transportation problem using column minima method Unbalanced supply and demand example

3. Unbalanced supply and demand example

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Home > Operation Research calculators > Column minima method example

5. column minima method example ( Enter your problem )

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Example-2
Unbalanced supply and demand example

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`D_(dummy)`

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 Column 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

In `1^(st)` column, The smallest transportation cost is 4 in cell `S_1 D_1`

The allocation to this cell is min(76,72) = 72.
This satisfies the entire demand of `D_1` and leaves 76 - 72 = 4 units with `S_1`

Table-1

In `2^(nd)` column, The smallest transportation cost is 8 in cell `S_1 D_2`

The allocation to this cell is min(4,102) = 4.
This exhausts the capacity of `S_1` and leaves 102 - 4 = 98 units with `D_2`

Table-2

In `2^(nd)` column, The smallest transportation cost is 16 in cell `S_3 D_2`

The allocation to this cell is min(77,98) = 77.
This exhausts the capacity of `S_3` and leaves 98 - 77 = 21 units with `D_2`

Table-3

In `2^(nd)` column, The smallest transportation cost is 24 in cell `S_2 D_2`

The allocation to this cell is min(82,21) = 21.
This satisfies the entire demand of `D_2` and leaves 82 - 21 = 61 units with `S_2`

Table-4

	`D_1`	`D_2`	`D_3`	`D_(dummy)`	Supply
`S_1`	4(72)	8(4)	8	0	0
`S_2`	16	24(21)	16	0	61
`S_3`	8	16(77)	24	0	0

Demand	0	0	41	20

In `3^(rd)` column, The smallest transportation cost is 16 in cell `S_2 D_3`

The allocation to this cell is min(61,41) = 41.
This satisfies the entire demand of `D_3` and leaves 61 - 41 = 20 units with `S_2`

Table-5

	`D_1`	`D_2`	`D_3`	`D_(dummy)`	Supply
`S_1`	4(72)	8(4)	8	0	0
`S_2`	16	24(21)	16(41)	0	20
`S_3`	8	16(77)	24	0	0

Demand	0	0	0	20

In `4^(th)` column, The smallest transportation cost is 0 in cell `S_2 D_(dummy)`

The allocation to this cell is min(20,20) = 20.
Table-6

	`D_1`	`D_2`	`D_3`	`D_(dummy)`	Supply
`S_1`	4(72)	8(4)	8	0	0
`S_2`	16	24(21)	16(41)	0(20)	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 (72)	8 (4)	8	0	76
`S_2`	16	24 (21)	16 (41)	0 (20)	82
`S_3`	8	16 (77)	24	0	77

Demand	72	102	41	20

The minimum total transportation cost `= 4 xx 72 + 8 xx 4 + 24 xx 21 + 16 xx 41 + 0 xx 20 + 16 xx 77 = 2712`

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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