Chapter 9: MANAGING FLOW VARIABILITY: Process Control and capability
9.1 Objective
This 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 process improvement such as Pareto charts. We then classify variability into “abnormal” and “normal” types. We introduce statistical process control over time as a device to identify and eliminate abnormal variability in the short run. Since our goal is to teach the fundamental framework of feedback control and control limit policy, our discussion of SPC is primarily conceptual, without getting into constructing charts from given data using ranges, constants, etc. However, these details can easily be covered by supplementing the chapter 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 in meeting external customer requirements, and how it can be improved. We conclude by discussing six sigma quality and some general principles of design for manufacturability and robust design. We have also discussed general TQM philosophy and Malcolm Baldrige award framework.
9.2 Additional Suggested Readings
We 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. Leonard
Suggested 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 include at least part of it in our course. Given its qualitative and fuzzy nature, it has been received with mixed success. More in line with this book, however, we have continued to teach the SPC tools. In the past we developed some data for the Hank Kolb case to illustrate the SPC tools that we have illustrated in this book using the MBPF example. We have also used the following service oriented case that one of us has developed 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 Set
Problem 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 /= 32.5 ± (3)(1.073)/= (31.57, 33.34)

Problem 9.2

a. In order to produce 98% of the boxes above 15.5 oz., the process mean must be z = 2.055 standard deviations 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 /= (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 individual boxes, 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.

Problem 9.3

From the 26 observations given, we can calculate the average number of errors per thousand transactions m = 3.3077, which is much better than the industry average = 15. a. We can then determine the control limits on the number of errors, assuming Poisson distribution, as m ± 3= (0, 8.75). b. Observe that three observations out of the 26 given exceed the UCL = 8.75. Hence, the process is not in control, even though on average it is better than the BAI standard! The process is not stable, and our estimate of m is not reliable. We first need to stabilize the process by removing assignable causes.

Problem 9.4

a. If the process mean  = 515, standard deviation  = 5 gms, and sample size n = 25, Control limits are LCL =  = 512 gms and UCL =  = 518 gms b. 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 decreased so that –3 = (500 - 503)/ or  = 1.

Problem 9.5

Given: process mean  = 6 cm, standard deviation  = 0.01 cm, and sample size n = 10,
Control limits are LCL =  = 5.9905 cm and UCL =  = 6.0095 cm
Note: customer specifications are irrelevant.

Problem 9.6
If the specifications do not change, improvement in the sigma capability means the standard deviation decreases. In that case, the control band should become narrower.

Problem 9.7
Currently, process mean  = 2.2 hours, and standard deviation  = 0.8 hours, so probability of processing within 4 hours is
P(T < 4) = Prob[Z < (4 – 2.2)/0.8 ] =Prob(Z < 2.25) = 0.9878 < 0.992, the bank’s requirement
With improved process mean  = 2 hours, and standard deviation  = 1.2 hours,
P(T < 4) = Prob[Z < (4 – 2)/1.2 ] =Prob(Z < 1.67) = 0.9616, which is worse than before.
To meet the bank’s requirement, the mean should be lowered to 2 hours without raising variability. In that case
P(T < 4) = Prob[Z < (4 – 2)/0.8 ] =Prob(Z < 2.5) = 0.9938 > 0.992, the bank’s requirement

Problem 9.8

If D is the diameter of basketballs produced, we need 95% of diameters to fall between 29.3 and 29.7. Given a centered process and the symmetry of the normal distribution, this means only 2.5% below lower spec and 2.5% above upper spec. Thus, from the Normal tables (single-sided at 2.5%), the specification limits should be z = 1.96 standard deviations on either side of the mean
So z = (29.7 – 29.5) /  = 1.96 or we need  = .2/1.96 = 0.102. inch .
Hence, Cp = (USL – LSL)/6  = (29.7 – 29.3)/6  = 0.65.
Please note that the Selected Answer at the end of the text erroneously corresponds to requiring that 98% of the ball diameters should fall between 29.3 and 29.7 inches.
Problem 9.9 a. False. Control limits only assure process stability. They have nothing to do with meeting customer specifications. b. True. As sample size increases, standard deviation of sample averages decreases, and the control bands becomes narrower. c. True. If control limits are 3 standard deviations from the mean, the probability of sample average falling outside control limits is 0.27%, regardless of the value of the standard deviation. d. If the process improves sigma capability, its standard deviation goes down, and the width of the control band decreases. e. False. A 6-sigma process may be out of control if an assignable cause leads to a mean shift (for instance). A 6-sigma process, however, is likely to produce very few defectives even when it is out of control. Thus, a 6-sigma process has the advantage of detecting small shifts even before they impact the customer.
Problem 9.10 a. R chart and X-bar chart b. p chart c. c chart d. c chart e. R chart and X-bar chart f. c chart

9.4 Test Questions
9.4.1. You are a manager with nine employees reporting directly to you. (You have full span of control.) All nine employees have essentially the same responsibilities. They all have numerous opportunities to make mistakes (of various types) in their jobs, but only a small chance of making any one particular mistake at any one time. In the past year you have recorded the following number of mistakes for each employee. Assume that all mistakes are equally critical, so that you cannot distinguish between employees based on the type of mistake they make.

Employee | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | No. of Mistakes | 10 | 15 | 11 | 5 | 17 | 22 | 11 | 12 | 10 | It is time for evaluations and merit raise recommendations. Suggest an approach (you do not have to show any computations) to decide who to reward and who to penalize? Assume that the total number of transactions performed by each employee are the same during the year.

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Answer: You should construct appropriate control limits based on the data available. For example, with this data, assuming Poisson distribution of errors, mean = variance = 12.55. So control limits are: 12.55 + 3 Sqrt (12.55) = (1.92, 23.18) Employees with error rates above the upper control limit or below the lower control limit would be asked for abnormal causes of variation. Effort should be made to eliminate assignable errors above the UCL and replicate assignable errors below the LCL. All employees with error rates within control limits should be treated equally. In this case, all employees seem to display normal variability, so take no action.
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In the long run, you should try to reduce the mean of 12.55 through better training, and mistake-proofing the process.

9.4.2 A customer wants delivery to be ensured between 10 am and 2 pm. Your truck leaves the factory at 4 am and the time taken to reach the customer is normally distributed with an average of 8 hrs and a standard deviation of 1 hr.

(a) What is the process capability ratio of the process?

0 0.67 1 0.50 2 1.00 3 2.00 4 None of the above

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Answer: Cp = (US – LS)/6 = 4/6 = 0.67

(b) What specific action(s) (in terms of mean and standard deviation targets to be achieved) would you take to improve the ability of the delivery process to be a 6- process (like Motorola)? Target mean =

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Answer: Same as before

Target standard deviation =

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Answer:  = 1/3 will yield Cp = 2, which corresponds to six sigma quality. To reduce the standard deviation, use interstate highways to avoid unforeseen traffic jams and lights. Get good, reliable trucks, reduce travel distance (which also affects the mean), improve driver training, etc.

9.4.3 An assembly process has twenty successive stages, each with 3- capability. The company has come up with a new design that will require only ten assembly stages. The capability at each stage still remains 3-. The overall assembly process capability with the new design will be

5 Higher than the original design. 6 Same as the original design. 7 Lower than the original design. -------------------------------------------------
Answer: Higher. Reducing the number of stages reduces the overall variability and improves process capability.

9.4.4 A machine produces drive shafts. Manufacturing currently takes samples of 10 shafts every half an hour to check if the process is in control. Control limits have been set at 3 standard deviations of the sample means and are set at 10 30.001 cm. A suggestion calls for control limits to be tightened to 2 standard deviations, i.e., 10 20.001 cm. As a result the proportion of shafts produced that are defective (outside specification limits) will

(i) Increase (ii) Decrease (iii) Remain unchanged

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Answer: Remain unchanged

Explain.

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Answer: Decreasing control limits increases the number of investigations to check whether the process is still behaving according to its distribution. This does not imply anything about meeting customer specification limits; we will simply be investigating more frequently.
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[As a second order effect, one can argue that a possible process change - getting out of control - is detected earlier, and hence, fewer defects would be produced.]

9.4.5 K-Log produces cereals that are sold in boxes labeled to contain 390 grams. If the cereal content is below 390 grams, K-Log may invite FDA scrutiny. Filling much more than 390 grams costs the company since it essentially means giving away more of the product. Accordingly, K-Log has set specification limits between 390 and 410 grams for the weight of cereal boxes. Currently the boxes are filled automatically by a filling machine and they have an average weight of 405 g with a standard deviation of 4 g. What process targets (in terms of mean and standard deviation of the filling process) are needed for the filling machine to have a six- capability?
Answer: We need to center the process at the middle of the specs, i.e., at 400 gms. To yield six sigma quality the standard deviation must be (400-390)/6 = 1.67 gms