Find solution using Two-Phase method
MIN Z = 5x1 + 2x2 + 10x3
subject to
x1 - x3 <= 10
x2 + x3 >= 10
and x1,x2,x3 >= 0;

Solution:
Problem is

Min `Z` `=` `` `5` `x_1` ` + ` `2` `x_2` ` + ` `10` `x_3`

subject to

`` `` `x_1` ` - ` `` `x_3` ≤ `10` `` `` `x_2` ` + ` `` `x_3` ≥ `10`

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 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

Min `Z` `=` `` `` `A_1`

subject to

`` `` `x_1` ` - ` `` `x_3` ` + ` `` `S_1` = `10` `` `` `x_2` ` + ` `` `x_3` ` - ` `` `S_2` ` + ` `` `A_1` = `10`

and `x_1,x_2,x_3,S_1,S_2,A_1 >= 0`

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

`` `` `x_1` ` - ` `` `x_3` ` + ` `` `S_1` = `10` `` `` `x_2` ` + ` `` `x_3` ` - ` `` `S_2` ` + ` `` `A_1` = `10`

Simplex tableau is
Tableau-0

`"Basis"`	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`A_1`	`RHS`
`R_1` `Z`	`0`	`0`	`0`	`0`	`0`	`-1`	`0`
`R_2` `S_1`	`1`	`0`	`-1`	`1`	`0`	`0`	`10`
`R_3` `A_1`	`0`	`1`	`1`	`0`	`-1`	`1`	`10`

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_3`(old)

	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`A_1`	`RHS`
`R_1`(old) =	`0`	`0`	`0`	`0`	`0`	`-1`	`0`
`R_3`(old) =	`0`	`1`	`1`	`0`	`-1`	`1`	`10`
`R_1`(new)`= R_1`(old) + `R_3`(old)	`0`	`1`	`1`	`0`	`-1`	`0`	`10`

Tableau-1

`"Basis"`	`x_1`	`x_2``darr`	`x_3`	`S_1`	`S_2`	`A_1`	`RHS`	`"Ratio"=(RHS)/(x_2)`
`R_1` `Z`	`0`	`1`	`1`	`0`	`-1`	`0`	`10`
`R_2` `S_1`	`1`	`0`	`-1`	`1`	`0`	`0`	`10`	`(10)/(0)` (ignore, denominator is 0)
`R_3` `A_1`	`0`	`(1)`	`1`	`0`	`-1`	`1`	`10`	`(10)/(1)=10``->`

Most Positive `Z` is `1`. So, the entering variable is `x_2`.

Minimum ratio is `10`. 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`	`x_3`	`S_1`	`S_2`	`A_1`	`RHS`
`R_3`(old) =	`0`	`1`	`1`	`0`	`-1`	`1`	`10`
`R_3`(new)`= R_3`(old)	`0`	`1`	`1`	`0`	`-1`	`1`	`10`

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

	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`A_1`	`RHS`
`R_2`(old) =	`1`	`0`	`-1`	`1`	`0`	`0`	`10`
`R_2`(new)`= R_2`(old)	`1`	`0`	`-1`	`1`	`0`	`0`	`10`

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

	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`A_1`	`RHS`
`R_1`(old) =	`0`	`1`	`1`	`0`	`-1`	`0`	`10`
`R_3`(new) =	`0`	`1`	`1`	`0`	`-1`	`1`	`10`
`R_1`(new)`= R_1`(old) - `R_3`(new)	`0`	`0`	`0`	`0`	`0`	`-1`	`0`

Tableau-2

`"Basis"`	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`A_1`	`RHS`
`R_1` `Z`	`0`	`0`	`0`	`0`	`0`	`-1`	`0`
`R_2` `S_1`	`1`	`0`	`-1`	`1`	`0`	`0`	`10`
`R_3` `x_2`	`0`	`1`	`1`	`0`	`-1`	`1`	`10`

Since all `Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=0,x_2=10,x_3=0`

Min `Z=0`


we eliminate the artificial variables and change the objective function for the original,
`Z - 5x_1 - 2x_2 - 10x_3=0`

Simplex tableau is
Tableau-0

`"Basis"`	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`RHS`
`R_1` `Z`	`-5`	`-2`	`-10`	`0`	`0`	`0`
`R_2` `S_1`	`1`	`0`	`-1`	`1`	`0`	`10`
`R_3` `x_2`	`0`	`1`	`1`	`0`	`-1`	`10`

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) + `2 R_3`(old)

	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`RHS`
`R_1`(old) =	`-5`	`-2`	`-10`	`0`	`0`	`0`
`R_3`(old) =	`0`	`1`	`1`	`0`	`-1`	`10`
`2 xx R_3`(new) =	`0`	`2`	`2`	`0`	`-2`	`20`
`R_1`(new)`= R_1`(old) + `2 R_3`(old)	`-5`	`0`	`-8`	`0`	`-2`	`20`

Tableau-1

`"Basis"`	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`RHS`
`R_1` `Z`	`-5`	`0`	`-8`	`0`	`-2`	`20`
`R_2` `S_1`	`1`	`0`	`-1`	`1`	`0`	`10`
`R_3` `x_2`	`0`	`1`	`1`	`0`	`-1`	`10`

Since all `Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=0,x_2=10,x_3=0`

Min `Z=20`

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