Find Solution using Least Cost method
| D1 | D2 | D3 | D4 | Supply |
| S1 | 11 | 13 | 17 | 14 | 250 |
| S2 | 16 | 18 | 14 | 10 | 300 |
| S3 | 21 | 24 | 13 | 10 | 400 |
| Demand | 200 | 225 | 275 | 250 | |
Solution:
TOTAL number of supply constraints : 3
TOTAL number of demand constraints : 4
Problem Table is
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply |
| `S_1` | 11 | 13 | 17 | 14 | | 250 |
| `S_2` | 16 | 18 | 14 | 10 | | 300 |
| `S_3` | 21 | 24 | 13 | 10 | | 400 |
| |
| Demand | 200 | 225 | 275 | 250 | | |
The smallest transportation cost is 10 in cell `S_3 D_4`
The allocation to this cell is min(400,250) = 250.
This satisfies the entire demand of `D_4` and leaves 400 - 250 = 150 units with `S_3`
Table-1
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply |
| `S_1` | 11 | 13 | 17 | 14 | | 250 |
| `S_2` | 16 | 18 | 14 | 10 | | 300 |
| `S_3` | 21 | 24 | 13 | 10(250) | | 150 |
| |
| Demand | 200 | 225 | 275 | 0 | | |
The smallest transportation cost is 11 in cell `S_1 D_1`
The allocation to this cell is min(250,200) = 200.
This satisfies the entire demand of `D_1` and leaves 250 - 200 = 50 units with `S_1`
Table-2
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply |
| `S_1` | 11(200) | 13 | 17 | 14 | | 50 |
| `S_2` | 16 | 18 | 14 | 10 | | 300 |
| `S_3` | 21 | 24 | 13 | 10(250) | | 150 |
| |
| Demand | 0 | 225 | 275 | 0 | | |
The smallest transportation cost is 13 in cell `S_3 D_3`
The allocation to this cell is min(150,275) = 150.
This exhausts the capacity of `S_3` and leaves 275 - 150 = 125 units with `D_3`
Table-3
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply |
| `S_1` | 11(200) | 13 | 17 | 14 | | 50 |
| `S_2` | 16 | 18 | 14 | 10 | | 300 |
| `S_3` | 21 | 24 | 13(150) | 10(250) | | 0 |
| |
| Demand | 0 | 225 | 125 | 0 | | |
The smallest transportation cost is 13 in cell `S_1 D_2`
The allocation to this cell is min(50,225) = 50.
This exhausts the capacity of `S_1` and leaves 225 - 50 = 175 units with `D_2`
Table-4
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply |
| `S_1` | 11(200) | 13(50) | 17 | 14 | | 0 |
| `S_2` | 16 | 18 | 14 | 10 | | 300 |
| `S_3` | 21 | 24 | 13(150) | 10(250) | | 0 |
| |
| Demand | 0 | 175 | 125 | 0 | | |
The smallest transportation cost is 14 in cell `S_2 D_3`
The allocation to this cell is min(300,125) = 125.
This satisfies the entire demand of `D_3` and leaves 300 - 125 = 175 units with `S_2`
Table-5
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply |
| `S_1` | 11(200) | 13(50) | 17 | 14 | | 0 |
| `S_2` | 16 | 18 | 14(125) | 10 | | 175 |
| `S_3` | 21 | 24 | 13(150) | 10(250) | | 0 |
| |
| Demand | 0 | 175 | 0 | 0 | | |
The smallest transportation cost is 18 in cell `S_2 D_2`
The allocation to this cell is min(175,175) = 175.
Table-6
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply |
| `S_1` | 11(200) | 13(50) | 17 | 14 | | 0 |
| `S_2` | 16 | 18(175) | 14(125) | 10 | | 0 |
| `S_3` | 21 | 24 | 13(150) | 10(250) | | 0 |
| |
| Demand | 0 | 0 | 0 | 0 | | |
Initial feasible solution is
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply |
| `S_1` | 11 (200) | 13 (50) | 17 | 14 | | 250 |
| `S_2` | 16 | 18 (175) | 14 (125) | 10 | | 300 |
| `S_3` | 21 | 24 | 13 (150) | 10 (250) | | 400 |
| |
| Demand | 200 | 225 | 275 | 250 | | |
The minimum total transportation cost `= 11 xx 200 + 13 xx 50 + 18 xx 175 + 14 xx 125 + 13 xx 150 + 10 xx 250 = 12200`
Here, the number of allocated cells = 6 is equal to m + n - 1 = 3 + 4 - 1 = 6
`:.` This solution is non-degenerate
This material is intended as a summary. Use your textbook for detail explanation.
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