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प्रश्न
A departmental head has three jobs and four subordinates. The subordinates differ in their capabilities and the jobs differ in their work
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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उत्तर
As the problem is unbalanced a dummy · job IV is introduced with zero performance as, the for:
| I | II | III | IV | |
| A | 7 | 3 | 5 | 0 |
| B | 2 | 7 | 4 | 0 |
| C | 6 | 5 | 3 | 0 |
| D | 3 | 4 | 7 | 0 |
As minimum element of each row is 'O' hence minimum element of each column is subtracted from all the elements of that column,
| I | II | III | IV | |
| A | 5 | 0 | 2 | 0 |
| B | 0 | 4 | 1 | 0 |
| C | 4 | 2 | 0 | 0 |
| D | 1 | 1 | 4 | 0 |
All the zeros of the above matrix a re covered with minimum number of lines is as below :
As the number of lines covering zeros is equal to the number of rows and columns. hence the optional solution is obtained it is as shown below :
As D gets a dummy job N, hence the assignment is as shown below :
A → Job II
B → Job I
C → Job III
and performance is 3 + 2 + 3 = 8 units
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| 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 |
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A natural truck-rental service has a surplus of one truck in each of the cities 1, 2, 3, 4, 5 and 6 and a deficit of one truck in each of the cities 7, 8, 9, 10, 11 and 12. The distance(in kilometers) between the cities with a surplus and the cities with a deficit are displayed below:
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| 7 | 8 | 9 | 10 | 11 | 12 | ||
| From | 1 | 31 | 62 | 29 | 42 | 15 | 41 |
| 2 | 12 | 19 | 39 | 55 | 71 | 40 | |
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|||
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| I | II | III | IV | |
| A | 3 | 11 | 10 | 8 |
| B | 13 | 2 | 12 | 2 |
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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 |
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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.
| 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.
