foreword
I have been picking up computer network principles for the past few days, but found that many IP calculation topics can’t be calculated, and then Baidu (Google) couldn’t find a better article for a long time, and it took me almost 3-4 hours to solve it. Come out, so just write an article to summarize it!
start
Before you start, you need to understand the basic concept of IP address
Basic concepts of IP addresses
1. IP address range, you need to know what type of IP address it is
A 0.0.0.0 - 127.255.255.255
B 128.0.0.0 - 191.255.255.255
C 192.0.0.0 - 223.255.255.255
D 224.0.0.0 - 239.255.255.255
E 240.0.0.0 - 255.255.255.2552. Combination of IP address, you need to know what combination it is made of
ps: Mainly look at the three types of addresses A, B, and C
They are composed of: network number + host number
3. To know what is and (&) operation, what is or (||) operation
And operation (&): that is, both sides must be 1 to get 1
1 1 = 1、1 0 = 0、0 1 = 0、0 0 = 0
Or operation (||): that is, if one of the two parties is 1, then 1 is obtained
1 1 = 1、1 0 = 1、0 1 = 1、0 0 = 0
4. Know what 11111111 stands for
255.255.255.255 = 11111111 11111111 11111111 11111111
So 8 add up = 255
1 1 1 1 1 1 1 1
128 64 32 16 8 4 2 1
topic one
It is known that the IP address of a host in a certain subnet is: 203.123.1.135, and the subnet mask is: 255.255.255.192.
(1). What is the subnet address of this subnet?
(2). What is the direct broadcast address of this subnet?
(3). What is the number of IP addresses in this subnet?
(4). What is the range of IP addresses that can be allocated?
(1). Subnet address = IP address and (&) operation subnet mask
203.123.1.135 11001011 01111011 00000001 10000111
255.255.255.192 11111111 11111111 11111111 11000000
203.123.1.128 11001011 01111011 00000001 10000000
Subnet address: 203.123.1.128
(2).Broadcast address = IP address or (||) operation The inverse of the subnet mask
The inverse code of the subnet mask: it is to turn all 1 into 0, and 0 into 1
255.255.255.192 11111111 11111111 11111111 11000000
0.0.0.127 00000000 0000000 0000000 00111111
Formal calculation: P address or (||) operation The inverse code of the subnet mask
203.123.1.128 11001011 01111011 00000001 10000000
0.0.0.127 000000 0000000 0000000 00111111
203.123.1.191 11001011 01111011 00000001 10111111
Broadcast address: 203.123.1.191
(3). Number of IP addresses = 2^host number (such as: )
Problem-solving idea: You need to memorize the network number and host number of the above-mentioned A, B, and C addresses, and then check how many 1s there are in the subnet mask, and then compare the network numbers. If there are fewer, use the host number to supplement. The rest of the host number is the number of IP addresses.
The slave host IP (203.123.1.135) is a class C address, the network number of the class C address is 24 bits, and the host number is 8 bits
Subnet mask to binary: 11111111 11111111 11111111 11000000, there are 26 1s, so: the subnet mask number is 26 bits
Subnet mask number - network number = 2, so the network number needs to borrow 2 from the host number, so 8-2=6, the host number is 6 digits
So the number of IP addresses = = 2*2*2*2*2*2 = 64
Number of IP addresses: 64
(4). IP assignable range = subnet address (+1) - broadcast address (-1)
Assignable IP address range: 203.123.1.129 - 203.123.1.190
Topic two
Known IP network 202.112.14.0/26
(1). Write down the number of its assignable IP addresses
(2). Suppose it is divided into three subnets, the first subnet has no less than 30 assignable IP addresses, the second subnet has no less than 14 assignable IP addresses, and the third subnet has no less than 14 assignable IP addresses. There are no less than 10 IP addresses, write their subnet address, subnet mask, and broadcast address respectively
(1). Number of IP addresses = 2^host number (such as: )
First write out the subnet mask , from 202.112.14.0/26, we know that the subnet mask number is 26 bits (IP/26: / is followed by the subnet mask number)
11111111 11111111 11111111 11000000
Subnet mask: 255.255.255.192
From the IP address (202.112.14.0), it is known that it is a class C address, the network number is 24 digits, the host number is 8 digits, the subnet mask number is 26 digits (see above), subnet mask number-network number= 2. The network number needs to borrow 2 from the host number, and the host number is 8-2=6 digits
So the number of IP addresses = = 2*2*2*2*2*2=64
The number of IP addresses is 64
(2).
No less than 30 problem-solving ideas
From the above we know how to calculate the number of IP addresses, let’s infer other cases from one instance, first calculate the lowest number of IP addresses, then look at the number of host numbers, and then deduce the subnet mask number, and then reverse the subnet address according to the subnet mask Perform an OR operation to obtain the broadcast address
2*2*2*2*2 = 32
Therefore, the host number is 5 digits, the network number is 24 digits, and the formula for calculating the subnet mask number is: original network number + (original host number - current host number)
24 + (8-5) = 24+3 = 27, so the subnet mask number is 27 bits
11111111 11111111 11111111 11100000
So the following result is obtained:
Subnet mask: 255.255.255.224
Subnet address: 202.112.14.0/27
Broadcast address = subnet mask inversion and subnet address for OR operation
11111111 11111111 11111111 11100000
00000000 00000000 00000000 00011111
Inverse subnet mask: 0.0.0.31
202.112.14.0 11001010 01110000 00001110 00000000
0.0.0.31 00000000 00000000 00000000 00011111
202.112.14.31 11001010 01110000 00001110 00011111
Subnet mask: 255.255.255.240
Subnet address: 202.112.14.0/28
Broadcast address: 202.112.14.31
No less than 14 problem-solving ideas
The logic of solving the problem is the same as above, but the subnet address = the broadcast address of the above problem + 1
2*2*2*2 = 16
24+(8-4) = 28
11111111 11111111 11111111 11110000
Subnet mask: 255.255.255.240
Subnet address: 202.112.14.32/28
Inverse subnet mask:
11111111 11111111 11111111 11110000
00000000 00000000 00000000 00001111
0.0.0.15
Broadcast address:
202.112.14.32 11001010 01110000 00001110 00100000
0.0.0.15 00000000 00000000 00000000 00001111
202.112.14.47 11001010 01110000 00001110 00101111
Subnet mask: 255.255.255.240
Subnet address: 202.112.14.32/28
Broadcast address: 202.112.14.47
No less than 10 problem-solving ideas
The logic of solving the problem is the same as above, but the subnet address = the broadcast address of the above problem + 1
2*2*2*2=16
24+(8-4) = 28
11111111 11111111 11111111 11110000
Subnet mask: 255.255.255.240
Subnet address: 202.112.14.48/28
Inverse subnet mask:
11111111 11111111 11111111 11110000
00000000 00000000 00000000 00001111
0.0.0.15
Broadcast address:
202.112.14.48 11001010 01110000 00001110 00110000
0.0.0.15 00000000 00000000 00000000 00001111
202.112.14.63 11001010 01110000 00001110 00111111
Subnet mask: 255.255.255.240
Subnet address: 202.112.14.48/28
Broadcast address: 202.112.14.63
Summarize
Did you see this and suddenly feel that solving the problem is very simple, hhh