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 > Simplex method example
2. Simplex method example ( Enter your problem )
( Enter your problem )
  1. Structure of Linear programming problem
  2. Algorithm (using `Z`-row method)
  3. Maximization example-1 (using `Z`-row method)
  4. Maximization example-2 (using `Z`-row method)
  5. Maximization example-3 (using `Z`-row method)
  6. BigM method Algorithm (using `Z`-row method)
  7. Minimization example-1 (using `Z`-row method)
  8. Minimization example-2 (using `Z`-row method)
  9. Minimization example-3 (using `Z`-row method)
  10. Degeneracy example-1 (Tie for leaving basic variable) (using `Z`-row method)
  11. Degeneracy example-2 (Tie first Artificial variable removed) (using `Z`-row method)
  12. Unrestricted variable example (using `Z`-row method)
  13. Multiple optimal solution example (using `Z`-row method)
  14. Infeasible solution example (using `Z`-row method)
  15. Unbounded solution example (using `Z`-row method)
  16. Algorithm (using `Z_j-C_j` method)
  17. Maximization example-1 (using `Z_j-C_j` method)
  18. Maximization example-2 (using `Z_j-C_j` method)
  19. Maximization example-3 (using `Z_j-C_j` method)
  20. BigM method Algorithm (using `Z_j-C_j` method)
  21. Minimization example-1 (using `Z_j-C_j` method)
  22. Minimization example-2 (using `Z_j-C_j` method)
  23. Minimization example-3 (using `Z_j-C_j` method)
  24. Degeneracy example-1 (Tie for leaving basic variable) (using `Z_j-C_j` method)
  25. Degeneracy example-2 (Tie first Artificial variable removed) (using `Z_j-C_j` method)
  26. Unrestricted variable example (using `Z_j-C_j` method)
  27. Multiple optimal solution example (using `Z_j-C_j` method)
  28. Infeasible solution example (using `Z_j-C_j` method)
  29. Unbounded solution example (using `Z_j-C_j` method)
  30. Algorithm (using `C_j-Z_j`method)
  31. Maximization example-1 (using `C_j-Z_j`method)
  32. Maximization example-2 (using `C_j-Z_j`method)
  33. Maximization example-3 (using `C_j-Z_j`method)
  34. BigM method Algorithm (using `C_j-Z_j`method)
  35. Minimization example-1 (using `C_j-Z_j`method)
  36. Minimization example-2 (using `C_j-Z_j`method)
  37. Minimization example-3 (using `C_j-Z_j`method)
  38. Degeneracy example-1 (Tie for leaving basic variable) (using `C_j-Z_j`method)
  39. Degeneracy example-2 (Tie first Artificial variable removed) (using `C_j-Z_j`method)
  40. Unrestricted variable example (using `C_j-Z_j`method)
  41. Multiple optimal solution example (using `C_j-Z_j`method)
  42. Infeasible solution example (using `C_j-Z_j`method)
  43. Unbounded solution example (using `C_j-Z_j`method)
 		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
6. BigM method Algorithm (using `Z`-row method)
(Previous example)		8. Minimization example-2 (using `Z`-row method)
(Next example)

7. Minimization example-1 (using `Z`-row method)


Find solution using Simplex method (BigM method)
MIN Z = x1 + x2
subject to
2x1 + 4x2 >= 4
x1 + 7x2 >= 7
and x1,x2 >= 0;

Solution:
Problem is
Min `Z``=``````x_1`` + ````x_2`
subject to
```2``x_1`` + ``4``x_2`≥`4`
`````x_1`` + ``7``x_2`≥`7`
and `x_1,x_2 >= 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 subtract surplus variable `S_1` and add artificial variable `A_1`

2. As the constraint-2 is of type '`>=`' we should subtract surplus variable `S_2` and add artificial variable `A_2`

After introducing surplus,artificial variables
Min `Z``=``````x_1`` + ````x_2`` + ``0``S_1`` + ``0``S_2`` + ``M``A_1`` + ``M``A_2`
subject to
```2``x_1`` + ``4``x_2`` - ````S_1`` + ````A_1`=`4`
`````x_1`` + ``7``x_2`` - ````S_2`` + ````A_2`=`7`
and `x_1,x_2,S_1,S_2,A_1,A_2 >= 0`


`Z`` - ````x_1`` - ````x_2`` - ``M``A_1`` - ``M``A_2`=`0`
```2``x_1`` + ``4``x_2`` - ````S_1`` + ````A_1`=`4`
`````x_1`` + ``7``x_2`` - ````S_2`` + ````A_2`=`7`


Simplex tableau is
Tableau-0
`"Basis"``x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_1` `Z``-1``-1``0``0``-M``-M``0`
`R_2` `A_1``2``4``-1``0``1``0``4`
`R_3` `A_2``1``7``0``-1``0``1``7`


Make the Z-row consistent with the rest of the table (set coefficient of basis variables to 0 in Z-row)
`R_1`(new)`= R_1`(old) + `M R_2`(old)
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_1`(old) = `-1``-1``0``0``-M``-M``0`
`R_2`(old) = `2``4``-1``0``1``0``4`
`M xx R_2`(new) = `2M``4M``-M``0``M``0``4M`
`R_1`(new)`= R_1`(old) + `M R_2`(old)`2M-1``4M-1``-M``0``0``-M``4M`
`R_1`(new)`= R_1`(old) + `M R_3`(old)
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_1`(old) = `2M-1``4M-1``-M``0``0``-M``4M`
`R_3`(old) = `1``7``0``-1``0``1``7`
`M xx R_3`(new) = `M``7M``0``-M``0``M``7M`
`R_1`(new)`= R_1`(old) + `M R_3`(old)`3M-1``11M-1``-M``-M``0``0``11M`


Tableau-1
`"Basis"``x_1``x_2``darr``S_1``S_2``A_1``A_2``RHS``"Ratio"=(RHS)/(x_2)`
`R_1` `Z``3M-1``11M-1``-M``-M``0``0``11M`
`R_2` `A_1``2``(4)``-1``0``1``0``4``(4)/(4)=1``->`
`R_3` `A_2``1``7``0``-1``0``1``7``(7)/(7)=1`


Most Positive `Z` is `11M-1`. So, the entering variable is `x_2`.

Minimum ratio is `1`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `4`.

Entering `=x_2`, Departing `=A_1`, Key Element `=4`
`R_2`(new)`= R_2`(old) `-: 4`
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_2`(old) = `2``4``-1``0``1``0``4`
`R_2`(new)`= R_2`(old) `-: 4``0.5``1``-0.25``0``0.25``0``1`
`R_3`(new)`= R_3`(old) - `7 R_2`(new)
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_3`(old) = `1``7``0``-1``0``1``7`
`R_2`(new) = `0.5``1``-0.25``0``0.25``0``1`
`7 xx R_2`(new) = `3.5``7``-1.75``0``1.75``0``7`
`R_3`(new)`= R_3`(old) - `7 R_2`(new)`-2.5``0``1.75``-1``-1.75``1``0`
`R_1`(new)`= R_1`(old) - `(11M-1) R_2`(new)
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_1`(old) = `3M-1``11M-1``-M``-M``0``0``11M`
`R_2`(new) = `0.5``1``-0.25``0``0.25``0``1`
`(11M-1) xx R_2`(new) = `5.5M-0.5``11M-1``-2.75M+0.25``0``2.75M-0.25``0``11M-1`
`R_1`(new)`= R_1`(old) - `(11M-1) R_2`(new)`-2.5M-0.5``0``1.75M-0.25``-M``-2.75M+0.25``0``1`


Tableau-2
`"Basis"``x_1``x_2``S_1``darr``S_2``A_1``A_2``RHS``"Ratio"=(RHS)/(S_1)`
`R_1` `Z``-2.5M-0.5``0``1.75M-0.25``-M``-2.75M+0.25``0``1`
`R_2` `x_2``0.5``1``-0.25``0``0.25``0``1``(1)/(-0.25)` (ignore, denominator is -ve)
`R_3` `A_2``-2.5``0``(1.75)``-1``-1.75``1``0``(0)/(1.75)=0``->`


Most Positive `Z` is `1.75M-0.25`. So, the entering variable is `S_1`.

Minimum ratio is `0`. So, the leaving basis variable is `A_2`.

`:.` The pivot element is `1.75`.

Entering `=S_1`, Departing `=A_2`, Key Element `=1.75`
`R_3`(new)`= R_3`(old) `-: 1.75`
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_3`(old) = `-2.5``0``1.75``-1``-1.75``1``0`
`R_3`(new)`= R_3`(old) `-: 1.75``-1.4286``0``1``-0.5714``-1``0.5714``0`
`R_2`(new)`= R_2`(old) + `0.25 R_3`(new)
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_2`(old) = `0.5``1``-0.25``0``0.25``0``1`
`R_3`(new) = `-1.4286``0``1``-0.5714``-1``0.5714``0`
`0.25 xx R_3`(new) = `-0.3571``0``0.25``-0.1429``-0.25``0.1429``0`
`R_2`(new)`= R_2`(old) + `0.25 R_3`(new)`0.1429``1``0``-0.1429``0``0.1429``1`
`R_1`(new)`= R_1`(old) - `(1.75M-0.25) R_3`(new)
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_1`(old) = `-2.5M-0.5``0``1.75M-0.25``-M``-2.75M+0.25``0``1`
`R_3`(new) = `-1.4286``0``1``-0.5714``-1``0.5714``0`
`(1.75M-0.25) xx R_3`(new) = `-2.5M+0.3571``0``1.75M-0.25``-M+0.1429``-1.75M+0.25``M-0.1429``0`
`R_1`(new)`= R_1`(old) - `(1.75M-0.25) R_3`(new)`-0.8571``0``0``-0.1429``-M``-M+0.1429``1`


Tableau-3
`"Basis"``x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_1` `Z``-0.8571``0``0``-0.1429``-M``-M+0.1429``1`
`R_2` `x_2``0.1429``1``0``-0.1429``0``0.1429``1`
`R_3` `S_1``-1.4286``0``1``-0.5714``-1``0.5714``0`


Since all `Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=0,x_2=1`

Min `Z=1`


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


6. BigM method Algorithm (using `Z`-row method)
(Previous example)		8. Minimization example-2 (using `Z`-row method)
(Next example)


Share this solution or page with your friends.
 
 
Home What's new College Algebra Games Feedback About us
 
Copyright © 2026. All rights reserved. Terms, Privacy
 
 

.