Given that the very last bit from a binary perspective, it cannot be a
network address, unless you wanna split hairs and call /32s networks.
Note that is a reserved for private lan ip, 172.16.0.0-172.31.0.0 is the
reserved block this is part of, that particular /21 would be
172.16.0.0 to 172.16.7.
172.16.0.1 is a private IP range and is not routable on the Internet. So an
ISP would not give you this range. See RFC's 1918 and RFC 1597.
Tim LeBrun
CCNA, CCDA
[EMAIL PROTECTED]
-Original Message-
From: CCIE TB [mailto:[EMAIL PROTECTED]]
Sent: Thursday, May 24, 2001 4:50 PM
To: [EMA
ack, before you nitpickers slam me, that entire private range block
ends in 172.31.255.255, not 172.31.0.0 as I stated previously.
Brian "Sonic" Whalen
Success = Preparation + Opportunity
On Thu, 24 May 2001, Brian wrote:
> Given that the very last bit from a binary perspective, it cannot be a
OK first of all 172.16.0.0/21 is actually 172.16.0.0 255.255.248.0
Which means the network address are:
172.16.8.0
172.16.16.0
172.16.24.0
172.16.32.0
172.16.40.0 etc. etc. going up by 8 in the 3rd octet until you hit
172.16.248.0
So the answer is no because 172.16.0.1 falls into the first invali
Unless you use "ip subnet-zero" Then it starts with 172.16.0.0. ;^)
I agree though, either way you slice this, it's not a network address.
D.
At 05:38 PM 05/24/2001 -0400, Michael Bambic wrote:
>OK first of all 172.16.0.0/21 is actually 172.16.0.0 255.255.248.0
>Which means the network addres
A network address can never end in an odd number. Given the IP address
172.16.0.1 and a /21 subnet mask, the network number in this case would be
172.16.0.0. The next network number would be 172.16.8.0 and so forth.
R/S, Dave Goldsmith
-Original Message-
From: CCIE TB [mailto:[EMAIL PR
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