ETSN01 Exam Solutions

Transcription

1 ETSN01 Exam Solutions March 014 Question 1 (a) See p17 of the cellular systems slides for a diagram and the full procedure. The main points here were that the HLR needs to be queried to determine the location of the mobile terminal, and then the terminal is paged by all base stations in the area controlled by its current MSC. (b) The home location register stores information (e.g. IMSI, telephone number, services availabe to that user) about all subscribers belonging to the network, including the current VLR for the user. The visitor location register stores information about the users in its location area (belonging to one MSC). Note that the VLR is not about roaming between networks even on the home network a user s information will be stored in the VLR responsible for the area they are currently in. Question (a) Soft handover is when a mobile terminal is connected to two or more base stations at the same time, rather than making a sharp transition from one to the other. The station can send and receive to/from all base stations simultaneously. (b) Soft handover prevents oscillating between different cells (as there is no sharp transition point). it also allows for better reception because of simultaneous transmission e.g. if a transmission is lost from one BS it may still be received from another. Question 3 (a) To determine whne to retransmit a segment. Since TCP relies on positive ACKs, the only way to determine if a segment was not received (or not received correctly) is by waiting to see if the sender receives an ACK. The RTO determines how long the sender will wait for the Ack before retransmitting the (presumably) lost segment. 1

2 (b) An adaptive retransmission timeout is an RTO which changes according to current network conditions, in partciular round trip time. (c) 1) If we receive an ACK for a retransmitted segment, we don t know whether it was for the original transmission or the retransmission, so we can t determine the correct round trip time. ) Since TCP uses block ACKs, an ACK only represents the round trip time for the latest segment it is acknowledging. Question 3 (a) Discrete-event simulations of networks rely on random events, controlled by the random number generators of the simulator. This means that a simulation will get different results each time it is run, even with the same parameters. We therefore need to run the simulation more than once to achieve statistical significance in our results. (b) Transient data is data collected during the start-up phase of the simulation, where the behaviour may be different than once it has reached steadystate. We can avoid transient data affecting simulation results by either running the simulation for a long time so that the effect of transient data is insignificant, or by not collecting any data until we have reached steady state. (c) Sample mean: 6.3, confidence interval: [5.73, 6.9] Question 4 (a) First find ρ: ρ = λ µ Using Poliazcek-Khinchine: = 50 packets/s s = 0.4 E[Q] = ρ + ρ + λ σ () = (1 0.4) = 4.7

3 (b) Using Little s Law: r = λt r T r = r λ = = Question 6 (a) See p3-7 of the statistics and probability review slides. Viewpoint 1 is about the behaviour in a small time interval, and is given by a Bernoulli distribution: did a packet arrive or not in t? Viewpoint is the behaviour over a long time interval and is given by a Poisson distribution: how many packets arrived in the time interval? Viewpoint 3 is about the behaviour between events and is given by a negative exponential distribution: how long was the time between two packet arrivals? (b) Using the Poisson formula: (c) Again, using the Poisson formula: P (no packets = e λt (λt) 0 0! = e λt = e = e = P (3 packets) = 1 P ( or fewer packets) = 1 (P (0 packets) + P (1 packet) + P ( packets)) ( e λt (λt) 0 = 1 + e λt (λt) 1 + e λt (λt) ) 0! 1!! ( ) = e + e 4 = =

4 Question 7 (a) The Kleinrock and Tobagi model of p-persistent CSMA derives metrics such as throughput and delay starting from the probability distribution of the packet arrival process. It assumes an infinite number of nodes collectively forming a Poisson packet arrival process, with propagation delay assumed to be equal between all nodes. The Bianchi model of the DCF uses a -dimensional Markov chain to derive performance metrics. It assumes a network at saturation, that is, all nodes always have a packet queued to send, and as such does not model the time taken to transmit each packet, instead only tracking which step of the DCF each node is up to. (b) The kleinrock and Tobagi model uses a Poisson packet arrival process collectively amongst all the nodes. The Bianchi model uses saturation: all nodes always have a packet queued to send, i.e. a packet always arrives at each node immediately after one has been sent. (c) The Bianchi model yields more tractable analysis because of the assumption of saturation conditions. This allows time to not be directly modelled, greatly simplifying the analysis. Question 8 (a) ALOHA is a medium access protocol for wireless packet networks in which when a node has a packet ready to send, it sends it, without regard to other nodes. There is no channel sensing or reservation. ALOHA uses separate channels for uplink and downlink and relies on positive acknowledgements for nodes to know whether their packets were received. (b) The utilisation for first period (TDMA) is 0.5, as each node uses its (entire) slot with probability 0.5, i.e. on average half of the slots, and hence half of the time, will be used. There are no collisions so all packets transmitted are useful data. The utilisation for the second period (ALOHA) follows the formula given: λe λ. The two periods occupy equal amounts of time, so the total utilisation is given by: U = λe λ = λe λ 4

5 (c) Take the derivative of the utilisation: du dt = 0 + λ e λ + 1 e λ = λ e λ + e λ = e λ ( 1 λ ) The maximum occurs when the derivative is zero. Since e λ 0 λ, we have: 1 λ = 0 λ = 1 λ = 1 (since λ > 0) Question 9 (a) Any diagram that showed an understanding of the system was accepted, including showing the two queues (with respective arrival rates) or showing the markov chains for each queue. (b) Queue 1 is an M/M/1/n system, with n=5. We can therefore use the blocking probability formula for M/M/1/n. First find ρ: ρ = λ µ = = 0.6 P B = ρn n+1 = =

6 (c) E[R] = n+1 = n iρ i i=0 5 i 0.6 i i=0 = ( ) = = 1.06 (d) Using Little s law, and noting that only non-blocked packets actually join the queue: E[R] = λ (1 P B ) E[T r ] 1.06 E[T r ] = 30 ( ) = s (e) Queue 1 is empty whenever it is in state 0. The proportion of time it will spend in this state is equivalent to the probability of being in the state. P (0) = ρ0 n+1 = = (f) Queue will only be served when Queue 1 is empty, therefore the effective service rate in Queue is: i.e. effective service time is s µ = = To determine ρ we first need the total arrival rate λ T, which includes all packets from Flow as well as blocked packets from Flow 1. λ t = λ + λ 1 P B = =

7 Then: ρ = = 0.5 Queue is an M/M/1 queue, so we can use the formula for average length: (g) Using Little s Law: E[R] = ρ = = r = λt r T r = r λ = = 0.1s (h) Packets from Flow 1 which go into Queue 1 experience the delay in Queue 1 (found in part (d)), whereas blocked packets experience the delay in Queue (found in part (g)). Hence the average delay is given by: ( ) = (i) To work out the average delay for all packets passing through the router, we can either take the average across the two flows, using the result from part (h) above, or the average across the two queues (making sure to include the blocked packets from Flow 1 in Queue s packets). Averaging across the two queues gives: 30 ( =