CS 5480/6480: Computer Networks Spring 2012 Homework 4 Solutions Due by 1:25 PM on April 11 th 2012

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1 CS 5480/6480: Computer Networks Spring 2012 Homework 4 Solutions Due by 1:25 PM on April 11 th 2012 Important: The solutions to the homework problems from the course book have been provided by the authors. CS5480 total points: 31 CS6480 total points: 40 Question 1 (Slotted Aloha) 5 points (1.25 points for each part): Suppose four active nodes nodes A, B, C and D are competing for access to a channel using slotted ALOHA. Assume each node has an infinite number of packets to send. Each node attempts to transmit in each slot with probability p. The first slot is numbered slot 1, the second slot is numbered slot 2, and so on. a. What is the probability that node A succeeds for the first time in slot 5? b. What is the probability that some node (either A, B, C or D) succeeds in slot 4? c. What is the probability that the first success occurs in slot 3? d. What is the efficiency of this four-node system? a) (1 p(a)) 4 p(a) where, p(a) = probability that A succeeds in a slot p(a) = p(a transmits and B does not and C does not and D does not) = p(a transmits) p(b does not transmit) p(c does not transmit) p(d does not transmit) = p(1 p) (1 p)(1-p) = p(1 p) 3 Hence, p(a succeeds for first time in slot 5) = (1 p(a)) 4 p(a) = (1 p(1 p) 3 ) 4 p(1 p) 3 b) p(a succeeds in slot 4) = p(1-p) 3 p(b succeeds in slot 4) = p(1-p) 3 p(c succeeds in slot 4) = p(1-p) 3 p(d succeeds in slot 4) = p(1-p) 3 p(either A or B or C or D succeeds in slot 4) = 4 p(1-p) 3 (because these events are mutually exclusive) c) p(some node succeeds in a slot) = 4 p(1-p) 3 p(no node succeeds in a slot) = 1-4 p(1-p) 3 Hence, p(first success occurs in slot 3) = p(no node succeeds in first 2 slots) p(some node succeeds in 3 rd slot) = (1-4 p(1-p) 3 ) 2 4 p(1-p) 3 d) efficiency = p(success in a slot) =4 p(1-p) 3

5 (a) (2 points) Suppose there are two ISPs providing WiFi access in a particular café, with each ISP operating its own AP and having its own IP address block. a. Further suppose that by accident, each ISP has configured its AP to operate over channel 11. Will the protocol completely break down in this situation? Discuss what happens when two stations, each associated with a different ISP, attempt to transmit at the same time. b. Now suppose that one AP operates over channel 1 and the other over channel 11. How do your answers change? a. The two APs will typically have different SSIDs and MAC addresses. A wireless station arriving to the café will associate with one of the SSIDs (that is, one of the APs). After association, there is a virtual link between the new station and the AP. Label the APs AP1 and AP2. Suppose the new station associates with AP1. When the new station sends a frame, it will be addressed to AP1. Although AP2 will also receive the frame, it will not process the frame because the frame is not addressed to it. Thus, the two ISPs can work in parallel over the same channel. However, the two ISPs will be sharing the same wireless bandwidth. If wireless stations in different ISPs transmit at the same time, there will be a collision. For b, the maximum aggregate transmission rate for the two ISPs is 11 Mbps. b. Now if two wireless stations in different ISPs (and hence different channels) transmit at the same time, there will not be a collision. Thus, the maximum aggregate transmission rate for the two ISPs is 22 Mbps for b. (b) (2 points) In Step 4 of the CSMA/CA protocol, a station that successfully transmits a frame begins the CSMA/CA protocol for a second frame at step 2, rather than at step 1. What rationale might the designers of CSMA/CA have had in mind by having such a station not transmit the second frame immediately (if the channel is sensed idle)? Suppose that wireless station H1 has 1000 long frames to transmit. (H1 may be an AP that is forwarding an MP3 to some other wireless station.) Suppose initially H1 is the only station that wants to transmit, but that while half-way through transmitting its first frame, H2 wants to transmit a frame. For simplicity, also suppose every station can hear every other station s signal (that is, no hidden terminals). Before transmitting, H2 will sense that the channel is busy, and therefore choose a random backoff value. Now suppose that after sending its first frame, H1 returns to step 1; that is, it waits a short period of times (DIFS) and then starts to transmit the second frame. H1 s second frame will then be transmitted while H2 is stuck in backoff, waiting for an idle channel. Thus, H1 should get to transmit all of its 1000 frames before H2 has a chance to access the channel. On the other hand, if H1 goes to step 2 after transmitting a frame, then it too chooses a random backoff value, thereby giving a fair chance to H2. Thus, fairness was the rationale behind this design choice. Question 6 (IP Mobility) 10 points: (a) (4 points) One proposed solution that allowed mobile users to maintain their IP addresses as they moved among foreign networks was to have a foreign network

6 advertise a highly specific route to the mobile user and use the existing routing infrastructure to propagate this information throughout the network. We identified scalability as one concern. Suppose that when a mobile user moves from one network to another, the new foreign network advertises a specific route to the mobile user, and the old foreign network withdraws its route. Consider how routing information propagates in a distance-vector algorithm (particularly for the case of interdomain routing among networks that span the globe). a. Will other routers be able to route datagrams immediately to the new foreign network as soon as the foreign network begins advertising its route? b. Is it possible for different routers to believe that different foreign networks contain the mobile user? c. Discuss the timescale over which other routers in the network will eventually learn the path to the mobile users. a. No. All the routers might not be able to route the datagram immediately. This is because the Distance Vector algorithm (as well as the inter-as routing protocols like BGP) is decentralized and takes some time to terminate. So, during the time when the algorithm is still running as a result of advertisements from the new foreign network, some of the routers may not be able to route datagrams destined to the mobile node. b. Yes. This might happen when one of the nodes has just left a foreign network and joined a new foreign network. In this situation, the routing entries from the old foreign network might not have been completely withdrawn when the entries from the new network are being propagated. c. The time it takes for a router to learn a path to the mobile node depends on the number of hops between the router and the edge router of the foreign network for the node. (b) (2 points) Suppose the correspondent in slide 6-43 were mobile. Sketch the additional network-layer infrastructure that would be needed to route the datagram from the original mobile user to the (now mobile) correspondent. Show the structure of the datagram(s) between the original mobile user and the (now mobile) correspondent, as in slide 6-48.

7 (c) (2 points) Consider two mobile nodes in a foreign network having a foreign agent. Is it possible for the two mobile nodes to use the same care-of address in mobile IP? Explain your answer. Two mobiles could certainly have the same care-of-address in the same visited network. Indeed, if the care-of-address is the address of the foreign agent, then this address would be the same. Once the foreign agent decapsulates the tunneled datagram and determines the address of the mobile, then separate addresses would need to be used to send the datagrams separately to their different destinations (mobiles) within the visited network.

8 Question 7 (required for cs6480, extra credit for cs5480) 9 points: Read the following paper: RSVP: A New Resource Reservation Protocol, by L. Zhang et al, in the IEEE Communications Magazine, May This paper is available from Answer the following questions that are based on this paper. (a) (2 points) Write two uses of reservation/path refresh messages. 1. Recover from an earlier lost reservation message. 2. Handle receiver/sender that goes away without teardown (caused by the lack of a refresh message). 3. Deal with path changes. (b) (1 point) What is soft state? Write one advantage of using soft state? Soft state lives only for a short amount of time unless refreshed. It is robust against equipment failure. For example, if resources are reserved and due to equipment failure they are not explicitly released, these would be wasted. In the case of soft state based resource reservation, these resources will be released unless the node that reserves resources receives a refresh. (c) (1 point) Does RSVP determine what routes packet will take? Explain in one sentence. No. That is the job of the routing protocols. (d) (1 point) What is the difference between the two filter types no filter and fixed filter? (e) (3 points) Consider Figure 3 and the no filter example in the paper. Show the tables at S1, S2, and S3 after the path messages from all H1 H5 hosts, and RESV messages from H1 and H2 have all been seen by these routers. Note that the table at S1 is given in the paper. S1: in L1, L2, L6 out L1(b), L2(b), L6 S2: in L5, L6, L7 out L5, L6(b), L7 S3: in L3, L4, L7 out L3, L4, L7(b)

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