IP 172.31.X.X Y is it a /12 not as /13 as it can use up to 5 bits

Shloima Berger

Y is it a /12 not as /13 as it can use up to 5 bits

Ronnie Wong

@Shloima-Berger,

You're looking at RFC 1918 address for Class B?

The range is from 172.16.0.1/12 to 172.31.255.254/12 is a complete network with a /12.

With a /13 determine what the network range would be 172.32.0.1/13 to 172.63.255.254/13.

Let me know if this isn't what your referring to.

Shloima Berger

yes I'm referring to the RFC 1918 class B, but if it goes until 172/.31 it should be a /13 (you got 8 bits from the 172 and then 5 bits from .31 16+8+4+2+1=31

Ronnie Wong

@Shloima-Berger,

Where does the /12 end then?

From my understanding the /12 network interval is

1st 172.0.0.0/12 (172.0.0.1 - 172.15.255.255/12)
2nd 172.16.0.0/12 (172.16.0.1 - 172.31.255.255/12)
3rd 172.32.0.0/12 (172.32.0.1 -172.47.255.255/12)

And the /13 network interval is

1st 172.0.0.0/13 (172.0.0.1 -172.7.255.255/13)
2nd 172.8.0.0/13 (17.8.0.1 -172.15.255.255/13)
3rd 172.16.0.0/13 (17.16.0.1 -172.23.255.255/13)
4th 172.24.0.0/13 (17.24.0.1 -172.31.255.255/13)
5th 172.32.0.0/13 (17.32.0.1 -172.39.255.255/13)
6th 172.40.0.0/13 (17.40.0.1 -172.47.255.255/13)

What what I can infer about your post clarification is the the following. What you should do in binary terms is determine the number of consecutive binary 1 from the LEFT (Highest Order to Lowest Order Bits) as follows, then convert to dotted decimal. This is how subnet masks are determined.

/12 = 255.240.0.0= 11111111.11110000.00000000.00000000
/13 = 255.248.0.0=11111111.11111000.00000000.00000000

When you clarified, you're counting up the number of host bits and converting to decimal...or if you're trying to determine the wildcard mask bits you would count the lowest bits to the highest bit orders.

I corrected the subnetting above, my original post, I completely did backward and wrong for the /13.