Hi,
Sure they are, do you need some help explaining why? We are here, just let us know .
As for your case, with /22 mask and prefix 10.56.4.x you have the whole range 10.56.4.0 - 10.56.7.255 available.
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Just a little bit of explanation:
This is your mask /22, written in decimal and binary respectively:
255. 255. 252. 0
11111111.11111111.11111100.00000000
Now, what does put hosts into same subnet? It's, as we all know, the identical prefix.
You have chosen the prefix 10.56.4.0 with mask of /22. When we apply the /22 mask on it, that means that the first 22 bits have to be exactly matched!
So, as for 10.56.4.0 we have to match 1st and 2nd octet precisely.
As for the 3rd octet, let's have a closer look at it. It is 4 in decimal. That is in binary 00000100.
Let's apply the subnet mask
11111111.11111111. 1111 1100 .00000000
1st octet.2nd octet. 0000 01xx .xxxxxxxx
The bold numbers are those, that need to be exactly matched. The others (marked with x) are identifiers of hosts in subnet (so called host portion of the address).
Now, we see that 1st and 2nd octet must be matched exactly and anything in the 4th octet would fit. But what about the possible combinations in the 3rd octet? Let's write them down.
1st octet.2nd octet. 0000 0100 .x = 10.56.4.x
1st octet.2nd octet. 0000 0101 .x = 10.56.5.x
1st octet.2nd octet. 0000 0110 .x = 10.56.6.x
1st octet.2nd octet. 0000 0111 .x = 10.56.7.x
where x = <0,255>, and 10.56.4.0 is the network address, 10.56.7.255 is the broadcast address.
If you need some more help, or I didn't explained it in an understandable manner, please feel free to ask or make a comment.
Best regards,
Jan
