pom chap 2 - 8

7/30/2019 POM chap 2 - 8

1/65

1

Linear Programming:Model Formulation

andGraphical Solution

7/30/2019 POM chap 2 - 8

2/65

2

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

7/30/2019 POM chap 2 - 8

3/65

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

7/30/2019 POM chap 2 - 8

4/65

4

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

7/30/2019 POM chap 2 - 8

5/65

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:

7/30/2019 POM chap 2 - 8

6/65

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).

6

7/30/2019 POM chap 2 - 8

7/657

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.

7/30/2019 POM chap 2 - 8

8/658

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

7/30/2019 POM chap 2 - 8

9/65

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.

9

7/30/2019 POM chap 2 - 8

10/65

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

7/30/2019 POM chap 2 - 8

11/6511

7/30/2019 POM chap 2 - 8

12/6512

7/30/2019 POM chap 2 - 8

13/6513

7/30/2019 POM chap 2 - 8

14/6514

7/30/2019 POM chap 2 - 8

15/6515

7/30/2019 POM chap 2 - 8

16/6516

7/30/2019 POM chap 2 - 8

17/65

17

7/30/2019 POM chap 2 - 8

18/65

18

7/30/2019 POM chap 2 - 8

19/65

19

7/30/2019 POM chap 2 - 8

20/65

20

7/30/2019 POM chap 2 - 8

21/65

21

7/30/2019 POM chap 2 - 8

22/65

22

7/30/2019 POM chap 2 - 8

23/65

23

7/30/2019 POM chap 2 - 8

24/65

24

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:

7/30/2019 POM chap 2 - 8

25/65

25


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:

7/30/2019 POM chap 2 - 8

26/65

26


Complete Linear Programming Model:

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40

4x2 + 3x2

120x1, x2 0

7/30/2019 POM chap 2 - 8

27/65

27

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

7/30/2019 POM chap 2 - 8

28/65

28

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

7/30/2019 POM chap 2 - 8

29/65

29

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

7/30/2019 POM chap 2 - 8

30/65

The graphic solution procedure is

one of the method of solving twovariable Linear programmingproblems. It consists of the following

steps:-

30

tep

7/30/2019 POM chap 2 - 8

31/65

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.

31

7/30/2019 POM chap 2 - 8

32/65

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

7/30/2019 POM chap 2 - 8

33/65

33

Coordinate AxesGraphical Solution of Maximization Model (1 of 12)

Maximize Z = $40x1

+ $50x2


4x2 + 3x2 120x1, x2 0

Figure 2.1Coordinates for Graphical Analysis

7/30/2019 POM chap 2 - 8

34/65

34

Labor ConstraintGraphical Solution of Maximization Model (2 of 12)

Maximize Z = $40x1

+ $50x2


4x2 + 3x2 120x1, x2 0

Figure 2.1Graph of Labor Constraint

7/30/2019 POM chap 2 - 8

35/65

35

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

Maximize Z = $40x1

+ $50x2


4x2 + 3x2 120x1, x2 0

Figure 2.3Labor Constraint Area

7/30/2019 POM chap 2 - 8

36/65

36

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

Maximize Z = $40x1

+ $50x2


4x2 + 3x2 120x1, x2 0

Figure 2.4Clay Constraint Area

7/30/2019 POM chap 2 - 8

37/65

37

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

7/30/2019 POM chap 2 - 8

38/65

38

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


4x2 + 3x2 120x1, x2 0

Figure 2.6Feasible Solution Area

7/30/2019 POM chap 2 - 8

39/65

39

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


4x2 + 3x2 120x1, x2 0

Figure 2.7Objection Function Line for Z = $800

7/30/2019 POM chap 2 - 8

40/65

40

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


4x2 + 3x2 120x1, x2 0

Figure 2.8Alternative Objective Function Lines

7/30/2019 POM chap 2 - 8

41/65

41

Optimal SolutionGraphical Solution of Maximization Model (9 of 12)


4x2 + 3x2 120x1, x2 0

Figure 2.9Identification of Optimal Solution

7/30/2019 POM chap 2 - 8

42/65

42

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


4x2 + 3x2 120x1, x2 0

Figure 2.10Optimal Solution Coordinates

7/30/2019 POM chap 2 - 8

43/65

43

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


4x2 + 3x2 120x1, x2 0

Figure 2.11Solution at All Corner Points

7/30/2019 POM chap 2 - 8

44/65

44

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


4x2 + 3x2 120x1, x2 0

Figure 2.12Optimal Solution with Z = 70x1 + 20x2

S

7/30/2019 POM chap 2 - 8

45/65

45

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

7/30/2019 POM chap 2 - 8

46/65

46

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

7/30/2019 POM chap 2 - 8

47/65

47

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

7/30/2019 POM chap 2 - 8

48/65

48

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

7/30/2019 POM chap 2 - 8

49/65

49

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

7/30/2019 POM chap 2 - 8

50/65

50

Feasible Solution AreaA Minimization Model Example (4 of 7)


subject to: 2x1 + 4x2 164x2 + 3x2 24

x1, x2 0

Figure 2.15Feasible Solution Area

O ti l S l ti P i t

7/30/2019 POM chap 2 - 8

51/65

51

Optimal Solution PointA Minimization Model Example (5 of 7)


subject to: 2x1 + 4x2 164x2 + 3x2 24

x1, x2 0

Figure 2.16Optimum Solution Point

S l V i bl

7/30/2019 POM chap 2 - 8

52/65

52

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

7/30/2019 POM chap 2 - 8

53/65

53

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

7/30/2019 POM chap 2 - 8

54/65

54

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

7/30/2019 POM chap 2 - 8

55/65

55

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

7/30/2019 POM chap 2 - 8

56/65

56

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

7/30/2019 POM chap 2 - 8

57/65

57

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


7/30/2019 POM chap 2 - 8

58/65

58


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

7/30/2019 POM chap 2 - 8

59/65

59

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

7/30/2019 POM chap 2 - 8

60/65

60

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

7/30/2019 POM chap 2 - 8

61/65

61

Step 1:Identify decision variables.

x1 = lb of chicken

x2 = lb of beef

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)

Solution

7/30/2019 POM chap 2 - 8

62/65

62

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)

7/30/2019 POM chap 2 - 8

63/65

63

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


7/30/2019 POM chap 2 - 8

64/65

64




6x1 + 6x2 36

x1 4

x1, x2 0

Step 2: Determine thefeasible solution space

Figure 2.22Feasible Solution Space and Extreme Points


7/30/2019 POM chap 2 - 8

65/65




6x1 + 6x2 36

x1 4

x1, x2 0

Step 3 and 4: Determine the

solution points and optimalsolution

Figure 2.22Optimal Solution Point

pom chap 2 - 8

Documents