z)) α αλzφ(z)(1 ω(z))] α λz [λ ω(z)(1 α z)(1 αφ(z)) α λ αzφ(z)(1 ω(z))]} 1. AccordingtotheL-Hospital rule and two lemmas which has given above, then the steady-state (or limiting) availability of the system is given by A = lim A(k) = lim k z 1 (1 z)ã(z) = α αe[τ]α. Remark 1 According to the operating characteristics of the discrete-time queueing system, no event occurs and the state of the system does not change in the period between k and (k 1). So the system availability at time epoch k could be considered as the system availability at outside observer s epoch. 4.2 The Failure Frequency of the System Let N wf (k) represent the number of the system failures that have occurred by time k. W f (k) = E [N wf (k) N wf (k 1)] is called the occurrence of failures in a slot. Then the expected number of failures from 0 to k is given by M f (k) = k W f (i). In view of model assumptions, we have W f (k) =αp 0 (k 1) αp 1 (k 1,m 1 ) i=1 αp 2 (k 1,m 2 ). From the explicit expressions obtained in Section 3, the Z transform of W f (k) isgivenby W f (z) =αz P 0 (z) αz P 1 (z,m 1 ) αz P 2 (z,m 2 ).

8 Geom/G 1,G 2/1/1 REPAIRABLE ERLANG LOSS SYSTEM 561 Using Lemmas 1 and 2, we give the steady-state failure frequency of the system as follows: M f (k) M = lim = lim k k z 1 (1 z) W αα f (z) = αe[τ]α. 4.3 The Steady-State Probabilities at Outside Observer s Epoch Similarly, because, in the period between k and (k 1), no event occurs and the state of the system does not change, we can get the steady-state probabilities at outside observer s epoch as follows. The steady-state probability for the server being idle: P 0 = lim k P 0(k) = lim z 1 (1 z) P 0 (z) = α αλ [λ(1 qψ 1 (α ) pψ 1 (α )ψ 2 (α )) αλ ](αe[τ]α ) ; The steady-state probability for the server in essential service period: P 1 = lim k P 1 (k, m 1 ) = lim z 1 (1 z) P 1 (z,m 1 ) = lim z 1 (1 z) P 1 (z,0) 1 ψ 1(α z) 1 α z = α λ(1 ψ 1 (α )) [λ(1 qψ 1 (α ) pψ 1 (α )ψ 2 (α )) αλ ](αe[τ]α ) ; The steady-state probability for the server in optional service period: P 2 = lim k P 2 (k, m 2 ) = lim z 1 (1 z) P 2 (z,m 2 ) = lim z 1 (1 z)pψ 1(α z) P 1 (z,0) 1 ψ 2(α z) 1 α z The steady-state probability for the server being busy: P busy = P 1 P 2 = = α λpψ 1 (α )(1 ψ 2 (α )) [λ(1 qψ 1 (α ) pψ 1 (α )ψ 2 (α )) αλ ](αe[τ]α ) ; α λ(1 qψ 1 (α ) pψ 1 (α )ψ 2 (α )) [λ(1 qψ 1 (α ) pψ 1 (α )ψ 2 (α )) αλ ](αe[τ]α ) ; The steady-state probability for the server being breakdown: F = lim k F (k, n) = lim z 1 (1 z) αe[τ] F (z,n) = lim F (z,0)(1 φ(z)) = z 1 αe[τ]α. Remark 2 Note that P 0 P busy F = 1, it confirms that our analysis technique is reasonable and the deduction process is right.

9 562 YINGHUI TANG MIAOMIAO YU CAILIANG LI 4.4 The Loss Probability First, we consider the probability that an arbitrary customer will be accepted (the accept probability). Under the four cases, an arriving customer will be accepted by the system in (k, k ): Case 1 The server is providing essential service to the customer at epoch (k 1) and the elapsed essential service time is m 1, m 1 =0, 1, 2,. Furthermore, the essential service will be completedin(k,k) and the catastrophe will not take place in (k, k ). Case 2 The server is providing optional service to the customer at epoch (k 1) and the elapsed optional service time is m 2, m 2 =0, 1, 2,. Furthermore, the optional service will be completedin(k,k) and the catastrophe will not take place in (k, k ). Case 3 The server is under repair at epoch (k 1) and the elapsed repair time is n, n =0, 1, 2,. Furthermore, the repair will be completed in (k,k) and the catastrophe will not take place in (k, k ). Case 4 Theserverisidleatepoch(k 1) and the catastrophe will not take place in (k, k ). Let P accept (k) represent the probability that a customer arrival occurring in (k, k )isaccepted by the system. So, we have P accept (k) = qα P 1 (k 1,m 1 )s 1 m 1 α F (k 1,n)r n α P 0 (k 1). α P 2 (k 1,m 2 )s 2 m 2 Taking the Z transform on both side of the above equation and using Lemmas 1 and 2, we can get the steady-state accept probability P accept = lim k P accept(k) = lim z 1 (1 z) P accept (z) = αα (1 α λ ) [λ(1 qψ 1 (α ) pψ 1 (α )ψ 2 (α )) αλ ](αe[τ]α ). So the steady-state loss probability of the queueing system is given by P loss =1 αα (1 α λ ) [λ(1 qψ 1 (α ) pψ 1 (α )ψ 2 (α )) αλ ](αe[τ]α ). 5 Numerical Results and Discussion To demonstrate the applicability of the analytical results obtained in the previous sections, some numerical results have been presented in self-explanatory table and graphs. Choose the system parameters as λ =0.25,α =0.075,E[τ] =2,p =0.5,q =0.5. In Table 1, the various steady-state and loss probabilities have been reported for three different essential and optional services times distributions, viz. deterministic (D 1 and D 2 ), geometric (Geom 1 and Geom 2 ) and general (General 1 and General 2 ). The service times distributions for essential and optional services are given below, respectively.

10 Geom/G 1,G 2/1/1 REPAIRABLE ERLANG LOSS SYSTEM 563 Essential service time distribution: Optional service time distribution: D 1 :Pr{η 1 =4} =1; D 2 :Pr{η 2 =2} =1; Geom 1 :Pr{η 1 = k} =0.25(0.75) k 1, Geom 2 :Pr{η 2 = k} =0.5(0.5) k 1, k =1, 2, ; k =1, 2, ; General 1 :Pr{η 1 =3} =0.25, General 2 :Pr{η 2 =1} =0.2, Pr {η 1 =4} =0.5, Pr {η 1 =5} =0.25. Pr {η 2 =2} =0.6, Pr {η 2 =3} =0.2. Remark 3 Three kinds of essential service time distributions have equal mean E[η 1 ]=4, and three kinds of optional service time distributions have also equal mean E[η 2 ] = 2. Under the assumption that two types of service time distributions have equal mean values, respectively, the loss probabilities for different service distributions have been compared in Table 1. Table 1 Steady state and loss probabilities under some typical service time distributions The distributions of Steady state probability essential and optinal and loss probability service times P 0 P 1 P 2 F P busy P loss P accept (D 1,D 2) (D 1,Geom 2) (D 1, General 2) (Geom 1,D 2) (Geom 1,Geom 2) (Geom 1, General 2) (General 1,D 2) (General 1,Geom 2) (General 1, General 2) The steady state repair probability F The steady state failure frequency M The steady state availability A The steady state repair probability F The steady state failure frequency M The steady state availability A E[τ]= α= The catastrophe arrvial rate α The mean repair time E[τ] Figure 2 α versus three kinds of steadystate performance measures Figure 3 E[τ] versus three kinds of steadystate performance measures From Table 1, it can be seen that when the essential and optional service times follow geometric distributions, the Geom/G 1,G 2 /1/1 queueing system has the smallest loss probability 16.33%. Furthermore, it should be noted that when the essential service time follows the geometric distribution, the queueing system has a smaller loss probability than other cases.

11 564 YINGHUI TANG MIAOMIAO YU CAILIANG LI Figure 2 describes the influence of the catastrophe arrival rate α on several performance measures. As to be expected, the steady-state availability decreases with increasing value of α, whereas the steady-state repair probability increases with increasing value of α. Figure 3 shows the effect of mean repair time on some steady-state performance measures. It can be seen that the influence of E[τ] on the steady-state repair probability and availability is less than the catastrophe arrival rate α. Also, from Figures 1 and 2, it can be observed that the influences of α and E[τ] on steady-state failure frequency are very little. 6 Conclusions In this paper, we provide a new analytic technique for discrete-time queueing system. We call our analysis method is discrete supplementary variable technique based on the elapsed time. It is very different from the discrete supplementary variable technique based on the remaining time (see [1 2]). In order to obtain the various state probabilities of the discrete-time queueing system, we introduce the hazard rate of unite slot and the tail probabilities of the service and the repair times. By our analysis, we establish a set of difference equations to govern the system, and through Z transform, we also derive some important steady-state performance measures. Furthermore, it should be noted that because the probability of several events that occur simultaneously in discrete-time case is not zero, the boundary conditions of our discretetime queueing system are more complex than the continuous-time counterpart. This is also the conspicuous difference between continuous-time and discrete-time queueing models. References [1] M. L. Chaudhry and U. C. Gupta, Queue-length and waiting time distributions of discrete-time GI x /Geom/1 queueing systems with early and late arrivals, Queuing Systems, 1997, 25: [2] M. L. Chaudhry, J. G. C. Templeton, and U. C. Gupta, Analysis of the discrete-time GI/Geom(n) /1/N queue, Computers and Mathematics with Applications, 1996, 31: [3] Y. H. Tang and X. W. Tang, Queueing Theories Foundations and Analysis Techniques, Science Press, Beijing, [4] K.C.Madan,AnM/G/1 queueing system with additional optional service and no waiting capacities, Microelectronics and Reliability, 1994, 34: [5] K.P.Sapna,AnM/G/1 type queueing system with non-perfect servers and no waiting capacity, Microelectronics and Reliability, 1996, 36: [6] N. S. Tian, X. L. Xu, andz. Y. Ma, Discrete-Time Queueing Theory, Science Press, Beijing, [7] D. R. Cox, The analysis of non-markovian stochastic processes by the inclusion of supplementary variables, Proc. Cambridge Philos. Soc., 1955, 51: [8] J. J. Hunter, Mathematical Techniques of Applied Probability, Vol. II, Discrete Time Models: Techniques and Applications, Academic Press, New York, 1983.