Re: Odd & Even ACLs

Hi Molomo,

Its just one of those shortcut steps to calculate the wildcard masks for
ACLs ( courtesy : IE Blog again ) . Therefore , you might be interested in
reading this

http://blog.internetworkexpert.com/2007/12/26/q-how-do-i-compute-complex-wildcard-masks-for-access-lists/

P.S - There is a difference in how I remember the output of a XOR operation
than the one specified on the blog, so whichever way you wanna remember it
..Do go through the comments in the blog..there is a nice explanation by a
gentleman called Paul ..

Regards,
Ravi

On Tue, Jul 7, 2009 at 12:09 AM, Molomo <letjedilakopa_at_gmail.com> wrote:

> Awesome, what a great binary lesson by Scott, it is just simply great...One
> just hopes it doesn't get that complicated in the lab.
>
> Ravi, great notes thanks a lot mate.. I still have one question though, the
> starting bit (0 for even & 1 for odd), you say "Therefore, the network
> address would become 1.1.0.0 ( which is an AND of the binary bits )" . Do
> you mind elaborating further? Everything else makes sense.
>
>
>
>
>
> On Tue, Jul 7, 2009 at 12:42 AM, Ravi Singh <way2ccie_at_googlemail.com>wrote:
>
>> As I was reviewing my sent mail I realised a very small typo in the odd
>> subnet section .. the line should say ..Converting them to binary shows that
>> for the octet to be "ODD" the last bit in the octet is always 1.
>>
>>
>> On Mon, Jul 6, 2009 at 11:38 PM, Ravi Singh <way2ccie_at_googlemail.com>wrote:
>>
>>> See if this helps ..Just a paste of some quick notes that I had prepared
>>>
>>> Filtering routes with an even subnet e.g for a subnet 1.1.0.0/16 , we
>>> need only routes with an even class C subnet for e.g 1.1.0.0/24,
>>> 1.1.2.0/24, 1.1.4.0/24 and so on . Converting them to binary shows that
>>> for the octet to be even the last bit in the octet is always 0
>>> 0 0 0 0 0 0 0 0 - 0
>>> 0 0 0 0 0 0 1 0 - 2
>>> 0 0 0 0 0 1 0 0 - 4
>>> 0 0 0 0 0 1 1 0 - 6 and so on
>>> Therefore, the network address would become 1.1.0.0 ( which is an AND of
>>> the binary bits ) and the wildcard mask will be 0.0.254.255 (which is an XOR
>>> of the binary values) In XOR an output is 1 only when all the bits are
>>> either 0's or 1's . Therefore , an XOR of the above values will be
>>> 0 0 0 0 0 0 0 1 . In another words this means that the values of the 7
>>> bits can be anything but the 8th bit has to be an exact match. This when
>>> represented in a wildcard mask gives a value of 254.
>>>
>>>
>>> Filtering routes with an odd subnet e.g for a subnet 1.1.0.0/16 , we
>>> need only routes with an even class C subnet for e.g 1.1.1.0/24,
>>> 1.1.3.0/24, 1.1.5.0/24 and so on . Converting them to binary shows that
>>> for the octet to be even the last bit in the octet is always 1.
>>> 0 0 0 0 0 0 0 1 - 1
>>> 0 0 0 0 0 0 1 1 - 3
>>> 0 0 0 0 0 1 0 1 - 5
>>> 0 0 0 0 0 1 1 1 - 7 and so on
>>> therefore an XOR for the above values will be the same i.e
>>> 0 0 0 0 0 0 0 1, what changes though is the AND of these values which
>>> means the network address of the subnet. An AND of the binaries is
>>>
>>> 0 0 0 0 0 0 0 1 , which therefore will make use of the value 1.1.1.0 and
>>> the wildcard value of 0.0.254.255 .
>>>
>>> Regards,
>>> Ravi
>>>
>>>
>>>
>>>
>>> On Mon, Jul 6, 2009 at 8:39 PM, Molomo <letjedilakopa_at_gmail.com> wrote:
>>>
>>>> Experts,
>>>>
>>>> I'm really battling to understand how the network and the wildcard bits
>>>> are
>>>> arrived at for odd and even octes on ACL.
>>>> Anyone cares to give brief explanation?
>>>>
>>>> Thanks in adavnce.
>>>>
>>>>
>>>> Molomo
>>>>
>>>>
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Received on Tue Jul 07 2009 - 01:09:27 ART