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Optimisation of cost function

Asked by SARNENDU JATI on 14 Jun 2018
Latest activity Commented on by SARNENDU JATI on 15 Jun 2018

I'm trying to optimise the cost function for a simple pendulum on a cart. I have written the following code that returns the value of J(cost function) fobs2.m

 function [J,u_candidate_fnc] =fobs2(lambda)
 sol=void(lambda(1),lambda(2),lambda(3),lambda(4));
 t=0:0.001:4.452;
 u_candidate_fnc=candidate_fnc(t,lambda);
 J=(max(sol.y(1,:))-pi)^2+ abs(max(candidate_fnc)) + abs(max(sol.y(3,:)));

Candidate_fnc.m defines candidate function for acceleration of the cart as the input to the system.

function u=candidate_fnc(t,lambda)
Tf=4.4520;
w=2*pi/Tf;
u1=lambda(1)*sin(w*t);
u2=lambda(2)*sin(2*w*t);
u3=lambda(3)*sin(3*w*t);
u4=lambda(4)*sin(4*w*t);
u5=(-5*lambda(1) -5/2*lambda(2)-5/3*lambda(3)-5/4*lambda(4))*sin(5*w*t);
u=u1+u2+u3+u4+u5;

I used BVP solver to give me better trajectories using initial guesses. I've written the following code to store the new trajectories in sol. void.m:

 function y=void(lambda1,lambda2,lambda3,lambda4)
 lambda=[lambda1 lambda2 lambda3 lambda4];
 t_opt=linspace(0,4452*0.001,4452)';
 solinit = bvpinit(t_opt,@xinit_fcn,lambda);
 sol = bvp4c(@system_odeset_fcn,@bc_function,solinit);
 % disp('Lambda values: ')
 y=sol;

when I run fobs2.m it says, not enough input arguments in sol=void(lambda(1),lambda(2),lambda(3),lambda(4)); and Undefined variable "sol".

  3 Comments

When you run fobs2, what are you passing to fobs2 on the command line?

At one point you had commented

 lambda= [0.1 0.075 0.1 0.25]

but did you then invoke

fobs2(lambda)

or did you just expect fobs2 to find the variable named lambda that you had assigned to?

I invoked fobs2(lambda)

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1 Answer

Answer by Walter Roberson
on 14 Jun 2018
 Accepted Answer

I had to get xinit_fcn from your https://www.mathworks.com/matlabcentral/answers/405238-how-to-resolve-interp1-error

I do not get any error about not enough input arguments when I invoke fobs2(lambda) . I do, however, run into problems with interp1 .

Your xinit_fcn is invoking interp1 with an x1_opt value derived from the t value that is passed into the function. However, bvpinit invokes the function with individual t values, not with the entire list of t values, so x1_opt will be a scalar. You are then trying to interp1(1 x 4452, 1 x 1, 1 x 1) . That fails because the second parameter must be the same length as the first parameter.

If you did have the entire original list of time values passed in, and were calculating x1_opt= -(pi/Tf)*All_t + pi and wanting to interpolate at the particular time, t, that was passed in to the xinit_fcn, then the result would always be the same as -(pi/Tf)*t + pi where t is the individual t. So it is not clear why you do not just have your xinit_fcn return -(pi/Tf)*t + pi .

  5 Comments

It is the same. I just took off the transpose. I plotted it and I'm getting the required result for x1_opt.

adjusted code:

 function xinit = xinit_fcn(t)
 t_opt=linspace(0,4452*0.001,4453);
 Tf=4.453;
 t=0:0.001:4.452;
 x1_opt= -(pi/Tf)*t + pi;
 figure,plot(x1_opt)
x1_opt= -(pi/Tf)*t + pi;
x2_opt=0;
x3_opt=0;
x4_opt=0;
xinit=[interp1(t_opt,x1_opt,t);       %x1*
       interp1(t_opt,x2_opt,t);       %x2*
       interp1(t_opt,x3_opt,t);       %x3*
       interp1(t_opt,x4_opt,t);] ;    %x4*

You have

x2_opt=0;
x3_opt=0;
x4_opt=0;
xinit=[interp1(t_opt,x1_opt,t);       %x1*
       interp1(t_opt,x2_opt,t);       %x2*
       interp1(t_opt,x3_opt,t);       %x3*
       interp1(t_opt,x4_opt,t);] ;    %x4*

Your x2_opt, x3_opt, and x4_opt are each scalar 0, not zeros the same size as t_opt. And with them being constant 0, there is little point doing interpolation: just output 0 without interpolation.

x2_opt=0;
x3_opt=0;
x4_opt=0;
xinit=[interp1(t_opt,x1_opt,t);       %x1*
       x2_opt;       %x2*
       x3_opt;       %x3*
       x4_opt;] ;    %x4*

Yeah, you're right. Thank you.

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