Chapter 3 Introduction to optimization models. Linear Programming The PCTech company makes and sells...
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Chapter 3 Introduction to optimization models
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Transcript of Chapter 3 Introduction to optimization models. Linear Programming The PCTech company makes and sells...
Chapter 3
Introduction to optimization models
Linear Programming• The PCTech company makes and sells two models for computers,
Basic and XP.• Profits for Basic is $80/unit and for XP is $129/unit. • Sales estimate is 600 Basics and 1200 XPs• Making the computers involves two operations:
Assembly: Basic requires 5 hours and XP requires 6 hours
Testing: Basic requires 1 hour and XP requires 2 hours
• Available labor hours:Assembly: 10000 hoursTesting: 3000 hours
Linear Programming• PC Tech wants to know how many of each model it should
produce (assemble and test) to maximize its net profit, but it cannot use more labor hours than are available, and it does not want to produce more than it can sell.
• The problem objective:– Use LP to find the best mix of computer models that maximizes
profit– Stay within the company’s labor availability– Don’t produce more than what can be sold
Graphical Method
x1 = Number of basic computer modelx2 = Number of XP computer model
Net profit = 80x1 + 129x2
x1
x2
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If x1 = 1290, x2 = 0, Net profit = 103,200
If x1 = 0, x2 = 800, Net profit = 103,200
Net profit = $103,200
x2
x1
Graphical Method
Net profit = 80x1 + 129x2
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1800Net profit = 80x1 + 129x2
x1 = Number of basic computer modelx2 = Number of XP computer model
If x1 = 1290, x2 = 0, Net profit = 103,200
If x1 = 0, x2 = 800, Net profit = 103,200
Net profit = $103,200
x2
x1
Net profit = $130,00
Net profit = $140,00
Graphical Method
Iso-profit line
Constraints
Basic Model XP Model Hours available
Assembly labor 5 hours/unit 6 hours/unit 10,000 hours
Testing labor 1 hour/unit 2 hours/unit 3,000 hours
Labor hours constraints
Basic Model XP Model
Maximum sales/month 600 1200
Sales constraints
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x1 = Number of basic computer modelx2 = Number of XP computer model
Assembly hours constraint:5x1 + 6x2 <= 10,000
If we make no XP model at all5(2000) + 6(0) = 10,000
If we make no Basic model at all5(0) + 6(1666.67) = 10,000
X1 = 2000, x2 = 0
Assembly Hours ConstraintsX1 = 0,
x2 = 1666.67
x1
x2
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x1 = Number of basic computer modelx2 = Number of XP computer model
Testing hours constraint:x1 + 2x2 <= 3,000
If we make no XP model at all(3000) + 2(0) = 3,000
If we make no Basic model at all(0) + 2(1500) = 3,000
Testing Hours Constraintsx2
x1
X1 = 3000, x2 = 0
X1 = 0, x2 = 1500
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x1 = Number of basic computer modelx2 = Number of XP computer model
Maximum sales Constraintsx2
x1
Maximum sales for basic model:x1 <= 600
X1 = 600, x2 = 0
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x1 = Number of basic computer modelx2 = Number of XP computer model
x2
x1
Maximum sales for XP model:x2 <= 1200
X1 = 0, x2 = 1200
Maximum sales Constraints
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x1 = Number of basic computer modelx2 = Number of XP computer model
Feasible regionx2
x1
x2 = 1200
x1 = 600
5x1 + 6x2 =10000
Feasible region
Redundant constraint
x1 + 2x2 <= 3000
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x1 = Number of basic computer modelx2 = Number of XP computer model
Optimum solutionx2
x1
Feasible region
Redundant constraintx1 + 2x2 <= 3000
Iso-profit line
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Optimum solutionx2
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Feasible region
Optimum solution
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Optimum solutionx2
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Feasible region
Optimum solution
Optimum Solution is the intersection between:x2 = 1200, and5x1 + 6x2 = 10000Solve and x1 = 560 and x2 = 1200Profit = 80(560) + 129(1200) = $199,600
5x1 + 6x2 =10000
x2 = 1200
The algebraic modelMaximize 80x1 + 129x2
subject to:5x1 + 6x2 < 10000
x1 + 2x2 < 3000
x1 < 600
x2 < 1200
x1, x2 > 0
Elements of LP model
• Decision variables– The variable whose values must be determined
• Objective function– A linear function of decision variables– The value of this function is to be optimized –
minimized or maximized• Constraints– Linear functions of the variables– Represents limited resources or minimum
requirements
LP requirements
• Proportionality of variables• Additivity of resources• Divisibility of variables• Non-negativity• Linear objective function• Linear constraints
Scaling in LP
• Poorly scaled model– model contains some very large numbers (e.g. 100,000 or
more) and some very small numbers (e.g. 0.001 or less)– Solver may erroneously give an error that the linearity
conditions are not satisfied• Three remedies for poorly scaled model– Use Automatic Scaling option in Solver/Options– Redefine the units in the model– Change the Precision setting in Solver's Options dialog
box to a larger number, such 0.00001 or 0.0001. (The default has five zeros.)
Solutions to LP problem
• Feasible solution• Feasible region• Optimal solution– Unique– Multiple– Unbounded
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Multiple Optimum solutionx2
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Multiple Optimum solutionx2
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Iso-profit line
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Unbounded Solution
Constraint 1
Constraint 2
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Unbounded Solution
Constraint 1
Constraint 2
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Infeasible Solution
Constraint 1
Constraint 2
Summary
• An LP model may result in– an unique optimum solution– multiple optimum solutions– unbounded feasible region– infeasible region
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