iSixSigma

How Can I Determine If My Rolled Throughput Yield (RTY) Is Acceptable?

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Let us postulate that a certain process is comprised of three progressive steps, say A, B, and C, respectfully. Further, we note the observed defects-per-unit (dpu) at the first step to be dpu.A = .10. At the second step, dpu.B = .05. The third step was found to be dpu.C = .008. Of course, we acknowledge dpu as a long-term measure of product quality – owing to its discrete nature.

At this juncture, we would seek to compute the rolled-throughput yield (Y.rt) of the process (probability of zero defects). Such an estimate can be realized by way of Poisson. Considering this function, we would recognize that Y = [(np)^r * e^-np] / r!, were n is the number of trials, p is the event probability, and r is the number of such events. By rational substitution, we declare that Y = [(dpu)^r * e^-dpu] / r!.  Thus, for the special case of r = 0 (zero defects), we recognize the throughput yield as Y.rt = e^-dpu.  Again, this is the probability of zero defects (for a given step in the process).

For process step A, we would estimate the throughput yield by computing the quantity Y.tp.A = e^-dpu.A = e^.10 = 0.904834, or about 90.48 percent. The throughput yield for step B would be computed as Y.tp.B = e^-dpu.B = e^.05 = 0.951229, or approximately 95.12 percent. Process step C would be given as Y.tp.C = e^-dpu.C = e^.008 = 0.992032, or roughly 99.20 percent.

The rolled-throughput yield (Y.rt) would then be determined by the cross product of all the subordinate throughput yields. Making this calculation discloses that Y.rt = 0.904834 * 0.951229 * 0.992032 = 0.853850, or about 85.39 percent. In equivalent form, we recognize that Y.rt = e^-tdpu = e^-(.010+.050+.008) = e^-.158 = 0.853850, where tdpu is the total defects per unit.

The normalized yield constitutes the “typical” yield when given a set of at least two yield values. Analogously, it is “kind of like” the average yield. Specifically, the normalized yield would be accepted as Y.norm = Y.tp^(1 / m), where m is the number of yield values that comprises the set of such values. For our example data, we would compute Y.norm = Y.tp^(1 / m) = 0.853850^(1 / 3) = 0.948696, or approximately 94.87 percent. This is to say that the “typical” yield per process step would be about 94.9 percent.  On the flip side of Y.norm is the idea of normalized dpu. This particular metric reports on the average dpu.  For our example, dpu.norm = – ln( Y.norm ) = 0.052667. Of course, this is equal to (.10 + .050 +.008) / 3 = .158 / 3 = 0.052667.

Once the normalized yield has been estimated, it is then possible to provide the “equivalent” capability per process step, but expressed in the form of a Z value (standard normal deviate). To facilitate this computation, we would likely employ Excel. Thus, we calculate the quantity Z.norm = normsivn(Y.norm) = 1.6323. So as to approximate the short-term capability, it would be necessary to add Z.shift = 1.5000. Doing so would reveal that Z.st.norm = Z.norm + Z.shift = 1.6323 + 1.5000 = 3.1323, or about 3.13 sigma. Classically speaking, it would be understood that Cp = Z.st / 3 = 1.044. 

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Of interest, actual benchmarking has revealed the typical process capability to be in the general range of 3.5 < Z.st < 4.5, with a modal condition of Z.st = 4.0. Based on this, we conclude that a capability of Z.st.norm = 3.13 would be considered below average. Thus, we conclude the example process would likely be considered an improvement candidate.