Example 4.1 by using M value = 1000 and Example 4.2 by using M value = M

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4.1) Find Solution of Transshipment Problem using vogel's approximation method

	3	4	5	6	Supply
1	1	4	M	M	100
2	3	2	M	M	200
3	0	1	6	M	0
4	1	0	5	8	0
5	M	M	0	1	0
Demand	0	0	150	150

Solution:
Problem Table is

	`3`	`4`	`5`	`6`		Supply
`1`	1	4	1000	1000		100
`2`	3	2	1000	1000		200
`3`	0	1	6	1000		0
`4`	1	0	5	8		0
`5`	1000	1000	0	1		0
Demand	0	0	150	150

`1,2` are pure supply nodes

`6` are pure demand nodes

`3,4,5` are transshipment nodes

Add Total value `=300` in supply and demand for transshipment nodes `3,4,5`

So again Problem Table is

	`3`	`4`	`5`	`6`		Supply
`1`	1	4	1000	1000		100
`2`	3	2	1000	1000		200
`3`	0	1	6	1000		300
`4`	1	0	5	8		300
`5`	1000	1000	0	1		300
Demand	300	300	450	150

Now, we solve this Transshipment problem
Table-1

	`3`	`4`	`5`	`6`		Supply	Row Penalty
`1`	1	4	1000	1000		100	`3=4-1`
`2`	3	2	1000	1000		200	`1=3-2`
`3`	0	1	6	1000		300	`1=1-0`
`4`	1	0	5	8		300	`1=1-0`
`5`	1000	1000	0	1		300	`1=1-0`
Demand	300	300	450	150
Column Penalty	`1=1-0`	`1=1-0`	`5=5-0`	`7=8-1`

The maximum penalty, 7, occurs in column `6`.

The minimum `c_(ij)` in this column is `c_54`=1.

The maximum allocation in this cell is min(300,150) = 150.
It satisfy demand of `6` and adjust the supply of `5` from 300 to 150 (300 - 150=150).

Table-2

	`3`	`4`	`5`	`6`		Supply	Row Penalty
`1`	1	4	1000	1000		100	`3=4-1`
`2`	3	2	1000	1000		200	`1=3-2`
`3`	0	1	6	1000		300	`1=1-0`
`4`	1	0	5	8		300	`1=1-0`
`5`	1000	1000	0	1(150)		150	`1000=1000-0`
Demand	300	300	450	0
Column Penalty	`1=1-0`	`1=1-0`	`5=5-0`	--

The maximum penalty, 1000, occurs in row `5`.

The minimum `c_(ij)` in this row is `c_53`=0.

The maximum allocation in this cell is min(150,450) = 150.
It satisfy supply of `5` and adjust the demand of `5` from 450 to 300 (450 - 150=300).

Table-3

	`3`	`4`	`5`	`6`		Supply	Row Penalty
`1`	1	4	1000	1000		100	`3=4-1`
`2`	3	2	1000	1000		200	`1=3-2`
`3`	0	1	6	1000		300	`1=1-0`
`4`	1	0	5	8		300	`1=1-0`
`5`	1000	1000	0(150)	1(150)		0	--
Demand	300	300	300	0
Column Penalty	`1=1-0`	`1=1-0`	`1=6-5`	--

The maximum penalty, 3, occurs in row `1`.

The minimum `c_(ij)` in this row is `c_11`=1.

The maximum allocation in this cell is min(100,300) = 100.
It satisfy supply of `1` and adjust the demand of `3` from 300 to 200 (300 - 100=200).

Table-4

	`3`	`4`	`5`	`6`		Supply	Row Penalty
`1`	1(100)	4	1000	1000		0	--
`2`	3	2	1000	1000		200	`1=3-2`
`3`	0	1	6	1000		300	`1=1-0`
`4`	1	0	5	8		300	`1=1-0`
`5`	1000	1000	0(150)	1(150)		0	--
Demand	200	300	300	0
Column Penalty	`1=1-0`	`1=1-0`	`1=6-5`	--

The maximum penalty, 1, occurs in row `4`.

The minimum `c_(ij)` in this row is `c_42`=0.

The maximum allocation in this cell is min(300,300) = 300.
It satisfy supply of `4` and demand of `4`.

Table-5

	`3`	`4`	`5`	`6`		Supply	Row Penalty
`1`	1(100)	4	1000	1000		0	--
`2`	3	2	1000	1000		200	`997=1000-3`
`3`	0	1	6	1000		300	`6=6-0`
`4`	1	0(300)	5	8		0	--
`5`	1000	1000	0(150)	1(150)		0	--
Demand	200	0	300	0
Column Penalty	`3=3-0`	--	`994=1000-6`	--

The maximum penalty, 997, occurs in row `2`.

The minimum `c_(ij)` in this row is `c_21`=3.

The maximum allocation in this cell is min(200,200) = 200.
It satisfy supply of `2` and demand of `3`.

Table-6

	`3`	`4`	`5`	`6`		Supply	Row Penalty
`1`	1(100)	4	1000	1000		0	--
`2`	3(200)	2	1000	1000		0	--
`3`	0	1	6	1000		300	`6`
`4`	1	0(300)	5	8		0	--
`5`	1000	1000	0(150)	1(150)		0	--
Demand	0	0	300	0
Column Penalty	--	--	`6`	--

The maximum penalty, 6, occurs in row `3`.

The minimum `c_(ij)` in this row is `c_33`=6.

The maximum allocation in this cell is min(300,300) = 300.
It satisfy supply of `3` and demand of `5`.


Initial feasible solution is

	`3`	`4`	`5`	`6`		Supply	Row Penalty
`1`	1(100)	4	1000	1000		100	3 |  3 |  3 | -- | -- | -- |
`2`	3(200)	2	1000	1000		200	1 |  1 |  1 |  1 | 997 | -- |
`3`	0	1	6(300)	1000		300	1 |  1 |  1 |  1 |  6 |  6 |
`4`	1	0(300)	5	8		300	1 |  1 |  1 |  1 | -- | -- |
`5`	1000	1000	0(150)	1(150)		300	1 | 1000 | -- | -- | -- | -- |
Demand	300	300	450	150
Column Penalty	1 1 1 1 3 --	1 1 1 1 -- --	5 5 1 1 994 6	7 -- -- -- -- --

The minimum total transportation cost `=1 xx 100+3 xx 200+6 xx 300+0 xx 300+0 xx 150+1 xx 150=2650`

Here, the number of allocated cells = 6, which is two less than to m + n - 1 = 5 + 4 - 1 = 8
`:.` This solution is degenerate

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4.2) Find Solution of Transshipment Problem using vogel's approximation method

	3	4	5	6	Supply
1	1	4	M	M	100
2	3	2	M	M	200
3	0	1	6	M	0
4	1	0	5	8	0
5	M	M	0	1	0
Demand	0	0	150	150

Solution:
Problem Table is

	`3`	`4`	`5`	`6`		Supply
`1`	1	4	M	M		100
`2`	3	2	M	M		200
`3`	0	1	6	M		0
`4`	1	0	5	8		0
`5`	M	M	0	1		0
Demand	0	0	150	150

`1,2` are pure supply nodes

`6` are pure demand nodes

`3,4,5` are transshipment nodes

Add Total value `=300` in supply and demand for transshipment nodes `3,4,5`

So again Problem Table is

	`3`	`4`	`5`	`6`		Supply
`1`	1	4	M	M		100
`2`	3	2	M	M		200
`3`	0	1	6	M		300
`4`	1	0	5	8		300
`5`	M	M	0	1		300
Demand	300	300	450	150

Now, we solve this Transshipment problem
Table-1

	`3`	`4`	`5`	`6`		Supply	Row Penalty
`1`	1	4	M	M		100	`3=4-1`
`2`	3	2	M	M		200	`1=3-2`
`3`	0	1	6	M		300	`1=1-0`
`4`	1	0	5	8		300	`1=1-0`
`5`	M	M	0	1		300	`1=1-0`
Demand	300	300	450	150
Column Penalty	`1=1-0`	`1=1-0`	`5=5-0`	`7=8-1`

The maximum penalty, 7, occurs in column `6`.

The minimum `c_(ij)` in this column is `c_54`=1.

The maximum allocation in this cell is min(300,150) = 150.
It satisfy demand of `6` and adjust the supply of `5` from 300 to 150 (300 - 150=150).

Table-2

	`3`	`4`	`5`	`6`		Supply	Row Penalty
`1`	1	4	M	M		100	`3=4-1`
`2`	3	2	M	M		200	`1=3-2`
`3`	0	1	6	M		300	`1=1-0`
`4`	1	0	5	8		300	`1=1-0`
`5`	M	M	0	1(150)		150	`M=M-0`
Demand	300	300	450	0
Column Penalty	`1=1-0`	`1=1-0`	`5=5-0`	--

The maximum penalty, M, occurs in row `5`.

The minimum `c_(ij)` in this row is `c_53`=0.

The maximum allocation in this cell is min(150,450) = 150.
It satisfy supply of `5` and adjust the demand of `5` from 450 to 300 (450 - 150=300).

Table-3

	`3`	`4`	`5`	`6`		Supply	Row Penalty
`1`	1	4	M	M		100	`3=4-1`
`2`	3	2	M	M		200	`1=3-2`
`3`	0	1	6	M		300	`1=1-0`
`4`	1	0	5	8		300	`1=1-0`
`5`	M	M	0(150)	1(150)		0	--
Demand	300	300	300	0
Column Penalty	`1=1-0`	`1=1-0`	`1=6-5`	--

The maximum penalty, 3, occurs in row `1`.

The minimum `c_(ij)` in this row is `c_11`=1.

The maximum allocation in this cell is min(100,300) = 100.
It satisfy supply of `1` and adjust the demand of `3` from 300 to 200 (300 - 100=200).

Table-4

	`3`	`4`	`5`	`6`		Supply	Row Penalty
`1`	1(100)	4	M	M		0	--
`2`	3	2	M	M		200	`1=3-2`
`3`	0	1	6	M		300	`1=1-0`
`4`	1	0	5	8		300	`1=1-0`
`5`	M	M	0(150)	1(150)		0	--
Demand	200	300	300	0
Column Penalty	`1=1-0`	`1=1-0`	`1=6-5`	--

The maximum penalty, 1, occurs in row `4`.

The minimum `c_(ij)` in this row is `c_42`=0.

The maximum allocation in this cell is min(300,300) = 300.
It satisfy supply of `4` and demand of `4`.

Table-5

	`3`	`4`	`5`	`6`		Supply	Row Penalty
`1`	1(100)	4	M	M		0	--
`2`	3	2	M	M		200	`M=M-3`
`3`	0	1	6	M		300	`6=6-0`
`4`	1	0(300)	5	8		0	--
`5`	M	M	0(150)	1(150)		0	--
Demand	200	0	300	0
Column Penalty	`3=3-0`	--	`M=M-6`	--

The maximum penalty, M, occurs in row `2`.

The minimum `c_(ij)` in this row is `c_21`=3.

The maximum allocation in this cell is min(200,200) = 200.
It satisfy supply of `2` and demand of `3`.

Table-6

	`3`	`4`	`5`	`6`		Supply	Row Penalty
`1`	1(100)	4	M	M		0	--
`2`	3(200)	2	M	M		0	--
`3`	0	1	6	M		300	`6`
`4`	1	0(300)	5	8		0	--
`5`	M	M	0(150)	1(150)		0	--
Demand	0	0	300	0
Column Penalty	--	--	`6`	--

The maximum penalty, 6, occurs in row `3`.

The minimum `c_(ij)` in this row is `c_33`=6.

The maximum allocation in this cell is min(300,300) = 300.
It satisfy supply of `3` and demand of `5`.


Initial feasible solution is

	`3`	`4`	`5`	`6`		Supply	Row Penalty
`1`	1(100)	4	M	M		100	3 |  3 |  3 | -- | -- | -- |
`2`	3(200)	2	M	M		200	1 |  1 |  1 |  1 | M | -- |
`3`	0	1	6(300)	M		300	1 |  1 |  1 |  1 |  6 |  6 |
`4`	1	0(300)	5	8		300	1 |  1 |  1 |  1 | -- | -- |
`5`	M	M	0(150)	1(150)		300	1 | M | -- | -- | -- | -- |
Demand	300	300	450	150
Column Penalty	1 1 1 1 3 --	1 1 1 1 -- --	5 5 1 1 M 6	7 -- -- -- -- --

The minimum total transportation cost `=1 xx 100+3 xx 200+6 xx 300+0 xx 300+0 xx 150+1 xx 150=2650`

Here, the number of allocated cells = 6, which is two less than to m + n - 1 = 5 + 4 - 1 = 8
`:.` This solution is degenerate

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