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

Jeff Bivin -- LZHS
3
Re-writing the Constraints

• x1 2x2 x3 lt 4
• x1 x2 gt 1 x3 gt 2

x1 2x2 s1 4
Jeff Bivin -- LZHS
4
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
9
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
Jeff Bivin -- LZHS
13
Function to minimize

• w -z 2x1 x2 x3

Re-writing
2x1 x2 x3 z 0
Jeff Bivin -- LZHS
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Initial Tableau
x1 x2 x3 s1 s2 s3 -z Values
1 2 1 1 0 0 0 4
1 1 0 0 -1 0 0 1
0 0 1 0 0 -1 0 2
2 1 1 0 0 0 1 0
Jeff Bivin -- LZHS
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Initial Tableau
contraints
x1 x2 x3 s1 s2 s3 -z Values
1 2 1 1 0 0 0 4
1 1 0 0 -1 0 0 1
0 0 1 0 0 -1 0 2
2 1 1 0 0 0 1 0
Jeff Bivin -- LZHS
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Initial Tableau
x1 x2 x3 s1 s2 s3 -z Values
1 2 1 1 0 0 0 4
1 1 0 0 -1 0 0 1
0 0 1 0 0 -1 0 2
2 1 1 0 0 0 1 0
function
Jeff Bivin -- LZHS
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Initial Tableau
x1 x2 x3 s1 s2 s3 -z Values
1 2 1 1 0 0 0 4
1 1 0 0 -1 0 0 1
0 0 1 0 0 -1 0 2
2 1 1 0 0 0 1 0
Jeff Bivin -- LZHS
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Initial Tableau Analysis
Jeff Bivin -- LZHS
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A negative basic variable in s2leads us to use
column x1 as the pivot column.
x1 x2 x3 s1 s2 s3 -z Values
1 2 1 1 0 0 0 4
1 1 0 0 -1 0 0 1
0 0 1 0 0 -1 0 2
2 1 1 0 0 0 1 0
Jeff Bivin -- LZHS
20
A negative basic variable in s2leads us to use
column x1 as the pivot column.
x1 x2 x3 s1 s2 s3 -z Values
1 2 1 1 0 0 0 4
1 1 0 0 -1 0 0 1
0 0 1 0 0 -1 0 2
2 1 1 0 0 0 1 0
s2 is a basic variable because it has only one
non-zero element in the column
Jeff Bivin -- LZHS
21
A negative basic variable in s1leads us to use
column x1 as the pivot column.
x1 x2 x3 s1 s2 s3 -z Values
1 2 1 1 0 0 0 4
1 1 0 0 -1 0 0 1
0 0 1 0 0 -1 0 2
2 1 1 0 0 0 1 0
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
22
Divide the value column by the pivot column.This
adds an additional column at the right.
x1 x2 x3 s1 s2 s3 -z Values
1 2 1 1 0 0 0 4 4
1 1 0 0 -1 0 0 1 1
0 0 1 0 0 -1 0 2 ??
2 1 1 0 0 0 1 0
Jeff Bivin -- LZHS
23
1 is the lowest non-negative quotient.
x1 x2 x3 s1 s2 s3 -z Values
1 2 1 1 0 0 0 4 4
1 1 0 0 -1 0 0 1 1
0 0 1 0 0 -1 0 2 ??
2 1 1 0 0 0 1 0
Jeff Bivin -- LZHS
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1 is the lowest non-negative quotient.Therefore,
the 1 in column x1 is determined to be the
pivot point.
x1 x2 x3 s1 s2 s3 -z Values
1 2 1 1 0 0 0 4 4
1 1 0 0 -1 0 0 1 1
0 0 1 0 0 -1 0 2 ??
2 1 1 0 0 0 1 0
Jeff Bivin -- LZHS
25
Determine the row operationsrow1 row1
row2row2 row2row3 row3row4 row4 2 x
row2
x1 x2 x3 s1 s2 s3 -z Values
1 2 1 1 0 0 0 4 4
1 1 0 0 -1 0 0 1 1
0 0 1 0 0 -1 0 2 ??
2 1 1 0 0 0 1 0
Jeff Bivin -- LZHS
26
Perform the row operations to determine Tableau 2
x1 x2 x3 s1 s2 s3 -z Values
r1r1-r2 0 1 1 1 1 0 0 3
r2 r2 1 1 0 0 -1 0 0 1
r3r3 0 0 1 0 0 -1 0 2
r4r4-2r2 0 -1 1 0 2 0 1 -2
Jeff Bivin -- LZHS
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2nd Tableau Analysis
Jeff Bivin -- LZHS
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A negative basic variable in s3leads us to use
column x3 as the pivot column.
x1 x2 x3 s1 s2 s3 -z Values
0 1 1 1 1 0 0 3
1 1 0 0 -1 0 0 1
0 0 1 0 0 -1 0 2
0 -1 1 0 2 0 1 -2
Jeff Bivin -- LZHS
29
Divide the value column by the pivot
column.This adds an additional column at the
right.
x1 x2 x3 s1 s2 s3 -z Values
0 1 1 1 1 0 0 3 3
1 1 0 0 -1 0 0 1 ??
0 0 1 0 0 -1 0 2 2
0 -1 1 0 2 0 1 -2
Jeff Bivin -- LZHS
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2 is the lowest non-negative quotient.
x1 x2 x3 s1 s2 s3 -z Values
0 1 1 1 1 0 0 3 3
1 1 0 0 -1 0 0 1 ??
0 0 1 0 0 -1 0 2 2
0 -1 1 0 2 0 1 -2
Jeff Bivin -- LZHS
31
1000 is the lowest non-negative
quotient.Therefore, 4 is determined to be the
pivot point.
x1 x2 x3 s1 s2 s3 -z Values
0 1 1 1 1 0 0 3 3
1 1 0 0 -1 0 0 1 ??
0 0 1 0 0 -1 0 2 2
0 -1 1 0 2 0 1 -2
Jeff Bivin -- LZHS
32
Determine the row operationsrow1 row1
row3row2 row2row3 row3 row4 row4 row3
x1 x2 x3 s1 s2 s3 -z Values
0 1 1 1 1 0 0 3
1 1 0 0 -1 0 0 1
0 0 1 0 0 -1 0 2
0 -1 1 0 2 0 1 -2
Jeff Bivin -- LZHS
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Perform the row operations to determine Tableau 3
x1 x2 x3 s1 s2 s3 -z Values
r1r1-r3 0 1 0 1 1 1 0 1
r2r2 1 1 0 0 -1 0 0 1
r3r3 0 0 1 0 0 -1 0 2
r4r4-r3 0 -1 0 0 2 1 1 -4
Jeff Bivin -- LZHS
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3rd Tableau Analysis
Jeff Bivin -- LZHS
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All basic variable are positive.But, we have a
negative value in the bottom row.
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1
1 1 0 0 -1 0 0 1
0 0 1 0 0 -1 0 2
0 -1 0 0 2 1 1 -4
Jeff Bivin -- LZHS
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This leads us to use column x2 as the pivot
column.
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1
1 1 0 0 -1 0 0 1
0 0 1 0 0 -1 0 2
0 -1 0 0 2 1 1 -4
Jeff Bivin -- LZHS
37
Divide the value column by the pivot
column.This adds an additional column at the
right.
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1 1
1 1 0 0 -1 0 0 1 1
0 0 1 0 0 -1 0 2 ??
0 -1 0 0 2 1 1 -4
Jeff Bivin -- LZHS
38
We have two lowest non-negative quotients.
Therefore, we may choose which to use.
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1 1
1 1 0 0 -1 0 0 1 1
0 0 1 0 0 -1 0 2 ??
0 -1 0 0 2 1 1 -4
Jeff Bivin -- LZHS
39
We have two lowest non-negative quotients.
Therefore, we may choose which to use. Lets
use the first row.
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1 1
1 1 0 0 -1 0 0 1 1
0 0 1 0 0 -1 0 2 ??
0 -1 0 0 2 1 1 -4
Jeff Bivin -- LZHS
40
Using the first row, 1 is determined to be the
pivot point.
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1 1
1 1 0 0 -1 0 0 1 1
0 0 1 0 0 -1 0 2 ??
0 -1 0 0 2 1 1 -4
Jeff Bivin -- LZHS
41
Determine the row operationsrow1 row1 row2
row2 row1row 3 row3row4 row4 row1
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1
1 1 0 0 -1 0 0 1
0 0 1 0 0 -1 0 2
0 -1 0 0 2 1 1 -4
Jeff Bivin -- LZHS
42
Perform the row operations to determine Tableau 4
x1 x2 x3 s1 s2 s3 -z Values
r1r1 0 1 0 1 1 1 0 1
r2r2-r1 1 0 0 -1 -2 -1 0 0
r3r3 0 0 1 0 0 -1 0 2
r4r4r1 0 0 0 1 3 2 1 -3
Jeff Bivin -- LZHS
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4th Tableau Analysis
Jeff Bivin -- LZHS
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Basic Variables x1, x2, and x3 are all positive
and the bottom row has all positive variable
coefficients.
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1
1 0 0 -1 -2 -1 0 0
0 0 1 0 0 -1 0 2
0 0 0 1 3 2 1 -3
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Basic Variables x1, x2, and x3 are all positive
and the bottom row has all positive variable
coefficients.
Process Complete
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1
1 0 0 -1 -2 -1 0 0
0 0 1 0 0 -1 0 2
0 0 0 1 3 2 1 -3
Jeff Bivin -- LZHS
46
Determine the Solutions
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1
1 0 0 -1 -2 -1 0 0
0 0 1 0 0 -1 0 2
0 0 0 1 3 2 1 -3
Jeff Bivin -- LZHS
47
Determine the Solutions
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1
1 0 0 -1 -2 -1 0 0
0 0 1 0 0 -1 0 2
0 0 0 1 3 2 1 -3
x1 0 / 1 0
Jeff Bivin -- LZHS
48
Determine the Solutions
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1
1 0 0 -1 -2 -1 0 0
0 0 1 0 0 -1 0 2
0 0 0 1 3 2 1 -3
x2 1 / 1 1
Jeff Bivin -- LZHS
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Determine the Solutions
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1
1 0 0 -1 -2 -1 0 0
0 0 1 0 0 -1 0 2
0 0 0 1 3 2 1 -3
x3 2 / 1 2
Jeff Bivin -- LZHS
50
Determine the Solutions
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1
1 0 0 -1 -2 -1 0 0
0 0 1 0 0 -1 0 2
0 0 0 1 3 2 1 -3
s1 0
Non-basic Variables 0
Jeff Bivin -- LZHS
51
Determine the Solutions
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1
1 0 0 -1 -2 -1 0 0
0 0 1 0 0 -1 0 2
0 0 0 1 3 2 1 -3
s2 0
Non-basic Variables 0
Jeff Bivin -- LZHS
52
Determine the Solutions
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1
1 0 0 -1 -2 -1 0 0
0 0 1 0 0 -1 0 2
0 0 0 1 3 2 1 -3
s3 0
Non-basic Variables 0
Jeff Bivin -- LZHS
53
Determine the Solutions
x1 x2 x3 s1 s2 s3 -z Values
0 1 0 1 1 1 0 1
1 0 0 -1 -2 -1 0 0
0 0 1 0 0 -1 0 2
0 0 0 1 3 2 1 -3
z -(-3 / 1) 3
Jeff Bivin -- LZHS
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Results

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

Jeff Bivin -- LZHS