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Design and analysis of experiments (2011)
From Statistics for Engineering
 
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Contents 
[edit] Course notes
 (PDF) Course notes
 Please print pages from Chapter 5.
 The full PDF is provided so that hyperlinks for crosssections will work as expected.
[edit] Projector overheads
[edit] Guest speaker overheads
[edit] Audio recordings of 2011 classes
Date  Material covered  Audio file 

07 March 2011  Single variable experiments, randomization, changing one single variable at a time (COST) approach  Class 22 
09 March 2011  Analysis of a 2 factor experiment: without interactions, and with interactions  Class 23 
10 March 2011  Factorial analysis using least squares; 3factor experiment example (contains an integer variable)  Class 24 
16 March 2011  Assess if factors are significant or not; blocking and confounding  Not available 
17 March 2011  Guest speaker, Emily Nichols, on DOEs  Class 26 
21 March 2011  Fractional factorials; generators and defining relationships  Class 27 
23 March 2011  Saturated designs; response surface methods  Class 28 
24 March 2011  Response surface methods  Class 29 
28 March 2011  Dealing with incorrect experiments and constraints; wrapping up DOEs; takehome review.  Class 30 
Thanks to the various students responsible for recording and making these files available
[edit] 3factor example
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The data are from a plastics molding factory which must treat its waste before discharge. The \(y\) variable represents the average amount of pollutant discharged [lb per day], while the 3 factors that were varied were:
 \(C\): the chemical compound added [A or B]
 \(T\): the treatment temperature [72°F or 100°F]
 \(S\): the stirring speed [200 rpm or 400 rpm]
 \(y\): the amount of pollutant discharged [lb per day]
Experiment Order \(C\) \(T\) [°F] \(S\) [rpm] \(y\) [lb] 1 5 A 72 200 5 2 6 B 72 200 30 3 1 A 100 200 6 4 4 B 100 200 33 5 2 A 72 400 4 6 7 B 72 400 3 7 3 A 100 400 5 8 8 B 100 400 4
We showed the cube plot for this system in the class on 10 March. From the cube plot we could already see the main factors, and even the CS interaction was noticeable.
C effect: There are 4 estimates of \(C = \displaystyle \frac{(+25) + (+27) + (1) + (1)}{4} = \frac{50}{4} = \bf{12.5}\)
T effect: There are 4 estimates of \(T = \displaystyle \frac{(+1) + (+3) + (+1) + (+1)}{4} = \frac{6}{4} = \bf{1.5}\)
S effect: There are 4 estimates of \(S = \displaystyle \frac{(27) + (1) + (29) + (1)}{4} = \frac{58}{4} = \bf{14.5}\)
CT interaction: There are 2 estimates of \(CT\). Recall that interactions are calculated as the half difference going from high to low. Consider the change in \(C\) when
 \(T_\text{high}\) (at \(S\) high) = 4  5 = 1
 \(T_\text{low}\) (at \(S\) high) = 3  4 = 1
 First estimate = [(1)  (1)]/2 = 0
 \(T_\text{high}\) (at \(S\) low) = 33  6 = +27
 \(T_\text{low}\) (at \(S\) low) = 30  5 = +25
 Second estimate = [(+27)  (+25)]/2 = +1
 Average CT interaction = (0 + 1)/2 = 0.5
 You can interchange \(C\) and \(T\) and still get the same result.
CS interaction: There are 2 estimates of \(CS\). Consider the change in \(C\) when
 \(S_\text{high}\) (at \(T\) high) = 4  5 = 1
 \(S_\text{low}\) (at \(T\) high) = 33  6 = +27
 First estimate = [(1)  (+27)]/2 = 14
 \(S_\text{high}\) (at \(T\) low) = 3  4 = 1
 \(S_\text{low}\) (at \(T\) low) = 30  5 = +25
 Second estimate = [(1)  (+25)]/2 = 13
 Average CS interaction = (13  14)/2 = 13.5
 You can interchange \(C\) and \(S\) and still get the same result.
ST interaction: There are 2 estimates of \(ST\): (1 + 0)/2 = 0.5, calculate in the same way as above.
CTS interaction: There is only a single estimate of \(CTS\):
 \(CT\) effect at high \(S\) = 0
 \(CT\) effect at low \(S\) = +1
 \(CTS\) interaction = [(0)  (+1)] / 2 = 0.5
 You can calculate this also by considering the \(CS\) effect at the two levels of \(T\)
 Or, you can calculate this by considering the \(ST\) effect at the two levels of \(C\).
 All 3 approaches give the same result.
Next, use computer software to verify that
The \(\mathbf{X}\) matrix and \(\mathbf{y}\) vector used to calculate the least squares model:
[edit] Visualizing a response surface
[edit] Some advice
You can easily create your own response surfaces and visualize them. Then practice on them before and during the takehome exam. Here are equations for the 4 examples given in class on Thursday:
 A simple optimum: \(y = 83 +9.4x_A +7.1x_B 6 x_A x_B  7.4 x_A^2 3.7 x_B^2\)
 A stationary ridge: \(y = 83 + 10.0x_A + 5.6x_B  7.6 x_A x_B  6.9 x_A^2  2.0 x_B^2\)
 A rising ridge: \(y = 83 + 8.8x_A + 8.2x_B  7.6 x_A x_B  7.0 x_A^2  2.4 x_B^2\)
 A saddle point: \(y = 83 + 11.1x_A + 4.1x_B  9.4 x_A x_B  6.5 x_A^2  0.4 x_B^2\)
Binary variables should be treated exactly like continuous variables in all respects during the optimization. The only difference is that it makes sense to visualize and use them at the two levels.
[edit] Plotting code
MATLAB code used in class on 23 March 2011 to do the first factorial:
X = [+1 +1 +1 +1; ... 1 +1 1 +1; ... 1 1 +1 +1; ... +1 1 1 +1]'; % Profit values recorded from the factorial experiments profit = [193 310 468 571]'; % Coefficient order = [Intercept, b_T, b_S, b_{TS}] b = inv(X'*X)*X'*profit % What does the model surface look like? [T, S] = meshgrid(2:0.1:2, 2:0.1:2); Y = b(1) + b(2).*T + b(3).*S + b(4).*T.*S; subplot(1, 2, 1) surf(T, S, Y) xlabel('T') ylabel('S') grid('on') subplot(1, 2, 2) contour(T, S, Y) xlabel('T') ylabel('S') grid('on') axis equal colorbar
MATLAB code used in class on 24 March 2011 to continue with the central composite design factorial:
% CCD design around point 6 % Put 4 factorial points first, then the center point % then the 4 axial (star) points a = sqrt(2); X = [1 1 1; ... 1 +1 1; ... 1 1 +1; ... 1 +1 +1; ... 1 0 0; ... 1 0 a; ... 1 +a 0; ... 1 0 a; ... 1 a 0]; X(:,4) = X(:,2) .* X(:,3); % TS = T x S X(:,5) = X(:,2) .* X(:,2); % TT = T x T X(:,6) = X(:,3) .* X(:,3); % SS = S x S y = [694 725 620 642 688 720 700 610 663]'; b = inv(X'*X)*X'*y [T,S] = meshgrid(2:0.1:3, 3:0.1:2); Y = b(1) + b(2).*T + b(3).*S + b(4).*T.*S + b(5).*T.*T + b(6).*S.*S; subplot(1, 2, 1) surf(T, S, Y) xlabel('T') ylabel('S') grid('on') subplot(1, 2, 2) contour(T, S, Y) xlabel('T') ylabel('S') grid('on') axis equal colorbar % Predicted optimum: $737, actual value at this point: $737.8 T = +2.25; S = 1.75; y_hat_opt = b(1) + b(2).*T + b(3).*S + b(4).*T.*S + b(5).*T.*T + b(6).*S.*S