Slides for Ch07

7/24/2019 Slides for Ch07

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Chapter 7

To accompanyQuantitative Analysis for Management, Eleventh Edition,by Render, Stair, and HannaPower Point slides created by Brian Peterson

Linear Programming Models:Graphical and Computer

Methods


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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 7-2

Introduction

Many management decisions involvetrying to make the most effective use oflimited resources.

Linear programming(LP) is a widely usedmathematical modeling techniquedesigned to help managers in planning anddecision making relative to resourceallocation. This belongs to the broader field of

mathematical programming.

In this sense, programmingrefers to modelingand solving a problem mathematically.


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Requirements of a LinearProgramming Problem

All LP problems have 4 properties in common:1. All problems seek to maximizeor minimizesome

quantity (the objective function).

2. Restrictions or constraintsthat limit the degree to

which we can pursue our objective are present.3. There must be alternative courses of action from which

to choose.

4. The objective and constraints in problems must beexpressed in terms of linearequations or inequalities.


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LP Properties and Assumptions

PROPERTIES OF LINEAR PROGRAMS

1. One objective function

2. One or more constraints

3. Alternative courses of action

4. Objective function and constraints are linearproportionality and divisibility

5. Certainty

6. Divisibility7. Nonnegative variables

Table 7.1


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Formulating LP Problems

One of the most common LP applications is theproduct mix problem.

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 Furniture 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 Furniture Company Data

The company wants to determine the bestcombination of tables and chairs to produce to reachthe maximum profit.

HOURS REQUIRED TOPRODUCE 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 are:T= 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 that:Carpentry time used Carpentry time available.

4T+ 3C 240 (hours of carpentry time)


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Similarly,

Painting 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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Graphical Solution to an LP Problem

The easiest way to solve a small LPproblems is graphically.

The graphical method only works when

there are just two decision variables. When there are more than two variables, a

more complex approach is needed as it isnot possible to plot the solution on a two-

dimensional graph. The graphical method provides valuable

insight into how other approaches work.


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Graphical Representation of aConstraint

100

80

60

40

20

C

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

0 20 40 60 80 100 T

N

umberofChairs

Number of Tables

This Axis Represents the Constraint T 0

This Axis Represents the

Constraint C 0

Figure 7.1

Quadrant Containing All Positive Values


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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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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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100

80

60

40

20

C

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

0 20 40 60 80 100 T

Nu

mberofChairs

Number of Tables

(T= 0, C= 80)

Figure 7.2

(T= 60, C= 0)

Graph of carpentry constraint equation


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100

80

60

40

20

C

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

0 20 40 60 80 100 T

NumberofChairs

Number of Tables

Figure 7.3

Any point on or belowthe constraint plot willnot violate the

restriction. Any point above the

plot will violate therestriction.

(30, 40)

(30, 20)

(70, 40)

Region that Satisfies the Carpentry Constraint


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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 (70, 40) lies above the plot anddoes not satisfy the constraint

4(70) + 3(40) = 400.


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100

80

60

40

20

C

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

0 20 40 60 80 100 T

Nu

mberofChairs

Number of Tables

(T= 0, C= 100)

Figure 7.4

(T= 50, C= 0)

Region that Satisfies the Painting andVarnishing Constraint


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To produce tables and chairs, bothdepartments must be used.

We need to find a solution that satisfies bothconstraints simultaneously.

A new graph shows both constraint plots.

The feasible region(or area of feasiblesolutions) 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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100

80

60

40

20

C

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

0 20 40 60 80 100 T

Nu

mberofChairs

Number of Tables

Figure 7.5

Feasible Solution Region for the FlairFurniture Company Problem

Painting/Varnishing Constraint

Carpentry ConstraintFeasibleRegion


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For the point (30, 20)

Carpentryconstraint

4T+ 3C 240 hours available

(4)(30) + (3)(20) = 180 hours used

Paintingconstraint


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

For the point (70, 40)

Carpentryconstraint


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

Paintingconstraint


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


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Isoprofit Line Solution 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

Nu

mberofChairs

Number of Tables

Figure 7.6

Profit line of $2,100 Plotted for the FlairFurniture Company

$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

Nu

mberofChairs

Number of Tables

Figure 7.7

Four Isoprofit Lines Plotted for the FlairFurniture Company

$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

Nu

mberofChairs

Number of Tables

Figure 7.8

Optimal Solution to the Flair Furniture problem

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

Maximum Profit Line

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



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100

80

60

40

20

C

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

0 20 40 60 80 100 T

N

umberofChairs

Number of Tables

Figure 7.9

Four Corner Points of the Feasible Region

1

2

3

4

Corner Point Solution Method


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To find the coordinates for Point accurately we have tosolve for the intersection of the two constraint lines.

Using the simultaneous equations method, we multiply thepainting equation by2 and add it to the 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 of T.

4T+ (3)(40) = 240 (carpentry line)4T+ 120 = 240

T= 30

3


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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, this isthe optimal solution.

3


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Slack and Surplus

Slackis the amount of a resourcethat is not used. For a less-than-or-equal constraint:Slack= Amount of resource available

amount of resource used.

Surplus is used with a greater-than-

or-equal constraint to indicate theamount by which the right hand sideof the constraint is exceeded.Surplus= Actual amountminimum

amount.


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Summary of Graphical SolutionMethods

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 7.4


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Using Excels Solver Command toSolve LP Problems

The Solver tool in Excel can be used to findsolutions to:

LP problems.

Integer programming problems.

Noninteger programming problems.

Solver is limited to 200 variables and 100constraints.


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Using Solver to Solve the FlairFurniture Problem

Recall the model for Flair Furniture is:

Maximize profit = $70T+ $50CSubject to 4T+ 3C 240

2T+ 1C 100

To use Solver, it is necessary to enter formulasbased on the initial model.


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1. Enter the variable names, the coefficients for theobjective function and constraints, and the right-hand-side values for each of the constraints.

2. Designate specific cells for the values of thedecision variables.

3. Write a formula to calculate the value of theobjective function.

4. Write a formula to compute the left-hand sides ofeach of the constraints.


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Program 7.2A

Excel Data Input for the Flair Furniture Example


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Program 7.2B

Formulas for the Flair Furniture Example


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Program 7.2C

Excel Spreadsheet for the Flair Furniture Example


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Once the model has been entered, the followingsteps can be used to solve the problem.

In Excel 2010, select DataSolver.

If Solver does not appear in the indicated place,see Appendix F for instructions on how toactivate this add-in.

1. In the Set Objective box, enter the cell addressfor the total profit.

2. In the By Changing Cells box, enter the celladdresses for the variable values.

3. Click Maxfor a maximization problem and Minfor

a minimization problem.


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4. Check the box for Make UnconstrainedVariables Non-negative.

5. Click the Select Solving Method button

and select Simplex LP from the menu thatappears.

6. Click Addto add the constraints.

7. In the dialog box that appears, enter the

cell references for the left-hand-sidevalues, the type of equation, and theright-hand-side values.

8. Click Solve.


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Figure 7.2E

SolverParametersDialog Box


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Figure 7.2G

Solver Results Dialog Box

S S


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Figure 7.2H

Solution Found by Solver


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Solving Minimization 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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Holiday 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 7.5


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f


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Feasible Region for the HolidayMeal Turkey Ranch Problem

20

15

10

5

0

X2

| | | | | |

5 10 15 20 25 X1

Po

undsofBrand2

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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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 8.4 poundsof brand 1 feed and 4.8 pounds of brand 2 feed for atotal cost of 31.2 cents per turkey.


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Graphical Solution to the Holiday Meal Turkey RanchProblem Using the Isocost Approach


20

15

10

5

0

X2

| | | | | |

5 10 15 20 25 X1

Pou

ndsofBrand2

Pounds of Brand 1

Figure 7.11

Feasible Region

(X1= 8.4, X2= 4.8)


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Solving the Holiday Meal Turkey Ranch ProblemUsing QM for Windows


Program 7.3


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Program 7.4A

Excel 2010 Spreadsheet for the Holiday Meal TurkeyRanch problem


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Program 7.4B

Excel 2010 Solution to the Holiday Meal TurkeyRanch Problem


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Four Special Cases in LP

No feasible solution This 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 Constraints

Figure 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 willbe open ended.

This usually means the problem has beenformulated improperly.


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A Feasible Region That is Unbounded to the Right

15

10

5

0

X2

| | | | |

5 10 15 X1Figure 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 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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Example of Alternate Optimal Solutions

8

7

6

5

4

3

2

1

0

X2

| | | | | | | |

1 2 3 4 5 6 7 8 X1

Figure 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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Sensitivity Analysis

Optimal solutions to LP problems thus far havebeen found under what are called deterministicassumptions.

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 sensitivea deterministicsolution is to changes in the assumptions of the

model. This is called sensitivity analysis, postoptimality

analysis, parametric programming, or optimalityanalysis.


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Sensitivity Analysis

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

Changes in the


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Changes in theObjective Function Coefficient

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.

Changes in the


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Changes in theObjective Function Coefficient

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

Old Profit Line for 50X1+ 120X2(Passes through Point a)

Profit Line for 50X1+ 150X2(Passes through Point a)

QM for Windows and Changes in


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QM for Windows and Changes inObjective Function Coefficients

Input and Sensitivity Analysis for High Note SoundData Using QM For Windows

Program 7.5B

Program 7.5A

Excel Solver and Changes in


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Excel Solver and Changes inObjective Function Coefficients

Excel 2010 Spreadsheet for High Note Sound Company

Program 7.6A


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Excel Solver and Changes in


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Excel Solver and Changes inObjective Function Coefficients

Excel 2010 Sensitivity Report for High Note SoundCompany

Program 7.6C

Changes in the


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Changes in theTechnological Coefficients

Changes in the technological coefficientsoftenreflect 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.

Changes in the


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Changes in theTechnological Coefficients

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

StereoReceivers

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

Changes in Resources or


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Changes in Resources orRight-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 muchshould be paid for additional resourcesand how much more of a resource wouldbe useful.



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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 priceor dualvalue.

The dual price for a constraint is the improvement

in the objective function value that results from aone-unit increase in the right-hand side of theconstraint.



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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.

Changes in the Electricians Time Resource


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Changes in the Electricians Time Resourcefor the High Note Sound Company

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 100Hoursof Electricians Time Resource

Figure 7.19



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60

40

20

15

| | |

0 20 40 60

|

30 X1

X2 (b)

a

b

c

Constraint Representing 60 Hours ofAudio Technicians Time Resource


Figure 7.19



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60

40

20

| | | | | |

0 20 40 60 80 100 120X1

X2 (c)

ConstraintRepresenting60 Hours of AudioTechniciansTime Resource


Figure 7.19


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Documents

Slides for Ch07