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 `z_j - c_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 Simplex method (BigM method)
MAX Z = 6x1 + 4x2
subject to
x1 + x2 <= 5
x2 >= 8
and x1,x2 >= 0;

Solution:
Problem is

Max `Z` `=` `` `6` `x_1` ` + ` `4` `x_2`

subject to

`` `` `x_1` ` + ` `` `x_2` ≤ `5` `` `` `x_2` ≥ `8`

and `x_1,x_2 >= 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 slack variable `S_1`

2. As the constraint-2 is of type '`>=`' we should subtract surplus variable `S_2` and add artificial variable `A_1`

After introducing slack,surplus,artificial variables

Max `Z` `=` `` `6` `x_1` ` + ` `4` `x_2` ` + ` `0` `S_1` ` + ` `0` `S_2` ` - ` `M` `A_1`

subject to

`` `` `x_1` ` + ` `` `x_2` ` + ` `` `S_1` = `5` `` `` `x_2` ` - ` `` `S_2` ` + ` `` `A_1` = `8`

and `x_1,x_2,S_1,S_2,A_1 >= 0`

Tableau-1	`C_j`	`6`	`4`	`0`	`0`	`-M`
`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`A_1`	`RHS`	`"Ratio"=(RHS)/(x_2)`
`R_1` `0`	`S_1`	`1`	`(1)`	`1`	`0`	`0`	`5`	`(5)/(1)=5``->`
`R_2` `-M`	`A_1`	`0`	`1`	`0`	`-1`	`1`	`8`	`(8)/(1)=8`
	`Z_j`	`0`	`-M`	`0`	`M`	`-M`	`Z=-8M`
	`Z_j-C_j`	`-6`	`-M-4``uarr`	`0`	`M`	`0`

Most Negative `Z_j-C_j` is `-M-4`. So, the entering variable is `x_2`.

Minimum ratio is `5`. So, the leaving basis variable is `S_1`.

`:.` The pivot element is `1`.

Entering `=x_2`, Departing `=S_1`, Key Element `=1`

`R_1`(new)`= R_1`(old)

	`x_1`	`x_2`	`S_1`	`S_2`	`A_1`	`RHS`
`R_1`(old) =	`1`	`1`	`1`	`0`	`0`	`5`
`R_1`(new)`= R_1`(old)	`1`	`1`	`1`	`0`	`0`	`5`

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

	`x_1`	`x_2`	`S_1`	`S_2`	`A_1`	`RHS`
`R_2`(old) =	`0`	`1`	`0`	`-1`	`1`	`8`
`R_1`(new) =	`1`	`1`	`1`	`0`	`0`	`5`
`R_2`(new)`= R_2`(old) - `R_1`(new)	`-1`	`0`	`-1`	`-1`	`1`	`3`

Tableau-2	`C_j`	`6`	`4`	`0`	`0`	`-M`
`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`A_1`	`RHS`	`"Ratio"`
`R_1` `4`	`x_2`	`1`	`1`	`1`	`0`	`0`	`5`
`R_2` `-M`	`A_1`	`-1`	`0`	`-1`	`-1`	`1`	`3`
	`Z_j`	`M+4`	`4`	`M+4`	`M`	`-M`	`Z=-3M+20`
	`Z_j-C_j`	`M-2`	`0`	`M+4`	`M`	`0`

Since all `Z_j-C_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=0,x_2=5`

Max `Z=20`

But this solution is not feasible (and also not optimal)
because the solution violates the `2^(nd)` constraint     `` `` `x_2` ≥ `8`.

and the artificial variable `A_1` appears in the basis with positive value `3`

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