From: John Edom (jedom123@gmail.com)
Date: Sat Dec 13 2008 - 12:38:47 ARST
hi
Ryan/Sachin thanks for link and explaination. I am clear about 3 octate now
just tell me what about 4 octate because we are matching bit by bit so since
in last octate all 0's are match it should be 0 in wildcard rather then 255?
Regards
On Sat, Dec 13, 2008 at 6:31 PM, Ryan DeBerry <rdeberry@gmail.com> wrote:
>
> http://blog.internetworkexpert.com/2007/12/26/q-how-do-i-compute-complex-wildcard-masks-for-access-lists/
>
>
> On Sat, Dec 13, 2008 at 9:31 AM, John Edom <jedom123@gmail.com> wrote:
>
>> Hi,
>>
>> Do anyone have some complex examples for batter understanding?
>>
>> Regards
>>
>>
>> On Sat, Dec 13, 2008 at 6:28 PM, John Edom <jedom123@gmail.com> wrote:
>>
>>> hi,
>>>
>>> i think wildcard will be 0.0.11.0 because in forth octet all networks
>>> have zeros so all match means 0.
>>>
>>> Third Octate
>>> 0100
>>> 0101
>>> 0110
>>> 0111
>>> 1100
>>> 1101
>>> 1110
>>> 1111
>>> --------
>>> 1011 - wildcard that is 11
>>>
>>> Correct me if i am wrong.......
>>>
>>> Regards
>>>
>>>
>>> On Sat, Dec 13, 2008 at 6:09 PM, Ryan DeBerry <rdeberry@gmail.com>wrote:
>>>
>>>> I think it should be
>>>>
>>>> 192.168.4.0 0.0.11.255
>>>>
>>>> Third octet has last 4 in common
>>>>
>>>> 8-4-2-1
>>>>
>>>> 4 is always on so wild card mask it would be off.
>>>>
>>>> 8+2+1=11
>>>>
>>>> On Sat, Dec 13, 2008 at 7:42 AM, John Edom <jedom123@gmail.com> wrote:
>>>>
>>>>> Hi,
>>>>>
>>>>> can we write these all network in single ACL line
>>>>>
>>>>> 192.168.4.0
>>>>> 192.168.5.0
>>>>> 192.168.6.0
>>>>> 192.168.7.0
>>>>>
>>>>> 192.168.12.0
>>>>> 192.168.13.0
>>>>> 192.168.14.0
>>>>> 192.168.15.0
>>>>>
>>>>> Regards
>>>>>
>>>>>
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>>>>>
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