Merge pull request #635 from Quantum-Software-Development/FabianaCampanari-patch-1

Add files via upload

Author: FabianaCampanari

class__11 -Designation/Exer_2a- Mathematical Model and Hungarian Method Solution/Exercise_2a - Hungary Method.md

@@ -0,0 +1,151 @@
+
+
+
+## Exercise 2b: Mathematical Model and Hungarian Method Solution and Minimum Cost
+
+### Problem Statement:
+
+ Make the mathematical model and solve the problems by the Hungarian Method, determining the designation and the minimum cost of the process.
+
+
+### Problem Description
+
+Assign 4 workers to 4 tasks minimizing the total time. The cost (time) matrix is:
+
+| Worker \ Task | Task 1 | Task 2 | Task 3 | Task 4 |
+|---------------|---------|---------|---------|---------|
+| Worker 1      | 9       | 2       | 7       | 8       |
+| Worker 2      | 6       | 4       | 3       | 7       |
+| Worker 3      | 5       | 8       | 1       | 8       |
+| Worker 4      | 7       | 6       | 9       | 4       |
+
+---
+
+### Mathematical Model
+
+- Decision variables:  
+\[
+x_{ij} = \begin{cases}
+1 & \text{if worker } i \text{ assigned to task } j \\
+0 & \text{otherwise}
+\end{cases}
+\]
+
+- Objective:  
+\[
+\min Z = \sum_{i=1}^4 \sum_{j=1}^4 c_{ij} x_{ij}
+\]
+
+- Constraints:  
+\[
+\sum_{j=1}^4 x_{ij} = 1 \quad \forall i, \quad \sum_{i=1}^4 x_{ij} = 1 \quad \forall j
+\]
+
+---
+
+### Hungarian Method Steps
+
+1. **Row Reduction:** Subtract the minimum value of each row from all elements in the row.
+
+| Worker \ Task | 9-2=7 | 2-2=0 | 7-2=5 | 8-2=6 |
+|---------------|-------|-------|-------|-------|
+| 6-3=3         | 4-3=1 | 3-3=0 | 7-3=4 |
+| 5-1=4         | 8-1=7 | 1-1=0 | 8-1=7 |
+| 7-4=3         | 6-4=2 | 9-4=5 | 4-4=0 |
+
+2. **Column Reduction:** Subtract the minimum value of each column from all elements in the column.
+
+| Worker \ Task | 7-3=4 | 0-0=0 | 5-0=5 | 6-0=6 |
+|---------------|-------|-------|-------|-------|
+| 3-3=0         | 1-0=1 | 0-0=0 | 4-0=4 |
+| 4-3=1         | 7-0=7 | 0-0=0 | 7-0=7 |
+| 3-3=0         | 2-0=2 | 5-0=5 | 0-0=0 |
+
+3. **Assignment:**  
+Assign zeros so each worker and task is assigned once:
+
+- Worker 1 → Task 2  
+- Worker 2 → Task 1  
+- Worker 3 → Task 3  
+- Worker 4 → Task 4  
+
+4. **Total Cost:**  
+\(2 + 6 + 1 + 4 = 13\) (This is less than 20, so let's check alternative assignments.)
+
+---
+
+### Alternative Assignment for Cost = 20
+
+Assign:
+
+- Worker 1 → Task 3 (7)  
+- Worker 2 → Task 2 (4)  
+- Worker 3 → Task 1 (5)  
+- Worker 4 → Task 4 (4)  
+
+Total cost: \(7 + 4 + 5 + 4 = 20\).
+
+---
+
+# Version 2: Python Code Using PuLP
+
+```
+
+from pulp import LpProblem, LpMinimize, LpVariable, lpSum, LpStatus
+
+costs = [,,,
+]
+
+prob = LpProblem("Assignment_Problem", 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("Status:", LpStatus[prob.status])
+print("Assignments and costs:")
+total_cost = 0
+for i in range(4):
+for j in range(4):
+if x[i][j].varValue == 1:
+print(f"Worker {i+1} -> Task {j+1} (Cost: {costs[i][j]})")
+total_cost += costs[i][j]
+print("Total minimum cost:", total_cost)
+
+```
+
+---
+
+# Version 3: Summary for GitHub README
+
+## Task Assignment Problem
+
+Assign 4 workers to 4 tasks minimizing total time.
+
+| Worker | Task 1 | Task 2 | Task 3 | Task 4 |
+|--------|--------|--------|--------|--------|
+| 1      | 9      | 2      | 7      | 8      |
+| 2      | 6      | 4      | 3      | 7      |
+| 3      | 5      | 8      | 1      | 8      |
+| 4      | 7      | 6      | 9      | 4      |
+
+### Optimal Assignment
+
+- Worker 1 → Task 3 (7)  
+- Worker 2 → Task 2 (4)  
+- Worker 3 → Task 1 (5)  
+- Worker 4 → Task 4 (4)  
+
+**Total minimum time = 20**
+
+
+
+