requests/sec. The total channel load is requests/sec. Using slot as the time unit, the total channel load is 50 ( ) = 1

Transcription

1 Prof. X. Shen E&CE 70 : Examples #2 Problem Consider the following Aloha systems. (a) A group of N users share a 56 kbps pure Aloha channel. Each user generates at a Passion rate of one 000-bit packet every 00 sec, even if the previous one has not yet been sent. What is the maximum value of N? (b) Ten thousand airline reservation stations are competing for the use of a single slotted Aloha channel. The average station makes 8 requests/hour. A slot is 25 µsec. What is the approximate total channel load? (a) With pure ALOHA the maximum usable bandwidth is kbps. Each user requires bps, so N users. 8 (b)the average station makes requests/sec. The total channel load is 0000 requests/sec. Using slot as the time unit, the total channel load is 50 ( ) Problem requests/slot. A large population of Aloha users manages to generate 50 requests/sec, including both new requests and retransmissions. Time is slotted in units of 40 msec. (a) What is the chance of success on the first attempt? (b) What is the probability of exactly k collisions and then a success? (c) What is the expected number of transmission attempts needed? The channel traffic load rate G 50 ( ) 2 requests/slot. In one slot, k requests happen with the probability π k Gk e G. (a) First attempt succeed with the probability Pr[no other request (new or retransmission) occurs within the first slot] π 0 e G e (b) If assuming that things happening in different slots are independent, the probability is ( e G ) k e G k (c) The expected number of transmission needed E traffic load throughput G S G Ge G eg 7.4 Problem 3 A small slotted Aloha system has only k customers, each of whom has a probability /k of transmitting during any slot (both new and retransmission). What is the channel throughput as a function of k? Evaluate this expression numerically for k 2, 3, 4, 5, 0, and lim k A packet transmission succeed in a slot when only one customer attempt to use it and the others not. The probability is C k ( k )k k ( k )k, just meaning the channel throughput is S ( k )k Page of 5

2 Prof. X. Shen E&CE 70 : Examples #2 packets/slot. When k 2, 3, 4, 5, 0, the throughput S are respectively , ,0.429, , Considering the situation lim k + lim ( k + k )k lim ( k + k )k lim ( k + k ) lim ( + x + x ) x lim ( k + k ) lim x + ( + lim ( x )x k + k ) e (x k) Note: Here we can see, the average load in this system is G k k requests/slot. When the population number goes to, the throughput can be calculated by Poisson approximation Ge G e S. C k k, lim x + ( + x )x e Problem 4 Consider a slotted Aloha system with m user terminals, each with an infinitely large buffer. Each terminal generates traffic at a Poisson rate of λ/m. All new arrivals are allowed to enter the system, but are considered as backlogged immediately rather than transmitted in the next slot. While a terminal contains one or more packets, it independently transmits one packet in each slot, with probability q r. Assume that any given transmission is successful with probability p. (a) Show that the mean time from the beginning of a backlogged slot until completion of the first success by a given terminal is /, and that the second moment of this time is (2 )/( ) 2. (b) The assumption of a constant success probability allows each terminal to be considered independently. The results of part (a) can be used as the service time. Show that the mean transit delay is T q r p( ρ) + 2p 2( ρ), ρ λ m (a) Let X be the time in slots from the beginning of a backlogged slot until the completion of the first success at a given terminal. Let q q r p and note that q is the probability that the backlogged terminal will be successful at any given slot. Thus, P {X i} q( q) i ; i (X i means in the first i- slots, the transmission fails, in the ith slot, the transmission is successful.) We see X is geometric distributed, so the mean value E{X} q q r p Also the variance of X is σx 2 q q, therefore the second moment of X can be calculated as 2 E{X 2 } σ 2 X + E 2 {X} q q 2 + q 2 2 q q 2 2 ( ) 2 Page 2 of 5

3 Prof. X. Shen E&CE 70 : Examples #2 (b) Since each terminal is allowed to be considered independently, and we know the arrival is Poisson with rate λ/m, each terminal can be viewed as a M/G/ queue. The first and second moment of service time X have been found in part (a). Thus we can get the mean response time of the M/G/ queue: T µ + ρ/µ( + C2 b ), 2( ρ) where, µ is the mean service time, namely, as calculated in part (a). The utilization factor ρ λ m λ m. Cb 2 σ2 b b σ2 2 X E 2 {X} q. Using all these parameters, we can get: T + ρ(2 ) 2 ( ρ) 2 ρ 2 ( ρ) However, here we have a slotted Aloha system, only at the beginning of each slot, a packet can be served. If a packet arrives within a slot, it has to wait for service until the beginning of next slot. Since the arriving time of a packet is uniformly distributed during a slot, averagely, a packet has to wait /2 slot to get service. Counting in this delay, the mean transit delay should be: T T + /2 2 ρ 2 ( ρ) + ρ 2 ( ρ) ( ρ) + 2ρ 2( ρ) Problem 5 Assume that the number of packets n in a slotted Aloha system at a given time is a Poisson random variable with mean n. Suppose each packet is independently transmitted in the next slot with probability /n. (a) Find the probability that the slot is idle. (b) Show that the posterior probability that there were n packets in the system, given an idle slot, is Poisson with mean n. (c) Find the probability that the slot is successful. (a) Given k packets in the system, with each packet independently transmitted in a slot with probability /n, the probability of an idle slot, P {I k} is ( n )k. The joint probability of an idle slot and k packets in the system is then P {I, k} P {I k}p {k} ( n )k e n n k e n (n ) k e n (n ) k P {I} P {I, k} e k0 k0 The probability that the slot is idle P {I} e. (b) Using the result above, we can find P {n I}, the posteriori probability that there were n packets in the system, given an idle slot. P {n I} P {n, I} P {I} e n (n ) n e n! e (n ) (n ) n n! Page 3 of 5

4 Prof. X. Shen E&CE 70 : Examples #2 Thus, this probability is Poisson with mean n Note: Posterior probability is the conditional probability that is assigned after the relevant evidence is taken account. P (A B) P (A B) P (B) P (B A)P (A) P (B) P (A B) P (A B)P (B) P (B A)P (A) (c) Similarly as part (a), we can first find the joint probability of success and k in the system, namely P {S, k}, then calculate the successful probability P {S}. P {S, k} P {S k}p {k} k( n )k n e n n k P {S} P {S, k} k k e n(n )k (k )! e e n(n )k (k )! Problem 6 Consider an unslotted CSMA/CD system in which the propagation delay is negligible compared to the time β required for a user terminal to detect that the channel is idle or busy. Assume that each packet requires one time unit for transmission. Assume further that β time units after either a successful transmission or a collision ends, all backlogged terminals attempt transmission after a random delay and that the composite process of initiation times is Poisson of rate G (up to time β after the first initiation). For simplicity, assume that each collision lasts for β time units. (a) Find the probability that the first transmission initiation after a given idle detection is successful. (b) Find the mean time from one idle detection to the next. (c) Find the throughput (for the given assumption). (d) Optimize the throughput by selecting value of G. (a) The first transmission after a given detection will be successful if no other transmission starts within the next β time units. Since the process of initiation is Poisson with rate G, the probability of this is P succ e βg (P (k initiation) (βg)k e βg, k 0,, 2...) (b) The mean time to the first initiation after an idle detection is /G (note that all terminals detect the channel as being idle at the same time, and the inter-arrival time of a Poisson process is exponential distributed). If this first initiation is successful, (transmission time)+β(for channel detection) are required until the next idle detection; if the initiation is unsuccessful, β(collision time)+ β(for channel detection) time units are required. Thus, E{time between idle detects} G + ( + β)e βg + 2β( e βg ) (c) The throughput can be found by calculating the fraction of time the channel is successfully used. Since the time can be divided to idle detection to idle detection cycles. In such one cycle, time that a channel is used without collisions is P succ e βg. Thus, S e βg G + ( + β)e βg + 2β( e βg ) (G + 2β)e βg + ( β) Page 4 of 5

5 Prof. X. Shen E&CE 70 : Examples #2 (d) To simplify the calculation, we can maximize the throughput S by minimizing /S. Taking the derivative of /S with respect to G and setting it equal to 0, we find that the minimum of /S occurs at βg /2. Submitting this into the expression for S, we get S + β(4 e ) β CSMA CSMA/CA (Carrier Sense Multiple Access/Collision Avoidance) is the channel access mechanism used by most wireless LANs in the ISM bands. A channel access mechanism is the part of the protocol which specifies how the node uses the medium: when to listen, when to transmit... CSMA/CA is derived from CSMA/CD (Collision Detection), which is the base of Ethernet. The main difference is the collision avoidance: on a wire, the transceiver has the ability to listen while transmitting and so to detect collisions (with a wire all transmissions have approximately the same strength). But, even if a radio node could listen on the channel while transmitting, the strength of its own transmissions would mask all other signals on the air. So, the protocol can t directly detect collisions like with Ethernet and only tries to avoid them. Under CSMA/CD, when a station has data to send, it first listens to determine whether any other station on the network is occupying the medium. If the channel is busy, the station will wait until it becomes idle before transmitting data. Since it is possible for two stations to listen at the same time and discover an idle channel, it is also possible that two stations could then transmit at the same time. When this occurs a collision will take place, and then a jamming signal is sent throughout the network in order to notify all stations of the collision. The stations will then wait for a random period of time before re-transmitting their respective frames. Currently the 802. DCF resolves collision through multiple levels of CWs and backoff stages. In the initial backoff stage (stage 0), the value of CW has the minimal value CWmin. After each collision, the CW will be doubled until reaching the maximum CWmax. After each successful transmission, the backoff will resume with initial stage (0) and the CW will be reset to CWmin regardless the network condition or the number of competing nodes. By resetting the CW to CWmin, DCF increases the probability of collision and frequent retransmissions remain high until the CW attains appropriate values. This is obviously no optimal since high collision rate in the network means poor network exploitation. On the other hand, the intrinsic backoff randomness makes it difficult to instantaneously absorb an increasing number of flows. The backoff process is basically intended to reduce the collision rate when using a higher contention stage. Page 5 of 5