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

0 0 0 0 0 1 0 1
0 0 0 0 1 1 0 1
---------------
10.0.5.0 0.0.10.255

0 0 0 0 1 1 0 0
---------------
10.0.12.0 0.0.0.0

HTH,
Mohammed Mahmoud.


Amit Kohli <amitkohli18@xxxxxxxxx> 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 <djerk@xxxxxxxx> 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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