Genetic Algorithm: Find best input combinations that will give global maximum of the output of a non-linear ODE mathematical model using Genetic algorithm

Question

Darshna Joshi 2022-1-12

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1627490-genetic-algorithm-find-best-input-combinations-that-will-give-global-maximum-of-the-output-of-a-no

编辑： Torsten 2024-1-8

在 MATLAB Online 中打开

The problem is to find the optimum(maximum) value of x3 in range of (-8e-4 to 2e-4) by varying kst,x1,x5 and xo)

x5=5 %Input 2 (Input 2 is a state variable and could vary in range of 4 to 15 while performing optimization)

kst=1 %Input 3 (Input 3 is in terms of rate constant, it could vary from 0.1 to 2)

xo=4 %Input 4 (Input 4 is a state variable and could vary in range of 4 to 10)

x1=1e-7 %Input 1 could vary from 1e-9 to 1e-6

%MATLAB CODE%

% Function file for ODE solution

options = odeset('InitialStep',0.0001,'RelTol',1e-09);

[T,Y]=ode15s(@Scrpt1,[0 60],[9e-13,0],options);

X3= Y(:,2);

plot(T,X3)

%Script file

function rest = Scrpt1(t,X)

x2 = X(1);

x3 = X(2);

%Parameters

if t<15

x1 = 1e-7; %Input 1 could vary from 1e-9 to 1e-6

else

x1 = 0;

end

x5=5; %Input 2 (Input 2 is a state variable and could vary in range of 4 to 15 while performing optimization)

kst=1; %Input 3 (Input 3 is in terms of rate constant, it could vary from 0.1 to 2)

xo=4; %Input 4 (Input 4 is a state variable and could vary in range of 4 to 10)

k1 = 6e7;

km1 = 0.20;

km4 = 0.003;

k3 = 2500.00;

k4 = km4/9;

km3 = km1;

LAP=1.5;

%Differential equations

dx2dt = km1*x3 + km3*LAP - k1*x1*x2 + km4*x3 - k4*x2;

dx3dt = k1*x1*x2 - km1*(x3+x5+xo) - k3*x3*kst;

rest = [dx2dt; dx3dt];

end

How to use Genetic Algorithm for this to solve the mentioned optimization problem of finding maximum value of x3. For which values of x5,kst,xo,x1 we get maximum x3?

My objective function is Maximize(x3).

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Torsten 2024-1-8

Maximum X3 after 60 time units ? Maximum value within the complete interval from 0 to 60 time units ? Maximum mean value in the interval from 0 to 60 time units ?

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Answer 1

VINAYAK LUHA 2024-1-8

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1627490-genetic-algorithm-find-best-input-combinations-that-will-give-global-maximum-of-the-output-of-a-no#answer_1385336

编辑：Torsten 2024-1-8

在 MATLAB Online 中打开

Hi Darshna,

I understand that you want to find the maximum value of the variable "x3" using Genetic Algorithm in MATLAB.

Go through the following steps to achieve the above objective.

Define the objective function that the GA will optimize
Set Up the GA Options and Constraints
Run the GA
Retrieve the Optimal Parameters and Maximum Value of x3

Please note that the GA inherently seeks to minimize the objective function. To find the maximum, we'll need to focus on the negative of the objective function. Additionally, you might need to adjust the GA settings or impose extra constraints to achieve optimal outcomes for your particular scenario.

Below is a sample code for implementing the steps mentioned:

% Define the bounds for the variables x1, x5, kst, xo
lb = [1e-9, 4, 0.1, 4]; 
ub = [1e-6, 15, 2, 10]; 
% Set up the GA options
options = optimoptions('ga','PopulationSize', 100,'MaxGenerations', 100,'FunctionTolerance', 1e-6,'Display', 'iter');
% Run the GA
[x_opt, fval] = ga(@objectiveFunction, 4, [], [], [], [], lb, ub, [], options);
Single objective optimization:
4 Variables

Options:
CreationFcn:       @gacreationuniform
CrossoverFcn:      @crossoverscattered
SelectionFcn:      @selectionstochunif
MutationFcn:       @mutationadaptfeasible

                                  Best           Mean      Stall
Generation      Func-count        f(x)           f(x)    Generations
    1              200              -0              -0        0
    2              295              -0              -0        1
    3              390              -0              -0        2
    4              485              -0              -0        3
    5              580              -0              -0        4
    6              675              -0              -0        5
    7              770              -0              -0        6
    8              865              -0              -0        7
    9              960              -0              -0        8
   10             1055              -0              -0        9
   11             1150              -0              -0       10
   12             1245              -0              -0       11
   13             1340              -0              -0       12
   14             1435              -0              -0       13
   15             1530              -0              -0       14
   16             1625              -0              -0       15
   17             1720              -0              -0       16
   18             1815              -0              -0       17
   19             1910              -0              -0       18
   20             2005              -0              -0       19
   21             2100              -0              -0       20
   22             2195              -0              -0       21
   23             2290              -0              -0       22
   24             2385              -0              -0       23
   25             2480              -0              -0       24
   26             2575              -0              -0       25
   27             2670              -0              -0       26
   28             2765              -0              -0       27
   29             2860              -0              -0       28
   30             2955              -0              -0       29

                                  Best           Mean      Stall
Generation      Func-count        f(x)           f(x)    Generations
   31             3050              -0              -0       30
   32             3145              -0              -0       31
   33             3240              -0              -0       32
   34             3335              -0              -0       33
   35             3430              -0              -0       34
   36             3525              -0              -0       35
   37             3620              -0              -0       36
   38             3715              -0              -0       37
   39             3810              -0              -0       38
   40             3905              -0              -0       39
   41             4000              -0              -0       40
   42             4095              -0              -0       41
   43             4190              -0              -0       42
   44             4285              -0              -0       43
   45             4380              -0              -0       44
   46             4475              -0              -0       45
   47             4570              -0              -0       46
   48             4665              -0              -0       47
   49             4760              -0              -0       48
   50             4855              -0              -0       49
   51             4950              -0              -0       50
ga stopped because the average change in the fitness value is less than options.FunctionTolerance.
% Retrieve the optimal parameters
optimal_x1 = x_opt(1);
optimal_x5 = x_opt(2);
optimal_kst = x_opt(3);
optimal_xo = x_opt(4);
% Maximum value of x3
max_x3 = -fval
max_x3 = 0
% Display the results
fprintf('Optimal x1: %g\n', optimal_x1);
Optimal x1: 1e-09
fprintf('Optimal x5: %g\n', optimal_x5);
Optimal x5: 4
fprintf('Optimal kst: %g\n', optimal_kst);
Optimal kst: 0.1
fprintf('Optimal xo: %g\n', optimal_xo);
Optimal xo: 4
fprintf('Maximum value of x3: %g\n', max_x3);
Maximum value of x3: 0
% Objective Function for the Genetic Algorithm
function x3_max = objectiveFunction(params)
    x1 = params(1);
    x5 = params(2);
    kst = params(3);
    xo = params(4);
    
    function rest = Scrpt1(t, X)
        x2 = X(1); 
        x3 = X(2); 
        if t < 15
            x1_val = x1;
        else
            x1_val = 0;
        end
        k1 = 6e7;
        km1 = 0.20;
        km4 = 0.003;
        k3 = 2500.00;
        k4 = km4 / 9;
        km3 = km1;
        LAP = 1.5;
        dx2dt = km1*x3 + km3*LAP - k1*x1_val*x2 + km4*x3 - k4*x2;
        dx3dt = k1*x1_val*x2 - km1*(x3+x5+xo) - k3*x3*kst;
        rest = [dx2dt; dx3dt];
    end
    options = odeset('InitialStep',0.0001,'RelTol',1e-09);
    [T,Y] = ode15s(@(t, X) Scrpt1(t, X), [0 60], [9e-13, 0], options);
    
    x3_max = -max(Y(:,2));
end