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 invalid range
which is 172.16.0.0 through 172.16.7.255.
Hope that helps.
Mike Bambic
CCDP, CCNP, CCIE Written
Looking for a job in Phoenix
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of
CCIE TB
Sent: Thursday, May 24, 2001 1:50 PM
To: [EMAIL PROTECTED]
Subject: IP Address [7:5786]
Hi Group members,
I came across a question in a Transcender exam in which they give a network
address as 172.16.0.1/21. This address is given by an ISP to your network.
Is that address a possible network address? It looks to me as a host
address. What I'm missing here?
Regards to all
Adia
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