curso nptel - optimization methods

Optimization Methods: Introduction and Basic Concepts - Learning Objectives

Module 1: Introduction and Basic Concepts

Learning Objectives

Optimization is the act of obtaining the best result under the given circumstances. In design

construction and maintenance of any engineering system, many technological and managerial

decisions have to be taken at several stages. The ultimate goal of all such decisions is either

to minimize the effort required or to maximize the desired benefit. Hence optimization can be

defined as the process of finding the conditions that give the minimum or maximum value of

a function, where the function represents the effort required or the desired benefit.

This module starts with a glance through the historical development of optimization methods.

Engineering applications of optimizations are scanned through from which one would get a

broad picture of the multitude of applications, the optimization methods have. The Art of

modeling is briefly explained with the various phases involved in modeling. In the second

lecture various components of the Optimization problem are discussed and summarized with

steps involved in formulating a mathematical programming problem. In the third lecture the

optimization problems are classified under various criteria to enable choosing an appropriate

model applicable to different types of optimization problems. In the final lecture a brief

introduction to the classical and advanced optimization techniques in use are given.

At the end of the module the reader will be able to

1. Understand the need and origin of the optimization methods.

2. Get a broader picture of the various applications of optimization methods used in engineering.

3. Define an optimization problem and its various components.

4. Formulate optimization problems as mathematical programming problems.

5. Classify optimization problems to suitably choose the method needed to solve the particular type of problem.

6. Briefly learn about classical and advanced techniques in optimizations.

D Nagesh Kumar, IISc, Bangalore M1LO

Optimization Methods: Introduction and Basic Concepts 1

Module – 1 Lecture Notes – 1

Historical Development and Model Building

Introduction

In this lecture, historical development of optimization methods is glanced through. Apart

from the major developments, some recently developed novel approaches, such as, goal

programming for multi-objective optimization, simulated annealing, genetic algorithms, and

neural network methods are briefly mentioned tracing their origin. Engineering applications

of optimization with different modeling approaches are scanned through from which one

would get a broad picture of the multitude applications of optimization techniques.

Historical Development

The existence of optimization methods can be traced to the days of Newton, Lagrange, and

Cauchy. The development of differential calculus methods for optimization was possible

because of the contributions of Newton and Leibnitz to calculus. The foundations of calculus

of variations, which deals with the minimization of functions, were laid by Bernoulli, Euler,

Lagrange, and Weistrass. The method of optimization for constrained problems, which

involve the addition of unknown multipliers, became known by the name of its inventor,

Lagrange. Cauchy made the first application of the steepest descent method to solve

unconstrained optimization problems. By the middle of the twentieth century, the high-speed

digital computers made implementation of the complex optimization procedures possible and

stimulated further research on newer methods. Spectacular advances followed, producing a

massive literature on optimization techniques. This advancement also resulted in the

emergence of several well defined new areas in optimization theory.

Some of the major developments in the area of numerical methods of unconstrained

optimization are outlined here with a few milestones.

• Development of the simplex method by Dantzig in 1947 for linear programming

problems

• The enunciation of the principle of optimality in 1957 by Bellman for dynamic

programming problems,

D Nagesh Kumar, IISc, Bangalore

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• Work by Kuhn and Tucker in 1951 on the necessary and sufficient conditions for the

optimal solution of programming problems laid the foundation for later research in

non-linear programming.

• The contributions of Zoutendijk and Rosen to nonlinear programming during the early

1960s have been very significant.

• Work of Carroll and Fiacco and McCormick facilitated many difficult problems to be

solved by using the well-known techniques of unconstrained optimization.

• Geometric programming was developed in the 1960s by Duffin, Zener, and Peterson.

• Gomory did pioneering work in integer programming, one of the most exciting and

rapidly developing areas of optimization. The reason for this is that most real world

applications fall under this category of problems.

• Dantzig and Charnes and Cooper developed stochastic programming techniques and

solved problems by assuming design parameters to be independent and normally

distributed.

The necessity to optimize more than one objective or goal while satisfying the physical

limitations led to the development of multi-objective programming methods. Goal

programming is a well-known technique for solving specific types of multi-objective

optimization problems. The goal programming was originally proposed for linear problems

by Charnes and Cooper in 1961. The foundation of game theory was laid by von Neumann in

1928 and since then the technique has been applied to solve several mathematical, economic

and military problems. Only during the last few years has game theory been applied to solve

engineering problems.

Simulated annealing, genetic algorithms, and neural network methods represent a new class

of mathematical programming techniques that have come into prominence during the last

decade. Simulated annealing is analogous to the physical process of annealing of metals and

glass. The genetic algorithms are search techniques based on the mechanics of natural

selection and natural genetics. Neural network methods are based on solving the problem

using the computing power of a network of interconnected ‘neuron’ processors.


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Engineering applications of optimization

To indicate the widespread scope of the subject, some typical applications in different

engineering disciplines are given below.

• Design of civil engineering structures such as frames, foundations, bridges, towers,

chimneys and dams for minimum cost.

• Design of minimum weight structures for earth quake, wind and other types of

random loading.

• Optimal plastic design of frame structures (e.g., to determine the ultimate moment

capacity for minimum weight of the frame).

• Design of water resources systems for obtaining maximum benefit.

• Design of optimum pipeline networks for process industry.

• Design of aircraft and aerospace structure for minimum weight

• Finding the optimal trajectories of space vehicles.

• Optimum design of linkages, cams, gears, machine tools, and other mechanical

components.

• Selection of machining conditions in metal-cutting processes for minimizing the

product cost.

• Design of material handling equipment such as conveyors, trucks and cranes for

minimizing cost.

• Design of pumps, turbines and heat transfer equipment for maximum efficiency.

• Optimum design of electrical machinery such as motors, generators and transformers.

• Optimum design of electrical networks.

• Optimum design of control systems.

• Optimum design of chemical processing equipments and plants.

• Selection of a site for an industry.

• Planning of maintenance and replacement of equipment to reduce operating costs.

• Inventory control.

• Allocation of resources or services among several activities to maximize the benefit.

• Controlling the waiting and idle times in production lines to reduce the cost of

production.

• Planning the best strategy to obtain maximum profit in the presence of a competitor.


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• Designing the shortest route to be taken by a salesperson to visit various cities in a

single tour.

• Optimal production planning, controlling and scheduling.

• Analysis of statistical data and building empirical models to obtain the most accurate

representation of the statistical phenomenon.

However, the list is incomplete.

Art of Modeling: Model Building

Development of an optimization model can be divided into five major phases.

• Data collection

• Problem definition and formulation

• Model development

• Model validation and evaluation of performance

• Model application and interpretation

Data collection may be time consuming but is the fundamental basis of the model-building

process. The availability and accuracy of data can have considerable effect on the accuracy of

the model and on the ability to evaluate the model.

The problem definition and formulation includes the steps: identification of the decision

variables; formulation of the model objective(s) and the formulation of the model constraints.

In performing these steps the following are to be considered.

• Identify the important elements that the problem consists of.

• Determine the number of independent variables, the number of equations required to

describe the system, and the number of unknown parameters.

• Evaluate the structure and complexity of the model

• Select the degree of accuracy required of the model

Model development includes the mathematical description, parameter estimation, input

development, and software development. The model development phase is an iterative

process that may require returning to the model definition and formulation phase.

The model validation and evaluation phase is checking the performance of the model as a

whole. Model validation consists of validation of the assumptions and parameters of the


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model. The performance of the model is to be evaluated using standard performance

measures such as Root mean squared error and R2 value. A sensitivity analysis should be

performed to test the model inputs and parameters. This phase also is an iterative process and

may require returning to the model definition and formulation phase. One important aspect of

this process is that in most cases data used in the formulation process should be different

from that used in validation. Another point to keep in mind is that no single validation

process is appropriate for all models.

Model application and implementation include the use of the model in the particular area

of the solution and the translation of the results into operating instructions issued in

understandable form to the individuals who will administer the recommended system.

Different modeling techniques are developed to meet the requirements of different types of

optimization problems. Major categories of modeling approaches are: classical optimization

techniques, linear programming, nonlinear programming, geometric programming, dynamic

programming, integer programming, stochastic programming, evolutionary algorithms, etc.

These modeling approaches will be discussed in subsequent modules of this course.


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D Nagesh Kumar, IISc Optimization Methods: M1L11

Introduction and Basic Concepts

(i) Historical Development and

Model Building


Objectives

Understand the need and origin of the optimization methods. Get a broad picture of the various applications of optimization methods used in engineering.


Introduction

Optimization : The act of obtaining the best result under the given circumstances.

Design, construction and maintenance of engineering systems involve decision making both at the managerial and the technological level

Goals of such decisions :– to minimize the effort required or

– to maximize the desired benefit


Introduction (contd.)

Optimization : Defined as the process of finding the conditions that give the minimum or maximum value of a function, where the function represents the effort required or the desired benefit.


Historical Development

Existence of optimization methods can be traced to the days of Newton, Lagrange, and Cauchy.

Development of differential calculus methods of optimization was possible because of the contributions of Newton and Leibnitz to calculus.

Foundations of calculus of variations, dealing with the minimizations of functions, were laid by Bernoulli, Euler, Lagrange, and Weistrass


Historical Development (contd.)

The method of optimization for constrained problems, which involve the inclusion of unknown multipliers, became known by the name of its inventor, Lagrange.

Cauchy made the first application of the steepest descent method to solve unconstrained optimization problems.


Recent History

High-speed digital computers made implementation of the complex optimization procedures possible and stimulated further research on newer methods.

Massive literature on optimization techniques and emergence of several well defined new areas in optimization theory followed.


Milestones

Development of the simplex method by Dantzig in 1947 for linear programming problems.

The enunciation of the principle of optimality in 1957 by Bellman for dynamic programming problems.

Work by Kuhn and Tucker in 1951 on the necessary and sufficient conditions for the optimal solution of problems laid the foundation for later research in non-linear programming.


Milestones (contd.)

The contributions of Zoutendijk and Rosen to nonlinear programming during the early 1960s

Work of Carroll and Fiacco and McCormick facilitated many difficult problems to be solved by using the well-known techniques of unconstrained optimization.

Geometric programming was developed in the 1960s by Duffin, Zener, and Peterson.

Gomory did pioneering work in integer programming. The most realworld applications fall under this category of problems.

Dantzig and Charnes and Cooper developed stochastic programming techniques.


Milestones (contd.)

The desire to optimize more than one objective or a goal while satisfying the physical limitations led to the development of multi-objective programming methods; Ex. Goal programming.

The foundations of game theory were laid by von Neumann in 1928; applied to solve several mathematical, economic and military problems, and more recently to engineering design problems.

Simulated annealing, evolutionary algorithms including genetic algorithms, and neural network methods represent a new class of mathematical programming techniques that have come into prominence during the last decade.


Engineering applications of optimization.

Design of structural units in construction, machinery, and in space vehicles. Maximizing benefit/minimizing product costs in various manufacturing and construction processes. Optimal path finding in road networks/freight handling processes. Optimal production planning, controlling and scheduling. Optimal Allocation of resources or services among several activities to maximize the benefit.


Art of Modeling : Model Building

Development of an optimization model can be divided into five major phases. – Collection of data – Problem definition and formulation– Model development– Model validation and evaluation or performance– Model application and interpretation of results


Data collection

Data collection – may be time consuming but is the fundamental

basis of the model-building process– extremely important phase of the model-building

process – the availability and accuracy of data can have

considerable effect on the accuracy of the model and on the ability to evaluate the model.


Problem Definition

Problem definition and formulation, steps involved: – identification of the decision variables; – formulation of the model objective(s);– the formulation of the model constraints.

In performing these steps one must consider the following. – Identify the important elements that the problem consists of.– Determine the number of independent variables, the number of

equations required to describe the system, and the number of unknown parameters.

– Evaluate the structure and complexity of the model– Select the degree of accuracy required of the model


Model development

Model development includes:– the mathematical description, – parameter estimation, – input development, and – software development

The model development phase is an iterative process that may require returning to the model definition and formulation phase.


Model Validation and Evaluation

This phase is checking the model as a whole. Model validation consists of validation of the assumptions and parameters of the model.The performance of the model is to be evaluated using standard performance measures such as Root mean squared error and R2

value. Sensitivity analysis to test the model inputs and parameters. This phase also is an iterative process and may require returning to the model definition and formulation phase.One important aspect of this process is that in most cases data used in the formulation process should be different from that used in validation.


Modeling Techniques

Different modeling techniques are developed to meet the requirement of different type of optimization problems. Major categories of modeling approaches are:

– classical optimization techniques, – linear programming, – nonlinear programming, – geometric programming, – dynamic programming, – integer programming, – stochastic programming, – evolutionary algorithms, etc.

These approaches will be discussed in the subsequent modules.


Thank You

Optimization Methods: Introduction and Basic concepts 1


Optimization Problem and Model Formulation

Introduction

In the previous lecture we studied the evolution of optimization methods and their

engineering applications. A brief introduction was also given to the art of modeling. In this

lecture we will study the Optimization problem, its various components and its formulation as

a mathematical programming problem.

Basic components of an optimization problem:

An objective function expresses the main aim of the model which is either to be minimized

or maximized. For example, in a manufacturing process, the aim may be to maximize the

profit or minimize the cost. In comparing the data prescribed by a user-defined model with the

observed data, the aim is minimizing the total deviation of the predictions based on the model

from the observed data. In designing a bridge pier, the goal is to maximize the strength and

minimize size.

A set of unknowns or variables control the value of the objective function. In the

manufacturing problem, the variables may include the amounts of different resources used or

the time spent on each activity. In fitting-the-data problem, the unknowns are the parameters

of the model. In the pier design problem, the variables are the shape and dimensions of the

pier.

A set of constraints are those which allow the unknowns to take on certain values but

exclude others. In the manufacturing problem, one cannot spend negative amount of time on

any activity, so one constraint is that the "time" variables are to be non-negative. In the pier

design problem, one would probably want to limit the breadth of the base and to constrain its

size.

The optimization problem is then to find values of the variables that minimize or maximize

the objective function while satisfying the constraints.

Objective Function

As already stated, the objective function is the mathematical function one wants to maximize

or minimize, subject to certain constraints. Many optimization problems have a single


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objective function. (When they don't they can often be reformulated so that they do) The two

exceptions are:

• No objective function. In some cases (for example, design of integrated circuit

layouts), the goal is to find a set of variables that satisfies the constraints of the model.

The user does not particularly want to optimize anything and so there is no reason to

define an objective function. This type of problems is usually called a feasibility

problem.

• Multiple objective functions. In some cases, the user may like to optimize a number of

different objectives concurrently. For instance, in the optimal design of panel of a

door or window, it would be good to minimize weight and maximize strength

simultaneously. Usually, the different objectives are not compatible; the variables that

optimize one objective may be far from optimal for the others. In practice, problems

with multiple objectives are reformulated as single-objective problems by either

forming a weighted combination of the different objectives or by treating some of the

objectives as constraints.

Statement of an optimization problem

An optimization or a mathematical programming problem can be stated as follows:

To find X = which minimizes f(X) (1.1)

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Subject to the constraints

gi(X) 0≤ , i = 1, 2, …., m

lj(X) 0= , j = 1, 2, …., p

where X is an n-dimensional vector called the design vector, f(X) is called the objective

function, and gi(X) and lj(X) are known as inequality and equality constraints, respectively.

The number of variables n and the number of constraints m and/or p need not be related in

any way. This type problem is called a constrained optimization problem.


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If the locus of all points satisfying f(X) = a constant c, is considered, it can form a family of

surfaces in the design space called the objective function surfaces. When drawn with the

constraint surfaces as shown in Fig 1 we can identify the optimum point (maxima). This is

possible graphically only when the number of design variables is two. When we have three or

more design variables because of complexity in the objective function surface, we have to

solve the problem as a mathematical problem and this visualization is not possible.

.

Optimum point

f = C3

f = C2

f= C4

f = C5

C1 > C2 > C3 >C4 …..> Cn

f = C1

Fig 1

Optimization problems can be defined without any constraints as well.

To find X = which minimizes f(X) (1.2)

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Such problems are called unconstrained optimization problems. The field of unconstrained

optimization is quite a large and prominent one, for which a lot of algorithms and software

are available.


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Variables These are essential. If there are no variables, we cannot define the objective function and the

problem constraints. In many practical problems, one cannot choose the design variable

arbitrarily. They have to satisfy certain specified functional and other requirements.

Constraints

Constraints are not essential. It's been argued that almost all problems really do have

constraints. For example, any variable denoting the "number of objects" in a system can only

be useful if it is less than the number of elementary particles in the known universe! In

practice though, answers that make good sense in terms of the underlying physical or

economic criteria can often be obtained without putting constraints on the variables.

Design constraints are restrictions that must be satisfied to produce an acceptable design.

Constraints can be broadly classified as:

1) Behavioral or Functional constraints: These represent limitations on the behavior

performance of the system.

2) Geometric or Side constraints: These represent physical limitations on design

variables such as availability, fabricability, and transportability.

For example, for the retaining wall design shown in the Fig 2, the base width W cannot be

taken smaller than a certain value due to stability requirements. The depth D below the

ground level depends on the soil pressure coefficients Ka and Kp. Since these constraints

depend on the performance of the retaining wall they are called behavioral constraints. The

number of anchors provided along a cross section Ni cannot be any real number but has to be

a whole number. Similarly thickness of reinforcement used is controlled by supplies from the

manufacturer. Hence this is a side constraint.


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D

Ni no. of anchors

W

Fig. 2

Constraint Surfaces

Consider the optimization problem presented in eq. 1.1 with only the inequality constraint

gi(X) . The set of values of X that satisfy the equation g0≤ i(X) 0≤ forms a boundary surface

in the design space called a constraint surface. This will be a (n-1) dimensional subspace

where n is the number of design variables. The constraint surface divides the design space

into two regions: one with gi(X) (feasible region) and the other in which g0< i(X) > 0

(infeasible region). The points lying on the hyper surface will satisfy gi(X) =0. The collection

of all the constraint surfaces gi(X) = 0, j= 1, 2, …, m, which separates the acceptable region is

called the composite constraint surface.

Fig 3 shows a hypothetical two-dimensional design space where the feasible region is

denoted by hatched lines. The two-dimensional design space is bounded by straight lines as

shown in the figure. This is the case when the constraints are linear. However, constraints

may be nonlinear as well and the design space will be bounded by curves in that case. A

design point that lies on more than one constraint surface is called a bound point, and the

associated constraint is called an active constraint. Free points are those that do not lie on any

constraint surface. The design points that lie in the acceptable or unacceptable regions can be

classified as following:

1. Free and acceptable point

2. Free and unacceptable point


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3. Bound and acceptable point

4. Bound and unacceptable point.

Examples of each case are shown in Fig. 3.

Fig. 3

Bound unacceptable

point.

Behavior constraint

g2 ≤ 0

.

Infeasible region

Feasible region

Behavior constraint

g1 ≤0

Side constraint

g3 ≥ 0

Bound acceptable point.

.

Free acceptable point Free unacceptable

point

Formulation of design problems as mathematical programming problems

In mathematics, the term optimization, or mathematical programming, refers to the study

of problems in which one seeks to minimize or maximize a real function by systematically

choosing the values of real or integer variables from within an allowed set. This problem can

be represented in the following way

Given: a function f : A R from some set A to the real numbers

Sought: an element x0 in A such that f(x0) ≤ f(x) for all x in A ("minimization") or such that

f(x0) ≥ f(x) for all x in A ("maximization").

Such a formulation is called an optimization problem or a mathematical programming

problem (a term not directly related to computer programming, but still in use for example,


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in linear programming – (see module 3)). Many real-world and theoretical problems may be

modeled in this general framework.

Typically, A is some subset of the Euclidean space Rn, often specified by a set of constraints,

equalities or inequalities that the members of A have to satisfy. The elements of A are called

candidate solutions or feasible solutions. The function f is called an objective function, or cost

function. A feasible solution that minimizes (or maximizes, if that is the goal) the objective

function is called an optimal solution. The domain A of f is called the search space.

Generally, when the feasible region or the objective function of the problem does not present

convexity (refer module 2), there may be several local minima and maxima, where a local

minimum x* is defined as a point for which there exists some δ > 0 so that for all x such that

;

and

that is to say, on some region around x* all the function values are greater than or equal to the

value at that point. Local maxima are defined similarly.

A large number of algorithms proposed for solving non-convex problems – including the

majority of commercially available solvers – are not capable of making a distinction between

local optimal solutions and rigorous optimal solutions, and will treat the former as the actual

solutions to the original problem. The branch of applied mathematics and numerical analysis

that is concerned with the development of deterministic algorithms that are capable of

guaranteeing convergence in finite time to the actual optimal solution of a non-convex

problem is called global optimization.

Problem formulation

Problem formulation is normally the most difficult part of the process. It is the selection of

design variables, constraints, objective function(s), and models of the discipline/design.

Selection of design variables

A design variable, that takes a numeric or binary value, is controllable from the point of view

of the designer. For instance, the thickness of a structural member can be considered a design

variable. Design variables can be continuous (such as the length of a cantilever beam),


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discrete (such as the number of reinforcement bars used in a beam), or Boolean. Design

problems with continuous variables are normally solved more easily.

Design variables are often bounded, that is, they have maximum and minimum values.

Depending on the adopted method, these bounds can be treated as constraints or separately.

Selection of constraints

A constraint is a condition that must be satisfied to render the design to be feasible. An

example of a constraint in beam design is that the resistance offered by the beam at points of

loading must be equal to or greater than the weight of structural member and the load

supported. In addition to physical laws, constraints can reflect resource limitations, user

requirements, or bounds on the validity of the analysis models. Constraints can be used

explicitly by the solution algorithm or can be incorporated into the objective, by using

Lagrange multipliers.

Objectives

An objective is a numerical value that is to be maximized or minimized. For example, a

designer may wish to maximize profit or minimize weight. Many solution methods work only

with single objectives. When using these methods, the designer normally weights the various

objectives and sums them to form a single objective. Other methods allow multi-objective

optimization (module 8), such as the calculation of a Pareto front.

Models

The designer has to also choose models to relate the constraints and the objectives to the

design variables. These models are dependent on the discipline involved. They may be

empirical models, such as a regression analysis of aircraft prices, theoretical models, such as

from computational fluid dynamics, or reduced-order models of either of these. In choosing

the models the designer must trade-off fidelity with the time required for analysis.

The multidisciplinary nature of most design problems complicates model choice and

implementation. Often several iterations are necessary between the disciplines’ analyses in

order to find the values of the objectives and constraints. As an example, the aerodynamic

loads on a bridge affect the structural deformation of the supporting structure. The structural

deformation in turn changes the shape of the bridge and hence the aerodynamic loads. Thus,

it can be considered as a cyclic mechanism. Therefore, in analyzing a bridge, the


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aerodynamic and structural analyses must be run a number of times in turn until the loads and

deformation converge.

Representation in standard form

Once the design variables, constraints, objectives, and the relationships between them have

been chosen, the problem can be expressed as shown in equation 1.1

Maximization problems can be converted to minimization problems by multiplying the

objective by -1. Constraints can be reversed in a similar manner. Equality constraints can be

replaced by two inequality constraints.

Problem solution

The problem is normally solved choosing the appropriate techniques from those available in

the field of optimization. These include gradient-based algorithms, population-based

algorithms, or others. Very simple problems can sometimes be expressed linearly; in that case

the techniques of linear programming are applicable.

Gradient-based methods

• Newton's method

• Steepest descent

• Conjugate gradient

• Sequential quadratic programming

Population-based methods

• Genetic algorithms

• Particle swarm optimization

Other methods

• Random search

• Grid search

• Simulated annealing

Most of these techniques require large number of evaluations of the objectives and the

constraints. The disciplinary models are often very complex and can take significant amount

of time for a single evaluation. The solution can therefore be extremely time-consuming.


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Many of the optimization techniques are adaptable to parallel computing. Much of the current

research is focused on methods of decreasing the computation time.

The following steps summarize the general procedure used to formulate and solve

optimization problems. Some problems may not require that the engineer follow the steps in

the exact order, but each of the steps should be considered in the process.

1) Analyze the process itself to identify the process variables and specific characteristics

of interest, i.e., make a list of all the variables.

2) Determine the criterion for optimization and specify the objective function in terms of

the above variables together with coefficients.

3) Develop via mathematical expressions a valid process model that relates the input-

output variables of the process and associated coefficients. Include both equality and

inequality constraints. Use well known physical principles such as mass balances,

energy balance, empirical relations, implicit concepts and external restrictions.

Identify the independent and dependent variables to get the number of degrees of

freedom.

4) If the problem formulation is too large in scope:

break it up into manageable parts, or

simplify the objective function and the model

5) Apply a suitable optimization technique for mathematical statement of the problem.

6) Examine the sensitivity of the result, to changes in the values of the parameters in the

problem and the assumptions.


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(ii) Optimization Problem and

Model Formulation


Objectives

To study the basic components of an optimization problem.Formulation of design problems as mathematical programming problems.


Introduction - Preliminaries

Basic components of an optimization problem :– An objective function expresses the main aim of the

model which is either to be minimized or maximized.

– A set of unknowns or variables which control the value of the objective function.

– A set of constraints that allow the unknowns to take on certain values but exclude others.


Introduction (contd.)

The optimization problem is then to: – find values of the variables that minimize or

maximize the objective function while satisfying the constraints.


Objective Function

As already defined the objective function is the mathematical function one wants to maximize or minimize, subject to certain constraints. Many optimization problems have a single objective function (When they don't they can often be reformulated so that they do). The two interesting exceptions are:

– No objective function. The user does not particularly want to optimize anything so there is no reason to define an objective function. Usually called a feasibility problem.

– Multiple objective functions. In practice, problems with multiple objectives are reformulated as single-objective problems by either forming a weighted combination of the different objectives or by treating some of the objectives by constraints.



To find X = which maximizes f(X)

Subject to the constraints gi(X) <= 0 , i = 1, 2,….,mlj(X) = 0 , j = 1, 2,….,p

⎟⎟⎟⎟⎟⎟

⎠

⎞

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⎝

⎛

nx

xx

.

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1


where– X is an n-dimensional vector called the design vector – f(X) is called the objective function, and – gi(X) and lj(X) are known as inequality and equality constraints,

respectively. This type of problem is called a constrained optimization problem. Optimization problems can be defined without any constraints as well. Such problems are called unconstrained optimization problems.



Objective Function Surface

If the locus of all points satisfying f(X) = a constant c is considered, it can form a family of surfaces in the design spacecalled the objective function surfaces. When drawn with the constraint surfaces as shown in the figure we can identify the optimum point (maxima). This is possible graphically only when the number of design variable is two. When we have three or more design variables because of complexity in the objective function surface we have to solve the problem as a mathematical problem and this visualization is not possible.


Objective function surfaces to find the optimum point (maxima)


Variables and Constraints

Variables – These are essential. If there are no variables, we cannot define the

objective function and the problem constraints.

Constraints – Even though Constraints are not essential, it has been argued that

almost all problems really do have constraints. – In many practical problems, one cannot choose the design

variable arbitrarily. Design constraints are restrictions that must be satisfied to produce an acceptable design.


Constraints (contd.)

Constraints can be broadly classified as :

– Behavioral or Functional constraints : These represent limitations on the behavior and performance of the system.

– Geometric or Side constraints : These represent physical limitations on design variables such as availability, fabricability, and transportability.


Constraint Surfaces

Consider the optimization problem presented earlier with only inequality constraints gi(X) . The set of values of X that satisfy the equation gi(X) forms a boundary surface in the design space called a constraint surface.

The constraint surface divides the design space into two regions: one with gi(X) < 0 (feasible region) and the other in which gi(X) > 0 (infeasible region). The points lying on the hyper surface will satisfy gi(X) =0.

The figure shows a hypothetical two-dimensional design space where the feasible region is denoted

by hatched lines.

Behavior constraint

g2 ≤ 0

.

Infeasible region

Feasible region Behavior constraint

g1≤ 0

Side constraint

g3 ≥ 0

Bound unacceptable point.

Bound acceptable point.

.

Free acceptable pointFree unacceptable point


Formulation of design problems as mathematical programming problems

The following steps summarize the procedure used to formulate and solve mathematical programming problems.

1. Analyze the process to identify the process variables and specific characteristics of interest i.e. make a list of all variables.

2. Determine the criterion for optimization and specify the objective function in terms of the above variables together with coefficients.

3. Develop via mathematical expressions a valid process model that relates the input-output variables of the process and associated coefficients. a) Include both equality and inequality constraintsb) Use well known physical principles c) Identify the independent and dependent variables to get the

number of degrees of freedom

4. If the problem formulation is too large in scope:a) break it up into manageable parts/ orb) simplify the objective function and the model

5. Apply a suitable optimization technique for mathematical statement of the problem.

6. Examine the sensitivity of the result to changes in the coefficients in the problem and the assumptions.


Thank You

Optimization Methods: Introduction and Basic Concepts


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Classification of Optimization Problems

Introduction

In the previous lecture we studied the basics of an optimization problem and its formulation

as a mathematical programming problem. In this lecture we look at the various criteria for

classification of optimization problems.

Optimization problems can be classified based on the type of constraints, nature of design

variables, physical structure of the problem, nature of the equations involved, deterministic

nature of the variables, permissible value of the design variables, separability of the functions

and number of objective functions. These classifications are briefly discussed below.

Classification based on existence of constraints.

Under this category optimizations problems can be classified into two groups as follows:

Constrained optimization problems: which are subject to one or more constraints.

Unconstrained optimization problems: in which no constraints exist.

Classification based on the nature of the design variables.

There are two broad categories in this classification.

(i) In the first category the objective is to find a set of design parameters that makes a

prescribed function of these parameters minimum or maximum subject to certain constraints.

For example to find the minimum weight design of a strip footing with two loads shown in

Fig 1 (a) subject to a limitation on the maximum settlement of the structure can be stated as

follows.

Find X =

db

which minimizes

f(X) = h(b,d)

Subject to the constraints (s X ) max ; b 0 ; d 0

where s is the settlement of the footing. Such problems are called parameter or static

optimization problems.



2

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It may be noted that, for this particular example, the length of the footing (l), the loads P1 and

P2 and the distance between the loads are assumed to be constant and the required

optimization is achieved by varying b and d.

(ii) In the second category of problems, the objective is to find a set of design parameters,

which are all continuous functions of some other parameter that minimizes an objective

function subject to a set of constraints. If the cross sectional dimensions of the rectangular

footings are allowed to vary along its length as shown in Fig 3.1 (b), the optimization

problem can be stated as :

Find X(t) =

)()(

tdtb

which minimizes

f(X) = g( b(t), d(t) )


(s X(t) ) max 0 t l

b(t) 0 0 t l

d(t) 0 0 t l

The length of the footing (l) the loads P1 and P2 , the distance between the loads are assumed

to be constant and the required optimization is achieved by varying b and d along the length l.

Here the design variables are functions of the length parameter t. this type of problem, where

each design variable is a function of one or more parameters, is known as trajectory or

dynamic optimization problem.

(a) (b)

Figure 1

l l

P1 P2

d

b

P2 P1

b(t)

d(t)

t



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Classification based on the physical structure of the problem

Based on the physical structure, optimization problems are classified as optimal control and

non-optimal control problems.

(i) Optimal control problems

An optimal control (OC) problem is a mathematical programming problem involving a

number of stages, where each stage evolves from the preceding stage in a prescribed manner.

It is defined by two types of variables: the control or design and state variables. The control

variables define the system and controls how one stage evolves into the next. The state

variables describe the behavior or status of the system at any stage. The problem is to find a

set of control variables such that the total objective function (also known as the performance

index, PI) over all stages is minimized, subject to a set of constraints on the control and state

variables. An OC problem can be stated as follows:

Find X which minimizes f(X) = ),(1

ii

l

ii yxf


1),( iiiii yyyxq i = 1, 2, …., l

0)( jj xg , j = 1, 2, …., l

0)( kk yh , k = 1, 2, …., l

Where xi is the ith control variable, yi is the ith state variable, and fi is the contribution of the

ith stage to the total objective function. gj, hk, and qi are the functions of xj, yj ; xk, yk and xi and

yi, respectively, and l is the total number of states. The control and state variables xi and yi

can be vectors in some cases.

(ii) Problems which are not optimal control problems are called non-optimal control

problems.

Classification based on the nature of the equations involved

Based on the nature of equations for the objective function and the constraints, optimization

problems can be classified as linear, nonlinear, geometric and quadratic programming

problems. The classification is very useful from a computational point of view since many



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predefined special methods are available for effective solution of a particular type of

problem.

(i) Linear programming problem

If the objective function and all the constraints are ‘linear’ functions of the design variables,

the optimization problem is called a linear programming problem (LPP). A linear

programming problem is often stated in the standard form :

Find X =

nx

xx

.

.2

1

Which maximizes f(X) = i

n

ii xc

1


ji

n

iij bxa

1, j = 1, 2, . . . , m

xi 0 , j = 1, 2, . . . , m

where ci, aij, and bj are constants.

(ii) Nonlinear programming problem

If any of the functions among the objectives and constraint functions is nonlinear, the

problem is called a nonlinear programming (NLP) problem. This is the most general form of

a programming problem and all other problems can be considered as special cases of the NLP

problem.

(iii) Geometric programming problem

A geometric programming (GMP) problem is one in which the objective function and

constraints are expressed as polynomials in X. A function h(X) is called a polynomial (with

m terms) if h can be expressed as

nman

mamam

nan

aanan

aa xxxcxxxcxxxcXh 22

11

2222

1212

1212

1111)(



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where cj ( mj ,,1 ) and aij ( ni ,,1 and mj ,,1 ) are constants with 0jc and

0ix .

Thus GMP problems can be posed as follows:

Find X which minimizes

f(X) = ,0

1 1

N

j

n

i

ijaij xc cj > 0, xi > 0

subject to

gk(X) = ,01 1

kN

j

n

i

ijkqijk xa ajk > 0, xi > 0, k = 1,2,…..,m

where N0 and Nk denote the number of terms in the objective function and in the kth constraint

function, respectively.

(iv) Quadratic programming problem

A quadratic programming problem is the best behaved nonlinear programming problem with

a quadratic objective function and linear constraints and is concave (for maximization

problems). It can be solved by suitably modifying the linear programming techniques. It is

usually formulated as follows:

F(X) =

n

i

n

jjiij

n

iii xxQxqc

1 11

Subject to

,1

j

n

iiij bxa

j = 1,2,….,m

xi 0 , i = 1,2,….,n

where c, qi, Qij, aij, and bj are constants.

Classification based on the permissible values of the decision variables

Under this classification, objective functions can be classified as integer and real-valued

programming problems.



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(i) Integer programming problem

If some or all of the design variables of an optimization problem are restricted to take only

integer (or discrete) values, the problem is called an integer programming problem. For

example, the optimization is to find number of articles needed for an operation with least

effort. Thus, minimization of the effort required for the operation being the objective, the

decision variables, i.e. the number of articles used can take only integer values. Other

restrictions on minimum and maximum number of usable resources may be imposed.

(ii) Real-valued programming problem

A real-valued problem is that in which it is sought to minimize or maximize a real function

by systematically choosing the values of real variables from within an allowed set. When the

allowed set contains only real values, it is called a real-valued programming problem.

Classification based on deterministic nature of the variables

Under this classification, optimization problems can be classified as deterministic or

stochastic programming problems.

(i) Stochastic programming problem

In this type of an optimization problem, some or all the design variables are expressed

probabilistically (non-deterministic or stochastic). For example estimates of life span of

structures which have probabilistic inputs of the concrete strength and load capacity is a

stochastic programming problem as one can only estimate stochastically the life span of the

structure.

(ii) Deterministic programming problem

In this type of problems all the design variables are deterministic.

Classification based on separability of the functions

Based on this classification, optimization problems can be classified as separable and non-

separable programming problems based on the separability of the objective and constraint

functions.

(i) Separable programming problems

In this type of a problem the objective function and the constraints are separable. A function

is said to be separable if it can be expressed as the sum of n single-variable functions,

nni xfxfxf ,..., 221 , i.e.



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n

iii xfXf

1)(

and separable programming problem can be expressed in standard form as :

Find X which minimizes

n

iii xfXf

1

)(

subject to

j

n

iiijj bxgXg

1

)( , j = 1,2,. . . , m

where bj is a constant.

Classification based on the number of objective functions

Under this classification, objective functions can be classified as single-objective and multi-

objective programming problems.

(i) Single-objective programming problem in which there is only a single objective function.

(ii) Multi-objective programming problem

A multiobjective programming problem can be stated as follows:

Find X which minimizes XfXfXf k,..., 21

Subject to

gj(X) 0 , j = 1, 2, . . . , m

where f1, f2, . . . fk denote the objective functions to be minimized simultaneously.

For example in some design problems one might have to minimize the cost and weight of the

structural member for economy and, at the same time, maximize the load carrying capacity

under the given constraints.



(iii) Classification of Optimization

Problems


Introduction

Optimization problems can be classified based on the type of constraints, nature of design variables, physical structure of the problem, nature of the equations involved, deterministic nature of the variables, permissible value of the design variables, separability of the functions and number of objective functions. These classifications are briefly discussed in this lecture.


Classification based on existence of constraints.

Constrained optimization problems: which are subject to one or more constraints.

Unconstrained optimization problems: in which no constraints exist.


Classification based on the nature of the design variables

There are two broad categories of classification within this classification First category : the objective is to find a set of design parameters that make a prescribed function of these parameters minimum or maximum subject to certain constraints.

– For example to find the minimum weight design of a strip footing with two loads shown in the figure, subject to a limitation on the maximum settlement of the structure.

The problem can be defined as follows

subject to the constraints

The length of the footing (l) the loads P1 and P2 , the distance between the loads areassumed to be constant and the required optimization is achieved by varying b and d.Such problems are called parameter or static optimization problems.


Classification based on the nature of the design variables (contd.)

Second category: the objective is to find a set of design parameters, which are all continuous functions of some other parameter, that minimizes an objective function subject to a set of constraints.

– For example, if the cross sectional dimensions of the rectangular footing is allowed to vary along its length as shown in the following figure.

The problem can be defined as follows


l

The length of the footing (l) the loads P1 and P2 , the distance between the loads areassumed to be constant and the required optimization is achieved by varying b and d.Such problems are called trajectory or dynamic optimization problems.


Classification based on the physical structure of the problem

Based on the physical structure, we can classify optimization problems are classified as optimal control and non-optimal control problems. (i) An optimal control (OC) problem is a mathematical

programming problem involving a number of stages, where each stage evolves from the preceding stage in a prescribed manner.

It is defined by two types of variables: the control or design variables and state variables.

The problem is to find a set of control or design variables such that the total objective function (also known as the performance index, PI) over all stages is minimized subject to a set of constraints on the control and state variables. An OC problem can be stated as follows:

Where xi is the ith control variable, yi is the ith state variable, and fi is the contribution of the ith stage to the total objective function. gj, hk, and qiare the functions of xj, yj ; xk, yk and xi and yi, respectively, and l is the total number of states. The control and state variables xi and yi can be vectors in some cases.

(ii) The problems which are not optimal control problems arecalled non-optimal control problems.

subject to the constraints:


Classification based on the nature of the equations involved

Based on the nature of expressions for the objective function and the constraints, optimization problems can be classified as linear, nonlinear, geometric and quadratic programming problems.


Classification based on the nature of the equations involved (contd.)

(i) Linear programming problemIf the objective function and all the constraints are linear functions of the design variables, the mathematical programming problem is called a linear programming (LP) problem.

often stated in the standard form :subject to



(ii) Nonlinear programming problemIf any of the functions among the objectives and constraint functions is nonlinear, the problem is called a nonlinear programming (NLP) problem this is the most general form of a programming problem.



(iii) Geometric programming problem– A geometric programming (GMP) problem is one in which the

objective function and constraints are expressed as polynomials in X.A polynomial with N terms can be expressed as

– Thus GMP problems can be expressed as follows: Find X which minimizes :

subject to:



where N0 and Nk denote the number of terms in the objective and kth

constraint function, respectively.(iv) Quadratic programming problem

A quadratic programming problem is the best behaved nonlinear programming problem with a quadratic objective function and linear constraints and is concave (for maximization problems). It is usually formulated as follows:

Subject to:

where c, qi, Qij, aij, and bj are constants.


Classification based on the permissible values of the decision variables

Under this classification problems can be classified asinteger and real-valued programming problems

(i) Integer programming problemIf some or all of the design variables of an optimization problem are restricted to take only integer (or discrete) values, the problem is called an integer programming problem.

(ii) Real-valued programming problemA real-valued problem is that in which it is sought to minimize or maximize a real function by systematically choosing the values of real variables from within an allowed set. When the allowed set contains only real values, it is called a real-valued programming problem.


Classification based on deterministic nature of the variables

Under this classification, optimization problems can be classified as deterministic and stochastic programming problems

(i) Deterministic programming problem• In this type of problems all the design variables are

deterministic.(ii) Stochastic programming problem

In this type of an optimization problem some or all the parameters (design variables and/or pre-assigned parameters) are probabilistic (non deterministic or stochastic). For exampleestimates of life span of structures which have probabilistic inputs of the concrete strength and load capacity. A deterministic value of the life-span is non-attainable.


Classification based on separability of the functions

Based on the separability of the objective and constraint functions optimization problems can be classified as separable and non-separable programming problems(i) Separable programming problems

In this type of a problem the objective function and the constraints are separable. A function is said to be separable if it can be expressed as the sum of n single-variable functions and separable programming problem can be expressed in standard form as :

subject to :

where bj is a constant.


Classification based on the number of objective functions

Under this classification objective functions can be classified as single and multiobjective programming problems.(i) Single-objective programming problem in which there is only a

single objective.(ii) Multi-objective programming problem

A multiobjective programming problem can be stated as follows:

where f1, f2, . . . fk denote the objective functions to be minimized simultaneously.

subject to :


Thank You



Classical and Advanced Techniques for Optimization

In the previous lecture having understood the various classifications of optimization

problems, let us move on to understand the classical and advanced optimization techniques.

Classical Optimization Techniques

The classical optimization techniques are useful in finding the optimum solution or

unconstrained maxima or minima of continuous and differentiable functions. These are

analytical methods and make use of differential calculus in locating the optimum solution.

The classical methods have limited scope in practical applications as some of them involve

objective functions which are not continuous and/or differentiable. Yet, the study of these

classical techniques of optimization form a basis for developing most of the numerical

techniques that have evolved into advanced techniques more suitable to today’s practical

problems. These methods assume that the function is differentiable twice with respect to the

design variables and that the derivatives are continuous. Three main types of problems can be

handled by the classical optimization techniques, viz., single variable functions, multivariable

functions with no constraints and multivariable functions with both equality and inequality

constraints. For problems with equality constraints the Lagrange multiplier method can be

used. If the problem has inequality constraints, the Kuhn-Tucker conditions can be used to

identify the optimum solution. These methods lead to a set of nonlinear simultaneous

equations that may be difficult to solve. These classical methods of optimization are further

discussed in Module 2.

The other methods of optimization include

• Linear programming: studies the case in which the objective function f is linear and

the set A is specified using only linear equalities and inequalities. (A is the design

variable space)

• Integer programming: studies linear programs in which some or all variables are

constrained to take on integer values.

• Quadratic programming: allows the objective function to have quadratic terms,

while the set A must be specified with linear equalities and inequalities.


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• Nonlinear programming: studies the general case in which the objective function or

the constraints or both contain nonlinear parts.

• Stochastic programming: studies the case in which some of the constraints depend

on random variables.

• Dynamic programming: studies the case in which the optimization strategy is based

on splitting the problem into smaller sub-problems.

• Combinatorial optimization: is concerned with problems where the set of feasible

solutions is discrete or can be reduced to a discrete one.

• Infinite-dimensional optimization: studies the case when the set of feasible solutions

is a subset of an infinite-dimensional space, such as a space of functions.

• Constraint satisfaction: studies the case in which the objective function f is constant

(this is used in artificial intelligence, particularly in automated reasoning).

Most of these techniques will be discussed in subsequent modules.

Advanced Optimization Techniques

• Hill climbing

Hill climbing is a graph search algorithm where the current path is extended with a

successor node which is closer to the solution than the end of the current path.

In simple hill climbing, the first closer node is chosen whereas in steepest ascent hill

climbing all successors are compared and the closest to the solution is chosen. Both

forms fail if there is no closer node. This may happen if there are local maxima in the

search space which are not solutions. Steepest ascent hill climbing is similar to best

first search but the latter tries all possible extensions of the current path in order,

whereas steepest ascent only tries one.

Hill climbing is used widely in artificial intelligence fields, for reaching a goal state

from a starting node. Choice of next node starting node can be varied to give a

number of related algorithms.


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• Simulated annealing

The name and inspiration come from annealing process in metallurgy, a technique

involving heating and controlled cooling of a material to increase the size of its

crystals and reduce their defects. The heat causes the atoms to become unstuck from

their initial positions (a local minimum of the internal energy) and wander randomly

through states of higher energy; the slow cooling gives them more chances of finding

configurations with lower internal energy than the initial one.

In the simulated annealing method, each point of the search space is compared to a

state of some physical system, and the function to be minimized is interpreted as the

internal energy of the system in that state. Therefore the goal is to bring the system,

from an arbitrary initial state, to a state with the minimum possible energy.

• Genetic algorithms

A genetic algorithm (GA) is a search technique used in computer science to find

approximate solutions to optimization and search problems. Specifically it falls into

the category of local search techniques and is therefore generally an incomplete

search. Genetic algorithms are a particular class of evolutionary algorithms that use

techniques inspired by evolutionary biology such as inheritance, mutation, selection,

and crossover (also called recombination).

Genetic algorithms are typically implemented as a computer simulation. in which a

population of abstract representations (called chromosomes) of candidate solutions

(called individuals) to an optimization problem, evolves toward better solutions.

Traditionally, solutions are represented in binary as strings of 0s and 1s, but different

encodings are also possible. The evolution starts from a population of completely

random individuals and occur in generations. In each generation, the fitness of the

whole population is evaluated, multiple individuals are stochastically selected from

the current population (based on their fitness), and modified (mutated or recombined)

to form a new population. The new population is then used in the next iteration of the

algorithm.


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• Ant colony optimization

In the real world, ants (initially) wander randomly, and upon finding food return to

their colony while laying down pheromone trails. If other ants find such a path, they

are likely not to keep traveling at random, but instead follow the trail laid by earlier

ants, returning and reinforcing it, if they eventually find any food.

Over time, however, the pheromone trail starts to evaporate, thus reducing its

attractive strength. The more time it takes for an ant to travel down the path and back

again, the more time the pheromones have to evaporate. A short path, by comparison,

gets marched over faster, and thus the pheromone density remains high as it is laid on

the path as fast as it can evaporate. Pheromone evaporation has also the advantage of

avoiding the convergence to a local optimal solution. If there was no evaporation at

all, the paths chosen by the first ants would tend to be excessively attractive to the

following ones. In that case, the exploration of the solution space would be

constrained.

Thus, when one ant finds a good (short) path from the colony to a food source, other

ants are more likely to follow that path, and such positive feedback eventually leaves

all the ants following a single path. The idea of the ant colony algorithm is to mimic

this behavior with "simulated ants" walking around the search space representing the

problem to be solved.

Ant colony optimization algorithms have been used to produce near-optimal solutions

to the traveling salesman problem. They have an advantage over simulated annealing

and genetic algorithm approaches when the graph may change dynamically. The ant

colony algorithm can be run continuously and can adapt to changes in real time. This

is of interest in network routing and urban transportation systems.


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References / Further Reading:

1. Deb K., Multi-Objective Optimization using Evolutionary Algorithms, First Edition, John

Wiley & Sons Pte Ltd, 2002.

2. Deb K., Optimization for Engineering Design – Algorithms and Examples, Prentice Hall

of India Pvt. Ltd., New Delhi, 1995.

3. Dorigo M., and T. Stutzle, Ant Colony Optimization, Prentice Hall of India Pvt. Ltd., New

Delhi, 2005.

4. Hillier F.S. and G.J. Lieberman, Operations Research, CBS Publishers & Distributors,

New Delhi, 1987.

5. Jain S.K. and V.P. Singh, Water Resources Systems Planning and Management, Elsevier

B.V., The Netherlands, 2003.

6. Loucks, D.P., J.R. Stedinger, and D.A. Haith, Water Resources Systems Planning and

Analysis, Prentice – Hall, N.J., 1981.

7. Mays, L.W. and K. Tung, Hydrosystems Engineering and Management, McGraw-Hill

Inc., New York, 1992.

8. Rao S.S., Engineering Optimization – Theory and Practice, Third Edition, New Age

International Limited, New Delhi, 2000

9. Ravindran A., D.T. Phillips and J.J. Solberg, Operations Research – Principles and

Practice, John Wiley & Sons, New York, 2001.

10. Taha H.A., Operations Research – An Introduction, Prentice-Hall of India Pvt. Ltd., New

Delhi, 2005.

11. Vedula S., and P.P. Mujumdar, Water Resources Systems: Modelling Techniques and

Analysis, Tata McGraw Hill, New Delhi, 2005.


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Classical and Advanced Techniques

for Optimization


Classical Optimization Techniques

The classical optimization techniques are useful in finding the optimum solution or unconstrained maxima or minima of continuous and differentiable functions.

These are analytical methods and make use of differential calculus in locating the optimum solution.

The classical methods have limited scope in practical applications as some of them involve objective functions which are not continuous and/ordifferentiable.

Yet, the study of these classical techniques of optimization form a basis for developing most of the numerical techniques that have evolved into advanced techniques more suitable to today’s practical problems


Classical Optimization Techniques (contd.)

These methods assume that the function is differentiable twice with respect to the design variables and the derivatives are continuous.

Three main types of problems can be handled by the classical optimization techniques:

– single variable functions

– multivariable functions with no constraints,

– multivariable functions with both equality and inequality constraints. In problems with equality constraints the Lagrange multiplier method can be used. If the problem has inequality constraints, the Kuhn-Tucker conditions can be used to identify the optimum solution.

These methods lead to a set of nonlinear simultaneous equations that may be difficult to solve. These methods of optimization are discussed in Module 2.


Numerical Methods of Optimization

Linear programming: studies the case in which the objective function f is linear and the set A is specified using only linear equalities and inequalities. (A is the design variable space)

Integer programming: studies linear programs in which some or all variables are constrained to take on integer values.

Quadratic programming: allows the objective function to have quadratic terms, while the set A must be specified with linear equalities and inequalities

Nonlinear programming: studies the general case in which the objective function or the constraints or both contain nonlinear parts.


Numerical Methods of Optimization (contd.)

• Stochastic programming: studies the case in which some of the constraints depend on random variables.

• Dynamic programming: studies the case in which the optimization strategy is based on splitting the problem into smaller sub-problems.

• Combinatorial optimization: is concerned with problems where the set of feasible solutions is discrete or can be reduced to a discrete one.

• Infinite-dimensional optimization: studies the case when the set of feasible solutions is a subset of an infinite-dimensional space, such as a space of functions.

• Constraint satisfaction: studies the case in which the objective function f is constant (this is used in artificial intelligence, particularly in automated reasoning).


Advanced Optimization Techniques

Hill climbing: it is a graph search algorithm where the current path is extended with a successor node which is closer to the solution than the end of the current path.In simple hill climbing, the first closer node is chosen whereas in steepest ascent hill climbing all successors are compared and the closest to the solution is chosen. Both forms fail if there is no closer node. This may happen if there are local maxima in the search space which are not solutions. Hill climbing is used widely in artificial intelligence fields, for reaching a goal state from a starting node. Choice of next node/ starting node can be varied to give a number of related algorithms.


Simulated annealing

The name and inspiration come from annealing process in metallurgy, a technique involving heating and controlled cooling of a material to increase the size of its crystals and reduce their defects.

– The heat causes the atoms to become unstuck from their initial positions (a local minimum of the internal energy) and wander randomly through states of higher energy;

– the slow cooling gives them more chances of finding configurations with lower internal energy than the initial one.

In the simulated annealing method, each point of the search space is compared to a state of some physical system, and the function to be minimized is interpreted as the internal energy of the system in that state. Therefore the goal is to bring the system, from an arbitrary initial state, to a state with the minimum possible energy.


Genetic algorithms

A genetic algorithm (GA) is a local search technique used to find approximate solutions to optimization and search problems.

Genetic algorithms are a particular class of evolutionary algorithms that use techniques inspired by evolutionary biology such as inheritance, mutation, selection, and crossover (also called recombination).


Genetic algorithms (contd.)

Genetic algorithms are typically implemented as a computer simulation, in which a population of abstract representations (called chromosomes) of candidate solutions (called individuals) to an optimization problem, evolves toward better solutions.

The evolution starts from a population of completely random individuals and occurs in generations.

In each generation, the fitness of the whole population is evaluated, multiple individuals are stochastically selected from the current population (based on their fitness), and modified (mutated or recombined) to form a new population.

The new population is then used in the next iteration of the algorithm.


Ant colony optimization

In the real world, ants (initially) wander randomly, and upon finding food return to their colony while laying down pheromone trails. If other ants find such a path, they are likely not to keep traveling at random, but instead follow the trail laid by earlier ants, returning and reinforcing it if they eventually find food

Over time, however, the pheromone trail starts to evaporate, thus reducing its attractive strength. The more time it takes for an ant to travel down the path and back again, the more time the pheromones have to evaporate.

A short path, by comparison, gets marched over faster, and thus the pheromone density remains high

Pheromone evaporation has also the advantage of avoiding the convergence to a locally optimal solution. If there were no evaporation at all, the paths chosen by the first ants would tend to be excessively attractive to the following ones. In that case, the exploration of the solution space would be constrained.


Ant colony optimization (contd.)

Thus, when one ant finds a good (short) path from the colony to a food source, other ants are more likely to follow that path, and such positive feedback eventually leaves all the ants following a single path.

The idea of the ant colony algorithm is to mimic this behavior with "simulated ants" walking around the search space representing the problem to be solved.

Ant colony optimization algorithms have been used to produce near-optimal solutions to the traveling salesman problem.

They have an advantage over simulated annealing and genetic algorithm approaches when the graph may change dynamically. The ant colony algorithm can be run continuously and can adapt to changes in real time.

This is of interest in network routing and urban transportation systems.


Thank You

Optimization Methods: Optimization using Calculus - Learning Objectives

Module 2: Optimization using Calculus

Learning Objectives

Optimization problems with continuous differentiable functions can be solved using the

classical methods of optimization. These analytical methods employ differential calculus to

locate the optimum points. The classical optimization techniques fail to have an application

where the functions are not continuous and not differentiable, and this happens with many

practical problems. However a study of these calculus based methods is a foundation for

development of most of the numerical techniques presented in later modules.

In this module a brief introduction to stationary points is followed by a presentation of the

necessary and sufficient conditions in locating the optimum solution of a single variable and

two variable functions. Convexity and concavity of these functions are explained. Then the

reader is introduced to the optimization of functions or single and multivariable functions

(with and without equality constraints). A few examples are discussed for each type. An

insight is also given to the Lagrangian function and Hessian matrix formulation. Finally we

take a look at the Kuhn-Tucker conditions with examples.

This module will help the reader to know about

1. Stationary points as maxima, minima and points of inflection

2. Concavity and convexity of functions

3. Necessary and sufficient conditions for optimization for both single and multivariable

functions

4. The Hessian matrix

5. Optimization of multivariable function with and without equality constraints

6. Kuhn-Tucker conditions


Optimization Methods: Optimization using Calculus-Stationary Points 1

Module - 2 Lecture Notes – 1

Stationary points: Functions of Single and Two Variables

Introduction

In this session, stationary points of a function are defined. The necessary and sufficient

conditions for the relative maximum of a function of single or two variables are also

discussed. The global optimum is also defined in comparison to the relative or local optimum.

Stationary points

For a continuous and differentiable function f(x) a stationary point x* is a point at which the

slope of the function vanishes, i.e. f ’(x) = 0 at x = x*, where x* belongs to its domain of

definition.

minimum inflection point maximum

Fig. 1

A stationary point may be a minimum, maximum or an inflection point (Fig. 1).

Relative and Global Optimum

A function is said to have a relative or local minimum at x = x* if * *( ) ( )f x f x h≤ + for all

sufficiently small positive and negative values of h, i.e. in the near vicinity of the point x*.

Similarly a point x* is called a relative or local maximum if * *( ) ( )f x f x h≥ +

*( ) ( )

for all values

of h sufficiently close to zero. A function is said to have a global or absolute minimum at x =

x* if f x f x≤ for all x in the domain over which f(x) is defined. Similarly, a function is


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said to have a global or absolute maximum at x = x* if *( ) ( )f x f x≥ for all x in the domain

over which f(x) is defined.

Figure 2 shows the global and local optimum points.

a b a bx x

f(x) f(x)

..

.. .

.A1

B1

B2

A3

A2 Relative minimum is also global optimum (since only one minimum point is there)

A1, A2, A3 = Relative maxima A2 = Global maximum B1, B2 = Relative minima B1 = Global minimum

Fig. 2

Functions of a single variable

Consider the function f(x) defined for a x b≤ ≤ . To find the value of x* ∈ such that x =

x

[ , ]a b* maximizes f(x) we need to solve a single-variable optimization problem. We have the

following theorems to understand the necessary and sufficient conditions for the relative

maximum of a function of a single variable.

Necessary condition: For a single variable function f(x) defined for x which has a

relative maximum at x = x

[ , ]a b∈* , x* ∈ if the derivative [ , ]a b ) /'( ) (f X df x dx= exists as a finite

number at x = x* then f ‘(x*) = 0. This can be understood from the following.


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Proof.

Since f ‘(x*) is stated to exist, we have

'

0

( * ) ( *)( *) limh

f x h f xf xh→

+ −= (1)

From our earlier discussion on relative maxima we have ( *) ( * )f x f x h≥ + for . Hence 0h →

( * ) ( *) 0f x h f xh

+ −≥ h < 0 (2)

( * ) ( *) 0f x h f xh

+ −≤ h > 0 (3)

which implies for substantially small negative values of h we have and for

substantially small positive values of h we have

( *) 0f x ≥

( *) 0f x ≤ . In order to satisfy both (2) and

(3), f ( *)x

(

= 0. Hence this gives the necessary condition for a relative maxima at x = x* for

)x . f

It has to be kept in mind that the above theorem holds good for relative minimum as well.

The theorem only considers a domain where the function is continuous and differentiable. It

cannot indicate whether a maxima or minima exists at a point where the derivative fails to

exist. This scenario is shown in Fig 3, where the slopes m1 and m2 at the point of a maxima

are unequal, hence cannot be found as depicted by the theorem by failing for continuity. The

theorem also does not consider if the maxima or minima occurs at the end point of the

interval of definition, owing to the same reason that the function is not continuous, therefore

not differentiable at the boundaries. The theorem does not say whether the function will have

a maximum or minimum at every point where f ‘(x) = 0, since this condition f ‘(x) = 0 is for

stationary points which include inflection points which do not mean a maxima or a minima.

A point of inflection is shown already in Fig.1


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f(x)

a b x

m2m1f(x*)

x*Fig. 3

Sufficient condition: For the same function stated above let f ’(x*) = f ”(x*) = . . . = f (n-1)(x*)

= 0, but f (n)(x*) 0, then it can be said that f (x≠ *) is (a) a minimum value of f (x) if f (n)(x*) > 0

and n is even; (b) a maximum value of f (x) if f (n)(x*) < 0 and n is even; (c) neither a

maximum or a minimum if n is odd.

Proof

Applying the Taylor’s theorem with remainder after n terms, we have

2 1( 1)( * ) ( *) '( *) ''( *) ... ( *) ( * )

2! ( 1)! !

n nn nh h hf x h f x hf x f x f x f x h

n nθ

−−+ = + + + + + +

− (4)

for 0<θ <1

since f ‘(x*) = f ‘’(x*) = . . . = f (n-1)(x*) = 0, (4) becomes

( * ) ( *) ( * )!

nnhf x h f x f x h

nθ+ − = + (5)

As f (n)(x*) 0, there exists an interval around x≠ * for every point x of which the nth derivative

f (n)(x) has the same sign, viz., that of f (n)(x*). Thus for every point x*+ h of this interval, f

(n)(x*+ h) has the sign of f (n)(x*). When n is even !n

nh

( * ) ( *)

is positive irrespective of the sign of h,

and hence f x h f x+ − will have the same sign as that of f (n)(x*). Thus x* will be a

relative minimum if f (n)(x*) is positive, with f(x) convex around x*, and a relative maximum if


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f (n)(x*) is negative, with f(x) concave around x*. When n is odd, !n

nh changes sign with the

change in the sign of h and hence the point x* is neither a maximum nor a minimum. In this

case the point x* is called a point of inflection.

Example 1.

Find the optimum value of the function 2( ) 3 5f x x x= + − and also state if the function

attains a maximum or a minimum.

Solution

'( ) 2 3 0f x x= + = for maxima or minima.

or x* = -3/2

''( *) 2f x = which is positive hence the point x* = -3/2 is a point of minima and the function

attains a minimum value of -29/4 at this point.

Example 2.

Find the optimum value of the function and also state if the function attains a

maximum or a minimum.

4( ) ( 2)f x x= −

Solution

3'( ) 4( 2) 0f x x= − = for maxima or minima.

or x = x* = 2 for maxima or minima.

2''( *) 12( * 2) 0f x x= − = at x* = 2

'''( *) 24( * 2) 0f x x= − = at x* = 2


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( ) 24* =′′′′ xf at x* = 2

Hence fn(x) is positive and n is even hence the point x = x* = 2 is a point of minimum and the

function attains a minimum value of 0 at this point.

Example 3.

Analyze the function and classify the stationary points as

maxima, minima and points of inflection.

5 4 3( ) 12 45 40 5f x x x x= − + +

Solution

4 3 2

4 3 2

'( ) 60 180 120 0 3 2 0or 0,1,2

f x x x xx x xx

= − + =

=> − + ==

Consider the point x =x* = 0

'' * * 3 * 2 *( ) 240( ) 540( ) 240 0f x x x x= − + =

−

at x * = 0

''' * * 2 *( ) 720( ) 1080 240 240f x x x= − + = at x * = 0

Since the third derivative is non-zero, x = x* = 0 is neither a point of maximum or minimum

but it is a point of inflection.

Consider x = x* = 1

'' * * 3 * 2 *( ) 240( ) 540( ) 240 60f x x x x= − + = at x* = 1

Since the second derivative is negative the point x = x* = 1 is a point of local maxima with a

maximum value of f(x) = 12 – 45 + 40 + 5 = 12


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Consider x = x* = 2

'' * * 3 * 2 *( ) 240( ) 540( ) 240 240f x x x x= − + = at x* = 2

Since the second derivative is positive, the point x = x* = 2 is a point of local minima with a

minimum value of f(x) = -11

Example 4.

The horse power generated by a Pelton wheel is proportional to u(v-u) where u is the velocity

of the wheel, which is variable and v is the velocity of the jet which is fixed. Show that the

efficiency of the Pelton wheel will be maximum at u = v/2.

Solution

K. ( )

0 K 2K

or 2

0

f u v uf v uu

vu

= −∂

= => − =∂

=

where K is a proportionality constant (assumed positive).

2

2

2

2Kvu

fu

=

∂= −

∂which is negative.

Hence, f is maximum at 2vu =

Functions of two variables

This concept may be easily extended to functions of multiple variables. Functions of two

variables are best illustrated by contour maps, analogous to geographical maps. A contour is a

line representing a constant value of f(x) as shown in Fig.4. From this we can identify

maxima, minima and points of inflection.


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Necessary conditions

As can be seen in Fig. 4 and 5, perturbations from points of local minima in any direction

result in an increase in the response function f(x), i.e. the slope of the function is zero at this

point of local minima. Similarly, at maxima and points of inflection as the slope is zero, the

first derivatives of the function with respect to the variables are zero.

Which gives us1 2

0; 0f fx x∂ ∂

= =∂ ∂

at the stationary points, i.e., the gradient vector of f(X), x fΔ

at X = X* = [x1, x2] defined as follows, must equal zero:

1

2

( *)0

( *)x

fx

ffx

∂⎡ ⎤Χ⎢ ⎥∂⎢ ⎥Δ = =∂⎢ ⎥

Χ⎢ ⎥∂⎣ ⎦

This is the necessary condition.

x2

x1

Fig. 4


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Global maxima Relative maxima

Relative minima

Global minima

Fig. 5

Sufficient conditions

Consider the following second order derivatives:

2 2 2

2 21 2 1

; ;2

f f fx x x x

∂ ∂ ∂∂ ∂ ∂ ∂

The Hessian matrix defined by H is made using the above second order derivatives.


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1 2

2 2

21 1 2

2 2

21 2 2 [ , ]x x

f fx x x

f fx x x

⎛ ⎞∂ ∂⎜ ⎟∂ ∂ ∂⎜ ⎟=⎜ ⎟∂ ∂⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠

H

a) If H is positive definite then the point X = [x1, x2] is a point of local minima.

b) If H is negative definite then the point X = [x1, x2] is a point of local maxima.

c) If H is neither then the point X = [x1, x2] is neither a point of maxima nor minima.

A square matrix is positive definite if all its eigen values are positive and it is negative

definite if all its eigen values are negative. If some of the eigen values are positive and some

negative then the matrix is neither positive definite or negative definite.

To calculate the eigen valuesλ of a square matrix then the following equation is solved.

0λ− =A I

The above rules give the sufficient conditions for the optimization problem of two variables.

Optimization of multiple variable problems will be discussed in detail in lecture notes 3

(Module 2).

Example 5.

Locate the stationary points of f(X) and classify them as relative maxima, relative minima or

neither based on the rules discussed in the lecture.

f(X) = + 5


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Solution

From 2

(X) 0fx∂

=∂

, 1 22 2x x= +

From 1

(X) 0fx∂

=∂

22 28 14 3x x 0+ + =

2 2(2 3)(4 1) 0x x+ + =

2 23 / 2 or 1/ 4x x= − = −

so the two stationary points are

X1 = [-1,-3/2]

and

X2 = [3/2,-1/4]

The Hessian of f(X) is

2 2 2 2

12 21 2 1 2 2 1

4 ; 4; 2f f f fxx x x x x x∂ ∂ ∂ ∂

= = = =∂ ∂ ∂ ∂ ∂ ∂

−


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14 22 4x −⎡ ⎤

= ⎢ ⎥−⎣ ⎦H

14 22 4

xλλ

λ−

=−

I - H

At X1= [-1,-3/2],

4 2( 4)( 4) 4

2 4λ

λ λ λλ

+= = + −

−I - H 0− =

2 16 4 0λ − − =

2λ = 12

1 212 12 λ λ= + = −

Since one eigen value is positive and one negative, X1 is neither a relative maximum nor a

relative minimum.

At X2 = [3/2,-1/4]

6 2( 6)( 4) 4

2 4λ

λ λ λλ

−= = − − −

−I - H 0=

2 10 20 0λ λ− + =

1 25 5 5 5λ λ= + = −

Since both the eigen values are positive, X2 is a local minimum.

Minimum value of f(x) is -0.375.


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Example 6

The ultimate strength attained by concrete is found to be based on a certain empirical

relationship between the ratios of cement and concrete used. Our objective is to maximize

strength attained by hardened concrete, given by f(X) = , where x2 21 1 2 220 2 6 3 / 2x x x x+ − + − 1

and x2 are variables based on cement and concrete ratios.

Solution

Given f(X) = ; where X = 21 1 2 220 2 6 3 / 2x x x x+ − + − 2 [ ]1 2,x x

The gradient vector 1 1

2

2

( *)2 2 06 3 0( *)

x

fx x

fxf

x

∂⎡ ⎤Χ⎢ ⎥∂ −⎡ ⎤ ⎡ ⎤⎢ ⎥Δ = = =⎢ ⎥ ⎢ ⎥−∂⎢ ⎥ ⎣ ⎦⎣ ⎦Χ⎢ ⎥∂⎣ ⎦

, to determine stationary point X*.

Solving we get X* = [1,2]

2 2 2

2 21 2 1 2

2; 3; 0f f fx x x x∂ ∂ ∂

= − = −∂ ∂ ∂ ∂

=

2 00 3−⎡ ⎤

= ⎢ ⎥−⎣ ⎦H

2 0( 2)( 3)

0 3λ

λ λλ

+= = + +

+I - H 0λ =

Here the values of λ do not depend on X and 1λ = -2, 2λ = -3. Since both the eigen values

are negative, f(X) is concave and the required ratio x1:x2 = 1:2 with a global maximum

strength of f(X) = 27 units.


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Optimization using Calculus

Stationary Points: Functions of Single and Two Variables


Objectives

To define stationary points

Look into the necessary and sufficient conditions for the

relative maximum of a function of a single variable and

for a function of two variables.

To define the global optimum in comparison to the

relative or local optimum


Stationary points

For a continuous and differentiable function f(x) a stationary point x* is a point at which the function vanishes, i.e. f ’(x) = 0 at x = x*. x* belongs to its domain of definition.

A stationary point may be a minimum, maximum or an inflection point


Stationary points

Figure showing the three types of stationary points (a) inflection point (b) minimum (c) maximum


Relative and Global Optimum

• A function is said to have a relative or local minimum at x = x* if for all sufficiently small positive and negative

values of h, i.e. in the near vicinity of the point x. • Similarly, a point x* is called a relative or local maximum if

for all values of h sufficiently close to zero. • A function is said to have a global or absolute minimum at x = x* if

for all x in the domain over which f(x) is defined. • Similarly, a function is said to have a global or absolute maximum at

x = x* if for all x in the domain over which f (x) is defined.

*( ) ( )f x f x h≤ +

*( ) ( )f x f x h≥ +

*( ) ( )f x f x≤

*( ) ( )f x f x≥


Relative and Global Optimum …contd.

a b a bx x

f(x)f(x)

.

..

..

.A1

B1

B2

A3

A2

Relative minimum is also global optimum

A1, A2, A3 = Relative maximaA2 = Global maximum

B1, B2 = Relative minimaB1 = Global minimum

Fig. 2


Functions of a single variable

Consider the function f(x) defined for

To find the value of x* such that x* maximizes f(x) we need to solve a single-variable optimization problem.

We have the following theorems to understand the necessary and sufficient conditions for the relative maximum of a function of a single variable.

a x b≤ ≤[ , ]a b∈


Functions of a single variable …contd.

Necessary condition : For a single variable function f(x) defined for xwhich has a relative maximum at x = x* , x* if the

derivative f ‘(x) = df(x)/dx exists as a finite number at x = x* thenf ‘(x*) = 0. We need to keep in mind that the above theorem holds good for relative minimum as well. The theorem only considers a domain where the function is continuous and derivative. It does not indicate the outcome if a maxima or minima exists at a point where the derivative fails to exist. This scenario is shown in the figure below, where the slopes m1 and m2 at the point of a maxima are unequal, hence cannot be found as depicted by the theorem.

[ , ]a b∈ [ , ]a b∈


Functions of a single variable …contd.

Some Notes:The theorem does not consider if the maxima or minima occurs at the end point of the interval of definition. The theorem does not say that the function will have a maximum or minimum at every point where f ’(x) = 0, since this conditionf ’(x) = 0 is for stationary points which include inflection points which do not mean a maxima or a minima.


Sufficient condition

For the same function stated above let f ’(x*) = f ”(x*) = . . . = f (n-1)(x*) = 0, but f (n)(x*) 0, then it can be said that f (x*) is

– (a) a minimum value of f (x) if f (n)(x*) > 0 and n is even

– (b) a maximum value of f (x) if f (n)(x*) < 0 and n is even

– (c) neither a maximum or a minimum if n is odd

≠


Example 1

Find the optimum value of the function and also state if the function attains a maximum or aminimum. Solution

for maxima or minima.

or x* = -3/2

which is positive hence the point x* = -3/2 is a point of minima and the function attains a minimum value of -29/4 at this point.

2( ) 3 5f x x x= + −

'( ) 2 3 0f x x= + =

''( *) 2f x =


Example 2

Find the optimum value of the function and also state if the function attains a maximum or a minimumSolution:

4( ) ( 2)f x x= −

3'( ) 4( 2) 0f x x= − = or x = x* = 2 for maxima or minima.

2''( *) 12( * 2) 0f x x= − = at x* = 2

'''( *) 24( * 2) 0f x x= − = at x* = 2

( ) 24* =′′′′ xf at x* = 2

Hence fn(x) is positive and n is even hence the point x = x* = 2 is a point of minimum and the function attains a minimum value of 0 at this point.


Example 3

Analyze the function and classify the stationary points as maxima, minima and points of inflection. Solution:

Consider the point x = x* = 0at x * = 0 at x * = 0

5 4 3( ) 12 45 40 5f x x x x= − + +

4 3 2

4 3 2

'( ) 60 180 120 0 3 2 0or 0,1,2

f x x x xx x xx

= − + =

=> − + ==

'' * * 3 * 2 *( ) 240( ) 540( ) 240 0f x x x x= − + =''' * * 2 *( ) 720( ) 1080 240 240f x x x= − + =


Example 3 …contd.

Since the third derivative is non-zero, x = x* = 0 is neither a point of maximum or minimum but it is a point of inflection

Consider x = x* = 1at x* = 1

Since the second derivative is negative the point x = x* = 1 is a point of local maxima with a maximum value of f(x) = 12 – 45 + 40 + 5 = 12

Consider x = x* = 2at x* = 2

Since the second derivative is positive, the point x = x* = 2 is a point of local minima with a minimum value of f(x) = -11

'' * * 3 * 2 *( ) 240( ) 540( ) 240 60f x x x x= − + = −

'' * * 3 * 2 *( ) 240( ) 540( ) 240 240f x x x x= − + =


Example 4

The horse power generated by a Pelton wheel is proportional to u(v-u) where u is the velocity of the wheel, which is variable and v is the velocity of the jet which is fixed. Show that the efficiency of the Pelton wheel will be maximum at u = v/2.Solution:

where K is a proportionality constant (assumed positive).

which is negative. Hence, f is maximum at

K. ( )

0 K 2K 0

or 2

f u v uf v uu

vu

= −∂

= => − =∂

=

2

2 2Kvu

fu

=

∂= −

∂2

2vu =



The concept discussed for one variable functions may be easily extended to functions of multiple variables. Functions of two variables are best illustrated by contour maps, analogous to geographical maps.

A contour is a line representing a constant value of f(x) as shown in the following figure. From this we can identify maxima, minima and points of inflection.


A contour plot


Necessary conditions

As can be seen in the above contour map, perturbations from points of local minima in any direction result in an increase in the response function f(x), i.e.

the slope of the function is zero at this point of local minima.

Similarly, at maxima and points of inflection as the slope is zero, the first derivative of the function with respect to the variables are zero.


Necessary conditions …contd.

Which gives us at the stationary points. i.e. the

gradient vector of f(X), at X = X* = [x1 , x2] defined as follows, must equal zero:

This is the necessary condition.

1 2

0; 0f fx x∂ ∂

= =∂ ∂

1

2

( *)0

( *)x

fx

ffx

∂⎡ ⎤Χ⎢ ⎥∂⎢ ⎥Δ = =∂⎢ ⎥

Χ⎢ ⎥∂⎣ ⎦

x fΔ



Consider the following second order derivatives:

The Hessian matrix defined by H is made using the above second order derivatives.

2 2 2

2 21 2 1 2

; ;f f fx x x x

∂ ∂ ∂∂ ∂ ∂ ∂

1 2

2 2

21 1 2

2 2

21 2 2 [ , ]x x

f fx x x

f fx x x

⎛ ⎞∂ ∂⎜ ⎟∂ ∂ ∂⎜ ⎟=⎜ ⎟∂ ∂⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠

H


Sufficient conditions …contd.

The value of determinant of the H is calculated and

if H is positive definite then the point X = [x1, x2] is a point of local minima.

if H is negative definite then the point X = [x1, x2] is a point of local maxima.

if H is neither then the point X = [x1, x2] is neither a point of maxima nor minima.


Example 5

Locate the stationary points of f(X) and classify them as relative maxima, relative minima or neither based on the rules discussed in the lecture.

Solution

3 21 1 2 1 2 2( ) 2 / 3 2 5 2 4 5f x x x x x x= − − + + +X


Example 5 …contd.

From ,

So the two stationary points are X1 = [-1,-3/2] and X2 = [3/2,-1/4]

1

(X) 0fx∂

=∂

22 28 14 3 0x x+ + =

2 2(2 3)(4 1) 0x x+ + =

2 23 / 2 or 1/ 4x x= − = −


Example 5 …contd.

The Hessian of f(X) is

At X1 = [-1,-3/2] ,

2 2 2 2

12 21 2 1 2 2 1

4 ; 4; 2f f f fxx x x x x x∂ ∂ ∂ ∂

= = = = −∂ ∂ ∂ ∂ ∂ ∂

14 22 4x −⎡ ⎤

= ⎢ ⎥−⎣ ⎦H

14 22 4

xλλ

λ−

=−

I - H

4 2( 4)( 4) 4 0

2 4λ

λ λ λλ

+= = + − − =

−I - H

2 16 4 0λ − − =

1 212 12 λ λ= + = −

Since one eigen value is positive and one negative, X1 is neither a relative maximum nor a relative minimum


Example 5 …contd.

At X2 = [3/2,-1/4]

Since both the eigen values are positive, X-2 is a local minimum. Minimum value of f(x) is -0.375

6 2( 6)( 4) 4 0

2 4λ

λ λ λλ

−= = − − − =

−I - H

1 25 5 5 5λ λ= + = −


Example 6

Maximize f(X) =

;

2 21 1 2 220 2 6 3 / 2x x x x+ − + −

1 1

2

2

( *)2 2 06 3 0( *)

x

fx x

fxf

x

∂⎡ ⎤Χ⎢ ⎥∂ −⎡ ⎤ ⎡ ⎤⎢ ⎥Δ = = =⎢ ⎥ ⎢ ⎥−∂⎢ ⎥ ⎣ ⎦⎣ ⎦Χ⎢ ⎥∂⎣ ⎦

X* = [1,2]

2 2 2

2 21 2 1 2

2; 3; 0f f fx x x x∂ ∂ ∂

= − = − =∂ ∂ ∂ ∂

2 00 3−⎡ ⎤

= ⎢ ⎥−⎣ ⎦H


Example 6 …contd.

Since both the eigen values are negative, f(X) is concave and

the required ratio x1:x2 = 1:2 with a global maximum strength

of f(X) = 27 MPa

2 0( 2)( 3) 0

0 3λ

λ λ λλ

+= = + + =

+I - H

1 22 3andλ λ= − = −


Thank you

Optimization Methods: Optimization using Calculus-Convexity and Concavity 1


Convexity and Concavity of Functions of One and Two Variables

Introduction

In the previous class we studied about stationary points and the definition of relative and

global optimum. The necessary and sufficient conditions required for a relative optimum in

functions of one variable and its extension to functions of two variables was also studied. In

this lecture, determination of the convexity and concavity of functions is discussed.

The analyst must determine whether the objective functions and constraint equations are

convex or concave. In real-world problems, if the objective function or the constraints are not

convex or concave, the problem is usually mathematically intractable.

Functions of one variable

Convex function

A real-valued function f defined on an interval (or on any convex subset C of some vector

space) is called convex, if for any two points a and b in its domain C and any t in [0,1], we

have

)()1()())1(( bftatfbttaf −+≤−+

Fig. 1

In other words, a function is convex if and only if its epigraph (the set of points lying on or

above the graph) is a convex set. A function is also said to be strictly convex if


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( (1 ) ) ( ) (1 ) ( )f ta t b tf a t f b+ − < + −

for any t in (0,1) and a line connecting any two points on the function lies completely above

the function. These relationships are illustrated in Fig. 1.

Testing for convexity of a single variable function

A function is convex if its slope is non decreasing or 2 2f / x∂ ∂ ≥ 0. It is strictly convex if its

slope is continually increasing or > 0 throughout the function.

Properties of convex functions

A convex function f, defined on some convex open interval C, is continuous on C and

differentiable at all or at most, countable many points. If C is closed, then f may fail to be

continuous at the end points of C.

A continuous function on an interval C is convex if and only if

( ) ( )2 2

a b f a f bf + +⎛ ⎞ ≤⎜ ⎟⎝ ⎠

for all a and b in C.

A differentiable function of one variable is convex on an interval if and only if its derivative

is monotonically non-decreasing on that interval.

A continuously differentiable function of one variable is convex on an interval if and only if

the function lies above all of its tangents: ( ) ( ) '( )( )f b f a f a b a≥ + − for all a and b in the

interval.

A twice differentiable function of one variable is convex on an interval if and only if its

second derivative is non-negative in that interval; this gives a practical test for convexity. If

its second derivative is positive then it is strictly convex, but the converse does not hold, as

shown by f(x) = x4.

More generally, a continuous, twice differentiable function of several variables is convex on

a convex set if and only if its Hessian matrix is positive semi definite on the interior of the

convex set.


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If two functions f and g are convex, then so is any weighted combination a f + b g with non-

negative coefficients a and b. Likewise, if f and g are convex, then the function maxf,g is

convex.

A strictly convex function will have only one minimum which is also the global minimum.

Examples

• The second derivative of x2 is 2; it follows that x2 is a convex function of x.

• The absolute value function |x| is convex, even though it does not have a derivative at

x = 0.

• The function f with domain [0,1] defined by f(0)=f(1)=1, f(x)=0 for 0<x<1 is convex;

it is continuous on the open interval (0,1), but not continuous at 0 and 1.

• Every linear transformation is convex but not strictly convex, since if f is linear, then

f(a + b) = f(a) + f(b). This implies that the identity map (i.e., f(x) = x) is convex but

not strictly convex. The fact holds good if we replace "convex" by "concave".

• An affine function (f (x) = ax + b) is simultaneously convex and concave.

Concave function

A differentiable function f is concave on an interval if its derivative function f ′ is decreasing

on that interval: a concave function has a decreasing slope.

A function that is convex is often synonymously called concave upwards, and a function

that is concave is often synonymously called concave downward.

For a twice-differentiable function f, if the second derivative, f ''(x), is positive (or, if the

acceleration is positive), then the graph is convex (or concave upward); if the second

derivative is negative, then the graph is concave (or concave downward). Points, at which

concavity changes, are called inflection points.

If a convex (i.e., concave upward) function has a "bottom", any point at the bottom is a

minimal extremum. If a concave (i.e., concave downward) function has an "apex", any point

at the apex is a maximal extremum.

A function f(x) is said to be concave on an interval if, for all a and b in that interval,


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[0,1], ( (1 ) ) ( ) (1 ) ( )t f ta t b tf a t f∀ ∈ + − ≥ + − b

Additionally, f(x) is strictly concave if

[0,1], ( (1 ) ) ( ) (1 ) ( )t f ta t b tf a t f b∀ ∈ + − > + −

These relationships are illustrated in Fig. 2

Fig. 2

Testing for concavity of a single variable function

A function is concave if its slope is non increasing or 2 2/f x∂ ∂ ≤ 0. It is strictly concave if its

slope is continually decreasing or 2 2/f x∂ ∂ <0 throughout the function.

Properties of a concave functions

A continuous function on C is concave if and only if

( ) ( )2 2

a b f a f bf + +⎛ ⎞ ≥⎜ ⎟⎝ ⎠

for any x and y in C.

Equivalently, f(x) is concave on [a, b] if and only if the function −f(x) is convex on every

subinterval of [a, b].


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If f(x) is twice-differentiable, then f(x) is concave if and only if f ′′(x) is non-positive. If its

second derivative is negative then it is strictly concave, but the opposite is not true, as shown

by f(x) = -x4.

A function is called quasiconcave if and only if there is an x0 such that for all x < x0, f(x) is

non-decreasing while for all x > x0 it is non-increasing. x0 can also be , making the

function non-decreasing (non-increasing) for all

±∞

x. The opposite of quasiconcave is

quasiconvex.

Example 1

Consider the example in lecture notes 1 for a function of two variables. Locate the stationary

points of and find out if the function is convex, concave or

neither at the points of optima based on the testing rules discussed above.

5 4 3( ) 12 45 40 5f x x x x= − + +

Solution 4 3 2

4 3 2

'( ) 60 180 120 0 3 2 0or 0,1,2

f x x x xx x xx

= − + =

=> − + ==

Consider the point x =x* = 0 3 2''( *) 240( *) 540( *) 240 * 0f x x x x= − + =

−

at x * = 0 2'''( *) 720( *) 1080 * 240 240f x x x= − + = at x * = 0

Since the third derivative is non-zero x = x* = 0 is neither a point of maximum or minimum

but it is a point of inflection. Hence the function is neither convex nor concave at this point.

Consider x = x* = 1 3 2''( *) 240( *) 540( *) 240 * 60f x x x x= − + = at x* = 1

Since the second derivative is negative, the point x = x* = 1 is a point of local maxima with a

maximum value of f(x) = 12 – 45 + 40 + 5 = 12. At this point the function is concave since 2 2/f x∂ ∂ < 0.

Consider x = x* = 2 3 2''( *) 240( *) 540( *) 240 * 240f x x x x= − + = at x* = 2

Since the second derivative is positive, the point x = x* = 2 is a point of local minima with a

minimum value of f(x) = -11. At this point the function is convex since 2 2/f x∂ ∂ > 0.


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A function of two variables, f(X) where X is a vector = [x1,x2], is strictly convex if

1 2 1( (1 ) ) ( ) (1 ) ( )f t t tf t fΧ + − Χ < Χ + − Χ2

2

where X1 and X2 are points located by the coordinates given in their respective vectors.

Similarly a two variable function is strictly concave if

1 2 1( (1 ) ) ( ) (1 ) ( )f t t tf t fΧ + − Χ > Χ + − Χ

Contour plot of a convex function is illustrated in Fig. 3

340

Fig. 3

Contour plot of a convex function is shown in Fig. 4

Fig. 4

450

70 x2

120

x1

x2

x1

110

210

305

40


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To determine convexity or concavity of a function of multiple variables, the eigenvalues of

its Hessian matrix are examined and the following rules apply.

(a) If all eigenvalues of the Hessian are positive the function is strictly convex.

(b) If all eigenvalues of the Hessian are negative the function is strictly concave.

(c) If some eigenvalues are positive and some are negative, or if some are zero, the

function is neither strictly concave nor strictly convex.

Example 2

Consider the example in lecture notes 1 for a function of two variables. Locate the stationary

points of f(X) and find out if the function is convex, concave or neither at the points of

optima based on the rules discussed in this lecture.

f(X) = 3 21 1 2 1 22 / 3 2 5 2 4 2x x x x x x− − + + + 5

Solution

21 1 2

1 2

2

( *)02 2 502 4 4( *)

x

fx x x

ff x xx

∂⎡ ⎤Χ⎢ ⎥∂ ⎡ ⎤− − ⎡ ⎤⎢ ⎥Δ = = =⎢ ⎥ ⎢ ⎥∂⎢ ⎥ − + + ⎣ ⎦⎣ ⎦Χ⎢ ⎥∂⎣ ⎦

Solving the above the two stationary points are

X1 = [-1,-3/2]

and

X2 = [3/2,-1/4]

The Hessian of f(X) is 2 2 2 2

12 21 2 1 2 2 1

4 ; 4; 2f f f fxx x x x x x∂ ∂ ∂ ∂

= = = =∂ ∂ ∂ ∂ ∂ ∂

−

14 22 4x −⎡ ⎤

= ⎢ ⎥−⎣ ⎦H

14 22 4

xλλ

λ−

=−

I - H

At X1

4 2( 4)( 4) 4

2 4λ

λ λ λλ

+= = + −

−I - H 0− =

2 16 4 0λ − − =


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2λ = 12

1 212 12 λ λ= + = −

Since one eigen value is positive and one negative, X1 is neither a relative maximum nor a

relative minimum. Hence at X1 the function is neither convex nor concave.

At X2 = [3/2,-1/4]

6 2( 6)( 4) 4

2 4λ

λ λ λλ

−= = − − −

−I - H 0=

2 10 20 0λ λ− + =

1 25 5 5 5λ λ= + = −

Since both the eigen values are positive, X2 is a local minimum, and the function is convex at

this point as both the eigen values are positive.


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Convexity and Concavity of Functions of One and

Two Variables


Objective

To determine the convexity and concavity of functions


Convex Function (Function of one variable)

A real-valued function f defined on an interval (or on any convex subset C of some vector space) is called convex, if for any two points x and y in its domain C and any t in [0,1], we have

In other words, a function is convex if and only if its epigraph (the set of points lying on or above the graph) is a convex set. A function is also said to be strictly convex if

for any t in (0,1) and a line connecting any two points on the function lies completely above the function.

( (1 ) ) ( ) (1 ) ( ))f ta t b tf a t f b+ − ≤ + −

( (1 ) ) ( ) (1 ) ( )f ta t b tf a t f b+ − < + −


A convex function


Testing for convexity of a single variable function

A function is convex if its slope is non decreasing or

0. It is strictly convex if its slope is continually

increasing or 0 throughout the function.

2 2/f x∂ ∂ ≥2 2/f x∂ ∂ >


Properties of convex functions

A convex function f, defined on some convex open interval C, is continuous on C and differentiable at all or at many points. If C is closed, then f may fail to be continuous at the end points of C.

A continuous function on an interval C is convex if and only if

for all a and b in C.• A differentiable function of one variable is convex on an interval if

and only if its derivative is monotonically non-decreasing on that interval.

( ) ( )2 2

a b f a f bf + +⎛ ⎞ ≤⎜ ⎟⎝ ⎠


Properties of convex functions (contd.)

A continuously differentiable function of one variable is convex on an interval if and only if the function lies above all of its tangents: for all aand b in the interval.

A twice differentiable function of one variable is convex on an interval if and only if its second derivative is non-negative in that interval; this gives a practical test for convexity.

More generally, a continuous, twice differentiable function of several variables is convex on a convex set if and only if its Hessian matrix is positive semi definite on the interior of the convex set.

If two functions f and g are convex, then so is any weighted combination af + b g with non-negative coefficients a and b. Likewise, if f and g are convex, then the function maxf, g is convex.


Concave function (function of one variable)

A differentiable function f is concave on an interval if its derivative function f ′ is decreasing on that interval: a concave function has a decreasing slope.A function f(x) is said to be concave on an interval if, for all a and bin that interval,

Additionally, f(x) is strictly concave if

[0,1], ( (1 ) ) ( ) (1 ) ( )t f ta t b tf a t f b∀ ∈ + − ≥ + −

[0,1], ( (1 ) ) ( ) (1 ) ( )t f ta t b tf a t f b∀ ∈ + − > + −


A concave function


Testing for concavity of a single variable function

A function is convex if its slope is non increasing or 0. It is strictly concave if its slope is continually

decreasing or 0 throughout the function.2 2/f x∂ ∂ ≤

2 2/f x∂ ∂ <


Properties of concave functions

A continuous function on C is concave if and only if

for any a and b in C.

• Equivalently, f(x) is concave on [a, b] if and only if the function −f(x) is convex on every subinterval of [a, b].

• If f(x) is twice-differentiable, then f(x) is concave if and only if f ′′(x) is non-positive. If its second derivative is negative then it is strictly concave, but the opposite is not true, as shown by f(x) = -x4.

( ) ( )2 2

a b f a f bf + +⎛ ⎞ ≥⎜ ⎟⎝ ⎠


Example

Consider the example in lecture notes 1 for a function of two variables. Locate the stationary points of and find out if the function is convex, concave or neither at the points of optima based on the testing rules discussed above.

5 4 3( ) 12 45 40 5f x x x x= − + +


Example …contd.



A function of two variables, f(X) where X is a vector = [x1,x2], is strictly convex if

where X1 and X2 are points located by the coordinates given in their respective vectors. Similarly a two variable function is strictly concave if

1 2 1 2( (1 ) ) ( ) (1 ) ( )f t t tf t fΧ + − Χ < Χ + − Χ

1 2 1 2( (1 ) ) ( ) (1 ) ( )f t t tf t fΧ + − Χ > Χ + − Χ


Contour plot of a convex function


Contour plot of a concave function



To determine convexity or concavity of a function of multiple variables, the eigen values of its Hessian matrix is examined and the following rules apply.

– If all eigen values of the Hessian are positive the function is strictly convex.

– If all eigen values of the Hessian are negative the function is strictly concave.

– If some eigen values are positive and some are negative, or if some are zero, the function is neither strictly concave nor strictly convex.


Example

Consider the example in lecture notes 1 for a function of two variables. Locate the stationary points of f(X) and find out if the function is convex, concave or neither at the points of optima based on the rules discussed in this lecture.

3 21 1 2 1 2 2( ) 2 / 3 2 5 2 4 5f x x x x x x= − − + + +X


Example (contd..)


Thank you

Optimization Methods: Optimization using Calculus - Unconstrained Optimization 1


Optimization of Functions of Multiple Variables: Unconstrained Optimization

Introduction

In the previous lectures we learnt how to determine the convexity and concavity of functions

of single and two variables. For functions of single and two variables we also learnt

determining stationary points and examining higher derivatives to check for convexity and

concavity, and tests were recommended to evaluate stationary points as local minima, local

maxima or points of inflection.

In this lecture functions of multiple variables, which are more difficult to be analyzed owing

to the difficulty in graphical representation and tedious calculations involved in mathematical

analysis, will be studied for unconstrained optimization. This is done with the aid of the

gradient vector and the Hessian matrix. Examples are discussed to show the implementation

of the technique.

Unconstrained optimization

If a convex function is to be minimized, the stationary point is the global minimum and

analysis is relatively straightforward as discussed earlier. A similar situation exists for

maximizing a concave variable function. The necessary and sufficient conditions for the

optimization of unconstrained function of several variables are given below.

Necessary condition

In case of multivariable functions a necessary condition for a stationary point of the function

f(X) is that each partial derivative is equal to zero. In other words, each element of the

gradient vector defined below must be equal to zero.

i.e. the gradient vector of f(X), x fΔ at X=X*, defined as follows, must be equal to zero:

1

2

( *)

( *)

( *)

x

n

fxfx

f

fdx

∂⎡ ⎤Χ⎢ ⎥∂⎢ ⎥∂⎢ ⎥

Χ⎢ ⎥∂⎢ ⎥Δ =⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥∂

Χ⎢ ⎥∂⎣ ⎦

M

M

= 0


M2L3


The proof given for the theorem on necessary condition for single variable optimization can

be easily extended to prove the present condition.


For a stationary point X* to be an extreme point, the matrix of second partial derivatives

(Hessian matrix) of f(X) evaluated at X* must be:

(i) positive definite when X* is a point of relative minimum, and

(ii) negative definite when X* is a relative maximum point.

Proof (Formulation of the Hessian matrix)

The Taylor’s theorem with reminder after two terms gives us

2

1 1 1 *

1( * ) ( *) ( *)2!

n n n

i i ji i ji i h

df ff h f h h hdx x x j θ= = = Χ=Χ +

∂Χ + = Χ + Χ +

∂ ∂∑ ∑∑

0<θ <1

Since X* is a stationary point, the necessary condition gives

0i

dfdx

= , i = 1,2,…,n

Thus

2

1 1 *

1( * ) ( *)2!

n n

i ji j i j h

ff h f h hx x

θ= = Χ=Χ +

∂Χ + − Χ =

∂ ∂∑∑

For a minimization problem the left hand side of the above expression must be positive.

Since the second partial derivative is continuous in the neighborhood of X* the sign of 2 *i jf x x hθ∂ ∂ ∂ Χ = Χ + is the same as the sign of 2 *i jf x x∂ ∂ ∂ Χ = Χ . And hence

will be a relative minimum, if ( * ) ( *)f h fΧ + − Χ

2

1 1 *

1( * ) ( *)2!

n n

i ji j i j

ff h f h hx x= = Χ=Χ

∂Χ + − Χ =

∂ ∂∑∑

is positive. This can also be written in the matrix form as

X=X*

TQ = h Hh

where

2

X=X**i j

fx x

Χ=Χ

⎡ ⎤∂= ⎢ ⎥

∂ ∂⎢ ⎥⎣ ⎦H

is the matrix of second partial derivatives and is called the Hessian matrix of f(X).


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Q will be positive for all h if and only if H is positive definite at X=X*. i.e. the sufficient

condition for X* to be a relative minimum is that the Hessian matrix evaluated at the same

point is positive definite, which completes the proof for the minimization case. In a similar

manner, it can be proved that the Hessian matrix will be negative definite if X* is a point of

relative maximum.

A matrix A will be positive definite if all its eigenvalues are positive. i.e. all values ofλ that

satisfy the equation

0λ− =A I

should be positive. Similarly, the matrix A will be negative definite if its eigenvalues are

negative. When some eigenvalues are positive and some are negative then matrix A is neither

positive definite or negative definite.

When all eigenvalues are negative for all possible values of X, then X* is a global maximum,

and when all eigenvalues are positive for all possible values of X, then X* is a global

minimum.

If some of the eigenvalues of the Hessian at X* are positive and some negative, or if some are

zero, the stationary point, X*, is neither a local maximum nor a local minimum.

Example

Analyze the function and classify the

stationary points as maxima, minima and points of inflection.

2 2 21 2 3 1 2 1 3 1 3( ) 2 2 4 5 2f x x x x x x x x x x= − − − + + + − +

Solution

11 2 3

2 12

3 1

3

( *)2 2 2 4 0

( *) 2 2 02 2 5 0

( *)

x

fx x x xff x xx

x xfx

⎡ ⎤∂Χ⎢ ⎥∂⎢ ⎥ − + + +⎡ ⎤ ⎡ ⎤

⎢ ⎥∂ ⎢ ⎥ ⎢ ⎥Δ = Χ = − + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥∂⎢ ⎥ ⎢ ⎥ ⎢ ⎥− + −⎣ ⎦ ⎣ ⎦⎢ ⎥∂Χ⎢ ⎥

∂⎣ ⎦

Solving these simultaneous equations we get X*=[1/2, 1/2, -2] 2 2 2

2 2 21 2 32 2

1 2 2 1

2; 2; 2

2

f f fx x x

f fx x x x

∂ ∂ ∂= − = − =

∂ ∂ ∂

∂ ∂= =

∂ ∂ ∂ ∂

−


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2 2

2 3 3 22 2

3 1 1 3

0

2

f fx x x x

f fx x x x

∂ ∂= =

∂ ∂ ∂ ∂

∂ ∂= =

∂ ∂ ∂ ∂

Hessian of f(X) is 2

i j

fx x

⎡ ⎤∂= ⎢ ⎥

∂ ∂⎢ ⎥⎣ ⎦H

2 2 22 2 02 0 2

−⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦

H

2 2 22 2 02 0 2

λλ λ

λ

+ − −0= − + =

− +I - H

or ( 2)( 2)( 2) 2( 2)(2) 2(2)( 2) 0λ λ λ λ λ+ + + − + + + = 2( 2)[ 4 4 4 4]λ λ λ+ + + − + = 0

0

3( 2)λ + =

or 1 2 3 2λ λ λ= = = −

Since all eigenvalues are negative the function attains a maximum at the point

X*=[1/2, 1/2, -2]


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Optimization of Functions of Multiple

Variables: Unconstrained Optimization


Objectives

To study functions of multiple variables, which are more difficult to analyze owing to the difficulty in graphical representation and tedious calculations involved in mathematical analysis for unconstrained optimization.

To study the above with the aid of the gradient vector and the Hessian matrix.

To discuss the implementation of the technique through examples


Unconstrained optimization

If a convex function is to be minimized, the stationary point is the

global minimum and analysis is relatively straightforward as

discussed earlier.

A similar situation exists for maximizing a concave variable function.

The necessary and sufficient conditions for the optimization of

unconstrained function of several variables are discussed.


Necessary condition

In case of multivariable functions a necessary condition for a stationary point of the function f(X) is that each partial derivative is equal to zero. In other words, each element of the gradient vector defined below must be equal to zero. i.e. the gradient vector of f(X), at X=X*, defined as follows, must be equal to zero:

x fΔ

*

1

*

2

*

( )

( )0

( )

x

n

fxfx

f

fdx

∂⎡ ⎤Χ⎢ ⎥∂⎢ ⎥∂⎢ ⎥Χ⎢ ⎥∂⎢ ⎥Δ = =

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥∂

Χ⎢ ⎥∂⎣ ⎦

M

M



For a stationary point X* to be an extreme point, the matrix of second partial derivatives (Hessian matrix) of f(X) evaluated at X* must be:

positive definite when X* is a point of relative minimum, andnegative definite when X* is a relative maximum point.

When all eigen values are negative for all possible values of X, then X* is a global maximum, and when all eigen values are positive for all possible values of X, then X* is a global minimum.If some of the eigen values of the Hessian at X* are positive and some negative, or if some are zero, the stationary point, X*, is neither a local maximum nor a local minimum.


Example

Analyze the function and classify the stationary points as maxima, minima and points of inflection Solution

2 2 21 2 3 1 2 1 3 1 3( ) 2 2 4 5 2f x x x x x x x x x x= − − − + + + − +

*

11 2 3

*2 1

23 1

*

3

( )2 2 2 4 0

( ) 2 2 02 2 5 0

( )

x

fx x x xff x xx

x xfx

⎡ ⎤∂Χ⎢ ⎥∂⎢ ⎥ − + + +⎡ ⎤ ⎡ ⎤

⎢ ⎥∂ ⎢ ⎥ ⎢ ⎥Δ = Χ = − + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥∂⎢ ⎥ ⎢ ⎥ ⎢ ⎥− + −⎣ ⎦ ⎣ ⎦⎢ ⎥∂Χ⎢ ⎥∂⎣ ⎦


Example …contd.


Thank you

Optimization Methods: Optimization using Calculus - Equality constraints 1


Optimization of Functions of Multiple Variables subject to Equality Constraints

Introduction

In the previous lecture we learnt the optimization of functions of multiple variables studied

for unconstrained optimization. This is done with the aid of the gradient vector and the

Hessian matrix. In this lecture we will learn the optimization of functions of multiple

variables subjected to equality constraints using the method of constrained variation and the

method of Lagrange multipliers.

Constrained optimization

A function of multiple variables, f(x), is to be optimized subject to one or more equality

constraints of many variables. These equality constraints, gj(x), may or may not be linear. The

problem statement is as follows:

Maximize (or minimize) f(X), subject to gj(X) = 0, j = 1, 2, … , m

where

1

2

n

xx

X

x

⎧ ⎫⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

M

with the condition that ; or else if m > n then the problem becomes an over defined one

and there will be no solution. Of the many available methods, the method of constrained

variation and the method using Lagrange multipliers are discussed.

m n≤

Solution by method of Constrained Variation

For the optimization problem defined above, let us consider a specific case with n = 2 and

m=1 before we proceed to find the necessary and sufficient conditions for a general problem

using Lagrange multipliers. The problem statement is as follows:

Minimize f(x1,x2), subject to g(x1,x2) = 0

For f(x1,x2) to have a minimum at a point X* = [x1*,x2*], a necessary condition is that the

total derivative of f(x1,x2) must be zero at [x1*,x2*].

1 21 2

0f fdf dx dxx x∂ ∂

= + =∂ ∂

(1)


M2L4


Since g(x1*,x2*) = 0 at the minimum point, variations dx1 and dx2 about the point [x1*, x2*]

must be admissible variations, i.e. the point lies on the constraint:

g(x1* + dx1 , x2* + dx2) = 0 (2)

assuming dx1 and dx2 are small the Taylor series expansion of this gives us

1 1 2 2 1 2 1 2 1 1 2 21 2

( * , * ) ( *, *) (x *,x *) (x *,x *) 0g gg x dx x dx g x x dx dxx x∂ ∂

+ + = + +∂ ∂

= (3)

or

1 21 2

0g gdg dx dxx x∂ ∂

= +∂ ∂

=

0

at [x1*,x2*] (4)

which is the condition that must be satisfied for all admissible variations.

Assuming (4)2/g x∂ ∂ ≠ can be rewritten as

12 1

2

/ ( *, *)/

g xdx x x dxg x 2 1∂ ∂

= −∂ ∂

(5)

which indicates that once variation along x1 (d x1) is chosen arbitrarily, the variation along x2

(d x2) is decided automatically to satisfy the condition for the admissible variation.

Substituting equation (5) in (1)we have:

1 2

11

1 2 2 (x *, x *)

/ 0/

g xf fdf dxx g x x

⎛ ⎞∂ ∂∂ ∂= − =⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

(6)

The equation on the left hand side is called the constrained variation of f. Equation (5) has to

be satisfied for all dx1, hence we have

1 21 2 2 1 (x *, x *)

0f g f gx x x x

⎛ ⎞∂ ∂ ∂ ∂− =⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

(7)

This gives us the necessary condition to have [x1*, x2*] as an extreme point (maximum or

minimum)

Solution by method of Lagrange multipliers

Continuing with the same specific case of the optimization problem with n = 2 and m = 1 we

define a quantityλ , called the Lagrange multiplier as

1 2

2

2 (x *, x *)

//

f xg x

λ ∂ ∂= −

∂ ∂ (8)

Using this in (6)


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1 21 1 (x *, x *)

0f gx x

λ⎛ ⎞∂ ∂

+ =⎜ ⎟∂ ∂⎝ ⎠ (9)

And (8) written as

1 22 2 (x *, x *)

0f gx x

λ⎛ ⎞∂ ∂

+ =⎜ ⎟∂ ∂⎝ ⎠ (10)

Also, the constraint equation has to be satisfied at the extreme point

1 21 2 ( *, *)

( , ) 0x x

g x x = (11)

Hence equations (9) to (11) represent the necessary conditions for the point [x1*, x2*] to be

an extreme point.

Note that λ could be expressed in terms of 1/g x∂ ∂ as well and 1/g x∂ ∂ has to be non-zero.

Thus, these necessary conditions require that at least one of the partial derivatives of g(x1, x2)

be non-zero at an extreme point.

The conditions given by equations (9) to (11) can also be generated by constructing a

function L, known as the Lagrangian function, as

1 2 1 2 1 2( , , ) ( , ) ( , )L x x f x x g x xλ λ= + (12)

Alternatively, treating L as a function of x1,x2 and λ , the necessary conditions for its

extremum are given by

1 2 1 2 1 21 1 1

1 2 1 2 1 22 2 2

1 2 1 2

( , , ) ( , ) ( , ) 0

( , , ) ( , ) ( , ) 0

( , , ) ( , ) 0

L f gx x x x x xx x xL f gx x x x x xx x xL x x g x x

λ λ

λ λ

λλ

∂ ∂ ∂= + =

∂ ∂ ∂∂ ∂ ∂

= +∂ ∂ ∂∂

= =∂

= (13)

The necessary and sufficient conditions for a general problem are discussed next.

Necessary conditions for a general problem

For a general problem with n variables and m equality constraints the problem is defined as

shown earlier


where

1

2

n

xx

x

⎧ ⎫⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

XM


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In this case the Lagrange function, L, will have one Lagrange multiplier jλ for each constraint

as (X)jg

1 2 , 1 2 1 1 2 2( , ,..., , ,..., ) ( ) ( ) ( ) ... ( )n m m mL x x x f g g gλ λ λ λ λ λ= + + + +X X X X

m

(14)

L is now a function of n + m unknowns, 1 2 , 1 2, ,..., , ,...,nx x x λ λ λ , and the necessary conditions

for the problem defined above are given by

1( ) ( ) 0, 1, 2,..., 1, 2,...,

( ) 0, 1, 2,...,

mj

jji i i

jj

gL f i n jx x xL g j m

λ

λ

=

∂∂ ∂= + = = =

∂ ∂ ∂

∂= = =

∂

∑X X

X

m (15)

which represent n + m equations in terms of the n + m unknowns, xi and jλ . The solution to

this set of equations gives us

1

2

**

*n

xx

x

⎧ ⎫⎪ ⎪⎪= ⎨⎪ ⎪⎪ ⎪⎩ ⎭

XM

⎪⎬ and

1

2

**

*

*m

λλ

λ

λ

⎧ ⎫⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

M (16)

The vector X corresponds to the relative constrained minimum of f(X) (subject to the

verification of sufficient conditions).

Sufficient conditions for a general problem

A sufficient condition for f(X) to have a relative minimum at X* is that each root of the

polynomial in ∈ , defined by the following determinant equation be positive.

11 12 1 11 21 1

21 22 2 12 22 2

1 2 1 2

11 12 1

21 22 2

1 2

00 0

0 0

n m

n m

n n nn n n mn

n

n

m m mn

L L L g g gL L L g g g

L L L g g g

g g gg g g

g g g

−∈−∈

−∈=

L L

M O M M O M

L L

L L L

M O M

M O M M M

L L L

(17)


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where 2

( *, *), for 1, 2,..., 1, 2,...,

( *), where 1, 2,..., and 1, 2,...,

iji j

ppq

q

LL i n jx x

gg p m

x

λ∂= =∂ ∂

∂= =∂

X

X

m

q n

=

=

2

(18)

Similarly, a sufficient condition for f(X) to have a relative maximum at X* is that each root of

the polynomial in ∈ , defined by equation (17) be negative. If equation (17), on solving yields

roots, some of which are positive and others negative, then the point X* is neither a

maximum nor a minimum.

Example

Minimize 2 21 1 2 2 1( ) 3 6 5 7 5f x x x x x= − − − + +X x

Subject to 1 2 5x x+ =

Solution

1 1 2( ) 5 0g x x= + − =X

1 2 , 1 2 1 1 2 2( , ,..., , ,..., ) ( ) ( ) ( ) ... ( )n m m mx x x f g g gλ λ λ λ λ λ= + + + +L X X X X

)

with n = 2 and m = 1

L = 2 21 1 2 2 1 2 1 1 23 6 5 7 5 ( 5x x x x x x x xλ− − − + + + + −

1 2 11

1 2 1

1

6 6 7

1 (7 )6

15 (7 )6

x xx

x x

λ

λ

λ

∂ 0= − − + + =∂

=> + = +

=> = +

L

or 1 23λ =

1 2 12

1 2 1

1 2 2 1

6 10 5 0

13 5 (5 )2

13( ) 2 (5 )2

x xx

x x

x x x

λ

λ

λ

∂= − − + + =

∂

=> + = +

=> + + = +

L


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21

2x −=

and, 1112

x =

Hence [ ]1 11* , ; * 22 2−⎡ ⎤= =⎢ ⎥⎣ ⎦

X λ 3

11 12 11

21 22 21

11 12

00

L L gL L gg g

−∈⎛ ⎞⎜ ⎟−∈ =⎜ ⎟⎜ ⎟⎝ ⎠

2

11 21 ( )

6Lx

∂= =∂

X*,λ*

L−

2

12 211 2 ( )

6L Lx x∂

= = = −∂ ∂

X*,λ*

L

2

22 22 ( )

10Lx

∂= =∂

X*,λ*

L−

111

1 ( )

112 21

2 ( )

1

1

ggx

gg gx

∂= =∂

∂= = =

∂

X*,λ*

X*,λ*

The determinant becomes

6 6 16 10 1

1 1 0

− −∈ −⎛ ⎞⎜ ⎟ 0− − −∈ =⎜ ⎟⎜ ⎟⎝ ⎠

or ( 6 )[ 1] ( 6)[ 1] 1[ 6 10 ] 0

2− −∈ − − − − + − + +∈ ==>∈= −

Since ∈ is negative, X*, correspond to a maximum. *λ


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Optimization of Functions of Multiple Variables subject to

Equality Constraints


Objectives

Optimization of functions of multiple variables subjected to equality constraints using

the method of constrained variation, and the method of Lagrange multipliers.


Constrained optimization

A function of multiple variables, f(x), is to be optimized subject to one or more equality constraints of many variables. These equality constraints, gj(x), may or may not be linear. The problem statement is as follows:

Maximize (or minimize) f(X), subject to gj(X) = 0, j = 1, 2, … , mwhere

1

2

n

xx

x

⎧ ⎫⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

XM


Constrained optimization (contd.)

With the condition that ; or else if m > n then the problem becomes an over defined one and there will be no solution. Of the many available methods, the method of constrained variation and the method of using Lagrange multipliers are discussed.

m n≤


Solution by method of Constrained Variation

For the optimization problem defined above, let us consider a specific case with n = 2 and m = 1 before we proceed to find the necessary and sufficient conditions for a general problem using Lagrange multipliers. The problem statement is as follows:

Minimize f(x1,x2), subject to g(x1,x2) = 0

For f(x1,x2) to have a minimum at a point X* = [x1*,x2

*], a necessary condition is that the total derivative of f(x1,x2) must be zero at [x1

*,x2*].

(1)1 21 2

0f fdf dx dxx x∂ ∂

= + =∂ ∂


Method of Constrained Variation (contd.)

Since g(x1*,x2

*) = 0 at the minimum point, variations dx1 and dx2 about the point [x1

*, x2*] must be admissible variations, i.e. the point lies on

the constraint:g(x1

* + dx1 , x2* + dx2) = 0 (2)

assuming dx1 and dx2 are small the Taylor series expansion of this gives us

(3)

*1 1 2 2

* * * * * *1 2 1 2 1 1 2 2

1 2

( * , )

( , ) (x ,x ) (x ,x ) 0

g x dx x dxg gg x x dx dxx x

+ +∂ ∂

= + + =∂ ∂



or at [x1*,x2

*] (4)

which is the condition that must be satisfied for all admissiblevariations.Assuming , (4) can be rewritten as

(5)

1 21 2

0g gdg dx dxx x∂ ∂

= + =∂ ∂

2/ 0g x∂ ∂ ≠

* *12 1 2 1

2

/ ( , )/

g xdx x x dxg x∂ ∂

= −∂ ∂



(5) indicates that once variation along x1 (dx1) is chosen arbitrarily, the variation along x2 (dx2) is decided automatically to satisfy the condition for the admissible variation. Substituting equation (5) in (1) we have:

(6)

The equation on the left hand side is called the constrained variation of f. Equation (5) has to be satisfied for all dx1, hence we have

(7)

This gives us the necessary condition to have [x1*, x2

*] as an extreme point (maximum or minimum)

* *1 2

11

1 2 2 (x , x )

/ 0/

f g x fdf dxx g x x

⎛ ⎞∂ ∂ ∂ ∂= − =⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

* *1 2

1 2 2 1 (x , x )

0f g f gx x x x

⎛ ⎞∂ ∂ ∂ ∂− =⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠


Solution by method of Lagrange multipliers

Continuing with the same specific case of the optimization problem with n = 2 and m = 1 we define a quantity λ, called the Lagrange multiplier as

(8)

Using this in (5)(9)

And (8) written as(10)

* *1 2

2

2 (x , x )

//

f xg x

λ ∂ ∂= −

∂ ∂

* *1 2

1 1 (x , x )

0f gx x

λ⎛ ⎞∂ ∂

+ =⎜ ⎟∂ ∂⎝ ⎠

* *1 2

2 2 (x , x )

0f gx x

λ⎛ ⎞∂ ∂

+ =⎜ ⎟∂ ∂⎝ ⎠


Solution by method of Lagrange multipliers…contd.

Also, the constraint equation has to be satisfied at the extreme point

(11)

Hence equations (9) to (11) represent the necessary conditions for the point [x1*, x2*] to be an extreme point.

Note that λ could be expressed in terms of as well andhas to be non-zero.

Thus, these necessary conditions require that at least one of the partial derivatives of g(x1 , x2) be non-zero at an extreme point.

* *1 2

1 2 ( , )( , ) 0

x xg x x =

1/g x∂ ∂ 1/g x∂ ∂


Solution by method of Lagrange multipliers…contd.

The conditions given by equations (9) to (11) can also be generated by constructing a functions L, known as the Lagrangian function, as

(12)

Alternatively, treating L as a function of x1,x2 and λ, the necessary conditions for its extremum are given by

(13)

1 2 1 2 1 2( , , ) ( , ) ( , )L x x f x x g x xλ λ= +

1 2 1 2 1 21 1 1

1 2 1 2 1 22 2 2

( , , ) ( , ) ( , ) 0

( , , ) ( , ) ( , ) 0

( , , ) ( , ) 0

L f gx x x x x xx x xL f gx x x x x xx x xL x x g x x

λ λ

λ λ

λ1 2 1 2λ

∂ ∂ ∂= + =

∂ ∂ ∂∂ ∂ ∂

= + =∂ ∂ ∂∂

= =∂


Necessary conditions for a general problem

For a general problem with n variables and m equality constraints the problem is defined as shown earlier


whereIn this case the Lagrange function, L, will have one Lagrange multiplier λj for each constraint as

(14)1 2 , 1 2 1 1 2 2( , ,..., , ,..., ) ( ) ( ) ( ) ... ( )n m m mL x x x f g g gλ λ λ λ λ λ= + + + +X X X X

1

2

n

xx

x

⎧ ⎫⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

XM


Necessary conditions for a general problem…contd.

L is now a function of n + m unknowns, , and the necessary conditions for the problem defined above are given by

(15)

which represent n + m equations in terms of the n + m unknowns, xi and λj. The solution to this set of equations gives us

(16)and

1 2 , 1 2, ,..., , ,...,n mx x x λ λ λ

1( ) ( ) 0, 1, 2,..., 1, 2,...,

( ) 0, 1, 2,...,

mj

jji i i

jj

gL f i n j mx x xL g j m

λ

λ

=

∂∂ ∂= + = = =

∂ ∂ ∂

∂= = =

∂

∑X X

X

*1

*2

*n

xx

x

⎧ ⎫⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

XM

*1

** 2

*m

λλ

λ

λ

⎧ ⎫⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

M


Sufficient conditions for a general problem

A sufficient condition for f(X) to have a relative minimum at X* is that each root of the polynomial in Є, defined by the following determinant equation be positive.

(17)

11 12 1 11 21 1

21 22 2 12 22 2

1 2 1 2

11 12 1

21 22 2

1 2

00 0

0 0

n m

n m

n n nn n n mn

n

n

m m mn

L L L g g gL L L g g g

L L L g g g

g g gg g g

g g g

−∈−∈

−∈=

L L

M O M M O M

L L

L L L

M O M

M O M M M

L L L


Sufficient conditions for a general problem…contd.

where

(18)

Similarly, a sufficient condition for f(X) to have a relative maximum at X*

is that each root of the polynomial in Є, defined by equation (17) be negative. If equation (17), on solving yields roots, some of which are positive and others negative, then the point X* is neither a maximum nor a minimum.

2* *

*

( , ), for 1,2,..., 1,2,...,

( ), where 1,2,..., and 1,2,...,

iji j

ppq

q

LL i n and j mx x

gg p m q

x

λ∂= = =∂ ∂

∂= = =∂

X

X n


Example

Minimize , Subject to

or

2 21 1 2 2 1 2( ) 3 6 5 7 5f x x x x x x= − − − + +X

1 2 5x x+ =

1 2 12

1 2 1

1 2 2 1

6 10 5 0

13 5 (5 )2

13( ) 2 (5 )2

x xx

x x

x x x

λ

λ

λ

∂= − − + + =

∂

=> + = +

=> + + = +

L

21

2x −=

1112

x =

[ ]1 11* , ; * 232 2−⎡ ⎤= =⎢ ⎥⎣ ⎦

X λ 11 12 11

21 22 21

11 12

00

L L gL L gg g

−∈⎛ ⎞⎜ ⎟−∈ =⎜ ⎟⎜ ⎟⎝ ⎠


Thank you

Optimization Methods: Optimization using Calculus – Kuhn-Tucker Conditions 1


Kuhn-Tucker Conditions

Introduction

In the previous lecture the optimization of functions of multiple variables subjected to

equality constraints using the method of constrained variation and the method of Lagrange

multipliers was dealt. In this lecture the Kuhn-Tucker conditions will be discussed with

examples for a point to be a local optimum in case of a function subject to inequality

constraints.


It was previously established that for both an unconstrained optimization problem and an

optimization problem with an equality constraint the first-order conditions are sufficient for a

global optimum when the objective and constraint functions satisfy appropriate

concavity/convexity conditions. The same is true for an optimization problem with inequality

constraints.

The Kuhn-Tucker conditions are both necessary and sufficient if the objective function is

concave and each constraint is linear or each constraint function is concave, i.e. the problems

belong to a class called the convex programming problems.

Consider the following optimization problem:

Minimize f(X) subject to gj(X) ≤ 0 for j = 1,2,…,p ; where X = [x1 x2 . . . xn]

Then the Kuhn-Tucker conditions for X* = [x1* x2

* . . . xn*] to be a local minimum are

10 1, 2,...,

0 1, 2,...,

0 1, 2,...,

0 1, 2,...,

m

jji i

j j

j

j

f g i nx x

g j

g j

m

m

j m

λ

λ

λ

=

∂ ∂+ = =

∂ ∂

= =

≤ =

≥ =

∑

(1)


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In case of minimization problems, if the constraints are of the form gj(X) 0, then ≥ jλ have

to be nonpositive in (1). On the other hand, if the problem is one of maximization with the

constraints in the form gj(X) ≥ 0, then jλ have to be nonnegative.

It may be noted that sign convention has to be strictly followed for the Kuhn-Tucker

conditions to be applicable.

Example 1

Minimize 2 21 22 3 2

3f x x x= + + subject to the constraints

1 1 2 3

2 1 2 3

2 12 3

g x x xg x x x= − − ≤= + − ≤

28

using Kuhn-Tucker conditions.

Solution:

The Kuhn-Tucker conditions are given by

a) 1 21 2 0

i i i

g gfx x x

λ λ∂ ∂∂+ + =

∂ ∂ ∂

i.e.

1 1 2

2 1 2

3 1 2

2 0 (2) 4 2 0 (3)6 2 3 0

xx

x

λ λλ λλ λ

+ + =− + =− − = (4)

b) 0 j jgλ =

i.e.

1 1 2 3

2 1 2 3

( 2 12) 0 (5)( 2 3 8) 0 (6)x x xx x x

λλ

− − − =+ − − =

c) 0 jg ≤


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i.e.,

1 2 3

1 2 3

2 12 0 (7)2 3 8 0 (8)

x x xx x x− − − ≤+ − − ≤

d) 0jλ ≥

i.e.,

1

2

0 (9)0 (10)

λλ≥≥

From (5) either 1λ = 0 or, 1 2 32 12x x x− − − = 0

Case 1: 1λ = 0

From (2), (3) and (4) we have x1 = x2 = 2 / 2λ− and x3 = 2 / 2λ .

Using these in (6) we get 22 2 28 0, 0 orλ λ λ+ = ∴ = −8

From (10), 2 0 λ ≥ , therefore, 2λ =0, X* = [ 0, 0, 0 ], this solution set satisfies all of (6) to (9)

Case 2: 1 2 32 12 0x x x− − − =

Using (2), (3) and (4), we have 1 2 1 2 1 22 2 3 12 02 4 3

λ λ λ λ λ λ− − − +− − − = or,

1 217 12 144λ λ+ = − . But conditions (9) and (10) give us 1 0λ ≥ and 2 0λ ≥ simultaneously,

which cannot be possible with 1 217 12 144λ λ+ = − .

Hence the solution set for this optimization problem is X* = [ 0 0 0 ]


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Example 2

Minimize 2 21 2 60 1f x x x= + + subject to the constraints

1 1

2 1 2

80 0120 0

g xg x x= − ≥= + − ≥

using Kuhn-Tucker conditions.

Solution

The Kuhn-Tucker conditions are given by

a) 31 21 2 3 0

i i i i

gg gfx x x x

λ λ λ ∂∂ ∂∂+ + + =

∂ ∂ ∂ ∂

i.e.

1 1 2

2 2

2 60 0 (11)2 0 (12)

xx

λ λλ

+ + + =+ =

b) 0 j jgλ =

i.e.

1 1

2 1 2( 120) 0 (14)x x( 80) 0 (13)xλ

λ + − =− =

80 0 (15)x − ≥

c) 0 jg ≤

i.e.,

1

1 2 120 0 (16)x x+ + ≥

d) 0jλ ≤

i.e.,


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1

2

0 (17)0 (18)

λλ≤≤

From (13) either 1λ = 0 or, 1( 80) 0x − =

Case 1: 1λ = 0

From (11) and (12) we have 21 302x λ= − − and 2

2 2x λ= −

Using these in (14) we get ; ( )2 2 150 0λ λ − = 2 0 150orλ∴ = −

Considering 2 0 λ = , X* = [ 30, 0].

But this solution set violates (15) and (16)

For 2 150 λ = − , X* = [ 45, 75].

But this solution set violates (15) .

Case 2: 1 80) 0x − =(

Using in (11) and (12), we have 1 80x =

2 2

1 2

22 220 (19)

xx

λλ

= −= −

Substitute (19) in (14), we have

( )2 22 40x x− − 0=

0=

.

For this to be true, either 2 20 40x or x= −

For , 2 0x = 1 220λ = − . This solution set violates (15) and (16)


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For ,2 40 0x − = 1 2140 80andλ λ= − = − . This solution set is satisfying all equations from

(15) to (19) and hence the desired. Therefore, the solution set for this optimization problem

is X* = [ 80 40 ].



International Limited, New Delhi, 2000.




Delhi, 2005.




M2L5


Introduction

Optimization with multiple decision variables and equality constraint : Lagrange Multipliers.

Optimization with multiple decision variables and inequality constraint : Kuhn-Tucker (KT) conditions

KT condition: Both necessary and sufficient if the objective function is concave and each constraint is linear or each constraint function is concave, i.e. the problems belongs to a class called the convex programming problems.


Kuhn Tucker Conditions: Optimization Model

Consider the following optimization problem Minimize f(X)

subject togj(X) 0 for j=1,2,…,p

where the decision variable vector

X=[x1,x2,…,xn]

≤


Kuhn Tucker Conditions

10 1, 2,...,

0 1, 2,...,

0 1, 2,...,

0 1, 2,...,

m

jji i

j j

j

j

f g i nx x

g j m

g j m

j m

λ

λ

λ

=

∂ ∂+ = =

∂ ∂

= =

≤ =

≥ =

∑

Kuhn-Tucker conditions for X* = [x1* x2

* . . . xn*] to be a local minimum are


Kuhn Tucker Conditions …contd.

jλ

In case of minimization problems, if the constraints are of the form gj(X) 0, then have to be non-positive

On the other hand, if the problem is one of maximization with the constraints in the form gj(X) 0, then have to be nonnegative.

≥ jλ

≥


Example (1)

Minimizesubject to

2 2 21 2 32 3f x x x= + +

1 1 2 3

2 1 2 3

2 122 3 8

g x x xg x x x= − − ≤= + − ≤


Example (1) …contd.

1 21 2 0

i i i

g gfx x x

λ λ∂ ∂∂+ + =

∂ ∂ ∂1 1 2

2 1 2

3 1 2

2 0 (2)4 2 0 (3)6 2 3 0 (4)

xxx

λ λλ λλ λ

+ + =− + =− − =

0 j jgλ = 1 1 2 3

2 1 2 3

( 2 12) 0 (5)( 2 3 8) 0 (6)x x xx x x

λλ

− − − =+ − − =

0 jg ≤1 2 3

1 2 3

2 12 0 (7)2 3 8 0 (8)

x x xx x x− − − ≤

+ − − ≤

0jλ ≥1

2

0 (9)0 (10)

λλ≥≥

Kuhn – Tucker Conditions



From (5) either = 0 or , Case 1From (2), (3) and (4) we have x1 = x2 = and x3 = Using these in (6) we get From (10), , therefore, =0, Therefore, X* = [ 0, 0, 0 ]This solution set satisfies all of (6) to (9)

1λ 1 2 32 12 0x x x− − − =

2 / 2λ− 2 / 2λ22 2 28 0, 0 8orλ λ λ+ = ∴ = −

2 0 λ ≥ 2λ



Case 2:

Using (2), (3) and (4), we have

or

But conditions (9) and (10) give us and simultaneously,

which cannot be possible with

Hence the solution set for this optimization problem is X* = [ 0 0 0 ]

1 2 32 12 0x x x− − − =1 2 1 2 1 22 2 3 12 02 4 3

λ λ λ λ λ λ− − − +− − − =

1 217 12 144λ λ+ = −

1 0λ ≥ 2 0λ ≥

1 217 12 144λ λ+ = −


Example (2)

Minimizesubject to

2 21 2 160f x x x= + +

1 1

2 1 2

80 0120 0

g xg x x= − ≥= + − ≥



1 21 2 0

i i i

g gfx x x

λ λ∂ ∂∂+ + =

∂ ∂ ∂

0 j jgλ =

0 jg ≤

0jλ ≥

Kuhn – Tucker Conditions

1 1 2

2 2

2 60 0 (11)2 0 (12)

xx

λ λλ

+ + + =+ =

1 1

2 1 2

( 80) 0 (13)( 120) 0 (14)xx x

λλ

− =+ − =

1

1 2

80 0 (15)120 0 (16)

xx x− ≥+ + ≥

1

2

0 (17)0 (18)

λλ≤≤



From (13) either = 0 or , Case 1From (11) and (12) we have and Using these in (14) we get

Considering , X* = [ 30, 0]. But this solution set violates (15)and (16)For , X* = [ 45, 75]. But this solution set violates (15)

1λ 1( 80) 0x − =

21 302x λ= − − 2

2 2x λ= −

( )2 2 150 0λ λ − =

2 0 150orλ∴ = −

2 0 λ =

2 150 λ = −



Case 2:

Using in (11) and (12), we have

Substitute (19) in (14), we have

For this to be true, either

1( 80) 0x − =

1 80x =

2 2

1 2

22 220 (19)

xx

λλ

= −= −

( )2 22 40 0x x− − =

2 20 40 0x or x= − =



For ,

This solution set violates (15) and (16)

For ,

This solution set is satisfying all equations from (15) to (19) and hence

the desired

Thus, the solution set for this optimization problem is X* = [ 80 40 ].

2 0x = 1 220λ = −

2 40 0x − = 1 2140 80andλ λ= − = −


Thank you

Optimization Methods: Linear Programming- Learning Objectives

Module 3: Linear Programming

Learning Objectives

It was discussed in module 2 that optimization methods using calculus have several

limitations and thus not suitable for many practical applications. Most widely used

optimization method is linear programming which is the main objective of this module. The

characteristics of linear programming problem (LPP) and also different techniques to solve

LPP are introduced.

The module starts with the definition of LPP, underlying assumptions and elementary

operations. Illustration of graphical method will help to conceive the idea behind the solution

of LPP. This will also help the reader to visualize the overall concept though explained for

only two decisions variables. Once the concept becomes clear, theoretical as well as logical

approach of most popularly used simplex method will be explained. Discussion on revised

simplex method, duality in LPP, Primal-Dual relation, Dual simplex method and sensitivity or

postoptimality analysis will help the reader to understand the practical application of LPP.

Among the other algorithms for solving LP problems, Karmakar’s projective scaling method

will be outlined.


1. Formulate the LPP.

2. Conceptualize the feasible region.

3. Solve the LPP with two variables using graphical method.

4. Solve the LPP using simplex method.

5. Formulate the dual problem from primal.

6. Analyse the sensitivity of a decision variable.

7. Run and analyse the results of user friendly software for LPP.


Optimization Methods: Linear Programming- Preliminaries


1

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Preliminaries

Introduction

Linear Programming (LP) is the most useful optimization technique used for the solution of

engineering problems. The term ‘linear’ implies that the objective function and constraints

are ‘linear’ functions of ‘nonnegative’ decision variables. Thus, the conditions of LP

problems (LPP) are

1. Objective function must be a linear function of decision variables

2. Constraints should be linear function of decision variables

3. All the decision variables must be nonnegative

For example,

Conditionity Nonnegativ0,Constraint3rd154Constraint2nd113

Constraint1st532tosubjectFunctionObjective56Maximize

πππππ

≥≤+≤+≤−

+=

yxyxyxyx

yxZ

is an example of LP problem. However, example shown above is in “general” form.

Standard form of LPP

Standard form of LPP must have following three characteristics:

1. Objective function should be of maximization type

2. All the constraints should of equality type

3. All the decision variables should be nonnegative

The procedure to transform a general form of a LPP to its standard form is discussed below.

Let us consider the following example.

1 2

1 2

1 2

1 2

1

2

Minimize 3 5subject to 2 3 15

34 2

0unrestricted

Z x xx x

x xx x

xx

= − −− ≤

+ ≤+ ≥≥



2

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The above LPP is violating the following criteria of standard form:

1. Objective function is of minimization type

2. Constraints are of inequality type

3. Decision variable 2x is unrestricted, i.e., it can take negative values also, thus

violating the non-negativity criterion.

However, a standard form for this LPP can be obtained by transforming it as follows:

Objective function can be rewritten as

21 53Maximize xxZZ +=−=′

The first constraint can be rewritten as: 1532 321 =+− xxx . Note that, a new nonnegative

variable x3 is added to the left-hand-side (LHS) to make both sides equal. Similarly, the

second constraint can be rewritten as: 3421 =++ xxx . The variables x3 and x4 are known as

slack variables. The third constraint can be rewritten as: 24 521 =−+ xxx . Again, note that a

new nonnegative variable 5x is subtracted form the LHS to make both sides equal. The

variable x5 is known as surplus variable.

Decision variable x2 can expressed by introducing two extra nonnegative variables as

222 xxx ′′−′=

Thus, 2x can be negative if 22 xx ′′<′ and positive if 22 xx ′′>′ depending on the values of

22 and xx ′′′ . 2x can be zero also if 22 xx ′′=′ .

Thus, the standard form of above LPP is as follows:

( )( )

( )( )

0,,,,,24

31532tosubject

53Maximize

543221

5221

4221

3221

221

≥′′′=−′′−′+

=+′′−′+=+′′−′−

′′−′+=−=′

xxxxxxxxxx

xxxxxxxx

xxxZZ

After obtaining solution for 22 and xx ′′′ , solution for 2x can be obtained as, 222 xxx ′′−′= .



3

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Canonical form of LPP

Canonical form of standard LPP is a set of equations consisting of the ‘objective function’

and all the ‘equality constraints’ (standard form of LPP) expressed in canonical form.

Understanding the canonical form of LPP is necessary for studying simplex method, the most

popular method of solving LPP. Simplex method will be discussed in some other class. In this

class, canonical form of a set of linear equations will be discussed first. Canonical form of

LPP will be discussed next.

Canonical form of a set of linear equations

Let us consider a set of three equations with three variables for ease of discussion. Later, the

method will be generalized.

Let us consider the following set of equations,

1023 =++ zyx (A0)

632 =+− zyx (B0)

12 =−+ zyx (C0)

The system of equations can be transformed in such a way that a new set of three different

equations are obtained, each having only one variable with nonzero coefficient. This can be

achieved by some elementary operations.

The following operations are known as elementary operations.

1. Any equation Er can be replaced by kEr, where k is a nonzero constant.

2. Any equation Er can be replaced by Er + kEs, where Es is another equation of the

system and k is as defined above.

Note that the transformed set of equations through elementary operations is equivalent to the

original set of equations. Thus, solution of the transformed set of equations will be the

solution of the original set of equations too.



4

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Now, let us transform the above set of equation (A0, B0 and C0) through elementary

operations (shown inside bracket in the right side).

310

31

32 =++ zyx (A1 =

31 A0)

38

38

380 =+− zy (B1 = B0 – A1)

317

35

310 −=−− zy (C1 = C0 – 2 A1)

Note that variable x is eliminated from equations B0 and C0 to obtain B1 and C1 respectively.

Equation A0 in the previous set is known as pivotal equation.

Following similar procedure, y is eliminated from A1 and C1 as follows, considering B1 as

pivotal equation.

40 =++ zx (A2 = A1 - 32 B2)

10 −=−+ zy (B2 = 83− B1)

6200 −=−+ z (C2 = C1 + 31 B2)

Finally, z is eliminated form A2 and B2 as follows, considering C2 as pivotal equation.

100 =++x (A3 = A2 – C3)

200 =++ y (B3 = B2 + C3)

300 =++ z (C3 = 21− C2)

Thus we end up with another set of equations which is equivalent to the original set having

one variable in each equation. Transformed set of equations, (A3, B3 and C3), thus obtained

are said to be in canonical form. Operation at each step to eliminate one variable at a time,

from all equations except one, is known as pivotal operation. It is obvious that the number of

pivotal operations is the same as the number of variables in the set of equations. Thus we did

three pivotal operations to obtain the canonical form of the set of equations having three

variables each.

It may be noted that, at each pivotal operation, the pivotal equation is transformed first and

using the transformed pivotal equation, other equations in the system are transformed. For



5

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example, while transforming, A1, B1 and C1 to A2, B2 and C2, considering B1 as pivotal

equation, B2 is obtained first. A2 and C2 are then obtained using B2. Transformation can be

obtained by some other elementary operations also but will end up in the same canonical

form. The procedure explained above is used in simplex algorithm which will be discussed

later. The elementary operations involved in pivotal operations, as explained above, will help

the reader to follow the analogy while understanding the simplex algorithm.

To generalize the procedure explained above, let us consider the following system of n

equations with n variables.

)(

)(

)(

2211

222222121

111212111

nnnnnnn

nn

nn

Ebxaxaxa

Ebxaxaxa

Ebxaxaxa

=+++

=+++

=+++

ΛΛΛΜΜΜΜ

ΛΛΛ

ΛΛΛ

Canonical form of above system of equations can be obtained by performing n pivotal

operations through elementary operations. In general, variable ( )nixi Λ1= is eliminated

from all equations except j th equation for which jia is nonzero.

General procedure for one pivotal operation consists of following two steps,

1. Divide j th equation by jia . Let us designate it as )( jE ′ , i.e., ji

jj a

EE =′

2. Subtract kia times of equation )( jE ′ from k th equation ( )njjk ,,1,1,2,1 ΛΛ +−= , i.e.,

jkik EaE ′−



6

M3L1

Above steps are repeated for all the variables in the system of equations to obtain the

canonical form. Finally the canonical form will be as follows:

)(100

)(010

)(001

21

2221

1121

cnnn

cn

cn

Ebxxx

Ebxxx

Ebxxx

′′=+++

′′=+++

′′=+++

ΛΛΛΜΜΜΜ

ΛΛΛ

ΛΛΛ

It is obvious that solution of the system of equations can be easily obtained from canonical

form, such as:

ii bx ′′=

which is the solution of the original set of equations too as the canonical form is obtained

through elementary operations.

Now let us consider more general case for which the system of equations has m equations

with n variables ( mn ≥ ). It is possible to transform the set of equations to an equivalent

canonical form from which at least one solution can be easily deduced.

Let us consider the following general set of equations.

)(

)(

)(

2211

222222121

111212111

mmnmnmm

nn

nn

Ebxaxaxa

Ebxaxaxa

Ebxaxaxa

=+++

=+++

=+++

ΛΛΛΜΜΜΜ

ΛΛΛ

ΛΛΛ



7

M3L1

By performing n pivotal operations (described earlier) for any m variables (say, mxxx Λ,, 21 ,

called pivotal variables), the system of equations reduced to canonical form will be as

follows:

)(100

)(010

)(001

11,21

22211,221

11111,121

cmmnmnmmmm

cnnmmm

cnnmmm

Ebxaxaxxx

Ebxaxaxxx

Ebxaxaxxx

′′=′′++′′++++

′′=′′++′′++++

′′=′′++′′++++

++

++

++

ΛΛΛΛΛΛΜΜΜΜ

ΛΛΛΛΛΛ

ΛΛΛΛΛΛ

Variables, nm xx ,,1 Λ+ , of above set of equations are known as nonpivotal variables or

independent variables. One solution that can be obtained from the above set of equations

is ii bx ′′= for mi Λ1= and 0=ix for ( ) nmi Λ1+= . This solution is known as basic

solution. Pivotal variables, mxxx Λ,, 21 , are also known as basic variables. Nonpivotal

variables, nm xx ,,1 Λ+ , are known as nonbasic variables.

Canonical form of a set of LPP

Similar procedure can be followed in the case of a standard form of LPP. Objective function

and all constraints for such standard form of LPP constitute a linear set of equations. In

general this linear set will have m equations with n variables ( mn ≥ ). The set of canonical

form obtained from this set of equations is known as canonical form of LPP.

If the basic solution satisfies all the constraints as well as non-negativity criterion for all the

variables, such basic solution is also known as basic feasible solution. It is obvious that, there

can be mnc numbers of different canonical forms and corresponding basic feasible solutions.

Thus, if there are 10 equations with 15 variables there exist 1015c = 3003 solutions, a huge

number to be inspected one by one to find out the optimal solution. This is the reason which

motivates for an efficient algorithm for solution of the LPP. Simplex method is one such

popular method, which will be discussed after graphical method.


Linear Programming

Preliminaries


Objectives

To introduce linear programming problems (LPP)

To discuss the standard and canonical form of LPP

To discuss elementary operation for linear set of

equations


Introduction and Definition

Linear Programming (LP) is the most useful optimization technique

Objective function and constraints are the ‘linear’ functions of ‘nonnegative’ decision variables

Thus, the conditions of LP problems are

1. Objective function must be a linear function of decision variables

2. Constraints should be linear function of decision variables

3. All the decision variables must be nonnegative


Example

Conditionity Nonnegativ0,

Constraint3rd154

Constraint2nd113

Constraint1st532tosubject

FunctionObjective56Maximize

p

p

p

p

p

≥

≤+

≤+

≤−

+=

yx

yx

yx

yx

yxZ

This is in “general” form


Standard form of LP problems

Standard form of LP problems must have following three characteristics:

1. Objective function should be of maximization type

2. All the constraints should be of equality type

3. All the decision variables should be nonnegative


General form Vs Standard form

Violating points for standard form of LPP:

1. Objective function is of minimization type

2. Constraints are of inequality type

3. Decision variable, x2, is unrestricted, thus, may take negative values also.

General form

edunrestrict0

243

1532tosubject53Minimize

2

1

21

21

21

21

xx

xxxx

xxxxZ

≥≥+≤+≤−−−=

How to transform a general form of a LPP to the standard form ?


General form Standard form

General form1. Objective function

2. First constraint

3. Second constraint

Standard form1. Objective function

2. First constraint

3. Second constraint

Transformation

21 53Minimize xxZ −−=

1532 21 ≤− xx 1532 321 =+− xxx

321 ≤+ xx 3421 =++ xxx

Variables and are known as slack variables3x 4x

21 53Maximize xxZZ +=−=′


General form Standard form

General form4. Third constraint

Standard form4. Third constraint

Transformation

24 521 =−+ xxx24 21 ≥+ xx

5. Constraints for decision variables, x1 and x2

Variable is known as surplus variable5x5. Constraints for decision

variables, x1 and x2

edunrestrict2x01 ≥x 01 ≥x

222 xxx ′′−′=0,and 22 ≥′′′ xx


Canonical form of LP Problems

The ‘objective function’ and all the ‘equality constraints’

(standard form of LP problems) can be expressed in canonical

form.

This is known as canonical form of LPP

Canonical form of LP problems is essential for simplex method

(will be discussed later)

Canonical form of a set of linear equations will be discussed

next.


Canonical form of a set of linear equations

Let us consider the following example of a set of linear equations

(A0)

(B0)

(C0)

The system of equation will be transformed through ‘Elementary Operations’.

1023 =++ zyx

632 =+− zyx12 =−+ zyx


Elementary Operations

The following operations are known as elementary operations:1. Any equation Er can be replaced by kEr, where k is a nonzero

constant.2. Any equation Er can be replaced by Er + kEs, where Es is

another equation of the system and k is as defined above.

Note: Transformed set of equations through elementary operations is equivalent to the original set of equations. Thus, solution of transformed set of equations is the solution of original set of equations too.


Transformation to Canonical form: An Example

Set of equation (A0, B0 and C0) is transformed through elementary operations (shown inside bracket in the right side)

⎟⎠⎞

⎜⎝⎛ ==++ 01 A

31A

310

31

32 zyx

( )101 ABB38

38

380 −==+− zy

( )101 2ACC3

1735

310 −=−=−− zy

Note that variable x is eliminated from B0 and C0 equations to obtain B1 and C1. Equation A0 is known as pivotal equation.


Transformation to Canonical form: Example contd.

Following similar procedure, y is eliminated from equation A1 and C1considering B1 as pivotal equation:

⎟⎠⎞

⎜⎝⎛ −==++ 212 B

32AA40 zx

⎟⎠⎞

⎜⎝⎛ −=−=−+ 12 B

83B10 zy

⎟⎠⎞

⎜⎝⎛ +=−=−+ 212 B

31CC6200 z


Transformation to Canonical form: Example contd.

Finally, z is eliminated form equation A2 and B2 considering C2 as pivotal equation :

( )323 CAA100 −==++x

( )323 CBB200 +==++ y

⎟⎠⎞

⎜⎝⎛ −==++ 23 C

21C300 z

Note: Pivotal equation is transformed first and using the transformed pivotal equation other equations in the system are transformed.

The set of equations (A3, B3 and C3) is said to be in Canonical form which is equivalent to the original set of equations (A0, B0 and C0)


Pivotal Operation

Operation at each step to eliminate one variable at a time, from all equations except one, is known as pivotal operation.

Number of pivotal operations are same as the number of variables in the set of equations.

Three pivotal operations were carried out to obtain the canonical form of set of equations in last example having three variables.


Transformation to Canonical form: Generalized procedure

Consider the following system of n equations with n variables

)(

)(

)(

2211

222222121

111212111

nnnnnnn

nn

nn

Ebxaxaxa

Ebxaxaxa

Ebxaxaxa

=+++

=+++

=+++

LLL

MM

MM

LLL

LLL



Canonical form of above system of equations can be obtained by performing n pivotal operations

Variable is eliminated from all equationsexcept j th equation for which is nonzero.

General procedure for one pivotal operation consists of following two steps,1. Divide j th equation by . Let us designate it as , i.e.,

2. Subtract times of equation from

k th equation , i.e.,

( )nix i L1=

jia

jia )( jE ′

kia )( jE ′

( )njjk ,,1,1,2,1 LL +−=

ji

jj a

EE =′

jkik EaE ′−



After repeating above steps for all the variables in the system of equations, the canonical form will be obtained as follows:

)(100

)(010

)(001

21

2221

1121

cnnn

cn

cn

Ebxxx

Ebxxx

Ebxxx

′′=+++

′′=+++

′′=+++

LLL

MM

MM

LLL

LLL

It is obvious that solution of above set of equation such as is the solution of original set of equations also.

ii bx ′′=


Transformation to Canonical form: More general case

Consider more general case for which the system of equations has m equation with n variables ( )

It is possible to transform the set of equations to an equivalent canonical form from which at least one solution can be easily deduced

mn ≥

)(

)(

)(

2211

222222121

111212111

mmnmnmm

nn

nn

Ebxaxaxa

Ebxaxaxa

Ebxaxaxa

=+++

=+++

=+++

LLL

MM

MM

LLL

LLL


Transformation to Canonical form: More general case

By performing n pivotal operations for any m variables (say, , called pivotal variables) the system of equations reduced to canonical form is as follows

Variables, , of above set of equations is known as nonpivotal variables or independent variables.

mxxx L,, 21

)(100

)(010

)(001

11,21

22211,221

11111,121

cmmnmnmmmm

cnnmmm

cnnmmm

Ebxaxaxxx

Ebxaxaxxx

Ebxaxaxxx

′′=′′++′′++++

′′=′′++′′++++

′′=′′++′′++++

++

++

++

LLLLLL

MM

MM

LLLLLL

LLLLLL

nm xx ,,1 L+


Basic variable, Nonbasic variable, Basic solution, Basic feasible solution

One solution that can be obtained from the above set of equations is

mxxx L,, 21

nm xx ,,1 L+

( ) nmixmibx

i

ii

,,1for0,,1forL

L

+===′′=

This solution is known as basic solution.

Pivotal variables, , are also known as basic variables.

Nonpivotal variables, , are known as nonbasic variables.

Basic solution is also known as basic feasible solution because it satisfies all the constraints as well as nonnegativity criterion for all the variables


Thank You

Optimization Methods: Linear Programming- Graphical Method 1


Graphical Method

Graphical method to solve Linear Programming problem (LPP) helps to visualize the

procedure explicitly. It also helps to understand the different terminologies associated with

the solution of LPP. In this class, these aspects will be discussed with the help of an example.

However, this visualization is possible for a maximum of two decision variables. Thus, a LPP

with two decision variables is opted for discussion. However, the basic principle remains the

same for more than two decision variables also, even though the visualization beyond two-

dimensional case is not easily possible.

Let us consider the same LPP (general form) discussed in previous class, stated here once

again for convenience.

5)(C&4)(C0,3)(C1542)(C1131)(C532tosubject

56Maximize

−−≥−≤+−≤+−≤−

+=

yxyxyx

yxyxZ

First step to solve above LPP by graphical method, is to plot the inequality constraints one-

by-one on a graph paper. Fig. 1a shows one such plotted constraint.

532 ≤− yx -2

-1

0

1

2

3

4

5

-2 -1 0 1 2 3 4 5

Fig. 1a Plot showing first constraint ( 532 ≤− yx )

Fig. 1b shows all the constraints including the nonnegativity of the decision variables (i.e.,

and ). 0≥x 0≥y

D Nagesh Kumar, IISc, Bangalore M3L2


113 ≤+ yx 154 ≤+ yx 0≥x 0≥y 532 ≤− yx

-2

-1

0

1

2

3

4

5

-2 -1 0 1 2 3 4 5

Fig. 1b Plot of all the constraints

Common region of all these constraints is known as feasible region (Fig. 1c). Feasible region

implies that each and every point in this region satisfies all the constraints involved in the

LPP.

-2

-1

0

1

2

3

4

5

-2 -1 0 1 2 3 4 5

Feasible region

Fig. 1c Feasible region

Once the feasible region is identified, objective function ( yxZ 56 += ) is to be plotted on it.

As the (optimum) value of Z is not known, objective function is plotted by considering any

constant, k (Fig. 1d). The straight line, kyx + =56 (constant), is known as Z line (Fig. 1d).

This line can be shifted in its perpendicular direction (as shown in the Fig. 1d) by changing

the value of k. Note that, position of Z line shown in Fig. 1d, showing the intercept, c, on the



y axis is 3. If, 55

6556 xykxykyx +6 k−

==>+−==>=+ , i.e., 5

=m 6− and

1535

==>== kc k .

-2

3

4

5

-2 0 1 2 3 4 5-1

0

1

2

-1

Z Line

Fig. 1d Plot of Z line and feasible region

-2

-1

0

1

2

3

4

5

-2 -1 4 50 1 2 3

Optimal Point

Z Line

Fig. 1e Location of Optimal Point

Now it can be visually noticed that value of the objective function will be maximum when it

passes through the intersection of 113+ =yx 154 and + =yx

*x 636.2* =y** 56 yx +=

(straight lines associated with

the second and third inequality constraints). This is known as optimal point (Fig. 1e). Thus

the optimal point of the present problem is and . And the optimal

solution is = 31.727

091.3=



Visual representation of different cases of solution of LPP

A linear programming problem may have i) a unique, finite solution, ii) an unbounded

solution iii) multiple (or infinite) number of optimal solutions, iv) infeasible solution and v) a

unique feasible point. In the context of graphical method it is easy to visually demonstrate the

different situations which may result in different types of solutions.

Unique, finite solution

The example demonstrated above is an example of LPP having a unique, finite solution. In

such cases, optimum value occurs at an extreme point or vertex of the feasible region.

Unbounded solution

If the feasible region is not bounded, it is possible that the value of the objective function

goes on increasing without leaving the feasible region. This is known as unbounded solution

(Fig 2).

-2

-1

0

1

2

3

4

5

-2 -1 0 1 2 3 4 5

Z Line

Fig. 2 Unbounded Solution



Multiple (infinite) solutions

If the Z line is parallel to any side of the feasible region all the points lying on that side

constitute optimal solutions as shown in Fig 3.

-2

-1

0

1

2

3

4

5

-2 -1 0 1 2 3 4 5

Parallel

Z Line

Fig. 3 Multiple (infinite) Solution

Infeasible solution

Sometimes, the set of constraints does not form a feasible region at all due to inconsistency in

the constraints. In such situation the LPP is said to have infeasible solution. Fig 4 illustrates

such a situation.

-2

-1

0

1

2

3

4

5

-2 -1 0 1 2 3 4 5

Z Line

Fig. 4 Infeasible Solution



Unique feasible point

This situation arises when feasible region consist of a single point. This situation may occur

only when number of constraints is at least equal to the number of decision variables. An

example is shown in Fig 5. In this case, there is no need for optimization as there is only one

solution.

-2

-1

0

1

2

3

4

5

-2 -1 0 1 2 3 4 5

Unique feasible point

Fig. 5 Unique feasible point



Linear Programming

Graphical method


Objectives

To visualize the optimization procedure explicitly To understand the different terminologies associated with the solution of LPP To discuss an example with two decision variables


Example

(c-1)

(c-2)

(c-3)

(c-4 & c-5)0,154113

532tosubject56Maximize

≥≤+≤+≤−+=

yxyxyxyx

yxZ


Graphical method: Step - 1

Plot all the constraints one by one on a graph paper



Identify the common region of all the constraints.

This is known as ‘feasible region’



Plot the objective function assuming any constant, k, i.e.

This is known as ‘Z line’, which can be shifted perpendicularly by changing the value of k.

kyx =+ 56



Notice that value of the objective function will be maximum when it passes through the intersection of and (straight lines associated with 2nd

and 3rd constraints).This is known as ‘Optimal

Point’

113 =+ yx154 =+ yx



Thus the optimal point of the present problem is

And the optimal solution is

31.726 56 ** =+ yx

091.3* =x636.2* =y


Different cases of optimal solution

A linear programming problem may have

1. A unique, finite solution (example already discussed)

2. An unbounded solution,

3. Multiple (or infinite) number of optimal solution,

4. Infeasible solution, and

5. A unique feasible point.


Unbounded solution: Graphical representation

Situation: If the feasible region is not bounded

Solution: It is possible that the value of the objective function goes on increasing without leaving the feasible region, i.e., unbounded solution


Multiple solutions: Graphical representation

Situation: Z line is parallel to any side of the feasible region

Solution: All the points lying on that side constitute optimal solutions


Infeasible solution: Graphical representation

Situation: Set of constraints does not form a feasible region at all due to inconsistency in the constraints

Solution: Optimal solution is not feasible


Unique feasible point: Graphical representation

Situation: Feasible region consist of a single point. Number of constraints should be at least equal to the number of decision variables

Solution: There is no need for optimization as there is only one feasible point


Thank You

Optimization Methods: Linear Programming- Simplex Method-I


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Simplex Method - I

Introduction

It is already stated in a previous lecture that the most popular method used for the solution of

Linear Programming Problems (LPP) is the simplex method. In this lecture, motivation for

simplex method will be discussed first. Simplex algorithm and construction of simplex

tableau will be discussed later with an example problem.

Motivation for Simplex method

Recall from the second class that the optimal solution of a LPP, if exists, lies at one of the

vertices of the feasible region. Thus one way to find the optimal solution is to find all the

basic feasible solutions of the canonical form and investigate them one-by-one to get at the

optimal. However, again recall the example at the end of the first class that, for 10 equations

with 15 variables there exists a huge number ( 1015c = 3003) of basic feasible solutions. In such

a case, inspection of all the solutions one-by-one is not practically feasible. However, this can

be overcome by simplex method. Conceptual principle of this method can be easily

understood for a three dimensional case (however, simplex method is applicable for any

higher dimensional case as well).

Imagine a feasible region (i.e., volume) bounded by several surfaces. Each vertex of this

volume, which is a basic feasible solution, is connected to three other adjacent vertices by a

straight line to each being the intersection of two surfaces. Being at any one vertex (one of

the basic feasible solutions), simplex algorithm helps to move to another adjacent vertex

which is closest to the optimal solution among all the adjacent vertices. Thus, it follows the

shortest route to reach the optimal solution from the starting point. It can be noted that the

shortest route consists of a sequence of basic feasible solutions which is generated by simplex

algorithm. The basic concept of simplex algorithm for a 3-D case is shown in Fig 1.



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Fig 1.

The general procedure of simplex method is as follows:

1. General form of given LPP is transformed to its canonical form (refer Lecture note 1).

2. A basic feasible solution of the LPP is found from the canonical form (there should

exist at least one).

3. This initial solution is moved to an adjacent basic feasible solution which is closest to

the optimal solution among all other adjacent basic feasible solutions.

4. The procedure is repeated until the optimum solution is achieved.

Step three involves simplex algorithm which is discussed in the next section.

Simplex algorithm

Simplex algorithm is discussed using an example of LPP. Let us consider the following

problem.

0,,

4225

024

622tosubject

24Maximize

321

321

321

321

321

≥

≤−−

≤+−

≤++

+−=

xxx

xxx

xxx

xxx

xxxZ



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Simplex algorithm is used to obtain the solution of this problem. First let us transform the

LPP to its standard form as shown below.

0,,,,,

4225

024

622tosubject

24Maximize

654321

6321

5321

4321

321

≥

=+−−

=++−

=+++

+−=

xxxxxx

xxxx

xxxx

xxxx

xxxZ

It can be recalled that 4x , 5x and 6x are slack variables. Above set of equations, including the

objective function can be transformed to canonical form as follows:

4100225

001024

600122

000024

654321

654321

654321

654321

=+++−−

=++++−

=+++++

=++++−+−

xxxxxx

xxxxxx

xxxxxx

Zxxxxxx

The basic solution of above canonical form is 64 =x , 05 =x , 46 =x , 0321 === xxx and

0=Z . It can be noted that, 4x , 5x and 6x are known as basic variables and 321 and, xxx are

known as nonbasic variables of the canonical form shown above. Let us denote each equation

of above canonical form as:

( )( )( )( ) 4100225

001024

600122

000024

6543216

6543215

6543214

654321

=+++−−

=++++−

=+++++

=++++−+−

xxxxxxx

xxxxxxx

xxxxxxx

ZxxxxxxZ

For the ease of discussion, right hand side constants and the coefficients of the variables are

symbolized as follows:

( )( )( )( ) 65665654643632621616

55565554543532521515

45465454443432421414

565544332211

bxcxcxcxcxcxcxbxcxcxcxcxcxcxbxcxcxcxcxcxcxbZxcxcxcxcxcxcZ

=+++++=+++++=+++++=++++++



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The left-most column is known as basis as this is consisting of basic variables. The

coefficients in the first row ( 61 cc Λ ) are known as cost coefficients. Other subscript notations

are self explanatory and used for the ease of discussion. For each coefficient, first subscript

indicates the subscript of the basic variable in that equation. Second subscript indicates the

subscript of variable with which the coefficient is associated. For example, 52c is the

coefficient of 2x in the equation having the basic variable 5x with nonzero coefficient (i.e.,

c55 is nonzero).

This completes first step of calculation. After completing each step (iteration) of calculation,

three points are to be examined:

1. Is there any possibility of further improvement?

2. Which nonbasic variable is to be entered into the basis?

3. Which basic variable is to be exited from the basis?

The procedure to check these points is discussed next.

1. Is there any possibility of further improvement?

If any of the cost coefficients is negative, further improvement is possible. In

other words, if all the cost coefficients are nonnegative, the basic feasible

solution obtained in that step is optimum.

2. Which nonbasic variable is to be entered?

Entering nonbasic variable is decided such that the unit change of this variable

should have maximum effect on the objective function. Thus the variable having

the coefficient which is minimum among all the cost coefficients is to be

entered, i.e., Sx is to be entered if cost coefficient Sc is minimum.

3. Which basic variable is to be exited?

After deciding the entering variable Sx , rx (from the set of basic variables) is

decided to be the exiting variable if rs

r

cb is minimum for all possible r, provided

rsc is positive.



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It can be noted that, rsc is considered as pivotal element to obtain the next

canonical form.

In this example, ( )41 −=c is the minimum. Thus, 1x is the entering variable for the next step

of calculation. r may take any value from 4, 5 and 6. It is found that 326

41

4 ==cb ,

010

51

5 ==cb

and 8.054

61

6 ==cb

. As, 51

5

cb

is minimum, r is 5. Thus 5x is to be exited and 51c is

the pivotal element and 5x is replaced by 1x in the basis. Set of equations are transformed

through pivotal operation to another canonical form considering 51c as the pivotal element.

The procedure of pivotal operation is already explained in first class. However, as a refresher

it is explained here once again.

1. Pivotal row is transformed by dividing it with the pivotal element. In this case, pivotal

element is 1.

2. For other rows: Let the coefficient of the element in the pivotal column of a particular

row be “l”. Let the pivotal element be “m”. Then the pivotal row is multiplied by l / m

and then subtracted from that row to be transformed. This operation ensures that the

coefficients of the element in the pivotal column of that row becomes zero, e.g., Z

row: l = -4 , m = 1. So, pivotal row is multiplied by l / m = -4 / 1 = -4, obtaining

00408164 654321 =+−+−+− xxxxxx

This is subtracted from Z row obtaining,

00406150 654321 =+++++− Zxxxxxx

The other two rows are also suitably transformed.

After the pivotal operation, the canonical form obtained is shown below.

( )( )( )( ) 415012180

0010241602129000406150

6543216

6543211

6543214

654321

=+−−−+=++++−=+−+−+=+++++−

xxxxxxxxxxxxxxxxxxxxx

ZxxxxxxZ

The basic solution of above canonical form is 01 =x , 64 =x , 46 =x , 0543 === xxx and

0=Z . However, this is not the optimum solution as the cost coefficient 2c is negative. It is



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observed that 2c (= -15) is minimum. Thus, 2=s and 2x is the entering variable. r may take

any value from 4, 1 and 6. However, ( )412 −=c is negative. Thus, r may be either 4 or 6. It is

found that, 667.096

42

4 ==cb , and 222.0

184

62

6 ==cb

. As 62

6

cb

is minimum, r is 6 and 6x is to

be exited from the basis. 62c (=18) is to be treated as pivotal element. The canonical form for

next iteration is as follows:

( )

( )

( )

( )92

181

1850

3210

98

92

910

3201

421

211400

310

65

610400

6543212

6543211

6543214

654321

=+−+−+

=+−+−+

=−++++

=++−+−+

xxxxxxx

xxxxxxx

xxxxxxx

ZxxxxxxZ

The basic solution of above canonical form is 98

1 =x , 92

2 =x , 44 =x , 0532 === xxx and

310=Z .

It is observed that 3c (= -4) is negative. Thus, optimum is not yet achieved. Following similar

procedure as above, it is decided that 3x should be entered in the basis and 4x should be

exited from the basis. Thus, 4x is replaced by 3x in the basis. Set of equations are

transformed to another canonical form considering 43c (= 4) as pivotal element. By doing so,

the canonical form is shown below.

( )

( )

( )

( )98

361

367

61010

914

365

361

61001

181

81

41100

322

31

311000

6543212

6543211

6543213

654321

=−−+++

=+−+++

=−++++

=++++++

xxxxxxx

xxxxxxx

xxxxxxx

ZxxxxxxZ



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The basic solution of above canonical form is 9

141 =x ,

98

2 =x , 13 =x , 0654 === xxx and

322=Z .

It is observed that all the cost coefficients are positive. Thus, optimum is achieved. Hence,

the optimum solution is

333.7322 ==Z

556.19

141 ==x

889.098

2 ==x

13 =x

The calculation shown above can be presented in a tabular form, which is known as Simplex

Tableau. Construction of Simplex Tableau will be discussed next.

Construction of Simplex Tableau

Same LPP is considered for the construction of simplex tableau. This helps to compare the

calculation shown above and the construction of simplex tableau for it.

After preparing the canonical form of the given LPP, simplex tableau is constructed as

follows.



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Variables Iteration Basis Z

1x 2x 3x 4x 5x 6x rb

rs

r

cb

Z 1 -4 1 -2 0 0 0 0 --

4x 0 2 1 2 1 0 0 6 3

5x 0 1 -4 2 0 1 0 0 0 1

6x 0 5 -2 -2 0 0 1 4 54

After completing each iteration, the steps given below are to be followed.

Logically, these steps are exactly similar to the procedure described earlier. However, steps

described here are somewhat mechanical and easy to remember!

Check for optimum solution:

1. Investigate whether all the elements in the first row (i.e., Z row) are nonnegative or

not. Basically these elements are the coefficients of the variables headed by that

column. If all such coefficients are nonnegative, optimum solution is obtained and no

need of further iterations. If any element in this row is negative, the operation to

obtain simplex tableau for the next iteration is as follows:

Operations to obtain next simplex tableau:

2. The entering variable is identified (described earlier). The corresponding column is

marked as Pivotal Column as shown above.

3. The exiting variable from the basis is identified (described earlier). The corresponding

row is marked as Pivotal Row as shown above.

4. Coefficient at the intersection of Pivotal Row and Pivotal Column is marked as

Pivotal Element as shown above.

5. In the basis, the exiting variable is replaced by entering variable.

Pivotal Column

Pivotal Element

Pivotal Row



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6. All the elements in the pivotal row are divided by pivotal element.

7. For any other row, an elementary operation is identified such that the coefficient in

the pivotal column in that row becomes zero. The same operation is applied for all

other elements in that row and the coefficients are changed accordingly. A similar

procedure is followed for all other rows.

For example, say, (2 x pivotal element + pivotal coefficient in first row) produce zero

in the pivotal column in first row. The same operation is applied for all other elements

in the first row and the coefficients are changed accordingly.

Simplex tableaus for successive iterations are shown below. Pivotal Row, Pivotal Column

and Pivotal Element for each tableau are marked as earlier for the ease of understanding.



rs

r

cb

Z 1 0 -15 6 0 4 0 0 --

4x 0 0 9 -2 1 -2 0 6 31

1x 0 1 -4 2 0 1 0 0 -- 2

6x 0 0 18 -12 0 -5 1 4 92

……continued to next page



10

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……continued from previous page



rs

r

cb

Z 1 0 0 -4 0 61−

65

310 --

4x 0 0 0 4 1 21

21− 4 1

1x 0 1 0 32− 0

91−

92

98 --

3

2x 0 0 1 32− 0

185−

181

92 --

Z 1 0 0 0 1 31

31

322

3x 0 0 0 1 41

81

81− 1

1x 0 1 0 0 61

361−

92

914

4

2x 0 0 1 0 61

367−

361−

98

Optimum value of Z

Value of 3x Value of 1x Value of 2x

All the coefficients are nonnegative. Thus optimum solution is achieved.



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As all the elements in the first row (i.e., Z row), at iteration 4, are nonnegative, optimum

solution is achieved. Optimum value of Z is 333.7322 = as shown above. Corresponding

values of basic variables are 556.19

141 ==x , 889.0

98

2 ==x , 13 =x and those of nonbasic

variables are all zero (i.e., 0654 === xxx ).

It can be noted that at any iteration the following two points must be satisfied:

1. All the basic variables (other than Z) have a coefficient of zero in the Z row.

2. Coefficients of basic variables in other rows constitute a unit matrix.

If any of these points are violated at any iteration, it indicates a wrong calculation. However,

reverse is not true.


Linear Programming

Simplex method - I


Introduction and Objectives

Simplex method is the most popular method used for the solution of Linear Programming Problems (LPP).

ObjectivesTo discuss the motivation of simplex method To discuss Simplex algorithm To demonstrate the construction of simplex tableau


Motivation of Simplex method

Solution of a LPP, if exists, lies at one of the vertices of the feasible region.All the basic solutions can be investigated one-by-one to pick up the optimal solution.For 10 equations with 15 variables there exists 15C10= 3003 basic feasible solutions!Too large number to investigate one-by-one.This can be overcome by simplex method


Simplex Method: Concept in 3D case

In 3D, a feasible region (i.e., volume) is bounded by several surfacesEach vertex (a basic feasible solution) of this volume is connected to the three other adjacent vertices by a straight line to each, being intersection of two surfaces.Simplex algorithm helps to move from one vertex to another adjacent vertex which is closest to the optimal solution among all other adjacent vertices.Thus, it follows the shortest route to reach the optimal solution from the starting point.


Simplex Method: Concept in 3D case

Optimum Solution

Starting Point Route followed

Thus there is no need to investigate all the basic feasible solutions.

Sequence of basic feasible solutions on the shortest route is generated by simplex algorithm

Pictorial representation


General procedure of simplex method

Simplex method involves following steps1. General form of given LPP is transformed to its canonical

form (refer Lecture notes 1).

2. Find a basic feasible solution of the LPP (there should exist atleast one).

3. Move to an adjacent basic feasible solution which is closest to the optimal solution among all other adjacent vertices.

4. Repeat until optimum solution is achieved.

Step three involves ‘Simplex Algorithm’


Simplex Algorithm

Let us consider the following LPP

0,,4225

024622tosubject

24Maximize

321

321

321

321

321

≥≤−−≤+−≤++

+−=

xxxxxx

xxxxxx

xxxZ


Simplex Algorithm … contd.

LPP is transformed to its standard form

0,,,,,

4225

024

622tosubject

24Maximize

654321

6321

5321

4321

321

≥

=+−−

=++−

=+++

+−=

xxxxxx

xxxx

xxxx

xxxx

xxxZ

Note that x4, x5 and x6 are slack variables



Set of equations, including the objective function is transformed to canonical form

Basic feasible solution of above canonical form is x4 = 6, x5 = 0, x6 = 4, x1 = x2 = x3 = 0 and Z = 0

4100225

001024

600122

000024

654321

654321

654321

654321

=+++−−

=++++−

=+++++

=++++−+−

xxxxxx

xxxxxx

xxxxxx

Zxxxxxx

x4, x5, x6 : Basic Variables; x1, x2, x3 : Nonbasic Variables



Symbolized form (for ease of discussion)

( )( )( )( ) 65665654643632621616

55565554543532521515

45465454443432421414

565544332211

bxcxcxcxcxcxcxbxcxcxcxcxcxcxbxcxcxcxcxcxcxbZxcxcxcxcxcxcZ

=+++++=+++++=+++++=++++++

• The left-most column is known as basis as this is consisting of basic variables

• The coefficients in the first row ( 61 ,, cc K ) are known as cost coefficients.

• Other subscript notations are self explanatory



This completes first step of algorithm. After completing each step (iteration) of algorithm, following three points are to be examined:

1. Is there any possibility of further improvement?2. Which nonbasic variable is to be entered into the basis?3. Which basic variable is to be exited from the basis?



Is there any possibility of further improvement?If any one of the cost coefficients is negative further improvement is

possible.

Which nonbasic variable is to be entered?Entering variable is decided such that the unit change of this

variable should have maximum effect on the objective function. Thus the variable having the coefficient which is minimum among all cost coefficients is to be entered, i.e., xs is to be entered if cost coefficient cs is minimum.



Which basic variable is to be exited?

After deciding the entering variable xs, xr (from the set of basic

variables) is decided to be the exiting variable if is

minimum for all possible r, provided crs is positive.

crs is considered as pivotal element to obtain the next canonical form.

rs

r

cb



Exiting variabler may take any value from 4, 5 and 6. It is found that ,

. As is minimum, r is 5. Thus, x5 is to be exited.

c51 ( = 1) is considered as pivotal element and x5 is replaced by x1 in the basis.

Thus a new canonical form is obtained through pivotal operation, which was explained in first class.

326

41

4 ==cb 0

10

51

5 ==cb

8.054

61

6 ==cb

Entering variablec1 is minimum (- 4), thus, x1 is the entering variable for the next step of calculation.

51

5

cb



Pivotal operation as a refresher

• Pivotal row is transformed by dividing it with the pivotal element. In this case, pivotal element is 1.

• For other rows: Let the coefficient of the element in the pivotal column of a particular row be “l”. Let the pivotal element me “m”. Then the pivotal row is multiplied by ‘l / m’ and then subtracted from that row to be transformed. This operation ensures that the coefficients of the element in the pivotal column of that row becomes zero, e.g., Z row: l = -4 , m = 1. So, pivotal row is multiplied by l / m = -4 / 1 = -4, obtaining

. This is subtracted from Z row, obtaining,

The other two rows are also suitably transformed.

00408164 654321 =+−+−+− xxxxxx00406150 654321 =+++++− Zxxxxxx



After the pivotal operation, the canonical form obtained as follows

The basic solution of above canonical form is x1 = 0, x4 = 6, x6 = 4, x2 = x3 = x5 = 0 and Z = 0.

( )

( )

( )

( ) 415012180

0010241

6021290

00406150

6543216

6543211

6543214

654321

=+−−−+

=++++−

=+−+−+

=+++++−

xxxxxxx

xxxxxxx

xxxxxxx

ZxxxxxxZ

Note that cost coefficient c2 is negative. Thus optimum solution is not yet achieved. Further improvement is possible.



Exiting variabler may take any value from 4, 1 and 6. However, c12 is negative (- 4). Thus, r

may be either 4 or 6. It is found that and .

As is minimum, r is 6. Thus x6 is to be exited. c62 ( = 18) is considered

as pivotal element and x6 is to be replaced by x2 in the basis.

Thus another canonical form is obtained.

667.096

42

4 ==cb

222.0184

62

6 ==cb

Entering variablec2 is minimum (- 15), thus, x2 is the entering variable for the next step of calculation.

62

6

cb



The canonical form obtained after third iteration

( )

( )

( )

( )92

181

1850

3210

98

92

910

3201

421

211400

310

65

610400

6543212

6543211

6543214

654321

=+−+−+

=+−+−+

=−++++

=++−+−+

xxxxxxx

xxxxxxx

xxxxxxx

ZxxxxxxZ

The basic solution of above canonical form is x1 = 8/9, x2 = 2/9, x4 = 4, x3 = x5 = x6 = 0 and Z = 10/3.



Exiting variable (Following the similar procedure)x4 is the exiting variable. Thus c43 ( = 4) is the pivotal element and x4 is to be replaced by x3 in the basis.

Thus another canonical form is obtained.

Entering variable (Following the similar procedure)x3 is the entering variable for the next step of calculation.

Note that cost coefficient c3 is negative. Thus optimum solution is not yet achieved. Further improvement is possible.



The canonical form obtained after fourth iteration

( )

The basic solution of above canonical form is x1 = 14/9, x2 = 8/9, x3 = 1, x4 = x5 = x6 = 0 and Z = 22/3.

( )

( )

( )98

361

367

61010

914

365

361

61001

181

81

41100

322

31

311000

6543212

6543211

6543213

654321

=−−+++

=+−+++

=−++++

=++++++

xxxxxxx

xxxxxxx

xxxxxxx

ZxxxxxxZ



Note that all the cost coefficients are nonnegative. Thus the optimum solution is achieved.

Optimum solution is333.7

322

==Z

556.19

141 ==x

889.098

2 ==x

13 =x


Construction of Simplex Tableau:General notes

Calculations shown till now can be presented in a tabular form, known as simplex tableau

After preparing the canonical form of the given LPP, first simplex tableau is constructed.

After completing each simplex tableau (iteration), few steps (somewhat mechanical and easy to remember) are followed.

Logically, these steps are exactly similar to the procedure described earlier.


Construction of Simplex Tableau:Basic steps

Check for optimum solution:1. Investigate whether all the elements (coefficients of the variables

headed by that column) in the first row (i.e., Z row) are nonnegative or not. If all such coefficients are nonnegative, optimum solution is obtained and no need of further iterations. If any element in this row is negative follow next steps to obtain the simplex tableau for next iteration.



Operations to obtain next simplex tableau:2. Identify the entering variable (described earlier) and mark that

column as Pivotal Column.

3. Identify the exiting variable from the basis as described earlier and mark that row as Pivotal Row.

4. Mark the coefficient at the intersection of Pivotal Row and Pivotal Column as Pivotal Element.



Operations to obtained simplex tableau…contd.:5. In the basis, replace the exiting variable by entering variable.

6. Divide all the elements in the pivotal row by pivotal element.

7. For any other row, identify the elementary operation such that the coefficient in the pivotal column, in that row, becomes zero. Apply the same operation for all other elements in that row and changethe coefficients.

Follow similar procedure for all other rows.


Construction of Simplex Tableau: example

Consider the same problem discussed before.Canonical form of this LPP is

4100225

001024

600122

000024

654321

654321

654321

654321

=+++−−

=++++−

=+++++

=++++−+−

xxxxxx

xxxxxx

xxxxxx

Zxxxxxx



Corresponding simplex tableau



Successive iterations



Successive iterations…contd.


Final Tableau


Final results from Simplex Tableau

All the elements in the first row (i.e., Z row), at iteration 4, are nonnegative. Thus, optimum solution is achieved.

Optimum solution is

333.73

22==Z

556.19

141 ==x

889.098

2 ==x

13 =x


It can be noted that at any iteration the following two points must be satisfied:

1. All the basic variables (other than Z) have a coefficient of zero in the Z row.

2. Coefficients of basic variables in other rows constitute a unit matrix.

Violation of any of these points at any iteration indicates a wrong calculation. However, reverse is not true.

Construction of Simplex Tableau: A note


Thank You

Optimization Methods: Linear Programming- Simplex Method - II


1

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Simplex Method – II

Introduction

In the previous lecture the simplex method was discussed with required transformation of

objective function and constraints. However, all the constraints were of inequality type with

‘less-than-equal-to’ ( ≤ ) sign. However, ‘greater-than-equal-to’ ( ≥ ) and ‘equality’ ( = )

constraints are also possible. In such cases, a modified approach is followed, which will be

discussed in this lecture. Different types of LPP solutions in the context of Simplex method

will also be discussed. Finally, a discussion on minimization vs maximization will be

presented.

Simplex Method with ‘greater-than-equal-to’ ( ≥ ) and equality ( = ) constraints

The LP problem, with ‘greater-than-equal-to’ ( ≥ ) and equality ( = ) constraints, is transformed

to its standard form in the following way.

1. One ‘artificial variable’ is added to each of the ‘greater-than-equal-to’ ( ≥ ) and equality

( = ) constraints to ensure an initial basic feasible solution.

2. Artificial variables are ‘penalized’ in the objective function by introducing a large

negative (positive) coefficient M for maximization (minimization) problem.

3. Cost coefficients, which are supposed to be placed in the Z-row in the initial simplex

tableau, are transformed by ‘pivotal operation’ considering the column of artificial

variable as ‘pivotal column’ and the row of the artificial variable as ‘pivotal row’.

4. If there are more than one artificial variable, step 3 is repeated for all the artificial

variables one by one.

Let us consider the following LP problem

1 2

1 2

2

1 2

1 2

Maximize 3 5subject to 2

63 2 18

, 0

Z x xx xxx x

x x

= ++ ≥≤+ =

≥

After incorporating the artificial variables, the above LP problem becomes as follows:



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1 2 1 2

1 2 3 1

2 4

1 2 2

1 2


63 2 18

, 0

Z x x Ma Max x x ax xx x a

x x

= + − −+ − + =+ =+ + =

≥

where 3x is surplus variable, 4x is slack variable and 1a and 2a are the artificial variables.

Cost coefficients in the objective function are modified considering the first constraint as

follows:

( )( )

1 2 1 2 1

1 2 3 1 2

3 5 0

2

Z x x Ma Ma E

x x x a E

− − + + =

+ − + =

Thus, pivotal operation is 21 EME ×− , which modifies the cost coefficients as follows:

( ) ( ) MMaaMxxMxMZ 2053 21321 −=++++−+−

Next, the revised objective function is considered with third constraint as follows:

( ) ( ) ( )( )4221

321321

1823

2053

Eaxx

EMMaaMxxMxMZ

=++

−=++++−+−

Obviously pivotal operation is 43 EME ×− , which further modifies the cost coefficients as

follows:

( ) ( ) MaaMxxMxMZ 20003543 21321 −=++++−+−

The modified cost coefficients are to be used in the Z-row of the first simplex tableau.

Next, let us move to the construction of simplex tableau. Pivotal column, pivotal row and

pivotal element are marked (same as used in the last class) for the ease of understanding.

Pivotal Column

Pivotal Row

Pivotal Column

Pivotal Row



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1x 2x 3x 4x 1a 2a rb

rs

r

cb

Z 1 M43 −− M35 −− M 0 0 0 M20− --

1a 0 1 1 -1 0 1 0 2 2

4x 0 0 1 0 1 0 0 6 -- 1

2a 0 3 2 0 0 0 1 18 6

Note that while comparing ( )M43 −− and ( )M35 −− , it is decided that

( ) ( )MM 3543 −−<−− as M is any arbitrarily large number.

Successive iterations are shown as follows:



rs

r

cb

Z 1 0 M+− 2 M33 −− 0 M43 + 0 M126 − --

1x 0 1 1 -1 0 1 0 2 --

4x 0 0 1 0 1 0 0 6 -- 2

2a 0 0 -1 3 0 -3 1 12 4

……continued to next page



4

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……continuing from previous page



rs

r

cb

Z 1 0 -3 0 0 M M+1 18 --

1x 0 1 32 0 0 0

13

6 9

4x 0 0 1 0 1 0 0 6 6 3

3x 0 0 31− 1 0 -1

31 4 --

Z 1 0 0 0 3 M M+1 36 --

1x 0 1 0 0 23

− 0 13

2 --

2x 0 0 1 0 1 0 0 6 -- 4

3x 0 0 0 1 31 -1

31 6 --

It is found that, at iteration 4, optimality has reached. Optimal solution is 36Z = with 1 2x =

and 2 6x = . The methodology explained above is known as Big-M method. Hope, reader has

already understood the meaning of the terminology!

‘Unbounded’, ‘Multiple’ and ‘Infeasible’ solutions in the context of Simplex Method

As already discussed in lecture notes 2, a linear programming problem may have different type

of solutions corresponding to different situations. Visual demonstration of these different types

of situations was also discussed in the context of graphical method. Here, the same will be

discussed in the context of Simplex method.



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Unbounded solution

If at any iteration no departing variable can be found corresponding to entering variable, the

value of the objective function can be increased indefinitely, i.e., the solution is unbounded.


If in the final tableau, one of the non-basic variables has a coefficient 0 in the Z-row, it

indicates that an alternative solution exists. This non-basic variable can be incorporated in the

basis to obtain another optimal solution. Once two such optimal solutions are obtained, infinite

number of optimal solutions can be obtained by taking a weighted sum of the two optimal

solutions.

Consider the slightly revised above problem,

1 2

1 2

2

1 2

1 2


63 2 18

, 0

Z x xx xxx x

x x

= ++ ≥≤+ =

≥

Curious readers may find that the only modification is that the coefficient of 2x is changed

from 5 to 2 in the objective function. Thus the slope of the objective function and that of third

constraint are now same. It may be recalled from lecture notes 2, that if the Z line is parallel to

any side of the feasible region (i.e., one of the constraints) all the points lying on that side

constitute optimal solutions (refer fig 3 in lecture notes 2). So, reader should be able to imagine

graphically that the LPP is having infinite solutions. However, for this particular set of

constraints, if the objective function is made parallel (with equal slope) to either the first

constraint or the second constraint, it will not lead to multiple solutions. The reason is very

simple and left for the reader to find out. As a hint, plot all the constraints and the objective

function on an arithmetic paper.

Now, let us see how it can be found in the simplex tableau. Coming back to our problem, final

tableau is shown as follows. Full problem is left to the reader as practice.



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Final tableau:



rs

r

cb

Z 1 0 0 0 0 M M+1 18 --

1x 0 1 32 0 0 0

13

6 9

4x 0 0 1 0 1 0 0 6 6 3

3x 0 0 31− 1 0 -1

31 4 --

As there is no negative coefficient in the Z-row the optimal is reached. The solution is 18Z =

with 1 6x = and 2 0x = . However, the coefficient of non-basic variable 2x is zero as shown in

the final simplex tableau. So, another solution is possible by incorporating 2x in the basis.

Based on the rs

r

cb , 4x will be the exiting variable. The next tableau will be as follows:



rs

r

cb

Z 1 0 0 0 0 M M+1 18 --

1x 0 1 0 0 23

− 0 13

2 --

2x 0 0 1 0 1 0 0 6 6 4

3x 0 0 0 1 31 -1

31 6 18

Thus, another solution is obtained, which is 18Z = with 1 2x = and 2 6x = . Again, it may be

noted that, the coefficient of non-basic variable 4x is zero as shown in the tableau. If one more

similar step is performed, same simplex tableau at iteration 3 will be obtained.

Coefficient of non-basic variable 2x is zero

Coefficient of non-basic variable 4x is zero



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Thus, we have two sets of solutions as 6

0

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

and 2

6

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

. Other optimal solutions will be obtained

as ( )6 2

10 6

β β⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪+ −⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

where, [ ]0,1β ∈ . For example, let 0.4β = , corresponding solution is

3.6

3.6

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

, i.e., 1 3.6x = and 2 3.6x = . Note that values of the objective function are not changed

for different sets of solution; for all the cases 18Z = .

Infeasible solution

If in the final tableau, at least one of the artificial variables still exists in the basis, the solution

is indefinite.

Reader may check this situation both graphically and in the context of Simplex method by

considering following problem:

1 2

1 2

1 2

1 2


3 2 18, 0

Z x xx xx x

x x

= ++ ≤+ ≥

≥

Minimization versus maximization problems

As discussed earlier, standard form of LP problems consist of a maximizing objective function.

Simplex method is described based on the standard form of LP problems, i.e., objective

function is of maximization type. However, if the objective function is of minimization type,

simplex method may still be applied with a small modification. The required modification can

be done in either of following two ways.

1. The objective function is multiplied by 1− so as to keep the problem identical and

‘minimization’ problem becomes ‘maximization’. This is because of the fact that

minimizing a function is equivalent to the maximization of its negative.

2. While selecting the entering nonbasic variable, the variable having the maximum

coefficient among all the cost coefficients is to be entered. In such cases, optimal



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M3L4

solution would be determined from the tableau having all the cost coefficients as non-

positive ( 0≤ )

Still one difficulty remains in the minimization problem. Generally the minimization problems

consist of constraints with ‘greater-than-equal-to’ ( ≥ ) sign. For example, minimize the price

(to compete in the market); however, the profit should cross a minimum threshold. Whenever

the goal is to minimize some objective, lower bounded requirements play the leading role.

Constraints with ‘greater-than-equal-to’ ( ≥ ) sign are obvious in practical situations.

To deal with the constraints with ‘greater-than-equal-to’ ( ≥ ) and = sign, Big-M method is to

be followed as explained earlier.


Linear Programming

Simplex method - II


Objectives

ObjectivesTo discuss the Big-M methodDiscussion on different types of LPP solutions in the context of Simplex method Discussion on maximization verses minimization problems


Big-M Method

Simplex method for LP problem with ‘greater-than-equal-to’ ( ) and ‘equality’ (=) constraints needs a modified approach. This is known as Big-M method.

The LPP is transformed to its standard form by incorporating a large coefficient M

≥


Transformation of LPP for Big-M method

1. One ‘artificial variable’ is added to each of the ‘greater-than-equal-to’ (≥) and equality (=) constraints to ensure an initial basic feasible solution.

2. Artificial variables are ‘penalized’ in the objective function by introducing a large negative (positive) coefficient for maximization (minimization) problem.

3. Cost coefficients, which are supposed to be placed in the Z-row in the initial simplex tableau, are transformed by ‘pivotal operation’ considering the column of artificial variable as ‘pivotal column’ and the row of the artificial variable as ‘pivotal row’.

4. If there are more than one artificial variables, step 3 is repeated for all the artificial variables one by one.


Example

Consider the following problem

1 2

1 2

2

1 2

1 2


63 2 18

, 0

Z x xx xxx x

x x

= ++ ≥≤+ =≥


Example

After incorporating the artificial variables

where x3 is surplus variable, x4 is slack variable and a1and a2 are the artificial variables

1 2 1 2

1 2 3 1

2 4

1 2 2

1 2


63 2 18

, 0

Z x x Ma Max x x ax xx x a

x x

= + − −+ − + =

+ =+ + =≥


Transformation of cost coefficients

Considering the objective function and the first constraint

( )( )

1 2 1 2 1

1 2 3 1 2

3 5 0

2

Z x x Ma Ma E

x x x a E

− − + + =

+ − + =

Pivotal Column

Pivotal Row

By the pivotal operation 21 EME ×− the cost coefficients are modified as

( ) ( ) MMaaMxxMxMZ 2053 21321 −=++++−+−


Transformation of cost coefficients

Considering the modified objective function and the third constraint

Pivotal ColumnPivotal Row

( )

By the pivotal operation the cost coefficients are modified as

( ) ( )( )

1 2 3 1 2 3

1 2 2 4

3 5 0 2

3 2 18

Z M x M x Mx a Ma M E

x x a E

− + − + + + + = −

+ + =

43 EME ×−

( ) ( ) MaaMxxMxMZ 20003543 21321 −=++++−+−


Construction of Simplex Tableau

Corresponding simplex tableau

Pivotal row, pivotal column and pivotal elements are shown as earlier


Simplex Tableau…contd.

Successive simplex tableaus are as follows:



Optimality has reached. Optimal solution is Z = 36 with x1 = 2 and x2 = 6


Simplex method: ‘Unbounded’, ‘Multiple’ and ‘Infeasible’ solutions

Unbounded solution

If at any iteration no departing variable can be found corresponding to entering variable, the value of the objective function can be increased indefinitely, i.e., the solution is unbounded.




If in the final tableau, one of the non-basic variables has a coefficient 0 in the Z-row, it indicates that an alternative solution exists.This non-basic variable can be incorporated in the basis to obtain another optimal solution. Once two such optimal solutions are obtained, infinite number of optimal solutions can be obtained by taking a weighted sum of the two optimal solutions.


Simplex method: Example of Multiple (indefinite) solutions

Consider the following problem

1 2

1 2

2

1 2

1 2


63 2 18

, 0

Z x xx xxx x

x x

= ++ ≥≤+ =≥

Only modification, compared to earlier problem, is that the coefficient of x2 is changed from 5 to 2 in the objective function. Thus the slope of the objective function and that of third constraint are now same, which leads to multiple solutions



Following similar procedure as described earlier, final simplex tableau for the problem is as follows:



As there is no negative coefficient in the Z-row optimal solution is reached.

Optimal solution is Z = 18 with x1 = 6 and x2 = 0

However, the coefficient of non-basic variable x2 is zero in the Z-row

Another solution is possible by incorporating x2 in the basis

Based on the , x4 will be the exiting variablers

r

cb



So, another optimal solution is Z = 18 with x1 = 2 and x2 = 6

If one more similar step is performed, previous simplex tableau will be obtained back



Thus, two sets of solutions are: and

Other optimal solutions will be obtained as where

For example, let β = 0.4, corresponding solution is

Note that values of the objective function are not changed for different sets of solution; for all the cases Z = 18.

6

0

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

2

6

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

( )6 2

10 6

β β⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪+ −⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭[ ]0,1β ∈

3.6

3.6

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭



Infeasible solution

If in the final tableau, at least one of the artificial variables still exists in the basis, the solution is indefinite.



Simplex method is described based on the standard form of LP problems, i.e., objective function is of maximization typeHowever, if the objective function is of minimization type, simplex method may still be applied with a small modification



The required modification can be done in either of following two ways.1. The objective function is multiplied by -1 so as to keep the

problem identical and ‘minimization’ problem becomes ‘maximization’. This is because minimizing a function is equivalent to the maximization of its negative

2. While selecting the entering nonbasic variable, the variable having the maximum coefficient among all the cost coefficients is to be entered. In such cases, optimal solution would be determined from the tableau having all the cost coefficients as non-positive ( ) 0≤



One difficulty, that remains in the minimization problem, is that it consists of the constraints with ‘greater-than-equal-to’ ( ) sign. For example, minimize the price (to compete in the market), however, the profit should cross a minimum threshold. Whenever the goal is to minimize some objective, lower bounded requirements play the leading role. Constraints with ‘greater-than-equal-to’( ) sign are obvious in practical situations.To deal with the constraints with ‘greater-than-equal-to’ ( ) and equality sign, Big-M method is to be followed as explained earlier.

≥

≥

≥


Thank You

Optimization Methods: Linear Programming- Revised Simplex Method


1

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Revised Simplex Method, Duality and Sensitivity analysis

Introduction

In the previous class, the simplex method was discussed where the simplex tableau at each

iteration needs to be computed entirely. However, revised simplex method is an improvement

over simplex method. Revised simplex method is computationally more efficient and accurate.

Duality of LP problem is a useful property that makes the problem easier in some cases and

leads to dual simplex method. This is also helpful in sensitivity or post optimality analysis of

decision variables.

In this lecture, revised simplex method, duality of LP, dual simplex method and sensitivity or

post optimality analysis will be discussed.

Revised Simplex method

Benefit of revised simplex method is clearly comprehended in case of large LP problems. In

simplex method the entire simplex tableau is updated while a small part of it is used. The

revised simplex method uses exactly the same steps as those in simplex method. The only

difference occurs in the details of computing the entering variables and departing variable as

explained below.

Let us consider the following LP problem, with general notations, after transforming it to its

standard form and incorporating all required slack, surplus and artificial variables.

( )( )( )

( )

1 1 2 2 3 3

11 1 12 2 13 3 1 1

21 1 22 2 23 3 2 2

1 1 2 2 3 3

0n n

i n n

j n n

l m m m mn n m

Z c x c x c x c x Z

x c x c x c x c x b

x c x c x c x c x b

x c x c x c x c x b

+ + + + + =

+ + + + =

+ + + + =

+ + + + =

L L L

L L L

L L L

M M MM M M

L L L

As the revised simplex method is mostly beneficial for large LP problems, it will be

discussed in the context of matrix notation. Matrix notation of above LP problem can be

expressed as follows:



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:with :subject toz Minimize T

0XBAXXC

≥=

=

where

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

nx

xx

Μ2

1

X ,

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

nc

cc

Μ2

1

C ,

1

2

m

b

b

b

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

BM

,

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

0

00

Μ0 ,

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

mnmm

n

n

ccc

cccccc

ΛΜΟΜΜ

ΛΛ

21

22221

11211

A

It can be noted for subsequent discussion that column vector corresponding to a decision

variable kx is

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

mk

k

k

c

cc

Μ2

1

.

Let SX is the column vector of basic variables. Also let SC is the row vector of costs

coefficients corresponding to SX and S is the basis matrix corresponding to SX .

1. Selection of entering variable

For each of the nonbasic variables, calculate the coefficient ( )cWP − , where, P is the

corresponding column vector associated with the nonbasic variable at hand, c is the cost

coefficient associated with that nonbasic variable and 1−= SCW S .

For maximization (minimization) problem, nonbasic variable, having the lowest negative

(highest positive) coefficient, as calculated above, is the entering variable.

2. Selection of departing variable

a. A new column vector U is calculated as BSU 1−= .

b. Corresponding to the entering variable, another vector V is calculated as PSV 1−= ,

where P is the column vector corresponding to entering variable.

c. It may be noted that length of both U and V is same ( m= ). For mi ,,1 Λ= , the

ratios, ( )( )ii

VU , are calculated provided ( ) 0>iV . ri = , for which the ratio is least, is

noted. The r th basic variable of the current basis is the departing variable.

If it is found that ( ) 0≤iV for all i , then further calculation is stopped concluding that

bounded solution does not exist for the LP problem at hand.



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3. Update to new basis

Old basis S , is updated to new basis newS , as [ ] 11 −−= ESS new

where

1

2

1

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

r

m

m

η

η

η

η

η

−

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

E

L L

L L

M M O M L M M

M M L L M M

M M L M O M M

L L

L L

and

( )( )

( )⎪⎪⎩

⎪⎪⎨

⎧

=

≠=

rirV

rirViV

i

for1

forη

S is replaced by newS and steps 1 through 3 are repeated. If all the coefficients calculated in

step 1, i.e., ( )cWP − is positive (negative) in case of maximization (minimization) problem,

then optimum solution is reached and the optimal solution is,

BSXS1−= and SCX=z

Duality of LP problems

Each LP problem (called as Primal in this context) is associated with its counterpart known

as Dual LP problem. Instead of primal, solving the dual LP problem is sometimes easier

when a) the dual has fewer constraints than primal (time required for solving LP problems is

directly affected by the number of constraints, i.e., number of iterations necessary to

converge to an optimum solution which in Simplex method usually ranges from 1.5 to 3

times the number of structural constraints in the problem) and b) the dual involves

maximization of an objective function (it may be possible to avoid artificial variables that

otherwise would be used in a primal minimization problem).

The dual LP problem can be constructed by defining a new decision variable for each

constraint in the primal problem and a new constraint for each variable in the primal. The

coefficients of the j th variable in the dual’s objective function is the i th component of the

primal’s requirements vector (right hand side values of the constraints in the Primal). The

dual’s requirements vector consists of coefficients of decision variables in the primal

objective function. Coefficients of each constraint in the dual (i.e., row vectors) are the

r th column



4

M3L5

column vectors associated with each decision variable in the coefficients matrix of the primal

problem. In other words, the coefficients matrix of the dual is the transpose of the primal’s

coefficient matrix. Finally, maximizing the primal problem is equivalent to minimizing the

dual and their respective values will be exactly equal.

When a primal constraint is less than equal to in equality, the corresponding variable in the

dual is non-negative. And equality constraint in the primal problem means that the

corresponding dual variable is unrestricted in sign. Obviously, dual’s dual is primal. In

summary the following relationships exists between primal and dual.

Primal Dual

Maximization Minimization

Minimization Maximization thi variable thi constraint thj constraint thj variable

0ix ≥ Inequality sign of thi Constraint:

≤ if dual is maximization

≥ if dual is minimization thi variable unrestricted thi constraint with = sign thj constraint with = sign thj variable unrestricted

RHS of thj constraint Cost coefficient associated with thj

variable in the objective function

Cost coefficient associated with thi variable in the objective

function

RHS of thi constraint

See the pictorial representation in the next page for better understanding and quick reference:



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M3L5

Coefficients of the 2nd constraint

Determine the sign of 1y

Determine the sign of 2y

1 1 2 2

11 1 12 2 1 1 1

21 1 22 2 2 2 2

1 1 2 2

1 2

Maximize

Subject to

0, unrestricted, , 0

n n

n n

n n

m m mn n m m

n

Z c x c x c x

c x c x c x b y

c x c x c x b y

c x c x c x b yx x x

= + + +

+ + + = ⎯⎯→

+ + + ≤ ⎯⎯→

+ + + ≤ ⎯⎯→≥ ≥

L L L

L L L

L L LM M

L L LL

M M M M

L L

1 1 2 2

11 1 21 2 1 1

12 1 22 2 2 2

1 1 2 2

1 2

MinimizeSubject to

unrestricted, 0, , 0

m m

m m

m m

n n mn m n

m

Z b y b y b yc y c y c y cc y c y c y c

c y c y c y cy y y

= + + ++ + + ≤+ + + =

+ + + ≤≥ ≥

L L LL L LL L L

M ML L L

L

Opposite for the Dual, i.e., Minimize

Cost coefficients for the Objective Function

Mark the corresponding decision variables in the dual

Coefficients of the 1st

constraint

Corresponding sign of the 1st

constraint is ≤

Right hand side of the 1st

constraint

Thus the Objective Function, 1 1 2 2Minimize m mb y b y b y+ + +L

Thus, the 1st constraint, 11 1 21 2 1 1m mc y c y c y c+ + + ≤L

Corresponding sign of the 2nd constraint is =

Right hand side of the 2nd

constraint

Thus, the 2nd constraint, 12 1 22 2 2 2m mc y c y c y c+ + + =L

Dual Problem

Determine the sign of my



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It may be noted that, before finding its dual, all the constraints should be transformed to ‘less-

than-equal-to’ or ‘equal-to’ type for maximization problem and to ‘greater-than-equal-to’ or

‘equal-to’ type for minimization problem. It can be done by multiplying with 1− both sides

of the constraints, so that inequality sign gets reversed.

An example of finding dual problem is illustrated with the following example.

Primal Dual

Maximize 21 34 xxZ += Minimize 321 400020006000 yyyZ +−=′

Subject to

600032

21 ≤+ xx

200021 ≥− xx

40001 ≤x

1x unrestricted

02 ≥x

Subject to

4321 =+− yyy

332

21 ≤+ yy

01 ≥y

02 ≥y

03 ≥y

It may be noted that second constraint in the primal is transformed to 1 2 2000x x− + ≤ −

before constructing the dual.

Primal-Dual relationships

Following points are important to be noted regarding primal-dual relationship:

1. If one problem (either primal or dual) has an optimal feasible solution, other problem

also has an optimal feasible solution. The optimal objective function value is same for

both primal and dual.

2. If one problem has no solution (infeasible), the other problem is either infeasible or

unbounded.

3. If one problem is unbounded the other problem is infeasible.



7

M3L5

Dual Simplex Method

Computationally, dual simplex method is same as simplex method. However, their

approaches are different from each other. Simplex method starts with a nonoptimal but

feasible solution where as dual simplex method starts with an optimal but infeasible solution.

Simplex method maintains the feasibility during successive iterations where as dual simplex

method maintains the optimality. Steps involved in the dual simplex method are:

1. All the constraints (except those with equality (=) sign) are modified to ‘less-than-

equal-to’ ( ≤ ) sign. Constraints with greater-than-equal-to’ ( ≥ ) sign are multiplied by

1− through out so that inequality sign gets reversed. Finally, all these constraints are

transformed to equality (=) sign by introducing required slack variables.

2. Modified problem, as in step one, is expressed in the form of a simplex tableau. If all

the cost coefficients are positive (i.e., optimality condition is satisfied) and one or

more basic variables have negative values (i.e., non-feasible solution), then dual

simplex method is applicable.

3. Selection of exiting variable: The basic variable with the highest negative value is

the exiting variable. If there are two candidates for exiting variable, any one is

selected. The row of the selected exiting variable is marked as pivotal row.

4. Selection of entering variable: Cost coefficients, corresponding to all the negative

elements of the pivotal row, are identified. Their ratios are calculated after changing

the sign of the elements of pivotal row, i.e., ⎟⎟⎠

⎞⎜⎜⎝

⎛×−

=rowpivotalofElements

tsCoefficienCostratio1

.

The column corresponding to minimum ratio is identified as the pivotal column and

associated decision variable is the entering variable.

5. Pivotal operation: Pivotal operation is exactly same as in the case of simplex

method, considering the pivotal element as the element at the intersection of pivotal

row and pivotal column.

6. Check for optimality: If all the basic variables have nonnegative values then the

optimum solution is reached. Otherwise, Steps 3 to 5 are repeated until the optimum is

reached.



8

M3L5

Consider the following problem:

1212342443

2tosubject2Minimize

21

21

21

1

21

≥+−≥+≤+

≥+=

xxxxxx

xxxZ

By introducing the surplus variables, the problem is reformulated with equality constraints as

follows:

121234

24432tosubject

2Minimize

621

521

421

31

21

−=+−−=+−−

=++−=+−

+=

xxxxxxxxxxx

xxZ

Expressing the problem in the tableau form:



Z 1 -2 -1 0 0 0 0 0

3x 0 -1 0 1 0 0 0 -2

4x 0 3 4 0 1 0 0 24

5x 0 -4 -3 0 0 1 0 -12

1

6x 0 1 -2 0 0 0 1 -1

Ratios 0.5 1/3 -- -- 0 --

Pivotal Column

Pivotal ElementPivotal Row



9

M3L5

Tableaus for successive iterations are shown below. Pivotal Row, Pivotal Column and Pivotal

Element for each tableau are marked as usual.

Variables Iteration Basis

Z 1x 2x 3x 4x 5x 6x rb

Z 1 -2/3 0 0 0 -1/3 0 4

3x 0 -1 0 1 0 0 0 -2

4x 0 -7/3 0 0 1 4/3 0 8

2x 0 4/3 1 0 0 -1/3 0 4

2

6x 0 11/3 0 0 0 -2/3 1 7

Ratios 2/3 -- -- -- -- --

Variables

Iteration Basis Z 1x 2x 3x 4x 5x 6x rb

Z 1 0 0 -2/3 0 -1/3 0 16/3

1x 0 1 0 -1 0 0 0 2

4x 0 0 0 -7/3 1 4/3 0 38/3

2x 0 0 1 4/3 0 -1/3 0 4/3

3

6x 0 0 0 11/3 0 -2/3 1 -1/3

Ratios -- -- -- -- 0.5 --

Variables

Iteration Basis Z 1x 2x 3x 4x 5x 6x rb

Z 1 0 0 2.5 0 0 -0.5 5.5

1x 0 1 0 -1 0 0 0 2

4x 0 0 0 5 1 0 2 12

2x 0 0 1 -0.5 0 0 -0.5 1.5

4

5x 0 0 0 -5.5 0 1 -1.5 0.5

Ratios



10

M3L5

As all the rb are positive, optimum solution is reached. Thus, the optimal solution is 5.5=Z

with 21 =x and 5.12 =x .

Solution of Dual from Final Simplex Tableau of Primal

Primal Dual

Final simplex tableau of primal:

As illustrated above solution for the dual can be obtained corresponding to the coefficients of

slack variables of respective constraints in the primal, in the Z row as, 11 =y , 31

2 =y and

31

3 =y and Z’=Z=22/3.

y1

y2

y3

0,,4225

024622tosubject

24Maximize

321

321

321

321

321

≥≤−−

≤+−≤++

+−=

xxxxxx

xxxxxx

xxxZ

0,,2222124

452tosubject406'Minimize

321

321

321

321

321

≥≥−+−≥−−

≥++++=

yyyyyy

yyyyyy

yyyZ

Z’



11

M3L5

Sensitivity or post optimality analysis

A dual variable, associated with a constraint, indicates a change in Z value (optimum) for a

small change in RHS of that constraint. Thus,

j iZ y b∆ = ∆

where jy is the dual variable associated with the thi constraint, ib∆ is the small change in the

RHS of thi constraint, and Z∆ is the change in objective function owing to ib∆ .

Let, for a LP problem, thi constraint be 1 22 50x x+ ≤ and the optimum value of the objective

function be 250. What if the RHS of the thi constraint changes to 55, i.e., thi constraint

changes to 1 22 55x x+ ≤ ? To answer this question, let, dual variable associated with the thi

constraint is jy , optimum value of which is 2.5 (say). Thus, 55 50 5ib∆ = − = and 2.5jy = .

So, 2.5 5 12.5j iZ y b∆ = ∆ = × = and revised optimum value of the objective function is

( )250 12.5 262.5+ = .

It may be noted that ib∆ should be so chosen that it will not cause a change in the optimal

basis.


Linear Programming

Revised Simplex Method, Duality of LP problems and Sensitivity analysis


Introduction

Revised simplex method is an improvement over simplex method. It is computationally more efficient and accurate.

Duality of LP problem is a useful property that makes the problem easier in some cases

Dual simplex method is computationally similar to simplex method. However, their approaches are different from each other.

Primal-Dual relationship is also helpful in sensitivity or post optimality analysis of decision variables.


Objectives

Objectives

To explain revised simplex method

To discuss about duality of LP and Primal-Dual relationship

To illustrate dual simplex method

To end with sensitivity or post optimality analysis


Revised Simplex method: Introduction

Benefit of revised simplex method is clearly comprehended in case of large LP problems.In simplex method the entire simplex tableau is updated while a small part of it is used.The revised simplex method uses exactly the same steps as those in simplex method. The only difference occurs in the details of computing the entering variables and departing variable.


Revised Simplex method

Consider the following LP problem (with general notations, after transforming it to its standard form and incorporating all required slack, surplus and artificial variables)

( )( )( )

( )

1 1 2 2 3 3

11 1 12 2 13 3 1 1

21 1 22 2 23 3 2 2

1 1 2 2 3 3

0n n

i n n

j n n

l m m m mn n m

Z c x c x c x c x Z

x c x c x c x c x b

x c x c x c x c x b

x c x c x c x c x b

+ + + + + =

+ + + + =

+ + + + =

+ + + + =

LLL

LLL

LLL

M M M

M M M

LLL

As the revised simplex method is mostly beneficial for large LPproblems, it will be discussed in the context of matrix notation.


Revised Simplex method: Matrix form

Matrix notation

:with :subject toz Minimize T

0XBAXXC

≥=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

nx

xx

M2

1

X

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

nc

cc

M2

1

C

1

2

m

b

b

b

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

BM ⎥

⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

0

00

M0

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

mnmm

n

n

ccc

cccccc

L

MOMM

L

L

21

22221

11211

A

where


Revised Simplex method: Notations

Notations for subsequent discussions:

Column vector corresponding to a decision variable is .

is the column vector of basic variables

is the row vector of cost coefficients corresponding to , and

is the basis matrix corresponding to

SX

SX

SC

S

SX

kx

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

mk

k

k

c

cc

M2

1


Revised Simplex method: Iterative steps

1. Selection of entering variable

For each of the nonbasic variables, calculate the coefficient (WP - c), where, P is the corresponding column vector associated with the nonbasic variable at hand, c is the cost coefficient associated with that nonbasic variable and W = CS S -1.

For maximization (minimization) problem, nonbasic variable, having the lowest negative (highest positive) coefficient, as calculated above, is the entering variable.



2. Selection of departing variablea) A new column vector U is calculated as U = S-1 Bb) Corresponding to the entering variable, another vector V is

calculated as V = S-1 P, where P is the column vector corresponding to entering variable.

c) It may be noted that length of both U and V is same (= m). For i = 1,…, m, the ratios, U(i)/V(i), are calculated provided V(i) > 0. i = r , for which the ratio is least, is noted. The r th basic variable of the current basis is the departing variable.If it is found that V(i) < 0 for all i, then further calculation is stopped concluding that bounded solution does not exist for the LP problem at hand.



3. Update to new BasisOld basis S, is updated to new basis Snew, as Snew = [ E S-1 ] -1

where

( )( )

( )⎪⎪⎩

⎪⎪⎨

⎧

=

≠=

rirV

rirViV

i

for1

forηand

r thcolumn

1

2

1

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

r

m

m

η

η

η

η

η

−

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

E

L L

L L

M M O M L M M

M M L L M M

M M L M O M M

L L

L L



S is replaced by Snew and steps1 through 3 are repeated.

If all the coefficients calculated in step 1, i.e., is positive (negative) in case of maximization (minimization) problem, then optimum solution is reached

The optimal solution is

XS=S-1B and z = CXS


Duality of LP problems

Each LP problem (called as Primal in this context) is associated with its counterpart known as Dual LP problem. Instead of primal, solving the dual LP problem is sometimes easier in following casesa) The dual has fewer constraints than primal

Time required for solving LP problems is directly affected by the number of constraints, i.e., number of iterations necessary to converge to an optimum solution, which in Simplex method usually ranges from 1.5 to 3 times the number of structural constraints in the problem

b) The dual involves maximization of an objective functionIt may be possible to avoid artificial variables that otherwise would be used in a primal minimization problem.


Finding Dual of a LP problem

Inequality sign of ith Constraint:if dual is maximization if dual is minimization

xi > 0

jth variablejth constraint ith constraint ith variable MaximizationMinimizationMinimizationMaximization

DualPrimal

≥≤

…contd. to next slide


Finding Dual of a LP problem…contd.

RHS of ith constraint constraintsCost coefficient associated with ith variable in the objective function

Cost coefficient associated with jth

variable in the objective functionRHS of jth constraint

jth variable unrestrictedjth constraint with = signith constraint with = signith variable unrestricted

DualPrimal

Refer class notes for pictorial representation of all the operations


Dual from a Primal


Finding Dual of a LP problem…contd.

Note: Before finding its dual, all the constraints should be transformed to ‘less-than-equal-to’ or ‘equal-to’ type for maximization problem and to ‘greater-than-equal-to’ or ‘equal-to’ type for minimization problem.

It can be done by multiplying with -1 both sides of the constraints, so that inequality sign gets reversed.


Finding Dual of a LP problem: An example

Primal DualMaximize Minimize

Subject to Subject to

600032

21 ≤+ xx

200021 ≥− xx

40001 ≤x

332

21 ≤+ yy

4321 =+− yyy

1 unrestrictedx02 ≥x

01 ≥y

02 ≥y

21 34 xxZ += 321 400020006000 yyyZ +−=′

Note: Second constraint in the primal is transformed to before constructing the dual.

1 2 2000x x− + ≤ −

03 ≥y


Primal-Dual relationships

If one problem (either primal or dual) has an optimal feasible solution, other problem also has an optimal feasible solution. The optimal objective function value is same for both primal and dual.If one problem has no solution (infeasible), the other problem is either infeasible or unbounded.If one problem is unbounded the other problem is infeasible.


Dual Simplex Method

Simplex Method verses Dual Simplex Method

1. Simplex method starts with a nonoptimal but feasible solution where as dual simplex method starts with an optimal but infeasible solution.

2. Simplex method maintains the feasibility during successive iterations where as dual simplex method maintains the optimality.


Dual Simplex Method: Iterative steps

Steps involved in the dual simplex method are: 1. All the constraints (except those with equality (=) sign) are

modified to ‘less-than-equal-to’ sign. Constraints with greater-than-equal-to’ sign are multiplied by -1 through out so that inequality sign gets reversed. Finally, all these constraints are transformed to equality sign by introducing required slack variables.

2. Modified problem, as in step one, is expressed in the form of a simplex tableau. If all the cost coefficients are positive (i.e., optimality condition is satisfied) and one or more basic variables have negative values (i.e., non-feasible solution), then dual simplex method is applicable.


Dual Simplex Method: Iterative steps…contd.

3. Selection of exiting variable: The basic variable with the highest negative value is the exiting variable. If there are twocandidates for exiting variable, any one is selected. The row of the selected exiting variable is marked as pivotal row.

4. Selection of entering variable: Cost coefficients, corresponding to all the negative elements of the pivotal row, are identified. Their ratios are calculated after changing the sign of the elements of pivotal row, i.e.,

The column corresponding to minimum ratio is identified as the pivotal column and associated decision variable is the entering variable.

⎟⎟⎠

⎞⎜⎜⎝

⎛×−

=rowpivotalofElements

tsCoefficienCostratio1


Dual Simplex Method: Iterative steps…contd.

5. Pivotal operation: Pivotal operation is exactly same as in the case of simplex method, considering the pivotal element as the element at the intersection of pivotal row and pivotal column.

6. Check for optimality: If all the basic variables have nonnegative values then the optimum solution is reached. Otherwise, Steps 3 to 5 are repeated until the optimum is reached.


Dual Simplex Method: An Example


1212342443

2tosubject2Minimize

21

21

21

1

21

≥+−≥+≤+

≥+=

xxxxxx

xxxZ


Dual Simplex Method: An Example…contd.

After introducing the surplus variables the problem is reformulated with equality constraints as follows:

121234

24432tosubject

2Minimize

621

521

421

31

21

−=+−−=+−−

=++−=+−

+=

xxxxxxxxxxx

xxZ



Expressing the problem in the tableau form:



Successive iterations:




As all the br are positive, optimum solution is reached. Thus, the optimal solution is Z = 5.5 with x1 = 2 and x2 = 1.5


Solution of Dual from Primal Simplex

0,,2222124

452tosubject406'Minimize

321

321

321

321

321

≥≥−+−≥−−

≥++++=

yyyyyy

yyyyyy

yyyZ

Primal

Dual0,,

4225024622tosubject

24Maximize

321

321

321

321

321

≥≤−−≤+−≤++

+−=

xxxxxx

xxxxxx

xxxZ y1

y2

y3

Z’



• Changes that can affect only Optimality• Change in coefficients of the objective function, C1, C2,..• Re-solve the problem to obtain the solution

• Changes that can affect only Feasibility• Change in right hand side values, b1, b2,..• Apply dual simplex method or study the dual variable values

• Changes that can affect both Optimality and Feasibility• Simultaneous change in C1, C2,.. and b1, b2,..• Use both primal simplex and dual simplex or re-solve



A dual variable, associated with a constraint, indicates a change in Z value (optimum) for a small change in RHS of that constraint.

j iZ y bΔ = Δ

where yj is the dual variable associated with the ith constraint,

Δbi is the small change in the RHS of ith constraint,

ΔZ is the change in objective function owing to Δbi.


Sensitivity or post optimality analysis: An Example

Let, for a LP problem, ith constraint be

and the optimum value of the objective function be 250.

RHS of the ith constraint changes to 55, i.e., ith constraint changes to

Let, dual variable associated with the ith constraint is yj , optimum value of which is 2.5 (say). Thus, Δbi = 55 – 50 = 5 and yj = 2.5

So, ΔZ = yj Δbi = 2.5x5 = 12.5 and revised optimum value of the objective function is 250 + 12.5 = 262.5.

1 22 50x x+ ≤

1 22 55x x+ ≤


Thank You

Optimization Methods: Linear Programming - Other Algorithms for Solving Linear Programming Problems

1


Other Algorithms for Solving Linear Programming Problems

Introduction

So far, Simplex algorithm, Revised Simplex algorithm, Dual Simplex method are discussed.

There are few other methods for solving LP problems which have an entirely different

algorithmic philosophy. Among these, Khatchian’s ellipsoid method and Karmarkar’s

projective scaling method are well known. In this lecture, a brief discussion about these new

methods in contrast to Simplex method will be presented. However, Karmarkar’s projective

scaling method will be discussed in detail.

Comparative discussion between new methods and Simplex method

Khatchian’s ellipsoid method and Karmarkar’s projective scaling method seek the optimum

solution to an LP problem by moving through the interior of the feasible region. A schematic

diagram illustrating the algorithmic differences between the Simplex and the Karmarkar’s

algorithm is shown in figure 1. Khatchian’s ellipsoid method approximates the optimum

solution of an LP problem by creating a sequence of ellipsoids (an ellipsoid is the

multidimensional analog of an ellipse) that approach the optimal solution.

Simplex Algorithm Karmarkar’s Algorithm

Optimal solution point

Feasible Region

Figure 1 Difference in optimum search path between Simplex and

Karmarkar’s Algorithm



2

Both Khatchian’s ellipsoid method and Karmarkar’s projective scaling method have been

shown to be polynomial time algorithms. This means that the time required to solve an LP

problem of size n by the two new methods would take at most where a and b are two

positive numbers.

ban

On the other hand, the Simplex algorithm is an exponential time algorithm in solving LP

problems. This implies that, in solving an LP problem of size n by Simplex algorithm, there

exists a positive number such that for any the Simplex algorithm would find its solution

in a time of at most . For a large enough n (with positive , b and ), . This

means that, in theory, the polynomial time algorithms are computationally superior to

exponential algorithms for large LP problems.

c nnc2 a c bn anc >2

Karmarkar’s projective scaling method

Karmarkar’s projective scaling method, also known as Karmarkar’s interior point LP

algorithm, starts with a trial solution and shoots it towards the optimum solution.

To apply Karmarkar’s projective scaling method, LP problem should be expressed in the

following form

:with1

:subject to Minimize T

0X1X

0AXXC

≥==

=Z

where , , , and . It is

also assumed that

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

nx

xx

M2

1

X

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

nc

cc

M2

1

C [ ] ( )n×= 1111 L1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

mnmm

n

n

ccc

cccccc

L

MOMM

L

L

21

22221

11211

A 2≥n

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

n

n

n

1

1

1

0M

X is a feasible solution and 0min =Z . The two other variables are

defined as ( )11−

=nn

r , ( )n

n3

1−=α .



3

Iterative steps are involved in Karmarkar’s projective scaling method to find the optimal

solution.

In general, iteration involves following computations: thk

a) Compute ( )[ ] T1TT CPPPPIC −−=p

where , ⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

ADP

k

kDCC T= and

( )( )

( )⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

nk

k

k

k

X

X

X

D

000

000

0020

0001

O

If , any feasible solution becomes an optimal solution. Further iteration is not

required. Otherwise, compute the following

0C =p

b) p

pnew r

C

CXY α−= 0 ,

c) newk

newkk Y1D

YDX =+1 . However, it can be shown that for 0k = , k new

newk new

=D Y Y

1D Y.

Thus, . 1 new=X Y

d) 1T

+= kZ XC

e) Repeat the steps (a) through (d) by changing as k 1+k .


0,,1

02 :subject to2 Minimize

321

321

321

32

≥=++=+−

−=

xxxxxxxxxxxZ

Thus, , , 3=n

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

=

1

2

0

C [ ]121 −=A ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

31

31

31

0M

X , ( ) ( ) 6

1133

11

1=

−=

−=

nnr ,

( ) ( )92

3313

31

=×−

=−

=n

nα .



4

Iteration 0 (k=0):

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

3/100

03/10

003/1

0D

[ ] [ 3/13/20

3/100

03/10

003/1

1200T −=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

×−== DCC ]

]

[ ] [ 3/13/23/1

3/100

03/10

003/1

1210 −=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

×−=AD

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

111

3/13/23/10

1

ADP

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−×⎥⎥⎦

⎤

⎢⎢⎣

⎡ −=

30

03/2

13/1

13/2

13/1

111

3/13/23/1TPP

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡=

−

3/10

05.11TPP

( )⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=−

5.005.0

010

5.005.01TT PPPP

( )[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

=−=−

6/1

0

6/1T1TT CPPPPIC p

62)6/1(0)6/1( 22 =++=pC



5

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

××

−

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=−=

3974.0

3333.0

2692.0

6/1

0

6/1

62

61

92

31

31

31

0Mp

pnew r

C

CXY α

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

==

3974.0

3333.0

2692.0

1 newYX

[ ] 2692.0

3974.0

3333.0

2692.0

1201T =

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

×−== XCZ

Iteration 1 (k=1):

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

3974.000

03333.00

002692.0

1D

[ ] [T1

0.2692 0 0

0 2 1 0 0.3333 0 0 0.6667 0.3974

0 0 0.3974

⎡ ⎤⎢ ⎥⎢ ⎥= = − × = −⎢ ⎥⎢ ⎥⎣ ⎦

C C D ]

]

⎤

⎦

3

⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦

[ ] [ 3974.06666.02692.03974.00003333.00002692.0

1211 −=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡×−=AD

1 0.2692 0.6667 0.3974

1 1 1

−⎛ ⎞ ⎡= =⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣

ADP

1

T

0.2692 10.2692 0.6667 0.3974 0.675 0

0.6667 11 1 1 0

0.3974 1

⎡ ⎤− ⎢ ⎥⎡ ⎤

⎢ ⎥= × −⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

PP

( ) 1T1.482 0

0 0.333

− ⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦

PP



6

( ) 1T T

0.441 0.067 0.492

0.067 0.992 0.059

0.492 0.059 0.567

−

⎡ ⎤⎢ ⎥⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦

P PP P

( ) 1T T T

0.151

0.018

0.132

p

−

⎡ ⎤⎢ ⎥

⎡ ⎤ ⎢ ⎥= − = −⎢ ⎥⎣ ⎦ ⎢ ⎥⎢ ⎥−⎣ ⎦

C I P PP P C

( )22 2(0.151) 0.018 ( 0.132) 0.2014p = + − + − =C

0

1 30.151 0.26532 1

1 3 9 6 0.018 0.34140.2014

0.132 0.39281 3

pnew

p

rα

⎡ ⎤⎢ ⎥ ⎡ ⎤ ⎡ ⎤×⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − = − × − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

CY X

C M

1

0.2692 0 0 0.2653 0.0714

0 0.3333 0 0.3414 0.1138

0 0 0.3974 0.3928 0.1561

new

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= × =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

D Y

[ ]1

0.0714

1 1 1 0.1138 0.3413

0.1561

new

⎡ ⎤⎢ ⎥⎢ ⎥= × =⎢ ⎥⎢ ⎥⎣ ⎦

1D Y

12

1

0.0714 0.20921 0.1138 0.3334

0.34130.1561 0.4574

new

new

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= = × =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

D YX1D Y

[ ]T2

0.2092

0 2 1 0.3334 0.2094

0.4574

Z

⎡ ⎤⎢ ⎥⎢ ⎥= = − × =⎢ ⎥⎢ ⎥⎣ ⎦

C X

So far, two successive iterations are shown for the above problem. Similar iterations can be

followed to get the final solution upto some predefined tolerance level.



7

It may be noted that, the efficacy of Karmarkar’s projective scaling method is more

convincing for ‘large’ LP problems. Rigorous computational effort is not economical for

‘not-so-large’ problems.







Delhi, 2005.



Linear Programming

Other Algorithms


Introduction & Objectives

Few other methods, for solving LP problems, use an entirely different algorithmic philosophy.– Khatchian’s ellipsoid method– Karmarkar’s projective scaling method

ObjectivesTo present a comparative discussion between new methods and Simplex methodTo discuss in detail about Karmarkar’s projective scaling method



Simplex AlgorithmKarmarkar’s Algorithm

Optimal solution point

Feasible Region

Khatchian’s ellipsoid methodand Karmarkar’s projective scaling method seek the optimum solution to an LP problem by moving through the interior of the feasible region.



1. Both Khatchian’s ellipsoid method and Karmarkar’s projective scaling method have been shown to be polynomial time algorithms. Time required for an LP problem of size n is at most anb , where a and b are two positive numbers.

2. Simplex algorithm is an exponential time algorithm in solving LP problems.Time required for an LP problem of size n is at most c2n, where c is a positive number



3. For a large enough n (with positive a, b and c), c2n > anb. The polynomial time algorithms are computationally superior to exponential algorithms for large LP problems.

4. However, the rigorous computational effort of Karmarkar’s projective scaling method, is not economical for ‘not-so-large’ problems.



Also known as Karmarkar’s interior point LP algorithm

Starts with a trial solution and shoots it towards the optimum solution

LP problems should be expressed in a particular form



LP problems should be expressed in the following form:

:with1

:subject to Minimize T

0X1X

0AXXC

≥==

=Z

where

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

nx

x

x

M

2

1

X

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

nc

cc

M2

1

C [ ] ( )n×= 1111 L1⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

mnmm

n

n

ccc

cccccc

L

MOMM

L

L

21

22221

11211

A 2≥nand



It is also assumed that is a feasible solution and

Two other variables are defined as: and .

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

n

n

n

1

1

1

0M

X 0min =Z

( )11−

=nn

r ( )n

n3

1−=α

Karmarkar’s projective scaling method follows iterative steps to find the optimal solution



In general, kth iteration involves following computations

a) Compute ( )[ ] T1TT CPPPPIC −−=p

where

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

ADP

k

kDCC T=

( )( )

( )⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

nk

k

k

k

X

X

X

D

000

000

0020

0001

O

If , ,any feasible solution becomes an optimal solution. Further iteration is not required. Otherwise, go to next step.

0C =p



b)p

pnew r

C

CXY α−= 0

c)newk

newkk Y1D

YDX =+1

However, it can be shown that for k =0, k newnew

k new

=D Y Y

1D YThus, 1 new=X Y

1T

+= kZ XCd)

Repeat the steps (a) through (d) by changing k as k+1e)


Karmarkar’s projective scaling method: Example

Consider the LP problem:

0,,1

02 :subject to2 Minimize

321

321

321

32

≥=++=+−

−=

xxxxxxxxxxxZ

3=nThus,⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

=

1

2

0

C [ ]121 −=A

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

31

31

31

0M

X

( ) ( ) 61

1331

11

=−

=−

=nn

r ( ) ( )92

3313

31

=×−

=−

=n

nαand also,



Iteration 0 (k=0):

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

3/100

03/10

003/1

0D

[ ] [ ]3/13/20

3/100

03/10

003/1

1200T −=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

×−== DCC

[ ] [ ]3/13/23/1

3/100

03/10

003/1

1210 −=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

×−=AD



Iteration 0 (k=0)…contd.:

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

111

3/13/23/10

1

ADP

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−×⎥⎥⎦

⎤

⎢⎢⎣

⎡ −=

30

03/2

13/1

13/2

13/1

111

3/13/23/1TPP

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡=

−

3/10

05.11TPP



( )⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=−

5.005.0

010

5.005.01TT PPPP


( )[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

=−=−

6/1

0

6/1T1TT CPPPPIC p

62)6/1(0)6/1( 22 =++=pC




⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

××

−

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=−=

3974.0

3333.0

2692.0

6/1

0

6/1

62

61

92

31

31

31

0Mp

pnew r

C

CXY α

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

==

3974.0

3333.0

2692.0

1 newYX [ ] 2692.0

3974.0

3333.0

2692.0

1201T =

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

×−== XCZ



Iteration 1 (k=1):

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

3974.000

03333.00

002692.0

1D

[ ] [ ]T1

0.2692 0 0

0 2 1 0 0.3333 0 0 0.6667 0.3974

0 0 0.3974

⎡ ⎤⎢ ⎥⎢ ⎥= = − × = −⎢ ⎥⎢ ⎥⎣ ⎦

C C D

[ ] [ ]3974.06666.02692.03974.00003333.00002692.0

1211 −=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡×−=AD




1 0.2692 0.6667 0.3974

1 1 1

−⎛ ⎞ ⎡ ⎤= =⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

ADP

1

T

0.2692 10.2692 0.6667 0.3974 0.675 0

0.6667 11 1 1 0 3

0.3974 1

⎡ ⎤− ⎢ ⎥⎡ ⎤ ⎡ ⎤

⎢ ⎥= × − =⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

PP

( ) 1T T

0.441 0.067 0.492

0.067 0.992 0.059

0.492 0.059 0.567

−

⎡ ⎤⎢ ⎥⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦

P PP P




( ) 1T T T

0.151

0.018

0.132

p

−

⎡ ⎤⎢ ⎥

⎡ ⎤ ⎢ ⎥= − = −⎢ ⎥⎣ ⎦ ⎢ ⎥⎢ ⎥−⎣ ⎦

C I P PP P C

( )22 2(0.151) 0.018 ( 0.132) 0.2014p = + − + − =C

0

1 30.151 0.26532 1

1 3 9 6 0.018 0.34140.2014

0.132 0.39281 3

pnew

p

rα

⎡ ⎤⎢ ⎥ ⎡ ⎤ ⎡ ⎤×⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − = − × − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

CY X

C M




1

0.2692 0 0 0.2653 0.0714

0 0.3333 0 0.3414 0.1138

0 0 0.3974 0.3928 0.1561

new

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= × =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

D Y

[ ]1

0.0714

1 1 1 0.1138 0.3413

0.1561

new

⎡ ⎤⎢ ⎥⎢ ⎥= × =⎢ ⎥⎢ ⎥⎣ ⎦

1D Y




12

1

0.0714 0.20921 0.1138 0.3334

0.34130.1561 0.4574

new

new

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= = × =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

D YX1D Y

[ ]T2

0.2092

0 2 1 0.3334 0.2094

0.4574

Z

⎡ ⎤⎢ ⎥⎢ ⎥= = − × =⎢ ⎥⎢ ⎥⎣ ⎦

C X

Two successive iterations are shown. Similar iterations can be followed to get the final solution upto some predefined tolerance level


Thank You

Optimization Methods: Linear Programming Applications - Learning Objectives

Module 4: Linear Programming Applications

Learning Objectives

In module 3, we discussed about how to represent a Linear Programming (LP) model in

various forms and also some of the most popular methods like simplex method to solve a LP

problem (LPP). In this module, an introduction to solve a LPP using software will be given.

This will help the reader to solve large LPP more easily avoiding the manual computations in

the form of simplex tables. Using this software, a LPP can be solved using graphical method

or simplex method.

This will be followed by the introduction of some bench mark problems that use LP for

solution. These include the classic transportation problem, assignment problem and some of

the examples from structural and water resources fields.


1. Use the software Mathematical Models for Optimization (MMO) and its application to

solve a LPP

2. Overview of MATLAB Optimization toolbox for solving LPP

3. Formulate and solve a classic transportation problem

4. Formulate and solve an assignment problem

5. Formulate and solve various structural and water resources problems


Optimization Methods: Linear Programming Applications – Software 1


Use of software for solving linear programming problems

Introduction

In this class, use of software to solve linear programming problem will be discussed. An MS-

Dos based software, known as MMO, will be discussed. Apart from MMO, simplex method

using optimization toolbox of MATLAB will be briefly introduced.

MMO Software

This is an MS-Dos based software to solve various types of problems. In this lecture notes,

only Graphical method and Simplex method for LP problem using MMO (Dennis and

Dennis, 1993) will be discussed. It may be noted that MMO can also solve optimization

problems related to integer programming, network flow models, PERT among others. For

more details of MMO, refer to ‘INFOFILE.TXT’ in the folder.

Installation

Download the “MMO.ZIP” file (in the ‘Module_4’ folder of the accompanying CD-ROM)

and unzip it in a folder in the PC. Open this folder and double click on the application file

named as “START”. It will open the MMO software. Opening screen can be seen as shown

in Fig. 1. Press any key to see Main menu screen of MMO as shown in Fig. 2. Use arrow

keys from keyboard to select different models.

Fig. 1. Opening Screen of MMO


M4L1


Fig. 2 Main Menu Screen of MMO

Select “Linear Programming” and press enter. Two options will appear as follows:

SOLUTION METHOD: GRAPHIC/ SIMPLEX

Graphical Method using MMO

Select GRAPHIC and press enter. You can choose a particular option using arrow keys from

the keyboard. It may be noted that graphical method can be used only for two decision

variables.

After waiting for a few moments screen for “data entry method” will appear (Fig. 3).

Fig. 3 Screen for “Data Entry Method”


M4L1


Data entry may be done by either of two different ways.

1. Free Form Entry: You have to write the equation when prompted for input.

2. Tabular Entry: Data can be input in spreadsheet style. Only the coefficients are to be

entered, not the variables.

Note that all variables must appear in the objective function (even those with a 0 coefficient);

if a variable name is repeated in the objective function, an error message will indicate that it

is a duplicate and allow you to change the entry. Constraints can be entered in any order;

variables with 0 coefficients do not have to be entered; if a constraint contains a variable not

found in the objective function, an error message indicates this and allows you to make the

correction; constraints may not have negative right-hand-sides (multiply by -1 to convert

them before entering); when entering inequalities using < or >, it is not necessary to add the

equal sign (=); non-negativity constraints are assumed and do not have to be entered.

However, this information can be made available by selecting “Information Screen”.

Let us take following problem

0,6

,52,532

21

21

21

1

21

≥≤+

−≥−≤+=

xxxxxx

xtoSubjectxxZMaximize

Thus, the second constraint is to be multiplied by -1 while entering, i.e., . In

the MMO software, let us select ‘Free Form Entry’ and ‘max’ while it asks about ‘TYPE OF

PROBLEM’ and press enter. After entering the problem the screen will appear as Fig. 4. Note

that at the last line of the constraints you have to write ‘go’ and hit the enter key from the

keyboard.

52 21 ≤+− xx

Next screen will allow checking the proper entry of the problem. If any mistake is found,

select ‘NO’ and correct the mistake. If everything is ok, select ‘YES’ and press the enter key.

The graphics solution will be displayed on the screen. Different handling options will be

shown on the right corner of the screen as follows:


M4L1


F1: Redraw

F2: Rescale

F3: Move Objective Function Line

F4: Shade Feasible Region

F5: Show Feasible Points

F6: Show Optimal Solution Point

F10: Show Graphical LP Menu (GPL)

We can easily make out the commands listed above. For example, F3 function can be used to

move the objective function line. Subsequent function keys of F4 and F5 can be used to get

the diagram as shown in Fig. 5.

Fig. 4 Screen after Entering the Problem.


M4L1


Fig. 5 Feasible Region (highlighted in white),

Feasible Points (white) and Optimal Solution Point (cyan)

The function key F10, i.e., Show Graphical LP Menu (GPL)’ will display the four different

options as shown in Fig. 6.

Fig. 6 Graphical LP Menu

‘Display Graphical Solution’ will return to the graphical diagram, ‘List Solution Values’ will

show the solution, which is with 67.15=Z 33.21 =x and 67.32 =x . ‘Show Extreme Points’

will show either ‘All’ or ‘Feasible’ extreme points as per the choice (Fig. 7)


M4L1


Fig. 7 List of ‘Extreme Points’ and ‘Feasible Extreme Points’

SIMPLEX Method using MMO

As we know, the graphical solution is is limited to two decision variables. However, simplex

method can be used for any number of variables, which is discussed in this section.

Select SIMPLEX in Linear Programming option of MMO software. As before, screen for

“data entry method” will appear (Fig. 3). The data entry is exactly same as discussed before.

Let us consider the earlier problem for discussion and easy comparison. However, we could

have taken a problem with more than two decision variables also.

0,6

,52,532

21

21

21

1

21

≥≤+

−≥−≤+=

xxxxxx


Once you run the problem, it will show the list of slack, surplus and artificial variables as

shown in Fig. 8. Note that there are three additional slack variables in the above problem.

Press any key to continue.

Fig. 8 List of slack, surplus and artificial variables


M4L1


It will show three different options (Fig. 9):

1. No Tableau: Shows direct solutions

2. All Tableau: Shows all simplex tableau one by one

3. Final Tableau: Shows only the final simplex tableau directly

Fig. 9 Different Options for Simplex Solution

Final simplex tableau for the present problem is shown in Fig. 10 and the final solution is

obtained as: Optimal with 667.15=Z 333.21 =x and 667.32 =x .

There is an additional option for ‘Sensitivity Analysis’. However, it is beyond the scope of

this lecture notes.


M4L1


Fig. 10 Final Simplex Tableau

MATLAB Toolbox for Linear Programming

Optimization toolbox of MATLAB (2001) is very popular and efficient. It includes different

types of optimization techniques. In this lecture notes, we will briefly introduce the use of

MATLAB toolbox for Simplex Algorithm. However, it is assumed that the users are aware of

basics of MATLAB.

To use the simplex method, you have to set the option as 'LargeScale' to 'off' and 'Simplex' to

'on' in the following way.

options = optimset('LargeScale', 'off', 'Simplex', 'on')

Then a function called ‘linprog’ is to be used. A brief MATLAB documentation is shown in

Fig. 11 for linear programming (linprog).


M4L1


Fig. 11 MATLAB Documentation for Linear Programming

Further details may be referred from the toolbox. However, with this basic knowledge, simple

LP problems can be solved. Let us consider the same problem as considered earlier.

0,6

,52,532

21

21

21

1

21

≥≤+

−≥−≤+=

xxxxxx



M4L1


Following MATLAB code will give the solution using simplex algorithm. clear all f=[-2 -3]; %Converted to minimization problem A=[1 0;-1 2;1 1]; b=[5 5 6]; lb=[0 0]; options = optimset('LargeScale', ff', 'Simplex', 'on'); 'o[x,fval]=linprog(f,A,b,[],[],lb); z=-fval %Multiplied by -1 x

Note that objective function should be converted to a minimization problem before entering

as done in line 2 of the code. Finally, solution should be multiplied by -1 to the optimized

(maximum) solution as done in last but one line. Solution will be obtained as

with and as in the earlier case.

667.15=Z

333.21 =x 667.32 =x

References

Dennis T.L. and L.B. Dennis, Microcomputer Models for Management Decision Making, West Publishing Company, 1993. MATLAB – User’s Manual, The Math Works Inc., 2001.


M4L1


Linear Programming Applications

Software for Linear Programming


Objectives

Use of software to solve LP problemsMMO Software with example

Graphical MethodSimplex Method

Simplex method using optimization toolbox of MATLAB


MMO SoftwareMathematical Models for Optimization

An MS-DOS based softwareUsed to solve different optimization problems

Graphical method and Simplex method will be discussed.

Installation– Download the file “MMO.ZIP” and unzip it in a

folder in the PC– Open this folder and double click on the application

file named as “START”. It will open the MMO software


Working with MMO

Opening Screen


Working with MMO

Starting Screen

SOLUTION METHOD: GRAPHIC/ SIMPLEX


Graphical Method

Data Entry


Data Entry: Few Notes

Free Form Entry: Write the equation at the prompted input.Tabular Entry: Spreadsheet style. Only the coefficients are to be entered, not the variables. All variables must appear in the objective function (even those with a 0 coefficient) Constraints can be entered in any order; variables with 0 coefficients do not have to be entered Constraints may not have negative right-hand-sides (multiply by -1 to convert them before entering) When entering inequalities using < or >, it is not necessary to add the equal sign (=) Non-negativity constraints are assumed and need not be entered


Example

Let us consider the following problem

Note: The second constraint is to be multiplied by -1 while entering, i.e.

0,6

,52,532

21

21

21

1

21

≥≤+

−≥−≤+=

xxxxxx


52 21 ≤+− xx


Steps in MMO Software

Select ‘Free Form Entry’ and Select ‘TYPE OF PROBLEM’ as ‘MAX’Enter the problem as shown

Write ‘go’ at the last line of the constraintsPress enterChecking the proper entry of the problem

– If any mistake is found, select ‘NO’ and correct the mistake– If everything is ok, select ‘YES’ and press the enter key


Solution

Z=15.67x1=2.33x2=3.67

F1: RedrawF2: RescaleF3: Move Objective

Function LineF4: Shade Feasible

Region F5: Show Feasible PointsF6: Show Optimal Solution

PointF10: Show Graphical LP

Menu (GPL)


Graphical LP Menu


Extreme points and feasibleextreme points


Simplex Method using MMO

Simplex method can be used for any number of variables

Select SIMPLEX and press enter.

As before, screen for “data entry method” will appear

The data entry is exactly same as discussed before.


Example

Let us consider the same problem. (However, a problem with more than two decision variables can also be taken)

0,6

,52,532

21

21

21

1

21

≥≤+

−≥−≤+=

xxxxxx



Slack, surplus and artificial variables

There are three additional slack variables


Different options for Simplex tableau

No Tableau: Shows direct solutions

All Tableau: Shows all simplex tableau one by one

Final Tableau: Shows only the final simplex tableau directly


Final Simplex tableau and solution

Final Solution

Z=15.67

x1=2.33

x2=3.67



Very popular and efficient

Includes different types of optimization techniques

To use the simplex method – set the option as

options = optimset ('LargeScale', 'off', 'Simplex', 'on')

– then a function called ‘linprog’ is to be used


Example

Let us consider the same problem as before

Note: The maximization problem should be converted to minimization problem in MATLAB

0,6

,52,532

21

21

21

1

21

≥≤+

−≥−≤+=

xxxxxx



Example… contd.

Thus,[ ] tscoefficienCost%32 −−=f

sconstraintoftsCoefficien%

1

2

0

1

1

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−=A

[ ] sconstraint ofsidehandRight%655=b

[ ] variablesdecisionofsLowerbound%00=lb


Example… contd.

MATLAB codeclear allf=[-2 -3]; %Converted to minimization problemA=[1 0;-1 2;1 1];b=[5 5 6];lb=[0 0];options = optimset ('LargeScale', 'off', 'Simplex', 'on');[x , fval]=linprog (f , A , b , [ ] , [ ] , lb );Z = -fval %Multiplied by -1x

SolutionZ = 15.667 with x1 = 2.333 and x2 = 3.667


Thank You

Optimization Methods: Linear Programming Applications – Transportation Problem 1


Transportation Problem

Introduction

In the previous lectures, we discussed about the standard form of a LP and the commonly

used methods of solving LPP. A key problem in many projects is the allocation of scarce

resources among various activities. Transportation problem refers to a planning model that

allocates resources, machines, materials, capital etc. in a best possible way so that the costs

are minimized or profits are maximized. In this lecture, the common structure of a

transportation problem (TP) and its solution using LP are discussed followed by a numerical

example.

Structure of the Problem

The classic transportation problem is concerned with the distribution of any commodity

(resource) from any group of 'sources' to any group of destinations or 'sinks'. While solving

this problem using LP, the amount of resources from source to sink will be the decision

variables. The criterion for selecting the optimal values of the decision variables (like

minimization of costs or maximization of profits) will be the objective function. And the

limitation of resource availability from sources will constitute the constraint set.

Consider a general transportation problem consisting of m origins (sources) O1, O2,…, Om and

n destinations (sinks) D1, D2, … , Dn. Let the amount of commodity available in ith source be

ai (i=1,2,….m) and the demand in jth sink be bj (j=1,2,….n). Let the cost of transportation of

unit amount of material from i to j be cij. Let the amount of commodity supplied from i to j be

denoted as xij. Thus, the cost of transporting xij units of commodity from i to j is c . ijij x×

Now the objective of minimizing the total cost of transportation can be given as

Minimize (1) ∑∑= =

=m

i

n

jijij xcf

1 1



Generally, in transportation problems, the amount of commodity available in a particular

source should be equal to the amount of commodity supplied from that source. Thus, the

constraint can be expressed as

∑ i= 1 ,2, … , m (2) =

=n

jiij ax

1,

Also, the total amount supplied to a particular sink should be equal to the corresponding

demand. Hence,

∑ j = 1 ,2, … , n (3) =

=m

ijij bx

1,

The set of constraints given by eqns (2) and (3) are consistent only if total supply and total

demand are equal.

∑ ∑= =

=m

i

n

jji ba

1 1

(4)

But in real problems this condition may not be satisfied. Then, the problem is said to be

unbalanced. However, the problem can be modified by adding a fictitious (dummy) source or

destination which will provide surplus supply or demand respectively. The transportation

costs from this dummy source to all destinations will be zero. Likewise, the transportation

costs from all sources to a dummy destination will be zero.

Thus, this restriction causes one of the constraints to be redundant. Thus the above problem

have m x n decision variables and (m + n - 1) equality constraints.

The non-negativity constraints can be expressed as

, i= 1 ,2, … , m , j = 1 ,2, … , n (5) 0≥ijx

This problem formulation is elucidated through an example given below.



Examples

Problem (1)

Consider a transport company which has to supply 4 units of paper materials from each of the

cities Faizabad and Lucknow to three cities. The material is to be supplied to Delhi,

Ghaziabad and Bhopal with demands of four, one and three units respectively. Cost of

transportation per unit of supply (cij) is indicated below in the figure. Decide the pattern of

transportation that minimizes the cost.

Solution:

Let the amount of material supplied from source i to sink j be xij. Here m =2; n = 3.

Total supply = 8 units and total demand = 4+1+3 = 8 units. Since both are equal, the problem

is balanced. The objective function is to minimize the total cost of transportation from all

combinations i.e.

Minimize ∑∑= =

=m

i

n

jijij xcf

1 1



Minimize f = 5 x11 + 3 x12 + 8 x13 + 4 x21 + x22 + 7 x23 (6)

subject to the constraints as explained below:

(1) The total amount of material supplied from each source city should be equal to 4.

∑ i= 1, 2 =

=3

14

jijx

i.e. x11 + x12 + x13 = 4 for i = 1 (7)

x21 + x22 + x23 = 4 for i = 2 (8)

(2) The total amount of material received by each destination city should be equal to the

corresponding demand.

∑=

=2

1,

ijij bx j = 1 ,2, 3

i.e. x11 + x21 = 4 for j = 1 (9)

x12 + x22 = 1 for j = 2 (10)

x13 + x23 = 3 for j = 3 (11)

(3) Non – negativity constraints

xij 0≥ i = 1, 2; j=1, 2, 3 (12)

Thus, the optimization problem has 6 decision variables and 5 constraints.

Since the optimization model consists of equality constraints, Big M method is used to solve.

The steps are shown below.



Since there are five equality constraints, introduce five artificial variables R1, R2, R3, R4 and

R5. Thus, the objective function and the constraints can be expressed as

Minimize

54321

232221131211 714835RMRMRMRMRM

xxxxxxf×+×+×+×+×+×+×+×+×+×+×=

subject to

x11 + x12 + x13 + R1 = 4

x21 + x22 + x23 + R2 = 4

x11 + x21+ R3 = 4

x12+ x22 + R4 = 1

x13+ x23+ R5 = 3

Modifying the objective function to make the coefficients of the artificial variable equal to

zero, the final form objective function is

54321

232221

131211

00000)27()21()24()28()23()25(

RRRRRxMxMxMxMxMxMf

×+×+×+×+×−×+−+×+−+×+−+×+−+×+−+×+−+

The solution of the model using simplex method is shown



Table 1

First iteration

Variables Basic variables x11 x12 x13 x21 x22 x23 R1 R2 R3 R4 R5

RHS Ratio

Z -5 +2M

-3 +2M

-8 +2M

-4 +2M

-1 +2M

-7 +2M 0 0 0 0 0 16M

R1 1 1 1 0 0 0 1 0 0 0 0 4 -

R2 0 0 0 1 1 1 0 1 0 0 0 4 4

R31 0 0

1 0

0 0 0 1 0 0 4 -

R4 0 1 0 0 1 0 0 0 0 1 0 1 1

R5 0 0 1 0 0 1 0 0 0 0 1 3 -

Table 2

Second iteration


RHS Ratio

Z -5+2M -1 -8+2M -4+2M 0 -7+2M 0 0 0 1-2M 0 1+14

M -

R1 1 1 1 0 0 0 1 0 0 0 0 4 -

R2 0 -1 0 1 0 0 0 1 0 -1 0 3 3

R3 1 0 0 1 0 0 0 0 1 0 0 4 4

X22 0 1 0 0 1 1 0 0 0 1 0 1 -

R5 0 0 1 0 0 1 0 0 0 0 1 3 -



Table 3

Third iteration


RHS Ratio

Z -5+2M -5+2M -8+2M 0 0 -7+2M 0 4-2M 0 -3 0 13+8M -

R1 1 1 1 0 0 0 1 0 0 0 0 4 4

X21 0 -1 0 1 0 0 0 1 0 -1 0 3 -

R3 1 1 0 0 0 0 0 -1 1 1 0 1 1

X22 0 1 0 0 1 1 0 0 0 1 0 1 -

R5 0 0 1 0 0 1 0 0 0 0 1 3 -

Table 4

Fourth iteration


RHS Ratio

Z 0 0 -8+2M 0 0 -7+2M 0 -1 5-2M

2-2M 0 18+6M -

R1 0 0 1 0 0 0 1 1 -1 -1 0 3 -

X21 0 -1 0 1 0 0 0 1 0 -1 0 3 -

X11 1 1 0 0 0 0 0 -1 1 1 0 1 -

X22 0 1 0 0 1 1 0 0 0 1 0 1 1

R5 0 0 1 0 0 1 0 0 0 0 1 3 3

Repeating the same procedure, we get the final optimal solution f = 42 and the optimum

decision variable values as : x11 = 2.2430, x12 = 0.00, x13 = 1.7570, x21 = 1.7570, x22 =

1.00, x23 = 1.2430.



Problem (2)

Consider three factories (F) located in three different cities, producing a particular chemical.

The chemical is to be transported to four different warehouses (Wh), from where it is

supplied to the customers. The transportation cost per truck load from each factory to each

warehouse is determined and are given in the table below. Production and demands are also

given in the table below.

Wh1 Wh2 Wh3 Wh4 Production

F1 523 682 458 850 60

F2 420 412 362 729 110

F3 670 558 895 695 150

Demand 65 85 80 70

Solution:

Let the amount of chemical to be transported from factory i to warehouse j be xij.

Total supply = 60+110+150 = 320 and total demand = 65+85+80+70 = 300. Since the total

demand is less than total supply, add one fictitious ware house, Wh5 with a demand of 20.

Thus, here m =3; n = 5

Wh1 Wh2 Wh3 Wh4 Wh5 Production

F1 523 682 458 850 0 60

F2 420 412 362 729 0 110

F3 670 558 895 695 0 150

Demand 65 85 80 70 20

The objective function is to minimize the total cost of transportation from all combinations.

Minimize f = 523 x11 + 682 x12 + 458 x13+ 850 x14 + 0 x15 + 420 x21 + 412 x22 + 362 x23

+ 729 x24 + 0 x25 + 670 x31 + 558 x32 + 895 x33 +695 x34 + 0 x35




x11 + x12 + x13 + x14+ x15 = 60 for i = 1

x21 + x22 + x23 + x24+ x25 = 110 for i = 2

x31 + x32 + x33 + x34 + x35 = 150 for i = 3

x11 + x21+ x31 = 65 for j = 1

x12 + x22+ x32 = 85 for j = 2

x13 + x23 + x23 =90 for j = 3

x14 + x24 + x24 =80 for j = 4

x15 + x25 + x25 =20 for j = 5

xij 0≥ i = 1, 2,3; j=1, 2, 3,4

This optimization problem can be solved using the same procedure used for the previous

problem.




Transportation Problem


Introduction

Transportation problem is a special problem of its own

structure.

Planning model that allocates resources, machines,

materials, capital etc. in a best possible way

In these problems main objective generally is to minimize

the cost of transportation or maximize the benefit.


Objectives

To structure a basic transportation problem

To demonstrate the formulation and solution with a

numerical example


Structure of the Problem

A classic transportation problem is concerned with the distribution of any commodity (resource) from any group of ‘sources’ to any group of destinations or ‘sinks’

The amount of resources from source to sink are the decision variables

The criterion for selecting the optimal values of the decision variables (like minimization of costs or maximization of profits) will be the objective function

The limitation of resource availability from sources will constitute the constraint set.


Structure of the Problem …contd.

Let there be m origins namely O1, O2, … , Om and n destinations D1, D2, … , Dn

Amount of commodity available in ith source is ai, i=1,2, … , m

Demand in jth sink is bj , j = 1, 2, … , n.

Cost of transportation of unit amount of material from i to j is cij

Amount of commodity supplied from i to j be denoted as xij

Thus, the cost of transporting xij units of commodity from i to j is

ijij xc ×



The objective of minimizing the total cost of transportation

Minimize (1)

Generally, in transportation problems, the amount of commodity available in a particular source should be equal to the amount of commodity supplied from that source. Thus, the constraint can beexpressed as

i= 1, 2, … , m (2)

∑ ∑= =

=m

i

n

jijij xcf

1 1

∑=

=n

jiij ax

1

,



Total amount supplied to a particular sink should be equal to the corresponding demand

j = 1, 2, … , n (3)

The set of constraints given by eqns (2) and (3) are consistent only if total supply and total demand are equal

i= 1, 2, … , m , j = 1 ,2, … , n (4)

∑=

=m

ijij bx

1,

∑ ∑= =

=m

i

n

jji ba

1 1



But in real problems constraint (4) may not be satisfied. Then, the problem is said to be unbalanced

Unbalanced problems can be modified by adding a fictitious (dummy) source or destination which will provide surplus supply or demand respectively

The transportation costs from this dummy source to all destinations will be zero

Likewise, the transportation costs from all sources to a dummy destination will be zero


Structure of the Problem… contd.

This restriction causes one of the constraints to be redundant

Thus the above problem have m x n decision variables and (m + n - 1)equality constraints.

The non-negativity constraints can be expressed as

i= 1, 2, … , m , j = 1, 2, … , n (5)0≥ijx


Example (1)

Consider a transport company which has to supply 4 units of paper materials from each of the cities Faizabad and Lucknow

The material is to be supplied to Delhi, Ghaziabad and Bhopal with demands of four, one and three units respectively

Cost of transportation per unit of supply (cij) is indicated in the figure shown in the next slide

Decide the pattern of transportation that minimizes the cost.


Example (1)… contd.

Solution:Let the amount of material suppliedfrom source i to sink jas xij

Here m =2; n = 3.



Total supply = 4+4 = 8 units Total demand = 4+1+ 3 = 8 unitsSince both are equal, the problem is balancedObjective function is to minimize the total cost of transportation from all combinations Minimize

i.e. Minimize

f = 5 x11 + 3 x12 + 8 x13 + 4 x21 + x22 + 7 x23 (6)

∑∑= =

=m

i

n

jijij xcf

1 1



The constraints are explained below: The total amount of material supplied from each source city should be equal to 4

; i= 1 ,2

i.e. x11 + x12 + x13 = 4 for i = 1 (7)x21 + x22 + x23 = 4 for i = 2 (8)

∑=

=3

14

jijx



The total amount of material received by each destination city should be equal to the corresponding demand

; j = 1, 2, 3

i.e. x11 + x21 = 4 for j = 1 (9)

x12 + x22 = 1 for j = 2 (10)

x13 + x23 = 3 for j = 3 (11)

Non-negativity constraints

xij ; i = 1, 2; j =1, 2, 3 (12)0≥

2

1ij j

ix b

=

=∑


Example (1)…contd.

Optimization problem has 6 decision variables and 5 constraints

Since the optimization model consists of equality constraints, Big M method is used to solve

Since there are five equality constraints, introduce five artificial variables viz., R1, R2, R3, R4 and R5.



The objective function and the constraints can be expressed as

Minimize

subject to x11 + x12 + x13 +R1 = 4

x21 + x22 + x23 + R2 = 4

x11 + x21 + R3 = 4

x12 + x22 + x23 + R4 = 1

x13 + x23 + R5 = 3

54321

232221131211 714835RMRMRMRMRM

xxxxxxf×+×+×+×+×+×+×+×+×+×+×=



Modifying the objective function to make the coefficients of theartificial variable equal to zero, the final form objective function is

The solution of the model using simplex method is shown in the

following slides

54321

232221

131211

00000)27()21()24()28()23()25(

RRRRRxMxMxMxMxMxMf

×+×+×+×+×−×+−+×+−+×+−+×+−+×+−+×+−+


Example (1)…contd.First iteration


Example (1)…contd.Second iteration


Example (1)…contd.Third iteration


Example (1)…contd.Fourth iteration



Repeating the same procedure

Minimum total cost,

f = 42

Optimum decision variable values:

x11 = 2.2430, x12 = 0.00, x13 = 1.7570,

x21 = 1.7570, x22 = 1.00, x23 = 1.2430.


Example (2)

Consider three factories (F) located in three different cities, producing a particular

chemical. The chemical is to get transported to four different warehouses (Wh),

from where it is supplied to the customers. The transportation cost per truck load

from each factory to each warehouse is determined and are given in the table

below. Production and demands are also given in the table below.



Solution:

Let the amount of chemical to be transported from factory i to warehouse j be

xij.

Total supply = 60+110+150 = 320

Total demand = 65+85+80+70 = 300

Since the total demand is less than total supply, add one fictitious ware

house, Wh5 with a demand of 20.

Thus, m =3; n = 5



The modified table is shown below



Objective function is to minimize the total cost of transportation from all combinations

Minimize f = 523 x11 + 682 x12 + 458 x13+ 850 x14 + 0 x15 + 420 x21

+ 412 x22 + 362 x23 + 729 x24 + 0 x25 + 670 x31

+ 558 x32 + 895 x33 +695 x34 + 0 x35


x11 + x12 + x13 + x14+ x15 = 60 for i = 1

x21 + x22 + x23 + x24+ x25 = 110 for i = 2

x31 + x32 + x33 + x34 + x35 = 150 for i = 3



x11 + x21+ x31 = 65 for j = 1

x12 + x22+ x32 = 85 for j = 2

x13 + x23 + x23 = 90 for j = 3

x14 + x24 + x24 = 80 for j = 4

x15 + x25 + x25 =20 for j = 5

xij i = 1, 2,3; j=1, 2, 3,4

This optimization problem can be solved using the same procedure used for

the previous problem.

0≥


Thank You

Optimization Methods: Linear Programming Applications – Assignment Problem 1


Assignment Problem

Introduction

In the previous lecture, we discussed about one of the bench mark problems called

transportation problem and its formulation. The assignment problem is a particular class of

transportation linear programming problems with the supplies and demands equal to

integers (often 1). Since all supplies, demands, and bounds on variables are integers, the

assignment problem relies on an interesting property of transportation problems that the

optimal solution will be entirely integers. In this lecture, the structure and formulation of

assignment problem are discussed. Also, traveling salesman problem, which is a special

type of assignment problem, is described.

Structure of assignment problem

As mentioned earlier, assignment problem is a special type of transportation problem in

which

1. Number of supply and demand nodes are equal.

2. Supply from every supply node is one.

3. Every demand node has a demand of one.

4. Solution is required to be all integers.

The goal of a general assignment problem is to find an optimal assignment of machines

(laborers) to jobs without assigning an agent more than once and ensuring that all jobs are

completed. The objective might be to minimize the total time to complete a set of jobs, or

to maximize skill ratings, maximize the total satisfaction of the group or to minimize the

cost of the assignments. This is subjected to the following requirements:



1. Each machine is assigned not more than one job.

2. Each job is assigned to exactly one machine.

Formulation of assignment problem

Consider m laborers to whom n tasks are assigned. No laborer can either sit idle or do

more than one task. Every pair of person and assigned work has a rating. This rating may

be cost, satisfaction, penalty involved or time taken to finish the job. There will be N2 such

combinations of persons and jobs assigned. Thus, the optimization problem is to find such

man- job combinations that optimize the sum of ratings among all.

The formulation of this problem as a special case of transportation problem can be

represented by treating laborers as sources and the tasks as destinations. The supply

available at each source is 1 and the demand required at each destination is 1.The cost of

assigning (transporting) laborer i to task j is cij.

It is necessary to first balance this problem by adding a dummy laborer or task depending

on whether m<n or m>n, respectively. The cost coefficient cij for this dummy will be zero.

Let ⎪⎩

⎪⎨⎧

=machineithetoassignedisjobjtheif

machineithetoassignednotisjobjtheifx

thth

thth

ij ,1,0

Thus the above model can be expressed as

∑∑= =

m

i

n

jijij xcMinimize

1 1

Since each task is assigned to exactly one laborer and each laborer is assigned only one

job, the constraints are



1

1

1 1,2,...

1 1,2,...

n

ijin

ijj

x for j n

x for i m

=

=

= =

= =

∑

∑

10 orxij =

Due to the special structure of the assignment problem, the solution can be found out using

a more convenient method called Hungarian method which will be illustrated through an

example below.

Example 1: (Taha, 1982)

Consider three jobs to be assigned to three machines. The cost for each combination is

shown in the table below. Determine the minimal job – machine combinations.

Table 1

Machine Job

1 2 3 ai

1 5 7 9 1

2 14 10 12 1

3 15 13 16 1

bj 1 1 1

Solution:

Step 1:

Create zero elements in the cost matrix (zero assignment) by subtracting the smallest

element in each row (column) from the corresponding row (column). After this exercise,

the resulting cost matrix is obtained by subtracting 5 from row 1, 10 from row 2 and 13

from row 3.



Table 2

1 2 3

1 0 2 4

2 4 0 2

3 2 0 3

Step 2:

Repeating the same with columns, the final cost matrix is

Table 3

1 2 3

1 0 2 2

2 4 0 0

3 2 0 3

The italicized zero elements represent a feasible solution. Thus the optimal assignment is

(1,1), (2,3) and (3,2). The total cost is equal to 60 (5 +12+13).

In the above example, it was possible to obtain the feasible assignment. But in more

complicated problems, additional rules are required which are explained in the next

example.

Example 2 (Taha, 1982)

Consider four jobs to be assigned to four machines. The cost for each combination is

shown in the table below. Determine the minimal job – machine combinations.



Table 4

Machine Job

1 2 3 4 ai

1 1 4 6 3 1

2 8 7 10 9 1

3 4 5 11 7 1

4 6 7 8 5 1

bj 1 1 1 1

Solution:

Step 1: Create zero elements in the cost matrix by subtracting the smallest element in each

row from the corresponding row.

Table 5

1 2 3 4

1 0 3 5 2

2 1 0 3 2

3 0 1 7 3

4 1 2 3 0

Step 2: Repeating the same with columns, the final cost matrix is

Table 6

1 2 3 4

1 0 3 2 2

2 1 0 0 2

3 0 1 4 3

4 1 2 0 0



Rows 1 and 3 have only one zero element. Both of these are in column 1, which means

that both jobs 1 and 3 should be assigned to machine 1. As one machine can be assigned

with only one job, a feasible assignment to the zero elements is not possible as in the

previous example.

Step 3: Draw a minimum number of lines through some of the rows and columns so that

all the zeros are crossed out.

Table 7

1 2 3 4

1 0 3 2 2

2 1 0 0 2

3 0 1 4 3

4 1 2 0 0

Step 4: Select the smallest uncrossed element (which is 1 here). Subtract it from every

uncrossed element and also add it to every element at the intersection of the two lines.

This will give the following table.

Table 8

1 2 3 4

1 0 2 1 1

2 2 0 0 2

3 0 0 3 2

4 2 2 0 0

This gives a feasible assignment (1,1), (2,3), (3,2) and (4,4) with a total cost of 1+10+5+5

= 21.

If the optimal solution had not been obtained in the last step, then the procedure of

drawing lines has to be repeated until a feasible solution is achieved.



Formulation of Traveling Salesman Problem (TSP) as an Assignment Problem

A traveling salesman has to visit n cities and return to the starting point. He has to start

from any one city and visit each city only once. Suppose he starts from the kth city and the

last city he visited is m. Let the cost of travel from ith city to jth city be cij. Then the

objective function is

∑∑= =

m

i

n

jijij xcMinimize

1 1


1

1

1 1,2,... , ,

1 1,2,... , ,

10 1

n

ijin

ijj

mk

ij

x for j n i j i m

x for i m i j i m

xx or

=

=

= = ≠

= = ≠

==

∑

∑

≠

≠

Solution Procedure:

Solve the problem as an assignment problem using the method used to solve the above

examples. If the solutions thus found out are cyclic in nature, then that is the final solution.

If it is not cyclic, then select the lowest entry in the table (other than zero). Delete the row

and column of this lowest entry and again do the zero assignment in the remaining matrix.

Check whether cyclic assignment is available. If not, include the next higher entry in the

table and the procedure is repeated until a cyclic assignment is obtained.

The procedure is explained through an example below.



Example 3:

Consider a four city TSP for which the cost between the city pairs are as shown in the

figure below. Find the tour of the salesman so that the cost of travel is minimal.

2

6 8 4

5 1 4

9 9

3

Table 9

1 2 3 4

1 ∞ 4 9 5

2 6 ∞ 4 8

3 9 4 ∞ 9

4 5 8 9 ∞

Solution:

Step 1: The optimal solution after using the Hungarian method is shown below.

Table 10

1 2 3 4

1 ∞ 0 5 0

2 2 ∞ 0 3

3 5 0 ∞ 4

4 0 3 4 ∞



The optimal assignment is 1→ 4, 2→ 3, 3→ 2, 4→ 1 which is not cyclic.

Step 2: Consider the lowest entry ‘2’ of the cell (2,1). If there is a tie in selecting the

lowest entry, then break the tie arbitrarily. Delete the 2nd row and 1st column. Do the zero

assignment in the remaining matrix. The resulting table is

Table 11

1 2 3 4

1 ∞ 0 4 0

2 2 ∞ 0 3

3 5 0 ∞ 4

4 0 0 0 ∞

Thus the next optimal assignment is 1→ 4, 2→1, 3→ 2, 4→ 3 which is cyclic. Thus the

required tour is 1→ 4→3→ 2→ 1 and the total travel cost is 5 + 9 + 4 + 6 = 24.




Assignment Problem


Introduction

Assignment problem is a particular class of transportation

linear programming problems

Supplies and demands will be integers (often 1)

Traveling salesman problem is a special type of

assignment problem


Objectives

To structure and formulate a basic assignment problem

To demonstrate the formulation and solution with a

numerical example

To formulate and solve traveling salesman problem as an

assignment problem


Structure of Assignment Problem

Assignment problem is a special type of transportation problem in

which

Number of supply and demand nodes are equal.

Supply from every supply node is one.

Every demand node has a demand for one.

Solution is required to be all integers.


Structure of Assignment Problem …contd.

Goal of an general assignment problem: Find an optimal assignment of machines (laborers) to jobs without assigning an agent more than once and ensuring that all jobs are completed

The objective might be to minimize the total time to complete a set of jobs, or to maximize skill ratings, maximize the total satisfaction of the group or to minimize the cost of the assignments

This is subjected to the following requirements:

Each machine is assigned no more than one job.

Each job is assigned to exactly one machine.


Formulation of Assignment Problem

Consider m laborers to whom n tasks are assigned

No laborer can either sit idle or do more than one task

Every pair of person and assigned work has a rating

Rating may be cost, satisfaction, penalty involved or time taken to finish the job

N2 such combinations of persons and jobs assigned

Optimization problem: Find such job-man combinations that optimize the sum of ratings among all.


Formulation of Assignment Problem …contd.

Representation of this problem as a special case of transportation problem

laborers as sources

tasks as destinations

Supply available at each source is 1

Demand required at each destination is 1

Cost of assigning (transporting) laborer i to task j is cij.

It is necessary to first balance this problem by adding a dummy laborer or task depending on whether m < n or m > n, respectively

Cost coefficient cij for this dummy will be zero.



Let xij be the decision variable

The objective function is

⎪⎩

⎪⎨⎧

=machineithetoassignedisjobjtheif

machineithetoassignednotisjobjtheifx

thth

thth

ij ,1,0

∑∑= =

m

i

n

jijij xcMinimize

1 1



Since each task is assigned to exactly one laborer and each laborer is

assigned only one job, the constraints are

Due to the special structure of the assignment problem, the solution can be found out using a more convenient method called Hungarian method.

1

1

1 1, 2,...

1 1, 2,...

n

ijin

ijj

x for j n

x for i m

=

=

= =

= =

∑

∑10 orxij =


Example (1)

Consider three jobs to be assigned to three machines. The cost for each combination is shown in the table below. Determine the minimal job – machine combinations



Solution:Step 1:

Create zero elements in the cost matrix (zero assignment) by subtracting the smallest element in each row (column) from the corresponding row (column). Considering the rows first, the resulting cost matrix is obtained by subtracting 5 from row 1, 10 from row 2 and 13 from row 3



Step 2:Repeating the same with columns, the final cost matrix is

The italicized zero elements represent a feasible solutionThus the optimal assignment is (1,1), (2,3) and (3,2)The total cost is equal to (5 +12+13) = 60


Example (2)

In the above example, it was possible to obtain the feasible assignment

But in more complicate problems, additional rules are required which are

explained in the next example.

Example 2 (Taha, 1982)

Consider four jobs to be assigned to four machines. Determine the minimal

job – machine combinations.



The cost for each combination is shown in the table below



Solution:

Step 1: Create zero elements in the cost matrix by subtracting the

smallest element in each row from the corresponding row.



Step 2: Repeating the same with columns, the final cost matrix is

Rows 1 and 3 have only one zero element

Both of these are in column 1, which means that both jobs 1 and 3 should be assigned to machine 1



As one machine can be assigned with only one job, a feasible assignment to the zero elements is not as in the previous exampleStep 3: Draw a minimum number of lines through some of the rows and columns so that all the zeros are crossed out



Step 4: Select the smallest uncrossed element (which is 1 here).Subtract it from every uncrossed element and also add it to every element at the intersection of the two lines. This will give the following table



This gives a feasible assignment (1,1), (2,3), (3,2) and (4,4)And the total cost is 1+10+5+5 = 21.If the optimal solution had not been obtained in the last step, then the procedure of drawing lines has to be repeated until a feasible solution is achieved.


Formulation of Traveling Salesman Problem (TSP) as an Assignment Problem

A traveling salesman has to visit n cities and return to the starting pointHe has to start from any one city and visit each city only once.Suppose he starts from the kth city and the last city he visited is mLet the cost of travel from ith city to jth city be cij.Then the objective function is

∑∑= =

m

i

n

jijij xcMinimize

1 1


Formulation of Traveling Salesman Problem (TSP) as an Assignment Problem …contd.


Solution Procedure: Solve the problem as an assignment problem using the method used to solve the above examplesIf the solutions thus found out are cyclic in nature, then that is the final solution

1

1

1 1,2,... , ,

1 1,2,... , ,

10 1

n

ijin

ijj

mk

ij

x for j n i j i m

x for i m i j i m

xx or

=

=

= = ≠ ≠

= = ≠ ≠

==

∑

∑


Formulation of Traveling Salesman Problem (TSP) as an Assignment Problem …contd.

Solution Procedure …contd. If it is not cyclic, then select the lowest entry in the table (other than zero)Delete the row and column of this lowest entry and again do the zero assignment in the remaining matrixCheck whether cyclic assignment is availableIf not, include the next higher entry in the table and the procedure is repeated until a cyclic assignment is obtained.


Traveling Salesman Problem (TSP) - Example

Consider a four city TSP for which the cost between the city pairs are as shown in the figure below. Find the tour of the salesman so that the cost of travel is minimal.

Cost matrix


Traveling Salesman Problem (TSP) – Example …contd.

Solution:Step 1: The optimal solution after using the Hungarian method isshown below.

The optimal assignment is 1→ 4, 2→ 3, 3→ 2, 4→ 1 which is not cyclic



Step 2:Consider the lowest entry ‘2’ of the cell (2,1)If there is a tie in selecting the lowest entry, then break the tie arbitrarilyDelete the 2nd row and 1st columnDo the zero assignment in the remaining matrix



The resulting table is

Next optimal assignment is 1→ 4, 2→1, 3→ 2, 4→ 3 which is cyclicRequired tour is 1→ 4→3→ 2→ 1 Optimal total travel cost is 5 + 9 + 4 + 6 = 24


Thank You

Optimization Methods: Linear Programming Applications – Structural & Water Resources Problems

1


Structural & Water Resources Problems

Introduction

In the previous lectures, some of the bench mark problems which use LP were discussed. LP

has been applied to formulate and solve several types of problems in engineering field also.

LP finds many applications in the field of water resources and structural design which include

many types like planning of urban water distribution, reservoir operation, crop water

allocation, minimizing the cost and amount of materials in structural design. In this lecture,

applications of LP in the plastic design of frame structures and also in deciding the optimal

irrigation allocation and water quality management are discussed.

Typical Example – Structural Design

(1) A beam column arrangement of a rigid frame is shown below. Moment in beam is

represented by Mb and moment in column is denoted by Mc. l = 8 units and h= 6 units and

forces F1=2 units and F2=1 unit. Assuming that plastic moment capacity of beam and

columns are linear functions of their weights; the objective function is to minimize the sum of

weights of the beam and column materials.

Fig. 1

F2=1

F1=2 3 4 52 6

h=6

1 7l=8l=8



2

Solution:

In the plastic limit design, it is assumed that at the points of peak moments, plastic hinges

will be developed. The points of development of peak moments are numbered in the above

figure from 1 through 7. The development of sufficient hinges makes the structure unstable

known as a collapse mechanism. Thus, for the design to be safe the energy absorbing

capacity of the frame (U) should be greater than the energy imparted by externally applied

load (E) for the various collapse mechanisms of the structure.

The objective function can be written as

Minimize f = weight of beam + weight of column

i.e. (2 2b )cf w lM hM= + (1)

where w is weight per unit length over unit moment in material. Since w is constant,

optimizing (1) is same as optimizing

( )2 216 12

b

b c

cf lM hMM M

= +

= + (2)

The four possible collapse mechanisms are shown in the figure below with the corresponding

U and E values. Constraints are formulated from the design restriction U for all the

mechanisms.

E≥

(a) (b)



3

(b) (d) Fig. 2

Hence, the optimization problem can be stated as

16 12b cMinimize f M M= + (2)

subject to

(3) 3≥cM

(4) 2≥bM

(5) 102 ≥+ cb MM

(6) 6≥+ cb MM

, 0≥bM 0≥cM (7)

Writing problem in standard format

16 12b cMinimize f M M= +



4

subject to -Mc ≤ -3

-Mb ≤ -2

-2Mb - M c ≤ -10

-Mb -Mc ≤ -6

Introducing slack variables X1, X2, X3, X4 all , the system of equations can be written in

canonical form as

0≥

-Mc+X1 = - 3

-Mb+ X2 = - 2

-2Mb -Mc +X3 = - 10

-Mb -Mc +X4 = - 6

16MB + 12MC – f = 0

This model can be solved using Dual Simplex algorithm which is explained below

Starting Solution:

Basic

Variables

Variables

MB MC X1 X2 X3 X4

br

f -16 -12 0 0 0 0 0

X1 0 -1 1 0 0 0 -3

X2 -1 0 0 1 0 0 -2

X3 -2 -1 0 0 1 0 -10

X4 -1 -1 0 0 0 1 -6

Ratio 8 12



5

Iteration 1:

Basic

Variables Variables

MB MC X1 X2 X3 X4

br

f 0 -4 0 0 -8 0 80

X1 0 -1 1 0 0 0 -3

X2 0 ½ 0 1 -½ 0 3

MB 1 ½ 0 0 -½ 0 5

X4 0 -½ 0 0 -½ 1 -1

Ratio 4

Iteration 2:

Basic

Variables Variables

MB MC X1 X2 X3 X4

br

f 0 0 -4 0 -8 0 92

MC 0 1 -1 0 0 0 3

X2 0 0 ½ 1 -½ 0 3/2

MB 1 0 ½ 0 -½ 0 7/2

X4 0 0 -½ 0 -½ 1 1

Ratio

The optimal value of decision variables are

MB =7/2; MC =3

And the total weight of the material required f = 92w units.



6

Typical Example – Water Resources

(2) Consider two crops 1 and 2. One unit of crop 1 produces four units of profit and one

unit of crop 2 brings five units of profit. The demand of production of crop 1 is A units and

that of crop 2 is B units. Let x be the amount of water required for A units of crop 1 and y be

the same for B units of crop 2. The amount of production and the amount of water required

can be expressed as a linear relation as shown below

A = 0.5(x - 2) + 2

B = 0.6(y - 3) + 3

Minimum amount of water that must be provided to 1 and 2 to meet their demand is two and

three units respectively. Maximum availability of water is ten units. Find out the optimum

pattern of irrigation.

Solution:

The objective is to maximize the profit from crop 1 and 2, which can be represented as

Maximize f = 4A + 5B;

Expressing as a function of the amount of water,

Maximize f = 4[0.5(x - 2) + 2] + 5[0.6(y - 3) + 3]= 2x + 3y + 10

subject to

; Maximum availability of water 10≤+ yx

; Minimum amount of water required for crop 1 2≥x

; Minimum amount of water required for crop 2 3≥y

The above problem is same as maximizing f’ = 2x + 3y subject to same constraints.



7

Changing the problem into standard form by introducing slack variables S1, S2, S3

Maximize f’ = 2x + 3y

subject to

x + y + S1 =10

-x + S2 = -2

-y + S3 = -3

This model is solved by forming the simplex table as below

Starting Solution:

Variables Basic Variables x y S1 S2 S3

RHS Ratio

f’ -2 -3 0 0 0 0

S1 1 1 1 0 0 10 10

S2 -1 0 0 1 0 -2 -

S3 0 -1 0 0 1 -3 3



8

Iteration 1:


RHS Ratio

f’ -2 0 0 0 -3 9 -

S1 1 0 1 0 1 7 7

S2 -1 0 0 1 0 -2 2

y 0 1 0 0 -1 3 -

Iteration 2:


RHS Ratio

f’ 0 0 0 -2 -3 13 -

S1 0 0 1 1 1 5 5

x 1 0 0 -1 0 2 -

y 0 1 0 0 -1 3 -3

Iteration 3:


RHS Ratio

f’ 0 0 3 1 0 28 -

S3 0 0 1 1 1 5 -

x 1 0 0 -1 0 2 -

y 0 1 1 1 0 8 -



9

Hence the solution is

x = 2; y = 8; f’ = 28

Therefore, f = 28+10 = 38

Thus, water allocated to crop A is 2 units and to crop B is 8 units and total profit yielded is 38

units.

Typical Example – Water Quality Management

Waste load allocation for water quality management in a river system can be defined as

determination of optimal treatment level of waste, which is discharged to a river; such that

the water quality standards set by Pollution Control Agency (PCA) are maintained through

out the river. Conventional waste load allocation involves minimization of treatment cost

subject to the constraint that the water quality standards are not violated.

Consider a simple problem, where, there are M dischargers, who discharge waste into the

river, and I checkpoints, where the water quality is measured by PCA. Let xj is the treatment

level and aj is the unit treatment cost for jth discharger (j=1,2,…,M). ci is the dissolved oxygen

(DO) concentration at checkpoint i (i=1,2,…,I), which is to be controlled. Therefore the

decision variables for the waste load allocation model are xj (j=1,2,…,M).

Thus, the objective function can be expressed as

1

M

j jj

Maximize f a x=

=∑

The relationship between the water quality indicator, ci (DO) at a checkpoint and the

treatment level upstream to that checkpoint is linear (based on Streeter-Phelps Equation)

when all other parameters involved in water quality simulations are constant. Let g(x) denotes

the linear relationship between ci and xj. Then,

( ) ,i jc g x i= ∀ j



10

Let cP be the permissible DO level set by PCA, which is to be maintained through out the

river. Therefore,

i Pc c≥ ∀i

Solution of the optimization model using simplex algorithm gives the optimal fractional

removal levels required to maintain the water quality of the river.




Structural & Water Resources Problems


Introduction

LP has been applied to formulate and solve several types of problems in engineering field

LP finds many applications in the field of water resources and structural design which include

Planning of urban water distribution

Reservoir operation

Crop water allocation

Minimizing the cost and amount of materials in structural design


Objectives

To discuss the applications of LP in the plastic design of frame

structures

To discuss the applications of LP in deciding the optimal pattern of

irrigation


Example – Structural Design

A beam column arrangement of a rigid frame is shown

Moment in beam is represented by Mb

Moment in column is denoted by Mc.

l = 8 units and h= 6 units

Forces F1=2 units and F2=1 unit.

Assuming that plastic moment capacity of beam and columns are linear functions of their weights; the objective function is to minimize weights of the materials.


Example - Structural Design …contd.

Solution:

In the limit design, it is assumed that at the points of peak moments, plastic hinges will be developed

Points of development of peak moments are numbered in the above figure from 1 through 7

Development of sufficient hinges makes the structure unstable known as a collapse mechanism

For the design to be safe the energy absorbing capacity of the frame (U) should be greater than the energy imparted by externally applied load (E) for the various collapse mechanisms of the structure




Minimize f = weight of beam + weight of column

where w is weight per unit length over unit moment in material

Since w is constant, optimizing (1) is same as optimizing

( )2 2 (1)b cf w lM hM= +

( )2 216 12

b c

b c

f lM hMM M

= +

= +



Four possible collapse mechanisms are shown in the figure below with the corresponding U and E values

(a) (b)



(c) (d)



The optimization problem can be stated as

subject to 16 12b cMinimize f M M= +

32

2 106

0; 0

c

b

b c

b c

b c

MM

M MM MM M

≥≥+ ≥+ ≥≥ ≥



Introducing slack variables X1, X2, X3, X4 all , the system of equations can be written in canonical form as

16MB + 12MC – f = 0

-Mc+X1 = - 3

-Mb+ X2 = - 2

-2Mb -Mc +X3 = - 10

-Mb -Mc +X4 = - 6

16MB + 12MC – f = 0



This model can be solved using Dual Simplex algorithm

The final tableau is shown below

The optimal value of decision variables areMB =7/2; MC =3

And the total weight of the material requiredf = 92w units


Example - Irrigation Allocation

Consider two crops 1 and 2. One unit of crop 1 produces four units of profit and one unit of crop 2 brings five units of profit. The demand of production of crop 1 is A units and that of crop 2 is B units. Let x be the amount of water required for A units of crop 1 and y be the same for B units of crop 2.

The amount of production and the amount of water required can beexpressed as a linear relation as shown below

A = 0.5(x - 2) + 2

B = 0.6(y - 3) + 3


Example - Irrigation Allocation …contd.

Consider two crops 1 and 2. One unit of crop 1 produces four units of profit and one unit of crop 2 brings five units of profit. The demand of production of crop 1 is A units and that of crop 2 is B units. Let x be the amount of water required for A units of crop 1 and y be the same for B units of crop 2.

The amount of production and the amount of water required can beexpressed as a linear relation as shown below

A = 0.5(x - 2) + 2

B = 0.6(y - 3) + 3



Solution:

Objective: Maximize the profit from crop 1 and 2

Maximize f = 4A + 5B;

Expressing as a function of the amount of water,

Maximize f = 4[0.5(x - 2) + 2] + 5[0.6(y - 3) + 3]

f = 2x + 3y + 10



subject to

; Maximum availability of water

; Minimum amount of water required for crop 1

; Minimum amount of water required for crop 2

The above problem is same as maximizing

f’ = 2x + 3y

subject to same constraints.

10≤+ yx

2≥x

3≥y



Changing the problem into standard form by introducing slack variables S1, S2, S3

Maximize f’ = 2x + 3y

subject to

x + y + S1 =10

-x + S2 = -2

-y + S3 = -3

This model is solved using simplex method



The final tableau

is as shown

The solution is x = 2; y = 8; f’ = 28

Therefore, f = 28+10 = 38

Water allocated to crop A is 2 units and to crop B is 8 units and total profit yielded is 38 units.


Example – Water Quality Management

Waste load allocation for water quality management in a river system can be defined as

Determination of optimal treatment level of waste, which is discharged to a river

By maintaining the water quality standards set by Pollution Control Agency (PCA), through out the river

Conventional waste load allocation involves minimization of treatment cost subject to the constraint that the water quality standards are not violated


Example - Waster Quality Management …contd.

Consider a simple problem of M dischargers, who discharge waste into the river, and I checkpoints, where the water quality is measured by PCA

Let xj is the treatment level and aj is the unit treatment cost for jth

discharger (j=1,2,…,M)

ci is the dissolved oxygen (DO) concentration at checkpoint i (i=1,2,…,I), which is to be controlled

Decision variables for the waste load allocation model are xj

(j=1,2,…,M).



Objective function can be expressed as

Relationship between the water quality indicator, ci (DO) at a checkpoint and the treatment level upstream to that checkpoint is linear (based on Streeter-Phelps Equation)

Let g(x) denotes the linear relationship between ci and xj.

Then,

1

M

j jj

Maximize f a x=

=∑

( ) ,i jc g x i j= ∀



Let cP be the permissible DO level set by PCA, which is to be maintained through out the river

Therefore,

This model can be solved using simplex algorithm which will give the optimal fractional removal levels required to maintain the water quality of the river

i Pc c i≥ ∀


Thank You

Optimization Methods: Dynamic Programming - Learning Objectives

Module 5: Dynamic Programming

Learning Objectives

It was discussed in modules 3 and 4 that most widely used optimization method is linear

programming technique. But this technique is not flexible in some complex problems and is

only restricted to linear optimization problems. Problems having a sequential optimization

nature can be solved easily using dynamic programming (DP), which is the main objective of

this module.

This module gives a brief description of sequential optimization and multistage decision

processes. The representation of multistage decision processes is given for a general problem

having N decision variables. This will enable the reader to formulate any multistage decision

problems. This will be followed by an explanation about the basic concepts on which a

problem is solved using dynamic programming. Formulation of recursive equations will be

discussed for a general problem. This will help the reader to develop recursive equations

using both backward and forward approach for any type of sequential optimization problem.

The computational procedure for DP will be discussed for the same general problem. At the

end, continuous DP, multiple state variable and curse of dimensionality will be described.


1. Represent a multistage decision process

2. Conceptualize the principle of optimality and sub-optimization

3. Formulate recursive equations for a DP

4. Acquire an idea about the computational procedure of DP

5. Differentiate between discrete DP and continuous DP

6. Deal with multiple state variables in DP


Optimization Methods: Dynamic Programming - Introduction 1


Introduction

Introduction

In some complex problems, it will be advisable to approach the problem in a sequential

manner in order to find the solution quickly. The solution is found out in multi stages. This is

the basic approach behind dynamic programming. It works in a “divide and conquer”

manner. The word "programming" in "dynamic programming" has no particular connection

to computer programming at all. A program is, instead, the plan for action that is produced. In

this lecture, the multistage decision process, its representation, various types and the concept

of sub-optimization and principle of optimality are discussed.

Sequential optimization

In sequential optimization, a problem is approached by dividing it into smaller subproblems

and optimization is done for these subproblems without losing the integrity of the original

problem. Sequential decision problems are those in which decisions are made in multiple

stages. These are also called multistage decision problems since decisions are made at a

number of stages.

In multistage decision problems, an N variable problem is represented by N single variable

problems. These problems are solved successively such that the optimal value of the original

problem can be obtained from the optimal solutions of these N single variable problems. The

N single variable problems are connected in series so that the output of one stage will be the

input to the succeeding stage. This type of problem is called serial multistage decision

process.

For example, consider a water allocation problem to N users. The objective function is to

maximize the total net benefit from all users. This problem can be solved by considering each

user separately and optimizing the individual net benefits, subject to constraints and then

adding up the benefits from all users to get the total optimal benefit.



Representation of multistage decision process

Consider a single stage decision process as shown in the figure below.

Net Benefits, NB1

Stage 1 Input S1 Output S2

Decision variable, X1 Fig 1.

Let S1 be the input state variable, S2 be the output state variable, X1 be the decision variable

and NB1 be the net benefits. The input and output are related by a transformation function

expressed as,

S2 = g(X1, S1)

Also since the net benefits are influenced by the decision variables and also the input

variable, the benefit function can be expressed as

NB1 = h(X1, S1)

Now, consider a serial multistage decision process consisting of T stages as shown in the

figure below.

Stage 1 Stage t Stage T St

NBt

Xt

St+1 S1 S2

NB1 NBT

X1 XT

ST ST+1

Fig 2.



Here, for the tth stage the, state transformation and the benefit functions are written as,

St+1 = g(Xt, St)

NBt = h(Xt, St)

The objective of this multistage problem is to find the optimum values of all decision

variables X1, X2,…, XT such that the individual net benefits of each stage that is expressed by

some objective function, f(NBt) and the total net benefit which is expressed by f(NB1,

NB2,…, NBT) should be maximized. The application of dynamic programming to a multistage

problem depends on the nature of this objective function i.e., the objective function should be

separable and monotonic. An objective function is separable, if it can be decomposed and

expressed as a sum or product of individual net benefits of each stage, i.e.,

either ( )∑ ∑= =

==T

t

T

tttt SXhNBf

1 1,

or ( )∏∏==

==T

ttt

T

tt SXhNBf

11

,

An objective function is monotonic if for all values of a and b for which the value of the

benefit function is ( ) ( )tttt SbxhSaxh ,, =≥= , then

( ) ( )1,211,21 ,,...,...,,,,...,...,, ++ =≥= tTttTt SxbxxxfSxaxxxf

should be satisfied.



Types of multistage decision problems

A serial multistage problem such as shown, can be classified into three categories as initial

value problem, final value problem and boundary value problem.

1. Initial value problem: In this type, the value of the initial state variable, S1 is given.

2. Final value problem: In this, the value of the final state variable, ST is given. A final

value problem can be transformed into an initial value problem by reversing the

procedure of computation of the state variable, St.

3. Boundary value problem: In this, the values of both the initial and final state

variables, S1 and ST are given.

Concept of sub-optimization and principle of optimality

Bellman (1957) stated the principle of optimality which explains the process of suboptimality

as:

“An optimal policy (or a set of decisions) has the property that whatever the initial state and

initial decision are, the remaining decisions must constitute an optimal policy with regard to

the state resulting from the first decision.”

Consider the objective function consisting of T decision variables x1, x2, …, xT,

( )∑ ∑= =

==T

t

T

tttt SXhNBf

1 1,

and satisfying the equations,

St+1 = g(Xt, St)

NBt = h(Xt, St) for t = 1,2,…,T

The concepts of suboptimization and principle of optimality are used to solve this problem

through dynamic programming. To explain these concepts, consider the design of a water

tank in which the cost of construction is to be minimized. The capacity of the tank to be

designed is given as K.



The main components of a water tank include (i) tank (ii) columns to support the tank and

(iii) the foundation. While optimizing this problem to minimize the cost, it would be

advisable to break this system into individual parts and optimizing each part separately

instead of considering the system as a whole together. However, while breaking and doing

suboptimization, a logical procedure should be used; otherwise this approach can lead to a

poor solution. For example, consider the suboptimization of columns without considering the

other two components. In order to reduce the construction cost of columns, one may use

heavy concrete columns with less reinforcement, since the cost of steel is high. But while

considering the suboptimization of foundation component, the cost becomes higher as the

foundation should be strong enough to carry these heavy columns. Thus, the suboptimization

of columns before considering the suboptimization of foundation will adversely affect the

overall design.

In most of the serial systems as discussed above, since the suboptimization of last component

does not influence the other components, it can be suboptimized independently. For the

above problem, foundation can thus be suboptimized independently. Then the last two

components (columns and foundation) are considered as a single component and

suboptimization is done without affecting other components. This process can be repeated for

any number of end components. The process of suboptimization for the above problem is

shown in the next page.



Tank Columns Foundation

Original System


Suboptimize design of Foundation component

Fig 3.



Suboptimize design of Foundation & Columns together

Optimize complete system



Dynamic Programming

Introduction



IntroductionComplex problems are sometimes solved quickly if approached in asequential mannerDynamic Programming : Sequential or multistage decision making processBasic approach behind dynamic programming: Solution is found out in multi stagesWorks in a “divide and conquer” manner

ObjectivesTo discuss the Sequential Optimization an Multistage Decision ProcessTo represent a Multistage Decision Process To describe the concept of sub-optimization and principle of optimality


Sequential Optimization

Problem is divided into smaller sub-problemsOptimize these sub-problems without losing the integrity of the original problemDecisions are made sequentiallyAlso called multistage decision problems since decisions are made at a number of stages A N variable problem is represented by N single variable problemsThese problems are solved successively to get the optimal value of the original problem


Sequential Optimization …contd.

Serial multistage decision process: Single variable problems are connected in series so that the output of one stage will be the input to the succeeding stageFor example, consider a water allocation problem to N usersThe objective function is to maximize the total net benefit fromall usersThe problem can be solved by considering each user separately and optimizing the individual net benefits, subject to constraints and then adding up the benefits from all users to get the total optimal benefit


Representation of Multistage Decision Process

Consider a single stage decision processHere

S1 is the input state variableS2 is the output state variableX1 is the decision variables and NB1 is the net benefit

The transformation function for the input and output isS2 = g(X1, S1)

Net benefits are expressed as a function of decision variables and input variable

NB1 = h(X1, S1)

Stage 1

Input S1

Net Benefit, NB1

Decision variable, X1

Output S2


Representation of Multistage Decision Process …contd.

Now, consider a serial multistage decision process consisting of T stages

For the tth stage the stage transformation and the benefit functions are St+1 = g(Xt, St)NBt = h(Xt, St)

Stage 1 Stage t Stage TSt

NBt

Xt

St+1S1 S2

NB1 NBT

X1 XT

ST ST+1



Objective of this multistage problem is to find the optimum values of all decision variables X1, X2,…, XT such that the individual net benefits of each stage that is expressed by some objective function, f(NBt) and the total net benefit which is expressed by f(NB1, NB2,…, NBT) should be maximizedDynamic programming can be applied to this multistage problem if the objective function is separable and monotonicAn objective function is separable, if it can be decomposed and expressed as a sum or product of individual net benefits of each stage

or

( )∑ ∑= =

==T

t

T

tttt SXhNBf

1 1,

( )∏∏==

==T

ttt

T

tt SXhNBf

11

,



An objective function is monotonic if for all values of a and b for which the value of the benefit function is ,

then,

is satisfied

( ) ( )tttt SbxhSaxh ,, =≥=

( ) ( )1,211,21 ,,...,...,,,,...,...,, ++ =≥= tTttTt SxbxxxfSxaxxxf


Types of Multistage Decision Process

A serial multistage problem can be divided into three categoriesInitial value problem

Value of the initial state variable, S1 is given

Final value problem Value of the final state variable, ST is givenFinal value problem can be transformed into an initial value problem by reversing the direction of the state variable, St

Boundary value problem.Values of both the initial and final state variables, S1 and STare given


Concept of Sub-Optimization

Consider the objective function consisting of T decision variables x1, x2, …, xT

satisfying the equations, St+1 = g(Xt, St)NBt = h(Xt, St) for t = 1,2,…,T

To solve this problem through dynamic programming, the concepts of sub-optimization and principle of optimality are used

( )∑ ∑= =

==T

t

T

tttt SXhNBf

1 1

,


Concept of Sub-Optimization …contd.

Consider the design of a water tank in which the cost of construction is to be minimized

Capacity of the tank to be designed is K

Main components of a water tank include (i) tank (ii) columns tosupport the tank and (iii) the foundation

Optimization can be done by breaking the system into individual parts and optimizing each part separately

While breaking and sub-optimizing, a logical procedure should be used; otherwise this approach can lead to a poor solution



For example, sub-optimization of columns without considering the other two components may cause the use of heavy concrete columns with less reinforcement, since the cost of steel is high

But while considering the sub-optimization of foundation component, the cost becomes higher as the foundation should be strong enough to carry these heavy columns

Thus, the sub-optimization of columns before considering the sub-optimization of foundation will adversely affect the overall design



In most of the serial systems sub-optimization can start from the last component (or first component) since it does not influence the other components

Thus, for the above problem, foundation can be suboptimizedindependently

Then, the last two components (columns and foundation) are considered as a single component and sub-optimization is done without affecting other components

This process can be repeated for any number of end components


Process of sub-optimizationTank Columns Foundation

Original System


Suboptimize design of Foundation component


Optimize complete system

Suboptimize design of Foundation & Columns together



Principle of Optimality

Belman stated the principle of optimality which

explains the process of multi stage optimization as:“An optimal policy (or a set of decisions) has the property that

whatever the initial state and initial decision are, the remaining

decisions must constitute an optimal policy with regard to the state

resulting from the first decision.”


Thank You

Optimization Methods: Dynamic Programming – Recursive Equations 1


Recursive Equations

Introduction

In the previous lecture, we have seen how to represent a multistage decision process and also

the concept of suboptimization. In order to solve this problem in sequence, we make use of

recursive equations. These equations are fundamental to the dynamic programming. In this

lecture, we will learn how to formulate recursive equations for a multistage decision process

in a backward manner and also in a forward manner.

Recursive equations

Recursive equations are used to structure a multistage decision problem as a sequential

process. Each recursive equation represents a stage at which a decision is required. In this, a

series of equations are successively solved, each equation depending on the output values of

the previous equations. Thus, through recursion, a multistage problem is solved by breaking it

into a number of single stage problems. A multistage problem can be approached in a

backward manner or in a forward manner.

Backward recursion

In this, the problem is solved by writing equations first for the final stage and then proceeding

backwards to the first stages. Consider the serial multistage problem discussed in the previous

lecture.


NBt

Xt

S1 S2

NB1 NBT

X1 XT

ST ST+1 St+1



Suppose the objective function for this problem is

( )

( ) ( ) ( ) ( ) ( ) )1...(,,...,...,,

,

111222111

1 1

TTTTTTttt

T

t

T

ttttt

SXhSXhSXhSXhSXh

SXhNBf

++++++=

==

−−−

= =∑ ∑

and the relation between the stage variables and decision variables are gives as

St+1 = g(Xt, St), t = 1,2,…, T. …(2)

Consider the final stage as the first subproblem. The input variable to this stage is ST.

According to the principle of optimality, no matter what happens in other stages, the decision

variable XT should be selected such that ( )TTT SXh ,∗

Tf

is optimum for the input ST. Let the

optimum value be denoted as . Then,

( )[ ] )3...(,)( TTTX

TT SXhoptSfT

=∗

Next, group the last two stages together as the second subproblem. Let be the optimum

objective value of this subproblem. Then, we have

∗−1Tf

( ) ( )[ ] )4...(,,)( 111,

111

TTTTTTXX

TT SXhSXhoptSfTT

+= −−−−∗−

−

From the principle of optimality, the value of should be to optimize for a given .

For obtaining , we need and . Thus, can be written as,

TX Th TS

TS 1−TS 1−TX )( 11 −∗− TT Sf

( )[ ] )5...()(,)( 111111

TTTTTX

TT SfSXhoptSfT

∗−−−−

∗− +=

−

By using the stage transformation equation, can be rewritten as, )( 11 −∗− TT Sf

( ) ( )( )[ ] )6...(,,)( 111111111

−−−∗

−−−−∗− +=

−

TTTTTTTX

TT SXgfSXhoptSfT



Thus, here the optimum is determined by choosing the decision variable for a given

input . Eqn (4) which is a multivariate problem (second sub problem) is divided into two

single variable problems as shown in eqns (3) and (6). In general, the i+1

1−TX

1−TSth subproblem (T-ith

stage) can be expressed as,

( ) ( ) ( )[ ] )7...(,,...,)( 111,,..., 1

TTTTTTiTiTiTXXX

iTiT SXhSXhSXhoptSfTTiT

+++= −−−−−−−∗−

−−

Converting this to a single variable problem,

( ) ( )( )[ ] )8...(,,)( )1( iTiTiTiTiTiTiTX

iTiT SXgfSXhoptSfiT

−−−∗

−−−−−−∗− +=

−

where denotes the optimal value of the objective function for the last i stages. Thus

for backward recursion, the principle of optimality can be stated as, no matter in what state of

stage one may be, in order for a policy to be optimal, one must proceed from that state and

stage in an optimal manner.

∗−− )1(iTf

Forward recursion

In this approach, the problem is solved by starting from the stage 1 and proceeding towards

the last stage. Consider the serial multistage problem with the objective function as given

below

( )

T T

( ) ( ) ( ) ( ) ( ) )9...(,,...,...,,

,

111222111

1 1

TTTTTTttt

t ttttt

SXhSXhSXhSXhSXh

SXhNBf

++++++=

==

−−−

= =∑ ∑

and the relation between the stage variables and decision variables are gives as

( ) TtSXgS ttt ,....,2,1, 11 =′= ++ …(10)

where St is the input available to the stages 1 to t.

Consider stage 1 as the first subproblem. The input variable to this stage is S1. The decision

variable X1 should be selected such that ( )SXh 111 , is optimum for the input S1.



The optimum value can be written as ∗1f

( )[ ] )11...(,)( 111111

SXhoptSfX

=∗

Now, group the first and second stages together as the second subproblem. The objective

function for this subproblem can be expressed as, ∗2f

( ) ( )[ ] )12...(,,)( 111222,

2212

SXhSXhoptSfXX

+=∗

But for calculating the value of , we need and . Thus, 2S 1S 1X

( )[ ] )13...()(,)( 11222222

SfSXhoptSfX

∗∗ +=

By using the stage transformation equation, can be rewritten as, )( 22 Sf ∗

( ) ( )( )[ ] )14...(,,)( 2221222222

SXgfSXhoptSfX

′+= ∗∗

Thus, here through the principle of optimality the dimensionality of the problem is reduced

from two to one. In general, the ith subproblem can be expressed as,

( ) ( ) ( )[ ]111222,...,,

,,...,)(21

SXhSXhSXhoptSf iiiXXX

iii

+++=∗ …(15)

Converting this to a single variable problem,

( ) ( )( )[ ]iiiiiiiX

ii SXgfSXhoptSfi

,,)( )1( ′+= ∗−

∗ …(16)

where denotes the optimal value of the objective function for the first i stages. The

principle of optimality for forward recursion is that no matter in what state of stage one may

be, in order for a policy to be optimal, one had to get to that state and stage in an optimal

manner.

∗if



Dynamic Programming

Recursive Equations



Introduction

Recursive equations are used to solve a problem in sequence

These equations are fundamental to the dynamic programming

Objectives

To formulate recursive equations for a multistage decision process

In a backward manner and

In a forward manner


Recursive Equations

Recursive equations are used to structure a multistage decision problem as a sequential processEach recursive equation represents a stage at which a decision is requiredA series of equations are successively solved, each equation depending on the output values of the previous equationsA multistage problem is solved by breaking into a number of single stage problems through recursionApproached can be done in a backward manner or in a forward manner


Backward Recursion

A problem is solved by writing equations first for the final stage and then proceeding backwards to the first stageConsider a serial multistage problem

Let the objective function for this problem is


NBt

Xt

S1 S2

NB1 NBT

X1 XT

ST ST+1St+1

( )

( ) ( ) ( ) ( ) ( ) 1...(,,...,...,,

,

111222111

1 1

TTTTTTttt

T

t

T

ttttt

SXhSXhSXhSXhSXh

SXhNBf

++++++=

==

−−−

= =∑ ∑


Backward Recursion …contd.

The relation between the stage variables and decision variables are St+1 = g(Xt, St), t = 1,2,…, T.

Consider the final stage as the first sub-problem. The input variable to this stage is ST. Principle of optimality: XT should be selected such that is optimum for the input ST

The objective function for this stage is

Next, group the last two stages together as the second sub-problem. The objective function is

( )TTT SXh ,

∗Tf

( )[ ]TTTX

TT SXhoptSfT

,)( =∗

( ) ( )[ ]TTTTTTXX

TT SXhSXhoptSfTT

,,)( 111,

111

+= −−−−∗−

−



By using the stage transformation equation, can be rewritten as

Thus, a multivariate problem is divided into two single variableproblems as shownIn general, the i+1th sub-problem can be expressed as

Converting this to a single variable problem

)( 11 −∗− TT Sf

( ) ( )( )[ ]11111111 ,,)(1

−−−∗

−−−−∗− +=

−

TTTTTTTX

TT SXgfSXhoptSfT

( ) ( ) ( )[ ]TTTTTTiTiTiTXXX

iTiT SXhSXhSXhoptSfTTiT

,,...,)( 111,,..., 1

+++= −−−−−−−∗−

−−

( ) ( )( )[ ]iTiTiTiTiTiTiTX

iTiT SXgfSXhoptSfiT

−−−∗

−−−−−−∗− +=

−

,,)( )1(



denotes the optimal value of the objective function for the last i stages

Principle of optimality for backward recursion can be stated as,

No matter in what state of stage one may be, in order for a policy to be optimal, one must proceed from that state and stage in an optimal manner sing the stage transformation equation

∗−− )1(iTf


Forward Recursion

The problem is solved by starting from the stage 1 and proceeding towards the last stageConsider a serial multistage problem

Let the objective function for this problem is


NBt

Xt

S1 S2

NB1 NBT

X1 XT

ST ST+1St+1

( )

( ) ( ) ( ) ( ) ( ) 1...(,,...,...,,

,

111222111

1 1

TTTTTTttt

T

t

T

ttttt

SXhSXhSXhSXhSXh

SXhNBf

++++++=

==

−−−

= =∑ ∑


Forward Recursion …contd.

The relation between the stage variables and decision variables are

where St is the input available to the stages 1 to tConsider the stage 1 as the first sub-problem. The input variable to this stage is S1

Principle of optimality: X1 should be selected such that is optimum for the input S1

The objective function for this stage is

( ) TtSXgS ttt ,....,2,1, 11 =′= ++

( )111 , SXh

∗1f

( )[ ]11111 ,)(1

SXhoptSfX

=∗



Group the first and second stages together as the second sub-problem. The objective function is

By using the stage transformation equation, can be rewritten as

In general, the ith sub-problem can be expressed as

( ) ( )[ ]111222,

22 ,,)(12

SXhSXhoptSfXX

+=∗

)( 22 Sf ∗

( ) ( )( )[ ]222122222 ,,)(2

SXgfSXhoptSfX

′+= ∗∗

( ) ( ) ( )[ ]111222,...,,

,,...,)(21

SXhSXhSXhoptSf iiiXXX

iii

+++=∗



Converting this to a single variable problem

denotes the optimal value of the objective function for the last i stagesPrinciple of optimality for forward recursion can be stated as,

No matter in what state of stage one may be, in order for a policy to be optimal, one had to get to that state and stage in an optimal manner

( ) ( )( )[ ]iiiiiiiX

ii SXgfSXhoptSfi

,,)( )1( ′+= ∗−

∗

∗if


Thank You

Optimization Methods: Dynamic Programming – Computational Procedure 1


Computational Procedure in Dynamic Programming

Introduction

The construction of recursive equations for a multistage program was discussed in the

previous lecture. In this lecture, the procedure to solve those recursive equations for

backward recursion is discussed. The procedure for forward recursion is similar to that of

backward recursion.

Computational procedure

Consider the serial multistage problem and the recursive equations developed for backward

recursion discussed in the previous lecture.

The objective function for this problem is

( )

( ) ( ) ( ) ( ) ( ) )1...(,,...,...,,

,

111222111

1 1

TTTTTTttt

T

t

T

ttttt

SXhSXhSXhSXhSXh

SXhNBf

++++++=

==

−−−

= =∑ ∑

Considering first subproblem i.e., the last stage, the objective function is

( )[ ] )2...(,)( TTTX

TT SXhoptSfT

=∗

NBT

ST ST+1 Stage T

XT

The input variable to this stage is ST. The decision variable XT and the optimal value of the

objective function depend on the input S∗Tf T. At this stage, the value of ST is not known. ST

can take a range of values depending upon the value taken by the upstream components. To



get a clear picture of the suboptimization at this stage, ST is solved for all possible range of

values and the results are entered in a graph or table. This table also contains the calculated

optimal values of , S∗TX ∗

TfT+1 and also . A typical table showing the results from the

suboptimization of stage 1 is shown below.

Table 1

Sl no ST ∗TX ∗

Tf ST+1

1 - - - -

- - - - -

- - - - -

Now, consider the second subproblem by grouping the last two components.

NBT-1 NBT

ST ST+1 Stage T Stage T-1

ST-1

The objective function can be written as XT-1 XT

( ) ( )[ ] )4...(,,)( 1TS − 111,

11

TTTTTTXX

T SXhSXhoptfTT

+= −−−∗−

−

As shown in the earlier lecture, can also be written as, )( 11 −∗− TT Sf

( )[ ] )5...()(,)( 111111

TTTTTX

TT SfSXhoptSfT

∗−−−−

∗− +=

−

Here also, a range of values are considered for S . All the information of first subproblem

can be obtained from Table 1. Thus, the optimal values of and are found for these

range of values. The results thus calculated can be shown in Table 2.

1−T

∗−1TX ∗

−1Tf



Table 2

Sl no ST-1 ∗−1TX ST )( TT Sf ∗ ∗

−1Tf

1 - - - -

- - - - -

- - - - -

In general, if suboptimization of i+1th subproblem (T-ith stage) is to be done, then the

objective function can be written as

( ) ( ) ( )[ ]

( )[ ] )7...(,

,,...,)(

)1(

111,,..., 1

∗−−−−−

−−−−−−−∗−

+=

+++=

−

−−

iTiTiTiTX

TTTTTTiTiTiTXXX

iTiT

fSXhopt

SXhSXhSXhoptSf

iT

TTiT

At this stage, the suboptimizaiton has been carried out for all last i components. The

information regarding the optimal values of ith subproblem will be available in the form of a

table. Substituting this information in the objective function and considering a range of

values, the optimal values of and can be calculated. The table showing the

suboptimization of i+1

iTS −

∗−iTf

∗−iTX

th subproblem can be shown below.

Table 3

Sl no ST-i ∗−iTX ST-(i-1) )( )1()1( −−

∗−− iTiT Sf ∗

−iTf

1 - - - -

- - - - -

- - - - -



This procedure is repeated until stage 1 is reached, i.e., Tth subproblem.


NBt

Xt

S1 S2

NB1 NBT

X1 XT

ST ST+1 St+1

Here, for initial value problems, only one value S1 need to be analyzed.

After completing the suboptimization of all the stages, we need to retrace the steps through

the tables generated to find the optimal values of X. In order to do this, the Tth subproblem

gives the values of and for a given value of S∗1X ∗

1f

∗2S

∗2X ∗

2f∗2S

∗3S

∗∗TXX ....,,2

1 (since the value of S1 is known for an

initial value problem). Use the transformation equation S2 = g(X1, S1), to calculate the value

of , which is the input to the 2nd stage ( T-1th subproblem). Then, from the tabulated results

for the 2nd stage, the values of and are found out for the calculated value of .

Again use the transformation equation to find out and the process is repeated until the 1st

subproblem or Tth stage is reached. Finally, the optimum solution vector is given by

. ∗ X,1



Dynamic Programming

Computational Procedure in Dynamic Programming


Objectives

To explain the computational procedure of solving the multistage decision process using recursive equations for backward approach


Computational Procedure

Consider a serial multistage problem and the recursive equations developed for backward recursion The objective function is

Considering first sub-problem (last stage),the objective function is

( )∑ ∑= =

==T

t

T

ttttt SXhNBf

1 1

,

( )[ ]TTTX

TT SXhoptSfT

,)( =∗

STStage T

NBT

XT

ST+1


Computational Procedure …contd.

The input variable is ST and the decision variable is XT

Optimal value of the objective function depend on the input ST

But at this stage, the value of ST is not knownValue of ST depends upon the values taken by the upstream components Therefore, ST is solved for all possible range of valuesThe results are entered in a graph or table which contains the calculated optimal values of , ST+1 and also .

∗Tf

∗TX ∗

Tf



The results are entered in a graph or table which contains the calculated optimal values of , ST+1 and also Typical table showing the results from the sub-optimization of stage 1

TABLE - 1

∗TX ∗

Tf



Consider the second sub-problem by grouping the last two components

The objective function is

( ) ( )[ ]TTTTTTXX

TT SXhSXhoptSfTT

,,)( 111,

111

+= −−−−∗−

−



From the earlier lecture,

The information of first sub-problem is obtained from the previous table

A range of values are considered for ST-1

The optimal values of and are found for these range of values

( )[ ])(,)( 111111

TTTTTX

TT SfSXhoptSfT

∗−−−−

∗− +=

−

∗−1TX

∗−1Tf



In general, consider the sub-optimization of i+1th sub-problem (T-ithstage)


( ) ( ) ( )[ ]

( )[ ] )7...(,

,,...,)(

)1(

111,,..., 1

∗−−−−−

−−−−−−−∗−

+=

+++=

−

−−

iTiTiTiTX

TTTTTTiTiTiTXXX

iTiT

fSXhopt

SXhSXhSXhoptSf

iT

TTiT



At this stage, the sub-optimization has been carried out for all

last i components

The information regarding the optimal values of ith sub-problem

will be available in the form of a table

Substituting this information in the objective function and

considering a range of values, the optimal values of and

can be calculated

∗−iTf

∗−iTX



The table showing the sub-optimization of i+1th sub-problem is shownTABLE - 2

This procedure is repeated until stage 1 is reachedFor initial value problems, only one value S1 need to be analyzed for stage 1



After completing the sub-optimization of all the stages, retrace the steps through the tables generated to find the optimal values of XThe Tth sub-problem gives the values of X1

* and f1* for a given value of

S1 (since the value of S1 is known for an initial value problem)Calculate the value of S2

* using the transformation equation S2 = g(X1, S1), which is the input to the 2nd stage ( T-1th sub-problem)From the tabulated results for the 2nd stage, the values of X2

* and f2* are

found outAgain use the transformation equation to find out S3

* and the process is repeated until the 1st sub-problem or Tth stage is reachedThe final optimum solution vector is given by ∗∗∗

TXXX ....,,, 21


Thank You

Optimization Methods: Dynamic Programming – Other Topics 1


Other Topics

Introduction

In the previous lectures we discussed about problems with a single state variable or input

variable St which takes only some range of values. In this lecture, we will be discussing about

problems with state variable taking continuous values and also problems with multiple state

variables.

Discrete versus Continuous Dynamic Programming

In a dynamic programming problem, when the number of stages tends to infinity then it is

called a continuous dynamic programming problem. It is also called an infinite-stage

problem. Continuous dynamic programming model is used to solve continuous decision

problems. The classical method of solving continuous decision problems is by the calculus of

variations. However, the analytical solutions, using calculus of variations, cannot be

generally obtained, except for very simple problems. The infinite-stage dynamic

programming approach, on the other hand provides a very efficient numerical approximation

procedure for solving continuous decision problems.

The objective function of a conventional discrete dynamic programming model is the sum of

individual stage outputs. If the number of stages tends to infinity, then summation of the

outputs from individual stages can be replaced by integrals. Such models are useful when

infinite number of decisions have to be made in finite time interval.

Multiple State Variables

In the problems previously discussed, there was only one state variable St. However there will

be problems in which one need to handle more than one state variable. For example, consider

a water allocation problem to n irrigated crops. Let Si be the units of water available to the

remaining n-i crops. If we are concerned only about the allocation of water, then this problem

can be solved as a single state problem, with Si as the state variable. Now, assume that L units

of land are available for all these n crops. We want to allocate the land also to each crop after



considering the units of water required for each unit of irrigated land containing each crop.

Let Ri be the amount of land available for n-i crops. Here, an additional state variable Ri is to

be included while suboptimizing different stages. Thus, in this problem two allocations need

to be made: water and land.

The figure below shows a single stage problem consisting of two state variables, S1 & R1.

Net Benefit, NB1

Input S1 & R1

Output S2 & R2

Stage 1


In general, for a multistage decision problem of T stages containing two state variables St and

Rt , the objective function can be written as

( )∑ ∑= =

==T

t

T

ttttt RSXhNBf

1 1,,

where the transformation equations are given as

St+1 = g(Xt, St) for t =1,2,…, T

& Rt+1 = g’(Xt, Rt) for t =1,2,…, T

Curse of Dimensionality

Dynamic programming has a serious limitation due to dimensionality restriction. As the

number of variables and stages increase, the number of calculations needed increases rapidly

thereby increasing the computational effort. If the number of stage variables is increased,

then more combinations of discrete states should be examined at each stage. For a problem



consisting of 100 state variables and each variable having 100 discrete values, the

suboptimization table will contain 100100 entries. The computation of this one table may take

10096 seconds (about 10092 years) even on a high speed computer. Like this 100 tables have

to be prepared, which explains the difficulty in analyzing such a big problem using dynamic

programming. This phenomenon is known as “curse of dimensionality” or “Problem of

dimensionality” of multiple state variable dynamic programming problems as termed by

Bellman.

References/ Further Reading:

1. Bellman, R., Dynamic Programming, Princeton University Press, Princeton, N.J, 1957.


New Delhi, 1987.


Analysis, Prentice-Hall, N.J., 1981.


International Limited, New Delhi, 2000


Delhi, 2005.





Dynamic Programming

Other Topics


Objectives

To explain the difference between discrete and continuous dynamic programming

To discuss about multiple state variables

To discuss the curse of dimensionality in dynamic programming


Discrete versus Continuous Dynamic Programming

Discrete dynamic programming problems: Number of stages is finite When the number of stages tends to infinity then it is called a continuous dynamic programming problemAlso called as infinite-stage problemContinuous dynamic problems are used to solve continuous decision problemThe classical method of solving continuous decision problems is by the calculus of variations


Discrete versus Continuous Dynamic Programming …contd.

However, the analytical solutions, using calculus of variations is applicable only for very simple problemsThe infinite-stage dynamic programming approach provides a very efficient numerical approximation procedure for solving continuous decision problems For discrete dynamic programming model, the objective function value is the sum of individual stage outputsFor a continuous dynamic programming model, summation is replaced by integrals of the returns from individual stages Such models are useful when infinite number of decisions have to be made in finite time interval


Multiple State Problems

Problems in which there are more than one state variableFor example, consider a water allocation problem to n irrigated cropsLet Si be the units of water available to the remaining n-i cropsConsidering only the allocation of water, the problem can be solved as a single state problem, with Si as the state variableNow, assume that L units of land are available for all these ncropsAllocation of land also to be done to each crop after considering the units of water required for each unit of irrigated land containing each crop


Multiple State Problems …contd.

Let Ri be the amount of land available for n-i cropsAn additional state variable Ri is included while sub-optimizing different stagesThus, in this problem two allocations need to be made: water and land.A single stage problem consisting of two state variables, S1 & R1 is shown below

Stage 1Input

S1 & R1

Net Benefit, NB1


Output

S2 & R2


Multiple State Problems …contd.

In general, for a multistage decision problem of T stages, containing two state variables St and Rt , the objective function can be written as

where the transformation equations are given as

St+1 = g(Xt, St) for t =1,2,…, T &

Rt+1 = g’(Xt, Rt) for t =1,2,…, T

( )∑ ∑= =

==T

t

T

ttttt RSXhNBf

1 1,,


Curse of Dimensionality

Limitation of dynamic programming: Dimensionality restrictionThe number of calculations needed increases rapidly as the number of variables and stages increaseIncreases the computational effortIncrease in the number of stage variables causes an increase in the number of combinations of discrete states to be examined at each stageFor a problem consisting of 100 state variables and each variable having 100 discrete values, the sub-optimization table will contain 100100 entries


Curse of Dimensionality …contd.

The computation of one table may take 10096 seconds (about 10092 years) even on a high speed computer

Like this 100 tables have to be prepared for which computation is almost impossible

This phenomenon as termed by Bellman, is known as “curse of dimensionality” or “Problem of dimensionality” of multiple state variable dynamic programming problems


Thank You

Optimization Methods: Dynamic Programming Applications - Learning Objectives

Module 6: Dynamic Programming Applications

Learning Objectives

The basic concepts of dynamic programming like concept of sub-optimization and principle

of optimality were discussed in module 5. The formulation of recursive equations for a

general problem and computational procedure was also discussed. In this module, the

application of these concepts in some common fields are discussed.

The module demonstrates the applications of dynamic programming in the fields of structural

engineering and water resources engineering. The problems discussed are design of

continuous beams, geometric layout of truss, water allocation, capacity expansion and

reservoir operation.


1. Formulate the dynamic programming problem.

2. Formulate the optimal design of continuous beams.

3. Formulate the optimal geometric layout of a truss.

4. Formulate and solve water allocation problem as a sequential process.

5. Formulate and solve capacity expansion problem.

6. Formulate and find steady state operating policies for a reservoir.


Optimization Methods: Dynamic Programming Applications – Design of Continuous Beam

1


Design of Continuous Beam

Introduction

In the previous lectures, the development of recursive equations and computational procedure

were discussed. The application of this theory in practical situations is discussed here. In this

lecture, the design of continuous beam and its formulation to apply dynamic programming is

discussed.

Design of continuous beam

Consider a continuous beam having n spans with a set of loadings W1, W2,…, Wn at the center

of each span as shown in the figure.

W1

0

L1

W2

1

L2

2

Wi

i-1

Li

Wi+1

i

Li+1

i+1

Wn

n-1

Ln

n

The beam rests on n+1 rigid supports. The locations of the supports are assumed to be

known. The objective function of the problem is to minimize the sum of the cost of

construction of all spans.

It is assumed that simple plastic theory of beams is applicable. Let the reactant support

moments be represented as m1, m2, …, mn. Once these support moments are known, the

complete bending moment distribution can be determined. The plastic limit moment for each

span and also the cross section of the span can be designed using these support moments.

The bending moment at the center of the ith span is -WiLi/4. Therefore, the largest bending

moment in the ith span can be computed as


Optimization Methods: Dynamic Programming Applications – Design of Continuous Beam

2

niforLWmm

mmM iiiiiii ,...2,1

42,,max 1

1 =⎭⎬⎫

⎩⎨⎧

−+

= −−

For a beam of uniform cross section in each span, the limit moment m_limi for the ith span

should be greater than or equal to Mi. The cross section of the beam should be selected in

such a way that it has the required limit moment. Since the cost of the beam depends on the

cross section, which in turn depends on the limit moment, cost of the beam can be expressed

as a function of the limit moments.

If represents the sum of the cost of construction of all spans of the beam where X

represents the vector of limit moments

∑=

n

ii XC

1)(

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

nm

mm

X

lim_

lim_lim_

2

1

M

then, the optimization problem is to find X so that ∑ is minimized while satisfying

the constraints

=

n

ii XC

1)(

niforMm ii ...,,2,1lim_ =≥ .

This problem has a serial structure and can be solved using dynamic programming.



Dynamic Programming Applications



Objectives

To discuss the design of continuous beams

To formulate the optimization problem as a dynamic

programming model



Consider a continuous beam having n spans with a set of loadings W1, W2,…, Wn

Beam rests on n+1 rigid supportsLocations of the supports are assumed to be known

W1

0

L1

W2

1

L2

2

Wi

i-1

Li

Wi+1

i

Li+1

i+1

Wn

n-1

Ln

n


Design of Continuous Beam …contd.

Objective function: To minimize the sum of the cost of construction of all spansAssumption: Simple plastic theory of beams is applicableLet the reactant support moments be represented as m1, m2, …, mn

Complete bending moment distribution can be determined once these support moments are knownPlastic limit moment for each span and also the cross section ofthe span can be designed using these support moments



Bending moment at the center of the ith span is -WiLi/4Thus, the largest bending moment in the ith span can be computed as

Limit moment for the ith span, m_limi should be greater than or equal to Mi for a beam of uniform cross sectionThus, the cross section of the beam should be selected such that it has the required limit moment

niforLWmmmmM iiiiiii ,...2,1

42,,max 1

1 =⎭⎬⎫

⎩⎨⎧ −+= −

−



The cost of the beam depends on the cross sectionAnd cross section in turn depends on the limit momentThus, cost of the beam can be expressed as a function of the limit momentsLet X represents the vector of limit moments

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

nm

mm

X

lim_

lim_lim_

2

1

Μ



The sum of the cost of construction of all spans of the beam is

Then, the optimization problem is to find X to

Minimize

satisfying the constraints .

This problem has a serial structure and can be solved using dynamic programming

∑=

n

ii XC

1)(

∑=

n

ii XC

1

)(

niforMm ii ...,,2,1lim_ =≥


Thank You

Optimization Methods: Dynamic Programming Applications – Optimum Geometric Layout of Truss

1


Optimum Geometric Layout of Truss

Introduction

In this lecture, the optimal design of elastic trusses is discussed from a dynamic programming

point of view. Emphasis is given on minimizing the cost of statically determinate trusses

when the cross-sectional areas of the bars are available.

Optimum geometric layout of truss

Consider a planar, pin jointed cantilever multi bayed truss. Assume the length of the bays to

be unity. The truss is symmetric to the x axis. The geometry or layout of the truss is defined

by the y coordinates (y1, y2, …, yn). The truss is subjected to a unit load W1. The details are

shown in the figure below.

h1

h2

h3

W1=1

y1

y2

y3

y4

x

1 1 1

Consider a particular bay i. Assume the truss is statically determinate. Thus, the forces in the

bars of bay i depend only on the coordinates yi-1 and yi and not on any other coordinates. The


Optimization Methods: Dynamic Programming Applications – Optimum Geometric Layout of Truss

2

cross sectional area of a bar can be determined, once the length and force in it are known.

Thus, the cost of the bar can in turn be determined.

The optimization problem is to find the geometry of the truss which will minimize the total

cost from all the bars. For the three bay truss shown above, the relation between y coordinates

can be expressed as

yi+1 = yi + di for i = 1,2,3

This is an initial value problem since the value y1 is already known. Let the y coordinate of

each node is limited to a finite number of values say 0.25, 0.5, 0.75 and 1. Then, as shown in

the figure below, there will be 64 different possible ways to reach y4 from y1.

y1 y4 y3 y2

0.25

0.50

0.75

1.00

This can be represented as a serial multistage initial value decision problem and can be

solved using dynamic programming.



Objectives

To discuss the design of elastic trusses

To formulate the optimization problem as a

dynamic programming model



Consider a planar, pin jointed cantilever multi bayed truss

Assume the length of the bays to be unity

The truss is symmetric to the x axis

Geometry or layout of the truss is defined by the y coordinates

(y1, y2, …, yn)

Truss is subjected to a unit load W1


Optimum Geometric Layout of Truss …contd.

h1

h2

h3

W1=1

y1

y2

y3

y4

x

1 1 1



Consider a particular bay I

Assume the truss is statically determinate

Forces in the bars of bay i depend only the coordinates yi-1 and

yi

Cross sectional area of a bar can be determined, once the

length and force in it are known

Cost of the bar can thus be determined.



The optimization problem is to find the geometry of the truss

which will minimize the total cost from all the bars

For the three bay truss, the relation between y coordinates can

be expressed as

yi+1 = yi + di for i = 1,2,3

This is an initial value problem since the value y1 is known



Let the y

coordinate of each

node is limited to a

finite number of

values say 0.25,

0.5, 0.75 and 1

y1 y4y3y2

0.25

0.50

0.75

1.00



As shown in the figure, there will be 64 different possible ways

to reach y4 from y1

This can be represented as a serial multistage initial value

decision problem and can be solved using dynamic

programming


Thank You

Optimization Methods: Dynamic Programming Applications – Water Allocation 1


Water Allocation as a Sequential Process – Recursive Equations

Introduction

As discussed in previous lecture notes, in dynamic programming, a problem is handled as a

sequential process or a multistage decision making process. In this lecture, we will explain

how a water allocation problem can be represented as sequential process and can be solved

using backward recursion method of dynamic programming.

Water allocation problem

Consider a canal supplying water to three fields in which three different crops are being

cultivated. The maximum capacity of the canal is given as Q units of water. The three fields

can be denoted as i=1,2,3 and the amount of water allocated to each field as xi.

Field 1

Field 2

Field 3

x1

x3

x2

The net benefits from producing the crops in each field are given by the functions below.

23333

22222

21111

7)(

5.18)(

5.05)(

xxxNB

xxxNB

xxxNB

−=

−=

−=



The problem is to determine the optimal allocations xi to each field that maximizes the total

net benefits from all the three crops. This type of problem is readily solvable using dynamic

programming.

The first step in the dynamic programming is to structure this problem as a sequential

allocation process or a multistage decision making procedure. The allocation to each crop is

considered as a decision stage in a sequence of decisions. If the amount of water allocated

from the total available water Q, to crop i is xi, then the net benefit from this allocation is

NBi(xi). Let the state variable Si defines the amount of water available to the remaining (3-i)

crops. The state transformation equation can be written as iii xSS = −+1

3

defines the state in

the next stage. The figure below shows the allocation problem as a sequential process.

Available Quantity, S1 = Q

Crop 1 Crop 2 Crop 3

x1 x2 x3

S2 = S1- x1 S3 = S2- x2

Net Benefits, NB1 (x1)



Remaining Quantity, S4 = S3- x3

The objective function for this allocation problem is defined to maximize the net benefits,

i.e., . The constraints can be written as ∑=1

)(maxi

ii xNB

3,2,10321

=≤≤≤++

iforQxQxxx

i



Let be the maximum net benefits that can be obtained from allocating water to crops

1,2 and 3. Thus,

)(1 Qf

⎥⎦

⎤⎢⎣

⎡= ∑

=≥≤++

3

10,,1 )(max)(

321321 i

iixxx

QxxxxNBQf

Transforming this into three problems each having only one decision variable,

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

++==−≤≤=−≤≤≤≤

)(max)(max)(max)( 330

220

110

1

32233

2122

11

xNBxNBxNBQfSxSx

xSxQx

xQx

x

Backward recursive equations

Considering the last term of this equation, let be the maximum net benefits from crop

3. The state variable for this stage is which can vary from 0 to Q. Therefore,

)( 33 Sf

3S

)( 22 Sf

)(max)( 330

33

333

xNBSfSx

x≤≤

=

Since , . Thus can be rewritten as 223 xSS −= )()( 22333 xSfSf −= )(1 Qf

⎥⎥⎦

⎤

⎢⎢⎣

⎡−++=

=−≤≤≤≤

)()(max)(max)( 223220

110

1

2122

11

xSfxNBxNBQfSxQx

xQx

x

Now, let be the maximum benefits derived from crops 2 and 3 for a given quantity

which can vary between 0 and Q. Therefore can be written as,

)( 22 Sf 2S

)()(max)( 223220

22

2122

xSfxNBSfSxQx

x−+=

=−≤≤

Again, since , which is the maximum total net benefit from the allocation to

the crops 1, 2 and 3, can be rewritten as

12 xQS −= )(1 Qf



[ ])()(max)( 12110

1

11

xQfxNBQfQx

x−+=

≤≤

Now, once the value of is calculated, the value of can be determined, from

which can be determined.

)( 33 Sf )( 22 Sf

)(1 Qf

Forward recursive equations

The procedure explained above can also be solved using a forward proceeding manner. Let

the function be the total net benefit from crops 1 to i for a given input of which is

allocated to those crops. Considering the first stage alone,

)( ii Sf iS

2S )( 22 Sf

)(max)( 1111

111

xNBSfSx

x≤

=

Since, the value of is not known (excepting that should not exceed Q), the equation

above has to be solved for a range of values from 0 to Q. Now, considering the first two crops

together, with units of water available to these crops, can be written as,

1S 1S

[ ])()(max)( 2212222

222

xSfxNBSfSx

x−+=

≤

This equation also should be solved for a range of values for from 0 to Q. Finally,

considering the whole system i.e., crops 1, 2 and 3, can be expressed as,

2S

)( 33 Sf

)( 33 Sf

[ ])()(max)( 3323333

333

xSfxNBSfQSx

x−+=

=≤

Here, if it is given that the whole Q units of water should be allocated, then the value of

can be taken as equal to Q. Otherwise, should be solved for a range of values from 0 to

Q.

3S



The basic equations for the water allocation problem using both the approaches are discussed.

A numerical problem and its solution will be described in the next lecture.




Water Allocation



Dynamic Programming : Sequential or multistage decision making processWater Allocation problem is solved as a sequential process using dynamic programming

ObjectivesTo discuss the Water Allocation ProblemTo explain and develop recursive equations for backward approach To explain and develop recursive equations for forward approach


Water Allocation Problem

Consider a canal supplying water for three different crops Maximum capacity of the canal is Q units of water.Amount of water allocated to each field as xi

Field 1

Field 2

Field 3

x1

x3

x2


Water Allocation Problem … contd.

Net benefits from producing the crops can be expressed as a function of the water allotted.

Optimization Problem: Determine the optimal allocations xi to each crop that maximizes the total net benefits from all the three crops

23333

22222

21111

7)(

5.18)(

5.05)(

xxxNB

xxxNB

xxxNB

−=

−=

−=


Solution using Dynamic Programming

Structure the problem as a sequential allocation process or a multistage decision making procedure.Allocation to each crop is considered as a decision stage in a sequence of decisions.Amount of water allocated to crop i is xi

Net benefit from this allocation is NBi(xi)Introduce one state variable Si :- Amount of water available to the remaining (3-i) cropsState transformation equation can be written as

defines the state in the next stageiii xSS −=+1


Sequential Allocation Process The allocation problem is shown as a sequential process


Backward Recursive Equations

Objective function: To maximize the net benefits

Subjected to the constraints

Let be the maximum net benefits that can be obtained from allocating water to crops 1,2 and 3

∑=

3

1)(max

iii xNB

3,2,10321

=≤≤≤++

iforQxQxxx

i

)(1 Qf

⎥⎦

⎤⎢⎣

⎡= ∑=≥

≤++

3

10,,1 )(max)(

321321 i

iixxx

QxxxxNBQf


Backward Recursive Equations … contd.

Transforming this into three problems each having only one decision variable

Now starting from the last stage, let be the maximum net benefits from crop 3. State variable for this stage can vary from 0 to Q Thus,

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

++==−≤≤=−≤≤≤≤

)(max)(max)(max)( 330

220

110

1

32233

2122

11

xNBxNBxNBQfSxSx

xSxQx

xQx

x

)( 33 Sf

3S

)(max)( 330

33

333

xNBSfSx

x≤≤

=



But . Therefore Hence,

Now, let be the maximum benefits derived from crops 2 and 3 for a given quantity which can vary between 0 and QTherefore,

223 xSS −= )()( 22333 xSfSf −=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−++=

=−≤≤≤≤

)()(max)(max)( 223220

110

1

2122

11

xSfxNBxNBQfSxQx

xQx

x

)( 22 Sf2S

)()(max)( 223220

22

2122

xSfxNBSfSxQx

x−+=

=−≤≤



Now since , can be rewritten as

Once the value of is calculated, the value of can be determined, from which can be determined.

12 xQS −= )(1 Qf

[ ])()(max)( 12110

1

11

xQfxNBQfQx

x−+=

≤≤

)( 33 Sf )( 22 Sf)(1 Qf


Forward Recursive Equations

Let the function be the total net benefit from crops 1 to i for a given input of which is allocated to those crops. Considering the first stage,

Solve this equation for a range of values from 0 to QConsidering the first two crops, for an available quantity of ,

can be written as

)( ii SfiS

)(max)( 1111

111

xNBSfSx

x≤

=

1S

2S)( 22 Sf

[ ])()(max)( 2212222

222

xSfxNBSfSx

x−+=

≤


Forward Recursive Equations … contd.

ranges from 0 to Q

Considering the whole system, can be expressed as,

If the whole Q units of water should be allocated then the value of can be taken as equal to Q

Otherwise, will take a range of values from 0 to Q

)( 33 Sf

[ ])()(max)( 3323333

333

xSfxNBSfQSx

x−+=

=≤

2S

3S

3S


Conclusion

The basic equations for the water allocation problem using both the approaches are discussed

A numerical problem and its solution will be described in the next lecture


Thank You



Water Allocation as a Sequential Process – Numerical Example

Introduction

In the previous lecture, recursive equations for a basic water allocation problem were

developed for both backward recursion and forward recursion. This lecture will further

explain the water allocation problem by a numerical example.

Numerical problem and solution

Consider the example previously discussed with the maximum capacity of the canal as 4

units. The net benefits from producing the crops for each field are given by the functions

below.

23333

22222

21111

7)(

5.18)(

5.05)(

xxxNB

xxxNB

xxxNB

−=

−=

−=

The possible net benefits from each crop are calculated according to the functions given and

are given below.

Table 1

ix )( 11 xNB )( 22 xNB )( 33 xNB

0 0.0 0.0 0.0

1 4.5 6.5 6.0

2 8.0 10.0 10.0

3 10.5 10.5 12.0

4 12.0 8.0 12.0

The problem can be represented as a set of nodes and links as shown in the figure below. The

nodes represent the state variables and the links represent the decision variables.



4

x1 x2 x3

4

3

2

1

0

4

3

2

1

0

0

1

2

3

4

0

1

0

1

2

4

3

2

1

0

0

1

0

1

2


The values inside the nodes show the value of possible state variables at each stage. Number

of nodes for any stage corresponds to the number of discrete states possible for each stage.

The values over the links show the different values taken by decision variables corresponding

to the value taken by state variables. It may be noted that link values for all links are not

shown in the above figure.

Solution using Backward Recursion:

Starting from the last stage, the suboptimization function for the 3rd crop is given as,

)(max)( 330

33

333

xNBSfSx

x≤≤

= with the range of from 0 to 4. 3S



The calculations for this stage are shown in the table below.

Table 2

)( 33 xNB State

3S3x : 0 1 2 3 4

)( 33 Sf ∗3x

0 0 0 0

1 0 6 6 1

2 0 6 10 10 2

3 0 6 10 12 12 3

4 0 6 10 12 12 12 3,4

Next, by considering last two stages together, the suboptimization function is

[ )()(max)( 2212222

222

xSfxNBSfSx

x− ]+=

≤

. This is solved for a range of values from 0 to 4.

The value of is noted from the previous table. The calculations are shown below.

2S

)( 223 xSf −

Table 3

State 2S 2x )( 22 xNB )( 22 xS − )( 223 xSf − )(

)()(

223

22

22

xSfxNB

Sf

−+

= )( 22 Sf ∗ ∗

2x

0 0 0 0 0 0 0 0

0 0 1 6 6 1

1 6.5 0 0 6.5 6.5 1

0 0 2 10 10

1 6.5 1 6 12.5 2

2 10 0 0 10

12.5 1

0 0 3 12 12

1 6.5 2 10 16.5

3

2 10 1 6 16

16.5 1



3 10.5 0 0 10.5

Table contd. on next page

0 0 4 12 12

1 6.5 3 12 18.5

2 10 2 10 20

3 10.5 1 6 16.5

4

4 8 0 0 8

20 2

Finally, by considering all the three stages together, the sub-optimization function is

[ ])()(max)( 12110

1

11

xQfxNBQfQx

x−+=

≤≤

. The value of 41 ==QS . The calculations are shown in

the table below.

Table 4

State

QS =11x )( 11 xNB )( 1xQ − )( 12 xQf −

)()(

)(

12

11

11

xQfxNB

Sf

−+

= )( 11 Sf ∗ ∗

1x

0 0 4 20 20

1 4.5 3 16.5 21

2 8 2 12.5 20.5

3 10.5 1 6.5 17

4

4 12 0 0 12

21 1

Now, backtracking through each table to find the optimal values of decision variables, the

optimal allocation for crop 1, = 1 for a value of 4. This will give the value of

as S . From Table 3, the optimal allocation for crop 2, for is 1. Again,

. Thus, from Table 2 is 2. The maximum total net benefit from all the

crops is 21. The optimal solution is given below

∗x S S

3112 =−= xS 2x 32 =S

S *3x

1 1 2

2223 =−= xS



2

1

1

21

3

2

1

=

=

=

=

∗

∗

∗

∗

x

x

x

f

Solution using Forward Recursion:

While starting to solve from the first stage and proceeding towards the final stage, the

suboptimization function for the first stage is given as,

)(max)( 1111

111

xNBSfSx

x≤

= . The range of values for is from 0 to 4. 1S

Table 5

State 1S 1x )( 11 xNB )( 22 Sf ∗ ∗1x

0 0 0 0 0

0 0 1

1 4.5 4.5 1

0 0

1 4.5 2

2 8

8 2

0 0

1 4.5

2 8 3

3 10.5

10.5 3

0 0

1 4.5

2 8

3 10.5

4

4 12

12 4



Now, considering the first two crops together, can be written as, )( 22 Sf

[ )()(max)( 2212222

222

xSfxNBSfSx

x− ]+=

≤

with ranging from 0 to 4. The calculations for this

stage are shown below.

2S

Table 6

State 2S 2x )( 22 xNB )( 22 xS − )( 221 xSf − )(

)()(

222

22

22

xSfxNB

Sf

−+

= )( 22 Sf ∗ ∗

2x

0 0 0 0 0 0 0 0

0 0 1 4.5 4.5 1

1 6.5 0 0 6.5 6.5 1

0 0 2 8 8

1 6.5 1 4.5 11 2

2 10 0 0 10

11 1

0 0 3 10.5 10.5

1 6.5 2 8 14.5

2 10 1 4.5 14.5 3

3 10.5 0 0 10.5

14.5 1,2

0 0 4 12 12

1 6.5 3 10.5 17

2 10 2 8 18

3 10.5 1 4.5 15

4

4 8 0 0 8

18 2

Now, considering the whole system i.e., crops 1, 2 and 3, can be expressed as, )( 33 Sf

[ ])()(max)( 3323333

333

xSfxNBSfQSx

x−+=

=≤

with the value of 43 =S .



The calculations are shown below.

Table 7

State 3S 3x )( 33 xNB 33 xS − )( 332 xSf − )(

)()(

332

33

33

xSfxNB

Sf

−+

= )( 33 Sf ∗ ∗

3x

0 0 4 18 18

1 6 3 14.5 20.5

2 10 2 11 21

3 12 1 6.5 18.5

4

4 12 0 0 12

21 2

In order to find the optimal solution, a backtracking is done. From Table 7, the optimal value

of is given as 2 for the value of 4. Therefore, *3x 3S 2332 = − =xSS

1=∗x 1

. Now, from Table 6,

the value of . Then, 2 221 =− xS = S and from Table 5, for 11 =S , the value of .

Thus, the optimal values determined are shown below.

1=∗x1

2

1

1

21

3

2

1

=

=

=

=

∗

∗

∗

∗

x

x

x

f

These optimal values are same as those we got by solving using backward recursion method.




Water Allocation –Numerical Example


Objectives

To demonstrate the water allocation problem through a numerical example using

Backward approach

Forward approach


Numerical Problem

Field 1

Field 2

Field 3

x1

x3

x2

Consider a canal supplying water for three different cropsMaximum capacity of the canal is 4 units of water.

Optimization Problem: Determine the optimal allocations xi to each crop that maximizes the total net benefits from all the three crops


Numerical Problem …contd.

Net benefits from producing the crops can be expressed as a function of the water allotted.

23333

22222

21111

7)(

5.18)(

5.05)(

xxxNB

xxxNB

xxxNB

−=

−=

−=

The net benefit values are calculated for each crop and are as shown



Representation of the problem as a set of nodes and links

4

x1 x2 x3

4

3

2

1

0

4

3

2

1

0

0

1

2

3

4

0

1

0

1

2

4

3

2

1

0

0

1

0

1

2




The values inside the node show the value of state variable at each stage

Number of nodes for any stage corresponds to the number of discrete states possible for each stage.

The values over the links show the different values taken by decision variables corresponding to the value taken by state variables


Numerical Problem: Solution by Backward recursion

Sub-optimization function for the 3rd crop: with the range of from 0 to 4. )(max)( 33

033

333

xNBSfSx

x≤≤

= 3S


Solution by Backward recursion … contd.

Next, by considering last two stages together, the sub-optimization function is

The calculations are shown belowTable 3

[ ])()(max)( 2212222

222

xSfxNBSfSx

x−+=

≤



Considering all the three stages togetherwith [ ])()(max)( 1211

01

11

xQfxNBQfQx

x−+=

≤≤

41 == QS



Backtrack through each table,Optimal allocation for crop 1, = 1 and = 4Thus, From 2nd stage, the optimal allocation for crop 2, = 1.Now, From 3rd stage calculations, = 2Maximum total net benefit from all the crops = 21

∗1x 1S

3112 =−= xSS

2x2223 =−= xSS

*3x


Numerical Problem: Solution by Forward recursion

Start from the first stage and proceed towards the final stage

Suboptimization function for the first stage

Range of values for is from 0 to 4

The calculations are shown in the table

)(max)( 1111

111

xNBSfSx

x≤

=

1S


Solution by Forward recursion … contd.


Solution by Forward recursion …contd.

⎥⎦

⎤⎢⎣

⎡−

+=

≤)(

)(max

)(

221

22

22

222 xSf

xNBSf

Sxx



with [ ])()(max)( 3323333

333

xSfxNBSfQSx

x−+=

=≤

43 =S



Backtrack through each table,Optimal allocation for crop 3, = 2 and = 4Then, The optimal allocation for crop 2,Now, From 1st stage calculations, Maximum total net benefit from all the crops = 21These solutions are the same as those we got from backward recursion method.

*3x 3S2332 =−= xSS

12 =∗x1221 =−= xSS

11 =∗x


Thank You

Optimization Methods: Dynamic Programming Applications – Capacity Expansion 1


Capacity Expansion

Introduction

The most common applications of dynamic programming in water resources include water

allocation, capacity expansion of infrastructure and reservoir operation. In this lecture,

dynamic programming formulation for capacity expansion and a numerical example are

discussed.

Capacity expansion

Consider a municipality planning to increase the capacity of its infrastructure (ex: water

treatment plant, water supply system etc) in future. The increments are to be made

sequentially in specified time intervals. Let the capacity at the beginning of time period t be St

(existing capacity) and the required capacity at the end of that time period be Kt. Let be

the added capacity in each time period. The cost of expansion at each time period can be

expressed as a function of and , i.e. C . The problem is to plan the time sequence

of capacity expansions which minimizes the present value of the total future costs subjected

to meet the capacity demand requirements at each time period. Hence, the objective function

of the optimization model can be written as,

tx

tS tx ),( ttt xS

tS

Minimize ∑=

T

tttt xSC

1),(

where is the present value of the cost of adding an additional capacity in the

time period t with an initial capacity . Each period’s final capacity or next period’s initial

capacity should be equal to the sum of initial capacity and the added capacity. Also at the end

of each time period, the required capacity is fixed. Thus, for a time period t, the constraints

can be expressed as

),( ttt xSC tx

TtforKSTtforxSS

tt

ttt

,...,2,1,...,2,1

1

1

=≥=+=

+

+



In some problems, there may be constraints to the amount of capacity added in each time

period i.e. can take only some feasible values. Thus,

tx

tx ttx Ω∈ .

The capacity expansion problem defined above can be solved in a sequential manner using

dynamic programming. The solution procedure using forward recursion and backward

recursion are explained below.

Forward Recursion

Consider the stages of the model to be the time periods in which capacity expansion is to be

made and the state to be the capacity at the end of each time period t, . Let be the

present capacity before expansion and be the minimum present value of total cost of

capacity expansion from present to the time t.

1+tS 1S

)( 1+tt Sf

TK )( 21 Sf

S 1K TK

Stage 1 Stage t

Stage T St

Ct

xt

St+1 S1 S2

C1 CT

x1 xT

ST ST+1

Considering the first stage, the objective function can be written as,

),(min),(min)(

1211

11121

SSSCxSCSf

−==

The values of can be between and where is the required capacity at the end of

time period 1 and is the final capacity required. In other words, should be solved

for a range of values between and . Then considering first two stages, the

suboptimization function is

2S 1K TK 1K

2



( )[ ]

( )[ ])(,min

)(,min)(

2312232

2122232

222

222

xSfxxSC

SfxSCSf

xx

xx

−+−=

+=

Ω∈

Ω∈

which should be solved for all values of ranging from to . Hence, in general for a

time period t, the suboptimization function can be represented as

3S 2K TK

( )[ ])(,min)( 1111 tttttttx

xtt xSfxxSCSftt

t

−+−= +−+

Ω∈

+

with constraint as Ttt KSK ≤≤ +1 . For the last stage, i.e. t=T, the function need to

be solved only for

)( 1+TT Sf

TT KS =+1 .

Backward Recursion

The expansion problem can also be solved using a backward recursion approach with some

modifications. Consider the state be the capacity at the beginning of each time period t.

Let be the minimum present value of total cost of capacity expansion in periods t

through T.

tS

)( TT Sf

]

1−tK TK

For the last period T, the final capacity should reach after doing the capacity expansions.

Thus, the objective function can be written as,

TK

( )[ ]TTTx

xTT xSCSfTT

T

,min)(Ω∈

=

This is solved for all values ranging from to . TS 1−TK TK

In general, for a time period t, the function can be expressed as )( tt Sf

( )[ )(,min)( 1 ttttttx

xtt xSfxSCSftt

t

++= +

Ω∈

which should be solved for all discrete values of

ranging from to .

tS



For period 1, the above equation must be solved only for the actual value of given. 1S

Numerical example (Loucks et al., 1981)

Consider a five stage capacity expansion problem. The minimum capacity to be achieved at

the end of each time period is given below.

Table 1

t Kt

1 5

2 10

3 20

4 20

5 25

The expansion costs for each combination of expansion for each stage are shown in the

corresponding links in the form of a figure below.



Solution Using Forward Recursion

The capacity at the initial stage is given as S1 = 0.

Consider the first stage, t =1. The final capacity for stage 1, S2 can take values between K1 to

K5. Let the state variable can take discrete values of 5, 10, 15, 20 and 25. The objective

function for 1st subproblem with state variable as S2 can be expressed as

),(min),(min)(

1211

11121

SSSCxSCSf

−==

The computations for stage 1 are given in the table below.

Table 2

Stage 1

State Variable, S2Added Capacity,

x1 = S2 – S1C1(S2) f1

*(S2)

5 5 9 9

10 10 11 11

15 15 15 15

20 20 21 21

25 25 27 27

Considering the 1st and 2nd stages together, the state variable S3 can take values from K2 to

K5. Thus, the objective function for 2nd subproblem is

( )[ ]

( )[ ])(,min

)(,min)(

2312232

2122232

222

222

xSfxxSC

SfxSCSf

xx

xx

−++−=

+=

Ω∈

Ω∈

The value of x2 should be taken in such a way that the minimum capacity at the end of stage 2

should be 10, i.e. S . 103 ≥



The computations for stage 2 are given in the table below.

Table 3

Stage 2

State

Variable,

S3

Added

Capacity, x2C2(S3)

S2= S3 -

x2f1

*(S2) f2(S3)=C2(S3)+f1*(S2) f2

*(S3)

0 0 10 11 11 10

5 8 5 9 17 11

0 0 15 15 15

5 8 10 11 19 15

10 10 5 9 19

15

0 0 20 21 21

5 7 15 15 22

10 10 10 11 21 20

15 13 5 9 22

21

0 0 25 27 27

5 7 20 21 28

10 9 15 15 24

15 14 10 11 25

25

20 20 5 9 29

24

Like this, repeat this steps till t = 5. For the 5th subproblem, state variable S6 = K5.



The computations for stages 3 to 5 are shown in tables below.

Table 4

Stage 3

State

Variable,

S4

Added

Capacity, x3C3(S4)

S3= S4 –

x3f2

*(S3) f3(S4)=C3(S4)+f2*(S3) f3

*(S4)

0 0 20 21 21

5 6 15 15 21 20

10 9 10 11 20

20

0 0 25 24 24

5 6 20 21 27

10 9 15 15 34 25

15 12 10 11 23

23

Table 5

Stage 4

State

Variable,

S5

Added

Capacity, x4C4(S5)

S4= S5 –

x4f3

*(S4) f4(S5)=C4(S5)+f3*(S4) f4

*(S5)

20 0 0 20 20 20 20

0 0 25 23 23 25

5 5 20 20 25 23



Table 6

Stage 5

State

Variable,

S6

Added

Capacity, x5C5(S6)

S5= S6 –

x5f4

*(S5) f5(S6)=C5(S6)+f4*(S5) f5

*(S6)

0 0 25 23 23 25

5 4 20 20 24 23

The figure below shows the solutions with the cost of each addition along the links and the

minimum total cost at each node.

From the figure, the optimal cost of expansion is 23 units. By doing backtracking from the

last stage (farthest right node) to the initial stage, the optimal expansion to be done at 1st stage

= 10 units, 3rd stage = 15 units and rest all stages = 0 units.



Solution Using Backward Recursion

The capacity at the final stage is given as S6 = 25. Consider the last stage, t =5. The initial

capacity for stage 5, S5 can take values between K4 to K5. The objective function for 1st

subproblem with state variable as S5 can be expressed as

( )[ ]55555 ,min)( xSfSfTT

Tx

xΩ∈

=

The computations for stage 5 are given in the table below

Table 7

Stage 5

State

Variable,

S5

Added

Capacity, x5

C5(S5) f5*(S5)

20 5 4 4

25 0 0 0

Following the same procedure for all the remaining stages, the optimal cost of expansion is

achieved. The computations for all stages 4 to 1 are given below.

Table 8

Stage 4

State

Variable,

S4

Added

Capacity, x4C4(S4) S5 = S4+ x4 f5

*(S5) f4(S4)=C4(S4)+f5*(S5) f4

*(S4)

0 0 20 4 4 20

5 5 25 0 5 4

25 0 0 25 0 0 0



Table 9

Stage 3

State

Variable, S3

Added

Capacity, x3C3(S3)

S4= S3 +

x3f4

*(S4) f3(S3)=C3(S3)+

f4*(S4)

f3*(S3)

10 9 20 4 13 10

15 12 25 0 12 12

5 6 20 4 10 15

10 9 25 0 10 10

0 0 20 4 4 20

5 6 25 0 5 4

25 0 0 25 0 0 0

Table 10

Stage 2

State

Variable, S2

Added

Capacity, x2C2(S2)

S3= S2 +

x2f3

*(S3) f2(S2)=C2(S2)+

f3*(S3)

f2*(S2)

5 8 10 12 20

10 10 15 10 20

15 13 20 4 17 5

20 20 25 0 20

17

0 0 10 12 12

5 8 15 10 18

10 10 20 4 14 10

15 14 25 0 14

12

0 0 15 10 10

5 7 20 4 11 15

10 9 25 0 9

9

0 0 20 4 4 20

5 7 25 0 7 4

25 0 0 25 0 0 0



Table 11

Stage 2

State

Variable, S1

Added

Capacity, x1C1(S1)

S2= S1 +

x1f2

*(S2) f1(S1)=C1(S1)+

f2*(S2)

f1*(S2)

5 9 5 17 26

10 11 10 12 23

15 15 15 9 24

20 21 20 4 25

0

25 27 25 0 27

23

The solution is given by the figure below with the minimum total cost of expansion at the

nodes.



The optimal cost of expansion is obtained from the node value at the first node i.e. 23 units

which is the same as obtained from forward recursion. The optimal expansions at each time

period can be obtained by moving forward from the first node to the last node. Thus, the

optimal expansions to be made are 10 units at the first stage and 15 units at the last stage.

Hence the final requirement of 25 units is achieved.

Although this type of expansion problem can be solved, the future demand and the future cost

of expansion are highly uncertain. Hence, the solution obtained cannot be used for making

expansions till the end period, T. It can be very well used to make decisions about the

expansion to be done in the current period. For this to be done, the final period T should be

selected far away from the current period, so that the uncertainty on current period decisions

is much less.

It may be note that, generally water supply projects are planned for a period of 25-30 years to

avoid undue burden to the present generation. In addition, change of value of money in time

(due to inflation and other aspects) is not considered in the examples above.




Capacity Expansion


Objectives

To discuss the Capacity Expansion ProblemTo explain and develop recursive equations for both backward approach and forward approachTo demonstrate the method using a numerical example


Capacity Expansion Problem

Consider a municipality planning to increase the capacity of itsinfrastructure (ex: water treatment plant, water supply system etc) in futureSequential increments are to be made in specified time intervalsThe capacity at the beginning of time period t be St

Required capacity at the end of that time period be Kt

Thus, be the added capacity in each time periodCost of expansion at each time period can be expressed as a function of and , i.e.

tx

tS tx ),( ttt xSC


Capacity Expansion Problem … contd.

Optimization problem: To find the time sequence of capacity expansions which minimizes the present value of the total futurecostsObjective function:

Minimize

: Present value of the cost of adding an additional capacity in the time period tConstraints: Capacity demand requirements at each time period

: Initial capacity

∑=

T

tttt xSC

1),(

),( ttt xSCtx

tS


Capacity Expansion Problem … contd.

Each period’s final capacity or next period’s initial capacity should be equal to the sum of initial capacity and the added capacity

At the end of each time period, the required capacity is fixed

Constraints to the amount of capacity added in each time period i.e. can take only some feasible values.

TtforxSS ttt ,...,2,11 =+=+

TtforKS tt ,...,2,11 =≥+

tx

ttx Ω∈


Capacity Expansion Problem: Forward Recursion

Stages of the model: Time periods in which capacity expansion to be madeState: Capacity at the end of each time period t,

: Present capacity before expansion : Minimum present value of total cost of capacity

expansion from present to the time t

1+tS

1S)( 1+tt Sf

Stage 1Stage t

Stage TSt

Ct

xt

St+1S1 S2

C1 CT

x1 xT

ST ST+1


Capacity Expansion Problem: Forward Recursion …contd.

For the first stage, objective function

Value of S2 can be between K1 and KT

where K1 is the required capacity at the end of time period 1 and KT is the final capacity required

Now, for the first two stages together,

Value of S3 can be between K2 and KT

),(min),(min)(

1211

11121

SSSCxSCSf

−==

( )[ ]

( )[ ])(,min

)(,min)(

2312232

2122232

222

222

xSfxxSC

SfxSCSf

xx

xx

−+−=

+=

Ω∈

Ω∈


Capacity Expansion Problem: Forward Recursion …contd.

In general, for a time period t,

Subjected to

For the last stage, i.e. t =T, need to be solved only for

( )[ ])(,min)( 1111 tttttttx

xtt xSfxxSCSftt

t

−+−= +−+Ω∈

+

Ttt KSK ≤≤ +1

)( 1+TT Sf

TT KS =+1


Capacity Expansion Problem: Backward Recursion

Stages of the model: Time periods in which capacity expansion to be madeState: Capacity at the beginning of each time period t,

: Minimum present value of total cost of capacity expansion in periods t through TFor the last period T, the final capacity should reach KT after doing the capacity expansions

Value of ST can be between KT-1 and KT

tS)( tt Sf

( )[ ]TTTx

xTT xSCSfTT

T

,min)(Ω∈

=


Capacity Expansion Problem: Backward Recursion …contd.

In general, for a time period t,

Solved for all values of St ranging from Kt-1 and Kt

For period 1, the above equation must be solved only for St = S1

( )[ ])(,min)( 1 ttttttx

xtt xSfxSCSftt

t

++= +Ω∈


Capacity Expansion: Numerical Example

Consider a five stage capacity expansion problem

The minimum capacity to be achieved at the end of each time period is given in the table below

255

204

203

102

51

Dtt


Capacity Expansion: Numerical Example …contd.

Expansion costs for each combination of expansion


Numerical Example: Forward Recursion

Consider the first stage, t =1The final capacity for stage 1, S2 can take values between D1 to D5

Let the state variable can take discrete values of 5, 10, 15, 20and 25Objective function for 1st subproblem with state variable as S2 can be expressed as

),(min),(min)(

1211

11121

SSSCxSCSf

−==


Numerical Example: Forward Recursion …contd. Computations for stage 1 are given in the table below


Numerical Example: Forward Recursion …contd.

Now considering the 1st and 2nd stages togetherState variable S3 can take values from D2 to D5

Objective function for 2nd subproblem is

The value of x2 should be taken in such a way that the minimum capacity at the end of stage 2 should be 10, i.e.

( )[ ]

( )[ ])(,min

)(,min)(

2312232

2122232

222

222

xSfxxSC

SfxSCSf

xx

xx

−++−=

+=

Ω∈

Ω∈

103 ≥S


Numerical Example: Forward Recursion …contd. Computations for stage 2 are given in the table below



Like this, repeat this steps till t = 5The computation tables are shown



For the 5th subproblem, state variable S6 = D5



Figure showing the solutions with the cost of each addition along the links and the minimum total cost at each node

Optimal cost of expansion is 23 unitsBy doing backtracking from the last stage (farthest right node) to the initial stage, the optimal expansion to be done at 1st stage = 10 units, 3rd stage = 15 units and rest all stages = 0 units


Numerical Example: Backward Recursion

Capacity at the final stage is given as S6 = 25

Consider the last stage, t =5

Initial capacity for stage 5, S5 can take values between D4 to D5

Objective function for 1st subproblem with state variable as S5

can be expressed as

The optimal cost of expansion can be achieved by following the same procedure to all stages

( )[ ]55555 ,min)( xSfSfTT

Tx

xΩ∈

=


Numerical Example: Backward Recursion …contd.

Computations for all stages



Optimal cost of expansion is obtained from the node value at the first node i.e. 23 units

Optimal expansions to be made are 10 units at the first stage and 15 units at the last stage


Capacity Expansion Problem: Uncertainty

The future demand and the future cost of expansion in this problem are highly uncertainHence, the solution obtained cannot be used for making expansions till the end period, TBut, decisions about the expansion to be done in the current period can be very well done through thisFor the uncertainty on current period decisions to be less , thefinal period T should be selected far away from the current period


Thank You

Optimization Methods: Dynamic Programming Applications – Reservoir Operation 1


Reservoir Operation

Introduction

In the previous lectures, we discussed about the application of dynamic programming in

water allocation and capacity expansion of infrastructure. Another major application is in the

field of reservoir operation, which will be discussed in this lecture.

Reservoir operation – Steady state optimal policy

Consider a single reservoir receiving inflow and making releases for each time period t.

The maximum capacity of the reservoir is K. The optimization problem is to find the

sequence of releases to be made from the reservoir that maximizes the total net benefits.

These benefits may be from hydropower generation, irrigation, recreation etc. Let and

be the initial and final storages for time period t. Expressing net benefits as a function of ,

and , the net benefit for period t is

ti tr

tS 1+tS

1+tS tr

tS

( )tttt rSSNB ,, 1+ .

If there are T periods in a year, then the objective function is to maximize the total net

benefits from all periods.

Maximize ( )∑=

+

T

ttttt rSSNB

11,,

This is subject to continuity and also capacity constraints. Neglecting all minor losses like

evaporation, seepage etc and assuming that there is no overflow, the continuity relation can

be written as,

TtforriSS tttt ,...2,11 =−+=+

The capacity constraint can be expressed as,

TtforKSt ,...2,1=≤



The above formulated problem can be solved as a sequential process either using forward or

backward approach. Here the stages are the time periods and the states are the storage

volumes.

Assume that there are T periods in a year. In order to find the steady state policy, select a

period in a particular year in the near future (to get steady solution). Usually in almost all

problems, the last period T is taken as the terminal period. At this stage, the optimal release

will be independent of the inflow and also the net benefit .

tr

ti tNB

Now, consider the terminal period as T of a particular year after which reservoir is no longer

useful. Solving this problem in a backward recursion method, let t represents the period in a

year from T to 1 and n represents the periods remaining from t till end. Thus, t will take

values starting from T, decreasing to 1 (which will complete one year) and then again taking

a value of T and repeating the values. The value of n starts from 1 (while considering the Tth

period of last year) and while moving backwards its value keeps on increasing i.e. at the

beginning of the last year, the value of n = T and at the beginning of second last year its

value will be equal to 2T and so on.

Starting from T of last year, which is at the far right, there is only one period remaining.

Thus, in this case t=T and n=1. Let ( )TT Sf 1

( )TT Sf 1

be the maximum net benefit in the last period of

the year considered. can be expressed as

( ) ( ) [ ]TTTTTT

KiSriSr

rTT rriSSNBSf

TTTTTT

T

,,max0

1 −+=

−+≥+≤

≥

which should be solved for all S values from 0 to K. T

Considering the last two stages together for which t=T-1 and n=2, the objective function can

be written as



( ) ( ) ( )[ ]1111

111111012

1 ,,max

111111

1−−−−−−−−−

−+≥+≤

≥−− −++−+=

−−−−−−

−TTTTTTTTTT

KiSriSr

rTT riSfrriSSNBSf

TTTTTT

T

This also is solved for all values from 0 to K. 1−TS

In general, for a period t of a particular year with n periods remaining, the function can be

written as

( ) ( ) ( )[ ]tttn

ttttttt

KiSriSr

rtn

t riSfrriSSNBSf

tttttt

t

−++−+= −+

−+≥+≤≥

110

,,max

where the index t decreases from T to 1 and then takes the value T again for the previous year

and the cycle repeats while the index n starts from 1 and increases at each successive stage.

This cycle can be repeated till the optimum values of for an initial storage will be the

same as the corresponding and of previous year. Such a solution is called stationary

solution. The maximum net benefit can be obtained as the difference of and

tr tS

tr tS

( )tTn

t Sf + ( )tn

t Sf

tS

for any and t.

Numerical example (Loucks et al., 1981)

Consider a reservoir for which the desirable constant storage is 20 units and the constant

release is 25 units. The capacity of the reservoir is 30 units and the inflows for three seasons

are given as 10, 50 and 20 units. The problem is to find the optimum and that

minimizes the total squared deviation from the release and storage targets given. Hence, the

objective function is

tS tr

( ) ( )[ ]22 2520 rS −+− S

tr

tt . Let take the discrete values of 0, 10, 20, 30

and take the values of 10, 20, 30, 40.

t

Solution:



Consider a year after which the reservoir is no longer useful. The problem can be expressed

as a sequential process as shown in the figure below

Here no. of seasons (periods), T = 3. Considering the last period for which t = 3 and n = 1,

the optimization function is

Minimize ( ) ( ) ( )[ ]23

233

13 2520 rSSf −+−=

Inflow for 3rd season, I3 = 20 units and capacity of the reservoir, K = 30 units.

The release constraints can be expressed as

203

333

+≤+≤

SISr

and

30203

333

−+≥−+≥

SKISr

The computation for the first subproblem (n = 1) is shown in the table below.



Table 1

State variable,

S3Release, r3 ( ) ( )3

23 2520 rS −+− ( )3

13 Sf

Optimal

release, r3*

10 625 0

20 425 425 20

10 325

20 125 10

30 125

125 20, 30

10 225

20 25

30 25 20

40 225

25 20, 30

10 325

20 125

30 125 30

40 325

125 20, 30

Now considering the last two periods (n =2), the optimization function is

Minimize ( ) ( ) ( )[ ] ( )2221

32

22

222

2 2520 rISfrSSf −++−+−=

Inflow for 2nd season, I2 = 50 units. The release constraints can be expressed as

5022 +≤ Sr and

305022 −+≥ Sr

The computation for the second subproblem (n = 2) is shown in the table below.

For S2=30, i.e. . Since r50..3050 222 ≥−+≥ reiSr 502 ≥r 2 can take values only of 10,

20, 30 and 40 only, the release cannot be made for S2=30.

Table 2



State

variable,

S2

Release,

r2

( )( ) ⎥

⎥⎦

⎤

⎢⎢⎣

⎡

−+

−2

2

22

25

20

r

SS2+

I2 -

r2

( )222

13

rISf

−+(5)+(3) ( )2

22 Sf

Optimal

release,

r2*

20 425 30 125 550

30 425 20 25 450 0

40 625 10 125 750

450 30

30 125 30 125 250 10

40 325 20 25 350 250 30

20 40 225 30 125 350 350 40

30 na na na na na na na

The same procedure is repeated for all stages till n = 7. The summarized solution for this

problem is given in the table below.

Table 3

n = 1 n = 2 n =3 Initial Storage, tS ( )3

13 Sf ∗

3r ( )22

2 Sf ∗2r ( )1

31 Sf ∗

1r

0 425 20 450 30 1075 10

10 125 20, 30 250 30 575 10, 20

20 25 20, 30 350 40 275 20

30 125 20, 30 -- na 375 30

Table 4

n = 4 n =5 n =6 Initial Storage, tS ( )3

43 Sf ∗

3r ( )25

2 Sf ∗2r ( )1

61 Sf ∗

1r

0 1200 10 725 30 1350 10

10 600 10 525 30 850 10, 20

20 300 20 625 40 550 20

30 400 30 -- na 650 30



Table 5

n = 7 Initial Storage, tS ( )3

73 Sf ∗

3r

0 1475 10

10 875 10

20 575 20

30 675 30

At this stage, the value of at n = 7 and n = 4 are exactly the same. Also the difference

- = 275 is same for all . This value is the minimum total squared deviations

from the target release and storage. Thus, the stationary policy obtained is given below.

∗3r

( )37

3 Sf ( )34

3 Sf tS

Table 6

Optimal Releases tS

∗1r ∗

2r ∗3r

0 10 30 10

10 10, 20 30 10

20 20 40 20

30 30 -- 30

A main assumption made in dynamic programming is that the decisions made at one stage is

dependent only on the state variable and is independent of the decisions taken in other stages.

In cases where decisions made at one stage are dependent on the earlier decisions, then

dynamic programming will not be an appropriate optimization technique.




1. Bellman, R., Dynamic Programming, Princeton University Press, Princeton, N.J, 1957.





4. Vedula, S., and P.P. Mujumdar, Water Resources Systems: Modelling Techniques and





Reservoir Operation


Objectives

To develop the steady state operational policy of a single reservoirTo develop backward recursive equations for this operational policy To demonstrate the method using a numerical example


Reservoir Operation – Steady State Policy

Consider a single reservoir receiving inflow it and making releases rt for each time period tMaximum capacity of the reservoir is KReservoir receives benefits from hydropower generation, irrigation, recreation etcOptimization problem: Find the sequence of releases to be made from the reservoir that maximizes the total net benefitsSt : initial storage for time period tSt+1 : final storage for time period tNet benefits can be expressed as a function of rt, St and St+1 as

( )tttt rSSNB ,, 1+


Reservoir Operation – Steady State Policy …contd.

Let there be T periods in a year, then the objective function is to maximize the total net benefits from all periods

Maximize

The constraints areContinuity Equation after neglecting all the minor losses and assuming no overflow

Capacity constraint

( )∑=

+

T

ttttt rSSNB

11,,

TtforriSS tttt ,...2,11 =−+=+

TtforKSt ,...2,1=≤


Steady State Policy : Backward Recursion

Take stages as the time periods and the states as the storage volumesAssume that there are T periods in a yearTo find the steady state policy, select a period in a particularyear, after which it is assumed that the reservoir is no longer useful (maximum lifetime of the reservoir)Usually in almost all problems, the last period T is taken as the terminal periodAt this stage, the optimal release will be independent of the inflow and also the net benefit.


Steady State Policy : Backward Recursion

Consider the terminal period as T of a particular year after which reservoir is no longer usefulSolving this problem in a backward recursion method, let t represents the period in a year from T to 1 and n represents the periods remaining from t till endt will take values starting from T, decreasing to 1 (which will complete one year) and then again taking a value of T and repeating the valuesValue of n starts from 1 (while considering the Tth period of last year) and while moving backwards its value keeps on increasing i.e. at the beginning of the last year, the value of n = T and at the beginning of second last year its value will be equal to 2T and so on.


Steady State Policy : Backward Recursion …contd.

Starting from the last period T of a particular year, there is only one period remaining

The maximum net benefit for this last period is

This is solved for all ST values from 0 to K

( ) ( ) [ ]TTTTTT

KiSriSr

rTT rriSSNBSf

TTTTTT

T

,,max0

1 −+=

−+≥+≤

≥



Considering the last two stages together, the objective function is

ST-1 also ranges from 0 to KIn general, for a period t with n periods remaining, the function can be written as

where the index t decreases from T to 1 and then takes the value T again and the cycle repeats for another year the index n starts from 1 and increases at each successive stage

( ) ( ) ( )[ ]tttn

ttttttt

KiSriSr

rtn

t riSfrriSSNBSf

tttttt

t

−++−+= −+

−+≥+≤

≥

110

,,max

( ) ( ) ( )[ ]1111

111111012

1 ,,max

111111

1−−−−−−−−−

−+≥+≤

≥−− −++−+=

−−−−−−

−TTTTTTTTTT

KiSriSr

rTT riSfrriSSNBSf

TTTTTT

T



The cycle is repeated till the optimum values of rt for an initial storage St will be the same as the corresponding rt and St of previous year

Such a solution is called stationary solution

Maximum net benefit is the difference of and for any St and t.

( )tTn

t Sf + ( )tn

t Sf


Steady State Policy : Numerical Example

Consider a reservoir for which the desirable constant storage is 20 units and the constant release is 25 units.

Capacity of the reservoir is 30 units.

Inflows for three seasons are 10, 50 and 20 units.

Optimization problem: To find the optimum rt and St that minimizes the total squared deviation from the release and storage targetsgiven.

Thus, the objective function is

Minimize ( ) ( )[ ]22 2520 tt rS −+−


Steady State Policy : Numerical Example …contd.

St take the discrete values of 0, 10, 20, 30 and rt take the values of 10, 20, 30, 40

Solution:Backward recursion is used to solve the problem Consider a year after which the reservoir is no longer useful.Expressing the problem as a sequential process as shown in the figure below



Considering the last period for which t = 3 and n = 1, the optimization function is

Minimize

Inflow for 3rd season, I3 = 20 units and capacity of the reservoir, K = 30 units

The release constraints can be expressed as

and

( ) ( ) ( )[ ]23

233

13 2520 rSSf −+−=

203

333

+≤+≤

SISr

30203

333

−+≥−+≥

SKISr



Computation for the first subproblem (n = 1)



Considering the last two periods (n =2), the optimization function is

Inflow for 2nd season, I2 = 50 units

Release constraints can be expressed as

and

While computing the optimal releases for S2=30,

Since r2 can take values only of 10, 20, 30 and 40 only, the release cannot be made for S2=30.

50..3050 222 ≥−+≥ reiSr

( ) ( ) ( )[ ] ( )2221

32

22

222

2 2520 rISfrSSf −++−+−=

5022 +≤ Sr 305022 −+≥ Sr


Steady State Policy : Numerical Example …contd.Computation for the second subproblem (n = 2)



The same procedure is repeated for all stages till n = 7. The summarized solution for this problem is given in the table below

Solution for the last year



Solution for the second last year



Solution for the last period of third last year



At this stage, the value of r3* at n = 7 and n = 4 are exactly the same.

Also the difference is same for all St

This value is the minimum total squared deviations from the target release and storage

Stationary optimal policy

obtained is shown

( ) ( ) 27534

337

3 =− SfSf


Steady State Policy: Limitations

Assumption in dynamic programming: Decisions made at one stage is dependent only on the state variable and is independent of the decisions taken in other stages

Therefore dynamic programming will not be an appropriate technique where decisions are dependent on the earlier decisions.


Thank You

Optimization Methods: Integer Programming - Learning Objectives

Module 7: Integer Programming

Learning Objectives

The previous modules discussed about the optimization methods using linear programming

and dynamic programming techniques with almost no limit in the values taken by decision

variables. But many practical problems require the constraint of integrality in decision

variable values. This can be solved using Integer programming which is the main objective of

this module.

The module gives a description of Integer programming, with emphasis on Integer Linear

Programming (ILP), its relation with linear programming and the various types of integer

programming,. Among the various techniques for solving ILP, a well known method

Gomory’s cutting plane method is explained. The concept of this method is explained

graphically to give the reader a better understanding. The generation of Gomory constraints

are also well explained for all- integer LP and also mixed ILP, which will in turn help the

reader to solve any type of integer programming.


1. Formulate the ILP.

2. Generate Gomory constraints.

3. Solve all-integer LP by Gomory cutting plane method.

4. Solve mixed Integer LP by Gomory cutting plane method.


Optimization Methods: Integer Programming – Integer Linear Programming 1


Integer Linear Programming

Introduction

In all the previous lectures in linear programming discussed so far, the design variables

considered are supposed to take any real value. However in practical problems like

minimization of labor needed in a project, it makes little sense in assigning a value like 5.6 to

the number of labourers. In situations like this, one natural idea for obtaining an integer

solution is to ignore the integer constraints and use any of the techniques previously

discussed and then round-off the solution to the nearest integer value. However, there are

several fundamental problems in using this approach:

1. The rounded-off solutions may not be feasible.

2. The objective function value given by the rounded-off solutions (even if some are

feasible) may not be the optimal one.

3. Even if some of the rounded-off solutions are optimal, checking all the rounded-off

solutions is computationally expensive ( possible round-off values to be considered

for an variable problem)

n2

n

Types of Integer Programming

When all the variables in an optimization problem are restricted to take only integer values, it

is called an all – integer programming problem. When the variables are restricted to take only

discrete values, the problem is called a discrete programming problem. When only some

variable values are restricted to take integer or discrete, it is called mixed integer or discrete

programming problem. When the variables are constrained to take values of either zero or 1,

then the problem is called zero – one programming problem.

Integer Linear Programming

Integer Linear Programming (ILP) is an extension of linear programming, with an additional

restriction that the variables should be integer valued. The standard form of an ILP is of the

form,



0

max

≥≤

XbAXtosubject

XcT

X must be integer valued

The associated linear program dropping the integer restrictions is called linear relaxation LR.

Thus, LR is less constrained than ILP. If the objective function coefficients are integer, then

for minimization, the optimal objective for ILP is greater than or equal to the rounded-off

value of the optimal objective for LR. For maximization, the optimal objective for ILP is less

than or equal to the rounded-off value of the optimal objective for LR.

For a minimization ILP, the optimal objective value for LR is less than or equal to the

optimal objective for ILP and for a maximization ILP, the optimal objective value for LR is

greater than or equal to that of ILP. If LR is infeasible, then ILP is also infeasible. Also, if LR

is optimized by integer variables, then that solution is feasible and optimal for IP.

A most popular method used for solving all-integer and mixed-integer linear programming

problems is the cutting plane method by Gomory (Gomory, 1957).

Gomory’s Cutting Plane Method for All – Integer Programming

Consider the following optimization problem.

0,4593


21

21

21

21

≥≤+≤−+=

xxxx

xxxxZ

x1 and x2 are integers

The graphical solution for the linear relaxation of this problem is shown below.



0 1 2 3 4 5 6 70

1

2

3

4

5

6

7

D C

B

A

x1

x2

It can be seen that the solution is 733,7

54 21 == xx and the optimal value of 7417=Z .

The feasible solutions accounting the integer constraints are shown by red dots. These points

are called integer lattice points. The original feasible region is reduced to a new feasible

region by including some additional constraints such that an extreme point of the new

feasible region becomes an optimal solution after accounting for the integer constraints.

The graphical solution for the example previously discussed taking x1 and x2 as integers are

shown below. Two additional constraints (MN and OP) are included so that the original

feasible region ABCD is reduced to a new feasible region AEFGCD. Thus the solution for

this ILP is and the optimal value is3,4 21 == xx 15=Z .

7417=Z

( )733,7

54

62 21

− xx

4593 21

≤

+ ≤xx



0 1 2 3 4 5 6 70

1

2

3

4

5

6

7

D C

B

A

Additional constraints

G

E

F

x1

x2

(4,3)

O

P

M

N

( )733,7

54 7

417=Z

15=Z

Gomary proposed a systematic method to develop these additional constraints known as

Gomory constraints.

Generation of Gomory Constraints:

Let the final tableau of an LP problem consist of n basic variables (original variables) and m

non basic variables (slack variables) as shown in the table below. The basic variables are

represented as xi (i=1,2,…,n) and the non basic variables are represented as yj (j=1,2,…,m).



Table 1

Variables Basis Z

1x 2x … ix … nx 1y 2y … jy … my rb

Z 1 0 0 0 0 c1 c2 cj cm b

1x 0 1 0 0 0 c11 c12 c1j c1m 1b

2x 0 0 1 0 0 c21 c22 c2j c2m 2b

…

ix 0 0 0 1 0 c31 c32 c3j c3m ib

…

nx 0 0 0 0 1 c41 c42 c4j c4m nb

Choose any basic variable with the highest fractional value. If there is a tie between two

basic variables, arbitrarily choose any of them as . Then from the i

ix

ix th equation of table,

∑=

−=m

jjijii ycbx

1 ……..(1)

Express both and as an integer value plus a fractional part. ib ijc

)3........()2........(

ijijij

iii

ccbb

αβ+=+=

where ib , ijc denote the integer part and iβ , ijα denote the fractional part. iβ will be a

strictly positive fraction ( 10 << i )β and ijα is a non-negative fraction ( )10 <≤ ijα .

Substituting equations (2) and (3) in (1), equation (1) can be written as

∑∑==

−−=−m

jjijiij

m

jiji ycbxy

11αβ …….(4)

For all the variables and to be integers, the right hand side of equation (4) should be

an integer.

ix jy

=−∑=

j

m

jiji y

1αβ integer ……(5)



Since ijα are non-negative integers and are non-negative integers, the term will

always be a non-negative number. Thus we have,

jy j

m

jij y∑

=1α

11

<≤⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−∑

=ij

m

jiji y βαβ …….(6)

Hence the constraint can be expressed as

01

≤−∑=

j

m

jiji yαβ …….(7)

By introducing a slack variable (which should also be an integer), the Gomory constraint

can be written as

is

ij

m

jiji ys βα −=−∑

=1 …….(8)

General procedure for solving ILP:

1. Solve the given problem as an ordinary LP problem neglecting the integer

constraints. If the optimum values of the variables are integers itself, then there is

nothing more to be done.

2. If any of the basic variables has fractional values, introduce the Gomory constraints

as discussed in the previous section. Insert a new row with the coefficients of this

constraint, to the final tableau of the ordinary LP problem (Table 1).

3. Solve this by applying the dual simplex method.

Since the value of 0=jy in Table 1, the Gomory constraint equation becomes

iis β−= which is a negative value and thus infeasible. Dual simplex method

is used to obtain a new optimal solution that satisfies the Gomory constraint.

4. Check whether the new solution is all-integer or not. If all values are not integers,

then a new Gomory constraint is developed from the new simplex tableau and the

dual simplex method is applied again.

5. This process is continued until an optimal integer solution is obtained or it shows that

the problem has no feasible integer solution.



Thus, the fundamental idea behind cutting planes is to add constraints to a linear program

until the optimal basic feasible solution takes on integer values. Gomory cuts have the

property that they can be generated for any integer program, but has the disadvantage that the

number of constraints generated can be enormous depending upon the number of variables.



Integer Programming

All Integer Linear Programming


Objectives

To discuss the need for Integer Programming (IP)

To discuss about the types of IP

To explain Integer Linear Programming (ILP)

To discuss the Gomory Cutting Plane method for solving ILP

Graphically

Theoretically


Introduction

In many practical problems, the values of decision variables areconstrained to take only integer values

For example, in minimization of labor needed in a project, the number of labourers should be an integer value

By rounding off a real value to an integer value have several fundamental problems like

Rounded solutions may not be feasibleEven if the solutions are feasible, the objective function given by the rounded off solutions may not be the optimal oneFinally, even if the above two conditions are satisfied, checking all the rounded-off solutions is computationally expensive (2n possible solutions to be considered for an n variable problem)

This demands the need for Integer Programming


Types of IP

All Integer Programming:All the variables are restricted to take only integer values

Discrete Programming:All the variables are restricted to take only discrete values

Mixed Integer or Discrete Programming: Only some variables are restricted to take integer or discrete values

Zero – One Programming:Variables are constrained to take values of either zero or 1


Integer Linear Programming (ILP)

An extension of linear programming (LP)Additional constraint: Variables should be integer valuedStandard form of an ILP:

X must be integer valued

Associated linear program, dropping the integer restrictions, is called linear relaxation (LR)

0

max

≥≤

XbAXtosubject

XcT


Checks for ILP:

Minimization: Optimal objective value for LR is less than or equal to the optimal objective for ILP

Maximization: Optimal objective value for LR is greater than or equal to that of ILP

If LR is infeasible, then ILP is also infeasible

If LR is optimized by integer variables, then that solution is feasible and optimal for IP


All – Integer Programming

Most popular method: Gomory’s Cutting Plane methodOriginal feasible region is reduced to a new feasible region by including additional constraints such that all vertices of the new feasible region are now integer pointsThus, an extreme point of the new feasible region becomes an optimal solution after accounting for the integer constraintsConsider the optimization problem

; x1 and x2 are integers0,4593


21

21

21

21

≥≤+

≤−+=

xxxx

xxxxZ


Graphical Illustration

Graphical solution for the linear approximation (neglecting the integer requirements) is shown in figure


Graphical Illustration …contd.

Optimal value of and the solution is Red dots in the figure show the feasible solutions accounting for the integer requirementsThese points are called integer lattice pointsNow to reduce the original feasible region to a new feasible region (considering x1 and x2 as integers) is done by including additional constraintsGraphical solution for the IP is shown in figure belowTwo additional constraints (MN and OP) are included so that the original feasible region ABCD is reduced to a new feasible region AEFGCD

7417=Z 7

33,754 21 == xx


Graphical Illustration …contd.

Optimal value of ILP is and the solution is15=Z 3,4 21 == xx


Generation of Gomory Constraints

Consider the final tableau of an LP problem consisting of nbasic variables (original variables) and m non basic variables (slack variables)The basic variables are represented as xi(i=1,2,…,n) and the non basic variables are represented as yj(j=1,2,…,m).


Generation of Gomory Constraints …contd.

Pick the variable having the highest fractional value. In case of a tie, choose arbitrarily any variable as

From the ith equation,

Expressing both bj and cij as an integer part plus a fractional part,

ix

ix

∑=

−=m

jjijii ycbx

1

ijijij

iii

ccbb

αβ

+=+=



, denote the integer part and

, denote the fractional part for which and

Thus, the equation becomes,

ib ijc

ijαiβ

∑∑==

−−=−m

jjijiij

m

jiji ycbxy

11αβ

( )10 << iβ ( )10 <≤ ijα



Considering the integer reuirements, the RHS of the equation also should be an integer. Thus, we have

Hence, the constraint can be expressed as,

After introducing a slack variable si, the final Gomory constraint can be written as,

11

<≤⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−∑

=ij

m

jiji y βαβ

01

≤−∑=

j

m

jiji yαβ

ij

m

jiji ys βα −=−∑

=1


Procedure for solving All-Integer LP

Solve the problem as an ordinary LP problem neglecting the integer requirements.

If the optimum values of the variables are not integers, then choose the basic variable which has the largest fractional value, and generate Gomory constraint for that variable.

Insert a new row with the coefficients of this constraint, to the final tableau of the ordinary LP problem.

Solve this by applying the dual simplex method


Procedure for solving All-Integer LP …contd.

Check whether the new solution is all-integer or not.

If all values are not integers, then a new Gomory constraint is developed for the non-integer valued variable from the new simplex tableau and the dual simplex method is applied again.

The process is continued until An optimal integer solution is obtained or

It shows that the problem has no feasible integer solution.


Thank You

Optimization Methods: Integer Programming – Mixed Integer Programming 1


Mixed Integer Programming

Introduction

In the previous lecture we have discussed the procedure for solving integer linear

programming in which all the decision variables are restricted to take only integer values. In a

Mixed Integer Programming (MIP) model, some of the variables are real valued and some are

integer valued. When the objective function and constraints for a MIP are all linear, then it is

a Mixed Integer Linear Program (MILP). Although Mixed Integer Nonlinear Programs

(MINLP) also exists, in this chapter we will be dealing with MILP only.

Mixed Integer Programming

In mixed integer programming (MIP), only some of the decision and slack variables are

restricted to take integer values. Gomory cutting plane method is used popularly to solve MIP

and the solution procedure is similar in many aspects to that of all-integer programming. The

first step is to solve the MIP problem as an ordinary LP problem neglecting the integer

restrictions. The procedure ends if the values of the basic variables which are constrained to

take only integer values happen to be integers in this optimal solution. If not, a Gomory

constraint is developed for the basic variable with integer restriction and has largest fractional

value which is explained below. The rest of the procedure is same as that of all-integer

programming.

Generation of Gomory Constraints:

Let be the basic variable which has integer restriction and also has largest fractional value

in the optimal solution of ordinary LP problem. Then from the i

ixth equation of table,

∑=

−=m

jjijii ycbx

1

As explained in all-integer programming, express as an integer value plus a fractional part. ib

iii bb β+=

ijc can be expressed as



−+ += ijijij ccc

where

⎪⎩

⎪⎨⎧

<

≥=

⎪⎩

⎪⎨⎧

<

≥=

−

+

0

00

00

0

ijij

ijij

ij

ijijij

cifc

cifc

cif

cifcc

Therefore,

( ) ( )iiij

m

jijij xbycc −+=+∑

=

−+ β1

Since is restricted to be an integer, ix ib is also an integer and 10 << iβ , the value of

( )iii xb −+β can be or <0. 0≥

Case I: ( )iii xb −+β 0≥

For to be an integer, ix ( ) ,....21 ++=−+ iiiiii ororxb ββββ

Therefore,

( ) ij

m

jijij ycc β≥+∑

=

−+

1

Finally it takes the form

ij

m

jij yc β≥∑

=

+

1

Case II: ( )iii xb −+β <0

For to be an integer, ix

( ) ,....321 iiiiii ororxb ββββ +−+−+−=−+

Therefore,

( ) 11

−≤+∑=

−+ij

m

jijij ycc β



Finally it takes the form

11

−≤∑=

−ij

m

jij yc β

Dividing this inequality by ( 1−i )β and multiplying with iβ , we have

ij

m

jij

i

i yc βββ

≥− ∑

=

−

11

Considering both cases I and II, the final form of the inequality becomes (since one of the

inequalities should be satisfied)

ij

m

jij

i

ij

m

jij ycyc β

ββ

≥−

+ ∑∑=

−

=

+

11 1

Then, the Gomory constraint after introducing a slack variable is is

ij

m

jij

i

ij

m

jiji ycycs β

ββ

−=−

−− ∑∑=

−

=

+

11 1

Generate the Gomory constraint for the variables having integer restrictions. Insert this

constraint as the last row of the final tableau of LP problem and solve this using dual simplex

method. MIP techniques are useful for solving pure-binary problems and any combination of

real, integer and binary problems.



Integer Programming

Mixed Integer Linear Programming


Objectives

To discuss about the Mixed Integer Programming (MIP)

To discuss about the Mixed Integer Linear Programming (MILP)

To discuss the generation of Gomory constraints

To describe the procedure for solving MILP


Introduction

Mixed Integer Programming:Some of the decision variables are real valued and some are integer valued

Mixed Integer Linear Programming:MIP with linear objective function and constraints

The procedure for solving an MIP is similar to that of All Integer LP with some exceptions in the generation of Gomory constraints.


Generation of Gomory Constraints

Consider the final tableau of an LP problem consisting of n basic variables (original variables) and m non basic variables (slack variables)Let xi be the basic variable which has integer restrictions



From the ith equation,

Expressing bj as an integer part plus a fractional part,

Expressing cij as where

∑=

−=m

jjijii ycbx

1

iii bb β+=

−+ += ijijij ccc

⎪⎩

⎪⎨⎧

<

≥=

⎪⎩

⎪⎨⎧

<

≥=

−

+

0

00

00

0

ijij

ijij

ij

ijijij

cifc

cifc

cif

cifcc



Thus,

Since xi and are restricted to take integer values and also the value of can be or <0

Thus we have to consider two cases.

( ) ( )iiij

m

jijij xbycc −+=+∑

=

−+ β1

ib 10 << iβ( )iii xb −+β 0≥



Case I: For xi to be an integer,

Therefore,

Finally it takes the form,

( ) 0≥−+ iii xbβ

( ) ,....21 ++=−+ iiiiii ororxb ββββ

( ) ij

m

jijij ycc β≥+∑

=

−+

1

ij

m

jij yc β≥∑

=

+

1



Case II: For xi to be an integer,

Therefore,

Finally it takes the form,

( ) 0<−+ iii xbβ

( ) ,....321 iiiiii ororxb ββββ +−+−+−=−+

( ) 11

−≤+∑=

−+ij

m

jijij ycc β

11

−≤∑=

−ij

m

jij yc β



Dividing this inequality by and multiplying with , we have

Now considering both cases I and II, the final form of the Gomory constraint after adding one slack variable si is,

( )1−iβ iβ

ij

m

jij

i

i yc ββ

β ≥− ∑

=

−

11

ij

m

jij

i

ij

m

jiji ycycs β

ββ

−=−

−− ∑∑=

−

=

+

11 1


Procedure for solving Mixed-Integer LP

Solve the problem as an ordinary LP problem neglecting the integrality constraints.

Generate Gomory constraint for the fractional valued variable that has integer restrictions.

Insert a new row with the coefficients of this constraint, to the final tableau of the ordinary LP problem.

Solve this by applying the dual simplex method

The process is continued for all variables that have integrality constraints


Thank You

Optimization Methods: Integer Programming - Examples 1


Integer Programming - Examples

Introduction

In the previous two lectures, we have discussed about the procedures for solving integer

linear programming and also mixed integer linear programming. In this lecture we will be

discussing a few examples based on the methods already discussed.

Example of Cutting plane method for ILP:

These steps are explained using the example previously discussed.

0,4593


21

21

21

21

≥≤+≤−+=

xxxx

xxxxZ

x1 and x2 are integers

The standard form for the problem can be written as

0,,,4593


2121

221

121

21

≥=++

=+−+=

yyxxyxx

yxxxxZ

areyandyxx 2121 ,, integers

Step 1: Solve the problem as an ordinary LP problem neglecting the integer requirements.

Table 1

Variables Iteration Basis

Z

1x 2x 1y 2y rb

rs

r

cb

Z 1 -3 -1 0 0 0 --

1y 0 2 -1 1 0 6 3

1

2y 0 3 9 0 1 45 15



Table 2


1x 2x 1y 2y rb

rs

r

cb

Z 1 0 25

− 23 0 9 --

1x 0 1 21

− 21 0 3 -- 2

2y 0 0 221

23

− 1 36 724

Table 3


1x 2x 1y 2y rb

rs

r

cb

Z 1 0 0 78

215

7123 --

1x 0 1 0 146

211

733 -- 3

2x 0 0 1 71

− 212

724 --

Optimum value of Z is 7

123 as shown above. Corresponding values of basic variables

are 754

733

1 ==x , 733

724

2 ==x and those of non-basic variables are all zero

(i.e., ). The values are not integers. 021 == yy

Step 2: Introduce Gomory constraints.

Gomory constraint is developed for which is having high fractional value. From the row

corresponding to in the last table, we can write,

1x

1x

211 211

146

733 yyx −−=



Here 2110,21

1,146,0,14

6,75,4,7

331212121111111 ========= ααβ andccccbb ii .

From Eqn. Gomory constraint can be expressed as

211

211

146.,. 211

12121111

−=−−

−=−−

yysei

yys βαα

By inserting a new row with the coefficients of this constraint to the last table, we get

Table 4

Variables rb Iteration Basis Z

1x 2x 1y 2y 1s

Z 1 0 0 78

215 0

7123

1x 0 1 0 146

211 0

733

2x 0 0 1 71

− 212 0

724

1s 0 0 0 146

− 211

− 1 75

−

Step 3: Solve using dual simplex method

Table 5


1x 2x 1y 2y 1s

Z 1 0 0 78

215 0

7123

1x 0 1 0 146

211 0

733

2x 0 0 1 71

− 212 0

724

1s 0 0 0 146

− 211

− 1 75

−

Ratio -- -- 38 5 --



Table 6


1x 2x 1y 2y 1s

Z 1 0 0 0 91

38

347

1x 0 1 0 0 0 1 4

2x 0 0 1 0 91

31

− 3

11 4

1y 0 0 0 1 91

37

− 35

Optimum value of Z from the present tableau is 3

47 as shown above. Corresponding values

of basic variables are ,41 =x3

112 =x ,

37

1 −=y and those of nonbasic variables are all zero

(i.e., ). Here also the values of and are not integers. 012 == sy 2x 1y

Step 4: Generation of new Gomory constraint for 2x

From the row corresponding to in the last table, we can write, 2x

122 31

91

311 syx +−=

Therefore Gomory constraint can be written as

32

31

91

122 −=+− sys

Step 5: Adding this constraint to the previous table and solving using dual simplex method



Table 7


1x 2x 1y 2y 1s 2s

Z 1 0 0 0 91

38 0

347

1x 0 1 0 0 0 1 0 4

2x 0 0 1 0 91

31

− 0 3

11

1y 0 0 0 1 91

37

− 0 35

2s 0 0 0 0 91

− 0 31

32

−

Ratio -- -- -- 1 -- -- --

Table 8


1x 2x 1y 2y 1s 2s

Z 1 0 0 0 0 38

31 15

1x 0 1 0 0 0 1 0 4

2x 0 0 1 0 0 31

− 31 3

1y 0 0 0 1 0 37

− 31 1

5

2s 0 0 0 0 1 0 -3 6

The solution from this tableau is 0,6,1,3,4 122121 ====== sysyxx and the value of Z =

15. These are satisfying the integer requirement and hence the desired solution.



Example of Cutting plane method for MILP

Consider the example previously discussed with integer constraint only on . 2x

The standard form for the problem is

0,,,4593


2121

221

121

21

≥=++

=+−+=

yyxxyxx

yxxxxZ

2x should be an integer

Step 1: Solve the problem as an ordinary LP problem. The final tableau is showing the

optimal solutions are shown below.

Table 9


1x 2x 1y 2y rb

rs

r

cb

Z 1 0 0 78

215

7123 --

1x 0 1 0 146

211

733 -- 3

2x 0 0 1 71

− 212

724 --

Optimum value of Z is 7

123 and corresponding values of basic variables are 754

733

1 ==x ,

733

724

2 ==x and those of non-basic variables are all zero (i.e., 021 == yy ). This is not

satisfying the constraint of to be an integer. 2x

Step 2: Generate Gomory constraint.

Gomory constraint is developed for .From the row corresponding to in the last table,

we can write,

2x 2x

122 212

71

724 yyx −+=



Here 212,7

1,724

22212 −=== ccb .

Since 222 β+= bb as per Eqn(), the value of 32 =b and 73

2 =β .

Similarly −+ += 212121 ccc and −+ += 222222 ccc .

Therefore,

71,0 2121 −== −+ cc since 21c is negative

0,212

2222 == −+ cc since 22c is positive

Thus Gomory constraint can be written as

ij

m

jij

i

ij

m

jiji ycycs β

ββ

−=−

−− ∑∑=

−

=

+

11 1

73

283

212.,. 122 −=−− yysei

Insert this constraint as the last row of the final tableau.

Table 10


1x 2x 1y 2y 2s

Z 1 0 0 78

215 0

7123

1x 0 1 0 146

211 0

733

2x 0 0 1 71

− 212 0

724

2s 0 0 0 283

−212

− 1 73

−



Step 3: Solve using dual simplex method

Table 11


1x 2x 1y 2y 2s

Z 1 0 0 78

215 0

7123

1x 0 1 0 146

211 0

733

2x 0 0 1 71

− 212 0

724

4

2s 0 0 0 283

−212

− 1 73

−

Ratio -- -- 3

32 2.5 --

Table 12


1x 2x 1y 2y 2s

Z 1 0 0 87 0 1

233

1x 0 1 0 83 0 1

29

2x 0 0 1 41

− 0 1 3 4

2y 0 0 0 89 1

221

− 29

Optimum value of Z from the present tableau is 2

33 as shown above. Corresponding values

of basic variables are , , 5.41 =x 32 =x 5.42 =y and those of non-basic variables are all zero

(i.e., ). This solution is satisfying all the constraints. 021 == sy



References/ Further Reading:


New Delhi, 1987.

2. Kasana H.S., and Kumar K.D., Introductory Operations Research – Theory and

Applications, Springer Pvt. Ltd, New Delhi, 2003.

3. Rao, S.S., Engineering Optimization – Theory and Practice, New Age Intl. Pvt. Ltd.,

New Delhi, 1996.




Delhi, 2005.



Integer Programming

Examples


Objectives

To illustrate Gomory Cutting Plane Method for solving

All Integer linear Programming (AILP)

Mixed Integer Linear Programming (MILP)


Example Problem (AILP)

Consider the problem

x1, x2 are integers Standard form of the problem can be written as

x1, x2, y1 and y2 are integers

0,4593


21

21

21

21

≥≤+

≤−+=

xxxx

xxxxZ

0,,,4593


2121

221

121

21

≥=++

=+−+=

yyxxyxx

yxxxxZ


Example Problem (AILP) …contd.

Optimum value of Z is and the values of basic variables are 7

123

;754

733

1 ==x 733

724

2 ==x

Solve the problem as an ordinary LP (neglecting integer requirements)The final tableau of LP problem is shown below



Since the values of basic variables are not integers, generate Gomory constraint for x1 which has a high fractional value. For this, write the equation for x1 from the table above

Here,

Thus, Gomory constraint can be written as

211 211

146

733 yyx −−=

2110,21

1

,146,0,14

6

,75,4,7

33

121212

111111

1

===

===

===

α

α

β

andcc

cc

bb ii

211

211

146.,. 211

12121111

−=−−

−=−−

yysei

yys βαα



Solve this using Dual Simplex method

Insert this constraint as a new row in the previous table



Optimum value of Z is and the values of basic variables are

and 3

47

;41 =x ;3

112 =x

37

1 −=y



Since the values of basic variable x2 from this table is not an integer, generate Gomory constraint for x2. For this, write the equation for x2 from the table above


Insert this constraint as a new row in the last table and solve using dual simplex method

122 31

91

311 syx +−=

32

31

91

122 −=+− sys



Optimum value of Z is 15 and the values of basic variables arex1 = 4, x2 = 3, y1 = 1, s2 = 6 and y2 = s1 = 0.These are satisfying the constraints and hence the desired solution.


Example Problem (MILP)

Consider the previous problem with integer constraint only on x2

; x2 is an integerStandard form of the problem can be written as

; x2 should be an integer

0,4593


21

21

21

21

≥≤+

≤−+=

xxxx

xxxxZ

0,,,4593


2121

221

121

21

≥=++

=+−+=

yyxxyxx

yxxxxZ


Example Problem (MILP) …contd.

Optimum value of Z is and the values of basic variables are 7

123

;754

733

1 ==x 733

724

2 ==x

Solve the problem as an ordinary LP (neglecting integer requirements)The final tableau of LP problem is shown below



Since the value of x2 is not an integer, generate Gomory constraint for x2. For this, write the equation for x2 from the table above

Here,

Thus, the value of and Since, and ,

since is negativesince is positive

122 212

71

724 yyx −+=

212,7

1,724

22212 −=== ccb

32 =b 73

2 =β−+ += 212121 ccc −+ += 222222 ccc

71,0 2121 −== −+ cc

0,212

2222 == −+ cc21c

22c




Insert this constraint as a new row to the previous table and solve it using Dual Simplex method

ij

m

jij

i

ij

m

jiji ycycs β

ββ

−=−

−− ∑∑=

−

=

+

11 1

73

283

212.,. 122 −=−− yysei


Optimum value of Z is and the values of basic variables are

and that of non-basic variables are zero.

This solution is satisfying all the constraints an hence the desired.

233

;5.41 =x ;32 =x 5.42 =y



Thank You

Optimization Methods: Advanced Topics in Optimization - Learning Objectives

Module 8: Advanced Topics in Optimization

Learning Objectives

In the previous modules, we discussed about almost all the major techniques used for

optimization. This module gives a brief outline of some additional methods often used to

solve some typical optimization problems.

The module starts with a description of piecewise linear approximation of non-linear

functions which will help the reader to approach the nonlinear problems to split up into linear

sub problems and solve it using the ordinary simplex method. This will be followed by a

discussion of multi-objective optimization problems, in which two methods will be explained

to deal the multiple objectives. The decomposition of larger problems into smaller ones is

illustrated in multilevel optimization. A brief outline of direct and indirect search methods

will be given next. The importance of evolutionary algorithms and steps involved will be

elaborated with special emphasis on genetic algorithms. The applications of these methods in

civil engineering will be outlined toward the end of this module.


1. Solve a nonlinear problem through its linear approximation

2. Solve a multi-objective problem through weighted and constrained methods

3. Acquire an idea about the various direct and indirect search methods

4. Understand evolutionary algorithms

5. Visualize advanced optimization applications in civil engineering


Optimization Methods: Advanced Topics in Optimization - Piecewise linear approximation of a nonlinear function

1


Piecewise Linear Approximation of a Nonlinear Function

Introduction

In the previous lectures, we have learned how to solve a nonlinear problem using various

methods. It is clear that unlike in linear programming, for nonlinear programming there exists

no general algorithm due to the irregular behavior of nonlinear functions. One commonly

used technique for solving nonlinear problems is to first represent the nonlinear function

(both objective function and constraints) by a set of linear functions and then apply simplex

method to solve this using some restrictions. In this lecture we will be discussing about a

method to approximate a nonlinear function using linear functions. This method can be

applied only to a single variable nonlinear function. For a nonlinear multivariable function

consisting of ‘n’ variables, this method is applicable only if the function is in separable form

i.e., can be expressed as a sum of ‘n’ single variable functions. In this lecture, only single

variable nonlinear functions are discussed.

Piecewise Linearization

Any nonlinear single variable function f(x) can be approximated by a piecewise linear

function. Geometrically, this can be shown as a curve being represented as a set of connected

line segments as shown in the figure below.

t1 t4

f(t4)

f(t3)

f(t2)

f(t1)

x

f(x)

t3t2



2

Method 1:

Consider an optimization function having only one nonlinear term f(x). Let the x-axis of the

nonlinear function f(x) be divided by ‘p’ breaking points t1, t2, t2, …, tp and the corresponding

function values be f(t1), f(t2),…, f(tp). Suppose if ‘x’ can take values in the interval

, then the breaking points can be shown as, Xx ≤≤0

Xttt p ≡<<<≡ ...0 21 .

Expressing ‘x’ as a weighted average of these breaking points,

pptwtwtwx +++= ...2211

∑=

=p

iii twxei

1

.,.

The function f(x) can be expressed as,

( ) ( ) ( ) ( )( )∑

=

=

+++=p

iii

pp

tfw

tfwtfwtfwxf

1

2211 ...

where ∑=

=p

iiw

1

1

Finally the model can be expressed as

Max or Min ( ) ( )∑=

=p

iii tfwxf

1

subject to the additional constraints

∑

∑

=

=

=

=

p

ii

p

iii

w

xtw

1

1

1

Linearly approximated model stated above can be solved using simplex method with some

restrictions. The restricting condition specifies that there should not be more than two ‘wi’ in

the basis and also two ‘wi’ can take positive values only if they are adjacent. This is because,

if the actual value of ‘x’ is between ti and ti+1, then ‘x’ can be represented as a weighted



3

average of wi and wi+1 only i.e., 11 +++= iiii twtwx . Therefore, the contributing weights wi and

wi+1 only will be positive, rest all weights be zero.

Similarly for an objective function consisting of ‘n’ variables (‘n’ terms) represented as

Max or Min ( ) ( ) ( ) ( )nn xfxfxfxf +++= ....2211

subjected to ‘m’ constraints

( ) ( ) ( ) mjforbxgxgxg jnnjjj ,...,2,1....2211 =≤+++

The linear approximation of this problem is

nkforw

mjforbtgwtosubjected

tfwMinorMax

p

iki

j

n

k

p

ikikjki

n

k

p

ikikki

,...,2,11

,...,2,1)(

)(

1

1 1

1 1

==

=≤

∑

∑∑

∑∑

=

= =

= =

Method 2:

Let ‘x’ be expressed as a sum, instead of expressing as the weighted sum of the breaking

points as in the previous method. Then,

∑−

=

−

+=

++++=1

11

1211 ....p

ii

p

ut

uuutx

where is the increment of the variable ‘x’ in the interval , i.e., the bound of is iu ),( 1+ii tt iu

iii ttu −≤≤ +10 .

The function f(x) can be expressed as

( ) ( ) i

p

iiutfxf ∑

−

=

+=1

11 α

where iα represents the slope of the linear approximation between the points and

given by

1+it it



4

( ) ( )ii

iii tt

tftf−−

=+

+

1

1α .


Max or Min ( ) ( ) i

p

iiutfxf ∑

−

=

+=1

11 α

subjected to the additional constraints

1,....,2,1,0 1

1

11

−=−≤≤

=+

+

−

=∑

pittu

xut

iii

p

ii

Example

The example below illustrates the application of method 1.

Consider the objective function

Maximize 231 xxf +=

subject to

0

401522

2

1

221

≥≤≤

≤+

xx

xx

The problem is already in separable form (i.e., each term consists of only one variable). So

we can split up the objective function and constraint into two parts

( ) ( )( ) ( )2121111

2211

xgxggxfxff

+=+=

where

( ) ( )( ) ( ) 2212

21111

2223111

2;2

;

xxgxxg

xxfxxf

==

==

Since and are in linear form, they are left as such and therefore is treated

as a linear variable. Consider five breaking points for .

( )22 xf ( )212 xg 2x

1x



5

Table 1

i it1 ( )ii tf 1 ( )ii tg 11 1 0 0 0 2 1 1 2 3 2 8 8 4 3 27 18 5 4 64 32

Thus, can be written as, ( )11 xf

( ) ( )

6427810 1514131211

5

111111

×+×+×+×+×=

=∑=

wwwww

tfwxfi

ii

and can be written as, ( )111 xg

( ) ( )

3218820 1514131211

5

1111111

×+×+×+×+×=

=∑=

wwwww

tgwxgi

iii

The linear approximation of the above problem thus becomes,

Maximize 215141312 64278 xwwwwf ++++=

subject to

5,...,2,101

152321882

1

1514131211

1215141312

=≥=++++

=+++++

iforwwwwww

sxwwww

i

This can be solved using simplex method in a restricted basis condition.

Table 2

Variables rb

rs

r

cb

Iteration Basis f

11w 12w 13w 14w 15w 2x 1s

f 1 0 -1 -8 -27 -64 -1 0 0 --

1s 0 0 2 8 18 32 2 1 15 1.87

1

11w 0 1 1 1 1 1 0 0 1 1



6

From the table above, it is clear that should be the entering variable. For that should

be the exiting variable. But according to restricted basis condition and cannot occur

together in basis as they are not adjacent. Therefore, consider the next best entering

variable . This also is not possible, since should be exited and and cannot occur

together. Again, considering the next best variable , it is clear that should be the

exiting variable.

15w 1s

15w 11w

14w 1s 14w 11w

13w 11w

Table 3

Variables rb

rs

r

cb

Iteration Basis f

11w 12w 13w 14w 15w 2x 1s

f 1 8 7 0 -19 -56 1 0 8 --

1s 0 -8 -6 0 10 24 2 1 7 3

1

13w 0 1 1 1 1 1 0 0 1 15

For the table 2 above, the entering variable is . Then the variable to be exited is and this

is not acceptable since is not an adjacent point to . Next variable can be admitted

by dropping .

15w 1s

15w 13w 14w

1s

Table 4

Variables rb

rs

r

cb

Iteration Basis f

11w 12w 13w 14w 15w 2x 1s

f 1 -7.2 -4.4 0 0 -10.4 4.8 1.9 21.3 --

14w 0 -0.8 -0.6 0 1 2.4 0.2 0.1 0.7 1

13w 0 1.8 1.6 1 0 -1.4 -0.2 -0.1 0.3



7

Now, cannot be admitted since cannot be dropped. Similarly and cannot be

entered as cannot be dropped. Thus, the process ends at this point and the last table gives

the best solution.

15w 14w 11w 12w

13w

7.0;3.0 1413 == ww

Now,

7.232 1413

5

1111

=×+×=

=∑=

ww

twxi

ii

02 =xand

The optimum value of . 3.21=f

This may be an approximate solution to the original nonlinear problem. However, the

solution can be improved by taking finer breaking points.



Advanced Topics in Optimization

Piecewise Linear Approximation of a Nonlinear Function



IntroductionThere exists no general algorithm for nonlinear programming due to its irregular behaviorNonlinear problems can be solved by first representing the nonlinear function (both objective function and constraints) by a set of linear functions and then apply simplex method to solve this using somerestrictions

ObjectivesTo discuss the various methods to approximate a nonlinear function using linear functionsTo demonstrate this using a numerical example


Piecewise Linearization

A nonlinear single variable function f(x) can be approximated by a piecewise linear function

Geometrically, f(x) can be shown as a curve being represented as a set of connected line segments


Piecewise Linearization: Method 1

Consider an optimization function having only one nonlinear term f(x)

Let the x-axis of the nonlinear function f(x) be divided by ‘p’breaking points t1, t2, t2, …, tpCorresponding function values be f(t1), f(t2),…, f(tp)

If ‘x’ can take values in the interval , then the breaking points can be shown as

Xx ≤≤0

Xttt p ≡<<<≡ ...0 21


Piecewise Linearization: Method 1 …contd.

Express ‘x’ as a weighted average of these breaking points

Function f(x) can be expressed as

where

pptwtwtwx +++= ...2211

∑=

=p

iiitwxei

1

.,.

( ) ( ) ( ) ( ) ( )∑=

=+++=p

iiipp tfwtfwtfwtfwxf

12211 ...

∑=

=p

iiw

11




subject to the additional constraints

( ) ( )∑=

=p

iii tfwxfMinorMax

1

∑

∑

=

=

=

=

p

ii

p

iii

w

xtw

1

1

1



This linearly approximated model can be solved using simplex method with some restrictions

Restricted condition: There should not be more than two ‘wi’ in the basis and

Two ‘wi’ can take positive values only if they are adjacent. i.e., if ‘x’ takes the value between ti and ti+1, then only wi and wi+1 (contributing weights to the value of ‘x’) will be positive, rest all weights be zero

In general, for an objective function consisting of ‘n’ variables (‘n’ terms)

represented as

( ) ( ) ( ) ( )nn xfxfxfxfMinorMax +++= ....2211



subjected to ‘m’ constraints

The linear approximation of this problem is

( ) ( ) ( ) mjforbxgxgxg jnnjjj ,...,2,1....2211 =≤+++

nkforw

mjforbtgwtosubjected

tfwMinorMax

p

iki

j

n

k

p

ikikjki

n

k

p

ikikki

,...,2,11

,...,2,1)(

)(

1

1 1

1 1

==

=≤

∑

∑∑

∑∑

=

= =

= =


Piecewise Linearization: Method 2

‘x’ is expressed as a sum, instead of expressing as the weighted sum of the breaking points as in the previous method

where ui is the increment of the variable ‘x’ in the interval i.e., the bound of ui is

The function f(x) can be expressed as

where represents the slope of the linear approximation between the

points and

∑−

=− +=++++=

1

111211 ....

p

iip utuuutx

),( 1+ii tt

iii ttu −≤≤ +10

( ) ( ) i

p

iiutfxf ∑

−

=

+=1

11 α

iα

it1+it ( ) ( )ii

iii tt

tftf−−=

+

+

1

1α




subjected to additional constraints

( ) ( ) i

p

iiutfxfMinorMax ∑

−

=

+=1

11 α

1,....,2,1,0 1

1

11

−=−≤≤

=+

+

−

=∑

pittu

xut

iii

p

ii


Piecewise Linearization: Numerical Example

The example below illustrates the application of method 1

Consider the objective function

subject to

The problem is already in separable form (i.e., each term consists of only one variable).

231 xxfMaximize +=

040

1522

2

1

221

≥≤≤

≤+

xx

xx


Piecewise Linearization: Numerical Example …contd.

Split up the objective function and constraint into two parts

where

and are treated as linear variables as they are in linear form

( ) ( )( ) ( )2121111

2211

xgxggxfxff

+=+=

( ) ( )( ) ( ) 2212

21111

2223111

2;2

;

xxgxxgxxfxxf==

==

( )22 xf ( )212 xg



Consider five breaking points for x1

can be written as, ( )11 xf

( ) ( )

6427810 1514131211

5

111111

×+×+×+×+×=

=∑=

wwwww

tfwxfi

ii



can be written as,

Thus, the linear approximation of the above problem becomes

subject to

( )111 xg

( ) ( )

3218820 1514131211

5

1111111

×+×+×+×+×=

=∑=

wwwww

tgwxgi

iii

215141312 64278 xwwwwfMaximize ++++=

5,...,2,101

152321882

1

1514131211

1215141312

=≥=++++

=+++++

iforwwwwww

sxwwww

i



This can be solved using simplex method in a restricted basis condition

The simplex tableau is shown below



From the table, it is clear that should be the entering variable

should be the exiting variable

But according to restricted basis condition and cannot occur together in basis as they are not adjacent

Therefore, consider the next best entering variable

This also is not possible, since should be exited and andcannot occur together

The next best variable , it is clear that should be the exiting variable

15w1s

15w 11w

14w

1s 14w 11w

13w 11w



The entering variable is . Then the variable to be exited is and this is not acceptable since is not an adjacent point to

Next variable can be admitted by dropping .


15w 1s15w 13w

14w1s



Now, cannot be admitted since cannot be dropped

Similarly and cannot be entered as cannot be dropped


15w 14w

11w 12w 13w



Since there is no more variable to be entered, the process ends

Therefore, the best solution is

Now,

The optimum value is

This may be an approximate solution to the original nonlinear problem

However, the solution can be improved by taking finer breaking points

7.0;3.0 1413 == ww

7.232 1413

5

1111 =×+×==∑

=

wwtwxi

ii

02 =xand

3.21=f


Thank You

Optimization Methods: Advanced Topics in Optimization - Multi-objective Optimization 1


Multi-objective Optimization

Introduction

In a real world problem it is very unlikely that we will meet the situation of single objective

and multiple constraints more often than not. Thus the rigidity provided by the General

Problem (GP) is, many a times, far away from the practical design problems. In many of the

water resources optimization problems maximizing aggregated net benefits is a common

objective. At the same time maximizing water quality, regional development, resource

utilization, and various social issues are other objectives which are to be maximized. There

may be conflicting objectives along with the main objective like irrigation, hydropower,

recreation etc. Generally multiple objectives or parameters have to be met before any

acceptable solution can be obtained. Here it should be noticed that the word “acceptable”

implicates that there is normally no single solution to the problems of the above type.

Actually methods of multi-criteria or multi-objective analysis are not designed to identify the

best solution, but only to provide information on the tradeoffs between given sets of

quantitative performance criteria.

In the present discussion on multi-objective optimization, we will first introduce the

mathematical definition and then talk about two broad classes of solution methods typically

known as (i) Utility Function Method (Weighting function method) (ii) Bounded Objective

Function Method (Reduced Feasible Region Method or Constraint Method ).

Multi-objective Problem

A multi-objective optimization problem with inequality (or equality) constraints may be

formulated as



Find (1)

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

=

nx

xx

X

.

.

.2

1

which minimizes f1(X), f2(X), … , fk(X) (2)

subject to

j= 1, 2 , … , m (3) ,0)( ≤Xg j

Here k denotes the number of objective functions to be minimized and m is the number of

constraints. It is worthwhile to mention that objective functions and constraints need not be

linear but when they are, it is called Multi-objective Linear Programming (MOLP).

For the problems of the type mentioned above the very notion of optimization changes and

we try to find good trade-offs rather than a single solution as in GP. The most commonly

adopted notion of the “optimum” proposed by Pareto is depicted below.

A vector of the decision variable X is called Pareto Optimal (efficient) if there does not exist

another Y such that for i = 1, 2, … , k with )()( XfYf ii ≤ )()( XfYf ij < for at least one j. In

other words a solution vector X is called optimal if there is no other vector Y that reduces

some objective functions without causing simultaneous increase in at least one other

objective function.



k

ij

Fig. 1

As shown in above figure there are three objectives i, j, k. Direction of their increment is also

indicated. The surface (which is formed based on constraints) is efficient because no

objective can be reduced without a simultaneous increase in at least one of the other

objectives.

Utility Function Method (Weighting function method)

In this method a utility function is defined for each of the objectives according to the relative

importance of fi.. A simple utility function may be defined as αifi(X) for ith objective where αi

is a scalar and represents the weight assigned to the corresponding objective. Then the total

utility U may be defined as weighted sum of objective functions as below

U = α,)(1∑=

k

iii Xfα i > 0, i = 1, 2, … , k. (4)

The solution vector X may be found by maximizing the total utility as defined above with the

constraint (3).

Without any loss to generality, it is customary to assume that although it is not

essential. Also α

11

=∑=

k

iiα

i values indicate the relative utility of each of the objectives.



The following figure represents the decision space for a given set of constraints and utility

functions. Here X = and two objectives are f⎥⎦

⎤⎢⎣

⎡

2

1

xx

1(X) and f2(X) with upper bound

constraints* of type (3) as in figure 2.

g1(X)

g2(X)

g3(X)

g4(X)

g5(X)

g6(X)

x2

A

BC

D

EO

x1

Decision Space

Fig. 2

*constraints g1(X) to g6(X) represent x1 , x2 ≥ 0

For Linear Programming (LP), the Pareto front is obtained by plotting the values of objective

functions at common points (points of intersection) of constraints and joining them through

straight lines in objective space.

It should be noted that all the points on the constraint surface need not be efficient in Pareto

sense as point A in the following figure.



A

BC

D

E

f2

f1

Objective Space

Fig. 3

By looking at Figure 3 one may qualitatively infer that it follows Pareto Optimal definition.

Now optimizing utility function means moving along the efficient front and looking for the

maximum value of utility function U defined by equation (4).

One major limitation is that this method cannot generate the complete set of efficient

solutions unless the efficiency frontier is strictly convex. If a part of it is concave, only the

end points of this can be obtained by the weighing technique.

Bounded objective function method

In this method we try to trap the optimal solution of the objective functions in a bounded or

reduced feasible region. In formulating the problem one objective function is maximized

while all other objectives are converted into constraints with lower bounds along with other

constraints of the problem. Mathematically the problem may be formulated as

Maximize fi(X)

Subject to ,0)( ≤Xg j j= 1, 2 , … , m (5)



fk(X) e≥ k ∀ ik ≠

here ek represents lower bound of the kth objective. In this approach the feasible region S

represented by ,0)( ≤Xg j j= 1, 2 , … , m is further reduced to S’ by (k-1) constraints

fk(X) e≥ k . ∀ ik ≠

e.g. let there are three objectives which are to be maximized in the region of constraints S.

The problem may be formulated as:

maximizeobjective-1

maximizeobjective-2

maximizeobjective-3

subject to X = ⎥⎦

⎤⎢⎣

⎡

2

1

xx

∈ S

In the above problem S identifies the region given by ,0)( ≤Xg j j= 1, 2 , … , m.

In the bounded objective function method, the same problem may be formulated as

maximizeobjective-1

subject to

objective-2 e≥ 1

objective-3 e≥ 2

X ∈ S

As may be seen, one of the objectives (objective-1) is now the only objective and all other

objectives are included as constraints. There are lower bounds specified for other objectives

which are the minimum values at least to be attained. Subject to these additional constraints,

the objective is maximized. Figure 4 illustrates the scheme.



x2

w1

w2

w3

S

S’A

BC

D

Ee1

e2P

x1F

Fig. 4

In the above figure w1, w2, and w3 are gradients of the three objectives respectively. If

objective-1 was to be maximized in the region S, without taking into consideration the

other objectives, then solution point is E. But due to lower bounds of the other objectives the

feasible region reduces to S’ and solution point is P now. It may be seen that changing e1

does not affect objective-1 as much as changing e2. This fact gives rise to sensitivity

analysis.

Exercise Problem

A reservoir is planned both for gravity and lift irrigation through withdrawals from its

storage. The total storage available for both the uses is limited to 5 units each year. It is

decided to limit the gravity irrigation withdrawals in a year to 4 units. If X1 is the allocation

of water for gravity irrigation and X2 is the allocation for lift irrigation, the two objectives

planned to be maximized are expressed as

Maximize Z1(X) = 3x1 - 2x2 and Z2(X) = - x1 + 4x2



For above problem, do the following

(i) Generate a Pareto Front of non-inferior (efficient) solutions by plotting Decision space and

Objective space.

(ii) Formulate multi objective optimization model using weighting approach with w1 and w2

as weights for gravity and lift irrigation respectively.

(iii) Solve it, for (i) w1=1 and w2=2 (ii) w1=2 and w2=1

(iv) Formulate the problem using constraint method

[Solution: (i) X1=0, X2=5; (ii) X1=4, X2=0 to 1 ]




Multi Objective Optimization


Introduction

IntroductionIn a real world problem it is very unlikely that we will meet the situation of single objective and multiple constraints more often than notThere may be conflicting objectives along with the main objective like irrigation, hydropower, recreation etc. Generally multiple objectives or parameters have to be met before any acceptable solution can be obtained. Multi-criteria or multi-objective analysis are not designed to identify the best solution, but only to provide information on the tradeoffs.


Multi-objective Problem

A multi-objective optimization problem with inequality (or equality) constraints may be formulated as

Find (1)

which minimizes f1(X), f2(X), … , fk(X) (2)

subject to j= 1, 2 , … , m (3)

Here k denotes the number of objective functions to be minimized and m is the number of constraints.

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

=

nx

xx

X

.

.

.2

1

,0)( ≤Xg j


Pareto Optimal Front

A vector of the decision variable X is called Pareto Optimal (efficient) if there does not exist another Y such that for i = 1, 2, … , k with for at least one jIn other words a solution vector X is called optimal if there is no other vector Y that reduces some objective functions without causing simultaneous increase in at least one other objective function.For the problems of the type mentioned above the very notion of optimization changes and we try to find good trade-offs rather than a single solution as in GP.

)()( XfYf ii ≤ )()( XfYf ij <


Pareto Optimal Front…

As shown in above figure there are three objectives i, j, k. Direction of their increment is also indicated. The surface (which is formed based on constraints) is efficient because no objective can be reduced without a simultaneous increase in at least one of the other objectives.


Utility Function Method (Weighting function method)

In this method a utility function is defined for each of the objectives according to the relative importance of fi.A simple utility function may be defined as αifi(X) for ithobjective where αi is a scalar and represents the weight assigned to the corresponding objective. Total utility U may be defined as weighted sum of objective functions as below

αi > 0, i = 1, 2, … , k. (4)Without any loss in generality it is customary to assume that

,)(1

∑=

k

iii Xfα

11

=∑=

k

iiα


Utility Function Method…

Figure represents decision space for a given set of constraints and utility functions. Here X = and two objectives are f1(X) and f2(X) with upper bound constraints* of type (3) as in figure 2. Pareto front is obtained by plotting the values of objective functions at common points (points of intersection) of constraints.

g1(X)

g2(X)

g3(X)

g4(X)

g5(X)

g6(X)

x2

2

A

BC

D

EO

Decision Space


Utility Function Method…

It should be noted that all the points on the constraint surface need not be efficient in Pareto sense as point A in the following figure.By looking at figure one may qualitatively infer that it follows Pareto Optimal definition. Now optimizing utility function means moving along efficient front and looking for the maximum value of utility function U defined by equation (4).

A

BC

D

E

f2


Bounded objective function method

In this method we try to trap the optimal solution of the objective functions in a bounded or reduced feasible region. In formulating the problem one objective function is maximized while all other objectives are converted into constraints with lower bounds along with other constraints of the problem. Mathematically the problem may be formulated as

Maximize fi(X)Subject to

j = 1, 2 , … , m (5)fk(X) ek

here ek represents lower bound on the kth objective.

,0)( ≤Xg j

≥ ∀ ik ≠


Bounded objective function method…

In this approach the feasible region S represented by j= 1, 2 , … , m is further reduced to S’ by (k-1) constraints fk(X) ek.e.g. let there are three objectives which are to be maximized inregion of constraints S.Problem may be formulated

maximizeobjective-1maximizeobjective-2maximizeobjective-3

subject toX = S

In above problem S identifies the region given by j= 1, 2 , … , m.

,0)( ≤Xgj

∀

,0)( ≤Xgj

∈

≥

⎥⎦

⎤⎢⎣

⎡

2

1

xx



In bounded objective function method, same problem may be formulated as

maximizeobjective-1subject to

objective-2 e1 objective-3 e2

X SAs may be seen that one of the objectives (objective-1) is now the only objective and all other objectives are included as the constraints.

∈≥≥

x2

w1

w2

w3

S

S’A

BC

D

E

e1

e2P



In previous figure w1, w2, and w3 are gradients of the three objectives respectively.If objective-1 was to be maximized in region S without taking into consideration other objectives then solution point had been E. But due to lower bound on other objectives the feasible region reduces to S’ and solution point is P now. It may be seen that changing e1 does not affect objective-1 as much as changing e2. This fact gives rise to sensitivity analysis.


Exercise Problem

A reservoir is planned both for gravity and lift irrigation through withdrawals from its storage. The total storage available for both the uses is limited to 5 units each year. It is decided to limit the gravity irrigation withdrawals in a year to 4 units. If X1 is the allocation of water to gravity irrigation and X2 is the allocation for lift irrigation, two objectives are planned to be maximized and are expressed as

Maximize Z1(X) = 3x1 - 2x2 and Z2(X) = - x1 + 4x2For above problem do the following

(i) Generate a Pareto Front of non-inferior (efficient) solutions by plotting Decision space and Objective space.

(ii) Formulate multi objective optimization model using weighting approach with w1 and w2 as weights for gravity and lift irrigation respectively.

(iii) Solve it, if (a) w1=1 and w2=2 (b) w1=2 and w2=1(iv) Formulate the problem using constraint method

[Solution: (i) x1=0, x2=5; (ii) x1=4, x2=0 to 1 ]


Thank You

Optimization Methods: Advanced Topics in Optimization - Multilevel Optimization 1


Multilevel Optimization

Introduction

The example problems discussed in the previous modules consist of very few decision

variables and constraints. However in practical situations, one has to handle an optimization

problem involving a large number of variables and constraints. Solving such a problem will

be quite cumbersome. In multilevel optimization, such large sized problems are decomposed

into smaller independent problems and the overall optimum solution can be obtained by

solving each sub-problem independently. In this lecture a decomposition method for

nonlinear optimization problems, known as model-coordination method will be discussed.

Model Coordination Method

Consider a minimization optimization problem consisting of n variables, . )(xF nxxx ,...,, 21

),...,,( 21 nxxxFMin

subjected to constraints

niuxxlx

mjxxxg

iii

nj

,...,2,1

,...,2,1,0),...,,( 21

=≤≤

=≤

where and represents the lower and upper bounds of the decision variable . ilx iux ix

Let be the decision variable vector. For applying the model coordination

method, the vector X should be divided into two subvectors, Y and Z such that Y contains the

coordination variables between the subsystems i.e., variables that are common to the

subproblems and Z vector contains the free or confined variables of subproblems. If the

problem is partitioned into ‘P’ subproblems, then vector Z can also be partitioned into ‘P’

variable sets, each set corresponding to each subproblem.

nxxxX ,...,, 21=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

PZ

ZZ

ZM

2

1

Thus the objective function can be partitioned into ‘P’ parts as shown, )(xF



( )∑=

=P

kkk ZYfxF

1,)(

where denotes the objective function of the k( kk ZYf , ) th subproblem. In this the coordination

variable Y will appear in all sub-objective functions and will appear only in kkZ th sub-

objective function.

Similarly the constraints can be decomposed as

PkforZYg kk ,...,2,10),( =≤

The lower and upper bound constraints are

PkforuZZlZuYYlY

kkk ,...,2,1=≤≤≤≤

The problem thus decomposed is solved using a two level approach which is described

below.

Procedure:

First level:

Fix the value of the coordination variables, Y at some value, say . Then solve each

independent subproblem and find the value of .

optY

kZ

( )kk ZYfMin ,

subject to

PkforuZZlZuYYlY

ZYg

kkk

kk

,...,2,1

0),(

=≤≤≤≤

≤

Let the values of obtained by solving this problem be . kZ optkZ

Second level:

Now consider the problem after substituting the values. optkZ

( )∑=

=P

koptkk ZYfYfMin

1,)(

subject to uYYlY ≤≤



Solve this problem to find a new . Again solve the first level problems. This process is

repeated until convergence is achieved.

optY




Multilevel Optimization


Objectives

To discuss about Multilevel Optimization

To describe a decomposition method for nonlinear

optimization problems, known as model-coordination

method


Introduction

In practical situations, an optimization problem involves a large

number of variables and constraints

In multilevel optimization, such large sized problems are

decomposed into smaller independent problems

The overall optimum solution is obtained by solving each sub-

problem independently


Model Coordination Method

Consider an minimization optimization problem consisting of n variables,

subjected to constraints

where and represents the lower and upper bound of the decision variable

Decision variable vector:

)(xFnxxx ,...,, 21

niuxxlxmjxxxg

iii

nj

,...,2,1

,...,2,1,0),...,,( 21

=≤≤

=≤

ilx iuxix

nxxxX ,...,, 21=


Model Coordination Method …contd.

For applying model coordination method, the vector X should be divided into two sub-vectors, Y and Z

Y contains the coordination variables between the subsystems i.e.,variables that are common to the sub-problems

Z vector contains the free or confined variables of sub-problems

If the problem is partitioned into ‘P’ sub-problems, then vector Z can also be partitioned into ‘P’ variable sets, each set corresponding to each sub-problem

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

PZ

ZZ

ZΜ

2

1



Thus the objective function can be partitioned into ‘P’

parts

where denotes the objective function of the kth sub-

problem

The coordination variable Y will appear in all sub-objective

functions and Zk will appear only in kth sub-objective function

)(xF

( )∑=

=P

kkk ZYfxF

1,)(

( )kk ZYf ,



Similarly the constraints are also decomposed as

The lower and upper bound constraints are

The problem is decomposed and solved using a two level approach

PkforZYg kk ,...,2,10),( =≤

PkforuZZlZuYYlY

kkk ,...,2,1=≤≤≤≤


Model Coordination Method: Procedure

First level:

Fix the value of the coordination variables, Y at some value, say Yopt

Solve each independent sub-problem and find the value of Zk

subject to

( )kk ZYfMin ,

PkforuZZlZuYYlY

ZYg

kkk

kk

,...,2,1

0),(

=≤≤≤≤

≤


Model Coordination Method: Procedure …contd.

Let the values of Zk obtained by solving this problem be Zkopt

Second level:

Now consider the problem after substituting the Zkopt values

subject to

( )∑=

=P

koptkk ZYfYfMin

1,)(

uYYlY ≤≤


Model Coordination Method: Procedure …contd.

Solve this problem to find a new Yopt

Again solve the first level problems

This process is repeated until convergence is achieved


Thank You

Optimization Methods: Advanced Topics in Optimization - Direct and Indirect Search Methods

1


Direct and Indirect Search Methods

Introduction

Most of the real world system models involve nonlinear optimization with complicated

objective functions or constraints for which analytical solutions (solutions using quadratic

programming, geometric programming, etc.) are not available. In such cases one of the

possible solutions is the search algorithm in which, the objective function is first computed

with a trial solution and then the solution is sequentially improved based on the

corresponding objective function value till convergence. A generalized flowchart of the

search algorithm in solving a nonlinear optimization with decision variable Xi, is presented in

Fig. 1.

Fig. 1 Flowchart of Search Algorithm

No

Start with Trial Solution Xi, Set i=1

Compute Objective function f(Xi)

Generate new solution Xi+1

Compute Objective function f(Xi+1)

Convergence Check

Set i=i+1

Optimal Solution Xopt=Xi

Yes



2

The search algorithms can be broadly classified into two types: (1) direct search algorithm

and (2) indirect search algorithm. A direct search algorithm for numerical search optimization

depends on the objective function only through ranking a countable set of function values. It

does not involve the partial derivatives of the function and hence it is also called nongradient

or zeroth order method. Indirect search algorithm, also called the descent method, depends on

the first (first-order methods) and often second derivatives (second-order methods) of the

objective function. A brief overview of the direct search algorithm is presented.

Direct Search Algorithm

Some of the direct search algorithms for solving nonlinear optimization, which requires

objective functions, are described below:

A) Random Search Method: This method generates trial solutions for the optimization model

using random number generators for the decision variables. Random search method includes

random jump method, random walk method and random walk method with direction

exploitation. Random jump method generates huge number of data points for the decision

variable assuming a uniform distribution for them and finds out the best solution by

comparing the corresponding objective function values. Random walk method generates trial

solution with sequential improvements which is governed by a scalar step length and a unit

random vector. The random walk method with direct exploitation is an improved version of

random walk method, in which, first the successful direction of generating trial solutions is

found out and then maximum possible steps are taken along this successful direction.

B) Grid Search Method: This methodology involves setting up of grids in the decision space

and evaluating the values of the objective function at each grid point. The point which

corresponds to the best value of the objective function is considered to be the optimum

solution. A major drawback of this methodology is that the number of grid points increases

exponentially with the number of decision variables, which makes the method

computationally costlier.

C) Univariate Method: This procedure involves generation of trial solutions for one decision

variable at a time, keeping all the others fixed. Thus the best solution for a decision variable



3

keeping others constant can be obtained. After completion of the process with all the decision

variables, the algorithm is repeated till convergence.

D) Pattern Directions: In univariate method the search direction is along the direction of co-

ordinate axis which makes the rate of convergence very slow. To overcome this drawback,

the method of pattern direction is used, in which, the search is performed not along the

direction of the co-ordinate axes but along the direction towards the best solution. This can be

achieved with Hooke and Jeeves’ method or Powell’s method. In the Hooke and Jeeves’

method, a sequential technique is used consisting of two moves: exploratory move and the

pattern move. Exploratory move is used to explore the local behavior of the objective

function, and the pattern move is used to take advantage of the pattern direction. Powell’s

method is a direct search method with conjugate gradient, which minimizes the quadratic

function in a finite number of steps. Since a general nonlinear function can be approximated

reasonably well by a quadratic function, conjugate gradient minimizes the computational time

to convergence.

E)Rosen Brock’s Method of Rotating Coordinates: This is a modified version of Hooke and

Jeeves’ method, in which, the coordinate system is rotated in such a way that the first axis

always orients to the locally estimated direction of the best solution and all the axes are made

mutually orthogonal and normal to the first one.

F) Simplex Method: Simplex method is a conventional direct search algorithm where the best

solution lies on the vertices of a geometric figure in N-dimensional space made of a set of

N+1 points. The method compares the objective function values at the N+1 vertices and

moves towards the optimum point iteratively. The movement of the simplex algorithm is

achieved by reflection, contraction and expansion.

Indirect Search Algorithm

The indirect search algorithms are based on the derivatives or gradients of the objective

function. The gradient of a function in N-dimensional space is given by:



4

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

=∇

Nxf

xf

xf

f

.

.

.2

1

(1)

Indirect search algorithms include:

A) Steepest Descent (Cauchy) Method: In this method, the search starts from an initial trial

point X1, and iteratively moves along the steepest descent directions until the optimum point

is found. Although, the method is straightforward, it is not applicable to the problems having

multiple local optima. In such cases the solution may get stuck at local optimum points.

B) Conjugate Gradient (Fletcher-Reeves) Method: The convergence technique of the steepest

descent method can be greatly improved by using the concept of conjugate gradient with the

use of the property of quadratic convergence.

C) Newton’s Method: Newton’s method is a very popular method which is based on Taylor’s

series expansion. The Taylor’s series expansion of a function f(X) at X=Xi is given by:

( ) ( ) ( ) ( ) [ ]( iiT

iiT

ii XXJXXXXfXfXf −−+−∇+=21 ) (2)

where, [Ji]=[J]|xi , is the Hessian matrix of f evaluated at the point Xi. Setting the partial

derivatives of Eq. (2), to zero, the minimum value of f(X) can be obtained.

( ) NjxXf

j

,...,2,1,0 ==∂∂ (3)



5

From Eq. (2) and (3)

[ ]( ) 0=−+∇=∇ iii XXJff (4)

Eq. (4) can be solved to obtain an improved solution Xi+1

[ ] iiii fJXX ∇−= −+

11 (5)

The procedure is repeated till convergence for finding out the optimal solution.

D) Marquardt Method: Marquardt method is a combination method of both the steepest

descent algorithm and Newton’s method, which has the advantages of both the methods,

movement of function value towards optimum point and fast convergence rate. By modifying

the diagonal elements of the Hessian matrix iteratively, the optimum solution is obtained in

this method.

E) Quasi-Newton Method: Quasi-Newton methods are well-known algorithms for finding

maxima and minima of nonlinear functions. They are based on Newton's method, but they

approximate the Hessian matrix, or its inverse, in order to reduce the amount of computation

per iteration. The Hessian matrix is updated using the secant equation, a generalization of the

secant method for multidimensional problems.

It should be noted that the above mentioned algorithms can be used for solving only

unconstrained optimization. For solving constrained optimization, a common procedure is the

use of a penalty function to convert the constrained optimization problem into an

unconstrained optimization problem. Let us assume that for a point Xi, the amount of

violation of a constraint is δ. In such cases the objective function is given by:

( ) ( ) 2δλ ××+= MXfXf ii (6)

where, λ=1( for minimization problem) and -1 ( for maximization problem), M=dummy

variable with a very high value. The penalty function automatically makes the solution

inferior where there is a violation of constraint.


http://en.wikipedia.org/wiki/Maxima_and_minima

http://en.wikipedia.org/wiki/Function_%28mathematics%29

http://en.wikipedia.org/wiki/Newton%27s_method

http://en.wikipedia.org/wiki/Hessian_matrix

http://en.wikipedia.org/wiki/Secant_method


6

Summary

Various methods for direct and indirect search algorithms are discussed briefly in the present

class. The models are useful when no analytical solution is available for an optimization

problem. It should be noted that when there is availability of an analytical solution, the search

algorithms should not be used, because analytical solution gives a global optima whereas,

there is always a possibility that the numerical solution may get stuck at local optima.




Direct and Indirect Search Methods


Introduction

Real World Problems

Nonlinear Optimization

Complicated Objective Function

Analytical Solution

Possible Solution: Search Algorithm


Flowchart for Search Algorithm

Start with Trial Solution Xi, Set i=1

Compute Objective function f(Xi)

Generate new solution Xi+1

Compute Objective function f(Xi+1)

Convergence Check

Optimal Solution, Xopt=Xi

Yes

Set i=i+1No


Classification of Search Algorithm

Search Algorithm


•Considers only objective function

•Partial derivatives of objective function: not considered

•Called nongradient or zeroth order method


•Partial derivatives of objective function are considered

•Also called descent method




Random Search Method

Grid Search Method

Pattern Directions

Univariate Method

Rosen Brock’s Method

Simplex Method


Direct Search Algorithm (Contd..)

Random Search Method:Generates trial solution for the decision variables.Classification: Random jump, random walk, and random walk with direction exploitation.Random Jump: generates huge number of data points assuming uniform distribution of decision variables and selects the best oneRandom Walk: generates trial solution with sequential improvements using scalar step length and unit random vector.Random Walk with Direction Exploitation: Improved version of random walk search, successful direction of generating trial solution is found out and steps are taken along this direction.



Grid Search MethodMethodology involves setting up grids in the decision space.Objective function values are evaluated at the grid points.The point corresponding to the best objective function value considered as optimum solution.Major Drawback: number of grid points increases exponentially with the number of decision variables.



Univariate Method:

Generates trial solution for one decision variable keeping all others fixed.Best solution for each of the decision variables keeping others constant are obtained.The whole process is repeated iteratively till convergence.


Pattern Directions:In Univariate method, search direction is same as coordinate axes directions.Makes the convergence process slow.In pattern directions, search is performed not along the coordinate directions, but along the direction towards the best solution.Popular Methods: Hooke and Jeeves’ Method, and Powellsmethod.Hooke and Jeeves’ Method: uses exploratory move and pattern move.Powells method: uses conjugate gradient




Rosen Brock’s Method of Rotating Coordinates:Modified version of Hooke and Jeeves’ Method.Coordinate system is rotated in such a way that the first axis

always orients to the locally estimated direction of the best solution and all the axes are made mutually orthogonal and normal to the first one.Simplex Method

Conventional direct search algorithm. The best solution lies on the vertices of a geometric figure in N-

dimensional space made of a set of N+1 points.The method compares the objective function values at the N+1

vertices and moves towards the optimum point iteratively. Movement of the simplex algorithm: reflection, contraction and

expansion.




Steepest Descent Method

Conjugate Gradient Method

Newton’s Method

Quasi-Newton Method

Marquardt Method


Indirect Search Algorithm (Contd..)

Steepest Descent Method:Search starts from an initial trial point X1Iteratively moves along the steepest descent direction until optimum point is found.Drawback: Solution may get stuck to a local optima.

Conjugate Gradient MethodConvergence technique uses the concept of conjugate gradientUses the properties of quadratic convergence


Indirect Search Algorithm (Contd..)Newton’s Method:

Based on Taylor series expansion.

[Ji]=[J]|xi is the Hessian MatrixSetting the partial derivatives of the objective function to zero, minimum f(x) can be obtained

From both the Eq.

The improved solution is given by

Process continues till convergence

( ) ( ) ( ) ( ) [ ]( )iiT

iiT

ii XXJXXXXfXfXf −−+−∇+=21

( ) NjxXf

j

,...,2,1,0 ==∂

∂

[ ]( ) 0=−+∇=∇ iii XXJff

[ ] iiii fJXX ∇−= −+

11


Indirect Search Algorithm (Contd..)

Marquardt Method:Combination of both steepest descent algorithm and Newton’s methodBy modifying the diagonal elements of Hessian matrix iteratively optimum solution is obtained

Quasi-Newton MethodBased on Newton’s methodThey approximate the Hessian matrix, or its inverse, in order to reduce the amount of computation per iteration.The Hessian matrix is updated using the secant equation, a generalization of the secant method for multidimensional problems


Constrained Optimization

Search algorithms cannot be directly applied to constrained optimization.Constrained optimization can be converted into unconstrained optimization using penalty function when there is a constraint violation.

λ=1( for minimization problem) and -1 ( for maximization problem), M=dummy variable with a very high value

( ) ( ) 2δλ ××+= MXfXf iiPenalty Function


Summary

Nonlinear optimization with complicated objective functions or constraints: Can be solved by search algorithm.Care should be taken so that the search should not get stuck at local optima.Search algorithms are not directly applicable to constrained optimizationPenalty functions can be used to convert constrained optimization into unconstrained optimization


Thank You

Optimization Methods: Advanced Topics in Optimization - Evolutionary Algorithms for Optimization and Search

1

Lecture Notes – 5

Evolutionary Algorithms for Optimization and Search

Introduction

Most real world optimization problems involve complexities like discrete, continuous or

mixed variables, multiple conflicting objectives, non-linearity, discontinuity and non-convex

region. The search space (design space) may be so large that global optimum cannot be found

in a reasonable time. The existing linear or nonlinear methods may not be efficient or

computationally inexpensive for solving such problems. Various stochastic search methods

like simulated annealing, evolutionary algorithms (EA) or hill climbing can be used in such

situations. EAs have the advantage of being applicable to any combination of complexities

(multi-objective, non-linearity etc) and also can be combined with any existing local search

or other methods. Various techniques which make use of EA approach are Genetic

Algorithms (GA), evolutionary programming, evolution strategy, learning classifier system

etc. All these EA techniques operate mainly on a population search basis. In this lecture

Genetic Algorithms, the most popular EA technique, is explained.

Concept

EAs start from a population of possible solutions (called individuals) and move towards the

optimal one by applying the principle of Darwinian evolution theory i.e., survival of the

fittest. Objects forming possible solution sets to the original problem is called phenotype and

the encoding (representation) of the individuals in the EA is called genotype. The mapping of

phenotype to genotype differs in each EA technique. In GA which is the most popular EA,

the variables are represented as strings of numbers (normally binary). If each design variable

is given a string of length ‘l’, and there are n such variables, then the design vector will have

a total string length of ‘nl’. For example, let there are 3 design variables and the string length

be 4 for each variable. The variables are 17,4 321 === xandxx . Then the chromosome

length is 12 as shown in the figure.

0 1 0 0 0 1 1 1 0 0 0 1

1x 2x 3x



2

An individual consists a genotype and a fitness function. Fitness represents the quality of the

solution (normally called fitness function). It forms the basis for selecting the individuals and

thereby facilitates improvements.

The pseudo code for a simple EA is given below

i = 0

Initialize population P0 Evaluate initial population

while ( ! termination condition)

i = i+1

Perform competitive selection

Create population Pi from Pi-1 by recombination and mutation

Evaluate population Pi

A flowchart indicating the steps of a simple genetic algorithm is shown below.



3

Generate Initial Population

Start

Encode Generated Population

Evaluate Fitness Functions

Meets Optimization

Criteria?

Best Individuals

Yes

Stop

Selection (select parents)

Mutation (mutate offsprings)

Crossover (selected parents)

No

REGENERAT ION

The initial population is usually generated randomly in all EAs. The termination condition

may be a desired fitness function, maximum number of generations etc. In selection,

individuals with better fitness functions from generation ‘i' are taken to generate individuals

of ‘i+1’th generation. New population (offspring) is created by applying recombination and

mutation to the selected individuals (parents). Recombination creates one or two new

individuals by swaping (crossing over) the genome of a parent with another. Recombined

individual is then mutated by changing a single element (genome) to create a new individual.



4

Finally, the new population is evaluated and the process is repeated. Each step is described in

more detail below.

Parent Selection

After fitness function evaluation, individuals are distinguished based on their quality.

According to Darwin's evolution theory the best ones should survive and create new offspring

for the next generation. There are many methods to select the best chromosomes, for example

roulette wheel selection, Boltzman selection, tournament selection, rank selection, steady

state selection and others. Two of these are briefly described, namely, roulette wheel

selection and rank selection:

Roulette Wheel Selection:

Parents are selected according to their fitness i.e., each individual is selected with a

probability proportional to its fitness value. In other words, depending on the percentage

contribution to the total population fitness, string is selected for mating to form the next

generation. This way, weak solutions are eliminated and strong solutions survive to form the

next generation. For example, consider a population containing four strings shown in the

table below. Each string is formed by concatenating four substrings which represents

variables a,b,c and d. Length of each string is taken as four bits. The first column represents

the possible solution in binary form. The second column gives the fitness values of the

decoded strings. The third column gives the percentage contribution of each string to the total

fitness of the population. Then by "Roulette Wheel" method, the probability of candidate 1

being selected as a parent of the next generation is 28.09%. Similarly, the probability that the

candidates 2, 3, 4 will be chosen for the next generation are 19.59, 12.89 and 39.43

respectively. These probabilities are represented on a pie chart, and then four numbers are

randomly generated between 1 and 100. Then, the likeliness that the numbers generated

would fall in the region of candidate 2 might be once, whereas for candidate 4 it might be

twice and candidate 1 more than once and for candidate 3 it may not fall at all. Thus, the

strings are chosen to form the parents of the next generation.



5

Candidate Fitness value Percentage of total fitness

1011 0110 1101 1001 109 28.09

0101 0011 1110 1101 76 19.59

0001 0001 1111 1011 50 12.89

1011 1111 1011 1100 153 39.43

Total 388 100

Rank Selection:

The previous type of selection may have problems when the fitnesses differ very much. For

example, if the best chromosome fitness is 90% of the entire roulette wheel then the other

chromosomes will have very few chances to be selected. Rank selection first ranks the

population and then every chromosome receives fitness from this ranking. The worst will

have fitness 1, second worst 2 etc. and the best will have fitness N (number of chromosomes

in population). By this, all the chromosomes will have a chance to be selected. But this

method can lead to slower convergence, because the best chromosomes may not differ much

from the others.

Crossover

Selection alone cannot introduce any new individuals into the population, i.e., it cannot find

new points in the search space. These are generated by genetically-inspired operators, of

which the most well known are crossover and mutation.

Crossover can be of either one-point or two-point scheme. In one point crossover, selected

pair of strings is cut at some random position and their segments are swapped to form new

pair of strings. In two-point scheme, there will be two break points in the strings that are

randomly chosen. At the break-point, the segments of the two strings are swapped so that



6

new set of strings are formed. For example, let us consider two 8-bit strings given by

'10011101' and '10101011'.

Then according to one-point crossover, if a random crossover point is chosen after 3 bits from

left and segments are cut as shown below:

100 | 11101

101 | 01011

and the segments are swapped to form

10001011

10111101

According to two-point crossover, if two crossover points are selected as

100 | 11 | 101

101 | 01 | 011

Then after swapping both the extreme segments, the resulting strings formed are

10001101

10111011

Crossover is not usually applied to all pairs of individuals selected for mating. A random

choice is made, where the probability of crossover being applied is typically between 0.6 and

0.9.

Mutation

Mutation is applied to each child individually after crossover. It randomly alters each gene

with a small probability (generally not greater than 0.01). It injects a new genetic character

into the chromosome by changing at random a bit in a string depending on the probability of

mutation.

Example: 10111011



7

is mutated as 10111111

It is seen in the above example that the sixth bit '0' is changed to '1'. Thus, in mutation

process, bits are changed from '1' to '0' or '0' to '1' at the randomly chosen position of

randomly selected strings.

Real-coded GAs

As explained earlier, GAs work with a coding of variables i.e., with a discrete search space.

GAs have also been developed to work directly with continuous variables. In these cases,

binary strings are not used. Instead, the variables are directly used. After the creation of

population of random variables, a reproduction operator can be used to select good strings in

the population.

Advantages and Disadvantages of EA:

EA can be efficiently used for highly complex problems with multi-objectivity, non-linearity

etc. It provides not only a single best solution, but the 2nd best, 3rd best and so on as required.

It gives quick approximate solutions. EA methods can very well incorporate with other local

search algorithms.

There are some drawbacks also in using EA techniques. An optimal solution cannot be

ensured on using EA methods, which are usually known as heuristic search methods.

Convergence of EA techniques are problem oriented. Sensitivity analysis should be carried

out to find out the range in which the model is efficient. Also, the implementation of these

techniques requires good programming skill.




Evolutionary Algorithms for Optimization and Search


Introduction

Real world optimization problems mostly involve complexities like

discrete-continuous or mixed variables, multiple conflicting objectives,

non-linearity, discontinuity and non-convex region

Global optimum cannot be found in a reasonable time due to the large

search space

For such problems, existing linear or nonlinear methods may not be

efficient

Various stochastic search methods like simulated annealing,

evolutionary algorithms (EA), hill climbing can be used in such

situations


Introduction

Among these techniques, the special advantage of EA’s are

Can be applied to any combination of complexities (multi-objective,

non-linearity etc) and also

Can be combined with any existing local search or other methods

Various techniques which make use of EA approach

Genetic Algorithms (GA), Evolutionary Programming, Evolution

Strategy, Learning Classifier System etc.

EA techniques operate mainly on a population search basis.


Objectives

To discuss the basic concept of Evolutionary Algorithms (EA)

To explain the basic steps for a simple EA like

Parent Selection

Cross over

Mutation

To discuss the advantages and disadvantages of EA


Basic Concept of EA

EAs start from a population of possible solutions (called individuals)

and move towards the optimal one by applying the principle of

Darwinian evolution theory i.e., survival of the fittest

Objects forming possible solution sets to the original problem is called

phenotype

The encoding (representation) or the individuals in the EA is called

genotype

The mapping of phenotype to genotype differs in each EA technique


Basic Concept of EA …contd.

In GA, variables are represented as strings of numbers (normally binary)

Let the number of design variables be n

Let each design variable is given by a string of length ‘l

Then the design vector will have a total string length ‘nl’

For example, if the string length be 4 for each design variable and there

are 3 design variables,

Then the chromosome length is 12

as shown

17,4 321 === xandxx



In GA, variables are represented as strings of numbers (normally binary)

Let the number of design variables be n

Let each design variable is given by a string of length ‘l

Then the design vector will have a total length ‘nl’

For example, if the string length be 4 and there are 3 design variables,

Then the chromosome of length 12 is

17,4 321 === xandxx



An individual consists of a genotype and a fitness function

Fitness Function:

Represents the quality of the solution

Forms the basis for selecting the individuals and thereby

facilitates improvements


Basic Concept of EA …contd.Pseudo code for a simple EA



Flowchart indicating the steps of a simple genetic algorithm



Initial population is randomly generated

Individuals with better fitness functions from generation ‘i' are taken to

generate individuals of ‘i+1’th generation

New population (offspring) is created by applying recombination and

mutation to the selected individuals (parents)

Finally, the new population is evaluated and the process is repeated

The termination condition may be a desired fitness function, maximum

number of generations etc


Parent Selection

Individuals are distinguished based on their fitness function value

According to Darwin's evolution theory the best ones should

survive and create new offspring for the next generation

Different methods are available to select the best chromosomes

Roulette wheel selection, Rank selection, Boltzman selection,

Tournament selection, Steady state selection


Parent Selection: Roulette wheel selection

Each individual is selected with a probability proportional to its

fitness value

In other words, an individual is selected depending on the

percentage contribution to the total population fitness

Thus, weak solutions are eliminated and strong solutions survive

to form the next generation


Parent Selection: Roulette wheel selection …contd.

Each string is formed by concatenating four substrings representing

variables a, b, c and d. Length of each string is taken as four bits

Consider a population containing four strings shown



First column represents the possible solution in binary form

Second column gives the fitness values of the decoded strings

Third column gives the percentage contribution of each string to the total

fitness of the population

Thus, the probability of selection of candidate 1, as a parent of the next

generation is 28.09%

Probabilities of other candidates 2, 3, 4 are 19.59, 12.89 and 39.43

respectively



These probabilities are represented on a pie chart

Then four numbers are randomly generated between 1 and 100

The likeliness of these numbers falling in the region of candidate 2 might be once, whereas for candidate 4 it might be twice and candidate 1 more than once and for candidate 3 it may not fall at all

Thus, the strings are chosen to form the parents of the next generation

The main disadvantage of this method is when the fitnesses differ very much

For example, if the best chromosome fitness is 90% of the entire roulette wheel then the other chromosomes will have very few chances to be selected


Parent Selection: Rank Selection

Population is ranked first

Every chromosome will be allotted with one fitness corresponding to this

ranking

The worst will have fitness 1, second worst 2 etc. and the best will have

fitness N (number of chromosomes in population)

By doing this, all the chromosomes have a chance to be selected

But this method can lead to slower convergence, because the best

chromosomes may not differ much from the others


Parent Selection

Through selection new individuals cannot get introduced into the

population

Selection cannot find new points in the search space

New individuals are generated by genetically-inspired operators

known are crossover and mutation.


Crossover

Crossover can be of either one-point or two-point schemeOne point crossover: Selected pair of strings is cut at some random position and their segments are swapped to form new pair of stringsConsider two 8-bit strings given by '10011101' and '10101011' Choose a random crossover point after 3 bits from left

Segments are swapped and the resulting strings are


Crossover …contd.

Two point crossover: There will be two break points in the strings that are randomly chosenAt the break-point the segments of the two strings are swapped so that new set of strings are formed If two crossover points are selected as

After swapping both the extreme segments, the resulting strings formed are


Mutation

Mutation is applied to each child individually after crossover

Randomly alters each gene with a small probability (generally not greater than 0.01)

Injects a new genetic character into the chromosome by changing at random a bit in a string depending on the probability of mutation

Example: 10111011 is mutated as

10111111

Sixth bit '0' is changed to '1‘

In mutation process, bits are changed from '1' to '0' or '0' to '1' at the randomly chosen position of randomly selected strings



Advantages

EA can be efficiently used for highly complex problems with

multi-objectivity, non-linearity etc

Provides not only a single best solution, but the 2nd best, 3rd

best and so on as required.

Gives quick approximate solutions

Can incorporate with other local search algorithms



Disadvantages

Optimal solution cannot be ensured on using EA methods

Convergence of EA techniques are problem oriented

Sensitivity analysis should be carried out to find out the range

in which the model is efficient

Implementation requires good programming skill


Thank You

Optimization Methods: Advanced Topics in Optimization - Applications in Civil Engineering

1


Applications in Civil Engineering

Introduction

In this lecture we will discuss some applications of multiobjective optimization and

evolutionary algorithms in civil engineering.

Water quality management, waste load allocation

Waste Load Allocation (WLA) in streams refers to the determination of required pollutant

treatment levels at a set of point sources of pollution, to ensure that water quality standards

are maintained throughout the stream. The stakeholders involved in a waste load allocation

are the Pollution Control Agency (PCA) and the dischargers (municipal and industrial) who

are discharging waste into the stream. The goals of the PCA are to improve the water quality

throughout the stream whereas that of the dischargers is to reduce the treatment cost of the

pollutants. Therefore a waste load allocation model can be viewed as a multiobjective

optimization model with conflicting objectives. If the fractional removal level of the pollutant

is denoted by x and the concentration of water quality indicator (e.g., Dissolved Oxygen) is

denoted by c, then the following optimization model can be formulated:

cMaximize (1)

xMinimize (2)

)(xfc = (3)

Eq. (3) represents the relationship between the water quality indicator and the fractional

removal levels of the pollutants. It should be noted that the relationship between c and x may

be nonlinear and therefore linear programming may not be applicable. In such cases the

applications of evolutionary algorithm is a possible solution. Interested readers may refer to

Tung and Hathhorn (1989), Sasikumar and Mujumdar (1998), and Mujumdar and Subbarao

(2004).



2

Reservoir Operation

In reservoir operation problems, to achieve the best possible performance of the system,

decisions need to be taken on releases and storages over a period of time considering the

variations in inflows and demands. The goals of a multipurpose reservoir operation problem

can be:

A) Flood control

B) Hydropower generation

C) Meeting irrigation demand

D) Maintaining downstream water quality

Therefore deriving the operation policy for a multipurpose reservoir can be considered as a

multiobjective optimization problem. A typical reservoir operation problem is characterized

by the uncertainty resulting from the random behavior of inflow and demand, incorporation

of which in terms of risk may lead to a nonlinear optimization problem. Application of

evolutionary algorithms is a possible solution for such problems. Interested readers may refer

to Janga Reddy and Nagesh Kumar (2007), Nagesh Kumar and Janga Reddy (2007).

Water Distribution Systems

The typical goals of water distribution systems problem in designing urban pipe system can

be:

A) Meeting the household demands.

B) Minimizing cost of pipe system.

C) Meeting the required water pressure at all nodes of the distribution system.

D) Optimal positioning of valves.

Therefore designing urban water distribution system is a multiobjective problem, which is

also characterized by nonlinearity resulting from the simulation model (e.g., Hardy Cross

method). Multiobjective techniques with search algorithm or evolutionary algorithm are

therefore used for solving such problems. Apart from these, determination of optimum



3

dosage of chlorine is also another important problem which is highly nonlinear because of the

nonlinear water quality simulation model. Evolutionary algorithms are successfully applied

for such problem in various case studies by different researchers.

Transportation Engineering

Evolutionary algorithm has become a useful tool for solving many problems in transportation

systems engineering. The problem, to efficiently move empty or laden containers for a

logistic company or Truck and Trailer Vehicle Routing Problem (TTVRP), is one among the

many potential research problems in transportation systems engineering. A general model for

TTVRP, consisting of a number of job orders to be served by trucks and trailers daily, is

constructed for a logistic company that provides transportation services for container

movement within the country. Due to the limited capacity of vehicles owned by the company,

the engineers of the company have to decide whether to assign the job orders of container

movement to its internal fleet of vehicles or to outsource the jobs to

other companies. The solution to the TTVRP consists of finding a complete routing schedule

for serving the jobs with - 1. minimum routing distance and 2. minimum number of trucks,

subject to a number of constraints such as time windows and availability of trailers.

Multiobjective evolutionary algorithm can be used to solve such models. Applications of

evolutionary algorithm in solving transportation problems can be found in Lee et al. (2003).


1. Deb K., Multi-Objective Optimization using Evolutionary Algorithms, First Edition, John

Wiley & Sons Pte Ltd, 2002.

2. Deb K., Optimization for Engineering Design – Algorithms and Examples, Prentice Hall

of India Pvt Ltd, New Delhi, 1995.

3. Dorigo M., and Stutzle T., Ant Colony Optimization, Prentice Hall of India Pvt Ltd, New

Delhi, 2005.







4



7. Taha H.A., Operations Research – An Introduction, Seventh Edition, Pearson Education,

New Delhi, 2004.

8. Vedula, S., and P.P. Mujumdar, Water Resources Systems: Modelling Techniques and


Journals - Suggested Reading

1. Janga Reddy, M., and Nagesh Kumar, D. (2007), Multi-objective differential evolution

with application to reservoir system optimization, Journal of Computing in Civil

Engineering, ASCE, 21(2), 136-146.

2. Lee, L.H., Tan, K. C., Ou, K. and Chew, Y.H. (2003), Vehicle capacity planning system

(VCPS): A case study on vehicle routing problem with time windows, IEEE Transactions

on Systems, Man and Cybernetics: Part A (Systems and Humans), 33 (2), 169-178.

3. Mujumdar, P.P. and Subbarao V.V.R (2004), Fuzzy waste load allocation model for river

systems: simulation – optimization approach, Journal of Computing in Civil Engineering,

ASCE, 18(2), 120-131.

4. Nagesh Kumar, D. and Janga Reddy, M. (2007), Multipurpose reservoir operation using

particle swarm optimization, Journal of Water Resources Planning and Management,

ASCE, 133(3), 192-201.

5. Sasikumar, K., and Mujumdar, P. P. (1998), Fuzzy optimization model for water quality

management of a river system, Journal of Water Resources Planning and Management,

ASCE, 124(2), 79-88.

6. Tung, Y. K., and Hathhorn, W. E. (1989), Multiple-objective waste load allocation, Water

Resources Management, 3, 129-140.


http://scitation.aip.org/cpo/

http://scitation.aip.org/cpo/

http://scitation.aip.org/wro



Applications in Civil Engineering


Introduction

Applications:Water Quality Management: waste load allocation.Reservoir Operation.Water Distribution SystemsTransportation Engineering.


Water Quality Management

Waste Load Allocation (WLA) in streams refers to the determination of required pollutant treatment levels at a set ofpoint sources of pollution to ensure that water quality standards are maintained throughout the stream. Stakeholders involved: Pollution Control Agency (PCA) and the dischargers (municipal and industrial) who are discharging waste into the stream Goals/ Objectives: goals of the PCA are to improve the water quality throughout the stream whereas that of dischargers is to reduce the treatment cost of the pollutants Optimization model with conflicting objectives


Water Quality Management (Contd..)

cMaximize

xMinimize

)(xfc =

Maximization of water quality indicator (Dissolved Oxygen)

Minimization of fractional levels to pollutant to reduce treatment cost

Multi-Objective

Water Quality Simulation Model: Nonlinear

Evolutionary Algorithm can be used

Suggested Readings: Tung and Hathhorn (1989), Sasikumar and Mujumdar (1998), and Mujumdarand Subbarao (2004)


Reservoir Operation

In reservoir operation problems, to achieve the best possible performance of the system, decisions need to be taken on releases and storages over a period of time considering the variations in inflows and demands. The goals can be:

1. Flood control 2. Hydropower generation 3. Meeting irrigation demand 4. Maintaining water quality

downstream.

Multiobjective in nature


Reservoir Operation (Contd..)

Characterized by the uncertainty resulting from the random behavior of inflow and demand, incorporation of which in terms of risk may lead to a nonlinear optimization problem. Application of evolutionary algorithms is a possible solution for such problems. Suggested Readings: Jangareddy and NageshKumar (2007), Nagesh Kumar and Janga Reddy (2007).


Water Distribution Systems

Objectives can be:1. Meeting the household

demands.2. Minimizing cost of pipe

system.3. Meeting the required water

pressure at all nodes of the distribution system.

4. Optimal positioning of valves.

Multiobjective in nature


Water Distribution Systems (Contd..)

Simulation model (e.g., Hardy Cross method): NonlinearDetermination of optimum dosage of chlorine is also another important problem which is highly nonlinear because of nonlinear water quality simulation model.Possible solution: Evolutionary Algorithm


Transportation Engineering

Efficiently moving empty or laden containers for a logistic company or Truck and Trailer Vehicle Routing Problem (TTVRP).Objectives:

1. Minimize routing distance 2. Minimize number of trucks Model is nonlinear due to complicated inter relationship between the components.Evolutionary Algorithm: Possible Solution.Suggested Reading: Lee et al. (2003)


Thank You

curso nptel - optimization methods

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