Awesome.  Thanks Tyson.  It is clear now.

Thanks,
Vince Mashburn
HP Americas Technology Services
Account Services Manager
FedEx On-site Office Phone: 901-263-6498
Mobile: 901-569-9734


-----Original Message-----
From: Tyson Scott [mailto:[email protected]] 
Sent: Tuesday, February 09, 2010 7:58 AM
To: Mashburn, Vince; [email protected]
Subject: RE: [OSL | CCIE_RS] Advanced ACL filtering

Vince,

let's say you have

172.16.18.0/24
172.16.82.0/24

As you can see there are only two variants in the third octet.  18 and 82.

Binary for these two numbers is
00010010
01010010

We can easily see that there is only one bit difference between the two
numbers, the 64 bit.

As a shortcut seeing that 18 and 82 are the only difference in the networks
above I could do
82-18 = 64.  64 is a power of 2 and it is a bit boundary.  So by subtracting
the two I quickly get my network and mask without taking it to binary.  The
network is the lowest common denominator, 18; and mask is the sum or 64.

Regards,
 
Tyson Scott - CCIE #13513 R&S, Security, and SP
Technical Instructor - IPexpert, Inc.
Mailto: [email protected]
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-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Mashburn, Vince
Sent: Tuesday, February 09, 2010 7:18 AM
To: [email protected]
Subject: [OSL | CCIE_RS] Advanced ACL filtering

Question on ACL filtering with binary operations. When referring to the
following shortcut:

With seemingly random numbers, if only two variants per octet, sibtract the
values.
If the answer is a power of 2, then only 1 bit in the ACL is different.

Can you clarify what "two variants per octet" is? Does it refer to the
number of 1's in the mask or the number of bits different in the network?
Also, when subtracting the two values, is that in decimal or binary?

Vince Mashburn
Sent from my Windows Mobile phone
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