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Laila Alzahrawi on 26 May 2023
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Commented: Walter Roberson on 26 May 2023
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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
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Answers (1)
Edited: Torsten on 26 May 2023
Open in MATLAB Online
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; FMINCONshifted 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 thanthe value of the step size tolerance but constraints are notsatisfied 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
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Laila Alzahrawi on 26 May 2023
Direct link to this comment
https://matlabcentral.mathworks.com/matlabcentral/answers/1974079-error-using-fmincon-line-619#comment_2760949
Edited: Laila Alzahrawi on 26 May 2023
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 on 26 May 2023
Direct link to this comment
https://matlabcentral.mathworks.com/matlabcentral/answers/1974079-error-using-fmincon-line-619#comment_2760959
Edited: Torsten on 26 May 2023
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 on 26 May 2023
Direct link to this comment
https://matlabcentral.mathworks.com/matlabcentral/answers/1974079-error-using-fmincon-line-619#comment_2761014
Edited: Laila Alzahrawi on 26 May 2023
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 on 26 May 2023
Direct link to this comment
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Edited: Torsten on 26 May 2023
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 on 26 May 2023
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Open in MATLAB Online
(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 on 26 May 2023
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Open in MATLAB Online
@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 on 26 May 2023
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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 ??
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