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Question
Solve the following minimal assignment problem :
| Machines | A | B | C | D | E |
| M1 | 27 | 18 | ∞ | 20 | 21 |
| M2 | 31 | 24 | 21 | 12 | 17 |
| M3 | 20 | 17 | 20 | ∞ | 16 |
| M4 | 21 | 28 | 20 | 16 | 27 |
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Solution
| Machines | A | B | C | D | E |
| M1 | 27 | 18 | __ | 20 | 21 |
| M2 | 31 | 24 | 21 | 12 | 17 |
| M3 | 20 | 17 | 20 | __ | 16 |
| M4 | 21 | 28 | 20 | 16 | 27 |
Step 1 : We introduce a dummy machine M2 with time zero for each job and replace '-' by ∞
| A | B | C | D | E | |
| M1 | 27 | 18 | ∞ | 20 | 21 |
| M2 | 31 | 24 | 21 | 12 | 17 |
| M3 | 20 | 17 | 20 | ∞ | 16 |
| M4 | 21 | 28 | 20 | 16 | 27 |
| M5 | 0 | 0 | 0 | 0 | 0 |
Step 2: Subtracting minimum element of each row from all its elememts
As minimum number of lines = 4 ≠ order of the matrix.
. . Optimal solution is not reached.
Step 3 : We subtract minimum element (from uncovered elements) from each uncovered element and add to intersection elements.
Minimum number of lines= 5
= order of matrix.
:. Optimum solution is reached.
Step 4 : Making assignment at single zero of the row and of the column.
. . The optimal assignment of jobs to machines is
M1 → B M2 → D M3 → E M4 → C M5 → A
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| I | II | III | IV | |
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Step I: Subtract the smallest element of each row from every element of that row:
| I | II | III | IV | |
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| B | 11 | 0 | 10 | 0 |
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Step II: Since all column minimums are zero, no need to subtract anything from columns.
Step III: Draw the minimum number of lines to cover all zeros.
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| A | 0 | 8 | 7 | 5 |
| B | 11 | 0 | 10 | 0 |
| C | 2 | 3 | 5 | 0 |
| D | 0 | 11 | 0 | 5 |
Since minimum number of lines = order of matrix, optimal solution has been reached
Optimal assignment is A →`square` B →`square`
C →IV D →`square`
Total minimum time = `square` hours.
