Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 1 of 54 Chapter 3 Linear Programming, A...

Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 1 of 54

Chapter 3

Linear Programming,

A Geometric Approach


Outline

3.1 A Linear Programming Problem

3.2 Linear Programming I

3.3 Linear Programming II


3.1 A Linear Programming Problem

1. The Problem2. Tabulate Data3. Translate the Constraints4. The Objective Function5. Linear Programming Problem6. Production Schedule7. No Waste8. Feasible Set


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?


Tabulate Data

It is helpful to tabulate data given in the problem.

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


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.


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


Translate the Constraints (3)

Similarly,

Finishing:

x + y < 18

Upholstery:

2x + 6y < 72

Number of chairs and sofas cannot be negative:

x > 0, y > 0


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


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.


Production Schedule

In the manufacturing problem, each pair of numbers (x,y) that satisfies the system of inequalities is called a production schedule.


Which of the following is a production schedule for

(11,6)? (6,11)?

Example Production Schedule

6 3 96

18

2 6 72

0, 0

x y

x y

x y

x y

Yes No


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


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.


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.


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


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


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.


3.2 Linear Programming I

1. Vertex

2. Fundamental Theorem of Linear Programming

3. Linear Programming Steps


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.


Example Vertex

Find the vertices ofy < -2x + 32

y < -x + 18

y < -x/3 + 12

x > 0, y > 0.

(0,0)

(0,12) (9,9)

(14,4)

(16,0)


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.


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).

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


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.


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.


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.


Example Linear Programming Steps

Rice 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. Find the optimal point that will minimize cost.


Example Step 1A

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

Organize the data.


Example Step 1B

B. Identify the unknown quantities and define corresponding variables.

x = number of cups of rice per day

y = number of cups of soybeans per day


Example Step 1C

C. 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


Example Step 1D

D. Form the objective function.

Minimize the cost in cents:

[Cost] = 21x + 14y


Example Step 2A

A. 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


Example Step 2B

B. 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


Example Step 2C

C. Determine the side of the line.

y > (-2/3)x + 4

y > -3x + 6

y > (-1/3)x + 3

x > 0

y > 0

feasible set


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)


Example Step 4

Determine the objective function at each vertex. Determine the optimal point.

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

The minimum cost is 66 cents for 6/7 cups of rice and 24/7 cups of soybeans.


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.


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.


3.3 Linear Programming II

1. Number of Variables2. Transportation Example


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.


Example Transportation

A TV dealer has stores in city A and B and warehouses in cities W and V. 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?


Step 1A

VStock: 40

B ANeeds: 30 Needs: 25

WStock: 45

$5

$3

$9

$6


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.


Step 1A&B

VStock: 40

B ANeeds: 30 Needs: 25

WStock: 45

$5

$3

$9

$6x y

25 - y30 - x


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


Step 1C - Simplified

x + y < 45

x + y > 15

x < 30

y < 25

0 < x

0 < y


Step 1D

The cost of transporting the TVs is to be minimized.

[cost] = 3x + 6y + 5(30 - x) + 9(25 - y)

[cost] = 375 - 2x - 3y


Step 2

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 3

y < 45 - x

y > 15 - x

x < 30

0 < x

y < 25

0 < y(0,15)

(0,25)

(30,15)

(30,0)(15,0)

(20,25)


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


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


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.

Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 1 of 54 Chapter 3 Linear Programming, A...

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