how to apply pso to formula

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kyungdoo lee
kyungdoo lee on 26 Oct 2021
Commented: Star Strider on 26 Oct 2021
i want to apply pso to function y = -0.0429x4 + 1.1527x3 - 11.178x2 + 43.153x + 19.63

Accepted Answer

Star Strider
Star Strider on 26 Oct 2021
The parameters have to be elements of a vector (and the appropriate operators must be provided, in this instance multiplication) —
y = @(x) -0.0429*x(4) + 1.1527*x(3) - 11.178*x(2) + 43.153*x(1) + 19.63
If the posted equation is actually something different, (such that ‘x4’ is actually ) those details — and a clear explanation of the actual problem — need to be provided. See the documentation on Anonymous Functions for details on them.
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  2 Comments
kyungdoo lee
kyungdoo lee on 26 Oct 2021
Are you saying the question has errors like vector parameters?
Star Strider
Star Strider on 26 Oct 2021
If the parameters to be optimised are ‘x’, they must be elements of a vector.
That is how the optimisation functions work.
Also, it is necessary to be clear about the objective of the optimisation.
For example —
y = @(x) -0.0429*x(4) + 1.1527*x(3) - 11.178*x(2) + 43.153*x(1) + 19.63;
x0 = rand(4,1);
[xe1,fval] = fminsearch(y, x0) % Absolute Minimum
Exiting: Maximum number of function evaluations has been exceeded - increase MaxFunEvals option. Current function value: -4553455067893434311976313508498416664992637329080320.000000
xe1 = 4×1
1.0e+49 * -8.3765 8.4337 0.3630 0.4905
fval = -4.5535e+51
fprintf('%23.15E\n',xe1)
-8.376509228679389E+49 8.433651206800148E+49 3.629666932350592E+48 4.905026065674677E+48
[xe2,fval] = fminsearch(@(x)norm(y(x)), x0) % Mean Square Minimum
xe2 = 4×1
-0.1648 1.1516 0.3351 0.7555
fval = 8.3061e-05
fprintf('%23.15E\n',xe2)
-1.647817685875986E-01 1.151648047524384E+00 3.351060012819946E-01 7.554828255085815E-01
Experiment to get the desired result.
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