Find solution using dual simplex method
MIN Z = 2x1 + 3x2 + 0x3
subject to
2x1 - x2 - x3 >= 3
x1 - x2 + x3 >= 2
and x1,x2,x3 >= 0;

Solution:

---

Problem is

Min `Z` `=` `` `2` `x_1` ` + ` `3` `x_2`

subject to

`` `2` `x_1` ` - ` `` `x_2` ` - ` `` `x_3` ≥ `3` `` `` `x_1` ` - ` `` `x_2` ` + ` `` `x_3` ≥ `2`

and `x_1,x_2,x_3 >= 0; `

In order to apply the dual simplex method, convert all `>=` constraint to `<=` constraint by multiply -1.

Problem is

Min `Z` `=` `` `2` `x_1` ` + ` `3` `x_2`

subject to

` - ` `2` `x_1` ` + ` `` `x_2` ` + ` `` `x_3` ≤ `-3` ` - ` `` `x_1` ` + ` `` `x_2` ` - ` `` `x_3` ≤ `-2`

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 add slack variable `S_2`

After introducing slack variables

Min `Z` `=` `` `2` `x_1` ` + ` `3` `x_2` ` + ` `0` `x_3` ` + ` `0` `S_1` ` + ` `0` `S_2`

subject to

` - ` `2` `x_1` ` + ` `` `x_2` ` + ` `` `x_3` ` + ` `` `S_1` = `-3` ` - ` `` `x_1` ` + ` `` `x_2` ` - ` `` `x_3` ` + ` `` `S_2` = `-2`

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

Tableau-1	`C_j`	`2`	`3`	`0`	`0`	`0`
`C_B`	`"Basis"`	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`RHS`
`R_1` `0`	`S_1`	`(-2)`	`1`	`1`	`1`	`0`	`-3``->`
`R_2` `0`	`S_2`	`-1`	`1`	`-1`	`0`	`1`	`-2`
	`Z_j`	`0`	`0`	`0`	`0`	`0`	`Z=0`
	`Z_j-C_j`	`-2`	`-3`	`0`	`0`	`0`
	`"Ratio"=(Z_j-C_j)/(S_1,j)` and `S_1,j<0`	`(-2)/(-2)` `=1``uarr`	`(-3)/(1)` `=`---	`(0)/(1)` `=`---	`(0)/(1)` `=`---	`(0)/(0)` `=`---

Most negative `RHS` is `-3` and its row index is `1`. So, the leaving basis variable is `S_1`.

Least positive ratio is `1` and its column index is `1`. So, the entering variable is `x_1`.

`:.` The pivot element is `-2`.

Entering `=x_1`, Departing `=S_1`, Key Element `=-2`

`R_1`(new)`= R_1`(old) `-: -2`

	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`RHS`
`R_1`(old) =	`-2`	`1`	`1`	`1`	`0`	`-3`
`R_1`(new)`= R_1`(old) `-: -2`	`1`	`-0.5`	`-0.5`	`-0.5`	`0`	`1.5`

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

	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`RHS`
`R_2`(old) =	`-1`	`1`	`-1`	`0`	`1`	`-2`
`R_1`(new) =	`1`	`-0.5`	`-0.5`	`-0.5`	`0`	`1.5`
`R_2`(new)`= R_2`(old) + `R_1`(new)	`0`	`0.5`	`-1.5`	`-0.5`	`1`	`-0.5`

Tableau-2	`C_j`	`2`	`3`	`0`	`0`	`0`
`C_B`	`"Basis"`	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`RHS`
`R_1` `2`	`x_1`	`1`	`-0.5`	`-0.5`	`-0.5`	`0`	`1.5`
`R_2` `0`	`S_2`	`0`	`0.5`	`(-1.5)`	`-0.5`	`1`	`-0.5``->`
	`Z_j`	`2`	`-1`	`-1`	`-1`	`0`	`Z=3`
	`Z_j-C_j`	`0`	`-4`	`-1`	`-1`	`0`
	`"Ratio"=(Z_j-C_j)/(S_2,j)` and `S_2,j<0`	`(0)/(0)` `=`---	`(-4)/(0.5)` `=`---	`(-1)/(-1.5)` `=0.6667``uarr`	`(-1)/(-0.5)` `=2`	`(0)/(1)` `=`---

Most negative `RHS` is `-0.5` and its row index is `2`. So, the leaving basis variable is `S_2`.

Least positive ratio is `0.6667` and its column index is `3`. So, the entering variable is `x_3`.

`:.` The pivot element is `-1.5`.

Entering `=x_3`, Departing `=S_2`, Key Element `=-1.5`

`R_2`(new)`= R_2`(old) `-: -1.5`

	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`RHS`
`R_2`(old) =	`0`	`0.5`	`-1.5`	`-0.5`	`1`	`-0.5`
`R_2`(new)`= R_2`(old) `-: -1.5`	`0`	`-0.3333`	`1`	`0.3333`	`-0.6667`	`0.3333`

`R_1`(new)`= R_1`(old) + `0.5 R_2`(new)

	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`RHS`
`R_1`(old) =	`1`	`-0.5`	`-0.5`	`-0.5`	`0`	`1.5`
`R_2`(new) =	`0`	`-0.3333`	`1`	`0.3333`	`-0.6667`	`0.3333`
`0.5 xx R_2`(new) =	`0`	`-0.1667`	`0.5`	`0.1667`	`-0.3333`	`0.1667`
`R_1`(new)`= R_1`(old) + `0.5 R_2`(new)	`1`	`-0.6667`	`0`	`-0.3333`	`-0.3333`	`1.6667`

Tableau-3	`C_j`	`2`	`3`	`0`	`0`	`0`
`C_B`	`"Basis"`	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`	`RHS`
`R_1` `2`	`x_1`	`1`	`-0.6667`	`0`	`-0.3333`	`-0.3333`	`1.6667`
`R_2` `0`	`x_3`	`0`	`-0.3333`	`1`	`0.3333`	`-0.6667`	`0.3333`
	`Z_j`	`2`	`-1.3333`	`0`	`-0.6667`	`-0.6667`	`Z=3.3333`
	`Z_j-C_j`	`0`	`-4.3333`	`0`	`-0.6667`	`-0.6667`
	Ratio

Since all `Z_j-C_j <= 0` and all `RHS >= 0`, thus the current solution is the optimal solution.

Hence, optimal solution is arrived with value of variables as :
`x_1=1.6667,x_2=0,x_3=0.3333`

Min `Z=3.3333`

---

---