Re: [PATCH] sched: Avoid side-effect of tickless idle onupdate_cpu_load (v2)

[Peter Zijlstra]

On Fri, 2010-05-14 at 18:21 -0700, Venkatesh Pallipadi wrote:


>  /*
> + * The exact cpu_load decay at various idx values would be
> + * [0] new = 0 * old + 1 * new
> + * [1] new = 1/2 * old + 1/2 * new
> + * [2] new = 3/4 * old + 1/4 * new
> + * [3] new = 7/8 * old + 1/8 * new
> + * [4] new = 15/16 * old + 1/16 * new

Would that be something like?

 load_i = ((2^i)-1)/(2^i) * load_i + 1/(2^i) * load_(i-1)

Where load_-1 == current load.

> + * Load degrade calculations below are approximated on a 128 point scale.
> + * degrade_zero_ticks is the number of ticks after which old_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).

So because we're in no_hz, current load == 0 and we could approximate
the thing by:

 load_i = ((2^i)-1)/(2^i) * load_i

Because for i ~ 1, there is no new input, and for i >> 1 the fraction is
small.

But why then do we precalculate these factors? It seems to me
((2^i)-1)/(2^i) is something that is trivial to compute and doesn't
warrant a lookup table?

> + * 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} };



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