Simplex Method Applications
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Simplex Method Applications
Transcript of Simplex Method Applications
Simplex Method Applications
A Business Application: Maximum Profit
ExampleA manufacturer produced three types of plastic fixtures. The timerequired for molding, trimming and packaging is given in theaccompanying table. Note: Times are given in hours per dozenfixtures.
Process Type A Type B Type C Total Time AvailableMolding 1 2 3
2 12,000Trimming 2
323 1 4,600
Packaging 12
13
12 2,400
Profit $11 $16 $15 —
How many dozens of each type of fixture should be produced toobtain a maximum profit?
A Business Application: Maximum Profit
First thing we need to do is assign variables.
Let x represent the number of dozens of type A fixtures.
Let y represent the number of dozens of type B fixtures.
Let z represent the number of dozens of type C fixtures.
What is our object function here?
M = 11x + 16y + 15z⇒ -11x-16y-15z+M=0
A Business Application: Maximum Profit
First thing we need to do is assign variables.
Let x represent the number of dozens of type A fixtures.
Let y represent the number of dozens of type B fixtures.
Let z represent the number of dozens of type C fixtures.
What is our object function here?
M = 11x + 16y + 15z⇒ -11x-16y-15z+M=0
A Business Application: Maximum Profit
First thing we need to do is assign variables.
Let x represent the number of dozens of type A fixtures.
Let y represent the number of dozens of type B fixtures.
Let z represent the number of dozens of type C fixtures.
What is our object function here?
M = 11x + 16y + 15z⇒ -11x-16y-15z+M=0
A Business Application: Maximum Profit
First thing we need to do is assign variables.
Let x represent the number of dozens of type A fixtures.
Let y represent the number of dozens of type B fixtures.
Let z represent the number of dozens of type C fixtures.
What is our object function here?
M = 11x + 16y + 15z⇒
-11x-16y-15z+M=0
A Business Application: Maximum Profit
First thing we need to do is assign variables.
Let x represent the number of dozens of type A fixtures.
Let y represent the number of dozens of type B fixtures.
Let z represent the number of dozens of type C fixtures.
What is our object function here?
M = 11x + 16y + 15z⇒ -11x-16y-15z+M=0
A Business Application: Maximum Profit
What is the system of constraints we get from this?
x + 2y + 3
2 z ≤ 1200023 x + 2
3 y + z ≤ 460012 x + 1
3 y + 12 z ≤ 2400
x ≥ 0, y ≥ 0, z ≥ 0
This becomes ...
x + 2y + 3
2 z + u = 1200023 x + 2
3 y + z + v = 460012 x + 1
3 y + 12 z + w = 2400
x ≥ 0, y ≥ 0, z ≥ 0
A Business Application: Maximum Profit
What is the system of constraints we get from this?
x + 2y + 3
2 z ≤ 1200023 x + 2
3 y + z ≤ 460012 x + 1
3 y + 12 z ≤ 2400
x ≥ 0, y ≥ 0, z ≥ 0
This becomes ...
x + 2y + 3
2 z + u = 1200023 x + 2
3 y + z + v = 460012 x + 1
3 y + 12 z + w = 2400
x ≥ 0, y ≥ 0, z ≥ 0
A Business Application: Maximum Profit
What is the system of constraints we get from this?
x + 2y + 3
2 z ≤ 1200023 x + 2
3 y + z ≤ 460012 x + 1
3 y + 12 z ≤ 2400
x ≥ 0, y ≥ 0, z ≥ 0
This becomes ...
x + 2y + 3
2 z + u = 1200023 x + 2
3 y + z + v = 460012 x + 1
3 y + 12 z + w = 2400
x ≥ 0, y ≥ 0, z ≥ 0
A Business Application: Maximum Profit
What is the system of constraints we get from this?
x + 2y + 3
2 z ≤ 1200023 x + 2
3 y + z ≤ 460012 x + 1
3 y + 12 z ≤ 2400
x ≥ 0, y ≥ 0, z ≥ 0
This becomes ...
x + 2y + 3
2 z + u = 1200023 x + 2
3 y + z + v = 460012 x + 1
3 y + 12 z + w = 2400
x ≥ 0, y ≥ 0, z ≥ 0
A Business Application: Maximum Profit
The initial tableau is
1 2 3
2 1 0 0 0 1200023
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
Where is our first pivot?
1 2 3
2 1 0 0 0 1200023
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
12000
2= 6000
460023
= 6900
240013
= 7200
A Business Application: Maximum Profit
The initial tableau is
1 2 3
2 1 0 0 0 1200023
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
Where is our first pivot?
1 2 3
2 1 0 0 0 1200023
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
12000
2= 6000
460023
= 6900
240013
= 7200
A Business Application: Maximum Profit
The initial tableau is
1 2 3
2 1 0 0 0 1200023
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
Where is our first pivot?
1 2 3
2 1 0 0 0 1200023
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
12000
2= 6000
460023
= 6900
240013
= 7200
A Business Application: Maximum Profit
The initial tableau is
1 2 3
2 1 0 0 0 1200023
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
Where is our first pivot?
1 2 3
2 1 0 0 0 1200023
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
120002
= 6000
460023
= 6900
240013
= 7200
A Business Application: Maximum Profit
The initial tableau is
1 2 3
2 1 0 0 0 1200023
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
Where is our first pivot?
1 2 3
2 1 0 0 0 1200023
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
12000
2= 6000
460023
= 6900
240013
= 7200
A Business Application: Maximum Profit
1 2 3
2 1 0 0 0 1200023
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
∼
12 1 3
412 0 0 0 6000
23
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
∼
12 1 3
412 0 0 0 6000
13 0 1
2 −13 1 0 0 600
13 0 1
4 −16 0 1 0 400
-3 0 -3 8 0 0 1 96000
Are we done?
A Business Application: Maximum Profit
1 2 3
2 1 0 0 0 1200023
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
∼
12 1 3
412 0 0 0 6000
23
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
∼
12 1 3
412 0 0 0 6000
13 0 1
2 −13 1 0 0 600
13 0 1
4 −16 0 1 0 400
-3 0 -3 8 0 0 1 96000
Are we done?
A Business Application: Maximum Profit
1 2 3
2 1 0 0 0 1200023
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
∼
12 1 3
412 0 0 0 6000
23
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
∼
12 1 3
412 0 0 0 6000
13 0 1
2 −13 1 0 0 600
13 0 1
4 −16 0 1 0 400
-3 0 -3 8 0 0 1 96000
Are we done?
A Business Application: Maximum Profit
1 2 3
2 1 0 0 0 1200023
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
∼
12 1 3
412 0 0 0 6000
23
23 1 0 1 0 0 4600
12
13
12 0 0 1 0 2400
-11 -16 -15 0 0 0 1 0
∼
12 1 3
412 0 0 0 6000
13 0 1
2 −13 1 0 0 600
13 0 1
4 −16 0 1 0 400
-3 0 -3 8 0 0 1 96000
Are we done?
A Business Application: Maximum Profit
12 1 3
412 0 0 0 6000
13 0 1
2 − 13 1 0 0 600
13 0 1
4 − 16 0 1 0 400
-3 0 -3 8 0 0 1 96000
∼
12 1 3
412 0 0 0 6000
13 0 1
2 − 13 1 0 0 600
1 0 34 − 1
2 0 3 0 1200-3 0 -3 8 0 0 1 96000
∼
0 1 3
834 0 −3
2 0 54000 0 1
4 −16 1 -1 0 200
1 0 34 −1
2 0 3 0 12000 0 −3
4132 0 9 0 99600
Now are we done?
A Business Application: Maximum Profit
12 1 3
412 0 0 0 6000
13 0 1
2 − 13 1 0 0 600
13 0 1
4 − 16 0 1 0 400
-3 0 -3 8 0 0 1 96000
∼
12 1 3
412 0 0 0 6000
13 0 1
2 − 13 1 0 0 600
1 0 34 − 1
2 0 3 0 1200-3 0 -3 8 0 0 1 96000
∼
0 1 3
834 0 −3
2 0 54000 0 1
4 −16 1 -1 0 200
1 0 34 −1
2 0 3 0 12000 0 −3
4132 0 9 0 99600
Now are we done?
A Business Application: Maximum Profit
12 1 3
412 0 0 0 6000
13 0 1
2 − 13 1 0 0 600
13 0 1
4 − 16 0 1 0 400
-3 0 -3 8 0 0 1 96000
∼
12 1 3
412 0 0 0 6000
13 0 1
2 − 13 1 0 0 600
1 0 34 − 1
2 0 3 0 1200-3 0 -3 8 0 0 1 96000
∼
0 1 3
834 0 −3
2 0 54000 0 1
4 −16 1 -1 0 200
1 0 34 −1
2 0 3 0 12000 0 −3
4132 0 9 0 99600
Now are we done?
A Business Application: Maximum Profit
12 1 3
412 0 0 0 6000
13 0 1
2 − 13 1 0 0 600
13 0 1
4 − 16 0 1 0 400
-3 0 -3 8 0 0 1 96000
∼
12 1 3
412 0 0 0 6000
13 0 1
2 − 13 1 0 0 600
1 0 34 − 1
2 0 3 0 1200-3 0 -3 8 0 0 1 96000
∼
0 1 3
834 0 −3
2 0 54000 0 1
4 −16 1 -1 0 200
1 0 34 −1
2 0 3 0 12000 0 −3
4132 0 9 0 99600
Now are we done?
A Business Application: Maximum Profit
0 1 3
834 0 −3
2 0 5400
0 0 14 −1
6 1 -1 0 200
1 0 34 −1
2 0 3 0 12000 0 −3
4132 0 9 0 99600
∼
0 1 3
834 0 −3
2 0 54000 0 1 −2
3 4 -4 0 8001 0 3
4 −12 0 3 0 1200
0 0 −34
132 0 9 0 99600
∼
0 1 0 1 −3
2 0 0 51000 0 1 −2
3 4 -4 0 8001 0 0 0 -3 6 0 6000 0 0 6 3 6 1 100200
Now are we done?
A Business Application: Maximum Profit
0 1 3
834 0 −3
2 0 5400
0 0 14 −1
6 1 -1 0 200
1 0 34 −1
2 0 3 0 12000 0 −3
4132 0 9 0 99600
∼
0 1 3
834 0 −3
2 0 54000 0 1 −2
3 4 -4 0 8001 0 3
4 −12 0 3 0 1200
0 0 −34
132 0 9 0 99600
∼
0 1 0 1 −3
2 0 0 51000 0 1 −2
3 4 -4 0 8001 0 0 0 -3 6 0 6000 0 0 6 3 6 1 100200
Now are we done?
A Business Application: Maximum Profit
0 1 3
834 0 −3
2 0 5400
0 0 14 −1
6 1 -1 0 200
1 0 34 −1
2 0 3 0 12000 0 −3
4132 0 9 0 99600
∼
0 1 3
834 0 −3
2 0 54000 0 1 −2
3 4 -4 0 8001 0 3
4 −12 0 3 0 1200
0 0 −34
132 0 9 0 99600
∼
0 1 0 1 −3
2 0 0 51000 0 1 −2
3 4 -4 0 8001 0 0 0 -3 6 0 6000 0 0 6 3 6 1 100200
Now are we done?
A Business Application: Maximum Profit
0 1 3
834 0 −3
2 0 5400
0 0 14 −1
6 1 -1 0 200
1 0 34 −1
2 0 3 0 12000 0 −3
4132 0 9 0 99600
∼
0 1 3
834 0 −3
2 0 54000 0 1 −2
3 4 -4 0 8001 0 3
4 −12 0 3 0 1200
0 0 −34
132 0 9 0 99600
∼
0 1 0 1 −3
2 0 0 51000 0 1 −2
3 4 -4 0 8001 0 0 0 -3 6 0 6000 0 0 6 3 6 1 100200
Now are we done?
A Business Application: Maximum Profit
0 1 0 1 − 3
2 0 0 51000 0 1 − 2
3 4 -4 0 8001 0 0 0 -3 6 0 6000 0 0 6 3 6 1 100200
x = 600
y = 5100
z = 800
u = 0
v = 0
w = 0
M = 100200
We have a maximum profit of$100,200 when we produce 600dozen of type A, 5100 dozen oftype B and 800 dozen of type C.
A Business Application: Maximum Profit
0 1 0 1 − 3
2 0 0 51000 0 1 − 2
3 4 -4 0 8001 0 0 0 -3 6 0 6000 0 0 6 3 6 1 100200
x = 600
y = 5100
z = 800
u = 0
v = 0
w = 0
M = 100200
We have a maximum profit of$100,200 when we produce 600dozen of type A, 5100 dozen oftype B and 800 dozen of type C.
A Business Application: Maximum Profit
0 1 0 1 − 3
2 0 0 51000 0 1 − 2
3 4 -4 0 8001 0 0 0 -3 6 0 6000 0 0 6 3 6 1 100200
x = 600
y = 5100
z = 800
u = 0
v = 0
w = 0
M = 100200
We have a maximum profit of$100,200 when we produce 600dozen of type A, 5100 dozen oftype B and 800 dozen of type C.
A Nutrition Problem
ExampleSuppose you wanted to make rice and soybeans a staple of your diet.The object is to design the lowest-cost diet that provides certainminimum levels of protein, calories and vitamin B12. One cup ofuncooked rice costs 21 cents and contains 15 grams of protein, 810calories and 1
9 mg of vitamin B12. One cup of uncooked soybeanscosts 14 cents and contains 22.5 grams of protein, 270 calories and 1
3mg of vitamin B12. The minimum daily requirements are 90 grams ofprotein, 1620 calories and 1 mg of vitamin B12. Design this lowestcost diet that meets these requirements.
The Table
Category Rice Soybeans RequirementProtein 15 22.5 90Calories 810 270 1620
Vitamin B1219
13 1
Cost 21 14
The System
Let x be the number of cups of rice and let y be the number of cups ofsoybeans.
Minimize C = 21x + 14y subject to the constraints
15x + 22.5y ≥ 90810x + 270y ≥ 162019 x + 1
3 y ≥ 1x ≥ 0, y ≥ 0
We can simplify some. Note that:We can multiply the first inequality by 2 to get rid of thefractions.All of the coefficients in the first one are now divisible by 15.All of the coefficients in the second one are divisible by 270.We can multiply the third one by 9 to get rid of the fractions.
The System
Let x be the number of cups of rice and let y be the number of cups ofsoybeans.
Minimize C = 21x + 14y subject to the constraints
15x + 22.5y ≥ 90810x + 270y ≥ 162019 x + 1
3 y ≥ 1x ≥ 0, y ≥ 0
We can simplify some. Note that:We can multiply the first inequality by 2 to get rid of thefractions.All of the coefficients in the first one are now divisible by 15.All of the coefficients in the second one are divisible by 270.We can multiply the third one by 9 to get rid of the fractions.
The System
Let x be the number of cups of rice and let y be the number of cups ofsoybeans.
Minimize C = 21x + 14y subject to the constraints
15x + 22.5y ≥ 90810x + 270y ≥ 162019 x + 1
3 y ≥ 1x ≥ 0, y ≥ 0
We can simplify some. Note that:We can multiply the first inequality by 2 to get rid of thefractions.
All of the coefficients in the first one are now divisible by 15.All of the coefficients in the second one are divisible by 270.We can multiply the third one by 9 to get rid of the fractions.
The System
Let x be the number of cups of rice and let y be the number of cups ofsoybeans.
Minimize C = 21x + 14y subject to the constraints
15x + 22.5y ≥ 90810x + 270y ≥ 162019 x + 1
3 y ≥ 1x ≥ 0, y ≥ 0
We can simplify some. Note that:We can multiply the first inequality by 2 to get rid of thefractions.All of the coefficients in the first one are now divisible by 15.
All of the coefficients in the second one are divisible by 270.We can multiply the third one by 9 to get rid of the fractions.
The System
Let x be the number of cups of rice and let y be the number of cups ofsoybeans.
Minimize C = 21x + 14y subject to the constraints
15x + 22.5y ≥ 90810x + 270y ≥ 162019 x + 1
3 y ≥ 1x ≥ 0, y ≥ 0
We can simplify some. Note that:We can multiply the first inequality by 2 to get rid of thefractions.All of the coefficients in the first one are now divisible by 15.All of the coefficients in the second one are divisible by 270.
We can multiply the third one by 9 to get rid of the fractions.
The System
Let x be the number of cups of rice and let y be the number of cups ofsoybeans.
Minimize C = 21x + 14y subject to the constraints
15x + 22.5y ≥ 90810x + 270y ≥ 162019 x + 1
3 y ≥ 1x ≥ 0, y ≥ 0
We can simplify some. Note that:We can multiply the first inequality by 2 to get rid of thefractions.All of the coefficients in the first one are now divisible by 15.All of the coefficients in the second one are divisible by 270.We can multiply the third one by 9 to get rid of the fractions.
The System
Minimize C = 21x + 14y subject to the constraints
2x + 3y ≥ 123x + y ≥ 6x + 3y ≥ 9x ≥ 0, y ≥ 0
Since this is not in standard form ...
−2x− 3y ≤ −12−3x− y ≤ −6−x− 3y ≤ −9x ≥ 0, y ≥ 021x + 14y + M = 0
The System
Minimize C = 21x + 14y subject to the constraints
2x + 3y ≥ 123x + y ≥ 6x + 3y ≥ 9x ≥ 0, y ≥ 0
Since this is not in standard form ...
−2x− 3y ≤ −12−3x− y ≤ −6−x− 3y ≤ −9x ≥ 0, y ≥ 021x + 14y + M = 0
The System
Minimize C = 21x + 14y subject to the constraints
2x + 3y ≥ 123x + y ≥ 6x + 3y ≥ 9x ≥ 0, y ≥ 0
Since this is not in standard form ...
−2x− 3y ≤ −12−3x− y ≤ −6−x− 3y ≤ −9x ≥ 0, y ≥ 021x + 14y + M = 0
The Matrix and Pivots
-2 -3 1 0 0 0 -12-3 -1 0 1 0 0 -6-1 -3 0 0 1 0 -921 14 0 0 0 1 0
∼
-2 -3 1 0 0 0 -12-3 -1 0 1 0 0 -6
-1 -3 0 0 1 0 -921 14 0 0 0 1 0
∼
-2 -3 1 0 0 0 -12-3 -1 0 1 0 0 -613 1 0 0 − 1
3 0 321 14 0 0 0 1 0
∼
-1 0 1 0 -1 0 -3−8
3 0 0 1 − 13 0 -3
13 1 0 0 − 1
3 0 3493 0 0 0 14
3 1 -42
The Matrix and Pivots
-2 -3 1 0 0 0 -12-3 -1 0 1 0 0 -6-1 -3 0 0 1 0 -921 14 0 0 0 1 0
∼
-2 -3 1 0 0 0 -12-3 -1 0 1 0 0 -6
-1 -3 0 0 1 0 -921 14 0 0 0 1 0
∼
-2 -3 1 0 0 0 -12-3 -1 0 1 0 0 -613 1 0 0 − 1
3 0 321 14 0 0 0 1 0
∼
-1 0 1 0 -1 0 -3−8
3 0 0 1 − 13 0 -3
13 1 0 0 − 1
3 0 3493 0 0 0 14
3 1 -42
The Matrix and Pivots
-2 -3 1 0 0 0 -12-3 -1 0 1 0 0 -6-1 -3 0 0 1 0 -921 14 0 0 0 1 0
∼
-2 -3 1 0 0 0 -12-3 -1 0 1 0 0 -6
-1 -3 0 0 1 0 -921 14 0 0 0 1 0
∼
-2 -3 1 0 0 0 -12-3 -1 0 1 0 0 -613 1 0 0 − 1
3 0 321 14 0 0 0 1 0
∼
-1 0 1 0 -1 0 -3−8
3 0 0 1 − 13 0 -3
13 1 0 0 − 1
3 0 3493 0 0 0 14
3 1 -42
The Matrix and Pivots
-1 0 1 0 -1 0 -3−8
3 0 0 1 − 13 0 -3
13 1 0 0 − 1
3 0 3493 0 0 0 14
3 1 -42
∼
1 0 -1 0 1 0 3− 8
3 0 0 1 − 13 0 -3
13 1 0 0 − 1
3 0 3493 0 0 0 14
3 1 -42
∼
1 0 -1 0 1 0 3− 7
3 0 − 13 1 0 0 -2
23 1 − 1
3 0 0 0 4353 0 14
3 0 0 1 -56
The Matrix and Pivots
1 0 -1 0 1 0 3
− 73 0 − 1
3 1 0 0 -223 1 − 1
3 0 0 0 4353 0 14
3 0 0 1 -56
∼
1 0 -1 0 1 0 37 0 1 -3 0 0 623 1 −1
3 0 0 0 4353 0 14
3 0 0 1 -56
∼
8 0 0 -3 1 0 97 0 1 -3 0 0 63 1 0 -1 0 0 6
-21 0 0 14 0 1 -84
The Matrix and Pivots
8 0 0 -3 1 0 97 0 1 -3 0 0 63 1 0 -1 0 0 6
-21 0 0 14 0 1 -84
∼
8 0 0 -3 1 0 91 0 1
7 − 37 0 0 6
73 1 0 -1 0 0 6
-21 0 0 14 0 1 -84
∼
0 0 − 8
737 1 0 15
71 0 1
7 − 37 0 0 6
70 1 − 3
727 0 0 24
70 0 3 5 0 1 -66
The Conclusion
0 0 −8
737 1 0 15
71 0 1
7 −37 0 0 6
70 1 −3
727 0 0 24
70 0 3 5 0 1 -66
We minimize the cost at $.66 when we use 67 cups of rice and 3 3
7 cupsof soybeans.
The Conclusion
0 0 −8
737 1 0 15
71 0 1
7 −37 0 0 6
70 1 −3
727 0 0 24
70 0 3 5 0 1 -66
We minimize the cost at $.66 when we use 67 cups of rice and 3 3
7 cupsof soybeans.
