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Final Exam: Mobile Networking
(Part II of the course “R´eseaux et mobilit´e”)
Prof. J.-P. Hubaux
February 12, 2004
Duration: 2 hours, all documents allowed
Please 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 of points (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 S
S1 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 other
a) When the basic scheme is used (no RTS/CTS):
- Describe a collision (what happens before, during, and after)
Data
Data
ACK
SIFS
DIFS
S
D
S
DIFS Time Data Data
SIFS
ACK
bkf = rand(doubled CW)
- What does the collision probability depend on? Contention window size Packet 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/ 2 S 2 → X/ 2
1.2 UDP and TCP over IEEE 802.11 (13 points)
Consider the following scenario:
2Mb/s MS1 MS
AP
S
10Mb/s
192.168.10.
192.168.12.
192.168.12.
192.168.10.
192.168.12.
where S0 is a station wire-connected to the access point (AP). MS1 and MS2 are wireless nodes associated to the AP, using IEEE 802.11 (DCF basic mode, i.e. no RTS/CTS). The cable capacity is 10 Mb/s, the total wireless channel capacity is 2 Mb/s. Note that all downlink flows share the same queue at the AP.
Scenario 1: Consider the channel to be clear (no noise). S0 sends 1 CBR/UDP flow to MS at 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 transmitting
Scenario 2: S0 sends 1 CBR/UDP flow to MS1 at 5 Mb/s, and 1 CBR/UDP flow to MS2 at 5 Mb/s, (both flows go through the AP). The channel between the AP and MS2 has a high error rate.
- Comment on the throughputs received at MS1 and MS2. Compare them to the result of Sce- nario 1. T hroughputS 2 << 1 M bps T hroughputS 1 < 1 M bps, because of sharing the queue, waiting for the retransmissions to MS2 to take place.
Scenario 3: The channel is clear, S0 sends TCP packets to MS1 only (via the AP).
- Using the notations in the legend for different TCP, IP and MAC packets, describe one cycle of packet exchange using the following figure:
S0 AP MS
IP[ TCP[data] ]
TCP[data]
Time
TCP−ACK IP[ ... ] MAC[ ... ] MAC−ACK
Legend
DIFS+backoff
DIFS+backoff
SIFS
MAC(IP(TCP−ACK))
MAC−ACK
SIFS
MAC(IP(...))
MAC−ACK
IP(TCP−ACK)
- What are the drawbacks? TCP will adapt to the wired link capacity, which has a higher data rate than what can be handled 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 methods in IEEE 802.11:
- The cheater sends his frame before the end of DIFS
- The cheater uses contention windows smaller than those specified in the standard
Questions:
- 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 ab- solute transmission success.
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) channel without interference. - Duplex allows both ends of a communication channel to send and receive signals at the same 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 communicate with 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 net- works. (2 points)
Near/far effect is a problem specific to CDMA-based cellular networks. If there is more than one active user, the interference power with respect to a given user (reference user) at the base station is the transmitted power of other users (non-reference users) suppressed by a factor depending on the code used by the CDMA system. However, if some non-reference users are closer to the base station than the reference user, it is possible that the interference caused by these non-reference users (however suppressed) has more power than the reference user. As a result, the SIR of the reference user becomes too lower to allow correct reception. Control the transmission power of each user such that the received power at the base station is equal for all user is a solution.
- Consider a cellular network with a given propagation loss exponent α; describe two ways to achieve a certain minimum required SIR? (3 points) - Determine a co-channel reuse ratio Q =
3 N. Note that a larger Q implies a larger cluster size and less channels assigned to each cell.
- Use sectoring technique to further divide a cell into multiple sectors. This means that the usage of directional antennae and a sacrifice of trunk efficiency within a cell.
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 N is 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 between randomly 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 total amount of data that can be simultaneously transmitted for one hop increases linearly with the total area of the ad hoc network. If node density is constant, this means that the total one-hop capacity is O(n), where n is the total number of nodes. However, as the network grows larger, the number of hops between each source and destination may also grow larger, depending on communication patterns. 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-end capacity 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 neighbors If the message sources and destinations are neighbors, all the communications are single-hop and the throughput is high.
(ii) The nodes send video traffic at a constant rate to randomly chosen destinations If destinations are chosen at random, the communications may be multi-hop, which decreases the end-to-end throughput. Furthermore, the video traffic assumes streaming, which induces more inter- ference, more collisions and therefore low end-to-end throughput.
(iii) The nodes communicate by occasionally sending messages (e.g., SMS) to randomly chosen destinations. As explained previously, multi-hop communications decrease end-to-end throughput. However, oc- casional traffic leads to less interference and collisions than in the previous case; the end-to-end throughput 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 the quantity of information exchanged is very important; a reactive routing protocol such as AODV and DSR is therefore not suitable for this kind of network.
b) Why is security important for routing? Describe one simple attack by which a single malicious node can disable all communication of its neighbors. If the routing protocol is not secure, an attacker can easily disrupt the communications in the ad hoc network. E.g., by sending forged routing packets, an attacker could route all packets for some destination to itself 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 pairwise keys, and if all mutual communication between nodes is encrypted with the keys that they share (both packet headers and payload), does this make the network resistant to attacks? If yes, explain how. If no, 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” described above is still possible.
