On 11/03/2012 12:44 AM, Hugo Mills wrote:
> 1 MiB stored in RAID-5 across 3 devices takes up 1.5 MiB -- multiplier ×1.5
> (1 MiB over 2 devices is 512 KiB, plus an additional 512 KiB for parity)
> 1 MiB stored in RAID-5 across 6 devices takes up 1.2 MiB -- multipler ×1.2
> (1 MiB over 5 devices is 204.8 KiB, plus an additional 204.8 KiB for parity)
>
> With the (initial) proposed implementation of RAID-5, the
> stripe-width (i.e. the number of devices used for any given chunk
> allocation) will be *as many as can be allocated*. Chris confirmed
> this today on IRC. So if I have a disk array of 2T, 2T, 2T, 1T, 1T,
> 1T, then the first 1T of allocation will stripe across 6 devices,
Interesting.
Let me simulate a possible output
> ./btrfs filesystem disk-usage -t /
Data Metadata Metadata System System
RAID5 Single DUP Single DUP Unallocated
/dev/dm-0 1.50TB 8.00MB - 4.00MB 16.00MB 500.00MB
/dev/dm-1 1.50TB - - 4.00MB 16.00MB 500.00MB
/dev/dm-2 1.50TB - - 4.00MB - 500.00MB
/dev/dm-3 1.00TB - 2x 100.00MB 4.00MB - 300.00MB
/dev/dm-4 1.00TB - 2x 100.00MB 4.00MB - 300.00MB
======= ======== =========== ====== ======= ===========
Total 5.00TB 8.00MB 200.00MB 4.00MB 8.00MB 2.10TB
Used 10.65GB 0.00 50.00MB 0.00 4.00KB
Would be it clear ? And what if we move the Total/used below the header ?
> ./btrfs filesystem disk-usage -t /
Data Metadata Metadata System System
RAID5 Single DUP Single DUP Unallocated
Total 5.00TB 8.00MB 200.00MB 4.00MB 8.00MB 2.10TB
Used 10.65GB 0.00 50.00MB 0.00 4.00KB
======= ======== =========== ====== ======= ===========
/dev/dm-0 1.50TB 8.00MB - 4.00MB 16.00MB 500.00MB
/dev/dm-1 1.50TB - - 4.00MB 16.00MB 500.00MB
/dev/dm-2 1.50TB - - 4.00MB - 500.00MB
/dev/dm-3 1.00TB - 2x 100.00MB 4.00MB - 300.00MB
/dev/dm-4 1.00TB - 2x 100.00MB 4.00MB - 300.00MB
GB
P.S:
The raid5 is composed by (4+1) x 1T and (2+1) x 0.5TB, supposing the
disks as 2x 1.5TB and 3x2Tb.
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