11.3 - Mixture Experiments

Example 11-2 Section

This is another class of response surface designs where the components are not just levels of factors but a special set where the \(x_1, x_2, \dots\) are coded and are the components of the mixture such that the sum of the \(x_i = 1\). So, these make up the proportions of the mixture.

Examples

If you are making any kind of product it usually involves mixtures of ingredients. A classic example is gasoline which is a mixture of various petrochemicals. In polymer production, polymers are actually mixtures of components as well. My favorite classroom example is baking a cake. A cake is a mixture of flour, sugar, eggs, and other ingredients depending on the type of cake. It is a mixture where the levels of x are the proportions of the ingredients.

This constraint that the sum of the x's sum to 1, i.e.,

\(0 ≤ x_i ≤ 1\)

has an impact on how we analyze these experiments.

Here we will write out our usual linear model:

\(Y_i=\beta_0 + \beta_1x_{i1}+\beta_2x_{i2}+\ldots +\beta_kx_{ik}+\varepsilon_i\)

where, \(1 =\sum\limits_{j=1}^k x_{ij}\)

If you want to incorporate this constraint then we can write:

\(Y_i= \beta_1x_{i1}+\beta_2x_{i2}+\ldots +\beta_kx_{ik}+\varepsilon_i\)

in other words, if we drop the \(\beta_0\), this reduces the parameter space by 1 and then we can fit a reduced model even though the x's are each constrained.

In the quadratic model:

\(Y_h=\sum\limits_{i=1}^k \beta_i x_{hi}+\mathop{\sum\sum}\limits_{i<j}\beta_{ij}x_{hi}x_{hj}+\varepsilon_h\)

This is probably the model we are most interested in and will use the most. Then we can generalize this into a cubic model which has one additional term.

A Cubic Model

\(Y_h=\sum\limits_{i=1}^k \beta_i x_{hi} +\mathop{\sum\sum}\limits_{i<j}\beta_{ij}x_{hi}x_{hj}
+\mathop{\sum\limits^k\sum\limits^k\sum\limits^k}\limits_{i<j<l}\beta_{ijl}x_{hi}x_{hj}x_{hl}
+\varepsilon_h\)

These models are used to fit response surfaces.

Let's look at the parameter of space. Let's say that k = 2. The mixture is entirely made up of two ingredients, \(x_1\) and \(x_2\). The sum of both ingredients is a line plotted in the parameter space below: An experiment made up of two components is either all of \(x_1\) or all of \(x_2\) or something in between, a proportion of the two. Use your mouse to click and drag the intersection point along the line that serves as a boundary to this region of experimentation.

Let's take a Look at the parameter space in three dimensions. Here we have three components: \(x_1, x_2 \text{ and } x_3\). As we satisfy our constraint that the sum of all the components equal 1 and then our parameter space is the plane that cuts the three-dimensional surface, intersecting these three points in the graph below scratch that in the plot below.

x 1 x 2 x 3 1 1 1

Plot showing the plane where the sum: x1+x2+x3 = 1

The triangle represents the full extent of the region of experimentation in this case with the points sometimes referred to as the Barycentric coordinates. The design question we want to address is where do we do our experiment? We are not interested in any one of the corners of the triangle where only one ingredient is represented, we are interested in some way on the middle where there is a proportion of all three of the ingredients included. We will restrict it to a feasible region of experimentation somewhere in the middle area.

Let's look at an example, for instance, producing cattle feed. The ingredients might include the following: corn, oats, hay, soybean, grass, ... all sorts of things.

In some situations it might work where you might have 100% of one component, but many instances of mixtures we try to partition off a part of the space in the middle where we think the combination is optimal.

In k = 4 the region of experimentation can be represented by the shape of a tetrahedron where each of the four sides of the tetrahedron is an equalateral triangle and has its own set of Barycentric coordinates. Each face of the tetrahedron corresponds to design region where one coordinate is zero, and the remaining three must sum to 1.