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Question
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?
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Solution
Step 1: Row minimum
Subtract the smallest element in each row from every element in its row.
The matrix obtained is given below:
| I | II | III | IV | V | |
| A | 30 | 0 | 55 | 60 | 80 |
| B | 15 | 0 | 10 | 40 | 55 |
| C | 30 | 0 | 45 | 60 | 70 |
| D | 0 | 0 | 30 | 30 | 60 |
| E | 20 | 0 | 35 | 45 | 70 |
Step 2: Column minimum
Subtract the smallest element in each column of assignment matrix obtained in step 1 from every element in its column.
| I | II | III | IV | V | |
| A | 30 | 0 | 45 | 30 | 25 |
| B | 15 | 0 | 0 | 10 | 0 |
| C | 30 | 0 | 35 | 30 | 15 |
| D | 0 | 0 | 20 | 0 | 5 |
| E | 20 | 0 | 25 | 15 | 15 |
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.
| I | II | III | IV | V | |
| A | 30 | 0 | 45 | 30 | 25 |
| B | 15 | 0 | 0 | 10 | 0 |
| C | 30 | 0 | 35 | 30 | 15 |
| D | 0 | 0 | 20 | 0 | 5 |
| E | 20 | 0 | 25 | 15 | 15 |
Step 4:
From step 3, minimum number of lines covering all the zeros are 3, which is less than order of matrix, i.e., 5.
∴ Select smallest element from all the uncovered elements, i.e., 15 and subtract it from all the uncovered elements and add it to the elements which lie at the intersection of two lines.
| I | II | III | IV | V | |
| A | 15 | 0 | 30 | 15 | 10 |
| B | 15 | 15 | 0 | 10 | 0 |
| C | 15 | 0 | 20 | 15 | 0 |
| D | 0 | 15 | 20 | 0 | 5 |
| E | 5 | 0 | 10 | 0 | 0 |
Step 5:
Draw minimum number of vertical and horizontal lines to cover all zeros.
| I | II | III | IV | V | |
| A | 15 | 0 | 30 | 15 | 10 |
| B | 15 | 15 | 0 | 10 | 0 |
| C | 15 | 0 | 20 | 15 | 0 |
| D | 0 | 15 | 20 | 0 | 5 |
| E | 5 | 0 | 10 | 0 | 0 |
Step 6:
From step 5, minimum number of lines covering all the zeros are 4, which is less than order of matrix, i.e., 5.
∴ 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:
| I | II | III | IV | V | |
| A | 15 | 0 | 30 | 15 | 10 |
| B | 15 | 15 | 0 | 10 | 0 |
| C | 15 | 0 | 20 | 15 | 0 |
| D | 0 | 15 | 20 | 0 | 5 |
| E | 5 | 0 | 10 | 0 | 0 |
Step 7:
The matrix obtained in step 6 contains exactly one assignment for each row and column.
∴ Optimal assignment schedule is as follows:
| Routes | Dairy Plant | Distance (kms) |
| A | II | 120 |
| B | III | 120 |
| C | V | 170 |
| D | I | 40 |
| E | IV | 70 |
| 520 |
∴ Minimum distance travelled
= 120 + 120 + 170 + 40 + 70
= 520 kms.
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RELATED QUESTIONS
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|
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|||
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P |
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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 |
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`
