# First Midterm for ECE374 03/24/11 Solution!!

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3 3 b. c. d... Problem 3: Consider sending a large file from a host to another over a TCP connection that has no loss. a. Suppose TCP uses AIMD for its congestion control without slow start. Assuming cwnd increases by 1 MSS every time a batch of ACKs is received and assuming approximately constant round- trip times, how long does it take for cwnd to increase from 5 MSS to 11 MSS (assuming no loss events)? b. What is the average throughput (in terms of MSS and RTT) for this connection up through time = 6 RTT? a. It takes 1 RTT to increase CongWin to 6 MSS; 2 RTTs to increase to 7 MSS; 3 RTTs to increase to 8 MSS; 4 RTTs to increase to 9 MSS; 5 RTTs to increase to 10 MSS; and 6 RTTs to increase to 11 MSS. b. In the first RTT 5 MSS was sent; in the second RTT 6 MSS was sent; in the third RTT 7 MSS was sent; in the forth RTT 8 MSS was sent; in the fifth RTT, 9 MSS was sent; and in the sixth RTT, 10 MSS was sent. Thus, up to time 6 RTT, = 45 MSS were sent (and acknowledged). Thus, we can say that the average throughput up to time 6 RTT was (45 MSS)/(6 RTT) = 7.5 MSS/RTT. Problem 4: Consider distributing a file of F = 15 Gbits to N peers. The server has an upload rate of us = 30 Mbps, and each peer has a download rate of di = 2 Mbps and an upload rate of u. For N = 10, 100, and 1,000 and u = 300 Kbps, 700 Kbps, and 2 Mbps, prepare a chart giving the minimum distribution time for each of the combination of N and u for both client- server distribution and P2P distribution. For calculating the minimum distribution time for client-server distribution, we use the following formula: D cs = max {NF/u s, F/d min } Similarly, for calculating the minimum distribution time for P2P distribution, we use the following formula: Where, F = 15 Gbits = 15 * 1024 Mbits u s = 30 Mbps d min = d i = 2 Mbps

4 Note, 300Kbps = 300/1024 Mbps. Client Server N Kbps u 700 Kbps Mbps Peer to Peer N Kbps u 700 Kbps Mbps Problem 5: Compare GBN, SR, and TCP (no delayed ACK). Assume that the timeout values for all three protocols are sufficiently long such that 5 consecutive data segments and their corresponding ACKs can be received (if not lost in the channel) by the receiving host (Host B) and the sending host (Host A) respectively. Suppose Host A sends 5 data segments to Host B, and the 2 nd segment (sent from A) is lost. In the end, all 5 data segments have been correctly received by Host B. a. How many segments has Host A sent in total and how many ACKs has Host B sent in total? What are their sequence numbers? Answer this question for all three protocols. b. If the timeout values for all three protocols are much longer than 5 * RTT, then which protocol successfully delivers all five data segments in shortest time interval? a. GoBackN: A sends 9 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later re-sent segments 2, 3, 4, and 5. B sends 8 ACKs. They are 4 ACKS with sequence number 1, and 4 ACKS with sequence numbers 2, 3, 4, and 5. Selective Repeat: A sends 6 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later re-sent segments 2. B sends 5 ACKs. They are 4 ACKS with sequence number 1, 3, 4, 5. And there is one ACK with sequence number 2. TCP: A sends 6 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later

5 5 re-sent segments 2. B sends 5 ACKs. They are 4 ACKS with sequence number 2. There is one ACK with sequence numbers 6. Note that TCP always send an ACK with expected sequence number. b. TCP. This is because TCP uses fast retransmit without waiting until time out. Problem 6: For this problem you should familiarize yourself with Figure 2 first. Now assume that in the network shown in Figure 2 two parallel TCP transmissions are performed. TCP1 is a transmission between Source A and Sink A that uses TCP Tahoe. TCP2 is a transmission between Source B and Sink B that uses TCP Reno. Initial ssthresh for both TCP transmissions is set to 32. In this specific scenario no additional delay through forwarding is introduced. Thus, the RTT is only composed of the sums of the delay indicated on each link, times two. a. For the TCP 1 transmission, draw the resulting congestion window, assuming that a packet loss (triple duplicate ACKs) is detected at time t=900ms. b. For the TCP 2 transmission, draw the resulting congestion window, assuming that a packet loss (triple duplicate ACKs) is detected at time t=650ms. c. Describe the benefit of TCP Reno over TCP Tahoe. Figure 2 Network layout for problem 5.

6 Figure 3 Solution

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