Discussion
Consider the following LP problem :
Max. : \(Z = x_{1} + x_{2/2}\)
Subject to the constraints :
\(3x_{1} + 2x_{2} \le 12\)
\(5x_{1} = 10\)
\(x_{1} + x_{2} \ge 8\)
\(-x_{1} + x_{2} \ge 4\) and
\(x_{1}, x_{2} \ge 0.\)
Then the LP problem has :
Answer: B
The LP has no feasible solution.
The equality \(5x_{1} = 10\) fixes \(x_{1} = 2\). Then:
\[ \begin{aligned} 3(2) + 2x_{2} \le 12 &\Rightarrow x_{2} \le 3 \\ 2 + x_{2} \ge 8 &\Rightarrow x_{2} \ge 6 \end{aligned} \]
No \(x_{2}\) can be both at most 3 and at least 6, so the constraints are inconsistent. (The last constraint, \(x_{2} \ge 6\), agrees with the third but cannot help.) The objective, printed with the slip \(x_{2/2}\), plays no part: with an empty feasible region there is no feasible solution, degenerate or not.
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