Solution 2.8 - Chemometrics: Data Analysis for the Laboratory and Chemical Plant
Education Article
- Published: Jan 1, 2000
- Channels: Chemometrics & Informatics
1. Calculate the partial derivatives, and set to 0, as follows:
giving, from the first equation
b = 5 – 2a
so that, substituting in the second equation
15 – 7 a = 0
or
a = 15/7 = 2.143, b = 5/7 = 0.714, y = 0.857.
2. The progress of the simplex is as follows
|
experiment |
a |
b |
y | |
|
initial |
0 |
0 |
8.00 | |
|
initial |
1 |
0 |
4.00 | |
|
initial |
0.5 |
0.866 |
3.35 | |
|
1 |
1.5 |
0.866 |
1.22 | |
|
2 |
1 |
1.732 |
3.07 | |
|
3 |
2 |
1.732 |
2.80 | |
|
4 |
2.5 |
0.866 |
1.08 | |
|
5 |
2 |
0 |
2.00 | |
|
6 |
2 |
1.732 |
2.80 | |
|
7 |
2 |
0 |
2.00 | |
|
8 |
2 |
1.732 |
2.80 |
3. The best conditions are found at experiment 4, with a = 2.5, b = 0.866, and y = 1.08. These conditions are not the same as the true best conditions as the step-size is quite large. There is oscillation after the minimum is found. There are a number of stopping rules, although it should be obvious from this example when the optimum has been reached.
4. The new optimisation is as follows:
|
experiment |
a |
b |
y | |||||
|
initial |
0 |
0 |
8.00 | |||||
|
initial |
0.5 |
0 |
5.75 | |||||
|
initial |
0.25 |
0.433 |
5.13 | |||||
|
1 |
0.75 |
0.433 |
3.35 | |||||
|
2 |
0.5 |
0.866 |
3.35 | |||||
|
3 |
1 |
0.866 |
2.04 | |||||
|
4 |
1.25 |
0.433 |
2.06 | |||||
|
5 |
1.5 |
0.866 |
1.22 | |||||
|
6 |
1.75 |
0.433 |
1.28 | |||||
|
7 |
2.25 |
0.433 |
1.00 | |||||
|
8 |
2 |
0.866 |
0.90 | |||||
|
9 |
2.75 |
0.433 |
1.21 | |||||
|
10 |
1.5 |
0.866 |
1.22 | |||||
|
11 |
2.75 |
0.433 |
1.21 | |||||
Now the oscillation is much gentler around the optimum, which gives a = 2.0, b = 0.866, and y = 0.90, closer to the true conditions both for y and a. The step-size in the original simplex was too large, as evidenced by the violent oscillation around the optimum, although even then, a reasonable answer was achieved quickly.
