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Drive Extender replacement due out in 2012. It's called Storage Spaces.

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skwire:
how does the math work out there?-superboyac (July 18, 2012, 03:42 PM)
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Here's a simple explanation:

Let's say you have four, 2 TB drives.  You would designate one of them as the parity drive, e.g.,:

D1 (2 TB)
D2 (2 TB)
D3 (2 TB)
P1 (2 TB)

This would give you 6 TB of data space with 2 TB of parity.  The simplest way to envision this is with math, i.e., 1+2+3=6 (D1 = 1, D2 = 2, D3 = 3, thus, P1 = 6).  If you lose any one of the "D" drives, you can easily calculate what you're missing.  Let's say you lost D2 and have replaced it.  The system starts to rebuild based data off the P1 parity drive information like this:

1+?+3=6

Obviously, ? = 2.  Yes, this is a 50,000 foot view, but does that help to explain things?  Also, if you were to lose two drives at one time, you wouldn't get all your data back (unless you had two parity drives).  In other words, you can simultaneously lose as many data drives as you have parity drives.  So goes the theory, anyway.  YMMV.   :P

tomos:
This would give you 6 TB of data space with 2 TB of parity.  The simplest way to envision this is with math, i.e., 1+2+3=6 (D1 = 1, D2 = 2, D3 = 3, thus, P1 = 6).  If you lose any one of the "D" drives, you can easily calculate what you're missing.  Let's say you lost D2 and have replaced it.  The system starts to rebuild based data off the P1 parity drive information like this:
-skwire (July 18, 2012, 04:28 PM)
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starting to get it - in conjunction with above explanation, this helped:
http://en.wikipedia.org/wiki/RAID#RAID_parity


very impressive functionality - so long as it works :D

superboyac:
Yes, very helpful explanation.  Thanks skwire!

40hz:
The math gets a little less obvious with something like RAID-5 since there isn't a designated parity drive. The parity data gets distributed among all the drives. But for RAID-5 you'll always need one additional drive to do an array.

Minimum number is three drives. Your data space is the capacity of all the drives in the array minus the capacity of one drive in the array. (Note: All the drives in a pure RAID-5 array must have identical capacity.)

Assuming 1TB drives, a three drive array would have a 2TB data space (i.e. 3TB-1TB). So 33% of the total capacity is 'lost' to parity. But a five drive array would have a 4TB data space (i.e 5TB-1TB) with only 20% total capacity lost to parity. And that percentage decreases with each drive added.

So as you can see, RAID-5 is least economical with only three drives - but it becomes increasingly economic with each drive added after that - up to the capacity of the RAID controller.

The tradeoff is that a standard RAID-5 array can only tolerate one drive failing at a time without experiencing data loss. As skwire pointed out earlier, the general rule is you can have as many drives simultaneously fail as you have parity drives. And in the case of RAID-5, you basically only have one "drive" for parity even though it isn't a specific drive.

In practice, it's a little more complicated than that. But not much. 8)

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