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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
8. Standard form-2 using Big M method : Example-2
(Previous example)

9. Standard form-2 using Big M method : Example-3


Find solution using Revised Simplex (BigM) 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`` - ``M``A_1`` - ``M``A_2`=`0`
`````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,-M,-M],[0,1,1,1,-1,0,1,0],[0,0,1,-1,0,1,0,0],[0,1,1,2,0,0,0,1]][[Z'],[x_1],[x_2],[x_3],[S_1],[S_2],[A_1],[A_2]]=[[0],[4],[2],[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,-M,0,-M],[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`
`A_1`		`beta_2`
`S_2`		`beta_3`
`A_2`		`y_1`
 		Min Ratio
`(X_B)/(y_1)`		`x_1``x_2``x_3``S_1`
`Z'``0``1``-M``0``-M`---`3``2``1``0`
`A_1``4``0``1``0``0`---`1``1``1``-1`
`S_2``2``0``0``1``0`---`0``1``-1``0`
`A_2``6``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"{(1^(st)" row of " B_1^(-1)) ("Columns " a_j " not in basis")}`

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

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

`=-3M+1` (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,-M,0,-M],[0,1,0,0],[0,0,1,0],[0,0,0,1]] [[1],[1],[-1],[2]]=[[-3M+1],[1],[-1],[2]]`

and `X_B = [[0],[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"{(4)/(1),(6)/(2)}`

`="Min"{4,3}`

`=3 ("correspnds to " x_(B3)/y_(33))`

Thus, vector `A_2` is selected to leave the basis, for `r=3`

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`
`A_1`		`beta_2`
`S_2`		`beta_3`
`A_2`		`y_3`
 		Min Ratio
`(X_B)/(y_3)`		`x_1``x_2``x_3``S_1`
`Z'``0``1``-M``0``-M``-3M+1`---`3``2``1``0`
`A_1``4``0``1``0``0``1``4``1``1``1``-1`
`S_2``2``0``0``1``0``-1`---`0``1``-1``0`
`A_2``6``0``0``0``1``2``3``1``1``2``0`


The table solution is now updated by replacing variable `A_2` 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``y_3`
`R_1``0``-M``0``-M``-3M+1`
`R_2``4``1``0``0``1`
`R_3``2``0``1``0``-1`
`R_4``6``0``0``1``2`


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


`R_1`(new)`= R_1`(old) + `(3M-1) R_4`(new)
`R_1`(old) = `0``-M``0``-M`
`R_4`(new) = `3``0``0``1/2`
`3M-1 xx R_4`(new) = `9M-3``0``0``(3M-1)/2`
`R_1`(new)`= R_1`(old) + `(3M-1) R_4`(new)`9M-3``-M``0``(M-1)/2`


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


`R_3`(new)`= R_3`(old) + `R_4`(new)
`R_3`(old) = `2``0``1``0`
`R_4`(new) = `3``0``0``1/2`
`R_3`(new)`= R_3`(old) + `R_4`(new)`5``0``1``1/2`


The improved solution is
Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`A_1`		`beta_2`
`S_2`		`beta_3`
`x_3`		`y_3`
 		Min Ratio
`(X_B)/(y_3)`		`x_1``x_2``A_2``S_1`
`Z'``9M-3``1``-M``0``(M-1)/2`---`3``2``M``0`
`A_1``1``0``1``0``-1/2`---`1``1``0``-1`
`S_2``5``0``0``1``1/2`---`0``1``0``0`
`x_3``3``0``0``0``1/2`---`1``1``1``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,-M,0,(M-1)/2]] [[3,2,M,0],[1,1,0,-1],[0,1,0,0],[1,1,1,0]]}`

`="Min"{[[(-M+5)/2,(-M+3)/2,(3M-1)/2,M]]}`

`=(-M+3)/2` (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,-M,0,(M-1)/2],[0,1,0,-1/2],[0,0,1,1/2],[0,0,0,1/2]] [[2],[1],[1],[1]]=[[(-M+3)/2],[1/2],[3/2],[1/2]]`

and `X_B = [[9M-3],[1],[5],[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"{(1)/(1/2),(5)/(3/2),(3)/(1/2)}`

`="Min"{2,10/3,6}`

`=2 ("correspnds to " x_(B1)/y_(12))`

Thus, vector `A_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`
`A_1`		`beta_2`
`S_2`		`beta_3`
`x_3`		`y_2`
 		Min Ratio
`(X_B)/(y_2)`		`x_1``x_2``A_2``S_1`
`Z'``9M-3``1``-M``0``(M-1)/2``(-M+3)/2`---`3``2``M``0`
`A_1``1``0``1``0``-1/2``1/2``2``1``1``0``-1`
`S_2``5``0``0``1``1/2``3/2``10/3``0``1``0``0`
`x_3``3``0``0``0``1/2``1/2``6``1``1``1``0`


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``9M-3``-M``0``(M-1)/2``(-M+3)/2`
`R_2``1``1``0``-1/2``1/2`
`R_3``5``0``1``1/2``3/2`
`R_4``3``0``0``1/2``1/2`


`R_2`(new)`= R_2`(old) `xx2`
`R_2`(old) = `1``1``0``-1/2`
`R_2`(new)`= R_2`(old) `xx2``2``2``0``-1`


`R_1`(new)`= R_1`(old) + `((M-3)/2) R_2`(new)
`R_1`(old) = `9M-3``-M``0``(M-1)/2`
`R_2`(new) = `2``2``0``-1`
`(M-3)/2 xx R_2`(new) = `M-3``M-3``0``(-M+3)/2`
`R_1`(new)`= R_1`(old) + `((M-3)/2) R_2`(new)`10M-6``-3``0``1`


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


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


The improved solution is
Basis Inverse `B_1^(-1)`Additional table
`B``X_B``beta_0`
`Z'`		`beta_1`
`x_2`		`beta_2`
`S_2`		`beta_3`
`x_3`		`y_2`
 		Min Ratio
`(X_B)/(y_2)`		`x_1``A_1``A_2``S_1`
`Z'``10M-6``1``-3``0``1`---`3``M``M``0`
`x_2``2``0``2``0``-1`---`1``1``0``-1`
`S_2``2``0``-3``1``2`---`0``0``0``0`
`x_3``2``0``-1``0``1`---`1``0``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,-3,0,1]] [[3,M,M,0],[1,1,0,-1],[0,0,0,0],[1,0,1,0]]}`

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

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.
Any bug, improvement, feedback then Submit Here


8. Standard form-2 using Big M method : Example-2
(Previous example)


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