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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 ) n1 Pn1 (t ) n1 Pn1 (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 ) Pn1 (t ) Pn1 (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 1t o(t )}{1 k 1t o(t )} + Pk 1 (t ){1 k 1t o(t )}{ k 1t o(t )} + o(t ) Pk (t t ) Pk (t ) (k k ) Pk (t )t Pk 1 (t )k 1t Pk 1 (t ) k 1t 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 1t o(t )}{ 1t 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 1t 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 ),..., PC1 (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 2P2 for n 1 (39) { (n 1) }Pn1 2Pn1 ( n ) Pn 2Pn for 1 n C (40) ( C ) PC 2PC 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 2P2 (45) ( ) P1 2P2 (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 2P2 (49) ( 2 2 ) P2 ( ) P1 2P3 (50) ( 2 ) P2 2P3 (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 2P2 (54) ( 2 2 ) P2 ( ) P1 2P3 (55) ( 3 2 ) P3 ( 2 ) P2 2P4 (56) ( 3 ) P3 2 P4 (57) From eq.(53), we have P0 k and P1 k From eq.(54) 2P2 ( ) P1 P0 17 ( ) k P2 k 2 2 ( ) k 2 From eq.(55), we get 2P3 ( 2 2 ) P2 ( ) P1 ( 2 2 )( )k 2P3 ( )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. 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