Exam (elaborations) TEST BANK FOR Data Communications and Networking 4TH Edition By Behrouz A. Forouzan (Solution Manual)

Exam (elaborations) TEST BANK FOR Data Communications and Networking 4TH Edition By Behrouz A. Forouzan (Solution Manual) CHAPTER 1 Introduction Solutions to Review Questions and Exercises Review Questions 1. The five components of a data communication system are the sender, receiver, transmission medium, message, and protocol. 2. The advantages of distributed processing are security, access to distributed databases, collaborative processing, and faster problem solving. 3. The three criteria are performance, reliability, and security. 4. Advantages of a multipoint over a point-to-point configuration (type of connection) include ease of installation and low cost. 5. Line configurations (or types of connections) are point-to-point and multipoint. 6. We can divide line configuration in two broad categories: a. Point-to-point: mesh, star, and ring. b. Multipoint: bus 7. In half-duplex transmission, only one entity can send at a time; in a full-duplex transmission, both entities can send at the same time. 8. We give an advantage for each of four network topologies: a. Mesh: secure b. Bus: easy installation c. Star: robust d. Ring: easy fault isolation 9. The number of cables for each type of network is: a. Mesh: n (n – 1) / 2 b. Star: n c. Ring: n – 1 d. Bus: one backbone and n drop lines 10. The general factors are size, distances (covered by the network), structure, and ownership. 2 11. An internet is an interconnection of networks. The Internet is the name of a specific worldwide network 12. A protocol defines what is communicated, in what way and when. This provides accurate and timely transfer of information between different devices on a network. 13. Standards are needed to create and maintain an open and competitive market for manufacturers, to coordinate protocol rules, and thus guarantee compatibility of data communication technologies. Exercises 14. Unicode uses 32 bits to represent a symbol or a character. We can define 232 different symbols or characters. 15. With 16 bits, we can represent up to 216 different colors. 16. a. Cable links: n (n – 1) / 2 = (6 × 5) / 2 = 15 b. Number of ports: (n – 1) = 5 ports needed per device 17. a. Mesh topology: If one connection fails, the other connections will still be working. b. Star topology: The other devices will still be able to send data through the hub; there will be no access to the device which has the failed connection to the hub. c. Bus Topology: All transmission stops if the failure is in the bus. If the drop-line fails, only the corresponding device cannot operate. d. Ring Topology: The failed connection may disable the whole network unless it is a dual ring or there is a by-pass mechanism. 18. This is a LAN. The Ethernet hub creates a LAN as we will see in Chapter 13. 19. Theoretically, in a ring topology, unplugging one station, interrupts the ring. However, most ring networks use a mechanism that bypasses the station; the ring can continue its operation. 20. In a bus topology, no station is in the path of the signal. Unplugging a station has no effect on the operation of the rest of the network. 21. See Figure 1.1 22. See Figure 1.2. 23. a. E-mail is not an interactive application. Even if it is delivered immediately, it may stay in the mail-box of the receiver for a while. It is not sensitive to delay. b. We normally do not expect a file to be copied immediately. It is not very sensitive to delay. c. Surfing the Internet is the an application very sensitive to delay. We except to get access to the site we are searching. 24. In this case, the communication is only between a caller and the callee. A dedicated line is established between them. The connection is point-to-point. 3 25. The telephone network was originally designed for voice communication; the Internet was originally designed for data communication. The two networks are similar in the fact that both are made of interconnections of small networks. The telephone network, as we will see in future chapters, is mostly a circuit-switched network; the Internet is mostly a packet-switched network. Figure 1.1 Solution to Exercise 21 Figure 1.2 Solution to Exercise 22 Station Station Station Repeat er Station Station Station Repeat er Station Station Station Repeater Hub Station Station Station Station Repeater Repeater 1 CHAPTER 2 Network Models Solutions to Review Questions and Exercises Review Questions 1. The Internet model, as discussed in this chapter, include physical, data link, network, transport, and application layers. 2. The network support layers are the physical, data link, and network layers. 3. The application layer supports the user. 4. The transport layer is responsible for process-to-process delivery of the entire message, whereas the network layer oversees host-to-host delivery of individual packets. 5. Peer-to-peer processes are processes on two or more devices communicating at a same layer 6. Each layer calls upon the services of the layer just below it using interfaces between each pair of adjacent layers. 7. Headers and trailers are control data added at the beginning and the end of each data unit at each layer of the sender and removed at the corresponding layers of the receiver. They provide source and destination addresses, synchronization points, information for error detection, etc. 8. The physical layer is responsible for transmitting a bit stream over a physical medium. It is concerned with a. physical characteristics of the media b. representation of bits c. type of encoding d. synchronization of bits e. transmission rate and mode f. the way devices are connected with each other and to the links 9. The data link layer is responsible for a. framing data bits b. providing the physical addresses of the sender/receiver c. data rate control 2 d. detection and correction of damaged and lost frames 10. The network layer is concerned with delivery of a packet across multiple networks; therefore its responsibilities include a. providing host-to-host addressing b. routing 11. The transport layer oversees the process-to-process delivery of the entire message. It is responsible for a. dividing the message into manageable segments b. reassembling it at the destination c. flow and error control 12. The physical address is the local address of a node; it is used by the data link layer to deliver data from one node to another within the same network. The logical address defines the sender and receiver at the network layer and is used to deliver messages across multiple networks. The port address (service-point) identifies the application process on the station. 13. The application layer services include file transfer, remote access, shared database management, and mail services. 14. The application, presentation, and session layers of the OSI model are represented by the application layer in the Internet model. The lowest four layers of OSI correspond to the Internet model layers. Exercises 15. The International Standards Organization, or the International Organization of Standards, (ISO) is a multinational body dedicated to worldwide agreement on international standards. An ISO standard that covers all aspects of network communications is the Open Systems Interconnection (OSI) model. 16. a. Route determination: network layer b. Flow control: data link and transport layers c. Interface to transmission media: physical layer d. Access for the end user: application layer 17. a. Reliable process-to-process delivery: transport layer b. Route selection: network layer c. Defining frames: data link layer d. Providing user services: application layer e. Transmission of bits across the medium: physical layer 18. a. Communication with user’s application program: application layer b. Error correction and retransmission: data link and transport layers c. Mechanical, electrical, and functional interface: physical layer 3 d. Responsibility for carrying frames between adjacent nodes: data link layer 19. a. Format and code conversion services: presentation layer b. Establishing, managing, and terminating sessions: session layer c. Ensuring reliable transmission of data: data link and transport layers d. Log-in and log-out procedures: session layer e. Providing independence from different data representation: presentation layer 20. See Figure 2.1. 21. See Figure 2.2. 22. If the corrupted destination address does not match any station address in the network, the packet is lost. If the corrupted destination address matches one of the stations, the frame is delivered to the wrong station. In this case, however, the error detection mechanism, available in most data link protocols, will find the error and discard the frame. In both cases, the source will somehow be informed using one of the data link control mechanisms discussed in Chapter 11. 23. Before using the destination address in an intermediate or the destination node, the packet goes through error checking that may help the node find the corruption (with a high probability) and discard the packet. Normally the upper layer protocol will inform the source to resend the packet. Figure 2.1 Solution to Exercise 20 Figure 2.2 Solution to Exercise 21 B/42 C/82 A/40 Sender Sender LAN1 LAN2 R1 D/80 42 40 A D Data T2 80 82 A D Data T2 B/42 C/82 A/40 Sender Sender LAN1 LAN2 R1 D/80 42 40 A D i j Data T2 80 82 A D i j Data T2 4 24. Most protocols issue a special error message that is sent back to the source in this case. 25. The errors between the nodes can be detected by the data link layer control, but the error at the node (between input port and output port) of the node cannot be detected by the data link layer. 1 CHAPTER 3 Data and Signals Solutions to Review Questions and Exercises Review Questions 1. Frequency and period are the inverse of each other. T = 1/ f and f = 1/T. 2. The amplitude of a signal measures the value of the signal at any point. The frequency of a signal refers to the number of periods in one second. The phase describes the position of the waveform relative to time zero. 3. Using Fourier analysis. Fourier series gives the frequency domain of a periodic signal; Fourier analysis gives the frequency domain of a nonperiodic signal. 4. Three types of transmission impairment are attenuation, distortion, and noise. 5. Baseband transmission means sending a digital or an analog signal without modulation using a low-pass channel. Broadband transmission means modulating a digital or an analog signal using a band-pass channel. 6. A low-pass channel has a bandwidth starting from zero; a band-pass channel has a bandwidth that does not start from zero. 7. The Nyquist theorem defines the maximum bit rate of a noiseless channel. 8. The Shannon capacity determines the theoretical maximum bit rate of a noisy channel. 9. Optical signals have very high frequencies. A high frequency means a short wave length because the wave length is inversely proportional to the frequency (λ = v/f), where v is the propagation speed in the media. 10. A signal is periodic if its frequency domain plot is discrete; a signal is nonperiodic if its frequency domain plot is continuous. 11. The frequency domain of a voice signal is normally continuous because voice is a nonperiodic signal. 12. An alarm system is normally periodic. Its frequency domain plot is therefore discrete. 13. This is baseband transmission because no modulation is involved. 14. This is baseband transmission because no modulation is involved. 15. This is broadband transmission because it involves modulation. 2 Exercises 16. a. T = 1 / f = 1 / (24 Hz) = 0.0417 s = 41.7 × 10–3 s = 41.7 ms b. T = 1 / f = 1 / (8 MHz) = 0. = 0.125 × 10–6 s = 0.125 μs c. T = 1 / f = 1 / (140 KHz) = 0. s = 7.14 × 10–6 s = 7.14 μs 17. a. f = 1 / T = 1 / (5 s) = 0.2 Hz b. f = 1 / T = 1 / (12 μs) =83333 Hz = 83.333 × 103 Hz = 83.333 KHz c. f = 1 / T = 1 / (220 ns) = Hz = 4.55× 106 Hz = 4.55 MHz 18. a. 90 degrees (π/2 radian) b. 0 degrees (0 radian) c. 90 degrees (π/2 radian) 19. See Figure 3.1 20. We know the lowest frequency, 100. We know the bandwidth is 2000. The highest frequency must be 100 + 2000 = 2100 Hz. See Figure 3.2 21. Each signal is a simple signal in this case. The bandwidth of a simple signal is zero. So the bandwidth of both signals are the same. 22. a. bit rate = 1/ (bit duration) = 1 / (0.001 s) = 1000 bps = 1 Kbps b. bit rate = 1/ (bit duration) = 1 / (2 ms) = 500 bps Figure 3.1 Solution to Exercise 19 Figure 3.2 Solution to Exercise 20 Frequency domain Bandwidth = 200 − 0 = 200 100 20 5 2100 Frequency domain