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533 CS - HOMEWORK 7 SOLUTION Page 1 ---------------------------- Version 4, May 7, 2000 Problem 1 (4 Points) --------- Short Answer: A has to wait 0.1 second, and B has to wait 2 seconds. Since A has higher priority than B, A only has to wait if a B packet gets on the wire (begins transmission) during a period when there are no A packets. On the other hand, B will always have to wait if there are any A packets (since the server is work conserving). We are given the following leaky bucket parameters: Flow beta rho ----------------------------- A 10 pkts 5 pps B 10 pkts 2 pps Since the link capacity is r = 10 pps (packets per second), there is enough capacity to handle the average aggregate traffic rate of 7 pps. The worst-case scenario for flow A is if B bursts and gets on the wire before A is able to send a packet. But in that case, A needs to wait at most for 1 B packet being transmitted at a link speed of 10 pps; i.e., A has to wait at most 0.1 second. The worst-case scenario for flow B is if A sends a maximum burst before B is able to send a packet. Even after the burst, A can continue to send at an average rate of 5 pps. In that case, B must wait for any link idle time during A's continuous average transmission phase. The link will drain at a rate r until such time t that the queue is depleted of A packets: 0 = rho_A t + beta_A - r t Solving for t: t = beta_A / (r - rho_A) = 10 pkts / (10 pps - 5 pps) = 2 sec. Thus, in the worst-case, B must wait for 2 sec. Page 2 Problem 2 (4 Points) --------- Short Answer (GPS Scheduling): 16 Kbits The weights indicate the guaranteed relative service rates for each flow. Since flow A is continuously backlogged in [0,1], there is no excess capacity from flow A that can be applied to flow B. Thus, flow B is guaranteed its rate which is proportional to its weight; i.e., 16 Kbits (= w_B x S_B = 4 x 4 Kbits). This follows from Equ. (9.5): g_i = r x w_i / (Sum w_i) which implies that between the two flows A and B the following holds: g_B / w_B = g_A / w_A or g_B = g_A x w_B / w_A or g_B x T = (g_A x T) x w_B / w_A = 4 Kbits x 4 / 1 where T is the length of any measurement period. Here, (g_i x T) is the service received by flow i in a period of length T when all flows are backlogged. Short Answer (FCFS Scheduling): No gurantee The weights have no meaning in FCFS scheduling, and there is no guarantee associated with FCFS scheduling. The packets are served in order of arrival. Page 2 Problem 3 (10 Points = 4 + 3 + 3) --------- [ <<< CAVEAT >>> In the computations below, we assume a byte-by-byte WFQ scheduler. Round and finish numbers for a bit-by-bit scheduler can be obtained by scaling by a factor of 8. ] We are given the following information about two connections and a WFQ scheduler: L_1 200 bytes Packet length on connection 1 w_1 2 Connection 1 weight L_2 400 bytes Packet length on connection 2 w_2 5 Connection 2 weight r 200 Bps Link rate a) Short Answer: Connection 1 finishes at real time 2 sec. and connection 2 finishes at time 3 sec. WFQ schedules packets according to finishing numbers which reflect an emulation of GPS. From Equ. (9.9), the finish numbers are: F (1,0,0) = L_1 (0) / w_1 = 200 / 2 = 100 F (2,0,0) = L_2 (0) / w_2 = 400 / 5 = 80 Since the packet from connection 2 will finish first, the WFQ scheduler will schedule it first. since But the link rate is 200 Bps, it will finish at real time: t = 400 bytes / 200 Bps = 2 sec. At this time the packet from connection 1 will begin transmission and finish at real time: t = 2 sec + 200 bytes / 200 Bps = 3 sec. b) Short Answer: Connection 2 will complete service when the round number is 57.1, and connection 1 will complete service when the round number is 100. The round number is a real-valued variable that increases at a rate inversely proportional to the sum of the active connection weights. In this problem, the sum of weights at time 0 is 7 so that in each round one byte is serviced from each of two queues and the round number (virtual time) is increased by 1/7. However, the actual computation of the round number needs to be done only when a packet arrives or departures. When the first packet on connection 2 finishes at real time 2 sec., the round number is 57.1 (= 400/7). But the slope of the service (emulated) curve will remain at 1/7 until the round number reaches 80 which happens at real time 2.8 sec (since 200 x 2.8 sec / 7 = 80). The packet from connection 1 is in service in the real time interval [2.0, 3.0]. In this interval, the slope of the round number curve will initially be 1/7 but will change to 1/2 (= 1/w_1) at real time 2.8 sec. At real time 3.0 sec, the packet from connection 1 will complete with the round number at 100. Note, however, that the round number needs to be computed only when the first and second packets complete service. Page 3 c) Short Answer: It's finish number would be 110. At real time 1.5 seconds, the packet from connection 2 is still in service, and the first packet from connection 1 is still queued for service. Thus, the finish number of the new packet is the finish number of the packet at the front of connection 1's queue plus the weighted length of the new packet: F(1, 1, 1.5) = F(1, 0, 1.5) + P(1, 1, 1.5) / w_1 = 100 + 20 / 2 = 110 This would occur at real time 3.1 sec. We have not included the round number in the above equation, but it is guaranteed to be less than the finish number of the queueed packet. The round number at real time 1.5 sec since: R(1.5) = (200 Bps x 1.5 sec) / 7 = 42.9 Problem 4 (4 Points) --------- Short Answer: 1.5 sec DRR scheduling attempts to approximate GPS scheduling but does so at a time scale of one frame time (the largest possible time taken to serve each of the backlogged connections). We assume that the quantum size has been set to the maximum packet length P_max. So, P_max = 8 KB. In the worst-case, all 500 connections are backlogged with maximum length packets and a packet arrives immediately after its queue has been examined for service. It must now wait one frame time before its last bit completes transmission. This frame time F in the worst case occurs when every connection counter is 1 byte under Pmax (e.g., there was a 1 byte packet in each queue; so the deficit counter is set to Q-1 after a round and then incremented to Q+P_max-1 in the next round): F = N x (Q + P_max -1) where Q is the quantum. So, F = 500 x (8 KB + 8 KB - 1) / 45 Mbps = 1456 msec = 1.5 sec This worst-case delay assumes that the packet arrives to an empty queue. If a burst of N packets arrive from a flow, each packet has to wait for the packets in front of it, and those packets are serviced in the worst-case at a rate of 1 packet every 1.5 sec.

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