From: Shine (shinepjoseph@iprimus.com.au)
Date: Thu Mar 22 2007 - 16:17:35 ART
This time it is more vague.
----- Original Message -----
From: "M S" <michaelgstout@hotmail.com>
To: "Jeff Mullan" <jmullan78@gmail.com>
Cc: <ccielab@groupstudy.com>
Sent: Friday, March 23, 2007 1:32 AM
Subject: RE: subnetmask problem
> No, sorry for not being clear. Three addresses
> 45.194.169.10561.202.173.24341.234.41.250I want the most specifcic one
> line
> access-list to permit these
>
>
> Date: Wed, 21 Mar 2007 21:36:15 -0700From: jmullan78@gmail.comTo:
> michaelgstout@hotmail.comSubject: Re: subnetmask problemCC:
> ccielab@groupstudy.comsorry, a bit confused. You have 2 examples here
> 194,202,234 and 2nd one 40,169,173,41 ?
> On 3/21/07, M S <michaelgstout@hotmail.com> wrote:
> Hello:I am looking at a summerization problem that requires the use of the
> mostefficient one line access-list to permit three ip addresses to telnet
> into
> a routerCan anybody verify my logic for these three numbers, please?194 =
> 1100
> 0010202 = 1100 1010234 = 1110 1010------------------------40 = 0010
> 1000So
> the mask for this Octet should be 40, is that corrrect? 169 = 1010 1001173
> =
> 1010 110141 = 0010 1001------------------------132 = 1000 0100So the
> mask
> for this octet should be 132, is that correct.Isthere a calculator
> shortcut
> that can be used to gain these mask values? Thank you very
> kindly.----------------------------------------------------------------------
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