Error using fmincon line 619

Hi,

I am trying to do a non-linear optimization using the "fmincon" function. I have got an error and couldn't find the problem. When I test my objective function on the initial conditions using

objective(x0) % Scalar?

I got the following answer

ans =

0 0 0 0 0 0 0 0 0 0

Your initial point x0 is not between bounds lb and ub; FMINCON

shifted x0 to satisfy the bounds.

The initial error obtained is this:

Error using fmincon (line 619)

Supplied objective function must return a scalar value.

Error in (line 41)

[x_opt, fval] = fmincon(objective, x0, [], [], [], [], lb, ub, nonlcon, options);

The code is the following:

R = [120.97, 462.26, 231.65, 140.94, 267.54, 95.49, 413.92, 361.17, 311.53, 107.76]; 
V = [5315.13, 11786.19, 8840.62, 15132.03, 23078.9, 7982.05, 11175.72, 7994.76, 10068.08, 11194.89]; 
IR = [60.67, 224.67, 177.67, 193.67, 239.33, 196.33, 220.58, 151.33, 169, 185.67]; 
Ky_initial =[0.2, 0.62, 0.2, 0.9, 0.4, 0.4, 0.2, 0.2]; 
Ky_flowering = [0.6, 0.9, 0, 1.1, 0, 1.5, 0.55, 0.8]; 
Ky_fruit = [0, 0.2, 0.2, 0.35, 0.25, 0.2, 0.2, 0]; 
Ky_total = [1, 1.1, 1.15, 0.9, 1.35, 1.05, 0.85, 1.25, 0.9, 0.85]; 
n = 10;
objective = @(A) -sum(A.*R); 
nonlcon = @(A) [
    sum(A.*V) - 34000000; 
    sum((10*A.*IR)/(86400*30)) - 3.78; 
    sum(A) - 3400; 
    min(A./sum(A))*100 - [20, 5, 3, 5, 3, 3, 10, 5, 5, 5]; 
    max(A./sum(A))*100 - [40, 15, 10, 15, 10, 10, 30, 30, 15, 15]; 
    ];

lb = [20, 5, 3, 5, 3, 3, 10, 5, 5, 5]; 
ub = [40, 15, 10, 15, 10, 10, 30, 30, 15, 15]; 

x0 = zeros(n, 1); 
options = optimoptions('fmincon', 'Display', 'iter', 'Algorithm', 'sqp');
[x_opt, fval] = fmincon(objective, x0, [], [], [], [], lb, ub, nonlcon, options);

disp('Optimized Solution:');
disp(x_opt);

obj_val = -sum(x_opt .*R); 
crop_obj_vals = x_opt .* R;

disp('Optimal Objective Value:');
disp(obj_val);
disp('Individual Objective Values:');
disp(crop_obj_vals);
disp(-fval); 

I would appreciate if somebody can help me find the error and fix it.

Thank you.

Laila

Answer 1

[Torsten]

Why do you solve a linear optimization problem with "fmincon" ?

Use "linprog" instead.

R = [120.97, 462.26, 231.65, 140.94, 267.54, 95.49, 413.92, 361.17, 311.53, 107.76]; 
V = [5315.13, 11786.19, 8840.62, 15132.03, 23078.9, 7982.05, 11175.72, 7994.76, 10068.08, 11194.89]; 
IR = [60.67, 224.67, 177.67, 193.67, 239.33, 196.33, 220.58, 151.33, 169, 185.67]; 
Ky_initial =[0.2, 0.62, 0.2, 0.9, 0.4, 0.4, 0.2, 0.2]; 
Ky_flowering = [0.6, 0.9, 0, 1.1, 0, 1.5, 0.55, 0.8]; 
Ky_fruit = [0, 0.2, 0.2, 0.35, 0.25, 0.2, 0.2, 0]; 
Ky_total = [1, 1.1, 1.15, 0.9, 1.35, 1.05, 0.85, 1.25, 0.9, 0.85]; 
n = 10;
objective = @(A) -sum(A.*R); 
nonlcon = @(A) deal([
    sum(A.*V) - 34000000; 
    sum((10*A.*IR)/(86400*30)) - 3.78; 
    sum(A) - 3400; 
    (min(A./sum(A))*100 - [20, 5, 3, 5, 3, 3, 10, 5, 5, 5]).'; 
    (max(A./sum(A))*100 - [40, 15, 10, 15, 10, 10, 30, 30, 15, 15]).' 
    ],[ ]);

lb = [20, 5, 3, 5, 3, 3, 10, 5, 5, 5]; 
ub = [40, 15, 10, 15, 10, 10, 30, 30, 15, 15]; 

x0 = zeros(1,n); 

options = optimoptions('fmincon', 'Display', 'iter', 'Algorithm', 'sqp');
[x_opt, fval] = fmincon(objective, x0, [], [], [], [], lb, ub, nonlcon, options);

Your initial point x0 is not between bounds lb and ub; FMINCON
shifted x0 to satisfy the bounds.
 Iter  Func-count            Fval   Feasibility   Step Length       Norm of   First-order  
                                                                       step    optimality
    0          11   -1.526094e+04     2.125e+01     1.000e+00     0.000e+00     4.623e+02  
    1          22   -4.937565e+04     1.105e+01     1.000e+00     4.441e+01     2.500e+01  
    2          34   -4.730441e+04     7.437e+00     7.000e-01     1.455e+01     2.500e+01  
    3          47   -4.811526e+04     8.388e+00     4.900e-01     6.863e+00     2.500e+01  
    4          62   -4.772028e+04     7.104e+00     2.401e-01     3.096e+00     2.500e+01  
    5          79   -4.786548e+04     7.541e+00     1.176e-01     1.203e+00     2.500e+01  
    6          97   -4.775070e+04     7.122e+00     8.235e-02     9.057e-01     2.500e+01  
    7         121   -4.773831e+04     7.091e+00     9.689e-03     9.777e-02     2.500e+01  
    8         146   -4.774656e+04     7.109e+00     6.782e-03     6.800e-02     2.500e+01  
    9         173   -4.774232e+04     7.093e+00     3.323e-03     3.344e-02     2.500e+01  
   10         203   -4.774087e+04     7.089e+00     1.140e-03     1.143e-02     2.500e+01  
   11         234   -4.774188e+04     7.091e+00     7.979e-04     7.993e-03     2.500e+01  
   12         266   -4.774117e+04     7.089e+00     5.585e-04     5.599e-03     2.500e+01  
   13         299   -4.774167e+04     7.091e+00     3.910e-04     3.919e-03     2.500e+01  
   14         333   -4.774132e+04     7.089e+00     2.737e-04     2.743e-03     2.500e+01  
   15         369   -4.774115e+04     7.089e+00     1.341e-04     1.344e-03     2.500e+01  
   16         405   -4.774132e+04     7.089e+00     1.341e-04     1.345e-03     2.500e+01  
   17         442   -4.774120e+04     7.089e+00     9.387e-05     9.406e-04     2.500e+01  
   18         480   -4.774129e+04     7.089e+00     6.571e-05     6.591e-04     2.500e+01  
   19         519   -4.774123e+04     7.089e+00     4.600e-05     4.609e-04     2.500e+01  
   20         560   -4.774120e+04     7.089e+00     2.254e-05     2.258e-04     2.500e+01  
   21         602   -4.774122e+04     7.089e+00     1.578e-05     1.582e-04     2.500e+01  
   22         646   -4.774121e+04     7.089e+00     7.731e-06     7.746e-05     2.500e+01  
   23         682   -4.774121e+04     7.089e+00     2.652e-06     2.657e-05     2.500e+01  

Converged to an infeasible point.

fmincon stopped because the size of the current step is less than
the value of the step size tolerance but constraints are not
satisfied to within the value of the constraint tolerance.

disp('Optimized Solution:');

Optimized Solution:

disp(x_opt);

   30.0000   15.0000   10.0000   15.0000   10.0000    5.5520   30.0000   30.0000   15.0000   15.0000

obj_val = -sum(x_opt .*R); 
crop_obj_vals = x_opt .* R;

disp('Optimal Objective Value:');

Optimal Objective Value:

disp(obj_val);

  -4.7741e+04

disp('Individual Objective Values:');

Individual Objective Values:

disp(crop_obj_vals);

   1.0e+04 *

    0.3629    0.6934    0.2316    0.2114    0.2675    0.0530    1.2418    1.0835    0.4673    0.1616

disp(-fval); 

   4.7741e+04

[Laila Alzahrawi]

I got this when I copied what you wrote.

"Your initial point x0 is not between bounds lb and ub; FMINCON

shifted x0 to satisfy the bounds."

What does this mean?

[Torsten]

It means that your x0 does not satisfy lb <= x0 <= ub, the lower and upper bounds you defined for your solution vector. That's the reason MATLAB had to shift your guess in order to satisfy your bound constraints.

[Laila Alzahrawi]

I just noticed that you asked this "Why do you solve a linear optimization problem with "fmincon" ?"

As I mentioned in the beginning of the question, I am trying to solve a nonlinear optimization problem.

Does this change your answer?

Also, I have provided numbers for the upper and lower bounds, but lets say they are not given, would I still be able to solve the problem by any way?

I appreciate your help and your fast responses.

[Torsten]

As I mentioned in the beginning of the question, I am trying to solve a nonlinear optimization problem. Does this change your answer?

The problem you posted is linear and can be solved using a linear optimizer, namely "linprog". If this is not your real problem and the real problem is non-linear, it's a different thing.

Also, I have provided numbers for the upper and lower bounds, but lets say they are not given, would I still be able to solve the problem by any way?

If your solution variables are not bounded from the application behind your optimization, you don't need to set the bounds. But usually, the problem needs to be constrained in order to get a finite value for the objective.

[Walter Roberson]

    (min(A./sum(A))*100 - [20, 5, 3, 5, 3, 3, 10, 5, 5, 5]).'; 

Your A is a vector, so sum(A) is a vector, and A./sum(A) is a vector. min() of a vector is a scalar, scalar times constant is scalar. So min(A./sum(A))*100 is a scalar.

What is the point of subtracting a vector from a scalar at that point? You are going for a <= 0 which is going to be most difficult to satisfy when the subtracted value is smallest -- so why not just subtract 3 there instead of the vector?

[Torsten]

@Walter Roberson

(min(A./sum(A))*100 - [20, 5, 3, 5, 3, 3, 10, 5, 5, 5]).'; 
(max(A./sum(A))*100 - [40, 15, 10, 15, 10, 10, 30, 30, 15, 15]).' 

Yes, these are strange conditions.

I would have guessed that the first condition has to be reversed:

- (min(A./sum(A))*100 + [20, 5, 3, 5, 3, 3, 10, 5, 5, 5]).';

so that they would read in a more comprehensible form

A(i) >= 3/100*sum_j(A(j)) for all i
A(i) <= 10/100*sum_j(A(j)) for all i

[Walter Roberson]

lb = [20, 5, 3, 5, 3, 3, 10, 5, 5, 5]; 
ub = [40, 15, 10, 15, 10, 10, 30, 30, 15, 15]; 

Those match the vectors in the odd condition. Are they a clumsy re-statement of the bounds ??