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
2Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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
3Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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
4Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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
5Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Summary of Model Formulation Steps
Step 1 : Clearly define the decision variables
Step 2 : Construct the objective function
Step 3 : Formulate the constraints
6Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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:
8Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
LP Model FormulationA Maximization Example (2 of 3)
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:
9Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
LP Model FormulationA Maximization Example (3 of 3)
Complete Linear Programming Model:
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40
4x1 + 3x2 120
x1, x2 0
10Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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
11Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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
12Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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
13Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Coordinate AxesGraphical Solution of Maximization Model (1 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Figure 2.2 Coordinates for Graphical Analysis
14Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Labor ConstraintGraphical Solution of Maximization Model (2 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Figure 2.3 Graph of Labor Constraint
15Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Labor Constraint AreaGraphical Solution of Maximization Model (3 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Figure 2.4 Labor Constraint Area
16Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Clay Constraint AreaGraphical Solution of Maximization Model (4 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Figure 2.5 Clay Constraint Area
17Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Both ConstraintsGraphical Solution of Maximization Model (5 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Figure 2.6 Graph of Both Model Constraints
18Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Feasible Solution AreaGraphical Solution of Maximization Model (6 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Figure 2.7 Feasible Solution Area
19Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Objective Function Solution = $800Graphical Solution of Maximization Model (7 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Figure 2.8 Objection Function Line for Z = $800
20Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Alternative Objective Function Solution LinesGraphical Solution of Maximization Model (8 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Figure 2.9 Alternative Objective Function Lines
21Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Optimal SolutionGraphical Solution of Maximization Model (9 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Figure 2.10 Identification of Optimal Solution
22Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Optimal Solution CoordinatesGraphical Solution of Maximization Model (10 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Figure 2.11 Optimal Solution Coordinates
23Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Extreme (Corner) Point SolutionsGraphical Solution of Maximization Model (11 of 12)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Figure 2.12 Solutions at All Corner Points
24Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Optimal Solution for New Objective FunctionGraphical Solution of Maximization Model (12 of 12)
Maximize Z = $70x1 + $20x2
subject to: 1x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Figure 2.13 Optimal Solution with Z = 70x1 + 20x2
25Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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
26Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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
27Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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 ?
29Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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)
30Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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
31Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Feasible Solution AreaA Minimization Example (4 of 7)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2 16 4x1 + 3x2 24
x1, x2 0
Figure 2.17 Feasible Solution Area
32Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Optimal Solution PointA Minimization Example (5 of 7)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2 16 4x1 + 3x2 24
x1, x2 0
Figure 2.18 Optimum Solution Point
33Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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)
34Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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)
35Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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
36Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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
37Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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
38Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Value of objective function increases indefinitely:
Maximize Z = 4x1 + 2x2
subject to: x1 4 x2 2 x1, x2 0
An Unbounded ProblemAn Unbounded Problem
Figure 2.22 Graph of an Unbounded Problem
39Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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.
40Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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
41Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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.
42Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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.
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)
43Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
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
44Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Solve the following model graphically:
Maximize Z = 4x1 + 5x2
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
45Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Example Problem No. 2 (2 of 3)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 the feasible solution space
Figure 2.24 Feasible Solution Space and Extreme Points
46Chapter 2 - Linear Programming: Model Formulation & Graphical Solution
Example Problem No. 2 (3 of 3)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 optimal solution
Figure 2.25 Optimal Solution Point
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