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Question
A department store has four workers to pack goods. The times (in minutes) required for each worker to complete the packings per item sold is given below. How should the manager of the store assign the jobs to the workers, so as to minimize the total time of packing?
| Workers | Packing of | |||
| Books | Toys | Crockery | Cutlery | |
| A | 3 | 11 | 10 | 8 |
| B | 13 | 2 | 12 | 12 |
| C | 3 | 4 | 6 | 1 |
| D | 4 | 15 | 4 | 9 |
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Solution
Step 1: Subtract the smallest element of each row from the every element of that row.
| Packing | Books | Toys | Crockery | Cutlery |
| Workers | ||||
| A | 0 | 8 | 7 | 5 |
| B | 11 | 0 | 10 | 0 |
| C | 2 | 3 | 5 | 0 |
| D | 0 | 11 | 0 | 5 |
Step 2: Since each column contains minimum element 0, subtracting it from every element of each column will given the same table as above.
Step 3: Drawing horizontal and vertical lines to cover all zeros.
| Packing | Books | Toys | Crockery | Cutlery |
| Workers | ||||
| A | 0 | 0 | 7 | 5 |
| B | 11 | 0 | 10 | 0 |
| C | 2 | 3 | 5 | 0 |
| D | 0 | 11 | 0 | 5 |
Step 4: Number of horizontal lines (4) = order of matrix (4), optimal solution has reached.
Assigning through zeroes, we get
| Packing | Books | Toys | Crockery | Cutlery |
| Workers | ||||
| A | `bb(0)` | 8 | 7 | 5 |
| B | 11 | `bb(0)` | 10 | `bb(cancel(0))` |
| C | 2 | 3 | 5 | `bb(0)` |
| D | `bb(cancel(0))` | 11 | `bb(0)` | 5 |
∴ Optimal Assignment schedule: A → Books
B → Bays, C → Cutlery, D → Crockery
Total minimum Time = 3 + 2 + 4 + 1 = 10 min
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Solution:
Step I: Subtract the smallest element of each row from every element of that row:
| I | II | III | IV | |
| A | 0 | 8 | 7 | 5 |
| B | 11 | 0 | 10 | 0 |
| C | 2 | 3 | 5 | 0 |
| D | 0 | 11 | 0 | 5 |
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.
| I | II | III | IV | |
| 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.
