CS/ECE 438: Communication Networks Fall 2006Problem Set 6 Due Friday, November 17Transport LayerNOTE: This assignment is due on a Friday, so you may ask for help on it after Wednesday’s class, unlikeprevious assignments. However, there is no automatic extension on this assignment, so any assignments nothanded in by the start of class on November 17th will receive no c redit.1. DemultiplexingOut of the packets below, show which ones will be delivered to the same socket.# Source IP Source Port Dest IP Dest Port Protocol1 1.2.3.4 1234 5.6.7.8 5678 UDP2 1.2.3.4 4321 5.6.7.8 5678 UDP3 1.2.3.4 1234 5.6.7.8 8765 UDP4 1.2.3.4 1234 5.6.7.8 5678 TCP5 1.2.3.4 4321 5.6.7.8 5678 TCP6 1.2.3.4 4321 5.6.7.8 8765 TCP7 1.2.3.5 4321 5.6.7.8 8765 TCP2. Reliable Data Transfer(a) Show how rdt3.0 can become unreliable if the network reorders packets. In particular, show asequence of messages, states, and network loss / delay events in a diagram similar to Figure 3.16(p213) of the textbook.(b) Notice that when the sender of rdt3.0 receives a “duplicate ACK” for the last packet, this ACKis ignored. Suppose that instead, rdt3.0 implemented a version of Fast Retransmit and resent thepacket whenever it received such an ACK. I.e. it would include the following transition rule instate ”Wait for ACK0”:rdtrcv(rcvpkt) && notcorrupt(rcvpkt) && isACK(rcvpkt,1)udtsend(sendpkt), start timerShow how with this modification, a single delayed ACK from the receiver can result in unnecessaryretransmissions of subsequent packets.3. Slow StartAssume a connection with RTT=100ms, MSS=1460 bytes. Ignoring overhead spent on headers, calcu-late the transfer time and the effective throughput for transferring a 500 KB file. Assume there are nolosses and that both the connection bandwidth and the receiver window are infinite (and in particular,not limited to 64K).(a) Calculate the transfer time for TCP using slow start(b) Calculate the transfer time supposing TCP used additive increase, starting with a window size of1 MSS and increasing it by 1 MSS every RTT.14. FairnessSupp ose there are two connections, A and B, sharing a 10 Kbit link. Let us simplify the TCP congestioncontrol protocol to include the following assumptions:• 1 MSS = 1000 bits• the two connections increase their window sizes by 1 MSS per RT T, in lock-step, whenever theiraggregate bandwidth is les s than or equal to 10 Kbits• when the aggregate bandwidth is greater than 10 Kbits, both connections simultaneously decreasetheir window size to 1/2 of the previous size. The w indow is rounded up to the next MSS size.(a) Suppose A’s window is currently 9000 bits and B’s window is 1000 bits, and both their RTTs are1s. Show how the window sizes change after 20s(b) Suppose A and B both have a window of 1000 bits, A has RTT of 1s and B has RTT of 2s. Showhow the window sizes change after 20s. (Assume that whenever the aggregate bandwidth exceeds10 Kbit, the next window of each of the connection will be cut in half.)5. RTT estimationYou may want to use a spreadsheet or a computer program to calculate the answers to this question.Recall that the RTT estimation algorithm is:RTTEstimate = RTTEstimate · (1 − α) + RTTSample · αSupp ose that a TCP connection has experienced a constant RTT of 500ms, so that RTTEstimate isnow 500ms. Now suppose the RTT drops to 200ms for the next N measurements, and then returns to500ms.(a) Consider the following strategy for setting the timeout:Timeout = 2 · RT T EstimateWhat is the smallest value of N so that a premature timeout will occur when the RTT returnsto 500ms? Let α = 0.1 in this case.(b) Now suppose the timeout is set using a deviation estimate:DeviationEstimate = DeviationEstimate · (1 − β) + |RTTSample − RTTEstimate| · βTimeout = RT T Estimate + 4 · DeviationEstimatewith β = 0.25. What will the timeout value be after N measurements of 200ms, using the valueof N from the previous question? (Assume that DeviationEstimate is 0 before the RTT drops
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