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Question
Solve the following maximal assignment problem :
| Branch Manager | Monthly Business ( Rs. lakh) | |||
| A | B | C | D | |
| P | 11 | 11 | 9 | 9 |
| Q | 13 | 16 | 11 | 10 |
| R | 12 | 17 | 13 | 8 |
| S | 16 | 14 | 16 | 12 |
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Solution
| Branch Manager | Monthly Business ( Rs. lakh) | |||
| A | B | C | D | |
| P | 11 | 11 | 9 | 9 |
| Q | 13 | 16 | 11 | 10 |
| R | 12 | 17 | 13 | 8 |
| S | 16 | 14 | 16 | 12 |
Step 1 : Since it is a maximization problem subtract each of the element in the table from the largest element.
| Branch Manager | Monthly Business ( Rs. lakh) | |||
| A | B | C | D | |
| P | 6 | 6 | 8 | 8 |
| Q | 4 | 1 | 6 | 7 |
| R | 5 | 0 | 4 | 9 |
| S | 1 | 3 | 1 | 5 |
Step 2 : Minimum element of each row is subtracted from every element in that row.
| Branch Manager | Monthly Business ( Rs. lakh) | |||
| A | B | C | D | |
| P | 0 | 0 | 2 | 2 |
| Q | 3 | 0 | 5 | 6 |
| R | 5 | 0 | 4 | 9 |
| S | 0 | 2 | 0 | 4 |
Step 3 : Minimum element of each column is subtracted from every element in that column.
| Branch Manager | Monthly Business ( Rs. lakh) | |||
| A | B | C | D | |
| P | 0 | 0 | 2 | 0 |
| Q | 3 | 0 | 5 | 4 |
| R | 5 | 0 | 4 | 7 |
| S | 0 | 2 | 0 | 2 |
Step 4 : Zero elements are covered With minimum number of straight lines.
Step 5 : Selecting the smallest element not covered by the lines and subtracting it from each uncovered element and add it to the intersection of the lines.
Step 6 : Covering all zeros by minimum number of straight lines.
Minimum number of lines = order of matrix. so optimal solution has reached.
Step 7 : Making assignment at single zero of the row and then at single zero of the column.
The optimal assignment is
P → D
Q → A
R → B
S → C
Maximum Profit = 9 + 13 + 17 + 16 = 55 lakhs.
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Solution:
Step 1: Subtract the smallest element in each row from every element of it. New assignment matrix is obtained as follows :
| Job | Machines (Processing cost in ₹) |
|||
| I | II | III | IV | |
| P | 6 | 0 | 8 | 4 |
| Q | 4 | 3 | 2 | 0 |
| R | 0 | 2 | 4 | 5 |
| S | 4 | 2 | 0 | 6 |
Step 2: Subtract the smallest element in each column from every element of it. New assignment matrix is obtained as above, because each column in it contains one zero.
Step 3: Draw minimum number of vertical and horizontal lines to cover all zeros:
| Job | Machines (Processing cost in ₹) |
|||
| I | II | III | IV | |
| P | 6 | 0 | 8 | 4 |
| Q | 4 | 3 | 2 | 0 |
| R | 0 | 2 | 4 | 5 |
| S | 4 | 2 | 0 | 6 |
Step 4: From step 3, as the minimum number of straight lines required to cover all zeros in the assignment matrix equals the number of rows/columns. Optimal solution has reached.
Examine the rows one by one starting with the first row with exactly one zero is found. Mark the zero by enclosing it in (`square`), indicating assignment of the job. Cross all the zeros in the same column. This step is shown in the following table :
| Job | Machines (Processing cost in ₹) |
|||
| I | II | III | IV | |
| P | 6 | 0 | 8 | 4 |
| Q | 4 | 3 | 2 | 0 |
| R | 0 | 2 | 4 | 5 |
| S | 4 | 2 | 0 | 6 |
Step 5: It is observed that all the zeros are assigned and each row and each column contains exactly one assignment. Hence, the optimal (minimum) assignment schedule is :
| Job | Machine | Min.cost |
| P | II | `square` |
| Q | `square` | 21 |
| R | I | `square` |
| S | III | 34 |
Hence, total (minimum) processing cost = 25 + 21 + 19 + 34 = ₹`square`
