diff options
| author | martin f. krafft <madduck@madduck.net> | 2008-10-15 10:13:23 +0200 |
|---|---|---|
| committer | martin f. krafft <madduck@madduck.net> | 2008-10-15 10:13:23 +0200 |
| commit | b8fab3955776b1d2d948a24351e726fc11ac5d13 (patch) | |
| tree | a0eb58b7174b0edd4ed80601f2b5b9f318cd251d /debian/FAQ | |
| parent | 6d6ba849967e924a8f612da4b414bcd9b321fe05 (diff) | |
fix FAQ 4b probability (#493577)
Diffstat (limited to 'debian/FAQ')
| -rw-r--r-- | debian/FAQ | 11 |
1 files changed, 7 insertions, 4 deletions
@@ -137,8 +137,8 @@ The latest version of this FAQ is available here: 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 + In two thirds 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. If A fails, the RAID6 can only survive if the second failing disk is either C or D; If B fails, your array is dead. @@ -151,8 +151,11 @@ The latest version of this FAQ is available here: See also question 18 further down. - 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/ + 0. it's actually (n-2)/(n-1), where n is the number of disks. I am not + a mathematician, see http://aput.net/~jheiss/raid10/, which gives the + chance of *failure* as 1/(n-1), so the chance of success is 1-1/(n-1), or + (n-2)/(n-1), or 2/3 in the four disk example. + (Thanks to Per Olofsson for clarifying this in #493577). 5. How to convert RAID5 to RAID10? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ |