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29th Paper11-09-51_ 23-10-11_ 0931 0951 0 Sunam11 295

VIEWS: 1 PAGES: 19

									      The Monitoring of The 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
Network traffic monitoring is an important way for network performance analysis and monitor.
The present analysis seeks to explore how to build the basic model of network traffic analysis
based on Queuing Theory [1]. Using this, we can obtain the network traffic forecasting ways and
the stable congestion rate formula, combining the general network traffic monitor parameters.
Consequently we can realize the estimation and monition process for the network traffic
rationally.
Queuing Theory, also called random service theory, is a branch of Operation Research in the
field of Applied Mathematics. It is a subject which analyze the random regulation of queuing
phenomenon, and builds up the mathematical model by analyzing the date of the network.
Through the prediction of the system, we can reveal the regulation about the queuing probability
and choose the optimal method for the system.
Adopting Queuing Theory to estimate the network traffic, it becomes the important ways of
network performance prediction, analysis and estimation and, through this way, we can imitate
the true network, it is useful and reliable for organizing, monitoring and defending the network.

2. The mathematical model of the queuing theory

In network communication, from sending, transferring to receiving data and the proceeding of
the data coding, decoding and sending to the higher layer, in all these process, we can find a
simple queuing model. According to the Queuing Theory, this correspond procedure can be
abstracted as Queuing theory model [2], like fig. 1. Considering this kind of simple data
transmitting system satisfies the queue model [3].


                                               1
                                           Nq                              Ts
                              TN
     λ'
                                                               TJ        TD        TC
                               λ

                               


                  Figure 1 The abstract model of communication process


From the above fig. 1,
         : Sending rate of the sender.
       TN : Transportation delay time.
        : Arriving speed of the data packets
       Nq : Quantity of data packets stored in the buffer (temporary storage).
        : Packets rate which have mistake in sending from receiver i.e., lost rate of the
       receiver.
        Ts : Service time of data packets in the server
where Ts=TJ+TD+TC
        TJ : Decoding time
       TD : Dispatching time
       TC : Calculating time or evaluating time or handling time.

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

In model M/M/1, the two M represent the sending process of the sender and the receiving
process of the receiver separately. They both follow the Markov Process [4], also keep to
Poisson Distribution, while the number 1 stands for the channel.

Let N(t)=n be the length of the queue at the moment of t. So the probability of the queue whose
length is n, be
                       Pn =Prob [N(t)= n]
In this model,
                       n = Rate of arrival into the state n
                       µn =Rate of departure from the state n

We have the transition rate diagram as follows



                                                 2
                                          Figure 2 State transition diagram

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

        And
                 d
                    P0 (t )  0 P0 (t )  1 P1 (t ) for n=0                        (2)
                 dt
In model M/M/1, we let
          n   And  n  
Where λ and µ are constants.
Then eqs.(1) and (2) reduces to
         d
            Pn (t )  Pn1 (t )  Pn1 (t )  (   ) Pn (t ) for n≥1              (3)
         dt

        And
                 d
                    P0 (t )  P0 (t )  P1 (t ) for n=0                            (4)
                 dt
Here, λ is considered as the arrival rate while μ as the service rate.
In the steady state equation
         Lt Pn (t )  Pn
         t 

                       d
        and Lt            {Pn (t )}  0
                t    dt
Hence, from eqs. (3) and (4) when t   we get
         0  Pn 1  Pn 1  (   ) Pn     for n≥1                                  (5)

        and 0  P0  P1




                                                           3
                            
This implies P1  ( ) P0
                            
From eq.(5) when n=1, we get
         (   ) P1  P0  P2

                               
Therefore,       P2  ( ) 2 P0
                               
                           
In general, Pn  ( ) n P0
                           
                                                       
         or, Pn   n P0 where  
                                                       
Here,  is called server utilization factor or traffic intensity.
                 
We know,        P
                n 0
                       n    1

         Also, Pn   n P0
                                   
Therefore,      P  
                n 0
                       n
                                   n 0
                                          n
                                              P0

                               
         or,    1  P0   n
                            n 0


Consequently, P0  1   , where  < 1

Hence,                     Pn   n (1   ) ,             n=0, 1, 2,… .          (6)


Suppose, L stands for the length of the queue under the steady state condition. It includes the
average volume of all the data packets which enter the processing module and store in the buffer.

                                   
         L   nPn   n n (1   )
               n 0                n 1
                                                  
                             (1   ) n n
                                               n 1


                                                   
                Hence                   L                                        (7)
                                              1 
                                              
         Also                  L                      Since,    
                                           


                                                                    4
If N q denotes the average volume of the buffers data packets then

                                                         2
                                     Nq  L                                    (8)
                                                        1 

                                                         2
                                 Also          Nq 
                                                       (   )
Using the Little’s law we have
                                     Ts and                                (9)
Using eq.(9), eq.(8) reduces to
                2
        Nq 
               1 

This implies (1  Ts ) N q  (Ts ) 2

        Or, ( Ts ) 2  Ts N q  N q  0 ,      (   ' )                      (10)

The above equation eq.(10) provides the relation between following parameters
        T s = Service time
           Sending rate
       N q  Quantity of data packets stored in the buffer
If we know any two variables, it is easy to gain the numerical value of the third one.
So, these three variables are key parameters for measuring the performance of the transmission
system.

4. Queuing theory and the network traffic monitor
4.1 Forecasting the network traffic using Queuing Theory
The network traffic is very common [5]. The system will be in worse condition, when the traffic
becomes under extreme situation, in which leads to the network 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 function of the router used in the network.

Considering a router’s arrival rate of data flow in groups is  , and the average time which the
                                      1
routers use to dispose each group is , the buffer of the routers is 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
                 1
waiting time is    . We identify Pi (t) to be the arrival probability of the queue length for the
                  
routers group at the moment of t, supposing the queue length is i:

                                                         5
P(t) = (P0 (t), P1 (t), . . . , Pn−1 (t) ), i = 0,1, . . . ,C+1.

Then the queuing system of the router’s date groups satisfies simple Markov Process [7],
according to Markov Process, we can find the diversion strength of matrix of model 1 as follow:

                     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

Network congestion rate is changing all the time [8]. The instantaneous congestion rate and the
stable congestion rate are often used to analysis the network traffic in network monitor. The
instantaneous rate AC (t ) is the congestion rate at the moment of t. The AC (t ) can be obtained by
solving the system length of the queue’s probability distributing, which is called PC 1 (t ) .

Let, Pk (t ) (k=0,1,. . .,C+1) to be the arrival probability of the queue length for the routers group
at the moment of t by considering the queue length is k.

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 differential difference equations.

Let,
                     Pk (t ) = Prob { k number of data packets present in the system in time t }
             and Pk 1 (t ) = Prob { k number of data packets present in the system in time (t + ∆t) }

Case 1:

For k ≥ 1

       Pk 1 (t  t ) = Prob { k number of data packets present in the system at time t }  Prob {

                     no data packet arrival in time ( ∆t) }  Prob { no data packet
                     departure in time ∆t }
                      + Prob { (k-1) number of data packets present in the system at time t } 

                                                           6
                      Prob { 1 data packet arrival in time ( ∆t) }  Prob { no data packet
                      departure in time ∆t }
                      + Prob { (k+1) number of data packets present in the system at time t } 
                      Prob {no data packet arrival in time ( ∆t) }  Prob { 1 data packet
                      departure in time ∆t }+…

 Pk (t  t )  Pk (t ){1  k t  o(t )}{1  k t  o(t )}
                    Pk 1 (t ){k 1t  o(t )}{1  k 1t  o(t )}

                   + Pk 1 (t ){1  k 1t  o(t )}{ k 1t  o(t )} + 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 )

Dividing both sides by t and taking limit as t  0
    d                                                                                         o(t )
      {Pk (t )}  (k   k ) Pk (t )  k 1 Pk 1 (t )   k 1 Pk 1 (t ) , since lim           0   (11)
    dt                                                                                 t    t




Here, in state k, data packets arrival is (  k )

     i.e.  k    k


Also, in state k, data packet departure is 
      i.e.  k  


Hence, eq.(11) reduces to
d
   {Pk (t )}  (  k   ) Pk (t )  {  (k  1) }Pk 1 (t )  Pk 1 (t )                           (12)
dt
where k=1,2,…,C

Case 2:

For k=0, we have
 P0 (t+∆t) = Prob { no data packet present in the system at time (t+∆t) }
            = Prob { no data packet present in time t }  Prob { no data packet arrival in
              time ∆t }
              + Prob {one data packet present in time t }  Prob { no data packet arrival

                                                              7
                in time ∆t }  Prob { one data packet departure in time ∆t } +…


            = P0 (t ){1  0 t  o(t )}  P1 (t ){1  1t  o(t )}{ 1t  o(t )}  o(t )

 P0 (t  t )  P0 (t )  0 P0 (t )  P (t ) 1  o(t )
                                          1




Dividing both sides by t and taking limit as t  0 , we obtain
                                 d
                                    {P0 (t )}  P0 (t )  P (t )
                                                              1                                    (13)
                                 dt
                         (since, k    k and  k   )


Case 3:
For k=C+1, we have
     PC 1 (t  t ) = Prob { (C+1 ) no. of data packets present in the system at time (t+∆t )}

                     = 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 }
                        + Prob { (C+1) no of data packets present in time t }  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 )

 PC 1 (t  t )  PC 1 (t )  PC (t )C t   C 1 PC 1 (t )t  o(t )


Dividing both sides by t and taking limit as t  0 we get
           d
              {PC 1 (t )}  PC (t )C   C 1 PC 1 (t )
           dt
    d
      {PC 1 (t )}  (  C ) PC (t )  PC 1 (t ) ,               since C    C            (14)
    dt


By solving this differential equation system, we get the instantaneous congestion rate
A0(t) as
                                                          
                                   A0 (t )  P1 (t )           (1  e (   )t )
                                                          


                                                                  8
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 the system. The so-called stable
congestion rate means, it will not change with the time changing, when the system works in a
stable operating condition. The definition of the stable congestion rate is

                                    AC  lim AC (t )
                                           t  



Considering, P  lim P(t ) as the distributing of the stable length of the queue and C as the buffer
                         t 
of the router, the stable congestion rate can be obtained in two ways: firstly, we obtain the
instantaneous congestion rate, then find its limit. According to its definition, it can 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 can be obtained through system of steady
state equations.

From eq.(12), eq.(13) and eq.(14), we have the system of differential difference equations as
follows

d
   {Pk (t )}  (  k   ) Pk (t )  {  (k  1) }Pk 1 (t )  Pk 1 (t )      (15)
dt
for k=1, 2, 3, … ,C
d
   {P0 (t )}  P0 (t )  P (t )
                             1             for k=0                                  (16)
dt
d
   {PC 1 (t )}  (  C ) PC (t )  PC 1 (t ) for k=C+1                         (17)
dt
According to some properties of Markov process, we know that
       Pi (t ) (i=0,1,2,…,C+1) satisfies the above differential equation.

Here, P(t )  [ P0 (t ), P1 (t ),..., PC1 (t )]

P(0)  [ P0 (0), P (0),..., PC 1 (0)]
                  1




P0 (0)  1, P (0)  0, P2 (0)  0,..., PC 1 (0)  0
             1




For steady state equation,
                                dPk (t )
lim Pk (t )  Pk and lim                 0
t                        t    dt




                                                             9
Under steady state condition, eqs.(15),(16) and (17) transform to following balance equations.
(  k   ) Pk (t )  {  (k  1) }Pk 1 (t )  Pk 1 (t ) for k=1, 2, 3, … ,C           (18)


0  P0 (t )  P (t ) for k=0
                  1                                                                           (19)

0  (  C ) PC (t )  PC 1 (t ) for k=C+1                                                 (20)
The above system of steady state equations can be written in matrix from as
                                                PQ  0
                                         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                                 0           0     
                                                                                               
  0       0                      (    3 )                           0            0     
Q                                                                                             
  0       0           0                                                    0           0     
                                                                                        
                                                                                               
  0       0           0               0                                 (    C )   C 
  0                                                                                        
          0           0               0                                                        

For C= 0,
From eq.(19), we have
                                                         P0  P1                            (21)

Also, P0  P1  1                                                                             (22)
Solving (21) and (22) we get
                                                     
                                           P1 
                                                    


Hence,

               
A0  P1                                                                                      (23)
            

                                                           10
For C=1
0  P0  P1                                                                           (24)
0  (     ) P1  P0  P2
                                                                                         (25)
0  (   ) P  P2
              1                                                                          (26)
Also, P0  P1  P2  1
From eq.(23), we get
P0       P
         1
              k
        
Therefore, P0  k , P1  k
From eq.(25), we have
                              
                    P2  (        )k , since, P  k                                    (27)
                                               1



Using eq.(26), we obtain
                                                                        
                   k[            ]  1           k 
                                                               (   )   (   )
From eq.(27) yields

                                                         
                              P2  (        )..
                                                 (   )   (   )

Hence,
                                                 (   )
                             A1  P2 
                                           (   )   (   )
For C=2,
0  P0  P1                                                                           (28)

0  (     ) P  P0  P2
                   1                                                                     (29)


0  (   ) P  (    2 ) P2  P3
              1                                                                          (30)

0  (  2 ) P2  P3                                                                   (31)

Also, P0  P1  P2  P3  1                                                              (32)

                                                          11
From eq.(28), we obtain
          P0  k and P1  k

From eq.(29), we have


        (   )
P2               k
           
From eq.(31), we have
       (  2 )
P3                P2
           
        (   )(   2 )
                           k
               2
From eq.(32), we have
                           2
k
     (   )  2   (   )   (   )(   2 )

                                      (   )(   2 )
Hence, A2  P2 
                         (   )   (   )   (   )(   2 )
                          2



For C=3, yielding
0  (     ) P  P0  P2
                   1                                                      (33)

0  (  2   ) P2  (   ) P1  P3                                 (34)

0  (  3   ) P3  (  2 ) P2  P4                                (35)

0  P0  P1                                                            (36)

0  (  3 ) P3  P4                                                    (37)
From eq.(36), we obtain
P0  k and P1  k

From eq.(33)
        (   )
P2               k
           
From eq.(34)
        (   )(   2 )k
P3 
                2

                                                         12
From eq.(37), we have
              (  3 )
      P4                 P3
                    
               (   )(   2 )(   3 )k
          
                            3
Also, P0  P1  P2  P3  P4  1

                                                  3
k 
          3 (   )   (   ) 2   (   )(  2 )   (   )(  2 )(  3 )
Hence ,
                                         (   )(   2 )(   3 )
A3  P4 
               (   )   (   )    (   )(   2 )    (   )(   2 )(   3 )
                3                       2




On the analogy of this, we conclude that, the stable congestion rate is


                                                  
AC  PC 1  1                                                                   , for C  2
                    (    C ) AC 1  {  (C  1) }(1  AC 1 ) AC 2  


5. The Queuing Model with additional one server (M/M/2) : ((C+1)/FCFS)

In this model, number of servers or channels is two and these are arranged in parallel. Here,
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  per unit time. The overall service rate can be obtained in
two situations. If there are n numbers of data packets are present in the system.
Case-1
For n < 2
There will be no queue. Therefore (2-n) server will remain idle and the combined service rate
will be

                                                 n  n ,        1 n  2

Case-2
For n  2

                                                             13
    Then, all the servers will be busy. So, maximum (n-2) (  C  1 ) number of data packets
present in the queue.

The combined service rate will be
                                 n  2 , n  2


Hence, combining Case-1 and Case-2, we have
                                  n   for all n  0
                                   n  n      for 1  n  2

                                  n  2 , n  2
                                  0  0 , n  0
                                  1   , n  1




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




                                                    14
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                                                        2           2    
          0            0                 0                                      


For C=0, we have

P0  P1                                                                       (42)

Also, P0  P1  1                                                               (43)

From eq.(43) yields P0  1  P1

Then, eq.(42) becomes

 (1  P1 )  P1

            
 P1 
          

                         
Hence, A0  P1 
                      

For C=1

P0  P1                                                                       (44)

(     ) P1  P0  2P2                                                    (45)

(   ) P1  2P2                                                              (46)

Also, P0  P1  P2  1                                                          (47)



                                                15
From eq.(44) we have

                           P0       P1
                                         k (say)
                                   

 and P0  k P1  k

From eq.(46) we get

                                (   )k
                         P2 
                                   2

Since, P0  P1  P2  1

                      
 k[     (            ) ]  1
                      2

              2
k 
       2  2   (   )
             2



Therefore,

                      2 2
     P0 
            2  2  2   (   )

                   2
and P1 
           2   2   (   )
                 2



                                (   )
Hence A1  P2 
                         (   )  2  (   )

For C=2

P0  P1                                                 (48)

(     ) P1  P0  2P2                              (49)

(  2  2 ) P2  (   ) P1  2P3                    (50)

(  2 ) P2  2P3                                       (51)

Also, P0  P1  P2  P3  1                               (52)



                                                     16
From eq.(48) yields P0  k and P1  k

From eq.(49), we obtain

       (   )
P2             k     since, P0  k , P  k
         2
                                         1



From eq.(51), we get

       (  2 )
P3              P2
          2

           (   )(   2 )
 P3                          k ,     Using the value of P2
                 4 2

Since, P0  P1  P2  P3  1

                (   )    (  2 )(    )
or, k[                                   ]  1
                  2              4 2

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

                                     (   )(   2 )
Hence A2  P3 
                      4 (   )  (   )2   (   )(   2 )
                         2




For C = 3
P0  P1                                                                (53)
(     ) P1  P0  2P2                                             (54)
(  2  2 ) P2  (   ) P1  2P3                                   (55)
(  3  2 ) P3  (  2 ) P2  2P4                                  (56)
(  3 ) P3  2 P4                                                     (57)

From eq.(53), we have

P0  k      and      P1  k

From eq.(54)

2P2  (     ) P1  P0


                                                        17
            (     )       k
 P2                    k 
                2             2

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

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

                             (  2  2 )
             (   )k [                   1]
                                  2

                 k (   )(   2 )
             
                         2

         k (   )(   2 )
 P3 
                 4 2
From eq.(57)
      (  3 )
P4             P3
         2
       k (   )(   2 )(   3 )
    
                    8 3

Also, P0  P1  P2  P3  P4  1

                        (   )  (   )(   2 )  (   )(   2 )(   3 )
  k[(    )                                                                     ] 1
                          2             4 2                       8 3
                                                  8 3
k 
           8 3 (   )  4 2 (   )  2(   )(  2 )   (   )(  2 )(  3 )


Hence,
                                        (   )(   2 )(   3 )
A3  P4 
              8 (   )  4 (   )  2 (   )(   2 )   (   )(   2 )(   3 )
                   3                 2




On the analogy of this, we conclude that, the stable congestion rate is
                                             2
AC  PC 1  1                                                                 , for C  2
                 (  2  C ) AC 1  {  (C  1) }(1  AC 1 ) AC  2  2


                                                          18
Conclusion

This research program cites the analysis of the network traffic model through Queuing Theory.
In the 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 the network traffic
through queuing theory models. In the present work two queuing models (M/M/1):
((C+1)/FCFS) and (M/M/2):((C+1)/FCFS) have been applied. These two models are used to
determine the forecast way for the stable congestion rate of the network traffic. Using the
Queuing Theory models, it is convenient and simple way for calculating and monitoring the
network traffic properly in the network communication system. We can monitor the network
efficiently, in the view of the normal, optimal and or even for the high overhead network
management, by monitoring and analyzing the network traffic rate.

Finally, we can say that network traffic rate can have an important role in the network
communication system.



References

[1] John N. Daigle, 2005, Queueing Theory with Applications to Packet Telecommunication,
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[2] Vern Paxson, Sally Floyd, 1997, Why We Don’t Know How To Simulate The Internet. In:
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    H. Withers, and B. L. Nelson, USA:ACM.
[3] Ren Xiangcai, Xiong Qibang, 2002, “An Application of Mobile Agent for IP Network Traffic
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[4] Li Da-Qi, Shen Jun-Yi, and Zhou Jiang-liang, 2007, “Queuing Theory Supervising K-Means
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[5] Wang Pei-Fa, Zhang Shi-wei, Li Jun, 2005, “The Application and Achievement of SVG in
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