REVIEW FOR CHAPTER 2
R1. List five nonproprietary Internet applications and
the application-layer protocols that they use.
Answer:
The Web: HTTP;
file transfer: FTP;
remote login: Telnet;
Network News: NNTP;
e-mail: SMTP.
Video conference: RTP
R2. For a communication session between a pair of
processes, which process is the client and which is the
sever?
• Ans:
• The process which initiates the communication is the
client; the process that waits to be contacted is the server.
R3:
What is the different between network architecture and
application architecture?
Answer:
Network architecture refers to the organization of the
communication process into layers (e.g., the five-layer
Internet architecture). Application architecture, on the
other hand, is designed by an application developer and
dictates the broad structure of the application (e.g., clientserver or P2P).
R4
What information is used by a process running on
one host to identify a process running on another host?
Answer:
The IP address of the destination host and the port
number of the destination socket.
R5
For a P2P file-sharing application, do you agree with
the statement, “There is no notion of client and server
sides of a communication session”? Why or why not?
Answer:
No. As stated in the text, all communication sessions
have a client side and a server side. In a P2P file-sharing
application, the peer that is receiving a file is typically the
client and the peer that is sending the file is typically the
server.
R6. Referring to Figure 2.4, we see that none of the
applications listed in Figure 2.4 requires both no data
loss and timing. Can you conceive of an application
that requires no data loss and that is also highly timesensitive.
• Ans:
network games
• network multimedia applications with high quality
requirements
•
R9.
Why do HTTP, FTP, SMTP, POP3 and IMAP run on top
of TCP rather than UDP?
• Answer:
• The applications associated with those protocols require
that all application data be received in the correct order
and without gaps. TCP provides this service whereas
UDP does not.
R10.
Recall that TCP can be enhanced with SSL to provide
process-to-process security services, including encryption.
Does SSL operate at the transport layer or the application
layer? If the application developer wants TCP to be enhanced
with SSL, what does the developer have to do?
• Ans:
• SSL operates at the application layer. The SSL socket
takes unencrypted data from the application layer,
encrypts it and then passes it to the TCP socket. If the
application developer wants TCP to be enhanced with
SSL, he/she has to include the SSL code in the
application.
R11
What is meant by a handshaking protocol?
Answer:
A protocol uses handshaking if the two communicating
entities first exchange
control packets before sending data to each other. SMTP
uses handshaking at the application layer whereas HTTP
does not.
R15
Suppose Alice, with a Web-based e-mail account (such as
Hotmail or gmail), sends a message to Bob, who accesses his
mail from his mail server using POP3. Discuss how the
message gets from Alice’s host to Bob’s host. Be sure to list
the series of application-layer protocols that are used to move
the message between the two hosts.
Answer:
Message is sent from Alice’s host to her mail server over
HTTP. Alice’s mail server then sends the message to
Bob’s mail server over SMTP. Bob then transfers the
message from his mail server to his host over POP3.
R16
Print out the header of an e-mail message you have
recently received. How many Received: header lines
are there? Analyze each of the header lines in the
message
. Answer:
skipped
R19.
Why is it said that FTP sends control information “outof-band”?
• Ans:
FTP uses two parallel TCP connections, one connection
for sending control information (such as a request to
transfer a file) and another connection for actually
transferring the file. Because the control information is not
sent over the same connection that the file is sent over,
FTP sends control information out of band.
R20
Consider a new peer Alice who joins BitTorrent without
possessing any chunks. Without any chunks, she cannot
become a top-four uploader for any of the other peers, since
she has nothing to upload. How then will Alice get her first
chunk?
Answer:
Alice will get her first chunk as a result of she being
selected by one of her neighbors as a result of an
“optimistic unchoke,” for sending out chunks to her.
P1
True or false?
A. A user requests a Web page that consists of some text and
three images. For this page, the client will send one request
message and receive four response messages.
B. Two distinct Web pages (for example,
www.mit.edu/research.html and
www.mit.edu/student.html )can be sent over the same
persistent connection.
C. With nonpersistent connection between browser and origin
server, it is possible for a single TCP segment to carry two
distinct HTTP request messages.
D. The Date: header in the HTTP response message indicates
when the object in the response was last modified.
Answer:
a) F
b) T
c) F
d) F
P4.
Obtain the HTTP/1.1 specification (RFC 2616). Answer
the following questions:
a. Explain the mechanism used for signaling between
the client and server to indicate that a persistent
connection is being closed. Can the client, the server,
or both signal the close of a connection?
b. What encryption services are provided by HTTP?
• ANS.:
• A. Persistent connections are discussed in section 8 of
RFC 2616 Sections 8.1.2 and 8.1.2.1 of the RFC indicate
that either the client or the server can indicate to the other
that it is going to close the persistent connection. It does
so by including the including the connection-token "close"
in the Connection-header field of the http request/reply.
•
• B. HTTP does not provide any encryption services
P7.
Suppose within your web browser you click on a link to obtain
a web page. Suppose that the IP address for the associated
URL is not cached in your local host, so that a DNS look up is
necessary to obtain the IP address. Suppose that n DNS
servers are visited before your host receives the IP address
from DNS; the successive visits incur a RTT of RTT1 , ...,
RTTN . Further suppose that web page associated with the
link contains exactly one object, a small amount of HTML text.
Let RTT0 denote the RTT between the local host and the
server containing the object. Assuming zero transmission time
of the object, how much time elapses from when the client
clicks on the link until the client receives the object
• ANS.:
• The total amount of time to get the IP address is
• RTT1 +RTT2 +… +RTTN .
• Once the IP address is known, RTT0 elapses to set up
the TCP connection and another
• RTT0 elapses to request and receive the small object.
The total response time is
• 2RTT0 +RTT1 +RTT2 +… +RTTN
P8.
Referring to question (7), suppose the page contains
three very small objects. Neglecting transmission times,
how much time elapses with
(a) nonpersistent HTTP with no parallel TCP
connections,
(b) nonpersistent HTTP with parallel connections,
(c) persistent HTTP with pipeline.
(d) persistent HTTP without pipeline
• Ans:
a)
RTT1 +… + RTTN +2RTT0 +3*2RTT0
=8*RTT0 +RTT1 +… + RTTN
b)
RTT1 +… + RTTN +2RTT0 + 2RTT0
= 4*RTT0 + RTT1 +… + RTTN
c)
RTT1 +… + RTTN +2RTT0 +RTT0
=3*RTT0 +RTT1 +… + RTT0
•
Find the total average
origin
response time
servers
b. Now suppose a cache is
installed in the institution
LAN. Suppose the hit rate
is 0.6. Find the total
response time.
a.
public
Internet
15 Mbps
access link
institutional
network
10 Mbps LAN
institutional
cache
a. Answer:
The time to transmit an object of size L over a link or rate R is
L/R.
6
  L / R  900,000bits / 15*10 bps  0.06( s )
The arrival rate of objects to the access link is 10 objects/second.
Thus the average access delay is 0.06/(1-0.06*10)=0.15 (S)
The total average response time is 0.15+2=2.15 seconds
Answer:
Since the hit rate is 0.6, the traffic on access link is reduced to
(1-0.6)=0.4. The arrival rate of objects to the access link is 4
objects/s.
Thus the average access delay is 0.06/(1-0.06*4)=0.079 (s). The
total average response time for the cache-missing objects is
2.079(s). Therefore we have the total average response time is
0.6*0+0.4*2.079=0.83(s)
b.
P16
Consider distributing a file of F=5 Gbits to N peers. The
server has an upload rate of us=20Mbps, and each peer has a
download rate of di=1Mbps and an upload rate of u. For
N=10,100, and 1,000 and u=100Kbps, 250Kbps,and
500Kbps,prepare a chart giving the minimum distribution time
for each of the combinations of N and u for both client-server
distribution and P2P distribution.
Answer:
For calculating the minimum distribution time for clientserver distribution, we use the following formula:
Similarly, for calculating the minimum distribution time for
P2P distribution, we use the following formula
N
DP 2 P  max{F / us , F / d min , NF / (us   ui )}
i 1
F  5Gbits  5*1024 Mbits
us  20 Mbits,
d min  d i  1Mbits
CS
N
u
10
100
1000
100 Kbps
5120
25600
256000
250 Kbps
5120
25600
256000
500 Kbps
5120
25600
256000
P2P
N
u
10
100
1000
100 Kbps
5120
17066.7
42666.7
250 Kbps
5120
11377.8
18963.5
500 Kbps
5120
7314.3
9846.2
P17
Consider distributing a file of F bits to N peers using
a P2P architecture. Assume a fluid model. For
simplicity assume that dmin is very large, so that peer
download bandwidth is never a bottleneck.
A. suppose that us  (us  u1  ...  uN ) / N . Specify a
distribution scheme that has a distribution time of F/us
B. suppose that us  (us  u1  ...  uN ) / N .Specify a
distribution scheme that has a distribution time of
F/(us+u1+…+uN)
• Ans:
• Assume that u=
• Thus:
1
⋯
,
DP 2 P  max{F / us , NF / u}
• A.
• By assumption: us  u / N
• Divide the file into N parts, with the ith part having size
(ui/(u-us))F. The server transmits the ith part to peer i at
rate ri = (ui/(u-us)) us. Note that r1 + r2 + ….. + rN = us, so
that the aggregate server rate does not exceed the link
rate of the server. Also have each peer I forward the bits it
receives to each of the N-1 peers at rate ri. The aggregate
forwarding rate by peer i is (N-1)ri. We have
ui
( N  1) ri  ( N  1)
us  ui
u  us
In this distribution scheme, peer i receives bits at an
aggregate rate of
ri   rj  us
j i
Thus, each peer receives the file in F/us.
• B.
• By assumption: us  u / N
ui
let : ri 
N 1
u  us
us 
N 1
rN 1 
N
• In this distribution scheme, the file is broken into N+1
parts. The server sends bits from the ith part to the ith peer
(i = 1, …., N) at rate ri. Each peer i forwards the bits
arriving at rate ri to each of the other N-1 peers.
Additionally, the server sends bits from the (N+1) st part at
rate rN+1 to each of the N peers. The peers do not forward
the bits from the (N+1)st part.
N
• The aggregate send rate of the server is  ri  NrN 1  u s
i 1
• Thus, the server’s send rate does not exceed its link rate.
The aggregate send rate of peer i is (N-1)ri=ui
• Thus, each peer’s send rate does not exceed its link rate.
In this distribution scheme, peer i receives bits at an
aggregate rate of
ri  rN 1   rj  u / N
j i
Suppose you open a startup company “starwar” and want to set
up your company network. Your network has the following serve
rs:
1. DNS server: “dns1.starwar.com” with IP as “128.119.12.40”
2. Web server: “starwar.com” with two IP as “128.119.12.55” a
nd “128.119.12.56”. The web server also has a name as “www.st
arwar.com”.
3. Email server: “galaxy.starwar.com” with IP as “128.119.12.60”
Your company’s email address is “username@starwar.com”.
1). What resource records (RRs) do you need to provide to the u
pper-level “.com” DNS server? (2 points)
2). What RRs do you need to put in your company’s DNS server
? (3 points)
• 1)
• （starwar.com, dns1.starwar.com, NS)
•
(dns1.starwar.com, 128.119.12.40, A)
• 2)
• (galaxy.starwar.com,128.119.12.60,A)
• (starwar.com, galaxy.starwar.com, MX)
• (www.starwar.com,starwar.com,CNAME)
• (starwar.com,128.119.12.55,A)
• (starwar.com,128.119.12.56,A)