From: Ryan DeBerry (rdeberry@gmail.com)
Date: Sat Dec 13 2008 - 12:46:00 ARST
If that were the case how many hosts are you gonna have on those networks?
NONE.
Make sense?
On Sat, Dec 13, 2008 at 9:38 AM, John Edom <jedom123@gmail.com> wrote:
> 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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