graphical method math statistics computation calculus algebra management science
TRANSCRIPT
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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
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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
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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
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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
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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
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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
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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
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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 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
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Flair Furniture Company
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
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Flair Furniture 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 actualdecisions we will make are
T= number of tables to be produced per week
C= number of chairs to be produced per week
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Flair Furniture Company
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)
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Flair Furniture Company
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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Flair Furniture Company
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)
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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
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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
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Graphical Representation of aConstraint
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
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Graphical Representation of aConstraint
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
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Graphical Representation of aConstraint
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
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Graphical Representation of aConstraint
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)
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Graphical Representation of aConstraint
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
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Graphical Representation of aConstraint
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
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Graphical Representation of aConstraint
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
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Graphical Representation of aConstraint
100
80
60
40
20
C
| | | | | | | | | | | |
0 20 40 60 80 100 T
NumberofChairs
Number of TablesFigure 7.5
Feasible solution region for Flair Furniture
Painting/Varnishing Constraint
Carpentry ConstraintFeasibleRegion
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Graphical Representation of aConstraint
For the point (30, 20)
Carpentryconstraint
4T+ 3C 240 hours available
(4)(30) + (3)(20) = 180 hours used
Paintingconstraint
2T+ 1C 100 hours available
(2)(30) + (1)(20) = 80 hours used
For the point (70, 40)
Carpentryconstraint
4T+ 3C 240 hours available
(4)(70) + (3)(40) = 400 hours used
Paintingconstraint
2T+ 1C 100 hours available
(2)(70) + (1)(40) = 180 hours used
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Graphical Representation of aConstraint
For the point (50, 5)
Carpentryconstraint
4T+ 3C 240 hours available
(4)(50) + (3)(5) = 215 hours used
Paintingconstraint
2T+ 1C 100 hours available
(2)(50) + (1)(5) = 105 hours used
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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
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Isoprofit Line Solution 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 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)
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100
80
60
40
20
C
| | | | | | | | | | | |
0 20 40 60 80 100 T
NumberofChairs
Number of TablesFigure 7.6
Isoprofit line at $2,100
$2,100 = $70T + $50C
(30, 0)
(0, 42)
Isoprofit Line Solution Method
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100
80
60
40
20
C
| | | | | | | | | | | |
0 20 40 60 80 100 T
NumberofChairs
Number of TablesFigure 7.7
Four isoprofit lines
$2,100 = $70T + $50C
$2,800 = $70T + $50C
$3,500 = $70T + $50C
$4,200 = $70T + $50C
Isoprofit Line Solution Method
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100
80
60
40
20
C
| | | | | | | | | | | |
0 20 40 60 80 100 T
NumberofChairs
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
Isoprofit Line Solution Method
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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
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100
80
60
40
20
C
| | | | | | | | | | | |
0 20 40 60 80 100 T
Numbe
rofChairs
Number of TablesFigure 7.9
Four corner points ofthe feasible region
1
2
3
4
Corner Point Solution Method
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Corner Point Solution Method
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
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Corner Point Solution Method
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
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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 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
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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
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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
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Using QM for Windows
Computer screen for input of data
Program 7.1A
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Using QM for Windows
Computer screen for input of data
Program 7.1B
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Using QM for Windows
Computer screen for output of solution
Program 7.1C
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Using QM for Windows
Graphical output of solution
Program 7.1D
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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
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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
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Holiday Meal Turkey Ranch
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
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Using the cornerpoint method
First we constructthe feasiblesolution region
The optimalsolution will lie aton of the cornersas it would in amaximizationproblem
Holiday Meal Turkey Ranch
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
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Holiday Meal Turkey Ranch
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
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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
Holiday 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 54cents, it is clearimprovement ispossible
Holiday Meal Turkey Ranch
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)
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QM for Windows can also be used to solve theHoliday Meal Turkey Ranch problem
Holiday Meal Turkey Ranch
Program 7.3
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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 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
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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 real
world It causes no particular problems, but
eliminating redundant constraints simplifiesthe 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 toselect 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 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
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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
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Sensitivity Analysis
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
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Sensitivity Analysis
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.
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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 + $120X2Subject to 2X1 + 4X2 80 (hours of electricians
time 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 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
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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
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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 60
X1
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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QM for Windows and Changes inObjective Function Coefficients
Input and sensitivity analysis for High Note Sounddata
Program 7.5B
Program 7.5A
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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
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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
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
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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?
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Changes in Resources orRight-Hand-Side Values
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
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Changes in Resources orRight-Hand-Side Values
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.
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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
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7/29/2019 graphical method math statistics computation calculus algebra management science
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Changes in the Electrician's Timefor 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 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
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Changes in the Electrician's Timefor High Note Sound
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
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Changes in the Electrician's Timefor High Note Sound
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
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QM for Windows and Changes inRight-Hand-Side Values
Input and sensitivity analysis for High Note Sounddata
Program 7.5B
Program 7.5A
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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
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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