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Defect per Unit, Correct Definition?

Six Sigma – iSixSigma Forums Old Forums General Defect per Unit, Correct Definition?

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  • #49528

    SickSigmaNewGuy
    Member

    All resources I can find give DPU as:
    DPU=(# defects)/(#units).  Note this formula counts defects, not defective units.
    My six sigma instructor gives the following definition for DPU:
    DPU=(PPM/1e6) x (# opportunities for defects);  where PPM=DPM=(#defectives)/(# units)*(1e6).  Note that this uses the # of defective units and not # of defects.
    First pass yield is then calculated from: FPY=exp(-DPU).
    Has anybody seen this definition for DPU and PPM before? Does the math work out for FPY? Or has my instructor been smoking something? 

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    #169403

    Anton Javier
    Participant

    Your instructor formula is correct.  DPU is the measurement of the number of defectives, not defects.  
    Anton Javier

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