written 3.8 years ago by | modified 18 months ago by |

Using simplex method solve the following L.P.P

Max. $Z =10x_1 +x_2 +x_3$, subject to $x_1 +x_2 -3x_3≤ 10 ; 4x_1 +x_2 +x_3≤20 $

$x_1 , x_2 , x_3 ≥ 0 $

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Using simplex method solve the following L.P.P

written 3.8 years ago by | modified 18 months ago by |

Using simplex method solve the following L.P.P

Max. $Z =10x_1 +x_2 +x_3$, subject to $x_1 +x_2 -3x_3≤ 10 ; 4x_1 +x_2 +x_3≤20 $

$x_1 , x_2 , x_3 ≥ 0 $

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written 3.7 years ago by |

We express the given problem in standard form, by taking slack variables $s_1$ & $s_2$

$z – 10x_1 – x_2 – x_3 +0s_1 +0s_2$

$x_1 + x_2 – 3x_3 +s_1 +0s_2 =10$

$4x_1 + x_2 + x_3 +0s_1 + s_2 =20$

Now we form the simplex table as:

Iteration no. | Basic Variable | Coefficient of $x_1$ | Coefficient of $x_2$ | Coefficient of $x_3$ | Coefficient of $s_1$ | Coefficient of $s_2$ | R.H.S. | Ratio |
---|---|---|---|---|---|---|---|---|

0 | z | -10 | -1 | -1 | 0 | 0 | 0 | |

$s_2$ leaves | $s_1$ | 1 | 1 | -3 | 1 | 0 | 10 | 10/1 = 10 |

$x_1$ enters | $s_2$ | 4 | 3 | 0 | 0 | 1 | 20 | 20/4 = 5 |

1 | z | 0 | 3/2 | 13/2 | 0 | 0 | 0 | 50 |

$s_1$ | 0 | 3/4 | -15/4 | 1 | -1/4 | 10 | 5 | |

$x_1$ | 1 | 1/4 | 3/4 | 0 | 1/4 | 20 | 5 |

In simplex method:

(i) We see that which one is the most minimum value in the row of z,here it is - 10.

(ii) Corresponding column is called key column

(iii) Now we divide each element of key column by corresponding element of R.H.S ,and form a new table called ratio

(iv) The variable corresponding to key column is called incoming & the variable corresponding to key row is called outgoing

(v) Neglecting negative & infinite ratios ,we see that which one is the most minimum positive ratio , here it is 5,corresponding row is called key row

(vi) Element common to key row & key column is called key element ,it is 4

(vii) Now we form a new table ,& convert key element in to 1 ,by dividing each element key row by 4

(viii) Using key element ,by row transformation convert other elements of key column into zero

(ix) We do the same procedure till all the elements in the row of z becomes positive

(x) Then R.H.S. gives the solution

Therefore solution is $x_1 = 5$ ,$x_2=0$, $x_3 = 0$ & $z_{max}=50$

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