MULTIPLE CHOICE QUESTIONS

Transcription

1 CHAPTER 7 2. Guided and unguided media 4. Twisted pair, coaxial, and fiber-optic cable 6. Coaxial cable can carry higher frequencies than twisted pair cable and is less sus-ceptible to noise. 8. a. The beam bends toward the horizontal axis, and passes into the less dense medium (R > I). b. The refracted beam travels along the horizontal axis (R = 90 degrees). c. The light beam will be reflected back into the original medium (I = R). 10. In multimode, multiple beams of light from one source travel through the core in different paths. In graded-index multimode, the core s density is not constant but is higher in the center and decreases gradually to a lower density at the edge. In single mode, a step-index fiber is used with a highly focused source of light. 12. Noise resistance, less signal attenuation, and higher bandwidth KHz to 300 GHz 16. Terrestrial microwaves require line of sight propagation, using repeaters (usually installed with each antenna) to increase the distance. 18. During a conversation a mobile phone may move from one cell to another and the signal may become weak. In this case, the MTSO seeks a new cell to accommo-date the communication. The MTSO changes the channel carrying the call. The process of handing the signal from the old channel to the new channel is called handoff. 20. Decibel measures the relative strength of two signals or a signal at two different points. 22. Propagation time = distance / propagation speed 24. The Shannon capacity determines the theoretical highest data rate for a channel. 26. See Figure 7.20 in the text. 28. The troposphere is the portion of the atmosphere extending outward approximately 30 miles from the earth surface and includes the stratosphere. It contains air and is used for the propagation of VLF, LF, MF, VHF and UHF signals. The ionosphere is the layer of the atmosphere above the troposphere. It contains free electrically charged particles (ions). The ionosphere is used to propagate HF signals. 30. The distance terrestrial microwaves can travel is limited by the curvature of the earth and surface obstacles because terrestrial microwaves require line of sight transmission. 32. b 33. b 34. a 35. b 36. a 37. b 38. b 39. d 40. c 41. c 42. b 43. a 44. a 45. d 46. c 47. b 48. a 49. b 50. a 51. b 52. a 53. a 54. a 55. c 56. b 57. a 58. d 59. a 60. b 61. c 62. c 63. a 64. b 65. d 66. c 67. b 68. a 69. b 70. c 71. d 72. b 73. a 74. c 75. d 76. b 77. d 78. d 79. b 80. c Kb / 5 s = 20 Kbps ,000 bits / 5 Kbps = 20 s ,000 km / 300,000 km/s = 1.33 s s 300,000 km/s = 144,000,000 km / = m = 0.3 mm The wavelength of infrared is longer than the wavelength of red because the frequency of the former is less than the frequency of the latter mm 5 = 5 mm 92. 2,000 km / 195,000 km/s = 0.01 s = 10 ms 93. 4,000 log 2 (1 + 1,000) = 40 Kbps (approximately) 94. 4,000 log 2 ( / 0.005) = 43,866 bps CHAPTER 8 2. In FDM each signal modulates a different carrier frequency. The modulated carriers are combined to form a new signal that is then sent across the link. 4. A demultiplexer uses a series of filters to decompose the multiplexed signal into its constituent component signals. 6. Synchronous and asynchronous 8. Synchronous and asynchronous. In synchronous TDM each frame contains at least one time slot dedicated to each device. The order in which each device sends its data is fixed. In asynchronous TDM the number of time slots is less than the number of devices and the slot order depends on which devices have data to send. Addressing of each time slot is necessary. 10. Inverse multiplexing splits a data stream from one high speed line onto multiple lower speed lines. 1

2 12. Voice channels (12 x 44 KHz) are multiplexed onto a higher bandwidth line to create a group (48 KHz). Up to five groups (5 x 48 KHz) can be multiplexed to create a super group (240 KHz). Ten super groups (10 x 240 KHz) are multiplexed to create a master group (2.52 MHz). Six master groups are multiplexed to create a jumbo group with MHz. 14. The DSU changes the rate of the digital data created by the subscriber s device to 56 Kbps and encodes it in the format used by the service provider. 16. DS is the name of the service, which is implemented by T-lines. The capacity of the lines precisely matches the data rate of DS-services. 18. ADSL divides the bandwidth of a twisted pair cable into three bands. The first band, 0-25 KHz is used for regular telephone service. The second band ( KHz) is for upstream communication and the third band (250 KHz 1 MHz) for downstream communication. 20. FTTC means fiber to the curb. Optical fiber is the medium from the central office of the telephone/cable company to the curb of the users premises. 22. A DSU is used in digital services and because the service is already digital a modem is not needed to transform analog data into digital. The DSU changes the rate of the digital data created by the subscriber s device to 56 Kbps and encodes it in the format used by the service provider. A modem takes a digital signal and changes it to an analog signal and vice versa. 24. For DS-0 the sampling rate is 8000 samples/second; with 8 bits per sample, this means bps or 64 Kbps. 25. d 26. a 27. d 28. a 29. a 30. b 31. a 32. a 33. b 34. d 35. a 36. a 37. b 38. c 39. b 40. d 41. b 42. d 43. d 44. a 45. c 46. b 47. a 48. d 49. d 50. c 51. b 52. a 53. c 54. c 55. a 56. c 57. c 58. d 60. (7900 Hz (200 x 2)) / 3 = 2.5 KHz bits per frame x 100 = 4.1 Kbps 64. See Figure us 68. n (n 1) / 2 = 500 (499)/2 = 124,750 lines; multiplexing can reduce the number of lines. 70. Nyquist theorem dictates that the sampling rate must be twice the highest fre-quency; 2 x 4000 Hz or 8000 Hz. 72. T1 line Þ (1,544, x 64000) / 24 = 333 bits /channel Þ 0.5% T2 line Þ (6,312,00 96 x 64000) / 96 = 1750 bits /channel Þ 2.7% T3 line Þ (44,736, x 64000) / 672 = 2571 bits /channel Þ 4.0% T4 line Þ (274,176, x 64000) / 4032 = 4000 bits /channel Þ 6.2% x 20 = 99 KHz 76. See Figure Thirty percent of the bandwidth is wasted. 80. See Figure 8.6 Output bit rate: 45 bits/second; duration; 22 ms; 45 slots per second Mbps; T1 is not appropriate in this case (1.544 Mbps) 84. See Figure

3 CHAPTER 9 2. Redundancy is a technique of adding extra bits into each data unit for the purpose of determining the accuracy of transmission. 4. A parity bit is added to every data unit so that the total number of 1s in the unit becomes even (in even-parity checking). If after transmission the number of 1s is odd, then there must be an error. 6. VRC is the least expensive and most common type of error detection. It can detect single bit errors as well as burst errors if the total number of bits changed is odd. 8. An LRC of n bits can detect single bit errors as well as most burst errors of n bits except for damage to corresponding bits of different data units. 10. The length of the divisor should be one bit more than the length of the CRC. 3

4 12. A polynomial should not be divisible by x and should be divisible by (x+1). 14. Checksum. 16. a. The checksum generator divides the data into equal segments. b. The segments are added together using one's complement arithmetic. c. The sum is complemented and sent to the receiver along with the data. 18. If a bit inversion in one data segment is balanced by an opposite bit inversion on the corresponding bit of another segment, the error can't be detected by checksum. 20. The purpose of the Hamming code is to correct one or more corrupted bits. 21. b 22. b 23. d 24. a 25. c 26. b 27. c 28. a 29. d 30. c 31. b 32. a 33. b 34. a 35. d 36. b 37. c 38. d 39. d 40. d 41. b 42. a 43. d 44. d 46. a. 0 b. 0 c. 1 d No error a. 5 b. 5 c. 5 d Bits 1, 5, 7, and 8 (from the right) are in error bits long 62. x 11 + x 6 + x 5 + x Two redundancy bits, one data bit. 66. No errors in the received code; the original code is the same as the received code. CHAPTER 10 REVIEW QUESTIONS 2. Line discipline, error control, and flow control. 4. The two methods for line discipline are ENQ/ACK and poll/select. The first method is used in peer-to-peer communication. The second method is used in pri-mary- secondary communication. 6. If the primary device wants to receive data it asks the secondary devices if they have anything to send. This is called polling. If the primary wants to send data, it tells the target secondary to get ready to receive the data. This is called selecting. 8. Polling is used by the primary device to receive transmissions from secondary devices. Selecting is used by the primary device if it wants to send something to secondary devices. 10. Each receiving device has a block of memory (buffer) that is used for storing the incoming data until they are processed. If the buffer starts to fill up, the receiver must be able to notify the sender. 12. In stop-and-wait flow control the sender waits for an ACK from the receiver after each frame sent. A new frame is only sent out if the previous frame has been acknowledged. 14. Error control refers to methods of error detection and retransmission and is based on automatic repeat request (ARQ), which means retransmission of the data in the case of damaged frames, lost frames, or lost acknowledgments. 16. The sender retransmits a packet if the packet was damaged or lost or if the acknowledgment of that packet was lost. 18. The two types of sliding window ARQ are go-back-n and selective-reject. In the first method, if one frame is lost or damaged, all frames since the last acknowledg-ment are retransmitted. In selective-reject only the specific damaged or lost frames will be retransmitted. 20. a. If a frame is damaged, the receiver sends a negative acknowledgment (NAK) indicating the last frame sent was damaged and needs to be retransmitted. b. If a frame is lost, the timer goes off and the frame is resent. 21. If a NAK gets lost, the sender timer expires. The sender automatically retransmits the last frame sent. The NAK frames are not numbered because the sender waits for an acknowledgment after each frame is sent. If the sender receives a NAK frame it automatically knows that the last frame sent was damaged and retransmits that frame. 22. In practice, go-back-n ARQ is more popular due to its simplicity of implementa-tion. 24. c 25. a 26. a 27. b 28. c 29. a 30. c 31. b 32. d 33. c 34. c 35. b 36. c 37. d 38. b 39. c 40. b 41. d 42. d 43. a 44. c 45. a 4

5 EXERCISES 46. See Figure a. Data frame or NAK b. ACK or NAK 50. a. The number refers to the receipt of an acceptable frame. b. The number refers to the next expected frame. c. The number refers to the next expected frame. 52. Four bits , 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, See Figure See Figure ms ms 66. See Figure See Figure

6 Chapter A DLE DLE pattern can be seen when the transparent region contains DLE as a part of the text. 4. In data communications protocol means a set of rules or specifications used to implement one or more layers of the OSI model. 6. Asynchronous protocols are mainly used in modems. 8. Synchronous protocols can be character-oriented or bit-oriented. In a character-oriented protocol, the frame is interpreted as a series of characters while in a bit-oriented protocol the frame is interpreted as a series of bits. 10. BSC is used in both point-to-point and multipoint configurations. It supports half-duplex transmission and uses stopand-wait ARQ for flow control and error control. 12. In a long BSC frame error detection is more difficult since changes in several bits can cancel each other. Therefore to decrease the probability of an error the message should be divided between several blocks. 14. Control frames are used to establish connections, maintain flow and error control during transmission, and terminate connections. 16. In an unbalanced configuration the primary device sends commands; the secondary devices send responses. In a symmetrical configuration, the primary part of one station sends commands to the secondary part of the other one and vice versa. In a balanced configuration both stations can send commands and responses. 18. Bit stuffing is the process of adding one extra 0 when there are five consecutive 1s in the data stream to distinguish data from a flag. 20. Piggybacking is combining data to be sent and acknowledgment of the received frame in one single frame. 22. Mode-setting, unnumbered-exchange, disconnection, initialization, and miscella-neous. 24. b 25. b 26. a 27. b 28. b 29. b 30. c 31. c 32. d 33. b 34. b 35. b 36. a 37. a 38. a 39. b 40. b 41. b 42. a figure 11-1 EXERCISES 43. SYN-SYN-EOT-SYN-H-E-L-L-O (The hyphens separate bytes). 44. SYN-SYN-EOT-SYN-DLE-B-Y-E (The hyphens separate bytes). 6

7 45. Only the control field is shown in Figure Ü Ü a (7) b. S-frame c. N/A d. 3 e. no f. no g. It is a poll frame. 49. a (7) b. S-frame c. N/A d. 3 e. no f. no g. It is a negative response to poll or positive response to select (depends on the previous packet). 50. a (3) b. S-frame c. N/A d. 3 e. no f. no g. Select 51. a (3) b. S-frame c. N/A d. 3 e. no f. no g. Negative response to select 52. a b. I-frame c. 2 d. 3 e f. no 53. a (3) b. I-frame c. 2 d. 3 e f. no 54. a (67) b. U-frame c. N/A d. N/A e. no f a. SYN-SYN-ENQ b. SYN SYN ACK0 c. SYN SYN STX 100bytes ITB BCC STX 100bytes ITB BCC STX 100bytes ITB BCC STX 100bytes ETB BCC SYN SYN STX 100bytes ITB BCC STX 100bytes ITB BCC STX 100bytes ITB BCC STX 100bytes ETB BCC SYN SYN STX 100bytes ITB BCC STX 100bytes ITB BCC STX 100bytes ITB BCC STX 100bytes ETX BCC d. SYN SYN ACK1 56. a. SYN SYN ENQ figure 11-2 b. SYN SYN STX 50 bytes ETX BCC c. SYN SYN ACK1 57. See Figure See Figure See Figure See Figure 11.5 figure 11-4 figure 11-3 figure 11-5 Chapter 12 CHAPTER CSMA/CD is the access mechanism used in Ethernet (802.3). If a station wants to send data over the network, it must first listen for existing traffic on the line. If no traffic is detected, the line is considered idle and transmission is initiated. Then the station continues to listen after sending the data. If a collision is detected, the station stops the current transmission and waits a certain amount of time for the line to clear and then starts all over again. 4. These fields are added by the MAC layer. 6. They are almost the same. The control field is moved to the LLC layer. The addresses are divided into two source and destination addresses. The type field is added to define the upper layer protocol using the frame. 7

8 8. Baseband refers to digital signal transmission (without modulation) and broadband refers to analog signal transmission (with modulation). 9. In 10Base5, the transceiver is placed onto the cable and connects the station NIC to the link via an AUI cable. In 10Base2, the transceiver circuit is in the NIC. In 10Base-T, instead of an individual transceiver, all of the networking operations are placed into an intelligent hub that has a port for every station. 10. Collision occurs in an Ethernet when two stations on the same link transmit data at the same time. A collision can be detected by an extremely high voltage on the line. 12. The AC field is a field in the Token Ring frame and is used to control the access on a Token Ring network. Since uses CSMA/CD as its access method, an AC field is not needed B/5B encoding transforms each four-bit data segment into a five-bit unit that contains no more than two consecutive 0s. Each of the 16 possible four-bit patterns is assigned a five-bit pattern to represent it. These patterns have been selected so that even sequential data units cannot result in a sequence of more than three 0s. None of the five-bit patterns start with more than one 0 or ends with more than two 0s. 16. A token circulates around the Token Ring network. This token travels from NIC to NIC in sequence, until it encounters a station that wants to send data. This station then keeps the token (if free) and sends its data. The data frame proceeds around the ring. When the frame reaches its destination, the receiving station copies the message, checks for errors and changes four bits in the last byte of the frame. The frame then continues around the ring. The original sender examines the frame and discards the used data frame and releases the token back on the ring. 18. The switch is a device that can recognize the destination address of a frame and can therefore route the frame to the port of the destination station. This way the rest of the media is not involved in the transmission Base-FX uses fiber optic cable which has less signal degradation. 22. b 23. b 24. a 25. c 26. c 27. c 28. d 29. c 30. b 31. b 32. d 33. b 34. d 35. c 36. d 37. c 38. d 39. a 40. a 41. b 42. a 43. b 44. d 45. a 46. d 47. d 48. d 49. b 50. a 51. b 52. c 53. b 54. d 55. b 56. c 57. b 58. Smallest Ethernet frame: 72 bytes (26 byte header + 46 bytes of data) Largest Ethernet frame: 1,526 bytes (26 byte header + 1,500 bytes of data) 60. Ratio for smallest Ethernet frame: 46 / 72 or Ratio for largest Ethernet frame: 1500 / 1526 or Average ratio: (46 / / 1526) / 2 or The Ethernet frame must have a minimum data size because a sending station must be able to sense a collision before the entire frame is sent. The minimum size of an Ethernet network is therefore determined by the minimum frame size. If there is no minimum data size, the minimum frame size would be only 27 bytes (26 byte header + 1 byte of data) which would make the minimum network size much shorter. 64. In the worst case, the collision would be sensed in = 27.8 microseconds. 66. If we call the minimum number of bits X: X bits / 10,000,000 bits/second = seconds X = 10,000, = bits bits / 8 = bytes The minimum frame size would have to be larger than 35 bytes for an Ethernet net-work of length 2500 meters to work properly. 68. The token size is 3 bytes. 3 bytes 8 bits / byte = 24 bits. If the data rate is 16 Mbps, 24 bits can be generated in: 24 bits / 16,000,000 bits/second = 1.5 microseconds See Table 12.1 Table 12.1 Exercise 72 Feature Ethernet Token Ring Preamble 56 bits None SFD None SD None One byte AC None One byte FC None One byte Destination address Six bytes Two to six bytes Source address Six bytes Two to six bytes Data size 46 to 1500 bytes Up to 4500 bytes CRC CRC-32 CRC-32 ED None One byte FS None One byte 8