Non Linear Constraint Optimization Problem Error

Question

0 个投票

Hello,

I am getting the same error as posted in the thread http://in.mathworks.com/matlabcentral/newsreader/view_thread/326736, i.e., " User supplied objective function must return a scalar value". However unlike the case stated in the thread, objective function i use returns a scalar value. I am pasting my objective function and the constraint function for your reference.

Objective Function *

function f = objfun(x)
global It
global Li
f=It*sqrt(Li/x(2))*(exp(-(sqrt(Li/((4*x(1)*x(1)*x(2))-Li)))*atan(sqrt(((4*x(1)*x(1)*x(2))-Li)/Li))));

Non Linear Constraint Function *

function [c,ceq] = confun(x)
global Li
global tmax
c=[1-(4*x(1)*x(1)*x(2)/Li);((pi - atan(sqrt(((4*x(1)*x(1)*x(2))-Li)/Li)))*2*x(1)*x(2))/(sqrt(((4*x(1)*x(1)*x(2))-Li)/Li)) + 2*x(1)*x(2) - tmax ];
ceq=[];

Minimization *

    x0=[0.1;1e-6];
    Li=1e-6;
    tmax=50e-6;
    It=1000;
    options=optimset('Algorithm','interior-point');
    [x,fval] = fmincon(@objfun,x0,[],[],[],[],[],[],@confun,options)

The objective function is returning a scalar value. But why such an error message comes is perplexing. Any suggestion is highly appreciated.

Thanks and Regards, Lekshman

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Follow Question

Answer 1

Torsten 2015-7-24

1 个投票

4*x(1)*x(1)*x(2))-Li < 0 at the start, thus taking the "sqrt" will result in a complex number.

Best wishes

Torsten.

3 个评论
显示 1更早的评论隐藏 1更早的评论

Matt J 2015-7-24

This ensures that sqrt((4*x(1)*x(1)*x(2) - Li)/Li) to be real.

At the solution, but not at the iterations. Here, it didn't seem to matter. The algorithm located a feasible point early on, but I think that might just have been lucky. It is good practice in your code to check for complex values and return Inf at those points.

JC 2015-7-27

Thanks Matt for clarifying that.

请先登录，再进行评论。

Answer 2

Matt J 2015-7-24

编辑：Matt J 2015-7-24

1 个投票

My suggestion would be to accept that there is an unintended event somewhere that is causing f to end up non-scalar. You should get familiar with MATLAB debugging tools for trapping this event. DBSTOP, for example, will stop the code when the error triggers, allowing you to inspect f. You can also use Conditional Breakpoints to force the code to stop specifically when f is non-scalar.

In any case, I cannot reproduce the error when I run your code. For me, fmincon completes with no errors, though of course I had to run without the "options" argument, which you did not provide. My guess would be that the global variables It and Li are not scalars as you suppose, or get changed by other code behind your back. That is why global variables are discouraged in favor of other ways to pass extra parameters, see

http://www.mathworks.com/help/optim/ug/passing-extra-parameters.html

4 个评论
显示 2更早的评论隐藏 2更早的评论

JC 2015-7-27

编辑：Matt J 2015-7-27

在 MATLAB Online 中打开

Thanks Matt for the suggestions. I followed the link you had shared and rewrote the function to take in parameter. I am posting the altered objective function and constraint functions below.

Objective Function

function f = parameterobjfun1(x,It,Li)
if ((4*x(1)*x(1)*x(2))-Li) < 0
  f=1e6; 
else
  f=It*sqrt(Li/x(2))*(exp(-(sqrt(Li/((4*x(1)*x(1)*x(2))-Li)))*atan(sqrt(((4*x(1)*x(1)*x(2))-Li)/Li))));
end
end

When (4*x(1)*x(1)*x(2))-Li) is < 0, I have set the function to default high value because if i return Inf for the complex value , the objective function reduces to zero

Constraint Function

function [c,ceq] = parameterconfun1(x,Li,tmax)
if ((4*x(1)*x(1)*x(2))-Li) < 0
  c=[1-(4*x(1)*x(1)*x(2)/Li); 2*x(1)*x(2) - tmax ];
else
  c=[1-(4*x(1)*x(1)*x(2)/Li);((pi - atan(sqrt(((4*x(1)*x(1)*x(2))-Li)/Li)))*2*x(1)*x(2))/(sqrt(((4*x(1)*x(1)*x(2))-Li)/Li)) + 2*x(1)*x(2) - tmax ];
end
ceq=[];
end

With

     options=optimset('Algorithm','sqp');
    f1=@(x)parameterobjfun1(x,It,Li);
    f2=@(x)parameterconfun1(x,Li,tmax);
    [x,fval] = fmincon(f1,x0,[],[],[],[],[0;0],[10;30e-6],f2,options)

I get the following message on the command window

    Local minimum found that satisfies the constraints.
  Optimization completed because the objective function is non-decreasing in 
  feasible directions, to within the default value of the function tolerance,
  and constraints are satisfied to within the default value of the constraint tolerance.
  <stopping criteria details>
  x =
      0.2266
      0.0000
  fval =
    431.8723

This however is not the minimum of the function. What can be done to fix this?

And Is it not possible to accept more than one answer for a thread? - Thanks, Lekshman

Walter Roberson 2015-7-27

fmincon() is not a global optimizer. You need a tool from the Global Optimization Toolbox to search for global minima.

JC 2015-7-27

编辑：JC 2015-7-27

Hi Torsten,

I had updated It value but forgot to invoke f1=@(x)parameterobjfun1(x,It,Li) after that. So was getting a different fval corresponding to x = [0.2266;0.00002]. The minimum corresponding to x = [0.2266;0.00002] is in fact 863.745. My mistake, sorry. Thanks for all the suggestions.

Hi Walter,

Yes the minima value is seen to be varying based on the initial guess.

Thanks and Regards,

Lekshman

请先登录，再进行评论。

Non Linear Constraint Optimization Problem Error

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

3 个评论
显示 1更早的评论隐藏 1更早的评论

更多回答（1 个）

4 个评论
显示 2更早的评论隐藏 2更早的评论

类别

标签

Community Treasure Hunt

Non Linear Constraint Optimization Problem Error

0 个评论 显示 -2更早的评论 隐藏 -2更早的评论

采纳的回答

3 个评论 显示 1更早的评论 隐藏 1更早的评论

更多回答（1 个）

4 个评论 显示 2更早的评论 隐藏 2更早的评论

类别

标签

另请参阅

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

3 个评论
显示 1更早的评论隐藏 1更早的评论

4 个评论
显示 2更早的评论隐藏 2更早的评论