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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
4. Standard form-2 using Two-Phase method : Example-1
(Previous example)		6. Standard form-2 using Two-Phase method : Example-3
(Next example)

5. Standard form-2 using Two-Phase method : Example-2


Find solution using Revised Simplex (Two-Phase) method
MIN Z = x1 + x2
subject to
x1 + 2x2 >= 7
4x1 + x2 >= 6
and x1,x2 >= 0

Solution:
Problem is
Min `Z``=``````x_1`` + ````x_2`
subject to
`````x_1`` + ``2``x_2`≥`7`
```4``x_1`` + ````x_2`≥`6`
and `x_1,x_2 >= 0; `
`:.` Max `Z``=`` - ````x_1`` - ````x_2`


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

After introducing surplus,artificial variables
`````x_1`` + ``2``x_2`` - ````S_1`` + ````A_1`=`7`
```4``x_1`` + ````x_2`` - ````S_2`` + ````A_2`=`6`


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

After introducing surplus,artificial variables
`````Z'`` + ````x_1`` + ````x_2`=`0`
` - ``5``x_1`` - ``3``x_2`` + ````S_1`` + ````S_2`` + ````x_5`=`-13`
`````x_1`` + ``2``x_2`` - ````S_1`` + ````A_1`=`7`
```4``x_1`` + ````x_2`` - ````S_2`` + ````A_2`=`6`


Now represent the new system of constraint equations in the matrix form
`[[1,1,1,0,0,0,0,0],[0,-5,-3,1,1,0,0,1],[0,1,2,-1,0,1,0,0],[0,4,1,0,-1,0,1,0]][[Z'],[x_1],[x_2],[S_1],[S_2],[A_1],[A_2],[x_5]]=[[0],[-13],[7],[6]]`

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],[0,1,0,0],[0,0,1,0],[0,0,0,1]]`

Then, `B_1^(-1)=[[1,C_B B^(-1)],[0,B^(-1)]]=1; B=[[1,0,0],[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`
`A_1`		`beta_3`
`A_2`		`y_1`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_1)`		`x_1``x_2``S_1``S_2`
`Z'``0``1``0``0``0`---`1``1``0``0`
`x_5``-13``0``1``0``0`---`-5``-3``1``1`
`A_1``7``0``0``1``0`---`1``2``-1``0`
`A_2``6``0``0``0``1`---`4``1``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"{(2^(nd)" row of " B_1^(-1)) ("Columns " a_j " not in basis")}`

`="Min"{[[0,1,0,0]] [[1,1,0,0],[-5,-3,1,1],[1,2,-1,0],[4,1,0,-1]]}`

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

`=-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=[[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]] [[1],[-5],[1],[4]]=[[1],[-5],[1],[4]]`

and `X_B = [[0],[-13],[7],[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)/(1),(7)/(1),(6)/(4)}`

`="Min"{0,7,3/2}`

`=3/2 ("correspnds to " x_(B3)/y_(31))`

Thus, vector `A_2` 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`
`A_1`		`beta_3`
`A_2`		`y_1`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_1)`		`x_1``x_2``S_1``S_2`
`Z'``0``1``0``0``0``1``0``1``1``0``0`
`x_5``-13``0``1``0``0``-5`---`-5``-3``1``1`
`A_1``7``0``0``1``0``1``7``1``2``-1``0`
`A_2``6``0``0``0``1``4``3/2``4``1``0``-1`


The table solution is now updated by replacing variable `A_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``beta_3``y_1`
`R_1``0``0``0``0``1`
`R_2``-13``1``0``0``-5`
`R_3``7``0``1``0``1`
`R_4``6``0``0``1``4`


`R_4`(new)`= R_4`(old)` -: 4`
`R_4`(old) = `6``0``0``1`
`R_4`(new)`= R_4`(old)` -: 4``3/2``0``0``1/4`


`R_1`(new)`= R_1`(old) - `R_4`(new)
`R_1`(old) = `0``0``0``0`
`R_4`(new) = `3/2``0``0``1/4`
`R_1`(new)`= R_1`(old) - `R_4`(new)`-3/2``0``0``-1/4`


`R_2`(new)`= R_2`(old) + `5 R_4`(new)
`R_2`(old) = `-13``1``0``0`
`R_4`(new) = `3/2``0``0``1/4`
`5 xx R_4`(new) = `15/2``0``0``5/4`
`R_2`(new)`= R_2`(old) + `5 R_4`(new)`-11/2``1``0``5/4`


`R_3`(new)`= R_3`(old) - `R_4`(new)
`R_3`(old) = `7``0``1``0`
`R_4`(new) = `3/2``0``0``1/4`
`R_3`(new)`= R_3`(old) - `R_4`(new)`11/2``0``1``-1/4`


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_1`		`y_1`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_1)`		`A_2``x_2``S_1``S_2`
`Z'``-3/2``1``0``0``-1/4`---`0``1``0``0`
`x_5``-11/2``0``1``0``5/4`---`0``-3``1``1`
`A_1``11/2``0``0``1``-1/4`---`0``2``-1``0`
`x_1``3/2``0``0``0``1/4`---`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"{[[0,1,0,5/4]] [[0,1,0,0],[0,-3,1,1],[0,2,-1,0],[1,1,0,-1]]}`

`="Min"{[[5/4,-7/4,1,-1/4]]}`

`=-7/4` (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,-1/4],[0,1,0,5/4],[0,0,1,-1/4],[0,0,0,1/4]] [[1],[-3],[2],[1]]=[[3/4],[-7/4],[7/4],[1/4]]`

and `X_B = [[-3/2],[-11/2],[11/2],[3/2]]`

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"{(-3/2)/(3/4),(11/2)/(7/4),(3/2)/(1/4)}`

`="Min"{-2,22/7,6}`

`=22/7 ("correspnds to " x_(B2)/y_(22))`

Thus, vector `A_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`
`A_1`		`beta_3`
`x_1`		`y_2`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_2)`		`A_2``x_2``S_1``S_2`
`Z'``-3/2``1``0``0``-1/4``3/4``-2``0``1``0``0`
`x_5``-11/2``0``1``0``5/4``-7/4`---`0``-3``1``1`
`A_1``11/2``0``0``1``-1/4``7/4``22/7``0``2``-1``0`
`x_1``3/2``0``0``0``1/4``1/4``6``1``1``0``-1`


The table solution is now updated by replacing variable `A_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``-3/2``0``0``-1/4``3/4`
`R_2``-11/2``1``0``5/4``-7/4`
`R_3``11/2``0``1``-1/4``7/4`
`R_4``3/2``0``0``1/4``1/4`


`R_3`(new)`= R_3`(old) `xx4/7`
`R_3`(old) = `11/2``0``1``-1/4`
`R_3`(new)`= R_3`(old) `xx4/7``22/7``0``4/7``-1/7`


`R_1`(new)`= R_1`(old) - `3/4 R_3`(new)
`R_1`(old) = `-3/2``0``0``-1/4`
`R_3`(new) = `22/7``0``4/7``-1/7`
`3/4 xx R_3`(new) = `33/14``0``3/7``-3/28`
`R_1`(new)`= R_1`(old) - `3/4 R_3`(new)`-27/7``0``-3/7``-1/7`


`R_2`(new)`= R_2`(old) + `7/4 R_3`(new)
`R_2`(old) = `-11/2``1``0``5/4`
`R_3`(new) = `22/7``0``4/7``-1/7`
`7/4 xx R_3`(new) = `11/2``0``1``-1/4`
`R_2`(new)`= R_2`(old) + `7/4 R_3`(new)`0``1``1``1`


`R_4`(new)`= R_4`(old) - `1/4 R_3`(new)
`R_4`(old) = `3/2``0``0``1/4`
`R_3`(new) = `22/7``0``4/7``-1/7`
`1/4 xx R_3`(new) = `11/14``0``1/7``-1/28`
`R_4`(new)`= R_4`(old) - `1/4 R_3`(new)`5/7``0``-1/7``2/7`


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_2`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_2)`		`A_2``A_1``S_1``S_2`
`Z'``-27/7``1``0``-3/7``-1/7`---`0``0``0``0`
`x_5``0``0``1``1``1`---`0``0``1``1`
`x_2``22/7``0``0``4/7``-1/7`---`0``1``-1``0`
`x_1``5/7``0``0``-1/7``2/7`---`1``0``0``-1`


`x_5=0`


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_2`
`C_k-Z_k`		Min Ratio
`(X_B)/(y_2)`		`S_1``S_2`
`Z'``-27/7``1``0``-3/7``-1/7`---`0``0`
`x_5``0``0``1``1``1`---`1``1`
`x_2``22/7``0``0``4/7``-1/7`---`-1``0`
`x_1``5/7``0``0``-1/7``2/7`---`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,-3/7,-1/7]] [[0,0],[1,1],[-1,0],[0,-1]]}`

`="Min"{[[3/7,1/7]]}`

Since all `Z_j-C_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=5/7,x_2=22/7`

Max `Z=-27/7`

`:.` Min `Z=27/7`

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


4. Standard form-2 using Two-Phase method : Example-1
(Previous example)		6. Standard form-2 using Two-Phase method : Example-3
(Next example)


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