If the problem $\min\{\mathbf{c}^{T}\mathbf{x}:A\mathbf{x}=\mathbf{b},\mathbf{x}\ge 0\}$ has a finite optimal solution, then the new problem $\min\{\mathbf{c}^{T}\mathbf{x}:A\mathbf{x}=\mathbf{b}′,\mathbf{x}\ge 0\}$ cannot be unbounded, no matter what value the vector $\mathbf{b}′$ might take. I believe this proof can be done using duality theorem? But I am not entirely sure how to go about it.
2026-03-24 20:32:00.1774384320
Proof that if an LP has a finite optimal solution then it cannot be unbounded for $\mathbf{b}'$
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If $$\min\{c^{\top}x\ :\ Ax=b, x \geq 0\}$$ has a finite optimal solution, then by strong duality also the dual $$\max\{b^{\top}x \ : \ A^{\top}y \leq c \}$$ has a finite optimal solution. Observe that for each vector $b^{\prime} \in \mathbb{R}^m$, the dual problem problem can not be unfeasible. Therefore it is either unbounded (and the primal is unfeasible), or it has a finite optimal solution (and the primal as well).