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rstaveleyFlag for United Kingdom of Great Britain and Northern Ireland

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RAID5 - childish understanding

I'm told that we have a RAID5 diskset. I am told that its capacity is equivalent to the capacity of the disks in the set minus one disk, which we can consider to be the capacity used for parity (and we ought to have minus another disk, which is reserved for hot swap, but let's ignore that).

I am told that the storage required for 1GB of data is a bit more than 1GB, but not as much as 2GB (1GB plus the parity bits - we can ignore all of the file system stuff, because my files are very large compressed data).

I am also told that if a catastrophe hit one of those disks, my 1G of data would not be lost.

Have I been misled?
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Richard Quadling
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I found that with Google, but I was too stupid to get what I needed from it.

If I have 1GB of data, does it occupy 1GB plus the parity or does it take ~2GB to store?

At least 2GB + parity for 1GB of data.

I think.

"Difficult to rebuild in the event of a disk failure (as compared to RAID level 1)" worries me though.

Oh my word... I get it now.

Here's where I found a good explanation cicuitously going via wikipedia:

http://www.pcguide.com/ref/hdd/perf/raid/concepts/genParity-c.html

That means that you can store 1G across (say) 5 disks (4 data and 1 parity) as 1.2G and your data would be safe in the event of a loss of one disk drive. The penny drops! How very neat.
Ditto the comment on adequate backups, otherwise the answer is that you are mostly correct.  However, it still just takes 1 GB of storage to store 1 GB of data.  The "loss" of storage space is factored in to your other statement about minus one drive for parity.  And a key part of your statement is that the RAID 5 only protects you against the most common failure of a single disk going bad.
Sorry masto, but I didn't get your response before accepting. Browser lag. It was the PC guide explanation I needed. I couldn't see how you could store data for free, but of course you are XOR-ing data bytes from all disks in the set to get the parity byte. Obvious when you know how.