Solution Manual for Computer Networking,8th
Edition by James Kurose
Chapter 1 Review Questions
1. There is no difference. Throughout this text, the words “host” and “end system” are
used interchangeably. End systems incl
...
Solution Manual for Computer Networking,8th
Edition by James Kurose
Chapter 1 Review Questions
1. There is no difference. Throughout this text, the words “host” and “end system” are
used interchangeably. End systems include PCs, workstations, Web servers, mail
servers, PDAs, Internet-connected game consoles, etc.
2. From Wikipedia: Diplomatic protocol is commonly described as a set of international
courtesy rules. These well-established and time-honored rules have made it easier for
nations and people to live and work together. Part of protocol has always been the
acknowledgment of the hierarchical standing of all present. Protocol rules are based on
the principles of civility.
3. Standards are important for protocols so that people can create networking systems and
products that interoperate.
4. 1. Dial-up modem over telephone line: home; 2. DSL over telephone line: home or
small office; 3. Cable to HFC: home; 4. 100 Mbps switched Ethernet: enterprise.
5. HFC bandwidth is shared among the users. On the downstream channel, all packets
emanate from a single source, namely, the head end. Thus, there are no collisions in the
downstream channel.
6. In most American cities, the current possibilities include: dial-up; DSL; cable modem;
fiber-to-the-home.
7. Ethernet LANs have transmission rates of 10 Mbps, 100 Mbps, 1 Gbps and 10 Gbps.
8. Today, Ethernet most commonly runs over twisted-pair copper wire. It also can run over
fibers optic links.
9. ADSL: up to 24 Mbps downstream and 2.5 Mbps upstream, bandwidth is dedicated;
HFC, rates up to 42.8 Mbps and upstream rates of up to 30.7 Mbps, bandwidth is
shared. FTTH: 2-10Mbps upload; 10-20 Mbps download; bandwidth is not shared.
10. There are two popular wireless Internet access technologies today:
a. Wifi (802.11) In a wireless LAN, wireless users transmit/receive packets
to/from an base station (i.e., wireless access point) within a radius of few tens
of meters. The base station is typically connected to the wired Internet and thus
serves to connect wireless users to the wired network.
b. 3G and 4G wide-area wireless access networks. In these systems, packets are
transmitted over the same wireless infrastructure used for cellular telephony,
with the base station thus being managed by a telecommunications provider.
This provides wireless access to users within a radius of tens of kilometers of
the base station.
11. At time t0 the sending host begins to transmit. At time t1 = L/R1, the sending host
completes transmission and the entire packet is received at the router (no propagation
delay). Because the router has the entire packet at time t1, it can begin to transmit the
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packet to the receiving host at time t1. At time t2 = t1 + L/R2, the router completes
transmission and the entire packet is received at the receiving host (again, no
propagation delay). Thus, the end-to-end delay is L/R1 + L/R2.
12. A circuit-switched network can guarantee a certain amount of end-to-end bandwidth
for the duration of a call. Most packet-switched networks today (including the Internet)
cannot make any end-to-end guarantees for bandwidth. FDM requires sophisticated
analog hardware to shift signal into appropriate frequency bands.
13. a) 2 users can be supported because each user requires half of the link bandwidth.
b) Since each user requires 1Mbps when transmitting, if two or fewer users transmit
simultaneously, a maximum of 2Mbps will be required. Since the available
bandwidth of the shared link is 2Mbps, there will be no queuing delay before the
link. Whereas, if three users transmit simultaneously, the bandwidth required
will be 3Mbps which is more than the available bandwidth of the shared link. In
this case, there will be queuing delay before the link.
c) Probability that a given user is transmitting = 0.2
d) Probability that all three users are transmitting simultaneously =
= (0.2)3 = 0.008. Since the queue grows when all the users are transmitting, the
fraction of time during which the queue grows (which is equal to the probability
that all three users are transmitting simultaneously) is 0.008.
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