On 03/17/2013 09:04 AM, Mel Gorman wrote:
Simplistically, the anon and file LRU lists are scanned proportionally
depending on the value of vm.swappiness although there are other factors
taken into account by get_scan_count(). The patch "mm: vmscan: Limit
the number of pages kswapd reclaims" limits the number of pages kswapd
reclaims but it breaks this proportional scanning and may evenly shrink
anon/file LRUs regardless of vm.swappiness.
This patch preserves the proportional scanning and reclaim. It does mean
that kswapd will reclaim more than requested but the number of pages will
be related to the high watermark.
Signed-off-by: Mel Gorman <mgorman@xxxxxxx>
---
mm/vmscan.c | 52 +++++++++++++++++++++++++++++++++++++++++-----------
1 file changed, 41 insertions(+), 11 deletions(-)
diff --git a/mm/vmscan.c b/mm/vmscan.c
index 4835a7a..182ff15 100644
--- a/mm/vmscan.c
+++ b/mm/vmscan.c
@@ -1815,6 +1815,45 @@ out:
}
}
+static void recalculate_scan_count(unsigned long nr_reclaimed,
+ unsigned long nr_to_reclaim,
+ unsigned long nr[NR_LRU_LISTS])
+{
+ enum lru_list l;
+
+ /*
+ * For direct reclaim, reclaim the number of pages requested. Less
+ * care is taken to ensure that scanning for each LRU is properly
+ * proportional. This is unfortunate and is improper aging but
+ * minimises the amount of time a process is stalled.
+ */
+ if (!current_is_kswapd()) {
+ if (nr_reclaimed >= nr_to_reclaim) {
+ for_each_evictable_lru(l)
+ nr[l] = 0;
+ }
+ return;
+ }
This part is obvious.
+ /*
+ * For kswapd, reclaim at least the number of pages requested.
+ * However, ensure that LRUs shrink by the proportion requested
+ * by get_scan_count() so vm.swappiness is obeyed.
+ */
+ if (nr_reclaimed >= nr_to_reclaim) {
+ unsigned long min = ULONG_MAX;
+
+ /* Find the LRU with the fewest pages to reclaim */
+ for_each_evictable_lru(l)
+ if (nr[l] < min)
+ min = nr[l];
+
+ /* Normalise the scan counts so kswapd scans proportionally */
+ for_each_evictable_lru(l)
+ nr[l] -= min;
+ }
+}
This part took me a bit longer to get.
Before getting to this point, we scanned the LRUs evenly.
By subtracting min from all of the LRUs, we end up stopping
the scanning of the LRU where we have the fewest pages left
to scan.
This results in the scanning being concentrated where it
should be - on the LRUs where we have not done nearly
enough scanning yet.
However, I am not sure how to document it better than
your comment already has...
Acked-by: Rik van Riel <riel@xxxxxxxxxx>
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