Solve the following assignment problem to maximization:
| I | II | III | IV | V | |
|---|---|---|---|---|---|
| 1 | 18 | 24 | 19 | 20 | 23 |
| 2 | 19 | 21 | 20 | 18 | 22 |
| 3 | 22 | 23 | 20 | 21 | 23 |
| 4 | 20 | 18 | 21 | 19 | 19 |
| 5 | 18 | 22 | 23 | 22 | 21 |
| Step - I | |||||
| Subtract the Smallest element of each row from every element of that row | |||||
| I | II | III | IV | V | |
| --- | --- | --- | --- | --- | --- |
| 1 | 0 | 6 | 1 | 2 | 4 |
| 2 | 1 | 3 | 2 | 0 | 3 |
| 3 | 2 | 3 | 0 | 1 | 3 |
| 4 | 2 | 0 | 3 | 1 | 1 |
| 5 | 0 | 4 | 5 | 4 | 3 |
| Step - II | |||||
| Subtract the smallest element of each column from every element of that column: | |||||
| I | II | III | IV | V | |
| --- | --- | --- | --- | --- | --- |
| 1 | 0 | 6 | 1 | 2 | 4 |
| 2 | 1 | 3 | 2 | 0 | 3 |
| 3 | 2 | 3 | 0 | 1 | 2 |
| 4 | 2 | 0 | 3 | 1 | 0 |
| 5 | 0 | 4 | 5 | 4 | 2 |
| Step - III | |||||
| Draw minimum number of lines covering all zeros. | |||||
| I | II | III | IV | V | |
| --- | --- | --- | --- | --- | --- |
| 1 | 0 | 6 | 1 | 2 | 4 |
| 2 | 1 | 3 | 2 | 0 | 3 |
| 3 | 2 | 3 | 0 | 1 | 2 |
| 4 | 2 | 0 | 3 | 1 | 0 |
| 5 | 0 | 4 | 5 | 4 | 2 |
| Step - 1V | |||||
| The smallest uncovered element is 1, which is to be subtracted from all uncovered elements and add it to all elements which lie at the intersection of two lines: | |||||
| I | II | III | IV | V | |
| --- | --- | --- | --- | --- | --- |
| 1 | 0 | 5 | 0 | 3 | |
| 2 | 2 | 3 | 2 | 0 | 3 |
| 3 | 3 | 3 | 0 | 2 | |
| 4 | 3 | 0 | 3 | 1 | 0 |
| 5 | 0 | 3 | 4 | 3 | |
| Step - V | |||||
| Draw minimum number of lines that are required to cover all zeros: | |||||
| I | II | III | IV | V | |
| --- | --- | --- | --- | --- | --- |
| 1 | 0 | 5 | 0 | 1 | 3 |
| 2 | 2 | 3 | 2 | 0 | 3 |
| 3 | 3 | 3 | 0 | 1 | 2 |
| 4 | 3 | 0 | 3 | 1 | 0 |
| 5 | 0 | 3 | 4 | 3 | 1 |
| Here minimum number of Lines ≠ order of matrix. | |||||
| Step - VI | |||||
| Find smallest uncovered element (1). Subtract this number from all uncovered elements and add it to all elements which lie at the intersection of two lines: | |||||
| I | II | III | IV | V | |
| --- | --- | --- | --- | --- | --- |
| 1 | 0 | 4 | 0 | 0 | 2 |
| 2 | 3 | 3 | 3 | 0 | 3 |
| 3 | 3 | 0 | 0 | ||
| 4 | 3 | 0 | 3 | 1 | 0 |
| 5 | 0 | 2 | 4 | 3 | 0 |
| Now minimium number of lines = order of matrix. | |||||
| The optimal assignment can be made. | |||||
| Optimal solution is | |||||
| 1 → I | |||||
| 2 → IV | |||||
| 3 → | |||||
| 4 → | |||||
| 5 → V | |||||
| Minimum value = |
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Start your 14-day free trial to unlock the full solution →Because the objective is to maximize, convert the profit matrix into a loss matrix using (subtract from the largest entry), then apply row reduction, column reduction and the Hungarian method. A complete set of five independent zeros gives the optimal assignment with maximum profit .
Given profit matrix (rows jobs –, columns machines I–V):
| I | II | III | IV | V | |
|---|---|---|---|---|---|
| 1 | 18 | 24 | 19 | 20 | 23 |
| 2 | 19 | 21 | 20 | 18 | 22 |
| 3 | 22 | 23 | 20 | 21 | 23 |
| 4 | 20 | 18 | 21 | 19 | 19 |
| 5 | 18 | 22 | 23 | 22 | 21 |
Step 1 — Convert to a minimization problem. The largest entry is . Replace each element by to get the opportunity-loss matrix:
| I | II | III | IV | V | |
|---|---|---|---|---|---|
| 1 | 6 | 0 | 5 | 4 | 1 |
| 2 | 5 | 3 | 4 | 6 | 2 |
| 3 | 2 | 1 | 4 | 3 | 1 |
| 4 | 4 | 6 | 3 | 5 | 5 |
| 5 | 6 | 2 | 1 | 2 | 3 |
Step 2 — Row reduction. Subtract the smallest entry of each row (row minima ):
| I | II | III | IV | V | |
|---|---|---|---|---|---|
| 1 | 6 | 0 | 5 | 4 | 1 |
| 2 | 3 | 1 | 2 | 4 | 0 |
| 3 | 1 | 0 | 3 | 2 | 0 |
| 4 | 1 | 3 | 0 | 2 | 2 |
| 5 | 5 | 1 | 0 | 1 | 2 |
Step 3 — Column reduction. Subtract the smallest entry of each column (column minima ):
| I | II | III | IV | V | |
|---|---|---|---|---|---|
| 1 | 5 | 0 | 5 | 3 | 1 |
| 2 | 2 | 1 | 2 | 3 | 0 |
| 3 | 0 | 0 | 3 | 1 | 0 |
| 4 | 0 | 3 | 0 | 1 | 2 |
| 5 | 4 | 1 | 0 | 0 | 2 |
Step 4 — Make the assignment (Hungarian method). Assign one zero per row and column:
- Row 1 has a single zero at II . …
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