3. Unbalanced supply and demand example
Unbalanced supply and demand
If the total supply is not equal to the total demand then the problem is called unbalanced transportation problem.
It's solution :
1. If the total supply is more than the total demand, then we add a new column, with transportation cost 0
2. If the total demand is more than the total supply, then we add a new row, with transportation cost 0
Example
Find Solution using Row minima method
| D1 | D2 | D3 | Supply | | S1 | 4 | 8 | 8 | 76 | | S2 | 16 | 24 | 16 | 82 | | S3 | 8 | 16 | 24 | 77 | | Demand | 72 | 102 | 41 | |
Solution:
TOTAL number of supply constraints : 3
TOTAL number of demand constraints : 3
Problem Table is
| | `D_1` | `D_2` | `D_3` | | Supply | | `S_1` | 4 | 8 | 8 | | 76 | | `S_2` | 16 | 24 | 16 | | 82 | | `S_3` | 8 | 16 | 24 | | 77 | | | | Demand | 72 | 102 | 41 | | |
Here Total Demand = 215 is less than Total Supply = 235. So We add a dummy demand constraint with 0 unit cost and with allocation 20.
Now, The modified table is
| | `D_1` | `D_2` | `D_3` | `D_(dummy)` | | Supply | | `S_1` | 4 | 8 | 8 | 0 | | 76 | | `S_2` | 16 | 24 | 16 | 0 | | 82 | | `S_3` | 8 | 16 | 24 | 0 | | 77 | | | | Demand | 72 | 102 | 41 | 20 | | |
In `1^(st)` row, The smallest transportation cost is 0 in cell `S_1 D_(dummy)`.
The allocation to this cell is min(76,20) = 20.
This satisfies the entire demand of `D_(dummy)` and leaves 76 - 20 = 56 units with `S_1`
Table-1
| | `D_1` | `D_2` | `D_3` | `D_(dummy)` | | Supply | | `S_1` | 4 | 8 | 8 | 0(20) | | 56 | | `S_2` | 16 | 24 | 16 | 0 | | 82 | | `S_3` | 8 | 16 | 24 | 0 | | 77 | | | | Demand | 72 | 102 | 41 | 0 | | |
In `1^(st)` row, The smallest transportation cost is 4 in cell `S_1 D_1`.
The allocation to this cell is min(56,72) = 56.
This exhausts the capacity of `S_1` and leaves 72 - 56 = 16 units with `D_1`
Table-2
| | `D_1` | `D_2` | `D_3` | `D_(dummy)` | | Supply | | `S_1` | 4(56) | 8 | 8 | 0(20) | | 0 | | `S_2` | 16 | 24 | 16 | 0 | | 82 | | `S_3` | 8 | 16 | 24 | 0 | | 77 | | | | Demand | 16 | 102 | 41 | 0 | | |
In `2^(nd)` row, The smallest transportation cost is 16 in cell `S_2 D_3`.
The allocation to this cell is min(82,41) = 41.
This satisfies the entire demand of `D_3` and leaves 82 - 41 = 41 units with `S_2`
Table-3
| | `D_1` | `D_2` | `D_3` | `D_(dummy)` | | Supply | | `S_1` | 4(56) | 8 | 8 | 0(20) | | 0 | | `S_2` | 16 | 24 | 16(41) | 0 | | 41 | | `S_3` | 8 | 16 | 24 | 0 | | 77 | | | | Demand | 16 | 102 | 0 | 0 | | |
In `2^(nd)` row, The smallest transportation cost is 16 in cell `S_2 D_1`.
The allocation to this cell is min(41,16) = 16.
This satisfies the entire demand of `D_1` and leaves 41 - 16 = 25 units with `S_2`
Table-4
| | `D_1` | `D_2` | `D_3` | `D_(dummy)` | | Supply | | `S_1` | 4(56) | 8 | 8 | 0(20) | | 0 | | `S_2` | 16(16) | 24 | 16(41) | 0 | | 25 | | `S_3` | 8 | 16 | 24 | 0 | | 77 | | | | Demand | 0 | 102 | 0 | 0 | | |
In `2^(nd)` row, The smallest transportation cost is 24 in cell `S_2 D_2`.
The allocation to this cell is min(25,102) = 25.
This exhausts the capacity of `S_2` and leaves 102 - 25 = 77 units with `D_2`
Table-5
| | `D_1` | `D_2` | `D_3` | `D_(dummy)` | | Supply | | `S_1` | 4(56) | 8 | 8 | 0(20) | | 0 | | `S_2` | 16(16) | 24(25) | 16(41) | 0 | | 0 | | `S_3` | 8 | 16 | 24 | 0 | | 77 | | | | Demand | 0 | 77 | 0 | 0 | | |
In `3^(rd)` row, The smallest transportation cost is 16 in cell `S_3 D_2`.
The allocation to this cell is min(77,77) = 77.
Table-6
| | `D_1` | `D_2` | `D_3` | `D_(dummy)` | | Supply | | `S_1` | 4(56) | 8 | 8 | 0(20) | | 0 | | `S_2` | 16(16) | 24(25) | 16(41) | 0 | | 0 | | `S_3` | 8 | 16(77) | 24 | 0 | | 0 | | | | Demand | 0 | 0 | 0 | 0 | | |
Initial feasible solution is
| | `D_1` | `D_2` | `D_3` | `D_(dummy)` | | Supply | | `S_1` | 4 (56) | 8 | 8 | 0 (20) | | 76 | | `S_2` | 16 (16) | 24 (25) | 16 (41) | 0 | | 82 | | `S_3` | 8 | 16 (77) | 24 | 0 | | 77 | | | | Demand | 72 | 102 | 41 | 20 | | |
The minimum total transportation cost `= 4 xx 56 + 0 xx 20 + 16 xx 16 + 24 xx 25 + 16 xx 41 + 16 xx 77 = 2968`
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.
Any bug, improvement, feedback then
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