dmc 1752 resource management techniques

RESOURCE MANAGEMENT TECHNIQUES

1 ANNA UNIVERSITY CHENNAI

NOTESUNIT I

CHAPTER 1

GENERAL FEATURES OFOPERATIONS RESEARCH

1.1 INTRODUCTION

Operations Research O.R.,

organisationsactivities

activities

The origins of the subject of often abbreviated as dateback to the Second World War (1939 - 1945) of the last century. During the war, the alliesU.K. and U.S.A. engaged groups of scientists belonging to different disciplines, in order tourgently evolve scientific approach to methodologies f allocation of scarce resourcesrequired for different activities in drawing up effective tactical and strategic operationalplans. It is claimed that the recommendations of these groups led to victories in severaltheatres of the war.

With the end of the war in the year 1945, these countr witnessed industrial boomwith the aid of newly invented technologies. The growi industries facing increasingcomplexities of their activities had to seek consultan y from such scientists who had workedin O.R. groups during the war. This was due to gradual awareness that industrial andbusiness activities, in essence, were of similar nature as were encountered in the militaryoperations. Actually during this period the subject of Operations Research acquired definitiveshape. Beginning with the pioneering work of the mathematician G.B.Dantzig during 1947- 1948, most of the methods of O.R. were in place by the end of the next decade. Insucceeding decades intensive development of the subjec has taken place with the discoveryof newer more efficient methods, treating large size a tivities and development of computercodes. Applications in new areas of agriculture, hospi s, communication networks andgovernment have also been found.

Operations Research is thus applicable to a variety of . An organisationoften has to deal with choice among alternative arising from conflict of interestamong its own components as well as those arising from other organisations with which ithas dealings. In some cases the choice has to be made n the backdrop of uncertainties.Thus the subject of O.R. is really “research” on opera ons or for most effectivelyrunning the organisation. Due to diverse nature of org nizations it is evident that unlike puresciences where answers can be deduced from a few fundamental laws, the answers in

DMC 1752


NOTES

deterministicprobabilistic

optimal

System Analysisteam

teamworkadvisory

Systems Analysis activitiesalternatives decisions

appropriate objectivesmathematical modeling

decision alternativesdecision variables

constraintsobjectives

O.R. are obtained in a variety of ways. Nevertheless, th the development of the subject,some standard types have emerged and the answers in the first instance may well be foundin these types. The scope of modification and improvement is always there. Whenuncertainties are absent the standard types are called , while in the contrarycase they are called . The standard types have the common running theme ofobtaining most effective (often called ) solution for the activities of the organisation.

In order to critically understand the activities of an organisation disregarding those inthe fringe (in other words performing of the organisation) and come upwith O.R. answers, a of individuals having specialized skills in different isciplines isrequired. Typically an O.R. team may contain those who are highly trained in O.R. methods,mathematics, probability and statistical theory, applied statistics, economics, businessadministration, computing, production, construction, process and communicationengineering, agriculture, hospital management and phys al sciences. Thus is anecessary ingredient of operations research. The role of the team is to themanagement of the organisation and must come up with t ir specific recommendations,including alternatives, but decisions are taken only b the management. It is clear that somemembers of the team must be in good communication with the management before beginningthe exercise and at the end present their recommendati s.

When an organisation is very large, an O.R. team is usually engaged to find solutionfor critically important segments of the organisation, which otherwise is not apparent. Fromthe recommended solutions, including alternatives, the management after taking intoconsideration the remaining segments of the whole organisation decides on the course ofthe activities, keeping in view the overall objective f stable profit, increased market share,product diversification, stable prices, improved worke morale, company prestige, socialresponsibilities while maintaining family control.

Because of its very nature, the management communicates to the O.R. team in avague, imprecise language about the segment requiring R. solution. From the input, theteam performs . This means that it has to identify the important involved, together with the associated which could be made andthe to be attained. For quantitative answers it is appare that

of the system is required in which the essence of the abovementioned ingredients are taken in to account. This procedure is true of most of puresciences as well, but the difference is that the O.R. dels are far more exact because ofthe inherent complexities of the system, losing minute details during abstract modeling.

Suppose that the identified are in number say 1, 2, …, n

(often the mathematical terminology is used for these). These areoften quantitative in nature and subject to which can be formulated asmathematical inequalities. The obviously depend on 1, 2, …, n and are

n x x x

x x x



NOTESobjective functions

programmingplanning of activities

Linear Programming Models

Nonlinear Programming Model

constants deterministicSensitivity Analysis

probabilisticexpectation or mean value

model validation

retrospective test

therefore mathematical functions of these variables. For O.R. solutions the values 1, 2,

…, n are required to be determined for which the are optimal, that is

to say, maximum or minimum. Thus a model may look like

Optimise: Objective Function of 1, 2… n

s.t.: Constraints involving 1, 2… n

(s.t. standing for “subject to”). In this book it is restricted to models in which there is onlyone objective function.

The solution methods are generally called (not to be confused with by which actually is implied. A well

known class of models are the in which the objectivefunction and the constraints are all linear functions of 1, 2, …, n . In the followingchapters we shall come across several models belonging to this class. When either theobjective function or any of the constraints is a nonl ear function of

1, 2…

n, the model

is called . In this book occasionally such models aredescribed.

The models alluded above generally contain several parameters associated with thedecision alternatives. These are supposed to be estima from the given system. If takenas , the models are . The effect of slight alterations in them due tosome uncertainty is an important question and is called . An O.R.team is required to perform this analysis too and inco orate appropriate recommendationsin their report. In some other models of a different kind, the decision alternatives dependon random parameters with some probabilities (which is not unnatural because of complexityof systems) and so the models become . In such cases the treatment is basedon the probability theory in which plays a central role. Adirect and more important class of probabilistic model is that of systems operating over alength of time, receiving and dispatching items at random. Historically, the mathematicaltheories of such problems were developed by noted math atical probability dating backto the beginning of the last century. Enriched by appl ation models the theories were lateron embodied in O.R. In this connection it will not be of place to mention that theliterature on the mathematics of the variety of problems constituting the deterministic andprobabilistic models are extremely rich.

Over the years several models have been standerdised, me of which form the basisof this book. But the application of a particular model to an organisational setup requiresadditional task of , it being presumed that gross errors in model buildingand trivial errors of computation are not present. A systematic approach when applicableis the . In this test, past data for the model parameters are used and thesolution compared with the past performance. If this solution is indicative of a degree ofagreement with the experience of possible good performance in the past, then the model

x xx

x x x

x x x

computer programming)

x x x

x x x

DMC 1752


NOTES

Implementation

1.2 FEATURES/CHARACTERISTICS OF O.R

Inter-disciplinary Team Approach

can be applied in the current situation as well, provided the system behaviour has notradically changed. In some cases past data may not be ailable. In such cases status quoin the level of activities can be maintained for some to generate data and theperformance compared with the solution. In case of major disagreement, the scope ofimprovement in the model structure opens and the O.R. m should be in a position toaddress such eventuality.

Over a length of time, due to complexity of real world, systems parameters mayundergo change and these require constant, monitoring. The solution is computer programsor special software packages, with flexibility to make changes in these parameters so thatrepeated application is possible.

of the solution of a validated model is the final task of an O.R.team.

It is one of the most important features of operations research. What it signifies is thatit is impossible for a single individual to have a tho ough and extensive knowledge of all theaspects of a particular problem which is to be analyze with the help of O.R. This wouldrequire a team of individuals having varied and divers skills and backgrounds. Such ateam should be inter-disciplinary and, it should include individuals having adequate degreeof proven skills in the fields of

a) statisticsb) engineeringc) computersd) economies ande) mathematics etc.

Every expert member of this team analyses each and eve aspect of the problem anddetermines if a similar problem has ever been undertak n previously by him or his colleagues.By functioning in such a manner, each member of the team suggests an approach whichmay be optimal for the problem under consideration, by utilizing his experience and skills.

Hence, operations research makes optimal and most effective utilization of peoplefrom diverse disciplines for developing latest tools a techniques applicable to the business.For example, while working on production planning and ontrol in an organization, onemay need the services of a production engineer, who kn ws the functions of the productionor assembly-line, a cost accountant and a statistician In this manner, every member of theteam so formed benefits from the views of other member of the team. Such a solutiondeveloped through team work and collaboration has rela vely, higher chance of acceptanceby the top management.



NOTESMethodological Approach

Objectivity Approach

Totalistic Approach

Continuous Process

O.R. is a highly systematic and scientific method. It llows a fixed and definite patternfrom the start to finish. The O.R. process starts with the careful observation and formulationwhere the problem is broken down into various workable and manageable parts and thencarefully observed by the O.R. team. The essence of th real problem is then sought to betranslated into the O.R. model. This model is examined solutions ascertained, most optimalones applied to the model and results examined. If the results are found to be satisfactory,these solutions are applied to the real or actual problem at hand.

The primary objective of O.R. is to find the best (optimal) solution to a problemunder consideration. To achieve this goal it is first necessary to define a measure ofeffectiveness that takes into consideration the main o tives of the organization. Thismeasure can then be used as a standard or bench-mark to compare and evaluate thealternative actions.

Any action or decision within an organization must first be analyzed and theirinteractions’ and the effect on the entire organizatio carefully considered. Under totalisticapproach, any action should not be seen in isolation. fore evaluating it, its impact on thewhole organization should always be kept in mind. For ample, to remove bottlenecks inproduction a production manger can store large quantity of inventories, raw materials andfinished goods. While this is important from the production manager’s view-point, it maylead to a situation where there may arise a direct conflict between the marketing and thefinance departments of the organization. In such a scenario, the O.R team examines all therelated factors such as cost of raw materials, holding & storage costs and competitor’sprices etc. Based on these, an appropriate O.R. model an be formulated for the solutionof the problem.

O.R. is a never-ending continuous process; it does not stop when a relevant O.R.model is applied to the problem since this may further create new problems in the relatedsectors as well as in the implementation’s of the decision taken. To implement the optimalsolutions (decisions), sometimes a change in the organizational structure might be needed.This is provided again by the O.R. After implementatio stage, the controlling of results isalso the primary function of operations research. Ther re, it can be considered to be acontinuous process.

DMC 1752


NOTES Broad Outlook

O.R. has a very great scope.

Economy of Operations

1.3 SCOPE OF OPERATIONS RESEARCH

Judgement Phase

Research Phase

Action Phase

It not only seeks to provide the most suitable optimalsolutions to a problem, but that it also uncovers new oblems by study methods.

The main function of O.R. is to provide solution to organizational problem and toassist in decision-making. In case of any conflict or mplexity in a situation, O.R. helps tominimize its costs and maximize profits resulting into “economy of operations”. Quite oftenthe problems are of such grave nature that they cannot be solved on the basis of intuition orpast experiences. E.g. New product development, product innovation, entry into newmarkets, product-mix etc.

The scope of operations research is very wide and depends upon the 3 key areas orphases of O.R. These are:-

The various components of this phase are as follows:

a) Ascertaining the operation

b) Establishing the objectives

c) Determining the appropriate effectiveness measures

d) Formulating the problem

The various components of this phase are as follows:

a) Observation and collection of data, to fully understan the various complex itemsthat form a par of the problem

b) Hypothesis or assumptions formulation

c) Model formulation

d) Analyzing the available data

e) Testing and verifying the accuracy of the hypothesis

f) Generalizing the results

g) Considering the alternative methods

After the above two phases are completed and an optimal solution has been obtained,in the action phase, recommendations are made to the t management for accepting theproposed solutions. Such a recommendation can be made y the persons who first presentedthe problem for consideration or by any other person who is in a position to make an



NOTES

Finance

Accounts

Marketing

effective decision. The decision, once accepted, is th n implemented in the implementationstage. Depending on these key areas, operations research can be applied in a very largenumber of real life complex situations. However, any t ols or techniques of OR should beutilized only after carefully considering the organisa onal requirements and should not befollowed blindly. Moreover the relative merits and demerits of a technique should alwaysbe kept in mind. It should be ensured that the basic postulates or assumptions on which itis based should not change. As and when the new requir nts emerge and the existingproblems become redundant, the technique in-use should either be mediated or replacedby another one immediately. In general, O.R. techniques can be applied extensively in thefollowing areas in an organisation:

i) Cash flow analysis and cash management models

ii) Investment portfolios

iii) Long range capital needs

iv) Financial planning models

v) Dividend policy

vi) Equipment replacement policy

vii) Capital allocation under various alternatives

i) Auditing and verification work

ii) Credit policy and credit risks

iii) Credit worthiness/Ratings

iv) Transfer pricing

v) Cost accounting - by-products and standard costs

i) Optimum product mix

ii) Advertising mix

iii) Sales promotion mix

iv) Product selection, timing and competitive action

v) Number of salesmen, frequency of calling and time spent on future customers

vi) Packaging

vii) Godown or warehousing

viii) Introducing a new product

DMC 1752


NOTES Purchases Management

Production Management

a. Physical Distribution

b. Distribution policy

c. Manufacturing

d. Maintenance and Project Scheduling

i) Terms and conditions for purchases

ii) Pricing of purchases

iii) Quantities to be purchased

iv) When to purchase

v) Bidding policies

vi) Searching for a new supplier or source

vii) Exploiting a new supplier or source

i) Location of godown or warehouse

ii) Size of godown or warehouse

iii) Channels of distribution

iv) Centres of distributions and retail outlets

i) Facilities Planning

ii) Number of factories/warehouses

iii) Location of factories/warehouses

iv) Facilities for loading/unloading for road as well as r lway transport schedule

v) Choosing sites with improved/better facilities

i) Production scheduling

ii) Production sequencing

iii) Minimising in-process inventory

iv) Minimising wastages/scraps/losses

v) Stabilizing production/lay-offs/product-mix

i) Maintenance policy

ii) Preventive maintenance

iii) Corrective maintenance

iv) Maintenance of crew size

v) Project scheduling

vi) Allocating the resources



NOTESe. Research and Development (R & D)

f. Organisational Development/Personnel Management

i) Areas interested in or key research areas

ii) Selecting the project

iii) Fixing the R & D budget

iv) Reliability and alternative design

v) Finding time - over runs

vi) Finding cost - over runs

vii) Controlling

i) Minimizing the need for temporary help through better cheduling

ii) Selecting the skilled personnel at lowest possible cost

iii) Recruitment policies

iv) Assigning the right job to the right man

v) Right blend of age and skills

DMC 1752


NOTESCHAPTER 2

LINEAR PROGRAM MING

2.1 LINEAR PROGRAM AND THE PRODUCT MIX MODEL

linearlinear

Linear Programming

Simplex Method

production planning

2.2 WHAT IS LINEAR PROGRAMMING

Many Operations Research mathematical models have the aracteristic feature thatthe objective function is some function of a certain number of decision alternatives

1, 2, …., n which are subject to a certain number of constraints. Alternativeresources to 1, 2, ….,

n at different levels satisfying the constraints results in different

values of the objective function and the problem is to determine the levels of the objectivefunction and the problem is to determine the levels for which the objective function becomesoptimal (maximal or minimal).

Historically, the problem in its general form was conc ved by the mathematician G.B.Dantzig and his associates in 1947, while working for States Department of AirForce. Beginning from the same year the economist T.C. Koopmans discovered that theformulation provided an excellent framework for the an lysis of classical economic theories.The term was coined by Koopmans and Dantzig in the summer of1948 during the course of their respective work. Dantzig’s group working on projectSCOOP (Scientific Computation of Optimum Programs) dev loped the theory and a rangeof applications. In 1949 Dantzig gave the of solution of the problemand put in place most of the theory. The development of the subject in the middle oftwentieth century is considered a landmark in the history of mathematics. The theory ishowever predated by some related work in 1939. L.V. Kantorovich considered a restrictedclass of problem occurring in the modeling of and proposed arudimentary method of solution. His work however remained neglected in the Soviet Unionand outside, but was much later awarded the Nobel Priz in Economics with Koopmans in1975. In the same year (1939), W.Karush investigated some aspects of optimality of afunction of variables subject to inequality constraints. These discoveries have proveduseful in linear programming theory. In subsequent years substantial developments havetaken place, notable by the discoveries of N.Karmarkar (1984). Adopting a radicallydifferent approach now called , substantial reduction in computingtime has been achieved for large dimensional problems.

Linear Programming is that branch of mathematical programming which is designedto solve optimization problems where all the constraints as well as the objectives areexpressed as linear function. It was developed by George B. Dentiz in 1947. Its earlierapplication was solely related to the activities of th Second World War. However soon itsimportance was recognized and it came to occupy a prominent place in the industry andtrade.

nx x x

x x x

n

Interior Point Method



NOTESLinear Programming is a technique for making decisions under certainty i.e., when all

the courses of options available to an organisation ar known and the objective of the firmalong with its constraints are quantified. That course of action is chosen out of all possiblealternatives which yield the optimal results. Linear P ing can also be used as averification and checking mechanism to ascertain the accuracy and the reliability of thedecisions which are taken solely on the basis of manag s experience without the aid of amathematical model.

Linear Programming is a method of planning and operati n involved in the constructionof a model of a real-life situation having the followi elements:

a. Variables which denote the available choices and

b. The related mathematical expressions which relate the riables to the controllingconditions reflect clearly the criteria to be employed for measuring the benefitsflowing out of each course of action and providing an ccurate measurement of theorganization’s objective. The method maybe so devised’ as to ensure the selectionof the best alternative out of a large number of alternative available to theorganization.

Linear Programming is the analysis of problems in whic a linear function of a numberof variables is to be optimized (maximized or minimize ) when whose variables are subjectto a number of constraints in the mathematical near inequalities.

From the above definitions, it is clear that:

i) Linear Programming is an optimization technique, where the underlying objectiveis either to maximize the profits or to minimize the c t function.

ii) It deals with the problem of allocation of finite limited resources amongst differentcompetiting activities in the most optimal manner.

iii) It generates solutions based on the feature and charac eristics of the actual problemor situation. Hence the scope of linear programming is very wide as it findsapplication in such diverse fields as marketing and production.

iv) Linear Programming has been highly successful in solving the following types ofproblems :

a. Product-mix problems

b. Investment planning problems

c. Blending strategy formulations and

d. Marketing and Distribution management.

v) Even though Linear Programming has wide and diversed applications, yet all LPproblems have the following properties in common:

a. The objective is always profit maximization or cost mi imization

b. Presence of constraints which limit the extent to whic the objective canbe pursued/achieved.

DMC 1752


NOTES

Decision or Activity Variables and their Inter-Relationship

Finite Objective Functions

Limiting Factors/Constraints

Presence of Different Alternatives

c. Availability of alternatives i.e., different courses o action to choose from.

d. The objectives and constraints can be expressed in the form of linearrelation.

Regardless of the size or complexity, all LP problems ake the same form i.e. allocatingscarce resources among various competiting alternatives. Irrespective of the manner inwhich one defines Linear Programming, a problem must h ve certain basic characteristicsbefore this technique can be utilized to find the opti l values.

The characteristics or the basic assumptions of linear programming are as follows:

The decision or activity variables refer to any activi es which are in competition withother variables for limited resources. Examples of such activity variables are: services,projects, products etc. These variables are most often inter-related in terms of utilization ofthe scarce resources and need simultaneous solutions. is important to ensure that therelationship between these variables be linear.

A Linear Programming problem requires a clearly defined, unambiguous objectivefunction which is to be optimized. It should be capable of being expressed as a linearfunction of the decision variables. The single-objective optimization is one of the mostimportant pre-requisites of linear programming. Examples of such objectives can be: cost-minimization, sales, profits or revenue maximization and the idle-time minimization etc.

These are the different kinds of limitations on the av ilable resources for e.g. importantresources like availability of machines, number of man hour’s available, production capacityand number of available markets or consumers for finished goods are often limited evenfor a big organization. Hence, it is rightly said that each and every organization functionswithin overall constraints of both internal and extern l. These limiting factors must be capableof being expressed as linear equations or in equations in terms of decision variables.

Different courses of action or alternatives should be vailable to the decision maker,who is required to make the decision which is the most effective or the o timal. For example,many grades of raw material may be available and the s e raw material can be purchasedfrom different supplier, the finished goods can be sol to various markets, production canbe done with the help of different machines.



NOTESNon-Negative Restrictions

Linearity Criterion

Additively

Mutually Exclusive Criterion

Divisibility

Certainty

Finiteness

Since the negative value of (any) physical quantity has no meaning, therefore all thevariables must assume nonnegative values. If some of he variables is unrestricted in sign,the non-negativity restriction can be enforced by the lp of certain mathematical toolswithout altering the original information contained in the problem

The relationship among the various decision variables st be directly proportionalto both the objective and the constraint and must be e pressed in terms of linear equationsor inequalities. For example, if one of the factor inputs (resources like material, labour,plant capacity etc.) increases, then it should result a proportionate manner in the finaloutput. These linear equations and in equations can graphically be presented as a straightline.

It is assumed that the total profitability and the tot amount of each resource utilizedwould be exactly equal to the sum of the respective in ividual amounts. Thus the functionor the activities must be additive - and the interaction among the activities of the resourcesdoes not exist.

All decision parameters and the variables are assumed o be mutually exclusive. Inother words, the occurrence of anyone variable rules out the simultaneous occurrence ofother such variables.

Variables may be assigned fractional values. i.e., the need not necessarily always bein whole numbers. If a fraction of a product cannot be produced, an integer programmingproblem exists.

Thus, the continuous values of the decision variables nd resources must be permissiblein obtaining an optimal solution.

It is assumed that conditions of certainty exist i.e., all the relevant parameters orcoefficients in the Linear Programming model are compl ely known and that they don’tchange during the period. However, such an assumption y not hold good at all times.

Linear Programming assumes the presence of a finite nu ber of activities and constraintswithout which it is not possible to obtain the best or the optimal solution.

DMC 1752


NOTES 2.3 ADVANTAGES AND LIMITATIONS OF LINEAR PROGRAMMING

Some of the advantages of Linear Programming approach

Scientific Approach to Problem Solving

Evaluation of all Possible Alternatives

Helps in Re-Evaluation

Quality of Decision

Focus on Grey Areas

Flexibility

Creation of Information Base

Now it is time to examine the advantages as well as the limitations of LinearProgramming.

Linear Programming is the application of scientific ap roach to problem solving. Henceit results in a better and true picture of the problems which can then be minutelyanalyzed and solutions ascertained.

Most of the problems faced by the present organizations are highly complicated whichcannot be solved by the traditional approach to decision making. The technique ofLinear Programming ensures that all possible solutions are generated out of which theoptimal solution can be selected.

Linear Programming can also be used in re-evaluation o a basic plan for changingconditions. Should the conditions change while the plan is carried out only partially,these conditions can be accurately determined with the help of Linear Programmingso as to adjust the remainder of the plan for best res s.

Linear Programming provides practical and better quality of decisions that reflectvery precisely the limitations of the system i.e., the various restrictions under whichthe system must operate for the solution to be optimal. If it becomes necessary todeviate from the optimal path, Linear Programming can easily evaluate theassociated costs or penalty.

Highlighting the grey area is the bottleneck of the production process and it is themost significant merit of Linear Programming. During the periods of bottlenecks,imbalances occur in the production department. Some of the machines remain idle forlong periods of time, while other machines are unable o meet the demand even at thepeak performance level.

Linear Programming is an adaptive and flexible mathematical technique and hencecan be utilized in analyzing a variety of multi-dimens onal problems quite successfully.

By evaluating the various possible alternatives in the light of the prevailing constraints,Linear Programming models provide an important databas from which the allocationof precious resources can be done rationally and judiciously.

§

§

§

§

§

§

§



NOTESMaximum optimal Utilization of Factors of Production.

2.4 LIMITATIONS OF LINEAR PROGRAMMING

Linear Relationship

Constant Value of objective and Constraint Equations

No Scope for Fractional Value Solutions

§

Linear Programming helps in optimal utilization of var ous existing factors of productionsuch as installed capacity labour and raw materials etc.

Although Linear Programming is a highly successful too having wide applications inbusiness and trade for solving optimization problems, t it has certain demerits or defects.Some of the important limitations in the application of Linear Programming are as follows:

Linear Programming models can be successfully applied nly in those situations wherea given problem can clearly be represented in the form of linear relationship betweendifferent decision variables. Hence it is based on the implicit assumption that the objectiveas well as all the constraints or the limiting factors can be stated in term of linear expressionswhich may not always hold well in real life situations In practical business problems, manyobjective function and constraints cannot be expressed linearly. Most of the businessproblems can be expressed quite easily in the form of quadratic equation (having apower 2) rather than in the terms of linear equation. ear Programming fails to operateand provide optimal solutions in all such cases.

E.g. A problem capable of being expressed in the form f:

ax2 + bx + c = 0 where a>0 can not be solved with the help of Linear Programmingtechniques.

Before a Linear Programming technique could be applied to a given situation, thevalues or the coefficients of the objective function as well as the constraint equations mustbe completely known. Further, Linear Programming assumes these values to be constantover a period of time. In other words, if the values w to change during the period ofstudy, the technique of LP would loose its effectiveness and may fail to provide optimalsolutions to the problem. However, in real life practi al situations often it is not possible todetermine the coefficients of objective function and t constraints equations with absolutecertainty. These variables in fact may, lie on a probability distribution curve and hence atbest, only the likelihood of their occurrence can be p cted. Moreover often the valuechanges due to extreme internal factors during the per d of study. Due to this, the actualapplicability of Linear Programming tools may be restricted.

There is absolutely no certainty that the solution to LP problem can always bequantified as an integer quite often. Linear Programmi g may give fractional-varied answers,

DMC 1752


NOTES

High Degree of Complexity

Multiplicity of Goals

Flexibility

which are rounded off to the next integer. Hence, the ution would not be the optimalone. For example, in finding out the number of men and machines required to perform aparticular job, a fractional solution would be meaning ss.

Many large-scale real life practical problems cannot be solved by employing LinearProgramming techniques even with the help of a compute , due to highly complex andlengthy calculations. Assumptions and approximations a required to be made so that thegiven problem can be broken down into several smaller roblems and, then solve separately.Hence, the validity of the final result, in all such cases, may be doubtful.

The long-term objectives of an organization are not confined to a single goal. Anorganization, at any point of time in its operations h a multiplicity of goals or the goalshierarchy - all of which must be attained on a priorit wise basis for its long term growth.Some of the common goals can be maximization or minimization, retainingmarket share, maintaining leadership position and prov ding quality service to the consumers.In cases where the management has conflicting, multipl goals, Linear Programming modelfails to provide an optimal solution. The reason being that under Linear Programmingtechniques, there is only one goal which can be expressed in the objective function. Hencein such circumstances, the situation or the given prob m has to be solved by the help of adifferent mathematical programming technique called th “Goal Programming”.

Once a problem has been properly quantified in terms of objective function and theconstraint equations and the tools of Linear Programming are applied to it, it becomes verydifficult to incorporate any changes in the system arising on account of any change in thedecision parameter. Hence, it lacks the desired operat nal flexibility.

Linear Programming is a mathematical technique for gen rating and selecting the optimalor the best solution for a given objective function. T chnically, Linear Programming may beformally defined as a method of optimizing (i.e., maxi izing or minimizing) a linear functionfor a number of constraints stated in the form of line equations.

Mathematically the problem of Linear Programming may be stated as that of theoptimization of linear objective function of the follo ng form:

Z = C1x2 + C

2x2 + ..........C

ix i+ ...........+ CnXn

profit cost



NOTESSubject to the Linear constrains of the form:

a 11

x1 + a

12x

2 + a

13x

3 ........+ a

1ix

i + .............. + a

lnx

n b

1

aj1x

1 + a

j2x

2 + a

j3 x

3 ........+ a

jix

i +.............. + a

jn x

n bm

1

am1x 1+ am 2

x 2+ am3 x3 + .........+ ami xi ..........+ amnxn bm

These are called the non-negative constraints. From the above, it is clear that a LPproblem has:

i) Linear objective function which is to be maximized or inimized.

ii) Various linear constraints which are simply the algebr ic statement of the limits ofthe resources or inputs at the disposal.

iii) Non-negatively constraints.

Linear Programming is one of the few mathematical tool that can be used to providesolution to a wide variety of large, complex managerial problems.

For example, an oil refinery can vary its product-mix y its choice among the differentgrades of crude oil available from various parts of the world. Also important is the processselected since parameters such as temperature would also affect the yield. As prices anddemands vary, a Linear Programming model recommends wh h inputs and processes touse in order to maximize the profits

Livestock gain in value as they grow, but the rate of in depends partially on the feedchoice of the proper combination of ingredients to maximize the net gain. This value can beexpressed in terms of Linear Programming. A firm which distributes products over a largeterritory faces an unimaginable number of different ch ces in deciding how best to meetdemand from its network of go down and warehouses. Each warehouse has a very limitednumber of items and demands often cannot be met from t nearest warehouse. If thereare 25 warehouses and 1,000 customers, there are 25,000 possible match ups betweencustomer and warehouse. LP can quickly recommend the s ipping quantities and destinationso as to minimize the cost of total distribution.

These are just a few of the managerial problems that h ve been addressed successfullyby LP. A few others are described throughout this text Project scheduling can be improvedby allocating funds appropriately among the most critical task so as to most effectivelyreduce the overall project duration. Production planni over a year or more can reducecosts by careful timing of the use of over time and in ntory to control changes in the sizeof the workforce. In the short run, personnel work sch ules must take into considerationnot only the production, work preferences for day offs and absenteeism etc.

Besides recommending solutions to problems like these, LP can provide usefulinformation for managerial decisions that can be solved by Linear Programming. Theapplication, however, rests on certain postulates and tions, which have to holdgood for the optimality of the solution to be effective during the planning period.

≤

≤

≤

DMC 1752


NOTES 2.5 APPLICATIONS OF LINEAR PROGRAMMING TECHNIQUES IN INDIAN ECONOMY

2.6 BUILDING LINEAR PROGRAMMING PROBLEMS

linear programming LP model

proportional objective function constraint

The various factors of productions such as

a. Skilled labour

b. Capital and

c. Raw material etc.,

are very precious and scarce for a country like India. The policy planner is, therefore facedwith the problem of scarce resource allocation to meet the various competiting demandsand numerous conflicting objectives. The traditional and conventional methods can nolonger be applied in the changed circumstances for solving this problem and are hence fastlosing their importance in the current economy. Hence, the planners in our country arecontinuously and constantly in search of highly objective and result oriented techniques forsound and proper decision making which can be effectiv at all levels of economic planning.Non-programmed decisions consist of:-

a. Capacity expansion

i) Plant location

ii) Product line diversification

iii) Expansion

iv) Renovation and

v) Modernization etc.

On the other hand, the programmed decisions consist of budgeting, replacement,procurement, transportation and maintenance etc. In these times, a number of new andbetter methods, techniques and tools have been developed by the economists all over theglobe. All these findings form the basis of operations research. Some of these well-knownoperations research techniques have been successfully pplied in Indian situations, suchas:

b. Business forecasting

i) Inventory models - deterministic and probabilistic

ii) Linear Programming

iii) Goal programming

iv) Integer programming and

v) Dynamic programming etc.

In building a (or briefly ) for a system, two crucialassumptions must be valid. The is that the activity level

j ( = 1, 2… ) has a

contribution to the as well as in a . If andfirst x j n

cjx

j



NOTESper unit

discounts additivity

linear functions

Product Mix Model productionplanning firm

resourcescapacities

return

Linear Programming Problem LP

aijxjith

cj

aij

second

simple additionx x x

n

n mmaximum available b b bm aij

ith unit jth

cj

nx

jjth

j n

c x c x cnxn

a x a x a nxn ba x a x a nxn b

am x am x amnxn bm

x x xn

x x xn divisible

x x xn

m

aa

am

aa

am

aa

amn

bb

bm

c c cn

are the respective contributions in the objective and the constraint, the contributions and are assumed constants for the system and must be known fromanalysis of the actual system. The proportionality ass mption precludes economics of returnto scale or . The assumption is that of of the contributions.The system must be such that the objective function and each of the constraint can beobtained from the contributions by . In this way, the objective function andeach of the constraints become 1, 2… .

As an example, consider the , applicable for . Suppose that a production unit (or in the terminology of economics)

produces number of products utilizing number of (of its own or of others)having 1, 2…, . Let be the amount consumedfrom the source for the production of a of the product which ensures (orprofit) . These data are supposed to be known from the cost an ysis for the productionunit and can be arranged as a table in the following m ner.

Given the data, it is required to find the level of pr ction of the products so thatthe total return Z is maximum. If we suppose that the level of production is for the product ( = 1, 2… ), then the mathematical problem of determining these ownsbecomes the (also written again in short as ):

Maximize : z = 1 1 + 2 2 + . . . +

s.t : 11 1 + 12 2 + . . . + 1 1

21 1 + 22 2 + . . . + 2 2

……………………………………..

1 1 + 2 2 + . . . +

1, 2, …, 0.

If the system allows 1, 2… to be numbers, then their values could beinteger, fractional and even theoretically any real number. This is obviously the case forexample, a paint manufacturer producing paints of diff ent colours and shades. There arehowever several instances, like manufacture of televis on sets, music systems etc. of differentmodels in an electronic manufacturing company, where it is required that 1, 2, …, must be integers, then this constraint also enters the list of constraints written above.

ProductResource

1 2 . . . nAvailable Capacity

12..

11

21

.

.1

12

22

.

.2

. . .

. . .

. . .

. . .

. . .

1n

2n

.

.

1

2

.

.

Unit Return 1 2 . . .

≤≤

≤

≥

DMC 1752


NOTES activities

Remark 1.

2.7 STEPS IN FORMULATINGA LINEAR PROGRAMMING MODEL

Step I. Identification of the Decision Variables

Step II. Identification of the Constraints

Step III. Identification of the Objective

2.8 EXAMPLES

2.8.1 Example 1

In general, instead of manufacturer of certain products one can have certain other of an organization which can be modeled on the linearity assumption. Evidently

the mathematical problem remains the same.

In many cases the available capacities are not simple constants but depend on and time with associated costs. In such cases the above model cannot

be used.

As we have so far seen that linear programming is one the most useful techniquesfor effective decision making. It is an optimization a proach with an emphasis on providingthe optimal solution for resource allocation.

How best to allocate the scarce organizational or national resources among differentcompeting and conflicting needs (or uses) forms the core of its working. The scope forapplication of linear programming is very wide and it cupies a central place in manydiversified decisional problems. The effective use and application of linear programmingrequires the formulation of a realistic model which represents accurately the objectives ofthe decision making subject to the constraints in which it is required to be made.

The basic steps in formulating a linear programming mo l are as follows:

The decision variables (parameters) having a bearing on the decision at hand shallfirst be identified, and then expressed or determined n the form of linear algebraic functionsor in equations.

All the constraints in the given problem which restric the operation of a firm at a givenpoint of time must be identified in this stage. Further these constraints should be brokendown as linear functions in terms of the pre-defined decision variables.

In the last stage, the objective which is required to optimized (i.e., maximized orminimized) must be clearly identified and expressed in terms of the pre-defined decisionvariables. With this background information, let us no try to formulate the given problemas a linear programming problem.

:- High Quality Furniture Ltd. manufactures two products,

i) Tables and

ii) Chairs.Both the products have to be processed through two mac nes:

i) Ml and

ii) M2

set-up takedown



NOTES

Solution.

Step I.

Step II.

Step III.

The total machine-hours available are:

i) 200 hours ofM1 and

ii) 400 hours of M2 respectively.

Time in hours required for producing a chair and a table on both the machines is asfollows:

Time in Hours

Profit from:-

i) Sale of table is Rs.50

and sale of chair is Rs.30

Determine optimal mix of tables and chair so as to maximized the total profitcontribution

Let x1

=no. of tables produced and

x2 =no. of chairs produced

The objective function for maximizing the profit is g en by

Maximize Z=50x1 +30x2 (objective function) (Since profit per unit from a table anda chair is Rs. 50 and Rs. 30 respectively).

List down all the constraints.

i) Total time on machine M1 cannot exceed 200 hours.

7x1+4x2 200

(Since it takes 7 hours to produce a table and 4 hours to produce a chair on machineM

1)

ii) Total time on machine M2 cannot exceed 400 hours

5x1+5x2 400

(Since it takes 5 hours to produce both a table and a on machine M2)

Presenting the problem. The given problem can now be ulated as a linearprogramming model as follows:

Maximize Z = 50x1 + 30x2

Subject to: 7x1+4x2 200

5x1+5x2 400

Futher; x1, x2 0

Machine Table ChairM1 7 4M2 5 5

∴ ≤

∴ ≤

≤

≤

≥

DMC 1752


NOTES2.8.2 Example 2

Solution.

Step 1

Step II

Step III Presenting the problem

2.8.3 Example 3.

(Since if x1 and x2 < 0 it means that negative quantities of products are beingmanufactured which has no meaning).

: Alpha Limited produces and sells 2 different products under the brandname black and white. The profit per unit on these products in Rs. 50 and Rs. 40respectively. Both black and white employ the same manufacturing process which has afixed total capacity of 50,000 man-hours. As per the e mates of the marketing researchdepartment of Alpha Limited, there is a market demand or maximum 8,000 units of blackand 10,000 units of white. Subject to the overall demand, the products can be sold in anypossible combination. If it takes 3 hours to produce one unit of black and 2 hours toproduce one unit of white, formulate the above as a li ar programming model.

Let x1, x2 denote the number of units produced of products black and whiterespectively.

: The objective function for maximizing the profit is ven by: maximize

Z= 50x1 + 40x2 (objective function)

: List down all the constraints.

i) Capacity or man-hours constraint:

(Since it takes 3 hours to produce one unit of x1 and 2 hours to produce 1 unit of x2

and the total available man - hours are 50,000)

ii) Marketing constraints:

(Since maximum 8,000 units of x1 can be sold) (Since maximum 10,000 units of x2 canbe sold).

: . Now, the given problem can be written as a linearprogramming model in the follows:

Maximize Z = 50x1 + 40x2

Subject to: 3x1+2x2 50,000

X1 8000

X2 10,000

Further; x1, x2 0

(Since if x1, x2< 0, it means that negative Quantities of products are being manufactured -which has no meaning)

Good Results Company manufactures and sells in the export marketthree different kinds of products:

i) P1,

ii) P2 and

iii) P3

≤

≤

≤

≥



NOTES

Solution.

Step I

Step II

The anticipated sales for the three products are:

i) 100 units of P1,

ii) 200 units of P2 and

iii) 300 units of P3.

As per the terms of the contract Good results must produce at least: -

i) 50 units of P1 and

ii) 70 units of P3.

Following is the break - up of the various production mes:

Management is free to establish the production schedule subject to the aboveconstraints.

Formulate as a linear programming model assuming profi maximization criterion forGood Results Company.

Let x1, x

2, x

3 denote the desired quantities of products P1, P2 and P

3 respectively.

. The objective function for maximizing total profits is given by:

Maximize Z =15x1 + 20x2 + 25x3 (objective function)

. List down all the constraints.

The available production hours for each of the products must satisfy the followingcriterion for each department:

i) 0.05x1+0.10x2+0.20x3 40.00

Total hours available on product - wise basis in Depar ent A

ii) 0.06x1+0.12x2+0.09x3 45.00

Total hours available on product - wise basis in Department B

iii) 0.07x1+0x2+0.07x3 50.00

Total hours available on product - wise basis in Department C

ProductProduction Hours

per UnitDepartment Department Department Department Unit(A) (B) (C) (D) Profit

P 1 0.05 0.06 0.07 0.08 15P 2 0.10 0.12 -- 0.30 20P 3 0.20 0.09 0.07 0.08 25

Available hours

40.00 45.00 50.00 55.00

≤

≤

≤

DMC 1752


NOTES

Step III:

2.8.4 Example 4

iv) 0.08x1+0.30x2+0.08x3 55.00

Total hours available on product wise basis in Department D

v) 50 x1 100

Minimum 50 units of P1 must be produced subject of maximum of 100 units.

vi) 0 x2 200

Maximum units of P2 that can be sold is 200 units.

vii) 70 x2 300

inimum 70 units of P3 must be produced subject to a ma mum of 300 units

viii.) Further, x1, x2, x3 0

Since negative values of P1, P2 and P3 has no meaning

Presenting the Problem

The given problem can be reduced as a LP model as under:

Maximize: Z=15x1 + 20x2 + 25x3

Subject to: 0.05x1+0.10x2+0.20x3

40.00

0.06x1+0.12x2+0.09x3 45.00

0.07x1+0x2+0.07x3 50.00

0.08x1+0.30x2+0.08x3 55.00

50 d x1 100

0 d x2 200

70 x2 300

x1, x2, x3 0

. The management of Surya Chemicals is considering the optimal mix oftwo possible processes. The values of input and output for both these process are given asfollows:

Maximum:

i) 500 units of Input I 1 and 300 units of I2,

ii) 300 units of I2

are available to Surya Chemicals in the local market.

Process Units Inputs Units OutputsI1 I2 O1 O2

X 2 6 3 7Y 4 8 5 9

≤

≤ ≤

≤ ≤

≤ ≤

≥

≤

≤

≤

≤

≤ ≤

≤ ≤

≤ ≤

≥



NOTES

Solution

Step I.

Step II

Step III

The forecasted demand for outputs O1 and O2 are at least:

i) 5,000 units and

ii) 7,000 units respectively.

The respective profits from process X and Y are:

i) Rs. 1,000 and

ii) Rs. 2,000 - per production run.

Formulate the above as a linear programming model.

. Let x1, x

2 represent the number of production runs of process x d y respectively.

The objective function for maximizing the total profits from both the process isgiven by:

Maximize Z 1000x1 + 2000x2

. List down all the constraints

i) 2x1 + 4x2 500

Maximum amount of input I 1 available for process x and y is 500 units

ii) 6x1 + 8x2 300

Maximum amount of input I 2 available for process x and y 300 units

iii) 3x1 + 5x2 5,000

Since market requirement is to produce at least 5000 units of O1

iv) 7x1 + 9x2 7,000

Since market requirement is to produce at least 7000 units of O2

Further: - x1,

x2

0 as always.

. Presenting the model.

Now, the current LP model can be presented as:

Maximize Z=1000x1+2000x2

Subject to: 2x1+4x2 500

6x1+8x2 300

3x1+5x2 5,000

7x1+9x2 7000

x1, x2 0

≤

≤

≤

≤

≥

≤

≤

≤

≤

≥

DMC 1752


NOTES 2.8.5 Example 5

Solution

Step 1

Step II

Inputs RequiredProduct Profit

: Chocolate India Ltd. produces three varieties of Chocolates –

i) Hard,

ii) Mild and

iii) Soft

from three different inputs I 1, I

2 and I

3.

One unit of Hard requires 2 units of I 1 and 4 unit of I 2. One unit of mild requires 5

units of I 1, 4 units of I 2 and 3 units of I 3 and one unit of soft requires 10 units of I 1 and 15

units of I 3. The total available inputs in the company’s warehouse is as under:

I3 - 50 units

I2 - 400 units

I1 - 100 units

The profit per unit for hard, mild and soft are

i. Rs. 30 and

ii. Rs. 20

iii. Rs. 40 respectively.

Formulate the problem so as to maximize the total prof t by using linear programming.

. At the beginning, it is better to present the problem in a tabular form

Let x1, x

2 and x3 denote the no. of units of the three varieties of cho olate - Hard, mild

and soft respectively.

The objective function for maximizing the profit is:

Maximize Z = 30x1 + 20x2 + 40x

3 (objective function)

List down all the constraints:

i.) 2x1+5x2+10x3 100

(Since Input I 1 required for the three products is 2, 5 and 10 units respectivelysubject to a maximum of 100 units).

ii.) 4x1+4x2+0x3 400

(Since input I 2 required for the three products is 4,4 and 0 units respectivelysubject to a maximum of 400 units).

I1 I2 I3 (Rs. Per unit )Hard 2 4 -- 20Mild 5 4 3 30Soft 10 -- 15 40

Maximum availability of inputs 100 400 50

≤

≤



NOTES

Step III

2.9 EXERCISES

Food value (gm) per 100 gmFood itemsCarbohydrates Proteins Fats

Cost per 100 gm (Rs.)

iii.) 0x1+3x2+15x3 50

(Since input I 2 required for the three products is 0,3 and 15 units respectivelysubject to a maximum of 50 units).

. Presenting the problem. The given problem can now be formulated as a linearprogramming model as follows:

Maximize Z = 30x1 + 20x2 + 40x3

Subject to: 2x1+5x2+10x3 100

4x1+4x2+0x3 400

0x1+3x2+15x3 50

Further:x1, x2, x3 0

1. A medical expert suggests that it is necessary for an dult to consume at least 200gm carbohydrate, 50 gm protein and 35 gm fat daily. Considering a person hasonly three food items, Rice, Vegetables and Fish with ood value and cost per 100gm is given in the following table:

The consumption of rice must not exceed 200 gm a day. ormulate the problem as alinear program to determine the requirement of the foo item at least cost.

[ 1 – Rice, 2 – Vegetables, 3 – Fish 100gm each.

Minimize: z = 2 1 + 1.5 2 + 7 3,

s.t: 65 1 + 40 2 + 12 3 200,

7 1 + 9 2 + 25 2 50,

3 1 + 42

+ 32 2 35,

1 2, 1, 2, 3 0]

2. The manufacturer of formulated medicines wants to prepare production plan fortwo medicines A and B every day. The market demand for A is at least one andhalf times that of B but the rate of bottling A to tha of B is 1 : 2 and there must bean inventory of 500 bottles of A at the beginning of e ery day. There is also storagelimitation of the bottles at 8000. The profits on selling medicines A and B arerespectively Rs. 6 and Rs. 8. Formulate the problem as L.P.

RiceVegetables Fish

654012

79

25

34

32

2.001.007.00

≤

≤

≤

≤

≥

≥

≥

≥

≤ ≥

x x x

x x x

x x x

x x x

x x x

x x x x

DMC 1752


NOTES

Machine Number of 200 ml bottles per min.

Number of jumbo bottles per min.

[ 1 – A, 2 – B (in thousands).

Maximize: z = 6000 1 + 8000 ,

s.t.: 1

(3/2) 2,

2 1 - 2 = const. 1,

1 + 2 8,

1, 2 0].

3. A soft drink manufacturer has two machines A and B for bottling 200ml and jumbosized bottles. The rates of bottling are

The machines can be run for 8 hours 5 days a week. The market demands for twotypes of bottles are at most 25000 and 8000 per week and the profits are 15 paiseand 75 paise per bottle. Formulate this as L.P for determining number of bottles tobe manufactured, of each type per week for maximum tot profit.

[ 1 – 200 ml bottles, 2 – jumbo bottles].

Maximize: z = .15 1+.75 2, s.t.: time constraints 1/100 + 2 /40 8 X 60 X 5 =2400, 1 / 60 + 2 / 75 2400, market demands 0 1 2500, 0 2 8000, 1,

2 integers 0].

4. A Television manufacturer produces BandW and color sets and sells them at aprofit of Rs. 500 and Rs. 1500 respectively on each set. The time required forassembling a color set is 5 hours and decorating it is 1 hour, while the respectivetimes for a BandW set are 2 hours and 45 minutes. The 12assemblers and 4 decorators working 8 hours per day. Formulate the problem asL.P. so that the manufacturer makes maximum profit by elling all models producedin a month.

[ 1 - BandW set, 2 – color set.

Maximize: z = 500 1 + 1500 2,

s.t.: 2 1 + 5 2 2880,

(3/4) 1 + 2 960,

1, 2 integers 0].

AB

10060

4075

x x

x x2

x x

x x

x x

x x

x x

x x x xx x x x x

x

x x

x x

x x

x x

x x

≥

≥

≤

≥

≤≤ ≤ ≤ ≤ ≤

≥

≤

≤

≥



NOTES

6.

Requirement per unitRaw Material

I II IIIAvailability

Activity Model I Model II Model III Available Time (hours)

5. A manufacturers produces three models I, II, III of a rtain product every week.He uses two types of raw materials A and B; the per unit requirement for the threemodels are

The labour time per unit of model I is twice that of I and three times that of III. Theentire labour force of the factory can produce the equ lent of 2000 units of modelI. The profit per unit of models I, II, III are Rs. 60, 40 and 100 respectively. Formulatethe problem.

[ 1 – I, 2 – II, 3 – III.

Maximize: z = 60 1 + 40 2 + 100 3 2000,

s.t.: 2 1 + 3 2 + 5 3 5000,

4 1 + 2 2 + 7 3 6000,

1 + (1/2) 2 + (1/3) 3 2000,

1 500, 2 500, 3 375, integer 1, 2, 3 0].

A manufacturer produces three models I, II, III of a p which require firstmachining and then assembling. The time (hours) required and the labour cost(Rs.) per machine are as follows:

The available cash to pay labour is Rs.1, 20,000. If t selling price of models I, II,III are Rs.4000, Rs.7500, and Rs.12600; find how the manufacturer can makemaximum return.

[ 1 - I, 2 - II,

3 - III.

Maximize z = (4000 - 2000) 1 + (7500 - 4500)

2 + (12600 - 8000) 3 = 2000 1

+ 3000 2 + 46003,

s.t.: 12 1 +15 2 + 163

3000, 3 1 + 4 2 + 6 3 1100, 2000 1 + 4500 2 +8000 3 120000,

1, 2, 3 integers 0].

AB

24

32

57

50006000

Minimum Demand 500 500 375

Machining Assembly

123

154

166

30001100

Labour cost 2000 4500 8000

x x x

x x x

x x x

x x x

x x x

x x x x x x

x x x

x x x xx x

x x x x x x x xx

x x x

≤

≤

≤

≤

≥ ≥ ≥ ≥

≤ ≤≤

≥

DMC 1752


NOTES 7. A woman worker of a cooperative has two types of job: spinning thread and(ii) dyeing the spun thread, producing a unit of each r hour. She is paid Rs.2 perunit for spinning and Rs.5 for dyeing and wants to ear not less than Rs.18 per dayworking for no more than 6 hours. The cooperative want that each day the spunthread should not exceed the thread consumed in dyeing by 2 units. If the sellingprofit is Rs.3 per unit for spinning and Rs.7 per unit for dyeing, formulate the L.Pfor determining the units the woman should spin and dye each day to makemaximum profit for the cooperative.

Try to solve the problem by trial.

[Spinning – 1 units, dyeing – 2 units.

Maximize z = 3 1 + 7 2,

s.t.: 2 1 + 5 2 18,

1 + 2 6,

0 1 - 2 2,

1, 2 integers 0.

Solution: 1 = 4, 2 =2, zmax = Rs.26].

8. A farmer has 100 acre land on which he wants to grow tomatoes, cabbages andradishes. The average yield per acre is 2000 kg of tomatoes, 3000 heads ofcabbage and 1000 kg of radishes. The price he can obtain is Rs.3 per kg oftomatoes, Rs.1 per head of cabbage and Rs.2 per kg of cost offertilizer is Rs.1.50 per kg and the amount required per acre is 100 kg for tomatoesand cabbage and 50 kg for radishes. Labour required for sowing, cultivating andharvesting are 5 man-days for tomatoes and radishes an 6 man-days for cabbages.A total of 400 man-days are available at Rs.40 per man day. Formulate the problemas an L.P for farmer’s maximum net profit.

[ 1 acres-tomatoes, 2 acres-cabbages, 3 acres-radishes. Maximum: z = 6000 1

+ 3000 2 + 2000 3 – 1.5 X (100 1 + 100 2 + 50 3) – 40 X (5 1 + 6 2 + 5 3) =5650 1 + 2610 2 + 1725 3, s.t.: acreage constraint 1 + 2 + 3 100, man-days constraint 5 1 + 6 2 + 5 3 400, 1, 2, 3 0].

9. A farmer wants to raise an orchard of mangoes and guav on three farms 1,2 and3 with usable acreage 400, 600 and 300 respectively. T water available in eachfarm is 2000 units and the consumption per acre are 5 ts for mangoes and 4units for guavas. In order to maintain a uniform work d among the farms thefraction of usable average must be kept the same at ea farm. If the expectedprofit per acre for mangoes is Rs. 400 and for guavas 300, the farmer wantsto know how many acres of each crop should be planted each of the farms inorder to have maximum profit.

x x

x x

x x

x x

x x

x x

x x

x x x xx x x x x x x x

x x x x x xx x x x x x

≥

≤

≤ ≤

≥

≤≤ ≥



NOTES

Holds Weight Capacity (tons) Volume Capacity (m3)

Commodity Weight (tons) Volume per ton (m3)

Profit per ton (Rs.)

Investment Alternatives

Return (percent)

Number of Years

Risk

∑∑==

+=

≤ ≤ ≤ ≤≥ ≥

≤ ≤ ≤ ≤≤ ≤ ≤

≤ ≤

≥

3

1

3

1gimi 300400z:Maximize trees.guava andmangosfor average x,[x

s.t.: m1 + g1 400, m2 + g2 600, m3 + g3 300, 5 mi + 4 gi 2000 for = 1, 2,3, ( m1 + g1) / 400 = ( m2 + g2) / 600 = ( m3 + g3) / 300, mi 0. gi 0 for =1,2,3].

10. Formulate the following LP. A ship has three cargo holds – forward, centre andaft. The capacity limits are

The following cargoes are offered: the ship owner may accept all or any part of eachcommodity

In order preserve the trim of the ship, the weight in ach hold must be proportional tothe capacity in tons. The cargo is to be distributed so as to maximize profit.

[ 1F, 1C, 1A – weight – (ton) loaded in forward, centre and aft etc., Maximize z =60 ( 1F + 1C + 1A) + 80( 2F + 2C + 2A) + 50( 3F + 3C + 3A), s.t.: 1F + 2F +

3F 2000, 1C + 2C + 3C 3000, 1A + 2A + 3A 1500, 1F + 1C + 1A

6000, 2F + 2C + 2A 4000, 3F + 3C + 3A 2000, 6 1F + 5 2F + 25 3F

10000, 6 1C + 5 2C + 25 3C 14000, 6 1A + 5 2A + 25 3A 3000, ( 1F + 2F +

3F) / 2000 = ( 1C + 2C + 3C) / 3000 = ( 1A + 2A + 3A) / 1500, 1F, 1C, x1A etc. 0].

11. (Investment Model) On retirement an employee wants to vest his retirementbenefits (provident fund, gratuity etc.) in four inves ment alternatives on a ten yearhorizon. The alternatives involve risks which he subjectively marks on a five pointscale. The data are given below:

ForwardCentreAft

200030001500

10000140003000

ABC

600040002000

65

25

608050

Term DepositsNSC

Equity SharesReal Estate

9.510.52025

265

10

10.572

igi

imi xx

x x x x x x x x ix x x x x x x x i

x x xx x x x x x x x x x x

x x x x x x x x x xx x x x x x x x x

x x x x x x x xx x x x x x x x x

DMC 1752


NOTES

(Investment Model)

A B C D

What percentage of his retirement benefits should he invest in the four schemes forhighest return if he decides not to take risks exceeding 3 and invest at least 30% inreal estate?

[ 1, 2, 3, 4 – respective percentages of total amount.

Maximize: z = 0.095 1 + 0.105 2 + 0.20 3 + 0.25 4,

s.t.: 2 1 + 6 2 + 5 3 +10 4 10,

1 + 0.5 2 + 73 + 2 4 3,

4 0.30,

1, 2, 3, 4 0].

12. A chartered accountant wants to determine best investmentportfolio in four proposals A, B, C, D. The portfolio ta are as follows:

The amount available is Rs 25 lakhs. He decides not to invest more than Rs 5 lakhsin A and B combined and for the sake of diversity at least 100 shares of each stockshould be purchased. Also, the weighted risk

esalternativtheallininvestmentTotal

)jealternativofRisk(x)jealternativinInvestment(4

1J

Should not exceed 0.10. How to maximize the rupee return (based on current price,annual growth rate and dividends per share) at the end of 1 year?

[1 – no. of shares of A etc.

Maximize z = (80 X 0.08 + 4) 1 + (100 X 0.07 + 4.50) 2 + (160 X 0.10 + 7.50) 3

+ (120 X 0.12 + 5.50) 4,

s.t.: 80 1 + 100 2 + 160 3 + 120 4 25,00,000,

80 1 + 100 2 5,00,000,

1 100, 2 100, 3 > 0, 4 > 0].

Current price per share (Rs.)Projected annual growth rate per yearProjected annual dividend per share (Rs.)

800.084.00

1000.074.50

1600.107.50

1200.125.50

Risk in turn 0.05 0.03 0.10 0.20

x x x x

x x x x

x x x x

x x x x

x

x x x x

x

x x xx

x x x x

x x

x x x x

≤

≤

≥

≥

∑=

≤

≤

≥ ≥



NOTES(Advertisement Model)

(Warehousing Model)

S

13. A firm wants to expend Rs. 10 lakhs for advertising oneof its products on radio, TV, and a newspaper. It can radio time forRs.2000 per spot, TV time for Rs.10,000 per spot and n paper advertisementfor Rs. 4000 per insertion. The payoff from the advert ement is a measure of theaudience reached. Based on past experience a radio spot is given 40 audiencepoints, a TV spot 160 points and a newspaper insertion 300 points. The firmwishes to determine the money to allocate to each medi so that the number oftotal audience points is maximized. Based on subjective consideration, the firmdecides not to spend more than Rs.2.5 lakhs on radio and at least Rs.4 lakhs onTV. Also, the firm wants to keep newspaper allocation exceed 50% ofallocation to TV. Formulate the problem as an LP.

[ 1 – radio, 2 – TV, 3 – newspaper (in lakhs of Rs.).

Maximize: z = (40/200) + (160/10000) + (300/4000) 3,

s.t.: 1 + 2 + 3 = 10,

1 2.5, 2 4,

3 – (1/2) 2 0,

1, 2, 3 0].

14. A warehouse has a fixed capacity of a commodity. Themanager buys and sells the stock at the intervals of periods (say weeks).Assuming that in the period constant unit and are and and unit is ; find how the manager should operate, assumingthat the warehouse is empty at the end.[ – level of stock at the beginning of period, , – number of units purchased

and sold. Maximize: z = ( – – ), s.t.: +1

= + – , ( = 1,2, …, -1),=1

, ( = 1, 2, …, ), + – = 0].

x x x

x1 x2 x

x x x

x x

x x

x x x

Cn n

jth selling purchasing prices pj

qj

holding price rj

xj

jth yj

zj

np

jz

jq

jy

jr

jx

jx

jx

jy

jz

jj n

j

xj

C j n xn yn zn

≤ ≥

≤

≥

≤

DMC 1752


NOTESCHAPTER 3

GRAPHICAL SOLUTION

3.1 GRAPHICAL APPROACH FOR GENERATING OPTIMAL SOLUTIONS TO A LP PROBLEM.

Step I

Step II

Step III

It is important to understand the Graphical Method of lution in Linear Programmingand find out as to how the graphical method of solution be used to generate optimal solutionto a Linear Programming problem.

Once the Linear programming model has been formulated the basis of the givenobjective and the associated constraint functions, the next step is to solve the problem andobtain the best possible or the optimal solution vario s mathematical and analytical techniquescan be employed for solving the Linear-programming mod .

The graphic solution procedure is one of the methods of solving two variable linearprogramming problems. It consists of the following steps:-

Defining the problem. Formulate the problem mathematically. Express it in terms ofseveral mathematical constraints and an objective func on. The objective function relatesto the optimization aspects such as maximization or mi imization criterion.

Plot the constraints graphically. Each inequality in the constraint equation has to betreated as an equation. An arbitrary value is assigned to one variable and the value of theother variable is obtained by solving the equation. In the similar manner, a different arbitraryvalue is again assigned to the variable and the corresponding value of other variable iseasily obtained.

These 2 sets of values are now plotted on a graph and ected by a straight line.The same procedure has to be repeated for all the constraints. Hence, the total straightlines would be equal to the total no.of equations, eac straight line representing one constraintequation.

Locate the solution space. Solution space or the feasi region is the graphical areawhich satisfies all the constraints at the same time. h a solution point alwaysoccurs at the corner. Points of the feasible region is determined as follows:

(x, y)



NOTES

Step IV

Important terms used in linear programming.Important Theorems

a. For “greater than” and “greater than or equal to” constraints (i.e.), the feasibleregion or the solution space is the area that lies abo the constraint lines.

b. For “Less than” and “Less than or equal to” constraint (i.e. ). The feasible regionor the solution space is the area that lies below the onstraint lines.

Selecting the graphic technique. Select the appropriate graphic technique to be usedfor generating the solution. Two techniques via: Corner Point Method and Iso-profit (orIso-cost) method may be used we give below both the te niques, however, it is easier togenerate solution by using the corner point method.

a. Corner Point Method.

i) Since the solution point (x,y) always occurs at the corner point of thefeasible or solution space, identify each of the ext e points or cornerpoints of the feasible region by the method of simulta eous equations.

ii) By putting the value of the corner point’s co-ordinates [e.g. (2, 3)] into theobjective function, calculate the profit (other cost) each of the cornerpoints.

iii) In a maximization problem, the optimal solution occurs at that corner pointwhich gives the highest profit.

iv) In a minimization problem, the optimal solution occurs at that corner pointwhich gives the lowest profit.

b. Iso-Profit (or Iso-Cost) method. The term Iso-profit signifies if is that anycombination of points produces the same profit as any other combination on thesame line. The various steps involved in this method a given below.

c. Selecting a specific figure of profit or cost, an iso-profit or iso-cost line is drawnup so that it lies within the shaded area.

d. This line is moved parallel to itself and farther or c ser with respect to the origin tillthat point after which any further movement would lead to this line falling totallyout of the feasible region.

e. The optimal solution lies at the point on the feasible region which is touched by thehighest possible iso-profit or the lowest possible iso-cost line.

f. The co-ordinates of the optimal point (x,y) are calculated with the help ofsimultaneous equations and the optimal profit or cost as curtained.

While obtaining the optimum feasible solution to the l near programming problem, thestatement of the following four important theorems are used:-

DMC 1752


NOTES Theorems I

Theorems II

Theorem III

Theorem IV

Important Terms

i. Solution

ii. Feasible Solution

iii. Basic Solution

iv. Basic Feasible Solution

The feasible solution space constitutes a convex set.

Within the feasible solution space, feasible solutions correspond to the extreme (orCorner) points of the feasible solution space.

There are a finite number of basic feasible solutions ith the feasible solution space.

The optimum feasible solution, if it exists. It will o ur at one, or more, of the extremepoints that are basic feasible solutions.

Note. Convex set is a polygon “Convex” implies that if any two points of the polygon areselected arbitrarily then straight line segment joinin these two points lies completely withinthe polygon. The extreme points of the convex set are basic solution to the linearprogramming problem.

Some of the important terms commonly used in linear programming are disclosed asfollows:

Values of the decision variable = 1, 2, 3, in) satisfying the constraints of a generallinear programming model is known as the solution to that linear programming model.

Out of the total available solution a solution that also satisfies the non-negativityrestrictions of the linear programming problem is call a feasible solution.

For a set of simultaneous equations in Q unknowns, a olution obtained by setting (P- Q) of the variables equal to zero and solving the remaining P equation in P unknowns isknown as a basic solution.

The variables which take zero values at any solution are detained as non-basic variablesand remaining are known as basic variables, often called basic.

A feasible solution to a general linear programming pr blem which is also basic solutionis called a basic feasible solution.

x ;( i



NOTESv. Optimal Feasible Solution

vi. Degenerate Solution

3.2 EXAMPLES

3.2.1 Example:1

Solution

Step I

Any basic feasible solution which optimizes (ie; maxim e or minimizes) the objectivefunction of a linear programming modes known as the op feasible solution to thatlinear programming model.

A basic solution to the system of equations is termed s degenerate if one or more ofthe basic variables become equal to zero.

X Ltd wishes to purchase a maximum of 3600 units of a product two types ofproduct ‘a’ and ‘b’ are available in the market. Product ‘a’ occupies a space of 3 cubicfeet and cost Rs. 9 whereas occupies a space of 1 cubic feet and cost Rs. 13 per unit. Thebudgetary constraints of the company do not allow to s more than Rs. 39,000. Thetotal availability of space in the company goes down is 6000 cubic feet. Profit margin ofboth the product and its Rs. is 3 and 4 respectively. ormulate as a linear programmingmodel and solve using graphical method. You are required to ascertain the best possiblecombination of purchase of a profit and so that the total profits are maximized.

Let x1

= no. of units of product a and

x2

= no. of units of product b

Then the problem can be formulated as a P model as follows:-Objective function,Maximize Z = 3x

1 + 4x2

Constraint equations: x1 + x2 3600(Maximum units constraints)

3x1 + x

2 6000 (Storage area constraint)

9x1 + 13x2 39000 (Budgetary constraint)

x1 , x

2 0

Treating all the constraints as equality, the first constraint is

x1 + x

2 = 3600

Put x1= 0 => x

2 =3600 and the point is (0, 3600)

≤

≤

≤

≥

DMC 1752


NOTES

Step II

Step III

Step IV

Step V

Step VI

Step VII

Step VIII

Finding Optimal Solution

Put x2= 0 => x

1 = 3600 and the point is (3600, 0)

Draw the graph with x1 on x-axis and x

2 on y-axis as shown in the figure.

Determine the set of the points which satisfy the cons aint:

x1 + x

2 = 3600

This can easily be done by verifying whether the origi (0, 0) satisfies the constraint. Here,0+0 = 3600. Hence all the points below the line will satisfy the constraint.

The 2nd

constraint is: 3x1 + x

2 6000

Put x1 =0 => x

2 = 6000 and the point is (0, 6000)

Put x 2 =0 => x

1 = 2000 and the point is (2000, 0)

Now draw its graph.

Similar to step II, determine the set of points which atisfy the constraint

3x1 + x

2 6000. At origin 0 +0 < 6000 Hence, all the points below the line will satisfy the

constraint.

The 3rd constraint is: 9x

1 + 13 x

2 39000

Put x1 = 0 => x

2 = 3000 and the point is (0, 3000) Put x

2 = 0 => x

1 =13000/3 and the point

is (13000/3, 0) Now draw its graph.

Again the point (0, 0) i.e. the origin satisfies the constraint 9x1 + 12x

2 39000. Hence,

all the points below the line will satisfy the constraint.

The intersection of the above graphic denotes the feasible region for the given problem.

Always keep in mind two things:

i. For constraint the feasible region will be the area, whic lies above the constraintlines, and for constraints, it will lie below the constraint lines. would beuseful in identifying the feasible region.

ii. According to a theorem on linear programming, an optimal solution to a problem(if it exists) is found at a corner point of the solut space.

≤

≤

≤

≤

≥≤



NOTES

Step IX

Corner Point W-Ordinates Objective Func Value

At corner points (O, A, B, C), find the profit value f the objective function thatpoint, which maximizes the profit is the optimal point.

For point B, solve the equation 9x1 + 13x 2

39000

And 3x1 + x

2 6000 to find point B

ie, 3x1 + x

2 = 6000 …(1)

9x1 + 13x

2 = 39000 … (2)

Multiply equ. (1) by 3 on both sides:

9x1 + 13x

2 = 39000 …(3)

9x1 + 3x

2 = 18000 …(4)

————————

-10x2 = -21000

x2 = 2100

Z=3x1+ 4x2

O (0,0) Z=0+0 0A (0,3000) Z=0+4x3000 12,000C (2000,0) Z=0+3x2000+0 6,000

(0, 0) 20003600 3900/9

6000

3000

3600

9X1+13X2=39,000

X1

B

3X1+X2=6000

X1+X2=3900

X2

O C

A

No.

of

units

≤

≤

∴

ð

ð

DMC 1752


NOTES

Result

3.2.2 Example:2

Solution

Put the Value of x2 in first equation:

=> x1 = 1300

At point (1300,2100)

Z = 3x1 + 4x

2

= 3x1300+ 4x2100

= 12,300 which is the maximum value.

The optimal solution is:

No of units of product a = 1300

No of units of product ß = 2100

Total profit = 12300 which is the maximum.

Let’s consider some more examples.

Great Well Ltd. produces and sells two different types of products P1 and P2 at aprofit margin of Rs. 4 and Rs. 3 respectively. The ava ability of raw materials the maximumno of production hours available and the limiting fact r of P2 can be expressed in terms ofthe following in equations:

4x1+ 2x2 10

2 x1+ 8 / 3 x2 8

x2 6

x1, x2 0

Formulate and solve the LP problem by using graphical hod so as to optimizeboth P

1 and P2

Objective:

Maximize Z = 4x 1 + 3x2

Since the origin (0, 0) satisfies each and every constraint, all points below the line willsatisfy the corresponding constraints.

≤

≤

≤

≥



NOTESConsider constraints as equations and plot then as under:

The area under the curve OABC is the solution space. The constraint x1=6 is not

considered since it does not contain the variable x2 .

Getting optimal solution.

Point B is the intersection of the curves 4x1 + 2x2= 10 and 2x

1 +8/3 x

2 = 8

Solving as system of simultaneous equation:

Point B is ( 8/5, 9/5)

Z = 4x1 + 3x

2

= 4(8/5)+ 3(9/5)= 59/5

4x1+2x2=10

2x1+8/3x2=8

5

A

X2

X1=6

X16C 5/2O(0, 0)

B

3

4

4x1+2x2 =10

put x1 = 0 =>x=5 and the point is (0,5) …(1)

Put x2 = 0 =>x=5/2 and the point is ( 5/2, 0)

2x1+8/3x2 =8 …(2)

put x1 = 0 => x2 =3 and the point is ( 0, 3)

put x2 = 0 =>x1 =4 and the point is ( 4, 0)

x1 = 6 …(3)

DMC 1752


NOTES Result

Remarks

3.2.3 Example:3

Solution

Step 1

Mathematical formulation of the problem

Nutrient Type Nutrient Constituents Minimum nutrient

Decision Variables

Number of Product

Type of Nutrient Constituent

Cost of Product

The optimal solution is:No. of units of P1= 8/5

No. of units of P2 = 9/5

Total profit = 59/5

. Since 8/5 and 9/5 units of a product cannot be produ hence these valuesmust be rounded off. This is the limitation of the lin r programming technique.

Fresh Products Ltd. is engaged in the business of breading cow’s quits farm. Since itis necessary to ensure a particular level of nutrients in their diet, Fresh Product Ltd. Buystwo products P1 and P2 the details of nutrient constituents in each of which are as follows:

The cost prices of both P1 and P2 are Rs. 20 per unit and Rs. 40 per unit respectively.

Formulate as a linear programming model and solve graphically to ascertain howmuch of the products P

1 and P2 must be purchased so as to provide the cow’s nutrients not

less then the minimum required?

Let x1 and x

2 be the number of units of product P

1 and P

2. The objective is to determine

the value of these decision variables which yields the minimum of total cost subject toconstraints. The data of the given problem can be summarized as below:

The above problem can be formulated and follows: Minimize Z =20x1 + 40x

2 subject

to the constraints:

in the product requirementsP1 P2

A 36 6 108B 3 12 36C 20 10 100

A B Cx1 1 36 3 20 20x2 2 6 12 10 40

Min. Requirement

108 36 100



NOTESStep 2

Step 3

Step 4

Corner points Co-ordinates of Objective function Value

Solution Space (Unbounded)

Graph the Constraints Inequalities. Next, we construct the graph by drawing horizontaland vertical axes. x

1 axes in the Cartesian X1 O X

2 plane. Since any point satisfying the

conditions x1 0 and x2

0 lies in the first quadrant only, search for or the desired pair(x

1, x

2) is restricted to the points of the first quadrant only.

The constraints of the given problem are plotted as de ribed earlier by treating themas equations: 36x

1 + 6x

2 = 108;3x 1+ 12x

2 = 36; and 20x1 + 10x

2 = 100

Since each of them happened to be ‘greater than or equ to type’, the points ( x1,x2)

satisfying them all will lie in the region that falls wards the right of each of these straightlines. The solution space is the intersection of all t se regions in the first quadrant.

Locate the solution point. The solution space is open B, P, Q and C as lowerpoints.

Value of objective function at corner points.

B (0, 18)

X2

18

16

14

12

10

8

6

4

2

0

2 4 6 8 10 12 14

A (0, 3) E (5, 0)

P (10, 0)

D (0, 3)

Q (4, 12)

C (12, 0)

X 1

Solution Space (Unbounded)

Corner Points (x1,x2) Z=20x1+40x2

B (0,18) 20(0) + 40(18) 720P (2,6) 20(2) + 40(6) 280Q (4,2) 20(4) + 40(2) 160C (12,0) 20(12) + 40(0) 240

≥ ≥

DMC 1752


NOTES Step 5

3.3 LIMITATIONS OF THE GRAPHICAL METHOD

3.4 EXERCISE

3.4.1

3.4.2

Raw Materials Product

Optimum value of the objective function. Here, we find that minimum cost of Rs. 160is found at point Q (4, 2), i.e., x

1 = 4 and x2 = 2

Hence the first should purchase 4 units of product P1 and 2 units of product P2 inorder to maintain a minimum cost of Rs. 160.

Once a Linear programming model has been constructed on the basis of the givenconstraints and the objective function, it can easily e solved by using the graphical methodand the optimal solution can be generated.

However, the applicability of the graphical method is ry limited in scope. This is dueto the fact that it is quite simple to identify all the corner points and then tests them foroptimality-in the case of a two-variable problem. As a result, the graphical method cannotbe always employed to solve the real-life practical Linear programming models whichinvolve more than two decision-variables.

A firm manufacturer sells two products P1 and P2 at a profit of Rs. 45 per unit andRs. 80 per unit respectively. The quantities of raw materials required for both P

1 and P2 are

given below

The maximum availability of both R1 and R2 is 400 and 450 units respectively.Formulate as a CP model and solve using the graphical ethod.

Fresh Products Ltd. is engaged in the business of breading cow’s quits farm. Sinceit is necessary to ensure a particular level of nutrients in their diet, Fresh Product Ltd. buystwo products P1 and P

2 the details of nutrient constituents in each of which are as follows:

P1 P2

R1 5 20

R2 10 15



NOTESNutrient Type Nutrient Constituents Minimum nutrient

in the product requirements

The cost prices of both P1 and P2 are Rs. 20 per unit and Rs. 40 per unit respectively.

Formulate as a linear programming model and solve graphically to ascertain, howmuch of the products P

1 and P2 must be purchased so as to provide the cow’s nutrients not

less then the minimum required?

P1 P2

A 36 6 108B 3 12 36C 20 10 100

DMC 1752


NOTESCHAPTER 4

SIM PLEX METHOD

4.1 STANDARD FORM OF LINEAR PROGRAM

minimization costmaximization return

The limitation of the graphical method is tackled by what is known as the simplexmethod. Developed in 1947 by George B-Dantizg, it remains a widely applicable methodfor solving complex LP problems. It can be applied to LP problem which can beexpressed in terms of a linear objective function subj t to a set of Linear Constraints. Assuch, no theoretical restrictions are placed on the nu ber of decision variables or constraintscontained in a linear programming problem.

In the examples and exercises of the preceding section we note that the optimizationof the objective function and the constraints occur in diverse form. For understanding theinherent mathematical nature of a linear program (or L in brief) and means of solving theproblem, it is necessary to devise a Standard Form of problem.

Suppose that in an LP there are decision variables 1, 2, 3……, and theobjective function is z - a linear function of the decision variables. The optimization of z issometimes a problem if it involves and sometimes it is a problem of

problem if is involved. Here the former is taken as standard; butthen require a means to convert a maximization proble to one of the minimization. Supposethat in the optimization: Maximize z, Max (z) = , then z and so –z – or Min (–z)= – . Hence,

Max (z) = = – Min (–z) (1)

Thus if z = c1 1 + c2 2 + ……. + cn n which is a linear function of j with coefficients

cj, ( = 1, 2, 3……….. ),

)}xc({Min}xc{Maxn

1jjj

n

1jjj

Then multiply the coefficients by -1 to convert a maximization problem into aminimization problem. Once the minimization of the new problem is found, then multiply by-1 in order to obtain the original maximum.

Next we form the constraints. If we have a linear equality constraints like

1

n

1jiij bxa

n x x x xn

K K KK

K

x x x xj n

cj

≤ ≥

∑∑==

−=

=∑=



NOTES

slack surplus variables

zero coefficients

c x

Ax = b, x 0.

then it is unlikely to cause much difficulty because such linear equations are common placein linear algebra. We would like to stick to such form as standard form. So, if we encounterinequality constraints like

3

n

jj3

n

1j2jj2 bxaandbxa

We convert them in to equalities by introducing extra ariables 2

0 and 3 0 such that

33

n

1jj3j2

n

1j2jj2 bSxaandb Sxa

S2 and S3 are respectively called in O.R. literature, and . We

note that such variables do not enter the objective fu tion z, or we may say that they enterinto the linear expression for z with . By this procedure we note howeverthat the dimension of the problem, that is to say the of variables is increased.Finally in all the examples, the decision variables by themselves are non-negative, that is tosay 0, ( = 1, 2… ). We retain these positive constraints as such and do not increasethe dimension any further by introducing more surplus ariables.

The standard form of a linear programming problem can herefore be formulated inthe following manner. Let the decision variables in an LP be

1, 2… n with an objective

function z, then determine them so as to

Minimize: z = 1 1 + 2 2 + … +

s.t.: 11 1 + 12 2 + … + 1 = 1

21 1 + 22 2 + … + 2 2 = 2 (3a)

——————————————————————

m1 1 + m2 2 + … + =

1, 2, …, 0.

More compactly we can write

Minimize: z = S =1

s.t.: S = , ( = 1, 2… ) (3b) =1

0, ( = 1, 2, …, )

Or, even more compactly, in the notations of matrix algebra

Minimize: S = T

s.t: (3c)

≥≤ ∑∑=

≥ ≥

=−=+ ∑∑==

≥

≥

≥

≥

S S

xj

j n

x x x

c x c x cnxn

a x a x a nxn b

a x a x a nx n b

a x a x amnxmn bm

x x xn

nc

jx

j

j

naijxj

bi i mj

xj

j n

DMC 1752


NOTES cost coefficientsunit cost vector c decision

technological coefficientstechnological constraints A

inconsistent right hand sidecoefficients right hand side vector b

c Ab

unrestricted

In the above 1, 2, …, are called associated with 1, 2, …, n

which constitute the components of = [ 1, 2, …, ] and = [ 1, 2, …, n] ; the superscript standing for transposition. The coefficients

, ( = 1, 2… and = 1, 2… ) are called the occurring inthe = 1, 2… . These coefficients define the x matrix = matrix of [ ], in other words

)4(

na...aa

............

na...aa

na...aa

A

m2m1m

22221

11211

must be , otherwise there would be more equations than the num r of unknownsand the equations would be . 1, 2, …, are called

, which constitute the components of the = [ 1, 2,

…, ] . Theoretically we assume the elements of and to have any sign, positive ornegative, but if necessary we can assume 0. This is possible because if any element is negative, we can multiply the technological constraint by –1 to obtain – > 0 on theright hand side.

In some models it sometimes happens that a variable > k > 0. In such a case wecan use the shift transformation = lk 0 in the problem. By this, the problem isconverted in to standard form involving the new variab e xk. On a few occasions, particularlyin the theory of LP we come across cases when is , that is, it can be > = or< 0. In such cases we can define 0 by

0xif,x

0xif,0

kx

0xif,0

0xif,x

kx

kk

k

k

kk(5)

Then evidently kx

k xxk and by this linear transformations in the LP we introd ce

two non-negative variables in place of one unrestricted variable When there is a group ofsuch then similar transformation can be used for each, but there will be twice as manyvariables 0 which would enter the problem.

c c cn x x xc c cn

T

x x x T Taij i m j n

i m m naij

m nb b bn

b bbm

T

bk

kth bk

xk lxk xk

xk

x

xk

vector x

=

>

≥

− ≥

>

<−

≥=

−

≤

>+=

+

−−

+=

≥



NOTES4.2 SOLVING LINEAR PROGRAMMING USING SIMPLEX METHOD

Remarks

The simplex method employed in solving LP problem is discussed as under:

The simplex method

i. The graphical method, is capable of solving problems h ing a maximum of twovariables. Hence, this method is used which can solve P problems with any numberof variable or constraints it is geared towards solving optimization problems whichhave constraints of less than or equal to type.

ii. This method utilizes the property of a LP problem of h ving optimal solution onlyat the corner point of the feasible solution space. It systematically generates cornerpoint solutions and evaluates them for optimality. The method stops when anoptimal solution is found. Hence, it is an iterative (repetitive) technique.

iii. If we get more variables and less equation, we can set extra variables equal tozero, to obtain a system of equal variables and equal ations. Such solution iscalled basic solution.

iv. The variables having positive values in a basic feasible solution are called basicvariable while the variables which are set equal to zero, so as to define a cornerpoint are called non-basic variables.

v. Slack variables are the fictitious variables which ind ate how much of a particularresource remains unused in any solution. These variabl s cannot be assigned negativevalues. A zero value indicates that all the resources fully used up in theproduction process.

vi. Cj column denotes the unit contribution margin.

vii. Cj row is simply a statement of the projective function.

viii. Zj row denotes the contribution margin lost if one uni is brought into the solution.Hence, it represents the opportunity cost. (Opportunit cost is the cost of sacrificei.e., the opportunity foregone by selecting a particular course of action out of anumber of different available alternatives).

ix. Cj - Z

j row denotes the Net potential contribution or the Net unit Margin

potential, per unit.

The rules used under simplex method, for solving a linear programming problemare as follows:

a. Select the largest value of Cj - Zj row. The column, under which this value falls, is the pivot-column.

b. Pivot-row selection rule. Find the ratio of quantity to the corresponding pivot-column co-efficient. The pivot-row selected is the var ble having the least ratio.

. Rows having negative or zero coefficients in the piv -column areto be neglected. Also, only the basic variables are examined.

c. The coefficient, which is in both, the pivot-row and the pivot-column is calledthe pivot-element or pivot-no.

DMC 1752


NOTES

4.3 EXAMPLES

4.3.1 Example: 1

Raw material type Price(Rs/lton) Quick Tuff

Raw material types Max available(Kg)

d. Updating Pivot-row. Pivot-row, also called replaced rows, is updated as under.All elements of old-row divided by Pivot-element now, n the basic activities column,write the pivot-column variable in place of the pivot-row variable. i.e. the pivot-row variable is to be replaced by the pivot-column variable.

e. Updating all other rows. Update all other rows by updating the formulae. (Old-row element) - (Corresponding pivot column element * u ted correspondingpivot row element) = ( New element)

f. Updating and - rows. Each ‘C’ is obtained as the sum of the products ofthe column coefficients multiplied by the corresponding coefficient in the Jthcolumn. (i.e.) the Quantity column). It is then subtracted from row values get C - values.

g. This pivoting is be repeated when no positive coefficients exist in the C - row, the optimal solution is known.

Let us consider some examples to test our understanding of the solution algorithmthat has been discussed so far.

Smart Limited manufactures two types of adhesives which are sold under the brandname quick and Tuff. Each product consumes the same raw materials but in varyingproportions. The following table depicts the amount of raw materials along with theirrespective cost.

Quick can be blended @ 1000 kg /hour whereas the blending rate for Tuff is 1250 kg/hour. Their respective selling prices are Rs. 1010 and Rs. 845. You may assume thevariable costs be Rs. 500 per hour of plant production time. The maximum availabilityof raw material is

N 600 350 200A 400 50 100P 400 50 100I 200 550 600

1000 kg 1000 kg

N 1000A 300P 250I 1800

Zj

Cj

Zj

Cj

Ci to

j Z

j

to j

Zi

to



NOTES

Solution

Step I

Step II

Step III

Step IV

Step V

4.3.2 Example: 2

Solution

Formulate as a linear programming model and find out the optimal units of quick andtuff to be produced so as to maximize the profits.

List the objective and constraint equations.

Introduce the slack variables

Arrange in the form of 1st tableau (table)

Find out the profit-margins from given sales price.

Generate solutions

The detailed solutions are as under:

X Ltd. produces two products P1, P2 having profit of Rs. 4 and Rs.3 each P1, P2

require 4 hrs and 2 hrs. Of machining respectively, th total available machining time is 10hours. P1, P2

consume 2 units and 8/3 units of raw material respectively subject to a totalof maximum 8 units. Any no. of P2 can be produced and sold but the no. of P

1 must not be

more than 6.

Formulate as a (LP model and solve by the simplex meth .)

Maximize = 4x1 + 3x2

Subject to:

4x1 + 2x2

10

2x1 +8/3x2 8 x1, x2 0

x1

≤

≤ ≥

≤6

DMC 1752


NOTESCj (Rs.) (Rs.) Basic Act. Qty. 4 x1 3 x2 0 S1 0 S2 0 S3

Up-dating S2 row Up-dating S3 row Up-dating Zj row

Cj (Rs.) Basic Act. Qty. 4 x1 3 x2 0 S1 0 S2 0 S3

I Tableau

Pivot – Column = x1

Pivot – Row =S1 S1 [S1 :10/4 = 5/4, S2:8/2=4, S3 : 6/1=6]

Pivot – Element = 4

Updating S2 row: 10/4 =2.5, 4/4=1, 2/4 =0.5, 1/4 =0.25, 0, 0

II Tableau:

0 0 0

S1

S2

S3

Zj (Rs.) Cj-Zj (Rs.)

10 8 6 0

42104

28/30 0 3

10 0 0 0

01 0 0 0

0 0 1 0 0

8 – (2 × 5/2) =3

2-(2× 1) =0

8/3 – (2× 1/2) = 5/3

0 – (2×1/4) = - 1/2

1 – (2×0) =1

0 – (2× 0) = 0

8 – (2 × 5/2) =7/2

1-(1× 1) =0

0 – (2×1/2) = 1/2

0 – (1× 1/4) = - 1/4

0 – (1× 1/4) = -1/4

0 – (1× 0) = 0

1 – (1×0) = 1

Qty: - 4× 5/2 = 10

X1 = 4

X2 = 4× 1/2=2

S1 = 4× 1/4 =2

S2 = 0

S3 = 0

400

X1

S2

S3

ZjCj-Zj

5/237/210

10040

½5/3-1/221

¼-1/2-1/41-1

01111

00100



NOTES

3 4 0 0 0CJ (Rs.)

Pivot row = S1 x [x1 + 5/2 + ½ =5, S2: 3+5/3 =9/5, S3: 7/2 + ½ = -7

9/5 -> Smallest and

7/2-> this will not be considered (read again pivot row selection rule).

Pivot element 5/3

Up dating pivot row:

Updating x1 row: Updating S3: Updating Zj row:

III Tableau:

Basic Act.Qty. X1 X2 S1 S2 S3

0 X1 8/5 1 0 2/5 -3/10 03 X2 9/5 0 1 -3/10 3/5 04 S3 22/5 0 0 -2/5 3/10 1

ZJ 59/5 4 3 7/10 3/5 0CJ - ZJ 0 0 -7/10 -3/5 0

053

51

,10

3

3521

,1,0,59

353

10x2

1100x

2

10

5

3

5

3x

2

10

10

3

5

3x

2

10

52

43

x21

41

52

x21

41

01x21

21

01x21

21

00x21

010x21

1

59x3

58

x4yQ522

59

x21

27

58

59

x21

25

==−

=−

=

=

−−=

−

=

−−−

=

−

−=

−−−−=

−

=

−−−=

−

=

−−=

−

+=+=

−−=

−

DMC 1752


NOTES

4.4 THE BIG M METHOD

The Big M Method to solve a linear programming problem

basicfeasible solution

all slack

The Big M Method

Steps

There are no +ve values in CJ - ZJ row, optional solution is reached.

Hence X1 = 8/5, X2 =9/5 and max. = 59/5 Ans.

In the previous discussions of the Simplex algorithm method starts with a . In the examples so far, when put into standard LP form, conveniently

have an starting solution. An all slack solution is only a possibility when all of theconstraints in the problem have <= inequalities. Here, we are going to look at methods fordealing with LPs having other constraint types.

Remember that simplex needs a place to start, it must from a basic feasiblesolution then move to another basic feasible solution o improve the objective value.

With these assumptions, One can obtain an initial basi feasible solution /dictionaryby letting all slack variables be basic, all original riables be non basic

Obviously, these assumptions do not hold for every LP.

When a basic feasible solution is not readily apparent, the Big M method or the two-phase simplex method may be used to solve the problem.

If an LP has any > or = constraints, a starting basic feasible solution may not bereadily apparent. The Big M method is a version of the Simplex Algorithm that first finds abasic feasible solution by adding “artificial” variabl to the problem. The objective functionof the original LP must, of course, be modified to ensure that the artificial variables are allequal to 0 at the conclusion of the simplex algorithm.

1. Modify the constraints so that the RHS of each constraint is nonnegative (Thisrequires that each constraint with a negative RHS be m tiplied by -1. Rememberthat if you multiply an inequality by any negative number, the direction of theinequality is reversed!). After modification, identify each constraint as a <, >, or =constraint.

2. Convert each inequality constraint to standard form (I constraint i is a < constraint,we add a slack variable s

i; and if constraint i is a > constraint, we subtract an

excess variable ei).

3. Add an artificial variable ai to the constraints identified as > or = constraints at

the end of Step 1. Also add the sign restriction ai> 0.



NOTES

If all artificial variables are equal to zero in the optimal solution, we have foundthe optimal solution to the original problem.

infeasible

4.5 EXAMPLES

4.5.1 Example 1

4. If the LP is a max problem, add (for each artificial v ble) -Mai to the objective

function where M denote a very large positive number.

5. If the LP is a min problem, add (for each artificial variable) Mai to the objective

function.

6. Solve the transformed problem by the simplex . Since ach artificial variable willbe in the starting basis, all artificial variables must be eliminated from row 0 beforebeginning the simplex. (In choosing the entering variable, remember that M is avery large positive number!).

If any artificial variables are positive in the optimal solution, the original problem is

Minimize z = 4x1 + x

2

Subject to:

3x1 + x

2 = 3

4x1 + 3x

2 >= 6

x1 + 2x

2 <= 4

x1, x

2 >= 0

By introducing a surplus in the second constraint and slack in the third we get the followingLP in standard form:

Minimize z = 4x1 + x2

Subject to:

3x1 + x

2 = 3

4x1 + 3x

2 – S

2 = 6

x1 + 2x

2 + s

3 = 4

x1, x

2, S

2, s

3 >= 0

Neither of the first two constraint equations has a slack variable or other variable thatwe can use to be basic in a feasible starting solution so we must use artificial variables. Ifwe introduce the artificial variables R

1 and R

2 into the first two constraints, respectively,

and MR1 + MR

2 into the objective function, we obtain:

DMC 1752


NOTES

Basic z x1

x2

S2

R1

R2

s3

Solution

Basic Z x1

x2

S2

R1

R2

s3

Solution

Minimize z = 4x1 + x2 + MR

1 + MR

2

Subject to:

3x1 + x

2 + R

1 = 3

4x1 + 3x

2 – S

2 + R = 6

x1 + 2x

2 + s

3 = 4

x1, x

2, S

2, s

3, R

1, R

2 >= 0

Set x1, x

2 and S

2 to zero and use R

1, R

2 and s

3 as the starting basic feasible solution.

In table form, we have:

At this point, we have our starting solution in place t we must adjust our z-row to

reflect the fact that we have introduced the variables R1 and R

2 with non-zero coefficients

(M).

We can see that if we substitute 3 and 6 into the objective function for R1 and R

2,

respectively, that z = 3M + 6M = 9M. In our tableau, wever, z is shown to be equal to0. We can eliminate this inconsistency by substitutin out R

1 and R

2 in the z-row. Because

each artificial variable’s column contains exactly one 1, we can accomplish this by multiplyingeach of the first two constraint rows by M and adding hem both to the current z-row.

New z-row = Old z-row + M*R1-row + M*R

2-row

The tableau now becomes

z 1 -4 -1 0 -M -M 0 0

R1

0 3 1 0 1 0 0 3

R2

0 4 3 -1 0 1 0 6

s3

0 1 2 0 0 0 1 4

z 1 -4+7M -1+4M -M 0 0 0 9M

R1

0 3 1 0 1 0 0 3

R2

0 4 3 -1 0 1 0 6

s3

0 1 2 0 0 0 1 4



NOTES

select R1 as our leaving variable

Two important considerations accompany use of the M method

Basic Z X1

X2

S2

R1

R2

S3

Solution

Now we have the expected form for our starting solution.

Apply the simplex method as before. Since this is a m imization problem we selectthe entering variable with the most objective row coefficient. In this case, that isx

1. Calculating the intercept ratios we get:

R1 – 3/3 = 1

R2 – 6/4 = 1.5

s3

4/1 = 4

So we .

Performing the Gauss-Jordan row operations, we obtain he new tableau:

In this tableau, we can see that x2 will be our next entering variable and R

2 will leave.

We can thus see that the simplex algorithm will quickly remove both R1 and R

2 from the

solution just as we intended when we assigned them the coefficient of M in the objectivefunction. If we continue to apply the simplex algorit m, we will find that the optimal solutionis:

x1 = 2/5

x2 = 9/5

S2 = 1

with z = 17/5

The use of the penalty M may not always force the artificial variable to zero level bythe final iteration. This can occur in the case where the given LP has no feasible solution.If any artificial variable is positive in the final iteration than the LP has no feasible solutionspace.

Theoretically, the application of the M technique requires that M approaches infinitybut to computerize the solution algorithm, M must be f nite while being “sufficiently large.”The pitfall in this case is, however, if M is too large it can lead to substantial round-off erroryielding an incorrect optimal solution. For this reason, most commercial LP solvers do notapply the M-method but use, rather, an artificial variable method called the two-phase

Z 1 0 (1+5M)/3 -M (4-7M)/3 0 0 4+2M

x1

0 1 1/3 0 1/3 0 0 1

R2

0 0 5/3 -1 -4/3 1 0 2

s3

0 0 5/3 0 -1/3 0 1 3

positive

−

DMC 1752


NOTES

4.5.2 Example 2

Minimize

Step1

Step2

Step3

Step4

Step5

Step6

method. For educational purposes, TORA allows the implementation of the M-methodwith a user selected value for M where M is sufficiently large to allow solution of theproblem. The definition of the term “sufficiently large” is dependent upon the problem inquestion and requires some judgment for implementation.

z = 2x1+3x

2

s.t 1/2x1+1/4x

2 4…………………1

x1+3x

2 20………………………2

x1+x

2=10…………………………3

x1,x

2 0

: Make the right hand side of all constraints positive We don’t have any negativeright hand side.

:Identify each constraint which is or =. Constraints 2 and 3 apply the aboveconditions.

: For each<= constraint add a slack variable and for each constraint subtract anexcess variable to make them equalities.

1…… 1/2x1+1/4x

2+s

1 =4

2…… x1+ 3x

2 -e

1=20

:For each>=or = constraint add an artificial variable ai(a

i, s>0),which is to be chosen

in the starting bfs.

2…… x1+3x

2-e

1+a

2=20

3…… x1+x

2+ +a

3=10

: If the LP is a min add +Mai to the objective function. If it is a max add Ma

i to the

objective function. Here M represents a very big number such that in the min problem+Mai is arbitrarily large so that a

i the artificial variable is best to be chosen as zero, ch

we require . Similar reasoning applies in the max prob m.

Min z =2x1+3x

2+Ma

2+Ma

3

: Choose those artificial variables in the starting bfs and proceed to find the OptimalTableau. If in the end, artificial variables are zero find the solution, but if they are notequal to zero then we don’t have a optimal solution. Thus, original LP is infeasible. Afterthese steps we have the LP:

Min Z = 2x1+3x

2+Ma

2+Ma

3

s.t 1/2x1+1/4x

2+s

1=4

≤

≥

≥

≥

−



NOTES x

1+3x

2-e

1+a

2=20

x1+x

2+ a3=10

After all, we have the table:

In the optimal solution, we have: {z,s1, x

2, x

1}={25,1/4,5,5}. Since we don’t have

any of artificial variables a1and a

2in the optimal solution, the solution is feasible. If y of

the artificial variables a 1, a

2were not equal to zero then we would have infeasibility as

described below:

Min z=x1+3x

2

s.t.

1/2x1+1/4x

2 4

x1+3x

2 10

x1+x

2 =10

After going through all the steps described above we e with the optimal tableau:

Entering variable

Basis z x1 x2 s1 e2 a2 a3 RHS

z 1 2M-2 4M-3 0 -M 0 0 30M

s1

Leaving var. a2

a3

0

0

0

1/4

1

1

1/4

3

1

1

0

0

0

-1

0

0

1

0

0

0

1

4

20

10

z

s1

x2

Leaving var. a3

1

0

0

0

(2M-3)/3

5/12

1/3

2/3

0

0

1

0

0

1

0

0

(M-3)/3

1/12

-1/3

1/3

(3-4M)/3

-1/12

1/3

-1/3

0

0

0

1

60+101

7/3

20/3

10/3

z

s1

x2

x1

1

0

0

0

0

0

0

1

0

0

1

0

0

1

0

0

-1/2

-1/8

-1/2

1/2

(1-2M)/2

1/8

-1/2

-1/2

(3-2M)/2

-5/8

-1/2

3/2

25

1/4

5

5

Basis z x1 x2 s1 e2 a2 a3 RHS

z

s1

a2

s2

1

0

0

0

1-2M

¼

-2

1

0

0

0

1

0

1

0

0

-M

0

-1

0

0

0

1

0

3-4M

-1/4

-3

1

30+6M

3/2

6

10

≤

≥

↓

←

←

DMC 1752


NOTES

4.5.3 Example 3

Maximize Z = x1 + 5x

2

In the above example, we have the optimal tableau since reduced costs of all nonbasic variables are non positive. However note that th optimal solution contains the verybig number M, which should not have been the case for min problem, thus we say thatour original LP was infeasible. We also say that from he fact that we have the artificialvariable a2 in the basic variables, which shouldn’t have been the case for a feasible LP.

Subject to:

3x1 + 4x

26

x1 + 3x

2 = 2

Where x1, x

2 0

Solution

Introducing slack and surplus variables

3x1 + 4x

2 + x

3 = 6

x1 + 3x

2 – x

4 = 2

Where: x3 is a slack variable x

4 is a surplus variable.

The surplus variable x4 represents the extra units.

Now if we let x1 and x

2 equal to zero in the initial solution, we will have x

3 = 6 and x

4= –2,

which is not possible because a surplus variable cannot be negative. Therefore, we needartificial variables

Maximize x1 + 5x

2 – MA

1

Subject to:

3x1 + 4x

2 + x

3 = 6

x1 + 3x

2 – x

4 + A

1 = 2

Where: x1

0, x2 0, x

3 0, x

4 0, AA

1 0

≤

≥

≥ ≥ ≥ ≥ ≥

.



NOTES

Table 2

Cj

1 5 0 0

CB Basic variables B x1

x2 x3 x4

Solution values b (= XB)

Zj– C

j 2 / 3 0 0 –5 / 3

Cj

1 5 0 0 M

CB

Basic variables B

x1

x2

x3

x4

A1

Solution values

b (= XB)

Zj

– Cj

–M – 1 –3M – 5 0 M 0

Here, a11

= 3, a12

= 4, a13

= 1, a14

= 0, a15

= 0, b1 = 6 a

21 = 1, a

22 = –3, a

23 = 0, a

24 = –1,

a25

= 1, b2 = 2

Calculating Zj – C

j

First column = 0 * 3 + (–M) * 1 – 1 = –M – 1 Second column = 0 * 4 + (–M) * 3 – 5 =–3M–5 Third column = 0 * 1 + (–M) * 0 – 0 = 0 Fourth column = 0 * 0 + (–M) * (–1)– 0 = M Fifth column = 0 * 0 + (–M) * 1 – (–M) = 0 Choose the smallest negative valuefrom Zj – Cj. Substitute M = 0 Smallest negative value is –5. So second column is the 3element column. Now find out the minimum positive value. Minimum (6 / 4, 2 / 3) = 2 / 3So second row is the element row. Here, the pivot (key) element = 3. Therefore, A1departs and x

2 enters.

Calculating values for table 2

Calculating values for first row

a11

= 3 – 1 * 4 / 3 = 5 /3 ;a12

= 4 – 3 * 4 / 3 = 0; a13

= 1 – 0 * 4 / 3 = 1 ;a14

= 0 – (–1) *4 / 3 = 4 / 3 ;b

1 = 6 – 2 * 4 / 3 = 10 / 3

Calculating values for key row

a21

= 1 / 3; a22

= 3 /3=1; a23 = 0 / 3 = 0; a24

= –1 / 3; b2 = 2 / 3

0 x3 5 / 3 0 1 4 / 3 10 / 3

5 x2 1 / 3 1 0 –1 / 3 2 / 3

0 x3 3 4 1 0 0 6

–M A1 1 3 0 1 1 2

DMC 1752


NOTES Table 3

.

4.6 SIMPLEX – BIG M METHOD

Solution:

negative

Cj

1 5 0 0

CB

Basic variables B x1

X2

x3

x4

Solution values b (= XB)

Zj

– Cj

11 / 4 0 5 / 4 0

Since all the values of Zj – C

j are positive, this is the optimal solution

x1 = 0, x

2 = 3 / 2

Z = 0 + 5 * 3 / 2 =15 / 2

Maximize 3X1 + 4X2

Subject to 2X1 + X

2 <= 600

X1 + X

2 <= 225

5X1 + 4X

2 <= 1000

X1 + 2X

2 >= 150

X1, X2 >= 0

Standard form:

Maximize 3X1 + 4X2

Subject to 2X1 + 3X2 + S1 = 600

X1 + X

2 + S

2 = 225

5X1 + 4X

2 + S

3 = 1000

X1 + 2X

2 - S4 = 150

X1, X2, S1, S2, S3

, S4 >= 0

Not in canonical form because there is no basic variable in the fourth equation.Therefore we add an artificial variable to that equation (R

1) and give it a large

coefficient in the objective function, to penalize it:

Maximize 3X1 + 4X2

Subject to 2X1 + 3 X

2 + S1 = 600

X1 + X

2 + S2 = 225

5X1 + 4X

2 +S

3 = 1000

X1 + 2X

2 - S

4 + R1 = 150

X1 , X

2 , S

1 , S

2 , S

3 , S

4 , R

1 >= 0

0 x4 5 / 4 0 3 / 4 1 5 / 2

5 x2 3 / 4 1 1 / 4 0 3/2



NOTESX

1X

2S

4S

1S

2S

3R

1B

X1

X2

S4

S1

S2

S3

R1

b

X1

X2

S4

S1

S2

S3

R1

b

Not in Canonical form because of +M entry on Z row for one basic variable (R1).

Pivot to replace +M on Z row by zero - Z row – M*R1 row:

Z -3 -4 0 0 0 0 +M

S1

2 3 0 1 0 0 0 600

S2

1 1 0 0 1 0 0 225

S3

5 4 0 0 0 1 0 1000

R1

1 2 -1 0 0 0 1 150

Z (-3-M) (-4-2M) M 0 0 0 0 -150M

S1

2 3 0 1 0 0 0 600

S2

1 1 0 0 1 0 0 225

S3

5 4 0 0 0 1 0 1000

R1

1 2 -1 0 0 0 1 150

X1

X2

S4

S1

S2

S3

R1

b

Z -1 0 -2 0 0 0 M 800

S1

½ 0 3/2 1 0 0 -3/2 375

S2

½ 0 ½ 0 1 0 -½ 150

S3

3 0 2 0 0 1 -2 700

X2

½ 1 -½ 0 0 0 ½ 75

Z -1/3 0 0 4/3 0 0 M 800 S

41/3 0 1 2/3 0 0 -1 250

S2

1/3 0 0 -1/3 1 0 0 25

S3

7/3 0 0 -4/3 0 1 0 200

X2

2/3 1 0 1/3 0 0 0 200

DMC 1752


NOTES

4.7 TWO PHASE METHOD

Motivation

Duality theory says

Remark

The Two-Phase Method

Artificial variables and auxiliary problem

X1

X2

S4

S1

S2

S3

R1

b

Optimal tableau: Solution: X1 = 75 X2 = 150 Z = 825

Consider the LP

Max cTx

(P) s.t Ax = b

X 0

We have assumed that a feasible basis is always given. But in practice, it is usually not easyto spot a feasible basis

: Optimal solutions to (P) and its dual are solutions to

Ax = b, X 0

ATy c

cTx – bTy = 0

So, is as hard as solving finding feasible solution LP. Two-phase method: an algorithmthat solves (P) in two phases, where

• In Phase 1, we solve an auxiliary LP problem to either get a feasible basis orconclude that (P) is infeasible.

• In Phase 2, we solve (P) starting from the feasible ba found in Phase 1.

: From Phase 1, we see that finding feasible basis is as easy as solving LP.

Consider the LP

Max cTx

(P) s.t Ax = b

X 0

Z 0 0 0 1 1 0 M 825 S

40 0 1 1 -1 0 -1 225

X1

1 0 0 -1 3 0 0 75

S3

0 0 0 1 -7 1 0 25

X2

0 1 0 1 -2 0 0 250

≥

≥

≥

≥



NOTES

4.8 EXAMPLE

4.8.1 Example 1

Assumption: b 0. (This is without loss of generality.) Suppose we relax the equalityconstraints to inequalities, and add slack variables u1, u2. . . um

Max cTx

s.t Ax + u = b

x , u 0

The basis having u1, u

2. . . u

m as basic variables is feasible; it determines the bfs *,

u*) = (0, b). These “slack” variables are called artificial variables. This new LP problem isNOT equivalent to (P). But, we can force all artificia variables to be zero, then the resultingsolution gives a feasible solution to (P).So, we change the objective function.

m

Max S ui

i=1

(A) s.t Ax + u =b

x , u 0

This is called an auxiliary problem.

Given the LP problem

Max (z =) –X1-X3+2X4

s.t X1+2X2+X4=4

(P) -X2+X3-X4 = -1

X1, X2, X3, X4 0

First, we make sure the right hand side is nonnegative.

Max (z =) –X1-X3+2X4

s.t X1+2X2+X4=4

(A) X2-X3+X4 = 1

X1, X2, X3, X4 0

Adding artificial variables u1, u2 gives the auxiliary problem

Max (w =) –u1-u2

s.t X1+2X2+X4 + u1=4

(A) X2-X3+X4 + u2= 1

X1, X2, X3, X4, u1, u2 0

≥

≥

≥

≥

≥

≥

DMC 1752


NOTES

Example (cont’d)

Any feasible solution of (A) has objective value 0.

=> (A) has optimal value 0.

[X*1, x*2, x*3 , x*4 ]T is feasible for (P),

=> [x*1, x*2, x*3, x*4, 0, 0] T is feasible for (A).

=> [x*1, x*2, x*3, x*4, 0, 0] T is optimal for (A) with value 0.

[X*1, x*2, x*3, x*4, u*1, u*2]T is optimal for (A) with value 0

=> u*1 = u*2 = 0

=> [x*1 , x*2 , x*3 , x*4 ]T is feasible for (P).

So (P) has a feasible solution () (A) has optimal value 0.

In general, the auxiliary problem is never unbounded; its optimal value is 0.Usingthe same argument as before, we can prove an LP problem (P) has a feasible solution itsauxiliary problem (A) has an optimal value 0.

The two-phase method constructs and solves the auxiliary problem (A) in the first phase.

• If (A) has optimal value < 0, we conclude that (P) is feasible.

• If (A) has optimal value = 0, we construct a feasible for (P) and solve it inthe second phase.

Max (z) = -X1 - X3 + 2X4

(P) s.t X1 + 2X2 + X4 = 4

X2 – X3 + X4 = 1

X1, X2, X3, X4 0

Max (w) = -X5 – X6

(A) s.t X1 + 2X2 + X4 + X5 = 4

X2 – X3 + X4 + X6 = 1

X1, X2, X3, X4, X5, X6 0

[Let x5 = u1 and x6 = u2.]

We solve the auxiliary problem starting from the obvious feasible basis B = {5, 6}.

The corresponding tableau is

W – X1 – 3X2 + X3 -2X4 = -5

X1 + 2X2 + X4 + X5 = 4

X2 – X3 + X4 + X6 =1

≤

≤

≤

≥

≥



NOTESExample (cont’d)

Example (cont’d)

Note: The w-row is obtained by subtracting x5-row and x6-row from w = “x5 “ x6.

c1 = 1 > 0, so x1 enters. t = min {4/1, -} = 4, so x5 leaves. Pivot on (5, 1) gives the tableau

W= – X2 + X3 – X4 + X5 = -1

X1 + 2X2 + X4 + X5 = 4

X2 – X3 + X4 + X6 =1

c2 = 1 > 0, so x2 enters. t = min {4/2, 1/1} = 1, so x6 leaves. Pivot on (6, 2) gives thetableau

W = X5 + X6 = 0

X1 + 2X3 - X4 + X5 - 2 X6 = 2

X2 – X3 + X4 + X6 =1

This tableau is optimal, and B = {1, 2} is an optimal is.

B = {1, 2} does not contain artificial variable => B = {1, 2} is a feasible basis for (P).

The tableau for (P) corresponding to B = {1, 2} is

Z = - X3 - X4 = -2

X1 + 2X3 - X4 = 2

X2 – X3 + X4 =1

Note: the z-row is obtained by eliminating the basic variables x1 and x2 from

z = “x1 “ x3 + 2x4.

c3 = 1 > 0, so x3 enters. t = ,22

min = 1, so x1 leaves. Pivot on (1, 3) gives the tableau

Z + ½ X1 – 3/2X4 = -1

½ X1 + X3 – ½ X4 = 1

½ X1 + X2 + ½ X4 = 2

c4 = 3/2 > 0, so x4 enters. t = 12

,min = 2, so x2 leaves.

Pivot on (2, 4) gives the tableau

Z + 2X1 + 3X2 = 5

X1 + X2 + X3 = 3

X1 + 2 X2 + X4 = 4

This tableau is optimal. The corresponding optimal solution is x* = [0, 0, 3, 4] T withoptimal value 5.

−

−

DMC 1752


NOTES 4.9 EXERCISES

Convert the following problems to standard form LP:

1. Maximize: z = 6 1 + 4 2 + 10

s.t.: 2 1 + 3 2 + 5 3 50

4 1 + 2 2 + 7 3 60

1 + 21

2 + 31

3 20

1 5, 2 4, 3 3.75

2. Minimize : z = 2 1 + 5 2 + 7 3

s.t.: 1 + 2 2 + 3 = 5

4 1 + 6 2 + 2 3 = 12

2 0, 3 0

And solve the problem.

3. Maximize: z = 1 + 2 2 – 3 3

s.t.: – 3 3 1 – 5 2 15, 2 0, 3 0.

4. Minimize: z = | 1| - 2| 2| + | 3|

s.t.: 1 + 2 – 3 10

1 – 3 2 + 2 3 5

5. Minimize: z = | 1 – 3| + | 2 + 4|

s.t.: 1 + 2 10

2 1 – 2 3

x x x3

x x x

x x x

x x x

x x x

x x x

x x x

x x x

x x

x x x

x x x x

x x x

x x x

x x x

x x

x x

x x

≤

≤

≤

≥ ≥ ≥

≥ ≥

≤ ≤ ≥ ≥

≤

≥

≤

≥



NOTESCHAPTER 5

VARIANTS OF THE SIMPLEX METHOD

5.1 INTRODUCTION

5.2 DUALITY OF LP PROBLEMS

By a of the Simplex Method it means, an algorithm consisting of a sequenceof pivot steps in the primal system using alternative for the selection of the pivot.Historically these variants were developed to take advantage of a situation where aninfeasible basic solution of the primal is available. n other applications there often occursa set of problems differing from one another only in their constant terms and cost factors.In such cases, it is convenient to omit Phase I and to use the optimal basis of one problemas the initial basis for the next.

Several methods have been proposed for varying the Sim x Algorithm to reducethe number of iterations. This is especially needed for problems involving many equationsin order to reduce the computation time. It is also needed for problems involving a largenumber of variables , for the number of iterations in practice appears to roughlyproportional to .

Each LP problem (called as Primal in this context) is ociated with its counterpartknown as dual LP problem. Instead of primal, solving t dual LP problem is sometimeseasier when a) the dual has fewer constraints then primal (time required for solving LPproblems is directly affected by the number of constra nts, i.e., number of iterations necessaryto converge to an optimum solution which in Simplex me od usually ranges from 1.5 to 3times the number of structural constraints in the problem) and b) the dual involvesmaximization of an objective function (it may be possible to avoid artificial variables thatotherwise would be used in a primal minimization problem).

The dual LP problem can be constructed by defining a new decision variable for eachconstraint in the primal problem and a new constraint r each variable in the primal. Thecoefficient of the jth variable in the dual’s objective function is the ith component of theprimal’s requirements vector (right hand side values o the constraints in the primal). Thedual’s requirement vector consists of coefficients of ecision variables in the primal objectivefunction. Coefficients of each constraint in the dual row vectors) are the columnvectors associated with each decision variables in the coefficients matrix of the primalproblem. In other words, the coefficients matrix of the dual is the transpose of the primal’scoefficient matrix. Finally, maximizing the primal pro em is equivalent to minimizing thedual and their respective values will be exactly equal.

variant

nn

DMC 1752


NOTES

Primal Dual

Primal Dual

When a primal constraint is less than equal to, the corresponding variable in the dualis non-negative. An equally constraint in the primal p oblem means that the correspondingdual variable is unrestricted in sign. Obviously, dual s dual I primal. In summary the followingrelationship exists between primal and dual.

It may be noted that, before finding its dual, all the constraints should be transformedto ‘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 si gets reversed.

An example of finding dual problem is illustrated with the following example.

Maximize Z=4x1 + 3x2 Minimize Z’=6000y1 – 2000y2 + 4000y3

Subject to Subject to x1 + 2/3 x2 6000 y1 –y2 + y3 = 4 x1 – x2 2000 2/3 y1 + y2 3

x1 4000 y1 0x1 0 y2 0

x2 0 y3 0

Maximization MinimizationMinimization Maximizationith variable jth constraintjth constraint jth variableXi 0 In equality sign of i th Constraint:

if dual is Maximization if dual is Minimization

ith variable unrestricted jth constraint with = signjth constraint with = sign ith variable unrestrictedRHS of jth constraint Cost coefficient associated with jth

variable in the objective functionCost coefficient associated with ith

variable in the objective functionRHS of ith constraint

≤≥ ≥

≤ ≥≥ ≥

≥ ≥

≥≤≥



NOTES

Primal-Dual relationships

Dual Simplex Method

Selection of exiting variable:

Primal

Dual

It may be noted that second constraint in the primal is transformed to –x1 + x2 -2000

before constructing the dual.

Following points are important to be noted regarding primal-dual relationship:

1. If one problem (either primal or dual) has an optimal easible solution, other problemalso has been optimal feasible solution. The optimal o ive function value issame for both primal and dual.

2. If one problem has no solution (infeasible), the other problem is either infeasible orunbounded.

3. If one problem is unbounded the other problem is infeasible.

Computationally, dual simplex method is same as simple method. However, theirapproaches are different from each other. Simplex method starts with a non optimal butfeasible solution where as dual simplex method starts th an optimal but infeasible solution.Simplex method maintains the feasibility during succes ive iterations whereas dual simplexmethod maintains the optimality. Steps involved in the dual simplex method are:

1. All the constraints (except those with equality (=) si ) are modified to ‘less-than-equal-to’ ( ) sign. Constraints with ‘greater-than-equal-to’ ( ) sign are multipliedby -1 through out so that inequality sign gets reversed. Finally, all these constraintsare transformed to equality (=) sign by introducing required slack variables.

2. Modified problem, as in step one, is expressed in the rm of a simplex tableau. Ifall the cost coefficients are positive (i.e., optimality condition is satisfied), then dualsimplex method is applicable.

3. The basic variable with highest negative value isthe exiting variable. If there are two candidates for variable, anyone isselected. The row of the selected exiting variable is arked as pivotal row.

Maximize Z= 4x1-x2 + 2x3

Subject to 2x1 + x2 +2x3? 6

x1 - 4x2 +2x3 0

5x1-2x2-2x3 ? 0

x1,x2,x3 0

Minimize Z’= 6y1 + 0y2 + 4y1

Subject to 2y1 + y2 + 5y3 4

y1- 4y2- 2y3 -1

2y1 + 2y2 -2y3 2

y1,y2,y3 0

≤

≤ ≥

≤

≥

≥

≥

≥

≥

DMC 1752


NOTES 4. Selection of entering variable:

Ratio = (Cost coefficients/ -1 × Elements of pivotal row)

5. Pivotal operation:

6. Check for optimality:

Cost coefficients, corresponding to all the negativeelements of the pivotal row, are identified. Their rat s are calculated after changingthe sign of the elements of pivotal row, i.e.,

The column corresponding to minimum ratio is identified as the pivotal column andassociated decision variable is the entering variable.

pivotal operation is exactly same as in the case of simplexmethod, considering the pivotal element as the intersection of pivotal row andpivotal column.

If all the basic variables have nonnegative values then theoptimum solution is reached, otherwise, Step 3 to 5 ar repeated until the optimum isreached.



NOTESUNIT II

CHAPTER 1

TRANSPORTATION PROBLEM

1. TRANSPORTATION PROBLEM

.

1.1 TRANSPORTATION MODELS

objective

1.2 MATHEMATICAL FORMULATION

The transportation problem is a special class of the linear programming problem. Itdeals with the situation in which a commodity is trans orted from to The objective is to determine the amount of commodity o be transported from each sourceto each destination so that the total transportation c t is minimum.

Transportation models or problems are primarily concerned with the optimal (bestpossible) way in which a product produced at different factories or plants (called supplyorigins) can be transported to a number of warehouses called demand destinations). The

in a transportation problem is to fully satisfy the destination requirements withinthe operating production capacity constraints at the m um possible cost. Wheneverthere is a physical movement of goods from the point o manufacture to the final consumersthrough a variety of channels of distribution (wholesa rs, retailers, distributors etc.), thereis a need to minimize the cost of transportation so as to increase the profit on sales.Transportation problems arise in all such cases. It aims at providing assistance to the topmanagement in ascertaining how many units of a particu ar product should be transportedfrom each supply origin to each demand destinations to that the total prevailing demand forthe company’s product is satisfied, while at the same me the total transportation costs areminimized.

Consider m-plant locations (origins) as and the n-retail shops(destination) as respectively. Let , be the amountavailable at the plant . Let the amount required at the shop Dj be bj

Let the cost of transporting one unit of soft drink fr ith origin to destination be If be the amount of soft drink to be transported from

origin to destination , then the problem is to determine so as to Minimize

Sources Destinations

O1 , O2 , …., Om

D1 , D2 , ….., Dn ai 0, i= 1,2, ….mith Oi jth 0, j=

1,2,….n.

jth Cij

, i= 1,2, ….m, j=1,2,….n. xij 0ith jth xij

³ ³

≥

DMC 1752


NOTES

THEOREM 1.1

Note.

The transportation table:

Supply

Requirement

∑∑= =

=

=∑=

=

∑ ∑= =

=

∑=

==∑=

m

i

n

jijij cxz

njm

ij

bij

x

m

i

n

jji ba

n

ji

aij

xandj

bm

iij

x

m+n equations mn

1 1

Subject to the constraint

and xij 0, for all i and j.

.,...2,1,1

This LPP is called a

A necessary and sufficient condition for the existence of a feasible solution to thetransportation problem is that

1 1

The set of constraints

11

Represents in non-negative variables. Each variable xij appearsin exactly two constraints, one is associated with the origin and the other is associated withthe destination.

If we are putting in the matrix from, the elements of A are either 0 or 1.

D1 D2 …… Dn O1 c11 c12 ….. c1n a1

O2 c21 c22 ….. …. c2n a2

… … … ….. ….. … .. … .. :

Om cm1 cm2 …. … cmn am

³

Transportation Problem.

Remark.



NOTES

1.2.1 Formulation of Transportation Problem as a Linea Programming Model

Objective-Function

Balanced Transportation Problem

Definition. (Loop). In a transportation table, an ordered set of four or more cells issaid to form a loop if:

I. Any two adjacent cells in the ordered set lie in the same row or in the samecolumn.

II. Any three or more adjacent cells in the ordered set do not lie in the same row or inthe same column.

A Transportation problem is simply a special case of Linear Programming problem.Hence to convert into a LP model, the same basics steps are employed which we havealready described in the previous section/chapters.

Let P denote the plant (factory) where the goods are b ng manufactured and denotethe warehouse (go down) where the finished products are stored by the company beforeshipping to various destinations. Further, Let, xij = quantity (amount of goods) shippedfrom plant P to the warehouse W

j, and C

ij = transportation cost per unit of shipping from

plant P to the Warehouse W j.

The objection function can be represented as:

Minimize

Z =C11X11 + C12X12 + C13X13+ C21X21 + C22X22 + C23X23 + C31X31 + C32X32 + C33X33

Either, xij for all values of i and j (ie; x

11, x12, … all such values are 0) It is further

assumed that: S1 + S2 + S3 = D1 + D2 + D3 i.e. the total supply available at the plantsexactly matches the total demand at the destinations. ence, there is neither excess supplynor excess demand.

Supply (from various sources) is written in the rows, ile a column is an expressionfor the demand of different warehouses. In general, if a transportation problem has m rowsand n column, then the problem is solvable if there are exactly (m + n–1) basic variables.

Supply constraints X11 + X12 + X13 = S1

X21 + X22 + X23 = S2

X31 + X32 + X33 = S3

Demand constraints X11 + X12 + X13 = D1

X11 + X12 + X13 = D1

X11 + X12 + X13 = D1

i

i

DMC 1752


NOTES Unbalanced Transportation Problem

unbalanced

Two situations are possible:-

Example.

Capacity constraints:

Requirement constraints:

A transportation problem is said to be if the supply and demand are notequal.

i. If supply < demand, a dummy supply variable is introduced in the equation tomake it equal to demand.

ii. If demand < supply, a dummy demand variable is introduced in the equation tomake it equal to supply. Formulate the following trans tation problem as a LPmodel:

Formulating as a Linear Programming model, we have:

Minimise

Z =11x11

+ 6x12 + 15x13 + 3x14 + 7x21 + 8x22 + 4x23 + 13x24+ 22x31

+ 17x32 + 8x33 + 31x34

Subject to:

x11 + x12 + x13 + x14 =21

x21 + x22 + x23 + x24 = 23

x31 + x32 + x33 + x34 = 29

Notation used:

xij = no. of units transported from source to destination J

x11

+ x21 + x31 = 6

X12 + x22 + x32 = 20

x13 + x23 + x33 = 22

x14 + x24 + x34 = 25Hence the transportation problem is successfully formulated as a LP model.

Designation W1 W2 W3 W4 (Supply)

Source Capacity 11 06 15 03

P1 X11 X12 X13 X14 2107 08 04 13

P2 X21 X22 X23 X24 2322 17 08 31

P3 X31 X32 X33 X34 29 Requirement

(Demand) 06 20 22 25 73

i



NOTES1.3 TRANSPORATION PROBLEM STEPS INVOLVED IN SOLVING A

TRANSPORTATION PROBLEM

Step I

Step II

Step III

Step IV

Remarks

Step V

Initial Feasible Solutions

a. The North-West Corner Rule (NWC Rule)

From the given problem, express the objective function which must be minimized along with the relevant constraints.

Set up a transportation table, where supply or sources (factories, plants) are denotedas rows and demand or destinations (warehouses, market are denoted by columns.

Find out an initial feasible solution by using one of he three rules.

A solution so obtained is said to be feasible if and o ly if, such a solution has allocationsin the total number of (m + n – 1) cells with independent positions.

When it is not possible to change an individual alloca on in any cell without violatingthe demand and/or supply constraints or changing the relative positions of the allocations,the position is said to be independent.

By ascertaining the opportunity costs, the solution obtained above can be checkedfor optimality. Opportunity cost may be defined as the cost of sacrifice. Here, it denotesthe reduction in cost due to the inclusion of a particular cell in the solution. An optimumsolution is obtained when there is a positive opportunity cost for each of the empty cells.

When the solution so obtained is not found to be optimal, that empty cell whichresults in the largest saving is included and the above steps are again repeated.

In a transportation problem, the initial feasible solu ion can be generated by a numberof methods. Three of the most common methods are discussed below:

As the name itself suggests, under this rule, first of all the upper-left or the north-westcorner of each cell is selected. It is one of the simp t methods which provides an initialfeasible solution to a transportation problem. The ste involved are as follows:

DMC 1752


NOTES Step I

Step II

Step III

Step IV

b. Least Cost Method (LCM)

Step I

Step II

Step III

Step IV

c. Vogel’s Approximation Method (VAM)

Upper-left corner cell of the transportation table is elected. Allocate as many units tothis cell as possible. This will be the least value of demand and supply. This process isrepeated for all the respective rows and columns.

After exhausting the supply for the first row, move down vertically to the Ist cell in2nd row and Ist column. Then repeat the Step I.

After exhausting the demand for the first column, move along horizontally to the nextcell in the IInd column & Ist row. Then repeat Step I.

Where demand is equal to supply for a cell, then further allocation is made in eitherthe next row or the next column cell. This procedure is continued till the total quantity thatis available is fully allocated to the various cells, required.

This is a time-saving method since it drastically reduces the numerous calculationrequired to be done under the northwest corner rule. The following steps are involved inthis method:

Out of all rows and columns in the transportation table, select that cell which has thelowest (minimum) transportation cost.

Where the smallest cost is not unique i.e., there are her cells having the same smallestcost, select any cell which has this smallest cost.

To the cell chosen in the above step, allocate maximum possible units. Eliminate thatrow or the column where either the demand is satisfied or the supply is exhausted.

For the reduced table so obtained, repeat the above st s till total demand and supplyare exhausted.

The Steps involved in this method are given below:-



NOTESStep I

Step II

Remarks

Step III

Step IV

Step V

Remarks

1.3.1 Methods Employed to Find the Optimal Solution to a Transportation Problem

a. Stepping Stone Method

Step I

Step II

Calculate the penalty for all rows and columns. (Penalty is the difference between thesmallest and the next smallest cost)

Select the row or column having maximum penalty. In th row or column, select thecell having the least cost. Allocate maximum possible its (quantity) to this lowest costcell.

In case of a tie, select the row/column having minimum cost. If, still, a tie persists,select the row/column having the maximum possible assignments or you may simply selectany row or column in case of a tie without considering minimum costs.

Reduce the demand or supply by the amount assigned to he cell.

If row supply is zero, eliminate it

If column demand is zero, eliminate it

If both are zero, eliminate both.

Again calculate penalty and repeat the same steps.

At the end, check that the total number of filled cell = M + N– 1. only then the initialsolution would be feasible. If filled cells are < M + – 1, an empty cell is to be filled in aparticular manner.

Basically 2 methods are widely used:-

The main steps involved in this method are as follows.

After determining the initial feasible solution, check that the number of occupied cellsis equal to M + N – 1 (M – rows, N –columns).

Test each of the unoccupied cells for evaluating its cost effectiveness as under:

DMC 1752


NOTES“turning points “stepping stones.”

Step III

Step IV

b. Modified Distribution (MODI) Method

Step I

Step II

Step III

Step IV

i. Select an empty cell. Use the most direct route horizontally or vertically whichpasses through 3 or more occupied cells to trace a closed path. The cells at the

: are known as

ii. Each corner cell of the path so traced is assigned a positive (+) and a negative( ) sign alternately. Start with the unoccupied cell which is to be evaluated andassign this cell a positive (+) sign.

iii. Sum all the unit cost figures in each cell having a po tive sign and then subtract allthe unit cost figures having a negative sign. This gives the net change in the cost.This step has to be repeated till the cost is calculated for all the unoccupied cells.

An optimum solution is reached, when all the net changes so calculated are 0.Wheresuch an optimum solution has not been reached, go to S IV.

The highest negative net cost unoccupied cell is selec ed. Find the maximum units thatcan be given to a cell having negative sign on the clo path relative to the unoccupiedcell. This number has to be added to the unoccupied an all other cells on the closed pathhaving a positive sign. Finally, this number is to be btracted from cells on the path whichhave a minus sign. The process must be repeated till a optimal solution is obtained.

The following are the steps involved in the Modified D tribution (MODI) method oftesting the optimality of a feasible solution in a transportation problem:-

By using any one of the three methods discussed above, obtain an initial feasiblesolution, having M + N – 1 allocation in independent p ition.

Assign an arbitrary value (zero) to one of the variabl s without violating the equations.(Since there are M + N – 1 occupied cells, there will M + N – 1 equation).

For every empty cell, calculate the improvement index .e.; its opportunity cost. Thishas to be calculated by adding the corresponding row and column number and thensubtracting the actual cost of this cell from it. The olution is optimal, if the opportunity costof all the empty cells > 0.

Where the solution is not optimal i.e.; we have empty lls with negative improvementindex (opportunity cost), select the empty cell having the largest value of negative opportunitycost.

−



NOTESStep V

Step VI

Remarks

1.4 EXAMPLES

1.4.1 Transportation Example 1:

Is this a BFS?

For the empty cell selected in Step IV, draw a closed – and assign alternatepositive (+) and negative (-) signs at the empty cell lying on the corner points of the path.The cell being evaluated i.e. as selected in Step IV will have a positive (+) sign.

Repeat this procedure till an optimal solution is achieved.

Both the stepping stone method and MODI differ in approach but provide the sameoptimal solution. i.e.; both give the solution (to the transportation problem) having thelowest shipping costs.

Initial solution: x11

= 5, x12

= 10, x23

= 15, x34

= 5, x35

= 10

Criterion for BFS: m + n - 1 basic variable, where m = 3 and n = 5. Seven basicvariables are required and need to identify 2 more variables as basic: let x

22 = 0, x

33 =0

45djSi:NoteN

1J

M

1i

=== ∑∑==

DMC 1752


NOTES Initial BFS:

Dependent Cells: Possible to trace a closed path through them.

In table below, the dependent cells are marked by × and form a closed path. Thevariables associated with those cells are 12, 15, 22, 23, 33 and 35.

Compute dual variables (ui, v

j) and reduced costs (w

ij):

Is the solution optimal?

If not, how do we improve it?

x x x x x x



NOTESAlgorithm

1.4.2 Example 2 (Using Simplex method)

Solution:

Consider a transport company which has to supply 4 units of paper materials fromeach of the cities Faizabad and Lucknow to three citie The material is to be supplied toDelhi, Ghaziabad and Bhopal with demands of four, one d three units respectively. Costof transportation per unit of supply ( ) is indicated below in the figure. Decide the patternof transportation that minimizes the cost.

Let the amount of material supplied from source to sink be . HereTotal supply = 8 units and total demand = 4+1+3 = 8 units. Since both are equal, theproblem is balanced. The objective function is to mini ze the total cost of transportationfrom all combinations i.e.

Minimize

Minimize

Subject to the constraints as explained below:

(1) The total amount of material supplied from each source city should be equal to 4.

4X3

1Jij

2,1I

i.e. x11 + x12 + x13 = 4 for i = 1

x21

+ x22

+ x23

= 4 for i = 2

Faizabad

Lucknow

Delhi

Ghaziabad

B hopal

C = 312

C = 421

C = 511

C = 122

C = 813

C = 723

cij

i j xij

m =2; n = 3.

f = 5 x11

+ 3 x12

+ 8 x13

+ 4 x21

+ x22

+ 7 x23

=∑∑==

DMC 1752


NOTES (2) The total amount of material received by each destination city should be equal to thecorresponding demand.

j = 1 ,2, 3

i.e. x11

+ x21

= 4 for j = 1

x12

+ x22

= 1 for j = 2

x13

+ x23

= 3 for j = 3

(3) Non – negativity constraints

Xij i = 1, 2; j=1, 2, 3

Thus, the optimization problem has 6 decision variables and 5 constraints. Since theoptimization model consists of equality constraints, Big M method is used to solve. Thesteps are shown below. Since there are five equality constraints, introduce five artificialvariables R

1, R

2, R

3, R

4 and R

5. Thus, the objective function and the constraints can be

expressed as

Minimize

subject to

Modifying the objective function to make the coefficients of the artificial variableequal to zero, the final form objective function is

The solution of the model using simplex method is shown

x11

+ x12

+ x13

+ R1

= 4

x21

+ x22

+ x23

+ R2 = 4

x11

+ x21

+ R3 = 4

x12

+ x22

+ R4 = 1

x13

+ x23

+ R5

= 3



NOTESTable 1

Table 2

Table 3

VariablesBasic

variablesRHS Ratio

VariablesBasic

variablesRHS Ratio

VariablesBasic

variablesRHS Ratio

First iteration

Second iteration

Third iteration

x11 x12 x13 x21 x2 2 x23 R1 R2 R3 R4 R5

x11 x1 2 x1 3 x2 1 x22 x23 R1 R2 R3 R4 R5

x11 x12 x1 3 x21 x2 2 x23 R1 R2 R3 R4 R5

Z-5

+2M

-3

+2M

-8

+ 2M

-4

+2M

-1

+2M

-7

+2M0 0 0 0 0 16M

R 1 1 1 1 0 0 0 1 0 0 0 0 4 -

R 2 0 0 0 1 1 1 0 1 0 0 0 4 4

R 3 1 0 0 1 0 0 0 0 1 0 0 4 -

R 4 0 1 0 0 1 0 0 0 0 1 0 1 1

R 5 0 0 1 0 0 1 0 0 0 0 1 3 -

Z-5

+2M-1

-8

+2M

-4

+2M0

-7

+2M0 0 0

1-

2M0

1+14

M-

R1 1 1 1 0 0 0 1 0 0 0 0 4 -

R2 0 -1 0 1 1 1 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 -

Z-5

+2M

-5

+2M

-8

+2M0 0

-7

+2M0

-4

2M0 -3 0 13+8M -

R1 1 1 1 0 0 0 1 0 0 0 0 4 4

X2 1 0 -1 0 1 0 0 0 1 0 -1 0 3 -

R3 1 0 0 0 0 0 0 -1 1 1 0 1 1

X2 2 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 -

DMC 1752


NOTES Table 4

1.5 EXERCISES

Solution:

Wh1 Wh2 Wh3 Wh4 Production

VariablesBasic variables RHS Ratio

Fourth iteration

Repeating the same procedure, we get the final optimal solution and theoptimum decision variable values as :

1. Consider three factories (F) located in three different cities, producing a particularchemical. The chemical is to be transported to four di ferent warehouses (Wh), from whereit is supplied to the customers. The transportation cost per truck load from each factory toeach warehouse is determined and are given in the table below. Production and demandsare also given in the table below.

Let the amount of chemical to be transported from factory to warehouse be .Total supply = 60+110+150 = 320 and total demand = 65+85+80+70 = 300. Since thetotal demand is less than total supply, add one fictitious ware house, Wh5 with a demandof 20. Thus, here

F1 523 682 458 850 60

F2 420 412 362 729 110

F3 670 558 895 695 150

Demand 65 85 80 70

Z 0 0-8

+2M0 0

-7

+2M0 -1

5-

2M

2-

2M0 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

f = 42x

11 = 2.2430, x

12 = 0.00, x

13 = 1.7570, x

21 =

1.7570, x22

= 1.00, x23

= 1.2430.

i j xij

m =3; n = 5

x11 x1 2 x13 x21 x22 x23 R1 R2 R3 R4 R5



NOTES

1.6 FINDING AN OPTIMAL SOLUTION

Steps

Wh1 Wh2 Wh3 Wh4 Wh5 Production

The objective function is to minimize the total cost of transportation from allcombinations.

After computing an initial basic feasible solution, we must now proceed to determinewhether the solution so obtained is optimal or not. Th next section, we will discuss aboutthe methods used for finding an optimal solution.

Stepping Stone Method:

It is a method for finding the optimum solution of a transportation problem.

1. Determine an initial basic feasible solution using any one of the following:

North West Corner Rule

Matrix Minimum Method

Vogel Approximation Method

2. Make sure that the number of occupied cells is exactly equal to m+n-1, where mis the number of rows and n is the number of columns.

3. Select an unoccupied cell. Beginning at this cell, tra a closed path, starting fromthe selected unoccupied cell until finally returning t that same unoccupied cell.The cells at the turning points are called “Stepping S es” on the path.

4. Assign plus (+) and minus (-) signs alternatively on each corner cell of the closedpath just traced, beginning with the plus sign at unoccupied cell to be evaluated.

5. Add the unit transportation costs associated with each of the cell traced in theclosed path. This will give net change in terms of cost.

6. Repeat steps 3 to 5 until all unoccupied cells are evaluated.

7. Check the sign of each of the net change in the unit transportation costs. If all thenet changes computed are greater than or equal to zero, an optimal solution hasbeen reached. If not, it is possible to improve the current solution and decreasethe total transportation cost, so move to step 8.

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

•

•

•

DMC 1752


NOTES

1.7 EXAMPLES

1.7.1 Example 3

Solution.

Depot

Factory D E F G Capacity

A

B

C

Requirement

DepotFactory D E F G Capacity

A

B

C

Requirement

8. Select the unoccupied cell having the most negative ne cost change and determinethe maximum number of units that can be assigned to th s cell. The smallest valuewith a negative position on the closed path indicates e number of units that canbe shipped to the entering cell. Add this number to the unoccupied cell and to allother cells on the path marked with a plus sign. Subtr t this number from cells onthe closed path marked with a minus sign.For clarity of exposition, consider thefollowing transportation problem.

A company has three factories A, B, and C with production capacity 700, 400, and600 units per week respectively. These units are to be shipped to four depots D, E, F, andG with requirement of 400, 450, 350, and 500 units per week respectively. Thetransportation costs (in Rs.) per unit between factories and depots are given below

The decision problem is to minimize the total transpor ation cost for all factory-depotshipping patterns.

An initial basic feasible solution is obtained by Matr Minimum Method and is shownbelow in table 1.

4 6 8 6 700

3 5 2 5 400

3 9 6 5 600

400 450 350 500 1700

4 8 700

5 5 400

9 6 600

400 450 350 500 1700



NOTESTable 1

Here, m + n - 1 = 6. So the solution is not degenerate.

Initial basic feasible solution

6 X 450 + 6 X 250 + 3 X 50 + 2 X 350 + 3 X 350 + 5 X 250 = 7350

The cell AD (4) is empty so allocate one unit to it. Now draw a closed path from AD. Theresult of allocating one unit along with the necessary adjustments in the adjacent cells isindicated in table 2.

Table2

Please note that the right angle turn in this path is ermitted only at occupied cells andat the original unoccupied cell.The increase in the tr portation cost per unit quantity ofreallocation is 4 – 6 + 5 – 3 =0.

This indicates that every unit allocated to route AD will neither increase nor decreasethe transportation cost. Thus, such a reallocation is necessary.

Choose another unoccupied cell. The cell BE is empty s allocate one unit to it. Nowdraw a closed path from BE as shown below in table 3.

Table 3

The increase in the transportation cost per unit quant of reallocation is +5 – 6 + 6– 5 + 3 – 3 = 0

This indicates that every unit allocated to route BE will neither increase nor decreasethe transportation cost. Thus, such a reallocation is necessary.

Evaluate all such unoccupied cells in this manner by f nding closed paths and calculatingthe net cost change as shown below.

DMC 1752


NOTES

Minimum transportation cost is:

1.8 DEGENERACY

1.9 EXAMPLES

1.9.1 Example 4

Unoccupied cells Increase in cost per unit of reallocation

Remarks

CECFAFBG

DealerFactory

1 2 3 4

Supply

A

B

C

Requirement

Since all the values of unoccupied cells are greater t n or equal to zero, the solutionobtained is optimal.

6 X 450 + 6 X 250 + 3 X 50 + 2 X 350 + 3 X 350 + 5 X 250 = Rs. 7350

If the basic feasible solution of a transportation pro lem with m origins and n destinationshas fewer than m + n – 1 positive xij (occupied cells), the problem is said to be a degener etransportation problem.

Degeneracy can occur at two stages:

At the initial solution

During the testing of the optimal solution

To resolve degeneracy, making use of an artificial quantity ( ),the quantity is assignedto that unoccupied cell, which has the minimum transpo tation cost. For calculation purposes,the value of is assumed to be zero.

The use of is illustrated in the following example.

+9 – 5 + 6 – 6 = 4 Cost Increases+6 – 3 + 3 – 2 = 4 Cost Increases+8 – 6 + 5 – 3 + 3 – 2 = 5 Cost Increases+5 – 5 + 3 – 3 = 0 Neither increase

nor decrease

2 2 2 4 1000

4 6 4 3 700

3 2 1 0 900

900 800 500 400

§

§

d d

d

d



NOTESSolution

Table 1

DealerFactory

1 2 3 4

Supply

A

B

C

Requirement

Dealer SupplyFactory

1 2 3 4

A

B

C

Requirement

An initial basic feasible solution is obtained by Matr x Minimum Method.

Number of basic variables =m+n–1=3+4–1=6. Since number of basic variables isless than 6,it is a degenerate transportation problem.

To resolve degeneracy, making use of an artificial qua ity (d),the quantity d is assignedto that unoccupied cell, which has the minimum transpo ation cost. The quantity d is sosmall that it does not affect the supply and demand co traints.

In the above table, there is a tie in selecting the smallest unoccupied cell. In thissituation, choose any cell arbitrarily.Select the cell C2 as shown in the following table.

2 4 1000

4 4 3 700

3 2 900

900 800 500 400

2 4 1000

4 4 3 700

3 900 + d

900 800 + d 500 400 2600 + d

DMC 1752


NOTES Table 2

Table 3

Unoccupied cells

Increase in cost per unit of reallocation

Remarks

A3A4B1B3B4C1

DealerFactory1 2 3 4

Supply

A

B

C

Requirement

Now, use the stepping stone method to find an optimal olution.Calculating opportunity cost

The cell B1 is having the maximum improvement potential, which is equal to -2. Themaximum amount that can be allocated to B1 is 700 and is will make the current basicvariable corresponding to cell B2 non basic. The impro ed solution is shown in the followingtable.

The optimal solution is

2 X 200 + 2 X 800 + 4 X 700 + 2 X d + 1 X 500 + 0 X 400 = 5300 + 2d.

Notice that is a very small quantity so it can be neglected in the optimal solution. Thus,the net transportation cost is Rs. 5300

+2 – 2 + 2 – 1 = 1 Cost Increases+4 – 2 + 2 – 0 = 4 Cost Increases+4 – 6 + 2 – 2 = –2 Cost Decreases+4 – 6 + 2 – 1 = –1 Cost Decreases+3 – 6 + 2 – 0 = –1 Cost Decreases+3 – 2 + 2 – 2 = 1 Cost Increases

2 4 1000

6 4 3 700

3 900

900 800 500 400 2600

d



NOTES1.10 EXERCISE

Consider the transportation problem having the followi parameter table

A) This problem has 3 special characteristics:

Number of sources= Number of destination

Each supply=1

Each destination=1

Transportation problem with 3 characteristics are of a special type called theassignment problem. Use the integer solution property explain why this type oftransportation problem can be interpreted as assigning sources to destination on a one-to-one basis.

B) How many basic variables are there in every BF solutio How many of these aredegenerate basic variable (=0)?

Destination

1 2 2 2 Supply

1 7 4 1 4

2 4 6 7 2 1

Source 3 8 5 4 6 1

4 6 7 6 3 1

Demand 1 1 1 1

•

•

•

DMC 1752


NOTESCHAPTER 2

VARIANTS OF THE ASSIGNMENT PROBLEM

2.1 ASSIGNMENT PROBLEM

Structure of assignment problem

2.1.1 Formulation of assignment problem

We have already discussed about one of the benchmark problems called transportationproblem and its formulation. The assignment problem is a particular class of transportationlinear programming problems with the supplies and demands equal to integers (often 1).Since all supplies, demands, and bounds on variables a integers, the assignment problemrelies on an interesting property of transportation pr lems that the optimal solution will beentirely integers. in this section, the structure and rmulation of assignment problem arediscussed. Also, traveling salesman problem, which is special type of assignment problem,is described.

As mentioned earlier, assignment problem is a special ype of transportation problemin which

1. Number of supply and demand nodes is 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 arecompleted. The objective might be to minimize the tota time to complete a set of jobs, orto maximize skill ratings, maximize the total satisfac ion of the group or to minimize the costof 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.

Consider labourers to whom tasks are assigned. No labourers can either sit idleor do more than one task. Every pair of person and assigned work has a rating. Thisratingmay be cost, satisfaction, penalty involved or time taken to finish the job. There willbe such combinations of persons and jobs assigned. Thus, he optimization problem isto find such man- job combinations that optimize the sum of ratings among all. Theformulation of this problem as a special case of trans rtation problem can be representedby treating as and the as . The supply available ateach source is 1 and the demand required at each destination is 1.The cost of assigning(transporting) laborer to task is

m n

N2

laborers sources tasks destinations

i j cij



NOTES

2.2 EXAMPLES

2.2.1 Example 1

Table 1

It is necessary to first balance this problem by adding a dummy laborer or taskdepending on whether m<n or m>n, respectively. The cos coefficient for this dummywill be zero.

Let

Thus the above model can be expressed as

Since each task is assigned to exactly one laborer and each laborer is assigned onlyone job, the constraints are

Due to the special structure of the assignment problem the solution can be found outusing a more convenient method called Hungarian method which will be illustrated throughan example below.

:

Consider three jobs to be assigned to three machines. he cost for each combinationis shown in the table below. Determine the minimal job – machine combinations.

MachineJob

1 2 3

1 5 7 9 12 14 10 12 13 15 13 16 1

1 1 1

cij

ai

bj

DMC 1752


NOTES Solution:

Step 1:

Table 2

Step 2:

Table 3

2.2.2 Example 2

Table 4

Create zero elements in the cost matrix (zero assignment) by subtracting the smallestelement in each row (column) from the corresponding ro (column). After this exercise,the resulting cost matrix is obtained by subtracting 5 from row 1, 10 from row 2 and 13from row 3.

Repeating the same with columns, the final cost matrix is

The italicized zero elements represent a feasible solu on. Thus the optimal assignmentis (1,1), (2,3) and (3,2). The total cost is equal to (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 t next example.

Consider four jobs to be assigned to four machines. Th cost for each combination isshown in the table below. Determine the minimal job – achine combinations.

1 2 3

1 0 2 4 2 4 0 2 3 2 0 3

1 2 3

1 2 2 2 4 0 3 2 3

MachineJob

1 2 3 4

1 1 4 6 3 12 8 7 10 9 13 4 5 11 7 14 6 7 8 5 1

1 1 1 1

0 0

0

ai

bj



NOTESSolution:

Step 1:

Step 2:

Table 5

Step 3:

Table 6

Step 4:

Table 7

Create zero elements in the cost matrix by subtracting the smallest element in eachrow from the corresponding row.

Repeating the same with columns, the final cost matri is

Rows 1 and 3 have only one zero element. Both of these are in column 1, whichmeans that both jobs 1 and 3 should be assigned to machine 1. As one machine can beassigned with only one job, a feasible assignment to t zero elements is not possible as inthe previous example.

Draw a minimum number of lines through some of the rows and columns so that allthe zeros are crossed out.

Select the smallest uncrossed element (which is 1 here). Subtract it from everyuncrossed element and also add it to every element at he intersection of the two lines. Thiswill give the following table.

1 2 3 4

1 0 3 5 2 2 1 0 3 2 3 0 1 7 3 4 1 2 3 0

1 2 3 4

1 2 1 1 2 2 0 2 3 0 3 2 4 2 2 0

1 2 3 4

1 0 3 2 2

2 1 0 0 2

3 0 1 4 3

4 1 2 0 0

0 0

0 0

DMC 1752


NOTES

Formulation of Traveling Salesman Problem (TSP) as an signment Problem

Solution Procedure

2.2.3 Example 3:

This gives a feasible assignment (1,1), (2,3), (3,2) and (4,4) with a total cost of1+10+5+5 = 21. If the optimal solution had not been obtained in the last step, then theprocedure of drawing lines has to be repeated until a easible solution is achieved.

A traveling salesman has to visit cities and return to the starting point. He has to st tfrom any one city and visit each city only once. Suppo he starts from the city and thelast city he visited is . Let the cost of travel from city to city be . Then theobjective function is

Subject to the constraints

Solve the problem as an assignment problem using the method used to solve theabove examples. If the solutions thus found out are cyclic in nature, then that is the finalsolution. If it is not cyclic, then select the lowest try in the table (other than zero). Deletethe row and column of this lowest entry and again do t zero assignment in the remainingmatrix. Check whether cyclic assignment is available. f not, include the next higher entry inthe table and the procedure is repeated until a cyclic assignment is obtained.

The procedure is explained through an example below.

Consider a four city TSP for which the cost between the city pairs are as shown in thefigure below. Find the tour of the salesman so that the cost of travel is minimal.

nk

th

m ith

jth

cij



NOTES

Table 8

Solution:

Step 1:

Table 9

Step 2:

The optimal solution after using the Hungarian method is shown below.

The optimal assignment is 1 4, 2 3, 3 2, 4 1 which is not cyclic.

Consider the lowest entry ‘2’ of the cell (2,1). If there is a tie in selecting the lowestentry, then break the tie arbitrarily. Delete the 2

nd row and 1

st column. Do the zero assignment

in the remaining matrix. The resulting table is

1 2 3 4

1 4 9 5

2 6 4 8

3 9 4 9

4 5 8 9

1 2 3 4

1 0 5

2 2 3

3 5 4

4 3 4

→ → → →

∝

∝

∝

∝

µ

µ

µ

µ

0

0

0

0

DMC 1752


NOTES Table 10

2.3 EXERCISE

Time–matrix (Time in hrs.)

Thus the next optimal assignment is 1 4, 2 1, 3 2, 4 3 which is cyclic. Thusthe required tour is 1 4 3 2 1 and the total travel cost is 5 + 9 + 4 + 6 = 24.

1. Study the following problem:-

Solve this assignment problem so as to minimize the time in hours.

2. A job has four men available for work on four separ te jobs. Only one man can work onany one job. The cost of assigning each man to each job is given in the following table. Theobjective is to assign men to jobs such that the total cost of assignment is minimum.

JobPerson1 2 3 4

1 6 12 3 72 13 10 12 83 2 5 15 204 2 7 8 13

JobsPerson 1 2 3 4

A 20 25 22 28B 15 18 23 17C 19 17 21 24D 25 23 24 24

→ → → →→ → → →



NOTES2.4 ASSIGNMENT MODELS-VARIANTS

minimizationtechnique

1. Maximization-Objective Problem

2. Unbalanced Problem

3. Multiple Solutions

4. Traveling Salesman Problem

•

•

Variants of assignment models

Method for solving complex problems

There are Important Variations although the assignment problem is a , being a special type of Linear Programming Problem. ransportation model,

can be applied in the following diverse situations:

The Hungarian model is fully capable of solving an assignment problem where theobjective is to maximize the total pay-offs. However, efore the successful application ofHungarian technique, such type of maximization problem should first to be converted intominimization problems.

Ideally, an assignment problem is easily solvable in the case of a square matrix (i.e.where the total number of rows and columns are equal). However, where the number ofrows and number of columns the assignment problem is not a square matrix. Such anassignment problem is known as an unbalanced assignment problem. In this case, beforeapplying the Hungarian method, the problem is balanced in the room of a square matrix byadding dummy rows or columns as required.

Sometimes, there may be multiple (more than 1) optimal solution to an assignmentproblem each having the same total pay-off. In Such ca s selecting a particular solutionout of equally suitable alternatives, depends upon discretion of the management. The optimalvalue best suited to the organizational requirements is thus selected.

The traveling salesman problem is peculiar to almost a l the manufacturing and marketingcompanies who employ salesmen in their channels of dis ribution to push the final productto the next channel i.e.; the whole-seller, a retailer or the end users. A traveling salesmanmust visit all the locations (cities, towns) in his as igned territory. The traveling plan is to becreated in such a manner that each and every location visited exactly once and thesalesman arrives back at the location from where he started his journey (i.e.back to thestarting or initiation point). The exact distances (also the time and related costs) betweenany two locations is determined before hand and the objective is to make an optimaltraveling plan which represents the least distance to covered in attaining the objective.

DMC 1752


NOTES Remarks

For number of locations to be visited, the total possible in which it can beachieved is (n-1)! Ways. For example, assume the total number of locations to be visitedis 4. Having started his journey, the traveling salesman can choose anyone of the 3destinations in 3 ways (i.e.; one out of three possibl choices). Having visited a particularlocation, he must now visit the remaining 2 locations. This can be done in 2 ways, bychoosing one location out of possible 2.Similarly there will be only one way in which thelast remaining location can be visited. Thus, the enti territory can be traveled in 3 x 2 x 1= 3! ways. The above problem can also be represented mathematically as under: - 1,where salesman travels from city i to city J xij = {0, in all other cases. Where:-

Cij = distance (also cost or time) between the city to = infinite, when =

This implies that a salesman can not travel from one l ation to the same location.

n

i j i j



NOTESUNIT III

CHAPTER 1

INTEGER LINEAR PROGRAMM ING

1.1 INTEGER LINEAR PROGRAMMING

Introduction

1.1.2 Types of Integer Programming

1.1.3 Integer Linear Programming

In linear programming, the design variables considered are supposed to take any realvalue. However in practical problems like minimization of labor needed in a project, itmakes little sense in assigning a value like 5.6 to the number of labourers. In this situation,one natural idea for obtaining an integer solution is ignore the integer constraints and useany of the techniques previously discussed and then round-off the solution to the nearestinteger value. However, there are several fundamental oblems in using this approach:

1. The rounded-off solutions may not be feasible.

2. The objective function value given by the rounded-off lutions (even if some arefeasible) 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 (2 possible round-off values to beconsidered for an variable problem)

When all the variables in an optimization problem are estricted to take only integervalues, it is called an . When the variables are restrictedto take only discrete values, the problem is called a .When only some variable values are restricted to take nteger or discrete, it is called

. When the variables are constrained to takevalues of either zero or one, then the problem is call .

Integer Linear Programming (ILP) is an extension of linear programming, with anadditional restriction that the variables should be in ger valued. The standard form of anILP is of the form,

n

n

all – integer programming problemdiscrete programming problem

mixedinteger or discrete programming problem

zero – one programming problem

X must be integer valued

DMC 1752


D C

B

A

NOTES

1.1.4 Gomory’s Cutting Plane Method for All – Integer Programming

The associated linear program dropping the integer restrictions is called

Thus, LR is less constrained than ILP. If the objectiv function coefficients areinteger, then for minimization, the optimal objective r ILP is greater than or equal to therounded-off value of the optimal objective for LR. For maximization, the optimal objectivefor 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 fo LR is less than or equal to theoptimal objective for ILP and for a maximization ILP, he optimal objective value for LR isgreater than or equal to that of ILP. If LR is infeasible, then ILP is also infeasible. Also, ifLR is optimized by integer variables, then that solution is feasible and optimal for IP. Amost popular method used for solving all-integer and mixed-integer linear programmingproblems is the cutting plane method by Gomory (Gomory 1957).

Consider the following optimization problem.

The graphical solution for the of this problem is shown below.

0X

bAXtosubject

Xcmax T

0x,x

45x9x3

6xx2tosubject

xx3ZMaximize

21

21

21

21

74177

33,754

62 21

4593 21

linearrelaxation

LR.

x1 and x2 are integers

linear relaxation

Z

xx

xx

≥

≤

≥

≤+

≤−

+=

=( )

≤−

≤+

0 1 2 3 4 5 6 70

1

2

3

4

5

6

7

x1

x2



(4,3)

NOTESIt can be seen that the solution is 733x,7

54x 21 and the optimal value of

7417Z . The feasible solutions accounting the integer constraints are shown by red

dots. These points are called integer lattice points. he original feasible region is reducedto a new feasible region by including some additional aints such that an extremepoint of the new feasible region becomes an optimal solution after accounting for the integerconstraints.

The graphical solution for the example previously discussed taking and asintegers are shown below. Two additional constraints (MN and OP) are included so thatthe original feasible region ABCD is reduced to a new easible region AEFGCD. Thus thesolution for this ILP is and the optimal value is.

Gomary proposed a systematic method to develop these a itional constraints known as. Generation of Gomory Constraints:

Let the final table of an LP problem consist of basic variables (original variables) and non basic variables (slack variables) as shown in the le below. The basic variables arerepresented as (i=1,2,…,n) and the non basic variables are represente as yj (j=1,2,…,m).

733,

754

15

7417

==

=

( )

=

=

x1 x2

Gomory constraints

n m

xi

Z

Z

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

O

P

M

N

DMC 1752


NOTES Table 1

Choose any basic variable ix with the highest fractional value. If there is a tie etween two

basic variables, arbitrarily choose any of them as ix . Then from the equation of table,

m

1jjijii ycbx …….. (1)

Express both ib and ijc as an integer value plus a fractional part.

)3........(cc

)2........(bb

ijijij

iii

Where ib , ijc denote the integer part and i , ij denote the fractional part i will be a

strictly positive fraction 10 i and is a non-negative fraction 10 ij .

Substituting equations (2) and (3) in (1), equation (1 can be written asm

1jjijiij

m

1jiji ycbxy ……. (4)

For all the variables ix and jy to be integers, the right hand side of equation (4) should

be an integer.

j

m

1jiji y integer ……(5)

Since ij are non-negative integers and jy are non-negative integers, the term j

m

1jij y

will always be a non-negative number. Thus we have,

VariablesBasis Z

1 2 … 1 2 … …

Z 1 0 0 0 0

1 0 1 0 0 0 1

2 0 0 1 0 0 2

…0 0 0 1 0

…

0 0 0 0 1

i th

x x ix nx y y jy my rb

c1 c2 cj cm bx c11 c12 c1j c1m b

x c21 c22 c2j c2m b

ix c31 c32 c3j c3m ib

nx c41 c42 c4j c4m nb

∑=

−=

α+=

β+=

β α β

( )<β< ( )<α≤

∑∑==

−−=α−β

=α−β ∑=

α ∑=

α



NOTES

1.1.5 General procedure for solving ILP

1y ij

m

1jiji ……. (6)

Hence the constraint can be expressed as

0y j

m

1jiji ……. (7)

By introducing a slack variable (which should also be n integer), the Gomory constraintcan be written as

ij

m

1jiji ys ……. (8)

1. Solve the given problem as an ordinary LP problem negl cting the integer constraints.If the optimum values of the variables are integers it lf, then there is nothing moreto be done.

2. If any of the basic variables has fractional values, i roduce the Gomory constraintsas discussed in the previous section. Insert a new row with the coefficients of thisconstraint to the final table of the ordinary LP probl (Table 1).

3. Solve this by applying the dual simplex method. Since e value of in Table 1, theGomory constraint equation becomes which is a negativ value and thus infeasible.Dual simplex method is used to obtain a new optimal solution that satisfies theGomory constraint.

4. Check whether the new solution is all-integer or not. all values are not integers,then a new Gomory constraint is developed from the new simplex table and thedual simplex method is applied again.

5. This process is continued until an optimal integer solution is obtained or it showsthat the problem has no feasible integer solution.

Thus, the fundamental idea behind cutting planes is to add constraints to a linearprogram until the optimal basic feasible solution takes on integer values. A Gomory cuthave the property that they can be generated for any integer program, but has thedisadvantage that the number of constraints generated n be enormous depending uponthe number of variables.

<β≤

α−β ∑

=

≤α−β ∑=

β−=α−∑=

DMC 1752


NOTESCHAPTER 2

M IXED INTEGER PROGRAMM ING

2.1 MIXED INTEGER PROGRAMMING

Introduction

Mixed Integer Programming

2.2 Generation of Gomory Constraints

In the previous chapter we have discussed the procedur for solving integer linearprogramming in which all the decision variables are re ricted to take only integer values.In a Mixed Integer Programming (MIP) model, some of th variables are real valued andsome are integer valued. When the objective function and constraints for a MIP are alllinear, then it is a Mixed Integer Linear Program (MILP). Although Mixed Integer NonlinearPrograms (MINLP) also exists, in this chapter we will dealing with MILP only.

In mixed integer programming (MIP), only some of the d cision and slack variablesare restricted to take integer values. Gomory cutting ne method is used popularly tosolve MIP and the solution procedure is similar in many aspects to that of all-integerprogramming. The first step is to solve the MIP problem as an ordinary LP problemneglecting the integer restrictions. The procedure ends if the values of the basic variableswhich are constrained to take only integer values happen to be integers in this optimalsolution. If not, a Gomory constraint is developed for the basic variable with integer restrictionand has largest fractional value which is explained below. The rest of the procedure issame as that of all-integer programming.

Let be the basic variable which has integer restricti n and also has largest fractionalvalue in the optimal solution of ordinary LP problem. hen from the equation of table,

m

1jjijii ycbx

As explained in all-integer programming, express as an integer value plus a fractionalpart.

iii bb ……….(1)

ijc Can be expressed as

ijijij ccc

i th

∑=

−=

β+=

−+ +=



NOTES

Case I

Case II

Where

0cifc

0cif0c

0cif0

0cifcc

ijij

ij

ij

ij

ijij

ij

Therefore,

iiij

m

1jijij xbycc

Since ix is restricted to be an integer ib , is also an integer and 10 i , the value of

iii xb can be 0 or <0.

: iii xb 0

For ix to be an integer, ,....2or1orxb iiiiii

Therefore,

ij

m

1jijij ycc

Finally it takes the form

ij

m

1jij yc

: iii xb <0

For to be an integer,

,....3or2or1xb iiiiii

Therefore,

1ycc ij

m

1jijij

<

≥=

<

≥=

−

+

( ) ( )−+β=+∑=

−+

<β<

( )−+β ≥

( )−+β ≥

( ) +β+ββ=−+β

( ) β≥+∑=

−+

β≥∑=

+

( )−+β

( ) β+−β+−β+−=−+β

( ) −β≤+∑=

−+

DMC 1752


NOTES

2.3 EXAMPLE

Example of Cutting plane method for MILP

Step 1

Finally it takes the form

1yc ij

m

1jij

Dividing this inequality by 1i and multiplying with, we have

ij

m

1jij

i

i yc1

Considering both cases I and II, the final form of the inequality becomes (since one of theinequalities should be satisfied)

ij

m

1jij

i

ij

m

1jij yc

1yc

Then, the Gomory constraint after introducing a slack ariable is

ij

m

1jij

i

ij

m

1jiji yc

1ycs

Generate the Gomory constraint for the variables havin integer restrictions. Insertthis constraint as the last row of the final table of problem and solve this using dualsimplex method. MIP techniques are useful for solving ure-binary problems and anycombination of real, integer and binary problems.

Consider the example previously discussed with integer constraint only on.Thestandard form for the problem is

0y,y,x,x

45yx9x3

6yxx2tosubject

xx3ZMaximize

2121

221

121

21

2x should be an integer

: Solve the problem as an ordinary LP problem. The fin tableau is showing theoptimal solutions which is shown below.

−β≤∑=

−

( )−β

β≥−β

β∑

=

−

β≥−β

β+ ∑∑

=

−

=

+

β−=−β

β−− ∑∑

=

−

=

+

≥

=++

=+−

+=



NOTESTable 9

Step 2

VariablesIteration Basis Z

Optimum value of Z is 7

123 and corresponding values of basic variables are

754

733

x1 , 733

724

x 2 and those of non-basic variables are all zero

(i.e. 0yy 21 ). This is not satisfying the constraint of 2x to be an integer..

: Generate Gomory constraint.

From the row corresponding to in the last table, we can write,

122 y212y7

17

24x

Here 212c,7

1c,724b 22212

Since 222 bb as per Eqn (), the value of 3b2

and 73

2 .

Similarly 212121 ccc and

222222 ccc .

Therefore, 71c,0c 2121 since is negative

0c,212c 2222 Since it is positive. The Gomory constraint can be written as

ij

m

1jij

i

ij

m

1jiji yc

1ycs

73y28

3y212s.,e.i 122

Insert this constraint as the last row of the final tableau.

1 2 1 2

Z 1 0 0

7

8

21

5

7

123 --

1 0 1 0

14

6

21

1

7

33 --

3

2 0 0 1

7

1

21

2

7

24 --

== ==

==

−+=

−===

β+= = =β

−+ += −+ +=

−== −+

== −+

β−=−β

β−− ∑∑

=

−

=

+

−=−−

−

x x y yrb

rs

r

cb

x

x

DMC 1752


NOTES Table 10

Step 3

Table 11


: Solve using dual simplex method


1 2 1 2 2

Z 1 0 07

8

21

50

7

123

1 0 1 014

6

21

10

7

33

2 0 0 17

1

21

20

7

24

2 0 0 028

3

21

21

7

3

1 2 1 2 2

Z 1 0 07

8

21

50

7

123

1 0 1 014

6

21

10

7

33

2 0 0 17

1

21

20

7

244

2 0 0 028

3

21

21

7

3

Ratio -- --3

322.5 --

rb

x x y y s

x

x

s

rb

x x y y s

x

x

s

−

− − −

−

− − −



NOTESTable 12


Optimum value of Z from the present tableau is 2

33 as shown above. Corresponding

values of basic variables are 5.4x1 , 3x 2 , 5.4y2 and those of non-basic variables

are all zero (i.e., 0sy 21 ). This solution is satisfying all the constraints.

1 2 1 2 2

Z 1 0 08

70 1

2

33

1 0 1 08

30 1

2

9

2 0 0 14

10 1 3

4

2 0 0 08

91

2

21

2

9

= = =

==

−

−

rb

x x y y s

x

x

y

DMC 1752


NOTESCHAPTER 3

BRANCH AND BOUND TECHNIQUE

3.1 BRANCH AND BOUND

Branch and Bound

branching

bounding

pruning

3.1.1 Effective subdivision

•

•

•

Branch and Bound is a general search method.

Starting by considering the root problem (the original problem with the completefeasible region), the lower-bounding and upper-bounding procedures are appliedto the root problem.

If the bounds match, then an optimal solution has been found and the procedureterminates.

(BB) is a general algorithm for finding optimal solutions of variousoptimization problems, especially in discrete and combinatorial optimization. It consists ofa systematic enumeration of all candidate solutions, where large subsets of fruitlesscandidates are discarded, by using upper and lower estimated bounds of the quantitybeing optimized.

For definiteness, we assume that the goal is to find the minimum value of a function( ) (e.g., the cost of manufacturing a certain product), where ranges over some set of

admissible or candidate solutions.

A branch-and-bound procedure requires two tools. The first one is a splitting procedurethat means, given a set of candidates, returns two or more smaller sets S1, S2, …whoseunion covers . Note that the minimum of ( ) over is min {?1, ?2

,…..}, where each isthe minimum of ( ) within . This step is called , since its recursive applicationdefines a tree structure (the search tree) whose nodes are the subsets of .

Another tool is a procedure that computes upper and lo r bounds for the minimumvalue of ( ) within a given subset . This step is called .

The key idea of the BB algorithm is’ if the lower bound for some tree node (set ofcandidates) is greater than the upper bound for some other node , then A may be safelydiscarded from the search. This step is called , and is usually implemented bymaintaining a global variable that records the minimum upper bound seen among all subregions examined so far. Any node whose lower bound is greater than can be discarded.

The recursion stops when the current candidate set is reduced to a single element orwhen the upper bound for set matches the lower bound. Either way, any element of willbe a minimum of the function within .

The efficiency of the method depends strongly on the node-splitting procedure andon the upper and lower bound estimators. All other things being equal, it is best to choosea splitting method that provides non-overlapping subse .

f x x S

SS f x S v

i

f x Si

S

f x S

A B

mm

SS S

S



NOTES

3.1.2 Applications

Ideally the procedure stops when all nodes of the search tree are either pruned orsolved. At that point, all non-pruned sub-regions will have their upper and lower boundsequal to the global minimum of the function. In practice the procedure is often terminatedafter a given time. At that point, the minimum lower bound and the maximum upper boundamong all non-pruned sections define a range of values that contains the global minimum.Alternatively, within an overriding time constraint, the algorithm may be terminated whensome error criterion such as (max - min) / (min - max) falls below a specified value.

The efficiency of the method depends critically on the effectiveness of the branchingand bounding algorithms used. Bad choices could lead t repeated branching without anypruning until the sub-regions become very small. In th case the method would be reducedto an exhaustive enumeration of the domain which is of n impractically large. There is nouniversal bounding algorithm that works for all problems, and there is little hope that onewill ever be found. Therefore the general paradigm needs to be implemented separatelyfor each application with branching and bounding algorithms, which are specially designedfor it.

Branch and bound methods may be classified according to the bounding methodsand according to the ways of creating/inspecting the s rch tree nodes.

The branch-and-bound design strategy is very similar to backtracking, in that a statespace tree is used to solve a problem. The differences are that the branch-and-boundmethod (1) does not limit us to any particular way of aversing the tree and (2) is used onlyfor optimization problems.

This method naturally lends itself for parallel and distributed implementations, e.g.,the traveling salesman problem article.

This approach is used for a number of NP-hard problems such as

Knapsack Problem (KP)

Integer Programming (IP)

Nonlinear Programming (NP)

Traveling Salesman Problem(TSP)

Quadratic Assignment Problem (QAP)

Maximum Satisfiability Problem (MAX-SAT)

Nearest Neighbor Search(NNS)

False Noise Analysis (FNA)

It may also be a base of various heuristics. For example, one may wish to stopbranching when the gap between the upper and lower bou becomes smaller than acertain threshold. This is used when the solution is “good enough for practical purposes”and can greatly reduce the computations required. This type of solution is particularly

•

•

•

•

•

•

•

•

DMC 1752


NOTES

Branch and Bound

3.1.3 The General Branch and Bound Algorithm

Definitions:

applicable when the cost function used is noisy or is he result of statistical estimates and sois not known precisely but rather only known to lie within a range of values with a specificprobability. An example of its application here is in iology when performing sadistic analysisto evaluate evolutionary relationships between organis where the data sets are oftenimpractically large without heuristics.

For this reason, branch-and-bound techniques are often used in game tree searchalgorithms most notably through the use of alpha-beta uning.

Although a number of algorithms have been proposed for the integer linear programmingproblem, the technique is used in almost all of the software. Thistechnique has proven to be reasonably efficient on pra ical problems, and it has the addedadvantage that it solves continuous linear programs as sub problems, that is, linearprogramming problems without integer restrictions.

Branch and bound is a systematic method for solving op imization problems

B&B is a general optimization technique that applies where the greedy methodand dynamic programming fail.

However, it is much slower. Indeed, it often leads to ponential time complexitiesin the worst case.

On the other hand, if applied carefully, it can lead t algorithms that run reasonablyfast on average.

The general idea of B&B is a BFS-like search for the optimal solution, but not allnodes get expanded (i.e., their children generated). Rather, a carefully selectedcriterion determines which node to expand and when, and another criterion tellsthe algorithm when an optimal solution has been found.

Each solution is assumed to be expressible as an array X [1: n] (as was seen inBacktracking).

A predictor, called an approximate cost function CC, is assumed to have beendefined.

A live node is a node that has not been expanded

A dead node is a node that has been expanded

The expanded node (or E-node for short) is the live no with the best CC value.

The general B&B algorithm is as follows:

Procedure B&B ()

begin

E: node pointer;

branch-and-bound

•

•

•

•

•

•

•

•

•

•



NOTES

3.1.4 Branch and Bound (tree search)

E: = new (node); — this is the root node which

— is the dummy start node

H: heap; — A heap for all the live nodes

— H is a min-heap for minimization problems,

— And a max-heap for maximization problems.

While (true) do

if (E is a final leaf) then

— E is an optimal solution

Print out the path from E to the root;

Return;

end if

Expand (E);

if (H is empty) then

Report that there is no solution;

Return;

End if

E: = delete-top (H);

End while

End

Procedure Expand (E)

begin

- Generate all the children of E;

- Compute the approximate cost value CC of each child;

- Insert each child into the heap H;

End

The most effective general purpose optimal algorithm is LP-based tree search (treesearch also being called branch and bound).

.

Where this method differs from the enumeration method that the feasiblesolutions are enumerated but only a fraction (hopefull a small fraction) of them. However

This is a way of systematically enumeratingfeasible solutions such that the optimal integer solution is found

not all

DMC 1752


NOTES

How can we rid ourselves of this troublesome fractionalvalue?

we can still that we will find the optimal integer solution. The method was firstput forward in the early 1960’s by Land and Doig.

Consider our example capital budgeting problem. What made this problem difficultwas the fact that the variables were restricted to be egers (zero or one). If the variableshad been allowed to be fractional (takes all values between zero and one for example)then we would have had an LP which we could easily solve. Suppose that we were tosolve this LP relaxation of the problem [replace x

j = 0 or 1 j=1,...,4 by 0 <= x

j <= 1

j=1,...,4]. Then using the package we get x2=0.5, x3=1, x1=x4=0 of value 0.65 (i.e. theobjective function value of the optimal linear program ng solution is 0.65).

As a result of this we now know something about the op imal integer solution, namelythat it is <= 0.65, i.e. this value of 0.65 is a (upper) on the optimal integer solution.This is because when we relax the integrality constraint we (as we are maximising) end upwith a solution value at least that of the optimal int r solution (and may be better).

Consider this LP relaxation solution. We have a variable x2 which is fractional when

we need it to be integer. To remove this troublesome fractional value we can ge rate two new problems:

original LP relaxation plus x2=0

original LP relaxation plus x2=1

Then we will claim that the optimal integer solution t the original problem is containedin one of these two new problems. This process of taki a fractional variable (a variablewhich takes a fractional value in the LP relaxation) and explicitly constraining it to each ofits integer values is known as . It can be represented diagrammatically as below(in a tree diagram, which is how the name arises).

Initial LP relaxation value 0.65

X2 fractional

X2 =1X2 =0

P1 P2

guarantee

bound

•

•

branchingtree search



NOTESWe now have two new LP relaxations to solve. If we do his we get:

P1 - original LP relaxation plus x2=0, solution x1=0.5, x3=1, x2=x4=0 of value 0.6

P2 - original LP relaxation plus x2=1, solution x2=1, x3=0.67, x1=x4=0 of value0.63

This can be represented diagrammatically as below.

To find the optimal integer solution we just repeat th process, choosing one of thesetwo problems, choosing one fractional variable and gen rating two new problems to solve.

Choosing problem P1 we branch on x1 to get our list of LP relaxations as:

P3 - original LP relaxation plus x2=0 (P1) plus x

1=0, solution x3=x

4=1, x1=x

2=0 of

value 0.6

P4 - original LP relaxation plus x2=0 (P1) plus x1=1, solution x1=1, x3=0.67,x2=x4=0 of value 0.53


This can again be represented diagrammatically as belo .Initial LP relaxation value 0.65

X2 fractionalX2 = 1

value 0.63

X3 fractional

P2Value 0.6

X1 fractional

X1 =1

X2 =0

P1

X1=0

P3

Value 0.6

integer feasible

Value 0.53

P4


X2 fractional

X2 =1

P2P1 Value 0.63

X3 fractionalX1 fractional

value 0.6

X2 = O

•

•

•

•

•

DMC 1752


NOTES At this stage we have identified an integer feasible solution of value 0.6 at P3. Thereare no fractional variables so no branching is necessary and P3 can be dropped from ourlist of LP relaxations.

Hence we now have new information about our optimal (b st) integer solution, namelythat it lies between 0.6 and 0.65 (inclusive).

Consider P4, it has value 0.53 and has a fractional variable (x3). However if we wereto branch on x

3 any objective function solution values we get after b nching can never be

better (higher) than 0.53. As we already have an integ feasible solution of value 0.6 P4can be dropped from our list of LP relaxations since b nching from it could never find animproved feasible solution. This is known as - using a known feasible solutionto identify that some relaxations are not of any inter t and can be discarded.

Hence we are just left with:


Branching on x3 we get

P5 - original LP relaxation plus x2=1 (P2) plus x

3=0, solution x1=x

2=1, x3=x

4=0 of

value 0.5

P6 - original LP relaxation plus x2=1 (P2) plus x3=1, problem infeasible

Neither of P5 or P6 lead to further branching so we are done, we have discoveredthe optimal integer solution of value 0.6 correspondin to x

3=x

4=1, x

1=x

2=0.

The entire process we have gone through to discover this optimal solution (and toprove that it is optimal) is shown graphically below.


X2 fractional

X2 =1X2 =0

P1Value 0.6

X1 fractional

X1 = 1 X3 = 0

P2 Value 0.63

X3 fractional

X3 = 1

P6

infeasiblevalue 0.5

P5P4

Value 0.53Value 0.6

integer feasible

X1=0

P3

bounding

•

•

•



NOTES

We do not need to examineas many LP’s as there are possible solutions.

3.1.5 Illustration on the Job Assignment Problem

2 4 52 7 105 3 7

You should be clear as to why 0.6 is the optimal integer solution for this problem,simply put if there were a better integer solution the above tree search process would(logically) have found it.

Note here that this method, like complete enumeration, also involves powers of twoas we progress down the (binary) tree. However also note that we did not enumerate allpossible integer solutions (of which there are 16). In d here we solved 7 LP’s. This isan important point, and indeed why tree search works at all.

While the computational efficiency oftree search differs for different problems, it is this basic fact that enables us to solve problemsthat would be completely beyond us were we trying comp te enumeration.

Input: n jobs, n employees, and an n x n matrix A wher Aij be the cost if person iperform job j.

Problem: find a one-to-one matching of the n employees to the n jobs so that thetotal cost is minimized.

formally, find a permutation f such that C(f), where

C (f) =A1f (1)

+ A2f (2)

+ ... + Anf(n)

is minimized.

A brute-force method would generate the whole solution tree, where every pathfrom the root to any leaf is a solution, then evaluate the C of each solution, andfinally choose the path with the minimum cost.

Illustration on this specific instance of the job-assi ment problem:

In informal terms, the problem is to choose a single number from each row suchthat (1) no two numbers are chosen from the same columns, and (2) the sum of thechosen numbers is minimized.

The first idea of B&B is to develop “a predictor” of the likelihood (in a loosesense) of a node in the solution tree that it will lead to an optimal solution. Thispredictor is quantitative.

With such a predictor, the B&B works as follows:

Which node to expand next: B&B chooses the live node with the best predictor valueB&B simply expands that node (i.e., generate all its children)

The predictor value of each newly generated node is computed. The just expandednode is now designated as a dead node and the newly ge ated nodes are designated aslive nodes.

•

•

•

•

•

•

•

•

•

•

DMC 1752


NOTES

Definition of the cost function C:

Theorem:

Proof:

2 4 52 7 105 3 7

2 4 52 7 105 3 7

2 4 52 7 105 3 7

Termination criterion: When the best node chosen for expansion turn out to be a finalleaf (i.e., at level n) when the algorithm terminates nd that node corresponds to the optimalsolution. The proof of optimality will be presented later on.

What could that predictor be?

In the case of minimization problem, one candidate predictor of any node is thecost so far. That is, each node corresponds to (partial) solution (from the root tothat node). The cost-so-far predictor is the cost of t partial solution.

Apply this preliminary algorithm on the above specific instance of the job assignmentproblem

A better predictor for the job assignment problem is:

(Cost so far) + (sum of the minimums of the remaining ws)

Apply B&B with the new predictor to the same instance the job assignmentproblem

A yet another predictor is cost so far + sumni=k+1p i

Where p i is the minimum value in row i of the cost matrix A, such that p is not in the

column of any of the terms chosen in the partial solution so far.

Apply B&B with the last predictor to the same instance of the job assignmentproblem

Criteria for the Choice of Approximate Cost Functions

For every node X in the solution tree, the costfunction C(X) is the cost of the best solution that go through node X.

In the case of minimization problems, if CC(X) <= C(X) for everynode X, and if CC(X)=C(X) for every final leaf node X, then the first expandingnode (best-CC node) that happens to be a final leaf corresponds to an optimalsolution.

Let E be the E-node that happens to be a final leaf.

•

•

•

•

•

•

•



NOTES

Implementation of the B&B Job Assignment Algorithm

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

Need to prove that C (E) <= C(X) for any live node X.

C(E)=CC(E) because E is a final leaf

CC(E) <= CC(X) for any live node X, because E is the e ing node, that is,the minimum-CC node

CC(X) <= C(X) by assumption.

Therefore, C (E) =CC (E) <=CC(X) <=C(X), leading to C <=C(X).

Therefore, the criteria for the approximate cost function CC for minimizationproblems are:

CC(X) <= C(X) for all live nodes X

CC(X) =C(X) for all final leaves.

The criteria for the approximate cost function CC for ximization problems are:

CC(X) >= C(X) for all live nodes X

CC(X) =C(X) for all final leaves.

Because of those criteria, CC is called an underestimate of C (in the case ofminimization), and an overestimate of C (in the case of maximization)

The closer CC is to C, the faster is the algorithm in reaching an optimal solution.

We need to define the full record of a node

We need to fully implement the Expand procedure

Every node corresponds to something like X[i] =j, whic signifies that the job X[i]assigned to person i is j.

Every node must store its CC value.

Each node must point to its parent so that when an opt l leaf is generated, thepath from that leaf to the root can be traced and printed out as the optimal solution.

Therefore, a node record structure should look like:

Record node

Begin

Parent: node pointer;

I: integer; — person I

J: integer; — Job J is assigned to person I

CC: real;

End

DMC 1752


NOTES •

•

•

•

Take the 2nd CC formula:

CC(X at level k) = cost so far + sumni=k+1mi

Where m i is the minimum of row i.

observe that if X is a pointer to a node, then

X->CC = X->Parent->CC + AX->I, A->J

- mX->I

Write a piece of code that computes the m is for i=1,2,...,n

Code for Expand(E):

Procedure Expand (E)

begin

/* generate all the children of E; */

I := E->I;

X,p: nodepointer;

S [1: n]: Boolean;

/* S is a bitmap set initialized to 0*/

/* S will contain all the jobs that have been

Assigned by the partial path from the root

to E */

p := E;

while (p is not the root) do

S[p->J] := 1;

p := p-> Parent;

End while

for job=1 to n do

if S[job] = 0 then

X: = new (node);

X->I := I + 1;

X->J := job;

X->Parent := E;

X->CC := E->CC + AX->I,X->J-mX->I;

Insert(X, H);

End if

End for

End



NOTESUNIT IV

CHAPTER 1

SCHEDULING BY PERT AND CPM

1.1 CRITICAL PATH METHOD (CPM)

Activities

The Critical Path Method (CPM) is one of several related techniques for doing projectplanning. CPM is for projects that are made up of a number of individual “activities.” Ifsome of the activities require other activities to fin h before they can start, then the projectbecomes a complex web of activities.

CPM helps to figure out:

How long the complex project will take to complete

Which activities are “critical,” meaning that they hav to be done on time or elsethe whole project will take longer time.

The information about the cost of each activity and ho much it costs to speed upeach activity, the following points are to be noted:

Whether to speed up the project, and, if so,

What is the least cost way to speed up the project?

An activity is a specific task. It gets something done. An activity can have theseproperties:

Names of any other activities that have to be completed before this one can start

A projected normal time duration

To do a speedup cost analysis, the following points about each activity are to benoted

A cost to complete

A shorter time to complete on a crash basis

The higher cost of completing it on a crash basis

CPM analysis starts after you have figured out all the individual activities in yourproject.

••

•

•

•

•

•

•

•

DMC 1752


NOTES CPM Analysis Steps

1.2 EXAMPLES

1.2.1 Example 1: Activities, precedence, and times

Step 1: List the activities

For example:

Step 2: Draw the diagram

Activity Destination Required Predecessor Duration

The steps for doing CPM analysis will be illustrated b two examples. It is recommendto work through these examples.

This first example involves activities, their precedence (which activities come beforeother activities), and the time the activities take. T objective is to identify the critical pathand figure out how much time the whole project will take.

CPM analysis starts with a table showing each activity for a project. For each activity,identify the other activities must be done before it starts, and how long the activity takes.

Draw a network diagram of the project that shows which activities follow the otherones. This can be tricky. The analysis method requires an “activity-on-arc” (AOA) diagram.An AOA diagram has numbered “nodes” that represent stages of project completion asshown in the given figure. Connect the nodes with arrows or “arcs” that represent theactivities that are listed in the above table.

A Product design (None) 5 monthsB Market research (None) 1C Production analysis A 2D Product model A 3E Sales brochure A 2F Cost analysis C 3G Product testing D 4H Sales training B,E 2I Pricing H 1J Project report F,G, I 1



NOTES

Example 1 Step 3: Set up the CPM spreadsheet

Some conventions about how to draw these diagrams:

All activities with no predecessor come off from node 1.

All activities with no successor points to the last no , which has to have a highestnode number.

In this example, A and B are the two activities that have no predecessor. They arerepresented as arrows leading away from node 1.J is th one activity that has no successor,in this example. It therefore points to the last node, which is node 8. If there were morethan one activity with no successor, all of those activitie’s arrows point to the highestnumber node.

The trickiest part for building the above diagram was guring what to do with activityH. First draw an arrow for activity B coming off node and going to node 3. Then drawan arrow for activity E coming off node 2 and going to node 6. Since H requires both Band E, erase the first E arrow and redraw it so it poi to the same node 3 that B did. Hthen comes off from node 3 and goes to node 6.

There are specialized commercial programs for doing CPM analysis. Rather thanpurchase and learn one of those. It is recommended to se the spreadsheet knowledge.Start up a new blank spreadsheet. In a blank spreadsheet, type the word “Activities” incell A1. In row 2, type the names of the activities, or their letters.

D

1

2

3

4

5

6

7 8

A

B

E

G

F

H

I

J

•

•

DMC 1752


NOTES

Step 4: Use Path find to get the paths

In row 3, type “Nodes”. In row 4, type in each activity’s start node — where the tail of itsarrow is. Below that, in row 5, type each activity’s end node — where the head of itsarrow is. Do this carefully. Mistakes here mess up everything that follows.

To the right, in K4 and K5, type the words “Start” and “End” to label those rows.

In cell A6, type “Times”. In row 7, type the time of each activity takes. Then, select therange of cells containing the node numbers

And copy it to the clipboard.

Path find is a computer program that helps to find and enter into the spreadsheet all ofthe possible paths through the diagram along the arrows from the first node to the last. Thisdiagram shows the four possible paths in this example.



NOTES

Step 5: Paste the path information into the spreadsheet

The four paths are A D G J, A C F J, A E H I J, and B J. To code them in thespreadsheet with a matrix of 0’s and 1’s. Rather than oing it manually, use path find to doit.

To use Path find, start up your Internet connection and your browser.

Loading the html file into the browser starts Path fin , which is a Java applet that runsinside the browser. When Path find is loaded:

Click in Path find’s upper text area.

Paste the range that is copied from the spreadsheet in that upper text area. Clickon Path find’s button. Path find’s lower text area wil give you a block of numbers.

Copy the highlighted numbers to the clipboard for past ng later into the spreadsheet.

Move the cell selector to cell A8. Type “Paths” in tha cell. Then move the cell selectorto A9, as shown here:

§

§

§

DMC 1752


NOTES

Example 1 Step 6: Calculate the paths’ times

Paste to that cell, to see this:

The pasted cells are all 0’s and 1’s. Each row represents a path. The 1’s indicatewhich activities are in that particular path. For example, row 9 (cells A9:J9) has 1’s underactivities A, C, F, and J. This says that this path includes activities A, C, F, and J. Thiscorresponds to the path through the middle of the diagram that goes:

1 A 2 C 7 J 8

The diagram above shows four paths from node 1 to node 8.Path find produces fourrows of 0’s and 1’s, one row for each path.

Move the cell selector to K9. Type in that cell.

− → − → − →

=SUMPRODUCT (A9:J9, $A$7:$J$7)



NOTES

Step 7: Identify the critical path

This formula multiplies each entry in row 9 by the corresponding entry in row 7.Because the entries in row 9 are all 0’s and 1’s, this has the effect of selecting the timesfrom row 7 that go with the activities represented in ow 7, and adding all those times.

When entering the formula, the number 11 should appear in K9. That’s the time itwould take to complete activities A, C, F, and J. (A takes 5 months, C takes 2, F takes 3,and J takes 1, for a total of 11).

To fill in the other paths’ times, copy cell K9, then ste it to K10:K12. The $ signsin the formula see to it that each path’s 1’s are mult plied by the corresponding numbers inrow 7.

The critical path is the path that takes the longest. this example, the critical path isthe one in row 10, which takes 13 months. The project therefore take 13 months, ifeverything is done on schedule with no delays. The tim a project takes is equal to the timeof its critical path.

The 1’s in row 10 tell that the critical path is 1 A 2 D 4 G 7 J 8. If any of

those activities is late, the project will be late. Ot r paths are not critical because they canwaste some time without slowing the project. For example, activity C, in row 9’s path, cantake up to two extra weeks and not hold up the project.

To make it easier to see what activities are in each path, go to cell A14. Type there. The letter A should appear in cell A14.

This =if (A9=1, a$2,”” function works this way: Inside the parentheses are threeexpressions separated by commas. The first expression =1) is something that can beeither true or false. If the expression is true, the s nd expression (A$2) is shown in thecell. Otherwise, the third expression (“”) is shown in the cell.

− → − → − → − →

=if(A9=1,A$2,””)

)

DMC 1752


NOTES In A14, the expression A9=1 is true, so the cell shows what is in A2, which is theletter “A”. If A9 had not contained a 1, then A14 woul have shown a blank. Copy A14 tothe clipboard. Then, starting in A14, select a range of cells that goes over to column J anddown four rows. The selected range should be the same e as the space that the paths’1’s and 0’s take up.

It shows which activities are in each path. If the results do not look like the above,make sure that there is one $ in the formula, and that it’s in front of the 2 and not in front ofthe A.

Go to cell J13 and type “Max”. Then go to cell K13. Type todisplay the longest path time.

=MAX (K9:K12)



NOTESMove to cell K14 and type =IF (K9=K$13,”Critical”,””) here.

This will put the word “Critical” next to a path whose time equals the maximum of allthe path times. Otherwise, it will put in a blank, as does here, because the 11 in K9 doesnot equal the 13 in K13.

Copy K14 to the clipboard. (It will seem strange to copy what appears to be anempty cell, but do it anyway.) Select cells K14 to K17, and paste.

DMC 1752


NOTES 4.2.2 Example 2: Costs and Crash Costs

Step 1: List the activities

Step 2: Draw the diagram

Activity Required Predecessor

Normal Time

Normal Cost

Crash Time

Crash Cost

This second example incorporates costs and the possibility of spending money tospeed up the project. The main objective is to determi how quickly to complete eachactivity, and thus how long the project as a whole should take. The presumption is thatthere is some reward for getting finished sooner. In o er to decide whether the reward isworth earning, and if so what is the best way to earn it. This example also shows how touse a dummy activity. A dummy activity is an activity at you add to the original activitieslist. A dummy activity takes no time, and it has no cost.

To start the network diagram, A, B, and C are the thre activities with no predecessor.They all come off from node 1. A can go from node 1 to node 2. B can go from node 1 tonode 3. C also starts at node 1. D requires A, so D starts at node 2.

Now for activity E. Activity E requires a special tric The problem is where E shouldstart. E requires A and B. A ends at node 2 and B ends at node 3. E is not allowed to startfrom both nodes 2 and 3. Activities can have only one tart node and only one end node.

What do we do about E? First choice is to consider connecting both A and B to node2, but that would mess up Activity D. If both A and B ere to run from node 1 to node 2

A (None) 3 weeks $3000 2 weeks $5000B (None) 4 $4000 2 $6000C (None) 5 $5000 3 $8000D A 8 $5000 6 $6000E A,B 3 $3000 2 $4000F C 5 $4000 3 $8000

2

3

4

1

A

B

C

D



NOTES

Step 3: Set up the CPM spreadsheet

and D came off from node 2, that would be saying that requires B as well as A. D issupposed to require A, not also B.

Here is the solution:

The solution is to add a dummy activity that runs from node 2 to node 3, as shown inthe diagram. Then start E at node 3.

If E starts at node 3, it means that E requires B and the dummy activity, the twoactivities that come in to node 3. The dummy activity, because it starts at node 2, requiresA. This makes E require B and A. The D starts at node so D only requires A. All therequirements are satisfied! Dummy activities add nothing to the time or the cost. Theirpurpose is to allow you to represent complex relations ps among activities.

Consider the CPM spreadsheet has 7 activities. That’s six lettered activities plusthe one dummy activity.

These instructions will create the spreadsheet from sc tch. Adapting the spreadsheetfrom example 1 is possible, but tricky, because example 2 has fewer columns and morerows than example 1.

To start with a blank spreadsheet, move the cell selector to A1 and type Activities”.In row 2, type the activities, letters or names. Put the Dummy at the right end of row 2, fortwo reasons:

The other activitie’s names matches will their column etters.

It will be easier to fill in the Solver Parameters box when we get to that stage.

In cell A3, type “Nodes”. In row 4, type the starting f the activities. In row 5, typethe ending of the activities.

2

3

4

51

A

B

CF

D

Dummy

E

•

•

DMC 1752


NOTES

Type “Times” in cell A6. In row 7, type each activitie normal time. The dummy timeis 0. In row 8, type each activities crash time. The d crash time is 0.

Copy row 7 and paste it into row 9. Row 9 will be the iable cells, while doing theoptimization later. Be sure to copy the numbers themselves from row 7 to row 9.

Label each of the Times rows by typing “Normal” in H7, “Crash” in H8, and “Actual”in H9.

To do the optimization, we’ll set a maximum time for t e project and tell the spreadsheetto find the combination of numbers in row 9 that compl es the project within that time forthe lowest possible cost.

To do that, it needs the costs. In cell A10, type “Cos ”. In row 11, put each activitiesnormal cost. In H11, type “Normal”. In row 12, put each activities crash cost. In H12,type “Crash”. In H13, type “Actual”. Type “0” in G13 for the Dummy actual cost.



NOTES

The diagram above also shows a formula to put in A13.

Row 13 will have formulas to calculate the actual cost for each activity. Each activitiesactual cost depends on how much it speed up or “crashed.” Assume a linear relationshipbetween speed-up and cost. For example if an activity can be shortened by 2 weeks atan added cost of $2000, assume that it can be shortened by 1 week for an added cost of$1000.

The formula to implement is at cell A13, A13 type =A1 (A7-A9)/ (A7-A8)*(A12-A11)

The logic of the formula:

(A7-A9) is the difference between the normal time and he actual time for activityA. This difference is how much time is reduced by speeding up activity A.

(A7-A8) is the difference between the normal time and he crash time. This is themost time that could save by speeding up activity A.

(A7-A9)/ (A7-A8) is how much time actually saved, as a fraction of how muchtime that could save, for activity A. In other words, is the proportion of thepossible time savings that are actually used.

(A12-A11) is the difference between the crash cost and the normal cost for activityA. This difference is how much cost would go up if activity A’s time is shortened asmuch as possible. Multiplying these, to get (A7-A9)/ (A7-A8)*(A12-A11), givesan additional cost that are incurring by shortening ac vity A’s time from its normaltime to the actual time chosen. This embodies the linearity assumption — that partway between the normal time and the crash time, the cost will be that same partway between the normal cost and the crash cost. The fu formula, A11+ (A7-A9)/(A7-A8)*(A12-A11), adds that additional cost to A11, the cost of doing the activityin normal time. This gives the cost of doing activity in the amount of time in A9.

§

§

§

§

DMC 1752


NOTES

Step 4: Use Path finds to get the paths

Step 5: Paste the path information into your spreadsheet

Don’t paste the formula to cell G13, because that would give a division-by-zeroerror. Similarly, if any of the other activities canno be speed up (the crash time equals thenormal time), put the normal cost number in the cell i row 13, not the formula.

Once that formula is in, copy cell A13 to the clipboard. Select cells A13:F13, andpaste.

Right now, it may look like the formula isn’t doing much, because the Actual costsmatch the Normal costs. This is because the Actual time matches the Normal time. It willvary, if the project is shortened.

Now, select the range of node numbers, being sure to i ude the dummy activity’snode numbers.

Copy this range to the clipboard.

Go to and paste in the node numbers that was just copied. Follow Path find’sinstructions for copying its reply.

Go back to the spreadsheet. In cell A14, type “Paths”. Then move the cell selector toA15, as shown here:



NOTES

Paste to that cell, it will produce the below results.

As with the first example, each row represents a path. The 1’s indicate which activitiesare in that particular path. For example, row 15 has 1’s under activities A and D. Thisrepresents the path at the top of the diagram. There are fourpossible paths from node 1 to node 5, so it have four ows of 0’s and 1’s.

1 -A-> 2 -D-> 5

DMC 1752


NOTES Step 6: Calculate the path times

Step 7: Identify the critical path

In cell H15, put .

Copy that cell and paste it to H15:H18.

It shows how long each path takes.

The critical path is in row 15, 1 -A-> 2 -D-> 5. It’s path with the longest time.

To make it easier to see which activities are in each go to cell A20 and type=if(A15=1,A$2,””) (Notice that the $ sign is before th 2, before the A.) This shouldput an “A” in A20.

Copy cell A20 to the clipboard. Select the range A20:G23 and then paste it.

It produces four rows showing the letters of the activ es that are in each path.

=SUMPRODUCT (A15:G15, $A$9:$G$9)

not



NOTES

Go back to cell G19 and type “Max”. Go to cell H19 and type =max(H15:H18).This shows how long the slowest path takes.

Go down to cell H20. Put in =IF (H15=H$19,”Critical”,””). In this example, “Critical”appears in cell H20, because this first path is the critical path.

Copy cell H20. Highlight H20:H23 and then paste it.

Sure enough, the first path with activities A and D, i the only one labeled “Critical”.

DMC 1752


NOTES

Step 8: Total cost formula

Go up to cell E14 and type “Total actual cost:” In cell H14, add up the actual costs ofall of the activities. The formula is =SUM (A13:G13)



NOTES

Step 9: Fill in the optimizer form

Upon completing the formula, if it uses all normal tim for all activities, the total costis $24,000.

Crash analysis is linear programming in disguise. To p form crash analysis, use theSolver tool. From ’s menu, select Tools, then Solver.

Fill in the Solver Parameters box as shown here.

The Target Cell is H14, the total cost.

Click on Min. to find the least-cost way to speed up the project.

The Changing Cells are the Actual times, in A9:G9.

The constraints, will be added by clicking the Add but n.

A9:G9 <= A7:G7 All the Actual times must be less than equal to the Normaltimes. Assume that, don’t save any money by going slower than the Normal time.

A9:G9 >= A8:G8 All the Actual times must be greater than or equal to the crashtimes. The crash times are, by definition, the fastest possible times for each activity.

H15:H18 <= 10 the slowest path can take no more than 1 weeks. 10 is chosenbecause it’s one less than 11, the normal completion t me. Later, by changing thisto 9, 8, etc., to see what happens while trying to finish the project in shorter andshorter times.

For using Excel 2000 or earlier, click on Options.

Click the checkbox for assuming a linear model.

Excel

•

•

•

•

•

•

•

DMC 1752


NOTES

Step 10: Solve

Click on OK in the Solver Options dialog box. Click on Solve in the Solver Parametersdialog box.

If an error message appears, bring back up the Solver rameters dialog box.

Make sure that Min is checked. A mistake here can caus an “unbounded solution”error.

Verify that the target cell and the changing cells are correct.

Verify that all the constraints are correct, with greater-than and less-than going inthe right directions. Mistakes here can cause “unbound solution,” “non-linear“and” no feasible solution” errors.

If it shows a message that there is no feasible solution, try changing the AssumeLinear Model option. That is, click Solvers Options bu n, then check AssumeLinear Model if it’s unchecked, or uncheck it if it’s cked.

If you still get the no-feasible-solution message, change the Solver Parameters asshown here to exclude the dummy activity from the Chan ng Cells and the Constraints.

•

•

•

•



NOTES

The range for the Changing Cells and the Constraints stop at column F. Column Ghas the dummy activity. Excel 2000 and later do not require this modification. If everythingis correct, a dialog box will appear and prompts for the report selection by the user. Theuser can requests whatever reports they like. The default result should be:

To get finished in 10 weeks (H19 now has 10), the total amount is$24,500 (H14 nowhas 24500).

There are two critical paths to worry about, 1 -A-> 2 5, and 1 -C-> 4 -F-> 5.If any of the four activities in those paths is late, e project will take more than 10 weeks.

Add three more rows, to make it easier to see, which a vities have speeded up andat an extra cost.

DMC 1752


NOTES

Step 11: Economic Analysis

WeeksProject costPenalty costTotal cost

Go to cell A24 and type “Crashed by how much” .Then go to A25 and put =A7-A9.This is the Actual time minus the Normal time. Go to cell A26. and put =A13-A11. This isthe difference between the current Actual cost and the Normal cost. Select A25:A26.Copy to the clipboard. Select A25:F25, and paste.

Use an economic problem to solve. Imagine that the Exa le 2 project is being doneon a contract, with a scheduled completion time of 8 weeks. There is a $2500 per weekpenalty for being late. There is also a $1000 per week bonus for being early. Here theresults so far. An explanation follows:

If it uses normal time, the project takes 11 weeks, wh h runs over the schedule by 3weeks. It makes to lose $7500 in penalties. The total cost is $24,000 + $7,500 =$31,500.

If it crash by 1 week, the project takes 10 weeks. The penalty cost is $5,000. Directproject cost is $24,500. The total is $29,500. This is less than $31,500, so ten weeks isbetter than eleven weeks.

11 $24,000 $7,500 $31,50010 24,500 5,000 29,500



NOTES

1.3 SUMMARY OF CPM STEPS

WeeksProject costPenalty costTotal cost

WeeksProject costPenalty cost Total cost

29,000

To try completing in 9 weeks, go back to the Solver (s 9). Change the H15:H18<= 10 constraint to H15:H18 <= 9. and solve. The cost es to $26,500. The penalty forbeing one week late is $2,500. Total cost is $26,500 + $2,500 = $29,000. This is lessthan $29,500, so nine weeks is better than ten weeks.

Notice that cutting the second week added more to the project cost (second column)than cutting the first week. Cutting one week increase project cost by $500. Cutting thesecond week increased project cost by $2,500 from $24,500 to $26,500. The law ofdiminishing returns is at work here.

Is it save three weeks and be on time? To try remove c pleting in 8 weeks, go backto the Solver. Change the H15:H18 <= 9 constraint to H15:H18 <= 8, then solve. Theresultant cost is $30,000. There is no penalty, but $30,000 is more than $29,500, so itloses money by being on time. It’s better off being one week late and paying the penalty.

What about saving four weeks and being early? As the law of diminishing returns,don’t need to consider shorter project durations than weeks. Going from 9 weeks to 8added $3,500 to cost. The law of diminishing returns i lies that going from 8 weeks to 7will add at least $3,500 more to cost. That is more than the $1000 bonus, so the 7 weeksis a loser.

If the bonus is linear then the law of diminishing returns implies that it can stop theanalysis as soon as the total cost, including the bonu starts to rise.

Here is the whole table. The bonus for being early is ted as a negative penalty.

The conclusion is that, the optimal production schedul is 9 weeks. It has the leasttotal cost.

CPM helps to identify a complex project’s critical pat , it enables to find how long aproject will take and which activities must be on time. If the information about costs, crashcosts and time is available then CPM helps to determin how long the project should take,and which activities should be speed up (“crashed”). The steps are:

11 $24,000 $7,500 $31,50010 24,500 5,000 29,5009 26,500 2,500 29,000

11 $24,000 $7,500 $31,50010 24,500 5,000 29,5009 26,500 2,5008 30,000 0 30,0007 Don't bother -1000 will be higher

DMC 1752


NOTES

1.4 EXERCISE

1.5 PROGRAM EVALUATION AND REVIEW TECHNIQUE (PERT)

Tasks Dependency Duration

§

§

§

§

§

§

§

§

§

§

Have a list of the activities.

Draw the network diagram.

Put activity names, node numbers, times, and costs in spreadsheet.

Use Path find to generate code for the paths.

Put the path information into the spreadsheet.

Calculate the paths’ times.

Identify the critical paths, and the activities in each path.

Set up the formula to calculate the project’s total cost.

Fill in the Tools | Solver... form.

Solve, and fix errors, if any.

For an economic analysis, change the maximum time cons raint and solve again. Repeatuntil costs, including penalties and bonuses, start to go up.

1. The software project consists of a list of tasks along with their estimated durationswhich are shown in the estimation table below. The project manager also knows thedependencies among those tasks. Please make a CPM gra and a Gantt chart to showhow to organize all the tasks so that the project can e down in the shortest time by Nancyand Julia.

Program Evaluation and Review Technique (PERT) and Cri ical Path Analysis (CPA)are tools that are used to schedule and manage large complex projects. These tools help tosimplify the planning and scheduling of projects to plan most importantly during the

A 4

B 3

C A 8

D A 7

E B and C 9

F B and C 12

G D and E 2

H D and E 5

I F and G 6



NOTES

1.5.1 Project Evaluation and Review Technique (PERT) and Critical PathMethod (CPM) Applications

or

Note: In the following examples read the diagrams from left to right.

1.6 EXAMPLE

1.6.1 Example 1

management of a project, It enable companies to monito achievement of project goalsand assist to identify where remedial action needs to e taken to ensure project is completedwithin the time frame.

PERT is used to estimate the time that each activity will take. PERT is intended forvery large-scale, one-time, complex, non-routine proje s. It is more of an event-orientedtechnique rather than start and completion-oriented an is used more in R&D-type projectswhere, time rather than cost is a major factor. A PERT diagram shows the relationshipbetween simultaneous steps in a project, after critica path analysis PERT diagram can alsobe used to show the critical path.

CPA formally identifies tasks which must be completed time for a project to becompleted on time, which tasks have the option of been delayed and the minimum lengthof time needed to complete a project. If a company would require accelerating a project,CPA analysis will helps to identify which project steps should accelerate to complete theproject within the available time. This helps to minim ze cost while still meeting the objective.

Two simple, yet interesting and important applications of partial ordering relations arethe PERT and CPM techniques in job scheduling. For an of the list of tasksneeded to build a house. let A = {set of tasks given below(include the exercise)} anddefine the relation R on A by: xRy if task x proceeds a prerequisite task of) task y task x = task y the Hasse diagram is given in the illu tration in the text. The above techniquesare explored through several examples as below.

The construction of a cottage requires the performance of certain tasks. The followingtable 1 lists the various tasks with their priority relationship. The start task ( ) and finishtask ( ) are added, is linked to task A and is linked to task K.

αω α ω

DMC 1752


NOTES

Figure 1

Code Of Task Duration(Weeks) Prerequisite tasks

Let S be the set of all tasks and consider the partial order relation R defined on S asfollows: for all tasks x and y in S,

xRy x = y or x precedes y.

The minimum time required to finish this job is calcul ed using Project Evaluation andReview Technique (PERT). The PERT diagram of the project is shown in the figure 1.

In this representation, the tasks are the vertices of e graph. The length of each taskis the duration of the corresponding task. The start a d finish of a task are two stages of theproject. Define an initial stage as start and a final tage as finish and a certain number ofintermediate stages. Each stage is defined by tasks already carried out.

In the beginning milestone of the project is start stage. The start followed by task Awhich is carried out in 7 weeks. After finishing stage A, it leads into two options. The taskB and task D can be started simultaneously. Task B takes 3 weeks to finish and task Dtakes 8 weeks to finish the job. Task B is immediately followed by task C, which takes oneweek to carry out. The minimum time required to start sk C is the total time required tofinish task B and A which is 10 weeks. Task C is followed by task E. To start task E,the

A Masonry 7 -B Carpentry for roof 3 AC Roof 1 BD Sanitary and electrical installations 8 AE Front 2 D,CF Windows 1 D,CG Garden 1 D,CH Ceiling 3 FJ Painting 2 HK Moving in 1 E,G,J

ó



NOTES

Table 2. below:

Figure 2

Task Earliest Start Time

task C and D should be finished. C is finished in 11 w s, and D is finished in 15 weeks,so it is very common that,the task E cannot be started before finishing task D which takes15 weeks to carry out. Same way, the earliest time to rt each task are identified, whichis shown in table 2 below.

The above analysis shows that at least 21 weeks is required to complete the project.The minimum duration of the project will be the length of the longest path between theinitial stage and the finish stage. This is shown in w h the darker line in the figure 2.

Start (a) 0

A 0

B 7

C 10

D 7

E 15

F 15

G 15

H 16

J 18

K 20

Finish ( ) 21ω

DMC 1752


NOTES

4.6.2 Example 2

If any work in this path, is delayed then the project ish time is also delayed. Thatmeans delay in performing any of these tasks in the path cause delay in the time required tofinish the project, and for these reason, this path is known as critical path. The tasksoutside the critical path can be delayed by certain amount of time. For example task E isnot on critical path and it is followed by task K. If E is delayed by 3 week than itsearliest start time, then the task K is started as per its earliest start time because K must bestarted after finishing J which takes 20 weeks. It makes the task E in 20 weeks by delaying3 weeks. In the opposition if any delay in work on the critical path, then it delays the timeto complete the project. For example 3 weeks delay in ask J cause 3 weeks delay to startK, and K start after 3 weeks than its earliest start time and these cause three week delayto complete the project. Now if task J finished earlier by one week then task K, can alsoreached earlier by one week and finish it off. And fin ly the overall project is also finishedone week before.

In order to construct a house, there are many jobs that need to be done at the sametime. Certain parts of the construction must be done b ore others can be started. Nobodycannot start wiring the house if the walls have not be n put up, and cannot put the walls uptill the foundation had been set. Before the foundation is set, one must choose a good plotto dig the foundation hole. However, there are certain jobs which can be donesimultaneously. While wiring the outlets, one can have someone to do the heating and airconditioning ducts or can have the landscaper working the outside layout. The partialorder diagram will allow one to see how long the whole process will take. Below, tableshows each task that needs to be done, and the amount time required for that task. Theamount of time is estimated.

The next table shows, what task precedes each task. This allows to see what taskscan be done at the same time. If one task does not depend on the other, then those twotasks can be done at the same time.

1 Find Plot 2 days

2 Excavate land 5 days

3 Dig for foundation 3days

4 Lay concrete foundation 5 days

5 Exterior firework 2 days

6 Exterior electrical 2 days

7 Exterior gas line 5 days



NOTESTask Percent Tasks Time

Partial Order Formula: X R Y <=> X = Y or X precedes Y

The following Hasse diagram from left to right to find the minimum time for the wholeprocess of constructing the house.

1 0 2 days

2 1 5 days

3 2 3days

4 3 5 days

5 1,2,3,4 2 days

6 1,2,3,4 2 days

7 1,2,3,4 5 days

8 4 2 days

9 8 2 days

10 8 2 days

11 8 2 days

12 8 2 days

13 8 2 days

14 10,11,12,13 2 days

15 14 2 days

16 8 2 days

17 15 2 days

18 17 2 days

19 17 2 days

20 19 2 days

21 20 2 days

DMC 1752


NOTES

1.7 EXERCISE

Duration in weeks Likely Optimistic Pessimistic

Partial Order Diagram: Construction of a House

Critical Path Method (CPM): 1, 2, 3,4,8,16,21 which is 35 days

The critical path time can be reduced if done certain asks differently. The time forlandscaping right now is 14 days. But if more workers re used, then the time will be less.If pre-fabricated materials are used, then it will take less time, because less people areneeded, and less time is used in making the material. so depending on the design of thehouse, the time can be reduced or even increased. If less room is needed by the family thatis living there, then it will take less time, then des gning at large size home with many rooms,and closets inside each room. What can also reduce the time is, experience of the constructionworkers. If the workers are new, then they will be a b slower, and making more mistakeswhich increases the time for each task.

1. Find the following problem using PERT model:

A) Calculate the expected duration per activity

i) 5+5*10+21/6 = 11

ii) 4+4*6+8/6 = 6

iii) 6+4*14+16/6 = 13

Activity A 10 5 21Activity B 6 4 8Activity C 14 6 16

Task1 2 Days

Task 2 5 Days

Task 3 3 Days

Task 4 5 Days

Task 5 2 Days

Task 6 2 Days

Task 7 2 Days

Task 8 5 Days

Task 13

Task 16

Task 12

Task 11

Task 10

Task 93 Days

Task 14

Task 15

Task 17



NOTES

1.8 NETWORK ANALYSIS: RESOURCE SCHEDULING

1.8.1 Resources and Networks

1.8.2 Project resources

1.8.3 Resource Scheduling Requirements

B) Determine the standard deviation

A 21-5 = 16/6 = 2.67

B 8-4 = 4/6 = 0.67

C 16-6 = 10/6 =1.67

((2.67)2 + (0.67)2 + (1.67)2)0.5 = 3.21

C) Assuming a normal distribution estimate the probabi ty that the project duration is

i) more than A days

ii) less than B days

iii) between C and D days

This section deals with the resources aspects of network analysis, and shows howresources of men, machines, etc could be scheduled over the project duration to be ableto deal with resource and/or time constraints.

The usefulness of networks is not confined only to the time and cost factors whichhave been discussed so far. Considerable assistance in planning and controlling the use ofresources can be given to management by appropriate development of the basic networktechniques.

The resources (men of varying skills, machines of all ypes, the required materials,finance, and space) used in a project are subject to varying demands and loadings as theproject proceeds. Management need to know what activities and what resources arecritical to the project duration, and if resource limi tions (eg shortage of materials, limitednumber of skilled personnel) might delay the project. They also wish to ensure, as far aspossible, constant work rates to eliminate paying overtime at one stage of a project andhaving short time working at another stage.

To be able to schedule the resource requirements for a project the following detailsare required.

DMC 1752


NOTES

1.8.4 Resources Scheduling Example Using a Gantt chart

PROJECT DATAACTIVITY

(DAYS)PRECEDING

ACTIVITYDURATION LABOUR

REQUIREMENT

a. The customary activity times, descriptions and sequenc s as previously described.

b. The resource requirements for each activity showing th classification of the resourceand the quantity required.

c. The resources in each classification that are available to the project. If variationsin availability are likely during the project life, th e must also be specified.

d. Any management restrictions that need to be considered eg which activities mayor may not be split or any limitations on labour mobility.

A simple project has the following time and resource d a (for simplicity, only the oneresource of labour is considered, but similar principles would apply to other types of inter-changeable resources).

Resource constraint - 2 men only available

Critical Path Activity D

Duration 5 Days (Without taking account of the resource constraint).

A - 1 2menB - 2 1manC A 1 1 manD - 5 1 manE B 1 1 manF C 1 1 man

0

0 0

1

1 3

3

2 4

2

2 4

4

5 5

A 1

C 1

D 5

F 1

E 1

B 2



NOTES1.8.5 Resource Scheduling Steps

Resource Aggregation Profile Based on EST’s

a. Draw the activity time of a Gantt or Bar Chart based on their ESTs

b. Based on the time bar chart prepares a Resource Aggregation Profile ie total resourcerequirements in each time period.

c. Examination of the above profile shows that at times more resources are required thanare available if activities commence at their EST’S. he ESTs/LSTs on the network showthat float is available for activities A, C, F, B and E. Having regard to these floats it isnecessary to ‘smooth out’ the resource requirements so that the resources required do notexceed the resource constraint, ie delay the commencement of activities (within their float)and if this procedure is still not sufficient then del the project as a whole. Carrying outthis procedure results in the following resource profi .

0 1 2 3 4 5

Time scale in days

D

A C F

B E

4

3

2

1

05

Time scale in days

0 1 2 3 4

Resource availabilityA

C F

B E

D

N o. of men required

DMC 1752


NOTES

Resource Allocation - with 2 man constraint

Note:

Resource Allocation Profile – with 5 day constraint

This procedure is sometimes termed RESOURCE LEVELLING.

d. Because of the resource constraint of 2 men it has een necessary to extend the projectduration by 1 day. Assume that management state that original project duration (5days) must not be extended and they require this to be achieved with the minimum extraresources. In such cases a similar process of varying activity start times within their float iscarried out, resulting in the following resource profi .

The above profile shows that to achieve the 5 day duration it is necessary tohave 3 men available from day 2 to day 4.

5

Time scale in days

0 1 2 3 4

A

C FB E

D

6

No. of men required

AB

D

E

C

F

0 1 2 3 4 5

4

3

2

1

0

No. of men required

Time scale in days



NOTES1.8.6 Summary

a. To enable resource scheduling to be carried out the resource requirements foreach activity must be specified.

b. In addition the various resources involved (men, machinery etc) must be classifiedand the availability and constraints specified.

c. After calculating the critical path in the usual manner a Resource AggregationProfile(s) is prepared ie the amount of the resource(s required in each time periodof the project based on the EST’s of each activity.

d. If the resource aggregation indicates that a constraint is being exceeded, and floatis available the resource usage is ‘smoothed’ i.e. the start of activities is delayed.

DMC 1752


NOTES



NOTESUNIT V

CHAPTER 1

QUEUEING MODELS

1.1 WAITING LINE MODELS

Introduction

Queueing theory

Basic Terminology

Queuing Model:

Waiting lines are the most frequently encountered problems in everyday life. Forexample, queue at a cafeteria, library, bank, etc. Common to all of these cases are thearrivals of objects requiring service and the attendant delays when the service mechanismis busy. Waiting lines cannot be eliminated completely but suitable techniques can be usedto reduce the waiting time of an object in the system. A long waiting line may result in lossof customers to an organization. Waiting time can be reduced by providing additionalservice facilities, but it may result in an increase i the idle time of the service mechanism.

is based on mathematical theories and deals with the roblems arisingdue to flow of customers towards the service facility.

The waiting line models help the management in balanci between the cost associatedwith waiting and the cost of providing service. Thus, ueing or waiting line models canbe applied in such situations where, decisions have to be taken to minimize the waiting timewith minimum investment cost.

The present section focuses on the standard vocabulary of Waiting Line Models.

It is a suitable model used to represent a service oriented problem,where customers arrive randomly to receive some service, where the service time is also arandom variable.

DMC 1752


NOTES Arrival:

Service Time:

Server:

Queue Discipline:

Poisson Process:

Queue:

Waiting time in queue:

Waiting time in the system:

Queue length:

Average length of line:

Average idle time:

FIFO:

Bulk Arrivals:

1.1.2 Components of Queuing System

1. Input Source:

2. Queue:

3. Service Discipline:

The statistical pattern of the arrival can be indicated through the probabilitydistribution of the number of the arrivals in an interval.

The time taken by a server to complete service is known as service time.

It is a mechanism through which service is offered.

It is the order in which the members of the queue are fered service.

It is a probabilistic phenomenon where the number of arrivals in aninterval of length t follows a Poisson distribution with parameter ?t, where ? is the rate ofarrival.

A group of items waiting to receive service, including those receiving the service,is known as queue.

Time spent by a customer in the queue before being served.

It is the total time spent by a customer in the system. It canbe calculated as follows:

Waiting time in the system = Waiting time in the queue + Service time

Number of persons in the system at any time.

The number of customers in the queue per unit of time.

The average time for which the system remains idle.

It is the first in first out queue discipline.

If more than one customer enter the system at an arri al event, it is knownas bulk arrivals.

The input source generates customers for the service mechanism.The most important characteristic of the input source s its size. It may be eitherfinite or infinite. Since the calculations are far easier for the infinite case, thisassumption is often made even when the actual size is elatively large. If the populationsize is finite, then the analysis of queueing model becomes more involved. Thestatistical pattern by which calling units are generated over time must also bespecified. It may be Poisson or Exponential probability distribution.

It is characterized by the maximum permissible number f units that it cancontain. Queues may be infinite or finite.

It refers to the order in which members of the queue areselected for service. Frequently, the discipline is first come, first served.



NOTESFollowing are some other disciplines:

4. Service Mechanism:

Unusual Customer/Server Behaviour

Customer’s Behavior

Balking.

o

o

o

LIFO (Last In First Out)

SIRO (Service In Random Order)

Priority System

A specification of the service mechanism includes adescription of time to complete a service and the number of customers who aresatisfied at each service event. The service mechanism also prescribes the numberand configuration of servers. If there is more than on service facility, the callingunit may receive service from a sequence of these. At given facility, the unitenters one of the parallel service channels and is completely serviced by that server.Most elementary models assume one service facility with either one or a finitenumber of servers. The following figure shows the phys cal layout of service facilities.

A customer may not like to join the queue due to long waiting line.

Single Channel, Single Phase

Service Facility

Single Channel, Multiple Phase

Service Facility

Service Facility

Multiple Channel, Single Phase

Service Facility

Service Facility

Multiple Channel, Multiple Phase

Service Facility

Service Facility

Service Facility

Service Facility

v

DMC 1752


NOTES Reneging.

Collusion.

Jockeying.

Server’s Behavior

Failure.

Changing service rates.

Batch processing.

What are the underlying assumptions?

1.2 THE M/M/1 ( /FIFO) SYSTEM

v

v

v

v

v

v

v

v

v

v

v

v

v

A customer may leave the queue after waiting for sometime due toimpatience.

Several customers may cooperate and only one of them y stand inthe queue.

When there are a number of queues, a customer may move from onequeue to another in hope of receiving the service quic y.

The service may be interrupted due to failure of a se er (machinery).

A server may speed up or slow down, depending onthe number of customers in the queue. For example, when the queue is long, aserver may speed up in response to the pressure. On the contrary, it may slowdown if the queue is very small.

A server may service several customers simultaneously thisphenomenon is known as batch processing.

The source population has infinite size.

The inter-arrival time has an exponential probability istribution with a mean arrivalrate of one customer arrivals per unit time.

There is no unusual customer behaviour.

The service discipline is FIFO.

The service time has an exponential probability distribution with a mean servicerate of m service completions per unit time.

The mean arrival rate is less than the mean service rate, i.e., l < m.

There is no unusual server behaviour.

It is a queueing model where the arrivals follow a Poi n process, service times areexponentially distributed and there is only one server In other words, it is a system withPoisson input, exponential waiting time and Poisson output with single channel.

Queue capacity of the system is infinite with first in first out mode. The first M in thenotation stands for Poisson input, second M for Poisson output, 1 for the number ofservers and a for infinite capacity of the system.

¥



NOTES

1.3 EXAMPLES

1.3.1 Example 1

Given

1.3.2 Example 2

-

Formulae

Students arrive at the head office of www.universaltea her.com according to a Poissoninput process with a mean rate of 40 per hour. The time required to serve a student has anexponential distribution with a mean of 50 per hour. Assume that the students are servedby a single individual find the average waiting time of a student.

Solution.

? = 40/hour, µ= 50/hour

Average waiting time of a student before receiving service (Wq)

8.440)-(5050

40(Wq) minutes

New Delhi Railway Station has a single ticket counter. During the rush hours, customersarrive at the rate of 10 per hour. The average number customers that can be served is 12per hour. Find out the following:

Probability that the ticket counter is free

Average number of customers in the queue

Probability of zero unit in the queue (Po) = 1 ? ----- µ

Average queue length (Lq ) = ?2

--------µ (µ- ? )

Average number of units in the system (Ls) = ?--------µ - ?

Average waiting time of an arrival (Wq) = ?----------µ(µ - ?)

Average waiting time of an arrival in the system (Ws) = 1--------- µ- ?

==

•

•

DMC 1752


NOTES Solution.

1.3.3 Example 3

Solution.

Given

? = 10/hour, µ = 12/hour

At Bharat petrol pump, customers arrive according to a Poisson process with anaverage time of 5 minutes between arrivals. The servic time is exponentially distributedwith mean time = 2 minutes. On the basis of this infor tion, find out

1. What would be the average queue length?

2. What would be the average number of customers in the q euing system?

3. What is the average time spent by a car in the petrol p?

4. What is the average waiting time of a car before receiving petrol?

Probability that the counter is free (Ws) = 1 -

10

-----

12

=1/6

Average number of customers in the queue (Lq ) =

(10)2

--------------

12 (12 - 10)

= 25/6

Average inter arrival time = 1

--- =?

1---hour =5 minutes12

?= 12/hour

Average service time = 1

--- =µ

1---hour =2minutes30

µ = 30/hour

Average queue length, Lq = (12)2

-----------30(30 - 12)

4 = ---- 15



NOTES

1.3.4 Example 4

Assuming Poisson arrivals and exponential service find

Solution.

Universal Bank is considering opening a drive in window for customer service.Management estimates that customers will arrive at the rate of 15 per hour. The tellerwhom it is considering to staff the window can service customers at the rate of one everythree minutes.

1. Average number in the waiting line.

2. Average number in the system.

3. Average waiting time in line.

4. Average waiting time in the system.

Given

? = 15/hour,

µ = 3/60 hour or 20/hour

Average number of custom ers, Ls = 12 -------30 - 12

2 = ---- 3

Average tim e spent at the petrol pum p = ? 1----------30 - 12

= 3.33 minutes

Average waiting tim e of a car before receiving petrol =µ

12------------30 (30 - 12)

= 1.33 minutes

Average number in the waiting line = (15)2

-------------20(20 - 15)

= 2.25 customers

Average number in the system = 15----------20 - 15

= 3 customers

Average waiting time in line = 15------------20(20 - 15)

= 0.15 hours

Average waiting time in the system = 1---------

20 - 15= 0.20 hours

DMC 1752


NOTES 1.3.5 Example 5

Solution.

1.4 THE M/M/1 (N/FIFO) SYSTEM

Chabra Saree Emporium has a single cashier. During the rush hours, customers arriveat the rate of 10 per hour. The average number of cust ers that can be processed by thecashier is 12 per hour. On the basis of this information, find the following:

Probability that the cashier is idle

Average number of customers in the queuing system

Average time a customer spends in the system

Average number of customers in the queue

Average time a customer spends in the queue

Given

? = 10/hour, µ = 12/hour

It is a queuing model where the arrivals follow a Pois process, service times areexponentially distributed and there is only one server Capacity of the system is limited toN with first in first out mode.

Po = 10 1- --- 12

=1/6

Ls =

10--------12 - 10

= 5 customers

Ws = 1----------12 - 10

= 30 minutes

Lq = (10)2

-------------12(12 - 10)

=25/6 customers

Wq =

10-------------12(12 - 10)

= 25 minutes

Ø

Ø

Ø

Ø

Ø



NOTES

1.5 EXAMPLES

1.5.1 Example

Solution.

-

-

The first M in the notation stands for Poisson input, econd M for Poisson output, 1for the number of servers and N for capacity of the system.?

Students arrive at the head office of www.universaltea her.com according to a Poissoninput process with a mean rate of 30 per day. The time required to serve a student has anexponential distribution with a mean of 36 minutes. As ume that the students are served bya single individual, and queue capacity is 9. On the b sis of this information, find the following:

The probability of zero units in the queue.

The average line length.

Po = 1-?--------1-? N + 1

Ls = ?--------1 - ?

–(N+1)?N +1

----------- 1-? N + 1

Lq = Ls – ?/µ

W q = Lq

---- ?

W s = Ls

----?

?= 30---------60 X 24

= 1/48 students per minute

µ = 1/36 students per minute

? = 36/48 = 0.75 N = 9

Po = 1 0.75-------------1- (0.75)9 + 1

= 0.26

Ls = 0.75 --------- 1 - 0.75

- (9+1)(0.75)9+1

------------------ 1 (0.75)9 + 1

= 2.40 or 2 students.

v

v

DMC 1752


NOTES 1.6 THE M/M/C (? /FIFO) SYSTEM

1.7 EXAMPLES

1.7.1 Example 1

Solution.

It is a queuing model where the arrivals follow a Pois process, service times areexponentially distributed and there are C servers.

The Silver Spoon Restaurant has only two waiters. Customers arrive according to aPoisson process with a mean rate of 10 per hour. The service for each customer isexponential with mean of 4 minutes. On the basis of th information, find the following:

Probability of waiting for a service.

The expected percentage of idle time for each waiter.

This is an example of M/M/C, where c = 2

? = 10 per hour or 1/6 per minute.

µ = 1/4 per minute

? = 1/3

1---- = P0

1

0

(?/µ)n

-------- n!

+(?/µ)c

-------- X c!

1----

1 - ?

Where ? = ?------

cµ

Lq = P0 X (?/µ)c

------ X c!

?--------(1 - ? )2

Wq = 1--- X

?Lq

Ws = Wq +

1----µ

Ls = Lq +?

----µ

•

•

∑−

=

c

n



NOTES

1.7.2 Example 2

Solution.

Case I - Treating depositors and withdrawers as unit of M/M/1 system.

Universal Bank has two tellers working on savings acco s. The first teller handleswithdrawals only. The second teller handles deposits o ly. It has been found that the servicetimes distributions for both deposits and withdrawals exponential with mean servicetime 2 minutes per customer. Deposits and withdrawals found to arrive in a Poissonfashion with mean arrival rate 20 per hour. What would be the effect on the averagewaiting time for depositors and withdrawers, if each teller could handle both withdrawersand depositors?

Given

? = 20 per hour or 1/3 per minute, = 1/2 per minute, c = 2

1---- = P0

1

0

(?/µ)n

-------- n!

+(?/µ)c

-------- c!

X 1 ----1 - ?

1---- = P0

1

0

(2/3)n

-------- n!

+(2/3)2

-------- 2!

X 1------1 - 1/3

1---- = P0

1 + 2------ 3

+1

---3

P0 = 1---2

The expected percentage of idle time for each waiter.1 - ? = 1 - 1/3 = 2/3 = 67%

Average waiting time of an arrival (Wq) = ?----------

µ (µ - ? )

Wq = 1/3 ---------------- = 4 minutes

1/2 (1/2 - 1/3)

µ

∑−

=

∑=

c

n

n

DMC 1752


NOTES Case II - If each teller handles both depositors and

1.8 THE M/EK/1 ( /FIFO) SYSTEM

An Erlang Family (Ek)

Hence, when both tellers handle both withdrawals and d posits, then expected waitingtime is reduced.

It is a queuing model where the arrivals follow a Pois n process, service time followsan Erlang (k) probability distribution and the number f server is one. Queue capacity ofthe system is infinite with first in first out mode. The first M in the notation stands forPoisson input, k for number of phases, 1 for the number of servers and a for infinitecapacity of the system.

Ek of a probability distribution is the probability distribution of a random variable,which can be expressed as the sum ? k µ independently, identically distributed exponentialvariables.

P0 = 1/2

Lq = 1/12

Wq = 1-----

?X Lq

Wq = 1/4 minutes

The expected numbers of customers in the queue, Lq = (1+k)--------

2kX

?2

-----------µ (µ - ? )

The expected waiting time before being served, Wq =(1+k)

-------- 2k

X ?-----------

( - ? )

The expected time spent in the system, Ws = Wq + 1----- µ

The expected numbers of customers in the system, Ls = ? Ws

¥

µ µ



NOTES1.9 EXAMPLES

1.9.1 Example 1

Solution.

1.9.2 Example 2

The registration of a student at www.universalteacher.com requires three steps to becompleted sequentially. The time taken to perform each step follows an exponentialdistribution with mean 30/3 minutes and is independent of each other. Students arrive atthe head office according to a Poisson input process with a mean rate of 25 per hour.Assuming that there is only one person for registration. On the basis of this information,find the following:

a. expected waiting time.

b. expected numbers of students in the queue.

This is an M/Ek/1 system.

Here k = 3, ? = 25 per hour.

Repair of a certain type of machine requires three ste to be completed sequentially.The time taken to perform each step follows an exponen l distribution with mean 20/3minutes and is independent of each other. The machine follows a Poissonprocess with rate of 1per 2 hours. Assuming that there is only one mechanic, find out

a. The expected idle time of a machine.

b. The average waiting time of a broken down machine in a queue.

c. The expected number of broken down machines in the queue.

d. The average number of machines which are not in operat n.

Service time per phase = 1

---3µ

=30----

3Therefore, µ = 30 per hour.

The expected numbers of students in the queue, Lq =

1+3------2 x 3

X(25)2

--------------30(30 - 25)

= 2.78 students or 3 students

The expected waiting time before being served, W q =

1+3 ------

2 x 3X

25-------------30(30 - 25)

=1/9 hour or 6.67

minutes

DMC 1752


NOTES Solution.

As k approaches to infinity, the expressions of Lq, W

q , W

s and L

s are given by

This is an M/Ek/1 system.

Here k = 3, ? = 1/2 per hour.

Here k =3, ? =1/2 per hour.

Service time per phase =

1---3µ

=20---- 3

Therefore, µ = 3 per hour.

The expected numbers of customers in the queue, Lq =

1+3------2 X 3

X(1/2)2

-----------3(3 - 1/2)

= 1.33 minutes

The expected waiting time before being served, W q =

1+3------2 X 3

X1/2

---------3(3 - 1/2)

=2 minutes40 seconds

The expected time spent in the system, W s =

2 ---- 45

+1

---- 3

= 22 minutes40 seconds

The expected numbers of customers in the system, Ls =

1----- 2

X17----- 45

= 11.33 minutes

Lq = ?2

----------- 2µ (µ - ? )

Wq = ? ------------ 2 ( - ? )

Ws = 1Wq + --- µ

Ls = ?Lq + ----- µ

µ µ



NOTES1.9.3 Example 3

Solution.

Summary

1.10 EXERCISES

At Indira Gandhi airport, it takes exactly 6 minutes to land an aeroplane, once it isgiven the signal to land. Although incoming planes hav scheduled arrival times, the widevariability in arrival times produces an effect which kes the incoming planes appear toarrive in a Poisson fashion at an average rate of 6 per hour. This produces occasionalstack-ups at the airport, which can be dangerous and c tly. Under these circumstances,how much time will a pilot expect to spend circling the field waiting to land?

Here, service time is fixed being equal to 6 minutes. service distribution is lastmember of Erlang family, i.e., for k = a Mean arrival rate of aeroplane, ? = 6 per hourMean landing rate of planes, = (1/6) X 60 = 10 per hour

Waiting lines, like birds, are omnipresent. For instance, queue at a cafeteria, library,bank, etc. The reason for waiting lines is that in any system, an item cannot be servicedimmediately on arrival and each new arrival has to wai for sometime before it is attended.Queuing or waiting line models can be applied in such tuations where decisions have tobe taken to minimize the waiting time with minimum inv stment cost.

In this chapter, we discussed the following waiting li models:

The M/M/1 ( /FIFO) system

The M/M/1 (N/FIFO) system

The M/M/C ( /FIFO) system

The M/Ek/1 ( /FIFO) system

1. In a work station orders arrive according to a Poisson arrival process with arrival rate?.An order consists of N independent jobs. The distribu n of N is given by

1kp)p1()kN(P

With k = 1,2,…. and 0 d” p < 1. Each job requires an exponentially distributedamount of processing time with mean 1/µ.

i. Derive the distribution of the total processing time o an order.

ii. Determine the distribution of the number of orders in e system.

Wq =

6

---------------------

2 X 10 X (10 - 6)

= 3/40 hours = 4.5 minutes

µ

Ø

Ø

Ø

Ø

∞

∞

∞

−−==

DMC 1752


NOTES 2. Consider a work station where jobs arrive according to a Poisson process with arrivalrate ?. The jobs have an exponentially distributed service time with mean 1/µ. So theservice completion rate (the rate at which jobs depart from the system) is equal to µ. Ifthe queue length drops below the threshold QL the serv e completion rate is loweredto _L. If the queue length reaches QH, where Q

H QL, the service rate is increased to

µH.(L stands for low, H for high.)Determine the queue le th distribution and the mean

time spent in the system.

3. A repair man fixes broken televisions. The repair time is exponentially distributed witha mean of 30 minutes. Broken televisions arrive at his repair shop according to aPoisson stream, on average 10 broken televisions per day (8 hours).

i. What is the fraction of time that the repair man has no work to do?

ii. How many televisions are, on average, at his repair shop?

iii. What is the mean throughput time (waiting time plus re ir time) of a television?

4. In a gas station there is one gas pump. Cars arrive at the gas station according to aPoisson process. The arrival rate is 20 cars per hour. Cars are served in order ofarrival. The service time (i.e. the time needed for pumping and paying) is exponentiallydistributed. The mean service time is 2 minutes.

i. Determine the distribution, mean and variance of the number of cars at the gasstation.

ii. Determine the distribution of the sojourn time and the waiting time.

iii. What is the fraction of cars that has to wait longer than 2 minutes?

An arriving car finding 2 cars at the station immediately leaves.

iv. Determine the distribution, mean and variance of the number of cars at the gasstation.

v. Determine the mean sojourn time and the mean waiting time of all cars (includingthe ones that immediately leave the gas station).

5. A gas station has two pumps, one for gas and the other for LPG. For each pumpcustomers arrive according to a Poisson process. On average 20 customers per hourfor gas and 5 customers for LPG. The service times are exponential. For both pumpsthe mean service time is 2 minutes.

i. Determine the distribution of the number of customers t the gas pump, and at theLPG pump.

ii. Determine the distribution of the total number of customers at the gas station.

6. Consider an M=M=1 queue with two types of customers. T mean service time of allcustomers is 5 minutes. The arrival rate of type 1 customers is 4 customers per hourand for type 2 customers it is 5 customers per hour. T e 1 customers are treated withpriority over type 2 customers.

i. Determine the mean sojourn time of type 1 and 2 custom rs under the preemptive-resume priority rule.

≥



NOTESii. Determine the mean sojourn time of type 1 and 2 customers under the non-

preemptive Priority rule.

7. Consider an M=M=1 queue with an arrival rate of 60 cus mers per hour and a meanservice time of 45 seconds. A period during which there are 5 or more customers inthe system is called crowded, when there are less than 5 customers it is quiet. What isthe mean number of crowded periods per day (8 hours) a how long do they last onaverage?

8. Consider a machine where jobs arrive according to a Po son stream with a rate of 20jobs per hour. The processing times are exponentially istributed with a mean of 1/µhours. The processing cost is 16µ dollar per hour, and the waiting cost is 20 dollar perorder per hour. Determine the processing speed µ minimizing the average cost perhour.

DMC 1752


NOTES NOTES



NOTESNOTES

DMC 1752


NOTES NOTES

dmc 1752 resource management techniques

Documents