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

(1)

210

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

Raminder Kaur Cheema

Research Scholar, Dept of Mathematics

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 .

(2)

Volume 3, Issue 1, January 2014

211

Number.Gupta,Deepak Gupta[11]studied the concept of steady state solutions of multiple parallel channels in series and non-serial multiple parallel channels both in balking and reneging.

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 , super-market 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 𝑆1,𝑆2and 𝑆3.

Where subsystem 𝑆1 consists of two biserial service channels 𝑠11 and 𝑠12 and also 𝑆3 contains two biserial channels 𝑠21 and 𝑠22. 𝑆1and 𝑆3 are linked with a common server 𝑆2 in between both. The service time at 𝑆𝑖𝑗 are distributed exponentially. For convenience, we assume the service rate 𝜇1,𝜇2,𝜇3,𝜇4,𝜇5.for 𝜇𝑖𝑗 at 𝑆𝑖𝑗 respectively. The customers initially join 𝑆𝑖𝑗 under poisson assumptions and the mean rate 𝜆1,𝜆2 respectively. Queues are formed in the front of the service channels 𝑆11and 𝑆12 if they are busy. Customers coming at the rate 𝜆1 after completion of phase service at 𝑆11 will join 𝑆12 or 𝑆2 i.e either they may go to the network of server 𝑆12 with the probability 𝑝12or go to the server 𝑆2 with the probabillty 𝑝13 such that 𝑝12+ 𝑝13=1.And the customers coming at the rate 𝜆1 after completion of phase service at 𝑆12 will join 𝑆11 or 𝑆2 i.e either they may go to the network of server 𝑆11 with the probability 𝑝21or go to the server 𝑆2 with the probabillty 𝑝23 such that 𝑝21+

𝑝23=1.After that customer will go from the server 𝑆2 either to the server 𝑆21 with the probability 𝑝34 or go to the server 𝑆22 with the probability

𝑝35 such that 𝑝34+𝑝35= 1

𝑆

1

𝑆

3

𝜆

₁

𝑛

₁

𝑆

₂

𝑝

₄₆

𝑝

₁₃

𝑛

₃

𝑝

57

𝜆

₂

𝑛

₂

𝑝

₂₃

Fig :1

IV. MATHEMATICAL ANALYSIS

𝑆11

𝑝12

𝑝21

𝑃

₂₃

𝑆12

𝜇1

𝜇₂

𝜇3

𝑛₄ 𝑆21

𝑝₃₄

𝑝45 𝑝54

𝑛5 𝑝45 𝑝₃₅

𝑝

₃₅

𝑆32

𝑆22 𝜇₄

(3)

212

Define 𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅ the probability ,there are 𝑛1 units wating in the queue 𝑄1 in front of 𝑆11 , 𝑛2 units waiting in the queue 𝑄2 in front of 𝑆12 ,𝑛3 units waiting in the queue 𝑄3 in front of 𝑆2,and 𝑛4 units wating in the queue 𝑄4 in front of 𝑆21 , 𝑛5 units waiting in the queue 𝑄5 in front of

𝑆22.In each case the waiting includes a unit in service ,if any (𝑛1,𝑛2,𝑛3,𝑛4,𝑛5> 0)

Differential Difference Equation in Transient form for the model is:

𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛5= ‒(λ1+λ2+μ1+μ2+μ3+μ4+μ5) 𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛5(t) + λ1𝑃𝑛₁−1,𝑛2,𝑛3,𝑛4,𝑛5(t)

+λ2𝑃𝑛₁,𝑛₂−1,𝑛3,𝑛4,𝑛5 (t)+μ1𝑝12(𝑛1+ 1)𝑃𝑛1+1,𝑛2−1,𝑛3,𝑛4,𝑛5(t)+μ1𝑝13(𝑛1+ 1)𝑃𝑛1+1,𝑛2,𝑛3−1,𝑛4,𝑛5(t)+μ2𝑝21(𝑛2+ 1)𝑃𝑛1−1,𝑛2+1,𝑛3,𝑛4,𝑛5(t)+μ2𝑝23(𝑛2+

1)𝑃𝑛₁,𝑛₂₊₁,𝑛₃−1,𝑛4,𝑛5(t)+μ3𝑝34(𝑛3+ 1)𝑃𝑛1,𝑛2,𝑛3+1,𝑛4−1,𝑛5(t)+μ3𝑝35(𝑛3+ 1)𝑃𝑛1,𝑛2,𝑛3+1,𝑛4,𝑛5−1(t)+μ4𝑝45(𝑛4+ 1)𝑃𝑛1,𝑛2,𝑛3,𝑛4+1,𝑛5−1(t)+μ4𝑝46(𝑛4+

1)𝑃𝑛1,𝑛2,𝑛3,𝑛₄₊₁,𝑛5(t)+μ5𝑝54(𝑛5+ 1)𝑃𝑛1,𝑛2,𝑛3,𝑛₄−1,𝑛₅₊₁(t)+μ5𝑝57(𝑛5+ 1)𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛₅₊₁(t) Equation in the steady state form:

(λ1+λ2+μ1+μ2+μ3+μ4+μ5)𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛5

=λ1𝑃𝑛₁−1,𝑛2,𝑛3,𝑛4,𝑛5+λ2𝑃𝑛1,𝑛2−1,𝑛3,𝑛4,𝑛5 +μ1𝑝12𝑃𝑛1+1,𝑛2−1,𝑛3,𝑛4,𝑛5+μ1𝑝13𝑃𝑛1+1,𝑛2,𝑛3−1,𝑛4,𝑛5+μ2𝑝21𝑃𝑛1−1,𝑛2+1,𝑛3,𝑛4,𝑛5+μ2𝑝23𝑃𝑛1,𝑛2+1,𝑛3−1,𝑛4,𝑛5+ μ₃𝑝34𝑃𝑛1,𝑛2,𝑛₃₊₁,𝑛₄−1,𝑛5+μ3𝑝35𝑃𝑛1,𝑛2,𝑛₃₊₁,𝑛4,𝑛₅−1+μ4𝑝45𝑃𝑛1,𝑛2,𝑛3,𝑛₄₊₁,𝑛₅−1+μ4𝑝46𝑃𝑛1,𝑛2,𝑛3,𝑛₄₊₁,𝑛5+μ5𝑝54𝑃𝑛1,𝑛2,𝑛3,𝑛₄−1,𝑛₅₊₁+μ5𝑝57𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛₅₊₁

(𝑛1,𝑛2,𝑛3,𝑛4,𝑛5> 0) (1) (λ1+λ2+μ₂+μ₃+μ₄+μ₅)𝑃0,𝑛₂,𝑛₃,𝑛₄,𝑛₅

=λ2𝑃0,𝑛₂−1,𝑛3,𝑛4,𝑛5 +μ1𝑝12𝑃1,𝑛2−1,𝑛3,𝑛4,𝑛5+μ1𝑝13𝑃1,𝑛2,𝑛3−1,𝑛4,𝑛5+μ2𝑝23𝑃0,𝑛2+1,𝑛3−1,𝑛4,𝑛5+μ3𝑝34𝑃0,𝑛2,𝑛3+1,𝑛4−1,𝑛5+μ3𝑝35𝑃0,𝑛2,𝑛3+1,𝑛4,𝑛5−1+ μ₄𝑝45𝑃0,𝑛2,𝑛3,𝑛₄₊₁,𝑛₅−1+μ4𝑝46𝑃0,𝑛2,𝑛3,𝑛₄₊₁,𝑛5+μ5𝑝54𝑃0,𝑛2,𝑛3,𝑛₄−1,𝑛₅₊₁+𝑝57𝑃0,𝑛2,𝑛3,𝑛4,𝑛₅₊₁

(𝑛1= 0,𝑛2,𝑛3,𝑛4,𝑛5> 0) (2) (λ1+λ2+μ3+μ4+μ5)𝑃𝑛1,0,𝑛3,𝑛4,𝑛5

=λ1𝑃𝑛₁−1,0,𝑛3,𝑛4,𝑛5+μ1𝑝13𝑃𝑛₁₊₁,0,𝑛₃−1,𝑛4,𝑛5+μ2𝑝21𝑃𝑛₁−1,1,𝑛3,𝑛4,𝑛5+μ2𝑝23𝑃𝑛1,1,𝑛₃−1,𝑛4,𝑛5+μ3𝑝34 𝑃𝑛1,0,𝑛₃₊₁,𝑛₄−1,𝑛5+μ3𝑝35𝑃𝑛1,0,𝑛₃₊₁,𝑛4,𝑛₅−1+ μ₄𝑝45𝑃𝑛₁,0,𝑛₃,𝑛₄₊₁,𝑛₅−1+μ4𝑝46𝑃𝑛1,0,𝑛3,𝑛4+1,𝑛5+μ5𝑝54𝑃𝑛1,0,𝑛3,𝑛4−1,𝑛5+1+μ5𝑝57𝑃𝑛1,0,𝑛3,𝑛4,𝑛5+1

(𝑛2= 0,𝑛1,𝑛3,𝑛4,𝑛5> 0) (3)

(λ1+λ2+μ₁+μ₂+μ₄+μ₅)𝑃𝑛₁,𝑛₂,0,𝑛₄,𝑛₅

=λ1𝑃𝑛₁−1,𝑛2,0,𝑛4,𝑛5+λ2𝑃𝑛1,𝑛2−1,0,𝑛4,𝑛5 +μ1𝑝12𝑃𝑛1+1,𝑛2−1,0,𝑛4,𝑛5+μ2𝑝21𝑃𝑛1−1,𝑛2+1,0,𝑛4,𝑛5+μ3𝑝34𝑃𝑛1,𝑛2,1,𝑛4−1,𝑛5+μ3𝑝35𝑃𝑛1,𝑛2,1,𝑛4,𝑛5−1+ μ₄𝑝45𝑃𝑛1,𝑛2,0,𝑛₄₊₁,𝑛₅−1+μ4𝑝46𝑃𝑛1,𝑛2,0,𝑛₄₊₁,𝑛5+μ5𝑝54𝑃𝑛1,𝑛2,0,𝑛₄−1,𝑛₅₊₁+μ5𝑝57𝑃𝑛1,𝑛2,0,𝑛4,𝑛₅₊₁

(𝑛3= 0,𝑛1,𝑛2,𝑛4,𝑛5> 0) (4) (λ1+λ2+μ1+μ2+μ3+μ5)𝑃𝑛1,𝑛2,𝑛3,0,𝑛5

=λ1𝑃𝑛₁−1,𝑛2,𝑛30,𝑛5+λ2𝑃𝑛1,𝑛₂−1,0,𝑛5 +μ1𝑝12𝑃𝑛₁₊₁,𝑛₂−1,𝑛30,𝑛5+μ1𝑝13𝑃𝑛₁₊₁,𝑛2,𝑛₃−1,0,𝑛5+μ2𝑝21𝑃𝑛₁−1,𝑛₂₊₁,𝑛3,0,𝑛5+μ2𝑝23𝑃𝑛1,𝑛₂₊₁,𝑛₃−1,0,𝑛5+ μ₃𝑝35𝑃𝑛1,𝑛2,𝑛₃₊₁,0,𝑛₅−1+μ4𝑝45𝑃𝑛1,𝑛2,𝑛3,1,𝑛₅−1+μ4𝑝46𝑃𝑛1,𝑛2,𝑛3,1,𝑛5+μ5𝑝57𝑃𝑛1,𝑛2,𝑛3,0,𝑛₅₊₁

(4)

213

=λ1𝑃𝑛₁−1,𝑛2,𝑛3,𝑛4,0+λ2𝑃𝑛1,𝑛₂−1,𝑛3,𝑛4,0+μ1𝑝12𝑃𝑛₁₊₁,𝑛₂−1,𝑛3,𝑛4,0+μ1𝑝13𝑃𝑛₁₊₁,𝑛2,𝑛₃−1,𝑛4,0+μ2𝑝21𝑃𝑛₁−1,𝑛₂₊₁,𝑛3,𝑛4,0+μ2𝑝23𝑃𝑛1,𝑛₂₊₁,𝑛₃−1,𝑛4,0+ μ₃𝑝34𝑃𝑛1,𝑛2,𝑛₃₊₁,𝑛₄−1,0+μ4𝑝46𝑃𝑛1,𝑛2,𝑛3,𝑛₄₊₁,0+μ5𝑝54𝑃𝑛1,𝑛2,𝑛3,𝑛₄−1,1+μ5𝑝57𝑃𝑛1,𝑛2,𝑛3,𝑛4,1

(𝑛5= 0,𝑛1,𝑛2,𝑛3,𝑛4, > 0) (6) (λ1+λ2+μ₃+μ₄+μ₅)𝑃0,0,𝑛₃,𝑛₄,𝑛₅

=μ₁𝑝13𝑃1,0,𝑛₃−1,𝑛4,𝑛5+μ2𝑝23𝑃0,1,𝑛₃−1,𝑛4,𝑛5+μ3𝑝34𝑃0,0,𝑛₃₊₁,𝑛₄−1,𝑛5+μ3𝑝35𝑃0,0,𝑛₃₊₁,𝑛4,𝑛₅−1+μ4𝑝45𝑃0,0,𝑛3,𝑛₄₊₁,𝑛₅−1+μ4𝑝46𝑃0,0,𝑛3,𝑛₄₊₁,𝑛5+ μ₅𝑝54𝑃0,0,𝑛₃,𝑛₄−1,𝑛₅₊₁+μ5𝑝57𝑃0,0,𝑛3,𝑛4,𝑛5+1

(𝑛1,𝑛2= 0,𝑛3,𝑛4,𝑛5> 0) (7) (λ1+λ2+μ2+μ4+μ5)𝑃0,𝑛2,0,𝑛4,𝑛5

=λ2𝑃0,𝑛₂−1,0,𝑛4,𝑛5 +μ1𝑝12𝑃1,𝑛2−1,0,𝑛4,𝑛5+μ2𝑝21𝑃𝑛1−1,𝑛2+1,𝑛3,𝑛4,𝑛5+μ3𝑝34𝑃0,𝑛2,1,𝑛4−1,𝑛5+μ3𝑝35𝑃0,𝑛2,1,𝑛4,𝑛5−1+μ4𝑝45𝑃0,𝑛2,0,𝑛4+1,𝑛5−1+

μ4𝑝46𝑃0,𝑛2,0,𝑛₄₊₁,𝑛5+μ5𝑝54𝑃0,𝑛2,0,𝑛₄−1,𝑛₅₊₁+μ5𝑝57𝑃0,𝑛2,0,𝑛4,𝑛₅₊₁

(𝑛1,𝑛3= 0,𝑛2,𝑛4,𝑛5> 0) (8)

(λ1+λ2+μ2+μ3+μ5)𝑃0,𝑛₂,𝑛₃,0,𝑛₅

=λ2𝑃,𝑛₂−1,𝑛3,0,𝑛5 +μ1𝑝12𝑃1,𝑛2−1,𝑛3,0,𝑛5+μ1𝑝13𝑃1,𝑛2,𝑛3−1,0,𝑛5+μ2𝑝23𝑃𝑛1,𝑛2+1,𝑛3−1,𝑛4,𝑛5+μ3𝑝35𝑃0,𝑛2,𝑛3+1,0,𝑛5−1+μ4𝑝45𝑃0,𝑛2,𝑛3,1,𝑛5−1+

μ4𝑝46𝑃0,𝑛2,𝑛3,1,𝑛5+μ5𝑝57𝑃0,𝑛2,𝑛3,0,𝑛₅₊₁

(𝑛1,𝑛4= 0,𝑛2,𝑛3,𝑛5> 0) (9) (λ1+λ2+μ2+μ3+μ4)𝑃0,𝑛₂,𝑛₃,𝑛₄,0

=λ2𝑃0,𝑛₂−1,𝑛3,𝑛4,0+μ1𝑝12𝑃1,𝑛2−1,𝑛3,𝑛4,0+μ1𝑝13𝑃1,𝑛2,𝑛3−1,𝑛4,0+μ2𝑃23𝑃0,𝑛2+1,𝑛3−1,𝑛4,0+μ3𝑃34𝑃0,𝑛2,𝑛3+1,𝑛4−1,0+μ4𝑃46𝑃0,𝑛2,𝑛3,𝑛4+1,0+

μ5𝑃54𝑃0,𝑛₂,𝑛₃,𝑛₄−1,1+μ5𝑃57𝑃0,𝑛2,𝑛3,𝑛4,1

(𝑛1,𝑛5= 0,𝑛2,𝑛3,𝑛4> 0) (10) (λ1+λ2+μ1+μ4+μ5)𝑃𝑛₁,0,0,𝑛₄,𝑛₅

=λ1𝑃𝑛₁−1,0,0,𝑛4,𝑛5+μ2𝑝21𝑃𝑛₁−1,1,0,𝑛4,𝑛5+μ3𝑝34𝑃𝑛1,0,1,𝑛₄−1,𝑛5+μ3𝑝35𝑃𝑛1,0,1,𝑛4,𝑛₅−1+μ4𝑝45𝑃𝑛1,0,0,𝑛₄₊₁,𝑛₅−1+μ4𝑝46𝑃𝑛1,0,0,𝑛₄₊₁,𝑛5+

μ5𝑝54𝑃𝑛₁,0,0,𝑛₄−1,𝑛₅₊₁+μ5𝑝57𝑃𝑛1,0,0,𝑛4,𝑛5+1

(𝑛2,𝑛3= 0,𝑛1,𝑛4,𝑛5> 0) (11)

(𝜆1+𝜆2+𝜇1+ 𝜇3+𝜇5)𝑃𝑛₁,0,𝑛₃,0,𝑛₅

=λ1𝑃𝑛₁−1,0,𝑛3,0,𝑛5+μ1𝑝13𝑃𝑛₁₊₁,0,𝑛₃−1,0,𝑛5+μ2𝑝21𝑃𝑛₁−1,1,𝑛3,0,𝑛5+μ2𝑝23𝑃𝑛1,1,𝑛₃−1,0,𝑛5+μ3𝑝35𝑃𝑛1,0,𝑛₃₊₁,0,𝑛₅−1+μ4𝑝45𝑃𝑛1,0,𝑛3,1,𝑛₅−1+

μ4𝑝46𝑃𝑛₁,0,𝑛₃,1,𝑛₅+μ5𝑝54𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄−1,𝑛₅₊₁+μ5𝑝57𝑃𝑛1,0,𝑛3,0,𝑛5+1

(𝑛2,𝑛4= 0,𝑛1,𝑛3,𝑛5> 0) (12) (λ1+λ2+μ1+μ3+μ4)𝑃𝑛₁,0,𝑛₃,𝑛₄,0

=λ1𝑃𝑛₁−1,0,𝑛3,𝑛4,0+μ1𝑝13𝑃𝑛₁₊₁,0,𝑛₃−1,𝑛4,0+μ2𝑝21𝑃𝑛₁−1,1,𝑛3,𝑛4,0+μ2𝑝23𝑃𝑛1,1,𝑛₃−1,𝑛4,0+μ3𝑝34𝑃𝑛1,0,𝑛₃₊₁,𝑛₄−1,0+μ4𝑝46𝑃𝑛1,0,𝑛3,𝑛₄₊₁,0+

(5)

214

(𝑛2 ,𝑛5=0,𝑛1,𝑛3,𝑛4> 0) (13) (λ1+λ2+μ1+μ2+μ5)𝑃𝑛1,𝑛2,0,0,𝑛5

=λ1𝑃𝑛₁−1,𝑛2,0,0,𝑛5+λ2𝑃𝑛1,𝑛₂−1,0,0,𝑛5 +μ1𝑝12𝑃𝑛₁₊₁,𝑛₂−1,0,0,𝑛5+μ2𝑝21𝑃𝑛₁−1,𝑛₂₊₁,0,0,𝑛5+μ3𝑝35𝑃𝑛1,𝑛2,1,0,𝑛₅−1+μ4𝑝45𝑃𝑛1,𝑛2,0,1,𝑛₅−1+μ4𝑝46𝑃𝑛1,𝑛2,0,1,𝑛5+

μ5𝑝57𝑃𝑛₁,𝑛₂,0,0,𝑛₅₊₁

(𝑛3 ,𝑛4=0,𝑛1,𝑛2,𝑛5> 0) (14) (λ1+λ2+μ1+μ2+μ4)𝑃𝑛1,𝑛2,0,𝑛4,0

=λ1𝑃𝑛₁−1,𝑛2,0,𝑛4,0+λ2𝑃𝑛1,𝑛₂−1,0,𝑛4,0+μ1𝑝12𝑃𝑛₁₊₁,𝑛₂−1,0,𝑛4,0+μ2𝑝21𝑃𝑛₁−1,𝑛₂₊₁,0,𝑛4,0+μ3𝑝34𝑃𝑛1,𝑛2,1,𝑛₄−1,0+μ4𝑝46𝑃𝑛1,𝑛2,0,𝑛₄₊₁,0+μ5𝑝54𝑃𝑛1,𝑛2,0,𝑛₄−1,1 +μ5𝑝57𝑃𝑛1,𝑛2,0,𝑛4,1

(𝑛3 ,𝑛5=0,𝑛1,𝑛2,𝑛4> 0) (15) (λ1+λ2+μ1+μ2+μ3)𝑃𝑛1,𝑛2,𝑛3,0,0

=λ1𝑃𝑛₁−1,𝑛2,𝑛3,0,0+λ2𝑃𝑛1,𝑛₂−1,𝑛3,0,0+μ1𝑝12𝑃𝑛₁₊₁,𝑛₂−1,𝑛3,0,0+μ1𝑝13𝑃𝑛₁₊₁,𝑛2,𝑛₃−10,0+μ2𝑝21𝑃𝑛₁−1,𝑛₂₊₁,𝑛3,0,0+μ2𝑝23𝑃𝑛1,𝑛₂₊₁,𝑛₃−1,0,0+

μ4𝑝46𝑃𝑛₁,𝑛₂,𝑛₃,1,0+μ5𝑝57𝑃𝑛₁,𝑛₂,𝑛₃,0,1

(𝑛4 ,𝑛5=0,𝑛1,𝑛2,𝑛3> 0) (16) (λ1+λ2+μ4+μ5)𝑃0,0,0,𝑛₄,𝑛₅

=μ3𝑝34𝑃0,0,1,𝑛₄−1,𝑛5+μ3𝑝35𝑃0,0,1,𝑛4,𝑛5−1+μ4𝑝45𝑃0,0,0,𝑛4+1,𝑛5−1+μ4𝑝46𝑃0,0,0,𝑛4+1,𝑛5+μ5𝑝54𝑃0,0,0,𝑛4−1,𝑛5+1+μ5𝑝57𝑃0,0,0,𝑛4,𝑛5+1

(𝑛1 ,𝑛2,𝑛3= 0𝑛4,𝑛5> 0) (17)

(λ1+λ2+μ3+μ5)𝑃0,0,𝑛₃,0,𝑛₅

=μ1𝑝13𝑃1,0,𝑛₃−1,0,𝑛5+μ2𝑝23𝑃0,1,𝑛3−1,0,𝑛5+μ3𝑝35𝑃0,0,𝑛3+1,0,𝑛5−1+μ4𝑝45𝑃0,0,𝑛3,1,𝑛5−1+μ4𝑝46𝑃0,0,𝑛3,1,𝑛5+μ5𝑝57𝑃0,0,𝑛3,0,𝑛5+1

(𝑛1 ,𝑛2,𝑛4= 0𝑛3,𝑛5> 0) (18) (λ1+λ2+μ3+μ4)𝑃0,0,𝑛3,𝑛4,0

=μ1𝑝13𝑃1,0,𝑛₃−1,𝑛4,0+μ2𝑝23𝑃0,1,𝑛₃−1,𝑛4,0+μ3𝑝34𝑃0,0,𝑛₃₊₁,𝑛₄−1,0+μ4𝑝46𝑃0,0,𝑛3,𝑛₄₊₁,0+μ5𝑝54𝑃0,0,𝑛3,𝑛₄−1,1+μ5𝑝57𝑃0,0,𝑛3,𝑛4,1

(𝑛1 ,𝑛2,𝑛5= 0𝑛3,𝑛4> 0) (19) (λ1+λ2+μ2+μ5)𝑃0,𝑛₂,0,0,𝑛₅

=λ2𝑃0,𝑛₂−1,00,𝑛5 +μ1𝑝12𝑃1,𝑛2−1,0,0,𝑛5+μ3𝑝35𝑃0,𝑛2,1,0,𝑛5−1+μ4𝑝45𝑃0,𝑛2,0,1,𝑛5−1+μ4𝑝46𝑃0,𝑛2,0,1,𝑛5+μ5𝑝57𝑃0,𝑛2,0,0,𝑛5+1 (𝑛1 ,𝑛3,𝑛4= 0𝑛2,𝑛5> 0) (20)

(λ1+λ2+μ2+μ4)𝑃0,𝑛₂,0,𝑛₄,0

=λ2𝑃0,𝑛₂−1,0𝑛4,0+μ1𝑝12𝑃1,𝑛2−1,0,𝑛4,0+μ3𝑝34𝑃0,𝑛2,1,𝑛4−1,0+μ4𝑝46𝑃0,𝑛20,𝑛4+1,0+μ5𝑝54𝑃0,𝑛2,0,𝑛4−1,1+μ5𝑝57𝑃0,𝑛2,0,𝑛4,1

(6)

215

(λ1+λ2+μ1+μ2+μ3+μ4+μ5)𝑃0,𝑛₂,𝑛₃,0,0

=λ2𝑃1,𝑛₂−1,𝑛3,0,0+μ1𝑝12𝑃1,𝑛2−1,𝑛3,0,0+μ1𝑝13𝑃1,𝑛2,𝑛3−1,0,0+μ2𝑝23𝑃0,𝑛2+1,𝑛3−1,0,0+μ4𝑝46𝑃𝑛1,𝑛2,𝑛3,1,0+μ5𝑝57𝑃0,𝑛2,𝑛3,0,1

(𝑛1 ,𝑛4,𝑛5= 0𝑛2,𝑛3> 0) (22)

(λ1+λ2+μ1+μ5)𝑃𝑛1,0,0,0,𝑛5

=λ1𝑃𝑛₁−1,0,0,0,𝑛5+μ2𝑝21𝑃𝑛1−1,1,0,0,𝑛5+μ3𝑝35𝑃𝑛1,0,1,0,𝑛5−1+μ4𝑝45𝑃𝑛1,0,0,1,𝑛5−1+μ4𝑝46𝑃𝑛1,0,0,1,𝑛5+μ5𝑝57𝑃𝑛1,0,0,0,𝑛5+1

(𝑛2 ,𝑛3,𝑛4= 0𝑛1,𝑛5> 0) (23) (λ1+λ2+μ1+μ4)𝑃𝑛1,0,0,𝑛4,0

=λ1𝑃𝑛₁−1,0,0,𝑛4,0+μ2𝑝21𝑃𝑛₁−1,1,0,𝑛4,0+μ3𝑝34𝑃𝑛1,0,1,𝑛₄−1,0+μ4𝑝46𝑃𝑛1,0,0,𝑛₄₊₁,0+μ5𝑝54𝑃𝑛1,0,0,𝑛₄−1,1+μ5𝑝57𝑃𝑛1,0,0,𝑛4,1

(𝑛2 ,𝑛3,𝑛5= 0𝑛1,𝑛4> 0) (24) (λ1+λ2+μ1+μ3)𝑃𝑛1,0,𝑛3,0,0

=λ1𝑃𝑛₁−1,0,𝑛3,0,0+μ1𝑝13𝑃𝑛₁₊₁,0,𝑛₃−1,0,0+μ2𝑝21𝑃𝑛₁−1,1,𝑛3,0,0+μ2𝑝23𝑃𝑛1,1,𝑛₃−1,0,0+μ4𝑝46𝑃𝑛1,0,𝑛3,1,0+μ5𝑝57𝑃𝑛1,0,𝑛3,0,1

(𝑛2 ,𝑛4,𝑛5= 0𝑛1,𝑛3> 0) (25)

(λ1+λ2+μ1+μ2)𝑃𝑛1,𝑛2,0,0,0

=λ1𝑃𝑛₁−1,𝑛2,0,0,0+λ2𝑃𝑛1,𝑛₂−1,0,0,0+μ1𝑝12𝑃𝑛₁₊₁,𝑛₂−1,0,0,0+μ2𝑝21𝑃𝑛₁−1,𝑛₂₊₁,0,0,0+μ4𝑝46𝑃𝑛1,𝑛2,0,1,0+μ5𝑝57𝑃𝑛1,𝑛2,0,01

(𝑛3 ,𝑛4,𝑛5= 0𝑛1,𝑛2> 0) (26) (λ1+λ2+μ5)𝑃0,0,0,0,𝑛₅

=μ3𝑝35𝑃0,0,1,0,𝑛₅−1+μ4𝑝45𝑃0,0,0,1,𝑛₅−1+μ4𝑝46𝑃0,0,0,1,𝑛5+μ5𝑝57𝑃0,0,0,0,𝑛₅₊₁

(𝑛1,𝑛2,𝑛3,𝑛4= 0,𝑛5> 0) (27) (λ1+λ2+μ1+μ2+μ3+μ4+μ5)𝑃0,0,0,𝑛₄,0

=μ3𝑝34𝑃0,0,1,𝑛₄−1,0+μ4𝑝46𝑃0,0,0,𝑛₄₊₁,0+μ5𝑝54𝑃0,0,0,𝑛₄−1,1+μ5𝑝57𝑃0,0,0,𝑛4,1

(𝑛1,𝑛2,𝑛3,𝑛5= 0,𝑛4> 0) (28) (λ1+λ2+μ3)𝑃0,0,𝑛3,0,0

=μ1𝑝13𝑃1,0,𝑛₃−1,0,0+μ2𝑝23𝑃0,1,𝑛₃−1,0,0+μ4𝑝46𝑃0,0,𝑛3,1,0+μ5𝑝57𝑃0,0,𝑛3,0,1

(𝑛1,𝑛2,𝑛4,𝑛5= 0,𝑛3> 0) (29) (λ1+λ2+μ2)𝑃0,𝑛2,0,0,0

(7)

216

(𝑛1,𝑛3,𝑛4,𝑛5= 0,𝑛2> 0) (30)

(λ1+λ2+μ1)𝑃𝑛₁,0,0,0,0 =λ1𝑃𝑛₁−1,0,0,0,0+ μ2𝑝21𝑃𝑛₁−1,1,0,0,0+μ4𝑝46𝑃𝑛1,0,0,1,0+μ5𝑝57𝑃𝑛1,0,0,0,1 (𝑛1,𝑛3,𝑛4,𝑛5= 0,𝑛2> 0) (31)

(λ1+λ2)𝑃0,0,0,0,0=μ4𝑝46𝑃0,0,0,1,0+μ5𝑝57𝑃0,0,0,0,1 (𝑛1,𝑛2,𝑛3,𝑛4,𝑛5= 0) (32)

Now Define Generating Function as F(X,Y,Z,R,S)= 𝑛∞1=0 𝑛∞2=0 ∞𝑛3=0 ∞𝑛4=0 𝑛∞5=0𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛5𝑋𝑛1𝑌𝑛2𝑍𝑛3𝑅𝑛4𝑆𝑛5 (33) Where 𝑋 , 𝑌 , 𝑍 , 𝑅 , 𝑆 =1 Also define partial generating function as 𝐹𝑛₂,𝑛₃,𝑛₄,𝑛₅(X) = ∞𝑛₁=0𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅ 𝑋𝑛1 (34)

𝐹𝑛3,𝑛4,𝑛5(X,Y) = 𝑛∞2=0𝐹𝑛2,𝑛3,𝑛4,𝑛5(X)𝑌𝑛2 (35)

𝐹𝑛4,𝑛5(𝑋,𝑌,𝑍)= ∞𝑛3=0𝐹𝑛3,𝑛4,𝑛5 (X,Y)𝑍𝑛3 (36)

𝐹𝑛₅(X,Y,Z,R) = 𝑛∞₄=0𝐹𝑛₄,𝑛₅(X,Y,Z)𝑅𝑛4 (37)

F(X,Y,Z,R,S)= ∞𝑛₅=0𝐹𝑛₅(X,Y,Z,R)𝑆𝑛5 (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- (λ1+λ2+μ1+μ2+μ3+μ4+μ5)F(X,Y,Z,R,S)−𝜇5F(X,Y,Z,R)−𝜇4F(X,Y,Z,S)− 𝜇3F(X,Y,R,S)−𝜇2F(X,Z,R,S)−𝜇1F(Y,Z,R,S)=𝜆1𝑋𝑓(𝑋,𝑌,𝑍,𝑅,𝑆)+ 𝜆2𝑋𝑓(𝑋,𝑌,𝑍,𝑅,𝑆)+ 𝜇1𝑝_𝑋12𝑌 𝐹(𝑋,𝑌,𝑍,𝑅,𝑆)− 𝐹(𝑌,𝑍,𝑅,𝑆)+ 𝜇1𝑝13𝑍 𝑋 𝐹(𝑋,𝑌,𝑍,𝑅,𝑆)− 𝐹(𝑌,𝑍,𝑅,𝑆) + 𝜇2𝑝21𝑋 𝑌 𝐹(𝑋,𝑌,𝑍,𝑅,𝑆)− 𝐹(𝑋,𝑍,𝑅,𝑆) + 𝜇2𝑝23𝑍 𝑌 𝐹(𝑋,𝑌,𝑍,𝑅,𝑆)− 𝐹(𝑋,𝑍,𝑅,𝑆)+ 𝜇3𝑝34𝑅 𝑍 𝐹(𝑋,𝑌,𝑍,𝑅,𝑆)− 𝐹(𝑋,𝑌,𝑅,𝑆)+ 𝜇3𝑝35𝑆 𝑍 𝐹(𝑋,𝑌,𝑍,𝑅,𝑆)− 𝐹(𝑋,𝑌,𝑅,𝑆) +𝜇4𝑝45𝑆 𝑅 𝐹(𝑋,𝑌,𝑍,𝑅,𝑆)− 𝐹(𝑋,𝑌,𝑍,𝑆) + 𝜇₄𝑝₄₆ 𝑅 𝐹(𝑋,𝑌,𝑍,𝑅,𝑆)− 𝐹(𝑋,𝑌,𝑍,𝑆)+ 𝜇₅𝑝₅₄𝑅 𝑆 𝐹(𝑋,𝑌,𝑍,𝑅,𝑆)− 𝐹(𝑋,𝑌,𝑍,𝑆)+ 𝜇5𝑝57 𝑆 𝐹(𝑋,𝑌,𝑍,𝑅,𝑆)− 𝐹(𝑋,𝑌,𝑍,𝑆) (39)

( 𝜆1(1− 𝑋) +𝜆2(1−𝑌)+𝜇1 1− 𝑝12_𝑋𝑌− 𝑝13_𝑋𝑍 +𝜇2 1− 𝑝21𝑋_𝑌− 𝑝23_𝑌𝑍 + 𝜇3(1−𝑝34_𝑍𝑅− 𝑝35_𝑍𝑆)+𝜇4(1−𝑝45𝑆_𝑅−𝑝_𝑅46)+𝜇5(1−𝑝54𝑅_𝑆− 𝑝57 𝑆 ))F(X,Y,Z,R,S)= 𝜇5F(X,Y,Z,R) (1−𝑝54 𝑅 𝑆− 𝑝57 𝑆 ) +𝜇4F(X,Y,Z,S) (1−𝑝45 𝑆 𝑅− 𝑝46 𝑅)+𝜇3F(X,Y,R,S) (1−𝑝34 𝑅 𝑍− 𝑝35 𝑆 𝑍)+𝜇2F(X,Z,R,S) 1− 𝑝21𝑋𝑌−𝑝23𝑍𝑌+𝜇1F(Y,Z,R,S)1−𝑝12𝑌𝑋−𝑝13𝑍𝑋 (40)

(8)

217

𝜇₅F X,Y,Z,R 1−𝑝54𝑅_𝑆−𝑝_𝑆57 +𝜇4F X,Y,Z,S 1−𝑝45_𝑅𝑆−𝑝_𝑅46 + 𝜇3F X,Y,R,S 1−𝑝34𝑅_𝑍−𝑝35𝑆_𝑍 + 𝜇₂F(X,Z,R,S) 1−𝑝₂₁𝑋_𝑌−𝑝₂₃𝑍_𝑌 +𝜇₁F(Y,Z,R,S) 1−𝑝₁₂𝑌_𝑋−𝑝₁₃𝑍_𝑋

( 𝜆₁(1−𝑋) +𝜆₂(1−𝑌)+𝜇₁ 1−𝑝₁₂𝑌_𝑋−𝑝₁₃𝑍_𝑋 +𝜇₂ 1−𝑝₂₁𝑋_𝑌−𝑝₂₃𝑍_𝑌 + 𝜇₃(1−𝑝₃₄𝑅_𝑍−𝑝₃₅𝑆_𝑍)+ 𝜇₄(1−𝑝₄₅_𝑅𝑆−𝑝_𝑅46)+𝜇₅(1−𝑝₅₄𝑅_𝑆−𝑝57_𝑆 ))

(41)

Let us denote :- F(Y,Z,R,S)=𝐹1

F(X,Z,R,S)=𝐹2

F(X,Y,R,S)=𝐹3

F(X,Y,Z,S)=𝐹4

F(X,Y,Z,R)=𝐹5

Also F(1,1,1,1,1)=1 Total probability ,

Let (X=1) as Y,Z,R,S→1

Now (41) becomes 0

0 form .Now by using L’Hospital rule ,we get by differentiating w.r.t.X

1= 𝜇1𝐹1 𝑝13+𝑝12 +𝜇2𝐹2(−𝑝21)

−𝜆1+𝜇1 𝑝12+𝑝13 +𝜇2(−𝑝21) −𝜆1+𝜇1− 𝜇2𝑝21=𝜇1𝐹1−𝜇2𝐹2𝑝21

where 𝑝13+𝑝12= 1 (42) Again differentiating w.r.t.Y gives Y=1

1=𝜇2𝐹2 𝑝23+𝑝21 +𝜇1𝐹1(−𝑝12) −𝜆₂+𝜇₂ 𝑝21+𝑝23 +𝜇1(−𝑝12)

⟹ −𝜆2− 𝜇1𝑝12+𝜇2=−𝜇1𝐹1𝑝12+𝜇2𝐹2 (43)

where 𝑝21+𝑝23=1 Again differentiating w.r.t.Z gives Z=1

1= 𝜇3𝐹3 𝑝34+𝑝35 +𝜇2𝐹2(−𝑝23)+𝜇1𝐹1(−𝑝13) 𝜇₃ 𝑝₃₄+𝑝₃₅ +𝜇₂(−𝑝₂₃)+𝜇₁(−𝑝₁₃)

⟹ −𝜇1𝑝13− 𝜇2𝑝23+𝜇3=−𝜇1𝑝13𝐹1− 𝜇2𝑝23𝐹2+𝜇3𝐹3 (44)

where 𝑝34+𝑝35=1

Again differentiating w.r.t.R gives R=1

1= 𝜇4𝐹4 𝑝45+𝑝46 +𝜇3𝐹3(−𝑝34)+𝜇5𝐹5(−𝑝54) 𝜇4 𝑝45+𝑝46 +𝜇3(−𝑝34)+𝜇5(−𝑝54)

⟹ −𝜇3𝑝34− 𝜇5𝑝54+𝜇4=−𝜇3𝑝34𝐹3− 𝜇5𝑝54𝐹5+𝜇4𝐹4 (45)

Where 𝑝45+𝑝46=1

(9)

218

1= 𝜇5𝐹5 𝑝54+𝑝57 +𝜇3𝐹3(−𝑝35)+𝜇4𝐹4(−𝑝45)

𝜇5 𝑝54+𝑝57 +𝜇3(−𝑝35)+𝜇4(−𝑝45)

𝜇5𝐹5− 𝜇4𝐹4(𝑝45)− 𝜇3𝐹3(𝑝35)=−𝜇3(𝑝35)− 𝜇4(𝑝45)+𝜇5 (46) After solving the above for 𝐹1,𝐹2,𝐹3,𝐹4,𝐹5 , we get

𝐹1=1−_(1−𝑝𝜆2+𝜆1𝑝12 21𝑝12)𝜇1 𝐹2=1−_(1−𝑝𝜆2+𝜆1𝑝12

21𝑝12)𝜇2

𝐹3=1−[ 𝜆2+𝜆1𝑝12_(1−𝑝 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13]

21𝑝12)𝜇3

𝐹4=1− 𝑝34+𝑝35𝑝54 [ 𝜆2+𝜆_(1−𝑝1𝑝12 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13] 21𝑝12)𝜇4(1−𝑝45𝑝54) ] 𝐹5=1− 𝜆2

+𝜆₁𝑝₁₂ 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13] (1−𝑝21𝑝12) ]

𝑝₄₅ 𝑝34+𝑝35𝑝54 +𝑝35 𝜇5(1−𝑝45𝑝54)

V. SOLUTION OF MODEL

𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛5=𝜌1 𝑛₁

𝜌₂𝑛2_𝜌 3 𝑛₃

𝜌₄𝑛4_𝜌 5 𝑛₅

(1−𝜌1) (1− 𝜌2)(1− 𝜌3)(1− 𝜌4)(1− 𝜌5) Where

𝜌1=

𝜆₁+𝜆₂𝑝₂₁ (1−𝑝21𝑝12 )𝜇1

𝜌2=_(1−𝑝𝜆2+𝜆1𝑝12 21𝑝12)𝜇2

𝜌3=[ 𝜆2+𝜆1𝑝12_(1−𝑝 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13]

21𝑝12 )𝜇3

𝜌4= 𝑝34+𝑝35𝑝54 [ 𝜆2+𝜆_(1−𝑝1𝑝12 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13] 21𝑝12)𝜇4(1−𝑝45𝑝54) ] 𝜌5= 𝜆2+𝜆1𝑝12_(1−𝑝 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13]

21𝑝12) ]

𝑝₄₅ 𝑝₃₄+𝑝₃₅𝑝₅₄ +𝑝₃₅ 𝜇5(1−𝑝45𝑝54)

The solution in steady state exists, if the conditions 𝜌1,𝜌2,𝜌3,𝜌4,𝜌5<1 are satisfied.

VI. VARIOUS QUEUE CHARACTERISITCS

A. Average number of customers

L= 𝑛𝑛∞₁ 𝑛∞₂ 𝑛∞₃ 𝑛∞₄ 𝑛∞₅ 1+𝑛2+𝑛3+𝑛4+𝑛5 𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅

= 𝑛𝑛∞1 𝑛∞2 𝑛∞3 ∞𝑛4 ∞𝑛5 1 𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛5 + 𝑛𝑛∞1 𝑛∞2 𝑛∞3 ∞𝑛4 𝑛∞5 2 𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛5

+ 𝑛𝑛∞1 𝑛∞2 𝑛∞3 ∞𝑛4 ∞𝑛5 3 𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛5+ 𝑛∞𝑛1 𝑛∞2 𝑛∞3 ∞𝑛4 𝑛∞5 4 𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛5

+ 𝑛𝑛∞1 𝑛∞2 𝑛∞3 ∞𝑛4 ∞𝑛5 5 𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛5

(10)

219

L= 𝐿1+𝐿2+𝐿3+𝐿4+𝐿5 L= 𝜆1+𝜆2𝑝21

(1−𝑝21𝑝12)𝜇1−(𝜆1+𝜆2𝑝21) +

𝜆₂+𝜆₁𝑝₁₂

(1−𝑝21𝑝12)𝜇2−(𝜆2+𝜆1𝑝12)+

𝜆2+𝜆1𝑝12 𝑝23+(𝜆1+𝜆2𝑝21)𝑝13

𝜇3(1−𝑝12𝑝21− 𝜆2+𝜆1𝑝12 𝑝23 +(𝜆1+𝜆2𝑝21 )𝑝13) +

𝑝34+𝑝35𝑝54 𝜆2+𝜆1𝑝12 𝑝23+(𝜆1+𝜆2𝑝21)𝑝13

(𝜇4(1−𝑝45𝑝54) 1−𝑝21𝑝12) − ( 𝑝34+𝑝35𝑝54 𝜆2+𝜆1𝑝12 𝑝23+(𝜆1+𝜆2𝑝21)𝑝13 + 𝜆2+𝜆1𝑝12 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13(𝑝45 𝑝34+𝑝35𝑝54 +𝑝35)

(𝜇₅(1−𝑝₄₅𝑝₅₄₎ 1−𝑝₂₁𝑝₁₂₎ − (𝑝₄₅ 𝑝34+𝑝35𝑝54 +𝑝35 𝜆2+𝜆1𝑝12 𝑝23+(𝜆1+𝜆2𝑝21)𝑝13 B. Variance of queue

V(𝑛1+𝑛2+𝑛3+𝑛4+𝑛5) = 𝑛𝑛∞₁ 𝑛∞₂ 𝑛∞₃ ∞𝑛₄ ∞𝑛₅ 1+𝑛2+𝑛3+𝑛4+𝑛5 − 𝐿 2𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅ = 𝑛𝑛∞1 ∞𝑛2 𝑛∞3 𝑛∞4 ∞𝑛5 1+𝑛2+𝑛3+𝑛4+𝑛5

2_𝑃

𝑛1,𝑛2,𝑛3,𝑛4,𝑛5+L

2 _𝑃

𝑛1,𝑛2,𝑛3,𝑛4,𝑛5 ∞ 𝑛5 ∞ 𝑛4 ∞ 𝑛3 ∞ 𝑛2 ∞

𝑛1 −2𝐿 𝑛𝑛∞1 𝑛∞2 ∞𝑛3 ∞𝑛4 ∞𝑛5 1+𝑛2+

𝑛3+𝑛4+𝑛5𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛5

= 𝑛𝑛∞₁ 𝑛∞₂ 𝑛∞₃ 𝑛∞₄ ∞𝑛₅ 1+𝑛2+𝑛3+𝑛4+𝑛5 2𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅− 𝐿2 = 𝑛𝑛∞1 𝑛∞2 𝑛∞3 𝑛∞4 𝑛∞5 1+𝑛2+𝑛3+𝑛4+𝑛5 𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛5

= 𝑛𝑛∞₁ ∞𝑛₂ ∞𝑛₃ ∞𝑛₄ 𝑛∞₅ 1 2𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅+ 𝑛∞𝑛₁ ∞𝑛₂ ∞𝑛₃ ∞𝑛₄ 𝑛∞₅ 2 2𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅+

𝑛∞𝑛₁ 𝑛∞₂ ∞𝑛₃ ∞𝑛₄ 𝑛∞₅ 3 2𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅+ 𝑛∞𝑛₁ ∞𝑛₂ ∞𝑛₃ ∞𝑛₄ 𝑛∞₅ 4 2𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅ + 𝑛∞𝑛₁ ∞𝑛₂ 𝑛∞₃ 𝑛∞₄ 𝑛∞₅ 5 2𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅+2

𝑛𝑛∞₁ 𝑛∞₂ 𝑛∞₃ 𝑛∞₄ 𝑛∞₅ 1𝑛2 𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅+2 𝑛𝑛∞₁ 𝑛∞₂ 𝑛∞₃ 𝑛∞₄ 𝑛∞₅ 1𝑛3 𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅+

2 𝑛𝑛∞₁ 𝑛∞₂ 𝑛∞₃ 𝑛∞₄ 𝑛∞₅ 1𝑛4 𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅+

2 𝑛𝑛∞1 𝑛∞2 𝑛∞3 𝑛∞4 𝑛∞5 1𝑛5 𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛5+

2 𝑛𝑛∞₁ 𝑛∞₂ 𝑛∞₃ 𝑛∞₄ 𝑛∞₅ 2𝑛3 𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅+………+

2 𝑛𝑛∞₁ 𝑛∞₂ 𝑛∞₃ 𝑛∞₄ ∞𝑛₅ 4𝑛5 𝑃𝑛₁,𝑛₂,𝑛₃,𝑛₄,𝑛₅ −𝐿2

After Substituting the values ,the result reduces to

[ 𝜆1+𝜆2𝑝21 (1−𝑝₂₁𝑝₁₂₎𝜇₁][ 1−

𝜆2+𝜆1𝑝12 (1−𝑝₂₁𝑝₁₂₎𝜇₁]

-2 _+[ 𝜆2+𝜆1𝑝12 (1−𝑝₂₁𝑝_{12 )}𝜇₂][1−

𝜆2+𝜆1𝑝12 (1−𝑝₂₁𝑝₁₂₎𝜇₂]

-2_+[[ 𝜆2+𝜆1𝑝12 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13] (1−𝑝₂₁𝑝_{12 )}𝜇₃ ][1−

[ 𝜆2+𝜆1𝑝12 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13] (1−𝑝₂₁𝑝₁₂₎𝜇₃ ]

-2

+[[ 𝜆2+𝜆1𝑝12 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13] (1−𝑝₂₁𝑝₁₂₎𝜇₃ ][1−

[ 𝜆2+𝜆1𝑝12 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13] (1−𝑝₂₁𝑝₁₂₎𝜇₃ ]

-2

+[ 𝜆2+𝜆1𝑝12 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13] (1−𝑝21𝑝12) ]

𝑝₄₅ 𝑝₃₄+𝑝₃₅𝑝₅₄ +𝑝₃₅ 𝜇5(1−𝑝45𝑝54) ][1−

𝜆₂+𝜆₁𝑝₁₂ 𝑝₂₃+(𝜆₂+𝜆₁𝑝₁₂)𝑝₁₃] (1−𝑝21𝑝12 ) ]

𝑝₄₅ 𝑝₃₄+𝑝₃₅𝑝₅₄ +𝑝₃₅

𝜇5(1−𝑝45𝑝54) ]

-2

C. Average waiting time for the customers in the system Where

E(W)= 𝜆1+𝜆2𝑝21

𝜆[(1−𝑝₂₁𝑝₁₂₎𝜇₁− 𝜆₁+𝜆₂𝑝₂₁ ] +

𝜆2+𝜆1𝑝12

𝜆[(1−𝑝₂₁𝑝₁₂₎𝜇₂− 𝜆₂+𝜆₁𝑝₁₂ ]+

𝜆2+𝜆1𝑝12 𝑝23+(𝜆1+𝜆2𝑝21)𝑝13

𝜆[𝜇₃(1−𝑝₁₂𝑝₂₁_{− 𝜆}₂₊_𝜆₁_𝑝₁₂_𝑝_{23 +(}_𝜆₁₊_𝜆₂_𝑝_{21 )}_𝑝₁₃_)] +

𝑝34+𝑝35𝑝54 𝜆2+𝜆1𝑝12 𝑝23+(𝜆1+𝜆2𝑝21)𝑝13

𝜆[(𝜇₄(1−𝑝₄₅𝑝₅₄₎ 1−𝑝₂₁𝑝₁₂₎ − ( 𝑝34+𝑝35𝑝54 𝜆2+𝜆1𝑝12 𝑝23+(𝜆1+𝜆2𝑝21)𝑝13] + 𝜆2+𝜆1𝑝12 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13(𝑝45 𝑝34+𝑝35𝑝54 +𝑝35)

(11)

220

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 𝑛1,𝑛2,𝑛3,𝑛4,𝑛5.

Step2. Obtain the values of mean service rate 𝜇1,𝜇2,𝜇3,𝜇4,𝜇5.

Step3.Obtain the values of mean arrival rate 𝜆1,𝜆2.

Step4.Obtain the values of the probabilities 𝑝12,𝑝21,𝑝13,𝑝23,𝑝34,𝑝35,𝑝45,𝑝46,𝑝54,𝑝57.

Step5. Calculate the value of (i) 𝐹1=1−

𝜆₂+𝜆₁𝑝₁₂ (1−𝑝21𝑝12)𝜇1

(ii) 𝐹2=1−

𝜆₂+𝜆₁𝑝₁₂ (1−𝑝21𝑝12)𝜇2

(iii) 𝐹3=1−[ 𝜆2+𝜆1𝑝12_(1−𝑝 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13]

21𝑝12)𝜇3

(iv)𝐹4=1− 𝑝34+𝑝35𝑝54 [ 𝜆2+𝜆_(1−𝑝1𝑝12 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13] 21𝑝12)𝜇4(1−𝑝45𝑝54) ] (v)𝐹5=1− 𝜆2+𝜆1𝑝12_(1−𝑝 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13]

21𝑝12) ]

𝑝45 𝑝34+𝑝35𝑝54 +𝑝35 𝜇₅(1−𝑝₄₅𝑝₅₄₎

Step 6.Calculate (i) 𝜌1=1−𝐹1 (ii) 𝜌2=1−𝐹2 (iii) 𝜌3=1−𝐹3 (iv) 𝜌4=1−𝐹4 (v) 𝜌5=1−𝐹5

Step 7.Check 𝜌1,𝜌2,𝜌3,𝜌4,𝜌5.<1

If so, then go to step (8) else steady state condition does not holds good.

Step 8. The joint probability

𝑃𝑛1,𝑛2,𝑛3,𝑛4,𝑛5=𝜌𝑛1𝜌𝑛2𝜌𝑛3𝜌𝑛4𝜌𝑛5(1−𝜌1) (1−𝜌2) (1−𝜌3) (1−𝜌4) (1−𝜌5)

Step9. Calculate average no. of customers (mean queue length) L= 𝜌1

(1−𝜌1)+ 𝜌₂ (1−𝜌2)+

𝜌₃ (1−𝜌3)+

𝜌₄ (1−𝜌4)+

𝜌₅ (1−𝜌5)

Step10. Calculate variance of queue V= 𝜌1

(1−𝜌₁)2+ 𝜌2 (1−𝜌₂)2+

𝜌3 (1−𝜌₃)2

𝜌4 (1−𝜌₄)2+

𝜌5 (1−𝜌₅)2

(12)

221

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. 𝑛1=3 𝜇1=9 𝜆1=2 𝑝12=0.6

2. 𝑛2=6 𝜇2=8 𝜆2=3 𝑝13=0.4

3. 𝑛3=7 𝜇3=7 𝑝21=0.7

4. 𝑛4=4 𝜇4=10 𝑝23=0.3

5. 𝑛5=5 𝜇5=15 𝑝45=0.5

𝑝46=0.5

𝑝34=0.6

𝑝35=0.4

𝑝54=0.7

𝑝57=0.3

SOLUTION:

𝜌1=_(1−𝑝𝜆1+𝜆2𝑝21

21𝑝12)𝜇1 =0.78

𝜌2=

𝜆₂+𝜆₁𝑝₁₂

(1−𝑝21𝑝12)𝜇2 =0.90

𝜌3=[ 𝜆2+𝜆1𝑝12_(1−𝑝 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13] 21𝑝12 )𝜇3 =0.71

𝜌4= 𝑝34+𝑝35𝑝54 [ 𝜆2+𝜆_(1−𝑝1𝑝12 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13]

21𝑝12)𝜇4(1−𝑝45𝑝54) ] =0.68 𝜌5= 𝜆2+𝜆1𝑝12_(1−𝑝 𝑝23+(𝜆2+𝜆1𝑝12)𝑝13]

21𝑝12) ]

𝑝45 𝑝34+𝑝35𝑝54 +𝑝35

𝜇₅(1−𝑝₄₅𝑝₅₄₎ =0.25

Average no. of customers

L= 𝐿1+𝐿2+𝐿3+𝐿4+𝐿5 = 3.66+9.54+2.5+2.50+0.8193

(13)

222

Average waiting time for customers

E(W)= 𝐿_𝜆 where 𝜆 =𝜆1+𝜆2 = 𝐿1+𝐿2+𝐿3+𝐿4+𝐿5

𝜆

= 0.732+1.908+0.5+0.5+0.16386

= 3.80386

Variance- 𝜌1 (1−𝜌1)2+

𝜌₂ (1−𝜌2)2+

𝜌₃ (1−𝜌3)2

𝜌₄ (1−𝜌4)2+

𝜌₅ (1−𝜌5)2

= 0.037752+90+8.442+6.6406+0.44

= 105.560352

REFERENCES

[1] Jackson R.R.P, “Queuing system with phase type service”, vol. 5. O.R Quart, 1954, pp.109-120.

[2]O`Brien , G.B, “The solution of some queuing problems”, vol.2. J.Soc.Ind. Appli. Math, 1954, pp. 133-142.

[3]Stephan ,F.F, “Two Queues under preemptive priority with pooisson Arrival band service Rates” , Vol.6. Opers .Res,1958, pp.399-418.

[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, 2005.

[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. 135-152.

[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 2222-2863, vol. 2 . No.3, 2007.

[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.