I have 6 devices: 4TB, 3TB, 2TB, 2TB, 1.5TB, 1TB (/dev/sda to /dev/sdf).
First question:
With RAID-1 I'd have:
My total size would be in that case (4 + 3 + 2 + 2 + 1.5 + 1) = 13.5/2 = 6.75TB
Will $ mkfs.btrfs --data raid1 --metadata raid1 /dev/sda /dev/sdb /dev/sdc /dev/sdd /dev/sde /dev/sdf
provide me with approximately 6.75TB? If yes, how many disks (and which?) can I afford losing?
Second question:
With the RAID-1 I can afford, for example, losing three disks:
without losing data.
How can I have the same freedom in losing the same disks with btrfs?
Thanks!
Btrfs distributes the data (and its RAID 1 copies) block-wise, thus deals very well with hard disks of different size. You will receive the sum of all hard disks, divided by two – and do not need to think on how to put them together in similar sized pairs.
If more than one disk fails, you're always in danger of losing data: RAID 1 cannot deal with losing two disks at the same time. In your example given above, if the wrong two disks die, you always lose data.
Btrfs can increase the chances of losing data if more than one disk fails, as it will distribute the blocks somewhat randomly, chances are higher that some blocks are only stored on the failed two devices. On the other hand, if you lose data, you probably lose less for the same reason. In average, it sums up to the same chance of losing n bits, but if you're interested in the chance of losing only a single bit you're worse of with btrfs.
And then again, you should also consider its advantage of using checksums which help against corrupted data on disk.