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Peter Zijlstra authored
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
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