"Computer Network" Final Review Notes

1. Labels for some English nouns (easy to remember)

ICMP : Reject PING sniffing, unreliable messages, TCP/IP errors need to be reported, errors in control transmission
FTP : Default number 21, file transfer
POP3 : take out mail, accept mail, obtain mail remotely
Router : multi-layer protocol support , not just point-to-point transmission
RTP : TCP/IP Vector Routing Protocol
VLAN : broadcast domain
DNS : domain name and IP conversion
RARP : MAC address to IP
ARP : IP to MAC address
ISP : Internet service provider
SDV : inter-adjacent layer
SMTP : electronic Protocol for mutual transfer of mails
IP : low-layer protocol, not in order, reorganization of destination host, packet switching, datagram
OSPF : intra-domain routing protocol
ARPANET : the earliest Internet
TCP : connection, communication regulations, Internet layer solves communication problems, data unit ( part)

2. OSI seven-layer protocol

Physical layer : actual communication, not involved in data encapsulation, the layer where the repeater is located,
data link layer : frame checksum, dividing the bit stream into frames, PPP protocol
network layer : routing, congestion control, interconnection, determining the path to pass Subnet, router selection
Transport layer : TCP/UDP, inter-process message transfer
Session layer :
Presentation layer :
Application layer : Users at both ends transfer files

3. Comprehensive questions

3.0 Knowledge point reserve

IP address type	binary	address range	Default subnet mask
Class A address	000	1-126	255.0.0.0
		127 is used for loopback test address
Class B address	100	128-191	255.255.0.0
Class C address	110	192-223	255.255.255.0
Class D address (multicast)	111	224-239	255.255.255.255

IP: network number + host number (network number is subnet mask 255).
If the host number is all 0 , it is calledNetwork segment address
If the host number is all 1 , it is calledbroadcast address
These two are not available

---

Regarding the calculation of CRC, the steps are:
Step 1: Determine based on the polynomial coefficient of G(x), such as $x^3+x^1+1$ , then the binary string is1011; if$x^3+1$ , then the binary string is,1001
at the end), the rule isStep 3, put the remainder at the end of the data sent, which is the answer3个010000101000相同为0，相异为1
10000101111

---

3.1 In the Internet network, the IP address of a certain computer is 11001010.01100000.00101100.01011000. Please answer the following questions

The first question: Convert binary to decimal, $2^0-2^7$
The second question: Since the IP address starts with110, it is concluded that it is a class C address.
The third question: Since it is a class C address, the network mask is.255.255.255.0
The fourth question: Since there is no division, the host number is the last segment of the IP address.

3.2 Assume that the data the sender wants to send is 10000101. The generator polynomial using CRC is G(X)=X3+X+1. Try to find the bit string actually sent by the sender?

3.3 The IP address of a certain host on a Class C network is 200.192.77.88. If the network needs to be divided into 14 available subnets, each subnet can access up to 14 hosts.

(1) What should be the subnet mask?
(2) In the subnet where the host is located, what is the range of IP addresses that can be assigned to the host?

The first question: Since it is a Class C address, it is 255.255.255.0
The second question:
First of all, if there are 14 stations, it must be taken up by the exponent of 2, 16that is, in the last 8-bit network address, the first 4 digits For the network address 11110000
00000000, the network segment address of the first subnet
11111111is the broadcast address of the sixteenth subnet.
The last four digits are the IP range of the host number of the current network segment.
For calculation, 16 5 = 80, 16 6 = 96
, so 88In the 5th subnet, i.e. between 80-95
due toNetwork segment address: host number all 0 and broadcast address: host number all 1Not available
, so the range of IP addresses that can be assigned to hosts is 81-94
i.e.200.192.77.81-200.192.77.94

3.4 Transmitting ASCII code information "Good morning!" through IEEE 802.3 LAN, if encapsulated into a MAC frame

(1) How many valid bytes are there in the data field of this frame? (2) How many bytes need to be filled?

3.5 The data field of a UDP user datagram is 9652 bytes and must be transmitted using Ethernet. Question: How many datagram fragments should be divided into? Specifies the length of the data field and the value of the fragment offset field for each datagram fragment.

4. Some words about this subject

I am majoring in electronic information, so our study of computer network will not be as in-depth as computers. We only need to be able to memorize and memorize, understand all the big topics, and clearly distinguish some Class A addresses, Class B addresses, etc. Subnet, what mask, etc. When memorizing multiple-choice questions, consciously summarize, so you need to summarize the knowledge points yourself. Just like many of my 一些英文名词的标签（方便记忆）knowledge points overlap, so sorting them out can help draw inferences.