1. South Dakota School of Mines & Technology Introduction to Probability & Statistics Industrial Engineering

2. South Dakota School of Mines & Technology Applications in Quality Industrial Engineering

3. South Dakota School of Mines & Technology Acceptance Sampling Industrial Engineering

4. Quality Methods Acceptance Sampling Process Control Design of Experiments

5. Quality Design & Process Variation 6080100120140 60140 60 Lower Spec Limit Upper Spec Limit Acceptance Sampling Statistical Process Control Experimental Design

6. Acceptance Sampling Suppose we buy parts in lot sizes of 5,000 and we wish to inspect to ensure that the incoming quality is sufficient. Obviously, we would not like to do 100% inspection. Idea: Select a portion, n, of the lot at random and determine the number of defects in the sample. If the number of defects, d, surpasses some critical value, c, assume the entire lot is defective and ship it back.

7. Example Let N = lot size = 5,000 n = sample size = 50 c = critical value = 1 d = number of defectives found in sample size n If d 1, accept the entire lot d 2, reject the entire lot  

8. Theory We assume that each part in the lot is either defective or “successful”. The probability of a defective part is constant with at some proportionate rate p. In our example we assume that an acceptable p is given by We call this the Acceptable Quality Level or AQL p c n a  1 50 022%.

9. Theory (cont.) Now, if p is constant from trial to trial in a sample of n, then the probability of accepting a lot given that the lot has acceptable quality level is given by P a = P{d c}  P n d pp a d c dnd            0 1()

10. Theory (cont.) Then, =.7358 P d a d dd           50 0298 0 1 50.(.)               50 0 0298 50 491 0298 050149 ! !!.(.) ! !!. )

11. Risks  P{reject lot | lot is good}  P{accept lot | lot is bad} HoHo HAHA HoHo HAHA Decision Truth no error no error  

12. Risks (cont.)  P{reject lot | lot is good} =1 - P(accept lot | lot is good} = 1 - P a = 1 -.7358 =.2642  P{accept lot | lot is bad} Problem: What constitutes a bad lot?

13. Computing  Let’s suppose that we consider a lot with 4% (p=.04) to be a bad lot. This is Lot Tolerance Percent Defective, LTPD. Then  P{accept lot | lot is bad} =P{d < c | p b }           50 0496 0 1 50 d d dd.(.)

14. Computing  Then =.4005 Suppose we wish to evaluate  for a number of different LTPD values (p b ).               50 0 0496 50 491 0496 050149 ! !!.(.) ! !!. )

15. Operating Characteristic

16. Acceptance Sampling u Uses u when testing is destructive u cost of 100% inspection is high u inspection error rate is high u cost of passing limited number of defective products is low u the process is stable

17. Acceptance Sampling u Limitations u does not estimate lot quality u does not provide for quality control u risks associated with accepting bad lots and rejecting good lots u Today’s Trends u high quality u small lots FRO’s find Acceptance Sampling of little use

19. South Dakota School of Mines & Technology Introductory SPC The “Magnificent 7” Industrial Engineering

20. Quality Design & Process Variation 6080100120140 60140 60 Lower Spec Limit Upper Spec Limit Acceptance Sampling Statistical Process Control Experimental Design

21. Statistical Process Control u i = random noise, unassignable causes of variation x i = assignable causes of variation u1u1 u2u2 u3u3 u4u4 x1x1 x2x2 x3x3 x4x4 Input Output

22. Statistical Process Control Objective: remove all assignable causes of variation smallest variation in output u1u1 u2u2 u3u3 u4u4 x1x1 x2x2 x3x3 x4x4 Input Output

23. Deming Wheel P D C A

24. The Magnificent “7” Problem Solving Step Understand Mess Find Facts Identify Problems Generate Ideas Develop Solutions Implementation Useful Tools Flow Charts Check Sheets Pareto Diagrams Histograms Cause-and-Effect Scatter Diagrams Control Charts

25. Problem Identification/Analysis Flow Chart Check Sheet Nominal Group Technique Brainstorming Histogram Scatter Diagram Process Capability Control Chart Force Field Analysis Pareto Chart Cause & Effect Run Chart Stratification Problem IdentificationProblem Analysis

26. Flow Charting Mix Dry Ingredients Mix Wet Ingredients Mix Wet & Dry Fold 10 Minutes Add Chips Fold 2 Minutes Extrude Dough Cut Dough Bake 10 Minutes OK Poor Qual. Pack and Ship Feed to Hogs

27. Check Sheet; Cookies TypeCheckTotal Burnt Crumbly Too Few Chips Poor Taste Other 11 5 14 6 3

28. Histograms Defects by Category 0 5 10 15 BurntCrumblyFew ChipsPoor TasteOther Type Defect Frequency

29. Pareto Chart

30. Histogram; Class Problem u Now that we’ve decided that too few chips is our quality problem. We need to determine how many chips are in each cookie. Count your chips.

31. Histogram; Class Problem

32. Cause & Effect Too Few Chips Automate Mix Chips in Dry Fold Longer Drop Liquid Chips Place Chips on by hand

33. Cause & Effect Effect PeopleMethodsHandling Design Tools

34. Idea; Cookies As a first step, let’s try Increasing Folding Time

35. Scatter Diagram Chips vs Fold Time 5 10 15 20 25 510152025 Fold Time # Chips

36. New Idea; Cookies Try Mixing Chips in With Dry Ingredients

37. Flow Charting Mix Dry Ingredients Mix Wet Ingredients Mix Wet & Dry Fold 10 Minutes Add Chips Fold 2 Minutes Extrude Dough Cut Dough Bake 10 Minutes OK Poor Qual. Pack and Ship Feed to Hogs

38. Evaluate; Control Chart Avg Chips Time 18 15 12

40. South Dakota School of Mines & Technology Control Charts Industrial Engineering

41. Control Charts  We assume that the underlying distribution is normal with some mean  and some constant but unknown standard deviation . Let x x n i i n    1

42. Distribution of x Recall that x is a function of random variables, so it also is a random variable with its own distribution. By the central limit theorem, we know that where, xN x  (,)  x n x  

43. Control Charts

44.   x UCL LCL UCL & LCL Set at Problem: How do we estimate  &  ?  3  x

45. Control Charts x x    m m i i 1    )( 1 f m R R m i

46. x x RALCL 2  x x RAUCL 2  RDLCL R 3  RDUCL R 4  x x    m m i i 1    )( 1 f m R R m i

47. Example u Suppose specialized o-rings are to be manufactured at.5 inches. Too big and they won’t provide the necessary seal. Too little and they won’t fit on the shaft. Twenty samples of 2 rings each are taken. Results follow.

48. Control Charts

49. Xbar charts can identify special causes of variation, but they are only useful if the process is stable (common cause variation).

50. Control Limits for Range UCL = D 4 R = 3.268*.002 =.0065 LCL = D 3 R = 0

51. pp The P-Chart p pUCL  3  p pLCL  3  where p d nm i i m    1 p  n   ()1

52. Example; P-Chart u Operators of a sorting machine must read the zip code on a letter and diver the letter to the proper carrier route. Over one month’s time, 25 samples of 100 letters were chosen, and the number of errors was recorded. Error counts for each of the 25 days follows.

53. Example

54. Example 25 03....00.01.03.   p

55. Example 25 03....00.01.03.   p p pp 070.0 100 )024.1(. 3. )1( 3      n UCL p

56. Example 25 03....00.01.03.   p p pp 070.0 100 )024.1(. 3. )1( 3      n UCL p p  pp 0.0022. )1(   n LCL p 3

57. Example; P-Chart

59. South Dakota School of Mines & Technology Process Capability Industrial Engineering

60. Process Capability

63. x LSLUSL x C LSL p x   6 

64. Process Capability x LSLUSL x C LSL pk xx         min,     33

65. Cookie Example u Suppose we have the cookie example and it is desired to have 15 + 3 chips.

66. Cookie Example 233.0 ).0,433.0min( )3(3 121.14, )3(3 1. 18 min             Cp

68. South Dakota School of Mines & Technology Funnel Experiment Industrial Engineering

69. Funnel Experiment u The Funnel Experiment is a famous experiment first developed by Dr. Deming. It was designed to show primarily two things: 1.All processes have statistical variation 2.Improving processes to reduce variation is infinitely better than trying to control process variation

70. Funnel Experiment

71. Rule 1: Do nothing Rule 2: Move the funnel in an equal but opposite direction Rule 3: Move the funnel in an equal but opposite direction from (0,0) Rule 4: Move the funnel to the last position where the last hit.

72. The Funnel Experiment Rule 1: (LTDTA) Set the funnel over the target at (0,0) and leave the funnel fixed through all 50 drops. Funnel Experiment; Rule 1 -4 -3 -2 0 1 2 3 4 -4-3-201234 Xk Yk

73. The Funnel Experiment Rule 2: Move the funnel in an equal but opposite direction from where the last marbel hit. The funnel should be moved the distance (-x k,-y k ) from its last resting point. Funnel Experiment; Rule 2 -4 -3 -2 0 1 2 3 4 -4-3-201234 Xk Yk

74. The Funnel Experiment Rule 2: Move the funnel in an equal but opposite direction from where the last marbel hit. The funnel should be moved the distance (-x k,-y k ) from its last resting point. Machine/Process adjustment rules using control chart limits. Automatic compensating machinery

75. The Funnel Experiment Rule 3: Move the funnel in an equal but opposite direction from target (0,0) to where the last marble hit. Rule 3 states that the funnel should be moved a distance (-x k,-y k ) from the target (0,0). Funnel Experiment; Rule 3 -4 -3 -2 0 1 2 3 4 -4-3-201234 Xk Yk

76. The Funnel Experiment Rule 3: Move the funnel in an equal but opposite direction from target (0,0) to where the last marble hit. Rule 3 states that the funnel should be moved a distance (-x k,-y k ) from the target (0,0). Management by Objectives Committees

77. The Funnel Experiment Rule 4: Move the funnel to the last position where the last marble hit. Rule 4 states that the funnel should be moved to the resting point, (x k,y k ). Funnel Experiment; Rule 4 -4 -3 -2 0 1 2 3 4 -4-3-201234 Xk Yk

78. The Funnel Experiment Rule 4: Move the funnel to the last position where the last marble hit. Rule 4 states that the funnel should be moved to the resting point, (x k,y k ). Operator matches color from batch to batch without reference to original swatch. Operator attempts to make every piece like the last one. On the job training.

79. Rule 2: Deadband Controlling oo X1X1  o  o

80. oo X1X1 X2X2  o  o

81. oo X1X1 X3X3 X2X2  o  o

82. oo X1X1 X3X3 X2X2 XnXn  o  o

83. oo X1X1 X3X3 X2X2 XnXn  o  o  )( no X 

84. oo X1X1 X3X3 X2X2 XnXn  o  o  )( no X  1. Deviation |X n -  o | is random; do nothing 2. Deviation |X n -  o | is shift; adjust by (  o - X n )

85. Infinitely Large Deadband  X1X1 X3X3 X2X2 XnXn 1. X n |X n-1 ~ N( ,  ) 2.  n =  n-1 = 

86. Infinitely Large Deadband  X1X1 X3X3 X2X2 XnXn X n ~ N( ,   )

87. Infinitesimally Small Deadband  X1X1 X3X3 X2X2 XnXn  )( n X  X nnn   1

88.  X1X1 X3X3 X2X2 XnXn  )( n X  ZnZn nn   )( 11 X nnn   1

89.  X1X1 X3X3 X2X2 XnXn  )( n X  ZnZn  ZnZn nn   )( 11 X nnn   1

90.  X1X1 X3X3 X2X2 XnXn  )( n X  nnn nn ZZX Z      1

91. nnn ZZX   1 1 ][][][][   nnn ZEZEEXE

92. nnn ZZX   1  1 ][][][][   nnn ZEZEEXE 0

93. Infinitesimally Small Deadband nnn ZZX   1  222 2][][)(  nnn XEXEXVar 1 ][][][][   nnn ZEZEEXE

94. Infinitesimally Small Deadband  X1X1 X3X3 X2X2 XnXn  )( n X  X n ~ N( , 2  2 )

95. Process Variation

97. South Dakota School of Mines & Technology Loss Functions Industrial Engineering

98. Cost of Quality Cost of Failure Cost of Control Total Cost Traditional View Quality Level Costs

99. Cost of Quality Cost of Failure Cost of Control Total Cost Contemporary View Quality Level Costs

100. Traditional Loss Function x LSLUSL T T LSL USL

101. Taguchi Loss Function x T

102. x T x T

103. 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.

104. Example x T U.S. Plant (  2 = 8.33) Japanese Plant (  2 = 2.78) LSLUSL

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

106. 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 milage. 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.

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

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

109. Expected Loss

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

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

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

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

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

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

120. Example (Piston Diam.)

121. Aside If we replace T by , L(x) looks similar to the variance function. Indeed, we can show that 2 )()(TxkxL  x))(()( 22 TkxL 

122. Example (Sony) x T U.S. Plant (  2 = 8.33) Japanese Plant (  2 = 2.78) LSLUSL E[L US (x)]= 0.16 * 8.33 = $1.33 E[L J (x)] = 0.16 * 2.78 = $0.44

123. Tolerance (Pistons) 10 L(x) 10.05 400 400 = k(10.05 - 10.00) 2 = k(.0025) k= 160,000 Recall,

124. Tolerance Suppose repair for an engine which will fail during warranty can be made for only $200 10 L(x) 10.05 400 LSLUSL 200

125. Tolerance Suppose repair for an engine which will fail during warranty can be made for only $200 200 = 160,000(tolerance) 2 10 L(x) 10.05 400 LSLUSL 200

126. Tolerance Suppose repair for an engine which will fail during warranty can be made for only $200 200 = 160,000(tolerance) 2 tolerance= (200/160,000) 1/2 =.0354 10 L(x) 10.05 400 LSLUSL 200