Help in understanding optimization problems and solving them in Matlab (choise of appropriate solver)

Hello

Maybe a classical phrase: I have been trying to understand this since several days but not succeed - really about me!

Lets take a simple constrained optimization problem from here maximazing revenu with bubget constraints. So the problem is

We can solve it in Matlab with fmincon

objective = @(x) -200*x(1)^(2/3)*x(2)^(1/3);
[X,FVAL,EXITFLAG,OUTPUT,LAMBDA,GRAD,HESSIAN] = fmincon(objective,[1,1],[],[],[20, 170],20000,[],[],[])
>> X =

  666.6669   39.2157
  LAMBDA = 

  struct with fields:

         eqlin: 2.5927

OR we can use Lagrangian cost function and rewrite these to unconstrained optimization problem (Am I right at this stage?)

Then solving this objective function with fminsearch or fminunc (even tried with ga)

lambda = 2.5927;
>> objective = @(x) -200*x(1)^(2/3)*x(2)^(1/3)-lambda*(20*x(1) + 170*x(2) - 20000);
x = fminuncfminunc(objective,[1, 1])

>> Problem appears unbounded.

fminunc stopped because the objective function value is less than
or equal to the value of the objective function limit.

<stopping criteria details>

x =

   1.0e+19 *

    0.2504    6.4423
    
>> x = fminsearch(objective,[1, 1])
 
Exiting: Maximum number of function evaluations has been exceeded
         - increase MaxFunEvals option.
         Current function value: -51855.290146 


x =

    0.1088    1.7458

gives different results and even not close to constrained solution with fmincon. I tried to change the sign, put lambda = 1....

So why its like this, where I am wrong or I dont understand something?

Answer 1

[Matt J]

You have to reformulate the problem with a convex objective to be certain that the Lagrangian minimization will behave as you're expecting (see Sufficient Conditions for Strong Duality). You can do this as follows:

objective = @(x) -2/3*log(x(1))-1/3*log(x(2));
[Xcon,FVAL,EXITFLAG,OUTPUT,LAMBDA,GRAD,HESSIAN] = fmincon(objective,[1,1],[],[],[20, 170],20000,[],[],[])

objectiveUC = @(x) -2/3*log(x(1))-1/3*log(x(2))+LAMBDA.eqlin*(20*x(1) + 170*x(2) - 20000);
Xunc = fminunc(objectiveUC,[1, 1])

This leads to

Xcon =

  666.6664   39.2157


Xunc =

  666.6533   39.2154

[Yurii Iotov]

Hi Matt!

Thank you so mach for your answer!

Yes I see my ''gap'' and thanks for referring to the topic! But is it the same that for Matlab I need to have analytic type function for optimization? Complex Numbers in Optimization Toolbox Solvers

I am writing my master's thesis about optimization, so other questions may appear about Matlab optimization, I would be grateful in a future...

[Yurii Iotov]

And Matt, please, what I dont understand also is why with these 3 different objective functions

"original"

objective_1 = @(x) -200*x(1)^(2/3)*x(2)^(1/3);

you showed me

objective_2 = @(x) -2/3*log(x(1))-1/3*log(x(2));

and with 200 term

objective = @(x) (-2/3*log(x(1))-1/3*log(x(2)))*200;

with constrained optimization:

[Xcon,FVAL,EXITFLAG,OUTPUT,LAMBDA,GRAD,HESSIAN] = fmincon(objective,[1,1],[],[],[20, 170],20000,[],[],[])

...Gives the same solution for Xcon, but others are differ, especially LAMBDA in wich I am also interested for results interpretation?

it blows my brain a little bit!

Is it because objective_1 and objective_2 have duality property?

And term 200 in objective is just a csaling factor?

...I feel I have some gaps in math...

[Matt J]

All 3 objectives differ by monotonic transformations, e.g.,

objective_2=-log(-objective_1/200)

is a monotonic function of objective_1. Therefore if one increases or decreases, so does the other, and they therefore have minima at the same locations.