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Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Chapter 3
Linear Programming,
A Geometric Approach
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Outline
3.1 A Linear Programming Problem
3.2 Fundamental Theorem of Linear Programming
3.3 Linear Programming
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 3.1
A Linear Programming
Problem
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Outline
1. The Problem2. Tabulate Data3. Translate the Constraints4. The Objective Function5. Linear Programming
Problem6. Production Schedule7. No Waste8. Feasible Set
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The Problem
A furniture manufacturer makes two types of furniture - chairs and sofas. The manufacture of a chair requires 6 hours of carpentry, 1 hour of finishing, and 2 hours of upholstery. Manufacture of a sofa requires 3 hours of carpentry, 1 hour of finishing, and 6 hours of upholstery. Each day the factory has available 96 labor hours for carpentry, 18 labor-hours for finishing, and 72 labor-hours for upholstery. The profit per chair is $80 and per sofa is $70. How many chairs and sofas should be produced each day to maximize the profit?
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Chair Sofa Available time
Carpentry
Finishing
Upholstery
Profit
6 hours 3 hours 96 labor-hours
1 hour 1 hour 18 labor-hours
2 hours 6 hours 72 labor-hours
$80 $70
Tabulate Data
It is helpful to tabulate data given in the problem.
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Translate the Constraints
Translate each of the constraints (restrictions on labor-hours available) into mathematical language.
Let x be the number of chairs and y be the number of sofas manufactured each day, respectively.
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Translate the Constraints (2)
Carpentry: [number of labor-hours per day]
= (number of hours required per chair) (number of chairs per day) + (number of hours required per sofa) (number of sofas per day)
= 6x + 3y
[number of labor-hours per day] < [maximum available]
6x + 3y < 96
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Translate the Constraints (3)
Similarly,
Finishing Constraint:
x + y < 18
Upholstery Constraint :
2x + 6y < 72
Number of chairs and sofas cannot be negative:
x > 0, y > 0
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The Objective Function
The objective of the problem is to optimize profit. Translate the profit (objective function) into mathematical language.
[profit] = [profit from chairs] + [profit from sofas]
= [profit per chair][number of chairs] + [profit per sofa][number of
sofas]
= 80x + 70y
Slide 11Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Linear Programming Problem
The manufacturing problem can now be written as a mathematical problem.
Find x and y for which 80x + 70y is as large as possible, and for which the following hold simultaneously: 6 3 96
18
2 6 72
0, 0.
x y
x y
x y
x y
This is called a linear programming problem.
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Production Schedule
In the manufacturing problem, each pair of numbers (x,y) that satisfies the system of inequalities is called a production schedule.
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Example Production Schedule
Which of the following is a production schedule for
(11,6)? (6,11)?
6 3 96
18
2 6 72
0, 0
x y
x y
x y
x y
Yes No
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No Waste
It seems clear that a factory will operate most efficiently when its labor is fully utilized (no waste).
This would require x and y to satisfy the system
6 3 96
18
2 6 72.
x y
x y
x y
Slide 15Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Example No Waste
Solve6 3 96
18
2 6 72.
x y
x y
x y
According to the graph of the three equations, there is no common intersection and therefore no solution.
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Feasible Set
The set of solutions to the system of inequalities is called the feasible set of the system. This represents all possible production schedules.
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Example Feasible Set
Find the feasible set for
6 3 96
18
2 6 72
0 , 0.
x y
x y
x y
x y
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Example Feasible Set (2)
Notice that (0,0) satisfies all the inequalities.
Graph the boundaries:
y < -2x + 32
y < -x + 18
y < -x/3 + 12
x > 0, y > 0 Feasible Set
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Summary Section 3.1
A linear programming problem asks us to find the point (or points) in the feasible set of a system of linear inequalities at which the value of a linear expression involving the variables, called the objective function, is either maximized or minimized.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 3.2
Fundamental Theorem of Linear
Programming
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Outline
1. Vertex
2. Fundamental Theorem of Linear Programming
3. Linear Programming Steps
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Vertex
The boundary of the feasible set is composed of line segments. The line segments intersect in points, each of which is a corner of the feasible set. Such a corner is called a vertex.
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Find the vertices ofy < -2x + 32
y < -x + 18
y < -x/3 + 12
x > 0, y > 0.
Example Vertex
(0,0)
(0,12) (9,9)
(14,4)
(16,0)
Find the vertices ofy < -2x + 32
y < -x + 18
y < -x/3 + 12
x > 0, y > 0.
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Fundamental Theorem of Linear Programming
Fundamental Theorem of Linear Programming The maximum (or minimum) value of the objective function in a linear programming problem is achieved at one of the vertices of the feasible set.
The point that yields the maximum (or minimum) value of the objective function is called an optimal point.
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Vertex Profit = 80x + 70y(0,0) 80(0) + 70(0) = 0
(0,12) 80(0) + 70(12) = 840(9,9) 80(9) + 70(9) = 1350
(14,4) 80(14) + 70(4) = 1400(16,0) 80(16) + 70(0) = 1280
Example Optimal Point
Find the point which maximizes Profit = 80x + 70y for the feasible set with vertices (0,0), (0,12), (9,9), (14,4) and (16,0).
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Linear Programming Steps - Step 1
Step 1 Translate the problem into mathematical language.A. Organize the data.B. Identify the unknown quantities and
define corresponding variables.C. Translate the restrictions into linear
inequalities.D. Form the objective function.
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Linear Programming Steps - Step 2
Step 2 Graph the feasible set.A. Put the inequalities in standard form.B. Graph the straight line corresponding
to each inequality.C. Determine the side of the line
belonging to the graph of each inequality. Cross out the other side. The remaining region is the feasible set.
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Linear Programming Steps - Steps 3 & 4
Step 3 Determine the vertices of the feasible set.
Step 4 Evaluate the objective function at each vertex. Determine the optimal point.
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Example Linear Programming Steps
Nutrition - PeopleRice and soybeans are to be part of a staple diet. One cup of uncooked rice costs 21 cents and contains 15 g of protein, 810 calories, and 1/9 mg of B2 (riboflavin). One cup of uncooked soybeans costs 14 cents and contains 22.5 g of protein, 270 calories, and 1/3 mg of B2. The minimum daily requirements are 90 g of protein, 1620 calories and 1 mg of B2. Design the lowest-cost diet meeting these specifications, (Find the optimal point that will minimize cost).
Slide 30Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Rice SoybeansRequired level
per day
Protein (g/cup) 15 22.5 90
Calories (per cup) 810 270 1620
B2 (mg/cup) 1/9 1/3 1
Cost (cents/cup) 21 14
Example Step 1A
Organize the data.
Slide 31Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Example Step 1B
Identify the unknown quantities and define corresponding variables.
x = number of cups of rice per day
y = number of cups of soybeans per day
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Example Step 1C
Translate the restrictions into linear inequalities.
Protein: 15x + 22.5y > 90
Calories: 810x + 270y > 1620
B2: (1/9)x + (1/3)y > 1
Nonnegative: x > 0, y > 0
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Example Step 1D
Form the objective function.
Minimize the cost in cents:
[Cost] = 21x + 14y
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Example Step 2A
Put the inequalities in standard form.
Protein: y > (-2/3)x + 4
Calories: y > -3x + 6
B2: y > (-1/3)x + 3
Nonnegative: x > 0
y > 0
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Example Step 2B
Graph the straight line corresponding to each inequality.
1. y = (-2/3)x + 4
2. y = -3x + 6
3. y = (-1/3)x + 3
4. x = 0
5. y = 0
1
2
3
4
5
Slide 36Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Example Step 2C
feasible set
Determine the side of the line.
y > (-2/3)x + 4
y > -3x + 6
y > (-1/3)x + 3
x > 0
y > 0
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Example Step 3
Determine the vertices of the feasible set.
x = 0 & y = -3x + 6: (0,6)
y = -3x + 6 & y = (-2/3)x + 4:
(6/7,24/7)
y = (-2/3)x + 4 & y = (-1/3)x + 3:
(3,2)
y = (-1/3)x + 3 & y = 0: (9,0)
(0,6)
(6/7,24/7)(3,2)
(9,0)
Slide 38Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Vertex Cost = 21x + 14y(0,6) 21(0) + 14(6) = 84
(6/7,24/7) 21(6/7) + 14(24/7) = 66
(3,2) 21(3) + 14(2) = 91
(9,0) 21(9) + 14(0) = 189
Example Step 4
Determine the objective function at each vertex. Determine the optimal point.
The minimum cost is 66 cents for 6/7 cups of rice and 24/7 cups of soybeans.
Slide 39Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Summary Section 3.2 - Part 1
The fundamental theorem of linear programming states that the optimal value of the objective function for a linear programming problem occurs at a vertex of the feasible set.
Slide 40Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Summary Section 3.2 - Part 2
To solve a linear programming word problem, assign variables to the unknown quantities, translate the restrictions into a system of linear inequalities involving no more than two variables, form a function for the quantity to be optimized, graph the feasible set, evaluate the objective function at each vertex, and identify the vertex that gives the optimal value.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 3.3
Linear Programming
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Outline
1. Number of Variables2. Transportation Example
Slide 43Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Number of Variables
On the surface some problems may appear to have more than two variables. However, sometimes they can be translated into mathematical language so that only two variables are required.
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Example Transportation - Shipping
A TV dealer has stores in city A (Annapolis) and B (Rockville) and warehouses in cities W (College Park) and V (Baltimore). The cost of shipping a TV from W to A is $6, from V to A is $3, from W to B is $9 and from V to B is $5. Store in A orders 25 TV sets and store in B orders 30 sets. The W warehouse has a stock of 45 sets and V warehouse has 40. What is the most economical way to supply the two stores the requested TV sets?
Slide 45Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Step 1A
VStock: 40
B ANeeds: 30 Needs: 25
WStock: 45
$5
$3
$9
$6
Slide 46Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Step 1B
The number of variables can be reduced by observing that what is not shipped from the warehouse in W must be shipped from the warehouse in V.
Let x be the number of TVs shipped from the W warehouse to the store in A and y be the number of TVs shipped from W to B. Then 30 - x is going from V to A and 25 - y from V to B.
Slide 47Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Step 1A&B
VStock: 40
A BNeeds: 30 Needs: 25
WStock: 45
$5
$3
$9
$6x y
25 - y30 - x
Slide 48Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Step 1C
Warehouse W:
x + y < 45
Warehouse V:
(30 - x) + (25 - y) < 40
Nonnegative restrictions:
0 < x and 30 - x > 0
0 < y and 25 - y > 0
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Step 1C - Simplified
Our restriction inequalities are these:
x + y < 45
x + y > 15
x < 30
y < 25
0 < x
0 < y
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Step 1D The Objective Function
The cost of transporting the TVs is to be minimized.
[cost] = 3x + 6y + 5(30 - x) + 9(25 - y)
[cost] = 375 - 2x - 3y
Slide 51Copyright © 2014, 2010, 2007 Pearson Education, Inc.
1
2 3
3
4
4
1. y < 45 - x
2. y > 15 - x
3. x < 30, 0 < x
4. y < 25, 0 < y
Step 2 Graph the System of Inequalities
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(0,15)
(0,25)
(30,15)
(30,0)(15,0)
(20,25)
Step 3 Determine the Vertices
y < 45 - x
y > 15 - x
x < 30
0 < x
y < 25
0 < y
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Vertex Cost = 375 - 2x - 3y
(0,15) 375 - 2(0) - 3(15) = 330(0,25) 375 - 2(0) - 3(25) = 300
(20,25) 375 - 2(20) - 3(25) = 260(30,15) 375 - 2(30) - 3(15) = 270(30,0) 375 - 2(30) - 3(0) = 315(15,0) 375 - 2(15) - 3(0) = 345
Step 4
Slide 54Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Step 4 - Solution
Optimum Point (20,25)V
Stock: 40
B A Needs: 30 Needs: 25
WStock: 45
$5
$3
$9
$6
20 25
010
Slide 55Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Summary Section 3.3
Sometimes it is necessary to use algebra to reduce the number of variables. Once the number of variables is reduced to two, the steps for solving a linear programming problem are followed.