Homework 1 - SOLUTION

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1 Homework - SOLUTION Problem M/M/ Queue ) Use the fact above to express π k, k > 0, as a function of π 0. π k = ( ) k λ π 0 µ 2) Using λ < µ and the fact that all π k s sum to, compute π 0 (as a function of λ and µ). = π k ( ) k λ µ λ/µ, because the sum converges as λ/µ <. Hence, we have π 0 = λ/µ. 3) Using the results above, compute the expected number of packets in the λ system at any given time. As you learnt in class, you should get µ λ. You may find it useful that ρ = λ/µ <. Observe that π 0 = ρ. Then, the expected number of packets in the system is kπ k = ρ( ρ) kρ k = rho( ρ) kρ k ρ ρk = ρ( ρ) ρ = ρ( ρ) ρ ρ = ρ ρ = λ µ λ. ( ) ρ k Notice that the sum converges for ρ <, and this allowed to swap the derivative and the sum.

2 4)What is the expected time T that a packet spends in the system (queue and service) if the arrival rate is λ and the departure rate is 3µ? M/M/c Queue T = 3µ λ. 5) What is π 0? As before, all the π s must sum up to. Therefore, It follows that = π k c ( c π 0 = ( c + π 0 (cρ c ) c + (cρ c) c + (cρ c) c ρ k ρ c ρ c ) ) The probability that all servers are occupied (and thus a new arrived packet has to wait in the queue) is π c+ = π k. This means that, with probability π c+ the new packet has to wait in line, while with probability π c+ it is served right away. 6) Compute π c+ as a function of c, ρ c, π 0. π c+ = (cρ c ) c π k k=c k=c ρ k (cρ c ) c. ρ c You could also (but, again, you don t have to) show that the expected time T c that a packet spends in this system (queue and service) is T c = ρ c λ( ρ c ) π c+. 7) What is the expected time T 3 that a packet spends in a M/M/3 queue with arrival rate λ and three servers each with service rate µ? (Do not write the complete expression for π 3+ ) First compute ρ 3 = λ 3µ 2

3 Then, T 3 = ρ 3 λ( ρ 3 ) π 3+ = 3µ λ π 3+ < 3µ λ, as π 3+ <. 8) Compare T 3 with T (computed above). For the same price, would you rather buy one outgoing link with rate 30Mbps or three outgoing links each with rate 0Mbps (assuming that you have a box that distributes your outgoing flow between them)? You would prefer the second choice as the expected delay is smaller. Problem 2: balancing the traffic. Consider the following M/M/ queues. Queue : There is an incoming flow of packets of constant length l = kbyte. The service time for a packet is T s () = 0.2ms (note that this value is a constant, as all packets have the same length), and the average time that a packet spends in the queue (waiting time in the buffer plus service time) is T () =.6ms. ) Compute the departure rate µ (). µ () = /T () s 2) Compute the arrival rate λ (). From the expression we get T () = 3) Compute the utilization factor ρ (). = 5000packets/s µ () λ (), λ () = µ () = 4375packets/s. T () ρ () = λ () /µ () = 87.5%. Queue 2: There is an incoming flow of packets of constant length l = kbyte, of rate R (2) = 0Mbps. The outgoing link is used at 75% of its capacity. 4) What is the utilization factor ρ (2)? 5) Compute the arrival rate. 6) Compute the departure rate µ (2). 7) Compute the service time T (2) s. ρ (2) = 75%. = R (2) /l = 250packets/s. µ (2) = /ρ (2) = 667packets/s. 3

4 T s (2) = /µ (2) = 0.6ms. 8) Compute the average time that a packet spends in the system T (2). T (2) = = 2.4ms. µ (2) λ (2) Consider an additional flow of packets, each of length l = kbyte, with arrival rate λ new = 400 packets per second. We want to split this flow among the two queue in a fair way, that is, in such a way that the two queues experience the same average packet delay. Writing λ new = λ () new + new, we mean that the new flow is split into two smaller flows, one with arrival rate λ () new added to Queue, and the other with arrival rate new added to Queue 2, resulting in a total flow to Queue with arrival rate λ () λ () + λ () new, and a total flow to Queue 2 with arrival rate + 9) What are the packet service time in Queue and Queue 2 once the new flows are added? Did they change? They don t change: the are still T s () = 0.2 and T s (2) = 0.6ms. 0) What is the average time T () tot a packet spends in Queue once the new flow is added? Write its expression as a function of µ (), λ (), λ () T () µ () λ () tot = µ () (λ () + λ () new). ) What is the average time T (2) tot a packet spends in Queue 2 once the new flow is added? Write its expression as a function of µ (2),, λ new, λ () new (not new). T (2) µ (2) tot = µ (2) ( + λ new λ () new). 2) To reach our goal of fairness, we want T () T (2) tot. Equate the two expressions obtained in parts 0) and ), and write an expression for λ () new in terms of µ (), λ (), µ (2),, λ new. We want that µ () (λ () + λ () new) = µ (2) ( + λ new λ () new), 4

5 that is, Therefore, µ () (λ () + λ () new) = µ (2) ( + λ new λ () new). λ () new = 2 (µ() λ () µ (2) + + λ new ). 3) Compute λ () new, λ () new = 2 (µ() λ () µ (2) + + λ new ) = 304packet/s. new = λ new λ () new = 96packet/s. 4) Compute T () tot and T (2) tot and check they are equal. T () T (2) µ () (λ () + λ () new) = 3.ms. µ (2) ( + new) = 3.ms. 5) Compute the resulting utilization factors of the queues (after the new flows are added). The resulting utilization factors are ρ () λ () tot/µ () = 4679/5000 = 93.58%, ρ (2) tot/µ () = 346/667 = 80.74%. 5