Using Spreadsheets for Linear Programming with The Simplex Method A sample problem By: Jeffrey Bivin...
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Using Spreadsheets for Linear Programming with The Simplex Method A sample problem By: Jeffrey Bivin Lake Zurich High School Last Updated: October 11, 2005
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Transcript of Using Spreadsheets for Linear Programming with The Simplex Method A sample problem By: Jeffrey Bivin...
Using Spreadsheets for
Linear Programming with
The Simplex Method
A sample problem
By: Jeffrey Bivin
Lake Zurich High School
Last Updated: October 11, 2005
A landscaper can buy three types of 100-pound bags of fertilizer, type A, type B, and type C. Each 100-pound bag of type A fertilizer costs $20 and contains 40 pounds of nitrogen, 30 pounds of phosphoric acid, and 10 pounds of potash. Each 100-pound bag of type B fertilizer costs $30 and contains 20 pounds of nitrogen, 20 pounds of phosphoric acid, and 55 pounds of potash. Each 100-pound bag of type C fertilizer costs $20 and contains no nitrogen, 30 pounds of phosphoric acid, and 40 pounds of potash. The landscaper requires 4000 pounds of nitrogen. 2000 pounds of phosphoric acid, and 2000 pounds of potash. How many bags of each type of fertilizer should the landscaper buy in order to minimize the cost? Also, find the minimum cost.
Jeff Bivin -- LZHS
Analyzing the problem
NitrogenPhos. Acid potash
Type A x1 $20 40 30 10
Type B x2 $30 20 20 55
Type C x3 $20 0 30 40
needs 4000 2000 2000
Jeff Bivin -- LZHS
Define the Constraints
40x1 + 20x2 > 4000
30x1 + 20x2 + 30x3 > 2000
10x1 + 55x2 + 40x3 > 2000
x1 > 0
x2 > 0
x3 > 0
Jeff Bivin -- LZHS
Function to minimize
z = 20x1 + 30x2 + 20x3
Jeff Bivin -- LZHS
Initial Tableau
x1 x2 x3 s1 s2 s3 z Values
40 20 0 -1 0 0 0 4000
30 20 30 0 -1 0 0 2000
10 55 40 0 0 -1 0 2000
20 30 20 0 0 0 1 0
Jeff Bivin -- LZHS
Initial Tableau Analysis
Jeff Bivin -- LZHS
A negative basic variable in s1
leads us to use column x1 as the pivot column.
x1 x2 x3 s1 s2 s3 z Values
40 20 0 -1 0 0 0 4000
30 20 30 0 -1 0 0 2000
10 55 40 0 0 -1 0 2000
20 30 20 0 0 0 1 0
Jeff Bivin -- LZHS
A negative basic variable in s1
leads us to use column x1 as the pivot column.
x1 x2 x3 s1 s2 s3 z Values
40 20 0 -1 0 0 0 4000
30 20 30 0 -1 0 0 2000
10 55 40 0 0 -1 0 2000
20 30 20 0 0 0 1 0
s1 is a basic variable because it has only one non-zero element in the
columnJeff Bivin -- LZHS
A negative basic variable in s1
leads us to use column x1 as the pivot column.
x1 x2 x3 s1 s2 s3 z Values
40 20 0 -1 0 0 0 4000
30 20 30 0 -1 0 0 2000
10 55 40 0 0 -1 0 2000
20 30 20 0 0 0 1 0
We use Column x1 as the pivot column because the 40 in column x1 is the left most positive number in the row with the
-1 value (negative basic variable).
Jeff Bivin -- LZHS
Divide the value column by the pivot column.This adds an additional column at the right.
x1 x2 x3 s1 s2 s3 z Values
40 20 0 -1 0 0 0 4000 100
30 20 30 0 -1 0 0 2000 66.667
10 55 40 0 0 -1 0 2000 200
20 30 20 0 0 0 1 0
Jeff Bivin -- LZHS
66.667 is the lowest non-negative quotient.
x1 x2 x3 s1 s2 s3 z Values
40 20 0 -1 0 0 0 4000 100
30 20 30 0 -1 0 0 2000 66.667
10 55 40 0 0 -1 0 2000 200
20 30 20 0 0 0 1 0
Jeff Bivin -- LZHS
66.667 is the lowest non-negative quotient.Therefore, 30 is determined to be the pivot point.
x1 x2 x3 s1 s2 s3 z Values
40 20 0 -1 0 0 0 4000 100
30 20 30 0 -1 0 0 2000 66.667
10 55 40 0 0 -1 0 2000 200
20 30 20 0 0 0 1 0
Jeff Bivin -- LZHS
Determine the row operations:row1 = 3 x row1 – 4 x row2
row3 = 3 x row3 – row2row4 = 3 x row4 – 2 x row2
x1 x2 x3 s1 s2 s3 z Values
40 20 0 -1 0 0 0 4000 100
30 20 30 0 -1 0 0 2000 66.667
10 55 40 0 0 -1 0 2000 200
20 30 20 0 0 0 1 0
Jeff Bivin -- LZHS
Perform the row operations to determine Tableau 2
x1 x2 x3 s1 s2 s3 z Values
r1=3r1-4r2 0 -20 -120 -3 4 0 0 4000
r2 30 20 30 0 -1 0 0 2000
r3=3r3-r2 0 145 90 0 1 -3 0 4000
r4=3r4-2r2 0 50 0 0 2 0 3 -4000
Jeff Bivin -- LZHS
2nd Tableau Analysis
Jeff Bivin -- LZHS
A negative basic variable in s1
leads us to use column s2 as the pivot column.
x1 x2 x3 s1 s2 s3 z Values
0 -20 -120 -3 4 0 0 4000
30 20 30 0 -1 0 0 2000
0 145 90 0 1 -3 0 4000
0 50 0 0 2 0 3 -4000
Jeff Bivin -- LZHS
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 -20 -120 -3 4 0 0 4000 1000
30 20 30 0 -1 0 0 2000 -2000
0 145 90 0 1 -3 0 4000 4000
0 50 0 0 2 0 3 -4000
Jeff Bivin -- LZHS
1000 is the lowest non-negative quotient.
x1 x2 x3 s1 s2 s3 z Values
0 -20 -120 -3 4 0 0 4000 1000
30 20 30 0 -1 0 0 2000 -2000
0 145 90 0 1 -3 0 4000 4000
0 50 0 0 2 0 3 -4000
Jeff Bivin -- LZHS
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 -20 -120 -3 4 0 0 4000 1000
30 20 30 0 -1 0 0 2000 -2000
0 145 90 0 1 -3 0 4000 4000
0 50 0 0 2 0 3 -4000
Jeff Bivin -- LZHS
Determine the row operations:row2 = 4 x row2 + row1row3 = 4 x row3 – row1row4 = 2 x row4 – row1
x1 x2 x3 s1 s2 s3 z Values
0 -20 -120 -3 4 0 0 4000
30 20 30 0 -1 0 0 2000
0 145 90 0 1 -3 0 4000
0 50 0 0 2 0 3 -4000
Jeff Bivin -- LZHS
Perform the row operations to determine Tableau 3
x1 x2 x3 s1 s2 s3 z Values
r1 0 -20 -120 -3 4 0 0 4000
r2=4r2+r1 120 60 0 -3 0 0 0 12000
r3=4r3-r1 0 600 480 3 0 -12 0 12000
r4=2r4-r1 0 120 120 3 0 0 6 -12000
Jeff Bivin -- LZHS
3rd Tableau Analysis
Jeff Bivin -- LZHS
A negative basic variable in s3
leads us to use column x2 as the pivot column.
x1 x2 x3 s1 s2 s3 z Values
r1 0 -20 -120 -3 4 0 0 4000
r2=4r2+r1 120 60 0 -3 0 0 0 12000
r3=4r3-r1 0 600 480 3 0 -12 0 12000
r4=2r4-r1 0 120 120 3 0 0 6 -12000
Jeff Bivin -- LZHS
Divide the value column by the pivot column.This adds an additional column at the right.
x1 x2 x3 s1 s2 s3 z Values
r1 0 -20 -120 -3 4 0 0 4000 -200
r2=4r2+r1 120 60 0 -3 0 0 0 12000 200
r3=4r3-r1 0 600 480 3 0 -12 0 12000 20
r4=2r4-r1 0 120 120 3 0 0 6 -12000
Jeff Bivin -- LZHS
66.667 is the lowest non-negative quotient.
x1 x2 x3 s1 s2 s3 z Values
r1 0 -20 -120 -3 4 0 0 4000 -200
r2=4r2+r1 120 60 0 -3 0 0 0 12000 200
r3=4r3-r1 0 600 480 3 0 -12 0 12000 20
r4=2r4-r1 0 120 120 3 0 0 6 -12000
Jeff Bivin -- LZHS
66.667 is the lowest non-negative quotient.Therefore, 30 is determined to be the pivot point.
x1 x2 x3 s1 s2 s3 z Values
r1 0 -20 -120 -3 4 0 0 4000 -200
r2=4r2+r1 120 60 0 -3 0 0 0 12000 200
r3=4r3-r1 0 600 480 3 0 -12 0 12000 20
r4=2r4-r1 0 120 120 3 0 0 6 -12000
Jeff Bivin -- LZHS
Determine the row operations:row1 = 30 x row1 + row3row2 = 10 x row2 – row3row4 = 5 x row4 – row3
x1 x2 x3 s1 s2 s3 z Values
0 -20 -120 -3 4 0 0 4000
120 60 0 -3 0 0 0 12000
0 600 480 3 0 -12 0 12000
0 120 120 3 0 0 6 -12000
Jeff Bivin -- LZHS
Perform the row operations to determine Tableau 4
x1 x2 x3 s1 s2 s3 z Values
r1=30r1+r3 0 0 -3120 -87 120 -12 0 132000
r2=10r2-r3 1200 0 -480 -33 0 12 0 108000
r3 0 600 480 3 0 -12 0 12000
r4=52r4-r3 0 0 120 12 0 12 30 -72000
Jeff Bivin -- LZHS
4th Tableau Analysis
Jeff Bivin -- LZHS
Now we can determine the solutions
x1 x2 x3 s1 s2 s3 z Values
0 0 -3120 -87 120 -12 0 132000
1200 0 -480 -33 0 12 0 108000
0 600 480 3 0 -12 0 12000
0 0 120 12 0 12 30 -72000
Basic Variables x1, x2, and s2 are all positive and the bottom row has all
positive variable coefficientsJeff Bivin -- LZHS
Determining the solutions:from row 1: s2 = 132000 / 120 = 1100from row 2: x1 = 108000 / 1200 = 90
from row 3: x2 = 12000 / 600 = 20from row 4: z = -(-72000 / 30) = 2400
also: x3 = s1 = s3 = 0
x1 x2 x3 s1 s2 s3 z Values
r1=30r1+r3 0 0 -3120 -87 120 -12 0 132000
r2=10r2-r3 1200 0 -480 -33 0 12 0 108000
r3 0 600 480 3 0 -12 0 12000
r4=52r4-r3 0 0 120 12 0 12 30 -72000
Jeff Bivin -- LZHS
Results
x1 = 90
x2 = 20
x3 = 0
s1 = 0
s2 = 1100
s3 = 0
z = 2400
Jeff Bivin -- LZHS
Process CompleteAnswer the Question
• The minimum cost is $2400 when you order • 90 bags of type A fertilizer • 20 bags of type B fertilizer• 0 bags of type C fertilizer
Jeff Bivin -- LZHS
A Quick Recap
Jeff Bivin -- LZHS
Initial Tableau
x1 x2 x3 s1 s2 s3 z Values
40 20 0 -1 0 0 0 4000 100
30 20 30 0 -1 0 0 2000 66.667
10 55 40 0 0 -1 0 2000 200
20 30 20 0 0 0 1 0
Jeff Bivin -- LZHS
Perform the row operations to determine Tableau 2
x1 x2 x3 s1 s2 s3 z Values
r1=3r1-4r2 0 -20 -120 -3 4 0 0 4000 1000
r2 30 20 30 0 -1 0 0 2000 -2000
r3=3r3-r2 0 145 90 0 1 -3 0 4000 4000
r4=3r4-2r2 0 50 0 0 2 0 3 -4000
Jeff Bivin -- LZHS
Perform the row operations to determine Tableau 3
x1 x2 x3 s1 s2 s3 z Values
r1 0 -20 -120 -3 4 0 0 4000 -200
r2=4r2+r1 120 60 0 -3 0 0 0 12000 200
r3=4r3-r1 0 600 480 3 0 -12 0 12000 20
r4=2r4-r1 0 120 120 3 0 0 6 -12000
Jeff Bivin -- LZHS
Perform the row operations to determine Tableau 4
x1 x2 x3 s1 s2 s3 z Values
r1=30r1+r3 0 0 -3120 -87 120 -12 0 132000
r2=10r2-r3 1200 0 -480 -33 0 12 0 108000
r3 0 600 480 3 0 -12 0 12000
r4=52r4-r3 0 0 120 12 0 12 30 -72000
Jeff Bivin -- LZHS
x1 x2 x3 s1 s2 s3 z Values
r1=30r1+r3 0 0 -3120 -87 120 -12 0 132000
r2=10r2-r3 1200 0 -480 -33 0 12 0 108000
r3 0 600 480 3 0 -12 0 12000
r4=52r4-r3 0 0 120 12 0 12 30 -72000
Determine the solutions from the 4th Tableau
from row 1: s2 = 132000 / 120 = 1100from row 2: x1 = 108000 / 1200 = 90
from row 3: x2 = 12000 / 600 = 20from row 4: z = -(-72000 / 30) = 2400
also: x3 = s1 = s3 = 0
Jeff Bivin -- LZHS
Process CompleteAnswer the Question
• The minimum cost is $2400 when you order • 90 bags of type A fertilizer • 20 bags of type B fertilizer• 0 bags of type C fertilizer
Jeff Bivin -- LZHS
