Quality and Performance Quality A term used by customers to describe their general satisfaction with a service or product Defect Any instance when a process fails to satisfy its customer Costs of Quality ID: 1019146
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1. Managing QualityChapter 3
2. Quality and PerformanceQualityA term used by customers to describe their general satisfaction with a service or productDefectAny instance when a process fails to satisfy its customer
3. Costs of QualityPrevention costsCosts associated with preventing defects before they happenAppraisal costsCosts incurred when the firm assesses the performance level of its processesInternal Failure costsCosts resulting from defects that are discovered during the production of a service or productExternal Failure costsCosts that arise when a defect is discovered after the customer receives the service or product
4. Total Quality Management and Six SigmaTotal Quality ManagementA philosophy that stresses three principles (customer satisfaction, employee involvement and continuous improvement) for achieving high level of process performance and qualitySix SigmaA comprehensive and flexible system for achieving, sustaining, and maximizing business success by minimizing defects and variability in processes
5. Total Quality ManagementFigure 3.1
6. Total Quality ManagementCustomer SatisfactionConformance to SpecificationsValueFitness for UseSupportPsychological Impressions
7. Total Quality ManagementEmployee InvolvementCultural ChangeQuality at the SourceTeamsEmployee EmpowermentProblem-solving teamsSpecial-purpose teamsSelf-managed teams
8. Total Quality ManagementContinuous ImprovementKaizenProblem-solving toolsPlan-Do-Study-Act Cycle
9. Six SigmaXXXXXXXXXXXXXXXXXProcess average OK;too much variationProcess variability OK;process off targetProcesson target withlow variabilityReducespreadCenterprocessXXXXXXXXXFigure 3.3
10. Six SigmaGoal of achieving low rates of defective output by developing processes whose mean output for a performance measure is +/- six standard deviations (sigma) from the limits of the design specifications for the service or product.
11. Acceptance SamplingAcceptance Sampling The application of statistical techniques to determine if a quantity of material from a supplier should be accepted or rejected based on the inspection or test of one or more samples.Acceptable Quality LevelThe quality level desired by the consumer.
12. Acceptance SamplingFirm A uses TQM or SixSigma to achieve internalprocess performanceSupplier uses TQM or SixSigma to achieve internalprocess performanceYesNoYesNoFan motorsFan bladesAcceptblades?SupplierManufacturesfan bladesTARGET: Firm A’s specsAcceptmotors?Motor samplingBlade samplingFirm AManufacturersfurnace fan motorsTARGET: Buyer’s specsBuyerManufactures furnacesFigure 3.4
13. Statistical Process Control (SPC)SPC The application of statistical techniques to determine whether a process is delivering what the customer wants.Variation of OutputsNo two services of products are exactly alike because the processes used to produce them contain many sources of variation, even if the processes are working as intended.
14. Statistical Process Control (SPC)Performance MeasurementsVariables - Service or product characteristics that can be measuredAttributes - Service or product characteristics that can be quickly counted for acceptable performance
15. Statistical Process Control (SPC)Complete InspectionInspect each service or product at each stage of the process for qualitySampling Sample SizeTime between successive samplesDecision rules that determine when action should be taken
16. Statistical Process Control (SPC)Figure 3.5
17. Statistical Process Control (SPC)The sample mean is the sum of the observations divided by the total number of observations.wherexi = observation of a quality characteristic (such as time)n = total number of observations = mean
18. Statistical Process Control (SPC)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
19. Statistical Process Control (SPC)Categories of VariationCommon cause - The purely random, unidentifiable sources of variation that are unavoidable with the current processAssignable cause - Any variation-causing factors that can be identified and eliminated
20. Statistical Process Control (SPC)Control Chart Time-ordered diagram that is used to determine whether observed variations are abnormalControls chart have a nominal value or center line, Upper Control Limit (UCL), and Lower Control Limit (LCL)
21. Statistical Process Control (SPC)Steps for using a control chartTake a random sample from the process and calculate a variable or attribute performance measure.If a statistic falls outside the chart’s control limits or exhibits unusual behavior, look for an assignable cause.Eliminate the cause if it degrades performance; incorporate the cause if it improves performance. Reconstruct the control chart with new data.Repeat the procedure periodically.
22. Control ChartsSamplesAssignable causes likely123UCLNominalLCLFigure 3.7
23. NominalUCLLCLVariationsSample numberControl Charts(a) Normal – No actionFigure 3.8
24. NominalUCLLCLVariationsSample numberControl Charts(b) Run – Take actionFigure 3.8
25. NominalUCLLCLVariationsSample numberControl Charts(c) Sudden change – MonitorFigure 3.8
26. NominalUCLLCLVariationsSample numberControl Charts(d) Exceeds control limits – Take actionFigure 3.8
27. Control ChartsType I error An error that occurs when the employee concludes that the process is out of control based on a sample result that fails outside the control limits, when it fact it was due to pure randomnessType II error An error that occurs when the employee concludes that the process is in control and only randomness is present, when actually the process is out of statistical control
28. Control ChartsVariable Control ChartsR-Chart – Measures the variability of the process-Chart – Measures whether the process is generating output, on average, consistent with a target valueAttribute Control Chartsp-chart – Measures the proportion of defective services or products generated by the processc-chart – Measures the number of defects when more than one defect can be present in a service or product
29. Control Charts for VariablesR-Chart UCLR = D4R and LCLR = D3R
30. Control Charts for VariablesUCL = + A2 and LCL = - A2 where = central line of the chart, which can be either the average of past sample means or a target value set for the process A2 = constant to provide three-sigma limits for the sample mean x-Chart
31. Control Charts for VariablesTable 3.1
32. Control Charts for VariablesCollect data.Compute the range.Use Table 3.1 to determine R-chart control limits.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.Calculate for each sample and determine the central line of the chart, Steps to Compute Control Charts:
33. Control Charts for VariablesUse Table 3.1 to determine control limits 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. Steps to Compute Control Charts:
34. Example 3.1The 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?Sample Number1234R 1.5014.5022.5009.5027.0018.50182.5021.5041.5024.5020.0021.50273.5018.5026.5035.5023.0017.50264.5008.5034.5024.5015.0026.50205.5041.5056.5034.5047.0022.5045OBSERVATIONSAverage .0021 .5027
35. Example 3.1Compute the range for each sample and the control limitsUCLR = D4 R =2.282(0.0021) = 0.00479 in.0(0.0021) = 0 in.LCLR = D3 R =
36. Example 3.1Process variability is in statistical control.Figure 3.9
37. Example 3.1Compute the mean for each sample and the control limits. = 0.5027 + 0.729(0.0021) = 0.5042 in.= 0.5027 – 0.729(0.0021) = 0.5012 in.LCLx = X – A2 RUCLx = X + A2 R
38. Example 3.1Process average is NOT in statistical control.Figure 3.10
39. Control Charts for Variableswhere σx = σ/ n σ = standard deviation of the process distribution n = sample size x = central line of the chart z = normal deviate numberIf the standard deviation of the process distribution is known, another form of the x -chart may be used:UCLx = x + zσx and LCLx = x – zσx
40. Example 3.2For 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 minutesStandard deviation of 1.5 minutesSample size of six customersDesign an -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?
41. Example 3.2x = 5 minutesσ = 1.5 minutesn = 6 customersz = 1.96The new process is an improvement. LCLx = x – zσ/n =5.0 – 1.96(1.5)/6 = 3.80 minutesThe process variability is in statistical control, so we proceed directly to the -chart. The control limits are UCLx = x + zσ/n =5.0 + 1.96(1.5)/6 = 6.20 minutes
42. Application 3.1Webster 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 NumberSample12345678AvgRange17.988.348.027.948.447.687.818.118.0400.7628.238.127.988.418.318.187.998.068.1600.4337.897.777.918.048.007.897.938.097.9400.3248.248.187.838.057.908.167.978.078.0500.4157.878.137.927.998.107.818.147.887.9800.3368.138.148.118.138.148.128.138.148.1300.03Avgs8.0500.38
43. Application 3.1Assuming that taking only 6 samples is sufficient, is the process in statistical control?UCLR = D4 R =LCLR = D3 R =1.864(0.38) = 0.7080.136(0.38) = 0.052The range chart is out of control since sample 1 falls outside the UCL and sample 6 falls outside the LCL. Conclusion on process variability given = 0.38 and n = 8:
44. Application 3.1Consider dropping sample 6 because of an inoperative scale, causing inaccurate measures.Tube NumberSample12345678AvgRange17.988.348.027.948.447.687.818.118.0400.7628.238.127.988.418.318.187.998.068.1600.4337.897.777.918.048.007.897.938.097.9400.3248.248.187.838.057.908.167.978.078.0500.4157.878.137.927.998.107.818.147.887.9800.33Avgs8.0340.45What is the conclusion on process variability and process average?
45. Application 3.1The resulting control charts indicate that the process is actually in control. Now = 0.45, = 8.034, and n = 8 UCLR = D4 R =1.864(0.45) = 0.839LCLR = D3 R =0.136(0.45) = 0.061UCL x = x + A2 R =8.034 + 0.373(0.45) = 8.2028.034 – 0.373(0.45) = 7.866LCL x = x – A2 R =
46. Control Charts for Attributesp-charts are used for controlling the proportion of defective services or products generated by the process.The standard deviation isp = the center line on the chartUCLp = + zσp and LCLp = – zσp
47. Example 3.3Hometown 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 recordedUsing three-sigma control limits, which will provide a Type I error of 0.26 percent, is the booking process out of statistical control?
48. Example 3.3Sample NumberWrong Account NumbersSample NumberWrong Account Numbers1 15 7242 12 873 19 9104 2 10175 19 11156 4 123Total147
49. Example 3.314712(2,500)= = 0.0049p =Total defectivesTotal number of observationsσp = (1 – )/n = 0.0049(1 – 0.0049)/2,500 = 0.0014Calculate the sample proportion defective and plot each sample proportion defective on the chart.UCLp = p + zσp= 0.0049 + 3(0.0014) = 0.0091= 0.0049 – 3(0.0014) = 0.0007LCLp = p – zσp
50. Example 3.3Fraction DefectiveSampleMeanUCLLCL.0091.0049.0007 | | | | | | | | | | | | 1 2 3 4 5 6 7 8 9 10 11 12XXXXXXXXXXXXThe process is NOT in statistical control.Figure 3.11
51. Application 3.2A 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 Tubes Sample Tubes Sample Tubes13 8615525 941603310917244112186521261926413520172141Total =72
52. Application 3.2Calculate 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.7220(144)= = 0.025p =Total number of leaky tubesTotal number of tubesp (1 – p )n0.025(1 – 0.025)144 = 0.01301σp = =UCLp = p + zσp= 0.025 + 3(0.01301)= 0.06403LCLp = p – zσp= 0.025 – 3(0.01301)= –0.01403 = 0
53. Control Charts for Attributesc-charts – A chart used for controlling the number of defects when more than one defect can be present in a service or product. The mean of the distribution is and the standard deviation is UCLc = c + z cLCLc = c – z c
54. Example 3.4The 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?
55. Example 3.4a. The average number of defects per roll is 20. Therefore= 20 + 2(20) = 28.94= 20 – 2(20) = 11.06UCLc = + z LCLc = - z
56. Example 3.4The process is technically out of control due to Sample 6. However, Sample 6 shows that the new supplier is a good one. b.Figure 3.12
57. Application 3.3At 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 #Lumps1656 95256410030711194486122Determine the c-chart two-sigma upper and lower control limits for this process.
58. Application Problem 3.3
59. Process CapabilityProcess Capability – The ability of the process to meet the design specification for a service or productNominal ValueA target for design specificationsTolerance An allowance above or below the nominal value
60. Process Capability202530MinutesUpperspecification LowerspecificationNominalvalue (a) Process is capableProcess distributionFigure 3.13
61. Process Capability202530MinutesUpperspecification LowerspecificationNominalvalue (b) Process is not capableProcess distributionFigure 3.13
62. Process CapabilityLowerspecificationMeanUpperspecification Nominal valueSix sigmaFour sigmaTwo sigmaFigure 3.14
63. Process CapabilityProcess Capability Index (Cpk)An index that measures the potential for a process to generate defective outputs relative to either upper or lower specifications. where σ = standard deviation of the process distributionCpk = Minimum of , – Lower specification3σ Upper specification – 3σ
64. Process CapabilityProcess Capability (Cp)The tolerance width divided by six standard deviations.Cp =Upper specification – Lower specification6σ
65. Example 3.5The 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 + 5 minutes.Is the lab process capable of four sigma-level performance? Upper specification = 30 minutesLower specification20 minutesAverage service 26.2 minutes =1.35 minutes
66. Example 3.5Cpk = Minimum of 26.2 – 20 , 30 – 26.2 3 ( 1.53) 3( 1.53)Cpk = Minimum of , – Lower specification3σ Upper specification – 3σ Process does not meets 4-sigma level of 1.33Cpk = Minimum of 1.53, 0.94Cpk = 0.94
67. Example 3.5Process did not meet 4-sigma level of 1.3330 – 20 6 (1.35)Cp = Upper specification – Lower specification 6σ= 1.23Cp =
68. Example 3.5New Data is collected:Upper specification = 30 minutesLower specification20 minutesAverage service 26.1 minutes = 1.20 minutesCp = Upper – Lower 6Cp = 30 – 20 = 1.39 6 (1.20) Process meets 4-sigma level of 1.33 for variability
69. Example 3.5Cpk = Minimum 26.1 – 20 , 30 – 26.1 3 ( 1.20) 3 ( 1.20)Cpk = 1.08Process does not meets 4-sigma level of 1.33 Cpk = Minimum ,– Lower specification 3σ Upper specification – 3σ
70. Application 3.4Webster 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.
71. Application 3.4The value of 0.948 is far below the target of 1.33. Therefore, we can conclude that the process is not capable. = Minimum = 1.135 = 0.9488.054 – 7.4003(0.192)8.600 – 8.0543(0.192)a. Process capability index: Cpk = Minimum ,– Lower specification 3σ Upper specification – 3σ
72. Application 3.4b. Process capability ratio: Cp =Upper specification – Lower specification6σ= = 1.04178.60 – 7.406(0.192)The value of Cp 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.
73. International Quality Documentation StandardsISO 9001:2008 – Documentation StandardsISO 14000:2004 – Environmental Management System
74. Benefits of Baldrige Performance Excellence ProgramBaldrige Performance Excellence ProgramApplication process is rigorous and helps organizations define what quality means to themInvestment in quality principles and performance excellence pays off in increased productivity
75. Benefits of Baldrige Performance Excellence ProgramSeven Major CriteriaLeadershipStrategic PlanningCustomer FocusMeasurement, Analysis, and Knowledge ManagementWorkforce FocusOperations FocusResults
76. Solved Problem 1The Watson Electric Company produces incandescent light bulbs. The following data on the number of lumens for 40-watt light bulbs were collected when the process was in control.ObservationSample123416046125886002597601607603358157058559246206055955885590614608604a. Calculate control limits for an R-chart and an -chart.b. Since these data were collected, some new employees were hired. A new sample obtained the following readings: 625, 592, 612, and 635. Is the process still in control?
77. Solved Problem 1SampleR1 601 242 602 103 582 224 602 325 604 24Total 2,991 112Averagex = 598.2R = 22.4604 + 612 + 588 + 6004= 601x =R =612 – 588 = 24a. To calculate , 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,
78. Solved Problem 1The R-chart control limits are2.282(22.4) = 51.12UCLR = D4 = LCLR = D3= The x-chart control limits are0(22.4) = 0UCLx = x + A2 R=598.2 + 0.729(22.4) = 614.53598.2 – 0.729(22.4) = 581.87LCLx = x – A2R=
79. Solved Problem 1b. First check to see whether the variability is still in control based on the new data. The range is 43 (or 635 – 592), which is inside the UCL and LCL for the R-chart. Since the process variability is in control, we test for the process average using the current estimate for . The average is 616 which is above the UCL for the chart. Since the process average is out of control, a service for assignable causes inducing excessive average lumens must be conducted.
80. Solved Problem 2The 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.
81. Solved Problem 2SampleNumber of Defective RecordsSampleNumber of Defective Records 1 7168 2 51712 3 19184 4 10196 5 112011 6 82117 7 122212 8 9236 9 6247 10 132513 11 182610 12 52714 13 16286 14 42911 15 11309Total 300
82. Solved Problem 2a. 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 RecordsTues17Wed15Thurs22Fri21What is the supervisor’s assessment of the data-entry process likely to be?
83. Solved Problem 2a. 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= 0.043007,500p =The control limits are
84. Solved Problem 2SampleNumber of Defective RecordsProportionTues170.068Wed150.060Thurs220.088Fri210.084b. 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.
85. Solved Problem 3The 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?
86. Solved Problem 3a. 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 whichThere 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.
87. Solved Problem 4Pioneer 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 -chart using the process standard deviation. Use three sigma limits. 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.
88. Solved Problem 4a. For the process standard deviation of 25 calories, the standard deviation of the sample mean is
89. Solved Problem 4Because 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 isb. The process capability index is= Minimum of = 1.60, = 1.07420 – 3003(25)500 – 4203(25)Cp =Upper specification – Lower specification6σ= = 1.33500 – 3006(25)Cpk = Minimum of ,– Lower specification3σ Upper specification – 3σ