From: Kenneth Wygand (KWygand@customonline.com)
Date: Sun May 02 2004 - 21:08:52 GMT-3
Alsontra,
Unfortunately not. If you actually map out the "don't care" bits in your masks, your first mask is "00000000.00000000.01111111.11111111". If we don't count the last octet (because these are host bits due to the fact that I said the networks were /24's), you have 7 "don't care" bits for your network portion. 2^7 is 128, so you are actually matching 128 networks with your first statement, much more than my question allowed. Likewise, your second statement includes 14 "don't care" bits, totaling 16384 networks (2^14).
Keep thinking... and think about how the "don't care" bits work. It might help to actual map out the networks I mentioned in binary...
Good luck!
Ken
-----Original Message-----
From: alsontra@hotmail.com [mailto:alsontra@hotmail.com]
Sent: Sun 5/2/2004 9:50 PM
To: Kenneth Wygand
Cc:
Subject: Re: Quiz Question of the Day 20040502
Sorry about that. I mean..
10.1.0.0 0.0.254.255
10.0.1.0 0.254.254.255
Thnaks- Here for lucky guesses...????
Alsontra
----- Original Message -----
From: "Kenneth Wygand" <KWygand@customonline.com>
To: <ccielab@groupstudy.com>
Sent: Sunday, May 02, 2004 4:36 PM
Subject: Quiz Question of the Day 20040502
> Write an access list (ACL 10) using the _fewest_ number of lines that
denies _all_ of the following networks _without_ oversummarization and
permits all other networks.
>
> 10.1.1.0 /24
> 10.1.2.0 /24
> 10.1.4.0 /24
> 10.1.8.0 /24
> 10.1.16.0 /24
> 10.1.32.0 /24
> 10.1.64.0 /24
> 10.1.128.0 /24
> 10.1.1.0 /24
> 10.2.1.0 /24
> 10.4.1.0 /24
> 10.8.1.0 /24
> 10.16.1.0 /24
> 10.32.1.0 /24
> 10.64.1.0 /24
> 10.128.1.0 /24
>
> If you really understand how access lists work, this one should be easy...
:)
>
> Ken
>
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