From: Joseph Brunner (joe@affirmedsystems.com)
Date: Thu Oct 25 2007 - 13:07:47 ART
That was quite awesome Scott!
Want to come to Vegas next weekend and play some blackjack?
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
Scott Vermillion
Sent: Thursday, October 25, 2007 11:29 AM
To: 'shiran guez'; 'Martin McGannon'
Cc: ccielab@groupstudy.com
Subject: RE: IPV6 Question
Hi Shiran,
Binary is certainly the most fool-proof method to be sure. But I like to be
able to do this sort of thing on the fly. It helps to start thinking in hex
and then you don't need to take so much time. As you noted, within a
quartet, each position represents 16 possibilities. Knowing this leads to
some pretty basic arithmetic. Using the below example, hex 0025 is decimal
37 (2 x 16 + 5). Knowing that and having memorized all of the powers of 2
that apply to the IP world allows you to quickly realize that you need six
bits of a mask to cover more than 32 but less than 64. And since the base
address was given, and it had a mask of /64, you can pretty quickly come to
the correct answer of /58. It can be done in your head almost on sight - no
binary conversions required. This sort of task takes no more than five
seconds to be solved and double-checked in your head if thought out in hex
and decimal with no (or very little) consideration for actual binary
numbers.
Regards,
Scott
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
shiran guez
Sent: Thursday, October 25, 2007 12:46 AM
To: Martin McGannon
Cc: ccielab@groupstudy.com
Subject: Re: IPV6 Question
The easyest way for this 2 network summary is to take the bit format
remember :
1) IPv6 is represented in Hex
2) each part contain 16 bit
So you have here contigues 3 parts so you have 3x16 = 48
now take the 4th part and convert it to bin
12
0000 0000 0001 0010
25
0000 0000 0010 0101
now you can see very clear how many bit you can take to the summary, 10 bit
now take the first 48 bits + the 10 from your summary and you got a /58
Realy not very hard just a lot of 0 and 1.
On 10/24/07, Martin McGannon <mmcgannon@pmis.ie> wrote:
>
> Hi,
>
> A few basic questions :-
>
>
> If requested to summarize the following prefixes in IPV6 what would the
> summary address be
>
> 2001:141:1:12/64
> 2001:141:1:25/64
>
> 2001:141:1::/58 or 59? and why.
>
> Also in a prevoius lab exexcise I was asked to implement Site Local
> Addressing
> with my existing IPV4 address in place. The interface address was
> 192.168.1.100 say. How would I create a Site Local address from this, give
> that the first 10 bits are always FEC0:X:X etc.. Any ideas.
>
> Lastly with RIPNG over NBMA do we need to add the broadcast keyword for
> the
> link locals. Can we use neighbor statements under the Router Process.
>
> Many Thanks,
>
> Martin
>
> _______________________________________________________________________
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-- Shiran Guez MCSE CCNP NCE1 http://cciep3.blogspot.com http://www.linkedin.com/in/cciep3
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