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I've recently read this post:

http://www.standalone-sysadmin.com/blog/2012/08/i-come-not-to-praise-raid-5/

tldr; It basically says that the error rate of a RAID system is tied to the hardware, and you need disks with a higher URE to make a RAID more stable.

We run a RAID 5 with these details:

  • 6 x 2TB drives = 10TB useable
  • URE <1 in 10^14 (each drive)

Understanding the article above I would like to check my numbers with someone else

  • 10^14 bits gives me 12,500GB's per chance of a URE
  • 12,500GBs / 10TBs = 1.25

Does that mean If one of my drives fails (lost 2 in the last 3 months) and I replace it, I have a 1 in 1.25 chance of getting a URE on one of my other drives?

I'm no expert so please feel free to tear this one apart and help me learn.

Thanks

outrunthewolf
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1 Answers1

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Everything on all of the remaining disks has to be read in order to reconstruct the data on the replacement drive.

So you are reading 5x2TB=10TB.

10^14 is 100 trillion bits, or 12.5 trillion bytes.

10/12.5, then, seems to be correct. You have approximately an 80% chance of having a URE occur during the rebuild. (For statistical reasons I don't quite fully understand, this is a bit wrong, but it's close enough to illustrate the risk. I welcome a good explanation of how the math really works.)

Michael Hampton
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  • Thank you Michael. I appreciate you taking the time to verify my math, if only that it is based on the workings from the post. I must note, the URE data of the hdd's is <1 so its obviously an upper limit. – outrunthewolf Jul 30 '14 at 21:23