Networking - Subnetting - Discussion

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?

SubnetMask CIDR NetBits HostBits #of Nets TotalHosts Hosts/Net.

255.255.224.0/19 3 13 6(8*) 8190 49140 (65520*).

Host is right 2^13 = 8192-2 = 8190.

Its class B address.

11111111 11111111 11100000 00000000.

Class A Class B Class C.

We have 2^3 8-sub networks. For 1 sub network we are calculating the how many hosts we have.

8192 hosts we have from these hosts we are subtracting 2 IP address network id and broadcast address.

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

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.

It is wrong answer because.

Number of hosts per subnet is ok, but 2^3 = 8 is wrong answer,

2^3 = 8-2 => 6.

Only 6 subnet are useful.

172.16.0.0---it is used for Network id.

And 172.16.255.255 is used for broadcast address, so we can't use 8 subnet id's even we have 3 bit's for subting.

19-16 = 3 (16 bit network address for class B).

2^3 = 8 subnet.

Total 32 bits.

32-19 = 13 bits available for host so.

2^13 = 80192 hosts for each subnet.

But 2 addresses are reserved for network and broadcast address therefore 80192-2 = 80190 hosts for each subnet are available.

@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.

Everything is correct, 3-bits are 8-subnets 2^3=8.

And then we have left 13 bits to work with which is:

2^13 = 80192-2 = 80190.

11111111.11111111.11100000.00000000.

1- = 19.

0 = 13---5+8 = 13.

I hope this is clear.

No. of network doesn't require to subtract anything.
Its only applicable for number of hosts that one is used for Broadcast and for the network.

So the formula for host is 2^n-2.
No. of subnets/networks is 2^n.

No actually I accept the answer of Rambabu because 2^3 = 8-2 = 6.

1 is for n/w and 1 for broadcast.

Host is right 2^13 = 8192-2 = 8190.