16:330:543 Communication Networks I Midterm Exam November 7, 2005

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1 l l l l l l l l 1 3 np n = ρ 1 ρ = λ µ λ. n= T = E[N] = 1 λ µ λ = 1 µ 1. 16:33:543 Communication Networks I Midterm Exam November 7, 5 You have 16 minutes to complete this four problem exam. If you know what you re doing, you should be able to complete the exam in half the allotted time. The point value of each problem is indicated. Put your name and your Rutgers netid (but no part of your SSN) on each exam book (1 points). Please read both sides of the exam carefully and ask the instructor if you have any questions points Data packets arrive at the input buffer of a transmission link as a Poisson process of rate 1 packets per second. The packets have exponential service requirements with mean 1/µ seconds.sketchamarkovchainforn, the number of buffered packets. Under what conditions on the service rate µ does the Markov chain have a stationary distribution? For those cases, find the stationary probabilities p n. What is the minimum service rate µ such that the average system time T is no more than msec? This problem is a gift. The system is an M/M/1 queue with arrival rate λ = 1 packets/sec, service rate µ. Here is the usual Markov chain for N: In terms of the offered load ρ = λ/µ = 1/µ, the stationary distribution is The average number of packets in the system is p n =(1 ρ)ρ n n =, 1,,... (1) E[N] = By Little s Law, the average system time is The requirement that T ms corresponds to µ 1 1/(.) = 5. Thus the system must have service rate µ 15 packets per second.. 5 points A collection of sensors transmit fixed unit-length packets over a shared multiaccess channel to data collector node. Packets are generated by the collection of sensors as a Poisson process of rate λ per unit time. The number of sensors is large enough so that the probability a sensor has more than one packet in its buffer is negligible. When a packet arrives, the sensor immediately begins to transmit that packet to the data collector. If all other sensors are silent during the transmission of a packet, then the transmitted packet is successfully received by the data collector. After a packet is transmitted, whether successful or not, the transmitting sensor discards that packet. At time t =, the system has been running for a long time as we begin to measure the performance of the system. 1

2 (a) Compare this system to unslotted Aloha. How is this system the same? How is this system different? The system is similar to slotted Aloha in that packets are transmitted immediately when they arrive; the system is unslotted; the model assumes that the number of terminals are high so each terminal has only one packet and thus a terminal never queues packets; overlapping transmission alays cause collisions. The system differs from slotted Aloha in that a packet is discarded when it collides with another packet transmission; there is no mechanism for backoff and retry; the system has no instability problem caused by backlogged packets; the system also has no need for feedback regarding transmitted packets. (b) What is the PMF of N(t), the number of packets generated in the interval [,t]? This is also a gift. For a Poisson process of rate λ, the number of packets in the interval [,t] has the Poisson PMF { (λt) P N(t) (n) = n e λt /n! n =, 1,,... otherwise (c) What is the probability q that a transmitted packet is successfully received by the data collector? A packet transmitted at time t is successfully received if there are no packets that start transmission either in the interval [t 1,t) or in the interval (t, t +1].Let N denote the number of arrivals over these two intervals. Since the probability of an arrival (really a second arrival) exactly at time t is zero, N is Poisson with expected value λ q = P {N =} =(λ) e λ /! = e λ. () (d) Let R(t) be the number of packets successfully received at the data collector over the interval [,t]. What is the expected success rate r = lim t E[R(t)]/t? What value of λ maximizes r? Suppose we divide the interval [,t] into t/ small intervals of size. Let I n denote an indicator variable that equals 1 if a packet is successfully transmitted in interval n. The probability of a successful transmission being initiated in interval n is E[I n ]=qλ. (3) since λ is the probability of a packet transmission in the interval and q is the probability the transmission is successful. The total number of successes is R(t) = t/ n=1 I n.

3 Note that the I n are dependent. When is small, I n =1implies I n 1 = I n+1 =. However, since the expected value of the sum equals the sum of the expeced values, no matter if they are dependent, t/ E[R(t)] = E[I n ]= n=1 t/ n=1 qλ = t qλ =qλt. It follows that R(t) r = lim = lim qλ = qλ = λe λ. t t t The maximum success rate is found by dr dλ = e λ λe λ =. This yields λ =1/, which balances setting λ high to increase the number of attempts against setting λ low to minimize the probability of a collision. (e) Is R(t) a Poisson process? Make sure to justify your answer. Successful packet transmissions represent arrivals for the R(t) counting process. However, the nature of the collisions is that if there is an arrival at time t, the next arrival cannot occur until after time t +1. Thus the process is not memoryless and is not Poisson points Packets arrive as a Poisson process of rate λ at a network. The packets proceed through a series of m unreliable communication links L 1,L,...,L m.oneach link, a transmission error can occur with probability q, independent of an error on any other link. However, error checking occurs only on an end-to-end basis; a packet error is detected at the output of link L m if an error occurs on any of the m links. If a packet is received in error, a feedback channel is used to transmit a NAK message back to the server at link L 1. Fortunately, the NAK signal is always received without error. Reception of the NAK at L 1 initiates a retransmission of the packet through links L 1,...,L m. Transmissions at each link (including the feedback link) require, on average, one unit of time for transmission. (a) Model the system consisting of links L 1,...,L m and the feedback link as a Jackson network. Sketch the network of queues. We model the system as a Jackson network with m +1 queues for the links L 1,...,L m and the feedback queue. For convenience, we call the feedback queue L. Since packets pass though each queue L j,andthen,withsomeprobability, require retrasmission: we model the system as the Jackson network: r L 1 L L m p L 1-p 3

4 The probability a packet is successfully transmitted through all m links and leaves the system is p =(1 q) m. The arrival rate at link L 1,satisfiesr = λ +(1 p)r, which implies r = λ p = λ (1 q) m. Moreover, since a packet that passes through link L 1 passes through links L,...,L m, the arrival rate is r at each of these links. (b) For what arrival rates λ is the network of queues stable? At each link L j, 1 j m, the arrival rate is r and the service rate is µ =1. Thus the offered load is ρ j = r µ = r, 1 j m. We conclude that each queue L j is stable if and only if r<1, orλ<(1 q) m. Also, we note that at the feedback link L, the arrival rate is r(1 p) <r.since the service rate on the feedback queue is also 1, the feedback queue has load ρ = r(1 p) =r[1 (1 q) m ]. The condition λ<(1 q) m is also sufficient to ensure that ρ < 1 and the feedback queue is stable. (c) Find the stationary probabilities when the queueing network is stable. Under the Jackson network model, the stationary distribution for the network has the form of m +1 independent M/M/1 queues: p(n,...,n m )= m (1 ρ i )ρ n i In this case, using the loads ρ i from the previous part, we obtain the stationary distribution i= p(n,...,n m )=[1 r(1 p)][r(1 p)] n (1 r) m r n 1+ +n m, n i. (4) (d) Identify any additional assumptions, whether physically reasonable or not, required for correctness the Jackson network model. For the Jackson network model, we make the assumption (unstated in the problem description) that the packets have exponential service times. In addition, we have to assume that the service times of a packet are independent from link to link. If the links have fixed data rates, this is equivalent to assuming the packet gets a new independent length at each queue. Note that this is not the same thing as the Kleinrock independence assumption in which we assume (or make the modeling assumption) that queues are independent. Rather, once we satisfy the assumptions of the Jackson network model, its a mathematical fact that the stationary distribution is the same as it would be if the queues were independent (despite the fact that the individual queue processes are not actually independent.) 4 i.

5 (e) Is this queueing network a time-reversible system? Explain. One approach to answering this question is to find the reverse time transition probabilities and show they are not the same as the transition probabilities of the ordinary forward time system. That s pretty tedious however. A simpler observation is to note that in reverse time, external packets will arrive at link L m and proceed through links m 1 down to link 1. At that point, a packet will either leave the system or go through the feedback queue back to the queue at link m. This process is easily distringuished from the forward time process in which the packet circulation is reversed. Hence the network is not reversible points Computation jobs arrive at a processor as a Poisson process of rate λ. For each job, the processor performs two calculations. These calculations require processing times X and Y with PDFs f X (x) = { 1/ x otherwise f Y (y) = { 1/4 y 4 otherwise The processing times required for each calculation, whether in the same job or different jobs, are independent. Given a job, a single processor performs the two calculations sequentially so the service time of a job is X + Y. A job is queued if it arrives when the processor is busy. (a) For what values of the arrival rate λ is the job queue stable? For the single processor, the service time of a packet is U = X + Y.theresulting queue is an M/G/1 queue with service time PDF f U (u) which is the convolution of the service time PDFs f X (x) and f Y (y). We can use the P-K formula to find the average waiting time E[W ]= λe[ U ] (1 λe[u]). In this problem, the appropriate definition for stability would be that E[W ] is finite since this would imply that the expected number of queued customers is finite. In this case, we would have two conditions: E [ U ] <, λe[u] < 1. Because X and Y 4, we know that U 6 and thus U 36. ThusE [ U ] is always finite. In addition, since X is uniform (, ) and Y is uniform (, 4), E[U] =E[X]+E[Y ]=1+=3. Thus the condition λe[u] < 1 simplifies to λ<1/3. (b) What is the expected waiting time E[W ]ofajob? To use the P-K formula, we need to find E [ U ]. We recall (or you can derive) that a uniform (a, b) random variable has variance (b a) /1. Since the calculation times X and Y are independent, σ U = σ X + σ Y = ( ) 1 + (4 ) 1 =

6 Since E[U] =3, U has second moment E [ U ] = σ U +(E[U]) = 5 3 Thus, by the P-K formula, for λ<3, +9= 3 3. E[W ]= λe[ U ] (1 λe[u]) = 16λ 3(1 3λ). (c) What is the expected number E[N] of jobs in the system? The expected system time of a job is E[T ]=E[W ]+E[U] = 3λ 9 11λ +3= 6(1 3λ) 3(1 3λ) By Little s Law, the expected number of jobs in the system is E[N] =λe[t ]= λ(9 11λ) 3(1 3λ). (d) points Suppose now that the processor is replaced by a dual processor system that can perform the two calculations in parallel. The processing times X and Y for each calculation remain independent and each has the same distribution as in the single processor system. A job is finished when both calculations are completed. For what range of arrival rates λ is the queue stable? What is the expected waiting time E[W ]ofajob? In this case, the service time is V =max(x, Y ) since the job is completed when both calculations are completed. The queue is still M/G/1, and by the P-K formula, the expected waiting time is E [ W ] = λe[ V ] (1 λe[v ]). However, we need to solve a probability problem to find E[V ] and E [ V ]. First we find the CDF F V (v) and then the PDF f V (v). Since V is the maximum of X and Y and since X and Y are independent, F V (v) =P {V v} = P {X x, Y v} = P {X v}p {Y v} = F X (v) F Y (v). Now we observe from the PDFs f X (x) and f Y (y) that x<, x<, F X (x) = x/ x, F Y (y) = y/4 y 4, 1 <x, 1 4 <y. This implies v< (v/)(v/4) v, F V (v) =F X (v) F Y (v) = v/4 v 4, 1 v>4. 6

7 Taking derivatives, V has PDF f V (v) = df V (v) dv = v/4 v, 1/4 v 4, otherwise. Thus the moments of V are and E[V ]= E [ V ] = v(v/4) dv + 4 v (v/4) dv + v(1/4) dv = v3 1 4 From the P-K formula, it follows that + v 8 v (1/4) dv = v4 16 E[W ]= λe[ V ] (1 λe[v ]) = 17λ 6 13λ. 4 = = v3 4 1 = We conclude that the queue is stable for arrival rates λ<6/13. Quite a few people guessed tht the queue was stable for rates λ<1/, since the longer calculation required expected time E[Y ]=. However, since the calculation times are random, sometimes X>Y. Averaged over all possible outcomes, this results in E[V ]=13/6 >. 7

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