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प्रश्न
A job production unit has four jobs A, B, C, D which can be manufactured on each of the four machines P, Q, R and S. The processing cost of each job for each machine is given in the following table:
| Jobs | Machines (Processing Cost in ₹) |
|||
| P | Q | R | S | |
| A | 31 | 25 | 33 | 29 |
| B | 25 | 24 | 23 | 21 |
| C | 19 | 21 | 23 | 24 |
| D | 38 | 36 | 34 | 40 |
Find the optimal assignment to minimize the total processing cost.
उत्तर
Step 1: Row minimum
Subtract the smallest element in each row from every element in its row. The matrix obtained is given below:
| Jobs | Machines (Processing Cost in ₹) |
|||
| P | Q | R | S | |
| A | 6 | 0 | 8 | 4 |
| B | 4 | 3 | 2 | 0 |
| C | 0 | 2 | 4 | 5 |
| D | 4 | 2 | 0 | 6 |
Step 2: Column minimum
Subtract the smallest element in each column of assignment matrix obtained in step 1 from every element in its column.
| Jobs | Machine (Processing Cost in ₹) |
|||
| P | Q | R | S | |
| A | 6 | 0 | 8 | 4 |
| B | 4 | 3 | 2 | 0 |
| C | 0 | 2 | 4 | 5 |
| D | 4 | 2 | 0 | 6 |
Step 3:
Draw minimum number of vertical and horizontal lines to cover all zeros.
First cover all rows and columns which have maximum number of zeros.
| Jons | Machines (Processing Cost in ₹) |
|||
| P | Q | R | S | |
| A | 6 | 0 | 8 | 4 |
| B | 4 | 3 | 2 | 0 |
| C | 0 | 2 | 4 | 5 |
| D | 4 | 2 | 0 | 6 |
Step 4:
From step 3, minimum number of lines covering all the zeros are 4, which is equal to order of the matrix, i.e., 4
∴ Select a row with exactly one zero, enclose that zero in () and cross out all zeros in its respective column.
Similarly, examine each row and column and mark the assignment ().
∴ The matrix obtained is as follows:
| Jobs | Machines (Processing Cost in ₹) |
|||
| P | Q | R | S | |
| A | 6 | 0 | 8 | 4 |
| B | 4 | 3 | 2 | 0 |
| C | 0 | 2 | 4 | 5 |
| D | 4 | 2 | 0 | 6 |
Step 5:
The matrix obtained in step 4 contains exactly one assignment for each row and column.
∴ Optimal assignment schedule is as follows:
| Jobs | Machines | Processing cost (₹) |
| A | Q | 25 |
| B | S | 21 |
| C | P | 19 |
| D | R | 34 |
∴ Total minimum processing cost = 25 + 21 + 19 + 34 = ₹ 99.
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संबंधित प्रश्न
Solve the following minimal assignment problem :
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| 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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| A | B | C | D | |
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| A | B | C | D | E | |
| 1 | 30 | 37 | 40 | 28 | 40 |
| 2 | 40 | 24 | 27 | 21 | 36 |
| 3 | 40 | 32 | 33 | 30 | 35 |
| 4 | 25 | 38 | 40 | 36 | 36 |
| 5 | 29 | 62 | 41 | 34 | 39 |
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| A | 7 | 25 | 26 | 10 |
| B | 12 | 27 | 3 | 25 |
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| I | II | III | IV | V | |
| A | 150 | 120 | 175 | 180 | 200 |
| B | 125 | 110 | 120 | 150 | 165 |
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| D | 40 | 40 | 70 | 70 | 100 |
| E | 45 | 25 | 60 | 70 | 95 |
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Choose the correct alternative:
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If the given matrix is ______ matrix, the assignment problem is called balanced problem
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| Machine | ||||
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| B | 10 | 25 | 16 | |
| C | 12 | 14 | 11 | |
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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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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 |
| B | 8 | 2 | 5 | 5 |
| C | 4 | 9 | 6 | 9 |
| D | 7 | 5 | 4 | 8 |
| E | 6 | 3 | 5 | 4 |
| F | 6 | 8 | 7 | 3 |
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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:
| To | |||||||
| 7 | 8 | 9 | 10 | 11 | 12 | ||
| From | 1 | 31 | 62 | 29 | 42 | 15 | 41 |
| 2 | 12 | 19 | 39 | 55 | 71 | 40 | |
| 3 | 17 | 29 | 50 | 41 | 22 | 22 | |
| 4 | 35 | 40 | 38 | 42 | 27 | 33 | |
| 5 | 19 | 30 | 29 | 16 | 20 | 33 | |
| 6 | 72 | 30 | 30 | 50 | 41 | 20 | |
How should the truck be dispersed so as to minimize the total 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?
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 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.
