Length of lower bounds is < length(x); filling in missing lower bounds with -Inf. Problem is unbounded

Hi,

I am trying to estimate $a_{j}$ that maximize the following objective function

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

p = 4 , t= 249 observations.

Update the question ::::

The idea is to sum the rows in $A_t $ then maximize the sum over the vector of a_j.

I used the following code :

A=readtable('4times6249datacsv');
A=table2array(A);
Aeq =ones(4,996);
lb = zeros(1,4) ;
beq =ones(1,4);
x = linprog(-A, [], [], Aeq, beq, lb, []);

I received the following :

Warning: Length of lower bounds is < length(x); filling in missing lower bounds with -Inf.

> In checkbounds (line 33)

In linprog (line 241)

Problem is unbounded. what does that mean ? Any suggestion to improve the code will be appreciated

[Torsten]

Your problem formulation is weird.

Multiplying a vector of dimension 1xp with a matrix of dimension pxt gives a vector of dimension 1xt.

So what do you mean by "maximize" if the object you want to maximize is a vector ?

[Az.Sa]

The idea is to maximize the sum of the rows in $A_t$ .So we sum the rows then maximize the sum over the vector of a_j. I hope this is clear now

Answer 1

[Matt J]

A=readtable('4times6249datacsv');
A=table2array(A);

f=sum(A,2);

Aeq =ones(1,4); beq = 1;
lb = zeros(4,1) ;

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

[Az.Sa]

The idea is to maximize the sum of the rows in $A_t$ then maximize the sum over the vector of a_j.

So I sum the rows same as what you provided f=sum(A,2); but I recived: Error using linprog (line 238)

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

I tried to change it to

A=readtable('all.reg.csv');
A=table2array(A);

f=sum(A,2);

Aeq =ones(1,249); beq = 1;
lb = zeros(1,4) ;

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

Warning: Length of lower bounds is < length(x); filling in missing lower bounds with -Inf.

> In checkbounds (line 33)

In linprog (line 241)

Problem is unbounded.

[Matt J]

You cannot "maximize the sum of the rows" of because is known and therefore the sum of its rows are known. There are no optimization variables controlling the sum over the rows.

[Az.Sa]

in the former code you provided , what the code was trying to do there ? the vector $a_j$ are unknown

[Matt J]

In the former code, I assumed you have only 4 unknowns only (a_1,...,a_4), which is what your post seems to convey. If you have only 4 unknowns, it does not make sense that your Aeq matrix has more than 4 columns.

[Az.Sa]

yes I have only 4 unknowns. but when I run the code where Aeq =ones(1,4)

I received: Error using linprog (line 238)

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

[Matt J]

In my posted answer, f has 4 elements, as it should. I don't know what you've done differently.

[Az.Sa]

$A_t$ is a matrix of p X t

p = 4 , t= 249 observations.

f = sum(A,2)

means f will be 249 x 1, am I write ?

[Az.Sa]

I think I found my mistake, my A_t was 249 x 4 , so I need to tranpose it first. then solve it. which is sucessfully solved. Unfortunatily I had a corner solution a=[0;0;0;1]

[Matt J]

That's no surprise. In fact, the whole exercise could have been done analytically, without linprog,

[~,i]=max(f);
a=zeros(1,4)
a(i)=1;

[Az.Sa]

Thank you,

if I want to remove and keep , so the change in my code will be : removing lb = zeros(4,1) only ? I tried to do that but the broblem will be unbounded. I am looking to rule out the corner solution

[Matt J]

You can't rule out the corner solution, because it is the only solution, assuming the f(j) have a unique maximal element f(jmax). The only reason to expect a different solution is if there are further requirements on a(j) that you haven't yet put in your constraints.