Advertisements
Advertisements
प्रश्न
Find the optimal solution for the assignment problem with the following cost matrix.
| Area | |||||
| 1 | 2 | 3 | 4 | ||
| P | 11 | 17 | 8 | 16 | |
| Salesman | Q | 9 | 7 | 12 | 6 |
| R | 13 | 16 | 15 | 12 | |
| S | 14 | 10 | 12 | 11 | |
Advertisements
उत्तर
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1: Select the smallest element in each row and subtract this from all the elements in its row.
| Area | |||||
| 1 | 2 | 3 | 4 | ||
| P | 3 | 9 | 0 | 8 | |
| Salesman | Q | 3 | 1 | 6 | 0 |
| R | 1 | 4 | 3 | 0 | |
| S | 4 | 0 | 2 | 1 | |
Step 2: Select the smallest element in each column and subtract this from all the elements in its column.
| Area | |||||
| 1 | 2 | 3 | 4 | ||
| P | 2 | 9 | 0 | 8 | |
| Salesman | Q | 2 | 1 | 6 | 0 |
| R | 0 | 4 | 3 | 0 | |
| S | 3 | 0 | 2 | 1 | |
Step 3: (Assignment)
Examine the rows with exactly one zero. Mark the zero by □ Mark other zeros in its column by X
| Area | |||||
| 1 | 2 | 3 | 4 | ||
| P | 2 | 9 | 0 | 8 | |
| Salesman | Q | 2 | 1 | 6 | 0 |
| R | 0 | 4 | 3 | 0 | |
| S | 3 | 0 | 2 | 1 | |
Thus all the four assignments have been made.
The optimal assignment schedule and total cost.
| Salesman | Area | Cost |
| P | 3 | 8 |
| Q | 4 | 6 |
| R | 1 | 13 |
| S | 2 | 10 |
| Total | 37 | |
The Optimum cost (minimum) = ₹ 37
APPEARS IN
संबंधित प्रश्न
Determine `l_92 and l_93, "given that" l_91 = 97, d_91 = 38 and q_92 = 27/59`
The assignment problem is said to be balanced if ______.
Choose the correct alternative :
In an assignment problem if number of rows is greater than number of columns then
State whether the following is True or False :
It is not necessary to express an assignment problem into n x n matrix.
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?
Choose the correct alternative:
When an assignment problem has more than one solution, then it is ______
If the given matrix is ______ matrix, the assignment problem is called balanced problem
A departmental head has four subordinates and four tasks to be performed. The subordinates differ in efficiency and the tasks differ in their intrinsic difficulty. His estimates of the time each man would take to perform each task is given below:
| 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 | |
How should the tasks be allocated to subordinates so as to minimize the total manhours?
Choose the correct alternative:
The purpose of a dummy row or column in an assignment problem is to
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.
