Simplex Method

1
Simplex Method

• Minimization

By Jeffrey Bivin Lake Zurich High School
Last Updated October 11, 2005
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The Constraints

• x1 2x2 x3 lt 4
• x1 x2 gt 1 x3 gt 2
• x1 gt 0
• x2 gt 0
• x3 gt 0

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3
Re-writing the Constraints

• x1 2x2 x3 lt 4
• x1 x2 gt 1 x3 gt 2

x1 2x2 s1 4
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Re-writing the Constraints

• x1 2x2 x3 lt 4
• x1 x2 gt 1 x3 gt 2

x1 2x2 x3 s1 4
Jeff Bivin -- LZHS
5
Re-writing the Constraints

• x1 2x2 x3 lt 4
• x1 x2 gt 1 x3 gt 2

x1 2x2 x3 s1 4
Jeff Bivin -- LZHS
6
Re-writing the Constraints

• x1 2x2 x3 lt 4
• x1 x2 gt 1 x3 gt 2

x1 2x2 x3 s1 4
Jeff Bivin -- LZHS
7
Re-writing the Constraints

• x1 2x2 x3 lt 4
• x1 x2 gt 1 x3 gt 2

x1 2x2 x3 s1 4 x1 x2 s2 1
Jeff Bivin -- LZHS
8
Re-writing the Constraints

• x1 2x2 x3 lt 4
• x1 x2 gt 1 x3 gt 2

x1 2x2 x3 s1 4 x1 x2 s2 1
Jeff Bivin -- LZHS
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Re-writing the Constraints

• x1 2x2 x3 lt 4
• x1 x2 gt 1 x3 gt 2

x1 2x2 x3 s1 4 x1 x2 s2 1 x2 s3
2
Jeff Bivin -- LZHS
10
Re-writing the Constraints

• x1 2x2 x3 lt 4
• x1 x2 gt 1 x3 gt 2

x1 2x2 x3 s1 4 x1 x2 s2 1 x3 s3
2
Jeff Bivin -- LZHS
11
Re-writing the Constraints

• x1 2x2 x3 lt 4
• x1 x2 gt 1 x3 gt 2

x1 2x2 x3 s1 4 x1 x2 s2 1 x3 s3
2
Jeff Bivin -- LZHS
12
Function to minimize

• w -z 2x1 x2 x3

We use a negative z because we want to minimize
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Function to minimize

• w -z 2x1 x2 x3

Re-writing
2x1 x2 x3 z 0
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Initial Tableau
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Initial Tableau
contraints
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Initial Tableau
function
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Initial Tableau
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Initial Tableau Analysis
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A negative basic variable in s2leads us to use
column x1 as the pivot column.
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A negative basic variable in s2leads us to use
column x1 as the pivot column.
s2 is a basic variable because it has only one
non-zero element in the column
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A negative basic variable in s1leads us to use
column x1 as the pivot column.
We use Column x1 as the pivot column because the
1 in column x1 is the left most positive number
in the row with the -1 value (negative basic
variable).
Jeff Bivin -- LZHS
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Divide the value column by the pivot column.This
adds an additional column at the right.
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1 is the lowest non-negative quotient.
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1 is the lowest non-negative quotient.Therefore,
the 1 in column x1 is determined to be the
pivot point.
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Determine the row operationsrow1 row1
row2row2 row2row3 row3row4 row4 2 x
row2
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Perform the row operations to determine Tableau 2
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2nd Tableau Analysis
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A negative basic variable in s3leads us to use
column x3 as the pivot column.
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Divide the value column by the pivot
column.This adds an additional column at the
right.
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2 is the lowest non-negative quotient.
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1000 is the lowest non-negative
quotient.Therefore, 4 is determined to be the
pivot point.
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Determine the row operationsrow1 row1
row3row2 row2row3 row3 row4 row4 row3
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Perform the row operations to determine Tableau 3
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3rd Tableau Analysis
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All basic variable are positive.But, we have a
negative value in the bottom row.
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This leads us to use column x2 as the pivot
column.
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Divide the value column by the pivot
column.This adds an additional column at the
right.
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We have two lowest non-negative quotients.
Therefore, we may choose which to use.
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We have two lowest non-negative quotients.
Therefore, we may choose which to use. Lets
use the first row.
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Using the first row, 1 is determined to be the
pivot point.
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Determine the row operationsrow1 row1 row2
row2 row1row 3 row3row4 row4 row1
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Perform the row operations to determine Tableau 4
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4th Tableau Analysis
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Basic Variables x1, x2, and x3 are all positive
and the bottom row has all positive variable
coefficients.
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Basic Variables x1, x2, and x3 are all positive
and the bottom row has all positive variable
coefficients.
Process Complete
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Determine the Solutions
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Determine the Solutions
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Determine the Solutions
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Determine the Solutions
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Determine the Solutions
Non-basic Variables 0
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Determine the Solutions
Non-basic Variables 0
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Determine the Solutions
Non-basic Variables 0
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Determine the Solutions
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Results

• x1 0
• x2 1
• x3 2
• s1 0
• s2 0
• s3 0
• z 3

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