Supernet divisibility 1

[SwinIT]

We are wanting to set up a supernet with a starting address of 192.168.0.0 - 192.168.7.0. I understand it has to be divisible by 8 in this case. Some posts suggest that a network starting with ZERO can be used. Does anyone have more insight into this?

 

[VinceWhirlwind]

255.255.255.0 gives you addresses 192.168.0.1 -> 192.168.0.254
/24

255.255.254.0 gives you addresses 192.168.0.1 -> 192.168.1.254
/23

255.255.252.0 gives you addresses 192.168.0.1 -> 192.168.3.254
/22

255.255.248.0 gives you addresses 192.168.0.1 -> 192.168.7.254
/21

For all of these, the network address is 192.168.0.0.
The broadcast address is the address ending in .255 immediately after the last IP address in the range.

I do not understand your question:
"posts suggest that a network starting with ZERO can be used".
What does that mean?

 

[SwinIT]

Vincewhirlwind - - Thank you for posting. I have read that the first networks third octet must be divisible by the number of networks you are creating. In my case 8 networks which would mean the 3rd octet would have to be 8, 16, 24, etc. But some posts elude to the fact that the first networks third octet can be ZERO in some cases. I'm trying to find out what those cases are? Sorry for any confusion.

 

[VinceWhirlwind]

Back in the dim mists of time, you couldn't use .0.

This hasn't been the case for a while now, but on some older kit you might see in the config a default command that says "IP subnet-zero". Changing that to "no ip subnet-zero" would render the advice you have received useful.
But you would never do that, so the advice you have received is unlikely to be useful unless you are using some really ancient equipment.