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

Kuan Hsun Chen kuan-hsun.chen at tu-dortmund.de
Fri Mar 10 23:05:28 UTC 2017


Hi,

The formulas both are correct.
More formally, we can explicitly define "n" as the maximum number of tasks
and "i" as the index of task.
Then the formulas will become more clear like this:

Utilization = \sum_{i=1}^{n} Time_i/Period_i
maximumUtilization = n * (2^{\frac{1}{n}} - 1)

Best,
Kuan-Hsun

2017-03-10 15:10 GMT-07:00 Joel Sherrill <joel at rtems.org>:

> 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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-- 
M.Sc. Kuan-Hsun Chen

TU Dortmund
Department of Computer Science 12
Design Automation of Embedded Systems
Otto-Hahn-Strasse 16, Room 102

44227 Dortmund
Germany

Phone:  *+49 231 755 6124 <+49%20231%207556124>*
Mail:   kuan-hsun.chen at tu-dortmund.de <kuan-hsun.chen at tu-dortmund.de>
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