From: Chris McGuire (chris.mcguire333@gmail.com)
Date: Wed May 21 2008 - 18:38:02 ART
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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