linear function important.ppt

7/30/2019 linear function important.ppt

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Linear Programming


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Objectives

Requirements for a linear programming model.

Graphical representation of linear models.

Linear programming results:

Unique optimal solution

Alternate optimal solutions

Unbounded models

Infeasible models

Extreme point principle.


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Objectives - continued

Sensitivity analysis concepts: Reduced costs

Range of optimality--LIGHTLY

Shadow prices Range of feasibility--LIGHTLY

Complementary slackness

Added constraints / variables

Computer solution of linear programming models WINQSB

EXCEL

LINDO


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A Linear Programming model seeks to maximizeor minimize a linear function, subject to a set oflinear constraints.

The linear model consists of the following

components: A set of decision variables. An objective function.

A set of constraints. SHOW FORMAT

3.1 Introduction to Linear

Programming


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The Importance of Linear Programming

Many real static problems lend themselves to linear

programming formulations.

Many real problems can be approximated by linearmodels.

The output generated by linear programs provides

useful whats best and what-if information.


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Assumptions of Linear

Programming (p. 48) The decision variables are continuous or divisible,

meaning that 3.333 eggs or 4.266 airplanes is an

acceptable solution The parameters are known with certainty

The objective function and constraints exhibitconstant returns to scale (i.e., linearity)

There are no interactions between decisionvariables


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Methodology of Linear

ProgrammingDetermine and define the decision variables

Formulate an objective function

verbal characterization

Mathematical characterization

Formulate each constraint


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3.2 THE GALAXY INDUSTRY PRODUCTIONPROBLEM - A Prototype Example

Galaxy manufactures two toy models:

Space Ray.

Zapper.

Purpose: to maximize profits

How: By choice of product mix

How many Space Rays?

How many Zappers?

A RESOURCE ALLOCATION PROBLEM


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Galaxy Resource Allocation

Resources are limited to

1200 pounds of special plastic available per week

40 hours of production time per week.

All LP Models have to be formulated in thecontext of a production period

In this case, a week


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Marketing requirement

Total production cannot exceed 800 dozens.

Number of dozens of Space Rays cannot exceed

number of dozens of Zappers by more than 450.

Technological input

Space Rays require 2 pounds of plastic and

3 minutes of labor per dozen.

Zappers require 1 pound of plastic and

4 minutes of labor per dozen.


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Current production plan calls for:

Producing as much as possible of the more profitableproduct, Space Ray ($8 profit per dozen).

Use resources left over to produce Zappers ($5 profit

per dozen).

The current production plan consists of:

Space Rays = 550 dozens

Zapper = 100 dozens

Profit = 4900 dollars per week


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Management is seeking aproduction schedule thatwill increase the companys

profit.


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MODEL FORMULATION

Decisions variables:

X1 = Production level of Space Rays (in dozens per week).

X2 = Production level of Zappers (in dozens per week).

Objective Function: Weekly profit, to be maximized


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The Objective Function

Each dozen Space Rays realizes $8 in profit.Total profit from Space Rays is 8X1.

Each dozen Zappers realizes $5 in profit.Total profit from Zappers is 5X2.The total profit contributions of both is

8X1 + 5X2

(The profit contributions are additive becauseof the linearity assumption)


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we have a plastics resource constraint, aproduction time constraint, and two marketing

constraints.

PLASTIC: each dozen units of Space Raysrequires 2 lbs of plastic; each dozen units of

Zapper requires 1 lb of plastic and within anygiven week, our plastic supplier can provide1200 lbs.


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The Linear Programming Model

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

2X1 + 1X2 < = 1200 (Plastic)

3X1 + 4X2 < = 2400 (Production Time)

X1 + X2 < = 800 (Total production)

X1 - X2 < = 450 (Mix)

Xj> = 0, j = 1,2 (Nonnegativity)


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3.4 The Set of Feasible Solutions

for Linear Programs

The set of all points that satisfy all the

constraints of the model is calleda

FEASIBLE REGION


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Using a graphical presentation

we can represent all the constraints,

the objective function, and the three

types of feasible points.


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1200

600

The Plastic constraint

Feasible

The plastic constraint:2X1+X2


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3.5 Solving Graphically for an

Optimal Solution


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600

800

1200

400 600 800

X2

X1

We now demonstrate the search for an optimal solutionStart at some arbitrary profit, say profit = $2,000...

Profit = $

000

2,

Then increase the profit, if possible...

3,4,

...and continue until it becomes infeasible

Profit =$5040


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600

800

1200

400 600 800

X2

X1

Lets take a closer look at

the optimal point

FeasibleregionFeasibleregion

Infeasible


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Summary of the optimal solution

Space Rays = 480 dozens

Zappers = 240 dozens

Profit = $5040 This solution utilizes all the plastic and all the production

hours.

Total production is only 720 (not 800).

Space Rays production exceeds Zapper by only 240

dozens (not 450).


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3.6 The Role of Sensitivity Analysis

of the Optimal Solution Is the optimal solution sensitive to changes in

input parameters?

Possible reasons for asking this question:

Parameter values used were only best estimates. Dynamic environment may cause changes.

What-if analysis may provide economical andoperational information.


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600

800

1200

400 600 800

X2

X1

The effects of changes in an objective function coefficienton the optimal solution


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600

800

1200

400 600 800

X2

X1

The effects of changes in an objective function coefficienton the optimal solution

Range of optimality


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Multiple changes

The range of optimality is valid only when a singleobjective function coefficient changes.

When more than one variable changes we turn to the

100% rule.

This is beyond the scope of this course


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Reduced costs

The reduced cost for a variable at its lower bound(usually zero) yields:

The amount the profit coefficient must change before

the variable can take on a value above its lower bound.

Complementary slackness

At the optimal solution, either a variable is at its lowerbound or the reduced cost is 0.


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3.8 Sensitivity Analysis of

Right-Hand Side Values Any change in a right hand side of a binding

constraint will change the optimal solution.

Any change in a right-hand side of a non-binding constraint that is less than its slackor surplus, will cause no change in the

optimal solution.


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In sensitivity analysis of right-hand sides of

constraints we are interested in the followingquestions:

Keeping all other factors the same, how much would the

optimal value of the objective function (for example, theprofit) change if the right-hand side of a constraintchanged by one unit?

For how many additional UNITS is this per unit changevalid?

For how many fewer UNITS is this per unit change valid?


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1200

600

X2

The Plastic constraint

FeasibleX1

600

800

Production timeconstraint

Maximum profit = 5040

The new Plastic constraint

Productionmix constraint

Infeasible extreme points


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Range of feasibility

The set ofright - hand side values for which the same set of

constraints determines the optimal extreme point.

The range over-which the same variables remain in solution

(which is another way of saying that the same extreme point

is the optimal extreme point)

Within the range of feasibility, shadow prices remain

constant; however, the optimal objective function value and

decision variable values will change if the corresponding

constraint is binding


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3.9 Other Post Optimality Changes

SKIP THIS Addition of a constraint.

Deletion of a constraint. Addition of a variable.

Deletion of a variable.

Changes in the left - hand side technology

coefficients.


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3.10 Models Without Optimal

Solutions

Infeasibility: Occurs when a model has no

feasible point.

Unboundedness: Occurs when the objective

can become infinitely

large.


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Infeasibility No point, simultaneously,

lies both above line andbelow lines and .

1

2

31

2 3


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Unbounded solution


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3.11 Navy Sea Ration

A cost minimization diet problem

Mix two sea ration products: Texfoods, Calration.

Minimize the total cost of the mix.

Meet the minimum requirements of

Vitamin A, Vitamin D, and Iron.


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Decision variables

X1 (X2) -- The number of two-ounce portions of

Texfoods (Calration) product used in a serving.

The ModelMinimize 0.60X1 + 0.50X2

Subject to

20X1 + 50X2 100 Vitamin A

25X1 + 25X2 100 Vitamin D

50X1 + 10X2 100 IronX1, X2 0

Cost per 2 oz.

% Vitamin Aprovided per 2 oz.

% required


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The Graphical solution

5

4

2

2 4 5

Feasible Region

Vitamin D constraint

Vitamin A constraint

The Iron constraint


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Summary of the optimal solution

Texfood product = 1.5 portions (= 3 ounces)

Calration product = 2.5 portions (= 5 ounces)

Cost =$ 2.15 per serving.

The minimum requirements for Vitamin D and iron are

met with no surplus.

The mixture provides 155% of the requirement for

Vitamin A.


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Linear programming software packages solve large

linear models. Most of the software packages use the algebraic

technique called the Simplex algorithm.

The input to any package includes: The objective function criterion (Max or Min).

The type of each constraint: .

The actual coefficients for the problem.

3.12 Computer Solution of Linear

Programs With Any Number ofDecision Variables

, ,


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The typical output generated from linearprogramming software includes:

Optimal value of the objective function.

Optimal values of the decision variables.

Reduced cost for each objective function coefficient.

Ranges of optimality for objective function coefficients.

The amount of slack or surplus in each constraint.

Shadow (or dual) prices for the constraints.

Ranges of feasibility for right-hand side values.


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WINQSB Input Data for theGalaxy Industries Problem

Variables arerestricted to >= 0No upper bound

Click to solve

Variableandconstraint

name canbechangedhere


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Simplex Algorithm Basics

Starting at a feasible extreme point, thealgorithm proceeds from extreme point to

extreme point until one is found that is betterthan all neighboring extreme points

The transition from one extreme point to the

next is called an iteration. The algorithm chooses which extreme point

to go to next based on the fastest rate of

improvement


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Basis and non-basis variables

The basis variable values are free to take onvalues other than their lower bounds

The non-basis variables are fixed at theirlower bounds (0)

THERE ARE ALWAYS AS MANY BASIS

VARIABLES AS THERE ARE CONSTRAINTS,ALWAYS


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Another problem--10 products

max 10x1 + 12 x2 + 15 x3 + 5 x4 + 8 x5 + 17x6+ 3 x7 + 9x8 + 11x10

s.t. 2x1 + x2 + 3x3 + x4 + 2x5 + 3x6 + x7 + 3x8 +

2x9 + x10 = 0

How many basis variables?

How many products should we be making?


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linear function important.ppt

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