Solution 2.13 - Chemometrics: Data Analysis for the Laboratory and Chemical Plant
Education Article
- Published: Jan 1, 2000
- Channels: Chemometrics & Informatics
1. The design table should look as follows
0.4 |
0.2 |
0.4 |
0.4 |
0.3 |
0.3 |
0 |
0.6 |
0.4 |
0.1 |
0.6 |
0.3 |
0 |
0.3 |
0.7 |
0.1 |
0.2 |
0.7 |
It is constructed as follows
- The first two experiments are at the high value of the first factor, and at the low value of one of the other factors. The third factor is simply the remainder.
- Similarly, the second two are at the high value of the second factor and the final two at the high value of the third factor.
The design is feasible because the sum of the high value for one factor, and the low values for the remaining two factors is £ 1. There are six vertices because, in fact, this sum is < 1 for all three factors.
2. There are a variety of ways of doing this, but a simple approach is as follows.
- Reorganise the vertices so that each successive vertex has one value of x in common. These vertices will lie on a straight-line in mixture space where the proportion of one component is constant. These are experiments 1, 3, 5, 7, 9 and 11 below, and are shaded for convenience.
- Then take the average of successive experiments to obtain conditions on the middle of the edges of the hexagon (experiments 2, 4, 6, 8, 10 and 12, the latter being the average of experiments 1 and 11.
Finally choose a centre point which is the average of all the experiments.
expt |
x_{1} |
x_{2} |
x_{3} | |
1 |
0.40 |
0.20 |
0.40 |
vertex |
2 |
0.40 |
0.25 |
0.35 | |
3 |
0.40 |
0.30 |
0.30 |
vertex |
4 |
0.20 |
0.30 |
0.50 | |
5 |
0.00 |
0.30 |
0.70 |
vertex |
6 |
0.00 |
0.45 |
0.55 | |
7 |
0.00 |
0.60 |
0.40 |
vertex |
8 |
0.05 |
0.60 |
0.35 | |
9 |
0.10 |
0.60 |
0.30 |
vertex |
10 |
0.10 |
0.40 |
0.50 | |
11 |
0.10 |
0.20 |
0.70 |
vertex |
12 |
0.25 |
0.20 |
0.55 | |
13 |
0.17 |
0.37 |
0.47 |
centre |
3. The mixture diagram is illustrated below
The six vertices of the irregular hexagon are illustrated. Note certain features. For example, the smaller the proportion the longer the line. Note also that the sum of the upper bounds of factors 1 and 2 equals 1, so the lines for these upper bounds meet on the left hand edge of the triangle, although outside the constrained mixture space.