solve function with two variables by ga toolbox

Question

nadia nadi 2023-5-21

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1969494-solve-function-with-two-variables-by-ga-toolbox

评论： nadia nadi 2023-5-21

hello,

I have a function with two variables ang i want to optimize the function by using ga. I used the toolbox long time ago and i forgot how to use it. Can any one help me please.

my function is

y= 0.25-4.27*x(1)+0.61*x(2)+13.34*x(1)*x(2)-4.69*x(2).^2;

and the range of the variables are

lb = [0.001,0.01];
ub = [0.045,0.1];;

I'm trying to use

numberOfVariables = 2;
[x,fval] = ga(FitnessFunction,numberOfVariables,lb,ub)

Thanks

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Answer 1

Walter Roberson 2023-5-21

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1969494-solve-function-with-two-variables-by-ga-toolbox#answer_1240789

在 MATLAB Online 中打开

format long g
FitnessFunction = @(x) 0.25-4.27*x(1)+0.61*x(2)+13.34*x(1)*x(2)-4.69*x(2).^2;
lb = [0.001,0.01];
ub = [0.045,0.1];
numberOfVariables = 2;
A = []; b = [];
Aeq = []; beq = [];
[x,fval] = ga(FitnessFunction, numberOfVariables, A, b, Aeq, beq, lb, ub)
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
x = 1×2
                     0.045                      0.01
fval = 
                  0.069484
%cross-check
[xf, fvalf] = fmincon(FitnessFunction, randn(numberOfVariables, 1), A, b, Aeq, beq, lb, ub)
Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.
xf = 2×1
        0.0449999033023298
        0.0100003582625156
fvalf = 
        0.0694847999986172