solve function with two variables by ga toolbox

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nadia nadi
nadia nadi 2023-5-21
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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采纳的回答

Walter Roberson
Walter Roberson 2023-5-21
Ran in:
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

更多回答（1 个）

Sulaymon Eshkabilov
Ran in:
Here is how it can be solved with GA:
lb = [0.001, 0.01];
ub = [0.045, 0.1];
numberOfVariables = 2;
y=@(x) 0.25-4.27*x(1)+0.61*x(2)+13.34*x(1)*x(2)-4.69*x(2).^2;
options = optimoptions('ga','PlotFcn', @gaplotbestf);
[x,fval] = ga(y,numberOfVariables,[],[],[],[],lb,ub, [], options)
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
x = 1×2
0.0450 0.0100
fval = 0.0695

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