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 <j...@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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