How to use lsqnonlin command for solving a cost function minimization problem which consists of optimization variable?

Hi,

I'm trying to solve an optimization problem by using 'lsqnonlin' comman in matlab. My objective function consists of optimization variables created by 'optimvar' command. I came across with this error,

"Error using lsqfcnchk (line 80)

If FUN is a MATLAB object, it must have an feval method."

when i tried to use that objective function with lsqnonlin. I read that lsqnonlin only accepts function handles or anonymous functions, not symbolic objects. In this case how can i use my objective function with lsqnonlin? I left a piece of code below to show my problem clearly. By the way, I'm new at optimization toolbox. So I'm open for suggestions to use proper methods.

C = [0	0	0	1
     0	1	0	0];
D = [0;0];

x = [zeros(4,1)];
u = optimvar('U',2);
f=0;

% for loop here to create an objective function
for i=1:2
    
    y_ara = C*x + D*u(i);
    error1 = 1-y_ara(1,1);
    error2 = 1-y_ara(2,1);
    
    f = f + error1^2 + error2^2; % Objective function including optimvar U(1) and U(2)
    
end

prob = optimproblem;
prob.Objective = f;

% initilization and constraits
x0 = [zeros(1,10)];
lb = [-10*ones(1,2)];
ub = [10*ones(1,2)];

options = optimoptions("lsqnonlin",'Algorithm','levenberg-marquardt');
[sol, fval, exitflag, output] = lsqnonlin(f, x0, lb, ub, options);

Error using lsqfcnchk
If FUN is a MATLAB object, it must have an feval method.

Error in lsqnsetup (line 46)
funfcn = lsqfcnchk(FUN,caller,lengthVarargin,funValCheck,flags.grad);

Error in lsqnonlin (line 207)
JACOB,OUTPUT,earlyTermination] = lsqnsetup(FUN,xCurrent,LB,UB,options,defaultopt, ...

[Matt J]

Your ub and lb bounds do not make sense. You have 10 bounds, but only 2 unknowns u(i).

[B. Berk]

Its just because i copied a part of whole code and little bit manipulated it to make proper here. It looks like i forgot to change that value, however, the error is still same.

Answer 1

[Matt J]

If you have an objective created with optimvar variables, you would solve the problem by using the solve command,

C = [0	0	0	1
     0	1	0	0];
D = [0;0];

...
    
prob = optimproblem;
prob.Objective = f;

sol=solve(prob);

In the most basic use case, you would not call lsqnonlin or any other solver directly. The solve() command will choose the appropriate solver for you.

[B. Berk]

Actually i was using the solve command at first, but resulting values of 'u' which makes objective function minimum was meaningles. Then i thought may its reason is solver that 'solve' command uses. I wasn't know that 'solve' choose it automatically according to objective function. What if my case is not basic one, should i call another solver like lsqnonlin directly?

If you have any suggestions for approach this problem or possible solution please let me know.

Thanks for your contribution.

[Matt J]

It depends what you mean by "not a basic one". There are different possible syntaxes for solve if you need to do something specific, e.g., set an initial guess x0 for the optimization

sol = solve(prob)

sol = solve(prob,x0)

sol = solve(prob,x0,ms)

sol = solve(___,Name,Value)

[B. Berk]

I admit that "not a basic one" is not a clear expression in this case. I just meant that how much 'solve' command is reliable if you have an objective function with high order (e.g. 10th order) polynomial. Should i try my chance with other solvers or trust 'solve's choosing.

[Torsten]

If you don't trust, you should use solver-based optimization, not problem-based :-)

[Matt J]

@Bekir Berk Sirim

Modeling with a 10th order polynomial sounds like a dangerous thing to do. At the very least, you will need a good initial guess, since polynomials tend to have lots of local minima. Thus, you might need to use.

sol = solve(prob,x0)

sol = solve(prob,x0,ms)

[B. Berk]

Thanks again for your contributions, i will consider your suggestions.

Answer 2

[Walter Roberson]

C = [0	0	0	1
     0	1	0	0];
D = [0;0];

x = [zeros(4,1)];
u = optimvar('U',2);
f=0;

% for loop here to create an objective function
for i=1:2
    
    y_ara = C*x + D*u(i);
    error1 = 1-y_ara(1,1);
    error2 = 1-y_ara(2,1);
    
    f = f + error1^2 + error2^2; % Objective function including optimvar U(1) and U(2)
    
end

prob = optimproblem;
prob.Objective = f;

% initilization and constraits
x0 = [zeros(1,10)];
lb = [-10*ones(1,2)];
ub = [10*ones(1,2)];

%options = optimoptions("lsqnonlin",'Algorithm','levenberg-marquardt');
[sol, fval, exitflag, output] = solve(prob)

Solving problem using lsqlin.

sol = struct with fields:
    U: [2×1 double]

fval = 4.0000

exitflag = 
    OptimalSolution

output = struct with fields:
         iterations: 0
          algorithm: 'mldivide'
      firstorderopt: []
    constrviolation: []
       cgiterations: []
       linearsolver: []
            message: ''
             solver: 'lsqlin'

prob

prob = 
  OptimizationProblem with properties:

       Description: ''
    ObjectiveSense: 'minimize'
         Variables: [1×1 struct] containing 1 OptimizationVariable
         Objective: [1×1 OptimizationExpression]
       Constraints: [0×0 struct] containing 0 OptimizationConstraints

  See problem formulation with show.

f

f = 
  Quadratic OptimizationExpression

    4

You have C*x but your x is all 0. You have D*u but your D is all 0. So your optimization expression becomes a constant.

[B. Berk]

Thank you for your response. 'x' values are updating according to (x_dot = A*x + B*u) equation at every loop in the original of code. I just copied and changed this part to make it proper here. This is why x values are zero and optimization expression becomes zero.