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Assignment Problem |
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ExamplesNow we will examine a few highly simplified illustrations. Later in the chapter, you will find more practical versions of assignment models like Crew assignment problem, Travelling salesman problem, etc.
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| Job | ||||
|---|---|---|---|---|
| Person | 1 | 2 | 3 | 4 |
| A | 20 | 25 | 22 | 28 |
| B | 15 | 18 | 23 | 17 |
| C | 19 | 17 | 21 | 24 |
| D | 25 | 23 | 24 | 24 |
Step 1Identify the minimum element in each row and subtract it from every element of that row. The result is shown in the following table.
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"A man has one hundred dollars and you leave him with two dollars, that's subtraction." -Mae West |
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Job
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|---|---|---|---|---|
| Person | 1 | 2 | 3 | 4 |
| A | 0 | 5 | 2 | 8 |
| B | 0 | 3 | 8 | 2 |
| C | 2 | 0 | 4 | 7 |
| D | 2 | 0 | 1 | 1 |
Identify the minimum element in each column and subtract it from every element of that column.
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Job
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|---|---|---|---|---|
| Person | 1 | 2 | 3 | 4 |
| A | 0 | 5 | 1 | 7 |
| B | 0 | 3 | 7 | 1 |
| C | 2 | 0 | 3 | 6 |
| D | 2 | 0 | 0 | 0 |
Make the assignments for the reduced matrix obtained from steps 1 and 2 in the following way:
An optimal assignment is found, if the number of assigned cells equals the number of rows (and columns). In case you have chosen a zero cell arbitrarily, there may be alternate optimal solutions. If no optimal solution is found, go to step 5.
| Job | ||||
|---|---|---|---|---|
| Person | 1 | 2 | 3 | 4 |
| A | |
5 | 1 | 7 |
| B | |
3 | 7 | 1 |
| C | 2 | |
3 | 6 |
| D | 2 | |
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Draw the minimum number of vertical and horizontal lines necessary to cover all the zeros in the reduced matrix obtained from step 3 by adopting the following procedure:
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You can also draw the minimum number of lines
by inspection.
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Select the smallest element (i.e., 1) from all the uncovered elements. Subtract this smallest element from all the uncovered elements and add it to the elements, which lie at the intersection of two lines. Thus, we obtain another reduced matrix for fresh assignment.
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Job
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|---|---|---|---|---|
| Person | 1 | 2 | 3 | 4 |
| A | 0 | 4 | 0 | 6 |
| B | 0 | 2 | 6 | 0 |
| C | 3 | 0 | 3 | 6 |
| D | 3 | 0 | 0 | 0 |
Now again make the assignments for the reduced matrix.
| Job | ||||
|---|---|---|---|---|
| Person | 1 | 2 | 3 | 4 |
| A | |
4 | |
6 |
| B | |
2 | 6 | |
| C | 3 | |
3 | 6 |
| D | 3 | |
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Since the number of assignments is equal to the number of rows (& columns), this is the optimal solution.
The total cost of assignment = A1 + B4 + C2 + D3
Substituting values from original table:
20 + 17 + 17 + 24 = Rs. 78.