Find solution using integer simplex method (Gomory's cutting plane method)
MAX Z = x1 + x2
subject to
3x1 + 2x2 <= 5
x2 <= 2
x1,x2 non-negative integers

Solution:
Problem is

Max `Z` `=` `` `` `x_1` ` + ` `` `x_2`

subject to

`` `3` `x_1` ` + ` `2` `x_2` ≤ `5` `` `` `x_2` ≤ `2`

and `x_1,x_2 >= 0; ``x_1,x_2` non-negative integers

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` `=` `` `` `x_1` ` + ` `` `x_2` ` + ` `0` `S_1` ` + ` `0` `S_2`

subject to

`` `3` `x_1` ` + ` `2` `x_2` ` + ` `` `S_1` = `5` `` `` `x_2` ` + ` `` `S_2` = `2`

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

`Z` ` - ` `` `x_1` ` - ` `` `x_2` = `0`

`` `3` `x_1` ` + ` `2` `x_2` ` + ` `` `S_1` = `5` `` `` `x_2` ` + ` `` `S_2` = `2`

Simplex tableau is
Tableau-0

`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`RHS`
`R_1` `Z`	`-1`	`-1`	`0`	`0`	`0`
`R_2` `S_1`	`3`	`2`	`1`	`0`	`5`
`R_3` `S_2`	`0`	`1`	`0`	`1`	`2`

Tableau-1

`"Basis"`	`x_1``darr`	`x_2`	`S_1`	`S_2`	`RHS`	`"Ratio"=(RHS)/(x_1)`
`R_1` `Z`	`-1`	`-1`	`0`	`0`	`0`
`R_2` `S_1`	`(3)`	`2`	`1`	`0`	`5`	`(5)/(3)=1.6667``->`
`R_3` `S_2`	`0`	`1`	`0`	`1`	`2`	`(2)/(0)` (ignore, denominator is 0)

Most Negative `Z` is `-1`. So, the entering variable is `x_1`.

Minimum ratio is `1.6667`. So, the leaving basis variable is `S_1`.

`:.` The pivot element is `3`.

Entering `=x_1`, Departing `=S_1`, Key Element `=3`

`R_2`(new)`= R_2`(old) `-: 3`

	`x_1`	`x_2`	`S_1`	`S_2`	`RHS`
`R_2`(old) =	`3`	`2`	`1`	`0`	`5`
`R_2`(new)`= R_2`(old) `-: 3`	`1`	`0.6667`	`0.3333`	`0`	`1.6667`

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

	`x_1`	`x_2`	`S_1`	`S_2`	`RHS`
`R_3`(old) =	`0`	`1`	`0`	`1`	`2`
`R_3`(new)`= R_3`(old)	`0`	`1`	`0`	`1`	`2`

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

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

Tableau-2

`"Basis"`	`x_1`	`x_2``darr`	`S_1`	`S_2`	`RHS`	`"Ratio"=(RHS)/(x_2)`
`R_1` `Z`	`0`	`-0.3333`	`0.3333`	`0`	`1.6667`
`R_2` `x_1`	`1`	`0.6667`	`0.3333`	`0`	`1.6667`	`(1.6667)/(0.6667)=2.5`
`R_3` `S_2`	`0`	`(1)`	`0`	`1`	`2`	`(2)/(1)=2``->`

Most Negative `Z` is `-0.3333`. So, the entering variable is `x_2`.

Minimum ratio is `2`. So, the leaving basis variable is `S_2`.

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

Entering `=x_2`, Departing `=S_2`, Key Element `=1`

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

	`x_1`	`x_2`	`S_1`	`S_2`	`RHS`
`R_3`(old) =	`0`	`1`	`0`	`1`	`2`
`R_3`(new)`= R_3`(old)	`0`	`1`	`0`	`1`	`2`

`R_2`(new)`= R_2`(old) - `0.6667 R_3`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`RHS`
`R_2`(old) =	`1`	`0.6667`	`0.3333`	`0`	`1.6667`
`R_3`(new) =	`0`	`1`	`0`	`1`	`2`
`0.6667 xx R_3`(new) =	`0`	`0.6667`	`0`	`0.6667`	`1.3333`
`R_2`(new)`= R_2`(old) - `0.6667 R_3`(new)	`1`	`0`	`0.3333`	`-0.6667`	`0.3333`

`R_1`(new)`= R_1`(old) - `-0.3333 R_3`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`RHS`
`R_1`(old) =	`0`	`-0.3333`	`0.3333`	`0`	`1.6667`
`R_3`(new) =	`0`	`1`	`0`	`1`	`2`
`-0.3333 xx R_3`(new) =	`0`	`-0.3333`	`0`	`-0.3333`	`-0.6667`
`R_1`(new)`= R_1`(old) - `-0.3333 R_3`(new)	`0`	`0`	`0.3333`	`0.3333`	`2.3333`

Tableau-3

`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`RHS`
`R_1` `Z`	`0`	`0`	`0.3333`	`0.3333`	`2.3333`
`R_2` `x_1`	`1`	`0`	`0.3333`	`-0.6667`	`0.3333`
`R_3` `x_2`	`0`	`1`	`0`	`1`	`2`

Since all `Z_j >= 0`

Hence, non-integer optimal solution is arrived with value of variables as :
`x_1=0.3333,x_2=2`

Max `Z=2.3333`

To obtain the integer valued solution, we proceed to construct Gomory's fractional cut, with the help of `x_1`-row as follows:

`0.3333= 1 x_1+0.3333 S_1 -0.6667 S_2`

`(0+0.3333)=(1+0) x_1+(0+0.3333) S_1+(-1+0.3333) S_2`

The fractional cut will become
`-0.3333 = Sg1 -0.3333 S_1 -0.3333 S_2 ->` (Cut-1)

Adding this additional constraint at the bottom of optimal simplex table. The new table so obtained is

Tableau-1

`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`Sg1`	`RHS`
`R_1` `0`	`Z`	`0`	`0`	`0.3333`	`0.3333`	`0`	`2.3333`
`R_2` `1`	`x_1`	`1`	`0`	`0.3333`	`-0.6667`	`0`	`0.3333`
`R_3` `1`	`x_2`	`0`	`1`	`0`	`1`	`0`	`2`
`R_4` `0`	`Sg1`	`0`	`0`	`(-0.3333)`	`-0.3333`	`1`	`-0.3333``->`
	`"Ratio"=(Z_j)/(Sg1,j)` and `Sg1,j<0`	`(0)/(0)` `=`---	`(0)/(0)` `=`---	`(0.3333)/(-0.3333)` `=-1``uarr`	`(0.3333)/(-0.3333)` `=-1`	`(0)/(1)` `=`---

Most negative `RHS` is `-0.3333`. So, the leaving basis variable is `Sg1`.

Least negative ratio is `-1`. So, the entering variable is `S_1`.

`:.` The pivot element is `-0.3333`.

Entering `=S_1`, Departing `=Sg1`, Key Element `=-0.3333`

`R_4`(new)`= R_4`(old) `-: -0.3333`

	`x_1`	`x_2`	`S_1`	`S_2`	`Sg1`	`RHS`
`R_4`(old) =	`0`	`0`	`-0.3333`	`-0.3333`	`1`	`-0.3333`
`R_4`(new)`= R_4`(old) `-: -0.3333`	`0`	`0`	`1`	`1`	`-3`	`1`

`R_2`(new)`= R_2`(old) - `0.3333 R_4`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`Sg1`	`RHS`
`R_2`(old) =	`1`	`0`	`0.3333`	`-0.6667`	`0`	`0.3333`
`R_4`(new) =	`0`	`0`	`1`	`1`	`-3`	`1`
`0.3333 xx R_4`(new) =	`0`	`0`	`0.3333`	`0.3333`	`-1`	`0.3333`
`R_2`(new)`= R_2`(old) - `0.3333 R_4`(new)	`1`	`0`	`0`	`-1`	`1`	`0`

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

	`x_1`	`x_2`	`S_1`	`S_2`	`Sg1`	`RHS`
`R_3`(old) =	`0`	`1`	`0`	`1`	`0`	`2`
`R_3`(new)`= R_3`(old)	`0`	`1`	`0`	`1`	`0`	`2`

`R_1`(new)`= R_1`(old) - `0.3333 R_4`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`Sg1`	`RHS`
`R_1`(old) =	`0`	`0`	`0.3333`	`0.3333`	`0`	`2.3333`
`R_4`(new) =	`0`	`0`	`1`	`1`	`-3`	`1`
`0.3333 xx R_4`(new) =	`0`	`0`	`0.3333`	`0.3333`	`-1`	`0.3333`
`R_1`(new)`= R_1`(old) - `0.3333 R_4`(new)	`0`	`0`	`0`	`0`	`1`	`2`

Tableau-2

`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`Sg1`	`RHS`
`R_1` `0`	`Z`	`0`	`0`	`0`	`0`	`1`	`2`
`R_2` `1`	`x_1`	`1`	`0`	`0`	`-1`	`1`	`0`
`R_3` `1`	`x_2`	`0`	`1`	`0`	`1`	`0`	`2`
`R_4` `0`	`S_1`	`0`	`0`	`1`	`1`	`-3`	`1`
	Ratio

Since all `Z_j >= 0`

Hence, integer optimal solution is arrived with value of variables as :
`x_1=0,x_2=2`

Max `Z=2`

The integer optimal solution found after 1-cuts.