On 09/12/17 23:36, Rich Freeman wrote: > you instead compute 5 sets of parity so that now you have 9 sets of > data that can tolerate the loss of any 5, then throw away the sets > containing the original 4 sets of data and store the remaining 5 sets > of parity data across the 5 drives. You can still tolerate the loss > of one more set, but all 4 of the original sets of data have been > tossed already.
Is that how ZFS works? Cheers, Wol
