d.given the following tableau find an alternative basic feasible optimal solution. (10 pts)

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Accepted Answer

# Explanation:
## Step1: Identify Non-Basic Variables with Zero Reduced Cost
In the tableau, the reduced costs (coefficients in the $ z $-row for non-basic variables) for $ x_5 $ and $ x_6 $ are $ 2 $ and $ 3 $? Wait, no, wait. Wait, the $ z $-row is $ [1, 0, 0, 0, 0, 2, 3] $? Wait, no, the columns are $ z, x_1, x_2, x_3, x_4, x_5, x_6 $, RHS. So the $ z $-row coefficients for non-basic variables ( $ x_3, x_4, x_5, x_6 $ since $ x_1, x_2 $ are basic) are $ 0, 0, 2, 3 $? Wait, no, the $ z $-row is $ z: [1, 0, 0, 0, 0, 2, 3] $, RHS 4. Wait, no, the reduced cost for a non-basic variable is the coefficient in the $ z $-row. For a maximization problem, if a non-basic variable has a reduced cost of zero, we can introduce it into the basis to get an alternative optimal solution. Wait, here, the $ z $-row for $ x_3 $: coefficient is $ 0 $, $ x_4 $: $ 0 $, $ x_5 $: $ 2 $, $ x_6 $: $ 3 $. Wait, no, maybe I misread. Let's check the tableau again:

Rows:

- $ z $: $ [1, 0, 0, 0, 0, 2, 3] $, RHS $ 4 $
- $ x_1 $: $ [0, 1, 0, 2, -1, -1, 1] $, RHS $ 2 $
- $ x_2 $: $ [0, 0, 1, -2, 1, 2, 3] $, RHS $ 2 $

So basic variables are $ x_1, x_2 $ (since their columns have 1 in their row and 0 elsewhere). Non-basic variables: $ x_3, x_4, x_5, x_6 $, with $ x_3 = 0 $, $ x_4 = 0 $, $ x_5 = 0 $, $ x_6 = 0 $.

Reduced cost for $ x_3 $: coefficient in $ z $-row is $ 0 $, $ x_4 $: $ 0 $, $ x_5 $: $ 2 $, $ x_6 $: $ 3 $. Wait, no, for a maximization problem, the reduced cost $ c_j - z_j $. Wait, maybe the tableau is in canonical form, so $ z_j = \sum c_i a_{ij} $, and reduced cost is $ c_j - z_j $. But in the $ z $-row, the coefficients for non-basic variables are $ c_j - z_j $. If $ c_j - z_j = 0 $, we can pivot to get an alternative solution. Here, $ x_3 $ and $ x_4 $ have $ c_j - z_j = 0 $ (since their coefficients in $ z $-row are 0). Let's pick $ x_3 $ (or $ x_4 $) to enter the basis.

Let's try $ x_3 $. Now, find the pivot row: for each basic variable row, compute $ \frac{\text{RHS}}{a_{ij}} $ where $ a_{ij} > 0 $ (since we want to maintain non-negativity).

For $ x_1 $ row: $ a_{13} = 2 $, RHS = 2, so $ \frac{2}{2} = 1 $

For $ x_2 $ row: $ a_{23} = -2 $, which is negative, so we ignore it.

So the pivot row is $ x_1 $ row, pivot element is $ a_{13} = 2 $.

## Step2: Perform Pivot Operation on $ x_3 $ and $ x_1 $
Divide $ x_1 $ row by 2:

$ x_1 $ row: $ [0, \frac{1}{2}, 0, 1, -\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}] $, RHS $ 1 $

Now, update $ z $ row and $ x_2 $ row to eliminate $ x_3 $.

$ z $ row: $ z - 0 \times x_1 $ row (since $ z $-row has 0 for $ x_3 $ coefficient, no change? Wait, original $ z $-row: $ z = 4 + 0x_3 + 0x_4 + 2x_5 + 3x_6 $. After pivoting, $ x_3 $ is basic, $ x_1 $ is non-basic? Wait, no, pivot is $ x_3 $ entering, $ x_1 $ leaving? Wait, no, when we pivot, the entering variable is $ x_3 $, leaving variable is the one with the minimum ratio. We had $ x_1 $ row with ratio 1, $ x_2 $ row had negative coefficient, so leaving variable is $ x_1 $.

So new basic variables: $ x_3, x_2 $

Update $ x_2 $ row: $ x_2 $ row = original $ x_2 $ row + 2 * new $ x_1 $ row (since $ a_{23} = -2 $, so to eliminate $ x_3 $, we do $ x_2 + 2x_1 $)

Original $ x_2 $ row: $ [0, 0, 1, -2, 1, 2, 3] $, RHS 2

New $ x_1 $ row: $ [0, \frac{1}{2}, 0, 1, -\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}] $, RHS 1

So $ x_2 $ row new: $ [0, 0 + 2*\frac{1}{2}, 1 + 2*0, -2 + 2*1, 1 + 2*(-\frac{1}{2}), 2 + 2*(-\frac{1}{2}), 3 + 2*\frac{1}{2}] $, RHS $ 2 + 2*1 = 4 $

Simplify: $ [0, 1, 1, 0, 0, 1, 4] $, RHS 4? Wait, no:

Wait, $ a_{21} = 0 + 2*(1/2) = 1 $

$ a_{22} = 1 + 2*0 = 1 $

$ a_{23} = -2 + 2*1 = 0 $

$ a_{24} = 1 + 2*(-1/2) = 1 - 1 = 0 $

$ a_{25} = 2 + 2*(-1/2) = 2 - 1 = 1 $

$ a_{26} = 3 + 2*(1/2) = 3 + 1 = 4 $

RHS: $ 2 + 2*1 = 4 $

Now, $ z $ row: $ z = 4 + 0x_3 + 0x_4 + 2x_5 + 3x_6 $. But now $ x_1 $ is non-basic, $ x_3 $ is basic. Wait, maybe I made a mistake. Alternatively, let's check the reduced cost for $ x_4 $. $ x_4 $ has $ c_j - z_j = 0 $ (coefficient in $ z $-row is 0). Let's try $ x_4 $ entering the basis.

For $ x_4 $, find pivot row: $ a_{ij} > 0 $

$ x_1 $ row: $ a_{14} = -1 $ (negative, ignore)

$ x_2 $ row: $ a_{24} = 1 $, RHS = 2, so ratio $ \frac{2}{1} = 2 $

So pivot row is $ x_2 $ row, pivot element $ a_{24} = 1 $

## Step3: Perform Pivot Operation on $ x_4 $ and $ x_2 $
Divide $ x_2 $ row by 1 (no change): $ x_2 $ row: $ [0, 0, 1, -2, 1, 2, 3] $, RHS 2

Update $ x_1 $ row: $ x_1 $ row = original $ x_1 $ row + 1 * $ x_2 $ row (since $ a_{14} = -1 $, so $ x_1 + x_2 $ to eliminate $ x_4 $)

Original $ x_1 $ row: $ [0, 1, 0, 2, -1, -1, 1] $, RHS 2

New $ x_1 $ row: $ [0, 1 + 0, 0 + 1, 2 + (-2), -1 + 1, -1 + 2, 1 + 3] $, RHS $ 2 + 2 = 4 $

Simplify: $ [0, 1, 1, 0, 0, 1, 4] $, RHS 4

Update $ z $ row: $ z $-row has $ x_4 $ coefficient 0, so no change (since $ c_j - z_j = 0 $)

Now, the new basic variables are $ x_1 $ and $ x_4 $? Wait, no, after pivoting, $ x_4 $ enters, $ x_2 $ leaves? Wait, $ x_2 $ row was the pivot row, so $ x_4 $ enters, $ x_2 $ leaves. Now basic variables are $ x_1 $ and $ x_4 $.

Wait, let's check the solution. Original solution: $ x_1 = 2 $, $ x_2 = 2 $, $ x_3 = 0 $, $ x_4 = 0 $, $ x_5 = 0 $, $ x_6 = 0 $, $ z = 4 $.

After pivoting with $ x_4 $ and $ x_2 $:

Basic variables: $ x_1 $ (row 1: $ x_1 + x_2 - 2x_3 + 0x_4 + x_5 + 4x_6 = 4 $? Wait, no, maybe I messed up the pivot. Let's start over.

Correct approach: In a linear programming tableau, an alternative optimal solution exists if there is a non-basic variable with reduced cost zero (for maximization, $ c_j - z_j = 0 $; for minimization, $ z_j - c_j = 0 $). Here, the $ z $-row (objective row) for non-basic variables $ x_3, x_4, x_5, x_6 $ has coefficients: $ x_3: 0 $, $ x_4: 0 $, $ x_5: 2 $, $ x_6: 3 $. So $ x_3 $ and $ x_4 $ have reduced cost zero (since $ c_j - z_j = 0 $), meaning we can introduce them into the basis to get an alternative solution.

Let's choose $ x_3 $ to enter the basis. The basic variables are $ x_1 $ (row 1) and $ x_2 $ (row 2). For $ x_3 $, the coefficients in the basic rows are $ a_{13} = 2 $ (row 1) and $ a_{23} = -2 $ (row 2). We only consider positive coefficients for the ratio test (to ensure non-negativity of RHS after pivoting). So only row 1 has $ a_{13} = 2 > 0 $, so pivot row is row 1, pivot element is 2.

### Pivot on $ x_3 $ (column 3) and $ x_1 $ (row 1):

1. Divide row 1 by 2:
   - New row 1: $ \frac{1}{2} \times [0, 1, 0, 2, -1, -1, 1] $ RHS $ \frac{2}{2} = 1 $
   - So row 1: $ [0, \frac{1}{2}, 0, 1, -\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}] $ RHS $ 1 $

2. Update row 2 ( $ x_2 $ row) to eliminate $ x_3 $:
   - Row 2: original row 2 + 2 × new row 1 (since $ a_{23} = -2 $, we add 2×new row 1 to make $ a_{23} = 0 $)
   - New row 2: $ [0, 0, 1, -2, 1, 2, 3] + 2×[0, \frac{1}{2}, 0, 1, -\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}] $
   - Calculate each element:
     - $ z $-column: $ 0 + 2×0 = 0 $
     - $ x_1 $-column: $ 0 + 2×\frac{1}{2} = 1 $
     - $ x_2 $-column: $ 1 + 2×0 = 1 $
     - $ x_3 $-column: $ -2 + 2×1 = 0 $
     - $ x_4 $-column: $ 1 + 2×(-\frac{1}{2}) = 1 - 1 = 0 $
     - $ x_5 $-column: $ 2 + 2×(-\frac{1}{2}) = 2 - 1 = 1 $
     - $ x_6 $-column: $ 3 + 2×\frac{1}{2} = 3 + 1 = 4 $
     - RHS: $ 2 + 2×1 = 4 $
   - So row 2: $ [0, 1, 1, 0, 0, 1, 4] $ RHS $ 4 $

3. Update the $ z $-row:
   - The $ z $-row has $ x_3 $-coefficient 0, so no change (since we're pivoting a variable with reduced cost 0, the objective value remains the same).

Now, the new basic variables are $ x_3 $ (row 1) and $ x_2 $ (row 2)? Wait, no, row 1 is now the $ x_3 $ row (since we pivoted $ x_3 $ into the basis, replacing $ x_1 $). Wait, basic variables are the ones with a single 1 in their column and 0 elsewhere. Now, column 3 ( $ x_3 $) has 1 in row 1, 0 in row 2. Column 2 ( $ x_2 $) has 1 in row 2, 0 in row 1. So basic variables are $ x_3 $ and $ x_2 $, non-basic variables are $ x_1, x_4, x_5, x_6 $.

Now, find the solution:

- $ x_3 = 1 $ (row 1 RHS)
- $ x_2 = 4 $ (row 2 RHS)
- $ x_1 = 0 $ (non-basic)
- $ x_4 = 0 $ (non-basic)
- $ x_5 = 0 $ (non-basic)
- $ x_6 = 0 $ (non-basic)
- $ z = 4 $ (from $ z $-row RHS)

Wait, but let's check the constraints. Original constraints (assuming standard form, $ x_1, x_2, x_3, x_4, x_5, x_6 \geq 0 $):

From row 1 (original): $ x_1 + 2x_3 - x_4 - x_5 + x_6 = 2 $

From row 2 (original): $ x_2 - 2x_3 + x_4 + 2x_5 + 3x_6 = 2 $

With $ x_1 = 0 $, $ x_2 = 4 $, $ x_3 = 1 $, $ x_4 = 0 $, $ x_5 = 0 $, $ x_6 = 0 $:

Row 1: $ 0 + 2(1) - 0 - 0 + 0 = 2 $ (satisfies)

Row 2: $ 4 - 2(1) + 0 + 0 + 0 = 2 $ (satisfies)

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QUESTION IMAGE

Question

Explanation:

Step1: Identify Non-Basic Variables with Zero Reduced Cost

Step2: Perform Pivot Operation on \( x_3 \) and \( x_1 \)

Answer:

Step1: Identify Non-Basic Variables with Zero Reduced Cost

Step2: Perform Pivot Operation on \( x_3 \) and \( x_1 \)

Step3: Perform Pivot Operation on \( x_4 \) and \( x_2 \)

Pivot on \( x_3 \) (column 3) and \( x_1 \) (row 1):