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| author | Peter Zijlstra <peterz@infradead.org> | 2023-05-22 13:46:30 +0200 |
|---|---|---|
| committer | Peter Zijlstra <peterz@infradead.org> | 2024-08-17 11:06:45 +0200 |
| commit | 857b158dc5e81c6de795ef6be006eed146098fc6 (patch) | |
| tree | f8732e2f0581f895e5f722e67590f6eeb11e6183 /lib/memory-notifier-error-inject.c | |
| parent | 85e511df3cec46021024176672a748008ed135bf (diff) | |
sched/eevdf: Use sched_attr::sched_runtime to set request/slice suggestion
Allow applications to directly set a suggested request/slice length using
sched_attr::sched_runtime.
The implementation clamps the value to: 0.1[ms] <= slice <= 100[ms]
which is 1/10 the size of HZ=1000 and 10 times the size of HZ=100.
Applications should strive to use their periodic runtime at a high
confidence interval (95%+) as the target slice. Using a smaller slice
will introduce undue preemptions, while using a larger value will
increase latency.
For all the following examples assume a scheduling quantum of 8, and for
consistency all examples have W=4:
{A,B,C,D}(w=1,r=8):
ABCD...
+---+---+---+---
t=0, V=1.5 t=1, V=3.5
A |------< A |------<
B |------< B |------<
C |------< C |------<
D |------< D |------<
---+*------+-------+--- ---+--*----+-------+---
t=2, V=5.5 t=3, V=7.5
A |------< A |------<
B |------< B |------<
C |------< C |------<
D |------< D |------<
---+----*--+-------+--- ---+------*+-------+---
Note: 4 identical tasks in FIFO order
~~~
{A,B}(w=1,r=16) C(w=2,r=16)
AACCBBCC...
+---+---+---+---
t=0, V=1.25 t=2, V=5.25
A |--------------< A |--------------<
B |--------------< B |--------------<
C |------< C |------<
---+*------+-------+--- ---+----*--+-------+---
t=4, V=8.25 t=6, V=12.25
A |--------------< A |--------------<
B |--------------< B |--------------<
C |------< C |------<
---+-------*-------+--- ---+-------+---*---+---
Note: 1 heavy task -- because q=8, double r such that the deadline of the w=2
task doesn't go below q.
Note: observe the full schedule becomes: W*max(r_i/w_i) = 4*2q = 8q in length.
Note: the period of the heavy task is half the full period at:
W*(r_i/w_i) = 4*(2q/2) = 4q
~~~
{A,C,D}(w=1,r=16) B(w=1,r=8):
BAACCBDD...
+---+---+---+---
t=0, V=1.5 t=1, V=3.5
A |--------------< A |---------------<
B |------< B |------<
C |--------------< C |--------------<
D |--------------< D |--------------<
---+*------+-------+--- ---+--*----+-------+---
t=3, V=7.5 t=5, V=11.5
A |---------------< A |---------------<
B |------< B |------<
C |--------------< C |--------------<
D |--------------< D |--------------<
---+------*+-------+--- ---+-------+--*----+---
t=6, V=13.5
A |---------------<
B |------<
C |--------------<
D |--------------<
---+-------+----*--+---
Note: 1 short task -- again double r so that the deadline of the short task
won't be below q. Made B short because its not the leftmost task, but is
eligible with the 0,1,2,3 spread.
Note: like with the heavy task, the period of the short task observes:
W*(r_i/w_i) = 4*(1q/1) = 4q
~~~
A(w=1,r=16) B(w=1,r=8) C(w=2,r=16)
BCCAABCC...
+---+---+---+---
t=0, V=1.25 t=1, V=3.25
A |--------------< A |--------------<
B |------< B |------<
C |------< C |------<
---+*------+-------+--- ---+--*----+-------+---
t=3, V=7.25 t=5, V=11.25
A |--------------< A |--------------<
B |------< B |------<
C |------< C |------<
---+------*+-------+--- ---+-------+--*----+---
t=6, V=13.25
A |--------------<
B |------<
C |------<
---+-------+----*--+---
Note: 1 heavy and 1 short task -- combine them all.
Note: both the short and heavy task end up with a period of 4q
~~~
A(w=1,r=16) B(w=2,r=16) C(w=1,r=8)
BBCAABBC...
+---+---+---+---
t=0, V=1 t=2, V=5
A |--------------< A |--------------<
B |------< B |------<
C |------< C |------<
---+*------+-------+--- ---+----*--+-------+---
t=3, V=7 t=5, V=11
A |--------------< A |--------------<
B |------< B |------<
C |------< C |------<
---+------*+-------+--- ---+-------+--*----+---
t=7, V=15
A |--------------<
B |------<
C |------<
---+-------+------*+---
Note: as before but permuted
~~~
From all this it can be deduced that, for the steady state:
- the total period (P) of a schedule is: W*max(r_i/w_i)
- the average period of a task is: W*(r_i/w_i)
- each task obtains the fair share: w_i/W of each full period P
Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org>
Tested-by: Valentin Schneider <vschneid@redhat.com>
Link: https://lkml.kernel.org/r/20240727105030.842834421@infradead.org
Diffstat (limited to 'lib/memory-notifier-error-inject.c')
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