From: Peter McCreesh (petermccreesh@gmail.com)
Date: Wed Dec 14 2005 - 11:41:49 GMT-3
Sorry to drag this one up again but I am just looking at the solution and
trying to fully understand it.
The ACL for R4 groups is fine (226.37.0.0 1.8.255.255).
The solution I get for the R3 groups is:
225.0.0.0 0.58.255.255
Some of you have stated that the answer is 225.10.0.0 0.48.255.255 but this
shouldn't include the 225.48.0.0 - 225.48.255.255 range (or should it??)
i.e. 10 --> 00001010
48--> 00110000
so a wildcard of 48 in the second octet will ignore the third and fourth
bits (from the left) and match on everything else so there should not be a
match.
Am I understanding this right.
Thanks,
Pete
On 11/13/05, Montiean <noktes@bellsouth.net> wrote:
>
> Additional comment on R3 groups,
> If we use wildcard 0.58.255.255 --> we will include some range that we
> don't need
> if we use 0.48.255.255 -- this is exactly macth.
>
> However, boths are fine
>
> -Montiean
>
> >
> > From: "Cisco Nuts" <cisconuts@hotmail.com>
> > Date: 2005/11/12 Sat PM 07:29:58 EST
> > To: ChoonHo.Loi@getronics.com
> > CC: ccielab@groupstudy.com
> > Subject: RE: Weird MCast ACL - ??
> >
> > Thank you very much!
> >
> > Hopefully, this clears up things for other people like me.
> >
> > Sincerely.
> >
> > --------------------------------------------------------------------
> >
> > From: "Loi, Choon Ho" <ChoonHo.Loi@getronics.com>
> > To: "Cisco Nuts" <cisconuts@hotmail.com>
> > Subject: RE: Weird MCast ACL - ??
> > Date: Sat, 12 Nov 2005 22:29:31 +0800
> >
> > R3 as RP for groups: Second Octet225.10.0.0 -
> > 225.10.255.255 0000 1010225.26.0.0 - 225.26.255.255 0001
> 1010225.42.0.0
> > - 225.42.255.255 0010 1010225.48.0.0 - 225.48.255.255 0011 0000
> > ------------------AND 0000 0000
> > ---> 0XOR 0011 1010 ---> 58 my
> > answer: 225.0.0.0 0.58.255.255
> > Solution:----------------> 225.10.0.0 0.48.25.255 R4 as RP for groups:
> > first & second octet226.37.0.0 - 226.37.255.255 1110
> 0010 0010
> > 0101226.45.0.0 - 226.45.255.255 1110 0010 0010 1101227.37.0.0 -
> > 227.37.255.255 1110 0011 0010 0101227.45.0.0 - 227.45.255.255 1110
> > 0011 0010
> 1101
----------------------------AND
> > 1110 0010 0010 0101 ---> 226 37XOR
> > 0000 0001 0000 1000 ---> 1 8 som Solution:-------------->
> > 226.37.0.0 - 1.8.255.255
> >
> > ------------------------------------------------------------------------
> >
> > From: Cisco Nuts [mailto:cisconuts@hotmail.com]
> > Sent: Saturday, November 12, 2005 6:45 PM
> > To: Loi, Choon Ho
> > Subject: RE: Weird MCast ACL - ??
> >
> > Would you kindly explain - if you don't mind?
> >
> > I just cannot get it !!
> >
> > :-(
> >
> > --------------------------------------------------------------------
> >
> > From: "Loi, Choon Ho" <ChoonHo.Loi@getronics.com>
> > To: "Cisco Nuts" <cisconuts@hotmail.com>
> > Subject: RE: Weird MCast ACL - ??
> > Date: Sat, 12 Nov 2005 18:36:40 +0800
> > >I agree with 226.37.0.0 1.8.255.255
> > >But, for the first one, my answer is 225.0.0.0 0.58.255.255
> > >
> > >-----Original Message-----
> > >From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf
> > Of
> > >Cisco Nuts
> > >Sent: Saturday, November 12, 2005 6:16 PM
> > >To: ccielab@groupstudy.com
> > >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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> >
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