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. > 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"). -- With respect, Roman ~~~~~~~~~~~~~~~~~~~~~~~~~~~ "Stallman had a printer, with code he could not see. So he began to tinker, and set the software free."
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