TEST BANK FOR Data Communications and Networking By Behrouz Forouzan (Solution Manual)

Introduction
Solutions to Odd-Numbered Review Questions and Exercises
Review Questions
1. The five components of a data communication system are the sender, receiver,
transmission medium, message, and protocol.
3. The three criteria are performance, reliability, and security.
5. Line configurations (or types of connections) are point-to-point and multipoint.
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.
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
11. An internet is an interconnection of networks. The Internet is the name of a specific
worldwide network
13. Standards are needed to create and maintain an open and competitive market for
manufacturers, to coordinate protocol rules, and thus guarantee compatibility of
data communication technologies.
Exercises
15. With 16 bits, we can represent up to 216 different colors.
17.
a. Mesh topology: If one connection fails, the other connections will still be working.
b. Star topology: The other devices will still be able to send data through the hub;
there will be no access to the device which has the failed connection to the hub.
c. Bus Topology: All transmission stops if the failure is in the bus. If the drop-line
fails, only the corresponding device cannot operate.
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d. Ring Topology: The failed connection may disable the whole network unless it
is a dual ring or there is a by-pass mechanism.
19. Theoretically, in a ring topology, unplugging one station, interrupts the ring. However,
most ring networks use a mechanism that bypasses the station; the ring can
continue its operation.
21. See Figure 1.1
23.
a. E-mail is not an interactive application. Even if it is delivered immediately, it
may stay in the mail-box of the receiver for a while. It is not sensitive to delay.
b. We normally do not expect a file to be copied immediately. It is not very sensitive
to delay.
c. Surfing the Internet is the an application very sensitive to delay. We except to
get access to the site we are searching.
25. The telephone network was originally designed for voice communication; the
Internet was originally designed for data communication. The two networks are
similar in the fact that both are made of interconnections of small networks. The
telephone network, as we will see in future chapters, is mostly a circuit-switched
network; the Internet is mostly a packet-switched network.
Figure 1.1 Solution to Exercise 21
Station
Station Station
Repeat er
Station
Station Station
Repeat er
Station
Station Station
Repeater
Hub
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CHAPTER 2
Network Models
Solutions to Odd-Numbered Review Questions and Exercises
Review Questions
1. The Internet model, as discussed in this chapter, include physical, data link, network,
transport, and application layers.
3. The application layer supports the user.
5. Peer-to-peer processes are processes on two or more devices communicating at a
same layer
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.
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
d. detection and correction of damaged and lost frames
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
13. The application layer services include file transfer, remote access, shared database
management, and mail services.
Exercises
15. The International Standards Organization, or the International Organization of
Standards, (ISO) is a multinational body dedicated to worldwide agreement on
international standards. An ISO standard that covers all aspects of network communications
is the Open Systems Interconnection (OSI) model.
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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
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
21. See Figure 2.1.
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.
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.
Figure 2.1 Solution to Exercise 21
B/42 C/82
A/40
Sender
Sender
LAN1 LAN2
R1
D/80
42 40 A D i j Data T2 80 82 A D i j Data T2
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CHAPTER 3
Data and Signals
Solutions to Odd-Numbered Review Questions and Exercises
Review Questions
1. Frequency and period are the inverse of each other. T = 1/ f and f = 1/T.
3. Using Fourier analysis. Fourier series gives the frequency domain of a periodic
signal; Fourier analysis gives the frequency domain of a nonperiodic signal.
5. Baseband transmission means sending a digital or an analog signal without modulation
using a low-pass channel. Broadband transmission means modulating a
digital or an analog signal using a band-pass channel.
7. The Nyquist theorem defines the maximum bit rate of a noiseless 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.
11. The frequency domain of a voice signal is normally continuous because voice is a
nonperiodic signal.
13. This is baseband transmission because no modulation is involved.
15. This is broadband transmission because it involves modulation.
Exercises
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
19. See Figure 3.1
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.
23.
a. (10 / 1000) s = 0.01 s
b. (8 / 1000) s = 0. 008 s = 8 ms
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c. ((100,000 × 8) / 1000) s = 800 s
25. The signal makes 8 cycles in 4 ms. The frequency is 8 /(4 ms) = 2 KHz
27. The signal is periodic, so the frequency domain is made of discrete frequencies. as
shown in Figure 3.2.
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
31. –10 = 10 log10 (P2 / 5) → log10 (P2 / 5) = −1 → (P2 / 5) = 10−1 → P2 = 0.5 W
33. 100,000 bits / 5 Kbps = 20 s
35. 1 μm × 1000 = 1000 μm = 1 mm
37. We have
4,000 log2 (1 + 10 / 0.005) = 43,866 bps
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
41. We have
SNR= (signal power)/(noise power).
However, power is proportional to the square of voltage. This means we have
Figure 3.1 Solution to Exercise 19
Figure 3.2 Solution to Exercise 27
0 20 50 100 200
Frequency domain
Bandwidth = 200 − 0 = 200
Amplitude
10 volts
Frequency
30
KHz
10
KHz
...
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SNR = [(signal voltage)2] / [(noise voltage)2] =
[(signal voltage) / (noise voltage)]2 = 202 = 400
We then have
SNRdB = 10 log10 SNR ≈ 26.02
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).
45. We have
transmission time = (packet length)/(bandwidth) =
(8,000,000 bits) / (200,000 bps) = 40 s
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
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CHAPTER 4
Digital Transmission
Solutions to Odd-Numbered Review Questions and Exercises
Review Questions
1. The three different techniques described in this chapter are line coding, block coding,
and scrambling.
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.
5. When the voltage level in a digital signal is constant for a while, the spectrum creates
very low frequencies, called DC components, that present problems for a system
that cannot pass low frequencies.
7. In this chapter, we introduced unipolar, polar, bipolar, multilevel, and multitransition
coding.
9. Scrambling, as discussed in this chapter, is a technique that substitutes long zerolevel
pulses with a combination of other levels without increasing the number of
bits.
11. In parallel transmission we send data several bits at a time. In serial transmission
we send data one bit at a time.
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
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.
17. See Figure 4.2. 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.