Roman Mamedov schrieb:
On Tue, 05 Jun 2012 19:27:53 +0200
"Joachim Otahal (privat)"<Jou@xxxxxxx> wrote:
Hi,
Debian 6.0.4 / superblock 1.2
sdc1 = 1.5 TB
sdd1 = 1.5 TB (cannot be used during --create, contains still data)
sde1 = 1 TB
sdf1 = 1 TB
sdg1 = 1 TB
Target: RADI5 with 4.5 TB capacity.
The normal case would be:
mdadm -C /dev/md3 --bitmap=internal -l 5 -n 5 /dev/sdc1 /dev/sdd1
/dev/sde1 /dev/sdf1 /dev/sdg1
What I expect: since the first and the second drive are 1.5 TB size the
third fouth and fifth drive are treated like 2*1.5 TB, creating a 4.5 TB
RAID.
Lolwhat.
Hey, there is a reason why I ask, no need to lol.
What would really be created: I know here are people that know and not
guess : ).
5x1TB RAID5. Lowest common device size across all RAID members is utilized in
an array.
But what you do after that, is you also create a separate 2x0.5TB RAID1 from
the 1.5TB drives' "tails", and join both arrays into a single larger volume using LVM.
The result: 4.5 TB of usable space, with one-drive-loss tolerance (provided by
RAID5 in the first 4 TB, and by RAID1 in the 0.5TB "tail").
Thanks for clearing that up. I probably would have noticed when trying
in a few weeks, but knowing beforehand helps.
To make you lol more, following would work too:
Use only 750GB partitions, use the 3*250 GB loss at the end of each 1 TB
drive for the fourth 750 GB, and RAID6 those 8*750. Result is 4.5 TB
with a one-drive-loss tolerance and really bad performance.
I spare you the 500 GB partitions example which result in 4.5 TB with a
one-drive-loss tolerance and really bad performance.
Jou
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