COE 431 Computer Networks. Welcome to Exam II Thursday May 15, Instructor: Dr. Wissam F. Fawaz

Transcription

1 1 COE 1 Computer Networks elcome to Exam II Thursday May 15, 01 Instructor: Dr. issam F. Fawaz Name: Student ID: Instructions: 1. This exam is Closed Book. Please do not forget to write your name and ID on the first page.. You have exactly 5 minutes to complete the 7 required problems.. Read each problem carefully. If something appears ambiguous, please write your assumptions.. Do not get bogged-down on any one problem, you will have to work fast to complete this exam. 5. Put your answers in the space provided only. No other spaces will be graded or even looked at. Good Luck!!

2 Problem I: Comparing terminologies (10 minutes) [10 Points] hat is the difference between each of the following pairs of concepts? 1. Congestion avoidance and slow start The TCP send rate starts slow but grows exponentially fast during the slow start phase. The congestion window increases linearly during the congestion avoidance phase.. EstimatedRTT and DevRTT The EstimatedRTT is a variable maintained by a TCP sender and it represents the best current estimate of the round trip time to a given destination. In addition to having an estimate of the RTT, it is also valuable to have a measure of the variability of the RTT. So, DevRTT defines the RTT variation as an estimate of how much a sample RTT typically deviates from EstimatedRTT.. SYN and FIN The SYN bit is set to 1 during the TCP connection establishment phase. FIN is set 1 when a TCP client application process issues a close command.. MSS and MTU The amount of data that a TCP sender can grab from its send buffer and place in a segment is limited by the Maximum Segment Size (MSS). Maximum Transmission Unit is the length of the largest link-layer frame that can be sent by a sending host. 5. Unchoked and optimistically unchoked The four peers to which a BitTorrent peer sends bits are said to be unchoked. Every 0 seconds, a peer randomly picks one additional neighbor and sends it chunks. This randomly chosen peer is said to be optimistically unchoked.

3 Problem II: DNS (10 minutes) [10 Points] Suppose you have just created a new startup company called starwar and you wish to set up your company s network. Your network has the following servers: 1. Authoritative DNS server: dns1.starwar.com with an IP address of eb server: starwar.com with two IPs, namely and The web server is assumed to have a name of server: galaxy.starwar.com with an IP of Your employees addresses have the form username@starwar.com. 1. hat resource records do you need to provide to the.com registrar? (starwar.com, dns1.starwar.com, NS) (dns1.starwar.com, , A). hat resource records need to be entered into your authoritative DNS server? (starwar.com, , A) (starwar.com, , A) ( starwar.com, CNAME) (galaxy.starwar.com, , A) (starwar.com, galaxy.starwar.com, MX)

4 Problem III: Socket Programming (10 minutes) [15 Points] Suppose a server with IP address 1... starts executing the following lines of code: ServerSocket sock = new ServerSocket(157); Socket connsock1 = sock.accept(); Socket connsock = sock.accept(); InetAddress x = connsock1.getinetaddress(); Now suppose a client with an IP address of...5 executes the following lines of code: Socket socka = new Socket( 1..., 157); InetAddress y = socka.getinetaddress(); At this point, how many sockets are there at the server? Justify your answer. sockets: 1 welcome socket and 1 connection socket. A short while later, another host with IP address..5.6 executes the following lines of code. Socket sockb = new Socket( 1..., 157); InetAddress z = sockb.getinetaddress(); At this point, how many port numbers are being used at the server? sockets: 1 welcome socket and connection sockets. hat are the values of the variables x, y, and z? x=...5; y = 1...; z = 1... Provide below a Java code fragment that causes the String who haha to be sent from the server to the second of the two clients. DataOutputStream outtoclient = new DataOutputStream(connSock.getOutputStream()); outtoclient.writebytes( who haha );

5 5 Problem IV: Peer to Peer networks (15 minutes) [15 Points] Suppose a movie studio wants to distribute a new movie as a digital file to 1000 movie theaters across the country using peer-to-peer file distribution. Assume that the studio and the theaters have DSL connections with an Mbs downstream rate and a Mbs upstream rate and that the file is 10 GBytes long. 1. Approximately, how much time is needed to distribute the file to all the theaters under ideal conditions? The total upstream bandwidth is about Gbs. Since the file must be delivered to 1,000 studios, we have 10 TB of data to be delivered. At Gbs, this takes 0,000 seconds, or roughly 6 hours.. Suppose that same studio wanted to use the client server method instead. hat is the smallest link rate that is required at the studio that will allow the file to be distributed in under 0000 seconds? This time period is twice the time used for the first part, so the server s upstream bandwidth must be half as large as the upstream bandwidth of the peers in the first part. So, Gbs is enough.

6 6 Problem V: TCP congestion control (10 minutes) [15 Points] Recall the macroscopic description of TCP throughput. In the period of time from when the connection s rate varies from ( x RTT) to RTT, only one packet is lost (at the very end of the period). Show that if a TCP connection has a loss rate of L, then its average throughput is approximately given by: L RTT MSS Throughput 1. The loss rate, L, is the ratio of the number of packets lost over the number of packets sent. In a cycle, 1 packet is lost. The number of packets sent in a cycle is 0 ) ( 1 n n 0 1 n n 1) ( 1 Thus the loss rate is L 1 For large,. Thus L or L. From the text, we therefore have Average Throughput RTT MSS L

7 7 Problem VI: TCP congestion control (15 minutes) [0 Points] Consider two TCP flows that pass through a common link served by a single queue. Assume that both flows have the same RTT. Suppose at time t 0, the queue is full, causing both flows to lose a packet. Suppose that the senders for both flows detect the lost packet at time t through the receipt of triple duplicate ACKs. If the first flow has a congestion window size equal to 100 MSS and the second has a congestion window size of 00 MSS at time t: 1. How many packets will the first flow send during the next RTT following t: a) if TCP Tahoe is used? and b) if TCP Reno is used? In the case of TCP Tahoe, it will go back to slow start and send one packet in the next RTT. In the case of TCP Reno, it will cut its congestion window in half, allowing it to send 50 packets in the next RTT.. How many packets will the second flow send during the next RTT following t: a) if TCP Tahoe is used? and b) if TCP Reno is used? In the case of Tahoe, it will send 1 packet. In the case of Reno, it will send 100 packets.. At time t+50 RTT, approximately what will be the size of the congestion window for the first flow: a) if TCP Tahoe is used? and b) if TCP Reno is used? For Tahoe, it will double cwnd for the first five RTTs, then add 1*MSS per RTT. So, it should be about 95*MSS at t+50*rtt. For Reno, it increases by 1*MSS every RTT, so it should be about 100*MSS. At time t+50 RTT, approximately what will be the size of the congestion window for the second flow: a) if TCP Tahoe is used? and b) if TCP Reno is used? For Tahoe, 15*MSS. For Reno, 150*MSS

8 Problem VII: Go-Back-N error recovery (15 minutes) [15 Points] The left hand side diagram below shows the state of the sending side of a communication session using a go-back-n protocol with cumulative acknowledgements, a window size of and a retransmission timeout of 5. The array represents the re-send buffer and each entry in the buffer represents a packet with its associated sequence number. As usual for go-back-n, we are using a single shared timer. Its expiration time is shown in the diagram. Assume that the channel may lose packets but never re-orders them. Suppose at time 7, the application passes us a new payload to be sent and at time 1, we receive an ACK with sequence number. 1. List all the packets sent by the sender between times 6 and 1, including repeats if any (in the form, p i, p j, p k, p i ) p5, p, p, p, p5. Show the state of the sender at time 1, in the right-hand side diagram, by giving proper values for the sendbase, timerexpiresat, and nextseqnum variables. sendbase= timerexpiresat=5 nextseqnum=6. At time 1, how many additional packets can be sent before an ACK is received? packets.