Difference between revisions of "Design and analysis of experiments (2014)"

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| 11A
| 11A
| [http://learnche.mcmaster.ca/media/2014-4C3-6C3-Class-11A.mp4 Video] (309 M)  
| [http://learnche.mcmaster.ca/media/2014-4C3-6C3-Class-11A.mp4 Video] (309 M)  
| [http://learnche.mcmaster.ca/media/2014-4C3-6C3-Class-11A.mp3 Audio] (42 M)  
| [http://learnche.mcmaster.ca/media/2014-4C3-6C3-Class-11A.mp3 Audio] (43 M)  
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Revision as of 15:43, 27 March 2014

Class date(s): 27 February to 27 March 2014
Nuvola mimetypes pdf.png (PDF) Course slides
Download video: Link (plays in Google Chrome) [227M]
Download video: Link (plays in Google Chrome) [299M]
Download video: Link (plays in Google Chrome) [296M]
Download video: Link (plays in Google Chrome) [286M]
Download video: Link (plays in Google Chrome) [311M]
Download video: Link (plays in Google Chrome) [296M]
Download video: Link (plays in Google Chrome) [280M]
Download video: Link (plays in Google Chrome) [305 M]
Download video: Link (plays in Google Chrome) [301 M]
Download video: Link (plays in Google Chrome) [309 M]
Download video: Link (plays in Google Chrome) [245 M]

Course notes and slides

Date Class number Video and audio files Other materials Reading (PID) Slides
27 February 07C Video (227 M) Audio (40 M) None Chapter 5 Nuvola mimetypes pdf.png Slides for class
03 March 08A Video (299 M) Audio (40 M) None
05 March 08B Video (296 M) Audio (42 M) None
06 March 08C Video (286 M) Audio (42 M) See the code and derivations below
10 March 09A Video (311 M) Audio (42 M)
13 March 09B Video (296 M) Audio (42 M)
17 March 10A Video (280 M) Audio (42 M) See the source code below.
19 March 10B Video (305 M) Audio (42 M) None
20 March 10C Video (301 M) Audio (42 M) None
24 March 11A Video (309 M) Audio (43 M) None
26 March 11B Video (245 M) Audio (42 M) See the source code below.
27 March 11C


Software source code

Class 08C, 06 March

Source code to estimate the DOE model 

T <- c(-1, +1, -1, +1)  # centered and scaled temperature
S <- c(-1, -1, +1, +1)  # centered and scaled substrate concentration
y <- c(69, 60, 64, 53)  # conversion is the response, y
mod <- lm(y ~ T + S + T * S)
summary(mod)

Call:
lm(formula = y ~ T + S + T * S)

Residuals:
ALL 4 residuals are 0: no residual degrees of freedom!

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)     61.5         NA      NA       NA
T               -5.0         NA      NA       NA
S               -3.0         NA      NA       NA
T:S             -0.5         NA      NA       NA

Residual standard error: NaN on 0 degrees of freedom
Multiple R-squared:     1,	Adjusted R-squared:   NaN 
F-statistic:   NaN on 3 and 0 DF,  p-value: NA

Class 08C, 06 March

3-factor example <rst> <rst-options: 'toc' = False/> The data are from a plastics molding factory which must treat its waste before discharge. The :math:`y` variable represents the average amount of pollutant discharged [lb per day], while the 3 factors that were varied were: 

	-	:math:`C`: the chemical compound added [A or B]

- :math:`T`: the treatment temperature [72°F or 100°F] - :math:`S`: the stirring speed [200 rpm or 400 rpm] - :math:`y`: the amount of pollutant discharged [lb per day]

.. tabularcolumns:: |l|l||c|c|c||c|

+-----------+-------+---------------+-----------------+-----------------+-----------------+ | Experiment| Order | :math:`C` | :math:`T` [°F] | :math:`S` [rpm] | :math:`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 :math:`C = \displaystyle \frac{(+25) + (+27) + (-1) + (-1)}{4} = \frac{50}{4} = \bf{12.5}`
  • **T effect**: There are 4 estimates of :math:`T = \displaystyle \frac{(+1) + (+3) + (+1) + (+1)}{4} = \frac{6}{4} = \bf{1.5}`
  • **S effect**: There are 4 estimates of :math:`S = \displaystyle \frac{(-27) + (-1) + (-29) + (-1)}{4} = \frac{-58}{4} = \bf{-14.5}`
  • **CT interaction**: There are 2 estimates of :math:`CT`. Recall that interactions are calculated as the half difference going from high to low. Consider the change in :math:`C` when

- :math:`T_\text{high}` (at :math:`S` high) = 4 - 5 = -1 - :math:`T_\text{low}` (at :math:`S` high) = 3 - 4 = -1 - First estimate = [(-1) - (-1)]/2 = 0 - :math:`T_\text{high}` (at :math:`S` low) = 33 - 6 = +27 - :math:`T_\text{low}` (at :math:`S` low) = 30 - 5 = +25 - Second estimate = [(+27) - (+25)]/2 = +1

- Average **CT** interaction = (0 + 1)/2 = **0.5** - You can interchange :math:`C` and :math:`T` and still get the same result.

  • **CS interaction**: There are 2 estimates of :math:`CS`. Consider the change in :math:`C` when

- :math:`S_\text{high}` (at :math:`T` high) = 4 - 5 = -1 - :math:`S_\text{low}` (at :math:`T` high) = 33 - 6 = +27 - First estimate = [(-1) - (+27)]/2 = -14 - :math:`S_\text{high}` (at :math:`T` low) = 3 - 4 = -1 - :math:`S_\text{low}` (at :math:`T` low) = 30 - 5 = +25 - Second estimate = [(-1) - (+25)]/2 = -13

- Average **CS** interaction = (-13 - 14)/2 = **-13.5** - You can interchange :math:`C` and :math:`S` and still get the same result.

  • **ST interaction**: There are 2 estimates of :math:`ST`: (-1 + 0)/2 = **-0.5**, calculate in the same way as above.
  • **CTS interaction**: There is only a single estimate of :math:`CTS`:

- :math:`CT` effect at high :math:`S` = 0 - :math:`CT` effect at low :math:`S` = +1 - :math:`CTS` interaction = [(0) - (+1)] / 2 = **-0.5**

- You can calculate this also by considering the :math:`CS` effect at the two levels of :math:`T` - Or, you can calculate this by considering the :math:`ST` effect at the two levels of :math:`C`. - All 3 approaches give the same result.


Next, use computer software to verify that

.. math::

y = 11.25 + 6.25x_C + 0.75x_T -7.25x_S + 0.25 x_C x_T -6.75 x_C x_S -0.25 x_T x_S - 0.25 x_C x_T x_S


The :math:`\mathbf{X}` matrix and :math:`\mathbf{y}` vector used to calculate the least squares model:

.. math::

\begin{bmatrix} 5\\30\\6\\33\\4\\3\\5\\4 \end{bmatrix} &= \begin{bmatrix} +1 & -1 & -1 & -1 & +1 & +1 & +1 & -1\\ +1 & +1 & -1 & -1 & -1 & -1 & +1 & +1\\ +1 & -1 & +1 & -1 & -1 & +1 & -1 & +1\\ +1 & +1 & +1 & -1 & +1 & -1 & -1 & -1\\ +1 & -1 & -1 & +1 & +1 & -1 & -1 & +1\\ +1 & +1 & -1 & +1 & -1 & +1 & -1 & -1\\ +1 & -1 & +1 & +1 & -1 & -1 & +1 & -1\\ +1 & +1 & +1 & +1 & +1 & +1 & +1 & +1\\ \end{bmatrix} \begin{bmatrix} b_0 \\ b_C \\ b_T \\ b_{S} \\ b_{CT} \\ b_{CS} \\ b_{TS} \\ b_{CTS} \end{bmatrix} \\ \mathbf{y} &= \mathbf{X} \mathbf{b} </rst>

Class 08C, 06 March

# Create the design matrix in a quick way in R
C <- T <- S <- c(-1, +1)
design <- expand.grid(C=C, T=T, S=S)
design
   C  T  S
1 -1 -1 -1
2  1 -1 -1
3 -1  1 -1
4  1  1 -1
5 -1 -1  1
6  1 -1  1
7 -1  1  1
8  1  1  1

C <- design$C
T <- design$T
S <- design$S
y <- c(5, 30, 6, 33, 4, 3, 5, 4)

# Full factorial model (you could make errors typing this in)
mod.full <- lm(y ~ C + T + S + C*T + C*S + S*T + C*T*S)

# This powerful notation will expand all terms up to the 3rd order interactions
mod.full <- lm( y ~ (C + T + S)^3 ) 

summary(mod.full)
Call:
lm(formula = y ~ C + T + S + C * T + C * S + S * T + C * T * S)

Residuals:
ALL 8 residuals are 0: no residual degrees of freedom!

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)    11.25         NA      NA       NA
C               6.25         NA      NA       NA
T               0.75         NA      NA       NA
S              -7.25         NA      NA       NA
C:T             0.25         NA      NA       NA
C:S            -6.75         NA      NA       NA
T:S            -0.25         NA      NA       NA
C:T:S          -0.25         NA      NA       NA

# Guide to estimating significant coefficients, ignoring the intercept (1st coefficient)
coeff.full <- coef(mod.full)[2:length(coef(mod.full))]

# Pareto plot of the absolute coefficients
library(lattice)
coeff <- sort(abs(coeff.full), index.return=TRUE)
barchart(coeff$x, 
         xlim=c(0, max(abs(coeff.full))+0.1),
         xlab=list("Magnitude of effect", cex=1.5), 
         ylab = list("Effect", cex=1.5),
         groups=(coeff.full>0)[coeff$ix], col=c("lightblue", "orange"),
         scales=list(cex=1.5)
)

# Eliminate some factors: use the regular linear model notation
# Eliminate C*T*S, C*T and S*T
mod.sub <- lm(y ~ C + T + S + C*S)
summary(mod.sub)
confint(mod.sub)
coeff.sub <- coef(mod.sub)[2:length(coef(mod.sub))]
coeff <- sort(abs(coeff.sub), index.return=TRUE)
barchart(coeff$x, 
         xlim=c(0, max(abs(coeff.sub))+0.1),
         xlab=list("Magnitude of effect", cex=1.5), 
         ylab = list("Effect", cex=1.5),
         groups=(coeff.sub>0)[coeff$ix], col=c("lightblue", "orange"),
         scales=list(cex=1.5)
)

Class 11B and 11C, 26 and 27 March

To find all the 2-factor and 3-factor coefficients in a linear model 

terms.formula( y ~ (A+B+C+D)^2 )
terms.formula( y ~ (A+B+C+D)^3 )

Class 11B, 26 March

Code for contour plots in response surface methods

R MATLAB 
# Initial set of 4 runs
T <- c(-1,    +1,   -1,   +1,    0)
S <- c(-1,    -1,   +1,   +1,    0)
y <- c(193,  310,  468,  571,  407)

# Visualize the surface
# ------------------------------------------------------
mod <- lm(y ~ T + S + T*S)
summary(mod)

N <- 50  # resolution of surface
         #  (higher values give smoother plots)

# The lower and upper bounds, in coded units, 
# over which we want
# to visualize the surface
bound <- 3 
T_plot <- seq(-bound, bound ,length=N)
S_plot <- seq(-bound, bound, length=N)
grd <- expand.grid(T=T_plot, S=S_plot)

# Predict directly from least squares model
grd$y <- predict(mod, grd)

library(lattice)
contourplot(y ~ T * S, 
            data = grd,
            cuts = 10, 
            region = TRUE,
            col.regions = terrain.colors,
            xlab = "Temperature",
            ylab = "Substrate",
            main = "Predicted response"
)

trellis.focus("panel", 1, 1, highlight=FALSE)
lpoints(T, S, pch="O", col="red", cex=3)
llines(c(-1, +1, +1, -1, -1), c(-1, -1, +1, +1, -1), 
    col="red", lwd=3)
lpoints(1, 2.44, col="red", cex=5, pch="*")
ltext(0.4, 2.44, "Expt 5", cex=2, col="red")

lpoints(2, 4.88, col="red", cex=5, pch="*")
ltext(1.2, 4.88, "Expt 6", cex=2, col="red")

trellis.unfocus()

X = [+1 +1 +1 +1 +1; ...
     -1 +1 -1 +1 0; ...
     -1 -1 +1 +1 0; ...
     +1 -1 -1 +1 0]';
 
% Profit values recorded from the factorial experiments
profit = [193 310 468 571 407]';
% 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
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