Risk and Uncertainty Reduction by Using Algebraic Inequalities [1 ed.] 0367898004, 9780367898007

Table of contents :
Cover
Half Title
Title Page
Copyright Page
Dedication
Table of Contents
Preface
Author
Chapter 1: Fundamental Concepts Related to Risk and Uncertainty Reduction by Using Algebraic Inequalities
1 1 Domain-Independent Approach to Risk Reduction
1 2 A Powerful Domain-Independent Method for Risk and Uncertainty Reduction Based on Algebraic Inequalities
1 2 1 Classification of Techniques Based on Algebraic Inequalities for Risk and Uncertainty Reduction
1 3 Risk and Uncertainty
Chapter 2: Properties of Algebraic Inequalities: Standard Algebraic Inequalities
2 1 Basic Rules Related to Algebraic Inequalities
2 2 Basic Properties of Inequalities
2 3 One-Dimensional Triangle Inequality
2 4 The Quadratic Inequality
2 5 Jensen Inequality
2 6 Root-Mean Square–Arithmetic Mean–Geometric Mean–Harmonic Mean (RMS-AM-GM-HM) Inequality
2 7 Weighted Arithmetic Mean–Geometric Mean (AM-GM) Inequality
2 8 Hölder Inequality
2 9 Cauchy-Schwarz Inequality
2 10 Rearrangement Inequality
2 11 Chebyshev Sum Inequality
2 12 Muirhead Inequality
2 13 Markov Inequality
2 14 Chebyshev Inequality
2 15 Minkowski Inequality
Chapter 3: Basic Techniques for Proving Algebraic Inequalities
3 1 The Need for Proving Algebraic Inequalities
3 2 Proving Inequalities by a Direct Algebraic Manipulation and Analysis
3 3 Proving Inequalities by Presenting Them as a Sum of Non-Negative Terms
3 4 Proving Inequalities by Proving Simpler Intermediate Inequalities
3 5 Proving Inequalities by Substitution
3 6 Proving Inequalities by Exploiting the Symmetry
3 7 Proving Inequalities by Exploiting Homogeneity
3 8 Proving Inequalities by a Mathematical Induction
3 9 Proving Inequalities by Using the Properties of Convex/Concave Functions
3 9 1 Jensen Inequality
3 10 Proving Inequalities by Using the Properties of Sub-Additive and Super-Additive Functions
3 11 Proving Inequalities by Transforming Them to Known Inequalities
3 11 1 Proving Inequalities by Transforming Them to an Already Proved Inequality
3 11 2 Proving Inequalities by Transforming Them to the Hölder Inequality
3 11 3 An Alternative Proof of the Cauchy-Schwarz Inequality by Reducing It to a Standard Inequality
3 11 4 An Alternative Proof of the GM-HM Inequality by Reducing It to the AM-GM Inequality
3 11 5 Proving Inequalities by Transforming Them to the Cauchy-Schwarz Inequality
3 12 Proving Inequalities by Segmentation
3 12 1 Determining Bounds by Segmentation
3 13 Proving Algebraic Inequalities by Combining Several Techniques
3 14 Using Derivatives to Prove Inequalities
Chapter 4: Using Optimisation Methods for Determining Tight Upper and Lower Bounds: Testing a Conjectured Inequality by a Simulation – Exercises
4 1 Using Constrained Optimisation for Determining Tight Upper Bounds
4 2 Tight Bounds for Multivariable Functions Whose Partial Derivatives Do Not Change Sign in a Specified Domain
4 3 Conventions Adopted in Presenting the Simulation Algorithms
4 4 Testing a Conjectured Algebraic Inequality by a Monte Carlo Simulation
4 5 Exercises
4 6 Solutions to the Exercises
Chapter 5: Ranking the Reliabilities of Systems and Processes by Using Inequalities
5 1 Improving Reliability and Reducing Risk by Proving an Abstract Inequality Derived from the Real Physical System or Process
5 2 Using Algebraic Inequalities for Ranking Systems Whose Component Reliabilities Are Unknown
5 2 1 Reliability of Systems with Components Logically Arranged in Series and Parallel
5 3 Using Inequalities to Rank Systems with the Same Topology and Different Component Arrangements
5 4 Using Inequalities to Rank Systems with Different Topologies Built with the Same Types of Components
Chapter 6: Using Inequalities for Reducing Epistemic Uncertainty and Ranking Decision Alternatives
6 1 Selection from Sources with Unknown Proportions of High-Reliability Components
6 2 Monte Carlo Simulations
6 3 Extending the Results by Using the Muirhead Inequality
Chapter 7: Creating a Meaningful Interpretation of Existing Abstract Inequalities and Linking It to Real Applications
7 1 Meaningful Interpretations of Abstract Algebraic Inequalities with Applications to Real Physical Systems
7 1 1 Applications Related to Robust and Safe Design
7 2 Avoiding Underestimation of the Risk and Overestimation of Average Profit by a Meaningful Interpretation of the Chebyshev Sum Inequality
7 3 A Meaningful Interpretation of an Abstract Algebraic Inequality with an Application for Selecting Components of the Same Variety
7 4 Maximising the Chances of a Beneficial Random Selection by a Meaningful Interpretation of a General Inequality
7 5 The Principle of Non-Contradiction
Chapter 8: Optimisation by Using Inequalities
8 1 Using Inequalities for Minimising the Deviation of Reliability-Critical Parameters
8 2 Minimising the Deviation of the Volume of Manufactured Workpieces with Cylindrical Shapes
8 3 Minimising the Deviation of the Volume of Manufactured Workpieces in the Shape of a Rectangular Prism
8 4 Minimising the Deviation of the Resonant Frequency from the Required Level for Parallel Resonant LC-Circuits
8 5 Maximising Reliability by Using the Rearrangement Inequality
8 5 1 Using the Rearrangement Inequality to Maximise the Reliability of Parallel-Series Systems
8 5 2 Using the Rearrangement Inequality for Optimal Condition Monitoring
8 6 Using the Rearrangement Inequality to Minimise the Risk of a Faulty Assembly
Chapter 9: Determining Tight Bounds for the Uncertainty in Risk-Critical Parameters and Properties by Using Inequalities
9 1 Upper-Bound Variance Inequality for Properties from Different Sources
9 2 Identifying the Source Whose Removal Causes the Largest Reduction of the Worst-Case Variation
9 3 Increasing the Robustness of Electronic Products by Using the Variance Upper-Bound Inequality
9 4 Determining Tight Bounds for the Fraction of Items with a Particular Property
9 5 Using the Properties of Convex Functions for Determining the Upper Bound of the Equivalent Resistance
9 6 Determining a Tight Upper Bound for the Risk of a Faulty Assembly by Using the Chebyshev Inequality
9 7 Deriving a Tight Upper Bound for the Risk of a Faulty Assembly by Using the Chebyshev Inequality and Jensen Inequality
Chapter 10: Using Algebraic Inequalities to Support Risk-Critical Reasoning
10 1 Using the Inequality of the Negatively Correlated Events to Support Risk-Critical Reasoning
10 2 Avoiding Risk Underestimation by Using the Jensen Inequality
10 2 1 Avoiding the Risk of Overestimating Profit
10 2 2 Avoiding the Risk of Underestimating the Cost of Failure
10 2 3 A Conservative Estimate of System Reliability by Using the Jensen Inequality
10 3 Reducing Uncertainty and Risk Associated with the Prediction of Magnitude Rankings
References
Index

Citation preview

Risk and Uncertainty Reduction by Using Algebraic Inequalities

Risk and Uncertainty Reduction by Using Algebraic Inequalities

Michael T. Todinov

First edition published 2020 by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 and by CRC Press 2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN © 2020 Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, LLC Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright. com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging-in-Publication Data Names: Todinov, M. T., author. Title: Risk and uncertainty reduction by using algebraic inequalities / Michael T. Todinov. Description: First edition. | Boca Raton, FL : CRC Press, [2020] | Includes bibliographical references and index. Identifiers: LCCN 2020005695 (print) | LCCN 2020005696 (ebook) | ISBN 9780367898007 (hardback) | ISBN 9780367504014 (paperback) | ISBN 9781003032502 (ebook) Subjects: LCSH: Risk assessment--Mathematics. | System failures--Prevention--Mathematics. | Risk management--Mathematics. | Inequalities (Mathematics) Classification: LCC TA169.55.R57 T63 2020 (print) | LCC TA169.55.R57 (ebook) | DDC 620/.004520151297--dc23 LC record available at https://lccn.loc.gov/2020005695 LC ebook record available at https://lccn.loc.gov/2020005696 ISBN: 978-0-367-89800-7 (hbk) ISBN: 978-1-003-03250-2 (ebk) Typeset in Times by Lumina Datamatics Limited

Dedication To Prolet

Contents Preface������������������������������������������������������������������������������������������������������������������� xiii Author��������������������������������������������������������������������������������������������������������������������xvii Chapter 1 Fundamental Concepts Related to Risk and Uncertainty Reduction by Using Algebraic Inequalities�����������������������������������������1 1.1 Domain-Independent Approach to Risk Reduction������������������ 1 1.2 A Powerful Domain-Independent Method for Risk and Uncertainty Reduction Based on Algebraic Inequalities���������� 4 1.2.1 Classification of Techniques Based on Algebraic Inequalities for Risk and Uncertainty Reduction��������� 8 1.3 Risk and Uncertainty��������������������������������������������������������������� 10 Chapter 2 Properties of Algebraic Inequalities: Standard Algebraic Inequalities����������������������������������������������������������������������������������������� 21 2.1 Basic Rules Related to Algebraic Inequalities������������������������ 21 2.2 Basic Properties of Inequalities����������������������������������������������� 21 2.3 One-Dimensional Triangle Inequality������������������������������������� 23 2.4 The Quadratic Inequality��������������������������������������������������������24 2.5 Jensen Inequality���������������������������������������������������������������������25 2.6 Root-Mean Square–Arithmetic Mean–Geometric Mean–Harmonic Mean (RMS-AM-GM-HM) Inequality������26 2.7 Weighted Arithmetic Mean–Geometric Mean (AM-GM) Inequality��������������������������������������������������������������� 27 2.8 Hölder Inequality���������������������������������������������������������������������28 2.9 Cauchy-Schwarz Inequality����������������������������������������������������� 29 2.10 Rearrangement Inequality������������������������������������������������������� 30 2.11 Chebyshev Sum Inequality������������������������������������������������������ 33 2.12 Muirhead Inequality����������������������������������������������������������������34 2.13 Markov Inequality������������������������������������������������������������������� 35 2.14 Chebyshev Inequality�������������������������������������������������������������� 36 2.15 Minkowski Inequality������������������������������������������������������������� 36 Chapter 3 Basic Techniques for Proving Algebraic Inequalities������������������������ 39 3.1 The Need for Proving Algebraic Inequalities������������������������� 39 3.2 Proving Inequalities by a Direct Algebraic Manipulation and Analysis���������������������������������������������������������������������������� 39

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3.3 Proving Inequalities by Presenting Them as a Sum of Non-Negative Terms����������������������������������������������������������� 42 3.4 Proving Inequalities by Proving Simpler Intermediate Inequalities������������������������������������������������������������������������������44 3.5 Proving Inequalities by Substitution��������������������������������������� 45 3.6 Proving Inequalities by Exploiting the Symmetry������������������ 45 3.7 Proving Inequalities by Exploiting Homogeneity������������������� 47 3.8 Proving Inequalities by a Mathematical Induction����������������� 49 3.9 Proving Inequalities by Using the Properties of Convex/Concave Functions������������������������������������������������� 52 3.9.1 Jensen Inequality�������������������������������������������������������� 54 3.10 Proving Inequalities by Using the Properties of Sub-Additive and Super-Additive Functions����������������������� 56 3.11 Proving Inequalities by Transforming Them to Known Inequalities������������������������������������������������������������������������������ 58 3.11.1 Proving Inequalities by Transforming Them to an Already Proved Inequality�������������������������������� 58 3.11.2 Proving Inequalities by Transforming Them to the Hölder Inequality��������������������������������������������� 59 3.11.3 An Alternative Proof of the Cauchy-Schwarz Inequality by Reducing It to a Standard Inequality������������������������������������������������������������ 60 3.11.4 An Alternative Proof of the GM-HM Inequality by Reducing It to the AM-GM Inequality�����������������60 3.11.5 Proving Inequalities by Transforming Them to the Cauchy-Schwarz Inequality����������������������������� 61 3.12 Proving Inequalities by Segmentation������������������������������������� 62 3.12.1 Determining Bounds by Segmentation����������������������64 3.13 Proving Algebraic Inequalities by Combining Several Techniques�������������������������������������������������������������������������������64 3.14 Using Derivatives to Prove Inequalities���������������������������������� 65 Chapter 4 Using Optimisation Methods for Determining Tight Upper and Lower Bounds: Testing a Conjectured Inequality by a Simulation – Exercises����������������������������������������������������������������������� 69 4.1 Using Constrained Optimisation for Determining Tight Upper Bounds�������������������������������������������������������������������������� 69 4.2 Tight Bounds for Multivariable Functions Whose Partial Derivatives Do Not Change Sign in a Specified Domain�������� 71 4.3 Conventions Adopted in Presenting the Simulation Algorithms������������������������������������������������������������������������������� 73 4.4 Testing a Conjectured Algebraic Inequality by a Monte Carlo Simulation��������������������������������������������������� 76 4.5 Exercises����������������������������������������������������������������������������������80 4.6 Solutions to the Exercises�������������������������������������������������������� 83

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Chapter 5 Ranking the Reliabilities of Systems and Processes by Using Inequalities.......................................................................................... 95 5.1 Improving Reliability and Reducing Risk by Proving an Abstract Inequality Derived from the Real Physical System or Process�������������������������������������������������������������������� 95 5.2 Using Algebraic Inequalities for Ranking Systems Whose Component Reliabilities Are Unknown���������������������� 95 5.2.1 Reliability of Systems with Components Logically Arranged in Series and Parallel������������������ 96 5.3 Using Inequalities to Rank Systems with the Same Topology and Different Component Arrangements...............100 5.4 Using Inequalities to Rank Systems with Different Topologies Built with the Same Types of Components.......... 103 Chapter 6 Using Inequalities for Reducing Epistemic Uncertainty and Ranking Decision Alternatives.................................................. 107 6.1 Selection from Sources with Unknown Proportions of High-Reliability Components������������������������������������������� 107 6.2 Monte Carlo Simulations........................................................ 110 6.3 Extending the Results by Using the Muirhead Inequality������� 113 Chapter 7 Creating a Meaningful Interpretation of Existing Abstract Inequalities and Linking It to Real Applications.............................. 119 7.1 Meaningful Interpretations of Abstract Algebraic Inequalities with Applications to Real Physical Systems������119 7.1.1 Applications Related to Robust and Safe Design...... 124 7.2 Avoiding Underestimation of the Risk and Overestimation of Average Profit by a Meaningful Interpretation of the Chebyshev Sum Inequality������������������ 126 7.3 A Meaningful Interpretation of an Abstract Algebraic Inequality with an Application for Selecting Components of the Same Variety���������������������������������������������������������������� 128 7.4 Maximising the Chances of a Beneficial Random Selection by a Meaningful Interpretation of a General Inequality����������� 130 7.5 The Principle of Non-Contradiction....................................... 134 Chapter 8 Optimisation by Using Inequalities................................................... 139 8.1 Using Inequalities for Minimising the Deviation of Reliability-Critical Parameters����������������������������������������� 139 8.2 Minimising the Deviation of the Volume of Manufactured Workpieces with Cylindrical Shapes����������������������������������������140

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8.3 Minimising the Deviation of the Volume of Manufactured Workpieces in the Shape of a Rectangular Prism�������������������������������������������������������������������������������������� 142 8.4 Minimising the Deviation of the Resonant Frequency from the Required Level for Parallel Resonant LC-Circuits�������������144 8.5 Maximising Reliability by Using the Rearrangement Inequality������������������������������������������������������������������������������� 146 8.5.1 Using the Rearrangement Inequality to Maximise the Reliability of Parallel-Series Systems��������������������������������������������������������������������� 146 8.5.2 Using the Rearrangement Inequality for Optimal Condition Monitoring����������������������������������������������� 150 8.6 Using the Rearrangement Inequality to Minimise the Risk of a Faulty Assembly����������������������������������������������� 151 Chapter 9 Determining Tight Bounds for the Uncertainty in Risk-Critical Parameters and Properties by Using Inequalities............................. 153 9.1 Upper-Bound Variance Inequality for Properties from Different Sources���������������������������������������������������������� 153 9.2 Identifying the Source Whose Removal Causes the Largest Reduction of the Worst-Case Variation�������������� 156 9.3 Increasing the Robustness of Electronic Products by Using the Variance Upper-Bound Inequality������������������� 157 9.4 Determining Tight Bounds for the Fraction of Items with a Particular Property������������������������������������������������������������� 158 9.5 Using the Properties of Convex Functions for Determining the Upper Bound of the Equivalent Resistance�����������������������������������������������������������������������������159 9.6 Determining a Tight Upper Bound for the Risk of a Faulty Assembly by Using the Chebyshev Inequality����������� 162 9.7 Deriving a Tight Upper Bound for the Risk of a Faulty Assembly by Using the Chebyshev Inequality and Jensen Inequality������������������������������������������������������������������� 164 Chapter 10 Using Algebraic Inequalities to Support Risk-Critical Reasoning.................................................................................. 169 10.1 Using the Inequality of the Negatively Correlated Events to Support Risk-Critical Reasoning��������������������������������������� 169 10.2 Avoiding Risk Underestimation by Using the Jensen Inequality������������������������������������������������������������������������������� 171 10.2.1 Avoiding the Risk of Overestimating Profit.............. 171 10.2.2 Avoiding the Risk of Underestimating the Cost of Failure................................................................... 173

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10.2.3 A Conservative Estimate of System Reliability by Using the Jensen Inequality�������������������������������� 174 10.3 Reducing Uncertainty and Risk Associated with the Prediction of Magnitude Rankings������������������������� 176 References������������������������������������������������������������������������������������������������������������� 181 Index���������������������������������������������������������������������������������������������������������������������� 185

Preface In the engineering design and the reliability and risk literature there is a surprising lack of discussion on the use of non-trivial algebraic inequalities to improve reliability and reduce uncertainty and risk. Consequently, the main objective of this book is to fill this significant gap in the application of algebraic inequalities. The application of algebraic inequalities for reliability improvement, uncertainty and risk reduction is associated with a number of big benefits. • Algebraic inequalities provide a powerful reliability improvement, uncertainty and risk reduction method that transcend engineering and can be applied in various unrelated domains of human activity. While reliability and risk assessment are truly domain-independent areas, this cannot be stated about the equally important areas of reliability improvement and uncertainty and risk reduction. For decades, the reliability and risk science failed to appreciate that reliability improvement and uncertainty and risk reduction are underpinned by general principles that work in many unrelated domains. Currently, reliability improvement and risk reduction almost solely rely on knowledge from a specific domain and are conducted exclusively by experts in that specific domain. This is the reason for the strong perception that effective risk reduction can be delivered solely by using methods offered by the specific domain, without resorting to general risk reduction principles and methods. As a result, in standard textbooks on mechanical engineering and design of machine components, for example, there is no mention of generic (domain-independent) methods for reliability improvement and risk reduction. This  resulted in ineffective reliability improvement and risk reduction across the entire industry. Valuable opportunities for improving reliability and reducing risk have been overlooked, which led to numerous accidents resulting in financial losses, fatalities and damage to the environment. Recently, the author formulated a number of domain-independent methods and principles for improving reliability and reducing risk. The proposed book extends this work by introducing a new, powerful domain-­independent method for improving reliability and reducing risk and uncertainty based on algebraic inequalities. • Algebraic inequalities provide a very effective tool for dealing with deep uncertainty related to key reliability-critical parameters of systems and processes. Reliability and risk assessments in the case of deep uncertainty related to the values of critical parameters present a great challenge to reliability engineering, risk management and decision making. Inequalities do not normally require any knowledge related to the values or the distributions of the controlling variables, which makes them ideal for handling systems and processes whose controlling variables are associated with deep uncertainty. Such applications are considered in the book, including applications related to decision making under deep uncertainty. For  example, xiii

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with the help of an algebraic inequality, a highly counter-intuitive result has been obtained. If no information about the component reliability characterising the individual suppliers is available, purchasing components from a single supplier or from the smallest possible number of suppliers maximises the probability that all purchased components will be of high reliability. • Algebraic inequalities permit a meaningful interpretation which yields new physical properties. While the proof of an inequality does not normally pose problems, the meaningful interpretation of an abstract inequality is not a straightforward process. The meaningful interpretation of the variables entering the inequality and the different parts of the inequality usually brings deep and nontrivial insights, some of which stand at the level of new physical properties. In this respect, the book introduces relevant meaning for the variables and provides a meaningful interpretation for a number of abstract inequalities. Different meaningful interpretations have also been made of the same abstract inequality. One such interpretation showed that the equivalent resistance of n resistors arranged in series is at least n2 times larger than the equivalent resistance of the same resistors arranged in parallel, irrespective of the individual values of the resistors. Similar interpretation has been created, relevant to the equivalent elastic constant of n elastic elements arranged in parallel and series and to the capacity n capacitors arranged in parallel and series. The meaningful interpretation of an abstract algebraic inequality led to a new theorem related to electrical circuits. The power output from a voltage source, on elements in series, is smaller than the total power output from the segmented source where each voltage segment is applied to a separate element. Finally, an important principle of non-contradiction has been formulated: If a correct algebraic inequality permits a meaningful interpretation related to a real phenomenon, the realisation of the phenomenon must yield results that do not contradict the abstract inequality. • Algebraic inequalities provide a very effective tool for determining tight bounds for the variation of risk-critical parameters and complying the design with these bounds. Algebraic inequalities are ideal for determining tight bounds of ­reliability-​ critical parameters associated with variation. In  many cases, the possible ­values of the reliability-critical parameters (e.g. component r­ eliabilities, ­mixing proportions from different suppliers, magnitudes of loads, ­damage, etc.) are unknown. However, if a tight bound for the variation of a critical parameter can be determined by using inequalities, the design could be complied with this bound and a number of failure modes avoided. As a result, inequalities producing such tight bounds can be used for reliability improvement, uncertainty reduction and risk reduction. • Algebraic inequalities are a very effective tool for revealing the intrinsic reliability of systems and processes with unknown reliability of their constituting parts and their ranking.

Preface

Suppose that two different system configurations are built by using the same set of components with unknown performance characteristics (e.g. reliabilities). If an algebraic inequality regarding the reliabilities of the competing systems could be proved, this would mean that the performance (e.g. reliability) of one of the systems is intrinsically superior to the performance of the other system. Then, the superior system can be selected and the risk of failure reduced in the absence of knowledge related to the reliabilities of the parts building the systems. The possibility of revealing the intrinsic reliability of two competing systems/processes and making a correct ranking in terms of reliability, in complete absence of knowledge about the reliabilities of their building parts, is a formidable advantage of the algebraic inequalities which is demonstrated in the book. In addition, • Algebraic inequalities are a very effective tool for optimising designs and processes by maximising performance and minimising the deviation of critical output parameters from their specified values. By using inequalities, the nominal design variables can be selected in such a way that the natural variation of the design variables results in a minimal variation of critical output parameters. By using the rearrangement ­inequality, components with similar functions but different ages (reliability) can be permuted in such a way that the reliability of a system is maximised. By using an abstract inequality, the power output from a voltage source can be maximised. These applications are demonstrated with case studies from ­mechanical engineering, manufacturing, electronics and process engineering. The word combination ‘algebraic inequalities’ and ‘risk/uncertainty reduction’ may give the incorrect impression of a niche subject with niche applications. The  proposed domain-independent method based on algebraic inequalities transcends mechanical engineering (where it originated) and has wide applications in diverse areas of human activity, for example in reliability engineering, systems engineering, risk management, economics, electrical engineering and electronics, manufacturing, materials science, process engineering, oil and gas production, project management, business planning, financial risk management, logistic supply and environmental science. This wide application of the algebraic inequalities is illustrated by the examples in the book which have been selected from the listed application areas. The book also fills a gap in the existing literature on techniques for proving algebraic inequalities. Despite the existence of substantial literature on algebraic inequalities and a large number of solved examples, there is clear absence of systematic exposition of the techniques through which a conjectured equality can be tested and proved. A good awareness of the available tools through which an inequality can be tested and proved significantly increases the benefit from using inequalities for uncertainty and risk reduction. It reduces the reliance on inspiration and tricks in conjecturing and proving algebraic inequalities and enhances the power of researchers.

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Preface

A number of powerful methods and techniques for testing and proving inequalities have not been discussed in the literature or have not received adequate coverage. The book equips readers with a powerful domain-independent method for reducing risk and uncertainty through (i) revealing the intrinsic reliability of systems and processes in the absence of knowledge related to the reliability of their building parts and ranking systems and processes in terms of reliability and risk; (ii) reducing epistemic uncertainty; (iii) reducing aleatory uncertainty; (iv) maximising system reliability; (v) minimising the risk of faulty assembly; (vi) providing tight upper and lower bounds for the variation of risk-critical parameters; (vii) providing support for risk-critical decisions; and (viii) optimisation and creating robust designs and processes. • Algebraic inequalities permit risk and uncertainty reduction to be conducted across all areas of human activity by combining the domain-­independent method based on inequalities with domain-specific knowledge. In  this respect, algebraic inequalities strengthen the credibility of the risk science as a source of valuable domain-independent methods not only for reliability and risk analysis but also for risk and uncertainty reduction. • Employing algebraic inequalities strengthens the reliability and risk science and creates immense opportunities for enhancing the reliability of products and processes across industries. • Algebraic inequalities deliver value to reliability improvement and uncertainty and risk reduction activities and deeply impact the current understanding of uncertainty and risk reduction. • Algebraic inequalities underpin the design for reliability efforts in all areas of mechanical engineering and demonstrate the value of the domain-­ independent methods for reducing uncertainty and risk. • Algebraic inequalities accelerate the development of a unified and powerful framework for uncertainty and risk reduction which transcends mechanical engineering and is applicable to diverse areas of human activity. The author is fully convinced that the method of algebraic inequalities will become an integral part of reliability and risk science and a necessary part of the education of every reliability and risk professional. In  conclusion, I thank the executive editor Cindy Carelli, the senior editorial assistant Erin Harris at CRC Press/Taylor & Francis Group and project management executive Dolarine Sonia Fonceca at Lumina Datamatics Ltd. for their excellent work and cooperation. Thanks also go to my academic colleagues for their useful comments related to various aspects of the content. Finally, I acknowledge the immense help and support from my wife Prolet during the preparation of this book. Michael T. Todinov Oxford, UK

Author Michael T. Todinov, PhD, has a background in mechanical engineering, applied mathematics and computer science. He  received his PhD and higher doctorate (DEng) from the University of Birmingham and is currently a professor in ­mechanical e­ ngineering at Oxford Brookes University, UK. Prof. Todinov pioneered research on reliability analysis based on the cost of failure, repairable flow networks and networks with disturbed flows, domain-independent methods for reliability improvement and risk reduction and reducing risk and uncertainty by using algebraic inequalities. In the area of reliability and risk, Prof. Todinov has authored five books with reputable academic publishers and numerous papers. In 2017, he received the prestigous award of the Institution of Mechanical Engineers (UK) in the area of risk reduction in mechanical engineering.

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Fundamental Concepts Related to Risk and Uncertainty Reduction by Using Algebraic Inequalities

1.1  DOMAIN-INDEPENDENT APPROACH TO RISK REDUCTION While reliability and risk assessment are truly domain-independent areas, this cannot be stated about the equally important areas of reliability improvement and risk reduction. For decades, the reliability and risk science failed to appreciate and emphasise that reliability improvement, risk reduction and uncertainty reduction are underpinned by general principles that work in many unrelated domains. The  reliability and risk research has been focused primarily on reliability and risk assessment and reliability and risk prediction rather than on reliability improvement and risk reduction. Similarly, methods for uncertainty quantification have been developed rather than methods for reducing uncertainty. As a consequence, the reliability and risk science developed primarily methods for measuring and assessing risk and uncertainty, not domain-independent methods for reducing risk and uncertainty which could provide direct input to the design process. Furthermore, the available methods for measuring and assessing reliability and risk cannot always be fully implemented in the design for the obvious reason that for new products and processes reliability data are simply unavailable. Even in cases where reliability data for the components and parts building the systems are available, they are relevant for a particular environment and duty cycle, and their mechanical application to another environment and duty cycle, as experience has shown, is questionable. The lack of predictive capability of the existing reliability tools caused many engineers to lose faith in the tools and discard them as not adding real value to their work. The direct consequence from this state of the risk and uncertainty reduction methods is that in standard textbooks on mechanical engineering and design of machine components, for example Collins (2003), Norton (2006), Pahl et al. (2007), Childs (2014), Budynas and Nisbett (2015), Mott et  al. (2018); and Gullo and Dixon (2018), there is no mention of generic (domain-independent) methods for reliability improvement and risk reduction. This is despite the circumstance that assuring the reliability of the designed product/process is of critical importance for their commercial success. Why is 1

2

Risk and Uncertainty Reduction by Using Algebraic Inequalities

the engineering design so slow in exploiting the achievements of the reliability and risk science to improve reliability and reduce risk? This is certainly not due to the complexity of the reliability improvement and risk reduction methods. In this respect, the contrast with the very complex generic mathematical methods for stress analysis, kinematic and dynamic analysis of solid bodies and fluids is striking. These generic mathematical modelling methods are penetrating all aspects of the engineering design. The problem is that the current approach to reliability improvement and risk and uncertainty reduction almost solely relies on knowledge from a specific domain and is conducted exclusively by experts in that specific domain. This created the damaging perception that effective risk reduction can be delivered solely by using methods offered by the specific domain, without resorting to a general risk reduction methodology. Another entrenched incorrect perception is that a good quality management is sufficient to deliver high-reliability products. These incorrect perceptions resulted in ineffective reliability improvement and risk reduction across the entire industry. Valuable opportunities for improving reliability and reducing risk have been overlooked which led to numerous accidents resulting in financial losses, fatalities and damage to the environment. Even riskreducing design methods that have been used for a very long period of time, such as reducing risk by introducing deliberate weaknesses, have not been acknowledged as valuable domain-independent methods for risk reduction and further developed and systematised in the reliability and risk literature. Current technology changes so fast that the domain-specific skills in reliability improvement and risk reduction are outdated almost as soon as they are learned. In contrast, domain-independent skills in reliability improvement and risk reduction are higher-order skills that permit applying strategically the same methods and principles in new, constantly changing situations and circumstances. On the other hand, domain-independent methods and principles cannot be effectively applied without domain-specific knowledge, just as it is not possible to communicate effectively on an unfamiliar topic by using only generic concepts. Domain-specific knowledge in the area of application is therefore a necessary prerequisite to the application of domain-independent methods. The  development of the physics-of-failure approach to reliability improvement (Pecht et al., 1990) has been prompted by the deficiencies of the data-driven approach: (i) the models based on data collected for a particular environment (temperature, humidity, pressure, vibrations, corrosive agents, etc.) often give poor predictions for the time to failure in a different environment; (ii) the data-driven approach measures the reliability performance of the product instead of improving the reliability of the product which could provide a direct input to the design process and (iii) the datadriven approach is critically dependent on the availability of past failure rates. According to the physics-of-failure approach, failures and decline in performance occur due to known underlying failure mechanisms. Failure mechanisms lead to accumulation of damage, and failure is initiated when the amount of accumulated damage exceeds the endurance limit. As a result, the time to failure of products can be physically modeled. The physics-of-failure approach was very successful in addressing the underlying causes of failure and eliminating failure modes, and contributed to the widespread

Fundamental Concepts Related to Risk and Uncertainty Reduction

3

view among reliability practitioners that only physics-of-failure models can deliver a real reliability improvement. However, it is necessary to point out that building accurate physics-of-failure models of the time to failure is not always possible because of the complexity of the physical mechanisms underlying the failure modes, the complex nature of the environment and the operational stresses. Physics-of-failure modelling certainly helps, for example, to increase the strength of a component by conducting research on the link between microstructure and mechanical properties of the material. However, this approach requires arduous and time consuming research, special equipment and human resources. Furthermore, despite their success and popularity, physics-of-failure models cannot transcend the narrow domain they serve and cannot normally be used to improve reliability and reduce risk in unrelated domains. A central theme in the new domain-independent approach for reliability improvement and risk reduction (Todinov, 2019a,b) is the concept that risk reduction is underlined by common domain-independent principles which, combined with knowledge from the specific domain, help to generate much more effective risk reducing solutions. With the exception of a few simple and well-known domain-independent methods for risk reduction such as implementing redundancy, eliminating vulnerabilities, upgrading with more reliable components, simplification of components, systems and operations and condition monitoring, the framework of the domain-independent methods for risk reduction is missing. Most of the listed methods for risk reduction are associated with significant cost of investment. Eleven new domain-independent risk reduction methods have been recently proposed in Todinov (2019b), and their application across unrelated domains was demonstrated through numerous examples ranging from mechanical engineering, electrical engineering, civil engineering, computer science, project management, health risk management, business planning, financial risk management and oil and gas production. The  domain-independent risk reduction principles and methods transcend the domain of mechanical engineering where they originated and can be applied in truly diverse areas of human activity. These methods and principles have been distilled from a large number of engineering solutions, analyzed for recurring reliability improvement patterns and invariants which were captured into categories, classes and individual techniques. The domain-independent methods do not rely on the availability of past failure data or detailed knowledge of the underlying mechanisms of failure. As a result, they are well suited for developing new designs, with unknown failure mechanisms and failure history. In many cases, these methods reduce risk at no extra cost or at a relatively small cost. Here it needs to be pointed out that the domain-independent reliability improving and risk reducing methods are not a substitute for the domain-specific methods. Rather, they are a powerful enhancement of the domain-specific risk reducing methods and, combined with knowledge from the specific domain, help to obtain superior solutions. The best results are obtained by combining domain-­ independent methods for risk reduction with domain specific knowledge (Todinov, 2019b). The lack of knowledge of the domain-independent method of segmentation, for example, is the reason for missing valuable opportunities for effective

4

Risk and Uncertainty Reduction by Using Algebraic Inequalities

risk reduction in such mature fields as stress analysis and kinematic analysis of mechanisms (Todinov, 2019c). The domain-independent principles and methods for risk reduction: • Provide a key input to the design process by improving the reliability of the designed product rather than measuring its performance only. • Show how risk reduction should be conducted across all areas of human activity: by combining domain-independent principles for risk and uncertainty reduction with domain-specific knowledge. • Significantly increase the standards of uncertainty and risk management, and create immense opportunities for enhancing the reliability of products and processes across various industries. • Deliver great value to all organisations involved in reliability improvement and risk reduction and deeply impact the current understanding of risk and uncertainty reduction. • Add value to processes aimed at improving reliability and reducing risk and uncertainty, irrespective of industry. • Underpin the design for reliability effort in all areas of engineering. • Accelerate the development of a unified and powerful framework for uncertainty and risk reduction, which transcends mechanical engineering and is applicable to diverse areas of the human activity. Establishing universally accepted theoretical principles for risk assessment requires a common definition of risk valid in unrelated domains of human activity (Aven, 2019). Similarly, establishing universally accepted theoretical principles for risk and uncertainty reduction goes through formulating domain-independent principles for reducing risk and uncertainty, valid in unrelated domains of human ­activity. Establishing risk research as a mainstream science requires solid and universally accepted theoretical principles for the two fundamental components of risk management: risk assessment and risk reduction.

1.2 A POWERFUL DOMAIN-INDEPENDENT METHOD FOR RISK AND UNCERTAINTY REDUCTION BASED ON ALGEBRAIC INEQUALITIES Reliability and risk assessments in the case of uncertainty related to the values of  risk-critical parameters still present a great challenge to reliability engineering, risk management and decision making. In many cases, the actual values of the risk-critical parameters (e.g. material properties, dimensions, loads, magnitudes of the consequences of failure) are unknown or are associated with large uncertainty. However, almost all existing reliability analysis tools require reliability data which are unavailable at the design stage, which makes it difficult to compare the performance of different design solutions. The aim of this book is to introduce a new, powerful domain-independent method for improving reliability and reducing uncertainty and risk, based on algebraic inequalities.

Fundamental Concepts Related to Risk and Uncertainty Reduction

5

This aim will be achieved by demonstrating the capabilities of algebraic inequalities for (i) revealing the inherent reliability of systems and processes and ranking them in terms of reliability and risk in the absence of knowledge related to the reliabilities of their building parts; (ii) reducing aleatory and epistemic uncertainty; (iii) obtaining tight upper and lower bounds of risk-critical parameters and properties; (iv) supporting risk-critical decisions and (v) optimisation to maximise reliability and achieve robust designs. Another important objective of this book is to demonstrate the benefits from combining the domain-independent method for risk reduction based on advanced algebraic inequalities and domain-specific knowledge in order to achieve effective uncertainty and risk reduction in diverse application domains. In this respect, the book demonstrates simple and effective solutions in such mature fields as electronics, stress analysis, mechanical design, manufacturing, economics and management. Most of these solutions have never been suggested in standard textbooks and research papers, which demonstrates that the lack of knowledge of the domain-independent method of inequalities for reducing risk and uncertainty made these simple solutions invisible to domain experts. A formidable advantage of algebraic inequalities is that they do not require knowledge related to the distributions of the variables entering the inequalities. This makes the method of algebraic inequalities ideal for handling deep uncertainty associated with components, properties and control parameters and for ranking designs in the absence of reliability data related to the separate components. The  method of algebraic inequalities does not  rely on reliability data or detailed knowledge of physical mechanisms underlying possible failure modes. This is why the method is appropriate for new designs, with no failure history and with unknown failure mechanisms. By proving algebraic inequalities related to the reliabilities of competing systems/processes, the proposed method has the potential to reveal their intrinsic reliability and rank them in terms of reliability in the absence of knowledge related to the reliabilities of their building parts. Suppose that two different system configurations are built by using the same set of n components with performance characteristics (e.g. reliabilities) x1, x2 ,..., xn that are unknown. Let the performance of the first configuration be given by the function f ( x1,..., xn ) while the performance of the second configuration is given by g( x1,..., xn ). If inequality of the type

f ( x1,..., xn ) > g( x1,..., xn )

could be proved, this would mean that the performance (e.g. reliability) of the first configuration is intrinsically superior to the performance of the second configuration. Then, the first system configuration can be selected and the risk of failure reduced in the absence of knowledge related to the reliabilities of the parts building the systems. The possibility of revealing the intrinsic reliability of competing systems/processes and making a correct ranking under a deep uncertainty related the reliabilities of their components is a formidable advantage of algebraic inequalities.

6

Risk and Uncertainty Reduction by Using Algebraic Inequalities

Similarly, if a particular order among the reliabilities of the separate components is present, the topology of the two competing systems may be the same but for a particular permutation (arrangement) of the components, the system reliability associated with that particular permutation may be superior to any other permutation. In both cases, the algebraic inequalities reveal the intrinsic reliability of a particular topology or arrangement. Another formidable advantage of the algebraic inequalities is their capacity to reduce aleatory and epistemic uncertainty and produce tight upper and lower bounds related to uncertain reliability-critical design parameters such as material properties, electrical parameters, dimensions, loads and component reliabilities. As a result, by establishing tight bounds related to properties and parameters, the method of algebraic inequalities can be applied to improve the robustness of designs, by complying them with the worst possible variation of the design parameters. As a result, a number of failure modes can be avoided. Yet another advantage of the method based on algebraic inequalities consists of its suitability for minimising deviations of key parameters from their required values. As a result, the proposed method can be applied to reduce the sensitivity of components and systems to variation of dimensions and other parameters thereby enhancing robustness and performance. Algebraic inequalities can also be used for maximising the reliability of systems and improving the robustness of manufacturing processes. These applications are demonstrated in the book with using algebraic inequalities to maximise the reliability of parallel-series systems and to find the optimal design parameters that guarantee the smallest deviation of a risk-critical parameter. Algebraic inequalities, can also be used for maximising the performance of engineering systems and processes and this application is demonstrated with maximising the power obtained from a voltage source. The method of algebraic inequalities reduces risk at no extra cost or at a low cost, unlike many other methods (e.g. ‘introducing redundancy’, ‘selecting better materials’, ‘strengthening weak links’ and ‘condition monitoring’). The  method of algebraic inequalities is domain-independent because of the domain-independent nature of mathematics. Consequently, the method is demonstrated in such diverse domains as mechanical design, electronic circuits design, project management, economics, decision-making under uncertainty, manufacturing and quality control. Finally, another formidable advantage of algebraic inequalities is that they admit a meaningful interpretation in terms of uncertainty, reliability and risk which can be attached to a real system or process and used to obtain new physical properties and bounds. To the best of our knowledge, creating meaningful interpretation for existing non-trivial abstract inequalities and attaching it to a real system or process has not yet been explored in the reliability and risk literature. Covering this gap constitutes an important objective of this book. Algebraic inequalities have been used extensively in mathematics. For  a long time, simple inequalities are being used to express error bounds in approximations and constraints in linear programming models (Figure 1.1). The  properties of a number of useful non-trivial algebraic inequalities, such as the arithmetic mean – geometric mean (AM-GM) inequality, Cauchy-Schwarz

Fundamental Concepts Related to Risk and Uncertainty Reduction

7

FIGURE 1.1  Simple inequality constraints used in linear programming.

inequality, the rearrangement inequality, the Chebyshev inequalities, Jensen inequality, Muirhead inequality, Hölder inequality, and so on, have also been well documented (Bechenbach and Bellman, 1961; Kazarinoff, 1961; Engel, 1998; Hardy et al., 1999; Steele, 2004; Pachpatte, 2005; Sedrakyan and Sedrakyan, 2010; Lugo, 2013). In reliability and risk research, inequalities have been used exclusively as a mathematical tool for reliability and risk evaluation and for characterisation of reliability functions (Berg and Kesten, 1985; Makri and Psillakis, 1996; Ebeling, 1997; Xie and Lai, 1998; Dohmen, 2006; Hill et  al., 2013; Kundu and Ghosh, 2017). It  is important to guarantee that the reliability of a system meets certain minimal expectations and inequalities have been used (Ebeling, 1997) for obtaining lower and upper bounds on the system reliability by using minimal cut sets and minimal path sets. Xie and Lai (1998), for example, used simple conditional inequalities to obtain more accurate approximations for system reliability, instead of the usual minimal cut and minimal cut bounds. By using improved Bonferroni inequalities, the lower and upper bounds of system reliability were derived by Makri and Psillakis (1996). Inequality-based reliability estimates for complex systems have also been proposed by Hill et al. (2013). In  the reliability and risk research, algebraic inequalities have also been used to express relationships between random variables and their transformations to generate insight into the structure of reliability distributions. Simple inequalities with relation to reliability prediction have been used by Berg and Kesten (1985); inequalities involving expectations have been used by Kundu and Ghosh (2017) for characterising some well-known reliability distributions. Well-known inequalities about a random variable X with unknown probability distribution are the Chebyshev inequality and Markov inequality (DeGroot, 1989). These inequalities are related to the probability that the distance of a random value from a specified number or from its mean will be larger than a specified quantity. An inequality related to the probability that the distances between the locations of a fixed number of uniformly distributed points on a segment will be greater than a specified quantity has been proved in Todinov (2002a).

8

Risk and Uncertainty Reduction by Using Algebraic Inequalities

Trivial inequalities, obtained from solving with respect to one of the variables, have been used for specifying the upper bound of the lineal density of Poissondistributed flaws to guarantee a probability of clustering below a maximum acceptable level (Todinov, 2005, 2006a, 2016). In reliability and risk research, inequalities have been used exclusively as a tool for reliability and risk evaluation and for characterisation of reliability functions. However, these reliability-related applications of inequalities are very limited and oriented towards measuring the reliability performance of the systems, instead of providing direct input to the design process by reducing uncertainty and improving the reliability of components, systems and processes. Despite that standard reliability textbooks (Ramakumar, 1993; Lewis, 1996; O’Connor, 2002; Dhillon, 2017; Modarres et al., 2017) do allocate some space on reliability improvement methods such as introducing redundancy, derating, eliminating common cause and condition monitoring, there is a clear lack of discussion related to reducing uncertainty and risk by using algebraic inequalities. To the best of our knowledge of existing approaches involving inequalities in reliability and risk, the method of algebraic inequalities has not yet been used as a domain-­independent uncertainty and risk reduction method. Covering this gap in the available literature is one of the objectives of this book. Applications of inequalities have also been considered in physics (Rastegin, 2012) and engineering (Cloud et  al., 1998; Samuel and Weir, 1999). However, in the mechanical engineering design literature (Cloud et  al., 1998; French, 1999; Thompson, 1999; Samuel and Weir, 1999; Collins, 2003; Norton, 2006; Pahl et al., 2007; Childs, 2014; Budinas and Nisbett, 2015; Gullo and Dixon, 2018; Mott et al., 2018) there is also a lack of discussion on the use of complex algebraic inequalities to improve reliability and reduce risk. In engineering design, the application of inequalities is mainly confined to inequalities linking design variables required to satisfy various design constraints in order to guarantee that the design will perform its required functions (Samuel and Weir, 1999). Recently, work related to applying advanced algebraic inequalities for uncertainty and risk reduction was published in Todinov (2019c), where a highly counter-intuitive result in decision making under deep uncertainty was obtained by using the Muirhead’s inequality. In Todinov (2019d), an advanced inequality known as ‘upper-bound variance theorem’ has been applied to create a sharp upper bound for the variance of properties from multiple sources. This book extends this work by introducing various applications of the new domain-­independent method based on algebraic inequalities for improving reliability and reducing risk and uncertainty.

1.2.1 Classification of Techniques Based on Algebraic Inequalities for Risk and Uncertainty Reduction The book introduces an important method based on algebraic inequalities for reliability improvement and uncertainty and risk reduction. The method consists of creating relevant meaning for the variables entering the algebraic inequalities and providing

Fundamental Concepts Related to Risk and Uncertainty Reduction

9

a meaningful interpretation of the different parts of the inequality. As a result, the abstract inequality is linked with a real physical system or process. By using this method, various meaningful interpretations have been made of the same abstract inequality. It has been shown that the equivalent resistance of n resistors arranged in parallel is at least n times smaller than the average resistance of the same resistors, irrespective of the individual values of the resistors. This upper bound is much stronger than the well-known upper bound: the equivalent resistance of n resistors arranged in parallel is smaller than the least resistance. Similar interpretation has been given to the equivalent elastic constant of n elastic elements arranged in series and parallel and to the capacity n capacitors arranged in series and in parallel. The  meaningful interpretation of an abstract algebraic inequality led to a new theorem related to electrical circuits. The power output from a voltage source, on elements arranged in series, is smaller than the total power output from the segmented source where each voltage segment is applied to a separate element. Finally, an important principle related to algebraic inequalities has been formulated: If a correct algebraic inequality permits a meaningful interpretation related to a real process/experiment, the realisation of the process/experiment must yield results that do not contradict the abstract inequality. As a domain-independent method, algebraic inequalities reduce risk by (i) revealing the intrinsic reliability of systems and processes and ranking them in terms of reliability and risk; (ii) reducing epistemic uncertainty; (iii) reducing aleatory uncertainty; (iv) maximising system reliability and system performance; (v) minimising the risk of faulty assembly; (vi) providing tight upper and lower bounds for the variation of risk-critical parameters; (vii) providing support for risk-critical decisions and (viii) creating the basis for robust designs and processes. A  classification has been presented in Figure  1.2,  which includes the different ways through which algebraic inequalities reduce uncertainty and risk. Despite the existence of a substantial amount of literature on algebraic inequalities and a significant amount of solved examples (Bechenbach and Bellman, 1961; Kazarinoff, 1961; Engel, 1998; Hardy et  al., 1999; Steele, 2004; Pachpatte, 2005; Sedrakyan and Sedrakyan, 2010), there is absence of a systematic exposition of the different techniques through which an advanced algebraic equality can be tested and proved. A good awareness of the most important tools through which an algebraic inequality can be tested and proved significantly increases the benefit from using inequalities in reliability improvement and risk reduction and enhances the power of researchers in conjecturing and proving inequalities. A number of powerful methods and techniques for testing and proving inequalities have not  been discussed in the literature. To the best of our knowledge the method of Monte Carlo simulation for testing a conjectured algebraic inequality has not been discussed in the literature. In some cases, discussion regarding a particular method or technique does exist, but it is insufficient or inadequate. Filling this gap in the current literature determined another objective of the book: to introduce systematically various important techniques through which an inequality can be conjectured and proved.

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Risk and Uncertainty Reduction by Using Algebraic Inequalities

FIGURE 1.2  Using inequalities for improving reliability and reducing risk and uncertainty.

The  application of algebraic inequalities has been demonstrated in various unrelated domains of human activity: reliability engineering, manufacturing, decision making, mechanical engineering design, electric engineering design, project management, economics and business planning, which shows that the method of algebraic inequalities is indeed a domain-independent method for improving reliability and reducing risk and uncertainty.

1.3  RISK AND UNCERTAINTY The purpose of risk analysis is to provide support in making correct management decisions. By evaluating the risk associated with a set of decision alternatives, risk analysis helps to identify the alternative which maximises the expected utility for the stakeholders while complying with a set of specified criteria and constraints. Key concepts in risk management are likelihood of failure, consequences from f­ailure, potential loss and conditional loss, whose various aspects have been discussed in Todinov (2019b). Risk is always associated with an adverse event (event with negative impact). Starting with the classical paper of Kaplan and Garrick (1981), a large number of definitions and interpretations of risk have been given. The  view adopted in this book is that risk is a unity of the following components: (i) an adverse event, (ii) likelihood of its occurrence, (iii) consequences given that the adverse event has occurred and (iv) uncertainty associated with the consequences.

Fundamental Concepts Related to Risk and Uncertainty Reduction

11

It needs to be pointed out that the likelihood of the exposure to a loss-generating factor and the uncertainty related to the consequences are conditional on the existing knowledge about the adverse event. Because risk is conditional on pre-existing knowledge, risk is relative (Kaplan and Garrick, 1981). The same action from the point of view of the subject and the observer can be a low-risk action or a high-risk action depending on the possession of additional information by the subject or by the observer. The likelihood of an adverse event is conditional on the existence of relevant knowledge ­(information) (Aven, 2019). Thus, the assessed likelihood of failure of a bearing is altered significantly, given that conditional information exists about increased vibration amplitudes and increased temperature of the bearing. Probability is often conditional, and the unconditional probability can be very different from the conditional probability. The condition can effectively be interpreted as knowledge about the system. Thus, according to the conditional probability formula, the probability P( A | B) of the event A given the condition B, P( A | B) = P(PA(∩B)B) , is very different from the unconditional probability P( A). For a market with n suppliers, the probability that a randomly purchased ­component will be a high-reliability component, P( A), is very different from the probability P( A | B) that the component will be a high-reliability component if it is known that the component has survived a particular test. Similarly, in the absence of relevant knowledge, the likelihood of an incident estimated by experts can be very different from the estimated likelihood if relevant knowledge is present. Often, increased uncertainty translates into increased risk, and this has created the misconception that risk is synonymous with uncertainty. It needs to be strongly emphasised that risk is not  synonymous with uncertainty. The  simplest example which exposes this misconception is the load-strength interference in Figure  1.3. Suppose for the sake of simplicity that the failure because of the load-strength interference is associated with a fixed loss. Then, the risk of failure is proportional to the probability of failure. In the load-strength configurations in Figure 1.3, failure is present if load (L) exceeds strength (S). In Figure 1.3a, the uncertainty associated with both the load and strength is significantly smaller than the uncertainty associated with the load and strength in Figure 1.3b, yet the risk of failure for the configuration in Figure 1.3a is significantly larger than the

FIGURE  1.3  Load-strength interference illustrating the fundamental difference between uncertainty and risk: (a) relatively small uncertainty in the load and strength distributions and large risk of failure and (b) large uncertainty in the load and strength distributions and small risk of failure.

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Risk and Uncertainty Reduction by Using Algebraic Inequalities

risk of failure for the configuration in Figure 1.3b. This is because the probability that the load (L) will exceed strength (S) in Figure 1.3a is significantly larger than the probability that the load (L) will exceed strength (S) in Figure 1.3b. Risk, in this case, is a result of the interaction of uncertainties, not a consequence of uncertainty magnitudes. Furthermore, risk is not  synonymous with the uncertainty associated with the consequences. Without exposure to an adverse event, the consequences will never materialise. Replacing ‘likelihood of occurrence of the adverse event’ with ‘uncertainty associated with the adverse event’ does not  correctly represent the risk. While often, larger uncertainty does translate into a larger risk, there are cases where even a proportional increase in the uncertainty of load and strength does not translate into a larger risk. Consider the two intervals in Figure 1.4a, depicting the independent variation of load (interval L) and strength (interval S). Each interval has length of 4 units and load and strength are uniformly distributed along their corresponding intervals (Figure 1.4a). Failure is present if load is greater than strength; therefore, the probability of failure is equal to the probability that load will be greater than strength. In order for this event to occur, both load and strength must belong to the intersection region (the probability of which is 0.5 × 0.5 = 0.25) and given that both load and strength are within the intersection region, load must exceed strength. The probability that load will exceed strength given that both are in the intersection region is 0.5 because, due to the symmetry, in half of the load-strength random configurations, load is greater than strength and in half of the configurations, strength is greater than load. The probability of the compound event that both load and strength are within the intersection region and load exceeds strength is therefore 0.25 × 0.5 = 0.125. In Figure 1.4b, the uncertainty intervals of both load and strength have been proportionally increased twice. Despite that the uncertainty has been increased proportionally, the likelihood that load will exceed strength remains the same as that in Figure 1.4a, equal to 0.125. Increasing proportionally the uncertainty associated with the load and strength had no effect on the risk of failure. Increasing the uncertainty (variation) in the load and strength does not necessarily increase the probability of failure. Further counterexamples to the erroneous view that increased uncertainty always translates into increased risk had been given in Todinov (2016) with the load-strength interference of asymmetric load and strength distributions. As can be seen, while larger likelihood of occurrence of the adverse event always means larger risk, larger uncertainty does not always mean larger risk. Therefore, the concept ‘likelihood of an adverse event’ permits risks to be ranked while the concept ‘uncertainty’ does not.

FIGURE 1.4  (a) A relatively small uncertainty associated with the load (L) and strength (S) and (b) A relatively large uncertainty associated with the load (L) and strength (S) which does not translate into a larger likelihood of failure.

Fundamental Concepts Related to Risk and Uncertainty Reduction

13

Another counterexample that increased variability (uncertainty) does not necessarily translate into increased risk was given in Todinov (2013) with the risk of a net loss from a sequence of statistically independent repeated good bets (bets with positive expected profit). Despite that with increasing the number of good bets, the variance (uncertainty) of the net profit increases, the probability of a net loss decreases significantly. In this case, the common belief that a larger uncertainty (variance) in the profits translates into a larger risk does not hold. In the case of repeated good bets, a larger uncertainty (variance) in the net profits actually coexists with a smaller risk of a net loss and the uncertainty of the net profit cannot be used to rank the risks of a net loss associated with different risky prospects. The risk can be measured in various ways. According to a popular classical definition (Henley and Kumamoto, 1981; Vose, 2000), the risk of failure is measured by the product of the probability of failure and the expected loss given failure:

K = p f C

(1.1)

where pf is the probability of failure and C is the expected loss given failure. To an operator of production equipment, for example, the expected loss given failure C may include several components: cost of lost production, cost of cleaning up polluted environment, medical costs, insurance costs, legal costs, costs of mobilisation of emergency resources, lost sales due to loss of reputation and low customer confidence, and so on. The expected loss to the manufacturer of production equipment may include warranty payment if the equipment fails before the agreed warranty time, loss of sales, penalty payments, compensation and legal costs. For a repairable system, the expected loss from failures in the finite time interval (0, a) is given by the equation (Todinov, 2007)

L = NC

(1.2)

where N is the expected number of failures in the interval (0, a) and C is the expected cost given failure. The problems with definitions of risk based on expected values have been analysed in Todinov (2016). These are one-dimensional definitions and, as a consequence, do not adequately reflect the existing risk. Indeed, consider a good bet where, with probability 0.8, a person receives their initial investment plus the size of the investment as profit or with probability 0.2 the person loses their investment. If a single bet with investment of $3000 is made, this investment is risky, with probability 0.2 that the person will lose their investment of $3000. However, if the investment is split into three sequential bets (each bet equal to $1000) instead of a single bet, the probability of a net loss is equal to the probability of no win in three bets. This probability is equal to = p 0= .23 0.008, which is significantly smaller than the probability of a net loss of 0.2 characterising the single bet of $3000. Despite that the expected loss in both cases is the same, 0.2 × 3000 = 600 and 0.2 × 1000 + 0.2 × 1000 + 0.2 × 1000 = 600, the single bet is associated with a significantly higher risk of a net loss.

14

Risk and Uncertainty Reduction by Using Algebraic Inequalities

Furthermore, the definition of risk based on the expected potential loss does not reflect the uncertainty associated with the consequences of the adverse event. This definition does not reflect the fact that for two events with a similar expected loss, the probability that the loss will exceed a particular critical point may differ significantly. Given that an accident/failure has occurred, for each identified set of initiating events, an assessment of the possible damage is made. In case of loss of containment, for example, depending on the release rate and the dispersion rate, the consequences can vary significantly. In case of a leak of toxic chemicals to the environment, the consequences are a function of the magnitude of the leak and the dispersion rate. Where possible, a distribution of the conditional losses (consequences given failure) should be produced. This distribution gives the likelihood that the consequences given failure will exceed a specified critical threshold. In case of n mutually exclusive failure scenarios, the conditional cumulative distribution C( x | f ) of the loss given failure/accident is described by the equation C( x | f ) = p1| f C1 ( x | f ) + p2| f C2 ( x | f ) + ... + pn| f Cn ( x | f )



(1.3)

where pi| f is the conditional probability that the i-th failure scenario will occur first i i | f = 1 and Ci ( x | f ) is the conditional cumulative distribution of the loss associated with the ith failure scenario. To reflect the likelihood of exceeding a critical value of the loss, the concept potential loss from an adverse event should be used. The cumulative distribution function C( x ) of the potential loss gives the probability that the potential loss X will not be greater than a specified value x. A loss is present only if an adverse event (failure) is present. Consequently, the unconditional probability C( x ) ≡ P( X ≤ x ) that the potential loss X will not be greater than a specified value x is equal to the sum of the probabilities of two mutually exclusive events: (i) failure will not occur and the loss will be not be greater than x and (ii) failure will occur and the loss will not be greater than x. The probability of the first compound event is (1− p f ) × H ( x ), where p f is the probability of the adverse event (failure) and H ( x ) is the conditional probability that the loss will not be greater than x given that no adverse event has occurred. This conditional probability can be presented by the Heaviside unit step function (Abramowitz and Stegun, 1964) H ( x ) = { 10,,xx≥ x a ), Figure 1.5b. The maximum potential loss xα at a specified level α is a risk measure that specifies the loss limit the probability of exceeding which is not greater than α . Probability is not the only measure of the likelihood of an adverse event. Very often, it is impossible to attach a number to the likelihood of an adverse event despite that the event may be unanimously perceived by large number of experts as ‘highly unlikely’, ‘unlikely’, ‘likely’, ‘very likely’, etc. Thus, for random demands for a unique piece of equipment, arriving randomly on a finite time interval, the likelihood of the adverse event ‘clustering of two or more random demands for the piece of equipment’ can be assessed by both: the probability of clustering of random demands and the expected time fraction of unsatisfied demand. There  are several fundamental ways of reducing risk. Risk can be reduced either by reducing the consequences (the loss) given failure or by reducing the likelihood of failure or by reducing both. Depending on how the risk is reduced, the risk reducing methods can be broadly divided into three major categories: preventive risk reducing methods, reducing the likelihood of failure, protective risk-reducing methods, reducing the loss given failure, or dual methods, reducing both (Todinov, 2007). As a rule, preventive risk-reducing methods are usually associated with smaller investment costs and maximise the profit for the business enterprise. Thus, between keeping and maintaining emergency resources which recover quickly the system from failure (protective measure) and redesigning the components to eliminate the possibility of failure (preventive measure), the second strategy should be preferred because it minimises the costs for the organisation and maximises the profit. This has led some reliability practitioners to label preventive risk-reduction methods as ‘proactive’ and

16

Risk and Uncertainty Reduction by Using Algebraic Inequalities

the protective risk reduction measures as ‘reactive’ and proclaim that preventive riskreduction methods should always be preferred to protective risk-reduction methods. This  erroneous view can be easily exposed by the method of reducing risk by deliberate weaknesses, which is a pure protective risk-reduction method. The essence of this risk reduction method is to deflect a potential failure into an inexpensive component, thereby protecting expensive equipment (Todinov, 2020). In many instances, a method based on introducing deliberate weaknesses works better than a costly preventive method which involves improving reliability of expensive components. The  common circuit breakers and fuses protecting electrical equipment is a good example. In other cases, there is little control over the hazards or a great deal uncertainty about the failure modes, hazards, working environment and environmental stresses. In these cases, reducing the likelihood of failure is difficult and the protective measures are vital to reducing the risk. They are designed to improve the resilience of the system to adverse events and reduce the cost of failure. In cases where the likelihood of failure is difficult to estimate, the principle of the risk-based design should be applied (Todinov, 2006b). This principle states that the reliability of a system/component should be proportional to the cost of its failure. Probability is the mathematical language of uncertainty, and the basic axioms and rules for manipulating probabilities are well understood and accepted (Winkler, 1996). Probability can be defined using a simple set of axioms known as the Kolmogorov’s axioms (Kolmogorov, 1933). The first axiom states that the probability P( A) of an event A is a number between zero and one (0 ≤ P( A) ≤ 1) . According to the first axiom, the probability of events is measured on a scale from 0 to 1, with ‘0’ being impossibility and ‘1’ being certainty. The second axiom states that the probability of the certain event is equal to one ( P(Ω) = 1) and the third axiom states that the probability of a union of mutually exclusive events equals the sum of the probabilities of the individual events:

A1, A2,… (Ai ∩ Aj = ∅, when i ≠ j )



P( A1 ∪ A2 ∪ ...) = P( A1 ) + P( A2 ) + ....

Uncertainty permeates all aspects of life. A proper understanding of uncertainty is vital to making correct decisions and conclusions in the absence of knowledge or under insufficient knowledge. There exist various approaches to describing uncertainty: theory of probability and mathematical statistics (DeGroot, 1989; Miller and Miller, 1999; Lee, 2012), imprecise probability (Walley, 1991), evidence theory (Shafer, 1976), possibility theory (Dubois and Prade, 2002) and fuzzy set theory (Ross, 1995). The most common approach to describing uncertainty is the theory of probability. In this respect, Bayesian inference has been widely used for reducing the uncertainty in reliability parameters (Ang and Tang, 2007). One of the major benefits of algebraic inequalities is their inherent capability to handle uncertainty. Two dimensions of uncertainty are usually distinguished in the risk literature: aleatory uncertainty and epistemic uncertainty (Hoffman and Hammonds, 1994; Oberkampf et al., 2004).

Fundamental Concepts Related to Risk and Uncertainty Reduction

17

The aleatory uncertainty is thought to be caused by the natural variation of the outcomes from the same experiment conducted under the same conditions. This type of uncertainty is thought to be caused by the intrinsic stochastic behaviour of the experiment, because of which, the outcomes of the experiments cannot be predicted. As a result, it is thought that the aleatory uncertainty (variation) cannot be reduced by gathering more knowledge about the phenomenon. This is why aleatory uncertainty is often referred to as inherent uncertainty, variability or irreducible uncertainty. A value of a mechanical property, for example (yield strength), is associated with aleatory uncertainty. Even conducted at the same conditions, the outcomes of the mechanical tests are subjected to a natural variation which cannot be reduced. The aleatory uncertainty is often modelled by a probability distribution or interpreted as relative frequency. The relative frequency interpretation of probability is the proportion of successful trials where a particular event occurs during a large number of trials. Thus, uncertainty of 0.7 for a positive experimental outcome indicates that 70% of the experiments had a positive outcome. The classical interpretation of probability is in terms of equally likely outcomes and consists of the ratio of the number of equally likely outcomes n A leading to event A and the total number of outcomes nT :

P( A) =

nA nT

Thus, in the classical sense, the probability of obtaining the sum of the points equal to 5 from rolling two unbiased dice (event A) is P = ( A) 4= / 36 1 / 9 because n A = 4 and nT = 36. The epistemic uncertainty is thought to be caused by missing knowledge about conditions, environment, probability distribution of a random variable, and so on. Epistemic uncertainty is not an inherent property of the system or process; it emerges from lack of knowledge or incomplete information about the system. An example of epistemic uncertainty is the lack of knowledge about a fixed but unknown characteristics of a system, for example the reliabilities of its components or the fractions of high-reliability components in a batch. Another example of epistemic uncertainty is the uncertainty about the model describing the behaviour of the system or process. Unlike the aleatory uncertainty, the epistemic uncertainty can be reduced by gathering more knowledge about the outcomes and their causes. Thus, the outcome of a game with existing winning strategy is initially associated with large epistemic uncertainty. With searching for more knowledge about the game and the winning strategy, the epistemic uncertainty can be eliminated completely to the level of predicting the outcome of the game with certainty. The  epistemic uncertainty correlates with the degree of belief, which increases with gathering more confirmatory evidence and is naturally measured with degree (level) of confidence. An interesting feature of epistemic uncertainty noted by Fox and Ülkümen (2011) is that as the knowledge about the possible outcome increases, the judged probability moves towards the extremes (towards 1 or 0). In the absence of any knowledge, uniform probabilities are assigned to potential outcomes. Thus, in the absence of

18

Risk and Uncertainty Reduction by Using Algebraic Inequalities

any knowledge about n potential outcomes, uniform (equal) probabilities of 1 / n are assigned to each outcome. Probability is also the common measure for epistemic uncertainties. According to Aven (2011), subjective probabilities used by experts in their assessments provide a consistent basis in the assessment and reporting of epistemic uncertainty. The probability is subjective in the sense that it is the probability of a particular expert and the probability varies among different experts. Despite these variations, these subjective uncertainty assessments are still considered valuable because they are based on strong knowledge and assumptions made by experienced experts in a specific domain. The  subjective probabilities are to be understood as knowledge-based probabilities expressing the degree of belief about an outcome of a random event and can be manipulated according to the usual mathematical rules of probability (Winkler, 1996). The two types of uncertainty are not mutually exclusive and can coexist. Often, no sharp distinction can be made between epistemic uncertainty and aleatory uncertainty. What often appears to be irreducible aleatory uncertainty can in fact be reduced by acquiring more knowledge about the system and its environment. For  example, the probability of landing of a projectile within a certain distance from a target appears to be measuring irreducible aleatory uncertainty. With precise research about the temperature and density of the air, the air resistance properties of the projectile, the initial velocity and its direction, the landing location of the projectile can be predicted with great accuracy. The landing location can be predicted by solving a differential equation of motion, to a very high degree of precision. Despite the blurred line between aleatory and epistemic uncertainty, the distinction between aleatory and epistemic uncertainty sometimes is very sharp and cannot be ignored. Here is an example. The  absence of knowledge about the relative fraction p (0 ≤ p ≤ 1) of reliable components in a very large batch of components is epistemic uncertainty. The probability that two randomly selected components from the batch will be reliable is p2, and this probability is associated with both epistemic uncertainty and aleatory uncertainty because both types of uncertainty are present. The epistemic uncertainty can be reduced by gathering more information about the batch. For example, by gathering more information it has been established that the relative share of the reliable components is p = 0.8 . Although the epistemic uncertainty has been eliminated, the aleatory uncertainty about whether both purchased components will be reliable still remains equal to= p2 0= .82 0.64 . The aleatory uncertainty in this case cannot be reduced by gathering more knowledge about the batch. No matter how much more information is gathered about the batch, the probability that two randomly selected will be reliable still remains 0.64. Uncertainty varies between complete certainty to the deepest level of uncertainty where the possible outcomes are not  known. This  type of uncertainty has been referred in Taleb (2007) as ‘black swan’ type of uncertainty. Between these extremes lie different levels of uncertainty. Thus, a relatively low level of uncertainty is present in cases where the possible outcomes and their associated probability distributions are known, and the parameters of the system and their

Fundamental Concepts Related to Risk and Uncertainty Reduction

19

associated distributions are known. This type of uncertainty can be easily resolved by using sensitivity analysis. Generating instances of the system parameters by sampling their distributions and combining them generates the system output with the associated uncertainty bounds. A higher level of uncertainty is present in cases where the possible outcomes are known, their probabilities are not known, but they can be ranked in terms of likelihood. Still a higher level of uncertainty is present where only the range of possible outcomes is known but information about vital structural parameters such as probability distributions, proportions, and so on is missing and it is not clear how the possible outcomes can be ranked in terms of likelihood. This  type of uncertainty will be referred to as ‘deep uncertainty’.

2

Properties of Algebraic Inequalities Standard Algebraic Inequalities

2.1  BASIC RULES RELATED TO ALGEBRAIC INEQUALITIES Inequalities are statements about expressions or numbers which involve the symbols ‘ ’ (greater than), ‘ ≤ ’ (less than or equal to) or ‘ ≥’ (greater than or equal to). The  basic inequalities introduced in this chapter are later used for proving inequalities related to reliability, risk and uncertainty. The basic rules related to handling algebraic inequalities can be summarised as follows: a. For any real numbers a and b, exactly one of the following holds:

a < b, a = b, a > b



b. If a > b and b > c then a > c c. If a > b , adding the same number c to both sides of the inequality does not alter its direction: a + c > b + c d. If a > b then −a < −b e. If a > 0 and b > 0 then ab > 0



By using these rules, a number of basic properties can be established.

2.2  BASIC PROPERTIES OF INEQUALITIES i. For any real number x, x 2 ≥ 0 . The equality holds if and only if x = 0. If x = 0 then x 2 = 0. If x > 0 applying rule (e) where a = x and b = x gives xx = x 2 > 0. If x < 0 then − x > 0 and applying rule (e) where a = − x and b = − x gives (− x ) × (− x ) = x 2 > 0. ii. If x > y , t > 0 then xt > yt and x /t > y /t. From rule (c) and x > y , it follows x − y > y − y = 0. From rule (e), it follows ( x − y )t > 0 or xt − yt > 0. Applying rule (c) where a = xt and c = yt results in xt − yt + yt > yt or xt > yt. The  property x /t > y /t follows if the setting c = 1/t > 0 is made. Then, according to what has been proved, xc > yc or x /t > y/t . From this property, it follows that if 0 < x < 1 then x 2 < x .

21

22











Risk and Uncertainty Reduction by Using Algebraic Inequalities

From this property and rule (d), it follows that if x > y , t < 0 then xt < yt and x / t < y / t . iii. If x > y > 0, u > v > 0 then xu > yv and x /v > y / u. From property (ii), it follows xu > yu and yu > yv . Applying basic rule (b), where = a xu = , b yu, c = yv, gives xu > yv . Since xu > yv, according to property (ii), dividing both sides of the inequality by the positive value uv > 0 will not  alter the direction of the inequality, therefore x /v > y / u holds. From this equality it follows that if a > b > 0 then a2 > b2. iv. If x > 0, y > 0, x ≠ y and x 2 > y2 then x > y . Suppose that 0 < x < y is true. According to property (iii), x 2 / x > y 2/ y which implies x > y . But this contradicts the assumption 0 < x < y. Therefore, x > y must be true. v. If a strictly increasing function is applied to both sides of an inequality, the inequality will still hold. Applying a strictly decreasing function to both sides of an inequality reverses the direction of the inequality. Thus, from x > y > 0, it follows that ln x > ln y, x n > yn where n > 0. From x > y > 0, it follows that, x − n < y − n where n > 0. Property (iv) follows from this property because u is a strictly increasing function which, applied to both sides of x 2 > y2, gives | x | > | y | and since x > 0, y > 0, the inequality x > y is obtained. vi. If a > b > 0 and p and q are positive real numbers, then a p / q > b p / q. Because the logarithm ln( x ) is a strictly increasing function (if x2 > x1 then ln( x2 ) > ln( x1 )), a > b > 0 and p / q is a positive real number, it follows that ln a > ln b

(2.1)

multiplying both sides of (2.1) by the positive value p/q gives ( p / q) ln a > ( p / q) ln b, which is equivalent to ln a p /q > ln b p /q. Using again the fact that ln ( x ) is a strictly increasing function gives the original inequality a p / q > b p / q. These rules and properties are sufficient to prove a number of useful inequalities. The first example involves proving the inequality

a2 + b2 ≥ 2ab

(2.2)

where a, b are arbitrary real numbers. Inequalities similar to inequality (2.2), where equality is attained for particular values of the variables, will be referred to as sharp (tight) inequalities. According to rule (c), proving this inequality is equivalent to proving the inequality

a2 − 2ab + b2 ≥ 0

(2.3)

23

Properties of Algebraic Inequalities

This is because inequality (2.2) can be obtained from inequality (2.3) by adding the quantity 2ab to both sides of the inequality. The left side of inequality (2.3) is a square: a2 − 2ab + b2 = (a − b)2 which, according to property (i), is non-negative. Equality is attained for a = b . The next example involves proving the inequality

a b + ≥ 2 b a

(2.4)

where a and b are both positive numbers. Proving this inequality is equivalent to proving the inequality ba + ba − 2 ≥ 0 or 2 2 the inequality a +bab−2 ab ≥ 0. The  left-hand side of the last inequality can be pre2 2 2 sented as a +bab−2 ab = ( a−abb ) . The  numerator is a square hence nonnegative; the 2 denominator is positive because a > 0 and b > 0. Consequently, ( a−abb ) ≥ 0 is true. Starting with the last inequality and conducting the operations in reverse yields ( a − b )2 a 2 −2 ab + b2 = ba + ba − 2 ≥ 0, which proves the original inequality (2.4). Equality = ab ab is attained for a = b . The third example involves proving the inequality

x ≥ 2 x 1− x

(2.5)

where 0 < x < 1. Because 1−xx ≥ 0 and 2 x ≥ 0 , following property (iv), to prove inequality (2.5) 2 is sufficient to prove that  x  ≥ (2 x )2, which is equivalent to proving 1−xx ≥ 4 x 2.  1− x  Proving the last inequality is equivalent to proving x  1−1x − 4 x  ≥ 0 or

 4x2 − 4x + 1  x  ≥ 0 1− x  

(2.6) 

2





2



The left side of inequality (2.6) can be presented as x  4 x 1−−4xx +1  = x  ( 21x−−x1) . Inequality 2 x  ( 21x−−x1)  ≥ 0 is true because x > 0; 1 − x > 0 and (2 x − 1)2 ≥ 0 . Starting with the last inequality and reversing the operations proves the original inequality (2.5). Equality is attained for x = 0.5.

2.3  ONE-DIMENSIONAL TRIANGLE INEQUALITY For any two real numbers a and b, the inequality

| a + b | ≤ | a | + | b |

(2.7)

holds, where | x | denotes the absolute value of x. The equality holds when both a and b have the same sign. Inequality (2.7) can be proved by noting that x 2 = | x |2. Accordingly, | a + b |2 = ( a + b)2 = a2 + 2ab + b2 = | a |2 + 2ab + | b |2 Because 2ab ≤ 2 | a || b | , the next inequality holds

| a |2 + 2ab + | b |2 ≤ | a |2 + 2 | a || b | + | b |2 = (| a | + | b |)2

24

Risk and Uncertainty Reduction by Using Algebraic Inequalities

Therefore, | a + b |2 ≤ (| a | + | b |)2



Since | a + b | ≥ 0 and (| a | + | b |) ≥ 0, according to property (iv): | a + b | ≤ | a | + | b |



From the one-dimensional triangle inequality, the following inequality can be deduced || a | − | b || ≤ | a − b |



(2.8)

Indeed, since | a | = | a − b + b | and | b | = | b − a + a |, applying the one-dimensional triangle inequality results in

| a | = | a − b + b | ≤ | a − b | + | b |

(2.9)



| b | = | b − a + a | ≤ | b − a | + | a |

(2.10)

From inequalities (2.9) and (2.10) it follows that

| a | − | b | ≤ | a − b |



| b | − | a | ≤ | a − b |

or || a | − | b || ≤ | a − b | is true. The one-dimensional triangle inequality for absolute values can be generalised for any set of real numbers a1, a2 ,..., an:

| a1 + a2 + ... + an | ≤ | a1 | + | a2 | + ... + | an |

A generalisation related to higher dimensions is presented in Section 2.15.

2.4  THE QUADRATIC INEQUALITY Under certain necessary conditions, the quadratic inequality states that

ax 2 + bx + c > 0

(2.11)

ax 2 + bx + c < 0

(2.12)

or for any real x, where a ≠ 0 .

Properties of Algebraic Inequalities

25

To derive the necessary conditions for the quadratic inequalities (2.11) or (2.12), the left-hand side of the inequalities will be presented as

2  b   b2 − 4ac   − ax 2 + bx + c = a   x +    2a   4a2   

(2.13)

by completing the square. From equation  (2.13), if the discriminant D = b2 − 4ac > 0, the equation 2 ax + bx + c = 0 has two different solutions, hence there will be values x for which ax 2 + bx + c > 0 and there will be values x for which ax 2 + bx + c < 0. If the discriminant D = b2 − 4ac < 0, the equation ax 2 + bx + c = 0 has no solutions 2  2  because both terms in the brackets,  x + 2ba  and −  b −42ac  , are positive and the sign of    4a  the trinomial ax 2 + bx + c = 0 is solely determined by the sign of the coefficient a. As a result, inequality (2.11) holds for any x if and only if D = b2 − 4ac < 0 and a > 0. The minimum value of the quadratic trinomial f ( x ) ≡ ax 2 + bx + c is attained at x = −b / (2a) and is positive. Inequality (2.12) holds if and only if D = b2 − 4ac < 0 and a < 0. The maximum value of the quadratic trinomial f ( x ) ≡ ax 2 + bx + c is attained at x = −b /(2a) and is negative. The properties of the discriminant of the quadratic trinomial will later be used to prove the Cauchy-Schwarz inequality in Section 2.9. Consider the quadratic trinomial f ( x ) ≡ ax 2 + bx + c, where x ∈[u1, u2 ]. If a > 0, the maximum of the quadratic trinomial is attained either at x = u1 or x = u2 . The maximum is found by comparing f (u1 ) and f (u2 ), whichever is greater. If a < 0, the minimum of the quadratic trinomial is attained either at x = u1 or x = u2 . The minimum is found by comparing f (u1 ) and f (u2 ), whichever is smaller. This provides a useful way of proving inequalities and determining tight upper and lower bounds. Thus, the tight lower bound of the expression

y = −3 x 2 + 2 x + 1

(2.14)

in the interval [−1, 2] is obtained by comparing the values of the quadratic trinomial at x = −1 and x = 2. Since y(−1) = −4 and y(2) = −7, the tight lower bound is attained at x = 2. For any real x in the interval −1 ≤ x ≤ 2 , the sharp inequality

−3 x 2 + 2 x + 1 ≥ −7

(2.15)

holds.

2.5  JENSEN INEQUALITY A function f ( x ) with a domain [a,b] is said to be convex if for all values x1 and x2 in its domain ( x1, x2 ∈ [ a, b]), the next inequality holds:

26



Risk and Uncertainty Reduction by Using Algebraic Inequalities

f (wx1 + (1 − w ) x2 ) ≤ wf ( x1 ) + (1 − w ) f ( x2 )

(2.16)

where 0 ≤ w ≤ 1. If the inequality

f (wx1 + (1 − w ) x2 ) < wf ( x1 ) + (1 − w ) f ( x2 )

(2.17)

holds, where 0 < w < 1, the function f ( x ) is said to be strictly convex. If f ( x ) is twice differentiable for all x ∈ (a, b), and if the second derivative is non-negative ( f "( x ) ≥ 0), the function f ( x ) is convex on (a, b). If f ( x ) is twice differentiable for all x ∈ (a, b), and if the second derivative is positive ( f ''( x ) > 0), the function f ( x ) is strictly convex on (a, b). For any convex function f ( x ), the Jensen’s inequality states that

f (w1 x1 + w2 x2 + ... + wn xn ) ≤ w1 f ( x1 ) + w2 f ( x2 ) + ... + wn f ( xn )

(2.18)

where wi (i = 1,..., n) are numbers (weights) that satisfy 0 ≤ wi ≤ 1 and w1 + w2 + ... + wn =1. If the function f ( x ) is strictly convex (second derivative is positive), equality in (2.18) is attained only for x= x= ... = xn . 1 2 A function f ( x ) with a domain [a,b] is said to be concave, if for all values x1 and x2 in its domain ( x1, x2 ∈ [ a, b]), the next inequality holds

f (wx1 + (1 − w ) x2 ) ≥ wf ( x1 ) + (1 − w ) f ( x2 )

(2.19)

where 0 ≤ w ≤ 1. For any concave function f ( x ), the Jensen inequality states that

f (w1 x1 + w2 x2 + ... + wn xn ) ≥ w1 f ( x1 ) + w2 f ( x2 ) + ... + wn f ( xn )

(2.20)

where wi (i = 1,..., n) are numbers (weights) that satisfy 0 ≤ wi ≤1 and w1 + w2 + ... + wn =1. If f ( x ) is twice differentiable for all x ∈ (a, b), and if the second derivative is not positive ( f ''( x ) ≤ 0 ) , the function f ( x ) is concave on (a, b). If f ( x ) is twice differentiable for all x ∈ (a, b), and if the second derivative is negative ( f ''( x ) < 0 ), the function f ( x ) is strictly concave on (a, b). If the function f ( x ) is strictly concave (second derivative is negative), equality in (2.20) is attained only for x= x= ... = xn . 1 2

2.6 ROOT-MEAN SQUARE–ARITHMETIC MEAN–GEOMETRIC MEAN–HARMONIC MEAN (RMS-AM-GM-HM) INEQUALITY For  a set of positive real numbers x1, x2 ,..., xn, the RMS-AM-GM-HM inequality states

x12 + x22 + ... + xn2 x1 + x2 + ... + xn n n ≥ ≥ x1 x2 ... xn ≥ 1 1 1 n n + + ... + x1 x2 xn

with equality attained only if x= x= ... = xn . 1 2

(2.21)

27

Properties of Algebraic Inequalities 2

2

2

The number x1 + x2 n+...+ xn is known as root-mean square; x1 + x2 n+...+ xn is the arithmetic n mean; n x1x2 ...xn is the geometric mean and 1 1 1 is the harmonic mean. + +...+ x1 x2 xn

For two non-negative numbers a, b, the inequality takes the form

a2 + b2 a + b 2 ≥ ≥ ab ≥ 1 1 2 2 + a b

(2.22)

Different techniques for proving the RMS-AM-GM-HM inequality will be demonstrated in the next chapter.

2.7 WEIGHTED ARITHMETIC MEAN–GEOMETRIC MEAN (AM-GM) INEQUALITY The weighted arithmetic mean–geometric mean (AM-GM) inequality is an important application of the Jensen inequality. For  a set of positive real numbers x1, x2 ,..., xn, and a set t1, t2 ,..., tn of positive weights t1 + t2 +  + tn = 1, the AM-GM inequality can be generalised with the weighted AM-GM inequality

t1 x1 + t2 x2 + ... + tn xn ≥ x1t1 x2t2 ... xntn

(2.23)

Introduce the values ai , i = 1,..., n as follows: a1 = ln x1,...,an = ln xn. Because e x is a convex function, from the Jensen inequality, the next inequality is obtained

e t1a1 +...+ tn an ≤ t1e a1 + ... + tne an

Since e t1a1 +...+ tn an = x1t1 x2t2 ... xntn and t1e a1 + ... + tne an = t1 x1 + ... + tn xn, the weighted AM-­ GM inequality (2.23) follows directly. An important corollary of the weighted AM-GM inequality is the Young inequality. If x, y are positive numbers and a, b are positive numbers such that 1/a + 1/ b = 1, the Young inequality states that

xy ≤

1 a 1 b x + y b a

(2.24)

The Young inequality follows directly from the weighted AM-GM inequality by noticing that xy = ( x a )1/ a × ( y b )1/ b . According to the weighted AM-GM inequality (2.23)

xy = ( x a )1/ a × ( y b )1/ b ≤

1 a 1 b x + x a b

28

Risk and Uncertainty Reduction by Using Algebraic Inequalities

2.8  HÖLDER INEQUALITY Let a11, a12 ,..., a1n ;a21, a22 ,..., a2 n ;...;am1, am 2 ,..., amn be m sequences of positive real numbers and λ1, λ 2,..., λ m be m positive real numbers adding up to 1 ∑im=1 λi = 1 . Then the Hölder inequality holds (Hardy et al, 1999):

(



( a11 + a12 ... + a1n )λ1 ( a21 + ... + a2 n )λ 2 ... ( am1 + am2 + ... + amn )λ m λ2 λm ≥ a1λ11a21 ...amλ1m + ... + a1λn1a2λn2 ...amn



)

(2.25)

Inequality (2.25) is also known as the elementary form of the Hölder inequality. For two sequences x1p , x2p ,..., xnp and y1q , y2q ,..., ynq of positive real numbers (m = 2, p > 0, q > 0, ) and weights λ1 = 1/ p; λ2 = 1/q, ( λ1 + λ2 = 1), the Hölder inequality yields

   

n

∑ i =1

1/ p

 x    p i

 ×  

n

∑ i =1

1/ q

 y    q i

n

≥

∑x y

(2.26)

i i

i =1

Inequality (2.26) is the Hölder inequality for sums which can be proved by using the Young inequality. Many inequalities can be proved by a direct application of the Hölder inequality. If a, b, c, x, y and z are positive real numbers, by a direct application of the Hölder inequality, for example, it can be shown that the following inequality

a3 b3 c3 (a + b + c)3 + + ≥ x y z 3( x + y + z)

holds. Indeed, note that the inequality

 a3 b3 c 3  (1 + 1 + 1) × ( x + y + z) ×  + +  ≥ (a + b + c)3 y z   x

is equivalent to the original inequality because the original inequality is obtained by dividing both sides by 3( x + y + z ). Taking λ1 = λ2 = λ3 = 1/3 ( λ1 + λ2 + λ3 = 1) and applying the Hölder inequality to the three sequences (1, 1, 1), ( x, y, z) and ( a3/ x, b3/ y , c3/ z ) gives: (1 + 1 + 1)1/ 3 ( x + y + z )1/ 3 (a3 / x + b3 / y + c3 /z )1/ 3 ≥

((1 ⋅ x ⋅ a /x ) 3

1/ 3

+ (1 ⋅ y ⋅ b3 / y)1/ 3 + (1 ⋅ z ⋅ c3 /z)1/ 3

Raising both sides of this inequality to the power of 3 gives

3( x + y + z )( a3/ x +b3/ y + c3/ z ) ≥ ( a + b + c)3

which completes the proof of the original inequality.

)

29

Properties of Algebraic Inequalities

2.9  CAUCHY-SCHWARZ INEQUALITY One of the most important algebraic inequalities is the Cauchy-Schwarz inequality (Cauchy, 1821), which states that for the sequences of real numbers a1, a2 ,..., an and b1, b2 ,..., bn , the following tight inequality holds:

(a1b1 + a2 b2 + ... + an bn )2 ≤ (a12 + a22 + ... + an2 )(b12 + b22 + ... + bn2 )

(2.27)

Equality holds if and only if for any i ≠ j , ai b j = a j bi are fulfilled. The Cauchy-Schwarz inequality is a powerful inequality. Many algebraic inequalities can be proved by reducing them to the Cauchy-Schwarz inequality by using an appropriate substitution. It has also a simple geometrical interpretation. The Cauchy-Schwartz inequality (2.27) is equivalent to the inequality

a1b1 + a2 b2 + ... + an bn a + a + ... + an2 × b12 + b22 + ... + bn2 2 1

2 2

≤ 1

(2.28)

If the sequences of real numbers a1, a2 ,..., an and b1, b2 ,..., bn are interpreted as the components of two vectors in the n-dimensional space, the numerator on the left side of (2.28) is the dot product of the two vectors and the denominator is the product of the magnitudes of the vectors. Consequently, the expression on the left side of inequality (2.28) is the cosine of the angle between these two vectors. In the threedimensional space (Figure  2.1) the cosine of the angle between two vectors with components a1, a2 , a3 and b1, b2 , b3 is given by

a1b1 + a2 b2 + a3b3 a + a22 + a32 × b12 + b22 + b32 2 1

= cos( γ )

and the Cauchy-Schwarz inequality expresses the fact that the cosine of an angle is a number which cannot exceed 1.

FIGURE 2.1  Geometric interpretation of the Cauchy-Schwartz inequality.

30

Risk and Uncertainty Reduction by Using Algebraic Inequalities

The Cauchy-Schwarz inequality is a special case of the Hölder inequality (2.26) for p= q= 2. An alternative way to prove the Cauchy-Schwarz inequality (2.27) is to use the properties of the quadratic trinomial introduced in Section  2.4. Consider the expression y = (a1t + b1 )2 + (a2t + b2 )2 + ... + (an t + bn )2



(2.29)

For  any sequences of real numbers a1, a2 ,..., an and b1, b2 ,..., bn , y is non-negative

the right-hand side of (2.29) and collecting the coefficients in ( y ≥ 0 ). Expanding 2 front of t and t, gives the quadratic trinomial

y = (a12 + a22 + ... + an2 )t 2 + 2(a1b1 + a2 b2 + ... + an bn )t + ... + b12 + b22 + ... + bn2

(2.30)

with respect to t. In equation (2.30), y is non-negative ( y ≥ 0 ) only if D ≤ 0 , where

D = (a1b1 + a2 b2 + ... + an bn )2 − (a12 + a22 + ... + an2 )(b12 + b22 + ... + bn2 )

is the discriminant of the quadratic trinomial. Therefore, the condition

(a1b1 + a2 b2 + ... + an bn )2 − (a12 + a22 + ... + an2 )(b12 + b22 + ... + bn2 ) ≤ 0

(2.31)

must hold for a non-negative y. The  condition (2.31) is identical to the CauchySchwarz inequality (2.27) which completes the proof. The Cauchy-Schwarz inequality will be illustrated by proving that ( a + b + c) (1/a + 1/b + 1/c ) ≥ 9



where a, b, c are positive real numbers. Proof. Consider the two sequences Cauchy-Schwarz inequality:



1 1  + b⋅ +  a⋅ a b  ≤  

c⋅

1   c 

a , b , c and 1/ a , 1/ b , 1/ c. From the 2



( a ) + ( b ) + ( c )  ×  (1 / a ) + (1 / b ) + (1 / c )  2

2

2

2

2

2

from which the inequality follows directly.

2.10  REARRANGEMENT INEQUALITY The  rearrangement inequality is a powerful yet underused inequality that can be applied to provide bounds for the uncertainty associated with reliability-critical parameters. Consider the two sequences a1, a2 ,..., an and b1, b2 ,..., bn of real numbers. It can be shown that:

31

Properties of Algebraic Inequalities





a. The  sum S = a1b1 + a2 b2 + ... + an bn is maximal if the sequences are sorted in the same way – both monotonically decreasing: a1 ≥ a2 ≥,..., ≥ an ; b1 ≥ b2 ≥,..., ≥ bn or both monotonically increasing: a1 ≤ a2 ≤,..., ≤ an ; b1 ≤ b2 ≤,..., ≤ bn. b. The  sum S = a1b1 + a2 b2 + ... + an bn is minimal if the sequences are sorted in the opposite way: one monotonically increasing and the other monotonically decreasing.

To prove the first statement, the extreme principle will be used. Suppose that there is a sum

S0 = a1b1 + ... + ar br + ... + as bs + ... + an bn

(2.32)

where the a-sequence and b-sequence are not both monotonically increasing or both monotonically decreasing and which is the largest possible sum. If the a-sequence and b-sequence are not  both sorted in ascending order or in a descending order, there will certainly be values ar, br and as , bs ( r < s ) for which either ar < as and br > bs is true or ar > as and br < bs is true. If no such pairs can be found, then the a-sequence and b-sequence are already either both increasing or both decreasing and the hypothesis that the a-sequence and b-sequence are not both sorted in ascending or in descending order does not hold. Suppose that ar < as and br > bs is true. Without loss of generality, it can be assumed that in equation (2.32) the terms ai bi ( i = 1,..., n ) have been sorted in ascending order of ai . This can always be done by a simple permutation of the terms. Now, consider the sum S1

S1 = a1b1 + ... + ar bs + ... + as br + ... + an bn

(2.33)

which has been obtained from the sum S0 by switching the positions of br and bs only. Subtracting S1 from S0 gives:

S0 − S1 = ar br + as bs − ar bs − as br = ar (br − bs ) − as (br − bs ) = (ar − as )(br − bs )

Because ar < as and br > bs is true, then S0 − S1 = (ar − as )(br − bs ) < 0 ; therefore the sum S1 is larger, which contradicts the assumption that S0 is the largest sum. Consequently, the hypothesis that the largest sum can be attained for sequences that are not both increasing or both decreasing is incorrect. The case ar > as and br < bs leads to a contradiction in a similar manner. In a similar fashion, statement (b) can also be proved. An important application of the rearrangement inequality is its use as a basis for generating new useful inequalities that can be used to produce bounds for the uncertainty in reliability-critical parameters. For two sequences a1, a2 ,..., an and b1, b2 ,..., bn of real numbers, the notation

 a1   b1

a2 b2

... ...

an   = a1b1 + a2 b2 + ... + an bn bn 

32

Risk and Uncertainty Reduction by Using Algebraic Inequalities

can be introduced (Engel, 1998). This is similar to the definition of a dot product of two vectors with components specified by the two rows of the matrix. Two important corollaries of the rearrangement inequality are the following: i. Given a set of real numbers a1, a2 ,..., an, for any permutation a1s , a2 s ,..., ans , the following inequality holds: a12 + a22 + ... + an2 ≥ a1a1s + a1a2 s + ... + ans



(2.34)

Proof. Without loss of generality, it can be assumed that a1 ≤ a2 ≤,..., ≤ an . Applying the rearrangement inequality to the pair of sequences ( a1, a2 ,..., an ), ( a1, a2 ,..., an ) and the pair of sequences ( a1, a2 ,..., an ), ( a1s , a2 s ,..., ans ) then yields



 a1   a1

a2 a2

an  a1 2 2 2  = a1 + a2 + ... + an ≥  an   a1s

... ...

a2 a2 s

... ...

an   ans 

= a1a1s + a2 a2 s + ... + an ans because the first pair of sequences are similarly ordered and the second pair of sequences are not. This completes the proof of inequality (2.34). ii. Given a set of positive real numbers a1, a2 ,..., an different from zero, for any permutation a1s , a2 s ,..., ans the following inequality holds: a1s a2 s a + + ... + ns ≥ n a1 a2 an



(2.35)

Proof. Without loss of generality, it can be assumed that a1 ≤ a2 ≤,..., ≤ an. The reciprocals can then be ranked as follows: 1/a1 ≥ 1 / a2 ≥,..., ≥ 1/an. Applying the rearrangement inequality to the pair of sequences (a1, a2 ,..., an ), (1/a1,1/a2 ,...,1/an ) and the pair of sequences (a1s , a2 s ,..., ans ), (1/a1,1/a2 ,...,1/an ) then yields  a1 1/a  1

a2 1/a2

 a1s = n ≤ 1/a1

... ... a2 s 1/a2

an  1/an  ... ...

ans  1/an 

= a1s /a1 + a2 s /a2 + ... + ans /an because the first pair of sequences are oppositely sorted (therefore associated with the smallest sum a1 ⋅ (1/a1 ) + a2 ⋅ (1/a2 ) + ... + an ⋅ (1/an ) = n) and the second pair of sequences are not oppositely sorted and therefore, the sum a1s /a1 + a2 s /a2 + ... + ans /an is not the smallest sum, which completes the proof of inequality (2.35).

33

Properties of Algebraic Inequalities

2.11  CHEBYSHEV SUM INEQUALITY Another important algebraic inequality is the Chebyshev sum inequality which states that for similarly sorted sequences of real numbers  – both monotonically decreasing: a1 ≥ a2 ≥,..., ≥ an and b1 ≥ b2 ≥,..., ≥ bn, or both monotonically increasing: a1 ≤ a2 ≤,..., ≤ an; b1 ≤ b2 ≤,..., ≤ bn, the following sharp inequality holds:

n(a1b1 + a2 b2 + ... + an bn ) ≥ (a1 + a2 + ... + an )(b1 + b2 + ... + bn )

(2.36)

a1b1 + ... + an bn a1 + ... + an b1 + ... + bn ≥ . n n n

(2.37)

or

If the sequences are oppositely sorted (for example, a1 ≥,..., ≥ an , b1 ≤,..., ≤ bn or a1 ≤ a2 ≤,..., ≤ an, b1 ≥ b2 ≥,..., ≥ bn ) the inequality is reversed. Equality is attained if a= a= .... = an or b= b= .... = bn holds. 1 2 1 2 Despite the existence of a number of alternative proofs of the Chebyshev’s sum inequality (Besenyei, 2018), probably the simplest proof is based on the rearrangement inequality. According to the rearrangement inequality, for similarly ordered sequences, the following inequalities are true: n

∑ a b = a b + a b + a b + ... + a b



1 1

i i

2 2

3 3

n n

i =1 n

∑ a b ≥ a b + a b + a b + ... + a b



1 2

i i

2 3

3 4

n 1

i =1 n

∑ a b ≥ a b + a b + a b + ... + a b



1 3

i i

2 4

3 5

n 2

i =1



… n

∑ a b ≥ a b + a b + a b + ... + a b



1 n

i i

2 1



n n −1

3 2

i =1

By adding these inequalities, the inequality n



n

∑ a b ≥ a (b + b + ... + b ) + a (b + b + ... + b ) + ... + a (b + b + ... + b ) i i

1

1

2

n

2

1

2

n

n

1

2

n

i =1

is obtained, which, after taking out the common factor (b1 + b2 + ... + bn ), leads to the Chebyshev inequality (2.36). Chebyshev inequality provides the valuable opportunity to segment an initial complex expression into simpler expressions. The complex terms ai bi in (2.36) are segmented into simpler terms involving ai and bi. The segmentation capability of the Chebyshev inequality will be illustrated by evaluating the lower bound of ∑in=1 xi2 if x1 + x2 + ... + xn = 1.

34

Risk and Uncertainty Reduction by Using Algebraic Inequalities

Without loss of generality, it can be assumed that x1 ≤ x2 ≤ ... ≤ xn . By setting = a1 x= x2 ,..., an = xn and = b1 x= x2 ,..., bn = xn , the conditions for applying 1, a2 1, b2 the Chebyshev inequality are fulfilled and x12 + ... + xn2 x1 + ... + xn x1 + ... + xn . ≥ n n n



(2.38)

holds. Substituting x1 + ... + xn = 1 in (2.38), gives the lower bound of ∑ in=1 xi2: n

∑x



2 i

≥ 1 / n

i =1

2.12  MUIRHEAD INEQUALITY Consider the two non-increasing sequences a1 ≥ a2 ≥,..., ≥ an and b1 ≥ b2 ≥,..., ≥ bn of positive real numbers. The sequence {a} is said to majorise the sequence {b} if the following conditions are fulfilled:

a1 ≥ b1; a1 + a 2 ≥ b1 + b2; ... ;a1 + a 2 + ... + an −1 ≥ b1 + b2 + ... + bn −1; a1 + a 2 + ... + an −1 + an = b1 + b2 + ... + bn −1 + bn

(2.39)

If the sequence {a} majorises the sequence {b} and x1, x2 ,..., xn are non-negative, the Muirhead inequality

∑x

x ...xnan ≥

a1 a2 1 2



sym

∑x

b1 b2 1 2

x ...xnbn

(2.40)

sym

holds (Hardy et al., 1999). For any set of non-negative numbers x1, x2 ,..., xn, the symmetric sum ∑ sym x1a1 x2a2 ...xnan, when expanded, includes n! terms. Each term of the symmetric sum is formed by a distinct permutation of the elements of the sequence a1, a2 ,..., an. Thus, if {a} = [2, 1, 0], the symmetric sum becomes

∑x x x

2 1 0 1 2 3

= x12 x2 + x12 x3 + x22 x1 + x22 x3 + x32 x1 + x32 x2

sym

If {a} = [2, 0, 0], the symmetric sum becomes

∑x x x

2 0 0 1 2 3

= 2 x12 + 2 x22 + 2 x32

sym

Here is an application example featuring an inequality that follows directly from the Muirhead inequality. Consider a set of real, non-negative numbers x1, x2 , x3. It can be shown that the following inequality holds:

x12 + x22 + x32 ≥ x1 x2 + x1 x3 + x2 x3

(2.41)

35

Properties of Algebraic Inequalities

Consider the set of non-negative numbers x1, x2 , x3 and the sequences {a} = [2, 0, 0] and {b} = [1, 1, 0]. Clearly, the sequence {a} = [2, 0, 0] majorises the sequence {b} = [1, 1, 0] because the conditions (2.39) are fulfilled: 2 ≥ 1; 2 + 0 ≥ 1 + 1 and 2 + 0 + 0 = 1 + 1 + 0.



According to the Muirhead inequality (2.40): 2 × ( x12 + x22 + x32 ) ≥ 2( x1 x2 + x1 x3 + x2 x3 )



which implies inequality (2.41).

2.13  MARKOV INEQUALITY Markov’s inequality is related to a random variable X with mean µ = E ( X ), which accepts only non-negative values. It states that the probability that the random variable will be greater than a positive constant a does not exceed the value µ /a:

µ P ( X ≥ a) ≤ a



(2.42)

This inequality is very general. It holds for any distribution of the random variable X. Proof. For a continuous probability density distribution f ( x ) of the random variable X, the expected value is given by

E( X ) =

∫ xf ( x) dx = ∫ xf ( x) dx + ∫ xf ( x) dx x

x 0 in (2.45) gives

P (| X − µ | ≥ tσ ) ≤

σ2 1 = t 2σ 2 t 2

(2.46)

The last formulation of the Chebyshev inequality states that for any positive constant t, the probability that the random variable X will take on a value more than t standard deviations from the mean µ = E ( X ) does not exceed 1/t 2 . The Chebyshev inequality is also very general. It holds for any distribution of the random variable X. Proof. Consider the auxiliary non-negative random variable Y = [ X − µ ]2. Because the random variable Y is non-negative, the Markov inequality (2.42) can be applied and

P(Y ≥ k 2 ) ≤

E(Y ) k2

Considering that E (Y ) = E[( X − µ )2 ] = V ( X ) and P (Y ≥ k 2 ) = P (| X − µ | > k ), the Chebyshev inequality (2.45) is finally obtained. Chebyshev’s inequality also provides a useful tight upper bound of the values of a random variable for values of the constant k greater than the standard deviation σ . The application of the inequality will be illustrated by an example. Consider loss given failure with mean µ . Suppose that the extreme value of the loss which triggers bankruptcy is 5σ , where σ is the standard deviation of the loss. According to the Chebyshev inequality, the probability that the loss will be larger than the extreme value of 5σ does not exceed 0.04:

P (| X − µ | ≥ 5σ ) ≤

σ2 = 0.04 52σ 2

The probability value of 0.04 is a tight upper bound for the extreme loss, which holds irrespective of the actual distribution of the loss given failure.

2.15  MINKOWSKI INEQUALITY The general Minkowski inequality for two sequences a1, a2 ,..., an and b1, b2 ,..., bn of non-negative numbers states (Hardy et al, 1999)

37

Properties of Algebraic Inequalities



( a1p + a2p + ... + anp )1/ p + (b1p + b2p + ... + bnp )1/ p ≥ [( a1 + b1 ) p + ( a2 + b2 ) p + ... + ( an + bn ) p ]1/ p



(2.47)

for p ≥ 1. For p < 1, the inequality direction is reversed (Bechenbach and Bellman, 1961). For p = 2, the general triangle inequality is obtained:

( a12 + a22 + ... + an2 )1/ 2 + (b12 + b22 + ... + bn2 )1/ 2 ≥ [( a1 + b1 )2 + ( a2 + b2 )2 + ... + ( an + bn )2 ]1/ 2



(2.48)

It  is not  difficult to see that the general triangle inequality is valid for arbitrary sequences a1, a2 ,..., an and b1, b2 ,..., bn of real numbers (not necessarily non-negative). If n = 3, from inequality (2.48) the three-dimensional triangle inequality is obtained

(a12 + a22 + a32 )1/ 2 + (b12 + b22 + b32 )1/ 2 ≥ [(a1 + b1 )2 + (a2 + b2 )2 + (a3 + b3 )2 ]1/ 2

which expresses geometrically that the sum of the two sides of the triangle with vertices (0, 0, 0), ( a1, a2 , a3 ) and ( a1 + b1, a2 + b2 , a3 + b3 ) is greater than or equal to the length of the third side (Figure 2.2). If n = 2, from inequality (2.48) the two-dimensional triangle inequality is obtained

(a12 + a22 )1/ 2 + (b12 + b22 )1/ 2 ≥ [(a1 + b1 )2 + (a2 + b2 )2 ]1/ 2

which expresses geometrically that the sum of the two sides of a triangle is greater than or equal to the length of the third side (Figure 2.3).

FIGURE 2.2  Geometric interpretation of the three-dimensional triangle inequality.

38

Risk and Uncertainty Reduction by Using Algebraic Inequalities

FIGURE 2.3  Geometric interpretation of the two-dimensional triangle inequality.

If n = 1, for arbitrary real numbers a1 and b1, from the general triangle inequality (2.48) the inequality

(a12 )1/ 2 + (b12 )1/ 2 ≥ [(a1 + b1 )2 ]1/ 2

is obtained, which is equivalent to

| a1 | + | b1 | ≥ | a1 + b1 |

which is the one-dimensional triangle inequality (2.7).

3

Basic Techniques for Proving Algebraic Inequalities

3.1  THE NEED FOR PROVING ALGEBRAIC INEQUALITIES A central part of the methodology underlying all applications of algebraic inequalities to reduce uncertainty and risk consists of techniques for proving algebraic inequalities. Despite the existence of extensive literature on algebraic inequalities and numerous solved examples involving algebraic inequalities, there is a clear absence of systematic exposition of the techniques through which an algebraic inequality can be proved. A good awareness of the available tools by which an inequality can be proved significantly increases the benefit from using the algebraic inequalities. The systematic exposition of such techniques reduces the dependence on inspiration and clever tricks, which enhances the power of researchers and practitioners in using inequalities to reduce uncertainty and risk. A number of techniques for proving inequalities have already been covered in the literature related to algebraic inequalities. However, there are also many powerful methods and techniques for proving algebraic inequalities which have not received adequate coverage in the available literature. Consequently, an important task of this chapter is compiling well-known methods and introducing little-known methods for proving algebraic inequalities.

3.2 PROVING INEQUALITIES BY A DIRECT ALGEBRAIC MANIPULATION AND ANALYSIS This technique is based on direct algebraic manipulation and analysis demonstrating that the inequality indeed holds. This technique will be illustrated by proving the arithmetic mean – harmonic mean (AM-HM) inequality

a1 + a2 + ... + an n ≥ 1 1 1 n + + ... + a1 a2 an

(3.1)

39

40

Risk and Uncertainty Reduction by Using Algebraic Inequalities

where ai are positive real numbers. To prove this inequality, it suffices to prove that 1 1 1 (a1 + a2 + ... + an )  + + ... + an  a1 a2



 2 ≥n 

(3.2)

Clearly, for two real numbers only, the inequality 1 1  (a1 + a2 )  +  ≥ 22  a1 a2 



(3.3)

is true. Indeed, expanding the left-hand side of (3.3) gives 1 1  a a (a1 + a2 )  +  = 1 + 1 + 2 + 1 a2 a1  a1 a2 



(3.4)

For any positive numbers a1 and a2 a1 a2 + ≥ 2 a2 a1



(3.5)

is fulfilled. This follows from the inequality a1 + a2 ≥ 2 a1a2



(3.6)

which follows directly from the inequality ( a1 − a2 )2 ≥ 0. Indeed, squaring both sides of inequality (3.6) (which are positive numbers) and dividing by a1a2 yields inequality (3.5), directly. Expanding the left-hand side of inequality (3.2) gives n 2 terms of the form aaij , 1 ≤ i, j ≤ n. From these n 2 terms, n terms have the same index i = j and for these terms aj ai 2 ai a j = 1. The rest of the n − n terms can be paired in the sums a j + ai . According to aj ai what has been proved earlier, for each of the paired sums a j + ai ≥ 2 holds. The num2 ber of these paired sums is ( n − n) / 2 and the left-hand side of (3.2) becomes

1 1 1 (a1 + a2 + ... + an )  + + ... + a a a 2 n  1

 n2 − n 2  ≥ n + 2 ×2 = n 

(3.7)

which proves inequalities (3.2) and (3.1). Equality is attained for a= a= ... = an . 1 2 Example 3.1 Let a1, a2 ,..., an be real numbers and b1, b2 , ..., b n be positive real numbers. Prove the Bergström inequality (Pop, 2009) by using a direct algebraic manipulation.



a12 a22 a2 (a + a + ... + an )2 + + ... + n ≥ 1 2 b1 b2 bn b1 + b2 + ... + bn

(3.8)

41

Basic Techniques for Proving Algebraic Inequalities Proof. Consider only two sequences a1, a2 and b1, b2 containing only two real numbers each. Proving that



a12 a22 (a1 + a2 )2 + ≥ b1 b2 b1 + b2

(3.9)

a12 a22 (a1 + a2 )2 + − ≥ 0 b1 b2 b1 + b2

(3.10)

is equivalent to proving that



Simplifying the left side gives (b1a2 − a1b2 )2 ≥ 0 which is always non-negative. Equality is present only if b1b2(b1 + b2) b1 / a1 = b2 / a2 . a2 Now, if the value 3 is added to both sides of inequality (3.9), the same reasonb3 (a + a )2 a2 ing can be applied to the two terms 1 2 , 3 and the inequality b1 + b2 b3 a12 a22 a32 (a1 + a2 )2 a32 [(a1 + a2 ) + a3 ]2 + ≥ + + ≥ b1 b2 b3 b1 + b2 b3 (b1 + b2 ) + b3





(3.11)

is obtained. Continuing this reasoning proves the inequality for a1, a2 ,..., an and b1, b2 ,..., bn. Equality is attained only if a= a= ... = an / bn . 1 / b1 2 / b2 The next example demonstrates proving an inequality by a direct analysis.

Example 3.2 For the four positive numbers x > 0, y > 0 and z > 0, t > 0 prove that



x × x+t

t − x+t

y 1 z × ≤ y+z y+z 2

(3.12)

Inequality (3.12) is equivalent to the inequality



2 xt 2 yz − ≤ 1 x+t y+z

(3.13)

Note that 2 xt ≤ x + t because ( x − t )2 ≥ 0 and 2 yz ≤ y + z because ( y − z )2 ≥ 0. The  left-hand part of inequality (3.13) is a difference of two 2 yz non-negative numbers 2x +xtt and y + z each of which is smaller than 1. Therefore their difference is smaller than 1, inequality (3.13) is true, and therefore inequality (3.12) is also true.

42

Risk and Uncertainty Reduction by Using Algebraic Inequalities

3.3 PROVING INEQUALITIES BY PRESENTING THEM AS A SUM OF NON-NEGATIVE TERMS This technique is also based on a direct algebraic manipulation following a powerful strategy. The inequality f ( x1,..., xn ) ≥ 0 is proved by transforming it into a sum of terms g j ( x1,..., xn ) ≥ 0, j = 1,..., m each of which is non-negative: f ( x1,..., xn ) = g1 ( x1,..., xn ) + ... + gm ( x1,..., xn )



(3.14)

To illustrate this important technique, consider the inequality x 2 + y2 + z 2 ≥ xy + yz + zx



(3.15)

where x, y, z are real numbers. The inequality is equivalent to the inequality x 2 + y2 + z 2 − xy − yz − zx ≥ 0



(3.16)

Starting with the identity (a − b)2 = a2 − 2ab + b2, a presentation for the left side of inequality (3.16) as a sum of non-negative terms is sought. The left-hand side of inequality (3.16) can be presented as a sum of squares which are guaranteed to be non-negative: x 2 + y 2 + z 2 − xy − yz − zx = (1 / 2)[2 x 2 + 2 y 2 + 2 z 2 − 2 xy − 2 yz − 2 zx ] = (1 / 2)[( x 2 − 2 xy + y 2 ) + ( y 2 − 2 yz + z 2 ) + ( z 2 − 2 zx + x 2 )] (3.17)

= (1 / 2)[( x − y )2 + ( y − z )2 + ( z − x )2 ]

and is therefore non-negative which completes the proof of inequalities (3.16) and (3.15). Equality in (3.15) is attained for x= y= z. A similar technique can also be used to prove the Cauchy-Schwarz inequality, (a1b1 + a2 b2 + ... + an bn )2 ≤ (a12 + a22 + ... + an2 )(b12 + b22 + ... + bn2 )



(3.18)

which holds for any two sequences of real numbers a1, a2 ,..., an and b1, b2 ,..., bn . Since the difference between the left and right side of inequality (3.18) can be presented as a sum of non-negative terms: n

(a + a + ... + a )(b + b + ... + b ) − (a1b1 + a2 b2 + ... + an bn ) = 2 1

2 2

2 n

2 1

2 2

2 n

2



∑ (a b − b a ) i j

i > j ; j =1

i j

2



inequality (3.18) and the conditions for attaining equality can be deduced directly from the equivalent inequality: n



∑ (a b − b a ) i j

i > j ; j =1

i j

2

≥ 0

(3.19)

43

Basic Techniques for Proving Algebraic Inequalities

As can be seen, equality holds for proportional ai and bi, if and only if for any i ≠ j , ai / bi = a j / b j are fulfilled. Presenting an inequality as a sum of non-negative terms is not necessarily confined to presenting it as a sum of squares. Non-negative terms can be formed by relentless factoring of the separate parts of the inequality. This powerful technique is illustrated by the next example. Example 3.3 If a, b are real numbers such that 0 ≤ a ≤ 1 and 0 ≤ b ≤ 1 , prove the inequality



a2b2 − 3ab + a + b ≥ 0

(3.20)

Proof. First note that the first term a2b2 in (3.20) can be presented as (1− a2 )(1− b2 ) − 1+ a2 + b2. As a result, the left side of inequality (3.18) can be presented as



a2b2 − 3ab + a + b = (1− a2 )(1− b2 ) + (a − b)2 − ab + a + b − 1

(3.21)

Factoring the expression −ab + a + b − 1 in the right side of (3.21) results in −ab + a + b − 1 = −(1− b)(1− a) and the right side of equality (3.21) becomes

(1− a2 )(1− b2 ) + (a − b)2 − ab + a + b − 1 = (1− a)(1+ a)(1− b2 ) + (a − b)2 − (1− b)(1− a) (3.22) Taking out the factor (1− a) in the right side of equality (3.22) gives

(1− a)(1+ a)(1− b2 ) + (a − b)2 − (1− b)(1− a) = (1− a) ((1+ a)(1− b2 ) − (1− b) ) + (a − b)2 (3.23) and taking out another factor (1− b) in the right side of (3.23) finally results in



(1− a) ((1+ a)(1− b)(1+ b) − (1− b) ) + (a − b)2 = (1− a)(1− b)(a + b + ab) + (a − b)2 (3.24) As a result, the left side of inequality (3.20) has been presented as a sum of two non-negative terms:



a2b2 − 3ab + a + b = (1− a)(1− b)(a + b + ab) + (a − b)2

(3.25)

because (1− a) ≥ 0 and (1− b) ≥ 0 . This completes the proof of the original inequality (3.20). Often, by exploiting the symmetry in the inequality, the variables in the inequality can be ordered, which permits presenting the inequality as a sum of non-negative terms. Section 3.6 demonstrates this approach.

44

Risk and Uncertainty Reduction by Using Algebraic Inequalities

3.4 PROVING INEQUALITIES BY PROVING SIMPLER INTERMEDIATE INEQUALITIES This is a powerful technique for proving inequalities, particularly for proving strict inequalities of the type f ( x ) > g( x ). The  inequality f ( x ) > g( x ) can be proved if simpler intermediate inequalities f ( x ) > h( x ) and h( x ) > g( x ) can be found and proved. From the properties of inequalities, it follows that f ( x ) > g( x ). An inequality f1 (a1,..., an ) + ... + fn (a1,..., an ) < C where C is a particular constant can be proved by proving g1 (a1,..., an ) + ... + gn (a1,..., an ) < C



(3.26)

where each term of inequality (3.26) majorises a corresponding term of the original inequality:

g1 (a1,..., an ) ≥ f1 (a1,..., an )



…



gn (a1,..., an ) ≥ fn (a1,..., an ) Example 3.4 Prove that



1 1 1 + + ... + 2 < 1 22 32 n Proof. Because for k ≥ 2 the inequality



1 k2

≤

1 ( k −1) k

(3.27)

holds, the next inequality also holds

1 1 1 1 1 1 + + ... + 2 < + + ... + 22 32 1× 2 2 × 3 n (n − 1)n

(3.28)

The left side of the original inequality (3.27) has been majorised by the expression 1 1 1 1×2 + 2×3 + ... + ( n−1) n . If the majorising inequality



1 1 1 + + ... + < 1 1× 2 2 × 3 (n − 1)n

(3.29)

can be proved, this will complete the proof of the original inequality (3.27). The presentation k ( k1−1) = ( k1−1) − k1 permits summing up the expression in the left side of (3.29):



1 1 1 1 1 1 1 1 1 n −1 + + ... + = − + − + ... − = 1− = 0, y > 0 and z > 0, t > 0 prove that

xy 1 zt 1 × − × ≤ 1 1 + xy 1 + zt 1 + zt 1 + xy

(3.32)

Noticing that 1+xyxy ≤ 1, 1+ztzt ≤ 1, 1+1xy = 1 − 1+xyxy ≤ 1, 1+1zt = 1 − 1+ztzt ≤ 1, inequality (3.32) can be proved by making the trigonometric substitutions 1+xyxy = sin 2 α and 1+ztzt = sin 2 β . From these substitutions, it follows:

1 xy = 1− = cos2 α 1 + xy 1 + xy

(3.33)

1 zt = 1− = cos2 β 1 + zt 1 + zt

(3.34)

and

Substituting back in the left side of inequality (3.32) gives

sin α cos β − cos a sin β = sin(α − β )

(3.35)

Since sin(α − β ) ≤ 1 is always fulfilled, this completes the proof of the original inequality (3.32).

3.6  PROVING INEQUALITIES BY EXPLOITING THE SYMMETRY The symmetry in the inequality permits introducing ordering of the variables without loss of generality. If f ( x1, x2 ,..., xn ) is a symmetric expression in x1, x2 ,..., xn, for any permutation z1, z2 ,..., zn of x1, x2 ,..., xn, the equality f ( x1, x2 ,..., xn ) = f ( z1, z2 ,..., zn )

46

Risk and Uncertainty Reduction by Using Algebraic Inequalities

holds. As a result, without loss of generality, a particular ordering of the variables in the inequality, for example, x1 ≥ x2 ≥,..., ≥ xn , can be assumed. The ordering can then be exploited to prove the inequality. Here is an example where ordering is exploited directly to prove the inequality: x 2 + y2 + z 2 ≥ xy + yz + zx



(3.36)

valid for any real x, y and z. Because inequality (3.36) is symmetric in x, y and z, without loss of generality, x ≤ y ≤ z can be assumed. Using this ordering, the original inequality (3.36) can be transformed as follows: x 2 + y 2 + z 2 − xy − yz − zx = z( z − y ) + y( y − x ) − x( z − x ) = z ( z − y ) + y( y − x ) − x( z − y + y − x ) = z ( z − y ) + y( y − x ) − x( z − y ) − x( y − x ) = ( z − y )( z − x ) + ( y − x )2





Since ( z − y)( z − x ) ≥ 0, the original inequality has been presented as a sum of nonnegative terms ( z − y)( z − x ) + ( y − x )2 ≥ 0, which completes the proof of the original inequality (3.36). The ordering can also be exploited through the rearrangement inequality. Because inequality (3.36) is symmetric in x, y and z, without loss of generality, x ≤ y ≤ z can be assumed. Applying the rearrangement inequality to the pair of sequences ( x, y, z ), ( x, y, z ) and the pair of sequences ( x, y, z ), ( y, z, x ) results in x  x



y y

z x 2 2 2  = x +y +z ≥ z y

y z

z  = xy + yz + zx x

because the sequences ( x, y, z ) and ( x, y, z ) are similarly ordered while the sequences ( x, y, z ), ( y, z, x ) are not. An example demonstrating the power of exploiting the symmetry through the rearrangement inequality is the next example, related to proving the inequality x 4 + y 4 + z 4 + t 4 ≥ 4 xyzt



where x, y, z and t are real numbers. Because the inequality is symmetric in x, y, z and t, without loss of generality x ≥ y ≥ z ≥ t can be assumed. From the rearrangement inequality:



x  x x  x

y y y

z z z

y

z

t x   t y = x 4 + y4 + z 4 + t 4 ≥  z t   t t

y z t

z t x

x

y

t  x = 4xyzt y  z

47

Basic Techniques for Proving Algebraic Inequalities

because the sequences ( x, y, z, t ) , ( x, y, z, t ) , ( x, y, z, t ) and ( x, y, z, t ) are similarly sorted while the sequences ( x, y, z, t ) , ( y, z, t, x ) and ( z, t, x, y ) and ( t, x, y, z ) are not. Example 3.5 Let a, b, c be non-negative real numbers. By exploiting symmetry, it can be shown that for any m > 0, the next inequality (Schur inequality) holds



am (a − b)(a − c) + bm (b − a)(b − c) + c m (c − a)(c − b) ≥ 0

(3.37)

Equality is attained only if a= b= c or if any two numbers are equal and the third is equal to zero. Clearly, this inequality is symmetric in the variables a, b, c and, without loss of generality, a ≤ b ≤ c can be assumed. Considering this ordering, the left-hand side of the inequality can be presented as am (a − b)(a − c) + bm (b − a)(b − c) + c m (c − a)(c − b)





(3.38)

= (c − b) × [c m (c − a) − b m (b − a)] + am (a − b)(a − c) Considering the ordering a ≤ b ≤ c, it can be shown that each term in the righthand side of equality (3.38) is non-negative. Indeed, am (a − b)(a − c) ≥ 0 because a − b ≤ 0 and a − c ≤ 0. To show that c m (c − a) − b m (b − a) ≥ 0, note that



c − a ≥ b − a

(3.39)

c m ≥ bm

(3.40)

and



Since c − a ≥ 0 and b − a ≥ 0, c m ≥ 0 and b m ≥ 0, multiplying inequalities (3.39) and (3.40), according to property (iii) from Section  2.2  results in c m (c − a) ≥ b m (b − a). Consequently, the right-hand side of equality (3.38) has been presented as a sum of non-negative terms and therefore inequality (3.37) is true.

3.7  PROVING INEQUALITIES BY EXPLOITING HOMOGENEITY This is a powerful yet underestimated method for proving inequalities. A function f ( x1, x2 ,..., xn ) is homogeneous of order k if f (λ x1, λ x2 ,..., λ xn ) = λ k f ( x1, x2 ,..., xn ) for every λ > 0. Consider the inequality f ( x1, x2 ,..., xn ) ≤ 0. The  homogeneity of the inequality permits assuming without loss of generality additional constraints, for example,

48

Risk and Uncertainty Reduction by Using Algebraic Inequalities

x1 + x2 + ... + xn = 1, x1m + x2m + ... + xnm = 1, m > 1; x1 x2 ... xn = 1. Indeed, the positive constant λ can always be selected to scale the variables x1, x2 ,..., xn so that, for example, they add up to 1: λ x1 + λ x2 + ... + λ xn = 1. If instead of the original variables x1, x2 ,..., xn, the scaled variables a1 = λ x1, a2 = λ x2 ,..., an = λ xn are substituted, the inequality will not be altered. As a result, the equivalent inequality f (a1, a2 ,..., an ) ≤ 0 is obtained, with a constraint a1 + a2 + ... + an = 1. This  technique will be illustrated with proving the special case of Hölder’s inequality:

(a1b1c1 + ... + an bn cn )3 ≤ (a13 + ... + an3 )(b13 + ... + bn3 )(c13 + ... + cn3 )

(3.41)

where ai > 0, bi > 0, ci > 0; i = 1,..., n. Constants λ , µ and ν can always be chosen such that

(λ a1 )3 + (λ a2 )3 + ... + (λ an )3 = 1



( µ b1 )3 + ( µ b2 )3 + ... + ( µ bn )3 = 1



(ν c1 )3 + (ν c2 )3 + ... + (ν cn )3 = 1

Substituting in (3.41) the original variables ai , bi , ci with the scaled variables λ ai , µ bi ,ν ci gives

λ 3 µ 3ν 3 (a1b1c1 + ... + an bn cn )3 ≤ λ 3 (a13 + ... + an3 )µ 3 (b13 + ... + bn3 )ν 3 (c13 + ... + cn3 )

which, after the cancellations, yields the original inequality (3.41). By setting xi = λ ai, yi = λ bi and zi = λ bi, the equivalent inequality

( x1 y1z1 + ... + xn yn zn )3 ≤ ( x13 + ... + xn3 )( y13 + ... + yn3 )( z13 + ... + zn3 )

(3.42)

is obtained, with constraints

x13 + x23 + ... + xn3 = y13 + y23 + ... + yn3 = z13 + z23 + ... + zn3 = 1

(3.43)

After substituting the constraints (3.43) in (3.42), the proof reduces to demonstrating that

x1 y1z1 + ... + xn yn zn ≤ 1

According to the AM-GM inequality, the following inequalities hold: xi3 + yi3 + zi3 ≥ xi yi zi , i = 1,..., n . 3 Adding the n inequalities and considering (3.43), yields x1 y1z1 + ... + xn yn zn ≤ 1, which proves inequality (3.42) and the original inequality (3.41).

49

Basic Techniques for Proving Algebraic Inequalities

3.8  PROVING INEQUALITIES BY A MATHEMATICAL INDUCTION Commonly, the method of mathematical induction for proving an inequality, stated for an arbitrary number n of variables, consists of the following steps.

1. Proof of a base case (for example, that the inequality holds for n = 2 variables). 2. Induction hypothesis: assuming that the inequality holds for n = k variables ( k ≥ 2 ). 3. Inductive step: Proof that whenever the inequality holds for n = k variables, it also holds for n = k +1 variables. 4. Conclusion: If steps 1, 2 and 3 have been proved, the inequality holds for n variables.

The method of mathematical induction will be illustrated by proving the inequality 1 1 1 (a1 + a2 + ... + an )  + + ... + a a a 2 n  1



 2 ≥n 

(3.44)

where ai are non-negative real numbers.

1. Proof that the inequality holds for n = 2 variables. For n = 2, the inequality becomes 1 1  (a1 + a2 )  +  ≥ 22  a1 a2 



Expanding the left-hand side gives 1 1  a a (a1 + a2 )  +  = 1 + 1 + 2 + 1 a a a a1 2  2  1



1 1  a1 a2 + ≥ 2, (a1 + a2 )  +  ≥ 4 = 22 and the base case has been a2 a1  a1 a2  proved. 2. Assume that the inequality holds for n = k (induction hypothesis).

Since



1 1 1  (a1 + a2 + ... + ak )  + + ... +  ≥ k 2 ak   a1 a2

(3.45)

3. It will be proven that if inequality (3.45) holds, the inequality also holds for n = k +1, (inductive step), which means proving the inequality 1 1 1 1  2 (a1 + a2 + ... + ak + ak +1 )  + + ... + +  ≥ (k + 1) a a a a 2 k k +1   1 Transforming the left-hand side of inequality (3.46) gives

(3.46)

50

Risk and Uncertainty Reduction by Using Algebraic Inequalities

 1 1 1  1  [(a1 + a2 + ... + ak ) + ak +1 ]  + + ... +  +  a a a ak +1  2 k   1 1 1 1   a a  a   a = (a1 + a2 + ... + ak )  + + ... +  +  1 + k +1  +  2 + k +1  (3.47) a2  a a a a a a 2 k   k +1 1   k +1  1  a a  + ... +  k + k +1  + 1 a  k +1 ak 



Because of the induction hypothesis and because aa1 + aka+1 ≥ 2; k +1 1 ak +1 ak ak +1 a2 + ≥ 2 ;...; + ≥ 2 , for equality (3.47), which is the left-hand a a a a k +1

2

k +1

k

side of inequality (3.46), we have

 1 1 1  1  2 2 [(a1 + a2 + ... + ak ) + ak +1 ]  + + ... +  +  ≥ k + 2k + 1 = (k + 1) a a a a 2 k  k +1   1

The inductive step has been proved, therefore inequality (3.44) holds for all of integer n. Example 3.6 This example is from Sedrakyan and Sedrakyan (2010). Suppose that for the real numbers ai , bi , ci , i = 1,..., n the following conditions hold:



a1 / c1 ≥ a2 / c2 ≥ ... ≥ an / cn , b1 / c1 ≥ b2 / c2 ≥ ... ≥ bn / cn and c1 ≥ 0, c2 ≥ 0,...,cn ≥ 0

(3.48) Then, it can be shown that the following general inequality (Sedrakyan inequality) holds: n



n

∑ i =1

ai bi ≥ ci

n

∑ ∑b ai ×

i =1

n

i

i =1

∑c



(3.49)

i

i =1

Inequality (3.49) will be proved by induction. The proof starts with the basic case corresponding to n = 2. For, n = 2, inequality (3.49) becomes



a1b1 a2b2 (a1 + a2 )(b1 + b2 ) + ≥ c1 c2 c1 + c2

(3.50)

To prove (3.50), it suffices to prove



a1b1 a2b2 (a1 + a2 )(b1 + b2 ) + − ≥ 0 c1 c2 c1 + c2

(3.51)

51

Basic Techniques for Proving Algebraic Inequalities The left-hand side of (3.51) can be transformed to: a1b1 a2b2 (a1 + a2 )(b1 + b2 ) (a1c2 − a2c1)(b1c2 − b2c1) + − = c1 c2 c1 + c2 c1c2 (c1 + c2 )



(3.52)

From the conditions (3.48), a1 / c1 ≥ a2 / c2 and b1 / c1 ≥ b2 / c2 ; therefore, (a1c2 − a2c1) ≥ 0 and (b1c2 − b2c1) > 0. As a result, the right side of (3.52) is non-­ (a c − a c )(b c − b c ) negative, 1 2 2 1 1 2 2 1 ≥ 0, which proves inequalities (3.51) and (3.50). c1c2 (c1 + c2 ) Now assume that inequality (3.49) is true for n = k ≥ 2 (induction hypothesis): k

k

∑



i =1

ai bi ≥ ci

k

∑ ∑b ai ×

i =1

i

i =1

k

∑c



(3.53)

i

i =1

It will be shown that inequality (3.53) is also true for n = k + 1 (induction step). a b Adding the term kc+1 k +1 to both sides of (3.53), (ak /ck >= ak+1/ck+1; bk /ck >= bk+1/ck+1), k +1 results in the inequality k

ak bk  a1b1  c + ... + c k 1 



 ak +1bk +1 ≥ + c k +1 

k

∑ ∑b ai ×

i =1

i

+

i =1

k

∑

ci

ak +1bk +1 ck +1

(3.54)

i =1

Denoting x = ∑ a , y = ∑ b , z = ∑ c , the right-hand side of (3.54) becomes k i =1 i

k i =1 i

k i =1 i

xy ak +1b k +1 + z c k +1 . It can be demonstrated that x/z>=ak+1/ck+1 and y/z>=bk+1/ck+1 (the proof is very similar to the one given in Section 9.4 and will not be duplicated here). Because inequality (3.49) holds for the basic case n = 2, the next inequality holds: k +1

xy ak +1bk +1 ( x + ak +1)( y + bk +1) + ≥ = z ck +1 z + ck +1

k +1

∑a ×∑b

i

i

i =1

k +1

i =1

∑c

i





i =1

Finally, k



ak bk  a1b1  c + ... + c k  1

 ak +1bk +1 = + c k +1 

i

i

i =1

k +1

k

∑a ×∑b k

∑ i =1

i =1

ci

+

ak +1bk +1 ≥ ck +1

k +1

∑a ×∑b

i

i

i =1

k +1

∑

i =1



(3.55)

ci

i =1

This proves the induction step and inequality (3.49) for arbitrary n. The inequality is sharp because equality is attained if ai / ci = aj / c j , i ≠ j and bi / ci = bj / c j, i ≠ j .

52

Risk and Uncertainty Reduction by Using Algebraic Inequalities

3.9 PROVING INEQUALITIES BY USING THE PROPERTIES OF CONVEX/CONCAVE FUNCTIONS The concept of convex/concave functions has already been introduced in formulating the Jensen inequality in Chapter  2. The  concept will be discussed in greater detail here because it is a very powerful tool for proving inequalities. A  function f ( x ) with a domain [a,b] is said to be convex if for all values x1 and x2 in its domain ( x1, x2 ∈ [ a, b]), the next inequality holds

f (wx1 + (1 − w ) x2 ) ≤ wf ( x1 ) + (1 − w ) f ( x2 )

(3.56)

where 0 ≤ w ≤ 1. The  function f ( x ) is said to be strictly convex if equality in (3.56) holds only when= w 0= ; w 1 or x1 = x2. The function f ( x ) is strictly convex if for 0 < w < 1, x1 ≠ x2 , the inequality

f (wx1 + (1 − w ) x2 ) < wf ( x1 ) + (1 − w ) f ( x2 )

(3.57)

holds. If f ( x ) is twice differentiable for all x ∈ (a, b), and if the second derivative is non-negative ( f ″( x ) ≥ 0 ), the function f ( x ) is convex on (a, b). If f ( x ) is twice differentiable for all x ∈ (a, b), and if the second derivative is positive ( f ″( x ) > 0 ), the function f ( x ) is strictly convex on (a, b). For  a strictly convex function f(x) of a single argument, the graph between any two points x1 < x2 remains below the line segment joining the values f ( x1 ), f ( x2 ) corresponding to these points (Figure 3.1a). An example of a convex function is f ( x ) = x m for m ≥ 1 and x ≥ 0. For concave functions, the inequalities are reversed. If f ( x ) is twice differentiable for all x ∈ (a, b), and if for the second derivative f ″( x ) ≤ 0 is fulfilled, the function f ( x ) is concave on (a, b). If f ( x ) is twice differentiable for all x ∈ (a, b), and if the second derivative is negative, f ″( x ) < 0, the function f ( x ) is strictly concave on (a, b). For a strictly concave function f(x) of a single argument, the graph between any two points x1 < x2 remains above the line segment joining

FIGURE 3.1  (a) A strictly convex function and (b) a strictly concave function of a single argument.

Basic Techniques for Proving Algebraic Inequalities

53

the values f ( x1 ), f ( x2 ) corresponding to these points (Figure 3.1b). An example of a concave function is f ( x ) = ln x . If a function f ( x ) is convex, the function − f ( x ) is concave and vice versa. Thus, f ( x ) = ln x and g( x ) = x are both concave functions but − ln( x ) and − x are both convex functions. A sum of convex functions is a convex function and a sum of concave functions is a concave function. A  convex function does not  mean that the function is increasing or decreasing. The absolute maximum of a convex function is never found in the interior of the domain where the arguments vary. It  occurs at the boundary of the domain. Similarly, the absolute minimum of a concave function is never found in the interior of the domain where the arguments vary. It occurs at the boundary of the domain. The concepts of convex/concave functions are very powerful and can be used to prove various inequalities. Example 3.7 Consider the quadratic trinomial f ( x) ≡ ax 2 + bx + c where x ∈[u1, u2 ]. If a > 0, the second derivative of the quadratic trinomial is positive, the function f ( x) ≡ ax 2 + bx + c is convex and its global maximum is attained either at x = u1 or x = u2. The maximum is found by comparing f (u1) and f (u2 ), whichever is greater. If a < 0, the second derivative of the quadratic trinomial is negative, the function f ( x) ≡ ax 2 + bx + c is concave and its global minimum is attained either at x = u1 or x = u2. The global minimum is found by comparing f (u1) and f (u2 ), whichever is smaller. This provides a useful technique for proving inequalities. Here is an example. For  the non-negative x, y , z for varying in the intervals 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 , 0 ≤ z ≤ 1 it can be shown that the inequality



x 2 (1− y ) + x 2 (1− z) + y 2 (1− z) + y 2 (1− x) + z 2 (1− x) + z 2 (1− y ) ≤ 2 holds. This inequality can be proved by using convex functions. Because the second partial derivatives of the function f ( x,y ,z)= x 2 (1− y )+ x 2 (1− z)+ y 2 (1− z)+ y 2 (1− x)+ z 2 (1− x)+ z 2 (1− y ) with respect to each of the variables x,  y  and    z are non-negative (∂ 2f ( x, y , z) / ∂x 2 = 2 × (2 − y − z) ≥ 0 ; ∂ 2f ( x, y , z) / ∂y 2 = 2 × (2 − x − z) ≥ 0 and ∂ 2f ( x, y , z) / ∂z 2 = 2 × (2 − x − y ) ≥ 0 ), the function f ( x, y , z) is convex and attains its global maximum at the boundary of its domain. Because of the symmetry, the combination of values to check for a maxi, , 0) = 2 and f (111 , , ) = 0 . The maximum value mum is f (0, 0, 0) = 0, f (1, 0, 0) = 2, f (11 of the function f ( x, y , z) is equal to 2, which proves the inequality.

Example 3.8 Determine the smallest constant C such that



2 x 2 − x − 2ln x ≤ C where 1 ≤ x ≤ 2.

54

Risk and Uncertainty Reduction by Using Algebraic Inequalities f ( x) = 2x 2 − x − 2ln x is a convex function because it is a sum of three convex functions: f1( x) = 2 x 2 , f2 ( x) = − x and f3 ( x) = −2ln( x) . Consequently, the maximum of f ( x) is obtained at the ends of the interval [1,2]. There  are two values to be compared: f (1) = 1 and f (2) = 5.2. The  maximum is attained at x = 2 and is equal to 5.2. Therefore, the smallest value of the constant is C = 5.2.

Example 3.9 In  the next example, for the non-negative numbers 0 ≤ a ≤ 1, 0 ≤ b ≤ 1 and 0 ≤ c ≤ 1 , by using the concept of convexity, it can be shown that



1 1 1 + + + a2b2 + b2c 2 + c 2a2 ≤ 15 / 4 (1+ a)2 (1+ b)2 (1+ c)2

(3.58)

The  function f (a, b, c) = (1+1a)2 + (1+1b)2 + (1+1c )2 + a2b2 + b2c 2 + c 2a2 is c­ontinuous on a closed domain 0 ≤ a ≤ 1, 0 ≤ b ≤ 1 and 0 ≤ c ≤ 1; therefore, according to the extreme value theorem from calculus, the function must attain its maximum and minimum in that domain. 1 1 1 The function f (a, b, c) = + + + a2b2 + b2c 2 + c 2a2 is strictly (1+ a)2 (1+ b)2 (1+ c)2 convex with respect to each of the variables a, b and c because



∂ 2f 6 = + 2b2 + 2c 2 > 0 ∂a2 (1+ a)4



∂ 2f 6 = + 2a2 + 2c 2 > 0 2 ∂b (1+ b)4



∂ 2f 6 = + 2a2 + 2b2 > 0 ∂c 2 (1+ c)4 Because the function is strictly convex and continuous on the bounded and closed domain 0 ≤ a ≤ 1, 0 ≤ b ≤ 1 and 0 ≤ c ≤ 1, it assumes its maximum at the boundaries of the domain. The boundary points are:



{0, 0, 0},{0, 0,1},{0,1, 0},{0,11 , },{1, 0, 0},{1, 0,1},{11 , , 0},{1,11 , }. The maximum is attained at f (111 , , ) = 15 / 4 , hence inequality (3.58) holds.

3.9.1  Jensen Inequality For any convex function f ( x ), the Jensen’s inequality states that

f (w1 x1 + w2 x2 + ... + wn xn ) ≤ w1 f ( x1 ) + w2 f ( x2 ) + ... + wn f ( xn )

(3.59)

where wi (i = 1,..., n ) are numbers that satisfy 0 ≤ wi ≤ 1 and w1 + w2 + ... + wn = 1. For concave functions, the direction of inequality (3.59) is reversed.

55

Basic Techniques for Proving Algebraic Inequalities

The function ln x is concave for positive values x > 0 because its second derivative (ln x )″ = −1 / x 2 is negative. According to the Jensen inequality for concave functions,  x + x + ... + xn  ln x1 + ln x2 + ... + ln xn ln  1 2 ≥ n n  



holds. Using the properties of the logarithms  x + x + ... + xn  1 ln  1 2  ≥ n ln( x1 x2 ... xn ) n  



(3.60)

Because both sides of (3.60) are positive and e x is an increasing function, it follows that   x + x + ... + xn exp  ln  1 2 n  



(

)

 n   ≥ exp ln x1 x2 ... xn 

or finally x1 + x2 + ... + xn n ≥ x1 x2 ... xn n



which is the AM-GM inequality. If x1, x2 ,..., xn are positive numbers, and α ≠ 0 is a real number, the number

(

x1α + x2α +...+ xαn n

)

1/ n

is known as a power mean of the positive numbers.

Applying the AM-GM inequality to the numbers x1α , x2α ,..., xnα yields x1α + x2α + ... + xnα ≥ x1α x2α ... xnα n

(

)

1/ n

= ( x1 x2 ... xn )

α /n

=

(

n

x1 x2 ... xn

) α

from which

x1α + x2α + ... + xnα ≥ n

(

n

x1 x2 ... xn

) α

(3.61)

is obtained. If α > 1 and xi > 0 then f ( x ) = x α is a convex function because f ″( x ) = α (α − 1) x α −2 > 0. According to the Jensen inequality for convex functions, α



x1α + x2α + ... + xnα  x1 + x2 + ... + xn  ≥  n n  

(3.62)

If 0 < α < 1 and xi > 0 then f ( x ) = x α is a concave function because f ″( x ) = α (α − 1) x α −2 < 0 . According to the Jensen inequality for concave functions

56

Risk and Uncertainty Reduction by Using Algebraic Inequalities α

x1α + x2α + ... + xnα  x1 + x2 + ... + xn  ≤  n n  



(3.63)

The next example illustrates another application of the Jensen inequality. Example 3.10 For the non-negative numbers x > 0, y > 0 and z > 0, prove that



1 1  1 1 1   ≤ + +  x + y + z 3 3  x y z 

(3.64)

Proof. The  function f (u) = u −1/ 2 is a strictly convex function for u > 0 because f ″(u) = (3 / 4)u −5/ 2 > 0. According to the Jensen inequality, for a convex function f ( x, y , z), the inequality



1 1 x+y+z 1 f ≤ f ( x) + f ( y ) + f ( z)  3 3 3  3  holds and for the convex function f (u) = u −1/ 2, the inequality



1 1 1 1 1   ≤  + + x + y + z 3  x y z  3 holds, from which inequality (3.64) follows. Because the function is strictly convex, the equality in (3.64) is attained only for x= y= z .

3.10 PROVING INEQUALITIES BY USING THE PROPERTIES OF SUB-ADDITIVE AND SUPER-ADDITIVE FUNCTIONS The properties of sub-additive and super-additive functions can be used for proving inequalities. A function f ( x, y) is sub-additive if for any real numbers x, y , the next inequality holds:

f ( x + y) ≤ f ( x ) + f ( y)

(3.65)

A function f ( x, y) is super-additive if for any real numbers x, y , the next inequality holds:

f ( x + y) ≥ f ( x ) + f ( y)

(3.66)

Multiplying the definition inequalities (3.65) and (3.66) by ‘–1’, reverses the sign of the inequality. Consequently, if f ( x, y) is sub-additive, − f ( x, y) is super-additive and if f ( x, y) is super-additive, − f ( x, y) is sub-additive.

57

Basic Techniques for Proving Algebraic Inequalities

An example of a super-additive function defined for −∞ < x < ∞ is the function e x . A key result related to sub-additive and super-additive functions can now be stated. If a function f ( x, y) with a domain [0, ∞) and range [0, ∞) is concave, then the function is sub-additive ( f ( x + y) ≤ f ( x ) + f ( y) ). If the function is convex, then it is super-additive (Alsina and Nelsen, 2010). Proof. Consider all x, y such that x + y ≠ 0. Because f ( x, y) is a concave function:

 x  x y y f ( x) = f  × ( x + y) + ×0 ≥ f ( x + y) + f (0) x y x y x y x y + + + +  

(3.67)



 x  y x y f (0) + f ( x + y) f ( y) = f  ×0 + × ( x + y)  ≥ x+y x+y  x+y  x+y

(3.68)

Adding inequalities (3.67) and (3.68) yields f ( x ) + f ( y) ≥ f ( x + y) + f (0)



Since f (0) ≥ 0 (the range of the function f ( x, y) is [0, ∞) ), f ( x ) + f ( y) ≥ f ( x + y) + f (0) ≥ f ( x + y)



Note that equality can only be attained if f (0) = 0. Indeed, for y = 0, f ( x + 0) = f ( x ) + f (0) = f ( x ). Similarly, for x = 0, f (0 + y ) = f (0) + f ( y) = f ( y). However, if f (0) > 0 the inequality is not sharp: f ( x ) + f ( y) ≥ f ( x + y) + f ( 0 ) > f ( x + y)



This proves inequality (3.66). Inequality (3.65) can be proved by using similar reasoning. Inequalities (3.65) and (3.66) can be generalised easily by induction for n variables x1,x2,....,xn: f(x1 + ... + xn) ≤  f(x1) + ... +  f(xn) holds for concave functions and f(x1 + ...+ xn) ≥  f(x1) + ... + f(xn) holds for convex functions. Example 3.11 Here is an example, illustrating the use of sub-additive functions. For any real x ≥ 0; y ≥ 0, prove the inequality



1− e −( x + y ) ≤ (1− e − x ) + (1− e − y )

(3.69)

The function f (u) = 1− e −u is a concave function because the second derivative is negative. Since f (0) = 1− e0 = 0 , the inequality follows from the key result stated earlier. Equality is attained for x = 0 or y = 0.

58

Risk and Uncertainty Reduction by Using Algebraic Inequalities

Inequalities (3.65) and (3.66) have very important potential applications. If the function f(z) measures the effect of a particular factor z, inequalities (3.65) and (3.66) provide the unique oppotunity to increase the effect of the factor by segmenting the factor (z = z1  + z2; f(z1) + f(z2) >= f(z)) or by aggregating the factor (z = z1  + z2); f(z) >= f(z1) + f(z2)), depending on whether the function f(z) is concave or convex.

3.11 PROVING INEQUALITIES BY TRANSFORMING THEM TO KNOWN INEQUALITIES Inequalities do not normally appear in the form of the standard inequalities discussed in Chapter 2. It is therefore very important to recognise what transformation of variables leads to standard inequalities or to inequalities which have already been proved.

3.11.1 Proving Inequalities by Transforming Them to an Already Proved Inequality The inequality is transformed until an inequality is reached which has already been proved. This technique will be illustrated by the next example. Let a1, a2 ,..., an be a set of positive real numbers and a1 + a2 + ... + an = s. Then, it can be shown that the inequality s s s n2 + + ... + ≥ s − a1 s − a2 s − an n − 1



(3.70)

holds. Multiplying the left-hand side by the sum of the reciprocals of the separate terms gives

 s s s  s − a + s − a + ... + s − a 1 2 n 

  s − a1 s − a2 s − an  ×  s + s + ... + s  

 2 ≥n 

(3.71)

which is the well-known inequality ( x1 + x2 + ... + xn )(1 / x1 + 1 / x2 + ... + 1 / xn ) ≥ n 2 , valid for any set of n positive real numbers xi . Since the second factor on the left side can be presented as s − an  s − a1 s − a2  s + s + ... + s 



 ns − s  = s = n − 1 

(3.72)

inequality (3.70) follows directly. Example 3.12 Prove the inequality x12 + x22 + ... + xn2 ≥ 1 / n, if 0 ≤ xi ≤ 1 and x1 + x2 + ... + xn = 1. The  inequality can be easily transformed into the already proved Bergström inequality (3.8):



x12 x22 x 2 ( x + x + ... + xn )2 + 2 + ... + 2n ≥ 1 2 = 1 / n 2 1 1 1 1+ 1+ ... + 1

59

Basic Techniques for Proving Algebraic Inequalities Many of the standard inequalities can be proved by reducing them to already proved inequalities. If, for example, the variables ci in the Sedrakian inequality (3.49) are selected such that ci = n; i = 12 , ,...,n, then ∑in=1 c i = n2 and inequality (3.49) yields the special-case sharp inequality a1b1 + a2b2 + ... + anbn a1 + a2 + ... + an b1 + b2 + ... + bn ≥ × n n n



which is the Chebyshev inequality.

3.11.2 Proving Inequalities by Transforming Them to the Hölder Inequality Transforming an inequality to the Hölder inequality introduced in Chapter  2  is a powerful technique for proving inequalities. The  method is often appropriate for cyclic inequalities and will be illustrated by the next two examples. Transformation to the Hölder inequality also works in the presence of constraints. Example 3.13 Let a, b, c and x, y , z be real positive numbers such that a + b + c = x + y + z . Then, the next inequality holds: a3 / x 2 + b3 / y 2 + c3 / z 2 ≥ a + b + c



(3.73)

Proof. Consider the three sequences ( x, y , z),( x, y , z), and (a3 / x 2 , b3 / y 2 , c3 / z 2 ). Applying the Hölder inequality with weights λ1 = λ2 = λ3 = 1 / 3 ∑3i=1λi = 1 gives

(

)

( x + y + z)1/ 3 ( x + y + z)1/ 3 (a3 / x 2 + b3 / y 2 + c3 / z 2 )1/ 3 ≥ a + b + c



Considering the constraint a + b + c = x + y + z , raising both sides into a power of 3 and dividing both sides by the positive value ( x + y + z)2 yields

(a3 / x 2 + b3 / y2 + c3 / z 2 ) ≥ (a + b + c)3 / ( x + y + z)2 = a + b + c



Example 3.14 Let a, b, c be real numbers such that a + b + c = 1. Then, the next inequality holds. 1 1 1 81 + + ≥ 4 a(b + c)2 b(a + c)2 c(a + b)2



(3.74)

Proof. Consider the four sequences (a, b, c) , (b + c, a + c, a + b), (b + c, a + c, a + b) and a( b +1 c )2 , b( a+1 c )2 , c( a+1b)2 . Applying the Hölder inequality with weights

(

)

4 λ1 = λ2 = λ3 = λ4 = 1 / 4 ( ∑i= 1λi = 1) gives

60

Risk and Uncertainty Reduction by Using Algebraic Inequalities

1 1 1   (a + b + c)(b + c + a + c + a + b)(b + c + a + c + a + b)  + + ≥ 2 2 b(a + c) c(a + b)2   a(b + c) 1/ 4 1/ 4 1/ 4 4 4 (1 + 1 + 1 ) = 3



Since a + b + c = 1 and b + c + a + c + a + b = 2, dividing both sides of the last inequality by 4 gives 1 1 1 + + ≥ 81 / 4 a(b + c)2 b(a + c)2 c(a + b)2



which completes the proof.

3.11.3 An Alternative Proof of the Cauchy-Schwarz Inequality by Reducing It to a Standard Inequality This alternative way of proving the Cauchy-Schwarz inequality has been given in Alsina and Nelsen (2010). Proving the Cauchy-Schwarz inequality (3.18) is equivalent to proving the next inequality. n

|

∑a b | k k

≤1

i =1

n

n

∑a × ∑b 2 k



2 k

i =1

Consider the numbers x =

| ak | n

∑a



i =1

and y =

2 k

| bk | n

∑b

. According to a standard

2 k

i =1

i =1

inequality (proved in Chapter 2), xy ≤ ( x + y ) / 2. Consequently, 2

n

|

∑a b | k k

n

k =1

n

n

∑a × ∑b 2 k



k =1

2 k

k =1

≤

∑ k =1

2

  | ak | | bk | 1 | bk |2  | ak |2 × ≤ + n n n 2 k =1  n 2  2 2 a bk2 k ak bk  k =1  k =1 k =1 k =1 n

∑

∑

∑

∑

∑

   1  = 2 (1 + 1) =1   

3.11.4 An Alternative Proof of the GM-HM Inequality by Reducing It to the AM-GM Inequality Let x1, x2 ,..., xn be a set of positive real numbers. The geometric mean – harmonic mean inequality states that

n

x1 x2 ... xn ≥

n 1 1 1 + + ... + x1 x2 xn

(3.75)

61

Basic Techniques for Proving Algebraic Inequalities

with equality attained only if x= x= ... = xn . Proving inequality (3.75) is equiva1 2 lent to proving n



x1 x2 ... xn  1 1 1  ×  + + ... +  ≥ 1 xn  n  x1 x2

the left-hand side of which can be transformed into n



n

x1 x2 ... xn  1 1 1  ×  + + ... +  = n x x x n  2  1

∑x

1

i =1

n

x1 x2 ... xn

i

n



1 The right side of the last equality is the average of the numbers xi i = 1,..., n . According to the AM-GM inequality n

∑x

1

i =1

n

x1 x2 ... xn

i

n



≥

n

n

x1 x2 ... xn ,

1 × n x1n x2n ... xnn = n 1 = 1 x1 x2 ... xn

This completes the proof of the original inequality (3.75).

3.11.5 Proving Inequalities by Transforming Them to the Cauchy-Schwarz Inequality This is a very important technique that can be applied to prove a vast range of inequalities. Let a1, a2 ,..., an be real numbers and b1, b2 ,..., bn be positive real numbers. Then, the inequality (Bergström inequality) a12 a22 a2 (a + a + ... + an )2 + + ... + n ≥ 1 2 b1 b2 bn b1 + b2 + ... + bn



(3.76)

holds (see Section 3.2). An alternative proof of inequality (3.76) can be obtained by transforming the inequality to the Cauchy-Schwarz inequality

( x1 y1 + x2 y2 + ... + xn yn )2 ≤ ( x12 + x22 + ... + xn2 )( y12 + y22 + ... + yn2 )

(3.77)

valid for any two sequences of real numbers x1, x2 ,..., xn and y1, y2 ,..., yn . Note that the substitutions xi = abi ( i = 1,..., n ) and yi = bi ( i = 1,..., n ) of the origii nal variables applied with the Cauchy-Schwarz inequality (3.77) leads to inequality (3.76):  a1   b1

( (a / 1

a b1 + 2 b2

a b2 + ... + n bn

)(

2

 bn  ≤  

b1 )2 + ... + (an / bn )2 ( b1 )2 + ... + ( bn )2

)

62

Risk and Uncertainty Reduction by Using Algebraic Inequalities

which shows that inequality (3.76) is a special case of the Cauchy-Schwarz inequality (3.77). The inequality x1 + ... + xn ≤ n



x12 + ... + xn2 n

(3.78)

where x1,..., xn is a set of real numbers is known also as arithmetic mean–root-mean square (AM-RMS) inequality. It can also be proved by showing that it is a special case of the Cauchy-Schwarz inequality. Indeed, consider the set of n real numbers x1, x2 ,..., xn and the set of n numbers 1, 1,..., 1. According to the Cauchy-Schwarz inequality:

x1 ⋅ 1 + x2 ⋅ 1 + ... + xn ⋅ 1 ≤ x12 + x22 + ... + xn2 × 12 + 12 + ... + 12

(3.79)

Dividing both sides of inequality (3.79) by the positive value n yields inequality (3.78). Example 3.15 For  any positive real numbers x, y and z such that x 2 + y 2 + z 2 = 1, prove the inequality



x 2 yz + y 2 zx + z 2 xy ≤ 1 / 3 Proof. Factoring xyz from the left hand side of the inequality results in x 2 yz + y 2 zx + z 2 xy = xyz ( x + y + z ). From the AM-GM inequality x 2 + y 2 + z2 3 2 2 2 ≥ x y z and from the constraint x 2 + y 2 + z 2 = 1  the next 3 inequality is obtained:



xyz ≤

1 3 3

From the Cauchy-Schwarz inequality



( x ⋅ 1+ y ⋅ 1+ z ⋅ 1) ≤

(x

2

)(

)

+ y 2 + z 2 12 + 12 + 12 = 3

1 1 Therefore, x 2 yz + y 2 zx + z 2 xy = xyz ( x + y + z ) ≤ × 3 = , which completes 3 3 3 the proof.

3.12  PROVING INEQUALITIES BY SEGMENTATION This is a powerful technique whose essence is to segment (split) the original inequality into simpler inequalities by using some of the standard inequalities (frequently the AM-GM inequality) and to sum the segmented inequalities in order to assemble the original inequality.

Basic Techniques for Proving Algebraic Inequalities

63

Here is an example of this technique demonstrated on the inequality

x 2 + y2 + z 2 ≥ xy + yz + zx

(3.80)

which is valid for any arbitrary real x, y, z . The left side of this inequality can be segmented into three inequalities by using the standard AM-GM inequality:



x 2 + y2 y2 + z 2 z2 + x 2 ≥ xy; ≥ yz; ≥ zx 2 2 2

Adding the three inequalities gives the original inequality (3.80) and completes the proof. Another example of this technique is proving the inequality

( x + y + z )2 ≥ x yz + y zx + z xy 3

(3.81)

valid for non-negative real numbers x, y and z. Since the terms x yz , y zx , z xy in the right-hand side can be obtained easily by applying the AM-GM inequality to xy + yz, yz + zx and zx + xy, correspondingly, the segmentation yields the following inequalities:

xy + yz ≥ 2 y zx

(3.82)



yz + zx ≥ 2 z xy

(3.83)



zx + xy ≥ 2 x yz

(3.84)

Adding inequalities (3.82)–(3.84) gives

2 xy + 2 yz + 2 xz ≥ 2 x yz + 2 y zx + 2 z xy

(3.85)

To make connection with the term ( x + y + z )2 on the left-hand side, note that the terms x yz , y zx , z xy on the right-hand side can also be obtained by applying the AM-GM inequality to x 2 + x 2 + y2 + z 2, y2 + y2 + x 2 + z 2 and z 2 + z 2 + x 2 + y2. Indeed, applying the AM-GM inequality yields:

x 2 + x 2 + y2 + z 2 ≥ 4 4 x 2 x 2 y2 z 2 = 4 x yz

(3.86)



y2 + y2 + x 2 + z 2 ≥ 4 4 y2 y2 x 2 z 2 = 4 y xz

(3.87)



z 2 + z 2 + x 2 + y2 ≥ 4 4 z 2 z 2 x 2 y2 = 4 z xy

(3.88)

Adding inequalities (3.86)–(3.88) results in

x 2 + y2 + z 2 ≥ x yz + y xz + z xy

(3.89)

64

Risk and Uncertainty Reduction by Using Algebraic Inequalities

Finally, adding inequalities (3.89) and (3.85) yields the inequality

( x + y + z)2 ≥ 3 x yz + 3 y xz + 3z xy

which is equivalent to inequality (3.81).

3.12.1 Determining Bounds by Segmentation Segmentation can also be used for determining bounds of various symmetric expressions. This  technique will be demonstrated by the next example determining the lower bound of the expression

f ( x, y, z) =

x 2 y2 y2 z 2 z 2 x 2 + 2 + 2 z2 x y

(3.90)

where x, y, z are positive real numbers that satisfy the constraint x 2 + y2 + z 2 = 1. The  lower bound will be found by segmenting expression (3.90) through the AM-GM inequality

x 2 y2 y2 z 2 + 2 ≥ 2 y 2 z2 x

(3.91)



y2 z 2 z 2 x 2 + 2 ≥ 2z 2 x2 y

(3.92)



x 2 y2 z 2 x 2 + 2 ≥ 2 x 2 z2 y

(3.93)

Adding inequalities (3.91)–(3.93) gives

 x 2 y2 y2 z 2 z 2 x 2  2  2 + 2 + 2  ≥ 2( x 2 + y2 + z 2 ) x y   z

which, after the substitution of the constraint x 2 + y2 + z 2 = 1 results in

f ( x, y, z) =

x 2 y2 y2 z 2 z 2 x 2 + 2 + 2 ≥ 1 z2 x y

(3.94)

for the lower bound. Equality is attained for x= y= z = 1 / 3 .

3.13 PROVING ALGEBRAIC INEQUALITIES BY COMBINING SEVERAL TECHNIQUES Often, a rigorous proof of an algebraic inequality may require combining several techniques. Consider the next example which involves substitution, the standard Cauchy-Schwarz inequality and finally, the AM-GM inequality.

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Basic Techniques for Proving Algebraic Inequalities

The example is related to proving the inequality (problem 2 from the International Mathematical Olympiad 1995)

1 1 1 3 + + ≥ a 3 ( b + c ) b 3 ( a + c ) c 3 ( a + b) 2

(3.95)

where a, b, c are positive real numbers for which abc = 1. Initially, the inequality is transformed by applying the substitution technique: Consider the substitution = x 1= / a; y 1 / b; z = 1 / c. Then xyz = 1 and substituting = a 1= / x; b 1 / y; c = 1 / z in the original inequality and by using yz = 1 / x , yx = 1 / z and xz = 1 / y , the original inequality (3.95) is transformed into the inequality

x2 y2 z2 3 + + ≥ y+z z+ x x+ y 2

(3.96)

Consider now the two sequences of real numbers  x y z  , ,  and  x + y   y + z z + x

{

y + z, z + x, x + y

}

Using the Cauchy-Schwarz inequality to these two sequences gives  x  y z × y+z + × z+x + × x+y   z+x x+y  y+z 



 x2 y2 z2  ≤ + +  × 2( x + y + z )  y+z z+ x x+ y

2



or

 x2 y2 z2  + + 2  ≥ ( x + y + z)  y+z z+ x x+ y

(3.97)

Using the AM-GM inequality for the right part of inequality (3.97) gives ( x + y + z) ≥ 3 3 xyz = 3. Hence, inequality (3.96) is obtained from inequality (3.97), which completes the proof of inequality (3.95).

3.14  USING DERIVATIVES TO PROVE INEQUALITIES Consider proving an inequality of the type f ( x ) ≥ g( x ) defined on an interval [a,b], where f ( x ) and g( x ) are differentiable functions. Given that f (a) ≥ g(a) and f ′( x ) ≥ g′( x ), it can be shown that f ( x ) ≥ g( x ) holds for any x ∈[ a, b]. To show why this is true, consider the function t ( x ) which is the difference t( x ) = f ( x ) − g( x ). This function is non-decreasing because t′( x ) = f ′( x ) − g′( x ) ≥ 0.

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Risk and Uncertainty Reduction by Using Algebraic Inequalities

Therefore, t ( x ) ≥ t (a) for any x ∈[ a, b]. Since t (a) = f (a) − g(a) ≥ 0, this means that t ( x ) ≥ 0 or f ( x ) ≥ g( x ) for any x ∈[ a, b]. This technique will be demonstrated by proving the Bernoulli inequality, (1 + x )n ≥ 1 + nx



(3.98)

for a non-negative x and any positive integer n ≥ 1. Proof. f ( x ) = (1 + x )n , g( x ) = 1 + nx . Clearly, f (0) = g(0) = 1 and f ′( x ) = n(1 + x )n −1 > g′( x ) = n , because (1 + x )n −1 ≥ 1. As a result, t ( x ) = (1 + x )n − (1 + nx ) is an increasing function over the interval [0, +∞). Since t (0) = f (0) − g(0) = 0 , and t ( x ) is an increasing function, it follows that t ( x ) ≥ 0, which means that f ( x ) ≥ g( x ) or (1 + x )n ≥ 1 + nx for any x ∈ [0, +∞]. Example 3.16 Another example is the proof of the inequality



 1  2 ln   ≥ 2x  1− x 

(3.99)

for x ∈[0,1. ) Proof. f ( x)=ln( 1−1x ), g( x) = 2 x 2. Clearly, f= (0) g= (0) 0 and f ′( x)= 1−1x ≥ g′( x)=4 x 2 because 1−1x −4 x = (21x−−x1) is non-negative for x ∈[0, 1). As a result, t( x)=ln( 1−1x ) −2 x 2 is an increasing function on the interval [0,1). Since t(0) = f (0) − g(0) = 0 and t ( x ) is an increasing function, it follows that t( x) ≥ 0, which means that f ( x) ≥ g( x) or ln( 1−1x ) ≥2 x 2 for any x ∈[0,1. )

Another technique employing derivatives for proving inequalities is the mean value theorem. The mean value theorem holds for a continuous function f ( x) on a closed interval [a, b] which is differentiable on the open interval (a, b). The theorem states that under these conditions, there exists a value ξ for which

f ( b) − f ( a) = f ′(ξ ) b−a

(3.100)

where f ′(ξ ) is the derivative of the function at a point ξ from the open interval (a, b). The mean value theorem can be used to prove the inequality

| sin( x2 ) − sin( x1 ) | ≤ x2 − x1

(3.101)

for any x1 < x2 . Applying the mean value theorem yields

sin x2 − sin x1 = cos(ξ ) x2 − x1

(3.102)

Basic Techniques for Proving Algebraic Inequalities

67

where ξ ∈ ( x1, x2 ). From (3.102) it follows

| sin x2 − sin x1 | = ( x2 − x1 ) × | cos(ξ ) |

(3.103)

Inequality (3.101) follows directly from equation (3.103) because | cos(ξ ) |≤ 1. Another useful inequality can be derived from inequality (3.101) by substituting x1 = 0 and x2 = x :

| sin( x ) | ≤ | x |

4

Using Optimisation Methods for Determining Tight Upper and Lower Bounds Testing a Conjectured Inequality by a Simulation – Exercises

4.1 USING CONSTRAINED OPTIMISATION FOR DETERMINING TIGHT UPPER BOUNDS An important problem in many applications is related to determining a tight upper bound M for a conjectured inequality of the type f ( x1, x2 ,..., xn ) ≤ U



where x1, x2 ,..., xn are arguments which vary within known limits:

x1 min ≤ x1 ≤ x1max ,..., xn min ≤ xn ≤ xn max

Suppose that f ( x1, x2 ,..., xn ) is a continuous function in the variables x1, x2 ,..., xn. The  tight upper limit U can be determined by finding the global maximum U of the function f ( x1, x2 ,..., xn ) by using a method for optimising multivariate functions. For  simple functions f ( x1, x2 ,..., xn ), the maximum can be obtained by exact analytical methods. For complex functions f ( x1, x2 ,..., xn ), the maximum can only be obtained by using heuristic optimisation, for example, by using particle swarm optimisation (Lazinica, 2009; Kiranyaz et al., 2014). Suppose that it is required to find an upper bound U (U is a particular constant) for a differentiable function of n variables f ( x1,..., xn ) under the constraint

g( x1, x2 ,..., xn ) = 0

which is also differentiable. 69

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Risk and Uncertainty Reduction by Using Algebraic Inequalities

For  relatively simple functions f ( x1,..., xn ) and g( x1,..., xn ), the bound can be found by using analytical methods for maximising the function f ( x1,..., xn ). From the theory of multivariable optimisation (Ellis and Gulick, 1991; McCallum et al., 2005), at the point of extremum, the equation:

grad ( f ( x1,..., xn ) − U ) = λ grad g ( x1,..., xn )

(4.1)

g( x1,..., xn ) = 0

(4.2)

and the equation

must be satisfied, where λ is a constant of proportionality (Lagrange multiplier). These conditions are then used to derive the points at which extremum is attained and also to evaluate the extremum. This approach will be illustrated by finding the value of the constant M which provides a tight upper bound U in the next inequality:

f ( x, y, z) = x 2 y + y2 z + z 2 x ≤ U

(4.3)

where x, y and z are non-negative real numbers for which x 2 + y2 + z 2 = 1. The constraint can be expressed as g( x, y, z) = x 2 + y2 + z 2 − 1. Because of the symmetry, without loss of generality, it can be assumed that x ≥ y ≥ z. Taking gradients results in

grad ( x 2 y + y2 z + z 2 x ) = (2 xy + z 2 )i + (2 yz + x 2 ) j + (2 zx + y2 )k

(4.4)



grad(g( x, y, z)) = 2 x i + 2 y j + 2 z k

(4.5)

where, i, j and k are the unit vectors. Equating the components of the gradient vectors from (4.4) and (4.5) gives the equations

2 xy + z 2 = λ 2 x

(4.6)



2 yz + x 2 = λ 2 y

(4.7)



2 zx + y 2 = λ 2 z

(4.8)

An additional equation is obtained by adding the constraint x 2 + y2 + z 2 = 1

Adding equations (4.6)–(4.8) gives

( x + y + z )2 = 2λ ( x + y + z )

from which λ = ( x + y + z )/ 2.

(4.9)

Using Optimisation Methods for Determining Tight Upper and Lower Bounds

71

Substituting λ in equation (4.6) and expanding results in

x ( x − y) + z( x − z ) = 0

(4.10)

Because x ≥ 0, z ≥ 0, x ≥ y and x ≥ z , the equality is only possible if x = y ≠ 0, z = 0; x= y= z = 0 or for x = y = z ≠ 0. The trivial solution x= y= z = 0 should be ignored because this solution does not satisfy the constraint (4.9). Finally, substituting λ = ( x + y + z ) / 2 in equation (4.8) and expanding results in

( y − z )( y + z) + z( x − y) = 0

(4.11)

Because x ≥ 0, z ≥ 0 and x ≥ y , y ≥ z hold, only the solution x = y = z ≠ 0 satisfies the equation (4.11). Substituting λ = ( x + y + z ) / 2 in equation (4.7) and expanding results in

( x − y)( x + y) − y( x − z) = 0

(4.12)

The solution x = y = z ≠ 0 also satisfies the equation (4.12). Consequently, the solution x = y = z ≠ 0 is the only solution that satisfies simultaneously equations (4.6)–(4.8); therefore, the extremum is attained at x = y = z ≠ 0. Considering the constraint x 2 + y2 + z 2 = 1, the solution is obtained from x 2 + x 2 + x 2 = 1 or for x= y= z = 1 / 3 . Evaluating the function f ( x, y, z) = x 2 y + y2 z + z 2 x yields f (1 / 3 , 1 / 3 , 1 / 3 ) = 3 × (1 / 3) (1 / 3 ) = 1 / 3 . The check shows that the obtained extremum is a maximum. As a result,

f ( x, y, z) = x 2 y + y2 z + z 2 x ≤ 1 / 3

(4.13)

which determines the value U = 1 / 3 as a tight upper bound.

4.2 TIGHT BOUNDS FOR MULTIVARIABLE FUNCTIONS WHOSE PARTIAL DERIVATIVES DO NOT CHANGE SIGN IN A SPECIFIED DOMAIN If a function of one variable is monotonic within a particular domain, the extreme values which serve as tight bounds are obtained at the boundary of the domain. To find the tight bounds, it is therefore sufficient to check the value of the function at the boundaries of the domain. Consider a multivariable function f ( x1, x2 ,..., xn ), where the arguments x1, x2 ,..., xn vary within the rectangular domain:

a1 ≤ x1 ≤ b1; a2 ≤ x2 ≤ b2 ;  ; an ≤ xn ≤ bn

(4.14)

and whose partial derivatives do not change sign in this domain. It has been shown (Todinov, 2015) that the global maximum and the global minimum for such functions are located at a corner of the rectangular domain. Despite this knowledge, determining the global extremum is not easy for functions of many variables because

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Risk and Uncertainty Reduction by Using Algebraic Inequalities

even checking the value of the function at the corners of the domain can be a formidable task. Thus, for a function f ( x1, x2 ,..., xn ), where the arguments x1, x2 ,..., xn vary within the rectangular domain (4.14) and whose partial derivatives do not  change sign in this domain, there are 2n different corners and therefore 2n checks need to be made in order to find the global maximum/minimum of the function. For a large number of variables n, the number of different corners 2n can be very big. To solve this predicament, the same sign local effect principle will be used (Todinov, 2015) for determining tight bounds of multivariable functions whose partial derivatives maintain the same sign in a specified domain. According to this principle, the coordinates of a corner point corresponding to a global minimum are specified, as follows. If the partial derivative with respect to xi is positive in the domain, the left end ai of the interval of variation of xi is selected as the ith coordinate of the corner point. If the derivative with respect to xi is negative in the domain, the right end bi of the interval of variation of xi is selected as the ith coordinate of the corner point. At the end of this process, a corner point with particular coordinates (e.g., a1, b2 , a3 , a4 ,...., bn ) will be specified. The as-specified corner point corresponds to a global minimum of the function in the specified domain. To find the coordinates of a corner point which correspond to a global maximum, the selection of the coordinates is reversed. If the partial derivative with respect to xi is positive in the domain, the right end bi of the interval of variation of xi is selected as the ith coordinate of the corner point. If the derivative with respect to xi is negative in the domain, the left end ai of the interval is selected as the ith coordinate of the corner point. At the end of this process, a corner point will be specified at which the global maximum is attained. As a result, the time complexity of finding the global extremum is reduced from exponential O(2n ) to linear O(n), which results in a significantly improved performance of the algorithm. Here is a simple illustrating example. Determine the lower bound L and the upper bound U of the function

f ( x, y) = x 3 y3 − (3 / 2) x 2 y2 + 4

(4.15)

where 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. The first partial derivatives are negative ∂∂fx = 3xy 2 ( xy − 1) ≤ 0 , ∂∂fy = 3x 2 y( xy − 1) ≤ 0 and do not change sign in the domain 0 ≤ x ≤ 1; 0 ≤ y ≤ 1. Following the procedure described earlier, the corner point where the global maximum is attained has coordinates (0,0), and = U f= (0, 0) 4 is obtained for the upper bound. The corner point where the global minimum is attained has coordinates (1,1) and for the lower bound = L f= (1, 1) 3.5 is obtained. As a result, the following tight inequalities hold:

3.5 ≤ x 3 y3 − (3 / 2) x 2 y2 + 4 ≤ 4

(4.16)

In the next example, it is required to determine the lower bound L and the upper bound U of the function

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Using Optimisation Methods for Determining Tight Upper and Lower Bounds



f ( x, y) = − x 2 + 4 x − y3 − sin y + cos y − 4

(4.17)

defined in the domain 0.1 ≤ x ≤ π /2, 0.1 ≤ y ≤ π/2 . The  partial derivative ∂∂fx = −2 x + 4 of the function f ( x, y) is with a positive sign in the domain 0.1 ≤ x ≤ π /2; 0.1 ≤ y ≤ π /2 while the partial derivative ∂f 2 ∂y = −3 y − cos y − sin y is with a negative sign in the domain. Following the procedure described earlier, the corner point where the global minimum is attained has coordinates (0.1, π /2 ). Evaluating the global minimum gives 

L = f (0.1, π /2) = −(0.1)2 + 4 × 0.1 − (π /2)3 − sin(π /2) + cos(π /2) − 4 = −8.486 (4.18)

The corner point where the global maximum is attained has coordinates (π /2,0.1) and for the upper bound

U = f (π /2, 0.1) = −(π /2)2 + 4 × (π /2) − 0.13 − sin(0.1) + cos(0.1) − 4 = 0.71

is obtained. As a result, the following inequalities hold:

−8.486 ≤ f ( x, y) = − x 2 + 4 x − y3 − sin y + cos y − 4 ≤ 0.71

An important application of the discussed technique is in improving the performance of heuristic algorithms determining the extrema of multivariable functions f ( x1, x2 , x3 ,..., xn ) in a large domain. A key step of a number of heuristic algorithms is a local search for the maximum/minimum value in the vicinity of a particular point. Such is, for example, the particle swarm optimisation algorithm (Lazinica, 2009; Kiranyaz et al., 2014), which searches in the vicinity of a given point ( x1, x2 ,..., xn) to determine the best value. Since in the small vicinity of the point ( x1, x2 ,..., xn), the partial derivative does not normally change sign, the largest or the smallest value in the vicinity of the specified point can be estimated very well by the method described earlier.

4.3 CONVENTIONS ADOPTED IN PRESENTING THE SIMULATION ALGORITHMS An array is a data structure that contains many data storage locations, each of which holds the same type of data. Each storage location is called an element of the array. The total number of storage locations is called size of the array. The statement ‘a[ ];’ defines an empty array; the statement a[0.1 0.4 0.9] defines an array of three elements with values a[1]=0.1, a[2]=0.4 and a[3]=0.9. The statement y=f(a,b,c) calls the function f (a,b,c) with parameters a, b, c and stores the result returned by the function in the variable y. The  statement a=rand() calls the built-in function rand(), which returns a uniformly distributed random number within the interval (0,1) and stores the

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Risk and Uncertainty Reduction by Using Algebraic Inequalities

random number in the variable a. To generate a random number x within the interval (a, b), the linear transformation x=a+(b-a)*rand() is used (Ross, 1997). As the built-in function rand() returns uniformly distributed numbers in the interval (0,1), the variable x receives random values uniformly distributed within the interval (a, b). A group of statements in braces are executed as a single block:    {     Statement_1;     Statement_2;     ......     Statement_n;    }

The conditional statements permit decisions to be made about which statements will be executed. The course of execution of the algorithm depends on whether a logical expression is true or false. In the next conditional statement, the logical expression can either evaluate to ‘true’ or ‘false’. The group of statements from 1, 2,..., n are executed as a single block, if the logical expression is true. If the logical expression is false, the statements 1, 2,..., n are not executed. if (logical_expression) then    {     Statement_1;     Statement_2;     ............     Statement_n;   }

In the next conditional statement, an else-clause is included. if  (logical_expression) then    {     Statement_1;     Statement_2;     ............     Statement_n;    }  else   {     Alt_Statement_1;     Alt_Statement_2;     ............     Alt_Statement_n;   }

If the logical expression is true, the block of statements Statement_1, Statement_2,...,  Statement_n is executed. If the logical  expression

Using Optimisation Methods for Determining Tight Upper and Lower Bounds

75

is false, the block of alternative statements Alt_Statement_1,  Alt_ Statement_2,..., Alt_Statement_n is executed instead. Repeated calculations in Monte Carlo simulations are organised in for-loops. The for-loop has the structure:   for k=1 to n do    {     Statement_1;     Statement_2;     ......     Statement_n;    }

The for-loop keeps repeating a set of statements enclosed in the body of the loop (defined by the braces in the loop) a number of times, equal to the specified end-value n. After each execution of the statements in the body of the loop, the control variable k is incremented. The number of repetitions (the number of times the loop will run the enclosed statements) is known in advance. If a statement break is executed anywhere in the for-loop, the loop is terminated immediately, before the control variable k reaches the end-value n. In the next forloop, if the condition in the if-statement is true, then Statement_3 and the statement break are executed. After the execution of the break-statement, the loop is exited immediately and the algorithm continues with the statement Statement_n+1;   for k=1 to n do    {     Statement_1;     Statement_2;     if (Condition) then         {          Statement_3;          break;         }     ......     Statement_n;    }     Statement_n+1;

The  use of the statement break saves a significant amount of computation time if there is no need to continue the simulation after a particular logical condition becomes true. The statement switch(m) branches the execution depending on the value of the variable ‘m’ switch (m)   {    case 1: {             Fist block of statements;            }

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Risk and Uncertainty Reduction by Using Algebraic Inequalities

   case 2: {             Second block of statements;            } ..........    case n: {             nth block of statements;            } }

If the value of the variable m is ′1′ then the first block of statements is executed and the rest of the blocks are skipped; if the value of the variable m is ′2′ then the second block of statements is executed and the rest of the blocks of statements are skipped and so on.

4.4 TESTING A CONJECTURED ALGEBRAIC INEQUALITY BY A MONTE CARLO SIMULATION Before attempting to prove a conjectured inequality rigorously, it is useful to confirm it first by a simulation testing. Only if the simulation provides support for the conjecture that the inequality is true is an attempt made to prove the inequality by using some combination of the techniques discussed in Chapter 3. In  some cases, n distinct components with unknown reliabilities  x1, x2 ,..., xn are used for  building two systems with reliabilities f ( x1,..., xn ) and g( x1,..., xn ). Because each of the variables x1, x2 ,..., xn represents reliability, the constraints 0 ≤ x1 ≤ 1; 0 ≤ x2 ≤ 1;...; 0 ≤ xn ≤ 1 hold for these variables. If the variables x1, x2 ,..., xn entering the inequality f ( x1,..., xn ) > g( x1,..., xn ) represent other quantities, they are usually subjected to the more general constraints a1 ≤ x1 ≤ b1; a2 ≤ x2 ≤ b2 ;...; an ≤ xn ≤ bn. To support the statement that the system topology with reliability f ( x1,..., xn ) is always superior to the system topology with reliability g( x1,..., xn ), irrespective of the specific values of the reliabilities x1, x2 ,..., xn of the components, it suffices to test the inequality f ( x1,..., xn ) > g( x1,..., xn ). This can be done by finding the global minimum L of the difference f ( x1,..., xn ) − g( x1,..., xn ) by using analytical or heuristic method for optimising a multivariate function and showing that the global minimum is a positive number ( L > 0 ) . A simple method to test an inequality of the type f ( x1,..., xn ) > g( x1,..., xn ) is to demonstrate that the inequality is not contradicted during a simulation involving various combinations of random values for the variables x1, x2 ,..., xn. This can be done by running an algorithm, the essence of which is repeated sampling from the intervals of variation of each variable xi , substituting the sampled values in the inequality and checking (millions of times) whether f ( x1,..., xn ) > g( x1,..., xn ) holds. Even a single combination of values for which the inequality does not hold disproves the inequality and shows that there is no point in searching for a rigorous proof because a counterexample has been found. If the inequality holds for millions of generated random values for the sampled variables, a strong support is provided for the conjecture that the inequality is true. However, such a support cannot serve as a substitute for proof. The  conjectured inequality must be proved by using analytical reasoning based on methods similar to the ones covered in Chapter 3.

Using Optimisation Methods for Determining Tight Upper and Lower Bounds

77

Here is the pseudo-code of an algorithm for testing a general conjectured inequality f ( x1,..., xn ) > g( x1,..., xn ), where the variables vary in the intervals a1 ≤ x1 ≤ b1; a2 ≤ x2 ≤ b2 ;...; an ≤ xn ≤ bn. Algorithm 4.1 x[];  a=[a1,a2,...,an]; b=[b1,b2,...,bn]; flag=0; for k=1 to num_trials do {   for k=1 to n do  x[k]=a[k]+(b[k]-a[k])*rand();    y1=f(x[1],x[2],...,x[n]);  y2=g(x[1],x[2],...,x[n]);    y=y1-y2;    if(y 0, and c > 0, prove the inequality b4 + c4 a4 + c4 a4 + b4 b + c a + c a + b + + ≥ + + a4 b4 c4 a b c



25. For the real numbers x, y, z varying in the intervals 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and 0 ≤ z ≤ 1, prove the inequality 1 + x 2 (1 − y ) + y 2 (1 − z ) + z 2 (1 − x ) ≤ 3/ 2 1+ x + y + z



26. For the real numbers x, y varying in the intervals 0 ≤ x ≤ π /2, 0 ≤ y ≤ π /2, prove the inequality sin( x + y ) ≤ sin x + sin y



27. For the positive real numbers x and y whose sum is equal to 1 ( x + y = 1), prove the inequality 1/ x 2 + 1/ y 2 ≥ 8



28. If a, b and c are positive real numbers, show that the following inequality (a 4 + 3)(b 4 + 3)(c 4 + 3)(d 4 + 3) ≥ (a + b + c + d )4

holds.

Using Optimisation Methods for Determining Tight Upper and Lower Bounds

83

29. If x, y and z are positive real numbers such that x + y + z = 1, show that the following inequality holds 1 1 1 9 + + ≥ x(3 y + 1) y(3z + 1) z (3x + 1) 2



30. If a ≥ b ≥ c and z ≥ y ≥ x are positive real numbers, prove the inequality  a+b+c  a b c + + ≥ 3  x y z  x + y + z 



4.6  SOLUTIONS TO THE EXERCISES

1. To prove this inequality it suffices to prove



1 + xy −1 > 0 x+y The left side of this inequality is transformed to 1+ xyx +− yx − y = (1− xx)(+1y− y ) . Since 0 < x < 1 and 0 < y < 1, it follows that 1 − x > 0 and 1 − y > 0 ; therefore, the left side of the inequality is positive.



2. From the identity ( x + y + z)2 = x 2 + y2 + z 2 + 2 xy + 2 yz + 2 xz, considering that x + y + z = 0, we get xy + yz + zx = −

Since

x2 + y2 + z2 2

≥ 0 , it follows that − x

2

x 2 + y2 + z 2 2

+ y2 + z2 2

≤ 0, which completes the proof.

3. To prove this inequality, it suffices to prove the equivalent inequality: 1 − 3xy − 3 yz − 3zx ≥ 0 Using the constraint x + y + z = 1 gives ( x + y + z )2 = 1 and, as a result, the left side of the inequality can be presented as



1 − 3 xy − 3 yz − 3zx = ( x + y + z)2 − 3 xy − 3 yz − 3zx = x 2 + y2 + z 2 − xy − yz − zx Since x 2 + y2 + z 2 − xy − yz − zx = (1 / 2)[( x − y)2 + ( y − z)2 + ( z − x )2 ], the left side of the inequality has been presented as a sum of non-negative terms, which completes the proof.



4. a. 1 − ( x + y + z ) + ( xy + yz + zy) − xyz = (1 − x )(1 − y)(1 − z ) Because 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ y ≤ 1, it follows that (1 − x ) ≥ 0, (1 − y) ≥ 0 and (1 − z) ≥ 0. Consequently, the product (1 − x )(1 − y)(1 − z) is non-negative. Equality is attained when any of the numbers x, y, z accepts value equal to 1.

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Risk and Uncertainty Reduction by Using Algebraic Inequalities

b. 1 − ( xy + yz + zx ) + ( x 2 yz + y2 zx + z 2 xy) − x 2 y2 z 2 = (1 − xy)(1 − yz )(1 − zx ) Because 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, it follows that (1 − xy) ≥ 0, (1 − yz ) ≥ 0 and (1 − zx ) ≥ 0. Consequently, the product (1 − xy)(1 − yz )(1 − zx ) is non-­ negative. Equality is attained when any of the terms xy, yz, zx are equal to 1. This happens when any of the following conditions are fulfilled = ( x 1,= y 1, z ≠ 1) or ( y = 1, z = 1, x ≠ 1) or ( z = 1, x = 1, y ≠ 1) or (= x 1= , y 1, z = 1).

5. The inequality will be proved by using induction. For n = 1, the inequality is true because 1 − x1 = 1 − x1. For n = 2, the lefthand side of the inequality becomes 2 ∏ i =1 (1 − xi ) = (1 − x1 )(1 − x2 ) = 1 − x1 − x2 + x1x2 . Since x1 x2 ≥ 0, dropping this term yields, 1 − x1 − x2 + x1 x2 ≥ 1 − x1 − x2 . The  right-hand part of the last inequality is 1 − ∑ i2=1 xi ; therefore, the inequality is also true for n = 2. Assume that the inequality is true for n = k ≥ 2. In other words, assume that k

∏



k

(1 − xi ) ≥ 1 −

∑x

(4.21)

i

i =1

i =1

holds. It will be shown that if the x1, x2 ,..., x k , xk +1, are real numbers in the interval [0,1], then k +1

∏



k +1

(1 − xi ) ≥ 1 −

∑x i

i =1

i =1

Since 0 ≤ x k +1 ≤ 1, if both sides of inequality (4.21) are multiplied by 1 − x k +1, the direction of the inequality will be preserved: k

∏

(1 − x k +1 )

i =1

 (1 − xi ) ≥ (1 − x k +1 )  1 −  



k

∑ x  i

i =1



Expanding both sides gives k +1



∏ i =1

k

(1 − xi ) ≥ 1 − x k +1 −

∑ i =1

 xi +  x k +1  

k

∑ i =1

 xi   

Since the term ( xk +1 ∑ ik=1 xi ) is non-negative, dropping this term from the right k +1 k +1 side can only strengthen the inequality; therefore, ∏ i =1 (1 − xi ) ≥ 1 −∏ i =1 xi holds. This completes the induction step. This, with the proved trivial case n = 2 n n means that the inequality ∏ i =1 (1 − xi ) ≥ 1 −∏ i =1 xi is valid for any integer n.



6. This inequality can be proved by using the segmentation method. Applying the AM-GM inequality six times yields the following relationships: x3 + x3 + y3 3 3 3 3 y3 + y3 + x3 ≥ ≥ x x y = x 2 y, 3 3

3

y3 y3x3 = y2 x



Using Optimisation Methods for Determining Tight Upper and Lower Bounds



y3 + y3 + z 3 3 3 3 3 z 3 + z 3 + y3 3 3 3 3 ≥ z z y = z2 y ≥ y y z = y2 z, 3 3



z3 + z3 + x 3 3 3 3 3 x 3 + x 3 + z3 3 3 3 3 ≥ x x z = x 2z ≥ z z x = z 2 x, 3 3 Adding the six inequalities gives the original inequality and completes the proof.



7. The  inequality can also be proved by using the segmentation method. Applying the AM-GM inequality gives



a + b ≥ 2 ab



b + c ≥ 2 bc



c + a ≥ 2 ca Because a, b and c are positive, a + b, b + c and c + a are also positive and multiplying the three inequalities will not alter the direction of the resultant inequality. Multiplying the three inequalities yields (a + b)(b + c)(c + a) ≥ 8abc.



8. Let us define the weights t1 = a / (a + b + c), t2 = b / (a + b + c) and t3 = c / (a + b + c), which add up to 1 ( t1 + t2 + t3 = 1). The  next inequality follows from the weighted AM-GM inequality:



t1b + t2c + t3a ≥ bt1 c t2 at3 Substituting back t1 = a / (a + b + c), t2 = b / (a + b + c), t3 = c / (a + b + c) and considering that a + b + c = 1, yields ab + bc + ca ≥ b ac b ac .





9. Consider the two sequences ( x, y, z ) and Cauchy-Schwarz inequality results in

(x

x+y y+z z

)

2

(

)

x , y , z . Applying the

≤ ( x 2 + y 2 + z 2 )( x + y + z )

Considering that x 2 + y2 + z 2 = 1 and taking square roots from both sides of the inequality (which are positive numbers) yields x x + y y + z z ≤ x + y + z . 10. The inequality is homogeneous and scaling of the variables can be made a := λ a, b := λ b , c := λ c such that (λ a) 4 + (λ b) 4 + (λ c) 4 = 1. Substituting the scaled variables into the inequality will not alter it. As a result, an equivalent inequality is obtained:

( a 5 + b 5 + c 5 ) 4 ≤ ( a 4 + b 4 + c 4 )5 with the constraint a 4 + b 4 + c 4 = 1. Because of the constraint a 4 + b 4 + c 4 = 1, for the variables a, b and c, the following relationships hold: 0 < a < 1; 0 < b < 1; 0 < c < 1. As a result, a5 ≤ a 4 , b5 ≤ b 4 and c5 ≤ c 4.

85

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Risk and Uncertainty Reduction by Using Algebraic Inequalities

Adding these three inequalities gives a5 + b5 + c5 ≤ a 4 + b 4 + c 4 = 1. Therefore, (a5 + b5 + c 5 )4 ≤ 1. Since (a 4 + b 4 + c 4 )5 = 1, (a5 + b5 + c 5 )4 ≤ 1 = (a 4 + b 4 + c 4 )5 which proves the original inequality. 11. Because of homogeneity, scaling of the variables can always be made: a := λ a, b := λ b , c := λ c such that a + b + c = 1. Substituting the scaled variables into the inequality will not alter it. As a result, an equivalent inequality is obtained a b c + + ≥2 b+c a+c a+b



with the constraint a + b + c = 1. From the constraint, it follows that b + c = 1 − a; a + c = 1 − b and a + b = 1 − c. From the constraint, it also follows that a < 1, b < 1 and c < 1. The original inequality transforms into the equivalent inequality a b c + + ≥2 1− a 1− b 1− c



In Chapter 2, it was shown that Therefore,

x 1− x

≥ 2 x , for 0 < x < 1.

a b c + + ≥ 2a + 2b + 2c = 2(a + b + c) = 2 1− a 1− b 1− c



This completes the proof of the original inequality. 12. The left side of the inequality can be presented as follows: a b c a+b+c a+b+c a+b+c + + = + + −3 b+c a+c a+b b+c a+c a+b 1  1  1 = (a + b + c)  + + −3  b+c a+c a+b 1 1   1 = (1 / 2) ( (b + c) + (a + c) + (a + b) )  + + −3 b + c a + c a + b  

Since



( (b + c) + (a + c) + (a + b) )  b + c + a + c + a + b  ≥ 32 = 9



a b c + + ≥ (1 / 2) × 9 − 3 = 3 / 2 b+c a+c a+b

1



which proves the Nesbitt inequality.

1

1





Using Optimisation Methods for Determining Tight Upper and Lower Bounds

13. The inequality can be proved by applying the segmentation technique. Note that the terms x 3 y, y3 z, z 3 x in the right-hand side can be obtained easily by applying the AM-GM inequality to x 4 + x 4 + x 4 + y 4 , y 4 + y 4 + y 4 + z 4 and z 4 + z 4 + z 4 + x 4, correspondingly:

x 4 + x 4 + x 4 + y 4 ≥ 4 4 x 4 x 4 x 4 y4 = 4 x 3 y



y 4 + y 4 + y 4 + z 4 ≥ 4 4 y 4 y 4 y 4 z 4 = 4 y3 z



z 4 + z 4 + z 4 + x 4 ≥ 4 4 z 4 z 4 z 4 x 4 = 4z3 x Adding the last three inequalities results in 4( x 4 + y 4 + z 4 ) ≥ 4( x 3 y + y3 z + z 3 x )



which proves the original inequality. 14. Consider the sequences {a} = [ 4, 0, 0] and {b} = [3, 1, 0]. Clearly, the sequence {a} majorises the sequence {b} because the following conditions are fulfilled: 4 > 3 and 4 + 0 ≥ 3 + 1 and 4 + 0 + 0 = 3 + 1 + 0. According to the Muir­head inequality (2.40): 2 × ( x 4 + y 4 + z 4 ) ≥ x 3 y + x 3 z + y3 x + y3 z + z 3 x + z 3 y



which completes the proof. 15. Denote the function to be maximised by f ( x, y, z) = xy + yz + zx and the constraint by g( x, y, z) = x + y + z −1 (g( x, y, z) = 0). The gradients of the functions f ( x, y, z) and g( x, y, z) are: grad f ( x, y, z ) = yi + zj + xk

and

grad g( x, y, z) = i + j + k where i, j and k are the unit vectors along the x, y and z axis. From grad f ( x, y, z ) = λ grad g ( x, y, z ), and the constraint x + y + z = 1, the following system of equations is obtained:



y = λ ; z = λ ; x = λ ; x + y + z = 1 Solving this system of equations  yields x= y= z = 1/3 as a critical point. The  check shows that the critical point is a global maximum. The tight upper bound M in the inequality xy + yz + zx ≤ M is therefore M = 1/3.

87

88

Risk and Uncertainty Reduction by Using Algebraic Inequalities

16. From the Cauchy-Schwarz inequality ( x 2 y + y2 z + z 2 x ) ≤ ( x 4 + y 4 + z 4 )( y2 + z 2 + x 2 )



Considering the constraint x 2 + y2 + z 2 = 1, the inequality transforms into x 2 y + y2 z + z 2 x ≤ x 4 + y 4 + z 4



The  function f ( x, y, z) = x 4 + y 4 + z 4 will be maximised under the con2 2 2 straint g( x, y, z) ≡ x + y + z − 1 = 0. The gradients of the functions f ( x, y, z) and g( x, y, z) are grad f ( x, y, z ) = 4 x 3i + 4 y3 j + 4 z 3k

and

grad g( x, y, z) = 2 x i + 2 yj + 2 zk



where i, j and k are the unit vectors along the x, y and z axis. From grad f ( x, y, z ) = λ grad g ( x, y, z ,) and the constraint x 2 + y2 + z 2 = 1, the following system of equations is obtained: x2 = λ / 2 y2 = λ / 2

z2 = λ / 2



x 2 + y2 + z 2 = 1 Solving this system of equations  yields x= y= z = 1 / 3 as a critical point. The  check shows that the critical point is a global maximum. The  tight upper limit M in the inequality x 4 + y 4 + z 4 ≤ M is therefore M = (1 / 3 )4 + (1 / 3 )4 + (1 / 3 )4 = 1 / 3 . Consequently, x 2 y + y 2 z + z 2 x ≤ x 4 + y 4 + z 4 ≤ 1 / 3 = 1 / 3 which proves the original inequality.

17. a. Because of the symmetry, without loss of generality, a ≥ b ≥ c can be assumed. From the rearrangement inequality, the following inequality follows:



a  a  a

b b b

c a   3 3 3 c  = a + b + c ≥ b  c c 

b c a

c  a  = 3abc b 

because in the left matrix, the rows are sorted in the same way while in the right matrix, they are not.

89

Using Optimisation Methods for Determining Tight Upper and Lower Bounds

b. From the rearrangement inequality, the following inequality follows



a  a a  a

b b b b

c a   c a = a4 + b4 + c4 ≥  b c   c c

b b c a

c  c = a2 bc + b2ca + c 2 ab a  b

because in the left matrix, the rows are sorted in the same way while in the second matrix, they are not. 18. Without loss of generality, suppose that a ≤ b ≤ c ≤ d . Then 1 / a ≥ 1 / b ≥ 1 / c ≥ 1 / d



According to the rearrangement inequality: 1 / a  1 / a 1 / a 1 / a  ≥ 1 / b 1 / c



1/ b 1/ b 1/ b 1/ b 1/ c 1/ d

1/ d  1 / d  = 1 / a3 + 1 / b3 + 1 / c 3 + 1 / d 3 1 / d 

1/ c 1/ c 1/ c 1/ c 1/ d 1/ a

1/ d 1 1 1 1  + + + 1 / a = abc bcd cda dab 1 / b 



Since

1 1 1 1 d a b c a+b+c+d + + + = + + + = abc bcd cda dab abcd abcd abcd abcd abcd the original inequality has been proved. This inequality can also be proved by using the Hölder inequality. Consider the three sequences (1/b3 ,1/c3 ,1/d 3 ,1/a3 ), (1/c3 ,1/d 3 ,1/a3 ,1/b3 ) and (1/d 3 ,1/a3 ,1/b3 ,1/c3 ). Applying the Hölder inequality with weights λ1 = λ2 = λ3 = 1 / 3 ∑ i3=1 λi = 1 gives

(

)

(1/b3 + 1/c3 + 1/d 3 + 1/a3 )(1/c3 + 1/d 3 + 1/a3 + 1/b3 )(1/d 3 + 1/a3 + 1/b3 +11/c3 )



  1 1/ 3  1 1/ 3  1 1/ 3  1 1/ 3  ≥  3 3 3  +  3 3 3  +  3 3 3  +  3 3 3    b c d  a c d  a b d   a b c   

3



which can be rewritten as 3



3

1 1 1   a+b+c+d   1 (1 / a3 + 1 / b3 + 1 / c3 + 1 / d 3 )3 ≥  + + +  =  abcd  bcd acd abd abc    Raising both sides to the power 1/3 completes the proof of the original inequality.

90

Risk and Uncertainty Reduction by Using Algebraic Inequalities

19. Note that the sequence (a / b, b / c , c / a ) and the identical sequence (a / b, b / c , c / a ) are similarly sorted while the sequence (a / b, b / c , c / a ) and the sequence (b / c , c / a , a / b) are not. According to the rearrangement inequality



a a b b c c a b b c c a a b c ⋅ + ⋅ + ⋅ ≥ ⋅ + ⋅ + ⋅ = + + b b c c a a b c c a a b c a b which completes the proof.

20. Consider the inequality ∑ in=1 ai bi ≥∑ in=1 ai qi . This inequality follows from the rearrangement inequality because the sequences a1, a2 ,..., an and b1, b2 ,..., bn are ordered in a similar way while the sequences a1, a2 ,..., an and q1, q2 ,..., qn are not. Multiplying both sides of the inequality by ‘–2’ reverses the sign of the inequality and the result is n

−

∑

n

2ai bi ≤

i =1

∑ 2a q i i

i =1

Note that ∑ in=1 bi2 = ∑ in=1 qi2 . Adding to the left side of the last inequality the term ∑ in=1 ai2 + ∑ in=1 bi2 and to the right side of the last inequality the identical term ∑ in=1 ai2 + ∑ in=1 qi2 results in the inequality n

n

∑ ∑ ai2 −



i =1

n

2ai bi +

i =1

∑

n

bi2 ≤

i =1

n

∑ ∑ ai2 −

i =1

n

2ai qi +

i =1

∑q 2 i

i =1

which is equivalent to the inequality n



∑

n

(ai − bi )2 ≤

i =1

∑ (a − q ) i

i

2

i =1

This completes the proof of the original inequality. 21. Applying the mean value theorem yields



arctan( x2 ) − arctan( x1 ) 1 = [arctan( x )]′x = ξ = x2 − x1 1+ξ 2 where ξ ∈( x1, x2 ). From the last equation it follows



arctan x2 − arctan x1 = ( x2 − x1 ) ×

1 1+ ξ 2

Because 1 2 ≤ 1, arctan x2 − arctan x1 ≤ ( x2 − x1 ). 1+ξ

91

Using Optimisation Methods for Determining Tight Upper and Lower Bounds

22. f ( x ) ≡ 3 x − x ln( x ) − x 5 − x 3 + 27 ≥ C is a concave function because it is a sum of four concave functions: f1 ( x ) = 3 x , f2 ( x ) = − x ln( x ) (the second derivative of f2 ( x ) is negative in the inteval [1,2]) f3 ( x ) = − x 5 and f4 ( x ) = 27 − x 3. Consequently, the absolute minimum of f ( x ) is obtained at the ends of the interval  [1,2]. There  are two values to be compared f (1) = 26.73 and f (2) = −11.94. The global minimum is attained at x = 2 and the constant C is equal to –11.94. 23. The  function f (u) = u is a concave function for u > 0 because f ''(u) = −1/( 4u u ) < 0. According to the Jensen inequality, for the concave function f (u), the following inequality holds: 1 1  x+ y+z  1 f ≥ f ( x ) + f ( y) + f ( z )  3 3 3  3 

or

( x + y + z) / 3 ≥ (1 / 3) x + (1 / 3) y + (1 / 3) z which transforms into the inequality



x + y + z ≥ ( 3 / 3)

(

x+ y+ z

)

24. To prove this inequality, a combination of the method of symmetry and the method of derivatives will be used. Because of the symmetry, without loss of generality, a ≥ b ≥ c can be assumed. Consider the function



f ( x) =

bx + cx ax + cx ax + bx + + ax bx cx

defined for × > = 0. The first derivative of the function is df ( x ) (b x ln b + c x ln c)a x − a x ln a(b x + c x ) = dx a2 x



+

( a x ln a + c x ln c)b x − b x ln b( a x + c x ) b2 x

+

( a x ln a + b x ln b)c x − c x ln c( a x + b x ) c2 x

After some algebra, the derivative becomes



2x x x 2x 2x 2x x x 2x 2x df ( x )  [ln a − ln b]c a b (a − b ) + [ln a − ln c]b a c (a − c )  =   dx + [ln b − ln c]a 2 x b x c x (b 2 x − c 2 x )  

(a

2x 2x 2x

b c

)

92

Risk and Uncertainty Reduction by Using Algebraic Inequalities

As can be seen, dfdx( x ) ≥ 0; therefore, the function f ( x ) is non-decreasing in the domain × > = 0. Consequently, f (4) ≥ f (1), which proves the original inequality. 25. It will be shown that the function f ( x, y, z ) = 1+ x +1 y + z + x 2 (1 − y ) + y 2 (1 − z ) + z 2 (1 − x ) is convex. Indeed, the function is convex with respect to each of the variables x, y and z because

∂2 f 2 = + 2(1 − y) > 0 2 ∂x (1 + x + y + z )3



∂2 f 2 = + 2(1 − z ) > 0 ∂y2 (1 + x + y + z )3



∂2 f 2 = + 2(1 − x ) > 0 ∂z 2 (1 + x + y + z )3 The function f ( x, y, z) is continuous on the closed domain 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 and 0 ≤ z ≤ 1; therefore, according to the extreme value theorem from calculus, the function must attain its maximum and minimum in that domain. Because the function is convex and continuous on the closed domain 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and 0 ≤ z ≤ 1, it assumes its maximum value at the boundaries of the domain. The boundary points are:



{0, 0, 0},{0, 0, 1},{0, 1, 0},{0, 1, 1},{1, 0, 0},{1, 0, 1},{1, 1, 0},{1,, 1, 1} and the maximum value is attained at



f= (1,0,0) f= (0,1,0) f (0,0,1) = 3/2 hence the inequality holds.

26. The inequality follows from the properties of sub-additive functions. The  function f (u) = sin(u) is a concave function in the interval [0, π / 2] because the second derivative f ''(u) = − sin(u) is negative in this interval. If a function is concave, the function is sub-additive, therefore sin( x + y) ≤ sin x + sin y, from the basic property of sub-additive functions. Since sin(0) = 0, the equality is attained for x = 0 or y = 0. 27. Because x + y = 1, note that

1 / x 2 + 1 / y2 = ( x + y)( x + y)(1 / x 2 + 1 / y2 ) Applying Hölder’s inequality gives 3

  x ⋅ x 1/ 3  y ⋅ y 1/ 3  ( x + y)( x + y)(1 / x + 1 / y ) ≥   2  +  2   = 23 = 8  x   y    2



which completes the proof.

2

93

Using Optimisation Methods for Determining Tight Upper and Lower Bounds 4 28. Consider the four sequences (a 4 , 1, 1, 1), (1, b , 1, 1), (1, 1, c 4 , 1) and (1, 1, 1, d 4 ) Taking λ1 = λ2 = λ3 = λ4 = 1 / 4 ( λ1 + λ2 + λ3 + λ4 = 1) and applying the Hölder’s inequality to the three sequences gives

( a 4 + 14 + 14 + 14 )(14 + b 4 + 14 + 14 )(14 + 14 + c 4 + 14 )(14 + 14 + 14 + d 4 )

≥ ( a ⋅1 ⋅11 ⋅1 + 1 ⋅ b ⋅1 ⋅1 + 1 ⋅1 ⋅ c ⋅1 + 1 ⋅1 ⋅1 ⋅ d ) 4 = ( a + b + c + d ) 4



which completes the proof of the inequality. 29. Consider the three sequences ( x, y, z), (3 y + 1, 3z + 1, 3 x + 1) and 1 1 . Applying the Hölder inequality gives y ( 3 z +1) , z ( 3 x +1)

)

(

1 x ( 3 y +1)

,

  1 1 1 ( x + y + z )(3 y + 1 + 3z + 1 + 3x + 1)  + + ≥  x(3 y + 1) y(3z + 1) z(3 x + 1) 

(11/ 3 + 11/ 3 + 11/ 3 )3 = 33



Since x + y + z = 1 and 3 y + 1 + 3z + 1 + 3 x + 1 = 6, dividing both sides by 6 gives



  1 1 1 + +  ≥9/2  x (3 y + 1) y(3z + 1) z(3 x + 1)  which completes the proof.

30. Consider the similarly ordered sequences a ≥ b ≥ c and 1 / x ≥ 1 / y ≥ 1 / z. Applying the Chebyshev inequality yields



a b c 3  + +  ≥ (a + b + c)(1 / x + 1 / y + 1 / z) x y z From the AM-GM inequality 3 (1 / x + 1 / y + 1 / z ) ≥ 3 xyz , therefore



 a b c  3(a + b + c) 3 + +  ≥ 3 xyz x y z From the AM-GM inequality 3 xyz ≤



x+ y+ z 3

, therefore

 a b c  9(a + b + c) 3 + +  ≥  x y z  ( x + y + z) which after dividing both sides by 3 completes the proof.

5

Ranking the Reliabilities of Systems and Processes by Using Inequalities

5.1 IMPROVING RELIABILITY AND REDUCING RISK BY PROVING AN ABSTRACT INEQUALITY DERIVED FROM THE REAL PHYSICAL SYSTEM OR PROCESS An important way of using inequalities to improve reliability and reduce risk is to start with the real system or a process and derive and prove an abstract inequality. This process includes several basic steps (Figure 5.1): (i) detailed analysis of the system (e.g., by using reliability theory); (ii) conjecturing an inequality about the competing alternatives or an inequality related to the bounds of a risk-critical parameter; (iii) testing the conjectured inequality by using Monte Carlo simulation and (iv) proving the conjectured inequality rigorously. This generic strategy can be followed, for example in comparing the reliabilities of competing systems. It starts with building the functional diagram of the system, creating the reliability network for the system, deriving expressions for the system reliability of the competing alternatives, conjecturing an inequality, testing the conjectured inequality by using Monte Carlo simulation and finishing with a rigorous proof by using some combination of the analytical techniques for proving inequalities (Figure 5.2).

5.2 USING ALGEBRAIC INEQUALITIES FOR RANKING SYSTEMS WHOSE COMPONENT RELIABILITIES ARE UNKNOWN Often, the reliabilities of the components building the system are unknown. The epistemic uncertainty associated with the reliabilities of the components building the system translates into epistemic uncertainty related to which system is superior. Algebraic inequalities can eliminate the uncertainty about which is the superior system or process. The first approach related to using inequalities for reliability improvement and risk reduction can be demonstrated with comparing the reliabilities of competing systems. It starts with building the functional diagram of the system, creating the reliability network for the system, deriving expressions for the system reliability of the competing alternatives, conjecturing inequalities ranking the competing alternatives, testing the conjectured inequalities and finishing with rigorous proofs based on some of the analytical techniques for proving inequalities.

95

96

Risk and Uncertainty Reduction by Using Algebraic Inequalities

FIGURE 5.1  A generic strategy for improving reliability and reducing risk by proving an abstract inequality derived from a real physical system/process.

FIGURE 5.2  Improving reliability and reducing risk by comparing the reliabilities of competing systems.

For two competing systems (a) and (b) built on components whose reliabilities are unknown, the steps which lead to establishing the system with intrinsically superior reliability can be summarised as follows. • For each of the competing systems, build the reliability network from its functional diagram. • By using methods from system reliability analysis, determine the system reliabilities Ra and Rb of the systems or the probabilities of system failure Fa and Fb. • Subtract the reliabilities of the competing systems or the probabilities of system failure and test and prove any of the inequalities: Ra − Rb > 0, Ra − Rb < 0, Fa − Fb > 0, Fa − Fb < 0. • Select the system with the superior reliability or the system with the smaller probability of failure.

5.2.1 Reliability of Systems with Components Logically Arranged in Series and Parallel This section covers the basics of evaluating the reliability of systems with components logically arranged in series and parallel.

Ranking the Reliabilities of Systems and Processes by Using Inequalities

97

FIGURE 5.3  A system with components (a) logically arranged in series and (b) logically arranged in parallel.

Consider a system including n independently working components. Let S denote the event ‘the system is in working state at the end of a specified time interval’ and Ck (k = 1, 2,..., n) denote the events ‘component k is in working state at the end of the specified time interval’. For components logically arranged in series (Figure 5.3a) the system is in working state at the end of the specified time interval only if all components are in working state at the end of the time interval. Reliability is the ability of an entity to work without failure for a specified time interval, under specified conditions and environment. The ability to work without failure within the specified time interval is measured by the probability of working without failure during the specified time interval. According to the reliability theory (Bazovsky, 1961), the probability of system success (system in working state at the end of the specified time interval) is a product of the probabilities that the components will be in working state at the end of the specified time interval:

P(S) = P(C1 ) × P(C2 ) × ... × P(Cn )

(5.1)

Denoting by R the probability P(S) that the system will be in working state at the end of the specified time interval and by rk = P(Ck ) the probability that the kth component will be in working state at the end of the specified time interval, equation (5.1) becomes

R = r1 × r2 × ... × rn

(5.2)

In equation (5.2), R will be referred to as the reliability of the system and rk as the reliability of the kth component related to the specified time interval. Now consider independently working components logically arranged in parallel (Figure  5.3b). According to the system reliability theory (Bazovsky, 1961; Hoyland and Rausand, 1994), the probability of system success (system in working state at the end of the specified time interval) is equal to the probability that at least a single component will be in working state at the end of the specified time interval. The event ‘at least a single component will be in working state at the end of the specified time interval’ and the event ‘none of the components will be in working state at the end of the specified time interval’ are complementary events. From

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probability theory (DeGroot, 1989), the probabilities of complementary events add up to unity. Therefore, the probability that at least a single component will be in working state at the end of the specified time interval can be evaluated by subtracting from unity the probability that none of the components will be in working state at the end of the specified time interval. The advantage offered by this inverse-thinking approach is that the probability that none of the components will be in working state at the end of the specified time interval is very easy to calculate. Indeed, if r1, r2 ,..., rn denote the reliabilities of the separate components (the probabilities that the components will be in working state at the end of the specified time interval), the probability P(S ) that none of the components will be in working state at the end of the specified time interval (the probability of system failure) is given by

P(S ) = (1 − r1 )(1 − r2 )...(1 − rn )

(5.3)

Consequently, the probability that the system will be in working state at the end of the specified time interval (the probability of system’s success) is given by

P(S) = 1 − P(S ) = 1 − (1 − r1 )(1 − r2 )...(1 − rn )

(5.4)

Note that for a logical arrangement of the components in series, the system reliability is a product of the reliabilities of the components, while for a logical arrangement of the components in parallel, the probability of system failure is a product of the probabilities of failure of the components. A system with components logically arranged in series and parallel can be reduced in complexity in stages, as shown in Figure 5.4. In the first stage, the components in Figure 5.4a, logically arranged in series, with reliabilities R1 and R2, are reduced to an equivalent component with reliability R12 = R1 R2. The components logically arranged in parallel with reliabilities R4 and R5 are reduced to an equivalent component with reliability R45 = 1 − (1 − R4 )(1 − R5 ) and the components in parallel with reliabilities R6 and R7 are reduced to an equivalent component with reliability R67 = 1 − (1 − R6 )(1 − R7 ). The resultant equivalent reliability network is shown in Figure 5.4b. In  the second stage, the components in parallel, with reliabilities R12 and R3 in Figure  5.4b, are reduced to an equivalent component with reliability R123 = 1 − (1 − R12 )(1 − R3 ) and the components in series with reliabilities R67 and R8 are reduced to an equivalent component with reliability R678 = R67 × R8. The resultant equivalent network is shown in Figure 5.4c. Next, the reliability network in Figure 5.4c is further simplified by reducing the equivalent components with reliabilities R123 and R45 to a single equivalent component with reliability R12345 = R123 × R45 . The final result is the reliability network in Figure 5.4d, whose reliability is R = 1 − (1 − R12345 )(1 − R678 ). It needs to be pointed out that there is a critical difference between a physical arrangement of components in a system and their logical arrangement. Thus, the valves in Figure 5.5a are physically arranged in series. If initially both valves are open, and the production fluid passes through the pipeline, with respect to stopping the production fluid through the pipeline, the valves are logically arranged in parallel (Figure 5.5b).

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FIGURE 5.4  Network reduction method for determining the reliability of a system including components logically arranged in series and parallel; (a), (b), (c) and (d) – stages of the network reduction method.

FIGURE 5.5  Difference between a physical arrangement (diagrams [a] and [c]) and logical arrangement (diagrams [b] and [d]) for valves on a pipeline with respect to the function ‘stopping the production fluid on command’.

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Risk and Uncertainty Reduction by Using Algebraic Inequalities

This is because at least one of the valves is necessary to work on command for the flow of production fluid through the pipeline to be stopped. In Figure 5.5c, the valves are physically arranged in parallel. If initially both valves are open, the production fluid passes through both valves. With respect to stopping the production fluid through the pipeline, the valves are now logically arranged in series (Figure 5.5d). This is because both valves must stop the flow in their branches for the flow of production fluid through the pipeline to be stopped. Methods for determining the reliability of systems with reliability networks different from networks with seriesparallel arrangement of the components are discussed in detail in (Todinov, 2016).

5.3 USING INEQUALITIES TO RANK SYSTEMS WITH THE SAME TOPOLOGY AND DIFFERENT COMPONENT ARRANGEMENTS Consider the functional diagrams of three systems built with pipes and four valves (A, B, C and D) with different ages (Figure 5.6). Valve A is a new valve, followed by valves B and C with intermediate age. Valve D is an old valve. The reliabilities of the valves are denoted by a, b, c and d. With respect to the function ‘valve closure on demand’, depending on their age, the reliabilities of the valves can be ranked as follows: a > b > c > d . The valves are working independently from one another, and initially, all of them are open. The  question of interest is which system is more reliable with respect to the function ‘stopping the fluid flow in the pipeline on command’. The signal for closing is issued to all valves simultaneously. It  has been conjectured that the arrangement in Figure  5.6c is superior to the arrangements in Figure 5.6a and b. The reliability networks (reliability block diagrams) of the systems with respect to the function ‘stopping the fluid in the pipeline on command’ are given in Figure 5.7. The reliability networks show the logical arrangement of the valves with respect to the function ‘stopping the fluid in the pipeline on command’.

FIGURE  5.6  Three possible arrangements involving four different types of valves: (a) valves A and B in the same branch and the rest of the valves in the other branch; (b) valves A and C in the same branch and the rest of the valves in the other branch; (c) valves A and D in the same branch and the rest of the valves in the other branch.

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FIGURE  5.7  Reliability networks (a), (b) and (c) corresponding to the three systems in Figure 5.6a–c with respect to the function ‘stopping the fluid in the pipeline on command’.

The reliabilities of systems (a), (b) and (c) in Figure 5.7 are given by Ra (a, b, c, d ) = [1 − (1 − a)(1 − b)] × [1 − (1 − c)(1 − d )] = (a + b − ab)(c + d − cd ) (5.5) Rb (a, b, c, d ) = [1 − (1 − a)(1 − c)] × [1 − (1 − b)(1 − d )] = (a + c − ac)(b + d − bd ) (5.6) Rc (a, b, c, d ) = [1 − (1 − a)(1 − d )] × [1 − (1 − b)(1 − c)] = (a + d − ad )(b + c − bc) (5.7) Proving that the configuration in Figure 5.7c is superior reduces to proving the inequalities

Rc (a, b, c, d ) > Ra (a, b, c, d )



Rc (a, b, c, d ) > Rb (a, b, c, d )

which are equivalent to the inequalities

(a + d − ad )(b + c − bc) − (a + b − ab)(c + d − cd ) > 0

(5.8)



(a + d − ad )(b + c − bc) − (a + c − ac)(b + d − bd ) > 0

(5.9)

Manipulating the left side of inequality (5.8) results in

(a + d − ad )(b + c − bc) − (a + b − ab)(c + d − cd )

= ab + dc − ad − bc = a(b − d ) − c(b − d ) = (a − c)(b − d )

Considering that a > b > c > d , it follows that (a − c)(b − d ) > 0. Therefore, the configuration in Figure 5.7c is more reliable than the configuration in Figure 5.7a. Manipulating the left-hand side of inequality (5.9) results in

(a + d − ad )(b + c − bc) − (a + c − ac)(b + d − bd )

= ac + db − ad − cb = a(c − d ) − b(c − d ) = (a − b)(c − d )

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Risk and Uncertainty Reduction by Using Algebraic Inequalities

FIGURE 5.8  The same system topology, built with new valves A and old valves B, in four distinct permutations (a), (b), (c) and (d).

Considering that a > b > c > d , it follows that (a − b)(c − d ) > 0. Therefore, the configuration in Figure 5.7c is more reliable than the configuration in Figure 5.7b. Consequently, the configuration in Figure 5.7c is the most reliable configuration. Consider now the functional diagrams of four possible configurations built with pipes and valves of the same type: new valves A and old valves B (Figure 5.8). With respect to the function ‘valve closure on command’, a new valve A is more reliable than an old valve B. If the reliabilities of the valves are denoted by a and b, the inequality a > b holds. The valves are working independently from one another and all of them are initially open. The  question of interest is which configuration is most reliable with respect to the function ‘stopping the fluid in the pipeline on command’. The signal for closure is issued to all valves simultaneously. The reliability block diagrams of the systems with respect to the function ‘stopping the fluid in the pipeline’, are given in Figure 5.9. The reliabilities of the systems in Figure 5.9a−d are given by

Ra (a, b) = [1 − (1 − a2 )(1 − ab)]b2 = ab2 (a + b − a2 b)

(5.10)



Rb ( a, b) = [1 − (1 − ab)(1 − b2 )]a2 = a2b( a + b − ab2 )

(5.11)



Rc (a, b) = [1 − (1 − ab)(1 − ab)]ab = a2 b2 (2 − ab)

(5.12)



Rd (a, b) = [1 − (1 − a2 )(1 − b2 )]ab = ab(a2 + b2 − a2 b2 )

(5.13)

Ranking the Reliabilities of Systems and Processes by Using Inequalities

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FIGURE  5.9  Reliability networks (a), (b), (c) and (d) corresponding to the systems in Figure 5.8a–d, with respect to the function ‘stopping the fluid in the pipeline on command’.

It is conjectured that the system in Figure 5.9b is the most reliable system. For the differences of the reliability of configuration (Figure 5.9b) and the reliabilities of the rest of the configurations, the following relationships hold:

Rb (a, b) − Ra (a, b) = ab(a2 − b2 ) > 0

(5.14)



Rb (a, b) − Rc (a, b) = a2 b(a − b) > 0

(5.15)



Rb (a, b) − Rd (a, b) = ab2 (a − b) > 0

(5.16)

The configuration in Figure 5.9b is indeed characterised by the highest reliability. In the examples considered in this section, the algebraic inequalities helped to reveal the intrinsic reliability of the competing design solutions and rank the systems in terms of reliability in the presence of significant uncertainty related to the reliabilities of their building parts.

5.4 USING INEQUALITIES TO RANK SYSTEMS WITH DIFFERENT TOPOLOGIES BUILT WITH THE SAME TYPES OF COMPONENTS Two systems with different topologies including the same types of valves (denoted by X, Y and Z) are shown in Figure 5.10 whose reliabilities (denoted by x, y and z) are unknown. The  valves are working independently from one another and all of them are initially open. The  question of interest is which system is more reliable with respect to the function ‘stopping the flow of fluid in the pipeline’. The signal for closing is sent to all valves simultaneously. The reliability networks of the systems from Figure 5.10 are shown in Figure 5.11. The  reliability values x, y and z characterising the separate valves are unknown. The only available information about the reliabilities of the valves are the obvious constraints: 0 < x < 1; 0 < y < 1; 0 < z < 1.

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Risk and Uncertainty Reduction by Using Algebraic Inequalities

FIGURE  5.10  Two competing systems (a) and (b) with different topology, built with the same type and number of components.

FIGURE  5.11  The  reliability networks (a) and (b), corresponding to the systems from Figure 5.10a and b.

Expressing the probabilities of failure characterising the competing systems as a function of the unknown reliabilities of the valves yields

Fa ( x, y, z ) = (1 − x 2 )(1 − y2 )(1 − z 2 ) and Fb ( x, y, z ) = (1 − xyz )2

Ranking the system reliabilities consists of proving Fa ( x, y, z ) − Fb ( x, y, z ) < 0 or Fa ( x, y, z ) − Fb ( x, y, z ) > 0. Proving Fa ( x, y, z ) − Fb ( x, y, z ) < 0, for example, is equivalent to proving the inequality

(1 − x 2 )(1 − y2 )(1 − z 2 ) < (1 − xyz )2

(5.17)

To prove inequality (5.17), it suffices to prove the equivalent inequality (1 − x 2 )(1 − y 2 )(1 − z 2 ) < (1 − xyz ) or the equivalent inequality

(1 − x 2 )(1 − y2 )(1 − z 2 ) + xyz < 1

(5.18)

Indeed, if inequality (5.18) is true, inequality (5.17) follows from it by squaring both sides of the inequality (1 − x 2 )(1 − y 2 )(1 − z 2 ) < 1 − xyz. The  squaring will not reverse the direction of the inequality because 0 < x < 1; 0 < y < 1; 0 < z < 1, and the following quantities are positive:

(1 − xyz) > 0, (1 − x 2 )(1 − y2 )(1 − z 2 ) > 0 .

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Ranking the Reliabilities of Systems and Processes by Using Inequalities

To prove inequality (5.18), a combination of the ‘substitution’ technique and proving a simpler, intermediate inequality will be used. Because the reliability ri of a component is a number between zero and unity, the trigonometric substitutions ri = sin α i where α ∈(0, π / 2) are appropriate. Making the substitutions: x = sin α ; y = sin β and z = sin γ transforms the left side of inequality (5.18) into

(1 − x 2 )(1 − y2 )(1 − z 2 ) + xyz = cos α × cos β × cos γ + sin α × sin β × sin γ

(5.19)

Next, the positive quantity cos α × cos β × cos γ + sin α × sin β × sin γ is replaced by the larger quantity cos α × cos β + sin α × sin β . Indeed, because 0 < cos γ < 1 and 0 < sin γ < 1, the inequality

cos α × cos β × cos γ + sin α × sin β × sin γ < cosα × cos β + sin α × sin β

(5.20)

holds. If the intermediate inequality cos α × cos β + sin α × sin β ≤ 1 could be proved, this would imply the inequality

cos α × cos β × cos γ + sin α × sin β × sin γ < 1

(5.21)

Since cos α × cos β + sin α × sin β = cos(α − β ), and cos(α − β ) ≤ 1, we finally get cos α × cos β × cos γ + sin α × sin β × sin γ < cos α × cos β + sin α × sin β

= cos(α − β ) ≤ 1



Inequality (5.18) has been proved and from it, inequality (5.17) follows. The system in Figure 5.10a is characterised by a smaller probability of failure than the system in Figure 5.10b; therefore, the system in Figure 5.10a is the more reliable system. In  the example considered in this section, the algebraic inequalities helped to reveal the intrinsic reliability of the competing design solutions and rank the systems in terms of reliability in the absence of any knowledge related to the reliabilities of their building parts.

6

Using Inequalities for Reducing Epistemic Uncertainty and Ranking Decision Alternatives

6.1 SELECTION FROM SOURCES WITH UNKNOWN PROPORTIONS OF HIGH-RELIABILITY COMPONENTS Consider a real-world example featuring a market where three suppliers A1, A2 and A3  produce high-reliability components of the same type, with proportions x1, x2 and x3, which are unknown. In the case of suspension automotive springs, for example, this means that only a fraction xi , i = 1, 2, 3 of the manufactured suspension springs can last for more than a specified number of cycles if tested on a specially designed test rig and the rest of the springs fail significantly below the specified limit. If two components are to be purchased and installed in a device, the question of interest is which strategy maximises the probability that both components will be highly reliable: (i) purchasing the two components from the same supplier or (ii) purchasing the two components from different suppliers. The same problem can be formulated for machine centres (instead of suppliers) producing high-reliability components with unknown proportions. This problem is mathematically equivalent to the following problem: Three boxes contain high-reliability components in unknown proportions x1, x2 and x3, characterising the boxes, correspondingly (Figure 6.1). Two components are to be installed in a device. Which strategy maximises the probability that both components will be highly reliable?

1. Taking the two components from the same, randomly selected box, or 2. Taking the two components from two different, randomly selected boxes.

At first glance, it seems that either of these strategies could be selected because the proportions x1, x2 and x3 of high-reliability components characterising the boxes/ suppliers are unknown. Surprisingly, this common-sense conclusion is incorrect. The probability of selecting two high-reliability components from the same box is

p1 = (1 / 3) x12 + (1 / 3) x22 + (1 / 3) x32

This is composed of the probabilities of three mutually exclusive events: (i) the probability (1/ 3) x12 that box 1 will be selected and both components taken from box 1 will 107

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FIGURE 6.1  Three boxes/suppliers containing high-reliability components with unknown proportions.

be highly reliable; (ii) the probability (1 / 3) x22 that box 2 will be selected and both components taken from box 2 will be highly reliable and (iii) the probability (1 / 3) x32 that box 3 will be selected and both components taken from box 3 will be highly reliable. Accordingly, the probability of selecting two high-reliability components from two randomly selected boxes is

p2 = (1 / 3) x1x2 + (1 / 3) x1x3 + (1 / 3) x2 x3

This is composed of the probabilities of three mutually exclusive events: (i) the probability (1 / 3) x1x2 that boxes 1 and 2 will be selected and both components taken from boxes 1 and 2 will be highly reliable; (ii) the probability (1 / 3) x1x3 that boxes 1 and 3 will be selected and both components taken from boxes 1 and 3 will be highly reliable and (iii) the probability (1 / 3) x2 x3 that boxes 2 and 3 will be selected and both components taken from boxes 2 and 3 will be highly reliable. The question of interest can be answered by comparing the probabilities p1 and p 2 , which is equivalent to proving the conjectured non-trivial algebraic inequality

(1 / 3) x12 + (1 / 3) x22 + (1 / 3) x32 ≥ (1 / 3) x1x2 + (1 / 3) x2 x3 + (1 / 3) x3 x1

(6.1)

This inequality can be proved rigorously by using a number of different techniques: by a direct manipulation, by using segmentation, by using the rearrangement inequality or by reducing it to a standard inequality (e.g., to the Muirhead inequality). The segmentation technique is a powerful technique whose essence is to segment (split) the original inequality into simple parts and express each part as an inequality by using some of the standard inequalities. Adding the separate parts assembles the original inequality and completes the proof. Proving inequality (6.1) is equivalent to proving the inequality

x12 + x22 + x32 ≥ x1x2 + x2 x3 + x3 x1

(6.2)

because inequality (6.1) can be obtained from inequality (6.2) by dividing both sides of (6.2) by the positive constant ′3′. Inequality (6.2) can be segmented into three parts by using the standard arithmetic mean – geometric mean (AM-GM) inequality:



x12 + x22 x 2 + x32 x2 + x2 ≥ x12 x22 = x1x2 , 2 ≥ x22 x32 = x2 x3 and 3 1 ≥ x32 x12 = x3 x1 2 2 2

Adding the three inequalities gives the original inequality (6.2) and completes the proof.

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109

This is a surprising and highly counter-intuitive result. After all, the proportions of high-reliability components characterising the boxes are unknown. Despite the total lack of knowledge related to the proportions of high-reliability components characterising the separate batches/suppliers and regarding existing interdependences (correlations) among the proportions of high-reliability components in the boxes, inequality (6.1) still holds. The reduced epistemic uncertainty allows an appropriate choice to be made, associated with a reduced risk of not selecting two high-reliability components. The same approach can be generalised for a larger number of selected components. Consider an analogous case of purchasing components from suppliers whose proportions of high-reliability components are unknown. If, for example, three components are to be purchased from three suppliers (n = 3) and installed in an assembly, the question of interest is to choose between several competing strategies: (a) purchasing the three components from a single, randomly selected supplier; (b) purchasing the three components from the three available suppliers or (c) purchasing the three components from two randomly selected suppliers. Suppose that three suppliers are characterised by unknown proportions x1, x2 and x3 of high-reliability components and unknown interdependences of the proportions of high-reliability components among the boxes. Proving that strategy ‘a’ is better than strategy ‘b’ reduces to proving the inequality

1 3 1 3 1 3 x1 + x2 + x3 ≥ x1x2 x3 3 3 3

(6.3)

The left side of inequality (6.3) is the probability of purchasing three high-reliability components from a randomly selected supplier. The right side of inequality (6.3) is the probability of purchasing three high-reliability components from the three available suppliers. Inequality (6.3) can be proved by a direct application of the AM-GM inequality:

x13 + x23 + x33 3 3 3 3 ≥ x1 x2 x3 = x1x2 x3 3

Proving that variant ‘a’ is better than variant ‘c’ reduces to proving the inequality

(1 / 3) x13 + (1 / 3) x23 + (1 / 3) x33 ≥ (1 / 6) x12 x2 + (1 / 6) x12 x3 + (1 / 6) x22 x1 + (1 / 6) x22 x3 + (1 / 6) x32 x1 + (1 / 6) x32 x2

(6.4)

The left side of inequality (6.4) is the probability of purchasing three high-reliability components from a randomly selected single supplier. The right side of inequality (6.4) is the probability of purchasing three high-reliability components from two randomly selected suppliers. Applying segmentation by using the AM-GM inequality six times yields the following inequalities:

x13 + x13 + x23 3 3 3 3 x 3 + x23 + x13 3 3 3 3 ≥ x2 x2 x1 = x22 x1, ≥ x1 x1 x2 = x12 x2 , 2 3 3

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x23 + x23 + x33 3 3 3 3 x 3 + x33 + x23 3 3 3 3 ≥ x3 x3 x2 = x32 x2 ≥ x2 x2 x3 = x22 x3 , 3 3 3 x33 + x33 + x13 3 3 3 3 x 3 + x13 + x33 3 3 3 3 ≥ x1 x1 x3 = x12 x3 ≥ x3 x3 x1 = x32 x1, 1 3 3

Adding the six inequalities and dividing both sides by the positive constant 3 gives inequality (6.4) and completes the proof.

6.2  MONTE CARLO SIMULATIONS All theoretical results from the application of the inequalities have been confirmed by Monte Carlo simulations, each of which involved 10 million trials. For fractions of high reliability components of 0.9, 0.55 and 0.35, characterising the three suppliers, the Monte Carlo simulation resulted in probabilities p1 = 0.41 and p2 = 0.33 of purchasing two high-reliability components from a randomly selected single supplier and from two randomly selected suppliers, correspondingly. The left and right parts of inequality (6.1) give p1 = 0.41 and p2 = 0.33 for the same probabilities, which confirms the validity of inequality (6.1). The simulation of purchasing two components from a randomly selected supplier and testing the components for high reliability has been done within a loop of ten million trials. The pseudo-code fragment related to evaluating the empirical probability of purchasing two high-reliability components from a single supplier is shown next. Algorithm 6.1 a=[0.9, 0.55, 0.35] n=3; num_trials=10000000; count1=0; for i=1 to num_trials do {   sup_no=[n*rand()]+1;  x=rand(); y=rand();  if(x n

(10.2)

This is because if at least two events Ai and Aj can occur simultaneously, at least one outcome must be counted twice: once for event Ai and once for event Aj . Dividing both sides of (10.2) by the positive value n does not alter the direction of inequality (10.2) and the result is the inequality

n1 / n + n2 / n + ... + nm / n > 1

(10.3)

which is inequality (10.1). The power of the simple inequality (10.1) can even be demonstrated on two events A and B only, where the event B is complementary to event B ( P ( B ) + P ( B) = 1) . 169

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If the non-occurrence of event B leads to the occurrence of event A and if the nonoccurrence of event A leads to the occurrence of event B, and if A and B can occur simultaneously, the conditions of inequality (10.1) are fulfilled and therefore

P( A) + P( B) > 1

(10.4)

Considering that P ( B ) = 1 − P ( B) , from inequality (10.4), the inequality

P ( A) > P ( B)

(10.5)

is obtained. Suppose that event A stands for ‘system A is working at the end of the time interval (0,t)’ and event B stands for ‘system B is not working at the end of the time interval (0,t)’. Suppose that the following conditions are fulfilled: • If system B is working, system A is also working. • If system A is not working, system B is not working. • It is possible that system A is working and system B is not working. If these conditions are fulfilled, according to the inequality of the negatively correlated events, inequality (10.5) is fulfilled, which means that system A is characterised by a higher reliability than system B. This conclusion will be illustrated by an example. System A and system B, shown in Figure 10.1 can be compared in terms of reliability, in the absence of any knowledge of the reliabilities of the components. Both systems are built on the same set of 6 components numbered from 1 to 6. The reliabilities r1, r2 , r3 , r4 , r5 , r6 of the components are unknown. Despite the deep uncertainty related to the components, the reliabilities of the systems can still be ranked by using the inequality of the negatively correlated events. Denote by A the event ‘the system in Figure 10.1a is working at the end of a specified time interval (0,t)’ and by A, the event ‘the system in Figure 10.1a is not working at the end of the specified time interval’. Similarly, denote by B the event ‘the system in Figure 10.1b is working at the end of a specified time interval (0,t)’ and by B the event ‘the system in Figure  10.1b is not  working at the end of the specified time interval’.

FIGURE 10.1  Ranking the reliabilities of two systems with reliability networks (a) and (b) in the absence of any information about the reliabilities of the components.

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171

Consider now the events A (system (a) is working at the end of the specified time interval) and B (system (b) is not working at the end of the specified time interval). The conditions of inequality (10.1) are fulfilled for these two events. Indeed, if event B does not occur, this means that system (b) is working. This can happen only if all components 4, 5 and 6 in Figure 10.1b are working, which means that system (a) is working. As a result, if event B does not occur then event A occurs. Conversely, if event A does not  occur, this means that at least one of the components  4, 5, 6 in Figure  10.1a does not  work, which means that system (b) does not work (the event B occurs). At the same time, both events can occur simultaneously ( P( A ∩ B) ≠ 0 ). This  is, for example, the case if components 1,2,3 are in working state at the end of the time interval (0,t) and component 5 is in a failed state. The conditions of inequality (10.1) are fulfilled, therefore

P ( A) + P ( B ) > 1

(10.6)

P ( A) > 1 − P ( B ) = P ( B)

(10.7)

holds, which is equivalent to

As a result, it follows that P ( A) > P ( B) irrespective of the reliabilities r1, r2 , r3 , r4 , r5 , r6 of the components building the systems. In  the example considered in this section, the algebraic inequalities helped to reveal the intrinsic reliability of the competing design solutions and rank the solutions in terms of reliability in the absence of any knowledge related to the reliabilities of their building parts.

10.2 AVOIDING RISK UNDERESTIMATION BY USING THE JENSEN INEQUALITY 10.2.1 Avoiding the Risk of Overestimating Profit Suppose that the demand for a particular product X is associated with variation (X is a random variable) and the variation of X cannot be controlled. The profits Y depend on the demand X through a particular function Y = f ( X ). The question of interest is which strategy is more conservative with respect to assessing the average profit in order to avoid an overestimation of the profits:

a. Averaging first different values of the demand X (x = (1/ n)∑in=1 xi ) by using n random values x1, x2 ,..., xn of the demand within the demand range, followed by evaluating the average profit from y = f ( x ), or b. Obtaining the average profit y = (1/ n)∑in=1 f ( xi ) by averaging the profits corresponding to n random demands x1, x2 ,..., xn within the demand range.

The choice of strategy which avoids overestimating the average profit depends on whether the dependence Y = f (X ) of the profit on the demand is concave or convex. Often, this dependence is concave, of the type in Figure 10.2, because after some

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FIGURE 10.2  Profit-demand dependencies are often concave functions.

continuous increase in the profits with increasing demand, a flat region follows due to limited production capacity. If the dependence Y = f (X ) is concave, the following Jensen inequality holds: f (w1 x1 + w2 x2 + ... + wn xn ) ≥ w1 f ( x1 ) + w2 f ( x2 ) + ... + wn f ( xn )



(10.8)

where wi ( i = 1,..., n ) are weights that satisfy 0 ≤ wi ≤ 1 and w1 + w2 + ... + wn = 1. If the weights are chosen to be equal wi = 1/ n, the Jensen inequality (10.8) becomes  f  (1/ n)  



n

∑ i =1

 xi  ≥ (1/ n)  

n

∑ f ( x ) i

(10.9)

i =1

In this case, the average of the profits at different levels of the demand can be significantly smaller than the profit calculated at an average level of the demand. To demonstrate this consider an example from the biotech industry, where the demand for a particular biochemical product varies uniformly from 0 to 300,000 kg per year and the capacity of the production plant for one year is only 200,000 kg of product per year. Suppose that the profit in USD generated from selling the product is given by y = 3.6 x, where x is the quantity of the product [in kg] sold. The profit function is therefore a concave function, defined in the following way: 3.6 x,  f ( x) =  3.6 × 200, 000,



0 ≤ x ≤ 200, 000

200, 000 ≤ x ≤ 300, 000

(10.10)

The average demand is obviously 300,000/2 = 150,000 kg per year. The profit corresponding to the average demand is y1 = 3.6 × 150, 000 = 540, 000. This is the value y1 = f (1/ n)∑in=1 xi on the left side of inequality (10.9).

(

)

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The average of the profits was calculated by using a simple Monte Carlo simulation whose algorithm in pseudo-code is given next.

Algorithm 10.1 n=100000; %number of simulation trials f(x): %the function that gives the profit at a random demand x   S=0; for i=1 to n do {   tmp=300000*rand();   y=f(tmp);   S=S+y;  } Average_profit=S/n 

Running the Monte Carlo simulation with 100,000 trials resulted in an average profit equal to 480,000. This is the value y2 = (1/ n)∑in=1 f ( xi ) on the right-hand side of inequality (10.9). The difference between the two values for the average profit is significant. Because of the significant difference, critical business decisions cannot be made on the basis of the simple calculation of the profit at the average demand. Instead, the average of the profits taken at different values of the random demand should be taken. This provides a realistic estimate of the real level of profits for the business. The correct decision which eliminates the risk of overestimating the average profit is averaging the profits at different levels of the demand rather than taking the profit at the average demand. Avoiding an overestimation of the profits avoids an optimistic valuation of the business.

10.2.2 Avoiding the Risk of Underestimating the Cost of Failure The downtime for repair X is always associated with variation (X is a random variable) and the variation of X cannot be controlled. An assessment of the average cost of failure Y (which depends on X through a particular dependence Y = f ( X )) can be made in two alternative ways. The question of interest is which strategy is more conservative with respect to assessing the average cost of failure Y:

a. Averaging n different downtimes x1, x2 ,..., xn , x = (1/ n)∑in=1 xi and assessing the average cost of failure y = f ( x ) with the average value x of the downtime, or b. Averaging the costs of= failure y1 f= ( x1 );...; yn f ( xn ) at the n different values x1, x2 ,..., xn of the downtimes: y = (1/ n)∑in=1 f ( xi ).

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FIGURE 10.3  The choice of a risk avoiding strategy is driven by whether the cost function is (a) concave or (b) convex.

Again, the choice of strategy depends on whether the cost-of-failure dependence Y = f (X ) is concave or convex. If the function Y = f (X ) is concave (Figure 10.3a), the following Jensen inequality holds:

 f  (1/ n)  

n

∑ i =1

 xi  ≥ (1/ n)  

n

∑ f ( x )

(10.11)

i

i =1

(

)

In this case, taking the cost at the average downtime from y = f (1/ n)∑in=1 xi gives a more conservative estimate of the cost of failure. If the function Y = f (X ) is convex (Figure 10.3b), the following Jensen inequality holds:

 f  (1/ n)  

n

∑ i =1

 xi  ≤ (1/ n)  

n

∑ f (x ) i

(10.12)

i =1

In this case, taking the average y = 1/n) ∑in=1 f ( xi ) of the cost at different downtimes gives a higher (more conservative) value of the cost of failure. These examples also show that the best results in eliminating profit overestimation and cost of failure underestimation are obtained from combining domain-specific knowledge and the domain-independent method based on algebraic inequalities. Domain-specific knowledge alone is not sufficient to achieve the risk reduction.

10.2.3 A Conservative Estimate of System Reliability by Using the Jensen Inequality Suppose that the reliability X of identical components from n separate batches is associated with variation (X is a random variable). The reliabilities of the components from the separate batches are x1, x2 ,..., xn and the batch-to-batch variation of X cannot be controlled. Suppose that an assembly needs to be made which includes m

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175

identical components ( m ≥ 2 ) logically arranged in parallel. The m identical components needed to build the assembly in parallel can be taken from any of the n available batches. The reliability of a system including m components logically arranged in parallel is given by the equation R = 1 − (1 − x ) m



(10.13)

where x is the component reliability characterising a particular batch. By using the particular values x1, x2 ,..., xn of the reliabilities of the components from the separate batches, an assessment of the reliability R of the assembly is made. The  question of interest is which approach gives a more conservative estimate of the average reliability of the assembly built with m components (m ≥ 2) arranged in parallel:



a. Averaging the values of the reliabilities of the components from the separate batches x = (1/ n)∑in=1 xi and performing a single calculation R = 1 − (1 − x )m of the system’s reliability with the average value x of the reliability of the components, or b. Taking the average R = (1/ n)∑in=1 Ri of the reliabilities Ri = 1 − (1 − xi )m ; i = 1,..., n of n assemblies built with components from the separate batches.

This question can be answered by investigating the system reliability function (10.13) whose second derivative with respect to x is negative:

R′′ = −m( m − 1)(1 − x ) m−2

Consequently, the reliability of the system is a concave function of the component reliability x, and for concave functions f ( x ), the Jensen inequality states

f ( w1x1 + w2 x2 + ... + wn xn ) ≥ w1 f ( x1 ) + w2 f ( x2 ) + ... + wn f ( xn )

(10.14)

where wi ( i = 1,..., n ) are weights that satisfy 0 ≤ wi ≤ 1 and w1 + w2 + ... + wn = 1. If the weights are chosen to be equal (wi = 1/ n), the Jensen inequality (10.14) becomes

 f  (1/ n)  

n

∑ i =1

 xi  ≥ (1/ n)  

n

∑ f ( x )

(10.15)

i

i =1

For the system reliability function (10.13), calculating with the average component reliability results in a higher and more optimistic value. n



R1 = 1 − (1 − x ) m ≥ R = (1/ n)

∑[1 − (1 − x ) ] i

i =1

m

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Risk and Uncertainty Reduction by Using Algebraic Inequalities

Consequently, taking R = (1/ n)∑in=1[1 − (1 − xi )m ] for the system reliability gives a conservative estimate.

10.3 REDUCING UNCERTAINTY AND RISK ASSOCIATED WITH THE PREDICTION OF MAGNITUDE RANKINGS This application has been prompted by the short note “Pick the Largest Number” in (Cover,1987). Consider a problem related to estimating the probability that the magnitude of a future random event will be exceeded by the magnitude of the next random event. This is a problem of predicting magnitudes ranking of future random events, and it is important in cases where it is necessary to decide on whether there is a need for extra resources to mitigate the consequences of future random events. Such random events are, for example, the leaks of crude oil into the marine water from a pipeline during subsea oil and gas production, the magnitudes of floods in consecutive years, the magnitudes of fires, the direction of movement of the price of a commodity on the stock market, and so on. Making correct predictions of this kind is not easy because of the deep uncertainty associated with the distribution of the magnitude of the random event. If no information is available about the distribution of the magnitude of the random event, it seems that the chance of guessing correctly whether the magnitude of the second random event will be larger than the magnitude of the first random event is only 50% and cannot be improved. Surprisingly, by introducing a random comparison threshold C before the occurrence of any random event (see Figure  10.4a), the uncertainty can be reduced by using an inequality and the chance of predicting correctly whether the magnitude of the second random event will exceed the magnitude of the first random event can be increased beyond 50%. The prediction strategy is simple (Cover, 1987). If the magnitude of the first random event is larger than the randomly pre-selected comparison threshold C (Figure 10.4b),

FIGURE 10.4  (a) Introducing a random threshold C reduces uncertainty and increases the likelihood of a correct prediction of the magnitude ranking of future random events to more than 50%; (b–d) represent the possible configurations of the magnitudes of the random events with respect to the random threshold C.

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a prediction is made that the second random event will have a smaller magnitude than the magnitude of the first event. If the magnitude of the first event is smaller than the comparison threshold C, a prediction is made that the second random event will have a larger magnitude than the magnitude of the first event (Figure 10.4c). Cover (2007) stated that the prediction strategy described earlier ensures correct prediction with probability strictly greater than 0.5. To test this assertion, an algebraic inequality will be employed. Since the distribution of the magnitude of the random event is unknown, denote by p the probability that a random event will have a larger magnitude M than the comparison threshold C; p = Pr( M > C ); 0 ≤ p ≤ 1. Correspondingly, the probability that a random event will not have a larger magnitude M than the comparison threshold C is 1− p = Pr( M ≤ C ). The  probability that both events will be with a larger magnitude than the pre-selected threshold C is p2 (Figure  10.4b); the probability that both events will not  have a larger magnitude than the pre-selected threshold C is (1 − p)2 (Figure 10.4c). Finally, the probability that one of the magnitudes will be greater than the pre-selected threshold C and the other magnitude will be smaller than the pre-selected threshold (Figure 10.4d) is p(1 − p) + (1 − p) p = 2 p(1 − p). In  the cases where both random events have larger magnitudes than the pre-selected threshold C or both random events have smaller magnitudes than the pre-selected threshold, the likelihood of a correct prediction related to the magnitude of the second random event is clearly 0.5. In the case where the first random event has a smaller magnitude than the threshold C and the second event has a larger magnitude, the likelihood of a correct prediction related to the magnitude of the second random event is 1.0. Similarly, in the case where the first random event has a larger than the threshold magnitude and the second event has a smaller magnitude, the likelihood of a correct prediction related to the magnitude of the second random event is also 1.0. Consequently, according to the total probability theorem, the likelihood L of a correct prediction related to the magnitude of the second random event is given by L = 0.5 p2 + 0.5(1 − p)2 + 1 × 2 × p(1 − p). It  can be shown that inequality (10.16) holds, irrespective of the value of the probability p:

L = 0.5 p2 + 0.5(1 − p)2 + 1× 2 × p(1 − p) ≥ 0.5

(10.16)

Indeed, expanding the left-hand side of inequality (10.16) reduces the inequality to L = 0.5 + p(1 − p) ≥ 0.5. Because 0 ≤ p ≤ 1, p(1 − p) ≥ 0, the quantity 0.5 + p(1 − p) is not smaller than 0.5, hence inequality (10.16) is true. The product of two positive values with a given sum has a maximum when the values are equal. Consequently, the product pq where p + q = 1 attains maximum when p= q= 1/ 2. As a result, the maximum possible value of the quantity p(1− p) is 0.25, and this value cannot be exceeded. Consequently, the most appropriate value of the comparison threshold C is a value for which the probability that the magnitude of the random event will be smaller than the threshold is equal to the probability that the magnitude will be larger than the threshold. In this case, a correct prediction will be made 75% of the time. In reality, the threshold position of equal probability is not known and the threshold is selected randomly within the physically possible range of values for the magnitude M.

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If p ≈ 0 or 1 − p ≈ 0, from inequality (10.16) it is clear that a probability of a correct guess strictly greater than 0.5 cannot be attained. This situation is present if, for example, the distribution of the magnitudes of the random events is characterised by a very small variance. In  this case, the likelihood that the magnitudes of two sequential events will fall on both sides of the threshold is extremely small. In the case where the variance of the magnitude distribution is not small but the selected threshold grossly overestimates or underestimates the expected magnitude, the probability p or the probability 1− p is very small and practically, a probability of a correct prediction strictly greater than 0.5 cannot be attained. To test the described prediction strategy, a Monte Carlo simulation experiment has been made. Suppose that the random event is a leak of crude oil from a pipeline with a maximum debit of 300 l/s. Because no oil leak can be negative and no oil leak can exceed the maximum debit of the pipeline, the absolute lower and upper limit of the range of possible leaks are abs _ lower_limit=0; and abs_ upper_ limit=300, correspondingly. The  future leak magnitudes within the physically possible range [0,300] are uncertain; this is why the distribution of the leak magnitudes is assumed to be uniform between randomly selected values a and b inside the physically possible range [0,300]. The  algorithm testing the described decision strategy in pseudo-code is given next.

Algorithm 10.2 num_trials=10000000;   abs_lower_limit=0; abs_upper_limit=300;     a=abs_lower_limit+(abs_upper_limit-abs_lower_limit)*rand(); b=abs_lower_limit+(abs_upper_limit-abs_lower_limit)*rand(); if(b