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Home > Operation Research calculators > Revised Simplex method example
9. Revised Simplex method example ( Enter your problem )
( Enter your problem )
  1. Standard form-1 : Example-1
  2. Standard form-1 : Example-2
  3. Standard form-1 : Example-3
  4. Standard form-2 using Two-Phase method : Example-1
  5. Standard form-2 using Two-Phase method : Example-2
  6. Standard form-2 using Two-Phase method : Example-3
  7. Standard form-2 using Big M method : Example-1
  8. Standard form-2 using Big M method : Example-2
  9. Standard form-2 using Big M method : Example-3
 		Other related methods
  1. Formulate linear programming model
  2. Graphical method
  3. Simplex method (BigM method)
  4. Two-Phase method
  5. Primal to dual conversion
  6. Dual simplex method
  7. Integer simplex method
  8. Branch and Bound method
  9. 0-1 Integer programming problem
  10. Revised Simplex method
5. Standard form-2 using Two-Phase method : Example-2
(Previous example)		7. Standard form-2 using Big M method : Example-1
(Next example)

6. Standard form-2 using Two-Phase method : Example-3


Find solution using Revised Simplex (Two-Phase) method
MIN Z = 3x1 + 2x2 + x3
subject to
x1 + x2 + x3 >= 4
x2 - x3 <= 2
x1 + x2 + 2x3 = 6
and x1,x2,x3 >= 0

Solution:
Problem is
Min `Z``=````3``x_1`` + ``2``x_2`` + ````x_3`
subject to
`````x_1`` + ````x_2`` + ````x_3`≥`4`
`````x_2`` - ````x_3`≤`2`
`````x_1`` + ````x_2`` + ``2``x_3`=`6`
and `x_1,x_2,x_3 >= 0; `
`:.` Max `Z``=`` - ``3``x_1`` - ``2``x_2`` - ````x_3`


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

After introducing slack,surplus,artificial variables
`````x_1`` + ````x_2`` + ````x_3`` - ````S_1`` + ````A_1`=`4`
`````x_2`` - ````x_3`` + ````S_2`=`2`
`````x_1`` + ````x_2`` + ``2``x_3`` + ````A_2`=`6`


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

After introducing slack,surplus,artificial variables
`````Z'`` + ``3``x_1`` + ``2``x_2`` + ````x_3`=`0`
` - ``2``x_1`` - ``3``x_2`` - ``2``x_3`` + ````S_1`` - ````S_2`` + ````x_5`=`-10`
`````x_1`` + ````x_2`` + ````x_3`` - ````S_1`` + ````A_1`=`4`
`````x_2`` - ````x_3`` + ````S_2`=`2`
`````x_1`` + ````x_2`` + ``2``x_3`` + ````A_2`=`6`


Now represent the new system of constraint equations in the matrix form
`[[1,3,2,1,0,0,0,0,0],[0,-2,-3,-2,1,-1,0,0,1],[0,1,1,1,-1,0,1,0,0],[0,0,1,-1,0,1,0,0,0],[0,1,1,2,0,0,0,1,0]][[Z'],[x_1],[x_2],[x_3],[S_1],[S_2],[A_1],[A_2],[x_5]]=[[0],[-10],[4],[2],[6]]`

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

where `e=beta_0,a_5=beta_1,a_6=beta_2,a_7=beta_3,a_8=beta_4`

Step-2 : The basis matrix `B_1` of order `(4+1)=5` can be expressed as

`B_1=[beta_0,beta_1,beta_2,beta_3,beta_4]=[[1,0,0,0,0],[0,1,0,-1,0],[0,0,1,0,0],[0,0,0,1,0],[0,0,0,0,1]]`

Then, `B_1^(-1)=[[1,C_B B^(-1)],[0,B^(-1)]]=1; B=[[1,0,-1,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]]=[beta_1,beta_2,beta_3,beta_4]; C_B=[0,0,0,0]`

Phase-1
Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`x_5`		`beta_2`
`A_1`		`beta_3`
`S_2`		`beta_4`
`A_2`		`y_1`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_1)`		`x_1``x_2``x_3``S_1`
`Z'``0``1``0``0``0``0`---`3``2``1``0`
`x_5``-10``0``1``0``-1``0`---`-2``-3``-2``1`
`A_1``4``0``0``1``0``0`---`1``1``1``-1`
`S_2``2``0``0``0``1``0`---`0``1``-1``0`
`A_2``6``0``0``0``0``1`---`1``1``2``0`


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"{(2^(nd)" row of " B_1^(-1)) ("Columns " a_j " not in basis")}`

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

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

`=-4` (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=[[1,0,0,0,0],[0,1,0,-1,0],[0,0,1,0,0],[0,0,0,1,0],[0,0,0,0,1]] [[2],[-3],[1],[1],[1]]=[[2],[-4],[1],[1],[1]]`

and `X_B = [[0],[-10],[4],[2],[6]]`

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"{(0)/(2),(4)/(1),(2)/(1),(6)/(1)}`

`="Min"{0,4,2,6}`

`=2 ("correspnds to " x_(B3)/y_(32))`

Thus, vector `S_2` is selected to leave the basis, for `r=3`

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`
`x_5`		`beta_2`
`A_1`		`beta_3`
`S_2`		`beta_4`
`A_2`		`y_2`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_2)`		`x_1``x_2``x_3``S_1`
`Z'``0``1``0``0``0``0``2``0``3``2``1``0`
`x_5``-10``0``1``0``-1``0``-4`---`-2``-3``-2``1`
`A_1``4``0``0``1``0``0``1``4``1``1``1``-1`
`S_2``2``0``0``0``1``0``1``2``0``1``-1``0`
`A_2``6``0``0``0``0``1``1``6``1``1``2``0`


The table solution is now updated by replacing variable `S_2` 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``beta_3``beta_4``y_2`
`R_1``0``0``0``0``0``2`
`R_2``-10``1``0``-1``0``-4`
`R_3``4``0``1``0``0``1`
`R_4``2``0``0``1``0``1`
`R_5``6``0``0``0``1``1`


`R_4`(new)`= R_4`(old)
`R_4`(old) = `2``0``0``1``0`
`R_4`(new)`= R_4`(old)`2``0``0``1``0`


`R_1`(new)`= R_1`(old) - `2 R_4`(new)
`R_1`(old) = `0``0``0``0``0`
`R_4`(new) = `2``0``0``1``0`
`2 xx R_4`(new) = `4``0``0``2``0`
`R_1`(new)`= R_1`(old) - `2 R_4`(new)`-4``0``0``-2``0`


`R_2`(new)`= R_2`(old) + `4 R_4`(new)
`R_2`(old) = `-10``1``0``-1``0`
`R_4`(new) = `2``0``0``1``0`
`4 xx R_4`(new) = `8``0``0``4``0`
`R_2`(new)`= R_2`(old) + `4 R_4`(new)`-2``1``0``3``0`


`R_3`(new)`= R_3`(old) - `R_4`(new)
`R_3`(old) = `4``0``1``0``0`
`R_4`(new) = `2``0``0``1``0`
`R_3`(new)`= R_3`(old) - `R_4`(new)`2``0``1``-1``0`


`R_5`(new)`= R_5`(old) - `R_4`(new)
`R_5`(old) = `6``0``0``0``1`
`R_4`(new) = `2``0``0``1``0`
`R_5`(new)`= R_5`(old) - `R_4`(new)`4``0``0``-1``1`


The improved solution is
Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`x_5`		`beta_2`
`A_1`		`beta_3`
`x_2`		`beta_4`
`A_2`		`y_2`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_2)`		`x_1``S_2``x_3``S_1`
`Z'``-4``1``0``0``-2``0`---`3``0``1``0`
`x_5``-2``0``1``0``3``0`---`-2``0``-2``1`
`A_1``2``0``0``1``-1``0`---`1``0``1``-1`
`x_2``2``0``0``0``1``0`---`0``1``-1``0`
`A_2``4``0``0``0``-1``1`---`1``0``2``0`


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"{(2^(nd)" row of " B_1^(-1)) ("Columns " a_j " not in basis")}`

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

`="Min"{[[-2,3,-5,1]]}`

`=-5` (correspnds to `z_3-c_3`)

Thus, vector `x_3` is selected to enter into the basis, for `k=3`

Step-4: To select a basic variable to leave the basis, we compute `y_k` for k=3, as follows


`y_3= B_1^(-1) a_3=[[1,0,0,-2,0],[0,1,0,3,0],[0,0,1,-1,0],[0,0,0,1,0],[0,0,0,-1,1]] [[1],[-2],[1],[-1],[2]]=[[3],[-5],[2],[-1],[3]]`

and `X_B = [[-4],[-2],[2],[2],[4]]`

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"{(-4)/(3),(2)/(2),(4)/(3)}`

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

`=1 ("correspnds to " x_(B2)/y_(23))`

Thus, vector `A_1` is selected to leave the basis, for `r=2`

The table with new entries in column `y_3` and the minimum ratio

Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`x_5`		`beta_2`
`A_1`		`beta_3`
`x_2`		`beta_4`
`A_2`		`y_3`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_3)`		`x_1``S_2``x_3``S_1`
`Z'``-4``1``0``0``-2``0``3``-4/3``3``0``1``0`
`x_5``-2``0``1``0``3``0``-5`---`-2``0``-2``1`
`A_1``2``0``0``1``-1``0``2``1``1``0``1``-1`
`x_2``2``0``0``0``1``0``-1`---`0``1``-1``0`
`A_2``4``0``0``0``-1``1``3``4/3``1``0``2``0`


The table solution is now updated by replacing variable `A_1` with the variable `x_3` 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``beta_3``beta_4``y_3`
`R_1``-4``0``0``-2``0``3`
`R_2``-2``1``0``3``0``-5`
`R_3``2``0``1``-1``0``2`
`R_4``2``0``0``1``0``-1`
`R_5``4``0``0``-1``1``3`


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


`R_1`(new)`= R_1`(old) - `3 R_3`(new)
`R_1`(old) = `-4``0``0``-2``0`
`R_3`(new) = `1``0``1/2``-1/2``0`
`3 xx R_3`(new) = `3``0``3/2``-3/2``0`
`R_1`(new)`= R_1`(old) - `3 R_3`(new)`-7``0``-3/2``-1/2``0`


`R_2`(new)`= R_2`(old) + `5 R_3`(new)
`R_2`(old) = `-2``1``0``3``0`
`R_3`(new) = `1``0``1/2``-1/2``0`
`5 xx R_3`(new) = `5``0``5/2``-5/2``0`
`R_2`(new)`= R_2`(old) + `5 R_3`(new)`3``1``5/2``1/2``0`


`R_4`(new)`= R_4`(old) + `R_3`(new)
`R_4`(old) = `2``0``0``1``0`
`R_3`(new) = `1``0``1/2``-1/2``0`
`R_4`(new)`= R_4`(old) + `R_3`(new)`3``0``1/2``1/2``0`


`R_5`(new)`= R_5`(old) - `3 R_3`(new)
`R_5`(old) = `4``0``0``-1``1`
`R_3`(new) = `1``0``1/2``-1/2``0`
`3 xx R_3`(new) = `3``0``3/2``-3/2``0`
`R_5`(new)`= R_5`(old) - `3 R_3`(new)`1``0``-3/2``1/2``1`


The improved solution is
Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`x_5`		`beta_2`
`x_3`		`beta_3`
`x_2`		`beta_4`
`A_2`		`y_3`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_3)`		`x_1``S_2``A_1``S_1`
`Z'``-7``1``0``-3/2``-1/2``0`---`3``0``0``0`
`x_5``3``0``1``5/2``1/2``0`---`-2``0``0``1`
`x_3``1``0``0``1/2``-1/2``0`---`1``0``1``-1`
`x_2``3``0``0``1/2``1/2``0`---`0``1``0``0`
`A_2``1``0``0``-3/2``1/2``1`---`1``0``0``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"{(2^(nd)" row of " B_1^(-1)) ("Columns " a_j " not in basis")}`

`="Min"{[[0,1,5/2,1/2,0]] [[3,0,0,0],[-2,0,0,1],[1,0,1,-1],[0,1,0,0],[1,0,0,0]]}`

`="Min"{[[1/2,1/2,5/2,-3/2]]}`

`=-3/2` (correspnds to `z_4-c_4`)

Thus, vector `S_1` is selected to enter into the basis, for `k=4`

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


`y_4= B_1^(-1) a_4=[[1,0,-3/2,-1/2,0],[0,1,5/2,1/2,0],[0,0,1/2,-1/2,0],[0,0,1/2,1/2,0],[0,0,-3/2,1/2,1]] [[0],[1],[-1],[0],[0]]=[[3/2],[-3/2],[-1/2],[-1/2],[3/2]]`

and `X_B = [[-7],[3],[1],[3],[1]]`

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"{(-7)/(3/2),(1)/(3/2)}`

`="Min"{-14/3,2/3}`

`=2/3 ("correspnds to " x_(B4)/y_(44))`

Thus, vector `A_2` is selected to leave the basis, for `r=4`

The table with new entries in column `y_4` and the minimum ratio

Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`x_5`		`beta_2`
`x_3`		`beta_3`
`x_2`		`beta_4`
`A_2`		`y_4`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_4)`		`x_1``S_2``A_1``S_1`
`Z'``-7``1``0``-3/2``-1/2``0``3/2``-14/3``3``0``0``0`
`x_5``3``0``1``5/2``1/2``0``-3/2`---`-2``0``0``1`
`x_3``1``0``0``1/2``-1/2``0``-1/2`---`1``0``1``-1`
`x_2``3``0``0``1/2``1/2``0``-1/2`---`0``1``0``0`
`A_2``1``0``0``-3/2``1/2``1``3/2``2/3``1``0``0``0`


The table solution is now updated by replacing variable `A_2` with the variable `S_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``beta_3``beta_4``y_4`
`R_1``-7``0``-3/2``-1/2``0``3/2`
`R_2``3``1``5/2``1/2``0``-3/2`
`R_3``1``0``1/2``-1/2``0``-1/2`
`R_4``3``0``1/2``1/2``0``-1/2`
`R_5``1``0``-3/2``1/2``1``3/2`


`R_5`(new)`= R_5`(old) `xx2/3`
`R_5`(old) = `1``0``-3/2``1/2``1`
`R_5`(new)`= R_5`(old) `xx2/3``2/3``0``-1``1/3``2/3`


`R_1`(new)`= R_1`(old) - `3/2 R_5`(new)
`R_1`(old) = `-7``0``-3/2``-1/2``0`
`R_5`(new) = `2/3``0``-1``1/3``2/3`
`3/2 xx R_5`(new) = `1``0``-3/2``1/2``1`
`R_1`(new)`= R_1`(old) - `3/2 R_5`(new)`-8``0``0``-1``-1`


`R_2`(new)`= R_2`(old) + `3/2 R_5`(new)
`R_2`(old) = `3``1``5/2``1/2``0`
`R_5`(new) = `2/3``0``-1``1/3``2/3`
`3/2 xx R_5`(new) = `1``0``-3/2``1/2``1`
`R_2`(new)`= R_2`(old) + `3/2 R_5`(new)`4``1``1``1``1`


`R_3`(new)`= R_3`(old) + `1/2 R_5`(new)
`R_3`(old) = `1``0``1/2``-1/2``0`
`R_5`(new) = `2/3``0``-1``1/3``2/3`
`1/2 xx R_5`(new) = `1/3``0``-1/2``1/6``1/3`
`R_3`(new)`= R_3`(old) + `1/2 R_5`(new)`4/3``0``0``-1/3``1/3`


`R_4`(new)`= R_4`(old) + `1/2 R_5`(new)
`R_4`(old) = `3``0``1/2``1/2``0`
`R_5`(new) = `2/3``0``-1``1/3``2/3`
`1/2 xx R_5`(new) = `1/3``0``-1/2``1/6``1/3`
`R_4`(new)`= R_4`(old) + `1/2 R_5`(new)`10/3``0``0``2/3``1/3`


The improved solution is
Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`x_5`		`beta_2`
`x_3`		`beta_3`
`x_2`		`beta_4`
`S_1`		`y_4`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_4)`		`x_1``S_2``A_1``A_2`
`Z'``-8``1``0``0``-1``-1`---`3``0``0``0`
`x_5``4``0``1``1``1``1`---`-2``0``0``0`
`x_3``4/3``0``0``0``-1/3``1/3`---`1``0``1``0`
`x_2``10/3``0``0``0``2/3``1/3`---`0``1``0``0`
`S_1``2/3``0``0``-1``1/3``2/3`---`1``0``0``1`


Iteration=4 : 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"{(2^(nd)" row of " B_1^(-1)) ("Columns " a_j " not in basis")}`

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

`="Min"{[[0,1,1,1]]}`

Since all `Z_j-C_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=0,x_2=10/3,x_3=4/3`

Max `Z=-8`

`:.` Min `Z=8`


Phase-2
We maximize `Z'` instead of `x_5`

Artificial variables will be removed from additional table
Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`x_5`		`beta_2`
`x_3`		`beta_3`
`x_2`		`beta_4`
`S_1`		`y_0`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_0)`		`x_1``S_2`
`Z'``-8``1``0``0``-1``-1`---`3``0`
`x_5``4``0``1``1``1``1`---`-2``0`
`x_3``4/3``0``0``0``-1/3``1/3`---`1``0`
`x_2``10/3``0``0``0``2/3``1/3`---`0``1`
`S_1``2/3``0``0``-1``1/3``2/3`---`1``0`


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,-1,-1]] [[3,0],[-2,0],[1,0],[0,1],[1,0]]}`

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

`=-1` (correspnds to `z_2-c_2`)

Thus, vector `S_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=[[1,0,0,-1,-1],[0,1,1,1,1],[0,0,0,-1/3,1/3],[0,0,0,2/3,1/3],[0,0,-1,1/3,2/3]] [[0],[0],[0],[1],[0]]=[[-1],[1],[-1/3],[2/3],[1/3]]`

and `X_B = [[-8],[4],[4/3],[10/3],[2/3]]`

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"{(4)/(1),(10/3)/(2/3),(2/3)/(1/3)}`

`="Min"{4,5,2}`

`=2 ("correspnds to " x_(B4)/y_(42))`

Thus, vector `S_1` is selected to leave the basis, for `r=4`

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`
`x_5`		`beta_2`
`x_3`		`beta_3`
`x_2`		`beta_4`
`S_1`		`y_2`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_2)`		`x_1``S_2`
`Z'``-8``1``0``0``-1``-1``-1`---`3``0`
`x_5``4``0``1``1``1``1``1``4``-2``0`
`x_3``4/3``0``0``0``-1/3``1/3``-1/3`---`1``0`
`x_2``10/3``0``0``0``2/3``1/3``2/3``5``0``1`
`S_1``2/3``0``0``-1``1/3``2/3``1/3``2``1``0`


The table solution is now updated by replacing variable `S_1` with the variable `S_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``beta_3``beta_4``y_2`
`R_1``-8``0``0``-1``-1``-1`
`R_2``4``1``1``1``1``1`
`R_3``4/3``0``0``-1/3``1/3``-1/3`
`R_4``10/3``0``0``2/3``1/3``2/3`
`R_5``2/3``0``-1``1/3``2/3``1/3`


`R_5`(new)`= R_5`(old) `xx3`
`R_5`(old) = `2/3``0``-1``1/3``2/3`
`R_5`(new)`= R_5`(old) `xx3``2``0``-3``1``2`


`R_1`(new)`= R_1`(old) + `R_5`(new)
`R_1`(old) = `-8``0``0``-1``-1`
`R_5`(new) = `2``0``-3``1``2`
`R_1`(new)`= R_1`(old) + `R_5`(new)`-6``0``-3``0``1`


`R_2`(new)`= R_2`(old) - `R_5`(new)
`R_2`(old) = `4``1``1``1``1`
`R_5`(new) = `2``0``-3``1``2`
`R_2`(new)`= R_2`(old) - `R_5`(new)`2``1``4``0``-1`


`R_3`(new)`= R_3`(old) + `1/3 R_5`(new)
`R_3`(old) = `4/3``0``0``-1/3``1/3`
`R_5`(new) = `2``0``-3``1``2`
`1/3 xx R_5`(new) = `2/3``0``-1``1/3``2/3`
`R_3`(new)`= R_3`(old) + `1/3 R_5`(new)`2``0``-1``0``1`


`R_4`(new)`= R_4`(old) - `2/3 R_5`(new)
`R_4`(old) = `10/3``0``0``2/3``1/3`
`R_5`(new) = `2``0``-3``1``2`
`2/3 xx R_5`(new) = `4/3``0``-2``2/3``4/3`
`R_4`(new)`= R_4`(old) - `2/3 R_5`(new)`2``0``2``0``-1`


The improved solution is
Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`x_5`		`beta_2`
`x_3`		`beta_3`
`x_2`		`beta_4`
`S_2`		`y_2`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_2)`		`x_1``S_1`
`Z'``-6``1``0``-3``0``1`---`3``0`
`x_5``2``0``1``4``0``-1`---`-2``0`
`x_3``2``0``0``-1``0``1`---`1``0`
`x_2``2``0``0``2``0``-1`---`0``0`
`S_2``2``0``0``-3``1``2`---`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,-3,0,1]] [[3,0],[-2,0],[1,0],[0,0],[1,1]]}`

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

Since all `Z_j-C_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=0,x_2=2,x_3=2`

Max `Z=-6`

`:.` Min `Z=6`

This material is intended as a summary. Use your textbook for detail explanation.
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