new qs 18-20

1 files changed, 81 insertions, 0 deletions

diff --git a/debian/FAQ b/debian/FAQ
index d35e1fda..e933b8f3 100644
--- a/debian/FAQ
+++ b/debian/FAQ
@@ -144,6 +144,8 @@ Also see /usr/share/doc/mdadm/README.recipes.gz
   use RAID6, or store more than two copies of each block (see the --layout
   option in the mdadm(8) manpage).
 
+  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/
 
@@ -402,6 +404,85 @@ Also see /usr/share/doc/mdadm/README.recipes.gz
   overridden with the --force and --assume-clean options, but it is not
   recommended. Read the manpage.
 
+18. How many failed disks can a RAID10 handle?
+~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+  (see also question 4b)
+
+  The following table shows how many disks you can lose and still have an
+  operational array. In some cases, you *can* lose more than the given number
+  of disks, but there is no guarantee that the array survives. Thus, the
+  following is the guaranteed number of failed disks a RAID10 array survives
+  and the maximum number of failed disks the array can (but is not guaranteed
+  to) handle, given the number of disks used and the number of data block
+  copies. Note that 2 copies means original + 1 copy. Thus, if you only have
+  one copy, you cannot handle any failures.
+
+                  1            2            3            4    (# of copies)
+        1        0/0          0/0          0/0          0/0
+        2        0/0          1/1          1/1          1/1
+        3        0/0          1/1          2/2          2/2
+        4        0/0          1/2          2/2          3/3
+        5        0/0          1/2          2/2          3/3                         
+        6        0/0          1/3          2/3          3/3
+        7        0/0          1/3          2/3          3/3
+        8        0/0          1/4          2/3          3/4
+  (# of disks)
+
+  Note: I have not really verified the above information. Please don't count
+  on it. If a disk fails, replace it as soon as possible. Corrections welcome.
+
+19. What should I do if a disk fails?
+~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+  Replace it as soon as possible:
+    
+    mdadm --remove /dev/md0 /dev/sda1
+    halt
+    <replace disk and start the machine>
+    mdadm --add /dev/md0 /dev/sda1
+
+20. So how do I find out which other disk(s) can fail without killing the
+    array?
+~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+  Did you read the previous question and its answer?
+  
+  For cases when you have two copies of each block, the question is easily
+  answered by looking at the output of /proc/mdstat. For instance on a four
+  disk array:
+
+    md3 : active raid10 sdg7[3] sde7[0] sdh7[2] sdf7[1]
+
+  you know that sde7/sdf7 form one pair and sdg7/sgh7 the other.
+
+  If sdh now fails, this will become
+
+    md3 : active raid10 sdg7[3] sde7[0] sdh7[4](F) sdf7[1]
+
+  So now the second pair is broken; the array could take another failure in
+  the first pair, but if sdg now also fails, you're history.
+
+  Now go and read question 19.
+
+  For cases with more copies per block, it becomes more complicated. Let's
+  think of a seven disk array with three copies:
+
+    md5 : active raid10 sdg7[6] sde7[4] sdb7[5] sdf7[2] sda7[3] sdc7[1] sdd7[0]
+
+  Each mirror now has 7/3 = 2.33 disks to it, so in order to determine groups,
+  you need to round up. Note how the disks are arranged in increasing order of
+  their indices (the number in brackes in /proc/mdstat):
+
+  disk:     -sdd7-  -sdc7-  -sdf7-  -sda7-  -sde7-  -sdb7-  -sdg7-
+  group:    [      one       ][      two       ][     three      ]
+
+  Basically this means that after two disk failed, you need to make sure that
+  the third failed disk doesn't destroy all copies of any given block. And
+  that's not always easy as it depends on the layout chosen: whether the
+  blocks are near (same offset within each group), far (spread apart in a way
+  to maximise the mean distance), or offset (offset by size/n within each
+  block).
+
+  I'll leave it up to you to figure things out. Now go read question 19.
+
  -- martin f. krafft <madduck@debian.org>  Thu, 26 Oct 2006 11:05:05 +0200
 
 $Id$