[PATCH 3/3] c-user/rate_monotonic_manager.rst: Use Latex math for utilization formulas

Joel Sherrill joel at rtems.org
Fri Mar 10 22:10:01 UTC 2017


I would appreciate someone double-checking that I translated the formulas
correctly.

Thanks.

--joel

On Fri, Mar 10, 2017 at 3:29 PM, Joel Sherrill <joel at rtems.org> wrote:

> ---
>  c-user/rate_monotonic_manager.rst | 14 ++++++--------
>  1 file changed, 6 insertions(+), 8 deletions(-)
>
> diff --git a/c-user/rate_monotonic_manager.rst b/c-user/rate_monotonic_
> manager.rst
> index 05eb59f..1309a88 100644
> --- a/c-user/rate_monotonic_manager.rst
> +++ b/c-user/rate_monotonic_manager.rst
> @@ -267,21 +267,19 @@ Processor Utilization Rule
>
>  The Processor Utilization Rule requires that processor utilization be
>  calculated based upon the period and execution time of each task.  The
> fraction
> -of processor time spent executing task index is ``Time(index) /
> -Period(index)``.  The processor utilization can be calculated as follows:
> +of processor time spent executing task index is ``Time(i) / Period(i)``.
> +The processor utilization can be calculated as follows:
>
> -.. code-block:: c
> +.. math::
>
> -    Utilization = 0
> -    for index = 1 to maximum_tasks
> -        Utilization = Utilization + (Time(index)/Period(index))
> +    Utilization = \sum_{i=1}^{maximumTasks} Time_i/Period_i
>
>  To ensure schedulability even under transient overload, the processor
>  utilization must adhere to the following rule:
>
> -.. code-block:: c
> +.. math::
>
> -    Utilization = maximum_tasks * (2**(1/maximum_tasks) - 1)
> +    maximumUtilization = maximumTasks * (2^{\frac{1}{maximumTasks}} - 1)
>
>  As the number of tasks increases, the above formula approaches ln(2) for a
>  worst-case utilization factor of approximately 0.693.  Many tasks sets
> can be
> --
> 1.8.3.1
>
>
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