k-501 lp class slide
TRANSCRIPT
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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