Karush–Kuhn–Tucker Conditions (MATLAB)

Code

%KKT with 2 variables, 2 linear constraints with one equality and one

%inequality constraint at once or 1 equality and no inequality constraint at once or no

%equality and one inequality constraint at once

clc;

syms x1 x2 g1 h1 u1 v1 s1

%input of a function like (x1 - 1.5).^2 + (x2 - 1.5).^2 for up to two variables x1, x2

fx=input('Enter the function of ''x'' to be minimized of the form (x - 4).^2 + (y - 6).^2 ');

clc;

C=input('Enter the number of constraints in the problem, e.g. 10 for one equality and no inequality constraint ');

clc;

if C==01

    %constraints input

    g1=input('Enter inequality constraint in the form x1+x2-10 ');

    clc;

    %Lagrange function calculation

    L=fx+(u1*(g1+s1.^2));

    %gradient conditions

    dx1=(diff (L,x1)); dx2=(diff (L,x2)); du1=(diff (L,u1)); ds1=(diff (L,s1));

    %simultaneous solution of gradient conditions

    [sols1,solu1,solx1,solx2]=solve(dx1==0,dx2==0,du1==0,ds1==0);

    a=zeros(3,4);

    for i=1:3

        a(i,1)=solx1(i,1);

        a(i,2)=solx2(i,1);

        a(i,3)=sols1(i,1);

        a(i,4)=solu1(i,1);

    end

    SET1=a(1,:); SET2=a(2,:); SET3=a(3,:);

    %feasibility and non-negativity of Lagrange multiplier check

    if SET1(1,3)>0 || SET1(1,4)>0

        disp('Set 1 '); disp ('     x1    x2    s    u'); disp (SET1)

    end

    if SET2(1,3)>0 || SET2(1,4)>0

        disp('Set 2 '); disp ('     x1    x2    s    u'); disp (SET2)

    end

    if SET3(1,3)>0 || SET3(1,4)>0

        disp('Set 3 '); disp ('     x1    x2    s    u'); disp (SET3)

    end

end

if C==10

    %constraints input

    h1=input('Enter equality constraint in the form x1+x2-10 ');

    clc;

    %Lagrange function calculation

    L=fx+(v1*h1);

    %gradient conditions

    dx1=(diff (L,x1)); dx2=(diff (L,x2)); dv1=(diff (L,v1));

    %simultaneous solution of gradient conditions

    [solv1,solx1,solx2]=solve(dx1==0,dx2==0,dv1==0);

    a=zeros(1,3);

    a(1,1)=solx1;

    a(1,2)=solx2;

    a(1,3)=solv1;

    disp('Set 1 '); disp ('     x1    x2    v'); disp (a);

end

if C==11

    %constraints input

    g1=input('Enter inequality constraint in the form x1+x2-10 ');

    clc;

    h1=input('Enter equality constraint in the form x1+x2-10 ');

    clc;

    %Lagrange function calculation

    L=fx+(v1*h1)+(u1*(g1+s1.^2));

    %gradient conditions

    dx1=(diff (L,x1)); dx2=(diff (L,x2)); du1=(diff (L,u1)); ds1=(diff (L,s1)); dv1=(diff (L,v1));

    %simultaneous solution of gradient conditions

    [sols1,solu1,solv1,solx1,solx2]=solve(dx1==0,dx2==0,du1==0,ds1==0,dv1==0);

    a=zeros(3,5);

    for i=1:3

        a(i,1)=solx1(i,1);

        a(i,2)=solx2(i,1);

        a(i,3)=sols1(i,1);

        a(i,4)=solu1(i,1);

        a(i,5)=solv1(i,1);

    end

    SET1=a(1,:); SET2=a(2,:); SET3=a(3,:);

    %feasibility and non-negativity of Lagrange multiplier check

    if SET1(1,3)>0 || SET1(1,4)>0

        disp('Set 1 '); disp ('       x1         x2         s         u         v'); disp (SET1)

    end

    if SET2(1,3)>0 || SET2(1,4)>0

        disp('Set 2 '); disp ('       x1         x2         s         u         v'); disp (SET2)

    end

    if SET3(1,3)>0 || SET3(1,4)>0

        disp('Set 3 '); disp ('       x1         x2         s         u         v'); disp (SET3)

    end

end

Example 4.29 Arora pg. 128

Step One

In this step the user enters the objective function:

Command Window

Enter the function of 'x' to be minimized (x1 - 1.5).^2 + (x2 - 1.5).^2

Step Two

In this step the user specifies the type(s) and quantity of constraints the objective function is subjected to.

01 means no equality and one inequality constraint, 10 means one equality and no inequality constraint, 11 means one equality and one inequality constraint.

Command Window

Enter the number of constraints in the problem, e.g. 10 for one equality and no inequality constraint 01

Step Three

In this step the user enter the constraint(s).

Command Window

Enter inequality constraint in the form x1+x2-10 x1 + x2 – 2

Step Four

Output of the program; feasible set with candidate minimum point(s).

Command Window

Set 1

     x1    x2    s    u

     1     1     0     1

Example 4.27 Arora pg. 122

Step One

Command Window

Enter the function of 'x' to be minimized (x1 - 1.5).^2 + (x2 - 1.5).^2

Step Two

Command Window

Enter the number of constraints in the problem, e.g. 10 for one equality and no inequality constraint 10

Step Three

Command Window

Enter equality constraint in the form x1+x2-10 x1 + x2 – 2

Step Four

Command Window

Set 1

     x1    x2    v

     1     1     1

Modifications

Extension to five or even more variables and more constraints can be made via simple modifications to the code. The structure of the new code will remain similar to the code mention above, though it will have many more lines of instructions.