We are give a linear programm in standard form. Given $A\in\mathbb{R}^{n}\times\mathbb{R}^{m}$ with Rank$(A)=m$ and $B$ a basis whos basis solution is optimal. Now the right side of the constraint $Ax=b$ is exchanged for $b+\lambda d$ where $\lambda\geq0$ and $0\neq\in\mathbb{R}^{m}$. What (implicit) condition does $d$ need to fulfill in order for $B$ to still induce a optimal basic solution for every $\lambda\geq 0$.
I conjecture that $d$ needs to be linearly dependent on $b$. Because if that is the case the original basis $B$ still induces a optimal basic solution for every $\lambda\geq 0$.
However I'm not quite sure how to show that this isn't the case anymore when $d$ and $b$ are linearlly independent. Am I on the right track tough ?
If $A_B$ is the $m \times m$ matrix formed by the basic columns of $B$, the basic solution for this basis is $x = A_B^{-1} (b + \lambda d)$. This is automatically optimal if it is feasible, because it satisfies complementary slackness with the basic solution of the dual problem, which is not affected by the change from $b$ to $b+\lambda d$. So the only requirement is that all entries of $x$ are nonnegative. This will be true for all $\lambda \ge 0$ iff all entries of $A_B^{-1} d$ are nonnegative. In particular, it is not required that $d$ and $b$ are linearly dependent.
For a simple example, consider
maximize $x_1 + x_2$
subject to
$$ \eqalign{x_1 + x_2 + x_3 & = 2\cr x_1 - x_2 + x_4 & = 1\cr x_1, x_2, x_3, x_4 \ge 0 &}$$ for which the optimal basis is $x_1, x_2$. We have $A_B = \pmatrix{1 & 1\cr 1 & -1\cr}$ and $A_B^{-1} = \pmatrix{1/2 & 1/2\cr 1/2 & -1/2}$. Since $A_B^{-1} \pmatrix{1\cr 0\cr} > 0$, the optimal solution to the problem with $\pmatrix{2\cr 1\cr}$ replaced by $\pmatrix{2+\lambda\cr 1}$ will have the same basis for all $\lambda \ge 0$.