1. Convert to Standard Form

First, we convert the given linear program into the standard form required for the simplex method. This involves three main steps:

Objective Function: The simplex method

Question

1. Convert to Standard Form

First, we convert the given linear program into the standard form required for the simplex method. This involves three main steps:

Objective Function: The simplex method is typically used for maximization problems. To minimize z, we can maximize its negative, Z=−z.

Original: minz=−3x1​+x2​+x3​

New: maxZ=3x1​−x2​−x3​

Constraints: We convert all inequality constraints to equality constraints by introducing slack, surplus, and artificial variables.

Constraint 1 (≤): Add a slack variable s1​.

x1​−2x2​+x3​+s1​=11

Constraint 2 (≥): Subtract a surplus variable s2​ and add an artificial variable a1​.

−4x1​+x2​+2x3​−s2​+a1​=3

Constraint 3 (=): Add an artificial variable a2​.

−2x1​+x3​+a2​=1

Penalize Artificial Variables: We modify the new objective function by subtracting a large positive number M for each artificial variable. This penalty ensures that artificial variables will be zero in the final solution.

Penalized Objective: maxZ=3x1​−x2​−x3​−Ma1​−Ma2​

The problem is now in standard form with all variables (x1​,x2​,x3​,s1​,s2​,a1​,a2​) being non-negative.

2. Initial Simplex Tableau

We set up the initial tableau. The objective function must be

VIPSolutions · Accepted Answer

Certainly! Here is your solution **rewritten as it is, with no changes to the calculations or logic, but laid out in a clean, clear format for readability**:

---

### 1. Convert to Standard Form

**Objective Function**  
The simplex method is typically used for maximization problems. To minimize $ z $, we can maximize its negative: $ Z = -z $.

- **Original:** $ \min z = -3x_1 + x_2 + x_3 $
- **New:** $ \max Z = 3x_1 - x_2 - x_3 $

**Constraints**  
We convert all inequality constraints to equality constraints by introducing slack, surplus, and artificial variables.

- **Constraint 1 ($\leq$)**: Add slack variable $ s_1 $:
  $$
  x_1 - 2x_2 + x_3 + s_1 = 11
  $$

- **Constraint 2 ($\geq$)**: Subtract surplus variable $ s_2 $, add artificial variable $ a_1 $:
  $$
  -4x_1 + x_2 + 2x_3 - s_2 + a_1 = 3
  $$

- **Constraint 3 ($=$)**: Add artificial variable $ a_2 $:
  $$
  -2x_1 + x_3 + a_2 = 1
  $$

**Penalize Artificial Variables**  
We modify the new objective function by subtracting a large positive number $ M $ for each artificial variable:
$$
\max Z = 3x_1 - x_2 - x_3 - M a_1 - M a_2
$$

All variables ($ x_1, x_2, x_3, s_1, s_2, a_1, a_2 $) are non-negative.

---

### 2. Initial Simplex Tableau

We set up the initial tableau. The objective function must be expressed in terms of non-basic variables only.  
We rewrite the objective function row ($ Z - 3x_1 + x_2 + x_3 + M a_1 + M a_2 = 0 $) by substituting out the artificial variables $ a_1, a_2 $ using the constraint equations:

From the constraints:
- $ a_1 = 3 + 4x_1 - x_2 - 2x_3 + s_2 $
- $ a_2 = 1 + 2x_1 - x_3 $

Substituting these into the objective function row gives us the initial Z-row for the tableau.

**Initial Tableau:**
$$
\begin{blockarray}{cccccccccc}
\text{Basis} & Z & x_1 & x_2 & x_3 & s_1 & s_2 & a_1 & a_2 & \text{RHS} \
\begin{block}{c(ccccccccc)}
Z   & 1 & 6M-3 & 1-M & 1-3M & 0 & M & 0 & 0 & -4M \
s_1 & 0 & 1 & -2 & 1 & 1 & 0 & 0 & 0 & 11 \
a_1 & 0 & -4 & 1 & 2 & 0 & -1 & 1 & 0 & 3 \
a_2 & 0 & -2 & 0 & 1 & 0 & 0 & 0 & 1 & 1 \
\end{block}
\end{blockarray}
$$

---

### 3. Simplex Iterations

**Iteration 1**

- **Entering Variable:** The most negative coefficient in the Z-row is $ 1-3M $, so $ x_3 $ enters the basis.
- **Leaving Variable:**  
  - Row $ s_1 $: $ 11/1 = 11 $
  - Row $ a_1 $: $ 3/2 = 1.5 $
  - Row $ a_2 $: $ 1/1 = 1 $ (Smallest Ratio)
- **Pivot:** The pivot element is 1 (row $ a_2 $, column $ x_3 $). Perform row operations.

**Tableau after Iteration 1:**
$$
\begin{blockarray}{cccccccccc}
\text{Basis} & Z & x_1 & x_2 & x_3 & s_1 & s_2 & a_1 & a_2 & \text{RHS} \
\begin{block}{c(ccccccccc)}
Z   & 1 & -1 & 1-M & 0 & 0 & M & 0 & 3M-1 & -M-1 \
s_1 & 0 & 3 & -2 & 0 & 1 & 0 & 0 & -1 & 10 \
a_1 & 0 & 0 & 1 & 0 & 0 & -1 & 1 & -2 & 1 \
x_3 & 0 & -2 & 0 & 1 & 0 & 0 & 0 & 1 & 1 \
\end{block}
\end{blockarray}
$$

---

**Iteration 2**

- **Entering Variable:** Most negative coefficient is $ 1-M $, so $ x_2 $ enters.
- **Leaving Variable:**  
  - Row $ s_1 $: $ 10/(-2) $ ⟹ invalid (negative)
  - Row $ a_1 $: $ 1/1 = 1 $ (Smallest Ratio)
  - Row $ x_3 $: $ 1/0 $ ⟹ invalid
- **Pivot:** Pivot element is 1 (row $ a_1 $, column $ x_2 $). Perform row operations.

**Tableau after Iteration 2:**  
Both artificial variables ($ a_1, a_2 $) have left the basis. This completes Phase I. Drop the $ a_1 $ and $ a_2 $ columns.

$$
\begin{blockarray}{cccccccc}
\text{Basis} & Z & x_1 & x_2 & x_3 & s_1 & s_2 & \text{RHS} \
\begin{block}{c(ccccccc)}
Z   & 1 & -1 & 0 & 0 & 0 & 1 & -2 \
s_1 & 0 & 3 & 0 & 0 & 1 & -2 & 12 \
x_2 & 0 & 0 & 1 & 0 & 0 & -1 & 1 \
x_3 & 0 & -2 & 0 & 1 & 0 & 0 & 1 \
\end{block}
\end{blockarray}
$$

---

**Iteration 3**

- **Entering Variable:** Only negative coefficient is -1, so $ x_1 $ enters.
- **Leaving Variable:**  
  - Row $ s_1 $: $ 12/3 = 4 $ (Smallest Ratio)
  - Row $ x_2 $: $ 1/0 $ ⟹ invalid
  - Row $ x_3 $: $ 1/(-2) $ ⟹ invalid
- **Pivot:** Pivot element is 3 (row $ s_1 $, column $ x_1 $). Perform row operations.

**Final Tableau:**  
(Correction in the final row operation for $ x_3 $ row: $ R_{x_3} \leftarrow R_{x_3} - (-2) R_{\text{pivot}} $. Calculation confirmed.)

$$
\begin{blockarray}{cccccccc}
\text{Basis} & Z & x_1 & x_2 & x_3 & s_1 & s_2 & \text{RHS} \
\begin{block}{c(ccccccc)}
Z   & 1 & 0 & 0 & 0 & 1/3 & 1/3 & 2 \
x_1 & 0 & 1 & 0 & 0 & 1/3 & -2/3 & 4 \
x_2 & 0 & 0 & 1 & 0 & 0 & -1 & 1 \
x_3 & 0 & 0 & 0 & 1 & 2/3 & -4/3 & 9 \
\end{block}
\end{blockarray}
$$

---

### 4. Conclusion ?

All coefficients in the Z-row are non-negative ($[0, 0, 0, 1/3, 1/3]$), which indicates that we have reached the optimal solution for the maximization problem.

From the final tableau, we read the values of the basic variables:
$$
x_1 = 4 \
x_2 = 1 \
x_3 = 9
$$
The maximum value of the objective function $ Z $ is the value in the RHS of the Z-row:
$$
\max Z = 2
$$
Since the original problem was to minimize \( z

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