1. Find solution using Revised Simplex (BigM) method
MAX Z = 2x1 + x2
subject to
3x1 + 4x2 <= 6
6x1 + x2 <= 3
and x1,x2 >= 0

Solution:
Problem is

Max `Z` `=` `` `2` `x_1` ` + ` `` `x_2`

subject to

`` `3` `x_1` ` + ` `4` `x_2` ≤ `6` `` `6` `x_1` ` + ` `` `x_2` ≤ `3`

and `x_1,x_2 >= 0; `

Step-1 :
The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

After introducing slack variables

`` `3` `x_1` ` + ` `4` `x_2` ` + ` `` `S_1` = `6` `` `6` `x_1` ` + ` `` `x_2` ` + ` `` `S_2` = `3`

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

After introducing slack variables

`` `` `Z'` ` - ` `2` `x_1` ` - ` `` `x_2` = `0` `` `3` `x_1` ` + ` `4` `x_2` ` + ` `` `S_1` = `6` `` `6` `x_1` ` + ` `` `x_2` ` + ` `` `S_2` = `3`

Now represent the new system of constraint equations in the matrix form
`[[1,-2,-1,0,0],[0,3,4,1,0],[0,6,1,0,1]][[Z'],[x_1],[x_2],[S_1],[S_2]]=[[0],[6],[3]]`

or
`[[1,-c],[0,A]][[Z],[x]]=[[0,b]]; x>=0`

where `e=beta_0,a_3=beta_1,a_4=beta_2`

Step-2 : The basis matrix `B_1` of order `(2+1)=3` can be expressed as

`B_1=[beta_0,beta_1,beta_2]=[[1,0,0],[0,1,0],[0,0,1]]`

Then, `B_1^(-1)=[[1,C_B B^(-1)],[0,B^(-1)]]=1; B=[[1,0],[0,1]]=[beta_1,beta_2]; C_B=[0,0]`

		Basis Inverse `B_1^(-1)`					Additional table
`B`	`X_B`	`beta_0` `Z'`	`beta_1` `S_1`	`beta_2` `S_2`	`y_1`	Min Ratio `(X_B)/(y_1)`	`x_1`	`x_2`
`Z'`	`0`	`1`	`0`	`0`		---	`-2`	`-1`
`S_1`	`6`	`0`	`1`	`0`		---	`3`	`4`
`S_2`	`3`	`0`	`0`	`1`		---	`6`	`1`

---

Iteration=1 : Repeat steps 3 to 5 to get new solution
Step-3: To select the vector corresponding to a non-basic variable to enter into the basis, we compute
`z_k-c_k="Min" {(z_j-c_j)<0;}`

`="Min"{(1^(st)" row of " B_1^(-1)) ("Columns " a_j " not in basis")}`

`="Min"{[[1,0,0]] [[-2,-1],[3,4],[6,1]]}`

`="Min"{[[-2,-1]]}`

`=-2` (correspnds to `z_1-c_1`)

Thus, vector `x_1` is selected to enter into the basis, for `k=1`

Step-4: To select a basic variable to leave the basis, we compute `y_k` for k=1, as follows


`y_1= B_1^(-1) a_1=[[1,0,0],[0,1,0],[0,0,1]] [[-2],[3],[6]]=[[-2],[3],[6]]`

and `X_B = [[0],[6],[3]]`

Now, calculate the minimum ratio to select the basic variable to leave the basis
`x_(Br)/y_(rk)= "Min" {x_(Bi)/y_(ik), y_(ik)>0}`

`="Min"{(6)/(3),(3)/(6)}`

`="Min"{2,1/2}`

`=1/2 ("correspnds to " x_(B2)/y_(21))`

Thus, vector `S_2` is selected to leave the basis, for `r=2`

The table with new entries in column `y_1` and the minimum ratio

		Basis Inverse `B_1^(-1)`					Additional table
`B`	`X_B`	`beta_0` `Z'`	`beta_1` `S_1`	`beta_2` `S_2`	`y_1`	Min Ratio `(X_B)/(y_1)`	`x_1`	`x_2`
`Z'`	`0`	`1`	`0`	`0`	`-2`	---	`-2`	`-1`
`S_1`	`6`	`0`	`1`	`0`	`3`	`2`	`3`	`4`
`S_2`	`3`	`0`	`0`	`1`	`6`	`1/2`	`6`	`1`

The table solution is now updated by replacing variable `S_2` with the variable `x_1` into the basis.

For this we apply the following row operations in the same way as in the simplex method

	`X_B`	`beta_1`	`beta_2`	`y_1`
`R_1`	`0`	`0`	`0`	`-2`
`R_2`	`6`	`1`	`0`	`3`
`R_3`	`3`	`0`	`1`	`6`

`R_3`(new)`= R_3`(old)` -: 6`

`R_3`(old) =	`3`		`0`	`1`
`R_3`(new)`= R_3`(old)` -: 6`	`1/2`		`0`	`1/6`

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

`R_1`(old) =	`0`		`0`	`0`
`R_3`(new) =	`1/2`		`0`	`1/6`
`2 xx R_3`(new) =	`1`		`0`	`1/3`
`R_1`(new)`= R_1`(old) + `2 R_3`(new)	`1`		`0`	`1/3`

`R_2`(new)`= R_2`(old) - `3 R_3`(new)

`R_2`(old) =	`6`		`1`	`0`
`R_3`(new) =	`1/2`		`0`	`1/6`
`3 xx R_3`(new) =	`3/2`		`0`	`1/2`
`R_2`(new)`= R_2`(old) - `3 R_3`(new)	`9/2`		`1`	`-1/2`

The improved solution is

		Basis Inverse `B_1^(-1)`					Additional table
`B`	`X_B`	`beta_0` `Z'`	`beta_1` `S_1`	`beta_2` `x_1`	`y_1`	Min Ratio `(X_B)/(y_1)`	`S_2`	`x_2`
`Z'`	`1`	`1`	`0`	`1/3`		---	`0`	`-1`
`S_1`	`9/2`	`0`	`1`	`-1/2`		---	`0`	`4`
`x_1`	`1/2`	`0`	`0`	`1/6`		---	`1`	`1`

---

Iteration=2 : Repeat steps 3 to 5 to get new solution
Step-3: To select the vector corresponding to a non-basic variable to enter into the basis, we compute
`z_k-c_k="Min" {(z_j-c_j)<0;}`

`="Min"{(1^(st)" row of " B_1^(-1)) ("Columns " a_j " not in basis")}`

`="Min"{[[1,0,1/3]] [[0,-1],[0,4],[1,1]]}`

`="Min"{[[1/3,-2/3]]}`

`=-2/3` (correspnds to `z_2-c_2`)

Thus, vector `x_2` is selected to enter into the basis, for `k=2`

Step-4: To select a basic variable to leave the basis, we compute `y_k` for k=2, as follows


`y_2= B_1^(-1) a_2=[[1,0,1/3],[0,1,-1/2],[0,0,1/6]] [[-1],[4],[1]]=[[-2/3],[7/2],[1/6]]`

and `X_B = [[1],[9/2],[1/2]]`

Now, calculate the minimum ratio to select the basic variable to leave the basis
`x_(Br)/y_(rk)= "Min" {x_(Bi)/y_(ik), y_(ik)>0}`

`="Min"{(9/2)/(7/2),(1/2)/(1/6)}`

`="Min"{9/7,3}`

`=9/7 ("correspnds to " x_(B1)/y_(12))`

Thus, vector `S_1` is selected to leave the basis, for `r=1`

The table with new entries in column `y_2` and the minimum ratio

		Basis Inverse `B_1^(-1)`					Additional table
`B`	`X_B`	`beta_0` `Z'`	`beta_1` `S_1`	`beta_2` `x_1`	`y_2`	Min Ratio `(X_B)/(y_2)`	`S_2`	`x_2`
`Z'`	`1`	`1`	`0`	`1/3`	`-2/3`	---	`0`	`-1`
`S_1`	`9/2`	`0`	`1`	`-1/2`	`7/2`	`9/7`	`0`	`4`
`x_1`	`1/2`	`0`	`0`	`1/6`	`1/6`	`3`	`1`	`1`

The table solution is now updated by replacing variable `S_1` with the variable `x_2` into the basis.

For this we apply the following row operations in the same way as in the simplex method

	`X_B`	`beta_1`	`beta_2`	`y_2`
`R_1`	`1`	`0`	`1/3`	`-2/3`
`R_2`	`9/2`	`1`	`-1/2`	`7/2`
`R_3`	`1/2`	`0`	`1/6`	`1/6`

`R_2`(new)`= R_2`(old) `xx2/7`

`R_2`(old) =	`9/2`		`1`	`-1/2`
`R_2`(new)`= R_2`(old) `xx2/7`	`9/7`		`2/7`	`-1/7`

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

`R_1`(old) =	`1`		`0`	`1/3`
`R_2`(new) =	`9/7`		`2/7`	`-1/7`
`2/3 xx R_2`(new) =	`6/7`		`4/21`	`-2/21`
`R_1`(new)`= R_1`(old) + `2/3 R_2`(new)	`13/7`		`4/21`	`5/21`

`R_3`(new)`= R_3`(old) - `1/6 R_2`(new)

`R_3`(old) =	`1/2`		`0`	`1/6`
`R_2`(new) =	`9/7`		`2/7`	`-1/7`
`1/6 xx R_2`(new) =	`3/14`		`1/21`	`-1/42`
`R_3`(new)`= R_3`(old) - `1/6 R_2`(new)	`2/7`		`-1/21`	`4/21`

The improved solution is

		Basis Inverse `B_1^(-1)`					Additional table
`B`	`X_B`	`beta_0` `Z'`	`beta_1` `x_2`	`beta_2` `x_1`	`y_2`	Min Ratio `(X_B)/(y_2)`	`S_2`	`S_1`
`Z'`	`13/7`	`1`	`4/21`	`5/21`		---	`0`	`0`
`x_2`	`9/7`	`0`	`2/7`	`-1/7`		---	`0`	`1`
`x_1`	`2/7`	`0`	`-1/21`	`4/21`		---	`1`	`0`

---

Iteration=3 : Repeat steps 3 to 5 to get new solution
Step-3: To select the vector corresponding to a non-basic variable to enter into the basis, we compute
`z_k-c_k="Min" {(z_j-c_j)<0;}`

`="Min"{(1^(st)" row of " B_1^(-1)) ("Columns " a_j " not in basis")}`

`="Min"{[[1,4/21,5/21]] [[0,0],[0,1],[1,0]]}`

`="Min"{[[5/21,4/21]]}`

Since all `Z_j-C_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=2/7,x_2=9/7`

Max `Z=13/7`