CS/ECE 438: Communication Networks for Computers Spring 2014 Midterm Study Guide

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1 CS/ECE 438: Communication Networks for Computers Spring 2014 Midterm Study Guide 1. Channel Rates and Shared Media You are entrusted with the design of a network to interconnect a set of geographically distributed hosts within your corporation. After some research, you narrow the options to two choices, a fiber-based token ring or a copper-based switched network. The pertinent statistics appear in the table below. type fiber-based token ring copper-based switched network signal bandwidth 10 MHz 1 MHz signal-to-noise ratio at transmitter 20 db 20 db attenuation rate 1 db/km 2 db/km The longest link in the network in either case is 10 km. (a) What link bandwidth is possible according to Shannon's Law i. for the fiber network? SNR = 20dB (10km*1dB/km) = 10dB 10dB = 10log 10 (S/N) S/N = 10 So, S/N at receiver is 10. C = B*log 2 (1+S/N) = 10MHz * log 2 (1+10) = Mbps (in fiber) ii. for the copper network? SNR = 20dB (10km*2dB/km) = 0dB 0dB = 10log 10 (S/N) S/N = 1 So, S/N at receiver is 1. C = B*log 2 (1+S/N) = 1MHz * log 2 (1+1) = 1Mbps (in copper) (b) Assuming that hosts in the copper network can all transmit at their link rate (the values found in part (a)) simultaneously, roughly how many hosts are necessary for the networks to provide equal aggregate bandwidth (the sum of bandwidth for all hosts)? we would need ~35 hosts on the copper network to equal the aggregate bandwidth of the fiber network. (c) Using the copper-based network with a 32-point QAM encoding, what modulation rate (baud) is necessary to obtain the bandwidth found in part (a)? 32 points = 5 bits/symbol 1Mbps / (5 bits/symbol) = 0.2 M symbols/sec = 200K baud

2 2. Medium Access Control This question concerns medium access control on a microwave network using carrier sense multiple access with collision detection (CSMA/CD, the algorithm used with Ethernet). The network consists of four hosts distributed as shown in the figure below. The microwaves are broadcast, and the signal travels directly along a line of sight from sender to all receivers. Assume that the signals propagate at the speed of light in a vacuum, m/sec. A 54 km 40 km 51 km B C 90 km 57 km 57 km D (a) If a transmitter sends at 1 Mbps, how long must packets be to guarantee collision detection by the transmitter? Longest link = 90 km, so maximum one-way delay is: (90km) / (3 x 108 m/s) = 30 x 10-5 sec = 300 µsec Therefore, maximum RTT is 600 µsec. So at a rate of 1 Mbps, we get: (1 Mbps) * 600 µsec = 600 bits = 75 bytes (b) Divide time into slots the length of the maximum round-trip propagation delay in the network. One packet may be transmitted each time slot. Assume that each of the four hosts attempts to transmit with probability p in each time slot. What is the probability of a successful transmission in any given slot if i. p = ¼? (4 C 1) * (¼) * (¾) 3 = 27 / 64 ii. p = ½? (4 C 1) * (½) * (½) 3 =1 / 4 iii. p = ¾? (4 C 1) * (¾) * (¼) 3 = 3 / 64 (c) Using the minimum transmission length from part (a) and the probability of successful transmission from part (b)(ii) (for p = ½), calculate the average throughput of the network if each packet requires 20 bytes of header/trailer and i. 10 bytes of data, and Given 30 bytes of information (data and headers), we need another 45 bytes to reach the minimum packet size (from part a). Therefore, we are transmitting only 10 bytes of actual data per 75 bytes transmitted. (10 bytes) / (75 bytes) = = 13.3% goodput (assuming no errors)

3 Now factor in the ¼ success rate (from part b), which lowers the goodput to 3.33% Therefore 3.33% of the 1 Mbps link is used for data throughput. This is equivalent to 33.3 Kbps. ii. 50 bytes of data. Given 70 bytes of information (data and headers), we need another 5 bytes to reach the minimum packet size (from part a). Therefore, we are now transmitting 50 bytes of actual data per 75 bytes transmitted. (50 bytes) / (75 bytes) = = 66.7% goodput (assuming no errors) Now factor in the ¼ success rate (from part b), which lowers the goodput to 16.67% Therefore 16.67% of the 1 Mbps link is used for data throughput. This is equivalent to 167 Kbps. Hint: Don't try to make part (c) too realistic. 3. Sliding Window Algorithms This question considers a sliding window implementation across a full-duplex point-to-point link. The link has a bandwidth of 327 kbps in each direction and a one-way propagation delay of 100 milliseconds. All packets sent across the link are 1,024 bytes long, including all headers and trailers. (a) How much data is required to fill the pipe for a RTT(contains 2 one-way delay and 1 transmission delay)on the network? 2 ( s) ( bps) +1024*8b = 65.4 Kb+8Kb = 73.4Kb (b) What send window size (SWS) is necessary to fully utilize the network? ( 73,400 bits) / (8,192 bits/packet) = [8.98 = 9 packets (c) For RWS= SWS/2, construct an example demonstrating that SWS+2 sequence numbers (e.g, from 0 to SWS+1, where SWS is your answer to part (b)) are not enough to guarantee correct operation of the sliding window algorithm. SWS = 9 RWS = 4 SWS+2 = 11 send receive all ACK s lost The receiver is now expecting frames 8, 9, 10, , but the sender times out and sends frames Unfortunately, the receiver is expecting the second incarnation of frames but gets the first incarnation of these frames. duplicate with what was previously received

4 (d) Given a go-back-n algorithm (with RWS=1), assume that data frames are received with probability p = 0.9 and that acknowledgements (ACK's) are always received (probability of 1). Further assume that the retransmission timeout used by the sender is a negligible amount of time longer than the round-trip time, implying that a packet is retransmitted as soon as the sender could possibly detect its loss on the previous transmission. Calculate the average transmission rate (in bits per second) achieved for long streams of data. transmission time = 8,192 b / bps = sec = 25 msec penalty for loss = 200 msec + 25msec(RTT) average number of transmissions = 1/p = 10/9 average time per frame = 25 msec + (225 * 1/9 msec) = 50 msec average throughput = 8, 192 b / (25 msec + (225 * 1/9 msec)) = Kbps 4. Token Ring Consider a token ring with latency of 500 µsec. Answer for both a single active host and for many active hosts. For the latter, assume that there are sufficiently many hosts transmitting that the time spent advancing the token can be ignored. Assume a packet size of 1500 B (a) Assuming that the delayed token release strategy is used, what is the effective throughput rate that can be achieved if the ring has a bandwidth of 3 Mbps? Single host: At 3 Mbps, it takes 4 ms to send a packet. A single active host would transmit for 4 ms and be idle for 0.5 ms until the frame was received back and idle for yet another 0.5 ms as the token went around the ring. So, efficiency = 4 / ( ) = 80 %. So, effective throughput = 2.4 Mbps Many hosts: Time spent advancing the token is ignored. So, the time between successive packet transfers is 4.5 ms. So, efficiency = 4 / 4.5 = %. So, effective throughput = 2.67 Mbps. (b) Assuming that the immediate token release strategy is used, what is the effective throughput rate that can be achieved if the ring has a bandwidth of 3 Mbps? Single host: A host sends a packet for 4 ms, and then remains idle for 0.5 ms as the token circulates. So, efficiency = 4 / ( ) = %. So, effective throughput = 2.67 Mbps. Many hosts: Hosts transmit one after the other. The time interval between successive packet transfers is the same as the transmission time. So, efficiency is close to 100% and effective throughput is close to 3 Mbps. (c) With delayed release, what is the effective throughout rate that can be achieved if the ring has a bandwidth of 100 Mbps? Single host: At 100 Mbps, transmission time = 120 µs. A single host would send for 120 µs and then wait for 500 µs for the packet to come around and then wait another 500 µs for the token to go around. So, efficiency = 120/( ) = 10.71%. So, effective throughput = Mbps.

5 Many hosts: Time between successive packet transfers = µs. So, efficiency = 120 / ( ) = %. So, effective throughput = Mbps. (d) With immediate release, what is the effective throughout rate that can be achieved if the ring has a bandwidth of 100 Mbps? Single host: A host would send for 120 µs and then wait 500 µs for the token to come back. So, efficiency = 120 / ( ) = %. So, effective throughput = Mbps. Many hosts: Hosts transmit one after the other. The time interval between successive packet transfers is the same as the transmission time. So, efficiency is close to 100% and effective throughput is close to 100 Mbps. 5. Name the OSI layer or layers in which medium access control (MAC) is addressed and state whether MAC is typically handled in hardware, in software, or in both in the Internet architecture. Data link layer Hardware 6. Explain one advantage of abstracting networked communication into multiple layers. Modularity Abstraction 7. Describe the problem solved by reliable transmission. Supports the illusion that a network neither loses nor reorders frames. Implemented in the transport layer in software. 8. Explain how a receiver detects the end of a frame with length-based framing. The length is indicated in the header of the frame. 9. Explain the main drawback of the stop-and-wait ARQ algorithm. By only allowing one outstanding packet, bandwidth is not used efficiently (i.e. stop-and-wait does not keep the pipe full) 10. Name the OSI layer or layers in which framing is addressed and state whether framing is typically handled in hardware, in software, or in both in the Internet architecture. Framing is addressed at the datalink layer in hardware 11. Define the Hamming distance of an encoding. The minimum number of bit flips to move from one to the other 12. Explain the effect of layering on end-to-end bandwidth.

6 The additional headers at each layer consume additional bandwidth. Each layer adds additional processing time. 13. Name the OSI layer or layers in which encoding is addressed and state whether encoding is typically handled in hardware, in software, or in both in the Internet architecture. Encoding is addressed in the OSI physical layer and handled in hardware. Physical Layer Hardware 14. Explain a drawback of forwarding packets with source routing. Header overhead because of storing the source route in each packet Nodes need to be aware of the topology of the entire network hosts must know topology and recognize failures 15. What delay is relevant and what bandwidth is relevant for computing the delay-bandwidth product of two links in series? Sum of link delays, minimum of link bandwidths 16. Describe a problem associated with communicating between heterogeneous architectures (e.g., a mixture of Sun and Intel hosts) on a network. They may not have the same data representation 17. What purpose do the four addresses in an IEEE packet serve? Wireless source and destination addresses for first and last (wireless) hops Source and destination AP addresses for wireline routing (forwarding) 18. Describe the problem solved by medium access control (MAC). Given multiple senders on some media, MAC aims to provide: Fair arbitration Good performance 19. (1)Describe the problem solved by framing. Separates continuous stream of bits into frames Marks start and end of each frame (2)Framing demarcates units of transfer. Name two benefits of framing for bit stream transmission. ability to packetize data multiplexing efficient/better error correction if error, easy retransmission of small sections of data sync recovery

7 20. Name the OSI layer or layers in which error detection is addressed in the Internet architecture and state whether error detection is typically handled in hardware, in software, or in both. Data link, Hardware

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