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प्रश्न
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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उत्तर
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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A job production unit has four jobs P, Q, R, S which can be manufactured on each of the four machines I, II, III and IV. The processing cost of each job for each machine is given in the following table :
| Job | Machines (Processing cost in ₹) |
|||
| I | II | III | IV | |
| P | 31 | 25 | 33 | 29 |
| Q | 25 | 24 | 23 | 21 |
| R | 19 | 21 | 23 | 24 |
| S | 38 | 36 | 34 | 40 |
Complete the following activity to find the optimal assignment to minimize the total processing cost.
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`
