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Can't run an equation in a function that works in a script

Daniel Harper さんによって質問されました 2019 年 2 月 18 日
最新アクティビティ Star Strider
さんによって コメントされました 2019 年 2 月 20 日
Hi,
I've used the following code for an equation in MATLAB and it works perfectly well in a script (cp1 and u1 have already been defined):
syms u4
eqn3 = cp1 == (u1+u4)*(u1.^2-u4.^2)/(2*u1.^3);
U4= double(solve(eqn3,u4));
u4 = real(U4( U4 > 0 ));
However, when I've tried to implement it into a function it doesn't work.
Can anyone help me with this please?

  2 件のコメント

madhan ravi
2019 年 2 月 18 日
show us your function
function [y] = test(x)
c1 = 0.5176;
c2 = 116;
c3 = 0.4;
c4 = 5;
c5 = 21;
c6 = 0.0068;
% equation 1 lamdai1
z1 = 1/(x(2) + 0.08*x(1)) - 0.035/((x(1).^3) +1);
y(1)=1/z1;
%equation 2 cp1
y(2) = c1*((c2/y(1))-c3*x(1)-c4)*exp(-c5/y(1))+c6*x(2);
%equation 3 u4 (z3)
syms y(3)
eqn3 = y(2) == (u1+y(3))*(u1.^2-y(3).^2)/(2*u1.^3);
Z3 = double(solve(eqn3,y(3)));
z3 = real(Z3( Z3>0 ));
y(3)=z3
%equation 4
y(4) = 0.5*(u1+y(3));
end
I'm using a genetic algorithm for inputs x(1) and x(2). cp1 and u4 etc. have been equated to y(1) etc.

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1 件の回答

回答者: Star Strider
2019 年 2 月 18 日
 採用された回答

Try this:
function [y] = test(x)
c1 = 0.5176;
c2 = 116;
c3 = 0.4;
c4 = 5;
c5 = 21;
c6 = 0.0068;
% equation 1 lamdai1
z1 = 1/(x(2) + 0.08*x(1)) - 0.035/((x(1).^3) +1);
y = sym('y', [1 4])
y(1)=1/z1;
%equation 2 cp1
y(2) = c1*((c2/y(1))-c3*x(1)-c4)*exp(-c5/y(1))+c6*x(2);
%equation 3 u4 (z3)
u1 = 3; % <— PROVIDE CORRECT VALUE
eqn3 = y(2) - ((u1+y(3))*(u1.^2-y(3).^2)/(2*u1.^3));
eqn3 = vpa(eqn3);
U4p = sym2poly(eqn3);
Z3 = roots(U4p);
z3 = real(Z3( Z3>0 ));
y(3)=z3;
%equation 4
y(4) = vpa(0.5*(u1+y(3)));
y = double(y);
end
It runs without error for me. You must determine if it produces the correct result.
Note that this cannot be your fitness function, since a fitness function is required to produce a scalar output.

  8 件のコメント

Star Strider
2019 年 2 月 20 日
Maximize it with respect to what?
The ‘y’ variable is a (1 x 12) vector that is not itself a function of any other variables. What do you intend by ‘maximizing’ it? You can get the largest value (and its index) with the max (link) function.
Star Strider
2019 年 2 月 20 日
Daniel Harper’s ‘Answer’ moved here:
Maximize it in terms of the inputs x. Like how do I maximize the fitness fuction rather than minimize it? I've seen you can do y = -x for the fuction however that doesn't seem to work
Star Strider
2019 年 2 月 20 日
If ‘test.m’ is your entire fitness function, it is not going to work. Fitness functions require scalar outputs (at least in my experience), and ‘y’ is a (1 x 12) vector.
Negating the fitness function output is the generally accepted way to maximimize a fitness (or other objective) function. That obviously works if there is a distinct (and preferably bounded) maximum. If your fitness function is for example a function that increases without bound in any direction, there is not going to be a distinct maximum.
So first, plot the function you want to optimize, if possible. (It initially appeared to be a function of two parameters, so that could work.) That will give you some idea of what it’s doing, and if finding a distinct maximum is even possible.
Meanwhile, I have gotten yuouir function to run, and run relatively efficiently. That was the initial objective.

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