Math Problem Statement

pivot the 9 to obtain a new tableau and corresponding solutions

pivot the 9 to obtain a new tableau and corresponding solutions

Solution

Let's go through the steps systematically. The problem involves a linear programming tableau that needs to be pivoted, and the goal is to determine the feasibility of solutions and their effect on maximizing MM.

Problem Breakdown:

  1. Identifying Groups I and II:

    • Group I typically contains the basic variables (the variables with pivot elements in the initial tableau).
    • Group II contains the non-basic variables.

  2. Pivoting Process:

    • We need to pivot around the entry "9" (in row 1, column 2) in the tableau.
    • Pivoting involves transforming this element to 1 and making all other entries in the same column zero.

  3. Reevaluating the Solution:

    • After pivoting, we re-calculate the values of the variables and determine the feasibility of the solution.

Tableau Analysis and Steps:

The initial tableau is:

& x & y & u & v & M & \text{RHS} \\ \hline \text{Row 1} & 2 & 9 & 1 & 0 & 0 & 180 \\ \text{Row 2} & 3 & 1 & 0 & 1 & 0 & 300 \\ \text{Row 3 (Objective)} & -10 & -8 & 0 & 0 & 1 & 0 \\ \end{array}$$ #### Step 1: Pivot the 9 (Row 1, Column 2) 1. **Make the pivot element (9) equal to 1** by dividing the entire first row by 9. Updated Row 1: $$\left[ \frac{2}{9}, 1, \frac{1}{9}, 0, 0, 20 \right]$$ 2. **Make all other elements in the pivot column (Column 2) zero** by adjusting the other rows: - Row 2: $$\text{New Row 2} = \text{Row 2} - 1 \times \text{New Row 1}$$ - Row 3: $$\text{New Row 3} = \text{Row 3} + 8 \times \text{New Row 1}$$ After applying these operations, the tableau should look as follows: $$\begin{array}{c|ccccc|c} & x & y & u & v & M & \text{RHS} \\ \hline \text{Row 1} & \frac{2}{9} & 1 & \frac{1}{9} & 0 & 0 & 20 \\ \text{Row 2} & \frac{25}{9} & 0 & -\frac{1}{9} & 1 & 0 & 280 \\ \text{Row 3 (Objective)} & -\frac{22}{9} & 0 & \frac{8}{9} & 0 & 1 & 160 \\ \end{array}$$ ### Feasibility and Grouping: - **Group I (Basic Variables)** now includes $$y$$ and $$v$$. - **Group II (Non-basic Variables)** includes $$x$$, $$u$$, and $$M$$. ### Feasible Solution Analysis: 1. **Feasibility** is determined by the non-negativity of the RHS values: - Both RHS values (20 and 280) are positive, so the solution is feasible. 2. **Maximizing $$M$$**: - The objective function value in Row 3 (160) indicates that this is the maximum value for $$M$$ under the current pivot operations. ### Summary: - **Feasible Solution**: Yes, this tableau represents a feasible solution. - **Maximizing $$M$$**: The solution here maximizes $$M$$ to 160. Would you like more details on each step or explanation of the pivoting process? --- Here are 5 related questions that could expand your understanding: 1. How do we select pivot elements in other situations for different objectives? 2. What is the purpose of dividing the pivot row by the pivot element? 3. How does changing the objective function impact the tableau? 4. How can we identify infeasible solutions during tableau transformations? 5. What are alternate methods for maximizing objectives in linear programming? **Tip:** When pivoting, always verify each transformed row to avoid calculation errors in large tableaux.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Programming
Simplex Method
Tableau Pivoting

Formulas

Pivoting Formula: Adjusting rows to make all other entries in the pivot column zero after dividing the pivot row by the pivot element

Theorems

Simplex Method for Linear Optimization

Suitable Grade Level

Undergraduate

Related Recommendation

Pivoting in Simplex Tableau: Pivot the 9 to Obtain New Solutions

Simplex Tableau and Pivot Operations in Linear Programming

Simplex Tableau Pivot and Solution Computation

How to Find the Pivot in a Simplex Tableau - Step-by-Step Guide

Simplex Method Pivoting for Maximizing Variable M in Linear Programming