03/15/22 Network Fundamentals IV Network Fundamentals IV Dr. Tim Lin ECE Department Cal Poly Pomona Add Add Corporate Corporate Logo Logo Here Here EXIT >
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04/18/23
Network Fundamentals IVNetwork Fundamentals IV
Dr. Tim Lin
ECE Department
Cal Poly Pomona
Add Add CorporateCorporate
LogoLogoHereHere
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Agenda• What is Computer Network• ISO / OSI model• Internet Organizations• IP Addresses
– v4• Classful• Classless
– V6• IP Protocol
– Header
• ICMP Protocol• ARP Protocol
• UDP Protocol• TCP Protocol
– Header– Flow Control– Error Control– Congestion Control
• FTP Protocol• HTTP Protocol• SMTP Protocol
• Technology (LAN)
Figure 21.2 Opening the control connection
Figure 21.3 Creating the data connection
Figure 21.6 Command processing
Figure 21.8 Example 21.1
220 (Service ready)
USER forouzan
LIST /usr/user/forouzan/reports
PASS xxxxxx
125 (Data connection OK)
331 (User name OK. Password?)
PORT 8888
150 (Data connection opens shortly)
230 (User login OK)
1
2
3
4
5
6
7
8
9
List of files or directories
List of files or directories
10
11
QUIT
226 (Closing data connection)
221 (Service closing)
12
13
14
DATATRANSFER
We show an example of anonymous FTP. We assume that some public data are available at internic.net.
ExampleExample 21.4
Figure 22.3 Browser
Figure 22.4 URL
Host can be a host name, an IP address or localhost
Figure 22.10 HTTP transaction
1
Request messageRequest line
Headers
Body
A blank line
2
Response message
Status line
Headers
Body
A blank line
Figure 22.13 Example 22.4
Figure 22.14 Example 22.5
HTTP uses ASCII characters. The following shows how a client can directly connect to a server using TELNET, which logs into port 80.
ExampleExample 22.6
Figure 23.1 First scenario
1
2
Figure 23.2 Second scenario
1
2 3 4
5
Figure 23.6 Format of an email
Figure 23.7 E-mail address
Figure 23.11 Message transfer
Figure 3.1 IEEE standard for LANs
Figure 3.2 Ethernet Frame
Figure 3.3 Maximum and minimum lengths
Figure 3.5 Unicast and multicast addresses
multicast: 1unicast: 0
Define the type of the following destination addresses: a. 4A:30:10:21:10:1A b. 47:20:1B:2E:08:EE c. FF:FF:FF:FF:FF:FF
SolutionTo find the type of the address, we need to look at the secondhexadecimal digit from the left. If it is even, the address is unicast. If it is odd, the address is multicast. If all digits are F’s, the address is broadcast. Therefore, we have the following:a. This is a unicast address because A in binary is 1010 (even).b. This is a multicast address because 7 in binary is 0111 (odd).c. This is a broadcast address because all digits are F’s.
ExampleExample 3.1
Figure 3.6 Ethernet evolution through four generations
10Base2, 10Base5, 10BaseT
Figure 3.10 Standard Ethernet implementation
Figure 3.7 Space/time model of a collision in CSMA
Time Time
BA C D
B startsat time t1
t1
Area whereA’s signal exists
C startsat time t2
t2
Area whereB’s signal exists
Area whereboth signals exist
Figure 3.8 Collision of the first bit in CSMA/CD
Figure 3.13 Basic service sets (BSSs)
Figure 3.14 Extended service sets (ESSs)
Figure 3.15 CSMA/CA flow diagram
Figure 3.17 Frame format
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