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Question
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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Solution
Step 1 : The problem is unbalanced. So, it is balanced by introducing a dummy machine M5 with O.
Step 2 : Minimum element of each row is subtracted from every element in that row.
Step 3 : Zero element are covered with minimum number of straight lines :
Minimum number of lines = order of matrix = 5
∴ Optimum solution is reached
Step 4 : Making assignment at single zero of the row and then at single zero of the column.
The optional assignment is
M1 → A
M2 → B
M3 → E
M4 → D
M5 → C
∴ Minimum Time= 9 + 9 + 7 + 7 + 0 = 32 hrs.
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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:
| To | |||||||
| 7 | 8 | 9 | 10 | 11 | 12 | ||
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| 3 | 17 | 29 | 50 | 41 | 22 | 22 | |
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A dairy plant has five milk tankers, I, II, III, IV and V. Three 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.
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|||
| I | II | III | IV | |
| P | 31 | 25 | 33 | 29 |
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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 |
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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`
A plant manager has four subordinates and four tasks to perform. The subordinates differ in efficiency and task differ in their intrinsic difficulty. Estimates of the time subordinate would take to perform tasks are given in the following table:
| I | II | III | IV | |
| A | 3 | 11 | 10 | 8 |
| B | 13 | 2 | 12 | 2 |
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Complete the following activity to allocate tasks to subordinates to minimize total time.
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.
