Re: Mainline kernel OLTP performance update

On Sat, 2009-01-24 at 09:36 +0200, Pekka Enberg wrote:

On Fri, 2009-01-23 at 10:22 -0500, Christoph Lameter wrote:

quoted

quoted

No there is another way. Increase the allocator order to 3 for the
kmalloc-8192 slab then multiple 8k blocks can be allocated from one of the
larger chunks of data gotten from the page allocator. That will allow slub
to do fast allocs.

On Sat, Jan 24, 2009 at 4:55 AM, Zhang, Yanmin
[off-list ref] wrote:

quoted

After I change kmalloc-8192/order to 3, the result(pinned netperf UDP-U-4k)
difference between SLUB and SLQB becomes 1% which can be considered as fluctuation.

Great. We should fix calculate_order() to be order 3 for kmalloc-8192.
Are you interested in doing that?

Pekka,

Sorry for the late update.
The default order of kmalloc-8192 on 2*4 stoakley is really an issue of calculate_order.


slab_size	order		name
-------------------------------------------------
4096            3               sgpool-128
8192            2               kmalloc-8192
16384           3               kmalloc-16384

kmalloc-8192's default order is smaller than sgpool-128's.

On 4*4 tigerton machine, a similiar issue appears on another kmem_cache.

Function calculate_order uses 'min_objects /= 2;' to shrink. Plus size calculation/checking
in slab_order, sometimes above issue appear.

Below patch against 2.6.29-rc2 fixes it.

I checked the default orders of all kmem_cache and they don't become smaller than before. So
the patch wouldn't hurt performance.

Signed-off-by Zhang Yanmin [off-list ref]

---

diff -Nraup linux-2.6.29-rc2/mm/slub.c linux-2.6.29-rc2_slubcalc_order/mm/slub.c

--- linux-2.6.29-rc2/mm/slub.c	2009-02-11 00:49:48.000000000 -0500
+++ linux-2.6.29-rc2_slubcalc_order/mm/slub.c	2009-02-12 00:08:24.000000000 -0500

@@ -1856,6 +1856,7 @@ static inline int calculate_order(int si
 	min_objects = slub_min_objects;
 	if (!min_objects)
 		min_objects = 4 * (fls(nr_cpu_ids) + 1);
+	min_objects = min(min_objects, (PAGE_SIZE << slub_max_order)/size);
 	while (min_objects > 1) {
 		fraction = 16;
 		while (fraction >= 4) {

@@ -1865,7 +1866,7 @@ static inline int calculate_order(int si
 				return order;
 			fraction /= 2;
 		}
-		min_objects /= 2;
+		min_objects --;
 	}
 
 	/*