The next three questions are related to the local area network shown below. In t

Question

The next three questions are related to the local area network shown below. In this network computers attached to inks 1 &2 send data to the server attached to link 3. Link 1 Link 2 Link 3 Following parameters of the network have been provided. Propagation delay in link 1-144 micro seconds Propagation delay in link 2-303 micro seconds Propagation delay in link 3-218 micro seconds Frame size 1428 bytes ACK frame size- 100 bytes Data rate 100 Mbps Link 1 uses a Go-back-N sliding window protocol with window size of 3 frames and link 2 uses a stop-and-wait flow control respectively What is the throughput on link 1? Consider only the throughput from the sender to the receiver. Express your answer in Mbps to two decimal points. Assume that the processing delay at all nodes are negligible

Explanation / Answer

Throughput of Link 1 :-

        Given Propogation delay of Link 1 is = 144 micro seconds

                 Frame size is = 1428 bytes

         ACK Frame size is = 100 bytes    

                 Date rate is = 100 Mbps

         Also Given that Link 1 uses Go-back-N sliding window protocol

                Window size is = 3 Frames

         We know Throughput.is = 1 Window / RTT

                                            = 3 Frames / RTT

         Here RTT is = Transmission delay + 2 * Propogation delay

         Here Transmission delay = Siize of the frame / Data Rate

                                              = 1428 * 8 bits / 100 * 106 bps

                                              = 114.24 micro secs

        Hence Transmission delay = 114.24 micro secs

        Now RTT is = 114.24 + 2 * 144 = 402 micro seconds (approx)

        Now Throughput.= 1 Window / RTT

                                = 3 * 1428 * 8 bits / 402 micro secs

                                = 85.25 * 106 bps

       Hence Throughput of Link 1 is = 85.25 Mbps

Throughput of Link 2 :-

      Given Propogation delay of Link 2 is = 303 micro seconds

                 Frame size is = 1428 bytes

         ACK Frame size is = 100 bytes    

                 Date rate is = 100 Mbps

         Also Given that Link 2 uses stop and wait protocol

          We know in stop andf wait Window size is = 1 Frame

             Hence Throughput.is = 1 Window / RTT

                                            = 1 Frame / RTT

         Here RTT is = Transmission delay + 2 * Propogation delay

         Here Transmission delay = Siize of the frame / Data Rate

                                              = 1428 * 8 bits / 100 * 106 bps

                                              = 114.24 micro secs

        Hence Transmission delay = 114.24 micro secs

        Now RTT is = 114.24 + 2 * 303 = 720 micro seconds (approx)

        Now Throughput.= 1 Window / RTT

                                = 1428 * 8 bits / 720 micro secs

                                = 15.87 * 106 bps

       Hence Throughput of Link 2 is = 15.87 Mbps

Minimum window size :-

          If there should not be any long term frame build up at the router

         in this case we need an optimal window size and it is given by

                Optimal window size = Maximum Throughput x RTT

Here RTT = Transmission delay + 2 * Propogation delay

= 114.24 + 2 * 218   = 550 micro secs

               Now Optimal window size = Maximum Throughput x RTT

                                                     = 100 Mbps * 550 micro secs

                                                     = 55000 bits = 6875 bytes

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