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Question
Solve the following problem :
A plant manager has four subordinates, and four tasks to be performed. The subordinates differ in efficiency and the tasks differ in their intrinsic difficulty. This estimate of the time each man would take to perform each task is given in the effectiveness matrix below.
| I | II | III | IV | |
| A | 7 | 25 | 26 | 10 |
| B | 12 | 27 | 3 | 25 |
| C | 37 | 18 | 17 | 14 |
| D | 18 | 25 | 23 | 9 |
How should the tasks be allocated, one to a man, as to minimize the total man hours?
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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 | |
| A | 0 | 18 | 19 | 3 |
| B | 9 | 24 | 0 | 22 |
| C | 23 | 4 | 3 | 0 |
| D | 9 | 16 | 14 | 0 |
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 | |
| A | 0 | 14 | 19 | 3 |
| B | 9 | 20 | 0 | 22 |
| C | 23 | 0 | 3 | 0 |
| D | 9 | 12 | 14 | 0 |
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 | |
| A | 0 | 14 | 19 | 3 |
| B | 9 | 20 | 0 | 22 |
| C | 23 | 0 | 3 | 0 |
| D | 9 | 12 | 14 | 0 |
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:
Assigning through zeroes, we get,
| I | II | III | IV | |
| A | 0 | 14 | 19 | 3 |
| B | 9 | 20 | 0 | 22 |
| C | 23 | 0 | 3 | 0 |
| D | 9 | 12 | 1 | 0 |
Step 5:
The matrix obtained in step 4 contains exactly one assignment for each row and column.
∴ Optimal assignment schedule is as follows:
| Job | Subordination | Time (hrs) |
| A | I | 7 |
| B | III | 3 |
| C | II | 18 |
| D | IV | 9 |
∴ Total minimum time = 7 + 3 + 18 + 9 = 37 hrs.
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|
Jobs
|
Machines |
|||
|
P |
Q |
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|
|
Processing Cost (Rs.)
|
||||
|
A |
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|
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|||
| I | II | III | IV | |
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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`
