Rene,

I have a doubt, all examples that you gave are “continuos” networks and an even number of networks.

And when we have networks like below? I just can solve them with binary method. Is there another form?

172.16.10.0/24

172.16.20.0/24

172.16.30.0/24

172.16.40.0/24

172.16.50.0/24

I choosed shortest and highest networks and convert them to binary, so the summary address will be

176.16.0.0/18 a block size 64 networks. I can’t solve it using CIDR notation and block size method.

Another example with an odd number of networks, I can solve it with binary method

192.168.0.0 / 24

192.168.1.0 / 24

192.168.2.0 / 24

192.168.3.0 / 24

192.168.4.0 / 24

The summary address will be 192.168.0.0/21, but when I was writing this example I saw that if I use block

size it’s able solve it too. It’s need always think in block size like “powers of 2”.

When you have a free time, please, detail below

8 + 8 + 6 = 24 bits

Hug and thanks for yours articles. It’s helping a lot