>Problem is that each raid1 block group contains two chunks on two >separate devices, it can't utilize fully three devices no matter what. >If that doesn't suit you then you need to add 4th disk. After >that FS will be able to use all unallocated space on all disks in raid1 >profile. But even then you'll be able to safely lose only one disk >since BTRFS still will be storing only 2 copies of data.
I hope I didn't say that I want to utilize all three devices fully. It was clear to me that there will be 2 TB of wasted space. Also I'm not questioning the chunk allocator for RAID1 at all. It's clear and it always has been clear that for RAID1 the chunks need to be allocated on different physical devices. If I understood Kai's point of view, he even suggested that I might need to do balancing to make sure that the free space on the three devices is being used smartly. Hence the questions about balancing. I mean in worst case it could happen like this: Again I have disks of sizes 3, 3, 8: Fig.1 Drive1(8) Drive2(3) Drive3(3) - X1 X1 - X2 X2 - X3 X3 Here the new drive is completely unused. Even if one X1 chunk would be on Drive1 it would be still a sub-optimal allocation. This is the optimal allocation. Will btrfs allocate like this? Considering that Drive1 has the most free space. Fig. 2 Drive1(8) Drive2(3) Drive3(3) X1 X1 - X2 - X2 X3 X3 - X4 - X4 >From my point of view Fig.2 shows the optimal allocation, by the time the disks Drive2 and Drive3 are full (3TB) Drive1 must have 6TB (because it is exclusively holding the mirrors for both Drive2 and 3). For sure now btrfs can say, since two of the drives are completely full he can't allocate any more chunks and the remaining 2 TB of space from Drive1 is wasted. This is clear it's even pointed out by the btrfs size calculator. But again if the above statements are true, then df might as well tell the "truth" and report that I have 3.5 TB space free and not 1.5TB (as it is reported now). Again here I fully understand Kai's explanation. Because coming back to my first e-mail, my "problem" was that df is reporting 1.5 TB free, whereas the whole FS holds 2.5 TB of data. So the question still remains, is it just that df is intentionally not smart enough to give a more accurate estimation, or is the assumption that the allocator picks the drive with most free space mistaken? If I continue along the lines of what Kai said, and I need to do re-balance, because the allocation is not like shown above (Fig.2), then my question is still legitimate. Are there any filters that one might use to speed up or to selectively balance in my case? or will I need to do full balance? On Sun, Sep 10, 2017 at 7:19 PM, Dmitrii Tcvetkov <demfl...@demfloro.ru> wrote: >> @Kai and Dmitrii >> thank you for your explanations if I understand you correctly, you're >> saying that btrfs makes no attempt to "optimally" use the physical >> devices it has in the FS, once a new RAID1 block group needs to be >> allocated it will semi-randomly pick two devices with enough space and >> allocate two equal sized chunks, one on each. This new chunk may or >> may not fall onto my newly added 8 TB drive. Am I understanding this >> correctly? > If I remember correctly chunk allocator allocates new chunks on device > which has the most unallocated space. > >> Is there some sort of balance filter that would speed up this sort of >> balancing? Will balance be smart enough to make the "right" decision? >> As far as I read the chunk allocator used during balance is the same >> that is used during normal operation. If the allocator is already >> sub-optimal during normal operations, what's the guarantee that it >> will make a "better" decision during balancing? > > I don't really see any way that being possible in raid1 profile. How > can you fill all three devices if you can split data only twice? There > will be moment when two of three disks are full and BTRFS can't > allocate new raid1 block group because it has only one drive with > unallocated space. > >> >> When I say "right" and "better" I mean this: >> Drive1(8) Drive2(3) Drive3(3) >> X1 X1 >> X2 X2 >> X3 X3 >> X4 X4 >> I was convinced until now that the chunk allocator at least tries a >> best possible allocation. I'm sure it's complicated to develop a >> generic algorithm to fit all setups, but it should be possible. > > > Problem is that each raid1 block group contains two chunks on two > separate devices, it can't utilize fully three devices no matter what. > If that doesn't suit you then you need to add 4th disk. After > that FS will be able to use all unallocated space on all disks in raid1 > profile. But even then you'll be able to safely lose only one disk > since BTRFS still will be storing only 2 copies of data. > > This behavior is not relevant for single or raid0 profiles of > multidevice BTRFS filesystems. -- To unsubscribe from this list: send the line "unsubscribe linux-btrfs" in the body of a message to majord...@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html
