Matlab Optimization Toolbox-Constrained Optimization problem

Started by zarith 7 years ago5 replieslatest reply 7 years ago316 views

I have a problem on solving a constrained optimization problem which has 5 variables and 3 constraints.

my code for the objective function is:
function y = objective_function(x)
y = 225*x(1)-45*x(2)-80*x(3)-1000*x(4)-90*x(5);

Which I saved in an m file.

The code for my constraint equations:
function [c,ceq] = constraint_equation (x)
c = [];
ceq = x(1)-4930-0.7548*x(3);

Which I saved in another m file.
The initial values were given as follow

After plugging them all in the optimization toolbox, I used the 'fmincon' solver and SQP as the algorithm. After running the system, I obtained a negative answer which is totally wrong. Really help anyone of you can help me on this!
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Reply by MichaelRWMarch 22, 2017

Out of curiosity, how do you know the answer is wrong?  Please discuss your rationale for reaching this conclusion.

[ - ]
Reply by zarithMarch 22, 2017

If I solve the problem by using the Newton Raphson method, the objective value is y=769036.96.

However, when I use the toolbox, I get a negative answer. Also, the question asked to maximize profit y. Sorry, its my bad for not clearly stating the problem.

I've also tried to write the code by using the 'fmincon' solver.

options = optimoptions('fmincon','Display','iter','Algorithm','sqp');

fun = @(x)-(225*x(1)-45*1000*x(4)-90*x(3)-1000*x(4)-80*x(5));

A = [];

b = [];

Aeq = [1,0,0,0,-0.7548];

beq = [4930];

lb = [];

ub = [];

nonlcon = @constraint_equation;

x0 = [12013.10423,189391.9536,-3.4,450,9384.08086];

x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,options);

[x,fval] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,options)

And the result is:

x =

   1.0e+15 *

    0.0000    0.0000   -0.0172   -8.7719    0.0000

fval =


Really need much help here. Thank you!

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Reply by zarithMarch 22, 2017

Also, please ignore the above constraint equation. Here's the right one.

function [c,ceq] = constraint_equation (x)

c = [];

ceq = x(1)-4930-0.7548*x(5);x(5)-((27)*(375-((217.5+0.039*x(5))/((548.8/x(3))+1)))*(1-exp(-0.005556*x(4))));[x(2)-45*(557.5+0.039*x(5)-((217.5+0.039*x(5))/((548.8/x(3))+1)))];


[ - ]
Reply by MichaelRWMarch 22, 2017

It looks like you've taken into account the change in sign necessary to change the maximization problem into a minimization problem that is suitable for the FMINCON function.  However, why did your objective function change?

You originally posted:

y = 225*x(1)-45*x(2)-80*x(3)-1000*x(4)-90*x(5);

Then you posted:

fun = @(x)-(225*x(1)-45*1000*x(4)-90*x(3)-1000*x(4)-80*x(5));

[ - ]
Reply by zarithMarch 22, 2017

y = 225*x(1)-45*x(2)-80*x(3)-1000*x(4)-90*x(5);

above equation was saved in an m.file for me to use in the toolbox.

As for fun = @(x)-(225*x(1)-45*1000*x(4)-90*x(3)-1000*x(4)-80*x(5));

This is a code I write, just an alternative to solve the problem. Also, I included the negative sign because the problem asked to maximize y. More details to maximize an objective function can be found here: