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					                                         VOL. 1, NO. 1, November 2011                                                       ISSN XXXX-XXXX
                                          ARPN Journal of Science and Technology
                                                   ©2011-2012 ARPN Journals. All rights reserved.


                                                           http://www.ejournalofscience.org

    Plant Stress Gene Database: A collection of plant genes responding to stress
                                    condition
                                  Ratna Prabha1,2, Indira Ghosh3 , Dhananjaya P. Singh 4

1                                              2                                                         3
  Project Trainee, School of Information        Dean, Professor                                           Senior Scientist,
Technology,                                    School of Information Technology,                         National    Bureau    of  Agriculturally
JNU, New Delhi- 110067, India.                 JNU, New Delhi- 110067, India                             Important Microorganisms,
Email: ratna.bioinfo@gmail.com                                                                           Kushmaur,
2
  Current Address: Senior Research Fellow,                                                               Mau Nath Bhanjan -275101, U. P., India
National     Bureau    of    Agriculturally
Important Microorganisms,
Kushmaur,
Mau Nath Bhanjan -275101, U. P., India




                                                                ABSTRACT
Network traffic monitoring is an important way for network performance analysis and monitor. The present article
explores how to build the basic model of network traffic analysis based on Queuing Theory. In the present work, two
queuing models (M/M/1): ((C+1)/FCFS) and (M/M/2): ((C+1)/FCFS) have been applied to determine the forecast way for
the stable congestion rate of the network traffic.
Using this we can obtain the network traffic forecasting ways and the stable congestion rate formula. Combining the
general network traffic monitor parameters, we can realize the estimation and monitor process for the network traffic
rationally.

Keywords: Network traffic, Queuing Theory, stable congestion rate

      1. INTRODUCTION                                                              Adopting Queuing Theory to estimate the network
                                                                              traffic, it becomes the important ways of network
    Network traffic monitoring is an important way for                        performance prediction, analysis and estimation and,
network performance analysis and monitor. The present                         through this way, we can imitate the true network, it is
analysis seeks to explore how to build the basic model of                     useful and reliable for organizing, monitoring and
network traffic analysis based on Queuing Theory [1].                         defending the network.
Using this, we can obtain the network traffic forecasting
ways and the stable congestion rate formula, combining                              2. THE MATHEMATICAL MODEL OF THE
the general network traffic monitor parameters.                                     QUEUING THEORY
Consequently we can realize the estimation and monition
process for the network traffic rationally.                                        In network communication, from sending, transferring
    Queuing Theory, also called random service theory, is                     to receiving data and the proceeding of the data coding,
a branch of Operation Research in the field of Applied                        decoding and sending to the higher layer, in all these
Mathematics. It is a subject which analyze the random                         process, we can find a simple queuing model. According
regulation of queuing phenomenon, and builds up the                           to the Queuing Theory, this correspond procedure can be
mathematical model by analyzing the date of the network.                      abstracted as Queuing theory model [2], like fig. 1.
Through the prediction of the system, we can reveal the                       Considering this kind of simple data transmitting system
regulation about the queuing probability and choose the                       satisfies the queue model [3].
optimal method for the system.

                                                                 Nq                                                 Ts
                                              TN
         λ'
                                                                                                    TJ           TD           TC
                                              λ

                                              

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                                         VOL. 1, NO. 1, November 2011                                                    ISSN XXXX-XXXX
                                          ARPN Journal of Science and Technology
                                              ©2011-2012 ARPN Journals. All rights reserved.


                                                      http://www.ejournalofscience.org


                                   Figure 1: The abstract model of communication process




From the above fig. 1,                                                               In model M/M/1, the two M represent the
             : Sending rate of the sender.                               sending process of the sender and the receiving process
          TN : Transportation delay time.                                  of the receiver separately. They both follow the Markov
           : Arriving speed of the data packets                           Process [4], also keep to Poisson Distribution, while the
          Nq : Quantity of data packets stored in the                      number 1 stands for the channel.
buffer (temporary storage).                                                          Let N(t)=n be the length of the queue at the
           : Packets rate which have mistake in sending                   moment of t. So the probability of the queue whose
                                                                           length is n, be
          from receiver i.e., lost rate of the receiver.
           Ts : Service time of data packets in the server                                              Pn =Prob [N(t)= n]
where Ts=TJ+TD+TC                                                          In this model,
           TJ : Decoding time                                                                        n = Rate of arrival into the state n
          TD : Dispatching time
                                                                                                     µn =Rate of departure from the state n
          TC : Calculating time or evaluating time or
handling time.
                                                                           We have the transition rate diagram as follows

3. Model-1: The Queuing model with one
server (M/M/1):((C+1)/FCFS)




                                                Figure 2: State transition diagram

         The system of differential difference equation is                         d
d                                                                                 dt
   {Pn (t )}  n Pn (t )   n Pn (t )  n1 Pn1 (t )   n1 Pn1 (t )
dt                                                                      Pn (t )  Pn1 (t )  Pn1 (t )  (   ) Pn (t )               for
 , for n≥1                                                                                                         (1)
                                                                           n≥1

         And
                d
                   P0 (t )  0 P0 (t )  1 P1 (t )     for
                                                                                         And
                                                                                                  d
                                                                                                     P0 (t )  P0 (t )  P1 (t )        for
                dt
                                                                                                  dt
n=0                                                                                                                (2)
                                                                           n=0
In model M/M/1, we let
                                                                           Here, λ is considered as the arrival rate while μ as the
         n      And   n                                              service rate.
Where λ and µ are constants.                                               In the steady state equation
Then eqs.(1) and (2) reduces to                                                          Lt Pn (t )  Pn
                                                                                         t 

                                                                                                     d
                                                                                         and Lt         {Pn (t )}  0
                                                                                                t  dt


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                                                       VOL. 1, NO. 1, November 2011                                                    ISSN XXXX-XXXX
                                                       ARPN Journal of Science and Technology
                                                            ©2011-2012 ARPN Journals. All rights reserved.


                                                                     http://www.ejournalofscience.org
Hence, from eqs. (3) and (4) when t   we get                                                                                     
                                                                                                              Hence      L                              (7)
             0  Pn 1  Pn 1  (   ) Pn                      for   n≥1                                                 1 
                                                                                                                             
                                                                                                                                                   
                                                              (5)
                                                                                                        Also           L                Since,
         and 0  P0  P1
                                                                                                                             

                            
This implies     P1  ( ) P0
                                                                                         If N q denotes the average volume of the buffers data

From eq.(5) when n=1, we get                                                              packets then

             (   ) P1  P0  P2
                                                                                                         2
                                                                                          Nq  L                                                     (8)
Therefore,       P2  ( ) 2 P0                                                                          1 
                       
                                                                                                                            Also
                     
In general,    Pn  ( ) n P0                                                                           2
                                                                                          Nq 
                                                                                                     (   )
                                                       
         or, Pn         n P0 where                                                    Using the Little’s law we have
                                                       
                                                                                                                                Ts       and        
Here,        is called server utilization factor or traffic
                                                                                                        (9)
intensity.
                                                                                          Using eq.(9), eq.(8) reduces to
                  
We know,        P
                 n 0
                        n   1
                                                                                                        Nq 
                                                                                                                 2
                                                                                                                1 
         Also, Pn             n P0
                                                                                          This implies (1  Ts ) N q  (Ts ) 2
                                
Therefore,       Pn    n P0
                n 0            n 0
                                                                                                        Or,            ( Ts ) 2  Ts N q  N q  0 ,

                                                                                          (   ' )                                         (10)
         or,      1  P0                 n

                              n 0                                                        The above equation eq.(10) provides the relation
                                                                                          between following parameters
Consequently, P0  1   , where                       < 1                                       T s = Service time
Hence,                      Pn   n (1   ) ,              n=0, 1, 2,… .                              Sending rate
                                                                                                    N q  Quantity of data (6)
                                                                                                                            packets stored in the
                                                                                          buffer
                                                                                          If we know any two variables, it is easy to gain the
Suppose, L stands for the length of the queue under the
                                                                                          numerical value of the third one.
steady state condition. It includes the average volume of
                                                                                          So, these three variables are key parameters for
all the data packets which enter the processing module
                                                                                          measuring the performance of the transmission system.
and store in the buffer.

                                                                                        4.      QUEUING              THEORY             AND         THE
         L   nPn   n (1   )             n

                n 0                n 1
                                                                                          NETWORK TRAFFIC MONITOR
                                    
                 (1   ) n n                                                          4.1.Forecasting the network traffic using
                                 n 1
                                                                                                Queuing Theory
                                                                                               The network traffic is very common [5]. The system
                                                                                          will be in worse condition, when the traffic becomes

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                                        VOL. 1, NO. 1, November 2011                                                  ISSN XXXX-XXXX
                                         ARPN Journal of Science and Technology
                                             ©2011-2012 ARPN Journals. All rights reserved.


                                                     http://www.ejournalofscience.org
under extreme situation, in which leads to the network                    be lost. When the arriving time of group timeouts, the
congestion [6]. There are a great deal of research about                  group has to resend. Suppose, the group’s average
monitoring the congestion at present ,besides, the                                            1
documents which make use of Queuing Theory to                             waiting time is       . We identify Pi (t) to be the arrival
research the traffic rate appear more and more. For                                           
forecasting the traffic rate, we often test the data disposal             probability of the queue length for the routers group at
function of the router used in the network.                               the moment of t, supposing the queue length is i:
                                                                          P(t) = (P0 (t), P1 (t), . . . , Pn−1 (t) ), i = 0,1, . . . ,C+1.
    Considering a router’s arrival rate of data flow in
groups is  , and the average time which the routers use                       Then the queuing system of the router’s date groups
                                                                          satisfies simple Markov Process [7], according to
                           1                                              Markov Process, we can find the diversion strength of
to dispose each group is     , the buffer of the routers is
                                                                         matrix of model 1 as follow:
C, if a certain group arrives, the waiting length of the
queue in groups has already reached, so the group has to
                                0               0                                               0               
                                                                                                                     0
                                                                                                                    
                (     )                  0                                             0           0     
             0              (    2 )       2                                           0           0     
                                                                                                                    
             0       0                      (    3 )                                     0            0     
           Q                                                                                                       
             0       0           0                                                              0           0     
                                                                                                             
                                                                                                                    
             0       0           0               0                                           (    C )   C 
             0                                                                                                  
                     0           0               0                                                                  
4.2 Network Congestion Rate
                                                                          For k ≥ 1
         Network congestion rate is changing all the
time [8]. The instantaneous congestion rate and the
stable congestion rate are often used to analysis the                       Pk 1 (t  t ) = Prob { k number of data packets
network traffic in network monitor. The instantaneous
                                                                          present in the system at time t } Prob {
rate AC (t ) is the congestion rate at the moment of t. The
                                                                           no data packet arrival in time ( ∆t) }  Prob { no data
AC (t ) can be obtained by solving the system length of
                                                                          packet departure in time ∆t }
the queue’s probability distributing, which is called
 PC 1 (t ) .                                                              + Prob { (k-1) number of data packets present in the
                                                                          system at time t }     
Let, Pk (t ) (k=0,1,. . .,C+1) to be the arrival probability              Prob { 1 data packet arrival in time ( ∆t) }        
of the queue length for the routers group at the moment                   Prob { no data packet            departure in time ∆t }            +
of t by considering the queue length is k.
                                                                          Prob { (k+1) number of data packets present in the
        Then, the queuing system of the router’s date                     system at time t }     
groups satisfies simple Markov Process. According to
                                                                          Prob {no data packet arrival in time ( ∆t) }         
Markov Process, Pk (t ) satisfies the following system of
differential difference equations.                                        Prob { 1 data packet           departure in time ∆t }+…

Let,                                                                        Pk (t  t )  Pk (t ){1  k t  o(t )}{1  k t  o(t )}
                  Pk (t ) = Prob { k number of data
packets present in the system in time t }
           and Pk 1 (t ) = Prob { k number of data
                                                                            Pk 1 (t ){k 1t  o(t )}{1  k 1t  o(t )}
packets present in the system in time (t + ∆t) }

Case 1:

                                                                                                                                     4
                                                         VOL. 1, NO. 1, November 2011                                                                    ISSN XXXX-XXXX
                                                          ARPN Journal of Science and Technology
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                                                                          http://www.ejournalofscience.org
                       +
                                                                                                    Dividing both sides by t and taking limit as t  0 ,
Pk 1 (t ){1  k 1t  o(t )}{ k 1t  o(t )} +
                                                                                                    we obtain
o(t )
 Pk (t  t )  Pk (t )  (k  k ) Pk (t )t  Pk 1 (t )k 1t  Pk 1 (t )k 1t  o(t )   d
                                                                                                       {P0 (t )}  P0 (t )  P (t )
                                                                                                                                 1
                                                                                                    dt
Dividing both sides by t and taking limit as t  0                                                                          (since,    k    k          and    k   )
      d
        {Pk (t )}  (k   k ) Pk (t )  k 1 Pk 1 (t ) 
     dt                                                                                             Case 3:
                                 o(t )
 k 1 Pk 1 (t ) , since lim           0                                                          For k=C+1, we have                            (11)
                          t    t
                                                                                                          PC 1 (t  t ) = Prob { (C+1 ) no. of data packets
Here, in state k, data packets arrival is                 (  k )                                 present in the system at time (t+∆t )}

      i.e.    k    k                                                                             = Prob { C no. of data packets present in time t }                          
                                                                                                    prob {1 data packet arrival in time ∆t }                  
                                                                                                    Prob { no data packet departure in time ∆t }
Also, in state k, data packet departure is 
                                                                                                    + Prob { (C+1) no of data packets present in time t }                         
        i.e.   k                                                                                 Prob { no data packet departure in time ∆t }+…

                                                                                                     PC (t ){C t  o(t )}{1  C t  o(t )}  PC 1 (t ){1  C 1t  o(t )}  o(t )
Hence, eq.(11) reduces to
d
   {Pk (t )}  (  k   ) Pk (t )  {  (k  1) }Pk 1 (t )  Pk 1 (t )                      PC 1 (t  t )  PC 1 (t )  PC (t )C t   C 1 PC 1 (t )t  o(t )
dt
                                        (12)
where k=1,2,…,C
                                                                                                    Dividing both sides by                     t and taking limit as
Case 2:                                                                                             t  0 we get
                                                                                                                 d
For k=0, we have                                                                                                    {PC 1 (t )}  PC (t )C   C 1 PC 1 (t )
                                                                                                                 dt
 P0 (t+∆t) = Prob { no data packet present in the system
                                                                                                         d
at time (t+∆t) }                                                                                           {PC 1 (t )}  (  C ) PC (t )  PC 1 (t ) ,
                                                                                                         dt
      = Prob { no data packet present in time t } 
                                                                                                    since    C    C
         Prob { no data packet arrival in time ∆t }
      + Prob {one data packet present in time t } 
       Prob { no data packet arrival in time ∆t }                                                  By solving this differential equation system, we get the
       Prob { one data packet departure in time                                ∆t }                 instantaneous congestion rate

+…                                                                                                  A0(t) as


                                                                                                                                 
                                                                                                    A0 (t )  P1 (t )                  (1  e (   )t )
                                                                                                                              
 P (t ){1  0t  o(t )}  P1 (t ){1  1t  o(t )}{ 1t  o(t )}  o(t )
= 0
                                                                                                             The instantaneous congestion rate can not be
 P0 (t  t )  P0 (t )  0 P0 (t )  P (t ) 1  o(t )
                                          1
                                                                                                    used to measure the stable operating condition of the
                                                                                                    system, so we must obtain the stable congestion rate of
                                                                                                    the system. The so-called stable congestion rate means, it


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                                                     VOL. 1, NO. 1, November 2011                                                                   ISSN XXXX-XXXX
                                                      ARPN Journal of Science and Technology
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                                                                   http://www.ejournalofscience.org
will not change with the time changing, when the system                                 Under steady state condition, eqs.(15),(16) and (17)
works in a stable operating condition. The definition of
                                                                                        transform            to       following              balance               equations.
the stable congestion rate is
                                                                                         (  k   ) Pk (t )  {  (k  1) }Pk 1 (t )  Pk 1 (t )
                                  AC  lim AC (t )                                       for k=1, 2, 3, … ,C                                  (18)
                                            t  


           Considering,       P  lim P(t ) as the distributing
                                     t                                                 0  P0 (t )  P (t ) for k=0
                                                                                                           1
of the stable length of the queue and C as the buffer of
the router, the stable congestion rate can be obtained in                                                                                     (19)
two ways: firstly, we obtain the instantaneous congestion
rate, then find its limit. According to its definition, it can                           0  (  C ) PC (t )  PC 1 (t ) for k=C+1
be obtained with the distributing of the length of the                                                                                                           (20)
queue. Secondly, according to the Markov Process, we
know that the distributing of the stable length of queue                                The above system of steady state equations can be
can be obtained through system of steady state                                          written in matrix from as
equations.
                                                                                                                                                                 PQ  0
From eq.(12), eq.(13) and eq.(14), we have the system of                                                                                     C 1
differential difference equations as follows                                                                         and                     P 1
                                                                                                                                             i 0
                                                                                                                                                      i


d
   {Pk (t )}  (  k   ) Pk (t )  {  (k  1) }Pk 1 (t )  Pk 1 (t )          where P  ( P0 , P ,..., PC 1 )
                                                                                                          1
dt
                                   (15)                                                 and

for k=1, 2, 3, … ,C                                                                         
                                                                                           
                                                                                                                 0               0                       0           
                                                                                                                                                                       
                                                                                                                                                                        0
                                                                                              (     )                  0                       0           0
d                                                                                          0              (    2 )       2               0           0     
   {P0 (t )}  P0 (t )  P (t )
                             1                       for k=0                               
                                                                                           0       0                      (    3 )         0            0
                                                                                                                                                                       
                                                                                                                                                                       
dt                                                                                       Q                                                                           
                                                                                           0       0           0                                  0           0     
                                                                                                                                                               
                                                            (16)                                                                                                      
                                                                                           0       0           0               0               (    C )   C 
                                                                                           0                                                                      
d                                                                                                  0           0               0                                      
   {PC 1 (t )}  (  C ) PC (t )  PC 1 (t ) for k=C+1
dt                                                                                      For C= 0,
                                                            (17)                        From eq.(19), we have

According to some properties of Markov process, we                                                                               P0  P1                (21)
know that
                                                                                        Also, P0  P  1
                                                                                                    1                                                     (22)
     Pi (t )      (i=0,1,2,…,C+1)            satisfies       the     above
                                                                                        Solving (21) and (22) we get
differential equation.
                                                                                                                                              
                                                                                                                                P1 
Here, P(t )  [ P0 (t ), P (t ),..., PC1 (t )]
                          1                                                                                                             

                                                                                        Hence,
P0 (0)  1, P (0)  0, P2 (0)  0,..., PC 1 (0)  0
             1                                                                                               
P(0)  [ P0 (0), P1 (0),..., PC 1 (0)]                                                  A0  P1                              (23)
                                                                                                          

                                                                                        For C=1
For steady state equation,
                                                                                         0  P0  P1
                          dP (t )                                                                                                                                (24)
lim Pk (t )  Pk and lim k        0
t                  t   dt                                                            0  (     ) P1  P0  P2
                                                                                                                                                                 (25)



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0  (   ) P  P2
              1                                   (26)
                                                                              From eq.(29), we have

Also, P0  P  P2  1
                                                                                        (   )
            1

                                                                               P2                k
                                                                                           
From eq.(23), we get                                                          From eq.(31), we have
P0
     
         P1
              k                                                                       (  2 )
                                                                               P3                    P2
                                                                                          
Therefore, P0  k , P  k
                      1
                                                                                        (   )(   2 )
                                                                                                           k
From eq.(25), we have                                                                          2
                 
     P2  (          )k ,   since, P  k (27)                               From eq.(32), we have
                                    1

                                                                                                           2
                                                                               k
                                                                                     (   )  2   (   )   (   )(   2 )
Using eq.(26), we obtain
                                                                              Hence,
                               
                   k[          ]  1                                                                     (   )(   2 )
                                                                              A2  P2 
                                                                                                 (   )   (   )   (   )(   2 )
                                                                                                  2

                      
k 
           (   )   (   )                                             For C=3, yielding
From eq.(27) yields                                                            0  (     ) P1  P0  P2                (33)

                                                                               0  (  2   ) P2  (   ) P1  P3

                           
P2  (        )..                                                                                                             (34)
                   (   )   (   )
                                                                               0  (  3   ) P3  (  2 ) P2  P4

Hence,                                                                                                                         (35)

                                                                               0  P0  P1                                  (36)

                      (   )                                                0  (  3 ) P3  P4                          (37)
A1  P2 
                (   )   (   )                                        From eq.(36), we obtain

For C=2,                                                                       P0  k and P1  k
0  P0  P1                           (28)                                 From eq.(33)

0  (     ) P1  P0  P2 (29)                                                   (   )
                                                                               P2                k
                                                                                           
                                                                              From eq.(34)
0  (   ) P  (    2 ) P2  P3
                                                                                        (   )(   2 )k
              1

                                                  (30)                         P3 
                                                                                                2
0  (  2 ) P2  P3                            (31)
                                                                              From eq.(37), we have
Also, P0  P  P2  P3  1                                                                    (  3 )
            1                                     (32)                                 P4                 P3
                                                                                                      
From eq.(28), we obtain
              P0  k and P1  k

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                  (   )(   2 )(   3 )k                                        per unit time. The overall service rate can be obtained in
            
                               3                                                       two situations. If there are n numbers of data packets are
                                                                                        present in the system.
Also, P0  P  P2  P3  P4  1
            1


                                            3                                        Case-1
k 
        (   )   (   )   (   )(  2 )   (   )(  2 )(  3 ) For n < 2
        3                      2


                                                                                        There will be no queue. Therefore (2-n) server will
Hence ,                                                                                 remain idle and the combined service rate will be

                                      (   )(  2 )(  3 )                                      n  n ,        1 n  2
A3  P4 
             (   )   (   )   (   )(  2 )   (   )(  2 )(  3 )
             3                      2



                                                                                        Case-2

On the analogy of this, we conclude that, the stable                                     For n  2
                                                                                            Then, all the servers will be busy. So, maximum (n-
congestion rate is                                                                      2) (  C  1 ) number of data packets present in the
                                                                                        queue.

                                                                                       The combined service rate will be
AC  PC 1  1 
                   (    C ) AC 1  {  (C  1) }(1  AC 1 ) AC 2                             n  2 , n  2
, for C  2
                                                                                        Hence, combining Case-1 and Case-2, we have

5. THE              QUEUING                  MODEL               WITH                                      n      for all n  0

     ADDITIONAL ONE SERVER (M/M/2)                                                                         n  n      for 1  n  2

     : ((C+1)/FCFS)                                                                                        n  2 , n  2

                       In this model, number of servers or
                                                                                                          0  0 , n  0
channels is two and these are arranged in parallel. Here,                                                  1   , n  1
arrival distribution is Poisson distribution with mean rate
   per unit time. The service time is exponentional with
mean rate  per unit time. Each server is identical i.e.

each server gives identically service with mean rate 




                                                         Figure 3: State transition rate diagram

The steady state equations are
                    P0  P1           for n=0                                                                  (38)

                    (     ) P1  P0  2P2                 for n  1                                       (39)


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                {  (n  1) }Pn1  2Pn1  (  n ) Pn  2Pn for 1  n  C                           (40)

                (  C ) PC  2PC 1 for k  C  1                                                              (41)


The above system of steady state balance equations can be written in matrix form as
                                                     PQ  0
                                      C 1
                       and            P 1
                                      i 0
                                              i



where P  ( P0 , P ,..., PC 1 )
                  1

and

                                    0               0                                             0                0
                                                                                                                  
                 (     )                      0                                     0             0    
              0       2        (  2  2 )        2                                    0            0    
                                                                                                                  
              0       0           2            (  2  3 )                               0             0    
            Q                                                                                                    
              0       0           0                 2                                        0             0    
                                                                                                           
                                                                                                                  
              0       0            0                 0                                    (  2  C )   C 
              0                                                                              2           2    
                      0            0                 0                                                            
From eq.(49), we obtain                                                        Hence
                                                                                                            (   )(   2 )
     (   )                                                                   A2  P3 
P2           k        since, P0  k , P  k                                              4 (   )  (   )2   (   )(   2 )
                                                                                                   2

       2
                                          1



From eq.(51), we get

        (  2 )                                                              For C = 3
P3               P2
           2                                                                   P0  P1
                                                                                                                                (53)
            (   )(   2 )                                                 (     ) P1  P0  2P2
 P3                           k ,          Using the value
                  4 2                                                                                               (54)
                                                                                (  2  2 ) P2  (   ) P1  2P3
of P2
                                                                                                                                (55)
                                                                                (  3  2 ) P3  (  2 ) P2  2P4
Since, P0  P  P2  P3  1
             1
                                                                                                                                (56)

                (   )    (  2 )(    )                                  (  3 ) P3  2 P4
or, k[                                   ]  1
                  2              4 2
                                                                                                                               (57)

                                 4 2                                          From eq.(53), we have
k 
          4 2 (   )  (   )2   (   )(  2 )
                                                                                P0  k          and   P1  k

                                                                               From eq.(54)


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2P2  (     ) P1  P0                                                                                                      2
                                                                                          AC  PC 1  1 
                                                                                                      (  2  C ) AC 1  {  (C  1) }(1  AC 1 ) AC  2  2
              (     )       k                                                          for C  2
 P2                      k                                                           ,
                  2             2
                                                                                         6. CONCLUSION
             (   )
                     k                                                                           This research paper cites the analysis of the
               2                                                                        network traffic model through Queuing Theory. In the
From eq.(55), we get                                                                     present analysis, we describe that how we can make a
                                                                                         queuing model on the basis of queuing theory and
                                                                                         subsequently we derive the estimation after analyzing
2P3  (  2  2 ) P2  (   ) P1                                                   the network traffic through queuing theory models. In
          (  2  2 )(    )k                                                      the present work two queuing models (M/M/1):
 2P3                                (   )k                                       ((C+1)/FCFS) and (M/M/2):((C+1)/FCFS) have been
                     2                                                                  applied. These two models are used to determine the
                                                                                         forecast way for the stable congestion rate of the
                                  (  2  2 )                                         network traffic. Using the Queuing Theory models, it is
              (   )k [                       1]
                                       2                                                convenient and simple way for calculating and
                                                                                         monitoring the network traffic properly in the network
                                                                                         communication system. We can monitor the network
                 k (   )(   2 )                                                   efficiently, in the view of the normal, optimal and or
             
                         2                                                              even for the high overhead network management, by
                                                                                         monitoring and analyzing the network traffic rate.
              k (   )(   2 )
 P3                                                                                    Finally, we can say that network traffic rate can have an
                     4 2                                                                important role in the network communication system.
From eq.(57)
       (  3 )                                                                         REFERENCES
P4              P3
          2                                                                             [1] John N. Daigle, 2005, Queueing Theory with
       k (   )(   2 )(   3 )                                                       Applications to Packet Telecommunication, ISBN:
                                                                                            0-387-22857-8, Springer, Boston, USA.
                    8 3
                                                                                         [2] Vern Paxson, Sally Floyd, 1997, Why We Don’t
                                                                                             Know How To Simulate The Internet. In
Also, P0  P  P2  P3  P4  1
            1
                                                                                               Proceedings of the 1997 Winter Simulation
                                                                                               Conference, ed. S. Andradóttir, K. J. Healy, D. H.
                                                                                               Withers, and B. L. Nelson, USA:ACM.
                  (   )  (   )(  2 )  (   )(  2 )(  3 )
 k[(   )                                                              ] 1        [3] Ren Xiangcai, Xiong Qibang, 2002, “An Application
                    2             4 2                     8 3                              of Mobile Agent for IP Network Traffic
                                                                                              Management,” Computer Engineering, 2002-11.
                                         8 3                                            [4] Li Da-Qi, Shen Jun-Yi, and Zhou Jiang-liang, 2007,
k  3                                                                                       “Queuing Theory Supervising K-Means Clustering
    8 (   )  4 2 (   )  2(   )(  2 )   (   )(  2 )(  3 )
                                                                                             Algorithm and ITS Appllication in Optimized
                                                                                             Design of TTC Network,” Journal of Astronautics,
                                                                                             28 (3), pp. 752-756.
Hence,                                                                                       [5] Wang Pei-Fa, Zhang Shi-wei, Li Jun, 2005, “The
                                     (   )(  2 )(  3 )                                 Application and Achievement of SVG in Network
A3  P4  3                                                                                      Netflow Monitor Field,” Chinese Journal of
         8 (   )  4 2 (   )  2 (   )(  2 )   (   )(  2 )(  3 )     Semiconductors, 22(4), pp. 162-165.
                                                                                         [6] Wang Ting, Wang Yu, 2007, “Survey on a Queue
                                                                                             Theory Based Handover Scheme for UAVS
On the analogy of this, we conclude that, the stable                                         Communication Network,” Chinese Journal of
congestion rate is                                                                           Sensors and Actuators, 2007, 04.
                                                                                         [7] Gunther, N., 1998, The Practical Performance
                                                                                         Analyst, McGraw-Hill Inc., New York.


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                                  VOL. 1, NO. 1, November 2011                                    ISSN XXXX-XXXX
                                   ARPN Journal of Science and Technology
                                       ©2011-2012 ARPN Journals. All rights reserved.


                                               http://www.ejournalofscience.org
[8] Han Jing, Guo Fang, Shi Jin-Hua, 2007, “Research                based on human immune algorithm,” Microcomputer
on the traffic monitoring of the distributed network                Information, 2007-18.




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