From: Bob Dixon (bobdixon@xxxxxxxxxxxx)
Date: Wed Apr 18 2001 - 22:37:00 GMT-3
Folks,
Thanks for your replies.
I was over-thinking the problem...I was trying to construct a wildcard mask
that allowed *ONLY* 192.168.2.0 and 192.168.4.0 using only ONE statement.
After looking at this for while and trying out lots of different combos, I
think that it is impossible to do what I was trying to do with just one
statement.
Thanks again for the quick replies.
Bob
----- Original Message -----
From: Bruce Williams <bruce@williamsnetworking.com>
To: 'Bob Dixon' <bobdixon@mediaone.net>; <ccielab@groupstudy.com>
Sent: Wednesday, April 18, 2001 8:59 PM
Subject: RE: wildcard mask problem
> The two network address have the first 5 bits of the third octet in
common.
> The wildcard mask is the sum of the bits that they do not have in common
in
> the third octet; 4, 2 ,1 and then the sum of all the bits in the last
octet.
>
> network 192.168.0.0 0.0.7.255 area 0
>
> Bruce
> mailto:bruce@williamsnetworking.com
>
>
>
> -----Original Message-----
> From: nobody@groupstudy.com [mailto:nobody@groupstudy.com]On Behalf Of
> Bob Dixon
> Sent: Wednesday, April 18, 2001 8:38 PM
> To: ccielab@groupstudy.com
> Subject: wildcard mask problem
>
>
> Folks,
>
> I should know how to do this, but I can't seem to get it.
>
> Question:
> take the following:
> network 192.168.2.0 0.0.0.255 area 0
> network 192.168.4.0 0.0.0.255 area 0
>
> and create one network statement that summarizes the two.
>
> Thanks,
> Bob
> **Please read:http://www.groupstudy.com/list/posting.html
> **Please read:http://www.groupstudy.com/list/posting.html
**Please read:http://www.groupstudy.com/list/posting.html
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