Merge pull request #777 from Quantum-Software-Development/FabianaCampanari-patch-1

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Author: FabianaCampanari

class__15-✍🏻 HandMade Excercise prep EXAM -LP -MathModels - Simplex Two-Stages + Transportation Problem/🇺🇸 Answers - English/Exercise_1-Answers/2-answer_1-MD.md

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+```markdown
+# Detailed Solution of the Problem Using the Two-Phase Simplex Method
+
+## Original Problem
+
+**Minimize:**
+
+$$
+Z = x_1 + x_2 + x_3
+$$
+
+**Subject to:**
+
+$$
+\begin{cases}
+2x_1 + x_2 - x_3 \leq 10 \\
+x_1 + x_2 + 2x_3 \geq 20 \\
+2x_1 + x_2 + 3x_3 = 60 \\
+x_1, x_2, x_3 \geq 0
+\end{cases}
+$$
+
+<br>
+
+## Phase I: Elimination of Artificial Variables
+
+### Step 1: Convert inequalities to equalities
+
+- For the $\leq$ constraint, add slack variable $s_1 \geq 0$:
+
+$$
+2x_1 + x_2 - x_3 + s_1 = 10
+$$
+- For the $\geq$ constraint, subtract excess variable $s_2 \geq 0$ and add artificial variable $a_1 \geq 0$:
+
+$$
+x_1 + x_2 + 2x_3 - s_2 + a_1 = 20
+$$
+- For the equality constraint, add artificial variable $a_2 \geq 0$:
+
+$$
+2x_1 + x_2 + 3x_3 + a_2 = 60
+$$
+
+
+### Step 2: Auxiliary objective function for Phase I
+
+$$
+\min W = a_1 + a_2
+$$
+
+<br>
+
+### Initial Tableau
+
+| **Base** | $x_1$ | $x_2$ | $x_3$ | $s_1$ | $s_2$ | $a_1$ | $a_2$ | **RHS** |
+| :--: | :--: | :--: | :--: | :--: | :--: | :--: | :--: | :--: |
+| $\mathbf{s_1}$ | 2 | 1 | -1 | 1 | 0 | 0 | 0 | 10 |
+| $\mathbf{a_1}$ | 1 | 1 | 2 | 0 | -1 | 1 | 0 | 20 |
+| $\mathbf{a_2}$ | 2 | 1 | 3 | 0 | 0 | 0 | 1 | 60 |
+| $\mathbf{W}$ | -3 | -2 | -5 | 0 | 1 | 0 | 0 | 80 |
+
+*Note: The $W$ row is obtained by substituting $a_1$ and $a_2$ from the constraints.*
+
+<br>
+
+### Step 3: Iteration 1 — Enter $x_3$, Leave $a_1$
+
+- **Pivot element:** $2$ (row $a_1$, column $x_3$).
+- **Row operations:** Divide row $a_1$ by $2$ and update other rows to zero out column $x_3$.
+
+<br>
+
+### Tableau after Iteration 1
+
+| **Base** | $x_1$ | $x_2$ | $x_3$ | $s_1$ | $s_2$ | $a_1$ | $a_2$ | **RHS** |
+| :--: | :--: | :--: | :--: | :--: | :--: | :--: | :--: | :--: |
+| $\mathbf{s_1}$ | 2.5 | 1.5 | 0 | 1 | -0.5 | 0.5 | 0 | 20 |
+| $\mathbf{x_3}$ | 0.5 | 0.5 | 1 | 0 | -0.5 | 0.5 | 0 | 10 |
+| $\mathbf{a_2}$ | 0.5 | -0.5 | 0 | 0 | 1.5 | -1.5 | 1 | 30 |
+| $\mathbf{W}$ | -0.5 | 0.5 | 0 | 0 | -1.5 | 2.5 | 0 | 130 |
+
+
+<br>
+
+### Step 4: Iteration 2 — Enter $s_2$, Leave $a_2$
+
+- **Pivot element:** $1.5$ (row $a_2$, column $s_2$).
+- **Row operations:** Divide row $a_2$ by $1.5$ and update other rows to zero out column $s_2$.
+
+<br>
+
+### Tableau after Iteration 2
+
+| **Base** | $x_1$ | $x_2$ | $x_3$ | $s_1$ | $s_2$ | $a_1$ | $a_2$ | **RHS** |
+| :--: | :--: | :--: | :--: | :--: | :--: | :--: | :--: | :--: |
+| $\mathbf{s_1}$ | 8/3 | 4/3 | 0 | 1 | 0 | 0 | 1/3 | 30 |
+| $\mathbf{x_3}$ | 2/3 | 1/3 | 1 | 0 | 0 | 0 | 1/3 | 20 |
+| $\mathbf{s_2}$ | 1/3 | -1/3 | 0 | 0 | 1 | -1 | 2/3 | 20 |
+| $\mathbf{W}$ | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 |
+
+
+<br>
+
+### ⚠️ **Attention: If Phase I ends with $W \neq 0$**
+
+If at the end of Phase I the auxiliary objective function $W$ **is not zero**, i.e., $W > 0$, this indicates that **there is no feasible solution to the original problem**.
+
+This happens because artificial variables could not be removed from the basis, signaling inconsistency in the original constraints.
+
+In this case, the Two-Phase Simplex method **stops and reports the problem as infeasible**.
+
+<br>
+
+## Phase II: Optimization of the Original Objective Function
+
+### Initial Tableau Phase II
+
+| **Base** | $x_1$ | $x_2$ | $x_3$ | $s_1$ | $s_2$ | **RHS** |
+| :--: | :--: | :--: | :--: | :--: | :--: | :--: |
+| $\mathbf{s_1}$ | 8/3 | 4/3 | 0 | 1 | 0 | 30 |
+| $\mathbf{x_3}$ | 2/3 | 1/3 | 1 | 0 | 0 | 20 |
+| $\mathbf{s_2}$ | 1/3 | -1/3 | 0 | 0 | 1 | 20 |
+
+
+<br>
+
+### Step 5: Write the original objective function in terms of the variables
+
+$$
+x_3 = 20 - \frac{2}{3}x_1 - \frac{1}{3}x_2
+$$
+
+$$
+Z = x_1 + x_2 + x_3 = x_1 + x_2 + 20 - \frac{2}{3}x_1 - \frac{1}{3}x_2 = 20 + \frac{1}{3}x_1 + \frac{2}{3}x_2
+$$
+
+---
+
+### Step 6: Optimality Check
+
+The coefficients of $Z$ for the non-basic variables $x_1$ and $x_2$ are positive $\left(\frac{1}{3}\right.$ and $\left.\frac{2}{3}\right)$, indicating the current solution is optimal.
+
+---
+
+## Optimal Solution
+
+$$
+\boxed{
+(x_1, x_2, x_3) = (0, 0, 20) \quad \text{with} \quad Z_{\min} = 20
+}
+$$
+
+<br>
+
+## Constraint Verification
+
+$$
+\begin{cases}
+2(0) + 0 - 20 = -20 \leq 10 \quad \checkmark \\
+0 + 0 + 2(20) = 40 \geq 20 \quad \checkmark \\
+2(0) + 0 + 3(20) = 60 = 60 \quad \checkmark
+\end{cases}
+$$
+
+<br>
+
+# **Summary**
+
+- The Two-Phase Simplex method was applied correctly.
+- Phase I eliminated artificial variables with $W = 0$, indicating the problem is **feasible**.
+- Phase II optimized the original objective function, finding the optimal solution.
+- If $W \neq 0$ at the end of Phase I, the problem would be **infeasible** (no feasible solution).
+````
+