# I Am Developing a Scorecard for a Project Level Sigma Score. How Do You Suggest a Scorecard Be Developed to Determine a Sigma Value for the Whole Project?

By Dr. Mikel Harry
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Creating a process scorecard is not an easy or straightforward task. It becomes even more difficult when the scorecard is comprised of continuous and discrete performance metrics, each having a unique level of complexity (defect opportunity space). Even more perplexing is when some of the baseline metrics are short-term in nature while others are inherently long-term.

Nonetheless, the easiest and perhaps simplest way to approach this problem is to first convert each baseline metric to an equivalent metric, such as DPU. For the ensuing example, we will use DPU as the base of conversion. Consider a certain critical-to-quality (CTQ) characteristic called “CTQ7.” In this case, CTQ7 was determined to have a short-term capability of Z.st = 3.2. Of course, such a level of short-term performance converts to an equivalent long-term capability of Z.lt = Z.st – 1.5 = 3.2 – 1.5 = 1.7.

At this point in our discussion, we should note that a post-mortem bias of 1.5 was employed to pragmatically degrade the reported short-term capability. This was done to compensate or otherwise account for long-term sources of variation (error) not reflected in, or intrinsic to, the Z.st metric. We employed the value of 1.5 since no other empirical information was available at the time of reporting. Certainly, if specific emperical data were available, the compensatory correction of 1.5 would not be needed; owing to our a priori knowledge of the long-term performance. In such instances, the shift factor (per se) would be estimate-able and, as a consequence, the 1.5 constant would not be needed as an approximation.

Meanwhile, we must convert the equivalent long-term standard normal deviate (Z.lt) to a yield value by way of the Excel equation: Yield = normsdist(Z.lt) = normsdist (1.7) = 0.955434. Following this task, we would compute the rolled-throughput yield as Y.rt = Y.ft^M = 0.955434^3 = 0.872173, where M is the number of defect opportunites.  Such a level of aggregate yield would then translate to DPU = -ln(Y.rt) = .136767. Normalizing to the opportunity level revels that DPMO = (DPU / M) * 10^6 = (.136767 / 3) * 10^6 = .045589 * 10^6 = 45,589. Thus, we determine the sigma capability to be about 3.2.