ECEN Final Exam Fall Instructor: Srinivas Shakkottai

Transcription

1 ECEN Final Exam Fall 2013 Instructor: Srinivas Shakkottai NAME: Problem maximum points your points Problem 1 10 Problem 2 10 Problem 3 20 Problem 4 20 Problem 5 20 Problem 6 20 total 100 1

2 2 Midterm #3 1. (10pt) Pick one and only one answer for each question (indicate by circling): (a) A router running distance-vector routing knows: i. Nothing ii. The entire network s topology iii. The distance to destinations through its immediate neighbors (b) A router running link-state routing knows: i. Nothing ii. The entire network s topology iii. Only the distance to destinations through its immediate neighbors (c) Which one of the following apply to MAC addresses? i. May be assigned dynamically ii. Used in the physical layer iii. Used in the data link layer iv. Used in the network layer v. Hierarchical (d) Which one of the following applies to IP addresses? i. May be assigned dynamically ii. Used in the physical layer iii. Used in the data link layer iv. Always globally unique (e) Once this test is over, you go to a take-out only McDonald s. While waiting you notice that, on average, a new customer goes in to the store every 5 minutes, and a customer receives their order and goes out on average 3 minutes after arriving. On average, how many customers are inside McDonald s: i. 2 customers ii. 5 customers iii. 0.6 customers iv. There is insufficient information provided December 6, 2013 Srinivas Shakkottai ECEN 424

3 Midterm # (10 pt) TCP. Complete the missing sequence numbers (Seq), acknowledgment numbers (ACK), and segment length (LEN) in the following TCP connection (Figure 1). We assume: No timeouts occur at the receiver. The connection is full duplex (bi-directional data flow in same connection). The sender and the receiver have always data to transmit. There are no delayed acknowledgements at the sender or the receiver (but packets can be lost as shown). See review 5 Figure 1: Sequence and ACK numbers for a TCP connection. ECEN 424 Srinivas Shakkottai December 6, 2013

4 4 Midterm #3 3. (20pt) TCP Congestion Control. Although slow start with congestion avoidance is an effective technique for coping with congestion, it can result in long recovery times in high-speed networks. (a) (10pt) Assume a roundtrip time delay of 100 ms (about what might occur across a continent) and a link with an available bandwidth of 600 Mbps and a segment size of 3000 bits. Determine the window size needed to keep the pipe full. Now, suppose that TCP has reached that window size and then a time out occurs. Compute the time that it will take to get back to that window size after the time out, assuming that there are no losses and that every packet is acked. (b) (10pt) As in part (a), assume that TCP has reached the maximum window size required to keep the pipe full, and then a time out occurs. Compute the minimum segment size required to guarantee that the time to reach the maximum window size after the time out is no more than 1.0 s. Again, assume that there are no losses after the time out and that every packet is acked. Hint: Remember to calculate the congestion threshold, and use it to divide the congestion window growth process into slow start and congestion avoidance phases. See review 5 December 6, 2013 Srinivas Shakkottai ECEN 424

5 Midterm # (20pt) Packet Timings. Consider a network with two links shown in Figure 2. Node X is trying to send data to node Z. The first link (X Y ) has capacity C and latency (i.e., propagation delay) L, and the second link (Y Z) has capacity 2C and latency L/2. Figure 2: Network with two links. (a) Suppose the network is circuit-switched. Assume that path setup has occurred and at time t =0 node X begins sending a 1MB file to Z. As soon as the last bit of the file has been put on the wire, X sends a 2MB file to Z. The last bit of the 1MB file arrives at Z at time t =0.8 sec, and the last bit of the 2MB file arrives at Z at time t =1.8 sec. What is L? What is C? (b) Suppose that it is packet-switched. Assume that at time t =0node X begins sending a 1KB packet to Z. As soon as this transmission is completely on the wire X sends a 2KB packet to Z. Assume there is no processing delay at any node. The first packet arrives at Z at time t =1.8 msec, and the second packet arrives at Z at time t =3.4 msec. What is L? What is C? Solution on next page ECEN 424 Srinivas Shakkottai December 6, 2013

6 Packet Timings I Consider a network with two links. Node X is trying to send data to node Z. X Y Z The first link (X-Y) has bandwidth B and latency (i.e., propagation delay) L, and the second link (Y-Z) has bandwidth 2B and latency L/2. Consider two cases: The network is circuit-switched: Assume that all path setup has already occurred and that at time t=0 node X begins sending a 1MB file to Z. As soon as the last bit of the file has been put on the wire, X sends a 2MB file to Z. The last bit of the 1MB file arrives at Z at time t=0.8sec, and the last bit of the 2MB file arrives at Z at time t=1.8sec. 1. What is L (in msec)? What is B (in Mbps)? 16 The circuit as a whole can offer service with end-to-end bandwidth B (we ll call this throughput T to avoid confusion with bytes), and latency of (L + L/2). 1MB / T + (L + L/2) =.8sec (1) (1MB + 2MB) / T + (L + L/2) = 1.8sec (2) 2MB / T = 1.8sec - 0.8sec = 1sec (3) from (1) + (2) T = 2MB/sec (4) from (3) 1MB / (2MB/sec) + (L + L/2) = 0.8sec (5) from (1) + (4) 0.5sec + (L + L/2) = 0.8sec (6) from (5) L + L/2 = 0.3sec (7) from (6) L = 0.2sec (8) from (7) File 1: Trans XZ 0.5s P XY 0.2 P YZ 0.1 P = propagation time File 2: Transmission XZ 1.0 P XY 0.2 P YZ 0.1 2

7 The network is packet-switched: Assume that at time t=0 node X begins sending a 1KB packet to Z. As soon as this transmission is completely on the wire X sends a 2KB packet to Z. Assume there is no processing delay at any node. The first packet arrives at Z at time t=1.8msec, and the second packet arrives at Z at time t=3.4msec. 3. What is L (in msec)? What is B (in Mbps)? 12.5 Note: since the second link is faster, there is no queueing delay. (1KB / T) + L + (1KB / 2T) + L/2 = 1.8ms (1) (1KB / T) + (2KB / T) + L + (2KB / 2T) + L/2 = 3.4ms (2) (5KB / 2T) = 3.4ms 1.8ms = 1.6ms (3) from (1) + (2) T = 5KB / 2 / 1.6ms = 12.5Mbps (4) from (3) 0.64ms + L ms + L/2 = 1.8ms (5) from (1) + (4) L + L/2 = 0.84ms (6) from (5) L = 0.56ms (7) from (6) File 1: Trans XY 0.64 P XY 0.56 Tr YZ 0.32 P YZ 0.28 File 2: Transmission XY 1.28 P XY 0.56 Tr YZ 0.64 P YZ

8 6 Midterm #3 5. (20pt) Statistical Multiplexing. Assume that we have a network technology that divides time into frames. Each frame contains two time slots, and the link capacity is such that a single packet can be served in each time slot. Assume there are two flows that generate packets at random, independent of each other. Thus, at the beginning of each frame, each flow generates packets with the following probabilities: 0 packets with probability packet with probability packets with probability 1 2 (a) (5pt) What is the average number of packets both flows generate in a frame? Remember, we are asking about the total generated by both flows, not the number generated by a single flow. (b) (5pt) Assume that each flow is assigned a single slot in each frame. That is, it can use one and only one slot to carry its packets in each frame; if it generates two packets in a frame, one of them must be dropped. For example, Flow 1 is always given slot 1 and Flow 2 is always given slot 2 in each frame. What is the average number of packets carried in each frame? Remember, we are asking about all the packets carried, not just about those for a particular flow, i.e, lots of cases need to be considered. (c) (10pt) Assume that the flows share the slots in a frame; if not all the packets fit, then the excess packets are dropped (ignore how we choose which packets to drop; it won t matter in the answer). That is, if two or fewer packets are generated by the flows (in aggregate), then all of them are carried in that frame, but if more than two are generated than only two are carried in a frame. How many packets are, on average, sent in each frame? Why is the throughput higher as compared to part (b)? (a) 2*(1* *0.5) = 2.5 (b) Cases: (0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2) Find the probability for each case, and the number of packets carried, assuming at most one packet for each flow. (c) Cases: (0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2) Find the probability for each case, and the number of packets carried, assuming at most two packets per frame. December 6, 2013 Srinivas Shakkottai ECEN 424

9 8 Midterm #3 6. (20pt) Token Bucket. Consider a traffic stream with rate indicated in Figure 3. Figure 3: Rate Diagram Suppose this stream is passed into a token bucket filter with token generation rate r (where one token is needed per bit) and the token buffer size is B. Assume that the token bucket is full at the beginning. Suppose r = 15 kbps (i.e., 15 kilo tokens per second). What is the minimum size of B required so that the filter lets the stream pass with no loss or delay? Hint: Divide your calculations into four parts corresponding to the four different arrival rates. In each part, the initial condition on the number of tokens in the token bucket is equal to the final condition of the previous part. Solution on next page December 6, 2013 Srinivas Shakkottai ECEN 424

10 9. Token Bucket (extra 10pt) Consider a traffic stream with rate indicated in the figure above. Suppose this stream is passed into a token bucket filter with token generation rate r (where one token is needed per bit) and the token buffer size is B. Assume that the token bucket is full at the beginning. Suppose r=15 kbps (i.e., 15 kilo tokens per second). What is the minimum size of B required so that the filter lets the stream pass with no loss or delay? Answer: Let R denote the bit rate of traffic stream. (1) In time interval (0, 5s), At t = 0, the filter starts with a full buffer; at t = 5 sec, the tokens left in the buffer are B + 15 kbps * 5s - 20 kbps * 5s = (B n 25) Kbits (2) In time interval (5s, 15s) At t = 5 sec, the tokens in the buffer are (B n 25) Kbits; at t = 15 sec, the tokens left in the buffer are (B n 25) Kbits + 15 kbps * (15 n 5)s - 10 kbps * (15 n 5)s = (B + 25) Kbits This means that the buffer is full, the tokens in the buffer are still B Kbits. (3) In time interval (15s, 25s) At t = 15 sec, the tokens in the buffer are B Kbits; at t = 25 sec, the tokens left in the buffer are B + 15 kbps * (25 n 15)s - 30 kbps * (25 n 15)s = (B n 150) Kbits (4) After t = 25 sec As r R, the number of tokens in the buffer will not decrease any more. Thus we do not need to worry about the rest time period. As the number of tokens in the buffer should 0 at any time, so both B n 25 0 and B n should be satisfied Thus the minimum requirement for B is 150 Kbits