Find Solution using Voggel's Approximation 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 | | |
Table-1
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply | Row Penalty |
| `S_1` | 11 | 13 | 17 | 14 | | 250 | `2=13-11` |
| `S_2` | 16 | 18 | 14 | 10 | | 300 | `4=14-10` |
| `S_3` | 21 | 24 | 13 | 10 | | 400 | `3=13-10` |
| |
| Demand | 200 | 225 | 275 | 250 | | | |
Column
Penalty | `5=16-11` | `5=18-13` | `1=14-13` | `0=10-10` | | | |
The maximum penalty, 5, occurs in column `D_1`.
The minimum `c_(ij)` in this column is `c_11` = 11.
The maximum allocation in this cell is min(250,200) = 200.
It satisfy demand of `D_1` and adjust the supply of `S_1` from 250 to 50 (250 - 200 = 50).
Table-2
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply | Row Penalty |
| `S_1` | 11(200) | 13 | 17 | 14 | | 50 | `1=14-13` |
| `S_2` | 16 | 18 | 14 | 10 | | 300 | `4=14-10` |
| `S_3` | 21 | 24 | 13 | 10 | | 400 | `3=13-10` |
| |
| Demand | 0 | 225 | 275 | 250 | | | |
Column
Penalty | -- | `5=18-13` | `1=14-13` | `0=10-10` | | | |
The maximum penalty, 5, occurs in column `D_2`.
The minimum `c_(ij)` in this column is `c_12` = 13.
The maximum allocation in this cell is min(50,225) = 50.
It satisfy supply of `S_1` and adjust the demand of `D_2` from 225 to 175 (225 - 50 = 175).
Table-3
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply | Row Penalty |
| `S_1` | 11(200) | 13(50) | 17 | 14 | | 0 | -- |
| `S_2` | 16 | 18 | 14 | 10 | | 300 | `4=14-10` |
| `S_3` | 21 | 24 | 13 | 10 | | 400 | `3=13-10` |
| |
| Demand | 0 | 175 | 275 | 250 | | | |
Column
Penalty | -- | `6=24-18` | `1=14-13` | `0=10-10` | | | |
The maximum penalty, 6, occurs in column `D_2`.
The minimum `c_(ij)` in this column is `c_22` = 18.
The maximum allocation in this cell is min(300,175) = 175.
It satisfy demand of `D_2` and adjust the supply of `S_2` from 300 to 125 (300 - 175 = 125).
Table-4
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply | Row Penalty |
| `S_1` | 11(200) | 13(50) | 17 | 14 | | 0 | -- |
| `S_2` | 16 | 18(175) | 14 | 10 | | 125 | `4=14-10` |
| `S_3` | 21 | 24 | 13 | 10 | | 400 | `3=13-10` |
| |
| Demand | 0 | 0 | 275 | 250 | | | |
Column
Penalty | -- | -- | `1=14-13` | `0=10-10` | | | |
The maximum penalty, 4, occurs in row `S_2`.
The minimum `c_(ij)` in this row is `c_24` = 10.
The maximum allocation in this cell is min(125,250) = 125.
It satisfy supply of `S_2` and adjust the demand of `D_4` from 250 to 125 (250 - 125 = 125).
Table-5
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply | Row Penalty |
| `S_1` | 11(200) | 13(50) | 17 | 14 | | 0 | -- |
| `S_2` | 16 | 18(175) | 14 | 10(125) | | 0 | -- |
| `S_3` | 21 | 24 | 13 | 10 | | 400 | `3=13-10` |
| |
| Demand | 0 | 0 | 275 | 125 | | | |
Column
Penalty | -- | -- | `13` | `10` | | | |
The maximum penalty, 13, occurs in column `D_3`.
The minimum `c_(ij)` in this column is `c_33` = 13.
The maximum allocation in this cell is min(400,275) = 275.
It satisfy demand of `D_3` and adjust the supply of `S_3` from 400 to 125 (400 - 275 = 125).
Table-6
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply | Row Penalty |
| `S_1` | 11(200) | 13(50) | 17 | 14 | | 0 | -- |
| `S_2` | 16 | 18(175) | 14 | 10(125) | | 0 | -- |
| `S_3` | 21 | 24 | 13(275) | 10 | | 125 | `10` |
| |
| Demand | 0 | 0 | 0 | 125 | | | |
Column
Penalty | -- | -- | -- | `10` | | | |
The maximum penalty, 10, occurs in row `S_3`.
The minimum `c_(ij)` in this row is `c_34` = 10.
The maximum allocation in this cell is min(125,125) = 125.
It satisfy supply of `S_3` and demand of `D_4`.
Initial feasible solution is
| | `D_1` | `D_2` | `D_3` | `D_4` | | Supply | Row Penalty |
| `S_1` | 11(200) | 13(50) | 17 | 14 | | 250 | 2 | 1 | -- | -- | -- | -- | |
| `S_2` | 16 | 18(175) | 14 | 10(125) | | 300 | 4 | 4 | 4 | 4 | -- | -- | |
| `S_3` | 21 | 24 | 13(275) | 10(125) | | 400 | 3 | 3 | 3 | 3 | 3 | 10 | |
| |
| Demand | 200 | 225 | 275 | 250 | | | |
Column
Penalty | 5
--
--
--
--
--
| 5
5
6
--
--
--
| 1
1
1
1
13
--
| 0
0
0
0
10
10
| | | |
The minimum total transportation cost `= 11 xx 200 + 13 xx 50 + 18 xx 175 + 10 xx 125 + 13 xx 275 + 10 xx 125 = 12075`
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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