We use cookies to improve your experience on our site and to show you relevant advertising. By browsing this website, you agree to our use of cookies. Learn more   

AtoZmath.com
Support us
(New) All problem can be solved using search box
I want to sell my website www.AtoZmath.com with complete code
Home What's new College Algebra Games Feedback About us
Algebra Matrix & Vector Numerical Methods Statistical Methods Operation Research Word Problems Calculus Geometry Pre-Algebra


Home > Operation Research calculators > Simplex method example
2. Simplex method example ( Enter your problem )
( Enter your problem )
  1. Structure of Linear programming problem
  2. Algorithm
  3. Maximization example-1
  4. Maximization example-2
  5. Maximization example-3
  6. BigM method Algorithm
  7. Minimization example-1
  8. Minimization example-2
  9. Minimization example-3
  10. Degeneracy example-1 (Tie for leaving basic variable)
  11. Degeneracy example-2 (Tie first Artificial variable removed)
  12. Unrestricted variable example
  13. Multiple optimal solution example
  14. Infeasible solution example
  15. Unbounded solution example
 		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
11. Degeneracy example-2 (Tie first Artificial variable removed)
(Previous example)		13. Multiple optimal solution example
(Next example)

12. 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`



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


11. Degeneracy example-2 (Tie first Artificial variable removed)
(Previous example)		13. Multiple optimal solution example
(Next example)


Share this solution or page with your friends.


Home What's new College Algebra Games Feedback About us
 
Copyright © 2024. All rights reserved. Terms, Privacy
 
 

.