genetic algorithm code with more than three variables

I need genetic algorithm code with more than three variables

Answer 1

[Sam Chak]

Hi @huda nawaf

Maybe this example would give you the basics of using the genetic algorithm (GA) to minimize a multivariate function. The problem to find the roots of a Cubic function given by

.

Since the cubic function has no global minima, and the GA only minimizes a given function, then the root-finding problem must be reformulated to become a convex optimization problem.

x = linspace(1, 6, 501);
y1 = x.^3 - 10*x.^2 + 31*x - 30;        % cubic function
y2 = abs(x.^3 - 10*x.^2 + 31*x - 30);   % absolute value of cubic function

subplot(2, 1, 1)                        % 1st subplot
plot(x, y1, 'linewidth', 1.5)
title('Roots of a Cubic function')

subplot(2, 1, 2)                        % 2nd subplot
plot(x, y2, 'linewidth', 1.5)
title('Absolute value of the Cubic function')

To setup the fitness function for GA, you can do as follows:

f     = @(x) abs(x(1).^3 - 10*x(1).^2 + 31*x(1) - 30) + abs(x(2).^3 - 10*x(2).^2 + 31*x(2) - 30) + abs(x(3).^3 - 10*x(3).^2 + 31*x(3) - 30);
nvars = 3;                  % 3 variables
A     = -eye(nvars);        % Constraints A*x <= b to search for solutions on the positive side
b     = zeros(nvars, 1);  
Aeq   = [];
beq   = [];
lb    = [1.0 2.5 4.0];      % bounds setup xlb < x < ub for x(1), x(2), x(3)
ub    = [2.5 4.0 6.0];
rootx = ga(f, nvars, A, b, Aeq, beq, lb, ub)

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

rootx = 1×3

2.0000 3.0000 5.0000

[huda nawaf]

thanks, but need to know if my problem has more than three variables, can use this code?

[Sam Chak]

Hi @huda nawaf

Can, just modify the above multivariate function in your problem and set the number of variables in nvars.

See example below:

fun   = @(x) (x(1) - 2).^2 + (x(2) - 3).^2 + (x(3) - 5).^2 + (x(4) - 7).^2 + (x(5) - 11).^2 + (x(6) - 13).^2;
nvars = 6;                  % 6 variables
A     = -eye(nvars);        % Constraints A*x <= b to search for solutions on the positive side
b     = zeros(nvars, 1);  
x     = ga(fun, nvars, A, b)

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

x = 1×6

2.0001 3.0001 5.0000 7.0006 11.0002 12.9998

If you find this tutorial on using ga() in the MATLAB code helpful, consider accepting ✔ and voting 👍 the Answer. Thanks, @huda nawaf.

[huda nawaf]

my function

f(x) = ((a + 2b + 3c + 4d) - 30).

all var are integers between 0 and 30

when I applied the above code such as

nvars=6;

x=ga(fun,nvars)

this message appears

Fitness function must be a function handle.

thanks in advance

[Sam Chak]

Good. You'll be fine.

1. Follow the format of how the fitness function is entered and
2. identify the number of variables correctly.
3. Also enter your bounds as shown in the example.

Post these 3 things later so that I can check.

[huda nawaf]

thanks for help

fun=@(x)((x(1) + 2*x(2) + 3*x(3) + 4*x(4)) - 30);

>> nvars=4;

>> A = -eye(nvars);

>> b== zeros(nvars, 1);

>> x=ga(fun,nvars,A,b,0,30)

Error using preProcessLinearConstr (line 64)

The number of columns in Aeq must be the same as the

length of X0.

Error in gacommon (line 104)

[Iterate.x,Aineq,bineq,Aeq,beq,lb,ub,msg,exitFlag] = ...

Error in ga (line 319)

[x,fval,exitFlag,output,population,scores,FitnessFcn,nvars,Aineq,bineq,Aeq,beq,lb,ub,

when apply thefunction below without make the constrains, i got:

x=ga(fun,nvars)

Optimization terminated: maximum number of generations exceeded.

x =

0.5246 -16.1315 -21.8449 -18.3718

In fact I do not know what A and b means, but i can guss lb and ub are the constrains (in my case 0-30)

[Sam Chak]

No worries, @huda nawaf. learn slowly.

If the MATLAB code solves your given problem, consider accepting ✔ and voting 👍 the Answer. Thanks!

f     = @(x) (x(1) + 2*x(2) + 3*x(3) + 4*x(4)) - 30;
nvars = 4;                  % 4 variables, x1, x2, x3, x4
A     = -eye(nvars);        % Constraints A*x <= b to force GA to search for solutions on the positive side
b     = zeros(nvars, 1);  
Aeq   = [];
beq   = [];
lb    = [0 0 0 0];          % bounds setup lb < x < ub for x1, x2, x3, x4
ub    = [30 30 30 30];
xsol  = ga(f, nvars, A, b, Aeq, beq, lb, ub)

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

xsol = 1×4

1.0e-04 * 0.6638 0.6049 0.2772 0.1609

[huda nawaf]

hi sam , thanks

the code now is running without problems and i made vote and accept the answer

but still I have query, I think f must equal zero, is not?

my example from a website, I wana to apply the code then apply my real problem.

in website the obj. fun is a + 2b + 3c + 4d=30

then he converted into f=a + 2b + 3c + 4d-30

but i think it is must be zero, why f = -29.99

in same time how know the code is minimize or maximize?

[Sam Chak]

Hi @huda nawaf

To make it easy for you to visualize a multivariate function in n-dimension, I have simplified the function to only two variables, a bivariate function given by

.

If you haven't noticed yet, is linear functional as shown by the flat surface in the 3D plot below:

[X, Y] = meshgrid(0:12/60:6);
Z      = X + 2*Y - 5;
surf(X, Y, Z)

Since the variables are constrained to non-negative values (in this example, from 0 to 6), naturally the minimum of the function must be at , as clearly shown in the plot.

Thus, the Genetic Algorithm will return the solution as close as possible to .

f     = @(x) x(1) + 2*x(2) - 5;
nvars = 2;                  % 2 variables, x1, x2
A     = -eye(nvars);        % Constraints A*x <= b to force GA to search for solutions on the positive side
b     = zeros(nvars, 1);  
Aeq   = [];
beq   = [];
lb    = [0 0];              % bounds setup lb < x < ub for x1, x2
ub    = [6 6];
xsol  = ga(f, nvars, A, b, Aeq, beq, lb, ub)

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

xsol = 1×2

0 0

You are advised to revisit your optimization problem and reformulate the objective function (if relevant, as a Convex function), perhaps together with meaningful constraints as well.

[huda nawaf]

thank u very much