# Raid 5

Nicholas Leippe nick at byu.edu
Fri Dec 2 13:57:21 MST 2005

```RAID 10 and RAID 0+1 offer significantly different failure tolerances.

RAID 0+1:
odds of failure:
RAID0                1/1000000 * 2 = 1/500000  (the RAID0)
RAID1   RAID1           (1/1000)^2    = 1/1000000 (per mirror)
A   B   C   D           1/1000                    (per drive)

is better than RAID 10:

RAID1                (1/500)^2  = 1/250000
RAID0   RAID0           1/1000 * 2 = 1/500
A   B   C   D           1/1000

RAID0           1/1000000 * 3 = 1/333333
RAID1   RAID1   RAID1   (1/1000)^2    = 1/1000000
A   B   C   D   E   F   1/1000

RAID1             (1/333)^2  = 1/110889
RAID0     RAID0        1/1000 * 3 = 1/333
A  B  C   D  E  F       1/1000

The failure odds of the RAID0+1 increase slower than the odds for RAID 10
as capacity is added.

IMO, just because the absolute probabilities may be small, the difference
between them (a factor of 2 or more) is still significant.  My reasoning
is thus:

I consider the odds of a drive failure to be 100%--not if, but when.  The odds
of the system failing then become the odds of the degraded system failing.
IOW, it can always sustain the loss of the first drive--that's the entire
purpose in the first place.  So the real issue is how well it handles a
double-fault--how much do you have to worry about getting the first drive
replaced.  (If I did this right, the relative odds should be the same--just
their magnitudes increased, but I think it also might make it easier to
visualize.)  This makes the odds become:

RAID1-Deg.              1/1000
-       B               1/1000

RAID0                1/1000 + 1/1000000 ~= 1/1000
Deg.   RAID1           1/1000, 1/1000000 respectively (the latter from above)
-   B   C   D           1/1000

RAID1-Deg.             1/500
_   RAID0             n/a, 1/500
-   B C   D             1/1000

RAID0           1/1000 + (1/1000000 * 2) ~= 1/1000
Deg.   RAID1   RAID1   1/1000, 1/1000000, 1/1000000
-   B   C   D   E   F   1/1000

RAID1-Deg.          1/333
-      RAID0         n/a, 1/1000 * 3 = 1/333
-  B  C   D  E  F       1/1000

RAID5-Deg.              1/1000 * 3 = 1/333
-  B  C  D              1/1000

RAID5-Deg.              1/1000 * 5 = 1/200
-  B  C  D  E  F        1/1000

So my final probabilities become:
RAID 1,   2 drives:  1/1,000 = 0.001
RAID 0+1, 4 drives: ~1/1,000 = 0.001
RAID 0+1, 6 drives: ~1/1,000 = 0.001
RAID 10,  4 drives:  1/500   = 0.002
RAID 10,  6 drives:  1/333   = 0.003
RAID 5,   4 drives:  1/333   = 0.003
RAID 5,   6 drives:  1/200   = 0.005

You can see from this that increasing capacity reduces the reliability of a
RAID 0+1 pretty much insignificantly, significantly reduces it for a RAID 10,
but still not as badly as for a RAID5.

> Basically, RAID 1, RAID 10, and RAID 5 differ little in probability of
> failure.  For any one of these configurations, you still have to lose
> two drives to fail, and the probability of that is small, as we see
> above.  One might compare it to buying four lottery tickets instead of
> one - you're still probably going to lose, even though your chances of
> winning are slightly higher than before.

If you consider as I do, that the first drive failure is a given,
then the odds of failure is magnitudes larger--back up to the same order
of magnitude as a single drive failing.

This makes the relative differences in failure odds become significant,
especially when you consider how the odds increase with additional capacity.

--
Respectfully,

Nicholas Leippe
Sales Team Automation, LLC
1335 West 1650 North, Suite C
Springville, UT  84663 +1 801.853.4090
http://www.salesteamautomation.com

```