From: Fahad Khan (fahad.khan@gmail.com)
Date: Sun Aug 31 2008 - 07:04:31 ART
Sorry for the typing mistake in my last post,
*For all odd*,
The IP address will be 1.1.1.1
With the wild card mask as *0.0.0.25* <http://0.0.0.25/>*4 <--------- its
254 not 25*
254 = 11111110
Here, 0 means DO CARE of the last bit in IP address (must be ONE)
Eventually, ACL will be "permit 1.1.1.1 0.0.0.254"
On 8/31/08, Fahad Khan <fahad.khan@gmail.com> wrote:
>
>
>
> Let say we have 1.1.1.0/24 as u mentioned,
>
> The game will be played over last octet, in fact the least significant bit
> of last octet,
>
> If it is 0, the IP address will be Even
>
> If it is 1, the IP address will be Odd
>
> For example,
>
> 1.1.1.00000001 = 1.1.1.1 - odd
>
> 1.1.1.00000011 = 1.1.1.3 - odd
>
> 1.1.1.00000010 = 1.1.1.2 even
>
> 1.1.1.00000100 = 1.1.1.4 even
>
> Etc
>
> Here the wild card bits must be used trickily
>
> *For all Even*,
>
> The IP address will be 1.1.1.0
>
> With the wild card mask as 0.0.0.254
>
> 254 = 11111110
>
> Here, 0 means DO CARE of the last bit in IP address (must be ZERO)
>
> Eventually, ACL will be "permit 1.1.1.0 0.0.0.254"
>
> *For all odd*,
>
> The IP address will be 1.1.1.1
>
> With the wild card mask as 0.0.0.25
>
> 254 = 11111110
>
> Here, 0 means DO CARE of the last bit in IP address (must be ONE)
>
> Eventually, ACL will be "permit 1.1.1.1 0.0.0.254"
>
>
>
> On 8/31/08, Ashish . <invent.xr@gmail.com> wrote:
>>
>>
>> Hi Farhad,
>>
>> I understood the logic u used in this particular prob....however i'm not
>> able to figure out the way to use the same logic in the age old questions
>>
>> a: permit all odd ip
>> b: permit all even ip
>>
>> can you help me understand the above two in say 1.1.1.0/24 .
>>
>> thanks in advance,
>> Ashish
>>
>>
>>
>>
>>
>>
>> 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
>>> >
>>> >
>>> > Blogs and organic groups at http://www.ccie.net
>>> >
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>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>>
>>>
>>> --
>>> Fahad Khan
>>>
>>>
>>> Blogs and organic groups at http://www.ccie.net
>>>
>>> _______________________________________________________________________
>>> Subscription information may be found at:
>>> http://www.groupstudy.com/list/CCIELab.html
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>
>
>
> --
> Fahad Khan
-- Fahad KhanBlogs and organic groups at http://www.ccie.net
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