A marketing manager has list of salesmen and territories. Considering the travelling cost of the salesmen and the nature of territory, the marketing manager estimates the total of cost per month (in thousand rupees) for each salesman in each territory. Suppose these amounts are as follows:
| Salesman | Territories | ||||
|---|---|---|---|---|---|
| I | II | III | IV | V | |
| A | 11 | 16 | 18 | 15 | 15 |
| B | 7 | 19 | 11 | 13 | 17 |
| C | 9 | 6 | 14 | 14 | 7 |
| D | 13 | 12 | 17 | 11 | 13 |
| Find the assignment of salesman to territories that will result in minimum cost. |
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Start your 14-day free trial to unlock the full solution →Balance the matrix with a dummy salesman (all zeros), apply the Hungarian method, and obtain , , , with unassigned; minimum cost .
Cost matrix (thousand ₹) with a dummy row added to make it square:
| I | II | III | IV | V | |
|---|---|---|---|---|---|
| A | 11 | 16 | 18 | 15 | 15 |
| B | 7 | 19 | 11 | 13 | 17 |
| C | 9 | 6 | 14 | 14 | 7 |
| D | 13 | 12 | 17 | 11 | 13 |
| E | 0 | 0 | 0 | 0 | 0 |
Step 1 — Row reduction (subtract each row's minimum; row is already ):
| I | II | III | IV | V | |
|---|---|---|---|---|---|
| A | 0 | 5 | 7 | 4 | 4 |
| B | 0 | 12 | 4 | 6 | 10 |
| C | 3 | 0 | 8 | 8 | 1 |
| D | 2 | 1 | 6 | 0 | 2 |
| E | 0 | 0 | 0 | 0 | 0 |
Step 2 — Column reduction. Every column already contains a , so the matrix is unchanged.
Step 3 — Cover the zeros. The zeros can be covered by lines (row ; columns I, II, IV), which is fewer than , so we improve. Repeatedly subtracting the smallest uncovered entry from all uncovered cells and adding it at line-intersections (standard Hungarian iterations) finally produces a reduced matrix admitting one independent zero per row and column:
| I | II | III | IV | V | |
|---|---|---|---|---|---|
| A | 0 | 2 | 3 | 2 | 0 |
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