MATLAB: Hi all; I try to use fmincon in order to find the optimal control force F for an active suspension system, every time I run fmincon, F and the objective function did not change. Kindly, can any on explain to me why I get this problem. Thanks

fmincon

Where the system is a dynamic system and it is subjected to the equation of motion. The objective function is determined based on Lagrange integral parameterfun(F).
Fo=0;
lb=0;
ub=3000;  
options = optimset('Display','iter','TolFun',1e-8,'algorithm','sqp');
[Fopt,fval,exitflag,output] = fmincon(@(F) parameterfun(F),Fo,...
                                 [],[],[],[],lb,ub,[],options);
where parameterfun(F) is the objective function which calculate the value of F
    function [val,F] = parameterfun(F)
    val = 0;
    [xu,t,z0dot]=xu_det(F);
    xu_frame=xu;
    input=z0dot;
    r1 = 1e+5;
    r2 = 0.5;
    q = 1e-5;
    R = diag([r1, 0, 0, 0, q, r2]);
    xu_frame_zs = zeros(6,length(t));
    for i=0:0.005:(t-0.005)
        dx = f(t(i),xu_frame(1:4,i), xu_frame(5,i),input(i));
        xu_frame_zs(:,i) = [xu_frame(:,i); dx(4)];
        val = val+ (t(i+0.005)-t(i))*xu_frame_zs(:,i)'*R*xu_frame_zs(:,i);
    end
    end
where xu_det(F) is a function which is:
      function [xu,t,z0dot] = xu_det(F)
      param.ms = 325;               % 1/4 sprung mass (kg)

      param.mus = 65;               % 1/4 unsprung mass (kg)

      param.kus = 232.5e3;          % tire stiffness (N/m)

      ks=2.36e4;
      cs=1.03e3;
      Aqcar = [0 1 0 0;-param.kus/param.mus -cs/param.mus ks/param.mus cs/param.mus;0 -1 0 1;0 cs/param.ms -ks/param.ms -cs/param.ms];
      Bqcar = [0 -1;-1/param.mus 0; 0 0; 1/param.ms 0];
      Cqcar = [1 0 0 0; 0 1 0 0; 0 0 1 0; 0 0 0 1];
      Dqcar = 0;
      qcar = ss(Aqcar,Bqcar,Cqcar,Dqcar);
      x0 = [0 0 0 0]';                        % initial state
      load IRI_737b                           % road profile data
      ddx = road_x(2) - road_x(1);             % spacial step for input data
      v = 10;                 % vehicle velocity (m/s)
      dt = 0.005;
      dt2 = ddx/v;                             % time step for input data
      z0dot = [0 diff(road_z)/dt];            % road profile velocity     
      tmax = 5;                               % simulation time length
      t = 0:dt:tmax;
      x = v*t;                 % time/space steps to record output
      z0dot_f=zeros(length(z0dot),2);
      z0dot_f(:,1)=z0dot;
      z0dot_f(:,2)=F;
      u = interp1(road_x,z0dot_f,x);%umf = 3;
      umf = 0.01;
   % Simulate quarter car model
      y = lsim(qcar,u*umf,t,x0);   
      yu=y;
      yu(:,5)=F;
      xu=yu';
    end
and f() is another function is used to determine the states variables derivative dx at each F value
function dx = f(~, x,F,z0dot )
J = jacobian();
A = J(1:4,1:4);
B = J(1:4, 5);
b_ramp = [-1;0; 0; 0];
    dx= A* x + B * F + b_ramp * z0dot; 
end
and jacobian is another function is used to determine A and B matrices
function J = jacobian()
param.ms = 325;               % 1/4 sprung mass (kg)
param.mus = 65;               % 1/4 unsprung mass (kg)
param.kus = 232.5e3;          % tire stiffness (N/m)
ks=2.36e4;
cs=1.03e3;
A = [ 0 1 0 0;
    -param.kus/param.mus -cs/param.mus ks/param.mus cs/param.mus;
    0 -1 0 1;
    0 cs/param.ms -ks/param.ms -cs/param.ms];
B = [0 -1/param.mus 0 1/param.ms]'; 
J = [A, B];
end
when I run fmincon the results as:
                                                           Norm of First-order
   Iter F-count            f(x) Feasibility  Steplength        step  optimality
      0       2   0.000000e+000  0.000e+000                          1.000e+000
      1       4   0.000000e+000  0.000e+000  1.000e+000  0.000e+000  0.000e+000
Local minimum found that satisfies the constraints.
Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the selected value of the function tolerance,
and constraints are satisfied to within the default value of the constraint tolerance.
<stopping criteria details>

Best Answer

Did you try plotting your objective function to make sure that it is not constant near the initial point? Are you sure that you calculated the gradient of your function correctly? If so, then tell fmincon that you are supplying the gradient by setting
options.GradObj = 'on';
If you are not supplying a gradient, then you might need larger finite difference steps sizes, as described in the documentation on optimizing simulations.
Good luck,
Alan Weiss
MATLAB mathematical toolbox documentation
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