Find Solution using North-West Corner 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 rim values for `S_1`=250 and `D_1`=200 are compared.
The smaller of the two i.e. min(250,200) = 200 is assigned to `S_1` `D_1`
This meets the complete demand of `D_1` and leaves 250 - 200=50 units with `S_1`
Table-1
| | `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 | | 400 |
| |
| Demand | 0 | 225 | 275 | 250 | | |
Move horizontally,
The rim values for `S_1`=50 and `D_2`=225 are compared.
The smaller of the two i.e. min(50,225) = 50 is assigned to `S_1` `D_2`
This exhausts the capacity of `S_1` and leaves 225 - 50=175 units with `D_2`
Table-2
| | `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 | 10 | | 400 |
| |
| Demand | 0 | 175 | 275 | 250 | | |
Move vertically,
The rim values for `S_2`=300 and `D_2`=175 are compared.
The smaller of the two i.e. min(300,175) = 175 is assigned to `S_2` `D_2`
This meets the complete demand of `D_2` and leaves 300 - 175=125 units with `S_2`
Table-3
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply |
| `S_1` | 11(200) | 13(50) | 17 | 14 | | 0 |
| `S_2` | 16 | 18(175) | 14 | 10 | | 125 |
| `S_3` | 21 | 24 | 13 | 10 | | 400 |
| |
| Demand | 0 | 0 | 275 | 250 | | |
Move horizontally,
The rim values for `S_2`=125 and `D_3`=275 are compared.
The smaller of the two i.e. min(125,275) = 125 is assigned to `S_2` `D_3`
This exhausts the capacity of `S_2` and leaves 275 - 125=150 units with `D_3`
Table-4
| | `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 | 10 | | 400 |
| |
| Demand | 0 | 0 | 150 | 250 | | |
Move vertically,
The rim values for `S_3`=400 and `D_3`=150 are compared.
The smaller of the two i.e. min(400,150) = 150 is assigned to `S_3` `D_3`
This meets the complete demand of `D_3` and leaves 400 - 150=250 units with `S_3`
Table-5
| | `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 |
| |
| Demand | 0 | 0 | 0 | 250 | | |
Move horizontally,
The rim values for `S_3`=250 and `D_4`=250 are compared.
The smaller of the two i.e. min(250,250) = 250 is assigned to `S_3` `D_4`
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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