Linear Programming The PCTech company makes and sells two models for computers Basic and XP Profits for Basic is 80unit and for XP is 129unit Sales estimate is 600 Basics and 1200 XPs ID: 641243
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Slide1
Chapter 3
Introduction to optimization modelsSlide2
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 hours
Testing:
3000 hoursSlide3
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 soldSlide4
Graphical Method
x
1
= Number of basic computer model
x
2
= Number of XP computer model
Net profit = 80x
1
+ 129x
2
x
1
x
2
Slide5
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3000
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x
1
= Number of basic computer model
x
2
= Number of XP computer model
If x
1
= 1290, x
2
= 0, Net profit = 103,200
If x
1
= 0, x
2
= 800, Net profit = 103,200
Net profit = $103,200
x
2
x
1
Graphical Method
Net profit = 80x
1
+ 129x
2Slide6
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3000
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1000
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Net profit = 80x
1
+ 129x
2
x
1
= Number of basic computer model
x
2
= Number of XP computer model
If x
1
= 1290, x
2
= 0, Net profit = 103,200
If x
1
= 0, x
2
= 800, Net profit = 103,200
Net profit = $103,200
x
2
x
1
Net profit = $130,00
Net profit = $140,00
Graphical Method
Iso
-profit lineSlide7
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
6001200
Sales constraintsSlide8
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3000
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800
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1200
1400
1600
1800
x
1
= Number of basic computer model
x
2
= Number of XP computer model
Assembly hours constraint:
5x
1
+ 6x
2
<= 10,000
If we make no XP model at all
5(2000) + 6(0) = 10,000
If we make no Basic model at all
5(0) + 6(1666.67) = 10,000
X
1
= 2000,
x
2
= 0
Assembly Hours Constraints
X
1
= 0,
x
2
= 1666.67
x
1
x
2
Slide9
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800
1800
1200
1400
1600
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x
1
= Number of basic computer model
x
2
= Number of XP computer model
Testing hours constraint:
x
1
+ 2x
2
<= 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 Constraints
x
2
x
1
X
1
= 3000,
x
2
= 0
X
1
= 0,
x
2
= 1500Slide10
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3000
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800
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1200
1400
1600
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x
1
= Number of basic computer model
x
2
= Number of XP computer model
Maximum sales Constraints
x
2
x
1
Maximum sales for basic model:
x
1
<= 600
X
1
= 600,
x
2
= 0Slide11
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3000
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800
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1200
1400
1600
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x
1
= Number of basic computer model
x
2
= Number of XP computer model
x
2
x
1
Maximum sales for XP model:
x
2
<= 1200
X
1
= 0,
x
2
= 1200
Maximum sales ConstraintsSlide12
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x
1
= Number of basic computer model
x
2
= Number of XP computer model
Feasible region
x
2
x
1
x
2
= 1200
x
1
= 600
5x
1
+ 6x
2
=10000
Feasible region
Redundant
constraint
x
1
+ 2x
2
<= 3000Slide13
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800
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x
1
= Number of basic computer model
x
2
= Number of XP computer model
Optimum solution
x
2
x
1
Feasible region
Redundant constraint
x
1
+ 2x
2
<= 3000
Iso
-profit lineSlide14
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800
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x
1
= Number of basic computer model
x
2
= Number of XP computer model
Optimum solution
x
2
x
1
Feasible region
Optimum solutionSlide15
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x
1
= Number of basic computer model
x
2
= Number of XP computer model
Optimum solution
x
2
x
1
Feasible region
Optimum solution
Optimum Solution is the intersection between:
x
2
= 1200, and
5x
1
+ 6x
2
= 10000
Solve and x
1
= 560 and x
2
=
1200
Profit = 80(560) + 129(1200) = $199,600
5x
1
+ 6x
2
=10000
x
2
= 1200Slide16
The algebraic model
Maximize 80
x
1
+ 129
x
2
subject to:
5
x
1
+ 6
x
2
<
10000
x
1
+ 2x2 < 3000
x1 < 600
x2 < 1200
x1, x2 > 0Slide17
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 maximizedConstraints
Linear functions of the variablesRepresents limited resources or minimum requirementsSlide18
LP requirements
Proportionality
of variables
Additivity
of resources
Divisibility of variables
Non-negativity
Linear objective function
Linear constraintsSlide19
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 modelUse Automatic Scaling option in Solver/Options
Redefine the units in the modelChange the Precision setting in Solver's Options dialog box to a larger number, such 0.00001 or 0.0001. (The default has five
zeros.)Slide20
Solutions to LP problem
Feasible solution
Feasible region
Optimal solution
Unique
Multiple
UnboundedSlide21
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800
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x
1
= Number of basic computer model
x
2
= Number of XP computer model
Multiple Optimum solution
x
2
x
1
Slide22
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x
1
= Number of basic computer model
x
2
= Number of XP computer model
Multiple Optimum solution
x
2
x
1
Iso
-profit lineSlide23
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x
1
= Number of basic computer model
x
2
= Number of XP computer model
x
2
x
1
Unbounded Solution
Constraint 1
Constraint 2Slide24
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x
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= Number of basic computer model
x
2
= Number of XP computer model
x
2
x
1
Unbounded Solution
Constraint 1
Constraint 2Slide25
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x
1
= Number of basic computer model
x
2
= Number of XP computer model
x
2
x
1
Infeasible Solution
Constraint 1
Constraint 2Slide26
Summary
An LP model may result in
an unique optimum solution
multiple optimum solutions
unbounded feasible region
infeasible region