12 Oct
2009
12 Oct
'09
6:47 p.m.
Hi,
On Monday 12 October 2009 19:44:47 Jörn Nettingsmeier wrote:
Arnold Krille wrote:
Please listen to what I said: Two (or more) disks of the same manufacturer and
same type and maybe even continuous id-numbers.
And, bang!, now the failure of one device is highly connected to the failure
of the other one. Because the error (or call it small mis-fit) happening in the
production of the first is repeated in the second with a high probability.
And I worked on a manufacturing band several times, I know how humans catch
slight errors only after doing them the third or fourth time. By which its too
late to sort out and check the last three products.
Unless I miss something on AVLinux:
On Saturday 03 October 2009 21:36:54 Jörn Nettingsmeier wrote:
that remark is bogus.
each disk design has a mean-time between failures
of X hours. that means of N disks, N/2 will have failed after time X for
a sufficiently high number of samples N.
the failure of each individual disk is a statistically independent
event. Jonathan E. Brickman wrote:
Loosing two disks means you got more then two coupled. If so, you have
definitely several of the same charge in that array. (Don't deny it, the
human laziness speaks against you.)
Recovering will still do heavy load on the disks. Now calculate the
probability of two disks of the same charge failing without a third disk
of that charge failing during recovery. Pretty low, huh? > You should at least use RAID6.
1. How does RAID6 differ from RAID5?
raid 6 uses a more elaborate algorithm than
the simple parity of raid5,
to calculate two redundant bits. so you can recover from the failure of
two disks out of a raid6 array. there is absolutely no synchronized self-destruct of
disks
belonging to the same batch.
Yes, there is! I know from bad experience, both my own (not very tragic) and
that of a big company.
Have fun and make backups,
Arnold