1. 5 – 1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Quality And Performance 5 For Operations Management, 9e by Krajewski/Ritzman/Malhotra © 2010 Pearson Education Homework: 5, 13, 14, 15

2. 5 – 2 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Costs of Quality A failure to satisfy a customer is considered a defect Prevention costs Appraisal costs Internal failure costs External failure costs Ethics and quality

3. 5 – 3 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Total Quality Management Figure 5.1 – TQM Wheel Customer satisfaction

4. 5 – 4 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Total Quality Management Customer satisfaction  Conformance to specifications  Value  Fitness for use  Support  Psychological impressions Employee involvement  Cultural change  Teams

5. 5 – 5 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Total Quality Management Continuous improvement  Kaizen  A philosophy  Not unique to quality  Problem solving process

6. 5 – 6 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. The Deming Wheel Plan Do Study Act Figure 5.2 – Plan-Do-Study-Act Cycle

7. 5 – 7 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Six Sigma X X X X X X X X X X X X X X X X X Process average OK; too much variation Process variability OK; process off target Process on target with low variability Reduce spread Center process X X X X X X X X X Figure 5.3 – Six-Sigma Approach Focuses on Reducing Spread and Centering the Process

8. 5 – 8 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. SPC, Causes of Variation Common causes  Random, unavoidable sources of variation Assignable causes (out of statistical control)  Can be identified and eliminated  Change in the mean, spread, or shape

9. 5 – 9 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Control Charts Time-ordered diagram of process performance

10. 5 – 10 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Control Chart Factors TABLE 5.1|FACTORS FOR CALCULATING THREE-SIGMA LIMITS FOR |THE x -CHART AND R -CHART Size of Sample ( n ) Factor for UCL and LCL for x -Chart ( A 2 ) Factor for LCL for R -Chart ( D 3 ) Factor for UCL for R -Chart ( D 4 ) 21.88003.267 31.02302.575 40.72902.282 50.57702.115 60.48302.004 70.4190.0761.924 80.3730.1361.864 90.3370.1841.816 100.3080.2231.777

11. 5 – 11 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using x - and R -Charts EXAMPLE 5.1 The management of West Allis Industries is concerned about the production of a special metal screw used by several of the company’s largest customers. The diameter of the screw is critical to the customers. Data from five samples appear in the accompanying table. The sample size is 4. Is the process in statistical control?

12. 5 – 12 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using x - and R -Charts Data for the x - and R -Charts: Observation of Screw Diameter (in.) Observation Sample Number 1234 Rx 10.50140.50220.50090.50270.00180.5018 20.50210.50410.50240.50200.00210.5027 30.50180.50260.50350.50230.00170.5026 40.50080.50340.50240.50150.00260.5020 50.50410.50560.50340.50470.00220.5045 Average0.00210.5027

13. 5 – 13 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using x - and R -Charts Figure 5.10 –Range Chart from the OM Explorer x and R-Chart Solver for the Metal Screw, Showing That the Process Variability Is in Control

14. 5 – 14 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using x - and R -Charts Figure 5.11 –The x -Chart from the OM Explorer x and R -Chart Solver for the Metal Screw, Showing That Sample 5 is out of Control

15. 5 – 15 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.1 Webster Chemical Company produces mastics and caulking for the construction industry. The product is blended in large mixers and then pumped into tubes and capped. Webster is concerned whether the filling process for tubes of caulking is in statistical control. The process should be centered on 8 ounces per tube. Several samples of eight tubes are taken and each tube is weighed in ounces. Tube Number Sample12345678AvgRange 17.988.348.027.948.447.687.818.118.0400.76 28.238.127.988.418.318.187.998.068.1600.43 37.897.777.918.048.007.897.938.097.9400.32 48.248.187.838.057.908.167.978.078.0500.41 57.878.137.927.998.107.818.147.887.9800.33 68.138.148.118.138.148.128.138.148.1300.03 Avgs8.0500.38

16. 5 – 16 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.1 Assuming that taking only 6 samples is sufficient, is the process in statistical control? UCL R = D 4 R = LCL R = D 3 R = 1.864(0.38) = 0.708 0.136(0.38) = 0.052 The range chart is out of control since sample 1 falls outside the UCL and sample 6 falls outside the LCL. This makes the x calculation moot. Conclusion on process variability given R = 0.38 and n = 8: Too much variability in the process!

17. 5 – 17 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Process Capability Process capability refers to the ability of the process to meet the design specification for the product or service Design specifications are often expressed as a nominal value and a tolerance Three-sigma quality, Ratio>1 Four-sigma quality, Ratio>1.33 Six-sigma quality, Ratio>2

18. 5 – 18 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Process Capability Index The process capability index measures how well a process is centered and whether the variability is acceptable C pk = Minimum of, x – Lower specification 3σ Upper specification – x 3σ where σ = standard deviation of the process distribution If C pk is less than the critical value, either the process average is too close to one of the tolerance limits and is generating defective output, or the process variability is too large.

19. 5 – 19 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Process Capability Ratio The process capability ratio tests whether process variability is the cause of problems C p = Upper specification – Lower specification 6σ

20. 5 – 20 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. In Stat Control, but meeting Design Specs? If a process is in statistical control, we may also want to determine if the process is meeting a design specification or standard. To do this first determine Cpk, if this metric is okay (1, 1.33, 2), the process is capable of meeting design specs to a high level of quality (three, four, or six- sigma). Otherwise, if Cpk is small (less than 1, 1.33, or 2), determine Cp. If the Cp ratio is not okay, we conclude the process has too much variability. If the Cp is okay (and Cpk is not), we conclude the process mean is not centered appropriately, i.e. to design spec.

21. 5 – 21 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Assessing Process Capability EXAMPLE 5.5 The intensive care unit lab process has an average turnaround time of 26.2 minutes and a standard deviation of 1.35 minutes The nominal value for this service is 25 minutes with an upper specification limit of 30 minutes and a lower specification limit of 20 minutes The administrator of the lab wants to have four-sigma performance for her lab Is the lab process capable of this level of performance?

22. 5 – 22 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Assessing Process Capability SOLUTION The administrator began by taking a quick check to see if the process is capable by applying the process capability index: Lower specification calculation = = 1.53 26.2 – 20.0 3(1.35) Upper specification calculation = = 0.94 30.0 – 26.2 3(1.35) C pk = Minimum of [1.53, 0.94] = 0.94 Since the target value for four-sigma performance is 1.33, the process capability index told her that the process was not capable. However, she did not know whether the problem was the variability of the process, the centering of the process, or both. The options available to improve the process depended on what is wrong.

23. 5 – 23 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Assessing Process Capability She next checked the process variability with the process capability ratio: The process variability did not meet the four-sigma target of 1.33. Consequently, she initiated a study to see where variability was introduced into the process. Two activities, report preparation and specimen slide preparation, were identified as having inconsistent procedures. These procedures were modified to provide consistent performance. New data were collected and the average turnaround was now 26.1 minutes with a standard deviation of 1.20 minutes. C p = = 1.23 30.0 – 20.0 6(1.35)

24. 5 – 24 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Assessing Process Capability She now had the process variability at the four-sigma level of performance, as indicated by the process capability ratio: However, the process capability index indicated additional problems to resolve: C p = = 1.39 30.0 – 20.0 6(1.20) C pk =, = 1.08 (26.1 – 20.0) 3(1.20) (30.0 – 26.1) 3(1.20) Minimum of

25. 5 – 25 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.4 Webster Chemical’s nominal weight for filling tubes of caulk is 8.00 ounces ± 0.60 ounces. The target process capability ratio is 1.33, signifying that management wants 4-sigma performance. The current distribution of the filling process is centered on 8.054 ounces with a standard deviation of 0.192 ounces. Compute the process capability index and process capability ratio to assess whether the filling process is capable and set properly.

26. 5 – 26 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.4 Recall that a capability index value of 1.0 implies that the firm is producing three-sigma quality (0.26% defects) and that the process is consistently producing outputs within specifications even though some defects are generated. The value of 0.948 is far below the target of 1.33. Therefore, we can conclude that the process is not capable. Furthermore, we do not know if the problem is centering or variability. C pk = Minimum of, x – Lower specification 3σ Upper specification – x 3σ = Minimum of = 1.135, = 0.948 8.054 – 7.400 3(0.192) 8.600 – 8.054 3(0.192) a.Process capability index:

27. 5 – 27 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.4 b.Process capability ratio: C p = Upper specification – Lower specification 6σ = = 1.0417 8.60 – 7.40 6(0.192) Recall that if the C pk is greater than the critical value (1.33 for four-sigma quality) we can conclude that the process is capable. Since the C pk is less than the critical value, either the process average is close to one of the tolerance limits and is generating defective output, or the process variability is too large. The value of C p is less than the target for four-sigma quality. Therefore we conclude that the process variability must be addressed first, and then the process should be retested.

28. 5 – 28 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 The Watson Electric Company produces incandescent lightbulbs. The following data on the number of lumens for 40- watt lightbulbs were collected when the process was in control. Observation Sample1234 1604612588600 2597601607603 3581570585592 4620605595588 5590614608604 a.Calculate control limits for an R -chart and an x -chart. b.Since these data were collected, some new employees were hired. A new sample obtained the following readings: 570, 603, 623, and 583. Is the process still in control?

29. 5 – 29 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 Sample R 160124 260210 358222 460232 560424 Total2,991112 Average x = 598.2 R = 22.4 x 604 + 612 + 588 + 600 4 = 601 x = R = 612 – 588 = 24 SOLUTION a.To calculate x, compute the mean for each sample. To calculate R, subtract the lowest value in the sample from the highest value in the sample. For example, for sample 1,

30. 5 – 30 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 1 The R -chart control limits are b.First check to see whether the variability is still in control based on the new data. The range is 53 (or 623 – 570), which is outside the UCL for the R -chart. Since the process variability is out of control, it is meaningless to test for the process average using the current estimate for R. A search for assignable causes inducing excessive variability must be conducted. 2.282(22.4) = 51.12 0(22.4) = 0 UCL R = D 4 R = LCL R = D 3 R = The x -chart control limits are UCL x = x + A 2 R = LCL x = x – A 2 R = 598.2 + 0.729(22.4) = 614.53 598.2 – 0.729(22.4) = 581.87

31. 5 – 31 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 4 Pioneer Chicken advertises “lite” chicken with 30 percent fewer calories. (The pieces are 33 percent smaller.) The process average distribution for “lite” chicken breasts is 420 calories, with a standard deviation of the population of 25 calories. Pioneer randomly takes samples of six chicken breasts to measure calorie content. a.Design an x -chart using the process standard deviation. b.The product design calls for the average chicken breast to contain 400 ± 100 calories. Calculate the process capability index (target = 1.33) and the process capability ratio. Interpret the results.

32. 5 – 32 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 4 SOLUTION a.For the process standard deviation of 25 calories, the standard deviation of the sample mean is UCL x = x + zσ x = LCL x = x – zσ x = 420 + 3(10.2) = 450.6 calories 420 – 3(10.2) = 389.4 calories

33. 5 – 33 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 4 Because the process capability ratio is 1.33, the process should be able to produce the product reliably within specifications. However, the process capability index is 1.07, so the current process is not centered properly for four-sigma performance. The mean of the process distribution is too close to the upper specification. The process capability ratio is b. The process capability index is C pk = Minimum of, x – Lower specification 3σ Upper specification – x 3σ = Minimum of = 1.60, = 1.07 420 – 300 3(25) 500 – 420 3(25) C p = Upper specification – Lower specification 6σ = = 1.33 500 – 300 6(25)