IP Addressing
It is a 32 bit logical address assigned to each each network. The responsibility of managing and assigning IP address has been given to INTERNIC. It is denoted in decimal dotted notation as shown below :
w.x.y.z
10.7.4.8
192.168.0.1
NOTE: Remember If you want to understand IP addressing and subnetting, you must revise your binary, decimal and hexadecimal conversions.
The 32 bit IP address is broken into 4 octet into a decimal dotted notation. One octet is a set of 8 bits and each bit represented by '1' and '0' due to the binary factor. The decimal range of value is 0 from 255.Each IP bit will start from 0 and end with 255.
When we write a number for example fifty, we write it in 50 but for the computer it is 110010 because the computer can only understand in binary.
Now here is the method that how you can convert a decimal into binary. Take a paper write values of 2^1 till 2^8 in a horizontal way like this
128 64 32 16 8 4 2 1
Now the on bit will be represented by 1 and the off bit by 0. from Values write 1 on those values which will add and make a sum of 50. The bits that will make a sum 50 are 32 ,16 and 2. Now we will write 1 on these bits.
128 64 32 16 8 4 2 1
0 0 1 1 0 0 1 0 = 50
So that's how we get the binary value. Now we write binary values for 0, 1, 255, 10.
128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 0 = 0
0 0 0 0 0 0 0 1 = 1
1 1 1 1 1 1 1 1 = 255
0 0 0 0 1 0 1 0 = 10
0 0 0 0 0 0 0 1 = 1
1 1 1 1 1 1 1 1 = 255
0 0 0 0 1 0 1 0 = 10
IP ADDRESS RANGE
The range of an IP address is classified by three classes
10 . 0 . 0 . 5
w . x . y . z
w is the network bit and x, y, z are the host bits.
Class B : The Class IP ranges from 128 to 191. In class B the first two bits will be the network and the remaining two will be the host bits.
190 . 10 . 0 . 5
w . x . y . z
w, x are the network bits and x,y,z are the host bits.
- Class A
- Class B
- Class C
10 . 0 . 0 . 5
w . x . y . z
w is the network bit and x, y, z are the host bits.
Class B : The Class IP ranges from 128 to 191. In class B the first two bits will be the network and the remaining two will be the host bits.
190 . 10 . 0 . 5
w . x . y . z
w, x are the network bits and x,y,z are the host bits.
Class C : The Class IP ranges from 192 to 223. In class C three bits will be the network and the remaining last will be the host bit.
200 . 0 . 10 . 5
w . x . y . z
w, x, y are the network bits and z is the host bit.
200 . 0 . 10 . 5
w . x . y . z
w, x, y are the network bits and z is the host bit.
Note: Seeing the first digit of the IP address will determine which class the IP belongs. The higher the number of bits increases the more will be the host connections.
No of hosts in each Classes
There is a formula through which we will derive the no of hosts of each class which is
2^n-2
n is the no of bits
We know that each bit of an IP address is equal to 8.
Class A :There are 3 host bits in class A which is equal to 24 so
2^24-2=17 million approximately
It means that 17 million IP addresses can be hosted
Class B :There are 2 host bits in class B which is equal to 16 so
2^16-2=65533
It means that 65533 IP addresses can be hosted .
Class C :There is only 1 bit in class C which is equal to 8 so
2^8-2=254
It means that 254 IP addresses can be hosted
Note: The network bit is controlled by INTERNIC which we cannot use .
Now there is a question that will be bothering you which is why are we deducting 2 address from each range for which we are using -2 in the formula.
The two address which we are deducting is for special purpose which you will learn later. From the range we are taking the first address and the last address. The first one will be call the Network ID and the last one will be called the broadcast ID. Here is an example
10.0.0.1 --> The Network ID
10.0.0.2
10.0.0.2
10.0.0.3
*
*
*
10.255.255.254
10.255.255.255 --> The Broadcast ID
10.255.255.255 --> The Broadcast ID
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