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
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?
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?
\(b = 0, x_6\)
\(x_1\) 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.
\(r_2 < \min{0, r_1, r_3}\) 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
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