03 graphical methods

7/28/2019 03 Graphical Methods

1/31

1

Lecture 3

The GraphicalMethod

Lecture 3

The GraphicalMethod


2/31

2

Our Very First ExampleThe Opti Mize Company manufactures two products that compete

for the same (limited) resources. Relevant information is:Product A B Available resources

Labor-hrs/unit 1 2 20 hrs/dayMachine hrs/unit 2 2 30 hrs/day

Cost/unit $6 $20 $180/day

Profit/unit $5 $15

Let X = number of units of product A to manufactureY = number of units of product B to manufacture

Max Profit = z = 5 X + 15 Y

subject to: X + 2 Y


3/31

3

The non-negativity constraints

X2

X1

Graphical Analysis the Feasible

Region


4/31

4

The GraphicalSolution

X

Y

5 10 15 2520 30

5

10

20

15

X + 2 Y = 20

2X + 2Y = 30

9

6X + 2Y = 180

The feasible region

X = number of units of product A to manufactureY = number of units of product B to manufactureMax Profitz =5 X +15 Y

subject to:X + 2Y


5/31

5

The Graphical SolutionAlternate Approach

X

Y

5 10 15 2520 30

5

10

20

15

9

(x = 5, y = 7.5; z = 137.5)

(x = 10, y = 5; z = 125)

Z = 5X + 15Y

(x = 15, y = 0; z = 75)(x = 0, y = 9; z = 135)

(x = 0, y = 0; z = 0)


6/31

6

The Graphical Solution(continued)

X

Y

5 10 15 2520 30

10

20

15

9 Z = 5X + 15Y = 30

2

6

4

Z = 5X + 15Y = 60

12

(5, 7.5)

Z = 5 (5) + 15 (7.5) = 137.5


7/31

7

The Graphical Analysis of Linear

ProgrammingThe set of all points that satisfy all theconstraints of the model is called

a

FEASIBLE REGIONFEASIBLE REGION

Using a graphical presentation

we can represent all the constraints,the objective function, and the three

types of feasible points.


8/31

8

This is powerful stuff!Can we see anotherexample followed by adescription of the

general model?

Sure can ..


9/31

9

The Galaxy Industries ProductionProblem

Galaxy manufactures two toy doll models: Space Ray.

Zapper.

Resources are limited to

1000 pounds of special plastic. 40 hours of production time per week.

Marketing requirements

Total production cannot exceed 700 dozens.

Number of dozens of Space Rays cannot exceed

number of dozens of Zappers by more than 350


10/31

10

The Galaxy Industries ProductionProblem

Technological input Space Rays requires 2 pounds of plastic and

3 minutes of labor per dozen.

Zappers requires 1 pound of plastic and

4 minutes of labor per dozen.

The current production plan calls for: Producing as much as possible of the more profitable

product, Space Ray ($8 profit per dozen).

Use resources left over to produce Zappers ($5 profit perdozen), while remaining within the marketing guidelines.

Management is seeking a production schedule that willincrease the companys profit.


11/31

11

The Galaxy LP Model Decisions variables:

X1 = Weekly production level of Space Rays (in dozens)

X2 = Weekly production level of Zappers (in dozens)

Objective Function: Weekly profit, to be maximized

Max 8X1 + 5X2 (Weekly profit)subject to

2X1 + 1X2 1000 (Plastic)3X1 + 4X2 2400 (Production Time)X1 + X2 700 (Total production)X1 - X2 350 (Mix)Xj> = 0, j = 1,2 (Nonnegativity)


12/31

12

1000

500

Feasible

X2

Infeasible

ProductionTime3X1+4X22400

Total production constraint:X1+X2700 (redundant)

500

700

The Plastic constraint

2X1+X2 1000

X1

700

Graphical Analysis the Feasible

Region


13/31

13

1000

500

Feasible

X2

Infeasible

ProductionTime3X1+4X22400

Total production constraint:X1+X2700 (redundant)

500

700

Production mixconstraint:

X1-X2350

The Plastic constraint

2X1+X21000

X1

700

Graphical Analysis the

Feasible Region

There are three types of feasible points

Interior points. Boundary points. Extreme points.


14/31

14

Solving Graphically for an

Optimal Solution


15/31

15

The search for an optimal

solutionStart at some arbitrary profit, say profit = $2,000...

Then increase the profit, if possible...

...and continue until it becomes infeasible

Profit =$4360

500

700

1000

500

X2

X1


16/31

16

Summary of the optimal solution

Space Rays = 320 dozen

Zappers = 360 dozen

Profit = $4360 This solution utilizes all the plastic and all the production

hours.

Total production is only 680 (not 700).

Space Rays production exceeds Zappers production by only

40 dozens.


17/31

17

If a linear programming problem has anoptimal solution, an extreme point is optimal.

Extreme points and optimal solutions


18/31

18

Other Post - Optimality

Changes Addition of a constraint.

Deletion of a constraint.

Addition of a variable.

Deletion of a variable.

Changes in the left - hand side

coefficients.


19/31

19

Infeasibility: Occurs when a model has no feasible

point. Unboundness: Occurs when the objective can become

infinitely large (max), or infinitely small (min).

Multiple solutions: Occurs when more than one pointoptimizes the objective function

Models Without Unique OptimalModels Without Unique Optimal

SolutionsSolutions


20/31

201

No point, simultaneously,

lies both above line and

below lines and

.

1

2 32

3

Infeasible Model


21/31

21

Unbounded

solution

Thefeasibleregion

Maximize

theObjectiveFunction


22/31

22

For multiple optimal solutions to exist, the objectivefunction must be parallel to one of the constraints

Multiple optimal solutions

Anyweighted average ofoptimal solutions is also anoptimal solution.


23/31

23

Lets Go to

The NextTopic !!


24/31

24

Exercise 1Suppose a company produces two types of widgets, manual and

electric. Each requires in its manufacture the use of three machines; A,B, and C. A manual widget requires the use of the machine A for 2hours, machine B for 1 hour, and machine C for 1 hour. An electricwidget requires 1 hour on A, 2 hours on B, and 1 hour on C.Furthermore, suppose the maximum numbers of hours available per

month for the use of machines A, B, and C are 180, 160, and 100,respectively. The profit on a manual widget is $4 and on electricwidget it is $6. See the table below for a summary of data. If thecompany can sell all the widgets it can produce, how many of eachtype should it make in order to maximize the monthly profit?

Manual Electr ic Hours

available

A 2 1 180

B 1 2 160

C 1 1 100

profit $4 $6


25/31

25

Exercise 1continued

Step I Identify decision variables:

x = number of manual widgets

y = number of electric widgets

Step II Identify constraints:

2x + y 180

x + 2y 160x + y 100

x 0

y 0


26/31

26

Step III Define objective function:

max P = 4x + 6y

Solving P for y gives

y = -2/3 + P/3.

This defines a so-called family of parallel lines,

isoprofit lines.

Each line gives all possible combinations of x and ythat yield the same profit.

Exercise 1 continued


27/31

27



28/31

28



29/31

29

The Classical Diet ProblemMr. U. R. Fattehas been placed on a diet by hisDoctor (Dr. ImaQuack) consisting of two foods:beer and ice cream. The doctor warned him to insureproper consumption of nutrients to sustain life. Relevant

information is:

Nutrients Beer Ice cream Weekly Requirement

I 2 mg/oz 3 mg/oz 3500 mgII 6 mg/oz 2 mg/oz 7000 mg

cost/oz 10 cents 4.5 cents

Exercise 2


30/31

30

The Mathematical ModelLet X = ounces of beer consumed per week

Y = ounces of ice cream consumed per week

Min cost = z = 10 X + 4.5 Y

subject to:2X + 3Y >= 3500

6X + 2Y >= 7000

X, Y >= 0



31/31

31

Graphical Solution to the Diet Problem

X

Y

1000

3000

2000

30002000

1000

40006x + 2y = 7000

2x + 3y = 3500

Z = 10x + 4.5y = 18000 cents

(x = 1000, y = 500; z = 122.50)


03 graphical methods

Documents