Applications of Operations Research and Management Science for Military Decision Making [1st ed. 2019] 978-3-030-20568-3, 978-3-030-20569-0

Table of contents :
Front Matter ....Pages i-xix
Mathematical Modeling, Management Science, and Operations Research for Military Decision-Making (William P. Fox, Robert Burks)....Pages 1-15
Statistics and Probability in Military Decision-Making (William P. Fox, Robert Burks)....Pages 17-99
Modeling by Fitting Data (William P. Fox, Robert Burks)....Pages 101-135
Mathematical Programming: Linear, Integer, and Nonlinear Optimization in Military Decision-Making (William P. Fox, Robert Burks)....Pages 137-191
Introduction to Multi-attribute Military Decision-Making (William P. Fox, Robert Burks)....Pages 193-249
Game Theory (William P. Fox, Robert Burks)....Pages 251-329
Modeling Change with Dynamical Systems Models (William P. Fox, Robert Burks)....Pages 331-393
MONTE CARLO Simulation and AGENT-BASED Modeling (ABM) in Military Decision-Making (William P. Fox, Robert Burks)....Pages 395-453
Logistics Network Modeling (William P. Fox, Robert Burks)....Pages 455-483
Back Matter ....Pages 485-487

Citation preview

International Series in Operations Research & Management Science

William P. Fox Robert Burks

Applications of Operations Research and Management Science for Military Decision Making

International Series in Operations Research & Management Science Volume 283

Series Editor Camille C. Price Department of Computer Science, Stephen F. Austin State University, Nacogdoches, TX, USA Associate Editor Joe Zhu Foisie Business School, Worcester Polytechnic Institute, Worcester, MA, USA Founding Editor Frederick S. Hillier Stanford University, Stanford, CA, USA

More information about this series at http://www.springer.com/series/6161

William P. Fox • Robert Burks

Applications of Operations Research and Management Science for Military Decision Making

William P. Fox Department of Mathematics College of William and Mary Williamsburg, VA, USA

Robert Burks Department of Defense Analysis Naval Postgraduate School Monterey, CA, USA

ISSN 0884-8289 ISSN 2214-7934 (electronic) International Series in Operations Research & Management Science ISBN 978-3-030-20568-3 ISBN 978-3-030-20569-0 (eBook) https://doi.org/10.1007/978-3-030-20569-0 © Springer Nature Switzerland AG 2019 This work is subject to copyright. All rights are reserved 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, express 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

Addressing the Current Needs In recent years of teaching mathematical modeling for decision-making coupled with conducting applied mathematical modeling research, we have found that (a) decision-makers at all levels must be exposed to the tools and techniques available to help them in the decision process, (b) decision-makers and analysts need to have and to use technology to assist in the analysis process, and (c) the interpretation and explanation of the results are crucial to understand the strengths and limitations of modeling. With this in mind, this book emphasizes and focuses on the model formulation and modeling building skills required for decision analysis, as well as the technology to support the analysis.

Audience This book would be best used for a senior-level discrete modeling course in mathematics, operations research, or industrial engineering or graduate-level discrete choice modeling courses or decision modeling courses offered in business schools offering business analytics. The book would be of interest to mathematics departments that offer mathematical modeling courses focused on discrete modeling or modeling for decision-making. The following groups would benefit from using this book: • Undergraduate students in quantitative methods courses in business, operations research, industrial engineering, management sciences, industrial engineering, or applied mathematics • Graduate students in discrete mathematical modeling courses covering topics from business, operations research, industrial engineering, management sciences, industrial engineering, or applied mathematics v

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• Junior analysts who want a comprehensive review of decision-making topics • Practitioners desiring a reference book

Objectives The primary objective of this book is illustrative in nature. It sets the tone in Chap. 1 through the introduction to mathematical modeling. In this chapter, we provide a process for formally thinking about the problem and illustrate many scenarios and examples. In these examples, we begin the setup of the solution process, and which will be covered in-depth in later chapters. Based on many years of applied research and modeling, we have considered which techniques should be included or excluded in a book of this nature. Finally, we decided on the main techniques that we cover in our three-course sequence in mathematical modeling for decision-making in the Department of Defense Analysis and the Naval Postgraduate School. We feel these subjects have served and prepared our students well, as they have all gone on as leaders and decision-makers for our nation.

Organization This book contains information that could easily be covered in a two-semester course or a one-semester overview of topics. This allows the instructors the flexibility to pick and choose topics consistent with their course and consistent with their current needs. In Chaps. 2–8, we present materials to solve the type of problems introduced in Chap. 1. The contexts of these problems are in military applications and related military processes. In Chap. 2, we describe statistical models in military decision-making. From modeling with basic statistical information of piracy through hypothesis tests, we show how to use and interpret these models. Case studies are used to highlight the use of statistical methods. Chapter 3 addresses the use of regression tools for analyzing from simple linear regression to advanced regression methods. Technology is an essential tool for regression analysis. Chapter 4 addresses the uses of mathematical programming (linear, integer, and nonlinear) to solve problems that help in military decision-making. We start with defining the mathematical programming methods and illustrate some formulation concepts. Technology is used to solve the formulated problems. Mathematical programming is used later in our chapters discussing data envelopment analysis and game theory.

Preface

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Chapter 5 discusses the use of multi-attribute decision-making. In the real world, there are always many criteria to consider in weighing alternatives and courses of actions. We discuss different weighting schemes, including entropy, rank order centroid, ratio, and pairwise comparison. We discuss MADM methods of data envelopment analysis, simple additive weights, analytical hierarchy process, and technique of order performance by similarity to ideal solutions. Chapter 6 covers game theory. Both total and partial conflict games are covered. Case studies are used to show the type of real decision problems and analysis for which game theory can be used. Chapter 7 discusses model of change, both discrete and continuous. Lanchester equations are described and examples given to combat modeling scenarios. From hand to hand, combat to today’s insurgency warfare examples will be provided and results interpreted. Chapter 8 discusses simple Monte Carlo simulations and an introduction to agentbased models. Examples are used to expand the modeling ability to include variables and situations for which analytical models cannot be adequately used. Chapter 9 describes supply chain network logistics and decisions and analysis related to logistics. In addition, covered in this chapter are network models as well as transportation, transshipment, and assignment optimization problems. This book shows the power and limitations for mathematical modeling to solve real-world military problems. The solutions shown might not be the best solution, but they are certainly solutions that are or could be considered in the decision analysis process. As evidenced by previous textbooks in mathematical modeling, such as A First Course in Mathematical Modeling, the scenarios are revisited to illustrate alternative techniques in solving these problems. As we have seen from many years of working with COMAP’s Mathematical Contest in Modeling, ingenuity and creativity in modeling methods and solution techniques are always present. In this book, we cannot address every nuance in modeling real-world problems. What we can do is provide a sample of models and possible appropriate techniques to obtain useful results. We can establish a process to “do modeling” and illustrate many examples of modeling and technique in order to solve the problem. In the technique chapters, we assume no or little background in mathematical modeling and spend a little time establishing the procedure before we return to provide examples and solution techniques. The data used in the examples presented in this book are unclassified in both nature and design as compared to the actual data that was used in the real world examples. This book can be applied to analysts to allow them to see the range and type of problems that fit into specific mathematical techniques understanding we did address all possible mathematics techniques. Because of space, we do leave out some important techniques such as differential equations. This book also applies to decision-makers. It shows the decision-makers the wide range of applications of quantitative approaches to aid in the decision-making process. As we say in our modeling classes every day, mathematics does not tell what to do, but it does provide insights and allows critical thinking into the decisionmaking process. In our discussion, we consider the mathematical modeling process

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as a framework for decision-makers. This framework has four key elements: the formulation process, the solution process, the interpretation of the mathematical answer in context of the actual problem, and the sensitivity analysis. At every step along the way in the process, the decision-maker should question procedures and techniques and ask for further explanations as well as assumptions used in the process. Two major questions could be as follows: “Did you use an appropriate technique” to obtain a solution? Why were the other techniques not considered or used? Another question could be the following: “Did you over simplify the process” so much that the solution does not really apply in this situation, or were the assumptions made fundamental to even be able to solve the problem? We thank all the mathematical modeling students that we have had over this time as well as all the colleagues who have taught mathematical modeling with us during this adventure. We particularly single out the following who helped in our threecourse mathematical modeling sequence at the Naval Postgraduate School over the years: Bard Mansger, Mike Jaye, Steve Horton, Patrick Driscoll, and Greg Mislick. We are especially appreciative of the mentorship of Frank R. Giordano over the past 30 plus years. Williamsburg, VA, USA Monterey, CA, USA

William P. Fox Robert E. Burks

Contents

1

2

Mathematical Modeling, Management Science, and Operations Research for Military Decision-Making . . . . . . . . . . 1.1 Introduction to Decision-Making . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Mathematical Modeling and Decision-Making Framework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Types of Decision Models . . . . . . . . . . . . . . . . . . . . . . 1.2.1.1 Deterministic Models . . . . . . . . . . . . . . . . . . 1.2.1.2 Stochastic Models . . . . . . . . . . . . . . . . . . . . . 1.2.2 Types of Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.2.1 Qualitative Data and Ordinal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.2.2 Quantitative Data and Cardinal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Steps in the Decision Process . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Illustrative Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Technology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8 Projects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suggested Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 4 6 12 13 13 14 15 15

Statistics and Probability in Military Decision-Making . . . . . . . . . . 2.1 Introduction to Statistics and Statistical Models . . . . . . . . . . . . . 2.2 Displaying Categorical Data . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Pie Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Displaying Quantitative Data . . . . . . . . . . . . . . . . . . . 2.2.2.1 Stem and Leaf . . . . . . . . . . . . . . . . . . . . . . 2.2.3 Symmetry Issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.4 Boxplot Used for Comparisons . . . . . . . . . . . . . . . . . .

17 17 20 20 21 21 23 24

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2.2.4.1 Comparisons . . . . . . . . . . . . . . . . . . . . . . . . Measures of Central Tendency or Location . . . . . . . . . . 2.2.5.1 Describing the Data . . . . . . . . . . . . . . . . . . . 2.2.5.2 The Mean . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.5.3 The Median . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.5.4 The Mode . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.6 Measures of Dispersion . . . . . . . . . . . . . . . . . . . . . . . . 2.2.6.1 Variance and Standard Deviation . . . . . . . . . . 2.2.6.2 Measures of Symmetry and Skewness . . . . . . 2.2.7 Section Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Classical Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1.1 The Law of Large Numbers . . . . . . . . . . . . . . 2.3.1.2 Properties of Probabilities . . . . . . . . . . . . . . . 2.3.1.3 Probability from Data . . . . . . . . . . . . . . . . . . 2.3.1.4 Intersections and Unions . . . . . . . . . . . . . . . . 2.3.1.5 The Addition Rule . . . . . . . . . . . . . . . . . . . . 2.3.2 Conditional Probability . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.3 Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.3.1 Definition of Independent Events . . . . . . . . . . 2.3.3.2 Independent Events . . . . . . . . . . . . . . . . . . . . 2.3.4 System Reliability in Series and Parallel of Independent Subsystems . . . . . . . . . . . . . . . . . . . . . . 2.3.5 Bayes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.5.1 Theorem of Total Probability . . . . . . . . . . . . . 2.3.5.2 Bayes’ Theorem . . . . . . . . . . . . . . . . . . . . . . Probability Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Discrete Distributions in Modeling . . . . . . . . . . . . . . . . 2.4.2 Poisson Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2.1 Section Exercises . . . . . . . . . . . . . . . . . . . . . 2.4.2.2 Chapter Projects Examples . . . . . . . . . . . . . . 2.4.3 Continuous Probability Models . . . . . . . . . . . . . . . . . . . 2.4.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.3.2 The Normal Distribution . . . . . . . . . . . . . . . . 2.4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.5 Exponential Distribution . . . . . . . . . . . . . . . . . . . . . . . . Military Applications of Distributions . . . . . . . . . . . . . . . . . . . . 2.5.1 Application of Probable Error in Targeting (Adapted from DA 3410 Course Notes, 2006) . . . . . . . . 2.5.1.1 Section Exercises . . . . . . . . . . . . . . . . . . . . . 2.5.2 Target Search Processes (Adapted from DA 3410 Course Notes, 2006) . . . . . . . . 2.5.2.1 Relative Motion . . . . . . . . . . . . . . . . . . . . . . 2.5.2.2 Cookie Cutter Sensor . . . . . . . . . . . . . . . . . . 2.2.5

2.3

2.4

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Searching Without Overlap Arbitrary Search . . . . . . . . . . . . . . . . . . . . . 2.5.2.4 Random Search . . . . . . . . . . . . . . . . . . . . . 2.6 Central Limit Theorem, Confidence Intervals, and Hypothesis Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . . 2.6.2 Confidence Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.3 Simple Hypothesis Testing . . . . . . . . . . . . . . . . . . . . . 2.6.3.1 Interpretation . . . . . . . . . . . . . . . . . . . . . . . 2.6.3.2 Excel Templates . . . . . . . . . . . . . . . . . . . . . 2.6.3.3 Section Exercises . . . . . . . . . . . . . . . . . . . . 2.6.3.4 Hypothesis Testing . . . . . . . . . . . . . . . . . . . 2.6.3.5 Notation and Definitions . . . . . . . . . . . . . . . 2.6.3.6 Hypothesis Testing . . . . . . . . . . . . . . . . . . . 2.7 Hypothesis Tests Summary Handout . . . . . . . . . . . . . . . . . . . . 2.7.1 Tests with One Sample Mean . . . . . . . . . . . . . . . . . . . 2.7.2 Tests with a Population Proportion (Large Sample) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.3 Tests Comparing Two Sample Means . . . . . . . . . . . . . 2.7.3.1 Section Exercises Hypothesis Test Problems . . . . . . . . . . . . . . . . . . . . . . 2.8 Case Studies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.1 Violence in the Philippines . . . . . . . . . . . . . . . . . . . . . 2.8.2 The Impact of Medical Support on Security Force Effectiveness (Adapted from a Course Project by LTC Ramey Wilson) . . . . . . . . . . . . . . . . . 2.8.2.1 Soldier Adaptation Model Framework . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.3 Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.3.1 Initial Limitations . . . . . . . . . . . . . . . . . . . . 2.8.3.2 Database . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.3.3 Definitions of Terms and Variables . . . . . . . 2.8.3.4 Results and Discussion . . . . . . . . . . . . . . . . 2.8.3.5 Limitations . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.3.6 Case Study Conclusions . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suggested Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Modeling by Fitting Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Introduction to Correlation and Simple Linear Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Correlation of Recoil Data . . . . . . . . . . . . . . . . . . . . . 3.2.2 Linear Regression of Recoil Data . . . . . . . . . . . . . . . . 3.2.2.1 Simple Least Squares Regression . . . . . . . . .

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88 89 90 91 92 94 97 98 98 99

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3.2.3 Linear Regression of Philippines SIGACTS . . . . . . . . . . Exponential Decay Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Introducing Hospital Recovery Data from a Military Hospital . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Linear Regression of Hospital Recovery Data . . . . . . . . 3.3.3 Quadratic Regression of Hospital Recovery Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.4 Exponential Decay Modeling of Hospital Recovery Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Sinusoidal Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Introducing Military Supply Shipping Data . . . . . . . . . . 3.4.2 Linear Regression of Shipping Data . . . . . . . . . . . . . . . 3.4.3 Sinusoidal Regression of Shipping Data . . . . . . . . . . . . 3.4.4 Introducing Sinusoidal Regression of Afghanistan Casualty . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Logistic Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Case Study: Dehumanization and the Outcome of Conflict with Logistic Regression . . . . . . . . . . . . . . . 3.5.1.1 Data Specifics . . . . . . . . . . . . . . . . . . . . . . . 3.5.2 A Binary Logistical Regression Analysis of Dehumanization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.3 Introducing International Alliance Data . . . . . . . . . . . . . 3.5.4 Logistic Regression of Alliance Data . . . . . . . . . . . . . . . 3.6 Poisson Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.1 Introducing SIGACTS Data . . . . . . . . . . . . . . . . . . . . . 3.6.2 Poisson Regression of SIGACTS Data . . . . . . . . . . . . . 3.7 Conclusions and Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suggested Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3

4

Mathematical Programming: Linear, Integer, and Nonlinear Optimization in Military Decision-Making . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Formulating Mathematical Programming Problems . . . . . . . . . . 4.2.1 Simple 3D Printing of Parts . . . . . . . . . . . . . . . . . . . . 4.2.2 Financial Planning Problem . . . . . . . . . . . . . . . . . . . . 4.2.2.1 Financial Planning Formulation . . . . . . . . . . 4.2.2.2 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2.3 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2.4 Objective Function . . . . . . . . . . . . . . . . . . . 4.2.3 Blending and Formulation Problem . . . . . . . . . . . . . . . 4.2.3.1 Blending Problem Solution . . . . . . . . . . . . . 4.2.4 Production Planning Problem . . . . . . . . . . . . . . . . . . . 4.2.4.1 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.4.2 Constraints . . . . . . . . . . . . . . . . . . . . . . . . .

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4.2.4.3 Objective Function . . . . . . . . . . . . . . . . . . . . Shipping Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.5.1 Integer Programming and Mixed-Integer Programming . . . . . . . . . . 4.2.5.2 Nonlinear Programming . . . . . . . . . . . . . . . . 4.2.5.3 Exercises 4.2 . . . . . . . . . . . . . . . . . . . . . . . . Graphical Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 The Feasible Region . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.2 Solving a Linear Programming Problem Graphically . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.3 Minimization Example . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.3.1 Exercises 4.3 . . . . . . . . . . . . . . . . . . . . . . . . 4.3.3.2 Projects 4.3 . . . . . . . . . . . . . . . . . . . . . . . . . Mathematical Programming with Technology . . . . . . . . . . . . . . . 4.4.1 Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1.1 Using EXCEL . . . . . . . . . . . . . . . . . . . . . . . 4.4.1.2 Answer Report . . . . . . . . . . . . . . . . . . . . . . . 4.4.1.3 Sensitivity Report . . . . . . . . . . . . . . . . . . . . . 4.4.1.4 Using EXCEL . . . . . . . . . . . . . . . . . . . . . . . 4.4.1.5 Solver . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1.6 Constraints into Solver . . . . . . . . . . . . . . . . . 4.4.1.7 Answer Report . . . . . . . . . . . . . . . . . . . . . . . 4.4.1.8 Sensitivity Report . . . . . . . . . . . . . . . . . . . . . 4.4.2 Alternate Optimal Solution Shadow Prices . . . . . . . . . . . 4.4.2.1 Using LINDO . . . . . . . . . . . . . . . . . . . . . . . 4.4.2.2 USING LINGO . . . . . . . . . . . . . . . . . . . . . . 4.4.2.3 Using MAPLE . . . . . . . . . . . . . . . . . . . . . . . 4.4.3 Integer and Nonlinear Programming with Technology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.3.1 Integer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.3.2 Nonlinear Programming . . . . . . . . . . . . . . . . 4.4.3.3 Section Exercises . . . . . . . . . . . . . . . . . . . . . Case Studies in Mathematical Programming . . . . . . . . . . . . . . . . 4.5.1 Example 1. Military Supply Chain Operations (from Fox and Garcia 2014) . . . . . . . . . . . . . . . . . . . . . 4.5.2 Example 2. Military Recruiting Raleigh Office (Modified from McGrath 2007) . . . . . . . . . . . . . . . . . . . 4.5.2.1 Section 4.5 Exercises . . . . . . . . . . . . . . . . . . Examples for Integer, Mixed-Integer, and Nonlinear Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.1 Example 1. Medical Emergency Services . . . . . . . . . . . . 4.6.2 Example 2. Optimal Path to Transport Hazardous Material . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.5

4.3

4.4

4.5

4.6

146 146 147 147 148 151 152 154 156 157 158 160 160 160 164 165 165 167 167 169 169 170 170 171 172 175 175 175 175 175 175 178 180 180 180 181

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4.6.3

Example 3. Minimum Variance of Expected Investment Returns in TSP . . . . . . . . . . . . . . . . . . . . . 4.6.4 Example 4. Cable Instillation . . . . . . . . . . . . . . . . . . . 4.6.5 Grid Coordinates of Five Departments . . . . . . . . . . . . . 4.6.5.1 The Model . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.6 Exercises 4.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Chapter Projects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Simplex Method in Excel . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8.1 Steps of the Simplex Method . . . . . . . . . . . . . . . . . . . 4.8.2 Section 4.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suggested Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

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Introduction to Multi-attribute Military Decision-Making . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Risk Analysis for Homeland Security . . . . . . . . . . . . . . . . . . . . . 5.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Data Envelopment Analysis (DEA) . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Description and Use . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.2 Methodology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.3 Strengths and Limitations to DEA . . . . . . . . . . . . . . . . . 5.3.4 Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.5 Illustrative Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.5.1 Exercises 5.3 . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Weighting Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Modified Delphi method . . . . . . . . . . . . . . . . . . . . . . . . 5.4.2 Rank Order Centroid (ROC) Method . . . . . . . . . . . . . . . 5.4.3 Ratio Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.4 Pairwise Comparison (AHP) . . . . . . . . . . . . . . . . . . . . . 5.4.5 Entropy Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Simple Additive Weighting (SAW) Method . . . . . . . . . . . . . . . . 5.5.1 Description and Uses . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.2 Methodology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.3 Strengths and Limitations . . . . . . . . . . . . . . . . . . . . . . . 5.5.4 Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.5 Illustrative Examples SAW . . . . . . . . . . . . . . . . . . . . . . 5.5.5.1 SAW Using Raw Data . . . . . . . . . . . . . . . . . 5.5.6 Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.7 Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.7.1 Exercises 5.5 . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Analytical Hierarchy Process (AHP) . . . . . . . . . . . . . . . . . . . . . 5.6.1 Description and Uses . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6.2 Methodology of the Analytic Hierarchy Process . . . . . . . 5.6.2.1 Methods to Solve for Decision-Maker Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . .

183 184 185 185 185 186 186 187 191 191 191 193 193 195 195 195 196 196 197 197 208 209 209 209 211 212 213 216 216 216 218 218 219 221 222 222 225 225 225 228 229

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5.6.3 5.6.4 5.6.5

Strengths and Limitations of AHP . . . . . . . . . . . . . . . . Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . Illustrative Examples with AHP . . . . . . . . . . . . . . . . . 5.6.5.1 Sensitivity Analysis . . . . . . . . . . . . . . . . . . 5.6.5.2 Pairwise Comparison Matrix . . . . . . . . . . . . 5.6.5.3 Sensitivity Analysis . . . . . . . . . . . . . . . . . . 5.6.5.4 Section 5.6 Exercises . . . . . . . . . . . . . . . . . 5.6.5.5 Section 5.6 Projects . . . . . . . . . . . . . . . . . . 5.7 Technique of Order Preference by Similarity to the Ideal Solution (TOPSIS) . . . . . . . . . . . . . . . . . . . . . . . . 5.7.1 Description and Uses . . . . . . . . . . . . . . . . . . . . . . . . . 5.7.2 Methodology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7.2.1 Normalization . . . . . . . . . . . . . . . . . . . . . . 5.7.3 Strengths and Limitations . . . . . . . . . . . . . . . . . . . . . . 5.7.4 Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7.5 Illustrate Examples with TOPSIS . . . . . . . . . . . . . . . . 5.7.5.1 Sensitivity Analysis . . . . . . . . . . . . . . . . . . 5.7.5.2 Comparison of Results for the Kite Network . . . . . . . . . . . . . . . . . . 5.7.5.3 Section 5.7 Exercises . . . . . . . . . . . . . . . . . 5.7.5.4 Section 5.7 Projects . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suggested Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Game Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Introduction to Game Theory . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 Two-Person Total Conflict Games . . . . . . . . . . . . . . . . 6.1.1.1 Linear Programming of Total Conflict Games . . . . . . . . . . . . . . . . . . . . . . 6.1.2 Two-Person Partial Conflict Games . . . . . . . . . . . . . . . 6.1.2.1 Discussion of Some Cooperative Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.2.2 Moving First or Committing to Move First . . . . . . . . . . . . . . . . . . . . . . . 6.1.2.3 Threats . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.2.4 Issuing a Promise . . . . . . . . . . . . . . . . . . . . 6.1.2.5 A Combination Threat and Promise . . . . . . . 6.1.2.6 Combination Threat and Promise . . . . . . . . . 6.1.3 Nash Arbitration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.3.1 Finding the Nash Arbitration Point . . . . . . . 6.1.4 Three-Person Games . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.4.1 Example of Three-Person Total Conflict . . . . . . . . . . . . . . . . . . . . . . .

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230 231 232 235 236 237 239 239

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240 240 240 242 243 243 243 245

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259 259 260 261 262 263 266 270

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6.2

6.3

6.4

Applied Game Theory to Improve Strategic and Tactical Military Decisions . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Proposed Update to the Methodology . . . . . . . . . . . . . 6.2.2 AHP Method for Pairwise Comparison . . . . . . . . . . . . 6.2.3 Proposed Application of AHP to the Military Decision-Making . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.4 Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . Two-Person Partial Conflict (Nonzero-Sum) Game . . . . . . . . . . 6.3.1 Nonlinear Programming Approach for Two or More Strategies for Each Player . . . . . . . . . 6.3.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1.2 Applying Game Theory . . . . . . . . . . . . . . . 6.3.1.3 Assigning Cardinal Values . . . . . . . . . . . . . 6.3.1.4 Converting the Original Game . . . . . . . . . . . 6.3.1.5 Nonlinear Programming Approach for Two or More Strategies for Each Player . . . . . . . . . . . . . . . . . . . . . . 6.3.1.6 Finding Prudential Strategies . . . . . . . . . . . . 6.3.1.7 Possible Conclusions and Interpretations . . . . . . . . . . . . . . . . . . . 6.3.1.8 Assumptions . . . . . . . . . . . . . . . . . . . . . . . 6.3.1.9 Case Study 4: A Model of Irregular Warfare . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1.10 Definition of Counterinsurgent Lines of Operations . . . . . . . . . . . . . . . . . . . . . . . 6.3.1.11 Definition of Insurgent Lines of Operations . . . . . . . . . . . . . . . . . . . . . . . 6.3.1.12 Setting up the GAME . . . . . . . . . . . . . . . . . 6.3.1.13 Nash Equilibrium (NE) . . . . . . . . . . . . . . . . 6.3.1.14 Prudential Strategies (P) . . . . . . . . . . . . . . . 6.3.1.15 Coin Prudential . . . . . . . . . . . . . . . . . . . . . 6.3.1.16 Insurgent Prudential . . . . . . . . . . . . . . . . . . 6.3.1.17 Nash Arbitration (NA) Point . . . . . . . . . . . . 6.3.2 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Three-Person Games . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.1 Solving the Three-Person Games in Game Theory using EXCEL or MAPLE . . . . . . . . . . . . . . . . 6.4.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 6.4.1.2 Three-Person Total Conflict Games . . . . . . . 6.4.1.3 Movement Diagram . . . . . . . . . . . . . . . . . . 6.4.1.4 Coalitions Possible . . . . . . . . . . . . . . . . . . . 6.4.1.5 Technology Assistant with EXCEL . . . . . . . 6.4.1.6 N-Person Games with Linear Programming . . . . . . . . . . . . . . . . . . . . . . .

. 272 . 274 . 274 . 278 . 281 . 284 . . . . .

285 287 288 291 291

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304 305 307 307 308 308 310 310 311

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6.4.1.7

A three-Person Game That Is a Strict Partial Conflict Game (Nonzero-Sum Game) Using Technology . . . . . . . . . . . . . . . 6.4.1.8 The Three-Person Game in MAPLE . . . . . . . . 6.4.1.9 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.2 Case Study: Korean Unification Game (Adapted from Olish and Spence Mathematics Class Project 2017) . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Sequential Game in Extensive Form . . . . . . . . . . . . . . . . . . . . . . 6.5.1 Cuban Missile Crisis Example . . . . . . . . . . . . . . . . . . . 6.5.2 North Korea Missile CRISIS . . . . . . . . . . . . . . . . . . . . . 6.5.3 Example 3: Kidnapping . . . . . . . . . . . . . . . . . . . . . . . . 6.5.3.1 Chapter Summary . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suggested Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Modeling Change with Dynamical Systems Models . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Discrete Lanchester Combat Models . . . . . . . . . . . . . . . . . . . . 7.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.2 Discrete Forms of Lanchester Equations . . . . . . . . . . . 7.2.3 Discrete Form of Lanchester’s Direct Fire Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.3.1 Eigenvalues and Eigenvectors . . . . . . . . . . . 7.2.4 Red and Blue Force Illustrative Example . . . . . . . . . . . 7.2.5 Defining a Fair Fight: Parity . . . . . . . . . . . . . . . . . . . . 7.2.6 Qualitative and Quantitative Approach . . . . . . . . . . . . 7.2.7 Illustrative “Direct Fire” Examples and Historical Perspective . . . . . . . . . . . . . . . . . . . . . . 7.2.7.1 The Battle of the Alamo . . . . . . . . . . . . . . . 7.2.7.2 The Battle of Trafalgar . . . . . . . . . . . . . . . . 7.2.8 Determining the Length of the Battle . . . . . . . . . . . . . . 7.2.8.1 Battle of Iwo Jima Example . . . . . . . . . . . . 7.2.9 Insurgency and Counter-Insurgency Operations . . . . . . 7.2.10 Comparison to Standard Lanchester Equations Via Differential Equations . . . . . . . . . . . . . . . . . . . . . 7.2.11 Conclusions for Lanchester Equations . . . . . . . . . . . . . 7.3 Insurgency Models with Discrete Dynamical Systems . . . . . . . . 7.3.1 Using EXCEL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.2 Results and Conclusion . . . . . . . . . . . . . . . . . . . . . . . 7.4 IRAQ Model for the Three Circles of War . . . . . . . . . . . . . . . . 7.4.1 Dynamic Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.2 Modeling with Shocks . . . . . . . . . . . . . . . . . . . . . . . . 7.4.3 Model with Responsive Actions and/or Policies . . . . . . 7.4.3.1 Model’s Algorithm . . . . . . . . . . . . . . . . . . .

318 319 320

320 322 322 323 325 326 326 328

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335 336 337 339 339

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349 349 350 352 354 354 356 357 358 359

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7.4.3.2 Model’s Variables and Q Matrix . . . . . . . . . 7.4.3.3 Illustrative Examples . . . . . . . . . . . . . . . . . 7.4.3.4 Conclusions and Recommendations . . . . . . . 7.5 Differential Equations (Optional) . . . . . . . . . . . . . . . . . . . . . . . 7.6 Systems of Differential Equations . . . . . . . . . . . . . . . . . . . . . . 7.6.1 Applied Systems of Differential Equations . . . . . . . . . . 7.6.2 Solving Homogeneous and Non-Homogeneous Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.3 Numerical Solutions to Systems of ODE . . . . . . . . . . . 7.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8 Chapter Projects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.1 Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.2 An Electrical Network . . . . . . . . . . . . . . . . . . . . . . . . 7.8.3 Interacting Species . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.4 Trapezoidal Method . . . . . . . . . . . . . . . . . . . . . . . . . . 7.9 Predator–Prey, SIR, and Combat Models . . . . . . . . . . . . . . . . . 7.9.1 Model Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . 7.9.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.9.2.1 The Battle of IWO JIMA . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suggested Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

MONTE CARLO Simulation and AGENT-BASED Modeling (ABM) in Military Decision-Making . . . . . . . . . . . . . . . . 8.1 Introduction to Monte Carlo Simulation . . . . . . . . . . . . . . . . . . 8.1.1 Random Number and Monte Carlo Simulation . . . . . . . 8.1.1.1 Random-Number Generators in Excel . . . . . 8.1.1.2 Examples in Excel . . . . . . . . . . . . . . . . . . . 8.1.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Probability and Monte Carlo Simulation Using Deterministic Behavior . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 Deterministic Simulation Examples . . . . . . . . . . . . . . . 8.2.2 Exercises 8.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Probability and Monte Carlo Simulation Using Probabilistic Behavior . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.1 Exercises 8.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.2 Projects 8.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Applied Military Simulations and Military Queuing Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.1 Exercises 8.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.2 Projects 8.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Case Studies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.1 New Metrics for the Detection of Suicide Bombers . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . .

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359 359 361 362 364 364

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8.5.1.2 8.5.1.3 8.5.1.4 8.5.1.5

Experimental Setup . . . . . . . . . . . . . . . . . . . . Results and Discussion . . . . . . . . . . . . . . . . . Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . Simulation Algorithm for Methodology Model for RCS, Radar, Video, and Thermal Imagery . . . . . . . . 8.5.2 Docking Two Models of Insurgency Growth (Jaye and Burks reference) . . . . . . . . . . . . . . . . 8.5.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.2.2 Epstein’s Civil Violence Model . . . . . . . . . . . 8.5.2.3 Insurrection as Epidemic . . . . . . . . . . . . . . . . 8.5.2.4 Docking and Model Validation . . . . . . . . . . . 8.5.2.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.3 Search and Rescue From COED (William P. Fox & Michael J. Jaye) . . . . . . . . . . . . . . . . 8.5.3.1 Military to the Rescue . . . . . . . . . . . . . . . . . . 8.5.4 Exercises 8.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suggested Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Logistics Network Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Transportation Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Transshipment Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Capacitated Flow Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Multi-Directional Capacitated Flow Models . . . . . . . . . . . . . . . . 9.6 Assignment Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.8 Acid Chemical Company Case Study . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suggested Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

420 425 428

429 431 431 432 436 438 441 442 442 450 451 453 455 455 458 464 470 473 476 477 479 483 483

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485

Chapter 1

Mathematical Modeling, Management Science, and Operations Research for Military Decision-Making

Objectives 1. 2. 3. 4.

Understand the process of mathematical modeling. Understand the process of decision modeling. Understand the types of models: deterministic and probabilistic. Understand models have both strengths and limitations.

Two military observation posts 5.43 miles apart pick up a brief radio signal. The sensing devices were oriented at 110
and 119
, respectively, when a signal was detected. The devices are accurate to within 2
(i.e., 2
of their respective angle of orientation). According to intelligence, the reading of the signal came from a region of active terrorist exchange, and it is inferred that there is a boat waiting for someone to pick up the terrorists. It is dusk, the weather is calm, and there are no currents. A small helicopter leaves a pad from Post 1 and is able to fly accurately along the 110
angle direction. This helicopter has only one detection device, a searchlight. At 200 ft, the searchlight can just illuminate a circular region with a radius of 25 ft. The helicopter can fly 225 miles in support of this mission due to its fuel capacity. Where do we search for the boat? How many search helicopters should you use to have a “good” chance of finding the target? (Fox and Jaye 2011).

1.1

Introduction to Decision-Making

We use the scientific approach to decision-making. We define this approach as the development of a mathematical model of a real-world problem to help inform the decision-maker. Decision-making is often referred to as quantitative analysis, management science, and operations research. In this text book, we will use a mathematical modeling approach to support the decision-making process. © Springer Nature Switzerland AG 2019 W. P. Fox, R. Burks, Applications of Operations Research and Management Science for Military Decision Making, International Series in Operations Research & Management Science 283, https://doi.org/10.1007/978-3-030-20569-0_1

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1 Mathematical Modeling, Management Science, and Operations Research. . .

2

This approach is not unique as many large Fortune 500 companies have analysts to examine and build mathematical models to aid in decision-making. The decision modeling presented in this book crosses the lines of decision-making for business, industry, and government (BIG). We will provide many government and militaryrelated examples throughout the book to help demonstrate the utility of the modeling concepts presented in this book. It is not enough to know the final mathematical model in a decision-making process. It is just as important to understand the process of mathematical modeling starting with the definition of the problem, to the development of the mathematical model, to ultimately the solution implementation. It is also important to know the strengths and limitations of these models. The correct use of good modeling tools and techniques usually results in solutions that are timely, useful, and easy to understand by those making the decisions. As far as this book is concerned, we will use mathematical modeling and operations research as the same terms.

1.2

Mathematical Modeling and Decision-Making Framework

1.2.1

Types of Decision Models

Decision models can be broadly classified into two categories based upon the assumptions made in the modeling framework. These are deterministic models and stochastic model. We discuss each in this section.

1.2.1.1

Deterministic Models

Deterministic models assume that all the relevant and important data used in the decision-making process are known with certainty to the mathematical modeler and decision-maker. With certainty implies that the data is readily available, accurate, and known or can be found. Examples abound in industry and the military where optimization, especially linear or integer programming, is used to help decision relative to product mix, blending of items, scheduling, facility location, resupply, and a like. The key is formulating the linear programming problem as we will discuss in Chap. 3. Solution techniques and analysis are also discussed in Chap. 3.

1.2.1.2

Stochastic Models

Stochastic models (also known as probabilistic models) assume that some or all input data are not known with certainty. They assume that values of some important

1.2 Mathematical Modeling and Decision-Making Framework

3

input information are not known before the decision has to be made. It is essential to incorporate this into any model developed. One way that we will examine how to do this is through the use of expected value which we define more formally later. Examples include reliability of weapon systems, reliability of sensors and other military systems, identifying a suicide bomber with radar, targeting options, etc. Since the results are based on stochastic inputs, the result merely suggests reasonable results to the decision-maker.

1.2.2

Types of Data

There are two types of data that we will be using quantitative and qualitative and two types of numbers ordinal and cardinal as each plays a role in decision-making. We define each of these below and provide examples.

1.2.2.1

Qualitative Data and Ordinal Numbers

Measurement or data are expressed not in terms of numbers, but rather by means of a natural language description. In statistics, it is often used interchangeably with “categorical” data. For example: favorite color ¼ “blue” height ¼ “tall”. Although we may have categories, the categories may have a structure to them. When there is not a natural ordering of the categories, we call these nominal categories. Examples might be gender, race, religion, or sport. When the categories may be ordered, these are called ordinal variables. Categorical variables that judge size (small, medium, large, etc.) are ordinal variables. Attitudes (strongly disagree, disagree, neutral, agree, strongly agree) are also ordinal variables; however, we may not know which value is the best or worst of these issues. Note that the distance between these categories is not something we can measure. Often we code these qualitative data in numbers: 1, 2, 3, 4, . . . for use in analysis.

1.2.2.2

Quantitative Data and Cardinal Numbers

Quantitative measurements are expressed not by means of a natural language description, but rather in terms of numbers. However, not all numbers are continuous and measurable—for example, social security number—even though it is a number it is not something that one can add or subtract. For example: favorite color ¼ “blue” height ¼ “1.8 m”. Quantitative data always are associated with a scale measure. Probably, the most common scale type is the ratio-scale. Observations of this type are on a scale that has a meaningful zero value but also have an equidistant measure (i.e., the difference between 10 and 20 is the same as the difference between 100 and 110). For example, a 10-year-old girl is twice as old as a 5-year-old girl. Since you can measure zero

1 Mathematical Modeling, Management Science, and Operations Research. . .

4

years, time is a ratio-scale variable. Money is another common ratio-scale quantitative measure. Observations that you count are usually ratio-scale (e.g., number of widgets). Numbers for which all mathematics has meaning within the numbers are cardinal numbers. A more general quantitative measure is the interval scale. Interval scales also have an equidistant measure. However, the doubling principle breaks down in this scale. A temperature of 50
C is not “half as hot” as a temperature of 100, but a difference of 10
indicates the same difference in temperature anywhere along the scale. The Kelvin temperature scale, however, constitutes a ratio-scale because on the Kelvin scale zero indicates absolute zero in temperature, the complete absence of heat. So one can say, for example, that 200 K is twice as hot as 100 K. Numbers that have meanings.

1.3

Steps in the Decision Process

We think the framework for the mathematical modeling process (see Giordano and Fox 2014) works very well in the decision-making framework with a few minor adjustments as shown in Fig. 1.1. Let’s discuss each of these nine steps in a little more depth. Step 1. Understand the problem or the question asked. To make a good decision, you need to understand the problem. Identifying the problem to study is usually difficult. In real life no one walks up to you and hands you an equation to be solved, usually, it is a comment like, “we need to make more money,” or “we need to improve our efficiency.” We need to be precise in our understanding of

Step 1. Define the problem Step 2. Make assumptions and choose variables Step 3. Acquire the data that is available Step 4. Construct a mathematical model Step 5. Solve the model Step 6. Perform model testing and sensitivity analysis Step 7. Perform a common sense test on the results Step 8. Consider both strengths and weaknesses to your modeling process. Step 9. Present the results to the decision maker. Fig. 1.1 Decision-making framework

1.3 Steps in the Decision Process

5

the problem if we will be precise in the formulation of the mathematics to describe the situation. Step 2a. Make simplifying assumptions. Start by brainstorming the situation making a list of as many factors, or variables, as you can. However, keep in mind that we usually cannot capture all these factors influencing a problem. The task is simplified by reducing the number of factors under consideration. We do this by making simplifying assumptions about the factors, such as holding certain factors as constants. We might then examine to see if relationships exist between the remaining factors (or variables). Assuming simple relationships might reduce the complexity of the problem. Once you have a shorter list of variables, classify them as independent variables, dependent variables, or neither. Step 2b. Define all variables and provide units. It is critical to clearly define all your variables and provide the mathematical notation and units for each variable. Step 3. Acquire the data. We note that acquiring the data is not an easy process. Step 4. Construct the model. Using the tools in this text and your own creativity build a mathematical model that describes the situation and whose solution helps to answer important questions. Step 5. Solve and interpret the model. We take the model we constructed in Steps 1–4 and solve it. Often this model might be too complex or unwieldy so we cannot solve it or interpret it. If this happens, we return to Steps 2–4 and simplify the model further. Step 6. Perform sensitivity analysis and model testing. Before we use the model, we should test it out. There are several questions we must ask. Does the model directly answer the question or does the model allow for the answer to the question(s) to be answered? During this step, we should review each of our assumptions to understand the impact on the mathematical model’s solution if the assumption is not correct. Step 7. Passing the common sense test. Is the model useable in a practical sense (can we obtain data to use the model)? Does the model pass the common sense test? We will say that we “collaborate the reasonableness” of our model. Step 8. Strengths and Weaknesses. No model is complete with self-reflection of the modeling process. We need to consider not only what we did right but we did that might be suspect as well as what we could do better. This reflection also helps in refining models. Step 9. Present results and sensitivity analysis to the Decision-Maker. A model is pointless if we do not use it. The more user-friendly the model is, the more it will be used. Sometimes, the ease of obtaining data for the model can dictate its success or failure. The model must also remain current. Often this entails updating parameters used in the model. In the mathematical design process, we must understand that there is a difference between the real-world and the mathematical world. Often, a mathematical model can help us understand an issue better, while allowing us to experiment mathematically with different conditions. For our purposes, we will consider a mathematical

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1 Mathematical Modeling, Management Science, and Operations Research. . .

model to be a mathematical representation designed to study a particular real-world system for which a decision needs to be made. The model allows us to use mathematical operations to reach mathematical conclusions about the system being modeled. We often study the models graphically to gain insight into the behavior under investigation. Through these activities we hope to develop a strong sense of the mathematical aspects of the problem, its physical underpinnings, and the powerful interplay between them. Often our own time table to obtain adequate results limits the continuation of model improvement, model refinement. Thus the better the initial model then the better off the modeling process becomes.

1.4

Illustrative Examples

We now use several examples to help demonstrate the types of problems we can use the modeling process that was presented in the previous section. Emphasis is placed on problem identification and choosing appropriate (useable) variables in this section. Example 1 Prescribed Drug Dosage mild for traumatic brain injuries Scenario. Consider a patient that needs to take a newly marketed prescribed drug for mild brain trauma. To prescribe a safe and effective regimen for treating the disease, one must maintain a blood concentration above some effective level and below any unsafe level. How is this determined? Understanding the Decision and Problem: Our goal is a mathematical model that relates dosage and time between dosages to the level of the drug in the bloodstream. What is the relationship between the amount of drug taken and the amount in the blood after time, t? By answering this question, we are empowered to examine other facets of the problem of taking a prescribed drug. Assumptions: We should choose or know the disease in question and the type (name) of the drug that is to be taken. We will assume in this example that the drug is called MBT, a drug taken to support better blood flow to the brain. We need to know or to find decaying rate of MBT in the blood stream. This might be found from data that has been previously collected in the study prior to the FDA’s approval. We need to find the safe and unsafe levels of MBT based upon the drug’s “effects” within the body. This will serve as bounds for our model. Initially, we might assume that the patient size and weight has no effect on the drug’s decay rate. We might assume that all patients are about the same size and weight. All are in good health and no one takes other drugs that affect the prescribed drug. We assume all internal organs are functionally properly. We might assume that we can model this using a discrete time period even though the absorption rate is a continuous function. These assumptions help simplify the model.

1.4 Illustrative Examples

7

Example 2 Emergency Military Medical Response The Emergency Service Coordinator (ESC) for the military is interested in locating the military base’s three ambulances to maximize the residents that can be reached within 8 min in emergency situations. The base is divided into six zones and the average time required to travel from one region to the next under semi-perfect conditions are summarized in Table 1.1. The population in zones 1, 2, 3, 4, 5, and 6 are given in Table 1.2: Understanding the Decision and Problem: We want better coverage and to improve the ability to take care of patients requiring to use an ambulance to go to a hospital. Determine the location for placement of the ambulances to maximize coverage within the predetermined allotted time. Assumptions: We initially assume that time travel between zones is negligible. We further assume that the times in the data are averages under ideal circumstances. Example 3 Military Credit Union Bank’s Service Problem The bank manager is trying to improve customer satisfaction by offering better service. The management wants the average customer to wait to be less than 2 min and the average length of the queue (length of the line waiting) to be 2 or fewer. The bank estimates about 150 customers per day. The existing arrival and service times are given in Tables 1.3 and 1.4. Determine if the current customer service is satisfactory according to the manager guidelines. If not, determine through modeling the minimal changes for servers required to accomplish the manager’s goal. We might begin by selecting a queuing model off the shelf to obtain some benchmark values. Understand the Decision and Problem: The bank wants to improve customer satisfaction. First, we must determine if we are or are not meeting the goal. Build a Table 1.1 Average travel times from Zone i to Zone j in perfect conditions

Table 1.2 Populations in each zone

1 1 2 3 4 5 6

1 2 3 4 5 6 Total

1 8 12 16 18 16

2 8 1 18 14 16 18

3 12 6 1.5 4 10 4

4 14 18 12 1 4 12

5 10 16 6 16 2 2

6 16 16 4 12 2 2

50,000 80,000 30,000 55,000 35,000 20,000 270,000

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1 Mathematical Modeling, Management Science, and Operations Research. . .

Table 1.3 Arrival times

Time between arrivals in minutes 0 1 2 3 4 5

Probability 0.10 0.15 0.10 0.35 0.25 0.05

Table 1.4 Service times

Service time in minutes 1 2 3 4

Probability 0.25 0.20 0.40 0.15

mathematical model to determine if the bank is meeting its goals and if not come up with some recommendations to improve customer satisfaction. Assumptions: Determine if the current customer service is satisfactory according to the manager guidelines. If not, determine through modeling the minimal changes for servers required to accomplish the manager’s goal. We might begin by selecting a queuing model off the shelf to obtain some benchmark values. Example 4 Measuring Efficiency of Units We have three major units where each unit has two inputs and three outputs as shown in Table 1.5. Understand the Decision and Problem: We want to improve efficiency of our operation. We want to be able to find “best practices” to share. First, we have to measure efficiency. We need to build a mathematic model to examine efficiency of a unit based upon their inputs and outputs and be able to compare efficiency to other units. Assumptions and Variable definitions: We define the following decision variables: ti ¼ value of a single unit of output of DMU i, for i ¼ 1, 2, 3, wi ¼ cost or weights for one unit of inputs of DMU i, for i ¼ 1, 2, efficiencyi ¼ (total value of i’s outputs)/(total cost of i’s inputs), for i ¼ 1, 2, 3. The following modeling initial assumptions are made: 1. No unit will have an efficiency more than 100 %. 2. If any efficiency is less than 1, then it is inefficient. Example 5 World War II Battle of the Bismarck Sea In February 1943, at a critical stage of the struggle for New Guinea, the Japanese decided to bring reinforcements from the nearby island of New Britain. In moving

1.4 Illustrative Examples

9

Table 1.5 Input and outputs Unit 1 2 3

Input #1 5 8 7

Input #2 14 15 12

Output #1 9 5 4

Output #2 4 7 9

Output #3 16 10 13

their troops, the Japanese could either route north where rain and poor visibility were expected or south where clear weather was expected. In either case, the trip would be 3 days. Which route should they take? If the Japanese were only interested in time, they would be indifferent to the two routes. Perhaps they wanted to minimize their convoy to attack by US bombers. For the United States, General Kenney also faced a difficult choice. Allied intelligence had detected evidence of the Japanese convoy assembling at the far side of New Britain. Kenney, of course, wanted to maximize the days the bombers could attack the convoy but he did not have enough reconnaissance planes to saturate both routes. What should he do? Understand the Decision and Problem: We want to build and use a mathematical model of conflict between players to determine the “best” strategy option for each player. Assumptions: Let’s assume that General Kenney can search only south or north. We will put these into rows. Let’s further assume that the Japanese can actually sail north or south and let’s put these in columns. Assume we get additional information from the intelligence community of the US Armed Forces, and that this information is accurate. This information states that if there is clear exposure then we bomb all 3 days. If we search south and do not find the enemy (then have to search north in the poorer weather will waste 2 days searching) and then have only 1 day to bomb. If we search north and Japanese sail north, the enemy will be exposed for 2 days. If we search north and the Japanese sail south, the enemy will be exposed for 2 days. Example 6 Risk Analysis for Homeland Security The Department of Homeland Security department only has so many assets and a finite amount of time to conduct investigations, thus priorities might be established to caseloads. The risk assessment office has collected the data for the morning meeting shown in Table 1.6. Your operations research team must analyze the information and provide a priority list to the risk assessment team for that meeting. Understand the Decision and Problem: There are more risks than we can possibly investigate. Perhaps if we rank these based upon useful criteria we can determine a priority for investigating these risks. We need to construct a useful mathematical model that ranks the incidents or risks in a priority order. Assumptions: We have past decision that will give us insights into the decisionmaker’s process. We have data only on reliability, approximate number of deaths, approximate costs to fix or rebuild, location, destructive influence, and number of intelligence gathering tips. These will be the criteria for our analysis. The data is accurate and precise and we can convert word data into ordinal numbers.

0.005 12 11

0.35

0.73 0.69

0.81 0.70

6. Major bank robbery 7. FAA Threat

0.0002 0.001

Approximate associated deaths (000) 10 0.8

Reliability of threat assessment 0.40 0.45

Threat alternatives \criterion 1. Dirty Bomb Threat 2. Anthrax-Bio Terror Threat 3. DC-Road and Bridge network threat 4. NY subway threat 5. DC Metro Threat

Table 1.6 Risk assessment priority

10 5

200 200

300

Cost to fix damages in (millions) 150 10

Urban dense Both urban dense and rural Urban dense Rural dense

Urban and rural

Location Urban dense Urban dense

Weak Moderate

Very strong Very strong

Strong

Destructive psychological influence Extremely intense Intense

16 15

5 5

8

Number of intelligence-related tips 3 12

10 1 Mathematical Modeling, Management Science, and Operations Research. . .

1.4 Illustrative Examples

11

Model: We could use multi-attribute decision-making techniques for our model. We decide on a hybrid approach of AHP-TOPSIS. We will use AHP with Saaty’s (1980) pairwise comparison to obtain the decision-maker weights. We will also use the pairwise comparison to obtain numerical values for the criteria: location and destructive influence. Then, we will use TOPSIS. Example 7 Discrete SIR Models of Epidemics or Weapons of Mass Destruction Consider a disease that is spreading throughout the United States such as the new deadly flu. The CDC is interesting in knowing and experimenting with a model for this new disease prior to it actually becoming a “real” epidemic. Let us consider the population being divided into three categories: susceptible, infected, and removed. We make the following assumptions for our model: • No one enters or leaves the community and there is no contact outside the community. • Each person is either susceptible, S (able to catch this new flu); infected, I (currently has the flu and can spread the flu); or removed, R (already had the flu and will not get it again that includes death). • Initially, every person is either S or I. • Once someone gets the flu this year they cannot get again. • The average length of the disease is 2 weeks over which the person is deemed infected and can spread the disease. • Our time period for the model will be per week. The model we will consider is an off-the-shelf model, the SIR model (see Allman and Rhodes 2004). Let’s assume the following definition for our variables. S(n) ¼ number in the population susceptible after period n. I(n) ¼ number infected after period n. R(n) ¼ number removed after period n. Let’s start our modeling process with R(n). Our assumption for the length of time someone has the flu is 2 weeks. Thus, half the infected people will be removed each week, Rðn þ 1Þ ¼ RðnÞ þ 0:5I ðnÞ The value, 0.5, is called the removal rate per week. It represents the proportion of the infected persons who are removed from infection each week. If real data is available, then we could do “data analysis” in order to obtain the removal rate. I(n) will have terms that both increase and decrease its amount over time. It is decreased by the number that are removed each week, 0.5I(n). It is increased by the numbers of susceptible that come into contact with an infected person and catch the disease, aS(n)I(n). We define the rate, a, as the rate in which the disease is spread or the transmission coefficient. We realize this is a probabilistic coefficient. We will

1 Mathematical Modeling, Management Science, and Operations Research. . .

12

assume, initially, that this rate is a constant value that can be found from initial conditions. Let’s illustrate as follows. Assume we have a population of 1000 students in the dorms. Our nurse found 3 students reporting to the infirmary initially in the first week. The next week, 5 students came in to the infirmary with flu-like symptoms. I (0) ¼ 3, S(0) ¼ 997. In week 1, the number of newly infected is 30. 5 ¼ aI ðnÞSðnÞ ¼ að3Þ  ð995Þ a ¼ 0:00167 Let’s consider S(n). This number is decreased only by the number that becomes infected. We may use the same rate, a, as before to obtain the model: Sðn þ 1Þ ¼ SðnÞ  aSðnÞI ðnÞ Our coupled SIR model is Rðn þ 1Þ ¼ RðnÞ þ 0:5I ðnÞ I ðn þ 1Þ ¼ I ðnÞ  0:5I ðnÞ þ 0:00167I ðnÞSðnÞ Sðn þ 1Þ ¼ SðnÞ  0:00167SðnÞI ðnÞ I ð0Þ ¼ 3,Sð0Þ ¼ 997,Rð0Þ ¼ 0 The SIR Model can be solved iteratively and viewed graphically. We will revisit this model again in Chap. 7. In Chap. 7, we determine the worse of the flu epidemic occurs around week 8, at the maximum of the infected graph. The maximum number is slightly larger than 400, from the table in Chap. 7 it is approximated as 427. After 25 weeks, slightly more than 9 people never get the flu. These examples will be among those solved in subsequent chapters.

1.5

Technology

Most real-world problem solving that we have been involved in modeling require technology to assist the analyst, the modeler, and the decision-maker. Microsoft Excel is available on most computers and represents a fairly good technological support for analysis of the average problems especially with Analysis ToolPak and the Solver installed. Other specialized software to assist analysts include: MatLab, Maple, Mathematica, LINDO, LINGO, GAMS, as well as some additional add-ins for Excel such as the simulation package, Crystal Ball. Analysts should avail themselves to have access to as many of these packages as necessary. In this book, we illustrate Excel and R although the other software may be easily substituted.

1.7 Exercises

1.6

13

Conclusions

We have provided a clear and simple process to begin mathematical modeling in applied situation requiring the stewardship of applied mathematics, operations research analysis, or risk assessment. We did not cover all the possible models but did highlight a few through illustrative examples. We emphasize that sensitivity analysis is extremely important in all models and should be accomplished prior to any decision being made. We show this in more detail in the chapters covering the techniques.

1.7

Exercises

Using Steps 1–3 of the modeling process above identify a problem from scenario 1–11 that you could study. There are no “right” or “wrong” answers just measures of difficulty. 1. The population of military and dependents in your community. 2. The economic impact of military and dependents in your community. 3. A new base exchange is being constructed. How should you design the illumination of the parking lot? 4. A new commander wants a successful command season with his unit. What factors make the command successful? What if the command is recruiting command, then what are the factors? 5. The military needs to purchase or lease a new fleet of sedans. What factors must be considered? 6. A new section in the Pentagon wants to go mobile with internet access and computers upgrades but cost might be a problem. 7. Starbucks has many varieties of coffee available at the Base Exchange. How can Starbucks make more money? 8. Navy Seal graduate student does not like math or math-related courses. How can a student maximize their chances for a good grade in a math class to improve their overall GPA? 9. Recruits don’t think they need basic training and that military occupational specialty training should be all that they need for success. 10. Troops are clamoring to fire the commanding general. 11. Some military bases would like to stock a fish pond with bass and trout. 12. Safety airbags in millions of cars are to be replaced at the factory. Can this be done in a timely manner?

14

1.8

1 Mathematical Modeling, Management Science, and Operations Research. . .

Projects

1. Are Robert E. Lee, Dwight D. Eisenhower, Norman Schwarzkopf, and William McCarther the greatest generals of the century? What variables and factors need to be considered? 2. What kind of vehicle should the military buy for everyday use? 3. What kind of vehicle should the military buy as a utility vehicle? 4. Recently, the United States and its allies fired missiles into Syria to destroy chemical weapons. The news media stated that the 106 missiles were fired to minimize the chance of escalation. How would you build a model for the Department of Defense for targeting to prevent escalation? 5. You are the commander of a large recruiting unit for the military. Recruiting has been off lately by not meeting quotas. What factors should be considered to improve the recruiting effort? 6. How would you go about building a model for the “best US general of all time”? 7. Should the logistics recommend the use of 3D printers for small and often used parts in a combat zone? What factors should be considered? 8. You are the military advisor to moving oil from off shore wells to a refinery plant located inland. How would you go about building such a model? 9. Recall the Military to the Rescue problem at the beginning of the chapter. How would you model the rescue? 10. Consider an upcoming insurgency between a country and insurgent forces. What factors should be considered in modeling this insurgency? 11. Insurgent forces have a strong foothold in the city of Urbania. Intelligence estimates they currently have a force of about 1000 fighters. Intelligence also estimates that around 120 new insurgents arrive from the neighboring country of Moronka each week. In conflicts with insurgent forces, the local police are able to capture or kill approximately 10% of the insurgent force each week on average. (a) Describe the behavior of the current system under the conditions stated: • Is there a stable equilibrium to the system under the current conditions? If so, is this an acceptable level? • How effective would an operation designed to slow (or stop) the influx of new insurgents be if the dynamics do not change? (b) What attrition rate does the police force need to achieve to drive the insurgent population to an equilibrium level below 500 in 52 weeks or less? (c) If the police force can, with advanced weapons, achieve a 30–40 % attrition rate, do they also have to engage in operations to stop the inflow of new insurgents? (d) What effects do changes in the external factor, change factor, and initial condition have on the system behavior curve? (e) What conditions are necessary to cause either case (1) or (2) to occur within the 52-week horizon?

References

15

References Allman, E., & Rhodes, J. (2004). Mathematical modeling with biology: An introduction. Cambridge: Cambridge University Press. Giordano, F. R., Fox, W., & Horton, S. (2014). A first course in mathematical modeling (5th ed.). Boston: Brooks-Cole Publishers. Fox, W. P. & Jaye, M. J. (2011). Search and Rescue Carry Through Mathematical Modeling Problem. Computers in Education Journal (COED), 2(3), 82–93. Saaty, T. (1980). The analytical hierarchy process. New York: McGraw Hill.

Suggested Reading Albright, B. (2010). Mathematical modeling with excel. Burlington: Jones and Bartlett Publishers. Bender, E. (2000). Mathematical modeling. Mineola: Dover Press. Fox, W. P. (2012). Mathematical modeling with maple. Boston: Cengage Publishing. Fox, W. (2013a). Mathematical modeling and analysis: An example using a catapult. Computer in Education Journal, 4(3), 69–77. Fox, W. P. (2013b). Modeling engineering management decisions with game theory. In F. P. García Márquez & B. Lev (Eds.), Engineering management. London: InTech. ISBN 978-95351-1037-8. Fox, W. (2014a). Chapter 17, Game theory in business and industry. In Encyclopedia of business analytics and optimization (Vol. V(1), pp. 162–173). Hershey: IGI Global and Sage Publications. Fox, W. (2014b). Chapter 21, TOPSIS in business analytics. In Encyclopedia of business analytics and optimization (Vol. V(5), pp. 281–291). Hershey: IGI Global and Sage Publications. Fox, W. P. (2016). Applications and modeling using multi-attribute decision making to rank terrorist threats. Journal of Socialomics, 5(2), 1–12. Fox, W. P., & Durante, J. (2015). Modeling violence in the Philippines. Journal of Mathematical Science, 2(4) Serial 5, 127–140. Meerschaert, M. M. (1999). Mathematical modeling (2nd ed.). New York: Academic Press. Myer, W. (2004). Concepts of mathematical modeling. New York: Dover Press. Washburn, A., & Kress, M. (2009). Combat modeling. New York: Springer. Winston, W. (1995). Introduction to mathematical programming (pp. 323–325). Belmont: Duxbury Press.

Chapter 2

Statistics and Probability in Military Decision-Making

Objectives 1. Understand concepts in basic statistics, displays, and measures of location and dispersion. 2. Understand the concepts of probability and solving probability problems. 3. Knowledge of basic probability distributions used in analysis. 4. Knowledge of the central limit theorem. 5. Understand hypothesis testing.

2.1

Introduction to Statistics and Statistical Models

In a 2010 statement, General McCaffrey stated that the casualties in Afghanistan would double in the next year and the United States should expect up to 500 casualties a month (Coughlan 2018). Figure 2.1 displays the Afghanistan data up until that time. Was their basis for his claim? Later, we will analyze this data and either support or refute his claim. In this chapter, we provide a review of topics from basic probability and statistics. For those individuals with a good statistical foundation, you can probably move directly to the case studies at the end of the chapter if desired. Statistics is the science of reasoning from data, so a natural place to begin is by examining what is meant by the term “data.” Data is information. The most fundamental principle in statistics is that of variability. If the world were perfectly predictable and showed no variability, there would be no need to study statistics. You will need to discover the notion of a variable and then first learn how to classify variables. Any characteristic of a person or thing that can be expressed as a number is called a variable. A value of that variable is the actual number that describes that person or © Springer Nature Switzerland AG 2019 W. P. Fox, R. Burks, Applications of Operations Research and Management Science for Military Decision Making, International Series in Operations Research & Management Science 283, https://doi.org/10.1007/978-3-030-20569-0_2

17

18

2 Statistics and Probability in Military Decision-Making

600

500

400

300

200

100

0 1

3

5

7

9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47

Fig. 2.1 US casualties in Afghanistan 2001–2009 Table 2.1 Heights of members of a squad 50 1000

60 200

50 500

50 200

60

50 900

50 400

50 1000

Table 2.2 Weights of members in a squad 135

155

215

192

173

170

165

142

thing. Think of the variables that might be used to describe you: height, weight, income, rank, branch of service, and gender. Data can be either quantitative or categorical (qualitative). We will explain each. Quantitative means that the data are numerical where the number has relative meaning. Examples could be a list of heights of soldiers in your platoon, number of targets hit for marksmanship by your unit, weights of soldiers in your unit, or IED fatalities (Tables 2.1, 2.2, and 2.3). These data elements provide numerical information and from it we can determine who is the tallest or shortest or which squad member weighs the most. We can also compare and contrast “mathematically” these values. Quantitative data can be either discrete (counting data) or continuous. These distinctions in data become important as we analyze the data and use it in models later in the book. Quantitative data allows us to “do meaningful mathematics,” such as addition, subtraction, multiplication, and division. Categorical (qualitative) data can describe objects, such as recording the people with a particular hair color as: blonde ¼ 1 or brunette ¼ 0. If we had four colors of

2.1 Introduction to Statistics and Statistical Models

19

Table 2.3 IED deaths from 2001 to 2014 (www.icasualties.org) Period 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014

IED 0 4 3 12 20 41 78 152 275 368 252 132 52 3

Total 4 25 26 27 73 130 184 263 451 630 492 312 117 13

Pct 0.00 16.00 11.54 44.44 27.40 31.54 42.39 57.79 60.98 58.41 51.22 42.31 44.44 23.08

Table 2.4 Display methods for univariate data Data display

Concern

Categorical Pie chart Bar chart Comparisons

Quantitative: continuous or discrete Stem and leaf Dot plot Histogram Shape and skewness

Comment

Often overlay the distribution of interest over the histogram

hair: blonde, brunette, black, and red; we could use as codes: brunette ¼ 0, blonde ¼ 1, black ¼ 2, and red ¼ 3. We certainly cannot have an average hair color from these numbers that would make sense. For example, if we had one individual with each hair color the average hair color would be 1.5. This value clearly has no significance in terms of hair color; it would not make sense. Another example is categories by gender: male ¼ 0 and female ¼ 1. In general, it may not make sense to do any arithmetic using categorical variables. Ranks: Lieutenant, Captain, Major, Lieutenant Colonel, etc. and services: Army, Navy, Air Force, Marine, Coast Guard, or International are additional examples of categorical data. Once you have learned to distinguish between quantitative and categorical data, we need to move on to a fundamental principle of data analysis: begin by looking at a visual display of the dataset. We will present five methods of displaying univariate data: pie chart, bar chart, stem and leaf (by hand only), histogram, and boxplot (by hand only). The displays should supply visual information to the decision-maker without them struggling to interpret the display (Table 2.4).

20

2.2 2.2.1

2 Statistics and Probability in Military Decision-Making

Displaying Categorical Data Pie Chart

The pie chart is useful to show the division of a total quantity into component parts. A pie chart, if done correctly, is usually safe from misinterpretation. The total quantity, or 100%, is shown as the entire circle. Each wedge of the circle represents a component part of the total. These parts are usually labeled with percentages of the total. A pie chart helps us see what part of the whole each group forms. Let’s review percentages. Let a represent the partial amount and b represent the total amount. Then P represents a percentage calculated by P ¼ a/b (100). A percentage is thus a part of a whole. For example, $0.25 is what part of $1.00? We let a ¼ 25 and b ¼ 100. Then, P ¼ 25/100 (100) ¼ 25%. Now, let’s see how Excel would create a pie chart for us in the following scenario. Consider soldiers choosing their Military Occupation Specialty (MOS). Out of the 632 new soldiers recruited in South Carolina that actually choose a MOS, the breakdown of selection is as follows. 1. Infantry 2. Armor 3. Artillery 4. Air Defense 5. Aviation 6. Signal 7. Maintenance Total

250 53 35 41 125 45 83 632

Figure 2.2 is a pie chart of the MOS breakdown. Each of the shaded regions displays the percentage (%) of soldiers out of 632 that chose that MOS. Clearly infantry has the largest percent of recruits, which MOS appears to have the least? What advantages and disadvantages can you see with using pie charts? Let’s view the data as a bar chart, Fig. 2.3: MOS Breakdown 1 6, 45, 7%

7, 83, 13%

2 1, 250, 40%

3 4 5

5, 125, 20% 4, 41, 6%

3, 35, 6%

2, 53, 8%

6 7

Fig. 2.2 Pie chart from Excel for MOS breakdown

2.2 Displaying Categorical Data

21

300 250 200 150 100 50

ce

l

an

na M

ai

nt en

Sig

on

e

Av ia

Ai r

De fe ns

ry le l Ar

Ar m or

In fa nt ry

0

Fig. 2.3 Bar chart of MOS breakdown

2.2.2

Displaying Quantitative Data

In quantitative data, we are concerned with the shape of the data. Shape refers to symmetry of data. “Is it symmetric?” “Is it skewed?” are questions we ask and answer.

2.2.2.1

Stem and Leaf

A stem-and-leaf plot uses the real data points in making a plot. The plot will appear strange because your plot is sideways. The rules are as follows: Step 1: Order the data Step 2: Separate according to the one or more leading digits. List stems in a vertical column. Step 3: Leading digit is the stem and trailing digit is the leaf. For example 32, 3 is the stem and 2 is the leaf. Separate the stem from the leafs by a vertical line. Step 4: Indicate the units for stems and leafs in the display. You will probably create these plots using technology. Example: Grades for 20 students in a course 53, 55, 66, 69, 71, 78, 75, 79, 77, 75, 76, 73, 82, 83, 85, 74, 90, 92, 95, 99

22

2 Statistics and Probability in Military Decision-Making

Stems are the leading digit: 5 6 7 8 9 Standing for 50s, 60s, 70s, 80s, and 90s. If there had been a score of 100, then the leading digit is in 100s. So we would need: 05 06 07 08 09 10 for 50s, 60s, 70s, 80s, 90s, and 100s Draw a vertical line after each stem. 5| 6| 7| 8| 9| Now add the leafs, which are the trailing digits, 53, 55, 66, 69, 71, 73, 74, 75, 75, 76, 77, 78, 79, 82, 83, 85, 90, 92, 95, 99 5| 3, 5 6| 6, 9 7| 1, 3, 4, 5, 5, 6, 7, 8, 9 8| 2, 3, 5 9| 0, 2, 5, 9 We can characterize this shape as almost symmetric. Note how we read the values from the stem and leaf. For example, we read the stem and leaf: 5| 3, 5 as data elements 53 and 55.

2.2 Displaying Categorical Data

23

Fig. 2.4 Shapes of distributions, symmetric, skewed right, and skewed left

2.2.3

Symmetry Issues

We look at these shapes as symmetric or skewed. Symmetric looks like a bell-shaped curve while skewed means that the plot appears lopsided. Three generic risk distribution shapes (symmetric, skewed right, and skewed left) shown in Fig. 2.4 Note: The shape of the distribution has important implications from a risk management standpoint. In Fig. 2.4a, the risk distribution is symmetric, and as a result, there are an equal number of people experiencing a high risk as there are a low risk. In Fig. 2.4b, the risk distribution is skewed to the right, with most people experiencing a low risk and a few experiencing a high risk, compared to Fig. 2.4c, where the distribution is skewed to the left translating to many people experiencing a higher risk and only a few people experiencing a lower risk. From a risk management or policy perspective, each of these situations would need to be assessed differently in light of the following considerations: the population (children, elderly, etc.) experiencing the higher risk; what the actual magnitude of higher risk is (high risk as defined in this context may not be very high when compared to other competing risks), if the higher risk is being borne as a result of voluntary or involuntary actions, and whether the people bearing the higher risk are in control of the risk situation, etc. A key aspect of the risk characterization stage is providing insight not only into the risk estimates, but also our confidence in the generated assessment. Such insights include: • The steps that could be taken to reduce the risk • Points in the process about which we have uncertainty and could benefit from more information • Points that have a significant impact on the risk, and as such would be ideal areas to focus more attention on so as to ensure they are under control In general, quantitative risk assessment models can be considered as contributing toward risk management decision-making by providing input along four avenues: • focusing attention on risk reducing areas • focusing attention on research areas

24

2 Statistics and Probability in Military Decision-Making

Histogram 45 40 35 30 25 Freq

20 15 10 5 0 1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20

Fig. 2.5 Example of a histogram.

• helping in the formulation of risk reduction strategies • providing a tool to test out formulated risk reduction strategies prior to implementation A histogram is provided in Fig. 2.5. Our examination shows the data displayed by the histogram appears to be skewed right.

2.2.4

Boxplot Used for Comparisons

We will present the information on how to construct and use a boxplot. Boxplots are a good way to compare datasets from multiple sources. For example, let’s look at violence in ten regions in Afghanistan. Putting the ten boxplots together allows us to compare many aspects such as medians, ranges, and dispersions. Boxplot Step 1. Draw a horizontal measurement scale that includes all data within the range of data. Step 2. Construct a rectangle (the box) whose left edge is the lower quartile value and whose right edge is the upper quartile value. Step 3. Draw a vertical line segment in the box for the median value. Step 4. Extend line segments from rectangle to the smallest and largest data values (these are called whiskers). 53, 55, 66, 69, 71, 73, 74, 75, 75, 76, 77, 78, 79, 82, 83, 85, 90, 92, 95, 99

2.2 Displaying Categorical Data

25

The values are in numerical order. What is needed are the range, the quartiles, and the median. Range is the smallest and largest values from the data: 53 and 99. The median is the middle value. It is the average of the 10th and 11th values as we will see later: (76 + 77)/2 ¼ 76.5 The quartile values are the median of the lower and upper half of the data. Lower quartile values: 53, 55, 66, 69, 71, 73, 74, 75, 75, 76. Its median is 72. Upper quartile values: 77, 78,79, 82, 83, 85, 90, 92, 95, 99. Its median is 84. You draw a rectangle from 72 to 84 with a vertical line at 76.5 Then draw a whisker to the left to 53 and to the right to 99. It would look something like the boxplot image in Fig. 2.6.

2.2.4.1

Comparisons

Consider our data for casualties in Afghanistan through the years 2002–2009. This is presented to you as a commander. What information can you interpret from this boxplot? We clearly see from Fig. 2.7 that the casualties increase over time indicating that the situation of the conflict is intensifying over time.

2.2.5

Measures of Central Tendency or Location

2.2.5.1

Describing the Data

In addition to plots and tables, numerical descriptors are often used to summarize data. Three numerical descriptors, the mean, the median, and the mode offer different ways to describe and compare datasets. These are generally known as the measures of location. Fig. 2.6 Boxplot

26

2 Statistics and Probability in Military Decision-Making Boxplots 2002-2009

600

# causalties

500 400

q1 min med max q3

300 200 100 0 1

2

3

4

5

6

7

8

years 2002-2009

Fig. 2.7 Comparative boxplots of casualties over time

2.2.5.2

The Mean

The mean is the arithmetic average, with which you are probably very familiar. For example, your academic average in a course is generally the arithmetic average of your graded work. The mean of a dataset is found by summing all the data and dividing this sum by the number of data elements. The following data represent ten scores earned by a student in a college algebra course: 55, 75, 92, 83, 99, 62, 77, 89, 91, 72. Compute the student’s average. The mean can be found by summing the ten scores 55 + 75 + 92 + 83 + 99 + 62 + 77 + 89 + 91 + 72 ¼ 795 and then dividing by the number of data elements (10), 795/10 ¼ 79.5 To describe this process in general, we can represent each data element by a letter with a numerical subscript. Thus, for a class of n tests, the scores can be represented by a1, a2, . . . , an. The mean of these n values of a1, a2, . . . , an is found by adding these values and then dividing this sum by n, the number of values. The Greek letter Σ (called sigma) is used to represent the sum of all the terms in a certain group. Thus, n P we may see this written as ai ¼ a1 þ a2 þ . . . þ an i¼1

2.2 Displaying Categorical Data

27 n P

ai

mean ¼ i¼1 n Think of the mean as the average. Notice that the mean does not have to equal any specific value of the original dataset. The mean value of 79.5 was not a score ever earned by our student. Batting average is defined as the total number of hits divided by the total number of official at bats. Is batting average a mean? Explain.

2.2.5.3

The Median

The median locates the true middle of a numerically ordered list. The hint here is that you need to make sure that your data is in numerical order listed from smallest to largest along the x-number line. There are two ways to find the median (or middle value of an ordered list) depending on n (the number of data elements): 1. If there is an odd number of data elements, then the middle (median) is the exact data element that is the middle value. For example, here are five ordered math grades earned by a student: 55, 63, 76, 84, 88. The middle value is 76 since there are exactly two scores on each side (lower and higher) of 76. Notice that with an odd number of values that the median is a real data element. 2. If there is an even number of data elements, then there is no true middle value within the data itself. In this case, we need to find the mean of the two middle numbers in the ordered list. This value, probably not a value of the dataset, is reported as the median. Let’s illustrate with several examples. (a) Here are six math scores for student one: 56, 62, 75, 77, 82, 85 The middle two scores are 75 and 77 because there are exactly two scores below 75 and exactly two scores above 77. We average 75 and 77. (75+77)/ 2 ¼ 152/2 ¼ 76 76 is the median. Note that 76 is not one of the original data values. (b) Here are eight scores for student two: 72, 80, 81, 84, 84, 87, 88, 89 The middle two scores are 84 and 84 because there are exactly three scores lower than 84 and three scores higher than 84. The average of these two scores is 84. Note that this median is one of our data elements. It is also very possible for the mean to be equal to the median.

2.2.5.4

The Mode

The value that occurs the most often is called the mode. It is one of the numbers in our original data. The mode does not have to be unique. It is possible for there to be

28

2 Statistics and Probability in Military Decision-Making

more than one mode in a dataset. As a matter of fact, if every data element is different from the other data elements then every element is a mode. For example, consider the following data scores for a mathematics class. 75, 80, 80, 80, 80, 85, 85, 90, 90, 100 The number of occurrences for each value is: Value 75 80 85 90 100

Number of occurrences 1 4 2 2 1

Since 80 occurred 4 times and that is the largest value among the number of occurrences, then 80 is the mode.

2.2.6

Measures of Dispersion

2.2.6.1

Variance and Standard Deviation

Measures of variation or measures of the spread of the data include the variance and standard deviation. They measure the spread in the data, how far the data are from the mean. The sample variance has notation S2 and the sample deviation has notation S. n  P

S2 ¼ i¼1

2 xi  x

n1

where n is the number of data elements.

S¼

vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi uP 2 un  u x  x ti¼1 i n1

where n is the number of data elements. Example 1 Consider the following ten data elements: 50, 54, 59, 63, 65, 68, 69, 72, 90, 90. The mean, x, is 68. The variance is found by subtracting the mean, 68, from each point, squaring them, add them up, and divide by n  1.

2.2 Displaying Categorical Data

S2

29

h ¼ ð50  68Þ2 þ ð54  68Þ2 þ ð59  68Þ2 þ ð63  68Þ2 þ ð65  68Þ2 þ ð68  68Þ2 þ ð69  68Þ2 þ ð72  68Þ2 þ ð90  68Þ2 þ ð90  68Þ2  =9 ¼ 180

S¼

pffiffiffiffiffi S2 ¼ 13:42:

Example 2 Consider a person’s metabolic rate at which the body consumes energy. Here are seven metabolic rates for men who took part in a study of dieting. The units are calories in a 24-h period. 1792

1666

1362

1614

1460

1867

1439

The researchers reported both x and S for these men. The mean: x ¼

1792 þ 1666 þ 1362 þ 1614 þ 1460 þ 1867 þ 1439 11,200 ¼ ¼ 1600 7 7

To see clearly the nature of the variance, start with a table of the deviations of the observations from the mean (Table 2.5). The variance, S2 ¼ 214,870/6p¼ 35,811.67 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi The standard deviation, S ¼ 35,811:67 ¼ 189:24 Some properties of the standard deviation are: • S measures spread about the mean. • S ¼ 0 only when there is no spread. • S is strongly influenced by extreme outliers.

Table 2.5 Table of deviations

Observations

Deviations

Xi 1792 1666 1362 1614 1460 1867 1439

xi  x 1792 – 1600 ¼ 192 1666 – 1600 ¼ 66 1362 – 1600 ¼ 238 1614  1600 ¼ 14 1460 – 1600 ¼ 140 1867  1600 ¼ 267 1439 – 1600 ¼ 161 Sum ¼ 0

Squared deviations  2 xi  x 36,864 4356 56,644 196 19,600 71,289 25,921 Sum ¼ 214,870

30

2 Statistics and Probability in Military Decision-Making

2.2.6.2

Measures of Symmetry and Skewness

We define a measure, the coefficient of skewness, Sk. Mathematically, we determine this value from formula:   3  X  X~ Sk ¼ S We use the following rules for skewness and symmetry. If Sk  0, the data is symmetric. If Sk > 0, the data is positively skewed (skewed right). If Sk < 0, the data is negatively skewed (skewed left). We use the bell-shaped curve to denote symmetry. Figure 2.8 provides an example of the classic symmetric bell-shaped curved (normal) distribution. Figure 2.9 provides examples of skewed distributions. Range is a measure that takes the maximum and minimum values of the data. Often, this is provided a single number. Assume we have the data in Table 2.6: The maximum value is 1867 and the minimum value is 1362. If you take the difference, 1867 – 1362 ¼ 505. What does 505 represent? I suggest you give the range as an interval [1362, 1867].

Fig. 2.8 Bell-shaped distribution

Normal_Distribution-Bell-Shaped Curve 0.025

0.02

0.015

0.01

0.005

0.0 40

60

80 x

100

120

2.2 Displaying Categorical Data

31

0.15

0.15

0.10

0.10

0.05

0.05

0.00

0.00 0

5

10

15

0

5

10

15

Fig. 2.9 Skewed distributions. (a) An example of positive skewness (skew to the right). (b) An example of negative skewness (skew to the left)

Table 2.6 Range data

2.2.7

1792 1666 1362 1614 1460 1867 1439

Section Exercises

1. The 1994 live birth rates per thousand population in the mountain states of Idaho, Montana, Wyoming, Colorado, New Mexico, Arizona, Utah, and Nevada were 12.9, 15.5, 13.5, 14.8, 16.7, 17.4, 20.1, and 16.4, respectively. What is the mean, variance, and standard deviation? 2. In five attempts, it took a soldier 11, 15, 12, 8, and 14 min to change a tire on a humvee. What is the mean, variance, and standard deviation? 3. A soldier is sent to the range to test a new bullet that the manufacturer says is very accurate. You send your best shooter with his weapon. He fires ten shots with each using the standard ammunition and then the new ammunition. We measure the distance from the bull’s eye to each shot’s location. Which appears to the better ammunition? Explain. Standard Ammunition: 3, 3, 1, 0, 0, 0, 1, 1, 1, 2 New Ammunition: 2, 1, 0, 0, 0, 0, 1, 1, 1, 2 4. AGCT Scores: AGCT-score AGCT stands for Army General Classification Test. These scores have a mean of 100, with a standard deviation of 20.0. Here are the AGCT scores for a unit: 79, 100, 99, 83, 92, 110, 149, 109, 95, 126, 101, 101, 91, 71, 93, 103, 134, 141, 76, 108, 122, 111, 97, 94, 90, 112, 106, 113, 114, 117 Find the mean, median, mode, standard deviation, variance, and coefficient of skewness for the data. Provide a brief summary to your S  1 about this data.

32

2.3

2 Statistics and Probability in Military Decision-Making

Classical Probability

2.3.1

Introduction

Consider a terrorist exploding an IED on a ship in the Mediterranean Sea with casualty results as presented in Table 2.7. One rule of disasters at sea is to rescue women and children first. Was this rule followed? Some basic calculation reveals the while only 19.6% survived, 70.4% of women and children survived. Such simple calculations can provide a lot of information. We discuss how we came up with these calculations in this section Probability is a measure of the likelihood of a random phenomenon or chance behavior. Probability describes the long-term proportion with which a certain outcome will occur in situations with short-term uncertainty. Probability deals with experiments that yield random short-term results or outcomes yet reveal long-term predictability. The long-term proportion with which a certain outcome is observed is the probability of that outcome.

2.3.1.1

The Law of Large Numbers

As the number of repetitions of a probability experiment increases, the proportion with which a certain outcome is observed gets closer to the probability of the outcome. In probability, an experiment is any process that can be repeated in which the results are uncertain. A simple event is any single outcome from a probability experiment. Each simple event is denoted ei. The sample space, S, of a probability experiment is the collection of all possible simple events. In other words, the sample space is a list of all possible outcomes of a probability experiment. An event is any collection of outcomes from a probability experiment. An event may consist of one or more simple events. Events are denoted using capital letters such as E. Example 1 Consider the probability experiment of flipping a fair coin twice (a) Identify the simple events of the probability experiment. (b) Determine the sample space. (c) Define the event E ¼ “have only one head.” Table 2.7 Terrorist IED causalities

Survived Died Total

Men 332 1360 1692

Women 318 104 422

Boys 29 35 64

Girls 27 18 45

Total 706 1517 2223

2.3 Classical Probability

33

Solution: (a) Events for two flips H¼head T¼tail (b) Sample space {HH, HT, TH, TT} (c) Having one head {HT, TH} The probability of an event, denoted P(E), is the likelihood of that event occurring.

2.3.1.2

Properties of Probabilities

1. The probability of any event E, P(E), must be between 0 and 1 inclusive. That is, 0  PðEÞ  1: 2. If an event is impossible, the probability of the event is 0. 3. If an event is a certainty, the probability of the event is 1. 4. If S ¼ {e1, e2, . . ., en}, then Pðe1 Þ þ Pðe2 Þ þ . . . þ Pðen Þ ¼ 1 where S is the sample space and ei are the events. P(only One head in two flips) ¼ Number of outcomes with only one head/total number of outcomes ¼ 2/4 ¼ 1/2 The classical method of computing probabilities requires equally likely outcomes. An experiment is said to have equally likely outcomes when each simple event has the same probability of occurring. An example of this is a flip of a fair coin where the chance of flipping a head is ½ and the chance of flipping a tail is ½. If an experiment has n equally likely simple events and if the number of ways that an event E can occur is m, then the probability of E, P(E), is PðE Þ ¼

Number of ways that E can occur m ¼ Number of Possible Outcomes n

So, if S is the sample space of this experiment, then PðE Þ ¼

N ðE Þ N ð SÞ

34

2 Statistics and Probability in Military Decision-Making

Example 2 Suppose a “fun size” bag of M&M’s contains nine brown candies, six yellow candies, seven red candies, four orange candies, two blue candies, and two green candies. Suppose that a candy is randomly selected. (a) What is the probability that it is brown? (b) What is the probability that it is blue? (c) Comment on the likelihood of the candy being brown versus blue. Solution: (a) P(brown) ¼ 9/30 ¼ 0.3. (b) P(blue) ¼ 2/30 ¼ 0.066666. (c) Since there are more brown candies than blue candies, it is more likely to draw a brown candy than a blue candy. These easily could be different ranks of soldiers preparing for a mission rather than colors of M&M’s.

2.3.1.3

Probability from Data

The probability of an event E is approximately the number of times event E is observed divided by the number of repetitions of the experiment. PðEÞ  relative frequency of E frequency of E ¼ number of trails of experiment Now, let’s return to our terrorist attack on the cruise ship (Table 2.7). We can use this method to compute the probabilities. PðSurvived the attackÞ ¼ 706=2223 ¼ 0:3176 PðDiedÞ ¼ 1517=2223 ¼ 0:6824

PðWomen and children survivedÞ ¼ ð318 þ 29 þ 27Þ=ð422 þ 64 þ 45Þ ¼ 374=531 ¼ 0:7043 PðMen survivedÞ ¼ 332=1692 ¼ 0:1962

2.3.1.4

Intersections and Unions

Now, let E and F be two events.

2.3 Classical Probability

35

E and F is the event consisting of simple events that belong to both E and F. The notation is \ (intersection), E \ F E or F is the event consisting of simple events that belong to either E or F or both. The notation is [(union), E [ F. Suppose that a pair of dice are thrown. Let E ¼ “the first die is a two” and let F ¼ “the sum of the dice is less than or equal to 5.” Find P(E \ F) and P(E [ F) directly by counting the number of ways E or F could occur and dividing this result by the number of possible outcomes shown in Fig. 2.10. Event E ¼ f2-1; 2-2; 2-3; 2-4; 2-5; 2-6g Event F ¼ f1-1; 1-2; 1-3; 1-4; 2-1; 2-2; 2-3; 3-1; 3-2; 4-1g There are 36 outcomes above. PðEÞ ¼ 6=36 ¼ 1=6 PðFÞ ¼ 10=36 ¼ 5=18 ðE \ F Þ ¼ f2-1; 2-2; 2-3g ðE [ FÞ ¼ f1-1; 1-2; 1-3; 1-4; 2-1; 2-2; 2-3; 3-1; 3-2; 4-1; 2-4; 2-5; 2-6g PðE \ F Þ ¼ 3=36 ¼ 1=12 PðE [ FÞ ¼ 13=36

Fig. 2.10 Outcomes for the roll of a pair of fair dice

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2 Statistics and Probability in Military Decision-Making

2.3.1.5

The Addition Rule

For any two events E and F, PðE or F Þ ¼ PðE Þ þ PðF Þ  PðE and F Þ PðE [ F Þ ¼ PðE Þ þ PðF Þ  PðE \ F Þ Let’s consider the following example. Let event A be the event a soldier on post takes the local newspaper and let event B be the event that a soldier on post takes the United States today. There are 1000 soldiers living on post and we know 750 take the local paper, and 500 take the United States today. We are told 450 take both papers. PðA \ BÞ ¼ 450=1000 ¼ 0:45 PðAÞ ¼ 0:75 PðBÞ ¼ 0:50 We can find the union, PðA [ BÞ ¼ PðAÞ þ PðBÞ  PðA \ BÞ PðA [ BÞ ¼ 0:75 þ 0:50  0:45 ¼ 0:8 Thus, 80% of the soldiers take at least one of the two newspapers. Venn diagrams represent events as circles enclosed in a rectangle as shown in Fig. 2.11. The rectangle represents the sample space and each circle represents an event. Consider the newspaper example, the Venn diagram would look like Fig. 2.12. E and F

Area of entire region = P(S) = 1

P(E)

Fig. 2.11 Venn diagram

P(F)

2.3 Classical Probability

37

Fig. 2.12 Newspaper example Venn diagram

0.45 0.2

0.3

.05

S=1

The following probabilities can be used or found from the Venn diagram. We always start filling in probabilities from inside the intersection of the events and move our way out. The sum total of all probabilities within the Venn diagram rectangle, S, the sample set is 1.0. P(A) ¼ 0.75 P(B) ¼ 0.5 P(A \ B) ¼ 0.45 P(A [ B) ¼ P(A) + P(B)  P(A \ B) ¼ 0.8 P (only A) ¼ 0.3 P(only B) ¼ 0.05 P( only take 1 paper) ¼ P(only A) + P(only B) ¼ 0.3 + 0.05 ¼ 0.35 P(a soldier does not take a paper) ¼ 0.2

2.3.2

Conditional Probability

The notation P(F | E) is read “the probability of event F given event E.” It is the probability of an event F given the occurrence of the event E. The idea in a Venn diagram here is if an event has happened then we only consider that circle of the Venn diagram and we look for the portion of that circle that is intersected by another event circle. Think of this formula as PðA \ BÞ PðBÞ PðA \ BÞ PðBjAÞ ¼ PðAÞ

PðAjBÞ ¼

In most cases, these conditional probabilities led to different probabilities as answers. Let’s return to our newspaper example. Find the P(A|B) and P(B|A).

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2 Statistics and Probability in Military Decision-Making

PðA \ BÞ ¼ 0:45 PðAÞ ¼ 0:75 PðBÞ ¼ 0:5 PðA \ BÞ :45 ¼ ¼ :9 PðBÞ :50 PðA \ BÞ :45 PðBjAÞ ¼ ¼ ¼ :60 PðAÞ :75 PðAjBÞ ¼

Notice that the probabilities increased as we obtained more information about the events occurring. The probabilities do not always increase, they could decrease, or remain the same. They do not have to be affected the same way.

2.3.3

Independence

Two events E and F are independent if the occurrence of event E in a probability experiment does not affect the probability of event F. Two events are dependent if the occurrence of event E in a probability experiment affects the probability of event F.

2.3.3.1

Definition of Independent Events

Two events E and F are independent if and only if PðF j E Þ ¼ PðF Þ or PðE j F Þ ¼ PðE Þ Another way to see this is if P(A \ B) ¼ P(A)  P(B) then the events A and B are independent. If P(A \ B) 6¼ P(A)  P(B), then the events are dependent.

2.3.3.2

Independent Events

If events E and F are independent, then the probability of E and F both occur is P(E \ F) ¼ P(E)  P(F) Example Are the events of getting the local newspaper and USA Today independent events?

2.3 Classical Probability

39

Solution: PðAÞ ¼ 0:75 PðBÞ ¼ 0:5 PðAÞ  PðBÞ ¼ ð0:75Þ  ð0:5Þ ¼ 0:375 PðA \ BÞ ¼ 0:45 Since P(A \ B) 6¼ P(A)  P(B), then these events are not independent. Example Given the following information: PðEÞ ¼ :2 PðFÞ ¼ :6 PðE [ FÞ ¼ 0:68 Are E and F independent events? Solution: PðEÞ  PðFÞ ¼ :12 P(E \ F) is not given and must be found first. We do not assume independence and use the product rule. We use the addition rule where PðE [ FÞ ¼ PðAÞ þ PðBÞ  PðE \ FÞ and solve for PðE \ FÞ: 0:68 ¼ 0:2 þ 0:6  PðE \ FÞ PðE \ FÞ ¼ 0:12 Since P(E \ F) ¼ 0.12 and P(A)  P(B) ¼ 0.12, then events E and F are independent.

2.3.4

System Reliability in Series and Parallel of Independent Subsystems

Given the military system in Fig. 2.13. System 1 consists of subsystems A and B in series. System 1 has a P(System 1) ¼ P(A)  P(B) ¼ 0.81. System 2 consists of subsystem C and D in parallel. P(System 2) ¼ P(C) + P (D)  P(C \ D) ¼ 0.9 + 0.9  (0.81) ¼ 0.99 Overall the system reliability is P(System 1)  P(System 2) ¼ 0.81  0.99 ¼ 0.7776.

40

2 Statistics and Probability in Military Decision-Making 0.8 C 0.9

0.9

A

B 0.8 D

Fig. 2.13 Military system with four subsystems {A, B, C, D}

2.3.5

Bayes’ Theorem

We begin with the theorem of total probability.

2.3.5.1

Theorem of Total Probability

Let E be an event that is a subset of a sample space S. Let A1, A2, . . ., An be a partition of the sample space, S. Then, PðE Þ ¼ PðA1 Þ  PðEjA1 Þ þ PðA2 Þ  PðEjA2 Þ þ . . . þ PðAn Þ  PðEjAn Þ This is illustrated in Fig. 2.14. If we define E to be any event in the sample space S, then we can write event E as the union of the intersections of event E with A1 and event E with A2. E ¼ ðE \ A1 Þ [ ðE \ A2 Þ If we have more events, we just expand the union of the number of events that E intersects with as in Fig. 2.15. PðEÞ ¼ PðA1 \ E Þ þ PðA2 \ E Þ þ PðA3 \ E Þ ¼ PðE \ A1 Þ þ PðE \ A2 Þ þ PðE \ A3 Þ ¼ PðA1 Þ  PðEjA1 Þ þPðA2 Þ  PðEjA2 Þ þ PðA3 Þ  PðEjA3 Þ This is easier to see in a tree diagram shown in Fig. 2.16.

2.3 Classical Probability

41

A1

A2 S E ∩ A1

E ∩ A2 E

Fig. 2.14 Illustration of the law of total probability

A1

A2

A3 S

E ∩ A1

E ∩ A2

E ∩ A3 E

Fig. 2.15 Three intersections depicted

Fig. 2.16 Tree diagram depicted

2.3.5.2

Bayes’ Theorem

Let A1, A2, . . ., An be a partition of a sample space S. Then for any event E that is a subset of S for which P(E) > 0, the probability of event Ai for i ¼ 1, 2, . . ., n given the event E, is PðAi jEÞ

¼ ¼

PðAi Þ  PðEjAi Þ PðEÞ

PðAi Þ  PðEjAi Þ PðA1 Þ  PðEjA1 Þ þ PðA2 Þ  PðEjA2 Þ þ . . . þ PðAn Þ  PðEjAn Þ

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2 Statistics and Probability in Military Decision-Making

Example 1 Recruiting Single Unemployed Women Problem: According to the United States Census Bureau 21.1% of American adult women are single, 57.6% of American adult women are married, and 21.3% of American adult women are widowed or divorced (other). Of the single women, 7.1% are unemployed; of the married women, 2.7% are unemployed; of the “other” women, 4.2% are unemployed. Suppose that a randomly selected American adult woman is determined to be unemployed. What is the probability that she is single? Approach: Define the following events: U: unemployed S: single M: married O: other We have the following probabilities: PðSÞ ¼ 0:211; PðM Þ ¼ 0:576; PðOÞ ¼ 0:213 PðU j SÞ ¼ 0:071; PðU j M Þ ¼ 0:027; PðU j OÞ ¼ 0:042 and from the Theorem of Total probability, we know P(U ) ¼ 0.039. We wish to determine the probability that a woman is single given the knowledge that she is unemployed. That is, we wish to determine P(S | U ). We will use Bayes’ Theorem as follows: PðSjU Þ ¼

PðS \ U Þ PðSÞ  PðUjSÞ ¼ P ðU Þ PðU Þ

Solution: PðSjU Þ ¼

0:211ð0:071Þ ¼ 0:384 0:039

There is a 38.4% probability that a randomly selected unemployed woman is single. We say that all the probabilities P(Ai) are a priori probabilities. These are probabilities of events prior to any knowledge regarding the event. However, the probabilities P(Ai | E) are a posteriori probabilities because they are probabilities computed after some knowledge regarding the event. In our example, the a priori probability of a randomly selected woman being single is 0.211. The a posteriori probability of a woman being single knowing that she is unemployed is 0.384. Notice the information that Bayes’ Theorem gives us. Without any knowledge of the employment status of the woman, there is a 21.1% probability that she is single. But, with the knowledge that the woman is unemployed, the likelihood of her being single increases to 38.4%. Let’s do one more example.

2.3 Classical Probability Table 2.8 Proportion of work-disabled Americans

43 Age

Event A1 A2 A3 A4 A5

18–24 25–34 35–44 45–54 55 and older

Proportion work-disabled 0.078 0.123 0.209 0.284 0.306

Source: United States Census Bureau

Example 2 Military Disability Problem: A person is classified as work-disabled if they have a health problem that prevents them from working in the type of work they can do. Table 2.8 contains the proportion of Americans that are 18 years of age or older that are work-disabled by age. If we let M represent the event that a randomly selected American who is 16 years of age or older is male, then we can also obtain the following probabilities: Pðmalej18  24Þ ¼ PðMjA1 Þ ¼ 0:471 Pðmalej35  44Þ ¼ PðMjA3 Þ ¼ 0:485 Pðmalej55 and older Þ ¼ PðMjA5 Þ ¼ 0:460

Pðmalej25  34Þ ¼ PðMjA2 Þ ¼ 0:496 Pðmalej45  54Þ ¼ PðMjA4 Þ ¼ 0:497

(a) If a work-disabled American aged 16 years of age or older is randomly selected, what is the probability that the American is male? (b) If the work-disabled American that is randomly selected is male, what is the probability that he is 25–34 years of age? Approach: (a) We will use the Theorem of Total Probability to compute P(M ) as follows: PðM Þ ¼ PðA1 Þ  PðMjA1 Þ þ PðA2 Þ  PðMjA2 Þ þ PðA3 Þ  PðMjA3 Þ þ PðA4 Þ  PðMjA4 Þ þ PðA5 Þ  PðMjA5 Þ (b) We use Bayes’ Theorem to compute P(25–34 | male) as follows: PðA2 jM Þ ¼ where P(M) is found from part (a).

PðA2 Þ  PðMjA2 Þ PðM Þ

44

2 Statistics and Probability in Military Decision-Making

Solution: (a) PðM Þ¼PðA1 ÞPðMjA1 ÞþPðA2 ÞPðMjA2 ÞþPðA3 ÞPðMjA3 ÞþPðA4 Þ PðMjA4 Þ þPðA5 ÞPðMjA5 Þ¼ ð0:078Þð0:471Þþð0:123Þð0:496Þþð0:209Þð0:485Þþð0:284Þ ð0:497Þþ ð0:306Þð0:460Þ¼0:481 There is a 48.1% probability that a randomly selected work-disabled American is male. ð0:496Þ (b) PðA2 jE Þ ¼ PðA2PÞPðEðÞEjA2 Þ ¼ 0:123 ¼ 0:127 0:481 There is a 12.7% probability that a randomly selected work-disabled American who is male is 25–34 years of age. Notice that the a priori probability (0.123) and the a posteriori probability (0.127) do not differ much. This means that the knowledge that the individual is male does not yield much information regarding the age of the work-disabled individual. Example 3 Terrorist Violence Victims The data presented in Table 2.9 represents the proportion of murder victims at the various age levels in 2017. If we let M represent the event that a randomly selected terrorist violence victim was male, then we can also obtain the following probabilities: PðMjA1 Þ ¼ 0:622 PðMjA2 Þ ¼ 0:843 PðMjA3 Þ ¼ 0:733 PðMjA4 Þ ¼ 0:730 PðMjA5 Þ ¼ 0:577 (a) What is the probability that a randomly selected murder victim was male? PðMÞ ¼

X5 i¼1

PðAi Þ  PðMjAi Þ ¼ 0:760179

(b) What is the probability that a randomly selected male murder victim was 17–29 years of age? P(M n A2) ¼ 0.3574 (c) What is the probability that a randomly selected male murder victim was less than 17 years of age? P(M n A1) ¼ 0.051

Table 2.9 Terrorist violence victims

Level Less than 17 years 17–29 30–44 45–59 At least 60 years

Event A1 A2 A3 A4 A5

Proportion 0.082 0.424 0.305 0.125 0.064

Source: Adapted from Federal Bureau of Investigation

2.3 Classical Probability

45

(d) Given that a victim was male, what is the probability that the victim between 17–29 years of age? P(A2 | M) ¼ 0.3574/0.760179 ¼ 0.4702 Example 4 Military/Government-Related Double Agents and Espionage Suppose that the CIA suspects that one of its operatives is a double agent. Past experience indicates that 95% of all operatives suspected of espionage are, in fact, guilty. The CIA decides to administer a polygraph to the suspected spy. It is known that the polygraph returns results that indicate a person is guilty 90% of the time if they are guilty. The polygraph returns results that indicate a person is innocent 99% of the time if they are innocent. What is the probability that this particular suspect is innocent given that the polygraph indicates that he is guilty? The question requires P(person is innocent given the polygraph says that they are guilty). PðPolygraph guiltyÞ ¼ 0:855 þ 0:0005 ¼ 0:8555 PðPolygraph not guiltyÞ ¼ 0:095 þ 0:0495 ¼ 0:1445 Pðperson is a double agent j polygraph says guiltyÞ ¼ 0:855=0:8555 ¼ 0:999415 Pðperson is a not a double agent j polygraph says guiltyÞ ¼ 0:0005=0:8555 ¼ 0:000585 This is quite small so we would feel comfortable testing in this manner (Fig. 2.17).

Fig. 2.17 Double agent decision tree

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2 Statistics and Probability in Military Decision-Making

2.4 2.4.1

Probability Distributions Discrete Distributions in Modeling

We will also use several probability distributions for discrete random variables. A random variable is a rule that assigns a number to every outcome of a sample space. A discrete random variable takes on counting numbers 0,1,2,3,. . .etc. These are either finite or countable. Then, a probability distribution gives the probability for each value of the random variable. Let’s return to our coin flipping example earlier. Let the random variable X be the number of heads of the two flips of the coin. The possible values of the random variable X are 0, 1, and 2. We can count the number of outcomes that fall into each category of X as shown in the probability mass function Table 2.10. Note that the ΣP(F) ¼ 1/4 + 2/4 + 1/4 ¼ 1. This is a rule for any probability distribution. Let’s summarize these rules: 1. P(each event) 0 2. Σ P(events) ¼ 1 Thus, the coin flip experiment is a probability distribution. All probability distributions have means, μ, and variances, σ2. We can find the mean and the variance for a random variable X using the following formulas: μ ¼ E½X ¼ ΣxPðX ¼ xÞ   σ2 ¼ E X2  ðE½XÞ2 For our example, we compute the mean and variance as follows: μ ¼ E½X ¼ ΣxPðX ¼ xÞ ¼ 0ð1=4Þ þ 1ð2=4Þ þ 2ð1=4Þ ¼ 1   σ2 ¼ E X2 -ðE½XÞ2 ¼ 0ð1=4Þ þ 1ð2=4Þ þ 4ð1=4Þ  12 ¼ :5 We can also find the standard deviation, σ. σ¼

pffiffiffiffiffi σ2

Thus, we find the variance first and then take its square root.

Table 2.10 Probability mass function

Random variable, X Occurrences Corresponding to events P(X ¼ x)

0 1 TT ¼

1 2 TH,HT 2/4

2 1 HH 1/4

2.4 Probability Distributions

47

σ¼

pffiffiffiffi :5

There will be several discrete distributions that will arise in our modeling: Bernoulli, Binomial, and Poisson. Consider an experiment made up of a repeated number of independent and identical trials having only two outcomes, like tossing a fair coin {Head, Tail}, or a {red, green} stoplight. These experiments with only two possible outcomes are called Bernoulli trials. Often they are found by assigning either a S (success) or F (failure) or a 0 or 1 to an outcome. Something either happened (1) or did not happen (0). A binomial experiment is found counting the number of successes in N trials. Binomial experiment: Consists of n trials where n is fixed in advance. Trials are identical and can result in either a success or a failure. Trials are independent. Probability of success is constant from trial to trail.   n x Formula: bðx; n; pÞ ¼ pðX ¼ xÞ ¼ p ð1  pÞnx 0 for x ¼ 0,1,2,. . .n x   x P n y p ð1  pÞny Cumulative binomial: pðX  xÞ ¼ Bðx; n; pÞ ¼ y y¼0 x ¼ 0,1,2,. . .n Mean: μ ¼ n p Variance: σ 2 ¼ n p (1  p)

(a) (b) (c) (d)

for

Example Flip of a fair coin Our coin flip experiment follows these above rules and is a binomial experiment. The probability that we got one head in two flips is:   2 P ð X ¼ 1Þ ¼ :51 ð1  :5Þ21 ¼ :50 1 If we wanted five heads in ten flips of a fair coin, then we can compute:  Pð X ¼ 5 Þ ¼

 10 :55 ð1  :5Þ105 ¼ 0:2461 5

Example 2 Munitions as a Binomial Experiment Munitions are manufactured in a small local plant. In testing the munitions, prior to packaging and shipping, they either work, S, or fail to work, F. The company cannot test all the munitions but does test a random batch of 100 munitions per hour. In this batch, they found 2% that did not work but all batches were shipped to distributors.

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2 Statistics and Probability in Military Decision-Making

As the unit supply officer, you are worried about past performance of these munitions that you distribute to the units. If a unit takes 20 boxs of munitions, what is the probability that all work? Problem ID: Predict the probability that x munitions out of N work. Assumptions: The munitions follow the binomial   distribution rules stated earlier. n x Model: Formula: bðx; n; pÞ ¼ pðX ¼ xÞ ¼ p ð1  pÞnx for x ¼ 0,1,2,. . .n x If we have discrete data that follows a binomial distribution, then its histogram might look as it does in Fig. 2.18. It is symmetric. The keys are the assumptions for the binomial as well as it being discrete. Example 3 Weapons firing at targets. A weapon has a 93% accuracy on average. If we fire 10 shots at a target, what is the probability that we hit the target 5 times, at most 5 times, at least 5 times? Solution: This is a binomial distribution because shots are fired independently, the probability of success is known (93%), and we know in advance the number of shots fired, n (ten shots fired). First, we use Excel to generate the PDF and the CDF given in Table 2.11. (a) P(X ¼ 5). This is a PDF value that we extract from n ¼ 5, under PDF. The value is 0.0003. P(X ¼ 5) ¼ 0.0003. Interpretation: If we fired ten shots at a target, the probability that exactly 5 of the 10 hit the target is 0.0003. (b) At most five hit the target ! P(X  5). This is a CDF value that we extract from n ¼ 5 since we include five under the CDF, P(X  5) ¼ 0.0003. Interpretation: if we fire ten shots at a target, the probability the five or fewer hit the target is P (X  5) ¼ 0.0003. (c) At least five hit the target ! P(X 5). This is NOT one of our known forms. We must convert the probability to its complement. P(X 5) ¼ 1  P 250

200

150

100

50

0 1

2

3

4

5

6

7

8

Fig. 2.18 Binomial distribution histogram

9 10 11 12 13 14 15 16 17 18 19 20 21

2.4 Probability Distributions Table 2.11 Excel generated PDF and CDF data

49 n 0 1 2 3 4 5 6 7 8 9 10

PDF 0.0000 0.0000 0.0000 0.0000 0.0000 0.0003 0.0033 0.0248 0.1234 0.3643 0.4840

CDF 0.0000 0.0000 0.0000 0.0000 0.0000 0.0003 0.0036 0.0283 0.1517 0.5160 1.0000

(X < 5) ¼ 1  P(X  4). We obtain P(X  4) from the CDF table and obtain 0.000 (to four decimal places). 1  0.0000 ¼ 1. Interpretation: we expect five or more rounds to hit the target with probability of 1. Example 4 IEDS as a Binomial Experiment An analysis of recent IED attacks in Afghanistan shows the following: 73% of IEDs are road side bombs using artillery rounds as the munitions and 27% are suicide bombs using other devices. The probability of disabling a target with a road side IED bomb is 0.80. The probability of disabling a target with a suicide IED is 0.45 (a) Construct a tree diagram of the above events showing their respective probabilities. (b) What is the probability that an IED is a suicide bomb and it disables the target? (c) What is the probability that a target is disabled? (d) Find the probability that a road side bombs was used given that the target was not disabled. (e) Find the probability that a road side bombs was used given that the target was disabled. Solution: What is the probability that an IED is a suicide bomb and it disables the target? 0.1215 (a) What is the probability that a target is disabled? 0.7055 (b) Find the probability that a road side bombs was used given that the target was not disabled. 0.4952 (c) Find the probability that a road side bombs was used given that the target was disabled. 0.8277 Example 5 Navy Seals Mission A SEAL platoon carries eight (8) shaped charges on an operation. The probability that one of the charges will fire properly is 0.98. All eight charges are fired

50

2 Statistics and Probability in Military Decision-Making

independent of each other. Assume that the shaped charges follow a Binomial distribution. Express your answers to three decimal place accuracy. (a) (b) (c) (d) (e)

What is the probability that six of the eight will fire properly? What is the probability that all eight fire properly? What is the probability that only one will misfire? What is the probably that between 4 and 6 (inclusive for both) will fire properly? What is the mean and standard deviation for the number of successful firings? Solution:

What is the probability that six of the eight will fire properly? 0.00992 What is the probability that all eight fire properly? 0.8508 What is the probability that only one will misfire? 0.1389 What is the probably that between 4 and 6 (inclusive for both) will fire properly? 0.01033 (e) What is the mean and standard deviation for the number of successful firings? μ ¼ 7.84, σ2 ¼ 0.1568, σ ¼ 0.39598

(a) (b) (c) (d)

Example 6 Missile Attack Military missiles have been used often in Afghanistan. The military commander has subdivided the entire region into 576 smaller regions (no region overlaps with another region). A total of 535 missiles hit the combined area of 576 regions, you are being assigned to a region, find the probability that a selected region was hit exactly twice with missiles, at least twice, at most twice. μ ¼ 535/576 ¼ 0.9288 P(X ¼ 2) ¼ 0.1806 P(X  2) ¼ 0.92294 P(X 2) ¼ 1  P(X  1) ¼ 0.25769

2.4.2

Poisson Distribution

A discrete random variable is said to have a Poisson distribution if the probability distribution function of X is: pðx; λÞ ¼

eλ λx 0, x!

for

x ¼ 0,1,2,3 . . .

for some λ > 0:

We consider λ as a rate per unit time or per unit area. A key assumption is that with a Poisson distribution the mean and the variance are the same. For example, let X represent the number of minor flaws on the surface of a randomly selected F-16. It has been found that on average, 5 flaws are found per F-16 surface. Find the probability that a randomly selected-16 has exactly 2 flaws.

2.4 Probability Distributions

51

pð X ¼ 2Þ ¼

e5 52 ¼ :084 2!

A Poisson distribution has a mean, μ, of λ and variance σ2 of λ. A Poisson process is a Poisson distribution that varies over time (generally its time). There exists a rate, called α for a short time period. Over a longer period of time, λ becomes αt. Example Suppose your pulse is read by an electronic machine at a rate of five times per minute. Find the probability that your pulse is read 15 times in a 4-min interval. λ ¼ α t ¼ 5 times 4 min ¼ 20 pulses in a 4-min period pðX ¼ 15Þ ¼

e20 2015 ¼ :052 15!

Poissonð15; 20; falseÞ ¼ 0:051648854 Poisson data usually at least is slightly positively skewed.

2.4.2.1

Section Exercises

1. If 75% of all purchases at the Base Exchange are made with a credit card and X is the number among ten randomly selected purchases made with a credit card, then find the following: (a) p(X ¼ 5) (b) p(X  5) (c) μ and σ2 2. Martin’s milling produces fine munitions and its known from experience that 10% of its munition lots have flaws and must be classified as “seconds.” (a) Among six randomly selected munition lots, how likely is it that one is a second? (b) Among the six randomly selected lots, what is the probability that at least two are seconds? (c) What is the mean and variance for “seconds?” 3. Consider the following TV ad for an exercise program: 17% of the participants lose 3 lb, 34% lose 5 lb, 28% lose 6 lb, 12% lose 8 lb, and 9% lose 10 lb. Let X ¼ the number of pounds lost on the program. (a) (b) (c) (d)

Give the probability mass function of X in a table. What is the probability that the number of pounds lost is at most 6? At least 6? What is the probability that the number of pounds lost is between 6 and 10? What are the values of μ and σ2?

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2 Statistics and Probability in Military Decision-Making

4. A military 5 KW generator fails on average 0.4 times a month (30 consecutive days). Determine the probability that there are ten failures in the next year.

2.4.2.2

Chapter Projects Examples

1. Iran Hostage Rescue Attempt. In 1979, President Carter authorized an attempt to rescue American hostages held in Iran. DoD estimated that at least six helicopters would have to complete the mission successfully, but that the total number of helicopters needed to be kept as small as possible for security reasons. Each helicopter was believed to have a 95% chance of completing the mission (based on historical maintenance records). DoD used eight helicopters. Three helicopters failed so the mission was aborted. Defend the use of the Poisson distribution over the Binomial distribution. Determine the minimum number of helicopters necessary to have successfully completed the mission. 2. Military Aircraft Accidents. In a 7-day period in September 1997, six military aircraft crashed, prompting the Secretary of Defense to suspend all training flights. There were 277 crashes in the previous 4 years. Show that this is a rare event. Was there anything special about this week (7-day period) other than the six crashes? How many 7-day period could have occurred in a 4-year period? What should the Secretary of Defense have done in this matter? Make some recommendations based upon sound probability analysis.

2.4.3

Continuous Probability Models

2.4.3.1

Introduction

Some random variables do not have a discrete range of values. In the previous section, we saw examples of discrete random variables and discrete distributions. What if we were looking at time, as a random event? Time has a continuous range of values and thus, as a continuous random variable can be continuous probability distribution. We define a continuous random variable as any random variable measured on continuous scale. Other examples include altitude of a plane, the percent of alcohol in a person’s blood, net weight of a package of frozen chicken wings, the distance a round misses a designated target, or the time to failure of an electric light bulb. We cannot list the sample space because the sample space is infinite. We need to be able to define a distribution as well as its domain and range. For any continuous random variable, we can define the cumulative distribution function (CDF) as F(b) ¼ P(X  b). For those that have seen calculus, the probability density function (PDF) of f(x) is Rb defined to be Pða  x  bÞ ¼ a f ðxÞdx. To be a valid probability density function (PDF): (a) f(x) must be greater than or equal to zero for all x in its domain.

2.4 Probability Distributions

(b) the integral

R1 1

53

x  f ðxÞdx ¼ 1 ¼ the area under the entire graph of f(x).

Expected value or average R 1value of a random variable x, with PDF defined as above, is defined as E½X  ¼ 1 x  f ðxÞdx. In this section, we will see some modeling applications using many continuous distributions such as the exponential distribution and the normal distribution. For each of these two distributions, we will not have to use calculus to get our answers to probability questions. Since we do not require calculus, we will discuss only a few of these distributions that we obtain results with Excel.

2.4.3.2

The Normal Distribution

A continuous random variable X is said to have a normal distribution with parameters μ and σ (or μ and σ2), where 1 < μ < 1 and σ > 0, if the PDF of X is 2

ðxμÞ 1 f ðx; μ; σ Þ ¼ pffiffiffiffiffiffiffiffi e ð2σ2 Þ , 2πσ

1x1

The plot of the normal distribution is our bell-shaped curve, see Fig. 2.19. To compute P(a < x < b) when X is a normal random variable, with parameters μ R 1 ð2σxμ2 Þ2 ffi e ð Þ dx. and σ, we must evaluate pffiffiffiffiffi 2πσ Since none of the standard integration techniques can be used to evaluate this integral, the standard normal random variable Z with parameters μ ¼ 0 and σ ¼ 1 has Fig. 2.19 Bell-shaped curve of the normal distribution

Normal_Distribution-Bell-Shaped Curve 0.025

0.02

0.015

0.01

0.005

0.0 40

60

80 x

100

120

54

2 Statistics and Probability in Military Decision-Making

been numerically evaluated and tabulated for certain values. Since most applied problems do not have parameters of μ ¼ 0 and σ ¼ 1, “standardizing” transformation can be used Z ¼ xμ σ . For example, the amount of fluid dispensed into a can of diet coke is approximately a normal random variable with mean 11.5 fluid ounces and a standard deviation of 0.5 fluid ounces. We want to determine the probability that between 11 and 12 fluid ounces, P(11 < x < 12) are dispensed. Z1 ¼ ð11  11:5Þ=:5 ¼ 1 Z2 ¼ ð12  11:5Þ=:5 ¼ 1 This probability statement P(11 < x < 12) is equivalent to P(1 < Z < 1). If we used the tables, we can compute this to be 0.8413  0.1587 ¼ 0.6826. However, we can easily use technology to compute the area between 11 and 12 (Fig. 2.20). Therefore, 68.26% of the time the cans are filled between 11 and 12 fluid ounces as shown in Fig. 2.20

2.4.4

Exercises

Find the following probabilities: 1. X ~ N (μ ¼ 10, σ ¼ 2), P(X > 6). 2. X ~ N (μ ¼ 10, σ ¼ 2), P(6 < x < 14). Fig. 2.20 Normal distribution area from 11 to 12

Normal_Distribution Area: .6826733603 0.75

0.5

0.25

0.0 10

11

12 x

Partitions: 100

Curve 1 f(x)

13

2.4 Probability Distributions

55

3. Determine the probability that lies within one standard deviation of the mean, two standard deviations of the mean, and three standard deviations of the mean. Draw a sketch of each region. 4. A tire manufacturer thinks that the amount of wear per normal driving year of the rubber used in their tire follows a normal distribution with mean ¼ 0.05 in. and standard deviation 0.05 in. If 0.10 in. is considered dangerous, then determine the probability that P(X > 0.10)

2.4.5

Exponential Distribution

Continuous distribution of a random variable X that has properties: μ ¼ 1/λ, variance ¼ σ2 ¼ 1/λ2, where λ is the rate. PDF ¼ λe-λx for x 0 CDF ¼ 1  eλx x 0 (represents the area under the curve). In probability theory and statistics, the exponential distribution (a.k.a. negative exponential distribution) is a family of continuous probability distributions. It describes the time between events in a Poisson process, i.e., a process in which events occur continuously and independently at a constant average rate. The exponential distribution occurs naturally when describing the lengths of the inter-arrival times in a homogeneous Poisson process. In real-world scenarios, the assumption of a constant rate (or probability per unit time) is rarely satisfied. For example, the rate of incoming phone calls differs according to the time of day. But if we focus on a time interval during which the rate is roughly constant, such as from 2 to 4 p.m. during work days, the exponential distribution can be used as a good approximate model for the time until the next phone call arrives. Similar caveats apply to the following examples which yield approximately exponentially distributed variables: • The time until a radioactive particle decays, or the time between clicks of a Geiger counter • The time it takes before your next telephone call • The time until default (on payment to company debt holders) given in a reduced form credit risk model. Exponential variables can also be used to model situations where certain events occur with a constant probability per unit length, such as the distance between mutations on a DNA strand, or between “road kills” on a given road. In queuing theory, the service times of agents in a system (e.g., how long it takes for a bank teller, etc. to serve a customer) are often modeled as exponentially distributed variables. (The inter-arrival of customers for instance in a system is typically modeled by the Poisson distribution in most management science textbooks.) Reliability theory and reliability engineering also make extensive use of the exponential distribution.

56

2 Statistics and Probability in Military Decision-Making

Reliability ¼ 1  Failure Series ---– (A) ---– (B) ---P(A and B) must work. A and B are independent so PðA n B Þ ¼ PðAÞ  PðBÞ Parallel events PðA or BÞ ¼ PðAÞ þ PðBÞ  PðA and BÞ PðA or BÞ ¼ PðAÞ þ PðBÞ  PðAÞ  PðBÞ Example 1 Let X ¼ amount of time (in minutes) a US postal clerk spends with his/her customer. The time is known to have an exponential distribution with the average amount of time equal to 4 min. The rate is 1 customer every 4 min or ¼ of a customer per minute. X is a continuous random variable since time is measured. It is given that μ ¼ 4 min. To do any calculations, you must know λ, the decay parameter. λ ¼ 1=μ Therefore, λ ¼ ¼ ¼ 0.25 The standard deviation, σ, is the same as the mean. μ ¼ σ. The distribution notation is X ~ Exp(λ). Therefore, X ~ Exp(0.25) The probability density function is f(X) ¼ λeλx. The number e ¼ 2.718 ... It is a number that is used often in mathematics. f ðXÞ ¼ 0:25  e0:25X

where X is at least 0 and λ ¼ 0:25:

CDF ¼ PðX < xÞ ¼ 1  eλx ¼ 1  e0:25x The graph is shown in Fig. 2.21: Notice the graph is a decreasing. Fig. 2.21 Exponential distribution with mean ¼ 4.

0.25

f(X)

m = 0.25

0.2 0.15 0.1 0.05 0 0

2

4

6

8

10 12 14 16 18 20 X μ=4

2.4 Probability Distributions

57

Probabilities: Find the P(X < 5), P(X > 5), P(2 < X < 6) PðX < 5Þ ¼ 1  e:255 ¼ 0:713495 PðX > 5Þ ¼ 1  PðX < 5Þ ¼ 1  0:713495 ¼ 0:2865 Pð2 < X < 6Þ ¼ PðX < 6Þ  PðX < 2Þ ¼ 0:7768698  0:393469 ¼ 0:3834008 Example 2 Service Times Find the probability that a clerk spends 4–5 min with a randomly selected customer. Pð4 < X < 5Þ Use CDF PðX < xÞ ¼ 1  eλx PðX < 5Þ ¼ 1  e0:255 ¼ 0:7135 PðX < 4Þ ¼ 1  e0:254 ¼ 0:6321 Pð4 < X < 5Þ ¼ PðX < 5Þ  PðX < 4Þ ¼ 0:7135  0:6321 ¼ 0:0814 Example 3 Finding percentiles of an exponential distribution. Half of all customers are finished within how long? (Find the 50th percentile) PðX < kÞ ¼ 0:50

PðX < kÞ ¼ 0:50

PðX < kÞ ¼ 1  e0:25k 0:50

¼ 1  e0:25k   ln e0:25k ¼ ln ð0:50Þ

e0:25k ¼ 1  0:50 ¼ 0:50  0:25  k ¼ ln ð0:50Þ k ¼ ln ð:50Þ=  0:25 ¼ 2:8

Which is larger, the mean or the median? Mean is 4 min (given), median is 2.8. The mean is larger. Example 4 Exponential Distribution Twenty units were reliability tested with the results presented in Tables 2.12 and 2.13 (Fig. 2.22): OK, now what. Assume an exponential distribution with μ ¼ 255 h or λ ¼ 1/255 ¼ 0.00392156: or 0.00392156 failures per hour. So the average lifetime is 255 h. PðX > 3Þ ¼ 1  PðX < 3Þ ¼ 1  expð0:0039  3Þ ¼ 0:0116957 Or about a 1.1% chance of having more than three failures in a given hour.

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2 Statistics and Probability in Military Decision-Making

Table 2.12 Unit time to failure

Number of units in group 7 5 3 2 1 2

Table 2.13 Descriptive statistics of group time to failure

Column1 Mean Standard error Median Mode Standard deviation Sample variance Kurtosis Skewness Range Minimum Maximum Sum Count

Fig. 2.22 Histogram of number of units is group

8

Time to failure 100 200 300 400 500 600

255 37.32856 200 100 166.9384 27,868.42 0.10518 0.959154 500 100 600 5100 20

7 6 5 4 3 2 1 0 1

2

3

4

So what if we want the following: P(more than 3 failures in a day) λ is now (0.00392156  24) ¼ 0.094117 per day PðX > 3Þ ¼ 1  PðX < 3Þ ¼ 1  0:754 ¼ 0:24599

5

6

2.5 Military Applications of Distributions

2.5 2.5.1

59

Military Applications of Distributions Application of Probable Error in Targeting (Adapted from DA 3410 Course Notes, 2006)

The purpose of this section is to show a practical military application of the normal distribution. Two measures of dispersion of considerable importance in gunnery and bombing are PROBABLE ERROR (PE) AND CIRCULAR ERROR PROBABLE (CEP). This section will discuss Probable Error. PROBABLE ERROR: Consider an artillery piece that has a fixed elevation and deflection. If rounds from this weapon fall in a horizontal impact area, the points of impact will tend to concentrate about a point called the Center of Impact (CI), located at the mean range and mean deflection. Any distance such as RD (see Fig. 2.23) is known as a Range Deviation, while a distance such as DD is called a Deflection Deviation. Let’s superimpose this description of the distribution pattern unto a rectangular coordinate system. Locate the origin at the Center of Impact and orient the x-axis parallel to the direction of fire. The range and deflection deviations are treated as normal random variable with means of zero. The values for the standard deviations of these random variables are considered properties of the particular gun being fired, provided that the weapon is fired at a fixed elevation using projectiles from the same lot with the same powder charge. First let us discus the range deviation. Knowing that this deviation is approximately normal with mean zero, we can use the standard normal table to find the percentage of rounds expected to land between any two multiples of its standard deviation, where these two distances (called them a and b) are measured from the CI parallel to the direction of fire. Let the random variable X denote range deviation and let its standard deviation be denoted asσ x. Positive values for σ x indicate that the distance is above the CI and negative numbers are short or below the CI. Then   X0  b  0 ¼ P ð a  Z  bÞ Pðaσ x  X  bσ x Þ ¼ P a  0  σx For example, if a ¼ 0 and b ¼ 1, we find

Fig. 2.23 Illustration of range deviation, RD

Round Impact DD

CI

RD

Direcon of fire

60

2 Statistics and Probability in Military Decision-Making

  X0 P ð0  σ x  X  1  σ x Þ ¼ P 0   1  0 ¼ Pð0  Z  1Þ ¼ 0:3413 σx which means that about 34% of the rounds fired may be expected to fall between the CI and a distance of one standard deviation above the CI. We can then also calculate the percentages of round that would fall beyond 1 standard deviation (or similarly those rounds that would fall below 1 standard deviation) as approximately PðZ > 1:0Þ ¼ 0:50  0:34 ¼ 0:16 The discussion for deflection deviation is similar (Fig. 2.24). Based upon the characteristics of the weapon, we would expect the dispersion pattern of rounds to resemble that depicted in the left diagram on Fig. 2.3. At the long range typical of artillery bombardments, however this pattern is nearly rectangular as shown on the right diagram of Fig. 2.25. Accordingly, if we now measure all distances perpendicular to the direction or fire and replace “beyond” or “below” the CI with to the “left” or “right” (left taking positive values and right taking on negative values) of the CI, then we may consider a new random variable Y to represent deflection deviation. Although the standard deviation for this random variable, σ y, would probably be different than σ x, the distribution would be similar to that shown in Fig. 2.24. f(x)

34

16

34

–σx

σx

Fig. 2.24 Bell-shaped curve with 1 and 2 σ probabilities

Deflecon Error

Range Error

Fig. 2.25 Deflection error

~

16

2.5 Military Applications of Distributions

61

An effective hit would require “hitting” the target in both range and deflection. Since the adjustments for range and deflection are controlled separately on the weapon, we can consider the event of hitting the target in range and the event of hitting the target in deflection to be independent of each other. Then using the multiplication principle of probability where P(HR) denotes the probability of hitting the target in range and P(HD) denotes hitting the target in deflection PðH R \ H D Þ ¼ PðH R Þ  PðH D Þ In artillery problems, it is not customary to measure deviation in units of standard deviation, but rather to use a more convenient measure called PROBABLE ERROR (PE). In Fig. 2.2, we note that about 68% of the rounds shot should fall within the interval [σ x, σ x] for range. A probable error may be defined as the distance such that exactly 50% of the rounds fall within the interval [PE, PE]. So we see that a probable error is a somewhat shorter distance than a standard deviation (See exercises Problem 1). Since P(1PE  X  + 1PE) ¼ 0.50, we can redraw the normal curve presented in Fig. 2.24 as Fig. 2.26 with areas expressed in terms of PEs. Note that P(1PE  X  + 1PE) ¼ 0.50 may be written as P(|X|  1PE) ¼ 0.50, from which it follows that P(|X| > 1PE) ¼ 0.50. In other words, it is just as likely that X will deviate (in absolute value) from its mean of zero, by more than one probable error as not. This leads to the following definition: DEFINITION: A PROBABLE ERROR is the distance in range or deflection from the center of impact such that P(1PE  X  + 1PE) ¼ 0.50. Next let’s examine how we can use this notion of probable error to calculate the probability of hitting a target. First, we must use the tables to find that the approximate probabilities for the area under the curve for P(1PE  X  2PE) ¼ 0.16; P (2PE  X  3PE) ¼ 0.07 and P(X > 3PE) ¼ 0.02. (See Exercises, Problem #2.) These probabilities can be displayed using a histogram of the given probabilities. In Fig. 2.27, a histogram showing these probabilities is presented. Notice that this is a probability distribution function that we shall call û(x). In solving problems, it may be useful to follow the below procedure: f(x)

Fig. 2.26 Probabilities in segments

0PE –1PE

25

+1PE

25

25

25

62

2 Statistics and Probability in Military Decision-Making Û(x)

.2

.25

.1

.1

.02

.02 .0

.07 –3PE

–2PE

–1PE

0

1PE

2PE

3PE

x

Fig. 2.27 Histogram

1. Sketch the target. 2. Locate the Center of Impact. 3. From the CI, Completely cover the targets with probable errors of range, PER, and probable errors of deflection, PED. 4. Calculate the probabilities of a hit in range P(HR), and a hit in deflection, P(HD). 5. A hit on the target is then a hit in range and deflection and is then P(HR \ HD) ¼ P (HR)  P(HD) Example 1 You are the gun commander of an 8-in. howitzer firing at an enemy bridge that is 10 m wide and 40 m long. The center of the bridge is determined to be the center of impact. At the range, elevation and type of projectile, the probable errors are PER ¼ 19 m and PED ¼ 6 m. Assuming that howitzer is correctly laid so that the center of impact of rounds fired coincides with the center of the bridge and that the direction of fire is along the longer axis of the bridge, compute the number of rounds you must fire in order to expect one hit. Following the procedure detailed above, first draw the target information as shown in Fig. 2.28: Here the 40 by 10 m bridge has been sketched and the center of impact corresponding to the center of the bridge is located. Next from the CI, completely cover the targets with probable errors of range, PER, and probable errors of deflection, PED. Notice from Fig. 2.29 that PER just about covers the entire length of the bridge with 1 m extending into the region between 1PER and 2PER (also between 1  PER and 2PER). Also notice that PED covers the entire width with 1 m to spare on each end. Now we calculate the probabilities of hitting the bridge in range and deflection. Use Figs. 2.30 and 2.31 to visualize and help with these calculations: P ðH R Þ ¼

1 1 ð:16Þ þ ð:25Þ þ ð:25Þ þ ð:16Þ ¼ 0:517 19 19

In a similar manner, the probability of a hit in deflection, P(HD), is

2.5 Military Applications of Distributions

63

CI

10 m

Direcon of fire

5m 20 m 40 m

Fig. 2.28 Target and CI

–1PE

0PE

19m

1PE 19m

–1PE 6m 0PE

10 m CI

6m

5m

1PE 20 m 40 m

Fig. 2.29 Target with PE added

5 5 PðH D Þ ¼ ð:25Þ þ ð:25Þ ¼ 0:416 6 6 Finally, we calculate the probability of hitting the bridge by multiplying PðH R Þ  PðH D Þ ¼ ð0:517Þð:417Þ ¼ 0:216: Since this is a binomial process (two possible independent outcomes, n repeated trials with a fixed probability), the expected number of hits is μ ¼ np. Therefore, in order to expect one hit (μ ¼ 1) you must fire n¼

1 1 ¼ ¼ 4:63 p 0:216

or rounding up 5 rounds. 2.5.1.1

Section Exercises

1. What would the z-value be for one PE? 2. Using the standard normal tables, find P(1PE  X  2PE) ¼ 0.16; P (2PE  X  3PE) ¼ 0.07 and P(X > 3PE) ¼ 0.02.

64

2 Statistics and Probability in Military Decision-Making Û(x)

.2

.2

.1

.1 .02

.02 .07

.0 –20m

x

20m

Fig. 2.30 Shaded error for range

Û(x)

.2

.2

.1

.1 .02

.02 .0

.07 –5m

5m

x

Fig. 2.31 Shaded error for deflection

3. For a certain artillery piece, the PER is 20 m and the PED is 10 m. The target is a bridge 50 m long (parallel to the gun-target line) and 10 m wide. The center of impact is the center of the bridge. What is the probability of a hit on the bridge? ANS: 0.145. 4. An artillery piece fires at a rectangular area target. The target is 100 m long (in the direction of fire) and 50 m wide (perpendicular to the direction of fire). The center of impact has been brought to a point on the center line of the target, but 25 m short of the target center. The PER is 35 m and the PED is 10 m Find: (a) The probability of a hit if one round is fired. ANS: 0.533. (b) The number of rounds which need to be fired in order to expect 3 hits on the target area. ANS: 6 rounds (c) The probability of getting at least one hit if four rounds are fired at the target area. ANS: 0.9525

2.5 Military Applications of Distributions

2.5.2

65

Target Search Processes (Adapted from DA 3410 Course Notes, 2006)

Often acquiring the target is not easy because the target is not in the observer’s field of view. Search models focus on the situation where a target is hidden in a large area of search. An observer searches for the target by moving a relatively small sensor field of view to examine different parts of the field of search. For most of the time, the target is not in field of view and thus, no detection is possible. When the field of view (FOV) overlaps with the target, the probability of detection can be analyzed. Search models try to describe the probability of the amount of time required to find the target. We will describe several search methods and their associated probability distributions. Koopman’s OEG56 report, “Search and Screening” is a good source. We make the following modeling assumptions: 1. 2. 3. 4. 5.

Assume a single target in the search area of size A. Assume we can compute the area, A, of the search region. Assume initially that the target is stationary. It does move to avoid detection. Assume that the target location in the region is random. Assume the searcher’s platform can move at a constant speed along any path in the search area. The sensor is carried on the platform and thus views various parts of the search area. 6. The sensor has a maximum range, RMAX, that is smaller than the search area dimensions. The search models provide answers to the questions: “What is the probability that our sensor covers the target with its field of view?” Or “What is the probability of target detection as a function of the search time?”

2.5.2.1

Relative Motion

As a sensor moves through the search area, it may, at some time, move within RMAX of the target and thus, have a chance to achieve a detection. This analysis is made easier by setting the coordinate system for (x,y) centered on moving sensor or its platform, see Fig. 2.32.

2.5.2.2

Cookie Cutter Sensor

Suppose a sensor has perfect coverage within a circle of radius RMAX. If the target ever gets inside the circle sensor pattern, then it is discovered. We use the lateral range curve, PBAR(X) to describe the probability.

66

2 Statistics and Probability in Military Decision-Making

X

y

RMAX

x

Fig. 2.32 Relative motion coordinates

PBAR(X) ¼ 1.0 is X < RMAX

2.5.2.3

0.0 otherwise

Searching Without Overlap Arbitrary Search

As the sensor platforms move through the search area, the sensor coverage pattern sweeps out a covered area of width, W ¼ 2 RMAX Suppose the searcher moves through the search area at a constant speed, V, for a total search time, T, then the path length is defined to be L ¼ VT. We set up the search so we do not overlap of the coverage. The total area covered is LW ¼ VTW. The probability of detection is just a fraction of the total area searched, PDET (T) ¼ P(detect in time T) ¼ L  W/A ¼ T  V  W/A ¼ S  T, where S ¼ (V  W/A). The value S is known as the search rate. This is valid for path length formula up to LMAX ¼ A/W and when the entire search area will be covered. The maximum coverage occurs when the search time is TMAX ¼ A/(V  W) ¼ 1/S and PDET(T) ¼ 1.0.

2.5.2.4

Random Search

A random search places a search path of length L ¼ V  T into a search area, A, in a random fashion. This means that the location and course at any time is independent of the location and course at other times that are not close to the first time. We first find the probability of detection in a short segment of N short segments: PDET ðT=N Þ ¼ ðT  V  W Þ=ðA  N Þ Assuming T/N is short enough not to overlap with itself. Then, the probability over the entire path is

2.5 Military Applications of Distributions

67

PDET ðT Þ ¼ 1  eST where S ¼ V  T  W/A is the search rate for an exponential process. Example 1 A patrol aircraft is searching a rectangular region of 40 Nm  80 Nm for an enemy submarine. The aircraft moves at a constant velocity of 200 knots and is using a sensor with lateral range curve shown in Fig. 2.33. W ¼ area under the curve. These are two right triangles, base is 2 units and the height is 0.9 units. W ¼ 2  2  .9 ¼ 3.6 (a) If the station time that the patrol aircraft remains in the search area is 2 h, find the probability of locating the submarine using an arbitrary search model. Arbitrary Search solution PDET(T) ¼ S  T S ¼ V  W/A and T ¼ 2 h S ¼ V  W/A ¼ 200 (3.6)/3200 ¼ 0.225 PDET(2) ¼ 2  0.225 ¼ 0.45 (b) Find TMAX for the arbitrary model. From the graph above RMAX is 2. The Area ¼ 3200. So LMAX ¼ A/(2RMAX) ¼ 3200/4 ¼ 800 TMAX ¼ LMAX/V ¼ 800/200 ¼ 4 h (c) What would be the probability if a random search method is used (Time is still 2 h). PDET(T) ¼ 1  eST PDET(2) ¼ 1  e(.2252) ¼ 1  0.6376 ¼ 0.36237

Fig. 2.33 Triangular distribution for the sensor

PDET 0.9

–2

2

68

2.6 2.6.1

2 Statistics and Probability in Military Decision-Making

Central Limit Theorem, Confidence Intervals, and Hypothesis Testing Central Limit Theorem

Often it is easier to model using the mean then the actual data especially if the real data is not symmetric. For example, given a large sample, with n larger (n > 30), regardless of the shape of the RV, X, the distribution of the mean, X is approximately normal with mean x, and standard deviation is s/sqrt(n). Thus to find probabilities, we assume we are more interested in X than X. X-exponential with a sample mean 0.55, and a sample standard deviation 0.547, n ¼ 49. X is approximately normal with mean .55 and s ¼ 0.547/7. P(X > .69) ¼ 1  0.96 ¼ 0.04.

2.6.2

Confidence Intervals

The basic concepts and properties of confidence intervals involve initially understanding and using two assumptions: 1. The population distribution is normal. 2. The standard deviation σ is known or can be easily estimated. In its simplest form, we are trying to find a region for μ (and thus a confidence interval) that will contain the value of the true parameter of interest. The formula for finding the confidence interval for an unknown population mean from a sample is X Z α2 pσffiffin The value of Z α2 is computed from the normality assumption and the level of confidence, 1  α, desired. Let’s consider a variation of the diet coke. For example, the amount of fluid dispensed into a can of diet coke is approximately a normal random variable with unknown mean fluid ounces and a standard deviation of 0.5 fluid ounces. We want to determine a 95% confidence interval for the true mean. A sample of 36 diet cokes was taken and a sample mean of x ¼ 11:35 was found. Now, 1 – α ¼ 0.95. Therefore, α ¼ 0.5 and since there are two regions then we need α2 ¼ 0:25 and Z α2 ¼ 1:96. This is seen in the Fig. 2.34. ffiffiffiffi or [11.1.8667 Our confidence interval for the parameter, μ, is 11:35 1:96  p0:5 36 11.51333] Let’s interpret this or any confidence interval. If we took 100 experiments of 36 random samples each, and calculated the 100 confidence intervals in the same manner, X Z α2 pσffiffin.

2.6 Central Limit Theorem, Confidence Intervals, and Hypothesis Testing Fig. 2.34 Confidence ffiffiffiffi interval 11:35 1:96  p0:5 36

69

Normal_Distribution Area: 2560763304 0.75

0.5

0.25

0.0 10

11

12

13

x Partitions: 100

Curve 1 f(x)

Thus, 95 of the 100 confidence intervals would contain the true mean, μ. We do not know which of the 95 confidence intervals contain the true mean. Thus, to a modeler, each confidence interval built will either contain the true mean or it will not contain the true mean. In EXCEL, the command is CONFIDENCE(alpha,st_dev,size) and it only proved the value of Z α2 pσffiffin, we must still combine to get the interval X Z α2 pσffiffin.

2.6.3

Simple Hypothesis Testing

A more powerful technique for interring information about a parameter is a hypothesis test. A statistical hypothesis test is a claim about a single population characteristic or about values of several population characteristics. There is a null hypothesis (which is the claim initially favored or believe to be true) and is denoted by H0. The other hypothesis, the alternate hypothesis, is denoted as Ha. We will always keep equality with the null hypothesis. The objective is to decide, based upon sample information, which of the two claims is correct. Typical hypothesis tests can be categorized by three cases: CASE 1: CASE 2: CASE 3:

H0: H0: H0:

μ ¼ μ0 μ  μ0 μ μ0

Versus Versus Versus

Ha: Ha: Ha:

μ 6¼ μ0 μ > μ0 μ < μ0

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2 Statistics and Probability in Military Decision-Making

Table 2.14 Type I and Type II errors

State of nature Test conclusion

Fail to reject H0 Reject H0

H0 true 1α α

Ha true β 1β

There are two types of errors that can be made in hypothesis testing, Type 1 errors called α error and Type II errors called β errors. It is important to understand these. Consider the information provided in Table 2.14. Some important facts about both α and β: 1. 2. 3. 4.

α ¼ P(reject H0jH0 is true) ¼ P(Type I error) β ¼ P(fail to reject H0jH0 is false) ¼ P(Type II error) α is the level of significance of the test 1  β is the power of the test

Thus, referring to the table we would like α to be small since it is the probability that we reject H0 when H0 is true. We would also want 1  β to be large since it represents the probability that we reject H0 when H0 is false. Part of the modeling process is to determine which of these errors is the costliest and work to control that error as your primary error of interest. The following template if provided for hypothesis testing: STEP 1: Identify the parameter of interest STEP 2: Determine the null hypothesis, H0 STEP 3: State the alternative hypothesis, Ha STEP 4: Give the formula for the test statistic based upon the assumptions that are satisfied STEP 5: State the rejection criteria based upon the value of α STEP 6: Obtain your sample data and substitute into your test statistic STEP 7: Determine the region in which your test statistics lies (rejection region or fail to reject region) STEP 8: Make your statistical conclusion. Your choices are to either reject the null hypothesis or fail to reject the null hypothesis. Insure the conclusion is scenario oriented You are a commander of a small aviation transport unit. You are tired of hearing higher headquarters complain that your crews rest too much during the day. Aviation rules require a crew to get around 9 h of rest each day. You collect a sample of 37 crew members and determine that their sample average, x, is 8.94 h with a sample deviation of 0.2 h. The parameter of interest is the true population mean, μ. H0 : μ 9

2.6 Central Limit Theorem, Confidence Intervals, and Hypothesis Testing

71

Ha : μ < 9 pffiffi. This is a one-tailed test. The test statistic is Z ¼ s=xμ n We select α to be 0.05. We reject H0 at α ¼ 0.05, if Z < 1.645. From our sample of 36 aviators, pffiffi ¼ 8:949 pffiffiffiffi ¼ 0:06ð6Þ=:2 ¼ 1:8. Z ¼ 1.8. Z ¼ s=xμ n :2= 36

2.6.3.1

we

find

Interpretation

Since 1.8 < 1.645, then we reject null hypothesis that aviators rest 9 or more hours per day and conclude the alternate hypothesis is true, that your aviators rest less than 9 h per day. Rejecting the null hypothesis is the better strategy because it is now concluded that we reject the null hypothesis that the aviator crews rest 9 or more hours a day. P-Value The P-Value is the appropriate probability related to the test statistic. It is written so that the result is the smallest alpha level in which we may reject the null hypothesis. It is normal probability. From above our test statistic is 1.8 and we are doing a lower tail test. P-Value is P(Z < 1.8) ¼ 0.0359. Thus, we reject the null hypothesis for all values of alpha >0.0359. Thus, we reject if alpha is 0.05 but fail to reject if alpha is 0.01. In statistical significance testing, the P-Value is the probability of obtaining a test statistic at least as extreme as the one that was actually observed, assuming that the null hypothesis is true. In this context, value a is considered more “extreme” than b if a is less likely to occur under the null. One often “rejects the null hypothesis” when the P-Value is less than the significance level α (Greek alpha), which is often 0.05 or 0.01. When the null hypothesis is rejected, the result is said to be statistically significant. The P-Value is a probability, with a value ranging from zero to one. It is the answer to this question: If the populations really have the same mean overall, what is the probability that random sampling would lead to a difference between sample means as large (or larger) than you observed? We usually use either a normal distribution directly or evoke the central limit theorem (for n > 30) for testing means. Let’s say we think our mean of our distribution is ½. We want to test if our sample comes from this distribution. H0 : μ ¼ 0:5 Ha : μ 6¼ 0:5 The test statistic is key. From our data with sample size n ¼ 49, we find that the mean is 0.41 and the standard deviation is 0.2.

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The test statistic for a one sample test of a proportion is p  p0 ffi z ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi po ð1  po Þ=n So we substitute p ¼ 1/2, po ¼ 0.41, (1  po) ¼ 0.59, n ¼ 49 0:41  0:5 z ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi :5ð1  :5Þ=49 We find z ¼ 1.26. Next we need to find the probability that corresponds to the statement P (Z  1.26) ¼ 0.010385. We compare this P-Value, p, to our level of significance. If p < α, then we have significance (α is usually either 0.05 or 0.01). Statistical calculations can answer this question: If the populations really have the same mean, what is the probability of observing such a large difference (or larger) between sample means in an experiment of this size? The answer to this question is called the P-Value. The P-Value is a probability, with a value ranging from zero to one. If the P-Value is small, you’ll conclude that the difference between sample means is unlikely to be a coincidence. Instead, you’ll conclude that the populations have different means.

2.6.3.2

Excel Templates

Given our hypothesis test above, the probability of a Type I error, α, is the area under the normal bell-shaped curve centered at μ0 corresponding to the rejection region. This value is 0.05 (Fig. 2.35).

2.6.3.3

Section Exercises

Discuss how to set up each of the following as a hypothesis test. 1. 2. 3. 4. 5. 6.

Does drinking coffee increase the risk of getting cancer? Does taking aspirin every day reduce the chance of a heart attack? Which of the two gauges is more accurate? Why is a person “innocent until proven guilty”? Is the drinking water safe to drink? Set up a fake trial for a suspected felon. Build a matrix for their innocence or guilt with an appropriate null hypothesis. Which error, Type I or Type II, is the worst error? 7. Numerous complaints have been made that a certain hot coffee machine is not dispensing enough hot coffee into the cup. The vendor claims that on average the

2.6 Central Limit Theorem, Confidence Intervals, and Hypothesis Testing

73

Hypothesis Test Template

2 Tail Test Mean Population Mean, hypoth Standard Deviation, S N, sample size Alpha Level Enter tail information Upper tail as 0 Lower tail as 1 Both tails as 2

Test Stat 8.94 9 0.2 36 0.05 2

Z=

Value

x −µ

–1.8

s / n\

–1.6449

Results Reject

User inputs are in yellow

Fig. 2.35 Screenshot Excel template for hypothesis test

machine dispenses at least 8 oz of coffee per cup. You take a random sample of 36 hot drinks and calculate the mean to be 7.65 oz with a standard deviation of 1.05 oz. Find a 95% confidence interval for the true mean. 8. Numerous complaints have been made that a certain hot coffee machine is not dispensing enough hot coffee into the cup. The vendor claims that on average the machine dispenses at least 8 oz of coffee per cup. You take a random sample of 36 hot drinks and calculate the mean to be 7.65 oz with a standard deviation of 1.05 oz. Set up and conduct a hypothesis test to determine if the vendors claim is correct. Use an α ¼ .05 level of significance. Determine the Type II error if the true mean were 7.65 oz. Further hypothesis needed. Simple means, simple proportions, two means, and two proportions.

2.6.3.4

Hypothesis Testing

Questions: to test the hypothesis—is the sample normally distributed? or is the sample large (n > 30) since the test concern means?

2.6.3.5

Notation and Definitions

H0 is the null hypothesis and is what we assume to be true. Ha is the alternative hypothesis and generally what is the worst case or what we want to prove. α ¼ P(Type I error) known as level of significance (usually 0.05 or 0.01).

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β ¼ P(Type II error). Type I error rejects the null hypothesis when it is true. Type II error fail to reject the null hypothesis when it is false. Power of test ¼ 1  β. We want this to be large. This is the probability that someone guilty is found guilty. Conclusions: reject H0 or fail to reject H0. One-tailed test from Ha. Two-tailed test from Ha. Test statistic, Ts, comes from our data and is found by z ¼ xμ . psffi n Rejected region: that area under the normal curve where we reject the null hypothesis. P-Values is the smallest level of significance at which H0 would be rejected when a specified test procedure is used on a given dataset. We compare P-Value to our given α. If P-Value  α ! we reject Ho at level α. If P-Value > a ! we fail to reject Ho at level α. It is usually thought of as the probability associated with the test statistic, P( x > TS)

2.6.3.6

Hypothesis Testing

Reject H0 Fail to Reject H0

H0 is true Type I error P(Type I error) ¼ α Correct decision

H0 is false Correct decision Type II error P(Type II error) ¼ β

Example: H0: The defendant is innocent HA: The defendant is guilty What is a Type I error: Someone who is innocent is convicted—we want that to be small. What is a Type II error: Someone who is guilty is cleared, we want that small also. Example: H0: The drug is not safe and effective. HA: The drug is safe and effective. What is a Type I error: Unsafe/ineffective drug is approved. What is a Type II error: Safe/effective drug is rejected. The reason we do it this way is we want to prove that the drug is safe and effective. Mathematically, when we examine hypothesis tests we always put the ¼ with H0!!!!! (Figs. 2.36 and 2.37)

2.6 Central Limit Theorem, Confidence Intervals, and Hypothesis Testing

Two-Tailed

Critical Region

–z α

75

Left-Tailed

Critical Region

Critical Region

zα

2

2

(critical value)

(critical value)

–zα (critical value)

Fig. 2.36 Two-tailed and one-tailed hypothesis test

Right-Tailed

Critical Region

zα (critical value)

Fig. 2.37 Right-tailed hypothesis test

Example 1 Two-tailed test. Being too big or too small is bad. A machine that produces rifle barrels is set so that the average diameter is 0.50 in. In a sample of 100 rifle barrels, it was found that xbar ¼ 0.51 in. Assuming that the standard deviation is 0.05 in., can we conclude at the 5% significance level that the mean diameter is not 0.50 in.? H0 : μ ¼ 0:50 HA : μ 6¼ 0:50 Rejection region: |z| > zα/2 ¼ z0.025 ¼ 1.9 Draw the picture and rejection region. Do problems without standard normal, just use normal.

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Test statistic: z ¼ (xbar  μ)/(σ/√n) ¼ (0.51  0.50)/(0.05/√100) ¼ 0.01/ 0.005 ¼ 2.0 Conclusion: Reject H0, Yes. P-Value: The probability of obtaining a sample result that is at least as unlikely as what is observed, or the observed level of significance. It is the probability in the tail associated with the value. P(Z > 2) or p( x > Xts) In the rifle barrel case: P-Value ¼ 0:5  0:4772 ¼ 0:0228 * z ¼ 2:00 ) Using EXCEL: ¼norm.dist(2,4,0,1) ¼ 0.022718 Example 2 Left-tailed test. Being too small is worst case. In the midst of labormanagement negotiations, the president of a company argues that the company’s blue-collar workers, who are paid an average of $30,000 per year, are well paid because the mean annual income of all blue-collar workers in the country is less than $30,000. That figure is disputed by the union, which does not believe that the mean blue-collar income is less than $30,000. To test the company president’s belief, an arbitrator draws a random sample of 350 blue-collar workers from across the country and asks each to report his or her annual income. If the arbitrator assumes that the blue-collar incomes are distributed with a standard deviation of $8000, can it be inferred at the 5% significance level that the company president is correct? H0 : μ 30; 000 HA : μ < 30; 000 Rejection region: z < zα ¼ z0:05 ¼ 1:645 * * One-tailedtest, draw picture again x ¼ 29; 120 Test statistic: z ¼ ( x  μ)/(σ/√n) ¼ (29,120  30,000)/(8000/√350) ¼ 880/ 427.618 ¼ 2.058 Conclusion: Reject H0, Yes. P-Value: the smallest value of α that would lead to rejection of the null hypothesis. P-Value ¼ Pðz < 2:058Þ ¼ 0:5  0:4803 ¼ 0:0197 z  2:06

2.6 Central Limit Theorem, Confidence Intervals, and Hypothesis Testing

77

Example 3 Left-tailed test. Being too large hurts unit performance and is worst case. We want to measure if the new regulations were effective so we look to the lefttailed test to prove they were effective. In an attempt to reduce the number of person-hours lost as a result of non-combat-related military accidents, the DOD has put in place new safety regulations. In a test of the effectiveness of the new regulations, a random sample of 50 units was chosen. The number of personhours lost in the month prior to and the month after the installation of the safety regulations was recorded. Assume that the population standard deviation is σ ¼ 5. What conclusion can you draw using a 0.05% significance level? x ¼ 1:2 H0 : μ 0 HA : μ < 0 Rejection region: z < zα ¼ z0:05 ¼ 1:644 * * One-tailed test Draw the picture Test statistic: z ¼ (xbar  μ)/(σ/√n) ¼ (1.20  0)/(5/√50) ¼ 1.2/ 0.707 ¼ 1.697 P-Value ¼ 0.5  0.4554 ¼ 0.0446 Conclusion: Reject H0, since 1.697 < 1.644 The new safety regulations are effective. Example 4 Right-Tailed Test Average, μ, of time spent reading newspapers: 8.6 min. Do people in military leadership positions spend more time than the national average time per day reading newspapers? H0 : μ  8:6 HA : μ > 8:6 We sampled 100 officers and found that they spend 8.66 min reading the paper (or from the web) with a standard deviation of 0.1 min. Z ¼ 8.66  8.6/(.1/10) ¼ 6 Reject if Z > Zα, assume α ¼ 0.01. Z α ¼ 2.32 Since 6 > 2.32, we reject Ho.

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2 Statistics and Probability in Military Decision-Making

(a) Type I Error: Rejecting the null hypothesis, H0, when it is true: Concluding that the mean newspaper-reading time for managers is greater than the national average of 8.6 min when in fact it is not. Possible consequences: Wasted money on campaigns targeting managers who are believed to spend more time reading newspapers than the national average. (b) Type II Error: Failing to reject the null hypothesis, H0, when it is false: Concluding that the mean newspaper-reading time for managers is less than or equal to the national average of 8.6 min when in fact it is greater than 8.6. Possible consequences: Missed opportunity to potentially access managers who may spend more time reading newspapers than the national average. Example 5 Mean filling weight: 16 oz/Container, σ ¼ 0.8 oz, Sample size: 30, α ¼ 0.05 H0 : μ ¼ 16 HA : μ 6¼ 16

Continue production Discontinue production

(a) Rejection Rule: Two-tailed test: z-value associated with alpha ¼ 0.05 is 1.96 ) Reject if z < 1.96 or if z > 1.96 (b) If xbar ¼ 16.32: (xbar  μ)/(σ/√n) ¼ (16.32  16)/(0.8/√30) ¼ 0.32/ 0.1460593 ¼ 2.19 Since 2.19 > |1.96| Reject H0; which means shut down production line. (c) If xbar ¼ 15.82: (xbar  μ)/(σ/√n) ¼ (15.82  16)/(0.8/√30) ¼ 0.18/ 0.1460593 ¼ 1.23 Do not reject H0; which means that no adjustment of production line is necessary. (d) P-Value (for case where sample mean was 16.32): (2) (0.5  0.4857) ¼ 0.0286. One-side is 0.01419. Excel (in its formula) always gives a P-Value based on a two-sided test. P-Value (for case where sample mean was 15.82): (2) (0.5  0.3907) ¼ 0.2186 Taking a second look at the interval estimation and hypothesis testing relationship:     μ0 zα=2 σ=√n ¼ 16 1:96 0:8=√30 ¼ 16 0:286 ) 15:714 . . . :16:286 Since 16.32 is outside of this range, we can conclude that we should reject H0; but since 15.82 is within the range, we fail to reject H0. Tests About A Population Mean:

2.6 Central Limit Theorem, Confidence Intervals, and Hypothesis Testing

79

Table 2.15 Mean earnings per share for financial service corporations 1.92

2.16

3.63

3.16

4.02

3.14

2.20

2.34

3.05

2.38

    Z or t ¼ x  μ = s=√n Example 6 Small sample from a normal is t. The population mean earnings per share for financial services corporations including American Express, ETrade Group, Goldman Sachs, and Merrill Lynch was $3 (Business Week, August 14, 2000). In 2001, a sample of ten financial service corporations provided the earnings per share in Table 2.15: Determine whether the population mean earnings per share in 2001 differ from $3 reported in 2000. α ¼ 0.05 H0 : μ ¼ 3 t0:025;9 ¼ 2:262

HA : μ 6¼ 3  EXCEL≔ TINVð0:05; 9Þ ¼ 2:262159

Reject if t < 2.262 or if t > 2.262

Sum: Mean:

Mean Earnings 1.92 2.16 3.63 3.16 4.02 3.14 2.20 2.34 3.05 2.38 28.00 2.8

(x  2.8)^2 0.77 0.41 0.69 0.13 1.49 0.12 0.36 0.21 0.06 0.18 4.42 0.4908 0.7006

¼ Variance ¼ St. Deviation

      t ¼ xbar  μ = s=√n ¼ ð2:8  3Þ= 0:7006=√10 ¼ 0:9027 p  value : EXCEL≔ðTDISTð0:9027; 9; 2ÞÞ ¼ 0:390

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2 Statistics and Probability in Military Decision-Making

Do not reject H0. We cannot conclude that the population mean earnings per share has changed. Again, utilizing a confidence interval to make a decision:     x tα=2 s=√n ¼ 2:8 2:262 0:7006=√10 ¼ 2:8 0:50 ) 2:30 . . . :3:30 As the claimed mean ($3) is within the range, we cannot reject the null hypotheses. Tests About a Population Proportion Example 7 In a television commercial, the manufacturer of a toothpaste claims that more than four out of five dentists recommend the ingredients in his product. To test that claim, a consumer-protection group randomly samples 400 dentists and asks each one whether he or she would recommend a toothpaste that contained certain ingredients. The responses are 0 ¼ No and 1 ¼ Yes. There were 71 No answers and 329 Yes answers. At the 5% significance level, can the consumer group infer that the claim is true or not true? p^ ¼ 329=400 ¼ 0:8225 p ¼ 0:8 H0 : p  0:8 HA : p > 0:8 Rejection region: z > zα ¼ z0.05 ¼ 1.645   Test statistic: z ¼ p^  p =√ðpq=nÞ ¼ ð0:8225  0:8Þ=√½ð0:8  0:2Þ=400 ¼ 0:0225=0:02 ¼ 1:125 Conclusion: Do not reject H0. The claim is likely to be true. If that (for some reason) remains 0.8225. How big would n have to be for us to be able to support the claim? 1:645 ¼ ð0:8225  0:8Þ=√ð0:8  0:2Þ=n



n ¼ 855:11 ¼ 856 Example 8 Alberta driving practices: 48% of drivers did not stop at stop signs on county roads. Two months and a serious information campaign later: Out of 800 drivers, 360 did not stop. (a) Has the proportion of drivers who do not stop changed? H0 : p ¼ 0:48

2.7 Hypothesis Tests Summary Handout

81

HA : p 6¼ 0:48 (b) Rejection region: zα/2 ¼ z0.025 ¼ 1.96 Reject if z < 1.96 or if z > 1.96 (c) p^ ¼ 360=800 ¼ 0:45 p ¼ 0:48  (d) ( p^  p =√ðpq=nÞ ¼ ð0:45  0:48Þ=√½ð0:48  0:52Þ=800 ¼ 0:03=0:0176635 ¼ 2 1:70 (e) Do not reject H0: We cannot conclude that the proportion of drivers who do not stop has not changed.

2.7

Hypothesis Tests Summary Handout H0 is true Type I error P(Type I error) ¼ α Correct decision

Reject H0 Do not reject H0

2.7.1

H0 is false Correct decision Type II error P(Type II error) ¼ β

Tests with One Sample Mean

H o: μ ¼ μ o Ha: This can be any of the following as required: μ 6¼ μo μ < μo μ > μo pffiffi 0 Test statistic: Z ¼ xbarμ σ= n

Decision: Reject the claim, Ho iff for μ 6¼ μo Either Z zα/2 or Z  zα/2 μ < μo Z  zα μ > μ o Z zα

2.7.2

Tests with a Population Proportion (Large Sample)

Null hypothesis: Ho: p ¼ po p1 p0 ffi Test statistic: z ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p0 ð1p0 Þ=n

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2 Statistics and Probability in Military Decision-Making

Alternative hypothesis Ha: p > p0 Ha: p < p0 Ha: ¼ p 6¼ p0

Rejection region z zα z  zα either z zα/2 or z  zα/2

These procedures are valid for np0 5 and n(1  po) 5.

2.7.3

Tests Comparing Two Sample Means

Ho: μ1 ¼ μ2 we write this as μ1  μ2 ¼ 0 Ha: This can be any of the following as required: Δμ6¼0, Δμ0 qffiffiffiffiffiffiffiffi Test statistic: Z ¼ xbar1xbar2 S2 S2 1 2 mþ n

Decision: Reject the claim, Ho iff for Δμ 6¼0 Either Z zα/2 or Z  zα/2 Δμ < 0 Z  zα Δμ > 0 Z zα

2.7.3.1

Section Exercises Hypothesis Test Problems

1. An intelligence agency claims that the proportion of the population who have access to computers in Afghanistan is at least 30%. A sample of 500 people is selected and 125 of these said they had access to a computer. Test the claim at a 5% level of significance. 2. A manufacturer of AA batteries claims that the mean lifetime of their batteries is 800 h. We randomly select 40 batteries and find their mean is 790 h with a standard deviation of 22 h. Test the claim at both a 5% and a 1% level of significance. 3. As a commander you are asked to test a new weapon in the field. This weapon is claimed to 95% reliable. You issue 250 of these weapons to your soldiers and of these 15 did not work properly, i.e., failed to meet military specifications. Perform a hypothesis test at a 5% level of significance. 4. You need steel cables for an upcoming mission that are at least 2.2 cm in diameter. You procure 35 of the cables and find through measurement that the mean diameter is only 2.05 cm. The standard deviation is .3 cm. Perform an hypothesis test of the cables at a 5% level. 5. For safety reason, it is important that the mean concentration of a chemical used to make a volatile substance does not exceed 8 mg/L. A random sample of 34 containers have a sample mean of 8.25 mg/L with a standard deviation of 0.9 mg/L. Do you conform to the safety requirements?

2.8 Case Studies

83

6. In a survey in 2003 adult Americans were asked which invention they hated the most but could not do without. 30% chose the cell phone. In a more recent survey, 363 of 1000 adult Americans surveyed stated that the cell phone was the invention they hated the most and could not do without. Test at the 5% level of significance if the proportion of adult Americans who hate and have a cell phone is the same as it was in 2003.

2.8 2.8.1

Case Studies Violence in the Philippines

In this case study, we examine how or if poverty levels in the community affect terrorist events using hypothesis testing based upon research by LTCOL J. Durante (Durante & Fox 2015). The population of the Philippines for 2010 is estimated to be 94 million. It has relatively grown from 76.9 million in the year 2000, with an annual growth rate of 2.36%, and 85.3 million in the year 2005, with an annual growth rate of 2.04% (National Statistics Coordination Board 2012). The high population growth, lack of jobs, and underemployment has contributed to a 33.7% poverty rate in 2003 (Abinales and Amoroso 2005). Income is distributed unevenly wherein the poorest 10% of the population only controls 1.7% of the national income while the top 10% of the population controls 38.4% (Abinales and Amoroso 2005). Many families rely on remittances of the seven million Filipinos living abroad which in recent years have sent home $6–7 billion annually (Abinales and Amoroso 2005). Following the reconstruction after World War II, the Philippines was one of the richest countries in Asia (Philippines 2012). However, economic mismanagement and political volatility during the Marcos regime, and the political instability during the Corazon Aquino administration contributed to economic stagnation and further dampened economic activity (Philippines 2012). A broad range of reforms were implemented by subsequent administrations to improve economic growth and attract foreign investments. Since the year 2000, the Gross Domestic Product (GDP) has been generally increasing except for 2009 where the GDP was at its lowest at 1.1%. This was mainly caused by consumer demand, a rebound in exports and investments, and election-related spending. However, it bounced back to 7.3% in 2010 and went down to 4% by 2011 (CIA Factbook 2012) (Fig. 2.38). From 2000 to 2011, the Philippine economy is considered to be stable. The economy was able to endure the 2008–2009 global recession compared to other countries in the region mainly due to minimal exposure to troubled international securities, lower dependence on exports, relatively resilient domestic consumption, large remittances from overseas Filipino workers, and a growing business process outsourcing industry (CIA Factbook 2012). Despite the stability, the country failed

84

2 Statistics and Probability in Military Decision-Making GDP 8 7.3

7 6

5.9 5.4

5 4.6 4

4.5

4.8 4

3.8

3.6 3

7.3

2.8

2 1

1.1

0 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012

Fig. 2.38 Philippines GDP from 2000 to 2011 (Index Mundi 2012)

to develop the domestic human capital. Not enough jobs were created and unemployment rate remained high. Other factors that restricted the growth of the economy are the huge deficit caused mainly by massive domestic and foreign debt, and the state’s inability to collect taxes. Due to limited government resources, social needs remained unmet which fueled political instability, which consequently discouraged foreign investment (Abinales and Amoroso 2005). Poverty is one of a number of factors that may contribute to violent conflict. It has been asserted that poverty is one of the main causes of insurgency. To analyze conflict and poverty in the Philippines, datasets on poverty and significant acts (Sigacts) are projected in a scatterplot. For 2003, 1355 violent incidents were recorded ranging from armed clashes, assassination, murder, kidnapping, arson, ambush, raid, bombing, shooting, and harassments. It can be observed from Fig. 2.39 that Sigacts increase as poverty index goes up. The correlation of 0.2315 reflects a weak linear relationship between these two variables. The linear regression equation only explains 5.36% of the data as depicted by R2. North Cotabato and Maguindanao are considered as outliers having considerably high Sigacts score of 217 and 225, respectively. Descriptive statistics shows that poverty has a mean of 31.77 and a median of 33.5, while Sigacts have a mean of 16.7 and a median of 8. In applying descriptive statistics, poverty index data for 2003 was partitioned into two groups. One group with a poverty index of less than 28 and the second group with more than 28. The hypotheses were formulated as follows: Ho: u1  u2 ¼ 0 Ha: u2 > u1 The null hypotheses (Ho) would state that both groups of the partitioned poverty index would have the same number of Sigacts with u1 being the group with lower

2.8 Case Studies

85

Sigacts

Poverty 03 vs Sigacts 2003 240 220 200 180 160 140 120 100 80 60 40 20 0 –20 0

Sigacts 2003 Linear (Sigacts 2003) y = 0.6038x -2.4557 R² = 0.0536 Correlaon = 0.2315 10

20

30 40 Poverty Index

50

60

70

Fig. 2.39 Scatterplot of poverty and Sigacts 2003

poverty index. Meanwhile, the alternate hypotheses would state that the group with higher poverty would have higher number of Sigacts. Descriptive statistics reveals the following values x ¼ 12.256 y ¼ 21.536 σ 2x ¼ 1177:936 σ 2γ ¼ 1333:305 m ¼ 39 n ¼ 41

Test statistics shows that the value of z ¼ 1.19. For a one-tailed test at 5% significance level, the value of the test statistic z reveals that it is not within the rejection region (Fig. 2.40). Since 1.19 > 1.65 and is not within the rejection region, the null hypothesis is not rejected and therefore conclude that the mean for sample 1 is equal to mean for sample 2 at α ¼ 0.05. As such, it is asserted that Sigacts is the same as poverty increase or decrease. For the year 2006, Sigacts declined with 1091 recorded incidents (Fig. 2.41). The linear trending only represents 4.38% of the data (R2). Moreover, the correlation coefficient attests that the relationship among the variables is only 0.2092, still a weak linear relationship between the variables. For descriptive statistics, poverty index data for 2006 was again partitioned into two groups. One group with a poverty index score of less than 37, and the second group with more than 37. The null hypotheses (Ho) would state that both groups of that of the partitioned poverty index would have the same number of Sigacts. Meanwhile, the alternate hypotheses would state that the group with higher rate of

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2 Statistics and Probability in Military Decision-Making

Fig. 2.40 Rejection region

Poverty 06 vs Sigacts 2006 70 60

Sigacts

50 40 Sigacts 2006

30

Linear (Sigacts 2006)

20

y = 0.2209x + 5.8604 R² = 0.0438 Correlaon = 0.2092

10 0 –10

10

30 50 Poverty Index

70

90

Fig. 2.41 Scatterplot of poverty and Sigacts 2006

poverty have higher number of Sigacts. Descriptive statistics reveals the following values: x ¼ 9.5 y ¼ 17.342 σ 2x ¼ 115:744 σ 2γ ¼ 338:880 m ¼ 40 n ¼ 41

Test statistics shows that the value of z ¼ 2.35. For a one-tailed test at 5% significance level, the value of z reveals that it is within the rejection region (Fig. 2.42). Since 2.347 > 1.65 and is within the rejection region, the null hypothesis is rejected and therefore conclude that the mean for sample 2 is greater than the mean sample 1 at α ¼ 0.05. As such, it is asserted that as poverty increases the number of Sigacts also increases.

2.8 Case Studies

87

Fig. 2.42 Rejection region

From a military prospective, we need to improve the welfare and wealth of the people in hopes of lowering the number of significant terrorist acts.

2.8.2

The Impact of Medical Support on Security Force Effectiveness (Adapted from a Course Project by LTC Ramey Wilson)

Introduction Many factors influence the effectiveness of security forces. While there has been significant inquiry and research on the impact of obvious factors, e.g., training, leadership, logistics, equipment, oversight, policies and legal institutions, the impact of medical support on security force effectiveness has received little attention. In fact, there is no evidence of any published quantitative or qualitative analysis on the role of medical support for security forces. It is undisputed that the delivery of security, especially in areas with active or latent instability, carries an inherent risk of injury for those tasked to provide it. For security to be effective and lasting, security forces must enter and control contested areas to establish order, apprehend criminals, and enforce peace. Establishing and maintaining security, however, exposes security forces to violence and the risk of injury. For the individual soldier or police officer, the risks are personal. For the state, the legitimacy of its governance often rests with establishing and maintaining order through the use of legitimate coercion and violence. Security, one of the pillars of development, remains a necessary condition for state development and economic progress. This case study explores the relationship between the effectiveness of police forces with varying levels of health support using a large-n quantitative approach. The results will demonstrate that security forces perform more effectively when they have a reasonable expectation of capable medical care in the event they are injured.

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Fig. 2.43 Soldier adaptation model (Bliese and Castro 2003: 188)

2.8.2.1

Soldier Adaptation Model Framework

The Soldier Adaptation Model (SAM) described by Bliese and Castro provides a framework to understand the factors influencing security personnel’s work motivation (Bliese and Castro 2003). The SAM uses a systems-based approach to describe performance outputs (See Fig. 2.43). The primary input, stressors, includes all aspects of the environment that “place a load or demand on the soldier” (Bliese and Castro 2003). These include weather, duty responsibilities, role ambiguity, workload, family separation, and danger. While some of these stressors will vary by location and time, such as danger, others are omnipresent. Moderators represent buffering or mitigating actions to decrease the impact of stressors. Training, unit cohesion, and leadership, for example, are moderators cultivated prior to stress exposure to reduce the impact of both anticipated and unanticipated stressors. As security personnel operate in the context of groups and organizations, moderators must be cultivated at each level to be mutually reinforcing in order to minimize the effects of stressors on performance. Bliese and Castro (2003) argue that “soldier well-being and performance is at its peak when moderation at each of the three levels, the individual, the group, and the organization, is maximal”(Bliese and Castro 2003). Stressors are mitigated by moderators and result in strains. At its basic form, “[strains] represent outcomes” (Bliese and Castro 2003). Categorized across the domains of health, attitudes, and performance, strains are analyzed through disease incidence, surveys, and performance metrics. Using the SAM framework, this analysis explores the relationship between the risk of injury (the stressor), medical support (the moderator), and security force performance (the strain) (see Fig. 2.44).

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Fig. 2.44 Study variables expressed in SAM system. The risk of injury and the presence of a pre-hospital medical system represented the independent variables influencing police performance, the dependent variable.

2.8.3

Hypothesis

In order to investigate this relationship, the following hypothesis was proposed: H0: Security forces will be more productive when medical care is readily available. To test this hypothesis, police productivity (dependent variable) will be analyzed with varying levels of medical support (independent variable) and risk (independent variable) (see Fig. 2.44). Though there may be disagreement on the use of police productivity as a metric of effectiveness, it appears to be the best metric available, as objective qualitative metrics of international security forces remain scarce. Few objective qualitative metrics on security forces and judicial systems have been applied globally. Existing performance metrics combine other factors influencing the legitimacy of the security force with performance, e.g., human rights violations and brutality. The Failed State Index produced by the Fund for Peace, for example, incorporates the influence of corruption, availability of weapons, professionalism, and the presence of private armies into their metric (The Failed State Index 2011). Speaking directly to the challenges of quantifying the quality of security forces, the United Nations Office on Drugs and Crime and European Institute for Crime Prevention and Control state that an overall assessment would necessarily mean an in-depth look at the criminal justice systems of the different countries in theory and practice. And even with sufficient knowledge on all criminal justice systems of the world, it would be a very ambitious task to translate this knowledge into a handy performance index, allowing for a ranking of countries based on the quality of criminal justice performance (Harrendorf et al. 2010). While the factors considered by the Failed State Index are helpful to establish a gestalt of a state’s security sector, the influence of health on the security system can be better evaluated through quantitative metrics that describe police performance as

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a behavior. Police productivity provides a metric quantifying security force behavior which can be analyzed in varying levels of medical support. The level of risk confronting security force personnel in the performance of their duties shapes the impact of efforts to mitigate risks. When risks are low, the perceived benefits of the moderator, in this case health support, may not be fully appreciated or factored into the individual’s behaviors. As risks increase, the perceived utility and impact of the moderator may emerge and directly impact behavior. If risks increase significantly, there exists a potential level of risk in which the moderator may not provide enough support to buffer the risks and lose its effect to modify behavior. For this analysis, a state’s level of violence represents the risk security officers must face in the performance of their duties. Levels of violence, as a metric, can be used in two different ways: as an independent variable or as a dependent variable. As an independent variable, violence creates strain on those working to reduce it. In areas with higher levels of violence, security forces face a higher risk of injury during the performance of their duties. As a dependent variable, violence is a by-product of delivered security and measures the quality of security delivery. As Nelson Mandela writes in the forward of the WHO’s 2002 “World Report on Violence and Health,” Nelson Mandela writes, “[violence] thrives in the absence of democracy, respect for human rights and good governance” (World Report on Violence and Health: Summary 2002). Used in this manner, violence could be used as a metric of a security force’s effectiveness. In this analysis, violence was used primarily as an independent variable to investigate the central hypothesis of this study. In specific sections, violence was used as a dependent variable to consider the quality of both security and medical support.

2.8.3.1

Initial Limitations

This analysis had several limitations that initially shaped the methodology and results. As no prior empirical or quantitative research has attempted to establish the relationship between security health support and police effectiveness, scarcity of prospective or empiric data required the use of reasonable indicators to quantify both dependent and independent variables, which allowed a reasonable appraisal of the hypothesis. While a causal relationship cannot be made with certainty, the goal of this paper is to illuminate the impact of health support on security force effectiveness and argue for additional emphasis on security force health development as the United Statespursues its strategic initiatives to strengthen the security forces of its partner nations.

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2.8.3.2

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Database

The majority of the data populating the study’s database was drawn from the World Health Organization’s (WHO) data registry and the International Statistics on Crime and Justice (ISCJ) report by the United Nations office on Drugs and Crime and the European Institute for Crime Prevention and Control (HEUNI). The WHO data registry collects and reports public health data on all WHO member countries and provides an extensive dataset for analysis (Global Health Observatory Data Repository 2012). All health-related information to include development metrics, economic factors, and disease rates are collected longitudinally. In 2008, the WHO published the Global Burden of Disease: 2004 Update (GBD) and included its data as part of the WHO data registry. As an update to data presented in 2002, the GBD summarized the impact of disease in its 192 member states in 2004. Drawing upon the resources of the WHO and international organizations to collect and verify this data, the report presented the data in normalized, age-adjusted metrics which adjusted for population distributions, allowing for comparison between states. As the wounds of conflict and instability consume health care resources, GBD provided metrics on the health impacts of war and violence throughout the world in 2004 (The Global Burden of Disease: 2004 Update 2008). Additional data from the WHO data registry data on health for 2004 was extracted to populate the database. When data was not available for 2004, data in close proximity to 2004 was used. The presence of a formal pre-hospital medical system (PHMS), for example, was only available for 2007. The ISCJ reports on the crime and criminal justice productivity of all United Nations member states (Harrendorf et al. 2010). Published in 2010, the report provides key metrics on police, prosecution, and detention capacity. From this dataset, information on police density and productivity was extracted and added to the database. The majority of the data reported for individual states covered the years 2004–2006. The data for some countries, however, falls outside this range or is not listed. As defined later, the metric for police productivity is a compilation of ratios that quantify the activity of the state’s criminal justice system. Missing information in the database was obtained, if available, through opensource documents, such as the US State Department website or the US embassy website responsible for the country of interest. The information gained from these sites, however, provided current evidence on the presence of a PHMS (information on the presence of a PHMS in these countries in 2004 was not available). In these situations, the current data represented the best available data and was used in the database. Once compiled, the database excluded states undergoing large armed conflicts from 2002 to 2004 to remove the bias of wars and post-conflict reconstruction on the data. Using the Uppsala Conflict Data Program/Centre for the Study of Civil Wars, International Peace Research Institute, Oslo (UCDP/PRIO) Armed Conflict Dataset v.4-2012, 1946–2011, countries with more than 1000 deaths/year from 2002 to 2004 were omitted (Themner and Wallensteen 2012). Counties with low-intensity conflict

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in 2004 remained in the dataset in order to evaluate the impact of increased violence on police performance with different levels of medical support. Using these criteria, nine countries were removed from the database: Nepal, Colombia, Sudan, Uganda, India, Liberia, Iraq, Russia, and Burundi. In addition, the following three states were excluded due to excessive war-related disability and death (greater than 1000 War Age-Standardized Disability Years per 100,000 people as reported in the GBD): Somalia, Democratic Republic of Congo, and the former Yugoslav Republic of Macedonia. The Democratic People’s Republic of Korea was excluded due to a paucity of data. Once completed, the database included 179 countries with populations ranging from 2000 to 1.3 billion and accounted for 4.9 billion people. Metrics on the quality of the PHMS, its penetration into rural areas, and the use of dedicated medical support outside of the civilian medical support system was unavailable. Summary information is provided in Table 2.16. Figure 2.45 graphs police productivity as a function of violence.

2.8.3.3

Definitions of Terms and Variables

Variables and metrics used in the database were defined as follows: Age-standardized, Disability-Adjusted Life Year (DALY): The DALY computes the burden of a disease process by computing the “. . .years of life lost from premature death and years of life lived in less than full health. . .” as a result of a specific disease (The Global Burden of Disease: 2004 Update 2008). While the GBD provided DALY in several formats, this study used the age-standardized metric. The Table 2.16 Database metrics

Item Total states included Total states excluded States with PMHS States without PMMS States with police productivity data States without police productivity data Item Median Population (thousands) 5799 All cause mortality DALY 19,032 (per 100,000) Violence DALY (per 236 100,000) War DAYL (per 100,000) 16 Burden of violence (%) 0.99% Police productivity (ratio) 0.077 Under five-mortality (death 25 per 1000 live)

Minimum 2 8013

Amount 179 14 140 39 91 88 Maximum 1,312,433 82,801

8

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0 0.09% 0 3

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Police Producvity (combined rao)

Police Producvity vs. Violence Levels 1 0.9 0.8 States with PHMS

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Fig. 2.45 Scatterplot graph of violence levels and police productivity for states

age-standardized DALY accounts for rates of disease by age and gender according to the WHO World Standard Population (The Global Burden of Disease: 2004 Update 2008). This standardization allows comparison of states with different population age densities (Ahmad et al. 2001). Violence (Age-standardized DALYs/100,000 of population): The number of DALYs attributed to violence per 100,000 people. War (Age-standardized DALYs/100,000 of population): The number of DALYs attributed to war conflict per 100,000 people. All-cause Mortality/Morbidity (Age-standardized DALYs/100,000 of population): The number of DALYs attributed to all diseases or disease processes to include conflict and violence, experienced by the population that affect their health and wellbeing. Burden of Violence (BOV): Burden of Violence represents the fraction of DALY attributed to violence in relation to the total disease burden (All-cause Mortality/ Morbidity). Pre-hospital Medical System (PHMS): Pre-hospital trauma care has begun to gain the attention of the international health community as an important element of essential medical services (Sasser et al. 2005). Pre-hospital medical systems connect the community to their medical system by responding to injuries and illness outside of the hospital, providing initial resuscitative care, stabilizing for transport, and moving the patients to the hospital for definitive care. Without a PHMS, patients must be brought to the hospital before receiving any medical treatment. While military forces often have expeditionary medical support to provide medical care in austere or deployed settings, police forces most commonly rely on the civilian medical system for emergency care. In 2007, the WHO collected self-reported data from member states on the presence of a formal publicly available pre-hospital care system in their country. Binomial data (yes or no) was provided by each state.

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For states without a reported value, evidence of a current PHMS found on current state department of US embassy websites populated the database. The metric is used in this analysis as the provision of medical care supporting security forces. With a PHMS, injured security force casualties receive expedited care and dedicated transport to the hospital if they are injured while performing their duties. Without PHMS, casualties may not be able to access emergency medical care in time to preserve their life or prevent permanent disability. Police Productivity Rate: The metrics of the United Nations Office on Drugs and Crime/European Institute for Crime Prevention and Control do not claim to measure quality or “. . . imply that a system with high productivity rates performs better than a system with low productivity rates” (Harrendorf et al. 2010). The metrics provided in their “International Statistics on Crime and Justice” provide a metric linking the willingness of security force personnel to make arrests and process them through the legal system. Police productivity is expressed as a metric averaging the ratios of three sub-metrics which evaluate security sector productivity: ratio of suspects per police officer, ratio of suspects brought before a court per prosecutor, and the ratio of convictions per prosecutor. These metrics quantified the output of security forces as they exposed, investigated, and supported the prosecution of criminals. In accordance with the SAM model, this metric represents a strain/outcome to evaluate the impact of medical support on the productivity of security forces. Under-five mortality rate: The under-five mortality rate represents the probability of dying by age 5 per 1000 live births. Commonly used as a metric for the effectiveness of a state’s health system, factors influencing its value include: the resources of health and nutrition services, food security, feeding practices, levels of hygiene and sanitation, access to safe water, female illiteracy, early pregnancy, access to health services, and gender equity. As an outcome metric, the under-five mortality provides feedback on how well a state’s health system operates in general and in coordination with other ministries (The State of the World’s Children 2007). In this study, under-five mortality was used as an indicator of the quality and development of the civilian medical sector which provided PHMS service. Population (1000s): The population metric represents the de facto population of each state. The WHO calculates population data from the most recent “World Population Prospects” report produced by the United Nations Population Division. Data was extracted for the 2004 time period (World Population Prospects: 2004 Revision 2005).

2.8.3.4

Results and Discussion

Hypothesis Testing H0: Security forces will be more productive when medical care is readily available.

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95

Analysis of the database revealed that security forces were significantly more productive when a PHMS was available to support their security operations (p ¼ 0.000). This finding supports the hypothesis that medical support influences the productivity of security forces. Additional analysis illuminated the impact of risk and violence to diminish the benefits gained by providing medical support, the role of medical support to improve the quality of the security force, the quality of the database, and the importance of fielding a quality PHMS.

Level of Risk The impact of medical support to strengthen security forces appears to be related to the relative risk security forces faced in the performance of their duties. Risk of injury, as indicated by the level of violence in a country, was normalized to the population size and reported as a rate per 100,000 people. Evaluating the risk as a percentage of the total disease burden of the country provided additional information on the amount of violence in a state. By dividing the DALY for violence by the DALY for All-cause Mortality/Morbidity (total disease burden), the Burden of Violence (BOV) was calculated as a percentage of total disease burden. Analysis revealed that once the BOV was above 2.25%, the increased risk could not be mitigated by the presence of a PHMS. Above BOV levels of 2.25%, the presence of a PHMS had no impact on risk (p ¼ 0.3097). Below BOV levels of 2.25%, a PHMS continued to have a significant impact on the level of risk (p ¼ 0.008). These findings suggest that when the BOV exceeded 2.25%, the magnitude of the relative risks diminishes the mitigating effects that a PHMS can provide to security operations. These findings were further strengthened when evaluating police productivity above and below the risk level associated with a BOV of 2.25%. At BOV levels up to 2.25%, police productivity continued to be significantly better with medical support (p ¼ 0.002). Once the level of risk increased above 2.25% BOV, the presence of medical support failed to improve the productivity of security forces (p ¼ 0.9512). These findings suggest that medical support significantly improved the productivity of security forces up to a certain level. Two likely causes of this ceiling effect are increased perceived risk by the security forces and the willingness of PHMS personnel to operate in areas of increased risk. The first cause suggests that the moderating effect of a PHMS diminishes once security personnel perceive the risk of injury as greater than the benefits of providing security. To counteract this change in productivity, security forces need to enhance other moderators, e.g., send a larger force, provide better armor, and improve training. The second cause proposes that the risk of violence also influences the reliability of the PHMS. Ambulance personnel must be willing to operate in areas of violence. If medical first responders are unwilling to enter areas of increased risk, medical support for security forces will be unavailable. As this analysis has shown, without the expectation of medical support, security forces will modify their security productivity. Dedicated medical support

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units to augment security personnel in regions with high levels of risk could increase the expectation of medical support.

Quality of Security Forces As previously discussed, this analysis predominantly used violence as an independent variable representing the risk of injury security forces must face when performing their duties. Violence, however, could also be considered a dependent variable that is influenced by the effectiveness of a security force. The following discussion considers the use of violence as a dependent variable representing the outcome of effective security operations. Data from the analysis provided evidence that the presence of a PHMS might improve the quality of security force effectiveness. In the subset of countries with reported police productivity metrics, levels of violence were significantly lower for those states with a PHMS (p ¼ 0.000). This difference persisted when comparing the levels of violence for all states in the database (p ¼ 0.025). To ensure consistency, these findings were compared to other measures of security quality. As previously mentioned, the Fund for Peace Failed State Index includes an indicator on each state’s security apparatus (The Failed State Index, 2011). Extracting the assigned security score for each state in the 2006 index, values were analyzed on the availability of a PHMS. The security apparatus indicator of states with a PHMS was significantly lower (better quality) than those without a PHMS (p ¼ 0.000). These findings suggest a relationship between the presence of medical support for security forces and the quality of the security they deliver.

Missing Data Data on police productivity were available for only 50% (91/179) of the WHO member states in the database. The lack of police productivity on such a large percentage of states could bias conclusions and limit their inferential power. To evaluate the impact of missing data, violence levels between countries with and without reported police metrics were compared. Countries without police productivity metrics experienced significantly higher levels of violence (p ¼ 0.002). If violence levels reflect the influence of security quality, higher levels of violence suggests that states without police productivity metrics are less effective and of lower quality. The very act of collecting and reporting productivity measures indicates a certain minimum level of security sector development. This finding suggests that the states with reported police productivity are, in general, more effective and of better quality than those that don’t report these metrics. If medical support improves the productivity of security forces of good quality, the effects would be expected in the lower quality security forces, as well. This assumes, though, that the PHMS will be of sufficient quality to create the expectation of care for the less developed security forces.

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Quality of Pre-Hospital Medical System While not a specific goal of this study, the data provide evidence on the effects of the quality of the medical support provided to security forces. For all countries without a PHMS, violence levels were not significantly different between those that reported their security productivity and those that didn’t (p ¼ 0.530). This suggests that the medical sector and the security sector were both underdeveloped. For those with a PHMS, countries that reported their productivity levels had a significantly lower level of violence compared to those that did not (p ¼ 0.001). This finding questions the impact or quality of the medical support provided by the PHMS in countries with less developed security. While appearing to have a PHMS, the medical system may fail to provide adequate support to security operations. By comparing the violence levels between states with reported productivity levels without a PHMS to those states without reported productivity levels with a PHMS, the lack of a significant difference between these metrics supports this conclusion (p ¼ 0.221). Further evidence of the impact of the quality of the medical sector was found when comparing the under-five mortality rates, a marker of health system output and effectiveness. The under-five mortality rates of those states with assumed lesser quality security forces (no report of productivity data) with a PHMS were compared to those of better quality security forces (reported productivity data) without a PHMS and showed no significant difference (p ¼ 0.7042). In order to impact security delivery, medical systems need both a quality PHMS and an adequately developed medical sector to strengthen the effectiveness of the security forces.

2.8.3.5

Limitations

Major limitations in this investigation extended from the quality and quantity of data. As this analysis relied on data collected for other studies or research efforts, its scope and conclusions were bounded. As a large-n study, the study’s descriptive power was, by nature, retrospective in order to demonstrate the relationship between risks, medical support, and security productivity. While the data allowed some discussion and investigation on issues of quality, quality was not specifically measured or quantified. Another major limitation of the analysis was the paucity of qualitative and quantitative metrics on the quality, reliability, and capabilities of each state’s PHMS. While the concept of PHMS quality was explored in regard to its relationship to provide reliable support to security forces, the binominal nature of the PHMS data restricted further analysis. Future efforts to establish both quantitative and qualitative measures of medical support for security operations would facilitate robust analysis on the key features of a PHMS that need to be developed in order to strengthen foreign security forces.

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Case Study Conclusions

Medical support for security forces plays a key role in strengthening the effectiveness of security operations. By mitigating the strain of potential injury, medical support encourages security productivity, a key aspect of effectiveness. These benefits, however, appear to plateau once the risk of injury exceeds a threshold where security personnel are overwhelmed by the risk or where the environment precludes effective and reliable medical support. As the United States seeks to strengthen the security forces of its partner nations, special emphasis and resourcing of medical support for those forces is essential. If training partner military forces for high operations, providing dedicated medical units able to endure the high risks are essential. If training internal security forces, all health or medical development endeavors must target interventions that strengthen civilian first responder skills, evacuation capacity, and hospital trauma capacity.

References Abinales, P., & Amoroso, D. (2005). State and society in the Philippines. Lanham: Rowland and Littlefield Publishers. Ahmad, O. B., et al. (2001). Age standardization of rates: A new WHO standard. Report, GPE Discussion Paper 31. Geneva: World Health Organization. Bliese, P. D., & Castro, C. (2003). The Soldier Adaptation Model (SAM): Applications to peacekeeping research. In T. W. Britt & A. B. Adler (Eds.), The psychology of the peacekeeper: Lessons from the field (pp. 185–203). Westport: Praeger. CIA Factbook. (2012). Retrieved March 13, 2012, from https://www.cia.gov/library/publications/ the-world-factbook/geos/rp.html Coughlan, T. (2018). US Forces in Afghanistan should expect up to 500 casualties a month. The Times. Retrieved November 10, 2018, from https://www.thetimes.co.uk/article/us-forces-inafghanistan-should-expect-up-to-500-casualties-a-month-gj0shwqdzr7 Durante, J, & Fox, W. P (2015). Modeling violence in the Philippines. Journal of Mathematical Science, 2(4) Serial 5, pp 127–140. Global Health Observatory Data Repository. (2012). Retrieved September 8, 2012. Harrendorf, S., Heiskanen, M., & Malby, S. (Eds.). (2010). International statistics on crime and justice. Report, HEUNI Publication 64. Helsinki: Institute for Crime Prevention and Control and United Nations Office on Drugs and Crime. Retrieved February 2, 2018, from http://www. heuni.fi/Etusivu/Publications/HEUNIreports/1266333832841 Index Mundi. (2012). Retrieved March 13, 2012, from http://www.indexmundi.com/g/g.aspx? c¼rp&v¼66 National Statistics Coordination Board. Population statistics. Retrieved March 13, 2012. Philippines. (2012). US Department of State. Retrieved April 7, 2012. Retrieved February 27, 2019, from http://www.icasualties.org/ (2019). Sasser, S., et al. (Eds.). (2005). Prehospital trauma care systems. Report. Geneva: World Health Organization. Retrieved from http://www.who.int/violence_injury_prevention/publications/ser vices/39162_oms_new.pdf The Failed State Index. (2011). Conflict assessment indicators: The fund for peace country analysis indicators and their measures. Report. Washington, DC: Fund for Peace. Retrieved January 1, 2018.

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The Global Burden of Disease: 2004 Update. (2008). Geneva: World Health Organization Press. Retrieved from http://www.who.int/healthinfo/global_burden_disease/2004_report_update/en/ index.html The State of the Worlds Children: 2008. (2007, December). Report. New York: United Nations Children Fund (UNICEF). Retrieved from http://www.unicef.org/sowc08/docs/sowc08.pdf World Population Prospects: 2004 Revision. (2005). United Nations. World Report on Violence and Health: Summary. (2002). Geneva: World Health Organization. Retrieved May 1, 2018, from http://www.who.int/violence_injury_prevention/violence/world_ report/en/summary_en.pdf

Suggested Reading Ahmad, O. B., Boschi-Pinto, C., Lopez, A. D., Murray, C. J. L., Lozano, R., & Inouem M. (2001). Age standardization of rates: A new WHO standard. Report. GPE Discussion Paper 31. Geneva: World Health Organization. Bliese, P. D., & Castro, C. A. (2003). The Soldier Adaptation Model (SAM): Applications to peacekeeping research. In T. W. Britt & A. B. Adler (Eds.), The psychology of the peacekeeper: Lessons from the field (pp. 185–203). Westport: Praeger. Sasser, S., et al. (Eds.). (2005). Prehospital trauma care systems. Report. Geneva: World Health Organization. Retrieved from http://www.who.int/violence_injury_prevention/publications/ser vices/39162_oms_new.pdf Themnér, L., & Wallensteen, P. (2012). Armed Conflict, 1946-2011. Retrieved September 8, 2012, from, http://www.pcr.uu.se/research/ucdp/datasets/ucdp_prio_armed_conflict_dataset/

Chapter 3

Modeling by Fitting Data

Objectives 1. Understand when to use simple regression analysis. 2. Understand what correlation means and how to determine it. 3. Understand the differences between exponential and sinusoidal regression models and when to use them.

3.1

Introduction

Often military analysis in data science requires analysis of the data and in many cases the use of regression techniques. Regression is not a one-method-fits-all approach; it takes good approaches and common sense to complement the mathematical and statistical approaches used in the analysis. This chapter discusses some simple and advanced regression techniques that have been used often in the analysis of data for business, industry, and government. We also discuss methods to check for model adequacy after constructing the regression model. We also believe technology is essential to good analysis and illustrate it in our examples and case studies. Often we might want to model the data in order to make predictions or explain what is occurring within the domain of the data. Besides the models, we provide insights into the adequacy of the model through various approaches including regression ANOVA output, residual plots, and percent relative error. In general, we suggest using the following steps in regression analysis. Step 1. Enter the data (x, y) and obtain a scatterplot of the data and note the trends. Step 2. If necessary, transform the data into “y” and “x” components. Step 3. Build or compute the regression Equation. Obtain all the output. Interpret the ANOVA output for R2, F-test, P-values for coefficients. Step 4. Plot the regression function and the data to obtain a visual fit. © Springer Nature Switzerland AG 2019 W. P. Fox, R. Burks, Applications of Operations Research and Management Science for Military Decision Making, International Series in Operations Research & Management Science 283, https://doi.org/10.1007/978-3-030-20569-0_3

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Step 5. Compute the predictions, the residuals, percent relative error as described later. Step 6. Insure the predictive results pass the common sense test. Step 7. Plot the residual versus prediction to determine model adequacy. We present several methods to check for model adequacy. First, we suggest your predictions pass the “common sense” test. If not, return to your regression model as we are shown with our exponential decay model in Sect. 3.3. The residual plot is also very revealing. Figure 3.1 shows possible residual plot results where only random patterns indicate model adequacy from the residual plot perspective. Linear, curve, or fanning trend indicates a problem in the regression model (Affi and Azen 1979) have a good and useful discussion on corrective action based upon trends found. Percent relative error also provides information about how well the model

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Fig. 3.1 Patterns for residuals. (a) No pattern (b) curved pattern (c) outliers (d) fanning pattern (e) linear trend

3.2 Introduction to Correlation and Simple Linear Regression

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approximates the original values and it provides insights into where the model fits well and where it might not fit well. We define percent relative error with Eq. (3.1),

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3.2 3.2.1

ð3:1Þ

Introduction to Correlation and Simple Linear Regression Correlation of Recoil Data

First, let’s define correlation. Correlation, ρ, measures the linearity between the datasets X and Y. Mathematically, correlation is defined as follows: The correlation coefficient, Eq. (3.2), between X and Y, denoted as ρxy, is ρxy ¼

COV ðX; Y Þ E ½XY   μx μy ¼ : σx σy σxσy

ð3:2Þ

The values of correlation range from 1 to +1. The value of 1 corresponds to perfect line with a negative slope and a value of +1 corresponds to a perfect line with a positive slope. A value of 0 indicates that there is no linear relationship. We present two rules of thumb for correlation from the literature. First, from Devore (2012), for math, science, and engineering data we have the following: 0.8 < |ρ|  1.0—Strong linear relationship 0.5 < |ρ|  0.8—Moderate linear relationship |ρ|  0.05—Weak linear relationship According to Johnson (2012) for non-math, non-science and non-engineering data, we find a more liberal interpretation of ρ: 0.5 < |ρ|  1.0—Strong linear relationship 0.5 < |ρ|  0.3—Moderate linear relationship 0.1 < |ρ|  0.3—Weak linear relationship |ρ|  0.1—No linear relationship Further, in our modeling efforts we emphasize the interpretation of |ρ|  0. This can be interpreted as either no linear relationship or the existence of a nonlinear relationship. Most students and many researchers fail to pick up on the importance of the nonlinear relationship aspect of the interpretation. Calculating correlation between two (or more) variables in Excel is simple. After loading in the recoil data (Table 3.1) in Excel, we can first visualize the data in tabular format. This lets us be sure that the data is in the proper format and that there

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3

Table 3.1 Spring-recoil system

Mass (g) 50 100 150 200 250 300 350 400 450 500 550

Modeling by Fitting Data Stretch (m) 0.1 0.1875 0.275 0.325 0.4375 0.4875 0.5675 0.65 0.725 0.80 0.875

are no oddities (missing values, characters entered instead of numbers) that would cause problems. Using either rule of thumb the correlation coefficient, |ρ| ¼ 0.999272, indicates a strong linear relationship. We obtain this value, look at Fig. 3.1, and we see an excellent linear relationship with a positive correlation very close to 1. To estimate the correlation between the two columns in this dataset, we simply find the correlation coefficient, ρ. The data’s correlation coefficient is 0.9993 that is very close to 1. Visualizing the data makes this relationship easy to see and we would expect to see a linear relationship with a positive slope as shown in Figs. 3.2 and 3.3.

3.2.2

Linear Regression of Recoil Data

3.2.2.1

Simple Least Squares Regression

The method of least squares curve fitting, also known as ordinary least squares and linear regression, is simply the solution to a model that minimizes the sum of the squares of the deviations between the observations and predictions. Least squares will find the parameters of the function, f(x) that will minimize the sum of squared differences between the real data and the proposed model, shown in Eq. (3.3). Minimize

SSE ¼

m  X

  y1  f x j 2

ð3:3Þ

j¼1

For example, to fit a proposed proportionality model y ¼ kx2 to a set of data, the least squares criterion requires the minimization of Eq. (3.4). Note in Eq. (3.3), k is estimated as follows.

3.2 Introduction to Correlation and Simple Linear Regression

105

Spring-Mass Data

0.8

0.6

0.4

0.2

100

200

300

400

500

Fig. 3.2 Plot of spring-mass data

Spring data scatterplot

y

0.75

Correlation coefficient: 0.999272

0.50

0.25

100

200

300

400

500

X

Fig. 3.3 Plot of recoil spring data with correlation value

Minimize

S¼

5 h X j¼1

yi  kx2j

i2

ð3:4Þ

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3

Table 3.2 Data for y ¼ kx2

x y

0.5 0.7

1.0 3.4

Modeling by Fitting Data

1.5 7.2

2.0 12.4

2.5 20.1

Minimizing Eq. (3.4) is achieved using the first derivative, setting it equal to zero, and solving for the unknown parameter, k.  X  ds ¼ 2 x2j y j  kx2j ¼ 0: Solving for k : dk

k¼

X

 X  x2j y j = x4j : ð3:5Þ

Given the dataset in Table 3.2, we will find the least squares fit to the model, y ¼ kx2. P  P  Solving for k: k ¼ x2j y j = x4j ¼ ð195:0Þ=ð61:1875Þ ¼ 3:1869 and the model y ¼ kx2 becomes y ¼ 3.1869x2. In Chap. 4, we will discuss more fully the optimization process. The use of technology: Excel, R, MINITAB, JUMP, MAPLE, MATLAB are bit a few software packages that will perform regression. Example 1 Regression of Recoil Data We can then perform simple linear regression on this recoil data and produce tables presenting coefficient estimates and a range of diagnostic statistics to evaluate how well the model fits the data provided. Estimate 0.001537 0.03245

x (Intercept)

Std. error 1.957e  05 0.006635

t value 78.57 4.891

Pr(>|t|) 4.437e  14 0.0008579

Fitting linear model: y ~ x Observations 11

R2 0.9985

Residual std. error 0.01026

Adjusted R2 0.9984

Analysis of Variance Table

x Residuals

Df 1 9

Sum Sq 0.6499 0.0009475

Mean Sq 0.6499 0.0001053

F value 6173 NA

Pr(>F) 4.437e  14 NA

We visualize this estimated relationship by overlaying the fitted line to the spring data plot. This plot shows that the trend line estimated by the linear model fits the data quite well as shown in Fig. 3.4. The relationship between R2 and ρ is that R2 ¼ (ρ)2.

3.2 Introduction to Correlation and Simple Linear Regression

107

Spring data scatterplot with fitted line

R2 = 0.998

y

0.75

0.50

0.25

100

200

300

400

500

X

Fig. 3.4 Regression plot of spring data

3.2.3

Linear Regression of Philippines SIGACTS

Here, we attempt to fit a simple linear model to the data from the Philippines case study in Chap. 2. Estimate 1.145 113

Literacy (Intercept)

Std. error 0.4502 37.99

t value 2.543 2.975

Pr(>|t|) 0.01297 0.003903

Fitting linear model: sigacts_2008 ~ literacy Observations 80

R2 0.07656

Residual std. error 25.77

Adjusted R2 0.06472

Analysis of Variance Table

Literacy Residuals

Df 1 78

Sum Sq 4295 51,805

Mean Sq 4295 664.2

F value 6.467 NA

Pr(>F) 0.01297 NA

Linear regression is not the answer to all analysis. As seen in this example between literacy and violent events the linear regression model, Fig. 3.5, is not helpful. We will return to this example later in this chapter.

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Modeling by Fitting Data

Number of violent events

Literacy and violent events in the Philippines

100

50

R2 = 0.077

0 75

80

85

90

95

Literacy %

Fig. 3.5 Regression with literacy and violent events data

3.3 3.3.1

Exponential Decay Modeling Introducing Hospital Recovery Data from a Military Hospital

We are given data from the VA to analyze to determine recovery information. The data is provided in Table 3.3. Plotting the table of recovery data shows that once again, the structure of the data is amenable to statistical analysis. We have two columns, T (number of days in the hospital) and Y (estimated recovery index) and we want to generate a model that predicts how well a patient will recover as a function of the time they spend in the hospital. Using Excel we can compute the correlation coefficient of ρ ¼ .941. Once again, creating a scatterplot, Fig. 3.6, of the data helps us visualize how closely the estimated correlation value matches the overall trend in the data. In this example, we will demonstrate linear regression, polynomial regression, and then exponential regression in order to obtain a useful model.

3.3.2

Linear Regression of Hospital Recovery Data

It definitely appears that there is a strong negative relationship: the longer a patient spends in the hospital, the lower their recovery index. Next, we fit an OLS model to the data to estimate the magnitude of the linear relationship.

3.3 Exponential Decay Modeling

109

Table 3.3 Patient recovery time T y

2 54

5 50

7 45

10 37

14 35

19 25

26 20

31 16

34 18

338 13

45 8

52 11

53 8

60 4

65 6

Recovery index scatterplot 50

Recovery index

40 Correlation coefficient: –0.941

30

20

0

0

20

40 Days in the hospital

60

Fig. 3.6 Scatterplot of days in the hospital and recovery index

Estimate 0.7525 46.46

T (Intercept)

Std. error 0.07502 2.762

t value 10.03 16.82

Pr(>|t|) 1.736e  07 3.335e  10

Fitting linear model: Y ~ T Observations 15

R2 0.8856

Residual std. error 5.891

Adjusted R2 0.8768

Analysis of Variance Table

T Residuals

Df 1 13

Sum Sq 3492 451.2

Mean Sq 3492 34.71

F value 100.6 NA

Pr(>F) 1.736e  07 NA

OLS modeling shows that there is a negative and statistically significant relationship between time spent in the hospital and patient recovery index. However, ordinary least squares regression may not be the best choice in this case for two reasons. First, we are dealing with real-world data: a model that can produce (for example) negative estimates of recovery index is not applicable to the underlying

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3

Modeling by Fitting Data

concepts our model is dealing with. Second, the assumption of OLS, like all linear models, is that the magnitude of the relationship between input and output variables stays constant over the entire range of values in the data. However, visualizing the data suggests that this assumption may not hold—in fact, it appears that the magnitude of the relationship is very high for low values of T and decays somewhat for patients who spend more days in the hospital. To test for this phenomenon, we examine the residuals of the linear model. Residuals analysis can provide quick visual feedback about model fit and whether the relationships estimated hold over the full range of the data. We calculate residuals as the difference between observed values Y and estimated values $Y& $, or Y i  Y i . We then normalize residuals as percent relative error between the observed and estimated values, which helps us compare how well the model predicts each individual observation in the dataset (Table 3.4): The residuals plotted, Fig. 3.7, show a curvilinear pattern, decreasing and then increasing in magnitude over the range of the input variable. This means that we can likely improve the fit of the model by allowing for nonlinear effects. Furthermore, the current model can make predictions that are substantively nonsensical, even if they were statistically valid. For example, our model predicts that after 100 days in the hospital, a patient’s estimated recovery index value would be 29.79. This has no common sense, as the recovery index variable is always positive in the real world. By allowing for nonlinear terms, perhaps we can also guard against these types of nonsense predictions.

Table 3.4 Residual analysis T 2 5 7 10 14 19 26 31 34 38 45 52 53 60 65

Y 54 50 45 37 35 25 20 16 18 13 8 11 8 4 6

Index 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Predicted 44.96 42.7 41.19 38.94 35.93 32.16 26.9 23.13 20.88 17.87 12.6 7.33 6.58 1.31 2.45

Residuals 9.04 7.3 3.81 1.94 0.93 7.16 6.9 7.13 2.88 4.87 4.6 3.67 1.42 2.69 8.45

Pct_Relative_Error 16.74 14.60 8.47 5.24 2.66 28.64 34.50 44.56 16.00 37.46 57.50 33.36 17.75 67.25 140.83

These data can also be plotted to visualize how well the model fits over the range of our input variable

3.3 Exponential Decay Modeling

111

Linear model residuals by time in hospital 10

Residuals

5

0

–5

–10 0

20

40 Time in hospital

60

Fig. 3.7 Residual plot for linear model

3.3.3

Quadratic Regression of Hospital Recovery Data

Including a quadratic term modifies the model formula: Y ¼ β0 + β1x + β2x2. Fitting this model to the data produces separate estimates of the effect of T itself as well as the effect of T2, the quadratic term.

T IT^2 Intercept

Estimate 1.71 0.01481 55.82

Std. error 0.1248 0.001868 1.649

t value 13.7 7.927 33.85

Pr(>|t|) 1.087e  08 4.127e  06 2.811e  13

Fitting the linear model Y ~ Intercept + B1T+B2T2 Observations 15

R2 0.9817

Residual std. error 2.455

Adjusted R2 0.9786

Analysis of Variance Table

T IT^2 Residuals

Df 1 1 12

Sum Sq 3492 378.9 72.34

Mean Sq 3492 378.9 6.029

F value 579.3 62.84 NA

Pr(>F) 1.59e  11 4.127e  06 NA

Including the quadratic term improves model fit as measured by R2 from 0.88 to 0.98—a sizable increase. To assess whether this new input variable deals with the

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3

Modeling by Fitting Data

Table 3.5 Residual analysis of quadratic model T 2 5 7 10 14 19 26 31 34 38 45 52 53 60 65

Y 54 50 45 37 35 25 20 16 18 13 8 11 8 4 6

Index 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Predicted 52.46 47.64 44.58 40.2 34.78 28.67 21.36 17.03 14.79 12.21 8.44 6.93 6.77 6.51 7.21

Residuals 1.54 2.36 0.42 3.2 0.22 3.67 1.36 1.03 3.21 0.79 0.44 4.07 1.23 2.51 1.21

Pct_Relative_Error 2.85 4.72 0.93 8.65 0.63 14.68 6.80 6.44 17.83 6.08 5.50 37.00 15.38 62.75 20.17

Quadratic model residuals by time in hospital 10

Residuals

5

0

–5

–10 0

20

40 Days in hospital

60

Fig. 3.8 Residual plot for polynomial regression model

curvilinear trend, we saw in the residuals from the first model, we calculate and visualize the residuals from the quadratic model (Table 3.5). Visually, Fig. 3.8, evaluating the residuals from the quadratic model shows that the trend has disappeared. This means that we can assume the same relationship holds whether T ¼ 1 or T ¼ 100. However, we are still not sure if the model produces numerical estimates that pass the common sense test. The simplest way to assess this

3.3 Exponential Decay Modeling

113

is to generate predicted values of the recovery index variable using the quadratic model, and plot them to see if they make sense. To generate predicted values in R, we can pass the quadratic model object to the predict() function along with a set of hypothetical input values. In other words, we can ask the model what the recovery index would look like for a set of hypothetical patients who spend anywhere from 0 to 120 days in the hospital. We can then plot these estimates to quickly gauge whether they pass the common sense test for real-world predictive value as shown in Fig. 3.9. The predicted values curve up toward infinity, Fig. 3.9; clearly, this is a problem. The quadratic term we included in the model leads to unrealistic estimates of recovery index at larger values of T. Not only is this unacceptable for the context of our model, but it is unrealistic on its face. After all, we understand that people generally spend long periods in the hospital for serious or life-threatening conditions such as severe disease or major bodily injury. As such, we can assess that someone who spends 6 months in the hospital probably should not have a higher recovery index than someone who was only hospitalized for a day or two.

3.3.4

Exponential Decay Modeling of Hospital Recovery Data

We may be able to build a model that both accurately fits the data and produces estimates that pass the common sense test by using an exponential decay model. This modeling approach lets us model relationships that vary over time in a nonlinear fashion—in this case, we want to accurately capture the strong correlation for lower Quadratic model: predicted recovery index

Predicted recovery index

60

40

20

0

25

50 75 Days in hospital

Fig. 3.9 Polynomial regression plot (quadratic polynomial)

100

125

114

3

Modeling by Fitting Data

ranges of T, but allow the magnitude of this relationship to decay as T increases, as the data seems to indicate. Generating nonlinear models in R is done using the nonlinear least squares or NLS function, appropriately labeled nls(). This function automatically fits a wide range of nonlinear models based on a functional form designated by the user. It is important to note that when fitting an NLS model in R, minimizing the sum of n P squares ðyi  aðexpðbxi ÞÞÞ2 is done computationally rather than mathematically. i¼1

That means that the choice of starting values for the optimization function is important—the estimates produced by the model may vary considerably based on the chosen starting values (Fox 2012). As such, it is wise to experiment when fitting these nonlinear values to test how robust the resulting estimates are to the choice of starting values. We suggest using a ln-ln transformation of this data to begin with and then transforming back into the original xy space to obtain “good” estimates. The model, ln(y) ¼ ln(a) + bx, yields ln(y) ¼ 4.037159  0.03797 x. This translates into the estimated model: y ¼ 56.66512e(.03797x). Our starting values for (a, b) should be (56.66512, 0.03797). This starting value can be found by performing linear regression on a ln-ln transformation of the model and converting back to the original space (Fox 2012). Fitting nonlinear regression model: Y ~ a  (e(b  T)) Parameter Estimates a 58.61

b .03959

Residual sum of squares: 1.951

The final model is y ¼ 58.61e0.03959x. Overlaying the trend produced by the model on the plot of observed values, Fig. 3.10, we see that the NLS modeling approach fits the data very well. Once again, we can visually assess model fit by calculating and plotting the residuals. The Fig. 3.11a, b show the same residuals plotted along both days in the hospital T and recovery index Y (Table 3.6). In both cases, Fig. 3.11a, b, we see that there is no easily distinguishable pattern in residuals. Finally, we apply the common sense check by generating and plotting estimated recovery index values for a set of values of T from 1 to 120. The predicted values generated by the exponential decay model make intuitive sense. As the number of days a patient spends in the hospital increases, the model predicts that their recovery index will decrease at a decreasing rate. This means that while the recovery index variable will continuously decrease, it will not take on negative values (as predicted by the linear model) or explosively large values (as predicted by the quadratic model). It appears that the exponential decay model not only fit the data best from a purely statistical point of view, but also generates values that pass the common sense test to an observer or analyst shown in Fig. 3.12.

3.4 Sinusoidal Regression

115

Exponential decay model: predicted recovery index

Recovery index

50 40 30 20 10

0

20

40 Days in hospital

60

Fig. 3.10 Exponential regression model and data

3.4 3.4.1

Sinusoidal Regression Introducing Military Supply Shipping Data

Consider a situation where we have shipping data that we need to model to estimate future results (Table 3.7). First, we obtain the correlation, ρ ¼ 0.6725644. Once again, we can visualize the data in a scatterplot to assess whether this positive correlation is borne out by the overall trend. Visualizing the data, Fig. 3.13, we see that there is a clear positive trend over time in shipping usage. However, examining the data in more detail suggests that a simple linear model may not be best-suited to capturing the variation in these data. One way to plot more complex patterns in data is through the use of a trend line using polynomial or non-parametric smoothing functions (Fig. 3.14). Plotting a trend line generated via a spline function shows that there seems to be an oscillating pattern with a steady increase over time in the shipping data.

3.4.2

Linear Regression of Shipping Data

As a baseline for comparison, we begin by fitting a standard OLS regression model using the lm() function in R.

116

3

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Exponential decay model residuals by days in hospital 5.0

Residuals

2.5

0.0

–2.5

–5.0 0

20

40

60

Time in hospital Exponential decay model residuals by recovery index 5.0

Residuals

2.5

0.0

–2.5

–5.0 10

20

30 Recovery index

40

50

Fig. 3.11 (a) Residual plot time in Hospital, (b) Residual plot recovery index ## Generate model shipping_model1 |t|) 0.001158 4.907e  06

3.4 Sinusoidal Regression

117

Table 3.6 Residual analysis of exponential model T 2 5 7 10 14 19 26 31 34 38 45 52 53 60 65

Y 54 50 45 37 35 25 20 16 18 13 8 11 8 4 6

Index 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Predicted 52.46 47.64 44.58 40.2 34.78 28.67 21.36 17.03 14.79 12.21 8.44 6.93 6.77 6.51 7.21

Residuals 0.14 1.92 0.58 2.44 1.34 2.62 0.93 1.17 2.75 0.01 1.86 3.52 0.81 1.45 1.53

Pct_Relative_Error 0.26 3.84 1.29 6.59 3.83 10.48 4.65 7.31 15.28 0.08 23.25 32.00 10.13 36.25 25.50

Predicted recovery index

Exponential decay model: predicted recovery index

40

20

0 0

25

50 75 Days in hospital

100

125

Fig. 3.12 Plot of exponential regression model

Fitting linear model: UsageTons ~ Month Observations 20

Residual std. error 5.079

R2 0.4523

Adjusted R2 0.4219

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3

Table 3.7 Usage Tons of shipping

Modeling by Fitting Data

Month 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

UsageTons 20 15 10 18 28 18 13 21 28 22 19 25 32 26 21 29 35 28 22 32

Shipping data scatterplot 40 Correlation coefficient: 0.673

Usage (tons)

30

20

10

0 5

10 Month

15

20

Fig. 3.13 Scatterplot of shipping data

Analysis of Variance Table

Month Residuals

Df 1 18

Sum Sq 383.5 464.3

Mean Sq 383.5 25.79

F value 14.87 NA

Pr(>F) 0.001158 NA

3.4 Sinusoidal Regression

119 Shipping data spline plot

40 Correlation coefficient: 0.673 Usage (tons)

30

20

10

0 5

10 Month

15

20

Fig. 3.14 Shipping data with data points connected show an oscillating trend

While the linear model, y ¼ 15.13 + 0.7954 x, fits the data fairly well, the oscillation identified by the spline visualization suggests that we should apply a model that better fits the seasonal variation in the data.

3.4.3

Sinusoidal Regression of Shipping Data

R, as well as other software, treats sinusoidal regression models as part of the larger family of nonlinear least squares (NLS) regression models. This means that we can fit a sinusoidal model using the same nls() function and syntax as we applied earlier for the exponential decay model. The functional form for the sinusoidal model we use here can be written as: Usage ¼ a  sin ðb  time þ cÞ þ d  time þ e This function can be expanded out trigonometrically as: Usage ¼ a  time þ b  sin ðc  timeÞ þ d  cos ðcðtimeÞÞ þ e This Equation can be passed to nls() and R will computationally assess best-fit values for the a, b, c, d, and e terms. It is worth stressing again the importance of selecting good starting values for this process, especially for a model like this one with many parameters to be simultaneously estimated. Here, we set starting values

120

3

Modeling by Fitting Data

based on pre-analysis of the data. It is also important to note that because the underlying algorithms used to optimize these functions differ between Excel and R, the two methods produce models with different parameters but nearly identical predictive qualities. The model can be specified in R as follows. ## Generate model shipping_model2 0.

4.2.2.3

Constraints

(a) limit on amount lent x1 þ x2 þ x3 þ x4  250 (b) policy condition 1 x1  0:55ðx1 þ x2 Þ (c) i.e., first mortgages 0.55(total mortgage lending) and also x1  0:25ðx1 þ x2 þ x3 þ x4 Þ (d) i.e., first mortgages >0.25(total loans) (e) policy condition 2 x2  0:25ðx1 þ x2 þ x3 þ x4 Þ

4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

144

(f) policy condition 3—we know that the total annual interest is 0.14x1 + 0.20x2 + 0.20x3 + 0.10x4 on total loans of (x1 + x2 + x3 + x4). Hence, the constraint relating to policy condition (3) is 0:14x1 þ 0:20x2 þ 0:20x3 þ 0:10x4  0:15ðx1 þ x2 þ x3 þ x4 Þ

4.2.2.4

Objective Function

To maximize interest income (which is given above), i.e., Maximize Z ¼ 0:14x1 þ 0:20x2 þ 0:20x3 þ 0:10x4

4.2.3

Blending and Formulation Problem

Consider the example of a manufacturer of animal feed who is producing feed mix for dairy cattle. In our simple example, the feed mix contains two active ingredients. One kilogram of feed mix must contain a minimum quantity of each of four nutrients as below: Nutrient Gram

A 90

B 50

C 20

D 2

The ingredients have the following nutrient values and cost:

Ingredient 1 (g/kg) Ingredient 2 (g/kg)

A 100 200

B 80 150

C 40 20

D 10 0

Cost/kg 40 60

What should be the amount of active ingredients in 1 kg of feed mix that minimizes cost?

4.2.3.1

Blending Problem Solution

Variables In order to solve this problem, it is best to think in terms of 1 kg of feed mix. That kilogram is made up of two parts—ingredient 1 and ingredient 2: x1 ¼ amount (kg) of ingredient 1 in 1 kg of feed mix

4.2 Formulating Mathematical Programming Problems

145

x2 ¼ amount (kg) of ingredient 2 in 1 kg of feed mix, where x1 > 0, x2 > 0 Essentially, these variables (x1 and x2) can be thought of as the recipe telling us how to make up 1 kg of feed mix.

Constraints • nutrient constraints 100x1 þ 200x2  90 ðnutrient AÞ 80x1 þ 150x2  50 ðnutrient BÞ 40x1 þ 20x2  20 ðnutrient CÞ 10x1  2 ðnutrient DÞ • balancing constraint (an implicit constraint due to the definition of the variables) x1 þ x2 ¼ 1

Objective Function Presumably to minimize cost, i.e., Minimize Z ¼ 40x1 þ 60x2 This gives us our complete LP model for the blending problem.

4.2.4

Production Planning Problem

A company manufactures four variants of the same table and in the final part of the manufacturing process there are assembly, polishing, and packing operations. For each variant, the time required for these operations is shown in Table 4.5 (in minutes) as is the profit per unit sold. • Given the current state of the labor force the company estimate that, each year, they have 100,000 min of assembly time, 50,000 min of polishing time, and 60,000 min of packing time available. How many of each variant should the company make per year and what is the associated profit?

146

4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

Table 4.5 Time and profit per variant

4.2.4.1

Variant 1 2 3 4

Assembly 2 4 3 7

Polish 3 2 3 4

Pack 2 3 2 5

Profit ($) 1.50 2.50 4.00 4.50

Variables

Let: xi be the number of units of variant i (i ¼ 1, 2, 3, 4) made per year, where xi > 0 i ¼ 1,2,3,4

4.2.4.2

Constraints

Resources for the operations of assembly, polishing, and packing 2x1 þ 4x2 þ 3x3 þ 7x4  100,000 ðassemblyÞ 3x1 þ 2x2 þ 3x3 þ 4x4  50,000 ðpolishingÞ 2x1 þ 3x2 þ 2x3 þ 5x4  60,000 ðpackingÞ

4.2.4.3

Objective Function Maximize Z ¼ 1:5x1 þ 2:5x2 þ 4:0x3 þ 4:5x4

4.2.5

Shipping Problem

Consider planning the shipment of needed items from the warehouses, where they are manufactured and stored to the distribution centers where they are needed as shown in the introduction. There are three warehouses at different cities: Detroit, Pittsburgh, and Buffalo. They have 250, 130, and 235 tons of paper accordingly. There are four publishers in Boston, New York, Chicago, and Indianapolis. They ordered 75, 230, 240, and 70 tons of paper to publish new books. Table 4.6 provides the costs in dollars of transportation of 1 ton of paper. Management wants you to minimize the shipping costs while meeting demand. We define xij to be the travel from city i (1 is Detroit, 2 is Pittsburg, 3 is Buffalo) to city j (1 is Boston, 2 is New York, 3 is Chicago, and 4 is Indianapolis).

4.2 Formulating Mathematical Programming Problems

147

Table 4.6 Transportation cost ($) per 1 ton of paper From/To Detroit (DT) Pittsburgh (PT) Buffalo (BF)

Boston (BS) 15 25 15

New York (NY) 20 13 15

Chicago (CH) 16 5 7

Indianapolis (IN) 21 11 17

Minimize Z ¼

15x11 þ 20x12 þ 16x13 þ 21x14 þ 25x21 þ 13x22 þ 5x23 þ 11x24 þ 15x31 þ 15x32 þ 7x33 þ 17x34

Subject to : x11 þ x12 þ x13 þ x14  250 ðavailability in Detroit Þ x21 þ x22 þ x23 þ x24  130 ðavailability in PittsburgÞ x31 þ x32 þ x33 þ x34  235 ðavailability in BuffaloÞ x11 þ x21 þ x31  75 ðdemand BostonÞ x12 þ x22 þ x32  230 ðdemand New York Þ x13 þ x23 þ x334  240 ðdemand ChicagoÞ x14 þ x24 þ x34  70 ðdemand IndianapolisÞ xij  0

4.2.5.1

Integer Programming and Mixed-Integer Programming

For integer and mixed integer programming, we will take advantage of technology. We will not present the branch and bound technique but we suggest that a thorough review of the topic can be found in Winston or other similar mathematical programming texts. Perhaps in Example 5, shipping, we decide that all shipment must be integer shipment and no partial shipments are allowed. That would cause us to solve Example 5 as an integer programming problem. Assignment problems, transportation problems, and assignments with binary constraints are among the most used integer and binary integer problems.

4.2.5.2

Nonlinear Programming

It is not our plan to present material on how to formulate or solve nonlinear programs. Often, we have nonlinear objective functions or nonlinear constraints. Suffice it to say, we will recognize these and use technology to assist in the solution. Excellent nonlinear programming information, methodology, and algorithms can be gained from our recommended suggested reading. Many problems are in fact, nonlinear. We will provide a few examples later in the chapter. We point out that

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4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

often numerical algorithms such as one-dimensional Golden section or two-dimensional gradient search methods are used to solve nonlinear problems.

4.2.5.3

Exercises 4.2

Formulate the following problems: 1. Modify the 3D printing problem as follows: A supply company wants use a 3D printer to produce parts as needed. It takes 3 h to print A, and it takes 2 h to label it correctly. It takes 3 h to print part B, and it takes 2.5 h to label it correctly. The supply company saves 15 h by printing A and 18 h by printing B in the field. Given that we have 40 h to devote to printing the parts and 35 h to devote to labeling the parts per day, how parts of each should be printed to maximize the time savings? 2. The Mariners Company wishes to repair make three models of ships to maximize their profits. They found that a model steamship takes the cutter 1 h, the painter 2 h, and the assembler 4 h of work; it produces a profit of $6.00. The sailboat takes the cutter 3 h, the painter 3 h, and the assembler 2 h. It produces a $4.00 profit. The submarine takes the cutter 1 h, the painter 3 h, and the assembler 1 h. It produces a profit of $2.00. The cutter is only available for 45 h per week, the painter for 50 h, and the assembler for 60 h. Assume that they sell all the ships that they make, formulate this LP to determine how many ships of each type that Mariners should produce. 3. In order to produce 1000 tons of non-oxidizing steel for engine valves, at least the following units of manganese, chromium, and molybdenum, will be needed weekly: 10 units of manganese, 12 units of chromium, and 14 units of molybdenum (1 unit is 10 lb). These materials are obtained from a dealer who markets these metals in three sizes small (S), medium (M), and large (L). One S case costs $9 and contains two units of manganese, two units of chromium, and one unit of molybdenum. One M case costs $12 and contains two units of manganese, three units of chromium, and one unit of molybdenum. One L case costs $15 and contains one unit of manganese, one units of chromium, and five units of molybdenum. How many cases of each kind (S, M, L) should be purchased weekly so that we have enough manganese, chromium, and molybdenum at the smallest cost? 4. The Recruiting headquarters hired an Advertising agency wishes to plan an advertising campaign in three different media—television, radio, and magazines. The purpose or goal is to reach as many potential customers as possible. Results of a marketing study are given in Table 4.7. The company does not want to spend more than $800,000 on advertising. It further requires (1) at least two million exposures take place among woman; (2) TV advertising be limited to $500,000; (3) at least three advertising units be bought on day time TV and two units on prime time TV, and (4) the number of radio and magazine advertisement units should each be between five and ten units.

4.2 Formulating Mathematical Programming Problems

149

Table 4.7 Advertising costs Day time TV $40,000 400,000

Prime time, TV $75,000 900,000

Radio $30,000 500,000

Magazines $15,000 200,000

300,000

400,000

200,000

100,000

Moody’s quality scale Aa

Bank’s quality scale 2

Years to maturity 9

Yield at maturity 4.3%

After-tax yield 4.3%

Aa Aaa Aaa Ba

2 1 1 5

15 4 3 2

5.4% 5% 4.4% 4.5%

2.7% 2.5% 2.2% 4.5%

Cost of advertising unit Number of potential customers reached per unit Number of woman customers reached per unit Table 4.8 Portfolio investment options Bond name A B C D E

Bond type MUNICIPAL AGENCY GOVT 1 GOVT 2 LOCAL

5. The mess hall is ordering food for the next month They orders meat for meatloaf (mixed ground beef, pork, and veal) for 1000 lb according to the following specifications: (a) (b) (c) (d)

Ground beef is to be no less than 400 lb and no more than 600 lb. The ground pork is to between 200 and 300 lb. The ground veal must weigh between 100 and 400 lb. The weight of the ground pork must be no more than one and one half (3/2) times the weight of the veal. The contract calls for the mess hall to pay $1200 for the meat. The cost per pound for the meat is: $0.70 for hamburger, $0.60 for pork, and $0.80 for the veal. How can this be modeled?

6. Portfolio Investments A portfolio manager in charge of a bank wants to invest $10 million. The securities available for purchase, as well as their respective quality ratings, maturate, and yields, are shown in Table 4.8. The Bank places certain policy limitations on the portfolios manager’s actions: (a) Government and Agency Bonds must total at least $4 million. (b) The average quality of the portfolios cannot exceed 1.4 on the Bank’s quality scale. Note a low number means high quality. (c) The average years to maturity must not exceed 5 years. Assume the objective is to maximize after-tax earnings on the investment. 7. Suppose a newspaper publisher must purchase three kinds of paper stock. The publisher must meet their demand but desire to minimize their costs in the process. They decide to use an Economic Lot Size model to assist them in their

4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

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decisions. Given an Economic Order Quantity Model (EOQ) with constraints where the total cost is the sum of the individual quantity costs: C ðQ1 ; Q2 ; Q3 Þ ¼ C ðQ1 Þ þ CðQ2 Þ þ CðQ3 Þ C ðQi Þ ¼ ai di =Qi þ hi Qi =2 where d is the order rate h is the cost per unit time (storage) Q/2 is the average amount on hand a is the order cost The constraint is the amount of storage area available to the publisher so that he can have the three kinds of paper on hand for use. The items cannot be stacked, but can be laid side by side. They are constrained by the available storage area, S. The following data is collected:

d a h s

TYPE I 32 rolls/week $25 $1/roll/week 4 sq ft/roll

TYPE II 24 $18 $1.5 3

TYPE III 20 $20 $2.0 2

You have 200 sq ft of storage space available. Formulate the problem 8. Suppose, you want to use the Cobb-Douglass function P(L,K) ¼ ALaKb to predict output in thousands, based upon amount of capital and labor used. Suppose you know the price of capital and labor per year is $10,000 and $7000 respectively. Your company estimates the values of A as 1.2, a ¼ 0.3 and b ¼ 0.6. Your total cost is assumed to be T ¼ PL  L + Pk  k, where PL and Pk are the price of capital and labor. There are three possible funding levels: $63,940, $55,060, or $71,510. Formulate the problem to determine which budget yields the best solution for your company. 9. The manufacturer of a new plant is planning the introduction of two new products, a 19-in. stereo color set with a manufacturer’s suggested retail price (MSRP) of $339 and a 21-in. stereo color set with a MSRP of $399. The cost to the company is $195 per 19-in. set and $225 per 21-in. set, plus an additional $400,000 in fixed costs of initial parts, initial labor, and machinery. In a competitive market in which they desire to sell the sets, the number of sales per year will affect the average selling price. It is estimated that for each type of set, the average selling price drops by one cent for each additional unit sold. Furthermore, sales of 19-in. sets will affect the sales of 21-in. sets and viceversa. It is estimated that the average selling price for the 19-in. set will be reduced by an additional 0.3 cents for each 21-in. set sold, and the price for the 21-in. set will decrease by 0.4 cents for each 19-in. set sold. We desire to provide

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151

them the optimal number of units of each type set to produce and to determine the expected profits. Recall Profit is revenue minus cost, P ¼ R  C. Formulate the model to maximize profits. Insure that you have accounted for all revenues and costs. Define all your variables. 10. Let us assume that a company has the potential to produce any number of TV sets per year. Now we realize that there is a limit on production capacity. Consideration of these two products came about because the company plans to discontinue manufacturing of its black-and-white sets, thus providing excess capacity at its assembly plants. This excess capacity could be used to increase production of other existing product lines, but the company feels that these new products will be more profitable. It is estimated that the available production capacity will be sufficient to produce 10,000 sets per year (about 200 per week). The company has ample supply of 19-in. and 21-in. color tubes, chassis, and other standard components; however, circuit assemblies are in short supply. Also the 19-in. TV requires different circuit assemblies than the 21-in. TV. The supplier can deliver 8000 boards per year for the 21-in. model and 5000 boards per year for the 19-in. model. Taking this new information into account, what should the company now do? Formulate this problem.

4.3

Graphical Linear Programming

Many applications in business and economics involve a process called optimization. In optimization problems, you are asked to find the minimum or the maximum result. This section illustrates the strategy in graphical simplex of linear programming. We will restrict ourselves in this graphical context to two-dimensions. Variables in the simplex method are restricted to positive variables (for example x > 0). A two-dimensional linear programming problem consists of a linear objective function and a system of linear inequalities called resource constraints. The objective function gives the linear quantity that is to be maximized (or minimized). The constraints determine the set of feasible solutions. Understanding the two-variable case helps the understanding of more complicated programming problems. Let’s illustrate a two-variable example. Example 1. Helping Victims of a Disaster or War Packages of food and clothing are being sent to assist victims in a disaster. Carriers will transport the packages, provided they fit in the available cargo space. Each 20-cu. ft. box of food weighs 40 lb, and each 30-cu. ft. box of clothing weighs 20 lb. The total weight cannot exceed 16,000 lb, and the total volume must not exceed 18,000 cu. ft. Each carton of food will feed ten people, while each carton of clothing will help put clothes on eight people. How many packages of food and how many packages of clothing should be sent in order to maximize the number of people assisted? How many people will be assisted? x1 ¼ number of boxes of food to send

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Table 4.9 Data for boxes Weight Space Benefit

Food 40 20 10

Clothes 20 30 8

Quantity available 16,000 18,000

x2 ¼ number of boxes of clothes to send The military expects a benefit of helping ten for each food box and eight for each clothes box. Table 4.9 has the technical data elements. The constraint information from the table becomes inequalities that are written mathematically as: 40x1 þ 20x2  16000ðweightÞ 20x1 þ 30x2  18000ðspace in cubic feetÞ x1  0,x2  0 The benefit equation is: Benefit Z ¼ 10x1 þ 8x2

4.3.1

The Feasible Region

We use the constraints of the linear program, 40x1 þ 20x2  16000ðweightÞ 20x1 þ 30x2  18000ðspace in cubic feetÞ x1  0,x2  0 The constraints of a linear program, which include any bounds on the decision variables, essentially shape the region in the x-y plane that will be the domain for the objective function prior to any optimization being performed. Every inequality constraint that is part of the formulation divides the entire space defined by the decision variables into two parts: the portion of the space containing points that violate the constraint, and the portion of the space containing points that satisfy the constraint. It is very easy to determine which portion will contribute to shaping the domain. We can simply substitute the value of some point in either half-space into the constraint. Any point will do, but the origin is particularly appealing. Since there’s only one origin, if it satisfies the constraint, then the half-space containing the origin will contribute to the domain of the objective function.

4.3 Graphical Linear Programming

153

When we do this for each of the constraints in the problem, the result is an area representing the intersection of all the half-spaces that satisfied the constraints individually. This intersection is the domain for the objective function for the optimization. Because it contains points that satisfy all the constraints simultaneously, these points are considered feasible to the problem. The common name for this domain is the feasible region. Consider our constraints: 40x1 þ 20x2  16000ðweightÞ 20x1 þ 30x2  18000ðspace in cubic feetÞ x1  0,x2  0 For our graphical work we use the constraints: x1 > 0, x2 > 0 to set the region. Here, we are strictly in the x1–x2 plane (the first quadrant). Let’s first take constraint #1 (weight) in the first quadrant: 40x1 + 20x2 < 16000 shown in Fig. 4.1 First, we graph each constraint as equality, one at a time. We choose a point, usually the origin to test the validity of the inequality constraint. We shade all the areas where the validity holds. We repeat this process for all constraints to obtain Fig. 4.2. Figure 4.2 shows a plot of (1) the assembly hour’s constraint and (2) the installation hour’s constraint in the first quadrant. Along with the non-negativity restrictions on the decision variables, the intersection of the half-spaces defined by these constraints is the feasible region shown in red. This area represents the domain for the objective function optimization. We region shaded in our feasible region. Fig. 4.1 Shaded inequality for weight

1000

800

600 x2 400

200

0 0

200

400

600 x1

800

1000

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4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

Fig. 4.2 Plot of (1) the weight constraint and (2) the space constraint in the first quadrant

1000

800

600 x2 400

200

0 0

200

400

600

800

1000

x1

4.3.2

Solving a Linear Programming Problem Graphically

We have decision variables defined and an objection function that is to be maximized or minimized. Although all points inside the feasible region provide feasible solutions the solution, if one exists, occurs according to the Fundamental Theorem of Linear Programming: If the optimal solution exists, then it occurs at a corner point of the feasible region. Notice the various corners formed by the intersections of the constraints in example. These points are of great importance to us. There is a cool theorem (didn’t know there were any of these, huh?) in linear optimization that states, “if an optimal solution exists, then an optimal corner point exists.” The result of this is that any algorithm searching for the optimal solution to a linear program should have some mechanism of heading toward the corner point where the solution will occur. If the search procedure stays on the outside border of the feasible region while pursuing the optimal solution, it is called an exterior point method. If the search procedure cuts through the inside of the feasible region, it is called an interior point method. Thus, in a linear programming problem, if there exists a solution, it must occur at a corner point of the set of feasible solutions (these are the vertices of the region). Note that in Fig. 4.2 the corner points of the feasible region are the four coordinates and we might use algebra to find these: (0,0), (0,600) (400, 0), and (150,500). How did we get the point (150,500)? This point is the intersection of the lines: 40x1 + 20x2 ¼ 16000 and 20x1 + 30x2 ¼ 18000. We use matrix algebra and solve for (x1,x2) from

4.3 Graphical Linear Programming Table 4.10 Coordinates and corresponding Z-values

155 Z ¼ 10x1 + 8x2. Z¼0 Z ¼ 4800 Z ¼ 5500 Z ¼ 4000 Z ¼ 5500

Coordinate of corner point (0,0) (0,600) (150,500) (400,0) Best solution is (150,500)



40 20 20 30



x1 x2



 ¼

16000 18000



Now, that we have all the possible solution coordinates for (x1,x2), we need to know which is the optimal solution. We evaluate the objective function at each point and choose the best solution. Our objective function is to Maximize Z ¼ 10x1 + 8x2. We can set up a table of coordinates and corresponding Z-values as shown in Table 4.10. Graphically, we see the result by plotting the objective function line, Z ¼ 10x1 + 8x2., with the feasible region. Determine the parallel direction for the line to maximize (in this case) Z. Move the line parallel until it crosses the last point in the feasible set. That point is the solution. The line that goes through the origin at a slope of 7/6 is called the ISO-Profit line. We have provided this in Fig. 4.3. Here is a short cut to sensitivity analysis using the KTC conditions. We set up the function, L, using the form,  L ¼ f ðxÞ þ l1 ðb1  g1 ðxÞÞ þ l2 b2  g2 ðxÞ þ . . . For our example this becomes, 140x1 þ 120x2 þ l1 ð1400  2x1 þ 4x2 Þ þ l2 ð1500  4x1 þ 3x2 Þ We take the partial derivatives of L with respect to x1,x2,l1,l2. For sensitivity analysis, we only care about the partial derivatives with respect to the l’s. Thus, we will solve the following two equations and two unknowns. 140  2l1  4l2 120  4l1  3l2 We find l1 ¼ 6 and l2 ¼ 32. We will see later with technology that these are shadow prices. We find here that a one unit increase in the second resource provides a larger increase to Z than a unit increase in the resource for the first constraint, (32Δ > 6Δ). We summarize the steps for solving a linear programming problem involving only two variables. 1. Sketch the region corresponding to the system of constraints. The points satisfying all constraints make up the feasible solution.

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Fig. 4.3 Iso-profit lines added

1000

800

600 x2 400

200

0 0

200

400

600

800

1000

x1

2. Find all the corner points (or intersection points in the feasible region). 3. Test the objective function at each corner point and select the values of the variables that optimize the objective function. For bounded regions, both a maximum and a minimum will exist. For an unbounded region, if a solution exists, it will exist at a corner.

4.3.3

Minimization Example Minimize Z ¼ 5x þ 7y Subject to : 2x þ 3y  6 3x  y  15 x þ y  4 2x þ 5y  27 x0 y0

The corner points in Fig. 4.4 are (0,2), (0,4,) (1,5), (6,3), (5,0), and (3,0). See if you can find all these corner points. If we evaluate Z ¼ 5x + 7y at each of these points, we find the values listed in Table 4.11. The minimum value occurs at (0, 2) with a Z value of 14. Notice in our graph that the blue ISO-Profit line will last cross the point (0,2) as it moves out of the feasible region in the direction that Minimizes Z.

4.3 Graphical Linear Programming

157 Feasible Region

10

8

6 y 4

2

–2

0

2

4

6

8

10

x –2

Fig. 4.4 Feasible region for minimization example Table 4.11 Solved corner points

4.3.3.1

Corner point (0,2) (1,5) (6,3) (5,0) (3,0) (0,4)

Z ¼ 5x + 7y (MINIMIZE) Z ¼ 14 Z ¼ 40 Z ¼ 51 Z ¼ 25 Z ¼ 15 Z ¼ 28

Exercises 4.3

Find the maximum and minimum solution. Assume we have x > 0 and y > 0 for each problem. 1. Z ¼ 2x þ 3y subject to : 2x þ 3y  6 3x  y  15 xþy4 2x þ 5y  27

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4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

2. Z ¼ 6x þ 4y subject to :  x þ y  12 x þ y  24 2x þ 5y  80 3. Z ¼ 6x þ 5y subject to : xþy6 2x þ y  9 4. Z ¼ x  y subject to : xþy6 2x þ y  9 5. Z ¼ 5x þ 3y subject to : 1:2x þ 0:6y  24 2x þ 1:5y  80

4.3.3.2

Projects 4.3

For each scenario (a) (b) (c) (d) (e) (f)

List the decision variables and define them. List the objective function. List the resources that constrain this problem. Graph the “feasible region”. Label all intersection points of the feasible region. Plot the Objective function in a different color (highlight the Objective function line, if necessary) and label it the ISO-Profit line. (g) Clearly indicate on the graph the point that is the optimal solution. (h) List the coordinates of the optimal solution and the value of the objective function. (i) answer all scenario-specific questions. 1. With the rising cost of gasoline and increasing prices to consumers, the use of additives to enhance performance of gasoline is being considered. Consider two additives, Additive 1 and Additive 2. The following conditions must hold for the use of additives: (a) Harmful carburetor deposits must not exceed 1/2 lb per car’s gasoline tank. (b) The quantity of Additive 2 plus twice the quantity of Additive 1 must be at least 1/2 lb per car’s gasoline tank. (c) 1 lb of Additive 1 will add 10 octane units per tank, and 1 lb of Additive 2 will add 20 octane units per tank. The total number of octane units added must not be less than six (6).

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159

(d) Additives are expensive and cost $1.53/lb for Additive 1 and $4.00/lb for Additive 2. We want to determine the quantity of each additive that will meet the above restrictions and will minimize their cost. (a) Assume now that the manufacturer of additives has the opportunity to sell you a nice TV special deal to deliver at least 0.5 lb of Additive 1 and at least 0.3 lb of Additive 2. Use graphical LP methods to help recommend whether you should buy this TV offer. Support your recommendation. • Write a one-page cover letter to your boss of the company that summarizes the results that you found. 2. A farmer has 30 acres on which to grow tomatoes and corn. Each 100 bushels of tomatoes require 1000 gallons of water and 5 acres of land. Each 100 bushels of corn require 6000 gallons of water and 2 1/2 acres of land. Labor costs are $1 per bushel for both corn and tomatoes. The farmer has available 30,000 gallons of water and $750 in capital. He knows that he cannot sell more than 500 bushels of tomatoes or 475 bushels of corn. He estimates a profit of $2 on each bushel of tomatoes and $3 of each bushel of corn. How many bushels of each should he raise to maximize profits? (a) Assume now that farmer has the opportunity to sign a nice contract with a grocery store to grow and deliver at least 300 bushels of tomatoes and at least 500 bushels of corn. Use graphical LP methods to help recommend a decision to the farmer. Support your recommendation. (b) If the farmer can obtain an additional 10,000 gallons of water for a total cost of $50, is it worth it to obtain the additional water? Determine the new optimal solution caused by adding this level of resource. (c) Write a one-page cover letter to your boss that summarizes the result that you found. 3. Fire Stone Tires headquartered in Akron, Ohio has a plant in Florence, SC which manufactures two types of tires: SUV 225 Radials and SUV 205 Radials. Demand is high because of the recent recall of tires. Each 100-SUV 225 Radials requires 100 gallons of synthetic plastic and 5 lb of rubber. Each 100 SUV 205 Radials require 60 gallons of synthetic plastic and 2 1/2 lb of rubber. Labor costs are $1 per tire for each type tire. The manufacturer has weekly quantities available of 660 gallons of synthetic plastic, $750 in capital, and 300 lb of rubber. The company estimates a profit of $3 on each SUV 225 radial and $2 of each SUV 205 radial. How many of each type tire should the company manufacture in order to maximize their profits? (a) Assume now that manufacturer has the opportunity to sign a nice contract with a tire outlet store to deliver at least 500 SUV 225 Radial tires and at least 300 SUV 205 radial tires. Use graphical LP methods to help recommend a decision to the manufacturer. Support your recommendation.

4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

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(b) If the manufacturer can obtain an additional 1000 gallons of synthetic plastic for a total cost of $50, is it worth it to obtain this amount? Determine the new optimal solution caused by adding this level of resource. (c) If the manufacturer can obtain an additional 20 lb of rubber for $50, should they do obtain the rubber? Determine the new solution caused by adding this amount. (d) Write a one-page cover letter to your boss of the company that summarizes the results that you found. 4. Consider a toy maker that carves wooden soldiers. The company specializes in two types: Confederate soldiers and Union soldiers. The estimated profit for each is $28 and $30, respectively. A Confederate soldier requires 2 units of lumber, 4 h of carpentry, and 2 h of finishing in order to complete the soldier. A Union soldier requires 3 units of lumber, 4.5 h of carpentry, and 3 h of finishing to complete. Each week the company has 100 units of lumber delivered. The workers can provide at most 120 h of carpentry and 90 h of finishing. Determine the number of each type wooden soldiers to produce to maximize weekly profits.

4.4

Mathematical Programming with Technology

4.4.1

Linear Programming

Technology is critical to solving, analyzing, and performing sensitivity analysis on linear programming problems. Technology provides a suite of powerful, robust routines for solving optimization problems, including linear programs (LPs). Technology that we illustrate t include Excel, LINDO, and LINGO as these appear to be used often in engineering. We also examined GAMS, which we found powerful but too cumbersome to discuss here. We tested all these other software packages and found them all useful. We show the computer chip problem first with technology. Profit Z ¼ 140x1 þ 120x2 Subject to : 2x1 þ 4x2  1400ðassembly timeÞ 4x1 þ 3x2  1500ðinstallation timeÞ x1  0,x2  0

4.4.1.1

Using EXCEL

(a) Put the problem formulation into Excel. Note, you must have formulas in terms of cells for the objective function and the constraints. Highlight the objective function, Open the Solver, select as the solution method.

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161

(b) SimplexLP (c) Insert the decision variables into the By Changing Variable Cells Enter the constraints by evoking the Add command. (d) Enter the constraints. (e) Solve. Save both the answer and sensitivity analysis worksheets. (f) View solution and analysis reports (Figs. 4.5, 4.6, 4.7, 4.8, 4.9, 4.10, and 4.11)

Fig. 4.5 Screenshot excel model of the problem

Fig. 4.6 Excel solver window

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4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

Fig. 4.7 Excel solver window changing variable cells

Fig. 4.8 Excel solver add constraint window

4.4 Mathematical Programming with Technology

Fig. 4.9 Excel solver window with constraints

Fig. 4.10 Excel solver solution window

163

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4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

Fig. 4.11 Screenshot excel model with solutions

4.4.1.2

Answer Report (Fig. 4.12)

Fig. 4.12 Screenshot excel solver answer report

4.4 Mathematical Programming with Technology

4.4.1.3

Sensitivity Report (Fig. 4.13)

Fig. 4.13 Screenshot excel solver sensitivity report

As expected, we have the same answers as we found earlier. We present the following example via each technology. Maximize Z ¼ 25x1 þ 30x2 Subject to : 20x1 þ 30x2  690 5x1 þ 4x2  120 x1,x2,  0

4.4.1.4

Using EXCEL (Figs. 4.14 and 4.15)

Fig. 4.14 Screenshot linear programming in excel

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4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

Fig. 4.15 Screenshot linear programming in excel—constraints

166

4.4 Mathematical Programming with Technology

4.4.1.5

167

Solver (Fig. 4.16)

Fig. 4.16 Excel solver window

4.4.1.6

Constraints into Solver (Fig. 4.17)

Full Set UP. Click Solve (Fig. 4.18). Obtain the answers as x1 ¼ 9, x2 ¼ 24, Z ¼ 972. Additionally, we can obtain reports from Excel. Two key reports are the answer report and the sensitivity report.

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4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

Fig. 4.17 Excel solver window with constraints

Fig. 4.18 Screenshot linear programming in excel—solutions

4.4 Mathematical Programming with Technology

4.4.1.7

169

Answer Report (Fig. 4.19)

Fig. 4.19 Screenshot linear programming in excel— answer Report

4.4.1.8

Sensitivity Report (Fig. 4.20)

Fig. 4.20 Screenshot linear programming in excel—sensitivity report

We find our solution is x1 ¼ 9, x2 ¼ 24, P ¼ $972. From the standpoint of sensitivity analysis Excel is satisfactory in that it provides shadow prices. Limitation: No tableaus are provided making it difficult to find alternate solutions. Further discussion:

4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

170

4.4.2

Alternate Optimal Solution Shadow Prices

4.4.2.1

Using LINDO

This is the format to type in the formulation directly into LINDO. MAX 25 X1 + 30 X2 SUBJECT TO 2) 20 X1 + 30 X2

We then call the optimization packages and in this case maximize the linear programming problem. There are two MAPLE approaches one with simplex and either maximize or minimize command and the other with LPSolve with either maximize or minimize as shown below to obtain our same answers.

> > > The basic LP package in MAPLE is not equipped to provide tableaus or sensitivity analysis directly. Fishback (2010) wrote a nice book on Linear Programming in Maple. This is a step by step process in which the user has to understand the Simplex procedure. Here are the commands for our problem and tableau are provided.

4.4 Mathematical Programming with Technology

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4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

174

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4.5 Case Studies in Mathematical Programming

4.4.3

Integer and Nonlinear Programming with Technology

4.4.3.1

Integer

175

Integer programming in Excel requires only that you identify the variables as integers in the constraint set. Your choices are binary integers {0, 1} or integers. We state that the Solver does not identify the methodology used.

4.4.3.2

Nonlinear Programming

There are many forms of nonlinear problems in optimization. MAPLE and EXCEL are both useful in obtaining solutions. We will illustrate the use of technology in the case studies examples later.

4.4.3.3

Section Exercises

Solve the exercises and projects in Sect. 4.3 using appropriate and available technology.

4.5

Case Studies in Mathematical Programming

4.5.1

Example 1. Military Supply Chain Operations (from Fox and Garcia 2014)

In our case study, we present linear programming for supply chain design. We consider producing a new mixture of gasoline. We desire to minimize the total cost of manufacturing and distributing the new mixture. There is a supply chain involved with a product that must be modeled. The product is made up of components that are produced separately as shown in Table 4.12. Demand information is contained in Table 4.13. Let i ¼ crude type 1, 2, 3 (X10, X20, X30 respectively) Let j ¼ gasoline type 1,2,3 (Premium, Super, Regular respectively) We define the following decision variables: Gij ¼ amount of crude i used to produce gasoline j For example, G11 ¼ amount of crude X10 used to produce Premium gasoline. G12 ¼ amount of crude type X20 used to produce Premium gasoline G13 ¼ amount of crude type X30 used to produce Premium gasoline G12 ¼ amount of crude type X10 used to produce Super gasoline G22 ¼ amount of crude type X20 used to produce Super gasoline

4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

176

Table 4.12 Supply chain gasoline mixture Crude oil type X10 X20 X30

Compound A (%) 35 50 60

Compound B (%) 25 30 20

Compound C (%) 35 15 15

Cost/ barrel $26 $32 $55

Barrel avail (000 of barrels) 15,000 32,000 24,000

Table 4.13 Supply chain gasoline demands

Gasoline Premium Super Regular

Compound A (%) >55

Compound B (%) 25

>40

Compound C (%) 22000 ðSuper Þ G13 þ G23 þ G33 > 25000 ðRegular Þ Availability of products G11 þ G12 þ G13 < 15000 ðcrude 1Þ G21 þ G22 þ G23 < 32000 ðcrude 2Þ G31 þ G32 þ G33 < 24000 ðcrude 3Þ Product mix in mixture format ð0:35G11 þ 0:50G21 þ 0:60G31Þ  =ðG11 þ G21 þ G31Þ > 0:55 ðX10 in PremiumÞ ð0:25G11 þ 0:30G21 þ 0:20G31Þ=ðG11 þ G21 þ G31Þ < 0:23ðX20 in PremiumÞ ð0:35G13 þ 0:15G23 þ 0:15G33Þ=ðG13 þ G23 þ G33Þ > 0:25ðX20 in Regular Þ ð0:35G13 þ 0:15G23 þ 0:15G33Þ=ðG13 þ G23 þ G33Þ < 0:35ðX30 in Regular Þ

4.5 Case Studies in Mathematical Programming Table 4.14 Supply chain solution

Decision variable G11 G12 G13 G21 G22 G23 G31 G32 G33

177 Z¼$1,940,000 0 0 14,000 15,000 7000 0 0 25,000 0

Z¼$1,940,000 1400 3500 9100 1100 20,900 0 12,500 7500 4900

ð0:35G12 þ 0:50G22 þ 0:60G23Þ=ðG12 þ G22 þ G32Þ < 0:40ðCompound X10 in Super Þ ð0:35G12 þ 0:15G22 þ 0:15G32Þ  =ðG12 þ G22 þ G32Þ < 0:25 ðCompound X30 in Super Þ The solution was found using LINDO and we noticed an alternate optimal solution: Two solutions are found yielding a minimum cost of $1,904,000 (Table 4.14). Depending on whether we want to additionally minimize delivery (across different locations) or maximize sharing by having more distribution point involved then we have choices. We present one of the solutions below with LINDO.LP OPTIMUM FOUND AT STEP 7 OBJECTIVE FUNCTION VALUE 1)

1904000.

VARIABLE VALUE P1 0.000000 R1 15000.000000 E1 0.000000 P2 0.000000 R2 7000.000000 E2 25000.000000 P3 14000.000000 R3 0.000000 E3 0.000000

REDUCED COST 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 9.000000 3) 0.000000 4.000000 4) 10000.000000 0.000000 5) 0.000000 -35.000000 6) 0.000000 -35.000000

4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

178 7) 8) 9) 10) 11) 12) 13)

0.000000 -35.000000 700.000000 0.000000 3500.000000 0.000000 1400.000000 0.000000 2500.000000 0.000000 2500.000000 0.000000 420.000000 0.000000

NO. ITERATIONS=

7

RANGES IN WHICH THE BASIS IS UNCHANGED:

VARIABLE P1 R1 E1 P2 R2 E2 P3 R3 E3

OBJ COEFFICIENT RANGES CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE 26.000000 INFINITY 0.000000 26.000000 0.000000 INFINITY 26.000000 INFINITY 0.000000 32.000000 0.000000 0.000000 32.000000 0.000000 0.000000 32.000000 0.000000 35.000000 35.000000 0.000000 4.000000 35.000000 INFINITY 0.000000 35.000000 INFINITY 0.000000

RIGHTHAND SIDE RANGES CURRENT ALLOWABLE ALLOWABLE RHS INCREASE DECREASE 2 15000.000000 4200.000000 0.000000 3 32000.000000 4200.000000 0.000000 4 24000.000000 INFINITY 10000.000000 5 14000.000000 10000.000000 14000.000000 6 22000.000000 0.000000 4200.000000 7 25000.000000 0.000000 4200.000000 8 0.000000 700.000000 INFINITY 9 0.000000 3500.000000 INFINITY 10 0.000000 INFINITY 1400.000000 11 0.000000 2500.000000 INFINITY 12 0.000000 INFINITY 2500.000000 13 0.000000 INFINITY 420.000000 ROW

4.5.2

Example 2. Military Recruiting Raleigh Office (Modified from McGrath 2007)

Although this is a simple model it was adopted by the US Army recruiting commend for operations. The model determines the optimal mix of prospecting strategies that a recruiter should use in a given week. The two prospecting strategies initially modeled and analyzed are phone and email prospecting. The data came from the Raleigh Recruiting Company United States Army Recruiting Command in 2006. On

4.5 Case Studies in Mathematical Programming

179

Table 4.15 Recruiter phone and email data Phone (x1) 60 min per lead $10 per lead

Prospecting time (minutes) Budget (dollars)

Email (x2) 1 min per lead $37 per lead

Table 4.16 Recruiter data sensitivity report Cell

Name

Variable cells $B$3 x1 $B$4 x2 Cell

Name

Constraints $C$10 $C$11 $C$12 $C$13

Final Value

Reduced Cost

Objective Coefficient

Allowable Increase

Decrease

294.2986425 1542.081448

0 0

0.041 0.142

8.479 0.0097

0.002621622 0.141316667

Final Value

Shadow Price

Constraint Allowable R.H. Side Increase

Decrease

19,200 60,000 294.2986425 1542.081448

4.38914E05 0.003836652 0 0

19,200 60,000 1 1

17518.64865 56764.16667 1E+30 1E+30

340,579 648,190 294.2986425 1541.081448

average each phone lead yields 0.041 enlistments and each email lead yields 0.142 enlistments. The forty recruiters assigned to the Raleigh recruiting office prospected a combined 19,200 minutes of work per week via phone and email. The company’s weekly budget is $60,000. The decision variables are: x1 ¼ number of phone leads x2 ¼ number of email leads Maximize Z ¼ 0:041x1 þ 0:142x2 Subject to 60x1 þ 1x2  19200 ðProspecting minutes availableÞ 10x1 þ 37x2  60000ðBudget dollars availableÞ x1 ,x2 > 0 ðnon-negativityÞ If we examine all the intersections point we find a sub-optimal point, x1 ¼ 294.29, x2 ¼ 154.082, achieving 231.04 recruitments. We examine the sensitivity analysis report in Tables 4.15 and 4.16, First, we see we maintain a mixed solution over a fairly large range of values for the coefficient of x1 and x2. Further the shadow prices provide additional information. A one unit increase in prospecting minutes available yields an increase of approximately 0.00004389 in recruits while an increase in budget of $1 yields an additional 0.003836652 recruits. At initial look at appears as though we might be better off with an additional $1 in resource.

4 Mathematical Programming: Linear, Integer, and Nonlinear Optimization. . .

180

Table 4.17 Supply chain gasoline mixture Crude oil type X10 X20 X30

Compound A (%) 45 60 70

Compound B (%) 35 40 30

Compound C (%) 45 25 25

Cost/ barrel $26.50 $32.85 $55.97

Barrel avail (000 of barrels) 18,000 35,000 26,000

Table 4.18 Supply chain gasoline demand Gasoline Premium Super Regular

Compound A (%) >55

Compound B (%) 25

>40

Compound C (%) 6).

4.8.2

Section 4.7 Exercises

Resolve exercises from Sect. 4.3 using the tableau method in Excel or Maple

References Fishback, P. E. (2010). Linear and nonlinear programming with Maple: An interactive, applications-based approach. Boca Raton: CRC Press. Retrieved February 2, 2019, from http://www.mathplace.org/C064X/main.pdf. Fox, W. P. (2012). Modeling with Maple, Cengage, Boston: MA. Fox, W., & Garcia, F. (2014). Modeling and linear programming in engineering management. In F. P. García (Ed.), Engineering management. London: Intech Open. McGrath, G. (2007). Email marketing for the U.S. Army and Special Operations Forces Recruiting. Master’s Thesis, Naval Postgraduate School.

Suggested Reading Albright, B. (2010). Mathematical modeling with EXCEL. Burlington: Jones and Bartlett Publishers, Chapter 7. Apaiah, R., & Hendrix E. (2006). Linear programming for supply chain design: A case on Novel protein foods. Ph.D. Thesis, Wageningen University, Netherlands. Balakrishnan, N., Render, B., & Stair, R. (2007). Managerial decision making (2nd ed.). Upper Saddle River: Prentice Hall. Bazarra, M. S., Jarvis, J. J., & Sheralli, H. D. (1990). Linear programming and network flows. New York: Wiley. Ecker, J., & Kupperschmid, M. (1988). Introduction to operations research. New York: Wiley. Fox, W. (2012). Mathematical modeling with maple. Boston: Cengage Publishers (Chapters 7–10). Giordano, F., Fox, W., & Horton, S. (2014). A first course in mathematical modeling (5th ed.). Boston: Cengage Publishers, Chapter 7. Hiller, F., & Liberman, G. J. (1990). Introduction to mathematical programming. New York: McGraw Hill. Márquez, & Lev B. (2013, March 3). ISBN 978-953-51-1037-8, InTech. Winston, W. L. (1994). Operations research: Applications and algorithms (3rd ed.). Belmont: Duxbury Press. Winston, W. L. (2002). Introduction to mathematical programming applications and algorithms (4th ed.). Belmont: Duxbury Press.

Chapter 5

Introduction to Multi-attribute Military Decision-Making

Objectives 1. 2. 3. 4. 5. 6.

Know the types of multi-attribute decision techniques. Know the basic solution methodologies. Know the weighting schemes. Know which technique or techniques to use. Know the importance of sensitivity analysis. Know the importance of technology is the solution process.

5.1

Risk Analysis for Homeland Security

The Department of Homeland Security (DHS) only has a limited number of assets and a finite amount of time to conduct investigations, thus DHS must establish priorities for its investigations. The risk assessment office has collected the data for the morning meeting shown in Table 5.1. Your operations research team must analyze the information and provide a priority list to the risk assessment team for that meeting with DHS. Problem: Build a model that ranks the list threats (Table 5.1) in a priority order. Assumptions: We have past decision that will give us insights into the decisionmaker’s thought process. We have data only on reliability, approximate number of deaths, approximate costs to fix or rebuild, location, destructive influence, and number of intelligence gathering tips. These will be the criteria for our analysis. The data is accurate and precise. This problem provides an example of what can be solved with mathematical modeling and we will solve this problem later in this chapter.

© Springer Nature Switzerland AG 2019 W. P. Fox, R. Burks, Applications of Operations Research and Management Science for Military Decision Making, International Series in Operations Research & Management Science 283, https://doi.org/10.1007/978-3-030-20569-0_5

193

Threat alternatives \criterion Dirty Bomb Threat Anthrax-Bio Terror Threat DC-Road and bridge network threat NY subway threat DC Metro Threat Major bank robbery FAA Threat

Approximate associated deaths (000) 10 0.8 0.005 12 11 0.0002 0.001

Reliability of threat assessment 0.40 0.45

0.35

0.73 0.69 0.81 0.70

Table 5.1 DHS risk assessment data

200 200 10 5

6.3 2.5 0.57 0.15

0.85

Location density in millions 4.5 3.2

7 7 2 4.5

6

Destructive psychological influence 9 7.5

5 5 16 15

8

Number of intelligence-related tips 3 12

5

300

Cost to fix damages in (millions) 150 10

194 Introduction to Multi-attribute Military Decision-Making

5.3 Data Envelopment Analysis (DEA)

5.2

195

Introduction

Multiple-attribute decision-making (MADM) concerns making decisions when there are multiple but a finite list of alternatives and decision criteria. This differs from analysis where we have alternatives and only one criterion such as cost. We address problems such as in the DHS scenario where we have seven alternatives and six criteria that impact the decision. Consider a problem where management needs to prioritize or rank order alternative choices such as identify key nodes in a business network, picking a contractor or sub-contractor, choosing an airport, ranking recruiting efforts, ranking banking facilities, ranking schools or colleges, etc. How does one proceed to accomplish this analytically? In this chapter, we will present four methodologies to rank order or prioritize alternatives based upon multiple criteria. These four methodologies include: Data Envelopment Analysis (DEA) Simple Average Weighting (SAW) Analytical Hierarchy Process (AHP) Technique of Order Preference by Similarity to Ideal Solution (TOPSIS) For each method, we will describe the method and provide a methodology, discuss some strengths and limitations to the method, discuss tips for conducting sensitivity analysis, and present several illustrative examples.

5.3 5.3.1

Data Envelopment Analysis (DEA) Description and Use

Data envelopment analysis (DEA) is a “data input-output driven” approach for evaluating the performance of entities called decision-making units (DMUs) that convert multiple inputs into multiple outputs (Cooper 2000). The definition of a DMU is generic and very flexible so that any entity to be ranked might be a DMU. DEA has been used to evaluate the performance or efficiencies of hospitals, schools, departments, US Air Force wings, US armed forces recruiting agencies, universities, cities, courts, businesses, banking facilities, countries, regions, SOF airbases, key nodes in networks, and the list goes on. According to Cooper (2000), DEA has been used to gain insights into activities that were not obtained by other quantitative or qualitative methods. Charnes et al. (1978) described DEA as a mathematical programming model applied to observational data. It provides a new way of obtaining empirical estimates of relationship among the DMUs. It has been formally defined as a methodology directed to frontiers rather than central tendencies.

196

5.3.2

5

Introduction to Multi-attribute Military Decision-Making

Methodology

The model, in simplest terms, may be formulated and solved as a linear programming problem (Winston 1995; Callen 1991). Although several formulations for DEA exist, we seek the most straightforward formulation in order to maximize an efficiency of a DMU as constrained by inputs and outputs as shown in Eq. (5.1). As an option, we might normalize the metric inputs and outputs for the alternatives if the values are poorly scaled within the data. We will call this data matrix, X, with entries xij. We define an efficiency unit as Ei for i ¼ 1, 2,. . .,nodes. We let wi be the weights or coefficients for the linear combinations. Further, we restrict any efficiency from being larger than 1. Thus, the largest efficient DMU will be 1. This gives the following linear programming formulation for single outputs but multiple inputs: Max E i subject to Pn w xij  Ei ¼ 0, j ¼ 1,2, . . . i i¼1 E i  1 for all i

ð5:1Þ

For multiple inputs and outputs, we recommend the formulations provided by Winston (1995) and Trick (2014) using Eq. (5.2). For any DMU0, let Xi be the inputs and Yi be the outputs. Let X0 and Y0 be the DMU being modeled. Min θ subject to Σλi X i  θX 0 Σλi Y i  Y 0 λι  0 Non-negativity

5.3.3

ð5:2Þ

Strengths and Limitations to DEA

DEA can be a very useful tool when used wisely according to Trick (1996). A few of the strengths that make DEA extremely useful are Trick 1996: (1) DEA can handle multiple input and multiple output models; (2) DEA does not require an assumption of a functional form relating inputs to outputs; (3) DMUs are directly compared against a peer or combination of peers; and (4) Inputs and outputs can have very different units. For example, X1 could be in units of lives saved and X2 could be in units of dollars without requiring any a priori tradeoff between the two.

5.3 Data Envelopment Analysis (DEA)

197

The same characteristics that make DEA a powerful tool can also create limitations to the process and analysis. An analyst should keep these limitations in mind when choosing whether or not to use DEA. A few additional limitations include: 1. Since DEA is an extreme point technique, noise in the data such as measurement error can cause significant problems. 2. DEA is good at estimating “relative” efficiency of a DMU but it converges very slowly to “absolute” efficiency. In other words, it can tell you how well you are doing compared to your peers but not compared to a “theoretical maximum.” 3. Since DEA is a nonparametric technique, statistical hypothesis tests are difficult and are the focus of ongoing research. 4. Since a standard formulation of DEA with multiple inputs and outputs creates a separate linear program for each DMU, large problems can be computationally intensive. 5. Linear programming does not ensure all weights are considered. We find that the value for weights is only for those that optimally determine an efficiency rating. If having all criteria weighted (inputs, outputs) is essential to the decision-maker, then do not use DEA.

5.3.4

Sensitivity Analysis

Sensitivity analysis is always an important element in analysis. According to Neralic (1998), an increase in any output cannot make a solution worse rating nor can a decrease in inputs alone worsen an already achieved efficiency rating. As a result in our examples, we only decrease outputs and increase inputs as just described (Neralic 1998). We will illustrate some sensitivity analysis, as applicable, in our illustrative examples next.

5.3.5

Illustrative Examples

Example 1 Manufacturing Consider the following manufacturing process (modified from Winston 1995), where we have three DMUs each of which has two inputs and three outputs as shown in Table 5.2. Table 5.2 Manufacturing output DMU 1 2 3

Input #1 5 8 7

Input #2 14 15 12

Output #1 9 5 4

Output #2 4 7 9

Output #3 16 10 13

198

5

Introduction to Multi-attribute Military Decision-Making

Since no units are given and the scales are similar, we decide not to normalize the data. We define the following decision variables: ti ¼ value of a single unit of output of DMU i, for i ¼ 1, 2, 3 wi ¼ cost or weights for one unit of inputs of DMU i, for i ¼ 1, 2 efficiencyi ¼ DMUi ¼ (total value of i’s outputs)/(total cost of i’s inputs), for i ¼ 1, 2, 3 The following modeling assumptions are made: 1. No DMU will have an efficiency of more than 100%. 2. If any efficiency is less than 1, then it is inefficient. 3. We will scale the costs so that the costs of the inputs equals 1 for each linear program. For example, we will use 5w1 + 14w2 ¼ 1 in our program for DMU1. 4. All values and weights must be strictly positive, so we use a constant such as 0.0001 in lieu of 0. To calculate the efficiency of DMU1, we define the linear program using Eq. (5.2) as Maximize DMU 1 ¼ 9t 1 þ 4t 2 þ 16t 3 Subject to 9t 1  4t 2  16t 3 þ 5w1 þ 14w2  0 5t 1  7t 2  10t 3 þ 8w1 þ 15w2  0 4t 1  9t 2  13t 3 þ 7w1 þ 12w2  0 5w1 þ 14w2 ¼ 1 t i  0:0001,i ¼ 1,2,3 wi  0:0001,i ¼ 1,2 Non-negativity To calculate the efficiency of DMU2, we define the linear program using Eq. (5.2) as Maximize DMU 2 ¼ 5t 1 þ 7t 2 þ 10t 3 Subject to 9t 1  4t 2  16t 3 þ 5w1 þ 14w2  0 5t 1  7t 2  10t 3 þ 8w1 þ 15w2  0 4t 1  9t 2  13t 3 þ 7w1 þ 12w2  0 8w1 þ 15w2 ¼ 1 t i  0:0001,i ¼ 1,2,3 wi  0:0001,i ¼ 1,2 Non-negativity To calculate the efficiency of DMU3, we define the linear program using Eq. (5.2) as

5.3 Data Envelopment Analysis (DEA)

199

Maximize DMU 3 ¼ 4t 1 þ 9t 2 þ 13t 3 Subject to 9t 1  4t 2  16t 3 þ 5w1 þ 14w2  0 5t 1  7t 2  10t 3 þ 8w1 þ 15w2  0 4t 1  9t 2  13t 3 þ 7w1 þ 12w2  0 7w1 þ 12w2 ¼ 1 t i  0:0001,i ¼ 1,2,3 wi  0:0001,i ¼ 1,2 Non-negativity The linear programming solutions show the efficiencies as DMU1 ¼ DMU3 ¼ 1, DMU2 ¼ 0.77303. Interpretation: DMU2 is operating at 77.303% of the efficiency of DMU1 and DMU3. Management could concentrate some improvements or best practices from DMU1 or DMU3 for DMU2. An examination of the dual prices for the linear program of DMU2 yields λ1 ¼ 0.261538, λ2 ¼ 0, and λ3 ¼ 0.661538. The average output vector for DMU2 can be written as: 2

3 2 3 2 3 9 4 5 0:2615384 4 5 þ 0:6615384 9 5 ¼ 4 7 5 16 13 12:785 and the average input vector can be written as  0:261538

     5 7 5:938 þ 0:661538 ¼ : 14 12 11:6

In our data, output #3 is 10 units. Thus, we may clearly see the inefficiency is in output #3 where 12.785 units are required. We find that they are short 2.785 units (12.785 – 10 ¼ 2.785). This helps focus on treating the inefficiency found for output #3. Sensitivity Analysis: Sensitivity analysis in a linear program is sometimes referred to as “what if” analysis. Let’s assume that without management engaging some additional training for DMU2 that DMU2 output #3 dips from 10 to 9 units of output while the input 2 h increases from 15 to 16 h. We find that these changes in the technology coefficients are easily handled in resolving the LPs. Since DMU2 is affected, we might only modify and solve the LP concerning DMU2. We find with these changes that DMU’s efficiency is now only 74% as effective as DMU1 and DMU3. Example 2 Social Networks and Ranking Nodes Consider the Kite Social Network (Krackhardt 1990 ) shown in Fig. 5.1. ORA (Carley 2011), a social network software, was used to obtain the metrics for this network. A subset of the output is shown in Table 5.3. We restricted the metrics presented: Total Centrality (TC), Eigenvector Centrality (EC), In-Closeness (IC),

200

5

Introduction to Multi-attribute Military Decision-Making

Fig. 5.1 Kite Network diagram from ORA (Carley 2011) Table 5.3 ORA metric measures as outputs for the Kite Network

TC 0.1806 0.1389 0.1250 0.1111 0.1111 0.0833 0.0833 0.0833 0.0556 0.0278

EC 0.1751 0.1375 0.1375 0.1144 0.1144 0.0938 0.0938 0.1042 0.0241 0.0052

IC 0.0920 0.0997 0.1107 0.0997 0.0997 0.0997 0.0997 0.0997 0.0997 0.0997

OC 0.1081 0.1003 0.0892 0.1003 0.1003 0.1003 0.1003 0.1003 0.1003 0.1003

INC 0.1088 0.1131 0.1131 0.1009 0.1009 0.0975 0.0975 0.1088 0.0885 0.0707

Betw 0.2022 0.1553 0.1042 0.0194 0.0194 0.0000 0.0000 0.3177 0.1818 0.0000

Out-Closeness (OC), Information Centrality (INC), and Betweenness (Betw), whose definitions can be found in recent social network literature (Fox and Everton 2013, 2014). We formulate the linear program from Eq. (5.1) to measure the efficiency of the nodes. We define the decision variables: ui ¼ efficiency of node i, i ¼ 1,2, 3,. . ., 10 wj ¼ weight of input j, j ¼ 1,2,3,4,5

5.3 Data Envelopment Analysis (DEA) Table 5.4 Social network solution

Susan Steven Sarah Tom Claire Fred David Claudia Ben Jennifer

201

DV DMU1 DMU2 DMU3 DMU4 DMU5 DMU6 DMU7 DMU8 DMU9 DMU10 w1 w2 w3 w4 w5 w6

1 0.785511 0.785511 0.653409 0.653409 0.535511 0.535511 0.59517 0.137784 0.02983 0 5.711648 0 0 0 0

Maximize u1 Subject to A¼0 ui  1 for i ¼ 1,2,3, . . . ,10 where

The linear programming solution is provided in Table 5.4. Interpretation: We interpret the linear programming solution as follows: Player 1, u1 ¼ Susan, is rated most influential followed closely by Sarah and Steven. Additionally, we see the most important criterion in solving the optimal problem was the eigenvector centrality, w2, of the network. The solution, translated back into the original variables is found as

202

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Introduction to Multi-attribute Military Decision-Making

Susan ¼ 1,Sarah ¼ 0:78551,Steven ¼ 0:78551,Claire ¼ 0:6534,Tom ¼ 0:6534,Fred ¼ 0:5355,David ¼ 0:5355,Claudia ¼ 0:5951,Ben ¼ 0:1377,and Jennifer ¼ 0:02983 while w1 ¼ w3 ¼ w4 ¼ w5 ¼ w6 ¼ 0 and w2 ¼ 5:7116: Since the output metrics are network metrics calculated from ORA, we do not recommend any sensitivity analysis for this type problem unless your goal is to improve the influence (efficiency) of another member of the network. If so, then the finding of the dual prices (shadow prices) would be required as shown in the first example. Example 3 Recruiting (Figueroa 2014) Data Envelopment Analysis to Obtain Efficiency’s in Recruiting Units is illustrated. Linear programming may be used to compare the efficiencies of units, known as DMUs. The data envelopment method uses the following linear programming formulation to calculate its efficiencies. We want to measure the efficiency of 42 recruiting companies that are part of a recruiting brigade in the United States. The model uses six input measures and two output measures created from data obtain directly from the sixth brigade in 2014. The outputs are the percent fill-todemand ratio for the unit and the percent language capability of the unit. The inputs are the number of recruiters and the percent of populations from which to recruit in a region. The main question was to determine if a larger percentage of recruiters’ ability to speak languages other than English improved their units’ ability to attract recruit. The goal is to identify those units that are not operating at the highest level so that improvement can be made to improve their efficiency. The data envelopment will calculate which of the companies, in this case, DMU1, DMU2, . . . , DMU42, are more efficient when compared to the others

5.3 Data Envelopment Analysis (DEA)

203

The linear formulation to implement the solutions of the DMUs is as follows : Objective Function : MaxDMU 1 , DMU 2 , . . . , DMU c Subject to : 2 3 2 3 W1 T1 Constraint 1 : 4 ⋮ 5  4 ⋮ 5  0; limits the resource of outputs to that of inputs 2W c 3 T c2 3 T1 DMU 1 5  4 ⋮ 5 ¼ 0; Constraint 2 : 4 ⋮ DMU c Tc cannot be more than the output values 2the efficiencies 3 DMU 1 5  1; limits the efficiency to values less than or equal to 1 Constraint 3 : 4 ⋮ 2 DMU 3 c w1 Constraint 4 : 4 ⋮ 5  0:001; wi limits   the input decision variables to values greater than zero t Constraint 5 : 1 > 0:001; t2 than zero 2limits the output decision variables to values 3 greater 2 3 X 1,input1 X 1,input 2 X 1,input 3    X 1, input i w1 6 X 2,input 1 X 2,input 2 X 2,input 3    X 2, input i 7 6 w2 7 7  6 7 ¼ 1; Constraint 6 : 6 4⋮ 5 4⋮ 5 ⋮ ⋮ ⋮ X c,input 1 X c,input 2 X c,input 3    X c, input i wi the multiplication of the input coefficients and decision variables must equal 1 The data needed for evaluating efficiency of the companies are displayed in Table 5.5: The weighted sum of the company’s populations, or the first five columns in the following table, must be equal to 1.00. It does not account for other ethnicities. The output matrix array is the set of coefficient vectors for the fill-to-demand and language-to-recruiter output variables. It also includes a portion of its output coefficients: In order to maximize the efficiency of the companies, or DMUs, the model formulation uses three set of decision variables. Excel Solver identifies the optimal values for the decision variables by solving a linear program, as shown in Fig. 5.2, of which the objective is to maximize the efficiency of the companies. Figure 5.2 shows how to implement the preceding DEA linear formulation using Excel Solver. The naming conventions in the Excel Solver screen (Fig. 5.2) represent the array of cells in which the data is found. Decision variables contain an array of cells in an Excel column that has all 42 decision variables assigned as DMU1, DMU2, . . . , DMUc; six values for the w1, w2, . . . , wi; and two values for the outputs t1 and t2. Each formulation, in the

%PopAPI 0.263722 0.087205 0.080011 0.157075 0.136177 0.036067 0.074656 0.035723 0.600513 0.926119 0.047907 0.040783 0.081499 0.113513 0.202628 0.086219 0.018162 0.043967 0.010439 0.017945 0.102264 0.006928 0.046855 0.201510

Company 6F2—SAN GABRIEL VL 6F3—LONG BEACH 6F5—SN FERNANDO VL 6F7—COASTAL 6F8—LOS ANGELES 6H1—EUGENE 6H2—VANCOUVER 6H3—WILSONVILLE 6H5—HONOLULU 6H7—GUAM 6I0—SIERRA NEVADA 6I1—REDDING 6I3—SACRAMENTO VL 6I4—SAN JOAQUIN 6I5—CAPITOL 6I6 —NORTH BAY 6J1—OGDEN 6J2—SALT LAKE 6J3—BUTTE 6J4—BOISE 6J6—LAS VEGAS 6J9—BIG HORN 6K1—REDLANDS 6K2—FULLERTON

%PopAA 0.031573 0.105755 0.035218 0.199102 0.055508 0.007294 0.038769 0.010366 0.014254 0.010042 0.019145 0.010726 0.036714 0.053919 0.107138 0.066036 0.008748 0.010922 0.002670 0.006657 0.107663 0.003742 0.080681 0.019564

Input coefficients

6th REC BDE %PopH 0.517156 0.672485 0.533119 0.295339 0.546700 0.093672 0.133544 0.173345 0.122297 0.000000 0.250423 0.151065 0.216618 0.445580 0.257041 0.319470 0.129949 0.149390 0.039882 0.165861 0.309920 0.067161 0.566912 0.492608

%PopW 0.185787 0.133068 0.350004 0.347086 0.260140 0.846818 0.745682 0.769427 0.260485 0.063839 0.664284 0.772217 0.657542 0.381530 0.427997 0.519875 0.833434 0.784530 0.885437 0.800868 0.474534 0.844465 0.300264 0.284349

%PopNative 0.001763 0.001486 0.001648 0.001397 0.001474 0.016149 0.007349 0.011140 0.002450 0.000000 0.018241 0.025210 0.007626 0.005457 0.005197 0.008401 0.009707 0.011190 0.061573 0.008668 0.005620 0.077704 0.005287 0.001969

Table 5.5 Input and output coefficients for the DEA approach (array named matrix (A))

#Recruiters 38 41 41 27 51 28 49 25 24 24 29 42 39 36 52 31 27 29 25 28 109 17 43 40

Output coefficients Fill-to-demand ratio Language-to-recruiter ratio 0.872152 0.2105 0.959648 0.2683 0.763623 0.0976 0.840470 0.2963 0.896214 0.3000 0.798541 0.0909 0.888734 0.2083 0.763134 0.1000 1.014687 0.0800 1.016260 0.0000 0.910788 0.0000 0.808599 0.0870 0.920515 0.0000 0.931590 0.0000 0.888965 0.0000 0.656557 0.0000 0.797701 0.1000 0.752606 0.0000 0.714416 0.0000 0.860633 0.0000 0.890533 0.2174 0.754591 0.0435 0.969034 0.1111 0.857143 0.1364

204 5 Introduction to Multi-attribute Military Decision-Making

6K4—LA MESA 6K5—NEWPORT BEACH 6K6—SAN MARCOS 6K7—RIVERSIDE 6K8—SAN DIEGO 6L1—EVERETT 6L2—SEATTLE 6L3—SPOKANE 6L4—TACOMA 6L5—YAKIMA 6L6—ALASKA 6L7—OLYMPIA 6N1—FRESNO 6N2—BAKERSFIELD 6N6—GOLD COAST 6N7—SOUTH BAY 6N8—EAST BAY 6N9—MONTEREY BAY

0.117789 0.139674 0.044696 0.086515 0.173149 0.086824 0.201467 0.023348 0.092312 0.016830 0.074101 0.048443 0.081141 0.036745 0.055439 0.321473 0.225293 0.213740

0.053590 0.013223 0.032839 0.090924 0.050862 0.018779 0.073000 0.012660 0.068854 0.008463 0.028043 0.018887 0.043657 0.072447 0.015520 0.038378 0.125870 0.023038

0.506028 0.270650 0.508044 0.563245 0.249331 0.109900 0.102603 0.055685 0.121083 0.338310 0.061478 0.094673 0.577334 0.546160 0.471056 0.254535 0.299244 0.451766

0.317863 0.574473 0.408251 0.256525 0.523374 0.768680 0.616732 0.891007 0.704523 0.614801 0.655457 0.814933 0.292158 0.338127 0.454545 0.383649 0.346959 0.309182

0.004730 0.001979 0.006170 0.002791 0.003284 0.015817 0.006198 0.017300 0.013227 0.021596 0.180921 0.023064 0.005710 0.006521 0.003440 0.001965 0.002633 0.002274

39 32 50 55 35 25 32 23 27 22 24 29 54 40 34 37 55 29

0.715259 0.743743 0.876591 0.855292 0.757979 0.837500 0.766444 0.796624 0.990457 0.775581 0.855114 0.901454 0.792119 0.836003 0.721532 0.638575 0.692921 0.726957

0.0690 0.0800 0.0000 0.1163 0.3500 0.3846 0.1364 0.1200 0.1818 0.3143 0.2000 0.1034 0.0238 0.0769 0.1111 0.1220 0.1290 0.0909

5.3 Data Envelopment Analysis (DEA) 205

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Fig. 5.2 Linear program for the DEA Problem Using Excel Solver

Subject to the Constraints block has similar naming conventions in order to simplify the location of the data in the Excel spreadsheet. The efficiency results in Table 5.6 provide an opportunity to determine whether the data envelopment method for ethnic populations correlates with the actual recruiting numbers by ethnicity. Note that the data envelopment analysis only uses the ethnic population distributions and the total number of recruiters; similarly, the outputs use the fill-to-demand and language-to-recruiter ratios. However, the actual recruiting data—the number of recruits by ethnicity—is neither part of the inputs nor the outputs of the DEA method. The DEA method accounts for the company’s performance in the form of the fill-to-demand ratio and indirectly, the P2P metrics. We will also show that the correlation between the recruiting efficiencies of the DEA and the P2P metrics suggests that the DEA model can be used to allocate recruiters with secondary languages. The decision-making criteria for allocating recruiters would be a bottom-up approach. In other words, the units at the bottom of the DEA ranking in Table 5.6 would be the ones to first receive new assignments of recruiters with secondary languages.

5.3 Data Envelopment Analysis (DEA)

207

Table 5.6 Optimal DEAs efficiencies for the sixth REC BDE’s companies DMU rankinga DMU-2 DMU-33 DMU-9 DMU-10 DMU-23 DMU-5 DMU-30 DMU-21 DMU-7 DMU-1 DMU-4 DMU-36 DMU-14 DMU-35 DMU-13 DMU-11 DMU-24 DMU-28 DMU-15 DMU-34 DMU-29 DMU-27 DMU-38 DMU-20 DMU-12 DMU-32 DMU-17 DMU-6 DMU-31 DMU-37 DMU-8 DMU-3 DMU-26 DMU-22 DMU-39 DMU-42 DMU-18 DMU-25 DMU-41 DMU-19 DMU-40 DMU-16 a

DEA efficiencies 1.0000 1.0000 0.9886 0.9631 0.9558 0.9506 0.9235 0.9173 0.9126 0.8976 0.8965 0.8892 0.8828 0.8779 0.8723 0.8631 0.8583 0.8498 0.8424 0.8411 0.8365 0.8307 0.8182 0.8156 0.7956 0.7954 0.7897 0.7874 0.7724 0.7587 0.757 0.7566 0.7318 0.7298 0.7213 0.7196 0.7132 0.7011 0.7002 0.677 0.6463 0.6222

DMUs rank from highest to lowest

Company 6F3—LONG BEACH 6L4—TACOMA 6H5—HONOLULU 6H7—GUAM 6K1—REDLANDS 6F8—LOS ANGELES 6L1—EVERETT 6J6—LAS VEGAS 6H2—VANCOUVER 6F2—SAN GABRIEL VALLEY 6F7—COASTAL 6L7—OLYMPIA 6I4—SAN JOAQUIN 6L6—ALASKA 6I3—SACRAMENTO VALLEY 6I0—SIERRA NEVADA 6K2—FULLERTON 6K7—RIVERSIDE 6I5—CAPITOL 6L5—YAKIMA 6K8—SAN DIEGO 6K6—SAN MARCOS 6N2—BAKERSFIELD 6J4—BOISE 6I1—REDDING 6L3—SPOKANE 6J1—OGDEN 6H1—EUGENE 6L2—SEATTLE 6N1—FRESNO 6H3—WILSONVILLE 6F5—SN FERNANDO VL 6K5—NEWPORT BEACH 6J9—BIG HORN 6N6—GOLD COAST 6N9—MONTEREY BAY 6J2—SALT LAKE 6K4—LA MESA 6N8—EAST BAY 6J3—BUTTE 6N7—SOUTH BAY 6I6—NORTH BAY

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Table 5.7 Hospital inputs and outputs

Introduction to Multi-attribute Military Decision-Making

Hospital 1 2 3

Table 5.8 Updated hospital inputs and outputs

Hospital 1 2 3

Inputs 1 5 8 7

Inputs 1 4 9 5

2 14 15 12

2 16 13 11

Outputs 1 9 5 4

Outputs 1 6 10 5

2 4 7 9

2 5 6 10

3 16 10 13

3 15 9 12

Analysis: The most efficient companies, those achieving a DEA score of 100%, are Long Beach, within the Los Angeles BN and Tacoma from the Seattle BN. The least efficient companies include North Bay from the Sacramento BN, achieving 62.2%, and South Bay from the Fresno BN, achieving 64.6%. There are many other factors for improving recruitment numbers, nevertheless, DEA can be used as a tool to assess changes in conditions such as evolving demographic data, to reallocate recruiting center areas of operation, to update rankings based on new recruiting production and fill-to-demand ratios, or to assess changes in the recruiter’s manning, or language-to-recruiter ratios.

5.3.5.1

Exercises 5.3

1. Given the input-output data in Table 5.7 for three hospitals where inputs are number of beds and labor hours in thousands per month and outputs, all measured in hundreds, are patient-days for patients under 14, patient-days for patients between 14 and 65, and patient-days for patients over 65. Determine the efficiency of the three hospitals. 2. Resolve problem 1 with the inputs and outputs in Table 5.8. 3. Consider ranking 4 bank branches in a particular city. The inputs are: Input 1 ¼ labor hours in hundred per month Input 2 ¼ space used for tellers in hundreds of square feet Input 3 ¼ supplies used in dollars per month Output 1 ¼ loan applications per month Output 2 ¼ deposits made in thousands of dollars per month Output 3 ¼ checks processed thousands of dollars per month The data in Table 5.9 is for the four bank branches. 4. What “best practices” might you suggest to the branches that are less efficient in problem 3?

5.4 Weighting Methods

209

Table 5.9 Bank branches inputs and outputs Branches 1 2 3 4

5.4 5.4.1

Input 1 15 14 16 13

Input 2 20 23 19 18

Input 3 50 51 51 49

Output 1 200 220 210 199

Output 2 15 18 17 21

Output 3 35 45 20 35

Weighting Methods Modified Delphi method

The Delphi method is a reliable way of obtaining the opinions of a group of experts on an issue by conducting several rounds of interrogative communications. This method was first developed by the US Air Force in the 1950's (Rand, 2019), mainly for market research and sales forecasting (Chan et al. 2001). This modified method is basically a way to obtain inputs from exerts and then average their scores. The panel consists of a number of experts chosen based on their experience and knowledge. As mentioned previously, panel members remain anonymous to each other throughout the procedure in order to avoid the negative impacts of criticism on the innovation and creativity of panel members. The Delphi method should be conducted by a director One can use the Delphi method for giving weights to the short-listed critical factors. The panel members should give weights to each factor as well as their reasoning. In this way, other panel members can evaluate the weights based on the reasons given and accept, modify, or reject those reasons and weights. For example, consider a search region in Fig. 5.3 that has rows A–G and columns 1–6 as shown. A group of experts then places an x in the squares. In this example, each of 10 experts place 5x’s in the squares. We then total the number of x’s in the squares and divide by the total of x placed, in this case 50. We would find the weights as shown in Table 5.10.

5.4.2

Rank Order Centroid (ROC) Method

This method is a simple way of giving weight to a number of items ranked according to their importance. The decision-makers usually can rank items much more easily than give weight to them. This method takes those ranks as inputs and converts them to weights for each of the items. The conversion is based on the following formula:  X 1 M 1 wi ¼ M n¼i n

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Fig. 5.3 Delphi example

Table 5.10 Modified Delphi to find weights

Selection A1 B3 B4 B6 C4 C5 D6 E5 F5 G3 G4 All others Total

Frequency 3 1 1 1 1 4 8 7 8 9 7 0 50

Relative frequency or weight 3/50 1/50 1/50 1/50 1/50 4/50 8/50 7/50 8/50 9/50 7/50 0/50 50/50 ¼ 1.0

1. List objectives in order from most important to least important 2. Use the above formulas for assigning weights where M is the number of items and Wi is the weight for the i item. For example, if there are four items, the item ranked first will be weighted (1 + 1/2 + 1/3 + 1/4)/4 ¼ 0.52, the second will be

5.4 Weighting Methods

211

Table 5.11 Ratio method Task/item Ranking Weighting Normalizing

Shorten schedule 1 50 41.7%

Project cost 2 40 33.3%

Agency control 3 20 16.7%

Competition 4 10 8.3%

weighted (1/2 + 1/3 + 1/4)/4 ¼ 0.27, the third (1/3 + 1/4)/4 ¼ 0.15, and the last (1/4)/4 ¼ 0.06. As shown in this example, the ROC is simple and easy to follow, but it gives weights which are highly dispersed (Chang 2004). As an example, consider the same factors to be weighted (shortening schedule, agency control over the project, project cost, and competition). If they are ranked based on their importance and influence on decision as (1) shortening schedule, (2) project cost, (3) agency control over the project, and (4) competition, their weights would be 0.52, 0.27, 0.15, and 0.06, respectively. These weights almost eliminate the effect of the fourth factor, i.e., among competitors. This could be an issue.

5.4.3

Ratio Method

The ratio method is another simple way of calculating weights for a number of critical factors. A decision-maker should first rank all the items according to their importance. The next step is giving weight to each item based on its rank. The lowest ranked item will be given a weight of 10. The weight of the rest of the items should be assigned as multiples of 10. The last step is normalizing these raw weights (see Weber and Borcherding 1993). This process is shown in the example below. Note that the weights should not necessarily jump ten points from one item to the next. Any increase in the weight is based on the subjective judgment of the decisionmaker and reflects the difference between the importance of the items. Ranking the items in the first step helps in assigning more accurate weights. Here is an example of the ratio method. There are four tasks listed in priority order from 1, most important, to 4, least important: 1-Shortening schedule, 2-Project Cost, 3-Agency Control, and 4-Competition. We assign the weights as 50, 40, 20, and 10 to these four tasks. We sum the weights (50 + 40 + 20 + 10 = 120) and we normalize the weights obtaining 41.7%, 33.3%, 16,7% and 8.3% for our four tasks. Normalized weights are simply calculated by dividing the raw weight of each item over the sum of the weights for all items. For example, normalized weight for the first item (shortening schedule) is calculated as 50/(50 + 40 + 20 + 10) ¼ 41.7%. The sum of normalized weights is equal to 100% (41.7 + 33.3 + 16.7 + 8.3 ¼ 100), see Table 5.11.

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Table 5.12 Saaty’s 9-point scale Intensity of importance in pairwise comparisons 1 3 5 7 9 2, 4, 6, 8 Reciprocals of above

5.4.4

Definition Equal Importance Moderate Importance Strong Importance Very Strong Importance Extreme Importance For comparing between the above In comparison of elements i and j if i is 3 compared to j, then j is 1/3 compared to i

Pairwise Comparison (AHP)

In this method, the decision-maker should compare each item with the rest of the group and give a preferential level to the item in each pairwise comparison (Chang 2004, Fox et al,. 2014, Fox et al. 2017). For example, if the item at hand is as important as the second one, the preferential level would be one. If it is much more important, its level would be ten. After conducting all of the comparisons and determining the preferential levels, the numbers will be added up and normalized. The results are the weights for each item. Table 5.2 can be used as a guide for giving a preferential level score to an item while comparing it with another one. The following example shows the application of the pairwise comparison procedure. Referring to the four critical factors identified above, let us assume that shortening the schedule, project cost, and agency control of the project are the most important parameters in the project delivery selection decision. Following the pairwise comparison, the decision-maker should pick one of these factors (e.g., shortening the schedule) and compare it with the remaining factors and give a preferential level to it. For example, shortening the schedule is more important than project cost; in this case, it will be given a level of importance of the 5. The decision-maker should continue the pairwise comparison and give weights to each factor. The weights, which are based on the preferential levels given in each pairwise comparison, should be consistent to the extent possible. The consistency is measured based on the matrix of preferential levels. The interested reader can find the methods and applications of consistency measurement in Temesi (2006). Table 5.12 provides the 9-point scale that we will use. Table 5.13 provides the rest of the hypothetical weights and the normalizing process, the last step in the pairwise comparison approach. Note that Column (5) is simply the sum of the values in Columns (1) through (4). Also note that if the preferential level of factor i to factor j is n, then the preferential level of factor j to factor i is simply 1/n. The weights calculated for this exercise are 0.6, 0.1, 0.2, and 0.1 which add up to 1.0. Note that it is possible for two factors to have the same importance and weight.

5.4 Weighting Methods

213

Table 5.13 Pairwise comparison example

Shorten the schedule Project cost Agency control Competition

Shorten the schedule (1) 1

Project cost (2) 5

Agency control (3) 5/2

Competition (4) 8

1/5

1

½

1

2.7

2/5

2

1

2

5.4

1/8

1

½

1

2.625

Total¼

5.4.5

Total (5) 16.5

27.225

Weights (6) 16.5/ 27.225 ¼ 0.60 2.7/27/ 225 ¼ 0.10 5.4/27/ 225 ¼ 0.20 2.625/27/ 225 ¼ 0.10 1

Entropy Method

Shannon and Weaver (1949) proposed the entropy concept, and this concept has been highlighted by Zeleny (1982) for deciding the weights of attributes. Entropy is the measure of uncertainty in the information using probability methods. It indicates that a broad distribution represents more uncertainty than does a sharply peaked distribution. To determine the weights by the entropy method, the normalized decision matrix we call Rij is considered. The Eq. (5.3), used is e j ¼ k

Xn

  R ln Rij ij i¼1

ð5:3Þ

where k ¼ 1/ln(n) is a constant that guarantees that 0  ej  1. The value of n refers to the number of alternatives. The degree of divergence (dj) of the average information contained by each attribute can be calculated as: d j ¼ 1  e j: The more divergent the performance rating Rij, for all i and j, then the higher the corresponding dj the more important the attribute Bj is considered to be. The weights are found by the Eq. (5.4),   1  ej : wj ¼ P  1  ej

ð5:4Þ

Let’s illustrate an example to obtain entropy weights. Example 1 Cars (a) The data (Table 5.14): (b) Sum the columns (Table 5.15) (c) Normalize the data. Divide each data element in a column by the sum of the column (Table 5.16).

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Table 5.14 Car performance data Cost 27.8 28.5 38.668 25.5 27.5 36.2

a1 a2 a3 a4 a5 a6

Safety 9.4 9.6 9.6 9.4 9.6 9.4

Reliability 3 4 3 5 5 3

Performance 7.5 8.4 8.2 7.8 7.6 8.1

MPG City 44 47 35 43 36 40

MPG HW 40 47 40 39 40 40

Interior/style 8.7 8.1 6.3 7.5 8.3 8

Table 5.15 Car performance sum of values sums

184.168

57

23

47.6

245

246

46.9

Table 5.16 Updated car performance values 0.150949 0.15475 0.20996 0.138461 0.14932 0.19656

0.164912 0.168421 0.168421 0.164912 0.168421 0.164912

0.13043478 0.17391304 0.13043478 0.2173913 0.2173913 0.13043478

0.157563 0.176471 0.172269 0.163866 0.159664 0.170168

0.17959184 0.19183673 0.14285714 0.1755102 0.14693878 0.16326531

0.162602 0.191057 0.162602 0.158537 0.162602 0.162602

0.185501066 0.172707889 0.134328358 0.159914712 0.176972281 0.170575693

(d) Use the entropy formula, where in the case k ¼ 6 (Table 5.17). e j ¼ k (e) (f) (g) (h) (i)

Xn i¼1

  Rij ln Rij

Find ej (Table 5.18), Compute weights by formula (Table 5.19) Check that weights sum to 1, as they did above. Interpret weights and rankings. Use these weights in further analysis. Let’s see the possible weights under another method. AHP: Use template for cars using pairwise comparison (Table 5.20) Results for the weights with a CR of 0.090 are:

Cost Safety Reliability Performance MPG City MPG HW Interior/style

0.3612331 0.2093244 0.14459 0.1166729 0.0801478 0.0529871 0.0350447

e1 0.28542 0.28875 0.32771 0.27376 0.28396 0.31976

e2 0.29723 0.30001 0.30001 0.29723 0.30001 0.29723

e3 0.2656803 0.3042087 0.2656803 0.3317514 0.3317514 0.2656803

e4 0.29117 0.30611 0.30297 0.29639 0.29293 0.30136

Table 5.17 Car performance updated with entropy formula e5 0.3083715 0.31674367 0.27798716 0.30539795 0.28179026 0.29589857

e6 0.29536 0.31623 0.29536 0.29199 0.29536 0.29536

e7 0.31251265 0.30330158 0.26965989 0.293142 0.3064739 0.3016761

k¼

0.558111

5.4 Weighting Methods 215

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Table 5.18 Car performance ej values 0.993081

0.999969

0.98492694

0.999532

0.99689113

0.998825

0.997213162

Ratio Method (Table 5.21).

5.5 5.5.1

Simple Additive Weighting (SAW) Method Description and Uses

This is a very straightforward and easily constructed process. Fisburn has referred to this also as the weighted sum method (Fishburn 1967). SAW is the simplest, and still one of the widest used of the MADM methods. Its simplistic approach makes it easy to use. Depending on the type of the relational data used, we might either want the larger average or the smaller average.

5.5.2

Methodology

Here, each criterion (attribute) is given a weight, and the sum of all weights must be equal to 1. If equally weighted criteria, then we merely need to sum the alternative values. Each alternative is assessed with regard to every criterion (attribute). The overall or composite performance score of an alternative is given simply by Eq. (5.5) with m criteria. Pi ¼

X m

w m =m j¼1 j ij

ð5:5Þ

It was previously thought that all the units in the criteria must be identical units of measure such as dollars, pounds, and seconds. A normalization process can make the values unit less. So, we recommend normalizing the data as shown in Eq. (5.6): Pi ¼

X m

w m =m j ijNormalized j¼1

ð5:6Þ

where (mijNormalized) represents the normalized value of mij, and Pi is the overall or composite score of the alternative Ai. The alternative with the highest value of Pi is considered the best alternative.

w

0.006919 0.234044

3.09E05 0.001046

Table 5.19 Car performance weight

0.01507306 0.50989363

0.000468 0.015834

0.00310887 0.1051674

0.001175 0.039742

0.002786838 0.094273533

0.029561 1

5.5 Simple Additive Weighting (SAW) Method 217

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Table 5.20 Car performance pairwise comparison

Cost Safety Reliability Performance MPG City MPG HW Interior/ style

Cost 1 1 1/2 1/3 1/3 1/4 1/5 1/7

Safety 2 2 1 1/2 1/2 1/3 1/4 1/6

Reliability 3 3 2 1 1/2 1/3 1/4 1/5

Performance 4 3 2 2 1 1/2 1/4 1/6

MPG City 5 4 3 3 2 1 1/3 1/6

MPG HW 6 5 4 4 4 3 1 1/3

Interior/ style 7 7 6 5 6 6 3 1

Table 5.21 Car performance ratio method Cost 70 0.25

5.5.3

Safety 60 0.214

Reliability 50 0.179

Performance 40 0.143

MPG City 30 0.107

MPG HW 20 0.714

Interior/ style 10 0.358

280 Sums to 1

Strengths and Limitations

The strengths are (1) the ease of use and (2) the normalized data allow for comparison across many differing criteria. Limitations include larger is always better or smaller is always better. There is no flexibility in this method to state which criterion should be larger or smaller to achieve better performance. This makes gathering useful data of the same relational value scheme (larger or smaller) essential.

5.5.4

Sensitivity Analysis

Sensitivity analysis should be applied to the weighting scheme employed to determine how sensitive the model is to the weights. Weighting can be arbitrary for a decision-maker or in order to obtain weights you might choose to use a scheme to perform pairwise comparison as we show in AHP that we discuss later. Whenever subjectivity enters into the process for finding weights, then sensitivity analysis is recommended. Please see later sections for a suggested scheme for dealing with sensitivity analysis for individual criteria weights.

5.5 Simple Additive Weighting (SAW) Method

219

Table 5.22 Raw data Cars Prius Fusion Volt Camry Sonata Leaf

Cost ($000) 27.8 28.5 38.668 25.5 27.5 36.2

MPG City 44 47 35 43 36 40

MPG HW 40 47 40 39 40 40

Performance 7.5 8.4 8.2 7.8 7.6 8.1

Interior and style 8.7 8.1 6.3 7.5 8.3 8.0

Safety 9.4 9.6 9.6 9.4 9.6 9.4

Reliability 3 4 3 5 5 3

Table 5.23 SAW using rank ordering of the data by criteria Cars Prius Fusion Volt Camry Sonata Leaf

5.5.5

Cost 3 4 6 1 2 5

MPG City 2 1 6 3 5 4

MPG HW 2 1 2 3 2 2

Perf. 6 1 2 4 5 2

Interior and style 1 3 6 5 2 4

Safety 2 1 1 2 1 2

Reliability 4 3 4 1 1 4

Value 2.857 2 3.857 2.714 2.572 3.285

Rank 4 1 6 1 2 5

Illustrative Examples SAW

Example 1 Car Selection (Data From Consumer’s Report and US News and World Report Online Data) We are considering six cars: Ford Fusion, Toyota Prius, Toyota Camry, Nissan Leaf, Chevy Volt, and Hyundai Sonata. For each car, we have data on seven criteria that were extracted from Consumer’s Report and US News and World Report data sources. They are cost, mpg city, mpg highway, performance, interior and style, safety, and reliability. We provide the extracted information in Table 5.22: Initially, we might assume all weights are equal to obtain a baseline ranking. We substitute the rank orders (first to sixth) for the actual data. We compute the average rank attempting to find the best ranking (smaller is better). We find our rank ordering is Fusion, Sonata, Camry, Prius, Volt, and Leaf (Table 5.23). Next, we apply a scheme to the weights and still use the ranks 1–6 as before. Perhaps we apply a technique similar to the pairwise comparison that we will discuss in the AHP Sect. 5.6. Using the pairwise comparison to obtain new weights, we obtain a new ordering: Camry, Sonata, Fusion, Prius, Leaf, and Volt. The changes in results of the rank ordering differ from using equal weights shows the sensitivity that the model has to be given criteria weights. We assume the criteria in order of importance are: cost, reliability, MPG City, safety, MPG HW, performance, interior, and style. We use pairwise comparisons to obtain a new matrix (Table 5.24):

COST 1 0.5 0.333333 0.25 0.2 0.166667 0.142857

Reliability 2 1 0.5 0.33333333 0.25 0.2 0.2

MPG City 3 2 1 0.333333 0.25 0.2 0.166667

Safety 4 3 3 1 0.5 0.333333 0.25

MPG HW 5 4 4 2 1 0.333333 0.25

Performance 6 5 5 3 3 1 1

Interior/style 7 5 6 4 4 1 1

5

COST Reliability MPG City Safety MPG HW Performance Interior/style

Table 5.24 Pairwise comparison of criteria

220 Introduction to Multi-attribute Military Decision-Making

5.5 Simple Additive Weighting (SAW) Method

221

The CR is 0.01862 and the new weights are: COST Reliability MPG City Safety MPG HW Performance Interior/style

0.38388 0.22224 0.15232 0.08777 0.06675 0.04612 0.04092

Using these weights and applying to the previous ranking, we obtain values that we average and we select the smaller average. We find the rank ordering is Fusion, Sonata, Camry, Prius, Leaf, and Volt. Prius Fusion Volt Camry Sonata Leaf

5.5.5.1

1.209897292 0.801867414 1.470214753 1.15961718 1.015172736 1.343230626

4 1 6 3 2 5

SAW Using Raw Data

We could also use the raw data directly from Table 5.25 except cost given that we using the ranks of the raw data. Now, only cost represents a value where smaller is better so we can replace cost with its reciprocal. So 1/cost represents a variable where larger is better. If we use the criteria weights from the previous results and our raw data replacing cost with 1/cost, we obtain a final ranking based upon larger values are better. Our rank ordering is Camry, Fusion, Sonata, Prius, Leaf, and Volt.

Table 5.25 SAW final ranking

Cars Prius Fusion Volt Camry Sonata Leaf

Value 0.16505 0.17745 0.14177 0.1889 0.1802 0.14663

Rank 4 2 6 1 3 5

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0.25 0.2 Prius Fusion

0.15

Volt Camry

0.1

Sonata Leaf

0.05 0 1

2

3

4

5

Fig. 5.4 Sensitivity analysis of SAW values for cars

5.5.6

Sensitivity Analysis

We suggest employing sensitivity analysis on the criteria weights as described earlier. We modified the weights in a controlled manner and resolved the SAW values. These are displayed in Fig. 5.4 where we see the top ranked cars (Fusion, Camry, and Prius) does not change over our range of sensitivity analysis. Example 2 Kite Network to rank nodes We revisit the Kite Network described earlier. Here, we present two methods that will work on the data from Example 2 from the previous section. Method I representing transforming the output data into rankings from first to last place. Then, we apply the weights and average all the values. We rank them smaller to larger to represent the alternative choices. We present only results using the pairwise compare criteria to obtain the weighted criteria (Table 5.26). Method I rankings: Steve, Susan, Claudia. Tom, Claire, Sarah, Ben Fred, David, and Jennifer. Method II uses the raw metrics data and the weights as above where larger values are better (Table 5.27). The results are Claudia, Susan, Steven, Sarah, Ben, Tom, Claire, Fred, David, and Jennifer. Although the top three are the same, their order is different. The model is sensitive both to the input format and the weights.

5.5.7

Sensitivity Analysis

We can apply sensitivity analysis to the weights, in controlled manner, and determine each changes impact on the final rankings. We recommend a controlled method

Weights Susan Steve Sarah Tom Claire Fred David Claudia Ben Jennifer Susan Steve Sarah Tom Claire Fred David Claudia Ben Jennifer

0.153209 1 2 3 4 4 6 6 6 9 10 0.153209 0.306418 0.459627 0.612835 0.612835 0.919253 0.919253 0.919253 1.37888 1.532089

0.144982 1 2 2 4 4 7 7 6 9 10 0.144982 0.289964 0.289964 0.579928 0.579928 1.014875 1.014875 0.869893 1.304839 1.449821

0.11944 10 2 1 2 2 2 2 2 2 2 1.194396 0.238879 0.11944 0.238879 0.238879 0.238879 0.238879 0.238879 0.238879 0.238879

Table 5.26 Kite Network method I pairwise rankings 0.067199 1 2 10 2 2 2 2 2 2 2 0.067199 0.134398 0.67199 0.134398 0.134398 0.134398 0.134398 0.134398 0.134398 0.134398

0.157688 3 1 1 5 5 7 7 3 9 10 0.473064 0.157688 0.157688 0.78844 0.78844 1.103816 1.103816 0.473064 1.419192 1.576879

0.357482 2 4 7 5 5 8 8 1 3 8 0.714965 1.42993 2.502377 1.787412 1.787412 2.859859 2.859859 0.357482 1.072447 2.859859 0.457969 0.426213 0.700181 0.690316 0.690316 1.04518 1.04518 0.498828 0.924772 1.298654

Steve Susan Claudia Tom Claire Sarah Ben Fred David Jennifer

0.426213 0.457969 0.498828 0.690316 0.690316 0.700181 0.924772 1.04518 1.04518 1.298654

5.5 Simple Additive Weighting (SAW) Method 223

Susan Steve Sarah Tom Claire Fred David Claudia Ben Jennifer

0.027663 0.021279 0.019151 0.017023 0.017023 0.012767 0.012767 0.012767 0.008512 0.004256

0.025384 0.019939 0.019939 0.016586 0.016586 0.013593 0.013593 0.015108 0.003497 0.000757

Table 5.27 Kite Network method II rankings 0.010938 0.011903 0.013226 0.011903 0.011903 0.011903 0.011903 0.011903 0.011903 0.011903

0.007262 0.006743 0.005994 0.006743 0.006743 0.006743 0.006743 0.006743 0.006743 0.006743

0.017164 0.017S33 0.017833 0.015916 0.015916 0.015381 0.015381 0.017164 0.013954 0.011146

0.0723 0.055503 0.037245 0.006938 0.006938 0 0 0.113562 0.064997 0

0.026793 0.0222 0.018898 0.012518 0.012518 0.010065 0.010065 0.029541 0.018268 0.005801

Claudia Susan Steve Sarah Ben Tom Claire Fred David Jennifer

0.029541 0.026793 0.0222 0.018898 0.018268 0.012518 0.012518 0.010065 0.010065 0.005801

224 5 Introduction to Multi-attribute Military Decision-Making

5.6 Analytical Hierarchy Process (AHP) Table 5.28 Hospital procedure Data Version I

225

Profit X-ray times Laboratory time

1 $200 6 5

Procedure 2 $150 5 4

3 $100 4 3

4 $80 3 2

Profit X-ray times Laboratory time

1 $190 6 5

Procedure 2 $150 5 4

3 $110 5 3

4 980 3 3

Table 5.29 Hospital procedure Data version II

to modify the weights. This is discussed later. You are asked in the Exercises set to perform sensitivity analysis to this problem.

5.5.7.1

Exercises 5.5

In each problem, use SAW to find the ranking under these weighted conditions: (a) All weights are equal. (b) Choose and state your weights. 1. 2. 3. 4.

For a given hospital, rank order the procedure using the data in Table 5.28. For a given hospital, rank order the procedure using the data in Table 5.29. Rank order the threats given in Table 5.30. Consider a scenario where we want to move to a new city. Table 5.31 provides our list of search characteristics. Rank the cities to determine the most desirable location. 5. Perform sensitivity analysis to the node ranking for the Kite example. 6. Use the entropy weight method for the cars example and determine the rankings. Compare our results shown in the text.

5.6 5.6.1

Analytical Hierarchy Process (AHP) Description and Uses

AHP is a multi-objective decision analysis tool first proposed by Saaty (1980). It is designed when either subjective and objective measures or just subjective measures are being evaluated in terms of a set of alternatives based upon multiple criteria, organized in a hierarchical structure, see Fig. 5.5.

Threat alternatives \criterion Dirty Bomb Threat Anthrax-Bio Terror Threat DC-Road and Bridge network threat NY subway threat DC Metro Threat Major bank robbery FAA Threat 0.005 12 11 0.0002 0.001

0.73 0.69 0.81 0.70

Approximate associated deaths (000) 10 0.8

200 200 10 5

300

Cost to fix damages in millions 150 10

6.3 2.5 0.57 0.15

0.85

Location density in millions 4.5 3.2

7 7 2 4.5

6

Destructive psychological influence 9 7.5

5 5 16 15

8

Number of intelligence-related tips 3 12

5

0.35

Reliability of threat assessment 0.40 0.45

Table 5.30 Threat assessment matrix

226 Introduction to Multi-attribute Military Decision-Making

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Table 5.31 City search characteristics

City 1 2 3 4 5 6 7 8 9

Affordability of housing (average home cost in hundreds of thousands) 250 325 676 1020 275 290 425 500 300

Cultural opportunities— events per month 5 4 6 10 3 4 6 7 8

Crime rate—number of reported # crimes per month (in hundreds) 10 12 9 6 11 13 12 10 9

Quality of Schools on average (quality rating between [0,1]) 0.75 0.6 0.81 0.8 0.35 0.41 0.62 0.73 0.79

Goal

Criterion 1

Criterion 2

Alternative 1

Criterion 3

Alternative 2

Criterion 4

Alternative 3

Fig. 5.5 Generic AHP hierarchy Table 5.32 Criteria data n RI

1 0

2 0

3 0.52

4 0.89

5 1.1

6 1.24

7 1.35

8 1.4

9 1.45

10 1.49

At the top level is the goal. The next layer has the criteria evaluated or weighted, and at the bottom level the alternatives are measured against each criterion. The decision-maker assesses their evaluation by making pairwise comparisons in which every pair is subjectively or objectively compared. This subjective method involves a 9-point scale that we present later in Table 5.32. We briefly discuss the elements in the framework of AHP. This process can be described as a method to decompose a problem into sub-problems. In most decisions, the decision-maker has a choice among many alternatives. Each alternative has a set of attributes or characteristics that can be measured, either subjectively or objectively. We will call these attributes or criteria. The attribute elements of the

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hierarchal process can relate to any aspect of the decision problem that either tangible or intangible, carefully measured or roughly estimated, well- or poorly understood—anything at all that applies to the decision at hand. We state simply that in order to perform AHP we need a goal or n objective and a set of alternatives, each with criteria (attributes) to compare. Once the hierarchy is built, the decision-makers systematically evaluate the various elements pairwise (by comparing them to one another two at a time), with respect to their impact on an element above them in the hierarchy. In making the comparisons, the decisionmakers can use concrete data about the elements or subjective judgments concerning the elements’ relative meaning and importance. Since we realize humans can easily change their minds, then sensitivity analysis will be very important. The AHP converts these subjective but numerical evaluations to numerical values that can be processed and compared over the entire range of the problem. A numerical weight or priority is derived for each element of the hierarchy, allowing diverse and often incommensurable elements to be compared to one another in a rational and consistent way. In the final step of the process, numerical priorities are calculated for each of the decision alternatives. These numbers represent the alternatives’ relative ability to achieve the decision goal, so they allow a straightforward consideration of the various courses of action. It can be used by individuals working on straightforward decision or teams working on complex problems. It has unique advantages when important elements of the decision are difficult to quantify or compare, or where communication among team members is impeded by their different specializations, terminologies, or perspectives. The techniques to do pairwise comparisons enable one to compare as will be shown in later examples.

5.6.2

Methodology of the Analytic Hierarchy Process

The procedure for using the AHP can be summarized as: Step 1. Build the hierarchy for the decision Goal Criteria Alternatives:

Select the best alternative c1, c2, c3, . . ., cm a1, a2, a3, . . ., an

Step 2. Judgments and Comparison Build a numerical representation using a 9-point scale in a pairwise comparison for the attributes criterion and the alternatives. The goal, in AHP, is to obtain a set of eigenvectors of the system that measures the importance with respect to the criterion. We can put these values into a matrix or table based on the values from Saaty’s 9-point scale, see Table 5.32.

5.6 Analytical Hierarchy Process (AHP)

229

We must ensure that this pairwise matrix is consistent according to Saaty’s scheme to compute the Consistency Ratio, CR. The value of CR must be less than or equal to 0.1 to be considered valid. Next, we approximate the largest eigenvalue, λ, using the power method (see Burden and Faires 2010). We compute the consistency index, CI, using the formula: CI ¼

ð λ  nÞ ð n  1Þ

Then, we compute the CR using: CR ¼

CI RI

If CR  0.1, then our pairwise comparison matrix is consistent and we may continue the AHP process. If not, we must go back to our pairwise comparison and fix the inconsistencies until the CR  0.1. In general, the consistency ensures that if A > B, B > C, that A > C for all A, B, and C all of which can be criteria o. Step 3. Finding all the eigenvectors combined in order to obtain a comparative ranking. Various methods are available for doing this.

5.6.2.1

Methods to Solve for Decision-Maker Weights

The use of technology is suggested to find the weights. We have found Excel a useful technology to assist. 1. Power method of estimated the dominant eigenvectors We suggest the method from Burden and Faires (2013) using the power method as it is straightforward to implement using technology. Definition of a dominant eigenvalue and dominant eigenvector: Let λ1, λ2, ..., λn be the eigenvalues of a n x n matrix A, λ1 is called the dominant eigenvalue of A if |λ1| > |λi|, for i ¼ 2,. . .,n. The eigenvectors corresponding to λ1 are called the dominant eigenvectors of A. The power method to find these eigenvectors is iterative. First, assume that the matrix A has a dominant eigenvalue with corresponding dominant eigenvectors. The choose an initial nonzero vector in Rn as the approximation, x0, of one of the dominant eigenvectors of A. Finally, form the iterative sequence.

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x1 ¼Ax0 x2 ¼Ax1 ¼A2 x0 x3¼Ax2 ¼A3 x0 ⋮ xk ¼Axk-1 ¼Ak x0 2. DDS approximation method (see Fox. W.P (2012). Mathematical modeling of the analytical hierarchy process using discrete dynamical systems in decision analysis, Computers in Education Journal, July-Sept. 27–34). Step 4. After the m  1 criterion weights are found and the n x m matrix for n alternatives by m criterion, we use matrix multiplication to obtain the n  1 final rankings. Step 5. We order the final ranking.

5.6.3

Strengths and Limitations of AHP

Like all modeling and MADM methods, the AHP has strengths and limitations. The main strength of the AHP is its ability to rank choices in the order of their effectiveness in meeting objectives. If the judgments made about the relative importance of criteria and those about the alternatives’ ability to satisfy those objectives have been made in good faith and effort, then the AHP calculations lead to the logical consequence of those judgments. It is quite hard, but not impossible, to manually change the pairwise judgments to get some predetermined result. A further strength of the AHP is its ability to detect inconsistent judgments in the pairwise comparisons using the CR value. If the CR value is greater than 0.1, then the judgments are deemed to be inconsistent. The limitations of the AHP are that it only works because the matrices are all of the same mathematical form. This is known as a positive reciprocal matrix. The reasons for this are explained in Saaty’s material (1980, 1990), so we will simply state that point that is the form that is required. To create such a matrix requires that, if we use the number 9 to represent “A is absolutely more important than B,” then we have to use 1/9 to define the relative importance of B with respect to A. Some people regard that as reasonable; others do not. Another suggested limitation is in the possible scaling. However, understanding that the final values obtained simply say that one alternative is relatively better than another alternative. For example, if the AHP values for alternatives {A, B, C} found were (0.392,0.406,0.204) then they only imply that alternatives A and B are about equally good at approximately 0.4, while C is worse at 0.2. It does not mean that A and B are twice as good as C. The AHP is a useful technique for discriminating between competing options in the light of arrange of objectives to be met. The calculations are not complex and,

5.6 Analytical Hierarchy Process (AHP)

231

while the AHP relies on what might be seen as a mathematical trick, you don’t need to understand the mathematics to use the technique. Be aware that it only shows relative values. Although AHP has been used in many applications in business, industry, and government as can be seen in literature searches of the procedure, Hartwich (1999) noted several limitations. First and foremost, AHP was criticized for not providing sufficient guidance about structuring the problem to be solved, forming the levels of the hierarchy for criteria and alternatives, and aggregating group opinions when team members are geographically dispersed or are subject to time constraints. Team members may carry out rating items individually or as a group. As the levels of hierarchy increase, so does the difficulty and time it takes to synthesize weights. One simple fix involves having the decision-making participants (the analysts and decision-maker) review the basics of the AHP methodology and work through examples so that concepts are thoroughly and easily understood (Hartwich 1999). Another critique of AHP is the “rank reversal” problem. Rank reversal involves the changing in the ordering of the alternatives when the procedure is changed, more alternatives are added, or the criteria changes. This implies that changes in the importance ratings whenever criteria or alternatives are added-to or deleted-from the initial set of alternatives being compared. Several modifications to AHP have been proposed to cope with this and other related issues. Many of the enhancements involved ways of computing, synthesizing pairwise comparisons, and/or normalizing the priority and weighting vectors. We mention the importance of rank reversal now because TOPSIS corrects this rank reversal issue.

5.6.4

Sensitivity Analysis

Since AHP, at least in the pairwise comparisons, is based upon subjective inputs using the 9-point scale then sensitivity analysis is extremely important. Leonelli (2012) in his master’s thesis outlines procedures for sensitivity analysis to enhance decision support tools including numerical incremental analysis of a weight, probabilistic simulations, and mathematical models. How often do we change our minds about the relative importance of an object, place, or thing? Often enough that we should alter the pairwise comparison values to determine how robust our rankings are in the AHP process. We suggest doing enough sensitivity analysis to find the “break point” values, if they exist, of the decision-maker weights that change the rankings of our alternatives. Since the pairwise comparisons are subjective matrices compiled using the Saaty’s method, we suggest as a minimum “trial-and-error” sensitivity analysis using the numerical incremental analysis of the weights. Chen and Kocaoglu (2008) grouped sensitivity analysis into three main groups that he called: numerical incremental analysis, probabilistic simulations, and mathematical models, The numerical incremental analysis, also known as one-at-a-time (OAT) or trial-and-error works by incrementally changing one parameter at a time, finding the new solution and showing graphically how the ranks change. There exist

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several variations of this method (Barker et al., 2011; Hurly 2001). Probabilistic simulations employ the use of Monte Carlo simulation (Butler et al. 1997) that allows random changes in the weights and simultaneously explores the effect on the ranks. Modeling may be used when it is possible to express the relationship between the input data and the solution results. We used Eq. (5.7) (Alinezhad and Amini 2011) for adjusting weights which falls under the incremental analysis: w0j ¼

1  w0p wj 1  wp

ð5:7Þ

where w0j is the new weight and wp is the original weight of the criterion to be adjusted and w0p is the value after the criterion was adjusted. We found this to be an easy method to adjust weights to reenter back into our model.

5.6.5

Illustrative Examples with AHP

Example 1 Car Selection Revisited We revisit Car Selection with our raw data presented in Table 5.22 to illustrate AHP in selecting the best alternative based upon pairwise comparisons of the decision criteria. Step 1. Build the hierarchy and prioritize the criterion from your highest to lower priority. Goal Criteria Alternatives

Select the best car c1, c2, c3, . . ., cm a1, a2, a3, . . ., an

For our cars example, we choose the priority as follows: Cost, MPG City, Safety, Reliability, MPG Highway, Performance, and Interior and Style. Putting these criteria in a priority order allows for an easier assessment of the pairwise comparisons. We used an Excel template prepared for these pairwise comparisons. Step 2. Perform the pairwise comparisons using Saaty’s 9-point scale. We used an Excel template created to organize the pairwise comparisons and obtain the pairwise comparison matrix. This yields the decision criterion matrix presented in Table 5.33, We check the CR, the consistency ratio, to ensure it is less than 0.1. For our pairwise decision matrix, the CR ¼ 0.00695. Since the CR < 0.1, we continue. We find the eigenvector as the decision weights (Table 5.34):

5.6 Analytical Hierarchy Process (AHP)

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Table 5.33 Decision criterion matrix

Cost MPG City MPG HW Safety Reliability Performance Interior and Style

Cost 1 0.5 0.5 0.3333 0.25 0.2 0.166

MPG City 2 1 0.5 0.333 0.25 0.2 0.2

MPG HW 2 2 1 0.5 0.5 0.333 0.333

Safety 3 3 2 1 0.5 0.333

Reliability 4 4 2 1 1 1 0.333

Performance 5 5 3 2 2 1 0.5

Interior and style 6 5 3 3 3 2 1

Table 5.34 Decision weights—eigenvector Cost City HW Safety Reliability Performance Interior and style

0.342407554 0.230887543 0.151297361 0.094091851 0.080127732 0.055515667 0.045672293

Step 3. For the alternatives, we either have the data as we obtained it for each car under each decision criterion or we can use pairwise comparisons by criteria for how each car fares versus its competitors. In this example, we take the raw data from before except now we will use 1/cost to replace cost before we normalize the columns. We have other options for dealing with a criteria and variable like cost. Thus, we have three courses of action, COA, (1) use 1/cost to replace cost, (2) use a pairwise comparison using the 9-point scale, or (3) remove cost from a criteria and a variable, run the analysis, and then do a benefit/cost ratio to re-rank the results. Step 4. We multiply the matrix of the normalized raw data from Consumer Reports and the matrix of weights to obtain the rankings. Using COA (1) from step 3, we obtain the results in Table 5.35. Camry is our first choice, followed by Fusin, Sonata, Leaf, and Volt. If we use method COA (2) in step 3, then within the final matrix we replace the actual costs with these pairwise results (CR ¼ 0.0576): Prius Fusion Volt Camry Sonata Leaf

Cost 0.139595 0.121844 0.041493 0.43029 0.217129 0.049648

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Table 5.35 AHP values and ranking

Introduction to Multi-attribute Military Decision-Making

Cars Prius Fusion Volt Camry Sonata Leaf

Values AHP 0.170857046 0.180776107 0.143888039 0.181037124 0.171051618 0.152825065

Rank 4 2 6 1 3 5

Then, we obtain the ranked results as: Cars Prius Fusion Volt Camry Sonata Leaf

Values AHP 0.14708107 0.152831274 0.106011611 0.252350537 0.173520854 0.113089654

Rank 4 3 6 1 2 5

If we do COA (3) in step 3, then this method requires us to redo the pairwise criterion matrix without the cost criteria. These weights are: City MPG HW MPG Safety Reliability Performance Interior/Style

0.363386 0.241683 0.159679 0.097 0.081418 0.056834

We normalize the original costs from Table 5.22, and divide these ranked values by the normalized cost to obtain a cost/benefit value. These are shown in ranked order: Camry Fusion Prius Sonata Leaf Volt

1.211261 1.178748 1.10449 1.06931 0.821187 0.759482

5.6 Analytical Hierarchy Process (AHP)

5.6.5.1

235

Sensitivity Analysis

We alter our decision pairwise values to obtain a new set of decision weights to use in COA (1) from step 3 to obtain new results: Camry, Fusion, Sonata, Prius, Leaf, and Volt. The new weights and model’s results are: Cost MPG City MPG HW Performance Interior Safety Reliability Alternatives Prius Fusion Volt Camry Sonata Leaf

0.311155922 0.133614062 0.095786226 0.055068606 0.049997069 0.129371535 0.225006578 Values 0.10882648 0.11927995 0.04816882 0.18399172 0.11816156 0.04357927

4 2 5 1 3 6

The resulting values have changed but not the relative rankings of the cars. Again, we recommend using sensitivity analysis to find a “break point,” if one exists. We systemically varied the cost weights using Eq. (5.5) with increments of () 0.05. We potted the results to show the approximate break point of the criteria cost as weight of cost +0.1 as shown in Fig. 5.6. Prius Fusion Volt Camry Sonata Leaf

0.170857 0.180776 0.143888 0.181037 0.171052 0.152825

0.170181 0.18119 0.145003 0.179903 0.170242 0.15395

0.169505 0.181604 0.146118 0.178768 0.169431 0.155074

0.16883 0.182018 0.147232 0.177634 0.168621 0.156198

We see that as cost decrease in weight and other criteria proportionally increase that Fusion overtakes Camry as number 1. Example 2 Kite Network Revisited with AHP Assume all we have are the outputs from ORA which we do not show here due to the volume of output produced. We take the metrics from ORA and normalize each column. The columns for each criterion are placed in a matrix X with entries, xij. We define wj as the weights for each criterion. Next, we assume we can obtain pairwise comparison matrix from the decisionmaker concerning the criterion. We use the output from ORA and normalize the results for AHP to rate the alternatives within each criterion. We provide a sample

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0.2 0.18 0.16 Prius

0.14

Fusion

0.12

Volt

0.1 0.08

Camry

0.06

Sonata

0.04

Leaf

0.02 0 1

2

3

4

Fig. 5.6 Camry overtakes fusion as the top alternative as we change the weight of Cost Table 5.36 Kite Network pairwise comparison matrix

Central Eigenvector In-degree Out-degree Information centrality Betweenness

Central 1 1= 3 1= 2 1= 2 2

Eigenvector 3 1 3 1= 2 2

Indegree 2 1= 3 1 1 4

Outdegree 2 1 1= 2 1 4

Information centrality 1= 2 2 1= 2 1= 4 1

1= 3 1= 2 1= 4 1= 4 1= 3

3

2

4

4

3

1

Betweenness

pairwise comparison matrix for weighting the criterion from the Kite example using Saaty’s 9-point scale. The CR is 0.0828, which is less than 0.1, so our pairwise matrix is consistent and we continue.

5.6.5.2

Pairwise Comparison Matrix (Table 5.36)

We obtain the steady-state values that will be our criterion weights, where the sum of the weights equals 1.0. There exist many methods to obtain these weights. The methods used here are the power method from numerical analysis (Burden et al. 2013) and discrete dynamical systems (Fox 2012; Giordano et al. 2014). 0.1532 0.1450 0.1194 0.0672 0.1577 0.3575

0.1532 0.1450 0.1195 0.0672 0.1577 0.3575

0.1532 0.1450 0.1194 0.0672 0.1577 0.3575

0.1532 0.1450 0.1194 0.0672 0.1577 0.3575

0.1532 0.1450 0.1194 0.0672 0.1577 0.3575

0.1532 0.1450 0.1194 0.0672 0.1577 0.3575

5.6 Analytical Hierarchy Process (AHP) Table 5.37 Kite Network rankings

Node Susan Steven Sarah Tom Claire Fred David Claudia Ben Jennifer

237 AHP Value 0.160762473 0.133201647 0.113388361 0.075107843 0.075107843 0.060386019 0.060386019 0.177251415 0.109606727 0.034801653

Rank 2 3 4 6 6 8 8 1 5 10

These values provide the weights for each criterion: centrality ¼ 0.1532, eigenvectors ¼ 0.1450, in-centrality ¼ 0.1194, out-centrality ¼ 0.0672, information centrality ¼ 0.1577, and betweenness ¼ 0.3575. We multiply the matrix of the weights and the normalized matrix of metrics from ORA to obtain our output and ranking (Table 5.37): For this example, with AHP Claudia, cl, is the key node. However, the bias of the decision-maker is important in the analysis of the criterion weights. The criterion, “Betweenness,” is two to three times more important than the other criterion.

5.6.5.3

Sensitivity Analysis

Changes in the pairwise decision criterion will cause fluctuations in the key nodes. We change our pairwise comparison so that “Betweenness” is not so dominant a criterion. With these slight pairwise changes, we now find Susan is ranked first, followed by Steven and then Claudia. The AHP process is sensitive to changes in the criterion weights. We vary betweenness in increments of 0.05 to find the break point (Table 5.38). With these slight pairwise changes, we now find Susan is ranked first, followed by Steven and then Claudia. The AHP process is sensitive to changes in the criterion weights. We vary betweenness in increments of 0.05 to find the break point (Table 5.39). Further, sensitivity analysis of the nodes is provided in Fig. 5.7. We varied the weight of the criterion Betweenness by lowering it by 0.05 each iteration and increasing the other weights using Eq. (5.1). We see the Claudia and Susan change as the top node when we reduce Betweenness by 0.1.

t c f s su st d cl b j 10 alternatives and 9 attributes or criterion Criterion weights

IN 0.111111 0.111111 0.083333 0.138889 0.166667 0.138889 0.083333 0.083333 0.055556 0.027778

0.034486 0.037178 0.045778 0.398079 0.055033 0.086323 0.135133 0.207991

Centrality 0.111111 0.111111 0.083333 0.125 0.180556 0.138889 0.083333 0.083333 0.055556 0.027778

w1 w2 w3 w4 w5 w6 w7 w8

Table 5.38 Kite Network with modified betweenness values OUT 0.111111 0.111111 0.083333 0.111111 0.194444 0.138889 0.083333 0.083333 0.055556 0.027778

Eigen 0.114399 0.114399 0.093758 0.137528 0.175081 0.137528 0.093758 0.104203 0.024123 0.005223

EIGENC 0.114507 0.114507 0.094004 0.137331 0.174855 0.137331 0.094004 0.104062 0.023985 0.005416

Close 0.100734 0.100734 0.097348 0.100734 0.122743 0.112867 0.097348 0.108634 0.088318 0.070542

IN-Close 0.099804 0.099804 0.09645 0.111826 0.107632 0.111826 0.107632 0.107632 0.087503 0.069891

Betw 0.019408 0.019408 0 0.104188 0.202247 0.15526 0 0.317671 0.181818 0

INFO Cen. 0.110889 0.108891 0.097902 0.112887 0.132867 0.123876 0.100899 0.110889 0.061938 0.038961

238 5 Introduction to Multi-attribute Military Decision-Making

5.6 Analytical Hierarchy Process (AHP) Table 5.39 Kite Network with modified betweenness values rankings

Tom Claire Fred Sarah Susan Steven David Claudia Ben Jennifer

239

0.098628 0.098212 0.081731 0.12264 0.161609 0.133528 0.083319 0.133428 0.0645 0.022405

Susan Steven Claudia Sarah Tom Claire David Fred Ben Jennifer

0.161609 0.133528 0.133428 0.12264 0.098628 0.098212 0.083319 0.081731 0.0645 0.022405

0.2 0.18

Susan

0.16

Steven

0.14

Sarah

0.12

Tom

0.1

Claire

0.08

Fred

0.06 0.04

David

0.02

Claudia Ben

0 1

2

3

4

Fig. 5.7 Sensitivity analysis for nodes varying only Betweenness

5.6.5.4

Section 5.6 Exercises

1. For the problems in Sect. 5.4, solve by AHP. Compare your results to your results using SAW. 2. Perform sensitivity analysis by changing the weight of your highest criteria weight until it is no longer the highest weighted criteria. Did it change the rankings?

5.6.5.5

Section 5.6 Projects

Construct a computer program to find the weights using AHP using the power method.

240

5.7 5.7.1

5

Introduction to Multi-attribute Military Decision-Making

Technique of Order Preference by Similarity to the Ideal Solution (TOPSIS) Description and Uses

The Technique for Order of Preference by Similarity to Ideal Solution (TOPSIS) is a multi-criteria decision analysis method, which was originally developed in a dissertation from Kansas State University (Hwang and Yoon 1981). It has been further developed by others (Yoon 1987; Hwang et al. 1993). TOPSIS is based on the concept that the chosen alternative should have the shortest geometric distance from the positive ideal solution and the longest geometric distance from the negative ideal solution. It is a method of compensatory aggregation that compares a set of alternatives by identifying weights for each criterion, normalizing the scores for each criterion and calculating the geometric distance between each alternative and the ideal alternative, which is the best score in each criterion. An assumption of TOPSIS is that the criteria are monotonically increasing or decreasing. Normalization is usually required as the parameters or criteria are often of incompatible dimensions in multi-criteria problems. Compensatory methods such as TOPSIS allow trade-offs between criteria, where a poor result in one criterion can be negated by a good result in another criterion. This provides a more realistic form of modeling than non-compensatory methods, which include or exclude alternative solutions based on hard cut-offs. We only desire to briefly discuss the elements in the framework of TOPSIS. TOPSIS can be described as a method to decompose a problem into sub-problems. In most decisions, the decision-maker has a choice among many alternatives. Each alternative has a set of attributes or characteristics that can be measured, either subjectively or objectively. The attribute elements of the hierarchal process can relate to any aspect of the decision problem whether tangible or intangible, carefully measured or roughly estimated, well or poorly understood information. Basically anything at all that applies to the decision at hand can be used in the TOPSIS process.

5.7.2

Methodology

The TOPSIS process is carried out as follows: Step 1 Create an evaluation matrix consisting of m alternatives and n criteria, with the intersection of each alternative and criterion given as xij, giving us a matrix (Xij)m x n.

5.7 Technique of Order Preference by Similarity to the Ideal Solution (TOPSIS)

D¼

A1 A2 A3 ⋮ Am

2 x1 x11 6 x21 6 6 x31 6 4⋮ xm1

x2 x12 x22 x32 ⋮ xm2

241

x3 . . . xn 3 x13 . . . x1n x23 . . . x2n 7 7 x33 . . . x3n 7 7 ⋮ ⋮ 5 xm3 . . . xmn

Step 2 The matrix shown as D above then is normalized to form the matrix R¼(Rij)m x n as shown using the normalization method xij r ij ¼ qffiffiffiffiffiffiffiffiffiffi P 2 xij for i ¼ 1,2. . .,m; j ¼ 1,2,. . .n Step 3 Calculate the weighted normalized decision matrix. First, we need the weights. Weights can come from either the decision-maker or by computation. Step 3a. Use either the decision-maker’s weights for the attributes x1, x2,.. xn or compute the weights through the use of Saaty’s (1980) AHP decision-maker weights method to obtain the weights as the eigenvector to the attributes versus attribute pairwise comparison matrix. n X

wj ¼ 1

j¼1

The sum of the weights over all attributes must be equal to 1 regardless of the method used. Use the methods described in Sect. 5.5.2 to find these weights. Step 3b. Multiply the weights to each of the column entries in the matrix from Step 2 to obtain the matrix, T.     T ¼ t ij m x n ¼ w j r ij m x n ,i ¼ 1,2, . . . , m Step 4 Determine the worst alternative (Aw) and the best alternative (Ab): Examine each attribute’s column and select the largest and smallest values appropriately. If the values imply larger is better (profit), then the best alternatives are the largest values, and if the values imply smaller is better (such as cost), then the best alternative is the smallest value.     Aw ¼ hmax t ij ji ¼ 1,2, . . . , m j j 2 J  i; hmin t ij ji ¼ 1,2, . . . , m j j 2 J þ i  t wj jj ¼ 1,2, . . . , n ,     Awb ¼ hmin t ij ji ¼ 1,2, . . . , m j j 2 J  i; hmax t ij ji ¼ 1,2, . . . , m j j 2 J þ i  t bj jj ¼ 1,2, . . . , n , where

242

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Introduction to Multi-attribute Military Decision-Making

J+ ¼ {j ¼ 1, 2, . . . n| j) is associated with the criteria having a positive impact. J ¼ {j ¼ 1, 2, . . . n| j) is associated with the criteria having a negative impact. We suggest that if possible make all entry values in terms of positive impacts. Step 5 Calculate the L2-distance between the target alternative i and the worst condition Aw d iw

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi Xn   t  t wj 2 ,i ¼ 1; 2, . . . , m ¼ j¼1 ij

and then calculate the distance between the alternative i and the best condition Ab dib ¼

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Xn   t  t bj 2 ,i ¼ 1; 2, . . . m j¼1 ij

where diw and dib are L2-norm distances from the target alternative i to the worst and best conditions, respectively. Step 6 Calculate the similarity to the worst condition: siw ¼

diw ,0  siw  1,i ¼ 1,2, . . . , m ðdiw þ dib Þ

Siw ¼ 1 if and only if the alternative solution has the worst condition. Siw ¼ 0 if and only if the alternative solution has the best condition. Step 7 Rank the alternatives according to their value from Siw (i¼1, 2,. . .,m) .

5.7.2.1

Normalization

Two methods of normalization that have been used to deal with incongruous criteria dimensions are linear normalization and vector normalization. Normalization can be calculated as in Step 2 of the TOPSIS process above. Vector normalization was incorporated with the original development of the TOPSIS method (Yoon 1987) and is calculated using the following formula: xij r ij ¼ qffiffiffiffiffiffiffiffiffiffi P 2 for i ¼ 1; 2 . . . ,m;j ¼ 1; 2, . . . n xij In using vector normalization, the nonlinear distances between single dimension scores and ratios should produce smoother trade-offs (Hwang and Yoon 1981). Let’s suggest two options for the weights in Step 3. First, the decision-maker might actually have a weighting scheme that they want the analyst to use. If not, we suggest using Saaty’s 9-point pairwise method developed for the Analytical Hierarchy Process (AHP) (Saaty 1980) to obtain the criteria weights as described in the previous section.

5.7 Technique of Order Preference by Similarity to the Ideal Solution (TOPSIS)

5.7.3

243

Strengths and Limitations

TOPSIS is based on the concept that the chosen alternative should have the shortest geometric distance from the positive ideal solution and the longest geometric distance from the negative ideal solution. TOPSIS is a method of many steps that compares a set of alternatives by identifying weights for each criterion, normalizing scores for each criterion and calculating the geometric distance between each alternative and the ideal alternative, which is the best score in each criterion.

5.7.4

Sensitivity Analysis

The decision weights are subject to sensitivity analysis to determine how they affect the final ranking. The same procedures discussed in Sect. 5.5 are valid here. Sensitivity analysis is essential to good analysis. Additionally, Alinezhad and Amini (2011) suggests sensitivity analysis for TOPSIS for changing an attribute weight. We will again use Eq. (5.6) in our sensitivity analysis.

5.7.5

Illustrate Examples with TOPSIS

Example 1 Car Selection Revisited (Table 5.22) We might assume that our decision-maker weights from the AHP section are still valid for our use. Weights from before: Cost MPG City MPGHW Performance Interior Safety Reliability

0.38960838 0.11759671 0.04836533 0.0698967 0.05785692 0.10540328 0.21127268

We use the identical data from the car example from AHP but we apply steps 3–7 from TOPSIS to our data (Table 5.40). We are able to keep the cost data and just inform TOPSIS that a smaller cost is better. We obtained the rank ordering of the cars: Camry, Fusion, Prius, Sonata, Volt, and Leaf (Table 5.41). It is critical to perform sensitivity analysis on the weights to see how they affect the final ranking. This time we work toward finding the break point where the order of cars actually changes. Since cost is the largest criterion weight, we vary it using

Cost 1 0.25 0.166667 0.2 0.166667 0.25 0.5 0

MPG_city 4 1 0.166667 0.333333 0.2 1 3 0

MPG_HW 6 6 1 2 2 3 4 0

Perf. 5 3 0.5 1 0.5 2 3 0

Interior 6 5 0.5 2 1 2 3 0

Safety 4 1 0.333333 0.5 0.5 1 2 0

Reliability 2 0.333333333 0.25 0.333333333 0.333333333 0.5 1 0

N/A 0 0 0 0 0 0 0 1

5

Cost MPG_city MPG_HW Perf. Interior Safety Reliability N/A

Table 5.40 Car pairwise comparisons

244 Introduction to Multi-attribute Military Decision-Making

5.7 Technique of Order Preference by Similarity to the Ideal Solution (TOPSIS) Table 5.41 Car decision criterion rankings

Car Camry Fusion Prius Sonata Leaf Volt

TOPSIS value 0.8215 0.74623 0.7289 0.70182 0.15581 0.11772

245 Rank 1 2 3 4 5 6

0.9 0.8 0.7

Prius

0.6

Fusion

0.5

Volt

0.4

Camry

0.3

Sonata

0.2

Leaf

0.1 0 1

2

3

4

Fig. 5.8 TOPSIS values of the cars by varying the weight for cost incrementally by 0.05 each of four increments along the x-axis

Eq. (5.5) in increments of 0.05. We see from Fig. 5.8, the Fusion overtakes Camry when cost is decreased by about 0.1, which allows reliability to overtake cost as the dominate-weighted decision criterion. Example 2 Social Networks We revisit the Kite Network with TOPSIS to find influences in the network. We present the extended output from ORA that we used in Table 5.42. We use the decision weights from AHP (unless a decision-maker gives us their own weights) and find the eigenvectors for our eight metrics (Table 5.43). We take the metrics from ORA and perform steps 2–7 of TOPSIS to obtain the results: We rank order the final output from TOPSIS as shown in the last column of Table 5.44. We interpret the results as follows: The key node is Susan followed by Steven, Sarah, Tom, and Claire.

5.7.5.1

Sensitivity Analysis

We used Eq. (5.7) and systemically altered the value of the largest criteria weight, EigenL and depict this in Fig. 5.9. We note that Susan remains the most influential node.

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Introduction to Multi-attribute Military Decision-Making

Table 5.42 Summary of extended ORA’s output for Kite Network IN 0.4 0.4 0.3 0.5 0.6 0.5 0.3 0.3 0.2 0.1

Tom Claire Fred Sarah Susan Steven David Claudia Ben Jennifer

OUT 0.4 0.4 0.3 0.4 0.7 0.5 0.3 0.3 0.2 0.1

Eigen 0.46 0.46 0.377 0.553 0.704 0.553 0.377 0.419 0.097 0.021

EigenL 0.296 0.296 0.243 0.355 0.452 0.355 0.243 0.269 0.062 0.014

Close 0.357 0.357 0.345 0.357 0.435 0.4 0.345 0.385 0.313 0.25

IN-Close 0.357 0.357 0.345 0.4 0.385 0.4 0.385 0.385 0.313 0.25

Betweenness 0.019 0.019 0 0.102 0.198 0.152 0 0.311 0.178 0

INF Centre 0.111 0.109 0.098 0.113 0.133 0.124 0.101 0.111 0.062 0.039

Table 5.43 Kite Network decision criterion rankings w1 w2 w3 w4 w5 w6 w7 w8

0.034486 0.037178 0.045778 0.398079 0.055033 0.086323 0.135133 0.207991

Table 5.44 Kite Network TOPSIS output S+ 0.0273861 0.0497878 0.0565358 0.0801011 0.0803318 0.10599 0.1112243 0.1115873 0.1714404 0.2042871

5.7.5.2

S– 0.181270536 0.148965362 0.14154449 0.134445151 0.133785196 0.138108941 0.12987004 0.128942016 0.113580988 0.130399883

C 0.86875041 0.749499497 0.714581437 0.626648721 0.624822765 0.565790826 0.538668909 0.536076177 0.398499927 0.389617444

SUSAN STEVEN SARAH TOM CLAIRE CLAUDIA DAVID ERED BEN JENNIFER

Comparison of Results for the Kite Network

We have also used the two other MADM methods to rank order our nodes in previous work in SNA (Fox and Everton 2013). When we applied data envelopment analysis and AHP to compare to TOPSIS, we obtained the results displayed in Table 5.45 for the Kite Network.

5.7 Technique of Order Preference by Similarity to the Ideal Solution (TOPSIS)

247

1 0.9

TOM

0.8

CLAIRE

0.7

FRED

0.6

SARAH

0.5

SUSAN

0.4

STEVEN

0.3 0.2

DAVID

0.1

CLAUDIA

0

1

2

3

4

5

BEN

Fig. 5.9 Sensitivity analysis plot as a function of varying EigenL weight in increments of 0.05 units Table 5.45 MADM applied to Kite Network Node Susan Sarah Steven Claire Fred David Claudia Ben Jennifer Tom

SAW 0.046 (1) 0.021 (4) 0.026 (3) 0.0115 (7) 0.0115 (7) 0.031 (2) 0.012 (8) 0.018 (5) 0.005 (10) 0.0143 (6)

TOPSIS value (rank) 0.862 (1) 0.675 (3) 0.721 (2) 0.649 (4) 0.446 (8) 0.449 (7) 0.540 (6) 0.246 (9) 0 (10) 0.542 (5)

DEA efficiency value (rank) 1 (1) 0.786 (2) 0.786 (2) 0.653 (4) 0.653 (4) 0.536 (8) 0.595 (6) 0.138 (9) 0.030 (10) 0.553 (7)

AHP value (rank) 0.159 (2) 0.113 (4) 0.133 (3) 0.076 (6) 0.061 (8) 0.061 (8) 0.176 (1) 0.109 (5) 0.036(10) 0.076 (6)

It might be useful to use this table as input for another round of one of these presented methods and then use sensitivity analysis.

5.7.5.3

Section 5.7 Exercises

1. For the problems in Sect. 5.4, solve by TOPSIS. Compare your results to your results using both SAW and AHP. 2. Perform sensitivity analysis by changing the weight of your highest criteria weight until it is no longer the highest weighted criteria. Did it change the rankings?

248

5.7.5.4

5

Introduction to Multi-attribute Military Decision-Making

Section 5.7 Projects

1. Write a program using the technology of your choice to implement any of all of the following: (a) SAW, (b) AHP, and (C) TOPSIS. 2. Enable your program in (1) to perform sensitivity analysis.

References Alinezhad, A., & Amini, A. (2011). Sensitivity analysis of TOPSIS technique: The results of change in the weight of one attribute on the final ranking of alternatives. Journal of Optimization in Industrial Engineering, 7, 23–28. http://onlinepubs.trb.org/onlinepubs/tcrp/tcrp_rpt_ 131AppF.pdf. Baker, T., & Zabinsky, Z. (2011). A multicriteria decision making model for reverse logistics using analytical hierarchy process. Omega, 39, 558–573. http://onlinepubs.trb.org/onlinepubs/tcrp/ tcrp_rpt_131AppF.pdf. Burden, R., & Faires, D. (2013). Numerical analysis (9th ed.). Boston: Cengage Publishers. http:// onlinepubs.trb.org/onlinepubs/tcrp/tcrp_rpt_131AppF.pdf. Butler, J., Jia, J., & Dyer, J. (1997). Simulation techniques for the sensitivity analysis of multi-criteria decision models. European Journal of Operations Research, 103, 531–546. Callen, J. (1991). Data envelopment analysis: Practical survey and managerial accounting applications. Journal of Management Accounting Research, 3, 35–57. Carley, K. M. (2011). Organizational risk analyzer (ORA). Pittsburgh: Center for Computational Analysis of Social and Organizational Systems (CASOS): Carnegie Mellon University. Charnes, A. W., Cooper, W., & Rhodes, E. (1978). Measuring the efficiency of decision making units. European Journal of Operations Research, 2, 429–444. Chen, H., & Kocaoglu, D. (2008). A sensitivity analysis algorithm for hierarchical decision models. European Journal of Operations Research, 185(1), 266–288. Cooper, W., Seiford, L., & Tone, K. (2000). Data envelopment analysis. Boston: Kluwer Academic Press. Figueroa, S. (2014). Improving recruiting in the 6th recruiting brigade through statistical analysis and efficiency measures. Master’s Thesis, Naval Postgraduate School. Fishburn, P. C. (1967). Additive utilities with incomplete product set: Applications to priorities and assignments. Operations Research Society of America (ORSA), 15, 537–542. Fox, W. P. (2012). Mathematical modeling of the analytical hierarchy process using discrete dynamical systems in decision analysis. Computers in Education Journal, 22, 27–34. Fox, W. P., Ormond, B., & Williams, A. (2014). Ranking terrorist targets using a hybrid AHPTOPSIS methodology. Journal of Defense Modeling and Simulation, 13(1), 77–93.. First published date: January-08-2015, http://onlinepubs.trb.org/onlinepubs/tcrp/tcrp_rpt_131AppF. pdf. https://doi.org/10.1177/1548512914563619. Fox, W. P., Greaver, B., Raabe, L., & Burks, R. (2017). CARVER 2.0: Integrating multi-attribute decision making with AHP in center of gravity vulnerability analysis. Journal of Defense Modeling and Simulation, 15(2), 154851291771705. Published on-line first, Retrieved June 29, 2017, from http://journals.sagepub.com/doi/abs/10.1177/1548512917717054; http:// onlinepubs.trb.org/onlinepubs/tcrp/tcrp_rpt_131AppF.pdf. Fox, W., & Everton, S. (2013). Mathematical modeling in social network analysis: Using TOPSIS to find node influences in a social network. Journal of Mathematics and System Science, 3(10), 531–541. Fox, W., & Everton, S. (2014). Mathematical modeling in social network analysis: Using data envelopment analysis and analytical hierarchy process to find node influences in a social network. Journal of Defense Modeling and Simulation, 2, 1–9. Giordano, F.R. , Fox, W. & . Horton S. (2014). A first course in mathematical modeling (5th ed.), Brooks-Cole Publishers. Boston

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Hartwich, F. (1999). Weighting of agricultural research results: Strength and limitations of the analytic hierarchy process (AHP). Universitat Hohenheim. Retrieved from https://entwicklungspolitik.unihohenheim.de/uploads/media/DP_09_1999_Hartwich_02.pdf. Hurly, W. J. (2001). The analytical hierarchy process: A note on an approach to sensitivity which preserves rank order. Computers and Operations Research, 28, 185–188. Hwang, C. L., & Yoon, K. (1981). Multiple attribute decision making: Methods and applications. Berlin: Springer. Hwang, C. L., Lai, Y., & Liu, T. Y. (1993). A new approach for multiple objective decision making. Computers and Operational Research, 20, 889–899. Krackhardt, D. (1990). Assessing the political landscape: Structure, cognition, and power in organizations. Administrative Science Quarterly, 35, 342–369. Leonelli, R. (2012). Enhancing a decision support tool with sensitivity analysis. Thesis, University of Manchester. Neralic, L. (1998). Sensitivity analysis in models of data envelopment analysis. Mathematical Communications, 3, 41–59. Saaty, T. (1980). The analytical hierarchy process. New York: McGraw Hill. http://onlinepubs.trb. org/onlinepubs/tcrp/tcrp_rpt_131AppF.pdf. Saaty, T. (1990). How to make decisions; the AHP. European Journal of Operation Research, 48 (1), 9–26. http://onlinepubs.trb.org/onlinepubs/tcrp/tcrp_rpt_131AppF.pdf. Shannon, C. E., & Weaver, W. (1949). The mathematical theory of communication. Champaign, IL: University of Illinois Press. ISBN 0-252-72548-4. http://onlinepubs.trb.org/onlinepubs/tcrp/ tcrp_rpt_131AppF.pdf. Temesi, J. (2006). Consistency of the decision-maker in pairwise comparisons. International Journal of Management and Decision Making, 7(2). http://onlinepubs.trb.org/onlinepubs/tcrp/ tcrp_rpt_131AppF.pdf. Trick, M. A. (1996). Multiple criteria decision making for consultants. Retrieved April 2014, from http://mat.gsia.cmu.edu/classes/mstc/multiple/multiple.html. Trick, M. A. (2014). Data envelopment analysis, Chapter 12. Retrieved April 2014, from http://mat. gsia.cmu.edu/classes/QUANT/NOTES/chap12.pdf. Weber, M., & Borcherding, K. (1993). Behavioral influences on weight judgments in multiattribute decision making. European Journal of Operational Research, 67, 1–12. http://onlinepubs.trb. org/onlinepubs/tcrp/tcrp_rpt_131AppF.pdf. Winston, W. (1995). Introduction to mathematical programming (pp. 322–325). Pacific Grove: Duxbury Press. Yoon, K. (1987). A reconciliation among discrete compromise situations. Journal of Operational Research Society, 38, 277–286. http://onlinepubs.trb.org/onlinepubs/tcrp/tcrp_rpt_131AppF. pdf. Zeleny, M. (1982). Multiple criteria decision making. New York: McGraw Hill. http://onlinepubs. trb.org/onlinepubs/tcrp/tcrp_rpt_131AppF.pdf.

Suggested Readings Baker, T., & Zabinsky, Z. (2011). A multicriteria decision making model for reverse logistics using analytical hierarchy process. Omega, 39, 558–573. Consumer’s Reports Car Guide. (2012). The editors of consumer reports. Cooper, W., Li, S., Seiford, L. M., Thrall, R. M., & Zhu, J. (2001). Sensitivity and stability analysis in DEA: Some recent developments. Journal of Productivity Analysis, 15(3), 217–246. Thanassoulis, E. (2011). Introduction to the theory and application of data envelopment analysis— A foundation text with integrated software. London: Kluwer Academic Press. Zhenhua, G. (2009). The application of DEA/AHP method to supplier selection. In: International Conference on Information Management, Innovation Management and Industrial Engineering (pp. 449–451).

Chapter 6

Game Theory

Objectives 1. 2. 3. 4.

Know the concept of formulating a two-person and three-person game. Understand total and partial conflict games. Understand solution methodologies for each type of game. Understand and interpret the solutions.

6.1

Introduction to Game Theory

According to Wasburn and Kress (2009), “military operations are conducted in the presence of uncertainty, much of which is due to the unpredictability of the enemy.” Further they state that there are two fundamental directions to go: game theory or wargaming. We discuss only game theory here in this report. According to Wasburn and Kress (2009) in their discussions, they limit analysis to the two-person zero-sum games for two reasons: (1) combat usually involves two opposing sides and (2) the two-person zero-sum solutions methods are more easily generalizable than the partial conflict (nonzero-sum) games. I think realism is essential in modeling and therefore cannot exclude partial conflict games from any of the analysis presented in this chapter. Military decision-making is a process that blends engineering, management, and business processes. As such the ability to make decision as well as model the decision-making process may be critical steps in the process. In game theory, we employ the process to gain insights into possible courses of action from each player assuming the players are rational, that is they want to maximize their gains. In many military situations, two or more decision-makers simultaneously and without communications choose courses of actions, and the action chosen by each affects the payoff or gains earned by all the other players. For example, consider a © Springer Nature Switzerland AG 2019 W. P. Fox, R. Burks, Applications of Operations Research and Management Science for Military Decision Making, International Series in Operations Research & Management Science 283, https://doi.org/10.1007/978-3-030-20569-0_6

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Table 6.1 Payoff Matrix, M, of a two-person total conflict game Player 1, Rose’s strategies Column 1 Row 1 M1,1,N1,1 Row 2 M2,1N2,1 . . . . . . Row m Mm,1,Nm,1

Column 2 M1,2N1,2 M22,N22 . . . Mm,2Nm,2

Player 2, Colin’s strategies ... Column n ... M1,nN1,n ... M2,n,N2,n . . . . . . ... Mm,n,Nm,n

fast food chain such as Burger King. If they choose an advertising strategy with pricing not only do they help their payoffs but their choices also affect all other fast food chains. Each company’s decision affects the revenues, profits, losses of the other fast food chains. Game theory is useful in analyzing decisions in cases where two or more decision-makers have conflicting interest. Most of what we present here concerns only the two-person game but we will also briefly examine the n-person game. In two-person games, each of the players has strategies or courses of action that they might choose. These courses of action lead to outcomes or payoffs to the decisionmaker and these payoffs might be any values (positive, negative, or zero). These payoffs are usually presented in a payoff matrix such as the general one presented in Table 6.1. In Table 6.1 player 1, whom we will call Rose, might have m course of actions available and player 2, whom we will call Colin, may have n courses of actions available. These payoff values might have come from ordinal utilities or cardinal utilities. For more information about obtaining payoff values, please see the additional reading (Straffin 2004; Von Neumann and Morgenstern 2004). Game theory is the branch of mathematics and decision theory concerned with strategic decisions when two or more players compete. The problems of interest involve multiple participants, each of whom has individual strategies related to a common system or shared resources. Because game theory arose from the analysis of competitive scenarios, the problems are called games and the participants are called players. But these techniques apply to more than just sport and are not even limited to competitive situations. In short, game theory deals with any problem in which each player’s strategy depends on what the other players do. Situations involving interdependent decisions arise frequently, in all walks of life. A few examples in which game theory might be used include: • • • • • • • •

Friends choosing where to go have dinner Couples deciding between going to ballet or a sporting event Parents trying to get children to behave Commuters deciding how best to travel to work Businesses competing in a fair market Diplomats negotiating a treaty Gamblers betting in a game of chance Military strategists weighing alternatives, such as attack or defend

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Table 6.2 Army versus Navy Recruiting Army

• • • •

Large City Small City

Navy Large City (60,40) (50,50)

Small City (75,25) (58,42)

Governmental diplomacy options for sanctions or actions Pitcher-batter dual in baseball or penalty kicker-goalies dual in soccer Manhunt situations: Searching for hiding terrorists Implementations of military or diplomatic sanctions

All of these situations call for strategic thinking, making use of available information to devise the best plan to achieve one’s objectives. Perhaps you are already familiar with assessing costs and benefits in order to make informed decisions between several options. Game theory simply extends this concept to interdependent decisions, in which the options being evaluated are functions of the players’ choices or their utility. Consider the situation where two military recruiting offices want to come into the same region. We will call these two major discount stores, Army and Navy. Each recruiting office can decide whether to build or place their station in the region’s larger city or in the region’s smaller city. The recruiting station desire the bigger market share of the consumers that yields more recruits for their respective services. Experts have estimated the market share in the region for the larger and smaller city building options based upon 100% of the consumer market and income of the region. Based upon this market research, Table 6.2, what decisions should each service make? As we will show later in this chapter, the best decision for each station is to locate in the larger city. Two types of games will be presented in this chapter: total conflict games and partial conflict games. Game theory then is the study of decisions where the outcome to the decision-maker depends not only on what he does, but the decision of one or more additional players. We classify the games depending upon whether the conflict between the players is total or partial. A total conflict game is a game where the sum of values in each cell of the payoff matrix, Mij + Nij either always equals 0 or always equals the same constant for each ij pair. In a partial conflict game, this sum does not always equals 0 or the same constant. We begin our discussion with the total conflict game described in Table 6.2. We also begin with simultaneous, non-cooperative games.

6.1.1

Two-Person Total Conflict Games

We begin with characteristics of the two-person total conflict game: 1. There are two persons (called the row player who we will refer to as Rose and the column player who we will refer to as Colin).

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2. Rose must choose 1 of m strategies and Colin must choose 1 of n strategies. 3. If Rose chooses the ith strategy and Colin the jth strategy, then Rose receives a payoff of aij and Colin loses an amount aij. 4. There are two types of possible solutions. Pure strategy solutions are when each player achieves their best outcomes by always choosing the same strategy in repeated games. Mixed strategy solutions are when players play a random selection of their strategies in order to obtain their best outcomes in repeated games. Games might be presented either in decision tree or payoff format. In a decision tree for sequential games, we look ahead and reason back. In simultaneous games, we use payoff matrices as shown in Table 6.1. This is a total conflict game if and only Mi,j+Ni,j equals either 0 or the same constant for all i and j. For example, if a player wins x when the other player loses x then their sum is zero or in business marketing strategy based upon 100% if one player get x% of the market then the other player gets y% such that their sum is x% + y% ¼ 100. Given a simple payoff matrix we look for the Nash equilibrium as the solution first with movement diagrams. Example 1: Navy Versus Army Recruiting Stations Placement Suppose Large City is located near Small City. Now assume the Department of the Navy recruiting would like to locate a franchise in either Large City or Small City. Further, the Department of the Army is making the same decision—they will locate either in Large City or Small City. Analysts have estimated the market shares of recruits and we place both sets of payoffs in a single game matrix. They both want to recruit as many new enlistees as possible. Listing the row player’s payoffs first, we have the payoff as shown in Table 6.3. We apply the movement diagram, were we draw arrows in each row (vertical arrow) and column (horizontal arrow) from the smaller payoff to the larger payoff. Note all arrows point into the payoff (60,40) at (Large City, Large City) strategies for both players and no arrow exits that outcome. This indicates that neither player can unilaterally improve their solution. This stable situation is called a Nash equilibrium. Often payoff matrices and movement diagrams may get convoluted or the arrows do not point to one or more points. In those more complex two-person games, we offer linear programming as the solution method.

Table 6.3 Payoff matrix for Example 1

Navy Large City Large City

60, 40

Small City 75, 25

Army Small City

50, 50

58, 42

6.1 Introduction to Game Theory

6.1.1.1

255

Linear Programming of Total Conflict Games

Every total conflict game may be formulated as a linear programming problem. Consider a total conflict two-person game in which maximizing player X has m strategies and minimizing player Y has n strategies. The entry (Mij,Nij) from the ith row and jth column of the payoff matrix represents the payoff for those strategies. We present the following formulation using the elements of M for the maximizing a player that provides results for the value of the game and the probabilities xi (Fox 2010; 2012a, b; Winston 2003). We note that if there are negative values in the payoff matrix then we need a slight modification to the formulation. We suggest the method by Winston (2003) to replace any variable that could take on negative values with the difference in two positive variables, Vj  V'j. We only assume that the value of the game could be positive or negative. The other values we are looking for are probabilities that are always non-negative. This is shown as Eq. (6.1). Maximize V

ð6:1Þ

Subject to : N 1,1 x1 þ N 2,1 x2 þ . . . þ N m,1 xn  V  0 N 2,1 x1 þ N 2,2 x2 þ . . . þ N m,2 xn  V  0 ... N m,1 x1 þ N m,2 x2 þ . . . þ N m, n xn  V  0 x1 þ x2 þ . . . þ xn ¼ 1 Non-negativity where the weights xi yields Rose’s strategy and the value of V is the value of the game to Colin. This is shown as Eq. (6.2). Maximize v

ð6:2Þ

Subject to : M 1,1 y1 þ M 2,1 y2 þ . . . þ M m,1 yn  v  0 M 2,1 y1 þ M 2,2 y2 þ . . . þ M m,2 yn  v  0 ... M m,1 y1 þ M m,2 y2 þ . . . þ M m, n yn  v  0 y1 þ y2 þ . . . þ yn ¼ 1 Non-negativity where the weights yi yield Colin’s strategy and the value of v is the value of the game to Rose. Our two formulations for this problem are for Rose and Colin, respectively: Maximize V c

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Subject to : 40x1 þ 50x2  V c  0 25x1 þ 42x2  V c  0 x1 þ x2 ¼ 1 x1 ,x2 , V c  0 Maximize V r Subject to : 60y1 þ 75y2  V r  0 50y1 þ 58y2  V r  0 y1 þ y2 ¼ 1 y1 ,y2 , V r  0 If we put our example into our two formulations and solve, we get the solution y1 ¼ 1, y2 ¼ 0 and Vr ¼ 60 from formulation (6.1) and x1 ¼ 1, x2 ¼ 0, and Vc ¼ 40 from formulation (6.2). The overall solution is (Large City, Large City) with value (60,40).

Constant-Sum to Zero-Sum The primal-dual only works in the zero-sum game format. We may convert this game to the zero-sum game format to obtain. Since this is a constant-sum game, all outcomes sum to 100. This can be converted to a zero-sum game through the positive linear function, y ¼ x  20. Use any two pairs of points and obtain the equation of the line and then make the slope positive. Using this transformation x1 ¼ x  20, we can obtain the payoffs for the row player in the zero-sum game. The new zero-sum payoff matrix may be written as presented in Table 6.4. For a zero-sum game, we can again look at movement diagrams, dominance, or linear programming. If one Rose’s information is present representing the zero-sum game, then only assume Colin’s values are the negative of Rose’s. We apply the movement diagram as before and place the arrows accordingly. The arrows point in and never leave 40. The large city strategy is the stable pure strategy solution. We define dominance as: Strategy A dominates a strategy B if every outcome in A is at least as good as the corresponding outcome in B, and at least one outcome in A is strictly better than the corresponding outcome in B. Dominance Principle: A rational player should never play a dominated strategy in a total conflict game.

In this case, the small city strategy payoffs for Rose are dominated by the large city strategy payoffs, thus we would never play small city. For Colin, the large city is better than the small city, so the large city dominates. Since large city is the dominated strategy, the solution is (40, 40). If we use linear programming, we only need a single formulation of the linear program. The row player maximizes and the column player minimizes with rows’

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Table 6.4 Payoff matrix for Example 1 Army

Large City

Navy Large City 40

Small City 55

Small City

30

38

values. This constitutes a primal and dual relationship. The linear program used for Rose in the zero-sum games is given as Eq. (6.3): Maximize V

ð6:3Þ

Subject to : a1,1 x1 þ a1,2 x2 þ . . . þ a1, n xn  V  0 a2,1 x1 þ a2,2 x2 þ . . . þ a2, n xn  V  0 : : : am,1 x1 þ am,2 x2 þ . . . þ am, n xn  V  0 x1 þ x2 þ . . . þ xn ¼ 1 V,xi  0 where V is the value of the game, am,n are payoff matrix entries, and xs are the weights (probabilities to play the strategies). We place these payoffs into our formulation: Max V r Subject to 40x1 þ 30x2  V r  0 55x1 þ 38x2  V r  0 x1 þ x2 ¼ 1 x1 ,x2 , V r  0 The optimal solution strategies found are identical as before with both players choosing Large City as their best strategy. This indicates that neither player can unilaterally improve a stable situation that we refer to as a Nash equilibrium. A Nash equilibrium is an outcome where neither player can benefit by departing unilaterally from its strategy associated with that outcome. We conclude our discussion of total conflict games with the analysis that linear programming may always be used all total conflict games but is most suitable for large games between two players each having many strategies (Fox 2010, 2012a, b).

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Table 6.5 Revised market shares

Navy Large City Army Small City

6.1.2

Large City 65, 25

Small City 50, 45

55, 40

62, 28

Two-Person Partial Conflict Games

In the previous example, the conflict between the decision-makers was total in the sense that neither player could improve without hurting the other player. If that is not the case, we classify the game as partial conflict as illustrated in the next example. Assume we have new market share analysis for our two stores as shown in Table 6.5 where the sums are not all equal to same constant. We begin with the movement diagram where the arrows do not find a stable point. In those cases, we need to find the equalizing strategies to find the Nash equilibrium. Other solution methods are found in the additional reading; we will describe only the use of linear programming as the method to use when the movement diagram fails to yield a stable point. Additionally, Gillman and Housman (2009) state that every partial conflict game also has equalizing strategy equilibrium even if it also has a pure strategy equilibrium. Since both players are maximizing their payoffs, we use the linear programming formulation presented as Eqs. (6.1) and (6.2). This yields two separate linear programs. Maximize V Subject to 65y1 þ 50y2  V  0 55y1 þ 62y2  V  0 y1 þ y2 ¼ 1 y1 ,y2 , V  0 The solution is y1 ¼ 6/11, y2 ¼ 5/11, and V ¼ 58.182. The other second LP formulation is Maximize v Subject to 25x1 þ 40x2  v  0 45 x1 þ 28x2  v  0 x1 þ x2 ¼ 1 x1 ,x2 , v  0

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The solution is x1 ¼ 17/32, y2 ¼ 15/32, and v ¼ 34.375. The results state that both the Army and the Navy must play their two strategies each a proportion of the time that they compete in order to obtain their best outcomes.

6.1.2.1

Discussion of Some Cooperative Methods

Another option available in partial conflict games is to consider allowing cooperation and communications between the game’s players. This allows for first moves, threats, promises, and combinations of threats and promises in order to obtain better outcomes. We call this strategic moves (Straffin 2004).

6.1.2.2

Moving First or Committing to Move First

We now assume both players can communicate their plans or their moves to the second player. If the Army can move first they can choose Large City or Small City. Examining the movement diagram, they should expect the Navy’s responses as follows: If Army plays Large City, Navy plays Small City resulting in the outcome (50, 45). If Army plays Small City, Navy plays Large City resulting in the outcome (55,40). Army prefers 55 to 50 so they would play Small City. If Army forces Navy to move first, then the choices are between (65,25) and (62,28) of which Navy prefers (62,28). Having Navy to move first gets Army a better outcome. How to get this to occur as well as the credibility of the first move is a concern.

6.1.2.3

Threats

In general, we describe the concept of issuing a threat. Rose may have a threat to deter Colin from playing a particular strategy. A threat must satisfy three conditions: Conditions for a Threat by Rose 1. Rose communicates that she will play a certain strategy contingent upon a previous action of Colin. 2. Rose’s action is harmful to Rose. 3. Rose’s action is harmful to Colin. In our game example, there is no valid threat. We present the classic game of chicken to show a valid threat (Table 6.6). In the game of Chicken, Rose wants Colin to play Swerve. Therefore, she makes the threat on Colin’s Not Swerve to deter him from choosing that strategy. Examining the movement diagram, if Colin plays Not Swerve, Rose plays Swerve yielding (2, 4). In order to harm herself, Rose must play Not Swerve. If Colin plays Not Swerve, then Rose plays Not Swerve yielding (1, 1). Is it a threat? It is

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Table 6.6 Conditions for a Threat by Rose Payoff Matrix

Colin Swerve Rose Not Swerve

Table 6.7 Updated Conditions for a Threat by Rose Payoff Matrix

Swerve (3, 3)

Not Swerve (2,4)

(4, 2)

(1, 1)

Colin Swerve Rose Not Swerve

Swerve (3, 3)

Not Swerve Eliminated by threat

(4, 2)

(1, 1)

contingent upon Colin choosing Not Swerve. Comparing (2, 4) and (1, 1), we see that the threat is harmful to Rose and is harmful to Colin. It is a threat and effectively eliminates the outcome (2, 4) updating the game in Table 6.7. Colin still has a choice of choosing Swerve or Not Swerve. Using the movement diagram, he analyzes his choices as follows: If Colin selects Swerve, Rose chooses Not Swerve yielding (4, 2). If Colin chooses Not Swerve, Rose chooses Not Swerve yielding (1, 1) (because of Rose’s threat). Thus, Colin’s choice is between a payoff of 2 and 1. He should choose Swerve yielding (4, 2). If Rose can make her threat credible, she can secure her best outcome.

6.1.2.4

Issuing a Promise

In our Army versus Navy game, there is no promise, so again we illustrate with the classic game of chicken. Again, if Colin has the opportunity to move first or is committed to (or possibly considering) Not Swerve, Rose may have a promise to encourage Colin to play Swerve instead. A promise must satisfy three conditions: Conditions for a Promise by Rose 1. Rose communicates that she will play a certain strategy contingent upon a previous action of Colin. 2. Rose’s action is harmful to Rose. 3. Rose’s action is beneficial to Colin. In the game of Chicken, Rose wants Colin to play Swerve. Therefore, she makes the promise on Colin Swerve to sweeten the pot so he will choose Swerve. Examining the movement diagram, normally, if Colin plays Swerve, Rose plays Not Swerve yielding (4, 2). In order to harm herself, she must play Swerve. Thus, the promise takes the form

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Table 6.8 Updated Conditions for a Threat by Rose Payoff Matrix Colin Swerve Rose Not Swerve

Swerve (3, 3)

Not Swerve (2, 4)

Eliminated by Promise

(1, 1)

If Colin plays Swerve, then Rose plays Swerve yielding (3, 3). Is it a promise? It is contingent upon Colin choosing Swerve. Comparing the normal (4, 2) with the promised (3, 3), we see that the promise is harmful to Rose and is beneficial to Colin. It is a promise and effectively eliminates the outcome (4, 2) updating the game in Table 6.8. Colin still has a choice of choosing Swerve or Not Swerve. Using the movement diagram, he analyzes his choices as follows: If Colin selects Swerve, Rose chooses Swerve yielding (3, 3) as promised. If Colin chooses Not Swerve, Rose chooses Swerve yielding (2, 4). Thus, Colin’s choice is between payoffs of 3 and 4. He should choose Not Swerve yielding (2, 4). Rose does have a promise. But her goal is for Colin to choose Swerve. Even with the promise eliminating an outcome, Colin chooses Not Swerve. The promise does not work. If both make a promise then perhaps (3, 3) is the outcome. In summary, the game of Chicken offers many options. If the players choose conservatively without communication, the maximin strategies yields (3, 3), which is unstable: both players unilaterally can improve their outcomes. If either player moves first or commits to move first, they can obtain their best outcome. For example, Rose can obtain (4, 2) which is a Nash equilibrium. If Rose issues a threat, she can eliminate (2, 4) and obtain (4, 2). A promise by Rose eliminates (4, 2) but results in (2, 4) which does not improve the (3, 3) likely outcome without communication.

6.1.2.5

A Combination Threat and Promise

Consider the game in Table 6.9: The movement diagram shows that (2,4) is the Nash equilibrium. Without communication, Colin gets his best outcome, but can Rose do better than (2, 4) with a strategic move? Rose First: If Rose moves R1, Colin should respond with C1 yielding (2, 4). If Rose play strategy R2, Colin responds with C1 yielding (1, 2). Rose’s best choice is (2, 4), no better than the likely conservative outcome without communication. Rose Threat: Rose wants Colin to play C2. Normally, if Colin plays C1, Rose plays R1 yielding (2, 4). To hurt herself, she must play R2 yielding (1, 2).

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Table 6.9 Combination Threat and Promise

Colin C1 R1

C2 (3, 3)

(2, 4)

Rose R2 Table 6.10 Updated Combination Threat and Promise

(1, 2)

(4, 1)

Colin R1 Rose R2

Table 6.11 Combination Threat and Promise

C1 Eliminated

C2 (3, 3)

(1, 2)

(4, 1)

Colin R1 Rose R2

C1 (2, 4)

C2 (3, 3)

(1, 2)

Eliminated

Comparing the normal (2, 4) and (1, 2), the threat is contingent upon Colin playing C1, hurts Rose and hurts Colin. It is a threat and effectively eliminates (2, 4) yielding the game in Table 6.10. Does the threat deter Colin from playing C1? Examining the movement diagram, if Colin plays C1 the outcome is (1, 2). If Colin plays C2, the outcome is (4, 1). Colin’s best choice is still C1. Thus there is a threat, but it does not work. Does Rose have a promise that works by itself? Rose Promise: Rose wants Colin to play C2. Normally, if Colin plays C2, Rose plays R2 yielding (4, 1). To hurt herself, she must play R1 yielding (3, 3) Comparing the payoffs (4, 1) with the promised (3, 3), the move is contingent upon Colin playing C2, hurts Rose and is beneficial to Colin. It is a promise and effectively eliminates (4, 1) yielding the game in Table 6.11. Does the promise motivate Colin to play C2? Examining the movement diagram, if Colin plays C1 the outcome is (2, 4). If Colin plays C2, the outcome is (3, 3). Colin’s best choice is still C1 for (2, 4). Thus there is a promise, but it does not work. What about combining both the threat and the promise?

6.1.2.6

Combination Threat and Promise

We see that Rose does have a threat that eliminates an outcome but does not work by itself. She also has a promise that eliminates an outcome but does not work by itself. In such situations, we can examine issuing both the threat and the promise to

6.1 Introduction to Game Theory Table 6.12 Updated Combination Threat and Promise

263

Colin R1 Rose R2

C1 Eliminated

C2 (3, 3)

(1, 2)

Eliminated

eliminate two outcomes to determine if a better outcome results. Rose’s threat eliminates (2, 4), and Rose’s promise eliminates (4, 1). If she issues both the threat and the promise, the outcomes in Table 6.12 are available. If Colin plays C1 the result is (1, 2), and choosing C2 yields (3, 3). He should choose C2, and (3, 3) represents an improvement for Rose over the likely outcome without communication (2, 4). Credibility Of course, commitments to first moves, threats, and promises must be made credible. If Rose issues a threat, and Colin chooses to Not Swerve anyway, will Rose carry out her threat and crash (1, 1) even though that action no longer promises to get her the outcome (4, 2)? If Colin believes that she will not carry through on her threat, he will ignore the threat. In the game of Chicken, if Rose and Colin both promise to Swerve and Colin believes Rose’s promise and executes Swerve, will Rose carry out her promise to Swerve and accept (3, 3) even though (4, 2) is still available to her? One method for Rose to gain credibility is to lower one or more of her payoffs so that it is obvious to Colin that she will execute the stated move. Or, if possible, she may make a side payment to Colin to increase his selected payoffs in order to entice him to a strategy that is favorable to her and is now favorable to him because of the side payment. An inventory of the strategic moves available to each player is an important part of determining how a player should act. Each player wants to know what strategic moves are available to each of them. For example, if Rose has a first move and Colin has a threat, Rose will want to execute her first move before Colin issues his threat. The analysis requires knowing the rank order of the possible outcomes for both players. Once a player has decided which strategy he wants the opposing player to execute, he can then determine how the player will react to any of his moves. As alluded earlier maybe the better option is to go to arbitration. We discuss that next.

6.1.3

Nash Arbitration

In the bargaining problem, Nash (1950) developed a scheme for producing a single fair outcome. The goals for the Nash arbitrations scheme are that the result will be at or above the status quo point for each player and that the result must be “fair.” Nash introduced the following terminology:

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6 Game Theory

Status Quo Point (We will typically use the intersection of Rose’s Security Level and Colin’s Security Level; the Threat positions may also be used). Negotiation Set: Those points in the Pareto Optimal Set that are at or above the “Status Quo” of both players. We use Nash’s four axioms that he believed that a reasonable arbitration scheme should satisfy rationality, linear invariance, symmetry, and invariance. A good discussion of these axioms and can be found in Straffin (2004, p. 104–105). Simply put the Nash Arbitration point is the point that follows all four axioms. This leads to Nash’s Theorem stated below: Nash’s Theorem: There is one and only one arbitration scheme which satisfies Axioms 1 through 4. It is this: if the status quo SQ ¼ (x0, y0), then the arbitrated solution point N is the point (x, y) in the polygon with x  x0 and y  y0 which maximizes the product: (x  x0) (y  y0).

Let’s examine this geometrically first as it will provide insights into using calculus methods. We produce the contour plot of our nonlinear function: (x  x0) (y  y0) when our status quo point is assumed to be (0,0). It is obvious that the northeast (NE) corner of quadrant 1 is where this function is maximized. This is illustrated in Fig. 6.1. We need a few more definition to use this Nash arbitration. In his theory for the arbitration and cooperative solutions, Nash (1950) stated the “reasonable” solution should be Pareto optimal and will be at or above the security level. The set of outcomes that satisfy these two conditions is called the negotiation set. The line segments that join the negotiation set must form a convex region as shown in Nash’s proof. Methodologies for solving for this point use basic calculus, algebra, and geometry. For any game theory problem, we next overlay the convex polygon onto our contour plot (Fig. 6.1). The most NE point in the feasible region is our optimal point and the Nash arbitration point. This will be where the feasible region is tangent to the hyperbola. It will always be on the line segment that joins the negotiation set. This is simply a constrained optimization problem. We can convert to a single variable problem as we will illustrate later in our example. In our example, we will use the security value as the status quo point to use in the Nash arbitration procedure. We additionally define the procedure to find the security value as follows: In a nonzero-sum game, Rose’s optimal strategy in Rose’s game is called Rose’s Prudential Strategy, the value is called Rose’s Security level. Colin’s optimal strategy in Colin’s game is called Colin’s Security level. We will illustrate this during the solution to find the Nash arbitration point in the Example 1. Finding the security levels in a nonzero-sum game

Rose

R1 R2

Colin C1 (2,6) (4,8)

C2 (10,5) (0,0)

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265

Fig. 6.1 Contour plot for (xy). We note that the direction of maximum increase is NE as indicated by the arrow

To find the security level (status quo point), we look at the following two separate games extracted from the original game and use movement diagrams, dominance, or our linear programming method to solve each game for those players’ values. In a prudential strategy, we allow a player to find their optimal strategy in their own game. For Rose, she would need to find her optimal solution in her own game. Rose’s game has a mixed strategy solution; V ¼ 10/3.

Rose

R1 R2

Colin C1 2 4

C2 10 0

For Colin, he would need to find his optimal solution in his own game. Colin’s game has a pure strategy solution, V ¼ 6.

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6 Game Theory

Rose

Colin C1 6 8

R1 R2

9

C2 5 0

(4.8)

8 7

(2,6)

6 5

Series1

4

(10,5)

3 2 1 0

0

2

4

6

8

10

12

(0,0)

Fig. 6.2 Payoff Polygon

The status quo point or security level from the prudential strategy is found to be (10/3, 6). We will use this point in the formulation of the Nash arbitration.

6.1.3.1

Finding the Nash Arbitration Point

We use the nonlinear programming method described by Fox (2010, 2012a, b). We set up the
convex  polygon (constraints) for the function that we want to maximize, which is x  10 3  ðy  6Þ. The convex polygon is the convex set from the values in the payoff matrix. Its boundary and interior points represent all possible combinations of strategies. Corner points represent pure strategies. All other points are mixed strategies. Occasionally, a pure strategy is an interior point. Thus, we start by plotting the strategies from our payoff matrix set of values {(2,6), (4,8), (10,5), (0,0)}, see Fig. 6.2. We note that our convex region has four sides whose coordinates are our pure strategies. We use the point-slope formula to find the equations of the line and then test points to transform the equations to inequalities. For example, the line form (4,8) to (10,5) is y ¼ .5x + 10. We rewrite as y + .5x ¼ 10. Our test point (0, 0) shows that inequality is 0.5x + y  10. We use this technique to find all boundary lines as well as add our security levels as lines that we need to be above.

6.1 Introduction to Game Theory

267

The convex polygon is bounded by the following inequalities: :5x þ y  10 3x þ y  0 0:5x  y  0 x þ y  4 x  x y  y where x and y are the security levels (10/3,6). The NLP formulation (Winston 2003; Fox 2012a, b) to find the Nash arbitration value following the format of equation is as follows shown as Eq. (6.4): Maximize Z ¼

  10 x  ð y  6Þ 3

ð6:4Þ

Subject to : 0:5x þ y  10 3x þ y  0 0:5x  y  0 x þ y  4 10 x 3 y6 We display the feasible region graphically in Fig. 6.3. The feasible region is the solid region. From the figure we can approximate the solution as the point of tangency between the feasible region and the hyperbolic contours in the north east (NE) region. Since we visually see that the solution must fall along the line segment y ¼ 0.5x + 10. We may use simple calculus.   10 Maximize x   ð y  6Þ 3 Subject to y ¼ :5x þ 10 We substitute to obtain a function of one variable, Maximize ðx  10=3Þð:5x þ 10  6Þ or

268

6 Game Theory

Opmal soluon is the point of tangency.

Fig. 6.3 Convex polygon and function contour plot

Maximize  :5x2 þ 34=6x  40=3 We find df dx ¼ 0 ¼ x þ 34=6. We find x¼17/3. 2 The second derivative test, ∂∂xf2 ¼ 1, which is less than 0, so this confirms we found a maximum. We substitute x ¼ 17/3 back into y ¼ .5x + 10 to obtain y ¼ 43/6. This point (17/3, 43/6) is the Nash arbitration point. Our optimal solution, the Nash arbitration point is found to be x ¼ 5.667 and y ¼ 7.167 and the value of the objective function payoff is 2.72. How do we obtain this value in a particle manner? An arbitrator plays the strategies BC (4,8) and AD (10,5) as follows described below. We can solve two equations and two unknowns from our strategies BC and AD equal to our Nash arbitration point. 

4 8

10 5

    x 5:667 ¼ y 7:167

We solve and find x ¼ 0.27777 or (5/18) y ¼ 0.72222 or 13/18. Example 2: Management-Labor Arbitration (Straffin 2004, p. 115–117) The convex polygon is graphed from the constraints (see the plots in Fig. 6.4):

6.1 Introduction to Game Theory

269

Fig. 6.4 The graphical NLP problem for the Management-Labor Arbitration

xþy0 0:5x þ y  0 0:25x þ y  1 xþy5 0:5x þ y  3:5 15 0:25x þ y  4 The status quo point (our security level) is (0,0), making the function to maximize simply xy. Our formulation is: Maximize x  y

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Table 6.13 Management-Labor Arbitration Payoff Matrix

(0,0) (2,2)

Eliminate Coffee Break (C) (4,1) (2,1)

Labor Concedes Automate checkpoint (A) (4,2) (2,0)

Both CA (8,3) (6,1)

(3,3) (5,5)

(1,2) (1,4)

(1,1) (1,3)

(5,0) (3,2)

Nothing Nothing Increase pension (P) $1 raise (R) Both PR

Management Concedes

Table 6.14 Three-person game

Rose

R1 R2

Larry, L1 Colin C1 (r1,c1,l1) (r2,c1,l1)

C2 (r1,c2,l1) (r1,c2,l1)

R1 R2

Larry, L2 Colin C1 (r1,c1,l2) (r2,c1,l2)

C2 (r1,c2,l2) (r2,c2,l2)

Subject to : xþy0 0:5x þ y  0 0:25x þ y  1 xþy5 0:5x þ y  3:5 15 0:25x þ y  4 The product is taken as xy ¼ 6.0 and the values are taken as x ¼ 3 and y ¼ 2. This optimal point is the point (3,2) on the line that is tangent to the contours in the direction of the NE increase shown in Fig. 6.4 and Table 6.13.

6.1.4

Three-Person Games

We restrict our discussion to the three-person games. We suggest placing the payoffs into payoff matrices as shown in Table 6.14. We will continue to use Rose and Colin but introduce Larry as our generic third player. We show with only two strategies each but the concept can be expanded. Again if ri+ci+li ¼ 0 or the same constant for all i we have a total conflict game otherwise we have a partial conflict game.

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271

Table 6.15 Updated ThreePerson Total Conflict Game

Rose

Larry L1 Colin C1

C2

R1

(2,2,4)

(1,3,2)

R2

(3,4,1)

(2,2,0)

Larry L2 Colin

Rose

Table 6.16 Updated ThreePerson Total Conflict Payoff Matrix

C2

R1

C1 (1,0,1)

R2

(2,3,1)

(2,1,3)

Rose R1 R2

C1L1 2,2 3,3

Colin and Larry C2L1 1,1 2,2

(2,1,3)

C1L2 1,1 2,2

C2L2 2,2 2,2

Movement diagram may again be used to examine the game for pure strategy solution. Arrow point from the small values to the larger values. The new arrows belong to Larry. Between Larry 1 and Larry 2 we draw arrows from smaller to larger by an arrow out from one matrix and an arrow into the other. We will illustrate with an example. Regardless if there is a pure solution or solutions or not, we will still consider coalitions. A coalition will be one or two players joining together to gain an advantage of a third player. We consider all such coalition in our analysis.

6.1.4.1

Example of Three-Person Total Conflict

Consider the three-person total conflict game presented in Table 6.15, between Rose, Colin, and Larry. We provide the payoffs and the movement diagram with all arrows. Our movement arrows indicate two stable pure strategies, R2C1L1 (3, 4,1) and R1C1L2 (1,0,1). These results are very different and not all players are satisfied at one or the other points. We now consider coalitions. We completely illustrate one coalition and provide the results for the others. Let’s assume that Larry and Colin form a coalition against Rose. Our new payoff matrix is now Table 6.16. As a zero-sum game, we may just list Rose’s values (Table 6.17).

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Table 6.17 Updated ThreePerson Total Conflict Payoff Matrix

Rose R1 R2

C1L1 2 3

Colin and Larry C2L1 1 2

C1L2 1 2

C2L2 2 2

We can use linear programming to obtain our solution for Rose. Since payoffs are negative so that we solution can negative we employ the transformation of Vr to Vr1  Vr2. Maximize V r1  V r2 Subject to 2y1  y2  y3  2y4  V r1  V r2r  0 3y1 þ 2y2  2y3 þ 2y4  V r1  V r2  0 y1 þ y2 þ y3 þ y4 ¼ 1 yi ,V r1  V r2  0 We find the optimal solution is Vr1 ¼ 0, V ¼ 1.2 so Vr ¼ 1.2 when y1 ¼ 0, y2 ¼ 0, y3 ¼ 4/5, and y4 ¼ 1/5. Thus, the coalition of Colin-Larry gains 1.2 units where we find Larry get 21/25 of the share and Colin gets 9/25 of the share. For the other coalitions, we may use the same procedures. Also, we may use the same procedures if we have a three-person partial conflict game. However, for those coalitions we must use the complete (M,N) formulations since M+N does not have to be always equal to zero.

6.2

Applied Game Theory to Improve Strategic and Tactical Military Decisions

In 1950, Haywood proposed the use of game theory for military decision-making while at the Air War College. This work culminated in an article, “Military Decisions and Game Theory” (Haywood 1954). Further work by Cantwell (2003) showed and presented a ten step-by-step procedure to assist analysts in comparing courses of action for military decisions. He illustrated his method using the Battle at Tannenberg between Russia and Germany in 1914 as his example (Schmitt 1994). Cantwell’s ten-step procedure (Cantwell 2003) was presented as follows: Step 1. Select the best-case friendly course of action for the friendly forces that achieves a decisive victory. Step 2. Rank order all the friendly courses of action from best effects possible to worse effects possible. Step 3. Rank order the enemy courses of action from best to worst in each row for the friendly player.

6.2 Applied Game Theory to Improve Strategic and Tactical Military Decisions

273

Table 6.18 Cantwell’s payoff matrix

Attack N, fix S Attack S, fix N Defend in place Defend along Vis. Minimax

Attack N 24

Attack S 23

Coordinated Att 22

ATT N, fix S 3

Attack S, fix N 15

Defend in depth 2

16

17

11

7

8

1

1

13

12

6

5

4

14

4

21

20

19

10

9

18

9

24

23

22

10

15

18

No saddle

Maximin 2

Step 4. Determine if the effect of the enemy courses of action result in a potential loss, tie, or win for the friendly player in every combination across each row. Step 5. Place the product of the number of rows multiplier by the number of column in the box representing the best case scenario for each player. Step 6–9. Rank order all combination for wins, tie, and losses descending down from the value of Step 5 to 1. Step 10. Put the matrix into a conventional format as a payoff matrix for the friendly player Now, the payoff matrix is displayed in Table 6.18 after executing all 10 steps. We can solve the payoff matrix for the Nash equilibrium. In Table 6.18, the saddle point method, Maximin, (Straffin 2004), illustrates that there is no pure strategy solution. When there is no pure strategy solution, there exists a mixed strategy solution (Straffin 2004). Using linear programming (Straffin 2004; Winston 1995; Giordano et al. 2014; Fox 2015), the game is solved obtaining the following results: V ¼ 9.462 when “friendly” chooses x1 ¼ 7.7%, x2 ¼ 0, x3 ¼ 0, x4 ¼ 92.3% while “enemy” best results come when y1 ¼ 0, y2 ¼ 0, y3 ¼ 0, y4 ¼ 46.2%, and y5 ¼ 53.8%. The interpretation, in military terms, appears to be that player one should feint an attack north and fix south while concentrating his maximum effort to defend along the Vistula River or they can leak misinformation slightly about the attack and maintain secrecy. Player two could mix their strategy: attack north—fix south or attack south—fix north. The value of the game, 9.462 is a relative value that has no real interpretation (Cantwell 2003). According to Cantwell, the results are fairly accurate as to the decisions.

274

6.2.1

6 Game Theory

Proposed Update to the Methodology

We propose a methodology change to obtain more representative preferences using multi-attribute decision-making, specifically AHP’s pairwise comparison method. The reason we make this recommendation is that ordinal numbers should not be used with mixed strategies. For example, if player A wins a race and player 2 finishes second, what does it mean to subtract the places? It makes more sense to have collected the times of the race and then subtract where the differences have real meaning and interpretation. Mixed strategies methods result in probabilities to play strategies that must be calculated utilizing mathematical principles. You cannot add, subtract, multiply, or divide ordinal numbers and make sense of the results.

6.2.2

AHP Method for Pairwise Comparison

AHP and AHP-TOPSIS hybrids have been used to rank order alternatives among numerous criteria in many areas of research in business industry, and government (Fox 2014) including such areas as social networks (Fox 2012b, 2014), dark networks (Fox 2014), terrorist phase planning (Fox and Thompson 2014), and terrorist targeting (Fox 2015). Table 6.19 represents the process to obtain the criteria weights using the Analytic Hierarchy Process used to determine how to weight each criterion for the TOPSIS analysis. Using Saaty’s 9-point reference scale (Saaty 1980), displayed in Table 6.19, we used subjective judgment to weight each criterion against all other criterion lower in importance. We begin with a simple example to illustrate. Assume that we have a zero-sum game where we might know preferences in an ordinal scale only. Table 6.19 Saaty’s 9-Point Scale Intensity of importance in pairwise comparisons 1 3 5 7 9 2,4,6,8 Reciprocals of above Rationale

Definition Equal importance Moderate importance Strong importance Very strong importance Extreme importance For comparing between the above In comparison of elements i and j if i is 3 compared to j, then j is 1/3 compared to i Force consistency; measure values available

6.2 Applied Game Theory to Improve Strategic and Tactical Military Decisions

275

Player 2 C1 C2 Player 1R1 R2 y z

w

x

Player 1’s preference ordering is x>y>w>z. Now we might just pick values that meet that ordering scheme, such as 10>8>6>4 yielding Player 2 C1 C2 Player 1R1 R2 8 4

6

10

There is no saddle point solution to this game. To find the mixed strategies, we could use the method of oddments. The method of oddment finds Player I plays R1 and R2 with probabilities ½ each and Player II plays ¾ C1 and ¼ C2. The value of the game is 7. The probabilities are function of the values chosen in the payoff matrix and not reflective of the utility the player has for each set of strategies. Therefore, rather than arbitrary values or even using the lottery method of von Neumann and Morgenstern (Winston 1995) we recommend using AHP to obtain the utility values of the strategies. We begin by numerating the strategies combinations in a subject priority order R1C2 > R2C1 > R1C1 > R2C2: Then, we use the pairwise values from Saaty’s 9-point scale in Table 6.19 to determine the relative utility. We prepared an Excel template to assist us in obtaining these utility values, as shown in Fig. 6.5. In this template, the prioritized strategies are listed so we can easily perform pairwise comparisons of the strategies We obtained the AHP pairwise comparison matrix in Table 6.20. The consistency ratio of this matrix, according to Saaty’s work (1980), must be less than 0.1. The consistency of this matrix was 0.0021, which is smaller than 0.1. We provide the formula and definition of terms. The Consistency Index for a matrix is calculated from (λmax  n)/(n  1) and, since n ¼ 4 for this matrix, the CI is 0.00019. The final step is to calculate the Consistency Ratio for this set of judgments using the CI for the corresponding value from large samples of matrices of purely random judgments using the data in Table 6.21, derived from Saaty’s book, in which the upper row is the order of the random matrix, and the lower is the corresponding index of consistency for random judgments. CR ¼ CI/RI In this example, the calculations give 0.00190/0.90 ¼ 0.0021. Saaty states that any CR < 0.1 indicates that the judgments are consistent. We obtain the weights, which are the eigenvector to the largest eigenvalue. They are presented here to three decimals accuracy.

276

6 Game Theory

Fig. 6.5 Screenshot Excel AHP Template Table 6.20 AHP Pairwise Comparison Matrix 1 2 3 4

x w y z

x 1 1 1/3 1/5 1/7

w 2 3 1 1/2 1/4

y 3 5 2 1 1/3

z 4 7 4 3 1

x ¼ 0:595 w ¼ 0:211 y ¼ 0:122 z ¼ 0:071 Thus, AHP can help obtain the relative utility values of the outcomes. These values are the cardinal utilities values based upon the preferences. The game with cardinal utilities is now Player 2 C1 C2 Player 1 R1 0.122 R2 0.211 0.071

0.595

N R

1 0

2 0

3 0.58

4 0.9

Table 6.21 AHP Consistency Matrix

5 1.12

6 1.24

7 1.32

8 1.41

9 1.45

10 1.49

11 1.51

12 1.48

13 1.56

14 1.57

15 1.59

6.2 Applied Game Theory to Improve Strategic and Tactical Military Decisions 277

278

6 Game Theory

Fig. 6.6 COA1-COA 4 Weighting Analysis

If we apply oddment to this game, we find Player I plays 22.8% of the time R1 and 77.2% of the time R2 while Play II plays C1 85.5% of the time and C2 14.5% of the time. The value for the revised game based on cardinal utility is 0.190.

6.2.3

Proposed Application of AHP to the Military Decision-Making

Example 1: Case Study: Two-Person Zero-sum Game with Cardinal Values The row player has four courses of action that might be compared initially. We provide an initial preference priority COA 4, COA 1, COA 2, COA 3 shown in Fig. 6.6. The consistency ratio is 0.002 which is less than 0.10 (Saaty, 1980). The weights calculated by the AHP template (Fox, 2016) are: COA4 COA1 COA2 COA3

0.59510881 0.2112009 0.12220096 0.07148933

6.2 Applied Game Theory to Improve Strategic and Tactical Military Decisions

279

Fig. 6.7 Enemies ECOA1-ECOA6 under player 1’s COA1

Now under each we will obtain weights as functions of the enemy COAs. For example, we display Fig. 6.7. The consistency ration is CR ¼ 0.03969 (less than 0.1 is acceptable). We find the sub-weights from the template. The sub-weights are COA1 COA2 COA3 COA4 COA5 COA6

0.431974 0.250029 0.162164 0.044169 0.0745 0.037163

To obtain the useable weights we form the product of COA1 times these sub-weight values.

280

6 Game Theory

Table 6.22 Military Decision-Making Course of Action Analysis Matrix

Major criteriarow player COA 4

Local weights 0.595

COA 1

0.211

COA 2

0.122

COA 3

0.0715

Sub criteria Player 2 COA 1 COA 2 COA 3 COA 4 COA 5 COA 6 COA 1 COA 2 COA 3 COA 4 COA 5 COA 6 COA 1 COA 2 COA 3 COA 4 COA 5 COA 6 COA 1 COA 2 COA 3 COA 4 COA 5 COA 6

Local weights 0.431974 .250 .162 0.404 0.0745 0.0372

Global decision weights (criteria weight  sub criteria weight) 0.091146 0.052756 0.034217 0.00932 0.01572 0.007841 0.033223 0.054067 0.016419 0.006478 0.007619 0.004395 0.017705 0.013371 0.005412 0.004151 0.003779 0.027081 0.235044 0.134041 0.079026 0.037179 0.030092 0.079718 Note that the SUM of all weights ¼ 1

0.091146 0.052756 0.034217 0.00932 0.01572 0.007841

We repeat the process for friendly COA 2 through friendly COA 4 for the enemies COA-1–COA 6 displayed in Table 6.22. These 24 entries are now the actual entries in the game matrix corresponding to R1–R4 for player 1 and C1–C6 for player 2 in this combat analysis. We developed a template to solve, via linear programming larger zero-sum games such as this game (Fox 2015).

6.2 Applied Game Theory to Improve Strategic and Tactical Military Decisions

281

Fig. 6.8 Excel Results using cardinal values in the combat analysis payoff matrix

Based upon these preference values, we enter our linear programming model template for game theory, displayed in Fig. 6.8. The results show a pure strategy solution that indicated Player 1 should defend the Vistula River and Player 2 should attack south, fix north to obtain their best outcomes. This is consistent with Cantwell’s results but perhaps more accurate since the values are based upon preferences not just ordinal rankings from 24 to 1.

6.2.4

Sensitivity Analysis

We used Eq. (6.5) [17] for adjusting weights of the primary COAs for player 1 and obtain new weights for the payoff matrix. 0

1  wp wj ¼ wj 1  wp 0

ð6:5Þ

where wj’ is the new weight and wp is the original weight of the criterion to be adjusted and wp’ is the value after the criterion was adjusted. We found this to be an easy method to adjust weights to reenter back into our model. We summarize some of the results in Table 6.23 that includes only the strategies for each player. We find the player 1 should always play strategy 4 either 100% or over 70%. Clearly that indicates a favorable strategy. If player 2 plays either a pure strategy

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6 Game Theory

Table 6.23 Course of Action Player Game and Analysis Summary Strategy played Player 1 COA 1 COA 2 COA 3 COA 4 Player 2 COA 1 COA 2 COA 3 COA 4 COA 5 COA 6

Ordinal preferences cantwell

Cardinal preferences

Sensitivity #1

Sensitivity #2

Sensitivity #3

0.077 0 0 0.923

0 0 0 1

0.28 0 0 0.72

0.25 0 0 0.75

0 0 0 1

0 0 0 0.462 0.538 0

0 0 0 0 1 0

0 0 0 0.567 0.433 0

0 0 0 0.77 0.23 0

0 0 0 0 1 0

with their COA 5 or a mixed strategy of COA 4 and COA 5 as indicated in the Table 6.23 to minimize their loss. Example 2: Case Study: Manhunting (Adapted from McCormick and Owen) In this example, we assume we have an entity (person, persons, etc.) that desire to hide and an opponent that desires to find them. We define this military game as follows. We consider a game in which a fugitive, who we will refer to as the hider (H), can hide in any of n cells. The authorities, who play the role of searcher(S), look for him in any one of the cells. If S looks in the cell, i, where H is hiding, there is a probability pi that S will find H. If he looks in a different cell, there is never the less a probability qi that H will be found, either because he might inadvertently give away his position or because he will be betrayed by those around him. We assume that for each cell i,0  qi < pi 1. We represent this game by an nn matrix A ¼ (aij), where aij ¼ p j qj

if i ¼ j

if i ¼ j:

Each row or column of the matrix is a pure strategy of the game. It is understood that S choses the row, j, while H chooses the column j. The payoff aij is the probability of finding H. S wishes to maximize this probability; H wishes to minimize this probability. We look here for optimal strategies of the two-person game that can be either in pure or mixed strategy. As mentioned before we can express this zero-sum game in terms of linear programming.

6.2 Applied Game Theory to Improve Strategic and Tactical Military Decisions Table 6.24 Hide & Seek Game S

R1 R2

Table 6.25 Updated Hide & Seek Game S

H C1 0.9 0.1

H C1 0.5 0.1 0.1

R1 R2 R3

283

C2 0.4 0.6

C2 0.1 0.4 0.1

C3 0.2 0.2 0.3

i j Consider a game with two cells. In the first cell, it is difficult to hide, but there is little chance of betrayal (perhaps there are few people around or those who might learn of H’s whereabouts are trustworthy). The second cell offers better hiding places if S mounts a search in that location, but there are many people around, some of whom cannot be trusted, so there is a chance of betrayal. A possible representation of this is the matrix in Table 6.24. It is easy to verify that the optimal search strategy for S in this game by the Methods of Oddments (William’s method) or linear programming is x ¼ (0.5,0.5) and the optimal hiding strategy for H is y ¼ (0.2,0.8). The value of the game is v ¼ (0.5, 0.5). Example 2: Consider a case with n ¼ 3 cells, and probabilities of discovery p1 ¼ 0:5, p2 ¼ 0:4, p3 ¼ 0:3,

q1 ¼ 0:1 q2 ¼ 0:1 q3 ¼ 0:2

The game matrix is then updated (Table 6.25) We solve this game, via linear programming, to obtain the optimal solution presented in Fig. 6.9. The value of this game is (0.2263, 0.2263) when S plays (0.316, 0.421, 0.263) while H plays 0.1578, 0.2105, 0.6315.We determine information relative to our search procedures that we might employ. For more information of the manhunt game, please see International Game Theory Review, Vol. 12, No. 4 (2010), pp. 293–308.

284

6 Game Theory

Fig. 6.9 Excel solution to Hide & Seek Game

6.3

Two-Person Partial Conflict (Nonzero-Sum) Game

There is no reason to assume that the game must be a zero-sum game. As a matter of fact, former President Clinton gave a talk in Dayton, Ohio where he discussed the need for nonzero-sum games. Therefore, we present some example using the methods shown to solve the games from Sect. 6.1.2. Example 1: Case Study: Revisit for COA Example from the Previous Section Cantwell’s method can be employed for the player 2 side to construct payoff that are in fact nonzero. Additionally, we might use the AHP method as we did to obtain player 1 values for player 2. We used the nonlinear programming approach presented in Barron (2003).

6.3 Two-Person Partial Conflict (Nonzero-Sum) Game

6.3.1

285

Nonlinear Programming Approach for Two or More Strategies for Each Player

For games with two players and more than two strategies each, we present the nonlinear optimization approach by Barron (2013). Consider a two-person game with a payoff matrix as before. Let’s separate the payoff matrix into two matrices M and N for players I and II. We solve the following nonlinear optimization formulation in expanded form, in Eq. (6.6). Maximiz

n X m X

xi aij y j þ

i¼1 j¼1

n X m X

xi bij y j þ p  q

i¼1 j¼1

Subject to m X

aij y j  p,

i ¼ 1,2, . . . , n,

xi bij  q,

j ¼ 1,2, . . . , m,

j¼1 n X

ð6:6Þ

i¼1 n X

xi ¼

i¼1

xi  0,

m X

yj ¼ 1

j¼1

yj  0

We developed a Maple procedure from Barron (2013) to perform our calculations.

286 > >

>

>

>

>

>

6 Game Theory

6.3 Two-Person Partial Conflict (Nonzero-Sum) Game

287

The NLP solution found was that Player 1 plays COA 4 and player 2 plays COA 4. Interpretation: The key result here is that after we analyzed this game as a nonzero game, player 1’s choice was still COA 4. Example 2: Case Study: An End-Game Strategy for Afghanistan: A Game Theory Approach (Adapted from Major Ryan Hartwig, Naval Postgraduate School, U.S. Army & Dr. William P. Fox, Naval Postgraduate School, Technical Report on Afghanistan Network Strategy)

6.3.1.1

Introduction

In recent United States (US) political discussion, the strategy for a way forward in Afghanistan has been little more than a debate over the rate and number of “stay behind” forces as bulk of US troops exit from the country. However, the withdrawal of US forces from Iraq in 2011, and Soviet withdrawal from Afghanistan in 1989 each demonstrate that leaving without a solid plan will likely lead to increased civil and intrastate strife (as in Iraq), or a slide backwards (the Taliban’s 1993–1995 resurgence in Afghanistan). Therefore, a well thought out end-game strategy is crucial to US policy makers for a way forward in Afghanistan. Clearly, the American people have practically turned their backs on and economy is no longer able to support a war that has cost trillions of dollars and thousands of lives. A desire for some sort of satisfactory conclusion, or minimax solution, so the amount of blood and treasure spent to date will have meaning and purpose. The purpose of this paper is to apply game theory while developing a solution in which both Afghanistan and the US receive the best outcome during the way forward. It is likely that the US will, in some capacity, continue to operate in Afghanistan. The primary traditional assets available to the US include powerful conventional forces and a withering financial support apparatus. There are also more unconventional resources available in the form of what are coined Village Stability Operations (VSO). In 2009, US Special Operations Forces (USSOF) embedded themselves into one village at a time, enabling governance; recruiting, training, and employing “home grown” security forces; and developing local infrastructure. By 2012, VSO was conducted in 100 Afghanistan villages, relatively even spread across the country. As a core competence of US Special Forces—Green Berets (and a few SEAL platoons and Marine Special Operations Teams—MSOTs), USSOF excelled at increasing village capacity (Gant 2009). To this point, no VSO end-game strategy during the US’s way forward in Afghanistan has surfaced. Afghanistan is often painted as a down trodden third world country with barren lands, with no assets beyond its yearly poppy harvests. Paradoxically, the situation in Afghanistan is not completely as it may seem. Between 2007 and 2011, the US Geological Survey (USGS) conducted research in Afghanistan by taking numerous soil samples throughout the country and reviewing similar research conducting by the Soviets during decades prior. The USGS concluded that extremely large amounts of strategic minerals, such as copper, gold, iron ore, oil, and natural gas are relatively

288

6 Game Theory

evenly distributed throughout the country (Peters et al. 2012). The Afghanistan Ministry of Mines and foreign investors are already in the hunt for the country’s strategic minerals, valued at an estimated at $1–3 trillion (Kral 2011; Lipow and Melese 2011; Global Data Ltd 2012). Afghanistan’s strategic mineral wealth opens new options for a better-funded Afghanistan military, a more prosperous economy, and a more influential government from the national to tribal levels.

6.3.1.2

Applying Game Theory

US Options Between now and 2015, as the US considers the draw-down options in Afghanistan, one of the largest concerns is that Taliban resurgence will destroy the future livelihood of Afghanistan and strategically damage the US. Four US options below are presented here with the future threat of the Taliban on the forefront of the Afghanistan Government’s future threats. Option 1: Fight Through the pain The US continues to engage in enemy-focused combat operations in Afghanistan. According to Powell Doctrine, this may be the most appropriate “common sense” approach to routing anti-Afghanistan forces and preventing Taliban resurgence. However, the costs to the America’s economy and military, frustrating effects of this approach make this a least desired option for the US President, Congress, and the American people. COA/Option 2: Show Me the Money! The US provides only financial aid to the Afghanistan Government, and encourages them to target the Taliban on their own. The biggest weakness to this option is the fact that the Afghanistan Government has expressed little to no interest (and even less effort) toward independently routing the Taliban. President Karzai has not appeared particularly willing to build a military or government capable of maintaining a post-2014 government and defeating the Taliban. But he would surely not balk at the opportunity to be given more money. COA/Option 3: Hybrid Lower Intensity Approach The US continues conventional combat operations (perhaps changing the name of the International Security Assistance Forces—ISAF to Nation Stability Forces) while also financially supporting the Afghanistan Government until they are strong enough to stand on their own. Implicit in this strategy is lower intensity, yet still conventional, fullspectrum combat operations. This commits a variety of resources to the problem, but continues to give the Afghanistan Government fish while hoping it can someday cast the bait on its own. This option does not hold a great deal of appeal to US leadership and America as a whole. COA/Option 4: VSO Focus Conventional forces, or ISAF, pack up and head home, but USSOF remain in Afghanistan villages conducting VSO with a clear end-game strategy. In order to maximize the Afghanistan Government’s ability to manage their economy, this option would include USSOF adjusting their disposition in Afghanistan to focus their VSO efforts toward the mineral rich terrain in the

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country. USSOF elements will influence and support tribal members during the negotiation of mineral rights contracts between the Afghanistan Government (local to national levels) and investors. Simultaneously, USSOF would continue enabling governance, training local security forces, and assisting the improvement of local infrastructure. This option should appeal to many tribal leaders and land-owners that do not hold the Afghanistan President Karzai in high regard, but are willing to work with organizations with goals of better their villages. The same appeal holds true in reverse. President Karzai doesn’t treat Afghanistan’s villages equally (mostly for tribal reasons), but he is willing to work with organizations with goals of bettering Afghanistan. This option does assume strategic risk since the “resource curse” will always loom (Jensen and Johnston 2011). However, Afghanistan’s short term ability to develop its own governance, security, and infrastructure likely outweighs the long term disadvantage of relying heavily on resources for economic advance. Each of these options has pros (benefits) and cons (risk), and the payoffs to each player are not diametrically opposed (as in a zero sum game). The result is a partial conflict game scenario for each course of action. The next step is to rank order the payoffs with ordinal numbers and then estimate and assign cardinal values for US and Afghanistan courses of action. Finally, a payoff matrix will result in a determination of the best course of action for the US and Afghanistan Government. 1. Assumptions To control for potential variables beyond the scope of this game, it is assumed that the US and Afghanistan are rational actors which make decisions to maximize their payoff. Furthermore, it is assumed that communications will take place during the game and the potential for cooperation exists. The motivations for the US are to initiate and maximize the potential for a VSO-based Resource Network End-Game Strategy in Afghanistan in a way that is less costly in terms of US blood and treasure, and is acceptable to the US President, Congress, and the American people. The motivations of the Afghanistan Government are to remain in power, increase the country’s financial wealth, and minimize US presence. 2. The Game with Ordinal Values Taking into account the aforementioned motivations of the US and Afghanistan, the following ordinal game is designed with the US player’s four Options or COAs versus the simpler Afghanistan Government’s options of high and low level of effort to combat the Taliban. Ordinal values are assigned to both players from one to four, with one being the least desirable, and four being the most desired. This is displayed in Fig. 6.10. COA/Option 1 US led enemy-focused combat operations to route the Taliban are the least desirable strategy for the US, no matter the level of effort by the Afghanistan Government to assist in the fight. The Powell Doctrine may seem the most “common sense” approach to ending the Taliban’s influence in Afghanistan, but this method hasn’t proven effective in practice and has placed a heavy burden on the US Soldier and tax payer. In addition, the Afghanistan Government and people in general do not want to conduct overt, high intensity, combat operations conducted on their soil, making this COA least desirable for Afghanistan.

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Fig. 6.10 Game with Ordinal Values

COA/Option 2 Financial aid is desirable for the US as long as it is combined with the Afghanistan Government taking aggressive action to defeat the Taliban. However, any US preference to provide purely financial aid to Afghanistan fades if the Afghanistan Government commits only minimal efforts to combating the Taliban. The Afghanistan Government, on the other hand, is happy to receive more funding from the US under either circumstance. However, they have slightly less interest in this option if they are held more responsible for taking the fight directly to the Taliban. The Afghanistan Government will make financial gains in either scenario. COA/Option 3 In many ways, this option resembles what the US is doing right now. The beginnings of withdrawal from Afghanistan have resulted in greatly reduced its efforts to conduct combat operations to target the Taliban, but US forces are still present and active. Many (including some members of USSOF) believe we should continue supporting the Afghanistan Special Forces (ASF) while continuing full-spectrum lower intensity operations. From Afghanistan’s perspective, they don’t want the US conducting large scale combat operations in their country, but have shown more tolerance for US operations as long as they are paired with financial assistance. The Afghanistan Government is especially happy with this option if they themselves are not required to commit to a high effort to fight the Taliban. COA/Option 4 VSO is a highly desirable option for the US, even more so if the Afghanistan Government increases its own efforts to combat the Taliban. Arguably, the best of all possible worlds would be for the Afghanistan Government to take more interest in asymmetric and full-spectrum combat, fully develop their security apparatus, support their governmental leaders from national to village levels, and produce their own capital to build further infrastructure. This option is desirable to Afghanistan at it magnifies their ability to increase their own security efforts with increased financial capabilities. This COA also enables the Afghanistan Government to develop leadership from the village to national levels, and ability to roll their mineral revenue into further mineral revenue. This COA is slightly more desirable to

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Table 6.26 US’s pairwise preference matrix Matrix

R4C2 R2C1 R4C1 R3C2 R1C1 R1C2 R3C1 R2C4

0 R4C2 1 1 1/2 1/2 1/2 1/7 1/7 1/7 1/7

R2C1 2 2 1 1 1/2 1/6 1/6 1/6 1/6

R4C1 3 2 1 1 1 1/5 1/5 1/5 1/5

R3C2 4 2 2 1 1 1/4 1/4 1/4 1/4

R1C1 5 7 6 5 4 1 1 1 1

R1C2 6 7 6 5 4 1 1 1 1

R3C1 7 7 6 5 4 1 1 1 1

R2C4 8 7 6 5 4 1 1 1 1

Afghanistan if they aren’t required to aggressively combat the Taliban considering it would allow the country increase to its mineral gains.

6.3.1.3

Assigning Cardinal Values

In order to more accurately assess the payoffs of each option, a weighted scale is designed based on the motivations of the US and Afghanistan Government. This will more accurately assess the payoffs of each COA. This changes the ordinal values form (Fig. 6.2) into cardinal values. The method employed in our analysis is based upon the method described by Fox (2015) using AHP to provided preference values using Saaty’s 9-point scale discussed earlier. We used the template devised for this analysis. We provide the matrices used to find the eigenvectors for the US and the Afghanistan Governments, Tables 6.26 and 6.27. The eigenvectors are found for each set and identified by strategy sets. This yields the game with the following cardinal payoffs.

6.3.1.4

Converting the Original Game

The resulting cardinal value totals for the US and Afghanistan Government were multiplied by the original ordinal numbers for each COA and player to obtain the following new game.

United States

R1: Enemy-focused combat operations R2: Financial aid R3: Combat Ops and Financial aid R4: VSO re-enforced

Afghanistan C1: Combat TB High (0.03876, 0.0379)

C2: Combat TB low (0.03876, 0.0511)

(0.1992, 0.1698)

(0.03876, 0.3110)

(0.03876, 0.0511) (0.1722,0.0511)

(0.1469,0.1680) (0.3265,0.1594)

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Table 6.27 Afghanistan’s pairwise preference matrix Matrix

0 R2C2 1 1 1/2 1/2 1/2 1/5 1/5 1/5 1/8

R2C2 R2C1 R3C2 R4C2 R1C2 R4C1 R3C1 R1C1

6.3.1.5

R2C1 2 2 1 1 1 1/4 1/4 1/4 1/6

R3C2 3 2 1 1 1 1/4 1/4 1/4 1/5

R4C2 4 2 1 1 1 1/3 1/3 1/3 1/4

R1C2 5 5 4 4 3 1 1 1 1

R4C1 6 5 4 4 3 1 1 1 1

R3C1 7 5 4 4 3 1 1 1 1

R1C1 8 8 6 5 4 1 1 1 1

Nonlinear Programming Approach for Two or More Strategies for Each Player

For games with two players and more than two strategies each, we present the nonlinear optimization approach by Barron (2013). Consider a two-person game with a payoff matrix as before. Let’s separate the payoff matrix into two matrices M and N for players I and II. We solve the following nonlinear optimization formulation in expanded form, in Eq. (6.6). Maximiz

n X m X

xi aij y j þ

i¼1 j¼1

n X m X

xi bij y j þ p  q

i¼1 j¼1

Subject to m X

aij y j  p,

i ¼ 1,2, . . . , n,

xi bij  q,

j ¼ 1,2, . . . , m,

j¼1 n X

ð6:6Þ

i¼1 n X

xi ¼

i¼1

xi  0,

m X

yj ¼ 1

j¼1

yj  0

We used the computer algebra system Maple to input the game and then solve. We let the matrix aij be labeled M and bij be labeled N in Maple.

6.3 Two-Person Partial Conflict (Nonzero-Sum) Game > >

>

>

>

>

>

>

>

>

293

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> >

>

>

>

>

The interpretation of the solution is x4 ¼ 1 and y2 ¼ 1 which represents the strategies R4 and C2, respectively. The US supports the VSO structure and the Afghanistan Government uses a low level approach to combating the Taliban. The Afghanistan Government will consider a more financially beneficial option, especially considering they are looking for the most financially beneficial solution now, the United States had vowed to withdrawal by 2015 and the Taliban’s threat is imminent. They understand that they are basically doomed without US support, after the US troop withdrawal. This is an indirect threat by the US toward the Afghanistan Government because they can just sit back and conduct drone strikes for the next 100 years. The Afghanistan Government doesn’t want this because of the potential collateral damage, and they’d undoubtedly settle for less payoff to avoid this.

6.3 Two-Person Partial Conflict (Nonzero-Sum) Game

295

Fig. 6.11 The Nash arbitration point for the end-game

6.3.1.6

Finding Prudential Strategies

We begin by using linear programming to determine the prudential strategies that each side should employ. We start by determining the optimal solution for the United States versus Afghanistan. We do this because the United States will have only four decision variables and only two constraints. We find the Prudential Strategy for the United States is to play half the time financial aid and half the time VSO never playing strategies involving combat operations yielding a security level value of 0.1822 for the United States. We find the Prudential Strategy for the Afghanistan Government is to always play C2 (Low combat operations against the Taliban) while the United States plays R4 with a security value of 0.1594. If the United States and Afghanistan go to arbitration, then we apply the Nash arbitration method and solve using the security levels that we just found. Figure 6.11 uses a template developed by Feix (2007) to find the Nash arbitration point. The Nask arbitration point is found as (0.254, .1972). Further how this arbitration game is played can be determined from the end point of the Pareto optimal line segment values of R2C2 and R4C2. The United States should play R2 1/3 of the time and R4 2/3 of the time in the negotiations while the Afghanistan Government always plays its C2 option.

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6.3.1.7

6 Game Theory

Possible Conclusions and Interpretations

As both sides attempt to obtain their best possible outcomes, we find many possible results. While assessing the payoff values based on strategic moves, it is most beneficial for the US Government to communicate and “sell” the value of VSO and Afghanistan mining its fortune in natural resources to fund their own combat operations targeting the Taliban, achieving a high payoff strategy. Although the Afghanistan Government cannot immediately conduct sustained-high intesity combat operations against the Taliban, it will attempt to achieve its Prudential Strategy/ Security level value of 0.1594. This will likely result in the United States opting for its own Prudential Strategy/Security level value of 0.1822. Each player choosing their own Prudential Strategy/Security level value will not work considering the United States will only opt to conduct VSO in Afghanistan villages achieving (0.1822, 0.2352) yielding no increase for the United States but an increase for Afghanistan. Afghanistan President Karzai’s remarks in March 2013 accused the US of secretly supporting and working with the Taliban. The same problem for each side exists in reverse for Afghanistan. Afghanistan eventually should take the offer of arbitration since the yields a values greater than their security level. It is very clear that if the Afghanistan Government will not take the US’s initial offer of the Nash equilibrium, the game or negotiations will be taken to NATO for Nash arbitration. If Nash arbitration does occur, and everything happens fairly, the United States would achieve a value of 0.254 and the Afghanistan Government a value of 0.1972. But, considering the values (0.254, 0.1972) are greater than the value Equilibrium Value, as well as other options NATO will need to better to sell the US’s VSO offer. An aspect of this game that the United States should consider are international laws followed by NATO that can be summarized as “if you broke it, you fix it or buy it.” The United States should remain weary of the Afghanistan Government’s willingness or the ability to justify the US forced troop or financial support post-2014 although the political bipartisanship and American tax payer’s opinion will likely only support the inexpensive and low risk methodology of VSO. Finally, while evaluating the game between the Afghanistan Government and the United States, the narrative accompanied by the values of each game will help determine the best course of action as communications and cooperation occur between the two parties. This is not a zero-sum game, therefore no player has to completely win, just as in reality. Conversely, each player can either accept a second and third best strategy, or allow Nash arbitration to occur by an outside organization, such as NATO, to handle mediation. Example 3: Case Study: Russia Annexation of Crimea (adapted from Wegersjoe, Fredrik FORNATL, SW [email protected], mathematical class project 2017) Many nations were surprised by Russia’s sudden annexation of Crimea. The explanation seems to be a response to the civil unrest in Kiev 2013, and that Ukraine wanted to become more “western friendly.” However, nations struggle to anticipate such crisis and to enhance their situational awareness, but apparently they are not very successful.

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297

This example will analyze if it would have been possible to anticipate Russia’s action in 2014 and if the outcome could have been different if Ukraine has chosen different options. Could we have anticipated Russia’s annexation of Crimea by the use of Game Theory? Three options will be analyzed: – Should Ukraine have deployed own forces? – Should Ukraine have become members of NATO? – Could United Nations have been an arbitrator that could have changed the outcome? By using game theory, this study will try to recreate Ukraine’s and Russia’s option and its value, with the actual outcome of this crisis in March 14, 2014. This example reveals that it could have been possible to anticipate Russia’s annexation of Crimea. It also explains how different options could have changed the outcome, or at least changed the levels of value that each stakeholder would have got. The arguments in this example are only hypothetical, but will reveal that the use of Game Theory, as support to decision-making, is worthwhile to adapt. This example will rely on rational actors and that every actor strives to achieve the best solution. However, the history reveals that quite often many decision-makers are not rational, and this is something you have to take into consideration when applying this theory. What is more common is that decision-makers more often try to achieve what is best for them. In 2013, Ukraine and EU initiated a dialogue about bilateral agreements between them. However, Ukraine president Viktor Yanukovich seemed to be more fascinated by the Russian side and changed the dialogue that they have started. This was the start of the civil unrest, later called Euromaidan that forced the Ukraine president to resign and flee to Russia. A new western friendly government were assigned and hope for new dialogue with EU aroused. Unfortunately, Russia wanted to protect their backyard from getting to tied to the west. In the spring of 2014, the annexation of Crimea was a fact, and the signals to the Ukraine’s was clear. The world could only observe how “little green men” walked around in Crimea and occupied government buildings. A definition of soldiers with a Russian uniform without any signs or flag.

6.3.1.8

Assumptions

Both actors will choose strategies rationally to maximize utility. NATO/EU cannot be involved in Ukraine because they are not NATO members or have bilateral agreements with EU. However, if Ukraine become members of NATO/EU, they will achieve protection and then become the stronger force.

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Ukraine The objectives for Ukraine are to protect national integrity and sovereignty. If there are indications of Russian operations, Ukraine will reinforce Crimea with necessary military capabilities to deter Russian operations. Russia Russia is the stronger military actor, with capacity to conduct a military operation in order to seize and control Crimea. The option for a military operation is a rapid and covert annexation intent to isolate a potential conflict in order to avoid NATO and EU military involvement. Objectives for Russian operation: Force Ukraine officials to negotiate with Russia instead of EU. To show that they are capable and have the will to launch operations abroad to protect their interests. Ukraine’s Options R1—No defense of Crimea R2—Proactive deployment of armed forces. Russia’s Options C1—No annexation C2—Annexation Ranked Outcomes for Ukraine R1C1—Status quo. Crimea still Ukraine’s property and no deployment. R2C1—Ukraine deploys and hopefully deterred Russia. Crimea still Ukraine’s property. R1C2—Russia annexes and Ukraine doesn’t deploy. Crimea goes Russian. R2C2—Russia annexes and Ukraine deploys. Most likely war and Ukraine loses. Ranked Outcomes for Russia R1C2—Russia annexes and Ukraine doesn’t deploy. Crimea goes Russian. R2C2—Russia annexes and Ukraine deploys. Most likely war and Ukraine loses. R1C1—Status quo. Crimea still Ukraine’s property and no deployment. R2C1—Ukraine deploys and hopefully deterred Russia (at least that is what the rest of the world will think). Crimea still Ukraine’s property. R1—No membership NATO/EU R2—Membership

6.3 Two-Person Partial Conflict (Nonzero-Sum) Game

299

Fig. 6.12 Option 1 Pure Strategy

Russia’s options: C1—No annexation C2—Annexation Ranked outcomes Ukraine: R2C1—Ukraine becomes members of NATO/EU and achieves protection. Crimea still Ukraine’s property. R1C1—Status quo. Crimea still Ukraine’s property. R2C2—Russia annexes and Ukraine becomes NATO/EU. Most likely war and Ukraine will win because of support from the west. R1C2—Russia annexes and Ukraine doesn’t become members. Crimea goes Russian. Ranked outcomes Russia: R1C2—Russia annexes and Ukraine doesn’t become members. Crimea goes Russian. R1C1—Status quo. Crimea still Ukraine’s property. R2C1—Ukraine becomes members of NATO/EU and achieves protection. Crimea still Ukraine’s property. R2C2—Russia annexes and Ukraine becomes NATO/EU. Most likely war and Ukraine will win because of support from the west (Figs. 6.12, 6.13 and 6.14 and Tables 6.28 and 6.29). We have pure strategies in both games without arbitration. However, in option 1 (Ukraine defend themselves). The outcome is annexation of Crimea without

300

Fig. 6.13 Option 2 Pure Strategy

Fig. 6.14 United Nations Arbitration

6 Game Theory

6.3 Two-Person Partial Conflict (Nonzero-Sum) Game

301

Table 6.28 Option 1 Payoff Matrix

Option 1 R1 R2

C1 4, 2 3, 1

C2 2, 4 1, 3

Table 6.29 Option 2 Payoff Matrix

Option 2 R1 R2

C1 3,3 4,2

C2 1,4 2, 1

deployment of own forces. This was what really happened and was ranked as number 3 out of 4 and a value of (13,56). More interesting is option 2, and whether Ukraine should become member of NATO/EU or not. Here, we also have a pure strategy revealing that Ukraine should have become a member and then Crimea would still be Ukraine’s property. The value of this game is the opposite of option 1 (54,13). However, by engaging United Nations as an arbitrator the Nash arbitrary point gives a value of (24,43). This is better for Ukraine than option 1, but it has lesser value for Russia. However, Russia can achieve other benefits if they go this way instead of invading Crimea. The world could eventually have a better assessment of Russia and how they engage in conflicts. The best solution for Ukraine was to become a member of NATO. This was not an option they chose. However, it is not too late to become a member of NATO, but Crimea will then be a more difficult issue to handle. The conclusion is that according to this game theory model, the annexation of Crimea might have been anticipated and might have been obstructed by membership in NATO.

6.3.1.9

Case Study 4: A Model of Irregular Warfare

Introduction Combat is chaotic in any form. Regardless of circumstances, the unknown and unknowable pose frightening questions to the participants and leaders in armed conflict. To mitigate this, militaries have built institutions and formal military processes to manage, to some degree, this chaos. The natural evolution of this process has resulted in doctrine. Doctrine is how the work of war is standardized and codified for instruction and implementation. While conventional combat maintains a wide margin for flexibility and creativity, the formalization of doctrine has created a common operating picture to direct the combatant toward a generally approved solution. Irregular warfare provides none of the standardization that conventional combat operations enjoy. Depending on an innumerable set of variables, the conditions that define the crux of one irregular conflict can differ wildly from the next. With irregular doctrine unable to provide prescriptive solutions, modern doctrine has

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shifted to a list of platitudes that may or may not apply to a given irregular warfare situation. With no definitive course of action prescribed by doctrine, leaders facing irregular challenges are presented with a host options. Selecting from these options becomes the art of military leadership. In an effort to facilitate the selection of appropriate efforts, and the weights to various options, game theory can be employed to facilitate decision-making. By limiting the scope of the inquiry to counterinsurgency, perhaps the most prevalent form of modern irregular warfare, the application of game theory becomes more defined. This project will demonstrate that common insurgent and counterinsurgent strategies can be captured in a two-player game that is type III (optimized against strategic agent) simultaneously, nonzero-sum.

Background The parameters of the game are set at a two-player scenario, but in the insurgentcounterinsurgent (COIN) environment a third party is present. That third party is the populace whose support or resistance becomes the measure of effectiveness and progress in COIN. As such the values of the game will be expressed in terms of popular perception of the given course of action. These values can be both positive and negative. As the players conduct efforts simultaneously with various degrees of competence and effectiveness, the skill and capability of each player in each course of action must be considered. Determining the optimal mix of strategies for each side of the conflict will be assessed in terms of effect on the populace, and the relationship between the two players will determine the priorities for each. Each side has unique resource requirements as the counterinsurgent faces a large expense over time that could affect popular support for the operation while the insurgent has a minimalist approach to expending just enough resources to stay in the fight in an attempt to prolong the conflict. In this manner, we can show that the different long-term strategies for each side effect the payoff values and their optimal solutions as well as their security values. Additionally, the nature of this relationship is adversarial, and optimal solutions for the conditions may lead the players to avoid options that maximize their own payoff if they can widen the margin between their payoff and that of their opponent. While the game remains nonzero-sum, the concept of maximizing payoffs for both sides is negated due to the nature of the relationship between the combatants. Each combatant has a different objective. The insurgents must maintain their existence first and foremost. While decisive victory is ideal, the reality is that insurgent or guerrilla elements typically are incapable of defeating their enemies through active and open combat. To simply maintain their standing with the populace and some degree of military viability will extend the fight, which is all they really need. Conversely, the stakeholders for the COIN forces want results and they measure success differently. When applied to game theory, payoff values of each option become expressive in two unique ways. First, each player has totally different

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objectives and that is reflected in their security values. Secondly, this contest is so adversarial that the optimal solution for a player isn’t always the highest payoff; rather, the optimal payoff may be the one that provides the widest margin of relative gain over the opponent.

6.3.1.10

Definition of Counterinsurgent Lines of Operations

COIN Governance (CG) The counterinsurgent (COIN) governance strategy involves focusing efforts toward establishing local and national level governance and administration. Counterinsurgent governance serves to uphold and protect the needs of the people and is not tarnished by rampant corruption and self-serving or ethnic interests. This strategy also involves establishing a justice system that promotes the rule of law and equality. The basis of this strategy is that with effective leadership and equal justice, the population will support the efforts of the counterinsurgent state. In our game, the value for counterinsurgent governance is determined by multiplying the effectiveness of the counterinsurgent governance (ECG) by the sum of the capability of counterinsurgent governance (CCG), the counterinsurgent governance effect on the population (PCG), and the effect of the insurgent’s selected strategy on counterinsurgent governance.
 CG ¼ ECG  CCG þ PCG þ OðxÞCG O(x)CG is the effect of the insurgent’s strategy (x) on the counterinsurgent governance strategy. Whatever strategy (column) the insurgent selects will determine the counterinsurgent governance (CG) value for that cell of the game. Thus, there are five different values for CG, one for each of the insurgent strategies (columns). The above equation is used for determining the values for all the counterinsurgent and insurgent strategies for our game. Each strategy has its own unique effectiveness (E), capability (C), and population effect (P). Capability and population effect are cardinal values ranging from 0 to 100, while effectiveness is a percentage ranging from 0 to 1. The effect of the opponent’s strategy, O(x), is also a cardinal value from 0 to 100. These values are selected based on the nature of the counterinsurgency and the environment and must be adjusted over time as the conflict changes. COIN Combat Operations (CC) Counterinsurgent combat operations strategy involves conducting direct military action against insurgent forces. Combat operations can take the form of precision capture/kill operations to broad clearing operations to regain large sections of territory. This strategy seeks to disregard the population and focus solely on defeating the insurgency through kinetic, military means. As a point of caution, this strategy can sometimes have a negative effect on the population if collateral damage is not minimized or if the population is disrupted by operations to remove insurgents.

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COIN Essential Services (CE) The counterinsurgent essential services strategy seeks to win population support through the provision and maintenance of essential services. Often in a counterinsurgency conflict, the people suffer from a breakdown of state control and infrastructure, which then translates into a shortage of essential services such as water, electricity, and food provisions. This strategy involves the counterinsurgent state focusing its efforts toward rebuilding control and infrastructure for the purposes of returning essential services to the population, and thus maintaining popular support. COIN Information Operations (CI) The counterinsurgent information operations strategy involves winning the support of the population by winning the battle of the narrative. The population must maintain positive perceptions of the counterinsurgent state through positive media, promotion of state efforts, and continuous highlighting of counterinsurgent successes. This strategy must be timely and must be able to counter the propaganda and misinformation strategy used by the insurgents. COIN Economic Development (CD) The counterinsurgent economic development strategy focuses on increasing the prosperity of the population through economic programs. These economic programs seek to improve personal wealth by enhancing agricultural operations, encouraging small businesses, and creating fair and legalized markets for economic growth. This strategy also seeks to ride the economy of unregulated gray and black markets, and illegal drug economies. The goal of this strategy is to win population support by fostering economic growth, job creation, and increased opportunities. COIN Host Nation Security Force Development (CH) The counterinsurgent host nation security force development strategy focuses on improving the security forces of the counterinsurgent state. The premise to this strategy is that if the state has a monopoly on the use of violence, then the insurgents will not be able to coerce and influence the population. The population will also then turn to the state for matters of security and justice. Security force development involves not only creating forces that can engage the insurgents, but also creating police forces that can ensure daily security and promote justice among the population. These security forces must be focused toward upholding a fair and equal rule of law in order to maintain legitimacy and support of the population.

6.3.1.11

Definition of Insurgent Lines of Operations

Insurgent Governance (IG) Insurgent governance is the ability of the insurgent forces to establish local level governance and administration contrary to the state’s administration. Insurgent governance efforts attempt to coerce population support by replacing local state leaders and administrators with insurgent members. This strategy is similar to the Viet Cong Infrastructure in South Vietnam that drove away local village leadership and state administrators through forced removal, assassinations, or

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intimidation. Insurgent governance also includes establishing an insurgent run justice system at the local levels to replace any state justice mechanisms. Insurgent Combat Operations (IC) Insurgent combat operations strategy involves taking direct action against the counterinsurgent security forces and the state’s control regime. Insurgent combat operations can take various forms from low level terrorism and harassment attacks, to guerrilla warfare, or to a conventional style war of movement. Like the other strategies, the success of the combat operations strategy depends on the effectiveness of the insurgent operations, the capabilities of the insurgents, the population’s effect on the insurgent’s operations, and the counterinsurgent state’s strategy. Insurgent Essential Services (IE) The insurgent essential services strategy involves winning population support by providing the essential services to the population that the counterinsurgent state cannot provide. Often in a counterinsurgency conflict, the people suffer from a breakdown in basic services such as water, electricity, and food provision. With this strategy, the insurgents arrive in the absence of the state to provide the much needed services and win population support. This strategy can also involve manipulating and controlling the population by having the ability to both provide essential services to cooperative population, and also withhold or disrupt essential services to non-supportive population. Insurgent Information Operations (II) The insurgent information operations strategy involves winning the support of the population by winning the battle of the narrative. This strategy seeks to use propaganda, political teachings, misinformation, and other information operations in order to sway public perception and direct population support toward the insurgency and away from the counterinsurgent state. Insurgent Economic Development (ID) The insurgent economic development strategy seeks to win population support by providing economic value to the populace. Insurgent economic development efforts can come in nefarious forms, such as the promotion of gray and black markets and the cultivation of the illegal drug market. However, insurgent economic development efforts can also take positive forms, such as providing cooperative populations with agricultural resources and the means to contribute their goods to markets. Insurgents can also foster economic development by providing paying jobs to population members within the insurgent organization.

6.3.1.12

Setting up the GAME

To demonstrate the game, we assigned values to each variable to determine the values for each Line of Operation (LOO). We obtain a two-person partial conflict game. The LOO values produced the matrix in Table 6.30. To conceptualize the region of possible solutions, the values are plotted as x and y

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Table 6.30 Example Game LOO Matrix

COIN FORCES

GOVERN COMBAT OPS ESS. SERVICES IO ECON.DEV. HN SEC. FORCES

INS FORCES GOVERN COMBAT OPS (48,45) (44,30) (27.5,35) (45,27)

ESS. SERVICES (40,51) (25,45)

(44,56) (17.5,56)

ECON. DEV. (40,39) (25,33)

(60,40)

(45,27)

(63,51)

(54,49)

(57,39)

(20,40) (52,40) (56,35)

(18,24) (44,27) (48,30)

(17,51) (52,51) (40,45)

(24,70) (52,56) (52,56)

(16,45) (60,39) (40,33)

IO

Fig. 6.15 Example Game Line of Operation

coordinates. A polygon is then created to determine the Pareto optimal region, which is the region where both players can maximize their returns. The plotting of the values above produces the graph in Fig. 6.15. Since there are multiple options, this game will be solved with non-linear programming (NLP). Each LOO will be assigned a variable. Let: COIN Governance ¼ x1 COIN Combat Operations ¼ x2 COIN Essential Services ¼ x3

6.3 Two-Person Partial Conflict (Nonzero-Sum) Game Fig. 6.16 Payoff polygon

80 70 60 50 40 30 20 10 0

307 INS

INS

0

10

20

30

40

50

60

70

COIN Information Operations ¼ x4 COIN Economic Development ¼ x5 COIN Host Nation Development ¼ x6 INS Governance ¼ y1 INS Combat Operations ¼ y2 INS Essential Services ¼ y3 INS Information Operations ¼ y4 INS Economic Development ¼ y5

6.3.1.13

Nash Equilibrium (NE)

First, NLP will be used to determine if there is a Nash equilibrium point. Nash equilibrium would mean that there is a point of balance where each side has chosen the best response to the other players’ choices. A limitation to this is that it supposes that each side correctly foresaw the others actions and correctly reacted to it. It is pointed out that there might be multiple equilibriums to this game. We use the NLP method outline using Eq. (6.6) using Maple©. We found two possible solutions as equilibrium values. Solution 1: (54,50) when x3¼0.846 and x6¼0.154 and y4¼1 Solution 2: (63,51) with the players playing x3¼1, y3¼1. Figure 6.16 provides the payoff polygon graph with the Nash equilibrium point. We find the Nash equilibrium at (63,51) is Pareto optimal.

6.3.1.14

Prudential Strategies (P)

Next, LP will be used to determine the prudential strategies for each player. A prudential strategy is a strategy that produces the highest value for the player independent of the other players’ actions. Since this is a value that the player can obtain without any interaction from the opponent, this is called a security level. To be enticed to cooperate with the opponent, a value higher than what could be obtained by the prudential strategy must be offered.

308

6.3.1.15

6 Game Theory

Coin Prudential

For the COIN player, we will maximize the value of the game for the COIN player in the COIN player's game. LP will be used to solve this, using the following: Maximize Vc Subject to : 48 x1 þ 27:5 x2 þ 60 x3 þ 20 x4 þ 52 x5 þ 56 x6  V  0 44 x1 þ 45 x2 þ 45 x3 þ 18 x4 þ 44 x5 þ 48 x6  V  0 40 x1 þ 25 x2 þ 63 x3 þ 17 x4 þ 52 x5 þ 40 x6  V  0 44 x1 þ 17:5 x2 þ 54 x3 þ 24 x4 þ 52 x5 þ 52 x6  V  0 40 x1 þ 25 x2 þ 57 x3 þ 16 x4 þ 60 x5 þ 40 x6  V  0 x1 þ x2 þ x3 þ x4 þ x 5 þ x6 ¼ 0 x1  1 x2  1 x3  1 x4  1 x5  1 x6  1 Vc ¼ 46:8 x3 ¼ 0:4 x6 ¼ 0:6

6.3.1.16

Insurgent Prudential

For the INS player, we will maximize the value of the game for the INS player in the COIN player’s game. LP will be used to solve this, using the following:

6.3 Two-Person Partial Conflict (Nonzero-Sum) Game

309

Room to improve

Population Control Value 80 70

Insurgent

60 50 40

LLO Values Nash Equilibrium Security Level

30 20 10 0 0

10

20

30

40

50

60

70

Coin

Fig. 6.17 LOO chart with security levels

Maximize Vi Subject to : 45y1 þ 30y2 þ 51y3 þ 56y4 þ 39y5  V  0 35y1 þ 27y2 þ 45y3 þ 56y4 þ 33y5  V  0 40y1 þ 27y2 þ 51y3 þ 49y4 þ 39y5  V  0 40y1 þ 24y2 þ 51y3 þ 70y4 þ 45y5  V  0 40y1 þ 27y2 þ 51y3 þ 56y4 þ 39y5  V  0 35y1 þ 30y2 þ 45y3 þ 56y4 þ 33y5  V  0 y1 þ y 2 þ y3 þ y4 þ y5 ¼ 0 y1  1 y2  1 y3  1 y4  1 y5  1 Vi ¼ 50:07692 y3 ¼ 0:538 y4 ¼ 0:462 The Security Level for the COIN player is 46.8 and the Security Level for the INS player is 50.08. To do this, the COIN player would use a mixed strategy of 40% COIN essential services and 60% COIN host nation security forces, while the INS player would use a mixed strategy of 53.8% INS essential services and 46.2% INS information operation. Figure 6.17 is the graph with the security levels added and the area where each side could improve their results through compromise and negotiation:

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6 Game Theory

6.3.1.17

Nash Arbitration (NA) Point

The Nash arbitration point is the highest value that each player could obtain through binding arbitration and compromising with the other player. This point is found by using the formulation expressed in Nash’s Theorem. We find the Nash arbitration point as ð55:91; 54:22Þ by playing about 64:4%of ð52; 56Þ and 35:6%of ð63; 51Þ: A Nash arbitration solution requires each side to cooperate with the other side in order to obtain a higher value. This cooperation could be obtained through incentives or coercion. In this scenario, the opponents are in an open conflict and neither side would want to see the other increase their return on the game. While unlikely, cooperation remains feasible. For example, threat of force escalation could be a coercive tactic to keep certain LOOs that are detrimental to one party off of the table. Also, outside parties could influence each party’s mixture of strategies, which could create the effect of cooperation between the two parties. In this example game, the COIN forces are in trouble. Their winning strategy is dependent on the insurgent forces being willing to negotiate to reach a Nash arbitration solution. In contrast, the winning strategy for the INS forces is the Nash equilibrium which assumes that each side has chosen the best response to the other side choices. Since the INS do not require any cooperation to achieve this result, it is likely that they will win this competition.

6.3.2

Conclusions

Although we presented methodologies to more accurately depict the use of game theory by using cardinal utilities, we only illustrated with the Tannenberg example from Cantwell. However, the results are promising enough to continue to employ these methodologies to assist military planners and decision-makers. Game theory does provide insights on how to play a game and therefore, we conclude that it does provide insights into military planning and strategy. Weighting can always be improved as shown by Khatwani and Kar (2016).

6.4 The Three-Person Games

311

Table 6.31 Generic three-person game between Rose, Colin, and Larry Larry L1

Rose

6.4

R1 R2

Larry L2 C1 (R1,C1,L1) (R2,C1,L1)

Colin C2 (R1,C2,L1) (R2,C2,L1)

R1 R2

C1 (R1,C1,L2) (R2,C1,L2)

C2 (R1,C2,L2) (R2,C2,L2)

The Three-Person Games

6.4.1

Solving the Three-Person Games in Game Theory using EXCEL or MAPLE

6.4.1.1

Introduction

In the three-person games, we find Nash equilibrium via movement diagrams and then break the game down into possible coalitions. This pits two players versus the third player. All possible coalitions game values are evaluated. From these results we look for any naturally forming coalitions. Next, we visit the partial conflict games. After covering the techniques for finding equilibrium and negotiated solution, we return to the three-person games. We cover the solution techniques for finding the Nash equilibrium and all the possible coalitions to attempt to determine what might happen. Let’s define a generic simultaneous three-person game theory payoff matrix as shown in Table 6.31. We give Larry two strategies {L1, L2}, Colin two strategies {C1, C2}, and Rose two strategies {R1, R2}. In a three-person total conflict game (zero-sum or constant-sum), the values in each triplet, (Ri,Ci,Li), sum to either zero or the same constant. In a three-person nonzero-sum game, the values in each triplet, (Ri,Ci,Li), do not all sum to zero or do they sum to the same constant. We also make the following assumptions about the game: Games are simultaneous. Players are rational meaning they want the best outcome possible versus their opponents. Games are repetitive. Players have perfect knowledge about their opponents.

6.4.1.2

Three-Person Total Conflict Games

The solution methodology of the three-person total conflict games involves several steps. First, we use the movement diagram as we describe how to find all the Nash

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6 Game Theory

Table 6.32 Three-person game example (Source: Straffin, Chap. 19) Rose

R1 R2

Rose

R1 R2

Larry L1 Colin C1 (1,1,2) (2,4,2) Larry L2 Colin C1 (3,2,1) (2,2,4)

C2 (4,3,1) (5,5,10)

C2 (6,6,12) (2,3,1)

equilibriums. The Nash equilibrium is defined when no player would unilaterally change their outcomes. Consider the three-person (total conflict) zero-sum game between Rose, Colin, and Larry (from Straffin, Chap. 19) shown in Table 6.32.

6.4.1.3

Movement Diagram

We define a movement diagram as follows for each player’s possible outcomes R1 or R2, C1 or C2, and L1 or L2, draw an arrow from the smallest to the largest value. For Rose, arrows are drawn vertically from smaller to larger. For example, under Larry L1 and Colin C1, the value 2 in R2 is greater than the value 1 in R1 so the arrow goes from R1 to R2. For Colin, arrows are drawn horizontally between C1 and C2 from smaller values to larger values. For Larry, arrows are drawn diagonally to represent the two games, L1 and L2 with arrows drawn from corresponding positions. This is illustrated in Table 6.33. We follow the arrows. If any set or sets of arrows bring us to a point where no arrow leaves that point or points, then we have an equilibrium point or points. Result: The movement diagram reveals two pure strategy Nash equilibriums at R1C1L2 (3, 2,1) and at R2C1L1 (2,4,2). These are not equivalent and not interchangeable. Going for one equilibrium point over another by either player may lead to a non-equilibrium outcome because of player’s preferences.

6.4.1.4

Coalitions Possible

Let’s consider communications with the ability to form coalitions. Assume first that Colin and Larry form a coalition against Rose. The following steps are helpful in the setting up and analysis of the coalition. Step 1. Build a payoff matrix for Rose against the Colin-Larry coalition using Rose’s values from the original payoffs in Table 6.34.

6.4 The Three-Person Games

313

Table 6.33 Movement diagram for three-person zerosum game

Larry L1 Colin

Rose

Rose

C1

C2

R1

(1,1,2)

(4,3,1)

R2

(2,4,2)

(5,5,10)

Larry L2 Colin C1

C2

R1

(3,2,1)

(6,6,12)

R2

(2,2,4)

(2,3,1)

Table 6.34 Three-person Coalition Values Rose

R1 R2

Colin-Larry C1L1 1 2

C2L1 4 5

C1L2 3 2

C2L2 6 2

Step 2. Find a solution for the Nash equilibrium using either (a) saddle point (maximin) or (b) mixed strategies methods. (a) No saddle point solution RowMin {6, 5} ColMax {2,4,3,2} (b) The graph, Fig. 6.18, shows that the Maximin solution is found by using the following values for Rose versus the Coalition. We can easily find the solution in Table 6.35. If the game has a saddle point solution that those values are the value of the game for all three players. Since we have a mixed strategy, then we must find the value for each of our three players. Step 3. Finding the values of the game for each player.

314

6 Game Theory Rose R1

Fig. 6.18 William’s graphical method to eliminate rows

Rose R2

Table 6.35 Three-person Coalition Values

Rose Oddments

R1 R2

Colin-Larry C2L1 4 5 1 4/5

C2L2 6 2 4 1/5

Oddments 2 3

3/5 2/5

Value

22/5 or 4.4

3 4 31 24 21  R1C2L1 þ R1C2L2 þ R2C2L1 þ R2C2L2 5 5 55 55 55 We now substitute the values from the original payoff matrix. 3 4 3 1 2 4  ð4; 3; 1Þ þ  ð6; 6; 12Þ þ  ð5; 5; 10Þ 5 5 5 5 5 5 2 1 þ  ð2; 3; 1Þ ¼ ð4; 4; 0:64; 5:04Þ 5 5 We find the payoffs are to Rose 4.4, to Colin .64, to Larry 5.04 Step 4. Redo steps 1–3 for Colin versus a coalition of Rose-Larry and then redo steps 1–3 for Larry versus a coalition of Rose-Colin. Results are as follows: Colin versus Rose-Larry: Value of (2,4,2) and this was the saddle point solution Larry versus Rose-Colin: (2.12,0.69,1.43) Rose versus Colin-Larry: (4.4,0.64, 5.04) from before Step 5. Determine which coalition, if any, yields the best payoff for each player. Rose: Max {2, 2.12, 4.4} is 2.12 so Rose prefers a coalition with Colin. Colin: Max {4,0.69,0.64} is 0.64 so Colin prefers a coalition with Larry. Larry: Max {2,1.43,5.04) is 5.04 so Larry prefers a coalition with Colin.

6.4 The Three-Person Games

315

Table 6.36 Three-person Coalition Payoff Larry

Fig. 6.19 Three-person with Coalition

L1 L2

Rose-Colin R1C1 2 1

R1C2 1 12

Larry L1

R2C1 2 4

R2C2 10 1

Larry L2

Maximin

In two of these cases, we find that Colin-Larry is the preferred coalition so we might expect that the Colin-Larry coalition will naturally be the coalition formed. We note that we may or may not be able to determine which coalition might be formed. We also note that there are both bribes and side payments allowed. These bribes or payments entice a coalition to either change or keep the coalition together. Characteristic function: The number v(S), called the value of S, is to be interpreted as the amount S would win if they formed a coalition. We assume that the empty coalition (none are formed) value is zero, v(∅) ¼ 0 Colin versus Rose-Larry: (2,4, 2) Larry versus Rose-Colin: (2.12,.69,1.43) Rose versus Colin-Larry: (4.4,.64, 5.04) We can build the functions: Empty set: v(∅) ¼ 0 Alone: v(Rose) ¼ 4.4, v(Colin) ¼ 4, v(Larry) ¼ 1.43 Coalition by two(s): v(Rose-Colin) ¼ 1.43 v(Rose-Larry) ¼ 4 v(Colin-Larry) ¼ 4.4 We add the payoff for the coalition’s partners in the associated games. Coalitions by three: These are zero-sum games so adding all payoff together ¼ 0 v(Rose-Colin-Larry) ¼ 0 Thus, Larry versus Rose-Colin (2.12, .69, 1.43) (Table 6.36) No saddle point exists since Max of {2,4} is 2 and Min of {1,12,2,10} is 1. We move on to find the mixed strategies (Fig. 6.19).

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6 Game Theory

Table 6.37 Three-person Coalition Payoff Subgame

Larry Oddments

L1 L2

R1C1 2 1 1 6/7

Rose-Colin R2C1 2 4 6 1/7

Oddments 4 3

3/7 4/7

Value is

10/7

Subgame (Table 6.37): (3/7)(6/7) (1,1,2) + (3/7)(1/7)(2,4,2)+(4/7)(6/7)(3,2,1)+(4/7)(1/7) (2,2,4) ¼ (104/49, -34/49 , 10/7) ¼(2.12,.069,1.43) rounded to two-decimal places. Although the mathematics is not difficult, the number of calculation is quite tedious. Therefore, we built a technology assistant for student use.

6.4.1.5

Technology Assistant with EXCEL

We developed a technology assistant to assist the students with the many calculations involved. Instructions are provided within the template, which is a macroenhanced Excel worksheet. These instructions include: 1. 2. 3. 4. 5.

Put the R,C,L entries into the blocks to the left Go to Coalition_R_CL and execute the Solver Go to Coalition_C_RL and execute the Solver Go to Coalition_L_RC and execute the Solver List the equilibrium values if the players play alone and the equilibriums in the three coalitions 6. Determine if any coalition naturally forms 7. Is there a legitimate bribe to change the coalition? In Fig. 6.20, we find the results or outcomes of the calculations made to find the pure strategies equilibrium and the results of the coalitions. The user must then interpret the results and make conclusion about those results as to what is likely to occur.

6.4.1.6

N-Person Games with Linear Programming

The coalition’s solution on each worksheet uses the Solver, specifically SimplexLP. We illustrate with a three-person zero-sum game that we just saw in the previous example. Recall, we created the game payoffs for the potential coalitions (Table 6.38):

6.4 The Three-Person Games

317

Fig. 6.20 Screenshot of three-person game template with instructions

Table 6.38 Three-person Coalition Payoff Rose

R1 R2

Colin-Larry C1L1 1 2

C2L1 4 5

C1L2 3 2

C2L2 6 2

This is a zero-sum game for solve for Rose and get the Colin-Larry coalition’s results come from the sensitivity column. Note there are some negative entries as payoffs so we let v ¼ V1  V2 (Winston 1995). We formulate the LP. Maximize v ¼ V1  V2 x1 þ 2x1  V1 þ V2  0 4x1  5x2  V1 þ V2  0 3x1 þ 2x2  V1 þ V2  0 6x1  2x2  V1 þ V2  0 x1 þ x2 ¼ 1 x1  1 x2  1 non  negativity We find the LP solution to this game to Rose is 4.4 when x1 ¼ 0.6 and x2 ¼ 0.4. We find from the reduced costs (the dual solution for Colin and Larry coalition),

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6 Game Theory

y1 ¼ y3 ¼ 0,

y2 ¼ 0:8,

and

y4 ¼ 0:2,

Vcl ¼ 4:4

Although this gives us a Coalition value, we must use all the probabilities for the players to obtain the values to each of our players separately. We only have to use the strategies with probabilities greater than 0: ð:6 Þð:8ÞR1C2L1 þ ð:4Þð:8ÞR2C2L1 þ ð:6 Þð:2ÞR1C2L2 þ ð:4Þð:2ÞR2C2L2 :48ð4; 3; 1Þ þ :32ð5; 5; 10Þ þ :12ð6; 6; 12Þ þ :08ð2; 3; 1Þ ¼ ð4:4; 0:64; 5:04Þ Rose loses 4.4 (as shown before) and the Coalitions 4.4 is broken down as 0.64 for Colin and 5.04 for Larry. We repeat this process for each Coalition to obtain these results: Colin vs: Rose  Larryð2; 4; 2Þ Larry vs: Rose  Colinð2:12; 0:69; 1:43Þ It is still up to the user to interpret and analyze these results. These procedures work for constant-sum games as well.

6.4.1.7

A three-Person Game That Is a Strict Partial Conflict Game (Nonzero-Sum Game) Using Technology

We also developed an assistant for the partial conflict game. This technology assistant requires the use of the Solver six times in the spreadsheet since each player or side in a coalition requires a linear programming solution. The instructions are listed inside the template (Fig. 6.21). The results here are as follows: Pure strategy by movement diagram finds an equilibrium at R1C1L2 with values (2,1,1)

R1C1L1 R1C2L1 R2C1L1 R2C2L1 R1C1L2 R1C2L2 R2C1L2 R2C2L2

Equilibrium No No No No Yes No No No

We easily see a better set of values as an output of (4,2,3) at R1C2L1. We analyze all coalitions to see if that solution rises from any coalitions.

6.4 The Three-Person Games

319

Fig. 6.21 Screenshot of three-person game results

Table 6.39 Three-person Coalition Payoff Coalition One Rose vs. Colin-Larry Coalition Two Colin vs. Rose-Larry Coalition Three Larry vs. Rose-Colin

Rose 1.5

To Players Colin 1

Larry 1

1.75

0.5

0.75

1.5

1.5

1

From the linear programming solution of the coalitions, we find (Table 6.39): Rose prefers a coalition with Larry, Colin prefers a coalition with Rose, and Larry prefers either a coalition with Colin or being alone. There is no preferred coalition and none gets us to the better value. Perhaps all the players should just all agree to play the strategies that provide the best solution.

6.4.1.8

The Three-Person Game in MAPLE

We provide the procedure in MAPLE and repeat these examples.

320

6.4.1.9

6 Game Theory

Conclusions

We have described the use of Excel templates and MAPLE to assist in the solution to the three-person games. We remark that users must still analyze the numerical values to determine what will most likely happen.

6.4.2

Case Study: Korean Unification Game (Adapted from Olish and Spence Mathematics Class Project 2017)

The fate of North and South Korean has been a hot political debate for the past 60 years, with the DPRK again stating it is “at war” with South Korea.(1) With International policy as a guiding force, what is the US’s best option for the future of Korea and which potential allies are in a position to either join or oppose a coalition? Who are the players and what is the game? We decide to examine the key players as: China, Russia, and the Unite States: To determine outcomes and possible coalitions, we first build a three-person game consisting of these three players with two strategies each: Unify(U) or Separate (S). Next, we assign ordinal values to each outcomes based upon the interpretation of the United States, China, and Russian Korean Unification polices. The following assumptions were made in this analysis: US 1. Wants peaceful unification 2. Current status quo is acceptable 3. Disfavor a China/Russian coalition to unify China 1. Wants to maintain a buffer 2. Views DPRK as a nuisance 3. Prefers Russia over the United States as an ally Russia 1. Fears a power dominance by anyone but themselves 2. Desires more power 3. Wants any agreement that benefits Russia Possible positions are identified in Table 6.40. Ordinal values are given based upon our experts (Table 6.41).

6.4 The Three-Person Games

321

Table 6.40 Three-person Unification Positions

Results Unification Unification Unification Unification Separate Separate Separate Separate

United States U U U S S U S S

China U U S U S S U S

Russia U S U U S S S U

Table 6.41 Expert Opinions on Three-person Unification Positions

Results Unification Unification Unification Unification Separate Separate Separate Separate

United States 10 7 8 1 5 6 4 2

China 1 3 2 4 10 8 9 6

Russia 5 2 7 8 5 9 6 4

Fig. 6.22 Movement Diagram three-person Unification Positions

We place the corresponding values into a three game template. Rose is the United States, Colin is China, and Larry is Russia (Fig. 6.22). The movement diagram shows a Nash equilibrium at (U, S, S), outcomes (6, 8, 9). Without coalitions, we find the equilibrium reflects the current international politics with respect to Korea. Next, we consider the ability of these countries to support coalitions if that enables them to achieve a better outcome. We set and solve the six associated coalition games using linear programming. We separate the results into values for each player in these games. We find the United States prefers a coalition with China against Russia, China prefers a coalition with Russia, and Russia prefers a coalition with China. A naturally

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6 Game Theory

forming coalition exists with China and Russia versus the United States. If we do not like this outcome what can we do? We might be able to economically influence the outcome. In 1972, President Nixon improved relations with China in order to balance Soviet international power. The situation is more complicated now, as both China and Russia present themselves as rising powers. This however is just one of the many games going on in the geopolitical world. Though the US prefers China in a coalition in this game, it might be best to slow friction between Russia and China or seek more gains in other diplomatic rows, as the status quo is acceptable to the United States.

6.5

Sequential Game in Extensive Form

Extensive games are sequential games. In a sequential game, each player takes a turn in implementing a strategy. A good example of this is the game of chess. These games seem to be more realistic in real conflict situations and the information about previous choices are available as the situation develops in the game. This is a good way to think about the game and these games might be able to be reduced to a matrix game. In this chapter, we provide multiple examples some of which we can easily collapse into a matrix game and one example where I cannot seem to find a good way to collapse the game.

6.5.1

Cuban Missile Crisis Example

In this case study adapted from Straffin (2004), we begin with an oversimplified description of the Cuban Missile crisis between the USSR and the United States. This was a game between President Kennedy (USA) and Premier Khrushchev (USSR) in 1963. Khrushchev began the game by deciding whether or not to place intermediate range ballistic missile in Cuba. If he places the missiles, Kennedy considers three options: do nothing, blockade Cuba, or eliminate the missile via a surgical airstrike or an invasion. If Kennedy chooses a more aggressive action such as the blockade or elimination strategies, Khrushchev can either give in to the demands and remove the missiles (acquiesce) or escalate the confrontation and perhaps risk nuclear war. The strategy definitions are provided and a tree diagram is shown in Fig. 6.23.

6.5 Sequential Game in Extensive Form

323

Fig. 6.23 Screenshot Excel Cuban Missile Crisis Game Table 6.42 Cuban Missile Crisis Game Payoff Matrix Kennedy

R1 R2 R3 C1 C2 C3 C4 C5

R1 R2 R3

C1 10 10 10

Khrushchev C2 10 8 6

C3 10 8 -20

C4 10 15 6

C5 10 15 20

No missiles Blockage Airstrike Don’t place missile Place missiles, acquiese Place missiles, acquiese to blockade, escalate airstrike Place missile, escalate to blockade, acquisies to airstrike Place missiles, always escalate

We can collapse this game into a matrix game where Kennedy has three strategies and Khrushchev has five strategies. We might have a game that looks like this (Table 6.42): This oversimplified game has a solution at R1C5, which represents do nothing for Kennedy and always escalate for Khrushchev. This is certainly not a favorable solution.

6.5.2

North Korea Missile CRISIS

Let’s consider a game where the leader in North Korea might act irrational or rational (Fig. 6.24). We can collapse and solve the game (Fig. 6.25). A more interesting question to consider is for what values of P are we happy with the results of the game and what values of P are we unhappy. This approach is very

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6 Game Theory

Fig. 6.24 Cuban Missile Crisis Game No Perfect Information

Fig. 6.25 Cuban Missile Crisis Strategy

similar to an approach shown by McCormick et al. in “Warlord Politics” having applications in dealing with warlord in various regions of concern.

6.5 Sequential Game in Extensive Form

325

Fig. 6.26 Kidnapping Decision Tree

6.5.3

Example 3: Kidnapping

Here is an example that we cannot collapse. It is based loosely on a terrorist type kidnapping scenario. A kidnapper takes a hostage and demands a ransom for the return of the hostage. The hostage may or may not get the ransom paid. The kidnapper may kill the hostage or not kill and release the hostage. If the hostage is released, he may or may not report the kidnapping to the police or FBI. We assume that the kidnapper gets +5 for getting paid, 2 for having the kidnapper report the crime, and 1 for killing the hostage. We assume these are additive at each node. The utilities for the hostage are 10 for getting killed, 2 for paying the kidnapper, and +1 for reporting the crime. The entries will be (kidnapper, hostage). For example, pay, release, report would be calculated as (+52, 2+1) or (3, 1). We can calculate the end values of the tree as: • • • • • •

Pay killed (4, 12) Pay, release, report (3, 1) Pay, release, no report (5, 2) No pay, killed (-1, 10) No pay, release, report (2, 1) No pay, release, no report (0, 0) We solve by backward methods and get Figs. 6.26 and 6.27.

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6 Game Theory

Fig. 6.27 Kidnapping Decision Tree Solution

The solution is (1, 10) which implies that we do not pay the ransom and allow the person kidnapped to be killed by the kidnapper. This implies why we should not deal or negotiate with terrorists.

6.5.3.1

Chapter Summary

Many example and case studies have been provided using game theory. We state that the use of game theory, in our opinion, is excellent for providing insights into the problems, strategies, and processes to play the game. The answer itself is not as essential and the methodology used and process development for the game.

References Barron, E. D. (2013). Game theory an introduction (pp. 156–158). Hoboken, NJ: Wiley. Cantwell, G. (2003). Can two person zero sum game theory improve military decision-making course of action selection? Fort Leavenworth, KS: School of Advanced Military Studies. Feix, M. (2007). Game theory: Toolkit and workbook for defense analysis students. MS Thesis, Naval Postgraduate School. Fox, W. (2016). Applied game theory to improve strategic and tactical military decision theory. J Def Manag, 6, 2. https://doi.org/10.4172/2167-0374.1000147.

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327

Fox, W. P. (2010). Teaching the applications of optimization in game theory’s zero-sum and non-zero sum games. International Journal of Data Analysis Techniques and Strategies, 2(3), 258–284. Fox, W. P. (2012a). Mathematical modeling of the analytical hierarchy process using discrete dynamical systems in decision analysis. Computers in Education Journal, 3(3), 27–34. Fox, W. P. (2012b). Mathematical modeling with Maple. Boston, MA: Cengage. Fox, W. P. (2014). Chapter 221, TOPSIS in business analytics. In Encyclopedia of business analytics and optimization (Vol. V(5), pp. 281–291). Hershey, PA: IGI Global and Sage. Fox, W. P. (2015). Two person zero-sum games as an application of linear programming using the excel solver. Computers in Education Journal, 6(3), 37–49. Fox, W., & Thompson, M. N. (2014). Phase targeting of terrorist attacks: simplifying complexity with analytical hierarchy process. International Journal of Decision Sciences, 5(1), 57–64. Gant, J. (2009). A strategy for success in Afghanistan: one tribe at a time. Los Angeles, CA: Nine Sisters Imports. Gillman, R., & Housman, D. (2009). Models of conflict and cooperation. Providence, RI: American Mathematical Society. Giordano, F., Fox, W., & Horton, S. (2014). A first course in mathematical modeling (5th ed.). Boston, MA: Cengage Publishing. Global Data Ltd. (2012, October). Afghanistan’s mining industry on the path of development. London, UK. Retrieved from http://search.proquest.com/ Haywood, O. (1954). Military decision and game theory. Journal of the Operations Research Society, 2(4), 365–385. Jensen, N., & Johnston, N. (2011). Political risk, reputation, and the resource curse. Comparative Political Studies, 44, 662–688. https://doi.org/10.1177/0010414011401208. Khatwani, G., & Kar, A. K. (2016). Improving the Cosine Consistency Index for the analytic hierarchy process for solving multi-criteria decision making problems. Applied Computing and Informatics, 13(2), 118–129. Kral, S. (2011, April). Minerals key to Afghanistan. Mining Engineering, 63(4), 136. Lipow, J., & Melese, F. (2011, September). Economic and security implications of Afghanistan’s newly discovered mineral wealth. Defense & Security Analysis, 27(3), 277. Retrieved from http://search.proquest.com/ Nash, J. (1950). The bargaining problem. Econometrica, 18, 155–162. Peters, S., Kalaly, S., Chirico, P., & Hubbard, B. (2012). Summaries of the important areas formineral investment and production opportunities of nonfuel minerals in Afghanistan: U.S. Geological Survey Open-File Report 2011-1204. Retrieved from http://pubs.usgs.gov/of/ 2011/1204/. Saaty, T. (1980). The analytic hierarchy process. New York: McGraw-Hill Book. Schmitt, J. (1994). Mastering tactics: A decision game workbook. Quantico, VA: Marine Corps Gazette. Straffin, P. D. (2004). Game theory and strategy. Washington, DC: Mathematical Association of America. Von Neumann, J., & Morgenstern, O. (2004). Theory of games and economic behavior (60th anniversary ed.). Princeton, NJ: Princeton University Press. Wasburn, A., & Kress, M. (2009). Combat modeling. (International Series in Operations Research & Management Science) 2009th Edition. New York: Springer. Winston, W. (1995). Introduction to mathematical programming; Applications and algorithms. Belmont, CA: Thomson. Winston, W. L. (2003). Introduction to mathematical programming (4th ed.). Belmont, CA, Duxbury Press.

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Suggested Reading Aumann, R. J. (1987). Game theory. In The New Palgrave: A dictionary of economics. London: Palgrave Macmillan. Bazarra, M. S., Sherali, H. D., & Shetty, C. M. (2006). Nonlinear programming (3rd ed.). New York: Wiley. Braun, S. J. (1994). Theory of moves. American Scientist, 81, 562–570. Camerer, C. (2003). Behavioral game theory: experiments in strategic interaction, Russell Sage Foundation, description and introduction (pp. 1–25). Chiappori, A., Levitt, S., & Groseclose, P. (2002). Testing mixed-strategy equilibria when players are heterogeneous: The case of penalty kicks in soccer. American Economic Review, 92(4), 1138–1151. Crawford, V. (1974). Learning the optimal strategy in a zero-sum game. Econometrica, 42(5), 885–891. Danzig, G. (1951). Maximization of a linear unction of variables subject to linear inequalities. In T. Koopman (Ed.), Activity analysis of production and allocation conference proceeding (pp. 339–347). New York: Wiley. Danzig, G. (2002). Linear programming. Operations Research, 50(1), 42–47. Dixit, A., & Nalebuff, B. (1991). Thinking strategically: The competitive edge in business, politics, and everyday life. New York: W.W. Norton. Dorfman, R. (1951). Application of the simplex method to a game theory problem. In T. Koopman (Ed.), Activity analysis of production and allocation conference proceeding (pp. 348–358). New York: Wiley. Dutta, P. (1999). Strategies and games: theory and practice. Cambridge, MA: MIT Press. Fox, W. P. (2008). Mathematical modeling of conflict and decision making: The writer’s guild strike 2007–2008. Computers in Education Journal, 18(3), 2–11. Fox, W. P., & Everton, S. (2013). Mathematical modeling in social network analysis: Using TOPSIS to find node influences in a social network. Journal of Mathematics and Systems Science, 3(2013), 531–541. Fox, W. P., & Everton, S. (2014a). Mathematical modeling in social network analysis: using data envelopment analysis and analytical hierarchy process to find node influences in a social network. Journal of Defense Modeling and Simulation, 2014, 1–9. Fox, W., & Everton, S. (2014b). Using mathematical models in decision making methodologies to find key nodes in the noordin dark network. American Journal of Operations Research, 2014, 1–13. Gale, D., Kuhn, H., & Tucker, A. (1951). Linear programming and the theory of games). In T. Koopman (Ed.), Activity analysis of production and allocation conference proceeding (pp. 317–329). New York: Wiley. Gintis, H. (2000). Game theory evolving: a problem-centered introduction to modeling strategic behavior. Princeton, NJ: University Press. Harrington, J. (2008). Games, strategies, and decision making. New York: Worth. Isaacs, R. (1999). Differential games: A mathematical theory with applications to warfare and pursuit, control and optimization. New York: Dover. Kilcullen, D. (2009, March 13). The accidental guerilla: fighting small wars in the midst of a big one. Oxford: Oxford University Press. Kim, H. (2014). China’s position on Korean unification and ROK-PRC relations. 전략연구 (Feb, 2014). pp 235–257. Retrieved from http://www.dbpia.co.kr/view/ar_view.asp?arid¼2501730 Kim, S. (2015). The day after: ROK-U.S. cooperation for Korean unification. The Washington Quarterly, 38(3), 37–58. https://doi.org/10.1080/0163660X.2015.1099024. Khatwani, G., Singh, S. P., Trivedi, A., & Chauhan, A. (2015). Fuzzy-TISM: A fuzzy extension of TISM for group decision making. Global Journal of Flexible Systems Management, 16(1), 97–112.

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Klarrich, E. (2009). The mathematics of strategy. Classics of the Scientific Literature. Retrieved October, 2009 from www.pnas.org/site/misc/classics5.shtml. Kuhn, H. W., & Tucker, A. W. (1951). Nonlinear programming. In J. Newman (Ed.), Proceedings of the second Berkley symposium on mathematical statistics and probability. Berkeley, CA: University of California Press. Leyton-Brown, K., & Shoham, Y. (2008). Essentials of game theory: A concise, multidisciplinary introduction. San Rafael, CA: Morgan & Claypool. Miller, J. (2003). Game theory at work: how to use game theory to outthink and outmaneuver your competition. New York: McGraw-Hill. Myerson, R. B. (1991). Game theory: Analysis of conflict. Cambridge, MA: Harvard University Press. Nash, J. (1951). Non-cooperative games. Annals of Mathematics, 54, 289–295. Nash, J. (2009). Lecture at NPS. Feb 19, 2009. Nash, J. (1950). Equilibrium points in n-person games. Proceedings of the National Academy of Sciences of the United States of America, 36(1), 48–49. North Korea Declares 1953 Armistice Invalid. (2013). CNN Wire, Mar 11, 2013. Osborne, M. (2004). An introduction to game theory. Oxford: Oxford University Press. Papayoanou, P. (2010). Game theory for business, e-book. Gainesville, FL: Probabilistic Publishing Retrieved from http://www.decisions-books.com/Links.html. Rasmusen, E. (2006). Games and information: An introduction to game theory (4th ed.). New York: Wiley-Blackwell. Shoham, Y., & Leyton-Brown, K. (2009). Multiagent systems: Algorithmic, game- theoretic and logical foundations. New York: Cambridge University Press. Smith, M. (1982). Evolution and the theory of games. Cambridge: Cambridge University Press. Smith, J. M., & Price, G. (1973). The logic of animal conflict. Nature, 246(5427), 15–18. Straffin, P. (1980). The prisoner’s dilemma. UMAP Journal, 1, 101–113. Straffin, P. (1989). Game theory and nuclear deterrence. UMAP Journal, 10, 87–92. Straffin, P. (2003). Game theory and strategy. Washington, DC: The Mathematical Association of America. Chapter 19. Williams, J. D. (1986). The compleat strategyst. New York: Dover. Webb, J. N. (2007). Game theory: decisions, interaction and evolution. London: Springer.

Chapter 7

Modeling Change with Dynamical Systems Models

Objectives 1. Formulate and solve a Discrete Dynamical System (DDS) 2. Formulate and solve a DE 3. Formulate and solve a system of DDS and/or DE

7.1

Introduction

In this chapter, we use the paradigm, Future ¼ Present þ Change in order to model systems that exhibit change. We choose to start with dynamical systems for several reasons. First, they are fairly easy to model following the paradigm and they can be solved through iteration on an Excel spreadsheet. We need to initially define a few terms. Let n be a counting number, 0, 1, 2, . . . representing the time steps to be modeled. Let A(n) represent the system at time period n. Let A(n+1) represent the system in the future, time period n+1. The model is Aðn þ 1Þ ¼ AðnÞ þ Change: We have to model the change that occurs to the system at each time step. A good method is to sketch a change diagram for the system, A(n). We illustrate with a prescribed drug problem. Example 1: Drug Dosage Problem for Mild Brain Trauma Suppose that a doctor prescribes that their patient takes a pill containing 100 mg of a certain drug every hour. Assume that the drug is immediately ingested into the bloodstream once © Springer Nature Switzerland AG 2019 W. P. Fox, R. Burks, Applications of Operations Research and Management Science for Military Decision Making, International Series in Operations Research & Management Science 283, https://doi.org/10.1007/978-3-030-20569-0_7

331

332

7 Modeling Change with Dynamical Systems Models

100 mg dose

25% removed by kidneys Amount of drug in the system aer

Fig. 7.1 Change Diagram for prescribed drugs in system

taken. Also, assume that every hour the patient's body eliminates 25% of the drug that is in his/her bloodstream. Suppose that the patient had 0 mg of the drug in his/her bloodstream prior to taking the first pill. How much of the drug will be in his/her bloodstream after 72 h? Problem Statement: Determine the relationship between the amount of drug in the bloodstream and time. Assumptions: The system can be modeled by a discrete dynamical system. The patient is of normal size and health. There are no other drugs being taken that will affect the prescribed drug. There are no internal or external factors that will affect the drug absorption rate. The patient always takes the prescribed dosage at the correct time. The change diagram is shown in Fig. 7.1. Variables: Define a(n) to be the amount of drug in the bloodstream after period n, n¼ 0,1,2,. . .hours. Model Construction: Let’s define the following variables: a(n+1) ¼ amount of drug in the system in the future a(n) ¼ amount currently in system We define change as follows: change ¼ dose – loss in system change ¼ 100  .25 a(n) So, Future¼Present + Change is a(n+1) ¼ a(n) .25 a(n) + 100 or a(n+1) ¼ .75 a(n) + 100 Since the body loses 25% of the amount of drug in the bloodstream every hour, there would be 75% of the amount of drug in the bloodstream remaining every hour. After 1 h, the body has 75% of the initial amount, 0 mg, to the 100 mg that is added every hour. So the body has 100 mg of drug in the bloodstream after 1 h. After 2 h, the body has 75% of the amount of drug that was in the bloodstream after 1 h (100 mg), plus an additional 100 mg of drug added to the bloodstream. So there would be 175 mg of drug in the bloodstream after 2 h. After 3 h, the body has 75% of the amount of drug that was in the bloodstream after 2 h (175 mg), plus an additional 100 mg of drug added to the bloodstream. So there would be 231.25 mg of drug in the bloodstream after 3 h and after a long time there will be 400 mg of the drug in the system. The table of iterated values and the scatterplot, Fig. 7.2, are given.

7.2 Discrete Lanchester Combat Models Fig. 7.2 Plot of drug build up in our system over time

333

400

300

200

100

0

10

20

30

40

drug_table := 0., 100.000, 175.00000, 231.2500000, 273.4375000, 305.0781250, 328.8085938, 346.6064453, 359.9548340, 369.9661255, 377.4745941, 383.1059456, 387.3294592, 390.4970944, 392.8728208, 394.6546156, 395.9909617, 396.9932213, 397.7449160, 398.3086870, 398.7315152, 399.0486364, 399.2864773, 399.4648580, 399.5986435, 399.6989826, 399.7742370, 399.8306777, 399.8730083, 399.9047562, 399.9285672, 399.9464254, 399.9598190, 399.9698643, 399.9773982, 399.9830487, 399.9872865, 399.9904649, 399.9928487, 399.9946365, 399.9959774, 399.9969830, 399.9977373, 399.9983030, 399.9987272, 399.9990454, 399.9992841, 399.9994630, 399.9995973

Interpretation If the patient requires 400 mg in their system, then this dosage and schedule will work. We now provided decision-making examples with system of dynamical systems.

7.2 7.2.1

Discrete Lanchester Combat Models Introduction

Since Lanchester’s initial combat models in 1914, differential equations have been the methodology to present and solve these combat models. James G. Taylor alluded

334

7 Modeling Change with Dynamical Systems Models

to difficulties in the solving “real” equations and suggested numerical methods in his work (Taylor 1983). The use of computers to analytically solve or numerically solve combat models is the standard method. We suggest using difference equations, the discrete form of Lanchester equations, in our combat models. We show the discrete forms and their solutions, where applicable. We also show a numerical solution. We compare several of these solutions with the differential equation form to show how close the discrete form matches. We also suggest uses of the equations in decisionmaking for our leaders.

7.2.2

Discrete Forms of Lanchester Equations

History is filled with examples of the unparalleled heroism and barbarism of war. Specific battles like Bunker Hill, the Alamo, Gettysburg, Little Big Horn, Iwo Jima, and the Battle of the Bulge are a part of our culture and heritage. Campaigns like the Cuban Revolution, Vietnam, and now the conflicts in Afghanistan and Iraq are a part of our personal history. Although combat is continuous, the models of combat usually employ discrete time simulation. For years, Lanchester equations were the norm for computer simulations of combat. The diagram of simple combat as modeled by Lanchester is illustrated in Fig. 7.1. We investigate the use of a discrete version of these equations. We will use models of discrete dynamical systems via difference equations to model these conflicts and gain insight into the different methods of “directed fire” conflicts like Nelson’s Battle at Trafalgar and the Alamo, and Iwo Jima. We employ difference equations which allow for a complete numerical and graphical solution to be analyzed and do not require the mathematical rigor of differential equations. We further investigate the analytical form of the “direct fire” solutions to provide a solution template to be used in modeling efforts. Lanchester model stated that “under conditions of modern warfare” that combat between two homogeneous forces could be modeled from the state condition of this diagram. We will call this diagram (Fig. 7.3) the change diagram. We will use the paradigm, Future ¼ Present þ Change to build our mathematical models. This will be paramount as eventually models will be built that cannot be solved analytically but can be solved by numerical (iteration) methods. We begin by defining the following variables: x(n) ¼ the number of combatants in the X-force after period n. y(n) ¼ the number of combatants in the Y-force after period n. Future is then x(n+1) and y(n+1), respectively. So, we have

7.2 Discrete Lanchester Combat Models

335 k2

Fig. 7.3 Change Diagram of Combat modeled by Lanchester

X

Y

k1

xðn þ 1Þ ¼ xðnÞ þ Change yðn þ 1Þ ¼ yðnÞ þ Change: Figure 7.1 provides the information of the change diagram that reflects change. Our dynamical systems of equations are: x ð n þ 1Þ ¼ x ð nÞ  k 1 y ð nÞ

ð7:1Þ

y ð n þ 1Þ ¼ y ð nÞ  k 2 x ð nÞ We define our starting conditions as the size of the combatant forces at time period zero: x ð 0Þ ¼ x0

and yð0Þ ¼ y0 :

Dynamical systems can always be solved by iteration, which make them quite attractive for use in computer modeling and simulations of combat. However, we can gain some powerful insights with those equations that have analytical solutions. This particular dynamical system of equations for Lanchester’s direct fire model does have an analytical solution.

7.2.3

Discrete Form of Lanchester’s Direct Fire Equations

We return to the typical system of equations of Lanchester’s direct fire equations in difference equation form from Eq. (7.1): x ð n þ 1Þ ¼ x ð nÞ  k 1 y ð nÞ yðn þ 1Þ ¼ k2 xðnÞ þ yðnÞ Now, we write these in matrix form in Eq. (7.2):

336

7 Modeling Change with Dynamical Systems Models




X nþ1

7.2.3.1

1 ¼ k 2


 k1 x Xn, X0 ¼ 0 1 y0

ð7:2Þ

Eigenvalues and Eigenvectors

Definition of eigenvectors and eigenvalues: Let A be a n  n matrix. The real number λ is called an eigenvalue of A if there exists a nonzero vector x in Rn such that Ax ¼ λx:

ð7:3Þ

The nonzero vector x is called an eigenvector of A associated with the eigenvalue λ. Equation (7.3) is written as Ax  λx ¼ 0,

or

ðA  λI Þx ¼ 0,

where I is a 2 x 2 identity matrix. The solution to finding λ comes from taking the determinant of the (AλI) matrix,
setting it equal to zero, and solving for λ. 1 k1 We know that matrix A is . k2 1 We
set up the form for the use of eigenvalues: 1  λ k1 ¼ 0 that yields the characteristic equation det k 2 1  λ ð1  λÞ  ð1  λÞ  k1 k 2 ¼ λ2  2λ þ 1  k1 k2 ¼ 0 We solve for λ. Although not intuitively obvious to the casual observer, the two eigenvalues are pffiffiffiffiffiffiffiffiffi λ1 ¼ 1 þ pkffiffiffiffiffiffiffiffi 1 k 2ffi : λ2 ¼ 1  k 1 k 2

ð7:4Þ

Therefore, we have the eigenvalues from the initial form of the equation in (7.4). We note that the eigenvalues are a function of the kill rates, k1 and k2. If you know the kill rates, then you can easily obtain the two eigenvalues. We also note two other characteristics of the eigenvalues: (1) λ1 + λ2 ¼ 2 and (2) λ1  λ2. For most of these combat models, one eigenvalue will be >1 and the other eigenvalue will be 1. Most literature on dynamical systems suggest that the dominant eigenvalue (that eigenvalue is the largest eigenvalue and greater than 1 in this case) will control the

7.2 Discrete Lanchester Combat Models

337

system. However, these combat models are observed to be controlled by the smaller eigenvalue. The general form of the solution is as follows in Eq. (7.5): X ðk Þ ¼ c1 V 1 ðλ1 Þk þ c2 V 2 ðλ2 Þk ,

ð7:5Þ

where the vector V1 and V2 are the corresponding eigenvectors. These eigenvectors, interestingly enough, are in a ratio of the attrition coefficients, k1 and k2. The vector for the dominant eigenvalue always has both a positive and a negative component as its eigenvector while the vector for the other smaller of the two eigenvalues always has two positive entries in this same ratio. This is because the equation for finding the eigenvector comes from Eq. (7.6): pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi k1 k2 c1  k1 c2 ¼ 0 and  k1 k2 d1  k1 d2 ¼ 0 So, pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi c1 ¼ k1 , c2 ¼ k1 k 2 and d1 ¼ k1 , d2 ¼ k1 k2

ð7:6Þ

Having simplified formulas for obtaining eigenvalues and eigenvectors allows us to quickly obtain the general form of the analytical solution. We can then use the initial conditions to obtain the particular solution.

7.2.4

Red and Blue Force Illustrative Example

For example, consider a battle between a Red force, R(n), and a Blue force, B(n), as given below: Bðn þ 1Þ ¼ BðnÞ  :1  RðnÞ, Bð0Þ ¼ 100 Rðn þ 1Þ ¼ RðnÞ  :05  BðnÞ, Rð0Þ ¼ 50 The ratio of B(0)/R(0) ¼ 100/50 ¼ 2. We are given the attrition coefficients, k1 ¼ 0.1 and k2 ¼ .050. Using the formulas that we just presented, we can quickly obtain the analytical solution. pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 1 k2 ¼ :1  :05 ¼ :0707 The eigenvalues are 1.0707 and 0.9293. We could build the closed form solution pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi k1 k 2 k 1 k2 with the ratio of the vectors as 1 and k1 . We find k1 ¼0.7070. Our general solution would be:

338

7 Modeling Change with Dynamical Systems Models

(a) Blue Force over time

(b) Red Force over time

95 90

40

85

30

80

20

75 10 70 0 0.0

2.5

5.0

7.5

10.0

12.5

0.0

2.5

x

5.0

7.5

10.0

12.5

x

Fig. 7.4 Solution Graphs for Blue and Red

 X ð k Þ ¼ c1

   1 1 ð1:0707Þk þ c2 ð:9293Þk :707 :707

With our initial conditions of (100,50) at period 0, we have the particular solution:    1 1 k ð:9293Þk X ðkÞ ¼ 14:64 ð1:0707Þ þ 85:36 :707 :707 

We can graph these separately and observe the behavior: These two graphs (Fig. 7.4a, b) of the analytical solution show as the y-force (initially at size 50) approaches 0 as the x-force (initially at 100) is slightly below 70. Thus, we know the x-force or the Blue force wins. Weffi can also develop a relationship for this “win” and quickly see that when pffiffiffiffiffiffiffiffi k1 k2  x0 > k1  y0 then the p X-force ffiffiffiffiffiffiffiffiffi wins. For our example, we find k 1 k2  x0 and k1  y0 . pffiffiffiffiffiffiffiffiffi x0  k 1 k2 ¼ 100  :0707 ¼ 7:07 k2  y0 ¼ :1  50 ¼ 5 7:07 > 5:0 Since 7.07 is greater than 5.0, then the X-force wins. In general, the relationship can be < , ¼, or >. So we state that

7.2 Discrete Lanchester Combat Models

8 9 pffiffiffiffiffiffiffiffiffi < > = k 1 k 2 x0 ¼ k 1 y0 : ;
, then X wins; when the relationship is 1:35 pffiffiffiffiffiffiffiffiffi Since k 1 k2  FS0 > k1  B0 then the French-Spanish Fleet win. The analytical solution can be easily developed as:  X ðkÞ ¼ 3

   1 1 ð:95Þk ð1:05Þk þ 30 1 1

In order for the British to win, we first find the values that provide then with a draw. We find the British would require 33 ships to have a draw. Additionally, we find that the British would have to increase their kill effectiveness to 0.07469 to obtain a draw. Increases just beyond these values, give the British the theoretical edge. However, there were no more ships and the armaments were in place on the ships already. The only option would be a change in strategy. We can also test this new strategy that was used by Admiral Nelson at the Battle of Trafalgar. Admiral Nelson decided to move away from the course of linear battle of the day and use a “divide and conquer” strategy. Nelson decided to break his fleet into two groups of size 13 and size 14. He also divided the enemy fleet into three groups: a force of 17 ships (called B), a force of 3 ships (called A), and a force of 13 ships (called C). We can assume these as the head, middle, and tail of the enemy fleet. His plan was to take the 13 ships and attack the middle 3 ships. Then have his reserve 14 ships rejoin the attack and attack the larger force B, and then turn to attack the smaller force C. How did Nelson’s strategy prevail? Assuming all other variables remain constant other than the order of the attacks against the differing size forces, we find the Admiral Nelson and the British fleet now win the battle sinking all French-Spanish ships and 13 to 14 ships remaining. How did we obtain these results? The easiest method was iteration and used three battle formulas. We stop each battle when one of the values gets close to zero (before going negative) shown in Fig. 7.10.

7.2.8

Determining the Length of the Battle

To determine the length of the battle, we need to see something in the solution system of equations ΔA ¼ A(n + 1)  A(n) ¼  k1B(n) and ΔB ¼ B(n + 1)  B (n) ¼  k2A(n) that should be obvious. Recall the solution to our first example:

346

7 Modeling Change with Dynamical Systems Models Trafalgar with Nelon's New Strategy

British, French-Spanish Fleets

30 25 20 15 10 5 0

1

3

5

7

9

11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 time

Fig. 7.10 British prevail with new strategy

 X ðkÞ ¼ 14:64

   1 1 ð1:0707Þk þ 83:36 ð:9293Þk :707 :707

This simplifies to     83:36 14:64 ð:9293Þk X ðk Þ ¼  ð1:0707Þk þ 58:9355 10:35 The graph shows that x wins (as our other analysis) so the time parameter we are interested in is “when does y go to zero?” If you try to use the x equation, we end up with trying to take the ln of a negative number, which is not possible. We use y(k) ¼  10.35  1.0707k + 58.9355  .9293k and set y(k)¼0. ln ð58:9335 10:35 Þ The solution for k (which is our time parameter) is k ¼ ln 1:0707 ¼ 12:28 time ð :9293 Þ periods. In general, the time parameter is either of the following two Eqs. (7.8):  ln ln

c1 v11 c2 v12



  or λ1 λ2

 ln ln

c1 v21 c2 v22



  λ1 λ2

ð7:8Þ

depending on which form yields the ln(positive number) in the numerator. If our Red–Blue combat data was in kills/hour, then the battle lasted for 12.28 h. Often we are interested in the approximate time or length of the battle. These formulas in Eq. (7.8) allow for a quick computation.

7.2 Discrete Lanchester Combat Models

7.2.8.1

347

Battle of Iwo Jima Example

At Iwo Jima in WWII, the Japanese had 21,500 soldiers and the United States had 73,000 soldiers. We assume that all forces were initially in place. The combatants engaged in conventional direct warfare, but the Japanese were fighting from reinforced entrenchments. The coefficient of effectiveness for the Japanese was assumed to be 0.0544 while that of the US side was assumed to be 0.0106 (based on data after the battle). If these values are approximately correct, which side should win? How many should remain on the winning side when the other side has only 1500 remaining? We move directly to both the winning conditions and an analytical solution to answer these questions. 8 9 pffiffiffiffiffiffiffiffiffi < > = k 1 k 2 x0 ¼ k 1 y0 : ; < 0:02401 ð73; 000Þ > ð0:0106Þ  ð21; 500Þ 1752:97 > 227:90, so we know that the United States wins decisively. The analytical solution is: 

   1 1 k X ðkÞ ¼ 12145:67 ð1:024Þ þ 60854:33 ð:976Þk :4414 :4414 or  X ðk Þ ¼

   60854,33 12145:67 ð:976Þk ð1:024Þk þ 26861:1 5361:1

We solve the equation for time it takes the Japanese to reach 1500 soldiers. We find that it takes 30.922 time periods for the Japanese to reach 1500 soldiers. Thus, the model shows that United States had approximately 53,999 soldiers remaining. The battle actually ended with 1500 Japanese survivors and 44,314 US survivors and took approximately 33–34 days. Our model’s approximations are not too bad. We are off by about 6% in the time and by 21.8% in the number of surviving US soldiers. The error in surviving soldiers should cause us to revisit the model’s assumptions for an explanation. The United States actually used a phased landing over 15 days of actual combat to reach their final force of 73,000 soldiers. We could have treated this like the Trafalgar battle with at least 15 different battles to be more accurate.

348

7.2.9

7 Modeling Change with Dynamical Systems Models

Insurgency and Counter-Insurgency Operations

Today’s warfare is different. The dynamics of today’s battlefield is quite different. Consider the later stages of the war in Iraq that has become a multi-ring conflict (Kilcullen’s view as seen in Fig. 7.11). Insurgency and counter-insurgency operations can be modeled in a simplified sense using the following discrete Lanchester model using a modified version of Brackney’s Mixed law (also called the Parabolic Law was developed in 1959). This can be used to represent Guerilla warfare and can now be used to represent insurgency and counter-insurgency operations. We define Y(n) to be the insurgent strength after period n. and we define X(n) to be the government troop strength after period n. Then, X ðn þ 1Þ ¼ X ðnÞ  k1  X ðnÞ  Y ðnÞ Y ðn þ 1Þ ¼ Y ðnÞ  k 2  X ðnÞ, where k1 and k2 are kill rates. Further if we model the total conflict with both growth and attrition, we could use the following models: X ðn þ 1Þ ¼ X ðnÞ þ a  ðK 1  X ðnÞÞ  X ðnÞ  k 1  X ðnÞ  Y ðnÞ Y ðn þ 1Þ ¼ Y ðnÞ þ b  ðK 2  Y ðnÞÞ  Y ðnÞ  k 2  X ðnÞ, where k1 and k2 are kill rates. a, b are positive constants. K1 and K2 are carrying capacities. Fig. 7.11 Kilcullen’s 2004 View of the Strategic problem in Iraq

Iraq - Nature of the Strategic Problem

Terrorism

Insurgency

Sectarian Conflict Iraq - Fragile State Underlying Problem: Nation-Building Divided Region - Sunni Arab versus Shi’a Persian @DJ Kilcullen 2004

7.2 Discrete Lanchester Combat Models

349

This is a combination of the growth model and the combat model and represents when conflict is ongoing and growth is still part of the insurgency operation. These type of equations can only be solved and analyzed using numerical iteration. Having laptops with Excel enable soldiers/decision-makers to characterize the situations and get quick “results.”

7.2.10 Comparison to Standard Lanchester Equations Via Differential Equations Let’s revisit the Red and Blue force illustrative example now as a differential equation. dxðt Þ ¼ :1  yðt Þ dt dyðt Þ ¼ :05  xðt Þ dt xð0Þ ¼ 100, yð0Þ ¼ 50 This system of differential equations yields the solution to three decimal places: xðt Þ ¼ 14:644  e0:0707t þ 85:355  e:707t yðt Þ ¼ 10:355  e0:0707t þ 60:355  e:707t We provide a plot of the solution via differential equations and the solution via difference equations in Fig. 7.12. Note how close they align.

7.2.11 Conclusions for Lanchester Equations The use of difference equations in combat modeling has practical value. Not only do analytical solutions allow analysts to provide decision-makers with quantitative information to quickly analyze potential results but every difference equation has a numerical solution that can be achieved easily. For the decision-maker in the field, a differential equation is an abstract concept and the tools for analysis are not available. However, an EXCEL spreadsheet is a powerful tool for decision-makers that are available in the field. The systems of difference equations, based upon “Future¼Present + Change” is an intuitive, non-evasive approach for which every combat model has a numerical solution and some combat models such as the direct fire models have analytical solutions that directly lend themselves to analysis and results. We are currently teaching this method to our military students in our modeling courses in Defense Analysis department.

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7 Modeling Change with Dynamical Systems Models

Fig. 7.12 Solution plots of Blue versus Red forces via DE and DDS methods

100

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50

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2.5

5.0

7.5

10.0

t

7.3

Insurgency Models with Discrete Dynamical Systems

Scenario 1 Insurgent forces have a strong foothold in the city of Urbania. Intelligence estimates they currently have a force of about 1000 fighters. Intelligence also estimates that around 120 new insurgents arrive from the neighboring country of Moronka each week. In conflicts with insurgent forces, the local police are able to capture or kill approximately 10% of the insurgent force each week on average. Problem Statement Determine the relationship between the size of the insurgent force and time. Assumptions The system can be modeled by a discrete dynamical system. The person is of normal size and health. There are no other factors that will affect insurgent force levels. The current force estimate occurs at time 0. The following questions are asked of the students as they explore their model and its solution in the lab. Questions that might be answered: 1. Describe the behavior of the current system under the conditions stated: (a) Is there a stable equilibrium to the system under the current conditions? If so, is this an acceptable level? (b) How effective would an operation designed to slow (or stop) the influx of new insurgents be if the dynamics do not change? 2. What attrition rate does the police force need to achieve to drive the insurgent population to an equilibrium level below 500 in 52 weeks or less? 3. If the police force can, with advanced weapons, achieve a 30–40% attrition rate, do they also have to engage in operations to stop the inflow of new insurgents?

7.3 Insurgency Models with Discrete Dynamical Systems

351

4. What effects do changes in the external factor, change factor, and initial condition have on the system behavior curve? 5. What conditions are necessary to cause either case (1) or (2) to occur within the 52-week horizon? We expect the students to obtain the following model: AðnÞ ¼ number of insurgents in the system after time period, n, where n ¼ 0,1,2, 3,. . . weeks. Aðn þ 1Þ ¼ AðnÞ  0:01 AðnÞ þ 120,

Að0Þ ¼ 1000

Note the sliders are built into the template to allow students the ability to easily change the parameters and watch the effects on the solution dynamics (Fig. 7.13). Scenario 2 Insurgent forces have a strong foothold in the city of Urbania, a major metropolis in the center of the country of Ibestan. Intelligence estimates they currently have a force of about 1000 fighters. The local police force has approximately 1300 officers, many of which have had no formal training in law enforcement methods or modern tactics for addressing insurgent activity. Based on data collected over the past year, approximately 8% of insurgents switch sides and join the police each week whereas about 11% of police switch sides and join the insurgents. Intelligence also estimates that around 120 new insurgents arrive from the neighboring country of Moronka each week Recruiting efforts in Ibestan yield about 85 new police recruits each week as well. In armed conflict with insurgent forces, the local police are able to capture or kill approximately 10% of the insurgent force each week on average while losing about 3% of their force. Problem Statement Determine the equilibrium state (if it exists) for this DDS. Assumptions The system can be modeled by a discrete dynamical system. Questions: 1. Build the DDS. 2. Determine who wins in the long run. 3. Find reasonable values that will alter the outcome. Explain how these values could be achieved? We define the variables P(n) ¼ the number of police in the system after time period n. I(n) ¼ the number of insurgents in the system after time period n. n ¼ 0,1,2, 3,. . . weeks Model: P(n+1) ¼ P(n)  0.03 P(n)  0.11 P(n) + 0.08 I(n) ¼ 85, P(0) ¼ 1300 I(n+1) ¼ I(n) + 0.11 P(n)  0.08 I(n)  0.01 I(n) + 120, I(0) ¼ 1000

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7 Modeling Change with Dynamical Systems Models

Fig. 7.13 Excel Screenshot of Insurgent Population Change

Again, an EXCEL template is provided to the student after they have developed the model. The template is available to provide a vehicle for analysis.

7.3.1

Using EXCEL

In building dynamical systems models in Excel, it is best to represent effects directly in the terms used in the model rather than simplifying the system’s representation first and then building the spreadsheet model. This is because (Fig. 7.14): (a) Each term and its coefficients have intuitive meaning for the problem based on a specific dynamic effect, which facilitates “what-if?” analysis. (b) Exploration is facilitated more easily using Scroll Bars, which need to be linked directly to single model parameters. Subsequent mathematical analysis, such as computing exact limiting behavior (e.g., equilibrium) follows directly by combining like terms and simplifying the system to identify the general form of the system and its solution. Two charts are of prime interest in analysis: individual population changes via a scatterplot, and force-on-force “state space” chart. The point of this article would be rapid exploration of alternative strategies. For this one we have:

7.3 Insurgency Models with Discrete Dynamical Systems

353

Fig. 7.14 Excel Screenshot of Insurgent Dynamical Model

1. Complete force-on-force where both sides “see” each other and the effects noted transpire. In this setting, the two strategies are: (a) Deploy all 1500 police at the start to fight the insurgency. (b) Deploy the best 500 police at the start to fight the insurgency, and establish a formal training program that graduates 500 police every 15 weeks, augmenting the field force. The major learning points in addition to the above would be: (a) Short-term effects are represented by “periodic bump-ups” in one side or the other. (b) Long-term effects are represented in the dynamics of the model, here represented by the coefficients in the system matrix. (c) Which is best to employ, separately or in combination, is difficult to determine without modeling the system behavior in a dynamic fashion?

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7 Modeling Change with Dynamical Systems Models

We do this in two examples in EXCEL: the simple insurgent model DDS in which our focus is only on tracking the insurgent population by modeling imposed effects on them only; and then introduce the two population model. In both models, we can illustrate the simplification of terms to a general form and then computing the equilibrium and examining the long-term behavior.

7.3.2

Results and Conclusion

In Scenario 2, the police lose to the insurgents. A modification that assumes we train and graduate 500 additional police every 15 weeks and adds them to the force improves our status but not our outcome. Students discussed that the surge in police might affect both the percentages of police and insurgents that switch sides. More insurgents will switch and less police as they all desire to be on the “winning side.” Taking this into account in the model reveal that the outcome will change and the police can defeat the insurgents. This tool does not tell us how to make this happen but suggest if we can make it happen that we can alter the final outcome for the police to win, our ultimate goal in decision-making assuming we are the police. The perfect partnership of technology and modeling allows us to “test” ideas in a non-threatening atmosphere to help us make better decisions.

7.4

IRAQ Model for the Three Circles of War

“The attack took place on American soil, but it was an attack on the heart and soul of the civilized world. And the world has come together to fight a new and different war, the first, and we hope the only one, of the 21st century. A war against all those who seek to export terror, and a war against those governments that support or shelter them.” —President George W. Bush, October 11, 2001

The strategic goal of the United States in Iraq is the creation of a unified, democratic, and federal Iraq that can govern, defend, and sustain it and is an ally in the global war on terrorism. The purpose of the modeling effort presented in this chapter is to provide a tool for leaders and decision-makers to measure the effects of proposed actions on all facets of the total mission in Iraq as well as to measure unforeseen events and their potential impact on Iraq’s stability. Mathematical models can not only provide insights into real-world behavior but also serve as a tool decision-makers can use to determine possible effects of “what-if” scenarios. Our Indicators of Force Multipliers (INFORM) model is a layered model, initially going to a depth of one or two layers. If our efforts provide fruitful results, the model can be expanded to more layers as needed. The INFORM model uses the circles-of-war approach described in Chap. 1. All three types of war—interstate war, insurgency, and civil war—have occurred during the conflict in Iraq, at times simultaneously. The conflict began as an interstate war

7.4 IRAQ Model for the Three Circles of War

355

Fig. 7.15 View of the strategic problem in Iraq

Conventional war

Civil War Insurgency

when the United States and allied forces invaded Iraq in 2003. After deposing the Saddam Hussein regime, US-led forces soon faced an insurgency led by Sunni factions—and including some al Qaeda elements—fighting the presence of coalition troops and the Shia-dominated Iraqi government and security forces. In 2006, the conflict began increasingly to take on the characteristics of a civil war, causing the government to split along Sunni/Shia lines, replete with reprisal attacks between Sunni and Shia militias. While the conflict in Iraq might currently be characterized as a civil war, it still contains elements of an insurgency and an interstate war, with threats of additional interstate involvement from Iran and Turkey. The three types of war overlap—incidents may involve elements of more than one dimension. For example, some insurgency is “pure” terrorist activity executed by al Qaeda in Iraq (AQI), while other acts are insurgent-motivated, and yet others exhibit a prominent sectarian dimension. Most incidents in fact include elements of two dynamics, or all three. The three types of war may be thought of in terms of a Venn diagram of overlapping circles (see Fig. 7.15), each constantly changing in size, in which any incident can be plotted somewhere within the interaction of the three dynamics. An underlying assumption of the model as applied to this conflict is the need for nation-building in Iraq. Pervasive security problems prevent us from getting at many of the underlying problems—crime, weak infrastructure, economic and social alienation, weak governance, and so on—that need to be addressed in order to deal with the nation-building requirement. The inability to get at this underlying problem perpetuates and exacerbates the security problems in Iraq as well as the three types of war. The three types of war can be mutually reinforcing—each can make the others worse. Terrorism provokes communal conflict, which in turn makes the insurgency more intractable, which in turn gives rise to terrorism, and so on.

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The solution sets to each problem also tend to be countervailing—the solution to one tends to make the others worse. For example, defeating the insurgency requires building indigenous security forces. But in a society that has weak national institutions and is divided along sectarian lines, a buildup of indigenous security forces can make the communal conflict worse. Resolving the communal conflict requires working with all community groups, including those groups who currently support terrorists. This may create animosity and increase the support of terrorist factions. Countering terrorist cells implies disrupting terrorist support, but that can make the insurgency worse—and so on, in an endless loop. We created the Indicators of Force Multipliers (INFORM) model of the war in Iraq as a dynamic systems model of both military and nonmilitary effects as applied to the war on terrorism in Iraq and Iraq’s own infrastructure resiliency. The basic form of this model stems from a Department of Homeland Security model. This model and the modified algorithm used in providing examples are explained in this chapter and its Appendix. The current effort is only the skeleton of a larger proposed stochastic model.

7.4.1

Dynamic Modeling

We used a discrete dynamic modeling structure to capture the effects over time as well as the results of interactions. We used the paradigm Future ¼ Present þ Change in order to build our dynamic model structure. The basic model without shocks is provided in the following equation:



pi ð t þ 1Þ Qxx ¼ s k ð t þ 1Þ Qsx

Qxs Qss

T


pi ð t Þ sk ðt Þ



Notation i ¼ 1, 2, . . ., I j ¼ 1, 2, . . ., J k ¼ 1, 2, . . .,K pi(t) sk(t) Q ¼ [Qik(t)]

major elements critical services related to elements set of i, j physical element-service pairs physical condition state of infrastructure element i at time t service process condition state for pair k at time t state interactivity level matrix at time t

Two additional vectors for degradation over time and maintenance over time can serve as multipliers in this model:

7.4 IRAQ Model for the Three Circles of War

di(t) dk(t) mi(t) mk (t)

357

degradation of the physical system by natural causes degradation of the service system by natural causes (between physical systems, services, and physical-service pairs) maintenance for physical layer for infrastructure i at time t maintenance for critical infrastructure service j for infrastructure i at time t

We note the following: • Q is assumed to be partially time invariant, asymmetric, and state independent in the basic model. • The block structure of Q represents the pairwise influence effects between state components as a result of linked interdependencies. • The scale chosen for Q components represents a complete effects cycle. • The state vector at time t ¼ 0 represents a conservative estimate of current state condition levels. • The degradation and maintenance vectors represent planned periodic resource investments for state conditions according to (possibly unique) periodic time sequence t. Examples of some Homeland Security layers that were used include physical layers such as energy, transportation, public health, finance, etc. and their respective service layers could be to provide residential electricity, provide rail and bus service, control disease, facilitate business transactions, etc. We modified these layers to fit our Iraq dynamic model. We begin with six variables, move to eight variables in our examples, and suggest a 30-variable model to be constructed. Our variables and their definitions are detailed in Section 2 of the Appendix, along with some brief discussion of data. We iterate the dynamic model and its interaction over time. We can affect the model with both degradation (natural degradation) and maintenance (upgrades) as necessary. We can “hit” the system with a shock, analyze which elements go out of control, and measure the net percentage change in these elements. In our group discussions, we discovered a need for our model to react to various stimuli, called “shocks” to the system. These shocks can be passive or aggressive stimuli. We built several scenarios based on shocks for preliminary analysis. Based on the results of our model and our analysis of these results, we believe this model is worthy of further development.

7.4.2

Modeling with Shocks

Let c(j) be a shock due to an outside force (insurgency attack, bombing, reduction of US troops, etc.). We can measure its effect throughout the elements (infrastructures) to determine courses of action to improve the situation. The model allows for experts to create a vector of scalars for the shock effect (or a distribution of the effect if not

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accurately known). The shock is applied after the period indicated for the shock. The new vector is then used for subsequent calculations within that scenario.



Qxx pi ð t þ 1Þ ¼ sk ðt þ 1Þ Qsx

Qxs Qss

T



 pi ð t Þ ci ðτÞ  , sk ðt Þ ck ð τ Þ

where


~ xx Q ~ sx Q

~ xs Q ~ ss Q

T


¼

Qxx Qsx

Qxs Qss

T




Cxx ðτÞ þ C sx ðτÞ

Cxs ðτÞ Css ðτÞ

T !

This modeling format can allow for the user (decision-maker) to consider responsive actions and policies.

7.4.3

Model with Responsive Actions and/or Policies




 


 
~ sx ~ xx Q pi ð t þ 1Þ pi ð t Þ mi ðtÞ Rxx ðδÞ Rsx ðδÞ Q ¼  ~ xs Q ~ ss þ Rxs ðδÞ Rss ðδÞ s k ð t þ 1Þ sk ðt Þ mk ðtÞ Q

 c ðτ Þ r ðδ Þ  i þ i ck ð τ Þ r k ðδ Þ
 


 
~ sx ~ xx Q pi ð t þ 1Þ pi ð t Þ mi ðtÞ Pxx ðδÞ Psx ðδÞ Q ¼  ~ xs Q ~ ss þ Pxs ðδÞ Pss ðδÞ s k ð t þ 1Þ sk ðt Þ mk ðtÞ Q

 c ðτ Þ x ðδ Þ  i þ i ck ð τ Þ xk ð δ Þ Dynamic pairwise effects imposed on the composite system by Q are proportional to the “health” of the “transmitting” system component. Specifically: pi ð t þ 1Þ ¼

pi ðt Þ þ

X

!

!

q ji p j ðt Þ mi ðt Þ envðt Þ

j6¼i

Since many of these variables stem from distributions, it is envisioned that a future simulation can be overlaid on the model and run thousands of times to capture the output statistics, which can then be analyzed. We illustrate these results below in scenario 5.

7.4 IRAQ Model for the Three Circles of War

7.4.3.1

359

Model’s Algorithm

The algorithm for the model in its current state is provided in step-by-step detail in Section 3 of the Appendix.

7.4.3.2

Model’s Variables and Q Matrix

Although our goal for this model is to include 15 physical variables and 15 service variables that capture most of the key aspects of Iraq’s infrastructure and the forces for security and combat (see the master list in Section 4 of the Appendix), in this report we limit our model to six and eight variables selected and modified from our master list. The values of the variables are derived from reports to Congress from 2003 to 2008, which we downloaded from the Internet and analyzed to extract the data. Many of the data elements were probability distributions, and we calculated the statistics to include means and standard deviations. In this version of the model, we used the means and then normalized them to values between 0 and 1. The Q matrix is a key element of the model. The values of the interactive effects of layers of the data among themselves must be provided by experts. These interactions are essential to the model, as they show the net effects of the shocks they propagate throughout the other layers of the model.

7.4.3.3

Illustrative Examples

We built and ran five scenarios in which we allowed a shock or multiple shocks to affect the dynamic system. The descriptions below provide only a summary. Scenario 1 In this example, we consider only three physical and three service layers: “electrical power” and “provide hours of power”; “Iraq police” and “provide local security”; and “terrorist activity” and “weekly attacks.” We watch the system dynamics operating until a shock hits the system directly at time period 6—in this case, car bombs are directed at police targets. This shock effect lowers the percentage levels of the police physical condition and their ability to provide local security directly, as we would expect. This is reflected in the graphs (Appendix, Section 5) showing a reduction over time of the police physical condition to less than 20% and provision of local security to less than 5%. The model shows the other effects as well: it predicts that the electrical power physical condition and the provision of power are diminished by more than 10% points and that the number of terrorist attacks increases by more than 20% points. Analysis of these factors suggests that we need to monitor the electrical power system in Iraq and concentrate additional efforts to deter terrorist activities. If we consider options, we can measure their impact on the system to try to restore the system as quickly as possible and even improve it.

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Scenario 2 In this example, we consider only three physical and three service layers: “economy” and “unemployment rate”; “Iraq police” and “provide local security”; and “terrorist activity” and “weekly attacks.” We watch the system dynamics operating until a shock hits the system directly at time period 6—here again, car bombs are directed at police targets. This shock effect lowers the percentage levels of the police and their ability to provide local security directly, as we would expect. This is reflected in the graphs (Appendix, Section 5) showing a reduction of the police physical condition to less than 20% and provision of local security to less than 5%. The model shows the other effects as well: it predicts that the economy is decreased by about 10% and provision of jobs by more than 15% and that the number of terrorist attacks increases more than 18%. Analysis of these factors suggests that we need to monitor the economic system in Iraq and concentrate additional efforts to deter terrorist activities. If we consider options, we can measure their impact on the system to try to restore the system as quickly as possible and even improve it. Scenario 3 In this example, we consider only four physical and four service layers: “US military” and “casualty rate”; “Iraq police” and “provide local security”; “insurgents” and “destabilize the government”; and “civil war” and “discontent.” We watch the system dynamics operating until a shock hits the system directly at time period 6—here again, car bombs are directed at police targets. This shock effect lowers the percentage levels of the police physical condition and their ability to provide local security directly, as we would expect. This is reflected in the graphs and table of values (Appendix, Section 5) showing a reduction of police effectiveness to less than 45% and provision of local security to less than 52.5%. The model shows the other effects as well: it predicts that US military effectiveness decreases by about 17% and casualties increase by 150%. The insurgency, which becomes the object of all coalition efforts, is temporarily decreased 72% and destabilization by only 10%. The civil war increases by 4% while discontent increases by 30%. Analysis of these factors suggests that we need to increase our efforts against the insurgents that will stop insurgent activities and maintain our level of effectiveness in Iraq. We should also concentrate additional efforts in other areas to curb the increase in the population’s discontent. If we consider options, we can measure their impact on the system to try to restore the system as quickly as possible and even improve it. Scenario 4 In this example, we consider only four physical and four service layers: “US military” and “casualty rate”; “Iraq military” and “provide security”; “insurgents” and “destabilize the government”; and “civil war” and “discontent.” We watch the system dynamics operating until a shock hits the system directly at time period 6—in this case, the US military withdraws substantial troops and the Iraqi military gains minor improvement in overall effectiveness. This shock effect lowers the percentage levels of the effectiveness of the US military and increases the casualty rate. The Iraqi military remains 93% effective, and their security is at 85%, both showing

7.4 IRAQ Model for the Three Circles of War

361

slight decreases. Insurgents and their ability to create instability are both decreased by these shocks, and the civil war and discontent levels both increase. We should recommend not having any rapid withdrawal of US troops until we can do so without causing any increase in casualties. Scenario 5: Stochastic Simulation Inputs and Outputs: In this model, we allowed the state conditions of the variables to be uniform distributions with the means as the values in scenario 4. We also allowed the shock multipliers to be uniform distributions with the means as the values in scenario 4. We captured the end state as distributions using 1000 runs of the simulated model from scenario 4. The simulated outputs are provided in Section 5 of the Appendix. In each case, the expected value or mean is established, but a probability distribution is shown for all possible other output values. Analysis: The model results show the possible range of values and the probabilities of obtaining those values. Sensitivity analysis shows some typical and atypical results. The variability in the lowering of US military effectiveness is a function of the force and its draw-down, which is typical. The same is true for the variable for the Iraqi military. The variability in the insurgency is atypical and is a function of the insurgency, but it is also 27.4% due to the Iraqi military and 16.6% due to civil war activities. The variability in the US casualty rate is affected by the civil war and the effectiveness of the Iraqi military.

7.4.3.4

Conclusions and Recommendations

The INFORM Iraq modeling effort should continue to be developed as a planning tool to help decision-makers make informed decisions about actions taken relative to Iraq. The model, in its infant stages, has been shown to be capable of capturing and explaining the dynamics of events occurring in Iraq and their effects on other components in Iraq. The larger model will incorporate more infrastructure variables since rebuilding is critical to the achievement of peace and stability in Iraq.

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Fig. 7.16 Graph of Mild Brain Trauma 300

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25

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7.5

Differential Equations (Optional)

All the modeling that we have presented can be solved via differential equations. In a DDS as presented earlier, Aðn þ 1Þ ¼ AðnÞ þ k  AðnÞ,

Að0Þ ¼ A0

Aðn þ 1Þ  AðnÞ ¼ k  AðnÞ,

Að0Þ ¼ A0

we can rearrange as

We divide by Δn and take the limit as n goes to infinity. This gives us our ordinary differential equation (ODE): dA ¼ kAðnÞ, dn

Að0Þ ¼ A0

We will take our mild brain trauma drug problem as an ODE, dA=dt ¼ -0:25 Aðt Þ þ 100,

Að0Þ ¼ 0:

The solution is found either in closed form or numerically in this case (Fig. 7.16),

7.5 Differential Equations (Optional) > > > >

> We plot the solution in MAPLE, >

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364

7.6 7.6.1

7 Modeling Change with Dynamical Systems Models

Systems of Differential Equations Applied Systems of Differential Equations

In this section, we introduce many mathematical models from a variety of disciplines. Our emphasis in this section is building the mathematical model, or expression, that will be solved later in the chapter. Recall previously that we discussed the modeling process. In this section, we will confine ourselves to the first three steps of the modeling process: (1) Identifying the problem, (2) Assumptions and variables, and (3) Building the model. Example 1: Economics: Basic Supply and Demand Models Suppose we are interested in the variation of the price of a specific product. It is observed that a high price for the product attracts more suppliers. However, if we flood the market with the product the price is driven down. Over time there is an interaction between price and supply. Recall the “tickle me Elmo” from Christmas a few years ago. Problem Identification: Build a model for price and supply for a specific product. Assumptions and variables: Assume the price is proportional to the quantity supplied. Also assume the change in the quantity supplied is proportional to the price. We define the following variables. PðtÞ ¼ the price of the product at time, t QðtÞ ¼ the quantity supplied at time, t We define two proportionality constants as a and b. The constant a is negative and represents a decrease in price as quantity increases. With our limited assumptions, the model could be dP ¼ aQ dt dQ ¼ bP dt Example 2: Competition Between Species Imagine a small fish pond supporting both trout and bass. Let T(t) denote the population of trout at time t and B(t) denote the population of bass at time t. We want to know if both can coexist in the pond. Although population growth depends on many factors we will limit ourselves to basic isolated growth and the interaction with the other competing species for the scarce life-support resources. We assume that the species grow in isolation. The level of the population of the trout or the bass, B(t) and T(t), depend on many variables such as their initial numbers, the amount of competition, the existence of predators, their individual species birth and death rates, and so forth. In isolation, we assume the following

7.6 Systems of Differential Equations

365

proportionality models (following the same arguments as the basic populations models that we have discussed before) to be true where the environment can support an unlimited number of trout and/or bass. Later, we might refine this model to incorporate the limited growth assumptions of the logistics model: dB ¼ mB dt dT ¼ aT dt Next, we modify the proceeding differential equations to take into account the competition of the trout and bass for living space, oxygen, and food supply. The effect is that the interaction decreases the growth of the species. The interaction terms for competition led to decay rate that we call n for bass and b for trout. This leads to following simplified model: dB ¼ mB  nBT dt dT ¼ aT  bBT dt If we have the initial stocking level, B0 and T0, we determine how the species coexist over time. If the model is not reasonable, we might try logistic growth instead of isolated growth. Logistic growth in isolation was discussed in first-order ODE models as a refinement. Example 3: Predator–Prey Relationships We now consider a model of population growth for two species in which one animal is hunted by another animal. An example of this might be wolves and rabbits where the rabbits are the primary food source for the wolves. Let R(t) ¼ the population of the rabbits at time t and W(t) ¼ the population of the wolves at time t. We assume that rabbits grow in isolation but are killed by the interaction with the wolves. We further assume that the constants are proportionality constants. dR ¼aRbRW dt We assume that the wolves will die out without food and grow through their interaction with the rabbits. We further assume that these constants are also proportionality constants. dW ¼ m  W þ n  R  W dt

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7 Modeling Change with Dynamical Systems Models

Example 4: Insurgencies Models As we look around the world, we see many conflicts involving insurgencies. We have the political faction (usually the status quo or the new regime) battling the insurgents or the rebels that are resisting the change or the political status. This also can be seen from history if we look at our own Revolutionary War. In Insurgency operations (IO), we follow the following assumption concerning IO that they are messy, grass root fights that are both confused and brutally contested. We find the definition of the enemy is loosely defined. We find that positive control of the forces is usually weak. There are few rules of engagement (they are often permissive). There are political divisions that are deep seated that leave little room for compromise. Further as we consider building a mobilization model, we assume that growth is subject to the same laws as any other natural or man-made population (basic growth or logistical growth as discussed before). Additionally, there are three considerations: pool of potential recruits, number of recruiters, and the transformation rate. We assume logistical growth and our systems could look like XðtÞ ¼ insurgency YðtÞ ¼ regime dX ¼ a  ðk 1  X Þ  X dt , dY ¼ b  ðk 2  Y Þ  Y dt where a measures insurgency growth rate. b measures regime growth rate. k1 and k2 are the respective carrying capacities.

7.6.2

Solving Homogeneous and Non-Homogeneous Systems

We can solve systems of differential equations of the form: dx ¼ ax þ by þ gðt Þ dt , dy ¼ mx þ ny þ hðt Þ dt where (1) a, b, m, n are constants and (2) the functions g(t) and h(t) can either be 0 or functions of t with real coefficients.

7.6 Systems of Differential Equations

367

When g(t) and h(t) are both 0 then the system of differential equations is called a homogeneous system, otherwise it is non-homogeneous. We will begin with homogeneous systems. The method we will use involves eigenvalues and eigenvectors. Example 1: Consider the following homogeneous system: with initial conditions 0 x ¼ 2x  y þ 0 0

y ¼ 3x  2y þ 0 xð0Þ ¼ 1,

y ð 0Þ ¼ 2

Basically, if we rewrite the system of differential equation in matrix form: 0

X ¼ Ax, where


 2 1
3 2
 0 dx dy x X ¼ , x¼ y dt dt A¼

then we can solve X' ¼ Ax. This form is highly suggestive of the first-order separable equation that we saw in the previous chapter. We can assume the solution to have a similar form: X¼Keλt, where λ is a constant and X and K are vectors. The values of λ are called eigenvalues and the components of K are the corresponding eigenvectors. We note that a full discussion of the theory and applications of eigenvalues and eigenvectors can be found in linear algebra textbooks as well as many differential equations textbooks. Since we have a 2 x 2 system, there are two linearly independent solutions that we call X1 and X2. The complementary solution or general solution X ¼ c1X1+c2X2, where c1 and c2 are arbitrary constants. We use the initial conditions to find specific values for c1 and c2. The following steps can be used when we have real distinct eigenvalues Step 1. Set up the system as a matrix, X’¼AX, X(0)¼X0 Step 2. Find the eigenvalues, λ1 and λ2. Step 3. Find the corresponding eigenvectors, K1 and K2. Step 4. Set up the complementary solution Xc ¼ c1 X1 + c2 X2, where

368

7 Modeling Change with Dynamical Systems Models

X 1 ¼ K 1 eλ 1 t X 2 ¼ K 2 eλ 2 t

Step 5. Solve for c1 and c2 and rewrite the solution for Xc. Step 2. Finding the eigenvalues. We set up the characteristic polynomial by finding the determinant of A – λI ¼ 0.

det




2λ 1 ¼0 3 2  λ

(2  λ)(2λ) + 3 ¼ 0 λ2 – 1 ¼ 0 λ ¼ 1, –1 Step 3. Finding the eigenvectors. We substitute each solution for l back in Ak ¼ 0 and solve the system of equations for k, the eigenvectors. Let λ ¼ 1. Let k1 and k2 be the components of eigenvector K1. k1 – k2 ¼ 0 3k1 – 3k2 ¼ 0 We arbitrarily make k1 ¼ 1 thus k2 ¼ 1. K1 ¼ [1,1] Let λ ¼ –1. Let k1 and k2 be the components of eigenvector K1. 3k1 – k2 ¼ 0 3k1 – k2 ¼ 0 We arbitrarily make k2 ¼ 3 thus k1 ¼ 1. K2 ¼ [1,3] Step 4. We set up the complementary solution. X c ¼ c1 X 1 þ c2 X 2 , where X 1 ¼ K 1 eλ 1 t X 2 ¼ K 2 eλ 2 t

7.6 Systems of Differential Equations

369

3.5 3.5

3

2.5

3

2

2.5

1.5 2 1 0

0.2 0.4 0.6 0.8

1 t

1.2 1.4 1.6 1.8

2

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 t

Fig. 7.17 Plot of x(t) and y(t)

X C ¼ c1



 1 t 1 t e þ c2 e 1 3

We find the complementary solution by setting Xc ¼ initial condition: Since we only had a homogeneous system, we will solve for c1 and c2 now using the initial conditions, x(0) ¼ 1, y(0) ¼ 2. We solve the system c1 þ c2 ¼ 1 c1 þ 3c2 ¼ 2 We find c1 and c2 both equal 0.5. The particular solution is X C ¼ 0:5



 1 t 1 t e þ 0:5 e 1 3

We might plot the solutions to the components X1 and X2, each a function of t. We note that both solutions grow without bound as t ! 1 (Fig. 7.17). Example 2: Complex Eigenvalues (eigenvalues of the form λ ¼ a  bi) We note here that we do not use the form ea  bi and that complex eigenvalues always appear in conjugate pairs. The key to finding two real linearly independent solutions from complex solutions is Euler’s identity:

370

7 Modeling Change with Dynamical Systems Models

eiθ ¼ cos θ þ i sin θ We can rewrite the solutions for X1and X2 using Euler’s identity. Keλt ¼ KeðaþbiÞ ¼ Keat ð cos bt þ i sin bt Þ Keλt ¼ K  eðabiÞ ¼ K  eat ð cos bt  i sin bt Þ Consider the following steps as a summary when we get complex eigenvalues. Step 1. Find the complex eigenvalues, λ ¼ a  bi Step 2. Find the complex eigenvector, K
K¼

u1 þ iv1 u2 þ iv2



Step 3. Form the real vectors


 u1 B1 ¼ u
2  v B2 ¼  1 v2

Step 4. Form the linearly independent set of real solutions: X 1 ¼ eat ðB1 cos bt þ B2 sin bt Þ X 2 ¼ eat ðB2 cos bt  B1 sin bt Þ  6 1 X Step 1. X ¼ 5 4 Step 2. We set up and solve for the eigenvalue. We solve the characteristic polynomial 0




(6-λ)(4--λ)+5¼0 29 – 10λ + λ2 ¼ 0 λ ¼ 5  2I We find the eigenvalues are 5+2I and 5–2I. Step 3. We find the eigenvectors we substituting λ as we did before. We then create the two vectors B1 and B2.

7.6 Systems of Differential Equations

371

Let λ ¼ –5 + 2I. Let k1 and k2 be the components of eigenvector K1. (1–2I)k1–k2 ¼ 0 5k1+(1–2I)k2 ¼ 0 We arbitrarily make k2¼(1-2I) thus k1¼1. K1 ¼ [1,1–2I] B1 ¼ real(K1) ¼ [1,1], B2 ¼ imaginary (K1) ¼ [0–2]. B1 ¼ [1,1] B2 ¼ [0,–2] By substitution, we find the complementary solution: 

  1 0 cos ð2t Þ  sin ð2t Þ 1 2 

  0 1 5t þ c2 e cos ð2t Þ þ sin ð2t Þ 2 1

X c ¼ c1 e5t

Since we only had a homogeneous system, we will solve for c1 and c2 now using the initial conditions, x(0) ¼1, y(0)¼2. We get two equations c1 ¼ 1 and c1  2c2 ¼ 2 whose solutions are c1 ¼ 1, c2 ¼ –0.5. 

  1 0 cos ð2t Þ  sin ð2t Þ 1 2 

  0 1 5t  0:5e cos ð2t Þ þ sin ð2t Þ 2 1

X p ¼ e5t

Again, we obtain plots of X1 and X2 as functions of t (Fig. 7.18). Example 3: Repeated Eigenvalues Solution When eigenvalues are repeated, we must find a method to obtain independent solutions. The following is a summary for repeated real eigenvalues. Step 1. Find the repeated eigenvalues, λ1 ¼ λ2,¼ λ Step 2. One solution is

372

7 Modeling Change with Dynamical Systems Models

80

2.4 2.2

60

2 1.8

40 1.6 1.4

20

1.2 1 0

0.1

0.2

0.3

0.4

0

0.5

0.2

0.4

0.6

0.8

t

t

Fig. 7.18 Plot of x(t) and y(t)

X 1 ¼ Keλt and the second linearly independent solution is given by X 2 ¼ Kteλt þ Peλt , where the components of P must satisfy the system ða  λÞp1 þ bp2 ¼ k1 cp1 þ ðd  λÞp2 ¼ k 2 and 
a b ¼A c d


 3 18 X 2 9 Step 2. Solve the characteristic equation (3λ)(9λ) + 36 ¼ 0. Step 1. X 0 ¼

We find we have repeated roots and λ ¼ 3,3. Step 3. We find K easily as the vector [3.1]. Then we solve for the vector P.


6 2

18 6




p1 p2

 ¼


 3 1

1

7.6 Systems of Differential Equations Fig. 7.19 Plot of Solution

373

1

0.5

0

0.2

0.4

t

0.6

0.8

1

–0.5

–1

We find p1 and p2 must solve p13p2¼1/2 or 2p16p2¼1. We select p1¼1 then p2 ¼ 1/6. Our complementary solution is " # !

 1 3 3t 3 3t 3t e þ c2 te þ 1 e X c ¼ c1 1 1 6

Since we only had a homogeneous system, we will solve for c1 and c2 now using the initial conditions, x(0)¼1, y(0)¼2. We obtain two equations: 3c1 þ c2 ¼ 1 c1 þ ð1=6Þc2 ¼ 2 c1 ¼ 5=3 c2 ¼ 4 " # !

 1 5 3 3t 3 3t 3t e 4 te þ 1 e Xc ¼ 1 3 1 6 We plot the solution in Fig. 7.19.

374

7.6.3

7 Modeling Change with Dynamical Systems Models

Numerical Solutions to Systems of ODE

In the previous chapter, we discussed the use of numerical solutions (Euler, Improved Euler, and Runge-Kutta methods) to first-order differential equations. In this chapter, we extend the use of numerical solutions to systems of differential equations. We show only Euler’s and Runge-Kutta methods. Our goal here is to provide a solution method for many models of systems of ODEs that do not have closed form analytical solutions. Throughout most of the chapter we have investigated and modeled autonomous systems of first-order differential equations. A more general form of systems of two ordinary first-order differential equations is given by: dx ¼ f ðt; x; yÞ dt dy ¼ gðt; x; yÞ dt

ð7:9Þ

If the variable t appears explicitly in one of the functions f or g, the system is not autonomous. In this section, we present numerical techniques for approximating solutions for x(t) and y(t) subject to initial conditions x(to)¼xo and y(to)¼yo. We will give the algorithm for each and show the Maple commands to execute a numerical solution. We also show how to obtain both the phase portraits and the plots of approximate numerical solutions. Euler’s method: Consider the iterative formula for Euler’s method for systems as xðnÞ ¼ xðn  1Þ þ f ðt ðn  1Þ; xðn  1Þ; yðn  1ÞÞΔt yðnÞ ¼ yðn  1Þ þ gðt ðn  1Þ; xðn  1Þ; yðn  1ÞÞΔt We illustrate a few iterations for the following initial value problem with a step size of Δt¼0.1: x0 ¼ 3x  2y,

x ð 0Þ ¼ 3

y0 ¼ 5x  4y, yð0Þ ¼ 6 xð0Þ ¼ 3, Given

y ð 0Þ ¼ 6

7.6 Systems of Differential Equations

375

xð1Þ ¼ 3 þ ð0:1Þ  ð3  3  2  6Þ ¼ 2:7 yð1Þ ¼ 6 þ ð0:1Þ  ð5  3  4  6Þ ¼ 5:1 and xð2Þ ¼ 2:7 þ ð0:1Þ  ð3  ð2:7Þ  2  ð5:1ÞÞ ¼ 2:49 yð2Þ ¼ 5:1 þ ð0:1Þ  ð5  ð2:7Þ  4  ð5:1ÞÞ ¼ 4:41

and so forth. In Excel, we enter the system and initial conditions and then iterate. Here are the numerical estimates using Euler’s method to our example (Table 7.2). We plot the estimates to see the solution in Fig. 7.20. The power of Euler’s method is two-fold. First, it is easy to use and second as a numerical method it can be used to estimate a solution to a system of differential equations that does not have a closed form solution. Assume we have the following predator–prey system that does not have a closed form analytical solution: dx ¼ 3x  xy dt dy ¼ xy  2y dt xð0Þ ¼ 1, yð0Þ ¼ 2 t 0 ¼ 1, Δt ¼ :1 We will obtain an estimated solution using Euler’s method (Fig. 7.21 and Table 7.3). We experiment and find that when we plot x(t) versus y(t) we have an approximately closed loop. We can use the improved Euler’s method and Runge-Kutta 4 methods to iterate solutions to systems of differential equations as well. The vector version of the iterative formula for Runge-Kutta is h X nþ1 ¼ X n þ ðK 1 þ 2K 2 þ 2K 3 þ K 4 Þ 6 where ðt n ; X n Þ K1 ¼ f   h h K 2 ¼ f tn þ ; X n þ K 1 2 2   h h K 3 ¼ f tn þ ; X n þ K 2 2 2 K 4 ¼ f ðt n þ h; X n þ hK 3 Þ

We repeat our example with Runge-Kutta (Fig. 7.22)

376 Table 7.2 Values Using Euler’s Method

7 Modeling Change with Dynamical Systems Models

7.6 Systems of Differential Equations

377

20 18 16 14 12 10 8 6 4 2 0

x y

0

0.5

1

1.5

2

Fig. 7.20 Plot of Estimates Using Euler’s Method

Fig. 7.21 Plot of Predator–Prey Model

2.5

3

3.5

378

7 Modeling Change with Dynamical Systems Models

Table 7.3 Iterated Values for Predator–Prey Model

7.7 Exercises

379

Fig. 7.22 Excel Screenshot Plot Using Runge-Kutta 4 Method

dx ¼ 3x  xy dt dy ¼ xy  2y dt xð0Þ ¼ 1, yð0Þ ¼ 2 t 0 ¼ 0, Δt ¼ 0:25

7.7

Exercises

1. Given the following system of Linear First-Order ODEs of species cooperation (symbiosis): dx1 =dt ¼ 0:5x1 þ x2 dx2 =dt ¼ 0:25x1  0:5x2 and

380

7 Modeling Change with Dynamical Systems Models

x1 ð0Þ ¼ 200 and x2 ð0Þ ¼ 500: (a) Perform Euler’s method with step size h ¼ 0.1 to obtain graphs of numerical solutions for x1(t) and x2(t) versus t and for x1 versus x2. You can put both x1(t) and x2(t) versus t on one axis if you want. (b) From the graphs discuss the long-term behavior of the system (discuss stability). (c) Analytically using Eigenvalues and Eigenvectors solve the system of DEs to determine the population of each species for t > 0. (d) Determine if there is a steady-state solution for this system. (e) Obtain real plots of x1(t) and x2(t) versus t and for x1(t) versus x2(t). Compare to the numerical plots. Briefly discuss. 2. Given a competitive hunter model defined by the system: dx=dt ¼ 15x  x2  2xy ¼ xð15  x  2yÞ dy=dt ¼ 12y  y2  1:5xy ¼ yð12  y  1:5xÞ (a) Perform a graphical analysis of this competitive hunter model in the x-y plane. (b) Identify all equilibrium points and classify their stability. (c) Find the numerical solutions using Euler’s method with step size h¼0.05. Try it from two separate initial conditions: first, use x(0)¼5 and y(0)¼4, then use x(0)¼ 3, y(0)¼9 . Obtain graphs of x(t), y(t) individually (or on the same axis) and then a plot of x versus y using your numerical approximations. Compare it to your phase portrait analysis. 3. Since bass and trout both live in the same lake and eat the same food sources, they are competing for survival. The rate of growth for bass (dB/dt) and for trout (dT/ dt) are estimated by the following equations: dB=dt ¼ ð10  B  T ÞB dT=dt ¼ ð15  B  3T ÞT Coefficients and values are in thousands. (a) Obtain a "qualitative" graphical solution of this system. Find all equilibrium points of the system and classify each as unstable, stable, or asymptotically stable. (b) If the initial conditions are B(0) ¼5 and T(0) ¼2, determine the long-term behavior of the system from your graph in part (a). Sketch it out. (c) Using Euler’s method, h¼0.1 and the same initial conditions as above, obtain estimates for B and T. Using these estimates determine a more accurate graph by plotting B versus T for the solution from t¼0 to t¼7. Euler’s method:

7.8 Chapter Projects

381

xnþ1 ¼ xn þ hf ðxn ; yn Þ and

ynþ1 ¼ yn þ hgðxn ; yn Þ

(d) Compare the graph in part (c) to the possible solutions found in (a) and (b). Briefly comment.

7.8 7.8.1

Chapter Projects Diffusion

Diffusion through a membrane leads to a first-order system of ordinary linear differential equations. For example, consider the situation in which two solutions of substance are separated by a membrane of permeability P. Assume the amount of substance that passes through the membrane at any particular time is proportional to the difference between the concentrations of the two solutions. Therefore, if we let x1 and x2 represent the two concentrations, and V1 and V2 represent their corresponding volumes, then the system of differential equations is given by: dx1 P ¼ ð x2  x1 Þ V1 dt , dx2 P ¼ ð x1  x2 Þ V2 dt where the initial amounts of x1 and x2 are given. Consider two salt concentrations of equal volume V separated by a membrane of permeability P. Given that P¼V, determine the amount of salt in each concentration at time t if x1(0)¼2 and x2(0)¼10. (a) Write out the system of differential equations that models this behavior. (b) Using the methods described in Chap. 7, solve this system. Clearly indicate your eigenvalues and eigenvectors. (c) Plot the solutions for x1 and x2 on the same axis and label each. Comment about the plots. (d) Use a numerical method (Euler or Runga-Kutta) and iterate a numerical solution to predict xi(4), use a step size of 0.5. Obtain a plot of your numerical approach. Compare it to the analytical plot. Comment about the plots. Diffusion through a double-walled membrane, where the inner wall has permeability P1 and the outer wall has permeability P2 with 0