sched/fair: Clean up the explanation around decaying load update misses

Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Cc: Mike Galbraith <efault@gmx.de>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Thomas Gleixner <tglx@linutronix.de>
Signed-off-by: Ingo Molnar <mingo@kernel.org>

kernel/sched/fair.c

@@ -4222,42 +4222,37 @@ static void dequeue_task_fair(struct rq *rq, struct task_struct *p, int flags)
  */
 
 /*
- * The exact cpuload at various idx values, calculated at every tick would be
- * load = (2^idx - 1) / 2^idx * load + 1 / 2^idx * cur_load
+ * The exact cpuload calculated at every tick would be:
  *
- * If a cpu misses updates for n-1 ticks (as it was idle) and update gets called
- * on nth tick when cpu may be busy, then we have:
- * load = ((2^idx - 1) / 2^idx)^(n-1) * load
- * load = (2^idx - 1) / 2^idx) * load + 1 / 2^idx * cur_load
+ *   load' = (1 - 1/2^i) * load + (1/2^i) * cur_load
+ *
+ * If a cpu misses updates for n ticks (as it was idle) and update gets
+ * called on the n+1-th tick when cpu may be busy, then we have:
+ *
+ *   load_n   = (1 - 1/2^i)^n * load_0
+ *   load_n+1 = (1 - 1/2^i)   * load_n + (1/2^i) * cur_load
  *
  * decay_load_missed() below does efficient calculation of
- * load = ((2^idx - 1) / 2^idx)^(n-1) * load
- * avoiding 0..n-1 loop doing load = ((2^idx - 1) / 2^idx) * load
+ *
+ *   load' = (1 - 1/2^i)^n * load
+ *
+ * Because x^(n+m) := x^n * x^m we can decompose any x^n in power-of-2 factors.
+ * This allows us to precompute the above in said factors, thereby allowing the
+ * reduction of an arbitrary n in O(log_2 n) steps. (See also
+ * fixed_power_int())
  *
  * The calculation is approximated on a 128 point scale.
- * degrade_zero_ticks is the number of ticks after which load at any
- * particular idx is approximated to be zero.
- * degrade_factor is a precomputed table, a row for each load idx.
- * Each column corresponds to degradation factor for a power of two ticks,
- * based on 128 point scale.
- * Example:
- * row 2, col 3 (=12) says that the degradation at load idx 2 after
- * 8 ticks is 12/128 (which is an approximation of exact factor 3^8/4^8).
- *
- * With this power of 2 load factors, we can degrade the load n times
- * by looking at 1 bits in n and doing as many mult/shift instead of
- * n mult/shifts needed by the exact degradation.
  */
 #define DEGRADE_SHIFT		7
-static const unsigned char
-		degrade_zero_ticks[CPU_LOAD_IDX_MAX] = {0, 8, 32, 64, 128};
-static const unsigned char
-		degrade_factor[CPU_LOAD_IDX_MAX][DEGRADE_SHIFT + 1] = {
-					{0, 0, 0, 0, 0, 0, 0, 0},
-					{64, 32, 8, 0, 0, 0, 0, 0},
-					{96, 72, 40, 12, 1, 0, 0},
-					{112, 98, 75, 43, 15, 1, 0},
-					{120, 112, 98, 76, 45, 16, 2} };
+
+static const u8 degrade_zero_ticks[CPU_LOAD_IDX_MAX] = {0, 8, 32, 64, 128};
+static const u8 degrade_factor[CPU_LOAD_IDX_MAX][DEGRADE_SHIFT + 1] = {
+	{   0,   0,  0,  0,  0,  0, 0, 0 },
+	{  64,  32,  8,  0,  0,  0, 0, 0 },
+	{  96,  72, 40, 12,  1,  0, 0, 0 },
+	{ 112,  98, 75, 43, 15,  1, 0, 0 },
+	{ 120, 112, 98, 76, 45, 16, 2, 0 }
+};
 
 /*
  * Update cpu_load for any missed ticks, due to tickless idle. The backlog