Bits needed for defining Net IDs and Host IDs of Class A, Class B and Class C

In classful addressing, an IP address of class A, B, and C are divided into netid and hostid. The netid determines the network address while the hostid determines the host connected to that network.

 

	Byte 1	Byte 2	Byte 3	Byte 4
Class A	Netid	Hostid
Class B	Netid		Hostid
Class C	Netid			Hostid
Class D	Multicast Address
Class E	Reserved for future use

 

Blocks and Hosts

Each class is divided into blocks. The number of blocks in a class can be calculated by the number of bits in the netid.

CLASS A

Class A has 1 byte (8 bits) netid and from the binary notation, we see that Class A address starts with 0 so there is a total of 7 bits that can be changed out of 8.

Therefore, total number of blocks in Class A = 27 = 128

There are 3 bytes (24 bits) for hostid in Class A so the total number of the host in each block = 224 = 16,777,216

So total number of addresses in Class A = No. of Blocks in Class A x No. of Hosts in each block of Class A

= 128 x 16,777,216

= 2,147,483,648

This is 50% of the total addresses in IPv4.

1st block of Class A has the netid 0

And the host id is between

0.0.0.0 ………….… 0.255.255.255

Similarly, 2nd block of Class A has the netid 1

And the host id is between

1.0.0.0

1.255.255.255

	Netid 0	Netid 1	…	Netid 127
Class A	0.0.0.0 to 0.255.255.255	1.0.0.0 to 1.255.255.255		127.0.0.0 to 127.255.255.255
	Block 1	Block 2		Block 128

 

CLASS B

Class B has 2 bytes (16 bits) netid and from the binary notation, we see that Class B address starts with 10, so there is a total of 14 bits that can be changed out of 16.

Therefore, total number of blocks in Class B = 214 = 16,384

There are 2 bytes (16 bits) for hostid in Class B so the total number of the host in each block = 216 = 65,536

So the total number of addresses in Class B = No. of Blocks in Class B x No. of Hosts in each block of Class B

= 16,384 x 65,536

= 1,073,741,824

This is 25% of the total addresses in IPv4. 

1st block of Class B has the netid 128.0

And the host id is between

128.0.0.0

…

128.0.255.255 

Similarly, 2nd block of Class B has the netid 128.1

And the host id is between

128.1.0.0

128.1.255.255

	Netid 128.0	Netid 128.1	…	Netid 191.255
Class B	128.0.0.0 to 128.0.255.255	128.1.0.0 to 128.1.255.255		191.255.0.0 to 191.255.255.255
	Block 1	Block 2		Block 16,384

 

CLASS C

Class C has 3 bytes (24 bits) netid and from the binary notation, we see that Class C address starts with 110, so there is a total of 21 bits that can be changed out of 24.

Therefore, total number of blocks in Class C = 221 = 2,097,152

There is 1 byte (8 bits) for hostid in Class C so the total number of the host in each block = 28 = 256

So total number of addresses in Class C = No. of Blocks in Class C x No. of Hosts in each block of Class C

= 2,097,152 x 256

= 536,870,912

This is 12.5% of the total addresses in IPv4.

 

1st block of Class C has the netid 192.0.0

And the host id is between

192.0.0.0

…

192.0.0.255

 

Similarly, 2nd block of Class B has the netid 192.0.1

And the host id is between

192.0.1.0

192.0.1.255

	Netid 192.0.0	Netid 192.0.1	…	Netid 223.255.255
Class C	192.0.0.0 to 192.0.0.255	192.0.1.0 to 192.0.1.255		223.255.255.0 to 223.255.255.255
	Block 1	Block 2		Block 2,097,152

(Source: https://dyclassroom.com/networking/netid-and-hostid)