pom chap 2 - 8
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Linear Programming:Model Formulation
andGraphical Solution
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Chapter Topics
Model Formulation
A Maximization Model Example
Graphical Solutions of Linear Programming Models
A Minimization Model Example
Irregular Types of Linear Programming Models
Characteristics of Linear Programming Problems
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Definition And Characteristics OfLinear Programming
Linear Programming is that branch ofmathematical programming which is designedto solve optimization problems where all theconstraints as will as the objectives areexpressed as Linear function .It wasdeveloped by George B. Denting in 1947. Itsearlier application was solely related to the
activities of the second World War. Howeversoon its importance was recognized and itcame to occupy a prominent place in the
industry and trade. 3
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Objectives of business firms frequently include maximizingprofit or minimizing costs.
Linear programming is an analysis technique in which linearalgebraic relationships represent a firms decisions given a
business objective and resource constraints.Steps in application:
Identify problem as solvable by linear programming.
Formulate a mathematical model of the unstructuredproblem.
Solve the model.
Linear ProgrammingAn Overview
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TERMINOLOGY
5
A LINEAR PROGRAMMING PROBLEM IS SAID TO BE IN
STANDARD FORM WHEN IT IS WRITTEN AS:
THE PROBLEM HAS MVARIABLES AND NCONSTRAINTS.IT MAY BE WRITTEN USING VECTOR TERMINOLOGY
AS:
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NOTE THAT A PROBLEM WHEREWE WOULD LIKE TO MINIMIZE THECOST FUNCTION INSTEAD OFMAXIMIZE IT MAY BE REWRITTEN IN
STANDARD FORM BY NEGATING THECOST COEFFICIENTS Cj(CT).
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Any vector x satisfying the constraints of the linear
programming problem is called a feasible solution of the problem.
Every linear programming problem falls into one of three categories:1. Infeasible. A linear programming problem is infeasible if a
feasible solution to the problem does not exist; that is, there is
no vectorx for which all the constraints of the problem are
satisfied.
2. Unbounded. A linear programming problem is unboundedif
the constraints do not sufficiently restrain the cost function so
that for any given feasible solution, another feasible solution
can be found that makes a further improvement to the cost
function.
3.Has an optimal solution. Linear programming problems that
are not infeasible or unbounded have an optimal solution; that
is, the cost function has a unique minimum (or maximum) cost
function value. This does not mean that the values of the
variables that yield that optimal solution are unique, however.
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Decision variables - mathematical symbols representinglevels of activity of a firm.
Objective function - a linear mathematical relationshipdescribing an objective of the firm, in terms of decisionvariables, that is maximized or minimized
Constraints - restrictions placed on the firm by theoperating environment stated in linear relationships of thedecision variables.
Parameters - numerical coefficients and constants used inthe objective function and constraint equations.
Model Components and Formulation
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STEPS FOR DEVELOPING ANALGEBRAIC LP MODEL
What decisions need to be made? Defineeach decision variable.
What is the goal of the problem? Write down
the objective function as a function of thedecision variables.
What resources are in short supply and/or
what requirements must be met?Formulate the constraints as functions of
the decision variables.
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Steps for Developing anLP Model in a Spreadsheet
Enter all of the data for the model. Make consistent use of rows andcolumns.
Make a cell for each decision to be made (changing cells). Follow thesame structure as the data. (Sometimes it is easier to do step 2 beforestep 1.)
What is the goal of the problem? Enter the equation that measures theobjective in a single cell on the worksheet (target cell). Typically:
SUMPRODUCT (Cost Data, Changing Cells)
or
SUMPRODUCT (Profit Data, Changing Cells)
What resources are in short supply and/or what requirements must bemet? Enter all constraints in three consecutive cells on thespreadsheet, e.g.,
amount of resource used , = , minimum requirement10
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Resource Requirements
ProductLabor
(hr/unit)Clay
(lb/unit)Profit
($/unit)
Bowl 1 4 40
Mug 2 3 50
Problem DefinitionA Maximization Model Example (1 of 2)
Product mix problem - Beaver Creek Pottery CompanyHow many bowls and mugs should be produced tomaximize profits given labor and materials constraints?
Product resource requirements and unit profit:
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Problem DefinitionA Maximization Model Example (2 of 3)
Resource 40 hrs of labor per dayAvailability: 120 lbs of clay
Decision x1 = number of bowls to produce per dayVariables: x2 = number of mugs to produce per day
Objective Maximize Z = $40x1 + $50x2Function: Where Z = profit per day
Resource 1x1 + 2x2 40 hours of labor
Constraints: 4x1 + 3x2 120 pounds of clayNon-Negativity x1 0; x2 0Constraints:
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Problem DefinitionA Maximization Model Example (3 of 3)
Complete Linear Programming Model:
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40
4x2 + 3x2
120x1, x2 0
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A feasible solution does not violate any of the constraints:
Example x1 = 5 bowls
x2 = 10 mugs
Z = $40x1 + $50x2 = $700
Labor constraint check:1(5) + 2(10) = 25 < 40 hours, within constraint
Clay constraint check:
4(5) + 3(10) = 70 < 120 pounds, within constraint
Feasible Solutions
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An infeasible solution violates at least one of theconstraints:
Example x1 = 10 bowls
x2 = 20 mugs
Z = $1400Labor constraint check:
1(10) + 2(20) = 50 > 40 hours, violates constraint
Infeasible Solutions
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Graphical solution is limited to linear programming modelscontaining only two decision variables (can be used withthree variables but only with great difficulty).
Graphical methods provide visualization of how a solutionfor a linear programming problem is obtained.
Graphical Solution of Linear Programming Models
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The graphic solution procedure is
one of the method of solving twovariable Linear programmingproblems. It consists of the following
steps:-
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tep
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tepDefining the problem. Formulate the problemmathematically. Express it in terms of several
mathematical constraints & an objectivefunction. The objective function relates to theoptimization aspect is, maximisation or
minimisation Criterion.
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Step IIPlot the constraints Graphically. Each inequality inthe constraint equation has to be treated as an
equation. An arbitrary value is assigned to onevariable & the value of the other variable is obtainedby solving the equation. In the similar manner, adifferent arbitrary value is again assigned to the
variable & the corresponding value of other variableis easily obtained.These 2 sets of values are now plotted on a graphand connected by a straight line. The sameprocedure has to be repeated for all the constraints.Hence, the total straight lines would be equal to thetotal no of equations, each straight line representing
one constraint equation. Q 32
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Coordinate AxesGraphical Solution of Maximization Model (1 of 12)
Maximize Z = $40x1
+ $50x2
subject to: 1x1 + 2x2 40
4x2 + 3x2 120x1, x2 0
Figure 2.1Coordinates for Graphical Analysis
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Labor ConstraintGraphical Solution of Maximization Model (2 of 12)
Maximize Z = $40x1
+ $50x2
subject to: 1x1 + 2x2 40
4x2 + 3x2 120x1, x2 0
Figure 2.1Graph of Labor Constraint
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Labor Constraint AreaGraphical Solution of Maximization Model (3 of 12)
Maximize Z = $40x1
+ $50x2
subject to: 1x1 + 2x2 40
4x2 + 3x2 120x1, x2 0
Figure 2.3Labor Constraint Area
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Clay Constraint AreaGraphical Solution of Maximization Model (4 of 12)
Maximize Z = $40x1
+ $50x2
subject to: 1x1 + 2x2 40
4x2 + 3x2 120x1, x2 0
Figure 2.4Clay Constraint Area
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Both ConstraintsGraphical Solution of Maximization Model (5 of 12)
Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 40
4x2 + 3x2 120x1, x2 0
Figure 2.5Graph of Both Model Constraints
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Feasible Solution AreaGraphical Solution of Maximization Model (6 of 12)
Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 40
4x2 + 3x2 120x1, x2 0
Figure 2.6Feasible Solution Area
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Objective Solution = $800Graphical Solution of Maximization Model (7 of 12)
Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 40
4x2 + 3x2 120x1, x2 0
Figure 2.7Objection Function Line for Z = $800
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Alternative Objective Function Solution LinesGraphical Solution of Maximization Model (8 of 12)
Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 40
4x2 + 3x2 120x1, x2 0
Figure 2.8Alternative Objective Function Lines
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Optimal SolutionGraphical Solution of Maximization Model (9 of 12)
Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 40
4x2 + 3x2 120x1, x2 0
Figure 2.9Identification of Optimal Solution
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Optimal Solution CoordinatesGraphical Solution of Maximization Model (10 of 12)
Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 40
4x2 + 3x2 120x1, x2 0
Figure 2.10Optimal Solution Coordinates
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Corner Point SolutionsGraphical Solution of Maximization Model (11 of 12)
Maximize Z = $40x1 + $50x2subject to: 1x1 + 2x2 40
4x2 + 3x2 120x1, x2 0
Figure 2.11Solution at All Corner Points
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Optimal Solution for New Objective FunctionGraphical Solution of Maximization Model (12 of 12)
Maximize Z = $70x1 + $20x2subject to: 1x1 + 2x2 40
4x2 + 3x2 120x1, x2 0
Figure 2.12Optimal Solution with Z = 70x1 + 20x2
S
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Standard form requires that all constraints be in the form of
equations.
A slack variable is added to a constraint to convert it to anequation (=).
A slack variable represents unused resources.A slack variable contributes nothing to the objective functionvalue.
Slack Variables
Li P i M d l
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Linear Programming ModelStandard Form
Max Z = 40x1 + 50x2 + s1 + s2subject to:1x1 + 2x2 + s1 = 40
4x2 + 3x2 + s2 = 120
x1, x2, s1, s2
0Where:
x1 = number of bowlsx2 = number of mugs
s1, s2 are slack variables
Figure 2.13Solution Points A, B, and C with Slack
P bl D fi iti
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Problem DefinitionA Minimization Model Example (1 of 7)
Chemical Contribution
Brand Nitrogen(lb/bag)
Phosphate(lb/bag)
Super-gro 2 4
Crop-quick 4 3
Two brands of fertilizer available - Super-Gro, Crop-Quick.Field requires at least 16 pounds of nitrogen and 24 poundsof phosphate.
Super-Gro costs $6 per bag, Crop-Quick $3 per bag.
Problem: How much of each brand to purchase to minimizetotal cost of fertilizer given following data ?
P bl D fi iti
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Problem DefinitionA Minimization Model Example (2 of 7)
Decision Variables:x1 = bags of Super-Grox2 = bags of Crop-Quick
The Objective Function:
Minimize Z = $6x1 + 3x2Where: $6x1 = cost of bags of Super-Gro$3x2 = cost of bags of Crop-Quick
Model Constraints:
2x1 + 4x2 16 lb (nitrogen constraint)4x1 + 3x2 24 lb (phosphate constraint)x1, x2 0 (non-negativity constraint)
M d l F l ti d C t i t G h
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Model Formulation and Constraint GraphA Minimization Model Example (3 of 7)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2 164x2 + 3x2 24
x1, x2 0
Figure 2.14Graph of Both Model Constraints
F ibl S l ti A
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Feasible Solution AreaA Minimization Model Example (4 of 7)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2 164x2 + 3x2 24
x1, x2 0
Figure 2.15Feasible Solution Area
O ti l S l ti P i t
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Optimal Solution PointA Minimization Model Example (5 of 7)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2 164x2 + 3x2 24
x1, x2 0
Figure 2.16Optimum Solution Point
S l V i bl
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Surplus VariablesA Minimization Model Example (6 of 7)
A surplus variable is subtracted from a
constraint toconvert it to an equation (=).
A surplus variable represents an excess above a constraintrequirement level.
Surplus variables contribute nothing to the calculated valueof the objective function.
Subtracting slack variables in the farmer problemconstraints:
2x1 + 4x2 - s1 = 16 (nitrogen)4x1 + 3x2 - s2 = 24 (phosphate)
Graphical Sol tions
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Minimize Z = $6x1 + $3x2 + 0s1 + 0s2
subject to: 2x1 + 4x2 s1 = 164x2 + 3x2 s2 = 24x1, x2, s1, s2 0
Figure 2.17Graph of Fertilizer Example
Graphical SolutionsA Minimization Model Example (7 of 7)
Irregular Types of Linear Programming Problems
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For some linear programming models, the general rules do
not apply.
Special types of problems include those with:
Multiple optimal solutions
Infeasible solutionsUnbounded solutions
Irregular Types of Linear Programming Problems
Multiple Optimal Solutions
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Objective function is parallelto a constraint line.
Maximize Z=$40x1 + 30x2
subject to: 1x1 + 2x2 404x2 + 3x2 120x1, x2 0
Where:x
1= number of bowls
x2 = number of mugs
Figure 2.18Example with Multiple Optimal Solutions
Multiple Optimal SolutionsBeaver Creek Pottery Example
An Infeasible Problem
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An Infeasible Problem
Every possible solutionviolates at least oneconstraint:
Maximize Z = 5x1 + 3x2subject to: 4x1 + 2x2 8
x1 4x2 6
x1, x2 0
Figure 2.19Graph of an Infeasible Problem
An Unbounded Problem
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Value of objective functionincreases indefinitely:
Maximize Z = 4x1 + 2x2subject to: x1 4
x2 2x1, x2 0
An Unbounded Problem
Figure 2.20Graph of an Unbounded Problem
Characteristics of Linear Programming Problems
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Characteristics of Linear Programming Problems
A linear programming problem requires a decision - a
choice amongst alternative courses of action.
The decision is represented in the model by decisionvariables.
The problem encompasses a goal, expressed as anobjective function, that the decision maker wants toachieve.
Constraints exist that limit the extent of achievement of the
objective.The objective and constraints must be definable by linearmathematical functional relationships.
Properties of Linear Programming Models
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Proportionality - The rate of change (slope) of the
objective function and constraint equations is constant.
Additivity - Terms in the objective function and constraintequations must be additive.
Divisability -Decision variables can take on any fractionalvalue and are therefore continuous as opposed to integerin nature.
Certainty - Values of all the model parameters are
assumed to be known with certainty (non-probabilistic).
Properties of Linear Programming Models
Problem Statement
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Problem StatementExample Problem No. 1 (1 of 3)
Hot dog mixture in 1000-pound batches.Two ingredients, chicken ($3/lb) and beef ($5/lb).
Recipe requirements:
at least 500 pounds of chicken
at least 200 pounds of beef
Ratio of chicken to beef must be at least 2 to 1.
Determine optimal mixture of ingredients that will minimize
costs.
Solution
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Step 1:Identify decision variables.
x1 = lb of chicken
x2 = lb of beef
Step 2:
Formulate the objective function.
Minimize Z = $3x1 + $5x2
where Z = cost per 1,000-lb batch$3x1 = cost of chicken$5x2 = cost of beef
SolutionExample Problem No. 1 (2 of 3)
Solution
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SolutionExample Problem No. 1 (3 of 3)
Step 3:Establish Model Constraints
x1 + x2 = 1,000 lbx1 500 lb of chickenx
2 200 lb of beef
x1/x2 2/1 or x1 - 2x2 0x1, x2 0
The Model: Minimize Z = $3x1 + 5x2subject to: x
1+ x
2= 1,000 lb
x1 50x2 200x1 - 2x2 0x1,x2 0
Example Problem No 2 (1 of 3)
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Solve the following modelgraphically:
Maximize Z = 4x1 + 5x2
subject to: x1 + 2x2 10
6x1 + 6x2 36
x1 4
x1, x2 0
Step 1: Plot the constraintsas equations
Example Problem No. 2 (1 of 3)
Figure 2.21Constraint Equations
Example Problem No 2 (2 of 3)
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Example Problem No. 2 (2 of 3)
Maximize Z = 4x1 + 5x2
subject to: x1 + 2x2 10
6x1 + 6x2 36
x1 4
x1, x2 0
Step 2: Determine thefeasible solution space
Figure 2.22Feasible Solution Space and Extreme Points
Example Problem No 2 (3 of 3)
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Example Problem No. 2 (3 of 3)
Maximize Z = 4x1 + 5x2
subject to: x1 + 2x2 10
6x1 + 6x2 36
x1 4
x1, x2 0
Step 3 and 4: Determine the
solution points and optimalsolution
Figure 2.22Optimal Solution Point