Chapter 9: MANAGING FLOW VARIABILITY: Process Control and capability9.1 Objective9.2 Additional Suggested Readings9.3 Solutions to the Problem SetProblem 9.1Problem 9.2Problem 9.39.4 Test QuestionsCHAPTER 9: MANAGING FLOW VARIABILITY: PROCESS CONTROL AND CAPABILITY9.1 ObjectiveThis chapter focuses primarily on product quality and process capability, although we try to position it more generally as dealing with variability in any product measure such as cost, availability and response time. We claim that they all vary from one flow unit to the next, and this variability leads to customer dissatisfaction. It is therefore necessary to measure this variability, track down its sources and eliminate them. In the process of measuring variability, we introduce some elementary tools of quality or processimprovement such as Pareto charts. We then classify variability into “abnormal” and “normal” types. Weintroduce statistical process control over time as a device to identify and eliminate abnormal variability inthe short run. Since our goal is to teach the fundamental framework of feedback control and control limitpolicy, our discussion of SPC is primarily conceptual, without getting into constructing charts from givendata using ranges, constants, etc. However, these details can easily be covered by supplementing thechapter with a one page handout on estimating the mean and standard deviation from ranges.Once the process is internally stabilized by removing abnormal variability, we can discuss its capability inmeeting external customer requirements, and how it can be improved. We conclude by discussing sixsigma quality and some general principles of design for manufacturability and robust design. We havealso discussed general TQM philosophy and Malcolm Baldrige award framework.9.2 Additional Suggested ReadingsWe have used in the past some HBR articles and HBS cases to discuss principles of TQM, when it was a hot topic in 1980’s. For example, the following two go well together.- “Incline of Quality” by F. Leonard and W. Sasser, Sept-Oct 1982.- “Hank Kolb: Director Quality Assurance”. HBS Case 681-083, 1981. Author: F. LeonardSuggested assignment questions:1. What can you say about the quality attitude in this company?”1. What seem to be the causes of the quality problem on the Greasex line?2. What can top management do to remedy the situation?As TQM lost its appeal, most of us have moved away from teaching it, although some of us still includeat least part of it in our course. Given its qualitative and fuzzy nature, it has been received with mixedsuccess. More in line with this book, however, we have continued to teach the SPC tools. In the past wedeveloped some data for the Hank Kolb case to illustrate the SPC tools that we have illustrated in thisbook using the MBPF example. We have also used the following service oriented case that one of us hasdeveloped for this purpose.- “Excel Logistics Services”, Kellogg Case 2001. Author: Sunil Chopra.- “Six Sigma Quality at Flyrock Tires”, Kellogg Case 2002. Author: Sunil Chopra.- “Quality Wireless (A and B), Kellogg Case 2005. Author: Sunil Chopra.9.3 Solutions to the Problem Set2Problem 9.1 a. Given the symmetric shape of normal distribution around its mean, maximum conformance of the output within the given specifications will be achieved by centering the process at the midpoint of the specifications, i.e., at  = 32.5 gms. Now if we desire 98% of the output to conform to the specifications, it means that only 2% can fall outside the specs. Given a centered process and the symmetry of the normal distribution, this means only 1% below lower spec and 1% above upper spec. Thus, from the Normal tables (single-sided at 1%), the specification limits should be z = 2.33 standard deviations on either side of the mean, i.e., (32.5 - 30)/ = 2.33, or  = 2.5/2.33 = 1.073 gms. The sigma-capability of this process is 2.33. The corresponding process capability ratio is Cp = (35- 30)/6 = (35-30)/(6*1.073) = 0.78.b. With n = 12, and  = 1.073 as above, we can now determine the ideal control limits on subgroup averages of 12 bottles as:Average control chart:  ± 3 /√n = 32.5 ± (3)(1.073)/= (31.57, 33.34)Problem 9.2a. In order to produce 98% of the boxes above 15.5 oz., the process mean must be z = 2.055 standarddeviations above 15.5, i.e.,  = 15.5 + (2.055)(0.5) = 16.53 oz.If  = 16.53, then the proportion of overweight boxes will be P(X > 16) = P[Z > (16-16.53)/0.5] = P(Z > -1.06) = 0.8554.b. With  = 16.53,  = 0.5, and n = 9, the control limits on the average weight in a sample of 9 boxes are:  + 3 /√9= (16.03, 17.03). The observed average of 15.9 is below the lower control limit of 16.03, which signals that the mean has shifted below 16.53 due to an assignable cause. They should stop the process and adjust the mean upward. That the average weight is above the minimum individual weight of 15.5 oz is irrelevant. The specifications are on weights of individualboxes, not on average weights.c. The FDA specifications require at least 15.5 ozs per box. To be a 6-sigma process, the filling process must produce output centered at a mean of μ and a standard deviation of σ, where (μ-15.5)/σ = 6. This can be achieved for many different values of μ and σ. In particular, μ = 16.1 and σ = 0.1 results in a 6-sigma process.3Problem 9.3From the 26 observations given, we can calculate the average number of errors per thousand transactionsm = 3.3077, which is much better than the industry average = 15. a. We can then determine the controllimits on the number of errors, assuming Poisson distribution, as m ± 3√m= (0, 8.75). b. Observe thatthree observations out of the 26 given exceed the UCL = 8.75. Hence, the process is not in control, eventhough on average it is better than the BAI standard! The process is not stable, and our estimate of m isnot reliable. We first need to stabilize the process by removing assignable causes.Problem 9.4a. If the process mean  = 515, standard deviation  = 5 gms, and sample size n = 25, Control limits are LCL =  √25= 512 gms and UCL =  √25= 518 gmsb. Proportion of underweight output is Prob(W < 500) = Prob [Z < (500 – 515) / 5] = Prob (Z < –3) = 0.0013 or 0.13%.c. If Prob (violation) = 0.13% and the mean is  = 503 gms, then  must be