Math Problem Statement

pivot the 3 to obtain a new tableau and corresponding solutions

Pivot the tableau about 2, 9, 3, and 1 (row 2, column 2) and compute the solution to the resulting tableau. Identify feasible solutions, the new group I and group II variables, and determine which feasible solution maximizes M.

Solution

Let's go through the question step by step:

(a) Identify Group I and Group II Variables

The tableau provided is a standard form in a linear programming problem setup. In this context:

  • Group I variables (basic variables) are those currently in the basis and have corresponding columns that act as identity matrix columns.
  • Group II variables (non-basic variables) are those that are currently out of the basis and usually have zero entries in their columns for the basic variable rows.

From the tableau:

x & y & u & v & M & \text{RHS} \\ \hline 2 & 9 & 1 & 0 & 0 & 180 \\ 3 & 1 & 0 & 1 & 0 & 300 \\ -10 & -8 & 0 & 0 & 1 & 0 \\ \end{array}$$ In this setup: - **Basic variables (Group I)**: `u`, `v`, and `M` (since they each have a column that resembles part of the identity matrix). - **Non-basic variables (Group II)**: `x` and `y`. ### (b) Pivoting the Tableau To perform a pivot, we need to: 1. Identify the pivot element, which is the element we want to make the new basis variable. 2. Adjust the tableau to reflect this new basis. #### Step 1: Choosing the Pivot Element We are instructed to pivot about the entry `3` in row 2, column 2 (associated with variable `x`). This means we will bring `x` into the basis, replacing `v`. #### Step 2: Performing Row Operations To make `x` a basic variable, we perform the following row operations: 1. **Make the pivot element (3) into 1** by dividing row 2 by 3. $$R_2 = \frac{1}{3}R_2 = \left[1, \frac{1}{3}, 0, \frac{1}{3}, 0, 100\right]$$ 2. **Zero out the other entries in the x-column** by adjusting rows 1 and 3. - For row 1: $$R_1 = R_1 - 2 \times R_2$$ - For row 3: $$R_3 = R_3 + 10 \times R_2$$ Let's calculate these adjustments: **Row 1 Adjustment**: \[ R_1 = [2, 9, 1, 0, 0, 180] - 2 \times \left[1, \frac{1}{3}, 0, \frac{1}{3}, 0, 100\right] \] $$R_1 = [0, 8\frac{2}{3}, 1, -\frac{2}{3}, 0, -20]$$ **Row 3 Adjustment**: \[ R_3 = [-10, -8, 0, 0, 1, 0] + 10 \times \left[1, \frac{1}{3}, 0, \frac{1}{3}, 0, 100\right] \] $$R_3 = [0, -\frac{2}{3}, 0, \frac{10}{3}, 1, 1000]$$ The new tableau is: \[ \begin{array}{ccccc|c} x & y & u & v & M & \text{RHS} \\ \hline 0 & 8\frac{2}{3} & 1 & -\frac{2}{3} & 0 & -20 \\ 1 & \frac{1}{3} & 0 & \frac{1}{3} & 0 & 100 \\ 0 & -\frac{2}{3} & 0 & \frac{10}{3} & 1 & 1000

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Simplex Method
Pivoting

Formulas

Pivoting operations for simplex method
Row operations for tableau transformation

Theorems

Simplex Algorithm
Basic Feasible Solution

Suitable Grade Level

Undergraduate

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