PENELITIAN OPERASIONAL I

(TIN 4109)

Lecture 3

LINEAR PROGRAMMING

Lecture 3

• Outline: – Graphical Method

– Mathematical Modeling

• References: – Frederick Hillier and Gerald J. Lieberman. Introduction

to Operations Research. 7th ed. The McGraw-Hill Companies, Inc, 2001.

– Hamdy A. Taha. Operations Research: An Introduction. 8th Edition. Prentice-Hall, Inc, 2007.

Graphical LP Solution

• 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

Example: The Galaxy Industries Production Problem

Graphical LP Solution Example: The Galaxy Industries Production Problem

• 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 per dozen), while remaining within the marketing guidelines.

Management is seeking a production schedule that will increase the company’s profit.

• 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)

Example: The Galaxy Industries Production Problem

Graphical LP Solution

1000

500

Feasible

X2

Infeasible

Production

Time

3X1+4X22400

Total production constraint:

X1+X2 700 (redundant) 500

700

Production mix

constraint:

X1-X2 350

The Plastic constraint

2X1+X2 1000

X1

700

• There are three types of feasible points

Interior points. Boundary points. Extreme points.

Graphical Analysis – the Feasible Region

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

Extreme points and optimal solutions

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.

• 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 point

optimizes the objective function

Models Without Unique Optimal Solutions

1

No point, simultaneously,

lies both above line and

below lines and

.

1

2 3 2

3

Infeasible Model

12

Unbounded solution

13

• For multiple optimal solutions to exist, the objective

function must be parallel to one of the constraints

Multiple optimal solutions

•Any weighted average of

optimal solutions is also an

optimal solution.

Latihan Soal

• A company produces two products, A and B. The sales volume for A is at least 80% of the total sales of both A and B. However, the company cannot sell more than 100 unit of A per day. Both products use one raw material whose maximum daily availability is limited to 240 lb a day. The usage rates of the raw material are 2 lb per unit of A and 4 lb per unit of B. The unit prices for A and B are $20 and $50, respectively.

Jawaban Latihan Soal

• Maximize Z = 20X1+50X2

• Subject to:

-0.2X1+0.8X2 ≥ 0

X1 100

2X1+4X2 240

Jawaban Latihan Soal

X1 50

150

100

200

50

200

2x1+4x2 = 240

-0.2x1+0.8x2 = 0

Z = 20x1 + 50x2 = $6000

(x1 = 100, x2 = 35; z = 2175)

25

120

100 150 240

X2

x1 = 100

Solution of Minimization Model

• Example: Diet Problem – Mr. U. R. Fatte has been placed on a diet by his Doctor,

consisting of two foods: beer and ice cream. The doctor

warned him to insure proper consumption of nutrients to

sustain life. Formulate the problem using this following

information.

Nutrients Beer Ice cream Minimum

Weekly Requirement

________________________________________________________

I 2 mg/oz 3 mg/oz 3500 mg

II 6 mg/oz 2 mg/oz 7000 mg

________________________________________________________

cost/oz 10 cents 4.5 cents

Solution of Minimization Model

• Example: Diet Problem

Let 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

Solution of Minimization Model

X

Y

1000

3000

2000

3000 2000

1000

4000 6x + 2y = 7000

2x + 3y = 3500

Z = 10x + 4.5y = 18000 cents

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

Example: Diet Problem

Graphical Sensitivity Analysis

• Sensitivity Analysis:

– the investigation of the effect of making changes in the model parameters on a given optimum LP solution.

• Changes in objective coefficients

• Changes in right-hand side of the constraints

Graphical Sensitivity Analysis

Example: Stereo Warehouse

Let x = number of receivers to stock

y = number of speakers to stock

Maximize 50x + 20y gross profit

Subject to 2x + 4y 400 floor space

100x + 50y 8000 budget

x 60 sales limit

x, y 0

Graphical Sensitivity Analysis • Example: Stereo Warehouse

0

50

100

150

200

0 50 100 150 200

Z=2000

Z=3000

Z=3600

Z=3800

A B

C

D

E

Optimal solution ( x = 60, y = 40)

Graphical Sensitivity Analysis Objective-Function Coefficients

0

50

100

150

200

0 50 100 150 200

z = 50x + 20y

x 60 (constraint 3 )

2 4 400x y (constraint 1)

100 50 8000x y (constraint 2)

A B

C

D

D(40, 80)

Graphical Sensitivity Analysis Right-Hand-Side Ranging

0

50

100

150

200

0 50 100 150 200

x 60 (constraint 3 )

100 50 8000x y (constraint 2)

A B I

C

D H

H(60, 280)

Lecture 4 - Preparation

• Read and Practice:

– Hamdy A. Taha. Operations Research: An Introduction. 8th Edition. Prentice-Hall, Inc, 2007. Chapter 3.