Simplex Method Questions and Answers

Answer: C

The problem has a unique optimal solution.

On the equality \(5x_{1} + 3x_{2} = 15\), write \(x_{1} = 3 - 0.6x_{2}\). The objective becomes

\[ x_{1} + 2.5x_{2} = 3 + 1.9x_{2} \]

which increases with \(x_{2}\). The constraints limit \(x_{2}\): \(-x_{1} + x_{2} \le 1\) gives \(x_{2} \le 2.5\), and \(2x_{1} + 5x_{2} \le 10\) gives \(3.8x_{2} \le 4\), so \(x_{2} \le \frac{20}{19}\). The optimum is the single point \(x_{2} = \frac{20}{19}\), \(x_{1} = \frac{45}{19}\), with value 5.

The trap is option B: the objective is parallel to \(2x_{1} + 5x_{2} \le 10\), which would give a whole edge of optima if the feasible region were two-dimensional. The equality constraint cuts the region down to a segment that meets that line at one point only. The region is non-empty and bounded, so A and D are wrong.

Answer: B

The optimal value is 4.

Since \(y \le x\),

\[ -4x + 5y \le -4x + 5x = x \le 4 \]

Both inequalities become equalities at \(x = y = 4\), which is feasible, so the maximum is exactly 4.

8 would need \(x = 8\), outside \(x \le 4\). 0 is the value at the origin, not the maximum. The bound above shows the objective cannot grow without limit, so an optimum does exist.

Answer: B

In the simplex method a negative value of the net evaluation \(\Delta_{j}\) means that bringing in that variable can improve the objective. If every element in its column is negative or zero, there is no row that limits how far the variable can be increased, since no basic variable is driven to zero, so the objective can be improved without limit. The solution is therefore unbounded, option (B). A degenerate solution has a basic variable equal to zero, and an alternative solution has a zero net evaluation.