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RE: Funky acl question-Check if this is correct posted 07/09/2007
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Mohamed
   
  See answer below
  

Mohamed M Moustafa <mmma@xxxxxxxxx> wrote:
  Hi All,

What about this one:

x x x x x x x x
0 0 0 0 0 1 0 1 - 5
0 0 0 0 1 0 1 0 - 10
0 0 0 0 1 1 0 0 - 12
0 0 0 0 1 1 0 1 - 13
0 0 0 0 1 1 1 1 - 15


0 0 0 0 1 0 1 0
0 0 0 0 1 1 1 1
---------------
10.0.10.0 0.0.5.255 Will match .10 .11 .14 .15 because as Djerk explained 2bits^2 = 4 addresses matched.

0 0 0 0 0 1 0 1
0 0 0 0 1 1 0 1
---------------
10.0.5.0 0.0.10.255 Will match .5 and .13            2^1 = 2addresses matched 

0 0 0 0 1 1 0 0
---------------
10.0.12.0 0.0.0.0  Will match .12 only                2^0= 1 address matched  

HTH,
Mohammed Mahmoud.


Amit Kohli wrote on 9 Jul 2007, 05:09 PM:
Subject: RE: Funky acl question-Check if this is correct
>Djerk
> 
> Very good and correct explanation. Thanks for pointing out the mistake in
>line 2.
> 
> Regards
> Amit
>
>Djerk Geurts wrote:
> Amit,
>
>My take is listed below...
>
>> 3 lines tho
>> 
>> permit 10.0.5.0 0.0.0.255 (Permit only .5)
>
>Correct
>
>> permit 10.0.10.0 0.0.5.255 (Permit only .10 and .15)
>
>A 2 bit mask (3rd octet) will allow 4 networks: .10 .11 .14 and .15
>
>> permit 10.0.12.0 0.0.1.255 (Permit only .12 and .13)
>
>Correct
>
>> Let me know if this is correct?
>
>You're close, don't forget that each bit is a don't care. Each bit in the
>network part of the address adds n2 options (aka matched networks).
>
>Regards,
>Djerk Geurts
>
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>Best Regards,
> Amit Kohli
> 
>
> 
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Best Regards,
  Amit Kohli
   

       
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