I need help with this problem. The exact problem is in this link http://d2vlcm61l7u1fs.cloudfront.net/media%2F959%2F959d289e-6f26-4e21-875e-bb71f3f5a49f%2Fphprimn1q.png
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Consider the linear programming problem of minimizing $c'x$ subject to $Ax=b, x \geq 0.$ Let $x^*$ be an optimal solution, assumed to exist , and let $p^*$ be an optimal solution to the dual.
(a) Let $\tilde{x}$ be an optimal solution to the primal, when $c$ is replaced by some $\tilde{c}$ . Show that $(\tilde{c} - c)' (\tilde{x} - x^* ) \leq 0.$
(b) Let the cost vector be fixed at c, but suppose that we now change $b$ to $\tilde{b}$, and let $\tilde{x}$ be a corresponding optimal solution to the primal. Prove that $(p^* )'(\tilde{b}-b) \leq c'(\tilde{x} - x^* ).$
(a) Both $x^*$ and $\tilde{x}$ are feasible solutions to $Ax=b, x \geq 0$, by definition of $\tilde{x}$, we have
$$\tilde{c}'\tilde{x}\leq \tilde{c} x^*$$
Similarly, by definition of $x^*$, we have
$$c'x^*\leq c'\tilde{x}.$$
Adding the two terms up, we have $$\tilde{c}'\tilde{x}+c'x^*\leq \tilde{c} x^*+c'\tilde{x}$$
which is equivalent to $$(\tilde{c}-c)'(\tilde{x}-x^*) \leq 0.$$
(b) $p^*$ is a feasible solution to $p'A \leq c$, by weak duality, we have
$$p^*\tilde{b} \leq c'\tilde{x}.$$
Also, by strong duality $$p^*b=c'x^*.$$
Hence, we have $$p^*\tilde{b}- p^*b \leq c'\tilde{x}-c'x^*.$$
$$p^*(\tilde{b}-b) \leq c'(\tilde{x}-x^*).$$