On Thu, Nov 2, 2017 at 1:21 AM koukou73gr <koukou73gr@xxxxxxxxx> wrote:
The scenario is actually a bit different, see:
Let's assume size=2, min_size=1
-We are looking at pg "A" acting [1, 2]
-osd 1 goes down
-osd 2 accepts a write for pg "A"
-osd 2 goes down
-osd 1 comes back up, while osd 2 still down
-osd 1 has no way to know osd 2 accepted a write in pg "A"
-osd 1 accepts a new write to pg "A"
-osd 2 comes back up.
bang! osd 1 and 2 now have different views of pg "A" but both claim to
have current data.
In this case, OSD 1 will not accept IO precisely because it can not prove it has the current data. That is the basic purpose of OSD peering and holds in all cases.
-Greg
-K.
On 2017-11-01 20:27, Denes Dolhay wrote:
> Hello,
>
> I have a trick question for Mr. Turner's scenario:
> Let's assume size=2, min_size=1
> -We are looking at pg "A" acting [1, 2]
> -osd 1 goes down, OK
> -osd 1 comes back up, backfill of pg "A" commences from osd 2 to osd 1, OK
> -osd 2 goes down (and therefore pg "A" 's backfill to osd 1 is
> incomplete and stopped) not OK, but this is the case...
> --> In this event, why does osd 1 accept IO to pg "A" knowing full well,
> that it's data is outdated and will cause an inconsistent state?
> Wouldn't it be prudent to deny io to pg "A" until either
> -osd 2 comes back (therefore we have a clean osd in the acting group)
> ... backfill would continue to osd 1 of course
> -or data in pg "A" is manually marked as lost, and then continues
> operation from osd 1 's (outdated) copy?
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