1. Statistical Process Control Operations Management Dr. Ron Tibben-Lembke

2. Designed Size 10 11 12 13 14 15 16 17 18 19 20

3. Natural Variation 14.5 14.6 14.7 14.8 14.9 15.0 15.1 15.2 15.3 15.4 15.5

4. Theoretical Basis of Control Charts 95.5% of all  X fall within ± 2  Properties of normal distribution

5. Theoretical Basis of Control Charts Properties of normal distribution 99.7% of all  X fall within ± 3 

6. Design Tolerances  Design tolerance: Determined by users’ needs UTL -- Upper Tolerance Limit LTL -- Lower Tolerance Limit Eg: specified size +/- 0.005 inches  No connection between tolerance and  completely unrelated to natural variation.

7. Process Capability and 6   A “capable” process has UTL and LTL 3 or more standard deviations away from the mean, or 3σ.  99.7% (or more) of product is acceptable to customers LTLUTL 33 66 LTLUTL

8. Process Capability LTLUTL LTL UTL CapableNot Capable LTLUTL LTLUTL

9. Process Capability  Specs: 1.5 +/- 0.01  Mean: 1.505 Std. Dev. = 0.002  Are we in trouble?

10. Process Capability  Specs: 1.5 +/- 0.01 LTL = 1.5 – 0.01 = 1.49 UTL = 1.5 + 0.01 = 1.51  Mean: 1.505 Std. Dev. = 0.002 LCL = 1.505 - 3*0.002 = 1.499 UCL = 1.505 + 0.006 = 1.511 1.499 1.511.491.511 Process Specs

11. Capability Index  Capability Index (C pk ) will tell the position of the control limits relative to the design specifications.  C pk >= 1.0, process is capable  C pk < 1.0, process is not capable

12. Process Capability, C pk  Tells how well parts produced fit into specs Process Specs 33 33 LTLUTL

13. Process Capability  Tells how well parts produced fit into specs  For our example:  C pk = min[ 0.015/.006, 0.005/0.006]  C pk = min[2.5,0.833] = 0.833 < 1 Process not capable

14. Process Capability: Re-centered  If process were properly centered  Specs: 1.5 +/- 0.01 LTL = 1.5 – 0.01 = 1.49 UTL = 1.5 + 0.01 = 1.51  Mean: 1.5 Std. Dev. = 0.002 LCL = 1.5 - 3*0.002 = 1.494 UCL = 1.5 + 0.006 = 1.506 1.4941.511.491.506 Process Specs

15. If re-centered, it would be Capable 1.4941.511.491.506 Process Specs

16. Packaged Goods  What are the Tolerance Levels?  What we have to do to measure capability?  What are the sources of variability?

17. Production Process Make Candy PackagePut in big bags Make Candy Mix Mix % Candy irregularity Wrong wt.

18. Processes Involved  Candy Manufacturing: Are M&Ms uniform size & weight? Should be easier with plain than peanut Percentage of broken items (probably from printing)  Mixing: Is proper color mix in each bag?  Individual packages: Are same # put in each package? Is same weight put in each package?  Large bags: Are same number of packages put in each bag? Is same weight put in each bag?

19. Your Job  Write down package # Weigh package and candies, all together, in grams and ounces Write down weights on form  Optional: Open package, count total # candies Count # of each color Write down Eat candies  Turn in form and empty complete wrappers for weighing

21. Peanut Color Mix website  Brown 17.7%20%  Yellow 8.2%20%  Red 9.5%20%  Blue15.4%20%  Orange26.4%10%  Green22.7%10%

22. Classwebsite  Brown12.1%30%  Yellow14.7%20%  Red11.4%20%  Blue19.5%10%  Orange21.2%10%  Green21.2%10% Plain Color Mix

23. So who cares?  Dept. of Commerce  National Institutes of Standards & Technology  NIST Handbook 133  Fair Packaging and Labeling Act

24. Acceptable?

26. Package Weight  “Not Labeled for Individual Retail Sale”  If individual is 18g  MAV is 10% = 1.8g  Nothing can be below 18g – 1.8g = 16.2g

27. Goal of Control Charts  collect and present data visually  allow us to see when trend appears  see when “out of control” point occurs

28. Process Control Charts  Graph of sample data plotted over time UCL LCL Process Average ± 3  Time X

29. Process Control Charts  Graph of sample data plotted over time Assignable Cause Variation Natural Variation UCL LCL Time X

30. Attributes vs. Variables Attributes:  Good / bad, works / doesn’t  count % bad (P chart)  count # defects / item (C chart) Variables:  measure length, weight, temperature (x-bar chart)  measure variability in length (R chart)

31. p Chart Example You’re manager of a 500- room hotel. You want to achieve the highest level of service. For 7 days, you collect data on the readiness of 200 rooms. Is the process in control (use z = 3)? © 1995 Corel Corp.

32. p Chart Hotel Data No.No. Not DayRoomsReady Proportion 12001616/200 =.080 2200 7.035 320021.105 420017.085 520025.125 620019.095 720016.080

33. p Chart Control Limits n n k i i k    1 1400 7 200

34. p Chart Control Limits 16 + 7 +...+ 16 p X n i i k i i k      1 1 121 1400 0864.n n k i i k    1 1400 7 200

35. p Chart Control Limits # Defective Items in Sample i Sample i Size UCLpz p n p X n p i i k i i k         (1 - p) 1 1

36. p Chart Control Limits # Defective Items in Sample i Sample i Size UCLpz pp) n p X n p i i k i i k        (1 1 1 z = 2 for 95.5% limits; z = 3 for 99.7% limits # Samples n n k i i k    1

37. p Chart Control Limits # Defective Items in Sample i # Samples Sample i Size z = 2 for 95.5% limits; z = 3 for 99.7% limits LCLpz n n k p X n p i i k i i k i i k       11 1 and pp) n  (1 UCLpz pp) n p   (1

38. p Chart p p  308643. n pp)  (1 200.0864 * (1-.0864) p X n i i k i i k      1 1 121 1400 0864.n n k i i k    1 1400 7 200 16 + 7 +...+ 16

39. p Chart  086405961460... or &.0268 p p  308643. n pp)  (1 200.0864 * (1-.0864) p X n i i k i i k      1 1 121 1400 0864.n n k i i k    1 1400 7 200 16 + 7 +...+ 16

40. p Chart UCL LCL

41. R Chart  Type of variables control chart Interval or ratio scaled numerical data  Shows sample ranges over time Difference between smallest & largest values in inspection sample  Monitors variability in process  Example: Weigh samples of coffee & compute ranges of samples; Plot

42. You’re manager of a 500- room hotel. You want to analyze the time it takes to deliver luggage to the room. For 7 days, you collect data on 5 deliveries per day. Is the process in control? Hotel Example

43. Hotel Data DayDelivery Time 17.304.206.103.455.55 24.608.707.604.437.62 35.982.926.204.205.10 47.205.105.196.804.21 54.004.505.501.894.46 610.108.106.505.066.94 76.775.085.906.909.30

44. R &  X Chart Hotel Data Sample DayDelivery TimeMeanRange 17.304.206.103.455.555.32 7.30 + 4.20 + 6.10 + 3.45 + 5.55 5 Sample Mean =

45. R &  X Chart Hotel Data Sample DayDelivery TimeMeanRange 17.304.206.103.455.555.323.85 7.30 - 3.45Sample Range = LargestSmallest

46. R &  X Chart Hotel Data Sample DayDelivery TimeMeanRange 17.304.206.103.455.555.323.85 24.608.707.604.437.626.594.27 35.982.926.204.205.104.883.28 47.205.105.196.804.215.702.99 54.004.505.501.894.464.073.61 610.108.106.505.066.947.345.04 76.775.085.906.909.306.794.22

47. R Chart Control Limits Sample Range at Time i # Samples From Exhibit 6.13

48. Control Chart Limits

49. R R Chart Control Limits R k i i k      1 385427422 7 3894.... 

50. R Chart Solution From 6.13 (n = 5) R R k UCLDR LCLDR i i k R R        1 4 3 385427422 7 3894 (2.11)(3.894)8232 (0)(3.894)0..... 

51. R Chart Solution UCL

52.  X Chart Control Limits UCLXAR X X k X i i k     2 1 Sample Range at Time i # Samples R R k i i i k    1

53.  X Chart Control Limits Sample Range at Time i # Samples Sample Mean at Time i From 6.13

54. Exhibit 6.13 Limits

55. R &  X Chart Hotel Data Sample DayDelivery TimeMeanRange 17.304.206.103.455.555.323.85 24.608.707.604.437.626.594.27 35.982.926.204.205.104.883.28 47.205.105.196.804.215.702.99 54.004.505.501.894.464.073.61 610.108.106.505.066.947.345.04 76.775.085.906.909.306.794.22

56.  X Chart Control Limits X X k R R k i i k i i k           1 1 532659679 7 5813 385427422 7 3894........  

57.  X Chart Control Limits From 6.13 (n = 5) X X k R R k UCLXAR i i k i i k X            1 1 2 532659679 7 5813 385427422 7 3894 5813058 *38948060............  

58.  X Chart Solution From 6.13 (n = 5) X X k R R k UCLXAR LCLXAR i i k i i k X X             1 1 2 2 532659679 7 5813 385427422 7 3894 5813(058) 5813(058) (3.894) = 3.566............   (3.894) = 8.060

59.  X Chart Solution* 0 2 4 6 8 1234567  X, Minutes Day UCL LCL

60. Thinking Challenge You’re manager of a 500- room hotel. The hotel owner tells you that it takes too long to deliver luggage to the room (even if the process may be in control). What do you do? © 1995 Corel Corp. N

61.  Redesign the luggage delivery process  Use TQM tools Cause & effect diagrams Process flow charts Pareto charts Solution MethodPeople Material Equipment Too Long

62. Dilbert’s View 11/27/06

63. Fortune Story  58 large companies have announced Six Sigma efforts  91% trailed S&P 500 since then, according to Qualpro, (which has its own competing system)  July 11, 2006

64.  Qualpro’s “Six Problems with Six Sigma” Six sigma novices get “low hanging fruit” “Without years of experience under the guidance of an expert, they will not develop the needed competence” Green belts get advice from people who don’t have experience implementing it Loosely organized methodology doesn’t guarantee results (and they do?) Six Sigma uses simple math – not “Multivariable Testing” (MVT) Six Sigma training for all is expensive, time-consuming Pressure to “do something” – low value projects

65. Six Sigma  Narrow focus on improving existing processes  Best and Brightest not focused on developing new products  Fortune July 11, 2006  Can be overly bureaucratic

66. Final Thought  Early 1980’s, IBM Canada, (Markham Ont.)  Ordered from new supplier in Japan.  Acceptable quality level 1.5% defects, a fairly high standard at the time.  The Japanese firm sent the order with a few parts packed separately, & the following letter... © 1995 Corel Corp.

67. Final Thought Dear IBM: We don’t know why you want 1.5% defective parts, but for your convenience we have packed them separately. Sincerely, © 1995 Corel Corp.