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Quality And Performance Total Quality Management

Quality And Performance Total Quality Management. 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. Service/product design. Process design. Continuous improvement.

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Quality And Performance Total Quality Management

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  1. Quality And PerformanceTotal Quality Management

  2. 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. Service/product design Process design Continuous improvement Employee involvement Customer satisfaction Problem-solving tools Purchasing Benchmarking Total Quality Management Figure 5.1 – TQM Wheel

  4. Total Quality Management • Customer satisfaction • Conformance to specifications • Value • Fitness for use • Support • Psychological impressions • Employee involvement • Cultural change • Teams

  5. Total Quality Management • Continuous improvement • Kaizen • A philosophy • Not unique to quality • Problem solving process

  6. Act Do Study The Deming Wheel Plan Figure 5.2 – Plan-Do-Study-Act Cycle

  7. X X X X X X X X X X X X X X X X X X X X X X X X X X Six Sigma Process average OK;too much variation Process variability OK;process off target Processon target withlow variability Reducespread Centerprocess Figure 5.3 – Six-Sigma Approach Focuses on Reducing Spread and Centering the Process

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

  9. Acceptance Sampling • Application of statistical techniques • Acceptable quality level (AQL) • Linked through supply chains

  10. fan motors Motor inspection fan blades Yes No Acceptmotors? Blade inspection Yes No Acceptblades? Acceptance Sampling Firm A uses TQM or SixSigma to achieve internalprocess performance Buyer Manufactures furnaces Supplier uses TQM or SixSigma to achieve internalprocess performance Firm A Manufacturersfurnace fan motors TARGET: Buyer’s specs Supplier Manufacturesfan blades TARGET: Firm A’s specs Figure 5.5 – Interface of Acceptance Sampling and Process Performance Approaches in a Supply Chain

  11. Statistical Process Control • Used to detect process change • Variation of outputs • Performance measurement – variables • Performance measurement – attributes • Sampling • Sampling distributions

  12. where xi= observation of a quality characteristic (such as time) n = total number of observations x = mean Sampling Distributions • The sample mean is the sum of the observations divided by the total number of observations

  13. Sampling Distributions • 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. Mean Process distribution 25 Time Sample and Process Distributions Distribution ofsample means Figure 5.6 – Relationship Between the Distribution of Sample Means and the Process Distribution

  15. 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. Average Time Assignable Causes (a) Location Figure 5.7 – Effects of Assignable Causes on the Process Distribution for the Lab Analysis Process

  17. Average Time Assignable Causes (b) Spread Figure 5.7 – Effects of Assignable Causes on the Process Distribution for the Lab Analysis Process

  18. Average Time Assignable Causes (c) Shape Figure 5.7 – Effects of Assignable Causes on the Process Distribution for the Lab Analysis Process

  19. Control Charts • Time-ordered diagram of process performance • Mean • Upper control limit • Lower control limit • Steps for a control chart • Random sample • Plot statistics • Eliminate the cause, incorporate improvements • Repeat the procedure

  20. UCL Nominal LCL Assignable causes likely 1 2 3 Control Charts Samples Figure 5.8 – How Control Limits Relate to the Sampling Distribution: Observations from Three Samples

  21. UCL Nominal Variations LCL Sample number Control Charts (a) Normal – No action Figure 5.9 – Control Chart Examples

  22. UCL Nominal Variations LCL Sample number Control Charts (b) Run – Take action Figure 5.9 – Control Chart Examples

  23. UCL Nominal Variations LCL Sample number Control Charts (c) Sudden change – Monitor Figure 5.9 – Control Chart Examples

  24. UCL Nominal Variations LCL Sample number Control Charts (d) Exceeds control limits – Take action Figure 5.9 – Control Chart Examples

  25. 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. UCLR = D4R and LCLR = D3R where R = average of several past R values and the central line of the control chart D3, D4 = constants that provide three standard deviation (three-sigma) limits for the given sample size SPC Methods • Control charts for variables • R-Chart

  27. Control Chart Factors

  28. Control charts for variables • x-Chart UCLx = x + A2R and LCLx= x–A2R 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 A2 = constant to provide three-sigma limits for the sample mean SPC Methods

  29. Calculate x for each sample Steps for x- and R-Charts • Collect data • Compute the range • Use Table 5.1 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.

  30. Use Table 5.1 to determine x-chart control limits Steps for x- and R-Charts • 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. 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. 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. Using x- and R-Charts

  33. UCLR = D4R = LCLR = D3R = 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 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.

  34. 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 Using x- and R-Charts

  35. 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. Using x- and R-Charts Step 5: Compute the mean for each sample. For example, the mean for sample 1 is

  36. 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 A2 from Table 5.1 for a sample size of 4 to construct the control limits: LCLx= x– A2R= UCLx= x+ A2R= Using x- and R-Charts 0.5027 – 0.729(0.0021) = 0.5012 in. 0.5027 + 0.729(0.0021) = 0.5042 in. 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.

  37. 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 Using x- and R-Charts

  38. If 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 where σx =σ/ n σ = standard deviation of the process distribution n = sample size x = central line of the chart z = normal deviate number An Alternate Form

  39. 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 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. • 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. x = 5 minutes σ = 1.5 minutes n = 6 customers z = 1.96 The process variability is in statistical control, so we proceed directly to the x-chart. The control limits are 5.0 + 1.96(1.5)/6 = 6.20 minutes UCLx = x + zσ/n = 5.0 – 1.96(1.5)/6 = 3.80 minutes LCLx = x – zσ/n = Using Process Standard Deviation SOLUTION

  41. 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. 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.

  43. Conclusion on process variability given R = 0.38 and n = 8: UCLR = D4R = LCLR = D3R = 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. Application 5.1 Assuming that taking only 6 samples is sufficient, is the process in statistical control? 1.864(0.38) = 0.708 0.136(0.38) = 0.052

  44. Application 5.1 Consider dropping sample 6 because of an inoperative scale, causing inaccurate measures. What is the conclusion on process variability and process average?

  45. Now R = 0.45, x = 8.034, and n = 8 UCLx = x + A2R = UCLR = D4R = LCLx = x–A2R = LCLR = D3R = Application 5.1 1.864(0.45) = 0.839 0.136(0.45) = 0.061 8.034 + 0.373(0.45) = 8.202 8.034 – 0.373(0.45) = 7.832 The resulting control charts indicate that the process is actually in control.

  46. p = the center line on the chart and UCLp = p + zσp and LCLp= p – zσp 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

  47. 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. 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. Using a p-Chart

  50. Step 1: Using this sample data to calculate p = = 0.0049 p = Total defectives Total number of observations UCLp = p + zσp = 0.0049(1 – 0.0049)/2,500 = 0.0014 147 12(2,500) LCLp = p–zσp Using a p-Chart σp = p(1 – p)/n = 0.0049 + 3(0.0014) = 0.0091 = 0.0049 – 3(0.0014) = 0.0007

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