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 North-West Corner 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 rim values for `S_1`=76 and `D_1`=72 are compared.

The smaller of the two i.e. min(76,72) = 72 is assigned to `S_1` `D_1`

This meets the complete demand of `D_1` and leaves 76 - 72=4 units with `S_1`

Table-1

	`D_1`	`D_2`	`D_3`	`D_(dummy)`		Supply
`S_1`	4(72)	8	8	0		4
`S_2`	16	24	16	0		82
`S_3`	8	16	24	0		77
Demand	0	102	41	20

Move horizontally,

The rim values for `S_1`=4 and `D_2`=102 are compared.

The smaller of the two i.e. min(4,102) = 4 is assigned to `S_1` `D_2`

This exhausts the capacity of `S_1` and leaves 102 - 4=98 units with `D_2`

Table-2

	`D_1`	`D_2`	`D_3`	`D_(dummy)`		Supply
`S_1`	4(72)	8(4)	8	0		0
`S_2`	16	24	16	0		82
`S_3`	8	16	24	0		77
Demand	0	98	41	20

Move vertically,

The rim values for `S_2`=82 and `D_2`=98 are compared.

The smaller of the two i.e. min(82,98) = 82 is assigned to `S_2` `D_2`

This exhausts the capacity of `S_2` and leaves 98 - 82=16 units with `D_2`

Table-3

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

Move vertically,

The rim values for `S_3`=77 and `D_2`=16 are compared.

The smaller of the two i.e. min(77,16) = 16 is assigned to `S_3` `D_2`

This meets the complete demand of `D_2` and leaves 77 - 16=61 units with `S_3`

Table-4

	`D_1`	`D_2`	`D_3`	`D_(dummy)`		Supply
`S_1`	4(72)	8(4)	8	0		0
`S_2`	16	24(82)	16	0		0
`S_3`	8	16(16)	24	0		61
Demand	0	0	41	20

Move horizontally,

The rim values for `S_3`=61 and `D_3`=41 are compared.

The smaller of the two i.e. min(61,41) = 41 is assigned to `S_3` `D_3`

This meets the complete demand of `D_3` and leaves 61 - 41=20 units with `S_3`

Table-5

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

Move horizontally,

The rim values for `S_3`=20 and `D_(dummy)`=20 are compared.

The smaller of the two i.e. min(20,20) = 20 is assigned to `S_3` `D_(dummy)`

Table-6

	`D_1`	`D_2`	`D_3`	`D_(dummy)`		Supply
`S_1`	4(72)	8(4)	8	0		0
`S_2`	16	24(82)	16	0		0
`S_3`	8	16(16)	24(41)	0(20)		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 (82)	16	0		82
`S_3`	8	16 (16)	24 (41)	0 (20)		77
Demand	72	102	41	20

The minimum total transportation cost `=4 xx 72+8 xx 4+24 xx 82+16 xx 16+24 xx 41+0 xx 20=3528`

Here, the number of allocated cells = 6 is equal to m + n - 1 = 3 + 4 - 1 = 6
`:.` This solution is non-degenerate

---