calculate appropriate level 0 chance

[sticnarf]

Problem

Currently the maximum level 0 chance is 0.98. Sometimes we still fail to achieve our target even with this maximum value. For example, if we have plenty of tasks of all levels to run and the average time spent in handling a level 0 task once is 50us while the average time spent in handling other tasks once is 1500us. We can calculate the final time proportion of level 0 is (50 * 0.98) / (50 * 0.98 + 1500 * 0.02) = 0.62, which is far lower than we expected.

Solution

Firstly, this PR raises the maximum level 0 chance to 0.9999, which should suffice in most cases. And the way we adjust the chance is improved. The new chance is now calculated based on the assumption that we have plenty of tasks in all levels. (It is OK when the assumption is false because then we actually don't need to balance tasks of different levels.

Formula

Say the average time spent in handling level 1 and 2 tasks once is k times of that for level 0 task, the current time proportion of level 0 tasks is m, the target proportion is m0 and the chance of level 0 tasks is p. We want to get an appropriate new chance of level 0 tasks, p'.

So, we have simultaneous equations below:

p/(p+k(1-p)) = m
p'/(p'+k(1-p')) = m0

Solving the equations we can get p' = m0(1-m)p/(m0(1-m0)-(m-m0)(p-(1-m0))).

In our implementation, we ensure 0 < p < 1, 0 <= m <= 1, 0 < m0 < 1. The denominator can be proved to be greater than zero so we needn't check it in our code.

[BusyJay]

the denominator can be proved to be greater than zero

How?

[sticnarf]

the denominator can be proved to be greater than zero

How?

if

then

Because m and m_0p are all > 0, p-(1-m0) > 0.
So we have

And also, we know m <= 1, then,

Multiply the denominator in both side, we get

But both (1-m0) and (1-p) > 0, so we get a contradiction.
QED

[BusyJay]

Cool.