# 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.

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