From: Montiean (noktes@bellsouth.net)
Date: Sat Nov 12 2005 - 11:21:46 GMT-3
Is the question really asking to get even number in wildcard??
From the solution, I think the question asking to use 1 statement in acl.
R3
10 = 0000 1010
26 = 0001 1010
42 = 0010 1010
48 = 0011 0000
= 00xx 0000 ==> we use wildcard bit 0-3 to be zero because we need it
to be 1010
so --> 225.10.0.0 0.48.255.255 ---> you get
225.(10,26,42,48).(0-255).(0-255)
R4
226 = 1110 0010
227 = 1110 0011
= 0000 000x
37 = 0010 0101
45 = 0010 1101
= 0000 x000
so --> 226.37.0.0 - 1.8.255.255 --> you get
(226-227).(37,45).(0-255).(0-255)
-Montiean
----- Original Message -----
From: "Cisco Nuts" <cisconuts@hotmail.com>
To: <ccielab@groupstudy.com>
Sent: Saturday, November 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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