Which-Is-Better (WIB): Problems in Reliability Theory 303127315X, 9783031273155

Table of contents :
Preface
Contents
1 Introduction
1.1 Maintenance Policies
1.2 Reliability Applications
References
2 Random Age Replacement Model
2.1 Random Replacement
2.2 Replacement First
2.3 Replacement Last
2.4 Replacement First and Last
2.5 Replacement Overtime
2.5.1 Replacement Overtime First
2.5.2 Replacement Overtime Last
2.6 Deviation Time
2.7 Problems
References
3 Replacement Model with Minimal Repair
3.1 Random Replacement
3.2 Replacement with Working Number
3.2.1 Replacement First
3.2.2 Replacement Last
3.3 Replacement First and Last
3.4 Replacement with Failure Number
3.4.1 Replacement First
3.4.2 Replacement Last
3.5 Replacement Overtime with Working Number
3.5.1 Replacement First
3.5.2 Replacement Last
3.6 Replacement Overtime with Failure Number
3.6.1 Replacement First
3.6.2 Replacement Last
3.7 Replacement with Working and Failure Numbers
3.7.1 Replacement First
3.7.2 Replacement Last
3.8 Problems
References
4 Periodic Replacement Models
4.1 Age Replacement
4.1.1 Replacement First
4.1.2 Replacement Last
4.2 Replacement with Working Number
4.2.1 Replacement First
4.2.2 Replacement Last
4.3 Replacement with Failure Number
4.3.1 Replacement First
4.3.2 Replacement Last
4.4 Replacement Over Failure Number
4.4.1 Replacement First
4.4.2 Replacement Last
4.5 Problems
References
5 Extended Replacement Models
5.1 Replacement with Two Failure Modes
5.1.1 Replacement First
5.1.2 Replacement Last
5.2 Undertime Replacement Model
5.2.1 Age Replacement
5.2.2 Replacement with Minimal Repair
5.3 Middle Replacement Models
5.3.1 Age Replacement
5.3.2 Comparison of Three Policies
5.3.3 Replacement with Minimal Repair
5.3.4 Comparisons of Three Policies
5.4 Problems
References
6 Which is Better for Standby or Parallel Systems
6.1 Reliability Measures
6.1.1 Constant Number of Units
6.1.2 Random Number of Units
6.2 Replacement Policies
6.2.1 MTTF for Replacement Time
6.2.2 Replacement for Number of Units
6.3 Scheduling of Random Works
6.3.1 Tandem and Parallel Works
6.3.2 Standby and Parallel Systems
6.3.3 Random Tandem Works
6.4 Modified Models
6.5 Problems
References
7 Which is Better Problems in Shock and Damage Models
7.1 Three Replacement Policies
7.2 Optimal Policies with Two Variables
7.3 Exponential Shock Times and Amount of Damage
7.4 Replacement Overtime Policies
7.5 Modified Replacement Models
References
8 Which is Better Problems in Backup Models
8.1 Database Backup Models
8.1.1 Optimal Backup Times
8.2 Overtime Backup First
8.2.1 Optimal Backup Times
8.3 Comparisons of Backup First and Overtime
8.4 Database Backup Last
8.4.1 Backup Last
8.4.2 Backup Overtime Last
8.5 Problem
References
9 Which is Better Problems in Checkpoint Models
9.1 Periodic Checkpoint Model
9.2 Random Checkpoint Models
9.2.1 Two Kinds of Tasks
9.2.2 Two Kinds of Checkpoints
9.3 Two Examples
9.4 Imperfect Checkpoint Model
9.4.1 Periodic Checkpoint
9.4.2 Random Checkpoint
9.5 Problem
References
Appendix Appendix A
A.1 Properties of Failure Rate f left parenthesis t right parenthesis divided by ModifyingAbove upper F With quotation dash left parenthesis t right parenthesisf(t)/overlineF(t) for Age Replacement
A.2 Properties of Failure Rate h left parenthesis t right parenthesish(t) for Replacement with Minimal Repair
A.3 Properties of Failure Rates f left parenthesis t right parenthesis divided by ModifyingAbove upper F With quotation dash left parenthesis t right parenthesisf(t)/overlineF(t) for Periodic Replacement
A.4 Properties of Failure Rates h left parenthesis t right parenthesish(t) for Periodic Replacement
A.5 Properties of Failure Rates h left parenthesis t right parenthesish(t) for Replacement Over Failure Number
A.6 Properties of Failure Rates for Extended Replacement
A.7 Properties of Failure Rates for Shock and Damage Models
Appendix Appendix B
Index

Citation preview

Springer Series in Reliability Engineering

Satoshi Mizutani Xufeng Zhao Toshio Nakagawa

Which-Is-Better (WIB): Problems in Reliability Theory

Springer Series in Reliability Engineering Series Editor Hoang Pham, Department of Industrial and Systems Engineering, Rutgers University, Piscataway, NJ, USA

Today’s modern systems have become increasingly complex to design and build, while the demand for reliability and cost effective development continues. Reliability is one of the most important attributes in all these systems, including aerospace applications, real-time control, medical applications, defense systems, human decision-making, and home-security products. Growing international competition has increased the need for all designers, managers, practitioners, scientists and engineers to ensure a level of reliability of their product before release at the lowest cost. The interest in reliability has been growing in recent years and this trend will continue during the next decade and beyond. The Springer Series in Reliability Engineering publishes books, monographs and edited volumes in important areas of current theoretical research development in reliability and in areas that attempt to bridge the gap between theory and application in areas of interest to practitioners in industry, laboratories, business, and government. Now with 100 volumes! **Indexed in Scopus and EI Compendex** Interested authors should contact the series editor, Hoang Pham, Department of Industrial and Systems Engineering, Rutgers University, Piscataway, NJ 08854, USA. Email: [email protected], or Anthony Doyle, Executive Editor, Springer, London. Email: [email protected].

Satoshi Mizutani · Xufeng Zhao · Toshio Nakagawa

Which-Is-Better (WIB): Problems in Reliability Theory

Satoshi Mizutani Department of Business Administration Aichi Institute of Technology Toyota, Aichi, Japan

Xufeng Zhao College of Economics and Management, Nanjing University of Aeronautics and Astronautics Nanjing, China

Toshio Nakagawa Department of Business Administration Aichi Institute of Technology Toyota, Aichi, Japan

ISSN 1614-7839 ISSN 2196-999X (electronic) Springer Series in Reliability Engineering ISBN 978-3-031-27315-5 ISBN 978-3-031-27316-2 (eBook) https://doi.org/10.1007/978-3-031-27316-2 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

Which is better, A or B? Different people have different answers due to their respective point of views. In the field of reliability, A or B problems are usually encountered such as standby or parallel systems, replacement or repair policies, maintenance first or last policies, discrete or continuous replacement policies, periodic or random inspection policies, incremental or differential backup policies, etc. in the viewpoint of reliability and maintainability, respectively. In this book, the above A or B problems are named as Which-Is-Better (WIB): Problems in Reliability Theory. For the last 10 years, the most appeared keywords in our research are first and last, which approach to trigger maintenance, replacement, and inspection policies. For example, replacement first means that the unit is replaced before failure at some events such as operating time, number of repairs, working number, etc. whichever occurs first, while replacement last means that the unit is replaced before failure at the above events, whichever occurs last. It would be natural that replacement first is more easier to prevent failures, while replacement last could let the unit operate work as longer as possible. However, only these findings are not enough for the first and last policies. In our publications, we not only surveyed the performability of the policies but also studied the comparative methods and found the comparative results in analytical ways. From the viewpoint of costs, the answers of WIB problems of replacement first and last polices have been obtained, and each policy comes with its own advantages and disadvantages. In addition, the modeling of first and last has been extended with overtime policies, e.g., inspection first, last, and overtime policies, and backup first, last, and overtime policies, which arisen new WIB problems. We have now already published six books that contain WIB problems in reliability theory, Maintenance Theory of Reliability, Shock and Damage Models in Reliability Theory, Advanced Reliability Models and Maintenance Policies, Random Maintenance Policies, Maintenance Overtime Policies in Reliability Theory, and Advanced Maintenance Policies for Shock and Damage Models from Springer Publisher. Based on our recent research papers, WIB problems will be surveyed systematically in this book. For example, the models of extended failure rates for replacement first, last, and overtime policies are studied, and their monotonic properties can be considered as the most important and difficult proofs in mathematics analysis. We believe that this v

vi

Preface

book would give great benefits for readers such as students, researchers, engineers, and managers who are usually troubled and worried with such WIB problems. This book is composed of nine chapters. Chapter 1 gives brief introduction of WIB problems in daily life and in the field of reliability theory. Chapters 2–5 are mainly devoted to the WIB problems for random age replacement policies; replacement policies with minimal repairs; periodic replacement policies; and replacement policies with two failure modes, undertime replacement policies and middle replacement policies, respectively. In Chaps. 6–9, the WIB problems in applications are introduced, such that which is better for standby or parallel systems, which is better for replacement policies in shock and damage environment, which is better for backup policies and which is better for checkpoint policies in computer science, respectively. Finally, Appendix A introduces detailed proofs of the expected failure rates for the above replacement policies and Appendix B gives answers for selected problems in chapters. We wish to thank Prof. Cunhua Qian, Prof. Kodo Ito, Prof. Mitsuhiro Imaizumi, Prof. Mitsutaka Kimura, Prof. Mingchih Chen, Prof. Syouji Nakamura, and Dr. Kenichiro Naruse, who are the co-authors of our research papers for chapters. We wish to express our special thanks to Prof. Tadashi Dohi, Prof. Krishna B. Misra, Prof. Shey-Huei Sheu, Prof. Won Young Yun, Prof. Hiroyuki Okamura, and Prof. Kazumi Yasui for their kind support in our research works. Finally, we would like to express our sincere appreciation to Prof. Hoang Pham, Rutgers University and Editor Anthony Doyle, Springer-Verlag, London for providing the opportunity for us to write this book. Toyota, Japan Nanjing, China Toyota, Japan November 2022

Satoshi Mizutani Xufeng Zhao Toshio Nakagawa

Contents

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Maintenance Policies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Reliability Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 2 3 5

2 Random Age Replacement Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Random Replacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Replacement First . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Replacement Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Replacement First and Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Replacement Overtime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 Replacement Overtime First . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2 Replacement Overtime Last . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Deviation Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7 7 10 13 16 18 20 23 26 28 28

3 Replacement Model with Minimal Repair . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Random Replacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Replacement with Working Number . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Replacement First . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Replacement Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Replacement First and Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Replacement with Failure Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Replacement First . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.2 Replacement Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Replacement Overtime with Working Number . . . . . . . . . . . . . . . . . . 3.5.1 Replacement First . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.2 Replacement Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Replacement Overtime with Failure Number . . . . . . . . . . . . . . . . . . . 3.6.1 Replacement First . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.2 Replacement Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Replacement with Working and Failure Numbers . . . . . . . . . . . . . . .

31 32 34 35 37 39 41 41 43 45 46 48 51 51 54 56 vii

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3.7.1 Replacement First . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.2 Replacement Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

56 58 60 61

4 Periodic Replacement Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Age Replacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Replacement First . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Replacement Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Replacement with Working Number . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Replacement First . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Replacement Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Replacement with Failure Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Replacement First . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.2 Replacement Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Replacement Over Failure Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Replacement First . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.2 Replacement Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

63 63 63 66 71 71 73 75 75 77 79 79 82 84 85

5 Extended Replacement Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Replacement with Two Failure Modes . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 Replacement First . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.2 Replacement Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Undertime Replacement Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Age Replacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 Replacement with Minimal Repair . . . . . . . . . . . . . . . . . . . . . . 5.3 Middle Replacement Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Age Replacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.2 Comparison of Three Policies . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.3 Replacement with Minimal Repair . . . . . . . . . . . . . . . . . . . . . . 5.3.4 Comparisons of Three Policies . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

87 87 88 91 97 97 100 103 103 107 110 113 115 116

6 Which is Better for Standby or Parallel Systems . . . . . . . . . . . . . . . . . . . 6.1 Reliability Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 Constant Number of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.2 Random Number of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Replacement Policies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 MTTF for Replacement Time . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.2 Replacement for Number of Units . . . . . . . . . . . . . . . . . . . . . . 6.3 Scheduling of Random Works . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 Tandem and Parallel Works . . . . . . . . . . . . . . . . . . . . . . . . . . . .

117 118 118 120 123 123 124 127 127

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6.3.2 Standby and Parallel Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.3 Random Tandem Works . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Modified Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

130 134 137 143 143

7 Which is Better Problems in Shock and Damage Models . . . . . . . . . . . 7.1 Three Replacement Policies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Optimal Policies with Two Variables . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Exponential Shock Times and Amount of Damage . . . . . . . . . . . . . . 7.4 Replacement Overtime Policies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Modified Replacement Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

145 146 150 154 158 163 173

8 Which is Better Problems in Backup Models . . . . . . . . . . . . . . . . . . . . . . 8.1 Database Backup Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 Optimal Backup Times . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Overtime Backup First . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 Optimal Backup Times . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Comparisons of Backup First and Overtime . . . . . . . . . . . . . . . . . . . . 8.4 Database Backup Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.1 Backup Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.2 Backup Overtime Last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

175 176 179 183 185 189 196 196 198 203 203

9 Which is Better Problems in Checkpoint Models . . . . . . . . . . . . . . . . . . 9.1 Periodic Checkpoint Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Random Checkpoint Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.1 Two Kinds of Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.2 Two Kinds of Checkpoints . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Two Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Imperfect Checkpoint Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.1 Periodic Checkpoint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.2 Random Checkpoint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

205 206 211 211 213 217 221 221 223 225 225

Appendix A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 Appendix B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277

Chapter 1

Introduction

It has been wellknown that Probability arose in the letter from Pascal to Fermat around 1654 on such questions as the fair division of the stakes in the game of change [1, p. 239]: In a game where Player A needs r points to win and Player B needs s points to win, and the correct division of the stakes between players A and B is given by Player A:

) s−1 ( r +s−1 Σ Σ (r + s − 1) r +s−1 , Player B: . k k k=0 k=s

(1) Which is more profitable, Player A or Player B ? In our daily life, we are frequently faced with the problems about Which is ∼er, A or B ? For examples: (2) When you buy 10 products of A, which is cheaper, 10 products with x1 free sales or 10 products with x2 % discounts ? (3) Product A has high cost with long life and Product B has low cost with short life, which is better to buy, Product A or Product B ? In the field of reliability, Barlow RE and Proschan F [2, p. 65] mentioned that we can achieve a specified mean time between failures by either changing the replacement time or increasing the number of redundant units. Not only that, A or B problems are usually encountered such as (1) standby or parallel systems, (2) replacement or repair policies, (3) maintenance first or last policies, (4) discrete or continuous replacement policies, (5) periodic or random inspection policies, (6) incremental or differential backup policies, and etc., in the viewpoint of reliability and maintainability, respectively. We call such A or B problems as Which-Is-Better (WIB) problems in reliability theory, which was firstly defined in the chapter of the book edited by Misra KB [3]. Not only that, we actually have published many research papers on A or B problems © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Mizutani et al., Which-Is-Better (WIB): Problems in Reliability Theory, Springer Series in Reliability Engineering, https://doi.org/10.1007/978-3-031-27316-2_1

1

2

1 Introduction

over the last ten years, one of the most appeared WIB problems is: Which is better, maintenance first or last policies ? In addition, when the first and last are extended overtime policies, how about WIB problems of first, last and overtime ? We aim to write this book from the viewpoint of WIB problems in reliability theory. In this chapter, we will survey WIB problems of maintenance policies and reliability applications, which are mainly based our research publications and, then, describe briefly further studies of WIB problems presented in this book. We believe that our efforts would give great benefits for students, researchers, engineers and managers who are usually troubled and worried with such WIB problems.

1.1 Maintenance Policies A very first appeared random maintenance was mentioned in the booked written by Nakagawa T [4, p. 245], in which a typical application is that maintenance can not be done in a strictly periodic fashion, but it should be performed at random times for the working completions. After that, a first WIB problem was studied for the mean times of random replacement policies, age replacement and random replacement, periodic inspection and random inspection [5]. However, it is more practical to plan age replacement with random replacement as alternatives, e.g., replacement policies based on the assumptions of whichever occurs first and whichever occurs last [6, 7], which were officially named as replacement first and replacement last and their typical WIB problems were obtained [8–11]. Other WIB problems of replacement first and last polices have also been studied, e.g., replacement with products update announcements [12], age and periodic replacement with two failure modes [13], replacement with time of operations, mission durations, minimal repairs and maintenance triggering approaches [14], and etc. Another WIB problem of age and random replacement polices is to find lower costs and its replacement policies for random replacement [15]. It would be nature that random replacement can save costs for replacement without interrupting working completions and it should be better than or at least equals to age replacement, which have also been summarized [16]. From such viewpoints, we find the best combination of age and random policies with different replacement costs. But, it is not easy to obtain the analytical solutions, so that in our most research papers, we had assumption that the costs of preventive replacement policies are the same with each other, e.g., replacement first, last and overtime policies[17], replacement policies with general models [18]. Regarding to the above overtime policies, it was assumed that replacement is done at the first completion of the working completion over planned replacement times such as age and working number, which were named as replacement overtime [6, 19]. Obviously, higher failure probability due to replacement delay for replacement overtime occurs, so that the answer of WIB problems should be age replacement is better than replacement overtime if we do not consider the penalty of working interruptions [19]. Replacement overtime have been extended for finite operating

1.2 Reliability Applications

3

intervals and random working cycles [20, 21]. In addition, the modeling of first and last have been extended with overtime policies, e.g., inspection first, inspection last and inspection overtime policies, which have arisen new WIB problems [22]. Various replacement overtime policies and their WIB problems have been summarized [23]. Which is better, which replacement is better, or comparisons of replacement policies, actually have been mentioned in the title of our research papers. For examples, Comparisons of periodic and random replacement policies [24], where WIB problems arise when periodic replacement, replacement first, replacement last and replacement overtime with minimal repairs are modeled; Which is better for replacement policies with continuous or discrete scheduled times [25], where WIB problems arise when replacement policies that are planned at continuous and discrete times; Comparisons of maintenance policies with periodic times and repair numbers [26], where WIB problems arise when replacement policies that are planned at periodic times or number of minimal repairs. Which replacement is better at working cycles or number of failures [27], where WIB problems arise when replacement policies that are planned at discrete working cycles or number of failures. Recently, WIB problems were defined in maintenance reliability policies [3], and maintenance policies for reliability systems have been compared. Anyhow, when the modelings of first, last and overtime are used, WIB problems should always be studied in our research, e.g., What is middle maintenance policy [28], WIB problems would be combining first, last and overtime modelings. Most recently, new concepts of deviation cost and deviation time have been modeled into age replacement, random replacement, replacement first and replacement last. One contribution of deviation cost modeling is to find optimal solutions for agebased replacement models with exponential failure distributions [29]. Actually, we had a very first breakthrough of age replacement model with an exponential failure distribution by introducing two kinds of costs, i.e., excess cost and shortage cost [5], which have also been taken up for periodic replacement policies [30]. Replacing deviation cost with excess cost and shortage cost, age replacement models for a parallel system have been studied when the lifetime of unit has an exponential distribution, however, the failure distribution of a whole system is not exponential [31, 32]. Similarly, optimal design and replacement policies for k-out-of-n systems with deviation time and cost have been studied [33]. Furthermore, we have found WIB problems of replacement policies with deviation costs or not, or their combinations with traditional assumptions of replacement costs.

1.2 Reliability Applications High system reliability can be achieved through redundancy and maintenance policies [34]. In most parallel systems, the number n of units is constant, and predetermined. However, for some complex and large systems, we might not know the exact number of units, but we could estimate the probability distribution of the number of units. A representative example is the fracture of an aircraft fuselage with multiple site

4

1 Introduction

damage around the large number of rivets [35]. Such a parallel system with random number of units is named as random parallel system [16]. For a K -out-of-n system, it can be operating if and only if at least number K of the total number of n units are operating. When a K -out-of-n system is designed, both K and n should be decided at early designing phase to confirm the principal specifications. However, when n is large, i.e., the system has enough redundancy, K may not be predefined constantly, but it is a random variable with an estimated probability function [36]. New WIB problems have raised for the above random systems when planning maintenance policies with constant number n or random number n, constant number K or random number K [35, 36]. In addition, the WIB problems of standby and parallel systems were studied in reliability, replacement, scheduling and application [37]. Cumulative damage models play an important role in reliability theory, which are considered as a sequence of shocks that occur randomly and give some amount of damage to the operating unit. New damage is accumulated to the current damage level, weakens the unit gradually, and makes it failure when the total damage exceeds a failure level K [38, 39]. Maintenance actions are generally grouped into time-based maintenance (TBM), and condition-based maintenance (CBM). For cumulative damage models, TBM is considered as maintenance policies made at a planned time T or at periodic times J T , and CBM includes maintenance policies made at a damage level Z or at a shock number N [38, p. 103]. Furthermore, WIB problems were shown in numerical examples [38, p. 52] that maintenance done at Z is better than those at T and N , and maintenance done at N is better than that at T in most cases, in other words, CBM is better than TBM under whichever occurs first, i.e., maintenance first. When we take up maintenance last, WIB problems of maintenance first and last have been obtained [40]. When the above overtime policies are applied cumulative damage models, maintenance policies with working completions and their WIB problems have been studied [21, 28, 41, 42]. A typical reliability application of cumulative damage models is backup policies for database systems, as random updates of database systems with random volumes form a compound stochastic process, which have similar formulations of shock and damage processes [38]. In our research, WIB problems arose among full backup, incremental backup and differential backup policies. It has been shown that full backup is a lazy but simple mode that exports all the data files updated since the last full backup, incremental backup is much smaller and quicker than full backups, and differential backup has quicker recovery time, as it requires only a full backup and the last differential backup to restore the updated data [43, 44]. When the above working completions are applied, WIB problems of backup policies arise among constant backup, random backup, backup first, backup last, and so on [43, 45]. Another WIB problem of reliability applications is random and periodic checkpoint policies for error detections and recoveries in a computer system, which will be surveyed in the last chapter of this book [34, p. 123], [46]. Actually, when random and periodic maintenance policies are used, we always have WIB problems to be discussed in reliability applications [47–49].

References

5

References 1. David FN (1962) Games, gods, and gambling. Griffin Press, London 2. Barlow RE, Proschan F (1965) Mathematical theory of reliability. Wley, New York 3. Mizutani S, Zhao X, Nakagawa T (2021) WIB (Which-Is-Better) problems in maintenance reliability policies. In: Misra KB (ed) Handbook of advanced performability engineering. Springer, Switzerland, pp 523–547 4. Nakagawa T (2005) Maintenance theory of reliability. Springer, London 5. Nakagawa T, Zhao X, Yun W (2011) Optimal age replacement and inspection policies with random failure and replacement times. Inter J Reliab Qual Saf Eng 18:405–416 6. Chen M, Mizutani S, Nakagawa T (2010) Random and age replacement policies. Inter J Reliab Qual Saf Eng 17:27–39 7. Chen, M, Nakamura S, Nakagawa T (2010). Replacement and preventive maintenance models with random working times. IEICE Trans Fund E93.A:500–507 8. Zhao X, Nakagawa T (2012) Optimization problems of replacement first or last in reliability theory. Euro J Oper Res 223:141–149 9. Zhao X, Nakagawa T, Zuo M (2014) Optimal replacement last with continuous and discrete policies. IEEE Trans Reliab 63:868–880 10. Zhao X, Liu H, Nakagawa T (2015) Where does whichever occurs first hold for preventive maintenance modelings. Reliab Eng Syst Saf 142:203–211 11. Zhao X, Al-Khalifa KN, Hamouda A, Nakagawa T (2016) First and last triggering event approaches for replacement with minimal repairs. IEEE Trans Reliab 65:197–207 12. Mizutani S, Dong W, Zhao X, Nakagawa T (2020) Preventive replacement policies with products update announcements. Comm Stat Theory Method 49:3821–3833 13. Mizutani S, Zhao X, Nakagawa T (2021) Age and periodic replacement policies with two failure modes in general models. Reliab Eng Syst Saf 214:107754 14. Zhao X, Cai J, Mizutani S, Nakagawa T (2021) Preventive replacement policies with time of operations, mission durations, minimal repairs and maintenance triggering approaches. J Manuf Syst 6:819–829 15. Nakagawa T, Zhao X (2013) Comparisons of replacement policies with constant and random times. J Oper Res Japan 56:1–14 16. Nakagawa T (2014) Random maintenance policies. Springer, London 17. Zhao X, Al-Khalifa KN, Hamouda A, Nakagawa T (2017) Age replacement models: A summary with new perspectives and methods. Reliab Eng Syst Saf 161:95–105 18. Chen M, Zhao X, Nakagawa T (2019) Replacement policies with general models. Ann Oper Res 277:47–61 19. Zhao X, Qian C, Nakamura S (2014) Age and periodic replacement models with overtime policies. Int J Reliab Qual Saf Eng 21:1450016 20. Mizutani S, Zhao X, Nakagawa T (2015). Overtime replacement policies with finite operating interval and number. IEICE Trans Fund E98-A:2069–2076 21. Zhao X, Nakagawa T (2016) Over-time and over-level replacement policies with random working cycles. Ann Oper Res 244:103–116 22. Zhao X, Nakagawa T (2015) Optimal periodic and random inspection with first, last, and overtime policies. Inter J Syst Sci 46:1648–1660 23. Nakagawa T, Zhao X (2015) Maintenance overtime policies in reliability theory. Springer, London 24. Zhao X, Nakagawa T (2013) Comparisons of periodic and random replacement policies. In: Frenkel I et al (eds) Applied reliability engineering and risk analysis: Probilistic models and statistical inference. Wiley, pp 193–204 25. Zhao X, Mizutani S, Nakagawa T (2015) Which is better for replacement policlies with continuous or discrete scheduled times. Euro J Oper Res 242:477–486 26. Zhao X, Qian C, Nakagawa T (2017) Comparisons of maintenance policies with periodic times and repair numbers. Reliab Eng Syst Saf 168:161–170

6

1 Introduction

27. Mizutani S, Zhao X, Nakagawa T (2020) Which replacement is better at working cycles or number of failures. IEICE Trans Fund E103. A:523–532 28. Zhao X, Al-Khalifa KN, Hamouda A, Nakagawa T (2016) What is middle maintenance policy. Qual Reliab Eng Inter 32:2403–2414 29. Zhao X, Li B, Mizutani S, Nakagawa T (2021) A revisit of age-based replacement models with exponential failure distributions. IEEE Trans Reliab. https://doi.org/10.1109/TR.2021. 3111682 30. Zhao X, Chen M, Nakagawa T (2022) Periodic replacement policies with shortage and excess costs. Ann Oper Res 311:469–487 31. Zhao X, Chen M, Nakagawa T (2016) Replacement policies for a parallel system with shortage and excess costs. Reliab Eng Syst Saf 150:89–95 32. Zhao X, Mizutani S, Chen M, Nakagawa T (2022) Preventive replacement policies for parallel systems with deviation costs between replacement and failure. Ann Oper Res 312:533–551 33. Wang J, Zhao X, Xiang J (2022) Optimum design and replacement policies for k-out-of-n systems with deviation time and cost. Ann Oper Res. https://doi.org/10.1007/s10479-02205043-1 34. Nakagawa T (2008) Advanced reliability models and maintenance policies. Springer, London 35. Nakagawa T, Zhao X (2012) Optimization problems of a parallel system with a random number of units. IEEE Trans Reliab 61:543–548 36. Ito K, Zhao X, Nakagawa T (2017) Random number of units for K -out-of-n systems. App Mathl Model 45:563–572 37. Zhao X, Chen M, Nakagawa T (2016) Comparisons of standby and parallel systems in reliability, replacement, scheduling and application. Proc Inst Mech Eng Part O: J Risk Reliab 230:101–108 38. Nakagawa T (2007) Shock and damage models in reliability theory. Springer, London 39. Zhao X, Nakagawa T (2018) Advanced maintenance policies for shock and damage models. Springer, Switzerland 40. Zhao X, Qian CH, Nakagawa T (2013) Optimal policies for cumulative damage models with maintenance last and first. Reliab Eng Syst Saf 110:50–59 41. Zhao X, Nakamura S, Nakagawa T (2013) Optimal maintenance policies for cumulative damage models with random working times. J Qual Maint Eng 19:25–37 42. Zhao X, Qian C, Sheu S (2014) Cumulative damage models with random working times. In: Nakamura S, Qian C, Chen M (eds) Reliability modeling with applications. World Scientific, pp 79–98 43. Nakamura S, Zhao X, Nakagawa T (2017) Constant and random full backup models with incremental and differential backup schemes. Inter J Reliab Qual Saf Eng 24:1750015 44. Zhao X, Nakamura S, Qian C, Sheu S (2017) Cumulative backup policies for database systems. In: Nakamura S, Qian C, Nakagawa T (eds) Reliability modeling with computer and maintenance applications. World Scientific, pp 235–254 45. Zhao X, Wang D, Mizutani S, Nakagawa T (2022) Data backup policies with failure-oblivious computing in reliability theory. Ann Oper Res. https://doi.org/10.1007/s10479-022-04941-8 46. Naruse K, Nakagawa T (2020) Optimal checking intervals, schemes and structures for computing modules. In: Pham H (ed) Reliability and statistical computing. Springer, London, pp 265–287 47. Zhao X, Nakamura S, Nakagawa T (2011) Two generational garbage collection models with major collection time. IEICE Trans Fund E94-A:1558–1566 48. Chen M, Zhao X, Nakamura S (2014) Periodic and random inspections for a computer system. In: Nakamura S, Qian C, Chen M (eds) Reliability modeling with applications. World Scientific, pp 249–267 49. Zhao X, Zheng Q, Jia X, Qian C, Mizutani S, Chen M (2022) Periodic and sequential inspection policies with mission failure probabilities. Qual Reliab Eng Inter 38:1539–1557

Chapter 2

Random Age Replacement Model

We consider the age replacement model in which the unit is replaced at time T or at failure, whichever occurs. The following five policies based on the age replacement are taken up: (1) Random replacement, (2) Age replacement, (3) Replacement first, (4) Replacement last, and (5) Replacement overtime. Comparing them, we propose Which-Is-Better (WIB) problems appeared in such models and make theoretical and numerical discussions for answering to these problems. Throughout this chapter, we make the following assumptions: Let X (0 < X ≤ ∞) be a random variable of the failure time of an operating ʃ ∞ unit which has a general distribution F(t) ≡ Pr{X ≤ t} with finite mean μ ≡ 0 F(t)dt, a density function f (t) ≡ dF(t)/dt, and failure rate h(t) ≡ f (t)/ F(t), where Φ(t) ≡ 1 − Φ(t) for a general function Φ(t). It is assumed that h(t) increases strictly with t from h(0) = 0 to h(∞) ≡ limt→∞ h(t) = ∞, except that the failure time is exponential. Furthermore, the unit operates for random working times Yn (n = 1, 2, . . . ) (0 < Yn < ∞). It is assumed that Yn are independent random variables of X and Y j ( j /= n), have an identical distribution G(t) ≡ Pr{Yn ≤ t} with finite mean 1/θ ≡ ʃand ∞ G(t)dt. n-fold Stieltjes convolution of G(t) is G (n) (t) ≡ Pr{Y1 + Y2 + · · · + 0 ʃThe ∞ (n−1) (t − u)dG(u) (n = 1, 2, . . . ) and G (0) (t) ≡ 1 for t ≥ 0. Yn ≤ t} = 0 G

2.1 Random Replacement Suppose that the unit is replaced at random Y (0 < Y ≤ ∞) or at failure, whichever occurs first, where Y is a random variable with a general distribution G(t) ≡ Pr{Y ≤ t} and is independent of the failure time X . This is called random replacement. Then, the probability that the unit is replaced at random time Y is

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Mizutani et al., Which-Is-Better (WIB): Problems in Reliability Theory, Springer Series in Reliability Engineering, https://doi.org/10.1007/978-3-031-27316-2_2

7

8

2 Random Age Replacement Model

ʃ

∞

Pr{Y ≤ X } =

ʃ

∞

F(t)dG(t) =

G(t)dF(t), 0

0

and the probability that it is replaced at failure is ʃ

∞

Pr{X ≤ Y } =

ʃ

∞

G(t)dF(t) =

F(t)dG(t). 0

0

Thus, the mean time to replacement is ʃ

∞

ʃ t F(t)dG(t) +

∞

ʃ

0

0

∞

t G(t)dF(t) =

G(t) F(t)dt. 0

Therefore, the expected cost rate is [1, p. 86], [2, p. 247], [3, p. 20] C(G) =

cF

ʃ∞ 0

ʃ∞ F(t)dG(t) + c R 0 F(t)dG(t) , ʃ∞ 0 G(t) F(t)dt

(2.1)

where c R = replacement cost at time Y and c F = replacement cost at failure with cF > cR . It has been shown [1, p. 87], [3, p. 21], [4] that (2.1) is ʃ∞ C(G) = ʃ0∞ 0

S1 (t)dG(t) S2 (t)dG(t)

,

where ʃ S1 (t) ≡ c F F(t) + c R F(t),

S2 (t) ≡

t

F(u)du. 0

Suppose that there exists a minimum value T of S1 (t)/S2 (t) for 0 < t ≤ ∞. We easily have ʃ ʃ ∞ S1 (T ) ∞ S2 (t)dG(t), S1 (t)dG(t) ≥ S2 (T ) 0 0 because

S1 (t) S1 (T ) ≥ . S2 (T ) S2 (t)

Thus, C(G) ≥ C(G T ) =

cT + (c F − cT )F(T ) S1 (T ) = ≡ C(T ), ʃT S2 (T ) 0 F(t)dt

(2.2)

2.1 Random Replacement

9

where cT = replacement cost at time T , and G T (t) is the degenerate distribution placing unit mass at T , i.e., G T (t) ≡ 1 for t ≥ T and 0 for t < T . If T = ∞ then the unit is replaced only at failure. This shows age replacement is better than random replacement. We find optimal T ∗ to minimize the expected cost rate C(T ) in (2.2) when failure rate h(t) increases strictly with t from 0 to h(∞) = ∞ [1, p. 87], [2, p. 74]. Differentiating C(T ) with respect to T and setting it equal to zero, ʃ

T

h(T ) 0

cT , or F(t)dt − F(T ) = c F − cT

ʃ

T

F(t)[h(T ) − h(t)]dt =

0

cT . c F − cT (2.3)

There exists a finite and unique T ∗ (0 < T ∗ < ∞) which satisfies (2.3), and the resulting cost rate is (2.4) C(T ∗ ) = (c F − cT )h(T ∗ ).

WIB 2.1: Age replacement with time T is better than random replacement.

It has been shown that when cT = c R , age replacement is better than random one. c R in which C(G) in (2.1) is the same as C(T ) in (2.2). When cT > c R , we compute ⌃ c R for given c F , cT , G(t) and We specify the computing procedure for obtaining ⌃ F(t) [3, p. 25], [4]: Setting C(G) = C(T ∗ ), i.e., ʃ∞ ⌃ c R + (c F − ⌃ c R ) 0 G(t)dF(t) = (c F − cT )h(T ∗ ), ʃ∞ G(t) F(t)dt 0 c R which satisfies the above equation. we compute ⌃ Example 2.1 Table 2.1 presents T ∗ , random cost ⌃ c R and cost difference rate 2 c R )/cT for cT when G(t) = 1 − e−t , F(t) = 1 − e−(t/10) and c F = 100. This (cT − ⌃ ∗ c R and (cT − ⌃ c R )/cT increase with cT , i.e., when cT are large, both indicates that T , ⌃ ⌃ c R and (cT − ⌃ c R )/cT are large. In other words, when cT become large, ⌃ c R become also large, however, do not become so large because the random time is constant, c R )/cT increase with cT . which follows that (cT − ⌃ 

10

2 Random Age Replacement Model

c R when G(t) = 1 − e−t , F(t) = 1 − e−(t/10) and c F = 100 Table 2.1 Optimal T ∗ , C(T ∗ ) and ⌃ cT −⌃ cR cT T∗ C(T ∗ ) ⌃ cR cT 2

2.304 3.365 4.264 5.107 6.790 8.646 10.908

5 10 15 20 30 40 50

4.378 6.056 7.248 8.171 9.506 10.376 10.908

2.451 4.129 5.321 6.244 7.578 8.448 8.981

0.510 0.587 0.645 0.688 0.747 0.789 0.820

WIB 2.2: When c R < ⌃ c R , random replacement is better than age replacement with time T .

2.2 Replacement First It is assumed that Yn (n = 1, 2, . . . ) is the working time of job n in Fig. 2.1, and is independent and has an identical distribution G(t) ≡ Pr{Yn ≤ t} with finite mean 1/θ (0 < 1/θ < ∞). Then, the probability that the unit works exactly n times in [0, t] is G (n) (t) − G (n+1) (t) (n = 0, 1, 2, . . . ). Suppose that the unit is replaced at a planned time T (0 < T ≤ ∞), at a planned number N (N = 1, 2, . . . ) of working times or at failure, whichever occurs first. This is called replacement first [3, p. 42]. Then, the probability that the unit is replaced at time T is [1 − G (N ) (T )] F(T ), the probability that it is replaced at work N is ʃ

T

F(t)dG (N ) (t),

0

Y1

Y2

0 Fig. 2.1 Process of working times Yn

Y3

Yn T

time

2.2 Replacement First

11

and the probability that it is replaced at failure is ʃ

T

[1 − G (N ) (t)]dF(t).

0

Thus, the mean time to replacement is ʃ

(N )

T

T [1 − G (T )] F(T ) + t F(t)dG 0 ʃ T [1 − G (N ) (t)] F(t)dt. =

(N )

ʃ

T

(t) +

t[1 − G (N ) (t)]dF(t)

0

0

Therefore, the expected cost rate is [3, p. 43] C F (T , N ) =

ʃT ʃT (N ) (t)]dF(t) + (c N −cT ) 0 F(t)dG (N ) (t) 0 [1−G , ʃT (N ) (t)]F(t)dt 0 [1 − G (2.5)

cT +(c F −cT )

where cT = replacement cost at time T , c N = replacement cost at work N and c F = replacement cost at failure with c F > cT and c F > c N . In particular, C F (T , ∞) ≡ lim N →∞ C F (T, N ) = C(T ) in (2.2). In particular, when the unit is replaced only at work N (N = 1, 2, . . . ), ʃ∞ c F − (c F − c N ) 0 G (N ) (t)dF(t) . C(N ) ≡ lim C F (T , N ) = ʃ∞ (N ) (t)]F(t)dt T →∞ 0 [1 − G

(2.6)

We find optimal N ∗ to minimize C(N ) when h(t) increases strictly with t to ∞ [3, p. 44]. Forming the inequality C(N + 1) − C(N ) ≥ 0, ʃ

∞

Q(N )

[1 − G (N ) (t)] F(t)dt −

0

ʃ

∞

[1 − G (N ) (t)]dF(t) ≥

0

cN , cF − cN

(2.7)

where ʃT Q 1 (T, N ) ≡ ʃ 0T 0

[G (N ) (t) − G (N +1) (t)]dF(t)

[G (N ) (t) − G (N +1) (t)]F(t)dt

< h(T ),

Q(N ) ≡ Q 1 (∞, N ) = lim Q 1 (T , N ). T →∞

It is easily proved that if Q(N ) increases strictly with N to Q(∞), then the lefthand side of (2.7) also increases strictly with N to μQ(∞) − 1. Thus, if Q(∞) > c F /[(c F − c N )μ], then there exists a finite and unique minimum N ∗ (1 ≤ N ∗ < ∞) which satisfies (2.7).

12

2 Random Age Replacement Model

In particular, when G(t) = 1 − e−θt , i.e., G (N ) (t) = 0, 1, 2, . . . ), from (1) of Appendix A.1, ʃT Q 1 (T , N ) = ʃ 0T 0

(θt) N e−θt dF(t)

(θt) N e−θt F(t)dt

∑∞

n −θt n=N [(θt) /n!]e

< h(T )

(N =

(2.8)

increases strictly with T from h(0) to Q(N ) and increases strictly with N from ʃT Q 1 (T ) ≡ Q 1 (T, 0) = ʃ 0T 0

e−θt dF(t) e−θt F(t)dt

to h(T ). Thus, noting that Q(N ) increases strictly with N to h(∞) = ∞, there exists a finite and unique minimum N ∗ (1 ≤ N ∗ < ∞) which satisfies (2.7). Next, when cT = c N , G(t) = 1 − e−θt and h(t) increases strictly with t to ∞, we find optimal TF∗ and N F∗ to minimize C F (T , N ) in (2.5). Differentiating C F (T , N ) with respect to T and setting it equal to zero, ʃ

T

h(T )

[1 − G (N ) (t)] F(t)dt −

0

ʃ

T

[1 − G (N ) (t)]dF(t) =

0

cT , c F − cT

or ʃ

T

[1 − G (N ) (t)] F(t)[h(T ) − h(t)]dt =

0

cT , c F − cT

(2.9)

which agrees with (2.3) when N = ∞. Noting that the left-hand side of (2.9) increases strictly with T from 0 to ∞, there exists a finite and unique TF∗ (0 < TF∗ < ∞) which satisfies (2.9), and the resulting cost rate is C(TF∗ , N ) = (c F − cT )h(TF∗ ).

(2.10)

Furthermore, because the left-hand side of (2.9) increases strictly with N to that of (2.3), TF∗ decreases with N to T ∗ given in (2.3), and TF∗ ≥ T ∗ . Forming the inequality C F (T , N + 1) − C F (T , N ) ≥ 0, ʃ

T

[1 − G (N ) (t)] F(t)[Q 1 (T , N ) − h(t)]dt ≥

0

Substituting (2.9) for (2.11), (2.11) is Q 1 (T , N ) ≥ h(T ),

cT . c F − cT

(2.11)

2.3 Replacement Last

13

Table 2.2 Optimal T ∗ and C(T ∗ )/cT , and optimal N ∗ and C(N ∗ )/c N when G(t) = 1 − e−t , 2 F(t) = 1 − e−(t/10) and c F = 100 cN cT N∗ C(N ∗ )/c N T∗ C(T ∗ )/cT or c F − cT cF − cN 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

3 5 6 7 8 9 10 11 12 13

0.73 0.49 0.40 0.34 0.31 0.28 0.26 0.24 0.23 0.22

3.19 4.55 5.62 6.54 7.38 8.15 8.89 9.58 10.26 10.91

0.64 0.45 0.37 0.33 0.30 0.27 0.25 0.24 0.23 0.22

which does not hold for any T . Thus, there does not exist any finite N F∗ , i.e., N F∗ = ∞ and TF∗ = T ∗ in (2.3), which follows that age replacement with time T is better than replacement first. In addition, noting that the left-hand side of (2.11) increases strictly with N to that of (2.3), if T > T ∗ , then there exists a finite and unique minimum N F∗ (1 ≤ N F∗ < ∞) which satisfies (2.11). Furthermore, because the left-hand side of (2.11) increases with T to that of (2.7), N F∗ decreases with T to N ∗ given in (2.7), and N F∗ ≥ N ∗ . Conversely, if T ≤ T ∗ , then N F∗ = ∞. Example 2.2 Table 2.2 presents optimal T ∗ which satisfies (2.3) and the resulting cost rate C(T ∗ )/cT in (2.4), and optimal N ∗ which satisfies (2.7) and C(N ∗ )/c N in (2.6). This indicates that both T ∗ and N ∗ increase with ci /(c F − ci ) (i = T , N ), and N ∗ > T ∗ except ci /(c F − ci ) = 0.1. Furthermore, C(N ∗ )/c N > C(T ∗ )/cT , however, their differences are smaller as ci /(c F − ci ) are larger, i.e., both T ∗ and N ∗ are larger.  WIB 2.3: Age replacement with time T is better than replacement with work N .

2.3 Replacement Last Suppose that the unit is replaced preventively at time T (0 ≤ T < ∞) or at work N (N = 0, 1, 2, . . . ), whichever occurs last. This is called replacement last [3, p. 46]. Then, the probability that the unit is replaced at time T is

14

2 Random Age Replacement Model

G (N ) (T ) F(T ), the probability that it is replaced at work N is ʃ

∞

F(t)dG (N ) (t),

T

and the probability that it is replaced at failure is ʃ

∞

F(T ) +

[1 − G (N ) (t)]dF(t).

T

Thus, the mean time to replacement is (N )

ʃ

T G (T ) F(T ) + ʃ ʃ T F(t)dt + = 0

∞

t F(t)dG T ∞

(N )

ʃ

T

(t) +

ʃ

∞

t dF(t) +

0

t[1 − G (N ) (t)]dF(t)

T

[1 − G (N ) (t)] F(t)dt.

T

Therefore, the expected cost rate is [3, p. 47] ʃ∞ cT + (c N − cT ) T F(t)dG (N ) (t) ʃ∞ ┌ ┐ + (c F − c N ) F(T ) + T [1 − G (N ) (t)]dF(t) C L (T , N ) = , ʃ∞ ʃT (N ) (t)] F(t)dt 0 F(t)dt + T [1 − G

(2.12)

where cT , c N and c F are given in (2.5). Note that C L (T , 0) = C(T ) in (2.2) and C L (0, N ) = C(N ) in (2.6). When cT = c N and h(t) increases strictly with t from 0 to ∞, we find optimal TL∗ and N L∗ to minimize C L (T , N ) in (2.12). Differentiating C L (T , N ) with respect to T and setting it equal to zero, ʃ 0

T

ʃ

∞

F(t)[h(T ) − h(t)]dt +

[1 − G (N ) (t)] F(t)[h(T ) − h(t)]dt ≥

T

cT , c F − cT (2.13)

which agrees with (2.3) when ʃ ∞ N = 0. Noting that the left-hand side of (2.13) increases strictly with T from − 0 [1 − G (N ) (t)]dF(t) to ∞. Thus, there exists a finite and unique TL∗ (0 < TL∗ < ∞) which satisfies (2.13), and the resulting cost rate is C(TL∗ , N ) = (c F − cT )h(TL∗ ).

(2.14)

Furthermore, because the left-hand side of (2.13) decreases with N from that of (2.3), TL∗ increases with N from T ∗ given in (2.3), and TL∗ ≥ T ∗ .

2.3 Replacement Last

15

Forming the inequality C L (T , N + 1) − C L (T , N ) ≥ 0, ┐ ┌ʃ T ʃ ∞ ʃ ∞ (N ) ˜ F(t)dt + [1−G (t)] F(t)dt − F(T )− [1−G (N ) (t)]dF(t) Q 1 (T , N ) 0

T

T

cT , ≥ c F − cT

(2.15)

where ʃ ∞ (N ) (t) − G (N +1) (t)]dF(t) T [G ˜1 (T, N ) ≡ ʃ ∞ > h(T ). Q (N ) (t) − G (N +1) (t)]F(t)dt T [G Substituting (2.13) for (2.15), (2.15) is ˜1 (T , N ) ≥ h(T ), Q which always holds for any T . Thus, C L (T , N + 1) − C L (T, N ) ≥ 0 for any N , i.e., N L∗ = 0 and TL∗ = T ∗ , which follows that age replacement with time T is better than replacement last. In particular, when G(t) = 1 − e−θt , from (2) of Appendix A.1, ʃ∞ N −θt dF(t) T (θt) e ˜ Q 1 (T, N ) ≡ ʃ ∞ > h(T ) N −θt F(t)dt T (θt) e

(2.16)

increases strictly with T from Q(N ) in (2.7) to h(∞) and increases strictly with N from ʃ ∞ −θt dF(t) T e ˜ ˜ Q 1 (T ) ≡ Q 1 (T , 0) = ʃ ∞ −θt F(t)dt T e ˜1 (T ) increases strictly with T from Q ˜1 (0) to h(∞). to h(∞), where Q In addition, note that the left-hand side of (2.15) increases strictly with N from ˜1 (T ) Q

ʃ

T 0

ʃ

T

F(t)dt − F(T ) > h(T )

F(t)dt − F(T ),

0

which agrees with that of (2.3). Thus, if T < T ∗ , then there exists a finite and unique minimum N L∗ (0 ≤ N L∗ < ∞) which satisfies (2.15). Furthermore, because the lefthand side of (2.15) increases with T from that of (2.7), N L∗ decreases with T from N ∗ given in (2.7) to 0, and N L∗ ≤ N ∗ . Conversely, if T ≥ T ∗ , then N L∗ = 0.

16

2 Random Age Replacement Model

WIB 2.4: When G(t) = 1 − e−θt , if T > T ∗ given in (2.3), then replacement first is better than replacement with time T , if T = T ∗ , then replacement with time T ∗ is better than replacement first and last, and if T < T ∗ , then replacement last is better than replacement with time T .

2.4 Replacement First and Last Suppose that the unit is replaced preventively at time T or time Y , where T is constant and Y has a general distribution G(t) ≡ Pr{Y ≤ t} with finite mean 1/θ ≡ ʃ∞ G(t)dt. When cT = c N and the unit is replaced preventively at time T or time 0 Y , whichever occurs first, the expected cost rate is [5], from (2.5), ʃT cT + (c F − cT ) 0 G(t)dF(t) . C F (T ) = ʃT 0 G(t) F(t)dt

(2.17)

Optimal TF∗ to minimize C F (T ) satisfies, from (2.9), ʃ

T

G(t) F(t)[h(T ) − h(t)]dt =

0

cT , c F − cT

(2.18)

and the resulting cost rate is C F (TF∗ ) = (c F − cT )h(TF∗ ).

(2.19)

When cT = c N and the unit is replaced preventively at time T or time Y , whichever occurs last, the expected cost rate is [5], from (2.12), C L (T ) =

ʃ∞ cT + (c F − cT )[F(T ) + T G(t)dF(t)] . ʃ∞ ʃT F(t)dt + G(t) F(t)dt 0 T

(2.20)

Optimal TL∗ to minimize C L (T ) satisfies, from (2.13), ʃ

T

ʃ

∞

F(t)[h(T ) − h(t)]dt +

G(t) F(t)[h(T ) − h(t)]dt =

T

0

cT , c F − cT

(2.21)

and the resulting cost rate is C L (TL∗ ) = (c F − cT )h(TL∗ ).

(2.22)

2.4 Replacement First and Last

17

We compare replacement first and last when h(t) increases strictly with t to ∞ [3, p. 38]. It is firstly noted from (2.19) and (2.22) that if TF∗ ≤ TL∗ , then replacement first is better than replacement last, and vice versa. Compare the left-hand side L F (T ) and L L (T ) of (2.18) and (2.21), respectively. Letting L(T ) ≡ L L (T ) − L F (T ) ʃ ʃ T G(t) F(t)[h(T ) − h(t)]dt − = 0

∞

G(t) F(t)[h(t) − h(T )]dt, (2.23)

T

ʃ∞ which increases strictly with T from − 0 G(t)dF(t) < 0 to ∞. Thus, there exists a finite and unique T A (0 < T A < ∞) which satisfies L(T ) = 0. Therefore, if L F (T A ) ≥ cT /(c F − cT ), then TF∗ ≤ TL∗ and if L F (T A ) < cT /(c F − cT ), then TL∗ < TF∗ . This means that if the ratio of cT /c F is larger than L(T A )/[1 + L(T A )], i.e., cost c F is nearly to cT , then TL∗ < TF∗ . Example 2.3 When F(t) = 1 − exp[−(t/10)2 ] and G(t) = 1 − e−θt , Table 2.3 presents optimal T ∗ , TF∗ , TL∗ and T A which satisfy (2.3), (2.18), (2.21) and L(T A ) = 0 in (2.23), respectively, and from (2.18), ʃ L F (T A ) =

TA

e−θt F(t)[h(T A ) − h(t)]dt.

0

This indicates that both TF∗ and TL∗ increase with cT /(c F − cT ), i.e., decrease with c F /cT . When c F /cT increase, the replacement times have to be smaller to prevent a high replacement cost. In other words, the unit can work longer as c F /cT become smaller. So that, replacement last is better than replacement first when c F /cT become Table 2.3 Optimal TF∗ , TL∗ , T A , T ∗ and L F (T A ) when F(t) = 1 − e−(t/10) and G(t) = 1 − e−θt cT 1/θ = 1 1/θ = 3 1/θ = 5 1/θ = 7 1/θ = 10 1/θ = ∞ c F − cT ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ TF TL TF TL TF TL TF TL TF TL∗ T∗ 2

0.1 0.3 0.5 0.7 1.0 3.0 5.0 7.0 10.0 TA L F (T A )

0.34 0.62 0.84 1.04 1.33 3.17 5.00 6.83 9.58

0.48 0.63 0.77 0.91 1.10 2.26 3.39 4.51 6.21 0.65 0.33

0.32 0.58 0.77 0.94 1.16 2.53 3.88 5.23 7.25

0.56 0.69 0.81 0.94 1.11 2.26 3.39 4.51 6.21 0.93 0.70

0.32 0.57 0.76 0.92 1.13 2.42 3.68 4.94 6.82

0.58 0.71 0.83 0.94 1.12 2.26 3.39 4.51 6.21 1.05 0.89

0.32 0.57 0.75 0.91 1.12 2.37 3.59 4.81 6.64

0.59 0.71 0.83 0.95 1.12 2.26 3.39 4.51 6.21 1.13 1.01

0.32 0.57 0.75 0.90 1.11 2.34 3.53 4.72 6.51

0.60 0.72 0.84 0.95 1.12 2.26 3.39 4.51 6.21 1.20 1.14

0.32 0.56 0.74 0.89 1.09 2.26 3.39 4.51 6.21

18

2 Random Age Replacement Model

smaller. For example, when cT /(c F − cT ) = 0.7, i.e., c F /cT = 17/7 and 1/θ = 1.0, TL∗ = 0.91 is much smaller than TF∗ = 1.04. When L F (T A ) ≥ cT /(c F − cT ), TF∗ ≤ TL∗ and replacement first is better than replacement last, and conversely, when L F (T A ) < cT /(c F − cT ), TF∗ > TL∗ and replacement last is better than replacement first. For example, when 1/θ = 1.0, L F (T A ) = 0.33, and TF∗ = 0.62 < TL∗ = 0.63 for cT /(c F − cT ) = 0.3, and TL∗ = 0.77 < TF∗ = 0.84 for cT /(c F − cT ) = 0.5. Similarly, TF∗ decrease with 1/θ to T ∗ and TL∗ increase with 1/θ from T ∗ . So that, replacement first is better than replacement last as 1/θ become larger. For example, when cT /(c F − cT ) = 0.7, if 1/θ ≤ 3, then replacement last is better than replacement first, and if 1/θ > 5, then replacement first is better than replacement last. 

WIB 2.5: Replacement first is better than replacement last as both c F /cT and 1/θ become larger, and replacement last is better than replacement first as both c F /cT and 1/θ become smaller.

2.5 Replacement Overtime It might be of advantage to replace an operating unit at the completion of its working time even if T comes because it continues to operate for some job [6]. Suppose that the unit is replaced preventively at the first completion of random working times Yn (n = 1, 2, . . . ) over time T (0 ≤ T < ∞). It is assumed that Yn is independent and has an identical distribution G(t) ≡ Pr{Yn ≤ t} with finite mean 1/θ in Fig. 2.1. This is called replacement overtime. The probability that the unit is replaced at the first completion of working times over time T is ∞ ʃ ∑ n=0

T

┌ʃ

∞ T −t

0

┐

F(t + u)dG(u) dG (n) (t),

and the probability that it is replaced at failure is F(T ) +

∞ ʃ ∑ n=0

0

T

┌ʃ

∞ T −t

Thus, the mean time to replacement is

┐ G(u)dF(t + u) dG (n) (t).

2.5 Replacement Overtime ∞ ʃ ∑

┌ʃ

T 0

n=0

∞ ʃ ∑

+ ʃ

n=0

=

T

0

∞ T −t T

19

┐ ʃ (t + u) F(t + u)dG(u) dG (n) (t) +

t dF(t)

0

┌ʃ

∞ T −t

F(t)dt +

0

T

┐ (t + u)G(u)dF(t + u) dG (n) (t)

∞ ʃ ∑ n=0

T 0

┌ʃ

∞ T −t

┐

G(u) F(t + u)du dG (n) (t),

where note that when n = 0, ʃ 0

T

┌ʃ

∞ T −t

┐ G(u) F(t + u)du dG

(n)

ʃ

∞

(t) =

G(u) F(u)du. T

Therefore, the expected cost rate is [3, p. 35] [6, p. 7], [7] { ┐ (n) } ∑ ʃ T┌ʃ ∞ c O T + (c F − c O T ) F(T ) + ∞ n=0 0 T −t G(u)dF(t +u) dG (t) C O (T ) = , ʃT ┐ ∑∞ ʃ T ┌ʃ ∞ (n) n=0 0 T −t G(u) F(t + u)du dG (t) 0 F(t)dt + (2.24) where c O T = replacement cost over time T and c F = replacement cost at failure with c F > c O T . Note that C O (0) = C(G) in (2.1) when c O T = c R . In particular, when G(t) = 1 − e−θt , ʃ∞ ┌ ┐ c O T + (c F − c O T ) F(T ) + T e−θ(t−T ) dF(t) C O (T ) = . ʃ∞ ʃT −θ(t−T ) F(t)dt F(t)dt + e T 0

(2.25)

Clearly, ʃ∞ c O T + (c F − c O T ) 0 e−θt dF(t) C O (0) = , ʃ∞ −θt F(t)dt 0 e

(2.26)

which agrees with C F (∞) in (2.17) and C L (0) in (2.20) when cT = c O T and G(t) = 1 − e−θt , and C O (∞) ≡ lim C O (T ) = T →∞

cF , μ

which agrees with C(∞) in (2.2). We find optimal TO∗ to minimize C O (T ) in (2.25) when h(t) increases strictly with t from 0 to ∞. Differentiating C O (T ) with respect to T and setting it equal to zero,

20

2 Random Age Replacement Model

ʃ

T

˜1 (T ) − h(t)]dt = F(t)[ Q

0

cO T , cF − cO T

(2.27)

˜1 (T ) is given in (2.16). Noting that the left-hand side of (2.27) increases where Q strictly with T from 0 to ∞, there exists a finite and unique TO∗ (0 < TO∗ < ∞) which satisfies (2.27), and the resulting cost rate is ˜1 (TO∗ ) = C O (TO∗ ) = (c F − c O T ) Q

c O T + (c F − c O T )F(TO∗ ) , ʃ TO∗ F(t)dt 0

(2.28)

˜ which agrees with C(TO∗ ) in (2.2) when ʃ ∞ c O T = cT . Furthermore, because Q 1 (T ) decreases strictly with θ from F(T )/ T F(t)dt to h(T ), TO∗ increases with θ to T ∗ given in (2.3) when cT = c O T , and TO∗ < T ∗ . Therefore, age replacement with time T is better than replacement overtime, because T ∗ minimizes the right-hand side of (2.28).

WIB 2.6: Age replacement with time T is better than replacement overtime.

2.5.1 Replacement Overtime First Suppose that the unit is replaced preventively at work N (N = 1, 2, . . . ) or at the first completion of working times over time T , whichever occurs first. This is called replacement overtime first. Then, the probability that the unit is replaced at work N is ʃ T F(t)dG (N ) (t), 0

the probability that it is replaced at the first completion of working times over time T is N −1 ʃ ∑ n=0

T 0

┌ʃ

∞ T −t

┐ F(t + u)dG(u) dG (n) (t),

and the probability that it is replaced at failure is ʃ 0

T

[1 − G

(N )

(t)]dF(t) +

N −1 ʃ ∑ n=0

0

T

┌ʃ

∞ T −t

┐ G(u)dF(t + u) dG (n) (t).

2.5 Replacement Overtime

21

Thus, the mean time to replacement is ʃ

T

t F(t)dG

(N )

(t) +

0

N −1 ʃ ∑ n=0

ʃ

T

T

n=0 T

=

[1 − G (N ) (t)] F(t)dt +

0

┐

(t + u) F(t + u)dG(u) dG (n) (t)

N −1 ʃ T{ʃ ∑

0

ʃ

∞ T −t

0

t[1 − G (N ) (t)]dF(t) +

+

┌ʃ

n=0

┌ʃ

T −t

0

N −1 ʃ T ∑

∞

┌ʃ

∞ T −t

0

t+u

┐ } y dF(y) dG(u) dG (n) (t)

T

┐ G(u) F(t + u)du dG (n) (t).

Therefore, the expected cost rate is [6, p. 9] C O F (T , N ) =

ʃT {ʃ T c O T + (c N − c O T ) 0 F(t)dG (N ) (t) + (c F − c O T ) 0 [1 − G (N ) (t)]dF(t) ┐ (n) } ∑ N −1 ʃ T ┌ʃ ∞ + n=0 0 T −t G(u)dF(t + u) dG (t) , ʃT ┐ ∑ N −1 ʃ T ┌ʃ ∞ (n) (N ) (t)]F(t)dt + n=0 0 T −t G(u) F(t + u)du dG (t) 0 [1 − G (2.29)

where c N = replacement cost at work N , and c O T and c F are given in (2.24). Clearly, lim T →∞ C O F (T , N ) = C(N ) in (2.6) and lim N →∞ C O F (T , N ) = C O (T ) in (2.24). When G(t) = 1 − e−θt and c O T = c N , the expected cost rate in (2.29) is c O T + (c F − c O T )

∑ N −1 {ʃ T n=0

0

[(θt)n /n!]e−θt dF(t)

ʃ∞ } + [(θT )n /n!] T e−θt dF(t) C O F (T , N ) = ∑ N −1 {ʃ T ʃ }. n −θt F(t)dt + [(θT )n /n!] ∞ e−θt F(t)dt n=0 0 [(θt) /n!]e T (2.30) Note that ʃ

ʃ (θt)n −θt (θT )n ∞ −θt e dF(t) + e dF(t) n! n! T 0 ⎧ʃ T ┐ ┌ʃ ∞ θ(θt)n−1 ⎪ −θu ⎪ ⎪ e dF(u) dt (n = 1, 2, . . . ), ⎨ 0 (n − 1)! t = ʃ ∞ ⎪ ⎪ ⎪ ⎩ e−θt dF(t) (n = 0), T

0

and

22

2 Random Age Replacement Model

ʃ

ʃ (θt)n −θt (θT )n ∞ −θt e F(t)dt + e F(t)dt n! n! 0 T ⎧ʃ T ┐ ┌ʃ ∞ θ(θt)n−1 ⎪ −θu ⎪ ⎪ e F(u)du dt (n = 1, 2, . . . ), ⎨ 0 (n − 1)! t = ʃ ⎪ ⎪ ∞ −θt ⎪ ⎩ (n = 0). e F(t)dt T

0

We find optimal TO∗ F and N O∗ F to minimize C O F (T , N ) in (2.30) when h(t) increases strictly with t from 0 to ∞. Differentiating C O F (T , N ) with respect to T and setting it equal to zero, N −1 ʃ ∑ n=0

T 0

(θt)n −θt cO T ˜1 (T ) − h(t)]dt = e F(t)[ Q , n! cF − cO T

(2.31)

whose left-hand side increases strictly with T from 0 to ∞. Thus, there exists a finite and unique TO∗ F (0 < TO∗ F < ∞) which satisfies (2.31) and the resulting cost rate is ˜1 (TO∗ F ) C O F (TO∗ F , N ) = (c F − c O T ) Q ∑ N −1 ʃ TO∗ F [(θt)n /n!]e−θt dF(t) c O T + (c F − c O T ) n=0 0 , = ∑ N −1 ʃ TO∗ F [(θt)n /n!]e−θt F(t)dt n=0 0

(2.32)

which agrees (2.5) when c O T = cT = c N and G(t) = 1 − e−θt . In addition, because the left-hand side of (2.31) increases strictly with N to that of (2.27), TO∗ F decreases with N to TO∗ given in (2.27), and TO∗ F ≥ TO∗ . Comparing (2.31) and (2.9) when c O T = cT = c N and G(t) = 1 − e−θt , we obtain ∗ TO F < TF∗ because the left-hand side of (2.31) is larger than that of (2.9). Thus, from (2.32) and (2.5), C O F (TO∗ F , N ) ≥ C F (TF∗ , N ), which follows that replacement first is better than replacement overtime first.

WIB 2.7: Replacement first is better than replacement overtime first.

Forming the inequality C O F (T , N + 1) − C O F (T , N ) ≥ 0 in (2.30), ┐ ʃ (θt)n −θt (θT )n ∞ −θt e F(t)dt + e F(t)dt Q 2 (T, N − 1) n! n! 0 T n=0 ┐ ʃ N −1 ┌ʃ T ∑ (θt)n −θt (θT )n ∞ −θt cN e dF(t) + e dF(t) ≥ , − n! n! c F − cN 0 T n=0 N −1 ┌ʃ ∑

where

T

(2.33)

2.5 Replacement Overtime

23

ʃT

Q 2 (T , N ) ≡

┌ʃ ∞ −θu ┐ N dF(u) dt 0 (θt) t e ʃT ┌ʃ ∞ ┐ N −θu F(u)du dt 0 (θt) t e

˜1 (T ) TO∗ given in (2.27), then there exists a finite and unique minimum N O∗ F (1 ≤ N O∗ F < ∞) which satisfies (2.33). Furthermore, because the left-hand side of (2.33) increases with T to that of (2.7), N O∗ F decreases with T to N ∗ given in (2.7), and N O∗ F ≥ N ∗ . Conversely, if T ≤ TO∗ , then N O∗ F = ∞.

2.5.2 Replacement Overtime Last Suppose that the unit is replaced preventively at work N (N = 0, 1, 2, . . . ) or at the first completion of working times over time T , whichever occurs last, i.e., it is replaced preventively at work N after time T or at overtime T after work N . This is called replacement overtime last [6]. Then, the probability that the unit is replaced at work N is ʃ ∞ F(t)dG (N ) (t), T

the probability that it is replaced at the first completion of working times over time T is ∞ ʃ ∑ n=N

0

T

┌ʃ

∞ T −t

┐

F(t + u)dG(u) dG (n) (t),

the probability that it is replaced at failure is ʃ

∞

F(T ) +

[1 − G

(N )

(t)]dF(t) +

T

Then, the mean time to replacement is

∞ ʃ ∑ n=N

0

T

┌ʃ

∞ T −t

┐ G(u)dF(t + u) dG (n) (t).

24

2 Random Age Replacement Model

ʃ

∞

t F(t)dG T

ʃ

+

0 ∞ ∑

+ ʃ =

T

(t) +

n=N T

T

∞ T −t

ʃ

n=N

T

┌ʃ

0

∞ T −t

0

┐

(t + u) F(t + u)dG(u) dG (n) (t)

┐ (t + u)G(u)dF(t + u) dG (n) (t)

F(t)dt +

∞ ʃ ∑

┌ʃ

T

┌ʃ

0

T

t[1 − G (N ) (t)]dF(t)

t dF(t) + ʃ

∞ ʃ ∑

n=N ∞

ʃ

0

+

(N )

T ∞ T −t

∞

[1 − G (N ) (t)] F(t)dt

┐ G(u) F(t + u)du dG (n) (t).

Therefore, the expected cost rate is [6, p. 13] C O L (T , N ) =

ʃ∞ { c O T + (c N − c O T ) T F(t)dG (N ) (t) + (c F − c O T ) F(T ) ʃ T ┌ʃ ∞ ʃ∞ ┐ (n) } ∑ + T [1 − G (N ) (t)]dF(t) + ∞ n=N 0 T −t G(u)dF(t + u) dG (t) . ʃ∞ ʃT F(t)dt + T [1 − G (N ) (t)] F(t)dt 0 ∑ ʃ T ┌ʃ ∞ ┐ (n) + ∞ n=N 0 T −t G(u) F(t + u)du dG (t) (2.34)

Clearly, C O L (0, N ) = C(N ) in (2.6) and C O L (T, 0) = C O (T ) in (2.24). When G(t) = 1 − e−θt and c O T = c N , the expected cost rate in (2.34) is { ∑ N −1 ʃ ∞ n −θt dF(t) c O T + (c F − c O T ) F(T ) + n=0 T [(θt) /n!]e ʃ } ∑∞ ∞ + n=N [(θT )n /n!] T e−θt dF(t) , C O L (T , N ) = ʃT ∑ N −1 ʃ ∞ F(t)dt + n=0 ʃT [(θt)n /n!]e−θt F(t)dt 0 ∑ ∞ −θt n + ∞ F(t)dt n=N [(θT ) /n!] T e (2.35) where note that C O L (T , 0) = C O L (T , 1). We find optimal TO∗ L and N O∗ L to minimize C O L (T, N ) when h(t) increases strictly with t from 0 to ∞. Differentiating C O L (T , N ) with respect to T and setting it equal to zero, ʃ

T

˜1 (T ) − h(t)]dt + F(t)[ Q

0

cO T = , cF − cO T

N −1 ʃ ∑ n=0

∞ T

(θt)n −θt ˜1 (T ) − h(t)]dt e F(t)[ Q n! (2.36)

2.5 Replacement Overtime

25

whose left-hand side increases strictly with T to ∞. Thus, there exists a finite and unique TO∗ L (0 < TO∗ L < ∞) which satisfies (2.36), and the resulting cost rate is ˜1 (TO∗ L ). C O L (TO∗ L , N ) = (c F − c N ) Q

(2.37)

Furthermore, because the left-hand side of (2.36) decreases strictly with N from that of (2.27), TO∗ L increases strictly with N from TO∗ given in (2.27), and TO∗ L ≥ TO∗ . Forming the inequality C O L (T , N + 1) − C O L (T , N ) ≥ 0, ┌ʃ

N −1 ʃ ∑

∞

(θt)n −θt e F(t)dt n! 0 n=0 T ┐ ʃ N −1 ʃ ∞ ∞ ∑ ∑ (θt)n −θt (θT )n ∞ −θt e F(t)dt − F(T ) − e dF(t) + n! n! T n=0 T n=N ʃ ∞ ∑ (θT )n ∞ −θt cO T e dF(t) ≥ , (2.38) − n! cF − cO T T n=N

˜2 (T , N − 1) Q

T

F(t)dt +

where ˜2 (T , N ) ≡ Q

ʃ∞

┌ʃ ∞ −θu ┐ N dF(u) dt T (θt) t e ʃ∞ ┌ʃ ∞ ┐ N −θu F(u)du dt T (θt) t e

˜1 (T ). >Q

˜2 (T, N ) increases strictly with T from Q(N + 1) to h(∞) and increases Note that Q ˜2 (T , 0) to h(∞) from (4) of Appendix A1. Substituting (2.36) strictly with N from Q for (2.38), ˜1 (T ), ˜2 (T, N − 1) ≥ Q Q which always holds for any T . Thus, C L (T , N + 1) − C L (T, N ) ≥ 0 for any N , i.e., N O∗ L = 0 or 1, and TO∗ L = TO∗ , which follows that replacement overtime with T is better than replacement overtime last. In addition, noting that the left-hand side of (2.38) increases strictly with N from that of (2.27), if T < TO∗ , then there exists a finite and unique minimum N O∗ L (0 ≤ N O∗ L < ∞) which satisfies (2.38). Furthermore, because the left-hand side of (2.38) increases with T from that of (2.7), N O∗ L decreases with T from N ∗ given in (2.7) to 0, and N O∗ L ≤ N ∗ . Conversely, if T ≥ TO∗ , then N O∗ L = 0.

26

2 Random Age Replacement Model

WIB 2.8: If T > TO∗ , replacement overtime first is better than replacement overtime with time T , if T = TO∗ , then replacement overtime with time TO∗ is better than replacement overtime first and last, and if T < TO∗ , then replacement overtime last is better than replacement overtime with time T .

2.6 Deviation Time We have considered optimization problems from the viewpoint of replacement costs before and after failure times. On the other hand, we are concerned about the deviation time between replacement time and failure time, which is defined by E|Y − X |, where Y is the replacement time and X is the failure time [8, 9]. Suppose that the unit is replaced at time Y or at failure, whichever occurs first, where Y has a general distribution G(t) ≡ Pr{Y ≤ t} with finite mean 1/θ. Then, the deviation time is ┐ ┐ ʃ ∞ ┌ʃ ∞ ʃ ∞ ┌ʃ t (t − u)dG(u) dF(t) + (u − t)dG(u) dF(t) D(G) = 0 0 0 t ┐ ┌ʃ ʃ ∞ ʃ ∞ t = G(u)du + G(u)du dF(t) 0 0 t ┐ ┌ʃ ʃ ∞ ʃ ∞ t + F(u)du dG(t) F(u)du + 0 0 t ʃ ∞ ʃ ∞ G(t)F(t)dt. (2.39) G(t) F(t)dt + = 0

0

Suppose there exists a minimum value T of We easily have ʃ

T

D(G) ≥ 0

ʃt 0

ʃ

F(u)du + ∞

F(t)dt +

ʃ∞ t

F(u)du for 0 ≤ t < ∞.

F(t)dt ≡ D(T ),

(2.40)

T

which follows that the deviation time of age replacement with time T is smaller than that of random replacement. It is clearly seen that D(0) = μ and D(∞) ≡ lim T →∞ D(T ) = ∞. Differentiating D(T ) with respect to T and setting it equal to zero, F(T ) =

1 . 2

(2.41)

2.6 Deviation Time

27

Therefore, optimal TM∗ to minimize D(T ) is given by (2.41), which is a median m of the failure distribution F(t). In particular, when F(t) = 1 − e−(λt) (m ≥ 1), ∗ 1/m TM = (log 2) /λ. Next, suppose that the unit is replaced at work N (N = 1, 2, . . . ) or at failure, whichever occurs first. Then, replacing G(t) in (2.39) with G (N ) (t), the deviation time is ʃ ∞ ʃ ∞ [1 − G (N ) (t)]F(t)dt. (2.42) D(N ) = G (N ) (t) F(t)dt + 0

0

We find optimal N ∗ to minimize D(N ). Forming the inequality D(N + 1) − D(N ) ≥ 0, ʃ

∞

θ[G (N ) (t) − G (N +1) (t)] F(t)dt ≤

0

1 . 2

(2.43)

In particular, when G(t) = 1 − e−θt , (2.43) is N ʃ ∑ j=0

∞

0

(θt) j −θt 1 e dF(t) ≥ , j! 2

(2.44)

ʃ ∞ −θt whose ʃ ∞ −θtleft-hand side increases strictly with N from 0 e dF(t) to∗ 1. Thus,∗ if dF(t) < 1/2, the there exists a finite unique minimum N (1 ≤ N < 0 e ʃ ∞and −θt ∞) which satisfies (2.44). Conversely, if 0 e dF(t) ≥ 1/2, then N ∗ = 0 and D(0) = μ, i.e., we should make no replacement until failure. When F(t) = 1 − e−λt , (2.43) is θ 1 [1 − G ∗ (λ)]G ∗ (λ) N ≤ , λ 2

(2.45)

where G ∗ (λ) is the LS (Laplace Stieltjes) transform of G(t). The left-hand side of (2.45) decreases strictly with N from (θ/λ)[1 − G ∗ (λ)] to 0. If (θ/λ)[1 − G ∗ (λ)] > 1/2, then there exists a finite and unique minimum N ∗ (1 ≤ N ∗ < ∞) which satisfies (2.45). Conversely, if (θ/λ)[1 − G ∗ (λ)] ≤ 1/2, then N ∗ = 0. Finally, when G(t) = 1 − e−θt and F(t) = 1 − e−λt , (2.44) and (2.45) are (

θ θ+λ

) N +1

≤

1 . 2

(2.46)

If θ/(θ + λ) > 1/2, i.e., 1/λ > 1/θ, then there exists a finite and unique minimum N ∗ (1 ≤ N ∗ < ∞) which satisfies (2.46), and if 1/λ ≤ 1/θ, then N ∗ = 0. Note that N ∗ decreases with 1/θ to 0 and increases with 1/λ to ∞. ∗ ∗ , and D(N M ), which satisfy Example 2.4 Table 2.4 presents optimal TM∗ , D(TM∗ ), N M −(t/10)m (2.41) and (2.45), respectively, when F(t) = 1 − e and G(t) = 1 − e−t . This

28

2 Random Age Replacement Model

∗ and D(T ∗ ), D(N ∗ ) when F(t) = 1 − e−(t/10) and G(t) = 1 − e−t Table 2.4 Optimal TM∗ , N M M M m

m

TM∗

D(TM∗ )

∗ NM

∗ ) D(N M

1.00 1.50 2.00 2.50 3.00

6.93 7.83 8.33 8.64 8.85

6.93 4.74 3.70 3.06 2.63

7 8 8 8 9

7.263 5.229 4.327 3.831 3.522

∗ indicates that both TM∗ and N M increase with m, are almost the same, and D(TM∗ ) < ∗ D(N M ). 

WIB 2.9: Age replacement with time T is better than random replacement for deviation time.

2.7 Problems 1. Sudden failure When the unit fails at an additional catastrophic failure with a general distribution K (t) ≡ Pr{Z ≤ t}, where Z is independent of failure time X and random working times Yn , reconsider the WIB problems of replacement first and last policies. 2. Finite interval When the unit has to be operating for a finite lifetime S in Sect. 2.5.1, obtain the expected cost rate [10]. 3. Mission time Suppose that the unit has to be operating for an interval [TO , TO + tx ] (0 ≤ tx < ∞), where TO is a random variable with a general distribution Y (t) ≡ Pr{TO ≤ t}. When the unit is replacement at time T (0 ≤ T < TO ), at TO + tx or at failure, whichever occurs first, obtain the expected cost rate [11]. 4. General models Consider a general replacement model in which the unit is replaced before failure at time T or at times Yi (i = 1, 2, . . . , n), whichever occurs first, and whichever occurs last [12].

References 1. Barlow RE, Proschan F (1965) Mathematical theory of reliability. Wiley, New York 2. Nakagawa T (2005) Maintenance theory of reliability. Springer, London

References

29

3. Nakagawa T (2014) Random maintenance policies. Springer, London 4. Nakagawa T, Zhao X (2013) Comparisons of replacement policies with constant and random times. J Oper Res Soci Japan 56:1–14 5. Zhao X, Nakagawa T (2012) Optimization problems of replacement first or last in reliability theory. Euro J Oper Res 223:141–149 6. Nakagawa T, Zhao X (2015) Maintenance overtime policies in reliability. Springer, London 7. Zhao X, Mizutani S, Nakagawa T (2015) Which is better for replacement policies with continuous or discrete scheduled times. Euro J Oper Res 242:477–486 8. Zhao X, Li B, Mizutani S, Nakagawa T (2021) A revisit of age-based replacement models with exponential failure distributions. IEEE Trans Reliab 71:1477–1487 9. Zhao X, Mizutani S, Chen M, Nakagawa T (2022) Preventive replacement policies for parallel systems with deviation costs between replacement and failure. Ann Oper Res 312:533–551 10. Zhao X, Mizutani S, Nakagawa T (2017) Replacement policies for age and working numbers with random life cycle. Commun Stat Theory Method 46:6791–6802 11. Zhao X, Cai J, Mizutani S, Nakagawa T (2021) Preventive replacement policies with time of operations, mission durations, minimal repairs and maintenance triggering approaches. J Manuf Syst 6:819–829 12. Chen M, Zhao X, Nakagawa T (2019) Replacement policies with general models. Ann Oper Res 277:47–61

Chapter 3

Replacement Model with Minimal Repair

We consider the replacement model in which the unit undergoes minimal repair at failures, and is replaced at a planned time T , at a working time N or at a failure number K . Then, the following five policies based on time T are taken up: (1) Random replacement, (2) Periodic replacement, (3) Replacement first, (4) Replacement last, and (5) Replacement overtime. Comparing the above policies, we propose Which-IsBetter (WIB) problems in such models and discuss for answering to these problems theoretically and numerically. Throughout this chapter, we make the following assumptions: Let F(t) ʃ ∞ be the failure distribution with a density function f (t) and finite mean μ ≡ 0 F(t)dt, where Φ(t) ≡ 1 − Φ(t) for a general function Φ(t). Failure rate h(t) ≡ f (t)/ F(t) increases strictly with t from h(0) = 0 to h(∞) ≡ limt→∞ h(t) = ∞. Furthermore, it is assumed that the unit undergoes minimal repair at failures, i.e., failure rate h(t) remains undistributed by minimal at failures [1, p. 96]. In other words, failures occur ʃ t at a nonhomogeneous Poisson Process with cumulative hazard function H (t) ≡ 0 h(u)du, i.e., h(t) = dH (t)/dt and F(t) = 1 − e−H (t) . Then, the probability that k (k = 0, 1, 2, . . . ) failures occur in [0, t] is pk (t) ≡

[H (t)]k −H (t) e , k!

which is called a Poisson distribution with mean value function H (t). ∑ ∑k−1 Letting Pk (t) ≡ ∞ j=k p j (t) and P k (t) ≡ 1 − Pk (t) ≡ j=0 p j (t), Pk (0) = 0, P k (0) = 1, Pk (∞) = 1, P k (∞) = 0, P0 (t) = 1, P 0 (t) = 0, limk→∞ Pk (t) = 0, limk→∞ P k (t) = 1, Pk (t) increases with t from 0 to 1 and decreases with k from 1 to 0. These probabilities have the following relations: For 0 < t < ∞ and k = 0, 1, 2, . . . ,

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Mizutani et al., Which-Is-Better (WIB): Problems in Reliability Theory, Springer Series in Reliability Engineering, https://doi.org/10.1007/978-3-031-27316-2_3

31

32

3 Replacement Model with Minimal Repair

ʃ Pk+1 (t) = ʃ

t

ʃ pk (u)h(u)du,

0 ∞

pk (u)h(u)du,

t

ʃ

∞

H (t)dPk (t) =

0

∞

P k+1 (t) =

P k (t)h(t)dt =

0

k−1 ʃ ∑ j=0

∞

p j (t)h(t)dt = k,

0

ʃ∞ ∑ ∑∞ where note that 0 pk (t)h(t)dt = 1, and H (t) ≡ ∞ k=0 kpk (t) = k=0 P k (t) represents the expected number of failures in [0, t].

3.1 Random Replacement An operating unit is replaced at time Y (0 < Y ≤ ∞) and undergoes minimal repair at each failure between replacements, where Y is a random variable with a general distribution G(t) ≡ Pr{Y ≤ t} and is independent of the failure times. Then, because failures occur at a nonhomogeneous Poisson process with mean value function H (t) [2, p. 96], [3, p. 27], the expected number of failures in [0, t] is H (t). Thus, the expected number of failures between replacement is ʃ

∞

ʃ

∞

H (t)dG(t) =

G(t)h(t)dt,

0

0

and the mean time to replacement is ʃ

∞

ʃ t dG(t) =

0

∞

G(t)dt. 0

Therefore, the expected cost rate is [2, p. 250] ʃ∞ c R + c M 0 H (t)dG(t) C(G) = , ʃ∞ 0 G(t)dt

(3.1)

where c R = replacement cost at time Y and c M = minimal repair cost at each failure. It has been shown [4, p. 55] that ʃ∞ C(G) = ʃ0∞ 0

S1 (t)dG(t) S2 (t)dG(t)

,

where S1 (t) ≡ c R + c M H (t),

S2 (t) ≡ t.

3.1 Random Replacement

33

Suppose that there exists a minimum value T of S1 (t)/S2 (t) for 0 < t ≤ ∞. We easily have ʃ

∞

S1 (t)dG(t) ≥

0

S1 (T ) S2 (T )

ʃ

∞

S2 (t)dG(t),

0

because S1 (T ) S1 (t) ≥ . S2 (T ) S2 (t) Thus, C(G) ≥ C(G T ) =

cT + c M H (T ) S1 (T ) = ≡ C(T ), T S2 (T )

(3.2)

where cT = replacement cost at time T and G T (t) is the degenerate distribution placing unit mass at T , i.e., G T (t) ≡ 1 for t ≥ T and 0 for t < T , and c R ≡ cT . If T = ∞ then the unit is not replaced and undergoes always only minimal repair at each failure. This shows that replacement with time T is better than random replacement. We find optimal T ∗ to minimize the expected cost rate C(T ) in (3.2) [1, p. 97], [2, p. 102], [4, p. 56] when failure rate h(t) increases strictly with t from h(0) = 0 to h(∞) = ∞. Differentiating C(T ) with respect to T and setting it equal to zero, T h(T ) − H (T ) =

cT , cM

ʃ

T

t dh(t) =

0

cT , or cM

ʃ

T

[h(T ) − h(t)]dt =

0

cT . cM (3.3)

Thus, there exists a finite and unique minimum T ∗ (0 < T ∗ < ∞) which satisfies (3.3), and the resulting cost rate is C(T ∗ ) = c M h(T ∗ ).

(3.4)

In particular, when F(t) = 1 − e−(λt) (m > 1), m

T∗ =

[ ]1/m 1 cT . λ (m − 1)c M

WIB 3.1: Replacement with time T is better than random replacement.

It has been shown that when cT = c R , replacement with time T is better than c R in which C(G) in (3.1) is the same as random one. When cT > c R , we compute ⌃

34

3 Replacement Model with Minimal Repair

c R when G(t) = 1 − e−t , H (t) = (t/10)2 and c M = 10 Table 3.1 Optimal T ∗ , C(T ∗ ) and ⌃ T∗ C(T ∗ ) ⌃ cR (cT − ⌃ c R )/cT cT 5 10 15 20 30 40 50

7.071 10.000 12.247 14.142 17.321 20.000 22.361

1.414 2.000 2.449 2.828 3.464 4.000 4.472

1.214 1.800 2.249 2.628 3.264 3.800 4.272

0.757 0.820 0.850 0.869 0.891 0.905 0.915

C(T ) in (3.2). We specify the computing procedure for obtaining ⌃ c R for given c M , cT , G(t) and F(t) [4, p. 27]: Setting C(G) = C(T ∗ ), i.e., ʃ∞ ⌃ c R + c M 0 H (t)dG(t) = c M h(T ∗ ), ʃ∞ G(t)dt 0 we compute ⌃ c R which satisfies the above equation. Example 3.1 Table 3.1 presents T ∗ , random cost ⌃ c R and cost difference rate (cT − ⌃ c R )/cT for cT when G(t) = 1 − e−t , H (t) = (t/10)2 and c M = 10. This shows c R and (cT − ⌃ c R )/cT increase with similar tendencies to Table 2.1, and T ∗ , C(T ∗ ), ⌃ cT . 

WIB 3.2: When random replacement cost c R < ⌃ c R , random replacement is better than replacement with time T .

3.2 Replacement with Working Number It is assumed that Yn (n = 1, 2, . . . ) is the working time of job n in Fig. ʃ ∞2.1 and has an identical distribution G(t) ≡ Pr{Yn ≤ t} with finite mean 1/θ ≡ 0 G(t)dt (0 < 1/θ < ∞). The n-fold Stieltjes convolution of G(t) is G (n) (t) ≡ Pr{Y1 + Y2 + · · · + Yn ≤ t} and G (0) (t) ≡ 1 for t ≥ 0 defined in Chap. 2. It is supposed that the unit is replaced at a planned time T or a planned number N of working times [4, p. 75], [5].

3.2 Replacement with Working Number

35

3.2.1 Replacement First Suppose that the unit is replaced at time T (0 < T ≤ ∞) or work N (N = 1, 2, . . . ), whichever occurs first. The probability that the unit is replaced at work N is G (N ) (T ), and the probability that it is replaced at time T is 1 − G (N ) (T ). Thus, the mean time to replacement is T [1 − G

(N )

ʃ

T

(T )] +

t dG

(N )

ʃ

T

(t) =

0

[1 − G (N ) (t)]dt,

0

and the expected number of failures until replacement is H (T )[1 − G

(N )

ʃ (T )] +

T

H (t)dG

(N )

ʃ (t) =

0

T

[1 − G (N ) (t)]h(t)dt.

0

Therefore, the expected cost rate is [4, p. 76] ʃT cT + (c N − cT )G (N ) (T ) + c M 0 [1 − G (N ) (t)]h(t)dt C F (T , N ) = , ʃT (N ) (t)]dt 0 [1 − G

(3.5)

where c N = replacement cost at work N , and cT and c M are given in (3.2). When N = ∞, C F (T , ∞) = C(T ) in (3.2), and when T =∞, N = 1 and c N = cT = c R , C F (∞, 1) = C(G) in (3.1). In particular, when the unit is replaced only at work N (N = 1, 2, . . . ), C(N ) ≡ lim C F (T , N ) =

cN + cM

T →∞

ʃ∞ 0

[1 − G (N ) (t)]h(t)dt . N /θ

(3.6)

We find optimal N ∗ to minimize C(N ) when h(t) increases strictly with t from 0 to ∞. Forming the inequality C(N + 1) − C(N ) ≥ 0, N R(N ) − θ

ʃ

∞

[1 − G (N ) (t)]h(t)dt ≥

0

cN , cM

where ʃT

[G (N ) (t) − G (N +1) (t)]h(t)dt < h(T ), ʃT (N ) (t) − G (N +1) (t)]dt 0 [G ʃ ∞ R(N ) ≡ lim R1 (T , N ) = θ[G (N ) (t) − G (N +1) (t)]h(t)dt.

R1 (T , N ) ≡

0

T →∞

0

(3.7)

36

3 Replacement Model with Minimal Repair

Thus, when R(N ) increases strictly with N to ∞, the left-hand side of (3.7) increases strictly with N to ∞. Thus, there exists a finite and unique minimum N ∗ (1 ≤ N < ∞) which satisfies (3.7). In particular, when G(t) = 1 − e−θt , from (1) of Appendix A.2 [4, p 77], ʃT R1 (T , N ) =

0

(θt) N e−θt h(t)dt < h(T ) ʃT N −θt dt 0 (θt) e

(3.8)

ʃ∞ increases strictly with T from 0 to R(N ) = 0 [θ(θt) N /N !]e−θt h(t)dt and increases strictly with N to h(T ), and R(N ) increases strictly with N to h(∞) from (1) of Appendix A.2. In this case, (3.7) is N −1 ʃ ∑ n=0

0

∞

(θt)n −θt cN e [R(N ) − h(t)]dt ≥ , n! cM

(3.9)

whose left-hand side increases strictly with N to ∞. Thus, there exists a finite and unique minimum N ∗ (1 ≤ N ∗ < ∞) which satisfies (3.9). Next, we find optimal TF∗ and N F∗ to minimize C F (T, N ) in (3.5) when cT = c N and h(t) increases strictly with t from 0 to ∞. Differentiating C F (T, N ) with respect to T and setting it equal to zero, ʃ

T

[1 − G (N ) (t)][h(T ) − h(t)]dt =

0

cT , cM

(3.10)

which agrees with (3.3) when N = ∞, and whose left-hand side increases strictly with T from 0 to ∞. Thus, there exists a finite and unique TF∗ (0 < TF∗ < ∞) which satisfies (3.10), and the resulting cost rate is C(TF∗ , N ) = (c F − cT )h(TF∗ ).

(3.11)

In addition, noting that the left-hand side of (3.10) increases strictly with N to that of (3.3), TF∗ decreases with N to T ∗ given in (3.3), and TF∗ ≥ T ∗ . Forming the inequality C F (T , N + 1) − C F (T , N ) ≥ 0, ʃ

T

[1 − G (N ) (t)][R1 (T, N ) − h(t)]dt ≥

0

Substituting (3.10) for (3.12), (3.12) is R1 (T , N ) ≥ h(T ),

cT . cM

(3.12)

3.2 Replacement with Working Number

37

which does not hold for any T . Thus, there does not exists any finite N F∗ , i.e., N F∗ = ∞, and TF∗ = T ∗ , which follows that replacement with time T is better than replacement first. In addition, when G(t) = 1 − e−θt , R1 (T, N ) is given in (3.8). Thus, noting that the left-hand side of (3.12) increases strictly with N to that of (3.3), if T > T ∗ given in (3.3), then there exists a finite and unique minimum N F∗ (1 ≤ N F∗ < ∞) which satisfies (3.12). Furthermore, because the left-hand side of (3.12) increases with T to that of (3.9), N F∗ decreases with T to N ∗ given in (3.9), and N F∗ ≥ N ∗ . Conversely, if T ≤ T ∗ , then N F∗ = ∞.

WIB 3.3: Replacement with time T is better than replacement first.

Example 3.2 Table 3.2 presents optimal T ∗ which satisfies (3.3) and the resulting cost rate C(T ∗ )/c M in (3.4), and optimal N ∗ which satisfies (3.9) and C(N ∗ )/c M in (3.6). This shows similar tendencies to Table 2.2. This indicates that both T ∗ and N ∗ increase with ci /c M (i = T , N ), T ∗ > N ∗ and C(T ∗ ) < C(N ∗ ). 

3.2.2 Replacement Last Suppose that the unit is replaced at time T (0 ≤ T < ∞) or work N (N = 0, 1, 2, . . . ), whichever occurs last. The probability that the unit is replaced at work N is 1 − G (N ) (T ), and the probability that it is replaced at time T is G (N ) (T ). Thus, the mean time to replacement is TG

(N )

ʃ

∞

(T ) +

t dG

(N )

T

ʃ

∞

(t) = T +

[1 − G (N ) (t)]dt,

T

and the expected number of failures until replacement is Table 3.2 Optimal T ∗ and C(T ∗ )/c M , and optimal N ∗ and C(N ∗ )/c M when G(t) = 1 − e−t , 2 F(t) = 1 − e−(t/10) and c M = 5 cT /c M or c N /c M T∗ C(T ∗ )/c M N∗ C(N ∗ )/c M 0.1 0.2 0.5 1.0 2.0 5.0

3.16 4.47 7.07 10.00 14.14 22.36

0.06 0.09 0.14 0.20 0.28 0.45

3 4 7 10 14 22

0.07 0.10 0.15 0.21 0.29 0.46

38

3 Replacement Model with Minimal Repair

H (T )G (N ) (T ) +

ʃ

∞

H (t)dG (N ) (t) = H (T ) +

ʃ

T

∞

[1 − G (N ) (t)]h(t)dt.

T

Therefore, the expected cost rate is [4, p. 79] ʃ∞ } { c N + (cT − c N )G (N ) (T ) + c M H (T ) + T [1 − G (N ) (t)]h(t)dt ʃ∞ C L (T , N ) = . T + T [1 − G (N ) (t)]dt (3.13) When N = 0, C L (T, 0) = C F (T, ∞) = C(T ) in (3.2), and when T = 0, C L (0, N ) = C F (∞, N ) = C(N ) in (3.6). We find optimal TL∗ and N L∗ to minimize C L (T , N ) in (3.13) when cT = c N and h(t) increases strictly with t from 0 to ∞. Differentiating C L (T , N ) with respect to T and setting it equal to zero, ʃ

T

ʃ

∞

[h(T ) − h(t)]dt +

0

[1 − G (N ) (t)][h(T ) − h(t)]dt =

T

cT , cM

(3.14)

which agrees ʃ ∞with (3.3) when N = 0, and whose left-hand side increases strictly with T from − 0 [1 − G (N ) (t)]h(t)dt < 0 to ∞. Thus, there exists a finite and unique TL∗ (0 < TL∗ < ∞) which satisfies (3.14), and the resulting cost rate is C(TL∗ ) = (c F − cT )h(TL∗ ).

(3.15)

In addition, noting that the left-hand side of (3.14) decreases strictly with N from that of (3.3), TL∗ increases with N from T ∗ given in (3.3), and TL∗ ≥ T ∗ . Forming the inequality C L (T , N + 1) − C L (T , N ) ≥ 0, [ ʃ ˜1 (T , N ) T + R

] ʃ [1 − G (N ) (t)]dt − H (T ) −

∞ T

∞

[1 − G (N ) (t)]h(t)dt ≥

T

cN , cM (3.16)

where ˜1 (T, N ) ≡ R

ʃ∞

[G (N ) (t) − G (N +1) (t)]h(t)dt ʃ∞ > h(T ). (N ) (t) − G (N +1) (t)]dt T [G

T

Substituting (3.14) for (3.16), (3.16) is ˜1 (T , N ) ≥ h(T ), R which always holds for any T , i.e., N L∗ = 0, which follows that replacement with time T is better than replacement last.

3.3 Replacement First and Last

39

WIB 3.4: Replacement with time T is better than replacement last.

In particular, when G(t) = 1 − e−θt , for N = 0, 1, 2, . . . , ˜1 (T , N ) = R

ʃ∞

(θt) N e−θt h(t)dt ʃ∞ N −θt dt T (θt) e

T

strictly with T from R(N ) to h(∞) and increases strictly with N from ʃincreases ∞ −θt θe h(T + t)dt to h(∞) from (2) of Appendix A.2. In this case, (3.16) is 0 ʃ

T

˜1 (T, N ) − h(t)]dt + [R

0

N −1 ʃ ∑ n=0

∞ T

(θt)n −θt ˜ cN e [ R1 (T , N ) − h(t)]dt ≥ , n! cM (3.17)

whose left-hand side increases strictly with N from ʃ T

∞

θe−θt h(T + t)dt − H (T ) > T h(T ) − H (T ),

0

which agrees with that of (3.3) to ∞. Thus, if T < T ∗ given in (3.3), then there exists a finite and unique minimum N L∗ (1 ≤ N L∗ < ∞) which satisfies (3.17). In addition, because the left-hand side of (3.17) increases with T from that of (3.9), N L∗ decreases with T from N ∗ given in (3.9) to 0, and N L∗ ≤ N ∗ . Conversely, if T ≥ T ∗ , then N L∗ = 0. WIB 3.5: When G(t) = 1 − e−θt , if T > T ∗ , replacement first is better than replacement with time T , if T = T ∗ , then replacement with time T ∗ is better than replacement first and last, and if T < T ∗ , then replacement last is better than replacement with time T .

3.3 Replacement First and Last Suppose that the unit is replaced at time T or at time Y , where T is constant and ʃ∞ Y has a general distribution G(t) ≡ Pr{Y ≤ t} with finite mean 1/θ ≡ 0 G(t)dt. When cT = c N and the unit is replaced at time T or time Y , whichever occurs first, the expected cost rate is, from (3.5),

40

3 Replacement Model with Minimal Repair

ʃT cT + c M 0 G(t)h(t)dt C F (T ) = . ʃT 0 G(t)dt

(3.18)

Optimal TF∗ to minimize C F (T ) satisfies, from (3.10), ʃ

T

G(t)[h(T ) − h(t)]dt =

0

cT , cM

(3.19)

and the resulting cost rate is C F (TF∗ ) = c M h(TF∗ ).

(3.20)

When cT = c N and unit is replaced at time T or time Y , whichever occurs last, the expected cost rate is, from (3.13), ʃ∞ [ ] cT + c M H (T ) + T G(t)h(t)dt . C L (T ) = ʃ∞ T + T G(t)dt

(3.21)

Optimal TL∗ to minimize C L (T ) satisfies, from (3.14), ʃ

T

ʃ

∞

[h(T ) − h(t)]dt +

0

G(t)[h(T ) − h(t)]dt =

T

cT , cM

(3.22)

and the resulting cost rate is C L (TL∗ ) = c M h(TL∗ ).

(3.23)

Compare the left-hand side L F (T ) and L L (T ) of (3.19) and (3.22), respectively. Letting L(T ) ≡ L L (T ) − L F (T ) ʃ ʃ T G(t)[h(T ) − h(t)]dt − = 0

∞

G(t)[h(t) − h(T )]dt,

T

ʃ∞ L(T ) increases strictly with T from − 0 G(t)h(t)dt < 0 to ∞. Thus, there exists a finite and unique TP (0 < TP < ∞) which satisfies L(T ) = 0. Therefore, if L F (TP ) ≥ cT /c M , then TF∗ ≤ TL∗ and if L F (TP ) < cT /c M , then TL∗ < TF∗ . In other words, if the ratio cT /c M is larger than L F (TP ), TL∗ < TF∗ and replacement last is better than replacement first, and vice versa. Example 3.3 When F(t) = 1 − exp[−(t/10)2 ] and G(t) = 1 − e−θt , Table 3.3 presents optimal T ∗ , TF∗ , TL∗ and TP for cT /c M and 1/θ and [4, p. 71]. This shows similar tendencies to Table 2.3, and TF∗ decrease with 1/θ to T ∗ and TL∗ increase with 1/θ from T ∗ . When L F (TP ) ≥ cT /c M , TF∗ ≤ TL∗ and replacement first is better

3.4 Replacement with Failure Number

41

Table 3.3 Optimal T ∗ , TF∗ , TL∗ , TP and L F (TP ) when F(t) = 1 − e−(t/10) and G(t) = 1 − e−θt 2

1/θ = 1

cT cM

TF∗

0.10 0.20 0.50 1.00 2.00 5.00 10.00 20.00 50.00 TP L F (TP )

6.00 3.18 11.00 4.47 26.00 7.07 51.00 10.00 101.00 14.14 250.99 22.36 500.90 31.62 1000.17 44.72 2488.17 70.71 1.30 0.182

TL∗

1/θ = 2 TF∗

TL∗

4.26 3.39 6.94 4.56 14.50 7.09 27.00 10.00 52.00 14.14 127.00 22.36 252.00 31.62 501.97 44.72 1251.59 70.71 2.60 0.349

1/θ = 5 TF∗

TL∗

3.53 5.25 5.25 5.94 9.21 7.78 14.74 10.31 24.97 14.24 55.00 22.37 105.00 31.62 205.00 44.72 505.00 70.71 6.50 0.759

1/θ = 10 TF∗

TL∗

3.34 9.39 4.83 9.76 8.01 10.85 11.98 12.53 18.41 15.56 34.69 22.81 59.98 31.75 110.00 44.75 260.00 70.71 13.00 1.145

1/θ = ∞ T∗ 3.16 4.47 7.07 10.00 14.14 22.36 31.62 44.72 70.71

than replacement last, and vice versa. Furthermore, because TP increase with 1/θ, replacement first is better than replacement last as 1/θ become larger. In addition, when cT /c M become larger, we should be longer to lessen its cost, and replacement last is better than replacement first. 

WIB 3.6: Replacement first is better than replacement last as 1/θ becomes larger, and replacement last is better than replacement first as cT /c M becomes larger.

3.4 Replacement with Failure Number We consider the replacement policies in which the unit is replaced at a planned time T or at a failure number K [7].

3.4.1 Replacement First Suppose that the unit is replaced at time T (0 < T ≤ ∞) or failure K (K = 1, 2, . . . ), whichever occurs first. The probability that the unit is replaced at failure K is PK (T ), and the probability that it is replaced at time T is P K (T ). Thus, the mean time to

42

3 Replacement Model with Minimal Repair

replacement is ʃ

T

T P K (T ) +

ʃ

T

t dPK (t) =

0

P K (t)dt,

0

and the expected number of failures until replacement is ʃ

T

H (T )P K (T ) +

ʃ

T

H (t)dPK (t) =

0

P K (t)h(t)dt.

0

Therefore, the expected cost rate is [2, p. 104] cT + (c K − cT )PK (T ) + c M C F (T , K ) = ʃT 0 P K (t)dt

ʃT 0

P K (t)h(t)dt

,

(3.24)

where cT = replacement cost at time T , c K = replacement cost at failure K and c M is given in (3.1). When K = ∞, C F (T , ∞) = C(T ) in (3.2). In particular, when the unit is replaced only at failure K (K = 1, 2, . . . ), cK + cM K . C(K ) ≡ lim C F (T , K ) = ʃ ∞ T →∞ 0 P K (t)dt

(3.25)

We find optimal K ∗ to minimize C(K ) when h(t) increases strictly with t from 0 to ∞. Forming the inequality C(K + 1) − C(K ) ≥ 0, ʃ

∞

H (K ) 0

P K (t)dt − K ≥

cK , cM

(3.26)

where ʃT

p K (t)h(t)dt < h(T ), ʃT 0 p K (t)dt 1 H (K ) ≡ lim H1 (T , K ) = ʃ ∞ . T →∞ p K (t)dt 0

H1 (T , K ) ≡

0

Note that from (3) of Appendix A.2, H1 (T , K ) increases strictly with T from h(0) ʃT to H (K ), and increases strictly with K from Q(T ) = F(T )/ 0 F(t)dt to h(T ). So that, H (K ) also increases strictly with K from 1/μ to h(∞). Therefore, there exists a finite and unique minimum K ∗ (1 ≤ K ∗ < ∞) which satisfies (3.26). Next, we find optimal TF∗ and K F∗ to minimize C F (T , K ) in (3.24) when cT = c K and h(t) increases strictly with t from 0 to ∞. Differentiating C F (T , K ) with respect to T and setting it equal to zero,

3.4 Replacement with Failure Number

ʃ

T

43

P K (t)[h(T ) − h(t)]dt =

0

cT , cM

(3.27)

which agrees with (3.3) when K = ∞, and whose left-hand side increases strictly with T from 0 to ∞. Thus, there exists a finite and unique TF∗ (0 < TF∗ < ∞) which satisfies (3.27), and the resulting cost rate is C F (TF∗ , K ) = c M h(TF∗ ).

(3.28)

In addition, noting that the left-hand side of (3.27) increases strictly with K to that of (3.3), TF∗ decreases with K to T ∗ given in (3.3), and TF∗ ≥ T ∗ . Forming the inequality C F (T , K + 1) − C F (T , K ) ≥ 0, ʃ

T

P K (t)[H1 (T , K ) − h(t)]dt ≥

0

cT . cM

(3.29)

Substituting (3.27) for (3.29), (3.29) is H1 (T, K ) ≥ h(T ), which does not hold for any T . Thus, there does not exist any finite K F∗ , i.e., K F∗ = ∞, and TF∗ = T ∗ given in (3.3), which follows that replacement with time T is better than replacement first.

WIB 3.7: Replacement with time T is better than replacement with failure K .

In addition, note that the left-hand side of (3.29) increases strictly with K to that of (3.3) and with T to that of (3.26). Thus, if T > T ∗ then there exists a finite and unique minimum K F∗ (1 ≤ K F∗ < ∞) which satisfy (3.29). Furthermore, because the left-hand side of (3.29) increases with T to that of (3.26), K F∗ decreases with T to K ∗ given in (3.26), and K F∗ ≥ K ∗ . Conversely, if T ≤ T ∗ , then K F∗ = ∞. Example 3.4 When F(t) = 1 − e−(t/10) , Table 3.4 presents optimal T ∗ given in (3.3) and the resulting cost rate C(T ∗ )/c M in (3.4), and optimal K ∗ which satisfies (3.26) and C(K ∗ )/c M in (3.25). This indicates that C(T ∗ ) < C(K ∗ ) and their differences are smaller as ci /c M (i = T, K ) are larger.  2

3.4.2 Replacement Last Suppose that the unit is replaced at time T (0 ≤ T < ∞) or failure K (K = 0, 1, 2, . . . ), whichever occurs last. The probability that the unit is replaced at failure

44

3 Replacement Model with Minimal Repair

Table 3.4 Optimal T ∗ and C(T ∗ )/c M , and optimal K ∗ and C(K ∗ )/c M when F(t) = 1 − e−(t/10)

2

cT cK or cM cM

T∗

C(T ∗ ) cM

K∗

C(K ∗ ) cM

0.1 0.2 0.5 1.0 2.0 5.0

3.16 4.47 7.07 10.00 14.14 22.36

0.06 0.09 0.14 1.20 0.28 0.45

1 1 1 2 3 6

0.12 0.14 0.17 0.23 0.30 0.46

K is P K (T ), and the probability that it is replaced at time T is PK (T ). Thus, the mean time to replacement is ʃ

∞

T PK (T ) +

ʃ

∞

tdPK (t) = T +

T

P K (t)dt,

T

and the expected number of failures until replacement is ʃ

∞

H (T )PK (T ) +

ʃ

∞

H (t)dPK (t) = H (T ) +

T

P K (t)h(t)dt.

T

Therefore, the expected cost rate is [7] ʃ∞ [ ] c K + (cT − c K )PK (T ) + c M H (T ) + T P K (t)h(t)dt C L (T , K ) = . ʃ∞ T + T P K (t)dt

(3.30)

When K = 0, C L (T , 0) = C F (T , ∞) = C(T ) in (3.2), and when T = 0, C L (0, K ) = C F (∞, K ) = C(K ) in (3.25). We find optimal TL∗ and K L∗ to minimize C L (T , K ) in (3.30) when cT = c K and h(t) increases strictly with t from 0 to ∞. Differentiating C L (T, K ) with respect to T and setting it equal to zero, ʃ

T 0

ʃ

∞

[h(T ) − h(t)]dt +

P K (t)[h(T ) − h(t)]dt =

T

cT , cM

(3.31)

which agrees with (3.3) when K = 0, and whose left-hand side increases strictly with T from −K to ∞. Thus, there exists a finite and unique TL∗ (0 < TL∗ < ∞) which satisfies (3.31), and the resulting cost rate is C L (TL∗ ) = c M h(TL∗ ).

(3.32)

In addition, noting that the left-hand side of (3.31) decreases strictly with K from (3.3), TL∗ increases with K from T ∗ given in (3.3), and TL∗ ≥ T ∗ .

3.5 Replacement Overtime with Working Number

45

Forming the inequality C L (T , K + 1) − C L (T , K ) ≥ 0, [ ʃ ˜ H1 (T , K ) T +

∞

]

ʃ

∞

P K (t)dt − H (T ) −

P K (t)h(t)dt ≥

T

T

cT , cM

(3.33)

where ˜1 (T , K ) ≡ H

ʃ∞

p K (t)h(t)dt ʃ∞ > h(T ) T p K (t)dt

T

ʃ ˜1 (0, K ) ≡ H (K ) = 1/ ∞ p K (t)dt to h(∞) and increases strictly with T from H 0ʃ ˜ ) = F(T )/ ∞ F(t)dt to h(∞) from ˜1 (T1 , 0) = Q(T increases strictly with K from H T (4) of Appendix A.2. Substituting (3.31) for (3.33), (3.33) is ˜1 (T, K ) ≥ h(T ), H which always holds for any T , i.e., K L∗ = 0, which follows that replacement with time T is better than replacement last. WIB 3.8: If T > T ∗ , then replacement first is better than replacement with time T , if T = T ∗ , then replacement with time T ∗ is better than replacement first and last, and if T < T ∗ , then replacement last is better than replacement with time T .

Furthermore, note that the the left-hand side of (3.33) increases strictly with K from ʃ T ˜ )T − H (T ) > [h(T ) − h(t)]dt, Q(T 0

˜ ) increases strictly with T from 1/μ which agrees with that of (3.3) to ∞, where Q(T to h(∞) from (4) of Appendix A.2. Thus, if T < T ∗ , then there exists a finite and unique minimum K L∗ (0 ≤ K L∗ < ∞) which satisfies (3.33). Furthermore, because the left-hand side of (3.33) increases with T from that of (3.26), K L∗ decreases with T from K ∗ given in (3.26) to 0, and K L∗ ≤ K ∗ . Conversely, if T ≥ T ∗ , then K L∗ = 0.

3.5 Replacement Overtime with Working Number We consider replacement overtime policies in which the unit is replaced at the first working time over a planned time T and at a number N of working times [8, p. 34].

46

3 Replacement Model with Minimal Repair

3.5.1 Replacement First Suppose that the unit is replaced at the first working time over time T (0 ≤ T < ∞) or at work N (N = 0, 1, 2, . . . ), whichever occurs first. Then, the probability that the unit is replaced at the first working time over time T is 1 − G (N ) (T ), and the probability that it is replaced at work N is G (N ) (T ). Thus, the mean time to replacement is ʃ

T

N −1 ʃ ∑

tdG (N ) (t) +

0

n=0

T

[ʃ

0

] N −1 1 ∑ (n) (t + u)dG(u) dG (n) (t) = G (T ), θ n=0 T −t ∞

and the expected number of failures until replacement is ʃ

T

H (t)dG (N ) (t) +

0

=

N −1 ʃ ∑ n=0

n=0

[ʃ

T 0

N −1 ʃ ∑

∞

T 0

[ʃ

∞ T −t

] H (t + u)dG(u) dG (n) (t)

]

G(u)h(t + u)du dG (n) (t).

0

Therefore, the expected cost rate is c O T + (c N − c O T )G (N ) (T ) ] (n) ∑ N −1 ʃ T [ʃ ∞ + c M n=0 0 0 G(u)h(t + u)du dG (t) , C O F (T , N ) = ∑ N −1 (n) G (T ) (1/θ) n=0

(3.34)

where c O T = replacement cost over time T , and c N and c M are given in (3.5). When T = ∞, C O F (∞, N ) = C(N ) in (3.6). In particular, when G(t) = 1 − e−θt and the unit is replaced only at the first working time over time T , the expected cost rate in (3.34) is ʃ∞ [ ] c O T + c M H (T ) + 0 e−θt h(T + t)dt . (3.35) C O (T ) ≡ lim C O F (T , N ) = N →∞ T + 1/θ We find optimal TO∗ to minimize C O (T ) when h(t) increases strictly with t from 0 to ∞. Differentiating C O (T ) with respect to T and setting it equal to zero, ʃ 0

∞

θe−θt [T h(T + t) − H (T )]dt =

cO T , cM

(3.36)

whose left-hand side increases strictly with T from 0 to ∞. Therefore, there exists a finite and unique TO∗ (0 < TO∗ < ∞) which satisfies (3.36), and the resulting cost rate is

3.5 Replacement Overtime with Working Number

C O (TO∗ ) = c M

ʃ 0

∞

47

θe−θt h(TO∗ + t)dt =

c O T + c M H (TO∗ ) , TO∗

(3.37)

which agrees with (3.2) when c O T = cT and TO∗ = T . When c O T = cT , from (3.3) and (3.36), ʃ

∞

θe−θt [T h(T + t) − H (T )]dt >

0

ʃ

T

[h(T ) − h(t)]dt,

0

which follows that TO∗ < T ∗ , where T ∗ minimizes C(T ) in (3.2). Thus, from (3.2) and (3.37) when c O T = cT , C O (TO∗ ) > C(T ∗ ), which follows that replacement with time T is better than replacement overtime with working number.

WIB 3.9: Replacement with time T is better than replacement overtime with working number.

Next, when c N = c O T and G(t) = 1 − e−θt , (3.34) is {ʃ T c O T + c M 0 [1 −ʃ G (N ) (t)]h(t)dt } ∞ + [1 − G (N ) (T )] 0 e−θt h(T + t)dt C O F (T , N ) = . ∑ N −1 (n) (1/θ) n=0 G (T )

(3.38)

We find optimal TO∗ F and N O∗ F to minimize C O F (T, N ) when h(t) increases strictly with t from 0 to ∞. Differentiating C O F (T, N ) with respect to T and setting it equal to zero, N ∑ n=1

G (n) (T )

ʃ

∞

e−θt h(T + t)dt −

0

ʃ

T

[1 − G (N ) (t)]h(t)dt =

0

cO T , cM

(3.39)

whose left-hand side increases strictly with T from 0 to ∞. Thus, there exists a finite and unique TO∗ F (0 < TO∗ F < ∞) which satisfies (3.39), and the resulting cost rate is C O F (TO∗ F , N ) = c M =

ʃ

∞ 0

θe−θt h(TO∗ F + t)dt ʃ TO∗ F

[1 − G (N ) (t)]h(t)dt , ∑N (1/θ) n=1 G (n) (TO∗ F )

cO T + cM

0

(3.40)

which agrees with (3.5) when c O T = cT = c N and TO∗ F = T , where note that ʃT ∑N (n) (N ) (t)]dt when G(t) = 1 − e−θt . In addition, because n=1 G (T ) = 0 θ[1 − G the left-hand side of (3.39) increases strictly with N to that of (3.36), TO∗ F decreases with N to TO∗ given in (3.36), and TO∗ F ≥ TO∗ .

48

3 Replacement Model with Minimal Repair

Comparing (3.39) and (3.10) when c O T = cT , we obtain TO∗ F < TF∗ because the left-hand side of (3.39) is larger than that of (3.10). Thus, from (3.5) and (3.40), C O F (TO∗ F , N ) ≥ C F (TF∗ , N ), which follows that replacement first is better than replacement overtime first.

WIB 3.10: Replacement first is better than replacement overtime first.

Forming the inequality C O F (T , N + 1) − C O F (T , N ) ≥ 0, R2 (T , N )

N −1 ∑

G (n) (T ) −

n=0

− [1 − G

(N )

ʃ

T

[1 − G (N ) (t)]h(t)dt

0

ʃ

∞

(T )]

e−θt h(T + t)dt ≥

0

cO T , cM

(3.41)

where ʃ T [ʃ ∞ R2 (T , N ) ≡

0

0

] ʃ ∞ e−θu h(t + u)du dG (N ) (t) < e−θt h(T + t)dt G (N ) (T ) 0

ʃ∞ increases strictly T from 0ʃ e−θt h(t)dt to R(N )/θ in (3.7) and increases strictly ʃ ∞ with ∞ with N from 0 e−θt h(t)dt to 0 e−θt h(T + t)dt from (5) of Appendix A.2. Substituting (3.39) for (3.41), (3.41) is ʃ

∞

R2 (T , N ) ≥

e−θt h(T + t)dt,

0

which does not hold for any T . Thus, there does not exist any finite N O∗ F , i.e., N O∗ F = ∞, and TO∗ F = TO∗ , which follows that replacement overtime with time T is better than replacement overtime first. In addition, noting that the left-hand side of (3.41) increases strictly with N to that of (3.36), if T > TO∗ given in (3.36), then there exists a finite and unique minimum N O∗ F (1 ≤ N O∗ F < ∞) which satisfies (3.41). Furthermore, because the left-hand side of (3.41) increases with T to that of (3.9), N O∗ F decreases with T to N ∗ given in (3.9), and N O∗ F ≥ N ∗ . Conversely, if T ≤ TO∗ , then N O∗ F = ∞.

3.5.2 Replacement Last Suppose that the unit is replaced at the first working time over time T (0 ≤ T < ∞) or at work N (N = 0, 1, 2, . . . ), whichever occurs last. Then, the probability that the unit is replaced at the first working time over time T is G (N ) (T ), and the probability

3.5 Replacement Overtime with Working Number

49

that it is replaced at work N is 1 − G (N ) (T ). Thus, the mean time to replacement is ʃ

∞

t dG

(N )

∞ ʃ T[ʃ ∑ (t) +

T

n=N

0

] [ ] ∞ ∑ 1 (n) (n) G (T ) , (t + u)dG(u) dG (t) = N+ θ T −t n=N ∞

and the expected number of failures until replacement is ʃ

∞

H (t)dG

(N )

(t) +

T

ʃ =

∞ ʃ ∑ n=N

∞

T

[ʃ

∞ T −t

0

[1 − G (N ) (t)]h(t)dt +

0

∞ ∑ n=N

ʃ

]

H (t + u)dG(u) dG (n) (t) T

[ʃ

0

∞

] G(u)h(t + u)du dG (n) (t).

0

Therefore, the expected cost rate is C O L (T , N ) = c O T + (c N − c O T )[1 − G (N ) (T )] ʃ T [ʃ ∞ {ʃ ∞ ] (n) } ∑ + c M 0 [1 − G (N ) (t)]h(t)dt + ∞ n=N 0 0 G(u)h(t + u)du dG (t) [ ] . ∑ (n) (1/θ) N + ∞ n=N G (T ) (3.42) When T = 0, C O L (0, N ) = C O F (∞, N ) = C(N ) in (3.6), and when N = 0, C O L (T , 0) = C O F (T , ∞). In particular, when c N = c O T and G(t) = 1 − e−θt , (3.42) is ʃ∞ { ) c O T + c M Hʃ (T ) + T [1 − G (N } (t)]h(t)dt ∞ −θt (N ) + G (T ) 0 e h(T + t)dt [ ] C O L (T , N ) = , ∑ (n) (1/θ) N + ∞ n=N G (T )

(3.43)

where note that C O L (T , 0) = C O L (T , 1) = C O (T ) in (3.35). We find optimal TO∗ L and N O∗ L to minimize C O L (T , N ) when h(t) increases strictly with t from 0 to ∞. Differentiating C O L (T, N ) with respect to T and setting it equal to zero, ʃ

[

∞

e

−θt

h(T + t)dt N +

0

− G (N ) (T )

∞ ∑

] G

(n)

∞ 0

∞

e−θt h(T + t)dt =

[1 − G (N ) (t)]h(t)dt

T

n=N

ʃ

ʃ

(T ) − H (T ) − cO T , cM

(3.44)

whose left-hand side increases strictly with T to ∞. Thus, there exists a finite and unique TO∗ L (0 ≤ TO∗ L < ∞) which satisfies (3.44), and the resulting cost rate is

50

3 Replacement Model with Minimal Repair

C O L (TO∗ L , N ) = c M

ʃ 0

∞

θe−θt h(TO∗ L + t)dt.

(3.45)

Noting that the left-hand side of (3.44) decreases strictly with N from that of (3.36), TO∗ L increases with N from TO∗ given in (3.36), and TO∗ L ≥ TO∗ . Forming the inequality C O L (T , N + 1) − C O L (T , N ) ≥ 0, [ ˜2 (T , N ) N + R − G (N ) (T )

∞ ∑

] G

n=N ∞ −θt

ʃ

e

(n)

ʃ

∞

(T ) − H (T ) −

[1 − G (N ) (t)]h(t)dt

T

h(T + t)dt ≥

0

cO T , cM

(3.46)

where ˜2 (T , N ) ≡ R

ʃ ∞ [ʃ ∞ T

0

] ʃ ∞ e−θu h(t + u)du dG (N ) (t) > e−θt h(T + t)dt 1 − G (N ) (T ) 0

increases strictly with T from R(N )/θ to h(∞) and increases strictly with N from ˜2 (T , 1) to h(∞) from (6) of Appendix A.2. R Substituting (3.44) for (3.46), (3.46) is ˜2 (T , N ) ≥ R

ʃ

∞

e−θt h(T + t)dt,

0

which always holds for any T , i.e., N O∗ L = 0 or 1, which follows that replacement overtime with T is better than replacement overtime last.

WIB 3.11: Replacement overtime with time T is better than replacement overtime first and last.

In addition, note that the left-hand side of (3.46) increases strictly with N from ʃ

∞

T

θe−θt h(T + t)dt − H (T ),

0

which agrees with that of (3.36). Thus, if T < TO∗ given in (3.36), then there exists a finite and unique minimum N O∗ L (1 ≤ N O∗ L < ∞) which satisfies (3.46). Furthermore, because the left-hand side of (3.46) increases strictly with T from ʃ 0

∞

[1 − G (N ) (t)][R(N ) − h(t)]dt,

3.6 Replacement Overtime with Failure Number

51

which agrees with that of (3.7), N O∗ L decreases with T from N ∗ given in (3.7), and N O∗ L ≤ N ∗ . Conversely, if T ≥ TO∗ , then N O∗ L = 0 or 1. WIB 3.12: When G(t) = 1 − e−θt , if T > TO∗ , then replacement overtime first is better than replacement overtime with time T , if T = TO∗ , then replacement overtime with TO∗ is better than replacement first and last, and if T < TO∗ , then replacement overtime last is better than replacement overtime with time T .

3.6 Replacement Overtime with Failure Number We consider replacement overtime policies in which the unit is replaced at the first failure over a planned time T or at a number of K of failures [8, p. 46].

3.6.1 Replacement First Suppose that the unit is replaced at the first failure over time T (0 ≤ T < ∞) or at failure K (K = 1, 2, . . . ), whichever occurs first. Then, the probability that the unit is replaced at the first failure over time T is P K (T ), and the probability that it is replaced at failure K is PK (T ). Thus, the mean time to replacement is ʃ

T 0

t dPK (t) +

ʃ

=

K −1 ʃ T ∑ k=0

T

e H (t)

0

∞

] u dF(u) dPk (t)

T

ʃ

∞

P K (t)dt + P K (T )

0

[ʃ

e−H (t)+H (T ) dt,

T

and the expected number of failures until replacement is K PK (T ) +

K −1 ∑ k=0

(k + 1) pk (T ) =

K −1 ∑

Pk (T ).

k=0

Therefore, the expected cost rate is ∑ K −1 Pk (T ) c O T + (c K − c O T )PK (T ) + c M k=0 , C O F (T , K ) = ʃ T ʃ∞ −H (t)+H (T ) dt 0 P K (t)dt + P K (T ) T e

(3.47)

52

3 Replacement Model with Minimal Repair

where c K is given in (3.24), and c O T and c M are given in (3.34). When T = ∞, C O F (∞, K ) = C(K ) in (3.25). In particular, when the unit is replaced only at the first failure over time T , the expected cost rate in (3.47) is c O T + c M [H (T ) + 1] ʃ∞ . T + T e−H (t)+H (T ) dt

C O (T ) ≡ lim C O F (T , K ) = K →∞

(3.48)

We find optimal TO∗ to minimize C O (T ) when h(t) increases strictly with t from 0 to ∞. Differentiating C O (T ) with respect to T and setting it equal to zero, ʃ

T

˜ ) − h(t)]dt = [ Q(T

0

cO T , cM

(3.49)

ʃ ˜ ) ≡ F(T )/ ∞ F(t)dt, and whose left-hand side increases strictly with where Q(T T T from 0 to ∞. Therefore, there exists a finite and unique TO∗ (0 < TO∗ < ∞) which satisfies (3.49), and the resulting cost rate is ∗ ˜ O∗ ) = c O T + c M H (TO ) , C O (TO∗ ) = c M Q(T TO∗

(3.50)

which agrees (3.2) when c O T = cT and TO∗ = T . When c O T = cT , from (3.3) and (3.49), ʃ

T

˜ ) − h(t)]dt > [ Q(T

0

ʃ

T

[h(T ) − h(t)]dt,

0

which follows that TO∗ < T ∗ , where T ∗ minimizes C(T ) in (3.2). Thus, from (3.2) and (3.50) when c O T = cT , C O (TO∗ ) > C(T ∗ ), which follows that replacement with time T is better than replacement overtime with failure.

WIB 3.13: Replacement with time T is better than replacement overtime with failure. Next, when c K = c O T , we find optimal TO∗ F and K O∗ F to minimize C O F (T, K ) in (3.47) when h(t) increases strictly with t from 0 to ∞. Differentiating C O F (T , K ) with respect to T and setting it equal to zero, ʃ

T 0

˜ ) − h(t)]dt = c O T , P K (t)[ Q(T cM

(3.51)

whose left-hand side increases strictly with T from 0 to ∞. Therefore, there exists a finite and unique TO∗ F (0 < TO∗ F < ∞) which satisfies (3.51), and the resulting cost

3.6 Replacement Overtime with Failure Number

53

rate is C O F (TO∗ F ,

˜ O∗ F ) = K ) = c M Q(T

ʃ T∗ c O T + c M 0 O F P K (t)h(t)dt , ʃ TO∗ F P K (t)dt 0

(3.52)

which agrees with (3.24) when c O T = c K = cT and TO∗ F = T . In addition, because the left-hand side of (3.51) increases strictly with K to that of (3.49), TO∗ F decreases with K to TO∗ given in (3.49), and TO∗ F ≥ TO∗ . Comparing (3.51) and (3.27) when c O T = cT , we obtain TO∗ F < TF∗ because the left-hand side of (3.51) is larger than that of (3.27). Thus, from (3.52) and (3.24), C O F (TO∗ F , K ) ≥ C F (TF∗ , K ), which follows that replacement first is better than replacement overtime first.

WIB 3.14: Replacement first is better than replacement overtime first.

Forming the inequality C O F (T , K + 1) − C O F (T, K ) ≥ 0, [ʃ

T

H2 (T , K )

ʃ

∞

P K (t)dt + P K (T ) T

0

] K∑ −1 cO T e−H (t)+H (T ) dt − Pk (T ) ≥ , cM k=0 (3.53)

where H2 (T , K ) ≡ ʃ T [ʃ ∞ 0

t

PK (T ) e−H (u)+H (t) du

]

dPK (t)

˜ ) < Q(T

ʃ∞ increases strictly with T from 1/μ to H (K ) = 1/ 0 p K (t)dt and increases strictly ˜ ) from (7) of Appendix A.2. with K from 1/μ to Q(T Substituting (3.51) for (3.53), (3.53) is ˜ ), H2 (T , K ) ≥ Q(T which does not hold for any T . Thus, there does not exist any finite K O∗ F , i.e., K O∗ F = ∞, and TO∗ F = TO∗ , which follows that replacement overtime with time T is better than replacement overtime first.

WIB 3.15: Replacement overtime with time T is better than replacement overtime first.

54

3 Replacement Model with Minimal Repair

In addition, noting that the left-hand side of (3.53) increases strictly with K to that of (3.49), if T > TO∗ given in (3.49), then there exists a finite and unique minimum K O∗ F (1 ≤ K O∗ F < ∞) which satisfies (3.53). Furthermore, because the left-hand side of (3.53) increases with T to that of (3.26), K O∗ F decreases with T to K ∗ given in (3.26), and K O∗ F ≥ K ∗ . Conversely, if T ≤ TO∗ , then K O∗ F = ∞.

3.6.2 Replacement Last Suppose that the unit is replaced at the first failure over time T (0 ≤ T < ∞) or at failure K , whichever occurs last. Then, the probability that the unit is replaced at the first failure over time T is PK (T ), and the probability that it is replaced at failure K is P K (T ). Thus, the mean time to replacement is ʃ

∞

t dPK (t) +

T

∞ ʃ ∑ k=K

ʃ

∞

=T+

0

T

[ʃ

∞

] u e H (t) dF(u) dPk (t)

T

ʃ

∞

P K (t)dt + PK (T )

e−H (t)+H (T ) dt,

T

T

and the expected number of failures until replacement is K P k (T ) +

∞ ∑

ʃ

∞

(k + 1) pk (T ) = H (T ) +

P K (t)h(t)dt + PK (T ).

T

k=K

Therefore, the expected cost rate is c O T +[ (c K − c OʃT )P K (T ) ] ∞ + c M H (T ) + T P K (t)h(t)dt + PK (T ) C O L (T , K ) = , ʃ∞ ʃ∞ T + T P K (t)dt + PK (T ) T e−H (t)+H (T ) dt

(3.54)

where note that C O L (T, 0) = C O L (T , 1). When T = 0, C O L (0, K ) = C O F (∞, K ) = C(K ) in (3.25), and when K = 0, C O L (T, 0) = C O F (T , ∞) = C O (T ) in (3.48). When c K = c O T , we find optimal TO∗ L and K O∗ L to minimize C O L (T , K ) when h(t) increases strictly with t from 0 to ∞. Differentiating C O L (T , K ) with respect to T and setting it equal to zero, ʃ 0

T

˜ ) − h(t)]dt + [ Q(T

ʃ

∞ T

˜ ) − h(t)]dt = c O T , P K (t)[ Q(T cM

(3.55)

whose left-hand side increases strictly with T to ∞. Thus, there exists a finite and unique TO∗ L (0 ≤ TO∗ L < ∞) which satisfies (3.55), and the resulting cost rate is

3.6 Replacement Overtime with Failure Number

55

˜ O∗ L ). C O L (TO∗ L , K ) = c M Q(T

(3.56)

Noting that the left-hand side of (3.55) decreases strictly with K from that of (3.49), TO∗ L increases with K from TO∗ given in (3.49), and TO∗ L ≥ TO∗ . Forming the inequality C O L (T , K + 1) − C O L (T , K ) ≥ 0, ] [ ʃ ∞ ʃ ∞ −H (t)+H (T ) ˜ P K (t)dt + PK (T ) e dt H2 (T , K ) T + T T [ ] ʃ ∞ cO T − H (T ) + P K (t)h(t)dt + PK (T ) ≥ , cM T

(3.57)

where P K (T ) ] −H e (u)+H (t) du dPK (t)

˜2 (T , K ) ≡ ʃ ∞ [ʃ ∞ H T

t

˜ ) > Q(T

ʃ∞ increases strictly with T from H (K ) = 1/ 0 p K (t)dt to h(∞) and increases strictly ˜2 (T , 1) to h(∞) from (8) of Appendix A.2. with K from H Substituting (3.55) for (3.57), (3.57) is ˜ ), ˜2 (T , K ) ≥ Q(T H which always holds for any T , i.e., K O∗ L = 0 or 1, which follows that replacement overtime with T is better than replacement overtime last.

WIB 3.16: Replacement overtime T is better than replacement overtime last.

In addition, note that left-hand side of (3.57) increases strictly with K from ʃ

T

˜ ) − h(t)]dt, [ Q(T

0

which agrees with that of (3.49). Thus, if T < TO∗ given in (3.49), then there exists a finite and unique minimum K O∗ L (1 ≤ K O∗ L < ∞) which satisfies (3.57). Furthermore, because the left-hand side of (3.57) increases strictly with T from ʃ H (K )

∞

P K (t)dt − K ,

0

which agrees with that of (3.26), K O∗ L decreases with T from K ∗ given in (3.26), and K O∗ L ≤ K ∗ . Conversely, if T ≥ TO∗ , then K O∗ L = 0 or 1.

56

3 Replacement Model with Minimal Repair

WIB 3.17: If T > TO∗ , then replacement overtime first is better than replacement overtime with time T , if T = TO∗ , then replacement overtime with time TO∗ is better than replacement first and last, and if T < TO∗ , then replacement overtime last is better than replacement overtime with time T .

3.7 Replacement with Working and Failure Numbers We consider the replacement policy in which the unit is replaced at a number N of working times and at a number K of failures [9].

3.7.1 Replacement First Suppose that the unit is replaced at work N (N = 1, 2, . . . ) or at failure K (K = 1, 2, . . . ), whichever occurs first. The probability that the unit is replaced at work N is ʃ ∞ P K (t)dG (N ) (t), 0

and the probability that it is replaced at failure K is ʃ

∞

[1 − G (N ) (t)]dPK (t).

0

Thus, the mean time to replacement is ʃ

∞

t P K (t)dG

(N )

0

ʃ

∞

(t) +

t [1 − G

(N )

ʃ ∞ (t)]dPK (t) = [1 − G (N ) (t)] P K (t)dt,

0

0

and the expected number of failures until replacement is ʃ 0

∞

H (t)P K (t)dG (N ) (t) +

ʃ

=

∞

ʃ

∞

H (t)[1 − G (N ) (t)]dPK (t)

0

[1 − G (N ) (t)] P K (t)h(t)dt.

0

Therefore, the expected cost rate is

3.7 Replacement with Working and Failure Numbers

57

ʃ∞ c K + (c N − c K ) 0 P K (t)dG (N ) (t) ʃ∞ + c M 0 [1 − G (N ) (t)] P K (t)h(t)dt . C F (N , K ) = ʃ∞ (N ) (t)]P (t)dt K 0 [1 − G

(3.58)

When K = ∞, C F (N , ∞) = C(N ) in (3.6), and when N = ∞, C F (∞, K ) = C(K ) in (3.25). We find optimal N F∗ and K F∗ to minimize C F (N , K ) when c K = c N , G(t) = 1 − e−θt and h(t) increases strictly with t from 0 to ∞. Forming the inequality C F (N + 1, K ) − C F (N , K ) ≥ 0, ʃ

∞

[1 − G (N ) (t)] P K (t)[R2 (N , K ) − h(t)]dt ≥

0

cN , cM

(3.59)

where ∑ K −1 ʃ ∞ R2 (N , K ) =

N −θt pk (t)h(t)dt k=0 0 (θt) e ∑ K −1 ʃ ∞ N −θt pk (t)dt k=0 0 (θt) e

increases strictly with N from R2 (0, K ) to h(∞) and increases strictly with K from R2 (N , 1) to R(N ) in (3.8) from (9) of Appendix A.3. Noting that the left-hand side of (3.59) increases strictly with N to ∞, there exists a finite and unique minimum N F∗ (1 ≤ N F∗ < ∞) which satisfies (3.59). In addition, because the left-hand side of (3.59) goes to that of (3.7) as K → ∞, N F∗ approaches to N ∗ given in (3) as K → ∞. Forming the inequality C F (N , K + 1) − C F (N , K ) ≥ 0, ʃ

∞

[1 − G (N ) (t)] P K (t)[H2 (K , N ) − h(t)]dt ≥

0

cK , cM

(3.60)

where ∑ N −1 ʃ ∞ H2 (K , N ) ≡

n −θt p K (t)h(t)dt n=0 0 [(θt) /n!]e ∑ N −1 ʃ ∞ n −θt p (t)dt K n=0 0 [(θt) /n!]e

increases strictly with K from H2 (0, N ) to h(∞) and increases strictly with N from H2 (K , 1) to H (K ) in (3.26) from (10) of Appendix A.2. Noting that the left-hand side of (3.60) increases strictly with K to ∞, there exists a finite and unique minimum K F∗ (1 ≤ K F∗ < ∞) which satisfies (3.60). In addition, because the left-hand side of (3.60) goes to that of (3.26) as N → ∞, K F∗ approaches to K ∗ given in (3.26) as N → ∞.

58

3 Replacement Model with Minimal Repair

3.7.2 Replacement Last Suppose that the unit is replaced at work N (N = 0, 1, 2, . . . ) or at failure K (K = 0, 1, 2, . . . ), whichever occurs last. The probability that the unit is replaced at work N is ʃ ∞ PK (t)dG (N ) (t), 0

and the probability that it is replaced at failure K is ʃ

∞

G (N ) (t)dPK (t).

0

Thus, the mean time to replacement is ʃ

∞

t PK (t)dG

(N )

ʃ

∞

(t) +

0

tG

(N )

ʃ

∞

(t)dPK (t) =

0

[1 − G (N ) (t)PK (t)]dt,

0

and the expected number of failures until replacement is ʃ

∞

0

H (t)PK (t)dG (N ) (t) +

ʃ

=

∞

ʃ

∞

H (t)G (N ) (t)dPK (t)

0

[1 − G (N ) (t)PK (t)]h(t)dt.

0

Therefore, the expected cost rate is c K + (c N − c K )

ʃ∞ 0

PK (t)dG (N ) (t)

ʃ∞ + c M 0 [1 − G (N ) (t)PK (t)]h(t)dt ʃ . C L (N , K ) = ∞ (N ) (t)P (t)]dt K 0 [1 − G

(3.61)

When K = 0, C L (N , 0) = C F (N , ∞) = C(N ) in (3.6), and when N = 0, C L (0, K ) = C F (∞, K ) = C(K ) in (3.25). We find optimal N L∗ and K L∗ to minimize C L (N , K ) when c K = c N , G(t) = 1 − e−θt and h(t) increases strictly with t from 0 to ∞. Forming the inequality C L (N + 1, K ) − C L (N , K ) ≥ 0, ʃ

∞ 0

where

˜2 (N , K ) − h(t)]dt ≥ [1 − G (N ) (t)PK (t)][ R

cN , cM

(3.62)

3.7 Replacement with Working and Failure Numbers

˜2 (N , K ) ≡ R

∑∞

59

ʃ

∞ N −θt pk (t)h(t)dt k=K 0 (θt) e ∑ ∞ ʃ∞ N −θt pk (t)dt k=K 0 (θt) e

˜2 (0, K ) to h(∞) and increases strictly with K from increases strictly with N from R R(N ) in (3.8) to h(∞) from (11) of Appendix A.2. Noting that the left-hand side of (3.62) increases strictly with N to ∞, there exists a finite and unique minimum N L∗ (0 ≤ N L∗ < ∞) which satisfies (3.62). In addition, because the left-hand side of (3.62) agrees with that of (3.9) when K = 0, N L∗ = N ∗ given in (3.9) when K = 0. Forming the inequality C L (N , K + 1) − C L (N , K ) ≥ 0, ʃ

∞

˜2 (K , N ) − h(t)]dt ≥ [1 − G (N ) (t)PK (t)][ H

0

cK , cM

(3.63)

where ˜2 (K , N ) ≡ H

ʃ∞ n −θt p K (t)h(t)dt n=N 0 [(θt) /n!]e ∑∞ ʃ ∞ n /n!]e−θt p (t)dt [(θt) K n=N 0

∑∞

˜2 (0, N ) to h(∞) and increases strictly with N from increases strictly with K from H H (K ) in (3.26) to h(∞) from (12) of Appendix A.2. Noting that the left-hand side of (3.63) increases strictly with K to ∞, there exists a finite and unique minimum K L∗ (0 ≤ K L∗ < ∞) which satisfies (3.63). In addition, because the left-hand side of (3.63) agrees with that of (3.26) when N = 0, K L∗ = K ∗ given in (3.26) when N = 0. Example 3.5 Table 3.5 presents optimal K ∗ , N ∗ , C(K ∗ )/c M and C(N ∗ )/c M for λ and ci /c M (i = N , K ). This indicates that for λ = 1, C(K ∗ )/c M < C(N ∗ )/c M , i.e., replacement with K ∗ is better than replacement with N ∗ . On the other hand, for λ = 0.1, C(K ∗ )/c M > C(N ∗ )/c M for K ∗ = c N /c M ≤ 5, and C(K ∗ )/c M < C(N ∗ )/c M for K ∗ ≥ 6. Optimal N ∗ decrease with λ. This means that when λ are large, interval times of failures become small and we should replace early to avoid its cost, and (λN ∗ )2 are almost the same as K ∗ . Table 3.6 presents (K F∗ ,N F∗ ), C F (K F∗ ,N F∗ )/c M , (K L∗ ,N L∗ ) and C L (K L∗ , N L∗ )/c M when c N = c K , G(t) = 1 − e−t and H (t) = (λt)2 for ci /c M (i = N , K ) when λ = 0.1. This indicates both Ci (K i∗ , Ni∗ ) (i = F, L) are smaller than C(K ∗ ) and C(N ∗ ), and C F (K F∗ , N F∗ )/c M < C L (K L∗ , N L∗ )/c M , however, they are almost the same. 

WIB 3.18: WIB problems for failure K and work N might depend on cost and ease of their replacements.

60

3 Replacement Model with Minimal Repair

Table 3.5 Optimal N ∗ , K ∗ , C(K ∗ )/c M and C(N ∗ )/c M when G(t) = 1 − e−t , H (t) = (λt)2 and cN = cK λ = 0.1 λ=1 cN cK ∗ ∗ or C(K ) C(N ) C(K ∗ ) C(N ∗ ) cM cM K ∗ N∗ N∗ cM cM cM cM 1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

10 14 17 20 22 22 24 26 28 32

0.226 0.301 0.361 0.413 0.459 0.500 0.539 0.575 0.608 0.640

0.210 0.293 0.357 0.410 0.457 0.503 0.542 0.578 0.611 0.643

1 1 2 2 2 2 3 3 3 3

2.257 3.009 3.611 4.127 4.585 5.002 5.387 5.746 6.084 6.404

3.000 4.000 4.500 5.000 5.500 6.000 6.333 6.667 7.000 7.333

Table 3.6 Optimal (K F∗ , N F∗ ), C F (K F∗ , N F∗ )/c M , (K L∗ , N L∗ ) and C L (K L∗ , N L∗ )/c M when G(t) = 1 − e−t , H (t) = (λt)2 , λ = 0.1 and c N = c K C F (K F∗ , N F∗ ) C L (K L∗ , N L∗ ) cN cK or (K F∗ , N F∗ ) (K L∗ , N L∗ ) cM cM cM cM 1 2 3 4 5 6 7 8 9 10

(3, 11) (4, 15) (5, 19) (6, 22) (7, 25) (8, 27) (9, 30) (10, 32) (11, 34) (12, 36)

(0, 9) (1, 13) (2, 16) (3, 18) (4, 20) (5, 22) (6, 24) (7, 25) (8, 26) (9, 28)

0.208 0.291 0.354 0.407 0.454 0.496 0.535 0.572 0.606 0.638

0.210 0.292 0.355 0.408 0.455 0.497 0.536 0.572 0.607 0.639

3.8 Problems 1. Over work N and failure K When the unit is replaced at the first failure time over work N , and it is replaced at the first working time over failure K , obtain two expected cost rates [9]. 2. Replacement with time T after failure When the unit is replaced at time T after failure K , obtain the expected cost rate. 3. Product update announcement (PUA) When the unit is replaced at PUA over time T or at time T after PUA, obtain the expected cost rate [10].

References

61

4. Block replacement and minimal repair interval [2, p. 117] Consider the combined replacement models with block replacement, no replacement and minimal repair, and obtain their expected cost rates [11].

References 1. Barlow RE, Proschan F (1965) Mathematical theory of reliability. Wiley, New York 2. Nakagawa T (2005) Maintenance theory of reliability. Springer, London 3. Nakagawa T (2011) Stochastic processes with applications to reliability theory. Springer, London 4. Nakagawa T (2014) Random maintenance policies. Springer, London 5. Zhao X, Mizutani S, Nakagawa T (2015) Which is better for replacement policies with continous or discrete scheduled times? Euro J Oper Res 242:477–486 6. Zhao X, Nakagawa T (2012) Optimization problems of replacement first or last in reliability theory. Euro J Oper Res 223:141–149 7. Zhao X, Al-Khalifa KN, Hamouda AM, Nakagawa T (2016) First and last triggering event approahes for replacement with minimal repairs. IEEE Trans Reliab 65:197–207 8. Nakagawa T, Zhao X (2015) Maintenance overtime policies in reliability. Springer, London 9. Mizutani S, Zhao X, Nakagawa T (2020) Which replacement is better at working cycles or numbers of failures. IEICE Trans Commun E103-A:523–532 10. Mizutani S, Dong W, Zhao X, Nakagawa T (2020) Preventive replacement policies with products update announcements. Comm Statist Theory Method. 49: 3821–3833 11. Zhang X, Cai J, Zhao X (2021) Optimizations of managerial maintenance policies. Inter J Reliab Qual Saf. 29(4):1–13

Chapter 4

Periodic Replacement Models

We consider the replacement model in which the unit is replaced at periodic times J T (J = 1, 2, . . . ) for a specified T [1, p. 236]. Then, we try to rewrite age replacement models in Chap. 2 and replacement models with minimal repair in Chap. 3 into periodic replacement models: We take up periodic policies of replacement first and last for age replacement in Sect. 4.1, and replacement with minimal repair with working and failure numbers in Sects. 4.2 and 4.3, respectively. In Sect. 4.4, we consider periodic replacement over failure number and compare their optimal policies. Furthermore, five modified periodic replacement models are given as Problems in Sect. 4.5. We make the following same assumptions as those of Chaps. 2 and 3: The unit has a failure distribution F(t) and failure rate h(t) which increases strictly with t from h(0) = 0 to h(∞) = ∞. Furthermore, the unit operates for the working times with an identical distribution G(t). When the unit undergoes minimal repair failures occur in [0, t] is pk (t) ≡ [H (t)k /k!]e−H (t) at failures, the probability that k∑ (k = 0, 1, 2, . . . ) and PK (t) ≡ ∞ k=K pk (t). Refer all notations used in this chapter to them in Chaps. 2 and 3.

4.1 Age Replacement It is assumed that Yn (n = 1, 2, . . . ) is the working time of job n in Fig. 2.1, and is independent with each other and has an identical distribution G(t) ≡ Pr{Yn ≤ t} with finite mean 1/θ (0 < 1/θ < ∞). Then, we rewrite replacement first and last of age replacement into periodic one.

4.1.1 Replacement First Suppose that the unit is replaced at a planned time J T (J = 1, 2, . . . ) for a specified T (0 < T < ∞), at a planned number N (N = 1, 2, . . . ) of working times, or at © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Mizutani et al., Which-Is-Better (WIB): Problems in Reliability Theory, Springer Series in Reliability Engineering, https://doi.org/10.1007/978-3-031-27316-2_4

63

64

4 Periodic Replacement Models

failure, whichever occurs first. Then, replacing T in (2.5) with J T , the expected cost rate is ʃ JT c J + (c F − c J ) 0 [1 − G (N ) (t)]dF(t) ʃ JT + (c N − c J ) 0 F(t)dG (N ) (t) , (4.1) C F (J, N ) = ʃ JT (N ) (t)]F(t)dt 0 [1 − G where c J = replacement cost at time J T , c N = replacement cost at work N and c F = replacement cost at failure with c F > c J and c F > c N . In particular, when the unit is replaced only at time J T (J = 1, 2, . . . ) [1, p. 236], C(J ) ≡ lim C F (J, N ) = N →∞

c J + (c F − c J )F(J T ) . ʃ JT F(t)dt 0

(4.2)

When the unit is replaced only at work N , ʃ∞ c N + (c F − c N ) 0 [1 − G (N ) (t)]dF(t) C(N ) ≡ lim C F ( J, N ) = , ʃ∞ (N ) (t)]F(t)dt J →∞ 0 [1 − G

(4.3)

which agrees with (2.6). Optimal N ∗ to minimize C(N ) has been already derived in Sect. 2.2. We find optimal J ∗ to minimize C( J ) in (4.2) when h(t) increases strictly with t from 0 to ∞. Forming the inequality C(J + 1) − C( J ) ≥ 0, ʃ

JT

Q1( J )

F(t)dt − F(J T ) ≥

0

cJ , cF − c J

(4.4)

where Q1( J ) ≡

F((J + 1)T ) − F(J T ) ʃ (J +1)T F(t)dt JT

ʃT increases strictly with J from Q 1 (0) ≡ Q(T ) = F(T )/ 0 F(t)dt to h(∞) from (1) of Appendix A.3. Thus, noting that h(J T ) < Q 1 (J ) < h((J + 1)T ), the left-hand side of (4.4) increases strictly with J from ʃ

T 0

ʃ F(t)[Q 1 (1) − h(t)]dt >

T

F(t)[h(T ) − h(t)]dt,

0

which agrees with that of (2.3) to ∞. Thus, if T < T ∗ given in (2.3), there exists a finite and unique minimum J ∗ (1 ≤ J ∗ < ∞) which satisfies (4.4). Conversely, if T ≥ T ∗ , then J ∗ = 1, i.e., the unit should be replaced only at time T .

4.1 Age Replacement

65

We find optimal JF∗ and N F∗ to∑minimize C F (J, N ) in (4.1) when c J = c N and n −θ t (N = 0, 1, 2, . . . ). Forming G(t) = 1 − e−θ t , i.e., G (N ) (t) = ∞ n=N [(θ t) /n!]e the inequality C F (J + 1, N ) − C F (J, N ) ≥ 0, ʃ

JT

[1 − G (N ) (t)] F(t)[Q 1 ( J, N ) − h(t)]dt ≥

0

cJ , cF − c J

(4.5)

where ʃ (J +1)T JT Q 1 ( J, N ) ≡ ʃ (J +1)T JT

[1 − G (N ) (t)]dF(t)

[1 − G (N ) (t)]F(t)dt

> h(J T )

increases strictly with J from Q 1 (0, N ) to h(∞) and increase strictly with N from Q 1 (J, 1) to Q 1 (J ), and h(J T ) < Q 1 (J, N ) < h(( J + 1)T ) from (2) of Appendix A.3. Thus, noting that the left-hand side of (4.5) increases strictly with J to ∞, there exists a finite and unique minimum JF∗ (1 ≤ JF∗ < ∞) which satisfies (4.5). In addition, because the left-hand side of (4.5) increases strictly with N to that of (4.4), JF∗ decreases with N to J ∗ given in (4.4), and JF∗ ≥ J ∗ . Forming the inequality C F (J, N + 1) − C F (J, N ) in (4.1), ʃ

JT

[1 − G (N ) (t)] F(t)[Q 2 ( J, N ) − h(t)]dt ≥

0

cJ , cF − c J

(4.6)

and forming the inequality C F (J, N − 1) − C F (J, N ) > 0, ʃ

JT

[1 − G (N ) (t)] F(t)[Q 2 (J, N − 1) − h(t)]dt
C F (J, N ). Thus, there does not exist any finite N F∗ , i.e., N F∗ = ∞ and JF∗ = J ∗ , which follows that periodic replacement with time J T is better than replacement first.

66

4 Periodic Replacement Models

WIB 4.1: Periodic replacement with J T is better than replacement first.

In addition, the left-hand side of (4.6) increases strictly with N to ʃ

JT

h(J T )

F(t)dt − F( J T ),

0

which agrees with that of (2.3) when J T = T . Thus, when c J = cT , if J T > T ∗ given in (2.3), there exists a finite and unique minimum N F∗ (1 ≤ N F∗ < ∞) which satisfies (4.6). Furthermore, because the left-hand side of (4.6) increases with J to that of (2.7), N F∗ decreases with J to N ∗ given in (2.7), and N F∗ ≥ N ∗ . Conversely, if J T ≤ T ∗ then N F∗ = ∞.

4.1.2 Replacement Last Suppose that the unit is replaced preventively at time J T (J = 0, 1, 2, . . . ) or at work N (N = 0, 1, 2, . . . ), whichever occurs last. Then, replacing T in (2.12) with J T , the expected cost rate is ʃ∞ { } c J + (c F − c J ) F(J T ) + J T [1 − G (N ) (t)]dF(t) ʃ∞ + (c N − c J ) J T F(t)dG (N ) (t) . C L ( J, N ) = ʃ∞ ʃ JT F(t)dt + J T [1 − G (N ) (t)] F(t)dt 0

(4.8)

Note that C L (J, 0) = C F ( J, ∞) = C( J ) in (4.2) and C L (0, N ) = C F (∞, N ) = C(N ) in (4.3). We find optimal JL∗ and N L∗ to minimize C L (J, N ) when c J = c N and G(t) = 1 − e−θt . Forming the inequality C L ( J + 1, N ) − C L ( J, N ) ≥ 0, ʃ

JT

˜1 (J, N ) − h(t)]dt + F(t)[ Q

ʃ

0

∞

˜1 ( J, N ) − h(t)]dt [1 − G (N ) (t)] F(t)[ Q

JT

cJ , ≥ cF − c J

(4.9)

and forming the inequality C L (J − 1, N ) − C L (J, N ) > 0, ʃ

ʃ ˜1 ( J −1, N ) − h(t)]dt + F(t)[ Q

JT

0

∞

˜1 (J −1, N ) − h(t)]dt [1 − G (N ) (t)] F(t)[ Q

JT

cJ , < cF − c J

(4.10)

4.1 Age Replacement

67

where ʃ (J +1)T (N ) G (t)dF(t) JT ˜ ≥ Q 1 (J ) Q 1 (J, N ) ≡ ʃ (J +1)T G (N ) (t)F(t)dt JT ˜1 (0, N ) to h(∞) and increases strictly with N increases strictly with J from Q ˜1 ( J, N ) < h((J + 1)T ) from (4) of from Q 1 (J ) to h((J + 1)T ), and h(J T ) < Q Appendix A.3. Thus, noting that the left-hand side of (4.9) increases strictly with J to ∞, there exists a finite and unique minimum JL∗ (0 ≤ JL∗ < ∞) which satisfies (4.9). In addition, because the left-hand side of (4.9) increases strictly with N from that of (4.4), JL∗ decreases with N from J ∗ given in (4.4), and JL∗ ≤ J ∗ . Forming the inequality C L (J, N + 1) − C L (J, N ) ≥ 0, ʃ

JT

˜2 (J, N ) − h(t)]dt + F(t)[ Q

0

ʃ

∞

˜2 (J, N ) − h(t)]dt [1 − G (N ) (t)] F(t)[ Q

JT

≥

cJ , cF − c J

(4.11)

where ʃ∞ N −θt dF(t) J T (θt) e ˜ > h(J T ) Q 2 ( J, N ) ≡ ʃ ∞ N −θt F(t)dt J T (θt) e ˜2 (J, 0) to h(∞) and increases strictly with J from increases strictly with N from Q ˜1 (J − 1, N ) < h(J T ) < Q ˜2 (J, N ) from ˜ Q 2 (0, N ) = Q(N ) in (2.7) to h(∞), and Q (5) of Appendix A.3. Substituting (4.10) for (4.11), (4.11) is ˜1 (J − 1, N ), ˜2 ( J, N ) > Q Q which always holds for any J , i.e., C L (J, N + 1) − C L (J, N ) ≥ 0. Thus, N L∗ = 0 and JL∗ = J ∗ given in (4.4), which follows that periodic replacement with time J T is better than replacement last.

WIB 4.2: Periodic replacement with J T is better than replacement last.

In addition, noting that the left-hand side of (4.11) increases strictly with N from ʃ 0

JT

˜2 (J, 0) − h(t)]dt > F(t)[ Q

ʃ 0

JT

F(t)[h( J T ) − h(t)]dt,

68

4 Periodic Replacement Models

if J T < T ∗ , then there exists a finite and unique minimum N L∗ (0 ≤ N L∗ < ∞) which satisfies (4.11). Furthermore, because the left-hand side of (4.11) increases with J from that of (2.7), N L∗ decreases with J from N ∗ given in (2.7) to 0, and N L∗ ≤ N ∗ . Conversely, if J T ≥ T ∗ , then N L∗ = 0. WIB 4.3: If J T > T ∗ , then replacement first is better than replacement with time J T , if J T = T ∗ , then replacement with time T ∗ is better than replacement first and last, and if J T < T ∗ , then replacement last is better than replacement with time J T .

Example 4.1 We present the√numerical examples when F(t) = 1 − e−(t/10) and G(t) = 1 − e−θ t , i.e., μ = 5 π  8.86, and h(t) = 0.02t, and c J /(c F − c J ) = c N /(c F − c N ) = 0.25 [3]. From (2.3), optimal T ∗ satisfies 2

ʃ

T

e−(t/10) dt − [1 − e−(T /10) ] = 0.25. 2

0.02T

2

(4.12)

0

Thus, T ∗ = 5.107. From (4.4), J ∗ satisfies e−(J T /10) − e−[(J +1)T /10] ʃ (J +1)T e−(t/10)2 dt JT 2

2

ʃ

JT

e−(t/10) dt − [1 − e−( J T /10) ] ≥ 0.25, 2

2

(4.13)

0

and from (4.2), the expected cost rate is ∗

0.25 + 1 − e−(J T /10) C(J ∗ ) = . ʃ J∗T cF − c J e−(t/10)2 dt 2

(4.14)

0

From (2.7), optimal N ∗ satisfies ʃ∞ 0

2 N −1 ʃ (θt) N e−θt (0.02t)e−(t/10) dt ∑ ∞ (θt)n −θ t −(t/10)2 ʃ∞ e e dt N −θt e−(t/10)2 dt n! 0 (θt) e n=0 0 N −1 ʃ ∞ ∑ (θt)n −θ t 2 e (0.02t)e−(t/10) dt ≥ 0.25, − n! n=0 0

(4.15)

and from (2.6), the expected cost rate is ∑ N ∗ −1 ʃ ∞ 2 n −θt 0.25 + n=0 (0.02t)e−(t/10) dt C(N ∗ ) 0 [(θ t) /n!]e = . ʃ ∑ N ∗ −1 ∞ n −θt e−(t/10)2 dt cF − c J n=0 0 [(θ t) /n!]e

(4.16)

4.1 Age Replacement

69

Table 4.1 Optimal J ∗ , N ∗ and C(J ∗ )/(c F − c J ), C(N ∗ )/(c F − c N ) when F(t) = 1 − e−(t/10) , G(t) = 1 − e−θ t and c J /(c F − c J ) = c N /(c F − c N ) = 0.25 1 C(J ∗ ) C(N ∗ ) T or J∗ N∗ θ cF − c J cF − cN 2

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.5 2.0

51 26 17 13 10 9 7 6 6 5 4 3 3

0.1021 0.1021 0.1021 0.1021 0.1022 0.1023 0.1022 0.1023 0.1023 0.1022 0.1023 0.1028 0.1032

52 26 18 13 11 9 8 7 6 6 5 4 3

0.1030 0.1038 0.1046 0.1053 0.1060 0.1067 0.1074 0.1080 0.1086 0.1093 0.1104 0.1119 0.1142

Table 4.1 presents J ∗ , C(J ∗ )/(c F − c J ), N ∗ and C(N ∗ )/(c F − c N ) for T or 1/θ < T ∗ = 5.107. This indicates that all of J ∗ and N ∗ decrease with T and 1/θ , and C(N ∗ ) > C(J ∗ ) as already shown in Sect. 4.1.1. This means that when T = 1/θ and T < T ∗ , periodic replacement with J is better than that with N . Furthermore, it is of interest that J ∗ T ∼ N ∗ /θ ∼ T ∗ = 5.107. In this case, if T < T ∗ , then a finite J ∗ exists, and if J ≤ J ∗ , then N F∗ = ∞. Conversely, if J > J ∗ , then a finite N F∗ exists, and from (4.6), N F∗ for J satisfies ʃ JT 0

2 N −1 ʃ t N e−t (0.02t)e−(t/10) dt ∑ J T t n −t −(t/10)2 e e dt ʃ JT N −t −(t/10)2 dt n! n=0 0 0 t e e N −1 ʃ J T n ∑ t −t 2 e (0.02t)e−(t/10) dt ≥ 0.25. − n! n=0 0

(4.17)

Table 4.2 presents optimal N F∗ for J and T where from Table 4.1, J ∗ = 10, 5, 3 for T = 0.5, 1.0, 1.5. Clearly, N F∗ decrease with J to 11, 6, 4 for T , respectively. Note that from Sect. 4.1, when T < T ∗ , if J > J ∗ , then N L∗ = 0, a finite N F∗ exists and satisfies (4.17), if J = J ∗ , then N F∗ = ∞ and N L∗ = 0, and if J < J ∗ , then N F∗ = ∞, and from (4.11), a finite N L∗ exists and satisfies

70

4 Periodic Replacement Models

Table 4.2 Optimal N F∗ when F(t) = 1 − e−(t/10) , G(t) = 1 − e−t and c J /(c F − c J ) = 0.25 2

T J

0.5

1.0

1.5

J∗ J∗ + 1 J∗ + 2 J∗ + 3 J∗ + 4 J∗ + 5 J∗ + 6 J∗ + 7 J∗

∞ 23 15 13 12 12 11 11 10

∞ 11 7 7 6 6 6 6 5

∞ 9 5 5 4 4 4 4 3

ʃ∞

t N e−t (0.02t)e−(t/10) dt ʃ∞ N −t −(t/10)2 dt JT t e e { N −1 ʃ ∑ 2 − 1 − e−( J T /10) + 2

{ʃ

JT

JT

n=0

e−(t/10) dt + 2

0 ∞

JT

N −1 ʃ ∑ n=0

∞ JT

t n −t −(t/10)2 e e dt n!

} t n −t −(t/10)2 e (0.02t)e dt ≥ 0.25. n!

}

(4.18)

Table 4.3 presents optimal N F∗ and N L∗ for J and T . This indicates that both N F∗ and N L∗ decrease with J and T , and when (T = 0.5, J = 10), (T = 1.0, J = 5), and (T = 1.5, J = 3), N F∗ = ∞ and N L∗ = 0. In addition, when T = 0.5, if J < 10, a positive N L∗ exists, when T = 1.0, if J < 5, a positive N L∗ exists and if J > 5, a finite Table 4.3 Optimal N F∗ and N L∗ when F(t) = 1 − e−(t/10) , G(t) = 1 − e−t and c J /(c F − c J ) = 0.25 T = 0.5 T = 1.0 T = 1.5 J N F∗ N L∗ N F∗ N L∗ N F∗ N L∗ 2

1 2 3 4 5 6 7 8 9 10 J∗

∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞

11 11 11 11 11 10 10 8 4 0 10

∞ ∞ ∞ ∞ ∞ 11 7 7 6 6

6 6 5 2 0 0 0 0 0 0 5

∞ ∞ ∞ 9 5 5 4 4 4 4

4 3 0 0 0 0 0 0 0 0 3

4.2 Replacement with Working Number

71

N F∗ exists, and when T = 1.5, if J < 3, a positive N L∗ exists and if J > 3, a finite N F∗ exists. 

4.2 Replacement with Working Number The unit undergoes minimal repair at failures in Chap. 3. When the unit is replaced at a planned time J T , we consider two replacement policies of replacement first and last in which it is replaced at a planned number N of working times in Sect. 3.2.

4.2.1 Replacement First Suppose that the unit is replaced at time J T (J = 1, 2, . . . ) for a specified T (0 < T < ∞) or at work N (N = 1, 2, . . . ), whichever occurs first. Then, replacing T in (3.5) with J T , the expected cost rate is ʃ JT c J + (c N − c J )G (N ) (J T ) + c M 0 [1 − G (N ) (t)]h(t)dt , C F (J, N ) = ʃ JT (N ) (t)]dt 0 [1 − G

(4.19)

where c M = minimal repair cost at each failure, and c J and c N are given in (4.1). In particular, when the unit is replaced only at time J T (J = 1, 2, . . . ), C(J ) = lim C F ( J, N ) = N →∞

c J + c M H (J T ) , JT

(4.20)

which agrees with (3.2) when J = 1 and c J = cT . When the unit is replaced only at work N , ʃ∞ c N + c M 0 [1 − G (N ) (t)]h(t)dt , (4.21) C(N ) ≡ lim C(J, N ) = J →∞ N /θ which agrees with (3.6), and optimal N ∗ is given in (3.7). We find optimal J ∗ to minimize C(J ) in (4.20) when h(t) increases strictly with t from 0 to ∞. Forming the inequality C(J + 1) − C(J ) ≥ 0, J H (( J + 1)T ) − (J + 1)H (J T ) ≥

cJ , cM

or J −1 ⎡ʃ ∑ j=0

( J +1)T JT

ʃ h(t)dt −

( j+1)T jT

⎤ cJ h(t)dt ≥ , cM

(4.22)

72

4 Periodic Replacement Models

whose left-hand increases strictly with J to ∞. Thus, there exists a finite and unique minimum J ∗ (1 ≤ J ∗ < ∞) which satisfies(4.22). Noting that when c J = cT , T ∗ given in (3.3) minimizes C(T ) in (3.2), C(T ∗ ) ≤ C(J ∗ ), and because the left-hand side of (4.22) increases strictly with T , J ∗ decreases with T to 1. Clearly, if T ≥ T ∗ , then J ∗ = 1. Next, we find optimal JF∗ and N F∗ to minimize C F (J, N ) in (4.19) when c N = c J , h(t) increases strictly with t from 0 to ∞ and G(t) = 1 − e−θt , i.e., G (N ) (t) = ∑ ∞ n −θ t . Forming the inequality C F ( J + 1, N ) − C F (J, N ) ≥ 0, n=N [(θ t) /n!]e ʃ

JT

[1 − G (N ) (t)][R1 (J, N ) − h(t)]dt ≥

0

cJ , cM

(4.23)

where ʃ (J +1)T R1 (J, N ) ≡

[1 − G (N ) (t)]h(t)dt > h(J T ) ʃ (J +1)T [1 − G (N ) (t)]dt JT

JT

increases strictly with J from R1 (0, N ) to h(∞) and increases strictly with N from R1 (J, 1) to [H ((J + 1)T ) − H (J T )]/T from (1) of Appendix A.4. Thus, noting that the left-hand side of (4.23) increases strictly with J to ∞, there exists a finite and unique minimum JF∗ (1 ≤ JF∗ < ∞) which satisfies (4.23). In addition, because the left-hand side of (4.23) increases strictly with N to that of (4.22), JF∗ decreases with N to J ∗ given in (4.22), and JF∗ ≥ J ∗ . Forming the inequality C F (J, N + 1) − C F (J, N ) ≥ 0, ʃ

JT

[1 − G (N ) (t)][R2 (J, N ) − h(t)]dt ≥

0

cJ , cM

(4.24)

and forming the inequality C F (J, N − 1) − C F (J, N ) > 0, ʃ

JT

[1 − G (N ) (t)][R2 ( J, N − 1) − h(t)]dt
C F (J, N ). Thus, there does not exist any finite N F∗ , i.e., N F∗ = ∞ and JF∗ = J ∗ given in (4.22), which follows that periodic replacement with J T is better than replacement first.

WIB 4.4: Periodic replacement with J T is better than replacement first.

In addition, the left-hand side of (4.24) increases strictly with N to J T h( J T ) − H (J T ), which agrees with that of (3.3) when J T = T . Thus, when c J = cT , if J T > T ∗ given in (3.3), then there exists a finite and unique minimum N F∗ (1 ≤ N F∗ < ∞) which satisfies (4.24). Furthermore, because the left-hand side of (4.24) increases with J to (3.7), N F∗ decreases with J to N ∗ given in (3.7), and N F∗ ≥ N ∗ . Conversely, if J T ≤ T ∗ then N F∗ = ∞.

4.2.2 Replacement Last Suppose that the unit is replaced at time J T (J = 0, 1, 2, . . . ) for a specified T (0 ≤ T < ∞) or at work N (N = 0, 1, 2, . . . ), whichever occurs last. Then, replacing T in (3.13) with J T , the expected cost rate is ʃ∞ { } c N + (c J − c N )G (N ) (J T ) + c M H (J T ) + J T [1−G (N ) (t)]h(t)dt ʃ∞ . C L (J, N ) = J T + J T [1 − G (N ) (t)]dt (4.26) When J = 0, C L (0, N ) = C F (∞, N ) = C(N ) in (4.21), and when N = 0, C L (J, 0) = C F (J, ∞) = C(J ) in (4.20). We find optimal JL∗ and N L∗ to minimize C L (J, N ) when c N = c J , h(t) increases strictly with t from 0 to ∞ and G(t) = 1 − e−θ t . Forming the inequality C L (J + 1, N ) − C L (J, N ) ≥ 0, ʃ 0

JT

˜1 (J, N ) − h(t)]dt + [R

ʃ

∞ JT

˜1 ( J, N ) − h(t)]dt ≥ [1 − G (N ) (t)] F(t)[ R

cJ , cM (4.27)

74

4 Periodic Replacement Models

and forming the inequality C L (J − 1, N ) − C L (J, N ) > 0, ʃ

JT

˜1 (J − 1, N ) − h(t)]dt + [R

ʃ

0

∞

˜1 (J −1, N ) − h(t)]dt [1 − G (N ) (t)] F(t)[ R

JT


h(J T )

increases strictly with N from ˜2 (J, 0) = R

ʃ∞

e−θt h(t)dt ʃ∞ = −θt dt JT e

ʃ

JT

∞

θe−θt h( J T + t)dt

0

ʃ∞ to h(∞), and increases strictly with J from R(N ) = 0 [θ (θ t) N e−θ t /N !]h(t)dt in ˜2 (J, N ) > h(J T ) > R ˜1 (J − 1, N ) from (4) of Appendix A.4. (3.8) to h(∞), and R Substituting (4.28) for (4.29), ˜1 (J − 1, N ), ˜2 ( J, N ) > R R which always holds for any J , i.e., C L ( J, N − 1) < C L ( J, N ). Thus, there does not exist for any positive N L∗ , i.e., N L∗ = 0 and JL∗ = J ∗ given in (4.22), which follows that periodic replacement with J T is better than replacement last.

4.3 Replacement with Failure Number

75

WIB 4.5: Periodic replacement with J T is better than replacement last.

In addition, note that the left-hand side of (4.29) increases strictly with N from ʃ JT

∞

θe−θ t h(J T + t)dt − H (J T ) > J T h(J T ) − H (J T ),

0

which agrees with that of (3.3) when J T = T . Thus, when c J = cT , if J T < T ∗ given in (3.3), then there exists a finite and unique minimum N L∗ (0 ≤ N L∗ < ∞) which satisfies (4.29). Furthermore, because the left-hand side of (4.29) increases with J from that of (3.9), N L∗ decreases with J from N ∗ given in (3.9) to 0, and N L∗ ≤ N ∗ . Conversely, if J T ≥ T ∗ , then N L∗ = 0. WIB 4.6: If J T > T ∗ , then replacement first is better than replacement with time J T , if J T = T ∗ , then replacement with time T ∗ is better than replacement first and last, and if J T < T ∗ , then replacement last is better than replacement with time J T .

4.3 Replacement with Failure Number The unit undergoes minimal repair at failures in Chap. 3. When the unit is replaced at a planned time J T , we consider two replacement policies in which it is replaced at a planned number K of minimal repair in Sect. 3.4 [2].

4.3.1 Replacement First Suppose that the unit is replaced at time J T (J = 1, 2, . . . ) for a specified T (0 < T < ∞) or at failure K (K = 1, 2, . . . ), whichever occurs first. Then, replacing T in (3.24) with J T , the expected cost rate is c J + (c K − c J )PK (J T ) + c M C F (J, K ) = ʃ JT P K (t)dt 0

ʃ JT 0

P K (t)h(t)dt

,

(4.30)

76

4 Periodic Replacement Models

where c K = replacement cost at failure K , and c J and c M are given in (4.19). When J = ∞, C F (∞, K ) = C(K ) in (3.25), and when K = ∞, C F (J, ∞) = C(J ) in (4.20). We find optimal JF∗ and K F∗ to minimize C F (J, K ) in (4.30) when c K = c J and h(t) increases strictly with t from 0 to ∞. Forming the inequality C F (J + 1, K ) − C F (J, K ) ≥ 0, ʃ

JT

P K (t)[H1 ( J, K ) − h(t)]dt ≥

0

cJ , cM

(4.31)

where ʃ (J +1)T H1 (J, K ) ≡

P K (t)h(t)dt > h( J T ) ʃ (J +1)T P (t)dt K JT

JT

increases strictly with J from H1 (0, K ) to h(∞) and increases strictly with K from H1 (J, 1) to [H ((J + 1)T ) − H (J T )]/T from (1) of Appendix A.5. Thus, noting that the left-hand side of (4.31) increases strictly with J to ∞, there exists a finite and unique minimum JF∗ (1 ≤ JF∗ < ∞) which satisfies (4.31). In addition, because the left-hand side of (4.31) increases strictly with K to that of (4.22), JF∗ decreases with K to J ∗ given in (4.22), and JF∗ ≥ J ∗ . Forming the inequality C F (J, K + 1) − C F (J, K ) ≥ 0, ʃ

JT

P K (t)[H2 (J, K ) − h(t)]dt ≥

0

cJ , cM

(4.32)

and forming the inequality C F (J, K − 1) − C F (J, K ) > 0, ʃ

JT

P K (t)[H2 (J, K − 1) − h(t)]dt
C F (J, K ). Thus, there does not exist any finite K F∗ , i.e., K F∗ = ∞, and JF∗ = J ∗ given in (4.22), which follows that periodic replacement with J T is better than replacement first.

WIB 4.7: Periodic replacement with J T is better than replacement first.

In addition, the left-hand side of (4.32) increases strictly with K to J T h(J T ) − H (J T ) which agrees with that of (3.3) when J T = T . Thus, when c J = cT , if J T > T ∗ given in (3.3), then there exists a finite and unique minimum K F∗ (1 ≤ K F∗ < ∞) which satisfies (4.32). Furthermore, because the left-hand side of (4.32) increases with J to that of (3.26), K F∗ decreases with J to K ∗ given in (3.26), and K F∗ ≥ K ∗ . Conversely, if J T ≤ T ∗ then K F∗ = ∞.

4.3.2 Replacement Last Suppose that the unit is replaced at time J T (J = 0, 1, 2, . . . ) for a specified T (0 ≤ T < ∞) or at failure K (K = 0, 1, 2, . . . ), whichever occurs last. Then, replacing T in (3.30) with J T , the expected cost rate is c J + (c K − c J )P K (J T ) + c M [H ( J T ) + C L (J, K ) = ʃ∞ J T + J T P K (t)dt

ʃ∞ JT

P K (t)h(t)dt]

. (4.34)

When J = 0, C L (0, K ) = C F (∞, K ) = C(K ) in (3.25), and when K = 0, C L (J, 0) = C F (J, ∞) = C(J ) in (4.20). We find optimal JL∗ and K L∗ to minimize C L (J, K ) in (4.34) when c K = c J and h(t) increases strictly with t from 0 to ∞. Forming the inequality C L (J + 1, K ) − C L (J, K ) ≥ 0, ʃ

JT 0

˜1 (J, K ) − h(t)]dt + [H

ʃ

∞

˜1 (J, K ) − h(t)]dt ≥ P K (t)[ H

JT

cJ , cM

(4.35)

and forming the inequality C L (J − 1, K ) − C K ( J, K ) > 0, ʃ

JT 0

˜1 (J − 1, K ) − h(t)]dt + [H

ʃ

∞ JT

˜1 (J − 1, K ) − h(t)]dt < c J , P K (t)[ H cM (4.36)

78

4 Periodic Replacement Models

where ˜1 (J, K ) ≡ H

ʃ (J +1)T

PK (t)h(t)dt < h(( J + 1)T ) ʃ (J +1)T PK (t)dt JT

JT

˜1 (0, K ) to h(∞) and increases strictly with K from increases strictly with J from H [H (( J + 1)T ) − H (J T )]/T to h(( J + 1)T ) from (3) of Appendix A.5. Thus, noting the left-hand side of (4.35) increases strictly with J to ∞, there exists a finite and unique minimum JL∗ (0 ≤ JL∗ < ∞) which satisfies (4.35). In addition, because the left-hand side of (4.35) increases strictly with K from that of (4.22), JL∗ decreases with K from J ∗ given in (4.22), and JL∗ ≤ J ∗ . Forming the inequality C L (J, K + 1) − C L (J, K ) ≥ 0, ʃ

JT

˜2 (J, K ) − h(t)]dt + [H

ʃ

0

∞

˜2 (J, K ) − h(t)]dt ≥ P K (t)[ H

JT

cJ , cM

(4.37)

where ˜2 (J, K ) ≡ H

ʃ∞

p K (t)h(t)dt ʃ∞ > h( J T ) J T p K (t)dt

JT

ʃ ˜ T ) in (3.33) ˜2 (J, 0) = F(J T )/ ∞ F(t)dt = Q(J increases strictly with K from H JT ʃ∞ to h(∞) and increases strictly with J from H (K ) = 1/ 0 p K (t)dt in (3.26) to ˜1 ( J − 1, K ) from (4) of Appendix A.5. ˜2 ( J, K ) > h( J T ) > H h(∞), H Substituting (4.36) for (4.37), (4.37) is ˜1 (J − 1, K ), ˜2 (J, K ) > H H which always holds for any J , i.e., C L (J, K + 1) ≥ C L (J, K ). Thus, there does not exist for any positive K L∗ , i.e., K L∗ = 0 and JK∗ = J ∗ given in (4.22), which follows that periodic replacement with J T is better than replacement last. In addition, the left-hand side of (4.37) increases strictly with K from ˜ T ) − H (J T ) > J T h( J T ) − H (J T ), J T Q(J which agrees with that of (3.3) when J T = T . Thus, when c J = cT , if J T < T ∗ then there exists a finite and unique minimum K L∗ (0 ≤ K L∗ < ∞) which satisfies (4.37). Furthermore, because the left-hand side of (4.37) increases with J from that of (3.26), K L∗ decreases with J from K ∗ given in (3.26) to 0, and K L∗ ≤ K ∗ . Conversely, if J T ≥ T ∗ then K L∗ = 0.

4.4 Replacement Over Failure Number

79

WIB 4.8: If J T > T ∗ , then replacement first is better than replacement with time J T , if J T = T ∗ , then replacement with time T ∗ is better than replacement first and last, and if J T < T ∗ , then replacement last is better than replacement with time J T .

4.4 Replacement Over Failure Number The unit is replaced at periodic times J T or at the first periodic time over failure K .

4.4.1 Replacement First Suppose that the unit is replaced at periodic times J T (J = 1, 2, . . . ) for a specified T (0 < T < ∞) or at the first periodic time over failure number K (K = 0, 1, 2, . . . ), whichever occurs first. We define that pk [H (( j + 1)T ) − H ( j T )] is the probability that k (k = 0, 1, 2, . . . ) failures occur in the interval [ j T , ( j + 1)T ] j T )] ≡ pk ( j T ) is the probability that k failures ( j = 0, 1, 2, . . . ), pk [H (∑ ∑ K −1 occur in p ( j T ), P ( j T ) ≡ 1 − P ( j T ) = [0, j T ], and PK ( j T ) ≡ ∞ K K k k=K k=0 pk ( j T ). Then, we have the relation K −1 ∑

pk [H ( j T )]

K∑ −k−1

k=0

=

pi [H (( j + 1)T ) − H ( j T )] =

i=0 K −1 ∑

K −1 ∑

pk [H (( j + 1)T )]

k=0

pk (( j + 1)T ) = P K (( j + 1)T ).

k=0

The probability that the unit is replaced at time J T is P K (J T ), and the probability that it is replaced over failure K is J −1 K −1 ∑ ∑

pk [H ( j T )]

j=0 k=0

=

−1 J −1 K ∑ ∑ j=0 k=0

=

−1 J −1 K ∑ ∑

∞ ∑

pi [H (( j + 1)T − H ( j T )]

i=K −k

pk [H ( j T )] −

−1 J −1 K ∑ ∑

pk [H ( j T )]

j=0 k=0

K∑ −k−1 i=0

[ pk ( j T ) − pk (( j + 1)T )] = PK ( J T ).

j=0 k=0

Thus, the mean time to replacement is

pi [H (( j + 1)T ) − H ( j T )]

80

4 Periodic Replacement Models K −1 ∑

(J T )

pk (J T ) +

k=0

=T

J −1 ∑

J −1 K −1 ∑ ∑ [( j + 1)T ] [ pk ( j T ) − pk (( j + 1)T )] j=0

k=0

P K ( j T ),

j=0

and the expected number of failures until replacement is K −1 ∑

k pk (J T ) +

k=0

−1 J −1 K ∑ ∑

pk [H ( j T )]

∞ ∑

(i + k) pi [H (( j + 1)T ) − H ( j T )]

i=K −k

j=0 k=0

J −1 ∑ = [H (( j + 1)T ) − H ( j T )] P K ( j T ). j=0

Therefore, the expected cost rate is c J + (c O K − c J )PK (J T ) ∑ −1 + c M Jj=0 [H (( j + 1)T ) − H ( j T )]P K ( j T ) C O F (J, K ) = , ∑ −1 T Jj=0 PK( jT)

(4.38)

where c O K = replacement cost over failure K , and c J and c M are given in (4.19). When K = ∞, C O F (J, ∞) = C(J ) in (4.20). When the unit is replaced only at the first periodic time j T over failure K , the expected cost rate is C O (K ) ≡ lim C O F (J, K ) J →∞ ∑ cO K + cM ∞ j=0 [H (( j + 1)T ) − H ( j T )]P K ( j T ) = , ∑ T ∞ j=0 P K ( j T )

(4.39)

where note C O (0) = [c O K + c M H (T )]/T which agrees with (3.2) when c O K = cT . We find optimal K O∗ to minimize C O (K ) (K = 0, 1, 2, . . . ) when h(t) increases strictly with t from 0 to ∞. Forming the inequality C O (K + 1) − C O (K ) ≥ 0, H3 (K )

∞ ∑ j=0

∞ ∑ cO K PK( jT) − [H (( j + 1)T ) − H ( j T )] P K ( j T ) ≥ , cM j=0

where ∑∞ H3 (K ) ≡

j=0 [H (( j

+ 1)T ) − H ( j T )] p K ( j T ) ∑∞ > T h(T ) j=0 p K ( j T )

(4.40)

4.4 Replacement Over Failure Number

81

increases strictly with K from H3 (0) to T h(∞) from (5) of Appendix A.5. Thus, noting that the left-hand side of (4.40) increases strictly with K from H3 (0) − H (T ) > T h(T ) − H (T ), which agrees with that of (3.3) to ∞, there exists a finite and unique minimum K O∗ (0 ≤ K O∗ < ∞) which satisfies (4.40). In addition, if T ≥ T ∗ given in (3.3), K O∗ = 0. We find optimal JO∗ F and K O∗ F to minimize C O F (J, K ) in (4.38) when c O K = c J and h(t) increases strictly with t from 0 to ∞. Forming the inequality C O F (J + 1, K ) − C O F ( J, K ) ≥ 0, [H ((J + 1)T ) − H ( J T )]

J −1 ∑

PK( jT) −

j=0

J −1 ∑ [H (( j + 1)T ) − H ( j T )] P K ( j T ) j=0

cO K ≥ . cM

(4.41)

Noting that the left-hand side of (4.41) increases strictly with J to ∞, there exists a finite and unique minimum JO∗ F (1 ≤ JO∗ F < ∞) which satisfies (4.41). Furthermore, because the left-hand side of (4.41) increases strictly with K to that of (4.22), JO∗ F decreases with K to J ∗ given in (4.22), and JO∗ F ≥ J ∗ . Forming the inequality C O F (J, K + 1) − C O F ( J, K ) ≥ 0, H3 (J, K )

J −1 ∑

PK( jT) −

j=0

J −1 ∑ cO K , (4.42) [H (( j + 1)T ) − H ( j T )] P K ( j T ) ≥ cM j=0

and forming the inequality C O F (J, K − 1) − C O F (J, K ) > 0, H3 (J, K − 1)

J −1 ∑ j=0

PK( jT) −

J −1 ∑ cO K [H (( j + 1)T ) − H ( j T )] P K ( j T ) < , cM j=0

(4.43) where ∑ J −1 H3 (J, K ) ≡

j=0 [H (( j +1)T ) − H ( j T )] p K ( j T ) ∑ J −1 j=0 p K ( j T )

< H (J T ) − H ((J −1)T )

increases strictly with K from H3 (J, 0) to H ( J T ) − H ((J − 1)T ) and increases strictly with J from H (T ) to H3 (K ) from (5) of Appendix A.5. Substituting (4.41) for (4.43), (4.43) is H3 (J, K − 1) < H ((J + 1)T ) − H (J T ),

82

4 Periodic Replacement Models

which always holds for any J , i.e., C O F (J, K − 1) > C O F ( J, K ). Thus, there does not exist any finite K O∗ F , and K O∗ F = ∞ and JO∗ F = J ∗ given in (4.22), which follows that periodic replacement with J T is better than replacement overtime first.

WIB 4.9: Periodic replacement with J T is better than replacement overtime first.

In addition, the left-hand side of (4.42) increases strictly with K to J [H (J T ) − H ((J − 1)T )] − H (J T ) < J [H ((J + 1)T ) − H (J T )] − H (J T ), which agrees with that of (4.22). Thus, when c O K = c J , if J ≥ J ∗ + 1 then there exists a finite and unique minimum K O∗ F (0 ≤ K O∗ F < ∞) which satisfies (4.42). Furthermore, because the left-hand side of (4.42) increases with J to that of (4.40), K O∗ F decreases with J to K O∗ given in (4.40), and K O∗ F ≥ K O∗ . Conversely, if J ≤ J ∗ then K O∗ F = ∞.

4.4.2 Replacement Last Suppose that the unit is replaced at periodic times J T (J = 0, 1, 2, . . . ) for a specified T (0 < T < ∞) or at the first periodic time over failure number K (K = 0, 1, 2, . . . ), whichever occurs last. Then, the probability that the unit is replaced at time J T is PK ( J T ), and the probability that it is replaced over failure K is P K (J T ). Thus, the mean time to replacement is (J T )

∞ ∑

pk ( J T ) +

k=K

∞ ∑

[( j + 1)T ]

j=J

= JT + T

∞ ∑

K −1 ∑

[ pk ( j T ) − pk (( j + 1)T )]

k=0

P K ( j T ),

j=J

and the expected number of failures until replacement is ∞ ∑

k pk (J T ) +

k=K

= H (J T ) +

∞ K −1 ∑ ∑

pk [H ( j T )]

j=J k=0 ∞ ∑

∞ ∑ (i +k) pi [H (( j +1)T )− H ( j T )]

i=K −k

[H (( j + 1)T ) − H ( j T )] P K ( j T ).

j=J

Therefore, the expected cost rate is

4.4 Replacement Over Failure Number

83

c J + (c K (J T ) { O K − c J )P∑ } + c M H (J T ) + ∞ j=J [H (( j + 1)T ) − H ( j T )] P K ( j T ) C O L (J, K ) = . ∑ JT + T ∞ j=J P K ( j T ) (4.44) When J = 0, C O L (0, K ) = C O F (∞, K ) = C O (K ) in (4.39), and when K = 0, C O L ( J, 0) = C O F (J, ∞) = C(J ) in (4.20). We find optimal JO∗ L and K O∗ L to minimize C O L (J, K ) in (4.44) when c O K = c J and h(t) increases strictly with t from 0 to ∞. Forming the inequality C O L (J + 1, K ) − C O L ( J, K ) ≥ 0, ⎡ [H ((J + 1)T ) − H ( J T )] ⎣ J +

∞ ∑

⎤ P K ( j T )⎦ − H (J T )

j=J

−

∞ ∑ cO K [H (( j + 1)T ) − H ( j T )] P K ( j T ) ≥ , cM j=J

(4.45)

and forming the inequality C O L (J − 1, K ) − C O L ( J, K ) > 0, ⎡ [H (J T ) − H ((J − 1)T )] ⎣ J +

∞ ∑

⎤ P K ( j T )⎦ − H (J T )

j=J ∞ ∑ cO K − [H (( j + 1)T ) − H ( j T )] P K ( j T ) < . cM j=J

(4.46)

Noting that the left-hand side of (4.45) increases strictly with J to ∞, there exists a finite and unique minimum JO∗ L (0 ≤ JO∗ L < ∞) which satisfies (4.45). Furthermore, because the left-hand side of (4.45) increases strictly with K from that of (4.22), JO∗ L decreases with K from J ∗ given in (4.22), and JO∗ L ≤ J ∗ . Forming the inequality C O L (J, K + 1) − C O L (J, K ) ≥ 0, ⎡ ˜3 ( J, K ) ⎣ J + H

∞ ∑

⎤ P K ( j T )⎦ − H (J T )

j=J

−

∞ ∑ cO K [H (( j + 1)T ) − H ( j T )] P K ( j T ) ≥ , cM j=J

(4.47)

where ˜3 (J, K ) ≡ H

∑∞

j=J [H (( j

+ 1)T ) − H ( j T )] p K ( j T ) ∑∞ > H ((J + 1)T ) − H (J T ) j=J p K ( j T )

84

4 Periodic Replacement Models

˜3 (J, 0) to h(∞) and increases strictly with J from increases strictly with K from H H3 (K ) to h(∞) from (6) of Appendix A.5. Substituting (4.46) for (4.47), (4.47) is ˜3 (J, K ) > H (J T ) − H ((J − 1)T ), H which always holds for any J , i.e., C O L (J, K + 1) > C O L (J, K ). Thus, there does not exist any positive K O∗ L , and K O∗ L = 0 and JO∗ L = J ∗ give in (4.22), which follows that periodic replacement with J T is better than replacement overtime last.

WIB 4.10: Periodic replacement with J T is better than replacement overtime last.

In addition, note that the left-hand side of (4.47) increases strictly with K from ˜3 (J, 0) − H (J T ) > J [H ((J + 1)T ) − H (J T )] − H (J T ), JH which agrees with that of (4.22). Thus, when c O K = c J , if J ≤ J ∗ − 1, then there exists a finite and unique minimum K O∗ L (0 ≤ K O∗ L < ∞) which satisfies (4.47). Furthermore, because the left-hand side of (4.47) increases with J from that of (4.40), K O∗ L decreases with J from K O∗ given in (4.40), and K O∗ L ≤ K O∗ . Conversely, if J ≥ J ∗ , then K O∗ L = 0. WIB 4.11: If J ≥ J ∗ + 1, then replacement overtime first is better than periodic replacement J T , if J = J ∗ , then periodic replacement with J ∗ T is better than replacement overtime first and last, and if J ≤ J ∗ − 1, then replacement overtime last is better than periodic replacement J T .

4.5 Problems 1. Work over J T When the unit is replaced at the first working time over time J T in age replacement, obtain the expected cost rate [3]. 2. Periodic time over work N When the unit is replaced at the first periodic time over work N in age replacement, obtain the expected cost rate. 3. Deviation time Derive the deviation time in Sect. 2.6 for periodic age replacement.

References

85

4. Work N over J T When the unit is replaced at the first working time over J T or work N in replacement with minimal repair, whichever occurs first, obtain the expected cost rate. 5. J T over work N When the unit is replaced at the first periodic time over work N or work N in replacement with minimal repair, whichever occurs first, obtain the expected cost rate.

References 1. Nakagawa T (2005) Maintenance theory of reliability. Springer, London 2. Zhao X, Qian CH, Nakagawa T (2017) Comparisons of replacement policies with periodic times and repair numbers. Reliab Eng Syst Saf 168:161–170 3. Mizutani S, Zhao X, Nakagawa T (2019) Random age replacement policies with periodic planning times. Reliab Qual Saf Eng 26(5):1950023. https://doi.org/10.1142/S0218539319500232

Chapter 5

Extended Replacement Models

The following three extended replacement models are considered: We take up age replacement with two failure modes such as major and minor failures in Sect. 5.1. Optimal policies for replacement first and last are discussed and are compared theoretically and numerically. We introduce a new policy in which the unit is replaced in the interval [TO , T ] before time T in Sect. 5.2. Optimal policies for age replacement and replacement with minimal repair are derived. Finally, we propose age replacement and replacement with minimal repair models with three kinds of replacement policies in which the unit is replaced at middle times in Sect. 5.3. These models are compared with replacement first and last, and the best policies are determined theoretically. Furthermore, four extended replacement models are given as Problems in Sect. 5.4.

5.1 Replacement with Two Failure Modes Consider the replacement model with the following two failure modes [1]: (a) Majorʃ failures occur according to a general distribution F(t) with finite mean ∞ μ ≡ 0 F(t)dt, a density function f (t) ≡ dF(t)/dt, and failure rate h(t) ≡ f (t)/ F(t) increases with t from h(0) = 0 to h(∞) = ∞, where Φ(t) ≡ 1 − Φ(t) for any function Φ(t). When major failure occurs, a failed system is replaced with a new one immediately. (b) Minor failures, which are independent of major failures, occur at a nonhomoʃt geneous Poisson process with cumulative hazard function R(t) ≡ 0 r (u)dt, where r (t) ≡ dR(t)/dt increases with t from r (0) = 0 to r (∞) = ∞. The system undergoes minimal repair at minor failures and can operate again without disturbing r (t). Then, the probability that k minor failures occur exactly in [0, t] is pk (t) ≡ [R(t)k /k!]e−R(t) (k = 0, 1, 2, . . . ).

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Mizutani et al., Which-Is-Better (WIB): Problems in Reliability Theory, Springer Series in Reliability Engineering, https://doi.org/10.1007/978-3-031-27316-2_5

87

88

5 Extended Replacement Models

The system operates ʃ ∞ for the working times with an identical distribution G(t) with finite mean 1/θ ≡ 0 G(t)dt, where G (n) (t) (n = 1, 2, . . . ) is the n-fold Stieltjes convolution of G(t) with itself and G (0) (t) ≡ 1 for t ≥ 0 in Sects. 2.2 and 3.2. We take up two policies of replacement first and last for the system with the above two failure modes.

5.1.1 Replacement First Suppose that the system is replaced at a planned time T (0 < T ≤ ∞), at a planned number N (N = 1, 2, . . . ) of working times or at major failure, whichever occurs first. Then, the probability that the system is replaced at time T is [1 − G (N ) (T )] F(T ), the probability that it is replaced at work N is ʃ

T

F(t)dG (N ) (t),

0

and the probability that it is replaced at major failure is ʃ

T

[1 − G (N ) (t)]dF(t).

0

Thus, the mean time to replacement is ʃ T ʃ T t F(t)dG (N ) (t) + t [1 − G (N ) (t)]dF(t) T [1 − G (N ) (T )] F(T ) + 0 0 ʃ T [1 − G (N ) (t)] F(t)dt, = 0

and the expected number of minor failures until replacement is ʃ T ʃ T R(T )[1 − G (N ) (T )] F(T ) + R(t) F(t)dG (N ) (t) + R(t)[1 − G (N ) (t)]dF(t) 0 0 ʃ T [1 − G (N ) (t)] F(t)r (t)dt. = 0

Therefore, the expected cost rate is

5.1 Replacement with Two Failure Modes

89

ʃT ʃT cT + (c N − cT ) 0 F(t)dG (N ) (t) + (c F − cT ) 0 [1−G (N ) (t)]dF(t) ʃT + c M 0 [1 − G (N ) (t)] F(t)r (t)dt , C F (T , N ) = ʃT (N ) (t)]F(t)dt 0 [1 − G (5.1) where cT = replacement cost at time T , c N = replacement cost at work N , c F = replacement cost at major failure, and c M = minimal repair cost at minor failure. When r (t) ≡ 0, and F(t) ≡ 0 and r (t) = h(t), C F (T , N ) agrees with (2.5) and (3.5), respectively. In particular, when the system is replaced only at time T or at major failure, whichever occurs first, the expected cost rate is C(T ) ≡ lim C F (T , N ) = N →∞

cT + (c F − cT )F(T ) + c M ʃT 0 F(t)dt

ʃT 0

F(t)r (t)dt

.

(5.2)

When the system is replaced only at work N or at failure, whichever occurs first, the expected cost rate is C(N ) ≡ lim C F (T , N ) T →∞ ʃ∞ ʃ∞ c N + (c F − c N ) 0 [1 − G (N ) (t)]dF(t) + c M 0 [1 − G (N ) (t)]F(t)r (t)dt . = ʃ∞ (N ) (t)]F(t)dt 0 [1 − G (5.3) When both h(t) and r (t) increase strictly with t from 0 to ∞, we find optimal T ∗ and N ∗ to minimize C(T ) and C(N ), respectively. Differentiating C(T ) in (5.2) with respect to T and setting it equal to zero, ʃ

T

(c F − cT )

ʃ F(t)[h(T ) − h(t)]dt + c M

T

F(t)[r (T ) − r (t)]dt = cT ,

(5.4)

0

0

whose left-hand side increases strictly with T from 0 to ∞. Thus, there exists a finite and unique T ∗ (0 < T ∗ < ∞) which satisfies (5.4), and the resulting cost rate is C(T ∗ ) = (c F − cT )h(T ∗ ) + c M r (T ∗ ).

(5.5)

Forming the inequality C(N + 1) − C(N ) ≥ 0, ʃ ∞ (c F − c N ) [1 − G (N ) (t)] F(t)[Q 1 (N ) − h(t)]dt ʃ ∞ 0 [1 − G (N ) (t)] F(t)[Q 3 (N ) − r (t)]dt ≥ c N , + cM 0

(5.6)

90

5 Extended Replacement Models

where ʃT Q 1 (T , N ) ≡ ʃ 0T Q 3 (T , N ) ≡

[G (N ) (t) − G (N +1) (t)]dF(t)

< h(T ),

(N ) (t) − G (N +1) (t)]F(t)dt 0 [G ʃ T (N ) (t) − G (N +1) (t)]F(t)r (t)dt 0 [G ʃT (N ) (t) − G (N +1) (t)]F(t)dt 0 [G

Q 1 (N ) ≡ lim Q 1 (T , N ), T →∞

< r (T ),

Q 3 (N ) ≡ lim Q 3 (T, N ). T →∞

∑ n −θ t (N = It is proved that when G(t) = 1 − e−θ t , i.e., G (N ) (t) = ∞ n=N [(θ t) /n!]e 0, 1, 2, . . . ), Q 1 (T , N ) increases strictly with T from h(0) to Q 1 (N ) and increases strictly with N from Q 1 (T , 0) to h(T ) from (1) of Appendix A.1, and Q 3 (T , N ) increases strictly with T from r (0) to Q 3 (N ) and increases strictly with N from Q 3 (T , 0) to r (T ), from (1) of Appendix A.6. Thus, Q 1 (N ) increases strictly with N from Q 1 (0) to h(∞) and Q 3 (N ) increases strictly with N from Q 3 (0) to r (∞). Then, noting that the left-hand side of (5.6) increases strictly with N to ∞, there exists a finite and unique minimum N ∗ (1 ≤ N ∗ < ∞) which satisfies (5.6). Next, when cT = c N and G(t) = 1 − e−θ t , we find optimal T ∗ and N ∗ to minimize C F (T , N ) in (5.1). Differentiating C F (T , N ) with respect to T and setting it equal to zero, ʃ T [1 − G (N ) (t)] F(t)[h(T ) − h(t)]dt (c F − cT ) 0 ʃ T [1 − G (N ) (t)] F(t)[r (T ) − r (t)]dt = cT , + cM

(5.7)

0

whose left-hand side increases strictly with T from 0 to ∞. Thus, there exists a finite and unique TF∗ (0 < TF∗ < ∞) which satisfies (5.7), and the resulting cost rate is C F (TF∗ , N ) = (c F − cT )h(TF∗ ) + c M r (TF∗ ).

(5.8)

Furthermore, because the left-hand side of (5.7) increases strictly with N to that of (5.4), TF∗ decreases with N to T ∗ given in (5.4), and TF∗ ≥ T ∗ . Forming the inequality C F (T , N + 1) − C F (T , N ) ≥ 0, ʃ T [1 − G (N ) (t)] F(t)[Q 1 (T , N ) − h(t)]dt (c F − cT ) 0 ʃ T [1 − G (N ) (t)] F(t)[Q 3 (T , N ) − r (t)]dt ≥ c N . + cM 0

Substituting (5.7) for (5.9), (5.9) is Q 1 (T , N ) + Q 3 (T , N ) ≥ h(T ) + r (T ),

(5.9)

5.1 Replacement with Two Failure Modes

91

which does not hold for any T . Thus, there does not exist any finite N F∗ , i.e., N F∗ = ∞ and TF∗ = T ∗ , which follows that replacement with time T ∗ is better than replacement first. WIB 5.1: Replacement with time T ∗ is better than replacement first.

In addition, noting that the left-hand side of (5.9) increases strictly with N to that of (5.4), if T > T ∗ given in (5.4), then there exists a finite and unique minimum N F∗ (1 ≤ N F∗ < ∞) which satisfies (5.9). Furthermore, because the left-hand side of (5.9) increases with T to that of (5.6), N F∗ decreases with T to N ∗ given in (5.6), and N F∗ ≥ N ∗ . Conversely, if T ≤ T ∗ , then N F∗ = ∞.

5.1.2 Replacement Last Suppose that the system is replaced preventively at time T (0 ≤ T < ∞) or work N (N = 0, 1, 2, . . . ), whichever occurs last. Then, the probability that the system is replaced at time T is G (N ) (T ) F(T ), the probability that it is replaced at work N is ʃ

∞

F(t)dG (N ) (t),

T

and the probability that it is replaced at major failure is ʃ

∞

F(T ) +

[1 − G (N ) (t)]dF(t).

T

Thus, the mean time to replacement is ʃ ∞ ʃ T ʃ ∞ T G (N ) (T ) F(T ) + t F(t)dG (N ) (t) + t dF(t) + t [1 − G (N ) (t)]dF(t) T 0 T ʃ ∞ ʃ T (N ) F(t)dt + [1 − G (t)] F(t)dt, = 0

T

92

5 Extended Replacement Models

and the expected number of minor failures until replacement is (N )

ʃ

∞

(N )

ʃ

T

R(t) F(t)dG (t) + R(t)dF(t) R(T )G (T ) F(T ) + T 0 ʃ ∞ R(t)[1 − G (N ) (t)]dF(t) + T ʃ ∞ ʃ T F(t)r (t)dt + [1 − G (N ) (t)] F(t)r (t)dt. = 0

T

Therefore, the expected cost rate is ʃ∞ (N ) cT + (c N − c{T ) T F(t)dG (t) ʃ∞ } + (c F − cT ) F(T ) + T [1 − G (N ) (t)]dF(t) ʃ∞ {ʃ T } + c M 0 F(t)r (t)dt + T [1 − G (N ) (t)] F(t)r (t)dt C L (T , N ) = . ʃ∞ ʃT (N ) (t)] F(t)dt 0 F(t)dt + T [1 − G

(5.10)

When r (t) = 0 and F(t) = 0, C L (T , N ) agrees with (2.12) and (3.13), respectively. Furthermore,C L (0, N ) = C F (∞, N ) = C(N )in(5.3)andC L (T , 0) = C F (T , ∞) = C(T ) in (5.2). When cT = c N and both h(t) and r (t) increase strictly with t from 0 to ∞, we find optimal TL∗ and N L∗ to minimize C L (T , N ) in (5.10). Differentiating C L (T , N ) with respect to T and setting it equal to zero, [ʃ

T

(c F − cT ) [ʃ + cM

ʃ F(t)[h(T ) − h(t)]dt +

0 T

∞

] (t)] F(t)[h(T ) − h(t)]dt

T

] [1 − G (N ) (t)] F(t)[r (T ) − r (t)]dt = cT ,

ʃ

∞

F(t)[r (T ) − r (t)]dt +

0

[1 − G

(N )

T

(5.11) whose left-hand side increases strictly with T from ʃ

∞

−(c F − cT ) 0

[1 − G (N ) (t)]dF(t) − c M

ʃ

∞

[1 − G (N ) (t)] F(t)r (t)dt < 0

0

to ∞. Thus, there exists a finite and unique TF∗ (0 < TF∗ < ∞) which satisfies (5.11), and the resulting cost rate is C L (TL∗ , N ) = (c F − cT )h(TL∗ ) + c M r (TL∗ ).

(5.12)

In addition, noting that the left-hand side of (5.11) decreases strictly with N from that of (5.4), TL∗ increases with N from T ∗ given in (5.4), and TL∗ ≥ T ∗ .

5.1 Replacement with Two Failure Modes

93

Forming the inequality C L (T , N + 1) − C L (T , N ) ≥ 0, ] ʃ ∞ (N ) ˜ ˜ F(t)[ Q 1 (T , N )−h(t)]dt + [1−G (t)] F(t)[ Q 1 (T , N )−h(t)]dt

[ʃ

T

(c F −cT ) 0 T

T

] ʃ ∞ ˜3 (T , N )−r (t)]dt + [1−G (N ) (t)] F(t)[ Q ˜3 (T , N )−r (t)]dt F(t)[ Q

[ʃ + cM

T

0

≥ cT ,

(5.13)

where when G(t) = 1 − e−θ t , ʃ∞

T ˜1 (T , N ) ≡ ʃ ∞ Q T

(θt) N e−θt dF(t)

(θt) N e−θt F(t)dt

> h(T )

increases strictly with T from Q 1 (N ) to h(∞) and increases strictly with N from ˜1 (T , 0) to h(∞), and Q ˜3 (T , N ) ≡ Q

ʃ∞

(θt) N e−θt F(t)r (t)dt > r (T ) ʃ∞ N −θt F(t)dt T (θt) e

T

increases strictly with T from Q 3 (N ) to r (∞) and increases strictly with N from ˜3 (T , 0) to r (∞), from (2) of Appendix A.6. Q Substituting (5.11) for (5.13), (5.13) is ˜3 (T , N ) ≥ h(T ) + r (T ), ˜1 (T , N ) + Q Q which always holds for any T . Thus, C L (T , N + 1) − C L (T , N ) ≥ 0 for any N , i.e., N L∗ = 0, and TL∗ = T ∗ , which follows that replacement with time T ∗ is better than replacement last. WIB 5.2: Replacement with time T ∗ is better than replacement last.

In addition, note that the left-hand side of (5.13) increases strictly with N from ʃ T ˜1 (T , 0) − h(t)]dt + c M ˜3 (T, 0) − r (t)]dt F(t)[ Q F(t)[ Q 0 0 ʃ T ʃ T F(t)[h(T ) − h(t)]dt + c M F(t)[r (T ) − r (t)]dt, > (c F − cT ) ʃ

(c F − cT )

T

0

0

which agrees with that of (5.4). Thus, if T < T ∗ given in (5.4), then there exists a finite and unique minimum N L∗ (0 ≤ N L∗ < ∞) which satisfies (5.13). Furthermore, because the left-hand side of (5.13) increases with T from that of (5.6), N L∗ decreases

94

5 Extended Replacement Models

with T from N ∗ given in (5.6) to 0, and N L∗ ≤ N ∗ . Conversely, if T ≥ T ∗ , then N L∗ = 0. WIB 5.3: If T > T ∗ given in (5.4), then replacement first is better than replacement with T , if T = T ∗ , then replacement with time T ∗ is better than replacement first and last, and if T < T ∗ , then replacement last is better than replacement with time T . ∑ n −θ t Example 5.1 When F(t) = 1 − e−λt , G(t) = ∞ for k = 2, 3, . . . , n=k [(θ t) /n!]e ∑ ∞ (N ) n −θ t i.e., G (t) = n=N k [(θ t) /n!]e (N = 0, 1, 2, . . . ), G (N ) (t) − G (N +1) (t) =

(N +1)k−1 ∑ n=N k

(θt)n −θ t e , n!

optimal T ∗ in (5.4) satisfies ʃ

T

e−λt [r (T ) − r (t)]dt =

0

cT , cM

and optimal N ∗ in (5.6) satisfies N∑ k−1 ʃ ∞ n=0

0

(θt)n −(θ +λ)t cN e [Q 3 (N ) − r (t)]dt ≥ , n! cM

where ∑(N +1)k−1 ʃ ∞ Q 3 (N ) =

n −(θ+λ)t r (t)dt n=N k 0 [(θ t) /n!]e . ∑(N +1)k−1 ʃ ∞ n −(θ+λ)t dt n=N k 0 [(θ t) /n!]e

Table 5.1 presents optimal T ∗ and N ∗ when cT = c N , F(t) = 1 − e−t/2 , G(t) = ∑∞ n −t and r (t) = 2α 2 t. This indicates that both T ∗ and N ∗ increase with n=2 (t /n!)e cT /c M , and decrease with α. This means that if the expected number R(t) = (αt)2 of minor failures increase and its cost c M is large, we should replace the system early. Optimal TF∗ in (5.7) satisfies N∑ k−1 ʃ T n=0

0

(θt)n −(θ +λ)t cT e [r (T ) − r (t)]dt = , n! cM

and optimal TL∗ in (5.11) satisfies

5.1 Replacement with Two Failure Modes

95

∑ n −t Table 5.1 Optimal T ∗ and N ∗ when cT = c N , F(t) = 1 − e−t/2 , k = 2, G(t) = ∞ n=2 (t /n!)e 2 and r (t) = 2α t α = 0.3 α = 0.5 α = 1.0 cT T∗ N∗ T∗ N∗ T∗ N∗ cM 1 2 3 4 5 6 7 8 9 10

4.57 7.51 10.32 13.11 15.89 18.67 21.44 24.22 27.00 29.78

3 5 8 10 12 14 16 18 20 22

2.40 3.68 4.82 5.90 6.94 7.96 8.98 9.99 10.99 12.00

2 3 3 4 5 6 7 7 8 9

1.09 1.60 2.02 2.40 2.74 3.07 3.38 3.68 3.98 4.26

1 1 2 2 2 2 2 3 3 3

∑ n −t Table 5.2 Optimal TF∗ and TL∗ when F(t) = 1 − e−t/2 , k = 2, G(t) = ∞ n=2 (t /n!)e and r (t) = 2 2α t α = 0.3 α = 0.5 α = 1.0 cT TF∗ TL∗ TF∗ TL∗ TF∗ TL∗ cM 1 2 3 4 5 6 7 8 9 10

4.58 7.52 10.36 13.19 16.02 18.84 21.67 24.50 27.32 30.15

ʃ 0

T

4.71 7.53 10.32 13.11 15.89 18.67 21.44 24.22 27.00 29.78

e−λt [r (T ) − r (t)]dt +

2.40 3.68 4.82 5.90 6.95 7.98 9.00 10.02 11.04 12.06

N∑ k−1 ʃ ∞ n=0

T

2.90 3.92 4.93 5.95 6.96 7.97 8.98 9.99 10.99 12.00

1.09 1.60 2.02 2.40 2.74 3.07 3.38 3.68 3.98 4.26

2.14 2.39 2.65 2.90 3.15 3.41 3.66 3.92 4.17 4.43

(θ t)n −(θ +λ)t cT e [r (T ) − r (t)]dt = . n! cM

Table 5.2 optimal TF∗ and TL∗ for cT /c M and α when F(t) = 1 − e−t/2 , ∑presents ∞ n G(t) = n=2 (t /n!)e−t , r (t) = 2α 2 t and N = 5. This indicates that both TF∗ and TL∗ increase with cT /c M , and decrease with α. Furthermore, replacement last is better than replacement first as α are small and cT /c M are large.

96

5 Extended Replacement Models

Optimal N F∗ in (5.9) satisfies N∑ k−1 ʃ T n=0

(θt)n −(θ +λ)t cN e [Q 3 (T , N ) − r (t)]dt ≥ , n! cM

0

where ∑(N +1)k−1 ʃ T Q 3 (T , N ) =

n −(θ+λ)t r (t)dt n=N k 0 [(θ t) /n!]e , ∑(N +1)k−1 ʃ T n −(θ +λ)t [(θ t) /n!]e dt n=N k 0

and N L∗ in (5.13) satisfies ʃ

T

˜3 (T , N ) − r (t)]dt + e−λt [ Q

0

N∑ k−1 ʃ ∞ n=0

≥

T

(θt)n −(θ +λ)t ˜ e [ Q 3 (T , N ) − r (t)]dt n!

cN , cM

where ˜3 (T , N ) = Q

∑(N +1)k−1 ʃ ∞

n −(θ+λ)t r (t)dt n=N k T [(θ t) /n!]e . ∑(N +1)k−1 ʃ ∞ n −(θ +λ)t dt n=N k T [(θ t) /n!]e

Table∑ 5.3 presents optimal N F∗ and N L∗ for c N /c M and α when F(t) = 1 − e−t/2 , n −t 2 G(t) = ∞ n=2 (t /n!)e , r (t) = 2α t and T = 5.0. This shows that optimal replacement first and last policies are given by the alternative of finite N F∗ or positive N L∗ .  Table 5.3 Optimal N F∗ and N L∗ when F(t) = 1 − e−t/2 , G(t) = α = 0.3

∑∞

n=2 (t

n /n!)e−t

α = 0.5

and r (t) = 2α 2 t

α = 1.0

cN cM

N F∗

N L∗

N F∗

N L∗

N F∗

N L∗

1 2 3 4 5 6 7 8 9 10

9 ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞

0 5 8 10 12 14 16 18 20 22

2 4 17 ∞ ∞ ∞ ∞ ∞ ∞ ∞

0 0 0 1 4 5 6 7 8 9

1 1 2 2 2 3 3 4 4 6

0 0 0 0 0 0 0 0 0 0

5.2 Undertime Replacement Model

97

5.2 Undertime Replacement Model We have introduced several replacement overtime policies in Chaps. 1–3 in which the unit is replaced at some works over time T . However, in such policies, the failure probability of units due to its performance delay might increase, because we have no opportunistic replacement before a planned time. From this viewpoint, we propose a new replacement policy in which the unit can be replaced preventively at the last working time before time T , which is called replacement undertime [2]. When the unit operates for successive jobs with a general distribution G(t) in Fig. 2.1, we consider the replacement undertime policies for age replacement in Chap. 2 and replacement with minimal repair in Chap. 3. It is assumed throughout this section that the unit has a failure distribution F(t) with mean μ and failure rate h(t) defined in Chap. 2.

5.2.1 Age Replacement Suppose that the unit is replaced at time T (0 < T ≤ ∞), at the completion of some working time in [TO , T ] (0 ≤ TO ≤ T ) or at failure, whichever occurs first. Then, the probability that the unit is replaced in [TO , T ] is ∞ ʃ ∑

TO

[ʃ

0

n=0

T

]

F(u)dG(u − t) dG (n) (t),

TO

the probability that it is replaced at time T is F(T )

∞ ʃ ∑ n=0

TO

G(T − t)dG (n) (t),

0

and the probability that it is replaced at failure is F(TO ) +

∞ ʃ ∑ n=0

TO 0

[ʃ

T TO

Thus, the mean time to replacement is

] G(u − t)dF(t) dG (n) (t).

98

5 Extended Replacement Models TO [ʃ T

∞ ʃ ∑ 0

n=0

ʃ

+

] ∞ ʃ ∑ u F(u)dG(u − t) dG (n) (t) + T F(T )

TO TO

t dF(t) +

0

ʃ =

TO

F(t)dt +

0

∞ ʃ ∑

n=0 0 ∞ ∑ ʃ TO n=0

TO

[ʃ

0

n=0

[ʃ

T

G(T − t)dG (n) (t)

0

] uG(u − t)dF(t) dG (n) (t)

TO T

TO

]

G(u − t) F(u)du dG (n) (t).

TO

Therefore, the expected cost rate is ] (n) ∑ ʃ TO [ʃ T cT + (c O T − cT ) ∞ n=0 0 TO F(u)dG(u − t) dG (t) { ] (n) } ∑ ʃ TO [ʃ T + (c F − cT ) F(TO ) + ∞ n=0 0 TO G(u − t)dF(t) dG (t) ] C(TO , T ) = , ʃ TO ∑∞ ʃ TO [ʃ T (n) (t) F(t)dt + G(u − t) F(u)du dG n=0 0 0 TO (5.14) where c O T = replacement cost in [TO , T ], cT = replacement cost at time T and c F = replacement cost at failure with c F > c O T and c F > cT . Note that C(TO , T ) includes the expected costs of the following three policies: When TO = 0, C(0, T ) =

cT + (c O T − cT )

ʃT 0

F(t)dG(t) + (c F − cT ) ʃT 0 G(t) F(t)dt

ʃT 0

G(t)dF(t)

, (5.15)

which agrees with C F (T , 1) in (2.5) when c O T = c N . When TO = T , C(T , T ) =

cT + (c F − cT )F(T ) , ʃT 0 F(t)dt

(5.16)

which agrees with C(T ) in (2.2). When T = ∞, ] (n) ∑ ʃ TO [ʃ ∞ c F − (c F − c O T ) ∞ n=0 0 TO F(u)dG(u − t) dG (t) C(TO , ∞) = ʃ TO , (5.17) ] ∑∞ ʃ TO [ʃ ∞ (n) (t) F(t)dt + G(u − t) F(u)du dG n=0 0 0 TO which agrees with C O (T ) in (2.24) when TO = T . In particular, when G(t) = 1 − e−θ t , the expected cost rate in (5.14) is C(TO , T ) = −(cT − c O T )θ ʃT ʃT [ ] cT + (cT − c O T )θ 0 O F(t)dt + (c F − cT ) F(TO ) + TO e−θ(t−TO ) dF(t) . + ʃT ʃ TO −θ(t−TO ) F(t)dt 0 F(t)dt + TO e (5.18)

5.2 Undertime Replacement Model

99

We find optimal TO∗ and T ∗ to minimize C(TO , T ). Differentiating C(TO , T ) with respect to T and setting it equal to zero, [ʃ (c F − cT )h(T )

TO

ʃ

T

F(t)dt +

0

[ ʃ − (c F − cT ) F(TO ) +

−θ(t−TO )

e

] F(t)dt

TO

T

] ʃ e−θ(t−TO ) dF(t) − (cT − c O T )θ

TO

F(t)dt = cT ,

0

TO

(5.19) or ʃ

TO

ʃ

T

F(t)[h(T ) − h(t)]dt +

0

cT + (cT − c O T )θ = c F − cT

ʃ TO 0

e−θ(t−TO ) F(t)[h(T ) − h(t)]dt

TO

F(t)dt

.

(5.20)

Differentiating C(TO , T ) with respect to TO and setting it equal to zero, (c F − c O T )

ʃT TO

e−θ t dF(t) + (cT − c O T )e−θ T F(T ) ʃ TO F(t)dt ʃT −θ t F(t)dt 0 TO e

− (c F − c O T )F(TO ) = c O T .

(5.21)

Substituting (5.20) for (5.21), (5.21) is ʃ

[

T −TO

e

−θ t

h(T ) − h(t + TO )

0

F(t + TO ) F(TO )

] dt =

cT − c O T . c F − cT

(5.22)

When cT = c O T , TO = T , i.e., the unit is replaced only at time T ∗ which satisfies, from (5.19), ʃ

T 0

F(t)[h(T ) − h(t)]dt =

cT , c F − cT

(5.23)

which agrees with (2.3). This shows that replacement with time T ∗ is better than replacement undertime. WIB 5.4: Replacement with time T ∗ is better than replacement undertime.

100

5 Extended Replacement Models

In addition, when cT = c O T , (5.21) is ʃ

TO

Q 1 (TO , T )

F(t)dt − F(TO ) =

0

cT , c F − cT

(5.24)

where ʃT T

e−θ t dF(t)

TO

e−θt F(t)dt

Q 1 (TO , T ) ≡ ʃ TO

increases strictly with TO from Q 1 (0, T ) = Q 1 (T ) in (2.8) to h(T ) and increases ˜1 (TO ) in (2.16), and h(TO ) ≤ strictly with T from h(TO ) to Q 1 (TO , ∞) = Q Q 1 (TO , T ) ≤ h(T ) from (3) of Appendix A.6. Thus, the left-hand side of (5.24) increases strictly with TO from 0 to that of (5.23). Therefore, if T > T ∗ , then there exists a finite and unique TO∗ (0 < TO∗ < T ) which satisfies (5.24), and the resulting cost rate is C(TO∗ , T ) = (c F − cT )Q 1 (TO∗ , T ).

(5.25)

Furthermore, because the left-hand side of (5.24) increases with T to that of (2.27) when TO = T , TO∗ decreases with T to a solution of (2.27). Conversely, if T ≤ T ∗ , then TO∗ = T .

5.2.2 Replacement with Minimal Repair Suppose that the unit undergoes minimal repair at failures in Chap. 3. Then, the unit is replaced at time T (0 < T ≤ ∞) or at the completion of some working time in [TO , T ] (0 ≤ TO ≤ T ), whichever occurs first. The probability that the unit is replaced in [TO , T ] is ∞ ʃ ∑ n=0

TO

[G(T − t) − G(TO − t)]dG (n) (t),

0

and the probability that it is replaced at time T is ∞ ʃ ∑ n=0

TO

G(T − t)dG (n) (t).

0

Thus, the mean time to replacement is

5.2 Undertime Replacement Model ∞ ʃ ∑ n=0

TO

[ʃ

0

]

T

u dG(u − t) dG (n) (t) + T

TO

= TO +

∞ ʃ ∑ n=0

101

TO

[ʃ

T

∞ ʃ ∑

G(T − t)dG (n) (t)

0

n=0

]

TO

G(u − t)du dG (n) (t),

TO

0

and the expected number of failures until replacement is TO[ʃ T

∞ ʃ ∑ n=0

0

TO

= H (TO ) +

] H (u)dG(u − t) dG

∞ ʃ ∑ n=0

TO 0

[ʃ

T

(n)

∞ ʃ ∑ (t) + H (T )

]

n=0

TO

G(T − t)dG (n) (t)

0

G(u − t)h(u)du dG (n) (t).

TO

Therefore, the expected cost rate is ∑ ʃ TO (n) c O T + (cT − c O T ) ∞ n=0 0 G(T − t)dG (t) ] { } ∑∞ ʃ TO [ʃ T (n) + c M H (TO ) + n=0 0 TO G(u − t)h(u)du dG (t) ] , C(TO , T ) = ∑ ʃ TO [ʃ T (n) TO + ∞ n=0 0 TO G(u − t)du dG (t) (5.26) where c M = minimal repair cost at each failure. Note that C(TO , T ) includes the expected costs of the following three policies: When TO = 0, c O T + (cT − c O T )G(T ) + c M C(0, T ) = ʃT 0 G(t)dt

ʃT 0

G(t)h(t)dt

,

(5.27)

which agrees with C F (T , 1) in (3.5) when c O T = c N . When TO = T , C(T , T ) =

cT + c M H (T ) , T

(5.28)

which agrees with C(T ) in (3.2). When T = ∞, ] { } ∑ ʃ TO [ʃ ∞ c O T + c M H (TO ) + ∞ G(u − t)h(u)du dG (n) (t) n=0 0 TO ] C(TO , ∞) = ∑ ʃ TO [ʃ ∞ (n) TO + ∞ n=0 0 TO G(u − t)du dG (t) ] (n) } ∑ ʃ TO [ʃ ∞ cO T + cM ∞ n=0 0 0 G(u)h(t + u)du dG (t) ∑ , (5.29) = (n) (1/θ ) ∞ n=0 G (TO ) which agrees with C O F (T , ∞) in (3.34) when TO = T .

102

5 Extended Replacement Models

In particular, when G(t) = 1 − e−θ t , the expected cost rate in (5.26) is C(TO , T ) =

ʃT [ ] c O T + (cT − c O T )e−θ(T −TO ) + c M H (TO ) + TOe−θ(t−TO ) h(t)dt TO + [1 − e−θ(T −TO ) ]/θ

.

(5.30) We find optimal TO∗ and T ∗ to minimize C(TO , T ) in (5.30). Differentiating C(TO , T ) with respect to T and setting it equal to zero, ʃ

TO

ʃ [h(T ) − h(t)]dt +

T

e−θ t [h(T ) − h(t)]dt =

TO

0

cT + (cT − c O T )θ TO . (5.31) cM

Differentiating C(TO , T ) with respect to TO and setting it equal to zero, (cT − c O T )e−θ T + c M

ʃT TO

e−θ t h(t)dt

(e−θ TO − e−θ T )/(θ TO )

− c M H (TO ) = c O T .

(5.32)

cT − c O T . cM

(5.33)

Substituting (5.31) and (5.32), (5.32) is ʃ

T

e−θ t [h(T ) − h(t)]dt =

TO

When cT = c O T , TO = T , i.e., the unit is replaced only at time T ∗ which satisfies, from (5.28), ʃ

T

[h(T ) − h(t)]dt =

0

cT , cM

(5.34)

which agrees with (3.3). This shows that replacement with T ∗ is better than replacement undertime. WIB 5.5: Replacement with time T ∗ is better than replacement undertime.

In addition, when cT = c O T , (5.32) is TO Q 2 (TO , T ) − H (TO ) =

cT , cM

where ʃT Q 2 (TO , T ) ≡

TO

e−θ t h(t)dt

ʃT

TO

e−θ t dt

(5.35)

5.3 Middle Replacement Models

103

increases strictly with TO from Q 2 (0, T ) = R1ʃ(T , 0) in (3.8) to h(T ) and increases ∞ strictly with T from h(TO ) to Q 2 (TO , ∞) = 0 θe−θ t h(TO + t)dt, and h(TO ) ≤ Q 2 (TO , T ) ≤ h(T ) from (4) of Appendix A.6. Thus, the left-hand side of (5.35) increases strictly with TO from 0 to that of (3.3) when TO = T . Therefore, if T > T ∗ given in (3.3), then there exists a finite and unique TO∗ (0 ≤ TO∗ < T ) which satisfies (5.35), and the resulting cost rate is C(TO∗ , T ) = c M Q 2 (TO∗ , T ).

(5.36)

Furthermore, because the left-hand side of (5.35) increases with T to that of (3.36), TO∗ decreases with T to a solution of (3.36). Conversely, if T ≤ T ∗ , then TO∗ = T .

5.3 Middle Replacement Models We consider three preventive replacement models and propose a new policy in which the unit is replaced at middle times. This is called replacement middle. We compare replacement middle with replacement first and last, and try to determine which policy is better [3]. Suppose that the unit is replaced at times T , Y1 or Y2 , where G 1 (t) ≡ Pr{Y1 ≤ t} and G 2 (t) ≡ Pr{Y2 ≤ t}. Then, we propose three policies of replacement first, last and middle for age replacement in Chap. 2 and replacement with minimal repair in Chap. 3. It is assumed throughout this section that the unit has a failure distribution F(t) with finite mean μ and failure rate h(t).

5.3.1 Age Replacement We consider three policies of replacement first, last and middle, obtain their expected cost rates and compare them analytically and numerically. (1) Replacement First Suppose that the unit is replaced preventively at time T (0 < T ≤ ∞), at time Y1 or at time Y2 , whichever occurs first. Then, the probability that the unit is replaced at time T is G 1 (T )G 2 (T ) F(T ), the probability that it is replaced at time Y1 is ʃ 0

T

F(t)G 2 (t)dG 1 (t),

104

5 Extended Replacement Models

the probability that it is replaced at time Y2 is ʃ

T

F(t)G 1 (t)dG 2 (t),

0

and the probability that it is replaced at failure is ʃ

T

G 1 (t)G 2 (t)dF(t).

0

Thus, the mean time to replacement is ʃ T ʃ T T G 1 (T )G 2 (T ) + t F(t)G 2 (t)dG 1 (t) + t F(t)G 1 (t)dG 2 (t) 0 0 ʃ T ʃ T t G 1 (t)G 2 (t)dF(t) = G 1 (t)G 2 (t) F(t)dt. + 0

0

Therefore, the expected cost rate is ʃT cT + (c F − cT ) 0 G 1 (t)G 2 (t)dF(t) C F (T ) = , ʃT 0 G 1 (t)G 2 (t) F(t)dt

(5.37)

where cT = replacement cost at time T , Y1 and Y2 , and c F = replacement cost at failure with c F > cT . We find optimal TF∗ to minimize C F (T ) when h(t) increases strictly with t from 0 to ∞. Differentiating C F (T ) with T and setting it equal to zero, ʃ

T

G 1 (t)G 2 (t) F(t)[h(T ) − h(t)]dt =

0

cT , c F − cT

(5.38)

whose left-hand side increases strictly with T from 0 to ∞. Thus, there exists a finite and unique TF∗ (0 < TF∗ < ∞) which satisfies (5.38), and the resulting cost rate is C F (TF∗ ) = (c F − cT )h(TF∗ ).

(5.39)

(2) Replacement Last Suppose that the unit is replaced preventively at time T (0 ≤ T < ∞), at time Y1 or at time Y2 , whichever occurs last. Then, the probability that the unit is replaced at time T is G 1 (T )G 2 (T ) F(T ),

5.3 Middle Replacement Models

105

the probability that it is replaced at time Y1 is ʃ

∞

F(t)G 2 (t)dG 1 (t),

T

the probability that it is replaced at time Y2 is ʃ

∞

F(t)G 1 (t)dG 2 (t),

T

and the probability that it is replaced at failure is ʃ

∞

F(T ) +

[1 − G 1 (t)G 2 (t)]dF(t).

T

Thus, the mean time to replacement is ʃ

∞

ʃ

T G 1 (T )G 2 (T ) F(T ) + t F(t)G 2 (t)dG 1 (t) + T ʃ ∞ ʃ T t dF(t) + t [1 − G 1 (t)G 2 (t)]dF(t) + 0 T ʃ ∞ ʃ T F(t)dt + [1 − G 1 (t)G 2 (t)] F(t)dt. = 0

∞

t F(t)G 1 (t)dG 2 (t)

T

T

Therefore, the expected cost rate is ʃ∞ { } cT + (c F − cT ) F(T ) + T [1 − G 1 (t)G 2 (t)]dF(t) C L (T ) = . ʃ∞ ʃT 0 F(t)dt + T [1 − G 1 (t)G 2 (t)] F(t)dt

(5.40)

We find optimal TL∗ to minimize C L (T ). Differentiating C L (T ) with respect to T and setting it equal to zero, ʃ

T 0

ʃ

∞

F(t)[h(T )−h(t)]dt +

[1−G 1 (t)G 2 (t)] F(t)[h(T )−h(t)]dt =

T

cT , c F − cT (5.41)

ʃ∞ whose left-hand side increases strictly with T from − 0 [1 − G 1 (t)G 2 (t)]dF(t) < 0 to ∞. Thus, there exists a finite and unique TL∗ (0 < TL∗ < ∞) which satisfies (5.41), and the resulting cost rate is C L (TL∗ ) = (c F − cT )h(TL∗ ).

(5.42)

106

5 Extended Replacement Models

(3) Replacement Middle Suppose that the unit is replaced preventively at middle times, i.e., at time T for Y1 < T < Y2 or Y2 < T < Y1 , at time Y1 for T < Y1 < Y2 or Y2 < Y1 < T , or at time Y2 for T < Y2 < Y1 or Y1 < Y2 < T , whichever occurs first. Then, the probability that the unit is replaced at time T is [G 1 (T )G 2 (T ) + G 1 (T )G 2 (T )] F(T ), the probability that it is replaced at time Y1 is ʃ

T

ʃ

∞

F(t)G 2 (t)dG 1 (t) +

F(t)G 2 (t)dG 1 (t),

T

0

the probability that it is replaced at time Y2 is ʃ

T

ʃ

∞

F(t)G 1 (t)dG 2 (t) +

0

F(t)G 1 (t)dG 2 (t),

T

and the probability that it is replaced at failure is ʃ

T

ʃ

∞

[1 − G 1 (t)G 2 (t)]dF(t) +

0

G 1 (t)G 2 (t)dF(t).

T

Thus, the mean time to replacement is [ ] T G 1 (T )G 2 (T ) + G 1 (T )G 2 (T ) F(T ) ʃ ∞ ʃ T t F(t)G 2 (t)dG 1 (t) t F(t)G 2 (t)dG 1 (t) + + T 0 ʃ ∞ ʃ T t F(t)G 1 (t)dG 2 (t) + t F(t)G 1 (t)dG 2 (t) + T 0 ʃ ∞ ʃ T t [1 − G 1 (t)G 2 (t)]dF(t) + t G 1 (t)G 2 (t)dF(t) + 0 T ʃ ∞ ʃ T G 1 (t)G 2 (t) F(t)dt. [1 − G 1 (t)G 2 (t)] F(t)dt + = 0

T

Therefore, the expected cost rate is ʃ∞ } {ʃ T cT + (c F −cT ) 0 [1−G 1 (t)G 2 (t)]dF(t) + T G 1 (t)G 2 (t)dF(t) C M (T ) = . ʃ∞ ʃT 0 [1 − G 1 (t)G 2 (t)]F(t)dt + T G 1 (t)G 2 (t) F(t)dt (5.43)

5.3 Middle Replacement Models

107

When T = 0, C M (0) = C F (∞), and when T = ∞, C M (∞) = C L (0). We find optimal TM∗ to minimize C M (T ). Differentiating C M (T ) with respect to T and setting it equal to zero, ʃ

ʃ

T

∞

[1 − G 1 (t)G 2 (t)] F(t)[h(T ) − h(t)]dt +

0

G 1 (t)G 2 (t) F(t)[h(T ) − h(t)]dt

T

=

cT , c F − cT

(5.44)

ʃ∞ whose left-hand side increases strictly with T from − 0 G 1 (t)G 2 (t)dF(t) to ∞. Thus, there exists a finite and unique TM∗ (0 < TM∗ < ∞) which satisfies (5.44), and the resulting cost rate is C M (TM∗ ) = (c F − cT )h(TM∗ ).

(5.45)

5.3.2 Comparison of Three Policies We compare the above three policies to find which policy is better. (1) Replacement First and Middle Letting L F (T ) and L M (T ) be the left-hand sides of (5.38) and (5.44), respectively, L M F (T ) ≡ L M (T ) − L F (T ) ʃ T ] [ G 1 (t)G 2 (t) + G 1 (t)G 2 (t) F(t)[h(T ) − h(t)]dt = 0 ʃ ∞ G 1 (t)G 2 (t) F(t)[h(t) − h(T )]dt, − T

which increases strictly with T from L M F (0) < 0 to ∞. Thus, there exists a finite and unique TM F (0 < TM F < ∞) which satisfies L M F (T ) = 0. Therefore, we have the following comparative results from (5.39) and (5.45): (i) If L M (TM F ) ≥ cT /(c F − cT ), then TF∗ ≤ TM∗ , i.e., replacement first is better than replacement middle. (ii) If L M (TM F ) < cT /(c F − cT ), then TF∗ > TM∗ , i.e., replacement middle is better than replacement first.

108

5 Extended Replacement Models

(2) Replacement Last and Middle Letting L L (T ) and L M (T ) be the left-hand side of (5.41) and (5.44), respectively, L M L (T ) ≡ L M (T ) − L L (T ) ʃ ∞ = [G 1 (t)G 2 (t) + G 1 (t)G 2 (t)] F(t)[h(t) − h(T )]dt T ʃ T G 1 (t)G 2 (t) F(t)[h(T ) − h(t)]dt, − 0

which decreases strictly with T from L M L (0) > 0 to −∞. Thus, there exists a finite and unique TM L (0 < TM L < ∞) which satisfies L M L (T ) = 0. Therefore, we have the following comparative results from (5.42) and (5.45): (iii) If L M (TM L ) ≤ cT /(c F − cT ), then TL∗ ≤ TM∗ , i.e., replacement last is better than replacement middle. (iv) If L M (TM L ) > cT /(c F − cT ), then TL∗ > TM∗ , i.e., replacement middle is better than replacement last. (3) Replacement First and Last From (5.38) and (5.41), L L F (T ) ≡ L L (T ) − L F (T ) ʃ T [1 − G 1 (t)G 2 (t)] F(t)[h(T ) − h(t)]dt = 0 ʃ ∞ [1 − G 1 (t)G 2 (t)] F(t)[h(t) − h(T )]dt, − T

which increases strictly with T from L L F (0) < 0 to ∞. Thus, there exists a finite and unique TL F (0 < TL F < ∞) which satisfies L L F (T ) = 0. Therefore, we have the following comparative results: (v) If L F (TL F ) ≥ cT /(c F − cT ), then TF∗ ≤ TL∗ , i.e., replacement first is better than replacement last. (vi) If L F (TL F ) < cT /(c F − cT ), then TF∗ > TL∗ , i.e., replacement last is better than replacement first. From the above results (i) ∼ (vi), we summarize the comparisons of replacement first, last and middle when C ≡ cT /(c F − cT ): Note that replacement middle is never last among three policies.

5.3 Middle Replacement Models

109

WIB 5.6: (i) If L M (TM F ) ≥ C and L M (TM L ) ≥ C, then the ranking of three policies is replacement first, middle and last. (ii) If L M (TM F ) < C < L M (TM L ) and L F (TL F ) ≥ C, the ranking is replacement middle, first and last. (iii) If L M (TM L ) < C < L M (TM F ) and L F (TL F ) < C, the ranking is replacement middle, last and first. (iv) If L M (TM F ) < C and L M (TM L ) ≥ C, then the ranking is replacement last, middle and first.

Example 5.2 When F(t) = 1 − e−(t/10) , G 1 (t) = 1 − e−0.15t and G 2 (t) = 1 −e−0.2t , Table 5.4 presents optimal TF∗ , TL∗ and TM∗ for C ≡ cT /(c F − cT ). This indicates that if C is small, replacement first is best, if C is middle, replacement middle is best, and if C is large, replacement last is best. Furthermore, it is of great interest that replacement middle is not the worst among the three policies.  2

Table 5.4 Optimal TF∗ , TL∗ and TM∗ when F(t) = 1 − e−(t/10) , G 1 (t) = 1 − e−0.15t and G 2 (t) = 1 − e−0.2t cT C≡ TF∗ TL∗ TM∗ c F − cT 2

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

1.06 1.54 1.93 2.27 2.58 2.87 3.15 3.41 3.66 3.91 6.14 8.20 10.21 12.19 14.18 16.16 18.14 20.12 22.10

7.20 7.26 7.31 7.37 7.43 7.48 7.53 7.58 7.61 7.69 8.22 8.73 9.24 9.73 10.22 10.71 11.20 11.68 12.16

2.25 2.41 2.57 2.72 2.87 3.02 3.16 3.31 3.45 3.58 4.85 5.98 7.02 8.01 8.96 9.89 10.80 11.69 12.58

110

5 Extended Replacement Models

5.3.3 Replacement with Minimal Repair When the unit undergoes minimal repair at each failure in Chap. 3, we consider three policies of replacement first, last and middle, obtain the expected cost rates and compare them analytically and numerically. (1) Replacement First Suppose that the unit is replaced at time T (0 < T ≤ ∞), at time Y1 or at time Y2 , whichever occurs first. Then, the probability that the unit is replaced at time T is G 1 (T )G 2 (T ), the probability that it is replaced at time Y1 is ʃ

T

G 2 (t)dG 1 (t),

0

the probability that it is replaced at time Y2 is ʃ

T

G 1 (t)dG 2 (t).

0

Thus, the mean time to replacement is ʃ

T

T G 1 (T )G 2 (T ) +

ʃ

T

t G 2 (t)dG 1 (t) +

0

ʃ

T

t G 1 (t)dG 2 (t) =

0

G 1 (t)G 2 (t)dt,

0

and the expected number of failures until replacement is ʃ T ʃ T H (T )G 1 (T )G 2 (T ) + H (t)G 2 (t)dG 1 (t) + H (t)G 1 (t)dG 2 (t) 0 0 ʃ T G 1 (t)G 1 (t)h(t)dt. = 0

Therefore, the expected cost rate is C F (T ) =

ʃT cT + c M 0 G 1 (t)G 2 (t)h(t)dt , ʃT G (t)G (t)dt 1 2 0

(5.46)

where c M = minimal repair cost at each failure and cT is given in (5.37). We find optimal TF∗ to minimize C F (T ) when h(t) increases strictly with t from 0 to ∞. Differentiating C F (T ) with respect to T and setting it equal to zero,

5.3 Middle Replacement Models

ʃ

T

111

G 1 (t)G 2 (t)[h(T ) − h(t)]dt =

0

cT , cM

(5.47)

whose left-hand side increases strictly with T from 0 to ∞. Thus, there exists a finite and unique TF∗ (0 < TF∗ < ∞) which satisfies (5.47), and the resulting cost rate is C F (TF∗ ) = c M h(TF∗ ).

(5.48)

(2) Replacement Last Suppose that the unit is replaced at time T (0 ≤ T < ∞), at time Y1 or at time Y2 , whichever occurs last. Then, the probability that the unit is replaced at time T is G 1 (T )G 2 (T ), the probability that it is replaced at time Y1 is ʃ

∞

G 2 (t)dG 1 (t),

T

and the probability that it is replaced at time Y2 is ʃ

∞

G 1 (t)dG 2 (t).

T

Thus, the mean time to replacement is ʃ

∞

T G 1 (T )G 2 (T ) + t G 2 (t)dG 1 (t) + T ʃ ∞ [1 − G 1 (t)G 2 (t)]dt, =T+

ʃ

∞

t G 1 (t)dG 2 (t)

T

T

and the expected number of failures until replacement is ʃ ∞ ʃ ∞ H (T )G 1 (T )G 2 (T ) + H (t)G 2 (t)dG 1 (t) + H (t)G 1 (t)dG 2 (t) T T ʃ ∞ [1 − G 1 (t)G 2 (t)]h(t)dt. = H (T ) + T

Therefore, the expected cost rate is ʃ∞ { } cT + c M H (T ) + T [1 − G 1 (t)G 2 (t)]h(t)dt ʃ∞ C L (T ) = . T + T [1 − G 1 (t)G 2 (t)]dt

(5.49)

112

5 Extended Replacement Models

We find optimal TL∗ to minimize C L (T ) when h(t) increases strictly with t from 0 to ∞. Differentiating C L (T ) with respect to T and setting it equal to zero, ʃ

T

ʃ

∞

[h(T ) − h(t)]dt +

0

[1 − G 1 (t)G 2 (t)][h(T ) − h(t)]dt =

T

cT , cM

(5.50)

ʃ∞ whose left-hand side increases strictly with T from − 0 [1 − G 1 (t)G 2 (t)]h(t)dt to ∞. Thus, there exists a finite and unique TL∗ (0 < TL∗ < ∞) which satisfies (5.50), and the resulting cost is C L (TL∗ ) = c M h(TL∗ ).

(5.51)

(3) Replacement Middle Suppose that the unit is replaced at time T for Y1 < T < Y2 or Y2 < T < Y1 , at time Y1 for T < Y1 < Y2 or Y2 < Y1 < T , or time Y2 for T < Y2 < Y1 or Y1 < Y2 < T , whichever occurs first. Then, the probability that the unit is replaced at time T is G 1 (T )G 2 (T ) + G 1 (T )G 2 (T ), the probability that it is replaced at time Y1 is ʃ

T

ʃ

∞

G 2 (t)dG 1 (t) +

0

G 2 (t)dG 1 (t),

T

and the probability that it is replaced at time Y2 is ʃ

T

ʃ

∞

G 1 (t)dG 2 (t) +

0

G 1 (t)dG 2 (t).

T

Thus, the mean time to replacement is [

]

ʃ

T

T G 1 (T )G 2 (T ) + G 1 (T )G 2 (T ) + t G 2 (t)dG 1 (t) + 0 ʃ ∞ ʃ T t G 1 (t)dG 2 (t) + t G 1 (t)dG 2 (t) + 0 T ʃ ∞ ʃ T [1 − G 1 (t)G 2 (t)]dt + G 1 (t)G 2 (t)dt, = 0

T

and the expected number of failures until replacement is

ʃ

∞ T

t G 2 (t)dG 1 (t)

5.3 Middle Replacement Models

113

ʃ

]

[

T

H (t)G 2 (t)dG 1 (t) H (T ) G 1 (T )G 2 (T ) + G 1 (T )G 2 (T ) + 0 ʃ T ʃ ∞ ʃ ∞ H (t)G 1 (t)dG 2 (t) + H (t)G 1 (t)dG 2 (t) H (t)G 2 (t)dG 1 (t) + + T 0 T ʃ ∞ ʃ T G 1 (t)G 2 (t)h(t)dt. [1 − G 1 (t)G 2 (t)]h(t)dt + = 0

T

Therefore, the expected cost rate is C M (T ) =

cT + c M

ʃ∞ } 0 [1 − G 1 (t)G 2 (t)]h(t)dt + T G 1 (t)G 2 (t)h(t)dt . ʃ∞ ʃT 0 [1 − G 1 (t)G 2 (t)]dt + T G 1 (t)G 2 (t)dt

{ʃ T

(5.52) When T = 0, C M (0) = C F (∞), and when T = ∞, C M (∞) = C L (0). We find optimal TM∗ to minimize C M (T ). Differentiating C M (T ) with respect to T and setting it equal to zero, ʃ

T

ʃ

∞

[1 − G 1 (t)G 2 (t)][h(T ) − h(t)]dt +

0

G 1 (t)G 2 (t)[h(T ) − h(t)]dt =

T

cT , cM (5.53)

ʃ∞ whose left-hand side increases strictly with T from − 0 G 1 (t)G 2 (t)h(t)dt to ∞. Thus, there exists a finite and unique TM∗ (0 < TM∗ < ∞) which satisfies (5.53), and the resulting cost rate is C M (TM∗ ) = c M h(TM∗ ).

(5.54)

5.3.4 Comparisons of Three Policies We compare the above three policies to find which policy is better. (1) Replacement First and Middle Letting L F (T ) and L M (T ) be the left-hand side of (5.47) and (5.53), respectively, L M F (T ) ≡ L M (T ) − L F (T ) ʃ T [ G 1 (t)G 2 (t) + G 1 (t)G 2 (t)[h(T ) − h(t)]dt = 0 ʃ ∞ G 1 (t)G 2 (t)[h(t) − h(T )]dt, − T

114

5 Extended Replacement Models

which increases strictly with T from L M F (0) < 0 to ∞. Thus, there exists a finite and unique TM F (0 < TM F < ∞) which satisfies L M F (T ) = 0. Therefore, we have the following comparative results from (5.48) and (5.54): (i) If L M (TM F ) ≥ cT /c M , then TF∗ ≤ TM∗ , i.e., replacement first is better than replacement middle. (ii) If L M (TM F ) < cT /c M , then TF∗ > TM∗ , i.e., replacement middle is better than replacement first. (2) Replacement Last and Middle Letting L L (T ) and L M (T ) be the left-hand side of (5.50) and (5.53), respectively, L M L (T ) ≡ L M (T ) − L L (T ) ʃ ∞ [ ] = G 1 (t)G 2 (t) + G 1 (t)G 2 (t) [h(t) − h(T )]dt T ʃ T G 1 (t)G 2 (t)[h(T ) − h(t)]dt, − 0

which decreases strictly with T from L M L (0) > 0 to −∞. Thus, there exists a finite and unique TM L (0 < TM L < ∞) which satisfies L M L (T ) = 0. Therefore, we have the following comparative results from (5.51) and (5.54): (iii) If L M (TM L ) ≤ cT /c M , then TL∗ ≤ TM∗ , i.e., replacement last is better than replacement middle. (iv) If L M (TM L ) > cT /c M , then TL∗ > TM∗ , i.e., replacement middle is better than replacement last. (3) Replacement First and Last From (5.47) and (5.50), L L F (T ) ≡ L L (T ) − L F (T ) ʃ ʃ T [1 − G 1 (t)G 2 (t)][h(T ) − h(t)]dt − = 0

∞

[1 − G 1 (t)G 2 (t)][h(t) − h(T )]dt,

T

which increases strictly with T from L L F (0) < 0 to ∞. Thus, there exists a finite and unique TL F (0 < TL F < ∞) which satisfies L L F (T ) = 0. Therefore, we have the following comparative results: (v) If L F (TL F ) ≥ cT /c M , then TF∗ ≤ TL∗ , i.e., replacement first is better than replacement last. (vi) If L F (TL F ) < cT /c M , then TF∗ > TL∗ , i.e., replacement last is better than replacement first. From the above results (i) ∼ (vi), we summarize the comparisons of replacement first, last and middle when C ≡ cT /c M : Note that the ranking moves to First, Middle and Last, and Middle is never last among three policies.

5.4 Problems

115

Table 5.5 Optimal TF∗ , TL∗ , TM∗ when F(t) = 1 − e−(t/10) , G 1 (t) = 1 − e−0.15t and G 2 (t) = 1 − e−0.2t cT TF∗ TL∗ TM∗ C= cM 2

0.01 0.02 0.05 0.10 0.20 0.50 1.00 2.00 5.00

1.06 1.54 2.57 3.87 6.01 11.56 20.36 37.87 90.40

6.71 6.76 6.90 7.15 7.62 8.98 11.05 14.58 22.44

2.74 2.87 3.26 3.87 4.98 7.73 11.48 17.89 35.31

WIB 5.7: (i) If L M (TM F ) ≥ C and L M (TM L ) ≥ C, then the ranking of three policies is replacement first, middle and last. (ii) If L M (TM F ) < C < L M (TM L ) and L F (TL F ) ≥ C, then the ranking is replacement middle, first and last. (iii) If L M (TM L ) < C < L M (TM F ) and L F (TL F ) < C, then the ranking is replacement middle, last and first. (iv) If L M (TM F ) < C and L M (TM L ) < C, then ranking is replacement last, middle and first.

Example 5.3 When F(t)=1−e−(t/10) , G(t) = 1 − e−0.15t and G 2 (t) = 1 − e−0.2t , Table 5.5 presents optimal TF∗ , TL∗ and TM∗ for C ≡ cT /c M . This shows similar tendencies to Table 5.4.  2

5.4 Problems 1. Suppose in Sect. 5.1 that the system is replaced preventively at time T (0 < T ≤ ∞) or at a planned number K (K = 1, 2, . . . ) of minor failures. Consider replacement first and last of this model [1]. 2. Suppose in Sect. 3.4 that the unit is replaced at time T or at failure during the interval [TO , T ] (0 ≤ TO ≤ T ), whichever occurs first, and obtain its expected cost rate.

116

5 Extended Replacement Models

3. Consider two modified replacement policies in Sect. 5.3.1 that the unit is replaced preventively at time T (0 < T ≤ ∞) or at max{Y1 , Y2 }, whichever occurs first, and at time T (0 ≤ T < ∞) or at min{Y1 , Y2 }, whichever occurs last [4]. 4. Consider replacement policies in Chap. 3 that the unit is replaced at time T , at work number N or at failure number K [5, p. 86].

References 1. Mizutani S, Zhao X, Nakagawa T (2021) Age and periodic replacement policies with two failure modes in general replacement models. Reliab Eng Syst Saf 214:107754 2. Zhao X, Al-Khalifa KN, Yun WY, Nakagawa T (2015) Optimal undertime replacement polices with random working cycles. MMR 2015, Tokyo 3. Zhao X, Nakagawa T, Al-Khalifa KN, Hamouda AMS (2015) What is middle maintenance policy? Qual Reliab Eng 32:7 4. Zhao X, Qian C, Nakagawa T (2018) Age replacement models with constant and random replacement times. Inter J Reliab Qual Saf Eng 25(6):1850029 5. Nakagawa T, Zhao X (2015) Maintenance overtime policies in reliability theory. Springer, London

Chapter 6

Which is Better for Standby or Parallel Systems

High system reliability can be achieved by redundancy and maintenance. It was shown by graph that the system can operate for a specified mean time by either the preventive replacement or increasing the number of redundant units [1, p. 64]. A variety of redundant systems and their optimization problems were discussed [2, p. 7]. Furthermore, several comparisons of reliability, replacement and scheduling for standby and parallel systems were made systematically [3–5]. Chapter 6 is written mainly based on the results [5]: Section 6.1 shows the reliability measures such as reliabilities, MTTFs, and failure rates of standby and parallel systems, compares them, and rewrites all results when the number of units is a random variable. Section 6.2 deals with age replacement for two systems and compares replacement times. Furthermore, we obtain the expected cost rates of two systems and make comparisons of their optimal policies. Section 6.3 introduces shortage and excess costs into scheduling models, and proposes comparative problems in which two systems operate for random works during a scheduling time. Finally, Sect. 6.4 takes up three modified models: (1) We apply two systems to partition problems for a data transmission model and compare them. (2) Using the deviation time introduced in Sect. 2.6, we obtain the expected cost rates of two systems and compare them. (3) Using a random K -out-of-n system, we derive a failure distribution of general redundant systems and show that all results in Sects. 6.1 and 6.2 are rewritten for any redundant systems. It is supposed throughout this chapter that both standby and parallel systems con. . . ) identical units. Each unit has a failure distribution F(t) with sist of n (n = 1, 2, ʃ∞ finite mean μ ≡ 0 F(t)dt, a density function f (t) ≡ dF(t)/dt, and failure rate h(t) ≡ f (t)/ F(t), where Φ(t) ≡ 1 − Φ(t) for any function Φ(t). In addition, h(t) increases from h(0) to h(∞) ≡ limt→∞ h(t), which might be infinity. Letting Φ ( j) (t) ( j = 1, 2, . . . ) be the j-fold Stieltjes convolution of Φ(t) with itself and Φ (0) (t) ≡ 1 for t ≥ 0, note that Φ ( j ) (t) ≤ Φ(t) j ( j = 0, 1, 2, . . . ) for any distribution Φ(t). © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Mizutani et al., Which-Is-Better (WIB): Problems in Reliability Theory, Springer Series in Reliability Engineering, https://doi.org/10.1007/978-3-031-27316-2_6

117

118

6 Which is Better for Standby or Parallel Systems

6.1 Reliability Measures 6.1.1 Constant Number of Units It is assumed that the number n (n = 1, 2, . . . ) of units for both standby and parallel systems can be constantly predefined. Then, reliability of a standby system at time t is R S (t) ≡ 1 − F (n) (t),

(6.1)

MTTF (Mean Time to Failure) is ʃ

∞

[1 − F (n) (t)]dt,

(6.2)

f (n) (t) . 1 − F (n) (t)

(6.3)

R P (t) = 1 − F(t)n ,

(6.4)

μS ≡ 0

and failure rate is h S (t) ≡ Reliability of a parallel system is

MTTF is

ʃ

∞

μP ≡

[1 − F(t)n ]dt,

(6.5)

n f (t)F(t)n−1 . 1 − F(t)n

(6.6)

0

and failure rate is

h P (t) ≡

In particular, when F(t) = 1 − e−λt (0 < λ < ∞), for a standby system, R S (t) =

n−1 ∑ (λt) j j=0

j!

e−λt ,

μS =

n , λ

λ(λt)n−1 /(n − 1)! h S (t) = ∑n−1 ≤ λ, j j=0 [(λt) /j!]

and for a parallel system, 1∑1 , λ j=1 j n

R P (t) = 1 − (1 − e−λt )n ,

h P (t) =

μP =

nλe−λt (1 − e−λt )n−1 nλ(1 − e−λt )n−1 = ≤ λ. ∑ n−1 −λt ) j 1 − (1 − e−λt )n j=0 (1 − e

6.1 Reliability Measures

119

We have the following comparative properties: (i) Both Ri (t) and μi (i = S, P) increase strictly with n. (ii) Both h i (t) (i = S, P) increase strictly with t from 0 to λ for n ≥ 2 and decrease strictly with n from λ to 0. When n = 1, Ri (t) = e−λt , μi = 1/λ and h i (t) = λ. (iii) R S (t) ≥ R P (t) and μ S ≥ μ P . (iv) h P (t) ≥ h S (t). The properties of (i)–(iii) can be easily shown. We prove (iv) as follows: When x ≡ λt for 0 < x < ∞, prove that n(1 − e−x )n−1 x n−1 /(n − 1)! ≥ ∑n−1 = h S (x). h P (x) = ∑n−1 −x j j j=0 (1 − e ) j=0 x /j! When n = 1, h P (x) = h S (x) = 1, and when n = 2, 2(1 − e−x )(1 + x) − x(2 − e−x ) = 1 − e−x + 1 − (1 + x)e−x > 0. Using a mathematical induction, supposing that −x n−1

n(1 − e )

n−1 j n−1 ∑ x x n−1 ∑ ≥ (1 − ex ) j , j! (n − 1)! j=0 j=0

we have n n ∑ xj xn ∑ (n + 1)(1 − e ) − (1 − e−x ) j j! n! j=0 j=0 [ ] ∑n−1 −x j n n−1 n (1 − e ) x x x j=0 ≥ (n + 1)(1 − e−x )n + − (1 − e−x ) j n! (n − 1)! n(1 − e−x )n−1 n! ⎡ ⎤ n n ∑ ∑ x n−1 ⎣ = (1 − e−x ) j − x (1 − e−x ) j ⎦ . (n + 1)x(1 − e−x )n + (n + 1) n! j=1 j=0 −x n

Letting L n (x) be the bracket of the above equation, L 0 (x) = 0 and L n (x) − L n−1 (x) =

n ∑ (1 − e−x ) j + nx(1 − e−x )n − nx(1 − e−x )n−1 + n(1 − e−x )n j=1

= n(1 − e−x )n−1 [1 − (1 + x)e−x ] > 0. Thus, L n (x) > 0 for any n ≥ 1, which completes the proof.

120

6 Which is Better for Standby or Parallel Systems

Table 6.1 Values of R S (t), R P (t), μ S , μ P , h S (t) and h P (t) when t = 1.0 and λ = 1.0 n R S (t) R P (t) μS μP h S (t) h P (t) 1 2 3 4 5 6 7 8 9 10

0.368 0.736 0.920 0.981 0.996 0.999 1.000 1.000 1.000 1.000

0.368 0.600 0.747 0.840 0.899 0.936 0.960 0.975 0.984 0.990

1.000 2.000 3.000 4.000 5.000 6.000 7.000 8.000 9.000 10.000

1.000 1.500 1.833 2.083 2.283 2.450 2.590 2.718 2.829 2.929

1.000 0.500 0.200 0.062 0.015 0.003 0.001 0.000 0.000 0.000

1.000 0.775 0.590 0.442 0.327 0.238 0.171 0.122 0.086 0.060

Example 6.1 Table 6.1 presents values of R S (t), R P (t), μ S , μ P , h S (t) and h P (t) for n at time t = 1.0 when λ = 1.0. When n become large, we can find obvious differences between standby and parallel systems. For MTTFs, μ S increase strictly with n to ∞, but μ P increase slowly and will be γ + log n when n is large enough, where γ is Euler’s constant and γ = 0.577215 . . . . For reliability and failure rate, it is much easier for a standby system to keep more reliable by increasing the number of units, e.g., when n increases to 7, R S (t) = 1.000 > R P (t) = 0.960 and h S (t) = ∎ 0.001 < h P (t) = 0.171.

6.1.2 Random Number of Units It is assumed that the system consists of N units, where N is a random variable with a probability function pn ≡ Pr{N = n} (n = 1, 2, . . . ) and its mean E{N } = ∑ ∞ n=1 npn [6]. Then, reliability of a standby system at time t is ˜S (t) = R

∞ ∑

pn [1 − F (n) (t)],

(6.7)

n=1

MTTF is ˜ μ S (t) =

∞ ʃ ∑ n=1

and failure rate is

∞

pn [1 − F (n) (t)]dt,

(6.8)

0

∑∞ pn f (n) (t) ˜ . h S (t) = ∑∞ n=1 (n) n=1 pn [1 − F (t)]

(6.9)

6.1 Reliability Measures

121

Reliability of a parallel system is ˜P (t) = R

∞ ∑

pn [1 − F(t)n ],

(6.10)

n=1

MTTF is ˜ μ P (t) =

∞ ʃ ∑ n=1

∞

pn [1 − F(t)n ]dt,

(6.11)

0

∑∞ pn n f (t)F(t)n−1 ˜ . h P (t) = ∑n=1 ∞ n n=1 pn [1 − F(t) ]

and failure rate is

(6.12)

(1) Geometric distribution When F(t) = 1 − e−λt and N has a geometric distribution, i.e., pn = pq n−1 (n = 1, 2, . . . ) for 0 < p < 1 and q ≡ 1 − p with mean 1/ p, for a standby system, ˜S (t) = e− pλt , R

˜ μS =

1 , pλ

˜ h S (t) = pλ,

which agrees with those of one-unit system with a failure distribution F(t) = 1 − e− pλt . For a parallel system, ˜P (t) = R

∞

e−λt , p + qe−λt

˜ μS =

1 ∑ qj , λ j=0 j + 1

˜ h P (t) =

pλ . p + qe−λt

We have the following comparative properties: (i) (ii) (iii) (iv)

˜i (t) and ˜ Both R μi (i = S, P) increase with q. Both ˜ h S (t) and ˜ h P (t) decrease with q from λ to 0. ˜P (t) and ˜ ˜S (t) ≥ R μS ≥ ˜ μP . R ˜ h S (t). h P (t) ≥ ˜

(2) Poisson distribution When F(t) = 1 − e−λt and N has a Poisson distribution, i.e., pn = [β n−1 /(n − 1)!]e−β (n = 1, 2, . . . ) for 0 < β < ∞ with mean β + 1, for a standby system, ˜S (t) = R

∞ ∑ (λt) j j=0

j!

e−λt

∞ ∑ βn n= j

n!

e−β ,

˜ μS =

∑ j j λ ∞ j=0 [(λt) /j!](β /j!) ˜ ∑ ≤ λ. h S (t) = ∑∞ ∞ j n j=0 [(λt) /j!] n= j (β /n!)

β+1 , λ

122

6 Which is Better for Standby or Parallel Systems

For a parallel system, ) ( ˜P (t) = 1 − (1 − e−λt ) exp −βe−λt , R

∞

1 ∑ β n −β ∑ 1 ˜ μP = e , λ n=0 n! j j=1 n+1

λ(1 + β)e−λt exp(−βe−λt ) ˜ ≤ λ. h P (t) = 1 − (1 − e−λt ) exp(−βe−λt ) We have the following comparative properties: (i) (ii) (iii) (iv)

˜i (t) and ˜ Both R μi (i = S, P) increase with β. ˜ Both h S (t) and ˜ h P (t) decrease with β from λ to 0. ˜P (t) and ˜ ˜S (t) ≥ R μS ≥ ˜ μP . R ˜ h S (t). h P (t) ≥ ˜

We prove (ii) as follows: Differentiating ˜ h S (t) with respect to β, ∞ ∞ ∞ ∑ (λt)i ∑ β n (λt) j+1 β j ∑ (i − j ) ( j + 1)! j! i=0 (i + 1)! n=i n! j=0 j ∞ ∞ ∑ ∑ (λt) j+1 β j ∑ (λt)i βn (i − j ) = ( j + 1)! j! i=0 (i + 1)! n! j=0 n=i ∞ ∞ ∞ ∑ ∑ βn (λt) j+1 β j ∑ (λt)i + (i − j ) ( j + 1)! j! i= j (i + 1)! n! j=0 n=i ⎡ ⎤ ∞ ∞ ∞ ∞ i ∑ n j ∑ n ∑ (λt) j+1 ∑ (λt)i β β β β ⎦ < 0, = ( j − i) ⎣ − ( j + 1)! i= j (i + 1)! i! n= j n! j! n=i n! j=0

∑ n −λt because (β j /j!) ∞ (0 ≤ x ≤ 1), difn= j (β /n!) decreases with j. When x ≡ e ˜ ferentiating h P (t) with respect to β, [ ] λxe−βx 1 − βx − eβx − x(1 − e−βx ) ≤ 0, which completes (ii). h S (t), which can be Unfortunately, we cannot prove mathematically that ˜ h P (t) ≥ ˜ shown numerically in the following example: ˜S (t), R ˜P (t), ˜ Example 6.2 Table 6.2 presents R μS , ˜ μP , ˜ h S (t) and ˜ h P (t) for β + 1 at time t = 1.0 when λ = 1.0. Compared with Table 6.1, similar comparative results between standby and parallel systems can be derived. However, for any redundant and parallel systems, it could be shown for most cases that reliabilities and MTTFs ˜S (t), R P (t) > R ˜P (t), μ S > ˜ μS , μ P > ˜ μ P , and become smaller such that R S (t) > R

6.2 Replacement Policies

123

˜S (t), R ˜P (t), ˜ h S (t) and ˜ h P (t) for β when t = 1.0 and λ = 1.0 Table 6.2 Values of R μS , ˜ μP , ˜ ˜S (t) ˜P (t) ˜ ˜ R R h S (t) h P (t) β+1 ˜ μS ˜ μP 1 2 3 4 5 6 7 8 9 10

0.368 0.654 0.817 0.906 0.953 0.977 0.989 0.995 0.997 0.999

0.368 0.562 0.697 0.790 0.855 0.900 0.930 0.952 0.967 0.977

1.000 2.000 3.000 4.000 5.000 6.000 7.000 8.000 9.000 10.000

1.000 1.429 1.752 2.006 2.213 2.386 2.536 2.666 2.782 2.886

1.000 0.472 0.259 0.145 0.080 0.043 0.023 0.012 0.006 0.003

1.000 0.905 0.759 0.618 0.494 0.390 0.304 0.235 0.181 0.137

˜P (t), ˜ ˜S (t) ≥ R μS ≥ ˜ μ P . Furthermore, failure rates become larger such that h S (t) < R ˜ h P (t), and ˜ h P (t). h S (t), h P (t) < ˜ h S (t) ≤ ˜ ∎

6.2 Replacement Policies 6.2.1 MTTF for Replacement Time It was shown that standby and parallel systems can operate for a specified mean time by either changing the preventive replacement time or increasing the number of redundant units [1, p. 64], [2, p. 7]. We first observe MTTFs for replacement in which the system is replaced preventively at time T (0 < T ≤ ∞). The mean time μ(T ) to the first system failure is given by a renewal function ʃ

T

μ(T ) =

t dF(t) + F(T )[T + μ(T )].

0

Solving the above renewal equation, 1 μ(T ) = F(T )

ʃ

T

F(t)dt.

(6.13)

0

Note that when F(t) = 1 − e−λt , μ(T ) = 1/λ = μ S = μ P for n = 1. When failure rate h(t) increases strictly with T from h(0) to h(∞), μ(T ) decreases strictly with T from 1/ h(0) to μ. Thus, there exists a finite and unique Ti∗ which satisfies μ(T ) = μi (i = S, P) for n ≥ 2. Clearly, when n = 1, Ti∗ = ∞.

124

6 Which is Better for Standby or Parallel Systems

For a standby system with n units, letting μ(T ) = μ S , from (6.2), TS∗ satisfies 1 F(T )

ʃ

T

ʃ

∞

F(t)dt =

0

[1 − F (n) (t)]dt,

(6.14)

0

and for a parallel system with n units, from (6.13), TP∗ satisfies 1 F(T )

ʃ

T

ʃ

∞

F(t)dt =

0

[1 − F(t)n ]dt.

(6.15)

0

We can compute replacement times Ti∗ which satisfy μ(Ti ) = μi (i = S, P). Example 6.3 When F(t) is a Weibull distribution, i.e., F(t) = 1 − e−(t/10) , 2

ʃ μ=

∞

√ 2 e−(t/10) dt = 5 π ,

√ μ S = 5n π ,

0

and Ti∗ (i = S, P) satisfy, respectively, 1 1−

e−(T /10)2

1−

e−(T /10)2

1

ʃ ʃ

T 0 T

√ 2 e−(t/10) dt = 5n π , ʃ ∞{ [ ] } 2 2 n e−(t/10) dt = 1 − 1 − e−(t/10) dt.

0

0

ʃ∞ √ 2 Table 6.3 presents TS∗ and TP∗ , μ S = 5n π and μ P = 0 [1−(1 − e−(t/10) )]dt 2 for standby and parallel systems with n (n ≥ 2) units when F(t) = 1 − e−(t/10) . Because both μ S and μ P increase with n and μ S > μ P as shown in Table 6.3, TS∗ and TP∗ decrease with n and TS∗ < TP∗ . From (6.13), this indicates that for a standby system with longer MTTF, F(T ) should be kept small enough, i.e., replacement should be done in much smaller times. For a parallel system with shorter MTTF, replacement could be done in longer times. ∎

6.2.2 Replacement for Number of Units Suppose that both systems operate for a single job with working time Y in Sect. 2.1, and fail when all of n units have failed. Let c A be an acquisition cost for one unit and c R be the replacement cost at each failure. Then, the total cost for a standby system until system failure is n(c A + c R ), and the expected cost rate is C S (n) =

n(c A + c R ) , μS

(6.16)

6.2 Replacement Policies

125

Table 6.3 TS∗ and TP∗ , and μ S and μ P when F(t) = 1 − e−(t/10)

2

n

TS∗

TP∗

μS

μP

2 3 4 5 6 7 8 9 10

5.99 3.86 2.86 2.28 1.89 1.62 1.42 1.26 1.13

10.48 8.83 8.02 7.52 7.17 6.91 6.70 6.53 6.39

17.72 26.59 35.45 44.31 53.17 62.04 70.90 79.76 88.62

11.46 12.90 13.89 14.62 15.20 15.68 16.09 16.45 16.76

where μ S is given in (6.2). Similarly, the total cost for a parallel system until system failure is nc A + c R , and the expected cost rate is C P (n) =

nc A + c R , μP

(6.17)

where μ P is given in (6.5). We compare the expected cost rates C S (n) and C P (n) numerically when F(t) = 1 − e−λt . Example 6.4 When F(t) = 1 − e−λt , the expected cost rates are, from (6.16) and (6.17), respectively, C S (n) = cA + cR , λ

C P (n) nc A + c R = ∑n . λ j=1 (1/j)

(6.18)

We find optimal n ∗P to minimize C P (n). From the inequality C P (n + 1) − C P (n) ≥ 0, n+1 ∑ 1 cA (n + 1) ≥ , (6.19) j c R j=2 whose left-hand side increases strictly with n from 1 to ∞. Thus, there exists a finite and unique minimum n ∗P (1 ≤ n ∗P < ∞) which satisfies (6.19). If c R ≥ c A , then n ∗P = 1 and C P (1) = C S (n). Furthermore, from (6.18) and (6.19), if c R < c A , i.e., n ∗P ≥ 2, and n∗

P 1 ∑ 1 cR ≥ , ∗ n P − 1 j=2 j cA + cR

(6.20)

126

6 Which is Better for Standby or Parallel Systems

then a parallel system would save more cost than a standby system does. However, (6.20) is ∑n ∗P cA j=2 (1/j) ≥ , (6.21) ∑n ∗P ∗ c R (1/j) n −1− P

j=2

and from (6.19), ⎤ ⎡ ∑n ∑n ) n n ( ∑ ∑ 1 1 j=2 (1/j) j=2 (1/j) ⎣n ∑n ∑n − = 1− − 1⎦≥ 0, n n −1− n−1− j (1/j) (/1 j) j j=2 j=2 j=2 j=2 which follows that there does not exit any n ∗P which satisfies (6.21). This shows that under the expected cost rates given in (6.18), a standby system saves more cost than a parallel system does. c A ≤ c A and its optimal number ⌃ n P for a parallel We next compute a modified cost ⌃ system, which might be the case where a parallel system is better than a standby n P )/λ = C S (n)/λ = c A + c R for given c A and c R , we have system. By setting C P (⌃ the following simultaneous equations: n P + 1) (⌃

⌃ n∑ P +1 j=2

1 ⌃ cA ≥ , j cR

⌃ n P⌃ c A + cR = cA + cR , ∑⌃n P j=1 (1/j )

i.e., we compute a minimum ⌃ n P which satisfies ⌃ n P (⌃ n P + 1)

⌃ n∑ P +1 j=2

1 ≥ j

(

)∑ ⌃ nP cA 1 +1 − 1. cR j j=1

Using ⌃ n P , we compute ⌃ c A /c R given by ⎤ ⎡ )∑ ( ⌃ nP ⌃ cA 1 1 ⎣ cA = +1 − 1⎦ . cR ⌃ nP cR j j=1 Table 6.4 presents original optimal n ∗P and its resulting cost C P (n ∗P )/(λc R ), and n P and ⌃ c A /c R . It is shown that ⌃ c A ≤ c A can be found, and we can provided modified ⌃ c A , as shown in Table 6.4 that ⌃ n P ≤ n ∗P . That is, less units for a parallel system under ⌃ c A for a parallel system is lower than that c A for a standby if the unit acquisition cost ⌃ system, then we could adopt a parallel system to save the total of acquisition and replacement costs, and otherwise, a standby system should be used. ∎

6.3 Scheduling of Random Works

127

Table 6.4 Optimal n ∗P , C P (n ∗P )/(λc R ), ⌃ n P and ⌃ c A /c R c A /c R

C S (n)/(λc R ) n ∗P

C P (n ∗P )/(λc R )

⌃ nP

⌃ c A /c R

1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0

2.000 3.000 4.000 5.000 6.000 7.000 8.000 9.000 10.000 11.000

2.000 3.333 5.455 7.091 10.080 12.000 15.766 17.956 22.449 24.898

1 2 2 3 3 3 4 4 4 4

1.000 1.750 2.500 3.250 3.333 3.944 4.556 4.437 4.958 5.479

1 2 3 3 4 4 5 5 6 6

6.3 Scheduling of Random Works We propose optimization problems of standby and parallel systems which operate for two kinds of tandem and parallel works [7, p. 176].

6.3.1 Tandem and Parallel Works (1) Single Work Suppose that a job has a working time Y which isʃa random time with a general dis∞ tribution G(t) ≡ Pr{Y ≤ t}, its mean time 1/θ ≡ 0 G(t)dt, and a density function g(t) ≡ dG(t)/dt. A job needs to be set up based on the scheduling time L. The problem is how to determine the scheduling time L for a job with a random working time Y [2, p. 83]. It is assumed that the scheduling time for a job is constant L (0 ≤ L < ∞). Next, we introduce the following two costs [8]: When the scheduling time is L, its cost is c O (L), If the work is accomplished up to time L, i.e., L ≥ Y , it requires the excess cost c E (L − Y ), and if it is not accomplished before time L and is completed after L, i.e., L < Y , it requires the shortage cost c S (Y − L). Then, the total cost until the work completion is ʃ

∞

C(L) = L

ʃ c S (t − L)dG(t) + 0

L

c E (L − t)dG(t) + c O (L).

(6.22)

128

is

6 Which is Better for Standby or Parallel Systems

In particular, when ci (t) = ci t and ci > 0 (i = O, S, E), the total expected cost ʃ L ʃ ∞ G(t)dt + c E G(t)dt + c O L . (6.23) C(L) = c S L

0

Note that C(0) = c S /θ and C(∞) = ∞. We find optimal L ∗ to minimize C(L) in (6.23). Differentiating C(L) with respect to L and setting it equal to zero, G(L) =

cS − cO . cS + c E

(6.24)

If c S ≤ c O , then L ∗ = 0, i.e., we should not set up the scheduling of a job. In particular, when c S = c E = 1 and c O ≡ 0, (6.23) agrees with (2.40) when L = T and G(t) = F(t), which corresponds to the problem of minimizing deviation times in Sect. 2.6. When L is not constant and is distributed exponentially with mean l, i.e., Pr{L ≤ t} = 1 − e−t/l , (6.24) is ʃ

∞

e−t/l dG(t) = G ∗ (1/l) =

0

cS − cO , cS + c E

(6.25)

ʃ∞ where G ∗ (s) ≡ 0 e−st dG(t) is LS (Laplace Stieltjes) transform of G(t) for Re(s) > 0. Thus, when G(t) = 1 − e−θt , optimal l ∗ is given by θl ∗ =

cS − cO . cE + cO

(6.26)

(2) Tandem Works Consider the scheduling problems of N works (N = 0, 1, 2, . . . ) in tandem in Fig. 2.1 when c S > c O : It is assumed that Y j ( j = 1, 2, . . . , N ) are the working times for a job, are independent and have an identical distribution G(t) ≡ Pr{Y j ≤ t} ∑ with finite mean 1/θ, where S N ≡ Nj=1 Y j . Then, the probability that N works are completed until time L is Pr{S N ≤ L} = G (N ) (L). Thus, replacing G(t) in (6.23) and (6.24) with G (N ) (t), the total expected cost until N work completion is ʃ

∞

C T (L , N ) = c S

[1 − G

(N )

ʃ (t)]dt + c E

L

L

G (N ) (t)dt + c O L ,

(6.27)

0

and optimal L ∗T to minimize C T (L , N ) for given N (N = 1, 2, . . . ) satisfies G (N ) (L) =

cS − cO . cS + c E

Note that L ∗T increases strictly with N from L ∗ given in (6.24) to ∞.

(6.28)

6.3 Scheduling of Random Works

129

Next, we find optimal N T∗ of works to minimize C T (L , N ) in (6.27) for given L (0 < L < ∞). Forming the inequality C T (L , N + 1) − C T (L , N ) ≥ 0, ʃ∞ L

[G (N ) (t) − G (N +1) (t)]dt

0

[G (N ) (t) − G (N +1) (t)]dt

ʃL ʃ

or

∞

θ

≥

[G (N ) (t) − G (N +1) (t)]dt ≥

L

cE , cS

cE . cS + c E

(6.29)

In particular, when G(t) = 1 − e−θt , N ∑ (θL) j

j!

j=0

e−θL ≥

cE . cS + c E

(6.30)

Thus, there exists a finite and unique minimum N T∗ (0 ≤ N T∗ < ∞) which satisfies (6.30). If 1 − e−θL ≤ c S /(c S + c E ), then N T∗ = 0, i.e., we should not set up any work for time L. In addition, because the left-hand side of (6.30) decreases strictly with L from 1 to 0, N T∗ increases with L from 0 to ∞. (3) Parallel Works Consider the scheduling problems of N works (N = 0, 1, 2, . . . ) in parallel when c S > c O . Then, the probability that N works are completed until time L is G(L) N . Replacing G(t) in (6.23) with G(t) N , the total expected cost until N work completion is ʃ ∞ ʃ L N C P (L , N ) = c S [1 − G(t) ]dt + c E G(t) N dt + c O L , (6.31) L

0

and optimal L ∗P to minimize C P (L , N ) satisfies G(L) N =

cS − cO . cS + c E

(6.32)

Note that L ∗P increases strictly with N from L ∗ given in (6.24) to ∞. In addition, because G(L) N ≥ G (N ) (L), L ∗P ≤ L ∗T . Next, we find optimal N P∗ to minimize C P (L , N ) in (6.31) for given L (0 < L < ∞). Forming the inequality C P (T , N + 1) − C P (T , N ) > 0, ʃ∞ L

G(t) N G(t)dt

0

G(t) N G(t)dt

ʃL

≥

cE . cS

(6.33)

130

6 Which is Better for Standby or Parallel Systems

Noting that ʃ

∞

ʃ

N +1

ʃ

L

∞

ʃ

G(t) G(t)dt G(t) G(t)dt − G(t) G(t)dt L 0 L [ʃ ∞ ʃ L G(t) N G(t)dt G(t) N G(t)dt ≥ G(L) N

L

ʃ

∞

−

ʃ

G(t) N G(t)dt

L

L

N

G(t) N +1 G(t)dt

0

0

L

] G(t) N G(t)dt = 0,

0

and ʃ∞

ʃ∞ G(L) N L G(t)dt ≥ lim ʃ L N N →∞ G(t) N G(t)dt 0 G(t) G(t)dt ʃ∞ L G(t)dt = lim ʃ L = ∞, N N →∞ 0 [G(t)/G(L)] G(t)dt

lim ʃLL N →∞ 0

G(t) N G(t)dt

the left-hand side of (6.33) increases strictly with N to ∞. Thus, there ʃ ∞ exists a finite and unique minimum N P∗ (1 ≤ N P∗ < ∞) which satisfies (6.33). If L G(t)dt/ ʃL ʃL ∗ 0 G(t)dt ≥ c E /c S , i.e., 0 θG(t)dt ≤ c S /(c S + c E ), then N P = 0. In addition, because the left-hand side of (6.33) decreases strictly with L from 1 to 0, N P∗ increases with L from 0 to ∞. In particular, when G(t) = 1 − e−θt , (6.33) is (1 − e−θL ) N +1 ≤

cS . cS + c E

Comparing (6.30) and (6.34), N P∗ ≤ N T∗ because (1 − e−θL ) N +1 ≥ j!]e−θL .

(6.34) ∑∞

j=N +1 [(θL)

j

/

6.3.2 Standby and Parallel Systems Suppose that standby and parallel systems with n units operate for a job with N tandem works.

(1) Standby System A standby system with n (n = 0, 1, 2, . . . ) units operates for a job with N (N = 0, 1, 2, . . . ) tandem works in Fig. 2.1. We introduce the same cost structure defined in Sect. 6.3.1: If the system fails at time t and N th work is completed at time u, then

6.3 Scheduling of Random Works

131

it requires the shortage cost c S (t − u) for t ≥ u and the excess cost c E (u − t) for u > t. Then, adding an operating cost c O n of n units, the total expected cost is ʃ

∞

[ʃ

t

(n)

]

C S (n, N ) = c O n + c S (t − u)dF (u) dG (N ) (t) 0 0 ] ʃ ∞ [ʃ t (N ) (t − u)dG (u) dF (n) (t) + cE 0 0 ʃ ∞ ʃ ∞ (N ) (n) = cO n + cS [1 − G (t)]F (t)dt + c E G (N ) (t)[1 − F (n) (t)]dt 0 0 ʃ ∞ N = n(c O + c E μ) + c S − (c S + c E ) [1 − G (N ) (t)][1 − F (n) (t)]dt. (6.35) θ 0 In particular, when G(t) = 1 − e−θt and F(t) = 1 − e−λt , ( cE ) N + cS C S (n, N ) = n c O + λ θ )j ( )i )( n−1 ∑ N −1 ( ∑ i+j cS + c E λ θ − . i θ + λ j=0 i=0 θ+λ θ+λ

(6.36)

We find optimal n ∗S for given N and N S∗ for given n to minimize C S (n, N ), respectively. Forming the inequality C S (n + 1, N ) − C S (n, N ) ≥ 0, (

λ θ+λ

)n+1 ∑ )( )j N −1 ( n+ j θ cE + cO λ ≤ . j θ + λ cS + c E j=0

(6.37)

Letting L S (n) be the left-hand side of (6.37), )N ( )n+1 ( )( λ θ n+N > 0, L S (n) − L S (n + 1) = θ+λ θ+λ N −1 which follows that L S (n) decreases strictly with n to 0. Thus, there exists a finite and unique minimum n ∗S (0 ≤ n ∗S < ∞) which satisfies (6.37) and increases with N . In addition, if [θ/(θ + λ)] N ≥ (c S − c O λ)/(c S + c E ), then n ∗S = 0. Furthermore, when N = 1, optimal n ∗S is given by an integer which satisfies (

cE + cO λ log cS + c E

)/ ( ) )/ ( ) ( λ cE + cO λ λ ∗ log − 1 ≤ n S < log log . θ+λ cS + c E θ+λ (6.38)

132

6 Which is Better for Standby or Parallel Systems

Table 6.5 Optimal n ∗S for N and N S∗ for n when c E /c S = 0.5 and c O λ/c S = 0.1 λ/θ 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

N

n

1

2

3

4

5

1

2

3

4

5

0 0 0 0 0 0 1 1 1

0 0 1 1 1 1 1 2 2

0 1 1 1 2 2 2 2 3

0 1 1 2 2 3 3 3 4

0 1 2 2 3 3 4 4 5

4 2 1 1 0 0 0 0 0

11 6 4 3 2 2 1 1 1

20 10 6 5 4 3 2 2 2

29 14 9 7 5 4 4 3 3

37 18 12 9 7 6 5 4 4

Forming the inequality C S (n, N + 1) − C S (n, N ) ≥ 0 in (6.36), (

θ θ+λ

) N +1 ∑ n−1 ( j=0

N+j j

)(

λ θ+λ

)j ≤

cS , cS + c E

(6.39)

whose left-hand side decreases strictly with N to 0. Thus, there exists a finite and unique minimum N S∗ (0 ≤ N S∗ < ∞) which satisfies (6.39) and increases with n. In addition, if [λ/(θ + λ)]n ≥ c E /(c S + c E ), then N S∗ = 0. Furthermore, when n = 1 optimal N S∗ is given by an integer which satisfies ( log

cS cS + c E

)/ ( log

θ θ+λ

)

− 1 ≤ N S∗ < log

(

cS cS + c E

)/ ( log

) θ . θ+λ (6.40)

Example 6.5 Suppose that F(t) = 1 − e−λt and G(t) = 1 − e−θt . Then, Table 6.5 presents optimal n ∗S and N S∗ for N and n for λ/θ, respectively, when c E /c S = 0.5 and c O λ/c S = 0.1. Optimal n ∗S increase with λ/θ and N , however, N S∗ decrease with λ/θ and increase with n. It is of interest that when λ/θ = 0.9, n ∗S = N , and n ∗S is almost equal to N λ/θ + 1 and N S∗ is almost equal to (n − 1)/(λ/θ) for n ≥ 2. In other words, the mean time to complete N works is almost the same mean failure time for a standby system with (n − 1) units. ∎

(2) Parallel System A parallel system with n units (n = 0, 1, 2, . . . ) operates for a job with N (N = 0, 1, 2, . . . ) tandem works. Then, replacing F (n) (t) in (6.35) with F(t)n formally,

6.3 Scheduling of Random Works

133

the total expected cost is ʃ C P (n, N ) = c O n + c S

∞

[1 − G

(N )

ʃ (t)]F(t) dt + c E

0

n

∞

G (N ) (t)[1 − F(t)n ]dt.

0

(6.41) In particular, when G(t) = 1 − e−θt and F(t) = 1 − e−λt , n N cE ∑ 1 + θ λ j=1 j ( ) [ )N ] ( n cS + c E ∑ θ j n 1 (−1) + 1− . j j λ jλ + θ j=1

C P (n, N ) = c O n + c S

(6.42)

We find optimal n ∗P for given N and N P∗ for given n to minimize C P (n, N ) in (6.42), respectively. Forming the inequality C P (n + 1, N ) − C P (n, N ) ≥ 0, [ ]N ] ( ) [ n ∑ 1 θ c E /(n + 1) + c O λ j n (−1) . 1− ≤ j j +1 ( j + 1)λ + θ cS + c E j=0

(6.43)

Letting L P (n) be the left-hand side of (6.43), [ ( ) ]N ] [ n ∑ n θ 1 L P (n) − L P (n + 1) = (−1) j ] 1− > 0, j j +2 ( j + 2)λ + θ j=0 which follows that L P (n) decreases strictly with n to 0, and the right-hand side of (6.43) decreases to c O λ/(c S + c E ). Thus, there exists a finite and unique minimum n ∗S (0 ≤ n ∗S < ∞) which satisfies (6.43). In addition, if [θ/(θ + λ)] N ≥ (c S − λc O /(c S + c E ), then n ∗P = 0. Forming the inequality C P (n, N + 1) − C P (n, N ) ≥ 0, n ∑ j=1

(−1) j+1

) N +1 ( )( n cS θ ≤ , j jλ + θ cS + c E

(6.44)

whose left-hand side decreases strictly with N to 0. Thus, there exists a finite and unique minimum N P∗ (0 ≤ N P∗ < ∞) which satisfies (6.44) and increases with n. In addition, if ( ) n ∑ n θ cS (−1) j+1 ≥ , j jλ + θ c S + cE j=1 then N P∗ = 0.

134

6 Which is Better for Standby or Parallel Systems

Table 6.6 Optimal n ∗P for N and N P∗ for n when c E /c S = 0.5 and λc O /c S = 0.1 N

n

λ/θ

1

2

3

4

5

1

2

3

4

5

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0 0 0 0 0 0 1 1 1

0 0 1 1 1 1 2 2 2

0 1 1 1 2 2 3 3 4

0 1 1 2 3 3 4 5 5

0 1 2 3 4 5 5 6 7

4 2 1 1 0 0 0 0 0

8 4 2 2 1 1 1 1 1

11 5 4 3 2 2 1 1 1

14 7 4 3 2 2 2 1 1

16 8 5 4 3 2 2 2 1

Example 6.6 Suppose that F(t) = 1 − e−λt and G(t) = 1 − e−θt . Then, Table 6.6 presents optimal n ∗P and N P∗ for N and n for λ/θ, respectively, when c E /c S = 0.5 and λc O /c S = 0.1 and shows a similar tendency to Table 6.5. Note that when λ/θ = 0.7, n ∗P = N . Comparing Table 6.6 with Table 6.5, n ∗P ≥ n ∗S and N P∗ ≤ N S∗ , because a parallel system needs more units than a standby system for the same job. It is of interest∑ that when λ/θ = 0.9, n ∗S = N , and n ∗S is almost equal to a minimum such that nj=1 (1/j ) ≥ N (λ/θ), and N S∗ is almost equal to a minimum such that ∑ N ≥ nj=1 (1/j )/(λ/θ). In other words, the mean time to complete N works is almost the same mean failure time for a parallel system with n units. ∎

6.3.3 Random Tandem Works It is assumed that the number N of works is not constant and is a random variable with a probability function pk ≡ Pr{N = k} with finite mean β + 1. We consider the scheduling problems of N tandem works when pk ≡ Pr{N = k} (k = 1, 2, . . . ). Then, the total expected cost until the completion of N works is, from (6.27), C T (L; p) = c O L + c S

∞ ∑

ʃ

∞

pk

k=1

(k)

[1 − G (t)]dt + c E

L

∞ ∑ k=1

ʃ

L

pk

G (k) (t)dt,

0

(6.45) and optimal L ∗T R satisfies ∞ ∑ k=1

pk G (k) (L) =

cS − cO . cS + c E

(6.46)

6.3 Scheduling of Random Works

135

(1) Standby System Suppose that a standby system operates for a job with a random number of N tandem works each of which has an identical probability function pk . Then, the total expected cost is, from (6.35), C S (n; p) = c O n + c S

∞ ∑

ʃ

+ cE

ʃ

[1 − G (k) (t)]F (n) (t)dt

0

k=1 ∞ ∑

∞

pk ∞

pk

G (k) (t)[1 − F (n) (t)]dt.

(6.47)

0

k=1

In particular, when G(t) = 1 − e−θt and F(t) = 1 − e−λt , ( β+1 cE ) + cS C S (n; p) = n c O + λ θ )( )j ( )i ∑ ∞ ( n−1 ∑ ∞ ∑ i+j λ θ cS + c E − pk . j θ + λ j=0 i=0 θ+λ θ + λ k=i+1

(6.48)

We find optimal n ∗S R to minimize C S (n; p). Forming the inequality C S (n + 1; p) − C S (n; p) ≥ 0, (

λ θ+λ

)n+1 ∑ ∞ ( j=0

n+ j j

)(

θ θ+λ

)j ∑ ∞

pk ≤

k= j+1

cE + cO λ . cS + c E

(6.49)

Letting L S (n) be the left-hand side of (6.49), ( L S (n) − L S (n + 1) =

λ θ+λ

)n+1 ∑ )( )j ∞ ( n+ j θ p j > 0, j −1 θ+λ j=1

which follows that L S (n) decreases strictly with n to 0. Thus, there exists a finite and unique minimum n ∗S R (0 ≤ n ∗S R < ∞) which satisfies (6.49), In addition, if ∞ ( ∑ k=1

then n ∗S R = 0.

θ θ+λ

)k pk ≥

c S − λc O , cS + c E

136

6 Which is Better for Standby or Parallel Systems

(2) Parallel System Suppose that a parallel system operates for a job with a random number of N tandem works with a probability function pk . Then, the total expected cost is, from (6.41), C P (n; p) = c O n + c S

∞ ∑

ʃ pk

+ cE

ʃ

pk

k=1

[1 − G (k) (t)]F(t)n dt

0

k=1 ∞ ∑

∞

∞

G (k) (t)[1 − F(t)n ]dt.

(6.50)

0

In particular, when G(t) = 1 − e−θt and F(t) = 1 − e−λt , n cE ∑ 1 β+1 + cS λ j=1 j θ [ ( ) ∑ )k ] ( n ∞ θ cS + c E ∑ j n 1 (−1) pk 1 − + . j j k=1 λ jλ + θ j=1

C P (n; p) = c O n +

(6.51)

We find optimal n ∗P R to minimize C P (n; p). Forming the inequality C P (n + 1; p) − C P (n; p) ≥ 0, [ ( ) ]k ] [ n ∞ ∑ 1 ∑ θ c E /(n + 1) + c O λ j n pk 1 − . (−1) ≤ ( j + 1)λ + θ cS + c E j j + 1 k=1 j=0 (6.52) Letting L P (n) be the left-hand side of (6.52), [ ]k ] ( ) [ ∞ n 1 ∑ θ L P (n) − L P (n + 1) = (−1) pk 1 − > 0, j j + 2 k=1 ( j + 2)λ + θ j=0 n ∑

j

which follows that L P (n) decreases strictly with n to 0, and the right-hand side of (6.52) decreases with n to c O λ/(c S + c E ). Thus, there exists a finite and unique minimum n ∗P R (0 ≤ n ∗P R < ∞) which satisfies (6.52). In addition, if ∞ ( ∑ k=1

then n ∗P R = 0.

θ θ+λ

)k pk ≥

c S − λc O , cS + c E

6.4 Modified Models

137

Table 6.7 Optimal n i∗ and n i∗R (i = S, R) when c E /c S = 0.5 and c O λ/c S = 0.1 λ/θ 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

n ∗S

n ∗S R

0 0 1 1 1 1 1 2 2

0 0 1 1 1 1 1 1 2

N =2 n ∗P 0 0 1 1 1 1 2 2 2

n ∗P R

n ∗S

n ∗S R

0 0 1 1 1 1 1 2 2

0 1 2 2 3 3 4 4 5

0 1 2 2 3 3 4 4 5

N =5 n ∗P 0 1 2 3 4 5 5 6 7

n ∗P R

n ∗S

N = 10 n ∗S R n ∗P

n ∗P R

0 1 2 3 3 4 5 5 6

1 2 3 4 5 6 7 9 10

1 2 3 4 5 6 7 9 10

1 3 5 6 7 8 8 9 9

1 3 5 7 8 8 9 9 9

Example 6.7 When pk = [β k−1 /(k − 1)!]e−β (k = 1, 2, . . . ) with mean β + 1, F(t) = 1 − e−λt and G(t) = 1 − e−θt , (6.49) is (

λ θ+λ

)n+1 ∑ ∞ ( j=0

n+ j j

)(

θ θ+λ

)j ∑ ∞ k= j

β k −β cE + cO λ e ≤ , k! cS + c E

(6.53)

and (6.52) is [ [ ]] ( ) n ∑ 1 θ ( j + 1)λ j n (−1) exp −β 1− ( j + 1)λ + θ ( j + 1)λ + θ j j +1 j=0 ≤

c E /(n + 1) + c O λ . cS + c E

(6.54)

Table 6.7 presents n ∗S in (6.37), n ∗S R in (6.53) and n ∗P in (6.43), n ∗P R in (6.54) when N = β + 1, c E /c S = 0.5 and λc O /c S = 0.1. All of n i∗ and n i∗R (i = S, P) increase with N and λ/θ, and n i∗ ≥ n i∗R . In addition, both n i∗ and n i∗R are almost the same, and n i∗ ≈ N λ/θ. ∎

6.4 Modified Models We apply standby and parallel systems into data transmission models and compare them. Next, we obtain the deviation times introduced in Sect. 2.6 for standby and parallel systems, and compare their expected cost rates. Finally, using a K -out-of-n system, we present a failure distribution of redundant systems and show that WIB problems are rewritten for any redundant systems.

138

6 Which is Better for Standby or Parallel Systems

(1) Data Transmission We apply standby and parallel systems to partition problems for a data transmission models [2, p. 110], [9]: Suppose that N (N = 1, 2, . . . ) units of data have to be transmitted from a sender to a receiver, and every transmissions of each unit of data fail according to an identical probability p (0 ≤ p < 1) due to some errors which have occurred independently. Firstly, we consider a standby transmission in which each of N units is sent to a receiver one by one successively. If this transmission fails, then we send it again until its success. Let t0 and t1 be the respective times needed for preparation and transmission of each unit. Then, the mean transmission time T1 of each unit until its success is given by a renewal equation: T1 = (1 − p)t1 + p(t1 + T1 ) + t0 , i.e., T1 =

t 0 + t1 . 1− p

(6.55)

Thus, the total time of N units until their successes is TS (N ) ≡ N T1 =

N (t0 + t1 ) . 1− p

(6.56)

Secondly, we consider a parallel transmission in which all of N units are sent to a receiver like one unit at the same time. If at least one of N transmissions fail, then we send all units again from the beginning until its success. Let t N be the time needed for transmission of N units. Then, the mean transmission time of N units until its success is given by a renewal equation: TP (N ) = (1 − p) N t N + [1 − (1 − p) N ][t N + TP (N )] + t0 , i.e., TP (N ) =

t0 + t N . (1 − p) N

(6.57)

Comparing (6.56) with (6.57), if t N + t0 > N (1 − p) N −1 , t1 + t0 then a standby transmission is better than a parallel one, and conversely, if t N + t0 < N (1 − p) N −1 , t1 + t0 then a parallel transmission is better than a standby one. In general, it would be t1 ≤ t N ≤ N t1 . Thus, if N (1 − p) N −1 < 1, then a standby transmission is better, and if N (1 − p) N −1 > (N t1 + t0 )/(t1 + t0 ), then a parallel

6.4 Modified Models

139

transmission is better. In addition, note that function N (1 − p) N −1 for 0 < p < 1 increases with N until N + 1 ≤ 1/ p and decreases to 0 after that. For example, when p = 0.1, N (0.9) N −1 increases with N until N = 9 and takes a maximum value 9(0.9)8 = 3.874, and decreases to 0 for N ≥ 10. Next, it is assumed that the number N of units is not constant and is a random variable with a probability function pk ≡ Pr{N = k} (k = 1, 2, . . . ) with finite mean β + 1. When t N = t1 , the respective mean transmission times of N units until its success for standby and parallel transmissions are, from (6.56) and (6.57), TS (N ; pk ) =

(t0 + t1 )(β + 1) , 1− p

t 0 + t1 . k k=1 pk (1 − p)

TP (N ; pk ) = ∑∞

In particular, when pk = [β k−1 /(k − 1)!]e−β (k = 1, 2, . . . ) with mean β + 1, TP (N ; pk ) =

t0 + t1 . (1 − p)e−β p

Thus, if eβ p > β + 1, then a standby transmission is better than a parallel one. For β = 36.15. So that if β < ⌃ β − 1, then example, when p = 0.1, e0.1β = β + 1 when ⌃ a standby transmission is better than a parallel one.

(2) Deviation Time We consider the deviation times introduced in Sect. 2.6 of standby and parallel systems with n (n = 1, 2, . . . ) units when they are replaced preventively at time T (0 < T ≤ ∞). Then, replacing F(t) with F (n) (t) in (2.40), the deviation time of a standby system is ʃ

T

D S (T ) =

F (n) (t)dt +

ʃ

0

∞

[1 − F (n) (t)]dt,

(6.58)

T

and optimal TS∗ to minimize D S (T ) satisfies F (n) (T ) =

1 , 2

(6.59)

and the deviation time of a parallel system is ʃ

T

D P (T ) = 0

ʃ

∞

F(t)n dt + T

[1 − F(t)n ]dt,

(6.60)

140

6 Which is Better for Standby or Parallel Systems

and optimal TP∗ to minimize D P (T ) satisfies 1 . 2

F(T )n =

(6.61)

Note that the above results are easily derived by setting G(t) with F (n) (t), F(t)n and L with T , and c S = c E = 1, c O = 0 in (6.23) and (6.24), respectively. Next, introduce the following costs: Let c O be an acquisition cost of each unit, c F be the replacement cost at system failure, and c D be the deviation cost per unit of time. Then, the expected cost rate for a standby system is [2, p. 10] {ʃ } ʃ∞ T nc O + c F F (n) (T ) + c D 0 F (n) (t)dt + T [1 − F (n) (t)]dt C S (T , n) = , ʃT (n) 0 [1 − F (t)]dt (6.62) and for a parallel system, {ʃ } ʃ∞ T nc O + c F F(T )n + c D 0 F(t)n dt + T [1 − F(t)n ]dt . C P (T , n) = ʃT n ]dt [1 − F(t) 0

(6.63)

We find optimal TS∗ and TP∗ to minimize C S (T , n) and C P (T , n) for given n, respectively, when failure rate h(t) of each unit increases with t. Differentiating C S (T , n) with respect to T and setting it equal to zero, [ ʃ c F h S (T ) [ + cD

T

[1 − F (n) (t)]dt − F (n) (T )

0 (n)

F (T ) 1 − F (n) (T )

ʃ

T

]

[1 − F (n) (t)]dt −

0

ʃ

T

] F (n) (t)dt = nc O + c D μ S ,

0

(6.64) and differentiating C P (T , n) with respect to T and setting it equal to zero, [

ʃ

T

c F h P (T ) [ + cD

] [1 − F(t) ]dt − F(T ) n

0

F(T )n 1 − F(T )n

ʃ

T

n

ʃ [1 − F(t) ]dt − n

0

T

] F(t) dt = nc O + c D μ P , n

(6.65)

0

where μ S and μ P in (6.2) and (6.5), h S (T ) and h P (T ) in (6.3) and (6.6), respectively. Thus, noting that both left-hand sides increase strictly with T from 0 to ∞, there exists finite and unique Ti∗ (i = S, P) which satisfy (6.64) and (6.65), respectively, and the resulting cost rates are

6.4 Modified Models

141

cD , 1 − F (n) (TS∗ ) cD C P (TP∗ , n) = c F h P (TP∗ ) − . 1 − F(TP∗ )n C S (TS∗ , n) = c F h S (TS∗ ) −

(6.66) (6.67)

Therefore, computing TS∗ and TP∗ from (6.64) and (6.65), and comparing (6.66) and (6.67), we can decide which standby or parallel system is better.

(3) Random K -out-of-n System Consider a K -out-of-n (K = 0, 1, 2, . . . , n; n = 1, 2, . . . ) system which can operate if and only if at least K units of the total n units are operable [1, p. 216], [2, p 12], [7, p. 155] when each unit has an identical failure distribution F(t) with failure rate h(t), reliability of the system at time t is Rn,K (t) =

n ( ) ∑ n i=K

i

F(t)i F(t)n−i .

(6.68)

Note that when K = n, Rn,n (t) = F(t)n , and when K = 1, Rn,1 (t) = 1 − F(t)n , which correspond to the respective reliabilities of series and parallel systems. Furthermore,∑ when K is a random variable with a probability pi ≡ Pr{K = i} (i = ∑ n pi = 1, and Pi ≡ Pr{K ≤ i} = ij=0 p j (i = 0, 1, 2, . . . , n), 0, 1, 2, . . . , n), i=0 where P0 ≡ 0 and Pn ≡ 1. Note that Pi represents the probability that when i units are operable, the system is operable, and increases with i from 0 to 1. Then, reliability of the system at time t is [10], [11] Rn,P (t) =

∞ ∑ j=0

pj

∞ ( ) ∑ n i= j

i

F(t)i F(t)n−i =

n ∑ i=0

Pi

( ) n F(t)i F(t)n−i , i

(6.69)

and its failure distribution is Fn,P (t) = 1 − Rn,P (t) =

n ∑ i=0

( ) n Pi F(t)i F(t)n−i , i

(6.70)

where P i = 1 − Pi . Thus, the mean time to failure (MTTF) is μn,P =

n ∑ i=0

( )ʃ ∞ n Pi F(t)i F(t)n−i dt, i 0

(6.71)

142

6 Which is Better for Standby or Parallel Systems

and failure rate is ( ) ∑n nh(t) i=1 pi n−1 F(t)i F(t)n−i = . h n,P (t) = (n )i−1 ∑n i n−i Rn,P (t) i=1 Pi i F(t) F(t) ' −Rn,P (t)

(6.72)

In particular, when F(t) = 1 − e−λt , (6.71) is 1 ∑ Pi , λ i=1 i n

μn,P =

(6.73)

and when F(t) = 1 − e−(λt) , it is approximately m

⌃ μn,P

( n )1/m 1 ∑ Pi = , λ i=1 i

(6.74)

Furthermore, using the entropy model [2, p. 199], a system complexity is defined as Hp ≡ −

n ∑

pi log2 pi ,

(6.75)

i=1

where pi log2 pi ≡ 0 when pi = 0. This shows that any redundant systems with identical units have the above reliability properties. Example 6.8 As an example of redundant systems, we give a bridge system with 5 units in Fig. 6.1. Then, P1 = 0,

P2 =

1 , 5

P3 =

4 , 5

P4 = P5 = 1,

and p1 = 0,

p2 =

Fig. 6.1 Bridge system with 5 units

1 , 5

p3 =

3 , 5

p4 =

1 , 5

p5 = 0.

1

2 3

4

5

References

143

Thus, reliability of the system is, from (6.69), ( ) ( ) ( ) ( ) 4 5 1 5 5 5 F(t)2 F(t)3 + F(t)3 F(t)2 + F(t)4 F(t) + F(t)5 5 2 5 3 4 5 [ ] = F(t)2 1 + 2F(t) + F(t)2 − 2F(t)3 ,

R5 (t) =

and when F(t) = 1 − e−λt , MTTF is, from (6.73), μ5 =

1 λ

(

1 1 4 1 1 1 × + × + + 5 2 5 3 4 5

) =

49 . 60λ

Furthermore, the complexity is, from (6.75), 1 1 3 3 1 1 H5 = − log2 − log2 − log2 ≈ 1.371. 5 5 5 5 5 5 Note that the complexity takes a maximum when pi = 1/n and is given by Hn = −

n ∑ 1 1 log2 = log2 n. n n i=1

In this case, log2 5 = 3.219, and the measure of complexity for a bridge system is H5 / log2 5 ≈ 0.426. ∎ Therefore, replacing F(t) with Fn,P (t) in (6.70), we could give WIB problems for any redundant systems with identical units and obtain their comparative results theoretically and numerically.

6.5 Problems 1. Consider replacement first and last of age replacement in Sects. 2.2 and 2.3 for a parallel system with constant and random number of n units. 2. Consider periodic replacement first and last of age replacement in Sect. 4.1 for a parallel system with constant and random number of n units.

References 1. Barlow RE, Proschan F (1965) Mathematical theory of reliability. Wiley, New York 2. Nakagawa T (2008) Advanced reliability models and maintenance policies. Springer, London

144

6 Which is Better for Standby or Parallel Systems

3. Safaei N, Tavakkoli-Moghaddam R, Sassani F (2009) A series-parallel redundant reliability system for cellular manufacturing design. J Risk Reliab 223:233–250 4. Chaturvedi SK, Basha SH, Amari SV, Zuo MJ (2012) Reliability analysis of generalized multistate k-out-of-n systems. J Risk Reliab 226:327–336 5. Zhao X, Chen M, Nakagawa T (2016) Comparisons of standby and parallel systems in reliability, replacement, scheduling and application. J Risk Reliab 230:101–108 6. Nakagawa T, Zhao X (2012) Optimization problems of a parallel system with a random number of units. IEEE Trans Reliab 543–548 7. Nakagawa T (2014) Random maintenance policies. Springer, London 8. Chen M, Zhao X, Nakagawa T (2012) Optimal scheduling of random works with reliability applications. Asia Pac J Oper Res 29:1–14 9. Yasui K, Nakagawa T, Sandoh H (2002) Reliability models in data communication systems. In: Osaki S (ed) Stochastic models in reliability and maintenance. Springer, Berlin, pp 281–301 10. Ito K, Zhao X, Nakagawa T (2017) Random number of units for k-out-of-n. Appl Math Comput 45:563–572 11. Ito K, Nakagawa T (2019) Reliability properties of K -out-of-n: G systems. In: Ram M, Dohi T (eds) System engineering reliability analysis using k-out-of-n structures. CRC Press, Boca Raton, FL, pp 25–40

Chapter 7

Which is Better Problems in Shock and Damage Models

We consider an operating unit suffered for the total damage due to random shocks which should operate for an infinite time span [1–5]. It is of great importance to adopt suitable maintenance plans introduced in Chaps. 2–5 to avoid serious failures when the total damage has exceeded a failure threshold level K (0 < K < ∞). We propose the following three preventive replacement policies [1, p. 40]: 1. The unit is replaced at a planned time T of its age, which is the easiest method. 2. Counting the number of shocks, the unit is replaced at a number N of shocks. 3. Investigating the amount of total damage at shock times, the unit is replaced at a damage level Z (0 ≤ Z ≤ K ). The above policies indicate that the unit is replaced correctively at failure K , and is replaced preventively at time T , shock N or damage Z . Note that the order of amount of information on three policies is damage Z , shock N and time T , and on the other hand, the order of readiness is time T , shock N and damage Z . Chapter 7 is written mainly based on the results [6]: In Sect. 7.1, we obtain the expected cost rates of three combined replacement policies with time T , shock N and damage Z , and derive analytically optimal policies to minimize the expected cost rates of each policy. In Sect. 7.2, comparing two policies with (T , N ), (T , Z ) and (N , Z ), we obtain their expected cost rates and determine which policy is better. In Sect. 7.3, when failure and shock times are exponential, we discuss optimal policies and give the solutions of which policy is better. In Sect. 7.4, we propose two replacement overtime with time T and shock N , and in Sect. 7.5, three modified models of replacement last, periodic replacement and replacement middle, and derive their expected cost rates. Furthermore, we obtain the deviation times of shock and damage models, using the method of Sect. 2.6 and discuss optimal policies [7].

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Mizutani et al., Which-Is-Better (WIB): Problems in Reliability Theory, Springer Series in Reliability Engineering, https://doi.org/10.1007/978-3-031-27316-2_7

145

146

7 Which is Better Problems in Shock and Damage Models

7.1 Three Replacement Policies Consider the standard shock and damage model [1, 2]: Shocks occur at a renewal process according to an identical ∫ ∞ distribution F(t) with a density function f (t) ≡ dF(t)/dt, finite mean μ ≡ 0 F(t)dt, and failure rate h(t) ≡ f (t)/ F(t) which increases from h(0) to h(∞) ≡ limt→∞ h(t), where Φ(t) ≡ 1 − Φ(t) for any distribution Φ(t). Thus, letting N (t) denote the number of shocks during the interval [0, t], the probability that n (n = 0, 1, 2, . . . ) shocks occur exactly in [0, t] is Pr{N (t) = n} = F (n) (t) − F (n+1) (t), where Φ (n) (t) is ∫ tthe n-fold Stieltjes convolution of any distribution Φ(t) with itself, i.e., Φ (n) (t) ≡ 0 Φ (n−1) (t − u)dΦ(u) (n = 1, 2, . . . ) and Φ (0) (t) ≡ 1 for t ≥ 0. Furthermore, an amount Wn (n = 1, 2, . . . ) of damage due to the nth ∫ ∞ shock has an identical distribution G(x) ≡ Pr{Wn ≤ x} with finite mean 1/ω ≡ 0 G(x)dx and the total damage W (t) at time t is additive to the current W0 ≡ 0. It is assumed that Σ N (t) Wn . Then, the distribution of the total damage W (t) damage level, i.e., W (t) ≡ n=0 at time t is ⎧ ⎫ ∞ n ⎨Σ ⎬ Σ Pr W j ≤ x Pr{N (t) = n} Pr{W (t) ≤ x} = ⎩ ⎭ n=0

j=0

∞ Σ [F (n) (t) − F (n+1) (t)]G (n) (x). =

(7.1)

n=0

In addition, the unit fails when the total damage has exceeded a failure level K (0 < K < ∞) at some shock, and its failure is detected immediately and is replaced with a new one. Suppose that the unit is replaced at time T (0 < T ≤ ∞), at shock number N (N = 1, 2, . . . ), at damage level Z (0 < Z ≤ K ), or at a failure level K , whichever occurs first. In addition, the unit is replaced at damage Z and K when the total damage has exceeded Z and K at shock N , respectively. Then, the probability that the unit is replaced at time T is N −1 Σ

[F (n) (T ) − F (n+1) (T )]G (n) (Z ),

n=0

the probability that it is replaced at shock N is F (N ) (T )G (N ) (Z ), the probability that it is replaced at damage Z is

7.1 Three Replacement Policies N −1 Σ

F (n+1) (T )

147

∫

Z

[G(K − x) − G(Z − x)]dG (n) (x),

0

n=0

and the probability that it is replaced at failure K is N −1 Σ

F

(n+1)

∫

Z

(T )

G(K − x)dG (n) (x).

0

n=0

Thus, the mean time to replacement is T

N −1 Σ

G (n) (Z )[F (n) (T ) − F (n+1) (T )] + G (N ) (Z )

N −1 ∫ Z Σ n=0

+

n=0

=

[G(K − x) − G(Z − x)]dG (n) (x)

0

t dF (N ) (t)

∫

T

t dF (n+1) (t)

0

N −1 ∫ Z Σ

N −1 Σ

T 0

n=0

+

∫

G(K − x)dG (n) (x)

0

∫

T

t dF (n+1) (t)

0

G (n) (Z )

∫

T

[F (n) (t) − F (n+1) (t)]dt.

(7.2)

0

n=0

Therefore, the expected cost rate is C F (T , N , Z ) = cT + (c N − cT )F (N ) (T )G (N ) (Z ) ∫Z Σ N −1 (n+1) + (c K − cT ) n=0 F (T ) 0 G(K − x)dG (n) (x) ∫Z Σ N −1 (n+1) + (c Z − cT ) n=0 F (T ) 0 [G(K − x) − G(Z − x)]dG (n) (x) , ∫T Σ N −1 (n) (n) (n+1) (t)]dt n=0 G (Z ) 0 [F (t) − F

(7.3)

where cT = replacement cost at time T , c N = replacement cost at shock N , c Z = replacement cost at damage Z and c K = replacement cost at failure K , where c K > ci (i = T , N , Z ). In particular, when the unit is replaced only at failure K , C ≡ C F (∞, ∞, K ) =

cK , μ[1 + MG (K )]

(7.4)

Σ (n) where MG (x) ≡ ∞ n=1 G (x) represents the expected number of shocks at damage level x. When the unit is replaced preventively at the first shock, C1 ≡ C F (∞, 1, K ) =

1 [c K − (c K − c N )G(K )]. μ

(7.5)

148

7 Which is Better Problems in Shock and Damage Models

When the unit is replaced preventively at time T (0 < T ≤ ∞) is C(T ) ≡ C F (T , ∞, K )

Σ (n+1) cT + (c K − cT ) ∞ (T )[G (n) (K ) − G (n+1) (K )] n=0 F . = ∫T Σ∞ (n) (n) (n+1) (t)]dt n=0 G (K ) 0 [F (t) − F

(7.6)

When the unit is replaced preventively at shock N (N = 1, 2, . . . ) is C(N ) ≡ C F (∞, N , K ) =

c K − (c K − c N )G (N ) (K ) , Σ N −1 (n) μ n=0 G (K )

(7.7)

which agrees with C1 when N = 1. When the unit is replaced preventively at damage Z (0 < Z ≤ K ) is ∫Z c K − (c K − c Z )[G(K ) − 0 G(K − x)dMG (x)] . μ[1 + MG (Z )] (7.8) We find optimal T ∗ , N ∗ and Z ∗ to minimize C(T ) in (7.6), C(N ) in (7.7) and C(Z ) in (7.8), respectively: C(Z ) ≡ C F (∞, ∞, Z ) =

(1) Optimal T ∗ Differentiating C(T ) with respect to T and setting it equal to zero, ∼1 (T ) Q −

∞ Σ

n=0 ∞ Σ

∫

T

[F (n) (t) − F (n+1) (t)]dt

0

F (n+1) (T )[G (n) (K ) − G (n+1) (K )] =

n=0

where

G (n) (K )

cT , c K − cT

(7.9)

Σ N −1 (n+1) (T )[G (n) (K ) − G (n+1) (K )] n=0 f ∼ , Q 1 (T , N ) ≡ Σ N −1 (n) (n+1) (T )]G (n) (K ) n=0 [F (T ) − F Σ∞ (n+1) (T )[G (n) (K ) − G (n+1) (K )] n=0 f ∼1 (T , ∞) = Σ ∼1 (T ) ≡ Q , Q ∞ (n) (n+1) (T )]G (n) (K ) n=0 [F (T ) − F

where f (n+1) (t) ≡ dF (n+1) (t)/dt. It has been shown that if h(t) increases with t, then f (n+1) (t)/[F (n) (t) − F (n+1) (t)] increases with t [8, p. 36]. In particular, when F(t) = 1 − e−λt , h(t) = f (n+1) (t)/[F (n) (t) − F (n+1) (t)] = λ for any n and t ≥ 0. ∼1 (∞), then the left-hand side of (7.9) ∼1 (T ) increases strictly with T to Q If Q ∼1 (∞)[1 + MG (K )] − 1. increases strictly to μ Q

7.1 Three Replacement Policies

149

Thus, if ∼1 (∞)[1 + MG (K )] > Q

cK , (c K − cT )μ

then there exits a finite and unique T ∗ (0 < T ∗ < ∞) which satisfies (7.9), and the resulting cost rate is ∼1 (T ∗ ). (7.10) C(T ∗ ) = (c K − cT ) Q Conversely, if ∼1 (∞)[1 + MG (K )] ≤ Q

cK , (c K − cT )μ

then T ∗ = ∞, i.e., the unit is replaced only at failure, and the resulting cost rate is given by C in (7.4). (2) Optimal N ∗ Forming the inequality C(N + 1) − C(N ) ≥ 0, r N (K )

N −1 Σ

G (n) (K ) − [1 − G (N ) (K )] ≥

n=0

cN , cK − cN

(7.11)

where for 0 < x ≤ K , r N (x) ≡

G (N ) (x) − G (N +1) (x) . G (N ) (x)

If r N (x) increases strictly with N to lim N →∞ r N (x) ≡ r∞ (K ), then the left-hand side of (7.11) increases strictly with N to r∞ (K )[1 + MG (K )] − 1. Thus, if r∞ (K )[1 + MG (K )] >

cK , cK − cN

then there exists a finite and unique minimum N ∗ (1 ≤ N ∗ < ∞) which satisfies (7.11). Conversely, if r∞ (K )[1 + MG (K )] ≤

cK , cK − cN

then N ∗ = ∞, i.e., the unit is replaced only at failure, and the resulting cost rate is given by C in (7.4). Noting that r N (K ) represents the probability that the unit surviving at the N th shock will fail at the next shock, it might increase with N to 1.

150

7 Which is Better Problems in Shock and Damage Models

(3) Optimal Z ∗ Differentiating C(Z ) with respect to Z and setting it equal to zero, ∫

K K −Z

[1 + MG (K − x)]dG(x) =

cZ , cK − cZ

(7.12)

whose left-hand side increases strictly with Z to MG (K ). Thus, if MG (K ) > c Z /(c K − c Z ), then there exists a finite and unique Z ∗ (0 < Z ∗ < K ) which satisfies (7.12), and the resulting cost rate is C(Z ∗ ) = (c K − c Z )

G(K − Z ∗ ) . μ

(7.13)

Conversely, MG (K ) ≤ c Z /(c K − c Z ), then Z ∗ = K , i.e., the unit is replaced only at failure, and the resulting cost rate is given by C in (7.4).

7.2 Optimal Policies with Two Variables To compare the above three preventive replacement policies when cT = c N = c Z , we obtain the expected cost rates with two variables, derive optimal policies and compare them analytically and numerically. (1) Optimal TF∗ and N F∗ When the unit is replaced preventively at time T or at shock N , whichever occurs first, the expected cost rate is, from (7.3), C F (T , N ) ≡ C F (T, N , K )

Σ N −1 (n+1) cT + (c K − cT ) n=0 F (T )[G (n) (K ) − G (n+1) (K )] . = ∫T Σ N −1 (n) (n) (n+1) (t)]dt n=0 G (K ) 0 [F (t) − F

(7.14)

We find optimal TF∗ and N F∗ to minimize C F (T, N ). Forming the inequality C F (T , N − 1) − C F (T , N ) > 0, ∫T 0

r N −1 (K )F (N ) (T ) [F (N −1) (t)

−

N −1 Σ n=0

−

F (N ) (t)]dt

N −1 Σ n=0

G (n) (K )

∫

T

[F (n) (t) − F (n+1) (t)]dt

0

F (n+1) (T )[G (n) (K ) − G (n+1) (K )]
∫ T 0

r N −1 (K )F (N ) (T ) [F (N −1) (t) − F (N ) (t)]dt

.

(7.17)

Thus, if the inequality (7.17) does not hold for any N , there does not exist any finite TF∗ which satisfies (7.17), i.e., optimal policy is TF∗ = ∞ and N F∗ = N ∗ given in (7.11). For example, when F(t) = 1 − e−λt and r N (x) increases strictly with N , (7.17) is Σ N −1 [(λT )n /n!][G (n) (K ) − G (n+1) (K )] > r N −1 (K ), Q 1 (T , N ) ≡ n=0 Σ N −1 n (n) n=0 [(λT ) /n!]G (K ) where note that Q 1 (T, N ) = G(K ) for N = 1 and increases strictly with T for N ≥ 2 from G(K ) to r N −1 (K ), and Q 1 (T, N ) ≤ r N −1 (K ). Thus, the above inequality does not hold for any finite T , which follows that if optimal N F∗ satisfies (7.15), then the left-hand side of (7.16) is less than cT /(c K − cT ), and TF∗ = ∞. This concludes that if inequality (7.17) does not hold for any N , then optimal policy is TF∗ = ∞ and N F∗ = N ∗ given in (7.11). Replacement with shock N is better than replacement with time T . Next, we derive optimal TF∗ for given N (1 ≤ N < ∞) when F(t) = 1 − e−λt and cT = c N . Then, (7.16) is Q 1 (T, N )

N −1 Σ

F

n=0

(n+1)

(T )G

(n)

(K ) −

N −1 Σ

F (n+1) (T )[G (n) (K ) − G (n+1) (K )]

n=0

cT = , c K − cT

(7.18)

whose left-hand side increases strictly with T from 0 to r N −1 (K )

N −1 Σ n=0

G (n) (K ) − [1 − G (N ) (K )] < r N (K )

N −1 Σ n=0

G (n) (K ) − [1 − G (N ) (K )],

152

7 Which is Better Problems in Shock and Damage Models

which agrees with that of (7.11). Thus, if N > N ∗ in (7.11), then there exists a finite and unique TF∗ (0 < TF∗ < ∞) which satisfies (7.18), and the resulting cost rate is C F (TF∗ , N ) = (c F − cT )λQ 1 (TF∗ , N ).

(7.19)

Conversely, if N ≤ N ∗ , then TF∗ = ∞. (2) Optimal TF∗ and Z ∗F When the unit is replaced preventively at time T or at damage Z , whichever occurs first, the expected cost rate is, from (7.3), C F (T , Z ) ≡ C F (T, ∞, Z )

∫Z Σ (n+1) cT + (c K − cT ) ∞ (T ) 0 G(K − x)dG (n) (x) n=0 F . = ∫T Σ∞ (n) (n) (n+1) (t)]dt n=0 G (Z ) 0 [F (t) − F

(7.20)

We find optimal TF∗ and Z ∗F to minimize C F (T, Z ). Differentiating C F (T , Z ) with respect to Z and setting it equal to zero, Q 2 (T , Z )

∞ Σ

G (n) (Z )

−

F (n+1) (T )

∫

[F (n) (t) − F (n+1) (t)]dt

Z

G(K − x)dG (n) (x) =

0

n=0

where

T 0

n=0 ∞ Σ

∫

cT , c K − cT

(7.21)

Σ (n) (n+1) G(K − Z ) ∞ (T ) n=0 g (Z )F Q 2 (T, Z ) ≡ Σ∞ , ∫T (n) (n) (n+1) (t)]dt n=0 g (Z ) 0 [F (t) − F

where g (n) (x) ≡ dG (n) (x)dx (n = 1, 2, . . . ) and g (0) (x) ≡ 0. Differentiating C F (T , Z ) with respect to T and setting it equal to zero, Q 3 (T , Z )

∞ Σ

G

(n)

∫

n=0

−

∞ Σ n=0

where

T

(Z )

F (n+1) (T )

[F (n) (t) − F (n+1) (t)]dt

0

∫

Z 0

G(K − x)dG (n) (x) =

cT , c K − cT

∫Z Σ∞ (n+1) (T ) 0 G(K − x)dG (n) (x) n=0 f . Q 3 (T , Z ) ≡ Σ∞ (n) (n+1) (T )]G (n) (Z ) n=0 [F (T ) − F

(7.22)

(7.23)

7.2 Optimal Policies with Two Variables

153

Substituting (7.21) for (7.22) is Q 3 (T, Z ) = Q 2 (T, Z ).

(7.24)

Thus, if Q 3 (T, Z ) < Q 2 (T , Z ) for any Z , then there does not exist any finite TF∗ which satisfies (7.22), i.e., optimal policy is TF∗ = ∞ and Z ∗F = Z ∗ given in (7.12). For example, when F(t) = 1 − e−λt , Q 2 (T, Z ) = λG(K − Z ) and Q 3 (T, Z ) = λ

∫Z n (n) n=0 [(λT ) /n!] 0 G(K − x)dG (x) Σ∞ n (n) n=0 [(λT ) /n!]G (Z )

Σ∞

< G(K − Z )

for any Z , i.e., Q 3 (T , Z ) < Q 2 (T , Z ) for any Z . Thus, if optimal Z ∗F satisfies (7.21), then the left-hand side of (7.22) is less than cT /(c K − cT ), and TF∗ = ∞. This concludes that if the equality (7.24) does not hold for any Z , then optimal policy is TF∗ = ∞ and Z ∗F = Z ∗ given in (7.12). Replacement with damage Z is better than replacement with time T . Next, we derive optimal TF∗ for given Z (0 < Z ≤ K ) when F(t) = 1 − e−λt and cT = c Z . Then, (7.22) is ∫ Z ∞ ∞ Σ Q 3 (T, Z ) Σ (n+1) (n) (n+1) F (T )G (Z ) − F (T ) G(K − x)dG (n) (x) λ 0 n=0 n=0 cT = . (7.25) c K − cT If Q 3 (T , Z )/λ increases strictly with T to G(K − Z ), then the left-hand side of (7.25) increases strictly with T from 0 to ∫

K K −Z

[1 + MG (K − x)]dG(x),

which agrees with that of (7.12). Thus, if Z > Z ∗ given in (7.12), then there exists a finite and unique TF∗ (0 < TF∗ < ∞) which satisfies (7.25). Conversely, if Z ≤ Z ∗ , then TF∗ = ∞. (3) Optimal N F∗ and Z ∗F When the unit is replaced preventively at shock N or at damage Z , whichever occurs first, the expected cost rate is, from (7.3), C F (N , Z ) ≡ C F (∞, N , Z ) =

c N + (c K − c N ) μ

Σ N −1 ∫ Z n=0

Σ N −1 n=0

0

G(K − x)dG (n) (x)

G (n) (Z )

.

(7.26)

154

7 Which is Better Problems in Shock and Damage Models

We find optimal N F∗ and Z ∗F to minimize C F (N , Z ). Differentiating C F (N , Z ) with respect to Z and setting it equal to zero, N −1 ∫ Σ n=0

Z

[G(K − Z ) − G(K − x)]dG (n) (x) =

0

cN . cK − cN

(7.27)

Forming the inequality C F (N + 1, Z ) − C F (N , Z ) ≥ 0, Σ N −1

(n) n=0 G (Z ) G (N ) (Z )

∫

Z

G(K − x)dG (N ) (x) −

0

N −1 ∫ Σ n=0

Z

G(K − x)dG (n) (x)

0

cN ≥ . cK − cN

(7.28)

Substituting (7.27) for (7.28) is ∫

Z

[G(K − Z ) − G(K − x)]dG (N ) (x) ≤ 0,

(7.29)

0

which does not hold for any Z . Thus, if optimal Z ∗F satisfies (7.27), then the lefthand of (7.28) is less than c N /(c K − c N ), and N F∗ = ∞. This concludes that if the inequality (7.29) does not hold for any Z , then optimal policy is N F∗ = ∞ and Z ∗F = Z ∗ given in (7.12). Replacement with damage Z is better than replacement with shock N . Next, we derive optimal N F∗ for given Z (0 < Z ≤ K ) when G(x) = 1 − e−ωx and c N = c Z . Then, (7.28) is [ e

−ω(K −Z )

r N (Z )

N −1 Σ

] G

(n)

(Z ) − 1 + G

(N )

n=0

(Z ) ≥

cN , cK − cN

(7.30)

whose left-hand increases strictly with N to ω Z e−ω(K −Z ) , which agrees with that of (7.12). Thus, if Z > Z ∗ in (7.12), then there exists a finite and unique minimum N F∗ (1 ≤ N F∗ < ∞) which satisfies (7.30). Conversely, if Z ≤ Z ∗ , then N F∗ = ∞.

7.3 Exponential Shock Times and Amount of Damage It is assumed that F(t) = 1 − e−λt and G(x) = 1 − e−ωx . Then,

7.3 Exponential Shock Times and Amount of Damage

F (n) (t) =

∞ Σ (λt) j j=n

j!

e−λt ,

G (n) (x) =

155

∞ Σ (ωx) j

j!

j=n

(ωx) N /N ! r N (x) = Σ∞ j j=N [(ωx) /j!]

e−ωx

(n = 0, 1, 2, . . . ),

(N = 0, 1, 2, . . . )

increases strictly with N from e−ωx to 1. The expected cost rate is, from (7.3), cT + (c N − cT )F (N ) (T )G (N ) (Z ) Σ N −1 (n+1) + (c K − cT ) n=0 F (T )[(ω Z )n /n!]e−ωK Σ N −1 (n+1) −ω Z −ωK + (c Z − cT )(e −e ) n=0 F (T )[(ω Z )n /n!] C F (T, N , Z ) = . Σ N −1 (n+1) λ (T )G (n) (Z ) n=0 F (7.31) We survey again optimal policies with one variable T ∗ , N ∗ and Z ∗ discussed in Sect. 7.1 and two variable (TF∗ , N F∗ ), (TF∗ , Z ∗F ) and (N F∗ , Z ∗F ) discussed in Sect. 7.2 for different replacement cost cT , c N and c Z . (1) Optimal T ∗ From (7.6) and (7.31), Σ cT + (c K − cT ) ∞ F (n+1) (T )[(ωK )n /n!]e−ωK C(T ) Σ∞ n=0 = . (n+1) (T )G (n) (K ) λ n=0 F

(7.32)

Differentiating C(T ) with respect T and setting it equal to zero, Q 1 (T )

∞ Σ

F (n+1) (T )G (n) (K ) −

n=0

where

∞ Σ n=0

F (n+1) (T )

(ωK )n −ωK cT e = , (7.33) n! c K − cT

Σ∞

n n n=0 [(λT ) /n!][(ωK ) /n!] Σ ∞ n j n=0 [(λT ) /n!] j=n [(ωK ) /j!]

Q 1 (T ) = Σ∞

increases strictly with T from e−ωK to 1. Thus, the left-hand side of (7.33) increases strictly T from 0 to ωK . Therefore, if ωK > cT /(c K − cT ), then there exists a finite and unique T ∗ (0 < T ∗ < ∞) which satisfies (7.33), and the resulting cost rate is C(T ∗ ) = (c K − cT )Q 1 (T ∗ ). λ

(7.34)

156

7 Which is Better Problems in Shock and Damage Models

(2) Optimal N ∗ From (7.7) and (7.31), C(N ) c N + (c K − c N )[1 − G (N ) (K )] = . Σ N −1 (n) λ n=0 G (K )

(7.35)

Forming the inequality C(N + 1) − C(N ) ≥ 0, r N (K )

N −1 Σ

G (n) (K ) − [1 − G (N ) (K )] ≥

n=0

cN , cK − cN

(7.36)

whose left-hand side of (7.36) increases strictly with N to ωK . Therefore, if ωK > c N /(c K − c N ), then there exists a finite and unique minimum N ∗ (1 ≤ N ∗ < ∞) which satisfies (7.36), and the resulting cost rate is C(N ∗ ) in (7.35). (3) Optimal Z ∗ From (7.8) and (7.31), c Z + (c K − c Z )e−ω(K −Z ) C(Z ) = . λ 1 + ωZ

(7.37)

Differentiating C(Z ) with respect to Z and setting it equal to zero, ω Z e−ω(K −Z ) =

cZ , cK − cZ

(7.38)

whose left-hand side increases strictly with Z from 0 to ωK . Therefore, if ωK > c Z /(c K − c Z ), then there exists a finite and unique Z ∗ (0 < Z ∗ < K ) which satisfies (7.38), and the resulting cost rate is C(Z ∗ ) ∗ = (c K − c Z )e−ω(K −Z ) . λ

(7.39)

It is of great interest that if ωK > ci /(c K − ci ) (i = T , N , Z ), then finite T ∗ , N ∗ and Z ∗ always exist uniquely. Conversely, if ωK ≤ ci /(c K − ci ), then T ∗ = N ∗ = ∞, Z ∗ = K , and C(K ) = C in (7.4), where note that MG (K ) = ωK represents the expected number of shocks until the total damage exceeds a failure level K . Example 7.1 We compute optimal λT ∗ , N ∗ and ω Z ∗ when F(t) = 1 − e−λt and G(x) = 1 − e−ωx . Tables 7.1–7.3 present optimal λT ∗ , N ∗ and ω Z ∗ , and their cost rates C(T ∗ )/(λcT ), C(N ∗ )/(λc N ) and C(Z ∗ )/(λc Z ) for ωK , c K /cT , c K /c N and c K /c Z . All of λT ∗ , N ∗ and ω Z ∗ are finite and exist uniquely because

7.3 Exponential Shock Times and Amount of Damage

157

Table 7.1 Optimal λT ∗ and C(T ∗ )/(λcT ) ωK = 5

cK cT

λT ∗

ωK = 10

C(T ∗ )/(λcT ) λT ∗

ωK = 20

C(T ∗ )/(λcT ) λT ∗

C(T ∗ )/(λcT )

5

3.328

0.617

5.750

0.260

11.835

0.106

10

2.221

0.910

4.378

0.334

9.971

0.124

15

1.804

1.130

3.805

0.382

9.137

0.134

20

1.567

1.317

3.460

0.420

8.617

0.142

30

1.292

1.637

3.040

0.479

7.962

0.153

50

1.019

2.162

2.594

0.564

7.235

0.168

Table 7.2 Optimal N ∗ and C(N ∗ )/(λc N ) ωK = 10

ωK = 5

ωK = 20

cK cN

N∗

C(N ∗ )/(λc N )

N∗

C(N ∗ )/(λc N )

N∗

5

3

0.508

6

0.213

13

0.089

10

2

0.684

5

0.253

12

0.100

C(N ∗ )/(λc N )

15

2

0.786

5

0.283

11

0.105

20

2

0.887

4

0.299

10

0.110

30

2

1.090

4

0.325

10

0.115

50

1

1.330

4

0.377

9

0.122

Table 7.3 Optimal ω Z ∗ and C(Z ∗ )/(λc Z ) ωK = 5

ωK = 20

ωK = 10

cK cZ

ωZ∗

C(Z ∗ )/(λc Z ) ω Z ∗

C(Z ∗ )/(λc Z ) ω Z ∗

5

2.642

0.378

6.710

0.149

15.850

0.063

10

2.073

0.482

6.009

0.166

15.089

0.066

C(Z ∗ )/(λc Z )

15

1.783

0.561

5.632

0.178

14.675

0.068

20

1.591

0.628

5.374

0.186

14.389

0.069

30

1.340

0.746

5.019

0.199

13.994

0.071

50

1.055

0.948

4.585

0.218

13.505

0.074

ωK > ci /(c K − ci ) (i = T , N , Z ). In addition, all of λT ∗ , N ∗ and ω Z ∗ increase with ωK and decrease with c K /ci , and their cost rates decrease with ωK and increase with c K /ci . When ωK = 5 and 10, optimal λT ∗ , N ∗ and ω Z ∗ are almost the same, and when ωK = 20, λT ∗ < N ∗ < ω Z ∗ . This means that when a failure level K becomes smaller or its cost c K becomes larger, preventive replacements should be done early and their cost rates become higher. When cT = c N = c Z , C(T ∗ ) > C(N ∗ ) > C(Z ∗ ), as shown previously. ∎

158

7 Which is Better Problems in Shock and Damage Models

7.4 Replacement Overtime Policies We consider the following two overtime policies: The unit is replaced at the first shock over time T and over damage level Z [6, p. 72].

(1) Replacement Overtime with T Suppose that the unit is replaced at failure level K or at the first shock over time T (0 ≤ T < ∞), whichever occurs first, i.e., it is replaced when the total damage has exceeded K before time T and is replaced at the first shock over time T . Then, the probability that the unit is replaced at the first shock over time T is ∞ Σ [F (n) (T ) − F (n+1) (T )]G (n+1) (K ), n=0

and the probability that it is replaced at failure K is ∞ Σ

F (n) (T )[G (n) (K ) − G (n+1) (K )].

n=0

Thus, the mean time to replacement is {∫ ∞ Σ [G (n) (K ) − G (n+1) (K )] n=0

]∫

0

∫

T

+

t dF 0

=μ

T

∞ Σ

(n+1)

} (t) +

∞ Σ

G

(n+1)

n=0

∞ T −t

] (t + u) dF(u) dF (n) (t) ∫

T

(K ) 0

]∫

∞ T −t

]

(t + u) dF(u) dF (n) (t)

F (n) (T )G (n) (K ).

n=0

Therefore, the expected cost rate is μC O (T ) =

Σ c O + (c K − c O ) ∞ F (n) (T )[G (n) (K ) − G (n+1) (K )] Σ∞n=0 (n) , (n) n=0 F (T )G (K )

(7.40)

where c O = replacement cost over time T where c O < c K . Clearly, C O (∞) = C in (7.4) and C O (0) = C1 in (7.5) when c O =Σc N . j −λt (n = 0, 1, 2, . . . ), we When F(t) = 1 − e−λt , i.e., F (n) (t) = ∞ j=n [(λt) /j!]e ∗ find optimal TO to minimize C O (T ). Differentiating C O (T ) with respect to T and setting it equal to zero,

7.4 Replacement Overtime Policies

Q 2 (T )

∞ Σ

F (n) (T )G (n) (K ) −

n=0

159 ∞ Σ

F (n) (T )[G (n) (K ) − G (n+1) (K )] =

n=0

where Q 2 (T ) ≡

cO , cK − cO (7.41)

Σ∞

n (n+1) (K ) − G (n+2) (K )] n=0 [(λT ) /n!][G Σ∞ . n (n+1) (K ) n=0 [(λT ) /n!]G

When r N (x) in (7.11) increases strictly with N to 1, Q 2 (T ) increases strictly with T from r1 (K ) to 1 from (1) of Appendix A.7. Thus, because the left-hand side of (7.41) increases strictly with T to MG (K ), if MG (K ) > c O /(c K − c O ) then there exists a finite and unique TO∗ (0 ≤ TO∗ < ∞) which satisfies (7.41), and the resulting cost rate is C O (TO∗ ) = (c K − c O )Q 2 (TO∗ ). (7.42) λ We compare C O (T ) in (7.40) with C(T ) in (7.6) when c O = cT , F(t) = 1 − e−λt and G(x) = 1 − e−ωx . Then, noting that Q 2 (T ) in (7.41) is greater than Q 1 (T ) in (7.33), i.e., Q 2 (T ) > Q 1 (T ), from (7.33) and (7.41), Q 2 (T )

∞ Σ

F (n) (T )G (n) (K ) − Q 1 (T )

n=0

−

∞ Σ

F (n+1) (T )G (n) (K )

n=0

∞ Σ (λT ) −λT (ωK ) −ωK e e = [Q 2 (T ) − Q 1 (T )] F (n) (T )G (n) (K ) > 0, n! n! n=0 n=0

∞ Σ

n

n

which follows that TO∗ < T ∗ . Furthermore, because Q 1 (T )

∞ Σ

F (n+1) (T )G (n) (K ) −

n=0

= Q 2 (T )

∞ Σ n=0

∞ Σ

F (n+1) (T )

n=0 ∞ Σ

F (n) (T )G (n) (K ) −

n=0

F (n) (T )

(ωK )n −ωK e n!

(ωK )n −ωK e , n!

(7.34) is Σ (n) ∗ n −ωK c O + (c K − c O ) ∞ C(T ∗ ) n=0 F (T )[(ωK ) /n!]e Σ = , ∞ (n) ∗ (n) λ n=0 F (T )G (K )

(7.43)

which agrees with (7.40) when c O = cT and T ∗ = T . Thus, from TO∗ < T ∗ , replacement overtime is better than replacement with time T . Next, we compare C O (T ) in (7.40) with C(N ) in (7.37) when c O = c N . For this purpose, we consider the following replacement overtime first: The unit is replaced at shock N (N = 1, 2, . . . ), at the first shock over time T (0 ≤ T < ∞) or at failure K , whichever occurs first. Then, the probability that the unit is replaced at shock N is

160

7 Which is Better Problems in Shock and Damage Models

F (N ) (T )G (N ) (K ), the probability that it is replaced at the first shock over time T is N −1 Σ

[F (n) (T ) − F (n+1) (T )]G (n+1) (K ),

n=0

and the probability that it is replaced at failure K is N −1 Σ

F (n) (T )[G (n) (K ) − G (n+1) (K )].

n=0

Thus, the mean time to replacement is G

(N )

∫

T

(K )

t dF

(N )

(t) +

0

+

N −1 Σ

T

+

G

(n+1)

[G (n) (K ) − G (n+1) (K )]

{∫ 0

}

t dF (n+1) (t) = μ

0

N −1 Σ

∫

T

(K ) 0

n=0

n=0

∫

N −1 Σ

T

]∫

∞ T −t

]∫

∞ T −t

]

(t + u)dF(u) dF (n) (t)

] (t + u)dF(u) dF (n) (t)

F (n) (T )G (n) (K ).

n=0

Therefore, the expected cost rate is μC O F (T , N ) =

Σ N −1 (n) c O + (c K − c O ) n=0 F (T )[G (n) (K ) − G (n+1) (K )] . (7.44) Σ N −1 (n) (n) n=0 F (T )G (K )

Clearly, when c O = c N , C O F (∞, N ) = C(N ) in (7.7), C O F (T, ∞) = C O (T ) in (7.40), and C O F (0, N ) = C O F (T , 1) = C O F (0, 1) = C1 in (7.5). We find optimal TO∗ F and N O∗ F to minimize C O F (T , N ) in (7.44) when F(t) = 1 − e−λt . Forming the inequality C O F (T , N − 1) − C O F (T, N ) > 0, N −1 Σ

F (n) (T )G (n) (K )[r N −1 (K ) − rn (K )]
r N −1 (K ), which does not hold for any N for N ≥ 2, i.e., C O F (T , N ) decreases with T and TO∗ F = ∞. Thus, replacement with shock N is better than replacement overtime with T . (2) Replacement Overtime with Z Suppose that the unit is replaced at the first shock over damage Z (0 ≤ Z ≤ K ) or at failure K , whichever occurs first. Then, the probability that the unit is replaced at the first shock over damage Z is ∞ ∫ Σ n=0

Z

]∫

] G(K − x − y)dG(y) dG (n) (x),

K −x Z −x

0

and the probability that it is replaced at failure K is ∞ {∫ Σ n=0

Z

G(K − x)dG (n) (x) +

∫

0

Z

]∫

K −x Z −x

0

] } G(K − x − y)dG(y) dG (n) (x) .

Thus, the mean time to replacement is {∞ ∫ Σ (n + 2) μ

+

∫

Z

(n + 1)

n=0 ∞ Σ n=0

] G(K − x − y)dG(y) dG (n) (x)

G(K − x)dG (n) (x)

0

∫

]∫

Z

(n + 2)

]

K −x Z −x

0

n=0 ∞ Σ

+

]∫

Z

0

∫

= μ 1 + G(K ) +

K −x Z −x

Z

]

}

G(K − x − y)dG(y) dG (n) (x) ]

G(K − x)dMG (x) .

0

Therefore, the expected cost rate is {∫ K c K − (c K − c O ) Z G(K − x)dG(x) ] } [ ∫ Z ∫ K −x + 0 G(K − x − y)dG(y) dM (x) G Z −x [ ] . C O (Z ) = ∫Z μ 1 + G(K ) + 0 G(K − x)dMG (x)

(7.47)

162

7 Which is Better Problems in Shock and Damage Models

We find optimal Z ∗O to minimize C O (Z ). Differentiating C O (Z ) with respect to Z and setting it equal to zero, ]

∫

r1 (K − Z ) 1 + G(K ) + ∫ + 0

Z

]∫

Z

]

Z −x

K

G(K − x)dMG (x) +

0

K −x

∫

] G(K − x − y)dG(y) dMG (x) − 1 =

G(K − x)dG(x)

Z

cO , cK − cO

(7.48)

where r1 (x) = [G(x) − G (2) (x)]/G(x). If r1 (K − Z ) increases strictly with Z to 1, then the left-hand side of (7.48) increases strictly with Z to MG (K ). Thus, if MG (K ) > c O /(c F − c O ), then there exists a finite and unique Z ∗O (0 ≤ Z ∗O < ∞) which satisfies (7.48), and the resulting cost rate is μC O (Z ∗O ) = (c K − c O )r1 (K − Z ∗O ).

(7.49)

In particular, note that if [G(K )2 − G (2) (K )]/G(K ) ≥ c O /(c K − c O ) then Z ∗O = 0. When F(t) = 1 − e−λt and G(x) = 1 − e−ωx , the expected cost rate is, from (7.47), c O + (c K − c O )[1 + ω(K − Z )]e−ω(K −Z ) C O (Z ) = , (7.50) λ 2 + ω Z − e−ω(K −Z ) and from (7.48), optimal Z ∗O satisfies e

−ω(K −Z )

]

] cO ω(K − Z )(1 + ω Z ) −1 = , −ω(K −Z ) 1−e cK − cO

(7.51)

whose left-hand side increases strictly with Z to ωK . Thus, if ωK > c O /(c K − c O ), then there exists a finite and unique Z ∗O (0 ≤ Z ∗O < K ) which satisfies (7.51), and the resulting cost rate is ∗

C O (Z ∗O ) ω(K − Z ∗ ) c O + (c K − c O )e−ω(K −Z O ) = (c K − c O ) ω(K −Z ∗ ) O = , O − 1 λ 1 + ω Z ∗O e

(7.52)

which agrees with (7.37) when c O = c Z and Z ∗O = Z . We compare C O (Z ) in (7.50) with C(Z ) in (7.37) when c O = c Z . Comparing the left-hand side of (7.51) and (7.38), ]

] ω(K − Z )(1 + ω Z ) − 1 − ω Z e−ω(K −Z ) 1 − e−ω(K −Z ) ] (1 + ω Z )e−ω(K −Z ) [ = ω(K − Z ) − 1 + e−ω(K −Z ) > 0, 1 − e−ω(K −Z )

e−ω(K −Z )

which follows that Z ∗O < Z ∗ . Thus, from (7.52), replacement with Z is better than replacement at the first shock over level Z O .

7.5 Modified Replacement Models

163

Table 7.4 Optimal λTO∗ and C O (TO∗ )/(λc O ) ωK = 5

ωK = 10

ωK = 20

cK cO

λTO∗

C(TO∗ )/(λc O )

λTO∗

C(TO∗ )/(λc O )

λTO∗

5

1.265

0.613

3.683

0.262

9.459

0.108

10

0.611

0.833

2.727

0.327

8.010

0.124

C(TO∗ )/(λc O )

15

0.349

0.969

2.311

0.368

7.341

0.134

20

0.194

1.066

2.056

0.398

6.918

0.141

30

0.009

1.194

1.741

0.444

6.378

0.152

50

0.000

1.981

1.402

0.507

5.772

0.165

Table 7.5 Optimal ω Z ∗O and C O (Z ∗O )/(λc O ) cK cO

ωK = 5 ω Z ∗O

C(Z ∗O )/(λc O )

ωK = 20

ωK = 10 ω Z ∗O

C(Z ∗O )/(λc O )

ω Z ∗O

C(Z ∗O )/(λc O )

5

1.527

0.445

5.248

0.166

14.136

0.067

10

0.876

0.611

4.423

0.191

13.245

0.071

15

0.565

0.745

3.991

0.207

12.770

0.073

20

0.369

0.866

3.700

0.220

12.444

0.075

30

0.124

1.087

3.306

0.241

11.997

0.078

50

0.000

1.662

2.831

0.271

11.450

0.081

Example 7.2 We compute optimal λTO∗ and ω Z ∗O when F(t) = 1 − e−λt and G(x) = 1 − e−ωx . Tables 7.4 and 7.5 present optimal λTO∗ , C O (TO∗ )/(λc O ), ω Z ∗O and C O (Z ∗O )/(λc O ) for ωK and c K /c O . Compared Table 7.4 with Table 7.1, λTO∗ < λT ∗ and C O (TO∗ )/(λc O ) < C(T ∗ )/(λcT ), which shows that when c O = cT , replacement overtime with time TO∗ is better than replacement with time T ∗ , as shown previously. Similarly, compared Table 7.5 with Table 7.3, ω Z ∗O < ω Z ∗ and C O (Z ∗O )/(λc O ) > C(Z ∗ )/(λc Z ), which shows that when c O = c Z , replacement with damage Z ∗ is better than replacement with overlevel Z ∗O . ∎

7.5 Modified Replacement Models We offer four modified replacement policies and give the deviation times for shock and damage models, and consider some other models.

(1) Replacement Last Suppose that the unit is replaced preventively at time T (0 ≤ T < ∞), at shock N (N = 0, 1, 2, . . . ) or at damage Z (0 ≤ Z ≤ K ), whichever occurs last [6, p. 50].

164

7 Which is Better Problems in Shock and Damage Models

Then, the probability that the unit is replaced at time T is ∞ Σ

[F (n) (T ) − F (n+1) (T )][G (n) (K ) − G (n) (Z )],

n=N

the probability that it is replaced at shock N is [1 − F (N ) (T )][G (N ) (K ) − G (N ) (Z )], the probability that it is replaced at damage Z is ∞ Σ

[1 − F (n+1) (T )]

∫

Z

[G(K − x) − G(Z − x)]dG (n) (x),

0

n=N

and the probability that it is replaced at failure K is ∞ Σ

F (n+1) (T )[G (n) (K ) − G (n+1) (K )]

n=0

+

∞ Σ

∫

[1 − F (n+1) (T )]

N −1 Σ

G(K − x)dG (n) (x)

0

n=N

+

Z

[1 − F (n+1) (T )][G (n) (K ) − G (n+1) (K )].

n=0

Thus, the mean time to replacement is T

∞ Σ

[F (n) (T ) − F (n+1) (T )][G (n) (K ) − G (n) (Z )]

n=N

+ [G (N ) (K ) − G (N ) (Z )] + +

∞ ∫ Σ n=N ∞ Σ

∫

∞

t dF (N ) (t)

T Z

[G(K − x) − G(Z − x)]dG (n) (x)

0

[G (n) (K ) − G (n+1) (K )]

+

N −1 Σ n=0

T

t dF (n+1) (t)

0

∞ ∫ Σ n=N

∞ T

∫

n=0

+

∫

Z

G(K − x)dG (n) (x)

∫

t dF (n+1) (t)

T

0

[G (n) (K ) − G (n+1) (K )]

∞

∫

∞ T

t dF (n+1) (t)

t dF (n+1) (t)

7.5 Modified Replacement Models

=

∞ Σ

G (n) (Z )

∫

+

G

(n)

∫

+

∞

(K )

[F (n) (t) − F (n+1) (t)]dt

T

n=0 ∞ Σ

[F (n) (t) − F (n+1) (t)]dt

T

n=N N −1 Σ

∞

165

G (n) (K )

n=0

∫

T

[F (n) (t) − F (n+1) (t)]dt.

0

Therefore, the expected cost rate is C L (T , N , Z ) =

Σ (n) (n+1) c K − (c K − cT ) ∞ (T )][G (n) (K ) − G (n) (Z )] n=N [F (T ) − F (N ) (N ) − (c K − c N )[1 − F (T )][G (K ) − G (N ) (Z )] ∫Z Σ (n+1) − (c K − c Z ) ∞ (T )] 0 [G(K − x) − G(Z − x)]dG (n) (x) n=N [1 − F ∫ ∞ (n) . Σ∞ G (n) (Z ) [F (t) − F (n+1) (t)]dt n=N Σ N −1 (n) T ∫ ∞ (n) + n=0 G (K ) T [F (t) − F (n+1) (t)]dt ∫ T (n) Σ (n) (n+1) (t)]dt + ∞ n=0 G (K ) 0 [F (t) − F (7.53)

Clearly, C L (0, 0, 0) = C F (∞, ∞, K ) = C in (7.4), C L (T , 0, 0) = C F (T, ∞, K ) = C(T ) in (7.6), C L (0, N , 0) = C F (∞, N , K ) = C(N ) in (7.7), C L (0, 0, Z ) = C F (∞, ∞, Z ) = C(Z ) in (7.8), and C L (0, 1, 0) = C F (∞, 1, K ) = C1 in (7.5). When cT = c N and the unit is replaced preventively at time T or at shock N , whichever occurs last, the expected cost rate is C L (T , N ) ≡ C L (T , N , 0) { } Σ (n+1) cT + (c K − cT ) 1 − ∞ (T )][G (n) (K ) − G (n+1) (K )] n=N [1 − F . = ∫∞ Σ N −1 (n) G (K ) T [F (n) (t) − F (n+1) (t)]dt n=0 ∫ T (n) Σ (n) (n+1) + ∞ (t)]dt n=0 G (K ) 0 [F (t) − F (7.54) When cT = c Z and the unit is replaced preventively at time T or at damage Z , whichever occurs last, the expected cost rate is C L (T , Z ) ≡ C L (T , 0, Z ) { Σ (n) (n+1) (n) cT + (c K − cT ) 1 − ∞ (T )][G (n) (K ) − G n=0 [F (T ) − F } (Z )] ∫ Σ∞ Z − n=0 [1 − F (n+1) (T )] 0 [G(K − x) − G(Z − x)]dG (n) (x) ∫∞ = . Σ∞ G (n) (Z ) T [F (n) (t) − F (n+1) (t)]dt n=0 ∫ Σ T (n) (n) (n+1) + ∞ (t)]dt n=0 G (K ) 0 [F (t) − F (7.55)

166

7 Which is Better Problems in Shock and Damage Models

When c N = c Z and the unit is replaced preventively at shock N or at damage Z , whichever occurs last, the expected cost rate is C L (N , Z ) ≡ C L (0, N , Z ) =

∫Z Σ (n) c N + (c K − c N )[1 − G (N ) (K ) + ∞ n=N 0 G (K − x)dG (x)] [Σ ] . Σ ∞ N −1 (n) (n) (Z ) + μ G G (K ) n=N n=0

(7.56)

By using the similar methods in Sect. 7.2, we can derive optimal policies to minimize the expected cost rates C L (T, N ), C L (T , Z ) and C L (N , Z ), and conclude that the best policy among three ones is replaced with damage Z , the second one is replaced with shock N and the third one is replacement with time T .

(2) Periodic Replacement Suppose that the unit is replaced preventively at time J T (J = 1, 2, . . . ) for a specified T > 0 in Chap. 4, at shock N or at damage Z , whichever occurs first [6, p. 218]. When c J is the replacement cost at time J T , replacing T with J T , the expected cost rate of replacement first is, from (7.3), c J + (c N − c J )F (N ) (J T )G (N ) (Z ) ∫Z Σ N −1 (n+1) + (c K − c J ) n=0 F (J T ) 0 G (K − x)dG (n) (x) ∫Z Σ N −1 (n+1) + (c Z −c J ) n=0 F (J T ) 0 [G(K−x)−G(Z −x)]dG (n) (x) C F (J, N , Z ) = , ∫ JT Σ N −1 (n) (n) (n+1) (t)]dt n=0 G (Z ) 0 [F (t) − F (7.57) and the expected cost rate of replacement last is, from (7.53), C L ( J, N , Z ) = c K − (c K − c J )

Σ∞

n=N [F

(n)

(J T ) − F (n+1) (J T )][G (n) (K ) − G (n) (Z )]

− (c K − c N )[1 − F (N ) (J T )][G (N ) (K ) − G (N ) (Z )] ∫Z Σ (n+1) − (c K − c Z ) ∞ (J T )] 0 [G(K −x) − G(Z −x)]dG (n) (x) n=N [1 − F ∫ ∞ (n) , Σ∞ G (n) (Z ) [F (t) − F (n+1) (t)]dt n=N Σ N −1 (n) J T∫ ∞ (n) + n=0 G (K ) J T [F (t) − F (n+1) (t)]dt ∫ J T (n) Σ (n) (n+1) + ∞ (t)]dt n=0 G (K ) 0 [F (t) − F (7.58) In particular, when the unit is replaced only at time J T (J = 1, 2, . . . ) or at failure K , whichever occurs first, the expected cost rate is, from (7.6),

7.5 Modified Replacement Models

167

C(J ) ≡ C F (J, ∞, K ) = C L (J, 0, 0) Σ (n) (n+1) c J + (c K − c J ) ∞ (K )]F (n+1) (J T ) n=0 [G (K ) − G . = ∫ JT Σ∞ (n) (n) (n+1) (t)]dt n=0 G (K ) 0 [F (t) − F

(7.59)

We find optimal J ∗ to minimize C(J ). Forming the inequality C(J + 1)− C(J ) ≥ 0, Q 1 (J )

∞ Σ

G (n) (K )

−

JT

[F (n) (t) − F (n+1) (t)]dt

0

n=0 ∞ Σ

∫

[G (n) (K ) − G (n+1) (K )]F (n+1) (J T ) ≥

n=0

cJ , cK − c J

(7.60)

where Q 1 (J ) =

Σ∞

n=0 [G

(n)

(K ) − G (n+1) (K )][F (n+1) ((J + 1)T ) − F (n+1) (J T )] . ∫ (J +1)T Σ∞ (n) [F (n) (t) − F (n+1) (t)]dt n=0 G (K ) J T

Thus, if Q 1 (J ) increases strictly with J to h(∞), then the left-hand side of (7.60) increases strictly J to μh(∞)[1 + MG (K )] − 1. Thus, if μh(∞)[1 + MG (K )] > c K /(c K − c J ), then there exists a finite and unique minimum J ∗ (1 ≤ J ∗ < ∞) which satisfies (7.60). In particular, when F(t) = 1 − e−λt , Q 1 (J ) =

λ

Σ∞

∫ (J +1)T (K ) − G (n+1) (K )] J T [(λt)n /n!]e−λt dt . ∫ (J +1)T Σ∞ (n) [(λt)n /n!]e−λt dt n=0 G (K ) J T

n=0 [G

(n)

It is easily shown that when rn (K ) = [G (n) (K ) − G (n+1) (K )]/G (n) (K ) increases strictly with n to 1, Q 1 ( J ) increases strictly with J to λ from (2) of Appendix A.7. Thus, if MG (K ) > c J /(c K − c J ), then there exists a finite and unique minimum J ∗ (1 ≤ J ∗ < ∞) which satisfies (7.60). Next, consider the periodic cumulative damage model: The damage caused by shocks is accumulated with time and is only inspected at periodic times j T ( j = 1, 2, . . . ) for given T (0 < T < ∞). Each amount W j of damage between [( j − 1)T ∫ ∞ , j T ] has an identical distribution G(x) ≡ Pr{W j ≤ x} with finite mean 1/ω ≡ 0 G(x)dx. Suppose that the unit is replaced preventively at time J T (J = 1, 2, . . . ) or at damage Z , whichever occurs first. Then, replacing T with J T and cT with c J , the expected cost rate is Σ −1 ∫ Z ( j) c Z + (c J − c Z )G (J ) (Z ) + (c K − c Z ) Jj=0 0 G (K − x)dG (x) . C F (J, Z ) = Σ J −1 ( j) T j=0 G (Z ) (7.61)

168

7 Which is Better Problems in Shock and Damage Models

Similarly, when the unit is replaced preventively at time J T (J = 0, 1, 2, . . . ) or at damage Z , whichever occurs last, the expected cost rate is c Z + (c J − c Z{)[G (J ) (K ) − G (J ) (Z )] } ∫Z Σ ( j) + (c K − c Z ) 1 − G (J ) (K ) + ∞ (K − x)dG (x) G j=J 0 [Σ ] . C L (J, Z ) = Σ J −1 ( j) ∞ ( j) T G (Z ) + G (K ) j=J j=0 (7.62)

(3) Replacement Middle We apply replacement middle introduced in Sect. 5.3 to shock and damage models with replacement time T , shock N and damage Z [6, p. 85]. Let t N , t Z and t K be the respective replacement times at shock N , damage Z and failure K . Suppose that the unit is replaced preventively at time T or at max{t N , t Z }, whichever occurs first. The probability that the unit is replaced at time T , i.e., T < t N < t Z , T < t Z < t N , t Z < T < t N or t N < T < t Z , is N −1 Σ

G (n) (K )[F (n) (T ) − F (n+1) (T )] +

n=0

∞ Σ

G (n) (Z )[F (n) (T ) − F (n+1) (T )],

n=N

the probability that it is replaced at shock N , i.e., t Z ≤ t N ≤ T , is F (N ) (T )[G (N ) (K ) − G (N ) (Z )], the probability that it is replaced at damage Z , i.e., t N ≤ t Z ≤ T , is ∞ Σ n=N

F (n+1) (T )

∫

Z

[G(K − x) − G(Z − x)]dG (n) (x),

0

and the probability that it is replaced at failure K , i.e, t K ≤ t N ≤ T or t N < t K ≤ T , is N −1 Σ

F (n+1) (T )[G (n) (K ) − G (n+1) (K )] +

n=0

Thus, the mean time to replacement is

∞ Σ n=N

∫ F (n+1) (T ) 0

Z

G(K − x)dG (n) (x).

7.5 Modified Replacement Models

T

{ N −1 Σ

169

}

G

(n)

(K )[F

(n)

(T )− F

n=0

+ [G (N ) (K ) − G (N ) (Z )]

(n+1)

n=N

∫

T 0

+

∞ Σ

[G (n) (Z ) − G (n+1) (Z )]

t dF (N ) (t) ∫

+

[G

(n)

(K ) − G

=

G (n) (Z )

∫

+

(n+1)

∫

T

(K )]

t dF (n+1) (t)

T

[F (n) (t) − F (n+1) (t)]dt

0

n=N N −1 Σ

t dF (n+1) (t)

0

n=0 ∞ Σ

T 0

n=N N −1 Σ

∞ Σ (T )] + G (n) (Z )[F (n) (T )− F (n+1) (T )]

G (n) (K )

∫

T

[F (n) (t) − F (n+1) (t)]dt.

0

n=0

Therefore, the expected cost rate is cT + (c N − cT )F (N ) (T )[G (N ) (K ) − G (N ) (Z )] ∫ Σ + (c Z {− cT ) ∞ F (n+1) (T ) 0Z [G(K − x) − G(Z − x)]dG (n) (x) Σ N −1 n=N (n+1) (T )[G (n) (K ) − G (n+1) (K )] + cK n=0 F } ∫ Σ∞ + n=N F (n+1) (T ) 0Z G (K − x)dG (n) (x) C M I (T , N , Z ) = . ∫ Σ∞ G (n) (Z ) 0T [F (n) (t) − F (n+1) (t)]dt n=N ∫ Σ N −1 (n) + n=0 G (K ) 0T [F (n) (t) − F (n+1) (t)]dt

(7.63) Next, suppose that the unit is replaced preventively at time T , at shock N or at damage Z , whichever occurs middle. The probability that the unit is replaced at time T , i.e., t Z < T < t N or t N < T < t Z , is N −1 Σ

[F (n) (T ) − F (n+1) (T )][G (n) (K ) − G (n) (Z )]

n=0

+

∞ Σ

G (n) (Z )[F (n) (T ) − F (n+1) (T )],

n=N

the probability that it is replaced at shock N , i.e., T < t N < t Z or t Z ≤ t N < T , is G (N ) (Z )[1 − F (N ) (T )] + F (N ) (T )[G (N ) (K ) − G (N ) (Z )], the probability that it is replaced at damage Z , i.e., T < t Z ≤ t N or t N < t Z ≤ T , is

170

7 Which is Better Problems in Shock and Damage Models N −1 Σ

[1 − F (n+1) (T )]

Z

[G(K − x) − G(Z − x)]dG (n) (x)

0

n=0

+

∫

∞ Σ

F (n+1) (T )

∫

Z

[G(K − x) − G(Z − x)]dG (n) (x),

0

n=N

and the probability that it is replaced at failure K , i.e., t K ≤ t N < T , t N < t K < T or T < t K ≤ t N , is N −1 Σ

∫ ∞ Σ F (n+1) (T )[G (n) (K ) − G (n+1) (K )] + F (n+1) (T )

n=0

+

N −1 Σ

[1 − F (n+1) (T )]

Z

G(K −x)dG (n) (x)

0

n=N

∫

Z

G(K − x)dG (n) (x).

0

n=0

Thus, the mean time to replacement is T

{ N −1 Σ

[F (n) (T ) − F (n+1) (T )][G (n) (K ) − G (n) (Z )]

n=0 ∞ Σ

+

} G

(n)

(Z )[F

n=N

+ G (N ) (Z )

∫

∞

(n)

(T ) − F

(n+1)

t dF (N ) (t) + [G (N ) (K ) − G (N ) (Z )]

T

+

∞ Σ

[G (n) (Z ) − G (n+1) (Z )]

+

[G

(n)

(Z ) − G

(n+1)

+ =

[G

(n)

(K ) − G ∫

+

n=N

∞

G (n) (K )

(n+1)

∫ (K )]

T

t dF (n+1) (t)

[F (n) (t) − F (n+1) (t)]dt

∫

T

[F (n) (t) − F (n+1) (t)]dt

0

n=0 ∞ Σ

t dF (n+1) (t)

T

n=0

+

∞

(Z )]

0

G (n) (Z )

N −1 Σ

t dF (n+1) (t)

T

n=0 N −1 Σ

T

∫

n=0 N −1 Σ

G

(n)

∫

T

(Z ) 0

∫

T 0

∫ 0

n=N N −1 Σ

(T )]

[F (n) (t) − F (n+1) (t)]dt.

t dF (N ) (t)

7.5 Modified Replacement Models

171

Therefore, the expected cost rate is C M I I (T , N , Z ) = { } cT + (c N −cT{) G (N ) (Z )[1 − F (N ) (T )] + F (N ) (T )[G (N ) (K ) − G (N ) (Z )] ∫Z Σ N −1 (n+1) + (c Z − cT ) (T )] 0 [G(K − x) − G(Z − x)]dG (n) (x) n=0 [1 − F } ∫Z Σ (n+1) (T ) 0 [G(K − x) − G(Z − x)]dG (n) (x) + ∞ n=N F {Σ N −1 (n+1) + (c K − cT ) (T )[G (n) (K ) − G (n+1) (K )] n=0 F ∫ Σ∞ Z + n=N F (n+1) (T ) 0 G (K − x)dG (n) (x) } ∫Z Σ N −1 + n=0 [1 − F (n+1) (T )] 0 G (K − x)dG (n) (x) . ∫∞ Σ N −1 (n) G (Z ) T [F (n) (t) − F (n+1) (t)]dt n=0 ∫T Σ N −1 (n) + n=0 G (K ) 0 [F (n) (t) − F (n+1) (t)]dt ∫ Σ∞ T + n=N G (n) (Z ) 0 [F (n) (t) − F (n+1) (t)]dt (7.64) Clearly, C M I I (∞, N , Z ) = C M I (∞, N , Z ). Using similar techniques in Sect. 7.4, we can discuss WIB problems of these models.

(4) Deviation Time We apply the deviation time introduced in Sect. 2.6 to shock and damage models [7]: Suppose that the unit is replaced at time T or at failure K , whichever occur first. The mean deviation time between T and t K when the total damage has exceeded failure K is ]∫ T ] ∫ ∞ ∞ Σ (n) (n+1) (n+1) (n+1) (K )] (T −t)dF (t) + (t −T )dF (t) D(T ) = [G (K )−G n=0 ∞ Σ

0

[G (n) (K ) − G (n+1) (K )]

=

n=0

T

{∫

} ∫ ∞ T F (n+1) (t) dt + [1 − F (n+1) (t)]dt . 0

T

(7.65) Σ (n) Clearly, D(0) = μ ∞ n=0 G (K ) and D(∞) = ∞. ∗ We find optimal T to minimize D(T ). Differentiating D(T ) with respect to T and setting it equal to zero, ∞ Σ 1 [G (n) (K ) − G (n+1) (K )]F (n+1) (T ) = , 2 n=0

(7.66)

whose left-hand side increases strictly with T from 0 to 1. Thus, there exists a finite and unique T ∗ (0 < T ∗ < ∞) which satisfies (7.66).

172

7 Which is Better Problems in Shock and Damage Models

Next, suppose that the unit is replaced at shock N or at failure K , whichever occurs first. The mean deviation time between t N and t K is N −1 Σ

D(N ) =

[G (n) (K ) − G (n+1) (K )]

n=0 ∞ Σ

+

∫

∞

0

[G

(n)

(K ) − G

(n+1)

{ N −1 Σ

∞

] u dF (N −n−1) (u) dF (n+1) (t)

0

∫

∞

(K )]

]∫

∞

u dF 0

n=N

=μ

]∫

+ =μ

(u) dF (N ) (t)

(n + 1 − N )[G (n) (K ) − G (n+1) (K )] }

(n + 1 − N )[G

(n)

(K ) − G

(n+1)

n=N

{ N −1 Σ

]

0

n=0 ∞ Σ

(n−N +1)

[1 − G

(n)

(K )] +

n=0

∞ Σ

(K )] }

G

(n)

(K ) .

(7.67)

n=N

Σ (n) Clearly, D(0) = μ ∞ n=0 G (K ) and D(∞) = ∞. ∗ We find optimal N to minimize D(N ). Forming the inequality D(N + 1) − D(N ) ≥ 0, 1 (7.68) 1 − G (N ) (K ) ≥ , 2 whose left-hand side increase strictly with N from 0 to 1. Thus, there exists a finite and unique minimum N ∗ (1 ≤ N ∗ < ∞) which satisfies (7.68). Example 7.3 When F(t) = 1 − e−λt and G(x) = 1 − e−ωx , (7.66) is ∞ Σ (ωK )n n=0

and (7.68) is

n!

e−ωK

n Σ (λT ) j j=0

N −1 Σ (ωK )n n=0

n!

j!

e−ωK ≥

e−λT =

1 , 2

1 . 2

Thus, using similar techniques in Sect. 2.6, we compute T ∗ , N ∗ and D(T ∗ ), D(N ∗ ), respectively, and can discuss WIB problems for deviation times of shock and damage models. ∎

References

173

References 1. Nakagawa T (2007) Shock and damage models in reliability theory. Springer, London 2. Zhao X, Nakagawa T (2010) Optimal replacement policies for damage models with the limit number of shocks. Inter J Reliab Qual Perf 2:13–20 3. Zhao X, Zhang H, Qian C, Nakagawa T, Nakamura S (2012) Replacement models for combining additive independent damage. Int J Perf Eng 8:91–100 4. Zhao X, Qian C, Nakagawa T (2013) Optimal maintenance policies for cumulative damage models with random working times. J Qual Mainte Eng 19:25–37 5. Zhao X, Qian C, Sheu S (2014) Cumulative damage models with random working times. In: Nakamura S, Qian C, Chen M (eds) Reliability modeling with applications. World Scientific, Singapore, pp 79–98 6. Zhao X, Nakagawa T (2018) Advanced maintenance policies for shock and damage models. Springer, London 7. Zhao X, Mizutani S, Chen M, Nakagawa T (2022) Preventive replacement policies for parallel systems with deviation costs between replacement and failure. Ann Oper Res 312:531–551 8. Barlow RE, Proschan F (1965) Mathematical theory of reliability. Wiley, New York

Chapter 8

Which is Better Problems in Backup Models

The data in a computer system are frequently updated by adding or deleting them, and are stored in a computer database system. However, data files in the database are often broken by several errors due to noises, human errors, software and hardware errors. In this case, we have to restore the same files from the beginning and recover the database. The most simple and recovery method to ensure the safety of data would be always to store up the backup copies of all files in other places, and to take them out when some files in the database are broken, which is called full backup [1]. However, this method would take hours and be costly as the size of files becomes larger. To take the backup copies quickly and efficiently, we might dump down only files which have changed since the last backup. This would reduce significantly both duration time, backup size and space [2], which is called incremental or differential backup [3]. This backup is mainly classified into three schemes [3], [4, p. 147]: (1) Full backup: All data files changed since the last full backup are updated, and the database returns to its initial state. (2) Incremental backup: Only data files changed since the last incremental backup are updated. To restore the entire updated data after failure, the total data of the last full backup and all incremental backups are needed. (3) Differential backup: Only data files changed since the last full backup are updated. To restore the entire updated data after failure, the total data of the last full backup and the last differential backup are needed. Suppose that backups are made at a planned time T or at a number N of updates. Then, introducing costs of backup and data restoration, the expected cost were obtained when we adopt incremental and differential backups between full backups. Optimal backup policies to minimize the expected costs were discussed deeply [5–10], using the techniques of random maintenance policies in Chap. 2 and cumulative damage models in Chap. 7. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Mizutani et al., Which-Is-Better (WIB): Problems in Reliability Theory, Springer Series in Reliability Engineering, https://doi.org/10.1007/978-3-031-27316-2_8

175

176

8 Which is Better Problems in Backup Models

Suppose in Sect. 8.1 that full backup is done at a number N of updates. Then, both expected costs of incremental and differential backups are obtained and compared. Furthermore, optimal update N ∗ to minimize the expected cost rates are discussed theoretically and numerically. In Sect. 8.2, full backup is done at the completion of the first update over a planned time T . Expected costs are obtained and similar discussions of minimizing them to Sect. 8.1 are made. In Sects. 8.3 and 8.4, full backup is done at the completion of the first update over time T or at a number N of updates, whichever occurs first or last, respectively. Two backup policies with overtime T and update N are compared theoretically and numerically. It is shown that backup policy with update N is better than that with time T .

8.1 Database Backup Models It is assumed that updates occur at a renewal process according to an identical F(t) with a density function f (t) ≡ dF(t)/dt and finite mean 1/λ ≡ distribution ∫∞ F(t)dt. A volume W j ( j = 1, 2, . . . ) of updated data due to the jth update has 0 ∫∞ an identical distribution G(x) ≡ Pr{W j ≤ x} with finite mean 1/ω ≡ 0 G(x)dx. It is denoted that Φ(t) ≡ 1 − Φ(t) for any function Φ(t), and Φ (n) (t) (n = 1, 2, . . . ) is the n-fold Stieltjes convolution of Φ(t) and Φ (0) (t) ≡ 1 for t ≥ 0. Furthermore, database failures occur randomly according to a general∫distribution D(t) with a ∞ density function d(t) ≡ dD(t)/dt and finite mean 1/μ ≡ 0 D(t)dt, and its failure rate is r (t) ≡ d(t)/D(t), which increases strictly with t from r (0) to r (∞). Suppose that full backup should be made immediately after database failures at renewal points of incremental and differential backups. To protect security of data and prevent the enormous restoration cost due to failures, full backup is scheduled preventively at a number N (N = 1, 2, . . . ) of updates. Then, the probability that full backup is done at failure is ∫

∞

[1 − F (N ) (t)]dD(t) =

0

∫

∞

D(t)dF (N ) (t),

0

and the probability that it is done at update N is ∫

∞

D(t)dF 0

(N )

∫ (t) =

∞

F (N ) (t)dD(t).

0

We introduce the following costs required for data backup and restoration schemes: c F is cost for full backup, c K + c O x is cost for incremental or differential backup when a total volume x of data has been updated, c R + c O x + nc N is data restoration cost after failure for a number n of backups, and c R + c O x is data restoration cost after failure for differential backup. It is denoted that for n = 1, 2, . . . ,

8.1 Database Backup Models

177

∫

∞

nc O , ω 0 ∫ ∞ nc O Nn = (c R + c O x)dG (n) (x) = c R + . ω 0

Mn ≡

(c K + c O x)dG (n) (x) = c K +

∑ Then, n M1 is the expected cost for n incremental backups, nj=1 M j is the expected cost for n differential backups, Nn + nc N is the data restoration cost for n incremental backups, and Nn is the data restoration cost for n differential backups. Thus, the expected cost until full backup is, for incremental backup, ∼I (N ) = (c F + N M1 ) C

∫

∞

F (N ) (t)dD(t)

0

+

N −1 ∑

∫

∞

(c F + n M1 + Nn + nc N )

[F (n) (t) − F (n+1) (t)]dD(t)

0

n=0

N ∫ ( c O ) ∑ ∞ (n) F (t)dD(t) = cF + cK + ω n=1 0 ∫ N −1 [ ( ∑ c O )] ∞ (n) cR + n cN + + [F (t) − F (n+1) (t)]dD(t), ω 0 n=0

(8.1)

and for differential backup, ( )∫ N ∑ ∼ Mn C D (N ) = c F + 0

n=1

+

N −1 ∑

⎛

⎝c F +

n=0

∞

n ∑ j=1

F (N ) (t)dD(t) ⎞

M j + Nn ⎠

∫

∞

[F (n) (t) − F (n+1) (t)]dD(t)

0

∫ nc O ) ∞ (n) cK + F (t)dD(t) = cF + ω 0 n=1 ∫ N −1 ( ∑ nc O ) ∞ (n) cR + + [F (t) − F (n+1) (t)]dD(t), ω 0 n=0 N ( ∑

(8.2)

∼D (1). The difference of C ∼I (1) = C ∼I (N ) and C ∼D (N ) is where note that C ∼I (N ) = ∼D (D) − C C ∫ ∞ N −1 N −1 ∫ ∞ ∑ cO ∑ (n+1) n F (t)dD(t) − c N n [F (n) (t) − F (n+1) (t)]dD(t), ω n=1 0 0 n=1 (8.3)

178

8 Which is Better Problems in Backup Models

which follows that if (

cO ) ∑ n ω n=1 N −1

cN +

∫

∞

F (n+1) (t)dD(t) > c N

0

N −1 ∫ ∑ n

∞

F (n) (t)dD(t),

0

n=1

then incremental backup is better than differential one. ∫∞ = F ∗ (μ)n , where F ∗ (μ) In particular, when D(t) = 1 − e−μt , 0 F (n) (t)dD(t) ∫ ∞ is Laplace-Stieltjes transform of F(t), i.e., F ∗ (μ) ≡ 0 e−μt dF(t). Thus, if cO ω

[

] F ∗ (μ) > cN , 1 − F ∗ (μ)

then incremental backup is better than differential one, independent of N . This means that because the expected number of updates before failure is ∫ ∞ ∑ n

∞

F (n) (t)dD(t) =

0

n=1

F ∗ (μ) , 1 − F ∗ (μ)

if the expected cost for the total volume of updated data before failure is larger than cost c N for one increment backup, then increment backup is better than differential one. The mean time to full backup is ∫

∞

t D(t)dF 0

(N )

∫ (t) + 0

∞

t[1 − F

(N )

∫

∞

(t)]dD(t) =

[1 − F (N ) (t)]D(t)dt.

0

(8.4) Thus, the expected cost rate for incremental backup is, from (8.1), ∑ N ∫ ∞ (n) c F + (c K + c O /ω) n=1 0 F ∫ (t)dD(t) ∑ N −1 ∞ + n=0 [c R + n(c N + c O /ω)] 0 [F (n) (t) − F (n+1) (t)]dD(t) , (8.5) C I (N ) = ∫∞ (N ) (t)]D(t)dt 0 [1 − F and the expected cost rate for differential backup is, from (8.2), ∫∞ ∑N c F + n=1 (c K + nc O /ω) 0 F (n) (t)dD(t) ∫∞ ∑ N −1 (c R + nc O /ω) 0 [F (n) (t) − F (n+1) (t)]dD(t) + n=0 . C D (N ) = ∫∞ (N ) (t)]D(t) dt 0 [1 − F

(8.6)

8.1 Database Backup Models

179

8.1.1 Optimal Backup Times We find optimal full backup times Ni∗ (i = I, D) to minimize C I (N ) in (8.5) and C D (N ) in (8.6), respectively.

(1) Incremental Backup Forming the inequality C I (N + 1) − C I (N ) ≥ 0, N ∫ ( c O ) ∑ ∞ (n) [1 − F (N ) (t)]D(t)dt − c K + F (t)dD(t) ω n=1 0 0 ∫ N −1 [ ( ∑ c O )] ∞ (n) [F (t) − F (n+1) (t)]dD(t) ≥ c F , cR + n cN + − ω 0 n=0

∫

Q I (N )

∞

(8.7)

where ∫∞ (c K + c O /ω) 0 F (N +1) (t)dD(t) ∫∞ + [c R + N (c N + c O /ω)] 0 [F (N ) (t) − F (N +1) (t)]dD(t) . Q I (N ) ≡ ∫∞ (N ) − F (N +1) (t)]D(t)dt 0 [F Let L I (N ) denote the left-hand side of (8.7) and M F (t) ≡ increases strictly with N to Q I (∞) and L I (∞) ≡

∑∞ n=1

F (n) (t). If Q I (N )

( )∫ ∞ Q I (∞) 2c O − cK + cN + M F (t)dD(t) − c R > c F , μ ω 0

then there exists a finite and unique minimum N I∗ (1 ≤ N I∗ < ∞) which satisfies (8.7). In particular, when D(t) = 1 − e−μt , ( Q I (N ) (c K + c O /ω)F ∗ (μ) cO ) = + N c + + cR N μ 1 − F ∗ (μ) ω increases strictly with N to ∞. Thus, there exists a finite and unique minimum N I∗ (1 ≤ N I∗ < ∞) which satisfies N ∑

[1 − F ∗ (μ)n ] ≥

n=1

When F(t) = 1 − e−λt , i.e., F (n) (t) =

cF . c N + c O /ω

∑∞

j=n [(λt)

j

/j!]e−λt , from (8.5),

(8.8)

180

8 Which is Better Problems in Backup Models

( cO ) C I (N ) = λ c K + ω ∫∞ ∑ N −1 c F + n=0 [c R + n(c N + c O /ω)] 0 [(λt)n /n!]e−λt dD(t) + . ∑ N −1 ∫ ∞ n −λt D(t)dt n=0 0 [(λt) /n!]e

(8.9)

Forming the inequality C I (N + 1) − C I (N ) ≥ 0, ] N −1 ∫ ∞ ∑ (λt)n −λt (λt)n −λt e D(t)dt − e dD(t) c R Q 1 (N ) n! n! n=0 0 n=0 0 } { N −1 ∫ ∞ N −1 ∫ ∞ ( ∑ ∑ cO ) (λt)n −λt (λt)n −λt e D(t)dt − n e dD(t) N Q 1 (N ) + cN + ω n! n! 0 n=0 0 n=0 [

N −1 ∫ ∑

∞

≥ cF ,

(8.10)

where Q 1 (N ) ≡ lim T →∞ Q 1 (T , N ) and ∫T Q 1 (T, N ) ≡ ∫ 0T 0

(λt) N e−λt dD(t)

(λt) N e−λt D(t)dt

,

which agrees with Q 1 (T , N ) in (2.8) when λ = θ and D(t) = F(t). It is easily proved from (1) of Appendix A.1 that when r (t) increases strictly with t from r (0) = 0 to r (∞) = ∞, Q 1 (T , N ) increases strictly with N from Q 1 (T, 0) to r (T ), and Q 1 (N ) increases strictly with N to ∞. Thus, the left-hand side of (8.10) increases strictly with N to ∞. In particular, when D(t) = 1 − e−μt , (8.10) is N [ ∑

( 1−

n=1

λ λ+μ

)n ] ≥

cF , c N + c O /ω

which agrees with (8.8) when F ∗ (μ) = λ/(λ + μ), and from (8.9), ( μ cO ) C I (N ) = λ c K + c R + λ ω ∑ N −1 c F + (c N + c O /ω)[μ/(λ + μ)] n=0 n[λ/(λ + μ)]n + . ∑ N −1 [1/(λ + μ)] n=0 [λ/(λ + μ)]n (2) Differential Backup Forming the inequality C D (N + 1) − C D (N ) ≥ 0 in (8.6),

(8.11)

8.1 Database Backup Models

∫

∞

Q D (N )

181

[1 − F (N ) (t)]D(t)dt −

0

−

N ( ∑

cK +

n=1

nc O ) ω

∫

∞

F (n) (t)dD(t)

0

∫ nc O ) ∞ (n) [F (t) − F (n+1) (t)]dD(t) ≥ c F , cR + ω 0

N −1 ( ∑ n=0

(8.12)

where ∫ ∞ (N +1) [c K + (N + 1)c O /ω] (t)dD(t) ∫ ∞ 0 (NF) + [c R + N c O /ω] 0 [F (t) − F (N +1) (t)]dD(t) . Q D (N ) ≡ ∫∞ (N ) (t) − F (N +1) (t)]D(t)dt 0 [F Let L D (T ) denote the left-hand side of (8.12). If Q D (N ) increases strictly with N to Q D (∞) and ∫ ∞ Q D (∞) ∑ ( nc O ) ∞ (n) − F (t)dD(t) cK + μ ω 0 n=1 ∫ cO ∞ − M F (t)dD(t) − c R > c F , ω 0

L D (∞) =

(8.13)

then there exists a finite and unique minimum N D∗ (1 ≤ N D∗ < ∞) which satisfies (8.12). In particular, when D(t) = 1 − e−μt , Q D (N ) [c K + (N + 1)c O /ω]F ∗ (μ) N cO = + + cR μ 1 − F ∗ (μ) ω increases strictly with N to ∞. Thus, there exists a finite and unique minimum N D∗ (1 ≤ N D∗ < ∞) which satisfies ∑ 1 cF [1 − F ∗ (μ)n ] ≥ . ∗ 1 − F (μ) n=1 c O /ω N

(8.14)

When F(t) = 1 − e−λt , (8.12) is ( cR + +

cO ω

N cO ω N −1 ∑ n=0

)[

∫

∞

Q 1 (N ) ∫

(N − n)

[1 − F (N ) (t)]D(t)dt −

0 ∞

F (n) (t)dD(t) ≥ c F ,

∫

∞

]

[1 − F (N ) (t)]dD(t)

0

(8.15)

0

where Q 1 (N ) is given in (8.10). Thus, when r (t) increases with t from 0 to ∞, the left-hand side of (8.15) increases strictly with N to ∞.

182

8 Which is Better Problems in Backup Models

In particular, when D(t) = 1 − e−μt , (8.15) is N −1 ∑

(

λ (N − n) λ + μ n=0

)n ≥

cF , c O /ω

which agrees with (8.14) when F ∗ (μ) = λ/(λ + μ), and from (8.6), ∑ N −1 n[λ/(λ + μ)]n μ c O ) c F + (c O /ω) n=0 . + C D (N ) = λ c K + c R + ∑ N −1 λ ω [1/(λ + μ)] n=0 [λ/(λ + μ)]n (

(8.16)

Example 8.1 When D(t) = 1 − e−μt , F(t) = 1 − e−t , c K = 0.1, c R = 0.2, c N = 0.5 and c F = 5.0, Table 8.1 presents optimal N I∗ , N D∗ and resulting cost rates C I (N I∗ )/μ, C D (N D∗ )/μ for μ and c O /ω. When c O /ω increase, the expected cost rates for all backup schemes increase. In order to prevent the full backup and restoration costs, optimal N I∗ and N D∗ decrease with c O /ω. Furthermore, N I∗ decrease with μ while N D∗ increase with μ. This means that incremental backup needs more costs after failure, and so that, we should decrease N I∗ to save the number of backups for restoration. For differential backup, the expected number of backups is limited by the increasing of μ, and so that, we should increase N D∗ as far as possible to lessen the restoration. This indicates that incremental backup is better when μ is small, and ∎ differential one is better when μ is large.

∗ , C (N ∗ )/μ and C (N ∗ )/μ when D(t) = 1 − e−μt , F(t) = 1 − e−t , Table 8.1 Optimal N I∗ , N D I D I D c K = 0.1, c R = 0.2, c N = 0.5 and c F = 5.0 c O /ω = 0.5 c O /ω = 1.0 μ ∗ ∗ )/μ N ∗ ∗ ∗ )/μ N I∗ C I (N I∗ )/μ N D C D (N D C I (N I∗ )/μ N D C D (N D I

0.01 0.02 0.05 0.10 0.20 0.50 1.00 2.00 5.00

35 26 18 14 12 11 12 13 13

94.334 55.229 29.342 19.615 14.364 11.619 11.892 13.200 13.173

6 6 6 6 6 8 12 17 21

313.843 159.303 66.684 35.990 20.987 12.927 11.892 12.881 12.767

28 21 14 11 9 8 9 9 9

151.724 85.549 42.832 27.199 18.799 13.927 13.308 14.087 13.780

4 4 4 4 5 5 7 9 11

480.286 243.388 101.340 54.142 30.619 17.330 14.199 14.087 13.476

8.2 Overtime Backup First

183

8.2 Overtime Backup First Suppose that full backup is made at the first update over a planned time T (0 ≤ T < ∞) or at database failure, whichever occurs first. Then, the probability that full backup is made at failure is ∞ ∫ ∑

D(T ) + =

∞ ∫ ∑

[∫

T

[∫

∞ T −t

0

n=0

∞ T −t

0

n=0

T

] F(u)dD(t + u) dF (n) (t) ]

D(t + u)dF(u) dF (n) (t),

and the probability that it is made at the first update over time T is ∞ ∫ ∑

[∫

T

T −t

0

n=0

∞

] D(t + u)dF(u) dF (n) (t).

Noting that ∞ ∫ ∑

[∫

0

n=0

=

T

0

∞ ∫ T ∑ n=0

] F(u)D(t + u)du dF (n) (t)

T −t

[F

(n)

−F

(n+1)

∫ (t)] D(t)dt =

T

D(t)dt, 0

0

the mean time to full backup is ∫

T

tdD(t) +

0

∫ =

∞ ∫ ∑ n=0

T

D(t)dt +

0

=

∞ ∫ ∑ n=0

0

T

T 0

∞ ∫ T ∑ n=0

[∫

∞

[∫

0

∞ T −t

[∫

] (t + u)D(t + u)dF(u) dF (n) (t) ∞

T −t

] F(u)D(t + u)du dF (n) (t) ]

F(u)D(t + u)du dF (n) (t).

0

Thus, the expected cost is, for incremental backup,

(8.17)

184

8 Which is Better Problems in Backup Models

∼O I (T ) = C

∞ ∑

∫ [c F + (n + 1)M1 ]

[F (n) (t) − F (n+1) (t)]dD(t)

∫ ∞ ∑ [c F + (n + 1)M1 ]

T

[∫

] F(u)dD(t + u) dF (n) (t)

] D(t + u)dF(u) dF (n) (t)

∞

∫

[∫

T

[c F + n M1 + Nn + nc N ] 0

n=0

∞ T −t

]

T −t

0

n=0 ∞ ∑

+

[∫

0

n=0 T 0

=

T

(c F + n M1 + Nn + nc N )

∫ +

[∫

∞ ∑

] D(t + u)dF(u) dF (n) (t)

∞ T −t

0

n=0

+

[∫

T

]

∞

F(u)dD(t + u) dF (n) (t)

0

[∫ ∞ ] ∞ ∫ ( cO ) ∑ T = cF + cK + D(t + u)dF(u) dF (n) (t) ω n=0 0 0 [∫ ∞ ] ∫ ∞ [ ( ∑ c O )] T + cR + n cN + [D(t + u) − D(t)]dF(u) dF (n) (t), ω 0 0 n=0 (8.18) and for differential backup, ∼O D (T ) = C

+

∞ ∑ n=0

⎛

∞ ∑

⎛ ⎝c F +

n=0

⎝c F +

n+1 ∑

⎞ Mj⎠

j=1 n ∑

⎞

M j + Nn ⎠

∫

T

[∫

T −t

0

∫

T

[∫

0

j=1

∞

∞

] D(t + u)dF(u) dF (n) (t) ]

F(u)dD(t + u) dF (n) (t)

0

] ∫ T [∫ ∞ ] (n + 1)c O D(t + u)dF(u) dF (n) (t) ω 0 0 n=0 [∫ ] ∫ ∞ ∞ ∑( nc O ) T + cR + [D(t + u) − D(t)]dF(u) dF (n) (t). ω 0 0 n=0

= cF +

∞ [ ∑

cK +

(8.19)

∼O I (T ) and C ∼O D (T ) is The difference of C ] ∫ T [∫ ∞ ∞ ∑ ∼O I (T ) = c O ∼O D (T ) − C D(t + u)dF(u) dF (n) (t) n C ω n=0 0 0 [∫ ] ∫ ∞ T ∞ ∑ − cN n [D(t + u) − D(t)]dF(u) dF (n) (t), n=0

0

0

8.2 Overtime Backup First

185

which follows that if (

] ∫ T [∫ ∞ ∞ cO ) ∑ cN + n D(t + u)dF(u) dF (n) (t) ω n=0 0 0 ∫ T ∞ ∑ > cN n D(t)dF (n) (t), n=0

0

then incremental backup is better than differential one. In particular, when D(t) = 1 − e−μt , if cO ω

[

] F ∗ (μ) > cN , 1 − F ∗ (μ)

then incremental backup is better than differential one. The expected cost rate for incremental backup is, from (8.17) and (8.18), C O I (T ) =

] (n) ∑ ∫ T [∫ ∞ c F + (c K + c O /ω) ∞ n=0 0 0 D (t + u)dF(u) dF (t) ∫ T {∫ ∞ } (n) ∑ + ∞ n=0 [c R + n(c N + c O /ω)] 0 0 [D(t + u) − D(t)]dF(u) dF (t) , ] ∑∞ ∫ T [∫ ∞ (n) n=0 0 0 F (u) D (t + u)du dF (t) (8.20)

and for differential backup, from (8.17) and (8.19). C O D (T ) = ∫ T [∫ ∞ ] (n) ∑ cF + ∞ n=0 [c K + (n + 1)c O /ω] 0 0 D (t + u)dF(u) dF (t) ∫ T {∫ ∞ } (n) ∑ + ∞ n=0 (c R + nc O /ω) 0 0 [D(t + u) − D(t)]dF(u) dF (t) . ] ∑∞ ∫ T [∫ ∞ (n) n=0 0 0 F (u) D (t + u)du dF (t)

(8.21)

8.2.1 Optimal Backup Times ∗ We find optimal TOi (i = I, D) to minimize C O I (T ) in (8.20) and C O D (T ) in (8.21), respectively, when D(t) = 1 − e−μt .

(1) Incremental Backup When D(t) = 1 − e−μt , the expected cost rate in (8.20) is

186

8 Which is Better Problems in Backup Models

∫ T −μt (n) ∑ c F + (c N + c O /ω)[1 − F ∗ (μ)] ∞ dF (t) C O I (T ) n=0 n 0 e = ∫T ∑ ∞ μ [1 − F ∗ (μ)] n=0 0 e−μt dF (n) (t) ( c O ) F ∗ (μ) + cK + + cR . (8.22) ω 1 − F ∗ (μ) Clearly, ( C O I (0) cF c O ) F ∗ (μ) = + c + c + , R K μ 1 − F ∗ (μ) ω 1 − F ∗ (μ) ) ( 2c O F ∗ (μ) C O I (∞) = cF + c R + cK + cN + . μ ω 1 − F ∗ (μ) Differentiating C O I (T ) with respect to T and setting it equal to zero, Q(T )

∞ ∫ ∑ n=0

T

e−μt dF (n) (t) −

∫ ∞ ∑ n

0

n=0

T

e−μt dF (n) (t)

0

cF = , (c N + c O /ω)[1 − F ∗ (μ)] where

(8.23)

∑∞ n f (n) (T ) , Q(T ) = ∑n=1 ∞ (n) (T ) n=1 f

where f (n) (t) ≡ dF (n) (t)/dt. If Q(T ) increases strictly with T to Q(∞) and Q(∞) −

F ∗ (μ) cF > , ∗ 1 − F (μ) c N + c O /ω

then there exists a finite and unique TO∗ I (0 ≤ TO∗ I < ∞) which satisfies (8.23), and the resulting cost rate is ( C O I (TO∗ I ) ( cO ) c O ) F ∗ (μ) Q(TO∗ I ) + c K + = cN + + cR . μ ω ω 1 − F ∗ (μ)

(8.24)

In particular, when F(t) = 1 − e−λt , Q(T ) = 1 + λT , and (8.23) is μ λ2 cF + λT − (1 − e−μT ) = , λ+μ (λ + μ)μ c N + c O /ω whose left-hand side increases strictly with T from μ/(λ + μ) to ∞. Thus, if cO ) μ ( ≥ cF , cN + λ+μ ω

(8.25)

8.2 Overtime Backup First

187

then TO∗ I = 0, i.e., full backup should be made at the first update, and ( C O I (0) cO ) λ = cF + c R + cF + cK + . μ ω μ

(8.26)

Next, when F(t) = 1 − e−λt , the expected cost rate in (8.20) is ∫∞ c F + c R [D(T ) +[ T e−λ(t−T ) dD(t)] ] ∫T ∫∞ + (c N + c O /ω) 0 λtdD(t) + λT T e−λ(t−T ) dD(t) C O I (T ) = ∫∞ ∫T −λ(t−T ) D (t)dt 0 D (t)dt + T e ( ) cO + λ cK + . (8.27) ω Clearly, C O I (0) c F + c R D ∗ (λ) cO = + cK + , λ 1 − D ∗ (λ) ω C O I (∞) (c F + c R )μ 2c O = + cK + cN + . λ λ ω Differentiating C O I (T ) with respect to T and setting it equal to zero, ] ∫ ∞ e−λt dD(t) ∫ T cO ) T cR + cN + D(t)dt (1 + λT ) ∫ ∞ −λt D(t)dt ω 0 T e ] [∫ T ∫ ∞ ( cO ) − c R D(T ) − c N + λt dD(t) − e−λ(t−T ) dD(t) = c F , (8.28) ω 0 T

[

(

whose left-hand side increases strictly with T from (c N + c O /ω)D ∗ (λ) to ∞. Thus, there exists a finite and unique TO∗ I (0 ≤ TO∗ I < ∞) which satisfies (8.28), and the resulting cost rate is ∫ ( ] T∞∗ e−λt dD(t) C O I (TO∗ I ) [ cO ) cO ∗ = cR + cN + . (1 + λTO I ∫ ∞O I + cK + −λt λ ω ω D(t)dt TO∗ I λe (8.29) In particular, when D(t) = 1 − e−μt , (8.28) agrees with (8.25).

(2) Differential Backup When D(t) = 1 − e−μt , the expected cost rate in (8.21) is

188

8 Which is Better Problems in Backup Models

∫ T −μt (n) ∑ c F + (c O /ω) ∞ dF (t) C O D (T ) n=0 n 0 e = ∫T ∑ ∞ μ [1 − F ∗ (μ)] n=0 0 e−μt dF (n) (t) ( c O ) F ∗ (μ) + cK + + cR . ω 1 − F ∗ (μ)

(8.30)

Clearly, ( C O D (0) cF c O ) F ∗ (μ) = + c + c + , R K μ 1 − F ∗ (μ) ω 1 − F ∗ (μ) [ [ ]] C O D (∞) cO 1 F ∗ (μ) = cF + c R + cK + . 1+ μ ω 1 − F ∗ (μ) 1 − F ∗ (μ) Differentiating C O D (T ) with respect to T and setting it equal to zero, Q(T )

∞ ∫ ∑ n=0

T

e−μt dF (n) (t) −

0

∫ ∞ ∑ n n=0

T

e−μt dF (n) (t) =

0

cF , c O /ω

(8.31)

whose left-hand side agrees with that of (8.23). Thus, if Q(T ) increases strictly with T and [ ] 1 F ∗ (μ) cF , Q(∞) − > 1 − F ∗ (μ) 1 − F ∗ (μ) c O /ω then there exists a finite and unique TO∗ D (0 < TO∗ D < ∞) which satisfies (8.31), and the resulting cost rate is ( C O D (TO∗ D ) c O Q(TO∗ D ) c O ) F ∗ (μ) = + c + + cR . K μ ω 1 − F ∗ (μ) ω 1 − F ∗ (μ)

(8.32)

In particular, when F(t) = 1 − e−λt , Q(T ) = 1 + λT , and (8.31) is 1 + λT +

λ2 T − μ

( )2 λ cF (1 − e−μT ) = , μ c O /ω

(8.33)

whose left-hand side increases strictly with T from 1 to ∞. Thus, if c O /ω ≥ c F , then TO∗ D = 0, and C O D (0) ( cO ) λ = cF + cK + + cF + cR . μ ω μ When F(t) = 1 − e−λt , the expected cost rate in (8.21) is ∫∞ [ ] c F + c R D(T ) + T e−λ(t−T ) dD(t) ∫T ( + (c O /ω) 0 λ(1 + λt)D (t)dt cO ) C O D (T ) = + λ cK + . ∫T ∫∞ −λ(t−T ) D (t)dt ω 0 D (t)dt + T e

(8.34)

8.3 Comparisons of Backup First and Overtime

189

Clearly, C O D (0) c F + c R D ∗ (λ) cO = + cK + , λ 1 − D ∗ (λ) ω ∫ C O D (∞) μ cO μ ∞ cO = (c F + c R ) + (1 + λt)D(t)dt + c K + . λ λ ω 0 ω Differentiating C O D (T ) with respect to T and setting it equal to zero, ∫∞

∫ e−λt dD(t) + (c O /ω)(1 + λT )e−λT D (T ) T D(t)dt ∫∞ −λt D (t)dt 0 T e ] [ ∫ T cO − c R D(T ) + λ(1 + λt)D(t)dt = c F , (1 + λT )D(T ) − ω 0

cR

T

(8.35)

whose left-hand side increases strictly with T from c O /ω to ∞. Thus, if c O /ω < c F , then there exists a finite and unique TO∗ D (0 < TO∗ D < ∞) which satisfies (8.35), and the resulting cost rate is ∫∞

∗

e−λt dD(t) + (c O /ω)(1 + λTO∗ D )e−λTO D D (TO∗ D ) ∫∞ −λt D (t)dt TO∗ D λe cO + cK + . (8.36) ω

cR C O D (TO∗ D ) = λ

TO∗ D

In particular, when D(t) = 1 − e−μt , (8.35) agrees with (8.33). Example 8.2 When D(t) = 1 − e−μt , F(t) = 1 − e−t , c K = 0.1, c R = 0.2, c N = 0.5 and c F = 5.0, Table 8.2 presents optimal TO∗ I , TO∗ D , and their cost rates C O I (TO∗ I )/μ and C O D (TO∗ D )/μ for c O /ω and μ. Comparing Tables 8.1 and 8.2, optimal TO∗ I and TO∗ D have the same properties as N I∗ and N D∗ for c O /ω and μ. We may also conclude that the backup model with update N is more economical than that with time T , because C I (N I∗ )/μ < C O I (TO∗ I )/μ and C D (N D∗ )/μ < C O D (TO∗ D )/μ, which will be approved analytically in the following section. ∎

8.3 Comparisons of Backup First and Overtime Suppose that full backup is made preventively at the first update over a planned time T (0 ≤ T < ∞) or at a number N (n = 0, 1, 2, . . . ) of updates, whichever occurs first. We discuss which policy is better at overtime T or update N analytically and numerically. Then, the probability that full backup is made at update N is ∫ 0

T

D(t)dF (N ) (t),

190

8 Which is Better Problems in Backup Models

Table 8.2 Optimal TO∗ I , TO∗ D , C O I (TO∗ I )/μ and C O D (TO∗ D )/μ when D(t) = 1 − e−μt , F(t) = 1 − e−t , c K = 0.1, c R = 0.2, c N = 0.5 and c F = 5.0 c O /ω = 0.5

μ

c O /ω = 1.0

TO∗ I

C O I (TO∗ I ) μ

TO∗ D

C O D (TO∗ D ) μ

TO∗ I

C O I (TO∗ I ) μ

TO∗ D

C O D (TO∗ D ) μ

0.01

33.556

94.757

4.505

338.239

27.122

152.381

3.179

532.282

0.02

24.419

55.619

4.540

171.465

19.645

86.168

3.196

269.194

0.05

16.474

29.674

4.646

71.474

13.123

43.385

3.248

111.405

0.10

12.688

19.888

4.832

38.275

9.980

27.671

3.338

58.919

0.20

10.371

14.571

5.250

21.951

7.987

19.182

3.533

32.903

0.50

9.481

11.880

6.937

13.305

6.937

14.305

4.263

18.187

1.00

10.989

12.788

10.989

12.788

7.662

14.293

5.996

15.291

2.00

15.376

16.875

20.219

16.414

10.461

17.940

10.461

17.940

5.00

26.434

27.753

38.735

24.161

18.818

30.147

22.771

28.946

the probability that it is made at the first update over time T is N −1 ∫ ∑ n=0

T

[∫

∞ T −t

0

] D(t + u)dF(u) dF (n) (t),

and the probability that it is made at failure is ∫

T

[1 − F

(N )

(t)]dD(t) +

0

N −1 ∫ ∑ n=0

T

[∫

∞ T −t

0

]

F(u)dD(t + u) dF (n) (t).

The mean time to full backup is ∫

T

t D(t) dF (N ) (t) +

0

∫ +

n=0 T

t[1 − F

(N )

0

=

N −1 ∫ ∑ n=0

N −1 ∫ ∑

0

T

[∫

T

[∫

0

(t)]dD(t) +

∞ T −t

N −1 ∫ ∑ n=0

∞

] (t + u)D(t + u)dF(u) dF (n) (t) T 0

[∫

∞ T −t

]

(t + u) F(u)dD(t + u) dF (n) (t)

]

F(u)D(t + u)du dF (n) (t),

0

which agrees with (8.4) when T = ∞ and with (8.17) when N = ∞. Therefore, the expected cost is, for incremental backup,

(8.37)

8.3 Comparisons of Backup First and Overtime

∼I (T, N ) = (c F + N M1 ) C

∫

T

D(t)dF (N ) (t)

0

+

N −1 ∑

∫

+

[∫

] D(t + u)dF(u) dF (n) (t)

∞ T −t

0

n=0 N −1 ∑

T

[c F + (n + 1)M1 ]

191

∫

T

(c F + n M1 + Nn + nc N )

[∫

0

n=0

∞

] F(u)dD(t + u) dF (n) (t)

0

] [∫ ∞ N −1 ∫ ( cO ) ∑ T D(t + u)dF(u) dF (n) (t) = cF + cK + ω n=0 0 0 ] ∫ [∫ ∞ N −1 [ ( ∑ c O )] T [D(t + u) − D(t)]dF(u) dF (n) (t), cR + n cN + + ω 0 0 n=0 (8.38) and for differential backup, ⎛ ∼D (T, N ) = ⎝c F + C

+

N −1 ∑ n=0

+

N −1 ∑ n=0

∫

T

+ 0

⎞ Mj⎠

∫

⎝c F +

n+1 ∑ j=1

⎛ ⎝c F +

n ∑ j=1

∞ T −t

⎞

Mj⎠

T

D(t)dF (N ) (t)

0

j=1

⎛

[∫

N ∑

∫

T 0

[∫ ⎞

M j + Nn ⎠

∞ T −t

[∫

] D(t + u)dF(t) dF (n) (t)

T

[F (n) (t) − F (n+1) (t)]dD(t)

0

] ] (n) F(t)dD(t + u) dF (t)

] ] ∫ T [∫ ∞ (n + 1)c O = cF + D(t + u)dF(u) dF (n) (t) cK + ω 0 0 n=0 ] ∫ [∫ ∞ N −1 ( ∑ nc O ) T [D(t + u) − D(t)]dF(u) dF (n) (t). cR + + ω 0 0 n=0 N −1 [ ∑

(8.39)

∼D (∞, N ) = C ∼I (N ) in (8.1), C ∼D (N ) in (8.2), and ∼I (∞, N ) = C Clearly, C ∼D (T , ∞) = C ∼O I (T ) in (8.18), C ∼O D (T ) in (8.19). ∼I (T , ∞) = C C ∼D (T , N ) is ∼I (T , N ) and C The difference of C

192

8 Which is Better Problems in Backup Models

] N −1 ∫ T [∫ ∞ ∑ ∼I (T, N ) = c O ∼D (T , N ) − C n D(t + u)dF(u) dF (n) (t) C ω n=0 0 0 ] [∫ ∫ N −1 T ∞ ∑ n [D(t + u)− D(t)]dF(t) dF (n) (t), − cN n=0

0

0

which follows that if (

] N −1 ∫ T [∫ ∞ cO ) ∑ n D(t + u)dF(u) dF (n) (t) ω n=0 0 0 N −1 ∫ T ∑ n D(t)dF (n) (t), > cN

cN +

n=0

0

then incremental backup is better than differential one. From (8.37), the expected cost rate is, for incremental backup, C I (T , N ) =

] ∑ N −1 ∫ T [∫ ∞ c F + (c K + c O /ω) n=0 D (t + u)dF(u) dF (n) (t) 0 0 ∫ T {∫ ∞ } (n) ∑ N −1 + n=0 [c R + n(c N + c O /ω)] 0 0 [D(t + u) − D(t)]dF(u) dF (t) , ] ∑ N −1 ∫ T [∫ ∞ (n) n=0 0 0 F (u) D (t + u)du dF (t) (8.40)

and for differential backup, C D (T , N ) = ∫ T [∫ ∞ ] ∑ N −1 c F + n=0 [c K + (n + 1)c O /ω] 0 0 D (t + u)dF(u) dF (n) (t) ∫ T {∫ ∞ } (n) ∑ N −1 + n=0 (c R + nc O /ω) 0 0 [D(t + u) − D(t)]dF(u) dF (t) . ] ∑ N −1 ∫ T [∫ ∞ (n) n=0 0 0 F (u) D (t + u)du dF (t)

(8.41)

Clearly, C I (∞, N ) = C I (N ) in (8.5), C D (∞, N ) = C D (N ) in (8.6), and C I (T , ∞) = C O I (T ) in (8.20), C D (T , ∞) = C O D (T ) in (8.21).

(1) Incremental Backup We find optimal TI∗N and N I∗T to minimize C I (T , N ) in (8.40) when D(t) = 1 − e−μt . Then, the expected cost rate in (8.40) is

8.3 Comparisons of Backup First and Overtime

193

∑ N −1 ∫ T −μt (n) c F + (c N + c O /ω)[1 − F ∗ (μ)] n=0 n 0 e dF (t) C I (T , N ) = ∫T ∑ N −1 μ [1 − F ∗ (μ)] n=0 0 e−μt dF (n) (t) ( c O ) F ∗ (μ) + cK + + cR . (8.42) ω 1 − F ∗ (μ) In particular, when N = 1, ( C I (T , 1) cF c O ) F ∗ (μ) = + c + + cR , K μ 1 − F ∗ (μ) ω 1 − F ∗ (μ) and TI∗N = ∞. Forming the inequality C I (T , N − 1) − C I (T, N ) < 0 and C I (T, N + 1) − C I (T , N ) ≥ 0 N −1 ∑

∫ (N − 1 − n)

T

cF (c N + c O /ω)[1 − F ∗ (μ)] ∫ T N ∑ ≤ (N − n) e−μt dF (n) (t),

e−μt dF (n) (t)
0. In addition, noting that the right-hand side increases strictly with T to ∑N

− F ∗ (μ)n ] , 1 − F ∗ (μ)

n=1 [1

optimal N I∗T decreases with T to N I∗ given in (8.8) and N I∗T ≥ N I∗ . Differentiating C I (T , N ) with respect to T and setting it equal to zero, Q(T, N )

N −1 ∫ ∑ n=0

T

e−μt dF (n) (t) −

0

N −1 ∫ ∑ n n=0

T

e−μt dF (n) (t)

0

cF = , (c N + c O /ω)[1 − F ∗ (μ)] where for N ≥ 2,

∑ N −1

Q(T , N ) ≡ ∑n=1 N −1 n=1

n f (n) (T ) f (n) (T )

(8.44)

≤ N − 1.

Substituting (8.43) for (8.44), Q(T , N ) > N − 1,

(8.45)

194

8 Which is Better Problems in Backup Models

which does not hold for any T . Thus, there does not exist any finite TI∗N which satisfies (8.44) for any N , and TI∗N = ∞, which follows that optimal policy is TI∗N = ∞, and N I∗N = N I∗ given in (8.8). (2) Differential Backup ∗ We find optimal TD∗ N and N DT to minimize C D (T , N ) in (8.41) when D(t) = 1 − −μt e . Then, the expected cost rate in (8.41) is

∑ N −1 ∫ T −μt (n) c F + (c O /ω) n=0 n 0 e dF (t) C D (T, N ) = ∫T ∑ N −1 μ [1 − F ∗ (μ)] n=0 0 e−μt dF (n) (t) ( c O ) F ∗ (μ) + cK + + cR . ω 1 − F ∗ (μ)

(8.46)

In particular, when N = 1, C D (T , 1) = C I (T , 1) and TD∗ N = ∞. Forming the inequality C D (T , N − 1) − C D (T, N ) < 0 and C D (T, N + 1) − C(T , N ) ≥ 0, N −1 ∑

∫ (N − 1 − n)

T

0

n=0

∑ cF ≤ (N − n) c O /ω n=0 N

e−μt dF (n) (t)
0. In addition, noting that the right-hand side increases ∗ decreases with T to strictly with T to the left-hand side of (8.14), i.e., optimal N DT ∗ ∗ ∗ N D given in (8.14), and N DT ≥ N D . Differentiating C D (T , N ) with respect to T and setting it equal to zero, Q(T , N )

N −1 ∫ ∑ n=0

T 0

e−μt dF (n) (t) −

N −1 ∫ ∑ n n=0

T

e−μt dF (n) (t) =

0

cF , c O /ω

(8.48)

whose left-hand side agrees with that of (8.44). Substituting (8.47) for (8.48) is Q(T , N ) > N − 1, ∗ which agrees with (8.45). Thus, optimal time is TD∗ N = ∞ and N DT = N D∗ given in (8.14). This concludes from the above discussions that when the full backup costs of time T and update N are the same, full backup with update N is better than that with time T for both incremental and differential backups.

8.3 Comparisons of Backup First and Overtime

195

Example 8.3 When D(t) = 1 − e−μt , F(t) = 1 − e−t , c O /ω = 0.5, c K = 0.1, ∗ for c R = 0.2, c N = 0.5 and c F = 5.0, Table 8.3 presents optimal N I∗T and N DT ∗ ∗ T and μ, and Table 8.4 presents optimal TI N and TD N for N and μ. ∗ Both N I∗T and N DT in Table 8.3 decrease with T to N I∗ and N D∗ in Table 8.1, and ∗ )/μ decrease with T to C I (N I∗ )/μ and their cost rates C I (T , N I∗T )/μ and C D (T , N DT ∗ C D (N D )/μ in Table 8.1. This means that backup with update N are more economical than those with overtime T . In Table 8.4, when failure rate μ become small, optimal TI∗N = ∞ that means full backup is made at the number N = 10 or N = 20 of incremental backup. In this case, N = 10 and N = 20 may be small enough to limit the data restoration costs, however, they are not optimal ones. In Table 8.4, when failure rate μ becomes high, optimal TD∗ N = ∞ and full backup is made at the number N = 10 of differential backups. In this case, N = 10 may be small enough to limit ∎ backup costs, however, they are not optimal ones.

∗ , C (T , N ∗ )/μ and C (T, N ∗ )/μ when D(t) = 1 − e−μt , F(t) = Table 8.3 Optimal N I∗T , N DT I D IT DT 1 − e−t , c O /ω = 0.5, c K = 0.1, c R = 0.2, c N = 0.5 and c F = 5.0 T = 10.0 T = 20.0 μ ∗ ∗ ) ∗ ∗ ) ∗ ∗ C D (T,N DT C D (T,N DT N I∗T C I (T,N I T ) N DT N I∗T C I (T,N I T ) N DT μ

0.01 0.02 0.05 0.10 0.20 0.50 1.00 2.00 5.00

59 34 19 14 12 11 12 17 31

119.184 64.168 31.127 20.052 14.497 11.857 12.790 16.947 30.902

μ

6 6 6 6 6 8 12 22 51

314.917 159.877 66.963 36.180 21.158 13.212 12.790 16.573 30.425

μ

39 26 18 14 12 11 12 17 29

98.726 55.977 29.387 19.642 14.420 11.846 12.760 16.790 29.217

μ

6 6 6 6 6 6 12 22 48

314.709 159.771 66.919 36.158 21.146 13.202 12.760 16.418 28.749

Table 8.4 Optimal TI∗N , TD∗ N , C I (TI∗N , N )/μ and C D (TD∗ N , N )/μ when D(t) = 1 − e−μt , F(t) = 1 − e−t , c O /ω = 0.5, c K = 0.1, c R = 0.2, c N = 0.5 and c F = 5.0 μ

N = 10

N = 20

TI∗N

C I (TI∗N ,N ) μ

0.01

∞

124.024

4.749

336.459

∞

99.191

4.505

338.239

0.02

∞

66.240

4.795

170.545

∞

55.840

4.540

171.465

TD∗ N

∗ ,N ) C D (TD N μ

TI∗N

C I (TI∗N ,N ) μ

TD∗ N

∗ ,N ) C D (TD N μ

0.05

∞

31.558

4.939

71.069

23.576 29.442

4.646

71.465

0.10

∞

19.991

5.204

38.083

13.300 19.858

4.832

38.275

0.20

∞

14.231

5.874

21.796

10.476 14.568

5.250

21.951

0.50

∞

10.975

11.707

13.236

9.525

6.939

13.305

11.880

1.00

∞

9.965

∞

9.965

11.172 12.788

11.172

12.788

2.00

180.935

9.454

∞

8.535

18.993 16.810

69.527

14.501

5.00

150.912

9.265

∞

6.920

62.595 18.939

111.488

11.607

196

8 Which is Better Problems in Backup Models

8.4 Database Backup Last We consider the backup last models in which full backup is made at a planned time T or at a number N of updates, whichever occurs last.

8.4.1 Backup Last Suppose that full backup is made preventively at time T (0 ≤ T < ∞) or at update N (N = 0, 1, 2, . . . ), whichever occurs last. Then, the probability that full backup is made at time T is F (N ) (T )D(T ), the probability that it is made at update N is ∫

∞

D(t)dF (N ) (t),

T

and the probability that it is made at failure is ∫

∞

D(T ) +

[1 − F (N ) (t)]dD(t).

T

Therefore, the expected cost until full backup is, for incremental backup, ∼I L (T , N ) = C ∞ ∑

(c F + n M1 )[F (n) (T ) − F (n+1) (T )]D(T ) + (c F + N M1 )

n=N

+ +

∞ ∑

(c F + n M1 + Nn + nc N )

n=0 N −1 ∑

∫ T

(c F + n M1 + Nn + nc N )

n=0

0

T

D(t)dF (N ) (t)

[F (n) (t) − F (n+1) (t)]dD(t)

∫ ∞ T

∫ ∞

[F (n) (t) − F (n+1) (t)]dD(t)

⎤ ⎡ ∞ ∫ N ∫ ∞ ( ∑ cO ) ⎣∑ T D(t)dF (n) (t) + D(t)dF (n) (t)⎦ = cF + cK + ω T 0 n=1 n=1 ⎫ ⎧ ∫ ∞ N −1 ⎬ ⎨ ) ( ∑ T ∑ ∫ ∞ cO + cN + F (n) (t)dD(t) + n [F (n) (t) − F (n+1) (t)]dD(t) ⎭ ω ⎩ T n=1 0 n=0 [ ] ∫ ∞ + c R D(T ) + [1 − F (N ) ]dD(t) , (8.49) T

8.4 Database Backup Last

197

and for differential backup, ∼DL (T , N ) = C

∞ ∑

⎛ ⎝c F +

n=N

( + cF +

N ∑

+

n=0

+

N −1 ∑

⎝c F +

n ∑

⎝c F +

n ∑

cO ω

M j + Nn ⎠ ⎞ M j + Nn ⎠

∫

T

[F (n) (t) − F (n+1) (t)]dD(t)

0

∫

∞

[F (n) (t) − F (n+1) (t)]dD(t)

T

j=1

∫ ∫ N ( ∑ nc O ) T nc O ) ∞ D(t)dF (n) (t) + D(t)dF (n) (t) cK + ω ω 0 T n=1 } ∫ N −1 ∞ ∑ (n) (n) (n+1) F (t)dD(t) + n [F (t) − F (t)]dD(t)

∞ ( ∑

cK +

n=1

+

⎞

j=1

⎛

M j ⎠ [F (n) (T ) − F (n+1) (T )]D(T )

D(t)dF (N ) (t)

T

n=0

= cF +

∞

Mn

⎛

⎞

j=1

)∫

n=1

∞ ∑

n ∑

{ [

∞ ∫ ∑ n=1

T

0

n=1

∫

∞

+ c R D(T ) +

T

] [1 − F (N ) (t)]dD(t) .

(8.50)

T

∼I L (T, N ) and C ∼I D (T , N ) is The difference of C ∼I L (T, N ) ∼DL (T , N ) − C C ] [∞ ∫ N −1 ∫ ∞ T ∑ cO ∑ (n+1) (n+1) n D(t)dF (t) + n D(t)dF (t) = ω n=1 0 T n=1 } {∞ ∫ N −1 ∫ ∞ ∑ T ∑ (n) (n) (n+1) F (t)dD(t) + n [F (t) − F (t)]dD(t) . − cN n=1

0

n=1

T

The mean time to full backup is ∫ ∞ ∫ T ∫ ∞ t D(t)dF (N ) (t) + t dD(t) + t[1 − F (N ) (t)]dD(t) TF (N ) (T )D(T ) + T 0 T ∫ ∞ ∫ T D(t)dt + [1 − F (N ) (t)] D(t) dt. (8.51) = 0

T

Therefore, the expected cost rate is, for incremental backup,

198

8 Which is Better Problems in Backup Models

C I L (T , N ) =

[∑ ∫ ] ∑N ∫ ∞ T ∞ (n) (n) c F + (c K +c O /ω) D (t)dF (t) + D (t)dF (t) n=1 T ∫ ∞ n=1 0 { } + c R D(T ) + {T [1 − F (N ) (t)]dD(t) ∑∞ ∫ T (n) + (c N + c O /ω) n=1 0 F (t)dD(t) } ∑ N −1 ∫ ∞ (n) + n=1 n T [F (t) − F (n+1) (t)]dD(t) , (8.52) ∫T ∫∞ (N ) (t)] D (t) dt. 0 D (t)dt + T [1 − F

and for differential backup, C DL (T , N ) = ∫T ∑ cF + ∞ (c + nc O /ω) 0 D (t)dF (n) (t) ∫∞ ∑ N n=1 K (n) + n=1 ∫ ∞O /ω) T (ND) (t)dF (t) { (c K + nc } + c R D(T{ ) + T [1 − F (t)]dD(t) } ∑∞ ∫ T (n) ∑ N −1 ∫ ∞ (n) (n+1) + (c O /ω) F (t)dD(t)+ n [F (t) − F (t)]dD(t) n=1 0 n=1 T . ∫T ∫∞ (N ) (t)] D (t) 0 D (t)dt + T [1 − F (8.53) Clearly, C I L (0, N ) = C I (N ) in (8.5) and C DL (0, N ) = C D (N ) in (8.6).

8.4.2 Backup Overtime Last Suppose that full backup is made preventively at the first update over a planned time T (0 ≤ T < ∞) or at a number N (N = 0, 1, 2, . . . ) of updates, whichever occurs last. Then, the probability that full backup is made at update N is ∫

∞

D(t)dF (N ) (t),

T

the probability that it is made at the first update over time T is ∞ ∫ ∑ n=N

0

T

[∫

∞ T −t

] D(t + u)dF(u) dF (n) (t),

and the probability that it is made at failure is ∫

∞

D(T ) + T

[1 − F

(N )

(t)]dD(t) +

∞ ∫ ∑ n=N

T 0

[∫

∞ T −t

]

F(u)dD(t + u) dF (n) (t).

8.4 Database Backup Last

199

The mean time to full backup is ∫

∞ T

∫

+

T

∫

[∫

[∫

∞ T −t

0

∞ T −t

∞

] ] y dD(y) dF(u) dF (n) (t)

t+u

T

[∫

0

n=N ∞

T

T

t[1 − F (N ) (t)]dD(t)

t dD(t) +

∞ ∫ ∑

=

n=N ∞

∫

0

+

∞ ∫ ∑

t D(t)dF (N ) (t) +

]

(t + u) F(u)dD(t + u) dF (n) (t)

[1 − F (N ) (t)]D(t) dt +

0

∞ ∫ ∑

T

[∫

0

n=N

∞

] F(u)D(t + u) du dF (n) (t).

0

(8.54) Therefore, the expected cost until full backup is, for incremental backup, ∼I L (T , N ) = C ∞ ∑

∫

[c F + (n + 1)M1 ]

n=N

T

0

∫

∞

+ (c F + N M1 )

[∫

∞ T −t

] D(t + u)dF(u) dF (n) (t)

D(t)dF (N ) (t)

T

+ +

N −1 ∑ n=0 ∞ ∑

∫

∞

(c F + n M1 + Nn + nc N )

[F (n) (t) − F (n+1) (t)]dD(t)

0

∫

T

(c F + n M1 + Nn + nc N ) 0

n=N

[∫

∞

]

F(u)dD(t + u) dF (n) (t)

0

{ N ∫ cO ) ∑ ∞ = cF + cK + D(t)dF (n) (t) ω 0 n=1 } ] ∞ ∫ T [∫ ∞ ∑ (n) D(t + u)dF(u) dF (t) + (

n=N

0

0

∫ ( c O )] ∞ (n) [F (t) − F (n+1) (t)]dD(t) cR + n cN + ω 0 n=0 ] ∫ T [∫ ∞ ∞ [ ( )] ∑ cO + [D(t + u) − D(t)]dF(u) dF (n) (t), cR + n cN + ω 0 0 n=N

+

N −1 [ ∑

(8.55) and for differential backup,

200

8 Which is Better Problems in Backup Models

∼DL (T , N ) = C ⎛ ⎞ ∫ ∞ n+1 ∑ ∑ ⎝c F + Mj⎠ n=N

N ∑

+ cF + +

N −1 ∑ n=0

+

∞ ∑

)∫

∞

n ∑

⎝c F +

⎞

M j + Nn ⎠

j=1

⎛

n ∑

⎝c F +

⎞ M j + Nn ⎠

cK +

n=1

∫

∞

nc O ) ω

∫

∞

[F (n) (t) − F (n+1) (t)]dD(t)

0

∫

T 0

j=1 N ( ∑

] D(t + u)dF(u) dF (n) (t)

D(t)dF (N ) (t)

T

n=N

= cF +

∞ T −t

Mn

n=1

⎛

[∫

0

j=1

(

T

[∫

∞ T −t

]

F(u)dD(t + u) dF (n) (t)

D(t)dF (n) (t)

0

] T [∫ ∞ (n + 1)c O + D(t + u)dF(u) dF (n) (t) cK + ω 0 0 n=N ∫ N −1 ( ∑ nc O ) ∞ (n) + [F (t) − F (n+1) (t)]dD(t) cR + ω 0 n=0 ] ∫ [∫ ∞ ∞ ( ∑ nc O ) T + [D(t + u) − D(t)]dF(u) dF (n) (t). cR + ω 0 0 n=N ∞ ∑

[

]∫

(8.56)

∼I L (0, N ) = C ∼DL (0, N ) = C ∼I (N ) in (8.1), C ∼D (N ) in (8.2), and Clearly, C ∼ ∼ ∼ ∼ C I L (T , 0) = C O I (T ) in (8.18), C DL (T , 0) = C O D (T ) in (8.19). ∼DL (T , N ) is ∼I L (T , N ) and C The difference of C ∼I L (T , N ) = ∼DL (T , N ) − C C ] [ N −1 ∫ ] ∫ T [∫ ∞ ∞ ∞ ∑ cO ∑ (n+1) (n) n D(t)dF (t) + n D(t + u)dF(u) dF (t) ω n=0 0 0 0 n=N ( N −1 ∫ ∞ ∑ − cN n [F (n) (t) − F (n+1) (t)]dD(t) 0

n=0

∫ ∞ ∑ n + n=N

T 0

[∫

∞

] [D(t + u) − D(t)]dF(u) dF

) (n)

0

When D(t) = 1 − e−μt , (8.57) is cO ∗ F (μ) − c N [1 − F ∗ (μ)]. ω

(t) .

(8.57)

8.4 Database Backup Last

201

From (8.54), (8.55) and (8.56), the expected cost rate is, for incremental backup, C I L (T , N ) =

{∑ ∫ ∞ N (n) c F + (c K + c O /ω) n=1 0 D(t)dF (t) } ∫ [∫ ] ∑ T ∞ (n) + ∞ n=N 0 0 D(t + u)dF(u) dF (t) ∫∞ ∑ N −1 [c R + n(c N + c O /ω)] 0 [F (n) (t) − F (n+1) (t)]dD(t) + n=0 ∫ T {∫ ∞ } (n) ∑ + ∞ n=N [c R + n(c N + c O /ω)] 0 0 [D(t + u) − D(t)]dF(u) dF (t) , ∫∞ ] ∑∞ ∫ T [∫ ∞ (N ) (t)]D(t) dt + (n) n=N 0 0 [1 − F 0 D(t + u) F(u)du dF (t) (8.58)

and for differential backup, C DL (T , N ) = ∫∞ ∑N c F + n=1 (c K + nc O /ω) 0 D (t)dF (n) (t) ∫ T [∫ ∞ ] ∑ + ∞ [c K + (n + 1)c O /ω] 0 0 D (t + u)dF(u) dF (n) (t) ∫ ∑n=N ∞ N −1 (c R + nc O /ω) 0 [F (n) (t) − F (n+1) (t)]dD(t) + n=0 ∫ T {∫ ∞ } (n) ∑∞ + n=N (c R + nc O /ω) 0 0 [D(t + u) − D(t)]dF(u) dF (t) . ∫∞ ] ∑∞ ∫ T [∫ ∞ (N ) (t)]D (t) dt + (n) n=N 0 0 [1 − F 0 D (t + u) F (u)du dF (t)

(8.59)

Clearly, C I L (0, N ) = C I (N ) in (8.5), C DL (0, N ) = C D (N ) in (8.6), and C I L (T , 0) = C O I (T ) in (8.20), C DL (T , 0) = C O D (T ) in (8.21).

(1) Incremental Backup We find optimal TI∗L and N I∗L to minimize C I L (T, N ) in (8.58) when D(t) = 1 − e−μt . Then, the expected cost rate in (8.58) is C I L (T , N ) =

∫ T −μt (n) ] ∑∞ ∗ n n F (μ) + n dF (t) n=0 n=N 0 e ∫ ∑ T −μt 1 − F ∗ (μ) N + [1 − F ∗ (μ)] ∞ dF (n) (t) n=N 0 e ( ) cO F ∗ (μ) + cK + + cR . (8.60) ω 1 − F ∗ (μ)

c F + (c N + c O /ω)[1 − F ∗ (μ)]

[∑ N −1

Forming the inequality C I L (T , N − 1) − C I L (T , N ) < 0 and C I L (T, N + 1) − C I (T , N ) ≥ 0,

202

8 Which is Better Problems in Backup Models N −1 ∑

(N − 1 − n)F ∗ (μ)n −

n=0

∞ ∑

∫

e−μt dF (n) (t)

0

n=N

cF ≤ < (ω + c O /ω)[1 − F ∗ (μ)] ∫ N ∞ ∑ ∑ (N − n)F ∗ (μ)n − (n − N ) n=0

T

(n + 1 − N )

T

e−μt dF (n) (t),

(8.61)

0

n=N

whose right-hand decreases with N at first, and after that, increases strictly to ∞. Thus, there exists a finite and unique minimum N I∗L (0 ≤ N I∗L < ∞) which satisfies (8.61) for any T . In addition, the right-hand side decreases strictly with T from the left-hand side of (8.43), i.e., optimal N I∗L increases with T from N I∗ given in (8.8), and N I∗L ≥ N I∗ . Differentiating C I L (T , N ) with respect to T and setting it equal to zero, ∼ , N) Q(T

[ N −1 ∑

∗

F (μ) + n

n=0

−

N −1 ∑ n=0

where

∞ ∫ ∑

T

e

dF

(n)

(t)

e−μt dF (n) (t) =

n

n=N

] −μt

0

n=N

∫ ∞ ∑ n F (μ) − n ∗

T

0

cF , (8.62) (c N + c O /ω)[1 − F ∗ (μ)]

∑∞ n f (n) (T ) ∼ > N. Q(T , N ) ≡ ∑n=N ∞ (n) (T ) n=N f

Substituting (8.61) for (8.62), (8.62) is ∼ , N ) > N − 1, Q(T which always holds for any T , i.e., TI∗L = 0. This shows that optimal policy is TI∗L = 0 and N I∗L = N I∗ given in (8.8). (2) Differential Backup ∗ ∗ We find optimal TDL and N DL to minimize C DL (T, N ) in (8.59) when D(t) = −μt 1 − e . Then, the expected cost rate in (8.59) is

∫ T −μt (n) ] ∑∞ ∗ n n F (μ) + n dF (t) n=N 0 e C DL (T , N ) = ∫ ∑ T −μt dF (n) (t) μ 1 − F ∗ (μ) N + [1 − F ∗ (μ)] ∞ n=N 0 e ( ) ∗ cO F (μ) + cK + + cR . (8.63) ω 1 − F ∗ (μ) c F + (c O /ω)

[∑

N −1 n=0

References

203

Forming the inequality C DL (T , N − 1) − C DL (T , N ) < 0 and C DL (T, N + 1) − C DL (T , N ) ≥ 0, N −1 ∑

∫ (N − 1 − n)

T

0

n=0

∑ cF ≤ (N − n) c O /ω n=0 N

e−μt dF (n) (t)
N − 1, Q(T ∗ which always holds for any T , i.e., TDL = 0. This shows that optimal policy is ∗ ∗ ∗ TDL = 0 and N DL = N D given in (8.15), which follows that full backup with N is better than backup last.

8.5 Problem 1. Consider incremental and differential backups in which full backup is made at volume K (0 ≤ K ≤ ∞) of update.

References 1. Siberschatz A, Korth HF, Sudarshan S (2010) Database system concepts, 6th edn. McGraw-Hill Education, New York 2. McDowall RD (2001) Computer system backup and recovery. Qual Assur J5:149–155 3. Velpuri R, Adkoli A (1998) Oracle 8 backup & recovery handbook. McGraw Hill, England 4. Nakagawa T (2007) Shock and damage models in reliability theory. Springer, London 5. Qian CH, Nakamura S, Nakagawa T (2002) Optimal backup policies for a database system with incremental backup. Electron Commun Japan, Part 3 85:1–9 6. Nakamuara S, Qian C, Funkumoto S, Nakamura T (2003) Optimal backup policy for a database system with incremental and full backup. Math Comput Model 38:1373–1379

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7. Qian CH (2010) Backup policies for a database system. In: Nakamura S, Nakagawa T (eds) Stochastic reliability modeling, optimaization and applications. World Scientific, Singapore, pp 177–204 8. Nakamura S, Zhao X, Nakagawa T (2017) Constant and random full backup models with incremental and differential backup schemes. Inter J Reliab Qual Saf Eng 24:3. https://doi.org/ 10.1142/S0218539317500152 9. Nakamura S, Zhao X, Qian CH (2019) Optimum backup policies with update failure at random times. Inter J Reliab Qual Saf Eng 26:1. https://doi.org/10.1142/S0218539319500025 10. Levitin L, Xing L, Dai Y (2017) Optimal distribution of nonperiodic full and incremental backups. IEEE Trans Syst Man Cybern 47:3310–3320

Chapter 9

Which is Better Problems in Checkpoint Models

It is of great importance to develop the design of computer systems with high reliability as human technologies have been rapidly growing up. Most systems consist of computing units which need high reliability and speed processes. Therefore, we have to design previously such computer systems with high quality. However, we cannot eliminate some computer errors in actual fields, and to mask these errors, we have to make a better planning for high reliable computer systems with multiple units and tolerant performances. Some errors may occur due to space radiation, electric-magnetic waves, low quality of hardware, element overheat and over current, and so on. These errors might lead to faults or failures and cause serious damage to systems. To prevent such faults, varied types of fault-tolerant technologies including system redundancies and configurations have been offered [1, 2]. Using such techniques, we can achieve high reliabilities and effective performances of objective computer systems. Failures due to errors are found and computer systems may lose their consistency. To protect such critical incidents, several recovery methods are required to restore a consistent state just before failures. The most standard method is to take copies of normal states at suitable times, which is called checkpoint times. When errors of the process occur and are detected, we perform the rollback operation to the nearest checkpoint time, restore a consistent state and reexecute the process. The simplest scheme for error detection recovery techniques was [3]: We execute two independent modules which compare two states at checkpoint times. If two states of each module do not match with each other, we go back to the previous checkpoint and make their retrials. We adopted a double modular structure and a majority decision structure as the checkpoint methods, and one and random tasks as the objective work of processes [4, 5]: One task is divided equally and checkpoints are placed at periodic times, and random tasks form tandem ones with an identical processing distribution and checkpoints are placed at each end of processing times. We obtained © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Mizutani et al., Which-Is-Better (WIB): Problems in Reliability Theory, Springer Series in Reliability Engineering, https://doi.org/10.1007/978-3-031-27316-2_9

205

206

9 Which is Better Problems in Checkpoint Models

the mean execution times until the success of all processes and discussed optimal checkpoint policies to minimize their mean times theoretically and numerically. This chapter is written based on the results [4, 5]: In Sect. 9.1, we take up a periodic checkpoint model for a double modular structure and a majority decision structure, obtain the mean execution times to complete the process and discuss optimal policies to minimize them. Using such results, we decide numerically which structure is better. In Sect. 9.2.1, we take up a random checkpoint model with N tandem and bulk tasks, and compare two kinds of tasks. In Sect. 9.2.2, supposing two kinds of checkpoints, we consider their three schemes, obtain the mean execution times to complete the process and compare them. In Sect. 9.3, two numerical examples are given and which scheme is better is determined numerically. Finally, in Sect. 9.4, we propose an imperfect checkpoint model and discuss optimal policies to minimize the mean execution times [6].

9.1 Periodic Checkpoint Model Suppose that the process has a native execution time S (0 < S < ∞) which does not include any overheads of retries and checkpoint generations. Then, we divide S equally into N (N = 1, 2, . . . ) intervals and place checkpoints at periodic times kT (k = 1, 2, . . . , N − 1), where S = N T [7, p. 123]. Furthermore, we take up the redundant techniques of error detection and masking for a finite execution time S, and adopt a double modular structure and a majority decision structure with (2n + 1) modules. For such periodic checkpoint models, we compute the mean execution times to complete the process and compare them theoretically and numerically.

(1) Double Modular Structure It is assumed that errors occur at constant rate λ. To detect errors, we provide two independent modules where they compare two states at periodic times kT (k = 1, 2, . . . , N ). When two states match with each other, two processes are correct and go forward. However, when two states do not match, it is judged that some errors have occurred. Then, we make a rollback operation to the previous checkpoint and reexecute the process. In this case, the probability that the process has no error during [0, t] is given by e−λt , and the probability that two modules have no error during [0, t] is F 1 (t) = e−2λt . The process completes and ends when all of two processes are successful for N intervals. Let introduce overhead c1 for the comparison of 2 states. The mean execution time l1 (N ) to complete the process is the summation of processing times and overhead c1 . When some errors have been detected at checkpoints, two processes are rolled back to the previous checkpoint. Then, the mean execution time ∼ l1 (1) for a checkpoint interval ((k − 1)T, kT ] is given by a renewal equation: ∼ l1 (1)(1 − e−2λT ), l1 (1) = T + c1 + ∼

9.1 Periodic Checkpoint Model

and solving it for ∼ l1 (1),

207

∼ l1 (1) = (T + c1 )e2λT .

Thus, the mean execution time to complete the process is L 1 (N ) ≡ N∼ l1 (1) = N (T + c1 )e2λT = (S + N c1 )e2λS/N

(N = 1, 2, . . . ). (9.1) We find optimal number N1∗ to minimize L 1 (N ) for a specified S. Noting that lim N →∞ L 1 (N ) = ∞, a finite N1∗ (1 ≤ N1∗ < ∞) exists. Rewriting (9.1) with the function of T (0 < T ≤ S), ( c1 ) 2λT L 1 (T ) = S 1 + e . T

(9.2)

Differentiating L 1 (T ) with respect to T and setting it equal to zero, T 2 + c1 T − Solving it with T , ∼1 = c1 T 2

c1 = 0. 2λ

(9.3)

(/

) 2 1+ −1 . λc1

(9.4)

Therefore, we have the following partition method of computing optimal N1∗ [7, p. 42]: ∼1 ] ≡ N , where [x] denotes the largest integer contained ∼1 < S, we put [S/T (i) If T in x, and compute L 1 (N ) and L 1 (N + 1) from (9.1). If L 1 (N ) ≤ L 1 (N + 1) then N1∗ = N and T1∗ = S/N1∗ , and conversely, if L 1 (N + 1) < L 1 (N ) then N1∗ = N + 1. ∼1 ≥ S then N1∗ = 1, i.e., we should place no checkpoint until time S, and the (ii) If T mean time is L 1 (S) = (S + c1 )e2λS . ∼1 in (9.3) increases with c1 , N1∗ decreases with c1 to 1. FurtherNote that because T ∼ more, T1 does not depend on S. Thus, when S is very large, is changed greatly or is ∼1 as an approximate checkpoint time. unclear, we could adopt T Example 9.1 Table 9.1 presents optimal N1∗ , λT1∗ and the resulting execution time λL 1 (N1∗ ) for λc1 when λS = 0.1. This indicates that N1∗ decrease to 1 and λT1∗ increase to λS = 0.1. For example, when λ = 0.01, c1 = 0.1 and S = 10.0, N1∗ = 5, T1∗ = 10/5 = 2.0, and L 1 (5) = 10.929, which is longer about 9.3% than S. ∎

(2) Triple Majority Decision Structure We provide a majority decision structure with three modules, i.e., a 2-out-of-3 structure as an error masking structure. When more than two states of three modules

208

9 Which is Better Problems in Checkpoint Models

Table 9.1 Optimal N1∗ , λT1∗ and λL 1 (N1∗ ) when λS = 0.1 ∼1 × 102 λc1 × 103 λT N∗ λL 1 (N ∗ ) × 102 1

0.1 0.2 0.3 0.4 0.5 1.0 1.5 2.0 3.0 4.0 5.0 10.0 20.0

0.702 0.990 1.210 1.394 1.556 2.187 2.665 3.064 3.726 4.277 4.756 6.589 9.050

1

14 10 8 7 6 5 4 3 3 2 2 2 1

10.286 10.406 10.499 10.578 10.650 10.929 11.143 11.331 11.715 11.936 12.157 13.435 14.657

λT1∗ × 102 0.714 1.000 1.250 1.429 1.667 2.000 2.500 3.333 3.333 5.000 5.000 5.000 10.000

match with each other, the process in this interval is correct, i.e., the structure can mask a single module error. Then, the probability that the process is correct during interval ((k − 1)T, kT ) (k = 1, 2, . . . , N ) is F 2 (T ) ≡ e−3λT + 3e−2λT (1 − e−λT ) = 3e−2λT − 2e−3λT .

(9.5)

Let c2 be the overhead for the comparison of a majority decision of 3 modules. By the similar method of obtaining (9.1), the mean time to compute the process is L 2 (N ) =

N (T + c2 ) F 2 (T )

=

S + N c2 − 2e−3λT

3e−2λT

Setting T ≡ S/N , L 2 (T ) =

(N = 1, 2, . . . ).

S(1 + c2 /T ) . 3e−2λT − 2e−3λT

(9.6)

(9.7)

∼2 to minimize (9.7). Differentiating L 2 (T ) with respect to T We find optimal T and setting it equal to zero, ( c2 ) c2 , = (eλT − 1) T 2 + c2 T − 2λ 6λ

(9.8)

whose left-hand side increases strictly with T from 0 to ∞. Thus, there exists a finite ∼2 (0 < T ∼2 < ∞) which satisfies (9.8). and unique T Therefore, using the partition method, we get optimal N2∗ and T2∗ which minimize L 2 (N ) in (9.6). In particular, when c1 = c2 , it is easily shown from (9.3) and (9.8) ∼1 , i.e., N2∗ ≤ N1∗ , which follows that a majority decision structure with 3 ∼2 > T that T

9.1 Periodic Checkpoint Model

209

Table 9.2 Optimal N2∗ , λT2∗ and λL 2 (N2∗ ) when λS = 0.1 ∼2 × 102 λc2 × 103 λT N∗ λL 2 (N ∗ ) × 102 2

0.1 0.2 0.3 0.4 0.5 1.0 1.5 2.0 3.0 4.0 5.0 10.0 20.0

2.605 3.298 3.787 4.178 4.510 5.721 6.579 7.265 8.358 9.232 9.974 12.675 16.087

3 3 2 2 2 1 1 1 1 1 1 1 1

2

10.050 10.091 10.101 10.129 10.158 10.191 10.270 10.345 10.492 10.634 10.773 11.448 12.762

λT2∗ × 102 3.333 3.333 5.000 5.000 5.000 10.000 10.000 10.000 10.000 10.000 10.000 10.000 10.000

modules is better than a double modular structure because 3e−2λT − 2e−3λT > e−2λT . Example 9.2 Table 9.2 presents optimal N2∗ , λT2∗ and the resulting mean time λL 2 (N2∗ ) for λc2 when λS = 0.1. This indicates that N2∗ decrease to 1 and λT2∗ increase to λS with c2 . Comparing Table 9.2 with Table 9.1, when c1 = c2 , N1∗ ≥ N2∗ and L 1 (N1∗ ) ≥ L 2 (N2∗ ), as shown above previously. For example, when c2 = 10c1 , L 1 (N1∗ ) > L 2 (N2∗ ) for λc1 × 103 < 0.3 and L 1 (N1∗ ) < L 2 (N2∗ ) for ∎ λc1 × 103 > 0.4.

(3) Majority Decision Structure We provide a majority decision structure with (2n + 1) modules, i.e., an (n + 1)out-of-(2n + 1) (n = 1, 2, . . . ) structure as an error masking structure. When more than (n + 1) states of (2n + 1) modules match with each other, the process in this interval is correct. Then, the probability that the process is correct during the interval ((k − 1)T , kT ) is F n+1 (T ) =

2n+1 ∑ j=n+1

(

) 2n + 1 −λT j ) (1 − e−λT )2n+1− j . (e j

(9.9)

210

9 Which is Better Problems in Checkpoint Models

Let cn+1 be the overhead for the comparison of a majority decision structure with (2n + 1) modules. By the similar method of obtaining (9.1), the mean time to complete the process is L n+1 (N ) =

N (T + cn+1 ) F n+1 (T )

S + N cn+1 = ∑2n+1 (2n+1) (e−λT ) j (1 − e−λT )2n+1− j j=n+1 j

(N = 1, 2, . . . ), (9.10)

which agrees with (9.6) when n = 1. Next, we consider the problem which majority structure is better, when the overhand of an (n + 1)-out-of-(2n + 1) is given by (

cn+1

) 2n + 1 = c1 2

(n = 1, 2, . . . ),

(9.11)

because we select two states and compare them from each of (2n + 1) modules, i.e., c2 = 3c1 , c3 = 10c1 , . . . . ∗ Example 9.3 Table 9.3 presents optimal Nn+1 and the resulting mean time λL n+1 ∗ ∗ (Nn+1 ) for n and λc1 when λS = 0.1. This indicates that L n+1 (Nn+1 ) increase with n, i.e., a 2-out-of-3 structure is the best among a majority decision structure. This means that overhead cn+1 might increase with n, however, increase less than that given in (9.11). The mean times are smaller for n = 1, 2 and are larger for n = 3, 4 when λc1 = 0.5 × 10−3 than 10.650 × 10−2 in Table 9.1 for a double modular structure. ∎

∗ ∗ ) when λS = 0.1 Table 9.3 Optimal Nn+1 and λL n+1 (Nn+1

n

λc1 = 0.05 × 10−3

λc1 = 0.1 × 10−3

λc1 = 0.5 × 10−3

∗ Nn+1

∗ ) × 102 λL n+1 (Nn+1

∗ Nn+1

∗ ) × 102 λL n+1 (Nn+1

∗ Nn+1

∗ ) × 102 λL n+1 (Nn+1

1

3

10.077

3

10.122

2

10.372

2

2

10.111

1

10.176

1

10.579

3

1

10.128

1

10.233

1

11.075

4

1

10.187

1

10.367

1

11.808

9.2 Random Checkpoint Models

211

9.2 Random Checkpoint Models 9.2.1 Two Kinds of Tasks Suppose that the process executes N tasks (N = 1, 2, . . . ), each of which has an identical processing time Yk (k = 1, 2, . . . , N ) with finite mean 1/θ. Then, we adopt a double modular structure with constant error rate λ for the process of each task, and consider the following two kinds of tasks in Figs. 9.1 and 9.2.

(1) Tandem Task Suppose that the process executes N tasks (N = 1, 2, . . . ) successively, which is called Tandem task in Fig. 9.1. Supposing T is a random variable with G(t) in Sect. 9.1, the mean execution time to complete the process is given by a renewal function: ∫ ∞ { } (t + N c1 )e−2λt + [t + N c1 + L T (N )](1 − e−2λt ) dG (N ) (t), L T (N ) = 0

(9.12) where G (N ) (t) is the N -fold Stieltjes convolution of G(t) with itself, and G (0) (t) ≡ 1 for t ≥ 0. Solving (9.12) with respect to L T (N ), the mean time to complete the process is Y1

YN

Y2 t

0 Fig. 9.1 Tandem task

Y1 Y2

YN −1 YN 0 Fig. 9.2 Bulk task

t

212

9 Which is Better Problems in Checkpoint Models

N (1/θ + c1 ) N (1/θ + c1 ) L T (N ) = ∫ ∞ −2λt (N ) = , G ∗ (2λ) N dG (t) 0 e

(9.13)

∫∞ where G ∗ (s) is Laplace-Stieltjes transform of G(t), i.e., G ∗ (s) ≡ 0 e−st dG(t) for Re(s) > 0, and L T (N ) agrees with (9.1) when 1/θ = T and G (N ) (t) = 0 for t ≤ T , G (N ) (t) = 1 for t > T .

(2) Bulk Task Suppose that the process executes N tasks (N = 1, 2, . . . ) simultaneously, which is called Bulk task in Fig. 9.2. When some errors are detected, the process returns to the previous checkpoint and reexecutes all N tasks, and ends when all N tasks have no error. Then, letting c N be the overhead for N tasks, the mean execution time to complete the process is ∫

∞

L B (N ) =

{

} (t + c N )e−2λt + [t + c N + L B (N )](1 − e−2λt ) dG(t) N . (9.14)

0

Solving (9.14) with respect to L B (N ), ∫∞ L B (N ) =

0

[1 − G(t) N ]dt + c N ∫∞ , −2λt dG(t) N 0 e

(9.15)

which agrees with L T (1) when N = 1. We compare L T (N ) in (9.13) and L B (N ) in (9.15) when c N = N c1 : From (9.13) and (9.15), N (1/θ + c1 ) − L T (N ) − L B (N ) = ∫ ∞ −2λt (N ) dG (t) 0 e

∫∞ 0

[1 − G(t) N ]dt + N c1 ∫∞ . −2λt dG(t) N 0 e

Noting that G(t) N ≥ G (N ) (t) (N = 1, 2, . . . ) and [1 − G(t) N ]/N decreases with N from G(t) to 0, we easily have ∫∞

−2λt dG(t) N 0 e ∫∞ −2λt dG (N ) (t) 0 e

≥

(1/N )

∫∞ 0

[1 − G(t) N ]dt + c1 . 1/θ + c1

Therefore, when c N = N c1 , Bulk task is better than Tandem one, because we can begin to execute N tasks simultaneously for Bulk one. However, overhead c N might be larger than N c1 . Example 9.4 When N = 2, G(t) = 1 − e−θt and F(t) = 1 − e−λt , the mean execution times are

9.2 Random Checkpoint Models

213

c2 and L B (2) when N = 2, L T (2) = L B (2), c1 = 1 and G(t) = 1 − e−θt Table 9.4 Overhead /\ −λt and F(t) = 1 − e θ = 1.00 θ = 0.75 θ = 0.50 θ = 0.25 λ /\ c2 L B (2) /\ c2 L B (2) /\ c2 L B (2) /\ c2 L B (2) 0.1 0.05 0.01 0.005 0.001 0.0005 0.0001

2.690 2.598 2.520 2.510 2.502 2.501 2.500

L T (2) =

4.840 4.410 4.080 4.040 4.009 4.004 4.001

2.958 2.817 2.698 2.682 2.670 2.668 2.667

2(1/θ + c1 ) , [θ/(θ + λ)]2

5.994 5.310 4.792 4.729 4.679 4.673 4.668

3.545 3.286 3.059 3.030 3.006 3.003 3.001

L B (2) =

8.640 7.260 6.242 6.121 6.024 6.012 6.002

5.667 4.909 4.196 4.099 4.020 4.010 4.002

19.600 14.400 10.816 10.404 10.080 10.040 10.008

3/(2θ) + c2 . + λ)(2θ + λ)]

2θ2 /[(θ

Thus, if 2θ(λ + 2θ)c2 − 8θ(θ + λ)c1 > 5λ + 2θ, then Tandem task is better than Bulk one. c2 and the mean time L B (2) for λ and θ when N = 2, Table 9.4 presents overhead /\ c2 = 2.598 L T (2) = L B (2) and c1 = 1. For example, when λ = 0.05 and θ = 1.00, /\ c2 and L B (2) increase with λ and 1/θ, and and L B (2) = L T (2) = 4.410. Values of /\ c2 , then Tandem task is better than Bulk one. This means that if 1/λ become if c2 > /\ smaller and 1/θ become larger, then Tandem task becomes better than Bulk one. ∎

9.2.2 Two Kinds of Checkpoints Suppose that we have to execute Tandem task with processing times Yk (k = 1, 2, . . . ) its mean time 1/θ ≡ ∫in∞Fig. 9.1 with an identical distribution G(t) ≡ Pr{Yk ≤ t} and −λt G(t)dt when a double modular structure with F(t) = e in (1) of Sect. 9.1 0 is adopted. Then, setting two kinds of checkpoints: Compare-and-store checkpoint (CSCP) and compare-checkpoint (CCP), we carry the following three checkpoint schemes: 1. CSCP is placed at each end of tasks. 2. CSCP is placed at N th (N = 1, 2, . . . ) end of tasks. 3. CCP is placed at each end of tasks and CSCP is placed at N th end of tasks in Fig. 9.3.

214

9 Which is Better Problems in Checkpoint Models

Y1

Y2

YN −1

Y3 CCP

YN

CSCP

Fig. 9.3 Task execution for Scheme 3

(1) Scheme 1 Suppose that CSCP is placed at each end of task k (k = 1, 2, . . . ). When two states of modules match with each other at the end of task k, the process of task k is correct and its state is stored. In this case, the process goes forward and executes task k + 1. However, when two states do not match, it is judged that some errors have occurred, and the process reexecutes task k again. Let c1 be the overhead for the comparison of two states and c S be the overhead for their store. Then, the mean execution time of task k is given by a renewal equation: ∫

∞

L 1 (1) = c1 +

{

} (c S + t)e−2λt + [t + L 1 (1)](1 − e−2λt ) dG(t).

(9.16)

0

Solving (9.16) for L 1 (1), L 1 (1) =

c1 + 1/θ + cS , G ∗ (2λ)

(9.17)

where G ∗ (s) is LS transform of G(t) defined in (9.13). Thus, the mean execution time per one task is given in (9.17).

(2) Scheme 2 Suppose that CSCP is placed only at the end of task N (N = 1, 2, . . . ): When two states of all task k (k = 1, 2, . . . , N ) match at the end of task N , its state is stored and the process goes forward and executes task N + 1. However, when at least two states of task k do not match, the process goes back to the first task 1 and reexecutes task 1 from the first one. By similar method of obtaining (9.16), the mean execution time l2 (N ) to complete the process of task k is ∫

∞

l2 (N ) = N c1 +

{ } (c S + t)e−2λt + [t + l2 (N )](1 − e−2λt ) dG (N ) (t), (9.18)

0

where G (N ) (t) is the N -fold Stieltjes convolution of G(t) defined in (9.12). Solving (9.18) for l2 (N ), N (c1 + 1/θ) l2 (N ) = + cS . [G ∗ (2λ)] N

9.2 Random Checkpoint Models

215

Thus, the mean execution time per one task is L 2 (N ) ≡

c1 + 1/θ cS l2 (N ) = ∗ + N G (2λ) N N

(N = 1, 2, . . . ),

(9.19)

which agrees with L 1 (1) in (9.17) when N = 1. We find optimal N2∗ to minimize L 2 (N ) in (9.19). Forming the inequality L 2 (N + 1) − L 2 (N ) ≥ 0, cS N (N + 1)[1 − G ∗ (2λ)] ≥ , (9.20) ∗ N +1 G (2λ) c1 + 1/θ whose left-hand side increases strictly with N to ∞. Thus, there exists a finite and unique minimum N2∗ (1 ≤ N2∗ < ∞) which satisfies (9.20). If cS 2[1 − G ∗ (2λ)] ≥ , G ∗ (2λ)2 c1 + 1/θ then N2∗ = 1, whose scheme corresponds to Scheme 1. In particular, when G(t) = 1 − e−θt , (9.20) is N (N + 1)

2λ θ

(

)N

2λ +1 θ

≥

cS , c1 + 1/θ

(9.21)

whose left-hand side increases strictly with λ, and so that, N2∗ decreases with λ to 1. Example 9.5 Table 9.5 presents optimal N2∗ , the resulting mean time θL 2 (N2∗ ) and θL 1 (1) in (9.17) for λ/θ and θc1 when θc S = 0.1. This indicates that N2∗ decrease with λ/θ and increase with θc1 . For example, when λ/θ = 0.005 and θc1 = 0.01, N2∗ = 3 and θL 2 (N2∗ ) = 1.077, which is about 4% smaller than θL 1 (1) = 1.120. In Table 9.5, if λ/θ ≤ 0.01, then Scheme 2 is better than Scheme 1. ∎

Table 9.5 Optimal N2∗ , θL 2 (N2∗ ) and θL 1 (N1∗ ) when θc S = 0.1 λ θ

θc1 = 0.005

θc1 = 0.01

θc1 = 0.05

N2∗

N2∗

N2∗

θL 2 (N2∗ ) θL 1 (1)

θL 2 (N2∗ ) θL 1 (1)

θL 2 (N2∗ ) θL 1 (1)

0.1

1

1.306

1.306

1

1.312

1.312

1

1.360

1.360

0.05

1

1.206

1.206

1

1.211

1.211

1

1.255

1.255

0.01

2

1.098

1.125

2

1.103

1.130

2

1.145

1.171

0.005

3

1.072

1.115

3

1.077

1.120

3

1.118

1.161

0.001

6

1.037

1.107

6

1.042

1.112

6

1.083

1.152

0.0005

8

1.028

1.106

8

1.033

1.111

7

1.074

1.151

0.0001

14

1.017

1.105

14

1.022

1.170

14

1.062

1.150

216

9 Which is Better Problems in Checkpoint Models

(3) Scheme 3 Suppose that CSCP is placed at the end of task N and CCP is placed at the end of task k (k = 1, 2, . . . , N − 1) between CSCPs in Fig. 9.3: When two states of task k (k = 1, 2, . . . , N − 1) match at the end of task k, the process executes task k + 1. When two states of task k (k = 1, 2, . . . , N ) do not match, the process goes back to the first task 1. When two states of all task N match, the process is completed and its state is stored, and the process executes task N + 1. Let l3 (k) be the mean execution time from task k to complete the process of task N . Using the similar method in (9.18), l3 (k) = c1 +

1 + θ

1 l3 (N ) = c1 + + θ

∫

∞

[ ] l3 (k + 1)e−2λt + l3 (1)(1 − e−2λt ) dG(t)

∞

[

0

∫

c S e−2λt

(k = 1, 2, . . . , N − 1), ] + l3 (1)(1 − e−2λt ) dG(t). (9.22)

0

Solving (9.22) for l3 (1), l3 (1) =

(c1 + 1/θ)[1 − G ∗ (2λ) N ] + cS . [1 − G ∗ (2λ)]G ∗ (2λ) N

(9.23)

Thus, the mean execution time per one task is L 3 (N ) ≡

l3 (1) (c1 + 1/θ)[1 − G ∗ (2λ) N ] c S = + (N = 1, 2, . . . ), N N [1 − G ∗ (2λ)]G ∗ (2λ) N N

(9.24)

which agrees with L 2 (1) in (9.17) when N = 1. Comparing (9.24) with L 2 (N ) in (9.19), Scheme 3 is better than Scheme 2, because N −1 ∑ 1 − G ∗ (2λ) N = G ∗ (2λ) j ≤ N . 1 − G ∗ (2λ) j=0

We find optimal N3∗ to minimize L 3 (N ) in (9.24). Forming the inequality L 3 (N + 1) − L 3 (N ) ≥ 0, N ∑ 1 cS [1 − G ∗ (2λ) j ] ≥ , ∗ N +1 G (2λ) c1 + 1/θ j=1

(9.25)

whose left-hand side increases strictly with N to ∞. Thus, there exists a finite and unique minimum N3∗ (1 ≤ N3∗ < ∞) which satisfies (9.25). If

9.3 Two Examples

217

Table 9.6 Optimal N3∗ and θL 3 (N3∗ ) when G(t) = 1 − e−θt and θc S = 0.1 λ θ 0.1 0.05 0.01 0.005 0.001 0.0005 0.0001

θc1 = 0.005

θc1 = 0.01

θc1 = 0.05

N3∗

θL 3 (N3∗ )

N3∗

θL 3 (N3∗ )

N3∗

θL 3 (N3∗ )

1 1 3 4 7 9 17

1.306 1.206 1.085 1.060 1.031 1.024 1.015

1 1 3 4 7 9 17

1.312 1.211 1.090 1.066 1.036 1.029 1.020

1 1 3 4 7 9 17

1.360 1.255 1.132 1.107 1.076 1.069 1.060

1 − G ∗ (2λ) cS ≥ , G ∗ (2λ)2 c1 + 1/θ then N3∗ = 1. Comparing (9.25) with (9.20), it can be seen that N3∗ ≥ N2∗ . When G(t) = 1 − e−θt , (9.25) is (

2λ + θ θ

[ ) N +1 ∑ N

( 1−

k=1

θ 2λ + θ

)k ] ≥

cS , c1 + 1/θ

(9.26)

whose left-hand side increases with λ from 0 to ∞. Thus, N3∗ decreases with λ to 1. If ) ( 2λ 2λ cS +1 ≥ , θ θ c1 + 1/θ then N3∗ = 1. Example 9.6 Table 9.6 presents optimal N3∗ and the resulting mean time θL 3 (N3∗ ) for λ/θ and θc1 when θc S = 0.1. For example, when λ/θ ≥ 0.05, N3∗ = 1 in all cases. Scheme 3 is better than Scheme 2, as shown previously. However, overhead ∎ c1 for Scheme 2 would be less than that for Scheme 3.

9.3 Two Examples Suppose that we have to execute Tandem tasks with processing times Yk and G(t) ≡ Pr{Yk ≤ t} = 1 − e−θt for a double modular structure with F(t) = 1 − e−λt in (1) of Sect. 9.1 It is assumed that c1 is the overhead for the comparison of two states and c S is the overhead for their store.

218

9 Which is Better Problems in Checkpoint Models

(1) (2) (3) Fig. 9.4 4 tasks with CSCPs

(1) Model 1 We carry Scheme i (i = 1, 2, 3) with 4 tasks in Fig 9.4, and discuss which schemes are better. Let lk (k = 1, 2, 3, 4) be the mean execution time from task k to complete 4 tasks. For Scheme 1, ∫ ∞ ∫ ∞ (c S + t + l2 )e−2λt dG(t) + (c1 + t + l1 )(1 − e−2λt )dG(t), l1 = 0 0 ∫ ∞ ∫ ∞ −2λt (3) (c S + t)e dG (t) + (c1 + t + l2 )(1 − e−2λt )dG (3) (t). l2 = 0

0

For Scheme 2, ∫ ∞ ∫ ∞ l1 = (c S + t + l3 )e−2λt dG (2) (t) + (c1 + t + l1 )(1 − e−2λt )dG (2) (t), 0 0 ∫ ∞ ∫ ∞ −2λt (2) (c S + t)e dG (t) + (c1 + t + l3 )(1 − e−2λt )dG (2) (t). l3 = 0

0

For Scheme 3, ∫ ∞ ∫ ∞ l1 = (c S + t + l4 )e−2λt dG (3) (t) + (c1 + t + l1 )(1 − e−2λt )dG (3) (t), 0 0 ∫ ∞ ∫ ∞ −2λt (c S + t)e dG(t) + (c1 + t + l4 )(1 − e−2λt )dG(t). l4 = 0

0

Solving the above equations with respect to l1 and setting L i (i = 1, 2, 3) of Scheme i, respectively, 1/θ + c1 [1 − G ∗ (2λ)] 3/θ + c1 [1 − G ∗ (2λ)3 ] L1 = L3 = + + 2c S , G ∗ (2λ) G ∗ (2λ)3 { } 2/θ + c1 [1 − G ∗ (2λ)2 ] L2 = 2 + cS . G ∗ (2λ)2 Comparing L 1 and L 2 , it is easily shown that L 2 < L 1 because

9.3 Two Examples

(1)

219

1

2

3

4

5

6

(2) (3) (4) (5) (6) (7) (8) (9) CCP

CSCP

Fig. 9.5 Task with CCP and CSCP

3/θ + c1 2/θ + c1 1/θ + c1 + ∗ > ∗ . G ∗ (2λ) G (2λ)2 G (2λ) This follows that we should place the checkpoint at the middle point of the number of tasks.

(2) Model 2 We carry Scheme i (i = 1, 2, 3, 4, 5, 6, 7, 8, 9) with 6 tasks in Fig. 9.5, and discuss numerically which schemes are better. The mean execution times L i of each scheme are obtained as follows [5]: L1 L3 L5 L6

[ ] c1 + 3/θ c1 + 6/θ + c S − c1 , L2 = 2 + c S − c1 , = ∗ G (2λ)6 G ∗ (2λ)3 [ [ ] ] c1 + 1/θ c1 + 2/θ =3 + c S − c1 , + c S − c1 , L4 = 6 G ∗ (2λ)2 G ∗ (2λ) (c1 + 3/θ)[1 + G ∗ (2λ)3 ] = + c S − c1 , G ∗ (2λ)6 (c1 + 2/θ)[1 + G ∗ (2λ)2 + G ∗ (2λ)4 ] = + c S − c1 , G ∗ (2λ)6

220

9 Which is Better Problems in Checkpoint Models

(c1 + 1/θ)[1 + G ∗ (2λ) + G ∗ (2λ)2 + G ∗ (2λ)3 + G ∗ (2λ)4 + G ∗ (2λ)5 ] G ∗ (2λ)6 + c S − c1 , } { (c1 + 1/θ)[1 + G ∗ (2λ) + G ∗ (2λ)3 ] L8 = 2 + c S − c1 , G ∗ (2λ)3 } { (c1 + 1/θ)[1 + G ∗ (2λ)] L9 = 3 + c S − c1 . G ∗ (2λ)2

L7 =

For example, L 1 and L 7 are computed as follows: Let lk (k = 1, 2, 3, 4, 5, 6) be the mean execution time for task k to complete task 6. For Scheme 1, ∫

∞

l1 =

−2λt

(c S + t)e

∫

(6)

dG (t) +

0

∞

(c1 + t + l1 )(1 − e−2λt )dG (6) (t)

0

= (c S − c1 )G ∗ (2λ)6 + c1 +

6 + l1 [1 − G ∗ (2λ)6 ]. θ

Solving the above equation for l1 , L 1 ≡ l1 =

c1 + 6/θ + c S − c1 . G ∗ (2λ)6

For Scheme 7, ∫ ∞ ∫ −2λt lk = (c1 + t + lk+1 )e dG(t) + 0

∫

(c1 + t + l1 )(1 − e−2λt )dG(t)

0 ∞

l6 =

∞

(c S + t)e−2λt dG(t) +

0

(k = 1, 2, 3, 4, 5),

∫

∞

(c1 + t + l1 )(1 − e−2λt )dG(t).

0

Solving the above equations for l1 , (c1 + 1/θ)[1 + G ∗ (2λ) + G ∗ (2λ)2 + G ∗ (2λ)3 + G ∗ (2λ)4 + G ∗ (2λ)5 ] G ∗ (2λ)6 + c S − c1 .

L7 =

Example 9.7 Table 9.7 presents L 1 ∼ L 9 for λ when c S = 0.1, c1 = 0.001 and G(t) = 1 − e−λt . Mean times L i increase with λ and the minimum L i among them move with λ = 0.1 ∼ 0.0005 as follows: L 4 → L 4 → L 8 → L 7 → L 7 → L 6 . ∎ When λ = 0, i.e., G ∗ (λ) = 1, the minimum L i is L 1 = c S + 6/θ = 6.100.

9.4 Imperfect Checkpoint Model

221

Table 9.7 Mean execution times L 1 ∼ L 9 when c S = 0.1, c1 = 0.001 and G(t) = 1 − e−t λ L1 L2 L3 L4 L5 L6 L7 L8 L9 0.1 0.05 0.01 0.005 0.001 0.0005

18.018 10.730 6.857 6.489 6.172 6.136

10.570 8.187 6.567 6.382 6.236 6.218

8.941 7.561 6.543 6.421 6.324 6.312

7.801 7.201 6.720 6.660 6.612 6.606

14.246 9.410 6.663 6.377 6.155 6.128

13.105 8.995 6.600 6.347 6.150 6.126

12.027 8.595 6.540 6.319 6.147 6.126

8.943 7.487 6.447 6.325 6.228 6.216

8.225 7.234 6.484 6.393 6.321 6.312

9.4 Imperfect Checkpoint Model Suppose that the process has a native execution time S (0 < S < ∞) for a double modular structure with periodic checkpoints in (1) of Sect. 9.1 and random checkpoint in (1) of Sect. 9.2. Then, we consider imperfect checkpoint models in which the process goes to forward recovery or makes reexecution with some probabilities when errors occur at constant rate λ, i.e., the probability that the process has no error in [0, t] is e−2λt [6].

9.4.1 Periodic Checkpoint We consider the same checkpoint model with periodic checkpoint times kT and its overhead c1 in (1) of Sect 9.1, and make the following assumptions: When errors occur in process k during the interval ((k − 1)T , kT ] (k = 1, 2, . . . , N ), where N T = S, the process (a) goes to (k + 1) process with probability p, (b) reexecutes process k again with probability q, (c) returns to process 1 with probability r , where p + q + r ≡ 1. The process completes when process N succeeds, i.e., no failure occurs or even if failures occur, the process succeeds with probability p, and its overhead is c S . The mean execution time l(k) from process k (k = 1, 2, . . . , N ) to complete the process N is given by renewal equations: l(k) = c1 + T + l(k + 1)e−2λT + [ p l(k + 1) + q l(k) + r l(1)](1 − e−2λT ) l(N ) = c1 + T + c S e

−2λT

+ [ p c S + q l(N ) + r l(1)].

(k = 1, 2, . . . , N − 1), (9.27)

222

9 Which is Better Problems in Checkpoint Models

Letting A ≡ 1 − q(1 − e−2λT ), we have

B ≡ e−2λT + p(1 − e−2λT ),

A − B = r (1 − e−2λT ).

Thus, (9.27) is A l(k) = c1 + T + B l(k + 1) + (A − B)l(1)

(k = 1, 2, . . . , N − 1),

A l(N ) = c1 + T + B c S + (A − B)l(1). Forming l(k) − l(k + 1), A [l(k) − l(k + 1)] = B[l(k + 1) − l(k + 2)]

(k = 1, 2, . . . , N − 1),

A [l(N − 1) − l(N )] = B[l(N ) − c S ].

(9.28)

Multiplying all equations in (9.28), A N −1 [l(1) − l(2)] = B N −1 l(N )[l(N ) − c S ].

(9.29)

Recalling that B[l(1) − l(2)] = c1 + T, B[l(1) − c S ] = c1 + T + A[l(1) − l(N )].

(9.30)

Solving (9.29) and (9.30) for l(1), c1 + T L(N ) ≡ l(1) = A−B

] [( ) A N − 1 + cS . B

(9.31)

In particular, when p = r = 0 and q = 1, A = B = e−2λT , L(N ) = N (T + c1 )e2λT = (S + N c1 )e2λS/N + c S , which agrees with (9.1) when c S = 0. Equation (9.31) is rewritten as c1 + T L(N ) = r (1 − e−2λT )

{[

1 − q(1 − e−2λT ) e−2λT + p(1 − e−2λT )

]N

} − 1 + cS .

We compute numerically optimal N ∗ to minimize L(N ) when N T = S.

(9.32)

9.4 Imperfect Checkpoint Model

223

Table 9.8 Optimal N ∗ and L(N ∗ ) when λ = 0.01, S = 10 and c S = 0.01 c1 = 0.001 c1 = 0.01 p q r N∗ L(N ∗ ) N∗ 0.9 0.8 0.7 0.6 0.6 0.5 0.5

0.07 0.15 0.2 0.3 0.2 0.4 0.3

0.03 0.05 0.1 0.1 0.2 0.1 0.2

13 19 22 27 24 30 28

10.0660 10.0976 10.1556 10.1641 10.2624 10.1713 10.2703

4 6 7 8 8 9 9

L(N ∗ ) 10.1215 10.1783 10.2528 10.2798 10.3700 10.3033 10.2954

Example 9.8 Table 9.8 presents optimal N ∗ and the mean time L(N ∗ ) in (9.32) for p, q, r and c1 when λ = 0.01, S = 10 and c S = 0.01. For example, when p = 0.7, q = 0.2 and r = 0.1, optimal policy is N ∗ = 22 and L(N ∗ ) = 10.1556 which takes 0.1556 larger than an original processing time S = 10. This indicates that L(N ∗ ) decrease with p and decrease with q and r . ∎

9.4.2 Random Checkpoint We consider the random checkpoint model with a processing time Pr{Yk ≤ t} = G(t) in Scheme 1 of Sect. 9.2.1 and make the same assumptions (a), (b), (c). Then, the mean execution time l(k) from process k (k = 1, 2, . . . , N ) to complete the process N is ∫ ∞ 1 e−2λt dG(t) l(k) = c1 + + l(k + 1) θ 0 ∫ ∞ (1 − e−2λt )dG(t), + [ p l(k + 1) + q l(k) + r l(1)] 0 ∫ ∞ 1 e−2λt dG(t) l(N ) = c1 + + c S θ 0 ∫ ∞ (1 − e−2λt )dG(t). (9.33) + [ p c S + q l(N ) + r l(1)] 0

Noting that G ∗ (s) ≡

∫∞ 0

e−st dG(t) and letting

A ≡ 1 − q[1 − G ∗ (2λ)], we have

B ≡ G ∗ (2λ) + p[1 − G ∗ (2λ)],

A − B = r [1 − G ∗ (2λ)].

224

9 Which is Better Problems in Checkpoint Models

Then, (9.33) is 1 + B l(k + 1) + ( A − B)l(1) (k = 1, 2, . . . , N − 1), θ 1 A l(N ) = c1 + + B c S + (A − B)l(1). (9.34) θ A l(k) = c1 +

Using the same method in Sect. 9.4.1 and solving (9.34) for l(1), c1 + 1/θ l(1) = A−B

] [( ) N −1 ( ) A N c1 + 1/θ ∑ A j − 1 + cS = + cS . B B B j=0

(9.35)

Therefore, the mean time per one task is L R (N ) ≡

N −1 ( ) l(1) c1 + 1/θ ∑ A j c S = + . N BN B N j=0

(9.36)

We find optimal N R∗ to minimize L R (N ) for 0 < r ≤ 1. Forming the inequality L R (N + 1) − L R (N ) ≥ 0, [ ( )j] N −1 ( ) N 1 ∑ A A cS − , ≥ B j=0 B B c1 + 1/θ

(9.37)

whose left-hand side increases strictly with N from A−B r [1 − G ∗ (2λ)] = B2 B2 to ∞. Thus, there exists a finite and unique minimum N R∗ (1 ≤ N R∗ < ∞) which satisfies (9.37). In particular, when G(t) = 1 − e−θt , A =1−

2λq , 2λ + θ

B=

2λ p + θ , 2λ + θ

and (9.37) is { ] ] } [ N −1 [ 1 ∑ 2λ(1 − q) + θ N 2λ(1 − q) + θ j cS − . ≥ B j=0 2λ p + θ 2λ p + θ c1 + 1/θ

(9.38)

References

225

Table 9.9 Optimal N R∗ and L R (N R∗ ) when θ = 1, λ = 0.01 and c S = 0.01 p 0.9 0.8 0.7 0.6 0.6 0.5 0.5

c1 = 0.001

c1 = 0.01

q

r

N R∗

L R (N R∗ )

N R∗

L R (N R∗ )

0.07 0.15 0.2 0.3 0.2 0.4 0.3

0.03 0.05 0.1 0.1 0.2 0.1 0.2

6 5 3 3 2 3 2

1.0061 1.0089 1.0122 1.0142 1.0159 1.0162 1.0179

5 5 3 3 2 3 2

1.0152 1.0180 1.0123 1.0213 1.0250 1.0254 1.0270

If

2λr (2λ + θ) cS ≥ , 2 (2λ p + θ) c1 + 1/θ

then N ∗ = 1. Example 9.9 Table 9.9 presents optimal N R∗ which satisfies (9.38) and the mean time L R (N R∗ ) in (9.36) for p, q, r and c1 when 1/θ = 1, λ = 0.01 and c S = 0.01. This indicates that N R∗ increase with p and decrease with q and r , which are contrary to Table 9.8, because N R∗ are number of tasks and N ∗ in Table 9.8 are the partition number of one task with S = 10.0. So that, L(N ∗ ) in Table 9.8 are almost the same ∎ as 10L R (N R∗ ).

9.5 Problem 1. Consider the imperfect checkpoint model in Sect. 9.4.2 for a 2-out-of-3 structure in (2) of Sect. 9.1 [6].

References 1. Lee PA, Anderson T (1990) Fault tolerance-principles and practice. Springer, Wien 2. Abd-El-Bar M (2007) Reliable and fault-tolerant. Imperial Colledge Press, London 3. Nakagawa S, Funkumoto S, Ishii N (2003) Optimal checking pointing intervals of three error detection schemes by a double modular redundancy. Math Comput Model 38:1357–1363 4. Naruse K, Nakagawa T (2020) Optimal checking intervals, schemes and structures for computing modules. In: Pham H (ed) Reliability and statistical computing. Springer, London, pp 265–287 5. Naruse K, Nakagawa T (2022) Optimal design of checkpoint systems with general structures, tasks and scheme. In: Ram H, Pham H (eds) Reliability and maintainability assesment of industrial systems. Springer, London, pp 73–91 6. Naruse K, Nakagawa T (2022) Optimal scheme models with imperfect checkpoint, 27th ISSAT International Conference on Reliability and Quality in Design, pp 96–100 7. Nakagawa T (2008) Advanced reliability models and maintenance policies. Springer, London

Appendix A

Suppose that the unit operates for a job with random working times. It ∫is assumed ∞ that the unit has a failure distribution F(t) for t ≥ 0 with finite mean μ ≡ 0 F(t)dt, where Φ(t) ≡ 1 − Φ(t) for any ∫ t function Φ(t). When F(t) has a density function f (t) ≡ dF(t)/dt, i.e., F(t) ≡ 0 f (u)du, failure rate h(t) ≡ f (t)/ F(t) for F(t) < 1, which increases strictly with t from h(0) to h(∞) ≡ limt→∞ h(t). In addition, the working times ∑ of a job have an identical exponential distribution G(t) = 1 − e−θt (N ) n −θt and G (t) = ∞ (N = 0, 1, 2, . . . ) for 0 < 1/θ < ∞ [1, p. 227], n=N [(θt) /n!]e [2, p. 105], [3]. We use the following notations: F(T )

Q(T ) = ∫ T 0

F(t)dt

∼1 (T ) = Q

,

∼ ) = ∫ ∞F(T ) , Q(T T F(t)dt

∫T Q 1 (T ) = ∫ 0T 0

∫∞

−θt dF(t) T e , ∫∞ −θt F(t)dt T e

∫

∞

R(N ) = 0

θ(θt) N −θt e h(t)dt, N!

∫∞ Q(N ) = ∫ 0∞ 0

e−θt dF(t) e−θt F(t)dt

(θt) N e−θt dF(t)

(θt) N e−θt F(t)dt

H (K ) = ∫ ∞ 0

,

,

1 , p K (t)dt

which increase strictly with i (i = T , N , K ) to h(∞).

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Mizutani et al., Which-Is-Better (WIB): Problems in Reliability Theory, Springer Series in Reliability Engineering, https://doi.org/10.1007/978-3-031-27316-2

227

228

Appendix A

A.1 Properties of Failure Rate f (t)/ F(t) for Age Replacement This appendix summarizes the properties of extended failure rates appeared in Chap. 2: (1) For 0 < T ≤ ∞ and N = 0, 1, 2, . . . , ∫T Q 1 (T , N ) = ∫ 0T 0

(θt) N e−θt dF(t)

(θt) N e−θt F(t)dt

increases strictly with T from h(0) to Q 1 (∞, N ) ≡ Q(N ) and increases strictly with ∫T ∫T N from Q 1 (T , 0) ≡ Q 1 (T ) = 0 e−θt dF(t)/ 0 e−θt F(t)dt to h(T ). Proof Note that ∫∞ lim Q 1 (T , N ) = h(0),

lim Q 1 (T , N ) = Q(N ) = ∫ 0∞

T →0

T →∞

0

lim Q 1 (T , N ) = h(T ),

(θt) N e−θt dF(t)

(θt) N e−θt F(t)dt

,

h(0) < Q 1 (T , N ) < h(T ).

N →∞

Differentiating Q 1 (T , N ) with respect to T , N −θT

(θT ) e

∫

T

F(T )

(θt) N e−θt F(t)[h(T ) − h(t)]dt > 0.

0

Forming Q 1 (T , N + 1) − Q 1 (T , N ), ∫

T

∫

T

dF(t) (θt) N e−θt F(t)dt 0 0 ∫ T ∫ T N −θt (θt) e dF(t) (θt) N +1 e−θt F(t)dt. −

L 1 (T ) ≡

(θt)

N +1 −θt

e

0

0

Then, we have L 1 (0) = 0 and L '1 (T ) = (θT ) N e−θT F(T )

∫

T

(θt) N e−θt F(t)(θT − θt)[h(T ) − h(t)]dt > 0,

0

which completes that proof. Thus, Q(N ) increases strictly with N from Q(0) to h(∞), and Q 1 (T ) increases strictly with T from h(0) to Q 1 (∞). (2)

For 0 ≤ T < ∞ and N = 0, 1, 2, . . . , ∫∞ N −θt dF(t) T (θt) e ∼ Q 1 (T , N ) = ∫ ∞ N −θt F(t)dt T (θt) e

Appendix A

229

∼1 (0, N ) = Q(N ) to h(∞) and increases strictly with increases strictly with T from Q ∫ ∫ ∼1 (T ) = ∞ e−θt dF(t)/ ∞ e−θt F(t)dt to h(∞). ∼1 (T , 0) ≡ Q N from Q T

T

Proof Note that ∼1 (T , N ) = lim Q ∼1 (T , N ) = h(∞), lim Q

T →∞

N →∞

∼1 (T , N ) < h(∞). h(T ) < Q

∼1 (T , N ) with respect to T , Differentiating Q (θT ) N e−θT F(T )

∫

∞

(θt) N e−θt F(t)[h(t) − h(T )]dt > 0.

T

∼1 (T , N + 1) − Q ∼1 (T , N ), Forming Q ∫

∫ ∞ (θt) N +1 e−θt dF(t) (θt) N e−θt F(t)dt T T ∫ ∞ ∫ ∞ N −θt (θt) e dF(t) (θt) N +1 e−θt F(t)dt. −

L 2 (T ) ≡

∞

T

T

Then, we have L 2 (∞) = 0 and ∫

L '2 (T ) = (θT ) N e−θT F(T )

∞

(θt) N e−θt F(t)(θt − θT )[h(T ) − h(t)]dt < 0,

T

∼1 (T ) increases strictly with T from Q ∼1 (0) to which completes the proof. Thus, Q h(∞). (3)

For 0 < T ≤ ∞ and N = 0, 1, 2, . . . , ∫T

Q 2 (T , N ) =

[∫ ∞ −θu ] N dF(u) dt 0 (θt) t e ∫T [∫ ∞ ] N −θu F(u)du dt 0 (θt) t e

∼1 (0) to Q 2 (∞, N ) = Q(N + 1) and increases increases strictly with T from Q ∼1 (T ). strictly with N from Q 2 (T , 0) to Q Proof Note that ∼1 (0), lim Q 2 (T , N ) = Q

T →0

∼1 (T ), lim Q 2 (T , N ) = Q

N →∞

∫∞

Q 2 (∞, N ) =

[∫ ∞ −θu ] N dF(u) dt 0 (θt) t e ∫∞ ] [∫ ∞ N −θu F(u)du dt 0 (θt) t e

= Q(N + 1).

∼1 (0) < Q 2 (T , N ) < Q ∼1 (T ), Q ∫∞

= ∫ 0∞ 0

(θt) N +1 e−θt dF(t)

(θt) N +1 e−θt F(t)dt

230

Appendix A

Differentiating Q 2 (T , N ) with respect to T , ∫ (θT )

∞

N

∫

−θt

e

(θt)

F(t)dt

T

{∫

T

∞

N

−θu

e

0

[

∼1 (T ) − Q ∼1 (t) F(u)du Q

} ]

dt > 0.

t

Forming Q 2 (T , N + 1) − Q 2 (T , N ), ∫

T

L 3 (T ) ≡

(θt)

N +1

0

∫

T

−

[∫

−θu

e

[∫ (θt) N

0

∞ t ∞

]

∫

T

dF(u) dt

] ∫ e−θu dF(u) dt

t

0 T

[∫ (θt)

N

∞

e

−θu

t

(θt) N +1

0

[∫

∞

] F(u)du dt

] e−θu F(u)du dt.

t

Then, we have L 3 (0) = 0 and L '3 (T )

∫

∞

= (θT ) e−θt F(t)dt T ] ∫ T [∫ ∞ [ ] −θu ∼1 (T ) − Q ∼1 (t) dt > 0, e F(u)du (θt) N (θT − θt) Q × N

0

t

which completes the proof. (4)

For 0 ≤ T < ∞ and N = 0, 1, 2, . . . , ∫∞ [∫ ∞ −θu ] N dF(u) dt T (θt) t e ∼ Q 2 (T , N ) = ∫ ∞ [∫ ∞ ] N −θu F(u)du dt T (θt) t e

increases strictly with T from Q(N + 1) to h(∞) and increases strictly with N from ∼2 (T ) to h(∞). ∼2 (T , 0) ≡ Q Q Proof Note that ∼2 (T , N ) = lim Q ∼2 (T , N ) = h(∞), lim Q

T →∞

N →∞

∼1 (T ) < Q ∼2 (T ) = Q ∼2 (T , 0) ≤ Q ∼2 (T , N ) < h(∞). Q Using similar methods of (2) and (3), we can prove (4).

A.2 Properties of Failure Rate h(t) for Replacement with Minimal Repair This appendix summarizes the properties of extended failure rates appeared in Chap. 3: (1) For 0 < T ≤ ∞ and N = 0, 1, 2, . . . ,

Appendix A

231

∫T R1 (T , N ) =

0

(θt) N e−θt h(t)dt ∫T N −θt dt 0 (θt) e ∫∞

increases strictly with T from h(0) to R(N ) = increases strictly with N from R1 (T , 0) to h(T ).

0

[θ(θt) N /N !]e−θt h(t)dt and

Proof Note that ∫ R1 (∞, N ) ≡ R(N ) =

lim R1 (T , N ) = h(0),

T →0

∞ 0

∫ T −θt θe h(t)dt , R1 (T , 0) = 0 1 − e−θT

lim R1 (T , N ) = h(T ),

N →∞

θ(θt) N −θt e h(t)dt, N!

h(0) < R1 (T , N ) < h(T ).

Differentiating R1 (T , N ) with respect to T , ∫

N −θT

T

(θT ) e

(θt) N e−θt [h(T ) − h(t)]dt > 0.

0

Forming R1 (T, N + 1) − R1 (T , N ), ∫

T

L 1 (T ) ≡

(θt)

N +1 −θt

e

∫

0

∫

T

−

N −θt

(θt) e

T

h(t)dt ∫

0 T

h(t)dt

0

(θt) N e−θt dt

(θt) N +1 e−θt dt.

0

Then, we have L 1 (0) = 0 and L '1 (T ) = (θT ) N e−θT

∫

T

(θt) N e−θt (θT − θt)[h(T ) − h(t)]dt > 0,

0

which completes the proof. Thus, R(N ) increases strictly with N from R(0) to h(∞). (2)

For 0 ≤ T < ∞ and N = 0, 1, 2, . . . , ∼1 (T , N ) = R

∫∞

(θt) N e−θt h(t)dt ∫∞ N −θt dt T (θt) e

T

increases strictly with T from R(N ) to h(∞) and increases strictly with N from ∫ ∼1 (T , 0) = ∞ θe−θt h(T + t)dt to h(∞). R 0

232

Appendix A

Proof Note that ∼1 (0, N ) = R(N ), R ∫ ∞ ∼ θe−θt h(T + t)dt, R1 (T , 0) =

∼1 (T , N ) = lim R ∼1 (T , N ) = h(∞), lim R

T →∞

N →∞

∼1 (T , N ) < h(∞). h(T ) < R

0

∼1 (T , N ) with respect to T , Differentiating R N −θT

∫

∞

(θT ) e

(θt) N e−θt [h(t) − h(T )]dt > 0.

T

∼1 (T, N + 1) − R ∼1 (T , N ), Forming R ∫ ∞ ∫ ∞ L 2 (T ) ≡ (θt) N +1 e−θt h(t)dt (θt) N e−θt dt T T ∫ ∞ ∫ ∞ N −θt (θt) e h(t)dt (θt) N +1 e−θt dt. − T

T

Then, we have L 2 (∞) = 0 and ∫ ∞ (θt) N e−θt (θt − θT )[h(T ) − h(t)]dt < 0, L '2 (T ) ≡ (θT ) N e−θT T

which completes the proof. (3)

For 0 < T ≤ ∞ and K = 0, 1, 2, . . . , ∫T H1 (T , K ) =

0

p K (t)h(t)dt ∫T 0 p K (t)dt

increases strictly with T from h(0) to H (K ) ≡ 1/ ∫T with K from Q(T ) ≡ F(T )/ 0 F(t)dt to h(T ).

∫∞ 0

p K (t)dt and increases strictly

Proof Note that H1 (∞, K ) = H (K ) = ∫ ∞

lim H1 (T , K ) = h(0),

T →0

0

1 , p K (t)dt

F(T ) = Q(T ), lim H1 (T , K ) = h(T ), h(0) < H1 (T ,K ) < h(T ). H1 (T , 0) = ∫ T K →∞ 0 F(t)dt Differentiating H1 (T , K ) with respect to T , ∫ T p K (t)[h(T ) − h(t)]dt > 0. p K (T ) 0

Appendix A

233

Forming H1 (T , K + 1) − H1 (T , K ), ∫

T

L 3 (T ) ≡

∫ p K +1 (t)h(t)dt

0

T

∫ p K (t)dt −

0

T

∫

T

p K (t)h(t)dt

0

p K +1 (t)dt.

0

Then, we have L 3 (0) = 0 and L '3 (T )

∫

T

≡

[ p K +1 (T ) p K (t) − p K (T ) p K +1 (t)][h(T ) − h(t)]dt

0

∫ =

T

0

H (T ) K H (t) K −H (T )−H (t) [H (T ) − H (t)][h(T ) − h(t)]dt > 0, e (K + 1)!K !

which completes the proof. Thus, H (K ) increases strictly with K from 1/μ to h(∞), and Q(T ) increases strictly with T from h(0) to 1/μ. (4)

For 0 ≤ T < ∞ and K = 0, 1, 2, . . . , ∼1 (T , K ) = H

∫∞

p K (t)h(t)dt ∫∞ T p K (t)dt

T

increases strictly with T from H (K ) to h(∞) and increases strictly with K from ∼ ) to h(∞). ∼1 (T , 0) = Q(T H Proof Note that ∼1 (0, K ) = ∫ ∞ 1 = H (K ), H 0 p K (t)dt

∼1 (T , K ) = lim H ∼1 (T , K ) = h(∞), lim H

T →∞

∼ ) > h(T ), ∼1 (T , 0) = ∫ ∞F(T ) = Q(T H F(t)dt T

K →∞

∼1 (T , K ) < h(∞). h(T ) < H

∼ ) increases Using similar methods of (2) and (3), we can prove (4). Furthermore, Q(T strictly with T from 1/μ to h(∞). (5)

For 0 < T ≤ ∞ and N = 0, 1, 2, . . . , ∫ T [∫ ∞ R2 (T , N ) =

0

0

] e−θu h(t + u)du dG (N ) (t) G (N ) (T )

∫ ∞ −θt increases∫ strictly with T from h(t)dt to R(N )/θ and increases strictly with 0 e ∫ ∞ ∞ N from 0 e−θt h(t)dt to 0 e−θt h(T + t)dt.

234

Appendix A

Proof Note that ∫

∞

lim R2 (T , N ) =

T →0

e

−θt

∫ R2 (∞, N ) =

∫0 ∞ e−θt h(t)dt, R2 (T , 0) =

N →∞

0

∫

∞

∞

(θt) N −θt R(N ) e h(t)dt = , N! θ 0 ∫ ∞ lim R2 (T , N ) = e−θt h(T + t)dt,

h(t)dt,

e−θt h(t)dt ≤ R2 (T , N )
0,

0

where g (N ) (t) ≡ dG (N ) (t)/dt. Forming R2 (T , N + 1) − R2 (T , N ), ∫

T

L 4 (T ) ≡

[∫

∞

−θu

e 0

]

h(t + u)du dG (N +1) (t)G (N ) (T )

0

∫

T

−

[∫

∞

−θu

e 0

]

h(t + u)du dG (N ) (t)G (N +1) (T ).

0

Then, we have L 4 (0) = 0 and L '4 (T )

∫

T

=

{∫

∞

}

−θu

[h(T + u) − h(t + u)]du ] × g (N ) (t)g (N +1) (T ) − g (N ) (T )g (N +1) (t) dt > 0, e

0

[

0

because g (N ) (t)g (N +1) (T ) − g (N ) (T )g (N +1) (t) =

θ2 (θt) N −1 (θT ) N −1 −θ(T +t) e (θT − θt) > 0, (N − 1)!N !

which completes the proof. (6)

For 0 ≤ T < ∞ and N = 1, 2, . . . , ∼2 (T , N ) = R

∫ ∞ [∫ ∞ T

0

] e−θu h(t + u)du dG (N ) (t) 1 − G (N ) (T )

increases strictly with T from R(N )/θ to h(∞) and increases strictly with N from ∼2 (T , 1) to h(∞). R

Appendix A

235

Proof Note that ∫ ∞ (θt) N −θt R(N ) ∼2 (0, N ) = ∼2 (T , N ) = h(∞), , lim R2 (T , N ) = lim R e h(t)dt = R θ N ! T →∞ N →∞ 0 ∫ ∞ ∼2 (T, N ) < h(∞). e−θt h(T + t)dt < R 0

Using similar methods of (3) and (5), we can prove (6). For 0 < T ≤ ∞ and K = 0, 1, 2, . . . ,

(7)

PK (T )

H2 (T , K ) = ∫ T [∫ ∞ 0

] e−H (u)+H (t) du dPK (t)

t

∼ ). increases strictly with T from 1/μ to H (K ) and increases with K from 1/μ to Q(T Proof Note that 1

lim H2 (T , K ) = ∫ ∞

T →0

e−H (t) dt

0

lim H2 (T , K ) = ∫ ∞

K →0

0

=

1 e−H (t) dt

=

1 , μ

T →∞

1 , μ

K →∞

lim H2 (T , K ) = H (K ) = ∫ ∞ 0

1 p K (t)dt

∼ ) = ∫ ∞F(T ) , lim H2 (T , K ) = Q(T T F(t)dt

1 ∼ ). < H2 (T , K ) < Q(T μ Differentiating H2 (T , K ) with respect to T , ∫

T

p K −1 (T )h(T ) 0

[∫

∞

e−H (u)+H (t) du −

t

∫

∞

] e−H (u)+H (T ) du dPK (t) > 0,

T

because ∫

PK' (t) = p K −1 (t)h(t),

∞

e−H (u)+H (t) du =

t

1 F(t)

∫

∞

F(u)du t

decreases strictly with t from μ to 1/ h(∞). Forming H2 (T , K + 1) − H2 (T , K ), ∫

T

L 5 (T ) ≡ PK +1 (T ) ∫ − PK (T )

0 T 0

[∫

∞ t

[∫

∞ t

] e−H (u)+H (t) du dPK (t)

] e−H (u)+H (t) du dPK +1 (t).

,

236

Appendix A

Then, we have L 5 (0) = 0 and L '5 (T )

∫

[∫

T

∞

∫

−H (u)+H (t)

∞

−H (u)+H (T )

= h(T ) h(t) e du − e 0 t T [ ] × p K (T ) p K −1 (t) − p K −1 (T ) p K (t) dt > 0,

] du

because p K (T ) p K −1 (t) − p K −1 (T ) p K (t) =

H (T ) K −1 H (t) K −1 [H (T ) − H (t)] > 0, K !(K − 1)!

which completes the proof. (8)

For 0 ≤ T < ∞ and K = 1, 2, . . . , ∼2 (T, K ) = ∫ ∞ [∫ ∞ H T

t

P K (T ) ] e−H (u)+H (t) du dPK (t)

increases strictly with T from H (K ) to h(∞) and increases strictly with K from ∼2 (T , 1) to h(∞). H Proof Note that ∼2 (T , K ) = lim H ∼2 (T , K ) = h(∞), ∼2 (0, K ) = ∫ ∞ 1 = H (K ), lim H H T →∞ K →∞ 0 p K (t)dt ∼ ) 0, k=0

0

because p K (T ) pk (t) − pk (T ) p K (t) =

H (T )k H (t)k −H (T )−H (t) [H (T ) K −k − H (t) K −k ] > 0, e K !k!

which completes the proof. (10)

For K = 0, 1, 2, . . . and N = 1, 2, . . . , ∑ N −1 ∫ ∞ H2 (K , N ) =

n −θt p K (t)h(t)dt n=0 0 [(θt) /n!]e ∑ N −1 ∫ ∞ n −θt p (t)dt K n=0 0 [(θt) /n!]e

238

Appendix A

increases strictly with K from ∫ ∞ H2 (0, N ) to h(∞) and increases strictly with N from H2 (K , 1) to H (K ) = 1/ 0 p K (t)dt. Proof Note that lim H2 (K , N ) = H (K ) = ∫ ∞

lim H2 (K , N ) = h(∞),

K →∞

N →∞

0

1 . p K (t)dt

For 0 < T < ∞, forming N −1 ∫ ∑

N −1 ∫ T ∑ (θt)n −θt (θt)n −θt L 8 (T , K ) = e p K +1 (t)h(t)dt e p K (t)dt n! n! n=0 0 n=0 0 N −1 ∫ T N −1 ∫ T ∑ ∑ (θt)n −θt (θt)n −θt e p K (t)h(t)dt e p K +1 (t)dt, − n! n! n=0 0 n=0 0 T

we have L 8 (0, K ) = 0 and L '8 (T , K ) = ×

n!

n=0

N −1 ∫ ∑ n=0

N −1 ∑ (θT )n

T 0

e−θT

(θt)n −θt e [h(T ) − h(t)][ p K +1 (T ) p K (t) − p K (T ) p K +1 (t)]dt > 0. n!

For 0 < T < ∞, forming N −1 ∫ T ∑ (θt) N −θt (θt)n −θt e p K (t)h(t)dt e p K (t)dt N! n! 0 n=0 0 ∫ T N −1 ∫ T ∑ (θt) N −θt (θt)n −θt e p K (t)dt e p K (t)h(t)dt, − N! n! 0 n=0 0

∫

L 9 (T , N ) =

T

we have L 9 (0, N ) = 0 and L '9 (T ,

N ) = p K (T ) [

N −1 ∫ ∑ n=0 N

T

p K (t)[h(T ) − h(t)]

0

] (θT ) −θT (θt)n −θt (θt) N −θt (θT )n −θT e e − e e × dt > 0, N! n! N! n! which completes the proof. (11)

For K = 0, 1, 2, . . . and N = 0, 1, 2, . . . ,

Appendix A

239

∼2 (N , K ) = R

∑∞

∫

∞ N −θt pk (t)h(t)dt k=K 0 (θt) e ∑ ∞ ∫∞ N −θt pk (t)dt k=K 0 (θt) e

∼2 (0, K ) to h(∞) and increases strictly with K from increases strictly with∫N from R ∞ ∼ R2 (N , 0) = R(N ) = 0 [θ(θt) N /N !]e−θt h(t)dt to h(∞). Proof Note that ∼2 (N , K ) = lim R ∼2 (N , K ) = h(∞), lim R

N →∞

K →∞

∼2 (N , K ) < h(∞). R(N ) < R

Using similar methods of (9), we can prove (11). (12)

For K = 0, 1, 2, . . . and N = 0, 1, 2, . . . , ∼2 (N , K ) = H

∫∞ n −θt p K (t)h(t)dt n=N 0 [(θt) /n!]e ∑∞ ∫ ∞ n −θt p K (t)dt n=N 0 [(θt) /n!]e

∑∞

∼2 (0, N ) to h(∞) and increases strictly with N from increases strictly with K from H H (K ) to h(∞). Proof Note that ∼2 (K , N ) = lim H ∼2 (K , N ) = h(∞), lim H

K →∞

N →∞

∼2 (K , N ) < h(∞). H (K ) < H

Using similar methods of (10), we can prove (12).

A.3 Properties of Failure Rates f (t)/ F(t) for Periodic Replacement This appendix summarizes the properties of extended failure rates appeared in Sect. 4.1. (1) For 0 < T < ∞ and J = 0, 1, 2, . . . , Q 1 (J ) =

F((J + 1)T ) − F(J T ) ∫ (J +1)T F(t)dt JT

increases strictly with J from Q 1 (0) = Q(T ) = F(T )/

∫T 0

F(t)dt to h(∞).

Proof Note that F(T ) , Q 1 (0) = Q(T ) = ∫ T 0 F(t)dt

lim Q 1 ( J ) = h(∞),

J →∞

240

Appendix A

h( J T ) < Q 1 ( J ) < h((J + 1)T ) < Q 1 (J + 1), which completes the proof. For 0 < T < ∞, J = 0, 1, 2, . . . and N = 1, 2, . . . ,

(2)

∑ N −1∫ (J +1)T

Q 1 (J, N ) =

[(θt)n /n!]e−θt dF(t) n=0 J T ∑ N −1∫ (J +1)T [(θt)n /n!]e−θt F(t)dt n=0 J T

∫ (J +1)T JT = ∫ (J +1)T JT

[1 − G (N ) (t)]dF(t)

[1 − G (N ) (t)]F(t)dt

increases strictly with J from Q 1 (0, N ) to h(∞) and increases strictly with N from Q 1 (J, 1) to Q 1 (J ). Proof Note that lim Q 1 ( J, N ) = h(∞),

J →∞

Q 1 (J, ∞) =

F((J + 1)T ) − F(J T ) = Q 1 (J ), ∫ (J +1)T F(t)dt JT

h( J T ) < Q 1 (J, N ) < h((J + 1)T ) < Q 1 (J + 1, N ), which proves that Q 1 (J, N ) increases strictly with J from Q 1 (0, N ) to h(∞). Forming Q 1 (J, N + 1) − Q 1 (J, N ), ∫

( J +1)T

[1 − G

(N )

{∫ (t)] F(t)

JT

( J +1)T JT

∫

( J +1)T

=

∫

( J +1)T

+ t

(θu) N −θu e F(u)[h(u) − h(t)]du N! JT } (θu) N −θu e F(u)[h(u) − h(t)]du dt. N!

[1 − G (N ) (t)] F(t)

JT

{∫

} (θu) N −θu e F(u)[h(t) − h(u)]du dt N!

t

Noting that ∫

( J +1)T

[1 − G (N ) (t)] F(t)

JT

∫

=

( J +1)T JT

(θt) N −θt e F(t) N!

{∫ {∫

t

} (θu) N −θu e F(u)[h(u) − h(t)]du dt N!

JT ( J +1)T

[1 − G

(N )

}

(u)] F(u)[h(t) − h(u)]du dt,

t

the above equation is ∫

(∫

( J +1)T

F(t) JT

t

( J +1)T

F(u)[h(u) − h(t)]

{ } ) (θu) N −θu (θt) N −θt − [1 − G (N ) (u)] × [1 − G (N ) (t)] e e du dt > 0, N! N!

Appendix A

241

because N −1 [ ∑ (θt)n n=0

=

n!

e−θt

(θu) N −θu (θu)n −θu (θt) N −θt − e e e N! n! N!

N −1 ∑ (θt)n (θu)n

n!N !

n=0

]

[ ] e−θ(t+u) (θu) N −n − (θt) N −n > 0,

which completes the proof. For 0 < T < ∞, J = 1, 2, . . . and N = 0, 1, 2, . . . ,

(3)

∫ JT Q 2 (J, N ) = ∫ 0J T 0

(θt) N e−θt dF(t)

(θt) N e−θt F(t)dt

increases strictly with J from Q 2 (1, N ) to Q 2 (∞, N ) = Q(N ) and increases strictly with N from Q 2 ( J, 0) = Q 1 (J T ) to h( J T ). Proof Note that lim Q 2 (J, N ) = h( J T ),

N →∞

h(0) < Q 2 (J, N ) < h(J T ).

Replacing J T with T in (1) of Appendix A.1, we can prove (3). For 0 < T < ∞, J = 0, 1, 2, . . . and N = 0, 1, 2, . . . ,

(4)

∫ (J +1)T (N ) ∑∞ ∫ (J +1)T [(θt)n /n!]e−θt dF(t) G (t)dF(t) n=N J T JT ∼ = ∫ (J Q 1 (J, N ) = ∑∞ ∫ (J +1)T +1)T n −θt [(θt) /n!]e F(t)dt G (N ) (t)F(t)dt n=N J T JT ∼1 (0, N ) to h(∞) and increases strictly with N from increases strictly with J from Q Q 1 (J ) to h((J + 1)T ). Proof Note that ∼1 ( J, N ) = h(∞), lim Q

J →∞

∼1 ( J, N ) = h(( J +1)T ), lim Q

N →∞

∼1 ( J, 0) = F((J∫ + 1)T ) − F(J T ) = Q 1 (J ), Q (J +1)T F(t)dt JT ∼1 (J, N ) < h((J +1)T ) < Q ∼1 (J+1, N ), h(J T ) < Q

∼ N ) to h(∞). Forming ∼1 (J, N ) increases strictly with J from Q(0, which proves that Q ∼1 (J, N ), ∼1 (J, N + 1) − Q Q

242

Appendix A

∫

( J +1)T

G

(N )

{∫ (t) F(t)

JT

∫

=

(J +1)T JT

( J +1)T

(θu) N −θu e F(u)[h(t) − h(u)]du N! JT } (θu) N −θu e F(u)[h(t) − h(u)]du dt. N!

G (N ) (t) F(t)

JT

∫

+

{∫

} (θu) N −θu e F(u)[h(t) − h(u)]du dt N!

( J +1)T

t

t

Noting that ∫

( J +1)T JT

∫

=

} (θu) N −θu e F(u)[h(t) − h(u)]du dt N! t {∫ t } N (θt) −θt G (N ) (u) F(u)[h(u) − h(t)]du dt, e F(t) N! JT

G (N ) (t) F(t)

( J +1)T JT

{∫

(J +1)T

the above equation is ∫

{∫

( J +1)T

t

F(u)[h(t) − h(u)]

F(t) JT

[

× G

(N )

JT N

] } (θu) −θu (θt) N −θt (N ) (t) − G (u) e e du dt > 0, N! N!

because ∞ [ ∑ (θt)n n=N

=

n!

e−θt

(θu) N −θu (θu)n −θu (θt) N −θt − e e e N! n! N!

]

∞ ∑ ] (θt) N (θu) N −θ(t+u) [ (θt)n−N − (θu)n−N > 0, e n!N ! n=N

which completes the proof. (5)

For 0 < T < ∞, J = 0, 1, 2, . . . and N = 0, 1, 2, . . . , ∫∞ N −θt dF(t) J T (θt) e ∼ Q 2 (J, N ) = ∫ ∞ N −θt F(t)dt J T (θt) e

∼2 (0, N ) = Q(N ) in (2) of Appendix A.1 to h(∞) increases strictly with J from Q ∼2 (J, 0) to h(∞). and increases strictly with N from Q

Appendix A

243

Proof Note that ∫∞ (θt) N e−θt dF(t) ∼ = Q(N ), Q 2 (0, N ) = ∫ 0∞ N −θt F(t)dt 0 (θt) e ∼2 (J, N ) = h(∞), lim Q N →∞

∼2 ( J, N ) = h(∞), lim Q

J →∞

∼2 ( J, N ) < h(∞). h(J T ) < Q

Replacing J T with T in (2) of Appendix A.1, we can prove (5).

A.4 Properties of Failure Rates h(t) for Periodic Replacement This appendix summarizes the properties of extended failure rates appeared in Sect. 4.2. (1) For 0 < T < ∞, J = 0, 1, 2, . . . and N = 1, 2, . . . , ∫ (J +1)T ∑ N −1 ∫ (J +1)T [(θt)n /n!]e−θt h(t)dt [1 − G (N ) (t)]h(t)dt n=0 J T = J∫T (J +1)T R1 (J, N ) = ∑ N −1 ∫ (J +1)T [(θt)n /n!]e−θt dt [1 − G (N ) (t)]dt n=0 J T JT increases strictly with J from R1 (0, N ) to h(∞) and increases strictly with N from R1 ( J, 1) to [H ((J + 1)T ) − H (J T )]/T . Proof Note that lim R1 (J, N ) = h(∞),

J →∞

lim R1 ( J, N ) =

N →∞

H ((J + 1)T ) − H (J T ) , T

h(J T ) < R1 (J, N ) < h(( J + 1)T ). Using similar methods of (2) of Appendix A.3, we can prove (1). (2)

For 0 < T < ∞, J = 1, 2, . . . and N = 0, 1, 2, . . . , ∫ JT R2 (J, N ) =

0

(θt) N e−θt h(t)dt ∫ JT N −θt dt 0 (θt) e

increases strictly with J from R2 (1, N ) to R(N ) and increases strictly with N from R2 (J, 0) to h( J T ). Proof Note that ∫

∞

lim R2 (J, N ) =

J →∞

0

θ(θt) N −θt e h(t)dt = R(N ), N!

lim R2 (J, N ) = h(J T ),

N →∞

h(0) < R2 ( J, N ) < h(J T ).

244

Appendix A

Using similar methods of (1) of Appendix A.1 and (3) of Appendix A.3, we can prove (2). (3)

For 0 < T < ∞, J = 0, 1, 2, . . . and N = 0, 1, 2, . . . , ∼1 (J, N ) = R

∫ (J +1)T [(θt)n /n!]e−θt h(t)dt n=N J T ∑∞ ∫ (J +1)T [(θt)n /n!]e−θt dt n=N J T

∫ (J +1)T

∑∞

G (N ) (t)h(t)dt ∫ (J +1)T G (N ) (t)dt JT

=

JT

∼1 (0, N ) to h(∞) and increases strictly with N from increases strictly with J from R ∼1 ( J, 0) = [H ((J + 1)T ) − H (J T )]/T to h(( J + 1)T ). R Proof Note that ∼1 (J, 0) = H ((J + 1)T ) − H (J T ) , R T ∼ h( J T ) < R1 (J, N ) < h(( J + 1)T ).

∼1 (J, N ) = h(∞), lim R

J →∞

∼1 (J, N ) = h((J + 1)T ), lim R

N →∞

Using similar methods of (4) of Appendix A.3, we can prove (3). (4)

For 0 < T < ∞, J = 0, 1, 2, . . . and N = 0, 1, 2, . . . , ∼2 (J, N ) = R

∫∞

N −θt h(t)dt J T (θt) e ∫∞ N −θt dt J T (θt) e

∼2 (0, N ) = R(N ) to h(∞) and increases strictly with increases strictly with R ∫ ∞J from −θt ∼ N from R2 (J, 0) = 0 θe h(J T + t)dt to h(∞). Proof Note that ∼2 (J, N ) = h(∞), lim R

J →∞

∼2 (J, N ) = h(∞), lim R

J →∞

∼2 ( J, 0) = R

∫∞

e−θt h(t)dt

∫∞

JT

JT

e−θt dt

∫ ∞ = θe−θth(t + J T )dt, 0

∼2 (J, N ) < h(∞). h(J T ) < R

Using (2) of Appendix A.1 and (5) of Appendix A.3, we can prove (4).

A.5 Properties of Failure Rates h(t) for Replacement Over Failure Number This appendix summarizes the properties of extended failure rates appeared in Sect. 4.3. (1) For 0 < T < ∞, J = 0, 1, 2, . . . and K = 1, 2, . . . ,

Appendix A

245

∫ (J +1)T

P K (t)h(t)dt ∫ (J +1)T P K (t)dt JT

H1 (J, K ) =

JT

increases strictly with J from H1 (0, K ) to h(∞) and increases strictly with K from H1 ( J, 1) to [H (( J + 1)T ) − H (J T )]/T . Proof Note that lim H1 (J, K ) = h(∞),

H ((J + 1)T ) − H (J T ) , T

lim H1 (J, K ) =

J →∞

K →∞

h(J T ) < H1 ( J, K ) < h(( J + 1)T ) < H1 (J + 1, K ), which proves that H1 (J, K ) increases strictly with J from H1 (0, K ) to h(∞). Forming H1 ( J, K + 1) − H1 (J, K ), ∫

( J +1)T

p K (t)

{ K −1 ∫ ∑

JT ( J +1)T

=

p K (t)

( K −1 {∫ ∑

JT

∫

+

pk (u)[h(t) − h(u)]du dt

t

pk (u)[h(t) − h(u)]du

JT

k=0 ( J +1)T

}

JT

k=0

∫

( J +1)T

})

pk (u)[h(t) − h(u)]du

dt.

t

Noting that ∫

{∫

( J +1)T

( J +1)T

p K (t)

JT

t

∫

{∫

( J +1)T

pk (t)

=

JT

t

} pk (u)[h(t) − h(u)]du dt

} p K (u)[h(u) − h(t)]du dt,

JT

the above equation is ∫

( J +1)T

p K (t)

JT ( J +1)T

−

JT

=

( J +1)T JT

because

( K −1 ∑

t

}) pk (u)[h(t) − h(u)]du

dt

JT

k=0

∫ ∫

( K −1 {∫ ∑

{∫ pk (t)

t JT

k=0

{ K −1∫ ∑

t

k=0

JT

}) p K (u)[h(t) − h(u)]du

dt }

[h(t) − h(u)][ p K (t) pk (u) − p K (u) pk (t)]du dt > 0,

246

Appendix A

p K (t) pk (u) − p K (u) pk (t) H (t)k H (u)k −H (t)−H (u) [H (t) K −k − H (u) K −k ] > 0, e K !k!

=

which completes the proof. (2)

For 0 < T < ∞, J = 1, 2, . . . and K = 0, 1, 2, . . . , ∫ JT H2 (J, K ) =

0

p K (t)h(t)dt ∫ JT p K (t)dt 0

increase strictly with J from H2 (1, K ) to H (K ) and increases strictly with K from ∫ JT H2 (J, 0) = Q( J T ) = F( J T )/ 0 F(t)dt to h( J T ). Proof Note that lim H2 ( J, K ) = ∫ ∞

J →∞

0

1 = H (K ), p K (t)dt

lim H2 (J, K ) = h(J T ),

K →∞

h(0) < H2 (J, K ) < h(J T ). Setting J T ≡ x and differentiating H2 (J, K ) with respect to x, ∫

x

p K (x)

p K (t)[h(x) − h(t)]dt > 0.

0

Forming H2 (J, K + 1) − H2 (J, K ), ∫

x

L 1 (x) ≡

∫

x

p K +1 (t)h(t)dt

0

∫ p K (t)dt −

0

x

∫

x

p K (t)h(t)dt

0

p K +1 (t)dt,

0

we have L 1 (0) = 0 and L '1 (x) =

∫

x

[h(x) − h(t)][ p K +1 (x) p K (t) − p K (x) p K +1 (t)]dt > 0,

0

which completes the proof. (3)

For 0 < T < ∞, J = 0, 1, 2, . . . and K = 0, 1, 2, . . . , ∼1 ( J, K ) = H

∫ (J +1)T

PK (t)h(t)dt ∫ (J +1)T PK (t)dt JT

JT

∼1 (0, K ) to h(∞) and increases strictly with K from increases strictly with J from H [H ((J + 1)T ) − H ( J T )]/T to h((J + 1)T ). Note that

Appendix A

247

∼1 ( J, 0) = H ((J + 1)T ) − H (J T ) , H T ∼1 ( J, K ) < h((J + 1)T ). h(J T ) < H

∼1 (J, K ) = h(∞), lim H

J →∞

∼1 (J, K ) = h(( J + 1)T ), lim H

K →∞

Using similar methods of (1), we can prove (3). (4) For 0 < T < ∞, J = 0, 1, 2, . . . and K = 0, 1, 2, . . . , ∼2 (J, K ) = H

∫∞

p K (t)h(t)dt ∫∞ J T p K (t)dt

JT

increases strictly with ∫ J from H (K ) to h(∞) and increases strictly with K from ∼ J T ) to h(∞). ∼2 (J, 0) = F( J T )/ ∞ F(t)dt = Q( H JT Proof Note that ∼2 (0, K ) = ∫ ∞ 1 H = H (K ), 0 p K (t)dt

∼2 (J, K ) = lim H ∼2 (J, K ) = h(∞), lim H

J →∞

F(J T ) ∼2 (J, 0) = ∫ ∞ ∼ T ), H = Q(J J T F(t)dt

K →∞

∼2 (J, K ) < h(∞). h(J T ) < H

Using similar methods of (2), we can prove (4). (5)

For 0 < T < ∞, J = 1, 2, . . . and K = 0, 1, 2, . . . , ∑ J −1 H3 ( J, K ) =

j=0 [H (( j

+ 1)T ) − H ( j T )] p K ( j T ) ∑ J −1 j=0 p K ( j T )

increases strictly with J from H (T ) to H3 (∞, K ) ≡ H3 (K ) and increases strictly with K from H3 ( J, 0) to H (J T ) − H ((J − 1)T ), and ∑∞ H3 (K ) ≡

j=0 [H (( j

+ 1)T ) − H ( j T )] p K ( j T ) ∑∞ j=0 p K ( j T )

increases strictly with K from H3 (0) to T h(∞). Proof Note that H3 (1, K ) = H (T ),

lim H3 ( J, K ) = H ( J T ) − H (( J − 1)T ),

K →∞

H (T ) ≤ H3 (J, K ) ≤ H (J T ) − H (( J − 1)T ), which proves that H3 (J, K ) increases strictly with J from H (T ) to H3 (K ). Forming H3 ( J, K + 1) − H3 (J, K ),

248

Appendix A

J −1 J −1 ∑ ∑ [H (( j + 1)T ) − H ( j T )] [ p K +1 ( j T ) p K (i T ) − p K ( j T ) p K +1 (i T )] j=0

=

i=0

J −1 ∑

{ j ∑

j=0

i=0

[H (( j + 1)T ) − H ( j T )]

[ p K +1 ( j T ) p K (i T ) − p K ( j T ) p K +1 (i T )]

⎫ J −1 ⎬ ∑ [ p K +1 ( j T ) p K (i T ) − p K ( j T ) p K +1 (i T )] . + ⎭ i= j

Noting that J −1 ∑

[H (( j + 1)T ) − H ( j T )]

j=0

J −1 ∑ [ p K +1 ( j T ) p K (i T ) − p K ( j T ) p K +1 (i T )] i= j

j J −1 {∑ ∑ = [H (( j + 1)T ) − H ( j T ) − H ((i + 1)T ) + H (i T )] j=0

i=0

}

[

]

× p K +1 ( j T ) p K (i T ) − p K ( j T ) p K +1 (i T )

> 0,

because H (( j + 1)T ) − H ( j T ) − H ((i + 1)T ) + H (i T ) > 0, p K +1 ( j T ) p K (i T ) − p K ( j T ) p K +1 (i T ) H ( j T )H (i T ) −H ( j T )−H (i T ) = [H ( j T ) − H (i T )] > 0, e K !(K + 1)! which completes the proof. Furthermore, H3 (K ) increases strictly with K from H3 (0) to ∫ ( J +1)T

lim

J →∞

h(t)dt = T h(∞),

JT

and H (T ) ≤ H3 (K ) ≤ T h(∞). (6)

For 0 < T < ∞, J = 0, 1, 2, . . . and K = 0, 1, 2, . . . , ∼3 (J, K ) = H

∑∞

j=J [H (( j

+ 1)T ) − H ( j T )] p K ( j T ) ∑∞ j=J p K ( j T )

increases strictly with J from H3 (K ) to h(∞) and increases strictly with K from ∼3 ( J, 0) to h(∞). H

Appendix A

249

Proof Note that ∼3 (J,K ) < h(∞). ∼3 (J,K ) = lim H ∼3 (J,K ) = h(∞), H (( J +1)T )− H ( J T ) < H lim H

J →∞

K →∞

Using similar methods of (5), we can prove (6).

A.6

Properties of Failure Rates for Extended Replacement

This appendix summarizes the properties of extended failure rates appeared in Chap. 5: (1) When r (t) increases strictly with t from r (0) to r (∞), for 0 < T ≤ ∞ and N = 0, 1, 2, . . . , ∫T (θt) N e−θt F(t)r (t)dt Q 3 (T , N ) = 0∫ T N −θt F(t)dt 0 (θt) e increases strictly with T from r (0) to Q 3 (∞, N ) ≡ Q 3 (N ) and increases strictly with N from Q 3 (T, 0) to r (T ). Proof Note that lim Q 3 (T , N ) = r (0),

lim Q 3 (T , N ) = r (T ), r (0) < Q 3 (T , K ) < r (T ).

T →0

N →∞

Differentiating Q 3 (T , N ) with respect to T , (θT ) N e−θT F(T ) R(T )

∫

T

(θt) N e−θt F(t) R(t)[r (T ) − r (t)]dt > 0.

0

Forming Q 3 (T , N + 1) − Q 3 (T , N ), ∫

T

L 1 (T ) =

(θt) N +1 e−θt F(t)r (t)dt

0

∫

−

T

N −θt

(θt) e

0

∫ ∫

F(t)r (t)dt

T

0 T

(θt) N e−θt F(t)dt

(θt) N +1 e−θt F(t)dt.

0

Then, we have L 1 (0) = 0 and L '1 (T ) = (θT ) N e−θT F(T ) ∫ T (θt) N e−θt F(t)(θT − θt)[r (T ) − r (t)]dt > 0, × 0

which completes the proof. Thus,

250

Appendix A

∫∞ Q 3 (N ) ≡ Q 3 (∞, N ) =

(θt) N e−θt F(t)r (t)dt ∫∞ N −θt F(t)dt 0 (θt) e

0

increases strictly with N from Q 3 (0) to r (∞). (2) When r (t) increases strictly with t from r (0) to r (∞), for 0 ≤ T < ∞ and N = 0, 1, 2, . . . , ∫∞ (θt) N e−θt F(t)r (t)dt ∼ Q 3 (T , N ) = T∫ ∞ N −θt F(t)dt T (θt) e ∼3 (0, N ) = Q 3 (N ) to r (∞) and increases strictly increases strictly with T from Q ∼ with N from Q 3 (T, 0) to r (∞). Proof Note that ∼3 (T, N ) = lim Q ∼3 (T , N ) = r (∞), lim Q

T →∞

N →∞

∼3 (T , N ) < r (∞). r (T ) < Q

Using similar methods of (1), we can prove (2). (3)

For 0 ≤ TO < T ≤ ∞, ∫T T

e−θt dF(t)

TO

e−θt F(t)dt

Q 1 (TO , T ) = ∫ TO

increases strictly with TO from Q 1 (0, T ) = Q 1 (T ) to h(T ) and increases strictly with ∼1 (TO ). T from h(TO ) to Q 1 (TO , ∞) = Q Proof Note that ∫T Q 1 (0, T ) = ∫ 0T ∫T

0

e−θt dF(t) e−θt F(t)dt

−θt dF(t) T e lim ∫ TO −θt T →TO F(t)dt TO e

= Q 1 (T ),

lim Q 1 (TO , T ) = h(T ),

TO →T

∫∞

= h(TO ),

T

Q 1 (TO , ∞) = ∫ ∞O

e−θt dF(t)

−θt F(t)dt TO e

h(TO ) < Q 1 (TO , T ) < h(T ). Differentiating Q 1 (TO , T ) with respect to TO , e−θTO F(TO )

∫

T

e−θt F(t) [h(t) − h(TO )]dt > 0,

TO

and differentiating Q 1 (TO , T ) with respect to T ,

∼1 (TO ), =Q

Appendix A

251 −θT

e

∫

T

F(T )

e−θt F(t)[h(T ) − h(t)]dt > 0,

TO

which completes the proof. (4)

For 0 ≤ TO < T < ∞, ∫T Q 2 (TO , T ) =

TO

e−θt h(t)dt

∫T

TO

e−θt dt

∫T −θT increases strictly with TO from Q 2 (0, T ) = R1 (T , 0) = 0 θe−θt h(t)dt/(1 ∫ ∞ −θt− e ) to h(T ) and increases strictly with T from h(TO ) to Q 2 (TO , ∞) = 0 θe h(TO + t)dt. Proof Note that ∫T

e−θt h(t)dt

∫T

θe−θt h(t)dt , lim Q 2 (TO , T ) = h(T ), −θt dt TO →T 1 − e−θT 0 e ∫ ∞ −θt h(t)dt ∫ ∞ −θt TO e lim Q 2 (TO , T ) = h(TO ), Q 2 (TO , ∞) = ∫ ∞ −θt = θe h(TO +t)dt, T →TO dt 0 TO e

Q 2 (0, T ) =

0

∫T

=

0

h(TO ) < Q 2 (TO , T ) < h(T ). Using similar methods of (3), we can prove (4).

A.7 Properties of Failure Rates for Shock and Damage Models This appendix summarizes the properties of shock and damage models appeared in Chap. 7. (1)

When rn (x) increases strictly with n to 1, for 0 < T < ∞ and 0 < K < ∞, Q 2 (T ) =

∑∞

n (n+1) (K ) − G (n+2) (K )] n=0 [(λT ) /n!][G ∑∞ n (n+1) (K ) n=0 [(λT ) /n!]G

increases strictly with T from r1 (K ) to 1. Proof Note that lim Q 2 (T ) = r1 (K ),

T →0

lim Q 2 (T ) = 1.

T →∞

252

Appendix A

Differentiating Q 2 (T ) with respect to T , ∞ ∞ ∑ ∑ (λT )n (λT ) j (n+2) (K )G ( j+1) (K )( j − n) G n! j! n=0 j=0 ⎧ ∞ ⎨∑ n ∑ (λT )n (λT ) j (n+2) (K )G ( j+1) (K )( j − n) G = ⎩ n! j! n=0 j=0 ⎫ ∞ ⎬ ∑ (λT )n (λT ) j (n+2) + (K )G ( j+1) (K )( j − n) . G ⎭ n! j! j=n

L(T ) ≡

On the other hand, n ∞ ∑ ∑ (λT )n (λT ) j (n+2) (K )G ( j+1) (K )( j − n) G n! j! n=0 j=0

=

∞ ∞ ∑ ∑ (λT )n (λT ) j (n+1) (K )G ( j+2) (K )(n − j ). G n! j! n=0 j=n

Thus, the above equation is ∞ ∞ ∑ ∑ (λT )n (λT ) j ( j − n)G (n+1) (K )G ( j+1) (K ) n! j! n=0 j=n [ (n+2) ] G (K ) G ( j+2) (K ) × − > 0, G (n+1) (K ) G ( j+1) (K )

L(T ) ≡

which proves that Q 2 (T ) increases strictly with T from r1 (K ) to 1. (2) When rn (x) increases strictly with n to 1, for J = 0, 1, 2, . . . and 0 < T < ∞, Q1( J ) =

λ

∑∞

∫ (J +1)T (K ) − G (n+1) (K )] J T [(λt)n /n!]e−λt dt ∫ (J +1)T ∑∞ (n) [(λt)n /n!]e−λt dt n=0 G (K ) J T

n=0 [G

(n)

increases strictly with J to λ. Proof Note that lim Q 1 ( J ) = λ.

J →∞

Using similar methods of (1), we can prove (2).

Appendix A

253

References 1. Nakagawa T (2014) Random maintenance policies. Springer, London 2. Nakagawa T, Zhao X (2015) Maintenance overtime policies in reliability theory. Springer, London 3. Zhao X, Cai J, Mizutani S, Nakagawa T (2020) Average failure rate and its applications of preventive replacement policies. In: Pham H (ed) Reliability and statistical computing. Springer, London, pp. 229–244

Appendix B

Chapter 2 2.1 Suppose that the unit is replaced at a planned time T (0 < T ≤ ∞), at a planned number N (N = 1, 2, . . . ) of working times, at time Z or at failure, whichever occurs first. Then, the probability that the unit is replaced at time T is [1 − G (N ) (T )] K (T ) F(T ), the probability that it is replaced at work N is ∫

T

K (t) F(t)dG (N ) (t),

0

the probability that it is replaced at failure is ∫

T

[1 − G (N ) (t)]K (t)dF(t),

0

the probability that is it replaced at time Z is ∫

T

[1 − G (N ) (t)] F(t)dK (t),

0

and the mean time to replacement is

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Mizutani et al., Which-Is-Better (WIB): Problems in Reliability Theory, Springer Series in Reliability Engineering, https://doi.org/10.1007/978-3-031-27316-2

255

256

Appendix B

∫

(N )

T

T [1 − G (T )]K (T ) F(T ) + t K (t) F(t)dG (N ) (t) 0 ∫ T ∫ T (N ) t[1 − G (t)]K (t)dF(t) + t[1 − G (N ) (t)] F(t)dK (t) + 0 0 ∫ T [1 − G (N ) (t)]K (t) F(t)dt. = 0

Therefore, the expected cost rate is C F (T , N ) =

∫T cT + (c F − cT ) 0 [1 − G (N ) (t)]K (t)dF(t) ∫T ∫T + (c N − cT ) 0 K (t) F(t)dG (N ) (t) + (c K − cT ) 0 [1 − G (N ) (t)] F(t)dK (t) , ∫T (N ) (t)]K (t) F(t)dt 0 [1 − G

where cT , c N and c F are given in (2.5), and c K = replacement cost at time Z . Similarly, when the unit is replaced preventively at time T or work N , whichever occurs last, the expected cost rate is C L (T, N ) =

∫∞ { } cT + (c F − cT∫) K (T )F(T ) + T [1 − G (N ) (t)]K (t)dF(t) ∞ + (c N − cT ) {T K (t) F(t)dG∫(N ) (t) } ∞ + (c K − cT ) K (T ) F(T ) + T [1 − G (N ) (t)] F(t)dK (t) . ∫∞ ∫T (N ) (t)]K (t) F(t)dt 0 K (t) F(t)dt + T [1 − G

Using similar techniques in Sects. 2.2, 2.3 and 2.4, we can discuss WIB problems. 2.2 The unit has to be operating for a specified S (0 < S < ∞) and is replaced at work N (N = 1, 2, . . . ), at the first completion of working times over time T (0 ≤ T ≤ S) or at failure, whichever occurs first [10]. The probability that it is replaced at work N is ∫ T F(t)dG (N ) (t), 0

the probability that it is replaced at the first completion of working times over time T is ] N −1 ∫ T [∫ S ∑ F(u)dG(u − t) dG (n) (t), n=0

0

T

the probability that it is replaced at time S is F(S)

N −1 ∫ ∑ n=0

0

T

G(S − t)dG (n) (t),

Appendix B

257

the probability that it is replaced at failure is ∫

T

[1 − G

(N )

N −1 ∫ ∑

(t)]dF(t) +

0

+ [F(S) − F(T )]

0

n=0

N −1 ∫ ∑

T

{∫

T

S

}

[F(u) − F(T )]dG(u − t) dG (n) (t)

T

G(S − t)dG (n) (t),

0

n=0

and the mean time to replacement is ∫

T

t F(t)dG

(N )

(t) +

0

+ S F(S) +

N −1 ∫ ∑ n=0

+ ∫

n=0 T

=

T

{∫

S T

T

[∫

0

T

[∫

0

S

]

u F(u) dG(u − t) dG (n) (t)

T

G(S − t)dG

(n)

∫

T

(t) +

0

n=0

0

N −1 ∫ ∑

n=0

N −1 ∫ ∑

T

N −1 ∫ ∑

t[1 − G (N ) (t)]dF(t)

0

[∫

u

]

}

y dF(y) dG(u − t) dG (n) (t)

T S

] y dF(y) G(S − t) dG (n) (t)

T

[1 − G (N ) (t)] F(t)dt +

0

N −1 ∫ ∑ n=0

0

T

[∫

S

] G(u − t) F(u)du dG (n) (t).

T

Thus, the expected cost rate is C O T (T , N ; S) =

] ∑ N −1 ∫ T [∫ S c F − (c F − c O T ) n=0 F(u)dG(u − t) dG (n) (t) 0 T ∫T ∑ N −1 ∫ T (n) − (c F − c N ) 0 F(t)dG (N ) (t) − (c F − c S ) F(S) n=0 0 G(S − t)dG (t) ] [ , ∫T ∫ ∫ ∑ N −1 T S (N ) (t)]F(t)dt + (n) n=0 0 0 [1 − G T G(u − t) F(u)du dG (t)

where c F , c O T and c N are given in (2.29) and c S = replacement cost at time S. When S is a random variable with a general distribution L(t) ≡ Pr{S ≤ t}, C O T (T , N ; S) is

258

Appendix B

∫T c F − (c F − c N ) 0 L(t) F(t)dG (N ) (t) ] (n) ∑ N −1 ∫ T [∫ ∞ − (c F − c O T{) n=0 0 T L(u) F(u)dG(u − t) dG (t) ∫T − (c F − c S ) 0 [1 − G (N ) (t)] F(t)dL(t) ] (n) } ∑ N −1 ∫ T [∫ ∞ G(u − t) F(u)dL(u) dG (t) + n=0 0 T C(T , N ; L) = . ∫T (N ) [1 − G (t)]L(t) F(t)dt 0 ∑ ] (n) N −1 ∫ T [∫ ∞ + n=0 0 T G(u − t) L(u) F(u)du dG (t) Using similar techniques in Sect. 2.5.1 when G(t) = 1 − e−θt , we can discuss WIB problems. 2.3 Suppose that the unit has to be operating during the interval [T0 , T0 + tx ] for a specified T0 (0 ≤ T0 < ∞) and tx (0 ≤ tx < ∞) [11]. When the unit is replaced at time T (0 < T < T0 ), at time T0 + tx or at failure, whichever occurs first, however, it is not replaced in [T0 , T0 + tx ]. Then, the probability that the unit is replaced at time T is F(T )Y (T ), the probability that it is replaced at T0 + tx is ∫

T

F(t + tx )dY (t),

0

the probability that it is replaced at failure is ∫

T

F(T )Y (T ) +

F(t + tx )dY (t),

0

and the mean time to replacement is ∫

T

T F(T )Y (T ) + (t + tx ) F(t + tx ) dY (t) 0 ] ∫ T ∫ T [∫ t+tx t dF(t) + u dF(u) dY (t) + Y (T ) 0 T

∫ = Y (T )

∫ F(t)dt +

0

0

0 0 T [∫ t+tx

] F(u)du dY (t).

0

Thus, the expected cost rate is [ ] ∫T cT + (c F − cT ) F(T )Y (T ) + 0 F(t + tx )dY (t) ] C F (T ; tx ) = , ∫T ∫ T [∫ t+t Y (T ) 0 F(t)dt + 0 0 x F(u)du dY (t)

Appendix B

259

where cT = replacement cost at time T or time T0 + tx , and C(T ; 0) agrees with C F (T , 1) in (2.5) when Y (t) = G(t) and cT = c N . When the unit is replaced preventively at time T (0 ≤ T < ∞) or at time T0 + tx , whichever occurs last, however, it is not replaced in [T0 , T0 + tx ]. Similarly, the expected cost rate is, for T ≥ tx , ∫∞ cT + (c F − cT )[F(T )Y (T − tx ) + T −tx F(t + tx )dY (t)] ] C L (T ; tx ) = ∫T ∫ ∞ [∫ t+t Y (T − tx ) 0 F(t)dt + T −tx 0 x F(u)du dY (t) ∫∞ cT + (c F − cT )[F(T ) + T Y (t − tx )dF(t)] = . ∫∞ ∫T 0 F(t)dt + T Y (t − tx ) F(t)dt Using similar techniques in Sect. 2.4, we can compare C F (T ; tx ) and C L (T ; tx ) when Y (t) = 1 − e−θt . 2.4 Suppose that the unit is replaced at random times Yi , where Yi (i = 1, 2, . . . , n) are independent and have the respective distribution G i (t) [12]. Denoting Ym ≡ min{Y1 , Y2 , . . . , Yn } and Y M ≡ max{Y1 , Y2 , . . . , Yn }, Pr{Ym ≤ t} = 1 −

n ∏

G i (t),

Pr{Y M ≤ t} =

i=1

n ∏

G i (t).

i=1

When the unit is replaced preventively at time Y F ≡ min{T , Ym }, i.e., it is replaced at time T , at time Ym or at failure , whichever occurs first, the expected cost rate is c R + (c F − c R ) C F (T ) = ∫ T ∏n 0

∫ T ∏n

i=1

0

i=1

G i (t)dF(t)

G i (t) F(t)dt

,

where c R = replacement cost at time Yi . When the unit is replaced preventively at time Y L ≡ max{T , Y M }, i.e, it is replaced at time T or at time Y M , whichever occurs last, the expected cost rate is ∫ ∞ ∏n c R + (c F − c R )[1 − T i=1 G i (t)dF(t)] C L (T ) = ∫ T . ∫∞ ∏n i=1 G i (t)] F(t)dt 0 F(t)dt + T [1 − Replacing G(t) with 1 − discuss WIB problems.

∏n i=1

G i (t) and G(t) with

∏n i=1

G i (t) in Chap. 2, we can

Chapter 3 3.1 Suppose that the unit is replaced at the first failure over working time N (N = 0, 1, 2, . . . ). Noting that F(t) = 1 − e−H (t) , f (t) ≡ dF(t)/dt = h(t)e−H (t) and the probability that the unit with age t fails in [u, u + du] for u > t is f (u)du/ F(t).

260

Appendix B

Thus, the mean time to replacement is ∫

∞

[∫

1

]

∞

u dF(u) dG (N ) (t) F(t) t [∫ ∞ ] ∫ ∞ (N ) −H (u)+H (t) [1 − G (t)] e du h(t)dt, =μ+ 0

0

t

and the expected number of failures until replacement is ∫

∞

[∫

1 F(t)

0

∞

] ∫ H (u)dF(u) dG (N ) (t) = 1 +

t

∞

[1 − G (N ) (t)]h(t)dt.

0

Therefore, the expected cost rate is C O (N ) =

∫∞ { } c O N + c M 1 + 0 [1 − G (N ) (t)]h(t)dt ∫∞ [∫ ∞ ] , μ + 0 [1 − G (N ) (t)] t e−H (u)+H (t) du h(t)dt

where c O N = replacement cost at the first failure over work N . Similarly, suppose that the unit is replaced at the first work over failure number K (K = 0, 1, 2, . . . ). The mean time to failure is

n=0

] } ∞ ∫ 1∑ ∞ y dG(y −u) dG (n) (u) dPK (t) = P K (t)dG (n) (t), θ n=0 0

∞{∫ t [∫ ∞

∞ ∫ ∑ 0

0

t

and the expected number of failures until replacement is ∞ ∫ ∑ 0

0

t

∞ ∫ ∑

∞[∫

∞

n=0

=

] } H (y)dG(y − u) dG (n) (u) dPK (t)

∞{∫ t [∫ ∞

n=0

0

]

G(u)h(t + u)du P K (t)dG (n) (t).

0

Thus, the expected cost rate is C O (K ) =

cO K + cM

∑∞ ∫ ∞[∫ ∞

]

0 G(u)h(t + u)du P K (t)dG ∫∞ ∑∞ (1/θ) n=0 0 P K (t)dG (n) (t)

n=0 0

(n)

(t)

,

where c O K = replacement cost at the first work over failure K . Using similar techniques in Sect. 3.7, we can derive optimal replacement policies to minimize C O (N ) and C O (K ), and can discuss WIB problems.

Appendix B

261

3.2 Suppose that the unit is replaced at time T (0 ≤ ∞) after failure K (K = 0, 1, 2, . . . ) failure, i.e., there is no replacement before failure K . The mean time to replacement is ∫ ∫ ∞

∞

(t + T ) dPK (t) = T +

0

P K (t)dt,

0

and the expected number of failures until replacement is ∫

∞

∫

∞

[K + H (t + T ) − H (t)]dPK (t) =

0

H (t + T ) dPK (t).

0

Thus, the expected cost rate is cT + c M

∫∞

H (t + T )dPK (t) , ∫∞ T + 0 P K (t)dt

C A (T , K ) =

0

which agrees with (3.2) when K = 0 and (3.25) when T = 0. Using similar techniques in Sect. 3.4, we can discuss WIB problems. 3.3 We introduce a notation of product update announcement (PUA) which arrives at random times with a distribution A(t). Suppose that if PUA has arrived before time T , the unit is replaced at time T , and conversely, if PUA has not arrived before time T , it is replaced at time PUA over time T . The mean time to replacement is ∫

∞

T A(T ) +

∫

∞

t dA(t) = T +

T

A(t)dt, T

and the expected number of failures until replacement is ∫

∞

H (T ) A(T ) +

∫

∞

H (t)dA(t) = H (T ) +

T

A(t)h(t)dt. T

Thus, the expected cost rate is ∫∞ [ ] cT + c M H (T ) + T A(t)h(t)dt C A (T ) = . ∫∞ T + T A(t)dt Differentiating C A (T ) with respect to T and setting it equal to zero, ∫

T 0

∫

∞

[h(T ) − h(t)]dt + T

A(t)[h(T ) − h(t)]dt =

cT , cM

which agrees with (3.31) when A(t) = P K (t). Using similar techniques in Sect. 3.4.2, we can discuss WIB problems.

262

Appendix B

3.4 Combining three kinds of block replacement, replacement with minimal repair and no replacement at failures [2, p. 126], we consider the following three replacement models for a finite interval S (0 < S < ∞): We divide [0, S] into two intervals [0, T ] and (T , S] (0 ≤ T ≤ ∞). (1) Block replacement and minimal repair Block replacement is done in [0, T ] and minimal repair is done at failures in (T , S], i.e., the unit is replaced with new ones at failures in [0, T ] and undergoes minimal in (T, S]. Then, the expected number of failures in [0, T ] is M(T ) ≡ repair ∑∞ at (failures j) j=1 F (T ) and the expected number of failures in (T , S] is ∫

S

[1 + H (S) − H (t)]dF(t) } ∫ T {∫ S [1 + H (S − t) − H (u − t)]dF(u − t) dM(t) +

T

0

T

∫

T

= [H (S) − H (T )] F(T ) +

[H (S − t) − H (T − t)] F(T − t) dM(t).

0

Thus, the total expected cost until time S is { C1 (T ) = c F M(T ) + c M [H (S) − H (T )] F(T ) ∫

T

+

} [H (S − t) − H (T − t)] F(T − t)dM(t) ,

0

where c F is replacement cost at each failure in [0, T ] and c M is cost of minimal repair at each failure in [T , S). (2) Block replacement and no maintenance Block replacement is done in [0, T ] and no maintenance is done in (T , S]. Then, the mean downtime from failure to time S is ] ∫ S ∫ T [∫ S (S − t) dF(t) + (S − u) dF(u − t) dM(t) T

0

∫

S

=

T

∫

[F(t) − F(T )]dt +

T

T

0

{∫

S

} [F(u − t) − F(T − t)] du dM(t).

T

Thus, the total expected cost until time S is (∫

S

C2 (T ) = c F M(T ) + c D ∫ + 0

T

T

{∫

S T

[F(t) − F(T )]dt }

)

[F(u − t) − F(T − t)]du dM(t) ,

Appendix B

263

where c D = downtime cost per unit of time elapsed time from failure to time S. (3) Minimal repair and no maintenance Minimal repair is done at failure in [0, T ] and no maintenance is done in (T , S]. Then, the mean downtime from failure to time S is ∫

1 F(T )

S

(S − t) dF(t) =

T

∫

1 F(T )

S

[F(t) − F(T )] dt.

T

Thus, the total expected cost until time S is ∫

cD

C3 (T ) = c M H (T ) +

F(T )

S

[F(t) − F(T )] dt.

T

Using similar techniques in [2, p. 126], we can derive optimal policies to minimize Ci (T ) (i = 1, 2, 3) and discuss WIB problems. Chapter 4 4.1 Suppose in age replacement that the unit is replaced at the first working time over time J T (J = 0, 1, 2, . . . ) or at failure, whichever occurs first. Then, the probability that the unit is replaced over time J T is ∞ ∫ ∑ n=0

[∫

JT

∞ J T −t

0

] F(t + u) dG(u) dG (n) (t),

and the probability that it is replaced at failure is ∞ ∫ ∑ n=0

[∫

JT

]

∞ J T −t

0

F(t + u) dG(u) dG (n) (t).

Thus, the mean time to replacement is ∞ ∫ ∑ n=0

+ =

JT

[∫

J T −t

0 ∞ ∫ ∑

JT

n=0 0 ∞ ∫ JT ∑ n=0

0

]

∞

(t + u) F(t + u) dG(u) dG (n) (t)

{∫

∞ J T −t

[∫

∞

[∫

t+u

] } y dF(y) dG(u) dG (n) (t).

0

]

G(u) F(t + u)du dG (n) (t).

0

Therefore, the expected cost rate is

264

Appendix B

] (n) ∑ ∫ J T [∫ ∞ c O J + (c F − c O J ) ∞ n=0 0 J T −t F(t + u) dG(u) dG (t) C O (J ) = , ] ∑∞ ∫ J T [∫ ∞ (n) n=0 0 0 G(u) F(t + u)du dG (t) where c O J = replacement cost over time J T and c F = replacement cost at failure. Using similar techniques in Sect. 4.1.1, we can discuss WIB problems. 4.2 Suppose in age replacement that the unit is replaced at the first periodic time over work N (N = 0, 1, 2, . . . ) or at failure, whichever occurs first. Then, the probability that the unit is replaced at the first periodic time over work N is ∞ ∑

F(( j + 1)T )[G (N ) (( j + 1)T ) − G (N ) ( j T )],

j=0

and the probability that it is replaced at failure is ∞ ∑

F(( j + 1)T )[G (N ) (( j + 1)T ) − G (N ) ( j T )].

j=0

Thus, the mean time to replacement is ∞ ∑ ( j + 1)T F(( j + 1)T )[G (N ) (( j + 1)T ) − G (N ) ( j T )] j=0

+

∞ ∫ ∑ j=0

=

∞ ∑ [

] u dF(u) dG (N ) (t)

( j+1)T [∫ ( j+1)T jT

0

]

G (N ) (( j + 1)T ) − G (N ) ( j T )

∫

( j+1)T

F(t)dt. 0

j=0

Therefore, the expected cost rate is [ (N ) ] ∑ c O N +(c F −c O N ) ∞ (( j +1)T )−G (N ) ( j T ) j=0 F(( j +1)T ) G C O (N ) = , ] ∫ ( j+1)T ∑∞ [ (N ) (( j + 1)T ) − G (N ) ( j T ) 0 F(t)dt j=0 G where c O N = replacement cost over work N . Replacing J T with T and using similar methods in Sect. 2.2, we can discuss WIB problems. 4.3 Replacing T with J T (J = 0, 1, 2, . . . ) in (2.40), the deviation time between J T and failure is ∫ ∞ ∫ JT F(t)dt + F(t)dt. D(J ) = 0

JT

Appendix B

265

Note that D(0) = μ and D(∞) = ∞. When TM is a solution of F(T ) = 1/2 in (2.41), optimal J ∗ is given: Set J ≡ [TM /T ], where [x] denotes the greatest integer in x, and compare D(J ) and D(J + 1). If D(J ) ≤ D(J + 1) then J ∗ = J , and if D( J ) > D(J + 1) then J ∗ = J + 1. 4.4 Suppose in replacement with minimal repair at failures that the unit is replaced at the first working time over time J T (J = 0, 1, 2, . . . ) or at work N , whichever occurs first. Then, the probability that the unit is replaced over time J T is 1 − G (N ) (J T ), and the probability that it is replaced at work N is G (N ) ( J T ). Thus, the mean time to replacement is N −1 ∫ ∑

[∫

1 θ

∞ J T −t

0

n=0

=

JT

N −1 ∑

] ∫ (t + u) dG(u) dG (n) (t) +

JT

t dG (N ) (t)

0

G (n) ( J T ),

n=0

and the expected number of failures until replacement is N −1 ∫ ∑ n=0

=

JT

[∫

J T −t

0

N −1 ∫ ∑ n=0

]

∞

JT 0

{∫

H (t + u) dG(u) dG

∞

(n)

∫ (t) +

JT

H (t) dG (N ) (t)

0

} [H (t + u) − H (t)] dG(u) dG (n) (t).

0

Therefore, the expected cost rate is c O J + (c N − c O J )G (N ) (J T ) } (n) ∑ N −1∫ J T{∫ ∞ + c M n=0 0 0 [H (t +u) − H (t)] dG(u) dG (t) , C O (J, N ) = ∑ N −1 (n) G (J T ), (1/θ) n=0 where c O J = replacement cost at the first work over J T . Using similar techniques in Sect. 4.2, we can discuss WIB problems. 4.5 Suppose in replacement with minimal repair that the unit is replaced at the first periodic time over work N or at time J T , whichever occurs first. Then, the mean time to replacement is

266

Appendix B J −1 ∑ ( j + 1)T [G (N ) (( j + 1)T ) − G (N ) ( j T )] + J T [1 − G (N ) (J T )] j=0

=T

J −1 ∑

[1 − G (N ) ( j T )],

j=0

and the expected number of failures until replacement is J −1 ∑

H (( j + 1)T )[G (N ) (( j + 1)T ) − G (N ) ( j T )] + H ( J T )[1 − G (N ) (J T )]

j=0 J −1 ∑ = [H (( j + 1)T ) − H ( j T )][1 − G (N ) ( j T )]. j=0

Therefore, the expected cost rate is c J + (c O N − c J )G (N ) (J T ) ∑ −1 [H (( j + 1)T ) − H ( j T )][1 − G (N ) ( j T )] + c M Jj=0 C O (N , T ) = , ∑ −1 [1 − G (N ) ( j T )] T Jj=0 where c O N = replacement cost at the first time over work N . Replacing G (N ) (t) with PK (t) in Sect. 4.4, we can discuss WIB problems. Chapter 5 5.1 Suppose that the unit is replaced at time T (0 < T ≤ ∞), at number K (K = 1, 2, . . . ) of minor failures or at major failure, whichever occurs first. Then, replacing G (N ) (t) with PK (t) formally, the probability that the unit is replaced at time T is P K (T ) F(T ), the probability that it is replaced at failure K is ∫

T

F(t) dPK (t),

0

and the probability that it is replaced at major failure is ∫

T

P K (t) dF(t),

0

∑ ∑ where Pk (t) ≡ kj=0 p j (t) = kj=0 [R(t) j /j!]e−R(t) (k = 0, 1, 2, . . . ). Then, the mean time to replacement is

Appendix B

267

∫

T

T P K (T ) F(T ) +

∫

T

t F(t) dPK (t) +

0

∫ t P K (t) dF(t) =

0

T

P K (t) F(t)dt,

0

and the expected number of minor failures until replacement is ∫ T ∫ T R(T )P K (T ) F(T ) + R(t) F(t) dPK (t) + R(t)P K (t) dF(t) 0 0 ∫ T P K (t) F(t)r (t) dt. = 0

Therefore, the expected cost rate is ∫T ∫T cT + (c K − cT ) 0 F(t)dPK (t) + (c F − cT ) 0 P K (t)dF(t) ∫T +c M 0 P K (t) F(t)r (t) dt C F (T , K ) = , ∫T 0 P K (t) F(t)dt where c K = replacement cost at failure K . Similarly, suppose that the unit is replaced preventively at time T (0 < T < ∞ ) or at number K (K = 0, 1, 2, . . . ) of minor failure, whichever occurs last. Then, the probability that the unit is replaced at time T is PK (T ) F(T ), the probability that it is replaced at failure K is ∫

∞

F(t) dPK (t),

T

and the probability that it is replaced at major failure is ∫

∞

F(T ) +

P K (t) dF(t).

T

Thus, the mean time to replacement is ∫

∞

T PK (T ) F(T ) + t F(t) dPK (t) + T ∫ ∞ ∫ T F(t) dt + P K (t) F(t) dt, = 0

∫ 0

T

∫

∞

t dF(t) + T

T

and the expected number of minor failures until replacement is

t P K (t) dF(t)

268

Appendix B

∫ ∞ ∫ T ∫ ∞ R(T )PK (T ) F(T ) + R(t) F(t)dPK (t) + R(t)dF(t) + R(t)P K (t)dF(t) T 0 T ∫ ∞ ∫ T F(t)r (t) dt + P K (t) F(t)r (t) dt. = 0

T

Therefore, the expected cost rate is ∫∞ cT + (c K − cT ) T F(t) ∫ ∞dPK (t) + (c F [− cT )[F(T ) + T P K (t) dF(t)] ] ∫∞ ∫T + c M 0 F(t)r (t) dt + T P K (t) F(t)r (t) dt C L (T , K ) = . ∫∞ ∫T F(t) dt + P (t) F(t) dt, K 0 T Using similar techniques in Sect. 5.1, we can discuss WIB problems. 5.2

The probability that the unit is replaced at time T is F(T ) F(T0 )

,

and the probability that it is replaced at failure in [T0 , T ] is F(T ) − F(T0 ) F(T0 )

.

Thus, the mean time to replacement is 1 F(T0 )

[

∫ T F(T ) +

T T0

] t dF(t) = T0 +

1 F(T0 )

∫

T

F(t) dt, T0

and the expected number of failures until replacement is H (TO ). Therefore, the expected cost rate is C(T0 , T ) =

(cT − c K )F(T ) + [c M H (T0 ) + c K ] F(T0 ) , ∫T T0 F(T0 ) + T0 F(t) dt

where c K = replacement cost at failure in [T0 , T ]. Using similar techniques in Sect. 5.2.2, we can discuss WIB problems. 5.3 Suppose that the unit is replaced at time T (0 < T ≤ ∞), at max{Y1 , Y2 } or at failure, whichever occurs first. Then, the probability that the unit is replaced at time T is F(T )[G 1 (T )G 2 (T ) + G 1 (T )G 2 (T ) + G 1 (T )G 2 (T )] = F(T )[1−G 1 (T )G 2 (T )],

Appendix B

269

the probability that it is replaced at time Y1 is ∫

T

F(t)G 2 (t) dG 1 (t),

0

the probability that it is replaced at time Y2 is ∫

T

F(t)G 1 (t) dG 2 (t),

0

and the probability that it is replaced at failure is ∫

T

[1 − G 1 (t)G 2 (t)] dF(t).

0

Then, the mean time to replacement is ∫ T T F(T )[1 − G 1 (T )G 2 (T )] + t F(t)G 2 (t) dG 1 (t) 0 ∫ T ∫ T t F(t)G 1 (t) dG 2 (t) + t [1 − G 1 (t)G 2 (t)] dF(t) + 0 0 ∫ T [1 − G 1 (t)G 2 (t)] F(t) dt. = 0

Therefore, the expected cost rate is ∫T cT + (c F − cT ) 0 [1 − G 1 (t)G 2 (t)] dF(t) C M F (T ) = . ∫T 0 [1 − G 1 (t)G 2 (t)]F(t) dt Next, suppose that the unit is replaced preventively at time T (0 ≤ T < ∞) or at min{Y1 , Y2 }, whichever occurs last. Then, the probability that the unit is replaced at time T is F(T )[1 − G 1 (T )G 2 (T )], the probability that it is replaced at time Y1 is ∫

∞

F(t)G 2 (t) dG 1 (t),

T

the probability that it is replaced at time Y2 is ∫

∞ T

F(t)G 1 (t) dG 2 (t),

270

Appendix B

and the probability that it is replaced at failure is ∫

∞

F(T ) +

G 1 (t)G 2 (t) dF(t).

T

Thus, the mean time to replacement is ∫ ∞ T F(T )[1 − G 1 (T )G 2 (T )] + t F(t)G 2 (t) dG 1 (t) T ∫ T ∫ ∞ ∫ ∞ t F(t)G 1 (t) dG 2 (t) + t dF(t) + t G 1 (t)G 2 (t) dF(t) + T 0 T ∫ ∞ ∫ T F(t) dt + G 1 (t)G 2 (t) F(t) dt. = 0

T

Therefore, the expected cost rate is ∫∞ cT + (c F − cT )[F(T ) + T G 1 (t)G 2 (t) dF(t)] C M L (T ) = . ∫∞ ∫T 0 F(t) dt + T G 1 (t)G 2 (t) F(t) dt Using similar techniques in Sect. 5.3.1, we can derive optimal policies to minimize C M F (T ) and C M L (T ). Furthermore, comparing C F (T ) in (5.46) and C L (T ) in (5.49), and C M F (T ) and C M L (T ), we can discuss WIB problems. 5.4 Suppose in Chap. 3 that the unit is replaced at time T (0 < T ≤ ∞), at working number N (N = 1, 2, . . . ) or at failure number K (K = 1, 2, . . . ), whichever occurs first. Then, the probability that the unit is replaced at time T is [1 − G (N ) (T )] P K (T ), the probability that it is replaced at work N is ∫

T

P K (t) dG (N ) (t),

0

and the probability that it is replaced at failure K is ∫

T

[1 − G (N ) (t)] dPK (t).

0

Thus, the mean time to replacement is

Appendix B

271

∫

(N )

T

T [1 − G (T )] P K (T ) + t P K (t) dG 0 ∫ T [1 − G (N ) (t)] P K (t) dt, =

(N )

∫

T

t [1 − G (N ) (t)] dPK (t)

(t) + 0

0

and the expected number of failures until replacement is ∫ T H (T )[1−G (N ) (T )] P K (T ) + H (t)P K (t) dG (N ) (t) 0 ∫ T ∫ T (N ) H (t)[1−G (t)] dPK (t) = [1 − G (N ) (t)] P K (t)h(t) dt. + 0

0

Therefore, the expected cost rate is C F (T , N , K ) =

∫T ∫T cT + (c N − cT ) 0 P K (t)dG (N ) (t) + (c K − cT ) 0 [1 − G (N ) (t)] dPK (t) ∫T + c M 0 [1 − G (N ) (t)] P K (t)h(t)dt . ∫T (N ) (t)]P (t) dt [1 − G K 0

Next, suppose that the unit is replaced at time T (0 ≤ T < ∞), at working number N (N = 0, 1, 2, . . . ) or at failure number K (K = 0, 1, 2, . . . ), whichever occurs last. Then, the probability that the unit is replaced at time T is G (N ) (T )PK (T ), the probability that it is replaced at work N is ∫

∞

PK (t) dG (N ) (t),

T

and the probability that it is replaced at failure K is ∫

∞

G (N ) (t) dPK (t).

T

Thus, the mean time to replacement is ∫

(N )

∞

(T )PK (T ) + t PK (t) dG T ∫ ∞ [1 − G (N ) (t)PK (t)] dt, =T+

TG

(N )

∫

∞

(t) + T

T

and the expected number of failures until replacement is

t G (N ) (t) dPK (t)

272

Appendix B

∫ ∞ ∫ ∞ H (T )G (N ) (T )PK (T ) + H (t)PK (t) dG (N ) (t) + H (t)G (N ) (t) dPK (t) T T ∫ ∞ = H (T ) + [1 − G (N ) (t)PK (t)]h(t) dt. T

Therefore, the expected cost rate is C L (T , N , K ) =

∫∞ ∫∞ cT + (c{ N − cT ) ∫T PK (t) dG (N ) (t) + (c K − cT}) T G (N ) (t) dPK (t) ∞ + c M H (T ) + T [1 − G (N ) (t)PK (t)]h(t) dt ∫∞ . T + T [1 − G (N ) (t)PK (t)] dt

Using techniques in Chap. 3, we can derive optimal policies to minimize C F (T , N , K ) and C L (T , N , K ). Furthermore, comparing their optimal policies, we can discuss WIB problems. Chapter 6 6.1 Consider a parallel system with n (n = 1, 2, . . . ) units which operates for a job with a random working time Y and G(t) ≡ Pr{Y ≤ t}, and suppose that it is replaced at time T , at work N or at failure, whichever occurs first. Then, replacing F(t) with F(t)n in (2.5), the expected cost rate is ∫T cT + (c F − cT ) 0 [1 − G (N ) (t)] dF(t)n C F (T , N ; n) = , ∫T (N ) (t)][1 − F(t)n ] dt 0 [1 − G where cT = replacement cost at time T or work N , and c F = replacement cost at failure. Furthermore, when the number of units is a random variable with a probability function pn in Sect. 6.1.2, the expected rate is ∫T ∑ (N ) cT + (c F − cT ) ∞ (t)] dF(t)n n=1 pn 0 [1 − G C F (T , N ; p) = . ∫T ∑∞ (N ) (t)][1 − F(t)n ] dt n=1 pn 0 [1 − G Similarly, suppose that the unit is replaced preventively at time T or at work N , whichever occurs last. From (2.12), ∫∞ ] [ cT + (c F − cT ) F(T )n + T [1 − G (N ) (t)] dF(t)n C L (T , N ; n) = ∫ T , ∫∞ n (N ) (t)][1 − F(t)n ] dt 0 [1 − F(t) ] dt + T [1 − G and for a parallel system with random units,

Appendix B

273

∫∞ { } ∑ n (N ) cT + (c F − cT ) ∞ (t)] dF(t)n n=1 pn F(T ) + T [1 − G {∫ }. C L (T , N ; p) = ∑ ∫∞ T ∞ n (N ) (t)][1 − F(t)n ] dt n=1 pn 0 [1 − F(t) ] dt + T [1 − G For a standby system with n units, we may replace F(t)n with F (n) (t). 6.2 Replacing T in Problem 6.1 with J T , we can easily obtain the expected cost rates C F ( J, N ; n), C F (J, N ; p), C L (J, N ; n) and C L (J, N ; p). Furthermore, replacing F(T ) in Chap. 2 with Fn,P (t) in (6.70), we can discuss WIB problems of any redundant systems with reliability Rn,P (t) in (6.69). Chapter 8 8.1 Suppose that full backup is made at a volume K (0 < K ≤ ∞) of update data. Then, the probability that full backup is made at volume K is ∫ ∞ ∑ [G (n) (K ) − G (n+1) (K )]

∞

D(t) dF (n+1) (t),

0

n=0

and the probability that it is made at failure is ∞ ∑

G (n) (K )

∫

∞

[F (n) (t) − F (n+1) (t)] dD(t).

0

n=0

The mean time to full backup is ∞ ∑

[G

(n)

(K ) − G

∞ ∑

∞ ∑ n=0

∫

G (n) (K )

(K )]

∞

t D(t) dF (n+1) (t)

G

∞

t [F (n) (t) − F (n+1) (t)] dD(t)

0

n=0

=

∫ 0

n=0

+

(n+1)

(n)

∫ (K )

∞

[F (n) (t) − F (n+1) (t)] D(t) dt,

0

and the expected number of backups until full backup is

274

Appendix B ∞ ∑

n[G (n) (K ) − G (n+1) (K )]

∞

D(t) dF (n+1) (t)

0

n=0

+

∞ ∑

nG (n) (K )

n=0 ∞ ∑

∫

∞

[F (n) (t) − F (n+1) (t)] dD(t)

0

G (n) (K )

=

∫

∫

∞

D(t) dF (n) (t),

0

n=1

and when full backup is made at (n + 1)th update of volume K , its cost is ∞ ∫ ∑ n=0

K

{∫

∞ K −x

0

} ∫ [c K + c O (x + y)]dG (n) (y) dG (n) (x)

∞

D(t)dF (n+1) (t).

0

Thus, denoting ∫ Mn (K ) ≡

K

∫

(c K + c O x) dG (n) (x),

K

Nn (K ) =

0

(c R + c O x) dG (n) (x),

0

the expected cost of differential backup is ∼D (K ) = c F + C +

∞ ∫ ∑ n=0

+

∞ ∑ n=0

= cF +

K

∞ ∑

{∫

n=1 ∞ K −x

0

∫

∫ Mn (K )

D(t) dF (n) (t)

0

} ∫ (n) [c K + c O (x + y)]dG(y) dG (x)

∞

Nn (K )

∞

∞

D(t)dF (n+1) (t)

0

[F (n) (t) − F (n+1) (t)]dD(t)

0 ∞ ∫ ∑ n=0

K

(c K + c O x)dG

(n)

∫

∞

(x)

0

D(t)dF (n+1) (t)

0

∫ ∞ ∞ c O ∑ (n) G (K ) D(t) dF (n+1) (t) + w n=0 0 ∫ ∞ ∞ ∫ K ∑ (n) + (c R + c O x)dG (x) [F (n) (t) − F (n+1) (t)]dD(t). n=0

0

Therefore, the expected cost rate is

0

Appendix B

275

∫∞ ∑ ∫K (n) cF + ∞ (x) 0 D(t)dF (n+1) (t) n=0∑0 (c K + c O x)dG ∫ ∞ + (c O /w) ∞ G (n) (K ) 0 D(t) dF (n+1) (t) ∫∞ ∑∞ ∫ K n=0 + n=0 0 (c R +c O x)dG (n) (x) 0 [F (n) (t)− F (n+1) (t)]dD(t). C D (K ) = . ∫∞ ∑∞ (n) (n) (n+1) (t)]D(t) dt, n=0 G (K ) 0 [F (t) − F Comparing C D (K ) with C D (N ) in (8.6), we can discuss WIB problems. Chapter 9 9.1 We apply a 2-out-of-3 structure in (2) of Sect. 9.1 to the random checkpoint model in Sect. 9.4.2. Replacing G ∗ (2λ) with 3G ∗ (2λ) − 2G ∗ (3λ), A = 1 − q[1 − 3G ∗ (2λ) + 2G ∗ (3λ)], B = 3G ∗ (2λ) − 2G ∗ (3λ) + p[1 − 3G ∗ (2λ) + 2G ∗ (3λ)], A − B = r [1 − 3G ∗ (2λ) + 2G ∗ (3λ)]. Thus, substituting A and B for (9.37), we can derive optimal N R∗ to minimize L R (N ) and the resulting mean time L R (N R∗ ).

Index

A Age replacement, 7, 9, 63, 97, 103, 228

B Backup first, 183, 189 Backup last, 196 Backup overtime last, 198 Backup policy, 176 Block replacement, 61, 262 Bulk task, 212 Bridge system, 142

C Checkpoint model, 205 Complexity, 142 Computer system, 175, 205 Cumulative hazard function, 31, 87

D Damage model, 145, 163, 175, 251 Database system, 175 Data transmission, 117, 138 Degenerate distribution, 9, 33 Deviation cost, 140 Deviation time, 26, 84, 128, 139, 163, 171, 264 Differential backup, 175, 176, 180, 187, 194, 202, 274

E Entropy model, 142 Euler’s constant, 120 Excess cost, 127, 131 Expected number of failures, 32, 262 Expected number of updates, 178

F Failure rate, 7, 9, 31, 63, 87, 103, 117, 146, 176, 227, 230, 239, 244, 249, 251 Finite interval, 28, 262 Full backup, 175, 189, 194, 199, 273

G Geometric distribution, 121

I Incremental backup, 175, 177, 179, 185, 192, 201

K k-out-of-n system, 117, 137, 141, 207, 209, 275

M Major decision structure, 206, 207, 209 Major failure, 87, 89, 91, 266 Mathematical induction, 119 Mean time to failure, 118, 141, 260

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Mizutani et al., Which-Is-Better (WIB): Problems in Reliability Theory, Springer Series in Reliability Engineering, https://doi.org/10.1007/978-3-031-27316-2

277

278 Mean time to replacement, 32, 35 Mean value function, 31 Median, 27 Middle replacement, 103 Minimal repair, 31, 32, 61, 71, 85, 89, 100, 110, 230, 262, 263, 265 Minor failure, 87, 115, 266 Mission time, 28 Modular structure, 206, 209, 213, 217, 221

N Nonhomogeneous poisson process, 31, 32, 87

O Opportunistic replacement, 97

P Parallel system, 117, 121, 130, 132, 136, 272 Parallel work, 127, 129 Partition method, 207, 208 Partition problem, 138 Periodic checkpoint, 206, 221 Periodic replacement, 31, 63, 143, 166, 239, 243 Poisson distribution, 31, 121 Preventive replacement, 103, 123, 150, 157 Product update announcement, 60, 261

R Random maintenance, 175

Index Random replacement, 7, 31, 32 Redundant system, 117, 137, 273 Renewal process, 146, 176 Replacement first, 7, 10, 16, 31, 35, 39, 41, 46, 51, 56, 63, 71, 75, 79, 88, 103, 107, 108, 110, 113, 114, 143 Replacement last, 7, 13, 31, 43, 48, 54, 58, 66, 73, 77, 82, 91, 104, 108, 111, 114, 143, 163 Replacement middle, 103, 106–108, 112– 114, 168 Replacement overtime, 7, 18, 20, 23, 31, 45, 51, 82, 84, 158, 161 Replacement undertime, 97, 102

S Scheduling problem, 128, 129, 134 Scheduling time, 117, 127 Shortage cost, 127, 131 Standby system, 118, 130, 135, 273 Sudden failure, 28

T Tandem task, 211 Tandem work, 128, 130, 132, 134, 136 Two failure modes, 87, 88

U Undertime replacement, 97

W Weibull distribution, 124