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When I use LVM to partition one or more storages I noticed it use 4 MiB' blocks (physical extent) by default, while the storage itself usually uses 512 bytes per sector.

I believed it should not be a problem if I align the LVM partitions to 4 MiB, but no matter what size they be, it will always show some "not usable" size when run pvdisplay:

--- Physical volume ---
PV Name               /dev/sda2
VG Name               xpto
PV Size               17.00 GiB / not usable 4.00 MiB
Allocatable           yes (but full)
PE Size               4.00 MiB
Total PE              4351
Free PE               0
Allocated PE          4351
PV UUID               xxxxxx-xxxx-xxxx-xxxx-xxxx-xxxx-xxxxxx

But even if I reallocate the PV to match the size minus the not usable, it decrease the total PE and still have remaining not usable size. Even using pvdisplay --unit B to see the exact number.

I'm wondering what is the exact size of the LVM header, I read some text arguing it would be 4 sectors of 512 bytes, other saying it is 180 KiB, but I have tried all combinations and was not able to discover how it is calculated.

Is there a way to align the PV partition in order to zero the not usable size, or at least reduce to minimum?

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  • I would swear storage has not been using 512 bytes for quite sometime now. The distribution/usage of files vs sectors is highly dependent also in what fs you are using on top of lvm Aug 6, 2018 at 17:59
  • @RuiFRibeiro I have read somewhere it could be 4096 bytes as well, but I believe it depend on vendor, or maybe could be customized, anyway in my case it is 512 bytes for sure. Aug 6, 2018 at 18:03
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    Exact partition size blockdev --getsize64 /dev/sda2? Perhaps some kind of rounding error? I get for example PV Size 401.00 MiB / not usable 0 so it works for me. The extra 1.00MiB is the default metadata size (1st PE Offset). If the size is correct, which distro, version of kernel / lvm2 / ...? Aug 6, 2018 at 18:11
  • Answer I wrote just yesterday to a similar question, may be of interest unix.stackexchange.com/a/460387/30851 Aug 6, 2018 at 18:13
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    That 4 MB is about 0.023 % of the size of the whole PV. Are you sure it's worth spending the effort to try to recover it?
    – ilkkachu
    Aug 6, 2018 at 18:24

1 Answer 1

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For those devices which sector size is 4K probably your device will never get unaligned, but for those with 512 bytes you should align the partition's first sector to match one divisible by 4096 bytes, in that case, divisible by 8 (4096/512), but remember that it starts with zero.

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