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

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1 MIT OpenCourseWare J / 6.780J / ESD.63J Control of Processes (SMA 6303) Spring 2008 For information about citing these materials or our Terms of Use, visit:

2 Control of Processes Subject 2.830/6.780/ESD.63 Spring 2008 Lecture #13 Modeling Testing and Fractional Factorial Designs April 1, J/6.780J/ESD.63J 1

3 Full Factorial Models Contrasts Extension to 2 k Outline Model Term Significance: ANOVA Checking Adequacy of Model Form Tests for higher order fits (curvature) Experimental Design Blocks and Confounding Single Replicate Designs Fractional Factorial Designs NB: Read Montgomery Chapter J/6.780J/ESD.63J 2

4 2 2 Model Based on Contrasts Two factor, two level experiments: y ŷ = β 0 + β 1 x 1 + β 2 + β 12 x 1 (Regression model) This defines a 3-D ruled surface x J/6.780J/ESD.63J 3

5 General Form for Contrasts Trial A B AB (1) a b ab A :[a + ab b (1)] B :[b + ab a (1)] AB :[ab + (1) a b] Contrast A = Trial Column A Contrast B = Trial Column B Contrast AB = Trial Column AB 2.830J/6.780J/ESD.63J 4

6 Extension to 2 k Consider 2 3 (3 factors, 2 levels each factor): Factor Levels Run Treatment x 1 x 3 Number Combination A B C 1 (1) y a y b y ab y c y ac y bc y abc y J/6.780J/ESD.63J 5

7 Generalization + B A + - C number of levels 2 k number of factors Courtesy of Dan Frey. Used with permission J/6.780J/ESD.63J 6

8 Contrasts as Surface Average Differences (b) + (bc) (ab) (abc) B (c) - (1) A (a) (ac) C + A 1 1 = bc 4 4 [( abc) + ( ab) + ( ac) + ( a) ] [( b) + ( c) + ( ) + (1) ] Courtesy of Dan Frey. Used with permission J/6.780J/ESD.63J 7

9 Contrasts for Main Effect (b) + (bc) (ab) (abc) B (c) - (1) A (a) (ac) C + B 1 1 = ac 4 4 [( abc) + ( ab) + ( bc) + ( b) ] [( a) + ( c) + ( ) + (1) ] Courtesy of Dan Frey. Used with permission J/6.780J/ESD.63J 8

10 Contrasts for Interaction Effect (b) + (bc) (ab) (abc) B (c) - (1) A (a) (ac) C AB = [(1) + ( ab) + ( c) + ( abc) ] [( a) + ( b) + ( ac) ( bc) ] Courtesy of Dan Frey. Used with permission J/6.780J/ESD.63J 9

11 Contrasts for 2 3 Contrast A :[a + ab + ac + abc b c bc (1)] Contrast ABC :[a + b + c + abc ab ac bc (1)] Effect = Contrast n2 k 1 Factorial Combination Treament Combination I A B AB C AC BC ABC (1) a b ab c ac bc abc A = 1 4n where n is the number of replicates at each treatment combination [a + ab + ac + abc b c bc (1)] 2.830J/6.780J/ESD.63J 10

12 Factorial Combinations Factorial Combination Treament Combination I A B AB C AC BC ABC (1) a b ab c ac bc abc Note: this is the scaled X matrix in the regression model 2.830J/6.780J/ESD.63J 11

13 Relationship to Regression Model β = (X T X) 1 X T y y ˆ = β + β x + β x + β x x A is the Effect of input 1 averaged over all other input changes (-1 to +1 or a total range of 2) B is the Effect of input 2 averaged over all other input changes,. or y is data from experimental design X β 0 = y β 1 = A 2 ; β 2 = B 2 ; β 12 = AB 2 regression model 2.830J/6.780J/ESD.63J 12

14 ANOVA for 2 k Now have more than one effect We can derive: SS Effect = (Contrast) 2 /n2 k And it can be shown that: SS Total = SS A + SS B + SS AB + SS Error 2.830J/6.780J/ESD.63J 13

15 ANOVA Table Source SS d.o.f. MS F 0 F crit A Contrast A 2 MS A MS E 2 2 n 1 SS A F 1,2n-4,α B Contrast B n 1 SS B MS B MS E AB Contrast AB n 1 SS C MS AB MS E Error SS E (2 2 n) 3 SS E (2 2 n) 3 Total ΣΣ( y ij -y) 2 (2 2 n) J/6.780J/ESD.63J 14

16 Alternative Form Source SS d.o.f. MS F mean nm β 0 2 x 1 nm β 1 2 nm β 2 2 x 12 nm β 12 2 ε total m n ε ij mn 4 i =1 j =1 m n y ij mn i =1 j =1 SS (β 0 ) 1 SS( β 1 ) 1 SS (β 2 ) 1 SS( β 12 ) 1 SS (ε) (mn 4) MS (β 0 ) MS (ε) MS ( β 1 ) MS (ε) MS (β 2 ) MS (ε) MS ( β 12 ) MS (ε) n = replicates m = 2 k SS Total includes the grand mean in this formulation For all terms F crit = F 1, mn 4,( 1 α ) 2.830J/6.780J/ESD.63J 15

17 Recall the Brakeforming Data (MIT 2002) ST AL 0.6 ST AL J/6.780J/ESD.63J 16

18 Inputs Punch Depth (x 1 ) 0.3 In (-1) 0.6 in (+1) Inputs and Levels Material Type/Thickness ( ) (e.g.. bending stiffness) Aluminum (-`1) Steel (+1) 2 Inputs 2 levels each Model Output: Angle (y) 2.830J/6.780J/ESD.63J 17

19 Data Table for 2 2 Model Test x1 x2 yi1 yi2 yi3 yi4 yi5 yi6 yi7 yi8 yi9 yi x 1 : Material : Depth 4 Tests 10 Replicates J/6.780J/ESD.63J 18

20 Looking only at Mean Response Test x1 x2 yibar x1 x2 x1x y = X= J/6.780J/ESD.63J 19

21 Solving β=x -1 y Model and Interpretation β= y = x x 1 + ε 2.830J/6.780J/ESD.63J 20

22 Residual Analysis y = β 0 + β 1 x 1 + β 2 + β 12 x 1 + h.o.t. + ε ŷ = β 0 + β 1 x 1 + β 2 + β 12 x 1 y ŷ = h.o.t. + ε = residual Properties of residual? if model is correct if model of error is ~N(0,σ 2 ) 2.830J/6.780J/ESD.63J 21

23 Residuals (ε) with Test J/6.780J/ESD.63J 22

24 Residual Distribution J/6.780J/ESD.63J 23

25 Aside: Use of All Data X η 1 x1 x2 x1x2 y β = (X T X) 1 X T η 55.1 β= Same as before! J/6.780J/ESD.63J 24

26 Response Surface in in -1 Al +1 St ,-1 Material Axis -1,+1 S1 S5 S9 S13 Depth Axis +1,+1 S21 S J/6.780J/ESD.63J 25

27 Side View of Surface in in -1 Al +1 St Degree of interaction? -1, Material Axis +1, J/6.780J/ESD.63J 26 S12-1,+1 S1 Depth Axis

28 Are the Model Terms Significant? Mean: β 0 Effect of Depth: 2β 1 Effect of Material: 2β 2 Contaminated by simultaneous change of modulus, thickness and yield Interaction of Depth and Material: 2β J/6.780J/ESD.63J 27

29 Look at Single Variable Plots Effect of Depth with Aluminum Only Variation between tests Variation within a test Depth 2.830J/6.780J/ESD.63J 28

30 Single Variable Plot: Material Effect Variation between tests Variation within a test J/6.780J/ESD.63J 29

31 Interaction Effect? 90 Interaction Plot 80 X1= in in -1 Al +1 St X1= ,-1 Material Axis +1,+1-1,+1 S12 S1 Depth Axis X J/6.780J/ESD.63J 30

32 ANOVA Test on Effects Test x1 x2 yi1 yi2 yi3 yi4 yi5 yi6 yi7 yi8 yi9 yi y = x x 1 + ε mean 2 nm β 0 1 x 1 2 nm β nm β 2 1 x 12 2 nm β 12 1 ε total m n ε ij mn 4 i =1 j =1 m n y ij mn i =1 j =1 SS (β 0 ) 1 SS( β 1 ) 1 SS (β 2 ) 1 SS( β 12 ) 1 SS (ε) (mn 4) MS (β 0 ) MS (ε) MS ( β 1 ) MS (ε) MS (β 2 ) MS (ε) MS ( β 12 ) MS (ε) ANOVA on Effects n=10 m=4 nm=40 SS DOF MS F F (0.05) mean E+05 1E X E+04 1E X X1X Error Total J/6.780J/ESD.63J 31

33 How to Test? Is Model Form Adequate? Consider additional experimental (center) points Is new data on or near surface? ,-1-1,+1 Intermediate points S1 S5 S9 S13 Depth Axis +1,+1 S21 S J/6.780J/ESD.63J 32

34 Questions and Hypotheses Lack of Fit Test: Is the Model Form Correct? H o : variance of lack of fit = pure (replicate) variance H 1 : variance of lack of fit pure (replicate) variance If H 0 the observed deviation from model prediction (e.g. at center point) could be explained by pure (replicate) error Not enough evidence to attribute to model structure error 2.830J/6.780J/ESD.63J 33

35 Testing for Quadratic Error Recall our Linear Model y = β 0 + β 1 x 1 + β 2 + β 12 x 1 + h.o.t. + ε Add a h.o.t.: β 11 x β 22 2 Check for deviation at center point x 1 = 0; = 0 What is our hypothesis H o :? H 1 :? y = β 0 + β 1 x 1 + β 2 + β 12 x 1 + β 11 x β ε 2.830J/6.780J/ESD.63J 34

36 Consider a Simple Test Full Factorial linear with interactions, no replicates (n=1) Cannot test significance Cannot test for model fit Add Central Point (A=0, B=0) with n replicates: ,-1 0-1,+1 S1 S5 0 S9 S21 S17 S13 Depth Axis +1, J/6.780J/ESD.63J 35

37 Use of Central Data Determine Deviation from Linear Prediction Quadratic Term, or Central Error Term Determine MS of that Error SS/dof Compare to Replication Error 2.830J/6.780J/ESD.63J 36

38 Definitions ŷ = β 0 + β 1 x 1 + β 2 + β 12 x 1 + β 11 + β 2 y = grand mean of all factorial runs F y = grand mean of all center point runs C SS = n F n C (y F y C )2 Quadratic n + n F C MS Quadratic = SS Quadratic n c 2.830J/6.780J/ESD.63J 37

39 Example: 2 2 Without Replicates; Replicated Intermediate Points +1 A B I A B AB (1) a b ab Contrasts Effect Model Coefficients Just using corner points: y = x x J/6.780J/ESD.63J 38

40 Use of Central Data (1) 40.3 a 40.5 b 40.7 ab 40.2 quad 40.6 Average SS Variance 0.04 A B ANOVA Source SS DOF MS F F(0.05) A B AB Quad Error Total J/6.780J/ESD.63J 39

41 Full Factorial Models Outline Contrasts Extension to 2 k Model Term Significance: ANOVA Checking Adequacy of Model Form Tests for higher order fits (curvature) Experimental Design Blocks and Confounding Single Replicate Designs Fractional Factorial Designs 2.830J/6.780J/ESD.63J 40

42 Experimental Design Issues Nuisance Factors Affect the output, but don t want the effect May not be able to run whole experiment holding that factor constant If Known but Uncontrollable Randomization: treat as random noise factor If Known and Controllable Separate data into block where nuisance is constant E.g., if replicating, run each replicate of full design with same block factor 2.830J/6.780J/ESD.63J 41

43 Replicated Block Design Hardness Test Test Coupon Tip Each Block is like a Factor Randomize Order Within Blocks y ij = μ + τ i + β j + ε ij i = 1, 2, 3...a j = 1, 2, 3, b 2.830J/6.780J/ESD.63J 42

44 Nonreplicated Block Design Suppose 2 2 design 2 factors, 2 levels each = 4 runs But we have to arrange to do 2 runs (block 1), then another 2 runs (block 2) Expect an unknown offset Δ between block 1 & 2 Question: How arrange runs? X 1 X 2 (1) a + b + ab + + Block 1 Block 1 Can t distinguish X 2 effect from Δ! 2.830J/6.780J/ESD.63J 43

45 Nonreplicated Block Design Better approach: Block 1 = (1) and ab Block 2 = a and b Contrast A = [a + ab b (1)] Contrast B = [b + ab a (1)] Contrast AB = [ab + (1) a b] Δ s within each block Δ s across each block 2.830J/6.780J/ESD.63J 44

46 Blocking and Confounding Contrast A = [a + ab b (1)] = (ab (1)) + (b a) Contrast B = [b + ab a (1)] = (ab (1)) + (a b) Contrast AB = [ab + (1) a b] = (ab + (1)) (a + b) Now assume the block effect is that block 2 has an offset of δ from what it would be if done in block 1 a = a + δ b = b + δ This gives Contrast A = (ab (1)) + (b + δ a δ ) Contrast B = (ab (1)) + (a + δ b δ ) Contrast AB = (ab + (1)) (a + δ + b + δ ) 2 δ δ s cancel δ s cancel δ s double! 2.830J/6.780J/ESD.63J 45

47 Fractional Factorial Experiments What if we do less than full factorial 2 k? From regression model for 3 inputs: y = β 0 + β 1 x 1 + β 2 + β 12 x 1 + β 3 x 3 +β 13 x 1 x 3 + β 23 x 3 + β 123z x 1 x 3 + ε We will not be able to find all 8 coefficients 2.830J/6.780J/ESD.63J 46

48 2 3-1 Experiment Consider doing 4 experiments instead of 8; e.g.: x 1 x This is a 2 2 array Could also be for 3 inputs if we define x 3 = x J/6.780J/ESD.63J 47

49 2 3-1 Experiment x 1 x But now we can only define 4 coefficients in the model: e.g.: ) y = β 0 + β 1 x 1 + β 2 + β 3 x 3 i.e. no interaction terms 2.830J/6.780J/ESD.63J 48

50 2 3-1 Experiment Or we could choose other terms: ) y = β 0 + β 1 x 1 + β 2 + β 13 x 1 x 3 or: ) y = β 0 + β 1 x 1 + β 12 x 1 + β 3 x 3 or: 2.830J/6.780J/ESD.63J 49

51 Confounding / Aliasing We actually have the following: ) y = β + β ' z + β ' z + β ' z where the z variable represent sums of the various input terms, e.g. z 1 = x 1 + x 3 z 2 = x 1 + x 3 L where the specific choice of the experimental array determines what these sums are 2.830J/6.780J/ESD.63J 50

52 Confounding / Aliasing 2 3 Array: (Our X matrix) Test I A B AB C AC BC ABC (1) a b ab c ac bc abc J/6.780J/ESD.63J 51

53 Consider upper half: Confounding / Aliasing Test I A B AB C AC BC ABC (1) a b ab c ac bc abc Look at columns for C - no change at all! or C = -I Also AC = -A and BC = -B, and ABC = -AB 2.830J/6.780J/ESD.63J 52

54 Confounding / Aliasing Test I A B AB C AC BC ABC (1) a b ab c ac bc abc Contrast A =[ -(1)+a-b+ab] Contrast AC =[ (1)-a+b-ab] Defining Relation I = -C AC is an alias of A Note that alias of A =A*(-C) 2.830J/6.780J/ESD.63J 53

55 Aliases Choice of Design? Must have one of the pair assumed negligible ( sparsity of effects ) Balance/Orthogonality Sufficient excitation of inputs 2.830J/6.780J/ESD.63J 54

56 Balance and Orthogonality Test I A B AB C AC BC ABC (1) a b ab c ac bc abc Note: All columns have equal number of + and - signs (Balance) Sum of product of any two columns = 0 (Orthogonality) -All combinations occur the same number of times 2.830J/6.780J/ESD.63J 55

57 Balance/Orthogonality in Test I A B C AB AC BC ABC a b c ab ac bc abc A, B and C are balanced but B and C are not orthogonal 2.830J/6.780J/ESD.63J 56

58 Design Resolution Test I A B C AB AC BC ABC a b c With this array: -balance for A, B, C -all but A B C are orthogonal -defining relation I=ABC e.g. aliases of A: A*ABC=A*I A*A = I BC aliased with A Main effects aliased with interactions only Aliases: A BC B AC C AB I ABC 2.830J/6.780J/ESD.63J 57

59 Resolution III No Main aliases Design Resolution Main - Interaction Aliases Resolution IV No Alias between main effects and 2 factor effects, but others exist Resolution V No Main and no 2 Factor Aliases J/6.780J/ESD.63J 58

60 p = 1 1/2 fraction p= 2 1/4 fraction p 1/2 p Smaller Fraction 2 k-p 2.830J/6.780J/ESD.63J 59

61 2 4-2 A B C D Four Main Effects Four tests? Suppose we want to alias A with BCD and ABC What are the defining relations? 2.830J/6.780J/ESD.63J 60

62 Full Factorial Models Summary Contrasts Extension to 2 k Model Term Significance: ANOVA Checking Adequacy of Model Form Tests for higher order fits (curvature) Experimental Design Blocks and Confounding Single Replicate Designs Fractional Factorial Designs 2.830J/6.780J/ESD.63J 61

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

2.830J / 6.780J / ESD.63J Control of Manufacturing Processes (SMA 6303) Spring 2008 MIT OpenCourseWare http://ocw.mit.edu 2.830J / 6.780J / ESD.63J Control of Processes (SMA 6303) Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

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