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1. Simplex (BigM) method example ( Enter your problem )
  1. Simplex method Algorithm
  2. BigM method Algorithm
  3. Maximization example
  4. Minimization example
  5. Degeneracy example-1 (Tie for leaving basic variable)
  6. Degeneracy example-2 (Tie - first Artificial variable removed)
  7. Unrestricted variable example
  8. Multiple optimal solution example
  9. Unbounded solution example
  10. Infeasible solution example
Other related methods
  1. Simplex (BigM) method
  2. Two-Phase method
  3. Graphical method
  4. Primal to dual conversion
  5. Dual simplex method
  6. Integer simplex method
  7. Branch and Bound method
  8. 0-1 Integer programming problem

7. Unrestricted variable example


Unrestricted variable
Sometimes decision variables are positive, negative or zero values, then they are called unrestricted variables.
In all such cases, the decision variables can be expressed as the difference between two non-negative variables.

For example, if `x_r` is unrestricted in sign, then
`x_r = x_r' - x_r''` ; `x_r' - x_r'' >= 0`
Example
Find solution using Simplex(BigM) method
MAX Z = 3x1 + 2x2 + x3
subject to
2x1 + 5x2 + x3 = 12
3x1 + 4x2 = 11
and x2,x3 >= 0 and x1 unrestricted in sign


Solution:
Problem is
Max `Z``=````3``x_1`` + ``2``x_2`` + ````x_3`
subject to
```2``x_1`` + ``5``x_2`` + ````x_3`=`12`
```3``x_1`` + ``4``x_2`=`11`
and `x_2,x_3 >= 0; ``x_1` unrestricted in sign


Since `x_1` is unrestricted in sign, introduce the non-negative variables `x_1',x_1''`

so that `x_1=x_1'-x_1''; x_1',x_1''>=0`.

The standard form of the LP problem becomes
Problem is
Max `Z``=````3``x_1'`` - ``3``x_1''`` + ``2``x_2`` + ````x_3`
subject to
```2``x_1'`` - ``2``x_1''`` + ``5``x_2`` + ````x_3`=`12`
```3``x_1'`` - ``3``x_1''`` + ``4``x_2`=`11`
and `x_1',x_1'',x_2,x_3 >= 0; `


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

1. As the constraint-1 is of type '`=`' we should add artificial variable `A_1`

2. As the constraint-2 is of type '`=`' we should add artificial variable `A_2`

After introducing artificial variables
Max `Z``=````3``x_1'`` - ``3``x_1''`` + ``2``x_2`` + ````x_3`` - ``M``A_1`` - ``M``A_2`
subject to
```2``x_1'`` - ``2``x_1''`` + ``5``x_2`` + ````x_3`` + ````A_1`=`12`
```3``x_1'`` - ``3``x_1''`` + ``4``x_2`` + ````A_2`=`11`
and `x_1',x_1'',x_2,x_3,A_1,A_2 >= 0`


Iteration-1 `C_j``3``-3``2``1``-M``-M`
`B``C_B``X_B``x_1'``x_1''` `x_2` Entering variable`x_3``A_1``A_2`MinRatio
`(X_B)/(x_2)`
 `A_1` Leaving variable`-M``12``2``-2` `(5)`  (pivot element)`1``1``0``(12)/(5)=2.4``->`
`A_2``-M``11``3``-3``4``0``0``1``(11)/(4)=2.75`
 `Z=0` `0=`
`Z_j=sum C_B X_B`
 `Z_j` `Z_j=sum C_B x_j` `-5M` `-5M=(-M)xx2+(-M)xx3`
`Z_j=sum C_B x_1'`
 `5M` `5M=(-M)xx(-2)+(-M)xx(-3)`
`Z_j=sum C_B x_1''`
 `-9M` `-9M=(-M)xx5+(-M)xx4`
`Z_j=sum C_B x_2`
 `-M` `-M=(-M)xx1+(-M)xx0`
`Z_j=sum C_B x_3`
 `-M` `-M=(-M)xx1+(-M)xx0`
`Z_j=sum C_B A_1`
 `-M` `-M=(-M)xx0+(-M)xx1`
`Z_j=sum C_B A_2`
`C_j-Z_j` `5M+3` `5M+3=3-(-5M)` `-5M-3` `-5M-3=(-3)-5M` `9M+2` `9M+2=2-(-9M)``uarr` `M+1` `M+1=1-(-M)` `0` `0=(-M)-(-M)` `0` `0=(-M)-(-M)`


Positive maximum `C_j-Z_j` is `9M+2` and its column index is `3`. So, the entering variable is `x_2`.

Minimum ratio is `2.4` and its row index is `1`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `5`.

Entering `=x_2`, Departing `=A_1`, Key Element `=5`

`R_1`(new)`= R_1`(old)`-: 5`

`R_2`(new)`= R_2`(old)`- 4 R_1`(new)

Iteration-2 `C_j``3``-3``2``1``-M`
`B``C_B``X_B` `x_1'` Entering variable`x_1''``x_2``x_3``A_2`MinRatio
`(X_B)/(x_1')`
`x_2``2` `12/5` `12/5=12-:5`
`R_1`(new)`= R_1`(old)`-: 5`
 `2/5` `2/5=2-:5`
`R_1`(new)`= R_1`(old)`-: 5`
 `-2/5` `-2/5=(-2)-:5`
`R_1`(new)`= R_1`(old)`-: 5`
 `1` `1=5-:5`
`R_1`(new)`= R_1`(old)`-: 5`
 `1/5` `1/5=1-:5`
`R_1`(new)`= R_1`(old)`-: 5`
 `0` `0=0-:5`
`R_1`(new)`= R_1`(old)`-: 5`
`(12/5)/(2/5)=6`
 `A_2` Leaving variable`-M` `7/5` `7/5=11-4xx12/5`
`R_2`(new)`= R_2`(old)`- 4 R_1`(new)
 `(7/5)` `7/5=3-4xx2/5` (pivot element)
`R_2`(new)`= R_2`(old)`- 4 R_1`(new)
 `-7/5` `-7/5=(-3)-4xx(-2/5)`
`R_2`(new)`= R_2`(old)`- 4 R_1`(new)
 `0` `0=4-4xx1`
`R_2`(new)`= R_2`(old)`- 4 R_1`(new)
 `-4/5` `-4/5=0-4xx1/5`
`R_2`(new)`= R_2`(old)`- 4 R_1`(new)
 `1` `1=1-4xx0`
`R_2`(new)`= R_2`(old)`- 4 R_1`(new)
`(7/5)/(7/5)=1``->`
 `Z=24/5` `24/5=2xx12/5`
`Z_j=sum C_B X_B`
 `Z_j` `Z_j=sum C_B x_j` `-(7M)/(5)+4/5` `-(7M)/(5)+4/5=2xx2/5+(-M)xx7/5`
`Z_j=sum C_B x_1'`
 `(7M)/(5)-4/5` `(7M)/(5)-4/5=2xx(-2/5)+(-M)xx(-7/5)`
`Z_j=sum C_B x_1''`
 `2` `2=2xx1+(-M)xx0`
`Z_j=sum C_B x_2`
 `(4M)/(5)+2/5` `(4M)/(5)+2/5=2xx1/5+(-M)xx(-4/5)`
`Z_j=sum C_B x_3`
 `-M` `-M=2xx0+(-M)xx1`
`Z_j=sum C_B A_2`
`C_j-Z_j` `(7M)/(5)+11/5` `(7M)/(5)+11/5=3-(-(7M)/(5)+4/5)``uarr` `-(7M)/(5)-11/5` `-(7M)/(5)-11/5=(-3)-((7M)/(5)-4/5)` `0` `0=2-2` `-(4M)/(5)+3/5` `-(4M)/(5)+3/5=1-((4M)/(5)+2/5)` `0` `0=(-M)-(-M)`


Positive maximum `C_j-Z_j` is `(7M)/(5)+11/5` and its column index is `1`. So, the entering variable is `x_1'`.

Minimum ratio is `1` and its row index is `2`. So, the leaving basis variable is `A_2`.

`:.` The pivot element is `7/5`.

Entering `=x_1'`, Departing `=A_2`, Key Element `=7/5`

`R_2`(new)`= R_2`(old)`xx5/7`

`R_1`(new)`= R_1`(old)`- 2/5 R_2`(new)

Iteration-3 `C_j``3``-3``2``1`
`B``C_B``X_B``x_1'``x_1''``x_2` `x_3` Entering variableMinRatio
`(X_B)/(x_3)`
 `x_2` Leaving variable`2` `2` `2=12/5-2/5xx1`
`R_1`(new)`= R_1`(old)`- 2/5 R_2`(new)
 `0` `0=2/5-2/5xx1`
`R_1`(new)`= R_1`(old)`- 2/5 R_2`(new)
 `0` `0=(-2/5)-2/5xx(-1)`
`R_1`(new)`= R_1`(old)`- 2/5 R_2`(new)
 `1` `1=1-2/5xx0`
`R_1`(new)`= R_1`(old)`- 2/5 R_2`(new)
 `(3/7)` `3/7=1/5-2/5xx(-4/7)` (pivot element)
`R_1`(new)`= R_1`(old)`- 2/5 R_2`(new)
`(2)/(3/7)=4.6667``->`
`x_1'``3` `1` `1=7/5xx5/7`
`R_2`(new)`= R_2`(old)`xx5/7`
 `1` `1=7/5xx5/7`
`R_2`(new)`= R_2`(old)`xx5/7`
 `-1` `-1=(-7/5)xx5/7`
`R_2`(new)`= R_2`(old)`xx5/7`
 `0` `0=0xx5/7`
`R_2`(new)`= R_2`(old)`xx5/7`
 `-4/7` `-4/7=(-4/5)xx5/7`
`R_2`(new)`= R_2`(old)`xx5/7`
---
 `Z=7` `7=2xx2+3xx1`
`Z_j=sum C_B X_B`
 `Z_j` `Z_j=sum C_B x_j` `3` `3=2xx0+3xx1`
`Z_j=sum C_B x_1'`
 `-3` `-3=2xx0+3xx(-1)`
`Z_j=sum C_B x_1''`
 `2` `2=2xx1+3xx0`
`Z_j=sum C_B x_2`
 `-6/7` `-6/7=2xx3/7+3xx(-4/7)`
`Z_j=sum C_B x_3`
`C_j-Z_j` `0` `0=3-3` `0` `0=(-3)-(-3)` `0` `0=2-2` `13/7` `13/7=1-(-6/7)``uarr`


Positive maximum `C_j-Z_j` is `13/7` and its column index is `4`. So, the entering variable is `x_3`.

Minimum ratio is `4.6667` and its row index is `1`. So, the leaving basis variable is `x_2`.

`:.` The pivot element is `3/7`.

Entering `=x_3`, Departing `=x_2`, Key Element `=3/7`

`R_1`(new)`= R_1`(old)`xx7/3`

`R_2`(new)`= R_2`(old)`+ 4/7 R_1`(new)

Iteration-4 `C_j``3``-3``2``1`
`B``C_B``X_B``x_1'``x_1''``x_2``x_3`MinRatio
`x_3``1` `14/3` `14/3=2xx7/3`
`R_1`(new)`= R_1`(old)`xx7/3`
 `0` `0=0xx7/3`
`R_1`(new)`= R_1`(old)`xx7/3`
 `0` `0=0xx7/3`
`R_1`(new)`= R_1`(old)`xx7/3`
 `7/3` `7/3=1xx7/3`
`R_1`(new)`= R_1`(old)`xx7/3`
 `1` `1=3/7xx7/3`
`R_1`(new)`= R_1`(old)`xx7/3`
`x_1'``3` `11/3` `11/3=1+4/7xx14/3`
`R_2`(new)`= R_2`(old)`+ 4/7 R_1`(new)
 `1` `1=1+4/7xx0`
`R_2`(new)`= R_2`(old)`+ 4/7 R_1`(new)
 `-1` `-1=(-1)+4/7xx0`
`R_2`(new)`= R_2`(old)`+ 4/7 R_1`(new)
 `4/3` `4/3=0+4/7xx7/3`
`R_2`(new)`= R_2`(old)`+ 4/7 R_1`(new)
 `0` `0=(-4/7)+4/7xx1`
`R_2`(new)`= R_2`(old)`+ 4/7 R_1`(new)
 `Z=47/3` `47/3=1xx14/3+3xx11/3`
`Z_j=sum C_B X_B`
 `Z_j` `Z_j=sum C_B x_j` `3` `3=1xx0+3xx1`
`Z_j=sum C_B x_1'`
 `-3` `-3=1xx0+3xx(-1)`
`Z_j=sum C_B x_1''`
 `19/3` `19/3=1xx7/3+3xx4/3`
`Z_j=sum C_B x_2`
 `1` `1=1xx1+3xx0`
`Z_j=sum C_B x_3`
`C_j-Z_j` `0` `0=3-3` `0` `0=(-3)-(-3)` `-13/3` `-13/3=2-(19/3)` `0` `0=1-1`


Since all `C_j-Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1'=11/3,x_1''=0,x_2=0,x_3=14/3`

Max `Z = 47/3`




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