# 1, Role of mask

## 1.2 the mask is used to divide the network part and the host part of the ip address

192.168.1.1 255.255.255.0

```11000000	10101000	00000001	00000001	IP address
11111111	11111111	11111111	00000000	Mask
11000000	10101000	00000001				network address
192			168			1
1
```

# 2, Network address

ip address and mask and result of operation
192.168.1.1 255.255.255.0

```11000000	10101000	00000001	00000001	IP address
11111111	11111111	11111111	00000000	Mask
And operation:
11000000	10101000	00000001	00000000	network address
192			168			1			0
```

# 3, Network segment

## 3.1 function of network segment

1. When the host communicates with other devices, use the network segment to judge whether the target host and itself are in the same network segment. If they are in the same network segment, directly find the target host; If you are not in the same network segment, find your own gateway first
2. In the router, when forwarding packets, the router should know which network segment the target host is in, and then check the routing table to find the route of this network segment
3. Neither the host nor the router knows the ip mask of other hosts, so they use their own mask and each other's ip to calculate the network segment

## 3.2 network segment composition

The network segment is composed of network address and mask. If 192.168.1.0/24

```1. Calculation 192.168.1.250/24 this ip Network segment of address
Network address: 192.168.1.250 And 255.255.255.0 And operation results=192.168.1.0

2. It's 192.168.1.250/26 this ip Network segment of address
Network address: 11000000	10101000	00000001	11111010 And 11111111	11111111	11111111	11000000 Doing and Computing=11000000	10101000	00000001	11000000(192.168.1.192)
Network segment=192.168.1.192/26

```

# 4, Class (main class) ip

Classified ip includes unicast ip, multicast ip and special ip

## 4.1 unicast ip address

```A Class:
First 8 bit Is 0 xxx xxxx
Minimum first 8 bit Address 0000 0000
Maximum first 8 bit Address 0111 1111
First paragraph ip Address range 0-127，The special addresses 0 and 127 shall be removed; 0.0.0.0/0 Represents all network segments, and represents the default route in the routing table; one hundred and twenty-seven.x.x.x Is the local loopback address; So the final available range is 1-126

B Class:
First 16 bit It's 10 xx xxxx
Minimum first 8 bit Address 1000 0000
Maximum first 8 bit Address 1011 1111
First paragraph ip Address range 128-191

C Class:
First 24 bit It's 110 x xxxx
Minimum first 8 bit Address: 11000000
Maximum first 8 bit Address 1101 1111
First paragraph ip Address range 192-223
```

## 4.2 special address

### 4.2.2 local loopback address 127.x.x.x

Full f of layer 2 broadcasting

The network bit address remains unchanged, and the host bit address is all 1

### 4.2.5 network address

The network bit address remains unchanged, and the host bit address is all 0

# 5, Private network address

```A Class:
10.0-255.0-255.0-255 Or 10.0.0.0/8

B Class:
172.16-31.0-255.0-255 Or 172.16.0.0/12

C Class:
192.168.0-255.0-255 Or 192.168.0.0/16
```

# 6, Classless address

## 6.1 in class addresses, the subnet mask of a group of eight bit s is fixed, either 0 or 255, but the mask without class address is not fixed

Therefore, the 8bit of the classless ip address mask can only be 012819222424825254255

## 6.2 formula for calculating the number of available host addresses in a network segment

```2^(32-Mask)-2
Minus 2 is the subtracted segment address and segment broadcast address
```

172.16.0.0/16
2^(32-16)-2 = 65536-2 = 65534

## 6.3 how many network segments are there after VLSM calculation

```2^(A-B)
A Is the mask after lengthening
B Is the class mask before getting longer

```

How many network segments are there after 172.16.0.0/16 becomes 172.16.0.0/24
2 ^ (24-16) = 2 ^ 8 = 256 network segments

## 6.4 calculate the information of each subnet after VLSM (network segment, network address, local broadcast address, available ip range)

Calculate the ip address 192.168.1.20/28 and find all subnet information divided by this mask

1. Calculate how many ip addresses this subnet has
2^(32-28)=16

2. Each network segment has 16 ip addresses, so the first ip address of each subnet from 0 to 255 is

```16	multiply	n	=
16		0	=	0
16		1	=	16
16		2	=	32
16		3	=	48
16		4	=	64
16		5	=	80
16		6	=	96
16		7	=	112
16		8	=	128
16		9	=	144
16		10	=	160
16		11	=	176
16		12	=	192
16		13	=	208
16		14	=	224
16		15	=	240
16		16	=	256
```
1. Now that you know the first ip address of each subnet, you know the information of each subnet
```Network segment					network address			Local broadcast address		available ip Range
192.168.1.0/28		192.168.1.0		192.168.1.15	192.168.1.1-14
192.168.1.16/28		192.168.1.16		192.168.1.31	192.168.1.17-30
192.168.1.32/28		192.168.1.32		192.168.1.47	192.168.1.33-46
192.168.1.48/28		192.168.1.48		192.168.1.63	192.168.1.49-62
192.168.1.64/28		192.168.1.64		192.168.1.79	192.168.1.65-78
192.168.1.80/28		192.168.1.80		192.168.1.95	192.168.1.81-94
192.168.1.96/28		192.168.1.96		192.168.1.111	192.168.1.97-110
192.168.1.112/28		192.168.1.112		192.168.1.127	192.168.1.113-126
192.168.1.128/28		192.168.1.128		192.168.1.143	192.168.1.127-142
192.168.1.144/28		192.168.1.144		192.168.1.159	192.168.1.145-158
192.168.1.160/28		192.168.1.160		192.168.1.175	192.168.1.161-174
192.168.1.176/28		192.168.1.176		192.168.1.191	192.168.1.177-190
192.168.1.192/28		192.168.1.192		192.168.1.207	192.168.1.193-206
192.168.1.208/28		192.168.1.208		192.168.1.223	192.168.1.209-222
192.168.1.224/28		192.168.1.224		192.168.1.239	192.168.1.225-238
192.168.1.240/28		192.168.1.240		192.168.1.255	192.168.1.241-254

```

## 6.5 different subnet masks may not be a network segment, because both parties have their own subnet masks to calculate which network segment the other party is

Are 192.168.1.100/28 and 192.168.1.101/29 a network segment?

1. 192.168.1.100/28 judge whether 192.168.1.101 is the same network segment as yourself
```ip	192.168.1.100
192.168.1.01100100
Network segment	192.168.1.96/28
Of this network segment ip Range	192.168.1.97-110
So at 192.168.1.100/28 It seems that he and 192.168.1.101 Is in the same network segment
```
1. 192.168.1.101/29 judge whether 192.168.1.100 is the same network segment as yourself
```3. 192.168.1.101/29 this ip Addressable vlsm The number of hosts per subnet is 2^(32-29)=8 individual
4. Therefore, each subnet segment is:
8	multiply	n	be equal to
8	0	0
8	1	8
8	2	16
.
.
.
8	12	96
8	13	104
5. So 192.168.1.101/29 this ip The subnet segment of the address is 192.168.1.96/29， ip The address range is 192.168.1.97-102
6. So at 192.168.1.101/29 It seems that he and 192.168.100 It is also the same network segment
```

## 6.6 the calculation uses CIDR to aggregate several small networks into a large network

Suppose a group of class C addresses are 192.168.8.0/24-192.168.15.0/24. If CIDR is used to aggregate these addresses into a network, what should the network address and subnet mask be?

1. First convert the third octet of class C address to binary:
```192.168.8.0 	192.168.00001 000.0
192.168.9.0 	192.168.00001 001.0
192.168.10.0 	192.168.00001 010.0
192.168.11.0 	192.168.00001 011.0
192.168.12.0 	192.168.00001 100.0
192.168.13.0 	192.168.00001 101.0
192.168.14.0 	192.168.00001 110.0
192.168.15.0 	192.168.00001 111.0
```
1. The first 5bit00001 is the same as everyone. It is the network address and mask of the third octet
192.168.00001000.0/21
192.168.8.0/21

## 6.7 special network segment

### 6.7.1 Internet segment

/30
Mask 30, with only two available ip addresses, is used for the interconnection of two routers

### 6.7.2 logical network segment

/32
Mask 32, loop, router ID, BGP, establish neighbors, manage addresses

### 6.7.3 business network segment

Network segments with more than 2 available ip addresses are used for various terminals

## 6.8 two algorithms for converting CIDR of several network segments into super network

There are three C network segments 192.168.1.0192.168.2.0192.168.3.0
Aggregate into a Supernet

### 6.8.1 method 1: take the same bit in the last 8bit network bit as the network bit; Different bits become host bits

```						192.168.1.0/24	192.168.0000 0001.0
192.168.2.0/24	192.168.0000 0010.0
192.168.3.0/24	192.168.0000 0011.0
Take the third group 8 in the network bit bit The same part is still a network bit				0000 00
Take the third group 8 in the network bit bit Different parts become host bits					   xx
The aggregated subnet mask is/22
The aggregated network address is the third group 8 bit Results with 192.168.0000 0000.0
The aggregated Supernet is 192.168.0.0/22

```

### 6.8.1 method 2 super network is to lend the original network address bit to the host bit for use

1. If 7 bits in the third 8bit bit are borrowed as host bits, and the mask is / 17, there will be 2 ^ 7 = 128 hosts
```128	multiply	n	be equal to
128		0	0
128		1	128
128		2	256

This division will have 2^1=2 The subnets are:
192.168.0.0/17
192.168.0.128/17
Network segment 192 of the first segment.168.0.0/17 Including subnets
192.168.0.0
192.168.1.0
.
.
.
192.168.127.0
```
1. If one bit of the third 8bit bit is borrowed as the master bit, and the mask is / 23, there will be 2 ^ 1 = 2
```2	multiply	n	be equal to
2		0	0
2		1	2
2		2	4
.
.
.
2		128	256
This division will have 2^7=128 The subnets are:
192.168.0.0/23
192.168.0.2
192.168.0.4
.
.
.

Network segment 192 of the first segment.168.0.0/23 It only contains two network segments, and there are no 2 and 3 network segments, so it is not satisfied
192.168.0.0
192.168.1.0
```
1. If 2 bits in the third 8bit bit are borrowed as host bits, and the mask is / 22, there will be 2 ^ 2 = 4
```4	multiply	n	be equal to
4		0	0
4		1	4
4		2	8
.
.
.
4		64	256
This division will have 2^6=64 The subnets are:
192.168.0.0/22
192.168.4.0
192.168.8.0
Network segment 192 of the first segment.168.0.0/22 Contains 192.168.0.0,192.168.1.0,192.168.2.0,192.168.3.0 So satisfied
```
1. Mask 22 satisfied

Keywords: Linux Operation & Maintenance udp

Added by kfir91 on Mon, 22 Nov 2021 22:49:48 +0200