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प्रश्न
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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उत्तर
| 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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संबंधित प्रश्न
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| I | II | III | IV | |
| A | 2 | 10 | 9 | 7 |
| B | 13 | 2 | 12 | 2 |
| C | 3 | 4 | 6 | 1 |
| D | 4 | 15 | 4 | 9 |
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Service Time ( in hrs.)
| Counters | Salesmen | |||
| A | B | C | D | |
| W | 41 | 72 | 39 | 52 |
| X | 22 | 29 | 49 | 65 |
| Y | 27 | 39 | 60 | 51 |
| Z | 45 | 50 | 48 | 52 |
Solve the following minimal assignment problem and hence find minimum time where '- ' indicates that job cannot be assigned to the machine :
| Machines | Processing time in hours | ||||
| A | B | C | D | E | |
| M1 | 9 | 11 | 15 | 10 | 11 |
| M2 | 12 | 9 | - | 10 | 9 |
| M3 | - | 11 | 14 | 11 | 7 |
| M4 | 14 | 8 | 12 | 7 | 8 |
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contents. With the help of the performance matrix given below, find out which of the four subordinates should be assigned which jobs ?
| Subordinates | Jobs | ||
| I | II | III | |
| A | 7 | 3 | 5 |
| B | 2 | 7 | 4 |
| C | 6 | 5 | 3 |
| D | 3 | 4 | 7 |
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The assignment problem is said to be balanced if ______.
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A dairy plant has five milk tankers, I, II, III, IV and V. These milk tankers are to be used on five delivery routes A, B, C, D and E. The distances (in kms) between the dairy plant and the delivery routes are given in the following distance matrix.
| I | II | III | IV | V | |
| A | 150 | 120 | 175 | 180 | 200 |
| B | 125 | 110 | 120 | 150 | 165 |
| C | 130 | 100 | 145 | 160 | 175 |
| D | 40 | 40 | 70 | 70 | 100 |
| E | 45 | 25 | 60 | 70 | 95 |
How should the milk tankers be assigned to the chilling center so as to minimize the distance travelled?
Choose the correct alternative:
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The assignment problem is said to be balanced if ______
If the given matrix is ______ matrix, the assignment problem is called balanced problem
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| Programmers | ||||
| P | Q | R | ||
| Programmers | 1 | 120 | 100 | 80 |
| 2 | 80 | 90 | 110 | |
| 3 | 110 | 140 | 120 | |
Assign the programmers to the programme in such a way that the total computer time is least.
Choose the correct alternative:
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A car hire company has one car at each of five depots a, b, c, d and e. A customer in each of the fine towers A, B, C, D and E requires a car. The distance (in miles) between the depots (origins) and the towers(destinations) where the customers are given in the following distance matrix.
| a | b | c | d | e | |
| A | 160 | 130 | 175 | 190 | 200 |
| B | 135 | 120 | 130 | 160 | 175 |
| C | 140 | 110 | 155 | 170 | 185 |
| D | 50 | 50 | 80 | 80 | 110 |
| E | 55 | 35 | 70 | 80 | 105 |
How should the cars be assigned to the customers so as to minimize the distance travelled?
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 |
Five wagons are available at stations 1, 2, 3, 4 and 5. These are required at 5 stations I, II, III, IV and V. The mileage between various stations are given in the table below. How should the wagons be transported so as to minimize the mileage covered?
| I | II | III | IV | V | |
| 1 | 10 | 5 | 9 | 18 | 11 |
| 2 | 13 | 9 | 6 | 12 | 14 |
| 3 | 7 | 2 | 4 | 4 | 5 |
| 4 | 18 | 9 | 12 | 17 | 15 |
| 5 | 11 | 6 | 14 | 19 | 10 |
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`
