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Question
State whether the following is True or False :
It is not necessary to express an assignment problem into n x n matrix.
Options
True
False
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Solution
It is not necessary to express an assignment problem into n x n matrix False.
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Solve the following minimal assignment problem and hence find the minimum value :
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| A | 2 | 10 | 9 | 7 |
| B | 13 | 2 | 12 | 2 |
| C | 3 | 4 | 6 | 1 |
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| A | B | C | D | |
| W | 41 | 72 | 39 | 52 |
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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 ?
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| I | II | III | IV | V | |
| 1 | 10 | 5 | 9 | 18 | 11 |
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The assignment problem is said to be balanced if it is a ______.
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| Programmers | ||||
| P | Q | R | ||
| Programmers | 1 | 120 | 100 | 80 |
| 2 | 80 | 90 | 110 | |
| 3 | 110 | 140 | 120 | |
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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 |
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Assign four trucks 1, 2, 3 and 4 to vacant spaces A, B, C, D, E and F so that distance travelled is minimized. The matrix below shows the distance.
| 1 | 2 | 3 | 4 | |
| A | 4 | 7 | 3 | 7 |
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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 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.
| I | II | III | IV | V | |
| A | 150 | 120 | 175 | 180 | 200 |
| B | 125 | 110 | 120 | 150 | 165 |
| C | 130 | 100 | 145 | 160 | 170 |
| 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?
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`
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 |
| C | 3 | 4 | 6 | 1 |
| D | 4 | 15 | 4 | 9 |
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.
