Network of Tandem and Bi-tandem Queueing Process with Reneging and Jockeying

The steady state behaviour of a queueing model where two bi-tandem channels are linked in tandem with a common channel has been studied using the concept of reneging and jockeying.

The concept of jockeying was first discussed by Glazer. Then using this concept in different models many researchers discussed as Keonigsberg [8], Disney and Mitchell [9] etc. The network of queues was studied by Finch [10], Kelly [11], Melamed [12] and recently Chandramouli [13] discussed a model in which two bi-tandem channels are linked with a common channel taking the concept of non-linear service growth rate. In the present paper the concept of reneging and jockeying has been introduced in this model when the service rates do not depend upon the queue length. The steady behaviour of this model has been discussed. The practical situation corresponding to this model can be realized in banks or in a publishing company etc. For example, consider a publishing company which has three types of machines say 1 2 , S S and 3 S .
Let 1 S print the matter in red ink and 2 S in blue ink. and 3 S denote the binding machine. We suppose that the arrivals (matters for printing) are printed in two colours (red or blue) and finally go to the binding process at 3 S . It has also been assumed that the binding machine 3 S undertakes outside printed matter for binding. Now in this situation, the reneging and jockeying at the arrivals may also be observed.

FORMULATION AND SOLUTION
Let 1 2 , S S and 3 S denote the three service channels in which it is supposed that 1 2 , S S that is 1 S and 2 S are in bi-tandem and 1 3 S S → or 2 3 S S → , that is, each is further linked in tandem with 3 S . An arriving unit for service at either 1 S or 2 S may follow one of the following routes for terminal services : This unit which arrives directly at 3 S departs from the system after servicing at 3 S . Let 1 Q , 2 Q and 3 Q be waiting line formed before 1 2 , S S and 3 S . If they are busy. It has been supposed that an arriving unit after intolerable waiting time in the queue 1 Q or 2 Q may renege (leave) at 1 S or 2 S without service. Also it has been assumed that units may jockey (move) from 1 for personal economic gains. Let λ i denote the Possion mean rate of arrivals at i Q before i S (i = 1, 2, 3), we assume that the input source is infinite. Let µ i denote the Possion mean departure rates at i S . Also let r b denote the constant rate of reneging from queues r Q (r = 1, 2). Further, let ir J (i ≠ r, i, r = 1, 2) denote the constant rates of jockeying from i r Q Q → .
Let 12 p and 13 p denote the probabilities that a unit after service at 1 S departs to join the respective queues 2 Q and 3 Q . Again let 21 p and 23 p denote the probabilities that a unit after service at 2 S join the respective queues 1 Q and 3 Q , where ij p ≥ 0 (i ≠ j, i = 1, 2, j = 1, 2, 3) and 12 13 p + p = 1, 21 23 p + p = 1. Let P (k, m, n) denote the steady-state probability that there are waiting k units in 1 Q , m units in 2 Q and n units in 3 Q . Each queue includes service also and k, m, n ≤ 0.  Again, if k, m, n are also zero. Then in this case for P (0, 0, 0), and negative of P (k, m, n) in zero and substitute these values in (1) we get one equation.
Hence, the above set of eight difference equations govern the model in steady state situation.
To solve the above set of difference equations we use the generating function technique and similar steps as Chandramouli [14] has taken in his paper. Now, define the generating function as Using L' Hospital's rule for indeterminate form %, and using F (x, 1, 1) = 1 as 1 x → and similarly other also we have the following set of equations : 1  where A = 0 (1,1) -  This result coincides with the result given by Jackson [2]. Case-III.
If we assume 2