Length of lower bounds is < length(x); filling in missing lower bounds with -Inf. Problem is unbounded
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Hi,
I am trying to estimate $a_{j}$ that maximize the following objective function
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1283421/image.png)
where ;
is unknown vector of 1 X p ,
is a matrix of p X t
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1283426/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1283431/image.png)
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
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 ?
采纳的回答
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);
11 个评论
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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