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ENGM 620: Quality Management

ENGM 620: Quality Management. 26 November 2012 Six Sigma. Problem Solving Quiz. I am a good problem solver because: My organization has no problems, so I must be good at solving them. I solve the same problems every day. I find the root cause and solve a problem once.

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ENGM 620: Quality Management

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  1. ENGM 620: Quality Management 26 November 2012 • Six Sigma

  2. Problem Solving Quiz • I am a good problem solver because: • My organization has no problems, so I must be good at solving them. • I solve the same problems every day. • I find the root cause and solve a problem once.

  3. Problem Solving Quiz • The people who work for me must be good problem solvers because: • I hear about no problems, so they must solve the problems. • They tell me they have no time for other things because they spend all their time solving problems. • Every member of my organization is trained in root-cause problem solving techniques.

  4. Problem Solving Fixing the symptoms, not the root cause!

  5. Six Sigma • The purpose of Six Sigma is to reduce variation to achieve very small standard deviations so that almost all of your products or services meet or exceed customer requirements.

  6. Reducing Variation Lower Spec Limit Upper Spec Limit 60 80 100 120 140 60 140 140 60

  7. Accuracy vs. Precision • Accuracy - closeness of agreement between an observed value and a standard • Precision - closeness of agreement between randomly selected individual measurements

  8. Six Ingredients of Six Sigma • Genuine focus on the customer • Data- and fact- driven management • Process focus, management, and improvement • Proactive management • Boundaryless collaboration • Drive for perfection, tolerate failure

  9. Key People in Six Sigma • Champion • Work with black belts to identify possible projects • Master Black Belts • Work with and train new black belts • Black Belts • Committed full time to completing cost-reduction projects • Green Belts • Trained in basic quality tools

  10. Six Sigma Problem Solving Process • Define the opportunity • Measure process performance • Analyze data and investigate causes • Improve the process • Control and process management

  11. Define • Four Phases (according to your text) • Develop the business case • Project evaluation • Pareto analysis • Project definition • Project Charter

  12. Some of the tools to “Define” • Project Desirability Matrix • Problem/objective statement • Primary/secondary metric • Change Management • Process Map • SIPOC, Flow chart, Value Stream, etc. • QFD Houses

  13. Project Assessment Stars ??? Return Low Hanging Fruit Dogs Risk

  14. Measure • Two major steps: • Select process outcomes • Verifying measurements

  15. Some of the tools to “Measure” • Magnificent 7 • Basic Statistics • FMEA • Time Series analysis • Process capability

  16. Analyze • Three major steps: • Define your performance objectives • Identify independent variables • Analyze sources of variability • Results of this step are potential improvements

  17. Some of the tools to “Analyze” • Graphic data analysis • Confidence intervals • Hypothesis tests • Regression/correlation • Process modeling / simulation

  18. Improve • Try your potential solutions • Off-line experiments • Pilot lines • Assure true improvement

  19. Some of the tools to “Improve” • Hypothesis tests • Multi-variable regression • Taguchi methods • Design of experiments

  20. Exercise: Anti-Solution • Objective: How do we best speed purchase order preparation? • Anti-Objective: How do we slow purchase order preparation down to a crawl? • Brainstorm the anti-objective • Examine each anti-objective for a positive idea • Record and add to the positive ideas

  21. Control • Sustain the improvements • Manage the process

  22. Some of the tools to “Control” • Implementation • Mistake proofing • Visible enterprise • Control Plan • Documentation • Training • Control Charts • Process Management Chart

  23. Taguchi Methods • The reduction of variability in processes and products Equivalent definition: • The reduction of waste • Waste is any activity for which the customer will not pay

  24. Traditional Loss Function LSL USL x T LSL T USL

  25. Example (Sony, 1979) Comparing cost of two Sony television plants in Japan and San Diego. All units in San Diego fell within specifications. Japanese plant had units outside of specifications. Loss per unit (Japan) = $0.44 Loss per unit (San Diego) = $1.33 How can this be? Sullivan, “Reducing Variability: A New Approach to Quality,” Quality Progress, 17, no.7, 15-21, 1984.

  26. Example LSL USL x T U.S. Plant (2 = 8.33) Japanese Plant (2 = 2.78)

  27. Taguchi Loss Function x T x T

  28. Taguchi Loss Function L(x) k(x - T)2 x T L(x) = k(x - T)2

  29. Estimating Loss Function Suppose we desire to make pistons with diameter D = 10 cm. Too big and they create too much friction. Too little and the engine will have lower gas mileage. Suppose tolerances are set at D = 10 + .05 cm. Studies show that if D > 10.05, the engine will likely fail during the warranty period. Average cost of a warranty repair is $400.

  30. Estimating Loss Function L(x) 400 10.05 10 400 = k(10.05 - 10.00)2 = k(.0025)

  31. Estimating Loss Function L(x) 400 10.05 10 400 = k(10.05 - 10.00)2 = k(.0025) k = 160,000

  32. Example 2 Suppose we have a 1 year warranty to a watch. Suppose also that the life of the watch is exponentially distributed with a mean of 1.5 years. The warranty costs to replace the watch if it fails within one year is $25. Estimate the loss function.

  33. Example 2 L(x) f(x) 25 1 1.5 25 = k(1 - 1.5)2 k = 100

  34. Example 2 L(x) f(x) 25 1 1.5 25 = k(1 - 1.5)2 k = 100

  35. Single Sided Loss Functions Smaller is better L(x) = kx2 Larger is better L(x) = k(1/x2)

  36. Example 2 L(x) f(x) 25 1

  37. Example 2 L(x) f(x) 25 1 25 = k(1)2 k = 25

  38. Expected Loss

  39. Expected Loss

  40. Expected Loss

  41. Expected Loss

  42. Expected Loss

  43. Expected Loss Recall, X f(x) with finite mean  and variance 2. E[L(x)] = E[ k(x - T)2 ] = k E[ x2 - 2xT + T2 ] = k E[ x2 - 2xT + T2 - 2x + 2 + 2x - 2 ] = k E[ (x2 - 2x+ 2) - 2 + 2x - 2xT + T2 ] = k{ E[ (x - )2 ] + E[ - 2 + 2x - 2xT + T2 ] }

  44. Expected Loss E[L(x)] = k{ E[ (x - )2 ] + E[ - 2 + 2x - 2xT + T2 ] } Recall, Expectation is a linear operator and E[ (x - )2 ] = 2 E[L(x)] = k{2 - E[ 2 ] + E[ 2x - E[ 2xT ] + E[ T2 ] }

  45. Expected Loss Recall, E[ax +b] = aE[x] + b = a + b E[L(x)] = k{2 - 2 + 2 E[ x - 2T E[ x ] + T2 } =k {2 - 2 + 22 - 2T + T2 }

  46. Expected Loss Recall, E[ax +b] = aE[x] + b = a + b E[L(x)] = k{2 - 2 + 2 E[ x - 2T E[ x ] + T2 } =k {2 - 2 + 22 - 2T + T2 } =k {2 + ( - T)2}

  47. Expected Loss Recall, E[ax +b] = aE[x] + b = a + b E[L(x)] = k{2 - 2 + 2 E[ x - 2T E[ x ] + T2 } =k {2 - 2 + 22 - 2T + T2 } =k {2 + ( - T)2} = k {2 + ( x - T)2} = k (2 +D2 )

  48. Example Since for our piston example, x = T, D2 = (x - T)2 = 0 L(x) = k2

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