Title Forouzan-Solutions-chapters-7-9 Computer Network Networks Routing Digital Subscriber Line Transmission Medium 165.6 KB 10
```                            Chapter 7
Transmission Media Solutions to Odd-Numbered Review Questions and Exercises
Review Questions
1. The transmission media is located beneath the physical layer and controlled by the physical layer.
3. Guided media have physical boundaries, while unguided media are unbounded.
5. Twisting ensures that both wires are equally, but inversely, affected by external influences such as noise.
7. The inner core of an optical fiber is surrounded by cladding. The core is denser than the cladding, so a light beam traveling...
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.
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
Table 7.2 Solution to Exercise 15
Distance
dB at 1 KHz
dB at 10 KHz
dB at 100 KHz
Table 7.3 Solution to Exercise 19
Distance
dB at 800 nm
dB at 1000 nm
dB at 1200 nm
SolStd08.pdf
Chapter 8
Switching Solutions to Odd-Numbered 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...
3. There are two approaches to packet switching: datagram approach and virtual- circuit approach.
5. The address field defines the end-to-end (source to destination) addressing.
7. In a space-division switch, the path from one device to another is spatially separate from other paths. The inputs and the ou...
9. In multistage switching, blocking refers to times when one input cannot be connected to an output because there is no path av...
Exercises
11. We assume that the setup phase is a two-way communication and the teardown phase is a one-way communication. These two phase...
3 [(5000 km)/ (2 ¥108 m/s)]+ 3 [(1000 bits/1 Mbps)] = 75 ms + 3 ms = 78 ms
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.
13.
15. In circuit-switched and virtual-circuit networks, we are dealing with connections. A connection needs to be made before the ...
17. Packet 1: 2 Packet 2: 3 Packet 3: 3 Packet 4: 2
19.
21.
23.
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.
25.
SolStd09.pdf
Chapter 9
Using Telephone and Cable Networks Solutions to Odd-Numbered Review Questions and Exercises
Review Questions
1. The telephone network is made of three major components: local loops, trunks, and switching 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 tele...
5. Telephone companies provide two types of services: analog and digital.
7. Telephone companies developed digital subscriber line (DSL) technology to provide higher-speed access to the Internet. DSL te...
9. To provide Internet access, the cable company has divided the available bandwidth of the coaxial cable into three bands: vide...
Exercises
11. Packet-switched networks are well suited for carrying data in packets. The end-to- end addressing or local addressing (VCI) ...
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.
15. See Figure 9.1.
17.
a. V.32
Æ
Time = (1,000,000 ¥ 8) /9600
ª 834 s
b. V.32bis
Æ
Time = (1,000,000 ¥ 8) / 14400
ª 556 s
c. V.90
Æ
Time = (1,000,000 ¥ 8) / 56000
ª 143 s
19. We can calculate time based on the assumption of 10 Mbps data rate: Time = (1,000,000 ¥ 8) / 10,000,000 ª 0.8 seconds
21. The cable modem technology is based on the bus (or rather tree) topology. The cable is distributed in the area and customers...
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