sched/eevdf: Use sched_attr::sched_runtime to set request/slice suggestion

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