graphical method math statistics computation calculus algebra management science

7/29/2019 graphical method math statistics computation calculus algebra management science

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2009 Prentice-Hall, Inc. 7 1

Linear Programming Models:Graphical and Computer Methods

Many management decisions involve trying tomake the most effective use of l imited resources Machinery, labor, money, time, warehouse space, raw

materials

Linear programmingLinear programming(LPLP) is a widely usedmathematical modeling technique designed tohelp managers in planning and decision makingrelative to resource allocation

Belongs to the broader field ofmathematicalmathematicalprogrammingprogramming

In this sense,programmingprogrammingrefers to modeling andsolving a problem mathematically


Requirements of a LinearProgramming Problem

LP has been applied in many areas overthe past 50 years

All LP problems have 4 properties incommon

1. All problems seek to maximizemaximize orminimizeminimizesome quantity (the objective functionobjective function)

2. The presence of restrictions orconstraintsconstraints thatlimit the degree to which we can pursue ourobjective

3. There must be alternative courses of action tochoose from

4. The objective and constraints in problemsmust be expressed in terms of linearlinearequationsorinequalitiesinequalities


Examples of SuccessfulLP Applications

Development of a production schedule that will satisfy future demands for a firms production

while minimizingminimizingtotal production and inventory costs

Determination of grades of petroleum products to yieldthe maximummaximum profit

Selection of different blends of raw materials to feedmills to produce finished feed combinations atminimumminimum cost

Determination of a distribution system that willminimizeminimize total shipping cost from several warehousesto various market locations


LP Properties and Assumptions

PROPERTIES OF LINEAR PROGRAMS

1. O ne 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


Basic Assumptions of LP

We assume conditions ofcertaintycertaintyexist and numbersin the objective and constraints are known withcertainty and do not change during the period beingstudied

We assumeproportionalityproportionalityexists in the objective andconstraints constancy between production increases and resource

utilization if 1 unit needs 3 hours then 10 require 30hours

We assume additivityadditivityin that the total of all activitiesequals the sum of the individual activities

We assume divisibilitydivisibilityin that solutions need not be

whole numbers All answers or variables are nonnegativenonnegative as we are

dealing with real physical quantities


Formulating LP Problems

Formulating a linear program involvesdeveloping a mathematical model to representthe managerial problem

The steps in formulating a linear program are1. Completely understand the managerial

problem being faced2. Identify the objective and constraints3. Define the decision variables4. Use the decision variables to write

mathematical expressions for the objectivefunction and the constraints


2/13


Formulating LP Problems

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

Two or more products are produced using limitedresources such as personnel, machines, and rawmaterials

The profit that the firm seeks to maximize isbased on the profit contribution per unit of eachproduct

The company would like to determine how manyunits of each product it should produce so as tomaximize overall profit given its l imitedresources


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 thepainting 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 of

painting 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 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 actualdecisions 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 ofTand 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)



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 forTand Cmust be nonnegative

T 0 (number of tables produced is greaterthan or equal to 0)

C 0 (number of chairs produced is greaterthan or equal to 0)

The complete problem stated mathematicallyMaximize profit = $70T+ $50C

subject to

4T+ 3C 240 (carpentry constraint)

2T+ 1C 100 (painting and varnishing constraint)

T, C 0 (nonnegativity constraint)


Graphical Solution 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 isnot possible to plot the solution on a two-dimensional graph

The graphical method provides valuableinsight into how other approaches work


Graphical Representation of aConstraint

100

80

60

40

20

C

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

0 20 40 60 80 100 T

NumberofChairs

Number of Tables

This Axis Represents the Constraint T 0

This Axis Represents theConstraintC 0

Figure 7.1



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 drawthe line



When Flair produces no tables, thecarpentry constraint is

4(0) + 3C= 2403C= 240C= 80

Similarly for no chairs4T+ 3(0) = 240

4T= 240T= 60

This line is shown on the following graph



100

80

60

40

20

C

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

0 20 40 60 80 100 T

NumberofChairs

Number of Tables

(T= 0, C= 80)

Figure 7.2

(T= 60, C= 0)

Graph of carpentry constraint equation


4/13



100

80

60

40

20

C

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

0 20 40 60 80 100 T

Numbe

rofChairs

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)



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 and

satisfies the constraint4(30) + 3(20) = 180

The point (30, 40) lies above the plot anddoes not satisfy the constraint

4(70) + 3(40) = 400



100

80

60

40

20

C

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

0 20 40 60 80 100 T

NumberofChairs

Number of Tables

(T= 0, C= 100)

Figure 7.4

(T= 50, C= 0)

Graph of painting and varnishingconstraint equation



To produce tables and chairs, bothdepartments must be used

We need to find a solution that satisfies bothconstraints simultaneouslysimultaneously

A new graph shows both constraint plots

The feasible regionfeasible region (orarea of feasiblearea of feasiblesolutionssolutions) is where all constraints are satisfied

Any point inside this region is a feasiblefeasiblesolution

Any point outside the region is an infeasibleinfeasiblesolution



100

80

60

40

20

C

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

0 20 40 60 80 100 T

NumberofChairs


Feasible solution region for Flair Furniture

Painting/Varnishing Constraint

Carpentry ConstraintFeasibleRegion



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


Carpentryconstraint


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

Paintingconstraint


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


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Carpentryconstraint


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

Paintingconstraint


(2)(50) + (1)(5) = 105 hours used


Isoprofit Line Solution Method

Once the feasible region has been graphed, we needto find the optimal solution from the many possiblesolutions

The speediest way to do this is to use the isoprofitline method

Starting with a small but possible profit value, wegraph the objective function

We move the objective function line in the direction ofincreasing profit while maintaining the slope

The last point it touches in the feasible region is theoptimal solution



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 the

graph This is obviously not the best possible solution Further graphs can be created using larger profi ts

The further we move from the origin whilemaintaining the slope and staying within theboundaries of the feasible region, the larger theprofit will be

The highest profit ($4,100) will be generated whenthe isoprofit line passes through the point (30, 40)


100

80

60

40

20

C

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

0 20 40 60 80 100 T

NumberofChairs


Isoprofit line at $2,100

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

(30, 0)

(0, 42)



100

80

60

40

20

C

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

0 20 40 60 80 100 T

NumberofChairs


Four isoprofit lines

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

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

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

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



100

80

60

40

20

C

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

0 20 40 60 80 100 T

NumberofChairs


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 methodcorner point method

It involves looking at the profit at everycorner point of the feasible region

The mathematical theory behind LP is that

the optimal solution must lie at one of thecorner pointscorner points, orextreme pointextreme point, in thefeasible region

For Flair Furniture, the feasible region is afour-sided polygon with four corner pointslabeled 1, 2, 3, and 4 on the graph

Corner Point Solution Method


100

80

60

40

20

C

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

0 20 40 60 80 100 T

Numbe

rofChairs


Four corner points ofthe feasible region

1

2

3

4




3

1

2

4

Point : (T= 0, C= 0) Profi t = $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 wehave to solve for the intersection of the twoconstraint lines

The details of this are on the following slide

3

3



Using the simultaneous equations methodsimultaneous equations method, wemultiply the painting equation by 2 and add it tothe carpentry equation

4T+ 3C= 240 (carpentry line)

4T 2C=200 (painting line)C= 40

Substituting 40 forC in either of the originalequations allows us to determine the value ofT

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

T= 30


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 thefeasible region is the optimal solution.

4. Find the values of the decision variables at this last point and compute theprofit (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 in

Step 3. This is the optimal solution.

Table 7.3


Solving Flair Furnitures LP ProblemUsing QM for Windows and Excel

Most organizations have access tosoftware to solve big LP problems

While there are differences betweensoftware implementations, the approacheach takes towards handling LP isbasically the same

Once you are experienced in dealing withcomputerized LP algorithms, you caneasily adjust to minor changes


7/13


Using QM for Windows

First select the Linear Programming module

Specify the number of constraints (non-negativityis assumed)

Specify the number of decision variables

Specify whether the objective is to be maximized

or minimized For the Flair Furniture problem there are two

constraints, two decision variables, and theobjective is to maximize profit



Computer screen for input of data

Program 7.1A



Computer screen for input of data

Program 7.1B



Computer screen for output of solution

Program 7.1C



Graphical output of solution

Program 7.1D


Solving Minimization Problems

Many LP problems involve minimizing anobjective such as cost instead ofmaximizing a profit function

Minimization problems can be solvedgraphically by first setting up the feasiblesolution region and then using either thecorner point method or an isocost lineapproach (which is analogous to theisoprofit approach in maximizationproblems) to find the values of the decisionvariables (e.g., X1 and X2) that yield the

minimum cost


8/13


The Holiday Meal Turkey Ranch is consideringbuying two different brands of turkey feed andblending them to provide a good, low-cost diet forits turkeys

Minimize cost (in cents) = 2X1 + 3X2subject to:

5X1 + 10X2 90 ounces (ingredient constraint A)

4X1 + 3X2 48 ounces (ingredient constraint B)0.5X1 1.5 ounces (ingredient constraint C)

X1 0 (nonnegativity constraint)

X2 0 (nonnegativity constraint)

Holiday Meal Turkey Ranch

X1 = number of pounds of brand 1 feed purchased

X2 = number of pounds of brand 2 feed purchased

Let



INGREDIENT

COMPOSITION OF EACH POUNDOF FEED (OZ.) MINIMUM MONTHLY

REQUIREMENT PERTURKEY (OZ.)BRAND 1 FEED BRAND 2 FEED

A 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


Using the cornerpoint method

First we constructthe feasiblesolution region

The optimalsolution will lie aton of the cornersas it would in amaximizationproblem


20

15

10

5

0

X2

| | | | | |

5 10 15 20 25 X1

PoundsofBrand2

Pounds of Brand 1

Ingredient C Constraint

Ingredient B Constraint

Ingredient A Constraint

Feasible Region

a

b

c

Figure 7.10



We solve for the values of the three corner points

Point a is 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 b with basic algebra we find X1 =8.4 and X2 = 4.8

Solving for point cwe find X1 = 18 and X2 = 0


Substituting these value back into the objectivefunction we find

Cost = 2X1 + 3X2Cost at point a = 2(3) + 3(12) = 42

Cost 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


Using the isocostapproach

Choosing aninitial cost of 54cents, it is clearimprovement ispossible


20

15

10

5

0

X2

| | | | | |

5 10 15 20 25 X1

PoundsofBrand2

Pounds of Brand 1Figure 7.11

Feasible Region

(X1 = 8.4, X2 = 4.8)


9/13


QM for Windows can also be used to solve theHoliday Meal Turkey Ranch problem


Program 7.3


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



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



A problem with no feasiblesolution

8

6

4

2

0

X2

| | | | | | | | | |

2 4 6 8 X1

Region Satisfying First Two ConstraintsRegion Satisfying First Two ConstraintsFigure 7.12

Region SatisfyingThird Constraint

X1+2X2


10/13



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 real

world It causes no particular problems, but

eliminating redundant constraints simplifiesthe model



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



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 toselect as the profit is the same at eachalternate solution



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 ofX1 and X2 AlongtheAB Segment

Isoprofit Line for $12Overlays Line SegmentAB

B

A

Maximize 3X1 + 2X2Subj. To: 6X1 + 4X2 < 24

X1 < 3

X1, X2 > 0


Sensitivity Analysis

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

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 sensitivesensitive a deterministicsolution is to changes in the assumptions of themodel

This is called sensitivity analysissensitivity analysis,postoptimalitypostoptimality

analysisanalysis,parametric programmingparametric programming, oroptimalityoptimalityanalysisanalysis



Sensitivity analysis often involves a series ofwhat-if? questions concerning constraints,variable coefficients, and the objective function What if the profit for product 1 increases by 10%? What if less advertising money is available?

One way to do this is the trial-and-error methodwhere 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 arange of changes in problem parameters that will notaffect the optimal solution or change the variables inthe solution without re-solving the entire problem


11/13



Sensitivity analysis can be used to deal notonly with errors in estimating inputparameters to the LP model but also withmanagements experiments with possiblefuture changes in the firm that may affect

profits.


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 + $120X2Subject to 2X1 + 4X2 80 (hours of electricians

time available)

3X1 + 1X2 60 (hours of audiotechnicians timeavailable)

X1, X2 0


The High Note Sound Company graphical solution

High Note Sound Company

b = (16, 12)

Optimal Solution at Point a

X1 = 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 theObjective Function Coefficient

In real-l ife problems, contribution rates in theobjective functions fluctuate periodically

Graphically, this means that although the feasiblesolution region remains exactly the same, theslope of the isoprofit or isocost line will change

We can often make modest increases ordecreases 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 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 60

X1

c

Figure 7.17

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

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


QM for Windows and Changes inObjective Function Coefficients

Input and sensitivity analysis for High Note Sounddata

Program 7.5B

Program 7.5A


12/13


Changes in theTechnological Coefficients

Changes in the technological coefficientstechnological coefficients oftenreflect changes in the state of technology

If the amount of resources needed to produce aproduct changes, coefficients in the constraintequations will change

This does not change the objective function, butit can produce a significant change in the shapeof the feasible region

This may cause a change in the optimal solution


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

StereoReceiv

ers

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 orRight-Hand-Side Values

The right-hand-side values of the constraintsoften represent resources available to the firm

If additional resources were available, a highertotal profit could be reali zed

Sensitivity analysis about resources will helpanswer questions such as: How much should the company be willing to pay

for additional hours?

Is it profitable to have some electricians workovertime?

Should we be willing to pay for more audiotechnician time?



If the right-hand side of a constraint ischanged, the feasible region will change(unless the constraint is redundant) andoften the optimal solution will change

The amount of change (increase ordecrease) in the objective function valuethat results from a unit change in one ofthe resources available is called the dualdualpriceprice ordual valuedual value



However, the amount of possible increase in theright-hand side of a resource is limited

If the number of hours increases beyond theupper bound (or decreases below the lowerbound), then the objective function would nolonger increase (decrease) by the dual price. There may be excess (slack) hours of a resource

or the objective function may change by anamount different from the dual price.

Thus, the dual price is relevant only within limits. If the dual value of a constraint is zero

The slack is positive, indicating unused resource Additional amount of resource will simply increase

the amount of slack. The upper limit of infinity indicates that addingmore hours would simply increase the amount ofslack.


Changes in the Electrician's Timefor 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

If the electricians hours are changed from 80 to100, the new optimal solution is (0,25) with profitof $3,000. The extra 20 hours resulted in anincrease in profit of $600 or $30/hour


13/13



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 6060 Hoursof Electricians Time Resource

Figure 7.19

If the hours are decreased to 60,the new solution is (0,15) and the

profit is $1,800. Reducing hours

by 20 results in a $600 decrease inprofit or $30/hour



60

40

20

| | | | | |

0 20 40 60 80 100 120X1

X2 (c)

ConstraintRepresenting60 Hours of AudioTechniciansTime Resource

Changed Constraint Representing240240 Hoursof Electricians Time Resource

Figure 7.19



If total electrician time was increased to 240,the optimal solution would be (0,60) with aprofit of $7,200. This is $2,400 (the originalsolution) + $30 (dual price)*160 hours(240-80)

If the hours increases beyond 240, then theoptimal solution would still be (0,60) and profitwould not increase. The extra time is slack duringwhich the electricians are not working


QM for Windows and Changes inRight-Hand-Side Values

Input and sensitivity analysis for High Note Sounddata

Program 7.5B

Program 7.5A


Flair Furniture SensitivityAnalysis

Dual/ValueDua l/ Va lue RHS changeRHS change Solut ion changeSolution change

C ar pe nt ry /1 .5 2 40241 30/40: 410

29.5/41: 411.5

Painting/. 5 100101 30/40: 410

31.5/38: 410.5

Profit per itemProf it per item Solut ionSolution RangeRange

Table 7 6.67 10

Chair 5 3.5 5.25


Personal Mini Warehouses

1. For the optimal solution, how much is spent onadvertising?

2. For the optimal solution, how much square footagewill be used?

3. Would the solution change if the advertising budgetwere only $300 instead of $400? Why?

4. What would the optimal solution be if the profit onthe large spaces were reduced from $50 to $45?

5. How much would earnings increase if the squarefootage requirement were increased from 8,000 to9,000?

1. 2*60 + 4*40 = 280 (note slack of 120 on advertising constraint)

2. 100*60 + 50*40 = 8,000 (no slack)

3. Advertising has a lower bound of 280 and upper bound of Infinity. 300 is within thebounds so the solution does not change.

4. The profit on large spaces has a range of 40 to Infinity, 45 is w ithin that range so thesolution would not change. However, earnings would be reduced from $3,800 to 45*60 +

20*40 = $3,500.5. The dual price for square footage is .4 and the upper bound is 9,500. By increasing thesquare footage by 1,000 we increase the earning by .4*1,000 = $400

graphical method math statistics computation calculus algebra management science

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