Chris Aitken <cs.aitken@???> wrote:
>>> RAID-5 is computationally expensive, and as it
>>> relies on at least three drives as opposed to
>>> RAID-1's two, it has a shorted MTBF.
>> Really? Do you have a reference for that?
> System MTBF = component MTBF/n
> where n is # of components.
>
> So for a component with MTBF of 50k hrs a 2 disk
> (RAID-1) system has an MTBF of 25k hours, and a
> 3 disk (RAID-5) system has an MTBF of ~16.7k
> hrs.
>
> During rebuilts the above RAID-1 has an MTBF of
> 50 k hrs again, and the above RAID-5 has 25k
> hour MTBF.
I'm replying to Chris' post, as Keith's is saying
much the same in a different way...
I'm still not following this reasoning. If the
above is true, RAID is pointless.
Here's an interesting, if rather brief, article on
Mean Time to Data Loss of RAID 5, 6, and - in the
comments - 1:
http://storageadvisors.adaptec.com/2005/11/01/raid-reliability-calculations/
That gives, for RAID 1:
MTTDL_R1 = MTBF ^ D
where MTBF is the time to failure of a single
disk, assuming they're all the same, and D is the
number of drives in the RAID 1 array.
For RAID 5, the calculation is:
MTBF ^ 2
MTTDL_R5 = -------------
D * (D - 1)
(That's ignoring time to rebuild after a failure,
reliability of the rest of the system, and common
factors affecting both disks. It also assumes that
failed drives are replaced.)
Those equations suggest that the reliability of
RAID 5 _is_ less than RAID 1 - for RAID 5 over
three drives and RAID 1 over 2, the MTTDL for the
former is a sixth of the latter.
However, they're both far better than the MTBF of
a single drive. For arrays with drives of 100k
Hours MTBF, two-drive RAID 1 gives an MTTDL of 10G
Hours, and three-drive RAID 5 gives an MTTDL of
1.7G Hours.
Nick.
--
Nick Chalk ................. once a Radio Designer
Confidence is failing to understand the problem.
