maximizing objective function with equality and inequality constraints

Hi,

I want to estimate x_1 ,...,x_4 by maximizing

subject to and ,

which function can help me to solve this problem ,

Also, how can I convert this objective function to be convex if that is possible.

Thanks in advance

[Torsten]

x1=x2=x3=0, x4=1

Should be obvious because the coefficient of x4 has the maximum value of all coefficients.

And your objective function is convex.

Answer 1

[Aditya]

Hi,

I understand that you want to solve this linear programming problem.

The solution for your example is trivial, as pointed out by @Torsten in comments.

In general, you can also use the linprog function to solve such problems. Here is an example to arrive at the trivial solution for your example.

Based on the documentation of linprog, I have defined the variables:

f = [4.22117991, 4.21111679, 4.22994893, 4.23060394];
Aeq = [1, 1, 1, 1];
lb = [0, 0, 0, 0];
beq = [1];
x = linprog(-f, [], [], Aeq, beq, lb, []);

You can see that the variable x is [0;0;0;1] which is the trivial solution to this problem.

The reason why I have passed negative f ( -f ) is because linprog minimizes the objective function. So, in order to maximize f, we minimize -f.

[Az.Sa]

Assume A_1,...,A_4 are vecotrs of 249 elements of each and my objective is to maximize :

I tried to write the code as :

clear all

data=readtable('utility.of.fitted.regressors.csv');

values=table2array(data);

A = values(:,[1]);

B= values(:,[2]);

C=values(:,[3]);

D=values(:,[4]);

f = [A , B, C , D ];

Aeq = ones(1,249);

lb = zeros(1,249);

beq = [1];

x = linprog(-f, [], [], Aeq, beq, lb, []);

I have an error Error using linprog (line 238)

The number of columns in Aeq must be the same as the number of elements of f.

length(f)= 249

[Torsten]

If A1,...,A4 are vectors , what do you try to maximize ? Vectors defined by A1x1+A2x2+A3x3+A4x4 cannot be maximized, only scalar values.

[Az.Sa]

I think I missrepresent my former question: I have the following miximization objective function

where ; is a vector of 1 X p , is a matrix of p X t

The idea is to maximize the sum of the rows . That is, we sum the values evaluated at each t ( in each row) and maximize the sum over the vector of a_j. I hope this makes more sense.

[Torsten]

Yes, that makes sense.

And the solution is as simple as in the case for t=1.

Take the row with the maximum sum, say row i, and set the corresponding a(i) to 1 and all other a(j)'s to 0.