The effective storage on RAID-6 depends on the implementation method and the drive count. In the “Adaptec RAID-6″ method shown above the overhead is exactly two if the drive count is even and therefore the effective storage is n-2. However if the drive count is odd then the overhead is slightly greater than two.

For this reason most RAID controllers are moving to a more standard Reed-Solomon implementation where the overhead is always two, regardless of drive count. In fact, R-S is extensible to more than two drive failures with an equal number of overhead disks.

TT

I am just re-entering the RAID world, after a several year hiatus (used to use it on DEC Alpha systems). Today is the first time I’ve read about RAID 6, and I appreciate the clear description. One thing I’m not clear about is the effective storage for the array. With striping, effective storage is (n/2), with RAID 5 its usually (n-1) [where n is the number of physical drives]. The descriptions seem to imply that in each case the RAID 6 effective storage is (2/3 * n), as each method seems to use one parity entitiy (row or column) for each 2 data entities. Is this correct?

Thanks,
Dave.

Great article! It’s good to see such clear information on how RAID 6 works.

Regarding RS and RAID-6 with n,t,k: In your example, n and k would be 10 and 8, and t is 1. Reed-Solomon error correction can detect and correct up to t data errors (corrupted bits), but 2t /erasures/, which is when you know which data is missing or incorrect (for errors, such as corrupted bits, you’re not necessarily sure which is corrupt).

Since a lost hard drive is an erasure, because you know which drive is not there, up to 2 drives could be corrected.

See http://en.wikipedia.org/wiki/Reed-Solomon_error_correction for more detail on erasures vs. errors.

Joe