From: Fahad Khan (fahad.khan@gmail.com)
Date: Sat Aug 30 2008 - 16:43:16 ART
For same number of inputs,output is the same in case of XOR.
On 8/31/08, syed hafiz <mail4hafiz@gmail.com> wrote:
>
> Thanks for tht reply anant tht solved my issue ,
> but continuing to tht question y have u made the last octet of the wild
> card mask '0'
> as i was stuck abt tht. is tht includes for XOR operation as well
>
> thanks
> ashhy
>
>
>
> On 8/30/08, Fahad Khan <fahad.khan@gmail.com> wrote:
>>
>> The actual game is of 3rd octat in this case.
>>
>> starting IP = x.x.4.0
>> ending IP = x.x.15.0
>>
>> Four most significant bits are common in the 3rd octat i.e 0000
>>
>> so we have 0100 (4) and 1111(15) for our play
>>
>> now for calculating IP in accesslist,
>>
>> perform AND operation in between, 0100 and 1111
>>
>> 0100
>> 1111
>> -------
>> 0100 = 4
>> -------
>> Hence, the IP is decided now as x.x.4.0
>>
>> Now for calculation of wildcard mask, perform XOR operation
>>
>> 0100
>> 1111
>> -------
>> 1011 = 11
>> -------
>>
>> Rules for XOR:
>>
>> - For same number of inputs output is 0 otherwise 1
>> - For even number of 1s output is 0 while for odd number of 1s output is
>> 1.
>>
>> hense you get the ACL as:
>>
>> permit x.x.4.0 0.0.11.0
>>
>>
>> HTH
>> On 8/30/08, ashhy <mail4hafiz@gmail.com> wrote:
>>>
>>> Hi all
>>> i want to know how to get this right acl for the below mentioned question
>>>
>>> i can use only one acl to do this ,i am getting 16 routes of class c
>>> from xxx.xxx.1.0 to xxx.xxx.16.0
>>> from backbone out of which i should allow only
>>>
>>> xxx.xxx.4.0
>>> xxx.xxx.5.0
>>> xxx.xxx.6.0
>>> xxx.xxx.7.0
>>> xxx.xxx.12.0
>>> xxx.xxx.13.0
>>> xxx.xxx.14.0
>>> xxx.xxx.15.0
>>>
>>> please guide me or else suggest if this is already disscussed.
>>>
>>> Thanks
>>> Asshy
>>>
>>>
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>>
>>
>> --
>> Fahad Khan
>
>
>
-- Fahad KhanBlogs and organic groups at http://www.ccie.net
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