btrfs: try to unlock parent nodes earlier when inserting a key
When inserting a new key, we release the write lock on the leaf's parent
only after doing the binary search on the leaf. This is because if the
key ends up at slot 0, we will have to update the key at slot 0 of the
parent node. The same reasoning applies to any other upper level nodes
when their slot is 0. We also need to keep the parent locked in case the
leaf does not have enough free space to insert the new key/item, because
in that case we will split the leaf and we will need to add a new key to
the parent due to a new leaf resulting from the split operation.
However if the leaf has enough space for the new key and the key does not
end up at slot 0 of the leaf we could release our write lock on the parent
before doing the binary search on the leaf to figure out the destination
slot. That leads to reducing the amount of time other tasks are blocked
waiting to lock the parent, therefore increasing parallelism when there
are other tasks that are trying to access other leaves accessible through
the same parent. This also applies to other upper nodes besides the
immediate parent, when their slot is 0, since we keep locks on them until
we figure out if the leaf slot is slot 0 or not.
In fact, having the key ending at up slot 0 when is rare. Typically it
only happens when the key is less than or equals to the smallest, the
"left most", key of the entire btree, during a split attempt when we try
to push to the right sibling leaf or when the caller just wants to update
the item of an existing key. It's also very common that a leaf has enough
space to insert a new key, since after a split we move about half of the
keys from one into the new leaf.
So unlock the parent, and any other upper level nodes, when during a key
insertion we notice the key is greater then the first key in the leaf and
the leaf has enough free space. After unlocking the upper level nodes, do
the binary search using a low boundary of slot 1 and not slot 0, to figure
out the slot where the key will be inserted (or where the key already is
in case it exists and the caller wants to modify its item data).
This extra comparison, with the first key, is cheap and the key is very
likely already in a cache line because it immediately follows the header
of the extent buffer and we have recently read the level field of the
header (which in fact is the last field of the header).
The following fs_mark test was run on a non-debug kernel (debian's default
kernel config), with a 12 cores intel CPU, and using a NVMe device:
$ cat run-fsmark.sh
#!/bin/bash
DEV=/dev/nvme0n1
MNT=/mnt/nvme0n1
MOUNT_OPTIONS="-o ssd"
MKFS_OPTIONS="-O no-holes -R free-space-tree"
FILES=100000
THREADS=$(nproc --all)
FILE_SIZE=0
echo "performance" | \
tee /sys/devices/system/cpu/cpu*/cpufreq/scaling_governor
mkfs.btrfs -f $MKFS_OPTIONS $DEV
mount $MOUNT_OPTIONS $DEV $MNT
OPTS="-S 0 -L 10 -n $FILES -s $FILE_SIZE -t $THREADS -k"
for ((i = 1; i <= $THREADS; i++)); do
OPTS="$OPTS -d $MNT/d$i"
done
fs_mark $OPTS
umount $MNT
Before this change:
FSUse% Count Size Files/sec App Overhead
0 1200000 0 165273.6 5958381
0 2400000 0 190938.3 6284477
0 3600000 0 181429.1 6044059
0 4800000 0 173979.2 6223418
0 6000000 0 139288.0 6384560
0 7200000 0 163000.4 6520083
1 8400000 0 57799.2 5388544
1 9600000 0 66461.6 5552969
2 10800000 0 49593.5 5163675
2 12000000 0 57672.1 4889398
After this change:
FSUse% Count Size Files/sec App Overhead
0 1200000 0 167987.3 (+1.6%) 6272730
0 2400000 0 198563.9 (+4.0%) 6048847
0 3600000 0 197436.6 (+8.8%) 6163637
0 4800000 0 202880.7 (+16.6%) 6371771
1 6000000 0 167275.9 (+20.1%) 6556733
1 7200000 0 204051.2 (+25.2%) 6817091
1 8400000 0 69622.8 (+20.5%) 5525675
1 9600000 0 69384.5 (+4.4%) 5700723
1 10800000 0 61454.1 (+23.9%) 5363754
3 12000000 0 61908.7 (+7.3%) 5370196
Reviewed-by: Josef Bacik <josef@toxicpanda.com>
Signed-off-by: Filipe Manana <fdmanana@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
1 file changed