第一部分：Multiple choice （共8道题）
1、 Which of the following is not Internet application layer Protocol?
A、SMTP
B、TCP
C、FTP
D、HTTP
答案：B
2 、DNS runs on top of ( ) and uses the port number 53.
A、 UDP
B、 HTTP
C、 TCP
D、 none of the above
答案：A
题目解析：
DNS---UDP--IP
3 FTP runs over ( ).
A、 UDP
B、 TCP
C、 HTTP
D、 none of the above
答案：B
4 Which service of the following can be provided by TCP？
A、 Reliable data transfer
B、 Throughput
C、 Timing
D、 Security
答案：A
5 Which of the following is the default port number of TCP control connection in FTP?
A、 20
B、 21
C、 25
D、 80
答案：B
题目解析：
20: data connection port using by FTP
6、Which tool can capture network packets and display that packet data?
A、Internet Explorer
B、 Wireshark
C、 FileZilla
D、 Firefox
答案：B
题目解析：
Wireshark: capture bidirectional packets on NIC
7 Which of the following port number is the default port number of HTTP? ( ).
A、 23
B、 21
C、 20
D、 80
答案：D
8 The application of ( ) is loss-tolerant.
A、 stored audio/video
B、 file transfer
C、 Web documents
D、 text messaging
答案：A
第二部分：True or False （共9道题）
1 The service of TCP is unreliable data transmission.
A、 True
B、 FALSE
答案：B
2 Status code 404 denotes that requested document does not exist on the server according the HTTP protocol.
A、 True
B、 FALSE
答案：A
3 Local DNS name servers cache resource records, but discard them after a period of time.
A、 True
B、 FALSE
答案：A
4 For the Web application, the user agent is the user’s keyboard.
A、 True
B、 FALSE
答案：B
5 The Date: header in the HTTP response message indicates when the object in the response was last modified.
A、 True
B、 FALSE
答案：B
6 Suppose that host A wants to send data over TCP to host B, and host B wants to send data to host A over TCP. Two separate TCP connections - one for each direction - are needed.
A、 True
B、FALSE
答案： B
7 For a P2P file-sharing application, There is no notion of client and server sides of a communication session.
A、True
B、FALSE
答案：B
8 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.
A、 True
B、FALSE
答案：B
9 Socket programming with UDP, the server process must run before the client process.
A、 True
B、 FALSE
答案：B
第三部分：Blank filling （共1道题）
1 [ 填空题 ]
The full text in English of HTTP is Hypertext Transfer Protocol.
The full text in English of FTP is ( ).
The full text in English of SMTP is ( ).
The full text in English of IMAP is ( ).
The full text in English of POP3 is ( ).
The full text in English of DNS is ( ).
参考答案：
File Transfer Protocol 或 file transfer protocol 、
Simple Mail Transfer Protocol 或 simple mail transfer protocol 、
Internet Mail Access Protocol 或 Internet mail access protocol 、
Post Office Protocol Version 3 或 post office protocol version 3 、
Domain Name System 或 domain name system
第四部分：comprehensive problems （共3道题）
1 [ 组合题 ]
Consider the following string of ASCII characters that were captured by Wireshark when the browser sent an HTTP request message (i.e., this is the actual content of an HTTP request message). The characters <cr><lf> are carriage return and line-feed characters. Answer the following questions, indicating where in the HTTP request message below you find the answer.
GET /HTML/intro/ HTTP/1.1<cr><lf>
Host: www.nsu.edu.cn <cr><lf>
User-Agent: Mozilla/5.0 (Windows;U; Windows NT 5.1; en-US; rv:1.7.2) <cr><lf>
Accept:ext/xml, application/xml, application/xhtml+xml, text/html;q=0.9, text/plain; q=0.8, image/png,*/*; q=0.5<cr><lf>
Accept-Language: en-us,en;q=0.5<cr><lf>
Accept-Encoding: zip,deflate<cr><lf>
Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.7<cr><lf>
Keep-Alive: 300<cr><lf>
Connection:close<cr><lf>
<cr><lf>
a. [ 填空题 ]
What is the method of the HTTP request message? ( )
参考答案：
GET 或 Get 或 get
b. [ 填空题 ]
What is the whole URL of the document requested by the browser? ( )
参考答案：
http://www.nsu.edu.cn/HTML/intro/|||www.nsu.edu.cn/HTML/intro/
c. [ 填空题 ]
What version of HTTP is the browser running?
参考答案：
HTTP/1.1 或1.1
d. [ 判断题 ]
The browser request a persistent connection.
A、 True
B、 false
参考答案：B
e. [ 判断题 ]
you can know the IP address of the host on which the browser is running.
A、 true
B、 false
参考答案：B
题目解析：
only see IP from IP packet.
2 [ 组合题 ]
Suppose within your Web browser you click on a link to obtain a Web page. The IP address for the associated URL is not cached in your local host, so a
DNS lookup 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 an RTT of RTT1, . . ., RTTn.
a. [ 单选题 ]
Further suppose that the Web page associated with the link contains exactly one object, consisting of 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?
A、 RTT1+RTT2+...+RTTn+RTT0
B、 RTT1+RTT2+...+RTTn+2RTT0
C、 RTT1+RTT2+...+RTTn+3RTT0
D、 RTT1+RTT2+...+RTTn+4RTT0
参考答案：B
题目解析：
ONLY html file: 2+0*n
b. [ 单选题 ] suppose the HTML file references 4 very small objects on the same server. Neglecting transmission times, how much time elapses with
a. Non-persistent HTTP with no parallel TCP connections?
A、 RTT1+RTT2+...+RTTn+2RTT0
B、 RTT1+RTT2+...+RTTn+4RTT0
C、 RTT1+RTT2+...+RTTn+6RTT0
D、 RTT1+RTT2+...+RTTn+10RTT0
参考答案：D
题目解析：
2+4*2=10
c. [ 单选题 ]
suppose the HTML file references 4 very small objects on the same server. Neglecting transmission times, how much time
elapses with
b. Non-persistent HTTP with the browser configured for 2 parallel connections?
A、RTT1+RTT2+...+RTTn+2RTT0
B、 RTT1+RTT2+...+RTTn+4RTT0
C、 RTT1+RTT2+...+RTTn+6RTT0
D、 RTT1+RTT2+...+RTTn+10RTT0
参考答案：C
题目解析：
2+2*(4/2)=6
d. [ 单选题 ]
suppose the HTML file references 4 very small objects on the same server. Neglecting transmission times, how much time
elapses with
c. Persistent HTTP without pipelining?
A、RTT1+RTT2+...+RTTn+2RTT0
B、 RTT1+RTT2+...+RTTn+4RTT0
C、 RTT1+RTT2+...+RTTn+6RTT0
D、 RTT1+RTT2+...+RTTn+8RTT0
参考答案：C
题目解析：
2+4*1=6
e. [ 单选题 ]
suppose the HTML file references 4 very small objects on the same server. Neglecting transmission times, how much time
elapses with
d. Persistent HTTP with pipelining?
A、 RTT1+RTT2+...+RTTn+RTT0
B、 RTT1+RTT2+...+RTTn+2RTT0
C、 RTT1+RTT2+...+RTTn+3RTT0
D、 RTT1+RTT2+...+RTTn+4RTT0
参考答案：C
题目解析：
2+1*1=3
3 [ 组合题 ]
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=(from the graph above, d1=d2=...=di, and u1=u2=...=u).
Notice, please use 1G = 1000 M = 1000000 K here, and directly remove the decimal part for the final result(for example, 100/3 = 33 and 200/3 = 66).
a. [ 填空题 ]
For N=10, u=100Kbps, the minimum distribution time for client-server distribution is ( )s.
For N=100, u=100Kbps, the minimum distribution time for client-server distribution is ( )s.
For N=1000, u=100Kbps, the minimum distribution time for client-server distribution is ( )s.
For N=1000, u=250Kbps, the minimum distribution time for client-server distribution is ( )s.
For N=1000, u=500Kbps, the minimum distribution time for client-server distribution is ( )s.
参考答案：
5000
、 25000
、 250000
、 250000
、 250000
题目解析：
For calculating the minimum distribution time for client-server distribution, we use the following formula:
Dcs = max {NF/us, F/dmin}
?Where, F = 5 Gbits = 5 * 1000 Mbits
us = 20 Mbps
dmin = di = 1 Mbps
b. [ 填空题 ]
For N=10, u=100Kbps, the minimum distribution time for P2P distribution is ( )s.
For N=100, u=100Kbps, the minimum distribution time for P2P distribution is ( )s.
For N=1000, u=100Kbps, the minimum distribution time for P2P distribution is ( )s.
For N=1000, u=250Kbps, the minimum distribution time for P2P distribution is ( )s.
For N=1000, u=500Kbps, the minimum distribution time for P2P distribution is ( )s.
参考答案：
5000
、 16666
、 41666
、 18518
、 9615
题目解析：
for calculating the minimum distribution time for P2P distribution, we use the following formula: (see text book):
Dp2p = max{F/us, F/dmin, NF/(us+N*u))
Where, F = 5 Gbits = 5 * 1000 Mbits
us = 20 Mbps
dmin = di = 1 Mbps, u = 100Kbps = 0.1Mbps