LINEAR PROGRAMMING PROJECT

VPAVITHRA SUKANYAH VK RIZWANA SULTANA SHILPA JAIN

INTRODUCTION

bull Modern technological advance growth ofscientific techniques

bull Operations Research (OR) recent additionto scientific tools

bull OR new outlook to many conventionalmanagement problems

bull Seeks the determination of best (optimum)course of action of a decision problem underthe limiting factor of limited resources

bull Operational Research can be considered asbeing the application of scientific method byinter-disciplinary teams to problemsinvolving the control of organized systems so as to provide solutionswhich best serve the purposes of theorganization as a whole

WHAT IS OR

CHARACTERISTIC NATURE OF OR

Inter-disciplinary team approachbull Systems approachbull Helpful in improving the quality of solutionbull Scientific methodbull Goal oriented optimum solutionbull Use of modelsbull Require willing executivesbull Reduces complexity

PHASES TO OR

bull Judgment phasendash Determination of the problemndash Establishment of the objectives and valuesndash Determination of suitable measures of effectiveness

bull Research phasendash Observation and data collectionndash Formulation of hypothesis and modelsndash Observation and experimentation to test the hypothesisndash Prediction of various results generalization consideration of alternative method

bull Action phasendash Implementation of the tested results of the model

METHODOLOGY OF OR

bull Formulating the problembull Constructing the model

bull Deriving the solutionndash Analytical methodndash Numerical methodndash Simulation methodbull Testing the validity

bull Controlling the solutionbull Implementing the result

PROBLEMS IN OR

bull Allocationbull Replacementbull Sequencingbull Routingbull Inventorybull Queuingbull Competitivebull Search

OR TECHINIQESbull Linear programming

bull Waiting line or queuing theorybull Inventory control planningbull Game theorybull Decision theorybull Network analysisndash Program Evaluation and Review

Techniquendash Critical Path Method (CPM) etcbull Simulationbull Integrated production models

SIGNIFICANCE OF OR

Provides a tool for scientific analysisbull Provides solution for various business problemsbull Enables proper deployment of resourcesbull Helps in minimizing waiting and servicing costsbull Enables the management to decide when to buy and how much to buybull Assists in choosing an optimum strategybull Renders great help in optimum resource allocationbull Facilitates the process of decision makingbull Management can know the reactions of the integrated

business systemsbull Helps a lot in the preparation of future managers

LIMITATIONS OF OR

The inherent limitations concerning mathematicalexpressions

bull High costs are involved in the use of OR techniquesbull OR does not take into consideration the intangible factorsbull OR is only a tool of analysis and not the complete decision-making processbull Other limitations ndash Bias ndash Inadequate objective functions ndash Internal resistance ndash Competence ndash Reliability of the prepared solution

INTRODUCTION TO LINEAR PROGRAMMING

Today many of the resources needed as inputs to operations are in limited supplyOperations managers must understand the impact of this situation on meeting their objectivesLinear programming (LP) is one way that operations managers can determine how best to allocate their scarce resources

Linear programmingWe use graphs as useful modeling abstractions to help us develop computational solutions for a wide variety of problemsA linear program is simply another modeling abstraction (tool in your toolbox)Developing routines that solve general linear programs allows us to embed them in sophisticated algorithmic solutions to difficult problems (eg like we did for TSP)The cutting edge algorithmic solutions to many problems use the ideas from mathematical programming linear programming forming the foundation

BASIC CONCEPT OF LP PROGRAM

Objective functionConstraintsOptimizationSolution of lppFeasible solutionOptimal solution

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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VPAVITHRA SUKANYAH VK RIZWANA SULTANA SHILPA JAIN

INTRODUCTION

bull Modern technological advance growth ofscientific techniques

bull Operations Research (OR) recent additionto scientific tools

bull OR new outlook to many conventionalmanagement problems

bull Seeks the determination of best (optimum)course of action of a decision problem underthe limiting factor of limited resources

bull Operational Research can be considered asbeing the application of scientific method byinter-disciplinary teams to problemsinvolving the control of organized systems so as to provide solutionswhich best serve the purposes of theorganization as a whole

WHAT IS OR

CHARACTERISTIC NATURE OF OR

Inter-disciplinary team approachbull Systems approachbull Helpful in improving the quality of solutionbull Scientific methodbull Goal oriented optimum solutionbull Use of modelsbull Require willing executivesbull Reduces complexity

PHASES TO OR

bull Judgment phasendash Determination of the problemndash Establishment of the objectives and valuesndash Determination of suitable measures of effectiveness

bull Research phasendash Observation and data collectionndash Formulation of hypothesis and modelsndash Observation and experimentation to test the hypothesisndash Prediction of various results generalization consideration of alternative method

bull Action phasendash Implementation of the tested results of the model

METHODOLOGY OF OR

bull Formulating the problembull Constructing the model

bull Deriving the solutionndash Analytical methodndash Numerical methodndash Simulation methodbull Testing the validity

bull Controlling the solutionbull Implementing the result

PROBLEMS IN OR

bull Allocationbull Replacementbull Sequencingbull Routingbull Inventorybull Queuingbull Competitivebull Search

OR TECHINIQESbull Linear programming

bull Waiting line or queuing theorybull Inventory control planningbull Game theorybull Decision theorybull Network analysisndash Program Evaluation and Review

Techniquendash Critical Path Method (CPM) etcbull Simulationbull Integrated production models

SIGNIFICANCE OF OR

Provides a tool for scientific analysisbull Provides solution for various business problemsbull Enables proper deployment of resourcesbull Helps in minimizing waiting and servicing costsbull Enables the management to decide when to buy and how much to buybull Assists in choosing an optimum strategybull Renders great help in optimum resource allocationbull Facilitates the process of decision makingbull Management can know the reactions of the integrated

business systemsbull Helps a lot in the preparation of future managers

LIMITATIONS OF OR

The inherent limitations concerning mathematicalexpressions

bull High costs are involved in the use of OR techniquesbull OR does not take into consideration the intangible factorsbull OR is only a tool of analysis and not the complete decision-making processbull Other limitations ndash Bias ndash Inadequate objective functions ndash Internal resistance ndash Competence ndash Reliability of the prepared solution

INTRODUCTION TO LINEAR PROGRAMMING

Today many of the resources needed as inputs to operations are in limited supplyOperations managers must understand the impact of this situation on meeting their objectivesLinear programming (LP) is one way that operations managers can determine how best to allocate their scarce resources

Linear programmingWe use graphs as useful modeling abstractions to help us develop computational solutions for a wide variety of problemsA linear program is simply another modeling abstraction (tool in your toolbox)Developing routines that solve general linear programs allows us to embed them in sophisticated algorithmic solutions to difficult problems (eg like we did for TSP)The cutting edge algorithmic solutions to many problems use the ideas from mathematical programming linear programming forming the foundation

BASIC CONCEPT OF LP PROGRAM

Objective functionConstraintsOptimizationSolution of lppFeasible solutionOptimal solution

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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• Slide 48
• Slide 49
• Slide 50

INTRODUCTION

bull Modern technological advance growth ofscientific techniques

bull Operations Research (OR) recent additionto scientific tools

bull OR new outlook to many conventionalmanagement problems

bull Seeks the determination of best (optimum)course of action of a decision problem underthe limiting factor of limited resources

bull Operational Research can be considered asbeing the application of scientific method byinter-disciplinary teams to problemsinvolving the control of organized systems so as to provide solutionswhich best serve the purposes of theorganization as a whole

WHAT IS OR

CHARACTERISTIC NATURE OF OR

Inter-disciplinary team approachbull Systems approachbull Helpful in improving the quality of solutionbull Scientific methodbull Goal oriented optimum solutionbull Use of modelsbull Require willing executivesbull Reduces complexity

PHASES TO OR

bull Judgment phasendash Determination of the problemndash Establishment of the objectives and valuesndash Determination of suitable measures of effectiveness

bull Research phasendash Observation and data collectionndash Formulation of hypothesis and modelsndash Observation and experimentation to test the hypothesisndash Prediction of various results generalization consideration of alternative method

bull Action phasendash Implementation of the tested results of the model

METHODOLOGY OF OR

bull Formulating the problembull Constructing the model

bull Deriving the solutionndash Analytical methodndash Numerical methodndash Simulation methodbull Testing the validity

bull Controlling the solutionbull Implementing the result

PROBLEMS IN OR

bull Allocationbull Replacementbull Sequencingbull Routingbull Inventorybull Queuingbull Competitivebull Search

OR TECHINIQESbull Linear programming

bull Waiting line or queuing theorybull Inventory control planningbull Game theorybull Decision theorybull Network analysisndash Program Evaluation and Review

Techniquendash Critical Path Method (CPM) etcbull Simulationbull Integrated production models

SIGNIFICANCE OF OR

Provides a tool for scientific analysisbull Provides solution for various business problemsbull Enables proper deployment of resourcesbull Helps in minimizing waiting and servicing costsbull Enables the management to decide when to buy and how much to buybull Assists in choosing an optimum strategybull Renders great help in optimum resource allocationbull Facilitates the process of decision makingbull Management can know the reactions of the integrated

business systemsbull Helps a lot in the preparation of future managers

LIMITATIONS OF OR

The inherent limitations concerning mathematicalexpressions

bull High costs are involved in the use of OR techniquesbull OR does not take into consideration the intangible factorsbull OR is only a tool of analysis and not the complete decision-making processbull Other limitations ndash Bias ndash Inadequate objective functions ndash Internal resistance ndash Competence ndash Reliability of the prepared solution

INTRODUCTION TO LINEAR PROGRAMMING

Today many of the resources needed as inputs to operations are in limited supplyOperations managers must understand the impact of this situation on meeting their objectivesLinear programming (LP) is one way that operations managers can determine how best to allocate their scarce resources

Linear programmingWe use graphs as useful modeling abstractions to help us develop computational solutions for a wide variety of problemsA linear program is simply another modeling abstraction (tool in your toolbox)Developing routines that solve general linear programs allows us to embed them in sophisticated algorithmic solutions to difficult problems (eg like we did for TSP)The cutting edge algorithmic solutions to many problems use the ideas from mathematical programming linear programming forming the foundation

BASIC CONCEPT OF LP PROGRAM

Objective functionConstraintsOptimizationSolution of lppFeasible solutionOptimal solution

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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• Slide 49
• Slide 50

bull Operational Research can be considered asbeing the application of scientific method byinter-disciplinary teams to problemsinvolving the control of organized systems so as to provide solutionswhich best serve the purposes of theorganization as a whole

WHAT IS OR

CHARACTERISTIC NATURE OF OR

Inter-disciplinary team approachbull Systems approachbull Helpful in improving the quality of solutionbull Scientific methodbull Goal oriented optimum solutionbull Use of modelsbull Require willing executivesbull Reduces complexity

PHASES TO OR

bull Judgment phasendash Determination of the problemndash Establishment of the objectives and valuesndash Determination of suitable measures of effectiveness

bull Research phasendash Observation and data collectionndash Formulation of hypothesis and modelsndash Observation and experimentation to test the hypothesisndash Prediction of various results generalization consideration of alternative method

bull Action phasendash Implementation of the tested results of the model

METHODOLOGY OF OR

bull Formulating the problembull Constructing the model

bull Deriving the solutionndash Analytical methodndash Numerical methodndash Simulation methodbull Testing the validity

bull Controlling the solutionbull Implementing the result

PROBLEMS IN OR

bull Allocationbull Replacementbull Sequencingbull Routingbull Inventorybull Queuingbull Competitivebull Search

OR TECHINIQESbull Linear programming

bull Waiting line or queuing theorybull Inventory control planningbull Game theorybull Decision theorybull Network analysisndash Program Evaluation and Review

Techniquendash Critical Path Method (CPM) etcbull Simulationbull Integrated production models

SIGNIFICANCE OF OR

Provides a tool for scientific analysisbull Provides solution for various business problemsbull Enables proper deployment of resourcesbull Helps in minimizing waiting and servicing costsbull Enables the management to decide when to buy and how much to buybull Assists in choosing an optimum strategybull Renders great help in optimum resource allocationbull Facilitates the process of decision makingbull Management can know the reactions of the integrated

business systemsbull Helps a lot in the preparation of future managers

LIMITATIONS OF OR

The inherent limitations concerning mathematicalexpressions

bull High costs are involved in the use of OR techniquesbull OR does not take into consideration the intangible factorsbull OR is only a tool of analysis and not the complete decision-making processbull Other limitations ndash Bias ndash Inadequate objective functions ndash Internal resistance ndash Competence ndash Reliability of the prepared solution

INTRODUCTION TO LINEAR PROGRAMMING

Today many of the resources needed as inputs to operations are in limited supplyOperations managers must understand the impact of this situation on meeting their objectivesLinear programming (LP) is one way that operations managers can determine how best to allocate their scarce resources

Linear programmingWe use graphs as useful modeling abstractions to help us develop computational solutions for a wide variety of problemsA linear program is simply another modeling abstraction (tool in your toolbox)Developing routines that solve general linear programs allows us to embed them in sophisticated algorithmic solutions to difficult problems (eg like we did for TSP)The cutting edge algorithmic solutions to many problems use the ideas from mathematical programming linear programming forming the foundation

BASIC CONCEPT OF LP PROGRAM

Objective functionConstraintsOptimizationSolution of lppFeasible solutionOptimal solution

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
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• Slide 25
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• Slide 30
• Slide 31
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• Slide 35
• Slide 36
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• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

CHARACTERISTIC NATURE OF OR

Inter-disciplinary team approachbull Systems approachbull Helpful in improving the quality of solutionbull Scientific methodbull Goal oriented optimum solutionbull Use of modelsbull Require willing executivesbull Reduces complexity

PHASES TO OR

bull Judgment phasendash Determination of the problemndash Establishment of the objectives and valuesndash Determination of suitable measures of effectiveness

bull Research phasendash Observation and data collectionndash Formulation of hypothesis and modelsndash Observation and experimentation to test the hypothesisndash Prediction of various results generalization consideration of alternative method

bull Action phasendash Implementation of the tested results of the model

METHODOLOGY OF OR

bull Formulating the problembull Constructing the model

bull Deriving the solutionndash Analytical methodndash Numerical methodndash Simulation methodbull Testing the validity

bull Controlling the solutionbull Implementing the result

PROBLEMS IN OR

bull Allocationbull Replacementbull Sequencingbull Routingbull Inventorybull Queuingbull Competitivebull Search

OR TECHINIQESbull Linear programming

bull Waiting line or queuing theorybull Inventory control planningbull Game theorybull Decision theorybull Network analysisndash Program Evaluation and Review

Techniquendash Critical Path Method (CPM) etcbull Simulationbull Integrated production models

SIGNIFICANCE OF OR

Provides a tool for scientific analysisbull Provides solution for various business problemsbull Enables proper deployment of resourcesbull Helps in minimizing waiting and servicing costsbull Enables the management to decide when to buy and how much to buybull Assists in choosing an optimum strategybull Renders great help in optimum resource allocationbull Facilitates the process of decision makingbull Management can know the reactions of the integrated

business systemsbull Helps a lot in the preparation of future managers

LIMITATIONS OF OR

The inherent limitations concerning mathematicalexpressions

bull High costs are involved in the use of OR techniquesbull OR does not take into consideration the intangible factorsbull OR is only a tool of analysis and not the complete decision-making processbull Other limitations ndash Bias ndash Inadequate objective functions ndash Internal resistance ndash Competence ndash Reliability of the prepared solution

INTRODUCTION TO LINEAR PROGRAMMING

Today many of the resources needed as inputs to operations are in limited supplyOperations managers must understand the impact of this situation on meeting their objectivesLinear programming (LP) is one way that operations managers can determine how best to allocate their scarce resources

Linear programmingWe use graphs as useful modeling abstractions to help us develop computational solutions for a wide variety of problemsA linear program is simply another modeling abstraction (tool in your toolbox)Developing routines that solve general linear programs allows us to embed them in sophisticated algorithmic solutions to difficult problems (eg like we did for TSP)The cutting edge algorithmic solutions to many problems use the ideas from mathematical programming linear programming forming the foundation

BASIC CONCEPT OF LP PROGRAM

Objective functionConstraintsOptimizationSolution of lppFeasible solutionOptimal solution

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
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PHASES TO OR

bull Judgment phasendash Determination of the problemndash Establishment of the objectives and valuesndash Determination of suitable measures of effectiveness

bull Research phasendash Observation and data collectionndash Formulation of hypothesis and modelsndash Observation and experimentation to test the hypothesisndash Prediction of various results generalization consideration of alternative method

bull Action phasendash Implementation of the tested results of the model

METHODOLOGY OF OR

bull Formulating the problembull Constructing the model

bull Deriving the solutionndash Analytical methodndash Numerical methodndash Simulation methodbull Testing the validity

bull Controlling the solutionbull Implementing the result

PROBLEMS IN OR

bull Allocationbull Replacementbull Sequencingbull Routingbull Inventorybull Queuingbull Competitivebull Search

OR TECHINIQESbull Linear programming

bull Waiting line or queuing theorybull Inventory control planningbull Game theorybull Decision theorybull Network analysisndash Program Evaluation and Review

Techniquendash Critical Path Method (CPM) etcbull Simulationbull Integrated production models

SIGNIFICANCE OF OR

Provides a tool for scientific analysisbull Provides solution for various business problemsbull Enables proper deployment of resourcesbull Helps in minimizing waiting and servicing costsbull Enables the management to decide when to buy and how much to buybull Assists in choosing an optimum strategybull Renders great help in optimum resource allocationbull Facilitates the process of decision makingbull Management can know the reactions of the integrated

business systemsbull Helps a lot in the preparation of future managers

LIMITATIONS OF OR

The inherent limitations concerning mathematicalexpressions

bull High costs are involved in the use of OR techniquesbull OR does not take into consideration the intangible factorsbull OR is only a tool of analysis and not the complete decision-making processbull Other limitations ndash Bias ndash Inadequate objective functions ndash Internal resistance ndash Competence ndash Reliability of the prepared solution

INTRODUCTION TO LINEAR PROGRAMMING

Today many of the resources needed as inputs to operations are in limited supplyOperations managers must understand the impact of this situation on meeting their objectivesLinear programming (LP) is one way that operations managers can determine how best to allocate their scarce resources

Linear programmingWe use graphs as useful modeling abstractions to help us develop computational solutions for a wide variety of problemsA linear program is simply another modeling abstraction (tool in your toolbox)Developing routines that solve general linear programs allows us to embed them in sophisticated algorithmic solutions to difficult problems (eg like we did for TSP)The cutting edge algorithmic solutions to many problems use the ideas from mathematical programming linear programming forming the foundation

BASIC CONCEPT OF LP PROGRAM

Objective functionConstraintsOptimizationSolution of lppFeasible solutionOptimal solution

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
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• Slide 48
• Slide 49
• Slide 50

METHODOLOGY OF OR

bull Formulating the problembull Constructing the model

bull Deriving the solutionndash Analytical methodndash Numerical methodndash Simulation methodbull Testing the validity

bull Controlling the solutionbull Implementing the result

PROBLEMS IN OR

bull Allocationbull Replacementbull Sequencingbull Routingbull Inventorybull Queuingbull Competitivebull Search

OR TECHINIQESbull Linear programming

bull Waiting line or queuing theorybull Inventory control planningbull Game theorybull Decision theorybull Network analysisndash Program Evaluation and Review

Techniquendash Critical Path Method (CPM) etcbull Simulationbull Integrated production models

SIGNIFICANCE OF OR

Provides a tool for scientific analysisbull Provides solution for various business problemsbull Enables proper deployment of resourcesbull Helps in minimizing waiting and servicing costsbull Enables the management to decide when to buy and how much to buybull Assists in choosing an optimum strategybull Renders great help in optimum resource allocationbull Facilitates the process of decision makingbull Management can know the reactions of the integrated

business systemsbull Helps a lot in the preparation of future managers

LIMITATIONS OF OR

The inherent limitations concerning mathematicalexpressions

bull High costs are involved in the use of OR techniquesbull OR does not take into consideration the intangible factorsbull OR is only a tool of analysis and not the complete decision-making processbull Other limitations ndash Bias ndash Inadequate objective functions ndash Internal resistance ndash Competence ndash Reliability of the prepared solution

INTRODUCTION TO LINEAR PROGRAMMING

Today many of the resources needed as inputs to operations are in limited supplyOperations managers must understand the impact of this situation on meeting their objectivesLinear programming (LP) is one way that operations managers can determine how best to allocate their scarce resources

Linear programmingWe use graphs as useful modeling abstractions to help us develop computational solutions for a wide variety of problemsA linear program is simply another modeling abstraction (tool in your toolbox)Developing routines that solve general linear programs allows us to embed them in sophisticated algorithmic solutions to difficult problems (eg like we did for TSP)The cutting edge algorithmic solutions to many problems use the ideas from mathematical programming linear programming forming the foundation

BASIC CONCEPT OF LP PROGRAM

Objective functionConstraintsOptimizationSolution of lppFeasible solutionOptimal solution

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
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• Slide 13
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• Slide 21
• Slide 22
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• Slide 25
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• Slide 30
• Slide 31
• Slide 32
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• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

PROBLEMS IN OR

bull Allocationbull Replacementbull Sequencingbull Routingbull Inventorybull Queuingbull Competitivebull Search

OR TECHINIQESbull Linear programming

bull Waiting line or queuing theorybull Inventory control planningbull Game theorybull Decision theorybull Network analysisndash Program Evaluation and Review

Techniquendash Critical Path Method (CPM) etcbull Simulationbull Integrated production models

SIGNIFICANCE OF OR

Provides a tool for scientific analysisbull Provides solution for various business problemsbull Enables proper deployment of resourcesbull Helps in minimizing waiting and servicing costsbull Enables the management to decide when to buy and how much to buybull Assists in choosing an optimum strategybull Renders great help in optimum resource allocationbull Facilitates the process of decision makingbull Management can know the reactions of the integrated

business systemsbull Helps a lot in the preparation of future managers

LIMITATIONS OF OR

The inherent limitations concerning mathematicalexpressions

bull High costs are involved in the use of OR techniquesbull OR does not take into consideration the intangible factorsbull OR is only a tool of analysis and not the complete decision-making processbull Other limitations ndash Bias ndash Inadequate objective functions ndash Internal resistance ndash Competence ndash Reliability of the prepared solution

INTRODUCTION TO LINEAR PROGRAMMING

Today many of the resources needed as inputs to operations are in limited supplyOperations managers must understand the impact of this situation on meeting their objectivesLinear programming (LP) is one way that operations managers can determine how best to allocate their scarce resources

Linear programmingWe use graphs as useful modeling abstractions to help us develop computational solutions for a wide variety of problemsA linear program is simply another modeling abstraction (tool in your toolbox)Developing routines that solve general linear programs allows us to embed them in sophisticated algorithmic solutions to difficult problems (eg like we did for TSP)The cutting edge algorithmic solutions to many problems use the ideas from mathematical programming linear programming forming the foundation

BASIC CONCEPT OF LP PROGRAM

Objective functionConstraintsOptimizationSolution of lppFeasible solutionOptimal solution

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
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• Slide 48
• Slide 49
• Slide 50

OR TECHINIQESbull Linear programming

bull Waiting line or queuing theorybull Inventory control planningbull Game theorybull Decision theorybull Network analysisndash Program Evaluation and Review

Techniquendash Critical Path Method (CPM) etcbull Simulationbull Integrated production models

SIGNIFICANCE OF OR

Provides a tool for scientific analysisbull Provides solution for various business problemsbull Enables proper deployment of resourcesbull Helps in minimizing waiting and servicing costsbull Enables the management to decide when to buy and how much to buybull Assists in choosing an optimum strategybull Renders great help in optimum resource allocationbull Facilitates the process of decision makingbull Management can know the reactions of the integrated

business systemsbull Helps a lot in the preparation of future managers

LIMITATIONS OF OR

The inherent limitations concerning mathematicalexpressions

bull High costs are involved in the use of OR techniquesbull OR does not take into consideration the intangible factorsbull OR is only a tool of analysis and not the complete decision-making processbull Other limitations ndash Bias ndash Inadequate objective functions ndash Internal resistance ndash Competence ndash Reliability of the prepared solution

INTRODUCTION TO LINEAR PROGRAMMING

Today many of the resources needed as inputs to operations are in limited supplyOperations managers must understand the impact of this situation on meeting their objectivesLinear programming (LP) is one way that operations managers can determine how best to allocate their scarce resources

Linear programmingWe use graphs as useful modeling abstractions to help us develop computational solutions for a wide variety of problemsA linear program is simply another modeling abstraction (tool in your toolbox)Developing routines that solve general linear programs allows us to embed them in sophisticated algorithmic solutions to difficult problems (eg like we did for TSP)The cutting edge algorithmic solutions to many problems use the ideas from mathematical programming linear programming forming the foundation

BASIC CONCEPT OF LP PROGRAM

Objective functionConstraintsOptimizationSolution of lppFeasible solutionOptimal solution

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
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• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

SIGNIFICANCE OF OR

Provides a tool for scientific analysisbull Provides solution for various business problemsbull Enables proper deployment of resourcesbull Helps in minimizing waiting and servicing costsbull Enables the management to decide when to buy and how much to buybull Assists in choosing an optimum strategybull Renders great help in optimum resource allocationbull Facilitates the process of decision makingbull Management can know the reactions of the integrated

business systemsbull Helps a lot in the preparation of future managers

LIMITATIONS OF OR

The inherent limitations concerning mathematicalexpressions

bull High costs are involved in the use of OR techniquesbull OR does not take into consideration the intangible factorsbull OR is only a tool of analysis and not the complete decision-making processbull Other limitations ndash Bias ndash Inadequate objective functions ndash Internal resistance ndash Competence ndash Reliability of the prepared solution

INTRODUCTION TO LINEAR PROGRAMMING

Today many of the resources needed as inputs to operations are in limited supplyOperations managers must understand the impact of this situation on meeting their objectivesLinear programming (LP) is one way that operations managers can determine how best to allocate their scarce resources

Linear programmingWe use graphs as useful modeling abstractions to help us develop computational solutions for a wide variety of problemsA linear program is simply another modeling abstraction (tool in your toolbox)Developing routines that solve general linear programs allows us to embed them in sophisticated algorithmic solutions to difficult problems (eg like we did for TSP)The cutting edge algorithmic solutions to many problems use the ideas from mathematical programming linear programming forming the foundation

BASIC CONCEPT OF LP PROGRAM

Objective functionConstraintsOptimizationSolution of lppFeasible solutionOptimal solution

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
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• Slide 35
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• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

LIMITATIONS OF OR

The inherent limitations concerning mathematicalexpressions

bull High costs are involved in the use of OR techniquesbull OR does not take into consideration the intangible factorsbull OR is only a tool of analysis and not the complete decision-making processbull Other limitations ndash Bias ndash Inadequate objective functions ndash Internal resistance ndash Competence ndash Reliability of the prepared solution

INTRODUCTION TO LINEAR PROGRAMMING

Today many of the resources needed as inputs to operations are in limited supplyOperations managers must understand the impact of this situation on meeting their objectivesLinear programming (LP) is one way that operations managers can determine how best to allocate their scarce resources

Linear programmingWe use graphs as useful modeling abstractions to help us develop computational solutions for a wide variety of problemsA linear program is simply another modeling abstraction (tool in your toolbox)Developing routines that solve general linear programs allows us to embed them in sophisticated algorithmic solutions to difficult problems (eg like we did for TSP)The cutting edge algorithmic solutions to many problems use the ideas from mathematical programming linear programming forming the foundation

BASIC CONCEPT OF LP PROGRAM

Objective functionConstraintsOptimizationSolution of lppFeasible solutionOptimal solution

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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• Slide 50

INTRODUCTION TO LINEAR PROGRAMMING

Today many of the resources needed as inputs to operations are in limited supplyOperations managers must understand the impact of this situation on meeting their objectivesLinear programming (LP) is one way that operations managers can determine how best to allocate their scarce resources

Linear programmingWe use graphs as useful modeling abstractions to help us develop computational solutions for a wide variety of problemsA linear program is simply another modeling abstraction (tool in your toolbox)Developing routines that solve general linear programs allows us to embed them in sophisticated algorithmic solutions to difficult problems (eg like we did for TSP)The cutting edge algorithmic solutions to many problems use the ideas from mathematical programming linear programming forming the foundation

BASIC CONCEPT OF LP PROGRAM

Objective functionConstraintsOptimizationSolution of lppFeasible solutionOptimal solution

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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• Slide 3
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• Slide 48
• Slide 49
• Slide 50

Linear programmingWe use graphs as useful modeling abstractions to help us develop computational solutions for a wide variety of problemsA linear program is simply another modeling abstraction (tool in your toolbox)Developing routines that solve general linear programs allows us to embed them in sophisticated algorithmic solutions to difficult problems (eg like we did for TSP)The cutting edge algorithmic solutions to many problems use the ideas from mathematical programming linear programming forming the foundation

BASIC CONCEPT OF LP PROGRAM

Objective functionConstraintsOptimizationSolution of lppFeasible solutionOptimal solution

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
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• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

BASIC CONCEPT OF LP PROGRAM

Objective functionConstraintsOptimizationSolution of lppFeasible solutionOptimal solution

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
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• Slide 4
• Slide 5
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• Slide 7
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• Slide 43
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• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

LP PROBLEMS IN OM PRODUCT MIX

ObjectiveTo select the mix of products or services

that results in maximum profits for the planning period

Decision VariablesHow much to produce and market of each

product or service for the planning period

ConstraintsMaximum amount of each product or

service demanded Minimum amount of product or service policy will allow Maximum amount of resources available

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
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• Slide 6
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• Slide 50

Objective function the linear functions which is to be optimized ie maximized or minimized this may be expressed in linear expressionSolution of Lpp

The set of all the values of the variable x1x2helliphellipxn which satisy the constraints is called the solution of Lpp Feasible solution The set of all the values of the variable x1x2helliphellipxn which satisy the constraints and also the non negative conditions is called the feasible solution of lpp

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
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• Slide 7
• Slide 8
• Slide 9
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• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

Recognizing LP Problems

Characteristics of LP Problems in OMA well-defined single objective must be statedThere must be alternative courses of actionThe total achievement of the objective must be constrained by scarce resources or other restraintsThe objective and each of the constraints must be expressed as linear mathematical functions

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
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• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
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• Slide 10
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• Slide 15
• Slide 16
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• Slide 19
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• Slide 24
• Slide 25
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• Slide 47
• Slide 48
• Slide 49
• Slide 50

Steps in Formulating LP Problems

Define the objective (min or max)Define the decision variables (positive) Write the mathematical function for the objectiveWrite a 1- or 2-word description of each constraintWrite the right-hand side (RHS) of each constraintWrite lt = or gt for each constraintWrite the decision variables on LHS of each constraintWrite the coefficient for each decision variable in each constraint

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
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• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

Linear Programming

An optimization problem is said to be a linear program if it satisfied the following properties There is a unique objective function Whenever a decision variable appears in

either the objective function or one of theconstraint functions it must appear only as a power term with an exponent of 1possibly multiplied by a constant

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
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• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

LP Problems in General

Units of each term in a constraint must be the same as the RHSUnits of each term in the objective function must be the same as ZUnits between constraints do not have to be the sameLP problem can have a mixture of constraint types

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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• Slide 2
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• Slide 50

No term in the objective function or in anyof the constraints can contain products ofthe decision variables

The coefficients of the decision variables inthe objective function and each constraintare constant

The decision variables are permitted toassume fractional as well as integer values

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

Examples of lpp

We are already familiar with the graphical representation of equations and inequations here we describe the application of linear equations and inequations in solving different kinds of problems The examples are stated belowExample 1

Find two positive numbers such that whose sum is atleast 15 and whose difference is at the most 7 such that the product is maximum

Step1 we have to choose the positive two numbers Let the 2 positive numbers be x and y this x and y are decision variables

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

Step 2 our objective is to minimize the product x y

Let z=xy we have to maximize zStep3 we have the following conditions on the variables as x and ystep 4 x+ygt=15 x-ylt=7 xygt0 as the linear constraints the mathemetical constraint of this equation is to maximize the objective function z=xy

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
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• Slide 9
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• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

PROBLEMS1 A producer wants to maximise revenues producing two goods x

1and x2 in the market Market prices of goods are 10 and 5 respectively Production of x 1and x2 requires 25 and 10 units of skilled labour and total endowment of skilled labour is1000 Similarly production of x1 and x2 also requires 20 and 50 units of unskilled labour and whose total endowment is 1500 How much should this firm produce x 1and x 2 in order to maximise the total revenue

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
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• Slide 7
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• Slide 41
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• Slide 45
• Slide 46
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• Slide 49
• Slide 50

SOLUTION

Max R =10x1 + 5x2

Subject toSkilled labour constraint 25x1 +10x2lt=1000 Unskilled labour constraint 20x1 +50x2 lt=1500 Non-negativity constraints x1 x2 gt=0

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

Pounds of each alloy needed per frameAluminum Alloy

Steel Alloy Deluxe 2 3 deluxe

Professional 4 2

Example LP Formulation

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
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• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

Define the objectiveMaximize total weekly profitDefine the decision variables

x1 = number of Deluxe frames produced weeklyx2 = number of Professional frames produced weeklyWrite the mathematical objective function

Max Z = 10x1 + 15x2

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
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• Slide 41
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• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

Write a one- or two-word description of each constraint

Aluminum availableSteel availableWrite the right-hand side of each constraint100 80Write lt = gt for each constraintlt 100lt 80

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
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• Slide 7
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• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

Write all the decision variables on the left-hand side of each constraint

x1 x2 lt 100x1 x2 lt 80Write the coefficient for each decision in each constraint+ 2x1 + 4x2 lt 100+ 3x1 + 2x2 lt 80

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
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• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

LP in Final FormMax Z = 10x1 + 15x2

Subject To2x1 + 4x2 lt 100 ( aluminum constraint)3x1 + 2x2 lt 80 ( steel constraint)x1 x2 gt 0 (non-negativity constraints

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
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• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
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• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
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• Slide 48
• Slide 49
• Slide 50

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
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• Slide 9
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• Slide 50

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
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• Slide 50

x

20

15

0

(55)

0

152

203

subject to

920 max

yx

yx

yx

yxyx

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
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• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
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• Slide 18
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• Slide 21
• Slide 22
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• Slide 25
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• Slide 27
• Slide 28
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• Slide 31
• Slide 32
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• Slide 47
• Slide 48
• Slide 49
• Slide 50

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
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• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
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• Slide 19
• Slide 20
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• Slide 24
• Slide 25
• Slide 26
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• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

20

15

0

(55)

Examplegraphical method

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
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• Slide 35
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• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

Examplegraphical method

0

152

203

subject to

920 max

yx

yx

yx

yxyx

x

y

20

15

0

(55)

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
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• Slide 15
• Slide 16
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• Slide 33
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• Slide 35
• Slide 36
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• Slide 38
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• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

THE SIMPLEX METHOD

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
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• Slide 13
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• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
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• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

So far we find an optimal point by searchingamong feasible intersection points1048698 The search can be improved by starting withan initial feasible point and moving to aldquobetterrdquo solution until an optimal one is found1048698 The simplex method incorporates bothoptimality and feasibility tests to find theoptimal solution(s) if one exists

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

An optimality test shows whether anintersection point corresponds to a value of theobjective function better than the best valuefound so far1048698 A feasibility test determines whether theproposed intersection point is feasible1048698 The decision and slack variables are separatedinto two nonoverlapping sets which we callthe independent and dependent sets

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

THE SIMPLEX METHOD

Transform Linear Program into a systemof linear equations using slack variables

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

0

152

203

subject to

920 max

yx

yx

yx

yxyx

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

0100920

1501012

2000113

0

0920

152

203

21

2

1

ssyx

Pyx

syx

syx

THE SIMPLEX METHOD

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34
• Slide 35
• Slide 36
• Slide 37
• Slide 38
• Slide 39
• Slide 40
• Slide 41
• Slide 42
• Slide 43
• Slide 44
• Slide 45
• Slide 46
• Slide 47
• Slide 48
• Slide 49
• Slide 50

Start from the vertex (x=0 y=0)Move to the next vertex that increases profit as much as possible

0100920

1501012

2000113

THE SIMPLEX METHOD

0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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0100920

1501012

2000113

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient

has larger magnitude than the y coefficient)bull Compute check ratios to find pivot row (smallest

ratio)

Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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Basic Idea Start from a vertex (x=0 y=0) Move to next vertex that increases profit as much as possible

bull At (00) P = 0bull Increasing x can increase P the most (x coefficient has

larger magnitude than the y coefficient)bull Compute check ratios to findpivot row (smallest ratio)bull Pivot around the element inboth pivot column and row

0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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0100920

1501012

2000113

x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

1

203

1

1

syx

syx

0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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0100920

1501012

3200031311x y s1 s2 P RHS

Pivoting means solve for that variableThen substitute into the other equations

3

20

3

1

3

11 syx

340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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340010320370

350132310

3200031311

x y s1 s2 P RHS

3

20

3

1

3

11 syx

Pivoting means solve for that variableThen substitute into the other equations

THE END

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THE END

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