Networking - Subnetting - Discussion

11111111 11111111 11100000 00000000 binary of class B.
16n bits+3host bits converted= 19 which is the CIDR,
therefore 3 will be use to calculate the subnet.
2^3= 8 which is the number of subnet.

32 bits-19 network bits= 13 host bits left,
2^13 = 8192.
8192-2 = 8190 which the number of host bits.

I agree with you, Thanks @James willie.

I can't understand it, Please explain.

I agree @Gajulapalli Naveen.

32 subnet and 8192 hosts. It's class b the values of third octet is 32.

How to calculate the number of host and number of the network for said IP-134.141.7.11 and subnet is 255.255.255.0

@All.

Always remember that subnet will calculate like this =2^given prefix - default prefix it will be =2^19-16 =2^3 =8 and valid host = count all 0 means here total host 13 then it will be (2^13)-2 means 8190 is the valid host.

172.16.0.0/19 means 19 bits for network ID part and 32-19 = 13 bits for host ID part. Now bits for subnet ID will be borrowed from host ID part or Network ID part?

Because generally, we borrow from host ID part, we have 2^3 = 8 subnets and 2^10-2 = 1046 hosts.

So, the maximum 8 subnets and maximum 1046 hosts are possible.

The answer provided by the website becomes correct only if we borrow the bits for subnet ID from Network ID part. How do we know that we should borrow bits from network ID part?

2^3 = 8 subnet.

2^13 -2= 8190 hosts each only Host need public IP & broadcast IP. Subnet don't need -2.

According to my knowledge sub networks can be;

172.16.32.0
172.16.64.0
172.16.96.0
172.16.128.0
172.16.160.0
172.16.192.0
172.16.224.0

As we know we cannot use all 0's and all 1's so ip address under 172.16.32.0 will be from 172.16.32.1 to 172.16.63.254. I just figure out 7 subnets. I don't know about 8th one. Can anyone help me where I am making mistake?

Thank you in advance