From owner-freebsd-questions  Fri Jul 30  7:56:20 1999
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From: "Bond, Jeffery" <Jeff.Bond@nectech.co.uk>
To: "'summoner@uswest.net'" <summoner@uswest.net>
Cc: "'FreeBSD questions'" <questions@FreeBSD.org>
Subject: RE: 172.16.0.0 private net question
Date: Fri, 30 Jul 1999 15:53:39 +0100
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Hi,

Since the netmask is 255.240.0.0, which is
11111111.11110000.00000000.00000000 in binary, 

and your private network is 172.16.0.0, which is
10101100.00010000.00000000.00000000 in binary

the entire range you have is (combining the address with the mask)
10101100.0001XXXX.XXXXXXXX.XXXXXXXX, where the X's can be anything you like,

This means that the second octet can range from 00010000 to 00011111 or 16
to 31 in decimal, and the 3rd and 4th octets can range from 0 - 255.

Hence the entire range is from 172.16.0.0 to 172.31.255.255

Hope this makes sense,

Jeff

Original message:
>Date: Thu, 29 Jul 1999 19:12:39 -0700
>From: Summoner <summoner@uswest.net>
>Subject: 172.16.0.0 private net question
>
>This isn't entirely FreeBSD related but I figure this would be one of
>the few good places to ask.
>
>Is the 172.16.0.0 private network range 172.16.0.0 to 172.31.255.255,
>or 172.0.0.0 to 172.16.255.255?  The netmask is 255.240.0.0, a 12-bit
>mask.  So it makes sense, mathematically, to me, at least, that the IP
>range would be 172.0 to 172.16.  But RFC1918 says it's 172.16 to
>172.31.  Does the 172.16 net just have the 12th bit always on or
>something?




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