**DPMO and RTY**

Hopefully I can explain this clearly.

total Defects =500

Total Opps =7787260

DPU = 0.41

m = number of opportunities/unit = 6383. I calculated this using the total # of defects (500) divided by the DPU to derive that the data was based on 1220 units then divided the total opps by the # of units.

RTY =e^-DPU=66.4%

DPMO = 64.21

Yna = Normalized yield = RTY ^ (1/m) = 0.664^(1/6383) =.99993585 which matches up to 99.99353% that you calculated

DPMO can be calculated from the Yna or from actual yield data

DPMO = (1-Yna)*10^6 = (1-0.9999353)*10^6 = 64.15

DPMO = # defects/# opportunities *10^6 = 500/7787260*10^6 =64.21

*So then, why does RTY not equal DPMO and DPMO not equal DPMO?*

DPMO = # defects/# opportunities *10^6 is an actual calculation based are units processed, number of defects found, and number of real opportunities per unit.

DPMO = (1-Yna)*10^6 is based on the Poisson distribution used to calculate RTY. RTY = e^-DPU assumes defects are distributed randomly and are assumed to follow a Poisson distrubtion. Yna, therefore, reflects the assumed Poisson distribution used to calculate RTY . As the defect rate falls below 10%, Yna and DPMO converge. The data that you have shows them converging quite nicely. *It appears that you were only missing the one step DPMO = (1-Yna)*10^6*

The benefit of the DPMO is that it normalizes the data so that assemblies of different complexities/defect opportunities can be compared objectively. The benefit of RTY is that if there is an issue with yields, it becomes immediately obvious.