the only way to make it safer, is put more devices on raid1
for example:
disks=6 (was wrong on last email)
raid0= 1-2(a) 3-4(b) 5-6(c)
raid1= a,b,c

or
raid1=1-2-3(a) 4-5-6(b)
raid0=a,b
now you can loose tree disks

2011/1/31 Roberto Spadim [off-list ref]:

rewriting..
using raid10 or raid01 you will have problems if you lose 2 drives too...
if you lose two raid 1 devices you loose raid 1...
see:

disks=4
RAID 1+0
raid1= 1-2(A)  ; 3-4(B); 5-6(C)
raid0= A-B-C
if you lose (A,B or C) your raid0 stop

RAID 0+1
raid0= 1-2-3(A)  ; 4-5-6(B)
raid1= A-B
if you lose (1,4 OR 1,5 OR 1,6 OR 2,4 OR 2,5 OR 2,6 OR 3,4 OR 4,5 OR
4,6) your raid0 stop

using raid1+0 or raid0+1 you can't lose two disks...



2011/1/31 Roberto Spadim [off-list ref]:

quoted

do you have a faster array using raid0+1 or raid1+0?

2011/1/31 Roberto Spadim [off-list ref]:

quoted

hum that's right,
but not 'increase' (only if you compare raid0+1 betwen raid1+0) using
raid1 and after raid0 have LESS point of fail between raid 0 and after
raid 1, since the number of point of fail is proportional to number of
raid1 devices.

2011/1/31 Robin Hill [off-list ref]:

quoted

On Mon Jan 31, 2011 at 01:00:13PM -0200, Roberto Spadim wrote:

quoted

i think make two very big raid 0
and after raid1
is better

Not really - you increase the failure risk doing this.  With this setup,
a single drive failure from each RAID0 array will lose you the entire
array.  With the reverse (RAID0 over RAID1) then you require both drives
in the RAID1 to fail in order to lose the array.  Of course, with a 4
drive array then the risk is the same (33% with 2 drive failures) but
with a 6 drive array it changes to 60% for RAID1 over RAID0 versus 20%
for RAID0 over RAID1.

Cheers,
   Robin
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Roberto Spadim
Spadim Technology / SPAEmpresarial

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Roberto Spadim
Spadim Technology / SPAEmpresarial

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Roberto Spadim
Spadim Technology / SPAEmpresarial

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Roberto Spadim
Spadim Technology / SPAEmpresarial
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