optimization for min and max values of a parameter

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gigi 2022-2-28

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https://ww2.mathworks.cn/matlabcentral/answers/1659980-optimization-for-min-and-max-values-of-a-parameter

评论： Torsten 2022-3-2

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)

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Matt J 2022-2-28

No you can't think of 61 values as 2 values. Matlab won't see it that way.

gigi 2022-2-28

l editted the description

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回答（2 个）

Matt J 2022-2-28

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1659980-optimization-for-min-and-max-values-of-a-parameter#answer_906195

编辑：Matt J 2022-2-28

在 MATLAB Online 中打开

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

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Matt J 2022-3-1

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.

gigi 2022-3-2

Thank you so much for your helpful suggests and supports, but this method doesnot solve the issue. Any different suggests will be appricated if you have.

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Torsten 2022-3-2

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1659980-optimization-for-min-and-max-values-of-a-parameter#answer_907780

编辑：Torsten 2022-3-2

在 MATLAB Online 中打开

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

8 个评论
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gigi 2022-3-2

Not separately, together...

Torsten 2022-3-2

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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