5 Quality And Performance PowerPoint Slides by Jeff Heyl For Operations Management, 9e by Krajewski/Ritzman/Malhotra © 2010 Pearson Education 5–1 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 5–2 Total Quality Management Customer satisfaction Figure 5.1 – TQM Wheel 5–3 Total Quality Management  Customer satisfaction  Conformance to specifications  Value  Fitness for use  Support  Psychological impressions Employee involvement  Cultural change  Teams 5–4 Total Quality Management  Continuous improvement  Kaizen A philosophy  Not unique to quality  Problem solving process 5–5 The Deming Wheel Plan Act Do Study Figure 5.2 – Plan-Do-Study-Act Cycle 5–6 Six Sigma Process average OK; too much variation Process variability OK; process off target X X X X XX XX X X X X X X X X X X Reduce spread Process on target with low variability Center process X XX X X X XX Figure 5.3 – Six-Sigma Approach Focuses on Reducing Spread and Centering the Process 5–7 Six Sigma Improvement Model Define Measure Analyze Improve Control Figure 5.4 – Six Sigma Improvement Model 5–8 Acceptance Sampling  Application of statistical techniques  Acceptable quality level (AQL)  Linked through supply chains 5–9 Acceptance Sampling Firm A uses TQM or Six Sigma to achieve internal process performance Buyer Manufactures furnaces fan Motor inspection Yes Accept motors? mo tors Firm A Manufacturers furnace fan motors TARGET: Buyer’s specs Supplier uses TQM or Six Sigma to achieve internal process performance fan bla des No Blade inspection Yes Accept blades? Supplier Manufactures fan blades TARGET: Firm A’s specs No Figure 5.5 – Interface of Acceptance Sampling and Process Performance Approaches in a Supply Chain – 10 The Baldrige Award  The seven categories of the award are Leadership Strategic Planning Customer and Market Focus Measurement, Analysis, and Knowledge Management Workforce Focus Process Management Results – 82 Solved Problem The Watson Electric Company produces incandescent lightbulbs The following data on the number of lumens for 40watt lightbulbs were collected when the process was in control Observation Sample 604 612 588 600 597 601 607 603 581 570 585 592 620 605 595 588 590 614 608 604 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? – 83 Solved Problem 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, x= 604 + 612 + 588 + 600 = 601 R = 612 – 588 = 24 x R 601 24 602 10 582 22 602 32 604 24 2,991 112 x = 598.2 R = 22.4 Sample Total Average – 84 Solved Problem The R-chart control limits are UCLR = D4R = 2.282(22.4) = 51.12 LCLR = D3R = 0(22.4) = The x-chart control limits are UCLx = x + A2R = 598.2 + 0.729(22.4) = 614.53 LCLx = x – A2R = 598.2 – 0.729(22.4) = 581.87 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 – 85 Solved Problem 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 – 86 Solved Problem Sample Number of Defective Records Sample Number of Defective Records 16 17 12 19 18 4 10 19 11 20 11 21 17 12 22 12 23 24 10 13 25 13 11 18 26 10 12 27 14 13 16 28 14 29 11 15 11 30 Total 300 – 87 Solved Problem a Based on these historical data, set up a p-chart using z = b Samples for the next four days showed the following: Sample Number of Defective Records Tues 17 Wed 15 Thurs 22 Fri 21 What is the supervisor’s assessment of the data-entry process likely to be? – 88 Solved Problem 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 p= 300 = 0.04 7,500 The control limits are UCL p = p + z p( 1− p ) 0.04(0.96) = 0.04 + = 0.077 n 250 p ( − p ) = 0.04 − 0.04(0.96) = 0.003 LCL p = p − z 250 n – 89 Solved Problem b Samples for the next four days showed the following: Sample Number of Defective Records Proportion Tues 17 0.068 Wed 15 0.060 Thurs 22 0.088 Fri 21 0.084 Samples for Thursday and Friday are out of control The supervisor should look for the problem and, upon identifying it, take corrective action – 90 Solved Problem 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? – 91 Solved Problem 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 UCLc = c + z c = + 3 = 8.20 LCLc = c – z c = – 3 = –2.196 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 – 92 Solved Problem 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 – 93 Solved Problem SOLUTION a For the process standard deviation of 25 calories, the standard deviation of the sample mean is σx = σ 25 = = 10.2 calories n UCLx = x + zσx = 420 + 3(10.2) = 450.6 calories LCLx = x – zσx = 420 – 3(10.2) = 389.4 calories – 94 Solved Problem b The process capability index is Cpk = Minimum of = Minimum of x – Lower specification Upper specification – x , 3σ 3σ 500 – 420 420 – 300 = 1.60, = 1.07 3(25) 3(25) The process capability ratio is Cp = Upper specification – Lower specification 500 – 300 = = 1.33 6σ 6(25) 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 – 95 – 96 ... distributions – 11 Sampling Distributions The sample mean is the sum of the observations divided by the total number of observations n x= ∑x i =1 i n where xi n x = observation of a quality characteristic... variance of a distribution An estimate of the process standard deviation based on a sample is given by σ= ∑( x − x) i n −1 or σ = x ∑ i− (∑ x ) n −1 i n where σ = standard deviation of a sample –... to be in control when it is actually out of statistical control  These errors can be controlled by the choice of control limits – 25 SPC Methods  Control charts for variables  R-Chart UCLR =