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Suppose that $A$ is an $m$ by $n$ real matrix, $c$ and $x$ are $n$-dimensional real vector, and $b$ is an $m$-dimensional real vector. Let '$\le$' be an elementwise partial order between two vectors. Suppose further that $m\le n$ and $A$ has its full rank. Then, consider a linear program(LP) ;

Minimize $c^Tx$ subject to $Ax\le b$.

It is elementary that if $D=\{x\in\mathbb R^n:Ax\le b\}$ is bounded and nonempty, then the LP has the optimal solution ; being closed and bounded, $D$ is compact. Since $x\mapsto c^Tx$ is continuous, it has its minimum.

Suppose now that $D$ is not necessarily bounded but suppose further that $c_i\ge0$ for each $i$ and that LP has $x_i\ge0$ as a constraint for each $i$. Does the LP have the optimal solution?

My trial 1

Yes. The LP always has the optimal solution. $L(x)=c^Tx\ge0$ by the hypotheses. Since $S=\{L(x)\in\mathbb R:Ax\le b\}$ is bounded below (by 0) and $S$ is not empty, $S$ has its infimum. $S$ is a continuous image of a closed set. A continuous image of a closed set is not closed in general. But if $S$ were closed, $S$ has its minimum and the LP has the optimal solution.

My trial 2

If $n=3$, the set $D$ resides in the first octant. $px+qy+rz=k$ ($p, q, r\ge0$) indicates a plane, whose intercept to each axis is nonnegative. Then, although $D$ is not bounded, $S=\{k\in\mathbb R:px+qy+rz=k,\text{constraints}\}$ is bounded below by 0. Therefore(?) $S$ has its minimum.

govindah
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  • What does element-wise total order mean? How does one compare $(0,1)$ and $(1,0)$ for instance? – AlvinL Jul 03 '24 at 04:50
  • Your claim that if $D$ is bounded an optimal solution exists is not quite right. The correct statement is that if $D$ is nonempty and bounded an optimal solution exists. You gave no reason to conclude that $D$ is nonempty. – Robert Israel Jul 03 '24 at 05:04
  • @RobertIsrael You're right. I forgot the nonempty condition. Now it's included. – govindah Jul 03 '24 at 05:10
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    @AlvinL Oh it's not a total order. it should be a partial order since one can't compare the two vector you suggested. Now I've edited. I wrote the term to justify the expression $Ax\le b$ – govindah Jul 03 '24 at 05:22

1 Answers1

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By the Fundamental Theorem of Linear Programming, a linear programming problem is either infeasible, unbounded (which for a minimization problem means it has feasible solutions with arbitrarily large negative objective values) or has an optimal solution. If $c \ge 0$ and all feasible solutions $x \ge 0$, then the objective value $c^T x \ge 0$ for all feasible solutions, so it can't be unbounded. Therefore if there are feasible solutions, there is an optimal solution.

Robert Israel
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