Mid Sweden University Dept. of IT and Media (ITM) Magnus Eriksson Tel 060-148740, 070-562 5502. Email magnus.eriksson@miun.se 9 Jan 2009 Page 1 of 6 Exam in DT024G, Computer Networks A Answers Time: 8:00-12:00 Permitted tools: Arbitrary pocket calculator. An English-Swedish dictionary. Preliminary requirement for approval: 30 out of 60 points. Only write on one side of each sheet. You may answer in Swedish or English. Your score will be presented at http://portal.miun.se via the LADOK service preliminary within three weeks, provided that you have concluded the course quizzes (assessments). Distance students: If you would like us to post you a copy of the marked exam, please contact our course expedition, e-mail itmexp.svl@miun.se. Good luck! THEORY PART 1. Draw a copy of the following table (with numbers but not all the text). Put a tic to mark if statement 1 to 10 is valid for networks of type A, B and C. There may be one or several tics in a column or in a row. (1 point per correct row. 0.5 points for a partly correct row.) (10 p) A: Circuit switched network B: Virtual circuit switched network 1. May use FDM or TDM multiplexing. 2. Based on packet switching. 3. The bit rate and delay may vary. 4. The data may be delivered in changed order because it does not follow a fix path. 5. One datastream or connection can not borrow unused capacity from another. 6. The resources are reserved in the network nodes during a connection phase. 7. Limited number of channels. Blocks new connections if too many simultaneous connections. 8. For example:.25, Frame relay and ATM 9. For example: ISDN and SONET/SDH 10. For example: Conventional IP networks C: Datagram network
Exam in Computer Networks A 9 January 2009 Page 2 2. Combine the words. One word in the left column corresponds to one description in the right column. State what number that corresponds to what letter in numerical order (not in alphabetical order). (12 p) 1. FDM A. Modulation technique that gives lower bit rate and lower bit error probability than 8PSK modulation for the same analogue bandwidth (in Hz) and SNR (in db 2. TDM B. Old name for IP router. May also mean application protocol converter. 3. Line coding C. Error control with sliding sliding window. A corrupt or lost frame and all subsequent frames are re-transmitted. Inefficient if retransmissions are common. 4. UTP D. Error control without sliding window. Inefficient for large delay-bandwidth product. 5. FSK E. Modulation technique that gives higher bit rate and larger bit error probability than 8PSK modulation for the same analogue bandwidth (in Hz) and SNR (in db). 6. Stop-and-wait ARQ F. Methods for transferring a digital bit stream over a low pass channel. Example: NRZ, Manchester, AMI. 7. Go-back-N ARQ G. One frequency band and carrier frequency per channel. 8. Gateway H. Connects LAN segments, and extends point-to-point links exposed to noise. Retransmits the data bit by bit instead of packet-by-packet. 9. 16PSK I. Based on start and stop bits, and arbitrary pause between the bytes. 10. Asynchronous serial J. One timeslot per channel. communications 11. Repeater K. Copper wires used in the telephone access network, and in switched and hubbed Ethernet. 12. Selective repeat ARQ L. Error control with sliding window. At retransmission, only the missing or corrupt frame is retransmitted, not subsequent correctly received frames. Drawback: Complex. Answer: 1G, 2J, 3F, 4K, 5A, 6D, 7C, 8B, 9E, 10i, 11H, 12L.
Exam in Computer Networks A 9 January 2009 Page 3 3. Draw a copy of the following table, and tick what functions and schemes (the columnes) that may be handled by what protocol (s) in the five TCP/IP model (rows). There should one or several ticks in each column, and one or several ticks in each row. (Marking: One point per correct row.) (12 p) A: Internet (network ) B. Physical C. Transport E. Application D. Data link 1. Automatic repeat request (ARQ), sliding window, flow control ( ) 2. Logical addressing and routing 3. Physical addressing (MAC addressing) 4. Port numbering. 5. PSK, FSK, QAM 6. TDMA, FDMA, CDMA/CD, ( ) CDMA/CA 7. TDM, FDM, WDM 8. Client-server protocols ( ) 9. Bit synchronization 10. Frame synchronization 11. Packet queuing, store-andforward ( ) 12. Standardization of radio frequencies, signal strengths, wave forms, electrical impedance Tics in parenthesis ( ) are not required but accepted. 4. What factors and mechanisms may affect the packet throughput measurements when Internet bandwidth capacity measurement utilities such as tptest and bredbandskollen.se are used? Why don t you always get the maximum bit rate offered by the ISP. (Hint: The communication may involve a user computer, a measurement server, routers, switches, wires, Ethernet hub/bus networks, xdsl modems or PSTN modems using the analog telephone access network, wireless mobile communication with varying signal-to-noise ratio, etc.) (6 p) Answer: Varying packet queue lengths in network nodes (routers, switches, bridges and hosts). Different packets may take different path due to rerouting. Collisions in non-switched Ethernet or WLANs may cause data link retransmissions. Long distance between modems or WLAN access points may cause that WLANs or modems negotiate on a robust but slow modulation scheme, resulting in low data rate
Exam in Computer Networks A 9 January 2009 Page 4 Packet errors due to interference in wireless links and modems may cause data link retransmissions, which causes a delay. Limited TCP window size in combination with large propagation delay, for example due to long distance may cause that the full bandwidth capacity can not be used (the long fat pipe syndrome). Missing packets due to congestion (the router skips packet when the queue is full) may cuase missing RTT data and congestion avoidance ( slow start ), i.e. reduced TCP window size. Processing delay due to other tasks in the user computers, or load from other users on the server computer. PROBLEM PART Show all calculcations. 5. What is the amplitude (in Volt) and the frequency (in Hertz) of the following sine wave? What is the phase in radians and degrees? Answer: 15V, 1MHz, π/8 radians, 22.5. v(t) = 15 sin (2л 1000000t + л/8) Volt (2 p) 6. What is the symbol rate in Mbaud (million symbols/s) of a 256 QAM modulator (using 256 different symbols), if the gross bit rate, inclusive of error-correcting codes, is 40 Mbit/s? (2 p) Answer:.40/8 = 5Mbaud. 7. The input power of a cable is 0.1 Watt. The cable has an attenuation of 0.1 db/meter. The output power is 0.001 Watt. How long is the cable? (2 p) Answer: 100 times attenuation equals 20 db. 20 db/(0.1 db/meter) = 200 meter. 8. Here follows two IP addresses. State for each of them what class they belong to (A/B/C/D/E), number of possible host adresses in the network, and the IP broadcast address of the network in case of classful addressing. (3 p) a) 16.12.128.1 b) 193.10.25.51 Answer: a) Class A. 2 24-2 host addresses. Broadcast IP address 16.255.255.255. b) Class C. 2 8-2 = 254 host addresses. Broadcast IP 193.10.25.255.-
Exam in Computer Networks A 9 January 2009 Page 5 9. You have got job at the Kiruna space center in the northern part of Sweden, with orders to design a system for datatransmission between earth and the international space station (ISS), which is at about 400 km above the surfice of the earth. You chose to utilize a radio link with the data rate 100 Mbit/s, which should send a sequence of data packets of size 4000 Byte data per packet. a) Calculate the packet transmission time, i.e. the time between that the first and the last bit in a packet has left the transmitter. b) Calculate the propagation delay from earth to the space station. The propagation speed of light is c = 3.00 10 8 m/s. c) If we use a Stop-And-Wait ARQ protocol, what is the waiting time between two packets, and what is the average throughput in bit/s? The ACK frames are very small. d) If we utilize a sliding window protocol, how large buffer memory in Byte is required in view to utilize the full capacity of the radio link, without any waiting time between the packet transmissions (provided that no retransmissions occur)? (7 p) Answer: a) Transmission time T t = 4000 *8 bit / 100 000 000 bit/s = 0.00032 s = 0.32 ms. (1p) b) Propagation time T p = 400 000 m / 300 000 000 m/s = 1.33 ms. (2p) c) The waiting time is 2T p = 2.67 ms. A new packet is sent every T t + 2T p = 3.0 ms (the total round-trip-time) if the ack packet transmission time is neglected. The throughput is 4000*8/0.030 = 10.7 Mbit/s. (2p) d) 100 Mbps * (T t +2T p ) / 8 = 39.000 Byte. Optional solution: The number of packets in the buffer is (T t +2T p ) / T t = 9.33, i.e. 10 full data packets, which is 10 * 4000 Byte = 40000 Byte. (2p)
Exam in Computer Networks A 9 January 2009 Page 6 Host C Host B Host D Private IP: 10.14.5.1 172.16.5.255 Public IP: 193.10.250.187 MAC addr: 0013020764AE Privat IP: 10.14.5.2 MAC: 02CB239B Router IP: 130.16.4.1 MAC: 015100212983 Host A 130.16.4.2 70DD35530178 130.16.4.3 BB26165274D3 Router + NAT server Private IP: 10.2.1.1 Public IP: 193.10.250.187 MAC addr: 31BE4A19273A 10.2.1.2 193.10.250.187 A0C11222F53B 10.2.1.3 001B55301781 10. An evil spy has installed a sniffer software on host computer A as well as on host B, in view to eavesdropp communication in the above network. Both sniffers detect an Ethernet frame (including an IP datagram) sent from host D, and destined to host C. Note that the right router includes a Network Address Translation (NAT) function, enabling several computers to share the same public IP address. What IP source address and destination address does the datagram have, and what physical source and destination address does the Ethernet frame have, a) according to the sniffer on host A? Answer: From 10.14.5.1 to 130.16.4.1. From 31BE to 001B b) according to the sniffer on host B? Answer: From 193.10.250.187 to 130.16.4.1. From BB26 to 0151. (Marking: 2p for correct IP addresses, 2p for correct MAC addresses.) (4 p)