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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?
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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|||
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I | II | III | IV | V | |
1 | 10 | 5 | 9 | 18 | 11 |
2 | 13 | 9 | 6 | 12 | 14 |
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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`