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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 

                Monitoring of Network Traffic based on Queuing Theory
                                                            S. Saha Ray, P. Sahoo
                                                      National Institute of Technology 
                                                        Department of Mathematics 
                                                          Rourkela‐769008, India 
                                     Email: santanusaharay@yahoo.com, saharays@nitrkl.ac.in



                                                            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

                                           

                                   Figure 1: The abstract model of communication process




                                                                                                                           1
                                          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 
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                    Here, λ is considered as the arrival rate while μ as the
d                                                                             service rate.
   {Pn (t )}   n Pn (t )   n Pn (t )   n 1 Pn 1 (t )   n 1 Pn 1 (t )
dt                                                                            In the steady state equation
 , for n≥1                                                                                                             (1)
                                                                                            Lt Pn (t )  Pn
                                                                                            t 
                 d
         And        P0 (t )  0 P0 (t )  1 P1 (t )      for
                                                                                                         d
                 dt                                                                         and Lt          {Pn (t )}  0
                                                                                                   t   dt
n=0                                                                                                                     (2)
In model M/M/1, we let                                                        Hence, from eqs. (3) and (4) when t   we get

         n       And   n                                                              0  Pn 1  Pn 1  (   ) Pn              for   n≥1

Where λ and µ are constants.                                                                                                         (5)
Then eqs.(1) and (2) reduces to                                                             and 0  P0  P1

           d                                                                                              
          dt                                                                  This implies         P1  ( ) P0
                                                                                                          
Pn (t )  Pn1 (t )  Pn1 (t )  (   ) Pn (t )        for
                                                                              From eq.(5) when n=1, we get
n≥1                                                                                                                    (3)
                                                                                            (   ) P1  P0  P2
                 d
         And        P0 (t )  P0 (t )  P1 (t )          for
                                                                                                          
                 dt                                                           Therefore,           P2  ( ) 2 P0
n=0                                                                                                                   (4)


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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 

                                                                                                                              Also
In general,     Pn  ( ) n P0
                                                                                                          2
                                                                                              Nq 
                                                                                                        (   )
          or, Pn            n P0         where     
                                                                                             Using the Little’s law we have

Here,     is called server utilization factor or traffic                                                                            Ts     and       
intensity.                                                                                                  (9)
                                                                                             Using eq.(9), eq.(8) reduces to
We know,         P
                 n 0
                        n    1
                                                                                                                    2
                                                                                                            Nq 
          Also, Pn              n P0                                                                             1 

                                                                                            This implies (1   Ts ) N q  (  Ts ) 2
Therefore,      P  
                n0
                        n
                                 n0
                                            n
                                                P0
                                                                                                            Or,           (  T s ) 2   T s N q  N q  0 ,
                                 
          or,     1  P0   n                                                                (   ' )                                        (10)
                                n 0
                                                                                              The above equation eq.(10) provides the relation
Consequently, P0  1   , where                        < 1                                  between following parameters
                                                                                                      Ts = Service time
Hence,                      Pn   n (1   ) ,                  n=0, 1, 2,… .                              Sending rate
                                                                                                        N q  Quantity of data (6)
                                                                                                                                packets stored in the
                                                                                              buffer
Suppose, L stands for the length of the queue under the                                       If we know any two variables, it is easy to gain the
steady state condition. It includes the average volume of                                     numerical value of the third one.
all the data packets which enter the processing module                                        So, these three variables are key parameters for
and store in the buffer.                                                                      measuring the performance of the transmission system.
                                      
         L   nPn   n n (1   )
                                                                                              4.      QUEUING             THEORY            AND         THE
                 n 0                n 1                                                     NETWORK TRAFFIC MONITOR
                                     
                  (1   ) n                 n
                                                                                              4.1.Forecasting the network traffic using
                                  n 1
                                                                                                    Queuing Theory
                                                      
                 Hence                      L                                (7)                  The network traffic is very common [5]. The system
                                                     1                                      will be in worse condition, when the traffic becomes
                                                                                              under extreme situation, in which leads to the network
                                                 
          Also                    L                         Since,                       congestion [6]. There are a great deal of research about
                                                                                           monitoring the congestion at present ,besides, the
                                                                                              documents which make use of Queuing Theory to
                                                                                              research the traffic rate appear more and more. For
                                                                                              forecasting the traffic rate, we often test the data disposal
If N q denotes the average volume of the buffers data
                                                                                              function of the router used in the network.
packets then
                                                                                                  Considering a router’s arrival rate of data flow in
                                                                                              groups is  , and the average time which the routers use
                         2                                                                                               1
Nq  L                                                                    (8)              to dispose each group is        , the buffer of the routers is
                        1                                                                                               
                                                                                              C, if a certain group arrives, the waiting length of the
                                                                                              queue in groups has already reached, so the group has to
                                                                                              be lost. When the arriving time of group timeouts, the
                                                                                              group has to resend. Suppose, the group’s average

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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 

                   1                                                                 Then the queuing system of the router’s date groups
waiting time is        . We identify Pi (t) to be the arrival                   satisfies simple Markov Process [7], according to
                                                                               Markov Process, we can find the diversion strength of
probability of the queue length for the routers group at                        matrix of model 1 as follow:
the moment of t, supposing the queue length is i:
P(t) = (P0 (t), P1 (t), . . . , Pn−1 (t) ), i = 0,1, . . . ,C+1.

                                 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                                                     no data packet arrival in time ( ∆t) }        Prob { no data
                                                                                packet departure in time ∆t }
         Network congestion rate is changing all the
time [8]. The instantaneous congestion rate and the                             + Prob { (k-1) number of data packets present in the
stable congestion rate are often used to analysis the                           system at time t }      
network traffic in network monitor. The instantaneous
rate AC (t ) is the congestion rate at the moment of t. The
                                                                                Prob { 1 data packet arrival in time ( ∆t) }       
                                                                                Prob { no data packet         departure in time ∆t }            +
AC (t ) can be obtained by solving the system length of
the queue’s probability distributing, which is called                           Prob { (k+1) number of data packets present in the
 PC 1 (t ) .                                                                   system at time t }      
                                                                                Prob {no data packet arrival in time ( ∆t) }       
Let, Pk (t ) (k=0,1,. . .,C+1) to be the arrival probability                    Prob { 1 data packet        departure in time ∆t }+…
of the queue length for the routers group at the moment
of t by considering the queue length is k.
                                                                                 Pk (t  t )  Pk (t ){1  k t  o(t )}{1  k t  o(t )}
        Then, the queuing system of the router’s date
groups satisfies simple Markov Process. According to
Markov Process, Pk (t ) satisfies the following system of
                                                                                 Pk 1 (t ){k 1t  o(t )}{1  k 1t  o(t )}
differential difference equations.
                                                                                                   + 
Let,
                    Pk (t ) = Prob { k number of data                           Pk 1 (t ){1  k 1t  o(t )}{k 1t  o(t )} + 
packets present in the system in time t }                                       o(t )
           and Pk 1 (t ) = Prob { k number of data
packets present in the system in time (t + ∆t) }                                P(t t)P(t) (k k)P(t)t P1(t)k1t P1(t)k1t o(t)
                                                                                  k        k              k       k             k


Case 1:
                                                                                Dividing both sides by t and taking limit as t  0
For k ≥ 1
                                                                                      d
                                                                                        {Pk (t )}   ( k   k ) Pk (t )   k 1 Pk 1 (t ) 
                                                                                     dt
 Pk 1 (t  t ) = Prob { k number of data packets
                                                                                                                 o(t )
                                                                                 k 1 Pk 1 (t ) , since lim           0
present in the system at time t }     Prob {                                                             t    t




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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 

Here, in state k, data packets arrival is (  k )                                 PC 1 (t  t ) = Prob { (C+1 ) no. of data packets
      i.e.   k    k                                                      present in the system at time (t+∆t )}
                                                                                = Prob { C no. of data packets present in time t }               
Also, in state k, data packet departure is                                   prob {1 data packet arrival in time ∆t }              
                                                                              Prob { no data packet departure in time ∆t }
       i.e.   k  
                                                                              + Prob { (C+1) no of data packets present in time t }              
                                                                              Prob { no data packet departure in time ∆t }+…
Hence, eq.(11) reduces to
                                                                              P (t){ C o()}{C o()}P1(t){ C1 o()}o()
                                                                                      t      t 1    t    t C 1            t    t     t
d                                                                               C
  {P (t)} (  k  )P (t) { (k 1)}P1(t)  P1(t)
    k                    k                  k         k
dt
                                 (12)                                          P 1 (t  t)  P 1 (t)  P (t)C t  C1P 1 (t)t  o(t)
                                                                                 C               C          C                C

where k=1,2,…,C

Case 2:                                                                       Dividing both sides by                t and taking limit as
                                                                              t  0 we get
For k=0, we have
                                                                                            d
 P0 (t+∆t) = Prob { no data packet present in the system                                       {PC 1 (t )}  PC (t )C   C 1 PC 1 (t )
                                                                                            dt
at time (t+∆t) }
                                                                                   d
      = Prob { no data packet present in time t }                                   {PC 1 (t )}  (  C ) PC (t )  PC 1 (t ) ,
                                                                                   dt
        Prob { no data packet arrival in time ∆t }
        + Prob {one data packet present in time t }                          since    C    C
         Prob { no data packet  arrival  in time ∆t } 
       Prob { one data packet departure in time             ∆t }              By solving this differential equation system, we get the
+…                                                                            instantaneous congestion rate
                                                                              A0(t) as


 P(t){ 0t o( )}P(t){  o( )}{1t o( )}o(t)
     1           t 1 1 1t          t          t                                                        
= 0
                                                                               A0 (t )  P1 (t )            (1  e (   ) t )
                                                                                                      
 P0 (t  t )  P0 (t )  0 P0 (t )  P1 (t ) 1  o(t )
                                                                                        The instantaneous congestion rate can not be
                                                                              used to measure the stable operating condition of the
                                                                              system, so we must obtain the stable congestion rate of
Dividing both sides by t and taking limit as t  0 ,                        the system. The so-called stable congestion rate means, it
                                                                              will not change with the time changing, when the system
we obtain                                                                     works in a stable operating condition. The definition of
                                                                              the stable congestion rate is

d                                                                                                             AC  lim AC (t )
   {P0 (t )}  P0 (t )  P (t )
                             1
                                                                                                                       t  
                                                                                                                       (13)
dt
                                                                                            Considering, P  lim P (t ) as the distributing
                       (since,   k    k   and   k   )                                                   t 
                                                                              of the stable length of the queue and C as the buffer of
                                                                              the router, the stable congestion rate can be obtained in
Case 3:                                                                       two ways: firstly, we obtain the instantaneous congestion
                                                                              rate, then find its limit. According to its definition, it can
For k=C+1, we have                                                            be obtained with the distributing of the length of the
                                                                              queue. Secondly, according to the Markov Process, we
                                                                              know that the distributing of the stable length of queue


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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 
can be obtained through system of steady state                                                                                                                          PQ  0
equations.
                                                                                                                                                C 1

From eq.(12), eq.(13) and eq.(14), we have the system of
                                                                                                                        and                     P 1
                                                                                                                                                i 0
                                                                                                                                                         i

differential difference equations as follows
                                                                                      where        P  ( P0 , P1 ,..., PC 1 )
d                                                                                     and
  {P (t)} (  k  )P (t) {  (k 1)}P 1(t)  P 1(t)
    k                   k                   k          k
dt
                                                                                                           0               0                                 0          0   
                                  (15)                                                  
                                                                                           (     )                  0                                  0          0
                                                                                                                                                                                 
                                                                                                                                                                                
                                                                                        0              (    2 )       2                                0          0   
for k=1, 2, 3, … ,C                                                                     
                                                                                                                                                  
                                                                                                                                                                                 
                                                                                        0       0                      (    3 )                          0          0   
                                                                                      Q                                                                                        
d                                                                                       0       0           0                                                  0          0   
   {P0 (t )}  P0 (t )  P1 (t )            for k=0                                                                                                                   
dt                                                                                      
                                                                                        0       0           0               0                   
                                                                                                                                                                                 
                                                                                                                                                          (     C )   C 
                                                                                        0       0           0               0                                               
                                                                                                                                                                                
                                                         (16)
d                                                                                     For C= 0,
   {PC 1 (t )}  (  C ) PC (t )  PC 1 (t ) for k=C+1                           From eq.(19), we have
dt
                                                         (17)
                                                                                                                                      P0  P1                  (21)

According to some properties of Markov process, we                                    Also,      P0  P1  1                                                 (22)
know that
                                                                                      Solving (21) and (22) we get 
       Pi (t )    (i=0,1,2,…,C+1)          satisfies     the      above                                                                          
                                                                                                                                       P                     
differential equation.
                                                                                                                                        1
                                                                                                                                                
Here,    P (t )  [ P0 (t ), P1 (t ),..., PC1 (t )]
                                                                                      Hence,
                                                                                                               
                                                                                       A0  P1                                     (23)
P0 (0)  1, P (0)  0, P2 (0)  0,..., PC 1 (0)  0
             1                                                                                             
P(0)  [ P0 (0), P1 (0),..., PC 1 (0)]
                                                                                      For C=1
                                                                                      0  P0  P1                                                                  (24)
For steady state equation,
                                  dPk (t )                                            0  (     ) P1  P0  P2                                                (25)
lim Pk (t )  Pk and lim                   0
t                          t    dt                                                0  (   ) P1  P2                                                           (26)
Under steady state condition, eqs.(15),(16) and (17)
transform         to      following         balance         equations.                Also,      P0  P1  P2  1

(  k   ) Pk (t )  {  (k  1) }Pk 1 (t )  Pk 1 (t )
for k=1, 2, 3, … ,C                           (18)                                    From eq.(23), we get
                                                                                       P0         P
                                                                                                  1
                                                                                                       k
0  P0 (t )  P1 (t ) for k=0                                                                 
                                              (19)
                                                                                      Therefore,         P0  k , P1  k

0  (  C ) PC (t )  PC 1 (t ) for k=C+1                                         From eq.(25), we have
                                                                                                            
                                                         (20)                                  P2  (           )k ,              since, P  k (27)
The above system of steady state equations can be
                                                                                                                                          1



written in matrix from as

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


From eq.(27) yields 
                                                                                                For C=3, yielding
                                                                                                0  (     ) P1  P0  P2                   (33)
                              
P2  (           )..                                                                           0  (  2   ) P2  (   ) P1  P3
                      (   )   (   )
                                                                                                                                                   (34)
Hence,
                                                                                                0  (  3   ) P3  (  2 ) P2  P4
                                                                                                                                                   (35)
                             (   )                                                          0  P0  P1                                     (36)
 A1  P2 
                       (   )   (   )
                                                                                                0  (  3 ) P3  P4                             (37)
For C=2,
                                                                                                From eq.(36), we obtain
0  P0  P1                                           (28)
                                                                                                P0  k and P1  k
0  (     ) P1  P0  P2 (29)
                                                                                                From eq.(33)
                                                                                                         (   )
0  (   ) P1  (    2 ) P2  P3                                                        P2                k
                                                                                                            
                                                                   (30)                         From eq.(34)
0  (  2 ) P2  P3                                             (31)                                  (   )(  2 )k
                                                                                                P3 
Also,      P0  P1  P2  P3  1                                   (32)                                          2
                                                                                                From eq.(37), we have
From eq.(28), we obtain
                                                                                                                   (  3 )
                P0  k and P1   k                                                                    P4                    P3
                                                                                                                          
From eq.(29), we have
                                                                                                                    (   )(  2 )(  3 )k
                                                                                                              
                                                                                                                                3
            (   )
P2                   k                                                                         Also,   P0  P1  P2  P3  P4  1
               
From eq.(31), we have
                                                                                                                                3
                                                                                                k 
           (  2 )                                                                                    3()()2 ()(2)()(2)(3)
P3                          P2
                  
           (   )(  2 )                                                                   Hence ,
                             k
                  2                                                                                                                    
                                                                                                                             ()( 2)( 3)
                                                                                                A P 
                                                                                                                                                        
                                                                                                               ()() ()(2)()( 2)( 3)
                                                                                                 3  4          3                    2
From eq.(32), we have




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On the analogy of this, we conclude that, the stable                          There will be no queue. Therefore (2-n) server will
                                                                              remain idle and the combined service rate will be
congestion rate is
                                                                                              n  n    ,     1 n  2
                                       
AC  P 1 1
                (  C )AC1 { (C1)}(  AC1)AC2 
      C
                                            1                                 Case-2
, for C  2
                                                                               For n  2
                                                                                   Then, all the servers will be busy. So, maximum (n-
                                                                              2) (  C  1 ) number of data packets present in the
5. THE           QUEUING           MODEL               WITH                   queue.
    ADDITIONAL ONE SERVER (M/M/2)
                                                                              The combined service rate will be
    : ((C+1)/FCFS)                                                                             n  2 ,   n  2

                   In this model, number of servers or                        Hence, combining Case-1 and Case-2, we have
channels is two and these are arranged in parallel. Here,                                         n       for all n  0
arrival distribution is Poisson distribution with mean rate
                                                                                                  n  n      for 1  n  2
 per unit time. The service time is exponentional with
mean rate  per unit time. Each server is identical i.e.                                          n  2 ,   n  2
each server gives identically service with mean rate                                            0  0 ,    n  0
per unit time. The overall service rate can be obtained in                                        1   , n  1
two situations. If there are n numbers of data packets are
present in the system.


Case-1
For n < 2




                                                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)

                 {  (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

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                                           C 1
                        and                P 1
                                            i 0
                                                    i



where     P  ( P0 , P1 ,..., PC 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       0            0                 0                                             2            2     
                                                                                                                            
From eq.(49), we obtain                                                               P0  P1
                                                                                                                                    (53)
     (   )
P2           k           since,       P0  k , P  k                              (     ) P1  P0  2 P2
       2
                                                   1
                                                                                                                            (54)
                                                                                      (  2  2 ) P2  (   ) P1  2P3
From eq.(51), we get
                                                                                                                                    (55)
         (  2 )                                                                    (  3  2 ) P3  (  2 ) P2  2P4
P3                P2                                                                                                               (56)
            2
                                                                                      (  3 ) P3  2P4
             (   )(  2 )
 P3                           k ,                Using the value
                                                                                                                                   (57)
                   4 2
of P2                                                                                 From eq.(53), we have

Since,    P0  P1  P2  P3  1                                                       P0  k          and   P1  k

                   (   )    (  2 )(   )                                      From eq.(54)
or, k[                                     ]  1
                     2              4 2                                             2P2  (     ) P1  P0

                                   4 2                                                             (     )      k
k                                                                                    P2                      k 
           4  2 (   )  (   ) 2   (   )(  2 )                                          2            2

                                                                                                    (   )
Hence                                                                                                       k
                           (   )(  2 )                                                         2
A2  P3                                                                              From eq.(55), we get
              4 (  )  (   )2  (   )(  2 )
                  2


                                                                                      2P3  (  2  2 ) P2  (   ) P1
                                                                                                (  2  2 )(   )k
                                                                                       2P3                                (   )k
                                                                                                           2
For C = 3


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                                         ARPN Journal of Science and Technology
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                            (  2  2 )                                 the network traffic through queuing theory models. In
           (    )k [                   1]                            the present work two queuing models (M/M/1):
                                 2                                        ((C+1)/FCFS) and (M/M/2):((C+1)/FCFS) have been
                                                                           applied. These two models are used to determine the
               k (   )(  2 )                                        forecast way for the stable congestion rate of the
                                                                          network traffic. Using the Queuing Theory models, it is
                       2                                                  convenient and simple way for calculating and
                                                                           monitoring the network traffic properly in the network
           k (   )(  2 )                                            communication system. We can monitor the network
 P3                                                                      efficiently, in the view of the normal, optimal and or
                  4 2
                                                                           even for the high overhead network management, by
From eq.(57)                                                               monitoring and analyzing the network traffic rate.
      (  3 )
P4             P3
         2                                                                Finally, we can say that network traffic rate can have an
                                                                           important role in the network communication system.
      k (   )(  2 )(  3 )
    
                   8 3                                                    REFERENCES

Also,    P0  P1  P2  P3  P4  1                                        [1] John N. Daigle, 2005, Queueing Theory with
                                                                               Applications to Packet Telecommunication, ISBN:
                                                                               0-387-22857-8, Springer, Boston, USA.
                                             
               () ()( 2) ()( 2)( 3)                      [2] Vern Paxson, Sally Floyd, 1997, Why We Don’t
   
k[( )                                          ] 1                      Know How To Simulate The Internet. In
                 2       42
                                          83
                                                                                 Proceedings of the 1997 Winter Simulation
                                                                                 Conference, ed. S. Andradóttir, K. J. Healy, D. H.
                         83                                                     Withers, and B. L. Nelson, USA:ACM.
 3
 k
                                         
   8 ()4()2)( 2 )()( 2 )( 3 )
              2
                       (                                                   [3] Ren Xiangcai, Xiong Qibang, 2002, “An Application
                                                                                of Mobile Agent for IP Network Traffic
                                                                                Management,” Computer Engineering, 2002-11.
                                                            [4] Li Da-Qi, Shen Jun-Yi, and Zhou Jiang-liang, 2007,
Hence,                                                          “Queuing Theory Supervising K-Means Clustering
                                   
                       ()( 2)( 3 )                       Algorithm and ITS Appllication in Optimized
A P  3
 3 4                                                            Design of TTC Network,” Journal of Astronautics,
                                               
      8 () 4() 2)( 2 ) ()( 2 )( 3 )
                  2
                             (
                                                                28 (3), pp. 752-756.
                                                                           [5] Wang Pei-Fa, Zhang Shi-wei, Li Jun, 2005, “The
On the analogy of this, we conclude that, the stable                           Application and Achievement of SVG in Network
congestion rate is                                                             Netflow Monitor Field,” Chinese Journal of
                              2                                               Semiconductors, 22(4), pp. 162-165.
A  P 1 1
 C   C                                                                     [6] Wang Ting, Wang Yu, 2007, “Survey on a Queue
            (2C)A 1 {(C1}(  A 1)A 2 2
                      C           ) 1 C C
                                                                               Theory Based Handover Scheme for UAVS
,   for C  2                                                                  Communication Network,” Chinese Journal of
                                                                               Sensors and Actuators, 2007, 04.
6. CONCLUSION
                                                                           [7] Gunther, N., 1998, The Practical Performance
                                                                           Analyst, McGraw-Hill Inc., New York.
         This research paper cites the analysis of the
network traffic model through Queuing Theory. In the                       [8] Han Jing, Guo Fang, Shi Jin-Hua, 2007, “Research
present analysis, we describe that how we can make a                       on the traffic monitoring of the distributed network
queuing model on the basis of queuing theory and                           based on human immune algorithm,” Microcomputer
subsequently we derive the estimation after analyzing                      Information, 2007-18.




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