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 PowerPoint Slides by Jeff Heyl

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. Six Sigma Improvement Model Control Improve Analyze Measure Define Figure 5.4 – Six Sigma Improvement Model

9. 5 – 9 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Acceptance Sampling Application of statistical techniques Acceptable quality level (AQL) Linked through supply chains

10. 5 – 10 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Acceptance Sampling Firm A uses TQM or Six Sigma to achieve internal process performance Supplier uses TQM or Six Sigma to achieve internal process performance YesNo YesNo fan motors fan blades Figure 5.5 –Interface of Acceptance Sampling and Process Performance Approaches in a Supply Chain Accept blades? Supplier Manufactures fan blades TARGET: Firm A’s specs Accept motors? Motor inspection Blade inspection Firm A Manufacturers furnace fan motors TARGET: Buyer’s specs Buyer Manufactures furnaces

11. 5 – 11 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Statistical Process Control Used to detect process change Variation of outputs Performance measurement – variables Performance measurement – attributes Sampling Sampling distributions

12. 5 – 12 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Sampling Distributions 1.The sample mean is the sum of the observations divided by the total number of observations where x i = observation of a quality characteristic (such as time) n = total number of observations x = mean

13. 5 – 13 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Sampling Distributions 2.The range is the difference between the largest observation in a sample and the smallest. The standard deviation is the square root of the variance of a distribution. An estimate of the process standard deviation based on a sample is given by where σ = standard deviation of a sample

14. 5 – 14 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Sample and Process Distributions Distribution of sample means 25Time Mean Process distribution Figure 5.6 –Relationship Between the Distribution of Sample Means and the Process Distribution

15. 5 – 15 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Causes of Variation Common causes  Random, unavoidable sources of variation  Location  Spread  Shape Assignable causes  Can be identified and eliminated  Change in the mean, spread, or shape  Used after a process is in statistical control

16. 5 – 16 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Assignable Causes (a) Location Time Average Figure 5.7 –Effects of Assignable Causes on the Process Distribution for the Lab Analysis Process

17. 5 – 17 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Assignable Causes (b) Spread Time Average Figure 5.7 –Effects of Assignable Causes on the Process Distribution for the Lab Analysis Process

18. 5 – 18 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Assignable Causes (c) Shape Time Average Figure 5.7 –Effects of Assignable Causes on the Process Distribution for the Lab Analysis Process

19. 5 – 19 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Control Charts Time-ordered diagram of process performance  Mean  Upper control limit  Lower control limit Steps for a control chart 1.Random sample 2.Plot statistics 3.Eliminate the cause, incorporate improvements 4.Repeat the procedure

20. 5 – 20 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Control Charts Samples Assignable causes likely 1 2 3 Figure 5.8 –How Control Limits Relate to the Sampling Distribution: Observations from Three Samples UCL Nominal LCL

21. 5 – 21 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Nominal UCL LCL Variations Sample number Control Charts Figure 5.9 –Control Chart Examples (a) Normal – No action

22. 5 – 22 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Nominal UCL LCL Variations Sample number Control Charts Figure 5.9 –Control Chart Examples (b) Run – Take action

23. 5 – 23 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Nominal UCL LCL Variations Sample number Control Charts Figure 5.9 –Control Chart Examples (c) Sudden change – Monitor

24. 5 – 24 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Nominal UCL LCL Variations Sample number Control Charts Figure 5.9 –Control Chart Examples (d) Exceeds control limits – Take action

25. 5 – 25 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Control Charts Two types of error are possible with control charts A type I error occurs when a process is thought to be out of control when in fact it is not A type II error occurs when a process is thought to be in control when it is actually out of statistical control These errors can be controlled by the choice of control limits

26. 5 – 26 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. SPC Methods Control charts for variables  R -Chart UCL R = D 4 R and LCL R = D 3 R where R =average of several past R values and the central line of the control chart D 3, D 4 =constants that provide three standard deviation (three-sigma) limits for the given sample size

27. 5 – 27 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

28. 5 – 28 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. SPC Methods UCL x = x + A 2 R and LCL x = x – A 2 R Control charts for variables  x -Chart where x =central line of the chart, which can be either the average of past sample means or a target value set for the process A 2 =constant to provide three-sigma limits for the sample mean

29. 5 – 29 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Steps for x - and R -Charts 1.Collect data 2.Compute the range 3.Use Table 5.1 to determine R -chart control limits 4.Plot the sample ranges. If all are in control, proceed to step 5. Otherwise, find the assignable causes, correct them, and return to step 1. 5.Calculate x for each sample

30. 5 – 30 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Steps for x - and R -Charts 6.Use Table 5.1 to determine x -chart control limits 7.Plot the sample means. If all are in control, the process is in statistical control. Continue to take samples and monitor the process. If any are out of control, find the assignable causes, correct them, and return to step 1. If no assignable causes are found, assume out-of-control points represent common causes of variation and continue to monitor the process.

31. 5 – 31 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? SOLUTION Step 1:For simplicity, we use only 5 samples. In practice, more than 20 samples would be desirable. The data are shown in the following table.

32. 5 – 32 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 Step 2:Compute the range for each sample by subtracting the lowest value from the highest value. For example, in sample 1 the range is 0.5027 – 0.5009 = 0.0018 in. Similarly, the ranges for samples 2, 3, 4, and 5 are 0.0021, 0.0017, 0.0026, and 0.0022 in., respectively. As shown in the table, R = 0.0021.

33. 5 – 33 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using x - and R -Charts Step 3:To construct the R -chart, select the appropriate constants from Table 5.1 for a sample size of 4. The control limits are UCL R = D 4 R =2.282(0.0021) = 0.00479 in. 0(0.0021) = 0 in. Step 4:Plot the ranges on the R -chart, as shown in Figure 5.10. None of the sample ranges falls outside the control limits so the process variability is in statistical control. If any of the sample ranges fall outside of the limits, or an unusual pattern appears, we would search for the causes of the excessive variability, correct them, and repeat step 1. LCL R = D 3 R =

34. 5 – 34 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

35. 5 – 35 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using x - and R -Charts Step 5:Compute the mean for each sample. For example, the mean for sample 1 is 0.5014 + 0.5022 + 0.5009 + 0.5027 4 = 0.5018 in. Similarly, the means of samples 2, 3, 4, and 5 are 0.5027, 0.5026, 0.5020, and 0.5045 in., respectively. As shown in the table, x = 0.5027.

36. 5 – 36 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using x - and R -Charts Step 7:Plot the sample means on the control chart, as shown in Figure 5.11. The mean of sample 5 falls above the UCL, indicating that the process average is out of statistical control and that assignable causes must be explored, perhaps using a cause-and-effect diagram. LCL x = x – A 2 R = 0.5027 + 0.729(0.0021) = 0.5042 in. 0.5027 – 0.729(0.0021) = 0.5012 in. UCL x = x + A 2 R = Step 6:Now construct the x -chart for the process average. The average screw diameter is 0.5027 in., and the average range is 0.0021 in., so use x = 0.5027, R = 0.0021, and A 2 from Table 5.1 for a sample size of 4 to construct the control limits:

37. 5 – 37 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

38. 5 – 38 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. An Alternate Form UCL x = x + z σ x and LCL x = x – zσ x If the standard deviation of the process distribution is known, another form of the x -chart may be used: where σ x = σ / n σ = standard deviation of the process distribution n = sample size x = central line of the chart z = normal deviate number

39. 5 – 39 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using Process Standard Deviation EXAMPLE 5.2 For Sunny Dale Bank the time required to serve customers at the drive-by window is an important quality factor in competing with other banks in the city. Mean time to process a customer at the peak demand period is 5 minutes Standard deviation of 1.5 minutes Sample size of six customers Design an x -chart that has a type I error of 5 percent After several weeks of sampling, two successive samples came in at 3.70 and 3.68 minutes, respectively. Is the customer service process in statistical control?

40. 5 – 40 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using Process Standard Deviation SOLUTION x = 5 minutes σ = 1.5 minutes n = 6 customers z = 1.96 UCL x = x + z σ/  n = LCL x = x – z σ/  n = 5.0 + 1.96(1.5)/  6 = 6.20 minutes 5.0 – 1.96(1.5)/  6 = 3.80 minutes The process variability is in statistical control, so we proceed directly to the x -chart. The control limits are

41. 5 – 41 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using Process Standard Deviation Obtain the value for z in the following way For a type I error of 5 percent, 2.5 percent of the curve will be above the UCL and 2.5 percent below the LCL From the normal distribution table (see Appendix 1) we find the z value that leaves only 2.5 percent in the upper portion of the normal curve (or 0.9750 in the table) So z = 1.96 The two new samples are below the LCL of the chart, implying that the average time to serve a customer has dropped Assignable causes should be explored to see what caused the improvement

42. 5 – 42 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

43. 5 – 43 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:

44. 5 – 44 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.1 Consider dropping sample 6 because of an inoperative scale, causing inaccurate measures. 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 Avgs8.0340.45 What is the conclusion on process variability and process average?

45. 5 – 45 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.1 UCL R = D 4 R = LCL R = D 3 R = 1.864(0.45) = 0.839 0.136(0.45) = 0.061 UCL x = x + A 2 R = LCL x = x – A 2 R = 8.034 + 0.373(0.45) = 8.202 8.034 – 0.373(0.45) = 7.832 Now R = 0.45, x = 8.034, and n = 8 The resulting control charts indicate that the process is actually in control.

46. 5 – 46 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Control Charts for Attributes p -charts are used to control the proportion defective Sampling involves yes/no decisions so the underlying distribution is the binomial distribution The standard deviation is p = the center line on the chart UCL p = p + zσ p and LCL p = p – zσ p and

47. 5 – 47 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using p -Charts Periodically a random sample of size n is taken The number of defectives is counted The proportion defective p is calculated If the proportion defective falls outside the UCL, it is assumed the process has changed and assignable causes are identified and eliminated If the proportion defective falls outside the LCL, the process may have improved and assignable causes are identified and incorporated

48. 5 – 48 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using a p -Chart EXAMPLE 5.3 Hometown Bank is concerned about the number of wrong customer account numbers recorded Each week a random sample of 2,500 deposits is taken and the number of incorrect account numbers is recorded The results for the past 12 weeks are shown in the following table Is the booking process out of statistical control? Use three-sigma control limits, which will provide a Type I error of 0.26 percent.

49. 5 – 49 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using a p -Chart Sample Number Wrong Account Numbers Sample Number Wrong Account Numbers 115724 21287 319910 42 17 5191115 64123 Total147

50. 5 – 50 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using a p -Chart 147 12(2,500) = = 0.0049 p = Total defectives Total number of observations σ p =  p (1 – p )/ n =  0.0049(1 – 0.0049)/2,500 = 0.0014 UCL p = p + zσ p LCL p = p – zσ p = 0.0049 + 3(0.0014) = 0.0091 = 0.0049 – 3(0.0014) = 0.0007 Step 1:Using this sample data to calculate p

51. 5 – 51 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using a p -Chart Step 2:Calculate the sample proportion defective. For sample 1, the proportion of defectives is 15/2,500 = 0.0060. Step 3:Plot each sample proportion defective on the chart, as shown in Figure 5.12. Figure 5.12 –The p -Chart from POM for Windows for Wrong Account Numbers, Showing That Sample 7 is Out of Control Fraction Defective Sample Mean UCL LCL.0091.0049.0007 |||||||||||| 123456789101112 X X X X X X X X X X X X

52. 5 – 52 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.2 A sticky scale brings Webster’s attention to whether caulking tubes are being properly capped. If a significant proportion of the tubes aren’t being sealed, Webster is placing their customers in a messy situation. Tubes are packaged in large boxes of 144. Several boxes are inspected and the following numbers of leaking tubes are found: Sample TubesSample TubesSample Tubes 13 86155 25 94160 33109172 44112186 52126192 64135201 72141Total =72

53. 5 – 53 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.2 Calculate the p -chart three-sigma control limits to assess whether the capping process is in statistical control. The process is in control as the p values for the samples all fall within the control limits.

54. 5 – 54 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Control Charts for Attributes c -charts count the number of defects per unit of service encounter The underlying distribution is the Poisson distribution The mean of the distribution is c and the standard deviation is  c UCL c = c + z  c and LCL c = c – z  c

55. 5 – 55 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using a c -Chart EXAMPLE 5.4 The Woodland Paper Company produces paper for the newspaper industry. As a final step in the process, the paper passes through a machine that measures various product quality characteristics. When the paper production process is in control, it averages 20 defects per roll. a.Set up a control chart for the number of defects per roll. For this example, use two-sigma control limits. b.Five rolls had the following number of defects: 16, 21, 17, 22, and 24, respectively. The sixth roll, using pulp from a different supplier, had 5 defects. Is the paper production process in control?

56. 5 – 56 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using a c -Chart SOLUTION a.The average number of defects per roll is 20. Therefore UCL c = c + z  c LCL c = c – z  c = 20 + 2(  20) = 28.94 = 20 – 2(  20) = 11.06 The control chart is shown in Figure 5.13

57. 5 – 57 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using a c -Chart Figure 5.13 –The c -Chart from POM for Windows for Defects per Roll of Paper b.Because the first five rolls had defects that fell within the control limits, the process is still in control. Five defects, however, is less than the LCL, and therefore, the process is technically “out of control.” The control chart indicates that something good has happened.

58. 5 – 58 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.3 At Webster Chemical, lumps in the caulking compound could cause difficulties in dispensing a smooth bead from the tube. Even when the process is in control, there will still be an average of 4 lumps per tube of caulk. Testing for the presence of lumps destroys the product, so Webster takes random samples. The following are results of the study: Tube #LumpsTube #LumpsTube #Lumps 1656 95 2564100 3071119 4486122 Determine the c -chart two-sigma upper and lower control limits for this process.

59. 5 – 59 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.3

60. 5 – 60 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

61. 5 – 61 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Process Capability 202530 Minutes Upper specification Lower specification Nominal value (a) Process is capable Process distribution Figure 5.14 –The Relationship Between a Process Distribution and Upper and Lower Specifications

62. 5 – 62 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Process Capability 202530 Minutes Upper specification Lower specification Nominal value (b) Process is not capable Figure 5.14 –The Relationship Between a Process Distribution and Upper and Lower Specifications Process distribution

63. 5 – 63 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Process Capability Figure 5.15 – Effects of Reducing Variability on Process Capability Lower specification Mean Upper specification Nominal value Six sigma Four sigma Two sigma

64. 5 – 64 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Process Capability 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

65. 5 – 65 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Process Capability The process capability ratio tests whether process variability is the cause of problems C p = Upper specification – Lower specification 6σ

66. 5 – 66 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining Process Capability Step 1.Collect data on the process output, and calculate the mean and the standard deviation of the process output distribution. Step 2.Use the data from the process distribution to compute process control charts, such as an x - and an R -chart.

67. 5 – 67 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining Process Capability Step 3.Take a series of at least 20 consecutive random samples from the process and plot the results on the control charts. If the sample statistics are within the control limits of the charts, the process is in statistical control. If the process is not in statistical control, look for assignable causes and eliminate them. Recalculate the mean and standard deviation of the process distribution and the control limits for the charts. Continue until the process is in statistical control.

68. 5 – 68 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Determining Process Capability Step 4.Calculate the process capability index. If the results are acceptable, the process is capable and document any changes made to the process; continue to monitor the output by using the control charts. If the results are unacceptable, calculate the process capability ratio. If the results are acceptable, the process variability is fine and management should focus on centering the process. If not, management should focus on reducing the variability in the process until it passes the test. As changes are made, recalculate the mean and standard deviation of the process distribution and the control limits for the charts and return to step 3.

69. 5 – 69 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?

70. 5 – 70 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.

71. 5 – 71 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)

72. 5 – 72 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

73. 5 – 73 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.

74. 5 – 74 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:

75. 5 – 75 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.

76. 5 – 76 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Quality Engineering Quality engineering is an approach originated by Genichi Taguchi that involves combining engineering and statistical methods to reduce costs and improve quality by optimizing product design and manufacturing processes. The quality loss function is based on the concept that a service or product that barely conforms to the specifications is more like a defective service or product than a perfect one.

77. 5 – 77 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Quality Engineering Loss (dollars) LowerNominalUpper specificationvaluespecification Figure 5.16 – Taguchi’s Quality Loss Function

78. 5 – 78 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. International Standards ISO 9000:2000 addresses quality management by specifying what the firm does to fulfill the customer’s quality requirements and applicable regulatory requirements while enhancing customer satisfaction and achieving continual improvement of its performance Companies must be certified by an external examiner Assures customers that the organization is performing as they say they are

79. 5 – 79 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. International Standards ISO 14000:2004 documents a firm’s environmental program by specifying what the firm does to minimize harmful effects on the environment caused by its activities The standards require companies to keep track of their raw materials use and their generation, treatment, and disposal of hazardous wastes Companies are inspected by outside, private auditors on a regular basis

80. 5 – 80 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. International Standards External benefits are primarily increased sales opportunities ISO certification is preferred or required by many corporate buyers Internal benefits include improved profitability, improved marketing, reduced costs, and improved documentation and improvement of processes

81. 5 – 81 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. The Baldrige Award The Malcolm Baldrige National Quality Award promotes, recognizes, and publicizes quality strategies and achievements by outstanding organizations It is awarded annually after a rigorous application and review process Award winners report increased productivity, more satisfied employees and customers, and improved profitability

82. 5 – 82 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. The Baldrige Award The seven categories of the award are 1.Leadership 2.Strategic Planning 3.Customer and Market Focus 4.Measurement, Analysis, and Knowledge Management 5.Workforce Focus 6.Process Management 7.Results

83. 5 – 83 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?

84. 5 – 84 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,

85. 5 – 85 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

86. 5 – 86 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 The data processing department of the Arizona Bank has five data entry clerks. Each working day their supervisor verifies the accuracy of a random sample of 250 records. A record containing one or more errors is considered defective and must be redone. The results of the last 30 samples are shown in the table. All were checked to make sure that none was out of control.

87. 5 – 87 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 Sample Number of Defective Records Sample Number of Defective Records 17168 251712 319184 410196 5112011 682117 7122212 89236 96247 10132513 11182610 1252714 1316286 1442911 1511309 Total 300

88. 5 – 88 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 a.Based on these historical data, set up a p -chart using z = 3. b.Samples for the next four days showed the following: SampleNumber of Defective Records Tues17 Wed15 Thurs22 Fri21 What is the supervisor’s assessment of the data-entry process likely to be?

89. 5 – 89 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 SOLUTION a.From the table, the supervisor knows that the total number of defective records is 300 out of a total sample of 7,500 [or 30(250)]. Therefore, the central line of the chart is = 0.04 300 7,500 p = The control limits are

90. 5 – 90 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 2 SampleNumber of Defective RecordsProportion Tues170.068 Wed150.060 Thurs220.088 Fri210.084 b.Samples for the next four days showed the following: Samples for Thursday and Friday are out of control. The supervisor should look for the problem and, upon identifying it, take corrective action.

91. 5 – 91 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 3 The Minnow County Highway Safety Department monitors accidents at the intersection of Routes 123 and 14. Accidents at the intersection have averaged three per month. a.Which type of control chart should be used? Construct a control chart with three sigma control limits. b.Last month, seven accidents occurred at the intersection. Is this sufficient evidence to justify a claim that something has changed at the intersection?

92. 5 – 92 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem 3 SOLUTION a.The safety department cannot determine the number of accidents that did not occur, so it has no way to compute a proportion defective at the intersection. Therefore, the administrators must use a c -chart for which There cannot be a negative number of accidents, so the LCL in this case is adjusted to zero. b.The number of accidents last month falls within the UCL and LCL of the chart. We conclude that no assignable causes are present and that the increase in accidents was due to chance. UCL c = c + z c LCL c = c – z c = 3 + 3 3 = 8.20 = 3 – 3 3 = –2.196

93. 5 – 93 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.

94. 5 – 94 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

95. 5 – 95 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)

96. 5 – 96 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.