Homework #4. Due: December 2, 4PM. CWND (#pkts)
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1 Homework #4 Due: December 2, 4PM EE122: Introduction to Communication Networks (Fall 2009) Department of Electrical Engineering and Computer Sciences College of Engineering University of California, Berkeley Please submit your solutions using BSpace. Assignment should be submitted in one of the following formats:.txt,.pdf, or.ps 1. TCP Congestion Control Consider a TCP connection that sends 12 packets (segments). During transmission, packet #4 is lost due to congestion. TCP successfully res this packet. No other packet is lost. For simplicity, assume: RTT is 1sec for the entire duration of transmission; TCP retransmission timeout (RTO) is 2sec for the entire duration of transmission. First packet is send at time 0, and the transmission time is 0. a) Fill the following table for vanilla TCP (no Fast Retransmission, no Fast Recovery) 3 <STALL> 4 SS SS 2 2 8, 9 6 AIMD , 11, 12 8
2 b) Fill the following table for TCP with Fast Retransmission. 3 SS SS 2 2 8, 9 5 AIMD , 11, c) Fill the following table for TCP- Reno (both Fast Retransmission and Fast Recovery are implemented). (Either answer is fine. The first one is TCP-Reno which does not inflate after 3dup ACKs, and the second one is TCP-new Reno which inflates after 3 dup ACKS.) 3 AIMD AIMD 3 2 8, 9, 10 5 AIMD , AIMD ACKs = 5 2 4, 8 (window full) 4 AIMD 6 2 9, 10, 11,
3 d) Compute throughput in packets/second for each scheme. The throughput is defined as (Number of packets successfully delivered)/((time when ack for the last lest packet is received by sender) (time of sending first packet)). What are the throughput improvements of (b) and (c) with respect to (a)? a. Vanilla TCP: totally seconds until the final ACK arrive. Thus, 12 / = 1.14 pkts/sec b. Fast Retransmission: Totally 6 seconds. Thus, 12 / 6 = 2 pkts/sec c. Fast Retransmission + Fast Recovery i. TCP- Reno: totally 6 seconds. So, 12/6 = 2pkts/sec ii. TCP- new Reno: totally 5 seconds. So,12 / 5 = 2.4 pkts/sec d. c) improves throughput by about 1% and d) does by 1% (or by 40% f or TCP- new Reno). (Below answer is not much correct, but it will also be given a full credit *only* for this homework since TA gave a wrong information.) Fill the following table for vanilla TCP (no Fast Retransmission, no Fast Recovery) 3 <STALL> 4 SS SS 2 2 5, 6 6 AIMD 3 2, 8, 9 AIMD , 11, 12 Fill the following table for TCP with Fast Retransmission. 3 SS SS 2 2 5, 6
4 5 AIMD 3 2, 8, 9 6 AIMD , 11, 12 Fill the following table for TCP- Reno (both Fast Retransmission and Fast Recovery are implemented). 3 SS 2 2 4, 5 4 AIMD 3 2 6,, 8 5 AIMD 4 2 9, 10, 11, 12 6 Compute throughput in packets/second for each scheme. The throughput is defined as (Number of packets successfully delivered)/((time when ack for the last lest packet is received by sender) (time of sending first packet)). What are the throughput improvements of (b) and (c) with respect to (a)? e. Vanilla TCP: totally 8 seconds until the final ACK arrive. Thus, 12 / 8 = 1.5 pkts/sec f. Fast Retransmission: Totally seconds. Thus, 12 / = 1.14 pkts/sec g. Fast Retransmission + Fast Recovery: Totally 6 seconds. Thus 12 / 6 = 2 pkts/sec h. c) improves throughput by about 14% and c) does by 33%. 2. TCP performance a) Consider two TCP flows, one over link X, and the other over link Y. X is a 100Mbps Ethernet link with a 10ms RTT, and Y is a 1Gbps satellite link with a 1000ms RTT. Assuming both links have the same loss probability p, and the flows always use the same packet size and never saturate the links, what s the link utilization ratio of X to Y?
5 Recall TCP throughput is proportional to the inverse of RTT. Thus, link utilization is proportional to the inverse of RTT * link bandwidth. Hence, link utilization ratio is 1/1000. b) Consider two TCP flows, one over wireless links X, and the other over wireless link Y. Both links are a and have 25ms RTT. If the interference causes a 1% packet loss on X, and a 4% packet loss on Y, respectively, what s the link throughput ratio of X to Y? (Assume there s no forward error correction, and there is no link layer retransmission.) TCP throughput is proportional to the inverse of square root of packet loss rate. Thus, throughput ratio is 2/1. c) From the above answers, we know that TCP does not perform very well in the presence of high RTT or lossy links. Give one solution to improve TCP throughput on links with high RTT, and one solution to improve TCP the throughput on lossy link s. i. For a link with high RTT, we can tweak. For example, we can start from a large and increase quickly. ii. For a wireless link, packet loss is no longer a good indication of congestion because link is already lossy. So, we better decouple congestion control signaling from packet loss or introduce explicit congestion bit in TCP header. d) Suppose a TCP flow and a UDP flow are sharing the same link, and the UDP sends at a rate equal to the link capacity. Assuming the flows stay for long enough, how would they split the bandwidth eventually? If they would equally share the bandwidth, rationalize it. If a particular one would monopolize the bandwidth, explain which one and why. i. UDP will use up the most of bandwidth because TCP keeps reducing its congestion window according to its congestion
6 control algorithm while UDP keeps taking the yielded bandwidth.!" (" #" )" $"," *" +" *" (" %" +" &" -" '" 3. Shortest- path routing a. Fill out the following table, for node B, assuming that Link- State routing is being used. S D(A),p(A) D(D),p(D) D(E),p(E) D(C),p(C) D(F),p(F) B 1, B 2, B, B BA 5, A 2, B, B BAE 4, E 5, E, E BAED 5, E, E BAEDC 6, C BAEDCF b. Fill in the forwarding table for node B. Destination Link A (B, A) C (B, E) D (B, E) E (B, E) F (B, E)!" #" $" %"
7 4. Wireless Media Access Control (MAC) protocol. Assume A & B can hear each other, B & C can hear each other, and C & D can hear each other. No other nodes can hear each other. For parts (a) and (b) also assume that RTS and CTS are not being used. a. If C wants to send to B while A is sending to B, will a collision occur? Why or why not? Will this be considered hidden or exposed terminal? Yes a collision will occur because C will not hear that A s transmission. This will be considered hidden terminal. b. If B wants to send to A while C is sending to D, will a collision occur? Why or why not? Will this be considered hidden or exposed terminal. No a collision will not occur because B will hear C s transmission and unnecessarily defer its own. This is considered exposed terminal. c. The Ethernet s CSMA/CD makes an assumption about the conditions being experienced by the sender and receiver that becomes invalid in the wireless domain. What is this assumption and how does RTS/CTS overcome it? The assumption is that the sender can accurately predict whether or not a collision would occur at the receiver simply by listening to the channel. This is not true for the wireless domain. RTS/CTS resolves this problem by providing the sender information about the conditions being experienced by the receiver. d. If the packets being sent by the wireless nodes are very short (few bytes each), would RTS/CTS be a useful mechanism if link- layer acks are already being used? Why or why not? No RTS/CTS would not be useful because the size of the packet will be comparable to the size of RTS/CTS messages. This means the overhead of RTS/CTS would be prohibitive
8 5. Interdomain Routing. Consider the following network of Autonomous Systems. Each circle represents an AS. Providers are higher than customers, and AS B and D are peers Assume an AS always prefers to route through customers, and assume that if an AS has multiple routes through customers, it will choose the one with least number of ASes. Customers won't carry traffic between providers. a) For destination E, which path would be advertised by B? B- C- E b) Which path will be used by A to reach E? Why this path may not be the shortest path for packet forwarding? A- D- E. Because inside one AS there may be many routers along the path. c) If D- E fails, what will D do? Which path will be used by A to reach E now? D withdraws path D- E and advertise D- B- C- E to A. A will use A- B- C- E. d) If both A- D and B- D fail, can C use the path C- E- D to reach D? Why?
9 No. Because customers will not carry traffic between its providers. 6. QoS. a) In DiffServ, how does a router recognize packets that require a better service? Those packets have DiffServ bits set. b) How are edge routers different from core routers in DiffServ, and why? Core routers have less per-packet processing time. So tagging packets at edge router reduces complexity of core routers. c) Name two differences between IntServ and DiffServ? 1) IntServ tries to provide per- flow guarantee while DiffServ is per- aggregate tiers of relative performance. 2) IntServ is end- to- end while DiffServ is per- domain.
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