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Question
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.
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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:
| 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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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`
