Hello. Need help with this practice question. 2. Use the Simplex Algorithm to solve the following 2-D Linear Programming Problem:Maximize x + 4y subject to:x + y < 8x -3y > 0x +3y < 10x > 0, y 2 0This LPP was solved geometrically in Homework III #3. Depict your elementary operations inthe usual manner (and in the usual place); don't forget to state clearly at the end the solutionto the LPP.

Given the LPP, Max x+4ysubject tox+y≤8x-3y≥0x+3y≤10x≥0,y≥0 Converting the problem into canonical…

Answered: Use the Simplex Algorithm to solve the…

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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Hello. Need help with this practice question.

2. Use the Simplex Algorithm to solve the following 2-D Linear Programming Problem:
Maximize x + 4y subject to:
x + y < 8
x -3y > 0
x +3y < 10
x > 0, y 2 0
This LPP was solved geometrically in Homework III #3. Depict your elementary operations in
the usual manner (and in the usual place); don't forget to state clearly at the end the solution
to the LPP.

Expert Solution

Step 1

$G i v e n t h e L P P, M a x x + 4 y subject to x + y \leq 8 x - 3 y \geq 0 x + 3 y \leq 10 x \geq 0, y \geq 0$

$Converting the problem into canonical form M a x z = x + 4 y + 0 s_{1} + 0 s_{2} + 0 s_{3} - M A_{1} subject to x + y + s_{1} = 8 x - 3 y - s_{2} + A_{1} = 0 x + 3 y + s_{3} = 10$

Step 2

Iteration-1		C_j	1	4	0	0	0	-M
B	C_B	X_B	x	y	s₁	s₂	s₃	A₁	Min ratio $\frac{x_{B}}{x}$
s₁	0	8	1	1	1	0	0	0	$\frac{8}{1} = 8$
A₁	-M	0	1	-3	0	-1	0	1	$\frac{0}{1} = 0 \to$
s₃	0	10	1	3	0	0	1	0	$\frac{10}{1} = 10$
z=0		z_j	-M	3M	0	M	0	-M
		z_j-c_j	-M-1 $↑$	3M-4	0	M	0	0

The most negative is -M-1 and hence the entering variable is x while the minimum ratio is 0 and hence the leaving variable is A₁.

We perform the following operations,

$R_{1} \to R_{1} - R_{2} R_{3} \to R_{3} - R_{2}$

Step 3

Iteration-2		C_j	1	4	0	0	0
B	C_B	X_B	x	y	s₁	s₂	s₃	Min ratio $\frac{x_{B}}{y}$
s₁	0	8	0	4	1	1	0	$\frac{8}{4} = 2$
x	1	0	1	-3	0	-1	0	$- - -$
s₃	0	10	0	6	0	1	1	$\frac{10}{6} = 1.667 \to$
z=0		z_j	1	-3	0	-1	0
		z_j-c_j	0	-7 $↑$	0	-1	0

The most negative is -7 and hence the entering variable is y while the least ratio is 1.667 and hence the leaving variable is s₃.

We perform the following operations,

$R_{3} \to \frac{R_{3}}{6} R_{1} \to R_{1} - 4 R_{3} R_{2} \to R_{2} + 3 R_{3}$

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