Solution 2.8 - Chemometrics: Data Analysis for the Laboratory and Chemical Plant

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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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