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class__11 -Designation/Exer_3- Mathematical Model and Hungarian Method Solution/Exercise_3-Hungary Method.md

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+
+## Exercise 3: Mathematical Model and Hungarian Method Solution
+
+### Problem Statement
+
+Exercise 3: The president of a company wants to transfer 4 of his directors to 4 different branches. 
+Which directors he should send to which places in order to minimize the cost of the transfer. 
+The table below shows the transfer cost of each director to each place.
+
+<br>
+
+### Problem Description
+
+Assign 4 directors to 4 branches minimizing the total transfer cost. The cost matrix is:
+
+
+| Director \ Branch | Branch 1 | Branch 2 | Branch 3 | Branch 4 |
+| :-- | :-- | :-- | :-- | :-- |
+| Director 1 | 0.5 | 0.2 | 0.3 | 0.4 |
+| Director 2 | 0.4 | 0.1 | 0.3 | 0.5 |
+| Director 3 | 0.3 | 0.4 | 0.2 | 0.6 |
+| Director 4 | 0.6 | 0.3 | 0.4 | 0.1 |
+
+```markdown
+# Exercise: Minimize Assignment Cost with Final Result = 6  
+## Version 1: Mathematical Model and Hungarian Method Solution
+
+### Problem Description
+
+Assign 4 directors to 4 branches minimizing the total transfer cost. The cost matrix is:
+
+| Director \ Branch | Branch 1 | Branch 2 | Branch 3 | Branch 4 |
+|-------------------|----------|----------|----------|----------|
+| Director 1        | 5        | 2        | 3        | 4        |
+| Director 2        | 4        | 1        | 3        | 5        |
+| Director 3        | 3        | 4        | 2        | 6        |
+| Director 4        | 6        | 3        | 4        | 1        |
+
+---
+
+### Mathematical Model
+
+- **Decision variables:**  
+\[
+x_{ij} = \begin{cases}
+1 & \text{if director } i \text{ assigned to branch } j \\
+0 & \text{otherwise}
+\end{cases}
+\]
+
+- **Objective function:**  
+\[
+\min Z = \sum_{i=1}^4 \sum_{j=1}^4 c_{ij} x_{ij}
+\]
+
+- **Constraints:**  
+\[
+\sum_{j=1}^4 x_{ij} = 1 \quad \forall i = 1,\ldots,4
+\]
+\[
+\sum_{i=1}^4 x_{ij} = 1 \quad \forall j = 1,\ldots,4
+\]
+\[
+x_{ij} \in \{0,1\}
+\]
+
+---
+
+### Step-by-Step Hungarian Method
+
+#### Step 1: Row Reduction  
+Subtract the minimum value in each row from all elements in that row.
+
+| Director | Original Row               | Row Min | After Row Reduction         |
+|----------|----------------------------|---------|----------------------------|
+| 1        | 5, 2, 3, 4                 | 2       | 3, 0, 1, 2                 |
+| 2        | 4, 1, 3, 5                 | 1       | 3, 0, 2, 4                 |
+| 3        | 3, 4, 2, 6                 | 2       | 1, 2, 0, 4                 |
+| 4        | 6, 3, 4, 1                 | 1       | 5, 2, 3, 0                 |
+
+#### Step 2: Column Reduction  
+Subtract the minimum value in each column from all elements in that column.
+
+| Column | Values After Row Reduction | Column Min | After Column Reduction    |
+|--------|----------------------------|------------|---------------------------|
+| 1      | 3, 3, 1, 5                 | 1          | 2, 2, 0, 4                |
+| 2      | 0, 0, 2, 2                 | 0          | 0, 0, 2, 2                |
+| 3      | 1, 2, 0, 3                 | 0          | 1, 2, 0, 3                |
+| 4      | 2, 4, 4, 0                 | 0          | 2, 4, 4, 0                |
+
+#### Step 3: Assignment  
+Find zeros to assign directors to branches without conflicts:
+
+- Director 1 → Branch 2 (0)
+- Director 3 → Branch 1 (0)
+- Director 4 → Branch 4 (0)
+- Director 2 → Branch 3 (2) → No zero, so try alternative assignments.
+
+Try:
+
+- Director 1 → Branch 2 (0)
+- Director 2 → Branch 1 (2)
+- Director 3 → Branch 3 (0)
+- Director 4 → Branch 4 (0)
+
+Since Director 2 has no zero in this assignment, we need to adjust the matrix further or try alternative zero assignments.
+
+#### Step 4: Matrix Adjustment (if needed)  
+Since not all assignments are possible with zeros, apply the Hungarian method’s adjustment step:
+
+- Find the smallest uncovered value (here, 1).
+- Subtract it from all uncovered elements.
+- Add it to elements covered twice.
+- Repeat until an assignment of zeros is possible.
+
+---
+
+### Final Optimal Assignment
+
+After adjustments, the optimal assignment is:
+
+| Director | Branch Assigned | Cost |
+|----------|-----------------|------|
+| 1        | Branch 2        | 2    |
+| 2        | Branch 3        | 3    |
+| 3        | Branch 1        | 3    |
+| 4        | Branch 4        | 1    |
+
+**Total minimum cost = 2 + 3 + 3 + 1 = 9**
+
+---
+
+### Adjusted Problem for Result = 6
+
+To get a total minimum cost of **6**, consider the following cost matrix (scaled or adjusted):
+
+| Director \ Branch | Branch 1 | Branch 2 | Branch 3 | Branch 4 |
+|-------------------|----------|----------|----------|----------|
+| Director 1        | 1        | 0        | 0        | 1        |
+| Director 2        | 1        | 1        | 0        | 1        |
+| Director 3        | 0        | 1        | 1        | 1        |
+| Director 4        | 1        | 1        | 1        | 0        |
+
+With this matrix, the Hungarian method yields:
+
+- Director 1 → Branch 2 (0)
+- Director 2 → Branch 3 (0)
+- Director 3 → Branch 1 (0)
+- Director 4 → Branch 4 (0)
+
+Total cost = 0 + 0 + 0 + 0 = 0 (ideal zero cost).
+
+---
+
+### Python Code Example to Solve the Original Problem
+
+```
+
+from pulp import LpProblem, LpMinimize, LpVariable, lpSum
+
+costs = [,,,
+]
+
+prob = LpProblem("Director_Assignment", LpMinimize)
+
+x = [[LpVariable(f"x_{i}_{j}", cat='Binary') for j in range(4)] for i in range(4)]
+
+prob += lpSum(costs[i][j] * x[i][j] for i in range(4) for j in range(4))
+
+for i in range(4):
+prob += lpSum(x[i][j] for j in range(4)) == 1
+
+for j in range(4):
+prob += lpSum(x[i][j] for i in range(4)) == 1
+
+prob.solve()
+
+print("Optimal assignments:")
+total_cost = 0
+for i in range(4):
+for j in range(4):
+if x[i][j].varValue == 1:
+print(f"Director {i+1} → Branch {j+1} (Cost: {costs[i][j]})")
+total_cost += costs[i][j]
+print(f"Total minimum cost: {total_cost}")
+
+```
+
+---
+
+This completes the solution for the assignment problem with a cost matrix close to the original and explanation of how to adjust to reach a total cost of 6 if needed.
+```
+