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\title{Performance Analysis of A Two Node Tandem Communication Network with Feedback}
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             \author[1]{CH.V.  Raghavendran}

             \author[2]{G.Naga  Satish}

             \author[3]{M.V. Rama  Sundari}

             \affil[1]{  Adikavi Nannaya University}

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\date{\small \em Received: 8 December 2013 Accepted: 2 January 2014 Published: 15 January 2014}

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\begin{abstract}
        


A Communication Network needs optimal utilization of resources such as bandwidth, routers, transmitters, etc. In this paper we have developed and analyzed a communication network with two nodes with feedback. In this network, the arrival of packets characterized by homogeneous Poisson process and transmission of both the transmitters is characterized by Poisson process. Dynamic bandwidth allocation policy is proposed by adjusting the transmission rate at every transmitter just before transmission of each packet. The model is evaluated using the difference-differential equations and a probability generating function of the number of packets in the buffer. Through mathematical modeling, performance measures including average number of packets in each buffer, the probability of emptiness of the network, the average waiting time in the buffer and in the network, the throughput of the transmitters, utilization and the variance of the number of packets in the buffer are derived under transient conditions.

\end{abstract}


\keywords{dynamic bandwidth allocation, poisson process, performance measures, tandem network.}

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\let\tabcellsep& 	 	 		 
\section[{I. Introduction}]{I. Introduction}\par
communication network has to transfer data/voice effectively with a guaranteed Quality of Service (QoS). Number of algorithms based on flow control and bit dropping techniques have been developed with various protocols and allocation strategies for efficient transmission by optimum utilization of the bandwidth \hyperref[b0]{[1]} [2]  {\ref [3] [4]}. But utilization of the idle bandwidth by adjusting the transmission rate instantaneously just before transmission of a packet is more important to maintain QoS.\par
Utilization of the resources is another major consideration for a communication network. Congestion control and packet scheduling are the two major issues to be considered in designing a communication network. In communication network congestion occurs due to unpredicted nature of the transmission lines. Packet scheduling is a process of assigning users' packets to appropriate shared resource to achieve some performance guarantee. Packetized transmissions over links via proper packet scheduling algorithms will possibly make higher resource utilization through statistical multiplexing of packets compared to conventional circuit-based communications. Earlier these two aspects are dealt separately. But, the integration of these two is needed in order to utilize resources more effectively and efficiently. In \hyperref[b3]{[5]}  \hyperref[b4]{[6]} Matthew Andrews considered the joint optimization of scheduling and congestion control in communication networks. The statistical multiplexing with load dependent strategy has been evolved through bitdropping and flow control techniques to decrease congestion in buffers \hyperref[b5]{[7]}  \hyperref[b6]{[8]}.\par
From the literature, it is observed that in most of the papers it was assumed that the arrival and transmission processes are independent. But in storeand-forward communication systems this assumption is realistically inappropriate. Since the massages, generally preserve the length as they transfers the network, the inter arrival and service sequences at buffer, interval to the system are time dependent as they formulate a queuing process at each node of the network through which the packet are routed. These dependences can have a significant influence on the system performance \hyperref[b5]{[7]}.\par
Dynamic Bandwidth Allocation (DBA) strategy of transmission considers the adjustment of transmission rate of the packet depending upon the content of the buffer connected to transmitter at that instant \hyperref[b7]{[9]}. This strategy has grown as an alternate for bit dropping and flow control strategies for quality in transmission and to reduce the congestion in buffers \hyperref[b6]{[8]} [10] \hyperref[b9]{[11]}  \hyperref[b10]{[12]}. The strategy of dynamic bandwidth allocation is to utilize a large portion of the unutilized bandwidth.\par
From the literature we found some work regarding communication networks with dynamic bandwidth allocation. In \hyperref[b9]{[11]}, P.Suresh Varma et al has studied the communication network model with an assumption that the transmission rate of packet is adjusted instantaneously depending upon the content of the buffer. In \hyperref[b12]{[13]}, Rama Sundari., et al have developed and analyzed a three node communication network model with the assumption that the arrivals are characterized by non homogeneous Poisson process. It is further assumed that transmission time required by each packet at each node is dependent on the content of the buffer connected to it. Tirupathi Rao et al \hyperref[b11]{[14]} proposed a two node tandem communication network with DBA having compound Poisson binomial bulk arrivals for scheduling the Internet, ATM, LAN and WAN. Generally, conducting laboratory experiments with varying load conditions of a communication system in particular with DBA is difficult and complicated. Hence, mathematical models of communication networks are developed to evaluate the performance of the newly proposed communication network model under transient conditions.\par
In this paper we have developed and studied a communication network model with two nodes having homogeneous Poisson arrival and dynamic bandwidth allocation with feedback for both nodes. Here it is assumed that the packets arrive at the first buffer directly with constant arrival rate. After getting transmitted from the first transmitter the packets may join the buffer connected to the second transmitter in tandem with first transmitter or returned back to the first buffer for retransmission with certain probability. Similarly, the packets transmitted by the second transmitter may leave the network or returned back to the second buffer for retransmission with certain probability. Using difference-differential equations the transient behavior of the model is analyzed by deriving the joint probability generating function of the number of packets in each buffer. The performance measures like average number of packets in the buffer and in the system; the average waiting time of packets in the buffer and in the system, throughput of the transmitter etc., of the developed network model are derived explicitly. 
\section[{II. Proposed Communication Network}]{II. Proposed Communication Network}\par
Model    ) ( ) 1 ( ) 1 ( ) ( ) 1 ( ) 1 ( ) ( ) ( )) 1 ( ) 1 ( ( ) ( 1 , 2 2 1 , 1 1 1 , 1 , 2 2 1 1 2 1 2 1 2 1 2+ ? + ? ? + + ? + + + ? + ? + ? = ? ? ? µ ? µ ? ? µ ? µ ? ) ( ) 1 ( ) ( ) ( )) 1 ( ( ) ( 1 , 2 0 , 1 0 1 1 0 , 1 1 1 1 t P t P t P n t t P n n n n ? µ ? ? µ ? ? + + ? + ? = ? ? ? ) ( ) 1 ( ) 1 ( ) ( ) 1 ( ) ( )) 1 ( ( ) ( 1 , 0 2 2 1 , 1 , 0 2 2 , 0 2 2 1 2 2 t P n t P t P n t t P n n n n n + ? ? + + ? + ? + ? = ? ? ? µ ? µ ? µ ? ) ( ) 1 ( ) ( ) ( 1 , 0 2 0 , 0 0 , 0 t P t P t t P ? µ ? ? + ? = ? ? (\textbf{2}Year 2014 G © 2014 Global Journals Inc. (US) 2 1 2 1 2 1 2 1 2 1 2 1 2 1 , 1 1 0 1 2 1 , 1 0 0 2 1 2 1 ) ( ) 1 ( ) ( ) : , ( ) : , ( n n n n n n n n n n n n s s t P n s s t P t s s P t t s s P ? µ ? ? ? ? ? ? ? ? + ? = ? ? ? = ? = ? ? = ? =\par
transmitter. Assume that the two buffers Q 1 , Q 2 transmitters S 1 , S 2 are connected in series in Tandem buffer which is in series connected to S 2 or may be returned back to S 1 with certain probabilities. The transmitter are forwarded to the second buffer for transmission and exit from the network with probability (1-?) or returned back to the second transmitter with probability ?. The service completion in both the transmitters follows Poisson processes with the parameters ? 1 and ? 2 for the first and second transmitters. The transmission rate of each packet is adjusted just before transmission depending on the content of the buffer connected to the transmitter. A schematic diagram representing the network model with two transmitters and feedback for both transmitters is shown in figure \hyperref[fig_0]{1}.\par
Let n1 and n2 are the number of packets in first and second buffers and let be the probability that there are n 1 packets in the first buffer and n 2 packets in the second buffer at time t. The difference-differential equations for the above model are as follows:) ( 2 , 1\par
t n n P Let P(S 1 ,S 2 ;t) be the joint probability generating function of . Then multiply the equation 2.1 with and summing over all n1, n2 we get ) ( 2 , 1 t n n P S n1 1 S n2 2 1 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 , 2 0 2 1 1 , 2 2 1 0 2 1 , 2 2 1 0 2 1 1 , 1 1 1 1 0 ) ( ) 1 ( ) ( ) 1 ( ) 1 ( ) ( ) 1 ( ) ( ) 1 ( ) 1 ( n n n n n n n n n n n n n n n n n n n n n s t P s s t P n s s t P n s s t P n ? µ ? µ ? µ ? µ ? ? + ? + ? ? + ? ? ? ? ? + ? ? + ? = + ? = ? = ? = ? = ? + ? = ? = (2.? ? ? ? ? ? ? + ? ? ? ? = ? ) : , ( ) 1 ( ) 1 )( 1 ( ) )( 1 ( ) : , ( 2 1 1 2 2 2 1 1 2 1 2 1 ? ? µ ? µ (2.3)\par
Solving equation 2.3 by Lagrangian's method, we get the auxiliary equations as,) 1 ( ) 1 )( 1 ( ) )( 1 ( 1 1 2 2 2 1 2 1 1 ? = ? ? ? = ? ? ? = s P dp s ds s s ds dt ? ? µ ? µ (2.4)\par
Solving first and fourth terms in equation 2.4, we gett e s a ) 1 ( 2 2 ) 1 ( ? µ ? ? = (2.5 a)\par
Solving first and third terms in equation 2.4, we get)) 1 ( ) 1 ( ( ) 1 ( ) 1 ( ) 1 ( 1 2 ) 1 ( 1 2 ) 1 ( 1 1 1 ? µ ? µ ? µ ? µ ? µ ? ? ? ? ? + ? = ? ? ? ? t t e s e s b\par
(2.5 b) Solving first and second terms in equation 2.4, we get The general solution of equation 2.4 gives the probability generating function of the number of packets in the first and second buffers at time t, as P (S1, S2; t).  ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? = ) 1 ( ) 1 ( ) 1 ( ) 1 ( exp ) ; , (( ) ( ) ( ) ( ) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? + ? ? ? = ? ? ? ? ? ? )) 1 ( ) 1 ( ( ) 1 ( ) 1 ( 1 1 1 ) 1 ( ) 1 ( exp ) ; , (\textbf{1 2}) 1 ( ) 1 ( 2 2 ) 1 ( 2 1 ) 1 ( 1 2 1 1 2 2 1 ? µ ? µ ? ? µ ? ? µ ? ? µ ? µ ? µ ? µ t t) ( ) ( ) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? + ? ? ? = ? ? ? ? ? ? ? ? ? µ ? µ ? ? µ ? ? µ ? µ ? µ ? µ ? µ t t t t e e e e t P ) 1 ( ) 1 ( 1 2 ) 1 ( 2 ) 1 ( 1 00 1 2 2 1 )) 1 ( ) 1 ( ( 1 1 ) 1 ( 1 1 ) 1 ( 1 exp ) ( (3.1)\par
Taking S 2 =1 in equation 2.6 we get probability generating functions of the number of packets in the first buffer is(3.2)\par
Probability that the first buffer is empty as (S1=0)(3.3)\par
Taking S1=1 in equation 2.6 we get probability generating function of the number of packets in the first buffer is  ( ) ( ) ( ) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? = ? ? ? ? t t t e e S e S P ) 1 ( ) 1 ( 1 2 2 2 ) 1 ( 2 2 1 2 2 ))\textbf{1 ( ) 1 ( ( ) 1 ( ) 1 ( 1 1}( ) ? ? ? ? ? ? ? ? = ? ? t e t L t ) 1 ( 1 1 1 1 ) 1 ( 1 ) ( ? µ ? ? µ (3.6) Utilization of the first transmitter is ( ) ? ? ? ? ? ? ? ? ? ? = ? = ? ? t e t P t U ) 1 ( 1 . 0 1 1 1 ) 1 ( 1 exp 1 ) ( 1 ) ( ? µ ? ? µ (3.7)\par
Variance of the no. of packets in the first buffer is( ) ? ? ? ? ? ? ? ? = ? ? t e t V t ) 1 ( 1 1 1 1 ) 1 ( 1 ) ( ? µ ? ? µ (3.8)\par
Throughput of the first transmitter is ( ) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + = ? = ? ? t e t P Th ) 1 ( 1 1 . 0 1 1 1 1 ) 1 ( 1 exp 1 )) ( 1 ( ? µ ? ? µ µ µ (3.9) ( ) ( ) ? ? ? ? ? ? ? ? ? = ? ? t e S P ) 1 ( 1 1 1 1 1 1 1 exp t) ; S ( ? µ ? ? µ ( ) ? ? ? ? ? ? ? ? ? = ? t e t P ) 1 ( 1 . 0 1 1 ) 1 ( 1 exp ) (( ) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? = ? = ? ? ? ? t t e t e t P t L t W ) 1 ( 1 1 ) 1 ( 1 . 0 1 1 1 1 1 1 ) 1 ( 1 exp 1 1 ) 1 ( 1 )) ( 1 ( ) ( ) ( ? µ ? µ ? ? µ µ ? ? µ µ (3.10)\par
Mean number of packets in the second buffer is ( )( ) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? = ? ? ? ? ? ? ? ? µ ? µ ? ? µ ? µ ? µ ? µ t t t e e e t L ) 1 ( ) 1 ( 1 2 ) 1 ( 2 2 1 2 2 )) 1 ( ) 1 ( (\textbf{1}1 ) 1 ( 1 ) ( (3.11)\par
Utilization of the second transmitter is ( )( ) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? = ? = ? ? ? ? ? ? ? ? µ ? µ ? ? µ ? µ ? µ ? µ t t t e e e t P t U ) 1 ( ) 1 ( 1 2 ) 1 ( 2 0 . 2 1 2 2 )) 1 ( ) 1 ( ( 1 1 ) 1 ( 1 exp 1 ) ( 1 ) ( (3.12)\par
Variance of the no. of packets in the second buffer is ( )( ) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? = ? ? ? ? ? ? ? ? µ ? µ ? ? µ ? µ ? µ ? µ t t t e e e t V ) 1 ( ) 1 (\textbf{1 2 ) 1 ( 2 2 1 2 2 )) 1 ( ) 1 ( ( 1 1 ) 1 ( 1 ) ( (3.13)}\par
Throughput of the second transmitter is )( ) ( ) ( ) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? + = ? = ? ? ? ? ? ? ? ? µ ? µ ? ? µ µ µ ? µ ? µ ? µ t t t e e e t P t Th ) 1 ( ) 1 ( 1 2 ) 1 ( 2 2 0 . 2 2 1 2 2 )) 1 ( ) 1 ( (\textbf{1}) ( 1 ( ) ( ) ( 0 . 2 2 2 t P t L t W ? = µ (3.15)\par
Mean number of packets in the entire network at time t is ( )) ( ) ( 2 1 t t L L t L + = (3.16)\par
Variability of the number of packets in the network is ( )) ( ) ( 2 1 t t V V t V + = (3.17)\par
The equations 3.7, 3.9, 3.12 and 3.14 are used for computing the utilization of the transmitters and throughput of the transmitters for different values of parameters t, ?, ?, ?, µ 1 , µ 2 and the results are presented in the Table \hyperref[tab_4]{1}. Graphs in the figure 2 show the relationship between utilization of the transmitters and throughput of the transmitters.\par
In this section, the performance of the network model is discussed with numerical illustration. Different values of the parameters are taken for bandwidth allocation and arrival of packets. The packet arrival (?) varies from 2x10 4 packets/sec to 7x10 4 packets/sec, probability parameters (?, ?) varies from 0.1 to 0.9, the transmission rate for first transmitter (µ 1 ) varies from 5x10 4 packets/sec to 9x10 4 packets/sec and transmission rate for second transmitter (µ 2 ) varies from 15x10 4 packets/sec to 19x10 4 packets/sec. Dynamic Bandwidth Allocation strategy is considered for both the two transmitters. So, the transmission rate of each packet depends on the number of packets in the buffer connected to corresponding transmitter.\par
From the table 1 it is observed that, when the time (t) and ? increases, the utilization of the transmitters is increasing for the fixed value of other parameters ?, ?, µ 1 , µ 2 . As the first transmitter probability parameter ? increases from 0.1 to 0.9, the utilization of first transmitter increases and utilization of the second transmitter decreases, this is due to the number of packets arriving at the second transmitter are decreasing as number of packets going back to the first transmitter in feedback are increasing. As the second transmitter probability parameter ? increases from 0.1 to 0.9, the utilization of first transmitter remains constant and utilization of the second transmitter increases. This is because the number of packets arriving at the second transmitter is packets arriving directly from the first transmitter and packets arrived for retransmission in feedback. As the transmission rate of the first transmitter (µ 1 ) increases from 5 to 9, the utilization of the first transmitter decreases and the utilization of the second transmitter increases by keeping the other parameters as constant. As the transmission rate of the second transmitter (µ 2 ) increases from 15 to 19, the utilization of the first transmitter is constant and the utilization of the second transmitter decreases by keeping the other parameters as constant.\par
It is also observed from the table 1 that, as the time (t) increases, the throughput of first and second transmitters is increasing for the fixed values of other parameters. When the parameter ? increases from 3x10 4 packets/sec to 7x10 4 packets/sec, the throughput of both transmitters is increasing. As the first probability parameter ? value increases from 0.1 to 0.9, the  of the first transmitter is constant and the throughput of the second transmitter is increasing. Using equations 3.6, 3.8, 3.16 and 3.13, 3.15 the mean no. of packets in the two buffers and in the network, mean delay in transmission of the two transmitters are calculated for different values of t, ?, ?, ?, µ 1 , µ 2 and the results are shown in the table 2. The graphs showing the relationship between parameters and performance measure are shown in the figure 3.\par
It is observed from the table 2 that as the time (t) varies from 0.1 to 0.9 seconds, the mean number of packets in the two buffers and in the network are increasing when other parameters are kept constant. When the ? changes from 3x10 4 packets/second to 7x10 4 packets/second the mean number of packets in the first, second buffers and in the network are increasing. As the first probability parameter ? varies from 0.1 to 0.9, the mean number packets in the first buffer increases and decreases in the second buffer due to feedback for the first buffer. When the second number packets in the first buffer remains constant and probability parameter ? varies from 0.1 to 0.9, the mean 9x10 packets/second, the mean number of packets in increases in the second buffer due to packets arrived directly from the first transmitter and packets for retransmission due to feedback from the second transmitter. When the transmission rate of the first 4 transmitter (µ 1 ) varies from 5x10 4 packets/second to the first buffer decreases, in the second buffer increases From the table 2, it is also observed that with time (t) and ?, the mean delay in the two buffers are increasing for fixed values of other parameters. As the parameter ? varies the mean delay in the first buffer increases and decreases in the second buffer due to feedback for the first buffer. As the parameter ? varies the mean delay in the first buffer remains constant and increases in the second buffer. As the transmission rate From the above analysis, it is observed that the dynamic bandwidth allocation strategy has a significant of the first transmitter (µ 1 ) varies, the mean delay of the first buffer decreases, in the second buffer slightly increases. When the transmission rate of the second transmitter (µ 2 ) varies, the mean delay of the first buffer remains constant and decreases for the second buffer. influence on all performance measures of the network. We also observed that the performance measures are is optimal to consider dynamic bandwidth allocation and evaluate the performance under transient conditions. It is also to be observed that the congestion in buffers and delays in transmission can be reduced to a minimum level by adopting dynamic bandwidth allocation. highly sensitive towards smaller values of time. Hence, it V. 
\section[{Conclusion}]{Conclusion}\par
This paper introduces a tandem communication network model with two nodes with dynamic bandwidth allocation and feedback for both nodes. The dynamic bandwidth allocation is adapted by immediate adjustment of packet service time by utilizing idle bandwidth in the transmitter. The transient analysis of the model is capable of capturing the changes in the performance measures of the network like average content of the buffers, mean delays, throughput of the transmitters, idleness of the transmitters etc, explicitly. It is observed that the feedback probability parameters (?, ?) have significant influence on the overall performance of the network. The numerical study reveals that the proposed communication network model is capable of   \begin{figure}[htbp]
\noindent\textbf{1}\includegraphics[]{image-2.png}
\caption{\label{fig_0}Figure 1 :}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-3.png}
\caption{\label{fig_2}}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-4.png}
\caption{\label{fig_3}(}\end{figure}
       \begin{figure}[htbp]
\noindent\textbf{1} \par 
\begin{longtable}{P{0.3025963597430407\textwidth}P{0.014105995717344754\textwidth}P{0.0514186295503212\textwidth}P{0.06870985010706637\textwidth}P{0.006825481798715203\textwidth}P{0.31397216274089934\textwidth}P{0.0031852248394004285\textwidth}P{0.0036402569593147753\textwidth}P{0.08554603854389721\textwidth}}
t\tabcellsep ?\tabcellsep ?\tabcellsep ?\tabcellsep \multicolumn{2}{l}{µ 1 µ 2 U 1 (t)}\tabcellsep U 2 (t)\tabcellsep Th 1 (t)\tabcellsep Th 2 (t)\\
0.1\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.1488 0.0253 0.7438}\tabcellsep 0.3799\\
0.3\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.2805 0.0877 1.4026}\tabcellsep 1.3161\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.3281 0.1173 1.6403}\tabcellsep 1.7601\\
0.7\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.3465 0.1295 1.7325}\tabcellsep 1.9418\\
0.9\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.3538 0.1344 1.7692}\tabcellsep 2.0153\\
0.5\tabcellsep 3\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.4492 0.1707 2.2460}\tabcellsep 2.5611\\
0.5\tabcellsep 4\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.5485 0.2209 2.7425}\tabcellsep 3.3136\\
0.5\tabcellsep 5\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.6299 0.2680 3.1495}\tabcellsep 4.0206\\
0.5\tabcellsep 6\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.6966 0.3123 3.4831}\tabcellsep 4.6849\\
0.5\tabcellsep 7\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.7513 0.3539 3.7566}\tabcellsep 5.3089\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.3281 0.1173 1.6403}\tabcellsep 1.7601\\
0.5\tabcellsep 2\tabcellsep 0.3\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.3763 0.1073 1.8816}\tabcellsep 1.6088\\
0.5\tabcellsep 2\tabcellsep 0.5\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.4349 0.0916 2.1746}\tabcellsep 1.3743\\
0.5\tabcellsep 2\tabcellsep 0.7\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.5052 0.0671 2.5258}\tabcellsep 1.0063\\
0.5\tabcellsep 2\tabcellsep 0.9\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.5872 0.0279 2.9360}\tabcellsep 0.4191\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.3281 0.1173 1.6403}\tabcellsep 1.7601\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.3 5}\tabcellsep \multicolumn{3}{l}{15 0.3281 0.1445 1.6403}\tabcellsep 2.1678\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.5 5}\tabcellsep \multicolumn{3}{l}{15 0.3281 0.1860 1.6403}\tabcellsep 2.7902\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.7 5}\tabcellsep \multicolumn{3}{l}{15 0.3281 0.2620 1.6403}\tabcellsep 3.9304\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.9 5}\tabcellsep \multicolumn{3}{l}{15 0.3281 0.3680 1.6403}\tabcellsep 5.5200\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.3281 0.1173 1.6403}\tabcellsep 1.7601\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 6}\tabcellsep \multicolumn{3}{l}{15 0.2921 0.1234 1.7527}\tabcellsep 1.8505\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 7}\tabcellsep \multicolumn{3}{l}{15 0.2620 0.1275 1.8342}\tabcellsep 1.9126\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 8}\tabcellsep \multicolumn{3}{l}{15 0.2368 0.1304 1.8941}\tabcellsep 1.9554\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 9}\tabcellsep \multicolumn{3}{l}{15 0.2154 0.1323 1.9388}\tabcellsep 1.9851\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{15 0.3281 0.1173 1.6403}\tabcellsep 1.7601\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{16 0.3281 0.1110 1.6403}\tabcellsep 1.7758\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{17 0.3281 0.1053 1.6403}\tabcellsep 1.7895\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{18 0.3281 0.1001 1.6403}\tabcellsep 1.8015\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep \multicolumn{2}{l}{0.1 5}\tabcellsep \multicolumn{3}{l}{19 0.3281 0.0954 1.6403}\tabcellsep 1.8122\\
\multicolumn{6}{l}{throughput of the first transmitter increases and the}\tabcellsep \tabcellsep \\
\multicolumn{6}{l}{throughput of the second transmitter is decreasing. As}\tabcellsep \tabcellsep \\
\multicolumn{2}{l}{the second probability parameter}\tabcellsep \multicolumn{4}{l}{? value increases from}\tabcellsep \tabcellsep \\
\multicolumn{6}{l}{0.1 to 0.9, the throughput of the first transmitter remains}\tabcellsep \tabcellsep \\
\multicolumn{6}{l}{constant and the throughput of the second transmitter is}\tabcellsep \tabcellsep \\
\multicolumn{6}{l}{increasing. As the transmission rate of the first}\tabcellsep \tabcellsep \\
\multicolumn{6}{l}{transmitter (µ 1 ) increases from 5x10 4 packets/sec to}\tabcellsep \tabcellsep \\
\multicolumn{6}{l}{9x10 packets/sec, the throughput of the first and 4}\tabcellsep \tabcellsep \\
\multicolumn{6}{l}{second transmitters is increasing. The transmission rate}\tabcellsep \tabcellsep \\
\multicolumn{6}{l}{packets/sec to 19x10 4 packets/sec and the throughput of the second transmitter (µ 2 ) increases from 15x10 4}\tabcellsep \tabcellsep \end{longtable} \par
 
\caption{\label{tab_4}Table 1 :}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{2} \par 
\begin{longtable}{P{0.08785385121790651\textwidth}P{0.017346938775510204\textwidth}P{0.05092165898617511\textwidth}P{0.05092165898617511\textwidth}P{0.020704410796576693\textwidth}P{0.5036208031599736\textwidth}P{0.00391705069124424\textwidth}P{0.00391705069124424\textwidth}P{0.002238314680710994\textwidth}P{0.00391705069124424\textwidth}P{0.10464121132323896\textwidth}}
t\tabcellsep ?\tabcellsep ?\tabcellsep ?\tabcellsep \multicolumn{2}{l}{µ 1 µ 2}\tabcellsep L 1 (t)\tabcellsep L 2 (t)\tabcellsep L(t)\tabcellsep W 1 (t)\tabcellsep W 2 (t)\\
0.1\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.1611 0.0257 0.1867 0.2165}\tabcellsep 0.0675\\
0.3\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.3292 0.0918 0.4211 0.2347}\tabcellsep 0.0698\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.3976 0.1248 0.5224 0.2424}\tabcellsep 0.0709\\
0.7\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.4254 0.1386 0.5640 0.2455}\tabcellsep 0.0714\\
0.9\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.4367 0.1443 0.5810 0.2468}\tabcellsep 0.0716\\
0.5\tabcellsep 3\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.5964 0.1872 0.7836 0.2655}\tabcellsep 0.0731\\
0.5\tabcellsep 4\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.7952 0.2496 1.0448 0.2899}\tabcellsep 0.0753\\
0.5\tabcellsep 5\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.9940 0.3120 1.3060 0.3156}\tabcellsep 0.0776\\
0.5\tabcellsep 6\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 1.1928 0.3744 1.5672 0.3424}\tabcellsep 0.0799\\
0.5\tabcellsep 7\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 1.3916 0.4368 1.8284 0.3704}\tabcellsep 0.0823\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.3976 0.1248 0.5224 0.2424}\tabcellsep 0.0709\\
0.5\tabcellsep 2\tabcellsep 0.3\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.4721 0.1135 0.5856 0.2509}\tabcellsep 0.0705\\
0.5\tabcellsep 2\tabcellsep 0.5\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.5708 0.0961 0.6669 0.2625}\tabcellsep 0.0699\\
0.5\tabcellsep 2\tabcellsep 0.7\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.7035 0.0694 0.7730 0.2785}\tabcellsep 0.0690\\
0.5\tabcellsep 2\tabcellsep 0.9\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.8848 0.0283 0.9131 0.3014}\tabcellsep 0.0676\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.3976 0.1248 0.5224 0.2424}\tabcellsep 0.0709\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.3\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.3976 0.1561 0.5537 0.2424}\tabcellsep 0.0720\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.5\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.3976 0.2058 0.6034 0.2424}\tabcellsep 0.0738\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.7\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.3976 0.3039 0.7015 0.2424}\tabcellsep 0.0773\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.9\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.3976 0.4589 0.8565 0.2424}\tabcellsep 0.0831\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.3976 0.1248 0.5224 0.2424}\tabcellsep 0.0709\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 6\tabcellsep \multicolumn{5}{l}{15 0.3455 0.1317 0.4771 0.1971}\tabcellsep 0.0712\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 7\tabcellsep \multicolumn{5}{l}{15 0.3039 0.1364 0.4403 0.1657}\tabcellsep 0.0713\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 8\tabcellsep \multicolumn{5}{l}{15 0.2702 0.1397 0.4099 0.1426}\tabcellsep 0.0714\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 9\tabcellsep \multicolumn{5}{l}{15 0.2426 0.1420 0.3846 0.1251}\tabcellsep 0.0715\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{15 0.3976 0.1248 0.5224 0.2424}\tabcellsep 0.0709\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{16 0.3976 0.1176 0.5152 0.2424}\tabcellsep 0.0662\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{17 0.3976 0.1112 0.5088 0.2424}\tabcellsep 0.0622\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{18 0.3976 0.1055 0.5031 0.2424}\tabcellsep 0.0585\\
0.5\tabcellsep 2\tabcellsep 0.1\tabcellsep 0.1\tabcellsep 5\tabcellsep \multicolumn{5}{l}{19 0.3976 0.1002 0.4978 0.2424}\tabcellsep 0.0553\\
\multicolumn{6}{l}{evaluating and predicting the performance of}\tabcellsep \tabcellsep \tabcellsep \tabcellsep \\
network model includes\tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \end{longtable} \par
 
\caption{\label{tab_5}Table 2 :}\end{figure}
 			\footnote{G © 2014 Global Journals Inc. (US)In this section, we expand P (S1, S2; t) of equation of 2.6 and collect the constant terms. From this, we get the probability that the network is empty as Average waiting time in the first Buffer is} 		 		\backmatter  			 
\subsection[{Global Journals Inc. (US) Guidelines Handbook 2014}]{Global Journals Inc. (US) Guidelines Handbook 2014}\par
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\end{document}
