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172.16.0.0/16 gives you 65534 possible hosts within 1 subnet.

By binary math, you would need to increase the subnet mask to divide the network into subnets.

Think of the additional subnet mask bits as a binary value.

With 0 extra bits in the masks, you have 1 subnet.

With 1 bit, you have 2.

With 2 bits, you have 4.

With 3 bits, you have 8.

With 4 bits, you have 16.

We need 8, so we should add 3 bits to the subnet mask.

This would be a base CIDR of 172.16.0.0/19.

Since there are 256 possibilities in the 3rd octet (Based on Class B network), divide by 8. The resultant number is the increment value for each network address.

256 / 8 = 32.

Therefore, your networks would be:

172.16.0.0/19

172.16.32.0/19

172.16.64.0/19

172.16.96.0/19

172.16.128.0/19

172.16.160.0/19

172.16.192.0/19

172.16.224.0/19

With maximum (65536 / 8 - 2) 8190 hosts per subnet.

Additional:

Since we have 19 network bits, there are 13 bits remaining for hosts (total, as always 32). An all 0 string of host bits would leave you with your base network number

ex>

172.16.0.0/19 , host with all 0's

10101100.00010000.00000000.00000000

The resultant would be 172.16.0.0, which is your network address. Conversely, having a host with all 1's:

10101100.00010000.00011111.11111111

This would result in 172.16.31.255, which is the broadcast address for the first subnet.

These aren't usable as assignable addresses, but are used for network and broadcast purposes.

Using all of the 3 borrowed bits in a 111 situation gives you a resultant of 172.16.224.0/19, which is a legitimate network address. Network bits in this case carry over from the /16, and are simply added on. An IP address of 172.16.255.255 would be the broadcast address for the final subnet.

ex>

10101100.00010000.11111111.11111111

If you were doing a class A network, this wouldn't be as applicable. If all 3 of the first bits of a network address are 1, this puts your IP into a class D (224.0.0.0/4) subnet which is reserved for multicasting. This is only applicable to the first bits of the network address (starting from the very beginning).

Hope this helped at least a little.

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Q: What is the second subnet of 172.16.0.0?
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