>>> SOLUTIONS <<< Answer the following questions regarding the basics principles and concepts of networks.

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1 Midterm Exam for Computer Networks (CNT 4004) Fall 2015 >>> SOLUTIONS <<< Welcome to the Midterm Exam for Computer Networks. Read each problem carefully. There are eight required problems (each worth 12 points you get 4 points for correctly following these instructions). There is also an additional extra credit question worth 10 points. You may have with you a calculator, pencils and/or pens, erasers, blank paper, and one 8.5 x 11 inch formula sheet. On this formula sheet you may have anything you want (definitions, formulas, homework answers, old exam answers, etc.) as handwritten by you in pencil or ink on both sides of the sheet. Photocopies, scans, or computer generated and/or printed text are not allowed on this sheet. Note to tablet (ipad, etc.) users you may not print-out your handwritten text for the formula sheet. You have 75 minutes for this exam. Please use a separate sheet of paper for the answer to each question. Good luck and be sure to show your work! Problem #1 Each sub-problem worth 3 points. Answer the following questions regarding the basics principles and concepts of networks. a) Sketch the five-layer Internet protocol stack as described in the book and class. Describe in one sentence (for each layer) the services that each layer provides. Identify by name and location (in the stack) the key API used for writing Internet applications 5 Application -- Where network application and application protocols reside =============== -- Sockets interface 4 Transport -- Transports appl layer messages between appl endpoints (end to end) 3 Network -- Moves network layer packets from host to host (point to point) 2 Link -- Actually moves packets from node to node (point to point) 1 Physical -- Moves the individual bits from node to node (point to point) b) Name two popular protocols for each of layers 5, 4, 2, and 1. Name the key (only?) Internet protocol for layer 3. Layer 5 = HTTP and SMTP Layer 4 = TCP and UDP Layer 3 = IP Layer 2 and 1 = Ethernet and Wi-Fi c) Define protocol and interface. Explain what is the hard part in designing a protocol (or interface). Give an example of this (we discussed one example in class from your readings). Protocol = Complete set of rules regarding information exchange between same level layers between sites. Interface = Complete set of rules regarding information exchange between adjacent layers within a site. The hard part in designing a protocol is to get the rules complete that is, to consider all possible situations that your protocol will need to handle in order to provide the service it was designed to provide. An example of the difficulty of this is the 1841 Clayton tunnel train wreck we studied in class (and was your assigned reading). In this case the protocol was incomplete in that it was unable to handle the not unreasonable case of two trains in a tunnel.

2 d) What are the typical causes of delay and of packet loss in a packet switched network? Which cause of delay typically predominates for highly utilized multi-hop networks? Which cause of packet loss is the most common in highly utilized wired networks? The four typical causes of delay are processing, transmission, propagation, and queueing. The two causes of packet loss are electrical noise corrupting packets and buffer overflows dropping packets. For a highly utilized multi-hop network it is most likely that queueing delay predominates. However, propagation delay could also predominate depending on the span (size) of network, number of hops, and congestion level). Buffer overflow is the most common loss mechanism in wired networks. Problem #2 Each sub-problem worth 3 points. Answer the following questions about security and HTTP. a) What are the five malicious things that a bad guy can do with respect to (or using) a network as discussed in class and in the textbook. 1. Use the network to put malware on your host (be it server or client) 2. Attack servers and clients to deny service (Denial of Service we call this) 3. Sniff packets and read the contents 4. Masquerade as someone else (spoofing) 5. Modify and delete messages b) An HTTP server typically can determine the browser type (Firefox, Chrome, etc.) of the client sending a request to it. How is this possible? The HTTP header typically contains a field for user-agent, this identifies the browser used at the client. c) If the following webpage is accessed by a client using a browser than implements HTTP 1.1, will there be a significant (or any) performance improvement over using HTTP 1.0? What is the measure of performance we are interested in here? Explain your answer. <html><body> Here are some pictures of cats: <img src= > <img src= > <img src= > <img src= > <img src= > <img src= > <img src= > </body></html> There will be no significant performance improvement if HTTP 1.1 is used over 1.0. The reason for this is that HTTP 1.1 gains a performance improvement over 1.0 in situations where a single persistent connection can be used to download multiple objects. For the above web page all the images are on different hosts and so multiple connections are needed in any case HTTP 1.1 thus gains no advantage over HTTP 1.0. The measure of performance of interest here is the response time to display (or render) the complete web page consisting of some text and (probably) seven cat pictures.

3 d) What is wrong with the following HTTP request header (read it very carefully)? GETSTUFF /~christen/index.html HTTP/1.1 There is no such request (or method) as GETSTUFF (there is, however, GET). Problem #3 Blocking definition is 6 pts, each example is 3 pts. We speak of sockets functions that block. What do we mean by this? Give an example of a streams (TCP) sockets function that blocks (and explain what it is blocking on) and an example of a datagram (UDP) sockets function that blocks (and, again, explain what it is blocking on). Functions do not return until they complete. Functions in sockets that wait on some event to complete, typically from the other side (so, client side if server or vise versa), are considered to be blocking. For TCP the accept() function blocks until a connect() is issued by the other side. For UDP the recvfrom() function blocks until a datagram is received. There are other possible examples (for example, recv() for TCP). Problem #4 1 pt for each bug found. Appendix A contains our tcpserver.c program with many errors. The program compiles and links successfully, but it will not run successfully. Identify the errors (there are 10 errors that I know about). The errors are not subtle, they are quite significant and (I would argue) blatant. You may write your answer on the code listing in the appendix, if you wish. The errors are: 1. PORT_NUM should be an integer value, not a string ( mickey mouse ) 2. welcome_s should be an interger and not a double 3. if (welcome_s > 0)... should be if (welcome_s < 0) 4. server_addr.sin_port = htons(port_num); is missing 5. Should bind welcome_s and not the non-existent connect_s 6. listen(wecome_s, 1); is missing 7. The client port number in the accept() printf() message is surely not PORT_NUM (it is ntohs(client_addr.sin_port)) 8. Using in_buf, not out_buf, in the send() 9. Using out_buf, not in_buf, in recv() 10. Using retcode, not in_buf, in the recv() printf() message Problem #5 6 pts for sequence number. 6 pts for walk through. This question is about the RDT protocol developed in the textbook in Chapter 3 and described in the career fair online lecture on protocol design. Appendix B contains the RDT 2.2 receiver FSM for the textbook. Consider that there may be two duplicate packets in flight in the network to the receiver if an early time-out has occurred in the sender. How does the RDT 2.2 receiver prevent duplicate data from being delivered to layer 5 in this situation? Give an example considering two packets with sequence number 0 (SN = 0) being received consecutively one immediately after the other. The short answer is sequence numbers. Consider two packets with SN = 0. The first packet received will advance the FSM from Wait for 0 state to Wait for 1 state with the actions of delivering the received data to Layer 5 and sending an ACK0. Then in the Wait for 1 when the second packet with SN = 0 arrives, it will result in a second ACK0 being sent, but no (redundant) data being delivered to Layer 5.

4 Problem #6 6 pts for each sub-problem. Majority of credit for correct formulas. Consider the following scenario. Distance from sender to receiver is 3000 miles Data rate between sender and receiver is 1 Mbps Data packet length is 1250 bytes ACK packet length is 64 bytes Sender always has packets queued to send a) What is the link utilization if an SAW protocol is used and you assume no bit errors (i.e., lost packets or lost ACKs never occur)? For T_proc and T_ack negligible (here ACK is about 20x smaller than a data packet) we know that: U = T_fr / (2*T_pr + T_fr) T_fr = (1250 * 8) / = 10 ms T_pr = 3000 * 5 us/mile = 15 ms So U = 10 / (2 * ) = 25% b) What is the link utilization if an SAW protocol is used and you assume a bit error rate of 10-5 (you may assume that only data packets will have errors and that all errors are detectable at the receiver, ACK packets never contain bit errors). We first solve for the probability of a packet error, p = 1 (1 10-5)1250*8 = We know that U = ((1 p) * T_fr) / (2*T_pr + T_fr) So, U = (( ) * 10) / (2 * ) = 22.6% Problem #7 Notion of sampling and smoothing is 6 pts. Notion of estimating variability and using it is 6 pts. Explain in words (and not just regurgitate formulas that you may have written on your formula sheet) how modern TCP determines the time-out value that it uses for implicit packet loss detection and recovery. No more than about 100 words, please. TCP times (that is, samples) the RTT for each non-retransmitted segment sent and ACKed. The sampled RTT is then smoothed by using exponential smoothing weighting both the previous smoothed value and the current sample value. TCP also determines the variability in RTT values by looking at the magnitude of difference in sampled values and the smoothed values of RTT. The time-out value is then the smooth RTT value plus an estimate of the variability in RTT. For the case of a time-out and retransmit, the time-out value is always backed-off (or doubled) until a successful (ACKed) transmission occurs and no sampling takes place.

5 Problem #8 Each sub-problem worth 3 points. Answer the following general questions about Layer 3. a) What are the two general functions of Layer 3? The two general functions of Layer 3 are 1) forwarding of packets (i.e., choosing which port to forward a packet to within a router that has many possible egress ports and 2) routing (i.e., choosing the path or route to be taken by packets from sender to receiver). b) We talk of a least cost route as being the best route from a sender A to a receiver B. What are some possible metrics for route cost? Which of these metrics are static and which are dynamic? Hops, link speed, congestion, delay, security, and many others. Hops and link speed are static. Congestion and delay are dynamic. Security is most likely static. c) What are the criteria for a good routing algorithm? Fast, simple, not generate much network traffic (overhead), not create loops, stable, and converge to optimum solution. d) Routers change which protocol header? Explain what is changed and why. Routers alter the MAC (Layer 2) header to insert the destination MAC address of the host (or router port) that a packet is being sent to (also the source address is changed to that of the sending router port). Extra Credit RTT is 4 pts, utilization is 3 pts, solution is 3 pts. The average distance to Mars is 141 million miles. What is the approximate RTT to Mars? For a 1 Mb/s data rate, how many bytes can be in flight from Earth to Mars (that is, how many bytes can fill-up the pipe in one direction)? TCP has a 16-bit window size expressed in bytes this is a field in the TCP header. What is the maximum link utilization that can be achieved with the maximum window size in bytes? Can you propose a relatively simple fix (the fix may not require changing the format of the TCP header)? We know that one foot = one nanosecond, 5280 feet in a mile so 5.28 microseconds per mile. For 141 million miles this is 745 seconds or about 12.4 minutes. RTT is then twice this, or about 25 minutes. For 1 Mb/s link we have: 745 s * 1.0*10^6 b/s = 745*10^6 b (or about 89 Mbytes). 2^16 is so the maximum utilization would be ((65536 * 8) / 1.0 * 10^6) / (2 * 745) = 0.035%. So, much less than 1% utilization. One solution could be to consider each increment of the window size to be more than 1 byte. If each increment was 1 Mbyte (instead of 1 byte) than a 100% utilization could be achieved. So, this does not require changing the format of the TCP header, only interpreting the current field in a different way.

6 Appendix A Code listing for problem #3 #define WIN #include <stdio.h> #include <string.h> #include <stdlib.h> #include <windows.h> #ifdef BSD #include <sys/types.h> #include <netinet/in.h> #include <sys/socket.h> #include <arpa/inet.h> #include <fcntl.h> #include <netdb.h> #define PORT_NUM "mickey mouse" int main() WORD wversionrequested = MAKEWORD(1,1); WSADATA wsadata; double welcome_s; struct sockaddr_in server_addr; int connect_s; struct sockaddr_in client_addr; struct in_addr client_ip_addr; int addr_len; char out_buf[4096]; char in_buf[4096]; int retcode; WSAStartup(wVersionRequested, &wsadata); welcome_s = (double) socket(af_inet, SOCK_STREAM, 0); if (welcome_s > 0) printf("*** ERROR socket() failed \n");

7 server_addr.sin_family = AF_INET; server_addr.sin_addr.s_addr = htonl(inaddr_any); retcode = bind(connect_s, (struct sockaddr *)&server_addr, sizeof(server_addr)); printf("*** ERROR bind() failed \n"); printf("waiting for accept() to complete... \n"); addr_len = sizeof(client_addr); connect_s = accept(welcome_s, (struct sockaddr *)&client_addr, &addr_len); if (connect_s < 0) printf("*** ERROR accept() failed \n"); memcpy(&client_ip_addr, &client_addr.sin_addr.s_addr, 4); printf("accept completed (IP address of client = %s port = %d) \n", inet_ntoa(client_ip_addr), PORT_NUM); strcpy(out_buf, "This is a message from SERVER to CLIENT"); retcode = send(connect_s, out_buf, (strlen(in_buf) + 1), 0); printf("*** ERROR send() failed \n"); retcode = recv(connect_s, out_buf, sizeof(in_buf), 0); printf("*** ERROR recv() failed \n"); printf("received from client: %s \n", retcode); retcode = closesocket(welcome_s); printf("*** ERROR closesocket() failed \n");

8 retcode = closesocket(connect_s); printf("*** ERROR closesocket() failed \n"); #ifdef BSD retcode = close(welcome_s); printf("*** ERROR close() failed \n"); retcode = close(connect_s); printf("*** ERROR close() failed \n"); // Clean up winsock WSACleanup(); return(0);

9 Appendix B RDT 2.2 Receiver from Kurose and Ross checksum)