Re: Q: Hard Disc Layout for Tjener? (Raid)

[Jonas Smedegaard <dr@jones.dk>]

Frank Weißer wrote:
> <zitiere wer="Jonas Smedegaard">
>
> > Frank Weißer wrote:
> >
> > > <zitiere wer="Maximilian Wilhelm">
> > >
> > > > RAID 5: (distributed space management)
> > > > =====================================>> * at least 3 disc (normally equal
> > > > size)
> > > > * If all disc have the same size
> > > >  size(RAID) = (n - 1 / n) * disc size.
> > > > * If one disc failes, replace it and be happy.
> > > >
> > > > So you would have ~108GB useable space.
> > >
> > >
> > > hmm, i always thought 3,75*36GB would be something like 135GB ;-)
> >
> > Maybe it's just me, but how do you end up using 3,75? Are you perhaps
> > setting up a RAID consisting of 4,75 disks?
> >
> > (more likely you are quoting a different formula than the one you are
> > using).
>
>
> Can't see that: (4 - 1/4) = 3.75 , isn't it, or the () have to be set in
> another way!

You are right (in interpreting the above) - I need my glasses fixed.

Nevertheless I agree with the later post by Max: The right calculation 
(and the way I wrongly read the above) is this:

  (n - 1) * [size of smallest disk]


 - Jonas


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--
* Jonas Smedegaard - idealist og Internet-arkitekt
* Tlf.: +45 40843136  Website: http://dr.jones.dk/

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