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Question
Find the optimal solution for the assignment problem with the following cost matrix.
| Area | |||||
| 1 | 2 | 3 | 4 | ||
| P | 11 | 17 | 8 | 16 | |
| Salesman | Q | 9 | 7 | 12 | 6 |
| R | 13 | 16 | 15 | 12 | |
| S | 14 | 10 | 12 | 11 | |
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Solution
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1: Select the smallest element in each row and subtract this from all the elements in its row.
| Area | |||||
| 1 | 2 | 3 | 4 | ||
| P | 3 | 9 | 0 | 8 | |
| Salesman | Q | 3 | 1 | 6 | 0 |
| R | 1 | 4 | 3 | 0 | |
| S | 4 | 0 | 2 | 1 | |
Step 2: Select the smallest element in each column and subtract this from all the elements in its column.
| Area | |||||
| 1 | 2 | 3 | 4 | ||
| P | 2 | 9 | 0 | 8 | |
| Salesman | Q | 2 | 1 | 6 | 0 |
| R | 0 | 4 | 3 | 0 | |
| S | 3 | 0 | 2 | 1 | |
Step 3: (Assignment)
Examine the rows with exactly one zero. Mark the zero by □ Mark other zeros in its column by X
| Area | |||||
| 1 | 2 | 3 | 4 | ||
| P | 2 | 9 | 0 | 8 | |
| Salesman | Q | 2 | 1 | 6 | 0 |
| R | 0 | 4 | 3 | 0 | |
| S | 3 | 0 | 2 | 1 | |
Thus all the four assignments have been made.
The optimal assignment schedule and total cost.
| Salesman | Area | Cost |
| P | 3 | 8 |
| Q | 4 | 6 |
| R | 1 | 13 |
| S | 2 | 10 |
| Total | 37 | |
The Optimum cost (minimum) = ₹ 37
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| M1 | 27 | 18 | ∞ | 20 | 21 |
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Solve the following maximal assignment problem :
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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 |
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| E | 45 | 25 | 60 | 70 | 95 |
How should the milk tankers be assigned to the chilling center so as to minimize the distance travelled?
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North – West Corner refers to ______
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If number of sources is not equal to number of destinations, the assignment problem is called ______
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The purpose of a dummy row or column in an assignment problem is to
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`
