Homework 3 Solution Guidelines – ECE/CSC 570-001, Fall 2007 1. (Tanenbaum, 4.3) Consider the delay of pure ALOHA versus slotted ALOHA at low load. Which one is less? Explain your answer. 1. At low load, no collisions are likely. In slotted ALOHA we still need to wait for the next slot beginning time to transmit, so delay is higher. 2. (Tanenbaum, 4.4) Ten thousand airline reservation stations are competing for the use of a single slotted ALOHA channel. The average station makes 18 requests/hour. A slot is 125 µs. What is the approximate total channel load? 2. Each terminal makes one request every 200 seconds, for a total load of 50 requests/second. This allows us to find the attempt rate, which is the quantity that determines channel load (not just original transmissions but the retransmissions as well). Thus G = 50/8000 = 1/160 is the answer. 3. (Tanenbaum, 4.5) A large population of ALOHA users manages to generate 50 requests/second, including both originals and retransmissions. Time is slotted in units of 40 ms. (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? 3. 50 requests per second boils down to (50 * 40/1000) = 2 requests per 40 msec (timeslot). Thus G = 2. This is slotted ALOHA (as per the reference to the timeslot). Therefore: (a) Chance of success on first try = α = e-G = 1/e 2 = 0.135 . (b) From the geometric argument, (1- e-G)k e-G. (c) Again from the geometric argument, expected number of transmissions until success = 1/α = eG = 7.39. 4. (Tanenbaum, 4.6) Measurements of a slotted ALOHA channel with an infinite number of users show that 10 percent of the slots are idle. (a) What is the channel load G ? (b) What is the throughput? (c) Is the channel underloaded or overloaded? 4. We are given that 10% of the slots are idle, that is the probability of an idle slot is 0.1 . We can only have an idle slot if the number of packet arrivals at the different stations during the last slot time was exactly zero. From the Poisson model, as given in the “ALOHA Throughput Study” slide of the lecture or Eq. 4-2 of Tanenbaum, the probability of zero arrivals during one timeslot is G0 e-G/0! = e-G .

(a) Equating this to 0.1, we have G = 2.3 . (b) Throughput S = G e-G = 0.23 packets per slot time. (c) Since G > 1, the channel is by definition overloaded. 5. Consider the following numerical values for the slotted ALOHA system. The arrival process at each station is an independent Poisson process, with rate 0.05 (arrivals/slot time). Backlogged stations attempt re-transmission with a probability of 0.02 in each frame slot. We use the model given for the rigorous analysis of slotted ALOHA, with all the assumptions listed there. Provide numerical answers for the questions below. (a) What is the numerical probability of the arrival of a new frame at the MAC layer of an unbacklogged station? (b) What is the numerical probability of the arrival of a new frame at a backlogged station, again at the MAC layer? (c) Assume that there are 20 stations on the medium, and at a given time, 5 of them are backlogged. What is the probability that the next slot will see a single successful retransmission? 5. (a) In this question, 0.05 takes the place of λ/m, the Poisson arrival rate at each station. This is an external arrival rate, that is, a function of the traffic process(es) which cause packets to arrive at the stations, and is unaffected by the backlogged state of the stations or the buffer capacity. Of course, according to our assumptions, the backlogged stations will not be able to accept new arrivals because the stations have no buffers, and unbacklogged stations will be able to only accept the first arrival (if there is one or more arrivals). In keeping with this, the probability of a new arrival at the MAC layer of an unbacklogged station is the same as the probability that at least one external arrival occurs at the station. This can be seen to be 1 – e-λ/m either by using the CDF of the exponential distribution (see Homework 3) to obtain the probability of “the next frame to arrive does so within the next one frame time”, or using the Poisson distribution to obtain the probability of “NOT zero frames arrive during the next one frame time”. Numerically, this comes out to nearly 0.05. (b) As already explained above, the external arrivals are at the same rate for all stations, but the MACs of backlogged nodes do not accept any new arrivals, so the probability is 0. (c) Clearly, all unbacklogged stations must have no new arrivals, the probability for this happening is (1 – 0.05)(20-5) = 0.472 . But exactly one out of the 5 backlogged nodes must choose to transmit. Since there are 5 ways to choose 1 node out of 5, (5C1 = 5), the probability of this is 5 * (0.02) (1-0.02)4 = 0.092 . The overall probability of a single

successful retransmission is thus around 0.043 . 6. Consider the more rigorous analysis of slotted ALOHA we performed in class. (a) We assume the original packet arrival rate at the medium to be λ , and this rate to be divided equally among the m stations on the medium. (Note: This is different from denoting the arrival rate at each station by λ.) We also denoted the probability that a single unbacklogged station transmits a packet in a given slot by qa . Express qa in terms of λ . (b) We denoted the probability that a backlogged station retransmits a packet in a given slot by qr . The probabilities Qa (i,n) and Qr (i,n) are as defined on slide 22 of the MAC lecture. Express these quantities in terms of qa and qr . I Write down the transition probabilities Pn,n+i for the Markov Chain on slide 23 for the four value ranges of i given on that slide, using the event descriptions provided on the slide. You should satisfy yourself first that the events in fact lead to the corresponding transitions. 6. (a) The Poisson rate of arrival of new packets to all the stations are assumed to be λ. As before, this is an external arrival rate, and backlogged stations will not be able to accept new arrivals because the stations have no buffers. Thus the effective arrival rate at the medium will decrease as the number of backlogged stations increase, as we saw in class. Thus the arrival rates at individual stations are of interest. We know that the Poisson rate is distributed evenly between the different stations. As we saw in class in our discussion on renewal processes, a split or join of independent Poisson processes leads to more Poisson processes, and the rates are nicely obtained as simple sums and differences of rates. Thus the arrival process at each (unbacklogged) station is also a Poisson process, with rate λ/m. In other words, packets arrive at each station randomly with an exponential inter-arrival time, arriving on the average at a rate of λ/m packets per time unit, leading to an overall arrival rate of λ packets per time unit. The only way a station can transmit a new packet at a slot is if it was unbacklogged at the last slot, and one or more packets arrived during the last slot time. Consider a given unbacklogged station. Again, we want the probability that more than zero packets arrived during the last slot time. From the definition of the Poisson distribution, the probability of zero arrivals during one slot time is ((λ/m)0/0!) e-λ/m , thus the probability that some number other than zero of packets will arrive is the probability of the complementary event and is e-λ/m . Once this event has been satisfied, it is certain that the station will attempt to transmit the newly arrived packet at the next slot, so we need not multiply further by a probability. (Equivalently, we multiply with a probability of 1). Thus the probability we denoted by qa is given by qa = 1 – e-λ/m .

(b) These are simple exercises in combination. Keeping in mind the fact that the behavior of each station is independent from the others,

Qa(i,n) = probability of i unbacklogged stations transmit packet at the next slot = (ways of choosing i nodes out of (m-n) unbacklogged ones) × (probability that each of i nodes transmits a new arrival with probability qa) × (probability that each of (m-n)-i nodes does not transmit a new arrival) = Ci

m-n × qai × (1 – qa)m-n-i

Qr(i,n) = probability that i backlogged nodes transmit packets at next slot = (ways of choosing i nodes out of n backlogged ones) × (probability that each of i nodes retransmits a packet with probability qr) × (probability that each of n-i nodes does not retransmit a packet) = Ci

n × qri × (1 – qr)n-i

(c) Pn,n+i is the probability that the system, while at state n, will transit to the state n+i at the next event that occurs; that is, the probability that an event occurs which causes the number of backlogged stations to change from n to n+i . Consider the events listed for the various cases of i on slide 19 of the first LAN lecture. Convince yourself that the events as listed there will lead to the corresponding transitions. Now, the events have been written down (though in imprecise, natural language) so that the clauses connected by a logical OR are disjoint (hence their probabilities can be simply added together), and the clauses themselves consist of subclauses which must be satisfied simultaneously, but which are independent (hence their probabilities may be multiplied together). This is a normal form for writing down events. Accordingly, we can write down the following transition probabilities, in exact correspondence with the events:

2 ≤ i ≤ (m-n), Pn,n+i = Qa(i,n) i = 1, Pn,n+i = Qa(1,n) (1 - Qr(0,n)) i = 0, Pn,n+i = Qa(1,n) Qr(0,n) + Qa(0,n) (1 - Qr(0,n) - Qr(1,n))

+ Qa(0,n) Qr(0,n) i = -1, Pn,n+i = Qa(0,n) Qr(1,n)

7. (Tanenbaum, 4.8) How long does a station, s, have to wait in the worst case before it can start transmitting its frame over a LAN that uses (a) the basic bit-map protocol? (b) Mok and Ward’s protocol with virtual station numbers? 7. (a) Assume each frame is d bits long. The worst case is: all stations want to send and s is the highest numbered station. Wait time is N + (N-1)*d bit times as follows: N bit contention period, and (N-1)*d bit (transmission of frames by all the other stations).

You might also validly consider that the worst case is when the frame arrives at the MAC layer of station s just after its contention slot passed. In this case, it has to wait (N-1)d bit times for the current sequence of frames (every other station has frame to send), plus the above answer. This answer is also valid. (b) The worst case is: all stations have frames to transmit and s has the lowest virtual station number at the time the frame arrives. Consequently, s will get its turn to transmit after the other N-1 stations have transmitted one frame each, and N contention periods of size log2N each. Wait time is thus (N-1)*d + N* log2 N bit times. 8. (Tanenbaum, 4.9) A LAN uses Mok and Ward’s version of binary countdown. At a certain instant, the ten stations have the virtual station numbers 8, 2, 4, 5, 1, 7, 3, 6, 9 and 0. The next three stations to send are 4, 3 and 9, in that order. What are the new station numbers after all three have finished their transmissions? 8. When station 4 sends, it becomes 0; and the stations numbered (at that time) 1, 2 and 3 are increased by 1. When station 3 sends, it becomes 0; and 0, 1, and 2 are increased by 1. Finally, when station 9 sends, it becomes 0 and all the other stations are incremented by 1. The result is 9, 1, 2, 6, 4, 8, 5, 7, 0 and 3. Diagrammatically: 8 2 4 5 1 7 3 6 9 0 8 3 0 5 2 7 4 6 9 1 8 0 1 5 3 7 4 6 9 2 9 1 2 6 4 8 5 7 0 3 9. (Tanenbaum, 4.20) Two CSMA/CD stations are each trying to transmit long (multiframe) files. After each frame is sent, they contend for the channel, using the binary exponential backoff algorithm. What is the probability that the contention ends on round k, and what is the mean number of rounds per contention period? 9. A “round” is each calculation of the backoff period followed by transmission attempt. Number rounds starting with 1 – at round 1, each of the two stations attempt to transmit in the immediately following slot (no backoff), thus collision is inevitable. At round 2, there are two slots at which each station might try – after waiting 0 slots (no wait) and after waiting 1 slot. Similarly the contention at round i is over 2i-1 slots. The “contention period” is the entire period between two successful frame transmissions. At round i, the first station might pick the slot 0 with probability 1/2i-1, so might the second station. Therefore the probability of a collision by both stations picking this slot is 1/22(i-1) . However, this is true not only of slot 0, but every slot from 0 to 2i-1, so the total probability of collision at the i-th round is 1/2i-1 . Clearly the probability of having exactly k rounds is the multiplication of the probabilities of collision at round 1, round 2, round 3, … round k -1, and the probability of not having a collision at round k. The second part of the question is a little difficult to interpret. A logical interpretation is that in the first part you found out the probability Pk of having exactly k backoff rounds

between two successful frame transmissions, in this part you are being asked to find the expectation of the same quantity. The answer is then simply Σ k Pk . Realistically, the probability Pk should be “frozen” after the k = 10, and the summation carried out only to k = 16, according to the details of the standard. If you took any other reasonable definition of the second part of the problem, that is also acceptable.