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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

  1. there's a non-negativity constraint.

  2. 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's gravatar image

accept rate: 0%

edited 23 Apr '16, 00:50

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Asked: 19 Apr '16, 14:06

Seen: 484 times

Last updated: 23 Apr '16, 00:50

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