This test is very approximately the "busy web server" workload. We set up a
number of processes each of which are reading many small files from different
parts of the disk.
Set up six separate copies of the 2.4.19 kernel tree, and then run, in
parallel, six processes which are reading them:
for i in 1 2 3 4 5 6
do
time (find kernel-tree-$i -type f | xargs cat > /dev/null ) &
done
With this test we have six read requests in the queue all the time. It's
what the anticipatory scheduler was designed for.
2.4.21-pre4:
6m57.537s
6m57.620s
6m57.741s
6m57.891s
6m57.909s
6m57.916s
2.5.61+hacks:
3m40.188s
3m51.332s
3m55.110s
3m56.186s
3m56.757s
3m56.791s
2.5.61+CFQ:
5m15.932s
5m16.219s
5m16.386s
5m17.407s
5m50.233s
5m50.602s
2.5.61+AS:
0m44.573s
0m45.119s
0m46.559s
0m49.202s
0m51.884s
0m53.087s
This was a little unfair to 2.4 because three of the trees were laid out by
the pre-Orlov ext2. So I reran the test with 2.4.21-pre4 when all six trees
were laid out by 2.5's Orlov allocator:
6m12.767s
6m12.974s
6m13.001s
6m13.045s
6m13.062s
6m13.085s
Not much difference there, although Orlov is worth a 4x speedup in this test
when there is only a single reader (or multiple readers + anticipatory
scheduler)
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This archive was generated by hypermail 2b29 : Sun Feb 23 2003 - 22:00:32 EST