From: shi bindong (bindong.shi@gmail.com)
Date: Sun Jul 02 2006 - 23:44:18 ART
i think the solution guide is correct. 51.1.0.1. 0.0.0.8 is simply derived
on the operation rule, but it also including subnet 51.1.0.1.(so it is not
correct). to be more acurate, we should creat 2 ACLs.
On 7/1/06, Sami <sy1977@gmail.com> wrote:
>
> Group,
>
> One of the task say use minimum amount of line necessary to comple this
> task.
>
> 51.3.0.1
> 51.5.0.1
> 51.7.0.1
> 51.3.0.9
> 51.5.0.9
> 51.7.0.9
>
> How can I combine in one access list ?
>
> Thanks
>
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-- your friendsbindong
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