Re: How does btrfs handle bad blocks in raid1?

[Hugo Mills <hugo@carfax.org.uk>]

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On Thu, Jan 09, 2014 at 06:34:23PM +0100, George Eleftheriou wrote:
> > claiming that RAID-10 (with 2-way mirroring) is guaranteed to survive
> > an arbitrary 2-device failure is incorrect.
> 
> Yes, you are right.  I didn't mean "any 2 devices". I should have
> added "from different mirrors" :)

   If you have an even number of devices and all the devices are the
same size, then:

 * the block group allocator will use all the devices each time
 * the amount of space on each device will always be the same

If the sort in the allocator is stable and resolves ties in free space
by using the device ID number, the above properties should guarantee
that the allocation is stable, so each new block group will have the
same functional chunk on the same device, and you get your wish.

   It's been a few months since I looked at that code, but I don't
recall seeing anything directly contradictory to the above
assumptions.

   Of course, if you have an odd number of devices, the allocator will
omit a different device on each block group, and you lose the ability
to survive (some) two-device failures. I suspect that the odds of
surviving a two-device failure are still non-zero, but less than if
you had an even number of devices. I'm not about to attempt an
ab-initio computation of the probabilities, but it shouldn't be too
hard to do either a monte-carlo simulation or a simple brute-force
enumeration of the possibilities for a given configuration.

   Hugo.

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