2 1 (20 points) Circuit switching vs datagram packet switching. (a)State two reasons why circuit switching can provide more predictable communication performance than datagram packet switching. (6 points) (b)Assume you wish to transfer an n-byte file along a path composed of the source, destination, 3 point-to-point links, and 2 switches. Suppose each link has a propagation delay of 20 ms, bandwidth of 2 Mbps, and that the switches support both circuit and datagram packet switching. Thus you can either break the file up into 1 KB packets, or set up a circuit through the switches and send the file as one contiguous bit stream. Suppose that packets have 24 bytes of packet header information and 1000 bytes of payload, that store-and-forward packet processing at each switch incurs a 1 ms delay after the packet has been completely received, that packets may be sent continuously without waiting for acknowledgements, and that circuit setup requires a 1 KB message to make one round- trip on the path incurring a 1 ms delay at each switch after the message has been completely received, and the delay at the destination is zero. Assume switches introduce no delay to data traversing a circuit. You should ignore the circuit teardown. You may also assume that file size is a multiple of 1000 bytes. For what file size n bytes is the total latency incurred before the entire file arrives at the destination less for circuits than for packets? (14 points) For packet, total transmission time is: Tp = (n/1000) * (1024*8/2M) + 3*20ms + 2*(1ms + 1024*8/2M) = n*4.096u + 70.192ms For circuit: Tc = (3*1000*8/2M + 3*20ms + 2*1ms)*2 + (n/1000) * (1000*8/2M) + 3 * 20ms = 208ms + n*4u So, for Tc < Tp, n*4u+ 208ms < n*4.096u+ 70.192ms n > 1.4355MB Guaranteed bandwidth or time slot, fixed network path.
3 2 (20 points) Broadcast and switched Ethernet. (a)State two reasons why broadcast Ethernet, where all hosts on the network share one single channel (i.e. wire) and CSMA/CD (carrier sense multiple access/collision detect) is used to arbitrate media access among the hosts, cannot support a large number of hosts spread across a large geographic area. (6 points) (b)A network based on Ethernet bridges (i.e. a switched Ethernet) does not use any routing protocol like distance vector or link-state to compute forwarding tables at the Ethernet bridges. Instead it uses two primitives: flooding and address learning. Briefly explain how these two mechanisms work together to let Ethernet bridges create forwarding tables. You may assume that the network has a tree topology. (14 points) If a bridge doesn’t know the destination address, flood packet, packet will reach destination. Each bridge notices the incoming port for each source address in the flooded packets and caches the port for subsequent forwarding to the previously seen source address. (1)To implement collision detection correctly, the maximum propagation delay of the network must be limited and related to the minimum packet size, thus as the physical length of the network increase, the minimum packet size must also increase, leading to inefficiency when sending small messages. (2)Since the medium is shared and has a limited bandwidth/capacity, performance will be very poor when a large number of hosts are sharing the medium.
4 3 (20 points) IP forwarding table aggregation. (a)Because IP packet forwarding is based on longest prefix matching of the packet destination address against forwarding table entries, it is possible to have redundant entries in the forwarding table. Forwarding table aggregation is a technique used to combine redundant entries to reduce the size of the forwarding table. In a few sentences, describe under what conditions two table entries can be combined and what the combined entry should contain. (10 points) (b)Perform forwarding table aggregation on the following IP forwarding table and write down the final table contents after all possible aggregations have been performed. Note that you may or may not receive any partial credit in this part if your answer to part (a) is incorrect. (10 points) Entry #Address PatternMaskNext Hop 122.214.171.124126.96.36.199 R1 2188.8.131.52184.108.40.206 R2 318.0.0.025220.127.116.11 R4 18.104.22.168 R4 522.214.171.124126.96.36.199 R1 6128.42.216.025188.8.131.52 R1 7184.108.40.206220.127.116.11 R3 (1)Both table entries have the same next hop output. (2)One entry’s address prefix defines an address space that is a superset of the other entry’s address space (3)The combined entry should contain the address pattern and the mask of the original entry that is the superset and the same next hop output Entries 1, 6 can be aggregated, so can entries 3, 4. Entry #Address PatternCIDR MaskNext Hop 118.104.22.16822.214.171.124 R2 126.96.36.199 R4 3188.8.131.52184.108.40.206 R1 4128.42.216.025220.127.116.11 R1 518.104.22.16822.214.171.124 R3 Aggregated entries
5 4 (20 points) Consider the simple network in the figure below, in which A and B exchange distance-vector routing information. All links have cost 1. Suppose the distance vector routing information has already converged, and now the A-E link fails. (a)Give a sequence of routing table updates that leads to a routing loop between A and B. (8 points) (b)Describe an addition to the basic distance-vector routing algorithm that will prevent the problem identified in (a). Show how it solves the problem in this specific case. (6 points) (c)Describe a generalization of the distance-vector routing algorithm that avoids the problem identified in (a) regardless of the network topology. (6 points) E E A A B B DestCostNext B1B E1E DestCostNext A1A E2A Node A Node B time A-E fails DestCostNext B1B Einfty- DestCostNext B1B E3B DestCostNext B1B E3E DestCostNext A1A E2A DestCostNext A1A E2A DestCostNext A1A E4A Split horizon, with or without poison reverse, eliminates the problem in this case. With split horizon, a node does not advertise a route to the node that it uses as a next hop to a given destination. With poison reverse, a node advertises a cost of infinity to the node that it uses as a next hop to a given destination. A path-vector algorithm eliminates the problem in any topology. With path-vector, nodes advertise the full path, thus allowing all nodes to check for routing loops.
6 5 (20 points) Sliding window based retransmission protocols. (a)In the stop and go protocol (i.e. the sliding window size is one packet), what is the minimum number of bits required to encode the sequence number for the protocol to work correctly? (3 points) (b)If sequence number is not used, the resulting “broken” stop and go protocol may fail to transfer data correctly. Describe a specific scenario that explains the problem of not using sequence number in stop and go. (7 points) (c)Assuming there is never any packet corruption or loss along a network path, therefore retransmission is never needed, what are the two primary factors that determine the overall throughput (in bits per second) of a data transfer using a sliding window protocol? (3 points) (d)State a mathematical formula to describe how the two factors determine the throughput.. (2 points) (e)What problems can arise if the retransmission timeout value is chosen arbitrarily? Be as specific as possible. (5 points) 1 bit Suppose the acknowledgement for the first packet is lost, and the sender retransmits the first packet after a timeout, then the receiver would mistake the retransmitted first packet as the second packet when no sequence number is used. The sliding window size and the round-trip delay. Window size in bits Round trip delay in seconds Throughput = If the RTO is chosen to be smaller than the RTT, then many unnecessary packet retransmission will result and waste network bandwidth. If the RTO is chosen to be much larger than the RTT, then the retransmission of a lost packet will be unnecessarily delayed, leading to poor performance.
10 COMP/ELEC 429 Spring 2006 Midterm Exam Exam instructions: 1.Once you open this exam, you have exactly 1.5 contiguous hour to work on the exam. It is your responsibility to time yourself and stop after 1.5 hour. No time credit for breaks allowed. 2.This exam is closed books, closed notes, closed computers. You must work on this exam on your own. 3.You are allowed to use a hand-held calculator for doing arithmetic. 4.You may not ask the instructors questions during the exam. The exam has been designed to eliminate ambiguities. When absolutely necessary, state your assumptions and proceed. 5.Be sure to fill out the front page with your name, email address, student ID number, and the time you started and finished the exam. 6.Sign the pledge in the space provided. 7.Notation: K=1000, M=1000000, G=1000000000, 1 byte (B) = 8 bits (b). 8.You may use the reverse side of a page to write your answers. The last few pages are scratch sheets you may use. 9.This exam is assigned on 3/2/2006 at 2:20pm. 10.This exam is due on 3/9/2006 at 1:00pm at Duncan Hall room 1042. Late exam turn-in will not be accepted. Early turn-in can be submitted to DH 3005. 11.This exam accounts for 15% of your final grade. 12.Points assigned to each problem roughly corresponds to the problem’s difficulty. 13.Be sure to show your work appropriately. 14.You should write all your answers in this provided exam, and only attach additional sheets of paper if necessary.