SOJOURN TIME DISTRIBUTIONS IN A MARKOVIAN G-QUEUE WITH BATCH ARRIVAL AND BATCH REMOVAL

We consider a single server Markovian queue with two types of customers; 
positive and negative, where positive customers arrive in batches and arrivals of negative customers remove positive customers in batches. Only 
positive customers form a queue and negative customers just reduce the 
system congestion by removing positive ones upon their arrivals. We derive the LSTs of sojourn time distributions for a single server Markovian 
queue with positive customers and negative customers by using the first 
passage time arguments for Markov chains.


Introduction
We consider a queue with two types of customers; positive and negative.Positive cus- tomers are ordinary ones who, upon arrival, join the queue with the intention of being served.In contrast to the positive customers, the arrival of a negative custom- er removes some of the positive customers from the system, if any available, and then disappears; otherwise the negative customer is lost.Only positive customers can form a queue and negative customers just reduce system congestion.Such queues have been called G-queue (Harrison and Pitel [12]).
Since Gelenbe [7] introduced the notion of negative customers to represent the inhi- bitator signal in neural networks and commands to delete some transactions in distri- buted computer systems or databases, there has been a growing interest not only in networks of queues ([3-9, 12, 14, 15]) but also in single node queues with negative cus- tomers ([1, 2,10,11,13,16]).Interest in time delays in the G-queue has increased re- cently.Harrison and Pitel [11] derived the LSTs of the sojourn time distributions for 2. Queue Length Distribution In this section, we describe the mathematical model in detail and derive the queue length distribution in equilibrium at the arrival instants of positive customers.We consider a single server queue in which positive customers arrive in batches according to a Poisson process with rate , +, and negative customers arrive according to Poisson process with rate ,-, which is independent of the arrival process of positive customers.We assume that each arrival of a negative customer removes a random number B of positive customers in the system.That is, upon a negative arrival, if there are k positive customers in the system, min(B,k) positive customers are remov- ed and the negative customer disappears.The service time distribution of all custom- ers is exponential with mean -.For the notational simplicity, we let -,-+ # and , ,+ + ,-.We assume that the batch size A of positive customers and the quota B of a negative customer take finite values to avoid calculations of infinite ma- trices.However, this assumption is not a strong restriction, since the supports of A and B may be arbitrarily large and one can apply our model to A and B taking infin- ite values by truncating the tail parts of the state spaces with sufficiently small tail probabilities.
Let P(A n) a n and P(B n) bn, n-l,2,.., with a n-O, n_>l+l and b n-0, n>_m+l for some l_<l, m<oe.We denote the means -a-E(A) and b-E(B) and generating functions A(z)-_,t n n= lan z and B(z) m bnz n.
Note that the stationary distribution of the queue length in this system is invari- ant under the service discipline and removal strategies and concern only positive customers.This model is equivalent to the MAIM B/1 queue where customers arrive in batches with batch size A according to a Poisson pross with rate A + and the customers are served in batches of maximum size B with b k P(B k), 1 <_ k <_ m, where A-b1+ # k-1 2<_k<_m, and the service time distribution is exponential with parameter .The necessary and sufficient condition for this system to be positive recurrent is given (e.g.Miller [17]) by A+ p-_<1.

#+A-b
We assume that p < 1 throughout.Now we turn our attention to the queue length distribution at the arrival epochs of positive customers, which will be imperative in the upcoming sections.Let {Nn} be the number of positive customers in the system at the epoch immediately before the arrival of the nth batch of positive customers.Let A n be the batch size of the nth arrival of positive customers with the same distributions as A and D n + 1, where Dn + 1 is the number of positive customers departed from the system during the (n + 1)st interarrival period of the batch of positive customers.Then it can be seen that Nn + max(Nn + An-Dn + 1,0) The probability d n that n positive customers potentially leave the system during the interarrival time of a batch of positive customers is given by where p ,x + +g q-1-p and b(j n) is the j-fold convolution of the probability mass function {bk, 0 <_ k <_ m}.Simple calculations yield m (z) E b'n zn 1 (#z + A-B(z)) n=l and hence the probability generating function d(z)-Y = odnz n is given by d(z) A + + #(1-z)+ A-(l-B(z))" Denoting d n -0 for n _< -1 and d n c= ndk, n >_ O, we deduce that the transi- tion probability matrix P (Pij) of {Nn} is given by l Following similar procedures as those in Miller [17], the stationary distribution r- {ri, i-0, 1,...} of {Nn} is given by Kn'-l(dj.. ) 0, (2.1) where ai, 1 <_ <_ K, is the solution of the equation ozl d(oz)alal-1 "t-a2oz l-2 +... + al with n being the multiplicity of a (l<_i<_K), such that l_<n i<_l and K lni l. cij, 0 <_ j <_ n 1, 1 <_ <_ K are arbitrary constants, which can be determined by the l-1 simultaneous equations: under the constraint rj E riPij' J 1,2,...,/-1, and C, the normalizing constant (in E ,'= ori-1), is given by t -a (2.4) After simple but tedious algebra, we have from (2.3) and (2.4) the following linear system of equations for {Cij 1 <_ <_ K, 0 <_ j <_ n 1}: Heel, where e--(c10, c11,...,Cl,n1_1,c2,0, c2,1...C2, n2_1,...CK, nK _1) t, e (0,0,...,0,1) is the /-unit vector and H is the xl matrix with its kth < < ro.
In this case, (2.2) becomes c 2-(1 + p)c + p-0 and the stationary distribution is given by p)p , n >_ o.

The First Passage Times
The sojourn times, which will be treated in the upcoming sections, can be considered as the first passage times of the corresponding Markov chains.So we need to investi- gate the first passage times for some Markov chains related to compound Poisson pro- cesses.
First, we consider the compound Poisson process N(t) x(t) i=1 where {N(t), t _> 0} is a Poisson process with rate and {Xi} is a sequence of indepen- dent and identically distributed (i.i.d.) random variables, which are independent of {N(t),t >_ 0} and have probability mass function x k P(X 1 k), k 1,2,... and probability generating function E(z)-] n =lXa z zl <1.LetUx(n be the first passage of time of X(t) to the state n, that is Ux(n --inf{t > 0:X(t) > n), and let Ux(n,t -P(Ux(n < t) be the probability distribution function of Ux(N).
By conditioning the first transition of the process {X(t)}, we have the following pro- position.
From the spatial homogeneity of the transition probability Qz(t) for states 'i(j,k), j > k, depends only on the difference of the states j-k and its distribution function is the same as that the Gz(k-j).Note that, by the Markovian property, T(0,1; n, j) and r_n(j,k), n >_ 1 are independent.By taking LST in (3.6), we have (3.4).By using the same arguments as in Chapter 2 of Neuts [18] we have that H*(s) is the minimal nonnegative solution of g*(s)-"1 + "2 -t-s E Cn [g*(s)]n" Hn+l(8)--"1+'2 +8 k=O For more details in calculation of H*(s), see Neuts [18].
2. Gz(n,t) in Proposition 2 can be considered as a busy period distribution in X1 1/ 1 the M M x2' /1 queue with arrival rate "1 and service rate "2 starting with n customers.

RCE With FCFS Discipline
Under the FCFS queueing discipline with RCE removal strategy, upon arrival of a negative customer, if the number of positive customers is fewer than B, then all the positive customers are removed; otherwise, B customers from the end of the queue are removed.Let W denote the time period during which the tagged customer spends in the system from the epoch of arrival to the epoch of its service completion.We assume that W is infinite if the tagged customer is removed from the system before its service completion.Let N a and N b be the numbers of customers ahead of and behind the tagged customer, respectively, immediately after its arrival instant, and let N be the number of customers in the system at ,L_. tagged customer's arrival.Let A* and A*_ be the batch size to which the tagged customer belongs and the number of customers in the preceding batch.Note that the probability mass functions of A* and A*__ are given by , 1 P(A* k) kak and P(A Ern E a_ Ep(W<_x]Na n=0 k=l j=O --n+j, Nb--k-j-1 ). (4.1) To calculate the conditional distributionP(W _< x lNa n,N b k), we define the Markov chain x(t) x + (t) x-(t), t >_ o, with X(0) 0, where X + (t) and X-(t) are the numbers of positive customers having arrived and potential removals by negative customers up to time t, respective- ly.Then the LST Gc -n, s) of the first passage time distribution function Gx( n) inf{t >_ 0"X(t) <_ n}, n >_ 1, can be obtained from (3.4) by replacing c in (3.3) by Let S n be the time needed to serve n consecutive customers.Since the service time distribution is exponential with parameter #, the probability density function sn(t of S n is given by n(t) (n-1)' t O. (4.2) Under the FCFS service discipline with RCE removal strategy, when N a -n, N b k for the tagged customer to complete its service without being removed, it must hold true that G x(k-1)> S n + 1" Hence, the conditional distribution of W given that N a n and N b k, is represented by Letting and P(W <_ x Na n,N b k) P(Gx( k-1) > S n + l,Sn + <_ x) x / P(Gx(-k 1) > t)s n + l(t)dt.0 Kj(a,s,t) E ane- ). 0 we have from (2.1) and (4.1)-( 4.3) the following proposition.Proposition 3: The LST W*(s) of W(x) is given by where i=1 j=O If/-1, that is, A-1, then we have from (2.6), (3.4) and (4.5) that W*(s) (1 a0)W*(a0, s) where (, s)-H*(s + #(1-c)) and 7(s)-H*(s + p).

RCH With the FCFS Discipline
Under the RCH removal strategy, when a negative customer finds n positive customers upon its arrival and n _< B, then he removes all the positive customers.If n > B, then B customers from the head containing the customer in service are removed, and hence the customers behind the tagged customer do not affect the sojourn time of the tagged customer.Thus, using the same argument as in (4.1), we have k-1 Letting Vn(x P(W <_xlN-n), n-0,1,2,.., and V(s) be the LST of Yn(x), the LST W*(s) of W(x)is given by oc ak k-1 : Now we derive V(s), n >_ O. Taking the conditional probability on the first departure of positive customers due to service completion or negative arrival and then using the total probability law, we have the following recursive relation" v;() (), where (s) We have from (2.1), (5.1) and (5.3) the following proposition.Proposition 4: The LST W*(s) of W(x) is given by '.w*(,) w*()-c ,C,o, i=1 j=O where and b k -0, k >_ n + 1. Simple calculation yields the generating (5.3) (5.4) (5.5) Special Cases .f A-, tn hv fom (.) tt W*(,)-V*(,)nd n, o (2.6) and (5.3) that (1 ao)(s (5.6) w*() If A-1 and B-1, then since c 0 -p and B(c)a, we have from (5.6) that W*(s) + + which is identical to the formula in Harrison and Pitel 11].

RCE With Preemptive LCFS Discipline
Here, the batch to which the tagged customer belongs goes immediately to the front of the queue and the removal of customers by a negative arrival is done at the end of the queue.Thus, when N a --n, N b --k, for the tagged customer to complete its service without being removed, the tagged customer must complete its service before the negative arrivals remove k + 1 positive customers.Note that when N a n, the time period that the tagged customer completes its service is the same as the busy period n+l in an ordinary MAIM/1 queue starting with n+ 1 customers.Let X-(t) be the numbers of potential removals by negative customers up to time t and U x (n) -inf{t _> 0, X-(t) _> n}, and (n, t)-P(n < t).Then we have P(W x N -n, Nb-k)-P(Ux-(k + l) > qn+l,n+ 1 X f P(Ux-( 0 k + 1) > t)(n + 1, dt).

RCH With Preemptive LCFS Discipline
Let {X + (t),t >_ O) and {r(/),/>_ 0) be independent compound Poisson processes with their one-dimensional distributions as those of the number of the arrived posi- tive customers and the number of positive customers potentially leaving the system due to service completion or negative arrivals up to time t, respectively.We define a Markov chain {Z(t), t _> 0} by z(t) x + (t)-Y(t).
Here the queue left behind the tagged customer is irrelevant throughout its sojourn time.Thus, using the same argument as (4.1), we have k-1 a k W(x) E --E P(W <-x Na j)" (7.1) k=l j=O Define Vn(x P(W <_ x N tional simplicity, we let -n) and denote its LST by V(s).

Acknowledgement
The author wishes to acknowledge the financial support of the Korean Research Foun- The m m matrix H*(s)in (3.5) can be calculated recursively by setting H(s) 0 in * (1+2) Ck[H:(s)] k.

Y
Cn's are of the same form as those in Section 3

Proposition 6 :
The LST W*(s) of W(x) is given by k