Defect per Unit, Correct Definition?

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

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    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? 


    Anton Javier

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

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