Quality And Performance Chapter 5 Quality And Performance Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Total Quality Management Process design Purchasing Service/product design Problem-solving tools Benchmarking Employee involvement Continuous improvement Customer satisfaction Figure 5.1 – TQM Wheel 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 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Sampling Distributions The sample mean is the sum of the observations divided by the total number of observations where xi = observation of a quality characteristic (such as time) n = total number of observations x = mean Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Sample and Process Distributions Mean 25 Time Distribution of sample means Process distribution Figure 5.6 – Relationship Between the Distribution of Sample Means and the Process Distribution Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Assignable Causes Average Time (a) Location Figure 5.7 – Effects of Assignable Causes on the Process Distribution for the Lab Analysis Process Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Assignable Causes Time Average (b) Spread Figure 5.7 – Effects of Assignable Causes on the Process Distribution for the Lab Analysis Process Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Assignable Causes Time Average (c) Shape Figure 5.7 – Effects of Assignable Causes on the Process Distribution for the Lab Analysis Process Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 Random sample Plot statistics Eliminate the cause, incorporate improvements Repeat the procedure Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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. Calculate x for each sample Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Steps for x- and R-Charts Use Table 5.1 to determine x-chart 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. If no assignable causes are found, assume out-of-control points represent common causes of variation and continue to monitor the process. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. SPC Methods 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. SPC Methods Control charts for variables R-Chart 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 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 (A2) Factor for LCL for R-Chart (D3) Factor for UCL for R-Chart (D4) 2 1.880 3.267 3 1.023 2.575 4 0.729 2.282 5 0.577 2.115 6 0.483 2.004 7 0.419 0.076 1.924 8 0.373 0.136 1.864 9 0.337 0.184 1.816 10 0.308 0.223 1.777 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Setting Chart Limits Data for the - and R-Charts: Ounces of Super Cola in each bottle Sample 1 2 3 4 5 R 11.90 11.83 11.99 11.71 12.00 11.886 0.29 12.10 12.19 12.136 0.09 12.20 12.01 12.29 12.180 0.28 11.80 11.898 0.19 11.70 11.900 0.40 Average = 12.000 0.25 1 = (11.90 + 11.83 + 11.99 + 11.71 + 12.00)/5 = 11.886 R1 = 12.00 – 11.71 = 0.290 Process average x = 12 ounces, Average range = .25 Sample size n = 5
Setting Chart Limits Process average x = 12 ounces Average range = .25 Sample size n = 5 UCL = 12.144 Mean = 12 LCL = 11.857 UCL = x + A2 = 12 + (.577)(.25) = 12 + .144 = 12.144 ounces LCL = x + A2 = 12 - (.577)(.25) = 12 - .144 = 11.857 ounces
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. An Alternate Form 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 z = 2 for 95.45% confidence and = 3 for 99.73% confidence Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Setting R-Chart Limits Upper control limit (UCLR) = D4 Lower control limit (LCLR) = D3 where = average range of the samples D3 and D4 = control chart factors from Table S6.1
Setting R-Chart Limits Average range = 0.25 ounces Sample size n = 5 From Table S6.1 D4 = 2.115, D3 = 0 UCL = 0.52875 Mean = 0.25 LCL = 0 UCLR = D4 = (2.115)(0.25) = 0.52875 ounces LCLR = D3 = (0)(0.25) = 0 ounces
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Control Charts UCL Nominal LCL 1 2 Assignable causes likely 3 Samples Figure 5.8 – How Control Limits Relate to the Sampling Distribution: Observations from Three Samples Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Run tests Nominal UCL LCL Variations Sample number (a) Normal – No action Figure 5.9 – Control Chart Examples Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Run tests Nominal UCL LCL Variations Sample number (b) Run – Take action Figure 5.9 – Control Chart Examples Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Run tests Upper control limit Target Lower control limit Ask the students to imagine a product, and consider what problem might cause each of the graph configurations illustrated. Trends in either direction, 5 plots. Investigate for cause of progressive change.
Run tests Upper control limit Target Lower control limit Ask the students to imagine a product, and consider what problem might cause each of the graph configurations illustrated. Two plots very near lower (or upper) control. Investigate for cause.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Control Charts Nominal UCL LCL Variations Sample number (c) Sudden change – Monitor Figure 5.9 – Control Chart Examples Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Control Charts Nominal UCL LCL Variations Sample number (d) Exceeds control limits – Take action Figure 5.9 – Control Chart Examples Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Run tests Upper control limit Target Lower control limit Ask the students to imagine a product, and consider what problem might cause each of the graph configurations illustrated. Erratic behavior. Investigate.
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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
(Don't stop production) Type I and Type II error Actual process state In control Out of control Decision based on SPC chart (Don't stop production) No error Type II error (Stop production) Type I error
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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 1 2 3 4 R x 0.5014 0.5022 0.5009 0.5027 0.0018 0.5018 0.5021 0.5041 0.5024 0.5020 0.0021 0.5026 0.5035 0.5023 0.0017 0.5008 0.5034 0.5015 0.0026 5 0.5056 0.5047 0.0022 0.5045 Average 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 UCLR = D4R = 2.282(0.0021) = 0.00479 in. LCLR = D3R = 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using x- and R-Charts 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 = 0.5027 – 0.729(0.0021) = 0.5012 in. UCLx = x + A2R = 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 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? 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 The process variability is in statistical control, so we proceed directly to the x-chart. The control limits are UCLx = x + zσ/n = 5.0 + 1.96(1.5)/6 = 6.20 minutes LCLx = x – zσ/n = 5.0 – 1.96(1.5)/6 = 3.80 minutes Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 Sample 1 2 3 4 5 6 7 8 Avg Range 7.98 8.34 8.02 7.94 8.44 7.68 7.81 8.11 8.040 0.76 8.23 8.12 8.41 8.31 8.18 7.99 8.06 8.160 0.43 7.89 7.77 7.91 8.04 8.00 7.93 8.09 7.940 0.32 8.24 7.83 8.05 7.90 8.16 7.97 8.07 8.050 0.41 7.87 8.13 7.92 8.10 8.14 7.88 7.980 0.33 8.130 0.03 Avgs 0.38 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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? Conclusion on process variability given R = 0.38 and n = 8: UCLR = D4R = 1.864(0.38) = 0.708 LCLR = D3R = 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 Sample 1 2 3 4 5 6 7 8 Avg Range 7.98 8.34 8.02 7.94 8.44 7.68 7.81 8.11 8.040 0.76 8.23 8.12 8.41 8.31 8.18 7.99 8.06 8.160 0.43 7.89 7.77 7.91 8.04 8.00 7.93 8.09 7.940 0.32 8.24 7.83 8.05 7.90 8.16 7.97 8.07 8.050 0.41 7.87 8.13 7.92 8.10 8.14 7.88 7.980 0.33 Avgs 8.034 0.45 What is the conclusion on process variability and process average? Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.1 Now R = 0.45, x = 8.034, and n = 8 UCLR = D4R = 1.864(0.45) = 0.839 LCLR = D3R = 0.136(0.45) = 0.061 UCLx = x + A2R = 8.034 + 0.373(0.45) = 8.202 LCLx = x – A2R = 8.034 – 0.373(0.45) = 7.832 The resulting control charts indicate that the process is actually in control. 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 UCLp = p + zσp and LCLp = p – zσp and Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Using a p-Chart Sample Number Wrong Account Numbers 1 15 7 24 2 12 8 3 19 9 10 4 17 5 11 6 Total 147 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Total number of observations Using a p-Chart Step 1: Using this sample data to calculate p p = Total defectives Total number of observations 147 12(2,500) = = 0.0049 σp = p(1 – p)/n = 0.0049(1 – 0.0049)/2,500 = 0.0014 UCLp = p + zσp = 0.0049 + 3(0.0014) = 0.0091 LCLp = p – zσp = 0.0049 – 3(0.0014) = 0.0007 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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. Fraction Defective Sample Mean UCL LCL .0091 .0049 .0007 | | | | | | | | | | | | 1 2 3 4 5 6 7 8 9 10 11 12 X Figure 5.12 – The p-Chart from POM for Windows for Wrong Account Numbers, Showing That Sample 7 is Out of Control Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 Tubes 1 3 8 6 15 5 2 9 4 16 10 17 11 18 12 19 13 20 7 14 Total = 72 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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. 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 UCLc = c + zc and LCLc = c – zc Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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? Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 UCLc = c + zc = 20 + 2(20) = 28.94 LCLc = c – zc = 20 – 2(20) = 11.06 The control chart is shown in Figure 5.13 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 # Lumps 1 6 5 9 2 4 10 3 7 11 8 12 Determine the c-chart two-sigma upper and lower control limits for this process. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application 5.3 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Six Sigma Process average OK; too much variation Process variability OK; process off target X X Process on target with low variability Reduce spread Center process X Figure 5.3 – Six-Sigma Approach Focuses on Reducing Spread and Centering the Process Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Process Capability 20 25 30 Minutes Upper specification Lower Nominal value Process distribution (a) Process is capable Figure 5.14 – The Relationship Between a Process Distribution and Upper and Lower Specifications Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Process Capability 20 25 30 Minutes Upper specification Lower Nominal value Process distribution (b) Process is not capable Figure 5.14 – The Relationship Between a Process Distribution and Upper and Lower Specifications Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Process Capability Lower specification Mean Upper Nominal value Six sigma Four sigma Two sigma Figure 5.15 – Effects of Reducing Variability on Process Capability Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
x – Lower specification Upper specification – x Process Capability The process capability index measures how well a process is centered and whether the variability is acceptable Cpk = Minimum of , x – Lower specification 3σ Upper specification – x where σ = standard deviation of the process distribution Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Upper specification – Lower specification Process Capability The process capability ratio tests whether process variability is the cause of problems Cp = Upper specification – Lower specification 6σ 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. 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. 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. 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? 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: = 1.53 26.2 – 20.0 3(1.35) Lower specification calculation = = 0.94 30.0 – 26.2 3(1.35) Upper specification calculation = Cpk = 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Assessing Process Capability She next checked the process variability with the process capability ratio: = 1.23 30.0 – 20.0 6(1.35) Cp = 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. 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: = 1.39 30.0 – 20.0 6(1.20) Cp = However, the process capability index indicated additional problems to resolve: , = 1.08 (26.1 – 20.0) 3(1.20) (30.0 – 26.1) Minimum of Cpk = Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
x – Lower specification Upper specification – x Application 5.4 a. Process capability index: Cpk = Minimum of , x – Lower specification 3σ Upper specification – x = Minimum of = 1.135, = 0.948 8.054 – 7.400 3(0.192) 8.600 – 8.054 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Upper specification – Lower specification Application 5.4 b. Process capability ratio: Cp = Upper specification – Lower specification 6σ = = 1.0417 8.60 – 7.40 6(0.192) Recall that if the Cpk is greater than the critical value (1.33 for four-sigma quality) we can conclude that the process is capable. Since the Cpk 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 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Quality Engineering Loss (dollars) Lower Nominal Upper specification value specification Figure 5.16 – Taguchi’s Quality Loss Function Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Acceptance Sampling Application of statistical techniques Acceptable quality level (AQL) Linked through supply chains Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
Acceptance Sampling An Accepting Sampling plan is defined by n,c n = Sample size c = Critical number of defectives in the sample N = Lot size Rejected lots are: Returned to the supplier Culled for defectives (100% inspection) Here again it is useful to stress that acceptance sampling relates to the aggregate, not the individual unit. You might also discuss the decision as to whether one should take only a single sample, or whether multiple samples are required.
Operating Characteristic (OC) Curve For a given (n,c), OC Curve is a plot of Pd on the x-axis and Pa on the y-axis Pd = Probability of defectives in the lot Pa = Probability of accepting the lot, i.e. P(X<=c), x = number of defectives in the sample Poisson probability tables can be used to develop OC-Curve Here again it is useful to stress that acceptance sampling relates to the aggregate, not the individual unit. You might also discuss the decision as to whether one should take only a single sample, or whether multiple samples are required.
Determining Pa Let the sampling plan be n = 60, and c = 2 For Pd = 0, Pa = 1.0 For Pd = 1%, compute nPd = (60 x .01) = 0.6 np 0 1 2 .40 .670 .938 .992 .45 .638 .925 .989 .50 .607 .910 .986 .55 .577 .894 .982 .60 .549 .878 .977 .65 .522 .861 .972 Here again it is useful to stress that acceptance sampling relates to the aggregate, not the individual unit. You might also discuss the decision as to whether one should take only a single sample, or whether multiple samples are required. Pa = .977
OC Curve for n=60, c=2 Pd nPd Pa 0% 0.0 1.000 1% 0.6 0.977 2% 1.2 0.879 3% 1.8 0.731 4% 2.4 0.570 6% 3.6 0.303 8% 4.8 0.143 10% 6.0 0.062 12% 7.2 0.025 14% 8.4 0.010 You can use this and the next several slides to begin a discussion of the “quality” of the acceptance sampling plans. You will find additional slides on “consumer’s” and “producer’s” risk to pursue the issue in a more formal manner in subsequent slides.
OC Curve Shows how well a sampling plan discriminates between good and bad lots (shipments) Shows the relationship between the probability of accepting a lot and its quality level You can use this and the next several slides to begin a discussion of the “quality” of the acceptance sampling plans. You will find additional slides on “consumer’s” and “producer’s” risk to pursue the issue in a more formal manner in subsequent slides.
AQL and LTPD AQL = Acceptable Quality Level The defective %, Pd, up to which quality the lot is said to be acceptable to the customer. LTPD =Lot Tolerance Percent Defective The defective %, Pd, above which quality the lot is said to be unacceptable to the customer. You can use this and the next several slides to begin a discussion of the “quality” of the acceptance sampling plans. You will find additional slides on “consumer’s” and “producer’s” risk to pursue the issue in a more formal manner in subsequent slides.
Producer’s and Consumer’s risks The lot state Pd <= AQL (The lot is good) Pd >= LTPD (The lot is bad) Decision based on n,c Sample defectives <= c Accept the lot No error Accept the lot Pa = Consumer’s risk (Type II error) Sample defectives > c Reject the lot 1- Pa = Producer’s risk (Type I error) You can use this and the next several slides to begin a discussion of the “quality” of the acceptance sampling plans. You will find additional slides on “consumer’s” and “producer’s” risk to pursue the issue in a more formal manner in subsequent slides.
Producer’s and Consumer’s risks Producer’s risk The probability of rejecting a good lot (i.e. Pd <= AQL) based on the acceptance sampling plan (Type I error = a) Consumer’s risk The probability of accepting a bad lot (i.e. Pd >= LTPD) based on the acceptance sampling plan (Type II error = b) You can use this and the next several slides to begin a discussion of the “quality” of the acceptance sampling plans. You will find additional slides on “consumer’s” and “producer’s” risk to pursue the issue in a more formal manner in subsequent slides.
An OC Curve Probability of Acceptance Percent defective | | | | | | | | | 0 1 2 3 4 5 6 7 8 100 – 95 – 75 – 50 – 25 – 10 – 0 – = 0.05 producer’s risk for AQL LTPD AQL = 0.10 Consumer’s risk for LTPD Bad lots Indifference zone Good lots
Drawing the OC Curve Application I.1
Drawing the OC Curve Application I.1 Finding (probability of rejecting AQL quality: Cumulative Poisson Probabilities p = .03 Pa = 0.965 np = 5.79 = 1 – .965 = 0.035
Drawing the OC Curve Application I.1 Finding (probability of accepting LTPD quality: Cumulative Poisson Probabilities p = .08 Pa = 0.10 np = 15.44 = Pa = 0.10
Drawing the OC Curve Application I.1
Drawing the OC Curve Application I.1
OC Curves (with c = 1) Producer’s Consumer’s Risk Risk n (p = AQL) (p = LTPD) 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – n = 60 60 0.122 0.126 80 0.191 0.048 n = 80 100 0.264 0.017 120 0.332 0.006 Probability of acceptance n = 100 n = 120 | | | | | | | | | | 1 2 3 4 5 6 7 8 9 10 (AQL) (LTPD) Proportion defective (hundredths)
OC Curves (with n = 60) Producer’s Consumer’s Risk Risk c (p = AQL) (p = LTPD) c = 1 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – c = 2 1 0.122 0.126 c = 3 2 0.023 0.303 c = 4 3 0.003 0.515 4 0.000 0.726 Probability of acceptance | | | | | | | | | | 1 2 3 4 5 6 7 8 9 10 (AQL) (LTPD) Proportion defective (hundredths)
Average Outgoing Quality AOQ = where, Pd = probability of defectives in the lot Pa = probability of accepting the lot N = Lot size n = sample size
Average Outgoing Quality Example I.2 Noise King example with rectified inspection for its single-sampling plan with n = 110, c = 3, N = 1000 Proportion Probability Defective of Acceptance (p) np (Pa) 0.01 1.10 0.974 0.02 2.20 0.819 0.03 3.30 0.581 = (0.603 + 0.558)/2 0.04 4.40 0.359 0.05 5.50 0.202 = (0.213 + 0.191)/2 0.06 6.60 0.105 0.07 7.70 0.052 = (0.055 + 0.048)/2 0.08 8.80 0.024
Average Outgoing Quality Example I.2 Proportion Probability Defective of Acceptance (p) np (Pa) AOQ 0.01 1.10 0.974 0.02 2.20 0.819 0.03 3.30 0.581 0.04 4.40 0.359 0.05 5.50 0.202 0.06 6.60 0.105 0.07 7.70 0.052 0.08 8.80 0.024 For p = 0.01, Pa = 0.974 AOQ = = 0.0087
Average Outgoing Quality Example I.2 Proportion Probability Defective of Acceptance (p) np (Pa) AOQ 0.01 1.10 0.974 0.0087 0.02 2.20 0.819 0.03 3.30 0.581 0.04 4.40 0.359 0.05 5.50 0.202 0.06 6.60 0.105 0.07 7.70 0.052 0.08 8.80 0.024
Average Outgoing Quality Example I.2 Proportion Probability Defective of Acceptance (p) np (Pa) AOQ 0.01 1.10 0.974 0.0087 0.02 2.20 0.819 0.0146 0.03 3.30 0.581 0.0155 0.04 4.40 0.359 0.0128 0.05 5.50 0.202 0.0090 0.06 6.60 0.105 0.0056 0.07 7.70 0.052 0.0032 0.08 8.80 0.024 0.0017
Average Outgoing Quality Example I.2 1.6 – 1.2 – 0.8 – 0.4 – 0 – | | | | | | | | 1 2 3 4 5 6 7 8 Defectives in lot (percent) Average outgoing quality (percent) Proportion Probability Defective of Acceptance (p) np (Pa) AOQ 0.01 1.10 0.974 0.0087 0.02 2.20 0.819 0.0146 0.03 3.30 0.581 0.0155 0.04 4.40 0.359 0.0128 0.05 5.50 0.202 0.0090 0.06 6.60 0.105 0.0056 0.07 7.70 0.052 0.0032 0.08 8.80 0.024 0.0017
Average Outgoing Quality Example I.2 1.6 – 1.2 – 0.8 – 0.4 – 0 – AOQL AOQL = Average Outgoing Quality Limit Average outgoing quality (percent) | | | | | | | | 1 2 3 4 5 6 7 8 Defectives in lot (percent)