Behrouz A. Forouzan - Data Communications and Networking Solution Manual PDF

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 s...

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 data- bases, 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 connec- tion) 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. 1 2 11. An internet is an interconnection of networks. The Internet is the name of a spe- cific 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 net- work. 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 differ- ent 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 work- ing. 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. How- ever, 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 sensi- tive 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 dedi- cated line is established between them. The connection is point-to-point.  3 Figure 1.1 Solution to Exercise 21 Hub Station Station Station Station Repeater Repeat er Station Station Station Station Repeat er Station Figure 1.2 Solution to Exercise 22 Station Station Repeater Repeater Station Station 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. 4 Sol-02.fm Page 1 Saturday, January 21, 2006 10:27 AM 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, net- work, 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 1 Sol-02.fm Page 2 Saturday, January 21, 2006 10:27 AM 2 d. detection and correction of damaged and lost frames 10. The network layer is concerned with delivery of a packet across multiple net- works; 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 data- base 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 corre- spond 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 com- munications 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 Sol-02.fm Page 3 Saturday, January 21, 2006 10:27 AM 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. Figure 2.1 Solution to Exercise 20 LAN1 LAN2 A/40 R1 Sender B/42 C/82 D/80 Sender 42 40 A D Data T2 80 82 A D Data T2 21. See Figure 2.2. Figure 2.2 Solution to Exercise 21 LAN1 LAN2 A/40 R1 Sender B/42 C/82 D/80 Sender 42 40 A D i j Data T2 80 82 A D i j Data T2 22. If the corrupted destination address does not match any station address in the net- work, the packet is lost. If the corrupted destination address matches one of the sta- tions, 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. Sol-02.fm Page 4 Saturday, January 21, 2006 10:27 AM 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. 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 fre- quency 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 modu- lation 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 nonperi- odic 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 dis- crete. 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. 1 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) = 0.000000125 = 0.125 × 10–6 s = 0.125 μs c. T = 1 / f = 1 / (140 KHz) = 0.00000714 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) = 4550000 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 Figure 3.1 Solution to Exercise 19 Frequency domain 0 20 50 100 200 Bandwidth = 200 − 0 = 200 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 Figure 3.2 Solution to Exercise 20 20 Frequency domain 5 100 2100 Bandwidth = 2100 − 100 = 2000 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  3 c. bit rate = 1/(bit duration) = 1 / (20 μs/10) = 1 / (2 μs) = 500 Kbps 23. a. (10 / 1000) s = 0.01 s b. (8 / 1000) s = 0. 008 s = 8 ms c. ((100,000 × 8) / 1000) s = 800 s 24. There are 8 bits in 16 ns. Bit rate is 8 / (16 × 10−9) = 0.5 × 10−9 = 500 Mbps 25. The signal makes 8 cycles in 4 ms. The frequency is 8 /(4 ms) = 2 KHz 26. The bandwidth is 5 × 5 = 25 Hz. 27. The signal is periodic, so the frequency domain is made of discrete frequencies. as shown in Figure 3.3. Figure 3.3 Solution to Exercise 27 Amplitude 10 volts... Frequency 10 30 KHz KHz 28. The signal is nonperiodic, so the frequency domain is made of a continuous spec- trum of frequencies as shown in Figure 3.4. Figure 3.4 Solution to Exercise 28 30 volts Amplitude 10 volts 10 volts Frequency 10 20 30 KHz KHz KHz 29. Using the first harmonic, data rate = 2 × 6 MHz = 12 Mbps Using three harmonics, data rate = (2 × 6 MHz) /3 = 4 Mbps Using five harmonics, data rate = (2 × 6 MHz) /5 = 2.4 Mbps 30. dB = 10 log10 (90 / 100) = –0.46 dB 31. –10 = 10 log10 (P2 / 5) → log10 (P2 / 5) = −1 → (P2 / 5) = 10−1 → P2 = 0.5 W 32. The total gain is 3 × 4 = 12 dB. The signal is amplified by a factor 101.2 = 15.85. 4 33. 100,000 bits / 5 Kbps = 20 s 34. 480 s × 300,000 km/s = 144,000,000 km 35. 1 μm × 1000 = 1000 μm = 1 mm 36. We have 4,000 log2 (1 + 1,000) ≈ 40 Kbps 37. We have 4,000 log2 (1 + 10 / 0.005) = 43,866 bps 38. The file contains 2,000,000 × 8 = 16,000,000 bits. With a 56-Kbps channel, it takes 16,000,000/56,000 = 289 s. With a 1-Mbps channel, it takes 16 s. 39. To represent 1024 colors, we need log21024 = 10 (see Appendix C) bits. The total number of bits are, therefore, 1200 × 1000 × 10 = 12,000,000 bits 40. We have SNR = (200 mW) / (10 × 2 × μW) = 10,000 We then have SNRdB = 10 log10 SNR = 40 41. We have SNR= (signal power)/(noise power). However, power is proportional to the square of voltage. This means we have SNR = [(signal voltage)2] / [(noise voltage)2] = [(signal voltage) / (noise voltage)]2 = 202 = 400 We then have SNRdB = 10 log10 SNR ≈ 26.02 42. We can approximately calculate the capacity as a. C = B × (SNRdB /3) = 20 KHz × (40 /3) = 267 Kbps b. C = B × (SNRdB /3) = 200 KHz × (4 /3) = 267 Kbps c. C = B × (SNRdB /3) = 1 MHz × (20 /3) = 6.67 Mbps 43. a. The data rate is doubled (C2 = 2 × C1). b. When the SNR is doubled, the data rate increases slightly. We can say that, approximately, (C2 = C1 + 1). 44. We can use the approximate formula C = B × (SNRdB /3) or SNRdB = (3 × C) /B We can say that the minimum SNRdB = 3 × 100 Kbps / 4 KHz = 75  5 This means that the minimum SNR = 10 SNRdB/10 = 107.5 ≈ 31,622,776 45. We have transmission time = (packet length)/(bandwidth) = (8,000,000 bits) / (200,000 bps) = 40 s 46. We have (bit length) = (propagation speed) × (bit duration) The bit duration is the inverse of the bandwidth. a. Bit length = (2 ×108 m) × [(1 / (1 Mbps)] = 200 m. This means a bit occupies 200 meters on a transmission medium. b. Bit length = (2 ×108 m) × [(1 / (10 Mbps)] = 20 m. This means a bit occupies 20 meters on a transmission medium. c. Bit length = (2 ×108 m) × [(1 / (100 Mbps)] = 2 m. This means a bit occupies 2 meters on a transmission medium. 47. a. Number of bits = bandwidth × delay = 1 Mbps × 2 ms = 2000 bits b. Number of bits = bandwidth × delay = 10 Mbps × 2 ms = 20,000 bits c. Number of bits = bandwidth × delay = 100 Mbps × 2 ms = 200,000 bits 48. We have Latency = processing time + queuing time + transmission time + propagation time Processing time = 10 × 1 μs = 10 μs = 0.000010 s Queuing time = 10 × 2 μs = 20 μs = 0.000020 s Transmission time = 5,000,000 / (5 Mbps) = 1 s Propagation time = (2000 Km) / (2 × 108) = 0.01 s Latency = 0.000010 + 0.000020 + 1 + 0.01 = 1.01000030 s The transmission time is dominant here because the packet size is huge. 6 CHAPTER 4 Digital Transmission Solutions to Review Questions and Exercises Review Questions 1. The three different techniques described in this chapter are line coding, block cod- ing, and scrambling. 2. A data element is the smallest entity that can represent a piece of information (a bit). A signal element is the shortest unit of a digital signal. Data elements are what we need to send; signal elements are what we can send. Data elements are being carried; signal elements are the carriers. 3. The data rate defines the number of data elements (bits) sent in 1s. The unit is bits per second (bps). The signal rate is the number of signal elements sent in 1s. The unit is the baud. 4. In decoding a digital signal, the incoming signal power is evaluated against the baseline (a running average of the received signal power). A long string of 0s or 1s can cause baseline wandering (a drift in the baseline) and make it difficult for the receiver to decode correctly. 5. When the voltage level in a digital signal is constant for a while, the spectrum cre- ates very low frequencies, called DC components, that present problems for a sys- tem that cannot pass low frequencies. 6. A self-synchronizing digital signal includes timing information in the data being transmitted. This can be achieved if there are transitions in the signal that alert the receiver to the beginning, middle, or end of the pulse. 7. In this chapter, we introduced unipolar, polar, bipolar, multilevel, and multitran- sition coding. 8. Block coding provides redundancy to ensure synchronization and to provide inher- ent error detecting. In general, block coding changes a block of m bits into a block of n bits, where n is larger than m. 9. Scrambling, as discussed in this chapter, is a technique that substitutes long zero- level pulses with a combination of other levels without increasing the number of bits. 1 2 10. Both PCM and DM use sampling to convert an analog signal to a digital signal. PCM finds the value of the signal amplitude for each sample; DM finds the change between two consecutive samples. 11. In parallel transmission we send data several bits at a time. In serial transmission we send data one bit at a time. 12. We mentioned synchronous, asynchronous, and isochronous. In both synchro- nous and asynchronous transmissions, a bit stream is divided into independent frames. In synchronous transmission, the bytes inside each frame are synchro- nized; in asynchronous transmission, the bytes inside each frame are also indepen- dent. In isochronous transmission, there is no independency at all. All bits in the whole stream must be synchronized. Exercises 13. We use the formula s = c × N × (1/r) for each case. We let c = 1/2. a. r = 1 → s = (1/2) × (1 Mbps) × 1/1 = 500 kbaud b. r = 1/2 → s = (1/2) × (1 Mbps) × 1/(1/2) = 1 Mbaud c. r = 2 → s = (1/2) × (1 Mbps) × 1/2 = 250 Kbaud d. r = 4/3 → s = (1/2) × (1 Mbps) × 1/(4/3) = 375 Kbaud 14. The number of bits is calculated as (0.2 /100) × (1 Mbps) = 2000 bits 15. See Figure 4.1. Bandwidth is proportional to (3/8)N which is within the range in Table 4.1 (B = 0 to N) for the NRZ-L scheme. Figure 4.1 Solution to Exercise 15 Average Number of Changes = (0 + 0 + 8 + 4) / 4 = 3 for N = 8 B (3 / 8) N Case a Case c 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1 Case b Case d 16. See Figure 4.2. Bandwidth is proportional to (4.25/8)N which is within the range in Table 4.1 (B = 0 to N) for the NRZ-I scheme. 17. See Figure 4.3. Bandwidth is proportional to (12.5 / 8) N which is within the range in Table 4.1 (B = N to B = 2N) for the Manchester scheme. 18. See Figure 4.4. B is proportional to (12/8) N which is within the range in Table 4.1 (B = N to 2N) for the differential Manchester scheme.  3 Figure 4.2 Solution to Exercise 16 Average Number of Changes = (0 + 9 + 4 + 4) / 4 = 4.25 for N = 8 B (4.25 / 8) N Case a Case c 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1 Case b Case d Figure 4.3 Solution to Exercise 17 Average Number of Changes = (15 + 15+ 8 + 12) / 4 = 12.5 for N = 8 B (12.5 / 8) N Case a Case c 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1 Case b Case d Figure 4.4 Solution to Exercise 18 Average Number of Changes = (16 + 8 + 12 + 12) / 4 = 12 for N = 8 B (12 / 8) N Case a Case c 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1 Case b Case d 4 19. See Figure 4.5. B is proportional to (5.25 / 16) N which is inside range in Table 4.1 (B = 0 to N/2) for 2B/1Q. Figure 4.5 Solution to Exercise 19 Average Number of Changes = (0 + 7 + 7 + 7) / 4 = 5.25 for N = 16 B (5.25 / 8) N Case a Case c 00 00 00 00 00 00 00 00 01 10 01 10 01 10 01 10 +3 +3 +1 +1 −1 −1 −3 −3 11 11 11 11 11 11 11 11 00 11 00 11 00 11 00 11 +3 +3 +1 +1 −1 −1 −3 −3 Case b Case d 20. See Figure 4.6. B is proportional to (5.25/8) × N which is inside the range in Table 4.1 (B = 0 to N/2) for MLT-3. Figure 4.6 Solution to Exercise 20 Average Number of Changes = (0 + 7 + 4 + 3) / 4 = 4.5 for N = 8 B (4.5 / 8) N Case a Case c 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 +V +V −V −V 1 1 1 1 1 1 1 1 0 0 0 1 1 0 0 0 +V +V −V −V Case b Case d 21. The data stream can be found as a. NRZ-I: 10011001. b. Differential Manchester: 11000100. c. AMI: 01110001. 22. The data rate is 100 Kbps. For each case, we first need to calculate the value f / N. We then use Figure 4.6 in the text to find P (energy per Hz). All calculations are approximations.  5 a. f /N = 0/100 =0 → P = 1.0 b. f /N = 50/100 = 1/2 → P = 0.5 c. f /N = 100/100 = 1 → P = 0.0 d. f /N = 150/100 = 1.5 → P = 0.2 23. The data rate is 100 Kbps. For each case, we first need to calculate the value f/N. We then use Figure 4.8 in the text to find P (energy per Hz). All calculations are approximations. a. f /N = 0/100 =0 → P = 0.0 b. f /N = 50/100 = 1/2 → P = 0.3 c. f /N = 100/100 = 1 → P = 0.4 d. f /N = 150/100 = 1.5 → P = 0.0 24. a. The output stream is 01010 11110 11110 11110 11110 01001. b. The maximum length of consecutive 0s in the input stream is 21. c. The maximum length of consecutive 0s in the output stream is 2. 25. In 5B/6B, we have 25 = 32 data sequences and 26 = 64 code sequences. The number of unused code sequences is 64 − 32 = 32. In 3B/4B, we have 2 3 = 8 data sequences and 24 = 16 code sequences. The number of unused code sequences is 16 − 8 = 8. 26. See Figure 4.7. Since we specified that the last non-zero signal is positive, the first bit in our sequence is positive. Figure 4.7 Solution to Exercise 26 a. B8ZS 1 1 1 0 0 0 0 0 0 0 0 0 0 0 B V V B 1 1 1 0 0 0 0 0 0 0 0 0 0 0 B V V b. HDB3 27. a. In a low-pass signal, the minimum frequency 0. Therefore, we have fmax = 0 + 200 = 200 KHz. → fs = 2 × 200,000 = 400,000 samples/s 6 b. In a bandpass signal, the maximum frequency is equal to the minimum fre- quency plus the bandwidth. Therefore, we have fmax = 100 + 200 = 300 KHz. → fs = 2 × 300,000 = 600,000 samples /s 28. a. In a lowpass signal, the minimum frequency is 0. Therefore, we can say fmax = 0 + 200 = 200 KHz → fs = 2 × 200,000 = 400,000 samples/s The number of bits per sample and the bit rate are nb = log21024 = 10 bits/sample N = 400 KHz × 10 = 4 Mbps b. The value of nb = 10. We can easily calculate the value of SNRdB SNRdB = 6.02 × nb + 1.76 = 61.96 c. The value of nb = 10. The minimum bandwidth can be calculated as BPCM = nb × Banalog = 10 × 200 KHz = 2 MHz 29. The maximum data rate can be calculated as Nmax = 2 × B × nb = 2 × 200 KHz × log24 = 800 kbps 30. We can first calculate the sampling rate (fs) and the number of bits per sample (nb) fmax = 0 + 4 = 4 KHz → fs = 2 × 4 = 8000 sample/s We then calculate the number of bits per sample. → nb = 30000 / 8000 = 3.75 We need to use the next integer nb = 4. The value of SNRdB is SNRdB = 6.02 × nb + 1.72 = 25.8 31. We can calculate the data rate for each scheme: a. NRZ → N=2 ×B=2 × 1 MHz = 2 Mbps b. Manchester → N=1 ×B=1 × 1 MHz = 1 Mbps c. MLT-3 → N=3 ×B=3 × 1 MHz = 3 Mbps d. 2B1Q → N=4 ×B=4 × 1 MHz = 4 Mbps 32. a. For synchronous transmission, we have 1000 × 8 = 8000 bits. b. For asynchronous transmission, we have 1000 × 10 = 10000 bits. Note that we assume only one stop bit and one start bit. Some systems send more start bits. c. For case a, the redundancy is 0%. For case b, we send 2000 extra for 8000 required bits. The redundancy is 25%. CHAPTER 5 Analog Transmission Solutions to Review Questions and Exercises Review Questions 1. Normally, analog transmission refers to the transmission of analog signals using a band-pass channel. Baseband digital or analog signals are converted to a complex analog signal with a range of frequencies suitable for the channel. 2. A carrier is a single-frequency signal that has one of its characteristics (amplitude, frequency, or phase) changed to represent the baseband signal. 3. The process of changing one of the characteristics of an analog signal based on the information in digital data is called digital-to-analog conversion. It is also called modulation of a digital signal. The baseband digital signal representing the digital data modulates the carrier to create a broadband analog signal. 4. a. ASK changes the amplitude of the carrier. b. FSK changes the frequency of the carrier. c. PSK changes the phase of the carrier. d. QAM changes both the amplitude and the phase of the carrier. 5. We can say that the most susceptible technique is ASK because the amplitude is more affected by noise than the phase or frequency. 6. A constellation diagram can help us define the amplitude and phase of a signal element, particularly when we are using two carriers. The diagram is useful when we are dealing with multilevel ASK, PSK, or QAM. In a constellation diagram, a signal element type is represented as a dot. The bit or combination of bits it can carry is often written next to it.The diagram has two axes. The horizontal X axis is related to the in-phase carrier; the vertical Y axis is related to the quadrature carrier. 7. The two components of a signal are called I and Q. The I component, called in- phase, is shown on the horizontal axis; the Q component, called quadrature, is shown on the vertical axis. 8. The process of changing one of the characteristics of an analog signal to represent the instantaneous amplitude of a baseband signal is called analog-to-analog con- 1 2 version. It is also called the modulation of an analog signal; the baseband analog signal modulates the carrier to create a broadband analog signal. 9. a. AM changes the amplitude of the carrier b. FM changes the frequency of the carrier c. PM changes the phase of the carrier 10. We can say that the most susceptible technique is AM because the amplitude is more affected by noise than the phase or frequency. Exercises 11. We use the formula S = (1/r) × N, but first we need to calculate the value of r for each case. a. r = log22 =1 → S = (1/1) × (2000 bps) = 2000 baud b. r = log22 =1 → S = (1/1) × (4000 bps) = 4000 baud c. r = log24 =2 → S = (1/2) × (6000 bps) = 3000 baud d. r = log264 =6 → S = (1/6) × (36,000 bps) = 6000 baud 12. We use the formula N = r × S, but first we need to calculate the value of r for each case. a. r = log22 =1 → N = (1) × (1000 bps) = 1000 bps b. r = log22 =1 → N = (1) × (1000 bps) = 1000 bps c. r = log22 =1 → N = (1) × (1000 bps) = 1000 bps d. r = log216 =4 → N = (4) × (1000 bps) = 4000 bps 13. We use the formula r = log2L to calculate the value of r for each case. a. log24 =2 b. log28 =3 c. log24 =2 d. log2128 =7 14. See Figure 5.1. a. We have two signal elements with peak amplitudes 1 and 3. The phase of both signal elements are the same, which we assume to be 0 degrees. b. We have two signal elements with the same peak amplitude of 2. However, there must be 180 degrees difference between the two phases. We assume one phase to be 0 and the other 180 degrees. c. We have four signal elements with the same peak amplitude of 3. However, there must be 90 degrees difference between each phase. We assume the first phase to be at 45, the second at 135, the third at 225, and the fourth at 315 degrees. Note that this is one out of many configurations. The phases can be at  3 Figure 5.1 Solution to Exercise 14 a. ASK b. BPSK Q Q I I 1 3 –2 2 Q Q 3 3 3 3 1 1 I I 3 3 1 1 3 3 c. QPSK d. 8-QAM 0, 90, 180, and 270. As long as the differences are 90 degrees, the solution is satisfactory. d. We have four phases, which we select to be the same as the previous case. For each phase, however, we have two amplitudes, 1 and 3 as shown in the figure. Note that this is one out of many configurations. The phases can be at 0, 90, 180, and 270. As long as the differences are 90 degrees, the solution is satisfac- tory. 15. See Figure 5.2 Figure 5.2 Solution to Exercise 15 a. b. Q Q I I 2 3 –3 3 Q Q 2 2 I I –2 2 –2 –2 c. d. a. This is ASK. There are two peak amplitudes both with the same phase (0 degrees). The values of the peak amplitudes are A1 = 2 (the distance between 4 the first dot and the origin) and A2= 3 (the distance between the second dot and the origin). b. This is BPSK, There is only one peak amplitude (3). The distance between each dot and the origin is 3. However, we have two phases, 0 and 180 degrees. c. This can be either QPSK (one amplitude, four phases) or 4-QAM (one ampli- tude and four phases). The amplitude is the distance between a point and the origin, which is (22 + 22)1/2 = 2.83. d. This is also BPSK. The peak amplitude is 2, but this time the phases are 90 and 270 degrees. 16. The number of points define the number of levels, L. The number of bits per baud is the value of r. Therefore, we use the formula r = log2L for each case. a. log22 =1 b. log24 =2 c. log216 =4 d. log21024 = 10 17. We use the formula B = (1 + d) × (1/r) × N, but first we need to calculate the value of r for each case. a. r = 1 → B= (1 + 1) × (1/1) × (4000 bps) = 8000 Hz b. r = 1 → B = (1 + 1) × (1/1) × (4000 bps) + 4 KHz = 8000 Hz c. r = 2 → B = (1 + 1) × (1/2) × (4000 bps) = 2000 Hz d. r = 4 → B = (1 + 1) × (1/4) × (4000 bps) = 1000 Hz 18. We use the formula N = [1/(1 + d)] × r × B, but first we need to calculate the value of r for each case. a. r = log22 = 1 → N= [1/(1 + 0)] × 1 × (4 KHz) = 4 kbps b. r = log24=2 → N = [1/(1 + 0)] × 2 × (4 KHz) = 8 kbps c. r = log216= 4 → N = [1/(1 + 0)] × 4 × (4 KHz) = 16 kbps d. r = log264= 6 → N = [1/(1 + 0)] × 6 × (4 KHz) = 24 kbps 19. First, we calculate the bandwidth for each channel = (1 MHz) / 10 = 100 KHz. We then find the value of r for each channel: B = (1 + d) × (1/r) × (N) → r = N / B → r = (1 Mbps/100 KHz) = 10 We can then calculate the number of levels: L = 2r = 210 = 1024. This means that that we need a 1024-QAM technique to achieve this data rate. 20. We can use the formula: N = [1/(1 + d)] × r × B = 1 × 6 × 6 MHz = 36 Mbps 21. a. BAM = 2 × B = 2 × 5 = 10 KHz  5 b. BFM = 2 × (1 + β) × B = 2 × (1 + 5) × 5 = 60 KHz c. BPM = 2 × (1 + β) × B = 2 × (1 + 1) × 5 = 20 KHz 22. We calculate the number of channels, not the number of coexisting stations. a. n = (1700 - 530) KHz / 10 KHz = 117 b. n = (108 - 88) MHz / 200 KHz = 100 6 CHAPTER 6 Bandwidth Utilization: Solutions to Review Questions and Exercises Review Questions 1. Multiplexing is the set of techniques that allows the simultaneous transmission of multiple signals across a single data link. 2. We discussed frequency-division multiplexing (FDM), wave-division multiplex- ing (WDM), and time-division multiplexing (TDM). 3. In multiplexing, the word link refers to the physical path. The word channel refers to the portion of a link that carries a transmission between a given pair of lines. One link can have many (n) channels. 4. FDM and WDM are used to combine analog signals; the bandwidth is shared. TDM is used to combine digital signals; the time is shared. 5. To maximize the efficiency of their infrastructure, telephone companies have tradi- tionally multiplexed analog signals from lower-bandwidth lines onto higher-band- width lines. The analog hierarchy uses voice channels (4 KHz), groups (48 KHz), supergroups (240 KHz), master groups (2.4 MHz), and jumbo groups (15.12 MHz). 6. To maximize the efficiency of their infrastructure, telephone companies have tradi- tionally multiplexed digital signals from lower data rate lines onto higher data rate lines. The digital hierarchy uses DS-0 (64 Kbps), DS-1 (1.544 Mbps), DS-2 (6.312 Mbps), DS-3 (44.376 Mbps), and DS-4 (274.176 Mbps). 7. WDM is common for multiplexing optical signals because it allows the multiplex- ing of signals with a very high frequency. 8. In multilevel TDM, some lower-rate lines are combined to make a new line with the same data rate as the other lines. Multiple slot TDM, on the other hand, uses multiple slots for higher data rate lines to make them compatible with the lower data rate line. Pulse stuffing TDM is used when the data rates of some lines are not an integral multiple of other lines. 9. In synchronous TDM, each input has a reserved slot in the output frame. This can be inefficient if some input lines have no data to send. In statistical TDM, slots are 1 2 dynamically allocated to improve bandwidth efficiency. Only when an input line has a slot’s worth of data to send is it given a slot in the output frame. 10. In spread spectrum, we spread the bandwidth of a signal into a larger bandwidth. Spread spectrum techniques add redundancy; they spread the original spectrum needed for each station. The expanded bandwidth allows the source to wrap its message in a protective envelope for a more secure transmission. We discussed frequency hopping spread spectrum (FHSS) and direct sequence spread spectrum (DSSS). 11. The frequency hopping spread spectrum (FHSS) technique uses M different car- rier frequencies that are modulated by the source signal. At one moment, the signal modulates one carrier frequency; at the next moment, the signal modulates another carrier frequency. 12. The direct sequence spread spectrum (DSSS) technique expands the bandwidth of the original signal. It replaces each data bit with n bits using a spreading code. Exercises 13. To multiplex 10 voice channels, we need nine guard bands. The required band- width is then B = (4 KHz) × 10 + (500 Hz) × 9 = 44.5 KHz 14. The bandwidth allocated to each voice channel is 20 KHz / 100 = 200 Hz. As we saw in the previous chapters, each digitized voice channel has a data rate of 64 Kbps (8000 sample × 8 bit/sample). This means that our modulation technique uses 64,000/200 = 320 bits/Hz. 15. a. Group level: overhead = 48 KHz − (12 × 4 KHz) = 0 Hz. b. Supergroup level: overhead = 240 KHz − (5 × 48 KHz) = 0 Hz. c. Master group: overhead = 2520 KHz − (10 × 240 KHz) = 120 KHz. d. Jumbo Group: overhead = 16.984 MHz − (6 × 2.52 MHz) = 1.864 MHz. 16. a. Each output frame carries 1 bit from each source plus one extra bit for synchro- nization. Frame size = 20 × 1 + 1 = 21 bits. b. Each frame carries 1 bit from each source. Frame rate = 100,000 frames/s. c. Frame duration = 1 /(frame rate) = 1 /100,000 = 10 μs. d. Data rate = (100,000 frames/s) × (21 bits/frame) = 2.1 Mbps e. In each frame 20 bits out of 21 are useful. Efficiency = 20/21= 95% 17. a. Each output frame carries 2 bits from each source plus one extra bit for syn- chronization. Frame size = 20 × 2 + 1 = 41 bits. b. Each frame carries 2 bit from each source. Frame rate = 100,000/2 = 50,000 frames/s. c. Frame duration = 1 /(frame rate) = 1 /50,000 = 20 μs. d. Data rate = (50,000 frames/s) × (41 bits/frame) = 2.05 Mbps. The output data rate here is slightly less than the one in Exercise 16.  3 e. In each frame 40 bits out of 41 are useful. Efficiency = 40/41= 97.5%. Effi- ciency is better than the one in Exercise 16. 18. a. Frame size = 6 × (8 + 4) = 72 bits. b. We can assume that we have only 6 input lines. Each frame needs to carry one character from each of these lines. This means that the frame rate is 500 frames/s. c. Frame duration = 1 /(frame rate) = 1 /500 = 2 ms. d. Data rate = (500 frames/s) × (72 bits/frame) = 36 kbps. 19. We combine six 200-kbps sources into three 400-kbps. Now we have seven 400- kbps channel. a. Each output frame carries 1 bit from each of the seven 400-kbps line. Frame size = 7 × 1 = 7 bits. b. Each frame carries 1 bit from each 400-kbps source. Frame rate = 400,000 frames/s. c. Frame duration = 1 /(frame rate) = 1 /400,000 = 2.5 μs. d. Output data rate = (400,000 frames/s) × (7 bits/frame) = 2.8 Mbps. We can also calculate the output data rate as the sum of input data rate because there is no synchronizing bits. Output data rate = 6 × 200 + 4 × 400 = 2.8 Mbps. 20. a. The frame carries 4 bits from each of the first two sources and 3 bits from each of the second two sources. Frame size = 4 × 2 + 3 × 2 = 14 bits. b. Each frame carries 4 bit from each 200-kbps source or 3 bits from each 150 kbps. Frame rate = 200,000 / 4 = 150,000 /3 = 50,000 frames/s. c. Frame duration = 1 /(frame rate) = 1 /50,000 = 20 μs. d. Output data rate = (50,000 frames/s) × (14 bits/frame) = 700 kbps. We can also calculate the output data rate as the sum of input data rates because there are no synchronization bits. Output data rate = 2 × 200 + 2 × 150 = 700 kbps. 21. We need to add extra bits to the second source to make both rates = 190 kbps. Now we have two sources, each of 190 Kbps. a. The frame carries 1 bit from each source. Frame size = 1 + 1 = 2 bits. b. Each frame carries 1 bit from each 190-kbps source. Frame rate = 190,000 frames/s. c. Frame duration = 1 /(frame rate) = 1 /190,000 = 5.3 μs. d. Output data rate = (190,000 frames/s) × (2 bits/frame) = 380 kbps. Here the output bit rate is greater than the sum of the input rates (370 kbps) because of extra bits added to the second source. 22. a. T-1 line sends 8000 frames/s. Frame duration = 1/8000 = 125 μs. b. Each frame carries one extra bit. Overhead = 8000 × 1 = 8 kbps 23. See Figure 6.1. 24. See Figure 6.2. 4 Figure 6.1 Solution to Exercise 23 O L E L Y I E B H H TDM Figure 6.2 Solution to Exercise 24 1 1 0 010 11 101 1 000 111 110 0 001 110 111 1 111 000 101 TDM 25. See Figure 6.3. Figure 6.3 Solution to Exercise 25 000000011000 101010100111 TDM 10100000 10100111 26. a. DS-1 overhead = 1.544 Mbps − (24 × 64 kbps) = 8 kbps. b. DS-2 overhead = 6.312 Mbps − (4 × 1.544 Mbps) = 136 kbps. c. DS-3 overhead = 44.376 Mbps − (7 × 6.312 Mbps) = 192 kbps. d. DS-4 overhead = 274.176 Mbps − (6 × 44.376 Mbps) = 7.92 Mbps. 27. The number of hops = 100 KHz/4 KHz = 25. So we need log225 = 4.64 ≈ 5 bits 28. a. 24 = 16 hops b. (64 bits/s) / 4 bits = 16 cycles 29. Random numbers are 11, 13, 10, 6, 12, 3, 8, 9 as calculated below: N1 = 11 N2 =(5 +7 × 11) mod 17 − 1 = 13 N3 =(5 +7 × 13) mod 17 − 1 = 10 N4 =(5 +7 × 10) mod 17 − 1 = 6  5 N5 =(5 +7 × 6) mod 17 − 1 = 12 N6 =(5 +7 × 12) mod 17 − 1 = 3 N7 =(5 +7 × 3) mod 17 − 1 = 8 N8 =(5 +7 × 8) mod 17 − 1 = 9 30. The Barker chip is 11 bits, which means that it increases the bit rate 11 times. A voice channel of 64 kbps needs 11 × 64 kbps = 704 kbps. This means that the bandpass channel can carry (10 Mbps) / (704 kbps) or approximately 14 channels. 6 CHAPTER 7 Transmission Media Solutions to Review Questions and Exercises Review Questions 1. The transmission media is located beneath the physical layer and controlled by the physical layer. 2. The two major categories are guided and unguided media. 3. Guided media have physical boundaries, while unguided media are unbounded. 4. The three major categories of guided media are twisted-pair, coaxial, and fiber- optic cables. 5. Twisting ensures that both wires are equally, but inversely, affected by external influences such as noise. 6. Refraction and reflection are two phenomena that occur when a beam of light travels into a less dense medium. When the angle of incidence is less than the crit- ical angle, refraction occurs. The beam crosses the interface into the less dense medium. When the angle of incidence is greater than the critical angle, reflection occurs. The beam changes direction at the interface and goes back into the more dense medium. 7. The inner core of an optical fiber is surrounded by cladding. The core is denser than the cladding, so a light beam traveling through the core is reflected at the boundary between the core and the cladding if the incident angle is more than the critical angle. 8. We can mention three advantages of optical fiber cable over twisted-pair and coax- ial cables: noise resistance, less signal attenuation, and higher bandwidth. 9. In sky propagation radio waves radiate upward into the ionosphere and are then reflected back to earth. In line-of-sight propagation signals are transmitted in a straight line from antenna to antenna. 10. Omnidirectional waves are propagated in all directions; unidirectional waves are propagated in one direction. 1 2 Exercises 11. See Table 7.1 (the values are approximate). Table 7.1 Solution to Exercise 11 Distance dB at 1 KHz dB at 10 KHz dB at 100 KHz 1 Km −3 −5 −7 10 Km −30 −50 −70 15 Km −45 −75 −105 20 Km −60 −100 −140 12. As the Table 7.1 shows, for a specific maximum value of attenuation, the highest frequency decreases with distance. If we consider the bandwidth to start from zero, we can say that the bandwidth decreases with distance. For example, if we can tol- erate a maximum attenuation of 50 dB (loss), then we can give the following list- ing of distance versus bandwidth. Distance Bandwidth 1 Km 100 KHz 10 Km 50 KHz 15 Km 1 KHz 20 Km 0 KHz 13. We can use Table 7.1 to find the power for different frequencies: 1 KHz dB = −3 P2 = P1 ×10−3/10 = 100.23 mw 10 KHz dB = −5 P2 = P1 ×10−5/10 = 63.25 mw 100 KHz dB = −7 P2 = P1 ×10−7/10 = 39.90 mw The table shows that the power is reduced 5 times, which may not be acceptable for some applications. 14. See Table 7.2 (the values are approximate). Table 7.2 Solution to Exercise 14 Distance dB at 1 KHz dB at 10 KHz dB at 100 KHz 1 Km −3 −7 −20 10 Km −30 −70 −200 15 Km −45 −105 −300 20 Km −60 −140 −400 15. As Table 7.2 shows, for a specific maximum value of attenuation, the highest fre- quency decreases with distance. If we consider the bandwidth to start from zero, we can say that the bandwidth decreases with distance. For example, if we can tol-  3 erate a maximum attenuation of 50 dB (loss), then we can give the following list- ing of distance versus bandwidth. Distance Bandwidth 1 Km 100 KHz 10 Km 1 KHz 15 Km 1 KHz 20 Km 0 KHz 16. We can use Table 7.2 to find the power for different frequencies: 1 KHz dB = −3 P2 = P1 ×10−3/10 = 100.23 mw 10 KHz dB = −7 P2 = P1 ×10−7/10 = 39.90 mw 100 KHz dB = −20 P2 = P1 ×10−20/10 = 2.00 mw The table shows that power is decreased 100 times for 100 KHz, which is unac- ceptable for most applications. 17. We can use the formula f = c / λ to find the corresponding frequency for each wave length as shown below (c is the speed of propagation): a. B = [(2 × 108)/1000×10−9] − [(2 × 108)/ 1200 × 10−9] = 33 THz b. B = [(2 × 108)/1000×10−9] − [(2 × 108)/ 1400 × 10−9] = 57 THz 18. a. The wave length is the inverse of the frequency if the propagation speed is fixed (based on the formula λ = c / f). This means all three figures represent the same thing. b. We can change the wave length to frequency. For example, the value 1000 nm can be written as 200 THz. c. The vertical-axis units may not change because they represent dB/km. d. The curve must be flipped horizontally. 19. See Table 7.3 (The values are approximate). Table 7.3 Solution to Exercise 19 Distance dB at 800 nm dB at 1000 nm dB at 1200 nm 1 Km −3 −1.1 −0.5 10 Km −30 −11 −5 15 Km −45 −16.5 −7.5 20 Km −60 −22 −10 20. The delay = distance / (propagation speed). Therefore, we have: a. Delay = 10/(2 × 108) = 0.05 ms b. Delay = 100/(2 × 108) = 0.5 ms c. Delay = 1000/(2 × 108) = 5 ms 4 21. See Figure 7.1. Figure 7.1 Solution to Exercise 21 Refraction a. 40 degrees Critical angle = 60 Critical angle Refraction b. 60 degrees Critical angle = 60 Critical angle Reflection c. 80 degrees Critical angle = 60 Critical angle a. The incident angle (40 degrees) is smaller than the critical angle (60 degrees). We have refraction.The light ray enters into the less dense medium. b. The incident angle (60 degrees) is the same as the critical angle (60 degrees). We have refraction. The light ray travels along the interface. c. The incident angle (80 degrees) is greater than the critical angle (60 degrees). We have reflection. The light ray returns back to the more dense medium. CHAPTER 8 Switching Solutions to Review Questions and Exercises Review Questions 1. Switching provides a practical solution to the problem of connecting multiple devices in a network. It is more practical than using a bus topology; it is more effi- cient than using a star topology and a central hub. Switches are devices capable of creating temporary connections between two or more devices linked to the switch. 2. The three traditional switching methods are circuit switching, packet switching, and message switching. The most common today are circuit switching and packet switching. 3. There are two approaches to packet switching: datagram approach and virtual- circuit approach. 4. In a circuit-switched network, data are not packetized; data flow is somehow a continuation of bits that travel the same channel during the data transfer phase. In a packet-switched network data are packetized; each packet is somehow an indepen- dent entity with its local or global addressing information. 5. The address field defines the end-to-end (source to destination) addressing. 6. The address field defines the virtual circuit number (local) addressing. 7. In a space-division switch, the path from one device to another is spatially separate from other paths. The inputs and the outputs are connected using a grid of elec- tronic microswitches. In a time-division switch, the inputs are divided in time using TDM. A control unit sends the input to the correct output device. 8. TSI (time-slot interchange) is the most popular technology in a time-division switch. It used random access memory (RAM) with several memory locations. The RAM fills up with incoming data from time slots in the order received. Slots are then sent out in an order based on the decisions of a control unit. 9. In multistage switching, blocking refers to times when one input cannot be con- nected to an output because there is no path available between them—all the possi- ble intermediate switches are occupied. One solution to blocking is to increase the number of intermediate switches based on the Clos criteria. 1 2 10. A packet switch has four components: input ports, output ports, the routing pro- cessor, and the switching fabric. An input port performs the physical and data link functions of the packet switch. The output port performs the same functions as the input port, but in the reverse order. The routing processor performs the function of table lookup in the network layer. The switching fabric is responsible for moving the packet from the input queue to the output queue. Exercises 11. We assume that the setup phase is a two-way communication and the teardown phase is a one-way communication. These two phases are common for all three cases. The delay for these two phases can be calculated as three propagation delays and three transmission delays or 3 [(5000 km)/ (2 ×108 m/s)]+ 3 [(1000 bits/1 Mbps)] = 75 ms + 3 ms = 78 ms We assume that the data transfer is in one direction; the total delay is then delay for setup and teardown + propagation delay + transmission delay a. 78 + 25 + 1 = 104 ms b. 78 + 25 + 100 = 203 ms c. 78 + 25 + 1000 = 1103 ms d. In case a, we have 104 ms. In case b we have 203/100 = 2.03 ms. In case c, we have 1103/1000 = 1.101 ms. The ratio for case c is the smallest because we use one setup and teardown phase to send more data. 12. We assume that the transmission time is negligible in this case. This means that we suppose all datagrams start at time 0. The arrival timed are calculated as: First: (3200 Km) / (2 × 108 m/s) + (3 + 20 + 20) = 59.0 ms Second: (11700 Km) / (2 × 108 m/s) + (3 + 10 + 20) = 91.5 ms Third: (12200 Km) / (2 × 108 m/s) + (3 + 10+ 20 + 20) = 114.0 ms Fourth: (10200 Km) / (2 × 108 m/s) + (3 + 7 + 20) = 81.0 ms Fifth: (10700 Km) / (2 × 108 m/s) + (3 + 7 + 20 + 20) = 103.5 ms The order of arrival is: 3 → 5 → 2 → 4 → 1 13. a. In a circuit-switched network, end-to-end addressing is needed during the setup and teardown phase to create a connection for the whole data transfer phase. After the connection is made, the data flow travels through the already-reserved resources. The switches remain connected for the entire duration of the data transfer; there is no need for further addressing. b. In a datagram network, each packet is independent. The routing of a packet is done for each individual packet. Each packet, therefore, needs to carry an end- to-end address. There is no setup and teardown phases in a datagram network (connectionless transmission). The entries in the routing table are somehow permanent and made by other processes such as routing protocols.  3 c. In a virtual-circuit network, there is a need for end-to-end addressing during the setup and teardown phases to make the corresponding entry in the switching table. The entry is made for each request for connection. During the data trans- fer phase, each packet needs to carry a virtual-circuit identifier to show which virtual-circuit that particular packet follows. 14. A datagram or virtual-circuit network handles packetized data. For each packet, the switch needs to consult its table to find the output port in the case of a datagram network, and to find the combination of the output port and the virtual circuit iden- tifier in the case of a virtual-circuit network. In a circuit-switched network, data are not packetized; no routing information is carried with the data. The whole path is established during the setup phase. 15. In circuit-switched and virtual-circuit networks, we are dealing with connections. A connection needs to be made before the data transfer can take place. In the case of a circuit-switched network, a physical connection is established during the setup phase and the is broken during the teardown phase. In the case of a virtual-circuit network, a virtual connection is made during setup and is broken during the tear- down phase; the connection is virtual, because it is an entry in the table. These two types of networks are considered connection-oriented. In the case of a datagram network no connection is made. Any time a switch in this type of network receives a packet, it consults its table for routing information. This type of network is con- sidered a connectionless network. 16. The switching or routing in a datagram network is based on the final destination address, which is global. The minimum number of entries is two; one for the final destination and one for the output port. Here the input port, from which the packet has arrived is irrelevant. The switching or routing in a virtual-circuit network is based on the virtual circuit identifier, which has a local jurisdiction. This means that two different input or output ports may use the same virtual circuit number. Therefore, four pieces of information are required: input port, input virtual circuit number, output port, and output virtual circuit number. 17. Packet 1: 2 Packet 2: 3 Packet 3: 3 Packet 4: 2 18. Packet 1: 2, 70 Packet 2: 1, 45 Packet 3: 3, 11 Packet 4: 4, 41 19. a. In a datagram network, the destination addresses are unique. They cannot be duplicated in the routing table. b. In a virtual-circuit network, the VCIs are local. A VCI is unique only in rela- tionship to a port. In other words, the (port, VCI) combination is unique. This means that we can have two entries with the same input or output ports. We can 4 have two entries with the same VCIs. However, we cannot have two entries with the same (port, VCI) pair. 20. When a packet arrives at a router in a datagram network, the only information in the packet that can help the router in its routing is the destination address of the packet. The table then is sorted to make the searching faster. Today’s routers use some sophisticated searching techniques. When a packet arrives at a switch in a virtual-circuit network, the pair (input port, input VCI) can uniquely determined how the packet is to be routed; the pair is the only two pieces of information in the packet that is used for routing. The table in the virtual-circuit switch is sorted based on the this pair. However, since the number of port numbers is normally much smaller than the number of virtual circuits assigned to each port, sorting is done in two steps: first according to the input port number and second according to the input VCI. 21. a. If n > k, an n × k crossbar is like a multiplexer that combines n inputs into k out- puts. However, we need to know that a regular multiplexer discussed in Chapter 6 is n × 1. b. If n < k, an n × k crossbar is like a demultiplexer that divides n inputs into k out- puts. However, we need to know that a regular demultiplexer discussed in Chapter 6 is 1 × n. 22. a. See Figure 8.1. Figure 8.1 Solution to Exercise 22 Part a 10 4 10 Crossbars Crossbars Crossbars 10 × 4 4 × 10 … n = 10 n = 10 … 10 × 10 … … … N = 100 n = 10 10 × 4 4 × 10 n = 10 N = 100 … … … … 10 × 10 … … 10 × 4 4 × 10 … n = 10 n = 10 … Stage 1 Stage 2 Stage 3 b. The total number of crosspoints are Number of crosspoints = 10 (10 × 4) + 4 (10 × 10) + 10 (4 × 10) = 1200 c. Only four simultaneous connections are possible for each crossbar at the first stage. This means that the total number of simultaneous connections is 40. d. If we use one crossbar (100 × 100), all input lines can have a connection at the same time, which means 100 simultaneous connections. e. The blocking factor is 40/100 or 40 percent.  5 23. a. See Figure 8.2. Figure 8.2 Solution to Exercise 23 Part a 10 6 10 Crossbars Crossbars Crossbars 10 × 6 6 × 10 … n = 10 n = 10 … 10 × 10 … … … N = 100 n = 10 10 × 6 6 × 10 n = 10 N = 100 … … … … 10 × 10 … … 10 × 6 6 × 10 … n = 10 n = 10 … Stage 1 Stage 2 Stage 3 b. The total number of crosspoints are Number of crosspoints = 10 (10 × 6) + 6 (10 × 10) + 10 (6 × 10) = 1800 c. Only six simultaneous connections are possible for each crossbar at the first stage. This means that the total number of simultaneous connections is 60. d. If we use one crossbar (100 × 100), all input lines can have a connection at the same time, which means 100 simultaneous connections. e. The blocking factor is 60/100 or 60 percent. 24. According to Clos, n = (N/2)1/2 = 7.07. We can choose n = 8. The number of cross- bars in the first stage can be 13 (to have similar crossbars). Some of the input lines can be left unused. We then have k = 2n − 1 = 15. Figure 8.3 shows the configura- tion. Figure 8.3 Solution to Exercise 24 Part a 13 15 13 Crossbars Crossbars Crossbars 8 × 15 15 × 8 … n=8 n=8 … 13 × 13 … … … N = 104 n = 8 8 × 15 15 × 8 … n=8 N = 104 … … … 13 × 13 … … 8 × 15 15 × 8 … n=8 n=8 … Stage 1 Stage 2 Stage 3 We can calculate the total number of crosspoints as 6 13 (8 × 15) + 15 (13 × 13) + 13 (15 × 8) = 5655 The number of crosspoints is still much less than the case with one crossbar (10,000). We can see that there is no blocking involved because each 8 input line has 15 intermediate crossbars. The total number of crosspoints here is a little greater than the minimum number of crosspoints according to Clos using the for- mula 4N[(2N)1/2 − 1], which is 5257. 25. a. Total crosspoints = N2 = 10002 = 1,000,000 b. Total crosspoints ≥ 4Ν[(2Ν)1/2 −1] ≥ 174,886. With less than 200,000 cross- points we can design a three-stage switch. We can use n = (N/2)1/2 =23 and choose k = 45. The total number of crosspoints is 178,200. 26. We give two solutions. a. We first solve the problem using only crossbars and then we replace the cross- bars at the first and the last stage with TSIs. Figure 8.1 shows the solution using only crossbars. We can replace the crossbar at the first and third stages with TSIs as shown in Figure 8.4. The total number of crosspoints is 400 and the total number of memory locations is 200. Each TSI at the first stage needs one TDM multiplexer and one TDM demultiplexer. The multiplexer is 10 × 1, but the demultiplexer is 1 × 4. In other words, the input frame has 10 slots and the output frame has only 4 slots. The data in the first slot of all input TSIs are directed to the first switch, the output in the second slot are directed to the sec- ond switch, and so on. Figure 8.4 First solution to Exercise 26 10 TSIs 10 TSIs 4 Crossbars 10 × 10 10 × 10 10 × 10 10 × 10 Stage 1 Stage 2 Stage 3  7 b. We can see the inefficiency in the first solution. Since the slots are separated in time, only one of the switches at the middle stage is active at each moment. This means that, instead of 4 crossbars, we could have used only one with the same result. Figure 8.5 shows the new design. In this case we still need 200 memory locations but only 100 crosspoints. Figure 8.5 Second solution to Exercise 26 10 TSIs 10 TSIs 1 Crossbars 10 × 10 Stage 1 Stage 2 Stage 3 8 CHAPTER 9 Using Telephone and Cable Networks Solutions to Review Questions and Exercises Review Questions 1. The telephone network is made of three major components: local loops, trunks, and switching offices. 2. The telephone network has several levels of switching offices such as end offices, tandem offices, and regional offices. 3. A LATA is a small or large metropolitan area that according to the divestiture of 1984 was under the control of a single telephone-service provider. The services offered by the common carriers inside a LATA are called intra-LATA services. The services between LATAs are handled by interexchange carriers (IXCs). These car- riers, sometimes called long-distance companies, provide communication services between two customers in different LATAs. 4. Signaling System Seven (SS7) is the protocol used to provide signaling services in the telephone network. It is very similar to the five-layer Internet model. 5. Telephone companies provide two types of services: analog and digital. 6. Dial-up modems use part of the bandwidth of the local loop to transfer data. The latest dial-up modems use the V-series standards such as V.32 and V.32bis (9600 bps), V.34bis (28,800 or 33,600 bps), V.90 (56 kbps for downloading and 33.6 kbps for uploading), and V.92. (56 kbps for downloading and 48 kbps for upload- ing. 7. Telephone companies developed digital subscriber line (DSL) technology to pro- vide higher-speed access to the Internet. DSL technology is a set of technologies, each differing in the first letter (ADSL, VDSL, HDSL, and SDSL). The set is often referred to as xDSL, where x can be replaced by A, V, H, or S. DSL uses a device called ADSL modem at the customer site. It uses a device called a digital sub- scriber line access multiplexer (DSLAM) at the telephone company site. 8. The traditional cable networks use only coaxial cables to distribute video infor- mation to the customers. The hybrid fiber-coaxial (HFC) networks use a combi- nation of fiber-optic and coaxial cable to do so. 1 2 9. To provide Internet access, the cable company has divided the available bandwidth of the coaxial cable into three bands: video, downstream data, and upstream data. The downstream-only video band occupies frequencies from 54 to 550 MHz. The downstream data occupies the upper band, from 550 to 750 MHz. The upstream data occupies the lower band, from 5 to 42 MHz. 10. The cable modem (CM) is installed on the subscriber premises. The cable modem transmission system (CMTS) is installed inside the distribution hub by the cable company. It receives data from the Internet and passes them to the combiner, which sends them to the subscriber. The CMTS also receives data from the sub- scriber and passes them to the Internet. Exercises 11. Packet-switched networks are well suited for carrying data in packets. The end-to- end addressing or local addressing (VCI) occupies a field in each packet. Tele- phone networks were designed to carry voice, which was not packetized. A cir- cuit-switched network, which dedicates resources for the whole duration of the conversation, is more suitable for this type of communication. 12. The setup phase can be matched to the dialing process. After the callee responds, the data transfer phase (here voice transfer phase) starts. When any of the parties hangs up, the data transfer is terminated and the teardown phase starts. It takes a while before all resources are released. 13. In a telephone network, the telephone numbers of the caller and callee are serving as source and destination addresses. These are used only during the setup (dialing) and teardown (hanging up) phases. 14. The delay can be attributed to the fact that some telephone companies use satellite networks for overseas communication. In these case, the signals need to travel sev- eral thousands miles (earth station to satellite and satellite to earth station). 15. See Figure 9.1. Figure 9.1 Solution to Exercise 15 60 kbps 56 kbps 50 kbps 40 kbps 30 kbps 20 kbps 14.4 kbps 10 kbps 9600 bps V.32 V.32bis V.90  3 16. See Figure 9.2. Figure 9.2 Solution to Exercise 16 55 Mbps 54 Mbps 25 Mbps 24 Mbps 6 Mbps 5 Mbps 4 Mbps 3 Mbps 2 Mbps 1 Mbps ADSL ADSL Lite HDSL SDSL VDSL Lower Higher 17. a. V.32 → Time = (1,000,000 × 8) /9600 ≈ 834 s b. V.32bis → Time = (1,000,000 × 8) / 14400 ≈ 556 s c. V.90 → Time = (1,000,000 × 8) / 56000 ≈ 143 s 18. a. ADSL → Time = (1,000,000 × 8) / 1,500,000 ≈ 5.3 s b. ADSL Lite → Time = (1,000,000 × 8) / 1,500,000 ≈ 5.3 s c. HDSL → Time = (1,000,000 × 8) / 1,500,000 ≈ 5.3 s d. SDSL → Time = (1,000,000 × 8) / 768,000 ≈ 10.42 s e. VDSL → Time = (1,000,000 × 8) / 25,000,000 ≈ 0.32 s 19. We can calculate time based on the assumption of 10 Mbps data rate: Time = (1,000,000 × 8) / 10,000,000 ≈ 0.8 seconds 20. The DSL technology is based on star topology with the hub at the telephone office. The local loop connects each customer to the end office. This means that there is no sharing; the allocated bandwidth for each customer is not shared with neigh- bors. The data rate does not depend on how many people in the area are transfer- ring data at the same time. 4 21. The cable modem technology is based on the bus (or rather tree) topology. The cable is distributed in the area and customers have to share the available band- width. This means if all neighbors try to transfer data, the effective data rate will be decreased. CHAPTER 10 Error Detection and Correction Solutions to Review Questions and Exercises Review Questions 1. In a single bit error only one bit of a data unit is corrupted; in a burst error more than one bit is corrupted (not necessarily contiguous). 2. Redundancy is a technique of adding extra bits to each data unit to determine the accuracy of transmission. 3. In forward error correction, the receiver tries to correct the corrupted codeword; in error detection by retransmission, the corrupted message is discarded (the sender needs to retransmit the message). 4. A linear block code is a block code in which the exclusive-or of any two code- words results in another codeword. A cyclic code is a linear block code in which the rotation of any codeword results in another codeword. 5. The Hamming distance between two words (of the same size) is the number of differences between the corresponding bits. The Hamming distance can easily be found if we apply the XOR operation on the two words and count the number of 1s in the result. The minimum Hamming distance is the smallest Hamming distance between all possible pairs in a set of words. 6. The single parity check uses one redundant bit for the whole data unit. In a two- dimensional parity check, original data bits are organized in a table of rows and columns. The parity bit is then calculated for each column and each row. 7. a. The only relationship between the size of the codeword and dataword is the one based on the definition: n = k + r., where n is the size of the codeword, k is the size of the dataword, and r is the size of the remainder. b. The remainder is always one bit smaller than the divisor. c. The degree of the generator polynomial is one less than the size of the divisor. For example, the CRC-32 generator (with the polynomial of degree 32) uses a 33-bit divisor. 1 2 d. The degree of the generator polynomial is the same as the size of the remainder (length of checkbits). For example, CRC-32 (with the polynomial of degree 32) creates a remainder of 32 bits. 8. One’s complement arithmetic is used to add data items in checksum calculation. In this arithmetic, when a number needs more than n bits, the extra bits are wrapped and added to the number. In this arithmetic, the complement of a number is made by inverting all bits. 9. At least three types of error cannot be detected by the current checksum calcula- tion. First, if two data items are swapped during transmission, the sum and the checksum values will not change. Second, if the value of one data item is increased (intentionally or maliciously) and the value of another one is decreased (intention- ally or maliciously) the same amount, the sum and the checksum cannot detect these changes. Third, if one or more data items is changed in such a way that the change is a multiple of 216 − 1, the sum or the checksum cannot detect the changes. 10. The value of a checksum can be all 0s (in binary). This happens when the value of the sum (after wrapping) becomes all 1s (in binary). It is almost impossible for the value of a checksum to be all 1s. For this to happen, the value of the sum (after wrapping) must be all 0s which means all data units must be 0s. Exercises 11. We can say that (vulnerable bits) = (data rate) × (burst duration) a. vulnerable bits = (1,500) × (2 × 10−3) = 3 bits b. vulnerable bits = (12 × 103) × (2 × 10−3) = 24 bits c. vulnerable bits = (100 × 103) × (2 × 10−3) = 200 bits d. vulnerable bits = (100 × 106) × (2 × 10−3) = 200,000 bits Comment: The last example shows how a noise of small duration can affect so many bits if the data rate is high. 12. a. (10001) ⊕ (10000) = 00001 b. (10001) ⊕ (10001) = 00000 c. (11100) ⊕ (00000) = 11100 d. (10011) ⊕ (11111) = 01100 Comment: The above shows three properties of the exclusive-or operation. First, the result of XORing two equal patterns is an all-zero pattern (part b). Second, the result of XORing of any pattern with an all-zero pattern is the original non-zero pattern (part c). Third, the result of XORing of any pattern with an all-one pattern is the complement of the original non-one pattern. 13. The codeword for dataword 10 is 101. This codeword will be changed to 010 if a 3-bit burst error occurs. This pattern is not one of the valid codewords, so the receiver detects the error and discards the received pattern.  3 14. The codeword for dataword 10 is 10101. This codeword will be changed to 01001 if a 3-bit burst error occurs. This pattern is not one of the valid codewords, so the receiver discards the received pattern. 15. a. d (10000, 00000) = 1 b. d (10101, 10000) = 2 c. d (1111, 1111) = 0 d. d (000, 000) = 0 Comment: Part c and d show that the distance between a codeword and itself is 0. 16. a. For error detection → dmin = s + 1 = 2 + 1 = 3 b. For error correction → dmin = 2t + 1 = 2 × 2 + 1 = 5 c. For error section → dmin = s + 1 = 3 + 1 = 4 For error correction → dmin = 2t + 1 = 2 × 2 + 1 = 5 Therefore dmin should be 5. d. For error detection → dmin = s + 1 = 6 + 1 = 7 For error correction → dmin = 2t + 1 = 2 × 2 + 1 = 5 Therefore dmin should be 7. 17. a. 01 b. error c. 00 d. error 18. We show that the exclusive-or of the second and the third code word (01011) ⊕ (10111) = 11100 is not in the code. The code is not linear. 19. We check five random cases. All are in the code. I. (1st) ⊕ (2nd) = (2nd) II. (2nd) ⊕ (3th) = (4th) III. (3rd) ⊕ (4th) = (2nd) IV. (4th) ⊕ (5th) = (8th) V. (5th) ⊕ (6th) = (2nd) 20. We show the dataword, the codeword, the corrupted codeword, and the interpreta- tion of the receiver for each case: a. Dataword: 0100 → Codeword: 0100011 → Corrupted: 0010011 This pattern is not in the table. → Correctly discarded. b. Dataword: 0111 → Codeword: 0111001 → Corrupted: 1111000 This pattern is not in the table. → Correctly discarded. 4 c. Dataword: 1111 → Codeword: 1111111 → Corrupted: 0101110 This pattern is in the table. → Erroneously accepted as 0101. d. Dataword: 0000 → Codeword: 0000000 → Corrupted: 1101000 This pattern is in the table. → Erroneously accepted as 1101. Comment: The above result does not mean that the code can never detect three errors. The last two cases show that it may happen that three errors remain unde- tected. 21. We show the dataword, codeword, the corrupted codeword, the syndrome, and the interpretation of each case: a. Datawor