Infeasible solution
If there is no any solution that satifies all the constraints, then it is called Infeasible solution

In the final simplex table when all `c_j - z_j` imply optimal solution but at least one artificial variable present in the basis with positive value. Then the problem has no feasible solution.

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Example
Find solution using Two-Phase method
MIN Z = X1 - 2X2 - 3X3
subject to
-2X1 + X2 + 3X3 = 2
2X1 + 3X2 + 4X3 = 1
and X1,X2,X3 >= 0;

Solution:
Problem is

Min `Z` `=` `` `` `X_1` ` - ` `2` `X_2` ` - ` `3` `X_3`

subject to

` - ` `2` `X_1` ` + ` `` `X_2` ` + ` `3` `X_3` = `2` `` `2` `X_1` ` + ` `3` `X_2` ` + ` `4` `X_3` = `1`

and `X_1,X_2,X_3 >= 0; `

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

1. As the constraint-1 is of type '`=`' we should add artificial variable `A_1`

2. As the constraint-2 is of type '`=`' we should add artificial variable `A_2`

After introducing artificial variables

Min `Z` `=` `` `` `A_1` ` + ` `` `A_2`

subject to

` - ` `2` `X_1` ` + ` `` `X_2` ` + ` `3` `X_3` ` + ` `` `A_1` = `2` `` `2` `X_1` ` + ` `3` `X_2` ` + ` `4` `X_3` ` + ` `` `A_2` = `1`

and `X_1,X_2,X_3,A_1,A_2 >= 0`

`Z` ` - ` `` `A_1` ` - ` `` `A_2` = `0`

` - ` `2` `X_1` ` + ` `` `X_2` ` + ` `3` `X_3` ` + ` `` `A_1` = `2` `` `2` `X_1` ` + ` `3` `X_2` ` + ` `4` `X_3` ` + ` `` `A_2` = `1`

Simplex tableau is
Tableau-0

`"Basis"`	`X_1`	`X_2`	`X_3`	`A_1`	`A_2`	`RHS`
`R_1` `Z`	`0`	`0`	`0`	`-1`	`-1`	`0`
`R_2` `A_1`	`-2`	`1`	`3`	`1`	`0`	`2`
`R_3` `A_2`	`2`	`3`	`4`	`0`	`1`	`1`

Make the Z-row consistent with the rest of the table (set coefficient of basis variables to 0 in Z-row)

`R_1`(new)`= R_1`(old) + `R_2`(old)

	`X_1`	`X_2`	`X_3`	`A_1`	`A_2`	`RHS`
`R_1`(old) =	`0`	`0`	`0`	`-1`	`-1`	`0`
`R_2`(old) =	`-2`	`1`	`3`	`1`	`0`	`2`
`R_1`(new)`= R_1`(old) + `R_2`(old)	`-2`	`1`	`3`	`0`	`-1`	`2`

`R_1`(new)`= R_1`(old) + `R_3`(old)

	`X_1`	`X_2`	`X_3`	`A_1`	`A_2`	`RHS`
`R_1`(old) =	`-2`	`1`	`3`	`0`	`-1`	`2`
`R_3`(old) =	`2`	`3`	`4`	`0`	`1`	`1`
`R_1`(new)`= R_1`(old) + `R_3`(old)	`0`	`4`	`7`	`0`	`0`	`3`

Tableau-1

`"Basis"`	`X_1`	`X_2`	`X_3``darr`	`A_1`	`A_2`	`RHS`	`"Ratio"=(RHS)/(X_3)`
`R_1` `Z`	`0`	`4`	`7`	`0`	`0`	`3`
`R_2` `A_1`	`-2`	`1`	`3`	`1`	`0`	`2`	`(2)/(3)=0.6667`
`R_3` `A_2`	`2`	`3`	`(4)`	`0`	`1`	`1`	`(1)/(4)=0.25``->`

Most Positive `Z` is `7`. So, the entering variable is `X_3`.

Minimum ratio is `0.25`. So, the leaving basis variable is `A_2`.

`:.` The pivot element is `4`.

Entering `=X_3`, Departing `=A_2`, Key Element `=4`

`R_3`(new)`= R_3`(old) `-: 4`

	`X_1`	`X_2`	`X_3`	`A_1`	`A_2`	`RHS`
`R_3`(old) =	`2`	`3`	`4`	`0`	`1`	`1`
`R_3`(new)`= R_3`(old) `-: 4`	`0.5`	`0.75`	`1`	`0`	`0.25`	`0.25`

`R_2`(new)`= R_2`(old) - `3 R_3`(new)

	`X_1`	`X_2`	`X_3`	`A_1`	`A_2`	`RHS`
`R_2`(old) =	`-2`	`1`	`3`	`1`	`0`	`2`
`R_3`(new) =	`0.5`	`0.75`	`1`	`0`	`0.25`	`0.25`
`3 xx R_3`(new) =	`1.5`	`2.25`	`3`	`0`	`0.75`	`0.75`
`R_2`(new)`= R_2`(old) - `3 R_3`(new)	`-3.5`	`-1.25`	`0`	`1`	`-0.75`	`1.25`

`R_1`(new)`= R_1`(old) - `7 R_3`(new)

	`X_1`	`X_2`	`X_3`	`A_1`	`A_2`	`RHS`
`R_1`(old) =	`0`	`4`	`7`	`0`	`0`	`3`
`R_3`(new) =	`0.5`	`0.75`	`1`	`0`	`0.25`	`0.25`
`7 xx R_3`(new) =	`3.5`	`5.25`	`7`	`0`	`1.75`	`1.75`
`R_1`(new)`= R_1`(old) - `7 R_3`(new)	`-3.5`	`-1.25`	`0`	`0`	`-1.75`	`1.25`

Tableau-2

`"Basis"`	`X_1`	`X_2`	`X_3`	`A_1`	`A_2`	`RHS`
`R_1` `Z`	`-3.5`	`-1.25`	`0`	`0`	`-1.75`	`1.25`
`R_2` `A_1`	`-3.5`	`-1.25`	`0`	`1`	`-0.75`	`1.25`
`R_3` `X_3`	`0.5`	`0.75`	`1`	`0`	`0.25`	`0.25`

Since all `Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`X_1=0,X_2=0,X_3=0.25`

Min `Z=0`

But this solution is not feasible (and also not optimal)
because the solution violates the `1^(st)` constraint  ` - ` `2` `X_1` ` + ` `` `X_2` ` + ` `3` `X_3` = `2`.

and the artificial variable `A_1` appears in the basis with positive value `1.25`

So phase-2 is not possbile.

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