From: Scott Vermillion (scott_ccie_list@it-ag.com)
Date: Thu Oct 25 2007 - 13:35:08 ART
Well, that's actually a very good point. That calc was not available during
the written of course, but in the lab I guess you would certainly be better
off to be 100% sure using the tools made available to you...
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
shiran guez
Sent: Thursday, October 25, 2007 9:37 AM
To: Scott Vermillion
Cc: Martin McGannon; ccielab@groupstudy.com
Subject: Re: IPV6 Question
call me old fashion but I prefer to be 2 sec slower and be accurate the bin
conversion to my opinion is easier then the basic arithmetic.
also relate to the test, even if you have problem with number conversion
then you have the calc and there is no simple then that.
On 10/25/07, Scott Vermillion <scott_ccie_list@it-ag.com> wrote:
>
> 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
> >
> > _______________________________________________________________________
> > Subscription information may be found at:
> > http://www.groupstudy.com/list/CCIELab.html
> >
>
>
>
> --
> Shiran Guez
> MCSE CCNP NCE1
> http://cciep3.blogspot.com
> http://www.linkedin.com/in/cciep3
>
> _______________________________________________________________________
> Subscription information may be found at:
> http://www.groupstudy.com/list/CCIELab.html
>
>
-- Shiran Guez MCSE CCNP NCE1 http://cciep3.blogspot.com http://www.linkedin.com/in/cciep3
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