optimization problem wont find any solutions

I have the following optimization problem

function [ Fx1, Fx2, Fx3, Fx4, alpha1, alpha2, alpha3, alpha4] = fcn(Fx_v, Fy_v, Mz_v, P_max_1, P_max_2, P_max_3, P_max_4, M_reg_1, M_reg_2, M_reg_3, M_reg_4, k, delta_T_aA, SW_1, SW_2, SW_3, SW_4, w_1, w_4, w_3, w_2, d_delay_1, d_delay_2, d_delay_3, d_delay_4, r_dyn_1, r_dyn_2, r_dyn_3, SOC_1, SOC_2, r_dyn_4, SOC_3, SOC_4, f_lag, n, j,w_vr, w_vl, w_hr, w_hl, l_h, l_v, c_w,d_min_vl, d_max_vl, d_min_vr, d_max_vr, d_min_hl, d_max_hl, d_min_hr, d_max_hr, A, B, FxB_max, SOC_krit)
fun =@(x)  (f_lag >= 0).*((A/SOC_1)*(x(1))^2+(A/SOC_2)*(x(2))^2+(A/SOC_3)*(x(3))^2+(A/SOC_4)*(x(4))^2) + (f_lag == -1).*((((B*(M_reg_1/(M_reg_1+M_reg_2+M_reg_3+M_reg_4)))/SOC_1)*(1/(x(1)^2)))+(((B*(M_reg_2/(M_reg_1+M_reg_2+M_reg_3+M_reg_4)))/SOC_2)*(1/(x(2)^2)))+(((B*(M_reg_3/(M_reg_1+M_reg_2+M_reg_3+M_reg_4))/SOC_3)*(1/(x(3)^2)))+(((B*(M_reg_4/(M_reg_1+M_reg_2+M_reg_3+M_reg_4)))/SOC_4)*(1/(x(4)^2)))));
x0 = [(Fx_v/4)*cos(SW_1),(Fx_v/4)*cos(SW_2),(Fx_v/4)*cos(SW_3),(Fx_v/4)*cos(SW_4),(d_delay_1-SW_1),(d_delay_2-SW_2),(d_delay_3-SW_3),(d_delay_4-SW_4)];
A = [];
b = [];
Aeq = [ 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0];
SOC_dif1=SOC_1-((SOC_1+SOC_2+SOC_3+SOC_4)/4);      %negative when 
SOC_dif2=SOC_2-((SOC_1+SOC_2+SOC_3+SOC_4)/4);
SOC_dif3=SOC_3-((SOC_1+SOC_2+SOC_3+SOC_4)/4);
SOC_dif4=SOC_4-((SOC_1+SOC_2+SOC_3+SOC_4)/4); 
if f_lag>=0
     
    if (SOC_dif1 < SOC_krit)          % state of charge of one actor is really low, actor is only used for steering %during driving mode
                                       %SOC_krit =-0.2(evaluate in tests)
                                                       
        Aeq(2,1) = 1;
    end
    
    if (SOC_dif2 < SOC_krit)          % state of charge of one actor is really low, actor is only used for steering %during driving mode
        Aeq(2,2) = 1;
   end
    
    if (SOC_dif3 < SOC_krit)         % state of charge of one actor is really low, actor is only used for steering %during driving mode
        Aeq(2,3) = 1;
    end
    
    if (SOC_dif4 < SOC_krit)          % state of charge of one actor is really low, actor is only used for steering %during driving mode
        Aeq(2,4) = 1;
    end
end

% jedes eine eigene Linie??
% Erfahrungswerte aus Testsimulation

beq = [ 0, 0]; 


% active stimulation for beta
%if k = 1 there is a stimulation
% k, j and n are Inputs, j and n are the actors stimulated
    if k == 1
       (Aeq(1,j) == 1)  && (Aeq(1,n) == -1);   % target moduls     
    end

    if k == 1
        beq(1) = delta_T_aA;            %target torque
    end
if f_lag == -1                    

%if it is possible the regenerative braking torque is the limit 
% sum of the Force from the regenerative braking torque in vehiclecoordinates is bigger than the demanded force for the whole vehicle
    % M_reg is always positive and Fx_v can be either positive or negative in the braking case 
         if (cos(SW_1)*(M_reg_1/r_dyn_1)+cos(SW_2)*(M_reg_2/r_dyn_2)+cos(SW_3)*(M_reg_3/r_dyn_3)+cos(SW_3)*(M_reg_4/r_dyn_4))> abs(Fx_v)  
         
             % max. regenerative braking torque
             lb = [-M_reg_1/r_dyn_1, -M_reg_2/r_dyn_2, -M_reg_3/r_dyn_3, -M_reg_4/r_dyn_4, d_min_vl-SW_1, d_min_vr-SW_2, d_min_hl-SW_3, d_min_hr-SW_4];    %d and betato the left side are positive, to the right side are negative  %regenerative braking torque must be positive
             ub = [M_reg_1/r_dyn_1, M_reg_2/r_dyn_2, M_reg_3/r_dyn_3, M_reg_4/r_dyn_4, d_max_vl-SW_1, d_max_vr-SW_2, d_max_hl-SW_3, d_max_hr-SW_4];
 
         else
            % mechanical braking
            %FxB_max is positive
            lb = [-FxB_max, -FxB_max, -FxB_max, -FxB_max, d_min_vl-SW_1, d_min_vr-SW_2, d_min_hl-SW_3, d_min_hr-SW_4];     %d and betato the left side are positive, to the right side are negative                                       %FxB_max is negative
            ub = [FxB_max, FxB_max, FxB_max, FxB_max, d_min_vl-SW_1, d_min_vr-SW_2, d_min_hl-SW_3, d_min_hr-SW_4];
         end                                                                         % close conditions for braking loop, max braking torque is positive

 %driving mode
else
    lb = [-P_max_1/(r_dyn_1*w_1), -P_max_2/(r_dyn_2*w_2), -P_max_3/(r_dyn_3*w_3), -P_max_4/(r_dyn_4*w_4), d_min_vl-SW_1, d_min_vr-SW_2, d_min_hl-SW_3, d_min_hr-SW_4];
    ub = [P_max_1/(r_dyn_1*w_1), P_max_2/(r_dyn_2*w_2), P_max_3/(r_dyn_3*w_3), P_max_4/(r_dyn_4*w_4), d_max_vl-SW_1, d_max_vr-SW_2, d_max_hl-SW_3, d_max_hr-SW_4];
end   
nlcon = @nonlnconstraints;

%options = optimoptions(@fmincon, 'MaxFunctionEvaluations',10.000000e+03);

options = optimoptions(@fmincon, 'Algorithm', 'sqp');


x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nlcon(Fx_v,Fy_v,Mz_v,SW_1,SW_2,SW_3,SW_4,w_vr, w_vl, w_hr, w_hl, l_h, l_v, c_w),options);

with these nonlinear constraints

function [c,ceq] = nonlnconstraints(Fx_v,Fy_v,Mz_v,SW_1,SW_2,SW_3,SW_4,c_w,l_v,l_h,w_hl,w_hr,w_vl,w_vr)
%following of the input trajectory
ceq =@(x) [Fx_v-(x(1)*cos(x(5)+SW_1)+x(2)*cos(x(6)+SW_2)+x(3)*cos(x(7)+SW_3)+x(4)*cos(x(8)+SW_4)-x(5)*sin(x(5)+SW_1)*c_w-x(6)*sin(x(6)+SW_2)*c_w-x(7)*sin(x(7)+SW_3)*c_w-x(8)*sin(x(8)+SW_4)*c_w); Fy_v-(x(1)*sin(x(5)+SW_1)+x(2)*sin(x(6)+SW_2)+x(3)*sin(x(7)+SW_3)+x(4)*sin(x(8)+SW_4)-x(5)*cos(x(5)+SW_1)*c_w-x(6)*cos(x(6)+SW_2)*c_w-x(7)*cos(x(7)+SW_3)*c_w-x(8)*cos(x(8)+SW_4)*c_w); Mz_v-(-x(1)*(cos(x(5)+SW_1)*w_vl+l_v*sin(x(5)+SW_1))+x(2)*(cos(x(6)+SW_2)*w_vr+sin(x(6)+SW_2)*l_v)+x(3)*(-cos(x(7)+SW_3)*w_hl-l_h*sin(x(7)+SW_3))+x(4)*(cos(x(8)+SW_4)*w_hr-sin(x(8)+SW_4)*l_h)+x(5)*(sin(x(5)+SW_1)*c_w*w_vl-cos(x(5)+SW_1)*c_w*l_v)+x(6)*(sin(x(6)+SW_2)*c_w*w_vr-cos(x(6)+SW_2)*c_w*l_v)+x(7)*(sin(x(7)+SW_3)*c_w*w_hl+cos(x(7)+SW_3)*c_w*l_h)+x(8)*(cos(x(8)+SW_4)*c_w*l_h-sin(x(8)+SW_4)*c_w*w_hl))];
c = [];
end

But matlab cant find any results for it.

Can anyone tell me whats wrong with it?

Thank you very much

Answer 1

[Matt J]

The problem is infeasible, or your initial guess x0 is too far from a feasible point.

[Sarah Kern]

Thank you very much! Now I changed my problem so its feasible, but it can only find a local solution.

Do you have any idea, how I could get a global solution?

Fx_v = 500;
Fy_v = 50;
Mz_v = 100;
P_max_1 = 10000;
P_max_2 = 10000;
P_max_3 = 10000;
P_max_4 = 10000;
M_reg_1 = 500;
M_reg_2 = 500;
M_reg_3 = 500;
M_reg_4 = 500;
k = 0;
delta_T_aA = 0;
SW_1 = 20;
SW_2 = 20;
SW_3 = 20;
SW_4 = 20;
w_1 = 80;
w_4 = 80;
w_3 = 80;
w_2 = 80;
d_delay_1 = 18;
d_delay_2 = 18;
d_delay_3 = 18;
d_delay_4 = 18;
r_dyn_1 = 0.28;
r_dyn_2 = 0.28;
r_dyn_3 = 0.28;
SOC_1 = 0.5;
SOC_2 = 0.5;
r_dyn_4 = 0.28;
SOC_3 = 0.5;
SOC_4 = 0.5;
f_lag = 1;
n = 0;
j = 0;
w_vr = 0.9;
w_vl = 0.9;
w_hr = 0.9;
w_hl = 0.9;
l_h = 2.8;
l_v = 2.8;
c_w = 0.2;
 d_min_vl = -60;   % in degree [°], positive against the clock
 d_max_vl = 90;
 d_min_vr = -90;
 d_max_vr = 60;
 d_min_hl = -90;
 d_max_hl = 60;
 d_min_hr = -60;
 d_max_hr = 90;
 A = 1;
 B = 1;
 FxB_max = 80000;
 SOC_krit = 0.15;
 C = 10000;
 
fun = @(x) C*((Fx_v-(x(1)*cos(x(5)+SW_1)+x(2)*cos(x(6)+SW_2)+x(3)*cos(x(7)+SW_3)+x(4)*cos(x(8)+SW_4)-x(5)*sin(x(5)+SW_1)*c_w-x(6)*sin(x(6)+SW_2)*c_w-x(7)*sin(x(7)+SW_3)*c_w-x(8)*sin(x(8)+SW_4)*c_w))^2+ (Fy_v-(x(1)*sin(x(5)+SW_1)+x(2)*sin(x(6)+SW_2)+x(3)*sin(x(7)+SW_3)+x(4)*sin(x(8)+SW_4)+x(5)*cos(x(5)+SW_1)*c_w+x(6)*cos(x(6)+SW_2)*c_w+x(7)*cos(x(7)+SW_3)*c_w+x(8)*cos(x(8)+SW_4)*c_w))^2+ (Mz_v-(x(1)*(-cos(x(5)+SW_1)*w_vl+l_v*sin(x(5)+SW_1))+x(2)*(cos(x(6)+SW_2)*w_vr+sin(x(6)+SW_2)*l_v)+x(3)*(-cos(x(7)+SW_3)*w_hl-l_h*sin(x(7)+SW_3))+x(4)*(cos(x(8)+SW_4)*w_hr-sin(x(8)+SW_4)*l_h)+x(5)*(sin(x(5)+SW_1)*c_w*w_vl+cos(x(5)+SW_1)*c_w*l_v)+x(6)*(-sin(x(6)+SW_2)*c_w*w_vr+cos(x(6)+SW_2)*c_w*l_v)+x(7)*(sin(x(7)+SW_3)*c_w*w_hl-cos(x(7)+SW_3)*c_w*l_h)+x(8)*(-cos(x(8)+SW_4)*c_w*l_h-sin(x(8)+SW_4)*c_w*w_hl)))^2) + (f_lag >= 0).*((A/SOC_1)*(x(1))^2+(A/SOC_2)*(x(2))^2+(A/SOC_3)*(x(3))^2+(A/SOC_4)*(x(4))^2+(d_delay_1-(SW_1+x(5)))^2+(d_delay_2-(SW_2+x(6)))^2+(d_delay_3-(SW_3+x(7)))^2+(d_delay_4-(SW_4+x(8)))^2) + (f_lag == -1).*((((B*(M_reg_1/(M_reg_1+M_reg_2+M_reg_3+M_reg_4)))/SOC_1)*(1/(x(1)^2)))+(((B*(M_reg_2/(M_reg_1+M_reg_2+M_reg_3+M_reg_4)))/SOC_2)*(1/(x(2)^2)))+(((B*(M_reg_3/(M_reg_1+M_reg_2+M_reg_3+M_reg_4)))/SOC_3)*(1/(x(3)^2)))+(((B*(M_reg_4/(M_reg_1+M_reg_2+M_reg_3+M_reg_4)))/SOC_4)*(1/(x(4)^2)))+(d_delay_1-(SW_1+x(5)))^2+(d_delay_2-(SW_2+x(6)))^2+(d_delay_3-(SW_3+x(7)))^2+(d_delay_4-(SW_4+x(8)))^2);


%setting the initial point
x0 = [(Fx_v/4)*cos(SW_1),(Fx_v/4)*cos(SW_2),(Fx_v/4)*cos(SW_3),(Fx_v/4)*cos(SW_4),(d_delay_1-SW_1),(d_delay_2-SW_2),(d_delay_3-SW_3),(d_delay_4-SW_4)];

%linear constraints
%inequality constraints
A = [];
b = [];

%equality constraints
% Aeq*x = beq
Aeq = [ 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0];

%steering priority
SOC_dif1=SOC_1-((SOC_1+SOC_2+SOC_3+SOC_4)/4);      %negative when 
SOC_dif2=SOC_2-((SOC_1+SOC_2+SOC_3+SOC_4)/4);
SOC_dif3=SOC_3-((SOC_1+SOC_2+SOC_3+SOC_4)/4);
SOC_dif4=SOC_4-((SOC_1+SOC_2+SOC_3+SOC_4)/4); 
%if f_lag>=0
     
   % if (SOC_dif1 < SOC_krit)          % state of charge of one actor is really low, actor is only used for steering %during driving mode
                                       %SOC_krit =-0.2(evaluate in tests)
                                                       
 %       Aeq(2,1) = 1;
  %  end
    
 %   if (SOC_dif2 < SOC_krit)          % state of charge of one actor is really low, actor is only used for steering %during driving mode
 %       Aeq(2,2) = 1;
 %  end
    
 %   if (SOC_dif3 < SOC_krit)         % state of charge of one actor is really low, actor is only used for steering %during driving mode
  %      Aeq(2,3) = 1;
   % end
    
%    if (SOC_dif4 < SOC_krit)          % state of charge of one actor is really low, actor is only used for steering %during driving mode
 %       Aeq(2,4) = 1;
  %  end
%end

% jedes eine eigene Linie??
% Erfahrungswerte aus Testsimulation

beq = [ 0, 0]; 



%case decision braking or driving backwards (decision input: flag= -1 oder 1)
%braking mode 
if f_lag == -1                    

%if it is possible the regenerative braking torque is the limit 
% sum of the Force from the regenerative braking torque in vehiclecoordinates is bigger than the demanded force for the whole vehicle
    % M_reg is always positive and Fx_v can be either positive or negative in the braking case 
         if (cos(SW_1)*(M_reg_1/r_dyn_1)+cos(SW_2)*(M_reg_2/r_dyn_2)+cos(SW_3)*(M_reg_3/r_dyn_3)+cos(SW_3)*(M_reg_4/r_dyn_4))> abs(Fx_v)  
         
             % max. regenerative braking torque
             lb = [-M_reg_1/r_dyn_1, -M_reg_2/r_dyn_2, -M_reg_3/r_dyn_3, -M_reg_4/r_dyn_4, d_min_vl-SW_1, d_min_vr-SW_2, d_min_hl-SW_3, d_min_hr-SW_4];    %d and betato the left side are positive, to the right side are negative  %regenerative braking torque must be positive
             ub = [M_reg_1/r_dyn_1, M_reg_2/r_dyn_2, M_reg_3/r_dyn_3, M_reg_4/r_dyn_4, d_max_vl-SW_1, d_max_vr-SW_2, d_max_hl-SW_3, d_max_hr-SW_4];
 
         else
            % mechanical braking
            %FxB_max is positive
            lb = [-FxB_max, -FxB_max, -FxB_max, -FxB_max, d_min_vl-SW_1, d_min_vr-SW_2, d_min_hl-SW_3, d_min_hr-SW_4];     %d and betato the left side are positive, to the right side are negative                                       %FxB_max is negative
            ub = [FxB_max, FxB_max, FxB_max, FxB_max, d_min_vl-SW_1, d_min_vr-SW_2, d_min_hl-SW_3, d_min_hr-SW_4];
         end                                                                         % close conditions for braking loop, max braking torque is positive

 %driving mode
else
    lb = [-P_max_1/(r_dyn_1*w_1), -P_max_2/(r_dyn_2*w_2), -P_max_3/(r_dyn_3*w_3), -P_max_4/(r_dyn_4*w_4), d_min_vl-SW_1, d_min_vr-SW_2, d_min_hl-SW_3, d_min_hr-SW_4];
    ub = [P_max_1/(r_dyn_1*w_1), P_max_2/(r_dyn_2*w_2), P_max_3/(r_dyn_3*w_3), P_max_4/(r_dyn_4*w_4), d_max_vl-SW_1, d_max_vr-SW_2, d_max_hl-SW_3, d_max_hr-SW_4];
end
%ub = (1:8);

    


nlcon = @nonlnconstraints;


%options = optimoptions(@fmincon, 'MaxFunctionEvaluations',10.000000e+03);

options = optimoptions(@fmincon, 'Algorithm', 'interior-point');


x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nlcon(Fx_v,Fy_v,Mz_v,SW_1,SW_2,SW_3,SW_4,w_vr, w_vl, w_hr, w_hl, l_h, l_v, c_w),options);


%x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nlcon(Fx_v,Fy_v,Mz_v,SW_1,SW_2,SW_3,SW_4,w_vr, w_vl, w_hr, w_hl, l_h, l_v, c_w));



Fx1 =  x(1);
Fx2 =  x(2);
Fx3 =  x(3);
Fx4 = x(4);
alpha1 = x(5);
alpha2 = x(6);
alpha3 = x(7);
alpha4 = x(8);

I would be very glad for your help.

Thank you

[Sarah Kern]

and the matlab function block tells me to change the options to 'sqp' not sure why.

With the different algorithm the problem is not solveable anymore

[Walter Roberson]

The parameter after ub needs to be empty or a function handle but instead you make a full function call to something that you do not provide the code for and seems unlikely to return a function handle.

[Sarah Kern]

The nonlinear constraints?

they are empty,

x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nlcon,options);
with 
function [c,ceq] = nonlnconstraints
ceq = [];
c = [];

end

I corrected it now, but it still wont accept interior point or find a global minimum

[Walter Roberson]

fmincon is only designed to find local minima. You should look at the tools in the Global Optimization Toolbox.

[Sarah Kern]

But my optimization works in a matlab file with fmincon and the interior point algorithm, but the matlab function block demands sqp, wich cant solve my problem.

Can you tell me why?

[Walter Roberson]

When you talk about MATLAB Function Block are you talking about using the code in Simulink?

[Sarah Kern]

yes

[Walter Roberson]

Please show the code version you use as the function block and please show the message that is produced about sqp

[Sarah Kern]

So my block input looks like

function [ Fx1, Fx2, Fx3, Fx4, alpha1, alpha2, alpha3, alpha4] = fcn(Fx_v, Fy_v, Mz_v, P_max_1, P_max_2, P_max_3, P_max_4, M_reg_1, M_reg_2, M_reg_3, M_reg_4, k, delta_T_aA, SW_1, SW_2, SW_3, SW_4, w_1, w_4, w_3, w_2, d_delay_1, d_delay_2, d_delay_3, d_delay_4, r_dyn_1, r_dyn_2, r_dyn_3, SOC_1, SOC_2, r_dyn_4, SOC_3, SOC_4, f_lag, n, j,w_vr, w_vl, w_hr, w_hl, l_h, l_v, c_w,d_min_vl, d_max_vl, d_min_vr, d_max_vr, d_min_hl, d_max_hl, d_min_hr, d_max_hr, A, B, FxB_max, SOC_krit, C, x0)
fun = @(x) C*((Fx_v-(x(1)*cos(x(5)+SW_1)+x(2)*cos(x(6)+SW_2)+x(3)*cos(x(7)+SW_3)+x(4)*cos(x(8)+SW_4)-x(5)*sin(x(5)+SW_1)*c_w-x(6)*sin(x(6)+SW_2)*c_w-x(7)*sin(x(7)+SW_3)*c_w-x(8)*sin(x(8)+SW_4)*c_w))^2+ (Fy_v-(x(1)*sin(x(5)+SW_1)+x(2)*sin(x(6)+SW_2)+x(3)*sin(x(7)+SW_3)+x(4)*sin(x(8)+SW_4)+x(5)*cos(x(5)+SW_1)*c_w+x(6)*cos(x(6)+SW_2)*c_w+x(7)*cos(x(7)+SW_3)*c_w+x(8)*cos(x(8)+SW_4)*c_w))^2+ (Mz_v-(x(1)*(-cos(x(5)+SW_1)*w_vl+l_v*sin(x(5)+SW_1))+x(2)*(cos(x(6)+SW_2)*w_vr+sin(x(6)+SW_2)*l_v)+x(3)*(-cos(x(7)+SW_3)*w_hl-l_h*sin(x(7)+SW_3))+x(4)*(cos(x(8)+SW_4)*w_hr-sin(x(8)+SW_4)*l_h)+x(5)*(sin(x(5)+SW_1)*c_w*w_vl+cos(x(5)+SW_1)*c_w*l_v)+x(6)*(-sin(x(6)+SW_2)*c_w*w_vr+cos(x(6)+SW_2)*c_w*l_v)+x(7)*(sin(x(7)+SW_3)*c_w*w_hl-cos(x(7)+SW_3)*c_w*l_h)+x(8)*(-cos(x(8)+SW_4)*c_w*l_h-sin(x(8)+SW_4)*c_w*w_hl)))^2); %(f_lag >= 0).*((A/SOC_1)*(x(1))^2+(A/SOC_2)*(x(2))^2+(A/SOC_3)*(x(3))^2+(A/SOC_4)*(x(4))^2+(d_delay_1-(SW_1+x(5)))^2+(d_delay_2-(SW_2+x(6)))^2+(d_delay_3-(SW_3+x(7)))^2+(d_delay_4-(SW_4+x(8)))^2) + (f_lag == -1).*((((B*(M_reg_1/(M_reg_1+M_reg_2+M_reg_3+M_reg_4)))/SOC_1)*(1/(x(1)^2)))+(((B*(M_reg_2/(M_reg_1+M_reg_2+M_reg_3+M_reg_4)))/SOC_2)*(1/(x(2)^2)))+(((B*(M_reg_3/(M_reg_1+M_reg_2+M_reg_3+M_reg_4)))/SOC_3)*(1/(x(3)^2)))+(((B*(M_reg_4/(M_reg_1+M_reg_2+M_reg_3+M_reg_4)))/SOC_4)*(1/(x(4)^2)))+(d_delay_1-(SW_1+x(5)))^2+(d_delay_2-(SW_2+x(6)))^2+(d_delay_3-(SW_3+x(7)))^2+(d_delay_4-(SW_4+x(8)))^2);
A = [];
b = [];
Aeq = [ 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0];
beq = [ 0, 0]; 
if f_lag == -1                    

%if it is possible the regenerative braking torque is the limit 
% sum of the Force from the regenerative braking torque in vehiclecoordinates is bigger than the demanded force for the whole vehicle
    % M_reg is always positive and Fx_v can be either positive or negative in the braking case 
         if (cos(SW_1)*(M_reg_1/r_dyn_1)+cos(SW_2)*(M_reg_2/r_dyn_2)+cos(SW_3)*(M_reg_3/r_dyn_3)+cos(SW_3)*(M_reg_4/r_dyn_4))> abs(Fx_v)  
         
             % max. regenerative braking torque
             lb = [-M_reg_1/r_dyn_1, -M_reg_2/r_dyn_2, -M_reg_3/r_dyn_3, -M_reg_4/r_dyn_4, d_min_vl-SW_1, d_min_vr-SW_2, d_min_hl-SW_3, d_min_hr-SW_4];    %d and betato the left side are positive, to the right side are negative  %regenerative braking torque must be positive
             ub = [M_reg_1/r_dyn_1, M_reg_2/r_dyn_2, M_reg_3/r_dyn_3, M_reg_4/r_dyn_4, d_max_vl-SW_1, d_max_vr-SW_2, d_max_hl-SW_3, d_max_hr-SW_4];
 
         else
            % mechanical braking
            %FxB_max is positive
            lb = [-FxB_max, -FxB_max, -FxB_max, -FxB_max, d_min_vl-SW_1, d_min_vr-SW_2, d_min_hl-SW_3, d_min_hr-SW_4];     %d and betato the left side are positive, to the right side are negative                                       %FxB_max is negative
            ub = [FxB_max, FxB_max, FxB_max, FxB_max, d_min_vl-SW_1, d_min_vr-SW_2, d_min_hl-SW_3, d_min_hr-SW_4];
         end                                                                         % close conditions for braking loop, max braking torque is positive

 %driving mode
else
    lb = [-P_max_1/(r_dyn_1*w_1), -P_max_2/(r_dyn_2*w_2), -P_max_3/(r_dyn_3*w_3), -P_max_4/(r_dyn_4*w_4), d_min_vl-SW_1, d_min_vr-SW_2, d_min_hl-SW_3, d_min_hr-SW_4];
    ub = [P_max_1/(r_dyn_1*w_1), P_max_2/(r_dyn_2*w_2), P_max_3/(r_dyn_3*w_3), P_max_4/(r_dyn_4*w_4), d_max_vl-SW_1, d_max_vr-SW_2, d_max_hl-SW_3, d_max_hr-SW_4];
end
nlcon = @nonlnconstraints;
y = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nlcon(Fx_v,Fy_v,Mz_v,SW_1,SW_2,SW_3,SW_4,w_vr, w_vl, w_hr, w_hl, l_h, l_v, c_w));
Fx1 =  x(1);
Fx2 =  x(2);
Fx3 =  x(3);
Fx4 = x(4);

alpha1 = x(5);
alpha2 = x(6);
alpha3 = x(7);
alpha4 = x(8);

with´the nonlinear constraints

ceq = [];
 
c = [];

If I try to run it like this the error

FMINCON requires 10 input arguments. Must set option Algorithm to 'sqp' and pass to solver.

Function 'CarMaker/VehicleControl/FTR1/Fahren/Aufteilung der Überaktuierung /Kräfteaufteilung/linearer Fahrbereich/MATLAB Function' (#84.7185.7298), line 116, column 5:

"fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nlcon(Fx_v,Fy_v,Mz_v,SW_1,SW_2,SW_3,SW_4,w_vr, "

Launch diagnostic report.

Comes up.

So if I change the algorithm to sqp

with

options = optimoptions(@fmincon, 'Algorithm', 'sqp');

there is no error anymore, but the output is random,

so if I copy it in a different file and try to solve it, this error comes up

Solver stopped prematurely.

fmincon stopped because it exceeded the function evaluation limit,

options.MaxFunctionEvaluations = 8.000000e+02.

also if I extend the iterations the solver is not able to solve it.

I dont know what else to try.

The guessed initial value is calculated before and should be really close to the x output already,

not sure whats the problem.

[Sarah Kern]

But if I try the extra file (not within simulink) with the 'interior-pint' algorithm it can solve my problem ( local minimum)

[Sarah Kern]

x = y in my upper code

[Walter Roberson]

I need a copy of your nonlnconstraints function, not just the knowledge that ceq = []; c = [];

Above you posted

function [c,ceq] = nonlnconstraints
ceq = [];
c = [];
end

but that is not your nonlnconstraints function that you are using: that function would die with a complaint about too many input arguments when you call it as

nlcon(Fx_v,Fy_v,Mz_v,SW_1,SW_2,SW_3,SW_4,w_vr, w_vl, w_hr, w_hl, l_h, l_v, c_w)

in the fmincon call.

[Sarah Kern]

I am calling this

nlcon = @nonlnconstraints;
options = optimoptions(@fmincon, 'Algorithm', 'sqp');
x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nlcon, options);

and this is my nonlinear constraints file

function [c,ceq] = nonlnconstraints()
ceq = [];
c = [];
end

[Walter Roberson]

I do not observe any difficulties when I use

options = optimoptions(@fmincon, 'Algorithm', 'sqp', 'MaxFunctionEvaluations', 1e6, 'MaxIterations', 1e6);
[xm,fval] = fmincon(fun,x0,[],[],[],[],lb,ub,[],options)

Your A and b are empty, your Aeq and beq are not empty but are all-zero and so are not doing anything so might as well be empty. Your nonlinear constraints are empty.

The optimum I get with this is not quite as good as you get with interior point, but is not far off in function value. x(5), x(6) and x(7) do differ by notable amounts, but due to non-linear contributions that doesn't always mean much.

The optimum you get with sqp is close to the optimum you get with fminsearch() when it is permitted enough iterations.

fmincon is not a global minimizer so if you need a global minima then you need to do something else.

[Sarah Kern]

allright, so I must have had some problems calling the nonlinear constraints.

Thank you very much for your help!

[Walter Roberson]

sqp gives the optimum near

-127.239597103982 128.835753420144 121.456350412052 125.969244882211 14.7285335181627 5.30599633554605 -19.9353352041771 5.15727654258455

with a function value of about 127529.373794526

Interior point gives the optimum near

-124.126884753518 129.235303885167 122.479502387116 127.655545239724 2.16075485377955 -0.98830809315363 -1.11150462680378 5.15802574308162

with a function value of about 126889.507492473

Using other techniques I find you can do slightly better, near

-124.113421979527 129.245561766186 122.471919627735 127.665321535654 2.16126732836261 -0.988860425787402 -1.11254769030373 5.15909384957014

with a function value of about 126889.418322864

[Sarah Kern]

I gonna look into all of them, thank you.