Solve assignment problem     C1  C2  C3  C4  C5  C6  C7
R1 35  22  60  41  27  52  44
R2 51  39  42  33  65  47  58
R3 25  32  53  41  50  36  43
R4 32  28  40  46   3  55  49
R5 43  36  45  63  57  49  42
R6 27  18  31  46  35  42  34
R7 48  50  72  59  43  64  58


The assignment problem can be solved using the Hungarian method (also known as the Kuhn-Munkres algorithm). This method finds the optimal assignment that minimizes the total cost. Here’s how you can solve the assignment problem step-by-step:

### Step 1: Create the Cost Matrix
The given matrix is:
\[
\begin{array}{ccccccc}
 & \text{C1} & \text{C2} & \text{C3} & \text{C4} & \text{C5} & \text{C6} & \text{C7} \\
\text{R1} & 35 & 22 & 60 & 41 & 27 & 52 & 44 \\
\text{R2} & 51 & 39 & 42 & 33 & 65 & 47 & 58 \\
\text{R3} & 25 & 32 & 53 & 41 & 50 & 36 & 43 \\
\text{R4} & 32 & 28 & 40 & 46 & 3 & 55 & 49 \\
\text{R5} & 43 & 36 & 45 & 63 & 57 & 49 & 42 \\
\text{R6} & 27 & 18 & 31 & 46 & 35 & 42 & 34 \\
\text{R7} & 48 & 50 & 72 & 59 & 43 & 64 & 58 \\
\end{array}
\]

### Step 2: Subtract Row Minimums
Subtract the minimum value of each row from all the elements of that row:

\[
\begin{array}{ccccccc}
 & \text{C1} & \text{C2} & \text{C3} & \text{C4} & \text{C5} & \text{C6} & \text{C7} \\
\text{R1} & 13 & 0 & 38 & 19 & 5 & 30 & 22 \\
\text{R2} & 18 & 6 & 9 & 0 & 32 & 14 & 25 \\
\text{R3} & 0 & 7 & 28 & 16 & 25 & 11 & 18 \\
\text{R4} & 29 & 25 & 37 & 43 & 0 & 52 & 46 \\
\text{R5} & 7 & 0 & 9 & 27 & 21 & 13 & 6 \\
\text{R6} & 9 & 0 & 13 & 28 & 17 & 24 & 16 \\
\text{R7} & 0 & 2 & 24 & 11 & 0 & 21 & 15 \\
\end{array}
\]

### Step 3: Subtract Column Minimums
Subtract the minimum value of each column from all the elements of that column:

\[
\begin{array}{ccccccc}
 & \text{C1} & \text{C2} & \text{C3} & \text{C4} & \text{C5} & \text{C6} & \text{C7} \\
\text{R1} & 13 & 0 & 24 & 8 & 5 & 9 & 16 \\
\text{R2} & 18 & 6 & -5 & -11 & 32 & -7 & 19 \\
\text{R3} & 0 & 7 & 14 & -4 & 25 & -3 & 12 \\
\text{R4} & 29 & 25 & 23 & 23 & 0 & 38 & 40 \\
\text{R5} & 7 & 0 & -5 & 7 & 21 & -1 & 0 \\
\text{R6} & 9 & 0 & -1 & 8 & 17 & 10 & 10 \\
\text{R7} & 0 & 2 & 10 & -3 & 0 & 7 & 9 \\
\end{array}
\]

### Step 4: Cover all zeros with a minimum number of horizontal and vertical lines
Use the minimum number of lines to cover all zeros in the matrix. We will find an optimal assignment by adjusting the matrix further if needed.

### Step 5: Find the Optimal Assignment
Using the Hungarian method, we continue to adjust the matrix by subtracting the smallest uncovered value from all uncovered elements and adding it to elements covered twice until an optimal assignment is found.

After performing these operations iteratively, the optimal assignment is:

- R1 -> C2 (Cost = 22)
- R2 -> C4 (Cost = 33)
- R3 -> C1 (Cost = 25)
- R4 -> C5 (Cost = 3)
- R5 -> C7 (Cost = 42)
- R6 -> C2 (Cost = 18)
- R7 -> C1 (Cost = 48)

The minimum total cost is \(22 + 33 + 25 + 3 + 42 + 18 + 48 = 191\).

Would you like more details or have any questions about this method?

Here are 5 related questions:
1. How does the Hungarian method ensure an optimal solution for the assignment problem?
2. Can the Hungarian method be applied to maximization problems?
3. What are the differences between the Hungarian method and other assignment problem-solving techniques?
4. How can the assignment problem be adapted for unbalanced matrices?
5. What are real-world applications of the assignment problem?

**Tip:** Practice solving smaller assignment problems by hand to understand the steps of the Hungarian method before applying it to larger problems.

Solve assignment problem     C1  C2  C3  C4  C5  C6  C7
R1 35  22  60  41  27  52  44
R2 51  39  42  33  65  47  58
R3 25  32  53  41  50  36  43
R4 32  28  40  46   3  55  49
R5 43  36  45  63  57  49  42
R6 27  18  31  46  35  42  34
R7 48  50  72  59  43  64  58

This detailed guide walks you through solving an assignment problem using the Hungarian method (Kuhn-Munkres algorithm). Learn how to reduce rows and columns, cover zeros with the minimum number of lines, and find an optimal assignment. Problem includes a 7x7 cost matrix, and the solution leads to a minimum total cost of 191.

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