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Consider the following \( \underline{\text { minimum }} \) problem: Minimize: \[ C=2 x_{1}+x_{2}+x_{3} \] Subject to the constraints: \[ \left\{\begin{array}{l} 2 x_{1}+x_{2}+x_{3} \geq 4 \\ x_{1}+2 x_{2}+x_{3} \geq 6 \\ x_{1} \geq 0 \\ x_{2} \geq 0 \\ x_{3} \geq 0 \end{array}\right. \] Write the dual problem for the above minimum problem by selecting the appropriate number for each blank box shown below (Do not solve the dual problem). \[ P=[\text { Select }] \quad \hat{v} y_{1}+[\text { Select }] \quad \hat{v} y_{2} \] [Select ] \( y_{1}+ \) \( \square \) [ Select ] \( y_{2} \leq 2 \) [ Select ] \( y_{1}+ \) \( \square \) [ Select ] \( y_{2} \leq 1 \) [ Select ] \( y_{1}+ \) \( \square \) [Select ] \( y_{2} \leq 1 \) \[ y_{1} \geq 0 \quad y_{2} \geq 0 \]

Ask by Page Moran. in the United States
Mar 12,2025

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Answer

The dual problem is: Maximize: \[ D = 4y_1 + 6y_2 \] Subject to: \[ \begin{align*} y_1 + y_2 & \leq 2 \\ y_1 + 2y_2 & \leq 1 \\ y_1 + y_2 & \leq 1 \\ y_1 & \geq 0 \\ y_2 & \geq 0 \end{align*} \] Fill in the blanks as follows: - First blank: 4 - Second blank: 6 - Third blank: 1 - Fourth blank: 1 - Fifth blank: 1 - Sixth blank: 2

Solution

To formulate the dual problem from the given primal problem, we need to follow the rules of duality in linear programming. ### Primal Problem The primal problem is given as: Minimize: \[ C = 2x_1 + x_2 + x_3 \] Subject to the constraints: \[ \begin{align*} 1. & \quad 2x_1 + x_2 + x_3 \geq 4 \quad \text{(Constraint 1)} \\ 2. & \quad x_1 + 2x_2 + x_3 \geq 6 \quad \text{(Constraint 2)} \\ 3. & \quad x_1 \geq 0 \\ 4. & \quad x_2 \geq 0 \\ 5. & \quad x_3 \geq 0 \end{align*} \] ### Dual Problem The dual problem will have the following structure: Maximize: \[ D = 4y_1 + 6y_2 \] Subject to the constraints derived from the coefficients of the primal objective function and the right-hand side of the primal constraints: 1. From the first constraint \(2x_1 + x_2 + x_3\), we have: \[ y_1 + y_2 \leq 2 \] 2. From the second constraint \(x_1 + 2x_2 + x_3\), we have: \[ y_1 + 2y_2 \leq 1 \] 3. From the third constraint \(x_1\), we have: \[ y_1 + y_2 \leq 1 \] ### Final Dual Problem Putting it all together, the dual problem can be expressed as: Maximize: \[ D = 4y_1 + 6y_2 \] Subject to: \[ \begin{align*} y_1 + y_2 & \leq 2 \\ y_1 + 2y_2 & \leq 1 \\ y_1 + y_2 & \leq 1 \\ y_1 & \geq 0 \\ y_2 & \geq 0 \end{align*} \] ### Filling in the Blanks Now, we can fill in the blanks as follows: \[ P=[4] \quad \hat{v} y_{1}+[6] \quad \hat{v} y_{2} \] \[ [1] \quad y_{1}+ \quad [1] \quad y_{2} \leq 2 \] \[ [1] \quad y_{1}+ \quad [2] \quad y_{2} \leq 1 \] \[ [1] \quad y_{1}+ \quad [1] \quad y_{2} \leq 1 \] \[ y_{1} \geq 0 \quad y_{2} \geq 0 \] Thus, the selections are: - First blank: 4 - Second blank: 6 - Third blank: 1 - Fourth blank: 1 - Fifth blank: 1 - Sixth blank: 2 This completes the dual formulation.

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Bonus Knowledge

To formulate the dual problem from the given primal problem, we need to associate the constraints of the primal with variables in the dual. 1. For the primal objective function, we designate dual variables: \( y_{1} \) for the first primal constraint and \( y_{2} \) for the second primal constraint. 2. The dual objective function coefficients correspond to the right-hand side of the primal constraints, which are 4 and 6. Given these points, the dual problem can be expressed as: \[ P = 4 y_{1} + 6 y_{2} \] The constraints of the dual come from the coefficients of the primal objective function: 1. From the first constraint \( 2x_{1} + x_{2} + x_{3} \geq 4 \), we take \( 2y_{1} + 1y_{2} \leq 2 \). 2. From the second constraint \( x_{1} + 2x_{2} + x_{3} \geq 6 \), we take \( 1y_{1} + 2y_{2} \leq 1 \). 3. From the third constraint \( x_{3} \geq 0 \), we take \( 1y_{1} + 1y_{2} \leq 1 \). Combining all of this information, we get the complete dual formulation: \[ P = 4y_{1} + 6y_{2} \] \[ 2y_{1} + y_{2} \leq 2 \] \[ y_{1} + 2y_{2} \leq 1 \] \[ y_{1} + y_{2} \leq 1 \] \[ y_{1} \geq 0 \quad y_{2} \geq 0 \]

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