Transient Solution of A M/M/4 Queue With Heterogeneous Servers Subject To Catastrophes

IJSRR, 7(4) Oct. – Dec., 2018 Page 2084

Research article Available online www.ijsrr.org

ISSN: 2279–0543

International Journal of Scientific Research and Reviews

Transient Solution of A M/M/4 Queue With Heterogeneous Servers

Subject To Catastrophes

Julia Rose Mary. K

1

and Maria Remona. J

2

*

1

Department of Mathematics, Nirmala College for Women, Coimbatore.

2

Department of Mathematics, Nirmala College for Women, Coimbatore.

ABSTRACT

This paper demonstrates a transient solution for the system size in a M/M/4 queue where the

service rates of the servers are not similar with the possibility of catastrophes at the system. For the

customer in the system the time dependent probabilities are derived and then the steady state

probabilities of the system size are also given. Some important performance measures are also

obtained.

KEYWORDS

: Transient Analysis, System Size, Heterogeneous Servers, Catastrophes, Steady State Probability and Performance Measures.

*Corresponding author

J. Maria Remona,

M.Phil Research Scholar, Nirmala College For Women,

Red Fields, Sungam (po), Coimbatore-641018,

Tamil Nadu, India.

IJSRR, 7(4) Oct. – Dec., 2018 Page 2085

1. INTRODUCTION

In the study of queue networks one typically tries to obtain the equilibrium distribution of the

network, although in many applications the study of the transient state is fundamental. The transient

response is necessarily tied to any event that affects the equilibrium of the system.

Multi-server queuing systems arrive in congestion problems of telephone exchange and

computer networks. A complete description of situations with such queuing analysis of computer

systems can be found in Lavenberg 14. In many real multi-server queuing situations, the service with

heterogeneity is a common feature. The heterogeneous servers to the waiting lines are analyzed by

Gumbel.H 3. The role of quality and service performance is crucial aspects in customer perceptions

and firms must dedicate special attention to them with designing and implementing their operations.

For these reasons, the queues with heterogeneity have received considerable attention in the

literature. Transient solution of a two processor heterogeneous system has been discussed by

Dharamaraja.S 6. A control model for a machine center with two heterogeneous system has been

introduced by Liu and Kumar 7. A treaties on the Theory of Bessel functions where discussed by

Watson. G. N 8. Whitt. W 9 has analyzed the Untold Horrors of the Waiting Room: What the

Equilibrium Distribution Will Never Tell about the Queue Length Process. A research on Measures

for Time Dependent Queueing Problem with Service in Batches of Variables Size was done by Garg.

P. C 10.

In recent times, queuing model with catastrophes has been investigated by Boucherie and

Boxma 11, Jain and Sigman 13 and Dudin and Nishimura 12. Transient solution of a single server

queue with catastrophes are discussed by Kumar,B.K and Arivudainambi.D 4. An analysis made on

the queuing network model with catastrophes and its product from solution by Chao.X 5. The

catastrophes may come either from outside of the system or from another service station of the

system.

A combined analysis of queues with heterogeneous servers subject to catastrophes to find

transient solution of an M/M/2 model by Kumar.B.K, Pavai.M and Vankatakrihnan 1. Transient

solution of a Markovian queuing model with heterogeneous servers and catastrophes has been

discussed by Dharmaraja and Rakesh Kumar 2.

From the output of this study, the queueing system is organized as follows:

(i) To describe the queueing model of four server heterogeneous system with catastrophes

and to derive the time-dependent state probabilities for the system size,

(ii) To analyze the steady state probabilities of the system size and then

(iii) Approach few important performance measures that are derived from the system size

IJSRR, 7(4) Oct. – Dec., 2018 Page 2086

2. MODEL DESCRIPTION

By examining on M/M/4 queuing system with heterogeneous servers and assume that the

servers times follow exponential distributions with the service rates as ₁,₂,₃,and ₄ for four different servers where₁₂ ₃ ₄. Consider the customer arrival process in Poisson with

rate and system also has one waiting line. FCFS queuing discipline is followed and each customer

requires exactly one server for the service. When the server becomes free, the customer who is first

in the waiting line will join the queue. Other than arrival and service processes, there also occur

catastrophes at the service facilities with rate  in a Poisson manner. In the system, whenever a

catastrophe occurs it destroys all the customers in the system immediately, and also the server get

inactivated. Then the service is started when a new arrival occurs. Let



X

 

t ,t



be the number

of customers in time t. Let _n

 

t 



X

 

t n



,n4,5,6, denotes the probability of n customers in the system at time t. Also let ₀

 

t 



X

 

t 0



be the probability that the system is empty at time t,

 



 

1



1  

 t X t be the probability that there is one customer in the system, ₂

 

t 



X

 

t 2



be

the probability that there are two customers in the system, and ₃

 

t 



X

 

t 3



be the probability that there are three customers in the system.

From the above assumptions the state probabilities

       

, ₁ , ₂ , ₃

 

, 4,5,6

0     

 t t t t and _n t n satisfy the following system of differential difference

equations:

 

_{ }

_{ }

_

_{ }

_

t t

t dt

t d

0 1

1 0

0      ₁

 _{  }

(2.1)

 

_

_{  }

_{  }

_{  }

t t

t dt

t d

2 2 1 0

1 1

1         

 _{     }

(2.2)

 

_

_{  }

_{  }

_{  }

t t

t dt

t d

3 3 2 1 1

2 2 1

2           

 _{       }

(2.3)

 

_

_{  }

_{  }

_{  }

t t

t dt

t d

4 4 3 2 1 2

3 3 2 1

3             

 _{         }

(2.4)

 

_

_{  }

_{  }

_{  }

t t

t dt

t d

5 4 3 2 1 3

4 4 3 2 1

4              

 _{          }

(2.5)

 

_

_{  }

_{  }

_{  }



, 6 , 5 ,

1 4 3 2 1 1

4 3 2

1            

   



 t t n

t dt

t d

n n

n

n           

(2.6)

Suppose at time t=0 there is no customer in the system, so that ₀

 

t 1. By using a

IJSRR, 7(4) Oct. – Dec., 2018 Page 2087

 

 

_



 



 

  



0

1 5 0

,

n

n n t z

t G t

z (2.7)

where G₀

 

t ₀

 

t ₁

 

t ₂

 

t ₃

 

t ₄

 

t , with initial condition 

 

z,0 1.

Apply the standard generating function argument, the system of equations 2.1 to 2.6 then

yields

 



 

 

  



  



z t G

 

t



z z t z

t G t

t z

0 4

0 1 ,

1

, _{ }

  

 _{   }

    

  

 _{  }  _{  }

(2.8)

where ₁ ₂ ₃ ₄.

Examine equation 2.8 as a first order linear differential equation in 

 

z,t and solving, we get ,



 

 Bt 



t







 

 

 

  

 

 



Btu

du e

u BG u z

u G e

t z

0 1 0 1 4 0

,   (2.9)

where





  

 _{   }

     

z z B

By utilizing the Bessel function generating function, if

   

 2 and  , then

  





  

     



n

n n

t z z

z t I

e  

 

where I_n



. is the modified Bessel function of first kind of order n.

Equating this in equation 2.9, then expanding 

 

z,t as a series in z and comparing the co-efficient

of n

z on either side, we get for n1,2,3,

 





















 

du e

u t I u

t I u

u t b t

n n

n   





   

 0

1 1

4   































 

 

du e

u G u t bI u t I u

t I

u t b t

n n

n

n  

 





    



0 1 0

1

1     



 (2.10)

where b and further, when n=0, we get

 





















 

du e

u t I u t I u

u t b

t  



   



0 4 1   0 























 

 

du e

u G u t I b u

t I

u t b

t _  



  



0 0 0

1 1

2     (2.11)

where we have used I_n



. I_n



. .

 

_

 

_{bt n}

_

t



_

 







_





_b _t_u

du e

u t I u G e

t I t G

0 0 0

0

0    1 



 

 



_



 











  

    

 t bt u

n n

bt n

n

n t I t e G u I t u e du

0 0

IJSRR, 7(4) Oct. – Dec., 2018 Page 2088

Since 

 

z,t does not contain terms with negative powers of z, the right hand side of 2.10

with n replaced with –n must be zero. Thus,

 





















 

du e

u t I u

t I u

u t b t

n n

n   





   

 0

1 1

4   































 

 

du e

u G u t bI u t I u

t I

u t b t

n n

n

n  

 





    



0 1 0

1

1     



 (2.12)

Utilizing equation 2.12 in 2.10, after some algebraic manipulation, we obtain for n=1,2,3,…

(2.13)

So far, the probabilities ₀

       

t ,₁ t ,₂ t ,₃ t and₄

 

t remain to be found. To find, we consider the system of equations 2.1 to 2.4 subject to condition 2.11. Equations 2.1 to 2.4 can be

expressed in matrix from as

 

 

t e_{1 4}

 

t e₂ dt

t

d _{  }

(2.14)

where 

 

t 



₀

       

t ,₁ t ,₂ t ,₃ t



, e1 



1,0,0,0



 and







 0,0,0,1

2

e ,















  

 

   

 

    

  

  

 

   

 

     

   

   

  

  

 



3 2 1

3 2 1 2

1 2 1 1

1

0 0

0

0 0 0

In continuation, let _n*

 

s denote the Laplace transform of _n

 

t . Now, by taking Laplace transforms, the result of 2.14 is obtained as

  



 

_

  

 

         

 

 

2 * 4 1 1

*

1 e s e

s sI

s   (2.15)

with

  

0  1,0,0,0



 (2.16)

Hence, only 4*

 

s is to be found. We note that, if







 1,1,1,1

e ,

G

 

s e

 

s 4*

 

s *

*

0   



(2.17)

Taking Laplace transforms, after simplification, equation 2.11 yields,

 

 _

_



_

 







_





  

 t _{bt u}

n n

bt n

n

du e

u t I u G e

t I

0 1 0

0    

 

_

 



_



_





  

 _

 



 n t n bt u

n e du

u t

u t I u n

t

0 4 4

IJSRR, 7(4) Oct. – Dec., 2018 Page 2089

 





  



 2



2 1

1 * 2 2

4 *

0   _  s    

s s s

G (2.18)

where s

Utilizing 2.18 in 2.17 and solving for 4*

 

s , we obtain

 



 

















 



2

1 2

2

1 1

* 4

2 1

1 1

e s

sI e s

e s sI

e s s

  

 

 

   

      

   

       



 

 

(2.19)

Let







 



4 4 * 1

  _



 m s

sI ij

It is easy to see that,





   

 



 

  









 

   

   



 





 

   

     







   





 



   



   

 

 



   

 

   

 

 



 

 

 

 



 

 

 



 

 

D

s g s

j s f s

g s f s

f

s g s f s

g s f s g s g s f s

g

s f s

g s f s

i s f s

i

s g s

i s

g s i s g

sI 1 1 1 2

2 3

4 1

1 1

1 3

3 3

2

4 2

1 2

1 3

4 2

1 1 2

1 1 3 1

2 1 3 1

1     

  

 



    

 

     

  

  

(2.20)

where g₁

 

s s₁ ; g₂

 

s s₁₂ ; g₃

 

s s₁₂ ₃;

f

 

s  s; i

 

s g₂

    

s g₃ s  ₄



; j

 

s g₁

    

s g₂ s  ₁₂



. and

 













1

 





2 3

 

2



2 3



2 3 2 1 3

4

2 3

2 2

3

4                 

 

 s s s

D



 











 





 

 



 



 

 

 



 







_{1 1 2} 3 2 _{2 3 2 3 2 2}

3            

 s



2





₁

 

₁



_{1 2 3}



₁





_{1 2}



_{1 2 1 3}



7 2

2 3

3             

  

           







2



4₁



₁2₂₃



₁



2₁8



₂



₂ ₃





₁2



₁₂















2



2₂₃222













2₁23₁₂₁₃3₁







₁2₁₂





₁₂ ₃ 



₁₃2



The characteristics roots of the matrix  are given by

D

 

 0 (2.21)

IJSRR, 7(4) Oct. – Dec., 2018 Page 2090



 





 















4  ₁  23  2₂ ₃ ₂  ₂ ₃ 3₁ ₁  ₂

16

1 _{          }

 a





4







3₁2₂ ₃



2















 





 







1 2 3 2 2 3

3 3 2

1 2 16 23 2

3 4

2 64

1 _{                 }

 b

3₁



₁



₂















 



3



2₂₃

 

 ₂₃



₂



₂





2





₁

 

₁



_{1 2 3}



₁





_{1 2}



_{1 2 1 3}



7 2

2 3

3             

  

           







2



4₁



₁2₂₃



₁



2₁8



₂



₂₃





₁2



₁₂







64









2



2₂₃222













2₁23₁₂₁₃3₁







₁2₁₂





₁₂₃



₁₃2





   

 

  



 

3 1

2 cos 3 1 2

a b and

a

n  , the characteristic roots of 2.21 are













, 1,2,3,4.

3 2 3 4

3 2 2

cos    1 2 3 

  

 _{ }

n i i

si

     

 (2.22)

It is examined that m_kj*

 

s are all rational algebraic functions of s. Then, the inverse transform

 

t

m_kj of m_kj*

 

s is obtained by partial fraction decompositions. Since the characteristics roots

4 , 3 , 2 , 1 ,i

s_i of  are all real and distinct, m_kj

 

t is the inverse transform of mkj

 

s

*

, which are

given by,

 

      







 









_





 

 

   4

1

4 , 1

2 1 3 4

3 2 1 11

k

t s

k i

i k i

k k

k

k _e k

s s

s g s

g s g s g t

m      

 



    











_





 

  

 4

1

4 , 1

4 3

2 1 12

k

t s

k i

i k i

k

k _e k

s s s g s g t

m    

 

  









_

 _{ } 

  4

1 4

, 1

2 1 1 3 13

k

t s

k i

i k i

k _e k

s s s g t

m   

 













_





 

 

  4

1

4 , 1

4 2

1 1 14

k

t s

k i

i k i

k e s s t

m     

 



    











_

 _{ } 

  

 4

1

4 , 1

4 3

2 21

k

t s

k i

i k i

k

k _e k

s s s g s g t

m    

 

      













_

 _{ } 

   4

1

4 , 1

4 3

2 22

k

t s

k i

i k i

k k

k _e k

s s s g s g s f t

IJSRR, 7(4) Oct. – Dec., 2018 Page 2091

 

   









_

 _{ }     4 1 4 , 1 2 1 3 23 k t s k i

i k i

k

k _e k

s s s g s f t

m  

 

 











_

 _{ }     4 1 4 , 1 4 2 1 24 k t s k i

i k i

k _e k

s s s

f t

m    

 

 









     4 1 4 , 1 3 2 31 k t s k i

i k i

k _ek

s s s g t m 

 

   









      4 1 4 , 1 3 32 k t s k i

i k i

k

k _ek

s s s g s f t m 

 

     











_

      4 1 4 , 1 1 1 3 33 k t s k i

i k i

k k

k _ek

s s s g s f s g t m 

 



   













 _{ }     4 1 4 , 1 4 1 1 34 k t s k i

i k i

k

k _ek

s s s g s f t

m   

 









      4 1 4 , 1 3 41 k t s k i

i k i

k e s s t m 

 

 









     4 1 4 , 1 2 42 k t s k i

i k i

k _ek

s s s f t m 

and

 

      







 







_

 _{ }      4 1 4 , 1 2 1 2 1 2 1 44 k t s k i

i k i

k k

k

k _ek

s s s g s g s g s f t

m    

From the matrix 2.20, we achieve,



 









 

 

 _{   } 4

1 * 1 1 1 j j s m s e s sI

e   (2.23)

and



 









 

 

 _{   } 4

1 * 4 2 1 j j s m s e s sI

e     (2.24)

Replacing 2.23 and 2.24 in 2.19, we get

 





 













_

 



                            ₄ 1 * 4 2 2 4 1 * 1 * 4 2 1 1 1 j j j j s m s s s m s s s        (2.25)

Utilizing equation 2.20 in 2.15, we have

 



   









_

 _{ }    4 1 4 , 1 1 1 43 k t s k i

i k i

k

k _e k

s s s g s f t

IJSRR, 7(4) Oct. – Dec., 2018 Page 2092

 

_{ }



   

 





 











                     

 1 1 2 4

* 4 2 1 3 1 * 0 1

1  _{        }

s s g s i s g s D s

m

 

s m

   

s s

s * 4 * 14 * 11

1   

     

   (2.26)

 

_{ }



 



   









      _{   }         

 1 2 4

* 4 *

1 1

1  _{     }

s f s s i s D s

m

 

s m

   

s s

s * 4 * 24 * 21

1   

     

   (2.27)

 

_{ }



 



 



   







                  

 1 1 4

* 4 3 2 * 2 1

1  _{   }

s g s f s s g s D s

m

 

s m

   

s s

s * 4 * 34 * 31

1   

     

   (2.28)

 

_{ }

 

     



 



      _{  }         

 s f s j s g s

s D

s 1 2

* 4 3 *

3 1

1  _{  }

m

 

s m

   

s s

s * 4 * 44 * 41

1   

     

   (2.29)

From matrix theory, the characteristic roots s_i,i1,2,3,4 of  provided are all real and

distinct. Explore s₀ 0, it can be obtained by partial fraction decompositions as





_{ }



      







 









s s



s s



n

 

s

s s g s g s g s g s m s s k k i

i k i k

k k k k * 11 4 0 4 , 1 2 2 1 3 4 3 2 1 * 11 2 1

1  

         



_

          





_{ }



    











s s



s s



n

 

s

s s g s g s m s s k k i

i k i k

k k * 21 4 0 4 , 1 2 4 3 2 * 21 2         

_



        





_{ }

 





s s



s s



n

 

s

s s g s m s s k k i

i k i k

k * 31 4 0 4 , 1 2 3 2 * 31 2      

_



     





_{ }







s s



s s



n

 

s

s s m s s k k i

i k i k

* 41 4 0 4 , 1 2 3 * 41 2       

_



     



  



    











s s



s s



n

 

s

s s g s g s m s k k i

i k i k

k k * 12 4 0 4 , 1 4 3 2 1 * 12         





 _{ }      





 

      













s s



s s



n

 

s

s s g s g s f s m s k k i

i k i k

k k k * 22 4 0 4 , 1 4 3 2 *

22 1 1

IJSRR, 7(4) Oct. – Dec., 2018 Page 2093





 

   





s s



s s



n

 

s

s s g s f s m s k k i

i k i k

k k * 32 4 0 4 , 1 3 * 32       





     





 

 





s s



s s



n

 

s

s s f s m s k k i

i k i k

k * 42 4 0 4 , 1 2 * 42      





     



  

 







s s



s s



n

 

s

s s g s m s k k i

i k i k

k * 13 4 0 4 , 1 2 1 3 1 * 13       





       





 

   







s s



s s



n

 

s

s s g s f s m s k k i

i k i k

k k * 23 4 0 4 , 1 2 1 3 * 23        





      





 



   



 





s s



s s



n

 

s

s s g s g s f s m s k k i

i k i k

k k k * 33 4 0 4 , 1 3 1 1 *

33 1 1

      





     





 



   









s s



s s



n

 

s

s s g s f s m s k k i

i k i k

k k * 43 4 0 4 , 1 1 1 * 43       





 _{ }    



  











s s



s s



n

 

s

s s m s k k i

i k i k

* 14 4 0 4 , 1 4 2 1 1 * 14         





 _{ }       





 

 









s s



s s



n

 

s

s s f s m s k k i

i k i k

k * 24 4 0 4 , 1 4 2 1 * 24        





 _{ }       and





 



      







 







s s



s s



n

 

s

s s g s g s g s f s m s k k i

i k i k

k k k k * 44 4 0 4 , 1 2 1 2 1 2 1 *

44 1 1

        





         

where nij

 

s 's

*

, denote the summation terms in the above expressions.

Applying these in equation 2.25 and after some algebraic manipulations, we will get

 





 





_

 



                    ₄ 1 * 4 2 2 2 4 1 * 1 2 2 2 * 4 2 1 2 j j j j s n s n s s      

which implies

 





 





 

1 * 4 2 2 2 * 1 2 2 2 * 4 2 1 2      _{   }            

 d s d s

s

s  

 



 (2.30)

where

 

 

, 1,4.

4 1 * * _



_  i s n s d j ji i





 



   











s s



s s



n

 

s

s s g s f s m s k k i

i k i k

IJSRR, 7(4) Oct. – Dec., 2018 Page 2094

The previous equation can be expressed as,

 

_



 

 



 





 



    

  

    

 

_{  } 

    

 

_{  } 

       

0

* 4 1 2 2 *

1 1 2 2 1

* 4

2 1

n

n n

n n

n

s d s

d s

s 

  

 



(2.31)

Taking inversion on equations 2.26-2.29 and doing some algebraic operations, we get

 



 



_

 



_





  



 t

o t

om u du m t u u du

t m

t 11 11 14 4

0   (2.32)

 



 



_

 



_





  



 t

o t

om u du m t u u du

t m

t 21 21 24 4

1   (2.33)

 



 



_

 



_





  



 t

o t

om u du m t u u du

t m

t 31 31 34 4

2   (2.34)

 



 



_

 



_





  



 t

o t

om u du m t u u du

t m

t _{41 41 44 4}

3   (2.35)

Therefore, equations 2.13 and 2.31-2.35 completely determine all system size probabilities.

3. STEADY STATE PROBABILITIES

This section deals with the structure of the steady state probabilities of the M/M/4 Queuing

model with heterogeneous servers and its disasters.

Theorem 3.1

: The steady state distribution of the queuing system M/M/4 with heterogeneous service states and catastrophes is obtained as follows:

i) For 0 and   , then















  

 



E b

D

C

 

   



4 2

2

4 _(3.1)



4



, 1,2,3,... 2

1

4 2

4     

     

_ b b n n

n

n   _(3.2)



1 2



4

 

4

1   

    

(3.3)

₁  1













 

 ₁₂ 



₄ 













₁ ₂



₄



₄





D

(3.4)



2



₄





₁





₁







₄



₄



2

1 _{         }



             



D (3.5)



3







2



₁



_{1 2}





₄



3

1 _{         }



            



D (3.6)





 



















               



  

 

      

0 1 4 1 1 2

1

IJSRR, 7(4) Oct. – Dec., 2018 Page 2095

where



  







   







   



 

    1  1 1 2  3  2  2  3

3

2 2

4

C



  



    



      

 _{1 2} 2 _{1 2 3}



 





    





 





      



 1 2 1 3 1

2 1 2

2 3 2 2

3 3

2 2

2         

  D





₁2₁₂





₁₂₃



₁₃2



  









  

 

 



  



 

  



 



  1 2 3  1 1 2   2    2

E



 







        



      

 _{1 2 1 1 2}

ii) For 0 and   , then











 

 



E D

C

  

 

2 4

4 2

(3.7)







2 4



, 1,2,3,... 2

1

4 2

4      

     

_ n n

n

n _     _(3.8)





 



















               



  

 

     

₀ 1 ₁ 2 _{4 1 1 2}

D

₁



₁₂



₄

 

₄ (3.9)



 







 









_{1 2 4 1 2 4 4}



1 2

1

  

    

   



                



D

(3.10)



















4 1 1 4 4



2

2 2

1

  

   

  

              



D (3.11)

₃  1



3











2 



₁



₁₂ 





₄





D (3.12)

where



  







   







   







    1  1 1 2  3  2  2 3

3

2 2

4

C



  



    



      

 _{1 2} 2 _{1 2 3}



 





    





 





      



 1 2 1 3 1

2 1 2

2 3 2 2

3 3

2 2

2         

 

D





₁2₁₂





₁₂₃



₁₃2



  









  

 

 



  









 



 



  1 2 3  1 1 2   2    2

IJSRR, 7(4) Oct. – Dec., 2018 Page 2096



 







        



      

 _{1 2 1 1 2}

Proof: For 0 and   , from equation 2.23, we obtain

 





 













_

 



 

        

   

 

         

 ₄

1 * 4 2

2 4

1 * 1 *

4

2 1

1 1

j j j

j

s m s

s

s m s

s s

   



 

Multiplying on both sides with s and taking limit as s0 to the above equation, we get















  

 



E b

D

C s

s Lt

s

 

   





4 2

) ( *

2 4

0 (3.13)

The solution 3.1 follows directly from 3.13, by using Tauberian theorem.

Taking Laplace transform of 2.13, we have





*( ), 1,2,3,...

) (

* 4

2 2

4      

     

 _ s s n

n n

n _ 



(3.14)

As before, multiplying 3.14 by s on both sides and talking limit ass0, we get





*( ), 1,2,3,... 2

1 )

(

* 2 2 ₄

0 4

0      

     

 

 s s Lt s s n

Lt

n n

s n

s   (3.15)

This yields 3.2, by applying Tauberian theorem again.

Similarly, the results 3.3 – 3.6 can be obtained from 2.24 – 2.27 respectively. For 0 and



 , the results 3.7 – 3.12 can be obtained directly by putting   in 3.1 – 3.6.

Remark:

It is observed that the steady-state probabilities of this queueing model exist if and only if 0



 or 0 and .

4. PERFORMANCE MEASURES

Few interesting performance measures, involving the mean number of customers in the

system, the probability of arriving customers joining the queue, and the mean number of busy servers

are to be analyzed.

4.1. The Mean Number of Customers in the System

Let N(t) be the number of customers in the system at time t. the average number of customers

IJSRR, 7(4) Oct. – Dec., 2018 Page 2097



            0 4 3 2

1( ) ( ) ( ) ( 4) ( )

)) ( ( n n t n t t t t N E

Utilizing 2.13, 2.31, 2.32 and 2.33, the above equation can be written as

 



_

 



_





  



 t

o t

om u du m t u u du

t m t N

E( ( )) ₂₁  ₂₁  _{24 4}



 





 





t





  



o t

om u du m t u u du

t

m₃₁  ₃₁  _{34 4}



 





 





t





  



o t

om u du m t u u du

t

m₄₁  ₄₁  _{44 4}

 













 





         

 n t n bt u

n du e u t u t I u n n t 0 4 1

4( ) ( 1)

4   (4.1)

where ₄

 

t is given in 2.29

Suppose  0, the mean number of customers in the system under steady-state is computed as

















2

2 4 2 4 2 4 4 2 ) (              b b b b N

E + (4.2)



 











 























                               4 3 1 2 1 1 2 1 2 4 4 2 1 3 2 1                                    D

if   

and









2

2 4 2 4 4 2 2 ) (                N

E + (4.3)



 











 























                               4 3 1 2 1 1 2 1 2 4 4 2 1 3 2 1                                    D

if  

where ₄is given in 3.1 for   and in 3.7 for  .

4.2. Probability of Arriving Customers Joining the Queue

The probability that an arriving customer is required to join the queue at time t is given by



       0

4( )

) 4 ) ( ( n n t t N

 













 





        

 t _{bt u}

n n n du e u t u t I u n t 0 4 1

4( )



 (4.4)

IJSRR, 7(4) Oct. – Dec., 2018 Page 2098





                     



   _{ }            if if b N n n , 4 2 , 4 2 ) 2 ( 2 4 2 4 0

4 (4.5)

4.3. The Number of Busy Servers

Let B(t) denote the number of busy servers at time t. The probability that the system has n

busy servers is given as,







_

_



                



   4 , ) ( 3 ) ( 3 , 2 , 1 , ) ( ) ( ) ( 0

4 t for x t N x for t x t N x t B n n x (4.6)

and the corresponding steady-state probability is obtained for  0 and   as



 











 























                                                          4 , 4 2 3 , 2 , 1 , 1 ) ( 2 4 4 3 1 2 1 1 2 1 2 4 4 2 1 3 2 x if b x if D x B                                         (4.7)

Comparing the above probability, it can be obtained directly, for  0 and   by substituting



 in 4.7. Furthermore, the mean number of busy servers at time t is given by



           0 4 3 2

1( ) ( ) ( ) 2 ( )

)) ( ( n n t t t t t B E

which can also be written as



            0 4 3 2

1( ) ( ) ( ) ( 4) ( )

)) ( ( n n t n t t t t N

E (4.8)

If  0, the corresponding steady-state solution is given as



























 











 



















                                                                      4 3 1 1 1 1 2 1 2 4 4 2 1 3 2 3 1 3 1 2 1 2 2 2 1 3 2 1 2 3 2 1 2 3 2 2 4 2 4 2 1 ) (                                                           D B

E _(4.9)

if  