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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 is given in the following table:
|
Jobs
|
Machines |
|||
|
P |
Q |
R |
S |
|
|
Processing Cost (Rs.)
|
||||
|
A |
31 |
25 |
33 |
29 |
|
B |
25 |
24 |
23 |
21 |
|
C |
19 |
21 |
23 |
24 |
|
D |
38 |
36 |
34 |
40 |
How should the jobs be assigned to the four machines so that the total processing cost is minimum?
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उत्तर
We can express the matrix form
`[(31,25,33,29), (25,24,23,21), (19,21,23,24), (38,36,34,40)]`
Subtracting the smallest element in each row from every element of it,
`[(6,0,8,4), (4,3,2,0), (0,2,4,5), (4,2,0,6)]`
Subtracting the smallest element In each column from very element of it.
`[(6,0,8,4), (4,3,2,0), (0,2,4,5), (4,2,0,6)]`
All the zeros of the above matrix are covered with minimum number of lines as below :
No. of lines = No. of rows/columns
Assignment of jobs :
A → Q, B → S, C → P, D → R
Minimum cost = 25 + 21 + 19 + 34
= Rs 99
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| Subordinates | Jobs | ||
| I | II | III | |
| A | 7 | 3 | 5 |
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| C | 6 | 5 | 3 |
| D | 3 | 4 | 7 |
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| I | II | III | IV | |
| A | 7 | 25 | 26 | 10 |
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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 |
| B | 125 | 110 | 120 | 150 | 165 |
| C | 130 | 100 | 145 | 160 | 175 |
| D | 40 | 40 | 70 | 70 | 100 |
| E | 45 | 25 | 60 | 70 | 95 |
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Three jobs A, B and C one to be assigned to three machines U, V and W. The processing cost for each job machine combination is shown in the matrix given below. Determine the allocation that minimizes the overall processing cost.
| Machine | ||||
| U | V | W | ||
| Jobs | A | 17 | 25 | 31 |
| B | 10 | 25 | 16 | |
| C | 12 | 14 | 11 | |
(cost is in ₹ per unit)
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| Tasks | |||||
| 1 | 2 | 3 | 4 | ||
| Subordinates | P | 8 | 26 | 17 | 11 |
| Q | 13 | 28 | 4 | 26 | |
| R | 38 | 19 | 18 | 15 | |
| S | 9 | 26 | 24 | 10 | |
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| 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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| 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 |
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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`
