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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
9. 0-1 Integer programming problem
(Previous method)		2. Standard form-1 : Example-2
(Next example)

1. Standard form-1 : Example-1


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`

This material is intended as a summary. Use your textbook for detail explanation.
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9. 0-1 Integer programming problem
(Previous method)		2. Standard form-1 : Example-2
(Next example)


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