Learning outcomes
 Learn the basic terminology of experiments: responses, factors, outcomes, realworld units vs coded units, confounding
 Analyze and interpret data from an experiment with 2, 3 or more factors by hand
 Use R to do the analysis, and interpret the various plots, such as the Pareto plot
 Analyze and interpret data from an experiment with 3 or more factors
 Recognize when to use a fractional factorial, to go mostly the same results as from a full factorial
 Understand when to use screening experiments
 Use the concepts of response surface methods to systematically reach an optimum
 What you can do when you make a mistake, or hit against constraints.
Resources
 Class notes 2015 This is a large file, at 219 Mb. You might prefer the smaller set of notes from 2014, but they are less comprehensive.
 Class notes 2014
 Textbook, chapter 5
 You must have completed steps 13, and 16 to 22 of the Software tutorial to successfully use R in this section.
 Take the Coursera course on this topic. That course was partially developed based on the Statistics for Engineers course.
 What experiments are you going to do for your course project? What are the factors, what are the levels? How did you chose your levels? What is your response? How will you measure your response?
 Quizzes (with solutions): attempt these after you have watched the videos
Tasks to do first

Quiz

Solution

Watch videos 1A, 1B, 1C and 1D

Quiz

Solution

Watch videos 2A, 2B, 2C, 3A and 3B (note!)

Quiz

Solution

Watch videos 2D, 3C and 3D and 4A (note!)

Quiz

Solution

Watch videos 4B, 4C, 4D, and 4E

Quiz

Solution

Watch videos 4F, 4G, 4H, 5A and 5B

Quiz

Solution

Watch videos 5C, 5D, 5E and 5F

Quiz

Solution

Extended readings/practice
Class videos from prior years
Videos from 2015
Watch all these videos in this YouTube playlist
 00  Introduction video for the Coursera online course [01:56]
 1A  Why experiments are so important [07:48]
 1B  Some basic terminology [06:37]
 1C  Analysis of your first experiment [09:00]
 1D  How NOT to run an experiment [03:07]
 2A  Analysis of experiments in two factors by hand [13:37]
 2B  Numeric predictions from twofactor experiments [07:25]
 2C  Twofactor experiments with interactions [15:15]
 2D  Indepth case study: analyzing a system with 3 factors by hand [17:28]
 3A  Setting up the least squares model for a 2 factor experiment [05:46]
 3B  Solving the mathematical model for a 2 factor experiment using software [08:46]
 3C  Using computer software for a 3 factor experiment [08:37]
 3D  Case study: a 4factor system using computer software [09:03]
 4A  The tradeoffs when doing halffraction factorials [13:20]
 4B  The technical details behind halffractions [09:38]
 4C  A case study with aliasing in a fractional factorial [06:38]
 4D  All about disturbances, why we randomize, and what covariates are [11:00]
 4E  All about blocking [09:21]
 4F  Fractional factorials: introducing aliasing notation [12:00]
 4G  Fractional factorials: using aliasing notation to plan experiments [10:45]
 4H  An example of an analyzing an experiment with aliasing [09:50]
 5A  Response surface methods  an introduction [06:13]
 5B  Response surface methods (RSM) in one variable [18:40]
 5C  Why changing one factor at a time (OFAT) will mislead you [05:33]
 5D  The concept of contour plots and which objectives should we maximize [03:40]
 5E  RSM in 2 factors: introducing the case study [19:20]
 5F  RSM case study continues: constraints and mistakes [13:45]
 5G  RSM case study continues: approaching the optimum [17:05]
 06  Wrapup: the course in review, multiple objectives, and references for the future [08:10]
Videos from 2014
See the webpage from 2014
Videos from 2013
See the webpage from 2013
Software codes for this section
Code to build a model for a 2factor system
# Centered and scaled temperature
T < c(1, +1, 1, +1)
# Centered and scaled substrate concentration
S < c(1, 1, +1, +1)
# Conversion is the response, y
y < c(69, 60, 64, 53)
# this works, but is more typing
mod < lm(y ~ T + S + T * S)
# preferred method
mod < lm(y ~ T * S)
summary(mod)
Detailed analysis of the 3 factor example
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 (see below) to verify that
\[y = 11.25 + 6.25x_C + 0.75x_T 7.25x_S + 0.25 x_C x_T 6.75 x_C x_S 0.25 x_T x_S  0.25 x_C x_T x_S\]
The \(\mathbf{X}\) matrix and \(\mathbf{y}\) vector used to calculate the least squares model:
\[\begin{split}\begin{bmatrix} 5\\30\\6\\33\\4\\3\\5\\4 \end{bmatrix} &= \begin{bmatrix} +1 & 1 & 1 & 1 & +1 & +1 & +1 & 1\\ +1 & +1 & 1 & 1 & 1 & 1 & +1 & +1\\ +1 & 1 & +1 & 1 & 1 & +1 & 1 & +1\\ +1 & +1 & +1 & 1 & +1 & 1 & 1 & 1\\ +1 & 1 & 1 & +1 & +1 & 1 & 1 & +1\\ +1 & +1 & 1 & +1 & 1 & +1 & 1 & 1\\ +1 & 1 & +1 & +1 & 1 & 1 & +1 & 1\\ +1 & +1 & +1 & +1 & +1 & +1 & +1 & +1\\ \end{bmatrix} \begin{bmatrix} b_0 \\ b_C \\ b_T \\ b_{S} \\ b_{CT} \\ b_{CS} \\ b_{TS} \\ b_{CTS} \end{bmatrix} \\ \mathbf{y} &= \mathbf{X} \mathbf{b}\end{split}\]
The above 3 factor example, but using computer software
# Create the design matrix in a quick way in R
x < c(1, +1)
design < expand.grid(C=x, T=x, S=x)
design
C < design$C
T < design$T
S < design$S
y < c(5, 30, 6, 33, 4, 3, 5, 4)
# Full factorial model (you could make errors typing this in)
mod.full < lm(y ~ C + T + S + C*T + C*S + S*T + C*T*S)
# This powerful notation will expand all terms up to the 3rd order interactions
mod.full < lm( y ~ C * T * S )
summary(mod.full)
Compare the result from the last line with the manual calculations above.
Quickly calculate, with software, the names of all the interactions terms
 Calculating all combinations of the interactions can be tedious byhand in larger models.
 Let the computer do the work for you.
 Run this code in a webbrowser
print('Twofactor interactions in a 4factor model')
attr(terms.formula( y ~ (A+B+C+D)^2 ), "term.labels")
print('Two and threefactor interactions in a 4factor model')
attr(terms.formula( y ~ (A+B+C+D)^3 ), "term.labels")
print('Two, 3, and 4factor interactions in a 4factor model')
attr(terms.formula( y ~ (A+B+C+D)^4 ), "term.labels")
Plot a Pareto plot of all coefficients in a leastsquares model
# First create a model to demonstrate.
x < c(1, +1)
design < expand.grid(C=x, T=x, S=x)
C < design$C
T < design$T
S < design$S
y < c(5, 30, 6, 33, 4, 3, 5, 4)
my.model < lm( y ~ C * T * S )
# The quick way to plot the coefficients: you can supply
# the "paretoPlot" function with any leastsquares model
#library(pid)
# The "PID" library is not part of this webbased version
# of R. So load it from this server:
source('https://yint.org/paretoPlot.R')
# And now we can generate the plot:
p < paretoPlot(my.model)
Fractionalfactorial example (saturated design) in R, including how to eliminate factors
 The code shows how the Pareto plot is used to find variables with low significance.
 These variables can be eliminated and the model rebuilt without them.
 Run this code in a webbrowser
A < B < C < c(1, +1)
design < expand.grid(A=A, B=B, C=C)
A < design$A
B < design$B
C < design$C
# Specify the other factors 4: up to 7 factors plus an
# intercept can be estimated from the 8 experiments.
D < A*B
E < A*C
F < B*C
G < A*B*C
# The results from the 8 experiments
y < c(77.1, 68.9, 75.5, 72.5, 67.9, 68.5, 71.5, 63.7)
mod < lm(y ~ A + B + C + D + E + F + G)
summary(mod)
# Visualize the Pareto plot using the PID library
#library(pid)
# The "PID" library is not part of this webbased version
# of R. So load it from this server:
source('https://yint.org/paretoPlot.R')
paretoPlot(mod)
# The factors B, F and D are not important.
# Remove and rebuild the model without them.
mod.update.1 < lm(y ~ A + C + E + G)
confint(mod.update.1)
paretoPlot(mod.update.1)
# Notice that the bars are the same length as before.
# Can you explain why?
# The confidence interval for factor E spans zero. It
# might be of practical relevance, but it is not of
# statistical relevance.
mod.update.2 < lm(y ~ A + C + G)
confint(mod.update.2)
Plot a contour plot of a leastsquares DOE model: a system with 2 factors
# First create a model to demonstrate.
# Note that this model has a centerpoint at (0,0)
T < c(1, +1, 1, +1, 0)
S < c(1, 1, +1, +1, 0)
y < c(193, 310, 468, 571, 407)
# Visualize the surface
my.model.round1 < lm(y ~ T*S)
# Note that the interaction is small;
# we should observe that in the contour plot
summary(my.model.round1)
# The PID library can draw contour plots for
# any least squares model where there are two factors:
# library(pid)
# The "PID" library is not part of this webbased version
# of R. So load it from this server:
source('https://yint.org/contourPlot.R')
# Confirms the weak interaction
contourPlot(my.model.round1)
# Now make the interaction more prominent
T < c(1, +1, 1, +1, 0)
S < c(1, 1, +1, +1, 0)
y < c(193, 310, 281, 571, 407)
my.model.round2 < lm(y ~ T*S)
# Note the stronger interaction
summary(my.model.round2)
# Confirms the strong interaction
contourPlot(my.model.round2)
Plot a contour plot of a leastsquares DOE model: a system with 3 factors
 When you have 3 (or more) factors in a DOE system, you can still use the contourPlot() function in R.
 Except, you need to tell the software which variables you want to see on the \(x\) and \(y\) axes.
 Run this code in a webbrowser
# First create a model to demonstrate.
x < c(1, +1)
design < expand.grid(C=x, T=x, S=x)
C < design$C
T < design$T
S < design$S
y < c(5, 30, 6, 33, 4, 3, 5, 4)
mod.full < lm( y ~ C*T*S )
# The summary shows a high C:S interaction, but
# the other interactions are weak. We should confirm
# this in the visualizations that follow next.
summary(mod.full)
# Visualize the surface using the PID library in R
# library(pid)
# The "PID" library is not part of this webbased version
# of R. So load it from this server:
source('https://yint.org/contourPlot.R')
# Any least squares model can be supplied as
# the first input into the contourPlot function.
# Start by visualizing the C and T factors
# (small interaction only)
contourPlot(mod.full, "C", "T", main='Weak interaction')
# Try the T and S factors
# (some interaction)
contourPlot(mod.full, "T", "S", main='Weak interaction')
# Try the C and S factor combination
# (high interaction)
contourPlot(mod.full, "C", "S", main='Strong interaction')
# Change the colour scheme and the plot resolution
contourPlot(mod.full, "C", "S", N=50,
colour.function=rainbow,
main='Different colour scheme')
# In which direction would you go to minimize
# the response surface?
Bringing it all together: a comprehensive response surface method (RSM) example)
The code is provided here for you to try: http://yint.org/rsmcode