Short answer (35 points)

Transcription

1 CPSC 360 Fall 2017 Exam 1 Version 2 Solutions (last updated 10/19/2017) This exam is closed book, closed notes, closed laptops. You are allowed to have one 8.5x11 sheet of paper with whatever you like written on the front and back. You may also use a calculator. Name: First and last warning. If I see you looking at someone s exam, you will get a 0 on this exam. Please sign below indicating you have read and understand this. Signature: Section 1 Short answer (35 points) 1 (5) What are the two most important items provided by a socket? Solution : Our socket discussion slides (socketsmain.pdf, slide 8) says a socket provides : Abstraction of an end-to-end communications pipe between two applications A programming API. 2 (4) HTML is the primary transfer protocol that a browser uses to interact with a web server. (True or False) False. HTML is the HyperText Markup Language- the language used to build web pages. The Hypertext Transfer Protocol (HTTP) is the transfer protocol used to handle transferring web pages from a web server to a client browser. 3. (4) A physical medium based on an optical fiber is able to carry energy encoded with data longer distances than a physical medium involving copper wiring. (True or False) True 4. (4) Frequency division multiplexing requires stations to share a physical medium by taking turns when sending. (True or False) False. FDM assigns different sub channels to each user 5 (6) We run our UDPEcho client on our VM and it interacts with the server running on a Linux machine on campus. We specify mode 0 (ECHO_MODE) with a message size of 1000 bytes. The figure lays out the frame that will get sent out the network. Assume a frame structure as described in class. Identify each section of the frame. To illustrate what we are after, we have filled in the last section with CRC. The last 4 octets of a frame contains data used to help the receiver detect if bit errors have occurred in the data 1

2 (it is a cyclic redundancy check or CRC, error detection method). You just need to identify by name what is contained in each section. CRC Frame Format MAC IP UDP Message (1000 bytes) MAC (CRC) HDR HDR HDR TRAILER 6. (6) Building off the previous question, assume that size of the frame is exactly 1048 bytes. What is the transmission time for the frame over a link with a data rate of 1.5 Mbps? Please show your answer in units of bits per second. 1048*8/ bps 7 (6) Provide a reasonable round trip time (RTT) estimate you would expect to observe when running UDPEcho or ping between your VM physically located on campus with each of the following partner Hosts. Please specify your answer in units of milliseconds.. HINT: There is a wide range of valid estimates for each partner Host. We are just looking to see if your answer is correct within order of magnitude. For example, the amount of time it takes to drive from my house to campus is 10 minutes. Answers of 1 minute or 100 minutes would be wrong. Your VM (using localhost) A Host in California (or some location on the West Coast) A Host located in Asia Answers for each should be within these bounds Less than several milliseconds (typical would be ms) Greater than several milliseconds but less than 100 ms (typical: tens of ms) Greater than 100 ms but less than 500 ms (typical: hundreds of ms) 2

3 Section 2 Problems (30 points) Two nodes communicate over the following network. One Router is in the path. The link between the router and Node 2 is the bottleneck as it operates over a link with a rate of 1.5 Mbps. Note that the maximum allowed size of the buffer for packets that arrive at Router 1while Link 2 is busy is 1024 packets. Link 1 Link 2 Node Gbps Ethernet ----Router Mbps Link Node 2 1 millisecond 10 millisecond Buffer size at Router 1 s link2: 1024 packets The only active application is the UDPEcho client and server program. The client operates at Node 1 in CBR mode configured with a message size of 1000 bytes and a target send rate of 1 Mbps. After accounting for all overhead, the size of a frame sent over a link is exactly 1048 bytes. This is the same for all links. Assume that any processing overhead is negligible. The server receives the message and just updates session counters. It does not send anything back. All questions assume ideal channels which means that bit errors will NOT occur. Packets could be dropped at Router 1 if it were to become congested and the queue fills. Further assume that if we tell the client to send at a rate of 1 Mbps,it will send at exactly that rate. Please have latency answers in units of seconds with microsecond precision, throughput in units of bits per second, and loss rate as a ratio (and not a percentage). Q2.1 (8) What is the one-way latency that each message sent by the client will experience? As a reminder, the one-way latency is the sum of all delays experienced by a message that is transmitted by Node 1 and successfully received by Node 2. Q2.2 (7) What is the application throughput you would expect the server to observe? Please specify in bits per second. Q2.3 (7) What is the expected loss rate observed by the server? Q2.4. (8) What happens if the client is told to send at a rate of 2 Mbps. Limit your answer to what you would expect the server to observe in terms of one-way latency, observed loss rate and throughput. Q2.1 The one-way delay consists of : Latency= Tx(link1) + Tprop(link1) + Tx(link2) + Tprop(link2) Tx(link1) = 1048*8/ = 8.38us = round down to 8 us Tx(link2) =1048*8/ = seconds 3

4 Latency = 8 10EXP = seconds Q2.2, 2.3 If the only traffic in the system is the client which sends at 1 Mbps, congestion will not occur. Therefore the throughput and loss observed by the server would be 1 Mbps and 0. Q2.4 If the client sends at 2 Mbps, this exceeds the capacity of Link 2. The buffer will fill to capacity and remain filled Traffic arrival rate 2 Mbps--- Queue (1024 pkts) Departure rate: 1.5 Mbps The figure simplifies the system to a queueing system with traffic arriving at a rate that exceeds the rate at which packets (which are in the queue) are being processed. The queue fills roughly at a rate of 500Kbps (2 Mbps- 1.5Mbps). This translates to a rate of queue growth of roughly 1 pkt every seconds (1048 *8 / ). The time for the queue to fill is 1024 * or 17.7 seconds. After this startup time the queue is always filled. When a packet arrives, there is a 0.75 chance the arriving packet is accepted and a 0.25 chance that the packet is dropped because the arrival happens to occur when the queue can accept one packet. Therefore the server will observe a loss rate of approximately Since the client does not react to the congestion the throughput observed by the receiver would be just under 1.5 Mbps. We know that the protocol headers adds 4.8% (48/1000) overhead and therefore we can approximate the application throughput as the bottleneck link speed minus 4.8% : Application throughput = *1,500,000 = Mbps. The average one way message latency extends the answer from Q2.1 which assumes no queue delay over the path. In this question, all packets that arrive at the queue when it is not full will need to wait 1024 packet transmission times before it has its turn to transmit. Therefore, the one way latency is: Latency = * = seconds. 4

5 Section 3 Programming (35 points) The following illustrates a greatly simplified pseudocode version of the UDPEcho client and server. Assume RTT_MODE (mode 0). We do not show or care about CBR_MODE for this problem. Assume that the routine gettimestamp() returns the current time as a double in units of seconds with microsecond precision. //Client: starttime=gettimestamp(); sock=socket(); //create the socket myaction.sa_handler = CatchAlarm; sigaction(sigalrm, &myaction, 0) rc=socketfunction1; rc=socketfunction2 //Server: starttime=gettimestamp(); sock=socket(); //create the socket rc=socketfunction3; rc=socketfunction4; rc=socketfunction5 Please answer the following questions. You do not need to get the syntax or the function parameters correct. Your answer should however clearly show that you know the answer. Identify the name of each socket function that makes the most sense (for example, when the socket is created, the answer is socket(). You do not need to show the parameters.) In the client, add the lines of code necessary to obtain an RTT sample each iteration. You can use the gettimestamp() to return the current time as a double. In the client, add the lines of code necessary to setup the retransmission timeout. You can assume the CatchAlarm has been successfully registered AND that you do NOT need to show the code for CatchAlarm. Just show what needs to be done to setup the timeout in the client pseudo code. o Hint: the man page for the alarm system call we use to set the timeout is: unsigned int alarm(unsigned int seconds); 5

6 //Client: double starttime=gettimestamp(); sock=socket(); //create the socket myaction.sa_handler = CatchAlarm; if (sigaction(sigalrm, &myaction, 0) < 0) { errorcount++; printf("udpechoclient(%s): HARD ERROR sigaction failed: errno:%d \n", argv[0],errno); exit(1); double RTTsample=0.0; alarm(2); double time1=gettimestamp(); rc=sendto(); rc=recvfrom(); double time2=gettimestamp(); alarm(0); RTTsample= time2- time1; //Server: sock=socket(); rc=bind() rc=recvfrom() rc=sendto() //create the socket Identify the name of each socket function that makes the most sense (for example, when the socket is created, the answer is socket(). You do not need to show the parameters.) 1. rc=sendto() 2. rc=recvfrom(); 3. rc=bind() 4. rc=recvfrom() 5. rc=sendto() In the client, add the lines of code necessary to obtain an RTT sample each iteration. You can use the gettimestamp() to return the current time as a double. Simply get a timestamp before the sendto(), and get a second timestamp upon successful return from a recvfrom(). By definition, a round trip time sample (RTT) is an estimate of the time it takes for one sendreceive iteration. Therefore, the sample: RTTsample= time2- time1; 6

7 In the client, add the lines of code necessary to setup the retransmission timeout. You can assume the CatchAlarm has been successfully registered AND that you do NOT need to show the code for CatchAlarm. Just show what needs to be done to setup the timeout in the client pseudo code. o Hint: the man page for the alarm system call we use to set the timeout is: unsigned int alarm(unsigned int seconds); The Retransmission Timer is set just before the client sends. If the echoed message is not received within the Retransmission Timeout (RTO) amount of time (the value used for the timer), the timer pops causing the client s timeout code to run. If this happens, the recvfrom unblocks with an error (we don t show this code). As we have seen in the UDPEcho client code, the alarm system call is used which is the program s interface to a single timer function that Unix makes available to all programs. The alarm(2) turns on the timer with a RTO of 2 seconds. After the recvfrom, in both the error and non-error cases, the alarm(0) turns off the alarm. 7