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

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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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x1 = Number of basic computer modelx2 = Number of XP computer model

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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x1 = Number of basic computer modelx2 = Number of XP computer model

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