Problem Max. Points Act. Points Grader

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1 Networks and Protocols Course: 00 Jacobs University Bremen Date: Dr. Jürgen Schönwälder Duration: 75 minutes Midterm Examination The Jacobs University s Code of Academic Integrity applies to this examination. Please fill in your name (please write readable) and sign below. Name: Signature: This exam is open book in the sense that you are allowed to use course slides, papers and your personal notes during the exam. You are not allowed to use any electronic equipment such as computers or cell phones. Please answer the questions on the problem sheets. If you need more space, feel free to write on the back of the pages. Please keep the papers stapled. Problem Max. Points Act. Points Grader M. 0 M. 0 M. 0 M.4 0 M.5 0 M.6 0 Total 00

2 Problem M.: fundamental concepts (++++ = 0 points) Indicate which of the following statements are correct or incorrect by marking the appropriate boxes. For every correctly marked box, you will earn two points. For every incorrectly marked box, you will loose one point. Statements which are not marked or which are marked as true and false will be ignored. The minimum number of points you can achieve is zero. true false When using Differential Manchester Encoding, it is possible to communicate even if the signal got inverted (for example by swapping wires). For an Ethernet interface using CSMA-CD, it is sufficient to listen for collisions only during the beginning of the transmission of a large Ethernet frame. The Internet checksum is able to detect more errors than the CRC- checksum. It is possible to design protocols that (a) only work with ACKs, (b) only work with NACKs, (c) use both ACKs and NACKs. Flow control deals with the sender s/receiver s finite buffers while congestion control deals with the finite buffers on forwarding devices (e.g., router, bridges) within the network. Solution: true false When using Differential Manchester Encoding, it is possible to communicate even if the signal got inverted (for example by swapping wires). For an Ethernet interface using CSMA-CD, it is sufficient to listen for collisions only during the beginning of the transmission of a large Ethernet frame. The Internet checksum is able to detect more errors than the CRC- checksum. It is possible to design protocols that (a) only work with ACKs, (b) only work with NACKs, (c) use both ACKs and NACKs. Flow control deals with the sender s/receiver s finite buffers while congestion control deals with the finite buffers on forwarding devices (e.g., router, bridges) within the network.

3 Problem M.: spanning tree protocol (0+0 = 0 points) Consider a Local Area Network (LAN) with the following topology: Segment A P. B P. P. Segment B B P. P. P. P. B P. Segment F Segment G Segment C P4. P5. Segment E Segment D B4 P4. P4. P5. B5 P5. Assume that the delay and cost on all segments is the same and that all systems use the same default priorities. a) Compute the spanning tree generated by the IEEE 80. spanning tree algorithm by marking the root bridge, the root port in all the other bridges, and the designated port for each LAN segment. If there is a tie, use the bridge/port with the lower ID. Which port(s) will be blocked? b) Assume bridge B4 fails. Compute the new spanning tree as in a). Solution: a) Bridge B has the lowest ID and becomes the root bridge. The root ports of the other bridges are P., P., P 4., P 5.. The designated ports of the segments A-G are P., P., P., P 4., P., P. and P 4.. The ports P., P 5. and P 5. will be blocked. b) Bridge B still has lowest ID and becomes the root bridge. The root ports of the other bridges are P., P., P 5.. The designated ports of the segments A,B,C,F are P., P., P., P.. The port P 5. will be blocked.

4 Problem M.: IPv4 over Ethernet (0 points) Consider the following Ethernet network topology. a b c A R B C e d E D The devices A, C, D, and E are IP hosts and R is an IP router and B a transparent bridge. The IP devices have static forwarding tables so that they can reach each other. The network just got initialized and no dynamic state information (except the static IP forwarding tables) is present in the devices. Let mac(x) denote the MAC address of node X and ip(x) denote the IP address of node X. The notation mac( ) denotes an Ethernet broadcast address an ip( ) an IP broadcast address. A first sends an IP datagram to C. Afterwards, D sends an IP datagram to E. Write down the messages that are exchanged over the various segments by filling out the following table: step proto mac-src mac-dst segments ip-src ip-dst description The proto column identifies the protocol contained in an Ethernet frame. The columns mac-src and mac-dst contain the Ethernet source and destination address while the columns ip-src and ip-dst contain the IP source and destination address where applicable. The column segments list the segments over which a frame is carried and the column description describes the contents of the frame.

5 Solution: step proto mac-src mac-dst segments ip-src ip-dst description ARP mac(a) mac(*) a who has ip(r)? ARP mac(r) mac(a) a ip(r) is at mac(r) IP mac(a) mac(r) a ip(a) ip(c) ip packet 4 ARP mac(r) mac(*) b,c,d who has ip(c)? 5 ARP mac(c) mac(r) c,b ip(c) is at mac(c) 6 IP mac(r) mac(c) b,c ip(a) ip(c) ip packet 7 ARP mac(d) mac(*) d,c,b who has ip(r) 8 ARP mac(r) mac(d) b,d ip(r) is at mac(r) 9 IP mac(d) mac(r) d,b ip(d) IP(E) ip packet 0 ARP mac(r) mac(*) e who has ip(e)? ARP mac(e) mac(r) e ip(e) is at mac(e) IP mac(r) mac(e) e ip(d) IP(E) ip packet

6 Problem M.4: IPv4 packet forwarding (4+8+8 = 0 points) a) During normal IPv4 packet forwarding by a router, which of the following packet fields are updated? Why? IPv4 header source address IPv4 header destination address IPv4 header time-to-live (TTL) IPv4 header checksum b) You are given the following network prefixes: / / / / / /4 Aggregate the prefixes together into the smallest possible number of shorter prefixes. What is the smallest number of prefixes, and what are they? prefix; / prefixes; /0, / 4 prefixes; /, /4, /4, / prefixes; /, /4 prefixes /0, /4, / Explain how you arrived at your answer. c) Explain how programs such as traceroute determine the route taken by packets on the Internet. Which protocol features on the networking layer are essential for traceroute to work? Solution: a) The time-to-live (TTL) field and the checksum field are updated while forwarding IPv4 datagrams. The source and destination address contained in an IPv4 datagram identify the communicating endpoints and never change during normal forwarding. (The exception to this rule are so called network address translators.) b) The two prefixes /4 and /4 can be aggregated to /. Similarly, the two prefixes /4 and /4 can be aggregated to /. A prefix combining / with /4 would include addresses from other prefixes and hence this combination does not work. A similar argument can be made for the prefix /4 and the aggregated prefix /. Hence, 4 prefixes are needed. c) Traceroute uses the following simplified algorithm: for (ttl = ; ttl < max; ttl++) { for (r = 0; r < retries; r++) { send-icmp-echo-request(dest, ttl); if (icmp-time-exceeded-received()) { print(icmp.source); continue } if (icmp-echo-reply-received()) { print(icmp.source); break; } } } Essential for traceroute to function is the ICMP protocol and a certain set of ICMP messages. Sites that filter ICMP messages will appear as black holes.

7 Problem M.5: open shortest path first (8+ = 0 points) Consider the following network topology: C 4 F A B E H D 4 G a) Determine the shortest-paths from node A to all destinations using Dijkstra s algorithm. For each step of the algorithm, fill in a row in a table structured as follows: step A B C D E F G H current permanent The column current indicates the current node and the column permanent indicates the set of nodes already marked as permanent. The columns A-H keep node specific information. After completing the algorithm, draw the shortest path spanning tree rooted at A. b) The OSPF protocol uses Dijkstra s algorithm to compute shortest paths. What is the motivation to support multiple areas in OSPF?

8 Solution: a) Computation of shortest paths using Dijkstra s algorithm: step A B C D E F G H current permanent A (A) B A (B) C A,B 4 4(B) D A,B,C 5 5(C) E A,B,C,D 6 5(C) F A,B,C,D,E 7 6(E) G A,B,C,D,E,F 8 7(E) H A,B,C,D,E,F,G C 4 F A B E H D 4 G b) Multiple areas (i) reduce the amount of traffic needed in the areas to distribute the link state information and (ii) reduce the effort to compute shortest pathes within an area.

9 Problem M.6: sockets (++++ = 0 points) Indicate which of the following statements are correct or incorrect by marking the appropriate boxes. For every correctly marked box, you will earn two points. For every incorrectly marked box, you will loose one point. Statements which are not marked or which are marked as true and false will be ignored. The minimum number of points you can achieve is zero. true false The socket API uses different C structs to represent IPv4 and IPv6 addresses. The accept() function must be used by a server to accept clients sending datagrams (SOCK DGRAM). The select() function can be used to implement server that can handle multiple clients in a single process. The getaddrinfo() function can map a given node name into multiple addresses and multiple address families. In order to communicate, a client and a server must use the same socket domain, type, and protocol. Solution: true false The socket API uses different C structs to represent IPv4 and IPv6 addresses. The accept() function must be used by a server to accept clients sending datagrams (SOCK DGRAM). The select() function can be used to implement server that can handle multiple clients in a single process. The getaddrinfo() function can map a given node name into multiple addresses and multiple address families. In order to communicate, a client and a server must use the same socket domain, type, and protocol.

Problem Max. Points Act. Points Grader

Problem Max. Points Act. Points Grader Networks and Protocols Course: 320301 Jacobs University Bremen Date: 2007-12-12 Dr. Jürgen Schönwälder Duration: 90 minutes Final Examination The Jacobs University s Code of Academic Integrity applies