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

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Example
Find Solution using Least Cost 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

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

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

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)	8	8	0(20)		0
`S_2`	16	24	16	0		82
`S_3`	8(16)	16	24	0		61
Demand	0	102	41	0

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)	8	8	0(20)		0
`S_2`	16	24	16	0		82
`S_3`	8(16)	16(61)	24	0		0
Demand	0	41	41	0

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)	8	8	0(20)		0
`S_2`	16	24	16(41)	0		41
`S_3`	8(16)	16(61)	24	0		0
Demand	0	41	0	0

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)	8	8	0(20)		0
`S_2`	16	24(41)	16(41)	0		0
`S_3`	8(16)	16(61)	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	24 (41)	16 (41)	0		82
`S_3`	8 (16)	16 (61)	24	0		77
Demand	72	102	41	20

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

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