Howdy, sorry for digging up this thread, but I've run into an issue again, and am looking for advice. Excerpts from Stan Hoeppner's message of 2014-02-04 03:00:54 -0500: > After a little digging and thinking this through... > > The default PE size is 4MB but up to 16GB with LVM1, and apparently > unlimited size with LVM2. It can be a few thousand times larger than > any sane stripe width. This makes it pretty clear that PEs exist > strictly for volume management operations, used by the LVM tools, but > have no relationship to regular write IOs. Thus the PE size need not > match nor be evenly divisible by the stripe width. It's not part of the > alignment equation. So in the course of actually going about this, I realized that this actually is not true (I think). Logical volumes can only have sizes that are multiple of the physical extent size (by definition, really), and so there's no way to have logical volumes end on a multiple of the array's stripe width, given my stripe width of 9216s, there doesn't seem to be an abundance of integer solutions to 2^n mod 9216 = 0. So my question is, then, does it matter if logical volumes (or, really, XFS file systems) actually end right on a multiple of the stripe width, or only that it _begin_ on a multiple of it (leaving a bit of dead space before the next logical volume)? If not, I'll tweak things to ensure my stripe width is a power of 2. Thanks again! -- Morgan Hamill _______________________________________________ xfs mailing list xfs@xxxxxxxxxxx http://oss.sgi.com/mailman/listinfo/xfs
