Math Problem Statement

Consider the following simplex tableau being used to maximize Z. The current solution is not optimal and needs at least one more pivot operation to complete the method. (A) Which variable is the entering variable? (B) Calculate the ratios, entered as decimals rounded to two decimal places: (Enter DNE in the last column if the ratio is not positive.) (C) Which variable is the exiting variable?

Solution

Let's go step-by-step to determine the answers.


(A) Entering Variable

The entering variable is determined by selecting the variable in the objective row (bottom row of the tableau) with the most negative coefficient. This indicates the direction to increase ZZ.

From the objective row:
1,1,0,0,2,0,11, -1, 0, 0, -2, 0, 1
Here, 2-2 is the most negative coefficient, and it corresponds to s2s_2.

Answer: s2s_2 is the entering variable.


(B) Calculate the Ratios

To determine the exiting variable, we calculate the ratios for each row using:

Ratio=Right-Hand Side (RHS)Coefficient of Entering Variable\text{Ratio} = \frac{\text{Right-Hand Side (RHS)}}{\text{Coefficient of Entering Variable}}

  1. Identify the column for s2s_2 (the entering variable). This is the 5th column in the tableau.
  2. For each row, divide the RHS (last column) by the coefficient in the s2s_2 column, but only if the coefficient is positive. If the coefficient is zero or negative, the ratio is DNE (Does Not Exist).

The tableau:

& x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & Z & \text{RHS} \\ \hline \text{Row 1} & 8 & 6 & 0 & 1 & 3 & 0 & 0 & 174 \\ \text{Row 2} & 0 & 1 & 1 & 0 & 5 & 0 & 0 & 196 \\ \text{Row 3} & 8 & 8 & 0 & 0 & 3 & 1 & 0 & 77 \\ \text{Objective} & 1 & -1 & 0 & 0 & -2 & 0 & 1 & 57 \end{array}$$ Now calculate the ratios: - **Row 1**: Coefficient of $$s_2 = 3$$ (positive). $$\text{Ratio} = \frac{174}{3} = 58.00$$ - **Row 2**: Coefficient of $$s_2 = 5$$ (positive). $$\text{Ratio} = \frac{196}{5} = 39.20$$ - **Row 3**: Coefficient of $$s_2 = 3$$ (positive). $$\text{Ratio} = \frac{77}{3} = 25.67$$ --- | Row | Basic Variable | Ratio as Decimal | |-------|----------------|------------------| | Row 1 | $$s_1$$ | $$58.00$$ | | Row 2 | $$x_2$$ | $$39.20$$ | | Row 3 | $$x_1$$ | $$25.67$$ | --- ### (C) **Exiting Variable** The exiting variable is the one with the **smallest positive ratio** calculated above. This ensures feasibility after the pivot. From the ratios: - Row 1: $$58.00$$ - Row 2: $$39.20$$ - Row 3: $$25.67$$ The smallest positive ratio is $$25.67$$, which corresponds to **Row 3**. The basic variable in Row 3 is $$x_1$$. **Answer:** $$x_1$$ is the exiting variable. --- ### Final Answers: 1. **Entering variable:** $$s_2$$ 2. **Ratios:** - Row 1: $$58.00$$ - Row 2: $$39.20$$ - Row 3: $$25.67$$ 3. **Exiting variable:** $$x_1$$ --- ### Related Questions: 1. What happens to the tableau after performing the pivot operation? 2. How do you determine optimality in the simplex method? 3. What is the significance of selecting the entering variable with the most negative coefficient? 4. What if no positive ratios exist for the exiting variable? 5. Can degeneracy occur in this tableau? Why or why not? **Tip:** Always verify calculations for the entering and exiting variables carefully, as mistakes here can affect the entire optimization process.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Simplex Method
Pivot Operations
Optimization

Formulas

Ratio = RHS / Coefficient of entering variable (when coefficient > 0)
Select entering variable: Most negative coefficient in the objective row
Exiting variable: Smallest positive ratio

Theorems

Simplex Method Optimality Condition
Feasibility Rule for Pivoting

Suitable Grade Level

Undergraduate (Optimization Methods, Operations Research)