2b gauss simplek dasar

8/2/2019 2b Gauss Simplek Dasar

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GOOD MORNING CLASS!

In Operation Research Class,WE MEET AGAINWITH A TOPIC OF :


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LINEAR PROGRAMMING

THE SIMPLEX METHOD :

GAUSSIAN ELIMINATION

SETTING UP THE INITIAL SOLUTION

DEVELOPING THE SECOND SOLUTION DEVELOPING THE THIRD SOLUTION


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Gaussian Elimination :

IS CHANGING :

5 4 -3

2 -4 3

4 -3 1

X

Y

Z

230

120

140= TO

1 0 0

0 1 0

0 0 1

X

Y

Z

50

40

60

=


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HOW TO CHANGE ?5 X+ 4 Y+ -3 Z = 230

2 X+ -4 Y+ 3 Z = 120

4 X+ -3 Y+ 1 Z = 140

5 4 -3 = 230 5

2 -4 3 = 120 2

4 -3 1 = 140 4

1 0,8 -0,6 = 46 0,8

0 -5,6 4,2 = 28 -5,6

0 -6,2 3,4 = -44 -6,2

1 0 0 = 50 0

0 1 -0,8 = -5 -0,75

0 0 -1,3 = -75 -1,25

1 0 0 = 50

0 1 0 = 40

0 0 1 = 60
http://gauss.xls/http://gauss.xls/http://gauss.xls/


5/14

LP as Three Stage Process :

Problem Formulation

Problem Solution

Solution Interpretation and Implementation

Maximize : Profit = 8 T + 6 C

Subject to the constraint :

4 T + 2 C = 0

2 T + 4 C = 0


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Problem Formulation

Maximize : Profit = 8 T + 6 C + 0 S1 + 0 S2

Subject to :

4 T + 2 C + 1 S1 + 0 S2 = 60

2 T + 4 C + 0 S1 + 1 S2 = 48

All variables >= 0


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Parts of the Simplex Tableau

Cj Product 8 6 0 0 Non-Negative

Mix Quantity T C S1 S2 Ratio

0 S1 60 4 2 1 0 150 S2 48 2 4 0 1 24

Cj column (profits per unit)

Productmix column

Constant Column (quantities

of product in the mix)

Variable columns

Real Product Slack Time

Cj row

Variable

row


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The Initial Simplex Tableau

Cj Product 8 6 0 0 Non-Negative

Mix Quantity T C S1 S2 Ratio

0 S1 60 4 2 1 0 15

0 S2 48 2 4 0 1 24

Zj 0 0 0 0 0

Cj - Zj 8 6 0 0

Pivot Point


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The Replacing Row & Second Simplex

Cj Product 8 6 0 0Mix Quantity T C S1 S2

8 T 15 1 0,5 0,25 0

0 S2 48 2 4 0 1

Zj

Cj - Zj

Cj Product 8 6,0 0,0 0,0

Mix Quantity T C S1 S2

8 T 15 1 0,5 0,3 0,0

0 S2 18 0 3,0 -0,5 1,0

Zj 120 8 4,0 2,0 0,0

Cj - Zj 0 2,0 -2,0 0,0


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Replacing Row of the Third Tableau

Cj Product 8 6,0 0,0 0,0 Non-NegativeMix Quantity T C S1 S2 Ratio

8 T 15 1 0,5 0,3 0,0 30

0 S2 18 0 3,0 -0,5 1,0 6

Zj 120 8 4,0 2,0 0,0

Cj - Zj 0 2,0 -2,0 0,0

Cj Product 8.0 6.0 0.0 0.0

Mix Quantity T C S1 S2

8 T 12 1.0 0.0 0.3 -0.2

6 C 6 0.0 1.0 -0.2 0.3Zj 132 8.0 6.0 1.7 0.7

Cj - Zj 0.0 0.0 -1.7 -0.7


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Summary Of Step InThe Simplex Maximization Procedure

Set up the inequalities describing the problem constraint

Convert the inequalities to equation by adding slack variables

Enter the equation in the simplex table

Calculate the Zj and Cj Zj values for this solution

Determine the entering variable (optimal column) by choosingthe one with highest Cj Zj value

Determine row to be replaced by dividing quantity columnvalues by their corresponding optimal column values andchoosing the smallest non negative ratio

Compute the values for the replacing rows

Compute the values for the remaining rows

Calculate Zj and Cj Zj values for this solution

If there is a positive Cj Zj value return to step 5.

If there is no positive Cj Zj value , the optimal solution hasbeen obtained


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Exercise for you :

The Tekno Fertilizer Company makes two types of fertilizerwhich are manufactured in two departments. Type Acontribution $3 per ton, and type B contributes $4 per ton

Department

Hours per ton Maximum hours

Type A Type B worked per week

1 2 3 40

2 3 3 75

Set up a linear programming problem to determine howmuch of the two fertilizer to make in order to maximizeprofits. Use simplex algorithm to solve your problem.


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THANK YOU !!!

SEE YOU NEXT WEEK

!!


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SOLUTION

Maximize : Profit = $3 A + $4 B

Subject to constraint :

2A + 3B

2b gauss simplek dasar

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