From: OhioHondo (ohiohondo@columbus.rr.com)
Date: Wed Apr 16 2003 - 14:40:30 GMT-3
Just for some fun info the
165.24.45.0 2.0.2.0 is the same as
167.24.45.0 2.0.2.0 or
167.24.47.0 2.0.2.0 or
165.24.47.0 2.0.2.0
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com]On Behalf Of
Brian McGahan
Sent: Wednesday, April 16, 2003 12:47 PM
To: 'Scott M. Livingston'; 'ccie_studying'; ccielab@groupstudy.com
Subject: RE: 1 aggregated ACL
Scott,
Close.
165.24.45.0 = 10100101.00011000.00101101.00000000
167.24.47.0 = 10100111.00011000.00101111.00000000
-------------------------------------------------
&& = 10100101.00011000.00101101.00000000
165.24.45.0 = 10100101.00011000.00101101.00000000
167.24.47.0 = 10100111.00011000.00101111.00000000
-------------------------------------------------
XOR = 00000010.00000000.00000010.00000000
165.24.45.0 2.0.2.0
HTH,
Brian McGahan, CCIE #8593
Director of Design and Implementation
brian@cyscoexpert.com
CyscoExpert Corporation
Internetwork Consulting & Training
Toll Free: 866.CyscoXP
Fax: 847.674.2625
> -----Original Message-----
> From: Scott M. Livingston [mailto:scottl@sprinthosting.net]
> Sent: Wednesday, April 16, 2003 9:42 AM
> To: 'Brian McGahan'; 'ccie_studying'; ccielab@groupstudy.com
> Subject: RE: 1 aggregated ACL
>
> And if we wanted to do it w/o leaking (OT) we would use the following;
>
> > > 165.24.45.0
> > > 167.24.47.0
> > > 175.28.65.0
>
> 165.24.45.0 0.0.2.0
> 175.28.65.0 0.0.0.0
>
> -check my math-
>
> scott
>
> -----Original Message-----
> From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf
Of
> Brian McGahan
> Sent: Tuesday, March 25, 2003 4:45 PM
> To: 'ccie_studying'; 'Scott M. Livingston'; ccielab@groupstudy.com
> Subject: RE: 1 aggregated ACL
>
> To find a network and wildcard pair, you need to use the AND and XOR
> logic gates.
>
> A AND B
> _____________
> | A | B | out |
> | 0 | 0 | 0 |
> | 0 | 1 | 0 |
> | 1 | 0 | 0 |
> | 1 | 1 | 1 |
> -------------
>
>
> A XOR B
> _____________
> | A | B | out |
> | 0 | 0 | 0 |
> | 0 | 1 | 1 |
> | 1 | 0 | 1 |
> | 1 | 1 | 0 |
> -------------
>
>
>
> Write the networks out in binary you are trying to find the list for:
>
> 165.24.45.0
> 167.24.47.0
> 175.28.65.0
>
>
> 10100101.00011000.00101101.00000000
> 10100111.00011000.00101111.00000000
> && 10101111.00011100.01000001.00000000
> ------------------------------------------
> 10100101.00011000.00000001.00000000 = 165.24.1.0
>
> ANDing them comes up with the network address.
>
>
> 10100101.00011000.00101101.00000000
> 10100111.00011000.00101111.00000000
> XOR 10101111.00011100.01000001.00000000
> ------------------------------------------
> 00001010.00000100.01101110.00000000 = 10.4.110.0
>
> XORing them comes up with the wildcard address.
>
> Therefore, the most specific match for these three networks is:
>
> 165.24.1.0 10.4.110.0
>
>
> Here's another of my threads on the same topic:
>
> http://www.groupstudy.com/archives/ccielab/200210/msg02503.html
>
>
> HTH
>
> Brian McGahan, CCIE #8593
> Director of Design and Implementation
> brian@cyscoexpert.com
>
> CyscoExpert Corporation
> Internetwork Consulting & Training
> Toll Free: 866.CyscoXP
> Fax: 847.674.2625
>
>
> > -----Original Message-----
> > From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf
> Of
> > ccie_studying
> > Sent: Tuesday, March 25, 2003 2:37 PM
> > To: Scott M. Livingston; ccielab@groupstudy.com
> > Subject: Re: 1 aggregated ACL
> >
> > I think if only summary to one network, it should be:
> >
> > 164.24.32.0 with wildcard 15.7.15.255 or subnet mask 240.248.240.0
> >
> > ----- Original Message -----
> > From: "Scott M. Livingston" <scottl@sprinthosting.net>
> > To: <ccielab@groupstudy.com>
> > Sent: Tuesday, March 25, 2003 11:26 AM
> > Subject: 1 aggregated ACL
> >
> >
> > > This was posted on another board so I wanted to check the answer
> that
> > > was given. It happens to be the same answer I came up with. Also,
> if
> > > someone has any other teasers maybe you can post them. I am using
> the
> > > formula Tim Fletcher taught those of us that were doing it another
> way
> > > (my wrong way :)).
> > >
> > > 165.24.45.0
> > > 167.24.47.0
> > > 175.28.65.0
> > >
> > >
> > > Answer:
> > > 165.24.1.0 mask 10.4.110.255
> > >
> > > thank you,
> > > scott
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