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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
1. Standard form-1 : Example-1
(Previous example)		3. Standard form-1 : Example-3
(Next example)

2. Standard form-1 : Example-2


Find solution using Revised Simplex (BigM) method
MAX Z = 3x1 + 5x2
subject to
x1 <= 4
x2 <= 6
3x1 + 2x2 <= 18
and x1,x2 >= 0

Solution:
Problem is
Max `Z``=````3``x_1`` + ``5``x_2`
subject to
`````x_1`≤`4`
`````x_2`≤`6`
```3``x_1`` + ``2``x_2`≤`18`
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
`````x_1`` + ````S_1`=`4`
`````x_2`` + ````S_2`=`6`
```3``x_1`` + ``2``x_2`` + ````S_3`=`18`


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

After introducing slack variables
`````Z'`` - ``3``x_1`` - ``5``x_2`=`0`
`````x_1`` + ````S_1`=`4`
`````x_2`` + ````S_2`=`6`
```3``x_1`` + ``2``x_2`` + ````S_3`=`18`


Now represent the new system of constraint equations in the matrix form
`[[1,-3,-5,0,0,0],[0,1,0,1,0,0],[0,0,1,0,1,0],[0,3,2,0,0,1]][[Z'],[x_1],[x_2],[S_1],[S_2],[S_3]]=[[0],[4],[6],[18]]`

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]`

Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`S_1`		`beta_2`
`S_2`		`beta_3`
`S_3`		`y_1`
 		Min Ratio
`(X_B)/(y_1)`		`x_1``x_2`
`Z'``0``1``0``0``0`---`-3``-5`
`S_1``4``0``1``0``0`---`1``0`
`S_2``6``0``0``1``0`---`0``1`
`S_3``18``0``0``0``1`---`3``2`


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

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

`=-5` (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],[0,1,0,0],[0,0,1,0],[0,0,0,1]] [[-5],[0],[1],[2]]=[[-5],[0],[1],[2]]`

and `X_B = [[0],[4],[6],[18]]`

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)/(1),(18)/(2)}`

`="Min"{6,9}`

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

Thus, vector `S_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`
`S_1`		`beta_2`
`S_2`		`beta_3`
`S_3`		`y_2`
 		Min Ratio
`(X_B)/(y_2)`		`x_1``x_2`
`Z'``0``1``0``0``0``-5`---`-3``-5`
`S_1``4``0``1``0``0``0`---`1``0`
`S_2``6``0``0``1``0``1``6``0``1`
`S_3``18``0``0``0``1``2``9``3``2`


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``y_2`
`R_1``0``0``0``0``-5`
`R_2``4``1``0``0``0`
`R_3``6``0``1``0``1`
`R_4``18``0``0``1``2`


`R_3`(new)`= R_3`(old)
`R_3`(old) = `6``0``1``0`
`R_3`(new)`= R_3`(old)`6``0``1``0`


`R_1`(new)`= R_1`(old) + `5 R_3`(new)
`R_1`(old) = `0``0``0``0`
`R_3`(new) = `6``0``1``0`
`5 xx R_3`(new) = `30``0``5``0`
`R_1`(new)`= R_1`(old) + `5 R_3`(new)`30``0``5``0`


`R_2`(new)`= R_2`(old)
`R_2`(old) = `4``1``0``0`
`R_2`(new)`= R_2`(old)`4``1``0``0`


`R_4`(new)`= R_4`(old) - `2 R_3`(new)
`R_4`(old) = `18``0``0``1`
`R_3`(new) = `6``0``1``0`
`2 xx R_3`(new) = `12``0``2``0`
`R_4`(new)`= R_4`(old) - `2 R_3`(new)`6``0``-2``1`


The improved solution is
Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`S_1`		`beta_2`
`x_2`		`beta_3`
`S_3`		`y_2`
 		Min Ratio
`(X_B)/(y_2)`		`x_1``S_2`
`Z'``30``1``0``5``0`---`-3``0`
`S_1``4``0``1``0``0`---`1``0`
`x_2``6``0``0``1``0`---`0``1`
`S_3``6``0``0``-2``1`---`3``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"{[[1,0,5,0]] [[-3,0],[1,0],[0,1],[3,0]]}`

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

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

and `X_B = [[30],[4],[6],[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"{(4)/(1),(6)/(3)}`

`="Min"{4,2}`

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

Thus, vector `S_3` 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`
`S_1`		`beta_2`
`x_2`		`beta_3`
`S_3`		`y_1`
 		Min Ratio
`(X_B)/(y_1)`		`x_1``S_2`
`Z'``30``1``0``5``0``-3`---`-3``0`
`S_1``4``0``1``0``0``1``4``1``0`
`x_2``6``0``0``1``0``0`---`0``1`
`S_3``6``0``0``-2``1``3``2``3``0`


The table solution is now updated by replacing variable `S_3` 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``30``0``5``0``-3`
`R_2``4``1``0``0``1`
`R_3``6``0``1``0``0`
`R_4``6``0``-2``1``3`


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


`R_1`(new)`= R_1`(old) + `3 R_4`(new)
`R_1`(old) = `30``0``5``0`
`R_4`(new) = `2``0``-2/3``1/3`
`3 xx R_4`(new) = `6``0``-2``1`
`R_1`(new)`= R_1`(old) + `3 R_4`(new)`36``0``3``1`


`R_2`(new)`= R_2`(old) - `R_4`(new)
`R_2`(old) = `4``1``0``0`
`R_4`(new) = `2``0``-2/3``1/3`
`R_2`(new)`= R_2`(old) - `R_4`(new)`2``1``2/3``-1/3`


`R_3`(new)`= R_3`(old)
`R_3`(old) = `6``0``1``0`
`R_3`(new)`= R_3`(old)`6``0``1``0`


The improved solution is
Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`S_1`		`beta_2`
`x_2`		`beta_3`
`x_1`		`y_1`
 		Min Ratio
`(X_B)/(y_1)`		`S_3``S_2`
`Z'``36``1``0``3``1`---`0``0`
`S_1``2``0``1``2/3``-1/3`---`0``0`
`x_2``6``0``0``1``0`---`0``1`
`x_1``2``0``0``-2/3``1/3`---`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"{[[1,0,3,1]] [[0,0],[0,0],[0,1],[1,0]]}`

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

Since all `Z_j-C_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=2,x_2=6`

Max `Z=36`

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


1. Standard form-1 : Example-1
(Previous example)		3. Standard form-1 : Example-3
(Next example)


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