Discussion
An objective function \(Z\) of primal variables (\(x_{1}\) and \(x_{2}\)) is described below:
Minimize \(Z = 0.07x_{1} + 0.05x_{2}\)
subject to \(0.1x_{1} \ge 0.4\)
\(0.1x_{2} \ge 0.6\)
\(0.1x_{1} + 0.2x_{2} \ge 2.0\)
\(0.2x_{1} + 0.1x_{2} \ge 1.8\)
\(x_{1}, x_{2} \ge 0\)
\(W\) is the objective function of the dual of \(Z\). \(k_{1}, k_{2}, k_{3}\), and \(k_{4}\) represent the corresponding dual variables. Which one of the following options represents the correct form of \(W\)?
Answer: A
The correct dual is the one with constraints \(0.1k_{1} + 0.1k_{3} + 0.2k_{4} \le 0.07\) and \(0.1k_{2} + 0.2k_{3} + 0.1k_{4} \le 0.05\).
For a minimisation primal with \(\ge\) constraints and non-negative variables, the dual maximises \(b^{T}k\) subject to \(A^{T}k \le c\), \(k \ge 0\). The objective uses the right-hand sides 0.4, 0.6, 2.0, 1.8, which all options share.
Each dual constraint comes from one primal variable's column. Column \(x_{1}\) has coefficients 0.1, 0, 0.1, 0.2 and cost 0.07; column \(x_{2}\) has 0, 0.1, 0.2, 0.1 and cost 0.05. That gives exactly the keyed pair.
The second option puts \(k_{2}\) in the \(x_{1}\) constraint in place of \(k_{3}\). The third and fourth swap the right-hand sides 0.07 and 0.05 between the columns.
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