diff options
| author | madduck <madduck@3cfab66f-1918-0410-86b3-c06b76f9a464> | 2006-10-26 18:40:26 +0000 |
|---|---|---|
| committer | madduck <madduck@3cfab66f-1918-0410-86b3-c06b76f9a464> | 2006-10-26 18:40:26 +0000 |
| commit | 577ef7c9ac951f20f6697a52ee25fc61f38df0c0 (patch) | |
| tree | 9f885fef40a0ea0ada8ab562b27a8f4e4626f1a2 /debian/FAQ | |
| parent | 767210c5fbd689906c7795cacae9904c24d0b113 (diff) | |
faq updates
Diffstat (limited to 'debian/FAQ')
| -rw-r--r-- | debian/FAQ | 6 |
1 files changed, 5 insertions, 1 deletions
@@ -129,6 +129,9 @@ Also see /usr/share/doc/mdadm/README.recipes.gz 4b. Can a 4-disk RAID10 survive two disk failures? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ + I am assuming that you are talking about a setup with two copies of each + block, so --layout=near2/far2/offset2: + In half of the cases, yes [0], and it does not matter which layout you use. When you assemble 4 disks into a RAID10, you essentially stripe a RAID0 across two RAID1, so the four disks A,B,C,D become two pairs: A,B and C,D. @@ -138,7 +141,8 @@ Also see /usr/share/doc/mdadm/README.recipes.gz Thus, if you see a disk failing, replace it as soon as possible! If you need to handle two failing disks out of a set of four, you have to - use RAID6. + use RAID6, or store more than two copies of each block (see the --layout + option in the mdadm(8) manpage). 0. it's actually 1/(n-1), where n is the number of disks. I am not a mathematician, see http://aput.net/~jheiss/raid10/ |