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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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