# What values make the solutions in the optimal? infeasible? degenerate? etc

 -1 Note that $$c_i$$'s in the $$z_j-c_j$$ row are not coefficients of the $$x_i$$'s. We can use instead $$r_1, r_2, r_3$$ (r for row). I'm assuming there's a non-negativity constraint. we need to state necessary conditions from the word 'required', but please give sufficient if you can. What I tried: The only amount we can deduce without further conditions is $$r_3 = 0$$ because $$x_3$$ is a basic variable Letter b Based on Case 1 here, I think: $$b < 0, x_6$$ Is this really the only way we have an infeasible solution? I know it is a sufficient condition but am not sure if it is a necessary condition. Infeasible means doesn't satisfy the constraints. What if solution satisfies non-negativity constraint but doesn't satisfy other constraints? Letter a Case 1: $$r_1 = 0, a_3 > 0, r_2 \ge 0, b \ge 0$$* Case 2: $$r_1 \ge 0, a_1 \in \mathbb R, r_2 = 0, b \ge 0$$ *Do we need $$a_3 > 0$$ instead of $$a_3 \in \mathbb R$$? If so, why? Avoiding unboundedness because unboundedness implies no optimal solutions? Letter c $$b = 0, x_6$$ Letter d $$x_1$$ is not in basis. If $$r_1 < 0$$ and $$a_3 \le 0$$, the LP is unbounded. If $$b \ge 0$$, the solution is feasible. Letter e $$r_2 < \min{0, r_1, r_3}$$ If $$a_1 > 0$$, we need $$b < \min{ 15/2, 3/a_1 }$$ If $$a_1 \le 0$$, we need $$b < 15/2$$ If $$b \ge 0$$, the solution is feasible. Anywhere I went wrong? From Chapter 2 here. asked 19 Apr '16, 14:06 BCLC 8●3●13 accept rate: 0%
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