Consider the primal problem : Maximize \(z=5 x_{1}+12 x_{2}+4 x_{3}\) Subject…
2022
Consider the primal problem :
Maximize \(z=5 x_{1}+12 x_{2}+4 x_{3}\)
Subject to
\(\begin{array}{l} \text {Subject to }x_{1}+2 x_{2}+x_{3}=10\\ 2 x_{1}-x_{2}+3 x_{3}=8 \\ x_{1}, x_{2}, x_{3} \geq 0 \end{array}\)
its dual problem is
Minimize \(w=10 y_{1}+8 y_{2}\)
Subject to
\(\begin{array}{l} y_{1}+2 y_{2} \geq 5\\ 2 y_{1}-y_{2} \geq 12 \\ y_{1}+3 y_{2} \geq 4 \end{array}\)
Which of the following is correct ?
- A.
\(𝑦_1≥0,𝑦_2\)unrestricted - B.
\(𝑦_1≥0,𝑦_2≥0\) - C.
\(𝑦_1\)is unrestricted,\(𝑦_2≥0\) - D.
\(𝑦_1\)is unrestricted.\(𝑦_2\)restricted
Attempted by 9 students.
Show answer & explanation
Correct answer: D
Answer: both y1 and y2 are unrestricted in sign.
Reason: For each equality constraint in the primal, the corresponding dual variable is unrestricted. The primal has two equality constraints, so both dual variables are unrestricted. The primal variables x1, x2, x3 are nonnegative, which determines the direction of the dual inequalities (≥).
From x1 ≥ 0: y1 + 2y2 ≥ 5
From x2 ≥ 0: 2y1 - y2 ≥ 12
From x3 ≥ 0: y1 + 3y2 ≥ 4
Conclusion: The correct sign restrictions are y1 unrestricted and y2 unrestricted. None of the provided options correctly state both dual variables as unrestricted.