Solve the below assignment problem : Task 1	Task 2	Task 3	Task 4
Agent 1	5 7 11 6
Agent 2	8 5 9 6
Agent 3	4 7 10 7
Agent 4	10 4 8 3

The assignment problem can be solved using the Hungarian Algorithm. This algorithm is used to find the optimal assignment that minimizes the total cost. Here’s a step-by-step solution for the given cost matrix:

### Cost Matrix:
\[
\begin{array}{c|cccc}
& \text{Task 1} & \text{Task 2} & \text{Task 3} & \text{Task 4} \\
\hline
\text{Agent 1} & 5 & 7 & 11 & 6 \\
\text{Agent 2} & 8 & 5 & 9 & 6 \\
\text{Agent 3} & 4 & 7 & 10 & 7 \\
\text{Agent 4} & 10 & 4 & 8 & 3 \\
\end{array}
\]

### Step 1: Subtract the row minimum from each row
- For Agent 1: Minimum = 5
\[
\begin{array}{cccc}
0 & 2 & 6 & 1 \\
\end{array}
\]
- For Agent 2: Minimum = 5
\[
\begin{array}{cccc}
3 & 0 & 4 & 1 \\
\end{array}
\]
- For Agent 3: Minimum = 4
\[
\begin{array}{cccc}
0 & 3 & 6 & 3 \\
\end{array}
\]
- For Agent 4: Minimum = 3
\[
\begin{array}{cccc}
7 & 1 & 5 & 0 \\
\end{array}
\]

The new matrix after row subtraction:
\[
\begin{array}{c|cccc}
& \text{Task 1} & \text{Task 2} & \text{Task 3} & \text{Task 4} \\
\hline
\text{Agent 1} & 0 & 2 & 6 & 1 \\
\text{Agent 2} & 3 & 0 & 4 & 1 \\
\text{Agent 3} & 0 & 3 & 6 & 3 \\
\text{Agent 4} & 7 & 1 & 5 & 0 \\
\end{array}
\]

### Step 2: Subtract the column minimum from each column
- For Task 1: Minimum = 0
\[
\begin{array}{c}
0 \\ 3 \\ 0 \\ 7 \\
\end{array}
\]
- For Task 2: Minimum = 0
\[
\begin{array}{c}
2 \\ 0 \\ 3 \\ 1 \\
\end{array}
\]
- For Task 3: Minimum = 4
\[
\begin{array}{c}
2 \\ 0 \\ 2 \\ 1 \\
\end{array}
\]
- For Task 4: Minimum = 0
\[
\begin{array}{c}
1 \\ 1 \\ 3 \\ 0 \\
\end{array}
\]

The new matrix after column subtraction:
\[
\begin{array}{c|cccc}
& \text{Task 1} & \text{Task 2} & \text{Task 3} & \text{Task 4} \\
\hline
\text{Agent 1} & 0 & 2 & 2 & 1 \\
\text{Agent 2} & 3 & 0 & 0 & 1 \\
\text{Agent 3} & 0 & 3 & 2 & 3 \\
\text{Agent 4} & 7 & 1 & 1 & 0 \\
\end{array}
\]

### Step 3: Cover all zeros with a minimum number of lines

Draw lines through rows and columns to cover all zeros:
\[
\begin{array}{c|cccc}
& \text{Task 1} & \text{Task 2} & \text{Task 3} & \text{Task 4} \\
\hline
\text{Agent 1} & \boxed{0} & 2 & 2 & 1 \\
\text{Agent 2} & 3 & \boxed{0} & 0 & 1 \\
\text{Agent 3} & \boxed{0} & 3 & 2 & 3 \\
\text{Agent 4} & 7 & 1 & 1 & \boxed{0} \\
\end{array}
\]

### Step 4: Optimal Assignment
By using the covered lines method and iteratively adjusting the matrix, we can assign tasks to agents such that:
- Agent 1: Task 1 (0)
- Agent 2: Task 2 (0)
- Agent 3: Task 3 (2)
- Agent 4: Task 4 (0)

So the optimal assignment minimizing the cost is:
- Agent 1 to Task 1
- Agent 2 to Task 2
- Agent 3 to Task 3
- Agent 4 to Task 4

The total cost is \(5 + 5 + 10 + 3 = 23\).

Would you like further details or have any questions?

Here are some related questions you might find interesting:
1. How does the Hungarian Algorithm compare to other optimization algorithms?
2. What are some real-world applications of the assignment problem?
3. Can the Hungarian Algorithm be used for maximization problems?
4. How can the Hungarian Algorithm be implemented in Python?
5. What modifications are needed to handle unbalanced assignment problems?
6. How does the initial reduction of the cost matrix aid in finding the optimal solution?
7. What are the computational complexities of the Hungarian Algorithm?
8. Are there any limitations to the Hungarian Algorithm?

**Tip:** When solving assignment problems, always ensure the cost matrix is square. If it's not, add dummy rows or columns with zero cost.

Learn how to solve the assignment problem using the Hungarian Algorithm with a step-by-step solution. This problem involves optimizing task assignments to minimize costs among agents. Suitable for advanced students and professionals.

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