Student Name: University of California at Berkeley College of Engineering Department of Electrical Engineering and Computer Science

Transcription

1 SI: University of alifornia at erkeley ollege of ngineering epartment of lectrical ngineering and omputer Science S Fall 00 MITRM XMINTION Monday, October 00 I. Stoica INSTRUTIONS R THM NOW! This examination is LOS OOK/LOS NOTS. There is no need for calculations, and so you will not require a calculator, Palm Pilot, laptop computer, or other calculation aid. Please put them away. You MY use one. by double-sided crib sheet, as densely packed with notes, formulas, and diagrams as you wish. The examination has been designed for 0 minutes/0 points ( point = minute, so pace yourself accordingly). ll work should be done on the attached pages. In general, if something is unclear, write down your assumptions as part of your answer. If your assumptions are reasonable, we will endeavor to grade the question based on them. If necessary, of course, you may raise your hand, and a T or the instructor will come to you. Please try not to disturb the students taking the examination around you. We will post solutions to the examination as soon as possible, and will grade the examination as soon as practical, usually within a week. Requests for regrades should be submitted IN WRITING, explaining why you believe your answer was incorrectly graded, within ON WK of the return of the examination in class. We try to be fair, and do realize that mistakes can be made during the regarding process. However, we are not sympathetic to arguments of the form I got half the problem right, why did I get a quarter of the points? (Name Please Print!) SI: (iscussion Section ay and Time) Instructional ccount: ee- (Signature) QUSTION POINTS SSIGN POINTS OTIN TOTL 0 Midterm Page October 00; -:0 PM

2 SI: Question. General Networking oncepts (0 points) For each of the following statements, indicate whether the statement is True or False, and provide a very short explanation of your selection ( points each). a. In packet switching networks, packets between two hosts always follow the same route. T F b. ccording to the nd-to-nd rgument the network should implement reliability. T F c. Link state routing protocols is more robust than distance vector protocols. T F d. etecting packet loss via UP cks is faster than waiting for a retransmission timeout. T F e. Shared bus routers have higher throughput than point-to-point routers. T F Midterm Page October 00; -:0 PM

3 SI: Question. Little s theorem (0 points) Murali wants to find the average number of people in his office hours, which are from :00-:00PM. He observes the following three people and the times that they arrive and leave: ngi: :00-:0PM en: :0-:0PM Greg: :0-:00PM a) Show this behavior on the following plot: b) What is the mean number of people in Murali s office hours? Solution: a) b) Mean arrival rate = #People/Time = /0 Mean service time = (0+0+0)/ = 0/ Mean # of people = (Mean arrival rate) * (Mean service time) = 0/0 = / Midterm Page October 00; -:0 PM

4 SI: Question. Store-and-forward (0 points) Hosts and are connected to each other via router R. R is a store-and-forward router. The bandwidth from to R is 0Mbps, and the bandwidth from R to is Mbps. The one-way latency of each link is ms. ssume host sends a 0K file to host. a) ssume the file is divided into two packets, p and p, where p has a length of 0K, and assume the packets are sent back-to-back. What is the difference between the arrival times of the first and the second packet at host? b) What is the effective throughput between and in part (a)? (The transmission time is the time interval from the time the first bit is sent at until the final bit is received at ). c) oes the throughput increase or decrease if we divide the file into smaller packets? Why? d) Now, assume each packet is acknowledged. The file is divided into packets of the same size. How long would it take to send the entire file assuming that the sender cannot send a new packet before it receives an acknowledgment for the previous packet? (The transfer time is the time interval measure at source from the time the first segment is sent until the acknowledgement of the last segment is received). Ignore the transmission time of the acknowledgements Solution: a) The difference is the transmission time of packet over the second link time: size/bw= *0/000 = 0/000 = ms b) Total transfer-time: calculate how long the second packet takes and add the wait for the first packet over link -R. -R wait is: *0/0000 = ms Now time to send is *0/ *0/000 + ms = ms + ms + ms=9ms Total time: 00ms THROUGHPUT: 0k/00ms = 0.Mps or. Mbits/s c) The throughput will increase because a smaller packet is received at R and can be forwarded immediately (don t have to wait till the entire original packet is transmitted over the first link to start transmitting over the slower second link) d) Size of packet = K transmit time /packet= + + ms + */0000+*/000= ms +ms + ms + ms = 00ms so for = 00ms Midterm Page October 00; -:0 PM

5 SI: Question. TP (0 Points) On any given Sunday, Sam initiates a connection with his girlfriend lice in order to plan out what movie they re going to watch on Tuesdays with Morrie. This Sunday, immediately after Sam initiates the connection, lice decides to send him a 0 K picture using TP. ssuming she sends the picture in 00 byte segments including a 0-byte header over a Mbps link with a propagation delay of 0 ms, what is the total time difference between when Sam initiates the connection to when lice receives the last K? The TP implements a slow-start mechanism with an initial congestion window of. Neglect queuing and processing delays at both the sender and receiver, and assume there are no losses. Solution: Let Tprop be one way propagation delay, Ttrans be transmission time of one 00 byte packet, and Twnd be the time for a single window to be transmitted and Ked. Tprop = 0 ms Ttrans = (00 bytes * bits/byte)/(0 bits/second) = 0 ms T-way handshake = * Tprop = 90ms Twnd = *Tprop+Ttrans = 0ms (pt) (pt) (pt) Size of application data in a TP packet = size of TP segment size of header = 0 bytes. Number of packets needed to transfer a 0 K file = (0*000)/0 = or packets. (this depends on whether you use K to be 000bytes or 0bytes. (pt) Hence, the window size grows as,,,, 0 or. (pt) Hence, the total time to transfer the packet after the -way handshake is: Tdata = *Tpkt + 0*Ttrans = 0ms or 0 Hence, total time to transfer = T-way-handshake + Tdata = 90 ms + Tdata ms = 0 or 0ms (pt) Strictly speaking, if the total transmission time of all the packets transmitted in a window exceeds the round trip time, then the time to issue a window of packets will be extended by that extra transmission time (i.e., the difference between the total transmission time and the round trip time). This extra time was not taken into account in the above solutions, but students who accounted for this extra time were not penalized. Midterm Page October 00; -:0 PM

6 SI: Question. Routing Protocols istance Vector ( points) ssume the following for this question: ll nodes implement distance vector routing. ll nodes have their time synchronized with each other. Route updates are exchanged in synchronized time step manner at fixed time intervals. Whenever an entry in the routing table of a node changes, that node sends its routing table to all of its immediate neighbors at the beginning of the next time step. Propagation time along all links takes a little less than time step unit (ie. an update sent by a node at the beginning of a time step reaches its neighbors before the end of the time step). y the end of a time step every node if it received any routing updates from neighbors, computes its latest routing table and updates it if needed, before the end of the time step. 9 onsider the network shown above. fter all the nodes converge, the routing tables in the nodes will look like the following: Node Node Node Node Node est ost est ost est ost est ost est ost The link between nodes and gets upgraded to have a small link cost of. oth nodes and immediately update their routing tables to reflect this change. fter several iterations the routing tables will eventually converge. Show step by step update propagation until all the tables converge. Note: In the next page we have drawn a couple of time step tables for your convenience. In your step by step answer, you should highlight the rows updated in each step by circling it. Midterm Page October 00; -:0 PM

7 SI: Time : Node Node Node Node Node est ost est ost est ost est ost est ost Time : Node Node Node Node Node est ost est ost est ost est ost est ost Time : Node Node Node Node Node est ost est ost est ost est ost est ost Time : Node Node Node Node Node est ost est ost est ost est ost est ost onverges in four iterations! Midterm Page October 00; -:0 PM

8 SI: Question. TP ( Points) Host sends a file consisting of 9 segments to a host using TP. The particular version of TP used by and does not implement fast recovery or fast retransmit. ssume that the th segment in the transmission is lost. ssume the retransmission timeout is T, the one-way latency is d, and that T > *d. Ignore the transmission time of the segments and of the acknowledgements. lso, ignore the three-way hand-shaking protocol. a) raw the time diagram showing each segment and acknowledgement until the entire file is transferred. Indicate on the diagram all changes of the cwnd and sstresh. How long does it take to transfer the file? b) nswer part (a) assuming TP Reno, i.e., the TP implements both fast retransmission and fast recovery. c) nswer part (b) assuming that only the th segment is dropped. Notes: a. When using TP Reno, cwnd does not increase upon receiving a UP ack. b. If the value of cwnd is fractional, you should round it to the closest larger integer. c. The transfer time is the time interval measure at source from the time the first segment is sent until the acknowledgement of the last segment is received Please use the following timing diagrams (which show the first packet) to answer this problem. Solution: (In this solution, the ck number represents the number of the last in-sequence received packet, not the number of the next in-sequence expected packet. oth notations are correct, and we have not deducted points for using the later notation.) cwnd= cwnd= cwnd= cwnd= sstresh= cwnd= cwnd= (a),,,,,,,,9,9 cwnd= cwnd= cwnd= cwnd= sstresh= cwnd= cwnd= (b),,,,,,,,9,9 cwnd= cwnd= cwnd= cwnd= cwnd= ssthresh= cwnd= (c),,,,,,,, 9, 9 Transfer time = *d + T Transfer time = 0*d Transfer time = 0*d Midterm Page October 00; -:0 PM