Hi! While investigating some compaction-related problems, I noticed, that many (even most) kernel objects are allocated on slabs with order 2 or 3. This behavior was introduced by commit 9b2cd506e "slub: Calculate min_objects based on number of processors." by Christoph Lameter. As I understand, the idea was to make kernel allocations cheaper by reducing the total number of page allocations (allocating 1 page with order 3 is cheaper than allocating 8 1-ordered pages). I'm sure, it's true for recently rebooted machine with a lot of free non-fragmented memory. But is it also true for heavy-loaded machine with fragmented memory? Are we sure, that it's cheaper to run compaction and allocate order 3 page than to use small 1-pages slabs? Do I miss something? Disabling this behavior dramatically reduces the number of 2- and 3-ordered allocations. Compaction is performed significantly rarer. This is especially noticeable on machines with intensive disk i/o. I do not see any performance degradation. But I'm not sure, that I'm not missing something. Any comments and/or ideas are welcomed. Thanks! Regards, Roman -- To unsubscribe, send a message with 'unsubscribe linux-mm' in the body to majordomo@xxxxxxxxx. For more info on Linux MM, see: http://www.linux-mm.org/ . Don't email: <a href=mailto:"dont@xxxxxxxxx";> email@xxxxxxxxx </a>