From: Schulz, Dave (DSchulz@dpsciences.com)
Date: Sat Nov 12 2005 - 11:22:52 GMT-3
I'll take a shot at the first one....looking at the bits on the second
octet...
0000 1010
0001 1010
0010 1010
0011 0000
...it appears that you would be able to summarize this in one acl if that last
number was 58, rather than 48, since this would make the second and the fourth
bits in common with the rest of the numbers....then you could summarize it
as....(as it, you would need more than one acl)...
225.10.0.0 0.48.255.255
on the odd and even numbers.....remember that the 1 is don't care and 0 is
remains the same in the mask.....
so, to summarize and even....you will always have a zero in the first bit with
the zero in the mask....
x.x.x.xxxxxxx0 x.x.x.xxxxxxx0
For the odd....you always have the 1 in the first bit....
x.x.x.xxxxxxx1 x.x.x.xxxxxxx0
....the zero in bit mask insures that the bit always matches.
Hope this helps you to give a shot at the second one. Please correct me if I
missed anything.
Dave
-----Original Message-----
From: nobody@groupstudy.com
To: ccielab@groupstudy.com
Sent: 11/12/2005 5:16 AM
Subject: Weird MCast ACL - ??
Hi Group: I have tried to understand this crazy acl. and just cannot get
it.How in the world do I get an even number in the wildcard mask?I wrote
down the 1's and 0's and tried to create a mask based on it but oh! boy
-
was I way off?If I get such an acl in the exam, I know I am sc****!!
Here
goes: R3 as RP for groups: 225.10.0.0 - 225.10.255.255 225.26.0.0 -
225.26.255.255 225.42.0.0 - 225.42.255.255 225.48.0.0 - 225.48.255.255
Solution:----------------> 225.10.0.0 0.48.25.255 R4 as RP for groups:
226.37.0.0 - 226.37.255.255 226.45.0.0 - 226.45.255.255 227.37.0.0 -
227.37.255.255 227.45.0.0 - 227.45.255.255 Solution:-------------->
226.37.0.0 - 1.8.255.255 Sincerely.
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