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

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
* 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

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

P.S. Please do not cc me when addressing this list. A default rule of Debian mailinglists is to not cc unless explicitly asked for.

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

 - Enden er nær: http://www.shibumi.org/eoti.htm

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