6/15/2015 Network Fundamentals IV Dr. Tim Lin ECE Department Cal Poly Pomona Add Corporate Logo Here...

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Transcript of 6/15/2015 Network Fundamentals IV Dr. Tim Lin ECE Department Cal Poly Pomona Add Corporate Logo Here...

04/18/23

Network Fundamentals IVNetwork Fundamentals IV

Dr. Tim Lin

ECE Department

Cal Poly Pomona

Add Add CorporateCorporate

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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