How to add a less than constraint condition to solve the transcendental equation

Question

Roy Francis 2018-1-7

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/376008-how-to-add-a-less-than-constraint-condition-to-solve-the-transcendental-equation

评论： Umair Naeem 2018-9-16

I have tried to solve the transcendental equation by Genetic algorithm. The fitness function used for the equations is given below:

function y = myFitnes(x)
y = ((cos(x(1)) + cos(x(2)) + cos(x(3)) + cos(x(4))-0.9*(pi/2)))^2...
   +(cos(3*x(1)) + cos(3*x(2)) + cos(3*x(3)) + cos(3*x(4)))^2...
  +(cos(5*x(1)) + cos(5*x(2)) + cos(5*x(3)) + cos(5*x(4)))^2 ...
  + (cos(7*x(1)) + cos(7*x(2)) + cos(7*x(3)) + cos(7*x(4)))^2;   
end

The main code written to solve the above equation using Genetic algorithm is given below:

objFcn =@myFitnes;
nvars = 4;
LB = [0 0 0 0];
UB = [pi/2 pi/2 pi/2 pi/2];
[x, fval] = ga(objFcn,nvars,[],[],[],[],LB,UB);

How to add the constraint condition 0<x(1)<x(2)<x(3)<x(4)<pi/2; to solve the above equations.

2 个评论
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Matt J 2018-1-7

Please indent your code so that it is fonted more readably (as I have done for you now)

Roy Francis 2018-1-8

Thanks a lot!!

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Answer 1

Matt J 2018-1-7

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https://ww2.mathworks.cn/matlabcentral/answers/376008-how-to-add-a-less-than-constraint-condition-to-solve-the-transcendental-equation#answer_299044

编辑：Matt J 2018-1-7

在 MATLAB Online 中打开

    A=[1,-1,0,0;
       0, 1,-1;0;
       0, 0, 1,-1];
    b=[0;0;0];
   objFcn =@myFitnes;
  nvars = 4;
  LB = [0 0 0 0];
  UB = [pi/2 pi/2 pi/2 pi/2];
  [x, fval] = ga(objFcn,nvars,A,b,[],[],LB,UB);

4 个评论
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Walter Roberson 2018-9-15

[1, -1, 0, 0]*[x1; x2; x3; x4] is x1-x2. Requiring that to be less than b(1)=0 is the condition x1-x2<=0. Add x2 to both sides to get x1<=x2

The A b matrix also encodes x2<=x3 the same way. Transitivity says you can then write x1 <= x2 <= x3 <= x4. The 0 at the beginning is expressed by lb 0. The const at the other end is ub const.

Umair Naeem 2018-9-16

thanks alot

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Answer 2

Walter Roberson 2018-1-7

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/376008-how-to-add-a-less-than-constraint-condition-to-solve-the-transcendental-equation#answer_299045

在 MATLAB Online 中打开

objFcn =@myFitnes;
nvars = 4;
A = [1 -1  0  0;
     0  1 -1  0;
     0  0  1 -1]
b = [0;
     0;
     0];
LB = [0 0 0 0] + realmin;    %disallow 0 exactly
UB = [pi/2 pi/2 pi/2 pi/2] * (1-eps);   %disallow pi/2 exactly
[x, fval] = ga(objFcn, nvars, A, b, [], [], LB, UB);

This implements 0 < x(1) <= x(2) <= x(3) <= x(4) < pi/2 which is not exactly what you had asked for. Coding strict inequalities would require adding the nonlinear constraint function, which is possible in this case (since you do not have any integer constraints), but is not as efficient.