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

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Az.Sa
Az.Sa 2023-2-2
评论： Matt J 2023-2-3
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
  2 个评论
Torsten
Torsten 2023-2-2
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
Az.Sa 2023-2-3
编辑：Az.Sa 2023-2-3
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

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采纳的回答

Matt J
Matt J 2023-2-2
编辑：Matt J 2023-2-2
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);
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Az.Sa
Az.Sa 2023-2-3
编辑：Az.Sa 2023-2-3
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
Matt J 2023-2-3
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.

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