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2. least cost method example ( Enter your problem )
Algorithm and examples
  1. Algorithm & Example-1
  2. Example-2
  3. Unbalanced supply and demand example
  4. Prohibited Routes example
Other related methods
  1. north-west corner method
  2. least cost method
  3. vogel's approximation method
  4. Row minima method
  5. Column minima method
  6. Russell's approximation method
  7. Heuristic method-1
  8. Heuristic method-2
  9. modi method (optimal solution)
  10. stepping stone method (optimal solution)

2. Example-2
(Previous example)
4. Prohibited Routes example
(Next example)

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 Least Cost method
D1D2D3Supply
S148876
S216241682
S38162477
Demand7210241


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`48876
`S_2`16241682
`S_3`8162477
Demand7210241


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`488076
`S_2`162416082
`S_3`81624077
Demand721024120


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`4880(20)56
`S_2`162416082
`S_3`81624077
Demand72102410


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)880(20)0
`S_2`162416082
`S_3`81624077
Demand16102410


The smallest transportation cost is 8 in cell `S_3 D_1`

The allocation to this cell is min(77,16) = 16.
This satisfies the entire demand of `D_1` and leaves 77 - 16 = 61 units with `S_3`

Table-3
`D_1``D_2``D_3``D_(dummy)`Supply
`S_1`4(56)880(20)0
`S_2`162416082
`S_3`8(16)1624061
Demand0102410


The smallest transportation cost is 16 in cell `S_3 D_2`

The allocation to this cell is min(61,102) = 61.
This exhausts the capacity of `S_3` and leaves 102 - 61 = 41 units with `D_2`

Table-4
`D_1``D_2``D_3``D_(dummy)`Supply
`S_1`4(56)880(20)0
`S_2`162416082
`S_3`8(16)16(61)2400
Demand041410


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-5
`D_1``D_2``D_3``D_(dummy)`Supply
`S_1`4(56)880(20)0
`S_2`162416(41)041
`S_3`8(16)16(61)2400
Demand04100


The smallest transportation cost is 24 in cell `S_2 D_2`

The allocation to this cell is min(41,41) = 41.
Table-6
`D_1``D_2``D_3``D_(dummy)`Supply
`S_1`4(56)880(20)0
`S_2`1624(41)16(41)00
`S_3`8(16)16(61)2400
Demand0000


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 24 (41)16 (41)0 82
`S_3`8 (16)16 (61)24 0 77
Demand721024120


The minimum total transportation cost `= 4 xx 56 + 0 xx 20 + 24 xx 41 + 16 xx 41 + 8 xx 16 + 16 xx 61 = 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.
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2. Example-2
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4. Prohibited Routes example
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