Steady State Behavior Of a Network Queue Model Comprised of Two Bi-serial Channels Linked with a Common Server

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1 Steady State Behavior Of a Network Queue Model Comprised of Two Bi-serial Channels Linked with a Common Server Deepak Gupta Prof. & Head, Dept of Mathematics Maharishi Markandeshwar University Mullana, Ambala,India Navjeet Kaur Research Scholar, Dept of Mathematics Maharishi Markandeshwar University Mullana, Ambala,India Raminder Kaur Cheema Research Scholar, Dept of Mathematics Maharishi Markandeshwar University Mullana, Ambala,India Abstarct-This paper is an attempt to study the steady state behavior of a network queuing model in which two bi-serial channels are linked with a common server.the arrivals of service pattern follow Poisson Law.The various queue characteristics such as mean queue length,variance of the queue length have been obtained explicitly by applying Generating function technique and laws of calculus.the model finds an application in decision making in the process industries,in banking and in many administrative steps. Keywords-Poisson Law, Steady State Behavior, Mean Queue Length, Variance and Bi-serial channels. I. INTRODUCTION Jackson [1]studied the behavior of a queuing system containing phase type service. O Brien[2] studied the transient solution of a queue model comprising of two queues in series in which the service parameter depends upon their queue lengths. Stephan[3] discussed two queues under pre-emptive priority with Poisson arrivals and service rate. Maggu[4] introduced the concept of bitendom in theory of queues which corresponds to a practical situation arises in production concern. Later on this idea was developed by various authors with different modifications in assumptions. Matoori Towfigh and Singh T.P[5] study a network queue model consisting two bi serial channels linked with a common server. Singh T.P. [6] studied the Transient analysis of feedback queue model under service parameter constraint. Recently, Singh T.P and Kumar Vinod et.al [7] studied the transient behavior of a queuing network with parallel bi serial queues. Further Singh T.P et al [8] studied steady state behavior of a queue model comprised of two subsystems with biserial channels linked with a common channel. Later on Gupta Deepak, Singh T.P et al [9] studied a network queue model comprised of biserial channel linked with a common server. Singh T.P.,Kusum, Deepak Gupta[10] introduced the Feedback Queue Model Assumed Service Rate Proportional to Queue 210

2 Number.Gupta,Deepak Gupta[11]studied the concept of parallel channels both in balking and reneging. steady state solutions of multiple parallel channels in series and non-serial multiple The present queue model differ the study made by Gupta Deepak, Singh T.P. et.al. in the sense that in this model the first and second both the systems consist of biserial channels and both the systems are linked with a common server. The differential difference equations formed in the model have been solved by using generating function technique, laws of calculus and statistical tools in order to find various queue characteristics such as mean queue length, variance of queue etc. II. PRACTICAL SITUATION Many practical situations of the model arise in industries, administrative setup, banking system, computer networks, office management, supermarket etc. For example: In a mall, there are three sections one is drink section second is for movie and third is food section.drink section consists of two subsections one is for soft-drink and other is for hot drink. A person may take the soft drink and directly go to watch the movie.or the person may also take the hot drink and then go to see the movie. After watching the movie the person may go to the food section to eat food. Similarly Food section consists of two subsections one is Indian food sections and other is Chinese food section.after movie,a person may either go to the Indian food section and leave the queue, or the person may go to the Chinese food section. Further this type of queue network is widely used for modeling and performance prediction in diverse areas as communication networks and teletraffic, multiprogramming systems, maintenance and repair facilities, production and assembly lines, steel making, air traffic control, urban transportation systems etc. III. THE PROBLEM The entire queue model is comprised of three service channels S 1, S 2 and S 3. Where subsystem S 1 consists of two biserial service channels s 11 and s 12 and also S 3 contains two biserial channels s 21 and s 22. S 1 and S 3 are linked with a common server S 2 in between both. The service time at S ij are distributed exponentially. For convenience, we assume the service rate μ 1, μ 2, μ 3, μ 4, μ 5.for μ ij at S ij respectively. The customers initially join S ij under poisson assumptions and the mean rate λ 1, λ 2 respectively. Queues are formed in the front of the service channels S 11 and S 12 if they are busy. Customers coming at the rate λ 1 after completion of phase service at S 11 will join S 12 or S 2 i.e either they may go to the network of server S 12 with the probability p 12 or go to the server S 2 with the probabillty p 13 such that p 12 p 13 =1.And the customers coming at the rate λ 1 after completion of phase service at S 12 will join S 11 or S 2 i.e either they may go to the network of server S 11 with the probability p 21 or go to the server S 2 with the probabillty p 23 such that p 21 p 23 =1.After that customer will go from the server S 2 either to the server S 21 with the probability p 34 or go to the server S 22 with the probability p 35 such that p 34 p 35 = 1 S 1 S 3 λ 1 n 1 S 11 S 2 S 21 p 46 μ 1 p 13 μ p 4 34 p 12 p 21 μ 3 p 45 p 54 p 57 λ 2 p 23 μ 2 p 45 p 35 μ 5 Fig :1 IV. MATHEMATICAL ANALYSIS 211

3 Define P n1,,,, the probability,there are n 1 units wating in the queue Q 1 in front of S 11, units waiting in the queue Q 2 in front of S 12, units waiting in the queue Q 3 in front of S 2,and units wating in the queue Q 4 in front of S 21, units waiting in the queue Q 5 in front of S 22.In each case the waiting includes a unit in service,if any (n 1,,,, > 0) Differential Difference Equation in Transient form for the model is: P n1,,,, = (λ 1 λ 2 μ 1 μ 2 μ 3 μ 4 μ 5 ) P n1,,,, (t) λ 1 P n1 1,,,, (t) λ 2 P n1, 1,,, (t)μ 1 p 12 (n 1 1) P n1 1, 1,,, (t)μ 1 p 13 (n 1 1) P n1 1,, 1,, (t)μ 2 p 21 ( 1) P n1 1, 1,,, (t)μ 2 p 23 ( 1) P n1, 1, 1,, (t)μ 3 p 34 ( 1) P n1,, 1, 1, (t)μ 3 p 35 ( 1) P n1,, 1,,n (t)μ p 45( 1) P n1,,, 1,n (t) μ p 46( 1) P n1,,, 1, (t)μ 5 p 54 ( 1) P n1,,, 1,n (t)μ p 57( 1) P n1,,,,n (t) 5 1 Equation in the steady state form: (λ 1 λ 2 μ 1 μ 2 μ 3 μ 4 μ 5 ) P n1,,,, =λ 1 P n1 1,,,, λ 2 P n1, 1,,, μ 1 p 12 P n1 1, 1,,, μ 1 p 13 P n1 1,, 1,, μ 2 p 21 P n1 1, 1,,, μ 2 p 23 P n1, 1, 1,, μ 3 p 34 P n1,, 1, 1, μ 3 p 35 P n1,, 1,,n μ p 45P n1,,, 1,n μ p 46 P n1,,, 1, μ 5 p 54 P n1,,, 1,n μ p 57 P n1,,,, 1 (n 1,,,, > 0) (1) (λ 1 λ 2 μ 2 μ 3 μ 4 μ 5 ) P 0,n2,,, =λ 2 P 0,n2 1,,, μ 1 p 12 P 1,n2 1,,, μ 1 p 13 P 1,n2, 1,, μ 2 p 23 P 0,n2 1, 1,, μ 3 p 34 P 0,n2, 1, 1, μ 3 p 35 P 0,n2, 1,,n 5 1 μ 4 p 45 P 0,n2,, 1,n μ p 46 P 0,n2,, 1, μ 5 p 54 P 0,n2,, 1,n p P 0,n2,,, 1 (n 1 = 0,,,, > 0) (2) (λ 1 λ 2 μ 3 μ 4 μ 5 ) P n1,0,,, =λ 1 P n1 1,0,,, μ 1 p 13 P n1 1,0, 1,, μ 2 p 21 P n1 1,1,,, μ 2 p 23 P n1,1, 1,, μ 3 p 34 P n1,0, 1, 1, μ 3 p 35 P n1,0, 1,,n 5 1 μ 4 p 45 P n1,0,, 1,n μ p 46 P n1,0,, 1, μ 5 p 54 P n1,0,, 1,n μ p 57 P n1,0,,, 1 ( = 0, n 1,,, > 0) (3) (λ 1 λ 2 μ 1 μ 2 μ 4 μ 5 ) P n1,,0,, =λ 1 P n1 1,,0,, λ 2 P n1, 1,0,, μ 1 p 12 P n1 1, 1,0,, μ 2 p 21 P n1 1, 1,0,, μ 3 p 34 P n1,,1, 1, μ 3 p 35 P n1,,1,,n 5 1 μ 4 p 45 P n1,,0, 1,n μ p 46 P n1,,0, 1, μ 5 p 54 P n1,,0, 1,n μ p 57 P n1,,0,, 1 ( = 0, n 1,,, > 0) (4) (λ 1 λ 2 μ 1 μ 2 μ 3 μ 5 ) P n1,,,0, =λ 1 P n1 1,, 0, λ 2 P n1, 1,0, μ 1 p 12 P n1 1, 1, 0, μ 1 p 13 P n1 1,, 1,0, μ 2 p 21 P n1 1, 1,,0, μ 2 p 23 P n1, 1, 1,0, μ 3 p 35 P n1,, 1,0,n μ p 45P n1,,,1,n μ p 46 P n1,,,1, μ 5 p 57 P n1,,,0, 1 ( = 0, n 1,,, > 0) (5) (λ 1 λ 2 μ 1 μ 2 μ 3 μ 4 ) P n1,,,,0 212

4 =λ 1 P n1 1,,,,0λ 2 P n1, 1,,,0μ 1 p 12 P n1 1, 1,,,0μ 1 p 13 P n1 1,, 1,,0μ 2 p 21 P n1 1, 1,,,0μ 2 p 23 P n1, 1, 1,,0 μ 3 p 34 P n1,, 1, 1,0 μ 4 p 46 P n1,,, 1,0μ 5 p 54 P n1,,, 1,1μ 5 p 57 P n1,,,,1 ( = 0, n 1,,,, > 0) (6) (λ 1 λ 2 μ 3 μ 4 μ 5 ) P 0,0,n3,, =μ 1 p 13 P 1,0,n3 1,, μ 2 p 23 P 0,1,n3 1,, μ 3 p 34 P 0,0,n3 1, 1, μ 3 p 35 P 0,0,n3 1,,n μ p 45P 0,0,n3, 1,n μ p 46 P 0,0,n3, 1, μ 5 p 54 P 0,0,n3, 1,n μ p P 0,0,n3,, 1 (n 1, = 0,,, > 0) (7) (λ 1 λ 2 μ 2 μ 4 μ 5 ) P 0,n2,0,, =λ 2 P 0,n2 1,0,, μ 1 p 12 P 1,n2 1,0,, μ 2 p 21 P n1 1, 1,,, μ 3 p 34 P 0,n2,1, 1, μ 3 p 35 P 0,n2,1,, 1 μ 4p 45 P 0,n2,0, 1, 1 μ 4 p 46 P 0,n2,0, 1, μ 5 p 54 P 0,n2,0, 1, 1 μ 5p 57 P 0,n2,0,, 1 (n 1, = 0,,, > 0) (8) (λ 1 λ 2 μ 2 μ 3 μ 5 ) P 0,n2,,0, =λ 2 P,n2 1,,0, μ 1 p 12 P 1,n2 1,,0, μ 1 p 13 P 1,n2, 1,0, μ 2 p 23 P n1, 1, 1,, μ 3 p 35 P 0,n2, 1,0, 1 μ 4p 45 P 0,n2,,1, 1 μ 4 p 46 P 0,n2,,1, μ 5 p 57 P 0,n2,,0, 1 (n 1, = 0,,, > 0) (9) (λ 1 λ 2 μ 2 μ 3 μ 4 ) P 0,n2,,,0 =λ 2 P 0,n2 1,,,0μ 1 p 12 P 1,n2 1,,,0μ 1 p 13 P 1,n2, 1,,0μ 2 P 23 P 0,n2 1, 1,,0μ 3 P 34 P 0,n2, 1, 1,0 μ 4 P 46 P 0,n2,, 1,0 μ 5 P 54 P 0,n2,, 1,1μ 5 P 57 P 0,n2,,,1 (n 1, = 0,,, > 0) (10) (λ 1 λ 2 μ 1 μ 4 μ 5 ) P n1,0,0,, =λ 1 P n1 1,0,0,, μ 2 p 21 P n1 1,1,0,, μ 3 p 34 P n1,0,1, 1, μ 3 p 35 P n1,0,1,,n μ 5 1 4p 45 P n1,0,0, 1,n μ 5 1 4p 46 P n1,0,0, 1, μ 5 p 54 P n1,0,0, 1,n μ 5 1 5p 57 P n1,0,0,, 1 (, = 0, n 1,, > 0) (11) (λ 1 λ 2 μ 1 μ 3 μ 5 )P n1,0,,0, =λ 1 P n1 1,0,,0, μ 1 p 13 P n1 1,0, 1,0, μ 2 p 21 P n1 1,1,,0, μ 2 p 23 P n1,1, 1,0, μ 3 p 35 P n1,0, 1,0,n μ 5 1 4p 45 P n1,0,,1,n 5 1 μ 4 p 46 P n1,0,,1, μ 5 p 54 P n1,,, 1,n μ 5 1 5p 57 P n1,0,,0, 1 (, = 0, n 1,, > 0) (12) (λ 1 λ 2 μ 1 μ 3 μ 4 ) P n1,0,,,0 =λ 1 P n1 1,0,,,0μ 1 p 13 P n1 1,0, 1,,0μ 2 p 21 P n1 1,1,,,0μ 2 p 23 P n1,1, 1,,0μ 3 p 34 P n1,0, 1, 1,0 μ 4 p 46 P n1,0,, 1,0 μ 5 p 54 P n1,0,, 1,1μ 5 p 57 P n1,0,,,1 213

5 (, =0, n 1,, > 0) (13) (λ 1 λ 2 μ 1 μ 2 μ 5 ) P n1,,0,0, =λ 1 P n1 1,,0,0, λ 2 P n1, 1,0,0, μ 1 p 12 P n1 1, 1,0,0, μ 2 p 21 P n1 1, 1,0,0, μ 3 p 35 P n1,,1,0,n μ 5 1 4p 45 P n1,,0,1,n μ 5 1 4p 46 P n1,,0,1, μ 5 p 57 P n1,,0,0, 1 (, =0, n 1,, > 0) (14) (λ 1 λ 2 μ 1 μ 2 μ 4 ) P n1,,0,,0 =λ 1 P n1 1,,0,,0λ 2 P n1, 1,0,,0μ 1 p 12 P n1 1, 1,0,,0μ 2 p 21 P n1 1, 1,0,,0μ 3 p 34 P n1,,1, 1,0 μ 4 p 46 P n1,,0, 1,0μ 5 p 54 P n1,,0, 1,1 μ 5 p 57 P n1,,0,,1 (, =0, n 1,, > 0) (15) (λ 1 λ 2 μ 1 μ 2 μ 3 ) P n1,,,0,0 =λ 1 P n1 1,,,0,0λ 2 P n1, 1,,0,0μ 1 p 12 P n1 1, 1,,0,0μ 1 p 13 P n1 1,, 10,0μ 2 p 21 P n1 1, 1,,0,0μ 2 p 23 P n1, 1, 1,0,0 μ 4 p 46 P n1,,,1,0μ 5 p 57 P n1,,,0,1 (, =0, n 1,, > 0) (16) (λ 1 λ 2 μ 4 μ 5 ) P 0,0,0,n4, =μ 3 p 34 P 0,0,1,n4 1, μ 3 p 35 P 0,0,1,n4, 1 μ 4p 45 P 0,0,0,n4 1, 1 μ 4p 46 P 0,0,0,n4 1, μ 5 p 54 P 0,0,0,n4 1, 1 μ 5p 57 P 0,0,0,n4, 1 (n 1,, = 0, > 0) (17) (λ 1 λ 2 μ 3 μ 5 ) P 0,0,n3,0, =μ 1 p 13 P 1,0,n3 1,0, μ 2 p 23 P 0,1,n3 1,0, μ 3 p 35 P 0,0,n3 1,0, 1 μ 4p 45 P 0,0,n3,1, 1 μ 4p 46 P 0,0,n3,1, μ 5 p 57 P 0,0,n3,0, 1 (n 1,, = 0, > 0) (18) (λ 1 λ 2 μ 3 μ 4 ) P 0,0,n3,,0 =μ 1 p 13 P 1,0,n3 1,,0μ 2 p 23 P 0,1,n3 1,,0μ 3 p 34 P 0,0,n3 1, 1,0 μ 4 p 46 P 0,0,n3, 1,0μ 5 p 54 P 0,0,n3, 1,1μ 5 p 57 P 0,0,n3,,1 (n 1,, = 0, > 0) (19) (λ 1 λ 2 μ 2 μ 5 ) P 0,n2,0,0, =λ 2 P 0,n2 1,00, μ 1 p 12 P 1,n2 1,0,0, μ 3 p 35 P 0,n2,1,0, 1 μ 4p 45 P 0,n2,0,1, 1 μ 4p 46 P 0,n2,0,1, μ 5 p 57 P 0,n2,0,0, 1 (n 1,, = 0, > 0) (20) (λ 1 λ 2 μ 2 μ 4 ) P 0,n2,0,,0 =λ 2 P 0,n2 1,0,0μ 1 p 12 P 1,n2 1,0,,0μ 3 p 34 P 0,n2,1, 1,0 μ 4 p 46 P 0,n2 0, 1,0μ 5 p 54 P 0,n2,0, 1,1μ 5 p 57 P 0,n2,0,,1 (n 1,, = 0, > 0) (21) 214

6 (λ 1 λ 2 μ 1 μ 2 μ 3 μ 4 μ 5 ) P 0,n2,,0,0 =λ 2 P 1,n2 1,,0,0μ 1 p 12 P 1,n2 1,,0,0μ 1 p 13 P 1,n2, 1,0,0μ 2 p 23 P 0,n2 1, 1,0,0 μ 4 p 46 P n1,,,1,0μ 5 p 57 P 0,n2,,0,1 (n 1,, = 0, > 0) (22) (λ 1 λ 2 μ 1 μ 5 ) P n1,0,0,0, =λ 1 P n1 1,0,0,0, μ 2 p 21 P n1 1,1,0,0, μ 3 p 35 P n1,0,1,0, 1 μ 4p 45 P n1,0,0,1, 1 μ 4p 46 P n1,0,0,1, μ 5 p 57 P n1,0,0,0, 1 (,, = 0n 1, > 0) (23) (λ 1 λ 2 μ 1 μ 4 ) P n1,0,0,,0 =λ 1 P n1 1,0,0,,0μ 2 p 21 P n1 1,1,0,,0μ 3 p 34 P n1,0,1, 1,0 μ 4 p 46 P n1,0,0, 1,0μ 5 p 54 P n1,0,0, 1,1μ 5 p 57 P n1,0,0,,1 (,, = 0n 1, > 0) (24) (λ 1 λ 2 μ 1 μ 3 ) P n1,0,,0,0 =λ 1 P n1 1,0,,0,0μ 1 p 13 P n1 1,0, 1,0,0μ 2 p 21 P n1 1,1,,0,0μ 2 p 23 P n1,1, 1,0,0 μ 4 p 46 P n1,0,,1,0μ 5 p 57 P n1,0,,0,1 (,, = 0n 1, > 0) (25) (λ 1 λ 2 μ 1 μ 2 ) P n1,,0,0,0 =λ 1 P n1 1,,0,0,0λ 2 P n1, 1,0,0,0μ 1 p 12 P n1 1, 1,0,0,0μ 2 p 21 P n1 1, 1,0,0,0 μ 4 p 46 P n1,,0,1,0μ 5 p 57 P n1,,0,01 (,, = 0n 1, > 0) (26) (λ 1 λ 2 μ 5 ) P 0,0,0,0,n5 =μ 3 p 35 P 0,0,1,0,n5 1 μ 4p 45 P 0,0,0,1,n5 1 μ 4p 46 P 0,0,0,1,n5 μ 5 p 57 P 0,0,0,0,n5 1 (n 1,,, = 0, > 0) (27) (λ 1 λ 2 μ 1 μ 2 μ 3 μ 4 μ 5 ) P 0,0,0,n4,0 =μ 3 p 34 P 0,0,1,n4 1,0 μ 4 p 46 P 0,0,0,n4 1,0μ 5 p 54 P 0,0,0,n4 1,1μ 5 p 57 P 0,0,0,n4,1 (n 1,,, = 0, > 0) (28) (λ 1 λ 2 μ 3 ) P 0,0,n3,0,0 =μ 1 p 13 P 1,0,n3 1,0,0μ 2 p 23 P 0,1,n3 1,0,0 μ 4 p 46 P 0,0,n3,1,0μ 5 p 57 P 0,0,n3,0,1 (n 1,,, = 0, > 0) (29) (λ 1 λ 2 μ 2 ) P 0,n2,0,0,0 =λ 2 P 0,n2 1,0,0,0μ 1 p 12 P 1,n2 1,0,0,0 μ 4 p 46 P 0,n2,0,1,0μ 5 p 57 P 0,n2,0,0,1 215

7 (n 1,,, = 0, > 0) (30) (λ 1 λ 2 μ 1 ) P n1,0,0,0,0 =λ 1 P n1 1,0,0,0,0 μ 2 p 21 P n1 1,1,0,0,0 μ 4 p 46 P n1,0,0,1,0μ 5 p 57 P n1,0,0,0,1 (n 1,,, = 0, > 0) (31) (λ 1 λ 2 ) P 0,0,0,0,0 = μ 4 p 46 P 0,0,0,1,0 μ 5 p 57 P 0,0,0,0,1 Now Define Generating Function as F(X,Y,Z,R,S)= n 1 =0 =0 =0 =0 =0 P n1,,,, X n 1Y Z R S Also define partial generating function as (n 1,,,, = 0) (32) F n2,,, (X) = n 1 =0 P n1,,,, X n 1 (34) F n3,, (X,Y) = =0 F n2,,, (X)Y (35) F n4, (X, Y, Z)= =0 F n3,, (X,Y)Z (36) F n5 (X,Y,Z,R) = =0 F n4, (X,Y,Z)R (37) (33) Where X, Y, Z, R, S =1 F(X,Y,Z,R,S)= =0 F n5 (X,Y,Z,R)S (38) Now proceeding on the lines of Maggu, Singh T.P. et.al. and following the standard technique,which after manipulation gives the final reduced results as- (λ 1 λ 2 μ 1 μ 2 μ 3 μ 4 μ 5 )F(X,Y,Z,R,S) μ 5 F(X,Y,Z,R) μ 4 F(X,Y,Z,S) μ 3 F(X,Y,R,S) μ 2 F(X,Z,R,S) μ 1 F(Y,Z,R,S)=λ 1 Xf(X, Y, Z, R, S) λ 2 Xf(X, Y, Z, R, S) μ 1p 12 Y X F(X, Y, Z, R, S) F(Y, Z, R, S) μ 1 p 13 Z X F(X, Y, Z, R, S) F(Y, Z, R, S) μ 2p 21 X Y F(X, Y, Z, R, S) F(X, Z, R, S) μ 2 p 23 Z Y μ 4p 45 S R F(X, Y, Z, R, S) F(X, Z, R, S) μ 3p 34 R F(X, Y, Z, R, S) F(X, Y, Z, S) Z F(X, Y, Z, R, S) F(X, Y, R, S) μ 3p 35 S Z F(X, Y, Z, R, S) F(X, Y, R, S) μ 4 p 46 R F(X, Y, Z, R, S) F(X, Y, Z, S) μ 5p 54 R S F(X, Y, Z, R, S) F(X, Y, Z, S) μ 5 p 57 S F(X, Y, Z, R, S) F(X, Y, Z, S) (39) Y ( λ 1 (1 X) λ 2 (1 Y)μ 1 1 p p Z 12 X 13 μ X X 2 1 p p Z 21 Y 23 μ R Y 3(1 p p S 34 Z 35 ) μ S Z 4(1 p p 46 )μ R 45 R R 5(1 p 54 S p 57 ))F(X,Y,Z,R,S)= μ R S 5F(X,Y,Z,R) (1 p p 57 ) μ S 54 S S 4F(X,Y,Z,S) (1 p p 46 ) μ R 45 R R 3F(X,Y,R,S) (1 p p S 34 Z 35 ) μ Z 2F(X,Z,R,S) 1 p21xy p23zyμ1f(y,z,r,s) 1 p12yx p13zx (40) F(X,Y,Z,R,S)= 216

8 R μ 5 F X,Y,Z,R 1 p 54 S p 57 S μ S 4 F X,Y,Z,S 1 p 45 R p 46 R μ R 3F X,Y,R,S 1 p 34 Z p 35 S Z X μ 2 F(X,Z,R,S) 1 p 21 Y p 23 Z Y μ Y 1F(Y,Z,R,S) 1 p 12 X p 13 Z X Y ( λ 1 (1 X) λ 2 (1 Y)μ 1 1 p 12 X p 13 Z X μ X 2 1 p 21 Y p 23 Z Y μ 3 (1 p 34 R Z p 35 S Z ) μ 4(1 p 45 S R p 46 R )μ 5(1 p 54 R S p 57 S )) (41) Let us denote :- F(Y,Z,R,S)=F 1 F(X,Z,R,S)=F 2 F(X,Y,R,S)=F 3 F(X,Y,Z,S)=F 4 F(X,Y,Z,R)=F 5 Also F(1,1,1,1,1)=1 Total probability, Let (X=1) as Y,Z,R,S 1 Now (41) becomes 0 form.now by using L Hospital rule,we get by differentiating w.r.t.x 0 1= μ 1F 1 p 13 p 12 μ 2 F 2 ( p 21 ) λ 1 μ 1 p 12 p 13 μ 2 ( p 21 ) λ 1 μ 1 μ 2 p 21 =μ 1 F 1 μ 2 F 2 p 21 Again differentiating w.r.t.y gives Y=1 where p 13 p 12 = 1 (42) 1= μ 2F 2 p 23 p 21 μ 1 F 1 ( p 12 ) λ 2 μ 2 p 21 p 23 μ 1 ( p 12 ) λ 2 μ 1 p 12 μ 2 = μ 1 F 1 p 12 μ 2 F 2 (43) where p 21 p 23 =1 Again differentiating w.r.t.z gives Z=1 1= μ 3F 3 p 34 p 35 μ 2 F 2 ( p 23 )μ 1 F 1 ( p 13 ) μ 3 p 34 p 35 μ 2 ( p 23 )μ 1 ( p 13 ) μ 1 p 13 μ 2 p 23 μ 3 = μ 1 p 13 F 1 μ 2 p 23 F 2 μ 3 F 3 (44) where p 34 p 35 =1 Again differentiating w.r.t.r gives R=1 1= μ 4F 4 p 45 p 46 μ 3 F 3 ( p 34 )μ 5 F 5 ( p 54 ) μ 4 p 45 p 46 μ 3 ( p 34 )μ 5 ( p 54 ) μ 3 p 34 μ 5 p 54 μ 4 = μ 3 p 34 F 3 μ 5 p 54 F 5 μ 4 F 4 (45) Where p 45 p 46 =1 Again differentiating w.r.t.s gives S=1 217

9 1= μ 5F 5 p 54 p 57 μ 3 F 3 ( p 35 )μ 4 F 4 ( p 45 ) μ 5 p 54 p 57 μ 3 ( p 35 )μ 4 ( p 45 ) μ 5 F 5 μ 4 F 4 (p 45 ) μ 3 F 3 (p 35 )= μ 3 (p 35 ) μ 4 (p 45 )μ 5 (46) After solving the above for F 1, F 2, F 3, F 4, F 5, we get F 1 =1 λ 2λ 1 p 12 (1 p 21 p 12) μ 1 F 2 =1 λ 2λ 1 p 12 (1 p 21 p 12) μ 2 F 3 =1 [ λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12) μ 3 F 4 =1 p 34 p 35 p 54 [ λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12) μ 4 (1 p 45 p 54) ] F 5 =1 λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12) ] p 45 p 34p 35 p 54 p 35 μ 5 (1 p 45 p 54) V. SOLUTION OF MODEL P n1,,,, =ρ 1 n 1 ρ 2 ρ 3 ρ 4 ρ 5 (1 ρ1 ) (1 ρ 2 )(1 ρ 3 )(1 ρ 4 )(1 ρ 5 ) Where ρ 1 = λ 1λ 2 p 21 (1 p 21 p 12 ) μ 1 ρ 2 = λ 2λ 1 p 12 (1 p 21 p 12) μ 2 ρ 3 = [ λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12 ) μ 3 ρ 4 = p 34 p 35 p 54 [ λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12) μ 4 (1 p 45 p 54) ] ρ 5 = λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12) ] p 45 p 34 p 35 p 54 p 35 μ 5 (1 p 45 p 54) The solution in steady state exists, if the conditions ρ 1, ρ 2, ρ 3, ρ 4, ρ 5 <1 are satisfied. A. Average number of customers VI. VARIOUS QUEUE CHARACTERISITCS L= n 1 n 1 P n1,,,, = n 1 n 1 P n1,,,, n 1 P n1,,,, n 1 P n1,,,, n 1 P n1,,,, n 1 P n1,,,, On Putting the values of P n1,,,,, ρ 1, ρ 2, ρ 3, ρ 4, ρ 5 and after simplication, We get 218

10 L= L 1 L 2 L 3 L 4 L 5 L= λ 1 λ 2 p 21 λ 2 λ 1 p 12 (1 p 21 p 12) μ 1 (λ 1 λ 2 p 21 ) (1 p 21 p 12) μ 2 (λ 2 λ 1 p 12 ) λ 2 λ 1 p 12 p 23 (λ 1 λ 2 p 21 )p 13 p 34 p 35 p 54 λ 2 λ 1 p 12 p 23 (λ 1 λ 2 p 21 )p 13 μ 3 (1 p 12 p 21 λ 2λ 1p 12 p 23 (λ 1λ 2p 21 )p 13 ) (μ 4 (1 p 45 p 54) 1 p 21 p 12) ( p 34 p 35 p 54 λ 2 λ 1 p 12 p 23 (λ 1 λ 2 p 21 )p 13 λ 2 λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 (p 45 p 34 p 35 p 54 p 35 ) (μ 5 (1 p 45 p 54) 1 p 21 p 12) (p 45 p 34 p 35 p 54 p 35 λ 2 λ 1 p 12 p 23 (λ 1 λ 2 p 21 )p 13 B. Variance of queue V(n 1 ) = n 1 L n 1 5 P n1,,,, = n n P n1,,,, L 2 n 1 P n1,,,, 2L n 1 n 1 n3n4n5pn1,n2,n3,n4,n5 = n n P n1,,,, L 2 = n 1 n 1 P n1,,,, = n 1 1 P n1,,,, n 1 2 P n1,,,, n 1 3 P n1,,,, n 1 4 P n1,,,, n 1 5 P n1,,,, 2 n 1 n 1 P n1,,,, 2 n 1 n 1 P n1,,,, 2 n 1 P n1,,,, n 1 2 n 1 P n1,,,, n 1 2 n 1 P n1,,,, 2 P n1,,,, n 1 L 2 After Substituting the values,the result reduces to λ [ 1 λ 2 p 21 ][ 1 λ 2λ 1 p 12 ] -2 λ [ 2 λ 1 p 12 ][1 λ 2λ 1 p 12 ] -2 [ [ λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] ][1 [ λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12) μ 1 (1 p 21 p 12) μ 1 (1 p 21 p 12 ) μ 2 (1 p 21 p 12) μ 2 (1 p 21 p 12 ) μ 3 [ [ λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12) μ 3 ][1 [ λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12) μ 3 ] -2 (1 p 21 p 12) μ 3 ] -2 [ λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12) ] p 45 p 34 p 35 p 54 p 35 μ 5 (1 p 45 p 54) Where E(W)= C. Average waiting time for the customers in the system λ 1 λ 2 p 21 λ 2 λ 1 p 12 λ[(1 p 21 p 12) μ 1 λ 1 λ 2 p 21 ] λ[(1 p 21 p 12) μ 2 λ 2 λ 1 p 12 ] ][1 λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12 ) ] p 45 p 34p 35 p 54 p 35 μ 5 (1 p 45 p 54) ] -2 λ 2 λ 1 p 12 p 23 (λ 1 λ 2 p 21 )p 13 p 34 p 35 p 54 λ 2 λ 1 p 12 p 23 (λ 1 λ 2 p 21 )p 13 λ[μ 3 (1 p 12 p 21 λ 2λ 1p 12 p 23 (λ 1λ 2p 21 )p 13 )] λ[(μ 4 (1 p 45 p 54) 1 p 21 p 12) ( p 34 p 35 p 54 λ 2 λ 1 p 12 p 23 (λ 1 λ 2 p 21 )p 13 ] λ 2 λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 (p 45 p 34 p 35 p 54 p 35 ) λ[(μ 5 (1 p 45 p 54) 1 p 21 p 12) (p 45 p 34 p 35 p 54 p 35 λ 2 λ 1 p 12 p 23 (λ 1 λ 2 p 21 )p 13 ] 219

11 VII. ALGORITHM The following algorithm provides the procedure to determine the joint probability and various queue characteristics of above discussed queue model: Step1.Obtain the number of customers n 1,,,,. Step2. Obtain the values of mean service rate μ 1, μ 2, μ 3, μ 4, μ 5. Step3.Obtain the values of mean arrival rate λ 1, λ 2. Step4.Obtain the values of the probabilities p 12, p 21, p 13, p 23, p 34, p 35, p 45, p 46, p 54, p 57. Step5. Calculate the value of (i) F 1 =1 λ 2λ 1 p 12 (1 p 21 p 12) μ 1 (ii) F 2 =1 λ 2λ 1 p 12 (1 p 21 p 12) μ 2 (iii) F 3 =1 [ λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12) μ 3 (iv) F 4 =1 p 34 p 35 p 54 [ λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12) μ 4 (1 p 45 p 54) ] (v) F 5 =1 λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12) ] p 45 p 34 p 35 p 54 p 35 μ 5 (1 p 45 p 54) Step 6.Calculate (i) ρ 1 =1 F 1 (ii) ρ 2 =1 F 2 (iii) ρ 3 =1 F 3 (iv) ρ 4 =1 F 4 (v) ρ 5 =1 F 5 Step 7.Check ρ 1, ρ 2, ρ 3, ρ 4, ρ 5. <1 If so, then go to step (8) else steady state condition does not holds good. Step 8. The joint probability P n1,,,, =ρ n 1ρ ρ ρ ρ (1 ρ 1 ) (1 ρ 2 ) (1 ρ 3 ) (1 ρ 4 ) (1 ρ 5 ) Step9. Calculate average no. of customers (mean queue length) L= ρ 1 (1 ρ 1 ) ρ 2 (1 ρ 2 ) ρ 3 (1 ρ 3 ) ρ 4 (1 ρ 4 ) ρ 5 (1 ρ 5 ) Step10. Calculate variance of queue V= ρ 1 (1 ρ 1 ) 2 ρ 2 (1 ρ 2 ) 2 ρ 3 (1 ρ 3 ) 2 ρ 4 (1 ρ 4 ) 2 ρ 5 (1 ρ 5 ) 2 Step11.Calculate average waiting time for customers 220

12 E(W)= L λ. VIII. NUMERICAL ILLUSTRATION Given customers coming to three servers out of which two servers consists biserial channels and one server is commonly linked in series with each of the two servers in biseries. The number of customers, mean service rate, mean arrival rate and associated probabilities are given as follows: S.No. No. of Customers Mean Service Rate Mean Arrival Rate Probabilities 1. n 1 =3 μ 1 =9 λ 1 =2 p 12 = =6 μ 2 =8 λ 2 =3 p 13 = =7 μ 3 =7 p 21 = =4 μ 4 =10 p 23 = =5 μ 5 =15 p 45 =0.5 p 46 =0.5 p 34 =0.6 p 35 =0.4 p 54 =0.7 p 57 =0.3 SOLUTION: ρ 1 = λ 1λ 2 p 21 (1 p 21 p 12) μ 1 =0.78 ρ 2 = λ 2λ 1 p 12 (1 p 21 p 12) μ 2 =0.90 ρ 3 = [ λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12 ) μ 3 =0.71 ρ 4 = p 34 p 35 p 54 [ λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12) μ 4 (1 p 45 p 54) ] =0.68 ρ 5 = λ 2λ 1 p 12 p 23 (λ 2 λ 1 p 12 )p 13 ] (1 p 21 p 12) ] p 45 p 34 p 35 p 54 p 35 μ 5 (1 p 45 p 54) =0.25 Average no. of customers L= L 1 L 2 L 3 L 4 L 5 = =

13 Average waiting time for customers E(W)= L λ where λ =λ 1 λ 2 Variance- = L 1L 2 L 3 L 4 L 5 λ = = ρ 1 (1 ρ 1 ) 2 ρ 2 (1 ρ 2 ) 2 ρ 3 (1 ρ 3 ) 2 ρ 4 (1 ρ 4 ) 2 ρ 5 (1 ρ 5 ) 2 = = REFERENCES [1] Jackson R.R.P, Queuing system with phase type service, vol. 5. O.R Quart, 1954, pp [2]O`Brien, G.B, The solution of some queuing problems, vol.2. J.Soc.Ind. Appli. Math, 1954, pp [3]Stephan,F.F, Two Queues under preemptive priority with pooisson Arrival band service Rates, Vol.6. Opers.Res,1958, pp [4]Maggu,P.L, Phase type service queue with two servers in Biserial, vol.13 No.1. J.O.P. RFS Soc Japan,1970. [5]Matoori Towfigh, On computer application of advanced level technique in sequencing and queuing to industrial and production problem,ph.d Thesis in Statstics, Agra University Agra,1989. [6]Singh T.P, Transient analysis of feedback queue model under service parameter constraint,1994. [7] Singh T.P., K. Vinod and K. Rajender On transient behavior of a queuing network with parallel biserial queues, JMASS vol. 1 No. 2., December, [8]Vinod Kumar, T.P Singh, Rajender Kumar, Steady-State behavior of a queue model comprised linked with a common channel, Reflection des ERA, vol.1.issue 2, 2006, pp [9]Gupta Deepak,Naveen Gulati,Sameer Sharma, Analysis of a network queue model comprised of biserial and parallel channel linked with a common server Computer Engineering and Intelligent systems ISSN , vol. 2. No.3, [10] Singh T.P.,Kusum,Deepak Gupta, Feed back Queue Model Assumed Service Rate Proportional to Queue Numbers Arya Bhatt Journal of Mathemetics and Informatics Vol.2 No.1,Jan-June,2010. [11].Meenu Gupta,Gupta Deepak, The Steady-State Solutions Of Multiple Parallel Channels In Series And Non-Serial Multiple Parallel Channels both With Balking & Reneging vol. 2.issue 2, International Journal of Computer Application (IJCA),