CS 455/655: Introduction to Computer Networks Fall 2013 Sample Final Examination

Transcription

1 CS 455/655: Introduction to Computer Networks Fall 2013 Sample Final Examination NAME: MATTA This exam is closed books and closed notes. Please write clearly and neatly. Be precise in your answers; do not just re-iterate what you know about the topic. Clearly state any assumptions you make. All questions are weighted equally. Answer all 5 questions. Problem 1 (IP addressing): Subnet B: 12 hosts Subnet A: 15 hosts Subnet C: 45 hosts R1 To public Internet via ISP Consider the network shown above, consisting of a single router, R1, with three subnets A, B and C, with 15, 12, and 45 hosts, respectively, on these subnets. Assign an address range to the hosts in subnets A, B, and C so that only a single aggregated address needs to be advertised by R1 to the public Internet, and that the size of the advertised aggregated address range is minimized. In a sentence or two, explain how you arrived at your answer. (Feel free to use variable names in addresses if the precise bit values are not important.) We need = 72 addresses, i.e. we need to 2 7 addresses or 7 bits. So we have a common prefix length of 32-7=25 bits. Aggregated CIDR address: X.Y.Z.0/25 Subnet A CIDR address: We need 2 4 addresses, or 4 bits. CIDR address: X.Y.Z.0/28 Subnet B CIDR address: We need 2 4 addresses, or 4 bits. CIDR address: X.Y.Z.16/28 Subnet C CIDR address: We need 2 6 addresses, or 6 bits. CIDR address: X.Y.Z.64/26 Note that other solutions are possible. 1

2 Problem 2 (Distance-vector routing): E A B Consider the simple network shown above, in which A and B periodically exchange distance-vector routing information. Assume routers do not implement split horizon. All links have cost 1. After reaching convergence, suppose the A E link fails. (a) Give a sequence of routing table updates that leads to a routing loop between A and B, for a at E. Before A advertises unreachability, node B advertises its distance of 2 to A, which A adopts. (b) Estimate the probability of the scenario in (a), assuming A and B send out routing updates at random times, each at the same average rate. Probability that B advertises before A = 0.5. (c) Estimate the probability of a loop forming if node A broadcasts an updated routing report after 1 second of discovering the A E failure, and B regularly broadcasts every 60 seconds. Probability that B advertises within 1 second = 1/60. Now, consider the following extended topology, and again suppose after reaching convergence, the A E link fails. Even if routers implement split horizon (without poison reverse), a routing loop between B and C, for a at E, can form consider the following scenario: node A advertises a non-reachability update to B and C; before this non-reachability update arrives at B and C, B and C had already exchanged their distances, but these routing messages arrive at B and C after the non-reachability message from node A had arrived. C E A B (d) Describe how the situation in this extended scenario will evolve. Then, describe how the situation would evolve if routers use split horizon with poison reverse. Without Poison Reverse, a loop forms between B and C. With Poison Reverse, B and C advertise unreachability to each other and the loop breaks. 2

3 Problem 3 (BGP routing): (a) Suppose P,, and R are network service providers, with respective CIDR address allocations C /8, C /8, and C /8 (using hexadecimal dotted notation with mask). Each provider s customers initially receive address allocations that are a subset of the provider s address space. P has the following customers: PA, with allocation C1.A3.0.0/16, and PB, with allocation C1.B0.0.0/12. has the following customers: A, with allocation C2.0A.10.0/20, and B, with allocation C2.0B.0.0/16. Assume there are no other providers or customers, and that each provider connects to both of the others. Suppose customer PA switches to provider and customer B switches to provider R, resulting in the AS topology shown below. Use the CIDR longest match rule to give the routing table for P (only as representative) that allow PA and B to switch without renumbering (i.e., keeping their initial address allocations). CIDR address next-hop (AS name) PB PA B A C /8 C /8 C1.A3.0.0/16 C2.0B.0.0/16 C1.B0.0.0/12 R // toward PA R // toward B PB P R (b) Suppose customer PA is multi-homed, in that it has two Internet connections from the two providers, P and, as shown in figure below. (PA is taking its address assignment from P.) has a CIDR longest match routing entry for PA. Which inbound (incoming) traffic to PA might flow on the PA connection, considering only traffic from domains P,, and H? Consider the two cases where does and does not advertise PA to the world using BGP. Recall shorter AS paths are preferred. does not advertise PA to the world: PA Only traffic from. P advertises PA to the world: Traffic from and H. X Y H 3

4 Problem 4 (Indirection and encapsulation): A à B TCP TCP header + Consider the IP gram carrying a TCP segment, shown above. Consider a situation where we want to route this gram through an intermediate router C by encapsulating it within another gram. (a) Draw the gram as it travels to C and then as it travels out of C toward the final B make sure to show the contents of the fields: source,,, and (payload). To router C From C, To B Aà C IP-in-IP original gram Aà B TCP TCP segment (b) Describe how the gram is processed at router C from when it arrives at C and until it comes out of C on its way toward the final B. At router C, IP demultiplexes the gram to the decapsulation software, which removes the header and forwards the original gram to the final. 4

5 Problem 5 (CSMA/CD ): (a) In CSMA/CD, after the fourth collision, what is the probability that the node chooses K = 3? What does the result K = 3 correspond to in terms of delay in microseconds on a 10 Mbps Ethernet? After the 4 th collision, K is chosen uniformly from 0,, 15. The probability that K=3 is then 1/16. K=3 corresponds to a delay of 512*3 bit times = (512*3)/10M= µsec. (b) Suppose nodes A and B are on the same 10 Mbps Ethernet segment, and the propagation delay between the two nodes is 225 bit times. Suppose at time t = 0, B starts to transmit a frame. Suppose A also transmits at some t = x, but before completing its transmission it receives bits from B (hence, a collision occurs at A). Assuming node A follows the CSMA/CD, what is the maximum value of x? x = 224 bit times /10M = 22.4 µsec. (c) Consider two nodes A and B on the same Ethernet segment, and suppose the propagation delay between the two nodes is 225 bit times. Suppose at time t = 0, both nodes A and B begin to transmit a frame. At what time do they detect the collision? Assuming both nodes transmit a 48-bit jam signal after detecting a collision 1, at what time (in bit times) do nodes A and B sense an idle channel? How many seconds is this for a 10 Mbps Ethernet? 225 bit times = 22.5 µsec. ( ) = 498 bit times = 49.8 µsec. 1 Note that once an adapter detects a collision, it stops transmitting its frame but continues to transmit a 48- bit jam signal to ensure that collision is detected by other adapters. 5