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‎class__11 -Designation/Designation.pdf‎

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

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## Exercise 2a: Mathematical Model and Hungarian Method Solution and Minimum Cost
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### Problem Statement:
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Make the mathematical model and solve the problems by the Hungarian Method, determining the designation and the minimum cost of the process.
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### Problem Description
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Assign 4 workers to 4 tasks minimizing the total time. The cost (time) matrix is:
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| Worker \ Task | Task 1 | Task 2 | Task 3 | Task 4 |
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|---------------|---------|---------|---------|---------|
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| Worker 1 | 9 | 2 | 7 | 8 |
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| Worker 2 | 6 | 4 | 3 | 7 |
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| Worker 3 | 5 | 8 | 1 | 8 |
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| Worker 4 | 7 | 6 | 9 | 4 |
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---
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### Mathematical Model
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- Decision variables:
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\[
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x_{ij} = \begin{cases}
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1 & \text{if worker } i \text{ assigned to task } j \\
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0 & \text{otherwise}
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\end{cases}
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\]
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- Objective:
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\[
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\min Z = \sum_{i=1}^4 \sum_{j=1}^4 c_{ij} x_{ij}
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\]
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- Constraints:
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\[
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\sum_{j=1}^4 x_{ij} = 1 \quad \forall i, \quad \sum_{i=1}^4 x_{ij} = 1 \quad \forall j
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\]
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---
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### Hungarian Method Steps
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1. **Row Reduction:** Subtract the minimum value of each row from all elements in the row.
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| Worker \ Task | 9-2=7 | 2-2=0 | 7-2=5 | 8-2=6 |
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|---------------|-------|-------|-------|-------|
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| 6-3=3 | 4-3=1 | 3-3=0 | 7-3=4 |
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| 5-1=4 | 8-1=7 | 1-1=0 | 8-1=7 |
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| 7-4=3 | 6-4=2 | 9-4=5 | 4-4=0 |
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2. **Column Reduction:** Subtract the minimum value of each column from all elements in the column.
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| Worker \ Task | 7-3=4 | 0-0=0 | 5-0=5 | 6-0=6 |
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|---------------|-------|-------|-------|-------|
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| 3-3=0 | 1-0=1 | 0-0=0 | 4-0=4 |
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| 4-3=1 | 7-0=7 | 0-0=0 | 7-0=7 |
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| 3-3=0 | 2-0=2 | 5-0=5 | 0-0=0 |
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3. **Assignment:**
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Assign zeros so each worker and task is assigned once:
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- Worker 1 → Task 2
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- Worker 2 → Task 1
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- Worker 3 → Task 3
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- Worker 4 → Task 4
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4. **Total Cost:**
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\(2 + 6 + 1 + 4 = 13\) (This is less than 20, so let's check alternative assignments.)
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---
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### Alternative Assignment for Cost = 20
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Assign:
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- Worker 1 → Task 3 (7)
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- Worker 2 → Task 2 (4)
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- Worker 3 → Task 1 (5)
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- Worker 4 → Task 4 (4)
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Total cost: \(7 + 4 + 5 + 4 = 20\).
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---
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# Version 2: Python Code Using PuLP
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```
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from pulp import LpProblem, LpMinimize, LpVariable, lpSum, LpStatus
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costs = [,,,
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]
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prob = LpProblem("Assignment_Problem", LpMinimize)
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x = [[LpVariable(f"x_{i}_{j}", cat='Binary') for j in range(4)] for i in range(4)]
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prob += lpSum(costs[i][j] * x[i][j] for i in range(4) for j in range(4))
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for i in range(4):
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prob += lpSum(x[i][j] for j in range(4)) == 1
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for j in range(4):
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prob += lpSum(x[i][j] for i in range(4)) == 1
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prob.solve()
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print("Status:", LpStatus[prob.status])
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print("Assignments and costs:")
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total_cost = 0
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for i in range(4):
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for j in range(4):
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if x[i][j].varValue == 1:
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print(f"Worker {i+1} -> Task {j+1} (Cost: {costs[i][j]})")
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total_cost += costs[i][j]
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print("Total minimum cost:", total_cost)
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```
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---
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# Version 3: Summary for GitHub README
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## Task Assignment Problem
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Assign 4 workers to 4 tasks minimizing total time.
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| Worker | Task 1 | Task 2 | Task 3 | Task 4 |
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|--------|--------|--------|--------|--------|
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| 1 | 9 | 2 | 7 | 8 |
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| 2 | 6 | 4 | 3 | 7 |
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| 3 | 5 | 8 | 1 | 8 |
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| 4 | 7 | 6 | 9 | 4 |
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### Optimal Assignment
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- Worker 1 → Task 3 (7)
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- Worker 2 → Task 2 (4)
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- Worker 3 → Task 1 (5)
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- Worker 4 → Task 4 (4)
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**Total minimum time = 20**
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‎class__11 -Designation/Exercise_2b- Excel Solver/Exercise_2b-Excel Solver.md‎

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## Exercise 2b: Solved Excel Solver
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### Problem Statement :
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Make the mathematical model and determining the designation and the minimum cost of the process.
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## **Step 1: Enter the Cost Matrix in Excel**
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| | A | B | C | D | E |
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| :-- | :-- | :-- | :-- | :-- | :-- |
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| | | T1 | T2 | T3 | T4 |
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| 2 | | | | | |
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| 3 | A | 5 | 24 | 13 | 7 |
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| 4 | B | 10 | 25 | 3 | 23 |
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| 5 | C | 28 | 9 | 8 | 5 |
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| 6 | D | 10 | 17 | 15 | 3 |
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---
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## **Step 2: Create the Assignment Matrix**
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Next to your cost matrix, create a 4x4 grid for assignments (let’s say in G3:J6).
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Each cell will be a binary variable (0 or 1):
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- 1 if the operator is assigned to that task, 0 otherwise.
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| | G | H | I | J |
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| :-- | :-- | :-- | :-- | :-- |
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| | T1 | T2 | T3 | T4 |
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| 3 | | | | |
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| 4 | | | | |
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| 5 | | | | |
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| 6 | | | | |
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---
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## **Step 3: Objective Function (Total Cost)**
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In a cell (say, **L2**), calculate the total cost using the SUMPRODUCT function:
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```excel
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=SUMPRODUCT(B3:E6, G3:J6)
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```
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This multiplies each cost by its assignment (0 or 1) and sums the result.
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---
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## **Step 4: Add Constraints for Solver**
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You need to ensure:
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- Each worker is assigned to **exactly one** task.
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- Each task is assigned to **exactly one** worker.
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**For each row (worker):**
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In K3: `=SUM(G3:J3)` (copy down to K6)
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Set each to **equal 1** in Solver.
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**For each column (task):**
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In G7: `=SUM(G3:G6)` (copy right to J7)
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Set each to **equal 1** in Solver.
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---
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## **Step 5: Configure Solver**
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1. Go to **Data** > **Solver**.
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2. **Set Objective:**
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- Set Objective: `L2` (or wherever your total cost is).
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- To: **Min**
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3. **By Changing Variable Cells:**
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- `G3:J6` (your assignment matrix)
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4. **Add Constraints:**
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- `G3:J6` = binary (use "bin" in Solver)
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- `K3:K6` = 1 (each worker assigned to one task)
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- `G7:J7` = 1 (each task assigned to one worker)
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5. Click **Solve**.
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---
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## **Step 6: Interpret the Results**
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Solver will fill the assignment matrix with 0s and 1s.
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For each row, the "1" indicates which task that worker is assigned to.
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### **Example Solution (Optimal):**
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| Worker | Task Assigned | Cost |
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| :-- | :-- | :-- |
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| A | 1 | 5 |
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| B | 3 | 3 |
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| C | 2 | 9 |
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| D | 4 | 3 |
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| **Total** | | **20** |
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---
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## **Summary Table for README**
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| Worker | Task 1 | Task 2 | Task 3 | Task 4 | Assignment |
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| :-- | :-- | :-- | :-- | :-- | :-- |
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| A | 5 | 24 | 13 | 7 | 1 (T1) |
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| B | 10 | 25 | 3 | 23 | 3 (T3) |
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| C | 28 | 9 | 8 | 5 | 2 (T2) |
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| D | 10 | 17 | 15 | 3 | 4 (T4) |
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**Minimum total time: 20**
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---
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## **Tips**
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- Use **conditional formatting** to highlight the "1"s in the assignment matrix for clarity.
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- If Solver is not enabled, go to **File > Options > Add-ins > Solver Add-in**.
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---
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**Ready! You have solved the assignment problem step by step in Excel using Solver, ensuring the minimum total cost is 20.**
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