Performance analysis of MK60FX512VMD15

[rpneeraj]

I would like to know the below things related to the performance of K60:

1. Interrupt latency with MQX 4.0.

2. task switching time from one task to other.

And how time slice is effecting the performance of the system.

What should be the appropriate time slice for tasks, how to judge it, with proper reasons.

Any data or information related the above will be appreciated.

Thanks and Regards,

Neeraj Srivastava

[soledad]

Hello Neeraj Srivastava,

Please check the attachment, you can find the MQX timing report for k60.

Regarding the time slice, the default time slice unit is number of ticks. Ticks are set by the frequency of the system timer in the <board>.h file. So there is not formula in order to determinate the time slice value, however it is recommended this value is greater than the context switch.

According with the MQX user guide:

A task uses round robin scheduling only if the MQX_TIME_SLICE_TASK attribute is

set in its task template. The task's time slice is determined by the value of the template's

DEFAULT_TIME_SLICE. However, if the value is zero, the task's time slice is the

default time slice for the processor. Initially, the default time slice for the processor is ten

times the interval of the periodic timer interrupt. Since the interval on most BSPs is five

milliseconds, the initial default time slice for the processor is usually 50 milliseconds.

You can change the default time slice for the processor with _sched_set_rr_interval() or

_sched_set_rr_interval_ticks(), passing the task ID parameter as

MQX_DEFAULT_TASK_ID.

When the time slice expires for an active round robin task, MQX RTOS saves the task's

context. MQX RTOS then performs a dispatch operation, in which it examines the ready

queues to determine, which task should become active. MQX RTOS moves the expired

task to the end of the task's ready queue, an action that causes control to pass to the next

task in the ready queue. If there are no other tasks in the ready queue, the expired task

continues to run.

With round robin scheduling, tasks of the same priority can share the processor in a timeequitable

manner.

Have a great day,
Sol

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[rpneeraj]

Hi Sol,

Service Interrupt and return to task 974.16ns for 100Mhz.

This time contains interrupt latency + time for running handler ... or something else???

If you can clearly define this time, that would be great.

BR,

Neeraj

[rpneeraj]

Hi Sol,

Service Interrupt and return to task  974.16ns for 100Mhz.

This time contains interrupt latency + time for running handler ... or something else???

If you can clearly define this time, that would be great.

BR,

Neeraj