1. Find Solution using North-West Corner method

	D1	D2	D3	D4	Supply
S1	19	30	50	10	7
S2	70	30	40	60	9
S3	40	8	70	20	18
Demand	5	8	7	14

Solution:
TOTAL number of supply constraints : 3
TOTAL number of demand constraints : 4
Problem Table is

	`D_1`	`D_2`	`D_3`	`D_4`		Supply
`S_1`	19	30	50	10		7
`S_2`	70	30	40	60		9
`S_3`	40	8	70	20		18
Demand	5	8	7	14

The rim values for `S_1`=7 and `D_1`=5 are compared.

The smaller of the two i.e. min(7,5) = 5 is assigned to `S_1` `D_1`

This meets the complete demand of `D_1` and leaves 7 - 5=2 units with `S_1`

Table-1

	`D_1`	`D_2`	`D_3`	`D_4`		Supply
`S_1`	19(5)	30	50	10		2
`S_2`	70	30	40	60		9
`S_3`	40	8	70	20		18
Demand	0	8	7	14

Move horizontally,

The rim values for `S_1`=2 and `D_2`=8 are compared.

The smaller of the two i.e. min(2,8) = 2 is assigned to `S_1` `D_2`

This exhausts the capacity of `S_1` and leaves 8 - 2=6 units with `D_2`

Table-2

	`D_1`	`D_2`	`D_3`	`D_4`		Supply
`S_1`	19(5)	30(2)	50	10		0
`S_2`	70	30	40	60		9
`S_3`	40	8	70	20		18
Demand	0	6	7	14

Move vertically,

The rim values for `S_2`=9 and `D_2`=6 are compared.

The smaller of the two i.e. min(9,6) = 6 is assigned to `S_2` `D_2`

This meets the complete demand of `D_2` and leaves 9 - 6=3 units with `S_2`

Table-3

	`D_1`	`D_2`	`D_3`	`D_4`		Supply
`S_1`	19(5)	30(2)	50	10		0
`S_2`	70	30(6)	40	60		3
`S_3`	40	8	70	20		18
Demand	0	0	7	14

Move horizontally,

The rim values for `S_2`=3 and `D_3`=7 are compared.

The smaller of the two i.e. min(3,7) = 3 is assigned to `S_2` `D_3`

This exhausts the capacity of `S_2` and leaves 7 - 3=4 units with `D_3`

Table-4

	`D_1`	`D_2`	`D_3`	`D_4`		Supply
`S_1`	19(5)	30(2)	50	10		0
`S_2`	70	30(6)	40(3)	60		0
`S_3`	40	8	70	20		18
Demand	0	0	4	14

Move vertically,

The rim values for `S_3`=18 and `D_3`=4 are compared.

The smaller of the two i.e. min(18,4) = 4 is assigned to `S_3` `D_3`

This meets the complete demand of `D_3` and leaves 18 - 4=14 units with `S_3`

Table-5

	`D_1`	`D_2`	`D_3`	`D_4`		Supply
`S_1`	19(5)	30(2)	50	10		0
`S_2`	70	30(6)	40(3)	60		0
`S_3`	40	8	70(4)	20		14
Demand	0	0	0	14

Move horizontally,

The rim values for `S_3`=14 and `D_4`=14 are compared.

The smaller of the two i.e. min(14,14) = 14 is assigned to `S_3` `D_4`

Table-6

	`D_1`	`D_2`	`D_3`	`D_4`		Supply
`S_1`	19(5)	30(2)	50	10		0
`S_2`	70	30(6)	40(3)	60		0
`S_3`	40	8	70(4)	20(14)		0
Demand	0	0	0	0

Initial feasible solution is

	`D_1`	`D_2`	`D_3`	`D_4`		Supply
`S_1`	19 (5)	30 (2)	50	10		7
`S_2`	70	30 (6)	40 (3)	60		9
`S_3`	40	8	70 (4)	20 (14)		18
Demand	5	8	7	14

The minimum total transportation cost `=19 xx 5+30 xx 2+30 xx 6+40 xx 3+70 xx 4+20 xx 14=1015`

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