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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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A plant manager has four subordinates and four tasks to perform. The subordinates differ in efficiency and task differ in their intrinsic difficulty. Estimates of the time subordinate would take to perform tasks are given in the following table:
| I | II | III | IV | |
| A | 3 | 11 | 10 | 8 |
| B | 13 | 2 | 12 | 2 |
| C | 3 | 4 | 6 | 1 |
| D | 4 | 15 | 4 | 9 |
Complete the following activity to allocate tasks to subordinates to minimize total time.
Solution:
Step I: Subtract the smallest element of each row from every element of that row:
| I | II | III | IV | |
| A | 0 | 8 | 7 | 5 |
| B | 11 | 0 | 10 | 0 |
| C | 2 | 3 | 5 | 0 |
| D | 0 | 11 | 0 | 5 |
Step II: Since all column minimums are zero, no need to subtract anything from columns.
Step III: Draw the minimum number of lines to cover all zeros.
| I | II | III | IV | |
| A | 0 | 8 | 7 | 5 |
| B | 11 | 0 | 10 | 0 |
| C | 2 | 3 | 5 | 0 |
| D | 0 | 11 | 0 | 5 |
Since minimum number of lines = order of matrix, optimal solution has been reached
Optimal assignment is A →`square` B →`square`
C →IV D →`square`
Total minimum time = `square` hours.
