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CHAPTER 1
Introduction
Solutions to Review Questions and Exercises
Review Questions
- The five components of a data communication system are the sender , receiver , transmission medium , message , and protocol.
- The advantages of distributed processing are security , access to distributed data- bases , collaborative processing , and faster problem solving.
- The three criteria are performance , reliability , and security.
- Advantages of a multipoint over a point-to-point configuration (type of connec- tion) include ease of installation and low cost.
- Line configurations (or types of connections) are point-to-point and multipoint.
- We can divide line configuration in two broad categories: a. Point-to-point : mesh , star , and ring. b. Multipoint : bus
- 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.
- 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
- 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
- The general factors are size , distances (covered by the network), structure , and ownership.
- An internet is an interconnection of networks. The Internet is the name of a spe- cific worldwide network
- 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.
- 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
- Unicode uses 32 bits to represent a symbol or a character. We can define 2 32 differ- ent symbols or characters.
- With 16 bits, we can represent up to 2 16 different colors.
- a. Cable links: n ( n – 1) / 2 = (6 × 5) / 2 = 15 b. Number of ports: ( n – 1) = 5 ports needed per device
- 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.
- This is a LAN. The Ethernet hub creates a LAN as we will see in Chapter 13.
- 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.
- 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.
- See Figure 1.
- See Figure 1.2.
- 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.
- 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.
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CHAPTER 2
Network Models
Solutions to Review Questions and Exercises
Review Questions
- The Internet model, as discussed in this chapter, include physical , data link , net- work , transport , and application layers.
- The network support layers are the physical , data link , and network layers.
- The application layer supports the user.
- 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.
- Peer-to-peer processes are processes on two or more devices communicating at a same layer
- Each layer calls upon the services of the layer just below it using interfaces between each pair of adjacent layers.
- 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.
- 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
- The data link layer is responsible for a. framing data bits b. providing the physical addresses of the sender/receiver c. data rate control
d. Responsibility for carrying frames between adjacent nodes: data link layer
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
- See Figure 2.1.
- See Figure 2.2.
- 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.
- 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/
A/
Sender
Sender
LAN1 LAN R D/
42 40 A DDataT2 80 82 A DData T
B/42 (^) C/
A/
Sender
Sender
LAN1 LAN R D/
42 40 A D i j Data T2 80 82 A D i j Data T
- Most protocols issue a special error message that is sent back to the source in this case.
- 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.
Exercises
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 μ s c. T = 1 / f = 1 / (140 KHz) = 0.00000714 s = 7.14 × 10 –6^ s = 7.14 μ s
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
a. 90 degrees (π/2 radian) b. 0 degrees ( 0 radian) c. 90 degrees (π/2 radian)
- See Figure 3.
- We know the lowest frequency, 100. We know the bandwidth is 2000. The highest frequency must be 100 + 2000 = 2100 Hz. See Figure 3.
- 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.
- 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
0 20 50 100 200
Frequency domain
Bandwidth = 200 − 0 = 200
100
20
5 2100
Frequency domain
Bandwidth = 2100 − 100 = 2000
c. bit rate = 1/(bit duration) = 1 / (20 μs/10) = 1 / (2 μs) = 500 Kbps
a. (10 / 1000) s = 0.01 s b. (8 / 1000) s = 0. 008 s = 8 ms c. ((100,000 × 8) / 1000) s = 800 s
- There are 8 bits in 16 ns. Bit rate is 8 / (16 × 10 −9^ ) = 0.5 × 10 −9^ = 500 Mbps
- The signal makes 8 cycles in 4 ms. The frequency is 8 /(4 ms) = 2 KHz
- The bandwidth is 5 × 5 = 25 Hz.
- The signal is periodic, so the frequency domain is made of discrete frequencies. as shown in Figure 3.3.
- The signal is nonperiodic, so the frequency domain is made of a continuous spec- trum of frequencies as shown in Figure 3.4.
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
- dB = 10 log 10 (90 / 100) = –0.46 dB
- –10 = 10 log 10 (P 2 / 5) → log 10 (P 2 / 5) = − 1 → (P 2 / 5) = 10−^1 → P 2 = 0.5 W
- The total gain is 3 × 4 = 12 dB. The signal is amplified by a factor 10 1.2^ = 15.85.
Figure 3.3 Solution to Exercise 27
Figure 3.4 Solution to Exercise 28
Amplitude
10 volts Frequency 30 KHz
10 KHz
......
Amplitude
30 volts
10 volts 10 volts Frequency 30 KHz
20 KHz
10 KHz
This means that the minimum SNR = 10 SNRdB/10^ = 107.5^ ≈ 31,622,
- We have
transmission time = (packet length)/(bandwidth) = (8,000,000 bits) / (200,000 bps) = 40 s
- We have
(bit length) = (propagation speed) × (bit duration) The bit duration is the inverse of the bandwidth. a. Bit length = (2 × 10 8 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 × 10 8 m) × [(1 / (100 Mbps)] = 2 m. This means a bit occupies 2 meters on a transmission medium.
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
- 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 × 10 8 ) = 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.
- 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.
- In parallel transmission we send data several bits at a time. In serial transmission we send data one bit at a time.
- 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
- 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
- The number of bits is calculated as (0.2 /100) × (1 Mbps) = 2000 bits
- 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.
- 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.
- 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.
- See Figure 4.4. B is proportional to (12/8) N which is within the range in Table 4. (B = N to 2N) for the differential Manchester scheme.
Figure 4.1 Solution to Exercise 15
0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1
0 1 0 1 0 1 0 1
Case a
Case b
Case c
Case d
Average Number of Changes = (0 + 0 + 8 + 4) / 4 = 3 for N = 8 B (3 / 8) N
Figure 4.2 Solution to Exercise 16
Figure 4.3 Solution to Exercise 17
Figure 4.4 Solution to Exercise 18
0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1
0 1 0 1 0 1 0 1
Case a
Case b
Case c
Case d
Average Number of Changes = (0 + 9 + 4 + 4) / 4 = 4.25 for N = 8 B (4.25 / 8) N
0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1
0 1 0 1 0 1 0 1
Case a
Case b
Case c
Case d
Average Number of Changes = (15 + 15+ 8 + 12) / 4 = 12.5 for N = 8 B (12.5 / 8) N
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 a
Case b
Case c
Case d
Average Number of Changes = (16 + 8 + 12 + 12) / 4 = 12 for N = 8 B (12 / 8) N
a. f /N = 0/100 = 0 → P = 1. b. f /N = 50/100 = 1/2 → P = 0. c. f /N = 100/100 = 1 → P = 0. d. f /N = 150/100 = 1.5 → P = 0.
- 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. b. f /N = 50/100 = 1/2 → P = 0. c. f /N = 100/100 = 1 → P = 0. d. f /N = 150/100 = 1.5 → P = 0.
- 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.
- In 5B/6B, we have 2 5 = 32 data sequences and 2 6 = 64 code sequences. The number of unused code sequences is 64 − 32 = 32. In 3B/4B, we have 2 3 = 8 data sequences and 2 4 = 16 code sequences. The number of unused code sequences is 16 − 8 = 8.
- See Figure 4.7. Since we specified that the last non-zero signal is positive, the first bit in our sequence is positive.
a. In a low-pass signal, the minimum frequency 0. Therefore, we have f (^) max = 0 + 200 = 200 KHz. → f (^) s = 2 × 200,000 = 400,000 samples/s
Figure 4.7 Solution to Exercise 26
1 1 1 0
V
a. B8ZS
b. HDB
B V
B
0 0 0 0 0 0 0 0 0 0
1 1 1 0
V
B V
0 0 0 0 0 0 0 0 0 0
b. In a bandpass signal, the maximum frequency is equal to the minimum fre- quency plus the bandwidth. Therefore, we have f (^) max = 100 + 200 = 300 KHz. → f (^) s = 2 × 300,000 = 600,000 samples /s
a. In a lowpass signal, the minimum frequency is 0. Therefore, we can say fmax = 0 + 200 = 200 KHz → f (^) s = 2 × 200,000 = 400,000 samples/s The number of bits per sample and the bit rate are nb = log 2 1024 = 10 bits/sample N = 400 KHz × 10 = 4 Mbps b. The value of n (^) b = 10. We can easily calculate the value of SNR (^) dB SNR (^) dB = 6.02 × nb + 1.76 = 61. c. The value of n (^) b = 10. The minimum bandwidth can be calculated as BPCM = nb × Banalog = 10 × 200 KHz = 2 MHz
- The maximum data rate can be calculated as
Nmax = 2 × B × nb = 2 × 200 KHz × log 24 = 800 kbps
- We can first calculate the sampling rate (fs) and the number of bits per sample (nb)
f (^) max = 0 + 4 = 4 KHz → f (^) s = 2 × 4 = 8000 sample/s We then calculate the number of bits per sample. → nb = 30000 / 8000 = 3. We need to use the next integer n (^) b = 4. The value of SNR (^) dB is SNRdB = 6.02 × nb + 1.72 = 25.
- We can calculate the data rate for each scheme:
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%.
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
