On 5/12/21 1:27 PM, David T-G wrote:
Phil, et al -- ...and then Phil Turmel said... % % I do this for my medium-speed read-mostly tasks. Raid10,n3 across 4 % or 5 disks gives me redundancy comparable to raid6 (lose any two) % without the CPU load of parity and syndrome calculations. I've been reading and I still need to catch up on the notation, but how much space do you get in the end?
One third of the total, since I'm using n3.
I'm hoping to grow our disk farm and end up with 8+ disks. I'm more than a bit nervous about RAID5 across a bunch of 6T (or bigger) disks, so I've been thinking of RAID6. That would give me 6x6 = 36T plus two parity. Putting 8 disks in RAID10 should give me 6x4 = 24T with mirroring. That's a pretty hefty space penalty :-( But ...
I wouldn't use only two copies, as you cannot lose any two. With three copies, 8x6T/3 = 16T usable space. Heftier space penalty, but necessary to have confidence surviving a rebuild after a disk replacement.
How does RAID10 across 5 disks as above 1) work and 2) work out?
1) Linux raid 10 does not layer raid 1 on top of raid 0, but implements "n" copies of each chunk striped across all devices. Which is why the number of devices doesn't have to be a multiple of "n".
If you had 8 disks with a huge need for space, how would y'all lay out everything?
raid6. I pretty much always layer LVM on top of mdraid.
Thanks in advance :-) :-D
Phil
