Operations Research [34 ed.] 9788182837218

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nPERATIONS

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(For Honours and Post-Graduate Students of Mathematics, Statistics, Commerce, Management and Engineering of all Universities)

(As per UGC Syll abus

By Dr. R.K. Gupta Rtd. Principal & Head Department of Mathematics S.S.V. College, Hapur (C.C.S. University, Meerut)

KRISHNA Prakashan Media (P) Ltd. rf

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KRISHNA HOUSE, 11, Shivaji Road, Meerut-250 001 (U.P.), India

16,4441

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pERATIONS

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First Edition: 1984 Twentieth Revised and Enlarged Edition: 2005 Twenty Eighth Revised Edition: 2010 Thirty Fourth Edition : 2015

Name, style or any part of this book thereof may not be reproduced in any form or by any means without the written permissior from the publishers and the author. Every effort has been made to avoid errors or omissions in this publication. In spite of this, some errors might have crept in. Any mistake, error or discrepancy noted may be brought to our notice which shall be token core of in the next edition. It is notified that neither the publisher nor the author or seller will be responsible for any damage or loss of action to anyone, of any kind, in any manner, therefrom. For binding mistakes, misprints or for missing pages, etc. the publisher's liability is limited to replacement within one month of purchase by similar edition. All expenses in this connection are to be borne by the purchaser.

I.S.B.N.

: 978-81-8283-721-8

Book Code No. : 241-34

Price

:

495.00 Only

Satyendra Rastogi "Mitre" for KRISHNA Prakashan Media (P) Ltd. 11, Shivaji Road, Meerut — 250 001 (U.P.) India Phones: 91.121.2644766, 2642946, 4026111/12; Fax: 91.121.2645855 Website: www. krishnaprakashan.com E-mail: [email protected] Chitf Editor : Sugam Rastogi Typesetting : Krishna Graphic Arts, Meerut (S.K.) Printed at : Vimal Offset Printers, Meerut.

Published by :

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Preface to the Present Edition In this edition, the book has been revised thoroughly according to the latest syllabus proposed by UGC for all Indian Universities. The book covers the syllabi of all Engineering and Professional institutes. We hope that this revised edition will be of great help for whom the book is written. For further improvement, the author will welcome all sorts of criticisms and comments from all readers. —R. K. Gupta

Preface to the First Edition The book entitled 'Operations Research', has been written to meet the requirements of Honours and Post-Graduate students of mathematics, statistics, commerce, management and engineering of all Universities on this subject. The book covers almost the entire syllabi of various Universities. Each topic in the book has been treated in an easy and lucid style without sacrificing any rigor. The language is simple and easily understandable. The book contains large number of important and interesting worked out examples. A set of exercises is given at the end of each chapter so that the students may have enough practice in the subject. Up-to-data references from question papers of various Universities are also given in the book. No claim to originality can be made but the treatment of the subject is in our own style. Though I have taken great care in eliminating the misprints, but if there are still any, I shall feel highly obliged to those who will take trouble of pointing them out. Suggestions for the improvement of the book will be gratefully acknowledged. The author is grateful to his publishers and printers for their full co-operation in bringing out the book in the present nice form. —R. K. Gupta 10, Ganga Nagar, Railway Road, Hapur

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Brief Contents Dedication

(iii)

Preface

(iv)

Brief Contents

(1)

Detailed Contents

(vi-xvi)

S.N.

Pages

Chapters

1.

Introduction to Operations Research

2.

Mathematical Preliminaries

3.

Inventory Theory

4.

Replacement Problems

115-160

5.

Waiting Line or Queuing Theory

161-234

6.

Allocation (General linear prog. problems)

235-276

7.

Convex Sets and Their Properties

277-292

8.

Simplex14-,thod

293-370

9.

Duality in Linear Programming

371-398

1-10 I 1-40 41-114

10. Sensitivity Analysis

399-434

11. Parametric Linear Programming

435-456

12. Integer Programming

457-488

13. Assignment Problem

489-538

14. Transportation Problem

539-602

15. Sequencing (Including Travelling Salesman Problem)

603-630

16. Dynamic Programming

631-658

17. Game Theory (Competitive Strategies)

659-724

18. Goal Programming

725-750

19. Network Analysis (PERT/CPM)

751-800

20. Information Theory

801-820

21. Non-Linear Programming

821-856

Index

857-862

I

•

•

Detailed Contents Pages

S.No. Chapters 1.

1.2

Nature and Definition of OR

1

1.3

Objective of OR

1.4

Phases of OR Method

1.5

Areas of Applications (Scope) of OR

1.6

Operations Research and Decision-Making

1.7

Scientific Method in OR

1.8

Characteristics of Operations Research

2 2

1.9

Modeling in OR

6

1.10

Types of Models

7

3

5 5

1.1 1 General Methods of Solution for OR Models 10 Exercise on Chapter 1 • 2.

1-10

Introduction to Operations Research Introduction (The Origin and the Development of OR) 1.1

8

- ID

Mathematical Preliminaries 2.1 Introduction

11

I. Elementary Probability Theory 2.2

Sample Space

2.3 Events

1I

11

11 12

2.4

Algebra of Events

2.5

Classical Definition of Probability

12

2.6

Odds in Favour and Odds Against

13

2.7

The Statistkal (or Empirical) Definition of Probability

2.8

Axiomatic Definition of Probability

13

2.9

Natural Assignment of Probabilities

14

13

2.10 Theorem of Total Probability or Additional Theorem of Probability 14

2.11

Compound Events

2.12

Independent and Dependent Events

2.13

Conditional Probability

2.14

Multiplication Theorem of Probability

15

15 15

14

2.15 I&ndom Variable ' 16 16 2.16 Discrete Probability Distributions 2.17 Expectation of a Random Variable 16 2.18 Special Discrete Probability Distributions 20 22 2.19 Continuous Probability Distributions 2.20 Special Continuous Probability Distributions 22 2. Matrices and Determinants 28 2.21 Definitions 28 2.22 Operations of Matrix Addition and Multiplication 28 2.23 Sub-Matrix 29 2.24 Minor of Order k 29 2.25 Determinant 30 2.26 Important Properties of Determinants 30 2.27 Minors 31 2.28 Cofactors 31 2.29 Rank of a Matrix 31 2.30 Adjoint of a Matrix 31 2.31 Singular and Non-Singular Matrices 32 2.32 Inverse of a Matrix 32 32 3. Vectors and Vector Spaces 2.33 Definitions 32 34 2.34 Euclidean Space 34 2.35 Linear Dependence and Independence of Vectors 34 2.36 Linear Combination (L.C.) of Vectors 34 2.37 Spanning Set 2.38 Basis Set 34 2.39 Some Useful Theorems of Linear Algebra 35 35 4. Simultaneous Linear Equations 2.40 Simultaneous Linear Equations 35 36 5. Finite Difference 36 2.41 First Difference off (x) 2.42 Second Difference of f(x) 37 37 2.43 Conditions for a Maximum or Minimum of f(x) 6. Differentiation of Integrals 37 38 7. Generating Functions 40 ♦ Exercise on Chapter 2

(WO

. Inventory Theory

41-114

41 3.1 Inventory 41 3.2 Variables in Inventory Problems 43 3.3 Need of Inventory 43 3.4 Inventory Problems 43 3.5 Advantages and Disadvantages of Inventory 43 3.6 Classification or Categories of Inventory Models 43 3.7 Some General Notations Used in Inventory Models 44 Deterministic Models 44 3.8 Economic Lot Size Models 3.9 Model I : Economic Lot Size Model with Uniform Rate of Demand 44 Infinite Production Rate and having no Shortages 3.10 Another form of Model I 46 3.11 Model II : Economic Lot-size Model with Different Rates of Demand in Different Production Cycles, Infinite Production Rate and having 47 no Shortages 3.12 Model III : Economic Lot-size Model with Uniform Rate of Demand, Finite Rate of Replenishment having no Shortages 48 Deterministic Models With Shortages 50 3.13 Model IV : Fixed Time Model 50 3.14 Model V : Economic Lot-size Model with Uniform Rate of Demand, Infinite Rate of Production and Having Shortages which are to be Fulfilled. 52 3.15 Model VI : Economic Lot-size Model with Uniform Rate of Demand, Finite Rate of Production and having Shortages which are to be Fulfilled 55 3.16 Multi Item, Deterministic Models with One Constant 59 77 3.17 Probabilistic Models 3.18 Model VII : Single Period Model with Discontinuous or Instantaneous Demand and Time Independent Costs (No Set up Cost Model) 77 3.19 Model VIII : Single Period Model with Uniform Demand. (No Set up Cost Model) 80 3.20 Model IX : The General Single Period Model of Profit Maximization with Time Independent Cost 85 87 3.21 Model X : Probabilistic Order Level System with Lead-Time 100 3.22 Purchase Inventory Models with Price Breaks 100 3.23 Model XI : Purchase Inventory Model 101 3.24 Model XII : Purchase Inventory Model with One Price Break

!3.25 Model XIII : Purchase Inventory MIode with Two Price Breaks 3.26 Model XIV : Purchase Inventory Model with Multiple Price Breaks • Exercise on Chapter 3 112

103 104

4.

Replacement Problems 1 1 5- 1 60 4.1 Introduction 115 4.2 Replacement of Major or Capital Item (Equipment) that Deteriorates with Time 115 4.3 To find the Best Replacement Age (Time) of a Machine when (i) Its Maintenance Cost is given by a Function Increasing with Time (ii) Its Scrap Value is Constant and (iii) The Money Value is not Considered 115 4.4 Few important Terms 124 4.5 To Determine the Best Replacement Age of Items whose Maintenance Costs Increase with Time and the Value of Money also Changes with Time 125 4.6 A Discounted Cost P(n) is Invested by taking Loan at the Interest Rate r; and the Loan is Repaid by Fixed Annual Payments say x, throughout the Life of the Machine. To Find the Minimum Value of x for Optimum Period n at which to Replace the Machine 128 4.7 Replacement of Items in Anticipation of Complete Failure the Probability of which increases with Time 137 4.8 To determine the Interval of Optimum Replacement 138 4.9 Problems in Mortality 139 4.10 Staffing Problem 142 4.11 Mortality Tables 142 • Exercise on Chapter 4 153

5.

Waiting Line or Queuing Theory 5.1 Introduction 161 5.2 Basic Queuing Process (system) and Its Characteristics 5.3 Customers Behaviour in a Queue 162 5.4 Important Definitions in Queuing Problem 163 5.5 The State of the System 163 5.6 Poisson Process 164 5.7 Poisson Arrivals 165 5.8 Theorem 166

161-234 161

168 5.9 Some Distributions 169 5.10 An Important Theorem 5.11 Notations 171 172 5.12 Classification of Queuing Models 5.13 Solution of Queue Models 173 5.14 Model I (M/M/1) : (oo/FCFS) (Birth and Death Model) 181 5.15 Relationship between 181 5.16 Model H (General Erlang Queuing Model) 196 5.17 Model III : (M/M/1) : (N/FCFS) 199 5.18 Model IV (M/M/S) : (./FCFS) 210 5.19 Model V : (M/E k /1) : (./FCFS) 217 5.20 Model VI (M/Ek /1) : (1/FCFS) 5.21 Machine Repair Problem 223 223 5.22 Model VII (M /M /R) : (k/GD), k < R 226 5.23 Model VIII Power Supply Model

♦ Exercise on Chapter 5

173

229

6. Allocation (General Linear Prog. Problems) .......

2. 5 - 2 7 6

235 6.1 Introduction 235 6.2 General Linear Programming Problems 237 6.3 Mathematical Formulation of a L.P.P. 242 6.4 Basic Solution (B.S.) 242 6.5 An Important Theorem 243 6.6 Some Important Definitions 247 6.7 Solution of a Linear Programming Problem 6.8 Geometrical (or Graphical) Method for the Solution of a Linear Programming. Problem 247 261 6.9 Analytic Method (Trial and Error Method) 6.10 Slack and Surplus Variables 262 266 6.11 Applications of Linear Programming Techniques 266 6.12 Advantages of Linear Programming Techniques 267 6.13 Limitations of Linear Programming

♦ Exercise on Chapter 6

268

-0.-- 277-292

7. Convex Sets and Their Properties 7.1 Introduction 277 7.2 7.3

Some Important Definitions Some important Theorems

♦ Exercise on Chapter 7

277 280 292

8. Simplex Method 8.1

Introduction : Simp ex Method

8.2

Some Definitions and Notations

8.3

Fundamental Theorem of Linear Programming

8.4

To obtain B.F.S. from a F.S.

8.5

To Determine Improved B.F.S.

8.6

Unbounded Solutions

8.7

Optimality Conditions

8.8

Alternative Optimal Solutions

8.9

Inconsistency and Redundancy in Constraint Equations

293 293 295

299 303

307 308 310

8.10 To Determine Starting B.F.S.

311

312

8.11 Computational Procedure of the Simplex Method for the Solution of a Maximization L.P.P.

315

8.12 Artificial Variables Technique

326•

8.13 Degeneracy in Simplex Method

334

8.14 Conditions for the Occurrence of Degeneracy in a L.P.P.

334

8.15 Computational Procedure to Resolve Degeneracy by Carner's Perturbation Method

334

8.16 Some Important Tips for Simplex Method

342

8.17 Linear Programming Problem (Special Cases)

343

8.18 Solution of System of Simultaneous Linear Equations by Simplex Method

356

8.19 Inverse of a Matrix by Simplex Method

♦ Exercise on Chapter 8

358

364

9. Duality in Linear Programming ................. —0...

...... 371-398

9.1

Duality in L.P. Problem

9.2

Standard Form of the Primal

9.3

Symmetric Dual Problem

9.4

Unsymmetric Dual Problem

372

9.5

The Dual of a Mixed System

372

9.6

Duality Theorems

9.7

Correspondence between Primal and Dual

9.8

To Read the Solution to the Dual from the Final Simplex Table

371 371

37 1

377

of the Primal and Vice-Versa

380

380

i -!----

9.9

1 !

A

Dual Simplek Method

388

9.10 Advantage of Dual Simplex Algorithm

388

9.11 Computational Procedure of the Dual Simplex Algorithm

•

Exercise on Chapter 9

397

399- 434

10. Sensitivity Analysis 399

10.1 Introduction

10.2 Variation in the Price Vector c

399

10.3 Variation in the Requirement Vector b

407

10.4 Variation in the Element ay of the Coefficient Matrix A 10.6 Addition of a New Constraint to the Problem

Exercise on Chapter 10

427

430

435-456

11. Parametric Linear Programming 11.1 Introduction

435

11.2 Linear Variations in c

435

11.3 Linear Variation in b

447

•

413

424

10.5 Addition of a New Variable to the Problem

•

389

Exercise on Chapter 11

454

12. Integer Programming ............... ..................... ............... 457-488 12.1 Introduction

457

457

12.2 Importance (or need) of I.P.P. 12.3 Solution of I.P.P.

458 458

12.4 Gomory's all I.P. Method

12.5 Construction of Gomory's Constraint and Gomory's Cutting Plane 461

12.6 All-Integer Cutting Plane Algorithm

474

12.7 Mixed-Integer Cutting Plane Algorithm 482

12.8 The Branch-and-Bound Technique 12.9 Branch-and-Bound Algorithm

•

483

Miscellaneous Exercise on Chapter 12

489

13.2 Assignment Problem 13.3 Important Theorem

488

489-538

13. Assignment Problem 13.1 Introduction

458

489 490

--I 3.4

509

13.5 Unbalanced Assignment Problems

513

13.6 Maximization Assignment Problem 520

13.7 Restrictions on Assignment

•

492

Hunnrian Method (Reduced Matrix Method)

Exercise on Chapter 13

522

14. Transportation Problem

539-602

539

14.1 Introduction

14.2 Transportation Problem

539 542

14.3 Difference between a Transportation and an Assignment Problem 14.4 Few Important Definitions

542

14.5 Some Important Theorems

543 545

14.6 Solution of a Transportation Problem

545

14.7 To Find an Initial Feasible Solution 550

14.8 Optimality Test 14.9 Theorem

551 553

14.10 Computational Procedure of Optimality Test

14.11 Transportation Algorithm or MODI (Modified Distribution) Method 565

14.13 Unbalanced Transportation Problems 576

14.14 Profit Maximization Problems 14.15 Prohibited Transportation Route

•

Exercise on Chapter 14

584

585

15. Sequencing (Including Travelling Salesman Problem) 15.1 Introduction

603-630

603 603

15.2 A Sequencing Problem 15.3 General Assumptions

604

15.4 Sequencing Decision Problem for n-jobs on two Machines

604

15.5 Sequencing Decision Problem for n-Jobs on Three Machines

605

15.6 Sequencing Decision Problem for n Jobs on m Machines 15.7 Processing Two Jobs Through m Machines 15.8 Graphical Method

Exercise on Chapter 15

612

612

15.9 Travelling Salesman (or Routing) Problem

•

553

565

14.12 Degeneracy in Transportation Problems

618

614

605

1

63 1 -658

16. DynaMic Programming 631 16.1 Introduction 16.2

63/

Bellman's Principle of Optimality in Dynamic Programming 631

16.3 Multistage Decision Problem

632

16.4 Characteristics of Dynamic Programming Problems

16.5 Solution of a Multi-stage Problem by Dynamic Programming with 632

Finite Number of Stages

16.6 Solution of Linear Programming Problem as a Dynamic 645

Programming Problem

16.7 Solution of an Inventory Problem as a Dynamic Programming Problem ♦

649

Exercise on Chapter 16

655

639-724

17. Game Theory (Competitive Strategies) 659

17.1 Introduction

659

17.2 Competitive Games

660

17.3 Finite and Infinite Games 660

17.4 Zero Sum Game

660

17.5 Two Person Zero Sum (or Rectangular) Games 660

17.6 Pay-off Matrix 17.7

Strategy

662 662

17.8 Solution of a Game

663

17.9 Maximin and Minimax Criterion of Optimality 17.10 Saddle Point

664 665

17.11 Solution of a Rectangular Game with Saddle Point

668

17.12 Solution of a Rectangular Game in terms of Mixed Strategies 670

17.13 Important Properties of Optimal Mixed Strategies

671

17.14 Solution of 2 x 2 Games Without Saddle Point 674

17.15 Dominance Property

17.16 Arithmetic Method (or the Method of Oddments or the Short Cut Method) for the Solution of 2 x 2 Game without Saddle Point 17.17 Graphical Method for the Solution of (2 x n) and (m x 2) Games 17.18 Algebraic Method for the Solution of a General Game 17.19 Iterative Method for Approximate Solution

682 684

692

697

17.20 Equivalence of the Rectangular Matrix Game and Linear Programming

700

17.21 Fundamental Theorem of Game Theory (Minimax Theorem)

702

17.22 Solution of a Redtangular Game by Simplex Method 17.23 Matrix Method for n x n (i.e, Square) Games

703

706

17.24 Summary of Methods for Solving the Rectangular

708

(Two Person Zero Sum) Games

709

17.25 Minimax and Maximin of a Function of Several Variables 17.26 Saddle Point of a Function of Several Variables

710

17.27 Necessary and Sufficient Condition for the Function E(x, y) to

710

Possess a Saddle Point (Existence of Saddle Point)

♦

Exercise on Chapter 17

18. Goal Programming 18.1 Introduction

712 725-750

725 725

18.2 Concept of Goal Programming

725

18.3 Formulation of a G.P. Problem as a L.P. Problem 18.4 Multiple Goals with Priorities and Weights 18.5 Solution of a Linear G.P. Problem

727

731

18.6 Graphical Method for the Solution of a Linear G.P. Problem

731

18.7 Simplex Method (Modified) for the Solution of a Linear G.P.

♦

Problem 734 Exercise on Chapter 18

746

19. Network Analysis (PERT/CPM) 19.1 Introduction 751 19.2 Theory of Graphs 751 751 19.3 Few Important Definitions 19.4 Network 752 19.5 Schedule Chart (Gantt Bar Chart) 752 19.6 Difference between CPM and PERT 753 19.7 Network Components 754 19.8 Construction of the Network Diagram 755 19.9 Dummy Activity 755 19.10 Errors in Drawing a Network 756 19.11 General Procedure for the Construction of a Network Diagram 19.12 Numbering the Events (Fulkeron's Rule) 758 19.13 CPM Computation (or Analysis) 768 19.14 The Float and Slack 772 19.15 Inequality relation between Floats of an Activity 773 19.16 Event Slacks 773 19.17 Critical, Event, Activity and Path 774

751- 800

757

-P 775 19.18 Procedure of determination of Critical Path 789 19. I 9 PERT 19.20 Estimate of Probability of Completing the Project by 789 Scheduled Time 796 Exercise on Chapter 19 • 20. Information Theory 801 20.1 Introduction 801 20.2 Communication Process 801 20.3 Description of a Communication System 802 20.4 A Quantitative Measure of Information 803 20.5 A Binary Unit of Information 804 20.6 Measure of uncertainty or Entropy 20.7 Properties of Average Measure of Uncertainty or Entropy 807 20.8 Important Relations for Various Entropies 812 20.9 Encoding 812 20.10 Few important Definitions 813 20.11 Unique Decipherability 20.12 Uses of Encoding; Efficiency and Redundancy 813 814 20.13 Shannon -Fano Encoding Procedure 20.14 Necessary and Sufficient Condition for Noiseless Coding Exercise on Chapter 20 820 • 21. Non-Linear Programming

Oil -8M))

804

816

821- 856

821 21.1 Introduction 21.2 General Non-linear Programming Problem 821 822 21.3 Mathematical Formulation of GNLPP 21.4 Solution of NLPP with ail Equally Constraints 825 21.5 Sufficient Conditions for Maximum or Minimum of the 828 Objective Function 837

21.6 Solution of NLPP when Constraints are Not all Equality Constraints 21.7 Kuhn-Tucker Necessary Conditions for the Optimality of the 837 Objective Function in a GNLP Problem 21.8 Kuhn-Tucker Sufficient Conditions for the Optimality of the Objective Function of a GNLPP with inequality Constraints 846 21.9 Graphical Solution 850 Exercise on Chapter 21 • Tables • Index

839

855-856 862

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Introduction to Operations Research

1.1 Introduction (The Origin and the Development of OR) The roots of operations research (abbreviated to OR) can be traced back many decades when early attempts were made to use a scientific approach in technical problems and in the management of organisations. During World War II, Britain was having very limited military resources, therefore there was an urgent need to allocate resources to the various military operations, and to the activities within each operation in an effective manner. Therefore, the British military executives and managers called upon a team of scientists to apply a scientific approach to study the strategic and tactical problems related to air and land defence of the country. Because the team was dealing with research on military operations, the work of this team of scientists was names as OR in Britain. Their efforts were instrumental in winning the "Air Battle of Britain", "Battle of the North Atlantic", etc. The success of this team of scientists in Britain encouraged United States, Canada and France to start with such teams. The work of this team was given various names in United States : operational analysis, operations evaluation, operations research, system analysis etc. The apparent success of OR in the military, attracted the attention of industrial management in the new field. In this way, OR began to creep into industry, business and governmental organisations. After the war, many of the scientists were motivated to pursue research relevant to these fields. The first technique in the field, called the Simplex Method for solving linear-programming problem, was developed by American mathematician George Dantzig in 1947. Since then many techniques called as the tools of OR, linear programming, dynamical-programming, inventory theory-queuing theory etc., are developed, Thus, the impact of OR can be felt in almost all walks of life.

1.2 Nature and Definition of OR [Rohilkhand 1995, 98, 2000; Meerut 2006, 08, 08 (B.P.), 09 (BP)]

What is operations research ? The answer to this question may be to give its definition. A definition should give a complete and clear picture of the subject matter of a discipline. As the name implies, operations research involves "research on operations". This says something about both the approach and the area of application. A number of definitions of OR have been formulated, but none of them is complete. Thus, it is important to study few definitions of OR. 1. OR is the application of the theories of probability, linear programming, queuing theory etc., to the problems of war and industry. This definition explain only the various disciplines used in OR.

Operations Research

2

2. 3.

4. 5.

6.

(Aurther Clarke) OR is the art of winning wars without actually fighting. This definition restricted OR to warfare only. OR is a management activity pursued in two complementary ways-one half by the free and bold exercise of common sense untrammelled by any routine and other half by the application of a repertoire of well establised pre-created methods and techniques. (Jagjit Singh 1968) OR is an art of giving bad answers to problems where otherwise worse answers are given: (T.L. Saaty 1958) OR is the application of scientific methods, techniques and tools to problems involving the operations of system so as to provide those. in control of the operations with optimum solutions to the problem. (Churchmann, Acoff; Annoff, 1957) OR is applied decision theory. It uses any scientific, mathematical or logical means to attempt to cope with the problems that confront the executive when he tries to achieve a thorough-going rationality in dealing with the decision problems. (D. W. Millar and M.W. Starr)

1.3 Objective of OR The objective of OR is to provide a scientific basis to the decision makers for solving the problems involving the operations of system to give a solution which is in the best interest of the organisation. The solution is called the optimum solution to the problem.

1.4 Phases of OR Method

[Rohilkhand 1997] The various phases of an operations research study are as follows :

1. Formulation of the problem To find the solution of a problem, one must be able to find the problem and formulate it so that it is susceptible to research. Formulation of a problem for research consists in identifying, defining and specifying the measure of the components of a decision model. For the formulation of a problem, the following information are needed : (i) Who will make the decision ? (ii) What are the objectives ? (iii) Controllable variables and their ranges. (iv) The uncontrolled variables that may effect the outcomes of the available choices. Since it is difficult to extract a right answer from the wrong problem, therefore, this phase should be executed with considerable care.

2. Construction of a mathematical model to represent the system under study After formulating the problem, the next phase is to reformulate this problem into a form that is convenient for analysis. The most convenient approach for this is to construct a mathematical model that represent the system under study. There are three types of models that are commonly used in OR and other sciences. They are (i) iconic, (ii) analogue and (iii) symbolic. (For detail see 1.10)

Introduction to Operations Research

3

3. Deriving a solution from the model After formulating the mathematical model for the problem under consideration, the next phase is to derive a solution from this model. Here in operations research we are always in the search for an optimal solution. A optimal solution-is one which maximise or minimise the objective function in the model in the best interest of the problem under consideration. Many procedures have been developed and will be discussed in detail. (See 1.11).

4. Testing the model and the solution derived from it When the model is complete, it should be tested again as a whole for obvious or oversight errors. This may be done by re-examining the formulation of the problem and comparing it with the model that may help to reveal any mistakes. The solution of an OR problem is supposed to yield better performance than that by some alternative procedure. A solution derived from the model of the problem should be further tested for its optimality.

5. Implementing and maintaining the solution The final phase of an operations research study is to implement the optimum solution derived by the OR team. As the conditions are constantly changing in the world, the model and the solution may not remain valid for a long time. Therefore, as the change occurs, it is to be detected as soon as possible so that the model, its solution and the resulting course of action can be modified accordingly.

1.5 Areas of Applications (Scope) of OR [Meerut 2007 (B.P.), 08, 09 (B.P.), 2010; Rohilkhand 1998, 2000]

OR has got wide scope. In general we can say that wherever there is a problem, there is OR for help. In addition to the military, operations research is widely used in many organisations including business and industry. Now we shall discuss the scope of OR in various important fields.

1. In Defence During World War II OR teams of Britain and America developed the techniques and strategies that helped them to win "Air Battle of Britain", "Island Campaign in the Pacific", "Battle of the North Atlantic" etc. In modern time war, the military operations are carried out by Airforce, Army and Navy. Therefore, there is a necessity to formulate optimum strategies that may give maximum benefit. OR helps the military executives and managers to select the best strategy (courses of action) to win the battle. Thus, OR has got great scope in defence.

2. In Industry Seeing the success of OR in the military, industry_became interested in this new field. As companies expanded, it became less and less possible for one man to manage them, therefore the job of one man was divided into parts and assigned to others. Thus, the company executives established different departments e.g. (i) Production Department : To minimize the cost of production. OR is useful to Production Specialist in (a) designing and selecting sites, (b) scheduling and sequencing the production run by proper allocation of machines and

Operations Research (c) calculating the optimum product mix. (ii) Marketing Department : To maximize the amount sold and to minimize the

cost of sales. OR is useful to the Marketing Managers in (a) determining, when to buy, how often to buy and what to buy to minimize the total costs, (b) calculating the miryimum sale price per unit and (c) knowing the customer's choice relating to colour, packing and size etc., for various products. (iii) Financial Department : To minimize the capital required to maintain any level of business. OR is useful to Financial Controller in (a) finding out the long term capital requirements (b) finding out a profit plan for the company and (c) determining the optimum replacement policies. These various departments came in conflict with each other as the policy of one department was against the policy of the other. The difficulty was solved by the application of OR techniques. Thus, OR has got great scope in industry. Many Indian industries including Delhi Cloth Mill, Indian Airlines, Indian Railways, Hindustan Lever, Defence Organisations, Fertilizer Corporation of India, Tata Iron and Steel Co., etc. have made wide spread use of operations research techniques to make optimum use of their limited resources to maximize their profit. 3.

In L.I.C.

OR techniques are also applicable to enable L.I.C. officers to decide the premium rates of various policies for the best interest of the corporation. 4.

In Agriculture

With the increase of population and consequent shortage of food, there is a need to increase agriculture output for a country. But there are many problems faced by the agriculture departments of a country e.g., (i) climatic conditions, (ii) problem of optimal distribution of water from the resources, etc. Thus, there is a need of best policy under the given restrictions. OR techniques may be fruitful to determine the best policies. Thus, OR has got great scope in agriculture. 5.

In Planning

Careful planning plays an important role for the economic development of a country. OR techniques may be fruitful for such planrying7. Planning commission made use of Operations Research Techniques for planning the optimum size of the caravelle fleet of Indian Airlines. Thus, OR has got great scope in planning also. In this way we can say that OR has got great scope almost in every walk of life.

1.6 Operations Research and Decision-Making [Meerut 2007 (B.P), 2010]

We daily take decisions about many major and minor issues/problems and a wrong decision may have an adverse effect. The management/decision-maker of an organisation have to make decisions most of the times in the best interest of the organisation. With the advancement of science and technology, the decision-making in business and industry have become extrernely difficult as the decision makers are

5

Introduction to Operations Research

faced with numerous variables and actions of various competitors over which he has no control. To overcome these difficulties the decision-maker remain in search of various methods and aids that may help him in decision making. The decisions by mere guess and experience are of no use for him. In these situations Operations Research comes to their help to make better decisions under uncertain future conditions. Operations Research may be regarded as a tool employed to increase the effectiveness of the decision maker. It enables the decision-maker to choose the best alternative from all available alternatives for some problem i.e. it helps the management/decision-maker to make better decisions. Operations Research which is mainly concerned with the techniques of applying scientific knowledge, provide an understanding which gives the decision-maker/ management new insights to take better decisions in his. decision making problem with great speed and confidence by working smartly but not by working hardly.

1.7 Scientific Method in OR The scientific method in OR study generally involve the following phases.

1. The Judgement Phase The judgement phase consists of the following : (i) The determination of the operation. (ii) The establishment of the objectives and the values related to the operation. (iii) The determination of suitable measures of effectiveness and (iv) The formulation of the problem relative to the objectives.

2. The Research Phase The research phase of scientific method is the largest and longest. The phase consists of the following : (i) Observations and data collection for a better understanding of the problem. (ii) Formulation of hypothesis and the models. (iii) Observation and experimentation to test the hypothesis on the bas& of other available data. (iv) Analysis of the available information. (v) Verification of the hypothesis using pre-established measures of effe'ctiveness. (vi) Predictions of various results from the hypothesis. (vii) Generalization of the result and consideration of alternative methods.

3. The Action Phase The action phase of scientific method in OR consists of making recommendation for decision process by any one in a position to make a decision influencing the operation in which the problem occurred.

1.8 Characteristics of Operations Research [Rohilkhand 1996, 2000, 2010]

The characteristics of OR are as follows :

1. OR is the inter-disciplinary team approach to find the optimum solution OR study is performed by a team of scientists rather than an individual. It has been recognised that scientists from different disciplines can produce better results,

Operations Research

6

than could be expected from the team of same number of scientists from the same discipline. Undoubtedly the scientists of different disciplines in a team will tackle the problem from different angles while the scientists of the same discipline cannot do so.

2.

OR emphasises on the overall approach to the system

The characteristics of OR is to consider all aspects of the problem under consideration. Often, it is found that the optimum decision from one aspect are very far from optimum decision from the other aspect. For example, in an industry the marketing department would like to have large inventory while the financial department will suggest for minimum possible inventory. Thus the decision will be taken by considering the problem from all aspects.

3.

OR tries to optimize the total output by maximizing the profit and minimizing the cost

The management of an industry will apply OR methods to maximize the profit and minimize the costs. Thus, OR tries to optimize the total output.

4.

OR uses scientific methods to arrive at an optimum solution Scientific methods are used in OR to solve a problem for optimum solution.

1.9 Modeling in OR

[Meerut 1998 (B.P.); Rohilkhand 2003] Models plays very important role in OR. They are representations of reality. Models provide distilled and economic descriptions and explanations of the operations of the system that they represent. By experimenting on them, we can determine how the changes in the relevant system will effect its performance. Models enables us-to experiment more effectively than on the system itself which is either impossible or too costly. A model in OR may be defined as an idealized representation of a real life system. ♦ Advantages of OR Models Models have many advantages over a verbal description of a problem. Some of them are as follows : (i) It describes a problem much more concisely. (ii) It provides some logical and systematic approach to the problem. (iii) It indicates the limitations and scope of the problem. (iv) It tends to make the over all structure of the problem more comprehensible. (v) It facilitates dealing with the problem in its entirety. (vi) It enables the use of high-powered mathematical techniques to analyse the problem. (vii) It helps in finding avenues for new research and improvements in a system. ♦ Disadvantages (or limitations) of OR Models Models have few disadvantages (limitations) also as follows : (i) Models are only an attempt in understanding an operation and should never be considered an absolute in any sense. (ii) The validity of any model with regard to the corresponding operation can only be verified by carrying on experiment and relevant data characteristics.

Introduction to Operations Research

7

Characteristics of a Good model

A model in OR should possess the following characteristics : (i) It should be capable of adjustments with new formulations without having any significant change in its frame. (ii) It should contain very few variables. (iii) A model should not take much time in its construction.

1.10 Types of Models

[Meerut 2006, 091

There are three types of models that are commonly used in OR as well as in most of sciences. 1. Iconic Models 2. Analogue Models 3. Symbolic Models 1. Iconic Models Iconic models represents the system as it is but in different size. Thus, Iconic models are obtained by enlarging or reducing the size of the system. In other words they are images. Some common examples are photographs, drawing, model aeroplanes, ships, engines, globes, maps etc. A toy aeroplanes is an iconic model of a real one. Iconic models of the sun and its planets are scaled down while the model of the atom is scaled up so as to make it visible to the naked eye. Iconic models have got some advantages as well as disadvantages as follows : ♦ Advantages (i) These are specific and concrete. (ii) These are easy to construct. (iii) These can be studied more easily than the system itself. ♦ Disadvantages (i) These are difficult to manipulate for experimental purposes. (ii) They cannot be used to study the changes in the operation of a system. (iii) It is not easy to make any modification or impiovement in these models. (iv) Adjustments with changing situations cannot be done in these models. 2. Analogue Models In analogue models one set of properties is used to represent another set of properties. After the problem is solved, the solution is re-interpreted

in terms of the original system. For example, contour lines on a map are analogues of elevation as they represent the rise and fall of heights. Graphs are analogues as distance is used to represent a wide variety of variables such as time, percentage, age, weight etc. ♦ Advantage Analogue models are easier to manipulate than iconic models. ♦ Disadvantage Analogue models are less specific and less concrete.

8

Operations Research

3. Mathematical (Symbolic) Models In symbolic models, letter, numbers, and other types of mathematical symbols are used to represent variables and the relationships between them. Thus, symbolic models are some kind of mathematical equations or inequalities reflecting the structure of the system they represent. Inventory models, queuing models etc., are the examples of symbolic models. ♦ Advantages (i) They are most abstract and most general. (ii) They are usually the easiest to manipulate experimentally. (iii) They usually yield more accurate results, under manipulation. Thus, in OR, symbolic models are used whenever possible.

1.11 General Methods of Solution for OR Models [Meerut 2009, 09 (B.P.); Rohilkhand 2003] Solving a model consists of finding the values of the controlled yariables that optimize the measure of performance, or of estimating them approximately. OR models are generally solved by the following three methods. 1. Analytic Methods In these methods all the tools of classical mathematics, such as differential calculus and finite differences are available for the solution of the model. Various inventory models are solved by the use of these analytic methods. 2. Numerical Methods Numerical methods concerns with the iterative or trial and error methods. Whenever the classical methods fail, we use iterative procedure. The classical methods may fail because of the complexity of the constraints or of the number of variables. In this procedure we start with a trial solution and a set of rules for improving it. The trial solution is improved by the given rules and is then replaced by this improved solution. This process of improvement is repeated until either no further improvement is possible or when the cost of further calculation cannot be justified. Iterative procedures are divided into three groups. (i) Here we know that each iteration will improve the solution and that after a finite number of repetitions no further improvement will be possible. (ii) Although successive iterations improve the solution, but here we are only guaranteed the solution as a limit of a infinite process. (iii) Here we include trial and error methods. We only know the successive trials tend to improve the result, without being sure of monotonic improvement. 3. Monte Carlo Technique Simulation The basis of Monte Carlo technique is random sampling of a variable's possible values. For this technique some random numbers are required which may be converted into random variates whose behaviour is known from past experience. Darker and Kac define Monte Carlo methods a combination of probability methods and sampling techniques providing solutions to complicated partial or integral differential equations. In short Monte Carlo technique is concerned with

Introduction to Operations Research

9

experiments on random numbers and it provide solutions to complicated OR

problems. Monte Carlo techniques are useful in the following situations. (i) Where one is dealing with a problem which have not yet arisen i.e., where it is not possible to gain any information from past experience. (ii) Where the mathematical and statistical problems are too complicated and some alternative methods are needed. (iii) To estimate parameters of a model. The main steps of Monte Carlo method are as follows : (i) To get the general idea of the system, a flow diagram is drawn. (ii) Then correct sample observations are taken to select some suitable model for the system. In this step some probability distribution for the variables of our interest is determined. (iii) Then the probability distribution is converted to a cumulative distribution function. (iv) Then a sequence of random numbers is selected with the help of random number tables. (v) Then a sequence of values of the variables of our interest is determined with the sequence of random numbers obtained in step (iv). (vi) Finally, some standard mathematical function is applied to the sequence of values obtained in step (v).

♦ Advantages of Monte Carlo Methods (i) These are helpful in finding solutions of complicated mathematical expressions which is not possible otherwise. (ii) By these methods difficulties of trial and error experimentation are avoided. ♦ Disadvantages of Monte Carlo Methods (i) These are costly way of getting a solution of any problem. (ii) These methods do not provide optimal answers to the problems. The answers are good only when the size of the sample is sufficiently large. •

Operations Research

10

+ Exercise on Chapter 1 + [Meerut 2006, UP TECH MBA 2002-03]

1.

What is Operations Research ?

2.

Explain the concept, scope and tools of OR as applicable to business and industry.

3.

Discuss the historical background of O.R. [UP TECH MBA 2002-03, 03-04, 05-06]

4. 5.

Discuss the various phases in solving an OR problem. Explain how and why operations research methods have been valuable in aiding executive decisions. Give a brief account of the methods used in model formulation. Discuss the three types of models with special emphasis on their important logical properties and the relationship they bear to each other. What is meant by a mathematical model of a real situation ? Discuss the importance of models in the solution of Operational Research problem.

6. 7. 8.

What are the advantages and disadvantages of Operations Research models ? Why is it necessary to test models and how would you go about testing models. 10. How can Operations Research model be classified ? Which is the best classification in terms of learning and understanding the fundamentals of OR ? 9.

11. Discuss the advantages and limitations of using results from a mathematical model to make decision about operations. 12. Define OR and discuss its scope. [Meerut 2007 (B.P), 08; UP TECH MBA 2003-04; Rohilkhand 2000] 13. Give the various general methods of solution for OR models. 14. Discuss scientific method in OR. 15. Discuss the scope of Operations Research. What is the role of Operations [Meerut 2007 (BP)] Research in Decision-Making. 16. What is Monto Carlo Simulation ? Explain this by an illustration. [UP TECH MBA 2004-05] 17. What is Operations Research ? State the areas of its applications. [UP TECH MBA 2006-07] 18. Outline broad features of the judgement phase and the research phase of scientific method in OR. Discuss fully any one of these phases. 19. Define OR. Give the main characteristics of OR. Also discuss the importance of OR in decision-making. 20. Write an essay on the scope, methodology and phases of Operational Research. 21. Explain the general nature of optimization problems and discuss their significance in relation ,to OR problems. Do all OR problems involve in optimization ? 22. Write an essay on the scope of Operations Research. 23. Explain the meaning and nature Of, .icdR. 24. Describe khe necessity of Operations Re\searc n industry. [UP TECH MBA 2003-04]

• ••

Mathematical Preliminaries

2.1 Introduction In this chapter we shall discuss some basic concepts and techniques of a number of mathematical topics that will facilitate the reader in the development and understanding of the further chapters of this book.

1. Elementary Probability Theory Before the definition of probability be given, it is necessary to understand the terms sample space, events etc., which are given here.

2.2 Sample Space Let el , e2, ...., en denote the outcomes of some particular experiment which are such that no two or more of them can occur simultaneously and exactly one of these must occur whenever the experiment is performed. The set S = {e 1 , e2,...., en } is called a sample space of an experiment satisfying the following conditions: (i) Each element of the set denotes a possible outcome of the experiment. and (ii) The outcome is one and only one element of the set whenever the experiment is performed. The sample space is called a finite sample space if the number of elements contained in the sample space S is finite and is called an infinite sample space if the number of elements contained in the sample space S is infinite.

2.3 Events The elements or the points of a sample space associated with an experiment are known as elementary events. Every subset of a sample space S of an experiment is called an event, generally denoted by E. ♦ Certain Event : The sample space S itself is called the certain event which is the subset of itself ♦ Impossible Event : An impossible event will be denoted by the empty subset I) of sample space S. ♦ Simple-Event : Any event that contains only one member of a sample space is called-a simple event in that space. „,„

Operations Research

12

♦ Equally Likely Events : Two events are considered equally likely if one of them cannot be expected in preference to the other. or

Two events are considered equally likely if both of them have equal chances to occur e.g., in the tossing of a fair coin, the two events, the head up or the tail up have the same chances of coming up. Thus these two events are equally likely events.

♦ Exhaustive Event : The set of all possible outcomes of an experiment is said to be an exhaustive event. ♦ Mutually Exclusive Events : Events are called mutually exclusive if the occurrence of any one of them excludes the occurrence of the others. e.g., in the tossing of a coin, the two events, the head up or the tail up are two mutually exclusive events; since both of them cannot come up together. If the head comes up, the tail is sure to remain downward i. e., the coming of head up prevents the coming of tail up and vice-versa.

2.4 Algebra of Events (i) Complementary of an Event : Let E be an event defined on a sample space S. Then the aggregate of all those sample points of S where E does not happen is known as complementary of an event E and is denoted by E e. g. , in the throw of a coin the sample space consists of two events, the head up (H) or the tail up (T). Obviously T is complementary event of H and vice-versa.

Fig. 2.1

(ii) Union of two Events : The set of all those points of S (the sample space), where either the event El happens, or the event E'2 happens or both of them happen together is known as the union of two events E1 and E 2 and is denoted by Ei u E 2 .

(iii) Intersection of two Events : The set of all those points of S (the sample space) where both the events El and E 2 happen together is known as the intersection of two events E1 and E 2 and is denoted by E1 n E 2 . e.g., in the throw of two coins the sample space consists of four points HH, TH, HT and TT. The set of all these points is the union and the set of points TH and HT is the intersection of the two events H and T.

Fig. 2.2

Fig. 2.3

2.5 Classical Definition of Probability If an event E can happen in m ways out of total of n possible equally likely ways, the probability of occurrence of the event E is denoted by

Mathematical Preliminaries

13

p = Pr {E} = and the probability of non-occurrence of the event E i.e., the probability of failure of the event E is denoted by n—m q = Pr {not — n

clearly

q=

n—m

p+q=1

n

—1

n

—1 p

i.e., Pr (E) + Pr (not E) = 1

2.6 Odds in Favour and Odds Against If p is the probability of occurrence of an event and q the probability of non-occurrence of the event, the odds in favour of the happening are p : q and the odds against its happening are q : p.

2.7 The Statistical (or Empirical) Definition of Probability The empirical probability of an event is taken as the relative frequency of occurrence of the event when the number of observations is very large. i.e., if the trials be repeated N times when N is large, under essentially the identical conditions, and an event E happens on pN occasions, then, pN Probability of the event E = lim — = p n -00 N This is more general definition. The statistical definition, although useful in practice, has difficulties from a mathematical point of view, since an actual limiting number may not really exist. For this reason, modern probability theory has been developed axiomatically in which probability is an undefined term.

2.8 Axiomatic Definition of Probability (Acceptable Assignment of Probabilities of Elementary Events) Let S be a sample space consisting of all n elementary Events el , e2,..., en . i. e.,

S=

e2,..., en 1.

To each elementary event ei e S, assign a real number called the probability of the elementary event el denoted by P (e1 ) such that

(i) The probability of each elementary event is a non-negative real number i.e., P(ei ) ?_ 0 i = 1,Z n and (ii) The sum of the probabilities assigned to all elementary events of the sample space is 1

i.e.,

P(e1 ) + P(e 2) +.... + P(en ) = 1

Such an assignment of real numbers to the elementary events of the sample space is called an acceptable assignment of probabilities.

Operations Research

14

♦ Important Results 1. Probability of an impossible event is zero : We define impossible event by the null set 4) on a sample space P (4)) = 0.

2. If E is complementary to E then P (E) = 1 — P (E) : Proof : Here the sample space S = {E, El P(E)+P(E)=1

P (E)=1— P (E)

2.9 Natural Assignment of Probabilities An assignment of probabilities is called natural (or equiprobable) assignment if each elementary event of a sample space S is assigned the same probability. Thus, if

S

then

P(ei )+ P(e 2 )+....+P(en )=1 P(e 1 )=P(e2 )=

=P(en ) =1 / n if P is natural assignment of probability.

2.10 Theorem of Total Probability or Additional Theorem of Probability If11'112, p3,..., pn are the probabilities of the happenings of n mutually exclusive events, and p the probability of the happening of any one of them, then P = +P2+••••+Pn or in other words. "The probability of the happening of any one of then mutually exclusive events is the sum of their separate probabilities".

Proof : If pi, p2,..., pn are the probabilities of the occurrence of n events El , E 2, ..., E n , respectively, and N is the total number of trials, then these events will occur p1N,p2N,...,pn N times respectively. Since the events are mutually exclusive, hence any one of them can happen on pi N + p2N + ....+ p n N occasions. The probability that one of the events happen pi N + p2N +....+ pn N N or

P Pi +P2+••••+Pn

2.11 Compound Events If two or more events occur simultaneously they are said to be compound

events. e.g., if two or more persons simultaneously draw one card each from a pack of cards, these events are compound events.

15

Mathematical Preliminaries

2.12 Independent and Dependent Events The compound events are of two types : (i) Independent Events : If the happening of one event does not effect the happening of the other events, these events are called independent events. e.g., In the throw of two dice, the coming up of 3 on one die does not effect the coming up of any number 1, 2, 3, 4, 5, 6, on the second die. Hence, these events are independent events. (ii) Dependent Events : If the happening of one event effects the happening of the other event, these events are called dependent events. e.g., If in the throw of two dice we require the sum of the two numbers on the upper faces to be 7 then if 3 comes up on the first dice then the face with number 4 must come up on the second dice. Thus the happening of the first dice effects the happening of the other dice. Hence, these events are dependent events.

2.13 Conditional Probability If El. and E2 are two events, the probability that E 2 occurs given that El has occurred is denoted by Pr {E 2 / El } or Pr {E 2 given El l and is called the conditional probability of E2 given that El has occurred. Thus, if we denote by E1E2 the event that both E1 and E2 occur simultaneously, then

Pr {EiE 2 } = Pr

. Pr {E 2 /

and for three compound events Ei, E 2, E 3 ; we have

Pr fE1E 2E 31 = Pr {El } Pr{E2 /Ei l Pr (E 3 /E1E2} i.e., the probability of occurrence of E1,E 2 and E3 is equal to the product of probability of El, probability of E 2 given that El has occurred and the probability of E 3 given that both E1 and E2 have occurred. In general, Pr {El E 2 E 3...En }=Pr IEOPr IE 2 /EOPi. {E 3 /E1E 2 }...Pr fE n /Ei E 2 ...En _i l Particular Case : In case of independent events 4{E2 /El} =Pr {E 2 } 4{E 3 / Ei E 2}=pr (E 3 ) Thus, in general, Pr {Ei E 2 ...En } =13,* {E l }Pr {E } {E3}

P,. {E n }.

Thus, we have the following theorem.

2.14 Multiplication Theorem of Probability (For Independent events) If pi , p2, p 3.... , p n be the probabilities of n independent events El, E then the probability of occurrence of El , E 2 , ,E n simultaneously is pl p 2

, En pn

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16

2.15 Random Variable To every point of the sample space S we can assign some real numbers known as random variable with the following properties : (i) The random variable takes some definite value at each sample point of the sample space. (ii) To each value of the random variable some probability is attached.

Thus, random variable is a numerical valued variable X defined . on a sample space. e.g., in the random experiment of finding the sum of the numbers on the upper faces in the throw of two fair dice, the random variable X will have numerical values from 2 to 12 and the sample space will consist of 36 points each having a probability equal to 1/36. A random variable is also known as chance variable or stochastic

variable. A random variable can be discrete as well as continuous.

Discrete Random Variable : Discrete random variables are those which take on a finite denumerable set of values. A denumerable set of values is a set whose elements can be put into one-to-one correspondence with the set of positive integers.

Continuous Random Variables : Continuous random variables are defined as those random variables which take on a continuum of values i.e., which takes all possible values between its limits.

2.16 Discrete Probability Distributions If variable X can assume a discrete set of values Xi, X2, ..., Xk with respective probabilities p i, 192, pk where pi + p2 + + pk = 1, then a discrete probability distribution for X has been defined. The function p (X) which has the respective values pi, p2,..., pk for X = Xi, X2,....Xk is called the probability function or frequency function for X. This variable X which assume certain discrete values with given probabilities is a discrete random variable. The mean (denoted by X) and variance (denoted`, by var (X)) of the distribution are defined by X =E, Xi P(Xi ) and

=

Xi 2 P

(Xi )—

var (X) =

(xi _ )2 P(Xi )

E Xi P(Xi)i _ ( k ) 2I p(xi) _,E xi2 .p xi) (x)2

2.17 Expectation of a Random Variable For a discrete random variable X, the expected value denoted by E(X) is just the sum of the products of the possible values the random variable X takes on and their respective probabilities.

Mathematical Preliminaries

17

Thus, if x1, x2, ..., xn are the only possible discrete values of the random variable X with respective probabilities p1. p 2, , p„ then E(X)=pl x1 + p2x2 + p3x3 +....+pn xn or

E(X)=1 pi i=i

Similarly for a continuous random variable X, the expected value is given by E (X) = 11 x f (x) d x where f (x) is the probability density function.

.91frattative eicamptett Example 1 : From a bag containing 10 black and 5 white balls, a ball is drawn at random. What is the probability that it is white ? Solution : Total number of balls in the bag = 10+5 = 15 Total number of ways of drawing one ball from the bag = 15C1 15. Total number of white balls in the bag = 5. Total number of ways of drawing one white ball from the bag = 5 Ci = 5. Probability of drawing one white ball from the bag = 5/15 = 1/3. Example 2 : From an urn containing 4 balls of different colours i.e., red (R), blue (B), yellow (Y), and green (G) any two balls are drawn : (i) simultaneously, (ii) one after another with replacement. Define the sample space in both the cases. Solution : (i) From the 4 balls, 2 balls can be drawn in 4C 2 = 6 ways. All possible combinations of two balls of different colours will constitute the sample space. Sample space is { RB, RY RG, BY, BG, YG} (ii) Here the balls are drawn one after another after replacement. Therefore, the combination of the same colour balls is also possible. Sample space is {RR, RB, RY, RG, BB, BR, BY, BG, YR, YB, YY, YG, GR, GB, GY, GG}. Example 3 : Find the probability of obtaining a total of 6 in a single throw of two dice. Solution : Total number of ways in which the total of the numbers with two ace-can come out = 6x 6=36. Total sum 6, on the two dice can be obtained in the following 5 ways 1+5, 2+4, 3+3, 4+2, 5+1 Probability of getting sum 6 from a single throw of two dice = 5/36.

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18

Example 4 : A coin is tossed three times in succession. If E is the event that there are at least two heads, and F is the event that first throw gives a head find P(E/F). Solution : Here sample space is {HHH, HHT, HTH, THH, HTT, THT, 7TH, TTT} Thus, there are 8 sample points and probability of each sample point is 1/8. Now sample space of event E is { HHH, HHT, HTH, THH} and sample space of event F is { HHH, HHT, HTH; H'TT} P(E) = (1/8) + (1/8) + (1/8) + (1/8) = 1/2 The event E and F happen simultaneously (i. e. , E n F) at the following points HHH, HTH, HHT P(E n F)= (1/8) + (1/8) + (1/8)= 3/8 P (E n F) 3/ 8 3 . — — P (E/F) — P(F), 1/2 4

Example 5 : Find the probability distribution in the throw of two fair dice where the discrete random variable X denote the sum of the numbers on the top of the two dice. Solution : Since the total seven can be obtained in the following 6 ways 1+6, 2+5, 3+4, 4+3, 5+2, 6+1. probability of getting sum 7 is 6/36. Similarly, all other probabilities may be obtained. Hence, the probability distribution is given by the following table. X

2

3

4

5

6

7

8

9

10

11

12

p(X)

1 36

2 36

3 36

4 36

5 36

6 36

5 36

4 36

3 36

2 36

1 36

Example 6 : Find the probability of boys and girls in families with 3 children, assuming equal probability of boys and girls. Solution : Let B and G represent the two events boy in the family and girl in the family respectively. Here Pr {B} =Pr {G} =1/ 2 In families of three children the following mutually exclusive events can occur. (i) 3 boys i.e., (BBB) Pr {BBB} = Pr {B} Pr (B)Pr (B)

= (1/2). (1/2). (1/ 2)=1/8. (ii) 2 boys i.e., (BBG + BGB + GBB) Pr {BBG + BGB+ GBB} =Pr {BBG} + Pr {BGB) + r {GBB} = Pr {B} . Pr {B} {G} + Pr {B} . Pr {G} Pr {B} + Pr {G} Pr {B}. Pr {B}

= (1/2). (1/2). (1/2)+ (1/2). (1/2). (1/2)+(1/2). (1/2). (1/2)=3/8

Mathematical Preliminaries

19

(iii) 1 boy i.e., (BGG + GBG + GGB) Pr {BGG +GBG + GGB} =(1/8)+ (1/8)+ (1/8),---. 3/8. (iv) No boy i.e., all the three girls or (GGG) Pr {GGG} = Pr {G). .{G} {G} =(1/2).(1/2).(1/2)=1/8. Representing the random variable, showing the number of boys in families with 3 children by X, the probability distribution is given by the following table. Number of boys X

0

1

2

3

Prob. P (X)

1 8

3 8

3 8

1 8

Example 7 : During the course of a day, a machine turns out either 0, 1 or 2 defective pens with probability 1/6, 2/3 and 1/6 respectively. Calculate the mean value and the variance of the defective pens produced by the machine in a day. Solution : Here the probability distribution is given by the following table : X

0 1 6

Prob. P (X)

.1 2 3

2 1 6

The mean value (average number of defective pens) is given by E (X) = P1 X1 + p2X2 + p3X3 = (1/6).0+ (2/3).1+ (1/6). 2 = 1 We have Var (X) = E (X 2 )—[E(X)] 2 y2 v 2 p3,. v 32 ( v2 I-I T P2!). 2 )

Now

=(1/6).02 +(2/3).12 + (1/6). 22 =4/3 From (1) Var (X) = (4/3) — (1)2 = 1/3 Example 8 : Given the following probability function : x

0

1

2

3

P (x)

0

X

2X,

2X,

4 3x

(i) Find 2 r (ii) Evaluate P (x 5) and P (x < 4). Solution : (i) We know that, EP (x) =1 0+X+2X+27,+37, or

10A,2 + 921 ./4 —

0

2 +2X2 +(7k2 +X)=1 or

(10A, —1) (X +1)= 0

5 x2

6 2X2

7 77 2 + x

Operations Research

20 1 10 2k, -1, since P (x) is always positive.

k =__ — ,-1.

But (ii)

2k. =1 / 10

P(x. 5) = P (5) + P(6)+ P (7) = x2 + 2x2 +012

2 4)- = 10)62 + = 10. — 1

10

and

1=1 + 10

5

P(x < 4) = P(0) + P(1) + P(2) + P(3) = 0+ X.+ 2X +3X =6X=6(1/10) =3/5

2.18 Special Discrete Probability Distributions Here we shall describe some special discrete probability distributions which are of great importance and are used in Operations Research work. (i) Binomial Distribution (Bernoulli Distribution)

If p is the probability that an event will happen in any single trial (called the probability of success) and q= 1 -p is the probability that it will fail to happen in any single trial (called the probability of failure) then the probability that the event will happen exactly X times in N trials i.e., X successes and N - X failures, is given by poo = N cx.px . e-X . ' or

P(X) -

N! X! (N - X)!

X N-X • P (I •

This discrete probability distribution is called the "Binomial distribution". For range of X = 0,1,2,..., N the probability distributions are N N N-1, N C 2p 2 q N-2, N C 3p 3 q N-3 pN q , Ci pq which are the successive terms in the binomial expansion of (q + p)N . The distribution (i) is also known as Bernoulli distribution. The mean of the Binomial distribution is given by n X= Xi P(Xi ) i =0 = 0 P(0) + 1. P(1) + 2. P(2) + 3.P(3) +... +N.P(N) = 0+ 1.NCip1qN -1 2. Nc2p2qN -2 3. N c3p3q N -3 +...+N.NCN PNqN-N = Np[q N-1 +N-1 N-2p+ N-1c2qN-3p2 4....+ c1g

= Np(q

_ Np

p

p=1

and the variance of the Binomial distribution is given by

1

21

Mathematical Preliminaries Var (X) =

Xi 2 P(X )- (5?) 2 i=0 = O. P(0) + 12. P(1)+22. P(2) + 32. P(3) +... + N 2P(N)- (Np) 2 =12 . N c1ple-1 + 22 . N c2p2 qN-2 + 32 . N c3p3q N-3 +... +N 2INI cNpNe-N _ N 2 p 2

=Nple + 2 N-1c N-2p + 3 N-1c2 qN-3p 2 +...+NpN-1] _ N 2p2 1g = NpRqN-1 +N-1 c1gN-2p +N-1 C2 (4N-3p2 +....+p N-11 ''2 p 2 +{N-1 C1o N-2 p + 2 N-1C2 q N-3 P 2 +... + (N - 1)pN-1)}] - iv = Np [ ((I p)N-1 + (N -1)p {qN-2 N-2c0N-3 p+...+ pN-2}] _ N 2 p2 +(N _ 1)p (q+ p)N-2] _ N 2 p 2

Np

=Np[1+ (N -1)A- N 2 p 2 *.• p + q = 1 = Np (1 - p)= Npq.

(ii) Poisson Distribution The discrete probability distribution.

P (X) -

X! where A, is positive parameter and X = 0, 1, 2, is called the Poisson distribution. The mean of the Poisson distribution is given by =yXi P(Xi ) i =0

=0.P(0)+1.P(1)+2P(2)+ 3.P(3)+.... ke-k x2e X +3. X =0+1.-+2. 1! 2! 3! 2

= Xe-2' + +=+.... = Xe-X e k =X, 1! 2! and the variance of the Poisson distribution is given by CO

2P(Xi )- (k)2

Var (X) = i= o

=0.P(0)+12.P(1)+ 22.P(2)+ 32.P(3)+....-A. -k _ =0+ 12 . Xe + 22 . k2e -X + 32 . X2e 3! 2! 1! = ke-k + 2k2Ck 1!

3k3e-X • • 2!

22

Operations Research = (1 + 0)X + (1+1)

1!

= ?LC?' [1. + X + X 2 1! 2! = Xe-x . +

x2 _x + (1+ 2) x 3 _x e e

2!

—k2

x 2 + A 2e-X [1 + — + 1! 2!

A,2

-a 2 =

Note : Poisson distribution may be derived as a limiting form of the binomial distribution when p is very small and n is very large.

2.19 Continuous Probability Distributions Let X be a continuous random variable, then it will have the infinite number of values in any interval however small. Thus, we can assign a probability to an interval of X. The probability, that the continuous random variable X lies in the infinitesimal in,terval (x, x + cbc) can be interpreted as fx (x) dx. Then fx (x) is known as the probability density function of the continuous random variable X. The distribution function for a continuous random variable can be written as Fx (b)= FIX (co)

=f

f x (x) dx.

When fx (x) is the density function of the continuous random variable X. The subscriptX is used to indicate the random variable that is under consideration. When there is no ambiguity, this subscript may be deleted, and f x (x) will be denoted by f (x). The knowledge of the density function enables us to calculate all sorts of probabilities. The probability that X lies in the interval a < X < b is given by P {a < X < b) = F (b) - F(a) = f b fx (x) dx a The probability that X lies in the interval x < X < x + dx is given by P{. < X < x + dx} = F (x + dx) - F(x) If F (x) is continuous and dx --> 0 then P (x = X) = 0 Thus, when we deal with continuous probability, it makes sense to talk of the probability that X is within an interval rather than at ooe point.

2.20 Special Continuous Probability Distributions Here we shall describe some special continuous probability distributions which are of great importance and are used in Operations Research work.

(i) The Exponential Distribution (Single parameter distribution) A continuous random variable whose density function is given by xe-xx for x> 0 f (x) = for x < 0 X. being some positive constant, is known as an exponentially distributed random variable.

Mathematical Preliminaries

23

1 1 The mean and variance of this distribution are - and — respectively. a2 a (ii) The Normal Distribution (Two parameter distribution) A continuous random variable whose density function is gil-ren by - (x-m )2 f (x) - 1 . e 202 -00 j or if all aij = 0 for i < j.

6. Null Matrix : A matrix is called a null matrix if all its elements are equal to zero.

7. Transposed Matrix : The transpose of a matrix A is a matrix obtained by interchanging its rows and columns and is denoted by A t or A'. If the matrix A is of order in x n then the order of A' is n x in.

8. Equality of Matrices : Two matrices A = [9] and B = [bii ] are said to be equal if

(i) they are of same order

and (ii) a 0 and

A f(a -1) = f (a) - f (a -1) < 0

i.e., if

A f (a-1) A f (a)

and A2 f (x) 0, for all x.

It is important to note that these conditions are sufficient but not necessary.

6, Differentiation of Integrals F(z) =

b(z) a(z)

f(x, z)clx

where x depends on z, a (z) and b (z) are the limits of integration expressed as the function of z.

Operations Research

38 Then the derivative of F (z), w.r.t., z is given by b (z) sb(z) a f(x, z) dx [foc,z). dx d F(z)= a(z) az dz dz x = a (z)

If a (z) and b (z) are constants, then df (z) dz

_ r b a f (x, z) dx

...(2)

J a az

Example 13 : Find the derivative of the function 2z w.r.t. z F(z) = f x 2 z 3 dx. z Solution : Here f (x, z) = x2z 3. dF(Z)

21'

dz

dX1 2z dz x = z 2z 3x 2z 2cbc [x2z 3 . .u.L‘] dz x = z

a f (X'Z) dX az L

Jz = 2z J x=z

2z 2 3 — (2z) — z 2 . Z 3 = [ x 3z 2 ] x = + z (2z) . z dz dz) = (203. z2 _ z 2 . 3 + 4z 5. 2— 5

142 5.

7. Generating Functions Generating functions technique is a very powerful method for solving the steady state difference equations in the analysis of queuing systems.

Definition Let Po, Pi, P2, .... be a sequence of real numbers. The function P(z) defined for some interval of real numbers containing 0 and 1, whose value at z is given by P(z) = Po + Piz + P2z 2 +.... 00

Or

P (z) =

P,Zn

n=0 is called the generating function of the sequence {Pn }. Note : In queuing models, Po, P1, P2, .... will denote the steady state probabilities. Example 14 : The generating function of the sequence 1, 2, 3, .... is 1 P(z) = 1+ 2z + 3z 2 +4z3+..... (1— z)2

39

Mathematical Preliminaries

Example 15 : Find the generating function of the sequence whose general term is k (k+1), k = 0, 1, 2, 3, .... Solution : Sequence is 0, 1.2, 2.3, 3.4, 4.5, .... 0, 2, 6, 12, 20, ....

i. e.,

The generating function is P(z) = 0+ 2z + 6z 2 +12z 3 + 20z 4 + 2z 2 + 6z 3 +12z 4 +

z P (z) = Subtracting,

(1- z)P(z)= 2z + 4z 2 + 6z 3 +8z 4 + = 2z(1 + 2z + 3z 9+ 4z 3 + = 2z. (1- z)-2 P(z) =

2z (1 - z)3 •

Operations Research

40

+ Exercise on Chapter 2 + Five salesmen of a company are considered for a three member trade delegation to represent the company in a conference. Find the sample space for selecting three salesmen for the delegation and the probabilities that (1) a particular salesman is selected in the delegation. (ii) either of any two particular salesmen (not both) are selected in the delegation. 2. Find the mean and standard deviation of a normal distribution of marks in an examination where 80% of the candidates obtained marks below 75, 4% get above 80 and the rest between 75 and 80. For a standard normal variable (X — m) z= , the area under the curve between z = ± 0.2 is 0.6 and between a z = ± 1. 8 is 0.92. 3. A car hire firm has two cars, which it hires out day by day. The number of demands for car on each day is distributed as a Poisson distribution with parameter 1:5. Calculate the portion of days on which either car is used and the proportion of days on which same demand is refused (e = 0.22). dF 4. Find —, where 1.

dy

2 (xy 2 + x 2y ) dx,

(1) F W) F f

1 2

log (xy 2 )dx,y >1

5. Find - the generating function of the sequence whose general term is A k , k = 0, 1, 2, 3

+ ANSWERS + 1. (1) 6/10, (ii) 6/10 7.44 approx., 3.12 2. 3. 0.22, 0.2025 2

4. (i) 5.

(2xY + x 2 ) clx, 1 (1. 7 Az)

f (2/y)dx

3

Inventory Theory ,

3.1 Inventory [Rohilkhand 1999, 2000; Meerut 2002 (B.P.) 03(B.P.)]

Definition : An inventory can be defined as a stock of goods which is held for the purpose of future production or sales. The stock of uods may be kept in the following forms. (a) raw-materials, (b) partly finished items and (c) finished (or prepared) goods, (d) spare parts etc. The objective of an inventory problem is to minimize total (actual or expected) cost or to maximize (actual or expected) profit.

3.2 Variables in Inventory Problems The variables associated with the inventory problems are classified into two categories. (a) The controlled variables and (b) The uncontrolled variables. (a) The controlled variables : The variables that may be controlled, separately or in combination are the following : 1. The quantity acquired (by purchase, production, or some other means) : The decision maker may have control over the purchase or production level. 2. The frequency or timing of acquisition : The decision maker may have control over how often Or when the inventory should be replenished. 3. The stage of completion of stocked items : The decision maker may have control over the stage at which the unfinished items be held so that there is no delay in supplying customers. (b) The uncontrolled variables : The variables that may not be controlled in an inventory problem are divisible into cost variables and others.

(c) Cost variables (or the costs) involved in inventory problems [Meerut 1995 (P), 96, 2002]

The main cost variables involved in inventory problems are as follows : 1.

Holding or storage costs.

[Rohilkhand 2001, 02]

The costs associated with the storage of the inventory until it is sold or used are known as the holding or storage costs. This cost is directly proportional to the increase in the inventory and the time for which the stocks are held.The various components of the holding costs are as follows :

Operations Research

42

(i) Handling cost : Which includes the cost of labour, transportation charges etc. (ii) Rent for the space or interest and the cost of depreciation on owned space. (iii) Costs of the staff required to keep records. (iv) Insurance and taxes etc. (v) Interest on the money locked for inventory. (vi) Deterioration costs etc. which arises in the case of fashion items or items that changes chemically during storage such as medicines , foods etc. 2.

Shortage or penalty costs

[Rohilkhand 2001, 02]

The unsatisfied demand or shortage penalty cost is incurred when the stock proves to be inadequate to meet the demand of the customers. The shortages may result in the cancellation of orders and loss of sales which may result in the loss of goodwill of the customers. The shortage costs are represented in two ways. One way is that the shortages can be fulfilled by back order, when the commodity next becomes available. This case is known as "back logging" of unsatisfied demand. These cost usually very directly with the shortage quantity and the delay time both. Another way is that the unsatisfied demand is lost. This case is known as "no back logging". This involve a fixed cost each time a shortage occurs. In this case the shortage costs are proportional to the shortage quantity only. Obviously the shortage costs reduces.with the increase in the inventory. 3.

Set up (or replenishment or ordering) costs

[Rohilkhand 2002]

This is the cost associated with the placing of an order for purchasing goods, or it is the cost of setting a machine before it starts production. This cost may depend on the quantity of goods purchased because of price breaks or quantity discounts. Besides these cost variables there are other variables that may not be controlled in any inventory problem. 1. Demand

Demand is the number of items required per period which is not necessarily equal to the amount sold as some demand may go unfulfilled because of shortage or delays. The demands may be of two types : (i) Deterministic demands : If the number of items required (i.e. demand) in a subsequent period of time is known exactly then such demands are called deterministic demands. (ii) Non deterministic or probabilistic demands : If the demands over a subsequent period of time is not known with certainty then such demands are called non deterministic or probabilistic demands. 2. Lead time

[Rohilkhand 2001]

The time gap between the time of placing an order or the starting of the production and the time of arrival or delivery of goods to the inventory is called "Lead time". Also the time gap between the time of demand and the time of filling the demand from the inventory is called lead time. If this time is known (constant) and not zero then one may order in advance by an amount of

43

Inventory Theory

time equal to the lead time. If it is a variable i. e., known only probabilistically then the question of when to order is difficult.

3. Amount delivered The supply of goods may be instantaneous or spread over a period of time.

3.3 Need of InVentory

[Meerut 2002]

The inventory in any business is maintained to decrease the set up costs and the shortage costs. If the demands of the customers are not fulfilled then it may result in the loss of their goodwills. If the orders are cancelled then it results in the loss of the business. Thus, there is always a need of inventory for the smooth running of any business.

3.4 Inventory Problems An inventory problem exists if the amount of the goods in stock (i. e., inventory) is subjected to control and if there is at least one cost that decreases as inventory increases.

3.5 Advantages and Disadvantages of Inventory [Meerut 2002, 03 (B.P.)]

The advantages : Associated with increased inventory are the following : (i) (ii) (iii) (iv) (v) (vi)

The economy of production with large run sizes. The smooth and efficient running of the business. The economy in transportation. The advantage of price discounts by bulk purchasing. Faster and adequate service to the customers, and Profit from speculation in the market where prices are expected to rise. The disadvantages associated with !pere-_•sP,1 inventory is that the holding of inventory, costs money, due to the following (i) Warehouse rent. (ii) Interest on invested capital. (iii) Physical handling. (iv) Accounting. and (v) Depreciation and deterioration.

3.6 Classification or Categories of Inventory Models [Rohilkhand 1999] The inventory problems (models) may be classified into two categories. (i) Deterministic models : These are the inventory models in which demand is assumed to be fixed for a subsequent period of time, and (ii) Probabilistic models : These are the inventory models in which the demand is a random variable having a known probabilistic distribution. Here the future demand is determined by collecting data from the past experience.

3.7 Some General Notations Used in Inventory Models We shall use the following general notations in inventory models of this chapter.

Operations Research

44

I = The cost of carrying one rupee in inventory for a unit time. C1 = Holding cost per unit per unit time. C2 = Shortage cost per unit per unit time. C3 = Set up cost per production run. q = Lot size per production run (i. e. quantity produced in one run). R = Total demand. r = Demand rate K = Production rate (i. e. rate of replenishment of inventory) C = Average total cost per unit time. t = Time interval between two consecutive replenishments of inventory. z = Order level or stock level. L = Lead time. q*, t*, z" = Optimal values of q, t, z respectively for which the cost C is minimum.

Deterministic Models First, we shall consider the deterministic models in which the demand is known and is fixed for a subsequent period of time.

3.8 Economic Lot Size Models The most common inventory problem faced by industry concerns the situation where stock levels are deplenished with time and then are replenished by the arrival of new items. The situation is given in the following economic lot size models. The inventory problems in which the demand is assumed to be fixed and completely predetermined are known as the Economic Lot size problem or Economic Order Quantity (EOQ) problem. The Economic Order Quantity (EOQ) is that size of order which minimizes the total costs that include carrying costs, set up costs and the shortage costs,under the assumption chat the demand is fixed and known.

3.9 Model I : Economic Lot Size Model with Uniform Rate of Demand Infinite Production Rate and having no Shortages To derive an economic lot size formula and the minimum average costs under the following assumptions. (i) demand is uniform at a rate of r units per unit time. (ii) production is instantaneous (i.e. production rate is infinite) (iii) lead time is zero. (iv) C1= holding cost per unit per unit time, (v) C3 = set up cost per production run, and (vi) shortages are not allowed. [Garhwal 1995, 98, 2001; Rohilkhand 2003; Meerut 2000, 01 (B.P.) 03 (B.P.), 07] Solution : Let q be the units of quantity produced (or ordered) per production run at intervals of time t.

Inventory Theory

45

The situation of the inventory is illustnited in fig. 3.1

0

t >i< A

t

>I< t -->k---- t --->i Time Fig. 3.1

Since the demand rate is r units per unit time. the total demand in one run of time interval t is rt. the quantity produced per production run (in the beginning) q = rt (Since shortages are not allowed) The cost of holding inventory (per production run) 1 =C1 (AreaofE OAB)=C1 — qt 2 and the set up cost (per production run) = C3 The tutal cost per production run of time t = — 1 Cl qt C 3 2 The average total cost per unit time C3 1 1 C(q) = — +— = 2 t 2

+

C 3r

._.(2)

q

This equation is known as cost equation. For minimum value of C(q) dC 1 dq 2 Since

3r —0 q2

d 2C 2C 3 r — is positive for q — dq 2 q 3

q -

[231)

(2 C3r Cl

C(q) given by (2) is minimum for q q* = 2

C31

...(3)

which is known as Harris or Wilson Economic Lot-size formula or simply Economic lot-size formula and q * given by (3) is called Economic order Quantity (EOQ) .

Operations Research

46 From (1) the optimum value of t is given by t = t* =

(2031 clr

...(4)

And from (2) the minimum cost per unit time is given by Cmin

1 C 4(2C 311

2

C1

+

3

I 2C3r

or C min = (2C1C3 r) ...(5) minimum cost per unit time is are constants then the If C1 and C3

proportional to the square root of the demand rate. 1. If the lead time is L (not zero), i.e. L is the time. r.7 betv.,c?n

!-4." " - of placing an order and the time of receiving of the goods to the inventory, then we have to place the order in advance by the time L so that we may get the total quantity q as soon as the inventory falls to zero. Thus at the time of placing an order we have the inventory equal to rL.This point is called the r?-nrder point p.

Thus, the re-order point (when lead time is L) is given by

p = rL. If we have to maintain the buffer stock B, then the re-order point (when lead time is L) is given by

p = rL + B. 2.

1 The total inventory in one cycle of time t is — qt. 2 1, t v,

=

Average inventory at any time = t

2

q Thus, throughout the period, average level of inventoryis—. 2

3.10 Another form of Model To derive an economic lot-size formula under the following assumptions : (i) 2 = the demand for product in one unit of time (say one year). (ii) production rate is infinite. (iii) lead time is zero. (iv) P = price of one item of product in Rupees. (v) I = cost of carrying one rupee to the inventory for one year. (v)i) C3 = Set up cost per order (per cycle). (vii) Shortages not allowed. [Meerut 1994, 97, 2004 (0)] Solution : Let q be the units of quantity produced (or ordered) per cycle (in time t). From model I total inventory in one cycle (in time t) =

1 2

qt.

the average inventory is q/2. the average inventory q/2 will continue throughout the year. Holding cost per item for one year = IP

Inventory Theory

47

[... C1 = IP]

The total holding cost per year = - 1P 2 Now the no. of orders (cycles) in one year = X/q. set up cost in one year =.C3 q 1 The total cost in one year, C(q) = -PqI + 2

C3 .

For C(q) to be minimum.

2LC C 1 = - PI - 3 - 0 2 q 2 q

l(2X,C 3 )

q-

i

2XE 3 , which is positive for q = 3 dq 2 q Thus, the economic lot-size formula is d 2C

=

(2X, C 3 j PI

(2 kC 3 ) q

-

PI

This may also be obtained from model I by replacing C1 by PI and r by X,.

3.11 Model II : Economic Lot-size Model with Different Rates of Demand in Different Production Cycles, Infinite Production Rate and having no Shortages To derive an economic lot-size formula and the minimum average cost under the same assumptions as in mode I except that the demand rates are different in different production cycles. [Meerut 2000] Solution Let q be the units of quantity produced (ordered) per production run.

0101011111111111,011111111110111111111i111011 0 k

t 1-->k— A

t3

>k

10111111111111i. t —>1 Time

Fig. 3.2 If R is the total demand in the total period t (fixed) then the number of production cycles.

= R/q = n (say) Let t1, t 2 ,

(Since shortages are not allowed)

t n be the times of the successive production cycles s.t.,

t = t1 + t 2 + t 3 +

+ to

...(1)

Operations Research

48

Thus, q, the quantity produced at the beginning of each production run is tn in the supplied with different uniform demand rates in times t1, t2, successive cycles. The situations of the inventory in different cycles is illustrated in fig. 3.2. The cost of holding inventory for the period t = C1 (sum of areas of n triangles in the fig. 3.2) =

(1 2- qti + Zgt 2 + 2gt 3 +... qt n )

= 4qCi. (ti + t 2 + t 3 +

+ tn )

qCit,

R

and the set up cost = nC3 = -.C3. q 1 R The total cost for the period t (fixed) = — gait — .C3. 2 1 RC 3 The average total cost per unit time, C (q) = — qC1 + 2 tq

...(2)

For minimum value of C(q) RC

dC 1 = C dq 2 1

3=0 0 tq 2

Sinced2C — = 2RC3 is positive for q" = dq 2 tq 3

q= q'=

211C 3 Ci t

(3) 1

2RC C1t

C(q) given by (2) is minimum for the value of q given by (3) Hence (3) gives the required economic lot-size formula. From (2) the minimum cost per unit time is given by _

or

(2R3 ) RC 3 1 01 2 Cit

Cit 2RC3

2C1C 3R) 1,

t

Note : It is clear that the relations (3) and (4), of model II may be obtained from the

corresponding relations (3) and (4) of model I replacing r (demand rate) by R/t (average demand rate).

3.12 Model III : Economic Lot-size Model with Uniform Rate of Demand, Finite Rate of Replenishment having no Shortages To derive an economic lot-size formula and minimum average cost under the same assumptions as in model I except that the replenishment rate (i.e., the production rate ) is finite. Solution : Let K > r be the number of items produced per unit time. If q is the number of items produced per production run then the production will continue for a time

Inventory Theory

49 tl=

q/K

... (1) and the time of one complete production run (i. e. the interval between two runs), t = q/r ...(2) (Since r is the demand rate and no shortages are allowed). The situation of the inventory in one production run is illustrated in fig. 3.3.

i i 11111111111111111110 t_ >I'

B Time

k Fig. 3.3 If Q is the inventory level at the moment the production is completed (i. e., at the end of time t1) then

Q = q - rti = q- K = q - — r

-.(3)

K

The cost of holding inventory for the period t (time interval between runs), . (Areas of ,6, ONB)

= Ci p t j= 1- qC1(12

2

from (3)

K

and the set up cost = C3. The total cost per run of period t 1 = C3 + — q -

t. lJ

The total cost per unit time C3 1 r C(q) = — + — qq. - — t 2 K)

C(q) =

1 + - qC1 (1 -1 q 2 K

C3 r

from (2)

For minimum value of C (q),

dC dq

_ C31. + 1 Ci 1 - r = 0 61 2

q = q* = 3r

Sinced — 2C

dq 2

q

3

2-

2C rK 11 Ci • K - r)}

is positive for this value of q given by (5).

...(4)

Operations Research

50 C (q)

given by (4) is minimum for the value of q given by (5).

Hence (5) gives the required economic lot-size formula. From (4) the minimum cost per unit time is given by Cmin = C 3r il

or

Ci

(K — r )1 1 (K — r ) I [2 C 3 (

2C 3

C min = [2C1C3 r

.

rK

+ Cl

2 K

•

I

Kr

C1 K—r

1

)

kIJ

Also time of one run is given by : t* —

q* r

—

2C 3 K (K — r)

Deterministic Models With Shortages Now we shall consider the deterministic models in which shortages are allowed but are to be fulfilled as soon as the items (goods) are received i. e., in these models back orders are permitted.

3.13 Model IV : Fixed Time Model Order level model 1. with uniform rate of demand Q to be fulfilled in constant time t infinite rate of production and having shortages which are to be fulfilled. (i.e. Backlogged case) To derive the optimal order level and the minimum average cost under the following assumptions. (i)

demand rate is uniform at a rate of r units per unit time.

(ii)

production is instantaneous (i. e. production rate of infinite),

(iii) lead time is zero, (iv)

C1 = holding cost per unit per unit time,

(v)

C = shortage cost per unit per unit time,

(vi) C3 = (vii)

set up cost per production run.

shortages are allowed and are backlogged

(viii) Q = total demand per production run of fixed time interval t. Let z be the order to which the inventory is raised in the beginning of a run of time interval t (constant). ... (1) This inventory is reduced to zero in time t1 = z/r. Then the shortages arise increase from 0 to Q— z in the remaining time (t — where Q is the total demand for one run. ... (2) Q = rt which is known exactly. t

The inventory problems involving z (the level of the inventory in the beginning) are known as order level problems.

Inventory Theory

51

The situation of the inventory and the shortages are illustrated in fig. 3.4.

I

C

1 111111„i=11101111„„1. ortage

Time

Fig. 3.4 The cost of holding inventory (per production run) = C1. (Area of A OAP) 1 C z2 =C1 2 z.t1 . 2r The shortage cost for one run = C 2 (Area of A ABC) 1 =C 2 - (Q - z). (r - t1 ) •2 1 = - C2 (Q - z) ( r r =

2r

from (1)

from (1) from (2)

(Q - Z) 2 .

and set up cost for one run = C3. The average total cost per unit time. C2(Q - Z)2]+C3 C (Z) 1=[C1 —Z2 — — t 2r 2r 21 r C3 [— z 2 +C2 (Q - z) = r- C1 + Q 2r 2r Q

c„ r C..1 (Q - z)-, + 2Q 2Q Q For the minimum value of C(z). C2 dC Ci — = — Z - -- (Q - z) = 0 dz Q Q C 2Q C 2 rt z = z-,— — Ci +C 2 C1 +C 2

or

G1 , C (z) = --= z - +

which is the required optimal order level.

-(3)

(... rC 3/Q is constant) ...(4)

Operations Research

52

The value z given by (4) gives the minimum value of C, since

dC 2 dz 2

—

+ C2

is

positive. ... From (3) the minimum cost per unit time is given by C ram —

or C mm —

C1 ( C2Q

2

,..., L. 2 ( 4 Q

2Q

2Q Ci +C2)

C 2Q

2

Ci +C2

+

rC 3 Q

rC 3 C1C 2Q rC3 C1C 2rt + — + 2(C 1 +C 2 ) Q 2 (Ci +C 2 ) Q

Note : If CI. 0 then Ci + C2 >C2 from (4) z r) units per unit time, (iii) lead time is zero, (iv) C1 = holding cost per unit per unit time, (v) C 2 = shortage cost per unit per unit time, (vi) C3 = set up cost per run, and

(vii) shortages are allowed and backlogged. Initially the stock is zero and the production starts with a finite rate K (> r) units per unit time while the demand is r units per unit time. Thus, the inventory (stock) increases with a rate (K — r) units per unit time. Let the production continue for a period

56

Operations Research The inventory level at the end of time t1 is ...(1)

Q= (K— r) t1

Now the production is stopped, and so this inventoryQ falls to zero with a rate r units per unit time in time t 2 (say) and then shortage arises with a rate r unit per unit time and continue for a period t3 (say). Let s be the maximum shortage at the end of time t 3.

and

Q = rt 2

...(2)

s= rt 3

(3)

At the end of time t 3, production starts again and the shortages are fulfilled in time t 4 (say) i. e. at the end of time t 4 shortages reaches to zero. The cycle then repeats itself, after the time t1 + t 2 + t 3 + t 4.

111111111111111111111111

t,

-t2

M t4

>I
0, a

> 0 and

a2c. a 2c .

at3

>0

59

Inventory Theory

for the values of t 2 and t3 given by (13). Hence for the values of t 2 and t 3 given by (13), the average cost per unit time given by (9) is minimum. From (9), the minimum cost per unit time is given by

2rCiC 2C 3 (1— C min

(C1 +C 2 )

and from (8) the optimum order quantity (EOQ) is given by

(2r:3 (C

+ C2

).( 1 — (1r/K)

ic2

j.

3.16 Multi Item, Deterministic Models with One Constant Here we shall consider the problems of storing several items under some restrictions or limitations. In some simple cases we may use Lagranges multiplier technique. First we consider the problem of inventory of n items, with

instantaneous production and no lead time. Let us suppose the following : (1) the demand is uniform and deterministic at a rate ri per unit time for the i-th item. (ii) C1i = carrying cost per unit of quantity of the i-th item. (iii) C3 i = set up cost per production run for the i-th item. (iv) no shortages are permitted. Cost per unit of time for the i-th item 1 rit Ci = — Cli 2

C 31

1 2

C3 i ri

=

(See Article 3.9 Model I)

qi +

(qi = rut)

...(1)

qi

where qi is the total quantity of i-th item produced at the beginning of the run. The total cost per unit of time C

1

C3iri

2

qi

I qi + i =1

ac 1 r, — — — lt 2 aqi for minimum value of C,

ac aqi

C3 i ri (112

=0

-

2C3i ri

C11

Operations Research

60

a 2c, for which

aqi 2

is positive.

the optimal value of qi is given by 12C 3 iri

_.(2)

q1 C li

1

Case I : Limitation on the amount to be invested on Inventory : i. e. there is a upper limit, say D, on the amount (in Rs.) to be invested on inventory. If C 4i is the unit price of the i-th item qi D

then

-.(3)

i=i Now we have to minimize the total cost per unit of time

C

I [-

qi +

{ =1 2

C3i ri qi

subject to the condition (3). If E C 4i qi *< D holds then optimal values qi *given by (2) are the required values. i =1 But if

C o qi *

D does not hold then optimal values qi * given by (2) are not

i =1 the required values. In this case we use the Lagranges multiplier technique. Let us consider the function 1

C 3i

i =1 _ 2

qi

F=I1--L,11 q1 +

E C4i —

where is a Lagranges multiplier. For F to be minimum C3i aF 1 —= + XC 4i = 0, 21 2 aqi qi and

aF

—=

-.(4)

=1

i = 1, 2,

n,

n r,

u 4i qi -D= 0

i =1

which give,

q1 -

-

2C3i ri C4i + 2XC4i

and

C4i q i =D i =1

, i =1, 2, . . . , n.

-(5)

...(6)

Inventory Theory

61

The values of qi * are obtained from (5) subject to the relation given in (6). The value of X is found by trial and error process. The process is well understood from Example 14, page 69. Case II : Limitations on the number of stocked units : i. e. there is a upper limit, say N, on the average number of all stocked units. 1" qi N 2 =1

or (Since

-.(7)

qi is the average number of units at any time).

Now we have to minimize the total cost per unit of time C 3 i ri 1 C= Cli qi + i=1 2 qi subject to the condition (7). 1 n If qi * N holds then optimal values qi * given by (2) are the required 2i

E

values. qi * But if 2 i= 1

N does not hold then optimal values qi * given by (2) are not

the required values. In this case we use the Lagranges multiplier technique. Let us consider the function F=

C311 [1 1 -Cii qi + + A. [21 =.ii_ qi qi i=1 2 in

N)

where X is a Largranges multiplier. . For F to be minimum C3i 3F 1 , —= —0 li 2 2 • , aqi 2 q and

DF 1 n —= a 2 I. =1

which gives qi = qi * -

and

-N= o

I 2C 3 ir I Cli + X

i = 1, 2, ..., n

1 n =1

=N

... (8)

-(9)

Thus, value of qi * are obtained from (8) subject to the relation given in (9). The value of ? is found by trial and error process.

Operations Research

62

Case III : Limitations on the warehouse floor space (in square feet). Let A be the upper limit of area of the floor space of a warehouse (in square feet). If A square feet floor space is required for one unit of i-th item, then

E Ai qi SA

...(10)

i Now we have to minimize the total cost per unit of time rt 1 C3i li C= Cli q, + qi i= 1 2 subject to the condition (10). If (10) holds then the optimal values qi * given by (2) are the required values. But if (10) does not hold then optimal values qi " given by (2) are not the required values. In this case we use Lagranges multiplier technique as in case I, to get the optimal value (it *

_

2C 3i ri C11

+ 2 XAi

,

= 1, Z

n

Ai qi = A ---1 Thus, values of qi * are obtained from (11) subject to the condition (12). and

...(12)

eXaMP/124 Example 1 : If in model I, the set up cost instead of being fixed is equal to C3 + bq, where b is the set up cost per item produced, then show that there is no change in the optimum order quantity produced due to change in the set up cost. Solution : Replacing C3 by C3 + bq in Model I eqn. (2). Average total cost per unit time. 1 2

C(t) = — Ci rt + or

C3 + bq C3

(... q = rt)

C(t) =— Ca rt + — + br 2

For minimum of C(t)

C1r C 3 d k.t) = —=0 2 t2 dt d2C(t) 2C 3

dt 2

t3

>0

t = t* —

63

Inventory Theory C (t) is minimum for t = t* =

2C0 and q* = rt * — Cl r

2C3r Cl

which is the same as in Model I. Example 2 : A manufacturer has to supply his customer with 600 units of his product per year. Shortages are not allowed and the storage cost amount to Rs. 0.60 per unit per year. The set up cost per run is Rs. 80.00. Find the optimum run size and the minimum average yearly cost. [Meerut 2001 (BP), 07] Solution : Considering one year as one unit of time, we have C1= Rs. 0.60, C3 = Rs. 80.00, r = 600 q

1(2C3r) ) = 11

From Article 3.9

(2 x 80 x 600) 0.60

— 400

2 q* 400 2 t=—= = — year = — x 12 = 8 months r 600 3 3 Thus, the manufacturer should produce 400 units of his product at an interval of 8 months. and

Minimum average cost = 4 (2C1C3r) = 4 (2 x 0.60 x 80 x 600) = Rs. 240 units/year Example 3 : The storage cost of one item is Rs.1 per month and the set up cost is Rs. 25 per run. If the production is instantaneous and the demand is 200 units per month, find the optimal size of the batch and the best time for the replenishment of inventory. [Meerut 2001 (BP), 02 (BP)] Solution : Considering one month as one unit of time, we have C1= Rs. 1.00, C3 = Rs. 25, r = 200 q* _

and

t =

(2C31

(2 x 25 x 200) 100 1 11

100 1 = — = — month = 15 days. r 200 2

Thus, he should produce 100 units of his product at interval of 15 days. Minimum average cost = (2C1C3r) = 4 (2 x 1 x 25 x 200) = Rs. 100/month. Note : Average inventory throughout the month = q/2 = 100. If he produces all 200 items in the beginning of a month then the cost = 25 x 1 + 100 = Rs. 125, which is more than the cost, if the items that are produced twice in a month.

Operations Research

64

Example 4 : Consider the inventoty system with the following data in usual notation: r = 100 units/year, I = 0.30, P = Rs. 0.50 per unit. C 3 =Rs. 10.00, L = 2yrs. (lead time). Determine the following (i) Optimal order quantity, (ii) Re-order point, (iii) Minimum average cost. Solution : Here C1 = IP = (0.30) x (0.50) = 0.15 per unit/year C3 = Rs. 10.00, r = 1000 units/year and L = 2 yrs. 3r)

(i) Optimal order quantity (EOQ), q* =

C1 2C 12 x 10 x 1000 = 365 units 0.15 1

(ii) Re-order point, p= rL = 1000 x 2 = 2000 units (iii) Minimum average cost = I (2C1C3r)

= 4{2 x (0.15) x 10 x 1000} = Rs. 54.80 Example 5 : In a central grain store, it takes about 15 days to get the stock after placing the order and daily 500 tons are despatched to neighbouring markets. On an adhoc basis safety stock is assumed to be 10 days stock. Calculate the recorder point p. Solution : Here safety stock B = 10 days stock =10 x SOO = 5000 tons L = 15 days and r = 500 tons. Reorder point p =. B + Lr = 5000 + 15 x 500 =12500 tons

Example 6 : You have to supply your customers with 100 units of a certain product every Monday and only then, you obtain the product from a local supplier at Rs. 60 per unit. The costs of ordering and transportation from the supplier are Rs. 150 per order. The cost of carrying inventory is estimated at 15% per year of the cost of the product carried. (i) Find the total size which will minimize the cost of the system. (ii) Determine the optimal cost. [Rohilkhand 1995] Solution : Here C1 = 15% per year of the cost of the product carried 15 = Rs. — x 60 = Rs. 2 per unit/year. 100 9 = Rs. — per unit/week 52 C3 = Rs. 150 and r = 100 units/week

Inventory Theory

65

2C r _ 42 (0 Optimal lot size, q* = 3F ) Ci 1

x 150 x 100 x 52) — 416 units 9

(ii) Optimal cost = 4 (2C1C3r) + 60 r =

9 2 x — x 150 x 1001+ 60 x 100 = Rs. 6072 52 (

Example 7 : ABC manufacturing company purchases 9000 parts of a machine for its annual requirements, ordering one month usage at a time. Each part costs Rs. 20. The ordering cost per order is Rs. 15, and the carrying charges are 15% of the average inventory per year. You have been asked to suggest a more economical purchasing policy for the company. What advice would you offer, and how much would it save per year ? [Meerut 2003] Solution : Here r = 9000 parts/year Cl = 15% of the average inventory per year 15 = Rs. 20 x100 — =Rs. 3 each part/year. C3 = Rs. 15 Economic lot size, q* — and

2C 3r C1

2 x 15 x 9000 = 300 units. 3

t* q* = 300 _ 1 year r 9000 30

minimum cost per year = 4 (2C1C3r) = 4 (2 x 3 x 15 x 9000) = Rs. 900 If the company follows the policy of ordering each month. Then, annual ordering cost = Rs. 12 x 15 = Rs. 180 and lot size each month — 9000 = 750 = q 12 q 750 average inventory at any time = — = — = 375 2 2 storage (or carrying) cost at any time = 375 C1 = 375 x 3 = Rs. 1125. Total annual cost = Rs. 1125 + Rs. 180 = Rs. 1305 1 Hence if the company purchase 300 parts at time intervals of — yrs. in place of 30 order 750 parts each month then the net saving of the company is Rs. 1305 — Rs. 900 = Rs. 405 per year

Operations Research

66

Example 8 : A company uses annually 24000 units of a raw material which costs Rs. 1.25/unit. Placing each order costs Rs. 22.5 and the carrying cost is 5.4% per year of the average inventory. Find the economic lost size and the total inventory cost (including the cost of material). [Meerut 1999] Solution : Here r = 24000 units/year. C1 = 5.4% of the average inventory per year 5.4 27 = Rs. 1.25 x — = Rs. — per unit/year 100 400 C3 = Rs. 22.5 2C3r 12 x 22.5 x 24000 x 400 — = 4000 units Economic lot size, q* — C1 27 1 q* 4000 1 — — year = 2 months. r 24000 6 The optimal lot size is 4,000 units, after every two months. Total annual inventory cost (including cost of material) = -,1(2C1C 3 r) + purchasing cost per year —

and

27 x 22.5 x 24000 + (1.25) x 24000 400 11 = Rs. 270 + Rs. 30000 = Rs. 30270 Example 9 : Neon lights are replaced at the rate of 100 units per day. The physical plant orders the neon lights periodically. It costs Rs. 100 to initiate a purchase order. A neon light kept in storage is estimated to cost Rs. 0.02 per day. The lead time between placing and receiving an order is 12 days. Determine the optimal inventory policy for ordering the neon lights. [Meerut 2002 (BP), 03 (BP)] Solution : Given that, C3 = Rs. 100, C1 = Rs. 0.02, r = 100 units per day 2C 3r 112 x 100 x 100 Optimal order quantity q* = — — — 1000 lights 0.02 =

2x

*

000

Length of the cycle, t * = 1 = - 10 days r 100 Here, lead time L = 12 days and the length of the cycle = 10 days i. e., the lead time is greater than the length of the cycle. the re-ordering will take place when the level of inventory is sufficient to satisfy the demands for 12 — 10 = 2 days. Hence the re-order point (ROP) = 2 x 100 = 200 lights. i. e., the quantity 100 lights is to be ordered when the level of inventory reaches 200 lights. Note : When the lead time L is greater then the length of the cycle t * , then the effective lead time is taken equal to L — t *. In this case effective lead time is taken as L — t * = 12 — 10 = 2 days instead of 12 days.

Inventory Theory

67

Example 10 : An item is produced at the rate of 50 items per day. The demand occurs at the rate of 25 items per day. If the set up cost is Rs. 100 per set up and holding cost is 0.01 per unit of item per day, find the economic lot size for one run, assuming that the shortages are not permitted. [Meerut 2004] Solution : It is the problem of Model III. Here K = 50 items/day, r = 25 items/day, C1 = Rs. 0.01/day, C3 = Rs. 100/run. lot size formula is 3

q* = and

rK —r)

12 x 100 25 x 50 — 1000 items 50 — 25 1 0.01

q* .1000 _ 40 days r 25

Minimum cost = 12C1C3r — 11 . 25 = 112 x 0.01 x 100 x 25 — — ) = Rs. 5 per day 50J Hence, total cost per run = Rs. 5 x 40 = Rs. 200. Example 11 : A contractor has to supply 10000 bearing per day to an automobile manufacturer. He finds that when he starts a production run, he can produce 25000 bearing per day. The cost of holding a bearing in stock for one year is 2 paise, and the set up cost of a production run is Rs. 18. How frequently should production run be made ? [Meerut 1996 (P), 2002; Rohilkhand 1995, 98] Solution : It is the problem of model III. Here K = 25,000 bearing/day, r = 10,000 bearing/day C1 = 2 paise per bearing per year. 2 1 = Rs. — Rs. per bearing per day. 100 x 365 50 x 365 C3 = Rs. 18 .

lot size formula is q* —

il

2C (_ rK ) C1 K — r

= il 2 x 18 x 50 x 365 10000 x 25000 25000 — 10000

and

= 104642 bearings. 104642 — — 10.46 days. 10000

Operations Research.

68

Example 12 : A contractor has to supply diesel engines to a truck manufacturer at the rate of 25 per day. There is a clause in the contract penalizing him Rs.10 per engine per day late for missing the scheduled delivery date. He finds that the cost of holding a completed engine in stock is Rs.16 per month. The production process is such that each month (30 days) he starts a batch of engines through the shops, and all these engines are available for delivery any time after the end of the month. What should his inventory level be at the beginning of each month ? [Meerut 2000] Solution : It is the problem of Model IV. Here

r = 25 engines/day Ci 16 8 = Rs. — per engine per day = Rs. — per engine per day. 15 30 C2 = Rs. 10 per engine per day

Optimal inventory level C rt 10 x 25 x 30 Z* 2 - 712 engines. C1 +C2 (8/15) + 10 Example 13 : The demand of an item is uniform at a rate of 25 units per month. The fixed cost is Rs. 15 each time a production run is made. The production cost is Rs. 1 per item, and the inventory carrying cost is Rs.0.30 per item per month. If the shortage cost is Rs.1.50 per item per month, determine how often to make a production run, and of what size it should be. [Meerut 1999; Rohllkhand 2003] Solution : It is a problem of model V. In this problem set up cost is a variable and is equal to C3 + Pq, where P is production cost of an item and q is the order quantity per production run. Proceeding as in model V, average total cost per unit time is given by z2 2r =

t

+

C2 (rt . 2r

—

z)2 C3 +Pq 2

z2

C2 (rt — z) —+—. + C3 +Pr

2r t 2r

t

Since Pr is constant as in model V, C (z, t) is minimum, when, t -

l2C3(Ci +C2 ) C1C 2 r

q* = rt* —

(Ci + C 2)C 3 C1C2

Here r = 25 units/month. C1 = Rs. 30/item/month, C2 = Rs. 1.50/item/month, C3 = Rs. 15 per production run.

(... q = rt)

Inventory Theory

69 q* -

and

12 x 15 x (0.30 + 1.50) x 25 (0.30) x (1.50) 1

-

55 items

t* = q* = 55 = 2.2 months r 25

Example 14 : For the following data, determine approximately the economic order quantities when the total value of average of inventory levels of three products is Rs.1000. Costs, Products —>

A

Holding cost

20

20

20

Cost per unit (Rs.)

6 50

7 42

5 60

10000

12000

7500

Set up cost (Rs.) Yearly demand rate

Solution : The optimal value qi of 'the i-th item produced at the beginning of the run is given by qi

12C3i .ri 1 di

(11 -

12 x 50 x 10000 - 100 4 5 = 223 20 1

(12 -

112 x 40 x 12000 20

- 40 4 30 = 219 1502

20

From equation, (6) article 3.16 amount to be invested on the average inventory =

=

4i

i =1

2

6 (5.1) + 7 (1 + 5 (q ) 2 2 2 qi Since average inventory at any time is — 2 qi in (6) qi is replaced by —

223 212 219 = Rs. 6 x — + 7 x — + 5 x —] = Rs. 1965.50 2 2 2 [ This amount Rs. 1965.50 is greater than Rs. 1000 available in the questions. So we try to find the value of from equation (5), article 3.16 to suit the given problem which is done by trial and error method.

Operations Research

70 2C 3 ri

*-

C11 + 2X,C4i

12 x 50 x 10000 — 121 If live take = 4, then qi* 1 20+ 2x 4x 6 (/2*

(13'

12 x 40 x 12000 — 112 1 20+ 2x 4x 7 _ 12x 60 x 7500 — 122 1 20+2x 4x5

Cost on average inventory

=

*

a„

1-:--J + 7 {2-i-} + 5 { 1= Rs. 1060> RS. 1000. 2 2 6 {a2 1

Again if we take, = 5, then qi

,

1 2 x 50 x 10000 . 112 11 20+2x 5x6

q2*

12 x 40 x 12000 1 20+2x 5 x7

q3

12 x 60 x 7500 _113 1 20 + 2 x 5 x 5

103

Now cost on average inventory = 6 {cli' + 7 I gq + 5 2 2

2

}= Rs. 979 < Rs. 1000

1120 1110 1090 1060 L 1030 (-) 1000 970 960 Fig. 3.6

Thus, it is clear that the value of X lies between 4 and 5. From the graph fig. 3.6, between cost on average inventory and the value of X, the appr. value of X obtained is X = 4.7.

Inventory Theory

71

For X = 4.7, ql* = 114, q2* = 106 and q3* = 116 Cost on average inventory = Rs. 1003 which is sufficiently close to Rs. 1000. Example 15 : A company, for one of the A-class item, placed 6 orders each of size 200 in a year. Giving ordering cost = Rs. 600, holding cost = 40% cost per unit = Rs. 40, find out the loss to the company in not operating scientific inventory policy? What are your recommendations for the future ? Solution : Considerii one year as one unit of time, we have 40 C1 = IP = Rs. — x 40 = Rs. 16 per unit/year. 100 C3 = Rs. 600. Total demand in one year, R = 6 x 200 = Rs. 1200 items (EOQ) q* =

2C3R

2 x 600 x 1200 — 300 units 16

Minimum cost per year = The cost of holding inventory + Ordering cost 1200 — x 6001= Rs. 4800.= —Co* 1 + (RC 3/q*) = Rs. ( x 16 x 300 +300 2 2 Total cost per year under the existing policy = The cost of holding inventory + Ordering cost 1 R = Ciq + — C3 -- Rs. (1x 16 x 200 + 1200 x 600) -- Rs. 5200 2 200 2 q Loss to the company in not operating scientific inventory policy = Rs. 5200— Rs. 4800 = Rs. 400 Recommendation for the future : The company should place four orders each of size 300 units in a year. Example 16 : A purchase manager has decided to place order for minimum quantity of 500 Nos. of a particular item in order to get a discount of 10%. From the records, it was found out that in the last year, 8 orders each of size 200 Nos. have been placed. Given ordering cost = Rs. 500 per order, Inventory carrying cost = 40% of the inventory value and the cost per unit = Rs. 400. Is the purchase manager justified of his decision ? What is the effect of his decision to the company ? Solution : Considering one year as one unit of time, we have 40 C1 = IP = Rs. — x 400= Rs. 160 per unit/year. 100 C3 = Rs. 500 Total demand in one year, R = 8 x 200 = 1600 Nos.

Operations Research

72 (EOQ)

,

(r

2C 3R Cit

2 x 500 x 1600 — 100 Nos: 160

Minimum cost per year = Purchasing cost + The cost of holding inventory + Ordering cost = R x 400+ 2CA* + (RC 3/q*)

= Rs. [1600 x 400 + x 160 x 100 + 1600 x 500/100] = Rs. 656000. For the existing last year's policy, the total cost per year = Purchasing cost + The cost of holding inventory + Ordering cost = 400 x R + 1Ci q + (RC 3/q) = Rs. [400 x 1600 + 1 160 x 200 + 1600 x 500/200) = Rs. 660000 When the manager decided to place order for minimum of 500 Nos. the discount on the cost Rs. 400 of an item. 10 = Rs. 400 x — = Rs.40. 100 . The discounted cost of an item = Rs. 360. ... In this case, the total cost per year = Purchasing cost + The cost of holding inventory + Ordering cost = 360 x R + iCiq + RC 3/q 40 = Rs. [360 x 1600I+x( x 360)x 500 + 1600 x 500/5001 2 100 = Rs.613600. From the above calculations it is obvious that the manager is justified in his decision. By placing an order for a minimum of 500 items to get a discount of 10%, the saving over the policy of placing an order of 200 Nos, = Rs. 660000 — Rs. 613600 = Rs. 46400. and the saving over the policy of economic lot size of 100 Nos, = Rs. 656000 — Rs. 613600 = Rs. 42400 Hence, the manger will save Rs. 46400 with his decision of purchasing a minimum of 500 Nos. to get a discount of 10% over that last year's purchasing policy of 8 orders each of size 200 Nos.

Inventory Theory

73

Example 17 : A company uses annually 50000 units of an item each costing Rs. 1.20. Each order costs Rs. 45 and inventory carrying costs 15% of the annual average inventory value. (i) Find EOQ (ii) If the company operates 250 days a years, the procurement time is 10 days and safety stock is 500 units, find reorder level, maximum, minimum and average inventory. Solution : We have 9 C1 = IP = Rs. 51 x 1.20 = Rs. — per unit/year. 100 50 C3 =Rs. 45 Total demand in one year R = 50000 units 2C R 2 x 45 x 50000 x 50 (i) EOQ — 3 C1 y 9 = 5000 units (ii) Number of days the company operates = 250 Lead time L = 10 day and safety stock B = 500 units Recorder level = Lead time demand + Safety stock 50000 n =LX— 4- D 250 = 10 x 200 + 500 = 2500 units Maximum inventory = EOQ + Safety stock = 5000 + 500 = 5500 units. Minimum inventory = Safety stock = 500 units (0 + EOQ) Average inventory = + Safety stock 2 =— 1 x 5000 + 500 2 = 3000 units. •

Operations Research

74

4 Exercise 3.1 f 1. A shop keeper has a uniform demand of an item at the rate of 50 items per month, he buys from a supplier at a cost of Rs. 6 per item and the cost of ordering is Rs. 10 each time. If the stock holding costs are 20% per year of stock value, how frequently should he replenish his stocks. [Meerut 1998 (BP); Rohilkhand 2002]

2.

3.

A contractor has to supply diesel engines to a truck manufacturer at a rate of 20 per day. The penalty in the contract is Rs. 10 per engine per day late for missing the scheduled delivery data. The cost of holding an engine in stock for one month is Rs. 15. His production process is such that each month (30 days) he starts a batch of engines through the agencies and all are available for supply after the end of the month, what should be the inventory level in the beginning of each month. A precision Engineering Factory consumes 50000 units of a component per year. The ordering receiving and handling costs are Rs. 3 per order while the trucking costs are Rs. 12 per order. Further details are as follows : Interest cost Rs. 0.06 per unit per year. Deterioration and observation cost Rs. 0.004 per unit per year. Shortage cost Rs. 1000 per year for 50000 units; calculate the economic order quantity. [Meerut 2006] [Hint : It is the problem on model I

Here C3 = Rs. 3 + Rs. 12 = Rs. 15/order, r = 50000, Cl= Rs. 0.06 + Rs. 0.004 + Rs. (1000/50000)

4.

5.

6.

= Rs. 0.084/unit. A company uses annually 12000 units of a raw material costing Rs. 1.25 per unit. Placing each order costs Rs. 0.45 and the carrying costs are 15% per year per unit of the average inventory. Find the Economic Order Quantity. A certain item costs Rs. 235 per ton. The monthly requirements are 5 tons and each time the stock is replenished there is a set up cost of Rs. 1000.The cost of carrying inventory has been estimated as 10% of the value of the stock per year. What is the optimal order quantity ? [Meerut 1997] An electric appliance manufacturer wishes to know what the economic quantity should be for a plastic impeller when the following information is available. The average daily requirement is 120 units and the company has 250 working days a year, so that the total yearly requirement is approximately 30000 units a year. The manufacturing cost is 50 paise per part. The sum of the annual rate for interest, insurance, taxes and so forth is 20% of the unit cost, and the cost of preparation is Rs. 50 per lot. [Hint : Model I, r = 3000 units, Ci = Rs. 0.50 x 0.20 = Rs. 0.1]

Inventory Theory

75

7.

A manufacturing company uses a certain part at a constant rate of 4000 units per year. Each unit costs Rs. 2 and the company personnel estimate that it cost Rs. 50 to place an order, and that the carrying cost of inventory is 20% per year. Find the optimal size of each order and minimum yearly cost. 8. A product is produced at the rate of 50 items per day. The demand occurs at the rate of 30 items per day. Given that, C3 = Rs. 100, C1 = Rs. 0.05, find the economic lot size and the associated total cost per cycle assuming that no shortage is allowed. 9. In question 7, suppose that shortage is allowed. If it is decided to use a lot-size of 600 items, what is the simplified shortage cost per unit under optimal conditions ? 10. An oil engine manufacturer purchases lubricants at the rate of Rs. 42 per piece from a vendor. The requirement of these lubricants is 1800 per year. What should be the order quantity per order, if the cost per placement of an order is Rs. 16 and inventory carrying charges per rupee per year is only 20 paise. 2C3 r [Hint : From Model I, r = 42 x 1800, q* — = 3478 appr.

11. 12.

13.

14.

15.

Optimum inventory quantity = 3478/42 = 83 lubricants] In Ex. 10, how frequently should production run be made if C1 = 20 paise/bearing/year and C3 = Rs. 180. A contractor has a requirement for a cement that amounts to 300 bags per day. No shortages are allowed Cement cost Rs. 12 per bag. Inventory carrying cost is 10% of the average inventory valuation per day and it costs Rs. 20 to purchase order. Find the minimum cost purchase quantity. [Hint : Mode I. Here r = 300, C1 = Rs. 0.10 x 12= Rs. 1.20, C3 = Rs. 20.] An aircraft company uses rivets at an approximate customer rate of 2500 kg per year. The rivets cost Rs. 30 per kg and the company personnel estimate that it costs Rs. 130 to place an order and the inventory carrying cost is 10% per year. How frequently should orders for rivets be placed and what quantities should be ordered ? [Meerut 2002] The XYZ manufacturing company has determined from an analysis of its accounting and production data for part number 625, that its cost to purchase is Rs. 36 per order and Rs. 2 Rer part. Its inventory carrying charge is 18% of the average inventory. The demand of this part is 10,000 units per annum. Find (a) What should the Economic order quantity be ? (b) What is the optimum number of day's supply per optimum order ? A manufacturer has to supply his customers 24000 units of his product per year. This demand is fixed and known. The customer has no storage space and so the manufacturer fails to ship a day's supply each day. If the manufacturer fails to supply, the penalty is Rs. 0.20 per unit per month. The inventory holding cost amounts to Rs. 0.10 per unit per month and the set up cost is Rs. 350.00 per production run. Find the optimum lot-size for the manufacturer.

Operations Research

76

16. The demand of an item is uniform at a rate of 20 units per month. The fixed cost is Rs. 10 each time a production run is made.The production cost is Rs. 1 per item and the inventory carrying cost is Rs. 0.25 per item per month. If the shortage cost is Rs. 1.25 per item per month, determine how often to make a production run and of what size it should be ? 17. The Anil manufacturing company has determined from an analysis of its accounting and production data for part no 100 that its cost to purchase is Rs. 35 per order and Rs. 2.20 per part. Its inventory carrying charge is 18% of average inventory. The firm currently purchases 22000 parts per year.

(a) What should be the economic order quantity ? (b) What is the optimum number of days supply per optimum order ? (c) What is the optimum number of order per year ? 18. A company producing three items has a limited storage space of averagely 750 items of all types. Determine the optimal production quantities for each item separately, when the following informations are given. Product —+

1

2

3

C1

0.05

0.02

0.04

C3

50

40

60

t

100

120

75

19. The uniform annual demands for two bulky items are 90 tons and 160 tons respectively. The carrying costs are Rs. 250, Rs. 200 per ton per year, and set up costs Rs. 50 and Rs. 40 per production respectively. No shortages are allowed. Space consideration restrict the average amount inventory of items to 4000 ft 3 . A ton of the first item occupies 1000 ft 3, and a ton of the second item 5000 ft 3 . Find the optimal lot-size. [Meerut 1997 (P)] 20. A small shop produces three machine parts 1, 2, 3, in lots. The shop has only 650 sq. ft. of storage space. The appropriate data for three items are presented in the following table.

1

2

Demand rate (units/yr.)

5000

2000

10000

Procurement cost (in Rs.)

100

200

75

10

15

5

0.70

0.80

0.40

Item

Cost per unit (in Rs.) Floor space required (sq. ft./unit)

3

The carrying charge on each item is 20% of average inventory valuation per annum. If no stock-outs are allowed determine the optimal lot-size for each item. •

Inventory Theory

77

+ ANSWERS + 1. Lot-size 100 items after every two months

2.

571 engines

3. q* = 4226 units

4.

240 units

5. 71.5 tons

6.

5477 units

7. 1000 units, Rs. 400

8.

13015, Rs. 11.00

9. C2 = 1/220 11. 10.46 days

10. 83 lubricants

13. 466 kg. appr., 5.3 orders per year

14. (a) 10001/5 units, (b) 42/10

15. 4578 units per run

16. 2.2 months

18. X = 0.00256 appr. 428, 628, 444

19. (11* = 2 tons, q2* = 12 tons

12. q* = 100 bags, Cali, = RS. 120

3.17 Probabilistic Models Now we shall discuss some models in which the demand is not known exactly. In these probabilistic models we shall minimize the expected costs instead of actual costs.

3.18 Model VII : Single Period Model with Discontinuous or Instantaneous Demand and Time Independent Costs (No Set up Cost Model) To determine the optimum order level 'z' which minimizes the total expected cost under the following assumptions. (i) t is the constant interval between orders. (So no need to consider set up cost in total cost). (ii) z is the stock level at the beginning of each period of time t. (iii) p(r) is the probability of requiring r units, in time t. (iv) lead time is zero. (v) C1 = holding cost per unit per unit time. (vi) C 2 = shortage cost per unit per unit time. Solution : (a) Discrete case (r has discrete values). [Garhwal MSc. (Stat) 1993, 95; Meerut 1996; Kanpur 1997] In this model the demand is instantaneous, therefore here we assume that the total demand is fulfilled at the beginning of the period.There are two cases, of the inventory. Case I When demand r does not exceed the stock z. L e., r < z. Here the total demand r is fulfilled at the beginning of the period and z — r is the inventory which is kept for time t.

Operations Research

78

In this case total cost is the holding cost. Cost of holding z — r units in stocks for unit time is equal to (z — r)Ci r5z for Since the prob. of requiring r units is p(r). The prob. of holding cost (z — r) C1 is also p(r). Total expected cost per unit time in this case is

Inventory

z-r

Time

Fig. 3.7 r z

(z — r) C1 p (r) r=0

Case II : When demand r exceeds the stock z L e., r > z. In this case the customers demand r is more than the stock z. Therefore, the total demand is not satisfied at the beginning of the period and r — z demand remains to be satisfied. Here the total cost is the shortage cost. Cost of shortage of r — z units for unit time is equal to (r — z) C2 Fig. 3.8 r > z for r > z. As in case I, total expected cost per unit time in this case is (r — z) C 2 p (r) r=z+1

The total expected cost per unit time for the model is given by z

C(z)=

(z — r) Ci p(r) + r=0

Cr — z) C2p(r) r=z+1

For minimum value of C(z), AC (z — 1) < 0< AC (z), Now, AC (z) = C (z + 1) C (z) z+1 = C1 f(z + 1) — p(r) + C 2 (r — z — 1) p(r) r=0

r= z + 2 z

00

(z — r) p(r) — C2

— C1 r=0

(r — z) p (r) r=z+1

79

Inventory Theory z

00

p(r) - C2

=C1

r=0

p(r)

r=z+1 z+1 Since (z + 1 - r) p (r) = (z + 1 - r) p (r) r=0 r=0 00

and

(r - z -1) p (r)

(r - z -1) p(r) =

r=z+2

z

r=z+1

-

z

=C1 I p (r) - C2 1p (r)

r=0

r=0

p (r)

r=0 ..

z = (CI. + C2 ) I p(r) -C2 r=0

...

I

p(r) = 1

r=0

z- 1 and similarly AC (z - 1) = (C1 + C2 ) 0.0

or

p (r) - C2. r=0 from (2), for minimum value of C(z), we have z-1 z (C1 + C2 ) p (r) - C2 < (Ci +C2 ) p(r) -C2 r=0 r=0 z-1 2 < p(r)< p(r) Ci +C2 r=0 r=0

...(3)

[Kanpur 1997; Garhwal MSc. (Stat) 1993, 95] The relation (3) will give the optimum value of stock level z. zo 1 Note : 1. If zo is such that

p (r) =

r=0

C2 +C
0 ... C(z) given by (4) is minimum for the value of z given by (6).

3.19 Model VIII : Single Period Model with Uniform Demand (No Set up Cost Model) To determine the optimum order level 'z' which minimises the total expected cost under the following assumptions. (i) t is constant interval between orders. (So no need ta_consider set up cost in total cost) (ii) z is the stock level at the beginning of each period of time t (iii) demand is uniform over the period and it is r units per period. (Note) (iv) p (r) is the probability of requiring r units in time t. (v) lead time is zero.

Inventory Theory

81

(vi) C1 = holding cost per unit per milt time. (vii) Shortages are allowed and back lodged. (viii) C 2 = shortage cost per unit per unit time. Solution : (a) Discrete case (where r has discrete values). (Meerut 1995) If z is the stock level at the beginning of each period of time t , then there are two cases of the inventory. (i) If r z, L e., when demand r does not exceed the stock z. The period t starts with a stock z and end with the stock z r. Total cost in this situation is the holding cost only and is equal to C1. Shaded area 1 = C1 — [z + (z — r)]t z

=C

r — — jt

1111111111111111111111111111111 t

Total expected cost in this situation Ci (z —

2

Time

Fig. 3.9 r z

is i• = o

z - r)

t p(r)

(ii) If r > z. i.e., when demand r exceed the stock z. In this case the period starts with a stock z which is supplied in time t1 with rate tit. For the remaining time t — t1 there is a shortage and which increase to r — z by the end of this period. r z = — . tl

(t-ti) Stock 10111111111111

0

Shorta e ti-->1

NONNMEINYIN Nominn•


z

Time (r - z)

Operations Research

82 so that

t = zt/r

t — ti = (r — z) t/r and Holding cost in this case is C1. (Area representing stock) zt1 = 1c z 2t =c 1. 2 2 1r and shortage cost in this case is C2. (Area representing shortage) —C (r — z) (t — t1) — C (r —z)2 t 2. 2 2 2r Total expected cost in this situation is — t [C1 z 2 + C 2(r — z)2 ] p (r) r = z +1

2r

Hence, the total expected cost per unit time for this model is given by C(Z)=Ci r=0

2 r — ) p(r) + I 1[clz +C2(r- z)2]p(r) —2 2r r=z +1

...(1)

For minimum value of C(z), we have AC(z —1) < 0 < AC (z)

...(2)

Now, AC(z) = C(z + 1) — C(z) z +1

=

1 + 1 — p (r) + — [Ci (z + 1) 2 + C 2 (r — z — 1)2] p (r) 2 2r r=0 ll r=z+2 (z —

—C

p

=

_ 1 [G,2z2 +C2 (r — z)2] p(r)

(r) —

r=0

r=z +1

E [(z +1_1_ _2 2

p(r) + C1 (z + 1

r=0

+ Ci

z+ 1) p(z + 1) 2

1 — 1 [(z + 1) 2 — z2]p(r) C1 (z + 1)2 p (z + 1) La 2r 2 (z +1) r=z +1 +C2

I p ( r) + C 1 (2z + 1) 2

• I- = 0

2r

1 —[ r=z +1 2r

v r=z+1 r

+C2

— z — 1)2 — (r — z)2] p(r)

.) p(► 1

—(1-2r + 2z) p (r) r=z +1 2r

Inventory Theory

83

1 1 = 1 I p(r) + Ci (z + 11 1 p(r) + C 2 (z. + "- p(r) 2 r=z +1 r 2 r=z +1 r r=0

y 2

-C 2

p(r) r= 7.

1) 1 p (r) + (C1 + C2 ) I z + p (r) 2 r=0 r= z+1

=C

p(r) -

p (r) r=0

r=0

v _ = (C1 + C2 ) [ i p(r) + (z + 1) 1 p (r)i C -7, 2) r=z +1 r r=0 Similarly, [z -1 AC (z - 1) = (Ci + C2 ) 1 p(r) + (z - -) r=0

)

j r=z

[

..

p(r) = r=0

1 p(r) - C 2

from (2) for minimum value of z, we have [z (C1 +C2 ) p(r) + - -1) . 1 p(r)] -C 2 6

0.01

0.01

0.01

0.00

Find the optimal number of spare parts which should be ordered with the order of the machine. The probability distribution of monthly sales of a certain item is as follows : Monthly sales (r) :

0

1

2

3

4

5

6

Probability p (r) :

0.02

0.05

0.30

0.27

0.20

0.10

0.06

The carrying cost is Rs.10.00 per unit per month. The current policy is to maintain a stock of four items at the beginning of each month, assuming that the cost of shortage is proportional to both time and quantity short, obtain the inputed cost of shortage of one item for one unit of time. [Hint : Use relation 3 r=0

4.

C2

4

p(r)« p(r) C1 + C 2 r= 0

17 .7 < C2 < 52.5 A news-paper boy buys papers for Rs. 2.40 each and sells them for Rs. 3.00 each. He cannot return unsold news-papers. Daily demand has the following distributions : No.of customers

23

24

25

26

27

28

29

30

31

32

(r) : Probability p(r): 0.01 0.03 0.06 0.10 0.20 0.25 0.15 0.10 0.05 0.05 If each day's demand is 'ndependent of previous day's demand how many papers should he order each day ? [Meerut 2009]

98

Operations Research

5.

Solve .Q. N. 4. If the news paper boys buys papers for Rs. 2.60 each and sells them for Rs. 3.60 each. [Meerut 2007 (B.P.)] Solve Q. 4. If the news paper boy buys papers for 13 paise each and sells them for 18 paise each. A news-paper boy buys papers for 5 paise each and sells them for 6 paise each. He cannot return unsold news-paper.Daily demand r for news-papers follows the distribution.

6. 7.

10

11

12

13

14

15

16

0.05

0.15

0.40

0.20

0.10

0.05

0.05

No.of customers (r) : Probability p (r) :

8.

If each days demand is independent of the previous day's how many papers should he order each day ? A news-paper boy buys papers for 25 paise each and sells them for 40 paise each. He cannot return unsold papers. Daily demand has the following distribution : No.of customers (r) :

23

24

25

26

27

28

29

30

0.01 0.04 0.05 0.10 0.15 0.25 0.15 0.10 0.08 0.07

Probability p (r) :

9.

22

21

If each day's demand is independent of the previous day's how many papers should he buy daily. Establish the formula you use. A news-paper boy buys papers for 40 paise and sells them for 70 paise each. He cannot return unsold news-papers. Daily demand has the following distribution. 25 No.of customers (r): Prob. p (r) :

26

27

28

29

30

31

32

33

34

35

36

0.03 0.05 0.05 0.10 0.15 0.15 0.12 0.10 0.10 0.07 0.06 0.02

If each day's demand is independent of the previous day's. How many papers he should order each day ? 10. A news-paper boy buys paper for 60 paise each and sells them for Rs. 1.40 each. He cannot return unsold news-papers. Daily demand has the following distribution. No.of customers

23

24

25

26

27

28

29

30

31

32

(r) :

Probability p (r) : 0.01 0.03 0.06 0.10 0.20 0.25 0.15 0.10 0.05 0.05

If each day's demand is independent of the previous day's, how may papers should be ordered each day ? [Meerut 2003] [Hint : Proceed as in Ex. 18 on page 92]. 11. A baking company sells one of its types of cakes by weight. If the product is not sold on the day it is baked, it can only be sold at a loss of 20 paise/kg. But there is unlimited market for one day old product. The cost of holding 1 kg of cake for one day is 15 paise; while the company makes a profit of one rupee/kg of cake

inventory Theory

99

sold on the day it is baked. If the probability density function of demand r is given as f(r) = 0.02 - 0.0002 r, find how many kgs of cake company should bake daily. [Meerut 1997(P)] [Hint : Proceed similarly as in Ex. 21 on page 94. Here Ci =0.20,C2 =1+0.15=1.15 C2 .% .1 f(r) dr z 2 - 200 z + 8518= 0 = 61.5] 0 Ci + C2 12. A baking company sells one of its type of cake by weight. It makes a profit of Rs. 2 per kg. on every kg. of cake sold on the day it is baked. It disposes of all cakes not sold on the day they are baked, at a loss of Rs. 0.50. If demand is known to have p.d.f. f (r) = 0.03 - 0.0003r. Find the optimum amount of cake the company should bake daily. [Meerut 2009 (BP)]

13. A shop owner places orders daily for goods which will be delivered 15 days later. On a certain day the owner has 10 items in stock and he has placed orders of 3 units for each of the previous 14 days. If C1 = Rs. 0.15 and C2 = Rs. 0.95, where C1 and C2 are carrying and shortage costs per unit of item. The distribution of requirements of r units for 1 r 15 is given by f(r) = 0.02 - 0.0002 r. How many items should be ordered for the 15th day hence ? [Hint Proceed similarly as in Ex. 21 on page 94. Here z = 63 appr. Items to be ordered for the 15th day hence = 63 - (10 + 3 x 14) = 11] 14. A baking company sells one of its types of cakes by weight. If the product is not sold on the day it is baked, it can only be sold at a loss of 15 paise per pound and there is unlimited market for one day old cake. The company makes a profit of 95 paise on every pound of cake sold on the day it is baked. Past daily orders form a triangular distribution with density function f(x) = 0.02 - 0.0002 x. 15. A baking company sells one of its types of cakes by weight. It makes a profit of 95 paise a pound on every pound of cake sold on the day it is baked. It disposes of all cakes not sold on the day it is baked at a loss of 15 paise a pound. If demand is known to he rectangular between 3000 and 4000 pounds, determine the optimum amount baked. 16. A baking company sells cake by the kg. It makes a profit of Rs. 5.00 a kg., on every kg sold on the day it is baked. It disposes of all cake not sold on the day it is baked at a loss of Rs. 1.20 a kg. If demand is known to be rectangular between 2000 and 3000 kg., determine the optimal daily amount baked. [Meerut 2008]

N + ANSWERS + 1.

4

2.

2

3.

Between 17.7 to 52.5 4.

27

5.

27

6.

27

7.

11

26

9.

30

10. 28

11. 61.5

12. 33.33

14. 63 or 136

15. 3864 pounds

16. 2806 appr

13. 11

8.

Operations Research

100

3.22 Purchase Inventory Models with Price Breaks It is a common practice that some discounts are offered by sellers to encourage large purchase orders. Thus, now we shall consider those problems of inventory in which the purchase price is not constant but decreases with the increase in the quantity or purchases. Let P. be the price of an item if the quantity purchased lies in between b j-1 to

b j. Price of an item Range of the purchased quantity ici ho 1 < Pl hi 5 q2 < b2

P2

3./-1

q j P2 > P3 > > Pn The price of item falls at the points bi , b2, are called the points of the price breaks.

....,bn

so they

3.23 Model XI : Purchase Inventory Model To find the optimum (economic) order quantity q which minimises the total cost, under the following assumptions. (i) P the price of an item, (ii) I the cost of carrying one rupee into the inventory for one year, (iii) r demand in one year (i. e., demand rate), (iv) C 3 set up cost per order, (v) q quantity ordered, (Optimum order quantity), (vi) shortages are not allowed. Here in this model the purchasing price of the inventory is also to be included in the total cost which is to be minimized. The total demand in one year is r the total number of runs in one year = n = r ,each of size q. q

If t is the time period for each run, then

nt = 1

1

— =q n r Now the average number of month inventories for each run 1 1 q2 = — . qt = — — 2 r 2 t=

Inventory Theory

101

and the number of lot month inventories per run 4t 1 2

q

1 q 2r

The associated annual costs are as follows : Purchase cost of item per year = rP r Set up cost per year = C3. n = C3. — q Annual inventory holding costs associated with the set up costs 4 )1 = 1C31 n.C3. (1`2r 2 Annual inventory holding costs associated with the purchasing costs = rp (1 1)1 = ipq r) 2 If C(q) is the total expected cost for one year, then r 1 1 C(q) = rP + C3 — + — I Pq + — C31 q 2 2 for C(q) to be minimum, Cr d — C(q) = — + —P = 0 2 2 dq 4 1(2rC3 1 q=q*,_1 IP

...(2)

2C 3r 3 •4 q* given by (2) is the optimal order level for minimum annual cost. Minimum annual cost is given by d2

dq2C(q) — — > 0

C(q*)

or

rP + C 3r.IP

C(q*) = rP + 2C 3I

2rC3

l

1p 2

2rC3 IP lI

+ C31

2

(2rIC 3P)}

...(3)

3.24 Model XII : Purchase Inventory Model with One Price Break Here we consider the purchase inventory problem with one price break bi, i. e., P, Price of an item Range of the purchased quantity '4

P1 P2

boq

Lot size Fig. 3.14 Economic lot-size curves • one price break

C (2) (q) are as shown in fig. 3.14. Here P2 < Pa , we have

rP2 +1 +2 1 C3I 2C 3 I rP1

Thus, from equations (1), (2) and the fig 3.14 it is obvious that C (1) ((ii*) > C (2) (q 2*) i.e., the minimum cost for curve II (corresponding to P2) is less than the minimum cost for curve I (corresponding to pi ). To find the economic lot size (optimum order quantity) for the model we proceed as follows.

103

Inventory Theory

Step I : First we calculate q2* from (4). If q2* > b1, as shown in fig. 3.15 then the optimum order quantity will be q2*.

" • /1

e

0 4.

Lot size

bo f_z q < 1)1

b1

>q ql q

* 2 b1 Lot size

bo 5_ qi < bl

4:17 .

b1 < q,

Fig. 3.15 Economic lot size curves : Fig 3.16 Economic lot-size curves :

412*

q2*

Step 2 : If q2*< b1 as shown in fig. 3.16. Then q2* is not the optimum order quantity i. e., quantity discounts are not applicable to the purchase quantity q2*. Since the cost is minimum at a point for at q 2* < b,, the total expected cost will be which the abscissa is less than monotonically increasing for q2 b1 and thus the minimum cost will be at q = b1. Hence, the optimum purchase quantity is obtained by comparing total expected cost for q = qi* with that q = bl From (1), and (2) C (1) (V) = C3.

+ q *

2

/Pi qi* a- /Pi + 2- C3/

(5)

i

r + /P214 + rP2 + 12.C3/ C (2) (bi ) =C3. — b 2 i

and

...(6)

If C (1) (q1*)< C (2) (k) then ql* is the optimum order quantity. Otherwise b1 is the optimum order quantity. • Working Method (i) Find q2* from (4). If q2* ?_ b1, then q2* is the optimum order quantity. (ii) If q2* < b1, then find ql* from (3), and C (1) (q1*) and. C (2) (b1 ) from (5) and (6) respectively. If c

(q 1

*)

< c

(2) c 100 then q1* is the optimum order quantity

otherwise b1 is the optimal order quantity.

3.25 Model XIII : Purchase Inventory Model with Two Price Breaks In this model we consider the purchase inventory problem with two price breaks b-i and b2, i. e.

Operations Research

104

P, Price of an item Range of purchased quantity '4 Pi

bo < gl < b1

P2

bl q2 < b2

P3

q3 h2

where P1 > 132 > P3 If C (i) (q), are the expected costs for one year when the purchased quantity q= qi (j =1, Z 3) lies in the range b 1

q(qi )< b j respectively, then

31 C ( j)(q) rPj + C3 — 1. + 2 IP.q + 1C 2

—(1)

q

d ( ) Now — C (q) = 0 give

dq

2rC 3 j„ , 3j

)= 1, 23

q1*, q 2*, q3* give the optimum order level for C (1) (q1 ), C respectively.

(2)

(2) (q2 ) and C (3) (q3 )

Here also with the same general discussions as in model XII, we arrive to the Following conclusion, to obtain the economic lot size for the model.

•

Working Method

(i)

Find q3* . If q3* ?_ b2, then q3* is the optimum order quantity.

(ii) If b1 < q3* < b2, then find q2' and if 1)1 q2* < b2, find C (2) (q2*) and C (3) (b2 ). If C (2) (q 2*) < C (3) (b2 ), then q2* is the optimum order quantity otherwise b2 is the optimum order quantity. (iii) If q2* < bi , then find q1* Which will satisfy ql* < b1 , find C (1) (q1'`) and C (2) (b1) If c, C (2) eh1 ), then (11* is the optimum order quantity otherwise b1 is the optimum order quantity. (q

* )
O. dq

(1

3

(2) give the optimal order quantity for C 0)(q) to be minimum. Minimum annual cost is given by

C*(i) (q) = C (i)(qt)= C 3r. or

ir

p,

- - IP;

C3

2rC3 2 '

1 + rP• + -C3I 2

C *(i) (q) = (2rIC 3Pj ) + rPi + ZC3 I

...(3)

Since Pi > P2 > P3 > > Pn

qi* < 42" < < qn* Thus, we see that corresponding to (n - 1) price breaks b1, b2.... , bn _ 1, there are n cost functions C*(-1) (q)and n economic lot sizes q j*, j = 1, 2, n given by (3) and (2) respectively. With the same general discussions as in model XII, we arrive to the following

conclusion to obtain the economic4ot size for the model. • Working Method (i) First Calculate qn*, from (2) If bn -1 0, gives 1 - vn

(1 - v) v n _vn+1

or

) (1 —vn)

1- v n 1 - v Cn+

or Cn +1 >

P(n)

(1 - v 1-v

1—V

C n+1 — P(n)i> 0

P(n) > 0 since

or C n +1

(1 - v) v n is positive as 0 < v < 1 (1 - v n+1)(1-vn )

> A +Ci + C2v +C3v 2 + 1 + v + v2 +

+C n v n + vn

...(6)

Operations Research

130 AC(n - 1) < 0, on simplification gives. 1 - v -1 Cn - P(n - 1) < 0 1-v + Cn A + 1 + C 2v + C 3v 2 + C„ < vn - 2 1 + v + v2

Similarly,

or

2 (7)

From (6) and (7) we conclude the same replacement policy as in article 4.5. (4) reduces to 1 - vn 1- v n- 1 P(n 1) < 0< C n+ 1 — P(n) ...(8) C 1- v 1- v which gives the best replacement age n of machine for which C (n) or P(n) is minimum. Then minimum value of x is obtained from (2).

gituttbudive Aaniptia Example 6 : The cost pattern for two machines A and B when money value is not considered, is given as follows : Year

Cost at the beginning of year in Rs. Machine A 900 600 700 2200

1 2 3 Total

Machine B 1400 100 700 2200

Find the cost pattern for each machine, when money is worth 10% per year, and hence find which machine is less costly. [Meerut 2002] Solution : The total expenditure for each machine in three years is equal to Rs. 2200 when the money value is not taken into consideration. Thus, the two machines are equally good if the money has no value over time. When money value is 10% per year, the discount rate 100 v- 0.9091 100 + 10 The present value of the maintenance costs in the years 1, 2, 3 for two machines are shown in the following table. Year

Machine A (Rs.)

Machine B (Rs.)

1

900

1400

2

600 x (0.9091) = 545.45

3

700 x (0.9091)2 = 578.52

700 x (0.9091)2 = 578.25

Total

Rs. 2023.97

Rs. 2069.43

100 x (0.9091) = 90.91

Replacement Problems

131

The present value of the total expenditure for machine A in three years is less than that for machine B. Hence, machine A is less costly. Example 7 : Let the value of money be assumed to be 10% per year and suppose that machine A is replaced after every 3 years whereas machine B is replaced after every six years. The :'early costs of both the machines are given as under : Year :

1.

2

3

4

5

6

Machine A :

1000

200

400

1000

200

400

Machine B :

1700

100

200

300

400

500

Determine which machine should be purchased. [Rohilkhand 2001] Solution : Since the value of money is 10% per year, the discount rate, 10 100 v— 100 + 10 11 The total discount cost (present worth) of costs of machine A for 3 years is = 1000 + 200 x (10/11) + 400 x (10/11)2 = Rs. 1512 approx. and the total discount cost (present worth) of costs of machine B for 6 years = 1700 + 100 x (10/11) + 200 x (10/11)2 + 300 x (10/11)3 + 400 x (10/11)4 + 500 x (10/11)5 = Rs. 2765 approx. Average yearly cost of machine A = 1512 / 3 = Rs. 504 and average yearly cost of machine B = 2765 / 6 = Rs. 460.83 From this machine B looks to be less costly than machine A. But it is not true since the periods of two machines considered are not the same. Now the total discount cost of machine A for 6 years = 1000+ 200 x (10 / 11) + 400 x (10 / 11)2 +1000 x (10/ 11)3 + 200 x (10/ 11)4 + 400 x (10 / 11)5 = Rs. 2647 approx. The present value of the total expenditure in 6 years on machine A is less than that for machine B. Thus, the machine A is less costly and hence machine A should be purchased. Example 8 : If you wish to have a return of 10% per annum on your investment, which of the following plans would you prefer ?

1st cost Scrap value after 15 years Excess of annual revenue over annual disbursement

Plan A (Rs.)

Plan B (Rs.)

200000 150000 25000

250000 180000 30000

Solution : Discount rate at 10% per annum, v = 100— 0.9091 100 + 10 Single payment present worth factor (pwf) at 10% for 15 years = V15 = (0.9091)15 = 0.2394

Operations Research

132

and annual payment present worth factor (pwf) at 10% for 15 years = v + v2 +

v15 =v (1 — v14 )/(1 —v)

= (0.9091) {1 — (0.9091)15} / (1 — 0.9091) = 7.6066 The present value of the total cost on the two plans for aperiod of 15 years are tabulated in the following table Present value of

Plan A in (Rs.)

Plan B in (Rs.)

200000

250000

1st cost

(1)

Scrap value after 15 years

(2) 150000 x 0.2394 = 35910 180000 x 0.2394 = 43092

(3) 25000 x 7.6066 = 190165 30000 x 7.6066 = 228198 Excess of annual revenue over annual disbursement 26075

Present value of total return in 15 years in Rs. (2) + (3) — (1)

21290

Since the present value of the total revenue in 15 years on plan A is more than that on plan B. Hence, the plan A should be preferred. Example 9 : The cost of new car is Rs. 10000. Compare the optimum moment of replacement assuming the following cost informations. Age of car, n Repair cost in n-th year Salvage value at the end of the n-th year 1

5000

8000

2

10000

6400

3

10000

5120

Assume that repairs are made at the end of each year only if the car is to be retained and are not necessary if the car to be sold for its salvage value. Also assume that the rate of discount is 10%. Solution : Here we shall compare the present value of the total investments on the car in 6 years with replacement policy after each year, two years and three years. Note that the period of 6 years is taken as the L.C.M. of 1, 2, 3 is 6. Since the rate of discount is 10%, discount rate, v

100 — 0.9091 100 + 10

Determination of discounted cost i. e., the present value of total investments.

Replacement Problems

133

Year Cost of the Repair cost Salvage new car at at the end value at n the end of of n-th the end of year n-th year n-th year in Rs. in Rs. in Rs. (iii) (ii) (iv) (i)

Total cost in Rs. (v) = (ii) + (iii) - (iv)

Discount factor v"

(v) x v n

Replacement after each year. 0

10000 -

1

10000

-

8000

2000

0.9091

1818.20

2

10000

-

8000

2000

0.8265

1653.00

3

10000

-

8000

2000

0.7513

1502.60

4

10000

-

8000

2000

0.6830

1366.08

5

10000

-

8000

2000

0.6209

1241.80

-

8000

- 8000

0.5645

- 4516.00

6

-

10000 1.0000 10000.00

4

Total = 13065.68

Replacement after 2 years 0 10000 -

-

10000 1.0000 10000.00

1

-

5000

-

5000

0.9091

4545.50

2

10000

-

6400

3600

0.8265

2975.40

3

-

5000

-

5000

0.7513

3756.50

4

10000

-

6400

3600

0.6830

2458.80

5

-

5000

-

5000

0.6209

3104.50

6

-

-

6400

- 6400

0.5645

-3612.80

Total = 23227.90

Replacement after 3 years. 0 10000 -

-

10000 1.0000 10000.00

1

-

5000

-

5000

0.9091

4545.50

2

-

10000

-

10000

0.8265

8265.00

3

10000

-

5120

4880

0.7513

3666.34

4

-

5000

5000

0.6830

3415.00

5

-

10000

10000

0.6209

6209.00

-

-

6

-

5120 -5120 0.5645 -2890.24 Total = 33210.60

Operations Research

134

Since the total discounted cost (1. e., the present value of costs) in 6 years with three replacement policies, that with replacement annually (i. e., every year) is minimum, hence yearly replacement policy is the optimum policy. Example 10 : The cost of a new machine is Rs. 5000. The maintenance cost of nth year is given by C = 500 (n - 1); n = 1, 2, .... . Suppose that the discount rate per year is 0.5. After how many years it will be economical to replace the machine by new one ? [Meerut 1996] = 500 (n 1), n = 1, 2, ....,1 - v= 0.5. Solution : Here v = 0.5, A = Rs. 5000 and Cr, If n years is the optimum replacement age, then from (3), article 4.5, we have l -v n 1 -v P(n) Cn P(n 1)< 0< 1 - v Cn + 1 -v where P(n) = A + C1 + C2v + C3v 2 +.... +C n V n -1 To find the value of n, satisfying (1), we compute the following table.

Year n (1)

v n-1 C r, a, n _ic, Rs. =(0.5)n -1 r '1 (2)

(3)

(4) = (2) x (3)

1 - vn 1-v

1 (1 - li n- ) .C n

(1 - v n ) 1-v

(5) = (2) - (4)

(6)

(7)

(8) = (6) - (7)

P(n)

c +1

•Cn+1 -P(n)

1

0

1.0000

00.00

00.00

500.00

5000.00

or


q 2 and

C1
X. If Rs. x is the running cost at time t. = 0 and the adjustment is made after time t hours then the running cost at time t (time of adjustment) = Rs. (x + bt). Since the running cost increases at a rate of b rupees per hour. ...(1) z = x + bt or t = (z - x)/b If C(z) is the total cost during the period of one adjustment to another adjustment, then C(z) = Cost of adjustment + Total running cost for time 0 to t. — x)/ b f t = (z - x)/ b 1 =K+ (x + bt) dt = K +[- (x + bt)2 =o 2 t=o 1 (z 2 - x 2 ) =K+— 2b

...(2)

Average cost per hour = C(z) / t = {[K + '

or

A(z) -

2b

(z 2 — X —

z-x

z+x Kb z x 2

Since, f(x) is the density function for the random variable x. Expected cost per hour z 2 X) b C E (z) = E (A(z)) = A(z) • f (x) dx = x = 0 (z + f (x) dx -x x=0 d Now C E CO is minimum for that value of z for which — CE (z) = 0 dz But

C (z) = LIT( = dz E dz

(_Kb z + x f (x) dx 2 0 Z - X

Operations Research

152

a ( Kb + z + x) f(x) dx 2 x = o az — x

fx

fX Kb + 1) f(x) dx J x 0 (z— x)2 2 X 1 = — Kb f(x) dx + —1 x f(x) dx 2 x=0 x = o (z — x) 2 x 1 1 f (x) dx = 1 f(x) dx + — = — Kb sx 2 Jx= x = 0 (z — x)2 for C E (z) to be minimum fx 1 1 CE (z) = — Kb f(x) dx + — = 0 dz 2 x = 0 (z — x)` fx 1 1 f(x) dx = 2Kb J x = 0 (z _ x)2

—(3)

From (3) we can find z. Case (ii). When z < X. In this case we cannot find minimum value of z. Thus, optimum time of adjustment is given by (3). Example 19 : Let p(t) be the probability that a machine in a group of 30 machines would breakdown in period t. The cost of repairing broken machine is Rs. 200.00. Preventive maintenance is performed by servicing all the 30 machines at the end of T units of time. Preventive maintenance cost is Rs. 15 per machine. Find optimal T which will minimise the expected cost per period of servicing, given that 0.03 , for t =1 p(t)= p(t — 1) + 0.01, for t = 2, 3, ...., 10

, for t = 11, 12, 13, ....

0.13 Hint : For t = 1 2

3

4

5

6

7

8

[Meerut 2004, 09] 9

10 11 12 13...

p1 = 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0

P12

...

= = P13 = • • • • •

Here, No = 30, Ni = Nopi = 1, N2 = N o p2 + N1 pi = 1, Similarly, N 3 = 2, N4 = 2, N 5 = 2, N 6 = 3, etc. Now average costs are, For T = 1, 30 x 15 + 200 x 1 = 650, For T = 2, (1/2) [30 x 15 + 200 x (1 + 1)] = 425, For T = 3, (1/3) [30 x 15 + 200 x (1 + 1 + 2)] = 416, For T = 4, 412, For T = 5, 410, For T = 6, 442 Since minimum is 410, For T = 5. Hence, optimal policy is to perform preventive maintenance to all machines at the end of 5th period. Thus, average cost per period = Rs. 410.00 •

153

Replacement Problems

+ Exercise on Chapter 4 + 1. 2.

State some of the simple replacement policies and give the average cost functions for the same, explaining your notations. [Meerut 2008 (BP)] (a) Write short note on replacement problems. (b) Discuss the importance of replacement models.

3.

[Meerut 2001, 02] (c) Discuss some important replacement situations. Write short note on different types of replacement models.

4.

Write short note on replacement of items that fails completely. [UP TECH MBA 2006-07]

5.

What is replacement ? Describe some important replacement situations and policies. [UP TECH MBA 2006-07]

6.

Explain how the theory of replacement .is used in the replacement of items whose maintenance cost varies with time ? [UP TECH MBA 2004-05]

7.

Write short note on Group replacement policy. [Meerut 2008]

8. 9.

Write short note on staffing problem. The cost of a machine is Rs. 5000. The running cost and the salvage value of the machine are given as under. Find the optimal replacement policy. 1

Year

2

3

4

5

6

7

8

Running cost in Rs.

1500 1600 1800 2100 2500 2900 3400 4000

Salvage value in Rs.

3500 2500 1700 1200

800

500

500

500

Salvage value means the cost at which the machine could be sold at the end of the year. 10. A firm is considering when to replace its machine whose price is Rs. 12200. The scrap value of the machine is Rs. 200 only. From past experience the maintenance costs of the machine are as under : Year Maintenance cost in Rs.

1

2

3

4

5

6

7

8

200

500

800

1200

1800

2500

3200

4000

Find when the new machine should be purchased. [Meerut 2001]

Operations Research

154

11. A truck owner finds from his past records that the maintenance costs per year,

of a truck whose purchase price is Rs. 8000, is as given below : 2

3

4

5

6

7

8

Maintenance 1000 cost

1300

1700

2200

2900

3800

4800

6000

4000

2000

1200

600

500

400

400

400

Year

Resale price

1

Determine at which time it is profitable to replace the truck. [Meerut 1999] 12. The cost per year of running a truck whose purchase price is Rs. 30000 are as

follows. Determine when is the replacement due. Year

1

2

3

4

5

Running cost (Rs.)

5000

6000

7000

9000

11500

Resale value (Rs.)

15000

7500

3750

1875

1000

6

7

14000 17000 1000

1000

13. The following table gives the running costs per year and resale values of a

certain equipment whose purchase price is Rs. 6500. At what year is the replacement due optimally, 1

Year

2

3

4

5

6

7

8

Running cost (Rs.)

1400 1500 1700 2000 2400 2800 3300 3900

Resale value (Rs.)

4000 3000 2200 1700 1300 1000 1000 1000

14. The cost of a truck is Rs. 3500. The salvage value and the running costs are

given below. Find the most economical age for replacement. Year

1

2

3

4

5

6

7

Running cost

600

700

800

900

1000

1200

1500

Resale value

2500

1833

1500

1250

1000

800

800

[Meerut 1994(P)] 15. The following table gives the running costs per year and resale prices of a

certain equipment whose purchase price is Rs. 10000. 1

2

3

4

5

6

7

Running cost (Rs.)

3000

3200

3600

4200

5000

5800

6800

Resale value (Rs.)

7000

5000

3400

2400

1600

1000

1000

Year

[Meerut 2000]

Replacement Problems

155

16. The maintenance cost and resale value per year of machine whose purchase

price is Rs. 7000 is given below : Year

1

2

3

4

Maintenance cost (In Rs.)

900

1200

1600

2100

4000

2000

1200

600

Resale value (In Rs.)

When should the machine be replaced ? [UP TECH MBA 2004-05] 17. A manufacturing firm has come to know from his past records that a machine costing Rs. 56000/- is not working satisfactorily in spite of its regular maintenance. With a view to replacing this machine the following facts were obtained : Year

Annual running Cost (Rs.)

Resale value (Rs.)

1

7,000

28,000

2

9,100

14,000

3

11,900

8,400

4

15,400

4,200

5

20,300

3,500

6

26,600

3,000

7

33,600

3,000

8

42,000

3,000

When should the.machine be replaced ? [UP TECH MBA 2006-07] 18. XYZ manufacturing company is using a machine whose purchase price is Rs. 65,000. The installation charges amount to Rs. 18,000 and the machine has a scrap value of only Rs. 8,000 because of firm has a monopoly of this type of work. The maintenance cost in various years is given in the following table : Year

Maintenance Cost (Rs.)

1

1250

2

3750

3

5000

4

7500

5

10500

6

14500

7

20000

8

24000

9

30000

Operations Research

156

Determine after how many years should the machine be replaced, assuming that the machine replacement can be done only at the years ends. [UP TECH MBA 2007-08]

19. The data of the operating costs per year and resale prices of equipment A whose purchase price is Rs. 10000 are given below : Year

1

2

3

4

5

6

7

Operating cost (Rs.)

1500

1900

2300

2900

3600

4500

5500

Resale value (Rs.)

5000

2500

1250

600

400

400

400

(0 What is the optimum period for replacement ?

[Meerut 1993, 2006]

(ii) When Equipment A is 2 years old, Equipment B, which is a new model for the

same usage is available. The optimum period for replacement is 4 years with an average cost of Rs. 3600. Should we change equipment A with that of B ? If so, when ? [Meerut 2006]

20. Assuming that the present value of one rupee to be spent in a year's time is Rs. 0.9 and c = Rs. 3000, capital cost of equipment and the running cost are given in the table. When should the machine be replaced ? Year

1

2

3

Running cost (Rs.)

500

600

800

4

5

6

1000 1300 1600

7 2000

21. The cost of a new machine is Rs. 500. The maintenance cost of n-th year is Suppose that the discount rate per given by Rn = 500 (n — 1), for n = 1, 2, year is 0.05 (or 5%). After how many years, it will be economical to replace the machine by new one ? [Meerut 2007]

22. A truck is priced at Rs. 60000 and running costs are estimated at Rs. 6000 for each of the first four years, increasing by Rs. 2000 per year in the fifth and subsequent years. If money is worth 10% per year, when should the truck be replaced ? Assume that the truck will eventually be sold for scrap at a negligible price.

[Meerut 1994, 98(B.P.)]

23. A machine cost Rs. 10000 operating costs Rs. 500 per year for the first five years. In the sixth and succeeding years operating cost increases by Rs. 100 per year. Assuming a 10% discount rate of money per year, find the optimum length of time to hold the machine before we replace it. 24. The following are the costs of running a particular car to date and the forecast into the future. Assuming that the car will be replaced by a similar car. When is the best time to replace it and what will be the average yearly running cost ?

157

Replacement Problems Year n

Resale value at the end of years in Rs.

0

-

Cost of petrol and Other maintenance tax during the cost during the n-th year in Rs. year in Rs.

700

—

—

1

625

90

10

2

575

90

30

3

550

90

50

4

500

90

70

5

450

90

90

6

400

90

110

7

350

90

130

300

90

150

25. A person is considering to purchase a machine for his own factory. Relevant

data about alternative machines are as follows : Machine A

Machine B

Machine C

Present investment (Rs.)

10000

12000

15000

Total annual cost (Rs.)

2000

1500

1200

10

10

10

500

1000

1200

Life (years) Salvage value

As an adviser to be buyer, you have been asked to select the best machine, considering 12% normal rate of return : You are given that : (i) Single payment present worth factor (Pwf) at 12% for 10 years = 0.322 (ii) Annual series present worth factor (Pwf) 12% for 10 years = 5.650. Hint : Proceed similarly as in Example 8 on page 131. 26. The mortality rates as obtained for an electric compound are given as under : Year

1

2

3

4

5

6

% failure at the end of the year

8

22

45

70

85

100

There are 1500 items in operation. It costs Rs. 20.00 to replace an individual item and Rs. 0.50 per items, if all are replaced simultaneously. If the policy is either to remove the items whenever they fail or a policy of group replacement along with the individuals. Suggest which should be better economically. [Meerut 1996(P)]

Operations Research

158

Hint :P1 = 0.08, P2 = 0.14, P3 = 0.23, P4 = 0.25, P5 = 0.15, P6 = 0.15 P1 + P2 +P3 +P4 +P5 +P6 = 1, P7 = 0 = P8 = N o = 1500, N1 = 120, N 2 = 306, N 3 = 386, N 4 = 476, N 5 = 410 etc. Average life of the component = E X. P = 3.70 Average number of replacement every year = 1500/3.70 = 405 27. The following failure rates have been-observed for certain items : 1

End of month

0.10

Prob. of failure of date

3

2

0.30 0.55

4

5

0.85

1.00

The cost of replacement an individual item is Rs. 1.25. The decision is made to replace all items simultaneously at fixed intervals and also to replace individual items as they fail. If the cost of group replacement is 50 paise, what is the best interval of group replacement ? At what group replacements price per item, would a policy of strictly individual replacement becomes preferable to the adopted policy ? 28. The following failure rates have been observed for a certain type of light bulb : End of week

1

Prob. of failure to date

2

3

5

4

6

7

8

0.05 0.13 0.25 0.43 0.68 0.88 0.96 1.00

The cost of replacing an individual failed bulb is Rs. 1.25. The decision is made to replace all bulbs simultaneously at fixed intervals, and also to replace individual bulbs as they fail in service. If the cost of group replacement is 30 paise per bulb and the total number of bulbs is 1000, what is the best intervals between group replacements ? At what group replacement price per bulb would a policy of strictly individual replacement becomes preferable to the adopted policy ? 29. A system has a large number of light bulbs, all of which we must keep in working order. If a bulb fails in service, it cost Rs. 1 to replace, but if we replace all the bulbs in the same operation we can do for only Rs. 0.35 a bulb. Find the optimum period for group replacement policy. The life distribution of the bulbs is as follows : 1

Week

2

3

4

5

6

Prob. of the bulb failing in the week 0.09 0.16 0.24 0.36 0.12 0.03 30. The prob. p,1 of failure just before age n are shown below. If individua replacement costs Rs. 1.25 and group replacement costs Rs. 0.50 per item, find the optimal group replacement policy (assuming that there are 1000 bulbs in use). it pn

1

2

3

4

5

6

7

8

9

10

11

0.01 0.03 0.05 0.07 0.10 0.15 0.20 0.15 0.11 0.08 0.05 [Meerut 1995]

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159

31. A computer contains 10000 resistors. When any resistor fails, it is replaced. The

cost of replacing a resistor individually is Rs. 1 only. If all the resistors are replaced at the same time, the cost of a resistor would be reduced to 35 paise. The percent surviving say S(t) by the end of month t and P(t) the probability of failure during the month t are given as below : Month t

0

S(t)

100

P(t)

—

1

2

3

4

5

6

97

90

70

30

15

0

0.03

0.07

0.20

0.40

0.15

0.15

What is the optimum replacement plan ? [Meerut 2005]

32. Find the intervals for the replacement of all the bulbs simultaneously in Ex.13 page 142 if the cost of a bulb is Rs. 10 if replaced individually and Rs. 4 if all the bulbs are replaced simultaneously. [Meerut 2010] 33. The following mortality rates have been found for a certain type of coal cutter motor Weeks

10

20

30

40

50

Total percentage failure up to the end of 10 weeks period

5

15 35

65

100

If the motors are replaced over the weekend the total cost is Rs. 20. If they fail during the week the total cost is Rs. 100 per failure. Is it better to replace the motors failure and if so when ? 34. Truck tyres which fail in service can cause expensive accidents. It is estimated that a failure in service results in an average cost of Rs. 1000 exclusive of the cost of replacing the burst tyre. New tyres cost Rs. 400 each and are subject to mortality as in table below. If the tyres are to be replaced after a certain fixed mileage or at failure (which ever occurs first), determine the replacement policy that minimizes the average cost per mile. Truck tyre mortality table Age of tyres at failure in miles

•

Proportion of tyres

5. 10000

0.000

10001 — 12000

0.020

12001 — 14000

0.035

14001 — 16000

0.063

16001 — 18000

0.100

18001 — 20000

0.220

20001 — 22000

0.345

22001 — 24000

0.205

24001 — 26000

0.012

Mention the assumptions you have made to arrive at the solution.

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35. A piece of equipment either can fail completely, so that it has to be scraped (no

salvage value) or may suffer a minor defect which can be repaired. The probability that it will not have be scraped before age t is f (t). The conditional probability that it will need a repair in the interval t to t + dt, given that it is in running order at age t, is r(t) dt. The probability of a repair or complete failures dependent only on the age of the equipment, and not on the previous repair history. Each repair costs C, and complete replacement costs K. For some considerable time the policy has been to replace only on failure. (a) Derive a formula for the expected cost per unit time of the present policy of replacing only on failure. [Meerut 1997] (b) It has been suggested that it might be cheeper to scrap equipment at some fixed age T. Show that the expected cost per unit time of such a policy is C0

[Meerut 1997]

f(u) • r(u) du + K f(u) du Jo

(c) By differentiating this expression, find a condition to be satisfied by T for minimum cost. 36. What are the situations which makes the replacement of items necessary? [Meerut 2010]

•

+ ANSWERS + 9. After 4 years 11. After 5 years

10. After 6 years

13. After 5 years

14. After 5 years

15. After 4 years

16. 4 years

17. 5 years

18. 6 years

19. (i) 5 years (ii) Change when

21. 2 years

22. 24. 26. 28. 30. 32. 35.

12. After 5 years

equipment A is 4 years old After 9 years 3 years, Rs. 170 Group replacement yearly 3 weeks; After every 5 weeks Two weeks (a) Total expected cost = K + C

23. After 19 years; 25. Machine B should be purchased 27. 3 weeks 29. 3 weeks 31. After every 3 months 33. After every 20 weeks 0

f(u) • r(u) du

(b) Minimum expected cost E(T) per unit time is given by E(T) = C • r(T) • • •

Waiting Lines or Queuing Theory im

t. 5.1 Introduction A group of items waiting to receive service, including those receiving the service is known as a waiting line or a queue. Queuing theory involves the mathematical study of "queues" or waiting lines. The formation of waiting lines (or queues) is a common phenomenon which occurs whenever the current demand for a service exceeds the current capacity to provide that service. The queues of peoples may be seen at a cinema ticket window, bus stop, reservation office, counters of super market etc. The person waiting in a queue or receiving the service is called the customer and the person by whom he is serviced is called a server. The customers arriving for service may form one queue and be serviced through only one station, as in a doctor's clinic; they may form one queue and be serviced through several stations, as in barber shop; or they may form several queues and be served through as many stations, as at the check out counters of a super market.

5.2 Basic Queuing Process (system) and Its Characteristics [Meerut 2002, 04; Rohilkhand 1996]

The basic queuing process can be described as a process in which the customers arrive for service at a service counter (or station), wait for their turn in the queue if the server is busy in the service of the other customer and are served when the server gets free. Finally the customer leave the system as soon as he is served.

Characteristics of Queuing System

[Meerut 2001, 2002]

The queuing system is described by the following basic characteristics. 1. The input (or arrival pattern). 2. Queue (or waiting line). 3. The service discipline (or queue discipline). 4. The service mechanism (or service pattern). Input

Queue

Service mechanism

Fig. 5.1 : The basic queuing process

Departure (Served units)

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1. The input (or arrival pattern)

[Meerut 2002]

The input describes the pattern in which the customers arrive for service. Since the units, for service (customers) arrive in a random fashion therefore, their arrival pattern can be described in terms of probabilities. Here in this chapter we assume that they arrive according to Poisson process i. e., the number of units arriving until any specific time has a Poisson distribution. This is the case where arrivals to the queuing system occur at random, but at a certain average rate.

2. Queue (or waiting line) The units requiring service enter the queuing system on their arrival and join a

"queue" which is characterized by the maximum permissible number of units that it can contain called the capacity of the system. A queue is called finite if the number of units in it is finite otherwise it is called infinite. 3. The service discipline (or queue discipline) [Meerut 2002; Rohilkhand 2003] The service discipline refers to the manner in which the members in the queue are chosen for service. The following service disciplines are seen in common practice. (1) First come, first served (FCFS) : According to this discipline the customers are served in the order of their arrival. This service discipline may be seen at a cinema ticket window, at a railway ticket window etc. Here in this chapter we shall deal with the queuing models in which the service discipline in FCFS. (ii) Last come, first served (LCFS) : This discipline may be seen in big godowns where the units (items) which come last are taken out (served) first.

(iii) Service in random order. (iv) Service on some priority-procedure : Some customers are served before the others without considering their order of arrival i. e., some customers are served on priority basis.

4. The service mechanism (or service pattern) The service mechanism refers to

(i) The pattern according to which the customers are served : Here in this chapter we shall deal with the queuing models in which the service time follows the 'Exponential' and `Erlang' probability distributions. (ii) Facilities given to the customers : The facilities given to the customers can be classified in two categories : (a) Single channel : Here the customers are served by one counter only i. e., only one customer can be served at a time e. g. , at cinema ticket window. (b) Multi-channel : Here the customers are served by several (more than one) counters i. e., more than one customer can be served at a time e. g. , in the barber shop.

5.3 Customers Behaviour in a Queue Customers generally behave in the following ways :

(i) Balking : A customer may not like to wait in a queue due to lack of time or space or otherwise. (ii) Reneging : A customer may leave the queue due to impatience.

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163

(iii) Collusion : Some customers may collaborate and only one of them may join the queue. As at the cinema ticket window one person may join the queue and purchase tickets for his friends.

(iv) Jockeying : If there are more than one queues then one customer may leave one queue and join the other. This occurs generally in the super market.

5.4 Important Definitions in Queuing Problem Here we give the definitions of various terms used in this chapter. (i) Queue Length i Queue length is defined by the number of persons (customers) waiting in the line at any time. (ii) Average Length of Line : Average length of line (or Queue) is defined by the number of customers in the queue per unit time. (iii) Waiting Time : It is time upto which a unit has to wait in the queue before it is taken into service. (iv) Servicing Time : The time taken for servicing of a unit is called its servicing time. (v) Busy Period : Busy-period of a server is the time during which he remains busy in servicing. Thus, it is the time between the start of service of the first unit to the end of service of the last unit in the queue. (vi) Idle Period : When all the units in the queue are served. The idle period of the server begins and it continues upto the time of arrival of the unit (customer). The idle period of a server is the time during which he remains free because there is no customer present in the system. (vii) Mean Arrival Rate : The mean arrival rate in a waiting-line situation is defined as the expected number of arrivals occuring in a time interval of length unity. [Meerut 2004(0)] (viii) Mean Servicing Rate : The mean servicing rate for a particular servicing station is defined as the expected number of services completed in a time interval of length unity, given that the servicing is going on throughout the entire time unit. [Meerut 2004(0)] (ix) Traffic Intensity (or utilization factor) : In case of a simple queue the traffic intensity (or utilization factor) is the ratio of mean arrival rate and the mean servicing rate. i. e.,Traffic intensity = Mean arrival rate Mean servicing rate [Meerut 2004(0)]

5.5 The State of the System A basic concept in the analysis of a queuing theory is that of a state of the system. It involves the study of a system's behaviour over time. The state of a system may be classified as follows : (i) Transient State : A system is said to be in transient state when its operating characteristics are dependent on time. Thus, a queuing system is in transient state when the probability distributions of arrivals, waiting time and servicing time of the customers are dependent on time. This state occurs at the beginning of the operation of the system.

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(ii) Steady State : A system is said to be in steady state when its operating characteristics becomes independent of time. Thus, a queuing system acquires steady state when the probability distributions of arrivals, waiting time and servicing time of the customers are independent of the time. This state occurs in the long run of the system. Let Pn (t) denote the probability that there are n units in the system at time t, then the system acquires steady state as t -p oo, i.e., if lim Pn (t) = I3n (independent of time t). -> 00 In most of the queuing problems, steady-state solutions exist independent of the initial state of the queue. (iii) Explosive State : If the arrival rate of the system is more than its servicing rate, the length of the queue will go on increasing with time and will tend to infinity as t ---> 0.. This state of the system is said to be explosive state.

5.6 Poisson Process

[Meerut 1995(P), 96, 2003] In Poisson process the probability of n arrivals during time intervals of length t is given by Ro n e-Xt (i) Pn (t) n! where Xt is the parameter. Case I: When n = 0

From (i), the probability of no arrival in time At is given by (At) - (X At)0 . e-for Po - e-X.At 0! k2 , ,) = 1 — XAt + — (Atr- ... = 1 — XAt + 0 (At) 2! where 0 (At) denotes a quantity which is of smaller order of magnitude than At. If At is very small then 0 (At) = 0 Po (At) ,--- 1 — X At i. e. , the probability of no arrival in time At = 1 - XAt. Case II : When n = 1 From (i), the probability of one arrival in time At is given by (At) - (2 At)1 . e-lAt Pi 1! 2 76 = X At . 1 / - XAt + — (At)2 ... = XAt + 0 (At) 2! P1(4t) = XAt. i. e., the probability of one arrival in time At = XAt.

(If A t is very small)

Case III : When n m > 1 From (i), the probability of m (more than one) arrivals in time At is given by

Waiting Lines or Queuing Theory Pm (At) -

165

(XAt)m . e-?,At m! Xrn (AO' {1 m!

m!

X At +

X2 (At ) 2

2!

-{(At) m - X (At)m +1

Pm (At) = 0.

(If A t is very small)

(m> 1) Thus, the probability of more than one arrival in time At = 0. (If the terms of second and higher powers of At are negligibly small). From the above discussion we arrive to the following result which is sometimes known as postulates for Poisson Process. [Rohilkhand 2003] Postulates for the Poisson Process Postulate I.

The number of arrivals in non-overlapping intervals .are statistically independent, i. e., the process has independent increments. Postulate 2 The probability that an arrival occurs between time t and time t + At is equal to + 0 (At). i. e.,

Pi (At) X At + 0 (At).

where X is a constant independent of Pn (t), At is small increment int and 0 (At) denotes 0 (At) = 0. a quantity which is of smaller order of magnitude than At s. t., lim At Postulate 3

The probability of more than one arrival between time t and time t + At is 0 (At), i. e. , the probability of two or more arrivals during the small time interval A t is negligible. Thus,

Po (At) + Pi (At) + 0 (At) = 1.

5.7 Poisson Arrivals In general, arrivals in a queuing system do not occur at regular intervals, but tend to be scattered in some fashion. The "Poisson assumption" specifies the behaviour of arrivals, by postulating the existence of constant X , which is independent of time, queue length, or any other random property of the queue, such that ... (1) P (an arrival occurs between time t and t + At) = X At (Probability) if the interval At is sufficient small. A waiting line for which arrivals occur in accordance with (1) is called a queue with Poisson arrivals.

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5.8 Theorem Let n be a random variable (discrete) representing the number of arrivals in some time interval of length t, then if equation (1) is satisfied, n obeys a Poisson distribution. okon e-Xt Pn (t) n! (Probability of n arrivals in time t) [Meerut 1996; Agra 2002] where Xt is a parameter. Proof : Here the arrivals are Poisson and follow the postulate for the Poisson Process (see 5.6). Let (0, t) and (t, t + At) be two consecutive time intervals where At is sufficiently small. Case I : When n 1 There are n arrivals in the interval (0, t + At) in the following ways. (i) There are n arrivals in the interval (0, t) and no arrival in the consecutive time interval (t, t + At). The probability in this case will be Pn (t) Po (At) (Since the events are mutually exclusive) (ii) There are (n - 1) arrivals in the interval (0, t) and one arrival in the consecutive time interval (t, t + At). The probability in this case will be P„ _ 1 (t) . Pi (At) . (iii) There are (n - k) arrival in the interval (0, t) and k > larrivals in the consecutive time interval (t, t + At). The probability in this case will be P„ _ k (t) . Pk (At). Thus, adding above independent probabilities the probability of n arrivals in the time interval (0, t + At) is given by ...(1) _ k (t) . Pk (At) Now P0 (At) = Probability that there is no arrival in time interval (t, t + At) = 1 - probability that there is an arrival in (t, t + At) P„ (t + At) = P„ (t) . Po (At ) + P„

(t) . Pi (At) +

1- XAt + 0(At) Also from the postulate of Poisson process Pi (At) = XAt + 0 (At) and Pk (At) = 0 (At) From (1), we have P„ (t + At) = (t) . [1 - At + 0(At)]

k>1

+ Pn _ 1 (t). [XAt + 0 (At)] + Pn _ k (t) . [0 (A t)] P„(t) - P„ (t) At + X Pn _ 1 (t). At + 0 (At) or

Pn (t + At) - Pn (t) At

XP, (t) + XP

(t)+

Now taking limit as At 0, we have P„' (t) = - X P„ (t) XPn - (t).

0 (At) At ...(2)

167

Waiting Lines or Queuing Theory

Case II : When n = 0 Po (t + At) = Probability of no arrival in time interval (0, t + At) = Probability of no arrival in time interval (0, t) x Probability of no arrival in consecutive time interval (t, t + At) = Po (t) . Po (At) = Po (t) . [1 - X . At + 0 (At)] Po (t + At) - Po (t) 0 (At) - P0 (t) + At At Taking limit as At --> 0, we have Po' (t) = - XP0 (t)

—(3)

Po' (t) =-X Po (t) Integrating, log Po (t) = - + A.

or

When t= 0, P0 (0) = 1, so that A = 0 log Po (t) = - Xt ... (4)

or

Po = e-Xt Now putting n = 1 in (2). We have Pi (t) = - X Pi (t) + X Po{t) Or

Pi

(t) + x

[Substituting Po (t) = e -at from (4)]

(t) = Xe-lt

which is linear differential of 1st degree. e

dt

=e

xt

Multiplying both sides by I.F. e Xt and then integrating, we have e Xt . 131 (1) = C + Xe-Xt . e A" t dt = C + Xt when

t = 0, P1 (0) = 0, so that C = 0 P1 (t) =?.t. e -Xt

(5)

Again putting n = 2, in the equation (2), we have P2 (t) = - XP2 (t) + XPi (t) = - XP2 (t) + X. Xte-xt or

P2 (t) + XP2(t) = X.2t e-Xt

which is L.D.E. proceeding as before, we have (x02 e-kt

x2t 2 P2(t) = 2!

.e

=

2!

[Substituting from (5)]

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Proceeding in the similar way for n = 3, 4, ...., we arrive to the result — at)n e —Xt Pn n! n arrivals in the system follows Poisson Process. i.e.,

_.(6)

Note : If the arrivals follow Poisson Process, from (6) Mean (expected number of arrivals in a time interval of length t) = Taking t =1 it follows that the mean arrival rate for Poisson arrivals is just 2. .

5.9 Some Distributions Here we give some distributions which are used in queuing theory.

(i) Exponential distribution : If T is a random variable representing the inter-arrival time (the rime between consecutive arrivals), then T obeys an exponential distribution i. e. , if f (T) denotes the probability density function of T, then f(T) = X e-XT where X is parameter.

(ii) Regular distribution : Here the inter-arrival time is assumed to be constant. This distribution is defined by f(T) = 0 =1 a is constant.

for T < a for T a.

(iii) Erlang Service time distribution with k phases (Ek ) [Meerut 2003] The Erlang distribution is a very important distribution in queuing theory. The probability density function for the Erlang distribution is given by k)k

f

.t

k

(k -1)1

k"

... (1)

0, k__1 where4t and k are positive parameters of the distribution. The Erlang family has close tie with the exponential distribution. In particular, if t1, t 2, t k are k independent random variables with an identical exponential 1 distribution whose mean is — , then their sum, t

t=

+ t 2 + ....+t k .

has an Erlang distribution with parameters 1.1 and k. Thus, a servicing station in which a unit goes through k independent service 1 phases, each exponential with mean time —, will have service times obeying the µk k-th Erlang distribution with parameter [1. The Erlang family has many interesting properties, some of which are as follows : (i) Each member of the family has the same mean Vv. i.e., E(t) = 1/µ (Independent of k, the number of phases)

169

Waiting Lines or Queuing Theory (ii) Variance is given by

V(t) 1(11, 2 (iii) The mode of t is located at t -

k I for general k. 1.t k

k=co

k=3 k=2 k-1 1/211 2/3[t

1/p. t (service time)

Fig. 5.2 1 A family or Erlang distribution with mean - (constant) 11 i. e., for k = 1, the mode is at t = 0 for k = 2, the mode is at t =1 /2 1.1 for k = 3, the mode is at t = 2 /3 II, etc. (iv) The exponential distribution is a special case of the Erlang distribution for k = 1. (v) The degenerate (constant) distribution is a special case of the , Erlang distribution for k = For k = co, the mode = 141 and variance, V(t) = 0. Thus, for k = oo, from (1) we get a family-for which the time is constant.

5.10 An Important Theorem If the arrival pattern in a queuing problem follows a poisson process, then the random variable T representing the inter arrival time follows the exponential distribution and vice-versa. [Meerut 1995(P), 96(P), 98(B.P.); Rohilichand 2003]

Proof : If n is the number of arrivals during time interval of length t, then Pn (t) -

(X,t)n .e-xt n

Let t o = instant of an arrival initially

...(i)

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The random variable T represents the inter arrival time, therefore, there is no arrival in the interval (t 0 , t o + T) and t + T + AT will be the instant of consecutive arrival. -+ T -> ÷AT 4--

I to+ T

to (First arrival)

to+ T+ AT (Second arrival)

Fig. 5.3

From (i), Po (T) = Probability of no arrival in time T (XT) 0 . e -XT - e -XT 0! and

Po (T + At) = Probability of no arrival in time T + AT (T + AT)} ° e -X(T + AT) 0! = e -X(T + AT) = e -XT e -X AT = Po (T) . [1 — AT + 0 (AT)2 ]

where 0 (AT)2 contain the terms of second or higher power of AT. P (T + AT) - Po (T) = - Po (T) AT + 0 (AT) 2 or

Po (T + AT) — Po (T) AT

—

Po (T) +

0 (AT) 2 AT

Taking limit as AT -+ 0, we have — Po (T)= - Po (T) —Po (T) f(T)= X e-XT dT dT

from (ii)

(Probability density function) (Leaving -ve sign since probability density function is always non-negative). Hence, T follows the exponential distribution law. Converse : The converse of the theorem may be proved in the similar way. Remark : In the similar way, we can show that, if the time t to complete service on a unit follows the Exponential distribution i. e., f (t) = the density function of t = µ elµt where 1.1 is the mean servicing rate for a particular servicing station then n the number of units the servicing station could serve in time t, if the servicing is going on throughout t will follow the Poisson distribution. i. e., 4 t (n) = Probability [n servicing in time t, if there were no enforced idle time] t)"

n!

Waiting Lines or Queuing Theory

171

As in article 5.6 (OM (°) = Probability of no service in time A t = 1 — µ At (1)&(1) = Probability of one service in time A t = µ At (I)At (m) = Probability or more than one service in time At = 0, m > 1

and

5.11 Notations The following notations will be used throughout the chapter. T

The inter arrival time between two successive customers.

n

Number of customers in the queuing system.

Pn(t)

Transient state-probability that there are exactly n units in the system at any time t.

Pn

Steady state probability of having n units in the system lim Pn (t ) Pn Mean arrival rate of customers when there are n units present in the system. Mean service rate when there are n units in the system.

n

Mean arrival rate of customer (independent of n).

E

s

Mean service rate (independent of n). Number of parallel service stations.

X, P=—

Traffic intensity (or utilization factor).

0 (At) FCFS

A quantity which is of smaller order of magnitude than At. First come first served (service discipline).

00) E(Ls )

Probability density function of waiting time in the system. Expected number of customers in the system (waiting and in service) i. e. , expected or mean line length.

E(Lq )

Expected number of customers in the queue i. e., expected or mean queue length (excluding the number of units in service).

E (Ws )

Expected waiting time per customer in the system (including service time).

E(Wq )

Expected waiting time per customer in the queue (excluding service time).

> 0) Expected length of the non-empty queue.

E(W/W > 0) Expected waiting time of a customer who has to wait. (m) = m-C n Note : The value of p = X/I.t must always be less than 1, so that the steady state condition may be obtained. If p > 1 i. e. , > µ, then extremely large queues will be obtained.

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5.12 Classification of Queuing Models A

B C d e

[Meerut 2001]

A queuing model is symbolically represented as (A/B/C) : (d/e), where Arrival pattern of the units which is given by the probability distribution of inter arrival time of unit. The probability distribution of servicing time of the unit (which is actually serviced). The number of servicing channels (stations) in the system. Capacity of the system i. e., maximum number allowed in the system (in service + waiting). The manner in which units are selected for service in the system (service discipline). The various types of the Queuing model discussed in this chapter are as follows.

Model I : (Erlang model); (M/M/1) : (./FCFS) (exponential inter arrival time) Here first M stands for Poisson arrival (exponential service time) second M stands for Poisson departure stands for single server. 1 stands for infinite capacity of the system. (service discipline). FCFS notation representing first come first served

Model II : (M/M/1) : (../FCFS) (General single station queuing model) , This model is same as Model I except that heie-the-rate-of_arrind service depend on the length of the line i. e., here some person interested in joining the queue may not join due to long queue and the servicing rate of the server (servicing station) is also effected by the length of the queue.

Model III : (M/1 l/1) : (N/FCFS) This is the model with Poisson arrival, Poisson departure, single servicing channel, capacity of the system is N (finite) and first come first served service discipline. Here, the number of arrival will not exceed N (capacity of the system).

Model IV : (M/M/S) : (./FCFS) This is the model with Poisson arrival, Poisson departure, number of servicing channels S (finite) with infinite capacity and service discipline is first come first served.

Model V : (M/Ek/1) : (./FCFS) This is the model with Poisson arrival, Erlangian service time with K phases, single servicing channel, infinite capacity and first come first served service discipline.

Model VI : (M/Ek/1) : (1/FCFS) This model is same as model V except that there is no queue in this model i.e. , the system contains only one unit.

Model VII : Machine Repair Problem (M/1VIJR) : (K/GD), K < R This is the model with Poisson arrival, Erlangian service time, R servicing channels (repairman), with capacity of K machines in the system and general queue discipline.

Model VIII : Power Supply Model.

Waiting Lines or Queuing Theory

173

5.13 Solution of Queue Models The solution of a queuing model consists of the following steps :

Step one : First of all we shall obtain the system of steady state equations governing the queue.

Step two : In step two we solve the equations obtained in step one and find the probability distribution of queue length.

Step three : Then we obtain various probability density functions and derive formulae for E(Ls), E(Lq), E(Ws), E(Wq), E(L/L > 0) and E (W/W > 0).

5.14 Model I (M/M/l) : (oo/FCFS) (Birth and Death Model) [Meerut 1993, 94, 96(B.P.), 97(P), 99, 2000, 02(B.P.), 03; Rohilkhand 1996, 97, 99, 2003; Agra 1999, 2002, 03] This is the queuing model, with Poisson arrival, Poisson service, single servicing channel, with infinite capacity. The service discipline is first come first serve. Here X = mean arrival rate of units and µ = mean service rate.

Step one : To find the system of steady state equations. There are n > 0 units in the system at time (t + At) in the following ways : (i) (n - 1) units in the system at time t with probability Pr, _ 1 (t) one arrival in time At with probability P1 (At) = X At . no service in time At with probability (I) At (0) = 1- µAt The probability in this case is Pn _ 1 (t). X At (1 - µ At). (ii) n units in the system at time t, with probability Pr, (t) no arrival in time At, with probability Po (At) = 1 - XAt no service in time At, with probability 4)A t (0) = 1 - µAt The probability in this case is Pn (t) . (1 - X At) . (J - µAt) and (iii) (n + 1) units in the system at time t, with probability Pn + 1(t) no arrival in time At, with probability Po (At) = 1 - X At one service in time At, with probability 41A t (1) = 1.1 A t. The probability in this case is Pn +1 (t) . (1 - At) .11 At. Since all the above three cases are mutually exclusive, therefore, Pn (t + At), the probability of n units in the system at time (t + At) is obtained by adding the independent probabilities in the above three cases. i.e., Pn (t + At) = Pn -1(t) XAt . (1 - µAt) + Pn (t). (1 - XAt) (1- µAt) Pn + 1 (t) • (1 - XAt) . µAt or

Pn (t + At) = [X.Pn

- (X + µ) Pn (t) + µPn

(t)] At + Pn (t) + 0 (At)

Pn (t + At) - Pn (t)

At

Pn — 1 a)

+ 1-1 ) Pn

+ 1-113n +

+

0 (At) At

(1)

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174

Similarly there is no unit in the system at time (t + At) in the following two ways : (i) no unit in the system at time t, with probability Po (t) no arrival in time At, with probability Po (At) = 1- X At The probability in this case is Po (t). (1 - X At) and (ii) one unit in the system at time t, with probability Pi (t) no arrival in time At, with probability Po (At) = 1- X At one service in time At, with probability (I)At (1) = 1.1 At. The probability in this case is Pi (t) . (1 - XAt). µAt Adding the independent probabilities in the above two cases, we get Po (t + At), the probability of no unit in the systnn at time (t + At). (t). (1 - XAt) . µAt

i. e.,

Po (t + At) = Po (t) . (1 - XAt) +

or

(01 At + 0 (At) . Po (t + At) = Po (t ) + [-XPO (t) + Po (t + At) - Po (t) 0 (At) XPO (t) + µ P1 (t) + At At

...(2)

Taking limit as At -# 0, from (1) and (2), we have dt

Pn

= XPr2 -

- (X ti) Pn

11Pn + l(t)

(t). — Po (t) = - A,P0 (t) + dt Now when the system reaches steady state, lim Pn (t) = Pn and lim Pn' (t) = 0 t-> t->

and

-.(3) ...(4)

Under steady state of the system, the equations (3) and (4) reduce to XPn -1 - (X + 11) Pn 11P0 + 1 =o (n > 0) and

-XPo 1-1131 = ° These equations (5) and (6) are the steady state equations of the system.

...(5) ...(6)

" 0) =

1-p

P _µ

... (13 )

p (e) To find probability density function of waiting time (excluding service) distribution. [Meerut 1994, 96(B.P.); Agra 2000]

At any time, let there be n units in the system, (n - 1) waiting and one in service. If one unit arrive at this time, then it has to wait for time between co and + dw only when (n - 1) waiting units are served in time co and one unit in service is served in time dw. Since servicing distribution is also Poisson. Probability [(n - 1) units waiting are served in time co] (wo n -1 (n - 1) ! and Probability [one unit in service is served in time do)] = p dco Alf n (a) ) do) = Probability that a new arrival is taken into service after a time

lying between to and to + do) . = Probability [(n - 1) units waiting are served in time to ] x Probability [one unit in service is served in time do) ] (10)) // -1 e Nw (n - 1) !

.1.1.c/C0

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178

Let W be the waiting time of a unit who has to wait s.t. o) W 5o) + do) . Since the queue length can vary between 1 and 00, therefore, the probability density function of the waiting time is given by CO

w(w) do) =

[Probability that there are n units in the system] x i n (03) do) n =1 Pri .vn (o))dco= E(1_ p) pn . n =1 n =1 „cor - 1 dco = )tp (1 - p) e-gm " n =1 (n - 1)!

=

uon -1 a µw p do) (n — 1)!

= pp (1 - p) e-µ" . ePR" do) = pp (1 - p) e-

P) (") . do)

X =µp l - — I. e- (IA - X) . do) C

=p

-

e- (11 - X) do) o > 0

...(14)

X Hence 1- w(co) do) = p = — 1. . This is because the case for which w = 0 is excluded. Probability [W > 0] =

111(o)) dw = p

Probability [W = 0] = Probability [no unit in the system] = Po =1-p. Since the sum of these probabilities for waiting time is 1. Complete distribution of the waiting time is given by (i) part continuous for o) 5_W .5 + do) with probability density function w (co) given by w(co) do) = 1.1p (1 - p) . e-(1- P)1") .do) = X (1 -

. e4P

AMo) 'd

and (ii) part discrete for W = 0 with probability (W = 0) = 1 - p =1

[Meerut 19991

Also Probability (W > to ) = the probability that an arrival has to wait at least for t units of time e -(µ - ?)to (p — e- (12 - (t) ci) to [1.

Waiting Lines or Queuing Theory

179

(f) To find busy period distribution I. e., it (co /co > 0) (conditional density function for waiting time, given that a person has to wait). [Meerut 1999; Agra 2000] w (w /co) > 0 = Probability density function for waiting time such that a person has to wait (w) (€0) Prob. (W > 0) w (co) dw = p (1

p) . e- - 1-1°)

NJ (co/0» 0) = R. (1 - p) e- - P ) P = (µ -

e-( µ

x)

which gives the conditional probability density function for waiting time, given that a person has to wait (p.d.f. for busy period). (g) To find E(Wq ) expected waiting time in the queue (excluded service time), i.e., average waiting time of an arrival in the queue. [Meerut 2000] We have

E(W )= q • —

w(co) dco

0

pp (1 _) e - (1 - p) 1.1 0) do)

= 1-IP el - p) ..I.

0

co. e- (1 - P1 11() dco

Integrating by parts taking w as first function = up (1 • p) [

co . e- (1 - P) 1-"° _

_

Irrt

Ca - P)1-1°) '''' _ 0211 2

0

co =-.. pp (1 - p) [ - lim + 03 ) - u (1 - p) e(1 - P) P-6) =

(1 - p) 0+

1 1 t 2 (1 p) 2 .

1 II 2

(1

p) 2

11 (1

13)

P -(16) (1 - P) 1-t (1-1 - X) • (h) To find probability density function of the time that an arriving unit has to spend in the system (waiting time + service time). Let O n (v) be the probability that an arriving unit spends time v in the system (waiting + service), when there are already n units in the system at the time of its arrival. as in (e), we have O il (v) = Probability [that n units are served in the time v] x Probability [one unit is served in time dv] E(W„ ) -

Operations Research

180 (i_tv)" . e-P . [tdv, n= 0, 1, 2, 3, ...., 00. n!

[Probability that there are it units in the system]

0 (v) dv = n=0

x [Probability that new arrival spends time v in the system] CO

00

„ (µv)" CI" . P„ . 0 „ (v) dv = 1(1 - p) p . n! n=0 n=0

=

p) e

_ [iv

µ (1 - p) e-

(p µv)n dv n! n=0 .

v dv

(v) dv = µ (1 - p) e-(1-P)[' v .dv = (11 - X,)e-(1-

or

dv

v dv

...(17)

(i) To find E (Ws ) expected waiting time in the system (including service time), i. e. , average time that an arrival spends in the system. [Meerut 2000; Agra 2003] We have,

v . 0 (v) dv

E (W) =

v 1.1 (1 - p) e -(1 P) P v dv

o

v • e- (1-p)µvdv -

1 µ (1 - p) .

[Integrating by parts taking v as first function as in (g)] 1 1 ... (18) s E(W) (1 - P) (j) To find E (W/ W > 0) expected waiting time of a customer who has to wait, i. e. , average waiting time of an arrival who waits. We have, E (W/ W > 0) = -w . tv (co /co > 0) dco = = E(W/ W> 0) -

co . (1 - p) e -(1 P)" do) 0

(1 - p) f 1 µ (1 - p)

o

e- - P) " do) 1 -

µ (1 - p) ...(19)

181

Waiting Lines or Queuing Theory

(k) To find. the probability of arrivals during the service time of any given customer.

We know that arrivals and service times follows Poisson and exponential distribution laws. The probability of m arrivals during the service time of any customer is e -Xt

=

f (t) dt P 0 m (t)

=

re-a+ " t'n dt m! o kni p m! (

F (m + 1) + +

(X.Om m!

Since

dt

e-at . t n dt - T (n + 1)1

jni

( A,+µ

(A.+11)

...(20)

5.15 Relationship between E(Ls ), E(LqE(Ws ) and E(Wq ).

In a steady state queuing process, it can be proved that,, under essentially general conditions

and

E (Ls ) = T,.E (Ws )

...(1)

E(Lq ) = k .E(Wq )

(2)

Also it is obvious that 1 E(Ws ) = E(Wq ) + —

—(3)

These relations (1) to (3) are very important since they enable us to determine all the four fundamental quantities E(Ls ), E(Lq ), E (Ws ) and E (Wq ) as soon as one of these is determined.

5.16 Model II (General Erlang Queuing Model) This model is same as model I with the difference that here mean arrival rate and mean service rate are not constant but depend on n, the number of units present in the system. i. e. Here „ = mean arrival rate of customers when there are n units present in the system and [i n = mean service rate when there are n units in the system. Step one : To find the system of steady state equations.

Proceeding similarly as in Model I, the probability of n units in the system at time (t + At) is given by

Operations Research

182 PTt

(t +Ot)= Pn- 1 (t).X n-1 At . (1 — [in +

1A t)

(t) . (1 — X n At) . (1 — „ At) Pn + 1 (t) . (1 — X n l At)

or

— n+µn)Pn (t)

13,(t + At) = [X n —1Pn —

+1-tn+1. Pn+i (t)] At + Pn (t) + 0 (At) (t + At) —

At

Pn

)

—

_ (t) —

n

n +I.t n ) Pn (t)

+ n + 1Pn + 1a) + °(A61) t

for n > 0 and probability of no unit in the system at time (t + At) is given by Po (t + At) = Po (t). (1 - X, oAt) + Pi (t). (1- ) i At). RiAt or

Po (t + At) = [-

oPo (t) + µ1Pl (t)] At + Po (t) + 0 (At)

Po (t + At) - Po (t) At

- X oPo(t) +1111A +

0(At) At

for n = 0 Taking limit as At -) 0, from (1) and (2), we have dt

PO)= X n -1Pn -1(t) - (X n

PO) I

n +1Pn + 1(0

—(3)

for n > 0 and

dt

Po

for n = 0

Pi (t)

X oPo(t)

...(4)

When the system reaches to steady state lira Pn (t) = Pn and lira P n (t) = 0 t

Under steady state of the system, the equations (3) and (4) reduce to X n - 1Pn -1 - (fin for n > 0 and

oPo for n = 0

=0

n ) Pn

n + 1Pn +1 = 0

-(5) ...(6)

These equations (5) and (6) are the steady state equations of the system.

Waiting Lines or Queuing Theory

183

Step two : To solve the steady state equations (5) and (6) obtained in step one. From (6), 1)1 Putting n = 1, in (5) X0 D

Pl. Po_,' (X1 + /1 1 ) P1 = - /11 P1 + X1 + ---) P1 112 /1 2 /1 2 [12 11 2 X0 X1n X1 = — P1 FO 112 111112

P2 =

Putting n = 2, in (5), P3 = =

X1 D _, (X2 +11 2) 2'1' 113 !I 3

li 2 [1- 3

2+

(X 2 ± il 2 3 P2 1-1 3 11-

2t, 2 p _ X 0 A,1 X 2 2 PO 111R 2113 113

Proceeding in this way, we have X0X1X2 -• r,

X n -1 • PO

rn -

-.(7)

11111 211 3 ••••Pn

00 But

Pn = 1 n=0

or

Po + PI + P2 + P3 + • • • • = 1 [1 + Xo +. XoXi + • • • • Po = 1

or

HI 2

or

P0 =

whe7e

S

=

JJ

1

-(8)

S

, 21/.4 0 + XoXi. +

-(9)

111

11•1112 If S is a divergent series then P0 = 0, which is meaningless.

we assume that S is convergent series. Particular Cases : Now we shall discuss the following particular cases : Case I : If = (Both independent of n) and II, = then

S = 1 + X + (X )2

11 1 11

11 1 =— 1-p

+ (X )3 +.... 11 whereX- = p < 1.

Operations Research

184 Po = (1 - p) and so, Pn = (--111 Po = pn - p). I-1 These correspond to the results of Model I. Note : In this case model II is the same as model I. Case II : If

„=

n+1

and Ix n = µ (constant).

In this case X,1 decreases as n increases, that is arrival rate decreases with the increase in queue length, and the rate of servicep„ =11 is independent of n. This is called the case of "Queue with discouragement." Substituting the values of n = 0, 1, 2, .... ) andll n in (9), we have X S=1++

3

1.2 µ 2 2.3 la 3 2 3 , P —-r, — P -r, = 1 + p-r 2! 3! = e P, which is always convergent for p to be finite.

...(10)

From (8), P0 = (1/S) = e —P and Pi =

Po = p e-P

P2 = X0X1 Po = 2! CP µiµ2

pn

Pn

eF n! Which implies that P„ follows the Poisson distribution. Case III : If n = X. (constant) and p. = In this case arrival rate X n = A, is independent of n while the service rate n = nix increases with the increase of it. This is known as a problem of "queue with infinite number of channels" i. e., if n customers arrive then n service stations will be available for all n = 1, 2, ...., o. Thus, each arrival will enter the service facility without waiting and so no queue will be formed. Substituting the values of „ = X, and Ix n = n ).1 (n = 1, 2, .... ) in (9), we have 2 , 3 S =1++ 11 .1.2 µ 2 1.2.3113 2

3

= 1 + p + P— + L + 2! 3! = eP

185

Waiting Lines or Queuing Theory 1 Po = — s = eP and

P1

P e-P 2 P -p P2 = e

nn

...(13) e P. n! Which implies thatPn follows the Poisson distribution in this case also. Step three : In case II or III p" . e-P Pn n! • (a) To find E(1.., ) expected line length (average number of customers) in the system. We have, — — n -p E(Ls ) = In P0 = I n . P e n! n= o n =o 00 nn - 1 00 n-1 = p . CP I v - pe-P . eP - eP (n 1)! I (n 1)! n --- 0 =1 X E (Ls) = p = — . Pn =

p

(b) Now with the help of relations (1), (2) and (3) of article 5.15, E(Lq ), E(Ws ) and E(Wq ) may be obtained.

.911tithadlue Actinide:ft Examples on Model I Example 1 : Customers arrive at a sales counter manned by a single person according to a Poisson process with a mean rate of 20 per hour. The time required to serve a customer has an exponential distribution with a mean of 100 seconds. Find the average waiting time of a customer and queue length. [UP TECH MBA 2007-08] Solution : It is (M/M/1) : (oo/FCFS) problem. Here, mean arrival rate A, = 20 per hour. 60 x 60 and mean service rate µ = — 36 per hour 100 The average waiting time to a customer in the queue 20 = E(W ) = hours = 125 seconds. 1.t(µ— X) 36 (36 — 20)

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186

The average waiting time of a customer in the system 1 = E (Ws ) = — — hours 1 = 225 seconds. — X 36 — 20 ,,2 2 25 = = 0.7 (appr.) customers. And average queue length = E (L ) = = q 1—p p.(p. — X) 36 Example 2 : Arrivals at a telephone booth are considered to be Poisson, with an average time of 10 minutes between one arrival and the next. The length of a phone call assumed to be distributed exponentially, with mean 3 minutes. Find the followings : (a) What is the probability that a person arriving at the booth will have to wait ? [Meerut 2002(B.P.), 03(B.P.), 04] (b) What is the average length of the queues that form from time to time ? [Meerut 2002 (B.P.), 03 (B.P.), 04] (c) The telephone company will instal a second booth when convinced that an arrival would expect to have to wait at least three minutes for the phone. By how much must the flow of arrivals be increased in order to justify a second booth ? [Meerut 2002(B.P.), 03(B.P.), 04] (d) Find the average number of units in the system. (e) What is the probability that it will take, an arrival, more than 10 minutes altogether to wait for the phone and complete his call ? (f) What is the probability that an arrival will have to wait more than 10 minutes before the phone is free ? (g) Estimate the fraction of a day that the phone will be in use. Solution : It is (M/M/1) : (00 /FCFS) problem. Here, mean arrival rate X = 1/10, and mean service rate 1.t = 1/3. (a) Probability (arrival will have to wait) = 1 — P0 = X./µ = 3/10 = 0.3 (b) Average length of the queue = E(L/L > 0) 1/3 10 — 1.43 customers — — 0/3) — 0/10) 7 (c) Average waiting time of an arrival in the queue = E(Wq ) =

µ (µ- X)

Here E(Wq ) = 3,11 = 1/3, to find mean arrival rate X = X' (say) 3— or

(1/3) [(1/3) —

1/3 — X' = X' or X' = 1/6

1 1 1 Increase in mean arrival rate = A. — X, = — — — = — = 0.067 arrivals per minute. 6 10 15 Increase in the flow of arrivals = 1x 60 = 4 per hour. 15

187

Waiting Lines or Queuing Theory

The second booth is justified if the increase in arrival rate is 0.067 customers per minute (or 4 customers per hour). (d) Average number of units in the system E (Ls ) -

(1/10) 3 - - 0.43 customers. (1/3) - (1/10) 7

-

(e) Probability [time that a unit spend in the system 10] =

• 0 (v) dv -(µ- X,) e- (1- x.)v dv from equation (19) of article 5.14-

=

E)

e -(µ- X) v r

e- - x)10

At)

IL

= e -[(1/3) - (1/10) .10

=e

- (7/3) = e -2.33 = 0.10

(f) Probability [Waiting time of an arrival in queue 10] • (o)) tho

= J 10 =

• — (1.1. - X.) e- (11 - X)' do)

10

-10 (A - X) = 3

=--e

10

- (1/10)] e 10.[(1/3)

=3 . e- 2.33 = 3 . (0. 10) = 0.03 10 10 (g) Expected fraction of a day that the phone will be in use = Expected fraction of any other time interval that the phone will be in use. = Expected fraction of a small time interval A t that the phone will be in use. (in particular) = Probability that the both is occupied at random instant = P [n > 0] = 1 - Po =1 - (1 - 1" = --. = 1 = 0.3

µ JJ

1.1

10

Example 3 : In a railway marshalling yard, goods trains arrive at a rate of 30 trains per day. Assuming that the inter-arrival time follows an exponential distribution and the service time distribution is also exponential with an average 36 minutes. Calculate the following : (a) The mean queue size (line length) and (b) The probability that the queue size exceeds 10. [UP TECH MBA 2005-06; Meerut 2000, 02; Agra 1998, 99]

Operations Research

188 Solution : It is (M/M/1) : (oo/FCFS) problem. Here, mean arrival rate, 30 1 — trains/minute — 60 x 24 48 and mean service rate, µ = p

trains/minute. 36 X3 g 4

(a) The mean queue size = E (L s) =

(3/4) — 3 trains — 1 — p 1 — (3/4) p

and (b) Probability [queue size 10] = p10

[From article 5.14 equation (9)]

= (I° =0.06 —4 Example 4 : A TV repairman finds that the time spent on his jobs has an exponential distribution with mean 30 minutes. If he repairs sets in the order in which they came in, and if the arrival of sets is approximately Poisson with an average rate of 10 per 8 hour day, what is repairman's expected idle time each day ? How many jobs are ahead of the average set just brought ? [Agra 1998] Solution : It is (M/M/1) : (FCFS) problem. 10 — 1 sets/minute Here, X= 8 x 60 48 µ =1/30 sets/minute and Probability that there is no unit in the system = P0 = 1 — = 1 — 5/8 = 3/8 Repairman's expected idle time in 8 hour day = (3/8) x 8 = 3 hours and Expected (average) number of jobs 2 — Jobs E (LS ) — — X. 3 Example 5 : On average 96 patients per 24 hour day require the service of an emergency clinic. Also on average, a patient requires 10 minutes of active attention. Assume that the facility can handle only one emergency at a time. Suppose that it costs the clinic Rs. 100 per patient treated to obtain an average serving time of 10 patients, and that each minute of decrease in the average time would cost by Rs. 10 per patient treated. How much would have to be budgeted by the clinic to decrease the average size of the queue from one and one-third patients to half a patient ? [Meerut 1997]

Waiting Lines or Queuing Theory

189

Solution : It is a (M/M/1) : (oo/FCFS) problem. = 96 1 Here, = patients/minute 24 x 60 15 = 1/10 patients/minute X/µ = 2/3. Expected (average) number of patients in the waiting line 2 P (2/3)2—4 — 11 1 — p 1 — (2/ 3) 3 3 Fraction of the time during which there are no patients =1 — p = 1/3. Average size of the queue 2 A,2 = E(L ) = — q 1 p (µ — X) Here,

E(Lq ) = 1/2 A = 1/15 1 (1/15)2 2 µ[µ— (1/15)] 2/15 patients/minute

average rate of treatment required 1 15 1 = — = — = 7 — minutes µ 2 2 So the decrease in the average rate of treatment 1 1 = 10 — 7- = 2- minutes 2 2 • Budget per patient = Rs. 100 + (5/2) x 10 = Rs. 125. • In order to get the required size of the queue the budget should be increased from Rs. 100 per patient to Rs. 125 per patient. Example 6 : A fertilizer company distributes its products by trucks loaded at its only loading station. Both, company trucks and contractor's trucks are used for this purpose. It was found out that on an average every 5 minutes one truck arrived and the average loading time was 3 minutes. 40% of the trucks belong to the contractors. Making suitable assumptions determine : (i) The probability that a truck has to wait. (ii) The waiting time of a truck that waits. (iii) The expected waiting time of contractor's trucks per day. Solution : The assumptions for solving the given problem are made as follows : 1. The arrival rate is randomly distributed according to Poisson distribution and mean arrival rate is X . 2. The service time follows Exponential distribution and mean service rate is g. 3. µ > X (i. e., service rate is greater than arrival rate).

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190

4. There is single service channel i. e., the trucks are loaded at one place only. 5. The number of trucks being loaded is infinite i. e., capacity of the system is infinite, and 6. The service discipline is FCFS. Here,

60 X. = — = 12 trucks/hour 5 60 = —= 20 trucks/hour. 3

(i) The probability that a truck has to wait = 1 - P0 = p =

12 = — = 0.6. 1.t 20

(ii) The waiting time of a truck that waits 1 1 1 = — = hours = 7.5 minutes. - X, 20 - 12 8 (iii) Expected number of contractor's trucks waiting per day = 40% of the number of trucks per day 40 = x 12 x 24 = 115. 2 100 Expected waiting time of contractor's trucks per day = Expected number of contractors truck per day x Expected waiting time of a truck = 115.2 x 11 (11

X.) 12 = 115.2 x - 8.64 hours per day 20 (20 - 12) Example 7 : A shipping company has single unloading berth with ships arriving in a Poisson fashion at an average rate of three per day. The unloading time distribution for a ship with n unloading crews is found to be exponential with average unloading time 1 — days. The company has a large labour supply without regular working hours, and to 2n avoid long waiting lines the company has a policy of using as many unloading crews as there are ships waiting in line or being unloading. Under these conditions, find (a) The average number of unloading crews working at any time, and (b) Probability that more than four crews will be needed. Solution : Here the mean service rate is not constant as in model I. = 3 ships/day Here and mean service rate with one unloading crew = µ = 2 ships/day Pn

e -)an ( x i II . n!

From model II.

Waiting Lines or Queuing Theory

191

(a) The expected number of crews working at any specified instant = E (Ls )

(According to the question)

00

=

00

nPn = n=o n=0 ( ‘.1

=e

.n

e-x./1.1 ( x )n n!

ot, 4.0n -1 n =1 (n - 1) I

= e -k [I (I e x, /IA = —X = 2- =1.5 crews. R 3 (b) Probability that a ship entering service will require more than four crews. = Probability that there are at least five ships in the system at any instant =

n=5

=

n=0

4 - EPii n=0

= 1 - (P0 + P1 + P2 + P3 + P4) [ = 1- 1 +

µ

+

(X 4)2 2!

3 (3/2)2 = 1 - [1 + + 2 2!

(A./11)3 3! (3/2)3 3!

(A. 4) 41 4!

-X/11.

(3/2)4 e-3/2 . 4!

= 0.019. Example 8 : Problems arrive at a computing centre in a Poisson fashion at an average rate of five per day. The rules of the computing centre are that man waiting to get his problem solved must aid the man whose problem is being solved. If the time to solve a problem with one man has an exponential distribution with mean time of 1/3 day, and if the average solving time is inversely proportional to the number of peoples working on the problem, approximate the expected time in the centre of a person entering the line. Solution : It is (M/M/1) : (oo/FCFS) problem (general) queuing model. Here,

= 5 problems/day

andi.t (mean service rate with one unsolved problem) = 3 problems/day and

e-X/ P Pn

n!

(71n • — •

(Case II of model II)

Operations Research

192

Expected 'number of persons working on the problem at any instant = E(Ls ) =

n . P„ n=0 ()"

=

n=0

n! /on -1}

= (i7

= e-kig

n

(n - 1)! 5 persons = = Ix 3

Average solving time (which is inversely proportional to the number of peoples working on the problem) = 1/5 day/problems. Expected time in the centre for a person entering the line. 1 1 5 1 = -. ) = - . - = - days = 8 hours. 5 5 3 3 Example 9 : Customers arrive at a one window derive in bank according to a Poisson distribution with mean 10 per hour. Service time per customer is exponential with mean 5 minutes. The car space in front of the window, including that for the service can accommodate maximum of 3 cars. Other cars can wait outside this space. (a) What is the probability that an arriving customer car drive directly to the space in front of the window ? (b) What is the probability that an arrival will have to wait outside the indicated space ? (c) How long is an arriving customer expected to wait before starting service ? [UP TECH MBA 2004-05] Solution : Here, X = 10 customers/hour = 60/5 = 12 customers/hour p = X/u. = 10/12 = 5/6 and Po = 1 - p 1/6. (a) Required probability = Po + P1 + = (1 + p + p2 ) Po = (1 + 5/6 + 25/36) (1/6) = 0.42 (b) Probability that an arrival will have to wait outside the indicated space = 1 - (P0 + P1 + P2 + P3 ) = 1 - (Po + + P2) - p3 Po

(c)

= 1- 0.42 - (5/6)3 . (1/6) = 0.48 10 X E (W, ) - 0.417 (u - X) 12 (12 - 10)

Waiting Lines or Queuing Theory

193

Example 10 : At what average rate must a clerk in a shop work in order to ensure a probability of 0.80 that the customer will not have to wait longer than 10 minutes. It is assumed that there is only one counter, to which customers arrive in a Poisson's process at an average rate of 15 per hour. The length of service by the clerk has exponential distribution. [Meerut 1996] Solution : Given that X, = 15/60 = 1/4 customers/minutes and

p (W > 10) = 1 - O. 80 = 0. 20 -k (11 -

k) do) = O. 20

J 10 1.1

e -io 41 -

or or

or

X)

= 0.20

1 e-10(1.1 -1/4) = 0.20

4µ

e -10p

- 4 x 0.20 x e-5/2

11 or or

µ . e10p -

e 5/2 (0.80)

-

12.1825 (0.80)

[t. e10µ = 15.228

To solve (1) for µ, we proceed as follows : _ eio [t Let

15.228 = 0

Since

f(0.3) = 0.3 x 20.0855 - 15.228 = - 9.2024 < 0

and

f(0.4) = 0.4 x 54.5981 - 15. 228 = 6.6112 > 0

...(2)

0.3 0, we have d — Po (t) = - X,P0 (t) + dt

(t) for n = 0

...(4)

197

Waiting Lines or Queuing Theory

dt dt

Pn (t) -F [1. Pn i (t) for 1 n < N

(t) = XPn _ i (t) - (X, +

(t) for n = N

-

PN (t) = XPN -

Under steady state of the system these equations reduce to (for n = 0)

- X.P0 + µP1 = 0, Pn

-1 -

-.(5)

11) Pn 11P/, +1 =° for 1 n < N

and

(for n = N)

_ -i1PN =0

-(7)

which are the steady state equations of the system. Step two : To solve the steady state equation (5), (6) and (7) obtained in step one. From (5), we have X, = Po = PP0 Putting n = 1 in (6), P2 =

2 + - ) =µP1 = [1. Po = P 2P0

-1 )

Putting n = 2 in (6), 3

3 k ) PO P Po 131 + 1 + ) P2 = () P2 = (-

P3 =

Proceeding in this way, we have 21" Pit = (PO = P n • PO

-(8)

1-t

From (7),

or

PN =

PN =

1-1

N-1

X N

µ

C

=

1n 1 servicing channels, service rate at each channel is the same (µ). The service discipline is first come first serve. In this model the length of the waiting line will depend on the number of occupied channels. If n < S, then there will be no customer waiting in queue as all of them will be served simultaneously and in this case S — n service channels will remain idle. Here the rate of service µ it = n If n = S, then all the service channels will be busy and the rate of serviceµ n = S If n > S, then all the service channels will be busy, while n — S customers will be waiting in queue and the rate of service II n = S Step one : To find the system of steady state equations :

In this model the arrival and service rates are as follows : for = 0, 1, 2, A' 11 {ni.t. 0 5.n 0, we have d — P0 (0= -XPo (t)+ µPi a)

for n = 0

dt

d P„ (t) = X Pn _ 1 (0 - (X + dt -

P„ (t) + (n + 1) i.tPn +1 (t),

for t_ S which are the steady state equations of the system. Step two : To solve the steady state equations obtained in step one. From (5),

Pi = Po

Putting n = 1 in (6), D

P2 =-

2u

+1.1) n

'

= 2u

2u

n

1

2

= 2! (Xu

n

Putting n = 2 in (6), P3

=

n

3µ

X 2 n -) 3u 3

X n (X = 3µ 3!

+(—

Proceeding in this way, we have n Pn = n

Pn - 1 - (—) ' PO for 0 < n < S n1 u

Putting n = S - 1 in (6) PS = — • Ps -1 =

sp

1 (2‘. S! [t

• Po

3

n ro

201

Waiting Lines or Queuing•Theory Putting n = S in (7), Ps -1 + Sp +1

Ps +1 =

1 C1 ,/4 S +1 • Po Ps = S . SI 1.! )

=

Putting n = S + 1 in (7), n

PS +2= S

+1=

)5 + 2

1 S 2 .S1

•

R

• PO

In general Prt = PS +. (n — S)

n-S . S1 (R)n

.P0, for n

S

But we have,

P„ = 1 n=0 S -1

00

1-Pn + 1Pn =1

or

n=S

n=0 S —1

V 1 (O n [711 C11I PO]

or

[ 1

or

SS pn S!

y,

Po

n=0n [s -1 (son PO

or

SS

—• s -1 (s,$)n

or

.".

Po

y,

r

S

S!

n = 0 n! +

n=0n

-1 + p S+1

S SS • p— 1 S! 1 — p

1 PO

PO =1

rt=s

n=0

[s -1 (son

(X)n

Since p =

x

is

)= 1 +...) =1

p S

Operations Research

202

Step three : (a) To find E(Lq ), expected queue length (average number of customers in the queue).

If n > S, then n — S customers will he waiting in queue. S S pn (n — S) = 1(n S) E(Lq ) = Po S! n=S n=S

(Sp)s S!

I (n —

Po

S) pn s

n=S

=Ps[0-Fp-F2p2 +3p3 +

Since Ps -

(sos Po S!

[from (9)]

= P Ps (1 E(L )= P PS q (1 - p)2

...(10)

(b) To find E(Ls )

From article 5.15, we have E(Ws ) = E(Wq ) + XE(147,)=XE(W„)+

2, 1-1

or

E(L, ) = E(L,, ) + [1

or

Ps

+ Sp E(Ls)- P (1 - p)2

(c) To find E(W5 ) 1 From article 5.15, we have E(Ws ) = E(Ls ) E(W5 ) = P Ps x.

p)

2

+

Sp

...(12)

(d) To find E(Wq ) 1 From article 5.15, we have E(W )= — E(L ) q

E(Wq)=

P

q

PS

X (1 - p) 2

(e) To find E(L / L > 0)

We have,

E(L/L > 0) —

E( n—S)Pn n=S+1 00 I Pn n=S+1

Waiting Lines or Queuing Theory

203

or

E(L /L> 0) = 1 (On simplification) ...(14) 1-p (f) Probability that some customer has to wait. = P[n > S] = P Ps ...(15) 1-p (g) To find Probability of Busy Period. All servicing channels will be busy if n S. Probability of Busy Period : (Probability that all the channels are occupied) = Probability (n > S) 1

p n

n=S

ri

(p S)S 1)0 S!

Las s n - S S.

1 1-p

(X Pr) 1

)

[As in step two on page 200]

Ps 1-p (h) E(W/ W > 0) -

(16)

1 -

...(17)

(i) Average number of items served. s -1 = n Pn Pn • n =1

n=5

gliuthiativ . 2 examplei Example on Model III Example 13 : If for a period of 2 hours in the day (8 -10 A.M.) trains arrive at the yard every 20 minutes but the service time continues to remain 36 minutes, then calculate for this period. (a) the probability that yard is empty. (b) average queue length, on the assumption that the line capacity of the yard is limited to 4 trains only. Solution : (a) It is (M/M/1) : (N/FCFS) problem. Here, = 1/20 trains/minute, = 1/36 trains/minute, N = 4 p = X./µ = 36/20 = 1. 8 > 1 Probability that the yard is empty = P0 -1 + p pN + 1 —1+1.8

+ (1.8)5

— 0.04

[From eqns. (10) of article 5.17]

Operations Research

204

and (b) Average queue length (units in the system) 4 n Pn = E (Ls ) = rr = o 4

= ID"Po

n P"

Po

n=0

From equation (8) article 5.17 =po[p+2p2 +3p3 +4p4 ] = (0.04) [(1.8) + 2 (1.8)2 + 3 (1.8)3 + 4 (1.8)4] = (0.072) [1 + 3.6 + 9.72 + 23.328] = (0.072)x (37.648) = 2.7 trains Examples on Model IV Example 14 : A telephone exchange has two long distance operators. The telephone company finds that, during the peak load, long distance calls arrive in Poisson fashion at an average rate of 15 per hour. The length of service on these calls is approximately exponentially distributed with mean length 5 minutes. (a) What is the probability that a subscriber will have to wai L for his long distance call during the peak hours of the day ? [Agra 2000] (b) If the subscribers will wait and serviced in turn, what is the expected waiting time ? [Agra 2000] Solution : It is (M/M/S) : (oo/FCFS) Problem. Here,

A, = 15/60 = 1/4 calls/minute. 1.1 = 1/5 calls/minute. S = 2. 1/4 5 X, P [tS (1/5) x 2 8

From equation (8) article 5.18, we have 1 PO

c [-l; 1. (so

n =0

n!

(soS

n

S !•(1 — p) 1

=

..=

1

(5/4)2 (5/4)n + n! 2! . [1 — (5/8)] n =o

3 1 = (5/4) 25.8 13 1+ + 1! 32.3

(a) Probability that a subscriber will have to wait for his long distance call 00

= Probability (n 2) =

Pn n= 2

Waiting Lines or Queuing Theory

205

1

=

Pn — n=0

Pn = 1 — Po — n=0

3 2 pS = — — — .Po •13 1!

Pn =

(S .p) n

n!

Po tor n < S or n

=1- 3 - 5 X 3 - 25 13 4 13 52 (b) Expected waiting time p . (Sp) s = E(W ) — • Po X . S ! (1 — p) 5 (5)2

2

From eqn. (13); article 5.18

3 125 — 3.2 minutes 2 13 39

8 (4) 1. 2! Cl1 — — 4 8

Example 15 : Four counters are being run on the frontier of a country to check the passports and necessary papers of the tourists. The tourists choose counter at random. If the arrivals at the frontier is Poisson at the rate X and the service time is exponential with parameter X / 2, what is the steady-stage average queue at each counter ? [Meerut 2002] Solution : It is (M/MS) : (oo/FCFS) problem. Here

X,=Xd.t =X/2. S= 4

p=k4tS=1/2

From eqn. (8), article 5.18, we have 1 PO =

(son

0

•4

(so S +

n!

2n nI

S ! (1 — p)

24

41 [1 — (1/ 2)] 1

3 23

2 2 2 23 1 + — + — + —+ 4 1! 2! 3! 3 Average queue length E(Ln ) =

p —

(Sp) s

S!

. Po From equation (10), article 5.18

(1/2) 24 3 4 [1 — (1/2)] 2 • 4! 23 = 23

Operations Research

206

Example 16 : A super market has two girls ringing sales at the counters. If the service time for each customer is exponential with mean 4 minutes and if people arrive in a Poisson fashion at the counter at the rate of 10 an hour : (a) What is the probability of having to wait for service ? [Meerut 1999, 2004; Agra 2001]

(b) What is the expected percentage of idle time for each girl ? [Meerut 1999, 2004; Agra 2001]

(c) If a customer has to wait, what is the expected length of his waiting time ? Solution : It is (M/M/S) : (oo/FCFS) problem Here,

A. = 10/60 = 1/6 people /minute = 1/4 people/minute. S = 2 p = /RS = 1/3

From eqn. (8) article 5.18, we have Po - [ s -1 (son (1 - p)1 (sos n = 0 n! +-S! 1 [ 1 (2/ 3)n + (2/3)2 =o n! n/

2! . [1 - (1/3)]

1 1 + (2/ 3) + 11 2 3] 1! LL (a) Probability of having to wait for service = P(n

n = 1 - Po - P1

2) = n=2

(Sp)n 1 (2 / 3) Po Po, for n 5_ S P n nI • 2 1! - 1 - 2 1 = 1 = 0.167 2 3 2 6 (b) Fraction of the time the service remains busy = p (traffic intensity) = 1/3 The fraction of the time the service remains idle = 1 - 1/3 = 2 /3 Expected percentage of idle time for each girl 2 = - x 100% = 67% 3 -1

Waiting Lines or Queuing Theory

207

(c) Expected length of the customers waiting time = E (W /W > 0) 1 Sp. — X 1 — 3 minutes. 2 . (1/4) — (1/6)

[Equation (17) article 5.18]

Example 17 : A bank has two tellers working on savings accounts. The first teller handles withdrawals only. The second teller handles deposits only. It has been found that the service time distributions for both deposits and withdrawals are exponential with mean service time 3 minutes per customer. Depositors are found to arrive in a Poisson fashion throughout the day with mean arrival rate 16 per hour. Withdrawers also arrive in a Poisson fashion with mean arrival rate 14 per hour. What would be the effect on the average waiting time for depositors and withdrawers if each teller could handle both withdrawals and deposits. What would be the effect if this could only be accomplished by increasing the service time to 3.5 minutes ? [Agra 1998] Solution : 1st Part : When two tellers handle deposits and withdrawals separately then it is (M/M/1) : (oo/FCFS) problem. —= — 4 per minute. Mean arrival rate of depositors = Xi = 16 60 15 14 Mean arrival rate of withdrawers = X 2 = - = 7 per minute. 60 30 Mean service rate for both tellers p. = 1/3 customers/minute. Average waiting time for depositors =

(Wq ) —

X1 )

(4/15) - 12 minutes (1/3) [(1/3) — (4/15)] and average waiting time for withdrawers X = E2 (Wq = — — 2) 11

(7/30) [(1/3) [(1/3) — (7/30)] — 7 minutes

2nd Part : If each teller could handle both withdrawals and deposits than it is (M/M/S) : (oo/FCFS) problem. In this case, mean arrival rate X —

16 + 14 1 — customers/minutes. 60 2

Operations Research

208 1 and mean service rate µ = — customers/minute and S = 2 3

(equation 13, article 5.1.8)

Average waiting time = E(W ) — P Ps X. (1 — p) 2

q

where

P

211.

—

3 (1/2) 2 . (1/2) 4 1

PO

[11 (son + (Sp) S n=0

1

Sql—p)

n!

(3/2)"

2../ +

(3/2) 2

I

1 + (3/2) + (9/2) 7

+

(3/2) 2 1 9 (Sp)S si Po — 2! 7 56 Average waiting time for a customer Ps =

PPs

= E(W )

(1 —p)

q

(3/4) (9/56)

2

—

27

(1/2) (1 — 3/4)2

— 3. 86minutes

7

Thus, when both the tellers do both works deposit and withdrawls then average waiting time of the customer is decreased. 3rd Part : If the service time is increased to 3.5 minutes, then mean service rate 2 customers/minutes. 31.5 = — =— 7 p _ (1/2) 7 2 Sp, 2. (2/7) 8 PO =

1

1 [2. (7/8)] n n! n=0

[2. (7/ 8)] 2 2! . [1 — (7/8)] 2

_1 1 1 + (7/4) (49/4) 15

1 (Sp)S 2 ';) •1 —5 49 .P = PS = 2! 480 S!

Waiting Lines or Queuing Theory

209

= E(W ) - P Ps q X. (1- p)2 7 49 8 480 - 343 - 11.43minutes. 2 30 2 (1 -8) Example 18 : An insurance company has three claims adjusters in its branch office. People with claims against the company are found to arrive in a Poisson fashion, at an / average rate of 20 per 8 hours. The amount of time that an adjuster spends with q, claimant is found-to have an exponential distribution, with mean service time 40) minutes. Claimants are processed in the order of their appearance. (a) How many hours a week can an adjuster expect to spend with claimants ? (Consider 5 working days each of 8 hours in a week). (b) How much time, on average does an claimant spend in the branch office ? [Rohillchand 1993] Solution : It is (M/M/S) : (.0 : FCFS) problem. 5 Here, = 20 = -2claiming/hour, 8 3 1 x 60 = - claimant/hour it = — 2 40 X, =5 S = 3, P= 9 1-1 [S -1 (son (Sp Average waiting time

Po = /

nn=0 = 0 n!

+ `

S !(1 - p)

\ 2 = NI 1 ( 5

1 (5)

1

[n.„i0 n! 3))11 + 3! 0; 1 - (5/9)1-1 ± 1 (5)2 +6(3)3 _9 1 .j. + 1 (5)1 !0) 1 2! 0) 4

.[

24 139 1 1 = — (Sp)° Po = — (5/3)n P0 n! n!

Since, P1

1 _(I 24 40 — 1! 3 139' 139

1 (5) 2 24 100 P.? — . = - 2! 139 417

Operations Research

210 Expected number of idle adjusters at any instant = 3 Po + 2 Pi + 1 P2 = 3 (24/139) + 2 (40/139) + (100/417) = 4/ 3 Adjusters. the probability that any adjuster is idle Expected number of idle adjusters Total number of adjusters = 4/3= 4 _ 9 3 the probability that no adjuster is idle = 1 — 4/9 = 5/9

(a) Expected time that an adjuster spends with claimants in one week = (5/9) x 40 = 22.2 hours. (b)

E (Ws ) =

p

.

X. (1—

(Sp)s

S

P+

(AA-)s Po

(S — 1)! (Sp. — 2)2

pS

See formula (12) of article 5.18

1

11

(3/2) (5/3)3. (24/139) 2 3 2! • [(9/2) — (5/2)12 125 24 2 x—+— 18 x 2 x 4 139 3 =

681 681 hours = — x 60 = 49 minutes 834 834

5.19 Model V : (M/Ek / 1) : (.3 / FCFS) This is the queuing model with Poisson arrival, Erlang service time with k phases, single server, with infinite capacity. The service discipline is first come first serve. In this model one unit is served in k phases. So the arrival or departure of one unit will mean the increase or decrease of k phases in the system. Here, in this model the completion of service of one phase of a unit will mean the decrease of one phase in the system. If at any instant m units are waiting in queue and one unit is in service which has to complete s phases, then the total number of phases n in the system at this instant is given by n = mk + s

Waiting Lines or Queuing Theory Here let

211

P„ (t) = Probability that there are n phases in the system

(waiting and in service) at any time t. If g is the number of units served per unit time, then kg will be the number of phases served per unit time. Thus, here we take, X n = X, arrivals (of units) per unit time. and pt. = µ k phases served per unit time. Step one : To find the system of steady state equations.

There are n > 0 phases in the system at time (t + At) in the following ways. (i) - k) phases in the system at time t, with probability P„ _ k (t) one arrival (of unit) in time At with probability X, A t no service (of any phase) in time At with probability 1 - kµ A t The probability in this case is Pn _ k (t) . X At . (1 - kg At). (ii) n phases in the system at time t, with probability P„ (t) no arrival of any unit in time At with probability 1 - X At no service of any phase in time At with probability 1 - kµ A t. The probability in this case is P0 (t) . (1 - X At) . (1 - kµ At) and (iii) (n + 1) phases in the system at time t, with probability Pn + 1(t) no arrival of any unit in time At, with probability 1 - X, At service of one unit (of k phase) in time At, with probability k µ At. The probability in this case is Pn 1(t). (1 - X At) . kg At. Therefore P„ (t + At) the probability that there are n phases in the system at time t + At is given by (t + At) = Pn - k (t) • X At . (1-kg At) + Pn (t). (1-2 At)

. (1- kg At) or

or

P„ (t + At) [X Pn — k(t ) (X• P„ (t + A t) — P„ (t)

At

11) Pn

+

Pn + 1(t) • (1- X At) kg At Pn + 1 (t)] At + P„ (t) + 0 (At)

— X, Pn _ k (t) — (X, + 1(0 Pn (t)

0 (At) Pn +

+

.

(1)

At

Similarly there is no phase in the system at time (t + At) in the following two ways : (i) No phase in the system at time t with probability Po (t) no arrival of any unit in time At with probability 1 - X, At The probability in the case is Po (t) (1 - X At) (ii) One phase in the system at time t with probability P1 (t) no arrival of any unit in time At, with probability 1 - XAt service of one unit (of k phase) in time At, with probability kp At. The probability in this case is P1 (t) . (1 - X At) . kg At. Po (t + At), the probability of no phase in the system at time (t + A t) is given by

Operations Research

212

or or

At Po (t + At) = Po (t) (1 - XAt) + (t) • (1 - X. At) Po (t + At) - Po (t) = [- XPo (t) + kp. (t)] A t + 0 (At) 0 (At) Po (t + At) - P0 (t) 2 P0(t) 10-1 Pi (t) + At At

...(2)

Taking limit as At -4 0, from (1) and (2), we have dt

Pn (t) = Pn _ k (t) -- (X +

Pn (t) +

Pn l (t)

—(3)

n>_1 d -Po (t) = — ?P0(t) + Icµ P1 (t) dt

and

(4)

When the system reaches the steady state, equations (3) and (4) reduce to -Pn + 1 - ° X Pn - k (X + lc[t) Pn + (1 + P) Pn = Pn + 1 + P13/1 - k n 1 for

or

-.(5)

-X, Po + 1(114 =0

and or

...(6)

P Po = Pl

x

where P

The equations (5) and (6) are the steady state equations of the system. Step two : To solve steady state equations (5) and (6) obtained in step one. So far we have used the analytic procedure to solve the steady state equations in Models Ito IV., But this procedure seems to be more complicated here. Therefore, we shall use the technique of generating function to solve the equations (5) and (6) obtained in step one. A generating function is defined as P(z) =

Pn . z n

—(7)

n=0

Multiplying both sides of (5) by z n and taking summation from 1 to oo (since the equation is true for n 1), we have 00

00

(1 + p)

Pn z n n =1

00

00

or

or

Pn - kz Pn + 1z n +P n =1 n =1

P„ + lz " + Pn z n + P Po =P1 + (1 + P) n =1 n =1

(1 + p)

Pr, z " -Po = n=0

n

CO

Pn - kz "

n =1 [Since p Po = P1, from (6)]

Pn - kZ n Pn + 12 n +P n=k n=0 [Since Pr; _ k =0forn- k 0) = 1 — Po = 0.833,

Req. % = 83.3°/0

19. Consider a box office ticket window being manned by a single individual.

Customers arrive , to purchase tickets according to a Poisson input process with a mean rate of 30 per hour. The time required to serve a customer has an exponential distribution with a mean of 90 seconds. Find the following : (i) Expected line length. (ii) Expected queue length. (iii) Expected waiting time in the system. (iv) Expected waiting time in the queue. [Meerut 2001(BP)] 20. In a Bank, every 15 minutes one customer arrives for cashing the cheque. The

staff in the payment counter takes 10 minutes for serving a customer on an average. State suitable assumptions and find : (1) The average queue length. [Meerut 1994(P)] (ii) Increase in the arrival rate in order to justify a second counter (when the

waiting time of a customer is at least 15 minutes the management will increase one more counter). 21. Custome

rive at a first class ticket counter of a theatre in a Poisson distribute. arrival rate of 25 per hour. Service time is constant at 2 minutes. Calculate (a) The mean number in the waiting line (b) The mean waiting time. [Hint : 2. = 25/60 = 5/12 customers/min, la = 1/2 customer/min, p = 5/6 (a) E (Ls ) = 5 customers, E(Lq ) = 25/6 = 4 customers (appr.). (b) E(W, ) —

— 10 min, E (Ws ) =

— 12 min]

22. The XYZ company's quality control deptt. is managed by a single clerk, who

takes on an average 5 minutes in checking parts of each of the machine coming for inspection. The machines arrive once in every 8 minutes on the average. One hour of the machine is valued at Rs. 15 and a clerk's time is valued at Rs. 4 per hour. What are the average hourly queuing system costs associated with the quality control departments ?

Waiting Lines or Queuing Theory

231

[Hint : = 60/8 = 15/2 machines /h, µ = 60/5 = 12 machines/h 1 2 — hours µ—A. —X 9 Total cost per hour = Average queuing cost per machine x number of machines arrived per hour + Average cost of the clerk per hour = Rs. [(15 x 2 / 9) x 15/ 2'+ 4] = Rs. 29] 23. A repairman is to be hired to repair machines which break down at an average rate of three per hour. Break downs are distributed in time in a manner that may be regarded as Poisson. Non-productive time on any one machine is considered to cost the company Rs. 5 per hour. The company has narrowed the choice down to two repairmen, one slow but cheap, the other fast but expensive. The slow cheap repairman demands Rs. 3 per hour; in return, he will service broken-down machines exponentially at an average rate of four per hour. The fast expensive repair man demands Rs. 5 an hour and will repair machines exponentially at an average rate of six per hour. Which repair man should be hired ? 24. A repairman is-to be hired to repair machines that break down following a Poisson process with an average rate of four per hour. The cost of non-productive machine is Rs. 9 per hour. The company has the option of choosing either a fast repairman or a slow repairman. The fast repairman _charges Rs. 6 per hour and will repair machines at an average rate of 7 per hour. The show repairman charges Rs. 3 per hour and will repair machines at an average rate of 5 per hour, which repairman should be hired ? 25. A firm has served machines and want to install its-own service facility for the repair of its machines. The average break down rate of the machines is 3 per day. The repair times has exponential distribution. The loss incurred due to the lost time of an inoperative machine is Rs. 40 per day. There are two repair facilities available. Facility A has on installation cost of Rs. 20000 and the facility B costs Rs. 40000. With facility A, the total labour cost is Rs. 5000 per year and with facility B, the total labour cost is Rs. 8000 per year. Facility A can repair 41machines per day and the facility B can repair 5 machines per day. 2 Both facilities have a life of 4 years. Which facility should be installed ? 1 [Hint : Total annual cost = — (total investment expenditure) + (annual labour cost) 4 + (annual cost due to inoperative machines) Total annual cost 1 For facility A = — (20000) + 5000 + (3 x 2/ 3 x 40 x 365) = Rs. 39200 4 For facility B = 1(40000) + 8000 + (3 x 1/ 2 x 40 x 365) = Rs. 39900 4 26. In a bank cheques are cashed at a single 'teller' counter. Customers arrive at the counter in a Poisson manner at an average rate of 30 customers per hour. The teller takes, on an average aiminute and a half to cash cheque. The service time has been shown to be exponentially distributed.

Operations Research

232

(i) Calculate the percentage of time the teller is busy. (ii) Calculate the average time a customer is expected to wait. [Hint : A. = 30 customer/h, µ = 40 customer/h, find (i)1 — Po = p (II) E(WS ) = 1/(µ, — X,)] 27. At what average rate must a clerk at a supermarket work in order to insure a probability of 0.90 that the customer will not have to wait longer than 12 minutes ? It is assumed that there is only one counter, to which customers arrive in a Poisson fashion at an average rate of 15 per hour. The length of service by the clerk has an exponential distribution. [Meerut 1999] [Hint : Proceed similarly as in Example 10] 28. A car park contains 5 cars. The arrival of the cars is Poissonian at a mean rate of 10 per hour. The length of time each car spends in the car park has negative exponential distribution with mean of 2 hours. How many cars are in the car park on average and what is the probability of a newly arriving customer finding the car park full and having to park his car elsewhere. [Hint : N = 5, X. = 10/60 cars/min, µ = 1/(2 x 60) cars/min, p = X/p. = 20 Find Po =

1— p , E(I,$ ) = Po _ pN + 1

n on ] n=0 At a one-man barber shop, let the customers arrive according to Poisson 29. fashion at an average rate of 5 per hour and they are served according to exponential distribution with an average service rate of 10 minutes. Suppose that the customers do not wait if they find no seat available (assuming that only 5 seats are available), then find the average number of customers in the system, average queue length and average waiting time a customer spends in the system. 30. At a railway station, only one train is handled at a time. The railway yard is sufficient Only for two trains to wait while other is given signal to leave the station. Trains arrive at the station at an average rate of 6 per hour and the railway station can handle them on an average of 12 per hour. Assuming Poisson arrivals and exponential service distribution, find the steady-state probabilities for the various number of trains in the system. Also find the average waiting time of a new train coming into the yard. [Hint : p =

= 0.5, Po — 1 p =0.53 1 — p3 +1

Use Pr, = PO pn

31.

(i) (ii) (iii)

E(Ls )=1P1 +2P2 + 3P3 =0.74 E(W) = (14t) E(LS ) = 3.8 m] Patients arrive at a clinic according to a Poisson distribution at the rate or 30 patients per hour. The waiting room does not accomodate more than 14 patients. Examination time per patient is exponential with mean rate 20 per hour. . Find the effective arrival rate at the clinic. What is the probability that an arriving patient will not wait ? Will you find a vacant seat in the room ? What is the expected waiting time until a patient is discharged from the clinic.

Waiting Lines or Queuing Theory

233

32. If the mean arrival rate is 24 per hour, consider, from the customers point of view, the time spent in the system, whether 3 channels in parallel with a mean service rate of 10 per hour is better or worse than a single channel with mean service rate of 30 per hour. 33. A petrol pump station has two pumps. The service time follow the exponential distribution with a mean of 4 minutes and cars arrive for service in a Poisson process at the rate ten cars per hour. Find the probability that a customer has to wait for service. What proportion of time the pumps remain idle ? 34. At a one man barber shop, customers arrive according the Poisson distribution with a mean arrival rate of 15 per hour and his hair cutting takes exactly 5/2 minutes for one hair cut, how long on the average a customer must wait for the service ? [Meerut 1998(BP)] [Hint : Proceed similarly as Ex. 16] 35. A barber with one man shop takes exactly 40 minutes to complete one hair cut. If the inter arrival time of the customer follow in exponential distribution with average one in every 50 minutes, how long a customer must wait for service ? [Meerut 1995 [P)] [Hint : Proceed similarly as in Ex. 16] A 36. supermarket has two girls ringing of sales. The people arrive at a rate of 15 an hour. The service time for each customer is exponential with mean 3 minutes. If arrival of units obey Poissons process, find (a) The probability to wait for service. (b) The expected waiting time of a customer who waits for his turn. [Meerut 1998] 37. In a car manufacturing plant, a loading crane takes exactly 10 minutes to load . a car into a wagon and again come back to the position to load another car. If the arrival of cars in a Poisson stream at an average of one every 20 minutes, calculate the average time of a car in a stationary state. 38. A colliery working one shift per day uses a large number of locomotives which break down at random intervals; on average one fails per 8 hour shift. The fitter carries out a standard maintenance schedule on each faulty loco. Each of the 1 five main parts of this schedule take on average — hour but the time varies 2 widely. How much time will the fitter have for the other tasks and what is the average time a loco is out of service ? 39. In a Bhawan cafeteria it was observed that there is only one bearer who takes exactly 4 minutes to serve a cup of coffee once the order has been placed with him. If the students arrive in the cafeteria at an average rate of 10 per hour, how much time one is expected to spend waiting for his turn to place the order ? .40. For a M/Ek/1 model with = 6/5 per hour and 1.1 = 3/2 per hour; find the average waiting time of a customer. Also find the average time he spends in the system. k2 1 1] A, [Hint : Here k ---> ...., W, = lim 1 k -> .0 k µ (1,1 — X.) ' Ws = Wq + — µ •

234

Operations Research

+ ANSWERS + 17,

(i) 3 (ii) 1/4

18.

(a) 5 customers in shop and 42- customers waiting for hair cut,

19.

(b) 16.7%, (c) 83.3% (i) 3, (ii) 2.25, (iii) 6 minutes/customer, (iv) 4.5 minutes/customer .

20. 21.

6

(i) 1.33 units (ii) 20 minutes (a) E (Ls ) = 5 customers, E (Lq ) = 4 customers (appr.) (b) E (Wq ) =10 minutes, E (Ws ) = 12 minutes

22. 23.

Rs. 29 The fast expensive repairmen should be hired as the total cost in his case is Rs. 10 per hour whereas the total cost in the case of cheap repairmen is Rs. 19 per hour. 24. The fast repairman should be hired as the total cost in his case is Rs. 18 per hour whereas the total cost in the case of slow repairman is Rs. 39 per hour. 25. Facility A should be installed. 26. (i) 75% (ii) 6 minutes 27. 2.48 minutes per service. 29. 1.97, 1.22 and 16.3 minutes 30. P1 = 0.27, P2 = 0.13, P3 = 0.07 etc. 3.8 minutes 31. (1) 19.98, (ii) 0.67, (iii) 0.65 hours 32. Single channel with mean rate of 30 per hour is better 33. 0.167, 67% for each pump 34. 25/12 minutes 35. 1 hour and 20 minutes 36. (a) 0.204, (b) 0.491 minutes 37. 5 minutes 38. 3.18 hours 39. 4 minutes 40. 4/3 hours, 2 hours •• •

6

Allocation (General Linear Programming Problems) •

6.1 Introduction A problem which involve the allocation (or allotment) of given number of resources to the job (or activities) is called an allocation problem. These problems occur when the available resources are not sufficient to allow each job to be carried out in the most efficient way. Therefi e, the objective of these problems is to optimize the total effectiveness i. e. to minimize the total cost or maximize the total return. Here in this book we shall discuss three types of the allocation problems. 1. General Linear programming problem. 2. Assignment problem. (chap. 13) 3. Transportation problem. (chap. 14)

6.2 General Linear Programming Problems [UP TECH MBA 2005-06]

The term "linear" means that all the relations is the particular problem are linear and the term "programming" refers to the process determining a particular programme or plan of action. A general linear programming problem includes a set of simultaneous linear equations which represent the conditions of the problem and a linear function which expresses the objective function of the problem. The linear function which is to be optimize, is called the objective function and the conditions of the problem expressed as simultaneous linear equations (or inequalities) are referred as constraints. A general linear programming problem can be stated as follows : Find xi, x 2,..., xn which optimize the linear function. Z = clxl+ c 2x2+....+cn xn ...(1) subject to the constraints xi + a lit Xit (5_ =?_) 1 auxi a12 X2 + a2j xj a21X1 a22 x2 a2n Xn ("=•-) b2 ...(2) aiixi

+

a- i2-2 Y

+

+

aii

Xi

+

+

aill

x11

(5' = ')

bi

... and xi

+ a-In 2 x2 +

+

ami xi

+

+

am„ xn

(.=_.) bm

Operations Research

236 and non-negative restrictions X j = 1,2,...,n. where all aij i s,b1 ' s and c j' s are constants and x j' s are variables. f

In the conditions given by (2) there may be any of the three signs 5 , =,?_ The function Z given by (1) is called the objective function and the conditions given by (2) are termed as the constraints of the linear programming problem. We shall always assume that all bi 0. If any one is negative, we make it positive by multiplying both sides of the corresponding inequality by — 1. By this multiplication the inequality is also reversed. The above linear programming problem may also be stated in matrix from as follows: Optimize Z = c.x subject to A x and x 0 where A = [9 ] is the matrix of coefficients of order in x n. c = 1, c2,..., cn ) is a row vector-known as price vector. _ x1 x2 x=

=.

x2,...,

* is column vector of variables

xn

b=

b2

, b Hi ]

= [b1, b 2 ,

is column vector called the requirement vector and 0 is n dimensional null column vector. The column vector formed by the coefficients of xi in all the constraints is denoted a J C111 a2i CC • = _[al j ,a2i ,..., amj] ami then

A=

).

Note : In a general linear programming problem it is assumed that the number of rows of

coefficient matrix A is less than its number of columns. *

We shall write n-dimensional row vector as (a1, a2, ..., an) while m-dimensional column vector will be written as [b1, b2 bm

Allocation (General Linear Programming Problems)

237

6.3 Mathematical Formulation of a L.P.P. It is important to recognize a problem which can be handled by linear programming and then to formulate its mathematical model. Several examples are given below where the mathematical model of given problem is formulated.

glitabLatim tiamplith Example 1 : A manufacturer of a line of patent medicines is preparing a production plan on medicines A and B. There are sufficient ingredients available to make 20000 bottles of A and 40000 bottles of B but there are only 45000 bottles into which either of the medicines can be put. Further more it takes 3 hours to prepare enough material to fill 1000 bottles of A, it takes one hour to prepare enough material to fill 1000 bottles of B and there are 66 hours available for this operation. The profit is Rs. 8.00 per bottle of A and Rs. 7.00 per bottle of B. (a) Formulate this problem as linear programming problem. (b) How should the manufacturer schedule production in order to maximise his profit ? Solution : (a) Let the manufacturer produce x1 and x2 bottles of medicines A and • B respectively. Total profit (in Rs.) Z = 8x1 + 7x2 ... (1) The time required to prepare xi bottles of medicine A 3xi hours. 1.000 x2 hours. and the time required to prepare x2 bottles of medicine B — 1000 Total time required to prepare xi bottles of medicine A and x2 bottles of 3x1 + hours. medicine B is 1000 1000 Since total time available for this operation is 66 hours. 3x1 x2 + - 5. 66 1000 1000 3x1 + x2 S 66000 or Since there are only 45000 bottles into which the medicines can be put x1+ x 2 S 45000 Also there are sufficient ingredients available to make 20000 bottles of A and 40000 bottles of B. 20000, x2 40000 Hence the linear programming problem of the given problem is as follows : Max. Z = 8x1 + 7x2 subject to constraints 3x1 + x2 66000 xi+

x2 45000 xl-L", 20000

x2 40000

Operations Research

238

0, x2 0 and (b) For the solution of the above problem see Ex. 10, Page No. 249. Example 2 : A resourceful home decorator manufactures two types of lamps say A and B. Both lamps go through two technicians first a cutter, second a finisher. Lamp A requires 2 hours of the cutter's time and 1 hour of the finisher's time. Lamp B requires 1 hour of cutter's and 2 hours of finisher's time. The cutter has 104 hours and finisher 76 hours of available time each month. Profit on one lamp A is Rs. 6.00 and on one B lamp is Rs. 11.00. Assuming that he can sell all that, he produces, how many of each type of lamps should he manufacturer to obtain the best return ? Solution : Formulation of the mathematical model. Let the decorator manufacture x1 and x2 lamps of types A and B respectively. Total profit (in Rs.) Z = 6x1 + 11x2. Total time of the cutter used in preparing x1 lamps of type A and x2 of type B is 2x1 + x2. Since cutter has 104 hours only for each month. 2x1 +x2 2 and the non-negativity restrictions xi, x2, x3 ?. 0. Example 5 : The objective of a diet problem is to ascertain the quantities of certain foods that should be eaten to meet certain nutritional requirement at a minimum cost. The consideration is limited to milk; beef and eggs, and to vitamins A, B, C. The number of milligrams of each of these vitamins contained within a unit of each food is given below. Table 6.3 Vitamin

Gallon of Milk

Pound of Beef

Dozen of Eggs

Minimum daily requirement

A.

1

1

10

1 mg

B C

100 10

10 100

10 10

50 mg 10 mg

Cost

Rs. 1.00

Rs. 1.10

Rs. 0.50

What is the Linear Programming formulation for this problem ? Solution : Let the daily diet consist of x1 gallons of milk, x2 pounds of beef and x3 dozens of eggs Total cost per day is Rs. Z = x1+1.10x2 +0.50x3 Total amount of Vitamin A in the daily diet is x1 + x2 + 10x3 mg which should be at least equal to 1 mg + x2 +10x3 .?_ 1 Similarly total amount of vitamin B and C in the daily diet are 100x1 + 10x2 +10x3 50 and 10x1 +100x2 +10x3?_10

Allocation (General Linear Programming Problems)

241

Hence, the linear programming formulation of this diet problem is as follows. Find xi , x2, x3 which minimize Z = x1 + 1.10x2 + 0.50x3 subject to the constrains x1 +x2 + 10x3 1 100x1 + 10x2 + 10x3 50 10x1 + 100x2 + 10x3 10 and

?_ 0, x2 >_0, x3 >_0.

Example 6 : A company has two grades of inspectors. I and II, who are to be assigned' for a quality control inspection. It is required that at least 2000 pieces be inspected per 8 hour day. Grade I inspectors can check pieces at the rate of 50/hour with the accuracy of 97%. Grade II inspectors can check pie: es at the rate of 40/hour with the accuracy of 95%. The wage rate of Grade I inspector is Rs. 4.50/hour and that of Grade II is Rs. 2.50/hour. Each time an error is made by an inspector, the cost to the company is one rupee. The company has available for the inspection job, 10 grade I and 5 grade II inspectors. Formulate the problem to minimize the total cost of inspection. Solution : Let x1 and x2 be the no. of inspectors of grade I and II respectively. They both will inspect (8 x 50)x1 + (8 x 40)x2 pieces daily. But the company requires at least 2000 items to be inspected daily, so we have (8 x 50)x1 + (8 x 40)x2 2000 i. e., 5x1 + 4x 2 25 The limitations on the available number of inspectors give the constraints 10, x2 55 ...(2) Further the company is bearing two types of costs, the wages of inspectors and the costs of inspection errors. The cost of each inspector per hour is as follows : Grade I : Rs. 14.50 + 1 x

3 x 50)1 = Rs. 6.00/hour 100

5 Grade II : Rs. 2.50 + 1 x (- x 40 1= Rs. 4.50/hour 100 )j [ Now the objective of the company is to minimize the total cost of inspection each day. If Z is the total cost of inspection, then the objective function is Minimize Z = 8 x 6.00 x x1 + 8 x 4.50 x x2 = 48x1 + 36x2 Hence, the required L.P.P. is Minimize Z = 48x1 + 36x2 subject to the constraints : 5x1 + 4x2 25, x1 510, x2 ni) and r (A) = r (Ab) = m. Then none of the equations is redundant.

If any m x m non-singular matrix (whose determinant is not zero) is chosen from A and if all the (n — in) variables not associated with the columns of this matrix are set equal to zero, the solution to the resulting system of equations is called a basic solution. In other words a solution obtained by setting any (n — tn)variables to zero is called a basic solution, provided the determinant of the coefficients of the remaining rit variables is not zero. Thus, in a basic solution at least (n — m) variables must vanish.

The in variables associated with the columns of the above non singular matrix which may be different from zero are called basic variables. It is important to note that the matrix formed by the coefficients of in basic variables is non-singular, hence the vectors associated to the basic variables are L.I.

and the Thus, a solution in which the vectors associated to m variables -are remaining (n — m) variables are zero, is called a basic solution. If B is the matrix of ni L.I. vectors of A and x B is the vector of the corresponding variables (basic variables), the basic solution is given by x B = 13-1 b. Since in vectors out of n (Columns of coefficient matrix A) can be selected in n

C,n

-

n!

I(n — m)!

ways,

Hence, maximum number of basic solution is n C m —

1. 2.

n!

in! (it — m)!. Basic solutions are of two types : Non-degenerate B.S. A B.S. is called non-degenerate B.S. if none of the basic variables is zero. In other words all the m basic variables are non-zero. Degenerate B.S. A basic solution is called degenerate B.S. if at least one of the basic variables is zero.

6.5 An Important Theorem A necessary and sufficient condition for the existence and non-degeneracy of all the basic solutions of Ax=b is that every set of m columns of the augmented matrix Ab = [A, b] is L.I.

Proof : Necessary Condition First we consider that all the basic solutions of Ax = b exist and are non-degenerate. Therefore every set of ni-column vectors of A are L.I. Let al , a 2 ..., a m be one set of m-column vectors of A, then from the given system we have aixt c(2x2+••••ainxm = b. But x # 0, since each solution is non-degenerate. Therefore the vector b can replace al in the basis a 2 ,...,am . (By replacement theorem of vectors). Thus, the vectors b, a 2, a3, ..., a m also form a basis and hence are L.I. In the similar way al, b, a3,...,am ;a1,a2, b, a 4,...,am , etc., are L.I.

Allocation (General Linear Programming Problems)

243

Thus, the vector b with (in - 1) vectors of A from a L.I. Set. Hence, every set of in columns of the augmented matrix Ab = [A, b] is L.I. Sufficient Condition : Here we consider that every set of m columns of the augmented matrix Ab = [A, b] is L.I. Obviously every set of m-column vectors ofA are L.I. which implies that the basic solution of the system Ax = b exist. Now to prove that the basic solution is non-degenerate, consider al, a 2, ..., a m be any in-columns vectors of A which are L.I. and let x1 , x2 , xm be the corresponding basic solutions (basic variables). + x2a 2 +. .+x. am= b or

-1. b+x1a1 + x2a 2 +....+Xn,

=0

Now if x1 = 0, then we. have -1.b +x2a 2 +. .+xm am= 0 which implies that the vectors b, a2 , a m are L.D. which is a contradiction since we have assumed that every set of in columns (vectors) of the augmented matrix Ab= [A,b] is L.I. Therefore, x1 # 0. Similarly we can prove that none of x2, x3,..., xn, are not zero. Hence, the basic solution is non-degenerate. In this way we can show that every basic solution is non-degenerate. Cor. : The necessary and sufficient condition for any given basic solution X B = B -ib to be non-degenerate is the linear independence of b and every set of (m - 1) columns from A. 6.6 Some Important Definitions 1. A Feasible Solution (F.S.) A feasible solution to a linear programming problem is the set of values of the variables which satisfies the set of constraints and the non-negative restrictions of the problem. 2. Optimum (or optimal) Solution

Afeasible solution to a L.P.P. is said to be optimum (or optimal) solution if it also optimizes the-objective function Z of the problem. 3. Basic Feasible Solution (B.F.S.)

In a L.P.P. a feasible solution which is also basic is called a basic feasible solution (B.F.S.). In other words a feasible solution to a L.P.P. in which the vectors associated to non-zero variables (non-zero variables are certainly positive in F.S.) are L.I. is called a basic feasible solution. Since at most m vectors of E n , Euclidean space of m-dimensions 'iere m is the number of constraints may be L.I., hence a B.F.S. cannot have more :han m non-zero (i.e., positive) variables. Thus for a F.S. to be a B.F.S., at least (n - m) variables must vanish. B.F. solutions are also finite in number and maximum number of them is n . 4. Basic Variables

The variables in a B.F.S. are called the basic variables.

Operations Research

244

5. Non-degenerate B.F.S A B.F.S. of a L.P. problem is said to be non-degenerate B.F.S. if none of the basic variables is zero. 6. Degenerate B.F.S A B.F.S. of a L.P. problem is said to be degenerate B.F.S. Vat least one of the basic variables is zero.

gliadAativRexamoth Example 7 : Is x1 = 1, x2 = 1/2 x3 = x4 = x5 = Oa basic solution of the following system, =2 x1 + 2x2 + x3 + x4 + x5 = 2 x1 + 2x2 + x3 Solution : The given system of equations can be expressed in matrix form as Ax = b where A = ,a2 ,a 3 , 4 ,ct 5 ), where r -I 12-1 [1 [1] = 101 = 121 a a ,a a ,b 4 0 5 [1] [2] 1 — 1_11 2 [2]' 3 — 1/2 and

x = [xi x2 x3 x4 x5] In the given solution x1 = 1, x2 = 1/2, x3 = x4 = x5 = 0, the basic variables are =2 1 x1 and x2 and the vectors corresponding to them are al = ,a2 1 2 1 =2—2=0 lal a21= 12 so the given solution is L.D., hence this solution is not the basic solution of the problem. Example 8 : Find all the basic solutions of the following system, x1 + 2x2 + x3 = 4 2x1 + x2 + 5x3 = 5 and prove that they are non-degenerate. Solution : In matrix form the given system of equations can be written as Ax b where A = (oc1,a2,c(3 ) xi. 1 1 al = [ , a2 = [ ], '13 = [ 1 x = x 2 , b = 2] 1 5 5 x3 This problem can have at most 3C2 = 3 basic solutions. Now the three sets of two vectors are [1 21 , 1 1 1 B1 = [cq ' a 2] = 2 li' B2 = Ectl'a3j = 2 5] B3 = EC(2' (131 = 12 5]. Here IBi l = —3# 0,11321= 3 # 0,IB31= 9 # 0.

Allocation (General Linear Programming Problems)

245

Since none of these is zero, therefore every set of two vectors of A are L.I. Hence, all the three basic solution exist. To compute all basic solutions If xBi,i = 1, 2, 3 are the vectors of the basic variables associated to the sets Bi,i = 1, 2, 3, respectively, then ix1] x 4]=[2] =_1[ _ B1 3 - 2 1] 5 1 x2

[

x3

= X B2

=

B2-lb = 3 5 -1 4 = _ 5 - 2 1. 1

[X21

= xr3 = B3-l b = 1- [ 1 5 -2 14 52 z/ 3 9 5 Hence the basic solutions are x1 = (2, 1, 0), x 2 = (5, 0, -1), x 3 = (0,31) x3

Now the determinants of matrices [at , b] and [a2,13] are not zero, i.e., b and every set of m -1 = 2 -1 = 1 column of Blare L.I. Hence by theorem of article 6.5, the basic solution xi is non-degenerate. Similarly we can prove that the other basic solutions x 2 and x 3 are also non-degenerate. Example 9 : Do all basic solutions of the following system of equations exist ? + 6x 2 +13x3 + X 4 + X 5 = 6 9x1 + x2 + 2x3 + 6x4 +10x5 =11 and 0 V`i = 1, 2, 3,4, 5. Find all existing basic solutions. Also find all basic feasible solutions. Solution : The given system of equations can be expressed in matrix form as A x = b, A = 8 6 13 1 1 where 1 9 1 2 6 10] = (at a2 a3 a4 a5), [8] [61 [13] [1] [ 1] b [ 6 where , , a2 a3 5 (x4 = , as [11]5 10 2 6 9 1 and x = [x1 x2 x3 x4 x5] Here n = 5, m = 2. Hence there can be at the most 5C2 = 10 basic solutions. Ten sets of two vectors out of al, a2, a3, a4, a5 are as follows : I-8 6 8 13 = [a1 az] = [9 1 5 B2 = [al a3] = 9 2 B3 = [a1 a4] =

[8 11

B4 = [al a5]

r8 1 1 [9 10Y

9 6.1 6 13] B5 = [a2 a3 ] - [6 B6 = [a 2 a4]=[ 1 2 1 6]' [13 11 [6 1 137 = [a2 a5] B8 = [0:3 a 4] = 1 10T L 6L '

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246 B9 =[a3 a 51=

[1

[13 q 2 10T B10 = [a4 a5]

11 6 10]

Here none of 1B11,1B 21, ... ,1B101is equal to zero i.e., every set of two vectors of A is L.I. Hence, all the ten basic solutions exist. To compute all basic solutions : Since in a basic solution at least n - m = 3 variables have their values equal to zero. giving zero-Values to three non-basic variables, the solutions are computed in the following table : Table 6.4 S.N.

Basic Variables

Non-Basic Variables

Equations to obtain B.S.

1.

xi, x2

x3 = 0 = x4 = x5

8x1 + 6x2 = 6

xi, x3

9x1 + x2 =11 8x1 + 13x3 = 6 x2 = 0 = x4 = x5

2.

9x1 + 2x3 =ii 3. 4. 5. 6.

xl, x4 xi, x5 x2, x3 x2, x4

x2 = 0 = x3 = x5 x2 = 0 = x3 = x4

8x1 + x4 = 6 9x1. + 6x4 = 11 8x1 + x5 = 6

9x1 + 10x5 = 11 xi =0= x4 = x5 6x2 +13x3 - 6 x2 + 2x3 = 11 6x2 + x4 = 6 xi = 0 = x3 = xs x2 + 6x4 = 11

7.

x2, x5

x1 = 0 = x3 = x4

8.

x3, x4

x1 = 0 = x2 = x5

9.

x3, x5

xi =0= x2 = x4

.10.

x4, x5

X1 = 0 = x2 = x3

+ x5 = 6 x2 + 10x5 =11 13x3 + x4 = 6 6x2

Basic Solution

Is it B.F.S.

xi = 30/23, x2 = -17/23 No x3 = 0 = x4 = x5 x_

_131

x

_-34

No

101' 3 101 X2 = 0 =-X4 = x5 x1 = 25/39, x4 = 34/39

Yes

x2 -0= x3 =x5 xi = 49/71, x5 = 34/71

Yes

x2 = 0 = x3 = x4 x2 =131, x3 =- 60

No

xi = 0 = x4 = x5 x2 = 5/7,x4 =12/7

Yes

xi = 0 = x3 = x5 x2 = 49/59, x5 = 60/59

Yes

1

xi = 0 = x3 = x4 x3 = 25/76, x4 =131/76 Yes 2 x3 + 6x4 = 11 xi = 0 = x2 = xs _ 131 13x3 4- xs = 6 x., = 49 Yes x 128 128' 5 i 2x3 + 6x5 =11 xi. = 0 = x2 = x4 x4 + x5 = 6 x4 = 49/4, x5 = -25/4 No

6x4 + 10x5 =11

xi = 0 = x2 =x3

Hence, all existing basic solutions of the system aie as follows : (30/23, - 17/23, 0, 0, 0), (131/101, 0, - 34/101, 0, 0), (25/39, 0, 0, 34/39, 0), (49/ 71, 0, 0, 0, 34/71), (0, 131,- 60, 0, 0), (0, 5/7, 0, 12/7, 0), (0, 49/59, 0, 0, 60/59), (0, 0, 25/76, 131/76, 0), (0, 0, 49/128, 0, 131/128), (0, 0, 0, 49/4, - 25/4).

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The basic feasible solutions (the basic solutions satisfying the non-negative restrictions xi 0) are as follows : (25/39, 0, 0, 34/39, 0), (49/71, 0, 0, 0, 34/71) (0, 5/7, 0, 12/7, 0) (0, 49/59, 0, 0, 60/59), (0, 0, 25/76, 131/76, 0), (0, 0, 49/128, 0, 131/128)

6.7 Solution of a Linear Programming Problem Iri general we use the following three methods for the solution of a L.P.P.

1. Geometrical (or Graphical) Method If the objective function Z is a function of two variables only then the problem can be solved by graphical method: A problem of three variables can also be solved by this method, but it is complicated enough.

2. Analytic Method (Trial and Error Method) The -L.P.P. having more than two variables cannot be solved by graphical method as even the problem of three variables becomes complicated enough. In such cases,-the analytic method (trial and error method) can be useful.

3. Simplex Method This is the most powerful tool of the linear programming as any problem can be solved by this method. This method is an algebraic protedure which progressively approaches the optimal solution. The procedure is straight forward and requires only time and patience to execute it manually. (This method is discussed in chapter 8).

6.8 Geometrical (or Graphical) Method for the Solution of a Linear Programming Problem In this method we proceed as follows :

Step one : First of all we consider all the constraints as equalities. Step two : Then we draw the lines in the plane corresponding to each equation (obtained in step one) and non-negative restrictions. Methods to draw lines. Putting xi. = 0 in the equation of a line find x2 and then putting x2 = 0 find xi. . Thus, we get the points of intersection of the line with the axes. The line is drawn by joining these points on the axes. Step three : Then we find the permissible region (feasible region) for the values of the variables which is the region bounded by the lines drawn in step two. For which we proceed as follows.

Working rule for finding Permissible Region (Feasible Region) Consider the constraint ax + by 5 or c, where c > 0. The line ax + by = c (drawn in step 2) divide the xy-plane in two regions, one containing and the. other not containing the origin. Since (0,0) satisfy the inequality ax + by c, so for the inequality ax + by c, the feasible region is the region which contains the origin. Also (0,0) does not satisfy the inequality ax + by c, so for the inequality ax + by c, the feasible region is the region which does not contain the origin. See fig. 6.1 (a). 0 , where m > 0. Again consider the constraint y — m x The line y — mx = 0 (drawn in step 2) divide the xy-plane in two regions, one containing the +ve x-axis and the other containing the +vey-axis. For the inequality

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y — mx < 0, the feasible region is the region which contains the positive x-axis and for the region y — rnx > 0, the feasible region is the region which contains the positive y-axis. See fig. 6.1 (b).

0

(a) ,

Fig. 6.1 Thus, we finfeasible regions corresponding to all inequalities. Then the region which is cOmmon to all these regions is the permissible region (i.e., feasible region) for the values of the variables. This permissible region is shaded. Step four : Her 1 we find the point in the permissible region (obtained in step 3 ) which gives the oiltiniurn value of the objective function Z. The point will be one of the extreme points (vertices) of -the convex polygon enclosing the permissible region. This pOint, can be attained by one of the following two methods. Corner Point Method : Determine the vertices of the convex polygon, which are the points of intersection of the straight lines passing through them. These vertices are called the extreme points of the set of all feasible solutions of the L.P.P. Then find the values of the objective function Z at all these points. The point where the objective function Z attains its optimum value (maximum or minimum value as the case may be) gives the optimum (or optimal) value of the L.P.P. If two vertices o the convex polygon give the same optimum value of the objective function Z, t en all poi is on the line segment joining these two vertices will give the optimum value of the objective function Z. In this case the L.P.P. is said to have infinite number of optimum solutions. Iso-Profit or Iso-Cost Methqd : Here, to find the vertex of the convex polygon, which gives the opti4m value ofithe objective function Z, draw a straight line in the feasible region correronding to!the equation obtained by giving some convenient value k to the objecti e function.1This line is called an iso-profit or iso-cost line, since every point on e line within the permissible region will yield the same value of Z. We can also t e k = 0. Which give a line passing through the origin and paralldl to iso-profit line. Thus, we draw the line (dotted line) through the origin corresponding to — 0. Then for the maximization problem the extreme point of the permissible region which is if[rthest away (i.e. at greatest distance) from this line and for the minimization prohlem the extreme point of the permissible region which is nearest to this line gives t e optimum values of Z. To obtain this extreme point of the permissible region, givin the optimum value of the objective function Z, we go on drawing li a parallel tot e line Z = 0. The farthest extreme point is the vertex of the permissibl re ion throu h which one of the parallel lines passes and after which it leaves this di anthe earest extreme point is the vertex of the permissible region through I lich the paflallel line enters this region.

249

Allocation (General Linear Programming Problems)

Note : If there is no permissible region in a problem then we say that the problem has no solution.

The procedure of the method may be understood by the following examples.

Ritualtaiiffe bcamplea Example 1.0: Solve the programming problem. Max. Z = 8x1 + 7x2 s.t. 3x1 + x2 66000 x1+ x2 45000 20000 x2 S 40000 and 0, x2 >_0. Solution : Step one : First we consider the constraints as equations 3x1 + x2 = 66000 + x2 = 45000 x1 = 20000 x2 = 40000 Step two : Now we draw the lines to each of the above equations in two dimensional plane. See Fig. 6.2 Step three.: The shaded region shown in fig. 6.2 is the permissible region for the values of the variables x1 and x2.

x2

60000

-

cia .'..;..v '3 O o

o o o c c-,1 II x2=40000

% •\r

34500\ Line for maximum value of Z 20000 . \ \ I i 8K \ —7K Z=8x1 + 7x2=0 xi —7 gives x2 =

10500

p 2 100 Fig 6.2

40000

60000

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250

i

Step four : By Corner-Point Method Solving simultaneously the equations of corresponding intersection lines, the coordinates of the vertices of the convex polygon OPQRST are 0(0, 0), P(20000, 0), Q(20000, 6000), R(10500, 34500), S(5000, 40000), T (0, 40000). fhe values of the objective function Z at these vertices (corner points) are as given in the following table.

Table 6.5 Point (xi, x2 ) O(0, 0) P(20000, 0) Q(20000, 6000) R(10500, 34500) S(5000, 40000) T(0, 40000)

Value of the objective function Z = 0+0 Z = 8.x 20000 + 0 Z = 8 x 20000 + 7 x 6000 Z = 8 x 10500 + 7 x 34500 Z = 8 x 5000 + 7 x 40000 Z = 0 + 7 x 40000

Z = 8x1 +7x 2 =0 = 160000 = 202000 = 325500 (Max.) = 320000 = 28000D

Obviously Z is maximum at R (10500, 34500). Hence Z is maximum for x1 = 10500, x2 = 34500, and maximum Z = Rs. 325500.

By Iso-Profit Method : Here to find xi , x2 for which Z is maximum, we draw the line (dotted line through 0), Z = 8x1 + 7x2 = 0x1 / x2 = (-7)/ 8. Now draw lines, parallel to this dotted line through 0, till we reach the point of the permissible region which is farthest away from 0. As shown in the figure 6.2, this point is R(10500, 34500), which is the point of intersection of the lines. 3x1 + x2 = 66000 and x1 + x2 = 45000 Hence, Z is maximum for x1 = 10500, x2 = 34500 and maximum Z = Rs. (8 x 10500 + 7 x 34500) = Rs. 325500. Example 11 : Solve the linear programming problem, Max.

Z = 6xi + 11x 2

s. t.

2x1+ x2 104 x1+ 2x2 5_ 76

and

xi 0, x 2 0

Solution : Step one First we consider the constraints as equations; 2x1 + x2 = 104 x1 + 2x 2 = 76 Step two : Here we draw lines in two dimensional plane,

Allocation (General Linear Programming Problems)

251

100

80

k

60

I

?!)‘\\

37'

20 16

Line for maximum value of Z

P (44, 16)

6K•,„ —11K

11111hkii, 20 40 44 Z=6x1 4-11x2=0 • gives

xl x.2.

>x,

—11

6

Fig. 6.3 Step three : The shaded region in the fig. 6.3 is the permissible region for the values of the variables x1 and x2. Step four : As in Ex. 10, we see that max. Z is obtained at the point P(44, 16) which is the point of intersection of lines. 2x1 + x2 = 104 and x1 + 2x2 = 76 for Max. Z, xi = 44,x2 = 16 and Max. Z = Rs. (6 x 44 + 11 x 16) = Rs. 440. Example 12 : Solve the following programming problem by graphical method; Max. Z = 5x1 + 7x2, s. t. xi + x2 4 + 8x2 24 10x1 + 7x2 35 and xi, X 2 O. Solution : Step one : First we consider the constraints as equalities. xi + x2 = 4

3x1 + 8x2 = 24 10x1 + 7x3 = 35

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252 Step two : Here we draw lines in two dimensional plane.

Line for maximum value of Z

5K x Z=5x1+7x2=0 —7 xl gives X9 — rJ

Fig. 6.4

Step three : The shaded region in fig. 6.4 is the permissible region for values of the variables xi and x2. Step four : From the fig. 6.4 it can be seen that Z is Max at the point (1.6, 2.4) which is the point of intersection of the lines xl+ x2 = 4 and 3x1 + 8x2 = 24 For Max. Z x1 = 1.6, and

x2 = 2.4 Max. Z = Rs. [5 x (1.6) + 7 x (2.4)] = Rs. 24.8

Example 13 : Solve the following linear programming problem : Maximize: Z = 40x + 35y Subject to: 2 x + 3y 60 4x+3yS 96 4 x + 3.5y 105 x, y 0 and

[UP TECH MBA 2002-03]

Solution : Step one : First we consider the constraints as equalities. 2x + 3y = 60, 4x + 3y = 96, 4x + 3.5y = 105 Step two : Here we draw lines in two dimensional plane :

Allocation (General Linear Programming Problems)

32

253

4x+3y=96

30

4x+3.5y=105

s(0, 20)' •

P(18, 8)

8k

-7k

30

A1?4, 0) Z =40x +35y =0 xi -7 gives X2 = 8

Fig. 6.5 Step three : The shaded region in the figure is the permissible region for the values of x and y. Step four : The extreme points of the permissible region are 0 (0, 0), A (24, 0) , B (0, 20) and P (18, 8) S. No.

Point

1 2 3 4

0 (0, 01 A (24, 0) B (0, 20) P (18, 8)

Z = 40x + 35y Z=0+0=0 Z = 40 x 24 + 0 = 960 Z = 0+35x 20 =700 Z = 40 x 18 + 35 x 8 = 1000 (Maximum)

Since Z is maximum at the point P(18,8), hence the required solution is x = 18, y = 8 and maximum Z = 1000 Note : The required solution can also he obtained by drawing the dotted line of objective function as shown in the fig. 6.5. Example 14 : Solve the following L.P.P. Minimize : Z = 40x + 24y Subject to:

20x + 50y 4,800 80x + 50y 7,200

and x, y 0 [UP TECH MBA 2006-07] Solution : Step one : First we consider the constraints as equalities : 20x + 50y = 4800 and 80x + 50y = 7200.

Operations Research

254 Step two : Here we draw lines in two dimensional plane.

— 3k 0 \ Z= 40x +24y =0 (90, 0) x 3 gives = 5 Y Fig. 6.6 Step three : The shaded region in the figure 6.6 is the permissible region for the values of x and y. Step four : The extreme points of the permissible region are A(240, 0), P(40, 80), B(0,144). S. No.

Point

1. 2. 3.

A (240, 0) B (40, 80) C (0, 144)

Z = 40x +24y Z = 40 x 240 + 0 = 9600 Z = 40 x 40 + 24 x 80 = 3520 Z = 0 + 24 x 144 = 3456 (Minimum)

Since Z is minimum at the point B (10,144) hence the required solution is x = 0, y = 144 and minimum Z = 3456. Note : The required solution can also be obtained by drawing the dotted line of objective function as shown in the fig. 6.6. Example 15 : Solve graphically the following linear programming problem. Min. Z = 3x1 + 5x2 s. t -3x1 + 4x2 •_ 12 2x1 - x2 -2 2x1 + 3x2 12 x1 54, x2 >_2 xi, x 2 0 and

255

Allocation (General Linear Programming Problems)

Solution : Step one : First we consider the constraints as equalities. —3x1 + 4x2 = 12 2x1 — x2 = —2 2x1 + 3x2 = 12 =4, X2 = 2 Step two : Here we draw lines in two dimensional plane. X2 A

x1=4

Line for minimum value of Z

31i

x,=2

z=3xi +5x2=0 xl -5 gives X2 -- = 3

Fig. 6.7

Step three : The shaded region in fig. 6.7 is the permissible region for the values of xi and x2 .

Step four : Z = 3x1 + 5x2 = 0 gives — = —5 x2

3

We draw this line through 0 (dotted line) and continue drawing lines parallel to it till we reach the point P of the permissible region which is nearest to the origin. Z is mini. at P (3, 2) which is the point of intersection of lines

x2 = 2 and 2x1 + 3x2 = 12. Z is Mini. when

and

x1 = 3,x2 = 2

Mini. Z = 3 x 3 + 5 x 2= 19.

Note : In this problem we need minimum value of

Z, therefore the point P of the' permissible-region which is nearest to 0 give mini. value of Z.

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Example 16 : Solve the following linear programming problem. Max. s. t.

Z = 0.75xi + x 2 xl — X2 — 0.5 xi + x2 1 x1, x2 0

Solution : Step one : First we consider the constraints as equalities x1 — X2 = 0, - 0. 5X1 + X2 = 1

Step two : Now we draw the lines to each of above equations in two-dimensional plane. See fig. 6.8 A x2 // A-`‘'

ti

\

A

xis 0 (vi) Max. Z = 5x1 + 3x, s.t. 3x1 + 5x2 _5 15 5x1 + 2x2 _5 10 xi, x2 > 0 (vii) Max. Z = 3x1 + 4x2 s.t. x1— x2< —1 x1 + x2 5_ 0 x1,x,>_0

= 3x1 + 2x7 2x1 + 3x2 5. 9 xi — 5x, —20 , x ?_ 0 (ix) Max. Z = x1 + x2 2x1. + x, 5 1 s.t. xi 5. 2 xi + x2 0 2. Solve graphically the following L.P. Problems. (viii) Max. s.t.

(i)

(ii)

Z

Mini Z = 4x1 + 2x2 s.t. x1 + 2x2 > 2 3x1 + x2 3 4x1 + 3x2 ?_ 6 and xi , x2 Mini Z = —x1 + 2x2. + 3x2 5_ 10 s.t. x1 + x2 5_ 6

— x2 2 xi,x2 (iii) Mini Z = 20x1 + 10x2 xi + 2x2 5 40 s.t. 3x1 + x2 30 4x1 + 3x2 60 xi, X2 _> 0 and (iv) Mini Z = 2x1 + 3x2 s.t. xi + x2 4

Allocation (General Linear Programming Problems) 6x1 + 2x2 =2 8 + 5x, ;_>. 4 x1 ° [UP TECH MBA 2004-05]

x2 4 7. Solve the following L.P. Problem : and xi, x2 -2. 0 [Agra 1999] Max. Z = 8x + 16y 3. Determine x 0 and y > 0 so as to subject to, x + y 200 y 5 125 maximize Z = 2x+3y subject to the constraints : x + y 30, y 5. 12, x520, y

3, x — y 0

and

2x + Oy .‘_•-• 900 x,y0 [UP TECH MBA 2001-02]

4. Does the following L.P.P. has a feasible solution ? Max. Z = x1+ x2 s.t.

x1— x2 3x1 — x2 5. —3 x2 -1>:. 0.

show with the help of a graph. 8. Mark the feasible region represented by constraint equations x1 x2 5 1, 3x1 + x2 3, xi 0, x2 > 0 of a linear optimizing function Z = xi + x2. 9.

A farmer has a 100—acre farm. He can sell all the tomatoes, lettuce, or radishes he can raise. The price he can obtain is Rs. 1 per kilogram for tomatoes, Rs. 0.75 a head for lettuce and Rs. 2 per kilogram for radishes. The average yield per acre is 2000 kilograms of tomatoes, 3000 heads of lettuce and 1000 kilograms of radishes. Fertilizer is available at Rs. 0.50 per kilogram and the amount

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required per acre is 100 kilograms each for tomatoes and lettuce and 50 kilograms for radishes. Labour required for sowing, cultivating and harvesting per acre is 5 man/days for tomatoes and radishes and 6 man/days for lettuce. A total of 400 man/days of labour are available at Rs. 20.00 per man-days. Formulate this problem as a L.P.P. to maximize the farmer's total profit. 10. A firm manufactures two types of products A and B and sells them at a profit of Rs. 2.00 on type A and Rs. 3.00 on type B. Each product is processed on two machines M1 and M2. Type A requires one minute of processing time on Ml and two minutes on M2; type B requires one minute on M1 and one minute on M2. The machine M1 is available for not more than 6 hours 40 minutes while machine M2 is available for 10 hours during any working day. Formulate the problem as a L.P.P. and find how many products of each type should the firm produce each day in order to get maximum profit. [Meerut 2008 (BP)]

11. Old hens can be bought for Rs. 2.00 each but young ones cost Rs. 5.00 each. The old hens lay 3 eggs per week and the young ones 5 eggs per week, each egg being worth 30 paise. A hen cost Rs.1.00 per week to feed. If I have only Rs. 80.00 to spend for hens, how many of each kind should I buy to give a profit of more than Rs. 6.00 per week, assuming that I cannot house more than 20 hens. 12. The postmaster of a local post office wishes to hire extra helpers during the Deepawli season, because of a large increase in the volume of mail handling and delivery. Because of the limited office space and the budgetary condition, the number of temporary helpers must not exceed 10. According to the past experience, men can handle 300 letters and 80 packages per day, on the average, and woman can handle 400 letters and 50 packages per day. The post master believes that the daily volume of extra mail and packages will be no less than 3400 and 680 respectively. A man receives Rs. 25 a day and a woman receives Rs. 22 a day. How many men and women helpers should be hired to keep the pay roll at a minimum ? 13. A soft drink plant has two bottling machines A and B. It produces and sells 8 ounce and 16 ounce bottles. The following data is available. Machine

8 Ounce

16 Ounce

A B

100/minute 60/minute

40/minute 75/minute

The machines can be run 8 hrs. per day 5 days per week. Weekly production of the drinks cannot exceed 300000 ounces and the market can absorb 25000 eight ounce bottles and 7000 sixteen ounce bottles per week. Profit on these bottles is 15 raise and 25 paise per bottle respectively. The planner wishes to maximize his profit subject to all the production and marketing restrictions. Formulate it as a L.P.P. and solve. [Meerut 1998]

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271

14. A machine producing either product A or B can produce A by using 2 units of chemicals and 1 unit of a compound and can produce B by using 1 unit of chemicals and 2 units of compound. Only 800 units of chemicals and 1000 units of the compound are available. The profits available per unit of A and B are respectively Rs. 30 and Rs. 20. Draw a suitable diagram to show the feasible region. Also find the optimum allocation of units between A and B to maximise the total profit. Find the maximum profit. 15. The ABC Electric Appliance company produces two products; refrigerators and ranges. Production takes place in two separate departments. Refrigerators are produced in Department I and ranges are produced in Department II. The company's two products are produced and sold on a weekly basis. The weekly production cannot exceed 25 refrigerators in Department I and 35 ranges in Department II, because of the limited available facilities in these two departments. The company regularly employs a total of 60 workers in the two departments. A refrigerator requires 2 man-week of labour, while a range requires 1 man-week of labour. A refrigerator contributes a profit of Rs. 60 and a range contributes a profit of Rs. 40. How many units of refrigerators and ranges should the company produce to realize a maximum profit ? 16. An automobile manufacturer makes automobile and trucks in a factory that is divided into two shops. Shop A, which performs the basic assembly operation, must work 5 man-days on each truck but only 2 man-days on each automobile. Shop B, which performs finishing operations, must work 3 man-days for each automobile or truck that it produces. Because of men and machine limitations shop A has 180 man-days per week available while shop B has 135 man-days per week. If the manufacturer makes a profit of Rs. 300 on each truck and Rs. 200 on each automobile, how many of each should he produce to maximize his profit ? 17. A manufacturer produces two types of models A and B. Each A model requires 4 hours of grinding and 2 hours of polishing; where as each B model requires 2 hours of grinding and 5 hours of polishing. The manufacturer has 2 grinders and 3 polishers. Each grinder works for 40 hours a week and each polisher works for 60 hours a week. Profit on an A model is Rs. 3.00 and on an B model is Rs. 4.00.Whatever is produced in a week is sold in the market. How should the manufacturer allocate his production capacity to the two types of models so that he may make the maximum profit in a week ? [Hint : Let xi and x2 be the number of model A and B respectively produced per week. Then L.P.P. is Max. Z = 3x1 + 4x2 s.t. 4xi + 2x2 _5 80; 2x1 + 5x2 5. 180; xi, x2 O. ]

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18. Solve O. N. 17 when the profit on an A model is Rs. 30 and on a B model is Rs. 40. 19. A firm manufacture 3 products A,13 and C. The profits are Rs. 3, Rs. 2 and Rs. 4 respectively. The firm has 2 machines and below is the required processing time in minutes for each machine on each product. Product

Machine A

M1 M2

2

B

C

3

5

2

4

Machines MI and M2 have 2000 and 2500 machine-minutes respectively. The firm must manufacturer 100 A's, 200 B's and 50C's but not more than 150 A's. Set up a L.P.P. to maximize profit. 20. A factory uses three different resources for the manufacture of two different products, 20 units of the resource A, 12 units of B and 16 units of C being available. 1 unit of the first product requires 2, 2 and 4 units of the respective resources and 1 unit of the second product requires 4, 2 and 0 units of the respective resources. It is known that the first product gives a profit of 2 monetary unit per unit and the second 3. Formulate the linear programming problem. How many units of each product should be manufactured for maximizing the profit ? Solve it graphically. 21. A pineapple firm produces two products canned pineapple and canned juice. The specific amounts of material, labour and equipment required to produce each product and the availability of each of these resources are shown in the table given below : Canned Juice

Canned Pineapple

Available Resources

Labour (man-hours)

3

2.0

12.0

Equipment (machine hours)

1

2.3

6.9

Material (unit)

1

1.4

4.9

Assuming one unit each of caned juice and canned pineapple has profit margins of Rs. 2 and Rs. 1 respectively; determine the produced mix that will maximize the profit.

Allocation (General Linear Programming Problems) 22.

273

Diet Problem : Consider two different types of food stuffs, say F1 and F2.

Assume that these food stuffs contain vitamin V1, V2 and V3 respectively. Minimum daily requirement of these vitamins are 1 mg of Vi , 50 mg of V2 and 10 mg of V3. Suppose that the food stuff F1 contains 1 mg of V1, No mg of V2 and 10 mg of V3; whereas the food stuff F2 contains 1 mg of V1, 10 mg of V2 and 100 mg of V3. Cost of one unit of food stuff is Rs. 1 and that of F2 is Rs. 1.5. Find the minimum cost diet that would supply the body at least the minimum requirements of each vitamin. [Meerut 1995] Hint : If x1 and x2 are the required units of Fl and F2, then L.P.P. is

Mini Z = x1 + 1.5x2 s.t. x1 + x2 1, 100 + 10x2 50 10x1 + 100x2 10 and x1, x2 >_0 23. A company sells two different products A and B. The company makes a profit of Rs. 40 and Rs. 30 per unit of products A and B respectively. The two products are produced in a common production process and are sold in two different markets. The production process has a capacity of 30000 man-hours. It takes 3 hours to produce one unit of A and one hour to produce one unit of B. The market has been surveyed and company officials feel that the maximum number of units of A that can be sold is 8000 and the maximum of B is 12000 units. Subject to these limitations, the products can be sold in any convex combination. Formulae the above problem as a L.P.P. and solve it by graphical method. 24. Show that the feasible solution x1 = 1, x2 = 0, x3 = 1, Z = 6 to the system of equations x1 + x2 + x3 = 2, — x2 + x3 = 2, xi >_0 which minimize Z = 2x1 + 3x2 + 4x3 is not basic. 25. Find all the basic solutions for the system of equations : 2x1 + 3x2 + 4x3 = 5 3x1 +

5x3__= 6

26. Determine all distinct basic solutions to the following equations : Xi 4- 2X2 + x3 + X4 = 2, x1 + 2x2 + 1x3 + xs = 2

2 27. Do all possible basic solutions of the following system of equations exist ? Find all existing basic solutions. x1 + 2x2 + 3x3 + 4x4 = 7 2x1 + x2 + x3 + 2x4 = 3

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28. Do the system xi + x2 = 1, x1 — x3 = 2, xi, x2, x3 0 have a feasible solution. [Hint : From the relation x1 + x2 = 1, x2 = 1— xl 1 x2 0 Also from the relation x1 — x3 = 2,x3 = x1 — 2 x3 >_0=x1 >_2. Thus, both given equation and the restriction x2 0, x3 0 are satisfied if xl S1andx1 >2. But no such value ofx1 0 exists. given problem has no F.S.] 29. Find all the B.F. Solutions of the following system : 8x1 + 6x2 + 13x3 + x4 + X5 = 6 9x1 + X2 + 2X3

6X4

10X5

10

xi 0, = 1, 2, 3, 4, 5 . and 30. Find an optimal solution of the following L.P.P. without using the simplex algorithm Z = + 2x2 Max. xl + x2 5. 10, 2x1 — x2 5 40 s.t. xi, x2 O. and •

Allocation (General Linear Programming Problems)

275

+ ANSWERS 41.

(i)

= 0, x2 = 1, Max. Z = 3

(ii) x1 = 3, x2 = 4, Max. Z = 10 (iii) xi = 8, x2 = 0, Max. Z = 24 (iv) x1 = 2, x2 = 1, Max. Z = 8 (v) xi = 400/13, x2 = 150/13, Max. Z = 1800/13 (vi) x1 = 20/19, x2 = 45/19, Max. Z = 235/19 (vii) No feasible solution (viii) No feasible solution (ix) Infinite number xi = Z x2 = 1 or xi = 2/3, x2 = 7/3, Max. Z = 3

2.

(x) Unbounded solution = 0.6, x2 = 1.2, Mini. Z = 4.8 (i) (ii) x1 = 2, x2 = 0, Mini. Z = -2 (iii) xi = 6, x2 = 12, Mini. Z = 240 (iv) xi = 8/7, x2 = 4/7, Mini. Z = 4 (v) x1= 1.2, x2 = 2.6, Mini. Z = 10.2

(vi) Infinite number 3. x = 18, y = 12, Max. Z = 72 4. No. 5. x1 = 21/2, x2 = 69/2, Max. Z = 651/2 6. xi = 9, x2 = 0, Max. Z = 9, 7. Infinite No. of Solutions xi = 50, x2 = 125 ; x1 = 100, x2 = 100, Max. Z = 2400 8. No feasible solution 9. If farmer sow tomatoes, lettuce and radishes in xi, x2, x3acres respectively, L.P.P. is Max Z = 1850x1 + 2080x2 + 1875x3 s.t. + x2 + x3 100, 5x1 + 6x2 + 5x3 5.. 400, xi, x2, x3 O. 10. If firm manufacture xi and x2 units of product A and B respectively, L.P.P. is Max. Z = 2x1 + 3x2 + x2 600, xi, x2 0 s.t. + x2 400, Optimum solution, x1 = 0, x2 = 400, Z = Rs.1200 11. If xi, x2 are the numbers of old and young lens respectively, then L.P.P is Max. Z = - 0.1x1 + 0.5x2 s.t. 2x1 + 5x2 80, + x2 20, xi, x2 >_0 Optimum solution, xi = 0, x2 = 16,Z = Rs. 8. 12. 6 men and 4 women.

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13. If x1 bottles of 8 ounces and x2 of 16 ounces are produced, then L.P.P. is Max. Z = 0.15x1 + 0.25x2 8x1 + 16x2 5 300000, 2x1 + 5x2 5 480000, 5x1 + 4x2 .5 720000 s.t. 5 25000,x2 5 7000, x1, x2 0. Optimum solution, x1 = 25000, x2 = 6250,Z = Rs. 5312.50. 14. 15. 16. 17. 18. 19.

200 units of A, 400 units of B, Max. Profit = Rs. 14000. Refrigerators = 12.5, Ranges = 35, Max. Profit = Rs. 2150. 30 Trucks and 15 Automobiles per week. A = 2.5, B = 35, Max. Profit = Rs. 147.50. A = 2.5,8 = 35, Max. Profit = Rs.1475. If x1 units of A, x2 units of B and x3 units of C are manufactured, then L.P.P. is Max. Z = 3x1 + 2x2 + 4x3 s.t. 4x1 + 3x2 + 5x3 5 2000, 2x1 + 2x2 + 4x3 5 2500, 100 5 x1 5. 150, 0 5 x2 5 200 and 0 5 x3 5 50.

20. If x1, x2 units of two products are manufactured, then L.P.P. is Max. Z = 2x1 + 3x2 s.t. x1 + 2x2 5 10, x1 + x2 5 6, xi 4, x1, x2 0 . Optimum solution, x1 = 2 units, x2 = 4 units. 21. Canned juice = 4 units, canned pineapple = 0 units, Profit = Rs. 8. 22. F1 = 1 unit, F2 = 0 unit, Minimum cost = Rs. 1. 23. If x1 units of product A and x2 unit's of B are produced, then L.P.P. is Max. Z = 40x1 + 30x2 3x1 + x2 5 30000, xi. 5 8000,x2 5. 12000, xi , x2 0 . s.t. Optimum solution, x1 = 6000, x2 = 12000 and Max. Z = Rs. 600000 . (-2, 3, 0), (-1/2, 0, 3/2) and (0, -1, 2). (2, 0, 0, 0, 0), (0, 1, 0, 0, 0), (0, 0, 4, -2, 0), (0, 0, 2, 0, 1) and (0, 0, 0, 2, 2). No. (-1/3, 11/3, 0, 0), (2/5, 0, 11/5, 0), (-1/3, 0, 0, 11/6), (0, 2, 1, 0) and (0, 0, 1, 1). No. (2/3, 0, 0, 2/3, 0), (50/71, 0, 0, 0, 26/71), (0, 26/35, 0, 54/35, 0), (0, 50/59, 0, 0, 54/59), (0, 0, 13/38, 59/38, 0) and (0, 0, 25/64, 0, 59/64). 30. xi = 0, x2 10, Max. Z = 20.

25. 26. 27. 28. 29.

• • •

ei

7

Convex Sets and Their Properties •

7.1 Introduction In this chapter we shall discuss convex set theory which has found many important applications in linear programming, games theory and other decision theories.

7.2 Some Important Definitions 1. Point Sets : Point sets are sets whose elements are points or vectors in E " (n-dimensional Euclidean space). For example, (i) A linear equation in two variables x1, x2 i. e., al x + a2x2 = b represents a line in two dimensions. This line may be considered as a set of those points (x1, x2 ) which satisfy ai + a2x2 = b. This set of points can be written as Sl = {(xi , x2 ) : + a2x2 =b}. (ii) Consider the set of points lying inside a circle of unit radius with centre at the orgin, in two dimensional space (E 2 ). Obviously the points (x1, x2 ) of this set satisfy the inequality x12 + x22 < 1. This set of point can be written as S2 = {(Xi, X2 ): x12+ x22 0 is defined to be the set of points X {x :1 x — al =

[Meerut LP 1993]

i. e., the equation of hypersphere in E n is (x1 — al )2 -1- (X2 — a2 )2 + where

a=

a2,

+

— a, ) 2 = a 2

an ), x = (x1, x2, ...., xn )

which represent a circle in E 2 and sphere in E 3 .

3. An a-neighbourhood : An a-neighbourhood about the point a is defined as the set of points lying inside the hypersphere with centre at a and radius E > 0. i. e., the E-neighbourhood about the point is the set of points, X ={X :lx— aka}

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278 4.

5.

6. 7. 8.

9.

An Interior Point : A point a is an interior point of the set S if there exists an a-neighbourhood about a which contains only points of the set S.

An interior point of S must be an element of S. A Boundary Point : A point a is a boundary point of the set S if every a-neighbourhood about a (a > 0 may be however small) contains points which are in the set and the points which are not in the set. A boundary point of S does not have to be an element of S. An Open Set : A set S is said to be an open set if it contains only interior points. A Closed Set : A set S is said to be a closed set if it contains all its boundary points. Lines : In E n, the line through the two points x1 and x 2, x1 x2 is defined to be the set of points X = fx : x = Xxi + (1 — X) x2, for all real Xl. Line Segments : In E n , the line segment joining two points x1 and x 2 is

defined to be the set of points X = fx : x = Xxi + (1 — X) x 2, 0 X 10. Hyperplane : A hyperplane is defined as the set of points satisfying

or

+ c 2x2 + • .

+ Cn Xn =z (not all ci =0) cx =z

for prescribed values of c1, c2,

cn and z,

The vector c is called vector normal to the hyper-plane and ± c I ci are called unit normals. A hyperplane divides the whole space E n into three mutually disjoint sets given by X1 = { x : cx > Z} X2 = {X : CX Z} X3 ={c : cx < z} The sets X1 and X3 are called open half spaces. The sets {x : cx z} and {x : cx z} are called closed half spaces. 11. Parallel Hyperplanes : Two hyperplanes eix = z and c2x= z are said to be parallel if they have the same unit normals i. e., if c1 = Xc2 for some ?, ? being

non zero. 12. Convex Combination : A convex combination of a finite number of

points x1, x2, ..., xn is defined as a point x =?.1x1 +? 2x2 + + X n xn

where Xi is real and

0,47` i

279

Convex Sets and Their Properties

and The convex combination of two points x1 and x2 is given by

x=

+ X 2x2, St. X.1, a'2>— 0, a1 +a2 =1

It can also be written as x = Xxi + (1 — X) x2, 0 X 5. 1. This show that the line segment joining two points xi and x2 is nothing but the set of all possible convex combinations of the points x1 and x2 .

13. Convex Set [Agra 1998] A set of points is said to be convex if for any two points in the set, the line segment joining these two points is also in the set. In other words a set is convex if the convex combination of any two points in the set, is also in the set.

Fig. 7.1 Convex sets

Fig. 7.2 Non-Convex sets

14. Extreme point of a convex set A point x in a convex set C is called an extreme point if x cannot be expressed as a nconvex combination of any two distinct points x1 and x 2 in C. In other words, a point x in a convex set C is an extreme point of Cif it does not lie on the line segment of any two points, different from x in the set. Mathematically, a point is an extreme point of a convex set if there do not exist other points x1, x2 (x1 # x2 ) in the set, such that x = Xxi + (1 — X) x2, 0 X < 1.

For example; The set C = { (x1 , x2} : x1. 2 + x22 5 1} is convex. The polygon which are convex sets have the extreme points as their vertices. Obviously, an extreme point is a boundary point of the set. It should be noted that all boundary points of a convex set are not necessarily extreme points. A point of C which is not an extreme point, is referred as an internal point of C.

280

Operations Research

15. Convex Hull : The convex hull C(x) of any given set of points Xis the set of all convex combinations of sets of points from X. Example : If x is just the eight vertices of a cube, then the convex hull C (x) is the whole cube. 16. Convex Function : A function f (x) is said to be strictly convex at x if for any two other distinct points x1 and x2, f {Xxi + (1 — 2)x 2 } < f (xi ) + (1 — X) f(x2 ), where 0 < X, < 1 On the other hand a function f (x) is strictly concave if —f (x) is strictly convex. 17. Convex Polyhedron : The set of all convex combinations of finite number of points is called he convex polyhedron generated by these points . Example : The set of the area of a triangle is a convex polyhedron of its vertices.

7.3 Some important Theorems Theorem I : A hyperplane is a convex set. Proof : Consider the hyper-plane X = {x : cx = z}. Let x1 and x2 be any two points in the hyper-plane X cx1 = z and cx2 = z If x3 =A,x1 + (1 — X) x2, 0 X 1 then cx3 =Xcx1 + (1 — X) cx2 = X2 + (1— X) z = z which implies that x3 = Xx1 + (1 — X)x2 is also point in the hyper-plane X. Hence, by definition, the hyperplane X is a convex set. Theorem II : The closed half spaces H1 = {x cx z} and H2 = : ex S z} are convex sets. [Meerut L.P. 1993] Proof : Let x1 and x2 be any two points of H1. Thencx1 z, cx2 z. Now for 0 _ 0 for all j = 1, 2 ,..., (n + m) brn are positive.

b1, b2,

The above I..P. problem may also be stated as Max. Z = cx subject to Ax = b and where c =

x>0 c2,..., cn , 0, 0,..., 0)1 x (,n+n) is a row vector of order 1 x

+ n)

A = [aii i mx(ni+ n) is the matrix of coefficients of order m x (m + ii) xI x2 • • • •

x=

X

X 2" •• •9

n + 1 • Xn+ml

xn-i 1 • •

(n +

b=

b2 „

m)xl

= [bp b 2,

, bm ]

mx1

If oc denotes the j-th column of the matrix A then i A = (apa'2, • • • ) (I n+m )• (1)

First we shall denote by B a m x m non-singular matrix whose column vectors are linearly independent columns of the matrix A. If these columns are denoted by pi , p 2, ,Pin then B = (131,13 2,••• , Pin )•

Matrix B is called the basis matrix. (ii) The variables corresponding to pi, p 2,...,13 n, called the basic variables will

be denoted by xifi , xB2,..., xBni respectively. The vector (column vector) of these in basic variables is denoted by x B or XB i.e., x B (or X B ) = [X BI , X B9 , • • • , XBin ]

where xB = B-1b, and is called the B.F.S. of the L.P.P.

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295

(iii) The coefficients of the basic variables xm xB2, xBni in the objective function Z will be denoted by c m, C B2 • " CBm, the row vector corresponding to these variables is denoted by cB or C B i.e., c B (or C B ) = (c Bly C- B25 • • • , CBm)' (iv) Since B = (1-1 ,r3 2, ... ) is non-singular matrix of order m x m, and the vectors 131,132, ..., p rn are linearly independent, they form the basis of E m . Therefore, each vector in E m can be expressed as a linear combination of vectors in B. Let a; =13iYi; +13 2Y 2j +.• •• • + 13 in Y rnj Ytj Y 2j = (P1

9, • • • • )13 m

ylj

= BY.

• •

Y111.1

Yj =

Y 2j

Y m}_rn x 1

Yt. Y 2j,• • • y nd , are scalar required to express each (xi, (j = 1, 2 , 3, ..., n + m) as L.C. of (31, ¢2,•••,13m• Y. = 13--1 a • (v) In the end we write Zj = CBYj = CB1Y1j

C B2Y 2j 4- • • • • + CBrnY rnj •

8.3 Fundamental. Theorem of Linear Programming If a L.P. problem s. t. Ax = b, x 0

Max. Z = c x,

where A is a m x (m + n) matrix of coefficients given by A = (a l , a 2,..., a optimal solution, then at least one feasible solution must be optimal.

), has an

[Meerut 1990, 95]

i. n ], be an optimal feasible solution of the given L.P.

Proof : Let x* = [x.1 , x2, ..., problem and Z* =

m+n ci xi be the maximum value of the objective function

i corresponding to this solution x*. Suppose that k components (variables) in x* are non-zero and remaining (m + n k) components are zero. We can assume these non-zero components as the first k components of x*. + rt — k) i.e.,

x* =

x2 ,

x k, 0, 0, ..., 0]

k Xi OGi = b

Operations Research

296 k

Z* =

and

1.

2.

I ci ai i

(2)

Now there are two possibilities. If a2,a2 ....,ak are L.I. : If the vectors a1'c(2,..., ak are L.I. then by definition x* is a B.F.S. which is also optimal. Hence the theorem is true in this case. This solution is degenerate if k < m and non-degenerate if k = tn. If al, ok are L.D.: This is the case when k > m. In this case we can reduce the number of non-zero variables step by step as follows.

Since at , a 2, ..., al, are L.D., therefore there exist scalars Xi, X 2,..., k, with at least one X i 0, s.t. =0

i Now suppose that at least on Xi is positive because if none is positive then we can multiply the equation by - 1 and get positive Xi . Let

v = Max ki i 5: k xi

, ...(4)

Clearly v > 0, since xi ?. 0, for i = 1, 2, ..., k and at least one X i is positive. Mulitklying (3) by 1/v and subtracting from (1), we have k

( xi

/ .---1

- ---L k ot•: = b v

(5)

from (5), it follows that

tn+n-k

Xk X-= [V1 --,--2 41 Y --) 42 • •••, xk — V

V

is also a solution of the system Ax = b. Also from (4), we have X• v> i = 1, 2,..., k, xi i i or x• > or xi -

...(6)

...„

V

0, i = 1,

k

i.e., all the components of the solution x' given by (6) are non-negative and thus x' is a F.S. to the given L.P.P. Also (4) implies that, at least for one value of i Xi X• v=— or x• - = 0 x• = — or xi Xi i.e. xi vanishes-at least for one value of i. Therefore, the F.S. x' given by-(6) V

cannot have more than (k - 1) non-zero components (variables).

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Simplex Method

Thus, we have derived a new F.S. from the given optimal solution, containing less number of non-zero variables. Now we shall prove that x' is also an optimal solution. Let Z ' be the value of the objective function corresponding to this solution. —X• Z cl x• 1=1 " v k

1 _k v i=1

1 --Eci Xi = Z * v

i xi

i=1

...(7)

Now if Z' is optimal then Z' = Z*, which is possible only if k

=0 i =1

To prove I c i Xi = 0, we shall-use contradiction method.

If possible, suppose that Ici X i 0 i k either (i)

k ci X i < 0 or

i =1

ci X > 0 •

(ii) i

We can find a real number r, negative in case (i) and positive in case (ii) such that k i=1 k

or

ci (rX ) > 0 k•

or

(r

)+ i =1

or

cixi >Ecixi i

(xi + rXi ) > Z* i =1

[from (2)] Now multiplying (3) by r and adding to (1), we have k

I (x1 + r7k,i )ai =

i =1

b

...(8)

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298

Which implies that n - k)

(in

[x1 +

1

xk + rk k , 0, 0, ..., 0 ], for all values of r is also a

x2 +

solution of the system Ax = b. If we choose r, s. t. x• + rX I• > - 0 At or

r?

i =1,2,...,k

—Xi

r>-

X.

if k • >0

' xi - — if ?• 0)

(9)

Also we have Max.

< 0 and

Mini. E xi i

>0

(Xi < 0)

(2L i > 0)

which implies that interval given by (9) is non-empty interval. Hence, when r lies in the non-empty interval given by (9) then the solution given by in n -k [x1 + rX1,x2 + rX 2,

xk + r X k , 0, 0,..., 0']

is also a F.S. of the system Ax = b and from (8), this solution gives a value Z which is greater than Z* (optimal value or greatest value). Which is a contradiction k ci Xi = 0 E i=0

i.e.

Z,=Z*

i.e., x' given by (6) is also an optimal solution. In this way from a given optimal solution we have constructed (derived) a new optimal solution containing less number of non-zero variables. This solution is optimal B.F.S. if the column vectors associated to non-zero variables in this new solution are L.I. Hence, the theorem is true. If these associated vectors are not L.I.

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Simplex Method

(i.e., solution not B.F.S.) then by repeating the above process we can get another optimal B.F.S. of the given system containing not more than (k — 2) non-zero variables. Continuing in this way for finite number of times we will certainly get an optimal B.F.S. of the system. Hence, the theorem is true. Note : In the F.S. x' the variable xr will become zero if

v.

fx,.1 xr

Max. 1=1 Lxi 1 5.i

xr.

8.4 To obtain B.F.S. from a F.S. Theorem : If a L.P.P. Max. Z = cx.

s.t. Ax = b, x 0

where A is a in x (in + n) matrix of coefficients given by A = a 9, • • ' Cit /11 n), has at least one feasible solution t then it has at least one basic feasible solution also. [Meerut 2007 (BP)]

Proof : Let x = [xl , x2 ..., xni+ n ] be a feasible solution of the given L.P.P. Suppose that, k components (variables) in x are non-zero and remaining (in + n k) components are zero. We can assume these non-zero components as the first k components of x i.e.,

x

x 2 ,..., xk ,1 0,0,...,01 ] (m + n — k) k

b,

...(1)

Now there are two possibilities, 1. If the vectors cq,a 2, ...,ak are L.I. If the vectors a.1 , a2 , , a k are L.I., then by definition x is a B.F.S. Hence the theorem is true in this case. This solution is degenerate if k < m and non-degenerate if k = m. 2. If the vectors ai,a2, ...,a k are L.D. : This is the case when k > m. In this case we can reduce the number of non-zero variables step by step, so that the feasible solution become B.F.S.. Proceeding as in article 8.3, we can reduce the F.S. x= [xi, x2 ,

xk, 10, 0, ..., (m + n

k)

to a new feasible solution 1 v X1 .,. "2 Aqc x' = E-1 — ---, .A.2 - -----,.•.,xk - —, ...., -,...• 01 1 v V ' v I i (in + n — k) t In L.P.P. any solution which satisfies x > 0 will be termed as feasible solution.

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300

which cannot contain more than (k — 1) non-zero components. Thus we have derived a new F.S. from the given F.S. containing less number of non-zero (positive) variables. If the column vectors associated to non-zero variables in this new solution are L.I. then this solution is B.F.S. and hence the theorem is true. But if these associated vectors are not L.I. then this solution is not B.F.S. In this case we repeat the above process again and get another B.F.S of the given system containing not more than (k — 2) non-zero variables. Continuing in this way for finite number of times we will certainly get an optimal B.F.S. of the system. Hence, the theorem is true. Note : The two theorems 8.3 and 8.4 in combined form may be stated as follows. If the I..P.P. Max. Z = c x, s.t. A x= b, x 0 has feasible solution, then at least one of the B.F. solution will be optimal.

.911liatAative EXanigeti Example 1 : If x1 = 2, x2 = 3, x3 = 1, be a feasible solution of the L.P.P. Max. Z = x1 + 2x2 + 4x3 2x1 + x2 + 4x3 = 11 3x1 + x2 + 5x3 = 14 xi , x2, x3 0 then find a B.F.S. Solution : The given L.P.P. can be written as s.t.

[Meerut 2007 (BP), 08]

Max. Z = x1 + 2x2 + 4x3 s.t.

where a1

aix1 +a2x2 +"3x3 = b xi. , x2, x3 0 1-21 = [41 = 11 b ,a2 = [3] [1] 1_5] 1_14]

Here in (number of constraints) = 2, so the B.F.S. of the L.P.P. cannot have more than 2 non-zero variables. The given feasible solution x1 = 2, x2 = 3, x3 = 1 is not B.F.S. In order to reduce this F.S. to a B.F.S. we have to make at least one variables zero. For this we proceed as follows. Since the vectors c1,a2,et3, associated with the non-zero variables are L.D., therefore one of these vectors may be expressed as a L.C. of the remaining two. Let a1 X 2a2 + X 3a3 [4] [X2 +4X 3] 111 [21 . or x 2 +x3 [1] 5 X 2 +5X 3 3 +4X 3 =2 and X 2 + 5X 3 =3 Solving, we have X = —2 and X 3 7-- 1 a1 =-2.a2 +1.a3

Simplex Method or

301 =° al. + 2(12 3 ai = 0, where Xi = 1, i

or

Now

_.(2) 2 = Z 7t, 3 = —1

v = Max. ( Xi j =Max. ( X1 , X 2 , X 3 ) i< i < 3 Xi Xi x2 x3 = Max.(1 2 —1 = 2 = X 2 2' 3' 1 3 x2 .

. The variable x2 should be zero or the vector a 2 should be eliminated. Substituting the given feasible solution xi. = 2, x2 = 3, x3 = 1 in (1), we have 2a1 +3a.2 + ct3 = b Substituting the value of a 2 from (2) in (3), we have ("3 —ai + a 3 = b 2a1 +3 2 JI 1 5 or a1 + "3 = b

...(3)

2

or

1 2

5 ai + "2 ± -2- a 3 =b

The new feasible solution is xi = 1/2, x2 = 0, x3 = 5/2 . Here the vectors a1 =, 2 a3 = 4 , associated to non-zero variables are L.I. Hence, this F.S. 3 L5 = 1/2, x2 = 0, x3 = 5/2 is B.F.S. Note : 1. Another B.F.S. of the L.P.P. is x1 = 3,x2 = 5, x3 = 0, obtained by writing (2) as —ai — 2a 2 +a3 = O. 2. The new values of the variables can also be found by using the formula • xi '= xi — X. Example 2 : Consider the set of equations 2x1 — 3x2 + 4x3 + 6x4 = 21 xi+ 2x2 + 3x3 — 3x4 + 5x5 = 9 If = 2,x2 = 1,x3 = 2, x4 = Z x5 = 1 is a feasible solution, then reduce it to two different B.F. solutions. Solution : The system of equations in matrix form, can be written as x1

[2 — 3 4 6 01 x2 =[ 21 Ll 2 3 — 3 5] x3 9 or

_x 4 _ a1x1 + a2x2 + a3x3 + a4x4 + a5x5 = b

...(1)

Operations Research

302 -3 a3 = [ ,a 4 where ai = [ ,a2 = [ 3 i

0] = [211 6 i,a5 =1 , D 5 3 29J

whose F.S. is x1 = 2, x2 = 1, x3 = 2, x4 = 2, x5 = 1 (Given) ...(2) 2a1 a2 + 2a3 + 2a 4 + a5 = b = 5 - 2 = 3 zero Here m = 2 and n = 5. So the B.F.S. will have at least n variables i.e., B.F.S. cannot have more than two non-zero variables. Thus, given F.S. is not basic F.S. In order to reduce this F.S. to B.F.S., we have to make at least three variables equal to zero. Thus, three variables will be reduced to zero one by one. (i) To reduce one of the variables to zero value. Since ai, , a3, a4, a 5 are L.D. so by careful inspection, we have 2a1 +2a, -a3 +a 4 + O. a5 = 0 Xi ai =0, where Xi =2, 2 = 2, 3 = -1,

or

-(3)

= 1, X 5 = 0

i -1 Now let v= Max. (Xi /xi ) = Max. (Xi /xi, X 2 /x2, X 3 /x3, X 4 /x4, X 5 /x5 ) 5 = Max. (2/2, 2/1, - 1/2, 1/2, 0/1) = 2/1 = 2 /x2. the variable x2 should be zero for which the vector a 2 should be eliminated between (2) and (3). Eliminating a, between (2) and (3), we have 1 2a1 +(- + - a3 - la 4) + 2a3 + 2a 4 + a5 = b 2 2 ...(4) or a1 + (5/2) a3 + (3/2)a4 +0(.5 = b Thus, xi = 1, x2 = 0, x3 = 5/ 2, x4 = 3/2, x5 = 1 is another feasible solution of the problem, but it is not B.F.S., as it contains more than two non-zero variables. (ii) To reduce other variable to zero value. Now since ai , 3, a4, a5 are L.D. so by their careful inspection, we have - 3.a3 + 2. a4 3.a5 = 0 -(5) =3 41 =O5X3 v =Max. (Xi/xi, X 3 /x3, X 4 /x4, X 5 /x5 ) Now, let = Max. (0/1, -3/(5/2), 2/(3/2), 3/1) = 3/1 = X 5/x5 the variable x5 should be zero, for which the vector a 5 should be eliminated between (4) and (5). Eliminating a5 between (4) and (5), we get 2 a1 + (5/2) a3 + (3/2) a4 + (a3 - - a4 ) = b ...(6) or a1 + (7/2) a3 + (5/ 6) (.1,4 = b Thus, x1 = 1, x2 = 0, x3 = 7/2, x4 = 5/6, x5 = 0 is another F.S. of the problem. It cannot be B.F.S. as it contains more than two non-zero variables.

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(iii) To reduce one more variables to zero. Now since al, a3, a 4 are L.D. let a = aa3 + ba 4 (2ll (4) ( 6) + 6b) or 1_ b (1) - a (3) (-3) 3a - 3b.) 4a + 6b = 2 and 3a - 3b 1 = a = 2/ 5, b = 1/15 (2/5)a3 + (1/15) a 4 15a1 - 6a3 -a 4 = 0 or ...(7) =15, X 3 = - 6, X 4 = -1 Now let v = Max. (XI /xi , X 3/x3, X 4/x4) = Max. [15/1,-6/(7/2),-1/(5/6)] = 15/1 = Xi/xi the variable x1 should be zero for which al should be eliminated between (6) and (7). Eliminating al between (6) and (7), we get {(2/5)a3 +(1/15)x4} + (7/2)a 3 + (5/6)a 4 = b or (39/10) a3 + (9/10) a 4 = b = 0, x2 = 0, x3 = 39/10, x4 = 9/10, x5 =0 which is another F.S. of the problem having only two non-zero variables. It is also B.F.S. since the column vector a3,a4 corresponding to non-zero (basic) variables x3, x4 are L.I. as 4 6 a a, a 4 I = 3 -3 -,- 30 0. To get another B.F.S., we can also write (7) as - 15a1 + 6a3 + a 4 =0, then Xi = -15, X 3 = 6, X 4 =1 Max. {Xi /x1, X 3 /x3, k 4 /x4} = Max. {- 15/1, 6/(7/2), 1/(5/6)1 - 12/7 = X.3 /x3 x3 should be zero for which a 3 should be eliminated between (6) and (7) Eliminating a3, between (6) and (7), we get (39/4) al + (1/4) a4 = b .•. F.S. is x1 = 39/4, x2 = 0,x3 = 0, x4 = 1/4,x5 = 0 which is also B.F.S. since column vectors ill a 4 corresponding to x1 and x4 (basic variables) are L.I. Hence, two different B.F.S. of the system are xl= 0,x2 = 0, x3 = 39/10, x4 = 9/10, x5 = 0 and x1= 39/4, x2 = 0, x3 = 0, x4 = 1/4, x5 = 0

8.5 To Determine Improved B.F.S. Now we shall prove the following important theorem which helps us to determine an improved B.F.S. from a given B.F.S. of a L.P.P.

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304

Theorem : Let xB = B-113 be a B.F.S. of a L.P.P. with Z = c B x B as the value of the objective function. If for any column 0,i in A, but not in B, the condition c 1• — Z 1• > 0 (or Z — c •J < 0) holds, and if at least one yij > 0, i = 1, 2, ... , in, then it is possible to obtain a new B.F.S. by replacing one of the columns in B by oti and if the new value of the objective function is Z' then Z Z. If the given B.F.S. is non-degenerate then Z'> Z. , X B2 , , X Bm ], be a B.F.S., of the given L.P.P., Let xB = Proof : 0 Max Z = cx, s.t. Ax = b, „ ) and basis matrix where A = (al , a 2, , B = (R1,[32,•••,Prn) ...(1) B. xB = b s.t. a • B a•EA Let Since Pi, p,,..,137, form the basic of A, therefore a j can be expressed as the L.C. of these vectors. i.e.

(xj = DiYij = PiYij +13 2Y2i +••••+

(2)

ritY mj

i =1

If y ri # 0, then we can replace p,. in B by a j and still B is a basis matrix. 0, from (2), we have

assuming that y Pr

1 "j Y rj

or

P1

Y

Yrj

1

Pr =

Y r-1,j

Y 2j

Yij

P2•••

in' y

Y rj

aj

a

yti

Yr+1,j

r-1

Y rj

Y mj r+1

13 m •

Yri

ij

-(3)

i=1Yri r

From (1), we have Pi X B1 or

V. 2 X B2 4-• • "413 r XBr +" • •-i- P rn XBin = b

Epi XBi +13 r X Br = b

...(4)

i =1 r

Substituting the value of p r from (3) in (4), we have Yij

rr`

E . =i Pi

XBr

XBi

XBr

„,

-(5)

j=b

Y rj Yrj it r Comparing (5) with (4), the new basic solution to Ax = b is given by B ={) B1 X' B2 • ' •

where

X 'Bi = X Bi -

Yj Y rj

Xfr Brn

Xs, , i = 1,

2, , m; i

r

...(6)

Simplex Method and

305

x' Br = XBr

Yrj This basic solution will be feasible if Yi Bi = X Bi --XBr 0, Yr j

and

X'Br =xBr > Yr j

(i = 1, 2,..., m, i r)

0

...(7) ...(8)

Since xB is a B.F.S. of L.P.P. xBr 0, i = 1, 2,..., m Thus, we see that (7) and (8) are satisfied only if yri > 0 and yu 0 (i =1, r). If y rj > 0 and yij > 0, then (7) and (8) are satisfied only if xBi XBr > Yij

or

Y rj Xgr < xBi

Yrj Or

i =1, 2, 3,...,

Yij

X B r =Mini

.

xBi

> 0}

Y rj

l YU Thus, we can obtain a new B.F.S. from the in'tial B.F.S. by removing the column vector f3 r of the basis matrix B , by a j , if y ri > 0 where r is to be selected such that XBr Yrj

xBi

= Mini. — ,y0 > 0 = v. Yii

(9)

The relation (9) gives the criterion, termed as simplex criterion for selecting the vector Or which is to be deleted (removed) from the basis matrix B. It is important to note that in the new B.F.S. the variable at the r-th position is different from that in the previous. B.F.S. while at all other positions there are the same variables, usually with different value. The variable at the r-th position in the previous B.F.S. is driven to zero. From (9) if v = 0 which is possible Only if xm. = 0, then we conclude that the initial (original) B.F.S. is degenerate B.F.S. To prove that Z' Z : We have m

Z=

m

cBi xBi and Z'= I c' Bi

i =1 i =1 where c'Bi are the coefficients of- the basic variable xBi (i = 1, 2, ..., m) in the objective function Z. It is obvious that (i = 1, 2,..., m; i r) C' Bi = C Bi and c'Br = cj •

Operations Research

306 Now using (6) we have m C 'hi X Bi —

Z =

Yii

Y

Y rj

i =1

XB r

+ C Br..

X

i

CBi

an —

Yii

„ xBr ' `-'). •

X Br.

Y rj

Yrj

C& X-•

xm.

Yii Yrj

CJ

XBr

•

Yrj

Yi i Since C Ri (XBi — ' Y rj nt

.I

Yrj

L=1 XBr

=0

for i = r.

m

XBr

C Bi X Bi + [C •

XBr

E C BiYi j I =1

c Bi yij = C B Yj i [Note : Z corresponds to the original B.F.S.] • ...(10) Z'= Z v(ci —Z j ) or Thus, the new value of the objective function is equal to the original value plus the quantity v(ci — Z j ). Z' >_Z if v(ci — Zi )?.. Now Since v>0 Z' >_Z only if c j — Z j >0 Z j — c j < 0. or Thus, by choosing the vector ot j for which c j — Z • > 0 and at least one yii > 0, we = Z

(Ci

- Zi )

where

Zj =

Yrj

obtain a new basic solution with Z' Z. If the original B.F.S. was non-degenerate then v > 0 and in this case we have Hence, the theorem is proved. Z' > Z Note :1. If v = 0, then xm., i. e., the original B.F.S. is degenerate B.F.S. In this case from (6), we have x'Bi =.scBi , r and km.= 0 = xik. Thus, we see that the values of the variables common to the two solutions (old and new) do not change, and the new variable is at zero level (at r-th position). Hence, the new B.F.S. is also degenerate B.F.S. 2. If yij < 0, for every xj3; which is zero in the original B.F.S. then none of these variable enter in the calculation of —Lbl x in (9) v > 0 in this case, and the new Y rj

B.F.S. will be non-degenerate B.F.S. 3. The vector 13,. to be deleted from the basic B is called the departing (outgoing) vector while the vector ai to be introduced into the basis is called the entering (incoming) vector and the element yr./ (in matrb A) is called the key element.

Simplex Method

307

8.6 Unbounded Solutions In theorem 8.5 we assumed that if we insert ai into the basis B and remove some column to form a new basis then at least one yi j > 0, i = 1, 2,...,m. Now we shall discuss the case when all yi j 5.. 0, i = 1, 2, ..., in. For this we prove the following theorem. Theorem : If for any basic feasible solution of a L.P.P., there is some column a j in A but not in B for which c j — Z j > 0 and yy 0, (i = 1, 2, ..., m) then the problem has an unbounded solution* the objective function is to be maximized. [Meerut L.P. 1994, 96, 97] Proof: Let xB= ---B1/ --B2) ' • XBrn be a B.F.S. of the given L.P.P. Max. Z = c x

s.t. Ax = b,

0

and with the basis matrix

where A = B = 031,[1 2,—,13 n1 )

771 BXB

=b

or

~XBiR j = b. i=1

...(1)

The value of the objective function for this B.F.S. is given by Z = LB X B = Cm X Bi • i =1

Let us insert the column a • in B, where a1• is in A but not in B and c • — Z > 0 .1

and yij 0, i = 1, 2, ... , m. If R be any scalar then adding and subtracting ka j in (1), we have ni X Bi f3;

j + Xa j =b

—(3)

i a1• EA, s.t. a • B. Since 131,132,...,13m form the basis of A, therefore a j can be expressed as the L.C. of these vectors. 177

i.e.,

j

ai = i =1

or

m Aaj = —A,iyij i =1

* A problem is said to have an unbounded solution if the objective function can be increased or decreased arbitrarily i. e., if there is no finite optimum of the objective function.

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Operations Research

Substituting in (3), we have XBiRi

-X

E Riyij + Xa j = b

i =1

=1. ni

... (4)

(xBi - Xyi j )13i +ra j = b

or i =1

When A, > 0, then xBi - Xyij ?_ 0 Since yi j from (4) we obtain a new feasible solution

0, i = 1, 2, ..., m.

Bm 1] x'B = [X' 131 132 i• • j1 = 1, 2,..., m, where X' Bi = X Bi in which (m + 1) variables can be different from zero. In the feasible solution x'B the number of positive variables is less than or equal to (m + 1). This may be less than (m + 1) since xBi - Xyij may be zero for some value of i. If the number of positive variables in this solution x.'B is equal to (m + 1) then this solution is a non-basic feasible solution. Now if Z' is the val4e of the objective function for this new feasible solution then

cBi (x

Z' i =1

Or

=

C Bi X Bi

(C j - C Bi yi j)

i =1

or Z'= Z + X(ci -Z j ) Since c.1 - Z I• > 0, we can make Z' as large as we please by taking X sufficiently large. Hence the problem has an unbounded solution if the objective function is to be maximized.

0, i =1, 2, ..., m, then the L.P.P. has an unbounded solution if the objective function is to be minimized.

Note : Similarly if for someai , Zi -ci > 0 and yi

8.7 Optimality Conditions Theorem : Let x B = B-lb be the basic feasible solution to the L.P.P. Max. Z = c:x, s.t. Ax = b,x 0 and Z* = cB xB the value of the objective function for this B. F. S. If c3 - Zi Ofor every column a j in A but not in B, then Z* is the optimum (maximum) value of objective function Z and this B.R.S. x B is an optimal B.F.S.

Proof : Let xB = [xffl , xB2, where

xBin ] be the B.F.S. of the given L.P.P

Max. Z = cx, s.t. Ax = b, x > 0 A = (ai , a 2, , m+ n ) and- with the basis matrix B=

Simplex Method

309

BxB = b

or

xB = B -1 b.

The value of the objective function for this B.F.S. is given by Z* = CB• XB =

C Bi XBi i =1

Also it is given that cj - Z If

x'=

...(1)

0 for every column aj in A but not in B.

x2',...,.x' n m] is any F.S. of the given problem then

B. xB =b = Ax'.

... (2)

If Z ' is the value of the objective function for this solution n+ ni then Z'= cx1 =

. -(3)

j =1

Now we proceed to prove that Z* ?_ Z'. From (2), we have XB = B-1113 = 8-1 (A x' ) = (B71A)x' = Yx'

Y2 , , Yni )

whei'e•Y =

= 07131721• • • YM+ n

X 2f

3.• "Xf ni+n

Yll

Y12

•• ••

Ylj

• •• •

.Y1,m+n

Y21

Y22

•• ••

Y 2j

• •• •

Y 2,m+ n

Yml

Ym 2

"•• Y mj

xi

" • Y itt,m+

711 +

or [xffl , XB2,..., XBi ..., XBIn ] m+n

m+n

m+n

j =1

j =1

= I y,..,•xf I Y 2j • xj , • • '' I Yi j x

+n mi . ;

3

j =1

j

m+n XBi = 1 yij . xi j =1 Now it is given that cj - Zi

...(4)

0, for all j for which oci 0 B.

Now if aj E B say a j = [3i then a j =(3i = 0.131 + 0.132 +.... +0.+1.+0.131 + +... + O. pm which shows that Yj = ei. -(a unit vector whose i-th component is unity)

ci = CBi

Since

ecj =

i.e.,

c; -Z i = ci - CB Yi = Ci -cBei = c j - CBI = 0

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310 a j =13i then c j Z j =

Thus, if Hence,

c — Z ./ 5 0 for all a • in A Ci Z

Or

Cjx1
0 for at least one i = 1, 2, ..., m then by replacing the column o r in B by the column oti which is in A but not in B, we obtain a new B.F.S., x'B given by X13 = [X' B1 I X' 132 , • • • I X' Brn

where

x Bi

Yij

= XBi —

X Br , i = 1,

2, ... , m, i # r

Yrj

and

X

XBr — Yr j

X Br

m i n. I XBi

Yr j

i

Yi j

y

> 0 . ij

The value of objective function for this new B.F.S. is given by z i = z 4_ X.Bi• Yrj If

(c

Z. )

c i — Z.1 = 0, then Z' = Z

i.e., the value of the objective function for this new B.F.S. is also equal to Z (optimal value). Hence, this new B.F.S. is an alternative optimal B.F.S.

8.9 Inconsistency and Redundancy in Constraint Equations Redundancy in Constraint Equations By redundancy in constraint equations we mean that the system has more than enough number of constraint equations, in other words it has more constraint equations than the number of variables. This is the situation when r(A) = r(A b) = k ._< i t < in. In this case there will be (in — k) redundant equations. • Inconsistency •

As already defined, the set of constraints (linear equations) is .said to be inconsistent if r (A) ;., r (A b). Before solving a L.P. problem by simplex method, we should have r (A) = r (A b) i.e., the constraint equations (after introducing the slack and artificial variables) should he consistent. Since in simplex method we always start with the basis matrix B = I, therefore we always have r(A) = r (A b) = in

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312

If the system Ax = b involves artificial variables, then we cannot say whether this system is consistent or therein any redundancy. Below we give the cases (without proof) to decide about the consistency and redundancy in such system. Case I

-Case II :

Case III :

If the basis B contains no artificial vector and the optimality condition is -satisfied (at any iteration) then the current solution is a B.F.S. of the problem. If one or more artificial vector appears in the basis B at a zero.leveli.e., the value of the artificial variables corresponding to artificial B are zero and the optimality condition is satisfied (at any iteration) and x Br = 0 then the system is consistent. Furthermore if yri = 0, and r corresponds to the row containing artificial vector, then the r-th constraint equation is redundant. If at least one artificial vector appear in the basis B at a positive level i.e., the value of at least one artificial variable corresponding to artificial vector in B is non-zero and the optimality condition is satisfied (at any iteration), then there exists no feasible solution of the problem.

8.10 To Determine Starting B.F.S. In the constraints of a general linear programming problem there may be any Assuming that the requirement vector b 0 (if any of the of the three signs 's is negative, multiply the corresponding constraint by —1) we shall discuss-the following three cases. Case I : To find initial B.F.S. when all the original constraints have sign. In this case all the constraints are converted into equations by inserting slack variables only. Inserting slack variables the equations obtained are as follows. . an an

+ an x2 + + ai„ xn + 1. x„+1 +1• x 0 + 2 + a22 X2 + . • • • + a2n •• •

•••

•••

= =

b2

•••

+1. x„4.,„ = bm

a ird am2X2 +... at„ xn Here x„+1, xn+ 2, xn+ ,„ are slack variables. In matrix form, these equations can be written as slack vectors

x1 a21

a12 a22

11

a21.„

0

0 1

x2 0

bi b2

X,

xi, +1 am1

amt • • • • antra

0

0

_b111

X II+ Ill

Here we take the initial basis matrix B = I . (unit matrix of order in x m)

• • •

313

Simplex Method

:.

the initial basic solution is given by x 8 =B- 1 b=lmb-=b~ 0

Thus, the initial B.F.S. is Xt = 0, X2 = 0, ... ,Xn = 0 Xn+l

= Xsi = bi,Xn+2 = Xs2

= b2,···,Xn+m

= Xsm = bm

Which can be obtained by writing all the non-basic variables (i . .!., given variables) x 1, x 2 , ... , Xn equal to zero and solving the equations for the remaining

variables (i.e., slack variables)

Xn+ 1, Xn+ 2, ... , Xn+m.

Case D : To find initial B.F.S. when all the original constraints have~ sign. In this case all the constraints are converted into equations by inserting surplus variables. Inserting surplus variables the equations obtained are as follows.

Om1X1 +am2x2 + .... +amnXn

Here xn + 1 , x n + 2 , ... , x 11 + 111 are surplus variables. In matrix form, these equations can be written as surplus vectors au

a12

aln

l_l

a21

a22

a2n

0

X1

01

0 -1

X2

b1

0 Xn

=

b2

Xn+l aml

0

am2

-1

0

Here taking the initial basis matrix B = - Im , We have

x8

= B- 1 b = -1 111 b = -b::; 0.

This basic solution is not B.F.S. In this case to avoid this difficulty we add one ,nore variables to each constraint. These variables are called "artificial variables". Adding surplus and artificial variables, the constraints of the given L.P.P. are converted to the following equations. au Xi a21X1

+ a12_:,.;2 +. ·· · + a111 Xn + a22X2 + .... +aznxn

Xn+I

+ Xn+ m +l +xn + m+2

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314

Here x n+ m+ 1) Xn+ m+ 2

•• • s

x„++ „, in are the artificial variables. In matrix form,

these equations can be written as xl

Artificial vectors all all

012

a22

al n 02n

—1 0

0 —1

0 0

0 0

....

0 0

X2

X, X n+1 • • • •

aml 'am t •• • •

a mn

0

0 ....

—1

0

0

X n+ m X n+ m+1

Xn+ m+ m -

62 bm

Here we can take the basis matrix B = I „ . XB = = / 7„13--b 0 which is a B.F.S. Thus, in this case the B.F.S. is and X n+ m +1 = X B1 = b 1) X n + 171 + 9 = X B2 = b 2,— ) Xn+m+m = X Bm = bm each other variable equal to zero which can be obtained by writing all the non-basic variables (i.e., given variables and surplus variables) xi , x2, , x x xn÷ 2, •• • • X n + m equal to zero and solving the equations for the remaining basic variables (i.e., artificial variables) x m +1 , x „+ ni+ 2, ..., X,+„, m and '=' Case III : To find initial B.F.S. when the constraints have 1 5.' , signs In this case the constraints are converted into equations by inserting slack, surplus and artificial variables. Here the basis matrix.13 = Im is obtained by taking the slack and artificial vectors. In this case also the initial B.F.S. is obtained by writing all the non-basic variables. (i.e., given variables and surplus variables) equal to zero and solving for remaining basic variables. Here also the initial B.F.S. x B = b ?_ 0. Note : 1. It is important to note that in all the three cases the initial B.F.S. consist of the R.H.S. constants b1, b2,..., b„, (note that b1, b2, , bm , are all positive). 2. Some times it happens that an identity is present in A without introducing the artificial variables. So we do not introduce the artificial variables in such case and get a B.F.S. with its basis matrix B = I,. (See mac. 9).

315

Simplex Method

8.1.1 Computational Procedure of the Simplex Method for the Solution of a Maximization L.P.P. Meerut 1995]

It consists of the following steps systematically. Step 1: If the problem is of minimization, convert it into the maximization problem.

The values of xi,

, x„ that minimize the function

Z = Clxl+02x2 +....+cn x„, maximize the function

Z'= —Z =

— c2x2 —....— c„xn .

Thus, in the problem. of minimization, we maximize the function Z' = —Z. If Z' = v is the maximum value of Z then — v will be the minimum value of Z. Step 2 : Make all the bi 'spositive : If any of the bi s is negative, multiply the

corresponding constraint by —1. Step 3 : (a) Convert the constraints into equations by introducing the non-negative slack or surplus variables. Also introduce artificial variables in t'ie constraints where surplus variables

are inserted and which do not form the columns of unit matrix. Step 4 : To find initial B.F.S.: Find the initial B.F.S. by the method of article 8.10. If in a L.P.P. artificial variables are also introduced in step 3 then we follow two-phase method. In phase I we proceed to get the starting B.F.S. in terms of non-artificial variables (see article 8.12). Big M-Method can also be used in such cases. Step 5 : Construction of starting simplex table : Now we construct the simplex table as shown in table 8.1.

Here the columns corresponding to the coefficients of x1, x2

are shown by

Yi$ Y2. • • Step 6 : Test of starting B.F.S. for optimality : This is done by computing an evaluation Ai for each variable xi by the formula A j = ci — C B . The value of A )• = 0, if x • is a basic variable. Insert the values of A • ' s for all variables as a row in the starting simplex table as shown in table 8.1. Here c • is the row of the coefficients of the variables in the objective function. (I) If

0 for each j (i. e. , no Ai > 0) the solution under test is optimal.

(a) If none of A • is positive, but any are zero (for non-basic variable), then other optimal solution exist with the same value of Z. (b) If all of Ai are negative (for all non-basic variables), the solution under test is unique optimal solution.

XB,•=br

xBm=brn

CBr

CBm

Yr/ -Fr

Yn I m

A i

Yr2=ar2

Yr1 =ar7

....

••••

--

••••

....

Al

iS.2 —

Ym1=aml Ym2=am2 ••••

Y22 =a22

yo—cti.)

yii=aii

Y21 = a21

2( =O'2)

Y1(=(11)

c2

Ck

1 1 /4

AR •"•

Yrnk'amP ••• • An

Yrn17 = an? n

YOZ =a171

Y2n=a2/7

Y2k=a2k --

....

Yln=aln

Yrk = ark

cn-I-1.

An i

0

0

1

Yrt-=((x.n.) Yn +1 (i3 l)

n

••.•

Yik=aik

Yk -=((xk) ••• •

1

An. r 2

0

1

0

Yn-1-2(f 32)

Cn I 2

••••

.. ' .

••• •

•"•

••••

••••

On +m.

1

0

0

Yn+m(13m)

cnd m

--->

Ratio XB/Yk

Mini

If corresponding to maximum positive Ai, all the elements in the column Yj are negative or zero, then the solution under test will be unbounded. (d) If the value of at least one artificial variable appearing in the basis is non-zero and the optimality condition is satisfied, then we shall say that the problem has no feasible solution. Step 7 : To Find incoming (or entering) and outgoing vectors. To improve the above solution (which is not optimal) we find the vector entering the basis matrix (called in-coming vector) and thevector to be removed from the basis matrix (called outgoing vector) by the following rules.

(c)

(ii) If Ai > 0 for any ji.e., if one or more of of are positive the solution under test is not optimal. If the solution under test is not optimal, we must proceed to the next step 7.

=0

Z=CBXB

c82 xj-32=b2

lin I .)

x131. = b1

CB1

In i 1

KB

Cr

B

CiC1

Table 8.1 Starting Simplex Table

timeasaN suolluiado

Simplex Method

317

To find in-coming vector : The incoming vector will be taken as a k if A k = Max. Ai .

If Max. value of A • occurs at more than one • ' then any one of these may be taken as an incoming vector. To find out-going vector : The out-going vector pg. is taken corresponding to that value of r for which X Br

mini. X Bi

y

> 0 ik

Y rk

i

Y ik

when a k is the incoming vector. If minimum is not unique i.e., the minimum occurs for more than one value of i, then more than one variable will vanish in the next solution. Therefore the next solution will become a degenerate B.F.S. for which out-going vector is selected in a different way discussed in article 8.13. Step 8 : When ak (= Yk ) is the incoming vector and Yr (= Q r ) the outgoing vector

then the element y rk (-= ark ) is called the key element or pivot element which is at the intersection of minimum ratio arrow (-4) and incoming vector arrow (T). We mark this element in 0 as shown in table 8.1 In order to bring a k (= Yk ) in place of Y,. (= (3,.) there should be unity at the position 1=1 i.e., the key element y rk (= ark ) should be equal to unity (= 1). If it is not 1 then divide all the elements of this row by this key element ark . Then subtract appropriate multipliers of this row (containing key element) from all the other rows and obtain zero (= 0) at all other positions of this column a k (= yk ). Now bring Yk (ak ) in place of Yr (a r i.e. r ) and construct new (revised) simplex table. In this way we get improved basic feasible solution. Step 9 : Now test the above improved B.F.S. for optimality as in step 6.

If this solution is not optimal then repeat steps (7) and (8) in succession, until an optimal solution is finally obtained. For the clear understanding of the procedure, few illustrative examples are given here.

gituathatio . .t? jxamplea Example 3 : Solve the L.P. Problem.

Maximize Z = 3x1 + 5x2 + 4x3 subject to 2x1 + 3x2 S. 8 2x2 + 5x3 5 10 3x1 + 2x2 + 4x3 5 15 and x1, x2, x3 0 [Meerut (LP) 1995, 2007 (BP), 08, (BP)] Solution : We shall solve this example step-wise, so that students may understand the procedure of the simplex method. Step 1 : The problem is a problem of the maximization. Step 2 : All the bi 's are already positive.

Operations Research

318

Step 3 : Now the inequalities are converted to equations by the introduction of slack variables x4, x5 and x6, as follows = 8 2x1 + 3x2 + Ox3 + x4 +X 5 = 10 0. + 2x2 + 5x3 + X6 = 15 3X1 + 2X2 + 4x3 Step 4 : Taking xi = 0,x2 = 0, x3 = Owe get x4 = 8,x5 = 10 and x6 =15 which is the starting B.F.S. Step 5 : Now we construct the starting simplex table 8.2 Step 6 : Here we compute A i for all zero variables (non-basic) x j , j=1, 2, 3 by the formula Ai = ci — C B Yi Al = — C H K = 3 — (0, 0, 0) (2, 0, 3) = 3

A 2 =c2 - C B Y2 = 5 — (0, 0, 0)(3, 2, 2) = 5 A3 =c3

C B Y3 = 4 7 (0, 0, 0) (0, 5, 4) = 4

A 4 = 0 = A 5 = A6, which corresponds to basic variables x4, x5, x6. Obviously Since all A are not less than or equal to zero, therefore the solution is not optimal. Note that the A i 's in the starting simplex table are noting but c -'s. Thus, there is na need to compute them here. So we proceed to, the next step. Step 7 : To find incoming vector. Since A 2 = 5 is max. of Al, A 2, A 3 :. a 2 (= Y2 ) is incoming vector. Table 8.2 : Starting Simplex Table

e)

3

5

4

0

0

0

Y1

Y2

Y-3

Y4

Y5

Y6

MM. Ratio XB

B

CB

XB

(ai )

(a2 )

(a3 )

(I31 )

(0 2 )

(133 )

Y2

Y4

0 0 0

8 10 15

2 0 3

fl 2 2

0 5 4

1 0 0

0 1 0

0 0 1

8/3 (Min.)--) 10/2 = 5 15/2

Ai

3

5

4

0

0

0

Y5

Y6 Z = CBXB =0

Incoming Vector Outgoing Vector

To find outgoing vector : Since 1'2 (a 2 ) is incoming vector therefore we consider

the ratio XB Y2 Since

XB,

( 8 10 151 A ( X B1 X112 ,-LX 33,Yi 2 > v = k• -i' 2 ' 2 Y12 Y22 Y32

XBi 3 )= 8 XB1 • = Mini. yi 2 > = .• • ( XB1 X62 XB — Yi9 3_ Y12 Y12 Y22 Y 32 3'r2 r = 1 i. e., pl (= y4 ) is the outgoing vector.

Simplex Method

319

Step 8 : Since Y2 (a2 ) is incoming vector and 131(Y4 ) is outgoing vector.

the key element is y12 (= a12) as shown in the table 8-.2 which is equal to 3 (not equal to 1) In order to bring Y2 in place of Y4 in the basis we make the following operations. Divide the first row containing key element a12 = 3 by 3, to get unity at this position and then subtract 2 tints of the first row (obtained after dividing by 3) from the second and the third rows and subtract 5 times of this row from the row of 's to get the new value of dl 's*. Thus, we construct the second simplex table in which 131 (= Y4 ) is replaced by a 2 (= Y2 ) as follows. Table 8.3 : Second Simplex Table

B

CB

Y2 Y5 Y6

5 0 0

Z = CBXB

= 40 / 3

ci

3

5

4

0

XB

Y1 (a1 )

Y2 (P)

Y3 (a3)

Y4 (a4)

Y5 (13 2)

Y6 (P 3 )

1 0 0

0 15 I 4

1/3 —2/3 —2/3

0 1 0

0 0 1

0

4

—5/3

0

0

8/3 2/3 14/3 — 4/3 29/3 5/3 Aj

—1/3

0,

0

Min. Ratio X B ! Y3

— 14/15 (Min. ) 29/12

i.

Incoming Vector Outgoing Vector

To check the values of di , we also compute Ai 's by using the formula = ci —CB Yi , for Yi , Y3 and Y4. 2 4 5 0 1 dl= — CB = 3 — (5, 0, 0) . ( — , — — , —) = 3 — -- — 3 3 3 3 3 A 3 =c3 — C B Y3 = 4 — (5, 0, 0) . (0, 5, 4) = 4 5 A 4 = C4 — C B Y4 0— (5, 0, 0) 1— 22 = ,3 3 3) • 3 The values of A 2, A s, A 6 will be zero as they correspond to unit column vectors. Since in table 8.3 all Ai are not less than or equal to zero, therefore this solution is also not optimal. Since A3 = 4 is max of the d i ' s. a3 (= Y3 ) is the incoming vector. The values of dl' s for the new simplex table are also calculated by row operations. Thus a lot of time is saved, Even then the students are advised to check them by computing from the formula dJ = cj CBYi.

Operations Research

320 IC Br

Also

xn •

= Mini. ---LL, yi3 > 0

Y r3

1 LYi 3 [XB2 , XB3 ,

xB1 not considered Y23 Y33 Y13 14= x/32 = Mini.[14 — 29]= 15' 12 15 y23

= Mini.

••• Y13 = °

r=2

i.e., 13 2 (= Y5) is the outgoing vector and y23 = a23 = 5 is the key element. In order to bring Y3 in place of 0 2 (= Y5), we divide the 2nd row containing the key element by 5 to get 1 at this position, then subtract 4 times of the second row thus obtained from third.row and also from the row of Ai's. . The third simplex table in which f32 ( = Y5 ) is replaced by Y3 is as follows. Table 8.4 [ B

CB

Y2 Ir3

Y6

c•

. 3

XB

Yi (a1)

S 8/3 2/3 4/15 4 14/15 0 89/15 41/15

Z = C B XB

A

11/15

5

4

0

0

0

Min. Ratio

Y2 (Pi)

173 (P2)

1/4 (a4)

Y5

Y6

X B /yi

(a's )

(P3)

1 0 0

0 1 0

1/3 0 -2/15 1/5 -2/15 - 4/5

0 0 1

0

0

-17/15 -4/5

0

4 89/41 (Min.)

= 256/15 Incoming Vector

4, Outgoing Vector

To check the values of Ai , we again compute o f by using the formula Ai = ci - CB Al =c1 -C B = 3 - (5, 4, 0)(2/3, - 4/15,41/15) = 3- (100 - 16/15) = 11/15 A4 =

c4 - CB Y4 = 0 - (5, 4, 0), (1/3, - 2/15, - 2/15) = -17/15

A 5 = c5 - C B Y5 = 0 — (5, 4, 0). (0, 1/ 5, - 4/5) = - 4/5 Also

02 = 0= A3 = A6.

Since all Ai are not less than or equal to zero therefore the solution is not optimal. Since Al = 11/15 is max. of all-the Ai's a1 (= Y1) is the incoming vector. [xBi xBr = Mini. Also —,yii > 0 Yo. i Yii

Simplex Method

321 [XD-I XDQI X Do

, not considered Yll Y31 Y21

=

y 21 is negative

= Mini. [ 4 " = 89 = XB3 41 41 y3i i.e., 13 3 (=

r=3

Y6 ) is the outgoing vector and y 31 = a31 = 41/15 is the key element.

Again in order to bring Yi in place of 13 3 (= Y6 ) we divide the 3rd row containing the key element by 41/15 to get 1 at this position, then subtract 2/3 times of the third row (thus obtained) from first row, add 4/15 times of third row to the second row and subtract 11/15 times of third row from the row of Ai 's. Thus, the fourth simplex table in which 133 (= Y6 ) is replaced by Y1 is as follows. Table 8.5 1

3

XB

Yi

c

CB

B

5 4

Y3

Y1 Z

3

50 41 62 41 89 41

= C B XB =

765

A1

0

0 Min. Ratio

(P3) Y2

0

4

5 Y2

Y3

Y4

(131) ( 3 2)

0

1

0

0

0

1

1

0

0

0

0

0

(Cf'4)

YS

(a5) (a o )

15 41 6 — 41 2 — 41 —

45 41

Y6

—

8 41 5 41 12 41

10 — 41 4 41 15 41

24 41

11 .711-

41

To check the value of A • 's we also compute A • by using the formula A = c • — C B • Y. A 4 = c4 — CB Y4 = 0 — (5, Similarly

A 0)

6 — -41 ' 41 4141)

A 5 = —24/41, A 6 = —11/41, Al = 0= A 2 = A 3

Since all the Ai 's for zero variables (non-basic variables) are negative so this solution is optimal. Optimal solution is x1 = 89/41, x2 = 50/41, x3 = 62/41 and

Max. Z = Rs. 765/41

Operations Research

322

All tables 8.2, 8.3, 8.4 and 8.5 can be computed in a single table as in table 8.6 Table 8.6 c•

3

5

4

0

0

0

Min. Ratio X B /Y2

B

CB

XB

Yi

Y2

Y3

14

Y5

Y6

Y4 Y5

0 0 0

8

2

3

0

1

0

0

8/4 (Min.)

10

0

2

5

0

1

0

10/2 =5

15

3

2

4

0

0

1

15/2

A•

3

5

4

0

0

0

XB /Y3

0

14/15 (Min.) —> 29/12

Y6

Z =CB XB

Y2 Y5 16

1

T

=0 5

8/3

2/3

1

0 0

14/3

- 4/3

29/3

5/3

Z =CB XB

0

1/3

0

0

5

-2/3

1

0

0

4

-2/3

0

1

-1/3

0

4 I

-5/3

0 si-

0

X B / Yi

8/3

2/3

1

0

1/3

0

4

A•

= 40/3 5

0

Y2 Y3

4

14/15

- 4/15

0

1

- 2/15

1/5

0

Y6

0

89/15 141/1.5j

0

0

-2/15

-4/5

1

0

0

-17/15

-4/5

0

0

1

0

15/41

89/41

0 1

0 0

1 0

- 6/41 - 2/41

A•

0

0

0

Z = C B XB

Aj

I

= 256/15 Y2 Y3 Y1

5 4 3

Z = CB X B

11/15

50/41 62/41

89/41 (Min.) )

T 8/41 -10/41 4/41 5/41 -12/41 15/41

-45/41 -24/41 -11/41

= 756/ 41 Example 4 : Solve by simplex method the following L.P. problem.

Minimize Z = x1 - 3x2 + 2x3 subject to 3x1 - x2 + 2x3 7 - 2x1 + 4x2 5 12 - 4x1 + 3x2 + 8x3 S lb xi , x2, x3 0. Solution: First we convert the problem of minimization to maximization problem

by taking the objective function as Z' = -Z Max. Z'= -Z = -x1 + 3x2 - 2x3 . Now the equation obtained by introducing slack variables x 4, x5, x6 are as follows :

323

Simplex Method

=7 3X1 — x2 + 2x3 x4 - 2x1 + 4x2 + 0. x3 +x 5 = 12 +x6 = 10 —4x1 + 3x2 + 8x3 Taking xi. = 0, x2 = 0,x3 = 0, we get x/i = 7,x5 = 12, x6 = 10, which is the starting B.F.S. Table 8.7 Starting Simplex Table ci 0 Min. Ratio —1 3 —2 0 0 B

CB

XB

Y6

0 0 0

7 12 10

Yl (a1) 3 —2 —4

Z'

— CB XB

Aj

—1

Y4

Y3

Y2 (a2) —1 {±] 3

Y3 ("3) 2 0 8

Y4 ((11) 1 0 0

Y5 (02) 0 1 0

116

( 33) 0 0 1

3

—2

0

0

0

4 / Y2 3 (Min.) —> 10/3

=0 Incoming Vector Outgoing Vector - C BY]. = -1 - (0, 0, 0). (3, — — 4) = — 1

Dl=

A 2 = c2

C B Y2 = 3 - (0, 0, 0). (-1, 11,

=

3

A 3 =C3 - C B Y3 = — 2 - (0, 0, 0). (2, 0, 8) = — 2 Since all A. are not less than or equal to zero, therefore the solution is not optimal. Here incoming vector is a 2 (= Y2 ) as max Aj = 3 = 02 , and by mini. ratio rule, outgoing vector is 13 2 (Y5 ) key element = v 22 = a 2 9 = 4. In order to bring a 2 (= Y2 ) in place off32 (= Y5 ), we divide the elements of 2nd row containing key element a22 = 4 by 4 and then add it to 1st row and subtract its 3 times from 3rd row and also from the row of Ai 's, the second simplex table is as follows : Table 8.8 : Second Simplex Table CB

B

0 3 Y2 0 Y6 Z' =CB X B

Y4

c• I

-1

3

-2

0

0

0

XB

Yl

Y2

Y3

Y4

Y5

(c'''1)

(I3 2)

("3)

(PI)

("5)

Y6 (132)

10 3 1

5/2 —1/2 —5/2

0 1 0

2 0 8

1 0 0

1/4 1/4 — 3/4

0 0 1

Aj

1/2

0

—2

0

—3/4

0

Min. Ratio X B / Y1

4 (Min.) —› — —

=9 Outgoing Vector Incoming Vector are not less than or equal to zero, therefore the solution is not optimal. Since all Ai Here Al = 1/2 is Max.

Operations Research

324

Yi is the incoming vector and by the minimum ratio rule we find (31(= Y4 ) as the outgoing vector. key element = Yii = 5/ 2 In order to bring Y1 in place of p1 (= Y4 ), we divide the 1st row containing the key element Yii = 5/Z by 5/2 and add 1/2, 5/2 and -1/2 times of the 1st row thus obtained in the 2nd, 3rd and the row of Ai's respectively. Thus, we get the third simplex table as follows : Table 8.9 : Third Simplex Table B

CB

cj XB

-1

3

-2

0

0

0

yi

Y2

113

Y4

Y5

Y6

(hi) ( 3 2) Y1 -1 4 3 5 Y2 0 11 Y6 Z'= C B XB =11 Ai

(a3)

(a4) (a5) (a6)

1 0 0

0 1 0

4/5 2/5 10

2/5 1/10 1/5 3/10 1 —1/2

0 0 1

0

0

—12/5

—1/5 —4/5

0

Min. Ratio

All tables 8.7, 8.8, 8.9 can be computed in a single table as in table 8.10. Table 8.10 cj

-1

3

-2

0

0

0

CB

XB

yi.

Y2

Y3

Y4

Y5

1'6

0 0 0 Y6 Z'= C B XB

7 12 10

3 -2 -4

-1 4 3

2 0 8

1 0 0

0 1 0

0 0 1

12/4= 3 (Min.) 10/3

A.1

-1

3

-2

0

0

0

XB /Yl

10 3 1

5/2 -1/2 -5/2

0 1 0

2 0 8

1 0 0

1/4 1/4 -3/4

0 0 1

Z' = CBXB =9

A.

1/2

0

—2

0

—3/4

0

Yi

4 5 11

1 0 0

0 1 0

4/5 2/5 10

2/5 1/5 1

3/10 -1/2

0 0' 1

A•

0

0

-12/5

-1/5

-4/5

0

B Y4 Y5

T

=0 Y4 Y2 Y6

0 3 0

—1 3 0

Y2 Y6 Z'= C B X B =11

1

T

1

1/10

MM. Ratio XB / Y2

4 (Min.) -

325

Simplex Method

Since all A • for all non-zero variables are negative, so this solution is optimal. Optimal solution is x1 = 4, x2 = 5, x3 = 0 and Min. Z = — Max. Z '= —11. Example 5 : Using simplex Algorithm solve the problem Max. Z = 2x1 + 5x2 + 7 x3 subject to 3x1 + 2x2 + 4x3 100 x1 + 4x2 + 2x3 100 xi + x2 + 3x3 100 xi, x2, x3 0. Solution : The equations obtained by introducing slack variables x4, x5, x6 are as follows : = 100 3x1 + 2x2 + 4x3 + x4 = 100 +x5 xi + 4x2 + 2x3 +x6 = 100 xi + x2 + 3x3 = 100, x5 = 100, x6 = 100, which is the = 0 we get x4 = 0, x2 = 0, x3 Taking x1 starting B.F.S. All computation work is done in the following table 8.11. Table 8.11 ci

2

5

7

0

0

0

Min. Ratio XB /Y3

B

CB

XB

Yi

Y2

Y3

Y4

Y5

Y6

Y4

0 0 0

100 100 100

3 1 1

2 4 1

F]

2 3

1 0 0

0 1 0

0 0 1

25 (Min.) —> 50 100/3

Z = C B XB

A. )

2

5

7 T

0 1

0

0

X3 /Y2

0 1 0

0 0 1

50 50/3 (Min. ) —

0

0

- Y5

Y6

=0 Y3 Y5 Y6

7 0 0

Z = CBXB =175 Y3 Y2 Y6

7 5 0

Z = C B XB

25 50 25 Al

50/3 50/3 100/3 Al

3/4 —1/2 —5/4

1/2 3 —1/2

1 0 0

1/4 —1/2 —3/4

—13/4

3/2 T

0

—7/4

1

5/6 —1/6 —4/3

0 1 0

1 0 0

1/3 —1/6 —5/6

—1/6 1/3 1/6

0 0 1

--3

0

0

—3/2

—1/2

0

= 200 Since all A • 's are zero or negative, so this solution is optimal Optimal solution is xi. = 0, x2 = 50/3, x3 = 50/3 and Max. Z = 200.

326

Operations Research

8.12 Artificial Variables Technique In L.P.P. some constraints may have the signs ?_ or = with all bi 's positive. In such problems we introduce surplus variables in the constraints with sign In these problems we cannot get the starting basic matrix B = I m . So to avoid this difficulty we add one more variable to each of such constraints. These variables are called 'artificial variables'. As the name implies these variables are fictitious and represent no physical entities. The artificial variables technique is merely a device to get the starting B.F.S. so that we may proceed with simplex method to get the optimal solution. Such problem are solved by two methods. (i) Two Phase Method (ii) Big M-Method or M-technique or "Method of Penalties" due to A. Charnes. Method 1 : Two Phase Method Here the solution procedure is separated into two phases. Phase I : In phase I we remove (i.e., eliminate) the artificial variables from the basic and introduce other variables (non-artificial variables). Phase II : In phase II, we use, the solution of phase I, as the initial basic solution and apply simplex method to determine the optimum solution. Computational Procedure of the Two Phase Method Phase Step 1 : After making all bi 's positive convert each of the constraints into equations by introducing slack, surplus and artificial variables, as required. Step 2 : Obtain the new objective function (say Z') by assigning costs —1 to each artificial variable and costs 0 to all other primary, slack and surplus variables. Thus, new objective function. Z' = (sum of the artificial variables). Step 3 : Using simplex method maximize the new objective function, Z', subject to the constraints of the original problem. If the original problem has a feasible solution then this new problem shall also have an optimal solution with optimal value of the objection function (new objective function Z') equal to _zero as each of the artificial variables will be equal to zero. 'niece are three cases that arise. (i) If max. Z' < 0 and at least one artificial variable appears in the optimal basis at a positive level, then the given problem does not posses:: any feasible solution, and so the procedure is terminated. (ii) If max. Z ' = 0 and at least one artificial variable appears in the optimal basis at zero level, then proceed to phase II. (iii) If max Z ' = 0 and no artificial variable appears in the optimal basis then also proceed to phase II. It is to be noted that the new objective function Z' is always of maximization type regardedless of whether the given (original) problem is of maximization or minimization type. Phase II : In phase II, start with the optimal basic feasible solution contained in the final simplex table of phase I and using simplex method maximize the original objective function Z (If original objective function is of minimization then change it to maximization) with the actual costs of the primal variables and zero costs corresponding to slack and surplus variables. For clear understanding of the method see following examples.

Simplex Method

327

glhaPtatilb2 EXIIMP1124 Example 6 : Solve the following L.P.P. by using the two phase method. Min. Z = + x 2 subject to 2x1 + x 2 4 xl+ 7x2 7 xi, x2 ?. 0 Solution : The given problem is of minimization. Converting it to maximization by taking the objective function as Z' ,—Z = — x2. Introducing the surplus variables x3, x4 to convert the constraint inequalities to equations and taking the artificial variables xai and xa2 the constraint inequalities reduces to the following equations : 2x1 + x2 — x3 =4 xl+ 7X2 — x4 + Xa2 = 7 xi ,x2,x3,x4,x,12 x,2 O. Phase I : Assigning costs —1 to artificial variables and cost 0 to all other variables, the new objective function of the auxiliary problem becomes Max. Z" = 0.x1 +0.x2 0. x2 + 0. x3 + 0. x4 — 1. xai — 1. xa, subject to the constraint equations given above. Taking xi = 0,x2 = 0,x3 = 0,x4 = 0, we get xa = 4, xa2 = 7 which is the initial B.F.S. Now applying the simplex method in the usual manner, we have the following table: Table 8.12 ci

0

0

0

0

—1

—1

Min. Ratio

B

CB

XB

Y1

Y2

Y3

Y4

A1

A2

X B /1'2

Al A2

-1 —1

4 7

2 1

1 Pi]

-1 0

0 —1

1 0

0 1

4/1 7/7 = 1 (Min.)

Z"= CB .XB

A

3

8

—1

—1

0

0

XB /Yi

1

-11 Al

T

—1 0

3 1

13/7 1/7

0 1

—1 0

1/7 —1/7

Z"= C B .XB

Ai

13/ 7

0

—1

1/7

Y2

I

=-3 Y1 Y2

0. 0

Z"= C B .XB =0

21/13 10/13

1 0

0 1

A3

0

0

—7/13 1/13 1/13 —2/13 0

0

0 0

1

21/13 (Min.) --> 7

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328

In the last table all A • < 0, and no artificial variable appears in the basis, therefore this solution is an optimal solution to the auxiliary problem. Phase II : Now the first simplex table of phase II is written as follows. Assigning the actual costs to the original variables, and cost zero to the surplus variables, the objective function becomes Max. Z'= — x2 + 0. x3 + 0. x4 Now replacing the ci row values by the costs in the above objective function and deleting the artificial values column from the last simplex table Phase I, we write the first simplex table Phase II. The solution of the problem, applying simplex method as usual is given in the following table : Table 8.13 ci

-- 1

—1

0

0

B

C El

XB

1'1

Y2

Y3

Y4

Y1 Y2

—1

21/13

1

0

—7/13

1/13

—1

10/13

0

1

1/13

—2/13

Ai

0

0

— 6/13

—1/13

Z = C B .XB

Min. Ratio

.— 31/13 Since Ai 5 0 for all j, therefore this solution is optimal. Hence the optimal solution of the given problem is x1 = 21/13, x2 = 10/13, and mini. Z = — Max. Z' = 31/13. Example 7 : Solve the following L.P.P. by using the two phase method. Max. Z= + 2x2 + X3 + 4x4 subject to 4x1 + 5x2 + x3 + 5x4 = 5 2x1 — 3x2 — 4x3 + 5x4 = 7 + 4x2 + 5x3 — 4x4 = 6 xi, x2, x3,x4 0 Solution : Introducing the artificial variables x a xa x a the constraint l 2 3 equations reduce to 4x1 + 5x2 + x3 + 5x4 + xai =5 2x1 3x2 — 4x3 + 5x4 +.xa 2 =7 + 4X2 5X3 - 4x4 +xa3 = 6 xi, x2, x3, x4, x , x , x 0 ai a2 a3 Phase I : Assigning costs —1 to artificial variables and cost 0 to all other variables, the new objective function becomes Max. Z' 0.x1 +0.x2 0. x2 + 0. x3 + 0. x4 — 1 • — 1. x — 1. x a2 a3 subject to the constraint equations given above.

329

Simplex Method

Taking xi = 0,x2 = 0,x3 = 0,x4 = 0, we get xai = 5, xa2 = 7, xa3 = 6, which is the starting B.F.S. Now applying the simplex method to this auxiliary problem, in the usual manner, we have the following table : Table 8.14 cj

0

0

0

0

-1

-1

-1

XB

Y1

Y2

Y3

Y4

Al

A2

A3

Min. Ratio XB /Y1

-1 -1 -1

5 7 6

F41 2 1

5 -3 4

1 5 5 -4 5 -4

1 0 0

0 1 0

0 0 1

5/4 (Min.)-÷ 7/2 6/1

Z' ---- CB•XB = - 18

A. 1

T7

0

0

0

5/4 9/2 19/4

1 0 0

5/4 1/4 -11/2 -9/2 11/4 19/4

5/4 5/2 -21/4

0 1 0

0 0 1

Ai

0

-11/4

1/4 i

-11/4

0

0 .l,

0 -1 0

1 9 1

1 0 0

21/19 -55/19 11/19

0 0 1

29/19 -47/19 -21/19

0 1 0

Z' = C B.X B

Ai

0

-55/19

0

- 47/19

0

B

CB

Al A2 A3

Yi A2 Y3

0 -1 -1

Z' = CB . XB

6

_-37/4 Y1 A2 Y3

2

6

XB /Y3 5 1 (Min)-›

=-9 Since all A i 5_ 0, therefore an optimal B.F.S. to the auxiliary problem has been attained. But the artificial variable vector (corresponding to xa, ) appears in the optimal basic (basic solution) at a positive level. Hence, the auxiliary as well as the original L.P.P. does not possess any feasible solution. Method II. Big M-Method (Carrier's M-Method) We have already discussed in article 8.9, that if the basis B (basis matrix) contains no artificial vector and the optimality condition is satisfied (at any interaction) then the current solution is a B.F.S. of the L.P. problem. On the other hand if the problem does not have a solution then one or more artificial variables will appear in the basis B at a positive level i.e., at least one artificial variable will appear in the final solution with positive value. Keeping this in mind a method known as Big M-Method or Carner's M-Method for the solution of L.P.P. involving artificial variables is given here. In this method a very large negative price say - M is assigned to each artificial variable in the objective function. Due to these negative prices, the objective function cannot be improved in the presence of the artificial variables. So we first remove these artificial vectors from the initial basis matrix. Once an artificial vector leaves the basis we forget about it forever and never consider it as a vector to enter into the basis at any iteration. For clear understanding of the procedure see the following examples.

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gittatkativRexampleA Example 8 : Solve the following L.P. problem. Max. Z = + 2x 2 + 3x 3 — x4 subject to xi + 2x2 + 3x3 = 15 2x1 + x2 + 5x3 = 20 xl+ 2x2 + x3 + x4 = 10 x1,x2,x3,x4 0 [Meerut 1996, 2008,09] Solution : Examining the constraints carefully we note that in order to get an identity matrix of order 3 x 3 we need two more unit vectors (columns of the unit matrix), as one unit vector (column of unit matrix, formed by coefficients of x4) is already present in the constraints. Thus, we need only two artificial variables in the first two constraints. Introducing the artificial variables xal and xa in the first two constraints, and assigning large negative price —M to the artificial variables, the problem reduces to the following form. Max. Z = xl + 2x2 + 3x3 — x4 — M. xai — M. x02 s.t + 2x2 + 3x3 + O. x4 + Xe =15 2x1 + x2 + 5x3 + 0. X4 +x =20 a2 xl+ 2x2 + x3 + x4 =10 xi 0, V i = 1, 2, ..., 4, xai , xa2 0 Taking xi = 0, x2 = 0, x3 = 0, we get x4 = 10, xai = 15, xa2 = 20, which is the starting B.F.S. All computation work is done in the following table 8.15 Table 8.15 cj

1

2

3

-1 -M -M Min. Ratio

B

CB

XB

Yi

Y2

Y3

Y4

Al

A2

XB /Y3

Al

-M

15

1

2

3

0

1

0

5

A2

-M

20

2

1

5

0

0

1

—1 Y4 Z = C B .XB

10

1

2

1

1

0

0

4 (Min) --> 10

3M+2 3M+4 8M+4 0

0

0

XB /Y2

A•I

T

=- 35M -10 Al

-M 3 —1

Y3 Y4 Z'= C B. XB

= — 3M + 6

3 4 6 6,

-1/5 2/5 3/5

7/5 1/5 9/5

0 1 0

0 0 1

1 0 0

15/7 (Min.) --) 4 10/3

(-M + 2)

7M + 16

0

0

0

XB /Yi

5

5 T

331

Simplex Method B

CB

XB

Yi.

Y2

Y3

Y4

MM. Ratio

Y2 Y3

2 3

15/7 25/7

—1/7 3/7

0 1

0 0

— 25/3

Y4

—1

15/7

6/7

1 0 0

0

1

5/2 (Min.)—>

Al

6/7 I

0

0

0

Z = C B .XB = 90/7 Y2

2

5/2

0

1

0

1/6

Y3

3

5/2

0

0

1

—1/2

Yi

1

5/2

1

0

0

7/6

A1

0

0

0

—1

Z = CB . X B =15

Here no Ai > 0, so this solution is optimal. Optimal solution is, x1 = 5/2, x2 = 5/2, x3 = 5/2 and Max. Z = 15. Example 9 : Using simplex algorithm solve the L.P. problem. Min. Z = 4x1 + 8x2 + 3x3 subject to x1 + x 2 2 2x1 + x3 5 xi , x2, x3 0 Solution : First we convert the problem of minimization to maximization problem by taking the objective function as Z'. Max. Z' = —Z = —4x1 8x2 — 3x3 Introducing the surplus variables x4, x5 the equations obtained are x1 +

x, + O. x3 — x4

2x1 + 0.x2 + x3

=2 —x5 = 5

Examining the equations carefully we note that the columns of the coefficients of x2 and x3 form a unit matrix. Therefore, there is no need to introduce the artificial variables. Taking x1 = 0, x4 = 0, x5 = 0, we have x2 = 2, x3 = 5 which is the starting B.F.S. All computation work is done in the following table 8.16

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332 Table 8.16

Min. Ratio XB /Yl

ci

-4

-8

-3

0

0

CB

XB

Y1

Y2

Y3

Y4

Ys

-8 -3

2 5

1 2

1 0

0 1

-1 0

0 -1

2 (Min.) -4 5/2

Z'= CBXB

A.

10 i

0

0

-8

-3

X B /Y4

2 1

1 0

1 -2

0 1

-1 2

0 -1

1/2 (Min.)

Ai

0

-10

0 1

T

2

-3

5/2 1/2

1 0

0 -1

1/2 1/2

0 1

-1/2 -1/2

Ai

0

-8

-1

0

-2

B Y2 Y3

1

=-31 Y1 Y3

-4 -3

Z'= C B XB

=- 11 Y1 -4 0 Y4 Z' = C B XB =-10

Here no Ai > 0, so this solution is optimal. Optimal solution is x1 = 5/2, x2 = 0, x3 = 0, and Min. Z = - Max. Z' = 10. Example 10 : Shri Ram Chemical Company produces two components A and B. The following table gives the units of ingredients C and D per kg of compounds A and B as well as minimum requirements of C and D and cost per kg of A and B. Using the simplex method, find the quantities of A and B which would give a supply of C and D at minimum cost Minimum Requirements

Cot-wounds Ingredient

A

B

C D

1 3

2 1

Cost Per kg

4

6

80 75

[UP TECH MBA 2003-04] Solution : Formulation of the Problem as L.P.P. Let the company produce x1 and x2 kg of two compounds A and B respectively for minimum cost of supply of ingredients C and D. Requirement of ingredient C, 1x1 + 2x2 80 and requirement of ingredient D, 3x1 +1. x2 75 and cost of ingredients C and D, Z = 4x1 + 6x2

333

Simplex Method

Thus the L.P.P. is as follows : Minimize, Z = 4x1 + 6x2 subject to, xl + 2x2 80 3x1 + x2 75 and x2 ?_ 0 Solution of the Problem : The given problem is of minimization, so we change it to maximization problem, Also introducing the surplus variables x3, x4 and artificial variables, xai , xa2 , the given problem reduces to the following form Maximize, Z' = -Z =- 4x1 - 6x2 + 0.x3 + 0.x4 - Mxai - Mxa2 s.t. + xa = 80 xl+ 2x2 - x3 + xa 2= 75 3x1 + x2 - x4 and xl , x2, x3, x4, xai.„ xa2 0 where M is very large positive price. Taking x1, = 0,x2 = 0, x3 = 0, x4 = 0, we get xat = 80, xa2 = 75, which is the starting B.F.S. All computation work is shown in the following table. Table 8.17 ci

—4

—6

0

0

—M

—M

MM. Ratio XB /Yi

B

CB

XB

Yl

Y2

Y3

Y4

Al

A2

Al

-M

80

1

2

-1

0

1

0

80/1 = 80

A2

-M

75

3

1

0

-1

0

1

75/3 = 25—> (Min.)

Z' = CBXB =- 140M

0- 4M - 4 3M - 6 -M -M I

0

0 I

XB /1/2

Al

-M

55

0

5/3

-1

1/3

1

li

-4

25

1

1/3

0

-1/3

0

Ai

0

5M-14

__A4

M-4

0

-4M+4

3

1

3

Z' = C B XB

3

=-55M+100

T

., Y2

—6

33 24

-4 li Z' = C B XB ), Ai =- 294

-1/3 55 = 33 (Min. ) 5/3 —> 1/3 25 = 75 1/3

a

0 1

1 -3/5 0 1/5

1/5 -2/5

0

0

_ 14 5

_2 5

3/5 -1/5

-1/5 2/5 -1)4+14 _ A4+ 2 5

Here no Ai > 0, so this solution is optimal The optimal solution is xl = 24, x2 = 33 and Mini Z = — Max. Z' = 294.

5

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334

Hence 24 kg and 33 kg of components A and B respectively should be produced for minimum cost of ingredients C and D. In this case the minimum cost is 294 units.

8.13 Degeneracy in Simplex Method

[Meerut 2001 (B.P.)] A B.F.S. of a L.P. problem is said to be degenerate B.F.S. if at least one of the basic variables is zero. So far we have considered the L.P. problems in which by minimum ratio rule we get only one vector to be deleted from the basis. But there are L.P. problems where we get more than one vector which may be deleted from, the basis. Thus, if min. { X B' , yik > 00, (a, k is incoming vector) occurs at i

, i 2, ..., is.

Y ik i. e., minimum occurs for more than one value of i, then the problem is to select the vector to be deleted from the basis. In such cases if we choose one vector say pi (i is one of 11 , i 9 , ..., ) ,nd delete it from the basis then the next solution (obtained at this iteration) may be a degenerate B.F.S.. Such problem is called the problem of degeneracy. It can be seen that when the simplex method is applied to a degenerate B.F.S., to get a new B.F.S., the value of the objective function may remain unchanged i.e., the value of the objective function is not improved. In some cases due to the presence of degeneracy the same sequence of simplex tables are repeated forever without ever reaching the optimal solution. This problem is called

cycling. The procedure which prevent cycling within the simplex routine and an optimal solution is obtained in finite number of steps is called the resolution of degeneracy problem.

8.14 Conditions for the Occurrence of Degeneracy in a L.P.P. [Meerut 2001 (BP)] In a L.P.P. the degeneracy may appear in the following two ways.

1.

The degeneracy appear in a L.P.P. at the very first iteration when some component of vector b i.e., some bi is zero.

2.

If none of the components b is zero at any iteration and choice of the outgoing vector P r at the same iteration is not unique, the next solution is bound to be degenerate. (For proofs see author's book of L.P.)

8.15 Computational Procedure to Resolve Degeneracy by Carner's Perturbation Method If Yk.(= { X D;

) is the incoming vector (i.e., vector entering the basis) and MM.

yik > 0 is not unique i.e., the same minimum ratio XBi /yik ,yik > 0 occurs for

Yik

more than one value of i, i.e. in more than one row. Say it occurs in the i1-th, i 2-th, i 3 -th, ... rows, then let the set , 2 , 13, ...1. Then to select the vector to be deleted from the basis (i.e., outgoing vector), proceed as follows :

335

Simplex Method

1.

Denote the columns of the unit matrix in proper order (i.e., the columns corresponding to the basic variables in proper order) in the simplex table by _ Y1 , Y 2, etc. Thus, -

= [1, 0, 0, ,

t, Y2 = [0, 1, 0, ,

Element if

2. Compute Mini.

t etc.

in i-th row

i E h Element of Yk in i-th row

11= {ii , i2 i 3 ,• • • •}

If this minimum is unique, say attained for i = ir , then the i1-th vector is taken, as the outgoing vector and the element where this i ,.-th row is intersected by column Yk is taken as the key element. If this minimum is not unique, then let / 2be the set of all those values of i E 11, for which there is a tie obviously / 2 c 11. Then proceed to the next step. Element if Y2 in i-th row

3. Compute Mini.

E i2 Element of Yk in i-th row

If this minimum is unique, say attained for i = is, then the is -th vector is taken as the outgoing vector and the element where this is -th row is intersected by column Yk is taken as the key element. If this minimum is also not unique, then let 13 be the set of those values of i E 12, for which there is a tie. Obviously 13 c 12 C 11. Then proceed to the next step.

4.

Compute Mini. E 13

Element if Y3 in i-th row

Element of Yk in i-th row

In this minimum is unique say attained for i = it E 13, then i t -th vector is taken as the outgoing vector and the element where this i t -th row is intersected by column Yk is taken as the key element. If this minimum is also not unique, then proceed similarly.

911.11.4.thatiM2 exampLezi Problems of Degeneracy at the Initial Stage Example 11 : Solve the following L.P.P. (problem with some XBi = 0 for which yik = 0). Z= 2x1 +3x2 +10x3 x1 + 2x3 = 0 s.t.

Mar.

x2 + x3 = 1

Solution Max. s.t.

Yc2, x3 0 : The given L.P.P. can be written as

Z = 2x1 + 3x3 + 10x3 x1+ O. x2 + 2x3 = 0 0.x1 + 1. x2 + 1. x3 =1

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336

Here the basis matrix I2 (formed by coefficients of x1 and x2) exists, so there is no need to introduce the artificial variables. Taking x3 (non-basic variable) = 0, we have x1 = 0, x2 = 1, so the starting B.F.S. is x1 = 0, x2 = 1, x3 = 0. This B.F.S. is degenerate as the basic variable'xi. = 0 The solution by simplex method is given in the following table. Table 8.18

B

CB

Yi. Y,

2 3

Z = C B XB

c I•

2

3

10

XB

Y1 Pi

y2 P2

y3 a3

0 1

1 0

0 1

2 1

0

0

3 is

A 1•

=3

1

10 3

0 1

1/2 —1/2

0 1

1 0

Z =C B XB = 3

Aj

—3/2

0

0

Y3 Y2

MM. Ratio XB /Y3

0/2 (Min.)(1/1)

Here no Ai > 0, so this solution is optimal. Hence, the optimal solution is x1 = 0, x2 = 1, x3 = 0 which is degenerate, as one basic variable x3 = 0 Note : Here in this example it is important to note that we have obtained an optimal degenerate solution from an degenerate solution without improving the value of Z. Example 12 : Solve the following L.P.P. (problem with some xBi = 0 for which Yik < 0). Max. Z = 2x1 + 3x2 + 10x3 s. t. X1 — 2x3 = 0 x2 + x3 =1 x2, x3 >_0. Solution : The given L.P.P. can be written as Max. Z = 2x1 + 3x2 + 10x3 s.t. 1. x1 + O. x2 — 2x3 = 0. x1 + 1. x2 + 1. x3 = 1. Here the basis matrix I 2 (formed by coefficients of x1 and x2) exists, so there is no need to introduce the artificial variables. Taking x3 (non-basic variable) = 0, we get x1 = 0, x2 = 1, so the starting B.F.S. is x1 = 0, x2 = 1, x3 = 0 which is degenerate as the basic variable x1 = 0.

Simplex Method

337

The solution by simplex method is given in the following table : Table 8.19 c1•

2

3

10

MM. Ratio

B

CB

XB

yi

Y2

Y3

xB/Y3

Yi. Y2

2 3

0 1

1 0

0 1

—2 1

— 1/1 (Min.)--)

Z = C B XB = 3

A•

0

0 1

11 T

2 10

2 1

1 0

2 1

f

0

—11

Yi Y3

Z = C B XB = 14

0 1

A.2 = C2 — CB Y2 = 3 — (2,10), (2,1)=— 11, Ai= 0 = A 3. This solution x1 = 2, x2 = 0, x3 = 1 is optimal solution (non-generate). Note : Here in this examples it may be noted that a non-degenerate optimal solution is obtained from a degenerate B.F.S. and the value of the objective function is also improved. Example 13 : Solve the L.P.P. Max. Z 5x1 + 3x2 s.t. x1 + x 2 5.. 2 5x1. + 2x2 10 3x1 + 8x2 12 xi, x 2 O. and Solution : Introducing the slack variables x3, x4, x5 the given L.P.P. can be written as Max. Z = 5x1 + 3x2 =2 s.t. x1 + x 2 + x3 + x4 = 10 5x1+ 2x2 3x1 8x2 -FX 5 = 12 Taking xi = 0, x2 = 0, we have x3 = 2, x4 = 10, x5 = 12 which is starting B.F.S. Table 8.20 Starting Simplex Table 5

3

0

0

0

XB

Y4

Y5

fl

Y2

123

0 0 0

2 10 12

Yi 1 5 3

Y2 1 2 8

Y3((31) 1 0 0

Y4((32) 0 1 0

Y5 433 ) 0 0 1

Z = C B XB

p.

5 T

3

0

0

0

ci B

CB

Y3 Y4 Y5 =0

Min. Ratio XB /Y1 2 2--> 4

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338

Here A1 = c1 — C BY1 = 5, A2 = c2 —CB Y2 = 3, A3 = 0= A4 = A 5 which are positive or zero, therefore this solution is not optimal. Since Max. Ai = 5 = 01. Entering (incoming) vector is Y1(= al ) By minimum ratio rule we find that minimum is not unique but occurs for i =1 and i = 2 both. Thus, the problem is the problem of degeneracy. to select the vector to be deleted from the basis, we proceed as explained in article 8.15. 1. First of all we renumber the columns associated to the identity matrix in the proper order by Y1, Y2 , Y3. Y3 =Y5 • Y1 = Y3, Y2 = Y4, 2. Since minimum ratio occurs for i = 1 and 2. 11 = {1, and incoming vector is Y1 Element of Yr in i- th row Compute Mini. i E I1 Element of Y1 in i-th row Element of Y1 in 1st row Element of Yr in 2nd row Element of Yl in 1st row Element of Y1 in 2nd row {1 01 0 = Mini. — =—. 15 5 This minimum is unique and correspond to i = 2 the vector in the second row i. e., Y4 is to be deleted and key element is Y 21 = 5. Further computation by simplex method is shown in the following table. Table 8.21 = Mini.

5

3

0

171

Y2

0 2 6

0 1 0

2/5

34/5

Y3 1 0 0

Y4 —1/5 1/5 -3/5

Y5 0 0 1

A

0

1 is

0 1

—1

0

0 2 6

0 1 0

1 0 0

5/3 -2/3 -34/3

—1/3 1/3 5/3

0 0 1

Ai

0

0

—5/3

—2/3

0

ci B

CB

Y3 0 Yi 5 Y5 0 Z = C B XB

XB

=10 Y2 3 yi 5 Ys 0 Z = CB XB

3/5

0

0

=10 Since no A • > 0 the solution is optimal: Optimal solution is, x1 = 2, x2 = 0 and Max. Z = 10.

Min. Ratio XB /Y2 0 (Min.) ----> 5 15/17

Simplex Method

339

Example 14 : Solve the L.P.P. Max. Z = 2x1 + X2, s.t. 4X1 + 3X2 12 4x1 + x2 8 4x1 - x2 8 and

x1, x2 ?_ 0.

Solution : Introducing the slack variables x3, x4 and x5 the given L.P.P. can be written as Max. Z = 2x1 + x 2 s.t. 4x1 + 3x2 + x 3 = 12 4x1 + x2 +x4 =8 4x1 - x., +x5 8 and Xi , X2, X3, X4, X5 > O. Taking xi. = 0,x2 = 0, we have x3 = 12, x4 = 8,x5 = 8, which is the starting B.F.S. The starting simplex table is Table 8.22 ci

B

CB

XB

Y3

0

12

Y4 Y5

0

2

1

0

0

0

Min. Ratio

Y4

Y5

Y11 72

T73

XB 1K

Yi

Y2

Y3

Y5

Y4

4

3

1

0

0

3

8

4 Y24

1

0 721

8

4 Y34

-1

°Y31

0 1

2

0

1 Y22 0 5i 23

Z = C B XB

Aj

1

0

0

0 1

=0

2 I

3 = 0 = A4 = As Here = - C B-Yi = Z A 2 = C2 - C B Y2 = positive or zero, therefore the solution is not optimal.

2-+

which are

Since Max Ai = 2 = O1. Entering (Incoming) vector is Y1 (= al ). By minimum ratio rule we find that minimum is not unique but occurs for i = 2 and i = 3 both Thus, the problem is the problem of degeneracy. to select the vector to be deleted from the basis, we proceed as explained in article 8.15 1. First of all we renumber the columns of _ the_ above table associated to the identity matrix in the proper order by Yl, Y2, Y3. • Let Yi =Y3 = 131, Y2 =Y4 = P 2) Y3 =Y5 =133

Operations Research

340

= {2, 3} and incoming vector is Since minimum ratio occurs for i = 2 and 3 Y1 Element of Y1 in i- th rowl Compute, Mini. i E il Element of Yii in i- th row

2.

00 = Mini. / —, — = Mini. {0, 0} ... i = 2, 3 4 4J Since the minimum is not unique and there is a tie at i = 2, 3. Let 12 = {2, 3} c Ii and proceed to the next step. Compute, Mini. I E 12

Element of 12 2 in i- th row Element of Y1 in i- th row

i = 2, 3 = Mini. {1/4, 0/4} = Min. {1, 0}, =0 i.e., minimum occurs at i=3. ... Vector in the 3rd row i.e., Y5 is the outgoing vector and key element =y 31 = 4. Thus, we enter Yj in place of Y5. The computational work by simplex method is shown in the following table. Table 8.23 c1-

2

1

0

0

0 Y5

Min. Ratio X B /Y2

B

CB

XB

Yi

Y2

Y3

Y4

Y3

0 0 2

4 0 2

0 0 1

4` 2 —1/4

1 0 0

0 1 0

—1 —1 1/4

1 0 (Min.) --> —

Z = C B XB

A. 1

0

3/2 T

0

0 1

—1/2

X B /Y5

0 1 2

4 0 2

0 0 1

0 1 0

1 0 0

—2 1/2 1/8

1 —1/2 1/8

4 (Min.) — 16

Z =C B XB

A.

0

0

0 1

—3/4

1/4 I

4 2 3/2

0 0 1

0 1 0

1 1/2 —1/8

—2 —1/2 3/8

1 0 0

Al

0

0

—1/4

—1/2

0

Y4 Y1 =4 Y3 Y2 Y1

=4 Y5

0 1 2

Y2 Yi

Z = C B XB =5

Since no 41 > 0, the solution is optimal. Optimal solution is x1 = 3/2, x2 = 2, and Max. Z = 5.

Simplex Method

341

Example 15 : Solve the following L.P.P. by simplex method. Max. Z = 5x1— 2x2+ 3x3 subject to 2x1 + 2x2 — x3 2, 3x1 — 4x2 3 x2 + 3x3 5 and xi, x2, x3 0 [Kanpur 1996] Solution : Introducing surplus variable x4, slack variables x5, x6 and artificial variable xa , the given L.P.P. can be written as Max.

Z = 5x1 - 2x2 + 3x3 + 0. x4 + 0. X5 + 0. X6 — M. X,

s .t.

2X1 2X2

x3 — X4

+ xR

3x1 — 4x2 and

=3

+ X5

x2

=2•

3x3

+ x6

=5

Xi, X2,..., X6, Xa 0

Taking x1 = 0 = x2 = x3 = x4, we get xa = 2, x5 = 3, x6 = 5 which is the starting B.F.S. The computation of the solution by simplex method is given in the following table. Table 8.24 1

5

-2

3

0

0

0

-M

Y3

Y4

15

Y6

A

—1 0 3

—1 0 0

0 1 0

0 0 1

1 0 0

A. 5 I-2M -2+2M 3-M -Al

0

0

C

B

CB

XB

yi

Y2

A Y5 Y6

—M

2 3 5

2 3 0

2 —4 1

0 0

Z= -2M

T

Min. Ratio XB / Yi

2/2=1 --> 3/3=1 —

0

1

Here we note that max. = Al so Y1 is the incoming vector and by minimum ratio rule we find the same minimum ratio in row 1 and row 2. So it is a case of degeneracy. But here in the first row we have the artificial vector A, so giving preference to A, to leave the basis we choose A as the outgoing vector. Here there is no need to apply the rule of resolving degeneracy. Thus, entering vector Y1 in place of A in the basis, y11 = 2 is the key element, further computations are shown in the following table :

Operations Research

342 Table 8.25

B

c1•

5

—2

3

0

0

0 —M

CB

XB

Yl

Y2

Y3

Y4

Y5

Y6

5 0

1 0

1 0

0

5

0

1 —7 1

—1/2 3/2 3

—1/2 3/2 0

0 1 0

0 0 1

— 0/ (3/2) Min. 5 5/3

A•

0

—7

11/2

5/2

0

0

X B Y2 — — 5/15 Min. --> XB/Y4

Y1 y 5 Y6 Z= 5

1

T Yi y 3 y6

0

— 4/3 —14/3 15

0 1 0

0 1 —3

1/3 2/3 —2

0 0 1

0

56/3

0

—3

—11/3

0

5 3

1 0

1 0

0

5 A•

Z= 5

A

Min. Ratio XB /Y3

T Y1

y3

y 2

5

13/9 3 14/9 1/3 _2

Z= 101/9

A•

1 0 0

0 0 1

0 — 4/15 1 1/15 0 —1/5

0

0

0

7/45 2/45 —2/15

4/45 14/45 1/15

— 70/3 Min. --) —

11/15 —53/45 — 56/45

T Y1 y4

Y2

5

23/3 0 70/3 5 —2

Z= 85/3

A•

1 0 0

0 0 1

4 15 3

0 1 0

1/3 2/3 0

4/3 14/3 1

0

0

—11

0

—5/3

—14/3

Since no Ai > 0, so this solution is optimal. Hence the optimal solution to the given problem is x1 = 23/3, x2 = 5, x3 = 0 and Max. Z = 85 / 3.

8.16 Some Important Tips for Simplex Method 1. The minimization problem may be converted to maximization one and vice versa only by reversing the sign of objection function i.e. if Z is the minimization objective function then —Z is the maximization one and vice-verse.. 2. When the objective function contains a constant term, then the simplex method • is applied by excluding (omitting) the constant term. In the end this constant term omitted is added to the optimal value of the objective function. 3. In a L.P.P. in which lower bounds are specified say xi ci , substitute xi = ci + yi and solve the problem in terms of yi with yi 0. In the end find xi = ci +yi.

Simplex Method

343

4. In a L.P.P. in which some or all variables are unrestricted in sign, substitute xi = xi' — xi" , where xi' , xi" 0, if xi is unrestricted in sign and solve the problem in terms of xi ', xi ". In the end find xi from xi = — xi" etc. 5. In the simplex table of a L.P.P. in which m is the number of constraints, there must be an identity matrix of order in x m. The columns, constituting this identity matrix need not be adjacent or in order. 6. In the solution of a L.P.P. by simplex method, as soon as an artificial variable is driven out of solution i.e., it become non-basic variable, the column corresponding to it may be omitted in the subsequent iterations. 7. If degeneracy occurs in the simplex method i.e., when the choice of outgoing (leaving) vector by minimum ratio rule is not unique and one of these leaving vectors is the artificial variable then give preference to the artificial variable to leave the basis. In such cases there is no need to apply the procedure-for resolving degeneracy.

8.17 Linear Programming Problem (Special Cases) -1. Problem having no Feasible Solution Example 16 : Solve the L.P.P. Max.

Z = —X1 — X 2

3X1 + 2X 2 30 —2x1 + 3x2 5 —30 + x2 5 5. x1, x2 >_0. Solution : Multiplying second constraint by — 1 (to make the right hand side positive), and adding slack and surplus variables, the given problem reduces to the form. Max. Z = —x1 — x2 + 0. x3 + 0. x4 + 0. xs s.t. 3x1 + 2x2 — x3 = 30 —x4 = 30 2x1 — 3x2 xi+ X 2 +X 5 = 5 In order to get a unit matrix /3 we have to add two artificial variables xai and xa2 in the first two constraints. Thus assigning a large negative price vector —M to the artificial variables the given problem reduces to the following form Max. Z = —x1 — x2 + 0. x3 + O. x4 + O. x5 — — Mxa2 s.t. 3x1 + 2x _ 2 — x3 + 0. x4 + 0. X5 + Xai + 0. Xa2 = 30 2x1 — 3x2 + 0. X3 — x4 + 0. x5 + 0. Xai + xa2 = 30 xi + x2 + 0. x3 + 0. x4 + x5 + 0. x + 0. x, = 5 -2 ai and xi, x2, x3, x4, x5, xai xa2 0 Taking xi = 0,x2 = 0,x3 = 0, x4 = 0, we get x5 = 5, ;LI = 30, xa2 = 30, which is the starting B.F.S. All computation work is done in the following table. s. t.

Operations Research

344 Table 8.26 C 1•

-1

-1

0

0

0

-M -M Min. Ratio XB /Yi A Al 2

B

CB

XB

Y1

Y2

Y3

Y4

Y5

A1

-M -M 0

30 30 5

3 2 1

2 -3 1

-1 0 0

0 -1 0

0 0 1

1 0 0

0 1 0

Z = C B XB = - 60M

A.

5M - 1 -M - 1 -M 1'

-M

0

0

0

Al

-M

A2 Yi

-M -1

15 20 5

0 0 1

-1 0 0

0 -1 0

-3 -2 1

1 0 0

0 1 0

Z = 35M - 5 A j

0

-M -5M 0

0

A2 Y5

-1 -5 1, -6M

-M

1

10 15 5 (Min.) -›

Here no Ai > 0. Hence the optimality condition is satisfied but here the artificial vectors A1, A 2 appear in the basis at positive level, which immediately indicates that the give- problem has no feasible solution.

2. Problem having an Unbounded Feasible Solution but Bounded Optimal Solution Example 17 : Solve the L.P.P. Max. Z = 6x1 - 2x2 s.t. 2x1 - x2 2 4 xi, x2 0 Solution : Adding slack variables x3 and x4, the inequalities reduces to the following equations 2x1 - x2 + x3 + 0.x2

=2 + x4 = 4

Taking x1 = 0, x2 = 0, we get x3 = 2, x4 = 4, which is the starting B.F.S. All computation work is done in the following table.

345

Simplex Method Table 8.27 c 1-

6

-2

0

0

MM. Ratio XB /Y1

B

CB

XB

y1

y2

y3

Y4

Y3

0 0

2 4

2 1

-1 0

1 0

0 1

1 (Min.) —> 4

pi

6 T

-2

0

0

X B /Y2

Y4 Z = C B XB =0 Y1

6

1

1

-1/2

1/2

0

-

Y4

0

3

0

1/2

-1/2

1

1 (Min.) ->

Z = C B XB

Ai

0

1 I

-3

0 1

=6 Yi

6

4

1

0

0

1

Y2

-2

6

0

1

-1

2

Z = C B XB

pi

0

0

-2

-2

=12 The optimal solution is x1 = 4, x2 = 6 and Max. Z = 12. From the starting simplex table we note that the elements of column vector Y2 are negative or zero which indicates that the feasible region is not bounded. Hence we conclude that a L.P.P. having unbounded feasible solution may have a bounded optimal solution. 3. Problem having Unbounded Solution Example 18 : Solve the following L.P.P Z— + 2x2 + X3 Max. - 3x1 + 2x2 + 2x3 = 8 S. t. - 3xi + 4x2 + x3 = 7 x1, x2, x3 0 and Solution : Introducing the artificial variables xai , xa2 , the given L.P.P. reduces to the following form Max. Z = 3x1 + 2x2 + x3 — Mxai — Mxa2 =8 —3x1 + 2x2 + 2x3 + xai =7 + X,2 —3 xi. + 4x2 + x3 Xi, X2, X3, , X,2 0 Taking x1 = 0, x2 = 0, x3 = 0, we get xai = 8, xa2 = 7 which is the starting B.F.S All computation work is done in the following table.

Operations Research

346 Table 8.28 ci

3

2

1

—M

—M

Y1 -3 -3

Y2

Y3

A2

B /Y2

2 4

2 1

Al A 1 0

0 1

0

0 1

4 7/4 (Min.) ---) XB/Y3

B

CB

)(13

Al A2

-M -M

8 7

Z = C B XB

Ai

=- 15M

-6M+3 6M + 2 3M+1 s

Min. Ratio

-3/2 -3/4

0 1

3/2 1/4

1 0

3 (nit ) 7

Z =C B XB A•• =7/2 -9M/2

(-3M + 9)/2

0

(3M + 1)/2 is

0 1

X B /Y1

1 3 Y3 2 1 Y2 z = C B X B = 5 AI

-1 -1/2

0 1

1 0

5

0

0

Al

-M 9/2 2 7/4

Y2

Here Yi is the incoming vector. All the elements of this column are negative. Thus, we cannot select the outgoing vector. Hence, in this case the solution is unbounded (see Article 8 .6). Example 19 : Solve the following L.P.P. Max. Z = 2x1 + x2 - x2 10 s.t. 2x1 - x 2 40 and xi 0, x2 O. Solution : Entering the slack variables, the inequalities reduces to the following equations. =10 — x2 + x3 2x1 — x2 +x4 = 40 Taking x1 = 0, x2 = 0, we get x3 = 10, x4 = 40 which is the starting B.F.S. The starting simplex table is as follows. Table 8.29 c.1• Min. Ratio 2 1 0 0 X B /Y1 y4 B CB XB Y1 Y2 Y3 al a2 a3 a4 0 0

10 40

1 2

—1 —1

1 0

0 1

Z = C B XB

Ai

2

1

0 1

0

Y3 Y4 =0

10/1 = 10 (Min.) ---> 40/2 = 20

347

Simplex Method

= c 2 — C B Y2 Here the column Y2 (a 2 ) E A but oB and for this column c1 = 1 > Oand yi 2 0, i = 1, 2. Hence, the solution of this problem is unbounded. If the computation work by simplex method is carried on without taking knowledge of 8.6, then we get the following table. Table 8.30 c•

2

1

0

0

B

CB

XB

Y1

Y2

Y3

y 4

Yi v

2 0

10 20

1 0

—1

1 —2

0 1

Z = C B XB

A.

0

3 T

—2

0 1

2 1

30 20

1 0

0 1

—1 —2

1 1

Z =C B XB

Ai

0

0

4

—3

'4

=20 111

Y2

1

T

= 80

Min. Ratio X B /Y2 — 20/1 (Min.)-3

Here Y3 is the incoming vector, but all the elements of this column are negative. Thus, we cannot select the outgoing vector. Hence, in this case the solution of the problem is unbounded.

4. L.P.P. with Unrestricted Variables If in a L.P.P. a variable say xi is unrestricted in sign then substitute 0, and solve the problem as usual. In case more than xi = x1' —x1" where xi' , one variable is unrestricted in sign then we replace all these variable by the difference of two non-negative variables and solve as usual L.P.P. method. For clear understanding of method see the following example. Example 20 : Max. Z = 2x1 + 3x2 s.t.—x1 + 2x2 5 4 x1 + x2 6 x1 + 3x2 9, x1, x2 unrestricted.

Solution : Taking xi = 1 — , x2 = — s.t. , x1" , x2' , x2" 0 and introducing the slack variables x3, x4 and x5 the given problem reduces to Max. s.t.

Z = 2x; — 2x1" + 3x2 —

+ O. x3 + O. x4 + 0.x5

+xi." + 2x2 — 2x2" + x3 xl -

xi" + x2' - x2"

— x1" + 3x2 —

=4

+ x4

=6

+x5 = 9

Taking x1' = 0 = xi" = x2 = , we get x3 = 4,x4 = 6,x5 = 9 which is the starting B.F.S. All computation work is done in the following table.

Operations Research

348 Table 8.31

c1•

2

—2

3

—3

0

0

0

Min. Ratio

CB

XB

Yi'

Yi"

17-2

Y2n

Y3

Y4

Y5

XB /172

Y3

0

4

—1

1

2

—2

1

0

0

Y4

0

6

1

—1

1

—1

0

1

0

2 (Min. ) -6

Y5

0

9

1

—1

3

—3

0

0

1

3

p•

2

—2

3

—3

0

0

0

XB /Yl'

B

Z = CBXB

i

=0

1

Y2'

3

2

—1/2

1/2

1

—1

1/2

0

0

—

Y4

0

4

3/2

—3/2

0

0

—1/2

1

0

8/3

Y5

0

3

5/2

—5/2

0

0

—3/2

0

1

6/5 (Min.) —>

A•

7/2 i

—7/2

0

0

—3/2

0

0

XB /Y3

0

1

—1

1/5

0

1/5

13

Z =•CB-XTB =6 Y2

3

13/5

Y4

0

11/5

0

0

0

0

2/5

1

Yi

2

6/5

1

—1

0

0

—3/5

0

2/5

Ai

0

0

0

0

3/5 I

0

—7/5

Z = C B XB

0

=51/5

—1/2

—3/5 11/ 2 (Min. )

1/2

Y2

3

3/2

0

0

1

—1

0

Y3

0

11/2

0

0

0

0

1

5/2 —3/2

Yi

2

9/2

1

—1

0

0 ,

0

3/2 —1/2

• Ai

0

0

0

0

— 3/2 —1/2

Z =C B X B

0

_

= 27/2 Since no A > 0, so this solution is optimal. The optimal solution to the given L.P.P. is x1' = 9/2, x1" = 0, x2' = 3/2, x2" = 0 and Max. Z = 27/2

i.e.,

x1 = x1'— x1" = 9/2, x2 = x2'— x2" = 3/2, Max Z = 27/2.

Example 21 : Solve the L.P.P.

Max. subject to

Z = 4x1 + 6x2 xl— 2x2 —4 2x1 + 4x2 5 12 + 3x2 5 9

x1, x 2 are unrestricted. [Meerut 1996 (BP)]

349

Simplex Method

Solution : Here x1 and x2 are unrestricted, taking x1 = xl - x1", x2 = x2 - .xT the given L.P.P. in the standard maximization form with x3, x4, x5 as slack variables can be written as Z = 4x11 — 4x11' + 6x2 - 6x2" Max. 2x2"+ x3 s.t. — x11 + x1"+ =4 = 12 2x11 — 2x1"+4x21— 4x2" + x4 x1' — xi" + 3x2' — 3x2" + x5 = 9 and , , x2 , x2" , x3, x4, x5 0 Taking xi' = 0 = x1" = x2' = x2" , x3 = 4,x4 = 12, x5 = 9, which is the starting B.F.S Proceeding by simplex method, we get the following table. Table 8.32 ci

4

-4

6

-6

0

0

0

B

CB

XB

Yi.'

Yi"

Y2

Y2"

Y3

Y4

Y5

Min. Ratio X B' /Y2

Y3 Y4 Y5

0 0 0

4 12 9

—1 2 1

1 -2 -1

2 4 3

—2 -4 -3

1 0 0

0 1 0

0 0 1

2 (Min.) —> 3 3

Z =C B XB

A•

4

-4

6

-6

0 1

0

0

X B Illi'

1 0 0

-1 0 0

1/2 -2 - 3/2

0 1 0

0 0 1

1 (Min.) --> 6/5 -

-3

0 1

0

=0 Y2 Y4 Y5

6 0 0

2 4 3

-1/2 1/2 -4 4 5/2 - 5/2

Z = CBXB

A 1•

7 i

-7

0

0

6 4 0

5/2 1 1/2

0 1 0

0 -1 0

1 0 0

-1 0 0

1/4 1/8 -1/2 1/4 -1/4 -5/8

0 0 1

Z = C B X. '-' =19

A..1

0

0

0 1

0

1/2 - 7/4 T

0

Y3 0 Yi.' 4 115 0 Z = C B XB

10 6 3

0 1 0

0 -1 0

4 2 1

-4 -2 -1

1 0 0

1/2 1/2 - 1/2

0 0 1

A

0

0

-2

2

0

-2

0

=12 Y2 Yi.' Y5

10 (Mini) -› -

= 24 Here Y2"is the incoming vector, but all the elements of this column are negative. Thus, we cannot select the outgoing vector. Hence, the solution of the problem seems to be unbounded. Now we try to solve the problem by graphical method.

Operations Research

350

Obviously Z is max. at the point P(6, 0) and Max. Z = 24. Thus, the optimal solution of the problem is x1 = 6, x2 = 0 and Max. Z = 24 Y

•116,

0)11 '4%1•‘`

x

,

Z = 4xi + 6x2 =0

Fig. 8.1 x1 —3 x2 2

Actually the value of Z can be decreased as much as we please. 5. L.P.P. with Some Constant Term in the Objective Function In a L.P.P. when the objective function contains a constant term, then the simplex method is applied by leaving this constant term in the beginning, and optimal solution is obtained. In the end the constant term (which was left initially) is added to the optimal value of the objective function. For clear understanding of the method see the following example. Example 22 : Solve the L.P.P. Max. Z = 2x1 — x 2 + x3 + 50 Subject to 2x1 + 2x2 — 6x3 16 12x1 — 3x2 + 3x3 6 — 2x1 — 3x2 + x3 4 Xi, X2, X3 0 and [Meerut 1997(P)] Solution : Leaving the constant term 50 from the objective function in the beginning, introducing slack, surplus and artificial variables, the given problem reduces to Max. Z'= 2x1 — x2 + x3 + 0. x4 + 0. x5 + 0. x6 — M. x al s.t. = 16 2x1 + 2x2 — 6x3 + x4 12x1 — 3x2 + 3x3 —x5 +xa = 6 -2x1 - 3x2 + x3 +x6 = 4 and xl , x _ 2, x _ 3,x _ 4,x5, .x 6 , x _ _ Taking x1 = 0 = x2 = x3 = x5, we get x4 = 16, x = 6,x6 = 4 which is the ai starting B.F.S. To solution of the problem by simplex method is given in the following table.

351

Simplex Method Table 8.33

B Y4 A Y6

c 1•

2

-1

1

0

0

0

-M

CB X B

yi

Y2

Y3

Y4

Y5

Y6

A

2 12 -2

2 -3 -3

-6 3 1

1 0 0

1 -1 0

0 0 1

0 1 0

16/2 6/12 (Min.) ---> -

0

-M

0

0 .1,

X B /Y3

0 -M 0

16 6 4

Z'= C B X B A•

2 +T12M -1 - 3M 1 + 3M

= - 6M Y4 Yi y 6

Min. Ratio X B /Y1

0

15

0

5/2

-13/2

1

1/6

0

2 o

1/2 5

1 0

- 1/4 - 7/2

1/4 3/2

0

-1/2

0

2 (Min.) -->

0

-1/6

1

10/3

Al

0 1,

-1/2

1/2 T

0

1/6

0

X B /Y5

0

-

Z'=C XB =1 Y4

0

1'3 Y6

1 0

28 2 2

Z' =C B XB A.

26 4

-4 -1

0 1

1

-2

0

-1/3

0

-

-6 *

-2

0

0

1/3

1

6 (Min.) --->

0

0

0

1/3 T

1

0

6

-2

=2 Y4

0

40

Y3 y 5

1 0

0

-10

-16

0

1

4

-2

-3

1

0

0

1

6

-18

-6

0

0

1

3

2

0

0

0

-1

- CB Xs A.3 Z' = _4

4

Here Y1 is the entering vector but all elements in column one are —ve, so we cannot select the outgoing vector. Hence, the solution of the problem is unbounded. Example 23 : Solve the following L.P.P., using simplex method :

Max. subject to,

6x1 + 3x2 + 5 x1 + 3x2 9, xj + x2 5

2x1 + x2 8 and xi, x2 0 [Meerut LP 1997 (P)] Solution : Let Z = 6x1 + 3x2 + 5 = Z ' + 5, where Z' = 6x1 + 3x2. Introducing the slack, surplus and artificial variables and assigning large negative cost —M to artificial variable xa in the objective function, the L.P.P.with objective function Z'

reduces to

Operations Research

352 Max. s.t.

Z'= 6x1 + 3x + 0. x3 + 0. x4 + 0. x5 — M. x„ x1 + 3x2 + x3 =9 + X2 -x4 + x, = 5 •=8 2x1 + x2 + x5 and xi, x2, x3, x4, x5, xa 0 Taking xi = 0 = x2 = xn, we get x3 = 9, xa = 5,x5 = 8, which is the starting B.F.S. The solution of the problem using simplex method is given in the following table: Table 8.34

B

CB

A

0 —M

Y5

0

Y3

Z = C B XB

c1•

6

3

0

0

0

-M

XB

Y1

Y2

Y3

Y4

Y5

A

9 5 8

1 1 2

3 1 1

1 0 0

0 —1 0

0 0 1

0 1 0

A.

6+iM

3 +M

0

—M

=- 5M

Y3 A

XB /Y 9/1 5/1 8/2 (Min.)--> XB /Y2

0

1

—

0 —M 6

Yi.

Min. Ratio

Z = C B XB

Yi (3 3 )

Y2

Y3 ( 71 )

Y4

Y5

A(Y2)

5 1 4

0 0 1

5/2 1/2 1/2

1 0 0

0 —1 0

—1/2 —1/2 1/2

0 1 0

Ai

0

0

1 —M —3 — — M 2

=—M+24

1

M

2 2 -* 8 r•

0 1 '

s4

T

Since minimum ratio is not unique but occurs for i =1 and 2 both. /1 = {1, 2}. So it is a problem of degeneracy. To resolve degeneracy we proceed as follows. Renumbering the columns in order corresponding to unit matrix. we have Y1 = Y3, Y2 = A, Y 3 = Y1

Since Y2 is the entering vector, so we compute Mini. i E ii

Element of Yi in i- th row Element of Y2 in i-th row

= mini. 2 , 0} = 0 = Mini. 1 , {5/2 1/2 5 i = 1 , i =2

For i = 2

i=1i=2

This minimum in unique and correspond to i = 2 ... the vector in the second row i.e., A is to be deleted and the key element is y 22 = 1/2. Entering vector Y2 into the basis in place of A, we get the following table.

Simplex Method

353 Table 8.35

B 1'3

ci

6

3

0

0

0

-M

CB

XB

Yt

Y9

Y3

Y4

Ys

A

0 3 6

0 2 3

0 0 1

0 1 0

1 0 0

5 -2 1

2 -1 1

-5 2 -1

A

0

0

0

0

-3

-M

Y2 Y1 Z = CB XB

j

MM. Ratio

=24 . Here no Ai > 0, so this solution is optimal. The optimal solution is xl = 3, x, = 2 and Max. Z' = 24. Hence, the optimal solution of the given problem is x1 = 3, x2 = Z Max. Z = Iviax. Z' + 5 = 24 + 5 = 29. Note : In this problem there is tie in minimum ratio in row 1 and 2 with artificial vector in row 2, so first we shall delete the artificial variable from the basis. 6. L.P.P. with Lower Bounds of Some or All Variables, Other than

Zero In some L.P.P. the lower bounds may be specified for examples, in a L.P.P. it may be stipulated that x1 >c1,x2 L. c2 etc. In such problems we substitute x1 = c1 + yi, x2 = c2 + y 2 etc., and then solve the problem in terms of yi and y2, where yi 0,y 2 0 etc. For clear understanding of the method see the following example. Example 24 : Solve the L.P.P. Max. Z = 3x1 + 5x2 + 4x3 s.t. 2x1 - 3x2 5. 8 2x2 + 5x3 5. 10 3x1 + 2x2 + 4x3 5, 15 xlL' 2 , x2 % 4,x3 O. [Meerut 1995] Taking xi = y1 Solution : + 2 , x2 = y + 4,x3 = y 3, the given problem reduces to Max. Z 3y1 + 5y2 + 4y 3 + 26 s.t. 2y1 - 3y 2 5 16, 2y 2 + 5y3 5. 2, 3y1 + 2y 2 + 4y3 5.. 1 and y1,y2,y3 0 Introducing the slack variables y4, y s, y 6 and leaving constant term 26 from Z in the beginning, the above L.P.P. reduces to Max. z' = 3y + 2 + ItY 3 where Z = Z' + 26 s.t.

2y1 - 3y 2 = 16 + Y4 2y2 + 5y 3 = 2 +y 5 3y1 + 2y 2 + 4y 3 +y6 = 1 and Y1, Y23Y3 Taking yi = 0 = y 2 = y 3, we get y 4 = 16, y 5 = 2, y6 = 1 which is the starting B.F.S. •

Operations Research

354

The solution by simplex method is shown in the following table. Table 8.36 ci

3

5

4

0

0

0

CB

XB

YY2

Y3

- Y4

Y5

Y6

0 0

16

Y5

2

2 0

—3 2

0 5

1 0

0 1

0 0

Y6

0

1

3

2

4

0

0

1

Z'= C B XB

Ai

3

5 I

4

0

0

0

B Y4

X B /Y2

=0

Y4

0

Y5

0

Y2

5

Z'=C B XB

35/2 1 1/2 A

i

Min. Ratio

13/2 —3 3/2

0 0 1

6 1 2

1 0 0

0 1 0

3/2 —1 1/2

—9/2

0

—6

0

0

—5/2

= 5/2

— 2/2 1/2 (Min.)--)

1

Since no Ai > 0, so this solution is optimal. Optimal solution is yi = 0,y 2 = 1/2, y3 = 0, Max. Z' = 5/2 x1 = + 2 = 2, x2 = y 2 + 4 = 9/2, Max. Z = + 26 = 57/2 i.e., x1 = 2, x2 = 9/2, Max. Z = 57/2.

7. L.P.P. having More than One Optimum Solution Some L.P.P. have more than one optimum solution. If in the final simplex table when an optimal solution is obtained we notice that, for a non-basic variable say xi the value of A = ci — CB Yi = 0, then an alternate optimal solution to the L.P.P. exists. This alternate optimal solution is obtained by taking the- corresponding column vector Y. as the entering vector and finding outgoing vector by the usual minimum ratio rule. For clear understanding of the procedure see the following example. Example 25 : Solve the following L.P.P. by simplex method. Max. Z = 4x1 +10x2 Subject to 2x1 + x2 5_ 50 2oxi + 5x2 5_ 100 2x1 + 3x2 90 and xi , x2 O. Also find the alternative optimum solution if exists. [Meerut 2002]

Simplex Method

355

Solution : Introducing the slack variables x3, x4, x5 the given problem becomes Max. Z = 4xt + 10x2 + 0. x3 + 0. x4 + 0. xs s.t. 2x1 + x2 + x3 = 50 2x1 + 5x2 +x4 = 100 2x1 + 3x2 +xs = 90 • and xi, x2, x3, x4, xs O. Taking xi = 0, x2 = 0, we get x3 = 50,x4 = 100, xs.„..= 90, which is the initial B.F.S. The solution to the problem using simplex method is given in the following table : Table 8.37 C•1

4

10

0

0

0

XB

1(1. 2 2 2

Y2

Y3

Y4

Y5

1 5 3

1 0 0

0 1 0

0 0 1

50/1 100/5 (Min.) --) 90/3

10 4 ,t

0

O. .

0

XB /Yi

1 0 0

-1/5 1/5 -3/5

0 0 1

75./4 (Min.) -4 50 • 75/2

0

-2

0

B

CB

Y3 Y4 Y5

0 0 0

50 100 90

= CB XB =0

A 1•

Z

Y3 Y2

0 10 0

Y5 Z = C B XB

30 20 30

8/5 2/5 4/5

41 — 0 T

0 1 0 0

Min. Ratio '22

1

•

= 200 In the last table all A • 5_ 0, therefore this solution is optimal. The optimal solution is x1 = 0, x2 = 20 and Max. Z = 200. Here corresponding to non-basic variable x1 in the last table Ai = 0 and Yi is not in the basis B, therefore an alternative solution also exists. Thus, the problem does not have unique solution. Now taking Yi as incoming vector and using minimum ratio rule we find that Y3 is the outgoing vector. Taking column Y1 in the basis in place of Y3 we obtain the next simplex table as follows : Table 8.38 c1

4

10

0

0

0

B

CB

XB

Yl

Y,

Y3

Y4

Ys

II.

4 10 0

75/4 25/2 15

1 0 0

0 1 0

5/8 -1/4 -1/2

-1/8 1/4 1/10

0 0 1

01

0

0

0

—2

0

Y2 Ys Z

= CB XB = 200

Min. Ratio

Operations Research

356

Here no Ai > 0, therefore this solution is also optimal having the same maximum value of Z. Second optimal solution is x1 = 75/4, x2 = 25/2 and Max. Z = 200 Hence, the two optimal solution of the problem are Max. Z = 200 x2 = 20, (i) x1 = 0, and (ii) x1 = 75/4, x2 = 25/2, Max. Z = 200 We know that the convex combination of B.F. solutions is also an optimal solution. S. 1, the following table gives For any arbitrary value of X. such that 0 different optimal solutions which are infinite in number. Variables

I Sol.

II Sol.

Gen. Sol.

x1 x2

0

75/4

0. X. + (75/ 4)(1 — X.)

20

25/2

20X + (25/ 2)(1 — X)

Taking a particular value of X. say A = 1/2, the third optimal solution is xi = 75/8, x, = 65/4 and Max. Z = 200.

8.18 Solution of System of Simultaneous Linear Equations by Simplex Method To solve a system of simultaneous linear equation by simplex method, wt introduce a dummy objective function Z with costs 0 to all given variables (non-basic variables) and cost —1 to each artificial variable (basic variables). If non-negative restriction is not given for the variables then each variable is replaced by the difference of two non-negative variables. For example variable xi is replaced by xi' — xi" where x1 ' , xi" 0, so that xi may be +ye, —ve, or zero. Then the problem is solved by simplex method as usual. For clear understanding of the method see the following example. Example 26 : Solve the following system of simultaneous lin)ear equations xj — x3 + 4x4 =3 2x1 —x2 =• 3 3x1 — 2x2 — x4 = 1 [Meerut 1995, 2001 (BP)] Solution : Simple method is applicable if the variables are non-negative. In this problem xi, x2, x3, x4 all are unrestricted in sign (may be +, — or zero). We take xi= xi — x1 x2 = xi — x2, x3 = x3 — X3, X4 = X4 — x4". such that x1 xi", xi, x2, x3, )6, x4;

0

Introducing the dummy objective function Z with costs zero to each given variable xi, x2, x3, x4 (Le, to xi, x1", xi, x2, )6, x;, x4, xii') and costs — 1 to artificial variables xai , x02 , x a3 , the given system of equations in the form of L.P.P. can be written as follows : Max. Z = 0.

+ xi" + O. xi + 0.

+ 0. xi + 0. )6' + 0. x + 0. x —1.

—1. x a, —1. xa3

Simplex Method

357 , — X3 +, X3 +

xi — Xi"

S.t.

=3

4x4 — 4x,"i + xai.

2xi.' — 2x1"— xi + x;

=

±Xa2

+ X43 --= 1

— x4 + x4"

3x1' — 3x1." — 2x2 + 2x;

3

x1, xi', xi, x;, x, x;, x4, x4, xal , x02 , x44 > 0

and

Taking xi.' = 0 = x1" = ,q = x2" = x3 = x; = x4' = x4", we get xal = 3 xa2 = 3, x43 = 1, which is the starting B.F.S. The solution of the problem using simplex method is given in the following table : Table 8.39 c.i000

0

0

B CB

XD

Y1' Y.1 "

Y2'

Y2"

Y3'

Al -1 A, -1 A3 -1

3 3 1

1 -1 2 -2 0 -3

0 -1 -2

0 1 2

-1 0 0

Z = -7

A•

6 -6 ,i,

-3

3

0 0 2/3 -2/3 0 0 1/3 -1/3 2/3 1 -1 -2/3

Al -1 8/3 A2' -i 7/3 Y1' 0 1/3

0

0

-1

0

-1 -1 Min. Ratio XE/Yi' A1 A, A. I

Y3" Y4 "

Y4 "

1 0 0

4 0 -1

-4 0 1

1 0 0 .3/1 0 1 0 3/2 0 0 1 1/3 (in.) ._)

-1

1

3

-3

0

-1 0 0

1 13/3 -13/3 0 2/3 -2/3 1/3 0 -1/3

-1

1

0

1 0 0 1 0 0

-5

Y1' 0 8/13 0 0 2/131-2/13 -3/13 3/13 A2 -1 25/13 0 0 3/13 -3/13 2/13 -2/13 Y1' 0 7/13 1 -1 -8/13 8/13 -1/13 1/13

1 0 0

-1 0 0

0 1 0

Z= -12/13

0

0

0

3/2 13/2 -13/2 1/2 -1/2 -3/2 3/2

0

Ai

A

'

0

0

1

0

0

3/13

-1

-3/13 2/13 -2/13

T

4

0

0

1

A2 -1

4

0

0

0

0

0

3

1 --1

0

• 0

-1

Z = -1

A

0

0

0

0

1/2

Y2' Y1'

0

Yip 0 25/3 0 0 Y4 " 0 2/3 0 0 Yi' 0 17/3 1 _1 Z =0

Ai

0

0

1 C 0

-1 0 0

1

0

0

4

-4(

-1/2 -3/2

3/2 T

2/3 -2/3 1/3 -1/3 1/3 -1/3 0

0

X6 /1/4 '

8/13(Min.)-4 7/2

XB /Y2'

T

-1 -3/2

1

0

0

5

7= -4

0 1

0 -1 0

0 1 0

0

0

( 0 0 1

-4,

8/2= 4 (Min.) 25/3 _ X8 / Y I "

7,/3 (Min.) -)

Operations Research

358 Since no. Ai> . 0,

this solution is optimal.

Optimal solution is

xl' = 17/3, x1" = 0, x,' = 25/3, x2" = 0, x3 = 0, xr = 0, x4" = 0, x4" = 2/3 • Hence, the optimal solution of the given system of equations is x1 =

-

= 17/3, X2 = X2' - X2" = 25/3, X3 = x3 - x3" = 0,

X4 =x4- v4"=- 2/3 i.e., xi = 17/3, x2 = 25/3, x3 = 0, x4 = -2/3

8.19 Inverse of a Matrix by Simplex Method LetA be an x n non-singular real matrix. Then to find the inverse of matrix A by simplex method proceed as. follows. Introducing a dummy n x 1 real matrix b, consider the following system of equations. Ax = b, x

0

Now introduce the artificial variables xa > 0 and a dummy objective function Z, with costs zero to variables in x and cost -1 to each artificial variable. Then find the solution of the following L.P.P. formed, by using simplex method : Max.

Z = O. x -1. xa

subject to Ax = b, x, xa > 0. Then in the final simplex table which gives the optimal solution (i.e. when all 0) and the columns of A becomes the columns of unit matrix I (i.e. when A is converted to an unit matrix or when all variables of vector z are in the basis), the inverse of matrix A is the matrix formed by the column vectors which were the column vectors of the initial basis. •

Ai

If in the final simplex table which gives the optimal solution (i.e. when all 0), the matrix A is not converted to an unit matrix (i.e. all variables of vector re

are not in the basis and artificial variable / variables appear in the basis), then we continuohe simplex method by droping the artificial variable and introducing the remaining variable (not in the basis) into the base:. By this operation (iteration), the solution obtained must remain optimal while it can be feasible-ur infeasible. The process is continued till A is converted to an unit matrix. Finally the inverse of A is obtained as above.

Method to find dummy n x 1 real matrix b Although there is no particular method to find dummy n x 1 real matrix b, even then in most of the cases it can be obtained as follows :

Simplex Method

359

b1

=

(Ili j=1.

b= b2 =

a2i

b,1 =

anj

For clear understanding of the method see the following examples. Example 27 : Apply simplex method to find the inverse of the matrix

3 2 4 —1 •

]

[Meerut 2001] Solution : Let A =

5 1 = Y-au = 3 + 2 [3 2] and b = = be a dummy real 4 -1 b2 = Xa 2j = 4— 1 3

column matrix. Consider the system of equations Ax =b i.e.

3x1 + 2 x2 =5 1-3 21 51 i. e. 4x1 - x2 = 3 4 —1_1 x2 — [3

Introducing the artificial variables xal 5a2 and the dummy objective function Z with cost 0 to each variable x1, x2 cost —1 to each artificial variable xai , xa2 , the resulting L.P.P. is Max. s.t.

Z = O.

+ O. x2 - 1.

3x1 + 2x2 + x

- 1. xa., =5

4x1 —

-1-X,-72 = 3 and xi, x2, x0 , xa2 O. Taking x1 = 0, x2 = 0, we get x01 = 5, x0, = 3 which is the starting B.F.S. The solution of the above L.P.P. by simplex method is given in the following table :

Operations Research

360 Table 8.40 .

B Ai A2

ci

0

0

-1

-1

. CB

XB

Yi

Y2

Al

A2

—1

5 3

3

2 —1

1 0

0 1.

5/3 3/4 (Min.)-->

1

0

0

XB /Y:),

-1

Z = C B XB

I 7

n,

T

1

Ai .

-1

11/4

0

11/4

1

-3/4

Yi

0

3/4

1

-1/4

0

1/4

Ai

0

11/4 I

0

0 1

1 0

4/11 1/11

-3/11 2/11

0

0

-1

-1

Z = C B XB = _ 11 4 Y2

0 0

Yi.

1 1

Z = C B XB

Min. Ratio K B /1).

1 (Min.)-> .

-

-7/4

=0 Since in the last table no Ai > 0. Therefore the solution is optimal. The last table in proper order (i.e. form) of unit matrix for written as B

CB

XB

171

Y2

Yi

0 0

1 1

1 0

0 1

11.2

Al

1/11 4/11

172 (i.e. A) can be

A2

2/11 —3/11-

Since the given inatrixA (given by columns Y1, Y2) has been converted to a unit matrix, therefore A -1 is given by the columns A1, A 2 of the initial basis. Hence A =

11/11 2/11]=1 [1 2 4/11 -73/1.1 11 4 -

Example 28 : Apply simplex method to find the inverse of the matrix

[4 3 3 2 11

Simplex Method

361

bi = aiii = 4 + 3 7 r4 3 Solution : Let A =1 ] and b = be a dummy real = L3 2 i b2 = Ert2i = 3 + 2 5 column matrix. Consider the system of equations Ax = b r4 L 3 2 j x2 i

i.e.

5

4x1 + 3x2 = 7 3x1 + 2x2 = 5

e.

Introducing the artificial variables xa , x„ and the dummy objective function Z with cost 0 to each variable xi, x2 cost --1 to each artificial variable x„ , x„,, the i resulting L.P.P. is Max. Z = 0.x1 +0.x2 0. x2 — 1. xal — 1. xa, s.t. 4x1 -F 3x7 + =7 3x1 -F 2x2

+Xe, =5 "2

and

x1, x 2, - cti xa2>—0. Taking x1 = 0, x, = 0, we get x = 7, x = 5, which is the starting B.F.S. 01. a2 The solution of the above L.P.P. by simplex method is given in the following table : Table 8.41 C.1

0

0

—1

—1

Al

A2

Min. Ratio X B /Y1

B

CB

XB

Yi

Y2

Al A2

—1 —1

7 5

4 3

3 _ 2

1 0

0 1

7/4 5/3 (Min.)—>

5

0

0

XB /Y2 1 (Min.)-4

Z = C B XB

a. 7

= — 12 Al Y

0

Z = CBXB

1/3

0

11/3

1

— 4/3

5/3

1

2/3

0

1/3

A I•

0

1/3

0

—7/3

-- 4 3

5/2

i

1

=— —

3 Y2

0

Yi Z = C B XB

1 1

0 1

1 0

3 —2

A•

0

0

—1 - -----==1

=0 Since in the last table no A • > 0, therefore the solution is optimal.

Operations Research

362

The last table in proper order (i.e. form) of unit matrix for Y1, Y2 (i.e. A) can be written as CB

XB

Y1

Y2

Al

A2

Y1

0

1

Y2

0

1

1 0

0 1

-2 3

3 -4

B

Since the given matrixA (given by colimms Y1, Y2) has been converted to a unit matrix, therefore A -1 is given by the columns A1, A2 of the initial basis 3 -2 Hence A 1 = [ 3-4 4 1 21 Example 29 : Apply Simplex method to find the inverse of the matrix- 0 1 0 8 4 5 2 =Eciii =4+1+2 7 41 Solution : Let A = 0 1 0 and b = b2 = Ect2i = 0 + 1 + 0 = 1 be a dummy 17_ 8 4 5 b3 = Ea3j = 8 + 4+ 5

real column matrix. Consider the system of equations Ax = b i.e.

7 fzi 1 21 xi. 0 1 0 x2 = 1 i. c. 17 8 4 5 ,..... 3

4 xi + x2 + 2 x3 = 7 0.x1 + x2 + n x3 = 1 8x1 +4 x2 + 5 x3 =17

Introducing the artificial variables xai , xa2 , xci3 and the dummy objective function Z with cost 0 to each variable x1 , x2, x3, cost -1 to each artificial variable , the resulting L.P.P. is xa , xa 2x„"3 Max.

Z = 0.

s.t.

4x1 + x2 + 2x3 + xai 0.

+ 0. x2 + 0. x3 — 1. !cal — 1. xa2 — 1..xa3

+x2 + 0. x3

X1, X 2, X3,

,X

=1

+X

(1 2

8x1 + 4x2 5x3 and

=7

+Xa a2

3

= 17

X 0. a3

Taking x1 = 0,x2 = 0,x3 = 0, we get xai = 7, xa2 = 1, xa3 = 17, which is the starting B.F.S.

Simplex Method

363

!

The solution Qfthe above L.P.P. bys~rnplexrnethod is given in the foilowing table:

·•

Table 8.42 ~

0

0

Yi

0

-1

-1

-1

Y2

Y3

A1

A2

--A3

2

1 0 0

0

0

1 0

0 1 17/8

0 J,

0

0

1/2

1/4

.,

0

l(Min.)➔

;t..

0 -2

0 1 0

0

0 1

1

3/2

3

1

-3

0

0

Xn!Yj

3

·•

-

B

Cn

XB

Ai A2 A3

-1 -1 -1

i

(TI

1 17

0

1 1

0

-s

4

5

A]-

12

6

7

1/4

Z =C 8 X 8

=-25

i

Yi

0

7/4

1

A2 A3

-1

1

-1

3

0 0

Aj

0

Z=C 8 X 8 =-4 0 0

Yi Y2

-1

A3

Z =C 8 X 8

3/2

l

1

IT] i

Mini. Ratio

XBIYi ➔

X 8 JY2

7

I

"+-

1/2

1

0 0

0 1 0

[TI

-2

-2

0 0 1

Aj

0

0

1

-3

-3

--0

3/4

-1/2

1

0

0

i

=-1

1/4 0

-1/4 1

0

1

1

0

0

5/4

0

1

0

1

0

0

Y3

0

1

0

0

1

-2

-2

1

Aj

0

0

0

-1

-1

-1

'

1_(Min.)-➔

J,

Yi ¥i

Z=C 8 Xn=O

7/4(Min.)

Since in the last table no _Aj > 0, therefore the solution is optimal. Since the gh1en matrix A (given by

Yi, Y2 ,-Y3 )

has been converted to a unit

matrix, therefore A- 1 is given by the columns A1 , A 2 , A 3 of the

..

Hence A- 1 =

[s;; -2

initial basis .

2 l [ 5 ; -11 0] =0 4 -8 -2 1

31

•

364

Operations Research

+

I 1.

2.

3.

4. 5.

'·

Exercise on Chapter 8

+

Write sho1t note on the simplex met.hod. Or Give computational procedure for simplex method in linear programming. What do you understand by optimization? State clearly a L.P.P. and give an account of the simplex method of solving this problem. Make a list of the type of errors which can be made during simplex calculations. Write short note on phase method. Expiain the term 'degeneracy' in the context of L.P.P. (i) What is degeneracy? Discuss a method to resolve degeneracy in L.P.P. (ii) Discuss earner's perturbation method for resolving degeneracy.

two

[Agra 2003]

(iii) Explain degeneracy. How it can harm to us? Explain a method to resolve

degeneracy.

6.

[Meel'ut 2009 (BP)]

If x1 = 1, x 2 = 0, x 3 = l, be a F.S. of the L.P.P. Min. Z = 2x1 + 3x 2 + 4x 3 subject to x1 + x 2 + x 3 = 2, x1 - x 2 + = 0, and x1 , 2 , x 3 :?: 0

x,;

7.

2x1 -

8.

x

then show that the given F.S. is not basic. (a) If x1 = l, x 2 = 2, x 3 = 1, x 4 = 3 be a F.S. tG the set of e_quations. Sx1 -4x 2 + 3x 3 + x 4 = 3 2x1 + x 2 + 5x 3 - 3x 4 = 0 x1 + 6x 2 - 4:: 3 + 2x 4 = 15 then find a B.F.S. (b) Let x1 = 2, x 2 = 4 and x 3 = 1 be a F.S. to the system of .f'quations

x 2 + 2x 3 = 2

x1 + 4x 2 = 18 Then find n-vo B.F. solutions of the problem. (1, 1, 1) is a fe asible solution to the system of equations =2

Reduce the given F.S. to a B.F.S. (2, 1., - 3) is a F.S. of the set of equations 4x1 + 2x 2 6x1 + 4x 2 - Sx 3 = l. Reduce this F.S. to a B.F.S. of the set. 10. x1 = l , x 2 =1, x 3 = 1, x 4 = 0 is a F.S. to the system of equations. Xl +2X2 +4X3 + X4 =7

9.

2x1 -

-

3x 3

= 1,

x 2 + 3x 3 - 2x 4 = 4

Reduce the F.S. to two different B.F. solutions. 11. Consider the system of equations 2x1 - 3x 2 + 4x 3 + 6x 4 = 25

[Meerut LP 1998]

x1 + 2x 2 + 3x 3 - 3x 4 + Sx 5 == 12 If x1 = 2, x 2 = 1, x 3 = 3,..x: 4 = 2, x 5 == l is a F.S. then reduce it to a B.F.S. of the syst.em.

365

Simplex Method 12. Solve the following problems by simplex method. (i) Max. Z = 2x1 + 4x2 + x3 + x4 + 3X2 + X4 *1--- 4 2x1 + X2 5_ 3 X2 -I- 4X3 -I- X4 5. 3 xi, x2, x3, x4 >_0 (ii) Max. Z, = 2x1 + 4x2 + x3 s.t.

s.t.

+ 2x2 5 4 2x1 + x2 _5 3 x, + 4x3 5. 3 xi, x2, x3 > 0 (iii) Max. Z = 3x1 + 5x2 + 4x3 s.t.

2x2 + 3x3 5 18 2x2 + 5x3 5 18 3x1 + 2x2 + 4x3 5_ 25

xpx2, x3, x4 (iv) Max. Z = 2x1 + 4x2 s.t. 2x1 + 3x2 5 48 xi + 3x2 5 42 x1 + x2 5 21 xi, x2 0 (v) Max. Z = 5x1 ± 3x2 s.t. 3x1 + 5x2 5 15 5x1 + 2x2 5. 10 Xi, X2 (vi) Max. Z = 3x1 — x2 s.t. 2x1 + x, 2 + 3x2 5 3 X2 5

4 , x2 .?. 0 (vii) Max. Z = x1 — x2 + 3x3 s.t. xi + x2 + x3 5 10 2x1 — x3 5 2 2x1 2x2 + 3x3 5 0 xi , x2, x3 >_0

(viii) Max. Z = 5x1 + 3x2

s.t.

x1 +x, 5- 2

5x1 + 2x2 5_ 10 3x1 + 8x2 _5 12 xi, x2 0 3 (ix) Max Z= — xi —150x2 + I 50 x3 — 6x4 4 1 s.t. — 4 xi — 60x2 — x3 + 9x4 _5/0 25 • 1 xi — 90x2 — x3 + 3x4 < 0 50 x3 _ 1,x2 >_ 3, x3, x4 0 [Meerut 1998 (BP)]

(ii)

Max. Z = 2x1 + 4x2 + x3 + x4 s.t. x1 + 3x2 + X4 4 2x1 + x2 53 X2 + 4X3 ± X4 5_ 3 xi > 2,x2 >_3,x3 >_2,x4 >_0

[Meerut 1994 (BP)]

Operations Research

368

18. A company produces two types of leather belts, say type A and B. Belt A is a

superior quality and belt B is a lower quality. Profits on the two types of belts are 40 and 30 paise per belt respectively. Each belt of type A requires twice as much time as required by a belt of type B. If all belts were of type B, the company would produce 1000 belts per day. But the supply of leather is sufficient only for 800 belts per day. Belt A requires a fancy buckle and only 400 fancy buckles are available for this, per day. For belt of type B, only 700 buckles are available per day. How should the company manufacture the two types of belts in order to have a maximum over all profit ? 19. A manufacturer of leather belts makes three types of belts A, B and C, which are processed on three machines M1, M2 and M3. Belt A requires 2 hours on machine M1 and 3 hours on machine M3. Belt B requires 3 hours on machine M1, 2 hours on machine M2 and 2 hours on machine M3 and Belt C requires 5 hours on machine M2 and 4 hours on machine M3. There are 8 hours of time per day available on machine M1, 10 hours of time per day available on machine M, and 15 hours of time per day available on machine M3. The profit gained from belt A is Rs. 3.00 per unit, from belt B is Rs. 5.00 per unit from belt C is Rs. 4.00 per unit. What should be the daily production of each type of belts so that the profit is maximum ? 20. Two products A and B are processed on three machines M1, M2 and M3. The processing times per unit, machine availability and profit per unit are as follows : Machine

Processing time (hours)

M2 M3

A 2 3 1

B 3 2 1

Profit per unit

Rs.10

Rs. 12

M1

Available (hours)

1500 1500 1000

Formulate the mathematical model, solve it by using the simplex method and also find the number of hours machine M3 remains unutilised. 21. A manufacturing firm has discontinued production of a certain unprofitable product line. This created considerable excess production capacity. Management is considering to devote this excess capacity to one or more of three products; call them products A, B and C. The available capacity on the machines which might limit output is summarized in the following table. Productivity (hour per unit)

Product —> Machine type1

Milling Machine Lathe Grinder

A 8 4 2

B 2 3 —

C 3 0 1

Available time (hours pm* week) r'

250 150 50

Simplex Method

369

The unit profit would be Rs. 20, Rs. 6 and Rs. 8 respectively for products A, and C. Find how much of each product the firm should produce in order to maximize profit.

B

22. A firm makes two types of furniture : Chairs and tables. The contribution for) each product as calculated by accounting department are Rs. 20 per chair and Rs. 30 per table. Both products are processed on three machines Mi. , M2, M3. The time required by each product and total time available per week on each machine are as follows : Machine Chair Table Available hours 3 M1 3 36 5 2 50 M2 M3 2 6 60 (i) Formulate this problem as a L.P.P. (ii) How should the manufacturer schedule his production in order to maximize contribution ? 23. Solve the following system of simultaneous linear equations by using simplex method. (i)

x1 + X2 =

(ii)

2X1 + X2 = 3 (iii) x1— x 3 + 4x4

2x1 — x2

=3 =3

3x1 — 2x2 — x4 = 1 Xi, X2, X3, X4 0 [Meerut L.P. 1995, 2001]

3x1 + 2x2 = 4 4x1 — x2 = 6

(iv)

3x1 + 2x2 +

x3 + 4x4 S 6

2x1 + x2 + 5x3 + x4 4 xl+

3x2 — 2x3 + 4x4 = 0 xi, x2, x3, x4 0 [Meerut 1997 (BP)]

24. Apply simplex method to find the inverses of the following matrices. 1 21 (i) [332_1 2] 6 6_1 •

Operations Research

370

+ ANSWERS + 7. (a) x1 = 225/139, x 2 = 234/139, 0 Z • = 0 = x2 = x3 = 0,x4 = 228/139 or x1 = 0,x2 = 216/86, x1= 0 = x2, x3 F 5/2, Min. Z= 5/2 = 447/86. x3 = 225/86, x4 • = 213/30, x2 = 0 = X3 = X5, (b) (26/9, 34/9, 0), (0, 9/2, 13/4) x4 = 13/10, x 6 = 2/5, 8. (0, 0: 2) or (2, 2, 0) Min. Z = 213/30 9. (1, 0, 1) (iv) • = 3, x2 = 18, Min. Z = 48 10. (0, 1/2, 3/2, 0), (3, 2, 0, 0) 11. xi = 147/2, x 2 = 0,x3 = 0, (v) x1 =0,x2 =5,Min. Z = 5 x4 = 1/12,x5 = 0 (vi) xi= 0, x2 = 21/5, x3 = 22/5, 12. Min. Z = 43/5 x1 = 1, x2 = 1, x3 = 1/2, x4 =0, (i) Max. Z = 13/2 14. x1 = 2/3,x2 = 5/3,x3 = 1/3, (ii) (i) x1= 3/5,x2 = 6/5, Max. . Z =12/5 Max. Z = 25/3 (ii) xi= 4500/13, x 2 = 2100/13, x1 = 7/3, x2 = 9, x3 = 0, (iii) Min. Z = 2700 Max. Z = 52 15. xi = 6, x2 =12, Max. Z = 60 (iv) (i) xi = 0,x2 = 5, Max. Z = 40 (v) x1 = 20/19, x 2 = 45/19, (ii) x1 = 2, x2 = 0, Max. Z = 6 Max. Z = 235/19 xi = 3,x2 = 0, Max. Z = 9 (iii) Unbounded (vi) 6 Z = = 4, Max. (iv) No. F.S. x1 = 0, x2 = 6, x3 (vii) 16. x1 = 11/2, x2 = 9/2,x3 = 0, (viii) x1 = 2, x2 = 0, Max. Z = 10 Max. Z = 189/2 xi = 1/25, x2 = 0 = x4, x3 = 1, (ix) (i) No. F.S. (ii) No F.S. 17. Max. Z = 1/20 18. 400 of type A, 600 of type B, (x) x1 = - 6/5, x2 = - 6/5, Max. Profit = Rs. 340 Max. Z = -48/5 19. Belt A-89/41, B-50/41, C-62/41, xi = 6, x2 = 0, Z = 126 (xi) Max. Z = Rs. 765/41 (xii) xi = 3, x2 = 0, Max. Z = 21 A = 300, B = 300 units, (xiii) xi = 0 = x2, x3 '4= 1, Max. Z = 3 20. 400 hours (xiv) xi = 50/7, x2 = 0, x3 = 55/7, A - 0 units, B - 50 units C-50 21. x4 = 0, Max. Z = 695/7 units per week, Profit = Rs. 700 (xv) . x1= 2, x2 = 5, x3 = 0, Max. Z = 7 22. 3 chairs, 9 tables (xvi) Unbounded 23. (i) xi = 2,x2 = -1 (xvii) No. F.S. (ii) xi = 16/11, x2 = -2/11 (xviii) xi = 5/3,x2 = 5,x3 = 0, (iii) xi = 5, x2 = 7, x3 = 2, x4 = 0 Max. Z = 30 (iv) x1 = 8/9, x2 = 0, x3 = 4/9, (xix) xi = 9/2,x2 = 3/2,x3 = 0, X4 = 0 Max. Z = 27/2 Unbounded (xx) (011--2 21 0011 6 -11 (xxi) xi = 100/3, x2 = 200/3, x3 = 0 24. 4 3 -1] 61_- 6 2.1 Max. Z = 22000/3 •• •

Duality in Linear Programming

9.1 Duality in L.P. Problem Every linear programming problem is associated with another linear programming problem called the dual of the problem. The original problem is called 'primal' while the other is called its 'dual'. The optimal solution of either problem reveals information concerning the optimal solution of the other. If the optimal solution of either problem (primal or its dual) is known then the optimal solution of the other is also available. This fact is important because sometimes it is easier to solve the dual than the primal.

9.2 Standard Form of the Primal To find the dual of a L.P.P., the primal (L.P.P) should be in the standard form, which are as follows. (i) All the constraints involve the sign < _ if it is a problem of maximization. or

(ii) All the constraints involve the sign if it is a problem of minimization.

9.3 Symmetric Dual Problem A symmetric relation between a primal and its dual problem exists. Consider a L.P. problem. Find xi, x2,..., xn , which Max. Z p = ci xi + c2x2 + +cn xn s. t.

+ anxi

x2 + a 22 X2

4-a2n Xn

anii + ant2x2 + and

xi, x2,

+a1n Xn b2

+ ainn xn

, xn 0

where the signs of all parameters a, b and c's are arbitrary. The dual problem of the above L.P. problem is obtained by (i) Transposing the coefficient matrix. (ii) Interchanging the role of constant terms and the coefficients of the objective function.

Operations Research

372

(iii) Reverting the inequalities and (iv) Minimizing the objective function instead of maximizing it. The dual problem is as follows : , wm , which Find w1) w 2) +bm wm Min. Z D = b1W1 + b2w 2+ s. t.a11W1 a21w2+ anwi.

+and win +ani2wm> C2

a 22W2

... (2)

aln W1 a2n W2 +

and

wl,w2,

+ mn aw — Cn m>

, Wrn

In matrix notation the primal and dual problems can be written as follows.

Primal Problem : Find a column vector x, which Max. s. t. and

Z = c.x Ax < _b x ?. 0, where x = [xi, x2, ..., xn ]

Dual Problem : Find a column vector w, which Min. Z D = b' w A' w>_ c' s.t. and w>0 where w = w2,..., wm ] and A, b', c' are the transposes of A, b and c respectively

9.4 Unsymmetric Dual Problem Primal Problem : Find a column vector x which Max. s.t.

ZP = C X Ax = b

and

x>_0

Dual Problem : Find a column vector w, which Min. Z D = b'w s.t. A' w e' It is important to note that the dual variables are unrestricted in sign.

9.5 The Dual of a Mixed System If a system consists of a mixture of equations, inequalities (in either direction), non-negative variables and unrestricted variables then the dual of the problem can be obtained by reducing the problem to the standard form (See article 9.2). For maximization primal the following procedure is adopted.

Duality in Linear Programming (i)

373

If a constraint has a sign , then multiply both sides by —1 and make the sign ,

(ii) If a constraint has a sign =, i.e., an equation, it is replaced by two constraints involving the inequalities going in opposite directions. For Ex. an equation

au xi = bi is replaced by two constraints (inequalities)

n• x5 bi

and

I ail x j ?. bi

j =1 j =1 4 in maximization L.P.P. The second inequality can be written as

— E au .xi 5_ — bi j=i while in minimization problem it will be replaced by two inequalities. a•• x • >— b1 j =1

and I 11xI• — b • i=1

The second inequality can be written as au x i > — bi

j=-1.

The above order should be noted carefully. (iii) Every unrestricted variable is replaced by the difference of two non-negative variables. Note : The dual variables which correspond to primal equality constraints, must be unrestricted in sign and those associated with the primal inequalities must be non-negative.

gitaishativ.42 bcamplia Example 1 : Write the dual of the problem. Min. Z = 3x1 + x2 s.

2x1 + 3x2 2 x1 + x2 1

and

xi, x2 0

Solution : The given L.P.P. is in the standard primal form. In matrix form the given problem can be written as MM.

Z = (3 1)[x1, x2] = -cx

s.t.

2 31 xi 2] [1 1.][x21- [1

i.e.,

Ax

-The dual of the given problem is

Operations Research

374 Max. s.t. or

Z D = b' w=(2, 1)[w1, w2] = 2 w1 + w2

A' w c' [2 < [3-1 3 1 j [14/2 Lli + w2 5 3

or 3w1

w2 1 w2 °

and Example 2 : Give the dual of the following L.P.P. Max. Z = 2x1 + 3x2 + x3 s.t. 4x1 + 3x2 + x3 = 6 + 2x2 + 5x3 = 4 [Kanpur 1995] x1, x2, x3 0 and Solution : First we shall write the, given problem in the standard primal form as follows, (i) Since it is a maximization problem, all the constraints must involve the sign 5— . (ii) The-first equation (constraint) is equivalent to 4x1 + 3:x2 + x3 5 6 and 4x1+ 3x2 + x3 6 the second can be written as — 4x1 — 3x2 — x3 5 —6 (iii) The second equation (constraint) is equivalent to x1 + 2x2 + 5x3 5 4 and x1 + 2x2 + 5x3 4 the second can be written as — 2x2 — 5x3 5 —4 Thus, the given L.P.P. in the standard primal form is Max. Z = 2x1 + 3x2 + x3 = (2, 3, 1) [x1, x2, x3] = cx s.t. 4x1 + 3x2 + x3 5 6 —4x1 — 3x2 — x3 5 —6 + 2x2 + 5x3 5 4 — xl— 2x2 — 5x3 5 —4 4 3 f —6 —4 —3 —1 or x2 < 4 1 2 5 —1 —2 —5 -X3- - 4 or Ax b and xi, x2, x3 O.

375

Duality in Linear Programming

The dual of the given problem is Min. Z D = b' w = (6, — 6, 4, — wi" , w2' ,w2" 7 = 6w1' — 6w1" + 4w2' — 4w2" s.t. A' w c'

or

[4 —4 1 —1 wi" 3 —3 2 —2 1 —1 5 —5 W2'

2

[3 1

:1-1) 2"

or

4w1' — ' + w2' — w2" 2 3w1' — 3w1" + 2w — 2w2" > 3

w1" + 5 wz' 5w2" 1 and wi" wz' w Writing wi — = wl , and w — w2 = w2, the dual of the given problem is Min. Z D = 6141 4w2 s.t 4w1 + w2 2 3w1 + 2w2 3 wl + 5w2 1 and w1, w2 are both unrestricted in sign. Example 3 : Find the dual of the following L.P.P. Min.

Z =

+ X2 -I- X 3

s.t. x1— 3x2 + 4x3 = 5

— 2x2 3 2x2 — x3 4 x1, x2 0, x3 is unrestricted in sign. [Meerut 1995 (P), Garhwal 1997] Solution : First we shall write the given problem in the standard primal form as

follows. (i) It is a minimization problem, all the constraints must contain the sign >_ . (ii) The variable x3 is unrestricted in sign. We write x3 = x3' — x3" where x3, x3" 0. (iii) The first constraint (equation) is equivalent to xl— 3x2 + 4 (x3' — x3") 5 and — 3x2 + 4 (x3' — x3" ) 5 The second can be written as — + 3x2 — 4x3' + 4x3" — 5

376

Operations Research

(iv) Multiplying second constraint by - 1, we have

+2x 2 +0.(x 3' - xt):2:·- 3

-=- Xi

Thus, the given problem in the standard primal form is

--: x{ = (1,

Min; Z = Xi + x 2 + x 3

1, 1, - l)[xi, x 2 , x 3', x 3"] ·

=CX

s.t.

x1 ,,,

-

3x 2 + 4x 3'

4xt ;::: 5

-

- x1 + 3,x 2 -4x 3' + 4x/' 2:: - 5 -· ..

- x1 +-2x 2 + 0.x 3' - 0.x{

0. xi+ 2x 2

x 3' +x{ 2:: 4

-

l _:3 4-4]

or

r- 1

3. - 4 2 0 . 2 -l

-1

0

3

:e:: -

4 0 1

5 X1

-5

Xz x 3'

-3

X ,, 3

4

:. The dual of the given problem is Max. Zv =b'w=(S, - 5, - 3, 4)[w1',w{,w 2 ,w 3 ]

= Sw/ s.t.

- Sw/' - 3w 2 +4w 3

A'w::; c'

-1 -1

[-!

or

3

2

-4 :...4 4

0 0

or

1

-fl

W2

1

W3

-1

- 3w1' + 3w/' +2w 2 + 2w 3

$

,1

4w{ + 0.w 2

$

1

-

w3

-

- 4w1' + 4wi" + 0. w 2 + w 3

w{, wt

and Writing w1'

s.t.

1

1

-Wz

'-

w1"

$

-1

,w 2, w3 2:: 0.

= w1 , the dual problem is

= Sw1 --: 3w 2 + 4w 3 w1

w2

::;

1

- 3w1 + 2w 2 + 2w 3

$

1

4wl

-

- W3

w2, w3

=1 ;:::

..

$

$

-w1"

~

1

+ 0 W3

Wi'

4w1'

Max. Zv

.

w' i w ,,

0, w1 unrestricted in sign.

.

Duality in Linear Programming

377

9.6 Duality Theorems Here we shall give a number of theorems to describe the relation between the primal and its dual. The relationship between the primal and the dual is extremely useful. The proofs of the theorems are out of scope of this book. For the proof see authors book of Linear Programming. Theorem 1 : Dual of the dual of a given primal, is the primal itself. [Meerut 1995, 1998, 2007, 09] Proof : Consider the L.P. problem Primal :

Max. s. t.

Z p = cl xl + c2x2 +....+c,xn an + ai2 x2 +....+ain xn bi a12x1 + a22x2 +• • ..+a2p xn S b2

ami

+

2x 2 +... .+a inn Xn

and

bm

xl , x2,....,

0.

Dual : The dual of the above primal (1) is given by

MM. s.t.

Z D = b114/1 b2W 2

w rn

an wi + a2114/ 2 +

+ ami wm

anwi a22w2 +.... +am2wm C2 ".(2) + ain

a2n w2

and

amnwm cn

wi,w2,••",wm 0.

Now to write the dual of the dual (2), we first write (2) in the standard primal form (1). In standard primal form the dual (2), can be written as Max.

— b2 W 2 —•

( —ZD) —

s.t.

wm

a21W 2 — • • • • — ant1W m anwi

Cl

a 22W2 — • • • •— ant 2Wm 5- — C 2 -.(3)

aln W1

and

a2n W 2 — • • • • — anin Wm 5- Cn Win 0.

Dual of the dual : Now (3) is of the form (1)

Operations Research

378 Considering (3) as primal its dual is given by Z D1 = — C1 / 2 — C 2V 2 — Min. S. t.

— Cn V n

a12v2 — • • " ain V n

bl

— a21vi — a22v2 — ••.. — a2n vn — b2 ...(4)

amivi am2v2 — • •• amnvn

bm

?_ 0.

and

or The dual (4) of the dual (2) can be written as Z D1 =C1V1 + C 2V 2 + Max.

s. t.

+ CnV n

any]. + a12 v2 + • • .. + ain v n 5 hi an y]. + a22v2 + ... • + a2n V n < b2 —(5)

a mi vi + am2v2 +

+ a mn vn bm

?_ 0.

and which is identical to (1). Hence, the dual of the dual is the primal.

Note : Instead of writing the primal [dual (2)] in the standard primal form (1), the dual of (2) can directly be written as (5). Thus, we conclude that in the sets (1) and (2) if either problem is considered to be the primal, the other is its dual. Miter :,Consider the following L.P.P. in matrix form. Z = cx Max. Primal:

b x> 0

Ax _ 0, then cx b' w i.e., Z p Z D . Theorem 3 : If x is a feasible solution to the primal Max. Z p = cx s. t. Ax b,

x> 0 and w is a feasible solution to its dual problem Min. Z D = b' w s. t. A'w c', w 0, such that c x = b' w then x and w are the optimal solutions of the primal and the dual problems respectively. Or Alternative Statement : The necessary and sufficient condition for any L.P.P.

and its dual to have optimal solution is that both have feasible solutions. Theorem 4 : Basic Duality Theorem : If xo is an optimum solution to the primal, then there exists a feasible solution w 0 to the dual such that cx = b' wo, where b' is the transpose of b. Theorem 5 : Fundamental Duality Theorem : If either the primal or the dual problem has a finite optimal solution, then the other problem also has a finite optimal solution and the values of the two objective functions are equal. Theorem 6 : If the primal problem has an unbounded solution, then the dual has either no solution or an unbounded solution. Theorem 7 : If any of the constraints in the primal ts. a perfect gquality, the corresponding dual variable is unrestricted in sign. Theorem 8 : If any variable of the primal is unrestricted in sign, the corresponding constraint in the dual will be strict equality. Theorem 9 : Complementary Slackness Theorem For the optimal feasible solutions of the primal and dual system. (a) Whatever inequality occurs in the i-th relation of either system, if the corresponding slack or surplus variable is positive, then the i-th variable of its dual is zero.

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380

(b) If the j-th variable is positive in either system, the j-th relation of its dual holds as a strict equality i.e., the corresponding slack or surplus variable w vanishes.

9.7 Correspondence between Primal and Dual The following correspondence rule between the primal and the dual problem hold. Dual

Primal 1. Objective function Max. a. Requirement vector. t 3. %' Coefficient matrix. 4. Constraints with sign. 5. Relation. 6. i-th inequality.

Z p .

Objective function Min. Z D. Price vector. Transpose of the coefficient matrix. Constraints with sign. Variable. i-th variable wi 0.

7.

i-th constraint an equality.

i-th variable wi unrestricted in sign.

8. 9.

Variable. i-th variable xi > 0.

Relation. i-th relation a strict equality.

10. 11. 12. 13.

i-th variable x unrestricted in sign. i-th slack variable positive. i-th variable zero. Finite optimal solution.

14.

Unbounded solution.

i-th constraint a strict equality. i-th variable zero. i-th surplus variable positive. Finite optimal solution with equal optimal value of objective function. No solution or an unbounded solution.

9.8 To read the solution to the Dual from _the Final Simplex Table of the Primal and Vice-Versa It is important to note that the final simplex table of the primal problem also contains, the optimal solution of the dual and vice-versa. For this observe the following rules. 1. The optimal value of primal objective function is equal to the optimal value of the dual objective function. i.e. Max. Z = Min. Z D 2.

The value of the optimal dual variables are the values of Ai with sign changed, under the columns of the corresponding slack variables (if present in the constraints), or under the columns of the corresponding surplus variables (if slack variable not present in the constraint). If no slack variable is present in the constraint then the value of Ai , leaving M, under the column Of the corresponding artificial variable also give the value of the corresponding optimal dual variable. For example : The value of the first optimal dual variable wl is the value of Ai with sign changed under the column corresponding to the slack variable (if present) or surplus variable (if slack variable not present) in the first constraint of the primal. Also the value of A leaving M under the column corresponding

38]

Duality in Linear Programming

3.

to the artificial variable (if slack variable not present) in the first constraint of the primal, is the value of the first optimal dual variable If either problem has unbounded solution, then the other will have no feasible solution or an unbounded solution.

Note : It is always advantageous to apply the simplex method to the problem having lesser number of constraints. Hence, it saves a lot of time, if the problem (primal or dual) with lesser number of constraints is solved by the simplex method then the solution _of the other problem is read from the final simplex table of the first.

gatuttatiol2 ixampfra Example 4 : Apply the simplex method to solve the following L.P.P. Max. Z = 30x1 + 23x 2 + 29x 3 S. t.

6x1 + 5x2 + 3x3 26 4x1 + 2x2 + 5x3 7

and

x1, x2, x3 ->

Also write the dual of the problem and solve. Read the solution of each problem from the final simplex table of the other. Solution : Introducing the slack variables x4, x5 the given problem reduces to, Max. Z = 30x1 + 23x2 + 29)(3 + 0.x4 + 0.x5 s.t. 6x1 + 5x2 + 3x3 + x4 = 26 4x1 + 2x2 + 5x3 + x5 = 7 Taking x1 = 0, x2 = 0,x3 = 0, we have x4 = 26, x5 = 7 which is the starting B.F.S. All computation work is done in the following table : Table 9.1 cj

30

23

29

0

0

MM. Ratio X B / Y]

B

CB

XB

Y1

Y2

Y3

Y4

Y5

Y4

0

Y5

0

26 7

6 4

5 2

3 5

1 0

0 1

13/3 7/4 (Min.)—>

Ai

30 T

23

29

0

0

XB /Y2

31/2 7/4

0 1

21/2

—9/2 5/4

1 0

—3/2 1/4

Ai

0

8 T

—17/2

0

—15/2

Z .C B XB =0 Y4

Yi.

0 30

Z = CBXB=105/2 Y4 Y2

0 23

Z = C B XB =161/2

, 1

17/2 7/2

—4 2

0 1

—19/2 5/2

1 0

— 5/2 1/2

Aj

—16

0

—57/2

0

— 23/2

31/4 7/2 (Min.)-9

Operations Research

382 Since No. Ai > 0, so this solution is optimal. Optimal solution is x1 = 0, x2 = 7/2, x3 = 0, Max. Z = 161/2. and To write the dual of the problem. The given problem can be written as Max. Z = (30, 23, 29) [xi. , x2, x3] s.t.

1-6 5 31 X1 < I-261 and [7 L4 2 5] 2 X3

xi, x2, x3 > 0

the dual of the given problem is given by Min. ZD = (26, 7)[w1, w21 s.t.

Or

30 6 41 w 5 2 [ 1 > 23 and 3 5 w2 29

Min.

Z D = 26w1 + 7w2-

s.t.

6wi + 4w2 30

W2 5-

—(2)

5w1 + 2w2 23

+ 5w2 29 _ w1, w2

and

° Solution of the dual (2) of the given problem (1); The dual problem (2), can be written as Max. s.t.

and

Z.' D = —Z D = —26w1 — 7w2 + 0. w3 + 0. w4 + 0. w5 —Mwai — Mwa2 — Mw„-3 6w1 + 4w2 — w3 5W1

2W2

3w1

5w2

W

= 30

ai

= 23

Wa

— W5

+ wa

3

= 29

W2, W3,W4,WS,Wai ,Wa2,Wa3 0.

Where w3, w4, w5 are surplus variables and wai , wa2 , wa3 are the artificial variables. Now proceeding as usual by simplex method (by Big. M-Method) we get the following simplex table.

Duality in Linear Programming

383

Table 9.2 c• J B

CB WB

Wai -M

30

-26

-7

0

0

0

-M

-M

-M

Min. Ratio

W1

W2

W3

W4

W5

Wai

W 02

Wa3

WB/W1

6

4

-1

0

0

1

0

0

30/6

0

-1

0

0

1

0

23/5(Min.)-)

0

0

-1

0

0

1

29/3 XB / W2

W '22 "M W -M a3

23

5

2

29

3

5

ZD= -82M

A.1

• -26+14M -7+11M -M t

-M

-M

0

0 !..

0

-M 12/5

0

8/5

-1

6/5

0

1

-6/5

0

12/8(Min.)-+

-26 23/5

1

2/5

0

-1/5

0

0

1/5

0

23/2

Wa3 -M 76/5

0

19/5

0

3/5

-1

0

-3/5

1

76/19

-(9M-26) -M 5

0

(-14M+26) 5

0

XH/W3

Wa

1

W1

A

ZD+ 598)/5 W2

i

-7 3/2

Wi -26

4

W M 19/2 a3

127M+17) -M 0 5 ,.. 0

1

-5/8

3/4

0

5/8

-3/4

0

1

0

1/4

'

-1/2

0

-1/4

1/2

0

16

0

0

19/8

-9/4

-1

-19/8

9/4

1

4 (Min.)-

0 (17 + 19M) _ (9M+ 31) -M _(27M+17) (5M+31) 8 4 8 4 t

0 1

)1B/W5

TD= (19M+249) 2

A. J

0

W2 Wi

-7 -26

4 3

0

1

0

3/19 -5/19

0

-3/19

1

0

0

-5/19 2/19

0

5/19

' W3

0

4

0

0

1

-18/19 -8/19

-1

18/19

ZD =-106

of

0 •.1•

0

0

_109 19

19 t

W2

-7 23/2

5/2

1

0

-1/2

0

W5

0 57/2

19/2

0

0

-5/2

W3

0

16

4

0

1

-2

161 2

A., -'

_ 17 2

0

0

-Z 2

7,

_

. n---

"

-M

5/19

-

-

-2/19 57/2(Min.)-4 8/19

-

_m4.10 _34 _17 19 19

0

1/2

0

1

0

5/2

-1

0

-1

2

0

0

0

-M+.72

-M

From the above table the solution of the dual problem is w1 = 0, w2 = 23/2 and Min. Z D = —Z' D = 161/2 To read the solution of dual problem from the final simplex table of the primal and vice-versa.

To read solution of dual problem from the final simplex table of the primal on page 381.

In this table, since x4 and xs are slack variable and A j below the corresponding columns Y4 and Y5 are 0 4 and 0 5,

Operations Research

384 w1.= —A 4 = 0 and w2 = —A 5 = 23/2 Z D = Max. Z = 161/2. and Min.

To read the solution of the primal problem from the final simplex table of the dual on page 383. In this table since w3, w4, w5 are the surplus variables (slack variables are not present in the constraints) and A j below the corresponding columns W3, W4, W5 are A 3/ A 4) A 5* = —A 3 = 0,X2 = —A 4 = 7/2, X3 —A 5 = 0 Max. Z = Min. Z D = 161/2. and Also note that leaving M in A j's below the columns Wai ,Wa2 , Wa3 corresponding to the artificial variables wai , wa2 , w a3 respectively also give the same values of_xi , x2 and x3. ExUmple 5 : Using the dual, solve the following L.P.P. Max. Z p = 3x1 — 2x2 s.t.

x1

4

x2 7200 [UP Tech. MBA 2006-07] and x, y 0 8. Use the simplex method to solve the following problem. Also verify your answers using the dual of the problem. Max. Z = 40x1 + 50x2 s.t. 2x1 + 3x2 3, 8x1 + 4x2 5 and xi, x2 0 9. Solve the following L.P.P. by simplex method. Min. Z = (15/2) xi — 2x2 s.t. 3X1 — X2 X3 3, xi — + X3 + 2 0, — X2 + X3 2 and xi, x2, x3 0 Write the dual of the above problem and solve. Read the solution of each problem from the final simplex table of the other. 10. Write the dual of the following problem and solve it. Max. Z = 4x1 + 2x2 s.t. —x1 — x2 5 —3,— xi + x2 —2 and xi, x2 0. Hence, or otherwise write down the solution of the primal.

7.

Duality in Linear Programming

387

11. Use principle of duality to solve the following L.P.P. Max. Z = 3x1 + 2x2 subject to xi + x2 1, xl+ x2 5_ 7, xi + 2x2 5 10 x2 5. 3 and xi, x2 0 12. Write the dual of the following problem. Max. Z = 5x1 — 2x2 + 3x3 subject to, 2x1 + 2x2 — x3 2, 3x1 — 4x2 5 3, x2 + 2x3 5 5 and xi, x2, x3 0 Hence, or otherwise find the solution of the given problem. 13. A company makes three products X, Y, Z out of three raw-materials A,B, C. The number of units of raw-materials required to produce one unit of the product is as given in the following table :

A B C

X 1 2 2

Y 2 1 5

Z 1 4 1

The unit profit contribution of the products X, Y and Z are Rs. 40, 25 and 50 respectively. The number of units of raw-materials available are 36, 60 and 45 respectively. (i) Determine the products mix that will maximize the total profit. (ii) Write the dual of the problem and through the final simplex table write the solution of the dual problem.

4* ANSWERS 40 1. Min. Z D = 6w1 + 9w2 + 10w3 s.t. 4w1 + 2w2 + w3 ?_ 1 + 3w2 + w3 1 —9w1 + 4w2 — 5w3 4 — 5w2 — 7w3 1 —2w1 + w2 + 11w3 9 and w,,w3 O.

3. Max. Z D = 2w1 — 6w2 + 4w3 s.t. w1 -2w2 +w3 -- CSI w3 is unrestricted.

2. 4. Max. Z D = —714,1 +12w2 +10w3 Max. Z D = 7w1 + 4w2 —10w3 + 3w4 + 2w5 s.t. —3w1 + 2w2 — 4w3 5 1 s.t. 3W1+ 6W 2-7W 3+ W 4+ W 5 3 + 4w2 — 3w3 3 5w1 + w2 + 2w3 — 2w4 + 7w5 -1/3

Z p = C B XB

A .1•

-7/3

0

0

-1/3

T

. - 2/3

The solution given in this table is x1 = 0, x2 = 2/3, x3 = -1/3, x4 = Owhich is infeasible but optimal hence it can be improved further for which we again find the leaving and entering vectors. To determine the leaving vector or ) Since xm. = Min. {xBi , xBi < 0} = Min. IxEn } = -1/3 = xBi r = 1, i. e. , 131 (= Y3 ) is the leaving vector. To determine the entering vector (ak ) A k _ Min.1A j Ak 4 -,, . yli < 0} = Min. { Al -Ai J1j Yll Y14 i Y rk Y1k A

= Min.{ -7/3 -1/3 Min. {7,1} =1 = -1/3 -1/3 Y14 k = 4 i. e. a4 (= Y4) is the entering vector. . Key element is y14 = -1/3 Proceeding as usual the third simplex table is as follows.

Operations Research

392 Table 9.6

B

CB

Y4 Y2

0

-1

Zp = -1

c .1•

-3

—1

0

0

xi, (= xB )

Yi

Y2(132)

Y3

Y4(R1)

1 1

0 1

—3 -1

1 0

-2

0

-1

0

1. 1 A 1•

The solution given in this table is xi = 0,x2 = 1, x3 = 0, (since all xi 0) and also optimal as all Ai 5 0.

= 1 which is feasible

Optimal solution of the given L.P.P. is x1 = 0, x2 = 1, Min. Z = - Max. Z p =1. Example 7 : Solve the following L.P.P. by the Simplex algorithm Min. Z = 3x1 + 2x2 + x3 + 4x4 s. t.

2x1 + 4x2 + 5x3 + x4 z 10 3x1 - x2 + 7x3 - 2x4 2 5x1 + 2.x2 + x3 + 6x4 z 15

and

Xi, X2, X3,

X4 (Meerut (LP) 1995]

Solution : Step I : The given L.P.P. in the standard primal form is Max. 3x1 - 2x2 - x3 — 4x4 s.t. —2x1 -4x2 -5x3 -x4

—4

0

0

0 1

T

=0

0

Step IV : Al = c1 —Z1 = c1 -C B K = -3, A2 =-2,A3 = -1,A 4 = — 4,A5 = 0= A6 = A7. Thus, the starting basic solution is infeasible but optimal. This solution can be improved further.

Step V : To determine the leaving vector (N.) xm. = Min. (xBi , xBi < 0)= Min. (-10, — 2, -15) = -15 = xB3 r = 3 i. e.,13 3 (= Y7 ) is the leaving vector To determine the entering vector (ak ) Since

Alc Min. A I Yrk Y3k Y3j

3j

n

l , 6'2 , A3 , =Min. A } Y31 Y32 Y33 Y34 = Min. f —3 -2 —1 —41 = —3 = Al -5' -2' —1' -61 5 — Y31 k = 1 i. e. ,

(= ) is the entering vector. Key element = y31 = -5 Proceeding as usual the second simplex table is as follows.

Table 9.8 c .1• CB

B

Y5 Y6 Y1

0

0 -3 Z p= CBXB = —9

-3

XB(xB) Yl$3)

-2

—1

-4

Y2

Y3

Y4

0

Y5 (131 ) Y6 0 2 )

0

Y7

7/5 28/5 6/5

1 0 0

0 1 0

-2/5 —> -3/5 -1/5

-2/5 -2/5

0 1

0

-3/5

—16/5 -23/5

7 3

0 0 1

A.

0

-4/5

—4

0

11/5 -32/5 2/5 1/5 i

Operations Research

394

The solution given in this table is xi = 3,x2 = 0 = x3 = x4, x5 = — 4 < 0,x6 = 7 and x7 = 0 which is infeasible and optimal. ... It can be improved further. To determine the leaving vector ((3, ) Since xBr = Min. (xBi , xBi < 0) = MM. (xBi ) = MM. (-4) = xffl r = 1 i.e. , [31 (= Y5 ) is the leaving vector. To determine the entering vector (a k ) 1 A 2 A, Ai A k = A k Min. {A j = : ,Yij < 0 =Min. __ , , ..... Y12 Y13 Y17 i Ylj Y rk Y1k — 4/5 —2/5 —3/5 =Min. {-16/ 5 '-23/5'-2/5 1 2 3 = 2 = 03 = min. { 5' 23' 2 23 y13 ... k = 3, ... a3 (= Y3 ) is the entering vector. ... Key element y13 = — 23/5 Proceeding as usual the third simplex table is as follows. Table 9.9 c.J

—3

—2

—1

—4

0

0

0

XB(xB)

Y1(133)

Y2

Y3(131)

Y4

Y5

Y602)

Y7

0 0 1

16/23 153/23 6/23

1 0 0

—7/23 84/23 29/23

—5/23 —32/23 1/23

0 1 0

2/23 —1/23 —5/23

0

—12/23

0

—12/23

—2/23

0

—13/23

B

CB

Y3 Y6 11.1

—1 ' 20/23 0 289/23 65/23 —3

Z p =CBXB

A.

= --215/23 The solution given in this table is xl= 65/23, x2 = 0, x3 = 20/23, X4 = 0= X5, x6 = 289/23, x7 = 0 which is feasible and optimal (since all Ai 5_ 0) Hence, the Dptimal feasible solution of the given L.P.P. is xl= 65/23, x2 = 0,x3 = 20/23, x4 = 0 and Min. Z = --Max. Z = 215/23. Example 8 : Use dual simplex method to solve the following L.P.P. Min. Z = 6x1 +7 x 2 + 3x3 + 5x 4 Subject to,

5x1 + 6x2 — 3x3 + 4x4 12

and

X2 + 5X3 --6X4 10 2x1 +5x2 +x3 + X4 8 xi, x2, x3, x4 = 0 [Meerut (LP) 1995 (BP), 98 (0), 2006 (B.P.)]

395

Duality in Linear Programming

Solution : Step I : The given L.P.P. in the standard primal form is Max. Zp= - 7x2 - 3x3 - 5x4 s.t. -5x1 -6x2 +3x3 -4x4 -12 -x2 - 5x3 + 6x4 -10 -2x1 — 5X2 — x3 — X4 — 8 and xi, x2, x3, x4 0 Since objective function is of maximization and all ci < 0, we can solve this L.P.P.by dual simplex algorithm. Step II : Introducing the slack variables x5, x6 and x7 the constraints of the above problem reduce to the following equalities = - 12 - 5x1 - 6x2 + 3x3 - 4x4 + x5 + x6 = — 10 —x2 — 5x3 + 6x4 -I x7 = — 8 - 2x1 - 5x2 - X 3 — x4 Taking x1 = 0 = x2 = x3 = x4, we get x5 = -1Z x6 = -10, x7 = -8 which is the starting basic solution to the primal and is infeasible. Step III : The starting simplex table is as follows : Table 9.10

B Y5 Y6 Y7

CB 0 0

0

ZP = 0

c 1•

-6

-7

-3

-5

XB(=xB)

Yl

Y2

Y3

Y4

-12 -10 -8

-5 0 -2

-6 -1 -5

3 -5 -1

-4 6 -1

A .1•

-6

-7

-3

-5

0 1 0 0

0 1 0

0 -p 0 1

0

0

0

1

A 4 = —5,A 5 = °= A6 = A7

Thus, the starting basic solution is infeasible but optimal. Step V : To determine the leaving vector or ). xBr = Min. (xBi , xBi < 0) = Min. (-12, -10, - 8) Since -12 = xffl r = 1,i. e., pi (= Y5 ) is the leaving vector. To determine the entering vector (ak ) Y rk

A k Mini.{ Ylk

j

Aj < 0 = Mini. Al '612 A4 Yll Y12 Y14

Ylj

-6 -7 -5} 7 42 —= = -5 -6 -4 6 Y12

= Mini. {—,

0

Y5 (131 ) Y6 (32 ) Y7 (133 )

Step IV : 01 = c1 - Zl = c1 - CB Yl = —6, A 2 = -7, 43 = -3,

Ak

0

Operations Research

396 k = 2, i.e. , a 2 (= Y2 ) is the entering vector. Key element is y12 = -6 Proceeding as usual the second simplex table is as follows.

Table 9.11 c .1•

-6

-7

-3

-5

0

XB(- X B )

Yi

Y2 oi )

Y3

Y4

Y5

-7 0 0

2 -8 2

5/6 5/6 13/6

1 0 0

Z P = -14

Aj

-1/6

0

B

Y2 Y6 Y7

CB

-1/2 2/3 -1/6 -11/2 20/3 -1/6 -7/2 7/3 -5/6 -13/2 - 1/3 -7/6

0

0

Y6 (R 2 ) Y7 0 3 ) 0 1 0

0 0 -> 1

0

0

1

i

The solution given in this table is x1 = 0 = x3 = x4 = x5, x2 = 2, x6 = -8, x7 = 2, which is infeasible and optimal. It can be improved further. To determine the leaving vector ((3,. ) . Since xm. = Min. (xBi ,xBi < 0) = Min. (xB2 ) = Min. (-8) = -8 = XB2 r = 2, i. e., [3 2 (= Y6 ) is the leaving vector. To determine the entering vector (a k ) . Ak = Ak = Mini. { ,y2i 13 Y 2j Here y 23 = —2/19 < 0, y 24 = 5/19 > 0, we cannot consider y 21 and y 22 as Y1(= ai. --- 132 ), Y2 (= a2 = R1) are in the basis. C4 — Z 4 Y24

< AC B2 < C3 - Z3

Y23

or

—5/19 —16/19 .., A ,.• LIcB2 5. — —2/19 5/19

or

—16/ 5 5 Aci. 5. 5/2 5 — 16/5 5 ci 5 5 + 5/2

or

[... Given ci = 5]

9/5 5. c15. 15/2

which give the variation in c1 without affecting the optimality of the solution. Example 3 : Solve the following L.P.P. Max. Z = -x2 + 3x3 - 2x5 s.t.

x1 + 3X2

- X3 + 2x5 = 7

-2X 2 + 4X 3 + X 4 = 12

— 4x2 + 3x3 + 8x5 + X6 = 10 and

x >_O, j= 1, 2 ,..., 6.

Find the variations of the costs c1, c 2, c 3, c 4, c5 and c6 for which the optimal solution remains optimal. [Meerut 2005 (BP)] Solution : The columns corresponding to the coefficients of x1, x4 and x6 form a unit matrix, so x1, x4, x6 may be taken as the basic variables and thus there is no need to consider artificial variables. Taking x2 = 0, x3 = 0,x5 = 0, we get x1 = 7,x4 = 12,x6 = 10, which is the starting B.F.S. The solution of the problem by simplex method is given in the following table :

405

Sensitivity Analysis Table 10.3 c•

0

—1

3

0

—2

0

Min. Ratio

Y1

Y2

Y3

Y4

Y5

Y6

XB /Y3

7 12 10

1 0 0

3 —2 —4

—1 4 3

0 1 0

2 0 8

0 — 0 12/4 = 3 (min.) 1 10/3

A•

0

—1

3

0

—2

0

1 0 0

1/4 1/4 — 3/4

2 0 8

0 0 1

—2

0

C B (CB ) XB (XB )

B Y1 Y4 Y6

0 0 0

Z = CB. XB = 0 Yi.

0

10

Y3 Y6

3 0

3 1

Z =cB. XB = 9

Y2 Y3 Y6

—1 3 0

Z =cB .XB =11

5/2 0 —1/2 1 —5/2 0

Ai

0 1,

1/2 T

0

— 3/4

4 5 11

2/5 1/5 1

1 0 0

0 1 0

1/10 3/10 —1/2

Ai

—1/5

0

0

—4/5 —12/5 0

XB /Y2 4 (min.) —> — —

4/5 0 2/5 0 10 1

In the last table all Ai _5 0, therefore this solution is optimal. Hence optimal solution of this L.P.P. is = 0, x2 = 4, x3 = 5, x4 = 0, x5 = 0, x6 =11 and Max. Z = 11 Here CB =- (CBri C B2, C B3 ) = (-1, 3, 0) .7-- (C2, C3, C6). To find variations in c1, c4, c5 0 CB . Since for Ck e CB, ACk

Operations Research

410 Optimal solution is xi. = 0, x2 = 4, x3 = 0, Max. Z = 8 Here B = (cc4,a2,a5 ),b = [10, 6, 4] X B = DC4, X 2, X 5 1 = [Xgi , XB2, X B3.1 = [6, 4, 10], [1 0 -1 t 1 B-1 = (31,32,33) = 0 0 0 -1 4

To find variation in b1: After changing b1 to bi + Abi, the new requirement vector is given by b*B = [10 + Abi , 6, 4] From relation (2), article 10.3 the range of Abi consistent with the optimal feasible solution is given by Max. [--xBi --i < Abi < Min. rxBi l Pii< 0 Pit

NI> ° Pa

Here (311 = 1 > 0. Since no Oil < 0 ... there is no upper bound to Ab1 x131 We have Max. [_ _ < AN < . Du Ab i o

Pil

i =1 in

Ak

and Aaik

for

C Bi Rif < 0 i =1

CBi Pi1

1=1

Hence, the range of Aaik (change in aik iB), so that the solution xB remains optimal and feasible, is given by

Ak

m C Bi Rif )> o I If

5- Aa/k

Ak

EC BiRif

... (2)

0, there is no upper bound to Aaik, 1=1 171

and if

c Bi

< 0 there is no lower bound to Aalk.

i Note : There is no change in the value of the objective function if Aaik satisfies (2). Case II : When au, is an element of the optimal basis B. Since aik is an element of the optimal basis B then if au( is changed to alk + Aaik

the optimal basis B will certainly be changed and hence B-1 will also be

changed. Consequently xB = 13-1b and Z3= cB 13-1ai will also change. A change in Z 3may disturb the optimality condition Ai = ci - Zi 0 while a change in xB may disturb the feasibility of the solution. Hence our aim is to find the range of variation of aik so that neither the optimality nor the feasibility of the solution is disturbed. When the element au( e B is changed to all( + Aaik then let optimal basis B, the solution xB and Zi be represented by B*, xB Let

B = (131 b 2, alk = b1 p

and

,b)

, and

Z; respectively.

and ak = b

all( Aaik = b1p + Abip

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416

(0 0 B* = B + 0

• 0

0 0

0 0

0

0 •• •

Abip ••

0

0

0 "l'

0 /

0

/-th row

p-th column where 0 ip is the null matrix except for the (1, p)-th element which equals to unity.

= B + Abi p .01p, = B(I + B-1 Abip .O/p )

B*-1 = {B (/ +13-1Abip .Oip )}-1 = (I + B-1 Abip .0 ip)

-1 . B-1

[.. (A/3)-1 = If B-1 = (131, R2,

. , Rnt), then 0 0 .... 0 0 ....

I + B-1 Ail ip .0 ip = I +B-1

0

0

0

0 0 .... Abip

• 0

0 0

0

1311 1312

Iii

Plm

13 21 1322

R21

13 2m

• =1+

p-th row

R pl Rp2 Rml

Pm 2

13 p1

R pm

ml

13 mm •

1-th column 0 0 .... 0 .... 0 .... 0 0 0 .... 0 0 .... Abip

0 0

... (1)

Sensitivity Analysis or

423

5/27 -5/27

n 56/27 Aa 25/27

or -1 Gall 56/25 Also D = 1 +13piAaik = 1 0, holds as Bpi = R 21 = 0.

...(A)

Again from (10) article 10.4, the range of variation of aik = an for the optimality of the solution is given by Max.

where

-A . ' < A a k < MM.[ Q• 0

Qi =13

plAi

(2)

,B,R ii y pi for all j not in the basis

+

=1, 3

Here Qj = .-B 21— A j+

IC

BiPil Y 2j

i =1

= 1321 A j + (CBI Pll CB2 P21 CB3 P31)Y2j = 0+ (3.5/9+10.0- 5.1/9)y 2j = (10/9)y zi

for all j = 3, 5, 6, 7 not in the basis. Q3 = 10/9. y 23 10/9.2/3 = 20/27 > 0 Q5 = 10/9. y25 = 10/9.0 = 0 Q6 = 10/9. y 26 = 10/9.1/3 = 10/27 > 0 Q7 = 10/9. y 27 = 10/9.0 = 0

Since no Qi < 0 there is no lower bound to au( from (2), we have A, A r, < Ach i - Q3

or or

Q61

< Min. [82/27 80/27 20/27' 10/27 -.< Gan 5_41/10

,..(B)

Since the range for Actii so that the solution remains optimal and feasible is given by (A) and (B) both, Both (A) and (B) are satisfied if -1 < A. an 56/25 Since all 1 limits of variation of all are or

_56/25+1 -1 + 1 an < 0 all 81/25

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424

To find limits of variation of a23 . Here a231 a a3 (Y3 ) which is not in B.

From (2), article 10.4 the range of Aaik (change in alk 0 B) so that the solution remains optimal and feasible is given by

Ak

I13„

I cm pi,)
0

ci3,

(i=4

Ak

Aaik 5_

0

f i=1

I = 2, k = 3

Here aik = a23 = 2

Ak = A 3 = — 82/27, fp_ = 3 (number of constraints) 3

EC Bi Pi1 =i =1C Bi Pi2 = CB1 P12 + CB2 P22 + CB3 P32 i =1 = 3. (— 5/27) + 10.(1/3) + 5(1/27) = 80/27 >-0 There is no upper bound to Aa23 from (3), we have A3 3 C Bi

or

a23 < 00

pi 2 >

i — 82/ 27 80/ 27

Aa23 0, then the solution xB is no more optimal for the new problem and can be improved by introducing a n+1 in the basis. Here, we can start with the last simplex table giving the optimal feasible solution of the original problem by introducing one more column corresponding to the variable xn+1.

glitatudiee Actinide Example 7 : Solve the following L.P.P. Max. Z = 3x1 + 5x2 S. t.

+X3 3x1 + 2x2

=4 +x 4 = 18

and Xi, X2, X3, X4 ?._ 0 If a new variable x5 is introduced in the above L.P.P. with price 7, then we have the following problem : Max. Z'= 3x7 + 5x 2 +7 x 5 s.t. x1 + X3 + X5 = 4 + 2x2 + x4 + 2x-5 =18 and xi, x2, x3, x4, x 5 0 Find the solution of the new L.P.P. Solution : First part: Taking x1 = 0, x2 =p, we ggt x3 = 4, x4 = 18 which is the starting B.F.S. (Here, the-columns of the coefficients of x3, x4 form a unit matrix, therefore ..x and x4 may be taken as-the basic variables). Proieeding as usual the successive simplex tables for the original L.P.P. are as follows : Table 10.8

B Y3 Y4

CB (C. B )

0 0

Z = cB.xB =0 Y3 Y2

0 5

Z = cB . xB = 45

c•

3

5

0

0

MM. Ratio

X B (X B )

Yi

Y2

Y3

Y4

xB /Y2

4 18

1 3

0 2

1 0

0 1

— 18/2 = 9(min.)--)

Ai

3

5

0

0

4 9

1 3/2

0 1

1. 0

0 1/2

Ai

— 9/2

0

0

—5/2

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426

In the last table all A i 0, therefore this solution is optimal. Optimal solution of the given L.P.P. is = 0, x2 = 9, x3 = 4, x4 = 0, Max. Z = 45 From the final table the solution of the dual of the given L.P.P. is w1

= (°, w2 = 5/ 2 Min. Z D = 45 Revised L.P.P. : The dual of the new L.P.P. is same as that the original L.P.P. with one more constraint w1 + 2w2 7 which corresponds to the new variable x5. The optimal solution of the dual of the original L.P.P. is wi = 0, w2 = 5/2 which does not satisfy this constraint. Thus, we see that the optimal solution of the dual of the given L.P.P. is not optimal solution of the dual of the revised L.P.P. Hence, the optimal solution of the given L.P.P. is not optimal for the revised problem and can be improved by introducing C1. 5 = [1, 2], (column of A corresponding to the new variable introduced) in the basis. Thus, consider one more column Y5 in the above table. 11 _1 = 11 Since B LO 1/2 ys =

B-lay =

11 0 1 111 = 111 21 LO 1/.

A =C 5 -CB Y5 = 7 - (0, 5), (1, 1) = 2 We get the following simplex table for revised L.P.P. Table 10.9

B

CB (C13 )

Y3 Y2

0 5

Z' = cB xB = 45 Y5

7 5

Y2 Z' = cB xB = 53

0

0

7

Min. Ratio

Yi

12 -)Y3

Y4

Y5

XB /Y5

4 9

1 3/2

0 1

1 0

0 1/2

1 1

4/1 (min.) --9/1

Ai

-9/2

0

0

-5/2

cJ

3

XB (XB )

5

i

2

I

4 5

1 1/2

0 1

1 -1

0 1/2

1 0

A•

-13/2

0

- 2 -5/2

0

In the last table all Al 0, therefore this solution is optimal. Optimal solution of revised L.P.P. is x1 = 0, x2 = 5,x3 = 0,x4 = 0,x5 = 4 and Max. Z' 53

427

Sensitivity Analysis

10.6 Addition of a New Constraint to the Problem Let a new constraint be introduced to L.P.P. whose optimal solution has been obtained. Here we assume that the additional constraint does not introduce any new variable with non-zero-price. Let Z* be the optimal (maximal) value of the objective function for the new L.P.P. while it was Z for the original problem. Let Z* > Z. Since the new optimal solution satisfies the first m constraints as well as the additional new constraint, therefore, it is also an optimal solution of the original problem which is contradiction to the fact that we already had an optimal solution to the original L.P.P. Here Z* cannot be more than Z. We conclude that Z* 5: Z. (Max.) Thus, we have the following two cases : Case I : If the opcimal solution of the original L.P.P. satisfies the new constraint, it is also an optimal solution of the new L.P.P. In this case the additional constraint is redundant. Case II : If the optimal solution of the original L.P.P. does not satisfy the new constraint, a new optimal solution of the new L.P. problem must be obtained as follows : To find the new Optimal solution of the new enlarged problem. Let B and Bi be the optimal basis of the original and the enlarged L.P. problem respectively. Clearly B1 is the square matrix of order (m + 1) if B is the square matrix of order m. We can write B1 =

B 01 [a +1

The last column of Bi corresponds to the slack, surplus or artificial vector associated with the additional new constraint and a is the row vector of the coefficients, in the new constraint, of the variables which correspond to the vectors in the optimal basis B. Since B-1 exists and is known therefore the inverse of B1 is given by 0 B-1 =[ -T- a13- ±1

_.(2)

Let am +1, i be the coefficient of xi in the new (m + 1)th constraint and ai the column vector of the coefficients of x i in the enlarged problem. Also if 11 and are Y•1 and Z • for the new problem, then we have

Operations Research

428 11

* [ B-1 01[ aj = Bji . a = T-c(B-1 + 1 am+i, Y• +aB_1a j ±a„,±1, j Y.

or

y:* -1

Now

Y.1 Z; = ein /7!= (cB, cB, in + 1 ) [+ a y +a

[-Tot Yi ±

j - ril+1, j]

or

Z; = CB Yj + cB, In i 1 • -CT - a Yj ± am+1, .0

-(3)

(i) If slack or surplus variable is introduced in the additional constraint. In this case c B, ,,,44 = 0 from (3), Z; = cB Yj = Z j c1• - Z! = c .1• - Z • 1.1 C•—e remains unchanged in this case. Since the optimal solution of the original problem does not satisfy the new constraint, therefore the slack or surplus variable introduced in the new constraint is negative. Hence, we can apply the dual simplex algorithm to find an optimal feasible solution of the new problem. (ii) If the artificial variable is introduces in the additional constraint i.e., if the additional constraint is a perfect equality. In this case the additional vector is an artificial vector. Now there are two possibilities : (a) If the artificial variable in the basic solution is negative, then assigning a price to the artificial variable we can use the dual simplex algorithm for the removal of the artificial variable from the basis. (b) If the artificial variable in the basic solution is positive, then assigning a price -M to the artificial variable we can use the standard simplex method for the removal of the artificial variable from the basis. It is important to note that in this case c I• - •Z .1• will be changed. Note : If the addition of new constraint alters the nature of the problem, then the new problem must be solved as a fresh problem.

Sensitivity Analysis

429

gituatitatthj2 example Example 8 : Consider the following table which presents an optimal solution to some

linear programming problem. c.1•

2

4

1

3

2

0

0

0

Y7

Y8

B

CB

XB

Y1

Y2

Y3

Y4

Y5

Y6

Yi

2 4

3 1 7

1 0 0

0 1 0

0 0 1

—1 2 —1

0 1 —2

0.5 0.2 —1 0 5 — 0.3

—1 0.5 2

Aj

0

0

0

—2

0

—2

—2

Y2 Y3

1

Z = cBxB =17

—0.1

If the additional constraint 2x1 + 3x 2 — x 3 + 2x4 — 4x 5 5were annexed to the system, would there be any change in the optimal solution ? Justify your answer. Solution : From the table the optimal solution of the given L.P. problem is

= 3, x2 = 1, X3 = 7, X4 = 0 = X8 =x6 = x7 = x8 which also satisfies the new additional constraint 2x1 + 3x2 — x3 +2~t4 — 4x5 5_ 5. Thus, the optimal solution of the given problem will not be changed if we introduce the above constraint to it. Hence, the additional constraint is redundant and the optimal sol. of the given L.P.P. is also the optimal solution of the new L.P.P. •

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+ Exercise on Chapter 10 4 1. Write short note on Sensitivity Analysis in L.P.P. [Meerut 2007 (BP), 08 (BP), 09, 09 (BP)]

2.

What is sensitivity analysis of linear programming problem and why it is needed ? Explain. [Meerut 2007 (BP)]

3.

Find an optimal solution to the following L.P.P. Max Z = 15x1 + 45x2 s.t.

x2 5 50 x1 + 1.6x2 5. 240

0.5x1 + 2.0x2 5 162 and x1 x2 O. If Max. Z = E ci xi (i = 1, 2) and c2 is kept fixed at 45, find how much can c1 be changed without affecting the above optimal solution. 4. Find an optimal solution to the following L.P.P. problem Max. Z = 3x1 + 5x2 s.t.

xi 5_ 4 x2 5 6

+ 2x2 518 and xi x2 0, what happens to this optimal solution if the objective function is changed to Z = 3x1 + x2. 5. Find an optimal solution to the following L.P.P. Max. Z = 15x1 + 45x2 s.t.

x1+ 16x2 5 240 5x1+

2x2 5 162

x2 5 50, xi x2 0. If max Z = E c j x j , j =1, 2 and c 2 is fixed at 45, determine how much c1 can be changed without affecting the above solution, 6. Solve the following L.P.P. and find how much can c1 be changed without affecting the optimal solution, Max. Z =15x1 + 45x2 = cl x1 + c2x2 s.t.

xl+ 1.6x2 5. 240, 0.5x1 + 2x2 + x3 =162 x2 + x4 = 50

and

X X X X4 0 i, 2, 3,

Sensitivity Analysis

431

7. Solve the L.P.P. Max. Z = 3x1 + 4x2 + x3 + 7x4 subject to 8x1 + 3x2 + 4x3 + x4 7 2x1 + 6x2 + x3 + 5x4 5_ 3 x1 + 4x2 + 5x3 + 2x4 5. 8, xi xz x3 x4 _>_. 0. Discuss the effect of discrete changes in c on the optimality of an optimum basic feasible solution to the above L.P.P. 8. For L.P.P. given in Example 4, determine the effect of discrete changes in c1- (j=1, 2, 3) on the optimal solution. 9.

For L.P.P. given Example 5, find the limits of variation of b2 without affecting the optimality of the solution. [Meerut 2006]

10. Discuss the effect of discrete changes in the requirements (on the right hand

sides of the inequalities) for the L.P.P. given in Q. 7. 11. (i) Solve the problem

Max. Z = 5x1 + 12x2 + 4x3 subject to, x1 + 2x2 + x3 5_ 5 2x1 — x2 + 3x3 = 2 and

x1 x 2 x3 ?_. O.

Also (ii) Discuss the effect of changing the requirement vectorfrom [5 to 2 optimum solution. [5] (iii) Discuss the effect of changing the requirement vector from to 2 optimum solution.

7 on the 2 3] 9

on the

12. Consider the following optimum simplex table for a maximization problem

(with all constraints of 5 type), where x4 is slack and al is an artificial variable. Let a new variable x5 0 be introduced in the problem with a cost 30 assigned to it in the objective function. Also suppose that the coefficients of x5 in the two constraints are 5 and 7 respectively. B

(cB)

Y2

12 5

171

Z = 141/5

XB

Y1

Y2

Y3

8/5 9/5

0 1

1 0

—1/5 7/5

A•

0

0

— 3/5 — 29/5

Y4 2/5 1/5

Al — 1/5 2/5 — M + 2/5

Discuss the effect of this new variable on the optimality of the given problem.

Operations Research

432 13. The following table gives the optimal solution to a L.P.P. Max. Z = 3x1 + 5x2 + 4x3, s.t.

2x1 + 3x2 8 2x2 + 5x3 10, 3x1 + 2x2 + 4x3 S 15. X2 X3 ?:

0

For the above L.P.P. calculate the following : (i) How much c3 and c4 can be increased before the present basic solution will no longer the optimal ? Also, find the change in the value of the objective function if possible. (ii) How much b2 can be changed maintaining the feasibility of the solution ? (iii) Find the limits for the changes in a14 and a24 so that the new solution remains optimal feasible solution. xB

Y1

YT-

Y3

Y4

Y5

Y6

50/41 62/41 89/41

0 0 1

1 0 0

0 1 0

15/41 — 6/41 —2/41.

8/41 5/41 —12/41

—10/41 4/41, i 15/41

A•

0

0

0

—45/41

—24/41

— 11/41

(CB )

B

Y2 Y3

5 4 3

Yi. Z = CB X B

= 765/41 14. Consider the following table which presents an optimal solution to some L.P.P.

B

(CB)

2 3

Y1 Y2 Z=8

ci

2

3

1

0

0

XB

Y11

Y2

Y3

Y4

Y5

1 2

1 0

0 1

0.5 1

4 —1

— 0.5 2

A 1•

0

0

—3

—5

—5

For the above problem assuming that Y4 and Ys were, in that order, in the initial identity matrix basis, calculate the followings : (i) How much can b1 and b2 be increased without effecting the optimality and feasibility of the solution ? (ii) How much c3 can be increased before the present basic solution will no longer be optimal ?

Sensitivity Analysis

433

15. Discuss the effect of adding a new non-negative variable x8 in the Q. 7 on the optimality of its optimum solution. It is given that the coefficient of x8in the constraints of the problem are 2, 7 and 3 respectively, the cost component associated with x8 being 5. Also explain the situation when we have c8 = 10 instead of 5. 16. Consider the L.P.P. Mill. Z = x2 — 3x3 2x5 s.t.

3x2 — x3 + 2x5 —2x2 +4x3 12 4x2 — 3x3 + 8x5 10

and

x 2, x3

0

The optimal tab e is given as follows :

B Y2 Y3 Y6

c•

0

1

—3

0

(CB )

XB

Yl

Y2

Y3

Y4

Y5 '

Y6

—1 3 0

4 5 11

2/5 1/5 1

1 0 0

0 1 0

1/10 3/10 —1/2

4/5 2/5 10

0 0 1

Aj

—1/5

0

0

—4/5

— 12/5

0

Z = cBxg = 11

2

0

Z = — 11

(i) Formulate the dual problem for this primal problem. (ii) What are the optimal values of dual variables ? (iii) How much c5 be decreased before Y5 goes into basis ? (iv) How much can the 7 in first constraint be increased before the basis would change? •

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434

4. ANSWERS + 3.

x1 = 184, x2 = 35, Max.

4.

x1 =

Z = 4335, 45/4 < cl _5 225/8

5.

2, x2 = 6, Max. Z = 36, x1 = 4, x2 = 3, Max. Z* = 15 x1 = 352/13, x2 = 173/13, Max. Z = 1005, 45/16 5 c1 5_ 225/2

6.

x1 = 184,

7.

x1 = 16/19,

x2 = 35, x3 = 0,x4 = 15, Max. Z = 4335, 45/4 5 c1 5 225/8 x2 = 0, x3 = 0, x4 = 5/19, Max. Z = 83/19, 14/5 5 ci 5 56,

-co 0 c' j —

If no > 0, then there is no upper bound to X and if no c' j - Z' 1 < 0, then there is no lower bound to X. Now let ci - Z j

Max. (c'i - z'i )‹ 0

c' j - Z' i

-.0 if no c' I• - Z' • < 0 C1 - Zi

Min. -z'i )> o

and

c' j

Z' i

if no c' i - Z' i > 0 The solution xB will remain optimal for those values of X for which ...(5) Now (i) ifT =

e., if no c' j - Z'1> 0, then the problem has optimal solution

for all X > X. Thus in this case the problem has an optimal solution at X I) also, and hence the problem is over in this case. (ii) If is finite then we want to improve the range of X beyond 7 ;%..

Improvement of the Range of X, When 2: is finite, then suppose that =

ci - Z•c k — Z k ' (c'1 —Z.'1 )> 0 [ c' i - Z' 1 c' k — Z'k Min.

...(6)

If at least one yik > 0, then we can improve the range of beyond 3: in the manner described below. If we imagine

_ 3„--

ck — Z k c' k — Zi k_

then from (2) ck* - Z k* = (ck — Z k ) + X(c'-k —

)=0

i. e., c - Z k* is the only one to vanish and for X > ck* - Z k* = (ck — Z k ) + (c' k Z' k ) = (c' k —

)(

ck — Zk ck—Zk

+x

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438

= (c' k -

k)

>T

+ >0 and

c' k - Z' k > 0

When A, > A„ we can introduce a k in the basis to get a better solution of the problem. Now if all corresponding yik 0 then the problem has no optimal solution for > A. and hence A, cannot be improved, beyond X. Thus, if at least one yik > 0, then we can improve the range of X. beyond A,. Let B1 be the new basis obtained by insertinga k in the basis B in place of outgoing vector a1 (selected by minimum ratio rule) in the usual manner. If x B1 is the corresponding solution and (c. j.* . - Z • *)(1) is (c • * - Z•*) for-the new basis B1, then we have (9* - Zi*)(1) = (c)* Zi*) - (c k* - Zk*). — for all j k Yik

-(7)

and (ck* - Zk*)(1) = 0 Now for A, = -3'0 •

.* 0 .* - z I.*)u) •, .* - z .* ) - (c* - Zk*). — = C .* — Z .1 Y jk

(as c k* - Z k* = 0 for 2. = Thus, the new solution x Bi is also an optimal (maximum) solution for X = 2. Since the vector a / is replaced by the vector a k from the basis we want to be sure that it cannot be further introduced in the basis in the next iteration. The vector a /cannot be further introduced in the basis if (c/* - Z1*)(1) 5_ 0. (c1* - Z1*)(1) = (c1* - Z 1*) - (ck* - Z k*). Yil - ck* Zk*

Ylk

Ylk (as yii = 1 and ci* -- Z 1'

[(Ck -Zk)-FA. (c'k - Z' k ] =1

0)

from (2)

Y lk if (G I* - Z1*)(1) < 0 then

or or

which is true.

-[(ck - Z k )+ Ck - Zk

(ci k - Zi k ] 0

Y lk > °

ck ' - Z k ' [from (6)]

Parametric Linear Programming

439

In this manner, we proceed from one range of values of A to the next until we include X = 4). To prove that this process is valid, we have only to see that no basis is repeated. To prove this let A be replaced by A + r, where E is an arbitrary small positive number. The steps for solving the problem for A + E. would be the same as we have already described. Since we have already assumed that the problem is non-degenerate, so we will either solve the problem for X = A + e or obtain the information that there is no finite optimal for that value of A. Hence, we cannot remain indefinitely (undetermined) at a value ,2%, such that A = X = A. After we have a basis, we cannot return to it or to any basis corresponding to lower values of X (since we have proved that A > A for the new solution). The various A and a A that arise are called the characteristic values of X, and the optimal solution corresponding to the various values of X are called characteristic solutions. Case II : When the problem has no finite optimal solution for A = S.

Since the problem has a basic feasible solution, hence in the procedure of simplex method for A = 8, a vector a k chosen to go into the basis cannot go to the basis because alltiik -O. Since a k is the vector chosen to enter the basis, ck* - Z k* = (ck Z k )+ X (c' k — Z' k )> 0 (1)

Now there are two possibilities : If c' k - Z' k 0 then ck* - Z k* = (ck — Z k ) + X(c'k — Z'k )> 0

,

for all values of A > Z. Hence, we cannot get a finite optimal solution of the problem for any A S. (ii) If c' k - Z' k < 0 then ck* - Z k* = (c k — Z k ) + (c'k - Z'k )> 0 for all X. such that X
co{ ci - Z'i

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440

A, 5. Xi, and thus we can Hence, the characteristic solution holds for adopt the procedure of case I. Again if not all (c3 — Z j ) +~.'1 (c'j — Z;) 0

can be introduced in the basis for the further improvement of the solution. The criterion is used in the following iterations, (i) until all the transformed (c j — Z j ) +

(9' Z'j ) 0 or (ii) until a vector ot p with (c p — Z p ) +

(ep — Z'p ) > 0

and all yip0 is obtained. The condition (i) can be handled by the method of case I. A.'1 and if In condition (ii) if VI; — Z'p 0, no finite optimal solution exist for

c p — Z'p < 0, no finite optimal solution exist for A < A'2 where _ cP— Z P ''' 2

and

c P— Z'p

X12 > Xi •

NoW we can proceed to determine whether a finite solution exists for A = The successive applications cf the above procedure will lead us to a finite optimal solution for some in which we can use case I or will give information that there is no finite optimal solution for A which is case II.

The above discussions may be summarized as follows : We find finite optimal solution for the value A, = S. Then we determine a set of and the corresponding characteristic solution distinct critical values Ai, X, 2, and a series of corresponding basis B1, B2, ..., Bp (Bp may not exist) is determined with the following properties (i) Each basis Bi differs from Bi _ 1 by a single vector if A i is determined by a unique minimum. (ii) The basis Bi is optimal basis for all values of A in the interval p

Xi Xi + 1 i = 1, 2, ..., p — 1, and

(iii) Bp is an optimal basis for every

p if exist any finite optimal solution for A ?_ A

Bp

exists (case I) or there does not

p •

gthatkative bcanipteA Example 1 : For the following L.P.P. Min. Z= — Xx2 — x3 + x4 s. t. 3x1 — 3x2 — x3 + x4 > 5 2x1 — 2x2 + x3 — x4 3

and

xi, x2, x3, x4 0

Find the range of A over which the solution remain basic feasible and optimal. [Meerut 2006]

441

.Parametric Linear Programming

Solution : The given problem can be written as +2x2 +x3 —x4 +O. x 5 +O. x6 —M x7 Max. Z' = —Z = + X, = 5 s.t. 3X1 — 3X2 — X3 + X4 — X5 + X6 =3 2x1 — 2X2 + x3 — x4 and xi, x2, ...., x7 0, where x5, x6 and xa are surplus variable, slack variable and artificial variable respectively and M is a large number. Taking x1 = 0 = X2 = X3 = X4 = X5, x6 = 3, xa = 5 The Starting Simplex Table is as follows. Table 11.1 B

cB*

xi)

Y1

Y2

Y3

Y4

Y5

Y6

A

MM. Ratio xB /Y1

A

—M

5

3

'-3

—1

.1

—1

0

1

5/3

CI

3

2

—2

1

—1

0

1

0

3/2 (Min.)—)

Z'= c*B XB

Cj

A,

1

—1

0

0

—M

c.1 — Zi 3M —3M —M+1 M-1 —M

0

0

0 1

0

Y6

—X

=—5M

c

/

— Z.* {

c'•.1 — Z' •I —1 T

1

ci* — Zi* = (ci — ) +

0

0

0

(ei z'i ) =

= - A, — (—M, 0) . (3, 2) = 3 M — c2* — Z2* = (C2 — Z2 ) -F (c'2 Z'2 ) = C2* — C B*. Y2 =

— (—M, 0). (— 3, — 2) = — 3 M + X

C3* — Z3* = (C3 —Z3 ),+

(C' 3 — Z'3 ) = C3* —

C B* Y3

= 1 — (— M, 0) . (-1, 1) = — M 4.1 c4* — Z4* = (c4 — Z4 ) + (c'4 — Z'4 ) = c4* — CB* Y4 = — 1 — ( — M, 0) . (1, — 1) = M — 1 c 5* — Z 5* = (c 5 Z 5 ) + (C' 5 — Z's ) = C5* — CB* Y5 = 0 — (— M, 0) . (— 1, 0) = — M

Entering Yi in place of Y6 in the basis, the second simplex table is as follows.

Operations Research

442 Table 11.2

Y3

Y4.

Y5

Y6

— 5/2

5/2

— 1 — 3/2

1/2

— 1/2

B

CB*

XB

Y1

Y2

A

—M

1/2

0

0

Y1

—X

3/2

1

—1

ci*

—X

ci —Z j

0

0 — 7-5,M + 1 -M — 1 — M 7 iM

ei — Vi

0

0

Z' = ie*B xB =—

M

C l* —

3k —

1

X

—1

0

1/2

-0

A

Min. Ratio XB/Y4

1

1/5 -->

0 —M

0

2

Zi*

—1/ 2

1/2

0

1/2

0 0

i

C2* — Z2* =(C2 — Z2 ) + (C'2 — Z'2 ) C2* — CB* Y2 = X — (—M, — A.) (0, — 1) = 0 + .0 C3* — Z3* =C3 — CB* Y3 = 1 — (—M, — A„) (— 5/2, 1/2) =1 — 5M/2+1X C4 — Z4 =C4* — CB* Y4 = — 1 — (— M, — X);(5/2, = — 1 + 5M/2 —1 X cs* — Zs* = Cs* — CB* Y5 = 0 — (—M,— X) (-1, 0) = — M + O. X C6*,..— Z 6* = C6* — CB* Y6 = 0 — (— M, — A.) (— 3/2, 1/2) = — 3M/2 + 1 X Entering Y4 in place of A in the basis, the third simplex table is as follows. Table 11.3

B

CB*

XB

171

Y2

Y3

Y4

Y6

Y5

Min.Ratio X B /Y-6

Y4

—1

1/5

0

0

—1

1

— 2/5

— 3/5

Y1

—A.

8/5

1

—1

0

0

—1/5

1/5

A.

1

—1

0

Z'= 4c*BxB ..-

8X+1

ci *

" —A,

0

5

CI* - Z J * 1

c• — Z•

0

0

0

0

— 2/5

— 3/5

C' 1 - ' Z. :1

0

0

0

0

— 1/5

1/5 T

1

— 8-3

Parametric Linear Programming

443

c 2* — Z 2* = 0 = c 3* — Z3* = C4* — Z4* c5* — Z5* = — (2 + X)/5, c6* - Z6* = —3)/5. Since in table 11.3, no c j - Z j > 0, the solution given in table is optimal for =0. For X— 0, the optimal solution is x1 = 8/5, x2 = 0 = x3, x4 = 1/5. From (4) article 11.2, this solution will remain optimal for those values of X for which X 5 ?. < X, where =

Max. - Vi )< 0

X=

Min. - z' j)> 0

C•— Z

' - Max. j —2' i

• — Z• ) - Min. C' J - Z' j C

and

ck — Zk c' k — Zak

[ cs — Zc - 2 c' _ Z. 5 5 C6 — Z6 —3 C' 6 -Z'6

k=6

i.e., the solution given in table 11.3 is optimal for those values of X for which - 2 5_ X 5_ 3. Since X, = = 3 is finite and y 26 = 1/5 > 0, we can try to improve the range of X beyond Xi = 3. When X. > 3, we can introduce Y6 in the basis. By min.-ratio rule Y1 is the outgoing vector. Hence, we introduce Y6 in place of Y1 in the basis, and the next optimal table is as follows : Table 11.4 B

CB*

XB

Yl

Y2

Y3

Y4

Y5

Y6

Y4

—1

5

3

-3

-1

1

-1

0

Y6

0

8

5

-5

0

0

-1

1

ci *

-A.

X.

1

--1

0

0

cj — Z j

3

-3

0

0

-1

0

c' j - Z' 3

_1

1

0

0

0

0

1

c1.* - Z 1 *

for X = 3 X=

Min. (c'1 - z'i)> 0 ck — Z k c'k — Zak

c • — Z• c' • — Z' 1•

-

cn -Z-2 c2 ' — Z2'

-3 -3 1

Operations Research

444

k=2 This correspond to Y2 (= c( 2 ). To improve X beyond 3 we shall have to introduce Y2 in the basis. Since all yi 2 5 0, hence Y2 cannot be introduced into the basis. Hence, no optimal sol. exist for X > 3. Hence , for the solution to remain optimal feasible X is given by — 2 5_ A, 5 3. Example 2 : For the following L.P.P. Max. Z = (3 — 6X)xi + (2 — 2?.)x2 + (5 + 5X)x3 s. t. xl + 2x2 + x3 5_ 430 + 3x1 460 xl+ 4x2 5_ 420 and xi, x2, x3 ?. 0 Find the range of X over which the solution remains basic feasible and optimal. [Meerut L.P. 1994, 95 (BP), 98 (0), 2007 (BP)] Introducing the slack variables x4, x5 and x6, the problem reduces to , Solution : Max. Z = (3 — 6X)x1 + (2 — 2,)x2 + (5 + 5X)x3 + 0.x4 + 0.x5 + 0.x6 s.t. = 430 x1 + 2x2 + x3 + x4 = 460 3x1 + 2x3 + x5 + x6 = 420 xi+ 4x2 and xi, x2, ...., x6 0 Taking xi = 0 = x3, x4 = 430, x5 = 460, x6 = 420, which is the starting B.F.S. The Starting Simplex Table is as follows. Table 11.5 12

Y3

Y4

Y5

VE,

Min. Ratio xB /Y3

1

2

1

1

0

0

430/1

460

3

0

2

0

1

0

460/2 -->

0

420

1

4

0

0

0

1

Z'= clixB

cj*

0

0

0

B

CB*

XB

Y4

0

430

Y5

0

Y6

Y1

3 — 6X 2 — 2X 5 + 5X

=0 c I•* — Z.* 1

I c.—Z.

3

2

5

0

0

0

C'i— Z' i

_6

—2

5 i

0

0 1

0

cl* — Zi* = (ci — ) + X (c'i —Z'1 ) = cl* — cB* = (3 — 6X) — (0, 0, 0) (1, 3, 1) = 3 — 6X Similarly c2* — Z 2* = 2 — 2X, c3* — Z3* = 5 + 5X Entering Y3 in place of Y5 in the basis, the second simplex table is as follows.

Parametric Linear Programming

445

Table 11.6

B

CB*

XB

Y1

Y2

Y3

Y4

Y5

Y6

Min. Ratio xB /Y2

0 Y4 Y3 5 + 5X

200

— 1/2

2

0

1

— 1/2

0

200/2 -->

230

3/2

0

1

0

1/2

0

0

420

1

.4

0

0

0

1

— 420/4

0

0

0

0

- 5/2

0

0

- 5/2

0

Y6

Z'= CB XB *

=1150

3 - 6X 2 - 2X 5 + 5X

x (1 + X)

,

c- Z 1 - 9/2 c3.* - Z3.* / c; - Z i - 27/2

2

0

—2

0

T

J.

ci* — Zi* = (ci — ) + (ci ' — Z1 ' ) = ci* — cB* . Y1 = (3 — 6A,) — (0, 5 + 5 A„, 0) (-1/2, 3/2, 1) = — 9/2 — 27X/2 C2* — Z2* = (C2 — Z2 ) + (C2' — Z2')= C2* — CB* . Y2 = (2 — 2X) — (0, 5 + 5 X, 0) (2, 0, 4) = 2 - 2X cs* - Z5* = (C5 — Z5 ) + (cs' — Zs' )= cs* — CB* Y5 = 0 — (0, 5 + 5X, 0) (-1/2, 1/2, 0) = 5/2 — 5X/2 Entering Y2 in place of Y4 in the basis, the third simplex table is as follows. Table 11.7

B

CB*

XB

Y1

Y2

Y3

Y4

Y5

Y6

Min. Ratio XB /Y4

I 1/2] — 1/4

Y2 2 — 2A,

100

— 1/4

1

0

Y3 5 + 5X

230

3/2

0

1

0

1/2

0

—

0

20

2

0

0

-2

1

1

-

0

0

\. 0

—2

0

Y6

0

Z'= c*BxB = 1350

c I.*

3 - 6X, 2 - 22k, 5 + 5X

+ 950 X

c1 -Z1 CI

— Zi* { c' i — Z' i•

-4

0

0

—1

- 14

0

0

1

1

T

—3.0

200 —>

Operations Research

446 Ci* — Z1* = C1* — CB* Yi = (3 - 6X) - (2 - 2?., 5 + 5?., 0) (1/4, 3/2, 2)

= - 4 - 14 X c4* - Z4* = C4* — CB* Y4 = 0 — (2 - 2X, 5+5X, 0) (1/ 2,, 0, - 2) =-1+X cs* - Z5* = C5* — CB* Y5 = 0 — (2 - 2A., 5 + 5?., 0) (-1/4, 1/2, 1) =-2-3X t,he solution is optimal for X. = 0 Since no ci - Zi > 0, The above solution xB given by table 11.7 will be optimal for the value of X for which,

x= Max. — (c'j - Vd< 0

where

c j - Zj

[

c'i — Z'i

—Max.

ci. - Zi. C'1 —Z'1 '

C5 — Z 5 C' 5 —Z'5

= max. [._ 4 _ 2] = _ 2 14' 3 7 and

X. =

c• - Z • [- —2--i I - Min. [ e4 - Z4 Min. (ej - z' j ) > o c'i - Z' j c'4 - Z'4

= Min. [1] - 1 -

ck - Z k ck ' zk '

k=4 The above solution remain optimal for - 2/7 5 1. Improvement of the range of X. Here 37, = 1 is finite and y14 (i.e.,Y14)> We can try to improve the range of A, beyond X = 1. Taking = = 1. •

When A, > A. (= 1), we can introduce Y4 in the basis.

By min. ratio rule Y2 is the outgoing vector. Entering Y4 in place of 172 in the basis, we get the next optimal table as table 11.6. Since in this table no c', -

•

>0

from (4) of article 11.2 the optimal solution exist for X. > 1.

Hence for the solution to remain optimal feasible the range of X, is given by - 2/7


01 then x3 is the variable which first become infeasible.

We delete Y3 = (0 2 ) column corresponding to (x3 ) from the optimal We apply the dual simplex method to find the feasible solution for 0 > 01. basis Bo. To find the entering vector (ak We have Ak

= L\ k - = Yrk Y2k

• {—, y21 < Y2j

Here we cannot find the entering vector as all the constraint coefficients corresponding to the second constraint (see table 11.8) are 0. From which it follows that the problem has no feasible solution for 0 > 2.3. Hence the required range for 0 is 0 0 5. 2.3.

•

Operations Research

454

+ Exercise on Chapter 11 + 1. Write a short not on parametric linear programming. 2. What are the types of linear variations in a linear programming problem and explain the variations in c vector. [Meerut 2007 (BP)] 3. Consider the parametric linear programming problem. Max. Z = (0 — 1)x1 + x2 s.t. x1 + 2x2 5. 10, 2x1 + x2 11, — 2x2 5 3, xi, x2 O. Perform a complete parametric programming analysis. Identify all critical values of the parameter 0 and all optimum basic solution. 4. Find the largest value of X for which the basic solution (x1 =1, x2 = 2, x3 = x4 = x5 = 0) of the problem Max. Z = 2,x1. + 3x2 + x3 s.t. 2X1 + X3 -V 8X4 — X5 = 2 3x2 + 3x3 — 3x4 + 6x5 = 6 and xi, X2, ...., X5 is optimal if c is replaced with c + X,c', c' (1, Z 1, 0, 0). 5. Solve the following parametric objective function problems for all values of X. Min. Z = 2X + (1 — X)X2 —3x3 +7v X4 -I- 2X5 — 3X X6 s.t. + 3x2 — x3 + 2x5 =7 = 12 —2x2 + 4x3 + x4 + 8x5 + x6 = 10 —4x2 + 3x3 and xi, x2, ...., x6 O. 6. For 0 5 (I) 5_ 3, find the various optimal basic feasible solutions of the following problem. Max. (1 + (h)x1 + (4 — (p)x2, s.t.—x1 + 2x2 4, 3x1 + 2x2 12 xi, x2 0. [Meerut (LP) 1996] 7. Find the set of optimal basic feasible solution of the problem : Max. x1 + (2 + x2 Subject to x1+ 3x2 5 12 x1 + 2x2 5_ 9 5 xi , x2 0 for I) E [0, 4] [Meerut (LP) 1997, 97 (BP)] 8. For 0 5 (131.5 1, find the various optimal basic feasible solution of the following problem. Max. (4 + 41))x1 + (7 — 24))x2 subject to 2x1 + 3x2 12, 2x1 — x2 4 xi, x2 0 [Meerut (LP) 1996 (BP)]

Parametric Linear Programming 9.

455

Given the L.P.P., Max Z = 3x1 + 4x2 s.t. 2x1 + 4x2 5 21 — 0 +x2 6 — 20, xi, x2 O. Determine two optimal solution for 0 = 0, and find the range of 0 > 0 under which the solution remain optimal.

10. The following L.P.P., Max. Z = x1 + 6x2 s.t. x1 < 4 + 80 3x1 + 2x2 18 — 240, xi, x2 >_0 has the optimal solution for 0 = 0 as follows : B

xB

x3 x2

4 9

Y2

Y3

Y4

1

0

1

0

3/2

1

0

1/2

Y1

Find the range of 0 for which the above solution remains basic feasible and optimal assuming 0 0. 11. Consider L.P.P., Max.

Z = 7x1 + 4x2+ 6x3+ 5x4

s.t.

2x1 + x2 + 2x3 + x4 5 10 + 0 2x1 — x2 + 4x3 + 2x4 26 — 0 3x1 + x2 — 2x3 + 3x4 45 — 0 x1, x2, x3, x4 >_0.

Perform a complete parametric programming analysis and identify all the critical values of the parameter 0. [Meerut LP 1995] 12. Find X such that for 0

X, the following L.P.P. has the same optimal basis

Max. 8x1 + x2 s.t. x1 + 3x2 12 + 24) 3x1 + x2 12 — X1, x2 > 0 13. Find X such that for 0 basis :

and

[Meerut 1996 (BP)]

?, the following problem has the same optimal

Max. 4x1 + 7x2 s.t. 2x1 +3x2 512-3 4 2x1 — x2 4 — 2 4) x1, x2 >_0.

[Meerut 1996

(B.P)] •

Operations Research

456

+ ANSWERS + 5.

()

= 0, x2 = 5 if 0

3/2

(ii) x1 = 4, x2 = 3 if 3/2 A, 5 3

7.

(iii)

= 5, x2 = 1 if X>_3

(i)

= 0 = x4 = x5, x2 = 4, x3 = 5, x6 = 11, — 1/27 5_ 4) 2

(ii)

= 0 = x3 = x5, x2 = 7/3, x4 = 50/3, x6 = 58/3, 2 5
_0 and some x are integers. There are two types of the Integer Programming Problems. 1.

2.

All Integer Programming Problem. (All I.P.P.) : An I.P.P. is termed as all I.P.P. (or pure I.P.P.) if all the variables in the optimal solution are restricted to assume non-negative integer values. [Agra 2000, 01] Mixed Integer Programming Problem. (Mixed I.P.P.) : An I.P.P. is termed as mixed I.P.P. if only some variables in the optimal solution are restricted to assume non-negative integer values while the remaining variables are free to take any non-negative values.

12.2 Importance (or need) of I.P.P.

[Meerut L.P. 1998 (0)]

Quite often, in business and industry we require the discrete nature (or values) of the variable involved in many decision-making situations. For example in a factory manufacturing trucks or cars etc. the quantity (or number) manufactured can be whole (discrete) number only as a fraction of truck or car is not required. In assignment problems and travelling salesman problems etc., the variables involved can assume integer values only. In allocation of goods, a shipment must involve a discrete number of trucks etc. In sequencing and routing decisions we require the

Operations Research

458

discrete values of variables. Thus, we come across many integer programming problems and hence need some systematic procedure for obtaining the exact optimal integer solution to such problems.

12.3 Solution of I.P.P. A symmetric procedure for solving an all I.P.P. was first developed by R. E .Gomory in 1958. He also extended the procedure to solve the mixed I.P.P. He derived algorithms to find the optimum solution of the given I.P.P in a finite number of steps, making use of familiar dual simplex method. After this several algorithms came up to solve I.P.P. An efficient method with relatively new approach developed by A. L. Lang and A. P. Doig is "Branch and Bound Technique".

12.4 Gomory's all I.P. Method

[Meerut L.P. 1995 (P) 2007]

In this method, the I.P.P. is first solved by the regular simplex method, ignoring the integer condition of the variables. If all the variables in the optimum solution thus obtained have integer values, the current solution is the desired optimum integer solution, otherwise the considered L.P.P is modified by inserting a new constraint known as "Gontoty's constraint" which reduces some non-integer values of variables to integer one but does not eliminate any feasible integer. Then an optimum solution to this modified I.P.P. is obtained by using standard algorithm. If all the variables in this solution obtained are integers then the optimum solution of the given I.P.P is attained otherwise another Gomory's constraint is inserted in the above L.P.P. and again this new problem is solved to get an integer valued optimum solution.This procedure is repeated interatively until the required integer valued optimum solution is obtained.

12.5 Construction of Gomory's Constraint and Gomory's Cutting Plane The construction of the Gomory's constraint is based on the fact that a solution which satisfies the constraints in the given I.P.P. also satisifies any other constraint derived by adding or subtracting two or more constraints or multiplying a constraint by a non-zero number. Now first we introduced two notions as follows. [a] = largest integral part of number a, i. e., the greatest integer less than or equal to a, and f = positive fractional part of number a, thus, we have a = [a] + f, clearly 0 f < 1. For example (0 if a = 4 3, then [a] = 4 and f = , so that 41= 4 + 1 3 3 3 and (ii) if a = — 43, then [ a] = — 5 and f =

3

so that — 41 = — 5 + 1 3. 3

459

Integer Programming

Now we proceed for the construction of the Gomory constraint, as follows. Let the optimum solution of the maximization L.P.P. (ignoring the condition of integer values of the variables) obtained by simplex method be expressed by the following table : Note that in this table the basic variables xffl, XB2. , XBm are arranged in order for convenience. Table 12.1 +1

••

cn

IT, (pi ) 1'2 (13 2 ) ... Yi (iii ) ... I'm (i3 n., ) Y. + 1

...

Yn

C 3• • C1 C2 ... Ci

...

Cm

Cm

B

CB

XB

Yl

c ffl

xffl

1

0

0

Yi, in + 1

•• •

Yln

Y2

CB2

XB2

0

1

0

Y 2, m + 1

•• •

Y 2ti

•

•

.

•

.

•

.

.

•

•

•

•

Yi

C Bi

X Bi

0

.0

...

1

...

0

Yi, m + 1

•• •

Yin

.

.

.

.

.

.

.

.

•

•

•

•

Ym

C Bm

XBm

0

0

1

Ym, m +1

•• •

Ymn

Xi

XIII

XB 2

XBm

0

...

0

.. .

X Bi

...

Let the i-th basic variable xBi be non-integer. Note that 1 i

in

Using i-th row of the above table 12.1, we have

.

xBi = O. xi 1O. x2 + = xi +

+ 1.

+

+ O.

+Yi, m + l X m+ 1 + +yin X„

yu xi j

=m+ Xi

Xi = XBi — j= +1

Let

xBi = [xffl ]+ fBi and yu = [yu] +

where

[xBi = Largest integral part of xBi , i.e., [xBi ] < xBi

and

[yu ] = Largest integral part of yu , i.e. ,[y y ] < yy fBi = positive fractional part of xBi , i.e., 0

and

fBi < 1

4 = positive fractional part of yu , i.e., 0 _c 4 < 1 Clearly [xBi xBi , [yu yip 0 5_ f Bi < 1 and 0 4 < 1 Thus, from (1), we have

Operations Research

460 xi

{[yo-Ffii}x;

= {[xBi ] fBi } — j =m

+1 n

n

or

.fsi —

./;)xj =

— [x/31] +

(2)

[yy] XJ j = m +1

j = tn +1

Now if the variables xi (i = 1, 2,...., m) and xi (j = m + 1, integers then the R.H.S of (2) is an integer and hence the L.H.S. fBi —

n) are all

E j=

fij

xi

+1

of (2) must also be an integer. • is positive; fi•x 1.1 I

Since

fij X j fBi

A J•

1

1

0

0

X B /Y3

B

Z =0

i

1

Y3

0

1

3

0

1

—2

1/3 (Min.) --->

12

1

2

0

1

0

—1

—

A/

1 'i‘

0

0 1

—1

Z=2

Yi

1

1/3

1

0

1/3

Y2

1

2

0

1

0

1

A I•

0

0

—1/3

—1/3

Z = 7/3

—2/3 '

Since A 5_ 0,At- j , so the optimal solution ignoring the integer condition is xl=

1/3, x2 = 2.

Max. Z = 7/3 Step 2 : In the above optimum solution, xi = 1/3 fractional, while the condition is that x1 is integer. So We proceed to the next step. Step 3 : It is a problem of mixed integer programming, so the Gomory's constraint is obtained by the formula. —

fBi

.Y1•.1• x1•

fat

E

1)

yi j x j — ,63; R_

Here xi = X Bi = X Bi = [X131 ] + fin = 1/3 = 0 + V3 f B1 = 1/3, i = 1 R+ = -(j= 3 : yi3 = 1/ 3 > 01, R_ = fj = 4 : yi4 =-2/3

0 1

Here xm = - 1/3 < 0, the solution given by the above table is not feasible. we proceed by using dual simplex algorithm.' r=3 Taking leaving vector as y i. e., xBr. = x B3 To determine the entering vector (a. k ) . A k , Ak

Here

=Min. { A .' Y3k i Y3i

Yrk

,

y

A• = mi n. { , A4 1 -_.= Y33 Y34

j< 0 3

in. — 1/3 -V3 =Min. {1.11 - 1/3' - 1/3 1

k = 3 or 4 Taking k = 4 i. e. , taking Y4 (= cc4 ) as the entering vector, and Ym as the leaving vector, the revised simplex table is Table 12.15

c1

1

1

0

0

0 YG1

B

GB

XB

Y1

Y2

Y3

Y4

Y1

1

1

1

0

1

0

—2

Y2

, 1 0

1

0

1

—1

0

3

1

0

0

1

1

—3

Ai

0

0

0

0

-1

Y4

Z =2

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478

so the solution is optimal which is also feasible as Here A• 5 xi 0, Vj • Thus, the optimal feasible solution is = 1, x2 = 1 and Max. Z = 2 Note Taking entering vector Y3 and leaving vector YGI, an alternate optimum solution is I = 0, x2 = 2 and Max. Z

2.

Example 4 : Solve the following mixed-integer programming problem, using Gomory's cutting plane method. Max. Z axi + x2 + 3x3 subject to the constraints - xt + 2x2 + x3 5 4, 4x2 - 3x3 5, 2, x1 - 3x2 + 2x3 5. 3, and xi, x2, x3 > Q where x1 and x3 are integers. Solution : Step 1 : Introducing the slack variables x4, xs and x6, the given L.P.P. into the standard maximization form is Max. Z = 3x1 + x2 + 3x3 -x1 + 2x2 + x3 + x4 s.t. =4 + x5 =2 4x2 - 3x3 -+ x6 = 3 x1 - 3x2 + 2x3 x1, x,, x3, x4, x5, x6 0 Ignoring the integer condition for variables xi and x3, and proceeding as usual, the final simplex table giving the optimum solution is as follows. Table 12.16 c•j

3

1

3

0

0

0 Y6

B

CB

XB

Y1

Y2

Y3

1'4

Y5

Y3

3

10/3

0

0

1

4/9

1/9

4/9

Y2

1

3

0

1

0

1/3

1/3

1/3

Yi

3

16/3

1

0

0

1/9

7/9

10/9

A•

0

0

0

-2

-3

-5

Z=29

j, so this solution, ignoring integer condition is Ai 5 0, xi = 16/3= 5 + 1/3, x2 = 3, x3 = 10/ 3= 3 + 1/3 and Max. Z = 29 Step 2 : Herexi and x3 are not integers, as required, so we proceed to the next step. Step 3 : Here x1 = 16/3= 5 + 1/3, x3 =10/ 3 =3 + 1/ 3 Since fractional parts in both x1 and x3 are equal, 1/3, 1/3. So we first consider x3 to change its value to integral value

Integer Programming

479

.6. X3 = XBi =X.81 = 1-°/ 3 = 3 + 1/ 3 = ()m + fin The Gomory's constraints is given by -

fin f ffi -

Yijxj R,

j

= 1, fm =1/3

2,, Ylixi

for i

1, i.e., in the first row in the above table, for non basic variables. R+ lj = 4, 5, 6 : yii > 01, R_ ={ j: yii < 0} = (i) i.e., Y14 = 4/ 9, 3/15 =1/9, y16 = 4/9 > 0 Gomory's constraint is. C1/31 1/3 - y14x4 y15x5 - Y16x6 /3 - 1 or - (4/9) x4 - (1/9) x5 - (4/9) x6 - 1/3 Introducing the slack variable xo., the Gommys constraint equation (i. e., Gomory's cutting plane) is - (4/ 9)x4 - (1/ 9)x5 - (4/9)x6 xed = - 1/3 Step 4 : Adding the Gomory's constraint equation at the bottom of the optimum simplex table given in step 1, we get the following table. Table 12.17 0 0 11 ci 3 1 3 0 0 I's

Y6

4/9

1/9

4/9

0

0

1/3

1/3

1/3

0

0

0

1/9

7/9

10/9

0

0

0

0 ' - 4/ 9

-} 1/9

- 4/9

1 -->

0

0

0

-3

-5

B

CB

XB

Yi.

Y2

Y3

Y3

3

10/3

0

0

1

Y2

1

3

0

1

Y1

3

16/3

1

YG1

0

- 1/3 A J-

Z = 29

l'4.

-2

1 YG1(1 4)

0

I

Since xm = -1/3 < 0, the solution given by the above table is not feasible. Hence, we proceed by using dual simplex algorithm. Taking leavingyector YGi i.e. , -Br = X84 6% r = 4 Yk (= "0 is the entering vector, where k

Ak

Y rk Y 4k

min.

Min.

H1

< 0}

Y 4j

A 4 A5 6 {Y44 Y45 Y461

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480

5 1 — Min.19/2, 5 - 27, 45/41= — 9 = 6'4 = Min. { — , _429 — 1/9 — 4/9J 2 Y44

k= 4 i. e. , Y4 is the entering vector and so y 44 is the key element. Now removing the vector YG1 and entering the vector Y4 in the basis, the revised simplex table is Table 12.18 c 1•

3

1

3

0

0

0

0

XB

Yi

Y2

Y3

Y4

Y5

Y6

Y G1

B

CB

n

3

3

0

0

1

0

0

0

1

Y2

1

11/4

0

1

0

0

1/4

0

3/4

li

3

21/4

1

0

0

0

3/4

1

1/4

Y4

0

3/4

0

0

0

1

1/4

1

— 9/4

A 1•

0

0

0

0

— 5/2

—3

— 9/2

Z= 43/2

Ati so this solution is optimal with x3 as integer. But xi. is A < 0" not integer, as required, so we proceed to the next step. Steps: ••• xi = xpi = XB3 = 21/4 = 5 + 1/4= [x1331 fB3 .• i = 3and fB3 = 1/4 So we construct second Gomory's constraint by using the formula y 3) xj —

fB3

E y 3j Xi 0}, R_

i.e.,

{j : y 3j
0 Second Gomory's constraint is, —Y35x5 Y36X6 Y 37 XGI

or

( 1/4 1/4 — 1

0 — 1/4

— (3/4)x5 — 1. x6 — (1/4)xm — 1/4

Introducing the slack variable xG2, the second Gomory's constraint equation (i. e., Gomory's cutting plane) is — (3/4)x5 — 1. x6 — (1/4)XG1 + XG2 = — 114 Step 6 : Adding the second Gomory's constraint equation at the bottom of the above optimal table, we get, the following table :

Integer Programming

.481 Table 12.19 L

c1

3

1

3

0

0

0

0

XB

Yi

Y2

n

Y4

Y5

Y6

YG1

0

B

CB

Y3

3

3

0

0

1

0

0

0

1

0

Y2

1

11/4

0

1

0

0

1/4

0

3/4

0

Y1

3

21/4

1

0

0

0

3/4

1

1/4

0

Y4

0

3/4

0

0

0

1

1/4

1

Y G2

0

-1/4

0

0

0

0

-3/4

A1

0

0

0

0

-5/2

Z = 43/2

1 YG2(13 5)

- 9/4 L

0

-1

-1/4

1-4

-3

-9/2 I

0

1‘

I

Since xG2 = —114 < 0, the solution given in the above table is not feasible. Hence we proceed by using dual simplex algorithm. Taking leaving vector y G2 i.e., xBi. = xB5 ...r = 5 Yk = (ak ) is the entering vector, where 0• A k = '6 ` k = Min. {-A5 A6 A 7 , Y S ' < 0} =Min. {— Y5j Y55 Y56 Y57 y1-1c Y 5k = Min.

A {9/21 — {10/3, 3, 18} = 3 =A - 3/4 - 1 - V4 Y56

... k = 6 i.e., Y6 is the entering and so y 56 is the key element. Now removing the vector YG2 and entering vector Y6 in the basis, the revised simplex table is Table 12.20 c 1•

3

1

3

0

0

0

0

0

XB

Yi

Y2

Y3

Y4,

Y5

y6

ym.

Y G2

B

GB

Y3

3

3

0

0

1

0

0

0

1

0

Y2

1

11/4

0

1

0

0

1/4

0

3/4

0

li

3

5

1

0

0

0

0

0

0

1

Y4

0

1/2

0

0

0

1

- 1/2

0

- 5/2

1

Y6

0

1/4

0

0

0

0

3/4

1

1/4

-1

A.1

0

0

0

0

- 1/4

0- 15/4

-3

Z= 107/4 =26.75

Operations Research

482

Since A j < 0, -"9`j so the solution is optimum with x1 and x3 having integral values. Hence, the required optimum solution is xl = 5, x2 = 11/4= 2.75, x3 = 3 and Max. Z = 26.75

Exercise 12.2 + Find the optimum solution to the following mixed-inter programming problems, by Gomory's cutting plane method. 2. Max. Z = 7x1 + 9x2

1. Max. Z = 3x1 + 4x3

subject to the constraints

subject to the constants

— + 3x2 5_ 6

3x1 — x2 12 3x1 +

7 xi + x2 35

66

x1, x2 0 and x2 is an integer. 3.

x1, x2 0 and x1 is an integer.

Max. Z = 4x1 + 6x2 + 2x3 subject to the constraints 4x1 — 4x2 5,

+ 6x2 5

— x1 + x2 + x3 5, x1, x2, x3 0, x1 and x3 are integers.

+ ANSWERS + 1.

x1 = 0, x2 =2orx1 = 1, x2 = 4, Z = 6

2.

x1 = 4, x2 = 10/3, Z = 58

3.

x1= Z x2 = 1, x3 = 6, Z = 26

12.8 The Branch-and-Bound Technique [Meerut 2006 (BP), 08; Agra 1999]

This technique is applicable to both all integer programming problems as well as to mixed integer programming problems. This is the most general technique for the solution of an I.P.P. i,n which only a few or all the variables are constrained by their upper or lower bounds, or by both. The technique, called the Branch-and-Bound technique, for a maximization problem is discussed below. Let the given I.P.P. be as follows : 11

Max. Z = j =1

•• cix1

...(1)

subject to the constraints aijxj j =1

bi , i = 1, 2, ...., in

...(2)

Integer Programming

483

xi is integral valued for j = 1, 2, ...., r n and x • > 0' for j = r + 1, r = 2, ...., n

(3) ... (4)

Also let there exist lower and upper bounds for the optimum values of each integef valued variable xi such that Li < xi

3 , j= 1, 2, ...., r.

-.(5)

Thus, any optimum solution of (1) to (5) must satisfy only one of the constraints xi [xi ]

and

xi > _[x3 ]+1

(6)

—(7)

Thus, ignoring the integer restriction (3) if xi * is the value of the variable xt in the optimum solution of the above L.P.P. given by (1) to (5),then in an integer valued solution we have either or

Lt < xt < [xt_*]

...(8)

[xt*] + 1 x t < Ut

—(9)

For example if xi. = 3.5 (ignoring integer constraint) then in integer valued solution either 3 or 4 Thus, the given I.P.P. given by (1) to (5) has two sub-I.P. problems : (i) given by (1), (2), (3), (4) and (8)

and (ii) given by (1), (2) (3), (4) and (9) In the above two sub-problems constraint (5) is modified only for xt (i. e., for xj , j= t) Now solve these two sub I.P. problems. If the two problems posses integer valued solution then the solution having the larger value of Z is taken as the optimum solution of the given problem. If either of these sub-problems does not have an integer valued solution then sub-divide this again into two sub-problems and proceed similarly till an optimum integral valued solution is obtained.

12.9 Branch-and-Bound Algorithm The systematic step by step solution of an I.P.P. by Branch-and-Bound technique is as follows : Step 1 : Solve the given I.P.P. ignoring the integer, valued constraint. Step 2 : Test the integrability of the optimum solution obtained in step 1. Now

there are two possibilities. (i) The optimum solution is integral valued then the required solution is obtained. (ii) The optimum solution is not integral valued then proceed to the next step 3.

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484

Step 3 : If the optimal value xt* of the variable xt is fractional then form two sub-

problems. Sub-problem 2 Sub-problem 1 Given problem with one Given problem with one more constraint more constraint xt [xt*] + 1 xt 5 [x t*] • Step 4 : Solve the two sub-problems 1 and 2 obtained in step 1. Now there are three possibilities. (i) If the optimal solutions of the two sub-problems are integral valued then the required solution is that which given large value of Z. If the optimal solution of one sub-problem is integer value and the (ii) other sub-problem has no feasible optimal solution, then the required solution is that of the sub-problem having integer valued solution. (iii) If the optimal solution of one sub-problem is integer valued while that of the other sub-problem is fractional valued then record the integer valued solution and repeat step 3 and 4 for the fractional valued sub-problem. Continue step 3 and 4 iteratively, till all integral valued solutions are recorded. From all the recorded integral valued solutions choose that solution Step 5 : which given the largest value of Z. This is the required optimal solution of the problem.

gituthwth'im bcample Example 5 : Use Branch-and-Bound technique to solve the following problem.

Max.

Z

= 7x1 9x2

Subject to — x1 + 3x 2 6 7x1 + x2 35 OS xi, x2 5_ 7 x1, x2 are integers.

[Agra 1998]

Solution : Step 1 : The given problem ignoring the integer value constraint can be written as

Max. s.t.

Z = 7x1 + 9x2 — x1 + 3x2 5 6 7x1 + x2 5 35 xl 7 x2 0 xi, x2 0 Step 4 : Solving the above sub-problems by graphical method (see fig. on page 486) we get the optimal solutions as follows. Sub-Problem 1. xl = 4, x2 = 10/ 3 = 33, Max. Z = 58 Sub-Problem 2.

x1 = 5, x2 = 0 and Max. Z = 35

which has integral values. Since the solution of Sub-Problem 1 is not integral as x2 = 31 3, i.e., [x2] = 3 we sub-divide sub-problem 1 into the following two sub-problems. Sub-Problem 3 Sub-Problem 4 Max. Z = 7x1 + 9x2 Max. Z = 7x1 + 9x2 s.t. — xl+ 3x2 5 6 s.t. + 3x2 5 6 7x1 + x2 5 35 7x1 + x2 5 35 x1 5 4 4 x2 >_4 x2 5_3 xi, x2 0 xi, x2 ?_ 0 Solving the above problems by graphical method (see the figure on page 486) we get the optimal solution as follows.

Operations Research

486 Sub-Problem 3. x1 = 4, x2 = 3, Max. Z = 55 Which is integral valued solution. Sub-problem 4. No-feasible solution.

Step 5 : In the solutions of the sub-problems we get the following integer values solutions. = 5, x2 = 0, Max. Z = 35 and (ii) = 4, x2 = 3, max. Z = 55 (i) larger of these two values of Z is 55. Hence, the required optimal solution is = 4, x2 = 4, Max. Z = 55 The entire procedure (Branch and Bound) is given in the following figure. Give'i Prob. = 9/2 = 4.5 x2 = 7/2 = 3.5 Max. Z = 63 5

x1 5_ 4

Sub- Problem 2

Sub- Problem 1 v

= 5x2 = 0 Max. Z = 35

=4 x2 = 10/3= 33 Max. Z = 58 x2 5_3 Sub-Problem 3 v

= 4, x2 = 3 Max. Z = 55

4 v Sub- Problem 4

No F.S.

487

Integer Programming

+ Exercise 12.3 + 1. Describe Branch and Bound Technique to solve an Integer Programming Problem. [Meerut 2006 (BP), 08 (BP)] 2. Write short note Branch and Bound Technique. [Meerut 2008] Use Branch-and-Bound technique to solve the following I.P.P.

8. 3. Max. Z = x1 + x2 s.t. 3x1 + 2x2 5 12 x2 5__2 xi, x2 0 and are integers. 9. 4. Max. Z = x1 + x2 s.t. 4x1 — x2 5 10 + 5x2 5 10 4x1 — 3x2 _5 6 x1, x2 = 0, 1, 2, 10. 5. Max. Z = 7x1 + 9x2 subject to the constraints — + 3x2 5 6 7x1 + x2 _5 36 x2 5 7 11. xi , x2 0 and x1, x2 are integers. [Agra 2002] 6. Min. Z = 4x1 + 3x2 subject to the constraints 5x1 + 3x2 30 xi 5_ 4, x2 5 6 12. x1, x2 0 and are integers. 7. Max. Z = x1 + x2 s.t. + 7x2 5 28 14x1 + 4x2 5 63 xi , x2 0 and integers.

Max. s.t.

= 2x1 + x2 x1+ 7x2 5. 28 14x1 + 4x2 5 63 x1, x2 0 and integers.

Z

[Meerut (L.P.) 1997]

Max . Z = 2x1 + 3x2 s.t. — 3x1 + 7x2 5 14 7x1 — 3x2 514 x1, x2 0 and integers. [Meerut 1996 (BP)]

Max. s.t.

= 2x1 + x2 x1 5 3/2 x2 5 3/2 x1, x2 0 integers Z

[Meerut 1996]

Max. Z = 3x1 + 3x2 13x3 s.t. — 3x1 -+ 6x2 + 7x3 5 8 5x1 — 3x2 + 7x3 5 8 0 5_ x j 55 and all x • are integers. for j = 1, 2, 3. Max. Z = 3x1 + 4x2 subject to the constraints 3x1 — x2 + x3 =12 3x1 + 11x2 + x4 = 66 xi 0, j = 1, 2, 3, 4

[Meerut (L.P.) 1997 (BP), 98 (P)]

+ ANSWERS + xi = 2, x2 =2;x1 = 3, x2 =1;x1 = 4, x2 = xi = 2, xi = 1, Z =3 xi = 4, x2 = 3, Z = 55 xi = 3, x2 = 5, Z = 27 7. xi = 0, x2 = 0, x3 = 1, Z = 13 12. xi = 5, x2 = 4, x3 = x4 = 0, Z = 31

.3. 4. 5. 6.

0, Z = 4

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488

+ Miscellaneous Exercise on Chapter 12 + 1. Explain all integer programming problem. Describe Gomory's constraint and [Agra 2000, 2001] explain its use in the solution of the problem. 2. Consider the problem Maximize, Z = 2x1 + 20x2 -10x3 subject to the constraints : 2x1 + 20x2 + 4x3 15, 6x1 + 20x2 + 4x3 20 x1, x2, x3 0 and are integers. Solve the problem as a (continuous) linear problem, then, show that it is impossible to obtain feasible integer solution by using simple rounding. Solve the problem using any integer program algorithm. 3. Suppose five items are to be loaded on the vessel. The weight W, volume V and price p are tabulated below. The maximum cargo weight and cargo volume are W = 112, V = 109 respectively. Determine the most valuable cargo load discrete unit of each item. 5 1 2 3 Item 4 W V

5 1

8 8

3 6

3 5

7 4

Price (in Rs.)

4

7

6

5

4

Formulate the problem as integer programming model and then solve. [Hint : If x1, x2, x3, x4, x5 are the cargo load then 1.P.P. is Max Z = 4x1 + 7x2 + 6x3 + 5x4 + 4x5 s.t. 5x1 + 8x2 + 3x3 + 2x4 + 7x5 112 xl+ 8x2 + 6x3 + 5x4 + 7x5 109 xi 0 and all integers for i = 1, 2, 3, ...., 5] Use Branch-and-Bound technique to solve

4. 5.

Max. s.t. Max s.t.

Z = 2x1 + 6x2 3x1 + x2 5, 4x1 + 4x2 9, x1, x2 0 and are integers. Z = 21x1 + 11x2 7x1 + 4x2 + x3 = 13, x2 3, x1, x2, x3 0 and are integers. [Agra 2001]

6.

Max. Z = 6x1 + 8x2 x1 + 4x2 5 8, 7x1 + 2x2 5_ 14, x1, x2 0 and are integers. s.t.

+ ANSWERS + 2. 3.

5. 6.

= 5/4, x2 = 5/8, x3 = 0, Max. Z = 15; xi = 2, x2 = 0, x3 = 2, Max. Z = -16 x1 = 15,x2 = 0, X3 - 1, X4 = 17, Xs = 0 or = 14, x2 = 0 = x3 = xs, x4 = 19, Max. Z = 151 x1 = 4, x2 =3, Max. Z = 55 xi = 0, x2 = 2, Max. Z = 16

• • •

13

Assignment Problem IN

13.1 Introduction It is a special type of linear programming problem in which the objective is to find the optimum allocation of a number of tasks (jobs) to an equal number of facilities (persons). Here we make the assumption that each person can perform each job but with varying degree of efficiency. For example, a departmental head may have four persons available for assignment and four jobs to fill. Then his interest is to find the best assignment which will be in the best interest of the department. Assignment problem finds many applications in allocation. For example in assigning men to works; truck or car or drivers to different roots; planes or pilots to different commercial flights etc. Although simplex method is powerful enough to solve all the L.P. problems, but the above type of the problems may be solved by special procedures which are described in the following sections.

13.2 Assignment Problem [UP TECH MBA 2002-03, 03-04, 05-06; Rohilkhand 2000] The assignment problem can be stated in the form of n x n, matrix [cu ] called the cost or effectiveness matrix, where cu is the cost of assigning i-th facility (person) to the j-th job. Effectiveness matrix Jobs n

1

2

3

1

C11

C12

C13

C1j

Cln

2

C21

C 22

C 23

C2j

C 211

C11

Ci 2

Ci 3

Cif

Cin

C n1

C n2

C,13

C ni

C nn

3 Persons

490

Operations Research

n persons can be assigned to n jobs in n! possible ways. One method may be to find all possible ri! assignments and evaluate total costs in all cases. Then the assignment with minimum cost (as required) will give the optimal assignment. But this method is extremely laborious. For example if n = 8, then the number of such possible assignments is 8! = 40320. The evaluation of costs for all these allocations will take a large time. Thus, there is a need to develop an easy computational technique for the solution of assignment problems. Mathematical Formulation of Assignment Problem [Meerut L.P. 1996 (B.P.); Rohilkhand 2000]

Mathematically an assignment problem can be stated as follows : Minimize the total cost n

Z=

n

>i=1 j=1 ij .x C ~

where xij =

1,

ij

if i- th person is assigned to the j- th job

0, if i- th person is not assigned to the j- th job Subject to the conditions = 1, j = 1, 2, ... , n

(i)

which means that only one job is done by the i-th person, i = 1, 2, ..., n Xij = 1, 1 = 1, 2, ..., n

(ii)

j=1 which means that only one-person should be assigned to the j-th job, j = 1, 2, ...., n . 13.3 Important Theorem Now we shall prove the following two important theorems on which the procedure of solution of an assignment problem is based. Theorem 1 : (Reduction theorem) : If, in an assignment problem, a constant is added or subtracted to every element of a row (or column) of the cost matrix [cti ], then an assignment which minimizes the total cost for one matrix, also minimizes the total cost for the other matrix. [Meerut 1996, 2005, 05(BP), 2010; Bhopal (Stat) 2002; Rohilkhand 1997, 99, 2001] or

Mathematically the theorem may be stated as follows : n

Ifx..=.-minimizes Z = 1=1 Xl•

=1=

n

I cij x.4 over all xij• s.t. j=1

Exi1 and x.. > j=i

0

491

Assignment Problem n

n

then xy = Xu also minimizes Z' =

c' 1 . xu 1=1 j=1

where c' = cij ± ai ±bi ai , b j are constants, i = 1, 2, ...., n; j = 1, 2, ...., n. Proof : We have n n V= i =1 j =1 n n

E E (cii ± ai ±b j ). xij

=

i =1 i=1 n n

I i=1j=1

=Z±

i =1

n n

n n

C-X.. 1.1 11

i=1j=1

b1• 11 i=1 j = 1

ai . I ± bi . j=i j=i i =1

=Z±Ifai.1±I 1).0. i =1 j =1 =Z±I ai f i =1

j =1

bi are independent of xiiit follows that Z' is minimized when ai , i =1 f =1 Z is minimized Hence, xu =Xij which minimizes Z also minimizes Z'. Since

Theorem 2 : If all cy _ > 0 and there exists a solution n xij = Xij

n

s. t.

cij X,i = 0, i =1 j =1

then this solution is an optimal solution (i.e. this solution minimizes Z). Proof : Since all cu 0 n n Z= cij Xu cannot be negative. i = 1 j =1 Thus, its minimum value is 0, when xu = Xu n n Hence, the solution xy =Xij for which I cl j x = 0 is an optimal i =1 j =1 solution.

492

Operations Research

13.4 Hungarian Method (Reduced Matrix Method) For solving a minimal assignment problem (Assignment algorithm). [UP TECH MBA 2001-02; Meerut 2000, 03]

From the 'above two theorems we get a powerful method termed as "assignment algorithm" for solving an assignment problem. The procedure for

the solution is as follows : Step 1 : Subtract the minimum element of each row in the cost matrix [cu ] from every element of the corresponding row. Step 2 : Subtract the minimum element of each column in the reduced matrix obtained in the step 1 from every element of the corresponding column. Step 3 : (a) Starting with row 1 of the matrix obtained in step 2, examine rows successively until a row with exactly one zero element is found. Mark (0), at this zero, as an assignment will be made there. Mark ( x ) at all other zeros in the column (in which we mark ❑ ) to show that they cannot be used to make other assignments. Proceed in this way until the last row is examined. (b) After examining all the rows completely, proceed similarly examining the columns. Examine columns starting with column 1 until a column containing exactly one unmarked zero is found. Mark (0) at this zero and cross (mark x) at all zeros of the row in which ❑ is marked. Proceed in this way until the last column is examined. (c) Continue these operations (a) and (b) successively until we reach to any of the two situations. (i) all the zeros are marked ❑ or crossed. or (ii) the remaining unmarked zeros lies at least two in each row and column. In case (i), we have a maximal assignment (assignment as much as we can) and in case (ii) still we have some zeros to be treated for which we use the trial and error method to avoid the use of highly complicated algorithm. Now there are two possibilities : (i) If it has an assignment in every row and every column (i.e., total number of marked ❑ zeros is exactly n), then the complete optimal assignment is obtained. (See example 1) (ii) If it does not contain assignment in every row and every column (i.e., the total number of marked ❑ zeros is less than n), then one has to modify the cost (effectiveness) matrix by adding or subtracting to create some more zeros in it. For this proceed to the next step 4. Step 4 : When the matrix obtained in step 3 does not contain assignment in every row and every column then we draw the minimum number of horizontal and vertical lines necessary to cover all zeros at least once. For this the following procedure is adopted. (i) Mark (I ) all rows for which assignment have not been made.

Assignment Problem

493

(ii) Mark (I ) column which have zeros is marked rows. (iii) Mark (I ) rows (not already marked) which have assignment in marked columns. (iv) Repeat step (ii) and (iii) until the chain of marking ends. (v) Draw minimum number of lines through unmarked rows and through marked columns to cover all the zeros. This procedure will yield the minimum number of lines (equal to the number of assignments in the maximal assignment obtained in step 3) that will pass through all zeros. Step 5 : Select the smallest of the elements that do not have a line through them, subtract it from all the elements that do not have a line through them, add it to every element that lies at the intersection of two lines and leave the remaining elements of the matrix unchanged. Step 6 : At the end of step 5, number of zeros are increased (never decreased) in the matrix than that in step 3. Now re-apply the step 3 to the modified matrix obtained in step 5, to obtain the desired solution. For the clear understanding of the procedure see Example 4.

glluAlhativp jxamplea Example 1 : Solve the following minimal assignment problem : Man --> Job

1

2

3

4

I

12

30

21

15

II

18

33

9

31

III

44

25

24

21

IV

23

30

28

14 [Agra 2000]

Solution : For the clear understanding, this example is solved step by step systematically. Step 1 : Subtracting the smallest element of each row from every element of the corresponding row, we get the following matrix : 1

2

3

4

I

0

18

9

3

II

9

24

0

22

III

23

4

3

0

IV

9

16

14

0

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Step 2 : Subtracting smallest element of each column from every element of the

corresponding column, we get the following matrix : 1

2

3

4

I

0

14

9

3

II

9

20

0

22

III

23

0

3

0

IV

9

12

14

0

Step 3 : Now we test whether it is possible to make an assignment using the zeros

by the method described in step 3 on page 492. Starting with row I, we mark ❑ (i.e., make assignment) in the row containing only one zero and cross (x) the zeros in the corresponding column in which D lies. Thus, we get the following table : 2

3

4

9

3

I

D O

14

II

9

20

III

23

0

3

IV

9

12

14

22 X

Again starting with column 1, we mark 0 (i.e., make assignment) in the column containing only one unmarked or uncrossed zero in the above table and cross out the zeros in the corresponding row in which this assignment D is marked. Thus, we get the following table : 3

4

14

9

3

1

I 11

9

20

CI

22

III

23

E

3

X

IV

9

12

14

El

Since in the last table, every row and every column have one assignment, so we have the complete optimal zero assignment. Job

I

II

III

IV

Man

I

3

2

4

which is the optimal assignment.

Assignment Problem

495

Example 2 : A department head has four subordinates, and four tasks to be performed. The subordinates differ in efficiency and the tasks differ in their intrinsic difficulty. His estimate of the times each man would take to perform each task is given in the effectiveness matrix below. How should the task be allocated, one to a man, so as to minimize the total man hour ? Subordinates

Tasks

I

II

III

IV

A

8

26

17

11

B

13

28

4

26

C

38

19

18

15

D

19

26

24

10

[Rohillchand 1997, 1999, 2003] Solution : Step 1 : Subtracting the smallest element in each row from every element of the corresponding row, we get the following matrix : I

II

III

IV

A

0

18

9

3

B

9

24

0

22

C

23

4

3

0

D

9

16

14

0

Step 2 : Subtracting the smallest element in each column of the above matrix from every element of the corresponding column, we get the following matrix : I

II

HI

IV

A

0

14

9

3

B

9

20

0

22

C

23

0

3

0

D

9

12

14

0

The above matrix is the same as obtained in step 3 in Example 1, therefore for minimum man hours the allotment should be as follows : Tasks Subordinates Man hours

A I 8

B III 4

and the total Man hours are 8 + 4 + 19 + 10 = 41.

C II 19

D IV 10

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Example 3 : Solve minimal assignment problem whose effectiveness matrix is

I

II

III

IV

A

2

3

4

5

B

4

5

6

7

C

7

8

9

8

D

3

5

8

4

[Meerut 2002 (BP), 03 (BP); L.P. 1995; Agra 2003] Solution : Step 1 : Subtracting the smallest element of each row from every element of the

corresponding row, we get the following matrix : I

H

III

IV

A

0

1

2

2

B

0

1

2

3

C

0

1

2

1

D

0

2

5

1

Step 2 : Subtracting the smallest element of each column of the above matrix

from the corresponding column, we get the following matrix : I

II

III

IV

A

0

0

0

2

B

0

0

0

2

C

0

0

0

0

D

0

1

3

0

Step 3 : Now we test whether it is possible to make an assignment using the zeros

of the above table by the method described in step 3 on page 492. Since none of the rows or columns contain exactly one zero, therefore the trial and error method is followed. Now we start searching two zeros. Starting with row 1 we find row 4 which contain two zeros. We make the assignment at the first zero as shown in table 1 and cross out the other zero in this row and all the other zeros of the first column in which we have made the assignment 0. Now starting with column 1 we find column 4 which contain one zero and make assignment ❑ at the zero of row 3 and cross out all other zeros of this row. Now again starting with row 1 we search row containing only one zero but we find no such row. The same is true when we check the columns for one zero. Again we start with row 1 searching two zeros and find the row (row 1) containing two unmarked zeros. We can make as assignment at any one

Assignment Problem

497

of these zeros and cross out the other zero and zeros of the corresponding column in which assignment is made (see table 1 and 2). If we made an assignment in first row, second column and-crossed other zero and zeros of second column then the second row contain only one t ero, in third column where we can make an assignment. Similarly if we made an assignment in first row, third column and crossed zeros of third column then the second rove contain only one zero irrsecond column where we can make an assignment (see table-2). Table 1

Table 2 IV

I II III

A 8

ID

III

II

I

El

X 2

S

1:1

X Q X 2

B

X X

c

X X X 1

2

X X X

El

3 X

2

n

1 3 X

Thus, we get the following two assignments. A ---> and

,B --> , C

A -->

IV, D— I

,B —> II,C —> IV,D ---> I

In both cases minimum cost = Rs. 20 Note : Other assignments also exist. Students may try to find them. Example 4 : Solve the assignment problem represented by the following matrix : I II III IV V VI A

9

22

58

11

19

27

B

43

78

72

50

63

48

41

28

91

37

45

33

D

74

42

27

49

39

32

E

36

11

57

22

25

18

F

3

56

53

31

17

28

[Meerut L.P. 1990; Rufulkhand 1996] Solution : Step 1 : Subtracting the smallest element of each row from every element of the corresponding row, we get the following reduced matrix :

Operations Research

498 1

A

II

III

VI

13 49 2 10 18

B

0 35 29 7 2C 5

C

13 0 63 9 17 5

D

4 15 0 22 12 5

E

25 0 46 11 14 7

F

0 53 50 28 14 25

Step 2 : Subtracting the smallest element of each column from every element of the corresponding column, we get the following reduced matrix :

A

I

11

III

IV

V

VI

0

13 49 0 0 13

B 0 35 29 5 10 0 C

13 0 63

D 47 15

0

7

7

0

20

2

0

25 0 46 9 4 2 F 0 53 50 26 41 20 Step 3 : Now we give the zero assignments in our usual manner and get the following matrix : X 13 49

X 13

35 29 5 10 13 X 63 7 7 X 47 15

20 2 X

25 I] 46 9 4 2

ri

53 50 26 4 20

Since row 3 and column 5 have no assignments so we proceed to the next step. Step 4 : In this step we draw minimum number of lines to cover all zeros at least once. For this we proceed as follows :

Assignment Problem

• 499

49

El

35

29

5

. 10

[11

13 X

63

7

7

X

47

15

0

20

2

25 MI

46

9

4

2

D 5:3

50

26

4

2p

L4.•

6 (i) We mark ) row 3 in which there is no assignment. (ii) Then we mark (I ) columns 2 and 6 which have zeros in marked row 3. (iii) Then we mark (✓ ) rows 5 and 2 which have_ assignments in the marked colunins 2 and 6. • •. (iv) Then we mark column 1 (not already marked) which has zero in the marked row 2. (v) • Then we mark row 6 which.has assignment in the marked column 1; • (vi) Now•we draw lines through all marked. columns .1, 2, 6:. Then we draw lines through unmarked row 1 and 4 having zeros through which there.is no line. ThuS, we get five lines (minimum number) to • cover all the zeros. Step 5 : Now the. smallest of the elements that do not.have a line through them is • 4. Subtracting this element 4 from all the elements that do not have aline through them and adding to every.element that lies at the intersection of two lines and leaving the remaining elements unchanged, we get the • following matrix : - I

II

III

4

17

49

El 35

25.

1

6

C

13

)6(

59

3

3

E

D

51

19

El

20

2

4

E

25 E]

42

5

X

2

F

53

46

22

Q

20

A

IV

V

VI

17

Step 6 : Again repeating the step 3 we make the zero assignments and get the

following matrix (see above table) : Thus, the optimal assignment is A •-> /V, B I,C VI, D

E

F V

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and minimum cost, Z =11+ 43 + 33 + 27 + 11 + 17 = 142. (from the given cost matrix) Another optimal solution of the assignment problem is as follows. A —) IV, B --) VI, C --> D E ---> V, F --> I. I II III IV V VI

A

4

17 49

X

35

17

25

1

6

0

59

3

3

X

20

2

4

C

13

D

51

19

25

X

42

53

46

B

X,

2

X

22

20

In this case minimum cost Z = 11 + 48 + 28 + 27 + 25 + 3 = 142 (from the given cost matrix) which is equal to the minimum cost for the first solution. Example 5 : Solve the following assignment problem. Man I

II

III

IV

V

A

1

3

2

3

6

B

2

4

3

1

5

Task C

5

6

3

4

6

D

3

1

4

2

2

E

1

5

6

5

4

[Meerut 2005; Rohilkhand 20001 Solution : Step 1 and 2 : Applying step 1 and 2 we obtain the following matrix : I

II

III

IV

V

A

0

2

1

2

4

B

1

3

2

0

3

C

2

3

0

1

2

D

2

0

3

1

0

0

4

5

4

2

Assignment Problem

501

Step 3 : Giving the zero assignment in our usual manner we get the following matrix : / L.1

II

III

IV

V

A

I=1

2

1

2

4

B

1

3

2

El

2 ..

-.3

2

E3

3

1

4

5

4

✓()

..........3

L2

2

L3

1

.L4

2

✓ Since row 5 and column 5 have no assignments, so we proceed to the next step. Step 4 : The minimum number of lines drawn in the usual manner are 4 (see above table). Step 5 : Now the smallest of the elements that do not contains line through them is 1. Subtracting this element 1 from the elements that do not have a line through them, adding to every element that lies at the intersection of two lines and leaving. the remaining elements unchanged, we get the following matrix II

L2 ! III

IV

V

A

1

lir,

1

3

B

3

2

C

3

II

3

1:3 3

E

3..

1

3 3

2 X

• L4

1

✓(ID

0 Step 6 : Again repeating the step 3 we make the zero assignments in matrix (above table) and see that even now the row 1 and column 5 do not contain any assignments. Therefore we again repeat step 4 of drawing lines. Step 7 According to our usual manner the minimum number of lines drawn are 4 (see table in step 5).

Operations Research

502 Step .8 : Again repeating step 5, we get following matrix

A El

II III . IV V X 2

B

3

3 2

C

3 .2

D

3 X

D 4 E] 4 1 X E 8 2 4 2 El Step 9

Repeating the step 3 we make the zero assignments and get the following optimal assignments, (see above table). D B V. A I, B ----> IV, C

The minimum cost, Z = 1 + 1 + 3 + 1 + 4 = 10 Example 6 : An air-line that operates seven days a week has time-table shown below. Crews must have a minimum:layover of 5 hours between flights. Obtain the pairing of flights that minimizes layover time away from home. For any given pairing the crew will he based at the city that results in the smaller layover. Delhi-Jaipur Flight No. Departure 1 7.00 A.M. 2 8.00 A.M. . 1.30 P.M. 3 6.30 P.M. . 4

Arrival 8.00 A.M. 9.00 A.M. 2.30 P.M. 7.30 P.M.

•

Jaipur-Delhi Flight No. 101 102 103 104

Departure 8.00 A.M. 8.30 A.M. 12.00 Noon 5.30 P.M.

Arriyal 9.15 A.M. 9.45 A.M. 1.15 P.M. 6.45 P.M.

For each pair also mention the town where the creu., should be based. [Meerut 2002 (BP), 03 (0); 09]

Solution : Step 1 : First we construct the table forlayover times between flights when crew is based at Delhi i.e., crew start from and come back to Delhi with halt of minimum time (layover time) not less than 5 hours, at Jaipur. Since the crew must have a minimum layover of 5 hours between flights, the layover time between flights 1 and 101 will be 24 hours. Also the layover times between flight 1 and 102, flights I and 103, flights 1 and 104 aret24.5 hours, 28 hours, 9.5 hours respectively. Similarly the layover times between other pairs of flights may be calculated which are shown in the following table :

503

Assignment Problem Layover times in hours when crew based at Delhi Table 1 Flights -4 101

102

103

104

1

24

24.5

28

9.5

2

23

23.5

27

8.5

3

17.5

18

21.5

27

4

12.5

13

16.5

22

Now we construct the table for layover times between tie pair of flights when the crew is based at Jaipur i.e., crew start from and come back to Jaipur with minimum halt time (layover time) not less than 5 hours, at Delhi. Since the plane arrive at Delhi at 9.15 A.M. by flight No. 101 and will depart to Jaipur at 7.00 A.M. by flight No. 1 after 21.75 hours. Therefore, layover time between pair of flight No. 101 and 1 is 21.75 hours. Similarly, the layover times between other pairs of flights may be calculated which are shown in table 2 : Layover times when crew based at Jaipur Table 2 Flights

1

2

3

4

101

21.75

22.75

28.25

9.25

102

21.25

22.25

27.75

8.75

103

17.75

18.75

24.25 .

5.25

104

12.25 13.25 18.75 23.75

Table 2 can also be written as follows : Flights -4 101

Table 3 102 103

104

1

21.75

21.25

17.75

12.25

2

22.75

22.25

18.75

13.25

3

28.25

27.75

24.25

18.75

4

9.25

8.75

5.25

23.75

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Step 2 : To avoid the fractions we consider either the layover times in terms of quarter hour as one unit of time or the layover times for four weeks. Thus, multiplying the matrices (tables 1 and 3) by 4, the modified matrices are as follows (table 4 and 5) : (Crew based at Delhi) Table 4 Flights —> 101 1

96

2

92

102

103

104

98

112

38

94

108

34

,

14

3

70

72

86

108

4

50

52

66

88

(Crew based at Jaipur) Table 5 Flights —>

101

102

103

104

87

85

71

49

91

89

75

53

113

111

97

75

37

35

21

95

Step 3 : Now we combine the table 4 and 5, choosing that base which gives a lesser layover time for each pairing. The layover times marked with 1"' denote crew based at Jaipur. Otherwise the crew in based at Delhi. Thus, we get the following table : Table 6 (Minimum layover time table) 87*

85*

71*

38

91*

89*

75*

34

70

72

86

75*

37*

35*

21*

88

Step 4 : Subtracting the smallest element of each row from every element of the corresponding row and then subtracting the smallest elenient of each column from every element of the corresponding column, we get the following matrix (table 7): :

505

Assignment Problem Table 7 L3

45*

33*

El ✓

53*

41*

X

X

'6

S*

L1

6;7

L2

Step 5 : Giving the zero assignments (see table 7) we find that there is no assignment in row 2 and column 2, so we draw minimum number of lines to cover all the zeros as shown in table 7 : Step 6 : Now subtracting the smallest uncovered element 33 from all uncovered elements and adding it to the elements which lie at the intersection of lines and leaving other elements as usual the matrix obtained is as follows : (table 8). Step 7 : Giving the zero assignments we can find that there is no assignment in row 1 and column 2, so we again draw minimum number of lines to cover all the zeros as shown in table 8 : Table 8 L2

16*

12* ral

24*

20* ski:

4-6*

L3

* 16

12*

100

0 Step 8 : Proceeding again as in step 6 the final matrix obtained is as follows (table 9) : Table 9 101

102

103

104

1

4*

0*

0*

0

2

12*

8*

8*

0

3

0

0

4

9"

0*

28 0"

50* 100

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Step 9 : Giving the zero assignments, we get the following two optimal assignments (table 10 and 11) :

Table 11

Table 10 101 102

103

104 X

101

102

103

104

1

4*

)3; *

0*

X

2

12k

8*

8*

E

4*

El*

X.*

2

12*

8*

8*

3

ID

X

28

50*

3

El

N's

28

50*

X*

0* 100

4

4*

a

:le

100

4*

Hence the required city for crew for pairing of flights for minimum layover time is as follows : (1) 1 102 (Crew at Jaipur), 2 > 104 (Crew at Delhi) 3 —> 101 (Crew at Delhi), 4 —> 103 (Crew at Jaipur) (ii) 1 103 (Crew at Jaipur), 2 —> 104 (Crew at Delhi) 3 101 (Crew at Delhi), 4 —> 102 (Crew at Jaipur) In both the cases minimum layover time is 210 hours for four weeks i.e., 52 hours 30 minutes per week,

Example 7 : A small aeroplane company, operating seven days a week, serves three cities A, B and C according to the schedule shown in the following table. The layover cost per stop is roughly proportional to the square of the layover time. How should planes be assigned the flights so as to minimize the total layover cost ? • Departure To Arrival Flight No. From Al B A2 B A3 B A4 C A5 C B1 A B2 A 23 A C1 A C2 A

-

A A A A A

B B B C C

09 A.M. 10 A.M. 03 P.M. 08 P.M. 10 P.M. 04 A.M. 11 A.M. • 03 P.M. 07 A.M. 03 P.M.

B B B C

C A . A

A A A

Noon 01 P.M. 06 P.M. Midnight 02 A.M. 07 A.M. 02 P.M. 06 P.M. 11 A.M. 07 P.M.

[Agra 1998] Solution : From the data given in the problem, we note that : (i) Five planes are operating at a time. (ii) Any plane flying from a station must come back within 24 hours for the scheduled trip.

Assignment Problem

507

(iii) A plane starting from A for B or C must come back to A at the earliest possible opportunity. (iv) A plane cannot make more than 2 trips in 24 hours i.e., it will just go from A and will come back to A within 24 hours. First we consider the layover.cost matrix for the flights between A and B. A plane from A to B through the B Noon B A 09 AM A,B route A1B may Come back to A through any of the routes BO, B2A or B3A. The plane A depart from A at 9 A.M. and arrives at B at noon i.e., at 12 P.M. It may go. back to A by flights Bi A, B2A, B3A at 04 A.M., 11 A.M., 03 P.M., respectively. So its layover times at B for these flights will be 16 hours, 23 hours, 3 hours respectively. The flights ill A, B2A, B3 A reaches A at 07 A.M., 02 P.M., 06 P.M. respectively and will depart from A for B at 09 A.M. Thus, the layover times of these flights Bi A, B2A, B3A at A will be 2 hours, 19 hours and 15 hours respectively. Thus the layover costs for the routes AT B to BT A, is 162 + 22 = 260 units. Note that the flights for the routes A1B to B2A is not completed within 24 hours. So this route is not allowed. Thus the cost of layover time for this route is considered to be very high, say co to avoid this route. The. layover costs for the route A1B to. B3A is 32 + 152 = 234 units. Similarly, the layover costs for the routes A2B to B1 A, A2B to B2A, A2B to B3A are 152 + 32 = 234 units, co and 22 + 162 = 260 units respectively. A 10 AM A2B

01 PM B

And the layover costs for the routes A3B to BO, A3B to B2A, A3B to B3A are 102 + 82 = 164 units, 172 + 12 = 290 units and co respectively. A

03 AM A3B

06 PM B

07 AM B1A

04 AM

02 PM

13,A

11 AM

B3A

03 PM

Thus, the layover costs for the three planes flying between stations A and B in different routes are as shown in the following table :

Operations Research

508 B1 A B2A B3A A1B

260

.

234

A2B

234

.

260

A3B

164

290

00

Solving this as an assignment problem the optimal assignment is given in the following table : BiA

B2A

B3A

A 1B

26

00

El

A2B

El

-

26

A3B

X

00

Now we construct the layover costs matrix for the flights between A and C, which is as follows :

A4C A5C

r

C1 A

C2A

72 + 92 = 130

152 + 12 = 226

52 + 112 =146

132 + 32 =178

C

2 AM

Solving this as an assignment problem, the optimal assignment is given in the following table : CI A C2A

64

A4C A5C

X

Assignment Problem

509

Hence, the optimal route schedule for minimum layover is as follows : Plane No. Departure Route Arrival Route

1 A1 B B3 A

3 A 3B B2A

,2 A2B Bi A

4 A 4C Ci A

5 A 5C C2 A

13.5 Unbalanced Assignment Problems An assignment problem is called an unbalanced assignment problem whenever the number of tasks (jobs) is not equal to - the number of facilities (persons). Thus, the cost matrix of an unbalanced assignment problem is not a square matrix. For the solution of such problems we add the dummy rows or columns to the given matrix to make it a square matrix. The costs in these dummy rows or columns are taken to be 0. Now the problem reduce to the balanced assignment problem and. can be solved by assignment algorithm 13.4.

MA&atiOR EXaMPICTA Example 8 : A company is faced with the problem of assigning six different machines

to five different jobs. The costs are estimated as follows (hundreds of rupees). Job 1

2

3

1

2.5

5

1

6

1

2

2

5

1.5

7

3

3

3

6.5

2

8

3

3.5

7

2

9

4.5

5

4

7

3

9

6

6

6

9

5

10

6

Machine 4

5

Solve the problem assuring that the objective is to minimise total cost. Solution : Since the given matrix is not a square matrix, we add one fictitious job

6 (sixth column) to make it a square matrix: Thus, the resulting matrix obtained is as follows : 1

2

3

4

5

6

1

2.5

5

1

6

1

0

2

2

5

1.5

7

3

0

3

3

6.5

2

8

3

0

4

3.5

7

2

9

4.5

0

5

4

7

3

9

6

0

6

6

9

5

10

6

0

Operations ReSearch

51Q

Step 1 and 2 : Subtracting the smallest element of each row from every element of the con esponding row and then subtracting smallest element of each column from every element of the corresponding column, we get the following matrix. • 6 2 3 1 4 5 0 0 0.5 0 0 0 0 0 0 0.5 1 2 2 3 4 5 6

1 1.5 2 4

1.5 2 2 4

1 1. 2 4

2 3 3 4

2 3.5 5 5

0 0 0 0

Step 3 : Giving zero assignments in the usual manner, we observe that thesows 3, 4, 5 and columns 4, 5, 6, have no zero assignments. So we proceed to the next step : L3

2

4

5

Li 1. 0:5

3 ;13:

L2-2.

0:5

1

2

1.5

1

2

2

4 1.5

2

1

3

3.5

5

2

2

2

•3

5

6

4

4

4

4

3

1

6

X

I• Step 4 : Here we draw minimum number of lines (horizontal and vertical) to cover all the zeros at least once. The number of such lines is 3. See table in step 3. Step 5 : Since the smallest element among all uncovered elements in the above table is 1, so subtracting this element 1 from all uncovered elements, adding to every elements that lies at the intersection of two lines and leaving remaining elements unchanged, the above table reduces to the following form. 1 2 3 6 5 4 1 0.5 0 0 0 0 1 0 0 0.5 1 2 1 2 1 3 0 0.5 0 1 0 1 0 0 4 0.5 2 2.5 5 1 1 1 2 •4 0 6 3 ' 3 3 3' 4 0

Assignment Problem

511

Step 6 : Giving zero assignments in the usual manner, we observe the row 6 and

column 5 have no zero assignments. So we proceed to the next step : L5

1

2

3

4

:ers

53:

p.8;

E

0:5

1

L3 3-

0:5

X

1

1

L4 4-

1

0

2

2.5

L1..1. 0:5

5

5

1

1

1

2

4

6

3

3

3

3

4

:6

X

I® Jo

Step 7 : We again draw minimum number of lines (horizontal and vertical) to

cover all zeros at least once. The number of such lines is 5. See table in step 6. Step 8 : Since the smallest element among uncovered elements in the above table

is 1, so subtracting this element 1 from all uncovered elements, adding to every element that lies at the intersection of two lines and leaving remaining elements unchanged, the above table reduces to the following form. 3

4

5

6

1

0.5

0

0

0

0

2

2

0

0

0.5

1

2

2

3

0

0.5

0

1

1

1

4

0.5

1

0

2

2.5

1

5

0

0

0

1

3

0

6

2

2

2

2

3

0

Step 9 : Giving zero assignments in the usual maniter, we observe that row 2 and

column 5 have no assignments. (see table on page 512) So we proceed to the next step.

Operations Research

512 L2 L3 L4

6

•

z. El

2

2

2

1

05

X

:14;

1

05 2

2.5

X

1

3

2

2

3

4 05 :E4( •

✓

I Step 10: Here we again draw minimum number of lines (horizontal and vertical) to cover all the zeroes at least once. The number of such zeroes is 5. See table in step 9. Step 11: Since the smallest element among all uncovered elements is 1, so subtracting this elements 1 from all uncovered elements, adding to every elements that lies at the intersection of two lines and leaving remaining elements unchanged, the above table reduces to the following form : 1

2

3

4

5

6

1.5

1

1

0

0

3

0

0

0.5

0

1

2

3

0

0.5

0

0

0

1

4

0.5

1

0

1

1.5

1

5

0

0

0

0

2

0

6

2

,, 2

2

1

2

0

1

Step 12 : Giving zero assignments in the usual manner we get the optimal assignments given by the following tables : 1

2

3

4

5

6

1

1

1.5

1

1

0

X

3

1.5

1

1

El

R

3

2

0

)3(

0.5

X

1

2

2

)3(

El

0.5

)3(

1

2

3

X

0.5

El

1

3

X

0.5

X

X

El

1

4

0.5

1

El

1

1,5

1

4

0.5

1

DO

1

1.5

1

5

XDX

X

2X

5

El

X

X

)3(

2

X

6

2

1

2

2

2

2

1

2

1:1

2

2

Assignment Problem 1 1 1.5

513

2

3

4

5

6

1

1

B

E

3

0.5

X

1

2

1

2

3

4

5

6

1

1.5

1

1

X

El)

3

2

2

:a

0.5

X.

1

2

X

1

3

CI

0.5

)3(B

:8;

1

0.5

1

1

1.5

1

XI

p

2

X

3

)31:

0.5

X

4

0.5

1

R-1 )

1

1.5

1

4

5

;$3(

p

X

X

2

X

5

6

2

2

2

1

2

rni

6

1

2

3

4

5

6

1

1.5

1

1

X

E

3

2

X

El

0.5

B

1

3

X

0.5

X

4

0.5

1

5

El

X

6

2

2

2

2

2

1

2

El

1

2

3

4

5

6

1 1.5

1

1

1

E]

3

2

2

B

B

0,5

Di

1

2

:€1:

1

3

D)

0.5

1

1.5

1

4

0.5

1

X

X

2

5

XOXX

2

1

2

6

2

11

2

1 1

2

1

1.5

1

2

X

2

g

Hence, the optimal solutions are 1—> 4, 2 —> 1, 3 —> 5, 4 —> 3, 5 —> 2, 1 —> 4, 2 —> 2, 3 —> 5, 4—> 3, 5 —> 1 1 —> 5, 2 —> 1, 3—> 4, 4—> 3, 5 —> 2, 1 —> 5, 2—> 2, 3—> 1, 4—> 3, 5 --4 4 1—> 5, 2 —> 2, 3 —> 4, 4—> 3, 5 —> 1, 1 --> 5, 2—> 4, 3 —> 1, 4 —> 3, 5 —> 2 In all the optimal solutions 6th machine do no job and the minimum total cost in all cases is Rs. 20 x 100 i.e., Rs. 2000.

13.6 Maximization Assignment Problem Some times the assignment problem deals with the maximization of the objective function i.e., the problem may be to assign persons to the jobs in such a way that the expected profit is maximized. Such maximization problem may be solved by converting it to minimization problem. This is done converting the profit_ matrix to the cost (i.e., loss) matrix in either of the following two ways. (1) Subtract each element of the given matrix (Profit matrix) from the greatest element of the matrix to get the equivalent cost (i.e., loss) matrix. or (ii) Place minus sign before each element of the profit matrix to get the equivalent cost matrix.

Operations Research

514

gliathtadog Exams Example 9 : Alpha corporation has four plants each of which can manufacture any of

the four products. Production costs differ from plant to plant as do sales revenue. From the following data, obtain which product each plant should produce to maximize profit ? Sales revenue (Rs. 1000) Product Plant

Production cost (Rs. 1000) Product

1

2

3

Plantl

1

2

3

4

A

50

68

49

62

A

49

60

45

61

B

60

70

51

74

B

55

63

45

69

C

55

67

53

70

C

52 '

62

49

68

D

58

65

54

69

D

55

64

48

66

[Bangalore B.E. 1990] Solution : Since, Profit = Sales revenue — Product cost, so the profit matrix is as

_-

follows : 2 A B C

D which is'Aia-kimization problem. We shall solve this problem by converting it to minimization problem in both way discussed in article 13.6. 1. By way (9_4 Subtracting each element of the above matrix from the greatest element the matrix, the equivalent loss matrix is 1

2

3

4

7

0

4

7

3

1

2

3

C

5

3

4

6

D

5

7

2

5

Step 1 and 2 : Subtracting the minimum element of each row from all the

elements of the corresponding row and then subtracting minimum element of each column from all the elements of the corresponding column, we get the following matrix.

Assignment Problem

515 2 A

5

0

3 4

4 5

B 0 0 1 0

C

0

0

1

1

D 1 5 0 1 Step 3 : Giving zero assignments in the usual manner, we get the following matrix. 1 2 3 4 A

B C D

In the above table there is an assignment in each row and each column. Hence the optimal assignment for maximum profit is A 2, B 4, C --> 1, D -> 3 and Max. Profit = Rs. (8 + 5 + 3 + 6) x 1000 = Rs. 22000 2. By way (ii) : Placing negative sign before each element of the profit matrix, the equivalent loss matrix is 2 A

3

4

—1 —8 —4 —1

B —5 —7 —6 —5

C

—3 —5 —4 —2

D —3 —1 —6 —3 Now subtracting the minimum element of each row from every elements of the corresponding row and then subtracting the minimum element of each column from every element of the corresponding column, we get the following matrix. 1

2

3

4

A

5

0

4

5

B

0

0

)1

0

C

0

0

1

1

D

1

5

0

1

Which is the same matrix as obtained in step 1 and 2 in way (0. Hence giving zero assignments we get the same optimal solution as by way (0.

516

Operations Rpseare

J

Example 10 : Suggest optimum solution to the following assignment problem and also the maximum sales : Salesmen Markets (Sales in Lakhs Rs.) A B C D

I

II

III

IV

44 60 36 52

80 56 60 76

52 40 48 36

60 72 48 40

[UP TECH MBA 2006-07] Solution : This is a maximization assignment problem. Placing negative sign before each elem,ent of the given matrix, the corresponding loss (or minimization) assignment problem is given by I

II

III

IV

— 80

— 52

— 60

B

— 44 — 60

— 56

— 40

—72

C

— 36

— 60

— 48

— 48

D

—52

—76

—36

—40

A

Step 1 and 2 : Subtracting the minimum element of each row from all the elements of the corresponding row and then subtracting the minimum element of each column from all the elements of the corresponding column, we get the following matrix : I II III IV A

24

0

16

20

B

0

16

20

0

C

12

0

0

12

D

12

0

28

36

Step 3 : Giving zero assignments in the usual manner, we get the following matrix : 0 )[I A 24 B L2...c. 12 D 12

III

IV

16 20 1-6

• .0

;€4;

$

0

12

28 36

I® ✓®

Assignment Problem

517

In the above table, we observe that the row IV and column IV has no zero assignment. So we proceed to the next step. Step 4 : Here we draw minimum number of lines (horizontal and vertical) to cover all the zeros at least once. The number of such lines is three. See table in step 3. Step 5 : Since the smallest element among all uncovered elements in the above table is 12, so subtracting 12 from all uncovered elements, adding it to every element that lies at the intersection of ts4o lines and leaving remaining elements unchanged, the above table reduces to the following form : II III IV A 12

DI 4 8

B X 28 20 C

12 12

D

EI

0

CI 12

X 16 24

Step 6 : Giving zero assignment in the usual manner, in the table obtained in step

5, we observe that each row and each column has assignment. Hence the required optimum solution is D —> I A B —> IV, C —> The maximum sales for the above assignments is Rs. (80+72+48+52) = Rs. 252. Example 11 : A company has four territories open, and four salesmen available for assignment. The territories are not equally rich in their sales potential. It is estimated that a typical salesman operating in each territory would bring in the following annual sales : Territory Annual sales (Rs.) :

I

II

III

IV

60000 50000 40000

30000

The four sales men are also considered to differ in ability; it is estimated that, working under the same conditions their yearty sales would be proportionally as follows : Salesmen : Proportion :

7

B C

D

5

4

5

If the criterion is maximum expected total sales, the intuitive answer is to assign the best salesman to the richest territory the next best salesman to the second richest, and so on. Verify this answer by the assignment technique. [Meerut 2004, 04 (0)] Solution : To construct the effectiveness matrix. The sum of proportions of sales of four salesmen = 7+5+5+4 = 21

518

Operations Research Taking the salesmen in the four territories, sales are as follows 7 — x3 21

for A, 2- x 6, 21

7 21

7 21

for B, 5 x6, 21

5 x5, 5 x4, 5 x3

5 for C, — x 6, 21

— x 5, 21

21

21

5

21

5

— x 4, 21 21 5"

4 x6, 4 x5, 4 x4,

forD,

21

21

21

-4-x3 21

To avoid fractions we consider the sales in 21 years which are as follows for A,

42,

35,

28,

21

for B,

30,

25,

20,

15

for C,

30,

25,_ 20,

15

for D,

24,

20,

12

16,

Thus, the problem is to determine the assignments (of salesmen to territories) which make the total sales maximum, and the effectiveness matrix is given by I

II

III

IV

A

42

35

28

21

B

30

25

20

15

C

30

25

20

15

D

24

20

16

12

To convert the maximization problem to a minimization problem. This is a maximization problem. By multiplying each element of this matrix by -1, it is converted to minimization problem. Thus, the resulting matrix is as follows : I

III

IV

A - 42 - 35

- 28

- 21

B -30 -25

-20

-15

-30 -25 -20

-15

C

II

D -24 -20

-16

-12

Assignment Problem

519

Solution of the minimization problem. Step 1, 2 and 3 : Applying step 1, 2 and 3 we obtain the following matrix. L1 -

3

6

9

X

1,

2

3

*

1

2

3

D

.......

L

Step 4 : Here only two lines (less than 4, the L1 number of rows or columns in matrix) cover all the zeros. So we 5 8 proceed to the next step. 1 2 I CD Step 5 : Subtracting 1, minimum of all uncovered elements by the lines, from 1 * 2 I 0 all the elements through which no line L3 passes, adding to the elements which I YK El lie at the intersection of two lines and leaving other elements as they are, we obtain, the following matrix. Step Giving the zero assignments in ' matrix we see that there is no assignment in row 3 and column 4. Therefore, we again proceed to draw minimum number of lines to cover all zeros. Step 7 : Proceeding in our usual manner we find that there are three lines (minimum number) to cover all zeros (see table in step 5). Step 8 : Now subtracting 1, the minimum of all uncovered elements by the lines, from all the elements through which no line passes, adding to the elements which lie at the intersection of two lines and leaving other elements as they are, we get the following matrix.

/0

0

2

4

7

0

0

0

1

0

0

0

1

2

1

0

0

Operations Research

520

Step 9 : Giving assignments in usual manner we get following two assignments. I II III IV

I II III IV

A

Q O

B

X :6; 0 1

C

x

x 1

2 1

X

D

2 4 7

11

Thus, we get the following two optimal solutions (assignments). (i) and

A

I, B -->

C —>

D —> IV

(ii) A —) I, B —>

C —)

D—>W

From both the solutions it is obvious that the best salesman A is assigned to the richest territory I and the worst salesman D to the poorest territory D.The salesmen B and C are equal in efficiency, so either of them may be assigned to territories II and M. This assignment verify the given problem.

13.7 Restrictions on Assignment Sometimes due to some restrictions the assignment of a particular facility to a particular job is not permitted. To overcome this difficulty a very high cost (infinite cost) is assigned to the corresponding cell which automatically exclude this activity from the optimal solution. For clear understanding, see the following example. Example 12 : Solve the following assignment problem : I II III IV A

9

11

B

12

9

C

14

15

10

11

10

9

11

14

11

7

8

12

7

8 [Meerut 2001 (BP)]

Solution : According to the given assignment table facility B cannot be assigned to job III and the facility C cannot be assigned to job I, so we assign a very high cost in

the cells (B, III) and (C, I). Also the matrix is not a square matrix i.e., it is a unbalanced assignment problem, so we add a dummy facility E with zero costs in all cells of this row. Thus, we get the following assignment table :

Assignment Problem

521 11

III

IV

V

15

10

11

9

00

10

9

11

14

11

7

A

9

B C D

12 14

8

12

7

8

E

0

0

0

0

0

Step 1 : Subtracting the minimum element of each row from every element of the

corresponding row and then subtracting the minimum element of each column from every element of the corresponding column, the reduced matrix is I

II

III

IV

V

A

0

2

6

1

2

B

3

0

00

1

0

C

00

4

7

4

0

D

7

1

5

0

1

E

0

0

0

0

0

Step 2 : Giving zero assignment in the usual manner, we get the following matrix. 11

III

IV

V

A

QO

2

1 2.

B

3

El

1

:€1;

C

co

4 7 4

El

D 7 1 5 13 1

E

X X

X

Since there is an assignnient in each row and each column, so the optimal assignment is A-3 I, B --> C -3 V, D IV Here the job HI remains undone. •

Operations Research

522

+ Exercise on Chapter 13 + [UP TECH MBA 2005-06] Define assignment problem. Write short note on "Assignment problem". [Meerut 2008 (BP)] Give an algorithm to solve an assignment problem. [Rohilkhand 2002] Give a mathematical formulation of the assignment problem. Show that an assignment problem does not change in structure if each row or column is reduced by a constant. Explain how this property can be used to find the solution of an assignment problem. 6. Show that if in an assignment problem, we add a constant to every element of row (or column) in the effectiveness matrix, then an assignment that minimizes the total effectiveness in one matrix also minimizes the total effectiveness in the other matrix. 7. Give in detail the computational procedure of solving the assignment problem. [Meerut 2000]

1. 2. 3. 4. 5.

8.

Explain Hungarian method for solving an assignment problem. [UP TECH MBA 2001-02; Meerut 2001, 02 (BP), 03, 03 (P)]

9.

There are five jobs to be assigned, one each to five machines and the associated cost matrix is as follows : Machine 'Job

1

2

3

4

5

•

11

17

8

16

20

B

9

7

12

6

15

C

13

16

15

12

16

D

21

24

17

28

26

E

14

10

12

11

15

Solve the following minimal assignment problem. [Meerut 1996 (P), 2001; Rohilkhand 1992] 10. Solve the following minimal assignment problem : I

II

III

IV

A

1

4

6

3

B

9

7

10

9 '

C

4

5

11

7

-D

8

7

8

5 [Meerut 2003]

523

Assignment Problem 11. Solve the following assignment problem. Time (In minutes) Works

Job 1

Job 2

A

4

2

B

8

5

3

C

4

5

6

Job 3

-,

7

[UP TECH MBA 2001-02] 12. A computer centre has got three expert programmers. The centre needs three

application programmes to be developed. The head of the computer centre, after studying carefully the programmes to be developed, estimates the computer time in minutes repaired by the experts to the application programmes as follows :

-- Programme A B

C

1

120 100 80

Programmer 2

70 90 110

3

110 140 120

Assign the programmers to the programmes in such a way that the total computer time is least. [Meerut 2007; Agra 2003] 13. An automobile dealer wishes to put four repairmen to four different jobs. The repairman have somewhat different kinds of skills and they exhibit different levels of efficiency from one job to the another. The dealer has estimated the number of man-hours that would be required for each job-man combination. This is given in matrix form in the following table : Job

ABCD Man

1

5 3 2 8

2

7 9 2 6

3

6 4 5 7

4

5 7 7 8

Find the optimal assignment that will result in minimum man-hours needed. [Rohilkhand 2002] 14. Find the optimal assignment for the problem with the following matrix :

1 11 111 W

5 3 1 8 B 7 9 2 6 C

6 4 5 7

D 5 7 7 6

Operations Research

524

15. Find the optimum solution to the assignment problem having the following cost matrix : Sales Territories (Cost in rupees thousands) I II III IV A 30 25 26 28 Salesman B 26 32 24 20 20 22 18 27 D 23 20 21 19, [Meerut 2002] 16. Solve the following assignment problem represented by the matrix : I II HI IV V A 6 5 8 11 16 10 B 1 13 16 1 C 16 11 8 8 8 16 D 9 14 12 10 8 E 10 13 11 16 [Rohilkhand 2001] 17. One car is available at each of the stations 1, 2, 3, 4, 5, 6, and one car is required at each of the stations 7, 8, 9, 10, 11, 12. The distance between the various stations are given in the matrix below. How should the cars be despatched so as to minimize the total mileage travelled ? 12 7 8 9 11 10 1 2 3 4 5 6

42 22 27 45 29 82

72 29 39 50 40 40

39 49 60 48 39 40

52 65 51 52 26 60

25 81 32 37 30 51

51 50 32 43 33 30

18. Solve the following assignment problem. Typist A B C -- , D

P 85 90 70 75

Job (Time) R Q 50 30 70 40 60 60 45 35

S 40 45 50 55 [UP TECH MBA 2002-03]

Assignment Problem

525

19. Consider the problem of assigning five jobs to five persons. The assignment

costs are given as follows : Job

A

B Person C D E

1 8 0 3 4 9

2 4 9 8 3 5

3 2 5 9 1 8

5 1 4 6 3 5

4 6 5 2 0 9

Determine the optimum assignment schedule. 20. Solve the following minimal assignment problem.

[Meerut 2006 (BP)]--

1

2

3

4

5

12

8

7

15

14

//

7

9

17

14

10

III

9

6

12

6

7

IV

7

6

14

6

10

V

9

6

12

10

6

Man Job

21. Solve the following minimal assignment problem.

1

2

3

4

A

10

12

19

11

B

5

10

7

8

C

12

14

13

11

D

8

15

11

9

22. Find the minimum cost solution for the 5 x 5 assignment problem whose

coefficients are as given below : 1 2 3 4 5

I/

III

IV

V

-2 0 -3

-4 -9 -8

-4 -9

-3 -5

-8 -5 -9 -1 -8

-6 -5 -2 0 -9

-1 -4 -6 -3 -5

526

Operations Research

23. Five men are available to do five different jobs. From past records, the time (in hours) that each man takes to do each job is known and is given in the following table. Job II

III

IV

V

1 2

2 6

9 8

2 7

7 6

1 1

Man 3 4 5

4 4 5

6 2 3

5 7 9

3 3 5

1 1 1

Find the assignments of men to jobs that will minimize the total time taken. 24. A national truck-rental service has a surplus of one truck in each of the cities 1, 2, 3, 4, 5 and 6; and a deficit of one truck in each of the cities 7, 8, 9, 10, 11 and 12. The distances (in kilometers) between the cities with a surplus and the cities with a deficit are displayed below : To 1 2

7 8 9 10 11 12 31 62 29 42 15 41 12 \19 39 55 71 40

From 3

17 `29 50 41 22 22

4 5 6

35 19 72

40 30 30

38 29 - 30

42 27 33 16 20 23 50 41 20

..1••=1=

How should the trucks be dispersed so as to minimize the total distance travelled ? 25. Six wagons are available at six stations A, B, C, D, E and F. These are acquired at stations I, II, III, IV, V and VI. The mileage between various stations is given by the following table I II— III IV V VI

A

E

20 50 60 6 18

23 20 30 7 19

18 17 40 10 28

10 16 55 20 17

F

9

10

20

30

B C D

16- 20 15 11 8 7 100 9 60 70 40

55

How should the wagons be transported in order to minimize the total mileage covered.

Assignment Problem

527

26. A car hire company has one car at each of five depots a, b, c, d and e. A customer requires a car in each town namely A, B, C, D and E. Distances (in Kms.) between depots (origins) and towns (destinations) are given in the-following, distance matrix. e a b C d A 160 130 ' 175 190 200 135 120 130 160 175 B 140 110 155 170 185 50 50 80 D 80 110 54 80 105 E 34 70 How should cars be assigned to customers so as to minimize the diskance travelled ? [IAS 1994] 27. Six men are available for six different jobs. From past records the time in hours taken by different persons for different jobs are given below : Jobs Men

1

2

3

4

6

5

1

2 9 2 7 9 1

2

6 8 7 6 14 1

3

4 6 5 3 8 1

4 5 6

4 2 7 3 10 1 5 3 9 -5 -12 1 9 8 12 13 9 1

Find out an allocation of men to different jobs which will lead to minimum operating time. 28. Solve the following assignment problem having the following cost matrix : 1 2 3 4 5 6 7 A

35 20 60 41 27 52 44

B

51 39 42 33 65 47 58

C

25 32 53 41 50 36 43

D

32

28

40

46

33

55

49

E

43

36

45

63

57

49

42

F

27

18

31

46

35

42

34

G

48

50

72

59

43

64

58 [Agra 1999]

Operations Research

528 29. Solve the following cost minimizing jobs problem : Jobs I

II

III

IV

V

11

10

18

5

9

14

13

12

19

6

5

3

4

2

4

D

15

18

17

9

12

E

10

11

19

6

14

A

Machine

[Delhi BSc (Maths) 1993] 30. Solve the following minimal assignment problem : I

II

HI

IV

V

1 15

21

6

4

9

2

3

40

21

10

7

3

9

6

5

8 10

4 14

8

6

9

3

5 21

16

18

7

4

[Meerut 2005 (BP)] 31. An air-line that operates seven days a week has time-table shown below. Crews must have a minimum layover of 5 hours between flights. Obtain the pairing of flights that minimizes layover time away from home. For any given pairing the crew will be based at the city that results in the smaller layover. Delhi-Srinagar Flight No. Departure 1 07.30 A.M. 2 08.15 A.M. 3 02.00 P.M. 4 05.45 P.M. 5 07.00 P.M.

Srinagar-Delhi

Arrival Flight No. Departure Arrival 09.00 A.M. 101 07.00 A.M. 10.00 A.M. 09.45 A.M. 102 07.45 A.M. 10.45 A.M. 03.30 P.M. 103 11.00 P.M. 02.00 P.M. 07.15 P.M. 104 06.00 P.M. 09.00 P.M. 08.30 P.M. 105 07.30 P.M. 10.30 P.M. [Meerut 1994 (P), 96 (BP), 98, 98 (BP)]

Assignment Problem

529

32. Solve Q. No. 31, when the time-table is as follows and minimum layover time is

6 hours. Delhi-Calcutta

Flight No. 1 2 3 4

Departure 07.00 A.M. 09.00 A.M. 01.30 P.M. 07.30 P.M.

Calcutta-Delhi

Arrival Flight No. 09.00 A.M. 101 11.00 A.M. 102 03.30 P.M. 103 09.30 P.M. 104

Departure Arrival 09.00 A.M. 11.00 A.M. 10.00 A.M. 12.00 Noon 03.30 P.M. 05.30 P.M. 08.00 P.M. 10.00 P.M.

[Meerut 1999] 33. XYZ airline operating 7 days a week has given the following time table. Crews

must have a minimum layover of 5 hours between flights. Obtain pairing flights that minimizes layover time away from home. For any given pairing the crew will be based at the city that results in the smaller layover : Chennai-Mumbai

Flight No. Al A2 A3 A4

Departure 06.00 A.M. 08.00 A.M. 02.00 P.M. 08.00 P.M.

Mumbai-Chennai

Arrival Flight No. Departure Arrival 08.00 A.M. Bi 08.00 A.M. 10.00 A.M. 10.00 A.M. B2 09.00 A.M. 11.00 A.M. 04.00 P.M. B3 02.00 P.M. 04.00 P.M. 10.00 P.M. B4 07.00 P.M. 09.00 P.M.

[C.A. (May) 2000] 34. Find the optimal assignment for the given assignment problem. Machine

1

2

3

1

5 7 9

Job 2

14 10 12

3

15 13 16 [IAS (Maths) 1999]

35. Find the optimal solution for the assignment problem with the following cost

matrix. II

III

IV

V

A

11

17

8

16

B

7

12

6

C

9 13

20 15

16

15

12

16

D E

21 14

24 10

17 12

28 11

26 15 [IAS (Maths) 2000]

Operations Research

530

36. ABC company is engaged in manufacturing 5 brands of packed snacks. It is having five manufacturing setups, each capable of manufacturing any of its brands, one at a time. The cost to make a brand on these setups vary according to the following table. S2 S3 S4 Si "5 B1

4

6

7

5

11

B2

7

3

6

9

5

B3

8

5

4

6

9

B4

9

12

7

11

10

B5

7

5

9

8

11

Assuming five setups are S1, S2, 53 , S4, S 5 and five brands are B1, B2, B3, B4,B5, find the optimum assignment of products on these setups resulting in the minimum cost.' [C.A. November 1998] 37. A Private firm employs typists on hourly piece rate basis for their daily work. Five typists are working in that firm and their charges and speeds are different. On the basis of some earlier understanding, one job is given to one typist and the typist is paid for full hours even when he or she works for a fraction of an hour. Find the least cost allocation for the following data: Typist A B C D E

Rate per No. of Pages hours (Rs.) typed per hour 5 6 3 4N 4

12 14 8 10 11

Job

No. of Pages

P

199 175 145 298 178

Q

R S

T

[C.A. (Nov.) 1996i Delhi MBA Nov. 19961 Hint : Time taken (in hours) 'by typist A to do the job P, Q R, S, T are respectively 199/12 =16.581 175/12 = 14.58, 145/12 = 12.08, 298/12 = 24.83, 178/12 = 14.83 When typist A do the job P, Q R, S, T then he will be paid for 17, 15, 13, 25, 15 hours respectively. i.e., he will be paid Rs. 5 x 17 = 85, 5 x 15 = 75, 5 x 13 = 65, 5 x 25 = 125, 5 x 15 = 75 respectively. Similarly, payment to other typists can be calculated when they do the jobs P. Q R, S or T. Thus, the matrix giving the cost (in Rs.) incurred, if the i-th Typist (i = A, B, C, D, E) do the j-th job ( j = P, Q R, S, T) is as follows :

Assignment Problem

531 Job P Q R S T A

85

.75

65

125

75

B

90 75

78

66

132

78

66

57

114

69

72

60

120

72

64

56

112

68

Typist C D

80 76

E

Now solve by Hungarian Method. 38. A department head has four tasks to be performed and three subordinates. The subordinates differ in efficiency.. The estimates of the time, each subordinate would take to perform, is given below in the matrix. How should he allocate the tasks, one to each man, so as to minimize the total man-hour ? Men 1 2 3 26

15

27 20

6 15

D 18 30 39. Solve the following assignment problem :

20

A

Tasks

B

9 13 35

VI

I

II

III

IV

12

10

15

22

18

B 10 C 11 6 8 E

18

25

15

16

10

3

8

5

14 12

10 11

13 7

13 13

A

8 12 9 12 10

[Meerut 2002] 40. Solve the following unbalanced assignment problem of minimizing total time for doing all the jobs. Job 1 2 3 4 5

Operator

1

6

2

5

2

6

2

2

5

8

7

7

3

7

8

6

9

8

4

6

2

3

4

5

5

9

3

8

9

0

6

4

7

5

6

8

Operations Research

532

41. Use the Hungarian Method to find which of the two jobs should be left undone

when each of the four persons will do only one job in the following cost minimizing assignment problem. Job

Person

J3

J4

J5

JG

P1

10

9

11

12

8

5

P2

12

10

9

11

9

4

P3

8

11

10

7

12

6

P4

10

7

8

10

10

5 [Meerut L.P. 1996]

42. A company has 4 machines to do 3 jobs. Each job can be assigned to one and

only one machine. The cost of each job on each machine is given in the following table : Machine

w

X

1(

A

18

24

28

32

Job B

8

13

17

19

C

10

15

19

22

What are the job assignments whichl^:ill minimize the cost ? [Gauhati MCA 1992] 43. There are 3 persons P1, P2 and P3 and 5 jobs Ji, Jz , J 5. Each persons can do

only one job and a job is to be done by one person only. Using Hungarian method, find which 2 jobs should be left undone in the following cost minimizing problem. J1

"2

J3

J5

P1

7

8

6

5

9

P2

9

6

7

6

10

P3

8

7

5

6 [Meerut (LP) 1997, 98]

44. In a machine shop, a supervisor wishes to assign five jobs among six machines.

Any one of the jobs can be processed completely by any one of the machines as given below :

Assignment Problem

533 Machines A

1

13

13

16

23

19

9

2

11

19

26

16

17

18

Job 3

12

11

4

9

6

10

4

7

15

9

14

14

13

5

9

13

12

8

14

11

The assignment of jobs to machines be on a one to one basis. Assign the jobs to machines so that the total is minimum. Find the minimum total cost. [IAS (Maths) 1998] 45. A marketing manager has 5 salesmen and sales-districts. Considering the capabilities of the salesmen and the nature of districts, the marketing manager estimates that sales per month (in hundred rupees) for each salesman in each district would be as follows : Districts A

B

C

D

E

1

32

38

40

28

40

Salesmen 2

40

24

28

21

36

3

41

27

33

30

37

4

22

38

41

36

36

5

29

33

40

35

39

,

Find the assignment of salesman to districts that will result in maximum sales. [UP TECH MBA 2005-06; Agra 1998; Meerut 2009 (BP)] 46. The owner of a small machine shop has four persons available to assign to jobs for the day. Five jobs are offered with the expected profit in rupees for each person on each job being as follows : Job A

B

1

6.20

7.80

5.00 10.10

8.20

2

7.10

8.40

6.10

7.30

5.90

3

8.70

9.20

11.10

7.10

8.10

4

4.80

6.40

8.70

7.70

8.00

C

D

E

Person

Operations Research

534

Find the assignments of persons to jobs that will result in a maximum profit. Which job should be declined ? [Meerut 1995 (BP), 2006] [Hint : Introduce 5th fictitious row with zero costs to make the unbalanced problem to balanced one. Multiply each element by 10 to change the numbers to complete numbers and also multiply each element by — 1 to change it to minimization problem. Now proceed by Hungarian Method] . 47. A company is faced with the problem of assigning 4 machines to 6 different jobs (one machine to one job only).The profits are estimated as follows : Machines Job

A

1

3

6

2 -

6

2

7

1

4

4

3

3

8

5

8

4

6

4

3

7

5

5

2

4

3

6

5

7

6

4

Solve the problem to maximize the total profits. [Hint : Change the problem to minimization assignment problem by multiplying each element by — 1, then change the unbalanced assignment problem to balanced one by adding two more columns with zero costs.] 48. A manufacturing company has four zones A, B, C, D, and four sales engineers P, Q R, S respectively for assignment. Since the zones are not equally rich in sales potential, it is estimated that a particular engineer operating in a particular zone will bring the following sales. Zone A Zone B Zond C Zone D

420000 336000 294000 462000

The engineers are having different sales ability. Working under the same conditions, their yearly sales are proportional to 14,'9, 11 and 8 respectively. The criteria of maximum expected total sales is to be met by assigning the best engineer to the richest zone, the next best to the second richest zone and so on. Find the optimum assignment and maximum sales. [CA (May) 1998] [Hint : Proceed similarly as in Ex. 11] 49.

A company has five jobs to be done. The following matrix shows the return in Rs. assigning i-th (i = 1, 2, ..., 5) machine to the j-th job (j = 1, 2, ..., 5). Assign the five jobs to the five machines so as to maximize the total return.

Assignment Problem

535

Job/Machine 1 2 3 4 5

1

2

3

5 4 5 11 10 12 4 2 4 6 3 5 3 12 5 14 6 6 14 4 11 7 8 12 5 7 9

[Meerut 2010; Agra 2002] 50. Four engineers are available to design four projects. Engineer 2 is not competent to design the project B. Given the following time estimates needed to each engineer to design a given project, find how should the engineers be assigned to projects so as to minimise the total design time projects. Project D A

Engineer

1

12

10

10

8

2

14

Not suitable

15

11

3 4

6 8

10 10

16 9

4 7

51. Four engineers are available to design four projects. Engineer 2 is not competent to design the project B. Given the following time estimates needed to each engineer to design a given project, find how should the engineers be assigned to projects so as to minimise the total design time of four projects. Project D

A

Engineer

1

16

2

16

3

11

4

8

14

14

12

17

13

15

21

9

10

9

7

[Meerut 1996] 52. Five operators have to be assigned to five machines. The assignment costs are given in the table below : Machine A

B Operator C D E

I

II

III

IV

V

5 7 9 7 6

5 4 3 2 5

— 2 5 6 7

2 3 — 7 9

6 4 3 2 1

Operator A cannot operate machine III and operator C cannot operate machine IV. Find the optimal assignment schedule.

536

Operations Research

53.

The Secretary of a school is taking bids on the city's four school bus routes. Four companies have made the bids as detailed in the following table. Route 1

Bids Route 2 Route 3 Route 4

1

Rs. 4000

Rs. 5000

-

-

2

--

Rs. 4000

-

Rs. 4000

3

Rs. 3000

-

Rs. 2000

-

4

-

-

Rs. 4000

Rs, 5000

Company

Suppose each bidder can be assigned only one route. Use the assignment model to minimize the school's cost of running the four bus routes. [C.A. Nov. 1995]

•

+ ANSWERS + 9. 10. 11. 12. 13.

Job --> Machine : A->1,B-> 4, C -> 5, D -› 3, E-4 2. = 60 A -> B -> C- > D > IV A-> Job 2, B -4 Job 3, C-> Job 1. Mini. Time = 9 min. Programmer -> Programme : 1- C, 2 -> B, 3 --> A. Mini. Time = 280m Man-*Job: 1-*B, 2->C, 3->D, 4->A or 1 —> 2 ->13, 3 -> B, 4 -> A Mini. Time = 17 h 14. A -> III, B /V, C II, D -> I. Mini. Cost = 16 15. Sales man --> Territory : A --> or A -> B -> IV, CSI,D-->II In both cases mini. cost = 86 16. A -->gB-41, C -> or A— > IV, C —> V, D Mini. Cost = 34 17. Station -> Station : 1 -> 11, 2 --> 8, 3 -> 7, 4 --> 9, 5 10, 6 --> 12 Mini. Mil = 185 18. A->S,B->Q,C->P,D->R 19. Person -÷ Job : A -> 5, B1,C---) 4, D 3, E--> 2. Mini. Cost = 9 20. Job --> Man : --> 3, 1, HI -> 2, IV-> 4, V -> 5 or 3, H 1,III-44,1V-->2,V—> 5 21. A —> 2, B —> 3, 4, D —> 1. Min. Cost = 38

537

Assignment Problem

22. 1 -> III, 2-> 11, 3 -> V, 4 ->/, 5 /Vorl-->/1,:, 2->//, 3-->///, 4->V, 5-4/ Mini. Cost = 36 23. Man --> Job : 1 ---> III, 2 -> V, 3 -> I, 4 -> IV, 5>II or or

I> /II, 2 -> V, 3>1V, 4 -9 1, 5 -4 11

24. 1->11, 2-98, 3-> 7, 4-> 9,5-910, 6-->12 Mini. Distance = 125 kms 25. AIV,B->V1,C->V,D-->III,E->l,F-H Mini. T. Mileage = 66 26. Depot -4 customer : a ---> D, b > C, C--4 B, d > E, e > A Mini. Dist. = 570 kms 27. Man-9 Job: 1-43, 2->6,3 Or

4

6->5

1 ->3, 2-96, 3-94, 4->2, 5 1, 6--45

Mini. time = 22 units 28. A.-92,13 -> 4,C-> 6,D-> 1,E7,F-> 3, G ->5 MM. Cost = 237 units 29. Machine -4 Job : A --> II, B --> V, C--> III, D > IV, E -->I Mini. Cost = 39 30. 1--->/V, 2 32. 1 ---> 103,

4-9111, 5->V 2 ->104,

3 -> 101,

4 -9 103

Delhi

Delhi

Calcutta

Crew at Delhi

33. Al -4 B3 A2 -> B4, A3 -4 B1, A4 B2 Mini. Layover time = 40 hours 34. Job -> Machine : 1 -9 1, 2 -> 3, 3 -> 2 35. A -> I, B --> IV, C -> V, D --> HI, E -> II Mini. Cost = 60 36.

-> St, B2

55, B3 -> S4, B4

S3,B5 -> S2

Mini. Cost = 27 37. Typist --> Job : A T, B -> R,

Q, D -> P, E--> S

T. Mini. Cost = Rs. 399 38. Task -> Man : A -> 1, B-> 3, C --> 2 Task D is undone, mini. total man hours = 35 39. A ->

not assigned

40. Operator -> Job : 1 -> 4, 2 --> 1, 4 -4 2, 5 -> 5, 6 > 3 Operator 3 do not job. Mini. total time = 11 units

538

Operations Research

41. Person —> Job : P1 —> J5,P2 —> J6, P3 —> J4,P4, —> J2 Jobs J1 and J3 are left undone 42. Job —> Machine :A--)W,BX,C-->YorA—>W,BY,CX

In both cases no job is assigned to machine Z 43. Person Job : P1 or P1 —> J4,P2

J3, P2 12,P3 —> J4; Jobs Ji and J5 left undone J2,P3 —> J5; Jobs J1 and J3 left undone

In both cases minimum cost is 17 Machine : 1 —> F, 2 —> A, 3 —> E, 4 —> C, 5 Mini. Cost = 43

D

44. Job

45. Salesman —> District : 1 Max. Sales = Rs. 19100

B, 2 —> A, 3 —> E, 4 —> C, 5

46. Person —> Job : 1 —> D, 2 ---> B, 3 C,.4

D

E

Job A should be declined 47. Job —> Machine : 2

A, 3 —> B, 4 —> D, 6 —> C

Jobs 1 and 5 are left undone. Max. Profit = 28 units 48. Engineer —> Zone : P —> D, Q —> B, R A, S —> C Max. Sales = 392000 units 49. Job Machine : 1-* 5, 2 4, 3 1, 4 —> 3, 5 —> 2 Max. Return = 23 units 50. Engineer —> Project : 1—> B, 2 —>D, 3

A, 4 —> C •

Mini. total time = 36 hours 51. Engineer —> Project : 1-* B, 2 —> D, 3 --> A, 4 Mini. total time = 47 hours

C

52. Operator —> Machine (i) A —> IV, B III, C —> II, D or

(ii) A -->

or

(iii) A ----> IV, B —>

C V, D

EV E —> I

53. Company-Route : 1- 1, 2 —> 2, 3 —> 3, 4 —> 4 Mini. Cost = Rs. 15000 • • •

C Chapter

14

Transportation Problem •

14.1 Introduction The transportation problem is a particular class of linear programming problem in which the objective is to transport a commodity or to provide services from several supply origins to different demand destinations at a minimum total cost. The transportation problem can easily be expressed by linear relations as a linear progrFnming problem and can be solved by simplex method. But it takes a long time due to the presence of a- large number of variables and constraints. To overcome this difficulty the transportation algorithms have been developed for the solution of a transportation problem.

14.2 Transportation Problem [Meerut 1995 (P), 2000, 02, 03 (BP): 06 (BP), 07; Rohilkhand 2000]

The transportation problem can be described as follows : Suppose that the factories Fi (i = 1, 2, 3, ..., m), called the origins or sources produce the non-negative quantities ai (1 of a product and the non-negative quantities bi (j = 1, 2, n) of the same product are required at other n places, called the destinations, such that the total quantity produced is equal to the total quantity required In

= Ebi i =1

j=1

Also suppose that cif is the cost of transportation of a unit from the i-th source to the j-th destination. Then the problem is to determine xif the quantity transported from the i-th source to the j-th destination, in such a way that the total transportation cost c.- x.• is minimized. i =1 j =1

The transportation problem as described above can be represented in a tabular form as follows :

540

Operations Research

W1 Destinations —> Sources 1 F1 F2

c11 C21

F.

... cil ...

•• •

W2

...

W3

...

Wn Capacities of the sources

C12 C22 ... ... Ci 2 ...

...

Cu

...

C1„

•• • ... ...

C2j ... ...

•• •

C2n

tj ..• ...

•• •

•-

... ... ...

... Cin ...

•• •

•• •

... ...

••

-•

Fiii

Cm1

Cm 2

•• •

Cmj

•• •

C nut

Requirements -4

b1

b2

...

b I•

...

bn

a1 a2

a,

a,, n m I ai = I bi i=i. j=i

The calculations are made directly on the 'transportation array' given below which gives the current trial solution.

Destinations —> W1 Sources 1 F1 F2

Fi

X11 x21 .. ... xii ... •

W2

...

Wi

...

Wn

Capacities of the sources

X12 x2, ... ... xi 2 ...

"• •• • ... ... .. • ...

X1j x 2i ... ... Xii ...

•" .. • ... ... ... ...

Xln X20 .. ... xin ...

al a2

• ••

•• •

•• •

Fm

Xml

X m2

•• •

-1Cmj

"•

Xmn

Requirements —>

b1

b2

...

b 1•

...

bn

at

am m

n ai = I bi i =1 j=1.

I

The above two tables can be combined together by writing the costs cif within the bracket ( ), as follows :

Transportation Problem Destinations Sources

541

11/1

W2

•—•

Wj

•—

Wn

Capacities of the sources

al a2

1 Fl

xi, (cll. ) X/2 (C12 ) ...

)

• ••

x1tt (C1tt )

F2

x21 (C 21 ) X22 (C 22 ) ... X 21 (C 2i )

•• • ... ... xii (cil ) xi 2 (Ci 2 ) ... ... ... ... x ml (C m1 -) x„,2 (c m 2 )

Fi

F,,,

Requirements

b1

b2

Xii (Cli

• ••

X2r, (C 2,1 )

• •• ...

... ...

...

... ...

...

Xti (Cy )

. ..

Xin (Cin )

... ...

... ...

...

• •• Xmj (C mi ) • ••

X mn (C mn )

...

b 1•

ai

•• •

...

bn

am 171

n

I ai = I bi i =1

j=1

Mathematical Formulation-of a Transportation Problem [Meerut 2003 (BP), 05 (BP); Rohilkhand 2000]

Mathematically : The transportation problem can be stated as follows : Find (i= 1, 2,... , m; j = 1, 2, . . . n) for which the total transportation cost m n Z=

...(2)

13 . ..i1

i=1 j =1

is minimized, subject to the restrictions = 1, 2, 3, ... , m

=a

j =1 tn 1 1

i =1

=b• 1,

= 1, 2, 3, ..., tt

E

ai = Ebi i =1 j= 1 and xy

0, for all i = 1, 2,

(4)

m; j = 1, 2, ..., n.

Thus the transportation problem is a L.P.P. of special type, where we are required to find the values of m.n variables that minimize the objective function Z given by (2), satisfying (in + n) restrictions (3), restrictions (4) and the non-negative restriction of variables. Note : The restriction (4) indicates that one of the m + n constraints given in (3) is redundant. Thus, in m x n transportation problem' or}ly m + n — 1 equations form a •. linearly independent set. See theorem 2 of article 14.5.

Operations Research

542

14.3 Difference between a Transportation and an [Meerut 1999, 2001(BP), 02(BP), Assignment Problem 03(BP), 04(0), 05; Agra 1999; Rohilkhand 1998] An assignment problem is a special case of the-transportation problem in which m = ri and all the ai 's and bi 's (i = 1, 2, ..., m; j = 1, 2, n) are unity and each xii is limited to one of the two values 0 and 1. In these circumstances, exactly n of xii can be non-zero, one in each row of the array and one in each column showing that only one source (person) can be assigned to each destination (job).

14.4 Few Important Definitions Now we shall define few terms that are used in the transportation problems.

1. A Feasible Solution (A.F.S.)

[Meerut 2001 (BP)] A feasible solution to a transportation problem is a set of non-negative individual allocation (xi.. >— 0) which satisfies the row and column sum restrictions [i.e., restrictions (3) article 14.2].

2. Basic Feasible Solution (B.F.S.) [Meerut 2001 (BP); Rohilkhand 1999] A feasible solution of m by n transportation problem is said to be a basic feasible solution if the total number of positive allocations xijis exactly equal to m + n — 1; i.e., one less than the sum of the number of rows and columns. 3. Optimal Solution [Rohilkhand 1999] A feasible solution (not necessarily basic) is said to be optimal if it minimizes the total transportation cost.

4. Non-Degenerate Basic Feasible Solution A feasible solution of m by n transportation problem is said to be

non-degenerate basic feasible solution if (i) Total number of positive allocations is exactly equal to (m + n — 1) and (ii)These allocations are in independent positions. In other words, if a F.S. involves exactly (m + n — 1) independent individual positive allocations, then it is known as non-degenerate B.F.S., otherwise it is said to be degenerate B.F.S. Here by independent positions of the allocation we mean that it is always impossible to form any closed circuit (loop) by joining these allocations by horizontal and vertical lines only. See the following tables in which the positions of allocations are indicated by *. Independent positions

Non-independent positions Closed circuit (loop)

*

* *

* *

*

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543

Non-independent positions

Non-independent positions

Closed circular loop

Note that in fourth table the lines joining allocations in cells (2, 1), (2, 4) and (3, 3), (1, 3) intersect each other at the point (3, 3), where there is no allocation. 14.5 Some Important Theorems Theorem 1 : Existence of Feasible Solution A necessary and sufficient condition for the existence of feasible solution of am x n transportation problem is

ai = i =1 j =1

[Meerut 2007]

Proof : The Condition is Necessary : Let there exist a feasible solution to the transportation problem, then xij = a- i =1, 2,...,mandIxii =bi ,j=1, 2,...,n

=1

i =1

Summing over all i and j respectively, we get m n

E

=Iai and

i =1 j =1

i =1

m n

But

j =1 i =1

n m

xij i =1 j =1 E

m

11

.1 ,

=

11

tt

=

.1 j =1

Eai = I

xij

j =1 i =1

i =1

j =1

The Condition is Sufficient i.e., if

ai = i =1

bi

then there exists a feasible solution of the transportation problem. Let i =1 If xij =

= k(say)

=

j=i

bi for all i = 1, 2, ..., m and j = 1, 2,

n where X i # 0 is any real

number, then = Xi j=1

j=1

1—i

b• = kX i

j=1

=1 — k

x-• = I

k

544

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bi

Thus, x, 1 - X b3 -

k

0, for all i and j. (As ai > 0, bi > 0 and k > 0)

Hence, a feasible solution of the transportation problem exists. Remark : A transportation problem in which I ai = E b j is called a balanced transportation problem. Hence, from the above theorem, we can say that A balanced transportation problem always has a feasible solution. Theorem 2 : Out of (in + n) equations, in am x n transportation equation, one (any) is redundant and remaining in + n- 1 equations form a linearly independent set. Solution : Consider the following m + n -1 equations of am x n transportation problem. Xij =

i = 1, a..., m m-row equations

-(1)

j =1 111

and i =1

x-.1.1 = b j j = 1, 2, . . (n - 1), (n - 1)column equations

... (2)

where

ai = bj -.(3) i =1 j =1 By these m + n -1 equations given in (1) and (2) and equation (3), we shall get the remaining n-th coltimn equation. Adding m row-equations given in (1), we get m n Xij = i =1 j =1 i =1 Again, adding (n - 1) column equations given in (2), we get n -1 m n -1 = Eb j -(5) j =1 i =1 j =1 Subtracting (5) from (4), we get n -1 m m n Exii i=i j=1 j=1 1=1 m {n or

xij i =1 j =1 m

or m

n -1

i =1

j =1

n -1 n n -1 xu = bj bi j =1 j =1 j =1

n -1 n-1 x..ti + x. v

i =1 _ j=1

or

m

=E,

•

IXin = bn i =1

which is the n-th column equation.

j=1

=

using (3)

n -1 n -1 b•+b b•j j n j=1

j

545

Transportation Problem

i.e., if m + n — 1 equations are satisfied then m + n equations will be automatically if Eai = satisfied Thus, out of in + n equations, one (any) is redundant equation i.e., only m + n —1 equations are linearly independent. Remark : From the above theorem it follows that a B.F.S. of am x n transportation equation will contain at most m + n — 1 positive variables while the remaining mn — (m + n — 1)variables will be zero. Theorem 3 : Existence of an Optimal Solution There always exists an optimal solution to a balanced transportation problem.

14.6 Solution of a Transportation Problem The solution (optimal) of a transportation problem consists of the following two steps Step 1 : To find an initial basic feasible solution. Step 2 : To obtain an optimal solution by making successive improvements to initial basic feasible solution (obtained in step 1) until no further decrease in the transportation cost is possible.

14.7 To Find an Initial Feasible Solution [Rohilkhand 2003]

There are several methods for finding the initial feasible solution of the given transportation problem. Here we describe the following three simple methods. Method 1 : North-West Corner Rule By this rule we allocate a set of allocations in the cells so that the row totals and column totals will be as indicated before each, as follows : (i) Start with the cell (1, 1) at the north-west corner i.e., the top most left corner and allocate it maximum possible amount. Then move to the right hand cell (1, 2) if there is still any available quantity (ii) left, otherwise move to the down cell (2, 1) and allocate it maximum possible amount. (iii) Repeat the step (ii) again and continue until all the available quantity is exhausted. The method is well explained by taking the following numerical example: To

g ra.,

Fi F2 F3 F4

Demand

W1

W2

W3

Supply

(2) (3) (5) (1)

(7) (3) (4) (6)

(4) (1) (7) (2)

5 8 7 14

7

9

18

34

Operations Research

546 To Wi

From

F1

5(2)

F2

2(3)

F3

W2

5=a1 6(3) 3(4)

F4 7 = b1

W3

9 = b2

8 = a2 4(7)

7 = a3

14(2)

14 = a4

18 = b3

(i) We start with the top-most left corner and allocate it maximum possible amount 5. (Since Mini of al = 5 and b1 = 7 is 5). (ii) Since there is no amount left available at source F1 so we move downwards to the cell (2, 1) in place of moving to right and allocate it maximum-possible amount. Since the column 1 still need the amount 2 and the amount 8 is available in row 2, so we allocate the maximum amount 2 to this cell (2, 1). Now the allocation for column 1 is complete, so we move to the right of this cell. Since the amount 6 is still available in row 2 and amount 9 is needed in column 2, so we allocate the maximum amount 6 in this cell (2, 2). Thus, allocation for row 2 is complete, so we move downwards to the cell (3, 2). Since the amount 3 is still needed in column 2 and amount 7 is available in row 3, so we allocate the maximum amount 3 in this cell (3, 2). Thus allocation for column 2 is complete, so we move to the right of this cell. Since the amount 4 is still available in row 3 and amount 18 is needed in column 3, so we allocate the maximunramount 4 to the cell (3, 3). Thus, the requirement of the row 3 is complete, so we move to the downward cell (4, 3). Since amount 14 is still available in row 3 and an equal amount 14 is needed in row 4, so we allocate this amount 14 to this cell (4, 3). This complete the allocation and the resulting feasible solution (allocations) are shown in the above table. In the end it may be checked that the sum of rows and columns are as needed. On multiplying each individual allocation by its corresponding cost in ( ), and adding, the total transportation cost to this F.S. is = Rs. (5x 2+ 2 x 3+ 6x 3+ 3 x 4+ 4 x 7 +14 x 2)= Rs. 102 In this north-west copier rule we move to the right or down so no loop (circuit) can be formulated here by drawing horizontal and vertical lines to the allocations. Also at each step (allocation) at least one row or column is discarded from further consideration and at the last allocation both rows and columns are discarded. So we cannot get more than (m+ n — 1) individual positive allocation by this rule. Thus, we always get a non-degenerate basic feasible solution by this north-west corner rule. Method 2 : Lowest Cost Entry Method (or Method of Matrix Minima)

In this method we write the cost and the requirement matrix. The cost are written within bracket ( ). Now we examine the cost matrix carefully and choose the cell with lowest cost and allocate there as much as possible. If such cell of lowest cost is not unique, we select the cell where we can allocate more amount. Again we

Transportation Problem

547

examine the cost matrix and select the cell will the lowest cost (the cell in which allocation has been made is not considered) and allocate there as much as possible. We continue the process until all the available quantity is exhausted. The method is well-explained by taking the same numerical example as in method 1. First we write the cost and requirement matrix as follows To w2

5

Fl

(2)

2 (7)

3 (4)

F2

(3)

(3)

8 (1)

8

F3

(5)

7(4)

(7)

7

F4

7 (1)

(6)

7 (2)

14

7

9

18

0 44

Examining the cost matrix we find that there is lowest cost 1 in cell (2, 3) and in (4, 1). We choose the cell (2, 3) as we can allocate the maximum amount 8 to this cell which is more than the max. amount 7 that can be allocated to cell (4, 1). Leaving this cell we find that there is lowest cost 1 in cell (4, 1) where we allocate the maximum amount 7. Continuing in this way we get the required feasible solution as shown in the above table. The total transportation cost to this F.S. is =Rs.(2x 7+ 3 x4+ 8 x1+7 x 4+7 x 1+7 x 2)=Rs.83 This cost is less than the cost associated with the F.S. obtained by north-west corner rule. -_ The initial feasible solution obtained by this method usually gives a lower transportation cost than that obtained is north-west corner rule.

Method 3 : Unit Cost-Penalty Method (Vogels Approximation Method i.e. V.A.M.) In this method we write the differences of the smallest and the second smallest costs in each column, below the corresponding column and write the similar differences of each row to the right of the corresponding row. These individual differences can be thought of a penalty for making allocations in the second lowest cost cell instead of lowest cost in each row or column. Now we select the row or column for which the penalty is the largest and allocate the maximum possible amount to the cell with lowest cost in that particular row or column. If there are more than one largest penalty rows or columns, then select that row or column in which we can allocate more amount in the lowest cost cell. Then we cross (or leave) that row or column in which the requirement (or demand) has been satisfied and construct the reduced matrix. We continue this process on the reduced matrices till all allocations have been made. The method is well explained by taking the same numerical example as in method 1.

Operations Research

548 First we write the cost and requirement matrix as follows : w2

W3

(2)

(7)

(4)

5

F2

(3)

(3)

8 (1)

8

(2) — a, a > 0 . i =1 (ii) xi 0 for all i, (a) Show that fN (a) satisfies the recursive relation. fN (a) = Min. { x P + fN _ (a — X)} (:).)c 1, p —1 a fN (a) = N(— )

Meerut 1997 (BP)]

Operations Research

658 25. Consider the following "fixed charge" problem. Maximize Z = g.1 (xi ) + g 2 (x2 ) 3x3, ± X2 s.t. 2x1 + X2 + 3x3 30 and xi, x2, x3 >_0 , 0 if xi = 0 where gi (xi ) = - 4 + 5x1 if x1 > 0 0 g2(x2) =

if x2 = 0

- 10 + 7x2 if x2 > 0

Use the dynamic programming to solve this problem. •

+ ANSWERS .4. 6.

(cvn cvn

dinN), Min. Z = nc2in

7.

A (a) = Min. lz }, fn (a) = Min. { za + _1(a - z)}. o< z a z 5_ a

8.

x1 = 5/3 = x2 = x3, Max. Z = 125/27

9.

xi = 2 x2 = x3. Min. Z = 20

10. fi (c)= Max. {gi (z)}, fn (c) = Max. {g n (z) fn -1 (c z)} o

It is clear that this game has a saddle point in the cell (2, 1). So the value of this game is 5/2 to A.

Example 13 : Solve the following game graphically: B

(yi ) I A

(y 2 )

I

2

II

3

7 5

III

11

2

Solution : This 3 x 2 game has no saddle point. Therefore it is reduced to 2 x 2 game by the graphical method. If yl, y2 are the probabilities with which the pl2yer B uses his pure strategies, then yi + y = 1, yi 0, y 2 0 y2 =1- Yl

B's expected pay-off for different pure strategies used by A may be tabulated as follows : Pure strategy used by A I II III

E(v), B's expected pay-off E(v) = 2y1 + 7y2 = 2y1 + 7(1 — .Y1) = —5.Y1 + 7 E(v) = 3y1 + 5y 2 = 3y1 + 5(1 — yl ) = —2y1 + 5 E(v) = 1 lyi + 2y 2 = 1 lyi + 2 (1 — yi ) = 9yi + 2

Operations Research

690

The three pay-off lines are plotted in usual manner in Fig. 17.4 It is obvious from the following figure that the minimax occurs at the point P (the lowest point on the upper boundary) of intersection of pay-off lines E(v) = — 5y1 + 7 and E(v) = 9yi + 2. Hence, the best strategies for player A are I and III. E(v) 11 10

11 10 9 8

e

9 Upper 8 boundary 7 I

Minimax point

5

4

E(v) — 2yi -I-5

Thus, the given game reduces to the following 2 x 2 game : B

I A

(xi)

III (x2 )

I (yi )

II (y 2 )

2

7

11

2

If x1 and x2 are the probabilities with which player A chooses strategy I and III, using formulae of article 17.4, we have 5 5 9 9 x1 = 14' x2 = 14; Yi = 14' Y2 = 14. Here all = 73 and v=— 14 Hence, an optimal solution of the given game is 5 Optimal strategy for player A (9, 0, — ) 14 14 Optimal strategy for player B (5— 9 14'14 and

73 value of the game to A is — 14

2 = 7,

a 21 =11, a22 = 2

Game Theory (Competitive Strategies)

691

Example 14 : The following matrix represents the pay-off to P1 in a rectangulargame between P1 and P2

P2 19 15 — 5 — 2 P1 19 15 17 16 0 20 15 5 By the notion of dominance sticurthat,the game is equivalent to one represented by a 2 x 4 matrix which is a sub-matrix of the above. Then obtain a solution of the game graphically. [Meerut 2002 (BP), 03 (BP)]

Solution : The game has no saddle point. Since every element of 2nd row is greater or equal to the corresponding element of 1st row therefore player A will never choose pure strategy I. Thus, by rule 1 of dominance, strategy I is dominated by strategy II. So the game reduces (is equivalent) to the following 2 x 4 game : P2 I

P1

///

H

III

IV

19 15 17 16 0 20 15 5

If x1, x2 (= 1— x1) are the probabilities with which player P1 chooses his pure strategies II and III, the expected pay-off to player Pl for different pure strategies used by P2 may be tabulated as follows : P21 s pure strategy

E(v) = Pi's expected pay-off

I

E(v) = 19x1 + O. x 2 = 19x1

H

E(v) = 15x1 + 20x 2 = 15x1 + 20(1— x1) = — 5x1 + 20

III

E(v) = 17 xi. + 15x2 = 17x1 + 15 (1 — xi ) = 2xi + 15

IV

E(v) = 16xi + Sx 2 = 16x1 + 5 (1 — xi ) .- ;11xi + 5

These four pay-off lines are plotted in usual manner in the fig. 17.5 (on the next page). It is clear from the figure that the maximum occurs at the point P and the best strategies for player P2 are II and IV. The game reduce to the following 2 x 2 game : P2 H P1

IV Y2

H

15

16

III x 2

20

5

Here an = 15, a12 = 16, a21 = 20, a22 = 5 If y1 y 2 are the probabilities with which the player P2 chooses his II and IV pure strategies then from article 17.14, we have

Operations Research

692 E(V) 22 20

I

4'(k-)

18 16 15 14 12

's" `I't

22 20 19 18 17 16 15

x-',,.)

10 8 6 5 4 2 0 I

II

Fig. 17.5 1 15 11 5 6 x2 = 16 xl = 1 Yi 16 Y2 = 16 245 v=— and 16 Hence, an optimal solution of the game is Optimal strategy for player P1(0, — 15, — 1 ') 16 16 Optimal strategy for player P2 (0, — 11 , 0, —5 6) 16 1 and the value of the game =

245 . 16

17.18 Algebraic Method for the Solution of a General Game This method known as algebraic method is a direct approach to solve any game. Here we first convert the game into a system of inequalities and then solve them. The method is quite lengthy when there are more strategies. Therefore in case of large 'games the problem is solved by transforming in into a L.P. problem (see article 17.20). Let ],,, be the pay-off matrix of a rectangular game between two persons. Suppose X = (x1, x2, , yn ) be the mixed xm. )5 Y = (yi , Y2, optimal strategies of player A and B respectively. ThenA's expected gains when B used his pure strategies 1, 2, ...., n respectively are In

aii xi ,

in

ai2x1,

,

air, xi

If v is the value of the game then, the minimum expected gain of A is v.

GamekTheory (Competitive Strategies)

693

1/1

ail xi i =1

V,

ai2 xi > v, i =1

a.t n -vI. > V

,

...(1)

i =1

Similarly considering B's expected losses and considering the fact that the maximum loss of B is v, we get the following system of inequalities : n n n Eaciii yi v, I a2)• y)•

v, then corresponding y. = 0

Theorem 2 : If A's optimal policy is a mixed strategy in which exactly r pure strategies have non-zero probabilities, then the optimal strategy of other player also involves exactly r pure strategies.

gthicifhatilb2 eXaMidea Example 15 : The pay-off matrix for 41in a two person zero-sum game is given below. Determine the value of the game and the optimal strategies for both players. B II III 1 1 2 -1 —2 2 3 —3 4

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694

_Solution : The game does not have a saddle point and cannot be reduced to 2 x 2 game by the dominance rules. Therefore we proceed to solve this game by algebraic method. Let (x1, x2, x3 ) and (yi, y2, y3 ) be the optimal mixed strategies of the two players A and B respectively and v the value of the game. Proceeding as usual we get the following relations : ...(1) —1.x1 + 1. x 2 + 3.x3 v For player A

For player B

2.x1 — 2. x2 + 4. x3 v 1.x1 +2.x2 2. x2 — 3. x3 ?. v — 1.y1 + 2. y 2 -I- 1.y3 v — 2.y 2 +2.y3

Also

and

v

341 + 442 — 343 5 v x1 + X2 + X3 = 1 Yi Y2 +Y3 = 1 xi, x2, x3 0

...(2) —(3) ...(4) —(5) ...(6) —(7) ...(8) -.(9)

Y1, Y2) Y 3 Now the problem is to find the values x1, x2, x3; yi, y2, y3 such that all the above relations are satisfied. For this first we consider the inequalities (1) to (6) as strict equations (10) to (15) as follows : ...(10) — x1 + x2+ 3x3 = v 2x1 — 2x2 + 4x3 = v + 2x2 — 3x3 = v —

+2y 2 +y3 =v

...(11) ...(12) ...(13)

2Y3 = v

...(14)

3Y1 + 4Y 2 — 3Y 3 = V Adding (10) and (12), we have,

...(15)

Y1 — 2Y2

3x2 = 2v

•

x2 = (2/3) v

Adding 2 times of (10) in (11), we have 10x3 = 3v

•

x3 = (3/10)v

Putting these values in (12), we have 17v =— 30 Substituting these values of x1, x2, x3 in (7), we have v = 15/23 x1= 17/46, x2 = 10/23, x3 = 9/46

which are all non-negative.

Game Theory (Competitive Strategies)

695

Again adding (13) and (14), we have 3y 3 = 2v y 3 = 10/23 Adding 3 times of (13) in (15), we have 10y2 = 2v y 2 = 6/23 Putting the values of y 2, y3 and v in (14), we have yl =7/23 These values of y1 y2, y 3 also satisfy (8) and also non-negative restriction. Thus, they constitute the solution of the given problem. Hence, the solution of the game is

and ,

Optimal mixed strategies for player A (17/46, 10/23, 9/46) Optimal mixed strategies for player B (7/23, 6/23, 10/23) the value of the game v = 15/23

Example 16 : Solve the game whose pay-off matrix is

B I (yi )

A

II (y 2 )

III (y 3 )

I (xi )

3

—2

4

II (x 2 )

—1

4

2

III (x3 )

2

2

6

Solution : The game does not have a saddle point and cannot be reduced to 2 X 2 by the dominance rules. Therefore we proceed to solve this game by algebraic method. Let (x1, x2, x3 ) and (yi, y2, y3 ) be the optimal mixed strategies of the two players A and B respectively and v the value of the game. Proceeding as usual we get the following relations for player A ...(1) 3x1 — 1x 2 + 2x3 v ...(2) —2x1 + 4x2 + 2x3 v

For player B

Also

4x1 + 2x2 + 6x3 v 3y — 2y 2 + 4y3 5 v

—(3) ...(4)

—lyi + 4y 2 + 2y3

—(5)

2yi + 2y 2 + 6y3 v x1 + x2 + x3 =1

...(6)

+ Y 2 +Y3 = 1 xi , x2, x3 0

...(8)

and

—(7)

...(9) Y2) Y3 °

Now to find the values of x1, x2, x3; y1, y 2, y3 satisfying all the above relations, first consider the inequalities (1) to (6) as strict equations (10) t‘,-) (15) as follows :

Operations Research

696 3x1 — x2 + 2x3 = v

...(10)

— 2x1 + 4x2 + 2x3 =v

...(11)

4x1 + 2x2 + 6x3 = v

...(12)

3y1 — 2Y2 + 4Y3 = v

...(13)

— Yt + 4Y2 + 2Y3 ..-= v 2y1 + 2y 2 + 6y 3 = v

...(14) ...(15)

Subtracting (11) from (10), we have . x1 =x2 . Putting x2 = x1 in (11) and (12), we-have 2 xi + 2x3 = v, 6x1 + 6x3 = v From these equations, we have V = 0, X2 = 0

x1 = 0, x3 = 0

But x1 = 0 = x2 = x3 does not satisfy (7). Hence, the system is inconsistent. Therefore we try to find a solution taking one or more of the inequalities as strict inequalities. Now we consider the following set of relations : 3x1 — x2 + 2x3 = v

...(16)

— 2x1 + 4x2 + 2x3 = v

...(17)

4x1 + 2x2 + 6x3 > v

...(18)

4y3 = v

...(19)

Yi 4Y2 2Y3 < v 2y1 + 2y 2 + 6y 3 = v

...(20)

3Y1

2Y2

...(21)

The relation (18) and (209, with the help of therm 1, article 17.18 imply that y 3 = 0 and x2 = 0. , -When x2 = 0, from (16) and (17), we have 3x1 + 2x3 = v and — 2x1 + 2x3 = v Solving we have xi = 0, x3 = 0 But these values of xi , x2, x3 do not satisfy (18) and (7). Therefore this system is also inconsistent. We again try another set of relation to get the optimal solution. Let us consider the following relations : ...(22) 3x1 — x2 + 2x3 = v ...(23) — 2x1. + 4x2 + 2x3 = v ...(24) 4x1 + 2x2 + 6x3 > v ...(25) 3Yi 2Y2 4y3 < ...(26) — yi + 4y 2 + 2y 3 =-4, 2y1 + 2y 2 + 6y 3 = v

...(27)

Game Theory (Competitive Strategies)

697

that The relations (24) and (25), witlt the help of them. 1, of article 17.18 imply y 3 = 0andx1 =0 When x1 = 0 from (22) and (23), we have —x2 + 2x3 = v and 4x2 + 2x3 = v Solving, we have x2 = 0, x3 = v/2 Putting these values x1 = 0, x2 = 0, x3 = v/2 in (7), we get v = 2.

x3 = 1

These values of x1, x2, x3 and v satisfy relation (24). Again when y 3 = 0, from (26J and (27), we have —Yl + 4y 2 = v and 2y1 + 2y 2 =-1) Solving, we have yi = (1/5)v = 2/5,y2 = (3/10)v = 3/5 These values yi = 2/5, y 2 = 3/5, y 3 = 0 also satisfy relations (25) and (8). Also the value of x1, x2, x3; yi, y 2, y 3 also satisfy the non-negative restrictions (2). Thus, we have got the optimal solution of the given game which is as follows : (i) Optimal strategy for player A is (0, 0, 1) (ii) Optimal strategy for player B is (2/5, 3/5, 0) and (iii) the value of the game = 2 Note : We can also solve this game by graphical method after deleting B's strategy III which is dominated by B' s I strategy. Also this game may be solved by L.P. Method which will be discussed in article 17.20.

17.19 Iterative Method for Approximate Solution By iterative method we can find the approximate value of the game to any desired degree of accuracy. In this method it is assumed that each player acts under the assumption that the past is the best guide to the future and will play in a manner to maximize his expected gain and tcl minimize his expected loss. This method is based on the following principle : "The two players are supposed to play the plays of the game iteratively and at each play the players choose a strategy which is best to himself or worst to opponent, in view of what the opponent has done upto that iteration". Let A and B be the maximizing and minimizing players of a game. If A starts the game by selecting arbitrarily his strategy Ai . Then the opponent player B will examine the strategy Ai of player A and choose that strategy Bj which is best to himself or most unfavorable to A i.e. B will choose strategy BB which correspond to the smallest element inA's strategy Ai . Now A examines this strategy B3 of B and will

698

Operations Research

respond to a strategy Al.which will maximize his average gain. For this A selects that strategy which corresponds to the pair of strategies Ai and Ar. By these strategies of A, player B will choose that strategy which will minimize his average loss. For this B adds both the strategies of A and choose the strategy which correspond to the least element. Now again A chooses a strategy which corresponds to the largest element of the sum of the two strategies of B. A and B continue choosing their strategies in the similar manner. This process is continued at least for 10 iterations. In the last, the mixed strategies of both players are obtained by dividing the number of times the pure strategies are used by the number of iterations. The lower and upper bonds of the game are found by dividing the smallest and largest elements of last strategies used by the players A and B respectively. For complete demonstration of the process see Example 17. Example 17: Solve the following game approximately.

A

1

—1

—1

II

—1

—1

3

III

—1

2

—1

[Meerut 2001 (BP), 02(BP), 03(BP)] Solution : To find the approximate solution of the given game we shall use iteration method. For this we shall put all necessary information in a systematic form as follows: Here we shall write the strategy of A (say II which is superior to others below the matrix and circle the smallest element — 1 which will indicate the B's next strategy to be selected. Now B will select the strategy/ corresponding to this smallest element —1 and write this on the right of the matrix and circle the largest element 1 which will indicate the A's next strategy. In case of tie in selecting the strategy for any player, the strategy different from the previous strategy used will be selected . PlayerA chooses the strategy I corresponding to the largest element in B's previous chooses strategy and add this strategy to his own previously chosen strategy. The sum of these two strategies is written below the matrix and its smallest element —2 is circled which will indicate the B's next strategy to be selected. Then player B chooses the strategy II corresponding to this smallest element 2 of A's strategy and add it in his own previously chosen strategy. The sum of these two strategies is written on the right of the matrix and its largest element 1 is circled which will indicate A's next strategy to be selected. We continue the process at least for 10 iterations (see the figure).

Game Theory (Competitive Strategies)

1 -1 -1 Ol All -1 -1 III -1

o

C)

699

o -1

®

C)-1. 0 C

o

3 -1-2 -3 -4 -1 -2 0 0 030 1 0-1 4/13

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

3

0

2 1

0

5/13

4/13

0®0

0

-1

3 -2 -1 2 03 0 0 1 -1 001

0-1

2

-2 5 -2 0 34 -3 0 7 6/13 4/13 3/13 Now the approximate strategies of the two players are found by dividing the number of circled elements in each row and column by the total number of iterations. Thus, here A's approximate strategy is (5/13, 4/13, 4/13) and B's approximate strategy is (6/13, 4/13, 3/13). The lower bound of the game is determined by dividing the smallest element - 4in the last row by 13 (the number of iterations) and similarly the upper bound of the game is determined by dividing the largest element -1 in the last column by 13 (the number of iterations). Therefore , if v is the value of the games then - 4/13 v 5_ - 1/13 Note : The approximate strategies of the two players can he made close to their optimal strategies by increasing the number of iterations.

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Alternate Method : Students are advised to solve this game by algebraic method and by linear programming also -By these methods solution of the problem will be as follows : A's optimal strategy (6/13, 3/13, 4/13) B's optimal strategy (6/13, 4/13, 3/13)

the value of the game = —1/13.

and

17.20 Equivalence of the Rectangular Matrix Game and Linear Programming [Meerut 2000, 02, 03, 04(0); Rohilkhand 1997] Consider a rectangular game played by two players A (maximizing player) and B (minimizing-player) with pay-off matrix [aij Let X =[x1,x2„ xm ], Y =[Y 1 ,y 2 ,Y n Ibe the mixed strategies of the two players A and B respectively by which A and B select their pure strategies. In article 17.12 we have shown that A selects his optimal mixed strategies which will an xi , i =1 i =1 x1 + X2 +

S.t.

ai„xi i =1

+ X„i = 1

= 1, 2 „ m . and xi 0, Also B selects his optimal mixed strategies which will n

n

n

Min. [ Max. { I al./ y j, I a2j y j,...., I ctini yi } Yj ' j =1 j =1 j =1 s.t.

and

Y1 + Y2 + •••• Yn = 1 j = 1, 2 , . . . . , n . y j 0, {171

If

ai Min. ~aii , ai2 ,...., i =1 i =1 i =1

=v

than A expects to gain at least v. The criterion for A to choose his strategy is to maximize his least gain v. Since v is minimum of all expected gaips therefore, we have ail Xi i =1

v,

I

ci• att 2x.t — > v3 . " .3 m x• > —v i =1 i =1

Thus, A's problem is to determine xi. „ Maximize Z = v subject to the constraints

, to

Game Theory (Competitive Strategies)

701

all xl + a21x2 +

+ and xm v

a12 Xl+ a22x2 +

+ am2 xm > v

a1n xl xl+

a2n x2 + • " • + a 17111

Xm

?- 1/

x2 +.... + xn, =1

and xi, x2, ....

0

We assume that v is positive. For v > 0, it is enough that all the elements of the pay-off matrix are positive. If all the elements are not positive then we can add a sufficient large quantify (say K) to every element of the pay-off matrix so that they

all become positive. By doing so the value of the game is also increased by K(see Prop. 3 article 17.13) but the solution remains the same.. Thus, we can take v to be positive. Dividing equations of (1) by v and taking xin X2 = X2, = — = X, , we have V

a11X1 a21 X2 +, a12X1

a22 X2 +.

+ and X, > 1 + a m 2 X rn >— 1

+ a2n X 2 +

+ am, X, ?.. 1 1 X1 + X2 + +X = In V Xl , X2, X„ 0

and

Since

Max. v = Min. (11

—1

= Min.

xi -I- X2 +

+ X,

i J

= Min. (Xi + X2 + + X„ ) Thus, the given rectangular game reduces to the following L.P. problem.

1 Min. x* = — = 1 + X2 + + X, V subject to the constraints a.11 Xl + a21 X2 + ....+ am] X m > 1 an Xi. + 0 22 X2 " "+ am 2 X,,,> 1 •••

and

...(2)

•• •

+ ainn Xrn a1mX1 + 0 2n X2 + 0, X2 > 0„ X„

0

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702

Now considering the problem from B's point of view who want to minimize v (because the gain of A is the loss of B), proceeding in the similar manner we get following L.P. problem : Maximize y* = 1= Yl + Y2 + subject to the constraints an Y1 + ai.2172 (122112 +

a21

ain Yr/

1

a2n Yn

1

—(3)

and Y1 + arn2Y2 + ....+ a„ Yr, 5_ 1 and where

?_ 0, Y2 0, Y^ =

1

=

, Yn 0 Yn V

Yi v Y2 2 = —) • • • •) Ytt V V

It can be seen that the L.P. problems given in (2) and (3) are the duals of each other i.e., B's problem is the dual of the A's problem and vice-versa. Therefore if one problem is solved then the other is solved automatically. After getting the values of Xi , Yj and min. of

Xi , which is equal to max. of

EYE , we can find the values of xi and yi from and

xi = vXt

=v Y3 .

17.21 Fundamental Theorem of Game Theory (Minimax Theorem) Theorem : Eveiy game can be solved in terms of mixed strategies i.e., if mixed

strategies are adopted there always exists a value of the game i.e., v = v = v where v and i7 are maximin and minimax values of v. Proof : It has been shown in article 17.20 that if X = (x1, x2, ..., xni )

y n ) are the mixed strategies of the two players where Y (yi, y 2, xl x2, ...., ; yi, y2, ...., y n are the probabilities with which they choose their pure strategies than Player A's problem is To minimize s.t.

x* =

+ X2

a11X1 + a21X2 +

Xrn + ctini Xin > 1

a12X1 a22X2 + • • • • + atn2Xm

ain Xi + a2n X 2 +

+

1

703

Game Theory (Competitive Strategies) and

Xi , X2 ,...., Xm 0

where

-V j = 1, 2,...., m.

=

And player B's problem is To maximize

+ Y2 4- • • • • + Yr/

s.t.

all Yl + an Y2 +

+ aln Yn < 1

a21Y1 + a22Y2 + •••• + a2r/ Yn

a in1Y1 + am 2172 + • • • • + amn and where

Y; =

1

5. 1

171) Y2) • —3 Yn ° Yj v

, v

=1, 2,.... , n

It is clear that B's problem is the dual of A's problem. But from duality theorem, we know that "if either the primal or the dual problem has a finite optimal solutions, then the other problem has a finite optimal solution and the optimal solution and the optimal values of the two objective functions are equal. i.e.,

Max, y" = Mini. x'

or

v = v = v, where v is the value of game.

17.22 Sal,_IL;on of a Rectangular Game by Simplex Method [Meerut 2005] If a problem has no saddle point then we convert it to a L.P. problem for player B by the method discussed in article 17.20. This L.P. problem in solved by simplex method and fr,-)in the final simplex table of the solution we can read the solution for player A by he dual method (see Chapter 9). For clear Inderstanding of the method see the following example 18. Example 18 : Find the best strategies and the value of the following game : B 1 —1 3 5 —3 6 2 —2 [Meerut 2003, 06] Solutioci Ffirst we construct the table for row minimums and column maximums as follows :

Operations Research

704 B

Row-Minimum

II III

1 A

II III

6

Column Maximum

0

C)

3

Tcl

0

2

1C)

5

3

Maximin Value (_v )

Minimax Value ( i7) It is clear from the above table that the value v of the game lies between —1 and 3. —1< v < 3 i.e. It is possible that the value of the game may be negative or-zero. Thus, we add a constant K= 4 to all the elements of the matrix so that all these elements of, the matrix becorre positive assuming that the value of the game represented by this new matrix is non-negative and non-zero. The transformed (reduced) matrix is as follows : B

A

I

II

III

I

5

3

7

II

7

9

1

III

10

6

2

Let X = (x1, x2, x 3 ) and Y = (yi, y 2, y3 ) be the mixed strategies of the two players respectively, where x1, x2, x3; Yl , y 2, y3 are the probabilities with which they choose their pure strategies. Therefore B's problem is To find (yi , y 2 , y 3 ) which minimize v' subject to 5.Y1 + 3Y2 + 7Y3

V/

7Y1 + 9Y2 + Y3

+ 6y 2 + 2y 3 5 v' and

Yi + Y2 + Y3 = 1 Yl , Y2, Y3

Here v' > 0, dividing above equations by v' an putting y1/v' = Yi , y2/v' = Y2, y 3/v' = Y3 the B's problem reduces to the following L.P. problem :

Game.Theory (Competitive Strategies)

1

Max.

Z - V'

s.t.

705

+ 2+ 3 V

- Y, + Y2 + Y3•'

5173. 3Y2 7Y3 1 7Y1 + 9Y2 + Y3 1 1 (WI+ 6/72

and

2Y3 5_ 1

li, Y2 Y3

Introducing the slack variables, the constraints reduce to the following equations : =1

5Y1 + 3Y2 + 7Y3 + Y4 7Y1 + 9Y2 + Y3

+ Y5

=1

10171 + 6Y2 + 2Y3

+ Y6 = 1 Taking Y1 = 0, Y2 = 0, Y3 = 0, we have Y4 = 1, Y5 = 1, Y6 = 1, which is the starting B.F.S. Computation work by simplex method is shown in the following table :

B

CB

cI

1

XB

°CI.

1 a2

1

0

0

0

Min. Ratio

a3

a4

a5

a6

X.13/ a3

a4 a5

0

1

5

3

7

1

0

0

1/7 (Min)-*

0

1

7

9

1

0

1

0

1/1

a6

0

1

10

6

2

0

0

1

1/2

Z =C B XB = 0

Ai

1

1

1 I

0 1

0

0

X /a 2

a3

1

1/7

5/ 7

3/7

1

1/ 7

0

0

1/3

a5

0

6/ 7

44/7

60/7

0

-1/7

1

0

1/10(Min.)->

a6

0

5/7

60/7

36/ 7

0

-2/7

0

1

5/36

Lj

2/7

4/7 "r

0

-1/7

0 1

0

Z= C B X B =1/7

a3

1

1/10

2/5

0

1

a2

1

1/10

11/15

1

0- 1/60

7/60

0

a6

0

1/5

24/5

0

0

-3/5

1

-2/15

0

0

- 2/15 -1/15

0

Z=C B XB = 1/5

Ai

3/20 -1/20

- 1/5

0

Yl =0, Y2 = 1/10, Y3 = 1/10 and Max. Z = 1/5 from Z = 1/v' we have Min. v' = 5 Yl =171 v' 0, y 2 = Y2 V' = 1/2 , y3 = Y3v' = 1 / 2

and

value of the original game v = v' - 4 = 1.

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Since A's strategy are the dual of the above problem. x3 xi 2 x2 1 = = ,A 2 = = , X3 = =0 v' 15 v' 15 v' (values of A 4 , A 5 , A 6 with sign changed) x1 =2/3,x2 =1/3,x3 =0. Hence the optimal solution is (i) Best strategy for A (2/3, 1/3, 0) (ii) Best strategy for B (0, 1/2 1/2) (iii) ValUe of the game v = 1.

17.23 Matrix Method for nxn (i.e, Square) Games If the pay-off matrix A of a two person zero-sum game is a square matrix without saddle point (original or obtained by using dominance rules), then the mixed optimal strategies for the two players can be obtained by matrix method. The step by step procedure of this method is as follows : Step 1 : Subtract each row of A from the row above it (i.e., subtract the 2nd row from the Ist row, 3rd row from the 2nd and so on) and write the resulting rows in order below matrix A. Thus, we shall get a new matrix R of order (n — 1) x n below A. Step 2 : Subtract each column of A from the column to its left (i.e., subtract the

2nd column from the 1st column, 3rd column from the 2nd column and so on) and write the resulting columns in order to the right to the matrix A. Thus, we shall get a now matrix C of order n x (n-1) to the right of A. Step 3 : Compute the magnitudes of oddments corresponding to each row and each column of A and write in the corresponding row and column of A as shown below on page 707. The oddment corresponding to i-th row, Ai ofA denoted by 0Ai is the determinant ICi I, where Ci is obtained from C by deleting its i-th row. Similarly, oddment corresponding to j-th column Bi of A, denoted by Ogi is the determinant I R • I , where Rj is obtained from R by deleting its j-th column. Step 4 : Find the sum of magnitudes of oddments for rows and columns. i.e., find E I 0Ai I and E 104i I. Step 5 : If the sum of magnitudes of oddments of rows and columns ofA are equal

i.e., if E I 0Ai I = I OBJ I = S (say) i.e. if both players use all their plays in their best strategies, then I0Ai l/S and I0Bi l/S are the probabilities with which the players A and B will choose their strategies. But if E 10A, I # I OBj I, then the method fails. All the entries are written as follows : (see on next page 707). Taking matrix A of order 3 X 3, (For simplification).

Game Theory (Competitive Strategies)

B1 Al [all

707

Matrix A B2 B3

Matrix C

a12 a13 A2 a21 a22 a23 A3 a31 a32 a 33 3 x 3 Matrix R I r11

Mag. of oddments

[C31 C32 3

10A21

10,ilI s I 0 A21 s

I °A31

I0A31

10 Ai l

C11 C12 C21 C22 x2

Prob.

s

r12 113 r22 1.23 2 x 3

Mag. of oddments I Off il, 10821) I °631 El 0Bi I = ZIOAi I = S (say) Probability :

1 0m I 101321 1 0B3I , , S S S

For clear understanding of the method see the following example : Example 19 : Solve the following game by Matrix method.

3 —1 1 2 —2 3 2 3 2 —2 —1 1 Solution : There is no saddle point of the given game. Applying dominance rule

4th column is dominated by the 3rd column. The size of the resulting 3 x 3 matrix cannot be reduced further by dominance rules. So we solve this 3 x 3 game by matrix method step by step. All computation work are shown in the table below. Step 1 and 2 : Subtracting each row of the matrix from the row above it we write the resulting rows below the matrix and subtracting each column from the column to its left, we write the resulting columns to the right of the matrix. Step 3 : Oddments corresponding to the rows and column of the given matrix are as follows : Matrix C Mag. of Oddments Bi B2 B3 I°A1I= 1 Al 3 — 1 1 4—2 1 —5 A2 — 2 3 2 1 0 A2I — 4 A3 2 — 2 — 1 4—1 10A3I = 6 Matrix R

[ 5—4—1 —4 5 3

]

Mag. of Oddments 10m. I = 7, I OB21 = 11, OB31 — 9 0A1 =

—5 1 = 1, 4 —1

Om =

—4 — 1 = 5 3

Here

°A2 =

i) LIB2

I °Bi I (= 27)

4 —2 = 4, 4 —1 5 —4

°A3

= 11, OB3 =

E I 0Ail(= 11)

4 —2 =—6 1 —5 5 —4 =9 5 -4

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Hence, the matrix method fails. In this case the method of linear programming may be used. Using the method of linear programming (i.e., simplex method) the optimal solution of the game is (left as an exercise for the students). A(5/9, 4/9, 0),B(4/9, 5/9,0, 0), v =7/9 Example 20 : Solve the following game by Matrix method —1 1 1 2 —2 2 3 3 —3 Solution : There is no saddle point of the game and the size of the game cannot be reduced by dominance rules. Also it is a square matrix, so it can be solved by matrix method. Step 1 and 2 : Subtracting each row of the matrix from the row above it we write the resulting rows below the matrix and subtracting each column from the column to its left, we write the resulting columns to the right of the matrix. (See table) Step 3 : Oddments corresponding to the rows and columns of the given matrix are as follows : Matrix C Magn. of Oddments Prob. B1 B2 B3 24/44 = 6/11 IOA1I = 24 Al —1 1 1 A2 2 — 2 2 12/44= 3/11 4—4 1 0'121 — 12 A3 3 3 — 3 r 0 26 0 8/44 = 2/11 °A31= 8 Matrix R

[— 3 3—1 -1 —5 5

]

Mag. of oddments I OBI I = 10, I OB2 I = 16, I OB3 I = 18 10 5 16 4 18 9 Prob. : = = = 44 22' 44 11' 44 22 Here I I OBi I = 44 = E I 0Ai I Hence, the optimal strategies of the two players are A(6/11, 3/11, 2/11), B(5/22, 4/11, 9/22) and

v = — 1 • (6/11) + 2. (3/11) + 3 • (2/11) = 6/11

17.24 Summary of Methods for Solving the Rectangular (Two Person Zero Sum) Games For solving a rectangular game the method discussed in this chapter are used systematically in the following order : 1. Search for saddle point. 2. If there is no saddle point then if possible reduce the size of the game by using the dominance rules. 3. If the reduced size of the game becomes 2 x 2 without saddle point then use the formulae or the method of article 17.14 to find its solution.

Game Theory (Competitive Strategies)

4. 5. 6. 7. 8.

709

If the game reduces to the size in x 2 or 2 x n then use graphical method to reduce it to the size 2 x 2 and then apply formulae or method of article 17.14. If all above methods fail then try to solve by a algebraic method. If all fails in case of large games then use simplex method by converting the problem into a L.P. Problem. Use arithmetic method (i.e. oddment method) of article 17.16 or the matrix method of article 17.23, if asked to solve by any of these methods. Use iterative method for approximate solution.

17.25 Minimax and Maximin of a Function of Several Variables In a m x n game between two players A and B let (x, y) be the expectation function, x€E n ,y E E n' . If x is kept fixed and y is varied then E(x, y) will have minimum value say 4) for some value of y. i.e., (I) = Min. E (x, y) y

The value of (I) will change with the change in the value of x. Thus, is a function of x and we write. (x) = Min. E (x, y) y

Now 4) (x) will be maximum for some value of x. 4) 1 = Max. ((x) = Max. Min. E(x, y) x y

x

In the similar way be keeping y fixed and varying x we can find maximum value of E (x, y) which will be a function of y. i.e. w(y) = Max. E(x, y) x

yr (y) will be minimum for some value of y. i.e., yrl = Min. w(y) = Min. Max. E(x, y) y x

y

Theorem : Let E (x, y) be such that both Max. Min. E(x, y) and Min. Max. E(x, y) x y

Y x

exists, then Max. Min. E(x, y) Min. Max. E(x, y) x y

y x

Proof : If xo and y 0 are two arbitrary chosen points in E" and E respectively, then we have ...(1) E (xo, y o ) Max. E(x, y o ) x

and

...(2)

E (xo, yo)'— Min.E(xo, y) y

From (1) and (2), we have Min. E (xo, y) S Max. E (x, y o ) y

—(3)

x

Now xo is any arbitrary chosen point in E" for which inequality (3) holds. Thus, if we choose xo to be that point in E n for which Min. (x, y) has the maximum y

value,

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710 then also the inequality (3) holds i.e. Max. Min. E(x, y) x y

Max. E(x, y0 )

...(4)

Again y 0 is any arbitrary point in E "1 for which inequality (4) holds. Thus, if we choose y 0 to be that point in E" for which Max. E(x, y) has the minimum value, then also the inequality (4) holds. i.e., Max. MM. E(x, y) x y

MM. Max E(x, y) y x

Cor. If [ay ] is a m x ti pay-off matrix of a two person zero sum game, then Max. Min. au Min. Max au .

17.26 Saddle Point of a Function of Several Variables Def. : Let E (x, y) be a function of two variables (vectors) x and y in E" and E m respectively. The point (x*, y*), x* E E" and y* E E m is said to be the saddle point of the function E (x, y) if E (x, y*) < E (x*, y*) < E (x*, y)

17.27 Necessary and Sufficient Condition for the Function E(x, y) to Possess a Saddle Point (Existence of Saddle Point) Let E (x, y) be a function of two variables (vectors) x E E" and y E E m, such that both Max. Mini. E (x, y) and Mini. Max. E (x, y) exist. Then necessary and sufficient condition for E(x, y) to possess a saddle point (x*, y*) is that Max. MM. E(x, y) =E(x*,y*) = Min. Max. E(x, y) x y

...(1)

y x

Proof : Necessary condition : i.e. if (x* y*) is the saddle point of the function E(x, y) then to prove the condition (1). Since (x*, y*) is the saddle point of the function E(x, y) from the definition of saddle point, we have and

E(x*, y*) E(x, y*),

49' x E E"

...(2)

E(x*, y*) 5 E(x*, y),

a Em

—(3)

But E(x, y*) will be max. for some value of xaE n , for which (2) holds E(x*, y*)

Max. E(x, y*)

...(4)

x

But

Max. E(x, y*)

Min. Max. E(x, y) y x From (4) and (5), we have x

E(x*, y*)

Min. Max. E(x, y)

...(5)

...(6)

y x

Again E(x*, y) will be Mini. for some value of y E E m , for which (3) holds.

Game Theory (Competitive Strategies)

711

E(x*, y*) 5 Min. E(x*, y)

—(7)

y

But

Min.E(x*, y)

Encoder

Channel

Decoder

--->

Receiver

Noise Fig. 20.2 This communication system (fig. 20.2) is composed of the following components. 1. Transmitter : It is the device that produces the information to be communicated. 2. Encoder : It is the device used to improve the efficiency of the medium through which the message is transformed. Encoder acts as a sort of set-up transformer. 3. Channel : It is the medium through which the coded message is transmitted. 4. Decoder : It is the device used to transform the encoded message in the original from which is acceptable to the receiver. Decoder acts as a sort of step-down transformer. 5. Receiver : It is destination of the information communicated. 6. Noise : It is general term for anything which tends to produce error in transmission, such as noise in radio and television:

20.4 A Quantitative Measure of Information Now we confine ourselves to models that are statistically defined. Here source, transmits at random any one of a set of prespecified message, we have no knowledge as to which message will be transmitted next but we know the probability of transmitting each message directly. Our search for an amount of information is virtually a search for a statistical •parameter associated with a probability scheme. The parameter should indicate a relative measure of uncertainty relevant to the occurrence of each particular message in the message essemble. To find the formula for the amount of information we consider the following example. Suppose we want to select a machine which have n distinct models {m1, m2 „ mn

Information Theory

803

The derived amount of information I(m k ) associated with the electron of a particular model mk must be a function of the probability of choosing mk . i.e., 1(ink ) = f(P{rnk }) ...(1) If each one of these models m1, m2, probability then P{rni } P{ m2} =

=

, mk is selected with an equal -

1

from (1), we have I(nik ) =f(1).

-.(2)

We also assume that each of the models m1, in 2, inn may be ordered in one of the in distinct colours ci, c2, .... . If each one of these colours of a model may be selected with an equal probability then the amount of information /(ci ) associated with the selection of a colour ci among all colours {c1, c m } is given by f(P{ci ) + f( 1 in)

-(3)

f (ci ) = f (c 2) =

= f(cm ) = 1 m where the function f must be the same as used in (2). Now we assume that the selection of a machine is done in two ways. 1. Firstly we select the machine and then select the colour. The two selections are independent of each other. Since

+ f (1) ...(4) In We select the machine and its colour simultaneously, (i.e., at the same time) as 1 one selection out of mn number of possible selection with equal probability —. mn / (mk and ci )=I (mk ) + / 2 (Ci )= fl

2.

i ) I(rnk and ci ) = fH n

-(5)

From (4) and (5), we get

f m n mn =f

...(6)

This eqn. (5) is a functional equation. This functional equation has several solution, but one of the solution is given by f(x) = log1) = - log x.

-(7)

It may be verified that this solution f(x)=- log x satisfy (6).

20.5 A Binary Unit of Information Let us consider a simple case of selection between two events E1 and E 2 with equal probability e.g., E1 and E 2 may be the head and tail in the throwing of an honest coin.

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Prob. of selection of any one of the two events E1 and E 2 = 1/2. The amount of information associated with the selection of one out of two equiprobable events is given by I(E1 )=I(E 2 )= — log (1/ 2) = log 2. If we take 2 as the base of the logarithm then I(E1 )= I(E 2 )= log 2 2 = 1 i.e., the amount of information associated with the selection of one out of the two equiprobable events is unit. This unit is known as a bit. If we take 10 as the base of the logarithms, a unit of information is called a decimal digit (in short a debit).

20.6 Measure of uncertainty or Entropy Consider a source with a finite number n of the messages m1, m2, ...., mn to be P(inn )be transmitted through some communication system. Let P(mi ),P(m2 ), , mn the probabilities associated with the selection of messages m1, m2, respectively. The source selects at random each one of the messages. The selections are assumed to be statistically independent. The amount of information associated with the transmission of message mk is defined by ...(1)

Ik = I k is called the amount of self information of the message mk . The average information per message for the source is given by I = statistical average of I k =

Plink 1. log P{mk }.

...(2)

k =1 The average information per message I is known as the entropy or the communication entropy of the source and is denoted by H. If the same information I is communicated through a communication system n times then let II , 12, ...., In be the amounts of information reaching the receiver pn respectively then the entropy of the each time with probabilities pi, P2, source is given by H(pi , p2,

pn ) =

E pi log pi . i =1

20.7 Properties' of Average Measure of Uncertainty or Entropy The entropy function H(pi , p2,...., pn ) given by H(pi, p2,

pn ) = —

pi .10g pi i =1

have the following basic properties.

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Information Theory

1. Continuity : i.e., the function H(p1, p2,

pn )should be continuous in pi 's,

i

1, 2, ...., n. That is if the probabilities of the occurrence of the events are slightly changed, then the measure of uncertainty associated with the system should adjust accordingly in a continuous manner. The property is self-evident, since a slight change in the probability of any event will not bring any significant change in value of the function. From (1), we have - H(pi, p2,

, pn ) =

pi log pi i= i + pi, log p

= pi log pi + p2 log p2 +

+ pn _ log p„ - 1 = P1 log P1 + P2 log p2 + - p„ _ 1 ) + (1 - - p2 - pn _ 1 ) log (1 - p2 [Since j Pi = i - p„ p„ _1 are all independent and 1- pi - p2 Since pi , P2, L will also be continuous. Also since being a function of =• D1, P 2'•P3, • •, Pn logarithm of a continuous function is continuous, therefore, log - pn -1) is continuous. (1 -

2. Symmetry : i.e., the entropy function H remain unchanged with the interchange of Pi' p2,

pn with one another.

or

H p2, ••••, pn ) = H(p2, pi, pn ) The proof of the property is self-evident by the formula (1).

3. External Property : i.e., the entropy function has a maximum value when all pi 's are equal. We have

H(p1, P2,

pi . log pi where

pn ) i =1

P2, depends on p1,

pi =1 i=1

pn _ are independent and pi, = 1- pl- p2 P2 , • • ' Pn -1. aH aH apl aH ap2 aH ap,

Pn -1

aP2 aPi aPn aPi d d = (- pi log pi ) + — (- pn log pn ). a Pn dpn a pi dpi

aPi aPi aPi

where i = 1, 2, ..., (n - 1)

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806

log2 e)- (log pn 422- log 2 ej a Pn Pn a Pi

(- log pi Pi

(using logarithm for binary system) =

log pi - log 2 e) - (log pn + log, e). (- 1)

Pi = log pn - log pi = - log -Pn But for maximum value of H. dH , = 0,1 = 1, 2, ...., (n - 1) dpi Pi

- log --= 0 Pn

which is true for pi =

i = 1, 2, ...., (n 1)

Hence for H to be minimum, 1 P2 = • • • • = Pn = 11

P1

4.

i.e., His maximum when all the events are equally likely to happen. If the event I n is divided into a number of mutually exclusive Additivity events J1, J2, ...., Jm each with probability q1, q 2, Then, WA., P2, • •

p„

q r., respectively.

q2. ...., qm ) = H

p .H( q1 , 412 , Pn

where

Pr! -1, pn)

P2, Pn

Lim ) Pn

Pr! =

j =1

Proof : From (1), we have MTh., P2, • • • • Pn -1, 42, • • • gm) n -1 rn = pi log pi q j log qi =.1 j= pi log pi - pn log pn i =1 =

q j log qi

— j

pi log pi + pn log pn i =1

a log qi j =1

...(1)

Information Theory

807

=—

E

tn

pi log pi +

111

E log p„ .J=1q;•

nr

Since p„

qi j

pi log pi —

—

qj (— log pn + log q i ) I j =1

i =1

tn =

I

(2L log q-i

pi log Pi -

)

= H(pi , p2 ,

Pn

Pn

j =1

i =1

pn H ( 41 42 Pn

-Pn

qin

Proved.

P

It should be noted that H(pi , 132, • • • • Pn -1,

q1 3

42, • • • •,

H (Pi, P2, • • •

Pn —1) Pn )

i.e., the partioning of events into subevents cannot decrease the entropy of the

system.

20.8 Important Relations for Various Entropies Now we shall give few expressions (relations) for various entropies without proof (which is out of scope of this book). Let, Average informations per character at the source, or the entropy of the source. Average information per character at the destination, or the entropy H(Y) = of the receiver. H(X, Y) = Average information per pairs of transmitted and received characters, or the average uncertainty (entropy of the communication system as a whole.) H(YI X) = Conditional entropy of the receiver or the measure of information about the receiving part, where it is known that X is transmitted. H(X I Y) = Conditional entropy of the source or the measure of information about the source, where it-is known that Y is received. H(X) =

X, Y are random variables associated with discrete sample spaces of the elementary events at the source and the receiver respectively. The messages transmitted at a source and received at the receiver are given by {x1, x2, ....,

and

Y2, ....,

respectively with their corresponding probabilities as p(xk ) and p(y )k = 1, 2, ..., ti; j= 1, 2, ...., m .i

pfxk ). log p{xk }

H(X) = k =1

...(1)

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808 nt

...(2)

p fy 1. log ply j }

H(Y) == -

.j=1 n

H(X, Y) —

nt

I p{xk , y i}. log

p{xk, y j}

...(3)

k =1 j =1 n m p {xk lyi )

...(4)

fxk , yi l log plyi I xk l

-.(5)

p{xk ,

H(XIY)= k =1 j =1

n m and

H(11 X) = k =1 j =1

where p{xkly j} —

pfxk, y

j}

P{Y j}

and p {k, j} is the point probability for the occurrence of two events E k (at the . source) and fi (at the receiver) simultaneously. The following fundamental mathematical relations exist among the above different entropies in a simple two parts communication system. (i) and (ii)

H(X, Y) = H(X1Y)+ H(Y) H(YIX)+ H(X) H(X IY) gitilAbkatab2 EXCN211241

Example 1 : A man is informed that when a pair of dice were rolled the result was seven. How much information is there in this message ? Solution : The sum seven in rolling a pair of dice can be obtained in the following ways : 1, 2, 3, 4, 5, 6 Number on 1st dice Number on 2nd dice 6, 5, 4, 3, 2, 1 Total number of ways of getting a sum of seven = 6 and the total number of ways (total number of events) in which we can get any sum = 36 Probability of getting a sum seven p(7) 6 36 6

The amount of information 1(7) associated with the sum seven on the two dice is given by 1(7) = f (N(7)}

= f(1/6) = — log (1/6)

from eqn. (7), article 20.4

809

Information Theory = log 6

If logarithm is taken to base 2

= log (2)2.58 = (2.58) log 2 = 2.58 bits.

Example 2 : In a certain community 25% of all girls are blondes and 75% of all blondes have blue eyes. Also 50% of all girls in the community have blue eyes. If you know that a girl has blue eyes, how much additional information do you get by being informed that she is blonde ? Solution : Let pl = Probability that a girl is blonde = 0.25 = 1/4 P2 = Probability that a blonde girl has blue eyes = 0.75 = 3/4 p3 = Probability that a girl has blue eyes = 0.50 = 1/2 p4 = Probability that a girl is blonde and has blue eyes. p x = Probability that a blue eyed girl is blonde. Therefore, we have P4 = P1P2 = P3Px or P x =

P1P2 P3

Now knowing that the girl has blue eyes, the additional information we get by being informed that she is blonde is f{px} = —log p = —log P1P2 P3 1 3 — x— = — log 4 4 1 2 3 = — log — = — log 3 + 3 log 2 23 = 31og 2 — log 21.58 = 3 — 1.58 If log is taken to base 2 = 1.42 bits. Example 3 : Evaluate entropy of the source associated with the probabilities of events shown below. Event : A B Probability :

1/3

1/4

1/8

1/8 1/12 1/12

Solution : Here pi = 1/3, p2 = 1/ 4, p3 = 1/8, p4 = 1/8, ps = 1/1Z p6 = 1/12 Entropy is given by 6 H(1)1)

P2, • • • • Pn ) =

pi log Pi i =1

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810

= -pi log p1 - p2 log p2 - p3 log p3 - p4 log p4 - p5 log p5 - p6 log p6 1 1 1 1 1 1 I 1 1 1 1 1 =_ -log - - -log- - -log- - tog - --log- - -log 3 3 4 4 8 8 8 8 12 12 12 12 1 = --Rilog 3 + 6 log 23 + 3 log 23 + 3 log 23 + 21og (23 x 3) 24 + 2Iog (22 x 3)] 1 = -[38log 2 + 12log 3] 24 1 4[38log 2 + 12 log 21'58] =2 + 12 x (1.58)] =24[38 1

If log is taken to base 2.

= 9.69 bits (appr). Example 4 : A transmitter has an alphabet consisting of five letters. {x1, x2, x3, x4 x51 and the receiver has an alphabet consisting of four letters {yl , y 2 , y3, y4}.

The joint probability of the communication are given below : Yi

Y2

Y3

x1 0.25 0 0 x 2 0.10 0.30 0

Y4

0 0

x3 0

0.05 0.10 0

x4 0 xs 0

0

0.05 0.10

0

0.05 0

_

Determine H(X), H(Y), H(X, Y), H(Y I X) and H(XIY). Solution : Here pfxi l = 0.25 + 0+ 0+ 0= 0.25 pfx21 = 0.10 + 0.30 + 0 + 0 = 0.40

p{x 3 } = 0 + 0.05 + 0.10 + 0= 0.15 p (x4 ) = 0 + 0 + 0.05 + 0.10 = 0.15 p (x 5 ) = 0+ 0+ 0.05+ 0= 0.05 131)9 = 0.25 + 0.10 + 0 + 0 + 0 =0.35 P{Y2} =-- + 0.30 + 0.05 + 0+ 0 = 0.35

ply 3 } = 0 + + 0.10 + 0.05 + 0.05 = 0.20 p{y 4 } =0+ 0+ 0+0.10+0=0.10 Pfxt I Yil Pfxi, pfyi 1 P {x2IY2} =

Y2} 0.30 6 ply 21 0.35 7

P {x2,

p{xi, yi }

0.25 5 0.35 7

1)-0(0 P{.Y2lx2}

-0.25 - 1 0.25

p{x 2 , y 21 -0.30 - 3 p{x2} 0.40 4

811

Information Theory 0.10

1

P{X3 iY3} = 0.20 2 0.10 P{x4IY4} = 0.10 Pfx2iYil =

0.10

-

0.35

P{x4IY3} =

- 0.10

1

P{Y4 I x4} P{Yi x2} =

7

1

0.20

4

P{Y2 A.31

0.05

1

0.40 = 4 0.05 = 1 ci

3

0.05

1

P{Y3 I x4} = 0.

3

0.05

P{x5IY3} = 0.20 - 1 4

P{Y3 I x5} = 0.05 - 1

p{x1/y3} = 0 etc. H(X) =

2 0.15 3 0.10

2

0.05

2

P Y3 I x3} = 0.15 - 3

0.10 1 0.35 -- 7

p(x3iY2}

0.10

{

P {•Y1/x3} = etc. 5 ylxkl.logp{xk} k =1

= - 0.25 log (0.25) - 0.40log (0.40) - 0.15 log (0.15)

H(Y)

- 0.15 log (0.15)- 0.051og (0.05) = 2.066 4 - Ep{yi}. log p{yi} = = - 0.35 log (0.35) - 0.35 log (0.35) - 0.20 log (0.20) 0.10 log (0.10) = 1.856

5 4 ' H(X, Y) = - I E p fxk , k =1 j=1 = - p {x1, yi}.logp fx1,

log pfxk , yi l - p {x2, y1}.1og {x2, yi}

- p {x2, y 2} log p {x2, y2}-- p {x3, y2} log p {x3, y2} - p{x3, y 3)-logp {x3, Y3} - p{x4, y3} loge {x4, y3} - p fx3, y 41 log p {x3, y4} - p{x5, y3} log p {x5 , Y3} -other terms are zero. = - 0.25log 0.25 - 0.101og 0.10 - 0.30 log 0.30 - 0.05..log 0.05- 0.10 log 0.10 - 0.05log 0.05 - 0.10log 0.10 - 0.051og 0.05 = 2.665 5 4 H(YI X)= I fxk,yi blogpfyi lXk l k =1 j =1 =

P{Xl , y1}log {yl I x1} p {x2,

log p {Yil x2}

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Operations Research

—p{x2 , Y2} logp {y 2 1 x2} —p{x3, Y2} logp {Y2 I x3} —Pfx3, Y3110gP{.Y3 lx3} — P{x4,Y3}10gP {Y3Ix4} — p{x4, y 4} log p{y4 lx4} —p{x5, y3} logp{y3 1x5} 1 = — 0.25 log 1 — 0.10 log — — 0.30 log (3/4)— 0.05 log (1/3) 4 —0.10 log (2/3) — 0.05 log (1/3) — 0.10 log (2/3) — 0.05 log 1 = 0.600. 5

4

and H(X/Y).

1){3, j , xk l, log p {.)ck I y k =1 j =I

= — 0.25 log (5/7) — 0.10 log (2/7)— 0.301og (6/7)— 0.05 log (1/7) 0.10log (1/2) — 0.05 log (1/4) — 0.10 log 1 — 0.05 log (1/4) = 0.809

20.9 Encoding Coding offers a most significant application of information theory. The main purpose of coding being, in general, to improve the efficiency of the communication system. If the message to be communicated over a communication system is too lengthy, then there is possibility of losing information. Therefore, to improve the efficiency of the system it is necessary to reduce the length of the message to be communicated. This is done by transforming the message in some other language. This procedure is known as encoding. Thus, encoding is a transformation procedure operating on the input prior to its entry into the communication channel. In the further discussion we shall assume that the successive messages are selected independently; that is, the source has no memory.

20.10 Few important Definitions 1.

Now we shall define few terms used in the further discussion of the chapter. Letter, Symbol or character : It is any individual member of the alphabet set e.g., if (a1, a2, ak ) is a given alphabet set then ai 's are letters.

2.

Message of Word : It is a finite sequence of letters of the alphabets e.g., if

3.

4.

5.

(a1, a2, , are the ak ) is a given alphabet sat then a1, a1a2a3, a1a2a4, messages. Length of a word : The number of letters in a word is known as the length of the word e.g., the length of the words a1 al a2 and a2 a2 a2 a2 are three and four symbols respectively. Encoding or Enciphering : It is one to one transformation of words in original message to words in some other language. That is encoding is a procedure for mapping a given set of messages (m1, m2, ...., ) onto a new set of encoded messages {z1, z 2 , z n } so that the transformation is not one to one. Decoding or Deciphering : It is the inverse process of encoding. That is, the transformation of the transmitted encoded message into original language.

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Information Theory

20.11 Unique Decipherability If we consider the following binary codes 0

m1

01 m3 10 010 M4 then each of the message m4, m1m3 or m2m1 can be represented by the same binary sequence 010. Thus, the sequence 010 cannot be decoded accurately. Therefore, some restriction must be placed on the assigning of code words. For this we define Unique Decipherability as follows : m2

A code is said to be uniquely decipherability (or separable) if every finite sequence of code characters corresponds to at most one message. Let us consider the following codes 0

m1 m2

10 110 111

m3 m4

If we write 00000100100011010 then this message may be uniquely decoded into ml mlml ml m1m2m1m2m1m1m3m2 .

20.12 Uses of Encoding; Efficiency and Redundancy Uses of Encoding : We note the following uses of encoding. 1. It increases the efficiency to transmission 2. It reduces the cost of transmission.

Efficiency : If ci is the cost, to each message mi , then the average cost per message is given by C =

ci p{mi }. i =1

where p{mi } is the probability of transmission of the message mi . The most of efficient transmission is one for which C is minimum. When all the symbols in a message have identical (same) cost of transmission, the average cost per message becomes proportional to the average number of symbols per message. If ni is the number of symbols in message mi then the average cost per message is given by

ni . p(mi ), • = ni , i = 1, 2, ...., N.

C=L=

i

•

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814

The efficiency of the encoding procedure can be defined if we know the lowest possible bound of L. It may be proved that subject to certain restrictions the lowest — H(x) bound for L is (log D) where H(x) = entropy of the original message essemble and

D = number of symbols in the encoding alphabet.

Defining the efficiency of an encoding procedure, as the ratio of the average information per symbol of the encoded language to the maximum possible average information per symbol, we have H(x) Efficiency _ L(log D) Redundancy : The redundancy of a code is defined as Redundancy =1 — efficiency i log D— H(X) L log D

20.13 Shannon-Fano Encoding Procedure In Shannon-Fano encoding procedure a sequence of binary numbers (0, 1) for encoding message through a memoryless channel* is used. Let [X] be the essemble of the messages to be transmitted and [P] their corresponding probabilities : [X] =

m2,

rnisi

[P] = [Pl, P2, Then we desire to associate a sequence of binary numbers (0, 1) of unspecified length ni to each message mi with the following properties. 1.

No sequences of employed binary number can be obtained from each other by adding more binary terms to the shorter sequence (prefix property). This property guarantees a one to one correspondence between any set of original message and the corresponding set of encoded messages.

2.

The transmission of the encoded message is reasonable efficient i.e., the alphabets in a binary sequence can be either 0 or 1 each appearing with equal probability!. 2 This property ensures the transmission of almost one bit of information per digit of the encoded message. The Shannon-Fano encoding procedure is as follows.

* A channel is said to be memoryless if the effect of the noise on the input letters is independent of the sequence of previously transmitted letters.

815

Information Theory

Step 1 : First write messages in descending order as regards their probabilities. Step 2 : Then divide the message set into two subsets of approximately equal probability. If {X1} and {X2} are subsets then the sum of probabilities associated with the messages of subset {X1} is approximately equal to the sum of probabilities associated with messages of the subset {X2}. Step 3 Assign 0 to each message contained in one subset say {{X1} and 1 to each message contained in the other subset say {X2}. Step 4 : Repeat the steps 1 to step 3 on subsets {X1) and {X2}. If subset {X1) is diiiided into two subsets {X11} and {X12} then the code word corresponding to a message {X11} will begin with 00 and that corresponding to chat in {X12} will begin with 01. This procedure is continued until each subset contains only one message. For clear understanding of the procedure see Ex. 5 on page 818. The procedure has got its advantages and disadvantages as follows.

Advantages 1. The efficiency of the transmission of information is 100%. 2. Average length of the message is minimum.

Disadvantages : If the sample space cannot be divided into two subsets of equal probabilities then this method may not be optimum. By this method the most efficient code can be derbied only if the message probability space can be repeatedly divided into two equiprobable subspaces so that we finally reach the situation where each message corresponds to only one partitioned subspace. The probability that the message mi is encoded in a sequence of binary numbers of length ni will be (1/2)m . The probability of the occurrence of each message mi is given by P{mi } = where tti is a positive integer and N y 2-n; = 2- ni +2-n2 +

2nN =

1

i =1 If encoding procedure is unambiguous (one-to-one), the prefix requirement is fulfilled, and the average length of encoded message is N

L=

ni .P{mi } =

P{mi } log{mi } =

which is equal to the entropy H(X) of the original message essemble P{mi log P{mi = E bits per message.

i.e., H(X) = =1

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816

20.14 Necessary and Sufficient Condition for Noiseless Coding Theorem : The necessary and sufficient condition for existence of an irreducible noiseless encoding procedure with specified word length n1, n2, n N is that a set of positive integers {n1, n2, ....nN } can be found such that D-ni 5_ 1 i =1

where D is the number of symbols in encoding alphabet. Proof : The Condition is Necessary In an encoded message there is always a possibility that two messages may have same length. Let the two messages mi and mi have the same length.

i. e.,

►ti = n

Let Wi be the number of encoded messages of length ni . Since the number of encoded messages with only one letter cannot be larger than D.

WI. SD Again because of the coding restriction, the number of encoded messages of length 2, cannot be larger than (D - W1 )D.

...(2)

W2 5_ (D - Wi ) D = D2 - Wi D Similarly,

W3

-

D - W2] D = -w1 D2 - W2D

(3)

Finally, if m is the maximum length of the encoded words, then we have Wm 5. Dm _ wiD m - 1 _ w2Dm - 2 _ _ 1D. ... (4) Dividing both sides of inequality (4) by Dm (which is positive), we have

Wm D-m < 1 - WiD-1 - W2D- 2 - - Wm -1D- (m -1) _or

VVID- + W2D -2 +.... + W D

51

rn 1

or

—(5)

i =1 L.H.S. of inequality (5) may be written as WiD-1 + W2D -1 + + Wm D-

= (D -1 + D -1 + .... + W1 times) + (D- 2 + +

+ (D- m + D-in +

2+

+ W2 times)

+ Wm times).

...(6)

Since expression within each bracket corresponds to a message ►ni , and therefore the total number of terms is N. From (6) there are W1 messages of length 1, W2 messages of length 2, ...., Wm messages of length m.

Information Theory

817

vvi +W2 + • • • • + VV. = N The term Wk correspond to the encoded message of length k. These later terms can be considered as ED' , when the summation takes place over all those terms with ni = k. Therefore, by a simple reassignment of terms, we may write 111

W-D —i =

D-11' < 1

from (5)

-.(7)

i =1

=1

which proves the necessity of the theorem. The condition is sufficient : Now we shall show that the condition

in = Wi D —1 + W2D — 2 + +

D

m _0

Taking the Langrangian multipliers Xi, X 2, the Lagrangian function X 2 ) is as follows : L(xi, x2, ...., xn ; L(xi, x2, ...., xn ; Xi, X 2 ) = f(xi , x2, x„ ) — Xi hi (xi, x2, ...., xn ) — X 2 /12 (Xi, X2, • • • • , X, ) which can be written as L(x; Xi, X 2 ) = f(x) — Xi (x) — X 2h2 (x)

Rn Assuming that L, f and h1 are all differentiable partially w.r.t., xi, x2, ...., xn and Xi , X 2 The necessary conditions for maximum or minimum of f(x) subject to h1 (x) = 0 (i = 1, 2) are given by aL, of an, ant -b`j = 1, 2, ...., n ) a.) — "1— a_Xi 2 _x; aL

=—

(x) = 0 and

= — h2 (x) = 0 a2

Thus, the necessary conditions for maximum or minimum of f = f(x), subject to hi. m. hl (x) = gi (x)— bi = 0 and h 2 -•-• h2 (X) = g 2 (X) — b2, an,(x) af(x) _ ax; 1 ax;

X= A2 , Dh2 (x) _v_i = ax;

hi (x)= 0 and h2 (x) = 0, x =

X2, ...., X, )E Rn are that n

x2,...., xn ).

4. n decision variables and m equality constraints, m < n Consider the NLPP Optimize Z = f (xi, x2, ...., xn ) = f(x)

Rn subject to the constraints gi (x)= bi and x, 0, i = 1, 2, ...., This problem can be re-written as Optimize Z = f(x)

n).

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828

s.t.

hi (x) = gi (x) — bi = 0, and xi

0, = 1, 2, .... , m( 0, so (5) is also satisfied. Thus, xi = 3.5, x2 = 3 are the stationary values. and give the maximum value of Z.

Max. Z = 3.6 x 3.5 - 0.4 x (3.5)2 + 1.6 x 3 - 0.2 x 2 =10.7 Example 7 : Solve the following NLPP Maximize Z = f(x) = (200x1 - 2x12 ) + (500x2 - 3x2 2 ) subject to the constraints 2x1 + x2 5 140, 2x1 + 3x2 5_ 180 and xi , x2 0 Solution : Here f(x) = (200x1 - 2x12 ) + (500x2 - 3 x22 ) h1 (x) = 2x1 + x2 -140 5. 0, h2 (x) = 2x1 + 3x2 - 180 5. 0

and

x = (xi , x2 ) 0

For the given function f(x), Hessian matrix is a 2f f ax1ax2 _ 1-4 1 01 a 2f L0—

DX2 1

HB

a 2f

ax2ax1

aX22 _

-4 0 = 24 , which are of 0 -6 alternate signs. So f(x) is concave function of x= (x1, x2 ). Whose principal minors are D1 = - 4, D2 =

Also h1 (x) and h2 (x) are convex functions. Thus the Kuch-Tucker necessary conditions for max. of f(x) will also be the sufficient conditions. The Kuhn-Tucker necessary conditions for maximum of f(x) are af(x)

(x) of

x

ax,

ax„

(x) a x2

=1

an, (x) ax2

(x) = 0, X 2h2 (x) = 0, h1 (x) 0, h2 (x) 5 0 and X1, X 2 0 i.e., the conditions are 200 - 4x1 =2X1 + 2X 2 ...(1), (2x1 + x2 -140) = 0 ...(3), 2x1 + x2 -1405.0 and Xi 0, X 2 0 There are four different cases.

...(5), -.(7)

...(2) 500 - 6x2 =a'1 + 3X 2 X, 2 (2X1 + 3X2 - 180) = 0 ...(4) 2x1 + 3x2 - 180.5 0

...(6)

843

Non-linear Programming

Case I. Xi = 0 and 2 = 0 : In this case from (1) and (2), we get 200 - 4x1 = 0 and 500 - 6x2 = 0 = x1 = 50 and x2 = 250/3. These values of x1 and x2 do not satisfy (5) and (6). So this solution is discarded. Case II. X # 0, X 2 = 0 : In this case from (1) and (2), we get ...(8) 200 - 4x1 = 2X1 and 500 - 6x2 = 1 = x1 - 3x2 + 200 = 0 If X1 # 0, then from (3), 2x1 + x2 - 140 = 0 -.(9) Solving (8) and (9), we get, xi = 220/7, x2 = 540/7 These values of x1 and x2 does not satisfy (6). So this solution is also discarded. Case III. Xi = 0, 2 # 0 : In this case from (1) and (2), we get 200 - 4x1 = 2X 2 and 500 - 6x2 = 2 = 3X1 - 3X2 + 100 = 0 ...(10) ...(11) 2x1 + 3x2 - 180 = 0 If X 2 # 0, then from (4), Solving (10) and (11), we get , x1 =16, x2 = 148/3 a2 = 68> 0 These values of xi and x2 satisfy (5). xl = 16, x2 = 148/3 is stationary point at which f(x) is maximum. Maximum value of f(x) i.e., of Z = Rs. 60160/3. Case IV. Xi # 0 and 2 2 # 0 : In this case from (3) and (4), we get = 60, x2 =20 2x1 + x2 - 140 = 0 and 2x1 +3x2 - 180 = 0 For these values of x1 and x2, from (1) and (2), we get Xi+ X 2 = - 20 and Xi +3X 2 = 380 Xi = - 220 < 0, 2 = 200> 0 Thus, (7) is not satisfied. So this solution is also discarded. Hence , the optional solution is x1 = 16, x2 = 148/3 and Max. Z = Rs. 60160/3 Example 8 : Determine x1, x2, x3 so as to maximize z = - xi2 - x22 - x32 4x1 + 6x2, subject to the constraints : x1 + x2 < 2, 2x1 + 3x2 5 12 and x1, x2 O. Solution: Here f(x) = Z = -.x12 - x 22 - x32 + 4x1 + 6x2, X = (x1, x2, x3) E R 3 2 0, h2 (x) = 2 xi + 3x2 - 12 0 and x= (x1, x2, x3 ) 0 For the given function f(x), Hessian matrix is h1 (x)=x1 +x2 -

Operations Research

844

a2f

a 2f

ax, ,a 2f HB =

a 2f

aXi aX2 ax1 aX3 a 2f a 2f

ax2ax3 a 2f a2 f axi aX 3 k ax3ax2

aX 2 axi.

ax22 a 2f

—2 0 0 —[ 0 —2 0 0 0 —2

Whose principal minors are Di = — 2, D2 =

—2 0 1 0 — 21

4 and .

0 —2 0 0 = — 8 which are of alternate signs. So f(x) is concave function 0 —2 D3 = 0 0 —2 of x= (x1, x2, x3 ). Also hi (x), h2 (x) are convex functions of x. Thus the Kuch-Tucker necessary conditions for max. of f(x) will also be the sufficient conditions. The Kuhn-Tucker necessary condition for maximum of f (x) are all' (x), af(x) — axi 1=1 axi

= 1, 2, 3.

(x) = 0, X 2h2 (x) = 0,111(x) 0, h2 (x) S 0 and Xi , X 2 O. i.e., the conditions are —2x1 + 4 = Xi +

2

—2x3 = 0

...(1),

— 2 X2 + 6 = Xi + 3X. 2

...(2)

...(3),

2`1.(x1 + x2 — 2) = 0 x1 + x2 — 2 0

...(4)

X. 2 (2 Xi + 3X2 —12) = 0 ...(5) 2 xi + 3x2 —12 5_ 0

...(7)

_0 ,A, 22 >

...(6) ...(8)

There are four different cases. Case I. Xi = 0 = X 2 = 0 : In this case from (1) and (2), we get + 4 = 0 and — 2 x2 + 6 = 0 = x1 = 2 and x2 =3. — These values do not satisfy (6) and (7). So this solution is discarded. Case H : Xi # 0, X 2 = 0 : In this case from (1) and (2), we get, — x2 + 1 = 0 —2x1 + 4 = Xi and — 2x2 + 6 = Xi If Xi # 0, then from (4), x1 + x2 — 2 = 0

...(9) ...(10)

Solving (9) and (10), we get x1 = 1/2, x2 = 3/2 These values of x1 and x2 satisfy (6) and (7). For these values of x1 and x2, from (1) and (2), we get Xi = 3 > 0, X 2 = 0 For this solution, x1 = 1/2, x2 = 3/Z x3 = 0, f(x) = Z = 17/2 .

845

Non-linear Programming Case III : Xi = 0, X 2 # 0 : In this case from (1) and (2), we get —2x1 + 4 = 2X 2 and-2x2 + 6 = 3X 2 =3x1 — 2 X2 =0

...(11) ...(12)

If?, 2 # 0, from (5) we get 2 xi + 3 x2 — 12 = 0

Solving (11) and (12), we get xi = 24/13, x2 = 36/13 also from (3) x3 = 0. These values does not satisfy (6). So this solution is discarded. Case IV : Xi 0, X 2 # 0 : In this case from (4) and (5), we get x1 + x2 — 2 = 0 and 2 xi + 3x2 — 12 = 0

= — 6, x2 = 8

x1 = — 6 < 0, So this solution is also discarded. Hence, the optimal solution giving the Max. value of f(x) is x1 = 1/2, x2 = 3/2, x3 = 0 and Max. Z = 17/2. Example 9 : Solve the following NLPP Minimize Z = (x1 — 2)2 + (x2 — 1)2 subject to the constraints X1 2 — X2 5_ Q + x2 2

and

xp x2 0

Solution : Here f(x) = Z = (x1 — 2)2+ (x2 — 1)2 (x) = — x12 + x2 o, h2 (x) = 2— x1 — x2 0 and x =(xi,x2 ) O. Note that it is a minimization problem so the signs in inequality constraints are taken as Here for the function f(x), Hessian matrix is a 2f a 2f

a 2f

axi

axiaX 2 a 2f

ax 2aX1

ax 22

HB =

[2 0] 0 2

Whose principal minors are Di = Z D2 =

20 =4 02

Which are both positive . So f(x) is convex function of x = (x1, x2 ) . Also hi (x) and h2 (x) are convex functions of x. Thus, the Kuhn-Tucker necessary conditions for mini. of f (x) will also be the sufficient conditions. The Kuhn-Tucker necessary conditions for minimum of f(x) are af(x)

Xi

—

2

X, i =1

ail. (X)

j =1 2 x. "

(x) = 0, X. 2 h2 (x) = 0, (x) 0, h2 (x) ?_ 0 and Xi, X. 2 0

Operations Research

846

i.e., the conditions are — X.2 ...(1), ...(3), X1(— x12 ± x2) = ° ...(5), — xi 2 + X2 0

2 (x1 — 2) = — 2 xi

2(x2 —1) =

...(2)

—

X 2 (2 — x1 — x2 ) = 0

...(4)

2—xi — x2 0

...(6)

...(7)

and Xi, X 2 0 There are four different cases.

Case I. Xi = 0 = 2 : In this case from (1) and (2), we get xi— 2 = 0 and x2 — 1 = 0 = x1 = 2, x2 =1 These values do not satisfy (5) and (6) and so are inadmissible. Thus, these values are discarded. = 0 and 2 # 0 : In this case from (1) and (2), we get Case II. ...(8) -x2 - 1 = 0 2(x1 — 2) = — 2 and (2 x2 — 1) = — 2 — xi — x2 + 2 = 0 Solving (8) and (9), we get xi = 3/2, x2 = 1/2

Also from (4),

—(9)

These values does not satisfy (5). So this solution is also discarded. Case HI. Xi and 2 = 0 : In this case from (1) and (2), we get — x1 + 2 x1 x2 — 2 = 0 xi— 2 = — x1 Xi and 2 (x2 — 1) = Also from (3)

...(10) ...(11)

— x12 + X2 = 0

From (10) and (11), we get 2x13 — — 2 = 0

= xi =1.52

x2 = 2.31

These values does not satisfy (6). So this solution is also discarded. Case IV. Xi 0 and 2 # 0 : In this case from (3) and (4), we get x2 = x12 and x12 + — 2 = 0 — xi 2 + x2 = 0 and 2— x1 — x2 = 0 = 1, x2 = 1 x2 = xi 2 and (xi. — 1) (xi + 2) = 0 From these values of x1 and x2, from (1) and (2), we get, 2X.1 + X 2 = 2 and Xi — X 2 = 0

= 2/3, X 2 = 2/3 0.

Hence the optimal solution of the problem is x1 = 1, x2 = 1, Min. Z =1

21.9 Graphical Solution In chapter 7 we have shown that the optimal solution of a linear programming problem is obtained at one of the extremities of the feasible region (permissible region). But in the case of NLP problems, the optimal solution is not necessarily obtained at the extreme point of the feasible region. The method of solution of NLP problem by graphical method is explained through the following examples.

847

Non-linear Programming

Example 10 : Solve the following non-linear programming problem graphically. Also verify the Kuhn-Tucker necessary conditions for maximum of the function. Maximize Z = 8x1 — x1 2 + 8x2 — x2 2 subject to the constraints [Agra 1999]

x1+ x2 5_ 12, xi — x2 ?. 4 and xi, x2 0

Solution : The constraints of the NLPP are + 5. 12, xi - x2 4 and xi, x2 0

0

8

12

Fig. 21.1 Considering these constraints as equalities and drawing the lines in the plane, the permissible region is DABD. The objective function Z = 8x1 — x12 + 8x2 — x22 is a circle with centre at (4, 4). The point giving the maximum value of Z is the point at which the feasible region is tangent to the circle given by the objection function Z = 8x1 x12 + 8x2 — X2 2 = 0. The centre of this circle is C(4,4). Differentiating Z = 8x1 — x1 2 + 8x2 — x22, w.r.t. x1, we get

Or

dx2— 2x2. dX 2 = 0 8 — 2x1 + 8.— dxi dx dx2 2x1 — 8 —4 — mi (say) cbci 8 — 2x2 4 — x2

Operations Research

848 For the line xi + x2 = 12,

dx2 = — 1 = m2 (say) dxi

The circle touch this line at the point where xi— 4 — 1=>x2 = x1 = rn2 4 — x2 Putting in xi + x2 = 12, we get xi = x2 = 6 i.e., the circle touches 'the line xi + x2 = 12 at the point P (6, 6). But the point P (6, 6) is not the point of the feasible region DAB. dx.) = 1 = m3 (say) Again for the line x1 — x2 = 4, dxi The circle touches this line at the point where —4 .X2 = 8— xl . In3 A — x2 Putting in x1 — x2 = 4, we get xi = 6

x2 = 2

i.e., the circle touches the line x1 — x2 = 4 atj the point Q(6, 2), which is the point in the feasible region DAB. For x1 = 6, x2 = 2, Z = 24. Hence, the optimal solution 9f the given problem is xi = 6, x2 = 2 and Mini. Z = 24. Verification of Kuhn-Tucker Conditions For the given problem, we have f(x) = 8x1 — x12 + 8x2 — x22, hi (x)=x1 +x2 —12 5_ 0, h2 (x) = 4— xi + x2 5 0 and xi, x2 0 The Kuhn-Tucker necessary conditions for max. of f (x) are

of (X) ax j Xi hi (x) = 0,

—

2

an, (x)

= 1.

ax3

X 2h2 (x) = 0,

, j= 1, 2

hi (x) 5_ 0, h2 (x) 0 and Xi , X 2

0 i.e., the

conditions are 8 — 2x1 = Xi 5X 2

...(1),

Xi (xi +x2 — 12) = 0 ...(3), + x2 — 12 5 0 and Xi, X 2

0

8 — 2x2 =X1 + X 2 + x2 )= 0 X 2 (4 —

...(2)

4 — x1 + x2 5. 0

...(6)

...(7)

The point (6, 2) satisfy conditions (5) and (6). .For these values from (1) and (2), we have

...(4)

Non-linear Programming

849

X1 — X 2 =— 4 and Xi + X 2 =

= 0, X 2 =4

i.e., conditions (3), (4), and (7) are also satisfied. Thus, the optimal solution x1 = 6)x2 = 2, obtained by graphical method satisfy all the Kuhn-Tucker conditions for max. of f(x) . Example 11 : Maximize Z = xl + 2x2 subject to the constraints : x1 2 + X2 2 1, 2X1 + X2 5_ 2 and x1, x2 Solution : The constraints of the NLPP are x1. 2 + x22 1, 2x1 + x2 2 and x1, x2 >_0. Considering these constraints as equalities and drawing in the plane, the permissible region is OABCO. The objective function is Z = x1 + 2x2.

x Z=

+ 2x2 = 0 _—2 X2 - 1

Fig. 21.2 Drawing the line Z = x1 + 2x2 = 0 through the origin, we draw lines I I to this line till we reach the extremity B of the permissible region. B is the most distant point of the permissible region, through with the line parallel to Z = 0 passes. Here B is the point of intersection of the circle x12 + x22 = 1 and line 2x1 + x2 = 2, so at B, xl = 3/5 = 0.6, x2 = 4/5= 0.8. And at this point Z = 11/ 5 = 2. 2 Hence the optimal solution of the NLPP is x1 = 0.6, x2 = 0.8 and max. Z = 2.2. •

Operations. Research

850

4. Exercise on Chapter 21 +. 1. Define general non-linear programming problem. 2. State and prove the necessary and sufficient conditions for the optimality of a NLPP with equality constraints, by the use of Lagrangian multipliers. 3. State and prove Kuhn-Tucker necessary and sufficient conditions in non-linear programming. [Agra 2000, 01, 03] 4. Obtain necessary conditions for the optimum solution of the following problem. Minimize f (x1, x2 ) = 3 e 2x1 1+ + 2e x, + 5 subject to the constraint : x1 + x2 —7 = 0 Solve the following non-linear programming problems, using the method of Lagrangian multipliers. 5. Minimize Z = 6x12 + 5x22 subject to the constraints : x1+ 5x2 = 3 and x1, X2 O. 6. Minimize Z = 2x12 — 24x1 + 2x22 — 8x2 + 2x32 — 12x3 +200 subject to the constraints : x1 + x2 + x3 = 11 and x1, x2, x3 0 7. Minimize Z = —2x12 + 5x1x2 — 4x12 + 18x1 subject to the constraints : x1 + x2 = 7 and x1, x2 0 8. Maximize Z =x12 + 4x1x2 + x2 2 subject to the constraint x12 + X2 2= , and x1, x2 0 9. Maximize Z = 7x1 — 0.3x12 + 8x2 — 0.4x22 subject to the constraints : 4x1 + 5x2 = 100 and Xi, X2 0 10. Maximize Z = 4x1 + 6x2 — 2x12 — 2x1 x2 — 2x2 2 subject to the constraints : x1+ 2x2 =2 and X2 0

[Agra 1998, 99]

[Agra 2000]

851

Non-linear Programming

11. Minimize Z = 4x12 + 2x22 + x32 — 4x1x2 subject the constraints : xi+ x2 + x3 = 15, 2x1 —x2 + 2x3 = 20, and xi, x2, x3 >_0. = 12. Minimize Z x1 2 + x2 2 + x3 2 subject to the constraints Xi + x2 +3x3 =2, 5x1 + 2x2 + x3 = 5 and xi, x2, x3 O. 13. Use method of Lagrangian multipliers to solve the following NLP problem. Does the solution maximize or minimize the objective function ? Optimize Z=x1 2 — 10x1 + x 22 — 6x 2 + x32 — 4x3 subject to the constraints : xi+x2 +x3 =7 and xi , x2, x3 O. Use the Kuhn-Tucker conditions to solve the following non-linear programming problems : 14. Maximize Z = 8x1 + 10x1 — x1 2 — x22 subject to the constraints : 3x1 + 2x2 6 and xi, x2 >_0. 15. Maximize Z = 2x12 + 12x1x2 — 7x22 subject to the constraints : 2x1 + 5x2 98 and xi, x2 O. 16. Maximize Z = 6x12 + 5x22 subject to the constraints : + 5x2 3 and xi , x2 0 17. Minimize Z = — log xi — log x2 subject to the constraints : + x2 2 and xi , x2 0 18. Maximize Z = 3 xi + x2, subject to the constraints : 2 2 x1 + x2

5

xi— x2 1

and

xi, x2>_0.

Operations Research

852 19.

Maximize Z = 10x1 — x12 + 10x2 — x22

Subject to the constraints : 14 • + + x2 6 — xi, x2 0 , and 20. Maximize Z = 12x1 + 21x2 + 2x1x2 — 2x12 — 2x22 subject to the constraints : • + x2 10 x2 58 xi, X2 >_ 0 and 21. Maximize Z = 2x1 — x12 + x2 subject to the constraints : 2x1 + 3X2 5_ 6 2x, + x2 4 xi, x2 0 and 22. Maximize Z = 10x1 — x12 + 10x2 — x22 subject to the constraints : xl+ x2 9 xl — x2 6 xi, x2 0 and 23. Maximize Z = 8x12 + 2x22 subject to the constraints : x1 2 + X22 _5_ 9 and

2 xi , x2 0

24. Maximize Z = 7x12 — 6x1 + 5x2 2 subject to the constraints : xl+ 2x2 10 x, — 3x2 9 xi, x2 0 and 25. Minimize Z = (x1 — 9/4)2 + (x2 —2) 2 subject to the constraints : _0 X2 — Xi 2 > x1 +

and

x2 6 x„, x2 0

853,

Non-linear Programming 26. Minimize Z = x12 — x1x2 + 2x22 — 4x1 — 5x2 subject to the constraints x1 + 2x2 6 52. x1

xi, x2 0

and

27. Minimize f(x1, x2 ) = (x1 — 1)2 + (x2 — 5)2 subject the constraints : —

+ X2 4

—(x1 — 2)2 + x2 4 and xi, x2 0 Solve the following NLP problems graphically. And verify that the Kuhn-Tucker necessary conditions hold at the point of optimal solution : Maximize Z = 2x1 + 3x2 subject to the constraints : 2 2 x1 + X2 5_ 20 and

xi x2 8 xi, x2 0

29. Maximize Z = 100x1 — x12 + 100x2 — x22 subject to the constraints : x1 + x2 80 x1 + 2x2 100 and xi , X2 0 30. Minimize Z = (x1 — 2)2 + (x2 — 1) 2 subject to the constraints : — 2 + X2 ?_ 0 x1 + x2 2

and

xi , x2 0

31. Minimize Z = (x1 — 1) 2 + (x2 — ) 2 subject to the constraints : 0 5_ xi. 5 2 0 x2 5.1. and Minimize the distance of the origin from the convex region bounded by the 32. constraints x1 + x2 4, 2x1 + x2 5 and x1, x2 0. Verify that the Kuhn-Tucker necessary condition hold at the point of minimum distance. •

854

Operations Research

+ ANSWERS + 4.

6e 2x1 +1 =

2e x2 + 5 =X, x1 +x2 = 7; 1 x1 = 1(11 — log 3), x2 = — (10 + log 3). 3 3 5. x1 = 3/31, x2 = 18/31 Mini. Z = 54/31. 6. xi = 6, x2 = 2, x3 = 3, Mini. Z = 102 . 8. xi = 1/4 2, x2 = 1/4 2, Max. Z = 3. 9. x1 = 12.06, x2 = 10.35, Max. Z = 80.73. 10. x1 = 1/3, x2 = 5/6, Max. Z = 25/6. 11. xi, = 11/3, x2 = 10/3, x3 = 8, Mini. Z = 820/9. 12. x1 =- 0.81,x2 = 0.35, x3 = 0.928, Mini. Z = 0.857. 13. x1 = 4, x2 = 2, x3 = 1, Mini. Z = — 35. 14. x1 = 4/13, x2 = 33/13, Max. Z = 21.3. 15. xi, = 44, x2 = 2, Max. Z = 4900. 16. x1 = 3/ 31, x2 = 18/31, Max. Z = 54/31. 17. x1 = 1, x2 = 1, Mini. Z = 0. 18. x1 = 1.43, x2 = 0.48, Max. Z = 4.77 19. x1 = 5, x2 = 5, Max. Z = 50. 20. x1 = 17/4, x2 = 23/4, Max. Z = 1734/16. 21. xi. = 2/3, x2 = 14/9, Max. Z = 22/9. 22. x1 = 7.5, x2 = 1.5, Max. Z = 31.50. 23. x1 = 2, x2 = 5, Max. Z = 42. 24. x1 = 48/5, x2 = 1/5, Max. Z = 587.72. 25. xi. = 3/2, x2 = 9/4, Mini. Z = Ai (10)/16 26. xi = 2, x2 = 7/4, Mini. Z = — 81/ 8 27. x1 = 1, x2 = 5, Mini. Z = 0. 28. x1 = 2, x2 = 4, Max. Z = 16. 29. x1 = 60, x2 = 20, Max. Z = 4000. 30. xi =1, x2 = land Mini. Z = 1. 31. x1 = 1, x2 = 1, Mini. Z = 1. 32. x1 = 2, x2 = 2, Mini. distance = 8. • • •

Tables

855

Table 1 Normal Distribution An entry in the table are the areas under the entire normal curve between normal variate Z = 0 and a positive value of Z. The curve is symmetrical about Z = 0. Z to Ftrst Decimal

•00

.01

.02

.03

.04

.05

•07

.08

.09

0.0 0.1 0.2 0.3 0.4 0.5

.0000 .0398 .0793 -1179 .1554 .1915

-0040 .0438 .0832 .1217 .1591 •1950

.0080 .0478 .0871 .1255 .1628 •1985

.0120 .0517 .0910 .1293 -1664 .2019

-0160 .0557 .0948 •1331 .1700 -2054

.0199 .0596 .0987 .1368 .1736 .2088

.0239 .0636 .1026 -1406 -1772 -2123

.0279 .0675 .1064 .1443 .1808 .2157

-0319 .0714 .1103 .1480 .1844 -2190

-0359 .0753 .1141 .1517 •1879 .2224

0.6 0.7 0.8 0.9 1.0

.2257 .2580 .2881 .3159 .3413

.2291 .2611 .2910 -3186 .3438

.2324 .2642 .2939 .3212 .3461

.2357 .2674 -2967 •3238 .3485

.2389 .2704 •2995 .3264 .3508

.2422 .2734 •3023 .3289 .3531

•2454 .2764 .3051 .3315 -3554

.2486 .2794 .3078 .3340 .3577

.2518 .2823 .3106 .3365 .3599

.2549 .2852 .3133 .3389 .3621

1.1 1.2 1.3 1.4 1.5

.3643 .3849 .4032 .4192 .4332

.3665 .3869 .4049 .4207 .4345

.3686 .3888 .4066 .4222 .4357

.3708 .3907 .4082 -4236 .4370

.3729 .3925 .4099 .4251 .4382

.3749 .3944 .4115 .4265 .4394

.3770 •3962 .4131 .4279 .4406

-3790 .3980 .4147 .4292 .4418

.3810 .3997 -4162 .4306 .4429

.3830 .4015 .4177 .4319 .4441

1.6 1.7 1.8 1.9 2.0

.4452 .4554 .4641 .4713 .4772

.4463 .4564 .4649 .4719 .4778

.4474 .4573 .4656 .4726 .4783

-4484 .4582 .4664 .4732 .4788

.4495 .4591 .4671 .4738 .4793

.4505 .4599 .4678 .4744 .4798

.4515 .4608 .4686 .4750 .4803

.4525 .4616 .4693 .4756 .4808

.4535 .4625 .4699 .4761 .4812

.4545 •4633 .4706 .4767 .4817

24 2.2 2.3 2.4 2.5

.4821 .4861 .4893 .4918 .4938

.4826 .4865 .4896 .4920 .4940

.4830 .4868 .4898 .4922 .4941

.4834 .4871 .4901 .4925 .4943

.4838 .4874 .4904 .4927 .4945

-4842 .4846 .4878 .4881 .4906 .4909 .4929 .4931 .4946 .4948

.4850 .4884 .4911 .4932 .4949

.4854 .4887 .4913 .4934 .4951

.4857 .4890 .4916 .4936 .4952 _

2.6 2.7 2.8 2.9 3.0

.4953 .4965 .4974 .4981 .4986

.4955 .4966 .4975 .4982 .4987 '

.4956 .4967 .4976 .4982 •4987'

.4957 .4968 .4977 .4983 .4988

.4959 .4969 .4977 .4984 .4988

.4960 .4970 .4978 .4984 .4989

.4962 .4972 .4979 .4985 .4989

.4963 .4973 .4980 .4986 .4990

.4964 .4974 .4981 .4986 .4990

Second Decimal

.06 ,.

.4961 .4971 .4979 .4985 .4989

Each of the Areas under the normal curve between Z = -00 to Z = 0 and that between Z = 0 to Z = 00 is equal to 0.5. Method to read Prob. from the table : Let Z = 1.21, then P (Z 1.21) = Area under the curve from Z = 00 to Z = 1.21 = 0.5 + 0.3869 = 0.8869 i.e. 88.69% Where, 0. 3869 = value opposite to 1.21 in the table and P (Z - 1. 21) = Area under the curve from Z = - oo to Z = -1.21 = 0.5- 0.3869 = 0.1131. i.e. 11.31%

856

Operations Research Table 2 Present Values

Year

1%

2%

3%

4%

5%

6%

7%

8%

9%

10%

1

.990

.980

.971

.962

.952

.943

.935

.926

.917

.909

2

.980

.961

.943

.925

.907

.890

.873

.857

.842

.826

3

.971

.942

.915

.889

.864

.840

.816

.794

.772

.751

4

.961

.924

.888

.855

.823

.792

.763

.735

.708

.683

5

.951

.906

.863

.822

.784

.747

.713

.681

.650

.621

6

.942

.888

.837

.790

.746

.705

.666

.630

.596

.564

7

.933

.871

.813

.760

.711

.665

.623

.583

.547

.513

8

.923

.853

.789

.731

.677

.627

.582

.540

.502

.467

9

.914

.837

.766

.703

.645

.592

.544

.500

.460

.424

10

.865

.820

.744

.676

.614

.558

.508

.463

.422

.386

11

.896

.804

.722

.650

.585

.527

.475

.429

.388

.350

12

.887

.789

.701

.625

.557

.497

.444

.397

.356

.319

13

.879

.773

.681

601

.530

.469

.415

.368

.326

.290

14

.870

.758

.661

.577

.505

.442

.388

.340

.299

.263

15

.861

.743

.642

.555

.481

.417

.362

.315

.275

.239

16

.853

.728

.623

.534

.458

.394

.339

.292

.252

.218

17

.844

.714

.605

.513

.436

.371

.317

.270

.231

.198

18

.836

.700

.587

.494

.416

.350

.296

.250

.212

.180

19

.828

.686

.570

.475

.396

.331

.277

.232

.194

.164

20

.820

.673

.554

.456

.377

.312

.258

.215

.178

.149

21

.811

.660

.538

.439

.359

.294

.242

.199

.164

.135

22

.803

.647

.522

.422

.342

.278

.226

.184

.150

.123

23

.795

.634

.507

.406

.326

.262

.211

.170

.138

.112

24

.788

.622

.492

.390

.310

.247

.197

.158

.126

.102

25

.780

.610

.478

.375

.295

.233

.184

.146

.116

.092

30

.742

.552

.412

.308

.231

.174

.131

.099

:075

.057

35

.706

.500

.355

.253

.181

.130

.094

.068

.049

.036

40

.672

.453

.307

.208

.142

.097

.067

.046

.032

.022

45

.639

.410

.264

.171

.111

.073

.048

.031

.021

.014

50

.806

.372

.228

•141

.087

.054

.034

.021

.013

.009

857

Index

+ Index + A Activity 754 Addition Law of Probability' 14 Adjoint of a matrix 31 Advantage of a model 6 Agriculture 4, 266 Algebraic method 692 Algebra events 12 All Integer prog. problem 457 Allocation 235 Analogue Model 8 Analytic method 247, 261 Arithmetic method 682 Arrival, pattern 161 rate 163 Artificial, Variables 326 Variable technique 326 Assigninent, algorithm 492 problem 266, 489 Augmented matrix 36 Average, inventory 46 waiting time 180 Axiomatic def. of Prob. 13 B Back, logged 50 Backward, difference 37 Computation Procedure 646 Pass method 770 Balking 162 Basic, degenerate solution 242 duality theorem 379 feasible solution 243, 542 Queuing Process 161 non-degenerate solution 242 solution 242 Variables 242, 243, 294 Basis, matrix 294 set 34 Bellman's principle of optimality in Dyn. prog. 631

Bernoulli Distribution 20 Big M-method 329 Binomial distribution 20, 228 Birth and death, model 173 Blending Prob. 266 Branch 751 Branch and Bound Algoth. 483 Branch Tech. 482 Buffer stock 46 Burst event 754 Busy period 163 C

Carrying cost 46 Garners M-method 329 Cell evaluations 553 Characteristics of, O.R. 5 Dyn. Prog. Problems 631, 632 Peturbation Method 334 Queuing System 161 Classification of Inv. models 43 172 Queuing model Classical definition of prob. 12 Closed set 278 Cofactor 31 Collusion 163 Communication Process 801 Competitive, games 659 situations 659 12 Complementary of an event Compound events 14 Concave 828 Conditional Probability 15 Connected graph 752 Constraints 235, 236 Consistent 36 Construction of Gomary's constraint 458 Continuous prob. distribution 22 Convex 828 Convex, combination 278 function 280

858 hull 280 polyhedron 280 set 277, 279 Cost, variables 41 equation 45 Critical activity 774 event 774 path 775 Critical path method (CPM) 751, 768 Current matainenance cost 116 Customers 161 Customers behaviour in a queue 162 Cycle 752 Cyclic error 756 Cycling in L.P.P. 334 D Danging error 756 Dantzig, G.B. 293 Decoder 802 Defence 3 Definition of O.R. 1 Degeneracy in L.P.P. 334 in T. Problem 565 Degenerate, B solution 242 B.F. solution 244, 542 Demand 42 Dependent Event 15 Depreciation ratio 125 Description of comm. sys 801 Determinant 28, 30 Deterministic, Inv. models 43, 44 demands 42 Deterministic models with shortages 50 Diagonal matrix 28 Difference interval 36 Difference between CPM & PERT 753 Trans. Prob. & Ass. prob. 542 Difference, operators 36 forward 36 backward 37

Operations Research Differentiation of intervals 37 Discontinuous demand 77 Discount rate 125 Discrete, prob. distributions 16 random variables 16 Dominance property 674 Duality in L.P. 371 Duality theorem 377 Dual problem 371, 372 Dual Simplex method 388 Dual variable 372 Dummy activity 755 Dynamic programming 631 Dynamic prog. problem 631 E Economic lot size, model formula 45 Efficiency 813 Encoder 802 End points 751 Entering vector 316 Entropy 804 Erlang, service time distribution 168 Model 172 Euclidean space 34 Event, 11, 754 Burst 754 certain 11 impossible 11 Equally likely 12 mutually exclusive 12 simple 11 Slacks 773 Exhaustive events 12 Expectation of a Random variable 16 Expected, gain 668 line length 175, 185, 198 queue length 176, 202 time 789 value of game 670, 671 waiting time in queue 179, 216

Index waiting time in system 180, 216 Explosive state 163 Exponential distribution 22, 168 Extreme point, 751 of a convex set 279 F Fair game 662 Feasible solution 243, 542 Finite, difference 36 game 660 Fixed time mode 50 Float 772 Flow capacity 752 Forward, difference 36 Past method 769 Free float 772 Fulkerson's Rule 758 Functional equation 633, 803 Fundamental duality theorem 379 Fundamental theorem of L. Progr. 295 G Games 659 theory 659 with saddle point 665 without saddle point 671 Gantt bar chart 752 Generating functions 38 Goal Programming 725 Gomory's, all I.P.P. Method 458 Constraints 460, 463 Cutting Plane Method 460 Graphical method for solution of game 684 Graphical method for solution of L.P.P 247 of Seq. Prob. 612 of N.L.P.P 846 Group replacement policy 137 H Harris Eco. lot size formula 45 Head event slack 774 Hessian Matrix 829

859 Holding cost 41 Hungarian Method 492 Hyper, plane 278 sphere 277 I Iconic model 7 Idle period 163 Incoming vector 316, 317 Inconsistent 36, 311 Individual replacement policy 137 Independent event 15 Independent float 773 Inferior 674 Infinite game 660 Industry 3 Information theory 801 Instantaneous production 44 Integer programming 457 Interior Point 278, 279 Intersection of events 12 161, 162 Input Inventory 41 Inventory theory 41 Inverse of a matrix by simplex method 358 Iso Profit (cost) method 248 Iterative method 697 J Jockeying in queues 163 Judgement phase 5 Johnson's Method 604 K Key element 317 Kuhn-Tucker Nec. condi. 837 Suff. condi 839 L (Lagrangian, function 825 multipliers 825 Lead time 42 Linear, Algebra 35 combination 34 Dependence 34, 296

860 equation 35 Independence 34, 296 programming, problem 235 prog. prob with unrestricted variables 347 Lines 278 Line segments 278 Loop 752 Looping error 756 Loops in T.P. ' 542 Lowest cost entry method 546 Lowest value of game 664 M Machine repair problem 172, 223 115, 117 Maintenance cost Marovian property 632 Math. Formulation of, Assign. problem 490 Goal Programming Problem 725 L.P. Problems 237 N.L.P Problems 822 Trans. Problems 541 Matrix 28 Matrix Method for solution of games 706 Maximin minimax criterion of optimality 663 Maximum Assign Problems 513 Mean arrival rate ' 163 Mean servicing rate 163 Merge Event 754 Merge & Burst event 754 Message 812 Method of oddments 682. Minors 29, 31 Mixed integer Programming Problems 457, 474 Cutting Plane Algo. 461 Mixed strategy 662 Model 6 Modeling in O.R. 6

Operations Research Modi method 550, 553 Money value 115, 124, Monte Carlo technique 8 Mortability Table 142 Most likely time 789 Move 660 Multi stage Decision Problem 631 Multi-item, deterministic model with one constantt 59 Multiple Goals 727 Multiplication thm. of prob. 15 N Nature and definition of O.R. 1 Network 751, 752 Components 754 News paper boy problem 85 Node 751 Noise 802 Non-degenerate B. Sol. 242 B.F. Sol. 244, 542 Non-linear programing problems 821 Normal distribution 23 North west corner rule 545 Notations 43, 171 Null matrix 28 Numbering the events 758 0 Objective function 235, 236 Oddment method 682 Open set 278 Operations Research 1 Nature and definition 1 Objective 2 2 Phases 3 Scope Optimal policy 631 243, 542 solution 550 test in T.P. Optimality condition in L.P. 308 Optimistic time 789 Oriented graph 751 316, 317 Outgoing vector

861

Index P Path 751 Pay of matrix 660 Parametric L.P. 435 PERT 751, 789 Permissible region 248 Penlity cost 42 Pessimistic time 789 Phases of O.R. 2 Pivot element 317 Planning 4 Point set 277 Poisson, distribution 21 process 164 arrival 1.65 Power supply model '172, 226 Predecessor (Preceding) activity 755 Present value 125 Price vector 236 Primal problem 372 Priorities 725, 727 Probabilistic elements 42 model 77 Processing 612 Probability 13 definition 13 Profit maximization T.P. 576 Prohibited Trans Route 584 Purchase Inventory model with price breaks 100, 101, 103, 104 Pure strategy 662 Q

Queue 161, 162 length 171, 176 theory 161 Que with discouragement 184 R Random, variables 16 Rank of matrix 31 Receiver 801

Recursive, equation 632, 633 relationship 632 Rectangular, distribution 24 game 660 Reduction Thm. 490 Redundancy 311, 312 Redundancy error 757 Redundant 36 236 Requirement Vector Regular distribution 168 Reneging 162 Re-order point 46 Replacement 115 Replacement, problems 115 35 Theorem Replenishment cost 42 Research phase 5 Restrictions on assignment 520 Routing problem 614 S

Saddle point 664 Sample space 11 752 Schedule chart Scientific method 5 Scrap value 115 Scope of O.R. 3 Second difference 37 Sensitivity Analysis 399 Sequencing 603 Server 161 Service, discipline 161 mechanism 161 Servicing time 163 Set up cost 42 Shannon-Fano Encoding Proc. 814 Shortage cost 42 Simplex, method 1, 247, 293 algorithm 293 Simultaneous linear eqn. 35 Sink 753, 801 Slack 772

Operations Research

862 Slack variables 262 Source 752, 801 Sol. of I.P.P. 458 Sol. off simult. linear eq n by

Upper value of a game 661 Unique Opt. Sol. 315 Unrestricted variables 347 Unsymmetric Dual Prob. 372

simplex method 356 Spanning set 34 Staffing problems 142 Standard derivation 789 State of the system 163 Statistical def. of Probability 13 Steady state 164 Stepping-Stone method 550 Storage cost 41 Strategy 662 Sub-matrix 29 Successor activity 755 Surplus variables 262 Symbolic models 8 Symmetric dual problem 371 T Terminal branch 751 Total float 772 Traffic intensity 163 Transportation algorithm 553 problem 266, 539 Transient State 163 Transposed matrix 28 Travelling salesman problem 603, 614 Tree 752 Triangular matrix 28 Two Finger Morra 661 Two person zero-sum games 660 Two-phase method 326 U Unbalanced, assignment problem 509 Transp. problem 565 Unbounded solution 307, 316 Unit cost penalty method 547 Unit matrix 28 Union of events 12

V Value of game 662, 664 Variables in Inv. Prob. 41 Variance 20, 21, 789 Variation in c, 399, 435 in b 399, 407, 447 in ctij 399, 413 Vectors 32 Vectors spaces 32 Vogel's appr. method (V.A.M.) 547 Waiting, lines 161 time 162, 163 weighted average cost 128 Weights 725 Wilson Eco. lot-size formula 45 Zero-sum game

660

• • •

Notes

863

Notes

Operations Research

864

Notes