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How to determine Supernetting/CIDR IP address range?

Last response: in Business Computing
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June 2, 2012 1:03:48 AM




Just when I thought I had an understanding on how to calculate the IP address ranges via Supernetting/CIDR I am stumped as to how did MS get 192.168.31.254 as the end of the range.specifically where did the ( .31 ) come from?
For example, to express the range of addresses for the address prefix 192.168.16.0/20:
The first IPv4 unicast address in the range is 11000000 10101000 0001000000000001 (host bits are bold), or 192.168.16.1.


The last IPv4 unicast address in the range is 11000000 10101000 0001111111111110 (host bits are bold), or 192.168.31.254.
June 2, 2012 1:19:14 AM

I don't get what you're trying to say
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June 3, 2012 1:05:29 PM

The range is correct for a /20.

Network Address = 192.168.1.0
First Usable Host Address = 192.168.1.1
Last Usable Host Address = 192.168.31.254
Broadcast Address = 192.168.31.255

The .31 is there because the 4th octet can only represent up to 255. To go higher, the 4th octet rolls over to 0 and the 3rd octet increments by 1. So since there is a total of 4096 addresses in a /20, the 4th octet has filled up to 255 and rolled over to 0 multiple times, and the 3rd octet has incremented each time, a total of 15 times from .16 to .31
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June 6, 2012 2:23:55 PM

Thank you for your response: I believe I got what you are saying but for clarity sakes what do you mean by the statements "rolled over to "0" multiple times

Simo606 said:
The range is correct for a /20.

Network Address = 192.168.1.0
First Usable Host Address = 192.168.1.1
Last Usable Host Address = 192.168.31.254
Broadcast Address = 192.168.31.255

The .31 is there because the 4th octet can only represent up to 255. To go higher, the 4th octet rolls over to 0 and the 3rd octet increments by 1. So since there is a total of 4096 addresses in a /20, the 4th octet has filled up to 255 and rolled over to 0 multiple times, and the 3rd octet has incremented each time, a total of 15 times from .16 to .31

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June 6, 2012 5:19:51 PM

For example:

Network address: 192.168.1.0 /23 - room for a total of 512 addresses.

Selected Example Host: 192.168.1.255

And you want to add another IP address.

It can't go to: 192.168.1.256. The limit is .255 for any octet.

Therefore you must increment the octet to the left by 1, then the 255 resets back to 0, ready to be filled up with another 256 addresses.

So it would look like: 192.168.2.0
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