Unbounded solution
In simplex table, if a variable should enter into the basis, but all the coefficients in that column are negative or zero. So this variable can not be entered into the basis, because for minimum ratio, negative value in denominator can not be considered and zero value in denominator would result `oo`.
Hence, the solution to the given problem is unbounded.

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Example
Find solution using Simplex method (BigM method)
MAX Z = 3x1 + 5x2
subject to
x1 - 2x2 <= 6
x1 <= 10
x2 >= 1
and x1,x2 >= 0;

Solution:
Problem is

Max `Z` `=` `` `3` `x_1` ` + ` `5` `x_2`

subject to

`` `` `x_1` ` - ` `2` `x_2` ≤ `6` `` `` `x_1` ≤ `10` `` `` `x_2` ≥ `1`

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 add slack variable `S_2`

3. As the constraint-3 is of type '`>=`' we should subtract surplus variable `S_3` and add artificial variable `A_1`

After introducing slack,surplus,artificial variables

Max `Z` `=` `` `3` `x_1` ` + ` `5` `x_2` ` + ` `0` `S_1` ` + ` `0` `S_2` ` + ` `0` `S_3` ` - ` `M` `A_1`

subject to

`` `` `x_1` ` - ` `2` `x_2` ` + ` `` `S_1` = `6` `` `` `x_1` ` + ` `` `S_2` = `10` `` `` `x_2` ` - ` `` `S_3` ` + ` `` `A_1` = `1`

and `x_1,x_2,S_1,S_2,S_3,A_1 >= 0`

Tableau-1	`C_j`	`3`	`5`	`0`	`0`	`0`	`-M`
`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`A_1`	`RHS`	`"Ratio"=(RHS)/(x_2)`
`R_1` `0`	`S_1`	`1`	`-2`	`1`	`0`	`0`	`0`	`6`	`(6)/(-2)` (ignore, denominator is -ve)
`R_2` `0`	`S_2`	`1`	`0`	`0`	`1`	`0`	`0`	`10`	`(10)/(0)` (ignore, denominator is 0)
`R_3` `-M`	`A_1`	`0`	`(1)`	`0`	`0`	`-1`	`1`	`1`	`(1)/(1)=1``->`
	`Z_j`	`0`	`-M`	`0`	`0`	`M`	`-M`	`Z=-M`
	`Z_j-C_j`	`-3`	`-M-5``uarr`	`0`	`0`	`M`	`0`

Most Negative `Z_j-C_j` is `-M-5`. So, the entering variable is `x_2`.

Minimum ratio is `1`. So, the leaving basis variable is `A_1`.

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

Entering `=x_2`, Departing `=A_1`, Key Element `=1`

`R_3`(new)`= R_3`(old)

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`A_1`	`RHS`
`R_3`(old) =	`0`	`1`	`0`	`0`	`-1`	`1`	`1`
`R_3`(new)`= R_3`(old)	`0`	`1`	`0`	`0`	`-1`	`1`	`1`

`R_1`(new)`= R_1`(old) + `2 R_3`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`A_1`	`RHS`
`R_1`(old) =	`1`	`-2`	`1`	`0`	`0`	`0`	`6`
`R_3`(new) =	`0`	`1`	`0`	`0`	`-1`	`1`	`1`
`2 xx R_3`(new) =	`0`	`2`	`0`	`0`	`-2`	`2`	`2`
`R_1`(new)`= R_1`(old) + `2 R_3`(new)	`1`	`0`	`1`	`0`	`-2`	`2`	`8`

`R_2`(new)`= R_2`(old)

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`A_1`	`RHS`
`R_2`(old) =	`1`	`0`	`0`	`1`	`0`	`0`	`10`
`R_2`(new)`= R_2`(old)	`1`	`0`	`0`	`1`	`0`	`0`	`10`

Tableau-2	`C_j`	`3`	`5`	`0`	`0`	`0`	`-M`
`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`A_1`	`RHS`	`"Ratio"=(RHS)/(S_3)`
`R_1` `0`	`S_1`	`1`	`0`	`1`	`0`	`-2`	`2`	`8`	`(8)/(-2)` (ignore, denominator is -ve)
`R_2` `0`	`S_2`	`1`	`0`	`0`	`1`	`0`	`0`	`10`	`(10)/(0)` (ignore, denominator is 0)
`R_3` `5`	`x_2`	`0`	`1`	`0`	`0`	`-1`	`1`	`1`	`(1)/(-1)` (ignore, denominator is -ve)
	`Z_j`	`0`	`5`	`0`	`0`	`-5`	`5`	`Z=5`
	`Z_j-C_j`	`-3`	`0`	`0`	`0`	`-5``uarr`	`M+5`

Variable `S_3` should enter into the basis, but all the coefficients in the `S_3` column are negative or zero. So `S_3` can not be entered into the basis.

Hence, the solution to the given problem is unbounded.

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