A Single Queue with Mutually Replacing m Servers

Abstract Customers arriving in a Poisson stream are served one by one exponentially by one of the m servers S 1, S 2, S 3, . . . , S m. A server who has just completed a service either continues the next service or is replaced by another server. The replacement of a server at each service completion is governed by a probabilistic criterion of availability of servers. Transient solutions in terms of Laplace transforms of the probability generating functions are obtained and among some special cases the earlier known results for the case of 2 servers are deduced. The steady state solutions and the average queue sizes have been obtained for some particular cases.


Introduction
( Madan, 1990) has studied a 2 server queue with correlated availability of servers.The idea was motivated by some earlier papers dealing with correlated arrivals, correlated departures and some other similar situations.cf (Chaudary, 1965), (Mohan, 1955) and (Murari, 1969).For an excellent account of the single server queue, we refer the reader to (Cohen, 1969), Gross and (Harris, 1985) and (Saaty, 1961).In this paper, we generalize the problem studied by (Madan, 1990) to m servers.The mathematical model is briefly described by the following assumptions: • Arrivals occur one by one in a Poisson stream with mean arrival rate λ(> 0).
• The system has m servers designated as S 1 , S 2 , S 3 , • • • , S m and only one of them serves the customers at a time.The service is provided on a first come, first served basis and the service times of are exponentially distributed with mean service time 1/μ j , j = 1, 2, 3, • • • , m.
• However, subsequently at the completion of each service, a server who has just completed a service either continues the next service or is replaced by another server.The availability criterion of servers is determined by the conditional probability p i j = probability that server S j replaces server S i , given that the server S i has just completed a service.Obviously, when j = i, it means that the server S i continues with the next service.Thus the m × m availability matrix is given by • It has been assumed that the replacement of servers is instantaneous.

Equations Governing the System
We define P ( j) n (t), n ≥ 0 as the probability that at time t there are n customers in the queue excluding the one being served by the jth server S j , j = 1, 2, 3, • • • , m and let Q(t) be the probability that at time t the queue length is zero and none of the m servers is providing service.The following set of forward equations govern the system for j = 1, 2, 3, • • • , m: We assume that initially there are no customers either waiting in queue or being served so that the initial condition is Taking Laplace transform of equations ( 1) to (3) and using equation ( 4), we have 3. The Time-Dependent Solution |z| ≤ 1 define the probability generating functions of queue length under various states of the system in terms of their Laplace transforms.Then multiplying equations ( 5) and ( 6) by suitable powers of z and simplifying, we obtain We solve the system of equations given by (8) for P ( j) (s, z), j = 1, 2, • • • , m and obtain Where Δ is the determinant of the m × m matrix and N ( j) is the m × m determinant which is obtained from the determinant Δ by replacing the jth column of Δ by the column vector ( We note that due to the vector (H 1 , H 2 , • • • , H m ) appearing in N ( j) , the numerator of each of the equations given by ( 9) involves m + 1 unknowns, namely Q(s), and We proceed to determine these unknowns.It is easy to see that each G j = (s + λ + μ j − λz) − p j j μ j has only one zero inside the unit circle |z| = 1 for j = 1, 2, 3, • • • , m and, for that matter, due to the product term G 1 G 2 • • • G m appearing in Δ, the denominator of the right side of (9) has m zeros inside |z| = 1.These zeros give rise to m equations in m + 1 unknowns mentioned above.In addition, equation ( 6) also involves the same m + 1 unknowns.Thus there are in all m + 1 equations in m + 1 unknowns.Hence all m probability generating functions P ( j) (s, z), j = 1, 2, • • • , m can be completely determined.

Case 1: (Sequential Service)
If we let p 12 = p 23 = p 34 = • • • = p m−1, m = 1 and all other transition probabilities are set to zero, then this essentially means that the servers are providing sequential service, one after the other.In this case, the corresponding results will be given by ( 9) where, now will have

Case 2: (each server completes his cycle)
Let p ii = 1 for i = 2, 3, 4, • • • , m and p i j = 0 for i j which means that S 1 , S 2 , S 3 , • • • , S m , whosoever starts service continues serving the customers until the queue becomes empty again.In this case, And then (9) yields The denominator of the right side of equation ( 10) has one zero inside the unit circle |z| = 1.Let z = z j be this zero.This zero yields λzπ j Q(s) − μ j P ( j) Using equation ( 11) in ( 7), we obtain And hence

Steady State Queue Length Probabilities for Various States of the System
Let P ( j) n , j = 1, 2, • • • , m and Q be the respective steady state probabilities corresponding to P ( j) n (t) and Q(t) and, for that matter, let P ( j) (z) be the steady state probability generating functions for the queue length corresponding to P ( j) (s, z).Then the steady state solution can be obtained by using the well known Tauberian property lim We proceed to derive the steady state results for this case as follows: Applying the Tauberian property, equation ( 10) yields Now, z = 1 is a zero of the denominator of the right hand side of equation ( 14).Therefore, its denominator must vanish for this zero, giving Using ( 15), equation ( 14) can be written as We see than for z = 1, equation ( 16) is indeterminate of the 0/0 form.Therefore, using L'Hopital's rule, we obtain Next, utilizing (17) in the normalizing condition m j=1 P ( j) (1) + Q = 1, we obtain We note that equation ( 18) gives the steady state probability that the system is empty and none of the servers is providing service.
Then using the value of Q from (18) in equation ( 16), we get Factoring (λ + μ j − λz)z − μ j as (z − 1)(μ j − λz) and canceling out the factor (z − 1), we can re-write equation ( 19) as This can be again written as Expanding the last factor of the right hand side of equation ( 21) and picking up the coefficients of nth power of z and simplifying, we have

Steady State Mean Queue Length
Let p n = n j=1 p ( j)  n be the steady state probability that the queue length is n ≥ 0, irrespective of whosoever server is providing service.Then the mean queue lengh L q is given by Carrying out the summations and simplifying, we have the mean queue length as 4.3 Case 3: (The Case of Two Servers) If the system has two servers, then m = 2 and, for that matter, π j = 0 = P ( j) (s, z), for j = 3, • • • , m.Also p i j = 0 for i, j = 3, Where (2) 0 (s) Solving ( 25) simultaneously, we have Results in ( 26) and ( 27) agree with known results of (Madan, 1990) Under the conditions of particular case 2 (each server completes his cycle), we have p 11 = 1 = p 22 and p 12 = 0 = p 21 , for j = 2. Consequently, the steady state results corresponding to equations ( 16), ( 18), ( 22) and ( 24) can be derived as follows: Again, all the above results given in equations ( 28), ( 29), ( 30) and (31) agree with the results of (Madan, 1990) except for notations.
• • • , m so that now we have 2 × 2 selection matrix given by ⎧ ⎩ p 11 p 12 p 21 p 22