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IntroductionIntroduction

We all face decision about how to useWe all face decision about how to uselimited resources such as:limited resources such as:

Oil in the earthOil in the earth TimeTime

MoneyMoney

WorkersWorkers

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Characteristics of OptimizationCharacteristics of Optimization

ProblemsProblems DecisionsDecisions

ConstraintsConstraints

ObjectivesObjectivesBasic Assumptions of LP Model

Certainty

Proportionality

Additivity

Divisibility

Nonnegativity

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An Example LP ProblemAn Example LP Problem

Blue Ridge Hot Tubs produces two types of hottubs: Aqua-Spas & Hydro-Luxes.

There are 200 pumps, 1566 hours of labor,and 2880 feet of tubing available.

Aqua-Spa Hydro-Lux

Pumps 1 1

Labor 9 hours 6 hours

Tubing 12 feet 16 feet

Unit Profit $350 $300

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5 Steps In Formulating LP Models:5 Steps In Formulating LP Models:

1. Understand the problem.1. Understand the problem.

2. Identify the decision variables.2. Identify the decision variables.

XX11=number ofAqua=number ofAqua--SpastoproduceSpastoproduceXX22=number of Hydro=number of Hydro--LuxestoproduceLuxestoproduce

3.3. State the objective functionasalinearState the objective functionasalinearcombination of the decision variables.combination of the decision variables.

MAX:350XMAX:350X11 + 300X+ 300X22

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5 Steps In Formulating LPModels5 Steps In Formulating LPModels(continued)(continued)

4. State the constraintsaslinear combinations4. State the constraintsaslinear combinationsof the decision variables.of the decision variables.

1X1X11 + 1X+ 1X22

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LPModel for Blue Ridge Hot TubsLPModel for Blue Ridge Hot Tubs

MAX: 350X1 + 300X2S.T.: 1X

1+ 1X

2

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Solving LPProblems:Solving LPProblems:An Intuitive ApproachAn Intuitive Approach

Idea: EachAquaIdea: EachAqua--Spa(XSpa(X11) generates the highest unit) generates the highest unitprofit ($350),soletsmake asmany of themaspossible!profit ($350),soletsmake asmany of themaspossible!

How many would that be?How many would that be?

Let XLet X22 = 0= 0

1st constraint:1st constraint: 1X1X11

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Solving LPProblems:Solving LPProblems:

A GraphicalApproac

hA Grap

hicalApproac

h

The constraints ofanLPproblemThe constraints ofanLPproblemdefines its feasible region.defines its feasible region.

The bestpoint inthe feasible region isThe bestpoint inthe feasible region isthe optimalsolutiontothe problem.the optimalsolutiontothe problem.

ForLPproblems with2variables, it isForLPproblems with2variables, it iseasy toplotthe feasible regionandeasy toplotthe feasible regionandfindthe optimalsolution.findthe optimalsolution.

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X2

X1

250

200

150

100

50

0

0 50 100 150 200 250

(0, 200)

(200, 0)

boundary line of pump constraint

X1 + X2 = 200

Plotting the FirstConstraint

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X2

X1

250

200

150

100

50

0

0 50 100 150 200 250

(0, 261)

(174, 0)

boundary line of labor constraint

9X1 + 6X2 = 1566

Plotting the SecondConstraint

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X2

X1

250

200

150

100

50

0

0 50 100 150 200 250

(0, 180)

(240, 0)

boundary line of tubing constraint

12X1 + 16X2 = 2880

Feasible Region

Plotting the ThirdConstraint

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X2 Plotting ALevelCurve of theObjective Function

X1

250

200

150

100

50

0

0 50 100 150 200 250

(0, 116.67)

(100, 0)

objective function

350X1 + 300X2 = 35000

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A SecondLevelCurve of theObjective FunctionX2

X1

250

200

150

100

50

0

0 50 100 150 200 250

(0, 175)

(150, 0)

objective function

350X1 + 300X2 = 35000

objective function350X1 + 300X2 = 52500

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Using ALevelCurve toLocatethe OptimalSolutionX2

X1

250

200

150

100

50

0

0 50 100 150 200 250

objective function

350X1 + 300X2 = 35000

objective function

350X1 + 300X2 = 52500

optimal solution

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Calculating the OptimalSolutionCalculating the OptimalSolution

The optimalsolution occurs w

here t

he pumps andT

he optimalsolution occurs w

here t

he pumps andlabor constraints intersect.labor constraints intersect.

This occurs where:This occurs where:

XX11 + X+ X22 = 200= 200 (1)(1)

andand 9X9X11 + 6X+ 6X22 = 1566= 1566 (2)(2) From(1) we have, XFrom(1) we have, X22 = 200= 200--XX11 (3)(3)

Substituting (3) for XSubstituting (3) for X22 in (2) we have,in (2) we have,

9X9X11 + 6 (200+ 6 (200--XX11) = 1566) = 1566

which reduces to Xwhich reduces to X11 = 122= 122 So the optimalsolution is,So the optimalsolution is,

XX11=122, X=122, X22=200=200--XX11=78=78

TotalProfit = $350*122 + $300*78 = $66,100TotalProfit = $350*122 + $300*78 = $66,100

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Enumerating The CornerPointsX2

X1

250

200

150

100

50

0

0 50 100 150 200 250

(0, 180)

(174, 0)

(122, 78)

(80, 120)

(0, 0)

obj. value = $54,000

obj. value = $64,000

obj. value = $66,100

obj. value = $60,900obj. value = $0

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Summary of GraphicalSolutionSummary of GraphicalSolution

toLPProblemstoLPProblems

1. Plot the boundary line of each constraint1. Plot the boundary line of each constraint

2. Identify the feasible region2. Identify the feasible region3.3. Locate the optimalsolution by either:Locate the optimalsolution by either:

a.a. Plotting levelcurvesPlotting levelcurves

b. Enumerating the extreme pointsb. Enumerating the extreme points

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SpecialConditions inLPModelsSpecialConditions inLPModels

A number ofanomaliescan occur inLPA number ofanomaliescan occur inLPproblems:problems:

Alternate OptimalSolutionsAlternate OptimalSolutions RedundantConstraintsRedundantConstraints

Unbounded SolutionsUnbounded Solutions

InfeasibilityInfeasibility

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Example ofAlternate OptimalSolutionsX2

X1

250

200

150

100

50

0

0 50 100 150 200 250

450X1 + 300X2 = 78300

objective function level curve

alternate optimal solutions

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Example ofaRedundantConstraintX2

X1

250

200

150

100

50

0

0 50 100 150 200 250

boundary line of tubing constraint

Feasible Region

boundary line of pump constraint

boundary line of labor constraint

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Example ofanUnbounded SolutionX2

X1

1000

800

600

400

200

0

0 200 400 600 800 1000

X1 + X2 = 400

X1 + X2 = 600

objective function

X1 + X2 = 800

objective function

-X1 + 2X2 = 400

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Example of InfeasibilityX2

X1

250

200

150

100

50

0

0 50 100 150 200 250

X1 + X2 = 200

X1 + X2 = 150

feasible region forsecond constraint

feasible regionfor firstconstraint

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EndEnd