1 chapter 2 - linear programming: model formulation & graphical solution chapter 2 linear...

1Chapter 2 - Linear Programming: Model Formulation & Graphical Solution

Chapter 2

Linear Programming: Model Formulation and Graphical Solution

Introduction to Management Science

9th Edition

by Bernard W. Taylor III

© 2007 Pearson Education


Chapter TopicsChapter Topics

Model FormulationModel Formulation

A Maximization Model ExampleA Maximization Model Example

Graphical Solutions of Linear Programming ModelsGraphical Solutions of Linear Programming Models

A Minimization Model ExampleA Minimization Model Example

Irregular Types of Linear Programming ModelsIrregular Types of Linear Programming Models

Characteristics of Linear Programming ProblemsCharacteristics of Linear Programming Problems


Objectives of business decisions frequently involve Objectives of business decisions frequently involve maximizing profit or minimizing costs. maximizing profit or minimizing costs.

Linear programming is an analytical technique in which linear Linear programming is an analytical technique in which linear algebraic relationships represent a firm’s decisions, given a algebraic relationships represent a firm’s decisions, given a business objective, and resource constraints.business objective, and resource constraints.

Steps in application:Steps in application:

Identify problem as solvable by linear programming.Identify problem as solvable by linear programming.

Formulate a mathematical model of the unstructured Formulate a mathematical model of the unstructured problem.problem.

Solve the model.Solve the model.

ImplementationImplementation

Linear Programming: An Overview


Decision variablesDecision variables - mathematical symbols - mathematical symbols representing levels of activity of a firm.representing levels of activity of a firm.

Objective functionObjective function - a linear mathematical relationship - a linear mathematical relationship describing an objective of the firm, in terms of decision describing an objective of the firm, in terms of decision variables - this function is to be maximized or minimized.variables - this function is to be maximized or minimized.

ConstraintsConstraints – requirements or restrictions placed on the – requirements or restrictions placed on the firm by the operating environment, stated in linear firm by the operating environment, stated in linear relationships of the decision variables.relationships of the decision variables.

ParametersParameters - numerical coefficients and constants used - numerical coefficients and constants used in the objective function and constraints.in the objective function and constraints.

Model ComponentsModel Components


Summary of Model Formulation Steps

Step 1 : Clearly define the decision variables

Step 2 : Construct the objective function

Step 3 : Formulate the constraints


Resource Requirements

Product Labor

(hr/unit) Clay

(lb/unit) Profit

($/unit)

Bowl 1 4 40

Mug 2 3 50

LP Model FormulationA Maximization Example (1 of 3)

Product mix problem - Beaver Creek Pottery CompanyProduct mix problem - Beaver Creek Pottery Company

How many bowls and mugs should be produced to How many bowls and mugs should be produced to maximize profits given labor and materials constraints?maximize profits given labor and materials constraints?

Product resource requirements and unit profit:Product resource requirements and unit profit:



Resource 40 hrs of labor per dayAvailability: 120 lbs of clay

Decision x1 = number of bowls to produce per day

Variables: x2 = number of mugs to produce per day

Objective Maximize Z = $40x1 + $50x2

Function: Where Z = profit per day

Resource 1x1 + 2x2 40 hours of laborConstraints: 4x1 + 3x2 120 pounds of clay

Non-Negativity x1 0; x2 0 Constraints:



Complete Linear Programming Model:

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40

4x1 + 3x2 120

x1, x2 0


A A feasible solutionfeasible solution does not violate any of the constraints: does not violate any of the constraints:

Example xExample x1 1 = 5 bowls= 5 bowls

xx2 2 = 10 mugs= 10 mugs

Z = $40xZ = $40x11 + $50x + $50x2 2 = $700= $700

Labor constraint check:Labor constraint check:

1(5) + 2(10) = 25 < 40 hours, within constraint1(5) + 2(10) = 25 < 40 hours, within constraint

Clay constraint check:Clay constraint check:

4(5) + 3(10) = 70 < 120 pounds, within constraint4(5) + 3(10) = 70 < 120 pounds, within constraint

Feasible SolutionsFeasible Solutions


An An infeasible solutioninfeasible solution violates at least one of the violates at least one of the constraints:constraints:

Example xExample x11 = 10 bowls = 10 bowls

xx22 = 20 mugs = 20 mugs

Z = $1400Z = $1400

Labor constraint check:Labor constraint check:

1(10) + 2(20) = 50 > 40 hours, violates constraint1(10) + 2(20) = 50 > 40 hours, violates constraint

Infeasible SolutionsInfeasible Solutions


Graphical solution is limited to linear programming Graphical solution is limited to linear programming models containing only two decision variables (can be models containing only two decision variables (can be used with three variables but only with great difficulty).used with three variables but only with great difficulty).

Graphical methods provide visualization of how a Graphical methods provide visualization of how a solution for a linear programming problem is obtainedsolution for a linear programming problem is obtained..

Graphical Solution of LP ModelsGraphical Solution of LP Models


Coordinate AxesGraphical Solution of Maximization Model (1 of 12)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.2 Coordinates for Graphical Analysis


Labor ConstraintGraphical Solution of Maximization Model (2 of 12)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.3 Graph of Labor Constraint


Labor Constraint AreaGraphical Solution of Maximization Model (3 of 12)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.4 Labor Constraint Area


Clay Constraint AreaGraphical Solution of Maximization Model (4 of 12)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.5 Clay Constraint Area


Both ConstraintsGraphical Solution of Maximization Model (5 of 12)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.6 Graph of Both Model Constraints


Feasible Solution AreaGraphical Solution of Maximization Model (6 of 12)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.7 Feasible Solution Area


Objective Function Solution = $800Graphical Solution of Maximization Model (7 of 12)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.8 Objection Function Line for Z = $800


Alternative Objective Function Solution LinesGraphical Solution of Maximization Model (8 of 12)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.9 Alternative Objective Function Lines


Optimal SolutionGraphical Solution of Maximization Model (9 of 12)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.10 Identification of Optimal Solution


Optimal Solution CoordinatesGraphical Solution of Maximization Model (10 of 12)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.11 Optimal Solution Coordinates


Extreme (Corner) Point SolutionsGraphical Solution of Maximization Model (11 of 12)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.12 Solutions at All Corner Points


Optimal Solution for New Objective FunctionGraphical Solution of Maximization Model (12 of 12)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.13 Optimal Solution with Z = 70x1 + 20x2


Standard form requires that all constraints be in the form of Standard form requires that all constraints be in the form of equations (equalities).equations (equalities).

A slack variable is added to a A slack variable is added to a constraint (weak constraint (weak inequality) to convert it to an equation (=).inequality) to convert it to an equation (=).

A slack variable typically represents an unused resource.A slack variable typically represents an unused resource.

A slack variable contributes nothing to the objective function A slack variable contributes nothing to the objective function value.value.

Slack VariablesSlack Variables


Linear Programming Model: Standard Form

Max Z = 40x1 + 50x2 + s1 + s2

subject to:1x1 + 2x2 + s1 = 40 4x1 + 3x2 + s2 = 120 x1, x2, s1, s2 0Where:

x1 = number of bowls x2 = number of mugs s1, s2 are slack variables

Figure 2.14 Solution Points A, B, and C with Slack


LP Model FormulationA Minimization 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.Two brands of fertilizer available - Super-Gro, Crop-Quick.

Field requires at least 16 pounds of nitrogen and 24 pounds Field requires at least 16 pounds of nitrogen and 24 pounds of phosphate.of phosphate.

Super-Gro costs $6 per bag, Crop-Quick $3 per bag.Super-Gro costs $6 per bag, Crop-Quick $3 per bag.

Problem: How much of each brand to purchase to minimize Problem: How much of each brand to purchase to minimize total cost of fertilizer given following data ?total cost of fertilizer given following data ?


LP Model FormulationA Minimization Example (2 of 7)

Decision Variables: x1 = bags of Super-Gro

x2 = bags of Crop-Quick

The Objective Function:Minimize Z = $6x1 + 3x2

Where: $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)


LP Model Formulation and Constraint GraphA Minimization Example (3 of 7)

Minimize Z = $6x1 + $3x2

subject to: 2x1 + 4x2 16 4x1 + 3x2 24

x1, x2 0

Figure 2.16 Graph of Both Model Constraints


Feasible Solution AreaA Minimization Example (4 of 7)


subject to: 2x1 + 4x2 16 4x1 + 3x2 24

x1, x2 0

Figure 2.17 Feasible Solution Area


Optimal Solution PointA Minimization Example (5 of 7)


subject to: 2x1 + 4x2 16 4x1 + 3x2 24

x1, x2 0

Figure 2.18 Optimum Solution Point


Surplus VariablesA Minimization Example (6 of 7)

A surplus variable is subtracted from a A surplus variable is subtracted from a constraint to constraint to convert it to an equation (=).convert it to an equation (=).

A surplus variable represents an excess above a constraint A surplus variable represents an excess above a constraint requirement level.requirement level.

Surplus variables contribute nothing to the calculated value Surplus variables contribute nothing to the calculated value of the objective function.of the objective function.

Subtracting slack variables in the farmer problem Subtracting slack variables in the farmer problem constraints:constraints:

2x2x11 + 4x + 4x22 - s - s11 = 16 (nitrogen) = 16 (nitrogen)

4x4x11 + 3x + 3x22 - s - s22 = 24 (phosphate) = 24 (phosphate)


Minimize Z = $6x1 + $3x2 + 0s1 + 0s2

subject to: 2x1 + 4x2 – s1 = 16 4x1 + 3x2 – s2 = 24

x1, x2, s1, s2 0

Figure 2.19 Graph of Fertilizer Example

Graphical SolutionsA Minimization Example (7 of 7)


For some linear programming models, the general rules For some linear programming models, the general rules do not apply.do not apply.

Special types of problems include those with:Special types of problems include those with:

Multiple optimal solutionsMultiple optimal solutions

Infeasible solutionsInfeasible solutions

Unbounded solutionsUnbounded solutions

Irregular Types of Linear Programming ProblemsIrregular Types of Linear Programming Problems


Objective function is parallel to to a constraint line.

Maximize Z=$40x1 + 30x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0Where:x1 = number of bowlsx2 = number of mugs

Figure 2.20 Example with Multiple Optimal Solutions

Multiple Optimal SolutionsBeaver Creek Pottery Example


An Infeasible ProblemAn Infeasible Problem

Every possible solution violates at least one constraint:

Maximize Z = 5x1 + 3x2

subject to: 4x1 + 2x2 8 x1 4 x2 6 x1, x2 0

Figure 2.21 Graph of an Infeasible Problem


Value of objective function increases indefinitely:


subject to: x1 4 x2 2 x1, x2 0

An Unbounded ProblemAn Unbounded Problem

Figure 2.22 Graph of an Unbounded Problem


Characteristics of Linear Programming ProblemsCharacteristics of Linear Programming Problems

A linear programming problem requires a decision - a A linear programming problem requires a decision - a choice amongst alternative courses of action.choice amongst alternative courses of action.

The decision is represented in the model by The decision is represented in the model by decision decision variablesvariables..

The problem encompasses a goal, expressed as an The problem encompasses a goal, expressed as an objective functionobjective function, that the decision maker wants to , that the decision maker wants to achieve.achieve.

ConstraintsConstraints exist that limit the extent of achievement of exist that limit the extent of achievement of the objective.the objective.

The objective and constraints must be definable by The objective and constraints must be definable by linearlinear mathematical functional relationships. mathematical functional relationships.


ProportionalityProportionality - The rate of change (slope) of the - The rate of change (slope) of the objective function and constraint equations is constant.objective function and constraint equations is constant.

Additivity Additivity - Terms in the objective function and - Terms in the objective function and constraint equations must be additive.constraint equations must be additive.

DivisibilityDivisibility -Decision variables can take on any -Decision variables can take on any fractional value and are therefore continuous as opposed fractional value and are therefore continuous as opposed to integer in nature.to integer in nature.

CertaintyCertainty - Values of all the model parameters are - Values of all the model parameters are assumed to be known with certainty (non-probabilistic).assumed to be known with certainty (non-probabilistic).

Properties of Linear Programming ModelsProperties of Linear Programming Models


Problem StatementExample Problem No. 1 (1 of 3)

Hot dog mixture in 1000-pound batches.Hot dog mixture in 1000-pound batches.

Two ingredients, chicken ($3/lb) and beef ($5/lb).Two ingredients, chicken ($3/lb) and beef ($5/lb).

Recipe requirements:Recipe requirements:

at least 500 pounds of chickenat least 500 pounds of chicken

at least 200 pounds of beefat least 200 pounds of beef

Ratio of chicken to beef must be at least 2 to 1.Ratio of chicken to beef must be at least 2 to 1.

Determine optimal mixture of ingredients that will minimize Determine optimal mixture of ingredients that will minimize costs.costs.


Step 1:

Identify decision variables.

x1 = lb of chicken in mixture (1000 lb.)

x2 = lb of beef in mixture (1000 lb.)

Step 2:

Formulate the objective function.


where Z = cost per 1,000-lb batch $3x1 = cost of chicken $5x2 = cost of beef

SolutionExample Problem No. 1 (2 of 3)


SolutionExample Problem No. 1 (3 of 3)

Step 3:

Establish Model Constraints x1 + x2 = 1,000 lb x1 500 lb of chicken x2 200 lb of beef x1/x2 2/1 or x1 - 2x2 0 x1, x2 0

The Model: Minimize Z = $3x1 + 5x2

subject to: x1 + x2 = 1,000 lb x1 50 x2 200 x1 - 2x2 0 x1,x2 0


Solve the following model graphically:


subject to: x1 + 2x2 10

6x1 + 6x2 36

x1 4

x1, x2 0

Step 1: Plot the constraints as equations

Example Problem No. 2 (1 of 3)Example Problem No. 2 (1 of 3)

Figure 2.23 Constraint Equations





6x1 + 6x2 36

x1 4

x1, x2 0

Step 2: Determine the feasible solution space

Figure 2.24 Feasible Solution Space and Extreme Points





6x1 + 6x2 36

x1 4

x1, x2 0

Step 3 and 4: Determine the solution points and optimal solution

Figure 2.25 Optimal Solution Point


End of Chapter

1 chapter 2 - linear programming: model formulation & graphical solution chapter 2 linear...

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objective function step

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constraints requirements