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9) The sender side of rdt30 simply ignores (ie, takes no action on) all received packets which are either in error, or have the wrong value in the acknum field of an acknowledgement packet Suppose that in such circumstances, rdt30 were to simply retransmit the current data packet Would the protocol still work (Hint: Consider what would happen in the case that there are only it errors; no packet losses and no premature timeouts occur Consider how many times the nth packet is sent, in the limit as n approaches infinity 10) Consider the cross-country example shown in Figure 34-10 How big would the window size have to be for the channel utilization to be greater than 90 % 11) Design a reliable, pipelined, data transfer protocol that uses only negative acknowledgements How quickly will your protocol respond to lost packets when the arrival rate of data ot the sender is low Is high 12) Consider transferring an enormous file of L bytes from host A to host B Assumme an MSS of 1460 bytes a) What us the maximum length of L such that TCP sequence numbers are not exhausted Recall that the TCP number field has four bytes b) For the L you obtain in (a), find how long it takes to transmit the file Assme that a total of 66 bytes of transport, network and data-link header are added to each segment before the resulting packet is sent out over a 10 Mbps link Ignore flow control and congestion control, so A can pump out the segments back-to-back and continuously 13) In Figure 35-5, we see that TCP waits until it has received three duplicate ACK before performing a fas retransmit Why do you think the TCP designers chose not to perform a fast retransmit after the first duplicate ACK for a segment is received 14) Consider the TCP procedure for estimating RTT Suppose that x = 1 Let SampleRTT1 be the most recent sample RTT, let SampleRTT2 be the next most recent sample RTT, etc (a) For a given TCP connection, suppose 4 acknowledgements have been returned with corresponding sample RTTs SampleRTT4, SampleRTT3, SampleRTT2, and SampleRTT1 Express EstimatedRTT in terms of the four sample RTTs (b) Generalize your formula for n sample round-trip times (c) For the formula in part (b) let n approach infinity Comment on why this averaging procedure is called an exponential moving average 15) Refer to Figure 37-3 that illustrates the convergence of TCP's additive increase, multiplicative decrease algorithm Suppose that instead of a multiplicative decrease, TCP decreased the window size by a constant amount Would the resulting additive increase additive decrease converge to an equal share algorithm Justify your answer using a diagram similar to Figure 37-3 16) Recall the idealized model for the steady-state dynamics of TCP In the period of time from when the connection's window size varies from (W*MSS)/2 to W*MSS, only one packet is lost (at the very end of the period) (a) Show that the loss rate is equal to L = loss rate = 1/[(3/8)*W2 - W/4] ( b) Use the above result to show that if connection has loss rate L, then its average bandwidth is approximately given by: average bandwidth of connection ~ 122 * MSS / (RTT * sqrt(L) ) 17) Consider sending an object of size O = 100 Kbytes from server to client Let S=536 bytes and RTT=100msec Suppose the transport protocol uses static windows with window size W a) For a transmission rate of 28 Kbps, determine the minimum possible latency Determine the minimum window size that achieves this latency b) Repeat (a) for 100 Kbps c) Repeat (a) for 1 Mbps d) Repeat (a) for 10 Mbps 18) Suppose TCP increased its congestion window by two rather than by one for each received acknowledgement during slow start Thus the first window consists of one segment, the second of three segments, the third of nine segments, etc For this slow-start procedure: a) Express K in terms of O and S b) Express Q in terms of RTT, S and R.



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6) Repeat the above question for the estimate of average delay deviation 7) Compare the procedure described in Section 63 for estimating average delay with the procedure in Section 5 for estimating round-trip time What do the procedures have in common How are they different 8) Consider the adaptive playout strategy described in Section 63 (a) How can two successive packets received at the destination have timestamps that differ by more than 20 msecs when the two packets belong to the same talkspurt (b) How can the receiver use sequence numbers to determine whether a packet is the first packet in a talkspurt Be specific 9) Recall the two FEC schemes for Internet phone described in Section 63 Suppose that the first scheme generates a redundant chunk for every four original chunks Suppose the second scheme uses a low-bit-rate encoding whose transmission rate is 25% the transmission rate of nominal stream (a) How much additional bandwidth does each scheme require How much playback delay does each scheme add (b) How do the two schemes perform if at most one packet is lost in every group of five packets Which scheme will have better audio quality (c) How do the two schemes perform if at most one packet is lost in every group of two packets Which scheme will have better audio quality .

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