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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
3. Standard form-1 : Example-3
(Previous example)		5. Standard form-2 using Two-Phase method : Example-2
(Next example)

4. Standard form-2 using Two-Phase method : Example-1


Find solution using Revised Simplex (Two-Phase) method
MAX Z = x1 + x2
subject to
2x1 + 5x2 <= 6
x1 + x2 >= 2
and x1,x2 >= 0

Solution:
Problem is
Max `Z``=``````x_1`` + ````x_2`
subject to
```2``x_1`` + ``5``x_2`≤`6`
`````x_1`` + ````x_2`≥`2`
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,surplus,artificial variables
```2``x_1`` + ``5``x_2`` + ````S_1`=`6`
`````x_1`` + ````x_2`` - ````S_2`` + ````A_1`=`2`


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

After introducing slack,surplus,artificial variables
`````Z'`` - ````x_1`` - ````x_2`=`0`
` - ``3``x_1`` - ``6``x_2`` - ````S_1`` + ````S_2`` + ````x_5`=`-2`
```2``x_1`` + ``5``x_2`` + ````S_1`=`6`
`````x_1`` + ````x_2`` - ````S_2`` + ````A_1`=`2`


Now represent the new system of constraint equations in the matrix form
`[[1,-1,-1,0,0,0,0],[0,-3,-6,-1,1,0,1],[0,2,5,1,0,0,0],[0,1,1,0,-1,1,0]][[Z'],[x_1],[x_2],[S_1],[S_2],[A_1],[x_5]]=[[0],[-2],[6],[2]]`

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

where `e=beta_0,a_4=beta_1,a_5=beta_2,a_6=beta_3`

Step-2 : The basis matrix `B_1` of order `(3+1)=4` can be expressed as

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

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

Phase-1
Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`x_5`		`beta_2`
`S_1`		`beta_3`
`A_1`		`y_1`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_1)`		`x_1``x_2``S_2`
`Z'``0``1``0``0``0`---`-1``-1``0`
`x_5``-2``0``1``-1``0`---`-3``-6``1`
`S_1``6``0``0``1``0`---`2``5``0`
`A_1``2``0``0``0``1`---`1``1``-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"{(2^(nd)" row of " B_1^(-1)) ("Columns " a_j " not in basis")}`

`="Min"{[[0,1,-1,0]] [[-1,-1,0],[-3,-6,1],[2,5,0],[1,1,-1]]}`

`="Min"{[[-5,-11,1]]}`

`=-11` (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],[0,1,-1,0],[0,0,1,0],[0,0,0,1]] [[-1],[-6],[5],[1]]=[[-1],[-11],[5],[1]]`

and `X_B = [[0],[-2],[6],[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"{(6)/(5),(2)/(1)}`

`="Min"{6/5,2}`

`=6/5 ("correspnds to " x_(B2)/y_(22))`

Thus, vector `S_1` is selected to leave the basis, for `r=2`

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`
`S_1`		`beta_3`
`A_1`		`y_2`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_2)`		`x_1``x_2``S_2`
`Z'``0``1``0``0``0``-1`---`-1``-1``0`
`x_5``-2``0``1``-1``0``-11`---`-3``-6``1`
`S_1``6``0``0``1``0``5``6/5``2``5``0`
`A_1``2``0``0``0``1``1``2``1``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``beta_3``y_2`
`R_1``0``0``0``0``-1`
`R_2``-2``1``-1``0``-11`
`R_3``6``0``1``0``5`
`R_4``2``0``0``1``1`


`R_3`(new)`= R_3`(old)` -: 5`
`R_3`(old) = `6``0``1``0`
`R_3`(new)`= R_3`(old)` -: 5``6/5``0``1/5``0`


`R_1`(new)`= R_1`(old) + `R_3`(new)
`R_1`(old) = `0``0``0``0`
`R_3`(new) = `6/5``0``1/5``0`
`R_1`(new)`= R_1`(old) + `R_3`(new)`6/5``0``1/5``0`


`R_2`(new)`= R_2`(old) + `11 R_3`(new)
`R_2`(old) = `-2``1``-1``0`
`R_3`(new) = `6/5``0``1/5``0`
`11 xx R_3`(new) = `66/5``0``11/5``0`
`R_2`(new)`= R_2`(old) + `11 R_3`(new)`56/5``1``6/5``0`


`R_4`(new)`= R_4`(old) - `R_3`(new)
`R_4`(old) = `2``0``0``1`
`R_3`(new) = `6/5``0``1/5``0`
`R_4`(new)`= R_4`(old) - `R_3`(new)`4/5``0``-1/5``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_2`		`beta_3`
`A_1`		`y_2`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_2)`		`x_1``S_1``S_2`
`Z'``6/5``1``0``1/5``0`---`-1``0``0`
`x_5``56/5``0``1``6/5``0`---`-3``0``1`
`x_2``6/5``0``0``1/5``0`---`2``1``0`
`A_1``4/5``0``0``-1/5``1`---`1``0``-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"{(2^(nd)" row of " B_1^(-1)) ("Columns " a_j " not in basis")}`

`="Min"{[[0,1,6/5,0]] [[-1,0,0],[-3,0,1],[2,1,0],[1,0,-1]]}`

`="Min"{[[-3/5,6/5,1]]}`

`=-3/5` (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=[[1,0,1/5,0],[0,1,6/5,0],[0,0,1/5,0],[0,0,-1/5,1]] [[-1],[-3],[2],[1]]=[[-3/5],[-3/5],[2/5],[3/5]]`

and `X_B = [[6/5],[56/5],[6/5],[4/5]]`

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"{(6/5)/(2/5),(4/5)/(3/5)}`

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

`=4/3 ("correspnds to " x_(B3)/y_(31))`

Thus, vector `A_1` is selected to leave the basis, for `r=3`

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`
`x_5`		`beta_2`
`x_2`		`beta_3`
`A_1`		`y_1`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_1)`		`x_1``S_1``S_2`
`Z'``6/5``1``0``1/5``0``-3/5`---`-1``0``0`
`x_5``56/5``0``1``6/5``0``-3/5`---`-3``0``1`
`x_2``6/5``0``0``1/5``0``2/5``3``2``1``0`
`A_1``4/5``0``0``-1/5``1``3/5``4/3``1``0``-1`


The table solution is now updated by replacing variable `A_1` 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``beta_3``y_1`
`R_1``6/5``0``1/5``0``-3/5`
`R_2``56/5``1``6/5``0``-3/5`
`R_3``6/5``0``1/5``0``2/5`
`R_4``4/5``0``-1/5``1``3/5`


`R_4`(new)`= R_4`(old) `xx5/3`
`R_4`(old) = `4/5``0``-1/5``1`
`R_4`(new)`= R_4`(old) `xx5/3``4/3``0``-1/3``5/3`


`R_1`(new)`= R_1`(old) + `3/5 R_4`(new)
`R_1`(old) = `6/5``0``1/5``0`
`R_4`(new) = `4/3``0``-1/3``5/3`
`3/5 xx R_4`(new) = `4/5``0``-1/5``1`
`R_1`(new)`= R_1`(old) + `3/5 R_4`(new)`2``0``0``1`


`R_2`(new)`= R_2`(old) + `3/5 R_4`(new)
`R_2`(old) = `56/5``1``6/5``0`
`R_4`(new) = `4/3``0``-1/3``5/3`
`3/5 xx R_4`(new) = `4/5``0``-1/5``1`
`R_2`(new)`= R_2`(old) + `3/5 R_4`(new)`12``1``1``1`


`R_3`(new)`= R_3`(old) - `2/5 R_4`(new)
`R_3`(old) = `6/5``0``1/5``0`
`R_4`(new) = `4/3``0``-1/3``5/3`
`2/5 xx R_4`(new) = `8/15``0``-2/15``2/3`
`R_3`(new)`= R_3`(old) - `2/5 R_4`(new)`2/3``0``1/3``-2/3`


The improved solution is
Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`x_5`		`beta_2`
`x_2`		`beta_3`
`x_1`		`y_1`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_1)`		`A_1``S_1``S_2`
`Z'``2``1``0``0``1`---`0``0``0`
`x_5``12``0``1``1``1`---`0``0``1`
`x_2``2/3``0``0``1/3``-2/3`---`0``1``0`
`x_1``4/3``0``0``-1/3``5/3`---`1``0``-1`


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"{[[0,1,1,1]] [[0,0,0],[0,0,1],[0,1,0],[1,0,-1]]}`

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

Since all `Z_j-C_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=4/3,x_2=2/3`

Max `Z=2`


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_2`		`beta_3`
`x_1`		`y_0`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_0)`		`S_1``S_2`
`Z'``2``1``0``0``1`---`0``0`
`x_5``12``0``1``1``1`---`0``1`
`x_2``2/3``0``0``1/3``-2/3`---`1``0`
`x_1``4/3``0``0``-1/3``5/3`---`0``-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"{[[1,0,0,1]] [[0,0],[0,1],[1,0],[0,-1]]}`

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

and `X_B = [[2],[12],[2/3],[4/3]]`

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"{(2/3)/(2/3)}`

`="Min"{1}`

`=1 ("correspnds to " x_(B2)/y_(22))`

Thus, vector `x_2` is selected to leave the basis, for `r=2`

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_2`		`beta_3`
`x_1`		`y_2`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_2)`		`S_1``S_2`
`Z'``2``1``0``0``1``-1`---`0``0`
`x_5``12``0``1``1``1``0`---`0``1`
`x_2``2/3``0``0``1/3``-2/3``2/3``1``1``0`
`x_1``4/3``0``0``-1/3``5/3``-5/3`---`0``-1`


The table solution is now updated by replacing variable `x_2` 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``y_2`
`R_1``2``0``0``1``-1`
`R_2``12``1``1``1``0`
`R_3``2/3``0``1/3``-2/3``2/3`
`R_4``4/3``0``-1/3``5/3``-5/3`


`R_3`(new)`= R_3`(old) `xx3/2`
`R_3`(old) = `2/3``0``1/3``-2/3`
`R_3`(new)`= R_3`(old) `xx3/2``1``0``1/2``-1`


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


`R_2`(new)`= R_2`(old)
`R_2`(old) = `12``1``1``1`
`R_2`(new)`= R_2`(old)`12``1``1``1`


`R_4`(new)`= R_4`(old) + `5/3 R_3`(new)
`R_4`(old) = `4/3``0``-1/3``5/3`
`R_3`(new) = `1``0``1/2``-1`
`5/3 xx R_3`(new) = `5/3``0``5/6``-5/3`
`R_4`(new)`= R_4`(old) + `5/3 R_3`(new)`3``0``1/2``0`


The improved solution is
Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`x_5`		`beta_2`
`S_2`		`beta_3`
`x_1`		`y_2`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_2)`		`S_1``x_2`
`Z'``3``1``0``1/2``0`---`0``0`
`x_5``12``0``1``1``1`---`0``0`
`S_2``1``0``0``1/2``-1`---`1``1`
`x_1``3``0``0``1/2``0`---`0``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"{(1^(st)" row of " B_1^(-1)) ("Columns " a_j " not in basis")}`

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

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

Since all `Z_j-C_j >= 0`

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

Max `Z=3`

This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here


3. Standard form-1 : Example-3
(Previous example)		5. Standard form-2 using Two-Phase method : Example-2
(Next example)


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