# Identify the netid and hostid of the IP Address given its subnet mask in binary. Show your solution. 192.168.15.187 255.255.255.0

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Identify the netid and hostid of the IP Address given its subnet mask in binary. Show your solution.

192.168.15.187

255.255.255.0

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Generally networks are divided into netid and hostid, the netid  of a network identifies the network, while the host id states the host connected to the network.

First thing needed in order to find the netid and the hostid of any given IP address, is to first figure out what class the IP address falls under, there are mainly 5 classes of IP addresses and these are the Class A,B,C,D,E. For the purpose of this question though, we would  focus on only class A and B and C

CLASS A:  Every IP address as we know is a dotted 32 bits octet, meaning that they are divided into 4 parts and each is made up of 8 bits. for example (AAA. BBB. CCC. DDD) each alphabet can be said to be 8 bits. now for class A networks the first octet ranges from 0 to 127. meaning that the IP address in the question (192.168.15.187) is not a class A network.

CLASS B: The class B networks range from 128 to 191. meaning the IP address is also not of class B.

CLASS C: The class C networks have a range from 192.0.0.0 to 233.255.255.255, so clearly the IP address above is a class C address.

So lets focus on the class C.

For the class C addressing scheme, the first 24 bits of the IP address is what defines the NetID , meaning that the net ID here is; 192.168.15 and the last 8 bits define the HostID so we have 187 as the HostID.

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