lp graphical for presentation
TRANSCRIPT
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2008 Prentice-Hall, Inc.
Chapter 7
To accompanyQuant i tat ive Analysis for Management, Tenth Edit io n,
by Render, Stair, and HannaPower Point slides created by Jeff Heyl
Linear Prog ramm ing Models:Graph ical and Compu ter
Methods
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Learn ing Ob ject ives
1. Understand the basic assumptions andproperties of linear programming (LP)
2. Graphically solve any LP problem that hasonly two variables by both the corner pointand isoprofit line methods
3. Understand special issues in LP such asinfeasibility, unboundedness, redundancy,
and alternative optimal solutions4. Understand the role of sensitivity analysis
After completing this chapter, students will be able to:
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Chapter Outl ine
7.1 Introduction
7.2 Requirements of a Linear ProgrammingProblem
7.3 Formulating LP Problems7.4 Graphical Solution to an LP Problem
7.5 Solving Flair Furnitures LP Problem usingQM for Windows and Excel
7.6 Solving Minimization Problems7.7 Four Special Cases in LP
7.8 Sensitivity Analysis
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In t roduct ion
Many management decisions involve trying tomake the most effective use of limited resources Machinery, labor, money, time, warehouse space, raw
materials
Linear programm ing(LP) is a widely usedmathematical modeling technique designed tohelp managers in planning and decision makingrelative to resource allocation
Belongs to the broader field of mathematicalprogramming
In this sense, programmingrefers to modeling andsolving a problem mathematically
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Requ irements of a Linear
Programm ing Problem
LP has been applied in many areas over the past50 years
All LP problems have 4 properties in common1. All problems seek to maximizeor minimizesome
quantity (the object ive func t ion)2. The presence of restrictions or constra in tsthat limit the
degree to which we can pursue our objective
3. There must be alternative courses of action to choosefrom
4. The objective and constraints in problems must beexpressed in terms of l inearequations or inequali t ies
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LP Propert ies and Assumpt ions
PROPERTIES OF LINEAR PROGRAMS
1. One objective function
2. One or more constraints
3. Alternative courses of action
4. Objective function and constraints are linear
ASSUMPTIONS OF LP
1. Certainty
2. Proportionality
3. Additivity
4. Divisibility
5. Nonnegative variables
Table 7.1
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Basic Assumpt ions of LP
We assume conditions of certaintyexist andnumbers in the objective and constraints areknown with certainty and do not change duringthe period being studied
We assume proport ional i tyexists in the objectiveand constraints
We assume addit iv i tyin that the total of allactivities equals the sum of the individual
activities We assume div is ib i l i tyinthat solutions need not
be whole numbers
All answers or variables are nonnegat ive
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Formulat ing LP Prob lems
Formulating a linear program involves developinga mathematical model to represent the managerialproblem
The steps in formulating a linear program are
1. Completely understand the managerialproblem being faced
2. Identify the objective and constraints
3. Define the decision variables
4. Use the decision variables to writemathematical expressions for the objectivefunction and the constraints
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Formulat ing LP Prob lems
One of the most common LP applications is theproduct m ix prob lem
Two or more products are produced usinglimited resources such as personnel, machines,and raw materials
The profit that the firm seeks to maximize isbased on the profit contribution per unit of eachproduct
The company would like to determine howmany units of each product it should produceso as to maximize overall profit given its limitedresources
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Flair Furn i ture Company
The Flair Furniture Company producesinexpensive tables and chairs
Processes are similar in that both require a certainamount of hours of carpentry work and in the
painting and varnishing department Each table takes 4 hours of carpentry and 2 hours
of painting and varnishing
Each chair requires 3 of carpentry and 1 hour ofpainting and varnishing
There are 240 hours of carpentry time availableand 100 hours of painting and varnishing
Each table yields a profit of $70 and each chair aprofit of $50
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Flair Furn i ture Company
The company wants to determine the bestcombination of tables and chairs to produce toreach the maximum profit
HOURS REQUIRED TO
PRODUCE 1 UNIT
DEPARTMENT(T)
TABLES(C)
CHAIRSAVAILABLE HOURSTHIS WEEK
Carpentry 4 3 240
Painting and varnishing 2 1 100
Profit per unit $70 $50
Table 7.2
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Flair Furn i ture Company
The objective is to
Maximize profit
The constraints are
1. The hours of carpentry time used cannotexceed 240 hours per week
2. The hours of painting and varnishing timeused cannot exceed 100 hours per week
The decision variables representing the actual
decisions we will make areT= number of tables to be produced per week
C= number of chairs to be produced per week
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Flair Furn i ture Company
We create the LP objective function in terms of Tand C
Maximize profit = $70T+ $50C
Develop mathematical relationships for the twoconstraints
For carpentry, total time used is(4 hours per table)(Number of tables produced)
+ (3 hours per chair)(Number of chairs produced)
We know thatCarpentry time used Carpentry time available
4T+ 3C 240 (hours of carpentry time)
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Flair Furn i ture Company
SimilarlyPainting and varnishing time used
Painting and varnishing time available
2 T+ 1C 100 (hours of painting and varnishing time)
This means that each table producedrequires two hours of painting andvarnishing time
Both of these constraints restrict productioncapacity and affect total profit
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Flair Furn i ture Company
The values for Tand Cmust be nonnegative
T 0 (number of tables produced is greaterthan or equal to 0)
C 0 (number of chairs produced is greater
than or equal to 0)
The complete problem stated mathematically
Maximize profit = $70T+ $50C
subject to4T+ 3C240 (carpentry constraint)
2T+ 1C100 (painting and varnishing constraint)
T, C0 (nonnegativity constraint)
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Graph ical Solut ion to an LP Problem
The easiest way to solve a small LPproblems is with the graphical solutionapproach
The graphical method only works whenthere are just two decision variables
When there are more than two variables, amore complex approach is needed as it is
not possible to plot the solution on a two-dimensional graph
The graphical method provides valuableinsight into how other approaches work
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Graph ical Represen tat ion o f a
Constraint
100
80
60
40
20
C
| | | | | | | | | | | |
0 20 40 60 80 100 T
NumberofCh
airs
Number of Tables
This Axis Represents the Constraint T 0
This Axis Represents theConstraint C 0
Figure 7.1
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Graph ical Represen tat ion o f a
Constraint
The first step in solving the problem is toidentify a set or region of feasiblesolutions
To do this we plot each constraintequation on a graph
We start by graphing the equality portionof the constraint equations
4T+ 3C= 240 We solve for the axis intercepts and draw
the line
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Graph ical Represen tat ion o f a
Constraint
When Flair produces no tables, thecarpentry constraint is
4(0) + 3C= 240
3C
= 240C= 80 Similarly for no chairs
4T+ 3(0) = 240
4T= 240T= 60
This line is shown on the following graph
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Graph ical Represen tat ion o f a
Constraint
100
80
60
40
20
C
| | | | | | | | | | | |
0 20 40 60 80 100 T
NumberofCh
airs
Number of Tables
(T= 0, C= 80)
Figure 7.2
(T= 60, C= 0)
Graph of carpentry constraint equation
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Graph ical Represen tat ion o f a
Constraint
100
80
60
40
20
C
| | | | | | | | | | | |
0 20 40 60 80 100 T
NumberofCh
airs
Number of TablesFigure 7.3
Any point on or below the constraintplot will not violate the restriction
Any point above the plot will violatethe restriction
(30, 40)
(30, 20)
(70, 40)
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Graph ical Represen tat ion o f a
Constraint
The point (30, 40) lies on the plot andexactly satisfies the constraint
4(30) + 3(40) = 240
The point (30, 20) lies below the plot andsatisfies the constraint
4(30) + 3(20) = 180
The point (30, 40) lies above the plot anddoes not satisfy the constraint
4(70) + 3(40) = 400
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Graph ical Represen tat ion o f a
Constraint
100
80
60
40
20
C
| | | | | | | | | | | |
0 20 40 60 80 100 T
NumberofCh
airs
Number of Tables
(T= 0, C= 100)
Figure 7.4
(T= 50, C= 0)
Graph of painting and varnishing
constraint equation
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Graph ical Represen tat ion o f a
Constraint
To produce tables and chairs, bothdepartments must be used
We need to find a solution that satisfies bothconstraints simul taneously
A new graph shows both constraint plots
The feasib le region(or area of feasib lesolut ions)is where all constraints are satisfied
Any point inside this region is a feasible
solution Any point outside the region is an infeasible
solution
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Graph ical Represen tat ion o f a
Constraint
100
80
60
40
20
C
| | | | | | | | | | | |
0 20 40 60 80 100 T
NumberofCh
airs
Number of TablesFigure 7.5
Feasible solution region for Flair Furniture
Painting/Varnishing Constraint
Carpentry Constraint
FeasibleRegion
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Graph ical Represen tat ion o f a
Constraint
For the point (30, 20)
Carpentry
constra in t
4T+ 3C 240 hours available
(4)(30) + (3)(20) = 180 hours used
Paint ing
constra in t
2T+ 1C 100 hours available
(2)(30) + (1)(20) = 80 hours used
For the point (70, 40)
Carpentry
constra in t
4T+ 3C 240 hours available
(4)(70) + (3)(40) = 400 hours used
Paint ing
constra in t
2T+ 1C 100 hours available
(2)(70) + (1)(40) = 180 hours used
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Graph ical Represen tat ion o f a
Constraint
For the point (50, 5)
Carpentry
constra in t
4T+ 3C 240 hours available
(4)(50) + (3)(5) = 215 hours used
Paint ing
constra in t
2T+ 1C 100 hours available
(2)(50) + (1)(5) = 105 hours used
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Isopro f i t Line Solu t ion Method
Once the feasible region has been graphed, weneed to find the optimal solution from the manypossible solutions
The speediest way to do this is to use the isoprofit
line method Starting with a small but possible profit value, we
graph the objective function
We move the objective function line in the
direction of increasing profit while maintaining theslope
The last point it touches in the feasible region isthe optimal solution
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Isopro f i t Line Solu t ion Method
For Flair Furniture, choose a profit of $2,100
The objective function is then
$2,100 = 70T+ 50C
Solving for the axis intercepts, we can draw thegraph
This is obviously not the best possible solution
Further graphs can be created using larger profits
The further we move from the origin, the larger theprofit will be
The highest profit ($4,100) will be generated whenthe isoprofit line passes through the point (30, 40)
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100
80
60
40
20
C
| | | | | | | | | | | |
0 20 40 60 80 100 T
NumberofCh
airs
Number of TablesFigure 7.6
Isoprofit line at $2,100
$2,100 = $70T+ $50C
(30, 0)
(0, 42)
Isopro f i t Line Solu t ion Method
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100
80
60
40
20
C
| | | | | | | | | | | |
0 20 40 60 80 100 T
NumberofCh
airs
Number of TablesFigure 7.7
Four isoprofit lines
$2,100 = $70T+ $50C
$2,800 = $70T+ $50C
$3,500 = $70T+ $50C
$4,200 = $70T+ $50C
Isopro f i t Line Solu t ion Method
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100
80
60
40
20
C
| | | | | | | | | | | |
0 20 40 60 80 100 T
NumberofCh
airs
Number of TablesFigure 7.8
Optimal solution to theFlair Furniture problem
Optimal Solution Point(T= 30, C= 40)
Maximum Profit Line
$4,100 = $70T+ $50C
Isopro f i t Line Solu t ion Method
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A second approach to solving LP problemsemploys the corner point method
It involves looking at the profit at every
corner point of the feasible region The mathematical theory behind LP is that
the optimal solution must lie at one of thecorner points, or extreme po int,in the
feasible region For Flair Furniture, the feasible region is a
four-sided polygon with four corner pointslabeled 1, 2, 3, and 4 on the graph
Corner Poin t Solu t ion Method
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100
80
60
40
20
C
| | | | | | | | | | | |
0 20 40 60 80 100 T
NumberofCh
airs
Number of TablesFigure 7.9
Four corner points ofthe feasible region
1
2
3
4
Corner Poin t Solu t ion Method
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Corner Poin t Solu t ion Method
3
1
2
4
Point : (T= 0, C= 0) Profit = $70(0) + $50(0) = $0
Point : (T= 0, C= 80) Profit = $70(0) + $50(80) = $4,000
Point : (T= 50, C= 0) Profit = $70(50) + $50(0) = $3,500
Point : (T= 30, C= 40) Profit = $70(30) + $50(40) = $4,100
Because Point returns the highest profit, thisis the optimal solution
To find the coordinates for Point accurately we
have to solve for the intersection of the twoconstraint lines
The details of this are on the following slide
3
3
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Corner Poin t Solu t ion Method
Using the simu l taneous equat ions method,wemultiply the painting equation by2 and add it tothe carpentry equation
4T+ 3C= 240 (carpentry line)4T2C=200 (painting line)
C= 40
Substituting 40 for Cin either of the original
equations allows us to determine the value ofT
4T+ (3)(40) = 240 (carpentry line)4T+ 120 = 240
T= 30
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Summary o f Graph ical Solut ion
Methods
ISOPROFIT METHOD
1. Graph all constraints and find the feasible region.
2. Select a specific profit (or cost) line and graph it to find the slope.
3. Move the objective function line in the direction of increasing profit (ordecreasing cost) while maintaining the slope. The last point it touches in the
feasible region is the optimal solution.4. Find the values of the decision variables at this last point and compute the
profit (or cost).
CORNER POINT METHOD
1. Graph all constraints and find the feasible region.
2. Find the corner points of the feasible reason.
3. Compute the profit (or cost) at each of the feasible corner points.
4. select the corner point with the best value of the objective function found inStep 3. This is the optimal solution.
Table 7.3
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Solv ing Minim izat ion Problems
Many LP problems involve minimizing an objectivesuch as cost instead of maximizing a profitfunction
Minimization problems can be solved graphically
by first setting up the feasible solution region andthen using either the corner point method or anisocost line approach (which is analogous to theisoprofit approach in maximization problems) tofind the values of the decision variables (e.g., X1
and X2) that yield the minimum cost
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Hol iday Meal Turkey Ranch
INGREDIENT
COMPOSITION OF EACH POUNDOF FEED (OZ.) MINIMUM MONTHLY
REQUIREMENT PER
TURKEY (OZ.)BRAND 1 FEED BRAND 2 FEEDA 5 10 90
B 4 3 48
C 0.5 0 1.5
Cost per pound 2 cents 3 cents
Holiday Meal Turkey Ranch data
Table 7.4
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Using the cornerpoint method
First we constructthe feasible
solution region The optimal
solution will lie aton of the cornersas it would in amaximizationproblem
Hol iday Meal Turkey Ranch
20
15
10
5
0
X2
| | | | | |
5 10 15 20 25 X1
PoundsofBrand
2
Pounds of Brand 1
Ingredient C Constraint
Ingredient B Constraint
Ingredient A Constraint
Feasible Region
a
b
c
Figure 7.10
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Hol iday Meal Turkey Ranch
We solve for the values of the three corner points
Point ais the intersection of ingredient constraintsC and B
4X1+ 3X2= 48
X1= 3
Substituting 3 in the first equation, we find X2= 12
Solving for point bwith basic algebra we find X1=8.4 and X2= 4.8
Solving for point cwe find X1= 18 and X2= 0
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Substituting these value back into the objectivefunction we find
Cost = 2X1+ 3X2
Cost at point a= 2(3) + 3(12) = 42Cost at point b= 2(8.4) + 3(4.8) = 31.2
Cost at point c= 2(18) + 3(0) = 36
Hol iday Meal Turkey Ranch
The lowest cost solution is to purchase 8.4pounds of brand 1 feed and 4.8 pounds of brand 2feed for a total cost of 31.2 cents per turkey
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Using the isocostapproach
Choosing aninitial cost of 54
cents, it is clearimprovement ispossible
Hol iday Meal Turkey Ranch
20
15
10
5
0
X2
| | | | | |
5 10 15 20 25 X1
PoundsofBrand
2
Pounds of Brand 1Figure 7.11
Feasible Region
(X1= 8.4, X2= 4.8)
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Four Special Cases in LP
Four special cases and difficulties arise attimes when using the graphical approachto solving LP problems
Infeasibility Unboundedness
Redundancy
Alternate Optimal Solutions
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Four Special Cases in LP
No feasible solution Exists when there is no solution to the
problem that satisfies all the constraintequations
No feasible solution region exists This is a common occurrence in the real world
Generally one or more constraints are relaxeduntil a solution is found
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Four Special Cases in LP
A problem with no feasible solution
8
6
4
2
0
X2
| | | | | | | | | |
2 4 6 8 X1
Region Satisfying First Two ConstraintsFigure 7.12
Region SatisfyingThird Constraint
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Four Special Cases in LP
Unboundedness Sometimes a linear program will not have a
finite solution
In a maximization problem, one or more
solution variables, and the profit, can be madeinfinitely large without violating anyconstraints
In a graphical solution, the feasible region will
be open ended This usually means the problem has been
formulated improperly
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Four Special Cases in LP
A solution region unbounded to the right
15
10
5
0
X2
| | | | |
5 10 15 X1
Figure 7.13
Feasible Region
X1 5
X2 10
X1+ 2X2 15
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Four Special Cases in LP
Redundancy A redundant constraint is one that does not
affect the feasible solution region
One or more constraints may be more binding
This is a very common occurrence in the realworld
It causes no particular problems, buteliminating redundant constraints simplifies
the model
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Four Special Cases in LP
A problem witha redundantconstraint
30
25
20
15
10
5
0
X2
| | | | | |
5 10 15 20 25 30 X1Figure 7.14
RedundantConstraint
FeasibleRegion
X1 25
2X1+ X2 30
X1+ X2 20
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Four Special Cases in LP
Alternate Optimal Solutions Occasionally two or more optimal solutions
may exist
Graphically this occurs when the objectivefunctions isoprofit or isocost line runsperfectly parallel to one of the constraints
This actually allows management greatflexibility in deciding which combination to
select as the profit is the same at eachalternate solution
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Four Special Cases in LP
Example ofalternateoptimalsolutions
8
7
6
5
4
3
2
1
0
X2
| | | | | | | |
1 2 3 4 5 6 7 8 X1Figure 7.15
FeasibleRegion
Isoprofit Line for $8
Optimal Solution Consists of AllCombinations of X1and X2Alongthe AB Segment
Isoprofit Line for $12Overlays Line Segment AB
B
A
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Sens i t iv i ty Analys is
Optimal solutions to LP problems thus far havebeen found under what are called determinist icassumpt ions
This means that we assume complete certainty inthe data and relationships of a problem
But in the real world, conditions are dynamic andchanging
We can analyze how sensi t ivea deterministicsolution is to changes in the assumptions of the
model This is called sensi t iv i ty analysis, postopt imal i ty
analysis, parametr ic prog ramm ing, or opt imal i tyanalysis
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Sens i t iv i ty Analys is
Sensitivity analysis often involves a series ofwhat-if? questions concerning constraints,variable coefficients, and the objective function
One way to do this is the trial-and-error method
where values are changed and the entire model isresolved
The preferred way is to use an analyticpostoptimality analysis
After a problem has been solved, we determine a
range of changes in problem parameters that willnot affect the optimal solution or change thevariables in the solution
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The High Note Sound Company manufacturesquality CD players and stereo receivers
Products require a certain amount of skilledartisanship which is in limited supply
The firm has formulated the following product mixLP model
High Note Sound Company
Maximize profit = $50X1+ $120X2
Subject to 2X1 + 4X2 80 (hours of electricianstime available)
3X1 + 1X2 60 (hours of audiotechnicians timeavailable)
X1, X2 0
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The High Note Sound Company graphical solution
High Note Sound Company
b= (16, 12)
Optimal Solution at Point aX1= 0 CD PlayersX2= 20 ReceiversProfits = $2,400
a= (0, 20)
Isoprofit Line: $2,400 = 50X1+ 120X2
60
40
20
10
0
X2
| | | | | |
10 20 30 40 50 60 X1
(receivers)
(CD players)c= (20, 0)Figure 7.16
Ch i th
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Changes in the
Object ive Func t ion Coeff ic ient
In real-life problems, contribution rates in theobjective functions fluctuate periodically
Graphically, this means that although the feasiblesolution region remains exactly the same, the
slope of the isoprofit or isocost line will change We can often make modest increases or
decreases in the objective function coefficient ofany variable without changing the current optimalcorner point
We need to know how much an objective functioncoefficient can change before the optimal solutionwould be at a different corner point
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Changes in the
Object ive Func t ion Coeff ic ient
Changes in the receiver contribution coefficients
b
a
Profit Line for 50X1+ 80X2(Passes through Point b)
40
30
20
10
0
X2
| | | | | |
10 20 30 40 50 60X1
c
Figure 7.17
Profit Line for 50X1+ 120X2(Passes through Point a)
Profit Line for 50X1+ 150X2
(Passes through Point a)
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Changes in the
Techno log ical Coeff ic ients
Changes in the technolog ical coeff ic ientsoftenreflect changes in the state of technology
If the amount of resources needed to produce aproduct changes, coefficients in the constraint
equations will change This does not change the objective function, but
it can produce a significant change in the shapeof the feasible region
This may cause a change in the optimal solution
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Changes in the
Techno log ical Coeff ic ients
Change in the technological coefficients for theHigh Note Sound Company
(a) Original Problem
3X1+ 1X2 60
2X1+ 4X2 80
OptimalSolution
X2
60
40
20
| | |0 20 40 X1
S
tereoReceivers
CD Players
(b) Change in CircledCoefficient
2 X1+ 1X2 60
2X1+ 4X2 80
StillOptimal
3X1+ 1X2 60
2X1+ 5 X2 80
OptimalSolutiona
d
e
60
40
20
| | |0 20 40
X2
X1
16
60
40
20
| | |0 20 40
X2
X1
|
30
(c) Change in CircledCoefficient
a
b
c
fg
c
Figure 7.18
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Changes in Resou rces or
Righ t-Hand-Side Values
The right-hand-side values of theconstraints often represent resourcesavailable to the firm
If additional resources were available, ahigher total profit could be realized
Sensitivity analysis about resources willhelp answer questions about how much
should be paid for additional resourcesand how much more of a resource wouldbe useful
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Changes in Resou rces or
Righ t-Hand-Side Values
If the right-hand side of a constraint is changed,the feasible region will change (unless theconstraint is redundant)
Often the optimal solution will change
The amount of change in the objective functionvalue that results from a unit change in one of theresources available is called the dual pr iceor dualvalue
The dual price for a constraint is the improvementin the objective function value that results from aone-unit increase in the right-hand side of theconstraint
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Changes in Resou rces or
Righ t-Hand-Side Values
However, the amount of possible increase in theright-hand side of a resource is limited
If the number of hours increased beyond theupper bound, then the objective function would no
longer increase by the dual price There would simply be excess (slack) hours of a
resource or the objective function may change byan amount different from the dual price
The dual price is relevant only within limits
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Changes in th e Electr ic ian's Time
for High Note Sound
60
40
20
25
| | |
0 20 40 60
|
50 X1
X2 (a)
a
b
c
Constraint Representing 60 Hours ofAudio Technicians Time Resource
Changed Constraint Representing 100 Hoursof Electricians Time Resource
Figure 7.19
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Changes in th e Electr ic ian's Time
for High Note Sound
60
40
20
15
| | |
0 20 40 60
|
30 X1
X2 (b)
a
b
c
Constraint Representing 60 Hours ofAudio Technicians Time Resource
Changed Constraint Representing 60Hoursof Electricians Time Resource
Figure 7.19
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Changes in th e Electr ic ian's Time
for High Note Sound
60
40
20
| | | | | |
0 20 40 60 80 100 120X1
X2 (c)
ConstraintRepresenting60 Hours of AudioTechnicians
Time Resource
Changed Constraint Representing 240Hoursof Electricians Time Resource
Figure 7.19