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TDA4VM: graph

jialin xie
Intellectual 260 points
Part Number: TDA4VM


Whether there will be a result output when vxGraphParameterEnqueueReadyRef is enqueued at a slower speed than the TIDL model calculation speed. For example, the queue time is 1 second, and the model calculation time is 0.1 seconds.

  • 0
    TI__Guru* 87291 points

    Hi,

    Please let me know if this some sample demo you are referring to?

    If not, could you tell me which graph parameter are you enqueuing here?

    Regards,
    Nikhil

  • 0 in reply to Nikhil Dasan
    Intellectual 260 points

    For example, in the multi-camera demo,
    1680.07> GRAPH: app_multi_cam_graph (#nodes = 3, #executions = 3934)
    1680.07> NODE: CAPTURE1: capture_node: avg = 33255 usecs, min/max = 32172 / 54692 usecs, #executions = 3934
    1680.07> NODE: DISPLAY1: DisplayNode: avg = 11348 usecs, min/max = 70 / 16930 usecs, #executions = 3934
    1680.09> NODE: VPAC_LDC1: ldc_node: avg = 13993 usecs, min/max = 13826 / 14570 usecs, #executions = 3934

    NODE: CAPTURE1 takes the most time,
    NODE: VPAC_LDC1 takes less time,
    If captureObj.raw_image_arr enqueued for 1 second, and LDC processing is much less than 1 second, will there be a result output in this case?

  • +1 in reply to jialin xie
    TI__Guru* 87291 points

    Hi,

    If a producer node is running slower than the consumer node, the consumer node would eventually get starved for buffers and be slowed down by the producer node.

    However, it would still produce outputs, it is just that the compute resource of the consumer node may be idle while the producer is creating outputs.

    It might be helpful to generate a timing diagram using the RT logging feature like the below to visualize:
    software-dl.ti.com/.../TIOVX_DEBUG.html