Answer :
To determine the optimal assignment of methods to operators and maximize weekly output, we can use the Hungarian algorithm for assignment problems. Here are the steps to solve this problem:
1. Create the Output Table:
We start with the given weekly output table where the rows represent operators and the columns represent methods.
| \ | M1 | M2 | M3 | M4 | M5 |
| -- | -- | -- | -- | -- | -- |
| A | 4 | 6 | 11 | 16 | 9 |
| B | 5 | 8 | 16 | 19 | 9 |
| C | 9 | 13 | 21 | 21 | 13 |
| D | 6 | 6 | 9 | 11 | 7 |
2. Set Up Cost Matrix:
To maximize the output, we can convert this assignment problem into a minimization problem by negating the values in the matrix. This step is necessary because the Hungarian algorithm finds the minimum cost assignment.
| \ | M1 | M2 | M3 | M4 | M5 |
| -- | -- | -- | -- | -- | -- |
| A | -4 | -6 | -11 | -16 | -9 |
| B | -5 | -8 | -16 | -19 | -9 |
| C | -9 | -13 | -21 | -21 | -13 |
| D | -6 | -6 | -9 | -11 | -7 |
3. Apply Hungarian Algorithm:
Using the Hungarian algorithm, we process the cost matrix to identify the optimal assignment of methods to operators that minimizes the total cost (or equivalently, maximizes the output).
4. Optimal Assignment:
The results from the Hungarian algorithm give us the following assignments:
- Operator A is assigned to Method M5.
- Operator B is assigned to Method M4.
- Operator C is assigned to Method M3.
- Operator D is assigned to Method M2.
5. Calculate the Maximum Output:
To determine the total weekly output for these assignments, we sum the outputs for each assigned operator-method pair:
- A (Operator 0) assigned to M5 (Column 4): 9 units
- B (Operator 1) assigned to M4 (Column 3): 19 units
- C (Operator 2) assigned to M3 (Column 2): 21 units
- D (Operator 3) assigned to M2 (Column 1): 6 units
Adding these values together:
[tex]\[ 9 + 19 + 21 + 6 = 55 \][/tex]
Thus, the optimal assignments are:
- Operator A to Method M5
- Operator B to Method M4
- Operator C to Method M3
- Operator D to Method M2
The maximum total weekly output achieved with these assignments is 55 units.
1. Create the Output Table:
We start with the given weekly output table where the rows represent operators and the columns represent methods.
| \ | M1 | M2 | M3 | M4 | M5 |
| -- | -- | -- | -- | -- | -- |
| A | 4 | 6 | 11 | 16 | 9 |
| B | 5 | 8 | 16 | 19 | 9 |
| C | 9 | 13 | 21 | 21 | 13 |
| D | 6 | 6 | 9 | 11 | 7 |
2. Set Up Cost Matrix:
To maximize the output, we can convert this assignment problem into a minimization problem by negating the values in the matrix. This step is necessary because the Hungarian algorithm finds the minimum cost assignment.
| \ | M1 | M2 | M3 | M4 | M5 |
| -- | -- | -- | -- | -- | -- |
| A | -4 | -6 | -11 | -16 | -9 |
| B | -5 | -8 | -16 | -19 | -9 |
| C | -9 | -13 | -21 | -21 | -13 |
| D | -6 | -6 | -9 | -11 | -7 |
3. Apply Hungarian Algorithm:
Using the Hungarian algorithm, we process the cost matrix to identify the optimal assignment of methods to operators that minimizes the total cost (or equivalently, maximizes the output).
4. Optimal Assignment:
The results from the Hungarian algorithm give us the following assignments:
- Operator A is assigned to Method M5.
- Operator B is assigned to Method M4.
- Operator C is assigned to Method M3.
- Operator D is assigned to Method M2.
5. Calculate the Maximum Output:
To determine the total weekly output for these assignments, we sum the outputs for each assigned operator-method pair:
- A (Operator 0) assigned to M5 (Column 4): 9 units
- B (Operator 1) assigned to M4 (Column 3): 19 units
- C (Operator 2) assigned to M3 (Column 2): 21 units
- D (Operator 3) assigned to M2 (Column 1): 6 units
Adding these values together:
[tex]\[ 9 + 19 + 21 + 6 = 55 \][/tex]
Thus, the optimal assignments are:
- Operator A to Method M5
- Operator B to Method M4
- Operator C to Method M3
- Operator D to Method M2
The maximum total weekly output achieved with these assignments is 55 units.
