LP Graphical for Presentation

8/10/2019 LP Graphical for Presentation

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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

2009 Prentice-Hall, Inc.


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2009 Prentice-Hall, Inc. 72

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


7/67 2009 Prentice-Hall, Inc. 77

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



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



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



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




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




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




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)




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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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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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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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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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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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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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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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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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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Constraint

100

80

60

40

20

C

| | | | | | | | | | | |

0 20 40 60 80 100 T

NumberofCh

airs


Feasible solution region for Flair Furniture

Painting/Varnishing Constraint

Carpentry Constraint

FeasibleRegion


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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


(2)(30) + (1)(20) = 80 hours used


Carpentry

constra in t


(4)(70) + (3)(40) = 400 hours used

Paint ing

constra in t


(2)(70) + (1)(40) = 180 hours used


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Constraint


Carpentry

constra in t


(4)(50) + (3)(5) = 215 hours used

Paint ing

constra in t


(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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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


Isoprofit line at $2,100

$2,100 = $70T+ $50C

(30, 0)

(0, 42)



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100

80

60

40

20

C

| | | | | | | | | | | |

0 20 40 60 80 100 T

NumberofCh

airs


Four isoprofit lines

$2,100 = $70T+ $50C

$2,800 = $70T+ $50C

$3,500 = $70T+ $50C

$4,200 = $70T+ $50C



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100

80

60

40

20

C

| | | | | | | | | | | |

0 20 40 60 80 100 T

NumberofCh

airs


Optimal solution to theFlair Furniture problem

Optimal Solution Point(T= 30, C= 40)

Maximum Profit Line

$4,100 = $70T+ $50C



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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


Four corner points ofthe feasible region

1

2

3

4



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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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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


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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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


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


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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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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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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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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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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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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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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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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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

Ch i th


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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)

Ch i th


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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

Ch i th


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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

C


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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

Ch i R


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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

Ch i th El t i i ' Ti


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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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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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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

Documents

LP Graphical for Presentation