From: Hardeep Bhutani (ccie172203@gmail.com)
Date: Wed May 21 2008 - 18:58:08 ART
Thank you very much Chris.
On 5/21/08, Chris McGuire <chris.mcguire333@gmail.com> wrote:
>
> The logic is straight forward. Take the following byte in binary:
>
> 11111111
>
> Each bit represents:
>
> 128 - 64 - 32 - 16 - 8 - 4 - 2 - 1
>
> Notice the only odd bit is 1.
>
> So if you have
>
> 0.0.0.0 254.255.255.255
>
> The 254 = 11111110
>
> This means the only thing that MUST match from the network address is the
> last bit of this octet.
>
> What is the last bit of the 1st octet of the network address 0.0.0.0?
>
> 00000000
>
> It is 0.
>
> So any possible option for this 1st octet that will match will have to be
> even because the addition of any other bit values is even.
>
> On the flip side, if you have:
>
> 1.0.0.0 254.255.255.255
>
> So now the last bit of the 1st octet has to be a 1 always. So if you add
> any
> other bit values to 1 they will always be odd.
>
> This is how you force an access-list to only match either odd or even
> routes.
>
> Hope this makes sense.
>
> Thanks,
>
> Chris
> 20701
>
> On 5/21/08 3:20 PM, "Hardeep Bhutani" <ccie172203@gmail.com> wrote:
>
> > Hi All
> >
> > Does any one can help me in understanding the calculation of even and odd
> > routes?
> >
> > Or it would be great if someone has any document.
> >
> > Thanks
> >
> > HB
> >
> >
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