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 Min Z to Max Z and all `>=` constraint to `<=` constraint by multiply -1.

Problem is

Max `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

Max `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`

Iteration-1		`C_j`	`-2`	`-3`	`0`	`0`	`0`
`B`	`C_B`	`X_B`	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`
`S_1`	`0`	`-3`	`(-2)`	`1`	`1`	`1`	`0`
`S_2`	`0`	`-2`	`-1`	`1`	`-1`	`0`	`1`
`Z=0`		`Z_j`	`0`	`0`	`0`	`0`	`0`
		`Z_j-C_j`	`2`	`3`	`0`	`0`	`0`
		`"Ratio"=(Z_j-C_j)/(S_1,j)` and `S_1,j<0`	`-1``uarr`	---	---	---	---

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

Maximum negative 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)`

`R_1`(old) =	`-3`	`-2`	`1`	`1`	`1`	`0`
`R_1`(new)`= R_1`(old) `-: (-2)`	`3/2`	`1`	`-1/2`	`-1/2`	`-1/2`	`0`

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

`R_2`(old) =	`-2`	`-1`	`1`	`-1`	`0`	`1`
`R_1`(new) =	`3/2`	`1`	`-1/2`	`-1/2`	`-1/2`	`0`
`R_2`(new)`= R_2`(old) + `R_1`(new)	`-1/2`	`0`	`1/2`	`-3/2`	`-1/2`	`1`

Iteration-2		`C_j`	`-2`	`-3`	`0`	`0`	`0`
`B`	`C_B`	`X_B`	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`
`x_1`	`-2`	`3/2`	`1`	`-1/2`	`-1/2`	`-1/2`	`0`
`S_2`	`0`	`-1/2`	`0`	`1/2`	`(-3/2)`	`-1/2`	`1`
`Z=-3`		`Z_j`	`-2`	`1`	`1`	`1`	`0`
		`Z_j-C_j`	`0`	`4`	`1`	`1`	`0`
		`"Ratio"=(Z_j-C_j)/(S_2,j)` and `S_2,j<0`	---	---	`-0.6667``uarr`	`-2`	---

Minimum negative `X_B` is `-1/2` and its row index is `2`. So, the leaving basis variable is `S_2`.

Maximum negative ratio is `-0.6667` and its column index is `3`. So, the entering variable is `x_3`.

`:.` The pivot element is `-3/2`.

Entering `=x_3`, Departing `=S_2`, Key Element `=-3/2`

`R_2`(new)`= R_2`(old) `xx(-2/3)`

`R_2`(old) =	`-1/2`	`0`	`1/2`	`-3/2`	`-1/2`	`1`
`R_2`(new)`= R_2`(old) `xx(-2/3)`	`1/3`	`0`	`-1/3`	`1`	`1/3`	`-2/3`

`R_1`(new)`= R_1`(old) + `1/2 R_2`(new)

`R_1`(old) =	`3/2`	`1`	`-1/2`	`-1/2`	`-1/2`	`0`
`R_2`(new) =	`1/3`	`0`	`-1/3`	`1`	`1/3`	`-2/3`
`1/2 xx R_2`(new) =	`1/6`	`0`	`-1/6`	`1/2`	`1/6`	`-1/3`
`R_1`(new)`= R_1`(old) + `1/2 R_2`(new)	`5/3`	`1`	`-2/3`	`0`	`-1/3`	`-1/3`

Iteration-3		`C_j`	`-2`	`-3`	`0`	`0`	`0`
`B`	`C_B`	`X_B`	`x_1`	`x_2`	`x_3`	`S_1`	`S_2`
`x_1`	`-2`	`5/3`	`1`	`-2/3`	`0`	`-1/3`	`-1/3`
`x_3`	`0`	`1/3`	`0`	`-1/3`	`1`	`1/3`	`-2/3`
`Z=-10/3`		`Z_j`	`-2`	`4/3`	`0`	`2/3`	`2/3`
		`Z_j-C_j`	`0`	`13/3`	`0`	`2/3`	`2/3`
		Ratio	---	---	---	---	---

Since all `Z_j-C_j >= 0` and all `X_(Bi) >= 0` thus the current solution is the optimal solution.

Hence, optimal solution is arrived with value of variables as :
`x_1=5/3,x_2=0,x_3=1/3`

Max `Z=-10/3`

`:.` Min `Z=10/3`

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