Errors in objective function

Question

Honey Adams 2018-7-12

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/410081-errors-in-objective-function

评论： Honey Adams 2018-7-17

data=[0;
0;
0;
37248;
58649;
85974;
132620;
200698;
266406;
325423;
383975;
443234;
498503;
558777;
612881;
659367;
710580;
749772;
794515;
847288;
910312;
963746;
1016473;
1060359;
1097173;
1135293;
1176867;
1212260;
1239208;
1267657]
function res=  Bass(x,v,data )
%objective function
F=[ -data+ (x(1)*x(2)*(1 -  exp(  x(2) + (( X(3)*x(4))/ x(3)-x(4))*exp(-x(5)*v)))/ ( 1+  (( x(3)*X(4))/ x(3)-x(4))*exp(-x(5)*v)))]
res=sum(F.^2)
%constriant 
function [c ceq]= nonlinc(x,v,data)
c=[];
ceq=[ -data +( x(1)*x(2)*(1 -  exp(  x(2) + (( x(3)*x(4))/ x(3)-x(4))*exp(-x(5)*v)))/ ( 1+  (( x(3)*x(4))/ x(3)-x(4))*exp(-x(5)*v)))]
fun=@Bass;
v=100;
nl=@nonlinc;
lb=[20,0,0,0,0]
ub=[inf,1,1,1,inf]
x0=[ 0,0,0,0,0]
j=fmincon(fun,x0,[],[],[],[],lb,ub,nl)

my aim was to find the parameters of a system on non_linear equations. I used the Res as my objective function and the constraints were the equations. I didn't use fsolve because I had boundaries.data is a vector . I have corrected this several times but results are always : Error using Bass (line 3) Not enough input arguments.

Error in fmincon (line 564) initVals.f = feval(funfcn{3},X,varargin{:});

Error in Untitled2 (line 41) x= fmincon(fun,x0,A,b,Aeq,beq,lb,ub);

Caused by: Failure in initial user-supplied objective function evaluation. FMINCON cannot continue.

>> I would be glad for a little assistance.Thank you :)

6 个评论
显示 4更早的评论隐藏 4更早的评论

Honey Adams 2018-7-13

I found the error in the original equation.its is quite complex

P*m*((1-e^(-(p+(q_m q_0)/(q_(0+) (q_m-q_0)e^(-vw) ))*t)))/(P+ ((q_m q_0)/(q_(0+) (q_m-q_0)e^(-vw) ))*e^(-(P+ ((q_m q_0)/(q_(0+) (q_m-q_0 ) e^(-vw) ))*t) )

I then tried to define the function using the general equation. When I attempted to test my function I was told Error using exp Not enough input arguments.

Error in Bass (line 5) F=[-data + (x(1)*x(2)*(1- exp*(-x(1)- ((x(3)*x(4))/(x(4)+(x(3)-x(4))*exp(-x(5)*v))))*t)/...

Honey Adams 2018-7-13

t was omitted in the first case and it is a vector from 1 to 30.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Torsten 2018-7-13

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/410081-errors-in-objective-function#answer_328700

在 MATLAB Online 中打开

function main
xdata=1:30;
ydata=[0,0,0,37248,58649,85974,132620,200698,266406,325423,383975,443234,498503,558777,612881,659367,710580,749772,794515,847288,910312,963746,1016473,1060359,1097173,1135293,1176867,1212260,1239208,1267657];
v=100;
fun=@(x,xdata)Bass(x,xdata,v);
lb=[20,0,0,0,0]
ub=[inf,1,1,1,inf]
x0=[0,0,0,0,0]
j=lsqcurvefit(fun,x0,xdata,ydata,lb,ub)
end
%objective function
function F = Bass(x,xdata,v)  
  F=x(1)*x(2)*(1-exp(-(x(1)+x(3)*x(4)/(x(4)+(x(3)-x(4))*exp(-x(5)*v)))*xdata))./...
  (x(1)+x(3)*x(4)/(x(4)+(x(3)-x(4))*exp(-x(5)*v))*exp(-(x(1)+x(3)*x(4)/(x(4)+(x(3)-x(4))*exp(-x(5)*v)))*xdata));
end

You can't specify every single equation as a constraint. Minimizing the objective function does the job.

Best wishes

Torsten.

32 个评论
显示 30更早的评论隐藏 30更早的评论

Honey Adams 2018-7-13

编辑：Matt J 2018-7-14

在 MATLAB Online 中打开

function main
t=1:30;
ydata=[0,0,0,37248,58649,85974,132620,200698,266406,325423,383975,443234,498503,558777,612881,659367,710580,749772,794515,847288,910312,963746,1016473,1060359,1097173,1135293,1176867,1212260,1239208,1267657];
v=1000;
fun=@(x,t)Bassmodel(x,t,v);
lb=[20,0,0,0,0];
ub=[inf,1,1,1,inf];
x0=[20,0,0,0,300];
x = lsqcurvefit(fun,x0,t,ydata,lb,ub)
end
%objective function
function F = Bassmodel(x,t,v)
F=[(x(1)*x(2)*((1-exp(-x(1)-((x(3)*x(4))/(x(4)+(x(3)-x(4))*exp(-x(5)*v))))*t)/...
      (x(1) + ((x(3)*x(4)) / (x(4)+((x(3)-x(4))*exp(-x(5)*v)))) * exp(-x(1) - ((x(3)*x(4)) /( x(4)+(x(3)-x(4))*exp(-x(5)*v))))*t)))]
end
Dear Torsten,
I corrected my objective function and tested it.it works fine but when I try to run the lsqcurvefit this is what I get
Warning: Rank deficient, rank = 0, tol = NaN. 
> In Bassmodel at 4
  In main>@(x,t)Bassmodel(x,t,v) at 5
  In lsqcurvefit at 198
  In main at 9 
F =
       0
Error using lsqcurvefit (line 247)
Function value and YDATA sizes are not equal.
Error in main (line 9)
x = lsqcurvefit(fun,x0,t,ydata,lb,ub)

Stephan 2018-7-16

编辑：Stephan 2018-7-16

在 MATLAB Online 中打开

Hi again,

see this "improved" version of the script:

%lsqcurvefit
t=1:30;
ydata=[0,0,0,37248,58649,85974,132620,200698,266406,325423,383975,443234,498503,558777,612881,659367,710580,749772,794515,847288,910312,963746,1016473,1060359,1097173,1135293,1176867,1212260,1239208,1267657];
v=1000;
fun=@(x,t)Bassmodel(x,t,v);
lb=[0,0,0,0,0];
ub=[1, Inf, 1, 1, Inf];
x0=[20,1,1,1,300];
options = optimoptions(@lsqcurvefit,'Display','off');
x = lsqcurvefit(fun,x0,t,ydata,lb,ub,options);
y_val = Bassmodel(x,t,v);
% lsqnonlin
t=1:30;
ydata=[0,0,0,37248,58649,85974,132620,200698,266406,325423,383975,443234,498503,558777,612881,659367,710580,749772,794515,847288,910312,963746,1016473,1060359,1097173,1135293,1176867,1212260,1239208,1267657];
v=1000;
lb=[0,0,0,0,0];
ub=[1, Inf, 1, 1, Inf];
x0=x;
fun = @(x)ydata-(x(1)*x(2)*(1-exp(-(x(1)+x(3)*x(4)/(x(4)+(x(3)-x(4))*exp(-x(5)*v))).*t))./...
  (x(1)+x(3)*x(4)/(x(4)+(x(3)-x(4))*exp(-x(5)*v))*exp(-(x(1)+x(3)*x(4)/(x(4)+(x(3)-x(4))*exp(-x(5)*v))).*t)));
options2 = optimoptions(@lsqnonlin,'Display','off');
x2 = lsqnonlin(fun,x0,lb,ub,options2);
y_val2 = Bassmodel(x2,t,v);
% plot results
plot(t,ydata,'x')
hold on
plot(t,y_val,'o')
hold on
plot(t,y_val2,'*')
hold off
%show results
disp(['x2(1): ', num2str(x2(1))])
disp(['x2(2): ', num2str(x2(2))])
disp(['x2(3): ', num2str(x2(3))])
disp(['x2(4): ', num2str(x2(4))])
disp(['x2(5): ', num2str(x2(5))])
%objective function for lsqcurvefit
function [F] = Bassmodel(x,t,v)
F=x(1)*x(2)*(1-exp(-(x(1)+x(3)*x(4)/(x(4)+(x(3)-x(4))*exp(-x(5)*v))).*t))./...
  (x(1)+x(3)*x(4)/(x(4)+(x(3)-x(4))*exp(-x(5)*v))*exp(-(x(1)+x(3)*x(4)/(x(4)+(x(3)-x(4))*exp(-x(5)*v))).*t));
end

What does it do? At first in calculates the same what you already got. Result is vector x. Then it uses lsqnonlin to calcualte x again as x2, but as x0 input for this we use the result x from the first run. Then it plots all 3 versions (original, lsqcurvefit & lsqnonlin) into one graph and writes the new x-values (x2) into the command window.

Note: Using lsqnonlin is equivalent to lsqcurvefit - the small differences are beause of the start Point x0. If you start both with the same x0 you will get the exactly same results. The same result would be possible to reach by running lsqcurvefit in a loop with updating x0 from x of the run before)

More interesting is what you get, when you look to the results for x(1)...x(5):

x2(1): 0.01122
x2(2): 1382939.8173
x2(3): 0.15215
x2(4): 1
x2(5): 300

You have no chance to get x(2) between [0...1] without getting a very poor result. But x(1), x(3) & x(4) are in the interval of [0...1]

--> Is it possible that by writing the equation to matlab code you confused x(1) with x(2)??? This would explain the problem.

I recommend you check this.

Best regards

Stephan

Stephan 2018-7-16

编辑：Stephan 2018-7-16

在 MATLAB Online 中打开

Hi,

just to check this again - you have:

P = x(1);
m = x(2);
q_m = x(3);
q_0 = x(4);
w = x(5);

with m, q_m and q_0 between [0...1]

???

I rewrote your objective function a Little bit:

function [F] = Bassmodel(x,t,v)
P = x(1);
m = x(2);
q_m = x(3);
q_0 = x(4);
w = x(5);
F = -(P.*m.*(exp(-t.*(P+(q_0.*q_m)./(q_0-exp(-v.*w).*(q_0-q_m))))-1.0))./(P+(q_0.*q_m.*exp(-t.*(P+(q_0.*q_m)./(q_0-exp(-v.*w).*(q_0-q_m)))))./(q_0-exp(-v.*w).*(q_0-q_m)));
end

I tested this function in the script and i get the same results we already know- so it works in principal. Now it is a little clearer which x value means which function parameter. With this setting you get the solution for x already mentioned above, which does not meet your expectations.

If you should find out, that it is P which should be x(2) and not m, then copy that code into your script and change the assignment from your Parameters to x(i).

I attached the way i "build" this function for transparency.

Best regards

Stephan

Stephan 2018-7-17

It was a kind of fun to help. I will now unfollow this question, since you seem to be happy. If you have further problems, come back with a new question.

Best regards

Stephan

Honey Adams 2018-7-17

Yes, thank you.

请先登录，再进行评论。

Answer 2

Stephan 2018-7-13

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/410081-errors-in-objective-function#answer_328699

在 MATLAB Online 中打开

Hi,

in addition to the issues that Torsten noted in his comments, note the following errors:

you have X(i) instead of x(i) in your objective function, which will cause problems. I found 2 locations in your Bass function where this error appears.
your functions should end with an end - this will also cause errors, when executing the code
functions in a script have to be at the end of the script - this will also cause Errors, when executing the code
the problem regarding the input arguments is described here. The way you do it, will not pass the required arguments to the function calls of nonlinc and Bass, when fmincon calls the functions to evaluate the actual values of vector x in every iteration. In the result you will get the error not enough input arguments. To resolve this you need a kind of outer function, like described in the link above.

Here is a corrected version that considered all the issues that i could find:

data=[0;
0;
0;
37248;
58649;
85974;
132620;
200698;
266406;
325423;
383975;
443234;
498503;
558777;
612881;
659367;
710580;
749772;
794515;
847288;
910312;
963746;
1016473;
1060359;
1097173;
1135293;
1176867;
1212260;
1239208;
1267657];
v=100;
x = outer_function(v, data);
disp(x)
function j = outer_function(v, data)
fun=@Bass;
nl=@nonlinc;
lb=[20,0,0,0,0];
ub=[inf,1,1,1,inf];
x0=[ 0,0,0,0,0];
j=fmincon(fun,x0,[],[],[],[],lb,ub,nl);
function res=  Bass(x)
%objective function
F= (- data + x(1)*x(2)*(1 -  exp(  x(2) + (( x(3)*x(4))/ x(3)-x(4))*exp(-x(5)*v)))/ ( 1+  (( x(3)*x(4))/ x(3)-x(4))*exp(-x(5)*v)));
res=sum(F.^2);
end
function [c, ceq]= nonlinc(x)
%constraint 
c=[];
ceq=- data + ( x(1)*x(2)*(1 -  exp(  x(2) + (( x(3)*x(4))/ x(3)-x(4))*exp(-x(5)*v)))/ ( 1+  (( x(3)*x(4))/ x(3)-x(4))*exp(-x(5)*v)));
end
end

If you run this script, it will work, but converges to an infeasible point, which may be because of what Torsten noted in his commentary:

Converged to an infeasible point.
fmincon stopped because the size of the current step is less than
the default value of the step size tolerance but constraints are not
satisfied to within the default value of the constraint tolerance.
<stopping criteria details>
     28.4354    0.0000    0.5313    0.5313    0.4404

So what you have to do is check what Torsten told you.

Best regards

Stephan

3 个评论
显示 1更早的评论隐藏 1更早的评论

Honey Adams 2018-7-13

编辑：Honey Adams 2018-7-13

Great. I understand now how the fmincon works even tough from Torsten, I have come to realize that using the fmincon isn't the right optimizer to use for my problem. Thank you for bearing with me and using part of ur precious time to go through the code and point out all the errors. I am grateful. I accepted Tortsen answers since I will be using the lsqcurvefit. But I gained a lot of insight from your short tutorial.