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2.830J / 6.780J / ESD.63J Control of Manufacturing Processes (SMA 6303)Spring 2008

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1Manufacturing

Control of Manufacturing Processes

Subject 2.830/6.780/ESD.63Spring 2008Lecture #4

Probability Models of Manufacturing Processes

February 14, 2008

2Manufacturing

Note: Reading Assignment

• May & Spanos– Read Chapter 4

• Montgomery– Skim/consult Chapters 2 & 3 if need additional

explanations or examples beyond May & Spanos

3Manufacturing

Turning Process

4Manufacturing

CNC Turning

0.6970

0.6980

0.6990

0.7000

0.7010

0.7020

Run Number

Shift changes

• Randomness + Deterministic Changes

random or unknown

Δα

Observations from Experiments

5Manufacturing

CNC Data

0.6235

0.6237

0.6239

0.6241

0.6243

0.6245

0.6247

0.6249

0.6251

0.6253

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

OuterMiddleInner

Operating Points:High Feed=1Low Feed = 2

6Manufacturing

Brake Bending of Sheet

M M

ε

σ

Output:

Angle

7Manufacturing

Bending Process

Springback

8Manufacturing

Angle changes with depthΔY ⇒ Δu

• Clear Input-Output Effects (Deterministic)• Also Randomness as well

Observations from Bending Process

20

30

40

50

60

70

80

90

Aluminum 0.3 In

Steel 0.3 In

Aluminum 0.6 In

Steel 0.6 In

9Manufacturing

Observations from Injection Molding

Run Chart for Injection Molded Part

40.60

40.65

40.70

40.75

40.80

40.85

40.90

40.95

41.00

0 10 20 30 40 50 60

Number of Run

Width (mm)Average

Holding Time = 5 secInjection Press = 40%

Holding Time = 10 secInjection Press = 40%

Holding Time = 5 secInjection Press = 60%

Holding Time = 10 secInjection Press = 60%

2/3/05 50

QuickTime™ and aGraphics decompressor

are needed to see this picture.

10Manufacturing

Observations from Data

• Clearly some measurement “noise”?

20

30

40

50

60

70

80

90

shift 1 shift 2

shift 3

11Manufacturing

Observations from Data

• Systematic/traceable “operator error”Sheet Shearing

0.85

0.87

0.89

0.91

0.93

0.95

0.97

0.99

1.01

shift change

shift changealuminum

steel

steel

12Manufacturing

Inputs

A Random Process + A Deterministic Process

How Model to Distinguish these Effects?

Process

Disturbances (Reducible)

Irreducible Disturbances

Outputs +

"Noise"

13Manufacturing

• Consider the Output-only, “Black Box” view of the Run Chart

• How do we characterize the process?– Using Y(t) only

• WHY do we characterize the process– Using Y(t) only?

Process Y(t)

Random Processes

14Manufacturing

The Why

• Did output really change?• Did the input cause the change?• If not, why did the output vary?• How confident are we of these answers?

• Can we model the randomness?

Process

15Manufacturing

Background Needed• Theory of Random Processes and Random

Variables• Use of Sample Statistics Based on

Measurements– SPC basis– DOE: use of experimental I/O data– Feedback control with random disturbances

16Manufacturing

How to Describe Randomness?

• Look at a Frequency Histogram of the Data• Estimates likelihood of certain ranges

occurring:

– Pr(y1 < Y <y2)

– “Probability that a random variable Y falls between the limits y1 and y2”

17Manufacturing

Example: Thermoforming Histogram (2000 data)

0

2

4

6

8

10

12

1.48 1.482 1.484 1.486 1.488 1.49 1.492 1.494 1.496 1.498 1.5 1.502

Thermoforming Run Chart

1.475

1.48

1.485

1.49

1.495

1.5

1.505

Run Number

Part Diameter

Shift Changes

18Manufacturing

How to Describe Continuous Randomness

• Process outputs Y are continuous variables• The Probability of Y(t) taking on any specific value for a

continuumProb(Y(t) = y*) = 0

• Must use instead a Cumulative Probability FunctionPr(Y(t)<y*)

– Look at Cumulative Frequency

19Manufacturing

Cumulative Frequency

0

10

20

30

40

50

60

1.48 1.482 1.484 1.486 1.488 1.49 1.492 1.494 1.496 1.498 1.5

Pr(Y<1.49)=14/50

= 0.28

0

2

4

6

8

10

12

1.48 1.482 1.484 1.486 1.488 1.49 1.492 1.494 1.496 1.498 1.5 1.502 A Continuous equivalent?

20Manufacturing

• Probability Function: (P(x))

• Probability Density Function pdf(x) = dP/dxx

P

x

p(x)

Continuous Equivalents

0

10

20

30

40

50

60

1.48 1.482 1.484 1.486 1.488 1.49 1.492 1.494 1.496 1.498 1.5

0

2

4

6

8

10

12

1.48 1.482 1.484 1.486 1.488 1.49 1.492 1.494 1.496 1.498 1.5 1.502

21Manufacturing

Process Outputs as a Random Variable

• The Histogram suggests a pdf– Parent or underlying behavior “sampled” by the

process• Standard Forms (There are Many)

– e.g. The Uniform and Normal pdf’s

0.1

0.2

0.3

0.4

-4 -3 -2 -1 0 1 2 3 4

Normal Probability Density Function for σ2 = 1, μ=0

x

p(x)

22Manufacturing

Analysis of Histograms

• Is there a consistent pattern?• Is an underlying “parent” distribution suggested?

23Manufacturing

Histogram for CNC Turning

0

5

10

15

20

25

0.697 0.698 0.699 0.7 0.701 0.702

Bin Dimension (in)

CNC Turning

0.6970

0.6980

0.6990

0.7000

0.7010

0.70201 3 5 7 9

11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47

Run Number

Shift changes

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0

2

4

6

8

10

12

14

134 134.5 135 135.5 136 136.5 137 137.5 138 138.5 139 139.5 140 140.5 141 141.5 142

0.3 in depth only

Histogram for Bending(MIT 2002 data)

Bending Data Sorted by Depth

80

90

100

110

120

130

140

150

1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93 97 101

105

109

113

117

121

125

129

133

Ang

le (D

eg.)

Steel0.6 in depth

Aluminum0.6 in depth

Steel0.3 in depth

Aluminum0.3in depth

Dashed Lines are at group changes

25Manufacturing

0

2

4

6

8

10

12

14

16

18

98 98.5 99 99.5 100 100.5 101 101.5 102 102.5 103 103.5 104 104.5

0.6 in depth only

Histogram for Bending(MIT 2002 data)

Bending Data Sorted by Depth

80

90

100

110

120

130

140

150

1 5 9

13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93 97

101

105

109

113

117

121

125

129

133

Ang

le (D

eg.)

Steel0.6 in depth

Aluminum0.6 in depth

Steel0.3 in depth

Aluminum0.3in depth

Dashed Lines are at group changes

Mean Shift?

26Manufacturing

Consider: No Intentional Changes (Δu = 0)

• Shearing during shift 1(‘02) , aluminum only

0

10

20

30

40

50

60

0.91 0.915 0.92 0.925 0.93 0.935 0.94 0.945 0.95

27Manufacturing

Consider: No Effective Changes (∂Y/∂u= 0)• Injection Molding Entire MIT Run (2002)

0

5

10

15

20

25

1.875 1.8775 1.88 1.8825 1.885 1.8875 1.89

1.865

1.87

1.875

1.88

1.885

1.89

1.895

Shift Change

28Manufacturing

Injection Molding (S’2003)

0

5

10

15

20

25

30

35

40

65.4 65.5 65.5 65.6 65.6 65.7 65.7 65.8 65.8 65.9 65.9 66 66

Fre

qu

en

cy

65.2

65.4

65.6

65.8

66

66.2

66.4

1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106 113 120 127

29Manufacturing

Conclusion?

• When there are no input effect (no Δu or ∂Y/∂u) a consistent histogram pattern can emerge

• How do we use knowledge of this pattern?– Predict behavior– Set limits on “normal” behavior

• Define analytical probability density functions

30Manufacturing

Underlying or “Parent” Probability• A model of the “true”, continuous behavior of the

random process• Can be thought of as a continuous random variable

obeying a set of rules (the probability function)• We can only glimpse into these rules by sampling the

random variable (i.e. the process output)• Underlying process can have

– Continuous values (e.g. geometry)– Discrete values (e.g. defect occurrence)

31Manufacturing

Continuous Probability Functions

• Recall Probability Function (Cumulative)P(x) = Prob(Y(t)<x)

• Define pdf = p(x) = dP/dx

Thus :P(x) = pdf (x)dx

−∞

x

∫

x

p(x)x

P

32Manufacturing

Use of the pdf : Expectation

E{x(t)} = expected value of x(t)

E{x(t)} = x(t) pdf (x,t )dx-∞

∞

∫μ(t) = E{x(t)} mean value of x

Note that pdf and μ (or any other expected value)can be functions of time.

In general, they may be non-stationary.

33Manufacturing

pdf(x,t) = pdf(x) = p(x) : stationary pdf

• For a stationary process μx is a constant

Stationary Processes

E{x} = x p(x)dx−∞

∞

∫ = μx

μx : theoretical or " true" mean

34Manufacturing

• For a stationary process σx is a constant

Stationary Processes

σ x2 = E{x 2} − μx

2

E{(x − μx )2} = (x − μx )2 p(x)dx−∞

∞

∫ = σ x2

= "true" variance

= mean square - square of mean

35Manufacturing

The Uniform Distribution

r x

p(x)

1/r

p(x) =1r

⇒ x1 < x < x2

p(x) = 0 ⇒ x < x1 x > x2

x1 x2

36Manufacturing

The Normal Distribution

p(x) =1

σ 2πe

− 12

x − μσ

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

z =x −

z

μσ

“Standard normal”

μz=0σz=1

0

0.1

0.2

0.3

0.4

-4 -3 -2 -1 0 1 2 3 4

37Manufacturing

-4 -3 -2 -1 0 1 2 3 4

P(z)

Cumulative Distribution

0

0.2

0.4

0.6

0.8

1

-4 -3 -2 -1 0 1 2 3 4

z

38Manufacturing

Properties of the Normal pdf

• Symmetric about mean• Only two parameters:

μ and σ

• Superposition Applies:– sum of normal random variables has a normal

distribution

p(x) =1

σ 2πe

− 12

x − μσ

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

39Manufacturing

Superposition of Random Variables

If we define a variabley = c1x1 + c2x2 + c3x3 + c4x4 + ...

• ci are constants • xi are independent random variables

μy = c1μ1 + c2μ2 + c3μ3 + c4μ4 + ...

σy2 = c1

2σι2 + c2

2σ22 + c3

2σ32 + c4

2σ42

From expectation operation, for any pdf.

40Manufacturing

Use of the PDF: Confidence Intervals

• How likely are certain values of the random variable?

• For a “Standard Normal” Distribution:

z = (x − μ)σ

N(0,1)μ = 0

σ =1

z = 1 ⇒ x = 1σz = 2 ⇒ x = 2σz = 3 ⇒ x = 3σ

0

0.1

0.2

0.3

0.4

-4 -3 -2 -1 0 1 2 3 4

41Manufacturing

Confidence Intervals

P(-1≤ z ≤ 1) = P(z ≤1) - P(z ≤-1) = 0.841 - (1-0.841) = 0.682(+ 1σ)

P(-2≤ z ≤ 2) = P(z ≤2) - P(z ≤-2) = 0.977 - (1-0.977) = 0.954(+ 2σ)

P(-3≤ z ≤ 3) = P(z ≤3) - P(z ≤-3) = 0.998 - (1-0.998) = 0.997(+ 3σ)

P(z) tabulated (e.g. p. 752 of Montgomery)

42Manufacturing

0.1

0.2

0.3

0.4

-4 -3 -2 -1 0 1 2 3 4

±σ

±2 σ

±3 σ

z

Confidence Intervals

68%95%

99.7%

• Probability that |x| >μ+3σ = 3/1000

43Manufacturing

Is the Process “Normal” ?

• Is the underlying distribution really normal?– Look at histogram– Look at curve fit to histogram– Look at % of data in 1, 2 and 3 σ bands

• Confidence Intervals– Probability (or qq) plots (see Mont. 3-3.7)– Look at “kurtosis”

• Measure of deviation from normal

44Manufacturing

Kurtosis: Deviation from Normal

k = E(x − μ x )4

σ 4

k =n(n + 1)

(n − 1)(n − 2)(n − 3)xi − x

s⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∑

4⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

−3(n − 1)2

(n − 2)(n − 3)

Or for sampled data:

k=1 - normalk>1 more “peaked”k<1 more “flat”

45Manufacturing

The Central Limit Theorem

– If x1, x2 ,x3 ...xN … are N independent observations of a random variable with “moments” μx and σ2

x,

– The distribution of the sum of all the samples will tend toward normal.

46Manufacturing

Example: Uniformly Distributed Data

Sum of 100 sets of 1000 points each

35 40 45 50 55 60 650

50

100

150

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

x1

x2

x100

+

+ . . .

y = xii =1

100

∑

47Manufacturing

Sampling: Using Measurements (Data) to Model the Random Process

• In general p(x) is unknown• Data can suggest form of p(x)

– e.g.. uniform, normal, weibull, etc.• Data can be used to estimate parameters of distributions

– e.g. μ and σ for normal distribution - p(x) = p(x, μ ,σ)

• How to Estimate– Sample Statistics

• Uncertainty in Estimates– Sample Statistic pdf’s

p(x) =1

σ 2πe

− 12

x − μσ

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

48Manufacturing

Sample Statistics

x =1n

x( j)j =1

n

∑ : Average or Sample Mean

S2 =1

n − 1

x( j) = samples of x(t ) taken n times

(x( j)j =1

n

∑ − x )2 : Sample Variance

S =1

n − 1(x( j)

j =1

n

∑ − x )2 : Sample Std.Dev.

49Manufacturing

Sample Mean Uncertainty

• If all xi come from a distribution with μx and σ2x, and

we divide the sum by n:

x =1n

xii =1

n

∑

μx = μx σ x2 =

1n

σ x2 or σ x =

1n

σ x

x = c1x1 + c2 x2 + c3x3 + K cn xn

ci =1n

Then: and

50Manufacturing

Conclusions

• All Physical Processes Have a Degree of Natural Randomness

• We can Model this Behavior using Probability Distribution Functions

• We can Calibrate and Evaluate the Quality of this Model from Measurement Data