# Computer Networks - CS132/EECS148 - Spring

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3 b - If the first segment arrives before the second segment, in the acknowledgment of the first arriving segments, what is the ACK number, the source port number and the destination port number? c - If the second segment arrives before the first segment, in the ACK of the first arriving segment, what is the ACK number? a) In the second segment from Host A to B, the sequence number is 207, source port number is 302 and destination port number is 80 b) If the first segment arrives before the second, in the acknowledgement of the first arriving segment, the acknowledgement number is 207, the source port number is 80 and the destination port number is 302 c) If the second segment arrives before the first segment, in the acknowledgement of the first arriving segment, the acknowledgement number is 127, indicating that it is still waiting for bytes 127 and onwards d) Problem 6 (Problem 31, Chapter 3-4 points) - Suppose that the five measured SampleRTT values are 106ms, 120ms, 140ms, 90ms, and 115ms Compute the EstimatedRTT after each of these SampleRTT values is obtained, using a value of alpha = 0125 and assuming that the value of EstimatedRTT was 100ms just before the first of these 5 samples were obtained Compute also the DevRTT after each sample is obtained, assuming a value of beta = 025 and assuming the value of DevRTT was 5ms just before the first of these five samples was obtained Last, Compute the TCP TimeoutInterval after each of these samples is obtained

4 EstimatedR TT = xsamplertt + (1 x ) EstimatedR TT DevRTT = y SampleRTT EstimatedRTT + (1 y ) DevRTT TimeoutInterval = EstimatedRTT + 4 * DevRTT After obtaining first samplertt is EstimatedRTT = 0125 * * 100 = 10075ms DevRTT = * 5 = 506ms TimeoutInterval = * 506 = 12099ms After obtaining second samplertt = 120ms: EstimatedRTT = 0125 * * = 10315ms DevRTT = * 506 = 8ms TimeoutInterval = * 8 = 13515ms After obtaining Third samplertt = 140ms: EstimatedRTT = 0125 * * = 10776ms DevRTT = * 8 = 1406ms TimeoutInterval = *1406 = 164ms After obtaining fourth samplertt = 90ms: EstimatedRTT = 0125 * * = 10554ms

5 DevRTT = *1406 = 1442ms TimeoutInterval = *1442 = 16322ms After obtaining fifth samplertt = 115ms: EstimatedRTT = 0125 * * = 10671ms DevRTT = *1442 = 1288ms TimeoutInterval = *1288 = 15823ms Problem 7 (Problem 40, Chapter 3-4 points) - Consider Figure 358 of your book, assuming TCP Reno is the protocol experiencing the behaviour shown in the figure, answer the following question In all cases, you should provide a short discussion justifying your answer a - Identify the intervals of time when TCP slow start is operating b - Identify the intervals of time when TCP congestion avoidance is operating a) TCP slowstart is operating in the intervals [1,6] and [23,26] b) TCP congestion avoidance is operating in the intervals [6,16] and [17,22] Problem 8 (Problem 44, Chapter 3-4 points) - 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 w/o slow start Assuming cwnd increases by 1MSS every time a batch of ACKs is received and assuming approximately constant roundtrip times, how long does it take for cwnd increase from 6MSS to 12MSS (Assuming no loss events)? b - What is the average throughout (in terms of MSS and RTT) for this connection up through time = 6RTT? 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; 6 RTTs to increase to 11 MSS; and 7 RTTs to increase to 12MSS 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 fourth 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

7 (in [0-50]msec] (in [0-50]msec) 50 5 the avg total 50msec is 300= ) 100 (in [50-100]msec] 100 (in [50-100]msec) the avg total 50msec is 200= ) 40 5 the avg total 100msec is 250= ( )/2 + ( )/2) the avg total 50msec is 90= (40+50) (no further decrease, as window size is already 1) 20 2 the avg total 100msec is 80= 20

8 (40+20)/2 + (50+50)/2) (no further decrease, as window size is already 1) (no further decrease, as window size is already 1) 20 1 the avg total 100msec is 40= (20+20)/2 + (20+20)/2) the avg total 50msec is 50= (40+10)

9 Based on the above table, we find that after 1000 msec, C1 s and C2 s window sizes are 1 segment each b) No In the long run, C1 s bandwidth share is roughly twice as that of C2 s, because C1 has shorter RTT, only half of that of C2, so C1 can adjust its window size twice as fast as C2 If we look at the above table, we can see a cycle every 200msec, eg from 850msec to 1000msec, inclusive Within a cycle, the sending rate of C1 is ( ) = 120, which is thrice as large as the sending of C2 given by ( ) = 40 Problem 10 (Problem 4, Chapter 4-4 points) - Consider the network shown in the figure of Problem 4 of chapter 4 in your book a - Suppose that this network is a datagram network Show the forwarding table in router A, such that all traffic destined to Host H3 is forwarded through interface 3 b - Suppose that this network is a datagram network Can you write down forwarding table in routed A, such that all traffic from H1 destined to Host H3 is forwarded through interface 3, while all traffic from H2 destined to Host H3 is forwarded through interface 4? (Hint : this is a trick question) a) Data destined to host H3 is forwarded through interface 3 Destination Address H3

11 Problem 13 (Problem 19, Chapter 4-3 points) - Consider sending a 2400-byte datagram into a link that has an MTU of 700 bytes Suppose the original datagram is stamped with the identification number 422 How many fragments are generated? what are the values in the various fields in the IP datagram(s) generated related to fragmentation? The maximum size of data field in each fragment = 680 (because there are 20 bytes IP header) Thus the number of required fragments [ / 680 ] = 4 Each fragment will have Identification number 422 Each fragment except the last one will be of size 700 bytes (including IP header) The last datagram will be of size 360 bytes (including IP header) The offsets of the 4 fragments will be 0, 85, 170, 255 Each of the first 3 fragments will have flag=1; the last fragment will have flag=0 Problem 14 (Problem 21, Chapter 4-4 points) - Consider the network setup in Figure 422 in your book, suppose that the ISP instead assigns the router the address and that the network address of the home network is /24 a - Assign addresses to all interfaces in the home network b - Suppose each host has two ongoing TCP connections, all to port 80 at host Provide the six corresponding entries in the NAT translation table

12 a) Home addresses: , , with the router interface being b) Problem 15 (Problem 22, Chapter 4-4 points) - Suppose you are interested in detecting the number of hosts behind a NAT You observe that the IP layer stamps an identification number sequentially on each IP packet The identification of the first IP packet generated by a host is a random number, and the identification numbers of the subsequent IP packets are sequentially assigned Assume all IP packets generated by hosts behind the NAT are sent to the outside world a - Based on this observation and assuming you can sniff all packets sent by the NAT to the outside, can you outline a simple technique that detects the number of unique hosts behind a NAT? justify your answer b - If the identification numbers are not sequentially assigned but randomly assigned, would your technique work? justify your answer a) Since all IP packets are sent outside, so we can use a packet sniffer to record all IP packets generated by the hosts behind a NAT As each host generates a sequence of IP packets with sequential numbers and a distinct (very likely, as they are randomly chosen from a large space) initial identification number (ID), we can group IP packets with consecutive IDs into a cluster The number of clusters is the number of hosts behind the NAT For more practical algorithms, see the following papers A Technique for Counting NATted Hosts, by Steven M Bellovin, appeared in IMW 02, Nov 6-8, 2002, Marseille, France Exploiting the IPID field to infer network path and end-system characteristics Weifeng Chen, Yong Huang, Bruno F Ribeiro, Kyoungwon Suh, Honggang Zhang, Edmundo de Souza e Silva, Jim Kurose, and Don Towsley PAM'05 Workshop, March 31 - April 01, 2005 Boston, MA, USA b) However, if those identification numbers are not sequentially assigned but randomly assigned, the technique suggested in part (a) won t work, as there won t be clusters in sniffed

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