Find KKT solution to a given problem

I'm trying to solve kkt problem with matlab but I'm having problem with my code, please, I want some one to assist me in correcting errors in my codes

problem:

Maximize R= 208*x1**0.323168*x2**0.172084*x3**0.085998*x4**0.188794*x5**0.229956

subject to:

2.8 * x1 + 22 * x2 + 4 * x3 + 0.272 * x4 + x5 <= 492

22 * x2 + 4 * x3 <= 212

x5 <= 20

Bounds

x1= (lb = 40, ub = 80)

x2= (lb = 5, ub = 12)

x3= (lb = 4, ub = 11)

x4= (lb = 20, ub = 70)

x5= (lb = 7, ub = 20)

X0 = [41,6,5,22,8]

Proposed solution:

fun = @(x)-208*x(1)^0.323168*x(2)^0.172084*x(3)^0.085998*x(4)^0.188794*x(5)^0.229956;

x0=[41,6,5,22,8]; %initial guess for the solution

%express first constraint in the form A*x<=b:

Aeq=[41,6,5,22,8];

beq = 1;

A=[2.8,22,4,0.272]; b=492;

C=[0,22,4,0,0]; d=212;

E=[0,0,0,0,1]; f=20;

%express constraints 2 and 3 by defining vectors lb and ub:

lb=[40,5,4,20,7]; %no lower blound constraints

ub=[80, 12,11,70,20]; %upper bounds on x(4),x(5)

[x,fval,exitflag,output,lambda]=fmincon(fun,x0,A,b,[],[],lb,ub);

disp(x) %value of x at the solution

disp(-fun(x)); %value of the function you wanted to maximize

disp(A*x') %check constraint 1

disp(lambda)

Answer 1

[Sai Teja G]

Hi Ezra,

You can refer to the below code to solve the problem.

% Objective function
fun = @(x)-208*x(1)^0.323168*x(2)^0.172084*x(3)^0.085998*x(4)^0.188794*x(5)^0.229956;

% Initial guess for the solution
x0 = [41, 6, 5, 22, 8];

% Express first constraint in the form A*x <= b
A = [2.8, 22, 4, 0.272, 1]; % Add the missing element for x5
b = 492;

% Combine the second and third constraints into A and b
A = [A; 0, 22, 4, 0, 0; 0, 0, 0, 0, 1]; % Add the second and third constraints
b = [b; 212; 20];

% Lower and upper bounds
lb = [40, 5, 4, 20, 7]; % Lower bounds
ub = [80, 12, 11, 70, 20]; % Upper bounds

% Run fmincon
[x, fval, exitflag, output, lambda] = fmincon(fun, x0, A, b, [], [], lb, ub);

Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.

% Display results
disp(x)         % value of x at the solution

   80.0000    7.6364   11.0000   70.0000   20.0000

disp(-fun(x));  % value of the function you wanted to maximize

   6.6388e+03

disp(A*x' <= b) % check constraints

   1
   1
   1

disp(lambda)    % display Lagrange multipliers

         eqlin: [0×1 double]
      eqnonlin: [0×1 double]
       ineqlin: [3×1 double]
         lower: [5×1 double]
         upper: [5×1 double]
    ineqnonlin: [0×1 double]

Hope this helps!

[Ezra]

Thank Dr Teja,

please, what are the values of lamda?

[John D'Errico]

This looks reasonable to me. I would probably have maximized the log of the objective though, as that is often a good way to avoid any potential numerical problems. However, in this case, the variables live in a small enough (well bounded) space that not taking the log will not be an issue. Fmincon seems happy with the result.

[Sai Teja G]

@Ezra, you can execute the code provided above and then examine the values of the Lagrange multipliers. To do this, enter 'lambda.upper' and 'lambda.lower' in the command window and run these commands to check the corresponding values.

[Ezra]

I have run the code in MATLAB R2013a and got almost all the results except for the values of lamda I got this error message :

Undefined variable "lamda" or function "lamda.upper".

Error in func (line 20)

display(lamda.upper)

thank you

[Star Strider]

Try:

disp(lambda.upper)

[Ezra]

Thank you all for your effort the matter of lamda is solved, it was an error in spelling from me. Please sorry for taking your time.

merci beaucoup