Final Exam: Mobile Networking (Part II of the course “Réseaux et ...

Final Exam: Mobile Networking(Part II of the course “Réseaux et mobilité”)Prof. J.-P. HubauxFebruary 12, 2004Duration: 2 hours, all documents allowedPlease write your answers on these sheets, at the end of each question;use extra sheets if necessary (put your name on them)You may write your answers in English or in French.The total number of points (65) corresponding to the proposed questions is higher than the number ofpoints (50) required to obtain the highest mark.Let p be the number of points you obtain; your final mark to this exam will be min(6, (1 + p/10)).This document contains 13 pages.Student First name:Last name:(answers to the questions are shown in italic and blue)

1 WLANs (30 points)1.1 IEEE 802.11 (10 points)Consider the following scenario:S1 D S2S1 and S2 send CBR / UDP traffic to the common destination D, using IEEE 802.11 (DCF).1.1.1 Scenario 1: Consider S1, S2 and D all within receive range of each othera) When the basic scheme is used (no RTS/CTS):• Describe a collision (what happens before, during, and after)S1DDIFSDataSIFSACKDIFSDatabkf = rand(doubled CW)DataSIFSACKTimeS2Data• What does the collision probability depend on ?Contention window sizePacket rate (but not packet size)Number of stations• Consider the total channel capacity to be X b/s. S1 and S2 send data flows at rates Y 1 = Y 2 >X. How is the channel capacity shared between S1 and S2?S 1 → X/2S 2 → X/22

) When RTS/CTS is used• What are the changes to the previous answers ? Comment also on the throughput and fairness.Channel is still equally sharedwith more overhead (throughput decreases)Shorter collision duration1.1.2 Scenario 2: Consider that S1 and S2 are hidden from each other, only D hears both (S1cannot even sense S2, and vice versa)a) When the basic scheme is used (no RTS/CTS):• What are the changes to the previous answers ?Collision probability depends also on the packet sizeThroughput decreases considerablythroughput S1 = throughput S2 = X ′ /2 X ′ /23

1.2 UDP and TCP over IEEE 802.11 (13 points)Consider the following scenario:S0192.168.10.110Mb/sAP192.168.10.2192.168.12.1MS1192.168.12.22Mb/sMS2192.168.12.3where S0 is a station wire-connected to the access point (AP). MS1 and MS2 are wireless nodesassociated to the AP, using IEEE 802.11 (DCF basic mode, i.e. no RTS/CTS). The cable capacity is10 Mb/s, the total wireless channel capacity is 2 Mb/s. Note that all downlink flows share the samequeue at the AP.Scenario 1: Consider the channel to be clear (no noise). S0 sends 1 CBR/UDP flow to MS1at 5 Mb/s, and 1 CBR/UDP flow to MS2 at 5 Mb/s. (Both flows go through the AP)• What is the throughput and packet drop rate for each flow ?Throughput = 1 Mbps, drop rate = 4 Mbps(or, more precisely) Throughput = 0.8 Mbps, drop rate = 4.2 Mbps• What is the probability of a collision on the radio channel ? Explain why.P(collision) = 0, i.e. no collisions, since only the AP is transmitting4

Scenario 2: S0 sends 1 CBR/UDP flow to MS1 at 5 Mb/s, and 1 CBR/UDP flow to MS2 at 5Mb/s, (both flows go through the AP). The channel between the AP and MS2 has a high errorrate.• Comment on the throughputs received at MS1 and MS2. Compare them to the result of Scenario1.T hroughput S2

• Compare the throughput to that of Scenario 2.T hroughput < 2Mbps, because of the overheadand because of the collisions between the AP (data) and MS1 (ACKs)• What is the packet drop rate ?≈ 0, because TCP adapts to the available data rateScenario 4: The channel is clear, S0 sends 1 TCP flow to MS1 and 1 TCP flow to MS2, (bothflows go through the AP).• Describe a collision (on the wireless channel) in this case, followed by a successful transmission,using the following figure:TimeS0APDataDataDataSIFSDIFSDatabkf = rand(doubled CW)DataMS1ACKTCP−ACKMS2Scenario 5: Since TCP (at S0) is unable to distinguish if a packet was dropped because ofcongestion at the AP or on the wireless channel, one possible solution is to install a “TCP proxy”at the AP, that sends back TCP-ACKs whenever a packet is successfully received from the wiredpart, and forwards the packet on the wireless channel.• What are the advantages ?TCP will not retransmit in case of channel packet loss, MAC will do it faster than TCP woulddo.TCP does not perform congestion control when loss is on the channel (not necessarily congestion).6

• What are the drawbacks ?TCP will adapt to the wired link capacity, which has a higher data rate than what can behandled at the wireless channel (which needs an infinite queue length)1.3 On the vulnerabilities of IEEE 802.11 (7 points)• Why is SIFS shorter than DIFS ?To give priority to ACKs, which must be transmitted before any new data packet.To increase his share of the throughput, a cheater may use the following cheating methodsin IEEE 802.11:• The cheater sends his frame before the end of DIFS• The cheater uses contention windows smaller than those specified in the standardQuestions:• Which one is more efficient ? Why ?DIFS. It gives the cheater absolute priority (like the priority SIFS gives to ACKs)• Which one is easier to detect ? Why ?DIFS. Small contention windows give statistical transmission successes, while DIFS gives absolutetransmission success.7

2 Cellular Networks (15 points)2.1 Cellular principles• What is the difference between multiple access (e.g., FDMA) and duplex (e.g., FDD)? (2 points)– Multiple access allows a number of users to access a single radio-frequency (RF) channelwithout interference.– Duplex allows both ends of a communication channel to send and receive signals at thesame time.• Is it possible to have soft handover in GSM? Why? (2 points)No. Because in GSM, which uses TDMA instead of CDMA, a handset cannot communicatewith more than one base station at the same time.• Explain in a few sentences what the near/far effect is and how it is coped with in cellular networks.(2 points)Near/far effect is a problem specific to CDMA-based cellular networks. If there is more thanone active user, the interference power with respect to a given user (reference user) at the basestation is the transmitted power of other users (non-reference users) suppressed by a factordepending on the code used by the CDMA system. However, if some non-reference users arecloser to the base station than the reference user, it is possible that the interference caused bythese non-reference users (however suppressed) has more power than the reference user. As aresult, the SIR of the reference user becomes too lower to allow correct reception. Control thetransmission power of each user such that the received power at the base station is equal forall user is a solution.• Consider a cellular network with a given propagation loss exponent α; describe two ways toachieve a certain minimum required SIR? (3 points)– Determine a co-channel reuse ratio Q = √ 3N. Note that a larger Q implies a largercluster size and less channels assigned to each cell.– Use sectoring technique to further divide a cell into multiple sectors. This means that theusage of directional antennae and a sacrifice of trunk efficiency within a cell.8

2.2 Capacity of Cellular SystemsAssume that in a FDMA cellular system, a minimum SIR of 17dB is required for satisfactory downlinkchannel performance. Given a propagation loss exponent α = 4, propose three frequency reusepatterns for the system. Note that the cluster size N should be the least integer that satisfies the minimumrequired SIR; choosing an unreasonably large N will not be considered as a solution (Hint:partitioning a cell into three 120 o sectors is not the only feasible way to do sectoring). (6 points)S• No sectoringI = ( D R )4 1 6(i.e., there are 6 co-channel interference cells).• 120 ◦ SsectoringI = ( D R )4 1 2(i.e., there are 2 co-channel interference cells).• 60 ◦ SsectoringI = ( D R )4 (i.e., there is only 1 co-channel interference cell).The requirement is that:S≥ 17dB = 50.12ISo, considering that ( D R )4 = 9N 2 and N can only take values of 1, 3, 4, 7, 9, · · ·, we have:⎧⎪⎨ 7 no sectoringN = 4 120 sectoring⎪⎩ 3 60 ◦ sectoring◦9

3 Ad hoc networks (20 points)3.1 Throughput (6 points)In static multi-hop ad hoc networks, the maximal throughput decreases approx. with √ N, where Nis the number of nodes in the network.a) Explain precisely which throughput decreases at that rate.The throughput that decreases at that rate is the throughput of the end-to-end traffic betweenrandomly chosen sources and destinations.b) Explain the intuition behind this throughput decrease.In an ad hoc network, nodes that are sufficiently distant can transmit concurrently. Therefore the totalamount of data that can be simultaneously transmitted for one hop increases linearly with the totalarea of the ad hoc network. If node density is constant, this means that the total one-hop capacityis O(n), where n is the total number of nodes. However, as the network grows larger, the numberof hops between each source and destination may also grow larger, depending on communicationpatterns. One might expect the average path length to grow with the spatial diameter of the network,equivalently the square root of the area, i.e., O( √ n). With this assumption, the total end-to-endcapacity is roughly O(n/ √ n), and the end-to-end throughput available to each node is O(1/ √ n).c) Explain how each of the following situations influences throughput:(i) The message sources and destinations are necessarily neighborsIf the message sources and destinations are neighbors, all the communications are single-hop and thethroughput is high.(ii) The nodes send video traffic at a constant rate to randomly chosen destinationsIf destinations are chosen at random, the communications may be multi-hop, which decreases theend-to-end throughput. Furthermore, the video traffic assumes streaming, which induces more interference,more collisions and therefore low end-to-end throughput.10

(iii) The nodes communicate by occasionally sending messages (e.g., SMS) to randomlychosen destinations.As explained previously, multi-hop communications decrease end-to-end throughput. However, occasionaltraffic leads to less interference and collisions than in the previous case; the end-to-endthroughput is therefore higher than for the previous case.(iv) Nodes are mobile, but the application is not latency-sensitive.If the application is not latency-sensitive, this means that buffering and retransmissions are possible.Therefore, the throughput is expected to be high.3.2 Routing (6 points)Today’s mainstream proposals for ad hoc network routing are DSR and AODV protocols.a) Would it be efficient to make use of AODV or DSR protocols for routing in today’s Internet ?If Yes, explain how. If No, explain why.No. Indeed, the Internet is a wired network where the majority of the nodes are static and where thequantity of information exchanged is very important; a reactive routing protocol such as AODV andDSR is therefore not suitable for this kind of network.11

) Assume that you need to implement a messaging service in a mobile ad hoc network in whicheach user is sending messages at a rate of 1 message/10 minutes. Would you use AODV or DSRfor message routing? Explain why and how. If Yes, explain how. If No, what would be your otherproposals?There is no need to establish a route between the source and the destination if the message sendingfrequency is low (1 message/10 minutes). Indeed, both AODV and DSR flood the network with requestsin order to get routes to a given destination. However, the returned route will be used to send onlyone packet (the route will most likely be stale after 10 minutes), which makes no sense. An alternativesolution would be to flood the network with the message to be sent; the message will eventually reachthe destination and the cost of the solution is lower than the cost of AODV or DSR.3.3 Security (8 points)a) How would you explain the fact that “Mobility breaks the security-routing dependency cycle”? Arethere other techniques by which the security-routing dependency cycle can be avoided?To establish a security association between nodes A and B, we don’t rely on multi-hop routes (whichleads to the security-routing dependency cycle), we rather establish it when A and B are within thepower range of each other; mobility increases the probability of encounter between the two entities.Other techniques can be used to avoid the security-routing dependency cycle (Preloading all pairs ofkeys into nodes, On-line authentication servers, ...).12

) Why is security important for routing? Describe one simple attack by which a single maliciousnode can disable all communication of its neighbors.If the routing protocol is not secure, an attacker can easily disrupt the communications in the ad hocnetwork.E.g., by sending forged routing packets, an attacker could route all packets for some destination toitself and then discard them. This attack is called “Black hole attack”.c) Consider an ad hoc network running a DSR protocol. If all nodes in the network share pairwisekeys, and if all mutual communication between nodes is encrypted with the keys that they share (bothpacket headers and payload), does this make the network resistant to attacks? If yes, explain how. Ifno, describe the attacks that you believe can harm the network routing protocol operation.No. Indeed, even if all nodes in the network share pairwise keys, the “Black hole attack” describedabove is still possible.13