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

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Yurii Iotov 2019-5-17

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评论： Matt J 2019-5-20

采纳的回答： Matt J

在 MATLAB Online 中打开

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?

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Answer 1

Matt J 2019-5-17

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https://ww2.mathworks.cn/matlabcentral/answers/462678-help-in-understanding-optimization-problems-and-solving-them-in-matlab-choise-of-appropriate-solver#answer_375516

编辑：Matt J 2019-5-17

在 MATLAB Online 中打开

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

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Yurii Iotov 2019-5-20

在 MATLAB Online 中打开

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 2019-5-20

在 MATLAB Online 中打开

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.

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