Design and analysis of experiments

Learning outcomes

• Learn the basic terminology of experiments: responses, factors, outcomes, real-world 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

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

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 two-factor experiments [07:25]
• 2C - Two-factor experiments with interactions [15:15]
• 2D - In-depth 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 4-factor system using computer software [09:03]
• 4A - The trade-offs when doing half-fraction factorials [13:20]
• 4B - The technical details behind half-fractions [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 - Wrap-up: the course in review, multiple objectives, and references for the future [08:10]

Software codes for this section

Code to build a model for a 2-factor system

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) # this works, but is more typing
mod <- lm(y ~ T*S)           # preferred method
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$S y <- c(5, 30, 6, 33, 4, 3, 5, 4) # Full factorial model (you could make errors typing this in) mod.full <- lm(y ~ C + T + S + C*T + C*S + S*T + C*T*S) # This powerful notation will expand all terms up to the 3rd order interactions mod.full <- lm( y ~ C*T*S ) summary(mod.full) Compare the result from the last line with the manual calculations above. Quickly calculate, with software, the names of all the interactions terms • Calculating all combinations of the interactions can be tedious by-hand in larger models. • Let the computer do the work for you. • Run this code in a web-browser print('Two-factor interactions in a 4-factor model') attr(terms.formula( y ~ (A+B+C+D)^2 ), "term.labels") print('Two- and three-factor interactions in a 4-factor model') attr(terms.formula( y ~ (A+B+C+D)^3 ), "term.labels") print('Two-, three-, and four-factor interactions in a 4-factor model') attr(terms.formula( y ~ (A+B+C+D)^4 ), "term.labels") Plot a Pareto plot of all coefficients in a least-squares model # First create a model to demonstrate. x <- c(-1, +1) design <- expand.grid(C=x, T=x, S=x) C <- design$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 least-squares model
library(pid)
paretoPlot(my.model)

Fractional-factorial 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 web-browser
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 + 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 in R library(pid) paretoPlot(mod) # The factors B, F and D are not important. Remove and rebuild 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 least-squares DOE model: a system with 2 factors # First create a model to demonstrate. # Note that this model has a center-point 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) summary(my.model.round1) # Note that the interaction is small; # we should observe that in the contour plot # The PID library can draw contour plots for # any least squares model where there are two factors: library(pid) contourPlot(my.model.round1) # confirms the weak interaction # 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) summary(my.model.round2) # note the stronger interaction contourPlot(my.model.round2) # confirms the strong interaction Plot a contour plot of a least-squares 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 web-browser # 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)

# Any least squares model can be supplied as the first input

# Start by visualizing the C and T factors (little interaction)
contourPlot(mod.full, "C", "T")

# Try the T and S factors (little interaction)
contourPlot(mod.full, "T", "S")

# Try the C and S factor combination (high interaction)
contourPlot(mod.full, "C", "S")

# Change the colour scheme and the plot resolution
contourPlot(mod.full, "C", "S", N=50, colour.function=rainbow)

# 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/rsm-code