optimization for min and max values of a parameter
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Im trying to optimize a kind of 4 bar mechanism (lenghts to minimize forces) which has to move within predefined min&max values of an angle as lower and upper bounds. l want my optimization to run according to taking those values into account in the objective function and nonlinear constraints where my system has to reach while working.(Defining the angle as a variable with upper/lower bounds leads to result of finding optimums of the angle too). creating a for loop to iterate every single movement is a good idea? how should l define those min max values? its simple version is below,
Angmin = -30;
Angmax = 30;
Ang = Angmin & Angmax;% optimizan process must take both values into account in the same process.
A =[];
b = [];
Aeq =[];
beq= [];
lb = [];
ub = [];
x0= [40,50,60,70];
function Fmin = minconfcn(x)
Fmin = (x(2)/x(1))*sqrt(x(3)^2+x(2)^2)*cos(x(4)+Ang)
end
function [c,ceq] = nlincnstfcn(x)
c(1) = asin(x(4)+Ang) + x(3)^2 - x(1)^2;
c(2) = sqrt((x(2)*asin(Ang))^2- x(2)^2)+x(1);
ceq = [];
end
[x,Fmin] = fmincon(@minconfcn,x0,A,b,Aeq,beq,lb,ub,@nlincnstfcn)
댓글 수: 8
답변 (2개)
Matt J
2022년 2월 28일
편집: Matt J
2022년 2월 28일
l think it should be a kind of passing variable extra parameter but l dont know how to do it.
One way is to use nested functions, as below. See also Passing Extra Parameters - MATLAB & Simulink.
function main()
Angmin = -30;
Angmax = 30;
A =[];
b = [];
Aeq =[];
beq= [];
lb = [];
ub = [];
x0= [40,50,60,70];
[x,Fmin] = fmincon(@minconfcn,x0,A,b,Aeq,beq,lb,ub,@nlincnstfcn)
function Fmin = minconfcn(x)
Fmin = (x(2)/x(1))*sqrt(x(3)^2+x(2)^2)*cos(x(4))
end
function [c,ceq] = nlincnstfcn(x)
c(1) = asin(x(4)+Angmin) + x(3)^2 - x(1)^2;
c(2) = sqrt((x(2)*asin(Angmax))^2- x(2)^2)+x(1);
ceq = [];
end
end
댓글 수: 7
Matt J
2022년 3월 1일
Again, all of the equations I have written are deliberately wrong. Only you can provide the correct equations. My role here was to show how to make other variables like Angmin avaialble to the objective function and constraints.
Torsten
2022년 3월 2일
편집: Torsten
2022년 3월 2일
function main
Angmin = -30*pi/180;
Angmax = 30*pi/180;
A =[];
b = [];
Aeq =[];
beq= [];
lb = [-Inf -Inf -Inf -Inf Angmin];
ub = [Inf Inf Inf Inf Angmax];
x0= [40,50,60,70*pi/180,10*pi/180];
[x,Fmin] = fmincon(@minconfcn,x0,A,b,Aeq,beq,lb,ub,@nlincnstfcn)
end
function Fmin = minconfcn(x)
Fmin = (x(2)/x(1))*sqrt(x(3)^2+x(2)^2)*cos(x(4)+x(5))
end
function [c,ceq] = nlincnstfcn(x)
c(1) = asin(x(4)+x(5)) + x(3)^2 - x(1)^2;
c(2) = sqrt((x(2)*asin(x(5)))^2- x(2)^2)+x(1);
ceq = [];
end
댓글 수: 8
Torsten
2022년 3월 2일
Then please write down what your objective function and the constraints are if you use Argmin and Argmax together in one run of fmincon. My phantasy has come to an end.
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