Use the Big M method to solve the following LPs: 4 min z = 3x1s.t.2x1 + x2 2 63x1 + 2x2 = 4X1, X2 2 0

The given LPP problem, min Z=3x1 Subject to the constraints, 2x1+x2≥63x1+2x2=4x1,x2≥0 Using the Big…

Answered: 4 min z = 3x1 s.t. 2x1 + x2 2 6 3x1 +…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Use the Big M method to solve the following LPs:

4 min z = 3x1
s.t.
2x1 + x2 2 6
3x1 + 2x2 = 4
X1, X2 2 0

Expert Solution

Step 1

The given LPP problem,

$m i n Z = 3 x_{1}$

Subject to the constraints,

$2 x_{1} + x_{2} \geq 6 3 x_{1} + 2 x_{2} = 4 x_{1}, x_{2} \geq 0$

Using the Big M method to solve the LPP.

Step 2

The given LPP problem,

$m i n Z = 3 x_{1}$

Subject to the constraints,

$2 x_{1} + x_{2} \geq 6 3 x_{1} + 2 x_{2} = 4 x_{1}, x_{2} \geq 0$

The problem is converted into canonical form by adding slack, surplus, and artificial variables as appropriate.

1. As the constraint- $1$ is of type $' \geq'$ we should subtract the surplus variable $S_{1}$ and add the artificial variable $A_{1}$ .

2. As the constraint- $2$ is of type $' ='$ we should add the artificial variable $A_{2}$ .

After introducing surplus, artificial variables.

$m i n Z = 3 x_{1} + 0 x_{2} + 0 S_{1} + M A_{1} + M A_{2}$

Subject to the constraints,

$2 x_{1} + x_{2} - S_{1} + A_{1} = 6 3 x_{1} + 2 x_{2} + A_{2} = 4 x_{1}, x_{2}, S_{1}, A_{1}, A_{2} \geq 0$

Step 3

First iteration:

Iteration-1		$C_{j}$	3	0	0	M	M
B	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$S_{1}$	$A_{1}$	$A_{2}$	Min Ratio $\frac{X_{B}}{x_{1}}$
$A_{1}$	M	6	2	1	-1	1	0	$\frac{6}{2} = 3$
$A_{2}$	M	4	3	2	0	0	1	$\frac{4}{3} = 1.3333$
$z = 10 M$		$Z_{j}$	5M	3M	-M	M	M
		$Z_{j} - C_{j}$	5M-3	3M	-M	0	0

Positive maximum $Z_{j} - C_{j}$ is $5 M - 3$ . So the entering variable is $x_{1}$ .

Minimum ratio is $1.3333$ . So the leaving basis variable is $A_{2}$ .

Therefore the pivot element is $3$ .

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