Michael,
Thanks a lto for mentioning my name, but
I've just looked through follow-ups and found out that John Neiberger was
the first to give correct answer.
RGRDS,
EVgeny
--
***********
Please DO NOT send replies just directly to me, post them to the Newsgroup
as well
***********
""Michael L. Williams"" <[EMAIL PROTECTED]> wrote in message
8jt2kl$2s4$[EMAIL PROTECTED]">news:8jt2kl$2s4$[EMAIL PROTECTED]...
> Okay. There have been alot of responses to this. And alot of people
> wanting the answer.
>
> SO, let me mention Evgeny Babanin and Chuck Hwang, since they gave correct
> responses to the question! Congrats!!
>
> Now, I'll everyone's agony........... =)
>
> The answer is:
>
> access-list 10 permit A.B.C.1 255.255.255.24
>
> Here's the explanation:
>
> Remeber in the wildcard mask, for a given bit:
> 0 = must match the same bit in the source address given
> 1 = doesn't matter what it is........
>
> I will use A.B.C.1 for the source address in the access-list command since
> (as we see below) the least significant should always be 1. Below I have
> the binary version of our address 0.0.0.1:
>
> aaa aaa.bbb bbb.ccc ccc.0000 0001 <- least significant bit = 1, thus
> the decimal value of that octect is 1. This is crutial because this is
the
> only non-zero bit in the source address. We'll use that later.
>
> Limiting ourselves to looking at the last octet of the target IPs we want
to
> let through,
> 0000 0001 = 1
> 0000 1001 = 9
> 0001 0001 = 17
> 0001 1001 = 25
> ---------------
> So since the last bit (xxxx xxx1) has to = 1 (like in the source address),
> it's corresponding wildcard bit must = 0
>
> so we know the wildcard mask so far has to be xxxx xxx0 (last octect
only)
>
> we also know that bits (from right to left) 2, 3, 6, 7, 8 HAVE to be = 0
(so
> they match the zeros in the source address), so therefore their
> corresponding bits in the wildcard mask must be = 0. So now we have:
>
> 000x x000 (last octect only)
>
> When looking at bit 4 and 5 we see that they could be either 0 or 1. Any
of
> the
> four combinations of 0 and 1 give us the four target addresses, so
> their bits in the wildcard mask are = 1. So our final wildcard mask will
be
>
> 0001 1000 (last octect only) = 24 (decimal)
>
> So...... the following command is the correct answer:
>
> access-list 10 permit A.B.C.1 255.255.255.24
>
> It's a tough question,...... That's why it's fun =) Thanks to everyone
that
> participated!
>
> Mike W.
>
> PS: Alrighty Chuck..... we're ready for the next question =)
>
> > Puzzle posed by Mike Williams:
> >
> > Make a (single line) access-list that will only allow traffic from
> addresses
> > matching the following model:
> >
> > x.x.x.1
> > x.x.x.9
> > x.x.x.17
> > x.x.x.25
> >
> > It's along the same lines as the question you posed, but a little more
> > challenging since you only want those 4 addresses to pass through.....
=)
>
>
>
> ___________________________________
> UPDATED Posting Guidelines: http://www.groupstudy.com/list/guide.html
> FAQ, list archives, and subscription info: http://www.groupstudy.com
> Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
> ---
___________________________________
UPDATED Posting Guidelines: http://www.groupstudy.com/list/guide.html
FAQ, list archives, and subscription info: http://www.groupstudy.com
Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
