edition- i meant to say that the wildcard bits will include 32 addresses,
with ip addresses A, B, included.



On Thu, Jul 7, 2011 at 7:18 AM, gaurav nunia <[email protected]> wrote:

> let say your both IPs as A, B.
>
> then
>
> A  'AND' B ===  8.0.1.0
>
> and
>
> A 'XOR' B = 194.0.2.4  =  5 ones, gives 32 ip addresses,
> so a ip address of 8.0.1.0 with a wildcard mask 'what u computed'  will
> include both as u suggested. :)
>
>
> .
>
>
>
>
> On Thu, Jul 7, 2011 at 5:13 AM, Alef <[email protected]> wrote:
>
>> Can anyone confirm that if i want to include both 10.0.1.4 and 200.0.3.0
>> in a single statement, that i would have to use a wildcard mask of 194.0.2.4
>> ?
>>
>> this would allow for 5 bits difference in total, 32 networks
>> and the following octet possibilities
>> [0,2,8,64,128,192,200,202]][0][1,3][4,0]
>>
>> 0.0.1.4            64.0.1.4          200.0.1.4
>> 0.0.1.0            64.0.1.0          200.0.1.0
>> 0.0.3.4            64.0.3.4          200.0.3.4
>> 0.0.3.0            64.0.3.0          200.0.3.0
>>
>> 2.0.1.4            128.0.1.4        202.0.1.4
>> 2.0.1.0            128.0.1.0        202.0.1.0
>> 2.0.3.4            128.0.3.4        202.0.3.4
>> 2.0.3.0            128.0.3.0        202.0.3.0
>>
>> 8.0.1.4            192.0.1.4
>> 8.0.1.0            192.0.1.0
>> 8.0.3.4            192.0.3.4
>> 8.0.3.0            192.0.3.0
>> _______________________________________________
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>> visit www.ipexpert.com
>>
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>>
>
>
>
> --
> thanks
> gaurav
>
> http://routing0sand1s.blogspot.com/
>
>


-- 
thanks
gaurav

http://routing0sand1s.blogspot.com/
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