# CS 5480/6480: Computer Networks Spring 2012 Homework 3 Due by 1:25 PM MT, Monday March 5 th 2012

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2 interfaces, Subnet 2 is to support up to 95 interfaces, and Subnet 3 is to support up to 16 interfaces. Provide three network addresses (of the form a.b.c.d/x) that satisfy these constraints / / /28 (b) (2 points) Consider a subnet with prefix /26. Give an example of one IP address (of form xxx.xxx.xxx.xxx) that can be assigned to this network. Suppose an ISP owns the block of addresses of the form /26. Suppose it wants to create four subnets from this block, with each block having the same number of IP addresses. What are the prefixes (of form a.b.c.d/x) for the four subnets? Any IP address in range to Four equal size subnets: /28, /28, /28, /28 Question 4 (NAT) 3 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 number 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. 2

3 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, 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 data. Question 5 (Routing Protocols) 18 points: (a) (3 points) Consider the following network. With the indicated link costs, use Dijkstra s shortest-path algorithm to compute the shortest path from x to all network nodes. Show how the algorithm works by computing a table similar to to the table on slide Step N D(t),p(t) D(u),p(u) D(v),p(v) D(w),p(w) D(y),p(y) D(z),p(z) 0 x 3,x 6,x 6,x 8,x 1 xv 7,v 6,v 3,x 6,x 6,x 8,x 2 xvu 7,v 6,v 3,x 6,x 6,x 8,x 3 xvuw 7,v 6,v 3,x 6,x 6,x 8,x 4 xvuwy 7,v 6,v 3,x 6,x 6,x 8,x 5 xvuwyt 7,v 6,v 3,x 6,x 6,x 8,x 6 xvuwytz 7,v 6,v 3,x 6,x 6,x 8,x (b) (3 points) Consider the three-node topology shown on slide Let the link costs be c(x,y) = 3, c(y,z) = 6, c(z,x) = 4. Compute the distance tables after the initialization step and after each iteration of the distance-vector algorithm as done for the example on slide

4 Node x table From y z Node y table x z Node z table x From y (c) (3 points) Consider the count-to-infinity problem in the distance vector routing. Will the count-to-infinity problem occur if we decrease the cost of a link? Why? How about if we connect two nodes which do not have a link? 4

7 Two limitations of flooding and pruning multicast routing: - When there are only a few members in the multicast group, flooding results in unnecessary transmissions on large part of the network without any group members. - Pruned state can delay transmission of multicast data to receivers that join the group later. CBT s does not use any flooding. Group members send join requests towards the wellknown core. Those senders that are also members of the group send the multicast data along the CBT. Those senders that are not members of the group send their multicast data straight to the core that in turn forwards the data along the CBT. Two limitations of CBT: - Multicast data transmission in CBTs can take longer paths. Cores will experience traffic concentration near themselves. Question 7 (required for CS6480, extra credit for CS5480) 7 points: Read the following paper: On Fast and Accurate Detection of Unauthorized Access Points Using Clock Skews by Suman Jana and Sneha Kumar Kasera, in the IEEE Transactions on Mobile Computing, March This paper is available from Answer the following questions that are based on this paper. (a) (1 point) Define clock skew. Clock skew is the rate of change of clock offset. (b) (2 points) Let us plot the observed clock offset, in microseconds, on the y-axis and the time since the start of the fingerprinting measurements, in seconds, at the fingerprinter, on the x-axis. Let (5, 60) and (7.5, 75) be two points at times 5s and 7.5s, where the clock offset is observed by the fingerprinter to be 60 and 75, respectively. Estimate the clock skew of the fingerprintee from these two points. You can assume that the network delays are negligible. The clock skew is the slope of the line connecting the two points = 15/2.5 = 6 ppm. (c) (2 points) What clock skew behavior do the authors observe in the case of virtual access points? They find that the clock skews of the virtual APs on the same physical AP hardware are the same. (d) (2 points) Why is it not easy to fabricate clock skews of access points? - unpredictable medium access control delay - floating point operations on wireless cards could take a few microseconds 7

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