Consider that only a single TCP(Reno) connection uses one 10Mbps link which does

Question

Consider that only a single TCP(Reno) connection uses one 10Mbps link which does not buffer any data. Suppose that this link is the only congested link between the sending and receiving hosts. Assume that the TCP sender has a huge file to send to the receiver, and the receiver's receive buffer is much larger than the congestion window. We also make the following assumptions: each TCP segment size is 1,500 bytes; the two-way propagation delay of this connection is 100msec; and this TCP connection is always in congestion avoidance phase, that is, ignore slow start.
a) What is the maximum window size(in segments) that this TCP connection can achieve?
b)What is the average window size(in segments) and the average throughput(in bps) of this TCP connection?
c) How long would it take for this TCP connection to reach its maximum window again after recovering from the packet loss?


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Explanation / Answer

Given data:

       Link capacity =10Mbps

       Each segment size=1500bytes

       Two-way propagation delay of this connection = 100msec

a)

The maximum window size measured in segments. Then W*MSS/RTT=10Mbps as packets will be dropped if the maximum sending rate exceeds link capacity.

                                   W*1500*8/ (100msec) = 10*10^6

                                   W*120000=10*10^6

                                    W= (10*10^6)/120000

                                        = 83.333

                                        =84segments                                  

b)

As congestion window size varies from W/2 to W,

Then the average window size =0.75*W

                                                  = 0.75*84

                                                  =63 segments.

Average throughput = (0.75*W)/RTT

                               = 63*(1500*8)/ (100*10^-3)

                                 =7.56Mbps

c)

It would take for this TCP connection to reach its maximum window again recovering from a packet loss=W/2*100msec

                              =84/2 *0.1

                              = 4.2 seconds, as the number of RTTs (this TCP connections needs in order to increase its window size from W/2 to W) is given by W/2. Recall the window size increases by one in each RTT.

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