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Computer Networking: A Top-Down Approach,
6th Edition
Solutions to Review Questions and Problems
Version Date: May 2012
This document contains the solutions to review questions and problems for the 5th
edition of Computer Networking: A Top-Down Approach by Jim Kurose and Keith Ross.
These solutions are being made available to instructors ONLY. Please do NOT copy or
Chapter 1 Review Questions
1. There is no difference. Throughout this text, the words “host” and “end system” are
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
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
5. HFC bandwidth is shared among the users. On the downstream channel, all packets
6. In most American cities, the current possibilities include: dial-up; DSL; cable
modem; fiber-to-the-home.
8. Today, Ethernet most commonly runs over twisted-pair copper wire. It also can run
over fibers optic links.
9. Dial up modems: up to 56 Kbps, bandwidth is dedicated; ADSL: up to 24 Mbps
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
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
12. A circuit-switched network can guarantee a certain amount of end-to-end bandwidth
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
14. If the two ISPs do not peer with each other, then when they send traffic to each other
they have to send the traffic through a provider ISP (intermediary), to which they
15. Google's private network connects together all its data centers, big and small. Traffic
between the Google data centers passes over its private network rather than over the
public Internet. Many of these data centers are located in, or close to, lower tier ISPs.
Therefore, when Google delivers content to a user, it often can bypass higher tier
ISPs. What motivates content providers to create these networks? First, the content
16. The delay components are processing delays, transmission delays, propagation
18. 10msec; d/s; no; no
19. a) 500 kbps
20. End system A breaks the large file into chunks. It adds header to each chunk, thereby
generating multiple packets from the file. The header in each packet includes the IP
21. The maximum emission rate is 500 packets/sec and the maximum transmission rate is
350 packets/sec. The corresponding traffic intensity is 500/350 =1.43 > 1. Loss will
22. Five generic tasks are error control, flow control, segmentation and reassembly,
23. The five layers in the Internet protocol stack are – from top to bottom – the
24. Application-layer message: data which an application wants to send and passed onto
the transport layer; transport-layer segment: generated by the transport layer and
25. Routers process network, link and physical layers (layers 1 through 3). (This is a little
26. a) Virus
Requires some form of human interaction to spread. Classic example: E-mail
27. Creation of a botnet requires an attacker to find vulnerability in some application or
system (e.g. exploiting the buffer overflow vulnerability that might exist in an
application). After finding the vulnerability, the attacker needs to scan for hosts that
28. Trudy can pretend to be Bob to Alice (and vice-versa) and partially or completely
modify the message(s) being sent from Bob to Alice. For example, she can easily
Chapter 1 Problems
Problem 1
There is no single right answer to this question. Many protocols would do the trick.
Here's a simple answer below:
Messages from ATM machine to Server
Msg name purpose
-------- -------
PASSWD Ask user for PIN (password)
OK last requested operation (PASSWD, WITHDRAWL)
OK
ERR last requested operation (PASSWD, WITHDRAWL)
in ERROR
BYE -------------->
<------------- BYE
In situation when there's not enough money:
HELO (userid) --------------> (check if valid userid)
<------------- PASSWD
PASSWD <passwd> --------------> (check password)
<------------- OK (password is OK)
BALANCE -------------->
<------------- AMOUNT <amt>
<------------- BYE
Problem 2
At time N*(L/R) the first packet has reached the destination, the second packet is stored
Problem 3
a) A circuit-switched network would be well suited to the application, because the
application involves long sessions with predictable smooth bandwidth requirements.
Problem 4
a) Between the switch in the upper left and the switch in the upper right we can have 4
connections. Similarly we can have four connections between each of the 3 other
Problem 5
Tollbooths are 75 km apart, and the cars propagate at 100km/hr. A tollbooth services a
car at a rate of one car every 12 seconds.
Problem 6
a)
smd prop /
seconds.
b)
RLd
tra ns
/
seconds.
Problem 7
Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the
packet must be generated. This requires
Problem 8
a) 20 users can be supported.
b)
1.0 p
.
c)
n
n
pp
n
¸
¸
·
¨
¨
§
120
1
120
.
Problem 9
¹
©
Problem 10
The first end system requires L/R1 to transmit the packet onto the first link; the packet
propagates over the first link in d1/s1; the packet switch adds a processing delay of dproc;
Problem 11
Because bits are immediately transmitted, the packet switch does not introduce any delay;
Problem 12
Problem 13
a) The queuing delay is 0 for the first transmitted packet, L/R for the second transmitted
packet, and generally, (n-1)L/R for the nth transmitted packet. Thus, the average delay
for the N packets is:
Problem 14
a) The transmission delay is
RL /
. The total delay is
RL
L
IL
/
Problem 15
Problem 16
The total number of packets in the system includes those in the buffer and the packet that
Problem 17
