In order to find network id (NID) of a Subnet, one must be fully acquainted with the Subnet mask.
It is used to find which IP address belongs to which Subnet. It is a 32 bit number, containing 0’s and 1’s. Here network id part and Subnet ID part is represented by all 1’s and host ID part is represented by all 0’s.
If Network id of a entire network = 22.214.171.124 (it is class C IP). For more about class C IP see Classful Addressing.
In the above diagram entire network is divided into four parts, which means there are four subnets each having two bits for Subnet ID part.
Subnet-1: 126.96.36.199 to 188.8.131.52 Subnet-2: 184.108.40.206 to 220.127.116.11 Subnet-3: 18.104.22.168 to 22.214.171.124 Subnet-4: 126.96.36.199 to 188.8.131.52
The above IP is class C, so it has 24 bits in network id part and 8 bits in host id part but you choose two bits for subnet id from host id part, so now there are two bits in subnet id part and six bits in host id part, i.e.,
24 bits in network id + 2 bits in subnet id = 26 (1's) and 6 bits in host id = 6 (0's)
Subnet Mask = 11111111.11111111.11111111.11000000 = 255.255.255.192
If any given IP address performs bit wise AND operation with the subnet mask, then you get the network id of the subnet to which the given IP belongs.
If IP address = 184.108.40.206 (convert it into binary form) = 11000001.00000001.00000010.10000001 Subnet mask = 11111111.11111111.11111111.11000000 Bit Wise AND = 11000001.00000001.00000010.10000000 Therefore, Nid = 220.127.116.11
Hence, this IP address belongs to subnet:3 which has Nid = 18.104.22.168
If IP address = 22.214.171.124 (convert it into binary form) = 11000001.00000001.00000010.01000011 Subnet Mask = 11111111.11111111.11111111.11000000 Bit Wise AND = 11000001.00000001.00000010.01000000 Therefore, Nid = 126.96.36.199
Hence, this IP address belongs to subnet:2 which has Nid = 188.8.131.52
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